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Solve the equation $x^6-2x^5+3x^4-3x^2+2x-1=0$ Solve the equation $$x^6-2x^5+3x^4-3x^2+2x-1=0$$ Let's divide both sides of the equation by $x^3\ne0$ (as $x=0$ is obviously not a solution, we can consider $x\ne0$). Then we have $$x^3-2x^2+3x-3\cdot\dfrac{1}{x}+2\cdot\dfrac{1}{x^2}-\dfrac{1}{x^3}=0\\\left(x^3-\dfrac{1}{x^3}\right)-2\left(x^2-\dfrac{1}{x^2}\right)+3\left(x-\dfrac{1}{x}\right)=0$$ What do we do now? If we say $y=x-\dfrac{1}{x}$, we won't be able to express $\left(x^2-\dfrac{1}{x^2}\right)$ in terms of $y$ because of the minus sign as $(a-b)^2=a^2-2ab\color{red}{+b^2}$. On the other side, $$y^3=x^3-\dfrac{1}{x^3}-3\left(x-\dfrac{1}{x}\right)\\x^3-\dfrac{1}{x^3}=y^3+3y$$ I think this is a traditional issue when solving reciprocal equations, but I can't figure out how to deal with it.
| Noticing that $x=1$ and $x=-1$ are roots and by long division, you'll get $x^6-2x^5+3x^4-3x^2+2x-1=(x-1)(x+1)(x^4-2x^3+4x^2-2x+1)$
Considering $x^4-2x^3+4x^2-2x+1=0$, you now divide by $x^2$ and you'll get $x^2+\frac{1}{x^2}-2(x+\frac{1}{x})+4=0$ which, using the substitution $y=x+\frac{1}{x}$, gives $y^2-2y+2=0$.
I believe you can finish this from here :)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Inscribed circle in a trapezoid BA = 3. The radius of the circle is 2. What is DC?
I know that AG = AF = 1. And i know that FC = CE. And DC = DE + EC. I drew a line from point A to line DC to create a right triangle, but I can’t figure out where to go from there
| Attach a coordinate system to the figure, and let its origin be at point $D$.
Then $Z = (2, 2)$, and $A = (3, 4) $, and $C = (x_1, 0) $
When want to solve for $x_1$ such that the distance between $Z$ and $AC$ is $2$.
The equation of $AC $ is $ y - 4 = \dfrac{-4}{x_1-3} (x - 3) $
We want to make sure the coefficients of $x$ and $y$ are negative.
Applying the distance formula using the above line equation and the point $Z = (2,2)$, we obtain
$ d = 2 = \dfrac{ -2 + 4 - \dfrac{4}{x_1 - 3} (2 - 3) }{ \
\sqrt{1 + \dfrac{16}{(x_1 - 3)^2}}} $
From which,
$ 4 \left( 1 + \dfrac{16}{(x_1 - 3) ^2 } \right ) = \left(2 + \dfrac{4}{x_1 - 3}\right)^2 = 4 + \dfrac{16}{(x_1 - 3)^2} + \dfrac{16}{(x_1 - 3)} $
So that,
$\dfrac{ 3 }{ ( x_1 - 3) ^ 2 } = \dfrac{1}{ (x_1 - 3) } $
And therefore,
$ (x_1 - 3) = 3 $
Hence $x_1 = 6 $
And this is equal to the the length $DC$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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The equation $x^{4}-3x^{3}-6x^{2}+ax+b=0$ has a triple root. Find $a$ and $b$, and hence all roots of this equation. The given question is:
The equation $x^{4}-3x^{3}-6x^{2}+ax+b=0$ has a triple root. Find $a$ and $b$, and hence all roots of this equation.
I am confused about how to work out this question, but I feel like it has something to do with using α,β,γ, δ as roots of the quartic.
I would greatly appreciate all help in a simple and efficient method to solving this question. Please note, according to the answers, there is only one value of $a$ and one value of $b$.
| We know that the equation has a triple root, so if let $\alpha$ be the triple root and $\beta$ be the single root, we get that
\begin{equation}
x^4 - 3x^3-6x^2 + ax + b = (x-\alpha)^3(x-\beta)
\end{equation}
By expanding the RHS, we see that
\begin{equation}
x^4 - 3x^3-6x^2 + ax + b = x^4 - (3\alpha+\beta)x^3 + (3\alpha\beta + 3\alpha^2)x^2 + \cdots
\end{equation}
Now equating the coefficients of the $x^2$ and $x^3$ term, we have the simultaneous equations
\begin{equation}
3 = 3\alpha+\beta \quad ; \quad -6 = 3\alpha\beta + 3\alpha^2
\end{equation}
By a simple substitution we conclude $(\alpha,\beta) = (2,-3)$ or $(\alpha,\beta) = (-1/2, 9/2)$. We can check that these indeed satisfy our conditions. It then follows that we have either $(a,b) = (28,-24)$ or $(a,b) = (-13/4,-9/16)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $xy+yz+zx-xyz \leq \frac{9}{4}.$ Currently I try to tackle some olympiad questions:
Let $x, y, z \geq 0$ with $x+y+z=3$. Show that
$$
x y+y z+z x-x y z \leq \frac{9}{4}.
$$
and also find out when the equality holds.
I started by plugging in $z=3-x-y$ on the LHS and got
$$
3y-y^2+3x-x^2-4xy+x^2y+xy^2 = 3y-(y^2+x^2)+3x-4xy+x^2y+xy^2\leq 3y-((y+x)^2)+3x-4xy+x^2y+xy^2
$$
But this got me nowhere.
Then I started again with the left hand side
$$
x y+y z+z x-x y z \Leftrightarrow yz(1-x)+xy+zx
$$
and $x+y+z=3 \Leftrightarrow y+z-2=1-x$ so
$$
yz(y+z-2)+x(y+z)
$$
But this also leaves no idea. Do I have to use a known inequality?
| In this answer I use Lagrange multipliers to find the critical points of the function within the region $x,y,z\geq0$, show that the only critical point is a minimum at $x=y=z=1$, and then argue it must achieve a maximum on the boundary and find that maximum.
Let $f=xy+yz+zx-xyz$ then $$\nabla f=(1-(y-1)(z-1),1-(z-1)(x-1),1-(x-1)(y-1))$$
Let $g=x+y+z-3$ then $\nabla g=(1,1,1)$ and set
$\nabla f-\lambda \nabla g=0$ then we have
$$
\begin{align}
(y-1)(z-1)&=1-\lambda\\
(x-1)(z-1)&=1-\lambda\\
(x-1)(y-1)&=1-\lambda\\
\end{align}
$$
Equating these easily gives us $x=y=z=1$.
By setting, say $x=\frac13, y=z=\frac43$, we get $f=56/27$, and we can see this is a minimum, and since there is no other critical point, the maximum value attained by $f$ must lie on the boundary.
Without loss of generality, set $x=0$ then $f=yz=(3-z)z$, which by symmetry attains a maximum at $z=3/2$. Set $x=y=0$ and $f\equiv 0$, so the maximum of $9/4$ is at $x=0, y=z=3/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Two different answers to $\lim_{x \to -\infty} \frac{8x^2-2x^3+1}{6x^2+13x+4}$ Let us evaluate
$$\lim_{x \to -\infty} \frac{8x^2-2x^3+1}{6x^2+13x+4}$$
Dividing the numerator and denominator by $x^3$ we will be left with
$$\lim_{x \to -\infty} \frac{\frac{8}{x}-2+\frac{1}{x^3}}{\frac{6}{x}+\frac{13}{x^2}+\frac{4}{x^3}}$$ and since all the terms containg $x$ term will go to $0$,our resultant limit becomes $-\frac{2}{0}=-\infty$.
But if we do it in another way substituting $x=-t$,then our limit becomes after dividing both numerator and denominator by $t$,
$$\lim_{t \to \infty} \frac{\frac{8}{t}+2+\frac{1}{t^3}}{\frac{6}{t}-\frac{13}{t^2}+\frac{4}{t^3}}$$
Here also since terms containing $t$ go to $0$,we are left with $\frac{2}{0}=+\infty$.
Why are we getting two different answers? Surely one of the method is invalid. In books,the first answer was marked correct. But i want to know what's wrong with the second approach.
| Note that
*
*for $x\to -\infty$
$$\frac{6}{x}+\frac{13}{x^2}+\frac{4}{x^3} = \frac{1}{x}\left(6+\frac{13}{x}+\frac{4}{x^2}\right)\to 0 \cdot6=0$$
with $\frac 1x <0$, then $0$ is reached from the left side ($0^-$).
*
*for $t=-x\to \infty$
$$\frac{6}{t}-\frac{13}{t^2}+\frac{4}{t^3} = \frac{1}{t}\left(6-\frac{13}{t}+\frac{4}{t^2}\right)\to 0 \cdot6=0$$
with $\frac 1t >0$, then $0$ is reached from the right side ($0^+$).
Therefore in both cases limit is $\infty$.
Note that the second approach is fine indeed from here
$$\lim_{x \to -\infty} \frac{8x^2-2x^3+1}{6x^2+13x+4}=\lim_{t \to \infty} \frac{8t^2+2t^3+1}{6t^2-13t+4}$$
we can proceed as follows
$$\frac{8t^2+2t^3+1}{6t^2-13t+4}=\frac{t^3}{t^2}\cdot \frac{\frac8t+2+\frac1{t^3}}{6-\frac{13}t+\frac 4{t^2}}=t\cdot \frac{\frac8t+2+\frac1{t^3}}{6-\frac{13}t+\frac 4{t^2}}$$
from wich we obtain $\infty$ as limit.
Note also that in a similar way for the first one
$$\frac{8x^2-2x^3+1}{6x^2+13x+4}=x\cdot \frac{\frac 8 x-2+\frac 1{x^3}}{6+\frac {13} x+\frac 4{x^2}}$$
wich also leads to $\infty$ as limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned}
$$
Then, we have
$$
\begin{aligned}
0<x<2\\
\implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned}
$$
Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.
This leads,
$$
\begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.
Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.
We have:
$$
\begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.
Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.
This case implies, $\left(2x-\frac a2\right)^2≥0$. This means, $f(x)≥\frac {3a^2}{4}-5a+4$
Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.
Finally, we have to combine all the solution sets we get.
I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?
| I've revised your work as follows.
The first step is fine
$$ 0<x<2 \implies -\frac a2<2x-\frac a2<4-\frac a2$$
As a minor issue we should use $\iff$ instead of $\implies$.
*
*Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$
The condition corresponds to $a\ge8$.
The inequality you reach $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0 \iff (a-4)(a-5)>0$ holds.
*
*Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$
The condition corresponds to $a\le0$.
The inequality you reach $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0 \iff (a-1)(a-4)>0$ holds.
*
*Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$
The condition corresponds to $0\le a\le8$. So we could also consider $0<a<8$ but it is not necessary.
The inequality you reach $\frac {3a^2}{4}-5a+4>0 \iff a<\frac{10}3-\frac{2\sqrt{13}}3\approx0.93 \land a>\frac{10}3+\frac{2\sqrt{13}}3\approx5.74 $ holds.
The method looks fine to me since you have covered all the cases!
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing $\int_1^\infty\left(\sqrt{\sqrt{x}-\sqrt{x-1}}-\sqrt{\sqrt{x+1}-\sqrt{x}}\right)dx=\frac4{15}\left(\sqrt{26\sqrt2-14}-2\right)$ A Putnam problem asked to show that some improper integral is convergent, but I was curious to see if it can be computed in closed form and Mathematica came up with this:
$$\int_1^{\infty} \left(\sqrt{\sqrt{x}-\sqrt{x-1}} -\sqrt{\sqrt{x+1}-\sqrt{x} } \right)dx=
\frac{4}{15} \left(\sqrt{26\sqrt{2}-14 }-2\right) \approx 0.739132$$
I did a few substitutions but it didn't turn out as an easy calculation. I remember that eliminating square roots required some Euler type substitutions. Any ideas of how one can arrive at such a surprising result?
| It's easy to show that the given integral comes down to evaluating:
$$I = \int_{0}^{1} \sqrt{\sqrt{x+1}-\sqrt{x}} \ dx$$
Define the following function on $[0,1]$:
$$f(x) = -\sqrt{\sqrt{x+1}-\sqrt{x}}$$
I claim that that this is invertible. Let's just find an explicit inverse for this. So:
$$y^2 = \sqrt{x+1}-\sqrt{x}$$
This implies that:
$$y^4+2\sqrt{x}y^2+x = x+1$$
$$\sqrt{x} = \frac{1-y^4}{2y^2}$$
which implies that $x = \frac{(1-y^4)^2}{4y^4}$. But this just means that:
$$f^{-1}(x) = \frac{1}{4x^4} -\frac{1}{2} + \frac{x^4}{4}$$
where the inverse is defined on $[-1,-\sqrt{\sqrt{2}-1}]$. Now, it follows that:
$$\int_{0}^{1} f(x) \ dx + \int_{-1}^{-\sqrt{\sqrt{2}-1}} f^{-1}(x) \ dx = -\sqrt{\sqrt{2}-1}$$
Let's evaluate the integral of the inverse function because that's easy. So, we have that:
$$\int_{-1}^{-\sqrt{\sqrt{2}-1}} f^{-1}(x) \ dx = \left[-\frac{1}{12x^3} -\frac{1}{2}x+\frac{x^5}{20}\right]_{-1}^{-\sqrt{\sqrt{2}-1}}$$
Now, I'm not going to include all the details here but you can use the above to show that:
$$I = \frac{4}{15} \sqrt{\sqrt{2}-1}(6+\sqrt{2})-\frac{8}{15}$$
This is just obtained by plugging in the bounds. But now, notice that:
$$\sqrt{\sqrt{2}-1}(6+\sqrt{2}) = \sqrt{(\sqrt{2}-1)(6+\sqrt{2})^2} = \sqrt{(\sqrt{2}-1)(38+12\sqrt{2})} = \sqrt{26\sqrt{2}-14}$$
and this gives the desired result. $\Box$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Rationalize the denominator of $\frac{1}{1+\sqrt[3]{3}-\sqrt[3]{9}}$ Rationalize the denominator of $$\dfrac{1}{1+\sqrt[3]{3}-\sqrt[3]{9}}$$ Usually we are supposed to use one of the formulas $$x^3\pm y^3=(x\pm y)(x^2\mp xy+y^2)$$ I don't think they will work here. We can say $\sqrt[3]{3}=t\Rightarrow t^3=3$ and the given expression is then $$\dfrac{1}{1+t-t^2}$$ I don't see anything else. What are the available approaches?
| You can proceed by identification
$(1+t-t^2)(a+bt+ct^2)=(b+c-a)t^2+(a+b-3c)t+(3c-3b+a)$
To get rid of the surds, solve $\begin{cases}b+c-a=0\\a+b-3c=0\end{cases}\implies\begin{cases}a=2c\\b=c\end{cases}$
We can set $c=1$ which correspond to the $(2+t+t^2)$ indicated in the other answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4530664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Integration Issue with Finding a Centroid I've been brushing up on my understanding of centroids in 2-dimensions, and I chose to try to find $M_x$ of the region bounded by $$f(x) = \sin(x-\frac{\pi}{2})+3$$ and the x-axis, $g(x)=0$, from $x=\frac{\pi}{2}$ to $x=\frac{7\pi}{2}$.
