Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the locus of the middle points of the chords of the hyperbola $3x^2-2y^2+4x-6y=0$ parallel to the line $y=2x$ .
Find the locus of the middle points of the chords of the hyperbola $3x^2-2y^2+4x-6y=0$ parallel to the line $y=2x$ .
Equation of the chords of the hyperbola bisected at $(h,k)$ is given by $S_1=T$ where $S_1=3xx_1-2yy_1+2(x+x_1)-3(y+y_1)=3x_1^2-2y_1^2+4x_1-6y_1$
where I have chosen $(x_1,y_1)$ to be any point of the line $y=2x$
But the answer is not coming.Is there any fault in my answer.
Please help.
| Let $y=2x+k$ be an equation of the chord and $(p,q)$ be a point on our locus.
Thus, for the $x$-coordinate of an intersect point we obtain:
$$3x^2-2(2x+k)^2+4x-6(2x+k)=0$$ or
$$5x^2+8(1+k)x+2k^2+6k=0,$$
which gives $$x_1+x_2=\frac{-8(1+k)}{5}$$ and
$$p=\frac{-4(1+k)}{5}$$ and $$q=2p+k=\frac{-8(1+k)}{5}+k=\frac{-3k-8}{5}.$$
Id est, $$k=\frac{5p+4}{-4}=\frac{5q+8}{-3},$$
which gives $$q=\frac{3}{4}p-1$$ and we got an equation of the locus:
$$y=\frac{3}{4}x-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to change a rectangle's area based on it's 4 coordinates? I have a rectangle with it's 4 bounds/coordinates on a Cartesian plane. Lets say I want to make the rectangle twice as big. How do i find the new coordinates?
Example, Dotted lines are the new lines after growing the area
In the example image above, the original coordinates are (5, 10) (6, 9) (5, 8) (4, 9). If i wanted to grow the rectangle an arbitrary number like 1.2 times, the new coordinates would be
(5, 11) (7, 9) (5, 7) (3, 9). How do i find these new coordinates mathematically?
| To double the area, multiply each of the four bounds by $\sqrt{2}$. That will multiply both side lengths by $\sqrt{2}$, and multiply the area by $\sqrt{2}\sqrt{2}=2$.
Example: $x=2, x=4$ and $y=5, y=10$. Current area = $(4-2)\times (10-5)=2\times 5=10$.
Multiply everything by $\sqrt{2}$. The new values will be $x=2\sqrt{2}, x=4\sqrt{2}$ and $y=5\sqrt{2}, y=10\sqrt{2}$. The new area will be $(4\sqrt{2}-2\sqrt{2})\times (10\sqrt{2}-5\sqrt{2})=(2\sqrt{2})\times (5\sqrt{2})=10\sqrt{2}\sqrt{2}=20$.
Second example: $(5, 10) (6, 9) (5, 8) (4, 9) \to (5\sqrt{2}, 10\sqrt{2}) (6\sqrt{2}, 9\sqrt{2}) (5\sqrt{2}, 8\sqrt{2}) (4\sqrt{2}, 9\sqrt{2})$
All four side lengths (in this example not horizontal/vertical) will be $\sqrt{2}$ longer, so the area will be twice as big.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2603791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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prove that $\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$
prove that $$\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx= \frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$$
My attempt:
C is semicircle in upper half complex plane
Simple poles = $e^{i\frac{\pi}{8}}, e^{i\frac{3\pi}{8}},e^{i\frac{5\pi}{8}},e^{i\frac{7\pi}{8}}$ lie in upper semi-circle C and real axis
Given integral value $= 2\pi i \cdot (\text{sum of residues}) = 2 \pi i \left(\frac{-1}{8}\right) \left[e^{i\frac{5\pi}{8}}+e^{i\frac{15\pi}{8}}+e^{i\frac{25\pi}{8}}+e^{i\frac{35\pi}{8}}\right] = 0.27059 \pi$
This is numerically equal to $\frac{\pi}{\sqrt 2} \sin \frac{\pi}{8}$. But without using calculator, how to get this expression.
| $$\int_{-\infty}^{\infty} \frac{x^4}{1+x^8} dx=2\int_{0}^{\infty} \frac{x^4}{1+x^8} dx=2\int_{0}^{\infty} \frac{(x^8)^{1/2}}{1+x^8} dx\\
=\frac{1}{4} \int_{0}^{\infty} \frac{u^{5/8-1}}{(1+u)^{1}} du=\frac{1}{4}B\left(\frac{5}{8}, 1-\frac{5}{8}\right) =\frac{1}{4}\frac{\Gamma(5/8)\Gamma(1-5/8)}{\Gamma(1) }\\=\color{blue}{\frac{1}{4}\frac{π}{\sin(\frac{5π}{8}) } =\frac{1}{4}\frac{π}{\sin(\frac{3π}{8}) }}\color{red}{=\frac{π}{\sqrt2}\sin(\frac{π}{8})} $$
where we made use of the beta and Gamma function and the Schwartz duplication formula Along with the the below relations
$$-\frac{\sqrt2}{2}= \cos(\frac{3π}{4}) = \left(\cos^2(\frac{3π}{8})-\sin^2(\frac{3π}{8})\right)=\left(1-2\sin^2(\frac{3π}{8})\right) \\\implies\color{blue}{\sin (\frac{3π}{8}) =\frac{\sqrt{2+\sqrt2}}{2}} $$
And by the same token,
$$\color{red}{\sin (\frac{π}{8}) =\frac{\sqrt{2-\sqrt2}}{2} = \frac{\sqrt{2}}{2\sqrt{2+\sqrt2}} =\frac{\sqrt{2}}{4\sin (\frac{3π}{8}) }.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving integral with Leibniz' rule I'm trying to prove the following integral with Leibniz' rule: $\int_{0}^{1}{\frac{x}{(1+50x)^2}dx}$. Firstly I wrote: $\int_{0}^{1}{\frac{x}{(1+bx)^2}dx}$ then I need to find a function $f$ that $\frac{\partial f}{\partial b}(x,b)= {\frac{x}{(1+bx)^2}}$ but now what?
| If you insist on using Leibniz' rule for differentiating under the integral sign it can be done as follows.
Let
$$I(b) = \int_0^1 \frac{dx}{1 + bx}, \quad b > 0.$$
Note that
$$I(b) = \frac{1}{b} \ln (1 + bx) \Big{|}_0^1 = \frac{1}{b} \ln (1 + b).$$
Also, on applying Leibniz' rule to $I(b)$ we have
$$I'(b) = -\int_0^1 \frac{x}{(1 + bx)^2} \, dx,$$
and we see that the required integral comes from setting $b = 50$ in the above integral. Now
$$I'(b) = \frac{d}{db} \left (\frac{\ln (1 + b)}{b} \right ) = \frac{1}{b(1 + b)} - \frac{1}{b^2} \ln (1 + b),$$
yielding
$$\int_0^1 \frac{x}{(1 + bx)^2} \, dx = \frac{1}{b^2} \ln (1 + b) - \frac{1}{b(1 + b)}.$$
Finally, setting $b = 50$ gives
$$\int_0^1 \frac{x}{(1 + 50 x)^2} \, dx = \frac{1}{2500} \ln (51) - \frac{1}{2550}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2606794",
"timestamp": "2023-03-29T00:00:00",
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Compute a division with integer and fractional part I have a problem that I don't know how to solve:
Compute $[\frac{\sqrt{7}}{frac(\sqrt{7})}]$
Here's what I've tried: $[\sqrt{7}]=2 \rightarrow frac(\sqrt{7}) = \sqrt{7}-2 \rightarrow [\frac{\sqrt{7}}{frac(\sqrt{7})}] =[\frac{\sqrt{7}}{\sqrt{7}-2}] = [\frac{\sqrt{7}(\sqrt{7}+2)}{7-4}] = [\frac{7+2\sqrt{7}}{3}]$
But from here I don't know what to do anymore.
| In some way you need to estimate $2\sqrt7$. First observe that $2\sqrt7=\sqrt{4\cdot 7}=\sqrt{28}$, thus $5=\sqrt{25}<\sqrt{28}<\sqrt{36}=6$, so
$$4=\left\lfloor\frac{7+5}3\right\rfloor\le\left\lfloor\frac{7+2\sqrt7}3\right\rfloor\le\left\lfloor\frac{7+6}3\right\rfloor=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2607814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Find the greatest and least values of $(\sin^{-1}x)^2+(\cos^{-1}x)^2$ Find the upper and lower limit of
$$
(\sin^{-1}x)^2+(\cos^{-1}x)^2
$$
My Attempt:
$$
\frac{-\pi}{2}\leq\sin^{-1}x\leq \frac{\pi}{2}\quad\&\quad0\leq\cos^{-1}x\leq\pi\\(\sin^{-1}x)^2\leq\frac{\pi^2}{4}\quad\&\quad(\cos^{-1}x)^2\leq\pi^2\\
0\leq(\sin^{-1}x)^2+(\cos^{-1}x)^2\leq\frac{\pi^2}{4}+\pi^2=\frac{5\pi^2}{4}
$$
Here, I can see the upper limit is $\frac{5\pi^2}{4}$ which is fine. But, $0$ is one lower limit not the lower limit.
Why am I not getting the lower limit in my approach ?
How do I approach similar problems involving max and min, when you don't get the lower or upper limits ?
| Let $\sin^{-1}x=a,\cos^{-1}x=b$
$a+b=\dfrac\pi2$
$$a^2+b^2=\left(\dfrac\pi2-b\right)^2+b^2=\dfrac{\pi^2}4+2\left(b-\dfrac\pi4\right)^2-2\left(\dfrac\pi4\right)^2$$
Again , $0\le b\le\pi\implies0\le\left(b-\dfrac\pi4\right)^2\le\left(\pi-\dfrac\pi4\right)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Find largest possible value of $x^2+y^2$ given that $x^2+y^2=2x-2y+2$
Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2$.
My attempt:
$x^2+y^2=2x-2y+2$
$(x^2-2x)+(y^2+1)=2$
$(x-1)^2+(y+1)^2=4$
I have no idea how to continue here. Any help?
| Thanks to StackTD, I have achieved my solution.
$x^2+y^2=2x-2y+2$
$(x^2-2x)+(y^2+1)=2$
$(x-1)^2+(y+1)^2=4$
We need to find the maximum distance from any point on the circle to the origin, squared(to find $x^2+y^2$).
By Pythagoras' theorem, the distance from the centre of the circle to the origin is $\sqrt{(-1)^2+1^2}=\sqrt2$. We then add $2$, the radius of the circle. We have $2+\sqrt2$, the maximum distance on any point on the circle to the origin. However, this is only $\sqrt{x^2+y^2}$, so we need to square $2+\sqrt2$, as mentioned above, and will get $6+4\sqrt2$ as answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
McLaurin expansions help The problem:
Study the convergence/divergence of this limit, as $a \in \mathbb{R}$ changes
$$\lim_{x\to 0^+} x^a \biggl( \frac{1}{\log(2x+\sqrt{1+x^3})} -\frac{1}{\log(2x+1)} \biggr)$$
I did that:
$\sqrt{1+x^3}=1+\frac{x^3}{2} + o(x^3)$ so
$\log\left(1 +2x +\frac{x^3}{2} + o(x^3) \right) = 2x - \frac{x^2}{2} +\frac{x^3}{2} +o(x^3)$
and $\log\left(1 +2x \right) = 2x - \frac{x^2}{2} +\frac{8x^3}{3} +o(x^3)$ .
Now how can I expand $\frac{1}{ 2x - \frac{x^2}{2} +\frac{x^3}{2} +o(x^3)}$ and $\frac{1}{ 2x - \frac{x^2}{2} +\frac{8x^3}{3} +o(x^3)}$ ?
I don't find on internet some expansion for these fractions(!)
| You can handle the reciprocals like $1/(2x-2x^2+8x^3/3+o(x^3))$ using binomial theorem as $$(1/2x)(1-x+4x^2/3+o(x^2))^{-1}=\frac{1}{2x}\left(1+x-x^2/3+o(x^2)\right)$$ which is the same as $(1/2x)+(1/2)-x/6+o(x)$. Similarly the other expression is $(1/2x)+(1/2)-7x/24+o(x) $ and thus the given expression is equal to $-x^{a+1}/8+o(x^{a+1})$ and you can now conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Chinese Remainder Theorem with 0 mod n I'm trying to get the least x from a system of congruences by applying the Chinese Remainder Theorem. Keep running into issues.
System of congruences:
$$
x \equiv 0 (_{mod} 7) \\
x \equiv 5 (_{mod} 6) \\
x \equiv 4 (_{mod} 5) \\
x \equiv 3 (_{mod} 4) \\
x \equiv 2 (_{mod} 3) \\
x \equiv 1 (_{mod} 2)
$$
I made the moduli relatively prime by removing 2nd and 6th congruence:
$1(_{mod} 2)$ is just a special case of $3 (_{mod} 4)$
and the 6th congruence because
$5 (_{mod} 6)$ splits into
$2 (_{mod} 3)$ and $1 (_{mod} 2)$, both of which are already represented in the system.
Product of moduli $m = 7 . 5 . 4 . 3 = 420$
Each respective $M_n$
$$
M_1 = 420/7 = 60 \\
M_2 = 420/5 = 84 \\
M_3 = 420/4 = 105 \\
M_3 = 420/3 = 140 \\
$$
Each respective modular inverse $y_n$
$$
y_1 = 0 \\
y_2 = 4 \\
y_3 = 3 \\
y_4 = 2 \\
$$
Trying to find solutions via CRT, I get
$x \equiv a_1 M_1 . y_1 + a_2 M_2 . y_2 + a_3 M_3 . y_3 + a_4 M_4 . y_4$
Plugging in the values:
$$
x \equiv 0 + 4 . 84 . 4 + 3 . 105 . 3 + 2 . 140 . 2 = 2849 \\
\equiv 329 (mod 42)
$$
And the "least" value being #329$ .
However, 329 doesn't satisfy the equation $329 (_{mod} 4) \equiv 3$.
What / where am I messing up?
| This is very similar to aid78's solution.
\begin{align}
x &\equiv 0 \pmod 7 \\
x &\equiv 5 \pmod 6 \\
x &\equiv 4 \pmod 5 \\
x &\equiv 3 \pmod 4 \\
x &\equiv 2 \pmod 3 \\
x &\equiv 1 \pmod 2
\end{align}
can be rewritten as
\begin{align}
x &\equiv \phantom{-}0 \pmod 7 \\
x &\equiv -1 \pmod 6 \\
x &\equiv -1 \pmod 5 \\
x &\equiv -1 \pmod 4 \\
x &\equiv -1 \pmod 3 \\
x &\equiv -1 \pmod 2
\end{align}
Since $\operatorname{lcm}(2,3,4,5,6)=60$, then the last five congruences are equivalent to $x \equiv -1 \pmod{60}$ So now you have to solve
\begin{align}
x &\equiv \phantom{-}0 \pmod 7 \\
x &\equiv -1 \pmod{60}
\end{align}
$x \equiv -1 \pmod{60}$ is equivalent to saying $x = 60n -1$ for some integer $n$.
This is one way to finish the problem.
\begin{align}
x &\equiv 0 \pmod{7} \\
60n - 1 &\equiv 0 \pmod{7} &\text{(Note $60\equiv 4 \pmod 7$)} \\
4n &\equiv 1 \pmod 7 \\
8n &\equiv 2 \pmod 7 \\
n &\equiv 2 \pmod 7 \\
n &= 7m + 2 \\
x &= 60n - 1 \\
x &= 60(7m + 2) - 1 \\
x &= 420m + 119 \\
x &\equiv 119 \pmod{420}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617244",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Solving $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$? (1983 AIME problem 3)
What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$?
I know it is a messy/bad idea, but I first started off by squaring both sides and moving everything to one side to get $$x^4 + 36x^3 + 384x^2 + 1080x + 900 - 4x^2 - 72x - 180 = x^4 + 36x^3 + 380x^2 + 1008x + 720 .$$
And by (generalisation) of Vieta's formula, the product of the real roots should be $\frac{720}{1} = 720$, but that is wrong, and I don't understand why.
| Use the following substitution. $$x^2+18x+45=t$$ and we obtain
$$t-2\sqrt{t}-15=0$$ ot
$$(\sqrt{t}-1)^2=16$$ and since $\sqrt{t}\geq0$, we obtain $$\sqrt{t}=5,$$
$$x^2+18x+20=0$$ or
$$(x+9)^2=61$$ and we get the answer:
$$\{-9\pm\sqrt{61}\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Evaluate the limit containing $\arctan{x}$ and $\arcsin{x}$ Evaluate:
$$\lim_{x\to{0}}\bigg(\frac{2}{x^3}.(\arcsin{x}-\arctan{x})\bigg)^{2/x^2}$$
I can just expand $\arcsin{x}$ and $\arctan{x}$ using their taylor expansions, but is there any other method?
| From https://math.stackexchange.com/a/198337/62967 we have
$$
\arcsin x = x + \frac{x^3}{6} + \frac{3x^5}{40} + o(x^5)
$$
and we know that
$$
\arctan x = x -\frac{x^3}{3} + \frac{x^5}{5} + o(x^5)
$$
Therefore
$$
\frac{2(\arcsin x-\arctan x)}{x^3}=1-\frac{1}{4}x^2+o(x^2)
$$
The logarithm of your function can therefore be written as
$$
\frac{2}{x^2}\log\left(1-\frac{1}{4}x^2+o(x^2)\right)
=
\frac{2}{x^2}\left(-\frac{1}{4}x^2+o(x^2)\right)=-\frac{1}{2}+o(1)
$$
Can you finish?
I see no other practical method.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Evaluating: $\int \frac {1+\sin (x)}{1+\cos (x)} dx$
Evaluate: $\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$
My Attempt:
$$=\int \dfrac {1+\sin (x)}{1+\cos (x)} dx$$
$$=\int \dfrac {(\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2}))^2}{2\cos^2 (\dfrac {x}{2})} dx$$
$$=\dfrac {1}{2} \int (\dfrac {\sin (\dfrac {x}{2}) + \cos (\dfrac {x}{2})}{\cos (\dfrac {x}{2})})^2 dx$$
$$=\dfrac {1}{2} \int (\tan (\dfrac {x}{2}) +1)^2 dx$$
How do I continue?
| The integration of
$$\frac{\sin x}{1+\cos x}$$ which is of the form $f'(x)/f(x)$ is immediate. The rest is
$$\int\frac{dx}{1+\cos x}=\int\frac{dx}{2\cos^2\frac x2}$$ and
$$I=-\log|1+\cos x|+\tan\frac x2+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 7
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Dynamical System is fixed point at origin hyperbolic or asymptotically stable and is the system Hamiltonian I am given a system
$\dot x = f(x,y) = (x^2 + y^2)(x^3 + y^2x -2y - x) \\
\dot y = g(x,y) = (x^2 + y^2)(y^3 + x^2y +2x - y) $
and I am asked if the fixed point at $(0,0)$ is hyperbolic or asymptotically stable. I am also asked whether the system is a Hamiltonian.
So I know that a hyperbolic fixed point is a fixed point that does not have a centre manifold, so this makes me think that it is a hyperbolic fixed point but I'm not really sure.
Then onto asymptotically stable, my understanding of this is that we require the system to be both Liapounov stable and quasi-asymptotically stable, but I really dont know how to go about this.
Finally about it being a Hamiltonian system. I know in order to be a Hamiltonian the system can be written in the form
$H(q_i,p_i)$ such that
$\dot q_i= \frac{\partial H}{\partial p_i} \\
\dot p_i= -\frac{\partial H}{\partial q_i}$
But I really dont know how to do this for this DS.
Also i it helps I converted the DS into polar coords, giving me
$\dot r = r^5 - r^3 \\
\dot \theta = 2r^2$
| This is about the stability part:
Asymptotic stability of the equilibrium means that it is stable and attractive.
The origin $(0,0)$ is an equilibrium.
Consider the Lyapunov function candidate $V(x,y) = \frac{1}{2}x^2 + \frac{1}{2}y^2$. The derivative along the trajectories of $\dot x = f(x,y)$ and $\dot y = g(x,y)$ yields
\begin{align}
\dot V = x\,\dot x + y\,\dot y &= x(x^2+y^2)(x^3+y^2x-2y-x) + y(x^2+y^2)(y^3+x^2y+2x-y) \\
& = (x^2+y^2)(x^4+y^2x^2-2yx-x^2) + (x^2+y^2)(y^4+x^2y^2+2xy-y^2) \\
& = (x^2+y^2)(x^4+y^4) + (x^2+y^2)(2y^2x^2) - (x^2+y^2)^2 \\
& = (x^2+y^2)\left((x^4+y^4) +(2y^2x^2) \right) - (x^2+y^2)^2 \\
& = (x^2+y^2)(x^2+y^2)^2 - (x^2+y^2)^2
\end{align}
Insert $2V = x^2+y^2$:
\begin{align}
\dot V & = 8\,V^3 - 4 V^2 = -8\,V^2\left(\frac{1}{2}-V\right)
\end{align}
(Note: this is qualitatively similar to that what could be derived from the representation in polar coordinates)
Accordingly, $\dot V$ is only negative in the set $\{(x,y)\in\mathbb R^2: x^2+y^2<1 \}$ which contains the origin. That is, the origin is (locally) asymptotically stable. The unit disc defines also the region of attraction.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Number of non negative integer solution of the equation $x+y+z=n$
Total number of non negative integer solution of the equation $x+y+z=n$ subjected to the condition $x\leq y \leq z.$
Try:if $x=0$ Then we have $y+z=n$
If $x=1$ Then we have $y+z=n-1$
.....if $x=n$ Then we have $y+z=0$ so $x=y=0$
Could some help me how to calculate for $x+y=n,n-1,n-2$ ect.
Or any nice way to find solution of original equation.Thanks
|
We transform the equation $x+y+z=n$ with integer solutions $0\leq x\leq y\leq z$ by putting
\begin{align*}
x&=a&a \geq 0\\
y&=x+b&b\geq 0\\
z&=y+c&c\geq 0
\end{align*}
and get the equivalent system
\begin{align*}
\color{blue}{3a+2b+c=n\qquad\qquad 0\leq a,b,c\leq n}\tag{1}
\end{align*}
We solve (1) with the help of generating functions. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of a series.
We obtain the number of valid solutions by calculating
\begin{align*}
\color{blue}{[z^n]}&\color{blue}{\left(1+z^3+z^6+\cdots\right)\left(1+z^2+z^4+\cdots\right)\left(1+z+z^2+\cdots\right)}\\
&=[z^n]\frac{1}{\left(1-z^3\right)\left(1-z^2\right)\left(1-z\right)}\tag{2}\\
&=[z^n]\left(\frac{1}{6}\cdot\frac{1}{(1-z)^3}+\frac{1}{4}\cdot\frac{1}{(1-z)^2}
+\frac{1}{4}\cdot\frac{1}{1-z^2}+\frac{1}{3}\cdot\frac{1}{1-z^3}\right)\tag{3}\\
&=\frac{1}{6}\binom{-3}{n}(-1)^n+\frac{1}{4}\binom{-2}{n}+\frac{1}{4}[[n\equiv 0(3)]]+\frac{1}{3}[[n\equiv 0(3)]]\tag{4}\\
&=\frac{1}{6}\binom{n+2}{2}+\frac{1}{4}\binom{n+1}{1}+\frac{1}{4}[[n\equiv 0(3)]]+\frac{1}{3}[[n\equiv 0(3)]]\tag{5}\\
&=\frac{1}{12}(n+1)(n+5)+\frac{1}{4}[[n\equiv 0(3)]]+\frac{1}{3}[[n\equiv 0(3)]]\tag{6}\\
&=\frac{1}{12}(n+3)^2-\frac{1}{3}+\frac{1}{4}[[n\equiv 0(3)]]+\frac{1}{3}[[n\equiv 0(3)]]\tag{7}\\
&\color{blue}{=\mathrm{round}\left(\frac{(n+3)^2}{12}\right)}\tag{8}\\
&\color{blue}{=(1, 1, 2, 3, 4, 5, 7, 8, 10, 12, 14, 16, 19, 21, 24, 27, 30, 33, 37,\ldots)}\tag{9}
\end{align*}
Comment:
*
*In (2) we use the geometric series expansion.