This gives:
$$M_x=\int_\frac{\pi}{2}^\frac{7\pi}{2}((\frac 12)(\sin(x- \frac{\pi}{2})+3))(\sin(x- \frac{\pi}{2})+3)dx$$
$$=\frac12\int_\frac \pi2^\frac{7\pi}{2}(\sin(x- \frac{\pi}{2})+3)(\sin(x- \frac{\pi}{2})+3)dx$$
$$=\frac12\int_\frac \pi2^\frac{7\pi}{2}(\sin(x- \frac{\pi}{2})+3)^2dx$$
$$=\frac12\int_\frac \pi2^\frac{7\pi}{2} (\sin^2(x- \frac{\pi}{2}) +6\sin (x- \frac{\pi}{2}) + 9) \ dx$$
$$=\frac12 \left[
{\sin^3(x- \frac{\pi}{2})\over 3\cos(x- \frac{\pi}{2})}
+ 6(-\cos (x- \frac{\pi}{2}) )
+ 9x
\right]_ \frac{\pi}{2} ^ \frac{7\pi}{2}
$$
$$=\frac12 \left[ \left(
0
+ 6
+ \frac{63\pi}{2}
\right)
- \left(
0
- 6
+ \frac{9\pi}{2}
\right)
\right]
$$
$$
=\frac12 (12 + \frac{54\pi}{2})
$$
$$
=6 + \frac{27\pi}{2}
$$
$$ \approx 48.4115$$
For the exact same integral Desmos gives: $M_x=50.7676953137$,
and Wolfram gives: $M_x \approx 50.768 $.
I can't tell what I'm doing wrong. Would someone be able to point it out for me?
| The integral $$M_x = \frac{1}{2} \int_{x=\pi/2}^{7\pi/2} \left(\sin \left(x - \frac{\pi}{2}\right) + 3 \right)^2 \, dx$$ is more easily evaluated by first performing the substitution $u = x - \pi/2$, $du = dx$ to obtain
$$\begin{align}
M_x &= \frac{1}{2} \int_{u=0}^{3\pi} \sin^2 u + 6 \sin u + 9 \, du \\
&= \frac{1}{4} \int_{u=0}^{3\pi} 1 - \cos 2u + 12 \sin u + 18 \, du \\&= \frac{1}{4} \left[ -\frac{1}{2} \sin 2u - 12 \cos u + 19u \right]_{u=0}^{3\pi} \\
&= \frac{1}{4} \left( -0 + 12 + 19(3\pi) + 0 + 12 - 19(0)\right) \\
&= 6 + \frac{57\pi}{4}.
\end{align}$$
As to why your calculation is incorrect, it appears that you are asserting $$\int \sin^2 x \, dx = \frac{\sin^3 x}{3 \cos x} + C.$$ This is not at all correct, as differentiation of the RHS will show that $$\frac{d}{dx}\left[\frac{\sin^3 x}{3 \cos x}\right] = \frac{2 \sin^2 x + \tan^2 x}{3} \ne \sin^2 x.$$ I'm not sure how you arrived at such a claim; perhaps you thought more generally
$$\int (f(x))^2 \, dx = \frac{(f(x))^3}{3 \int f(x) \, dx}$$ in some kind of "reverse chain rule" sense? This is not a valid identity. To integrate $\sin^2 x$, you should employ the trigonometric identity $$\sin^2 x = \frac{1- \cos 2x}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all real solutions of the equation $x^{10} - x^8 + 8x^6 - 24x^4 + 32x^2 - 48 = 0$ I have been able to factorize the polynomial as follows: $$(x^2 - 2)(x^8 + x^6 + 10x^4 - 4x^2 + 24)$$ from which $\sqrt2$ and $-\sqrt2$ are obvious solutions. My guess is that $x^8 + x^6 + 10x^4 - 4x^2 + 24 = 0$ does not have any real solutions, but how would I show that?
| Try to write $x^8 + x^6 + 10x^4 - 4x^2 + 24 = 0$ as sum of squares and positive constant.
There is only one negative term, and else are all square already. So I try to factorize the negative term into a square, like
$x^8 + x^6 + 9x^4 + x^4 - 4x^2 + 4 + 20 =...>0$
You can try to finish the rest.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Inequality with trigonometric relations performing a Von Neumann analysis, I arrived to the relation
\begin{equation}
E_m(t+\Delta t)=\underbrace{\left[1-\frac{r}{4}\left(3-4e^{-i\theta}+e^{-2i\theta}\right)\right]}_{G}E_m(t),
\end{equation}
where $\theta\in[-\pi,\pi]$ and $r>0$. The question is to find the largest $r$ such that the factor $G$ has absolute value less or equal than 1, i.e., $|G|\leq 1$.
I know that I can estimate $r$ numerically, but the idea is to find some analytical relation. I have spent quite some time looking to that expression without success, so, any help will be appreciated.
Update: The inequality $|G|\leq 1$ has to be valid for all values of $\theta$.
| Note that for $z=e^{-i\theta}$ one has $G(z)=1-\frac{r}{4}(z^2-4z+3)$ so by the polar C-R the maximum of $|G(z)|$ when $|z|=1$ is attained at points where $\frac{zG'(z)}{G(z)} \ge 0$ (which of course means $\frac{zG'(z)}{G(z)}$ real)
Now $G'(z)=\frac{r}{2}(2-z)$ is non zero on $|z|=1$ so the above is equivalent to $\frac{G(z)}{zG'(z)} > 0$ and simplifying $r/2 >0$ one gets $\frac{G(z)}{2z-z^2}>0$ hence one needs $\Im (G(z) (2\bar z-\bar z^2)) =0$ so using $z\bar z=1$ one needs
$$\Im (2\bar z-\bar z^2-\frac{r}{4}(2z+6\bar z+4\bar z-3\bar z^2))=0$$
But $z=e^{-i\theta}$ so $\Im (2z+10\bar z)=8\sin \theta$ hence we get:
$2\sin \theta -\sin 2\theta-2r\sin \theta +\frac{3r}{4} \sin 2\theta=0$ which means $\sin \theta =0$ or $2-2\cos \theta-2r+\frac{3r}{2}\cos \theta=0$
In particular for $\theta =\pi$ one gets that $z=-1$ and $G(-1)=1-2r$ hence we need $r \le 1$.
For $r=1$ one gets $\cos \theta =0$ as the other roots of the maximum modulus equation so $z =\pm i$ and $G(\pm i)=1-\frac{1}{4}(2 \pm 4i)$ has modulus greater than $1$ so $r<1$ and we need to study the points where $\cos \theta =\frac{4-4r}{4-3r}=f(r)$ so $\sin^2 \theta =1-f(r)^2, \cos 2\theta =2f(r)^2-1, \sin 2\theta=2f(r)\sin \theta$
Now (separating in real and imaginary parts) $$|G(r,\theta)|^2=(1-\frac{r}{2}(1-\cos \theta)^2)^2+\frac{r^2(2-\cos \theta)^2\sin^2 \theta}{4}$$ so substituting as above, one gets
$$|G_{\max}(r)|^2 \ge (1-\frac{r^3}{2(4-3r)^2})^2+\frac{r^2}{4}(2-\frac{4-4r}{4-3r})^2(1-\frac{(4-4r)^2}{(4-3r)^2})$$ or by further simplifications
$$|G_{\max}(r)|^2 \ge \frac{(18r^2+32-r^3-48r)^2+4r^3(2-r)^2(8-7r)}{4(4-3r)^4}$$
But (with the help of Wolfram Alpha) we note that:
$$\frac{(18r^2+32-r^3-48r)^2+4r^3(2-r)^2(8-7r)}{4(4-3r)^4}=\frac{(2-r)^2(r+4)}{4(4-3r)}=\frac{r^3}{4(4-3r)}+1$$
which shows that $$|G(r, \theta)|^2 \ge \frac{r^3}{4(4-3r)}+1$$ for $0 \le r \le 1$ and actually the analysis above shows that is the maximum for $0 \le r \le 1$ and it is bigger than $1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4536282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $|z| \leq 1$, then $|\text{Im}(\bar{z} + 1 )| \leq \sqrt{3-x^2}$ If $|z| \leq 1$, then $|\text{Im}(\bar{z} + 1 )| \leq \sqrt{3-x^2}$ $\quad $ for $0 \leq x \leq1.$
Attempt:
Let $z = x + iy$ such that $\bar{z} = x -iy$.
Thus,
$$|z|^2 = x^2 + y^2 \leq 1 \quad \quad \quad (1)$$ and
$$\bar{z} = x - iy$$
$$\bar{z} + 1 = (x+1) - iy$$
$$\text{Im}(\bar{z} + 1) = -y$$
$$|\text{Im}(\bar{z} + 1 )| = |-y| = |y| \leq \sqrt{1-x^2} \leq \sqrt{3-x^2}\quad \quad \text{(because of the inequality in (1))}$$
which is true for $0 \leq x \leq1.$
Are my arguments until the final answer valid and logical?
| Your way is fine and correct, more simply, as an alternative, from here $|y| \leq \sqrt{3-x^2}$ we can square both side to obtain
$$y^2\le 3-x^2 \iff x^2+y^2 \le 3$$
which is true since $x^2 + y^2 \leq 1$.
Another way, by polar coordinates with $|r|\le 1$
$$|y| \leq \sqrt{3-x^2} \iff r|\sin \theta|\le\sqrt{3-r^2\cos^2 \theta}\le\sqrt 2$$
which is true since $r|\sin \theta|\le 1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For what positive integers $A$ is the sequence eventually constant?
For each integer $n \ge 0$ let $S(n)=n-m^2$, where $m$ is the greatest integer such that $m^2\le n$. Define $(a_k)_{k=0}^\infty$ as $a_0=A$ and $a_{k+1}=a_k+S(a_k)$ for $k \ge0$. For what positive integers $A$ is the sequence eventually constant?
I'm trying to investigate this sequence to find some pattern, but I don't find any... If $A=1$, then $$\begin{align*} a_0 &=1 \\ a_1&= a_0+S(a_0)= 1+(1-m^2) = 1+(1-1)=1 \\ a_2&= a_1+S(a_1)= 1+S(1)=1+(1-1)=1\end{align*}$$ so it would seem that for $A=1$ the sequence is always constant.
For $A=2$ $$\begin{align*} a_0 &=2 \\ a_1&=2+S(2)=2+(2-m^2)=2+(2-1)=3 \\ a_2 &=3+S(3)=3+(3-m^2)=3+(3-1)=5 \\ a_3&=5+S(5)=5+(5-m^2)=5+(5-4)=6 \\ a_4&=6+S(6)=6+(6-m^2)=6+(6-4)=8 \\a_5&=8+S(8)=8+(8-m^2)=8+(8-4)=12 \\ a_6&=12+S(12)= 12+(12-m^2)=12+(12-9)=15\end{align*}$$
so the sequence would seem to be just increasing If $A=3$, then $$\begin{align*} a_0 &=3 \\ a_1&=3+S(3)=5 \\a_2&=5+S(5)=6 \\ a_3&=6+S(6)=8 \end{align*}$$ and this will just be the same as when $A=2$ except were skipping the first term.
Now if $A=4$, then $$\begin{align*} a_0 &=4 \\ a_1&=4+S(4)=4+(4-m^²)=4+(4-2)=6 \\a_2&= 6+S(6)=8 \\a_3&= 8+S(8)=12 \\ a_4&=12+S(12)=15 \\ a_5&=15+S(15)=15+(15-9)=21 \\ a_6&=21+S(21) = 21+(21-16) = 26\end{align*}$$
but this also seems to be increasing? What is the pattern here?
| Suppose $a_k=n=m^2+e$ is a nonsquare term, where $m$ is the largest integer such that $m^2\le n$. Then $1\le e\le 2m$ since the next square is $(m+1)^2=m^2+2m+1$, and $S(n)=e$ so $a_{k+1}=m^2+2e$. We have
$$m^2<m^2+2e\le m^2+4m<m^2+4m+4=(m+2)^2$$
and $m^2+2e\ne(m+1)^2$ since the parities of both sides differ (subtract $m^2$). Thus $a_{k+1}$ is also a nonsquare, and since $e\ge1$ at every step the sequence increases forever after $a_k$.
This reasoning applies at $a_0=A$ as well, so the sequence is eventually constant iff $A$ is a perfect square itself.
| {
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Find the second derivative of $y=\left(1-2\sqrt{x}\right)^3$ Find the second derivative of $$y=\left(1-2\sqrt{x}\right)^3$$ Let's find the first derivative: $$y'=3(1-2\sqrt{x})^2(1-2\sqrt{x})'=3\left(0-2\dfrac{1}{2\sqrt{x}}\right)(1-2\sqrt{x})^2=-3\dfrac{1}{\sqrt{x}}(1-2\sqrt{x})^2$$ The second derivative of a function is the derivative of the derivative of that function, so $$y''=\left(-3\dfrac{1}{\sqrt{x}}(1-2\sqrt{x})^2\right)'=-3\left(\dfrac{1}{\sqrt{x}}\cdot(1-2\sqrt{x})^2\right)'=-3T$$ I am really having troubles with finding that derivative as I get confused (too many things going on). $$T=\left(\dfrac{1}{\sqrt{x}}\right)'(1-2\sqrt{x})^2+\dfrac{1}{\sqrt{x}}\left((1-2\sqrt{x})^2\right)'=\\-\dfrac12x^{-\frac32}(1-4\sqrt{x}+4x)-\dfrac{2}{x}(1-2\sqrt{x})$$ Is there an easier approach?
| Your calculations are correct. After simplification, the result is $$y'' = \frac{3(1-4x)}{2x^{3/2}}.$$
A simpler approach that works in this case is to expand the cube:
$$y = (1 - 2x^{1/2})^3 = 1 - 6x^{1/2} + 12x - 8x^{3/2}.$$ Then no product or chain rule is required:
$$y' = -3x^{-1/2} + 12 - 12x^{1/2},$$
and
$$y'' = \frac{3}{2}x^{-3/2} - 6x^{-1/2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show that $\ln(2) > \frac{2}{3}$ I'm trying to show that $\ln(2) > \frac{2}{3}$. However, I'm pretty stuck on how to proceed.
I tried to show that this is the case using a similar technique to
$$
e < 4 \iff \sqrt{e} < 2 \iff \frac{1}{2} < \ln(2)
$$
However, while I can show that $\sqrt{e} < 2$ (because $2^2 = 4 > e$), I don't think it's easy to do this when $\frac{2}{3}$ is the exponent (at least it's not easy for me).
The other approach I can think of is to use a series expansion for $\ln(2)$, but the only one I know is $\ln(1+x)$ for $\lvert x\rvert < 1$.
| $\ln(2) > \frac{2}{3}$ is equivalent to $e^2 < 8$, so we need an upper bound for $e$. A standard trick is to replace the “tail” of the exponential series by a geometric series, which can be computed explicitly:
$$
e = \sum_{n=0}^\infty \frac{1}{n!}= 1 + 1 + \frac 12 \left(
1 + \frac{1}{3} + \frac{1}{3\cdot 4} + \cdots\right) \\
< 1 + 1 + \frac 12 \left(
1 + \frac{1}{3} + \frac{1}{3^2} + \cdots\right) \\
= 2 + \frac 12 \cdot \frac {1}{1-1/3} = \frac{11}{4} \, .
$$
It follows that
$$
e^2 < \left(\frac{11}{4} \right)^2 = \frac{121}{16} < 8
$$
and taking logarithms we get $2/3 < \ln(2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$\cos(105°)$ using sum and difference identities I was trying to solve $\cos(105°)$ using sum and difference identities.
My solution:
$\cos(105°) = \cos(60°+45°)$
$\cos(60°)\cos(45°) - \sin(60°)\sin(45°)$
so,
$\frac{1}{2} \cdot \frac{\sqrt{2}}{2}$ - $\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}$
then,
$\frac{\sqrt{2}}{4} - \frac{\sqrt{6}}{4}$
last,
$\frac{\sqrt{2}-\sqrt{6}}{4}$
but when I try to use calculator it says,
$\frac{-\sqrt{6}+\sqrt{2}}{4}$
What part of the solution I am doing it wrong?
| $a - b = a + (-b) = (-b) + a = -b + a$.
Thus,
$$\frac{\sqrt{2} - \sqrt{6}}{4} = \frac{-\sqrt{6} + \sqrt{2}}{4}$$
and both your answers are correct (they are equivalent).
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Possible "clever" ways to solve $x^4+x^3-2x+1=0$, with methodological justification
Solve the quartic polynomial :
$$x^4+x^3-2x+1=0$$
where $x\in\Bbb C$.
Algebraic, trigonometric and all possible methods are allowed.
I am aware that, there exist a general quartic formula. (Ferrari's formula). But, the author says, this equation doesn't require general formula. We need some substitutions here.
I realized there is no any rational root, by the rational root theorem.
The harder part is, WolframAlpha says the factorisation over $\Bbb Q$ is impossible.
Another solution method can be considered as the quasi-symmetric equations approach. (divide by $x^2$).
$$x^2+\frac 1{x^2}+x-\frac 2x=0$$
But the substitution $z=x+\frac 1x$ doesn't make any sense.
I want to ask the question here to find possible smarter ways to solve the quartic.
| You can easily observe that, the expression $(x-1)^2$ is almost included in the polynomial $P(x):=x^4+x^3-2x+1$.
Let's rewrite your polynomial as follows:
$$
\begin{align}P(x)=x^4+x^3-\color{red}{x^2}+\color{red}{x^2}-2x+1\end{align}
$$
Based on this observation, we will represent the polynomial $P(x)$ as a "quadratic" polynomial:
$$
\begin{align}P(x):&=x^4+x^3-x^2+(x-1)^2\\
&=x^2(x^2+x-1)+(x-1)^2\\
&=x^4+x^2(x-1)+(x-1)^2\\
&=\color{red}{(x-1)^2}+\color{blue}{x^2}\color{red}{(x-1)}+\color{blue}{x^4}\end{align}
$$
Letting $x-1=u$ and writing $P(x)=0$ leads to:
$$
\begin{align}&\color{red}{u^2}+\color{blue}{x^2}\color{red}{u}+\color{blue}{x^4}=0\\
&\Delta_{\color{red}{u}}=x^4-4x^4=-3x^4\\
\implies &u_{1,2}=\frac{-x^2\pm i\sqrt 3x^2}{2}\\
\implies &u_{1,2}=x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)\\
\implies &x-1=x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)\\
\implies &x^2\left(\frac{-1\pm i\sqrt 3}{2}\right)-x+1=0.\end{align}
$$
Now, this is obvious that the given polynomial has no real roots.
Finally, you can complete the solution by using the quadratic formula to find all complex roots.
| {
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"timestamp": "2023-03-29T00:00:00",
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2022 Gr11 Fermat math contest question #20 Question: A sequence of numbers $t_1, t_2, t_3$,... has its terms defined by $t_n = \frac{1}{n} - \frac{1}{n+2}$ for every integer $n \geq 1.$ For example, $t_4 = \frac{1}{4} - \frac{1}{6}$. What is the largest positive integer k for which the sum of the first k terms (that is, $t_1 + t_2 +\cdots+t_{k-1} + t_k$) is less than 1.499?
Correct answer: 1998
My solution: First, to find the sum of all numbers from $t_1$ to $t_k$, I wrote the equation $\frac{(\frac{2}{3} + \frac{1}{k} - \frac{1}{k+2}) * k}{2}$, note: I got $\frac{2}{3}$ from $\frac{1}{1} - \frac{1}{1+2}$. Next, I just substituted the 5 options this question gave me into my equation (this is a multiple choice question), but I got around $666$ when I substituted the different options which is no where near $1.499$. This means the problem is somewhere in my equation for the sum of all numbers from $t_1$ to $t_k$, but I do not know where I went wrong in my equation, or even if it was the right first step to solving this question.
| Given, $$t_n=\frac1n-\frac{1}{n+2}$$ we get $$S_k=\sum_{n=1}^k t_n=\sum_{n=1}^k \frac1n-\frac{1}{n+2}$$$$= \sum_{n=1}^k\frac1n-\sum_{n=1}^k\frac{1}{n+2}$$
Reindexing the second sum and replacing $n+2=i$, we get $$S_k= \sum_{n=1}^k\frac1n- \sum_{i=3}^{k+2}\frac1k$$$$=1+\frac12-\frac{1}{k+1}-\frac{1}{k+2}=\frac{k(3k+5)}{2(k+1)(k+2)}$$ Now, given that $S_k<1.499$ we infer that $$1.5-\frac{1}{k+1}-\frac{1}{k+2}<1.499$$ or that $$\frac{1}{k+1}+\frac{1}{k+2}>0.001$$
To better visualise the reindexing, we can write $$S_k=\left(\color{red}{1+\frac12}+ \frac13+ …+ \frac{1}{k}\right)-\left(\frac13+ \frac14+…+ \frac{1}{k}+\color{red}{\frac{1}{k+1}+\frac{1}{k+2}}\right)$$ where the $\displaystyle \frac13+ …+ \frac{1}{k}$ gets cancelled.
| {
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"timestamp": "2023-03-29T00:00:00",
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show that $f(x)=x$ where $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$
Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be a function such that $\,\forall x,y\in\mathbb{R}^+, f(x+f(y+xy)) = (y+1)f(x+1)-1$. Prove that $f(x)=x$ for all real numbers $x>0$.
I think it might be useful to prove that f is injective and then make a substitution. It should be possible to prove $f$ is surjective too. It might also help to prove additional properties of f such as multiplicativity, that f has infinitely many fixed points, that f satisfies the functional equation $f(x+y) = f(x)+f(y)$ for all positive real numbers x,y, etc. First substituting $x=y$ gives $f(x+f(x+x^2)) = (x+1)f(x+1)-1.$ Swapping $x$ and $y$ gives $f(y+f(x+xy)) = (x+1)f(y+1)-1.$ Substituting $x=1$ into the latter equation gives $f(1+f(2)) = 2 f(2) - 1.$ In general for any positive integer $n, f(n+f(n+1)) = 2f(n+1)-1$. Suppose $f(x)=f(y).$ Then It might be useful to make the substitution $u=y+1, v= x+1$. Then we have $f(u-1 + f(u(v-1))) = uf(v) - 1$ for all $u,v > 1.$
If $f(\dfrac{x+y}2) = \dfrac{f(x)+f(y)}2$ for $x,y \in \mathbb{R}_{\ge 0}$, then letting $g(x) = f(x) - f(0)$ we get $g((x+y)/2) = (g(x)+g(y))/2$ for all $x\in \mathbb{R}^+$ so we may assume $f(0) = 0$. Then $f(x) = f(2x)/2$ for all nonnegative x and so $f(x+y) = (f(2x)+f(2y))/2 = f(x)+f(y)$, implying that $f(nx) = nf(x)$ for all rationals $n > 0$ and $x \ge 0$.
| Let $P(x,y)$ be the claim $$f(x+f(y+xy)) = (y+1)f(x+1)-1.$$
Notice that $P(x,\frac{y}{x+1})$ gives
$$
f(x+f(y))=(y+x+1)\frac{f(x+1)}{x+1}-1.
$$
To prove injectivity, assume $f(a)=f(b)$ for some $a,b \in \mathbb{R}^+$. Then comparing $P(x,\frac{a}{x+1})$ and $P(x,\frac{b}{x+1})$ we get
$$
(a+x+1)\frac{f(x+1)}{x+1}-1=(b+x+1)\frac{f(x+1)}{x+1}-1
$$
and hence $a=b$, i.e. $f$ is injective. Next $P(x,\frac{1}{f(x+1)})$ gives
$$
f(x+f(\frac{x+1}{f(x+1)}))=f(x+1)
$$
and by injectivity we get $f(\frac{x+1}{f(x+1)})=1$. So let $a$ be any such that $f(a)=1$, then $P(x,\frac{a}{x+1})$ gives
$$
f(x+1)=(a+x+1)\frac{f(x+1)}{x+1}-1
$$
and so solving for $f(x+1)$ we have
$$
f(x+1)=\frac{1}{a}(x+1).
$$
Plugging to the original equation we have
$$
f(x+f(y+xy)) = \frac{1}{a}(y+1)(x+1)-1.
$$
RHS is symmetric in $x,y$, so LHS must be too, hence we get a new claim $Q(x,y)$:
$$
x+f(y+xy)=y+f(xy+x).
$$
Now $Q(\frac{x}{2},1)$ gives
$$
\frac{x}{2}+f(\frac{x}{2}+1)=1+f(x).
$$
Since from previous $f(\frac{x}{2}+1)=\frac{1}{a}(\frac{x}{2}+1)$, we have found
$$
\frac{x}{2}+\frac{1}{a}\left(\frac{x}{2}+1\right)=1+f(x).
$$
Thus $f(x)$ is a linear function. Let $f(x)=cx+d$, then plugging into the original equation and comparing the coefficients we get $c=1,d=0$. So $f(x)=x$ is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computing the eigenvalues and eigenvectors of a $ 3 \times 3$ with a trick The matrix is: $
\begin{pmatrix}
1 & 2 & 3\\
1 & 2 & 3\\
1 & 2 & 3
\end{pmatrix}
$
The solution says that
$ B\cdot
\begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}
=
\begin{pmatrix}
6 \\
6 \\
6\end{pmatrix}$
$ B\cdot
\begin{pmatrix}
1 \\
1 \\
-1
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$
$ B\cdot
\begin{pmatrix}
3 \\
0 \\
-1
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$
Thus the eigenvalues are $ \lambda_{1}=6,\lambda_{2}=0 $
My question is, how can I easily find
$\begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$ and $\begin{pmatrix} 6 \\
6 \\
6\end{pmatrix}$?
Is there any way to see it "quickly"?
| One way to instantly see that $0$ is an eigenvalue is that the three columns are clearly not linearly independent --- in fact, they're all constant multiples of each other! Next, any dependency vector of those columns gives you an eigenvector for the eigenvalue $0$. For example, clearly the expression
$$\text{column 1 + column 2 - column 3}
$$
evaluates to the zero column vector. Putting the three coefficients $1, 1, -1$ of that expression into a column vector you therefore get the following eigenvector for the eigenvalue $0$ namely $$ \begin{pmatrix}
1 \\
1 \\
-1
\end{pmatrix}$$
I'm sure, without looking back at your own post, you can easily "see" another dependency vector to get another eigenvector for the eigenvalue $0$ which is linearly independent of the one just given.