*In (3) we do a (nice) partial fraction expansion.
*In (4) we select the coefficients of the binomial series. We also use the Iversion brackets $[[P]]$ which is $1$ iff $P$ is true and $0$ otherwise.
*In (5) we use the binomial identities $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$ and $\binom{p}{q}=\binom{p}{p-q}$.
*In (6) we do a simplification.
*In (7) we do a small trick which is indicated in the note at the end.
*In (8) we find a nice representation as rounded expression which can be easily justified by checking the remainders modulo $6$ of the terms in (7).
Note: The numbers listed in (9) are stored in OEIS as A001399. The nice representation (8) is stated there in the formula section.
| {
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Number of non-negative integral solutions of $2x+3y+z=19$ The question is to find the number of solutions of $2x+3y+z=19$.
There are posts like this with answers arrived by counting the possibilities. But I decided to approach it in a different way.
My Approach:
I learnt a theorem stating that
The number of integral solutions of $$a_1x_1+a_2x_2+a_3x_3...+a_nx_n=r$$ is the coefficient of $x^r$ in $$(1+{x}^{a_1}+{x}^{2\cdot{a_1}}+...+{x}^{\left \lfloor{\frac{r}{a_1}}\right \rfloor\cdot{a_1}})\cdot(1+{x}^{a_2}+{x}^{2\cdot{a_2}}+...+{x}^{\left \lfloor{\frac{r}{a_2}}\right \rfloor\cdot{a_2}})...(1+{x}^{a_n}+{x}^{2\cdot{a_n}}+...+{x}^{\left \lfloor{\frac{r}{a_n}}\right \rfloor\cdot{a_n}})$$ (I don't know the name of this theorem)
Substituting and reducing the equation, I landed up with finding coefficient of $x^{19}$ in $${\frac{1-x^{20}}{1-x^2}}\cdot{\frac{1-x^{21}}{1-x^3}}\cdot{\frac{1-x^{20}}{1-x}}$$
But, I had no clue of finding that.
My doubts:
*
*Is there any other relatively elegant way of finding it? (Other than counting or stars and bars - as in here)
*If no, how to solve further to get the answer?
Thanks in advance...
| Finally, I figured it out.
From this:
The number of integral solutions of $$a_1x_1+a_2x_2+a_3x_3...+a_nx_n=r$$ is the coefficient of $x^r$ in $$(1+{x}^{a_1}+{x}^{2\cdot{a_1}}+...+{x}^{\left \lfloor{\frac{r}{a_1}}\right \rfloor\cdot{a_1}})\cdot(1+{x}^{a_2}+{x}^{2\cdot{a_2}}+...+{x}^{\left \lfloor{\frac{r}{a_2}}\right \rfloor\cdot{a_2}})...(1+{x}^{a_n}+{x}^{2\cdot{a_n}}+...+{x}^{\left \lfloor{\frac{r}{a_n}}\right \rfloor\cdot{a_n}})$$
We substitute and get to find coefficient of $x^{19}$ in
$${\frac{1-x^{20}}{1-x^2}}\cdot{\frac{1-x^{21}}{1-x^3}}\cdot{\frac{1-x^{20}}{1-x}}$$
But, any term other than $1$ cannot give $x^{19}$ on multiplying with$\frac{1}{(1-x^2)(1-x^3)(1-x)}$
So, take the numerator as $1$
Now, we multiply and divide the function by $(1+x^3)(1+x^2+x^4)^2(1+x)$ to get$$\frac{(1+x^3)(1+x^2+x^4)^2(1+x)}{(1-x^6)^3}$$
This results in $$\frac{(1+x^3)(1+x^4+x^8+2x^2+2x^4+2x^6)(1+x)}{(1-x^6)^3}$$
Now, we have the following constraints:
*
*The maximum power of numerator is $12$
*Powers in expansion of $(1-x^6)^{-3}$ are $1, 6, 12, 18, 24... $
*For multiplication of $(1-x^6)^{-3}$ to numerator to yield $x^{19}$, there must be $x$ raised to $18, 13, 7,$ or $1$
*But, $x$ raised to $18$ and $13$ are not possible
Manipulating the numerator, we get the coefficients of $x$ and $x^7$ to be $1$ and $5$ respectively
Now we represent the expression as $$(x+x^7+P)(1-x^6)^{-3}$$
So, $x^{12}$ and $x^{18}$ terms of $(1-x^6)^{-3}$ are required to get $x^{19}$
So, we express the expression as $$5\cdot{4\choose2}(x^{12})(x^7)-{5\choose3}(x)(x^{18}) + Q$$
$$=40x^{19}+Q$$
So, coefficient of $x^{19}$ is $40$
$\therefore$ No. of non- negative integral solutions of $2x+3y+z=19$ is $40$
| {
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Prove the convergence of $\sum\limits_{k=2}^\infty \frac 1 {(2k+1)(2k+5)}$ Hints preferred:
I need to prove convergence and determine the limit of following series: $$ \sum_{k=2}^\infty \frac 1 {(2k+1)(2k+5)}$$
I need to use partial fraction decomposition to simplify the term, which I attempted with following:
$$
\frac 1 {(2k+1)(2k+5)} = \frac a {(2k+1)} + \frac b {(2k+5)} \\~\\
\frac 1 {(2k+1)(2k+5)} = \frac {a(2k+5)} {(2k+1)(2k+5)} + \frac {b(2k+1)} {(2k+1)(2k+5)} \\~\\$$
Then we have:
$$
k: 2a + 2b = 0 \\
k^0: 5a + b = 1 \\~\\
a = \frac 1 4 ~ \land b = -\frac 1 4
$$
So we have:
$$
\frac 1 {(2k+1)(2k+5)} = \frac { \frac 1 4 } {(2k+1)} - \frac { \frac 1 4} {(2k+5)} = \\ \frac { 1 } {4(2k+1)} - \frac { 1 } {4(2k+5)}
$$
Which can be defined as:
$$
\frac 1 4 (\frac { 1 } {2k+1} - \frac { 1 } {2k+5} )
$$
I'm not sure how to turn the result into a telescopic series (if the partial fraction decomposition is right in the first place).
| Set $u_{k} =\frac { 1 } {2k+1}\implies u_{k+1} =\frac { 1 } {2k+3}~~ãnd ~~~u_{k+2} =\frac { 1 } {2k+5} $ then by telescopic sum we have
$$\begin{align}\sum_{k=2}^n \frac 1 {(2k+1)(2k+5)}&=
\frac 1 4 \sum_{k=2}^n\left(\frac { 1 } {2k+1} - \frac { 1 } {2k+5} \right)\\
&=\frac 1 4 \left(\sum_{k=2}^nu_k-u_{k+2}\right)\\
&=\frac 1 4 \left(\sum_{k=2}^nu_k-u_{k+1}+u_{k+1}-u_{k+2}\right)\\
&=\frac 1 4 \left(\sum_{k=2}^nu_k-u_{k+1}+\sum_{k=2}^nu_{k+1}-u_{k+2}\right)\\
&=\frac 1 4 \left(u_2-u_{n+1}+u_{3}-u_{n+2}\right)\to \frac 1 4 \left(u_2+u_{3}\right)\\&= \frac{3}{35}\end{align}$$
| {
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New Golden Ratio (phi) Sequences This is the golden ratio (phi):
$\phi = \sqrt{5}/2 - 1/2 $
If we perform alternate subtraction and addition to ascending powers of phi and its reciprocal we get the sequence of numbers known as the 'Lucas numbers'.
$1 = 1/\phi - \phi $
$3 = 1/\phi^2 + \phi^2 $
$4 = 1/\phi^3 - \phi^3 $
$7= 1/\phi^4 + \phi^4 $
$11 = 1/\phi^5 - \phi^5 $
...
If we perform alternate addition and subtraction to ascending powers of phi and its reciprocal we get this strange sequence of numbers.
$ \sqrt{5} = 1/\phi + \phi $
$ \sqrt{5} = 1/\phi^2 - \phi^2 $
$ (2)\sqrt{5} = 1/\phi^3 + \phi^3 $
$ (3)\sqrt{5} = 1/\phi^4 - \phi^4 $
$ (5)\sqrt{5} = 1/\phi^5 + \phi^5 $
...
As you can see in the following examples we can calculate the square or square-root of phi to any power using simple arithmetic.
If we set the value of x to equal phi, we get the following sequence:
$ x^2 = 1 - x
\\ x^3 = x - x^2
\\ x^4 = x^2 - x^3
\\ x^5 = x^3 - x^4
\\ x^6 = x^4 - x^5
\\ x^7 = x^5 - x^6
\\ x^8 = x^6 - x^7 $
...
$ x^2 = 3/2 - \sqrt{5}/2
\\ x^3 = 4/2 - (2) \sqrt{5}/2
\\ x^4 = 7/2 - (3) \sqrt{5}/2
\\ x^5 = 11/2 - (5) \sqrt{5}/2
\\ x^6 = 18/2 - (8) \sqrt{5}/2
\\ x^7 = 29/2 - (13) \sqrt{5}/2
\\ x^8 = 47/2 - (21) \sqrt{5}/2 $
...
If we set the value of x to equal the reciprocal of phi, we get this second sequence:
$ x^2 = 1 + x
\\ x^3 = x + x^2
\\ x^4 = x^2 + x^3
\\ x^5 = x^3 + x^4
\\ x^6 = x^4 + x^5
\\ x^7 = x^5 + x^6
\\ x^8 = x^6 + x^7 $
...
$ x^2 = 3/2 + \sqrt{5}/2
\\ x^3 = 4/2 + (2) \sqrt{5}/2
\\ x^4 = 7/2 + (3) \sqrt{5}/2
\\ x^5 = 11/2 + (5) \sqrt{5}/2
\\ x^6 = 18/2 + (8) \sqrt{5}/2
\\ x^7 = 29/2 + (13) \sqrt{5}/2
\\ x^8 = 47/2 + (21) \sqrt{5}/2 $
...
Notice in both cases the similarity to the Fibonacci sequence.
Do theses sequence have a name?
| Observe these two closed formulas for the Lucas and Fibonacci numbers, known as Binet's formulas:
$$L_n = \phi^{n} + (-\phi)^{-n}$$
$$F_n = \frac{1}{\sqrt{5}}\left(\phi^{n} - (-\phi)^{-n}\right)$$
Now your observation is:
$$\phi^n = \frac{1}{2}(L_n + F_n\sqrt{5})$$
Plug in the formulas above to verify its correctness. And this is a well-known relationship between the Lucas and Fibonacci numbers and the golden ratio.
| {
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Computing $\lim_{x \to 1}\frac{x^\frac{1}{5}-1}{x^\frac{1}{6} -1}$ I cannot figure out how to
get around the zero numerator and denominator in order to compute the limit below:
$$\lim_{x \to 1}\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}$$
I tried:
$$ \lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{1}{6} - 1) (x^\frac{1}{6} + 1) } $$
$$\lim_{x \to 1} \frac{ (x^\frac{1}{5} - 1) (x^\frac{1}{6} + 1) }{ (x^\frac{2}{6} - 1) } $$
| By L'Hopital's rule:\begin{align}\lim_{x\to1}\frac{x^\frac15-1}{x^\frac16-1}&=\lim_{x\to1}\frac{\frac15x^{-\frac45}}{\frac16x^{-\frac56}}\\&=\frac65.\end{align}
| {
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How do I solve this fractional indices equation $\frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}$? Solve the equation $\frac{3^{5x+2}}{9^{1-x}}=\frac{27^{4+3x}}{729}$
I thought that the best way of approaching this would be to rewrite everything using $3$ as the base of the exponents, hence creating an equivalence which would allow me to equate numerators to numerators, and denominators to denominators. Doing this yields:
$$\frac{3^{5x+2}}{3^{2(1-x)}}=\frac{3^{3(4+3x)}}{3^6}$$
Equating the exponents of each numerator:
$$ 5x+2=3(4+3x) $$
$$ 5x+2=12+9x $$
$$ -4x=10 $$
$$ x = \frac{10}{-4}=-\frac{5}{2} $$
Doing this for the denominator yields a different value of $x$:
$$ 2(1-x)=6 $$
$$ 2-2x=6 $$
$$ -2x=4 $$
$$ x = -2 $$
Why is that I'm obtaining two different values of $x$?
Further to this, the solution in the book states the answer as $x=-3$, what am I doing wrong?
| First subtract the exponents both side and then equate.
$$3^{5x+2-2(1-x)}=3^{3(4+3x)-6}$$
$$5x+2-2(1-x)=3(4+3x)-6 \implies x=-3$$
| {
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The sum of two positive numbers is 1. The sum of their cubes is a maximum. What are the numbers? I set this up and end up finding the minimum (the two numbers would both be $1/2$). To find a maximum value, I could reflect the functions and use $y^3-x^3$ but I still end up finding $1/2$ as the two numbers. It does make sense that two two values to produce a max would be 0 and 1 but I can't figure out how to set up the problem from the start. What I have is...
$x+y=1 \\
x^3 + y^3 = max$
Subsitution...
$x^3 - (1-x)^3 = max \\
x^3 - 1 + 3x - 3x^2 + x^3 = max \\
2x^3 - 3x^2 + 3x - 1 = max \\
6x^2 - 6x + 3 = 0 \\
x = 1/2 \text{ which makes }y = 1/2 \\
$
That is where my issue is. How do I set it up to find the sum of the cubes to be a max?
| Well, at first:
$$x^3+y^3=(x+y)(x^2-xy+y^2)$$
Since you want to maximize this and $x+y=1$, we get that:
$$x^3+y^3=x^2-xy+y^2$$
Now, you can see the above as a quadratic with respect to $x$ - a parabola, and calculate:
$$\Delta=y^2-4y^2=-3y^2<0$$
So, this parabola attains its minimum value - it has a positive leading coefficient - at $x_0=\frac{y}{2}$ and it maximum value on the boundaries of the interval that $x$ belong to. In our case, since we have $x\in(0,1)$, it is clear that this parabola cannot have a maximum value. However, if we let - as proposed by many - $x,y\in[0,1]$, we see that the so wanted maximum value can be attained for:
$$x=0,y=1$$
or, symmetrically, for $$x=1,y=0$$
Alternatively, you can substitute $y=1-x$ in the above quadratic and get that:
$$x^3+y^3=x^2-x(1-x)+(1-x)^2=x^2-x+x^2+x^2-2x+1=3x^2-3x+1$$
which, again has a global minimum and attains its maximum value either on $0$ or $1$ - actually, exactly on both of them.
| {
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How to solve that? I have no idea about how to solve the following:
$$\sqrt[4]{13x+1} + \sqrt[4]{4x-1} = 3\sqrt[4]{x}$$
Could somebody help me, please?
| Let $\sqrt[4]{13x+1}=a$ and $ \sqrt[4]{4x-1}=b$.
Thus, $a\geq0$ and $b\geq0$ and $$a+b=3\sqrt[4]{\frac{a^4+b^4}{17}},$$ which is homogeneous and
$$\frac{a^4-1}{13}=\frac{b^4+1}{4}.$$
The first gives after squaring of the both sides:
$$(a-2b)(2a-b)(16a^2+23ab+16b^2)=0.$$
Since $16a^2+23ab+16b^2=0$ gives $a=b=0,$ which is impossible,
we get $(a-2b)(2a-b)=0$ only. I think now you can end it.
| {
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Inequality with $x^2+y^2+z^2=3$ If x,y,z are positive real numbers and $x^2+y^2+z^2=3$ prove that
$$\sum_\text{cyc}\frac{1}{x^3+y^2+z}\le\frac{1}{xyz}.$$
I have tried to prove it this way:
$$\sum_\text{cyc}\frac{1}{x^3+y^2+z}\le\frac{1}{xyz}$$
$$\Leftrightarrow \sum_\text{cyc}\frac{xyz}{x^3+y^2+z}\le1$$
$$\Leftrightarrow \sum_\text{cyc}\frac{1}{\frac{x^2}{yz}+\frac{y}{xz}+\frac{1}{xy}}\le1$$
But we know that $$\frac{x^2}{yz}+\frac{y}{xz}+\frac{1}{xy}\ge3\sqrt[3]{\frac{x^2}{yz}\cdot\frac{y}{xz}\cdot\frac{1}{xy}}=\frac{3}{\sqrt[3]{z^2y}}$$
$$\Rightarrow \sum_\text{cyc}\frac{1}{\frac{x^2}{yz}+\frac{y}{xz}+\frac{1}{xy}}\le\sum_{cyc}\frac{\sqrt[3]{z^2y}}{3}$$
We now have to prove that $\sum_\limits\text{cyc}\frac{\sqrt[3]{z^2y}}{3}\le1$.
$$\frac{\sqrt[3]{z^2y}}{3}+\frac{\sqrt[3]{x^2z}}{3}+\frac{\sqrt[3]{y^2x}}{3}=\frac{\sqrt[3]{1\cdot z^2\cdot y}}{3}+\frac{\sqrt[3]{1\cdot x^2\cdot z}}{3}+\frac{\sqrt[3]{1\cdot y^2\cdot x}}{3}$$
$$\le\frac{\frac{1+ z^2+ y}{3}+\frac{1+ x^2+ z}{3}+\frac{1+ y^2+ x}{3}}{3}=\frac{3+3+x+y+z}{9}$$
And the last part: $\;\dfrac{3+3+x+y+z}{9}\le1 \Leftrightarrow x+y+z\le3$ or
$$\frac{x+y+z}{3}\le \sqrt{\frac{x^2+y^2+z^2}{3}}$$ which is true.
I hope that it's correct. Any suggestions are welcome.
| I think your proof is right and very nice!
Also, we can use C-S:
$$\sum_{cyc}\frac{1}{x^3+y^2+z}=\sum_{cyc}\frac{x+y^2+z^3}{(x^3+y^2+z)(x+y^2+z^3)}\leq\sum_{cyc}\frac{x+y^2+z^3}{(x^2+y^2+z^2)^2}.$$
| {
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Given $\frac{1}{1^4}+\frac{1}{2^4}...\infty=\frac{\pi^4}{90}$ If $$\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}...\infty=\frac{\pi^4}{90}$$
then, $$\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}...\infty=?$$
Please provide a hint instead of the complete answer. Thanks. [Please edit the tags for me as I have no idea which sector this problem lies in.]
| hint: For the even terms: $\dfrac{1}{(2n)^4} = \dfrac{1}{16}\cdot \dfrac{1}{n^4}$, and you get to solve $S + \dfrac{S}{16} = \dfrac{\pi^4}{90}$, with $S$ is the desire sum.
| {
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Solving a biquadratic.
If $x$ is a real number and satisfies $$x+ \sqrt[4] {5-x^4}=2$$ then find the value of $$x\sqrt[4] {5-x^4}$$
My try :
The question is significantly asking for the value of $-x(x-2)$ if we get the root of the equation $$2x^4-8x^3+24x^2-32x+11=0$$
Using this I have reached till
$$-2x(x-2)(x^2-2x+8)=11$$
$$-x(x-2)=\frac {11}{2(x^2-2x-8)}$$
but I couldn't manipulate it further.
Also upon some second thought I want to ask whether could it be possible to form a quadratic polynomial with two roots $\alpha$ and $\beta$ such that $\alpha=x$ and $\beta = \sqrt[4] {5-x^4}$
but I still couldn't proceed further. Somebody please share some hints.
| hint: You have: $x+y = 2, x^4+y^4 = 5\implies 4 = (x+y)^2 = x^2+y^2 + 2xy \implies 4t^2 =4(xy)^2 = (x^2+y^2-4)^2 = x^4+y^4+16+2(xy)^2 - 8(x^2+y^2) = 5+16+2t^2 - 8(4-2t) $. Can you take it from here ? You got a quadratic equation in $t = xy$.
| {
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"question_score": "11",
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Solving $A_{n+1}=3A_n+2^n$ Let's say I want to find a formula for the following expression given $n$ number of threes
$$\ldots(3(3(3(3(3(3+1)+2)+4)+8)+16)+\ldots$$
If $A_0=1$, then
$$A_{n+1}=3A_n+2^n$$
Plugging in values to see the pattern,
$$A_2 = 3+1$$
$$A_3 = 3^2+3+2^1$$
$$A_4 = 3^3+3^2+3\cdot2+2^2$$
But I don't know how to condense something like this into an explicit formula.
|
Solution of the recurrence
Given sequences $g(n) \neq 0$ and $b(n)$, we have that $f(n)$
the solution of the recurrence
$$f(n+1)=g(n).f(n)+b(n)$$
is given by
$$f(n)= \bigg(\sum^{n-1}_{p=1}\frac{b(p)}{\prod\limits^{p}_{k=1}g(k)}+f(1) \bigg)\prod^{n-1}_{k=1}g(k). $$
See the proof here
Now taking $g(n)= 3$ and $b(n) =2^n.$ One obtains
$$A_n= \prod^{n-1}_{k=1}3\bigg(\sum^{n-1}_{p=1}\frac{2^p}{\prod\limits^{p}_{k=1}3}+A_1 \bigg)= 3^{n-1}\bigg(\sum^{n-1}_{p=1}\frac{2^p}{3^p}+A_1 \bigg)\\=3^{n-1}\bigg(\frac{2}{3}\frac{\left(\frac{2}{3}\right)^{n-1}-1}{\frac{2}{3}-1}+A_1 \bigg)=3^{n-1}\bigg(2\left[1-\left(\frac{2}{3}\right)^{n-1}\right]+A_1 \bigg)\\=\left(2\cdot 3^{n-1}-2^n+ 3^{n-1}\cdot A_1\right).$$
Finally, With $A_1= 4$ since $A_0=1$ $$A_n=\left(2\cdot 3^{n-1}-2^n+ 3^{n-1}\cdot 4\right) = 2\cdot 3^{n}-2^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Proof of the generating function of $1 -2x +3x^2 -4x^3+5x^4-6x^5+\cdots$ The ordinary generating function for the sequence $\{a_n\}_{n\geq0}$ where $a_n = (-1)^n\,n$ is
$$1 -2x +3x^2 -4x^3+5x^4-6x^5+\cdots = \frac{1}{(x+1)^2}$$
I can see from the geometric formula, and even from long division that
$$1 -x +x^2 -x^3+x^4-x^5+\cdots = \frac{1}{1+x}$$
but I'm not seeing the coefficients to explain $\frac{1}{(x+1)^2}.$
| Just to be different.
$\dfrac1{1-x}
=\sum_{n=0}^{\infty} x^n
$.