For the remaining eigenvector, you could notice that every column has constant coefficients, and therefore the expression
$$\text{column 1 + column 2 + column 3}
$$
evaluates to a column vector that also has constant coefficients. Since the three coefficients $1, 1, 1$ of that expression are constant, putting those three coefficients into column vector therefore gives you an eigenvector
\begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}
And since the actual sum of the columns is
\begin{pmatrix}
6 \\
6 \\
6
\end{pmatrix}
you immediately conclude that $6$ is the eigenvalue.
| {
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Creating a complex number so that its norm equals to 1 I would like to create a complex number c so that its norm is equal to some number a (for the purpose of this question let's assume a = 1) if I already have either its real or imaginary part. I know that:
$$\lVert \mathbf{c} \rVert = \sqrt{\sum_{i=1} ^{n} c_i \overline{c_i}}.$$
In this case:
$$\lVert \mathbf{c} \rVert = \sqrt{(a+bi)(a-bi)} = \sqrt{a^2+abi-abi-b^2i^2} = \sqrt{a^2-b^2\cdot{-1}} = \sqrt{a^2+b^2}.$$
As I said, we already know either a or b, let's say we know that a = 2. How can I find what must b be equal to if the norm must be 1? Some simple equations:
$$ \sqrt{2^2+b^2} = 1 \iff 2^2+b^2 = 1 \iff 4+b^2 = 1 \iff b^2 = 1 - 4 \iff b^2 = -3 \iff b = \pm\sqrt{-3} \iff $$
$$ \iff b = \pm\sqrt{3} * \sqrt{-1} \iff b = \pm\sqrt{3}i $$
Okay, so based on the above eqations, the imaginary part can be $\pm\sqrt{3}i$. So let's say $c=2+\sqrt{3}i$. But clearly my equations are wrong, because:
$$\sqrt{(2+\sqrt{3}i)(2-\sqrt{3}i)} = \sqrt{7},$$ which does not equal to 1. Please, help me with my confusion and lack of knowledge.
| Take any nonzero complex number $c=a+bi$. Divide by it's length, which is $\sqrt{a^2+b^2}$ to get a complex number of length $1$.
$$z = \frac{c}{||c||}= \frac{a}{\sqrt{a^2+b^2}} + \frac{b}{\sqrt{a^2+b^2}}i.$$
Compute the length of $z$ to see how it works.
| {
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The solution of differential equation $2xy+6x+(x^2-4)y'=0$ When I solved the above DE, I reached to the step and confused how to continue
$2x(y+3)+(x^2-4)dy/dx$
$(x^2-4) dy/dx=-2x(y+3)$
$dy/(y+3)=-2x/(x^2-4) dx$
$∫dy/(y+3)=∫-2x/(x^2-4) dx$
$\ln|y+3|=-ln|x^2-4|+C$
$\ln|y+3|+ln|x^2-4|=C$
$\ln|(y+3)(x^2-4)|=C$
$|(y+3)(x^2-4)|=e^C$
then how can I get the function $y$ for the last equation and when I checked the solution by W.F, I found it as
$y=c/(x^2-4)-3x^2/(x^2-4)$
plz can anyone help me to show how happed that
thanks
| From this step you use exponential
$$\ln\vert y+3\vert=-\ln\vert x^2-4\vert +C$$
$$\Rightarrow e^{\ln\vert y+3\vert}=e^{-\ln\vert x^2-4\vert +C}$$
$$\Rightarrow \vert y+3\vert=\frac{e^C}{e^{\ln\vert x^2-4\vert}}$$
$$\Rightarrow \vert y+3\vert=\frac{K}{\vert x^2-4\vert}$$
For some positive number $K$
$$\Rightarrow y+3=\frac{K}{x^2-4}\text{ or }y+3=-\frac{K}{x^2-4}$$
Which by allowing $K$ to be positive or negative can be consolidated to become
$$\Rightarrow y+3=\frac{K}{x^2-4}$$
$$\Rightarrow y=\frac{K}{x^2-4}-3; K\in\mathbb{R}.$$
Now the solution that you saw is equivalent to this one since when you take a common denominator you get
$$\Rightarrow y=\frac{-3x^2+12+K}{x^2-4}$$
And taking $12+K$ to be a constant
$C’$ you get
$$\Rightarrow y=\frac{-3x^2+C’}{x^2-4}$$
Which is the solution you saw in the answer sheet.
Hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the closed formula for the following recusive sequence. Find the closed formula for the recursive sequence defined by $a_0 = 4$, $a_1 = 12$, $a_n = 6a_{n-1}-a_{n-2}$ for $n>1$. This question is stumping me. I don't know any methods besides guess and check.
Current progress:
The first five terms are $4, 12, 68, 396, 2308$.
Each term is divisible by four, yielding $1, 3, 17, 99, 577$
Each of these is off of a perfect square by one or negative one $0-1, 4-3, 16-17, 100-99, 576-577$ = $-1,1,-1,1,-1$
Other than that I'm lost. Any help would be appreciated.
|
Using the following fact,
$$\begin{cases}a_{n}=ba_{n-1}+ca_{n-2},\\a_{0}=c,\quad a_{1}=d \end{cases}$$ then $$\begin{cases} a_{n}=\alpha r_{1}^{2}+\beta r_{2}^{2}, \quad \text{if} \quad r_{1}\not=r_{2},\quad r^{2}-br-c=0,\\a_{n}=br_{1}^{n}+dnr_{2}^{n},\quad \text{if}\quad r_{1}=r_{2},\quad r^{2}-br-c=0,\end{cases}$$
where $\alpha$ and $\beta$ can be found using the initial condition $a_{0}=c$ and $a_{1}=d$ by substitution in $a_{n}$.
Then, we have the recursive sequence with initial condition, $$\begin{cases}a_{n}=6a_{n-1}+(-1)\cdot a_{n-2},\\a_{0}=4,\quad a_{1}=12\end{cases}$$
Since $$r^{2}-6r-(-1)=0,\quad \text{then}\quad r=\frac{6\pm \sqrt{6^{2}-4}}{2},\quad \text{then}\quad r=3\pm \sqrt{8}, \quad \text{then}\quad r_{1}\not=r_{2}. $$
Then, the general solution is given by
$$a_{n}=\alpha\left(3+\sqrt{8}\right)^{n}+\beta\left(3-\sqrt{8}\right)^{n}.$$
Setting $a_{0}=4$ and $a_{1}=12$ we have
$$\begin{cases}4=\alpha\left(3+\sqrt{8}\right)^{0}+\beta\left(3-\sqrt{8}\right)^{0},\quad n=0,\\ 12=\alpha\left(3+\sqrt{8}\right)^{1}+\beta\left(3-\sqrt{8}\right)^{1},\quad n=1.\end{cases},\quad \text{then}\quad \begin{cases}\alpha=2,\\ \beta=2.\end{cases}$$
Therefore, the particular solution for the recursive sequence is given by
$$\color{blue}{\boxed{a_{n}=2\left(3+\sqrt{8}\right)^{n}+2\left(3-\sqrt{8}\right)^{n},\quad n=0,1,\ldots}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Can we make the inequality $A-B \leq \frac{A^2}{4}$ strict? I have an interesting problem:
Given that $$
A=\frac{1}{2022}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2022}\right)$$
$$B=\frac{1}{2023}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2023}\right)
$$
I got an upper bound on $A-B$.The Lower bound is zero, that is $A-B>0$ is easy to prove.For the upper bound , here is my analysis:
$$\text{Let}\:H=\sum_{k=1}^{2022}\frac{1}{k}$$
So we have $$2022=\frac{H}{A} \implies 2023=\frac{H+A}{A}$$
So $$B=\frac{H+\frac{1}{2023}}{2023}=\frac{\left(H^2+A H+A\right) A}{(H+A)^2}$$
Finally we get a quadratic in $H$ as:
$$(A-B) H^2+\left(A^2-2 A B\right) H+A^2(1-B)=0$$ and since $H \in \mathbb{R}$ the discriminant should be non negative.
So we have
$$(A^2-2AB)^2-4(A-B)A^2(1-B)\geq 0$$
Simplifying we get
$$A-B \leq \frac{A^2}{4}$$
Now is it possible to make it strict?
EDIT: After deep thinking i came to the conclusion that inequality should be strict:
It is evident that $A<2, B<1$
Let us assume equality holds true, that is $$A-B=\frac{A^2}{4}$$
So we get
$$(A-2)^2=4(1-B) \implies A=2-2\sqrt{1-B}$$
Now i am pretty sure that $\sqrt{1-B}$ is irrational. So that makes $A$ irrational, which contradicts that $A$ is rational. Hence the inequality is strict.
| If $\;A-B=\dfrac{A^2}4\;,\;$ then , by solving the quadratic equation in $H$, we get that $\;H=2-A\;,\;$ hence ,
$H=\dfrac{4044}{2023}<2<1+\dfrac12+\dfrac68<H_8<H_{2022}=H\;.$
So, $\;A-B\neq\dfrac{A^2}4\;,\;$ consequently ,$\;\;A-B<\dfrac{A^2}4\,.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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How to evaluate $\sum_{n=1}^\infty \frac{n^5}{e^{2\pi n}-1}$? I recently found the following result on Twitter.
$$\sum_{n=1}^\infty \frac{n^5}{e^{2\pi n}-1}=\frac{1}{504}$$
I know that $\int_0^\infty \frac{x^5}{e^{2 \pi x}-1} dx = \frac{5!}{(2\pi)^6}\zeta(6)=\frac{1}{504}$
How to show that the sum is also equal to the same number?
Can this be generalized to a class of functions where the sum (over positive integers) and the definite integral (from 0 to $\infty$) are the same?
| I'm not sure if it's more appropriate to post this here or elsewhere, but we can integrate the function $$f(z) = \frac{z^{5}e^{iz}}{\cosh(z) - \cos(z)} $$ around the contour $$\left[-\frac{(2N+1) \pi \sqrt{2}}{2}, \frac{(2N+1) \pi \sqrt{2}}{2}\right] \cup \frac{(2N+1) \pi \sqrt{2}}{2} e^{i[0, \pi]}. $$
(For any positive integer $N$, the semicircular part of the above contour passes halfway between two adjacent poles of $f(z)$.)
As $N \to \infty$ through the positive integers, the magnitude of $f(z)$ decays exponentially to zero on the semicircle. This is due to the fact that the magnitude of $e^{iz}$ decays exponentially to zero as $\Im(z) \to + \infty$, while the magnitude of $\cosh(z)$ grows exponentially as $\Re(z) \to \pm \infty$.
We therefore have $$\begin{align} \int_{-\infty}^{\infty} f(x) \, \mathrm dx &= 2 \pi i \left(\sum_{n=1}^{\infty}\operatorname{Res}\left[f(z), n \pi(1+i) \right]+ \sum_{n=1}^{\infty}\operatorname{Res}\left[f(z), n \pi(-1+i) \right] \right) \\ &= 2 \pi i \left(\sum_{n=1}^{\infty}\lim_{z \to n\pi(1+i)} \frac{z^{5}e^{iz}}{\sinh(z) + \sin (z)} + \sum_{n=1}^{\infty} \lim_{z \to n \pi(-1+i)} \frac{z^{5}e^{iz}}{\sinh(z) + \sin (z)} \right) \\&= 2 \pi i \left(\sum_{n=1}^{\infty} \frac{n^{5} \pi^{5}(1+i)^{5}(-1)^{n} e^{- n \pi}}{(-1)^{n}(1+i)\sinh(n \pi)} + \sum_{n=1}^{\infty} \frac{n^{5} \pi^{5}(-1+i)^{5}(-1)^{n}e^{-n \pi}}{(-1)^{n}(-1+i) \sinh(n \pi)} \right) \\ &= -16 \pi^{6} i \sum_{n=1}^{\infty} \frac{n^{5}e^{-n \pi}}{\sinh(n \pi)} \\ &=-32 \pi^{6} i \sum_{n=1}^{\infty} \frac{n^{5}}{e^{2 \pi n}-1}. \end{align} $$
Equating the imaginary parts on both sides of the equation, we get $$ \begin{align} \sum_{n=1}^{\infty} \frac{n^{5}}{e^{2 \pi n}-1} &= -\frac{1}{32 \pi^{6}} \int_{-\infty}^{\infty} \frac{x^{5} \sin(x)}{\cosh(x) - \cos(x)} \, \mathrm dx \\ &= -\frac{1}{16 \pi^{6}} \int_{0}^{\infty}\frac{x^{5} \sin(x)}{\cosh(x) - \cos(x)} \, \mathrm dx \\ &= -\frac{1}{8 \pi^{6}} \, \Im \int_{0}^{\infty}x^{5} \sum_{n=1}^{\infty} e^{-(1-i)nx} \, \mathrm dx \\ &= -\frac{1}{8 \pi^{6}} \, \Im \sum_{n=1}^{\infty} \int_{0}^{\infty} x^{5} e^{-(1-i)nx} \, \mathrm dx \\ &= -\frac{1}{8 \pi^{6}} \, \Im \sum_{n=1}^{\infty} \frac{\Gamma(6)}{(1-i)^6n^{6}} \\ &= \frac{15}{8 \pi^{6}} \sum_{n=1}^{\infty} \frac{1}{n^{6}} \\ &= \frac{1}{504}. \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4559942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Is it possible for two non real complex numbers a and b that are squares of each other? ($a^2=b$ and $b^2=a$)? Is it possible for two non real complex numbers a and b that are squares of each other? ($a^2=b$ and $b^2=a$)?
My answer is not possible because for $a^2$ to be equal to $b$ means that the argument of $b$ is twice of arg(a) and for $b^2$ to be equal to $a$ means that arg(a) = 2.arg(b) but the answer is it is possible.
How is it possible when arg(b) = 2.arg(a) and arg(a) = 2.arg(b) contradict each other?
| The arguments $\arg(a)$ and $\arg(b)$ are only defined up to a multiple of $2\pi$. Or, if we require that $\arg(z) \in (-\pi,\pi]$ for all $z$, then it's not necessarily the case that $\arg$ is multiplicative: we might end up having to add or subtract $2\pi$.
If we go with the first option, we know that $\arg(a) \equiv 2\arg(b) \pmod{2\pi}$ and $\arg(b) \equiv 2\arg(a) \pmod{2\pi}$, but this is not a problem. Substituting, we get $\arg(a) \equiv 4 \arg(a) \pmod{2\pi}$, or $3\arg(a) \equiv 0 \pmod{2\pi}$; this is possible and requires $\arg(a)$ to be a multiple of $\frac{2\pi}{3}$.
Or, if we go with the second option, we should consider alternatives to $\arg(a) = 2\arg(b)$ and $\arg(b) = 2\arg(a)$. For example, if we have $\arg(a) = 2\arg(b) + 2\pi$ and $\arg(b) = 2\arg(a) - 2\pi$, then we get $\arg(a) = 4\arg(a) - 2\pi$, which leads to $\arg(a) = \frac{2\pi}{3}$ as the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $a+b+c+d$ is a composite number Let $a,b,c,d$ be positive integers such that $a^2+3ab+b^2=c^2+3cd+d^2$. Prove that $a+b+c+d$ is a composite number.