Therefore
$\begin{array}\\
\dfrac1{(1-x)^2}
&=\sum_{n=0}^{\infty} x^n\sum_{m=0}^{\infty} x^m\\
&=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty} x^{m+n}\\
&=\sum_{k=0}^{\infty}\sum_{m=0}^{k} x^{k}\\
&=\sum_{k=0}^{\infty} x^{k}\sum_{m=0}^{k}1\\
&=\sum_{k=0}^{\infty}(k+1) x^{k}\\
\text{so}\\
\dfrac1{(1+x)^2}
&=\sum_{k=0}^{\infty}(k+1)(-1)^k x^{k}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is $\int \frac{ \sqrt{(a^2 - x^2)}}{x^2}$? I tried the following trig substitution:
$x = a\sin \theta$
$$ \int \frac{ \sqrt{(a^2 - x^2)}}{x^2} = \int \frac{\sqrt{a^2 - a^2 \sin^2 \theta}}{a^2 \sin^2 \theta} = \int \frac{a \cos \theta}{a^2 \sin^2}$$
Setting $u = \sin x$ yields:
$$\frac{1}{a} * \int u^{-2} = - \frac{1}{a \sin \theta}$$
However my textbook states that the answer should really be:
$$ -\frac{\sqrt{a^2 - x^2}+x\tan^{-1}(\frac{x}{\sqrt{a^2-x^2}})}{x}$$
Where did I mess up?
| You need to replace $dx$ with $d\theta$ and then $du$. You don't seem to have done that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Exponentiation of a symmetric $2 \times 2$ matrix Why does the following hold?
$$\left[\begin{matrix}a & b\\b & a\end{matrix}\right]^k=\frac{1}{2}\left[\begin{matrix}\left(a - b\right)^{k} + \left(a + b\right)^{k} & - \left(a - b\right)^{k} + \left(a + b\right)^{k}\\- \left(a - b\right)^{k} + \left(a + b\right)^{k} & \left(a - b\right)^{k} + \left(a + b\right)^{k}\end{matrix}\right]$$
| Note that
$$
\pmatrix{a&b\\b&a} = \frac 12 \pmatrix{1&1\\1&-1}\pmatrix{a+b&0\\0&a-b} \pmatrix{1 & 1\\1&-1}
$$
where we observe that
$$
\pmatrix{1 & 1\\1&-1}^2 = 2I
$$
With that, we compute
$$
\begin{align*}
\pmatrix{a&b\\b&a}^k &= \left[\frac 12 \pmatrix{1&1\\1&-1}\pmatrix{a+b&0\\0&a-b} \pmatrix{1 & 1\\1&-1}\right]^k\\
&=
\frac 12 \pmatrix{1&1\\1&-1}\left[\pmatrix{a+b&0\\0&a-b} \underbrace{\left(\frac 12 \pmatrix{1 & 1\\1&-1}\pmatrix{1&1\\1&-1}\right)}_I\right]^{k-1}
\\ & \qquad \cdot \pmatrix{a+b&0\\0&a-b} \pmatrix{1&1\\1&-1}
\\ & = \frac 12 \pmatrix{1&1\\1&-1}\pmatrix{a+b&0\\0&a-b}^{k}\pmatrix{1&1\\1&-1}
\end{align*}
$$
I'll let you figure out the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Evaluating $\int x \cdot \sqrt{x-1} \cdot \mathrm{d}x$ using substitution I know that my result is wrong. What did I do wrong?
Let $u = \sqrt{x-1}$. Then we find first:
$$ u^2 = x - 1 \Rightarrow x = u^2 + 1$$
$$ \Rightarrow \mathrm{d}x = 2u \cdot \mathrm{d}u $$
So:
$$\int x \cdot \sqrt{x-1} \cdot \mathrm{d}x = \int (u^2 + 1) \cdot u \cdot 2u \cdot \mathrm{d}u = 2 \cdot \int (u^4 + u^2) \mathrm{d}u$$
$$ = 2 \cdot (\frac{1}{5} u^5 + \frac{1}{2} u^2) + c = \frac{2}{5} u^5 + u^2 + c$$
$$ = \frac{2}{5} (\sqrt{x-1})^5 + x + c$$
Differentiating this does not give me $ x \cdot \sqrt{x-1}$, so it must be wrong. Which step went wrong and why?
| $$\int u^2\,\mathrm du=\frac13u^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Lipschitz constant for $f(x) = 1/(1+x^2)$ from the definition of Lipschitz Lipschitz constant for $f(x) = 1/(1+x^2)$ from the definition of Lipschitz
Bit confused on how to do this from the definition instead of using any derivatives
So far I've reduced it to
$$|f(x) - f(y)| \leqslant |(x+y)(x-y)|$$ by simplification and noting that $f(x) \le 1$.
| Correct me if wrong:
$|f(x)-f(y)| =|\dfrac{y^2-x^2}{(1+x^2)(1+y^2)}|$
$= \dfrac{|y-x||x+y|}{(1+x^2)(1+y^2)}.$
Note: $|x+y| \le |x|+|y| \le$
$ \sqrt{x^2+1} + \sqrt{y^2+1}\le$
$(1+ x^2) +(1+y^2).$
Hence:
$|f(x)-f(y)| \lt$
$ |x-y|( \dfrac{1}{1+y^2} +\dfrac{1}{1+x^2}) \le$
$2 |x-y|$.
Note :$ |x| =\sqrt{x^2} \lt \sqrt{x^2+1} \le x^2+1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find Matrix $B$ s.t $Rowspan(A)=Rowspan(B)$ And $(1,1,1)\in Columnspan(B)$
Let $A=\begin{pmatrix}
1 & 0 & 3 \\
2 & 1 & 5 \\
-1 & 2 &-5
\end{pmatrix}$
Find a matrix $B$ such that $Rowspan(A)=Rowspan(B)$ and $\begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}\in Columnspan(B)$
So to find $Rowspan(A)$ I row reduced $\begin{pmatrix}
1 & 0 & 3 \\
2 & 1 & 5 \\
-1 & 2 &-5
\end{pmatrix}$ to $\begin{pmatrix}
1 & 0 & 3 \\
0 & 1 & -1 \\
0 & 0 &0
\end{pmatrix}$ and because row operations do not change the row span so $B$ should span the same space of $CF(A)$ and $\begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}\in Columnspan(B)$
But where should I start?
| From here
$$A^{RREF}=\begin{pmatrix}
1 & 0 & 3 \\
0 & 1 & -1 \\
0 & 0 &0
\end{pmatrix}$$
you are done, indeed just consider
$$B=\begin{pmatrix}
1 & 0 & 3 \\
0 & 1 & -1 \\
1 & 0 &3
\end{pmatrix}$$
and note that $$B\cdot (1,1,0)=(1,1,1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Projection of the vector $\vec{b}=\left(\begin{smallmatrix}3\\-1\\4\end{smallmatrix}\right)$ in some given direction
Find the projection of the vector $\vec{b}=\begin{pmatrix}
3 \\ -1\\ 4
\end{pmatrix}$ in the direction
$(a)$ perpendicular to the plane: $$2x-2y+z=5.$$
$(b)$ parallel to the plane: $$2x-2y+z=5.$$
For the part $(a)$ projecting $\vec{b}=\begin{pmatrix}3 \\ -1\\ 4\end{pmatrix}$ to $\vec{a}=\begin{pmatrix}2 \\ -2\\ 1\end{pmatrix}$ will give me the perpendicular projection right?
For part $(b)$ I know I have to subtract what I found in a from a vector to find the projection of the vector parallel to the plane, but I'm kinda lost.
| Perpendicular
The plane $E = 2x-2y+z=5 $ is defined by the normal vector $\vec{n}=\begin{pmatrix}{2 \\ -2\\ 1}\end{pmatrix}$ and the projection of $\vec{b}$ on $\vec{n}$, we define it as $\vec{l}$, is obtained by $\,|l|\cdot \left(\frac{1}{|n|}\vec{n}\right)\,$ (See unit vector)
It follows:
$$|l| = |b|\cdot \cos(\alpha) = |b|\frac{\vec{b}\cdot \vec{n}}{|b|\cdot |n|} =
\frac{\vec{b}\cdot \vec{n}}{|n|} = 4,\quad\text{ and }\quad \frac{1}{|n|}\vec{n} = \frac{1}{3}\cdot\begin{pmatrix}{2 \\ -2\\ 1}\end{pmatrix}$$
$$\vec{l} = |l|\cdot \left(\frac{1}{|n|}\cdot\vec{n}\right) = 4\cdot\frac{1}{3}\cdot \begin{pmatrix}{2 \\ -2\\ 1}\end{pmatrix} = \begin{pmatrix}{\frac{8}{3} \\ -\frac{8}{3}\\ \frac{4}{3}}\end{pmatrix}$$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
Parallel
And the vector $\vec{t}$ ist parallel to the plane $E$. This is also the projection of $\vec{b}$ on $E$:
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
$$\vec{t} = \vec{b}-\vec{l} = \begin{pmatrix}{\quad 3 - \frac{8}{3}\\ -1 + \frac{8}{3}\\\quad 4 - \frac{4}{3}}\end{pmatrix} = \begin{pmatrix}{\frac13\\\frac53\\\frac83 }\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Factor $x^{80} - 1$ over $F_3$ Factor $x^{80} - 1$ over $F_3$
Effort
Factor $x^{80} - 1$ over $F_3$
$$\begin{aligned}
x^{40}-1&=(x^{20}+1)(x^{20}-1)
\\&=(x^{20}+1)(x^{10}+1)(x^{10}-1 )
\\&=(x^{20}+1)(x^{10}+1)(x^{5}+1 )(x^{5}-1 )
\end{aligned} $$
since we are in $F_3$ , $(x^{20}+1),(x^{10}+1),(x^{5}+1 )$ are factorable by $(x-2)$ and $x^5-1$ by $x-1$
so $$x^5-1= (x-1)(x^4+x^3+x^2+x+1) $$
what computer is saying that it factors to
(x + 1) * (x + 2) * (x^2 + 1) * (x^2 + x + 2) * (x^2 + 2*x + 2) * (x^4 +
x + 2) * (x^4 + 2*x + 2) * (x^4 + x^2 + 2) * (x^4 + x^2 + x + 1) * (x^4
+ x^2 + 2*x + 1) * (x^4 + 2*x^2 + 2) * (x^4 + x^3 + 2) * (x^4 + x^3 +
2*x + 1) * (x^4 + x^3 + x^2 + 1) * (x^4 + x^3 + x^2 + x + 1) * (x^4 +
x^3 + x^2 + 2*x + 2) * (x^4 + x^3 + 2*x^2 + 2*x + 2) * (x^4 + 2*x^3 + 2)
* (x^4 + 2*x^3 + x + 1) * (x^4 + 2*x^3 + x^2 + 1) * (x^4 + 2*x^3 + x^2 +
x + 2) * (x^4 + 2*x^3 + x^2 + 2*x + 1) * (x^4 + 2*x^3 + 2*x^2 + x + 2)
Been trying to find examples. I do not see the factors just how one would hypothetically find them. I think finding them takes a lot of work for a human. maybe it factors to something simple,beautiful and elegant but it does not seems to be the case unless I am wrong.
| Note that $x^{81}-x = x(x^{80}-1)$. $81 = 3^4$, so $x^{81}-x$ factors into all the irreducible polynomials of degree $1,2,4$ over $\mathbb{F}_3$. So we could do this by listing polynomials of degrees $1,2,4$ and discovering which are irreducible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
If $(a,b)=1$, then $ \gcd(a^2+b^2,a^3+b^3)\mid (a-b)$
If $(a,b)=1$, then $\gcd(a^2+b^2,a^3+b^3)\mid (a-b)$
The only way that can help is to find some common factor of $a^2+b^2$ and $a^3+b^3$.
That does not seem obvious enough, so will directly try to divide $a^3+b^3$ by $a^2+b^2$.
This leads to nowhere too.
It seems that need to use the fact that $a,b$ are relatively prime, but am unable to use that.
Let for some suitable integers $x, y$, have $ax +by =1$.
| Say $$d=\gcd(a^2+b^2,a^3+b^3)$$ then $d\mid (a+b)(a^2+b^2-ab)$ and $d\mid a^2+b^2$ so, we get: $$d\mid ab(a+b) = (a+b)(a^2+b^2)- (a+b)(a^2+b^2-ab)$$
Now suppose there is prime $p$ such that $p\mid d$ and $p\mid a$. Then $p\mid a^2$ and so $p\mid b^2 = (a^2+b^2)-a^2$. A contradiction, since $a,b$ are relatively prime. So $a,d$ are relatively prime (and $b,d$ also) and $$d\mid a+b\Longrightarrow d\mid a(a+b)=a^2+ab$$
so $$ d\mid (a^2+ab)- (a^2+b^2)= b(a-b)$$
By Euclid lemma we have $d\mid a-b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Volume of an ellipsoid using cylindrical polar coordinates I've been working on a question about finding the volume of an ellipsoid
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1.$$
This is fine if I consider rescaling the axes to give a sphere, but I wanted to try to solve the problem specifically using polar coordinates, $(\rho, \Phi, z)$ in a triple integral. My thoughts (to find the limits of integration) were as follows:
*
*$z$ varies between $-c$ and $c$; by fixing $z$, we may consider the ellipse given by
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} =1- \frac{z^2}{c^2}$$
*$\Phi$ varies between $0$ and $2\pi$; by further fixing the angle $\Phi$, we have that $\rho$ varies between $0$ and $\sqrt{(1-\frac{z^2}{c^2})(a^2cos^2\Phi + b^2sin^2\Phi)}$, the horrible square root just being $\sqrt {x^2+y^2}$ in a different form.
However, evaluating $\iiint \rho\, dz\,d\Phi\, d \rho$ with these limits gives me $\frac{2\pi}{3}c(a^2+b^2)$ which is clearly wrong. I'm guessing my mistake is with my limits in the $\rho$ integral and would appreciate some guidance.
| From the equation you should have
$$\rho^2=x^2+y^2=b^2-\frac{b^2}{c^2}z^2\left(1-\frac{b^2}{a^2}x^2\right)$$
thus $\rho$ varies from $0$ to
$$\sqrt{b^2-\frac{b^2}{c^2}z^2+\left(1-\frac{b^2}{a^2}\cos^2\theta\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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I would like to find the center of this group
Show that the center of $SL(2, \mathbb{C})$ is $\pm Id.$ Hint: Use $
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
$, $
\begin{pmatrix}
1 & 0 \\
1 & 1 \\
\end{pmatrix}
$.
The center of a group is $C(G) = \{ g \in G : g\ast h = h \ast g \quad \forall h \in G \}.$ And it is true that $
\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
$, $
\begin{pmatrix}
-1 & 0 \\
0 & -1 \\
\end{pmatrix}
$, $
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
$, $
\begin{pmatrix}
1 & 0 \\
1 & 1 \\
\end{pmatrix}
$, have determinant $1$, hence, they are in the group, and particularily, the suggested matrices commute with both $\pm Id.$ But this doesn't prove that only two matrices, $Id, -Id$, are in the center. And It doesn't show that, $Id, -Id$ commutes with all the matrices in the group. So why would I pick those matrices in the first place? What am I missing?
| Suposse that for all $B\in\text{PSL}(2,\mathbb{C})$ holds $AB=BA$. We want to prove $A=\pm\text{Id}$. Let $A=\left(\begin{array}{cc} a & b\\ c & d\end{array}\right)$. Because holds for all matrix, then, in particular $B=\left(\begin{array}{cc} 1 & 0\\ 1 & 1\end{array}\right)$. thus, we have that $$\left(\begin{array}{cc} a+b & b\\ c+d & d\end{array}\right)=\left(\begin{array}{cc} a & b\\ c & d\end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 1 & 1\end{array}\right)=\left(\begin{array}{cc} 1 & 0\\ 1 & 1\end{array}\right)\left(\begin{array}{cc} a & b\\ c & d\end{array}\right)= \left(\begin{array}{cc} a & b\\ a+c & b+d\end{array}\right)$$So we have that $a+b=a$, $b=b$, $c+d=a+c$, $b+d=d$. Then, we conclude that $b=0$. In other way, also the above is $B=\left(\begin{array}{cc} 1 & 1\\ 0 & 1\end{array}\right)$. So, we have that $$\left(\begin{array}{cc} a & b+a\\ c & d+c\end{array}\right)=\left(\begin{array}{cc} a & b\\ c & d\end{array}\right)\left(\begin{array}{cc} 1 & 1\\ 0 & 1\end{array}\right)=\left(\begin{array}{cc} 1 & 1\\ 0 & 1\end{array}\right)\left(\begin{array}{cc} a & b\\ c & d\end{array}\right)=\left(\begin{array}{cc} a+c & b+d\\ c & d\end{array}\right)$$Thus, $a+c=a$, $b+d=b+a$, $c=c$ y $d+c=d$. We conclude that $c=0$ and $d=a$. Well then, we have that $b=c=0$ and that $d=a$. Also, $ad-bc=ad=1$. Thus, $a=d=1$ or $a=d=-1$. Therefore, $A=\left( \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)$ or $A=\left(\begin{array}{cc} -1 & 0\\ 0 & -1\end{array}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$A^n = I$ only if $n\equiv 0 \pmod{4}$ Is the following proof correct? In addition could you pleases suggest a more cleaner or shorter proof.
Theorem. Given any $n\in\mathbb{Z^+}$ then
$$\begin{pmatrix}
0&0&-1\\
0&1&0\\
1&0&0\\
\end{pmatrix}^n
=
\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{pmatrix}
$$
only if $4 \mid n$.
Proof. Assume that $n\in\mathbb{Z^+}$ and let $A$ denote the matrix
$$
\begin{pmatrix}
0&0&-1\\
0&1&0\\
1&0&0\\
\end{pmatrix}
$$
We prove the contrapositive. Given the standard choice of basis for $\mathbb{R}^3$ we can see $A$ represents a $90^{\circ}$ clockwise rotation about the $y$-axis consequently we may write out $A^2$, $A^3$ and $A^4$ as follows:
$$
A^2 =
\begin{pmatrix}
-1&0&0\\
0&1&0\\
0&0&-1\\
\end{pmatrix}
$$
$$
A^3 =
\begin{pmatrix}
0&0&1\\
0&1&0\\
-1&0&0\\
\end{pmatrix}
$$
$$
A^4 =
\begin{pmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{pmatrix}
$$
Now assume that $4\not\mid n$ then the division theorem implies that $n$ is of the form $4q+1$, $4q+2$ or $4q+3$. Considering the case where $4q+1$ we can see that
$$A^{4q+1} = (A^{4})^q\cdot A = I^q\cdot A = I\cdot A\neq I.$$
By similar reasoning we can make the same conclusion in the other two cases.
$\blacksquare$
| Yes, it is correct, but I would not use a geometric argument here, since a very simple computation is enough to obtain $A^n$ for $n\in\{2,3,4\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the maximum value of $ax+ by$ $$ x^2 + xy + y^2 = t^2 $$
Find the maximum value of $ax + by$
One way of doing this is substituting
$ x = r \cos w $ and $ y= r \sin w $
Then using calculus we can find the maximum value but this is a very lengthy process
So I wanted to know if there is a shorter way of doing this
| $$(ax+by)^2\leq\frac{4}{3}(a^2-ab+b^2)(x^2+xy+y^2)$$ it's
$$((a-2b)x+(2a-b)y)^2\geq0.$$
The equality occurs for $(a-2b)x+(2a-b)^2y=0$ and $x^2+xy+y^2=t^2.$
Thus, $$\max\limits_{x^2+xy+y^2=t^2}(ax+by)=\frac{2}{\sqrt3}|t|\sqrt{a^2-ab+b^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2640403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $\tan^{-1} \left(\frac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$, then prove $x^2=\sin(2\alpha)$ If $\tan^{-1} \left(\dfrac {\sqrt {1+x^2} - \sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) = \alpha$ then prove that: $x^2= \sin (2\alpha) $
My Attempt:
$$\tan^{-1} \left(\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}\right) =\alpha$$
$$\dfrac {\sqrt {1+x^2}-\sqrt {1-x^2}}{\sqrt {1+x^2} + \sqrt {1-x^2}}=\tan (\alpha )$$
$$\dfrac {1+x^2-2\sqrt {1+x^2}.\sqrt {1-x^2}+ 1 - x^2}{1+x^2-1+x^2}=\tan (\alpha)$$
$$\dfrac {1-\sqrt {1+x^2}.\sqrt {1-x^2}}{x^2}=\tan (\alpha)$$
| Let $A=\sqrt{1+x^2}$ and $B=\sqrt{1-x^2}$, then
$$\tan(\alpha) = \frac{A-B}{A+B}, \quad A^2+B^2=2,\quad A^2-B^2 = 2x^2$$
and also note
$$\tan(a) = \underbrace{\frac{\sin(2a)}{2\sin(a)\cos(a)}}_{=\,1} \cdot \frac{\sin(a)}{\cos(a)}= \sin(2a)\cdot\frac{1}{2}\sec^2(a) = \sin(2a) \frac{1}{2} \left(1+\tan^2(a)\frac{}{}\right)$$
hence,
\begin{align}
\sin(2a)
&= \frac{2 \tan(a)}{1+\tan^2(a)}
= \frac{2 \tan(a)}{1+\tan^2(a)} \cdot \frac{\big(A+B\big)^2}{\big(A+B\big)^2}\\
&= \frac{2 (A+B)(A-B)}{(A+B)^2 + (A-B)^2}
= \frac{2 (A^2-B^2)}{2A^2+2B^2} = x^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2644104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Laurent series of complex function So I want to calculate the Laurent series of this function
$$ f: \mathbb{C} \to \mathbb{C}, \quad f(z) = \frac{1}{z^{2}+1}.$$
The Laurent series has to be in this form:
$$\sum_{n=- \infty }^{ \infty } a_{n} (z-i)^n$$
for a circular disc
$$ 0<| z-i|<p,$$
where $p$ has to be found.
With partial fraction expansion I am getting $$ f(z) =\frac{i}{2}\left( \frac{1}{z+i} - \frac{1}{z-i}\right).$$
For the first summand, $$\frac{1}{z+i} = \frac{1}{2i} \frac{1}{1+\frac{z-i}{2i}} = \frac{1}{2i} \sum_{n= 0 }^{ \infty }\left(\frac{-(z-i)}{2i}\right)^n = \frac{1}{2i} \sum_{n= 0 }^{ \infty } \left(\frac{i}{2}\right)^n (z-i)^n $$
for $$\left|\frac{-(z-i)}{2i}\right| < 1 \Longrightarrow \left| z-i \right| < 2.$$
Now I don't know how to continue with $$\frac{1}{z-i} .$$
| Note: Despite the question at the end regarding how to continue note that already all calculations were done in order to solve the problem.
The function
\begin{align*}
f: \mathbb{C} \to \mathbb{C}, \quad f(z) &= \frac{1}{z^{2}+1}\\
&=\frac{i}{2}\left( \frac{1}{z+i} - \frac{1}{z-i}\right)\tag{1}
\end{align*}
is to expand in a Laurent series at $z=i$.
We observe in (1) that
\begin{align*}
-\frac{i}{2}\cdot\frac{1}{z-i}\tag{2}
\end{align*}
is the principal part of the Laurent series of $f$. On the other hand we know that
\begin{align*}
\frac{i}{2}\cdot\frac{1}{z+i}=\frac{1}{4} \sum_{n= 0 }^{ \infty } \left(\frac{i}{2}\right)^n (z-i)^n\tag{3}
\end{align*}
is the power series representation of $f$ at $z=i$ with region of convergence $|z-i|<2$.
We conclude from (2) and (3) the Laurent series expansion of $f$ around $z=i$ is
\begin{align*}
f(z)&=\frac{1}{z^{2}+1}\\
&\color{blue}{=-\frac{i}{2}\cdot\frac{1}{z-i}+\frac{1}{4}\sum_{n= 0 }^{ \infty } \left(\frac{i}{2}\right)^n (z-i)^n}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A person rolls five fair dice simultaneously. Find the probability that the total number of spots showing is less than or equal to $11$ I am having trouble with this problem. When they say spot I think they are essentially saying the sum, so its the probability that the sum of dice is $11$ or less.
I understand that there are $6^5$ combinations.
I found 6 ways that it can equal to $11$ $(2,3,2,2,2)(3,3,1,1,3),(4,4,1,1,1),(5,2,2,1,1),(6,2,1,1,1)(7,1,1,1,1)$, but I know there has to be an easier way than just counting. Is it like $\binom{6}{5}$? Thanks for the help.
| We can solve the problem using generating function.
\begin{align}\left(\frac16\sum_{i=1}^6x^6\right)^5=\frac{1}{6^5}\left( \frac{x(1-x^6)}{1-x}\right)^5 \end{align}
Our goal is to find the sum of coefficients of $x^5$ to $x^{11}$.
$$(1-x^6)^5 =1-5x^6+\text{higher order terms} $$
By negative binomial series
$$(1-x)^{-5}=\sum_{k=0}^\infty \binom{4+k}{k}x^k$$
Hence,
\begin{align}\left(\frac16\sum_{i=1}^6x^6\right)^5=\frac{1}{6^5}x^5(1-5x^6+\text{higher order terms})\sum_{k=0}^\infty \binom{4+k}{k}x^k\end{align}
and the sum of coefficients is
$$\frac1{6^5}\left(\sum_{k=0}^{6}\binom{4+k}{k}-5\binom{4+0}{0} \right)=\frac1{6^5}\left(\sum_{k=0}^{6}\binom{4+k}{4}-5 \right)=\frac{462-5}{6^5}=\frac{457}{7776}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integration with Cos function in denominator I need to solve following integration:
$$\int_0^{\frac{\pi }{2}} \frac{1}{\left(a \cos ^2(x)+1\right)^m} \, dx$$
In Mathematica, the solution is given with $\text{Hypergeometric2F1}$ function. However, I want to know how to derive this step-by-steps. Can someone please guide me?
I started by applying the negative binomial expansion, however it works only when $a \cos ^2(x)<1$. I am wondering that if there is any general expression !!!
| So as @Kiryl Pesotski suggests, we need to rewrite your integral in the form
$$_2 F_1(a,b;c;x) = \frac{\Gamma (c)}{\Gamma (b) \Gamma (c- b)} \int_0^1 \frac{t^{b - 1} (1 - t)^{c - b - 1}}{(1 - xt)^a} \, dt.\tag1$$
For convergence, I will assume the $a$ appearing in your integral is greater than negative one $(a > -1)$.
Since $\cos^2 x = 1 - \sin^2 x$, we begin by writing the integral as
$$I = \int_0^{\pi/2} \frac{dx}{(a + 1 - a \sin^2 x)^m} = \frac{1}{(a + 1)^m} \int_0^{\pi/2} \frac{dx}{\left (1 - \frac{a}{a + 1} \sin^2 x \right )^m}.$$
Setting $t = \sin^2 x$, since $\cos x = \sqrt{1 - t}$ then $dx = dt/(2 \sqrt{t} \sqrt{1 - t})$ while for the limits of integration $(0,\pi/2) \mapsto (0,1)$. Thus
\begin{align*}
I &= \frac{1}{2(a + 1)^m} \int_0^1 \frac{t^{-1/2} (1 - t)^{-1/2}}{\left (1 - \frac{a}{a + 1} t \right )^m} \, dt\\
&= \frac{1}{2 (a + 1)^m} \cdot \frac{\Gamma (1/2) \Gamma (1/2)}{\Gamma (1)} \cdot \frac{\Gamma (1)}{\Gamma (1/2) \Gamma (1/2)} \int_0^1 \frac{t^{1/2 - 1} (1 - t)^{1 - 1/2 - 1}}{\left (1 - \frac{a}{a + 1} t \right )^m} \, dt.
\end{align*}
Now the above integral is exactly in the form of (1) where $a = m, b = 1/2, c = 1$, and $x = a/(a+1)$. So in terms of the hypergeometric function the integral we started out with can be written as
$$\int_0^{\pi/2} \frac{dx}{(a \cos^2 x + 1)^m} = \frac{\pi}{2(a + 1)^m}\ _2 F_1 \left (m,\frac{1}{2};1; \frac{a}{a + 1} \right ).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Limit of the sequence $\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$ As in the title, we have to calculate the limit $$\frac{1}{n^2+1}+\frac{2}{n^2+2}+...+\frac{n}{n^2+n}$$ when $n\rightarrow+\infty$. It is easy to see that $$\lim_{n\to\infty} \frac{1}{n^2+1}+\frac{1}{n^2+2}+...+\frac{1}{n^2+n}=0$$ and $$\lim_{n\to\infty} \frac{n}{n^2+1}+\frac{n}{n^2+2}+...+\frac{n}{n^2+n}=1$$
I don't know how to calculate the limit. I am pretty sure that we somehow apply the sandwich rule, but I am afraid that there is something similary with $$\lim_{n\to\infty} \frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{n^2}=\frac{\pi^2}{6}$$
| Replace all the denominators by $n^2+n$, and then by $n^2+1$, to use squeeze theorem and obtain the answer to be $\frac 12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Strange Trig Substitution in Integral Compute the following integral
$$\int _0^{0.6}\:\frac{x^2}{\sqrt{9-25x^2}}\mathrm dx$$
So as I have been doing in the past, I thought the required substitution would have been $x = 3\sin\theta$. But in this case, it is actually $x = \frac{3}{5}\sin\theta$. I am not entirely sure why this is. Is it because the coefficient of the $x^2$ in the denominator is also a square number so we can square root it and place it in the denominator of the substitution? Is this a universal rule?
| Maybe to make things clear for you, you need to perform two substitutions.
$$\begin{align}
J=\int_0^{0.6}\frac{x^2}{\sqrt{9-25x^2}}\,dx
&=\int_0^{0.6}\frac{x^2}{3\sqrt{1-\left(\frac{5}{3}x\right)^2}}\,dx
\end{align}$$
Perform the change of variable $y=\frac{5}{3}x$,
$$\begin{align}
J&=\frac{3}{5}\int_0^{1}\frac{\left(\frac{3}{5}y\right)^2}{3\sqrt{1-y^2}}\,dy\\
&=\frac{9}{125}\int_0^{1}\frac{y^2}{\sqrt{1-y^2}}\,dy
\end{align}$$
Since $0.6\times \frac{5}{3}=1$.
The integral bounds belong to $[-1;1]$ and there is the factor $\dfrac{1}{\sqrt{1-y^2}}$ therefore you know the substitution $y=\sin u$ is a way to complete your computation.
Perform the substitution $y=\sin u$,
$$\begin{align}
J&=\frac{9}{125}\int_0^{1}\frac{\sin^2 u}{\sqrt{1-\sin^2 u}}\times \cos u\,du\\
&=\frac{9}{125}\int_0^{1}\sin^2 u\,du\\
\end{align}$$
The substitution works because,
$$\begin{align} \frac{1}{\sqrt{1-y^2}}\,dy&=\frac{\cos u}{\sqrt{1-\sin^2 u}}\,du\\
&=\,du
\end{align}$$
(the derivative of the sinus function is the cosinus function)
No more denominator.
To achieve the computation you need to linearize $\sin^2 u$.
If you know that
$$\begin{align}\cos(2u)&=\cos^2 u-\sin^ 2 u\\
&=\left(1-\sin^2 u\right) -\sin^2 u\\
&=1-2\sin^2 u\\
\end{align}$$
you are done.
But, there is an algorithmic way to tackle the question.
$$\begin{align}\sin^2 u&=\left(\frac{\text{e}^{iu}-\text{e}^{-iu}}{2i}\right)^2\\
&=-\frac{1}{4}\left(\text{e}^{i2u}+\text{e}^{-i2u}-2\times \text{e}^{iu}\times \text{e}^{-iu}\right)\\
&=-\frac{1}{4}\left(\text{e}^{i2u}+\text{e}^{-i2u}-2\right)\\
&=\frac{1}{2}-\frac{1}{2}\cos(2u)
\end{align}$$
Since,
$$\displaystyle \sin u=\frac{\text{e}^{iu}-\text{e}^{-iu}}{2i},\cos u=\frac{\text{e}^{iu}+\text{e}^{-iu}}{2},i^2=-1$$
Therefore,
$$\begin{align}J&=\frac{9}{125}\int_0^1 \left(\frac{1}{2}-\frac{1}{2}\cos(2u)\right)\,du\\
&=\frac{9}{125}\Big[\frac{1}{2}u-\frac{\sin(2u)}{4}\Big]_{0}^{\frac{\pi}{2}}\\
&=\frac{9}{125}\times \frac{\pi}{4}\\
&=\boxed{\frac{9}{500}\pi}
\end{align}$$
NB:
$\sqrt{1-x^2}$ does exist if and only if $x\in [-1;1]$
You can compute integral $\displaystyle \int_0^1 \frac{P(x)}{\sqrt{1-x^2}}\,dx$ , $P$ is a polynomial function.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $(x,y) \in \Bbb R^+$ Prove that $\Bigr(1+\frac{1}{x}\Bigl)\Bigr(1+\frac{1}{y}\Bigl)\ge \Bigr(1+\frac{2}{x+y}\Bigl)^2$ Let $(x,y) \in \Bbb R^+$ Prove that $$\Bigr(1+\frac{1}{x}\Bigl)\Bigr(1+\frac{1}{y}\Bigl)\ge \Bigr(1+\frac{2}{x+y}\Bigl)^2$$
My try
Well, i didn't see a way to factorize this, so i put it in WolframAlpha and i got that we can rewrite it like this, which is obviously true for $(x,y) \in \Bbb R^+$:
$$\frac{(y-x)^2(x+y+1)}{xy(x+y)^2}\ge 0$$
But i don't see the way to get there manually, im stuck here:
$$\frac{(x+y)(y+1)}{xy}\ge \frac{(x+y+2)^2}{(x+y)^2}$$
| I give a proof using Cauchy-Schwartz for fun.
$$\Bigr(1+\frac{1}{x}\Bigl)\Bigr(1+\frac{1}{y}\Bigl)\ge \Bigr(1+\frac{2}{x+y}\Bigl)^2$$
$$\iff \Bigr((x+y)(1+\frac{1}{x})\Bigl)\Bigr((x+y)(1+\frac{1}{y})\Bigl)\ge \Bigr(x+y+2\Bigl)^2$$
$$\iff \Bigr(x+y+1+\frac{y}{x}\Bigl) \Bigr(x+y+1+\frac{x}{y}\Bigl) \ge \Bigr(x+y+2\Bigl)^2$$
$$\iff \Bigr(\sqrt x^2+\sqrt y^2+1^2+\sqrt{\frac{y}{x}}^2\Bigl) \Bigr(\sqrt x^2+\sqrt y^2+1^2+\sqrt{\frac{x}{y}}^2\Bigl) \\
\ge (\sqrt x \cdot \sqrt x + \sqrt y \cdot \sqrt y + 1\cdot1 + \sqrt{\frac{y}{x}} \cdot \sqrt{\frac{x}{y}})^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is this answer to this probability question correct?
A class of $400$.
• $220$ women.
• $180$ men.
Divided into study groups of 10 students each.
Q. What is the probability that a certain group will have exactly $4$ men?
$$ P(x) = \frac{n!}{x! \cdot (n-x)!} \cdot p^x \cdot (1-p)^{(n-x)}$$
$ n=4 \qquad x=4 \qquad p=0.45 $
$ P(x) = \frac{10!}{4! \cdot (10-4)!} \cdot 0.45^4 \cdot (1-0.45)^{(10-4)} $
$ \qquad \,\, = 23.84\%$
This is from the Binomial Distributions topic of a Udemy course I'm taking. That's the given answer. This seems like how to find the probability of a certain group having four men if each person was chosen from $10$ different groups where probability of choosing a man is $0.45$.
So, is that answer correct? What am I missing?
| Lets make a small example: You have a group of 2 men and 3 woman. Now you want divide into a study group of 3 students with 2 women ($Y$). The ways to do that is
$\texttt{wwm, wmw, mww}$
Each group has the same probability. Let us focus on the first group. The probability to choose a woman first is $\frac{3}{2+3}=\frac{3}{5}$
The probability to choose woman again is $\frac{3-1}{2+3-1}=\frac{2}{4}=\frac{1}{2}$
Here you can see that you have to use a distribution which regard, that the woman are not replaced.
The probability to choose a man is $\frac{2}{2+3-2}=\frac{2}{3}$
Thus the probability to choose $\texttt{wwm}$ is $\frac{3}{5}\cdot \frac{1}{2}\cdot \frac{2}{3}=\frac1{5}=20\%$.
Since we have three ways to choose $2$ woman, the probability to choose $2$ woman is $P(Y=2)=60\%$. This is equal to
$$P(Y=2)=\frac{\binom{3}{2}\cdot \binom{2}{1}}{\binom{5}{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654348",
"timestamp": "2023-03-29T00:00:00",
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An Infinite Series Contradiction Let A represent $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+......$ and B represent $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+......$
$$A+B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+......=2A-1$$
and then
$$A-B=1$$
thus
$$\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+......=1$$
Now set N as $\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{7}-\frac{1}{8}\right)+......$, a positive number
$$N+(A-B)=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+......=\frac{1}{2}$$
$$N+1=\frac{1}{2}$$
$$N=-\frac{1}{2}$$
What is wrong with this logic?
| I will try to point some things out, but I would strongly advise the OP to read about absolute convergence, conditional convergence and finally, summation of divergent series which they seem to be interested in.
First things first, as the comments pointed out, we can't manipulate with divergent series. They do not represent any real number and thus can not be added or subtracted to give a real number. We'd have the same luck adding the color blue and an apple and getting $\pi$ as a result.
A probably more striking example of such sensless games is this:
$$S=1+2+4+\dots+2^n+\dots \\ 2S=S-1 \\ S = - 1$$
This doesn't make sense, because adding all these terms we will never get the number on the right hand side (though again, I urge the OP to look up summation of divergent series).
Much more interesting topic is conditionally convergent series.
The most well known is the alternating harmonic series:
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+ \dots = \log 2$$
This one is not just formal, it really works. If we add the terms exactly in the same order as they are written. If we change the order of a finite number of terms, it's ok. Rules of arithmetic will hold. But if we change the order of the infinte number of terms, the value will change as well.
This can be seen by considering the general terms:
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$$
Since we are going in order, we can group the successive even and odd terms in pair, and that will not change the sum:
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=\sum_{k=1}^\infty \left(\frac{1}{2k-1}-\frac{1}{2k} \right)=\sum_{k=1}^\infty \frac{1}{2k(2k-1)} $$
The general term is now always positive and the series we obtained converges absolutely, as can be seen from the denominator. Indeed, we can write:
$$\left(1-\frac{1}{2} \right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+ \dots = \log 2$$
In the same way we can group every $4$ terms, every $6$ terms and such, the value of the series will stay the same.
So what happens if we group the terms out of order? We will either get another value, or even get the series to diverge. Which is why they are called 'conditionally convergent'.
Let's now group the terms in another way:
$$\left(1-\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{6}-\frac{1}{8}\right)+\left(\frac{1}{5}-\frac{1}{10}-\frac{1}{12}\right)+ \dots = ?$$
To find the value of this new series (made of exactly the same terms with the same signs) we need to find the formula for the general term:
$$\frac{1}{2k-1}-\frac{1}{4k-2}-\frac{1}{4k}=\frac{1}{4k(2k-1)}$$
But we can see from the previous formula that:
$$\sum_{k=1}^\infty \frac{1}{4k(2k-1)}= \frac{\log 2}{2}$$
Somehow, we have obtained half the value of the original series.
Let's now try another way:
$$\left(1+\frac{1}{3}-\frac{1}{2}\right)+\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{4}\right)+\left(\frac{1}{9}+\frac{1}{11}-\frac{1}{6}\right)+ \dots = ?$$
$$\frac{1}{4k-3}+\frac{1}{4k-1}-\frac{1}{2k}=\frac{8k-3}{2k(4k-3)(4k-1)}$$
Looking at the general term, we can see that the new series converges absolutely. Its value now:
$$\sum_{k=1}^\infty \frac{8k-3}{2k(4k-3)(4k-1)}= \frac{3\log 2}{2}$$
Unlike the OP's manipulations, all three series I obtained are absolutely convergent and really give us the values on the right hand side (or approximations to them, for a finite number of terms).
| {
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Expressions that equate to a square $4x+1 = a^2$
Where a and x is an integer, what are all possible value of x?
I always encounter these types of question even in the form of
$ax^2 + bx + c$
How do I approach these kinds of questions?
| You are looking for $a$ such that the remainder of dividing $a^2$ by $4$ is $1$ (then $x$ is the quotient). The possible remainders when dividing by $4$ are $0$, $1$, $2$, and $3$. Suppose the remainder of dividing $a$ by $4$ is $r$. Then the remainder for $a^2$ is directly related to $r^2$: $a=4x+r$ implies $$a^2=(4x+r)^2=16x^2+8x+r^2=4(4x^2+2x)+r^2,$$
where if $r^2>4$ then we should extract factors of 4 from it (and add them to the quotient) until it turns below 4, in order to get a true remainder.
If $r=0$ then $r^2=0$; if $r=1$ then $r^2=1$; if $r=2$ then $r^2=4$, which changes to $0$ (by adding 1 to the quotient) since $r^2<4$. If $r=3$ then similarly $r^2=9$ changes to $1$. Therefore the remainder of $a^2$ is $1$ if and only if the remainder of $a$ is either 1 or 3, $a=4c+1$ or $a=4c+3$ for some $c$: 1 and 3 (c=0), 5 and 7 (c=1), etc. So the solutions are all odd integers, since $a=4c+1$ or $a=4c+3$ is equivalent to $a=2d+1$ for some $d$.
By the way, if you are interested in this kind of reasoning, you sould learn about modular arithmetic, which studies the algebra of remainders. In this example we would say that $a^2\equiv 1(\mod 4)$ implies $a\equiv 1,3(\mod 4)$ and also that $r\equiv 9\equiv 1(\mod 4)$.
| {
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Find the coefficient of $x^{18}$ in $(x + x^2 + x^3 + x^4 + x^5)(x^2 + x^3 + x^4 +\cdots)^5 $
Find the coefficient of $x^{18}$ in $$(x + x^2 + x^3 + x^4 + x^5)(x^2 + x^3 + x^4 +\cdots)^5 $$
This is the first time coming across a generating function question and am not quite sure how to solve this and am looking for some help, thanks!
| You can rewrite your expression as $x^{11}(1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+\ldots)^5$ so you are looking for the coefficient of $x^7$ in $$(1+x+x^2+x^3+x^4)(1+x+x^2+x^3+x^4+\ldots)^5$$ This is the number of ways to write $7$ as the sum of six nonnegative integers with the first being at most four. Stars and bars gives you the number without the restriction on the first as ${12 \choose 5}$. Now subtract off the number of ways to express $2,1,0$ as five nonnegative integers, so we have $${12 \choose 5}-{6 \choose 4}-{5 \choose 4}-{4 \choose 4}$$
| {
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How to show that for each $n\in\mathbb{N}$, $\gcd(3^{n}+5^{n+1},3^{n+1}+5^{n})=2$ or $ 14$? For each $n\in\mathbb{N}$ show that
$$
\gcd(3^{n}+5^{n+1},3^{n+1}+5^{n})=2\text{ or } 14.
$$
I tried induction but I got stuck.
| They are both odd, so their gcd is even. Moreover
\begin{align}
\gcd(3^n+5^{n+1},3^{n+1}+5^n)&=\gcd(3^n+5^{n+1}-5(3^{n+1}+5^n),3^{n+1}+5^n)\\
&=\gcd(3^n (1-3\cdot 5),3^{n+1}+5^n) \\
&=\gcd(3^n \cdot 14,3^{n+1}+5^n) \\
&=\gcd(14,3^{n+1}+5^n)
\end{align}
divides $14$. So it can be only $2$ or $14$.
| {
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Determinant using factor theorem
Prove
$$\Delta=\begin{vmatrix}
(y+z)^2 & x^2 & x^2 \\
y^2 & (z+x)^2 & y^2 \\
z^2 & z^2 & (x+y)^2 \\
\end{vmatrix} = 2xyz(x+y+z)^3$$
using factor theorem.
This is solved in Demonstrate using determinant properties that the determinant of A is equal to $2abc(a+b+c)^3$ using factor theorem.
My Attempt:
$$
x=0\text{ or }y=0\text{ or }z=0\implies\Delta=0\text{ , So $x,y,z$ are factors of }\Delta.\\
(x+y+z)=0\implies \Delta=\begin{vmatrix}x^2&x^2&x^2\\y^2&y^2&y^2\\z^2&z^2&z^2\end{vmatrix}=0\text{ , So $(x+y+z)$ is a factor of $\Delta$.}
$$
$\color{black}{\text{But how do i extract the remaining term $(x+y+z)^2$ to prove $\Delta=2xyz(x+y+z)^3$ }\color{red}{ ?}}$
Similar Example:
Please check answer of @user348749 in How to solve this determinant,
$$
\Delta'=\begin{vmatrix}
(b+c)^2&ab&ca\\
ab&(a+c)^2&bc\\
ac&bc&(a+b)^2
\end{vmatrix}=2abc(a+b+c)^3
$$
it is said that
$$
(a+b+c)=0\implies\Delta'=\begin{align*}
\begin{vmatrix}
c^2 & ca & bc \\
ca & a^2 & ab \\
bc & ab & b^2 \\
\end{vmatrix}
=abc\begin{vmatrix}
c & a & b \\
c & a & b \\
c & a & b \\
\end{vmatrix}
\end{align*}=0
$$
"Since all rows are identical, $(a+b+c)^2$ is a factor. The determinant is a polynomial of degree 6 and hence the remaining factor is linear and since it is symmetric, the factor must be $k(a+b+c)$."
$\color{black}{\text{How can we say this }\color{red}{ ?}}$
My Understanding:
If the problem was similar to this, answer of @Saibal in Factorise a matrix using the factor theorem,
$$\Delta''=
\begin{vmatrix}
x&y&z\\
x^2&y^2&z^2\\
x^3&y^3&z^3\\
\end{vmatrix}$$
I could without doubt do as below:
$$
x=0\text{ or }y=0\text{ or }z=0\implies\Delta''=0\\
x=y\text{ or }y=z\text{ or }z=x\implies\Delta''=0
$$
Thus, $x,y,z,(x-y),(y-z),(z-x)$ are factors of $\Delta''$. ie. $\Delta''=kxyz(x-y)(y-z)(z-x)$
| Your attempt is fine. And all the further explanations you need are already given by user348749 in How to solve this determinant
I could just rephrase this here: Since for $x+y+z=0$, all three columns are identical (same argument as all three rows), $(x+y+z)^2$ is a factor (see user348749's explanation). So the remaining factor is linear. Since $\Delta$ does not change under $x \leftrightarrow y$ or any other permutation, the remaining linear factor must be symmetric under these exchanges, and only $C \cdot (x+y+z)$ is such a factor. Again, not my arguments so far, but user348749's.
The thing that remains to be done is to calculate the constant. Choose e.g. $x=y=z=1$ and you get
$$
\Delta=\begin{vmatrix}
4 & 1 & 1 \\
1 & 4 & 1 \\
1 & 1 & 4 \\
\end{vmatrix} = 4\cdot 15 - 2\cdot 3 =54 = C \cdot 3^3 = C \cdot 27
$$
which clearly gives $C =2$.
| {
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Limit question using limit as a sum
$$\lim_{n\rightarrow \infty}\frac{1}{n}\bigg(\sum^{n}_{i=1}\left\lfloor \frac{2n}{i}\right\rfloor-2\left\lfloor\frac{n}{i}\right\rfloor \bigg)$$
Where $\lfloor x \rfloor$ is a floor function of $x$
Trial: using limit as a sum put $\displaystyle \frac{i}{n}=x$ and $\displaystyle \frac{1}{n}=dx$
$$\int^{1}_{0}\bigg(\left\lfloor\frac{2}{x}\right\rfloor -2\left\lfloor\frac{1}{x}\right\rfloor \bigg)dx$$
Let $\displaystyle \frac{1}{x}=t$ and $dx=-\frac{1}{t^2}dt$
So $\displaystyle \int^{1}_{0}\bigg(\lfloor 2t \rfloor - 2\lfloor t\rfloor \bigg)\cdot\frac{1}{t^2}dt$
Could some help me to solve it, Thanks
| Write $I$ for the integral and notice that
$$
I
= \int_{0}^{1} \left( \left\lfloor \frac{2}{x} \right\rfloor - 2 \left\lfloor \frac{1}{x} \right\rfloor \right) \, dx
= \int_{0}^{1} \left( \left\lfloor \frac{2}{x} \right\rfloor - \frac{2}{x} \right) \, dx + 2\int_{0}^{1} \left( \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor \right) \, dx
$$
(Without adding and subtracting the intermediate term $\frac{2}{x}$, you have $\infty-\infty$ indeterminate which is never good.) Now by the substitution $x \mapsto x/2$, the second integral becomes
$$ 2\int_{0}^{1} \left( \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor \right) \, dx
= \int_{0}^{2} \left( \frac{2}{x} - \left\lfloor \frac{2}{x} \right\rfloor \right) \, dx $$
Plugging this back, we find that
$$
I
= \int_{1}^{2} \left( \frac{2}{x} - \left\lfloor \frac{2}{x} \right\rfloor \right) \, dx
= \int_{1}^{2} \left( \frac{2}{x} - 1 \right) \, dx
= 2\log 2 - 1.
$$
| {
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How to cancel an irrational power We all learned in 9th/10th grade that problems like:
$$\sqrt{x+2}=4$$
$$(x+2)^\frac 3 4=4$$
Can be solved by exponentiation
$$\text{If}: \sqrt{x+2}=4, \text{ then}:x+2=4^2$$
$$If: (x+2)^\frac 3 4=4,\text{ then}:(x+2)^3=4^4$$
And we can then expand the second problem via binomial expansion or whatever we like.