I think this statement is correct, at least using the program I came across only composite numbers. An approximate solution plan is as follows: The equality $(2a+3b)^2-5b^2=(2c+d)^2-5d^2$ holds. We can factor it in the ring $\mathbb Z[\sqrt5]$. It is not a UFD, but it is reasonable to assume that there is a further factorization in a wider area - involving what Kummer called "ideal numbers". From here, maybe some information about the coefficients can be extracted. Any ideas?
| Suppose $p:= a+b+c+d$ is a prime. Since $d=p-a-b-c$ we have:
$$a^2+3ab+b^2 = c^2+3pc-3ac-3bc-3c^2+p^2-2p(a+b+c)+a^2+b^2+c^2+2ab+2bc+2ca$$
i.e.
$$ab+ ac+bc+c^2=p^2-p(2a+2b-c)$$
or $$(a+c)(b+c) = p(p-2a-2b+c)$$
Last equation implies $p\mid a+c$ or $p\mid b+c$. In each case we have a contradiction, since $p$ is greater then $a+c$ and $b+c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Solve the equation $\frac{\sqrt{4+x}}{2+\sqrt{4+x}}=\frac{\sqrt{4-x}}{2-\sqrt{4-x}}$ Solve the equation $$\dfrac{\sqrt{4+x}}{2+\sqrt{4+x}}=\dfrac{\sqrt{4-x}}{2-\sqrt{4-x}}$$ The domain is $4+x\ge0,4-x\ge0,2-\sqrt{4-x}\ne0$.
Note that the LHS is always positive, so the roots must also satisfy: $A:2-\sqrt{4-x}>0$.
Firstly, I decided to raise both sides of the equation to the power of $2$. I came at $$\dfrac{4+x}{8+x+4\sqrt{4+x}}=\dfrac{4-x}{8-x-4\sqrt{4-x}}$$
Another thing I tried is to let $\sqrt{4+x}=u\ge0$ and $\sqrt{4-x}=v\ge0$. Then $$\begin{cases}\dfrac{u}{2+u}=\dfrac{v}{2-v}\\4+x=u^2\\4-x=v^2\end{cases}$$ Adding the second to the third equation, we get $8=u^2+v^2$.
And the last thing: I cross-multiplied $$2\sqrt{4+x}-\sqrt{16-x^2}=2\sqrt{4-x}+\sqrt{16-x^2}\\\sqrt{4+x}=\sqrt{4-x}+\sqrt{16-x^2}$$ Raising to the power of 2 gives $$4+x=4-x+16-x^2+2\sqrt{(4-x)(16-x^2)}\\2\sqrt{(4-x)(16-x^2)}=x^2+2x-16$$ Is there something easier?
| Your analysis is good.
Starting in the middle:
*
*$u = \sqrt{4+x}.$
*$v = \sqrt{4-x}.$
*$\dfrac{u}{2+u} = \dfrac{v}{2-v}.$
So,
$$2u - uv = 2v + uv \implies 2u - 2v = 2uv \implies $$
$$2\sqrt{4+x} - 2\sqrt{4-x} = 2\sqrt{16 - x^2} \implies $$
$$4(4 + x) + 4(4-x) - 8\sqrt{16 - x^2} = 4(16 - x^2) \implies $$
$$32 - 8\sqrt{16 - x^2} = 4(16 - x^2).$$
Let $t = \displaystyle \sqrt{16 - x^2} \implies t \geq 0.$
Then
$$32 - 8t = 4t^2 \implies t^2 + 2t - 8 = 0 \implies$$
$$(t - 2)(t + 4) = 0.$$
$(t + 4) = 0$ must be rejected, since $t \geq 0.$
Therefore,
$$\sqrt{16 - x^2} = t = 2 \implies 16 - x^2 = 4 \implies x = \pm 2\sqrt{3}.$$
These two roots of $x$ may be checked as follows:
$$\left(\sqrt{3} + 1\right)^2 = 4 + 2\sqrt{3} \implies \sqrt{4 + 2\sqrt{3}} = \sqrt{3} + 1.$$
$$\left(\sqrt{3} - 1\right)^2 = 4 - 2\sqrt{3} \implies \sqrt{4 - 2\sqrt{3}} = \sqrt{3} - 1.$$
Plugging $x = 2\sqrt{3}$ back into the original equation gives
$$\frac{\sqrt{3} + 1}{2 + [\sqrt{3} + 1]} = \frac{\sqrt{3} - 1}{2 - [\sqrt{3} - 1]}. \tag1 $$
Plugging $x = -2\sqrt{3}$ back into the original equation gives
$$\frac{\sqrt{3} - 1}{2 + [\sqrt{3} - 1]} = \frac{\sqrt{3} + 1}{2 - [\sqrt{3} + 1]}. \tag2 $$
The assertion in (1) above is easily determined to to be true and the assertion in (2) above is easily determined to be false.
Therefore, of the two candidate roots of
$~\displaystyle x = \pm 2\sqrt{3}~$, the first root of $~\displaystyle x = 2\sqrt{3}~$ is confirmed and the second root of $~\displaystyle x = -2\sqrt{3}~$ must be rejected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4566671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove that $\int_0^{2\pi} \frac{ab}{a^2\cos^2t+b^2\sin^2t}\mathrm dt=2\pi$. Probably equation $\int_0^{2\pi} \frac{ab}{a^2\cos^2t+b^2\sin^2t}\mathrm dt=\int_0^{2\pi} \frac{ab}{a^2\sin^2t+b^2\cos^2t}\mathrm dt$ is useful. Double integration is also a tool, but I don't know next step.
| Substitute $y = \tan t$
\begin{align}
\int_0^{2\pi} \frac{ab}{a^2\cos^2t+b^2\sin^2t}dt=&
\ 4ab \int_0^{\pi/2} \frac{ \sec^2 t}{a^2+b^2\tan^2t}dt\\
=& \ 4ab\int_0^\infty \frac {1}{a^2+b^2y^2}dy\\
=& \ 4\arctan \left(\frac{b\tan y}a\right) \bigg|_0^\infty = 2\pi
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Trick of selecting fundamental matrix in nonhomogeneous differential equations to simplify calculations Problem:
I came across the following differential equation set:
$$
\begin{cases}
\dot{x}&=3x-2y+t\\
\dot{y}&=4x-y+t^2\\
\end{cases}
$$
My solution
I followed normal practices and found the general solution to the homogeneous equation $$
\left[ \begin{array}{c}
x\\
y\\
\end{array} \right] '=\left[ \begin{matrix}
3& -2\\
4& -1\\
\end{matrix} \right] \cdot \left[ \begin{array}{c}
x\\
y\\
\end{array} \right] \Rightarrow \left[ \begin{array}{c}
x\\
y\\
\end{array} \right] =\left[ \begin{array}{c}
e^t\left( c_1\cos 2t+c_2\sin 2t \right)\\
e^t\left( \left( c_1-c_2 \right) \cos 2t+\left( c_1+c_2 \right) \sin 2t \right)\\
\end{array} \right]
$$
I selected $(c_1, c_2)=(1,0)$ and $(0,1)$ to form a fundamental matrix $X=\left[ \begin{matrix}
e^t\cos 2t& e^t\sin 2t\\
e^t\left( \cos 2t+\sin 2t \right)& e^t\left( \sin 2t-\cos 2t \right)\\
\end{matrix} \right] $, and plugged it into the variation of constants formula to get a particular solution $y(t)=X(t)\int_{0}^{t}\left(X^{-1}(\tau)\cdot \left[ \begin{array}{c}
\tau\\
\tau ^2\\
\end{array} \right]\right)\mathrm{d}\tau $, and after a one-page long calculation and the help of Wolframalpha, I got the following solution of $y$:
Question
I wonder what's wrong with my selection since the what wolfram gave me if I typed the original equation set would be like this:
So I guess there must be something in my selection of the fundamental matrix that makes the calculations really complex, but how can I improve it?
Also, what's your favourite way of solving this kind of nonhomogeneous linear differential equation sets?
| We want to solve for $x,y$ such that
$$
\frac{d}{dt}\left[\begin{array}{c}x \\ y\end{array}\right]
= \left[\begin{array}{cc}3 & -2 \\ 4 & -1\end{array}\right]\left[\begin{array}{c}x \\ y\end{array}\right] +\left[\begin{array}{c}t \\ t^2\end{array}\right].
$$
Let $C$ be the constant coefficient matrix on the right. Then the above may be written as
$$
\frac{d}{dt}\left[\begin{array}{c}x\\y\end{array}\right]-C\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c} t \\ t^2\end{array}\right] \\
\frac{d}{dt}\left(e^{-tC}\left[\begin{array}{c} x \\ y\end{array}\right]\right) = e^{-tC}\left[\begin{array}{c} t \\ t^2\end{array}\right]
$$
Integrating over $[0,t]$ and rearranging terms gives
$$
e^{-tC}\left[\begin{array}{c} x \\ y\end{array}\right] = \left[\begin{array}{c}x_0\\ y_0\end{array}\right]+\int_0^t e^{-sC}\left[\begin{array}{c} s \\ s^2 \end{array}\right]ds \\
\left[\begin{array}{c} x \\ y\end{array}\right]= e^{tC}\left[\begin{array}{c}x_0 \\ y_0\end{array}\right]+\int_0^t e^{(t-s)C}\left[\begin{array}{c}s \\ s^2\end{array}\right]ds
$$
Now the task is reduced to finding $e^{uC}$ for a real parameter $u$. You can use complex variables to write
$$
e^{uC} = \frac{1}{2\pi i}\oint_{\gamma} \frac{e^{uz}}{zI-C}dz,
$$
where the contour $\gamma$ is a positively-oriented, simple, closed, rectifiable curve enclosing all eigenvalues of $C$ in its interior. The coefficient matrix $C$ has characteristic polynomial
$$
p(z)= (z-3)(z+1)+8=z^2-2z+8=(z-4)(z+2)
$$
Therefore, $p(C)=0$, which gives
\begin{align}
p(z)I&=p(z)I-p(C)\\
&=(z^2 I-C^2)-2(z I-C) \\
&=(z I-C)(z I+C)-2(z I-C) \\
&=(z I-C)(z I+C-2I)
\end{align}
Hence, $z I-C$ is invertible for all $z$ for which $p(z)\ne 0$, and the inverse of $z I-C$ is given by
$$
(z I-C)^{-1}=\frac{1}{p(z)}((z-2)I+C).
$$
And that gives an integral which may be directly computed by residues at $z=-2,4$, assuming that the contour $C$ encloses $z=2$ and $z=4$ in its interior:
$$
e^{uC} = \frac{1}{2\pi i}\oint_{\gamma}\frac{e^{uz}}{(z-4)(z+2)}((z-2)I+C)dz.
$$
This is a relatively straightforward integral to evaluate by residues because the integrand has only poles of order $1$ at $z=-2,4$. Hence, $e^{uC}$ is a linear combination of matrices $I$ and $C$:
\begin{align}
e^{uC}&= \frac{e^{-2u}}{-6}(-4I+C)+\frac{e^{4u}}{6}(2I+C) \\
&= \left(\frac{2}{3}e^{-2u}+\frac{1}{3}e^{4u}\right)I+\left(-\frac{1}{6}e^{-2u}+\frac{1}{6}e^{4u}\right)C.
\end{align}
Therefore, using the above, the solution is known at this point, even though the fully expanded expression is messy:
$$
\left[\begin{array}{c} x \\ y\end{array}\right]= e^{tC}\left[\begin{array}{c}x_0 \\ y_0\end{array}\right]+\int_0^t e^{(t-s)C}\left[\begin{array}{c}s \\ s^2\end{array}\right]ds
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
I have a circle with $n$ equal tangent circles inscribed, what is the diameter of the largest circles that could be inscribed in the space left? Hello I know that the question is very long and maybe confused, I'll do my best to explain it step by step.
The problem is relevant in the cable design industry. The first step is to define the diameter of a circle and its numerosity n: from that I calculate the diameter of a supercicle that contains n tangent circles. To calculate that I cheated a little bit: I used a factor called "lay up factor" (that factor it's defined in an international ISO norm), so for instance if I have $3$ circles of diameter d the diameter s of the supercircle will be $2.13*d$. After that I have to calculate the diameter of the largest circle that can be inscribed between the supercircle and the base circles.
Maybe some images will make the problem clearer.
My final goal is to calculate the diameter of the red circles:
First step is to put n circles of diameter d tangent to each other:
I draw the supercicle (yellow) and calculate its diameter s:
It would be nice to understand how to calculate the red circles diameter for any numerosity, but I'm interested in the case of $3$ (so $3$ base circles).
| Let's put the diagram on the coordinate plane, with the supercircle centered at the origin, and the $+x$-axis going through the tangent point of two blue circles and the center of one red circle. Call $S$ the radius of the supercircle, $R$ the radius of the blue circles, and $r$ the radius of the red circle.
All the blue circles have centers on a circle around the origin with radius $S-R$. So the two blue circles touching the $+x$-axis have centers at $((S-R) \cos \frac{\pi}{n}, \pm (S-R) \sin \frac{\pi}{n})$. But the distance of either center from the $x$-axis is also $R$, so if $n \geq 3$,
$$ (S-R) \sin \frac{\pi}{n} = R $$
$$ R = \frac{\sin \frac{\pi}{n}}{1+\sin \frac{\pi}{n}} S $$
$$ S = R\left(1+\csc \frac{\pi}{n}\right) $$
And we can rewrite the centers of the two blue circles touching the $+x$-axis as $(R \cot \frac{\pi}{n}, \pm R)$.