I recently encountered a problem that looked like this:
$$\frac{a(ax^2+c-d)}{a^\sqrt2}=(\frac{a^\sqrt2}{a}x^\sqrt2+d)^\sqrt2$$
(attempting to solve for d)
Is this solveable?
Before we could take advantage of the fact that the exponent was rational and so we could take any expression
$$x^\frac n m=y^\frac p q$$
and transform it to
$$x^{nq}=y^{mp}$$
No trick like this will work in solving problems with irrational exponents. Is there an alternate strategy that allows for solving problems like this?
| Ignoring the nasty thornbush of how you define what $b^x$ can mean if $x$ is not rational and assuming we got that all worked out, we still have $(b^x)^y = b^{xy}$ if $b > 0$. (That's actually not at all trivial and stops being true for complex numbers and why this is true for reals is a significant result. But it is a question for another time.)
So for a problem like $x^{\sqrt 2} = 7$ we'd just say $(x^{\sqrt 2})^{\frac 1{\sqrt 2} } = x = 7^{\frac 1{\sqrt 2}}$.
Or we could do $(x^{\sqrt 2})^{\sqrt 2} = x^2 = 7^{\sqrt 2}$ so ($x \ge 0$ for $x^{\sqrt 2}$ to b defined) so $x = 7^{\frac {\sqrt 2}{2}}=7^{\frac 1{\sqrt 2}}$
You can do either of those on:
$\frac{a(ax^2+c-d)}{a^\sqrt2}=(\frac{a^\sqrt2}{a}x^\sqrt2+d)^\sqrt2$
$a(ax^2+c-d) = (\frac{a^\sqrt2}{a}x^\sqrt2+d)^\sqrt2 a^{\sqrt 2}$
$[a(ax^2 + c -d)]^{\sqrt 2} = ({a^\sqrt2}x^\sqrt2+da)^2$
$[a(ax^2 + c -d)]^{\frac {\sqrt 2}2} = {a^\sqrt2}x^\sqrt2+da$
which doesn't make it any easier. We'll have to come up with something else.
I'm not entirely sure what.
$\frac{a(ax^2+c-d)}{a^\sqrt2}=(\frac{a^\sqrt2}{a}x^\sqrt2+d)^\sqrt2$
$a^{1-\sqrt 2}(ax^2 + c +d) = (a^{\sqrt 2 - 1}x^\sqrt2 +d)^{\sqrt 2}$
$ax^2 + c + d = a^{\sqrt 2-1}(a^{\sqrt 2 - 1}x^\sqrt2 +d)^{\sqrt 2}$
$ax^2 + c + d = (a^{\sqrt 2-1 + \frac {\sqrt 2-1}{\sqrt 2}}x^\sqrt2 + da^{ \frac {\sqrt 2-1}{\sqrt 2}} )^{\sqrt 2}= (a^{\frac 1{\sqrt 2}}x^\sqrt2 +da^{ \frac {\sqrt 2-1}{\sqrt 2}} )^{\sqrt 2}$
Yeah, I got nothing.
In general $x^2 + k = w(x^{\sqrt{2}} + j)^{\sqrt 2}$ doesn't seem to be simplifiable.
We can transpose to $(x^2 + k)^{\frac 1 {\sqrt 2}} = w^{\frac 1{\sqrt 2}}(x^\sqrt{2} + j)$
or even $x = (w(x^{\sqrt{2}} + j)^{\sqrt 2} -k)^{\frac 12}$ but... I just don't see how we can distribute through powers.
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Compute $\lim_{x\to\frac{\pi}{2}} \frac{(1-\sin x)(1-\sin^2x)\dots(1-\sin^nx)}{\cos^{2n}x}$
$$\lim_{x\to\frac{\pi}{2}} \frac{(1-\sin x)(1-\sin^2x)\dots(1-\sin^nx)}{\cos^{2n}x}=?$$
Here's what I have done so far:
$y=\frac{\pi}{2}-x$ The limit becomes:$$\lim_{y\to 0}\frac{(1-\cos y)(1-\cos^2y)\dots(1-\cos^ny)}{\sin^{2n}y}=\lim_{y\to 0}\frac{1-\cos y}{\sin^2y}\frac{1-\cos^2y}{\sin^2y}\dots\frac{1-\cos^ny}{\sin^2y}$$
Also $1-\cos y=2\sin^2 \frac{x}{2}$
How should i write $1-\cos^2y ,1-\cos^3y,\dots1-\cos^ny$ in order to get the final answer.
| Look into this
$\lim_{x\rightarrow 0} \frac{(1-\cos x)}{\sin^2 x} \cdot \frac{1-\cos^2 x}{\sin^2 x} \cdots \frac{1-\cos^n x}{\sin^2 x}$ the lim distribute over the fractions and each can be calculated using Lohpital rule
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Number of real $x$ satisfying the trigo equation $1+\sin 2x=\sin x+\sin^2 x$ Number of real $x$ satisfying the trigo equation $1+\sin 2x=\sin x+\sin^2 x$ in $x\in(0,2\pi)$
solution i try $1+\sin 2x =\sin x+\sin^2 x$
$5+4\sin 2x=4\sin^2 x+4\sin x+1$
$5+\sin 2x=(2\sin x+1)^2$ How i solve it after that point
| Assume $y = \sin x$
$$1 + 2y\sqrt{1-y^2} = y + y^2 $$
$$2y\sqrt{1-y^2} = (y + y^2 -1) \implies 4y^2(1-y^2) = (y+ y^2 -1)^2$$
Simplifying, we get $5y^4 + 2y^3 - 5y^2 -2y + 1 = 0$
Solve somehow to get $y = 0.311$ or $y = 0.913$ and so $x = 0.316$ or $x = 1.15$
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Find the remainder when $3^{29}$ is divided by $12$.
Find the remainder when $3^{29}$ is divided by $12$.
a) $2$ ; b) $3$ ; c) $7$ ; d) $9$ ; e) $12$
Since $\dfrac{3^{29}}{12} = \dfrac{3^{28}}{4} = \dfrac{9^{14}}{4}$, and $9 \equiv 1 \mod 4$, I thought I could do $9^{14} \equiv 1^{14} \mod 4$, so that the answer is $1$. However, that is not one of the answer choices... Where did I go wrong?
| Modular arithmetic does not work well with division. So you cannot say that $\frac{3^{29}}{12} = \frac{3^{28}}{4}$ above, and a collection of similar arguments means your attempt is incorrect.
For the answer, try to find ways to simplify your calculation. For example, note that $3^3 = 27 \equiv 3 \mod 12$, so $$3^{29} \equiv (3^{3})^9 \times 3^2 \equiv 3^9 \times 3^2 \equiv (3^{3})^{3} \times 3^2 \equiv 3^3 \times 3^2 \equiv 3 \times 3^2 \equiv 27 \equiv 3 \mod 12$$
Where I basically used the fact noted many times in a row. Alternately, you could have also said that $3^k$ alternates modulo $27$, and this comes down to the same noted fact.
| {
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If $A+B+C=0$, then prove that the value of the determinant is $0$. I'll state the question from my textbook below:
If $A+B+C=0$, the prove that $\begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix} = 0$.
This is how I tried solving the problem:
$LHS = \begin{vmatrix}1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix}$
$= 1(1- \cos^2 A) - \cos C (\cos C - \cos A \cos B) + \cos B (\cos A \cos C - \cos B)$
$= 1 + 2 \cos A \cos B \cos C - (\cos^2 A +\cos^2 B + \cos^2 C)$
I don't know how to proceed further. I tried using the fact that $A+B+C=0$ but it didn't lead to anything I could solve. I don't know where is it supposed to be used.
Also, I read a solution to this problem somewhere in which the term $(\cos^2 A +\cos^2 B + \cos^2 C)$ was replaced by $1 + 2 \cos A \cos B \cos C$ as $A+B+C=0$. Are these two terms equal for the given condition? Also, is there a way to prove the statement without using this fact?
Any help would be appreciated.
| If one of the angles, say $A=0$, then two rows become identical and the determinant becomes zero.
Else we multiply the rows in order by $\sin A, \sin B, \sin C$ and we get
\begin{vmatrix}\sin A & \sin A\cos C & \sin A\cos B \\ \sin B\cos C & \sin B & \sin B\cos A \\ \sin C\cos B & \sin C\cos A & \sin C \end{vmatrix}
Now with $R_1 \rightarrow R_1+R_2+R_3$ we see for instance that $$a_{11} = \sin A +\sin B \cos C+\sin C \cos B = \sin A+\sin (B+C) = \sin A - \sin A =0$$
i.e. the first row becomes null and hence the determinant equals zero.
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Determinant of Union Jack matrix without Scottish diagonal Let $n \ge 1$ be an integer , and let $A_n$ be the matrix in $M_{2n-1}\mathbb(F)$ with entries $(a_{ij})$ where $a_{ij}=1 $ if $i+j=2n$ or $i=n$ or $j=n,$ and $a_{ij}=0$ otherwise . Find det$(A_n)$
my idea:
for $n=4$ the matrix of the form from the given data
$$\begin{bmatrix}
& & & 1 & & &1 \\
& & & 1& & 1& \\
& & & 1& 1 & & \\
1&1 & 1 & 1 & 1 &1 &1 \\
& & 1& 1& & & \\
& 1& & 1& & & \\
1& & &1 & & &
\end{bmatrix}$$
How to find genearl formula for determent :
| Take the mirror image of the matrix, so you get a factor $(-1)^{n-1}$. Now the mirror image is
$$
I
+\pmatrix{
&&&1\\
&&&\vdots\\
&&&1\\
1&\cdots&1&0&1&\cdots&1\\
&&&1\\
&&&\vdots\\
&&&1}=I+B\quad \text{(say)}.
$$
Clearly, the matrix $B$ has rank $2$ and $(1,\ldots,1,\pm\sqrt{2n-2},1,\ldots,1)^T$ is an eigenvector for the eigenvalue $\pm\sqrt{2n-2}$. Hence $\det A=(-1)^{n-1}(1+\sqrt{2n-2})(1-\sqrt{2n-2})=(-1)^n(2n-3)$.
Edit. Alternatively, subtract the middle row by the sum of all other rows. Then subtract the middle column by the sum of all other columns. What remains is a matrix with $(1,\ldots,1,-(2n-3),1,\ldots,1)$ on its anti-diagonal and zeroes elsewhere. Therefore $\det(A)=-(2n-3)\times(-1)^{n-1}=(-1)^n(2n-3)$.
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Jordan basis unique? Is Jordan basis unique? I have a 4x4 matrix, find eigenvectors and one generalized eigenvector, also trying different linearly independent eigenvectors but the matrix P so that PJP^-1 =A only works for certain vector.
\begin{pmatrix}-\frac{1}{2}&0&-\frac{1}{2}&0\\ \frac{3}{2}&0&\frac{1}{2}&0\\ -\frac{3}{2}&0&-\frac{3}{2}&-1\\ 1&1&1&1\end{pmatrix}\begin{pmatrix}1&-1&0&-1\\ 0&2&0&1\\ -2&1&-1&1\\ 2&-1&2&0\end{pmatrix}\begin{pmatrix}1&1&0&0\\ -1&0&1&1\\ -3&-1&0&0\\ 3&0&-1&0\end{pmatrix}=\begin{pmatrix}-1&0&0&0\\ 0&1&0&-1\\ 0&0&1&1\\ 0&0&0&1\end{pmatrix}
| No, unless your matrix is diagonalisable and has four distinct eigenvalues. To see that, note that if $v$ and $w$ are generalised eigenvectors for $A$ with the same eigenvalue $\lambda$, then so is $v+w$: indeed, $(A-\lambda I)^4(v+w)=(A-\lambda I)^4v+(A-\lambda I)^4w=0+0=0$.
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If $\cos2\theta=0$, then $\Delta^2=$? I'll state the question from my textbook here:
If $\cos2\theta=0$, then $\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2=$?
This is how I solved the problem:
$\begin{vmatrix} 0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta \end{vmatrix}^2$
$= (\cos^3 \theta + \sin^3 \theta)^2$
$= (\cos \theta + \sin \theta)^2(\cos^2 \theta - \cos \theta \sin \theta + \sin^2 \theta)^2$
$= (1+ \sin2\theta)(1-\sin2\theta + \sin^2 \theta \cos^2 \theta)$
$= (1+ \sin2\theta)(1-\sin2\theta) + (1 + \sin2\theta)\sin^2\theta \cos^2 \theta$
$= \cos^2 2\theta + \frac 14 (1 + \sin2\theta)\sin^2 2\theta$
$= \frac 14 (1 + \sin2\theta)\sin^2 2\theta$
Now since $\cos2\theta=0$, $\sin2\theta = \pm 1$.
Therefore the above expression can take the values 0 and $\frac12$.
My textbook gives the answer as $\frac12$. I don't see any grounds on rejecting the other answer of 0. Have I made a mistake somewhere? Or am I forgetting something?
| If we compute the square of the matrix, we get
$$
\begin{bmatrix}
1 & \sin\theta\cos\theta & \sin\theta\cos\theta \\
\sin\theta\cos\theta & 1 & \sin\theta\cos\theta \\
\sin\theta\cos\theta & \sin\theta\cos\theta & 1
\end{bmatrix}=
\frac{1}{2}
\begin{bmatrix}
2 & \sin2\theta & \sin2\theta \\
\sin2\theta & 2 & \sin2\theta \\
\sin2\theta & \sin2\theta & 2
\end{bmatrix}
$$
whose determinant is
$$
\frac{1}{8}(8+2\sin^32\theta-6\sin^22\theta)
$$
From $\cos2\theta=0$, we have $\sin^22\theta=1$, so finally we get
$$
\Delta^2=\frac{1}{4}(1+\sin2\theta)=
\begin{cases}
1/2 & \text{if $\sin2\theta=1$}\\[4px]
0 & \text{if $\sin2\theta=-1$}
\end{cases}
$$
both possibilities being allowed by the hypothesis that $\cos2\theta=0$.
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Finding Jordan-normal form of a special $n\times n$ matrix Let $$A := \begin{pmatrix}
0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \cdots & \vdots \\
-a_n & -a_{n - 1} & -a_{n - 2} & \cdots & -a_1
\end{pmatrix}$$
I have to find $B$, $J$ such that $A = BJB^{-1}$ such that $J$ is a Jordan matrix. I already proved that $$\det(\lambda E - A) = \lambda^n + a_1\lambda^{n - 1} + ... + a_n$$ (just simple induction). I know that this gives the eigenvalues, and if $a_n, ..., a_1$ are known, I can find the eigenvalues and generalised eigenvectors, and write down the Jordan normal form, but I don't see how there is a general solution to this problem, as $(A - \lambda E)^n$ does not have a nice closed form. This is homework for a differential equation course, so maybe we can use some of that theory here?
| Let $\lambda$ be a root of $x^n+a_1x^{n-1}+\cdots a_n=0$ then with
$$A=\begin{pmatrix}
0&1&0\\
0&0&1\\
-a_3&-a_2&-a_1
\end{pmatrix} $$
we have
$$A-\lambda I=\begin{pmatrix}
-\lambda&1&0\\
0&-\lambda&1\\
-a_3&-a_2&-a_1-\lambda
\end{pmatrix} $$
and
$$\begin{pmatrix}
-\lambda&1&0\\
0&-\lambda&1\\
-a_3&-a_2&-a_1-\lambda
\end{pmatrix} \begin{pmatrix}
1\\
\lambda\\
\lambda^2
\end{pmatrix}=
\begin{pmatrix}
0\\0\\-(a_3+a_2\lambda+a_1\lambda^2+\lambda^3)\end{pmatrix}
=\begin{pmatrix}
0\\0\\0\end{pmatrix} $$
Note further that
$$\begin{pmatrix}
-\lambda&1&0\\
0&-\lambda&1\\
-a_3&-a_2&-a_1-\lambda
\end{pmatrix} \begin{pmatrix}
0\\
1\\
2\lambda\\
\end{pmatrix}
=\begin{pmatrix}
1\\\lambda\\-a_2-2\lambda a_1-2\lambda^2\end{pmatrix}
=\begin{pmatrix}
1\\\lambda\\\lambda^2\end{pmatrix} $$
provided that
$$p^{\prime}(\lambda)=0$$
| {
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"url": "https://math.stackexchange.com/questions/2674971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Bertrand's Box Paradox Generalized Original Problem
There are three boxes, each with a different combination of gold and silver bars in them: One has two gold bars, one has a gold and a silver bar, and one has two silver bars. If a bar from a box is chosen at random, if that bar is a gold bar, what is the chance that the other bar in the box is gold?
Although the solution may seem to be $\frac{1}{2}$, it is actually $\frac{2}{3}$. There are three gold bars that could've been chosen:
There are two gold bars that could've been in the same box, and one silver bar, therefore the probability is $\frac{2}{3}$. (This problem is very similar to the Monty Hall problem.)
Generalization
There are $n+1$ boxes, each with $n$ bars that are either gold or silver, so that the first one has $n$ gold bars, the second has $n-1$ gold bars and $1$ silver bar, and so on. If a bar chosen at random is gold, what is the probability that another gold bar is chosen at random from the same box?
Arranging all the boxes side by side, we see an $n$ by $n+1$ array of bars, so there are $n(n+1)$ bars total. Half of them are gold, so there are $\frac{n(n+1)}{2}$ gold bars. The number of gold bars is also equal to $n+(n-1)+...+2+1$, which is also equal to $\frac{n(n+1)}{2}$. This means that the chance of each gold bar being chosen is $\frac{1}{\frac{n(n+1)}{2}}=\frac{2}{n(n+1)}$.
The probability for the gold bar chosen being in the first box is $\left(\frac{2}{n(n+1)}\right)(n)$, because there are $n$ gold bars there, and a $\frac{2}{n(n+1)}$ chance of each. If the gold bar chosen is in the first box, the probability for the other bar chosen later being gold is $\frac{n-1}{n-1}$.
the probability for the gold bar being in the second box is $\left(\frac{2}{n(n+1)}\right)(n-1)$, and the probability for the other bar chosen later being gold is $\frac{n-2}{n-1}$.
Continuing the pattern, we get the sum
$$\left(\frac{2}{n(n+1)}\right)(n)\left(\frac{n-1}{n-1}\right)+\left(\frac{2}{n(n+1)}\right)(n-1)\left(\frac{n-2}{n-1}\right)+\left(\frac{2}{n(n+1)}\right)(n-2)\left(\frac{n-3}{n-1}\right)+...+\left(\frac{2}{n(n+1)}\right)(2)\left(\frac{1}{n-1}\right)+\left(\frac{2}{n(n+1)}\right)(1)\left(\frac{0}{n-1}\right)+\left(\frac{2}{n(n+1)}\right)(0)\left(\frac{-1}{n-1}\right).$$
The last two terms in the sum can be canceled, and we can factor out the $\left(\dfrac{2}{n(n+1)}\right)$ to get
$$\left(\frac{2}{n(n+1)}\right)\left((n)\left(\frac{n-1}{n-1}\right)+(n-1)\left(\frac{n-2}{n-1}\right)+(n-2)\left(\frac{n-3}{n-1}\right)+...+(2)\left(\frac{1}{n-1}\right)\right).$$
Now we can also factor out $\frac{1}{n-1}$ to get
$$\left(\frac{2}{(n-1)n(n+1)}\right)\left((n)(n-1)+(n-1)(n-2)+(n-2)(n-3)+...+(2)(1)\right).$$
Because $(1)(2)+(2)(3)+(3)(4)+\cdots+(n)(n+1)=\dfrac{(n)(n+1)(n+2)}{3}$ (enter link description here), $(n)(n-1)+(n-1)(n-2)+(n-2)(n-3)+...+(2)(1)=\dfrac{(n-1)n(n+1)}{3}$. Therefore the answer becomes
\begin{align*}
\left(\frac{2}{(n-1)n(n+1)}\right)((n)(n-1)+(n-1)(n-2)+(n-2)(n-3)+...+(2)(1)) &= \left(\frac{2}{(n-1)n(n+1)}\right)\left(\frac{(n-1)n(n+1)}{3}\right)
&= \boxed{\frac{2}{3}}.
\end{align*}
This is the same answer as the original problem. Why is this the case?
Further Generalization
How would you solve the problem for $k$ different metals? Also, how many different boxes with $k$ different metals and $n$ bars per box are possible, without order mattering?
Note: drhab below has found the answer to another generalization; can anybody explain why the answer is so neat?
| I will set up a further generalization.
Let it be that there are $k$ different metals and that for every tuple $\langle n_1,\dots,n_k\rangle$ of non-negative integers with $n_1+\cdots+n_k=n$ there is exactly one box containing $n_i$ bars of metal $i$ (among them is metal gold).
Randomly a box is picked out. Then $m$ bars are picked out without replacement and all $m$ bars appear to be gold.
From the same box again a bar is picked out and the question is now: what is the probability that again the bar is of gold?
Let $G$ denote the number of gold bars that are in the selected box, and for $i=1,\dots,m+1$ and let $E_i$ denote the event that the first $i$ bars that are picked out are all gold.
Then to be found is:$$P(E_{m+1}\mid E_m)=P(E_{m+1})/P(E_m)$$
Applying stars and bars we find that total there are $\binom{n+k-1}{k-1}$ boxes and that $\binom{n+k-r-2}{k-2}$ of them contain exactly $r$ gold bars.
Then with total probability we find:
$$\begin{aligned}P\left(E_{i}\right) & =\sum_{r=i}^{n}P\left(E_{i}\mid G=r\right)P\left(G=r\right)\\
& =\sum_{r=i}^{n}\binom{r}{i}\binom{n-r}{0}\binom{n}{i}^{-1}\binom{n+k-r-2}{k-2}\binom{n+k-1}{k-1}^{-1}\\
& =\binom{n}{i}^{-1}\binom{n+k-1}{k-1}^{-1}\sum_{r=i}^{n}\binom{r}{i}\binom{n+k-r-2}{k-2}\\
& =\binom{n}{i}^{-1}\binom{n+k-1}{k-1}^{-1}\binom{n+k-1}{i+k-1}\\
& =\frac{i!\left(k-1\right)!}{\left(i+k-1\right)!}\\
& =\binom{i+k-1}{i}^{-1}
\end{aligned}
$$
Then it follows that: $$P(E_{m+1}\mid E_m)=\frac{m+1}{m+k}$$
In your original case we had $m=1$ and $k=2$ leading to probability $\frac23$.
Again the nice equality $P(E_i)=\binom{i+k-1}{i}^{-1}$ strikes me.
There must(!) be another, more elegant route to this result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that square of number of form $2k+1$ is $8n+1$ Need show that $\forall k, \exists n \in \mathbb{Z}, (2k+1)^2 =8n+1$.
$4k^2 + 4k +1 = 4k(k+1) +1$.
Although, $2k+1$ is an odd number, but $k$ can be any integer. So, $k+1$ can be an odd or an even number. Hence, not sure of getting the number of the form $8n+1$.
| Recursively By induction : if $n$ is odd then $$\begin{align}n^2 &= (n-2 +2)^2 \\ &= (n-2)^2+ 4(n-2) + 4 \\ &= (n-2) ^2 + 4(n-1)\end{align}$$ And since $n-1$ is even, $4(n-1)$ is a multiple of $8$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $1+20+20^2+20^3+....+20^{21} \equiv 0\bmod 23$ I am in search of a correct and shortcut techniques to prove this.
Otherwise I have calculated each remainder according to the power of $20$ to prove this :-
$1 \equiv 1 \bmod 23$
$\Rightarrow 20 \equiv -3\bmod 23$
$\Rightarrow 20^2 \equiv (-3)^2 \equiv 9\bmod 23$
$20^{2n}\equiv (-3)^{2n}\bmod 23$
$20^{2n-1}\equiv (-3)^{2n-1}\bmod 23$.
Therefore
$1+20+20^2+20^3+....+20^{21} \equiv 0\bmod 23$
For $\sum_{n=1}^{11} 20^{2n-1} \equiv r \bmod 23$
we have $$-3-4+10-2+5-1-9+11+7-6-8 \equiv 0 \bmod 23$$
For $\sum_{n=0}^{10} 20^{2n} \equiv s \bmod 23$
we have $$1+9+12+16+6+8+3+4+13+2-5 \equiv 0\bmod 23$$
Therefore $$\sum_{n=1}^{11} 20^{2n-1}+\sum_{n=0}^{10} 20^{2n} \equiv r+s \bmod 23$$
$$\Rightarrow 1+20+20^2+20^3+....+20^{21} \equiv 0+0 \equiv 0\bmod 23$$
If possible just show any short-cut correct way to prove this. Any help is appreciated.
| Bracket expansion and Fermat's little theorem gives
$$
(1-20)(1+20+\cdots + 20^{21}) = 1-20^{22}\equiv 1-1 = 0\pmod{23}
$$
And since $-19$ multiplied by your sum is congruent to $0$ modulo $23$, and $-19$ is coprime to $23$, that means the sum itself must also be congruent to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove an estimation of $\int_{2}^{x} \frac{dt}{\ln t}$ The problem is:
Let $k > 0$ be an integer. Prove that
$$
\int_{2}^{x} \frac{dt}{\ln t} = \frac{x}{\ln x} + \frac{1!x}{\ln^2 x} + \cdots + \frac{(k-1)!x}{\ln^k x} + \mathcal{O}\left(\frac{x}{\ln^{k+1} x}\right)
$$
I use integral by part:
\begin{align*}
\int_{2}^{x} \frac{dt}{\ln t}
&= \left. \frac{t}{\ln t} \right|_2^x - \int_{2}^{x} t d\frac{1}{\ln t}\\
&=
\frac{x}{\ln x} + \int_{2}^{x} \frac{1}{\ln^2 t} dt - \frac{2}{\ln 2}\\
&=
\frac{x}{\ln x} + \left. \frac{t}{\ln^2 t} \right|_2^x - \int_{2}^{x} t d\frac{1}{\ln^2 t} - \frac{2}{\ln 2}\\
&=
\frac{x}{\ln x} + \frac{x}{\ln^2 x} + \int_{2}^{x} \frac{2!}{\ln^3 t} dt - \frac{2}{\ln 2} - \frac{2}{\ln^2 2}\\
&=
\cdots\\
&=
\frac{x}{\ln x} + \frac{1!x}{\ln^2 x} + \cdots + \frac{(k-1)!x}{\ln^k x} + \int_{2}^{x} \frac{k!}{\ln^{k+1} t} dt - \frac{2}{\ln 2} - \frac{2}{\ln^2 2} - \cdots - \frac{2}{\ln^k 2}
\end{align*}
so finally I need to prove $\int_{2}^{x} \frac{k!}{\ln^{k+1} t} dt - \frac{2}{\ln 2} - \frac{2}{\ln^2 2} - ... - \frac{2}{\ln^k 2} = \mathcal{O}\left(\frac{x}{\ln^{k+1} x}\right)
$
But I cannot proceed any more. How to estimate the inequality here? Thank you for any help!
| Note that $$\begin{align}\int_2^x\frac{dt}{\ln^nt}&=\int_2^\sqrt x\frac{dt}{\ln^nt}+\int_\sqrt x^x\frac{dt}{\ln^nt}\\&\le\frac{\sqrt x}{\ln^n2}+\frac{x}{\ln^n\sqrt x}=\frac{\sqrt x}{\ln^n2}+\frac{2^nx}{\ln^nx}\\&=\mathcal{O}\left(\frac x{\ln^nx}\right)\end{align}$$
so
$$\frac{x}{\ln x} + \frac{1!x}{\ln^2 x} + ... + \frac{(k-1)!x}{\ln^k x} + \int_{2}^{x} \frac{k!}{\ln^{k+1} t} dt - \frac{2}{\ln 2} - \frac{2}{\ln^2 2} - \cdots- \frac{2}{\ln^k 2}$$ is the same as $$\sum_{n=1}^k\frac{(n-1)!x}{\ln^nx}+k!\cdot \mathcal{O}\left(\frac x{\ln^{k+1}x}\right)-2\sum_{n=1}^k\frac1{\ln^n2}$$ or just $$\frac{x}{\ln x} + \frac{1!x}{\ln^2 x} + \cdots+ \frac{(k-1)!x}{\ln^k x} + \mathcal{O}\left(\frac{x}{\ln^{k+1} x}\right)$$ since the rest are constants.
| {
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"url": "https://math.stackexchange.com/questions/2680416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\int \sqrt{25-x^2}x^2\,dx$ with cos-substitution I have tried this integral with cos substitution, but I don't understand why it's wrong. So could you please tell me which step is wrong? Here are my steps:
$$
x=5\cos\theta\\
dx=-5\sin\theta\,d\theta\\
$$
therefore
\begin{align}
&(1) = \int \sqrt{25-25\cos^2θ} \, 25\cos^{2}\theta \, (-5\sin\theta) \, d\theta\\
&(2) = -5^4\int \sin^{2}\theta \, \cos^{2}\theta\, d\theta \\
&(\sin 2\theta=2\sin\theta\cos\theta \implies \sin^22\theta=4\sin^2\theta\cos^2\theta)\\
&(3) =\frac{-5^4}{4}\int\sin^{2}2\theta\,d\theta\\
&(\sin^2\theta=\frac{1}{2}(1-\cos2\theta) \implies \sin^22\theta=\frac{1}{2}(1-\cos4\theta)\\
&(4) =\frac{-5^4}{8}\int (1-\cos4\theta)\,d\theta\\
&(5) = \frac{-5^4}{8} \left(\theta-\frac{1}{4}\sin4\theta\right) + C \\
&(\sin4\theta=2\sin2\theta\cos2\theta=4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta))\\
&(6)=\frac{-5^4}{8}(\theta-\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)) + C\\
&(\cos\theta=\frac{x}{5} \implies \sin\theta=\frac{\sqrt{25-x^2}}{5})\\
&(7)= \frac{-5^4}{8}\left(\arccos\frac{x}{5}-\frac{\sqrt{25-x^2}}{5}\cdot\frac{x}{5}\cdot\frac{25-2x^2}{5^2}\right) + C\\
&(8)=\frac{-5^4}{8}\left(\arccos\frac{x}{5}-\frac{x\sqrt{25-x^2}}{25}\cdot\frac{25-2x^2}{25}\right) + C\\
\end{align}
| It all looks fine, except for a little mistake here in line (7):
$$\frac{-5^4}{8}\left(\arccos\frac{x}{5}-\frac{\sqrt{25-x^2}}{5}\cdot\frac{x}{5}\cdot\frac{\color{red}{25-2x^2}}{5^2}\right)+C,$$
where your sign is flipped. Instead that fraction must be
$$\cos^2\theta-\sin^2\theta=\left(\frac{x}{5}\right)^2-\left(\frac{\sqrt{25-x^2}}{5}\right)^2=\frac{x^2}{25}-\frac{25-x^2}{25}=\frac{2x^2-25}{25}.$$
As far as I can tell, this is the only mistake.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trigonometric substitution hint I don't know how to get the following problem started and would appreciate a hint:
$$\int \frac{1}{x^2 \sqrt{9x^2+4}}dx$$
| $\int \dfrac{1}{3x^{2}\sqrt{x^{2}+\dfrac{4}{9}}}$
let $x=\dfrac{2}{3}tan(u)$ so $dx=\dfrac{2}{3}(\sec^2 u)du$
$$I=\int \dfrac{\frac{2}{3}(\sec^{2}u)}{3\frac{4}{9} \tan^2u\sqrt{\frac{4}{9} \tan^2u + \frac{4}{9}}}du$$
$$=\int \frac{\sec^2 u }{\frac{4}{3} \tan^2u \sqrt{\sec^2u}}$$
$$=\frac{3}{4} \csc u\cot u du = -\frac{3}{4} \csc u + c$$
since $x= \frac{2}{3}\tan u$ it follows (draw a triangle) that $\csc(u)=\frac{\sqrt{9x^2+4}}{3x}$ and so the integral reduces to
$$-\frac{\sqrt{9x^2+4}}{4x} +C$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there a quick way to arrive to this partial fraction decomposition? I'm reading a book where the author claims without showing the work that after we go through with the algebra of partial fractions we arrive to the formula (note $|z|<1$):
$$\frac{1}{(1-z)(1-z^2)(1-z^3)} = \frac{1}{6}\frac{1}{(1-z)^3}+\frac{1}{4}\frac{1}{(1-z)^2}+\frac{1}{4}\frac{1}{(1-z^2)}+\frac{1}{3}\frac{1}{(1-z^3)}$$
Naturally, I'm trying to reproduce the result, but I'm taking a very naive approach, namely I've expressed the LHS as a product of irreducible factors over $\mathbb{R}$ and I am trying to determine the coefficients:
$$\frac{1}{(1-z)^3}\frac{1}{(1+z)}\frac{1}{(z^2+z+1)} = \frac{A}{(1-z)}+\frac{B}{(1-z)^2}+\frac{C}{(1-z)^3}+\frac{D}{(1+z)}+\frac{Ex+F}{(z^2+z+1)}$$
This seems to have some drawbacks though, because on top of being long winded, once I do obtain all the coefficients I will have to recombine some of the terms to arrive at the author's answer. My question is then: is there some sort of trick which we can use here which I am not aware of, or do we have to suffer through the algebra patiently?
| Here's a bit of a shortcut (if you already know the coefficients).
Multiply the whole thing by $(1-z)(1-z^2)(1-z^3)$, to get
\begin{align}
& \frac{1}{6}\frac{(1-z)(1-z^2)(1-z^3)}{(1-z)^3}+\frac{1}{4}\frac{(1-z)(1-z^2)(1-z^3)}{(1-z)^2}+\frac{1}{4}\frac{(1-z)(1-z^2)(1-z^3)}{(1-z^2)}+\frac{1}{3}\frac{(1-z)(1-z^2)(1-z^3)}{(1-z^3)}\\ \ \\
&= \frac{(1+z)(1+z+z^2)}{6}+\frac{(1+z)(1-z^3)}{4}+\frac{(1-z)(1-z^3)}{4}+\frac{(1-z)(1-z^2)}{3}\\ \ \\
&= \frac{2(1+z)(1+z+z^2)+3(1+z)(1-z^3)+3(1-z)(1-z^3)+4(1-z)(1-z^2)}{12}\\ \ \\
&=\frac{2(1+z)(1+z+z^2)+6(1-z^3)+4(1-z)(1-z^2)}{12}\\ \ \\
&=\frac{2z^3+4z^2+4z+2+6-6z^3+4-4z-4z^2+4z^3}{12}\\ \ \\
&=\frac{12}{12}=1.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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I need help with integrating $\int\frac{1}{(x^2-1)^2}dx$ The problem is:
$$\int\frac{1}{(x^2-1)^2}dx$$
I tried using substitution $u=x^2-1$ but that does not bring me got results. I get an integral:
$$\frac{1}{2}\int\frac{1}{u^2\sqrt{u+1}}$$
And that does not really make things any more simple. From here I tried using partial decomposition but didn't really get anywhere.
Any help with this would be much appreciated.
| The standard way consists in decomposing the integrand into simple fractions:
$$\frac1{(x^2-1)^2}=\frac1{(x-1)^2(x+1)^2}=\frac A{x-1}+\frac B{(x-1)^2}+\frac C{x+1}+\frac D{(x+1)^2}.$$
Noting the integrand is an even function, we deduce that $C=-A$ and $D=B$.
Now multiply both sides by $(x^2-1)^2$: we obtain
$$1= A(x-1)(x+1)^2+B(x+1)^2+C(x-1)^2(x+1)+D(x-1)^2,$$
and setting $x=1$ yields the relation $\;1=4B$, whence $B=D=\frac14$.
Next set $x=0$: we now get $1=-A+B+C+D=-2A+\frac12$, whence $A=\frac14$, $C=-\frac14$.
Can you proceed from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that ${n \choose k} \ge (\frac{n}{k})^k$ for any given $n \ge k$ using induction I need to prove that ${n \choose k} \ge (\frac{n}{k})^k$ for any given $n \ge k$ using induction.
I started working on this but got stuck with the inductive step, namely I want to prove that
$${n \choose k} \ge \left(\frac{n}{k}\right)^k \Longrightarrow {n \choose k+1} \ge \left(\frac{n}{k+1}\right)^{k+1}$$
This is how I started:
$$\left( \frac{n}{k+1}\right)^{k+1} = \left(\frac{n}{k+1} \right) \left(\frac{n}{k+1}\right)^k \le \left(\frac{n}{k+1} \right) \left(\frac{n}{k} \right)^k \\ \le \underbrace{\left( \frac{n}{k+1}\right) {n \choose k}}_\text{hypothesis}$$
Right now, I don't see any reasonable way to continue with this. Any hints will be most appreciated. Thank you in advance for your answers.
| Following @gimusi hint,
$\binom{n+1}{k} = \frac{n+1}{n+1-k} \binom{n}{k} \ge \frac{n+1}{n+1-k} \left(\frac{n}{k}\right)^k = \left(\frac{n+1}{k}\right)^k \left(\frac{n}{n+1}\right)^k \frac{n+1}{n+1-k} = \left(\frac{n+1}{k}\right)^k \frac{1}{\left(1+\frac{1}{n}\right)^k} \frac{1}{\left(1 - \frac{k}{n+1}\right)}$
So if we can show that $\frac{1}{\left(1+\frac{1}{n}\right)^k} \frac{1}{\left(1 - \frac{k}{n+1}\right)} \ge 1$, then we are done.
To prove:$\left(1+\frac{1}{n}\right)^k \left(1 - \frac{k}{n+1}\right) \le 1$
Even to a first degree, the above $= \left(1 + \frac{k}{n}\right)
\left(1 - \frac{k}{n+1}\right) = 1 + \frac{k}{n(n+1)} - \frac{k^2}{n(n+1)}$
The case for $k = n+1$ is trivially correct. Assume $k = n$.
So even then
$1 + \frac{n}{n(n+1)} - \frac{n^2}{n(n+1)} = 1 + \frac{1}{n+1}- \frac{n}{n+1} = \frac{2}{n+1}$ and we know $0 < \frac{2}{n+1} \lt 1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Plot $|z^2-1|=1$ $$|z^2-1|=1$$
$|x^2-y^2-1+2xyi|=1$
$\sqrt{(x^2-y^2-1)^2+(2yx)^2}=1$
${(x^2-y^2-1)^2+4x^2y^2}=1$
$x^4+y^4+2x^2y^2+2x^2-2y^2=1$
$(x^2+y^2)^2=2(x^2-y^2)+1$
I am trying to bring it to the form of $(x^2+y^2)^2=2a(x^2-y^2)$ but cant get rid of the $1$
| Hint
$$(x^2-y^2-1)^2+4x^2y^2=1\to (x^2-y^2)^2-2(x^2-y^2)+1+4x^2y^2=1$$
$$(x^2+y^2)^2-2(x^2-y^2)=0\to (x^2)^2+x^2(2y^2-2)+y^4+2y^2=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Mutual information between two dependent variables Let $X$ be a discrete random variable such that
\begin{equation}
X = \begin{cases}
1 & \text{with p = 1/3} \\
-1 & \text{with p= 1/3} \\
0 & \text{with p = 1/3}
\end{cases}
\end{equation}
and let $Y = X^2$ whose distribution is
\begin{equation}
Y = \begin{cases}
1 & \text{with p = 2/3} \\
0 & \text{with p = 1/3}.
\end{cases}
\end{equation}
Note that it is clear that $Y$ is completely determined by $X$.
However, $X$ can not be completely determined by $Y$.
It follows from
$$ P(X=a,Y=b) = P(Y=b|X=a)P(X=a)$$
that we have the following joint distribution of $(X,Y)$,
$$P(X=1,Y=1) = P(X=-1,Y=1) = P(X=0,Y=0) = 1/3.$$
Note that $P(Y=b|X=a)$ is either 1 or 0, as $Y$ is completely determined by $X$.
Therefore, we compute the mutual information of $X$ and $Y$ and obtain
$$ I(X;Y) = \sum_{(x,y)} p(x,y)\log \left(\frac{p(x,y)}{p(x)p(y)}\right)
= \frac{1}{3}\log(3^3/4) > 0 $$
[Editted] Previously, wrong computation.
Based on the wikipedia
"Intuitively, mutual information measures the information that X and Y
share: It measures how much knowing one of these variables reduces
uncertainty about the other."
Question
In this example, knowing $X$ completely reduces uncertainty in $Y$.
Then how should I interpret this value $I(X;Y) = \frac{1}{3}\log(3^3/2) > 0$
in order to draw a conclusion that
"$Y$ can be completely determined by $X$"?
My thought
I think the mutual information of $X, Y=X^2$, doesn't make sense,
as knowing $X$ completely determines $Y$, however,
knowing $Y$ does not.
This is because $I(X;Y) = I(Y;X)$.
That is by changing the role of $X$ and $Y$, we obtain the same value even knowing the fact that $X$ completely determines $Y$ but $Y$ doesn't.
This doesn't sound right to me.
It would be very appreciated if someone gives some comments or suggestions or any thoughts.
Thanks in advance.
| Let's take the case $ x = 1 $ and $ y = 1 $. The fraction inside the log is $ \frac{ \frac{1}{3} }{ \frac{1}{3} \frac{2}{3} } = 1.5 $.
It is the same for $ x = -1 $ and $ y = 1 $.
For $ x = 0 $ and $ y = 0 $ we have $ \frac{ \frac{1}{3} }{ \frac{1}{3} \frac{1}{3} } = 3 $.
So the $ \log \left( \cdot \right) $ argument is always bigger than 1 and all is OK.
| {
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Why does $A^8=I_3$ implies $A^4=I_3$?
Let $A\in M_3(\mathbb{Q})$ such that $A^8=I_3$. Prove that $A^4=I_3$
(We denote by $m_X$ the minimal polynomial of matrix X and by $p_X$ its characteristic polynomial and by $gr \space X$ its degree)
The solution is the following:
$1)$ $m_a$ divides $P\in\mathbb{Q}[X]$, $$P(x)=x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)$$
$2)$ Since $A^4\neq I_3$, then $p_A$ would have at least one common root with $x^4+1$
$3)$ Since $x^4+1$ is irreducible over $\mathbb{Q}[X]$ we have that $x^4+1|m_A$
$4)$ $A\in M_3(\mathbb{Q})$ so $gr\space m_A\leq3$ and $gr\space x^4+1=4$ , contradiction
I understood only points $1)$ and $4)$. Can you please help me to understand $2)$ and $3)$? Are they even right?
| 2) Since $m_A\mid(x^4-1)(x^4+1)$, every root of $p_A$ is a root of $(x^4-1)(x^4+1)$. If $A^4\neq\operatorname{Id}_3$, then no root of $x^4-1$ is a root of $p_A$. Therefore, $p_A$ must have at least one root of $x^4+1$.
3) The polynomial $x^4+1$ is irreducible in $\mathbb{Q}[x]$ and it has at least a common root with $m_A$. Since $x^4+1$ and $m_A$ have a common factor, $\gcd(x^4+1,m_A)\neq1$. Since $\gcd(x^4+1,m_A)\mid x^4+1$ and this polynomial is irreducible, $x^4+1\mid\gcd(x^4+1,m_A)$. In particular, $x^4+1\mid m_A$.
| {
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Alternative for calculating the nth of quadratic sequence Given the quadratic sequence
$$f(n)=1, 7, 19, 37, \cdots$$
To calculate the $f(n)$ for $n\ge1$. $$f(n)=an^2+bn+c$$
We start with the general quadratic function, then sub in for $n:=1,2$ and $3$
$$f(1)=a+b+c$$
$$f(2)=4a+2b+c$$
$$f(3)=9a+3b+c$$
Now solve the simultaneous equations
$$a+b+c=1\tag1$$
$$4a+2b+c=7\tag2$$
$$9a+3b+c=19\tag3$$
$(2)-(1)$ and $(3)-(2)$
$$3a+b=6\tag4$$
$$5a+b=12\tag5$$
$(5)-(4)$
$$a=3$$
$$b=-3$$
$$c=1$$
$$f(n)=3n^2-3n+1$$
This method is very long. Is there another easy of calculating the $f(n)$?
| Take a better basis. Namely, $\{(n-1)(n-2),(n-1)(n-3),(n-2)(n-3)\}$. If $$f(n) = \alpha(n-1)(n-2) + \beta(n-1)(n-3) + \gamma(n-2)(n-3),$$
then:
$$f(1) = 0 + 0 + \gamma(1-2)(1-3),$$
$$f(2) = 0 + \beta(2-1)(2-3) + 0,$$
$$f(3) = \alpha(3-1)(3-2) + 0 + 0.$$
| {
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Prove this Hard inequality to CS or AM-GM?
Let $x,y,z\ge 0$ such $x+y+z=1$, show that
$$\sqrt{\dfrac{x(x+1)}{1-x}}+\sqrt{\dfrac{y(y+1)}{1-y}}+\sqrt{\dfrac{z(z+1)}{1-z}}\ge\sqrt{6}$$
This inequality is creat by wangyongxi,and when $x=y=z=\dfrac{1}{3}$ or $x=y=\dfrac{1}{2},z=0$ is $=$.
and we found this $f(x)=\sqrt{\dfrac{x(x+1)}{1-x}},0\le x\le 1$,not Convex function,because we get
$$f''(x)=-\dfrac{x^4-4x^3-6x^2-4x+1}{4(x-1)^4\left(-\dfrac{x(x+1)}{x-1}\right)^{3/2}}$$
so when $x<0.2$then $f''(x)<0$ and other $f''(x)>0$
Now I wonder to look such $AM-GM$ or Cauchy inequality to solve it? Thanks
| Write a tangent line at $T(\sqrt{2}-1,1)$ for $f(x)= \sqrt{\dfrac{x(x+1)}{1-x}} $
$$ y= (\sqrt{2}+1)x$$
and see that it is entirely under the graph of $f$ on $[0,1]$:
$$\sqrt{\dfrac{x(x+1)}{1-x}}\geq (\sqrt{2}+1)x$$
So $$f(x)+f(y)+f(z)\geq \sqrt{2}+1$$
| {
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how to write the polar form of $x^2/(x^2 + y^2)^2 - y^2/(x^2 + y^2)^2$? $$x^2/(x^2 + y^2)^2 - y^2/(x^2 + y^2)^2\\
r=(x^2+y^2)^{1/2}\\
x^2/r^4-y^2/r^4$$
Convert $x^2-y^2$ to polar form
$$x=r\cos(\theta)$$
$$y=r\sin(\theta)$$
$$(r\cos(\theta))^2-(r\sin(\theta))^2=r^2 \cos(2 θ)$$
$$r^2 \cos(2 θ)/r^4= \cos(2\theta)/r^2$$
Is that correct?
| HINT: it is $$\frac{r^2\cos^2(\theta)-r^2\sin^2(\theta)}{(r^2)^2}$$
| {
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Calculate a quartic monic polynomial given that $P(1)=10,P(2)=20$ and $P(3)=30$. Let
$$P(x)=x^4+ax^3+bx^2+cx+d$$
where $a,b,c$ and $d$ are constants.