The center of the red circle on the $+x$-axis is $(S-r,0)$. The distance between the center of this red circle and the center of a tangent blue circle must be $R+r$:
$$ \sqrt{\left(S-r- R \cot \frac{\pi}{n}\right)^2 + R^2} = R+r $$
$$ \left(R + R \csc \frac{\pi}{n} - R \cot \frac{\pi}{n} - r\right)^2 + R^2 = R^2+2Rr+r^2 $$
$$ \left(R + R \tan \frac{\pi}{2n} -r\right)^2 = 2Rr + r^2 $$
$$ R^2 \left(1+ \tan \frac{\pi}{2n}\right)^2 - 2Rr \left(1+\tan\frac{\pi}{2n}\right) + r^2 = 2Rr + r^2 $$
$$ R^2 \left(1+\tan \frac{\pi}{2n}\right)^2 = 2Rr \left(2+\tan\frac{\pi}{2n}\right) $$
$$ R \left(\cos\frac{\pi}{2n} + \sin\frac{\pi}{2n}\right)^2 = 2r \left(2\cos^2 \frac{\pi}{2n} + \cos\frac{\pi}{2n} \sin\frac{\pi}{2n}\right) $$
$$ R \left(\cos^2 \frac{\pi}{2n}+2\cos\frac{\pi}{2n}\sin\frac{\pi}{2n} + \sin^2 \frac{\pi}{2n}\right) = 2r\left(1+\cos \frac{\pi}{n}+\frac{1}{2} \sin \frac{\pi}{n}\right) $$
$$ R \left(1+\sin\frac{\pi}{n}\right) = r\left(2+2\cos \frac{\pi}{n}+\sin\frac{\pi}{n}\right) $$
$$ r = \frac{1+\sin\frac{\pi}{n}}{2+2\cos\frac{\pi}{n}+\sin\frac{\pi}{n}} R $$
$$ r = \frac{\sin \frac{\pi}{n}}{2+2\cos\frac{\pi}{n}+\sin\frac{\pi}{n}} S $$
For $n=3$, we have $\frac{S}{R} = 1+\frac{2}{\sqrt{3}} \approx 2.155$, $\frac{r}{R} = 2-\sqrt{3} \approx 0.4827$, and $\frac{r}{S} = \frac{1}{1+2\sqrt{3}} \approx 0.2240$. The result for $\frac{S}{R}$ is a bit larger than your quoted $2.13$; I'm not sure why. The result for $\frac{r}{R}$ in my notation matches the result for $\frac{u}{r}$ in the notation of @Nathalieissweetandawesome's answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proportions of ornamental gables in gothic architecture I'm trying to figure out mathematically precise proportions for gothic architecture-style gables.
The gable has the shape of a triangle. There is a central incircle, as well as a smaller incircle in the remaining space at the top. The legs of the triangle are tangent to two additional circles of the same radius as the smaller circle at the top, which are also tangent to the central incircle. If the base of the triangle is drawn so that it is tangent to the central incircle, then the two neighboring circles at the bottom are cut exactly in half.
I hope the image makes it more clear.
Now, my question is: for any given base or height of the triangle, or for any given radius of the central incircle, what is the radius of the smaller circles?
Since the legs of the triangle are tangent to 3 circles on either side, which also touch, and since the smaller circles are of the same size, I feel like there should be exactly one answer where the proportions exactly match.
I've tried a few things but my maths is a bit rusty and I can't seem to make a lot of progress. :/ I'd be grateful even for a suggestion from which direction to tackle this problem.
|
We have isosceles $\triangle ABC$ with $AB = AC$. $O$ is the midpoint of $BC$, and angle $ t = \angle OAC $. Further the height is $h = AO$
From this, it follows that
$OB = OC = h \ \tan t $
The semi-perimeter is given by
$ s = OB + AB = h ( \tan t + \sec t ) $
The radius of the incircle (shown in red) is
$ R = \dfrac{\text{Area}}{s} = \dfrac{h \tan t }{\tan t + \sec t } $
Let $f = \dfrac{1 - \sin t }{1 + \sin t } $, then the radius of the small circle on top is $r$ which is given by
$ r = f R $
This is also the radius of each of the small circles on either side on the bottom. Since their centers is on the base, then the distance of the center of the bottom right circle $G$ from the base midpoint $O$ is
$ d = h \tan t - r \sec t = h ( \tan t - f \dfrac{\tan t }{\tan t + \sec t } )$
Applying Pythagorean theorem to $\triangle ONB $
$ d^2 + R^2 = (R + r)^2 $
which simplifes to
$ d^2 = 2 R r + r^2 = (f^2 + 2 f ) R^2$
Substituting $d$ and $R$, and dividing by $h^2$ gives us
$ ( \tan t - f \dfrac{\tan t }{\tan t + \sec t } )^2 = (f^2 + 2 f) \dfrac{\tan^2 t }{(\tan t + \sec t )^2} $
Multiplying through by $ (\tan t + \sec t )^2 $, gives
$ ( \tan t (\tan t + \sec t ) - f \tan t )^2 = (f^2 + 2 f) \tan^2 t $
Dividing through by $\tan^2 t $, gives
$ ( \tan t + \sec t - f)^2 = f^2 + 2 f $
Substituting $f$ and multiplying through by $(1 + \sin t )^2 $, gives us
$ ( (\tan t + \sec t) (1 + \sin t ) - (1 - \sin t ) )^2 = (1 - \sin t )^2 + 2 (1 - \sin t)(1 + \sin t) $
Multiplying through by $ \cos^2 t $, simplifies the above expression to,
$ ( (\sin t + 1)^2 - 1 + \sin t)^2 = (1 - \sin^2 t) ( 3 - 2 \sin t - \sin^2 t )$
Simplifying,
$ ( \sin^2 t + 3 \sin t)^2 = \sin^4 t + 2 \sin^3 t - 4 \sin^2 t - 2 \sin t + 3 $
And further,
$ \sin^4 t + 6 \sin^3 t + 9 \sin^2 t =\sin^4 t + 2 \sin^3 t - 4 \sin^2 t - 2 \sin t + 3 $
And finally,
$ 4 \sin^3 t + 13 \sin^2 t + 2 \sin t - 3 = 0 $
which is a cubic polynomial equation in $\sin t$. Feeding this to wolframalpha.com gives only one valid solution which is
$ \sin t = \dfrac{\sqrt{17} - 1} { 8 } $
This is the golden trigonometric ratio for this problem. It follows that
$t= \sin^{-1} \bigg(\dfrac{\sqrt{17} - 1} { 8 }\bigg) \approx 22.9786^\circ $
So, given $h$ we can now compute all the required radii using this value of $t$.
Finally, note the ratio
$\dfrac{r}{R} = f = \dfrac{1 - \sin t }{1 + \sin t } = \dfrac{ 9 -\sqrt{17} }{ \sqrt{17} + 7 } = \dfrac{ (9 - \sqrt{17})(7 - \sqrt{17}) }{49 - 17} \\= \dfrac{ 80 - 16 \sqrt{17} }{32} = \dfrac{5 - \sqrt{17}}{2} $
The inverse ratio is
$ \dfrac{R}{r} = \dfrac{2}{5 - \sqrt{17}} = \dfrac{ 2 (5 + \sqrt{17} ) }{ 8 } = \dfrac{5 + \sqrt{17}}{4} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find a, real number for the following limit Find a, real number, such that $\displaystyle{\lim_{n\to\infty}{(n^3+an^2)^\frac{1}{3}-(n^2-an)^\frac{1}{2}}}=1$
I noted x=$(n^3+an^2)^\frac{1}{3}$
and y=$(n^2-an)^\frac{1}{2}$
I applied with the conjugate ($x^2-xy+y^2$) but I do not know how to continue. Any ideas?
Thank you for your help!
| Let $\alpha=n^3+an^2$ and $\beta=n^2-an$. Introduce a difference of $6^{\rm th}$ powers:
$$\begin{align*}
\alpha^{1/3} - \beta^{1/2} &= (\alpha^2)^{1/6} - (\beta^3)^{1/6} \\[1ex]
&= \frac{\left((\alpha^2)^{1/6}\right)^6 - \left((\beta^3)^{1/6}\right)^6}{(\alpha^2)^{5/6} + (\alpha^2)^{4/6}(\beta^3)^{1/6} + (\alpha^2)^{3/6} (\beta^3)^{2/6} + (\alpha^2)^{2/6} (\beta^3)^{3/6} + (\alpha^2)^{1/6} (\beta^3)^{4/6} + (\beta^3)^{5/6}} \\[1ex]
&= \frac{\alpha^2 - \beta^3}{\alpha^{5/3} + \alpha^{4/3} \beta^{1/2} + \alpha \beta + \alpha^{2/3} \beta^{3/2} + \alpha^{1/3} \beta^2 + \beta^{5/2}} \\[1ex]
&= \frac{5an^5 - 2a^2 n^4 + a^3 n^3}{(1+\gamma_+^{5/3}+\gamma_+^{4/3}\gamma_-^{1/2}+\gamma_+^{2/3}\gamma_-^{3/2} + \gamma_+^{1/3}\gamma_-^2 + \gamma_-^{5/2})n^5 - an^4 + an^3 - a^2n^2}
\end{align*}$$
where $\gamma_{\pm}=1\pm\frac an$. Both $\gamma_{\pm}\to1$ as $n\to\infty$, so the overall expression converges to $\frac{5a}{6}$, hence $a=\frac65$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Seeking for other methods to evaluate $\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx$ for $n\geq 2$. Inspired by my post, I go further to investigate the general integral and find a formula for
$$
I_n=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx =-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[\gamma+\psi\left(1-\frac{1}{n}\right)\right] \tag*{}
$$
Let’s start with its partner integral
$$
I(a)=\int_0^{\infty}\left(x^n+1\right)^a d x
$$
and transform $I(a)$, by putting $y=\frac{1}{x^n+1}$, into a beta function
$$
\begin{aligned}
I(a) &=\frac{1}{n} \int_0^1 y^{-a-\frac{1}{n}-1}(1-y)^{-\frac{1}{n}-1} d y \\
&=\frac{1}{n} B\left(-a-\frac{1}{n}, \frac{1}{n}\right)
\end{aligned}
$$
Differentiating $I(a)$ w.r.t. $a$ yields
$$
I^{\prime}(a)=\frac{1}{n} B\left(-a-\frac{1}{n}, \frac{1}{n}\right)\left(\psi(-a)-\psi\left(-a-\frac{1}{n}\right)\right)
$$
Then putting $a=-1$ gives our integral$$
\begin{aligned}
I_n&=I^{\prime}(-1) \\&=\frac{1}{n} B\left(1-\frac{1}{n}, \frac{1}{n}\right)\left[\psi(1)-\psi\left(1-\frac{1}{n}\right)\right] \\
&=-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[\gamma+\psi\left(1-\frac{1}{n}\right)\right]\end{aligned}
$$
For examples,
$$
\begin{aligned}& I_2=-\frac{\pi}{2} \csc \frac{\pi}{2}\left[\gamma+\psi\left(1-\frac{1}{2}\right)\right]=\pi \ln 2,\\ & I_3=-\frac{\pi}{3} \csc \left(\frac{\pi}{3}\right)\left[\gamma+\psi\left(\frac{2}{3}\right)\right]=\frac{\pi \ln 3}{\sqrt{3}}-\frac{\pi^2}{9} ,\\ &I_4=-\frac{\pi}{4} \csc \left(\frac{\pi}{4}\right)\left[\gamma+\psi\left(\frac{3}{4}\right)\right]=\frac{3 \pi}{2 \sqrt{2}}\ln 2-\frac{\pi^2}{4 \sqrt{2}},\\
& I_5=-\frac{\pi}{5} \csc \left(\frac{\pi}{5}\right)\left[\gamma+\psi\left(\frac{4}{5}\right)\right]=-\frac{2 \sqrt{2} \pi}{5 \sqrt{5-\sqrt{5}}}\left[\gamma+\psi\left(\frac{4}{5}\right)\right], \\
& I_6=-\frac{\pi}{6} \csc \left(\frac{\pi}{6}\right)\left[\gamma+\psi\left(\frac{5}{6}\right)\right]=\frac{2 \pi}{3} \ln 2+\frac{\pi}{2} \ln 3-\frac{\pi^2}{2 \sqrt{3}},
\end{aligned}
$$
Furthermore, putting $a=-m$, gives
$$\boxed{I(m,n)=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{(x^n+1)^m} dx = \frac{1}{n} B\left(m-\frac{1}{n}, \frac{1}{n}\right)\left[\psi(m)-\psi\left(m-\frac{1}{n}\right)\right] }$$
For example,
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(x^6+1\right)}{(x^6+1)^5} dx & =\frac{1}{6} B\left(\frac{29}{6}, \frac{1}{6}\right)\left[\psi(5)-\psi\left(\frac{29}{6}\right)\right] \\
& =\frac{1}{6} \cdot \frac{21505 \pi}{15552} \cdot\left(-\frac{71207}{258060}-\frac{\sqrt{3} \pi}{2}+\frac{3 \ln 3}{2}+2 \ln 2\right) \\
& =\frac{21505 \pi}{93312}\left(-\frac{71207}{258060}-\frac{\sqrt{3} \pi}{2}+\frac{3 \ln 3}{2}+2 \ln 2\right)
\end{aligned}
$$
Are there any other methods?
Your comments and alternative methods are highly appreciated.