If $P(1)=10,P(2)=20$ and $P(3)=30$, then what is the value of $$\frac{P(12)+P(-8)}{10}$$
I tried substituting $x=1,2,3$ in $P(x)$ to get three linear equations in $a,b,c$ and $d$, but it ended up being quite difficult.
$$a+b+c+d=9$$
$$8a+4b+2c+d=4$$
$$27a+9b+3c+d=-51$$
Is there a simpler method to this question?
| Define $Q(x) = P(x) - 10x$, then$$
P(1) = 10,\ P(2) = 20,\ P(3) = 30 \Longrightarrow Q(1) = Q(2) = Q(3) = 0.
$$
Since $\deg Q = 4$ and $Q$ is monic, there exists $x_0 \in \mathbb{R}$ such that$$
Q(x) = (x - 1)(x - 2)(x - 3)(x - x_0).
$$
Therefore,\begin{align*}
&\mathrel{\phantom{=}}{} P(12) + P(-8)\\
&= Q(12) + Q(-8) + 40\\
&= 11 × 10 × 9 × (12 - x_0) + (-9) × (-10) × (-11) × (-8 - x_0) +40\\
&= 11 × 10 × 9 × (12 - x_0) + 11 × 10 × 9 × (8 + x_0) +40\\
&= 11 × 10 × 9 × ((12 - x_0) + (8 + x_0)) +40\\
&= 11 × 10 × 9 × 20 +40\\
&=19840.
\end{align*}
| {
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Find all solutions to the congruence $x^3\equiv 1 {\pmod{77}}$ Find all solutions to the congruence $x^3\equiv 1 \pmod{77}$
I was able to do this for $x^2 \equiv 1\pmod{77}$ since $x$ would either be $1$ or $-1$, in each of $\bmod 7$ and $\bmod 11$. I don't know what $x$ would be in this case.
I can do the Chinese remainder theorem to find the final answers.
| Presumably you know that Fermat's little theorem (FLT) says that for prime modulus $p$ and $a$ coprime to $p$ we have $a^{p-1}\equiv 1 \bmod p$.
This also means that if we have $a^k \equiv 1 \bmod p$, and $k$ is the smallest positive integer for which this is true ($k$ is the order of $a \bmod p$), then $k$ divides $p{-}1$ - since otherwise FLT would not hold.
This means that we cannot have any such elements for $p=11$ and $k=3$, since $3 \nmid 10$. So $x^3\equiv 1 \bmod 11$ has only one solution, $x\equiv 1 \bmod 11$, since only $1$ divides both $3$ and $10$.
By constrast $3\mid (7-1) $ so there should be three solutions (including $1$), and $7$ is small enough to search case-by-case even if $7+1=8=2^3$ doesn't occur to you. The other cube root is $2^2 = 4$, which is apparent without further multiplying-out since $4^3=2^6\equiv 1^2\equiv 1 \bmod 7$.
| {
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Prove that the sequence converges to using the $\varepsilon$-$N$ definition of a limit Question: Prove that the sequence $\{a_n\}_{n=1}^\infty$ defined by $a_n=\frac{2n}{3n+1}$ converges to $\frac{2}{3}$ using the $\varepsilon$-$N$ definition of a limit.
I'm still a beginner when it comes to these types of questions. I've had a go at it but I'm pretty sure this is nonsense. Would appreciate some feedback and suggestions. Thanks!
Solution: Fix $\varepsilon>0$. We need to find $N\in\mathbb{N}$ such that $n>N \implies|a_n-\frac{2}{3}|<\varepsilon.$
We have $|a_n-\frac{2}{3}| < \varepsilon.$
$\iff |\frac{2n}{3n+1}-\frac{2}{3}| < \varepsilon$
$\iff |\frac{2n}{3n+1}-\frac{2}{3}| < \varepsilon$
Now $2n\ge2$ for all $n\ge1$ and $3n+1\ge3n$ for all $n$, so we have
$\iff |\frac{2}{3n}-\frac{2}{3}| < \varepsilon$
$\iff |\frac{2}{3}(\frac{1}{n}-1)| < \varepsilon$
The sequence is positive for all $n$ so the absolute values are redundant
$\iff \frac{2}{3}(\frac{1}{n}-1) < \varepsilon$
$\iff \frac{1}{n}-1 < \frac{3\varepsilon}{2}$
$\iff \frac{1}{n} < \frac{3\varepsilon}{2}+1$
$\iff \frac{1}{n} < \frac{3\varepsilon+2}{2}$
$\iff n> \frac{2}{3\varepsilon+2}$
So choose any $N>\frac{2}{3\varepsilon+2}$ and the definition is satisfied. Q.E.D.
| You're certainly making this complicated. And the following step is wrong, and I gave up on trying to figure out why you thought it was true:
$\iff |\frac{2n}{3n+1}-\frac{2}{3}| < \varepsilon$
Now $2n\ge2$ for all $n\ge1$ and $3n+1\ge3n$ for all $n$, so we have
$\iff |\frac{2}{3n}-\frac{2}{3}| < \varepsilon$
Remember that the way to subtract fractions is by using a common denominator:
$$
\left| \frac{2n}{3n+1} - \frac 2 3 \right| = \left| \frac{3\cdot2n}{3(3n+1)} - \frac{2(3n+1)}{3(3n+1)} \right| = \left| \frac{6n - (6n+2)}{3(3n+1)} \right| = \frac 2 {3(3n+1)}
$$
The problem now is how to make $n$ so big that that last fraction is less than $\varepsilon.$
| {
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Differentiation using product rule I'm having trouble simplifying these questions, particularly when they involve square roots of $x$.
Differentiate the following with respect to $x$ and simplify:
$y=(x+2)x^\frac{3}{2}$
My attempt:
Using product rule: $u=x^\frac{3}{2}, v=(x+2)$ therefore $\frac{du}{dx}=\frac{3}{2}x^\frac{1}{2}, \frac{dv}{dx}=1\\$
$\frac{dy}{dx}= \frac{3\sqrt{x}}{2}(x+2)+(\sqrt{x})^3$
Factorise:$\sqrt{x}[\frac{3}{2}(x+2)+(\sqrt{x})^2]\\\sqrt{x}[\frac{3}{2}x+3+x]$
The given answer is $\frac{\sqrt{x}}{2}(5x+6)$, which I can't achieve and I can't understand why the denominator of 2 is a common factor.
| You derivation is correct indeed note that
$$\sqrt{x}\left(\frac{3}{2}x+3+x\right)=\frac{\sqrt{x}}{2}\left(3x+6+2x\right)=\frac{\sqrt{x}}{2}\left(5x+6\right)$$
| {
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Find $f'(a)$ if $f(x) = \frac {x^n - a^n}{x-a}$ I'll state the multiple choice question from my textbook below:
If $f(x) = \frac {x^n - a^n}{x - a}$ for some constant '$a$', then $f'(a)$ is
(A) $1$
(B) $0$
(C) does not exist
(D) $\frac 12$
Here's what I tried:
By using quotient rule we get:
$f'(x) = \frac {(n-1)x^n - nax^{n-1} + a^n}{(x-a)^2}$
Clearly,
$f'(a) = \frac 00$
And the correct choice according to my book is (C). Does getting $\frac 00$ for a particular value of the variable (in this case, for $x=a$) mean that the derivative of the function doesn't exist at that point?
Not satisfied by the answer I factorised the numerator of $f(x)$ and cancelled the factor $(x-a)$ as below:
$f(x) = \frac {(x-a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 +..........+ xa^{n-2} + a^{n-1})}{x-a}$
$\implies f(x) = x^{n-1} + x^{n-2}a + x^{n-3}a^2 +..........+ xa^{n-2} + a^{n-1}$
Differentiating with respect to $x$ we get,
$f'(x) = (n-1)x^{n-2} + (n-2)x^{n-3}a + (n-3)x^{n-4}a^2 +..........+ a^{n-2}$
$\implies f'(a) = \Big[(n-1) + (n-2) + (n-3) + .......... + 1\Big]a^{n-2}$
$\implies f'(a) = \frac {n(n-1)}2 a^{n-2}$
Now this is a completely different answer.
So what have I done wrong? Is there some problem with cancelling the factor $(x-a)$? Does it have something to with the continuity and differentiabilty of $f(x)$ at $a$? What if the function was $f(x) = \begin{cases} \frac {x^n - a^n}{x - a}, & \text{if $x \ne a$} \\ na^{n-1}, & \text{if $x = a$}
\end{cases}$? How do I find the derivative at $x = a$ in this case?
| For the first case the derivative $f'(a)$ doesn't exist since $f(x)$ is not defined for $x=a$ and the derivative is defined by
$$\lim_{x\to a} \frac{f(x)-\color{red}{f(a)}}{x-a}$$
and the existence of $f(a)$ is required.
In the second case we are allowed to apply the definition by limit.
| {
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If $x\sqrt{1+y}+y\sqrt{1+x}=0$ find $y'$
Find $\frac{dy}{dx}$ if $x\sqrt{1+y}+y\sqrt{1+x}=0$ for $-1\leq x\leq 1$
My Attempt
$$
x\sqrt{1+y}=-y\sqrt{1+x}\implies x^2(1+y)=y^2(1+x)\implies x^2+x^2y=y^2+xy^2\\
2x+2xy+x^2\frac{dy}{dx}=2y\frac{dy}{dx}+y^2+2xy\frac{dy}{dx}\\
\frac{dy}{dx}\Big[ x^2-2y-2xy \Big]=y^2-2x-2xy\\
\frac{dy}{dx}=\frac{y^2-2x-2xy}{x^2-2y-2xy}
$$
How do I proceed further and find the derivative ?
| $x^2+x^2y = y^2 + xy^2 $
$x^2+x^2y -y^2-xy^2 = 0 $
$(x-y)(x+y) + xy(x-y) = 0 $
$(x-y)(x+xy+y) = 0 $
$(x-y) = 0 , or, x + xy + y = 0 $
$x \neq y$ since $x\sqrt{1+y} = -y\sqrt{1+x}$ unless (0,0)
$x + xy + y = 0 $
$x + y(x + 1) = 0 $
$y = \frac{-x}{x+1} = -1 + \frac{1}{x+1}$
$\frac{dy}{dx} = -(x+1)^2$
| {
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Fubini's Theorem contradiction Why does the fact that
$\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \,dy \,dx = \frac{\pi}{4}$ and $ \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \,dx \,dy = -\frac{\pi}{4}$ doesn't contradict Fubini's Theorem? Does the tangent function have something to do with that matter?
| Fubini's theorem says that if $\displaystyle \int_{X \times Y} \vert f(x,y) \vert d(x,y) < \infty$ then $$\int_X \left( \int_Y f(x,y) dy \right) dx = \int_Y \left( \int_X f(x,y) dx \right) dy = \int_{X \times Y} f(x,y) d(x,y)$$
But here, $\displaystyle \int_{X \times Y} \vert f(x,y) \vert d(x,y) = \infty$ so we should not expect the integrals to be equal.
Specifically,
$$\begin{align*} \int_{I \times I} \left\vert \frac{x^2-y^2}{(x^2+y^2)^2} \right\vert d(x,y) &\ge \int_0^1 \left(\int_0^ x\frac{x^2-y^2}{(x^2+y^2)^2} dy \right) dx\\ &=\int_0^1 \left[ \frac{y}{x^2+y^2} \right]_{y=0}^x dx\\ &= \int_0^1 \frac{1}{2x} dx = + \infty. \end{align*}$$
| {
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Roots of $3z^2+3z+b$ and $0$ lie on equilateral triangle Given that roots of $3z^2+3z+b$ and $0$ lie on equilateral triangle, find condition on $b$.
If $a$ is a root then $a(\frac{1}{2}+i\frac{\sqrt 3 }{2})$ i also a root. Sum of root is $-1 = a(\frac{3}{2}+i\frac{\sqrt3}{2})$ and product is $a^2(\frac{1}{2}+i\frac{\sqrt 3 }{2}) = \frac{b}{3}$
I am asking is this correct and also does veeta's formula work in complex numbers? What is your method and how to proceed using this, find $a$ and put in second equation to get $b$ is complicated.
Thanks a lot!!
| Let the roots of the quadratic be $p\pm iq$
Looking at the sum of roots, we have $p=-\frac 12$
Since the triangle is equilateral, $$\sqrt{p^2+q^2}=2|q|$$
So $$q=\frac{1}{2\sqrt{3}}$$
Looking at the product of roots, $$\frac b3=p^2+q^2=\frac 13$$
Hence $b=1$
| {
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Evaluating infinite series with $a^{(2^n)} + 1$ form denominator Evaluate:
$$\sum_{n = 0}^\infty \frac{2^n}{a^ \left( 2^n \right) + 1 } $$
This doesn't seem to have factors in the denominator. I tried stuff like expanding this series and then multiplying by a factors like $ \frac{1}{x-1} $ and then shifting the terms, subtracting the series.
(Edit: Sorry for not telling that a $ \gt $ 1).
| Alternative approach: for any $a>0$ we have
$$ 1-a^2 = (1+a)(1-a),\quad 1-a^4=(1+a^2)(1+a)(1-a),\quad 1-a^8 = (1+a^4)(1+a^2)(1+a+1)(1-a),\qquad \frac{1-a^{2^N}}{1-a}=\prod_{k=0}^{N-1}\left(1+a^{2^k} \right)\tag{1}$$
and
$$ \log(1-a^{2^N})-\log(1-a) = \sum_{k=0}^{N-1} \log\left(a^{2^k}+1\right)\tag{2} $$
so by applying $\frac{d}{da}$ to both sides of $(2)$, then multiplying by $a$:
$$ \frac{2^N a^{2^N}}{a^{2^N}-1}-\frac{a}{a-1}=\sum_{k=0}^{N-1}\frac{2^k a^{2^{k}}}{a^{2^k}+1}\tag{3}$$
then mapping $a$ into $\frac{1}{a}$:
$$ \frac{2^N}{1-a^{2^N}}+\frac{1}{a-1}=\sum_{k=0}^{N-1}\frac{2^k}{a^{2^k}+1}.\tag{4}$$
If $a>1$, by considering the limit of both sides of $(4)$ as $N\to +\infty$ we get
$$ \sum_{k=0}^{N-1}\frac{2^k}{a^{2^k}+1} = \frac{1}{a-1}.\tag{5} $$
Notice that this is equivalent to the fact that every natural number has a unique representation in base $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2731886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the inflection points of $f(x)={1 +\ln^2 x \over x}$
Given $$f(x)={1+\ln^2x\over x}, x>0$$
• Find the inflection points of $C_f.$
Personal work:
In order to find the inflection points of $C_f$ we first need to find the second derivative and then at which "$x$", $f''(x)=0.$
$${d\over dx}({1+\ln^2 x \over x})={2lnx-1-\ln^2x\over x^2}$$
$${d\over dx}({2lnx-1-\ln^2x\over x^2})=\cdots={2\ln^2-6\ln x+4 \over x^3}$$
or
$${d\over dx}({d\over dx}({1+\ln^2 x \over x}))={2\ln^2-6\ln x+4 \over x^3}$$
I'm struggling on finding the sign of $f'(x)$ and the points that $f''(x)=0.$
| We have $$f''(x)=\frac{\left(\frac{2}{x}-2\ln(x)\cdot \frac{1}{x}\right)x^2-\left(2\ln(x)-1-\ln(x)^2\right)\cdot 2x}{x^4} \\ = \frac{\left(2-2\ln(x)\right)+\left(-2\ln(x)+1+\ln(x)^2\right)\cdot 2}{x^3} \\ = \frac{2-2\ln(x) -4\ln(x)+2+2\ln(x)^2}{x^3} \\ = \frac{2\ln(x)^2 -6\ln(x)+4}{x^3}$$
$f''(x)=0 \Longleftrightarrow 2\ln(x)^2 -6\ln(x)+4=0$
Now, leting $\ln(x) = u$, we have: $$2u^2 -6u+4=0 \implies u=1,2 \\ \implies x=e,e^2$$ which are your possible points of inflection.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solve $x+\frac{1}{y-x}=1$, $y+\frac{1}{x-y}=2$ I've got this problem:
Solve for pairs of reals,
$$
\left \{
\begin{array}{rcl}
x+\dfrac{1}{y-x} & = & 1 \\
y+\dfrac{1}{x-y} & = & 2
\end{array}
\right.
$$
I've tried many different approaches and I tried on Wolfram Mathematica, which gives me the solutions, but I don't understand how to prove that there are no other.
Solution pairs $(x,y)=(2,1)$ and $(\frac{1}{2}, \frac{5}{2})$.
| $y-x= a $ ⇒ $y=x+a$
$y-1/a=2$
$x+a -1/a=2 $
$x+1/a=1$
Subtracting two relations we get:
⇒$x+1/a-x-a+1/a=1-2 ⇒ a^2-a-2=0 ⇒ a=2,.. a=-1$
$a=2$ gives $y-x=2$ and summing two initial relations gives $x+y=3$. The solution of this system of equation is $x=0,5 $ and $y=2.5$
$a=-1$ gives $y-x=-1$this with $x+y=3$give another system of equation which solution is $x=2$ and $y=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find the coefficient of $x^{203}$ in the expansion of $(x-1)(x^2 - 2)(x^3-3)\dots(x^{20} - 20)$?
How to find the coefficient of $x^{203}$ in the expansion of $(x-1)(x^2 - 2)(x^3-3)\dots(x^{20} - 20)$?
I took $x$ as common from each bracket making $x^{190}$ but I don't understand what to do next. Please help.
| The ways of writing $7$ as a sum of distinct positive integers are $7,1+6,2+5,3+4,1+2+4$, so the coefficient of $x^{203}$ will be $-7+1\cdot 6 + 2\cdot 5 + 3 \cdot 4 - 1\cdot 2 \cdot 4 = 13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How can I calculate this limit? ideas? How can I calculate this limit?
$$\lim_{n\rightarrow\infty} \frac{7^{\sqrt{n+1}-\sqrt{n}}\cdot(\frac{n+1}{2})!\cdot(\frac{n+1}{2})!}{(n+1)\cdot(\frac{n}{2})!\cdot(\frac{n}{2})!}$$
I don't have idea and I will be happy for help.
| Let apply Stirling’s approximation
$$n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n$$
that is
$$\left[\left(\frac{n+1}{2}\right)!\right]^2\sim 2 \pi \frac{n+1}2\left(\frac{n+1}{2e}\right)^{n+1}=\pi\frac{(n+1)^{n+2}}{2^{n+1}e^{n+1}}$$
$$\left[\left(\frac{n}{2}\right)!\right]^2\sim 2 \pi \frac n 2\left(\frac{n}{2e}\right)^n=\pi\frac{n^{n+1}}{2^ne^n}$$
and
$$\sqrt{n+1}-\sqrt{n}=\frac1{\sqrt{n+1}+\sqrt{n}}\sim \frac1{2\sqrt n}$$
then
$$\frac{7^{\sqrt{n+1}-\sqrt{n}}\cdot(\frac{n+1}{2})!\cdot(\frac{n+1}{2})!}{(n+1)\cdot(\frac{n}{2})!\cdot(\frac{n}{2})!}\sim 7^{\frac1{2\sqrt n}}\frac{(n+1)^{n+1}}{2^{n+1}e^{n+1}}\frac{2^ne^n}{n^{n+1}}=\frac{7^{\frac1{2\sqrt n}}}{2e}\left(1+\frac1n\right)^{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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How do you do change of variables for triple integrals? I am evaluating the function over the following bounds.$$\int_0^2\int_0^{\sqrt{4-x^2}}\int_0^{\sqrt{4-x^2-y^2}}z\sqrt{4-x^2-y^2}\,\mathrm dz\,\mathrm dy\,\mathrm dx$$
I'm not sure how to combine triple integrals and change of variables. Can someone run me through the steps for this problem?
thanks
| \begin{align}
& \int_0^2 \left( \int_0^{\sqrt{4-x^2}} \left( \int_0^{\sqrt{4-x^2-y^2}}z\sqrt{4-x^2-y^2}\,\mathrm dz\right) \,\mathrm dy \right)\,\mathrm dx \\[10pt]
& \text{The innermost integral is easy:} \\
& \int_0^{\sqrt{4-x^2-y^2}} z \underbrace{\sqrt{4-x^2-y^2}}_\text{No $z$ appears here.} \,\, \mathrm dz \\[10pt]
= {} & \sqrt{4-x^2-y^2} \int_0^{\sqrt{4-x^2-y^2}} z\,\mathrm d z \\
& \text{This can be done because the factor that} \\
& \text{was pulled out does not depend on $z$.} \\[10pt]
= {} & \sqrt{4-x^2-y^2} \cdot \frac{4-x^2-y^2} 2 = \frac 1 2 (4-x^2-y^2)^{3/2}. \\[10pt]
& \text{So now we have} \\
& \frac 1 2 \int_0^2 \int_0^{\sqrt{4-x^2}} (4-x^2-y^2)^{3/2} \, \mathrm dy \, \mathrm dx \\[10pt]
= {} & \frac 1 2 \iint\limits_{\{\,(x,y)\,:\, x^2+y^2\,\le\,4 \, x,y\,\ge\,0 \,\}} (4-x^2-y^2) \, \mathrm d(x,y) \\[10pt]
= {} & \frac 1 2 \int_0^{\pi/2} \underbrace{\left( \int_0^2 (4-r^2)^{3/2} r\, dr \right)}_\text{No $\theta$ appears here.} \, d\theta \\[10pt]
= {} & \frac 1 2 \cdot \frac \pi 2 \int_0^2 (4-r^2)^{3/2} r\,dr \quad \text{This works because no $\theta$ was in $\int_0^{\pi/2}\cdots\,d\theta$.} \\[10pt]
= {} & \frac \pi 4 \int_4^0 u^{3/2} \left( \frac{-du} 2\right) = \cdots
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof similarity of complex power series Let be the functions
$$
f(x)=\sqrt{2}\frac{\cos(\frac{\pi}{4}-\frac{1}{2}\arctan(x))}{(1+x^2)^{1/4}} \qquad g(x)=\frac{1}{\sqrt{1-x}}
$$
Obviously, they are holomorphic functions in a neighborhood of $0$. The series of $g(z)$ is well known and using Mathematica we have
$$
f(x)= 1+\frac{x}{2}- \frac{3x^2}{8}-\frac{5x^3}{16}+\frac{35x^4}{128}+\frac{63x^5}{256}-\frac{231x^6}{1024}-\frac{429x^7}{2048}+\dotsc
$$
$$
g(x)= 1+\frac{x}{2}+ \frac{3x^2}{8}+\frac{5x^3}{16}+\frac{35x^4}{128}+\frac{63x^5}{256}+\frac{231x^6}{1024}+\frac{429x^7}{2048}+\dotsc
$$
Exercise. Proof the absolute value of the coefficients of the two series are the same.
Attemp. With trigonometric manipulation I get
$$
f(x)=\frac{1+x+\sqrt{1+x^2}}{\sqrt{2}(1+x^2)^{3/4}\sqrt{1+\frac{1}{\sqrt{1+x^2}}}}
$$
My idea is use the Cauchy Integral Theorem to calculate the coefficients
$$
a_n = \frac{1}{2\pi i}\int_C \frac{f(z)}{z^{n+1}}dz
$$
and with changes of variables get something that seems to
$$
\frac{1}{2\pi i}\int_C \frac{g(z)}{z^{n+1}}dz
$$
But if I work in $\mathbb{C}$ I need to define $\log(1+z^2)$ and work with it in changes of variables doesn't look so friendly.
Can you give me a hint to prove it?
| Hint
Let $\phi(z)=\sum_k a_k z^k$.
The first step is to see how to express $a_0+a_1 z-a_2 z^2 -a_3 z^3 +a_4 z^4+\ldots$ using $\phi$.