| The integral admits elementary close-form as evaluated below
\begin{align}
I_n=&\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} \overset{x\to 1/x}{dx}\\
=& \ \frac12 \int_0^{\infty} \frac{(1+x^{n-2})\ln \left(x^n+1\right)}{x^n+1}dx-\frac n2 \int_0^{\infty} \frac{x^{n-2}\ln x}{x^n+1}dx
\end{align}
where $\int_0^{\infty} \frac{x^{n-2}\ln x}{x^n+1}dx=\frac{\pi^2}{n^2}\csc\frac\pi n\cot\frac \pi n$
\begin{align}
&\int_0^{\infty} \frac{(1+x^{n-2})\ln \left(x^n+1\right)}{x^n+1}dx\\
=& \int_0^{\infty}\int_0^1 \frac{nt^{n-1}(1+x^{n-2})}{(x^n+1)(t^n+x^n)}dt\ dx\\
=& \int_0^1 \frac{2t^{n-1}-t^{n-2}-1}{t^n-1}\int_0^\infty\frac n{x^n+1}dx \ dt \\
=&\ \frac\pi2\csc\frac\pi n \int_0^1 \frac{2t^{n-1}-t^{n-2}-1}{t^n-1}dt\\
=& \ \frac{4\pi}n \csc\frac\pi n
\bigg(\frac n4\ln2 - \sum_{k=1}^{[\frac n2]}\frac{\ln\csc\frac{k\pi}n}{\csc^2\frac{k\pi}n}\bigg)
\end{align}
Substitute above results into $I_n$ to obtain
$$I_n= \frac{4\pi}n \csc\frac\pi n
\bigg(\frac n4\ln2 -\frac\pi8 \cot\frac\pi n- \sum_{k=1}^{[\frac n2]}\frac{\ln\csc\frac{k\pi}n}{\csc^2\frac{k\pi}n}\bigg)
$$
In particular, the close-form yields
\begin{align}
I_2&=\pi \ln2\\
I_3 &=\frac\pi {\sqrt3}\ln3-\frac{\pi^2}9\\
I_4 &=\frac{3\pi}{2\sqrt2}\ln2-\frac{\pi^2}{4\sqrt2}\\
I_5 &= \frac{\pi\sqrt{\phi}}{\sqrt[4]5}\left(\frac12\ln5+\frac1{\sqrt5}\ln\phi\right)-\frac{\pi^2\phi^2}{5\sqrt5}\\
I_6&= \frac{2\pi}3\ln2+\frac\pi2\ln3 -\frac{\pi^2}{2\sqrt3}\\
I_7&= \frac{4\pi}7\csc\frac\pi7\bigg( \frac74\ln2-\frac\pi8\cot\frac\pi7 - \frac{\ln\csc\frac{\pi}7}{\csc^2\frac{\pi}7}- \frac{\ln\csc\frac{2\pi}7}{\csc^2\frac{2\pi}7}-\frac{\ln\csc\frac{3\pi}7}{\csc^2\frac{3\pi}7}\bigg)\\
I_8&= \pi\sqrt{1+\frac1{\sqrt2}}\left(\ln2+\frac1{2\sqrt2}\ln(\sqrt2+1)-\frac\pi8(\sqrt2+1)\right)\\
I_9&= \frac{4\pi}9\csc\frac\pi9\bigg( \frac32\ln2+\frac38\ln3-\frac\pi8\cot\frac\pi9\\
&\hspace{3cm} - \frac{\ln\csc\frac{\pi}9}{\csc^2\frac{\pi}9}- \frac{\ln\csc\frac{2\pi}9}{\csc^2\frac{2\pi}9}-\frac{\ln\csc\frac{4\pi}9}{\csc^2\frac{4\pi}9}\bigg)\\
I_{10}&= \frac{\pi \phi }{10}\bigg(\frac5{2}\ln5+4\ln2 +3\sqrt5\ln\phi-\pi\sqrt[4]5 \ \phi^{3/2}\bigg)\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Pairs of eigenvectors that are not multiple of each other I have the following matrix
\begin{align}
J
&= \begin{bmatrix}
0 & \lambda a \\
a & \delta \\
\end{bmatrix}
\end{align}
For which I find two eigenvalues ($e_-$ and $e_+$):
\begin{align}
\det \left( J - e I \right) &= 0 \\
\det \left(
\begin{bmatrix}
-e & \lambda a \\
a & \delta -e \\
\end{bmatrix} \right) &= 0 \nonumber \\
e^2 - \delta e - \lambda a^2 &= 0 \nonumber \\
e_\pm = \frac{\delta \pm \sqrt{\delta^2 + 4 \lambda a^2}}{2}
\end{align}
Then I find the eigenvectors (denoted with v):
\begin{align}
\left( J - e_\pm I \right) v &= 0 \\
\begin{bmatrix}
- e_\pm & \lambda a \\
a & \delta - e_\pm \\
\end{bmatrix}
\begin{bmatrix}
v^1 \\
v^2 \\
\end{bmatrix} &=
\begin{bmatrix}
0 \\
0 \\
\end{bmatrix} \nonumber \\
\begin{bmatrix}
- e_\pm v^1 + \lambda a v^2 \\
a v^1 + (\delta - e_\pm) v^2 \\
\end{bmatrix}
&=
\begin{bmatrix}
0 \\
0 \\
\end{bmatrix} \nonumber
\end{align}
For the eigenvalue $e_+$:
\begin{align}
v^2 &= \frac{e_+}{\lambda a} v^1
\end{align}
\begin{align}
v^1 &= \frac{e_+ - \delta}{a} v^2
\end{align}
For the eigenvalue $e_-$:
\begin{align}
v^2 &= \frac{e_-}{\lambda a} v^1
\end{align}
\begin{align}
v^1 &= \frac{e_- - \delta}{a} v^2
\end{align}
Therefore, setting $v^1 = 1$:
\begin{align}
\left\{ e_+ \ ; \ [ 1, \frac{e_+}{\lambda a}]^\top \right\} \quad \left\{ e_- \ ; \ [ 1, \frac{e_-}{\lambda a}]^\top \right\}
\end{align}
Setting $v^2 = 1$:
\begin{align}
\left\{ e_+ \ ; \ [ \frac{e_+ - \delta}{a}, 1]^\top \right\} \quad \left\{ e_- \ ; \ [ \frac{e_- - \delta}{a}, 1]^\top \right\}
\end{align}
Could you clarify why I obtain two pairs of eigenvectors that are not multiple of each other? How can the number of eigenvectors be greater than the dimension of the matrix?
| They are scalar multiples of each other:
$$
\left(\frac{e_+-\delta}{a},1\right)
=\frac{e_+-\delta}{a}\left(1,\frac{e_+}{\lambda a}\right).
$$
Note that $\dfrac{e_+-\delta}{a}\dfrac{e_+}{\lambda a}=1$ because $e_+$ is a root of the characteristic equation $x^2-\delta x-\lambda a^2=0$ of $J$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$ Inspired by my post, I decided to investigate the integral in general
$$
I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$$
by the powerful substitution $x=\frac{1-t}{1+t} .$
where $n$ is a natural number greater $1$.
Let’s start with easy one
\begin{aligned}
I_1 &=\int_0^1 \frac{\ln \left(\frac{2 t}{1+t}\right)}{1+t^2} d t \\ &=\int_0^1 \frac{\ln 2+\ln t-\ln (1+t)}{1+t^2} d t \\&=\frac{\pi}{4} \ln 2+\int_0^1 \frac{\ln t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2} d t \\&=\frac{\pi}{4} \ln 2-G-\frac{\pi}{8} \ln 2 \\
&=\frac{\pi}{8} \ln 2-G\end{aligned}
By my post
$$I_2= \frac{\pi}{4} \ln 2-G $$
and
$$\begin{aligned}I_4 &=\frac{3 \pi}{4} \ln 2-2 G
\end{aligned}
$$
$$
\begin{aligned}
I_3=& \int_0^1 \frac{\ln (1-x)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1+x+x^2\right)}{1+x^2} d x \\=& \frac{\pi}{8} \ln 2-G+\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+x+x^2\right)}{1+x^2} d x-G
\\ =& \frac{\pi}{8} \ln 2-\frac{4G}{3} +\frac{\pi}{6} \ln (2+\sqrt{3})
\end{aligned}
$$
where the last integral refers to my post.
Let’s skip $I_5$ now.
$$
I_6=\int_0^1 \frac{\ln \left(1-x^6\right)}{1+x^2} d x=\int_0^1 \frac{\ln \left(1+x^3\right)}{1+x^2} d x+I_3\\
$$
$$\int_0^1 \frac{\ln \left(1+x^3\right)}{1+x^2} d x = \int_0^1 \frac{\ln (1+x)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1-x+x^2\right)}{1+x^2} d x\\=\frac{\pi}{8}\ln 2+ \frac{1}{2} \int_0^{\infty} \frac{\ln \left(1-x+x^2\right)}{1+x^2} d x-G \\= \frac{\pi}{8}\ln 2+ \frac{1}{2}\left( \frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G \right)- G \\= \frac{\pi}{8} \ln 2+\frac{\pi}{3} \ln (2+\sqrt{3})-\frac{5}{3} G $$
Hence $$I_6= \frac{\pi}{4} \ln 2+\frac{\pi}{2} \ln (2+\sqrt{3})-3 G$$
How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x?$
| For even cases, apply
\begin{align}
&1-x^{4m}=(1-x^4) \prod_{k=1}^{m-1} \left(1+2x^2\cos\frac{k\pi}{m}+x^4 \right)\\
& 1-x^{4m+2}= (1-x^2)\prod_{k=0}^{m-1} \left(1+2x^2\cos\frac{(2k+1)\pi}{2m+1}+x^4\right)
\end{align}
and
$$\int_0^1 \frac{\ln(1+2x^2\cos \theta +x^4)}{1+x^2}dx
=\pi \ln\left(2\cos\frac{\theta}4\right)-2G
$$
to obtain
\begin{align}
&\int_0^1\frac{\ln(1-x^{2n})}{1+x^2}dx
=-nG+\frac{(2n-1)\pi}4\ln2+\pi \sum_{k=1}^{[\frac{n-1}2]}\ln \cos\frac{(n-2k)\pi}{4n}
\end{align}
In particular
\begin{align}
\int_0^1\frac{\ln(1-x^{2})}{1+x^2}dx
=& -G+\frac{\pi}4\ln2 \\
\int_0^1\frac{\ln(1-x^{4})}{1+x^2}dx
=& -2G+\frac{3\pi}4\ln2 \\
\int_0^1\frac{\ln(1-x^{6})}{1+x^2}dx
=& -3G-\frac{\pi}4\ln2 +\pi \ln(1+\sqrt3)\\
\int_0^1\frac{\ln(1-x^{8})}{1+x^2}dx
=& -4G+\frac{3\pi}4\ln2 +\frac\pi2\ln(2+\sqrt2)\\
\int_0^1\frac{\ln(1-x^{10})}{1+x^2}dx
=&-5G-\frac{3\pi}4\ln2 +\pi \ln\left(1+\sqrt5+\sqrt{2(5+\sqrt5)}\right)\\
\int_0^1\frac{\ln(1-x^{12})}{1+x^2}dx
=& -6 G+\frac{\pi}4\ln2 +\pi\ln(3+\sqrt3)\\
\int_0^1\frac{\ln(1-x^{14})}{1+x^2}dx
=& -7G+ \frac{13\pi}4\ln2 +\pi \ln\left(\cos\frac\pi{28} \cos\frac{3\pi}{28} \cos\frac{5\pi}{28} \right)\\
\int_0^1\frac{\ln(1-x^{16})}{1+x^2}dx
=&-8G+\frac{5\pi}4\ln2 +\pi\ln\left(1+\sqrt2+\sqrt{2+\sqrt2}\right)\\
\int_0^1\frac{\ln(1-x^{18})}{1+x^2}dx
=& -9G+ \frac{11\pi}4\ln2 +\pi \ln(1+\sqrt3)\\
&\ +\pi\ln\left(\cos\frac\pi{36} \cos\frac{5\pi}{36} \cos\frac{7\pi}{36} \right)\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
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A lower bound for upper Darboux sums of $\int_0^1 x^3 \, dx$
Find the largest $\lambda$ such that $$\sum_{k=1}^n x_k^3\left(x_{k}-x_{k-1}\right)\ge \frac{1}{4}+\frac{\lambda}{n}$$ holds for any $n \in \mathbb{N}$ and any $x_0,x_1,x_2,\cdots, x_n\in\mathbb{R}$ satisfying $$0=x_0\le x_1\le x_2\le \cdots\le x_n=1.$$
It's obvious that the problem is to estimate the Darboux sum of $f(x)=x^3.$
Note that for any $a,b$, it holds that $$a^3(a-b)-\frac{1}{4}(a^4-b^4)=\frac{1}{4}(a-b)^2((a+b)^2+2a^2)\ge0,$$namely $$a^3(a-b)\ge \frac{1}{4}(a^4-b^4).$$
Therefore $$\sum_{k=1}^n x_k^3\left(x_{k}-x_{k-1}\right)\ge \frac{1}{4}\sum_{k=1}^n \left(x_{k}^4-x_{k-1}^4\right)=\frac{1}{4}(x_n^4-x_0^4)=\frac{1}{4}.$$
But how to make the inequality more precisely?
| For every partition $0 = x_0 \le x_1 \le x_2 \le \cdots \le x_n = 1$ of $[0, 1]$ is
$$
\sum_{k=1}^n x_k^3(x_k - x_{k-1}) \ge \frac 14 + \frac 38 \frac 1n
$$
and the factor $\boxed{\lambda = \frac 38}$ is best possible.
Proof: The identity
$$
a^3 (a-b) = \frac 14 (a^4 - b^4) + \frac 38 (a^2-b^2)^2 + \frac 18 (a-b)^3(3a+b)
$$
can be verified easily. It follows that
$$
S_n =\sum_{k=1}^n x_k^3(x_k - x_{k-1}) = A_n + B_n + C_n
$$
where
$$
A_n = \frac 14 \sum_{k=1}^n (x_k^4 - x_{k-1}^4) = \frac 14 \, ,
$$
then, using the Cauchy-Schwarz inequality,
$$
B_n = \frac 38 \sum_{k=1}^n (x_k^2 - x_{k-1}^2)^2
\underset{(*)}{\ge} \frac 38 \frac 1n \left( \sum_{k=1}^n (x_k^2 - x_{k-1}^2)\right)^2
= \frac{3}{8n} \, ,
$$
and finally,
$$
C_n = \frac 18 \sum_{k=1}^n (x_k-x_{k-1})^3(3 x_k + x_{k-1}) \ge 0 \, .
$$
Adding these inequalities gives the desired estimate.
In order to show that the factor $3/8$ is best possible we consider the partitions $x_k = \sqrt{k/n}$, $0 \le k \le n$. Then equality holds at $(*)$ in the estimate of $B_n$, so that
$$
S_n = \frac 14 + \frac 38 \frac 1n + C_n \, .
$$
Now
$$
0 \le C_n \le \frac 12 \sum_{k=1}^n \left( \frac kn \right)^{1/2} \left(
\left(\frac kn \right)^{1/2} - \left(\frac {k-1}n \right)^{1/2}\right)^{3} \\
= \frac{1}{2n^2} \sum_{k=1}^n k^2 \left( 1 - \left( 1 - \frac 1k \right)^{1/2}\right)^{3} \, .
$$
Using $\sqrt{1-x} \ge 1-x$ for $0 \le x \le 1$ it follows that
$$
0 \le C_n \le \frac{1}{2n^2} \sum_{k=1}^n \frac 1k \le \frac{1 + \ln(n)}{2 n^2} \, .
$$
We have therefore shown that for these partitions
$$
S_n = \frac 14 + \frac{3}{8n} + o\left( \frac{1}{n}\right)
$$
and that concludes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Find all the real solutions of $(1+x^2)(1+x^4)=4x^3$. As title suggests, the question is to find all the real roots to the polynomial:
$$(1+x^2)(1+x^4)=4x^3$$
This problem was asked in the Kettering University Math Olympiad a few years back, it's an interesting problem with many different ways to approach it. I'm going to share my own approach here, please make sure to let me know if there are any mistakes in mine or if anything can be improved, and share your own methods too!