The four roots of $X^4=1$ being $\lbrace \pm 1, \pm i \rbrace$. You have to solve the system:
$$\begin{pmatrix}1&1&1&1\\1&i&-i&-1\\1^2&i^2&(-i)^2&(-1)^2\\1^3&i^3&(-i)^3&(-1)^3 \end{pmatrix} \begin{pmatrix} a\\ b \\c \\d \end{pmatrix}=\begin{pmatrix} 1\\ 1 \\-1 \\-1 \end{pmatrix}$$
As the solutions are $a=d=0$, $b=\frac{1-i}{2}, c=\frac{1+i}{2}$ you obtain:
$$a_0+a_1 z-a_2 z^2 -a_3 z^3 +a_4 z^4+\ldots = \frac{1}{2} \left((1-i)\phi(ix)+(1+i)\phi(-ix) \right)$$
Here you then have to prove that near $0$:
$$f(x)=\frac{1}{2} \left(\frac{1-i}{\sqrt{1-ix}}+\frac{1+i}{\sqrt{1+ix}} \right)$$
Using your expression of $f$ this is the same to prove that near $0$:
$$(1+i) \sqrt{1-ix}+(1-i) \sqrt{1+ix}=\sqrt{2}\frac{1+x+\sqrt{1+x^2}}{\sqrt{1+\sqrt{1+x^2}}}=2\sqrt{x+\sqrt{1+x^2}}$$
which seems easier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Number of solutions to a polynomial I need to find the number of solution to the polynomial
$$(x-2y-1)^2+(4x-3y-4)^2+(x-2y-1)(4x-3y-4)=0$$
Clearly, $x=1$ and $y=0$ is a solution. How to prove it has no other solutions or more solutions?
| Considering the previous answers, I could be totally wrong.
If the equation is
$$(x-2y-1)^2+(4x-3y-4)^2+(x-2y-1)(4x-3y\color{red}{+}4)=0$$
expand it as a polynomial of $x$ to get
$$21 x^2- (39 y+34)x+(19 y^2+23 y+13)=0$$ Solve the quadratic to get
$$x_\pm=\frac{1}{42} \left(39 y+34\pm\sqrt{-75 y^2+720 y+64}\right)$$ and notice that
$$-75 y^2+720 y+64 \geq 0 \qquad \text{if}\qquad \frac{8}{15} \left(9-2 \sqrt{21}\right)\leq y \leq \frac{8}{15} \left(9+2 \sqrt{21}\right)$$ which seems to mean that there is an infinite number of solutions in the given range for $y$ which seems normal since the equation corresponds to a rotated ellipse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Does this prove that no sequential squares have a ratio of 2? The goal:
Prove that there is no integer $k$ such that ${(k+1)^2\over{k^2}}=2$.
My proof:
If ${(k+1)^2\over{k^2}}=2$, then ${{k+1}\over{k}}=\sqrt2$, and if $k$ is an integer, $k+1$ is also an integer. This implies that $\sqrt2$ is a rational number, which is also provably false.
| Just to be different:
$\frac {(k+1)^2}{k^2} = \frac {k^2 + 2k + 1}{k^2} = 1 + \frac {2k + 1}{k^2}$
So $\frac {(k+1)^2}{k^2} = 2\implies \frac {2k + 1}{k^2} = 1 \implies$
$2k + 1 =k^2$ and as $k \ne 0$ (otherwise $2 = 0$ which isn't true)
$2 + \frac 1k = k$. But if $k \ne \pm 1$ then $\frac 1k$ is not an integer.
And if $k = \pm 1$ then $\frac 1k = k$ and $2 = k -\frac 1k = 0$ which is impossible.
....
But seriously...
$\frac {(k+1)^2}{k^2} = (\frac {k+1}{k})^2 = 2$ implies there is a rational number whose square is $2$ which is well-known to be false, if a perfect and irrefutable proof. And probably the absolute easiest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 9,
"answer_id": 7
} |
Solving a system of equations: $\begin{cases}\frac xy-\frac yx=\frac{15}4\\2x-5y=9\end{cases}$ How should I approach this type of a system equation?
$$
\begin{cases}
\dfrac xy-\dfrac yx=\dfrac{15}4\\
2x-5y=9
\end{cases}
$$
I tried to multiply the first equation by $4xy$ and divide the second one by $2$. After that I got this system:
$$
\begin{cases}
4x^2 - 4y^2 = 15xy\\
x - 2.5y = 4.5 \Longrightarrow x = 4.5 + 2.5y
\end{cases}
$$
Then I put $x$ from the second equation in the first one:
$$4(4.5 + 2.5y)^2 - 4y^2 = 15y(4.5 + 2.5y)$$
When I solved it I got these results:
*
*$$x_1 = \frac{189}{22},\ y_1 = \frac{18}{11}.$$
*$$x_2 = -3,\ y_2 = -3.$$
But these results are incorrect.
These are the answers from my book:
*
*$$x_1 = 12,\ y_1 = 3.$$
*$$x_2 = \frac{9}{22},\ y_2 = -\frac{18}{11}.$$
| Multiply the first equation by $4xy$ and rewrite as
$$4x^2-15xy-4y^2=0,$$ which factors as
$$(4x+y)(x-4y)=0.$$
(you can obtain this be solving the quadratic equation in $x$).
Now you have two easy linear systems
$$\begin{cases}4x+y=0,\\2x-5y=9,\end{cases}$$
$$\begin{cases}x-4y=0,\\2x-5y=9.\end{cases}$$
For the first, $y=-4x$ and $22x=9$. For the second, $x=4y$ and $3y=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Find the Taylor series about $z=0$ and the Laurent series about $z=-3$
*
*Let $f(z)=\frac{10z}{z^2+z-6}$, find the coefficient of $z$ in the Taylor series of $f$ expanded about $z=0$ and state the open set in $\mathbb C$ where the series converges.
*Find the Laurent series of $f$ about $z=-3$ and state the open set where this series converges.
*
*For the first part:
$$\begin{align}f(z)&=\frac{6}{z+3}+\frac{4}{z-1}\\
&=\frac{6}{3}\frac{1}{1+\frac{z}{3}}-4\frac{1}{2-z}\\
&=2\frac{1}{1-(-\frac{z}{3})}-2\frac{1}{1-\frac{z}{2}}\end{align}$$
So Taylor series would be
$$2\sum_{n=0}^{\infty}\left(-\frac{z}3\right)^n-2\sum_{n=0}^{\infty}\left(\frac{z}2\right)^n$$
Where $|-\frac{z}{3}|<1$ and $|\frac{z}{2}|<1$, so $|z|<3$ and $|z|<2$, is this correct so far and how do I deduce the open set from this? Would it be $B_3(0)\setminus B_2(0)$?
*
*For the second part:
Let $z=-3+w$ and input into $f$,
$$\begin{align}f(z)&=\frac{6}{(-3+w)+3}+\frac{4}{(-3+w)-2} \\
&=\frac{6}{w}+\frac{4}{-5+w}\\
&=\frac{6}{1-1+w}+\frac{-4}{5-w}\end{align}$$
....and then I get a little lost from there.
Any corrections, tips or solutions would be great!!
| We consider the function
\begin{align*}
f(z)&=\frac{10z}{z^2+z-6}\\
&= \frac{6}{z+3} +\frac{4}{z-2}\\
\end{align*}
First part: The function $f$ is to expand around the center $z=0$ as Taylor series.
Since there are simple poles at $z=-3$ and $z=2$ we have to distinguish three regions of convergence
\begin{align*}
D_1:&\quad 0\leq |z|<2\\
D_2:&\quad 2<|z|<3\\
D_3:&\quad |z|>3
\end{align*}
*
*The first region $D_1$ is a disc with center $z=0$, radius $2$ and the pole at $z=2$ at the boundary of the disc. It admits for both fractions a representation as power series.
*The region $D_2$ is an annulus containing all points outside the closure of $D_1$ and the closure of $D_3$. It admits for the fraction with pole at $z=2$ a representation as principal part of a Laurent series and for the fraction with pole at $z=3$ a power series.
*The region $D_3$ contains all points outside the disc with center $z=0$ and radius $3$. It admits for both fractions a representation as principal part of a Laurent series.
We are interested in an expansion as Taylor series and observe the region $D_1$ is the corresponding open set of convergence with the representation of $f$ as Taylor series as stated by OP:
\begin{align*}
f(z)&= \frac{6}{z+3} + \frac{4}{z-2}\\
&= 2\sum_{n=0}^\infty\left(-\frac{z}{3}\right)^n-2\sum_{n=0}^\infty\left(-\frac{z}{2}\right)^n
\end{align*}
Second part: The function $f$ is to expand around the center $z=-3$ as Laurent series and the region of convergence is to determine.
Since there are simple poles at $z=-3$ and $z=2$ we have to distinguish two regions of convergence when expanding around $z=-3$:
\begin{align*}
D_4:&\quad 0< |z+3|<5\\
D_5:&\quad |z+3|>5
\end{align*}
*
*The region $D_4$ is a punctured disc with center $z=-3$, radius $5$ and the pole at $z=2$ at the boundary of the disc. It admits for the fraction with pole at $z=-3$ a representation as principal part of a Laurent series and for the fraction with pole at $z=2$ a power series.
*The region $D_5$ contains all points outside the disc with center $z=-3$ and radius $5$. It admits for both fractions a representation as principal part of a Laurent series.
The expansion of $f$ as Laurent series in $D_4$ is
\begin{align*}
f(z)&= \frac{6}{z+3} + \frac{4}{z-2}\\
&= \frac{6}{z+3} +\frac{4}{(z+3)-5}\\
&= \frac{6}{z+3} -\frac{4}{5}\cdot\frac{1}{1-\frac{z+3}{5}}\\
&= \frac{6}{z+3} -\frac{4}{5}\sum_{n=0}^\infty\frac{1}{5^{n}}(z+3)^{n}\\
\end{align*}
The expansion of $f$ as Laurent series in $D_5$ can be calculated similarly.
| {
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"url": "https://math.stackexchange.com/questions/2746373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Evaluate without L'Hopital: $\lim_{x\to1}\left\{\frac{9}{x^9-1}-\frac{5}{x^5-1}\right\}$ Evaluate the following limit without using L'Hospital method
$$\lim_{x\to1}\left\{\frac{9}{x^9-1}-\frac{5}{x^5-1}\right\}$$
My turn is
$$L=\lim_{x\to1}\left\{\frac{9(x^5-1)-5(x^9-1)}{x^{14}-x^9-x^5+1}\right\}$$
$$L=\lim_{x\to1}\left\{\frac{9(x^5-1)-5(x^9-1)}{(x^{14}-1)-(x^9-1)-(x^5-1)}\right\}$$
then divide the numerator and the denominator by $$x-1$$
But I got again $$\frac{0}{0}$$
| Note that
$$ x^n - 1 = (x-1)(x^{n-1}+x^{n-2}+\cdots + x + 1). $$
Thus we have
\begin{align}
\frac{9}{x^9-1} - \frac{5}{x^5-1} &= \frac{1}{x-1}\left[\frac{9}{\sum_{i=1}^9 x^{i-1}} - \frac{5}{\sum_{i=1}^5 x^{i-1}} \right] \\
&=\frac{1}{x-1}\left[\frac{9\sum_{i=1}^5 x^{i-1} - 5\sum_{i=1}^9 x^{i-1}}{(\sum_{i=1}^9 x^{i-1})(\sum_{i=1}^5 x^{i-1})}\right].
\end{align}
And note that
\begin{align}
9\sum_{i=1}^5 x^{i-1} - 5\sum_{i=1}^9 x^{i-1}
&=4\sum_{i=1}^5 x^{i-1} - 5x^5\sum_{i=1}^4 x^{i-1} \\
&=4\sum_{i=1}^4 x^{i-1}(1-x^5) + x^4(4-x-x^2-x^3-x^4) \\
&=4(1-x)\left(\sum_{i=1}^4 x^{i-1}\right)\left(\sum_{i=1}^5 x^{i-1}\right) \\
&\hspace{0.5cm} +x^4(1-x)(1+(1+x)+(1+x+x^2)+(1+x+x^2+x^3))
\end{align}
Therefore,
\begin{align}
\frac{9}{x^9-1} - \frac{5}{x^5-1} &= -\frac{4(\sum_{i=1}^4 x^{i-1})(\sum_{i=1}^5 x^{i-1}) + x^4(4+3x+2x^2+x^3)}{(\sum_{i=1}^9 x^{i-1})(\sum_{i=1}^5 x^{i-1})}
\end{align}
and by letting $x \to 1$, we obtain
$$
\lim_{x \to 1} \left[\frac{9}{x^9-1} - \frac{5}{x^5-1}\right]
= -2.
$$
| {
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"url": "https://math.stackexchange.com/questions/2746728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sum_{k=1}^n\binom{2n-1}{n-k}=4^{n-1}$. Let $S_n=\sum_{k=1}^n\binom{2n-1}{n-k}$, I tried to prove that $S_{n+1}=4S_n$:
\begin{align}
S_{n+1}&=\sum_{k=1}^{n+1}\binom{2n+1}{n+1-k}\\
&=\sum_{k=0}^{n}\binom{2n+1}{n-k}=\text{stuck}
\end{align}
Maybe some $\Gamma$ functions?
$$S_n=\Gamma(2n)\sum_{k=1}^n\frac{1}{\Gamma(n-k+1)\Gamma(n+k)}=?$$
Any hint will be appreciated.
| Since you are aware of the gamma function and its existence, I'm going to assume that you are also well aware of the hockey stick identity.$$\binom {n-1}{r-1}+\binom {n-1}{r}=\binom nr\tag1$$Which can be proven algebraically by expanding both sides and seeing that they're both equal. Starting from where you left off, we use the (1) to break the sums up in terms of $\binom {2n-1}k$ where $1\leq k\leq n$. The first expansion yields$$\begin{align*}S_{n+1} & =\color{blue}{\binom {2n+1}n}+\color{red}{\binom {2n+1}{n-1}}+\cdots+\color{brown}{\binom {2n+1}1}+\color{green}{\binom {2n+1}0}\\ & =\color{blue}{\binom {2n}{n-1}+\binom {2n}n}+\color{red}{\binom {2n}{n-2}+\binom{2n}{n-1}}+\cdots+\color{brown}{\binom{2n}0+\binom{2n}1}+\color{green}{\binom {2n+1}0}\\ & =\binom {2n}n+2\binom {2n}{n-1}+2\binom {2n}{n-2}+\cdots+2\binom {2n}1+2\binom {2n}0\end{align*}$$Note that the last term appears because $\binom {2n}0=\binom{2n+1}0=1$. With that out of the way, we again use (1) to expand $S_{n+1}$. The reason behind this is because we have $S_n$ defined as$$S_n=\sum\limits_{k=1}^n\binom {2n-1}{n-k}\tag2$$So if we can get the top term of the binomial coefficient to equal $2n-1$, then we can use (2) to write a recurrence relation. From which we can easily solve for $S_{n+1}$. Hockey stick identity gives us$$\begin{align*}S_{n+1} & =\binom {2n}n+\color{blue}{2\binom {2n}{n-1}}+\color{red}{2\binom {2n}{n-2}}+\cdots+\color{green}{2\binom{2n}1}+2\binom {2n}0\\ & =\binom {2n}n+\color{blue}{2\left[\binom {2n-1}{n-2}+\binom {2n-1}{n-1}\right]}+\color{red}{2\left[\binom {2n-1}{n-3}+\binom {2n-1}{n-2}\right]}+\\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\cdots+\color{green}{2\left[\binom {2n-1}0+\binom {2n-1}1\right]}+2\\ & =\binom {2n}n+4\sum\limits_{k=1}^n\binom {2n-1}{n-k}-2\binom {2n-1}{n-1}\\ & =\binom {2n}n-2\binom {2n-1}{n-1}+4S_n\end{align*}$$The right-hand side can be reduced to $4S_n$ for all $n>0$ and $n$ is an integer. To prove that the two binomials reduce to zero, we can show that they are equal to each other. Using the basic definition of the binomial coefficient that we all learned in kindergarden, the simplification is as follows$$\begin{align*}\binom {2n}n & =\frac {(2n)!}{(n!)^2}\\ & =\frac {2n}n\frac {(2n-1)!}{n!(n-1)!}\\ & =2\left[\frac {(2n-1)!}{n!(n-1)!}\right]\\ & =2\binom {2n-1}{n-1}\tag3\end{align*}$$Hence, by (3), we have$$\begin{align*}S_{n+1} & =4S_n\qquad\implies\qquad S_n=4^{n-1}\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Sum to $n$ terms of the given series Find the sum to $n$ terms of the given series:
$$0.3+0.33+0.333+0.3333+\cdots$$
My Attempt:
Let
$$S=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$
$$=\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots$$
$$=\frac {3}{10} \left[1+\frac {11}{10}+\frac {111}{100}+\frac {1111}{1000}+\cdots \text{ to $n$ terms}\right]$$
How do I continue from here?
| Here's different flavor that's more elementary and explicit (maybe).
At the $k$-th term among summing $k=1$ to $k = n$, we have the multiplicative factor you got:
\begin{align*}
\frac{ 111\cdots11 }{ 100\cdots00} &= 1 + \frac1{10} + \frac1{100} + \cdots \frac1{10^{k-1}} \tag*{, totally $k$ terms}\\
&= \frac{ 1 - (1/10)^k}{1 - 1/10} \\
&= \frac{10}9 \left( 1 - \frac1{10^k}\right)
\end{align*}
Thus the original summation of $n$ terms is
\begin{align*}
S = \frac3{10}\frac{10}9 \sum_{k = 1}^n \left( 1 - \frac1{10^k}\right) &= \frac13 \cdot \left(n - \frac{ \frac1{10} - \frac1{10^{n+1}} }{1 - \frac1{10}} \right) \\
&=\frac13 \cdot \left(n - \frac19 \left(1 - \frac1{10^n} \right) \right) \\
&= \frac13 \cdot \left(n - \frac19 \frac{ \,\,\overbrace{ 999\cdots 99 }^{n~\text{digits} }}{ \underbrace{ 100\cdots 000 }_{n+1~\text{digits} } } \right) \\
&= \frac13 \left( n - \frac{ \,\,\overbrace{ 111\cdots 11 }^{n~\text{digits} }}{ \underbrace{ 100\cdots 000 }_{n+1~\text{digits} } }\right) = \frac13 \cdot (n - 0.111\cdots 11)
\end{align*}
There are $n$ digits of $1$ following the decimal point. Or if you prefer,$$S = \frac13 \bigl(n - 1 + 0.888\cdots 89 \bigr)$$where there are $n-1$ digits of $8$ followed by a single $9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Evaluate the integral$ \int_{-\infty}^{\infty}\frac{b\tan^{-1}\big(\frac{\sqrt{x^2+a^2}}{b}\big)}{(x^2+b^2)(\sqrt{x^2+a^2})}\,dx$. I am attempting to evaluate
$$\int_{-\infty}^{\infty}\dfrac{b\tan^{-1}\Big(\dfrac{\sqrt{x^2+a^2}}{b}\Big)}{(x^2+b^2)(\sqrt{x^2+a^2})}\,dx. $$
I have tried using the residue formula to calculate the residues at $\pm ib,\pm ia,$ but it got messy very quickly. Then I tried to use a trigonometric substitution $x=a\tan(\theta)$; $dx=a\sec^{2}(\theta)\,d\theta$ which led me to the integral $$\int_{-\infty}^{\infty}\dfrac{b\tan^{-1}\Big(\dfrac{a\sec(\theta)}{b}\Big)\sec(\theta)}{(a^2\tan^{2}(\theta)+b^2)}\,d\theta.$$ The bounds for this integral seem incorrect, but I am more worried about the actual expression before I deal with the bounds, which may have to be changed into a double integral where $0\leq\theta\leq2\pi$ and the second bound would range from $-\infty$ to $\infty$. I am wondering if there is some kind of substitution I have missed, but I have hit the wall. The OP of this problem said there were cases that would come into play, but when I asked him whether or not those cases arose from $b<0$ and $b>0$ he told me they did not. The cases most likely arise from whether $a$ and $b$ are positive or negative, because the case where $b=0$ is trivial, and in the case where $a=0$ I used wolframalpha and the integral evaluates to $\dfrac{\pi\ln(2)\lvert b \rvert}{b^2}$ for $\Im(b)=0 \land \Re(b)\neq0.$ Contour integration may be necessary. I am stuck on this problem and I would greatly appreciate the help. Thank you for your time.
| In the following we shall assume that $a$, $b$, and $c$ are real valued and that $a>b>0$.
Let $F(c)$ be represented by the integral
$$F(c)=b\int_{-\infty}^\infty \frac{\arctan\left(\frac{\sqrt{x^2+a^2}}{c}\right)}{(x^2+b^2)\sqrt{x^2+a^2}}\,dx\tag 1$$
Differentiating $(1)$ reveals
$$\begin{align}
F'(c)&=-b\int_{-\infty}^\infty \frac{1}{(x^2+b^2)(x^2+a^2+c^2)}\,dx\\\\
&=-\frac{\pi}{c^2+a^2+b\sqrt{c^2+a^2}}\tag2
\end{align}$$
Integrating $(2)$ and using $\lim_{c\to \infty}F(c)=0$, we find that
$$\begin{align}
F(c)&=\pi\,\left(\frac{\arctan\left(\frac{bc}{\sqrt{a^2-b^2}\sqrt{a^2+c^2}}\right)-\arctan\left(\frac{b}{\sqrt{a^2-b^2}}\right)+\pi/2-\arctan\left(\frac{c}{\sqrt{a^2-b^2}}\right)}{\sqrt{a^2-b^2}}\right)
\end{align}$$
Setting $c=b$ yields the coveted result
$$\int_{-\infty}^\infty \frac{b\arctan\left(\frac{\sqrt{x^2+a^2}}{b}\right)}{(x^2+b^2)\sqrt{x^2+a^2}}\,dx=\pi\,\left(\frac{\arctan\left(\frac{b^2}{\sqrt{a^2-b^2}\sqrt{a^2+b^2}}\right)+\pi/2-2\arctan\left(\frac{b}{\sqrt{a^2-b^2}}\right)}{\sqrt{a^2-b^2}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2751743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Is the function is differentiable at $0$? Is the function given by
$\displaystyle f(x) = \begin{cases} \dfrac{1}{x\log(2)} - \dfrac{1}{2^x -1}, \quad &x \neq 0, \\ \dfrac{1}{2} , &x = 0 \end{cases}$
differentiable at $0$ ?
My attempt : $f'(0) = \frac {f(x) -f(0)}{x-0}$
$$\lim _{x\to 0}f'(0) = \lim _{x\to 0}\frac {\frac{1}{x\log(2)} - \frac{1}{2^x -1} - \frac{1}{0\log(2)} - \frac{1}{2^0 -1}}{x- 0}$$
$$\lim _{x\to 0}\frac {\frac{1}{x\log(2)} - \frac{1}{2^x -1} - \frac{1}{0} - \frac{1}{0}}{x- 0}$$
Now I am not able to proceed further.
Any hints /solution will be appreciated.
Thanks in advance
| The idea is to use an expansion near $x=0$:
$$\frac{1}{2^x-1}=\frac{1}{e^{x \ln(2)}-1}=\frac{1}{1+x \ln(2) +\frac{x^2 \ln(2)^2}{2}+\frac{x^3 \ln(2)^3}{6}-1+o(x^3)}=\frac{1}{x \ln(2) \left(1+\frac{x \ln(2)}{2}+\frac{x^2 \ln(2)^2}{6}+o(x^2) \right)}$$
so near $x=0$:
\begin{align}
f(x)&=\frac{1}{x \ln(2)}-\frac{1}{x \ln(2) \left(1+\frac{x \ln(2)}{2}+\frac{x^2 \ln(2)^2}{6}+o(x^2) \right)}\\
&=\frac{1}{x \ln(2)}-\frac{1}{x \ln(2)} \left(1-\frac{x \ln(2)}{2}-\frac{x^2 \ln(2)^2}{6}+\left(\frac{x \ln(2)}{2}\right)^2+o(x^2) \right)\\
&=\frac{1}{2}+x \left(\frac{\ln(2)}{6}-\frac{\ln(2)}{4} \right)+o(x)
\end{align}
so, as $f(0)=\frac{1}{2}$:
$$\frac{f(x)-f(0)}{x-0}=\frac{\frac{1}{2}-x \frac{\ln(2)}{12}+o(x)}{x}=-\frac{\ln(2)}{12}+o(1)$$
so $f$ is differentiable at $x=0$ and:
$$f'(0)=-\frac{\ln(2)}{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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