Here's my approach for the problem:
$$(1+x^2)(1+x^4)=4x^3$$
Divide by $x^3$ on both sides:
$$\frac{1+x^2}{x}\cdot\frac{1+x^4}{x^2}=4$$
$$\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right)=4$$
Now we can set $y=x+\frac{1}{x}$, thus, $y^2-2=x^2+\frac{1}{x^2}$
$$y(y^2-2)=4$$
$$y^3-2y-4=0$$
$$y^3-8-2y+4=0$$
$$(y-2)(y^2+2y+4)-2(y-2)=0$$
$$(y-2)(y^2+2y+2)=0$$
Now obviously, since we need to find real solutions, the term $(y^2+2y+2)$ will need to be rejected, as it can easily be checked that it yields no real roots.
[Alternatively, we can set $y^2+2y+2=0$ and observe that $(y+1)^2=-1$, thus this term yields no real solution]
Therefore:
$$y=2$$
$$x+\frac{1}{x}=2$$
$$\frac{x^2+1}{x}=2$$
$$x^2-2x+1=0$$
$$(x-1)^2=0$$
Therefore, $x=1$ is our only real solution.
| Nice answer.
An amusing way is to use AM/GM. First, note that any solution is positive, since the left side is positive, and thus $4x^3$ must be positive, so $x$ must be positive.
Then AM/GM says $$\frac{1+x^2}{2}\geq x\\\frac{1+x^4}{2}\geq x^2$$
So $$\frac{(1+x^2)(1+x^4)}{4}\geq x^3$$ for $x>0,$ with equality only when $1=x^2=x^4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Solution for $\Delta u = 0$ for which $u(\cos(t), \sin(t)) = t(2\pi - t) \ , \ 0 \leq t \leq 2\pi$ I have the following problem: Find the solution for $\Delta u = 0$ for which $u(\cos(t), \sin(t)) = t(2\pi - t), \\ 0 \leq t \leq 2\pi$.
My attempt:
I rewrite $u(\cos(t), \sin(t)) = u(x(t), y(t))$ with $x(t) = \cos(t), \ y(t) = \sin(t)$. The Laplace equation is $\frac{d^2 u}{dx^2} + \frac{d^2 u}{dy^2} = 0$, where $\frac{d^2 u}{dx^2} = \frac{d}{dx} (\frac{du}{dx})$ and $\frac{d^2 u}{dy^2} = \frac{d}{dy} (\frac{du}{dy})$. Since $x(t) = \cos(t), \ y(t) = \sin(t)$, $u$ is basically a function of $t$, so I use the chain rule to get $\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt}$.
So we have $\frac{dx}{dt} = \frac{d}{dt} \cos(t) = -\sin(t)$ and $\frac{dy}{dt} = \frac{d}{dt} \sin(t) = \cos(t)$. Here I run into some problems.
*
*How do I calculate $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ ?
Because, as far as I understand it, I should rewrite the part $t(2\pi - t)$ in terms of $x$ and $y$ (or maybe in terms of $\cos(t)$ and $\sin(t)$ ?). But I don't know how to do it (if it's even necessary)
*How would the Laplace equation then look in terms of $u$ being dependent on $t$?
| I’ve made some mistakes in the original answer and is easier to redo everything. I can’t delete the answer so I’m compelled to at another one.
As I understand the problem is this: solve $\Delta u=0$ for $u(\cos t, \sin t)=t(2\pi-t), 0\le t\lt 2\pi$
There is a lot of differentiating from now on.
$$\frac{du}{dx}=\frac{\frac{du}{dt}}{\frac{dx}{dt}}=\frac{2\pi-2t}{-\sin t}=2\frac{t-\pi}{\sin t}$$
$$\frac{d^2u}{dx^2}=\frac{d}{dx}\left(\frac{du}{dx}\right)=\frac{\frac{d}{dt}\left(\frac{du}{dx}\right)}{\frac{dx}{dt}}=\frac{2\frac{\sin t-(t-\pi)\cos t}{\sin ^2 t}}{-\sin t}=-2\frac{\sin t-(t-\pi)\cos t}{\sin ^3 t}$$
In similar manner,
$$\frac{du}{dy}=\frac{2\pi-2t}{\cos t}=-2\frac{t-\pi}{\cos t}$$
$$\frac{d^2u}{dy^2}=\frac{-2\frac{\cos t-(t-\pi)(-\sin t)}{\cos ^2 t}}{\cos t}=-2\frac{\cos t+(t-\pi)\sin t}{\cos^3 t}$$
Now replacing $\sin t =y, \cos t=x$:
$$\Delta u =0\Leftrightarrow \frac{y-(t-\pi)x}{y^3}+\frac{x+(t-\pi)y}{x^3}=0, x^2+y^2=1$$
Solving for t:$$t=\pi+\frac{xy}{x^2-y^2}$$
Replacing this expression of t in the definition of u:
$$u(\cos t, \sin t)=t(2\pi-t)\Leftrightarrow u(x,y)= \pi^2-\left(\frac{xy}{x^2-y^2}\right)^2, x^2+y^2=1$$
Now the function u could be differentiated for x and y:
$$\frac{du}{dx}=-2\frac{xy}{x^2-y^2}\cdot \frac{y(x^2-y^2)-xy\cdot 2x}{(x^2-y^2)^2}=2\frac{xy^2}{(x^2-y^2)^3}$$
$$\frac{d^2u}{dx^2}=2\frac{y^2(x^2-y^2)^3-xy^2\cdot 3(x^2-y^2)^2\cdot 2x}{(x^2-y^2)^6}=-2y^2\frac{1+4x^2}{(x^2-y^2)^4}$$
$$\frac{du}{dy}= -2\frac{xy}{x^2-y^2}\cdot \frac{x(x^2-y^2)-xy\cdot (-2y)}{(x^2-y^2)^2}=-2\frac{x^2y}{(x^2-y^2)^3} $$
$$\frac{d^2u}{dy^2}=-2\frac{x^2(x^2-y^2)^3-x^2y\cdot 3(x^2-y^2)^2\cdot (-2y)}{(x^2-y^2)^6}=-2x^2\frac{1+4y^2}{(x^2-y^2)^4}$$
Solving $\Delta u=0$:
$$\Delta u=0\Leftrightarrow-y^2(1+4x^2)-x^2(1+4y^2)=0\Leftrightarrow 1+8x^2y^2=0$$
This equation has no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Find the largest real root of the equation $2x^2+6x+9=7x\sqrt{2x+3}$
Find the largest real root of the equation $2x^2+6x+9=7x\sqrt{2x+3}$
Source: https://www.hkage.edu.hk/uploads/file/202207/6cda89c718b674f6ac3aa2c19049abe5.pdf
I tried substituting $y = 2x+3$ and ended up with $\frac{y^2}{2}-\frac{7}{2}y^{\frac 3 2} + \frac {21}{2} y^{\frac 1 2} + \frac 9 2 = 0$ but I'm still not sure how to find the largest $y$ from this.
I also tried calculating the derivative, ending up with $(4x+6)\sqrt{2x+3}-21x-21=0$. However, solving for $x$ is not trivial after simplifying the equation.
Also, I believe that calculus is not needed in this contest so I am looking for an elementary solution to this problem. Thanks in advance.
| Notice that
\begin{align*}
2x^2+6x+9-7x\sqrt{2x+3}&=2x^2-7x\sqrt{2x+3}+3(2x+3)\\
&=\left(x-3\sqrt{2x+3}\right)\left(2x-\sqrt{2x+3}\right).
\end{align*}
The real solution of $x-3\sqrt{2x+3}=0$ is $x=9+6\sqrt 3$ and the real solution of $2x-\sqrt{2x+3}=0$ is $x=\frac{1+\sqrt{13}}4$. Finally,
$$9+6\sqrt 3>9>\frac{1+4}4>\frac{1+\sqrt{13}}4.$$
Therefore, the desired answer is $9+6\sqrt 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4588986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of walks of length k between 2 vertices in path graph Let $P_n$ define a path graph with $n$ vertices (example $P_5$ shown on image). Let $f_{n, k}(a, b)$ define a number of walks of length $k$ in $P_n$ starting from vertex $a$ and ending in vertex $b$.
Is there a formula for $f_{n, k}(a, b)$? Even after some research, I couldn't really find any resources about that problem (even though path graphs seem to be pretty simple objects so they should've already been studied very well). Thanks in advance!
| We can use the adjacency matrix: $$f_{n,k}(a,b) = (A(P_n)^k)_{ab}$$ where $A(P_n)$ is the adjacency matrix of $P_n$. That adjacency matrix has the form $$\begin{bmatrix}0 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 1 & 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & 1 & \cdots & 0 & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & 0 & 0 & \cdots & 1 & 0\end{bmatrix}$$ with $1$'s above and below the main diagonal, and zeroes everywhere else.
To get a closed form, we need to diagonalize $A(P_n)$. The eigenvalues are going to have the form $\lambda_j = 2 \cos \frac{j\pi}{n+1}$ for $j=1, \dots, n$, with the eigenvector $v^{(j)}$ of $\lambda_j$ proportional to $(\sin \frac{j\pi}{n+1}, \sin \frac{2j\pi}{n+1}, \dots, \sin \frac{n j \pi}{n+1})$. (This form of the eigenvector has norm $\sqrt{\frac{n+1}{2}}$; we could normalize it, but it's easier to just divide through by $\frac{n+1}{2}$ at the end.) Verifying these boils down to checking the trigonometric identity that $$\sin\left(\theta - \frac{j\pi}{n+1}\right) + \sin\left(\theta + \frac{j\pi}{n+1}\right) = 2 \cos \frac{j\pi}{n+1} \sin \theta$$ which tells us that the $i^{\text{th}}$ component of $A(P_n) v^{(j)}$ is $v^{(j)}_{i+1} + v^{(j)}_{i-1} = \lambda_j v^{(j)}_i$.
In the eigenvector basis, the standard basis vector $e^{(b)}$ is written as:
$$e^{(b)} = \frac{2}{n+1} \sum_{j=1}^n (v^{(j)} \cdot e^{(b)}) v^{(j)} = \frac{2}{n+1} \sum_{j=1}^n \sin \frac{bj\pi}{n+1} v^{(j)}$$
Multiplying through by $A(P_n)^k$, we get
$$
A(P_n)^k e^{(b)} = \frac{2}{n+1} \sum_{j=1}^n \sin \frac{bj\pi}{n+1} \lambda_j^k v^{(j)} = \frac{2}{n+1} \sum_{j=1}^n \sin \frac{bj\pi}{n+1} \left(2 \cos\frac{j\pi}{n+1}\right)^k v^{(j)}.
$$
We want to take the $(a,b)$ entry of $A(P_n)^k$, which is equal to $e^{(a)} \cdot A(P_n)^k e^{(b)}$. So we get
\begin{align}
f_{n,k}(a,b) &= e^{(a)} \cdot A(P_n)^k e^{(b)} \\
&= \frac{2}{n+1} \sum_{j=1}^n \sin \frac{bj\pi}{n+1} \left(2 \cos\frac{j\pi}{n+1}\right)^k (e^{(a)} \cdot v^{(j)}) \\
&= \frac{2}{n+1} \sum_{j=1}^n \sin \frac{aj\pi}{n+1} \sin \frac{bj\pi}{n+1} \left(2 \cos\frac{j\pi}{n+1}\right)^k.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How might I have anticipated that $\frac14(\sqrt{5+2\sqrt5}+\sqrt{10+2\sqrt{5}})$ simplifies to a single surd (namely, $\frac14\sqrt{25+10\sqrt{5}}$)? This is perhaps a silly question related to calculating with surds. I was working out the area of a regular pentagon ABCDE of side length 1 today and I ended up with the following expression :
$$\frac{\sqrt{5+2\sqrt5}+\sqrt{10+2\sqrt{5}}}{4}$$
obtained by summing the areas of the triangles ABC, ACD and ADE.
I checked my solution with Wolfram Alpha which gave me the following equivalent expression :
$$\frac{\sqrt{25+10\sqrt{5}}}{4}$$
I was able to show that these two expressions are equivalent by squaring the numerator in my expression, which gave me
$$15+4\sqrt5+2\sqrt{70+30\sqrt5},$$
and then "noticing" that
$$\sqrt{70+30\sqrt5}=\sqrt{25+30\sqrt5+45}=5+3\sqrt5.$$
My question is the following : how could I have known beforehand that my sum of surds could be expressed as a single surd, and is there a way to systematize this type of calculation ? I would have liked to find the final, simplest expression on my own without the help of a computer.
Thanks in advance !
| $10 + 2 \sqrt 5$ has norm $100 - 5 \cdot 4 = 80.$ $5 + 2 \sqrt 5$ has norm $25 - 5 \cdot 4 = 5.$ The ratio of the norms is $\frac{80}{5} = 16,$ which is an integer and a square, so the ratio might be very nice.
$$ \frac{10+2 \sqrt 5}{5 + 2 \sqrt 5} \cdot \frac{5-2 \sqrt 5}{5 - 2 \sqrt 5}
= \frac{30-10 \sqrt 5}{5 } = 6 - 2 \sqrt 5 $$
Next, $36 - 5 \cdot 4 = 16$ so $ 6 - 2 \sqrt 5 $ might be a square. Indeed, by inspection it is $\left( 1 - \sqrt 5 \right)^2 = \left( \sqrt 5 - 1 \right)^2$ So $10 + 2 \sqrt 5 = \left( \sqrt 5 - 1 \right)^2 \left( 5 + 2 \sqrt 5 \right) $ and
$ \sqrt{10 + 2 \sqrt 5} = \left( \sqrt 5 - 1 \right) \sqrt { 5 + 2 \sqrt 5 } $ Thus
$$ \sqrt{10 + 2 \sqrt 5} + \sqrt { 5 + 2 \sqrt 5 } = \left( \sqrt 5 \right) \sqrt { 5 + 2 \sqrt 5 } = \sqrt { 25 + 10 \sqrt 5 }$$
$$ \color{red}{ \sqrt{10 + 2 \sqrt 5} + \sqrt { 5 + 2 \sqrt 5 } = \sqrt { 25 + 10 \sqrt 5 } } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4590677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Find all three complex solutions of the equation $z^3=-10+5i$ Let $z\in \mathbb{C}$. I want to calculate the three solutions of the equation $z^3=-10+5i$. Give the result in cartesian and in exponential representation.
Let $z=x+yi $.
Then we have $$z^2=(x+yi)^2 =x^2+2xyi-y^2=(x^2-y^2)+2xyi$$
And then $$z^3=z^2\cdot z=[(x^2-y^2)+2xyi]\cdot [x+yi ] =(x^3-xy^2)+2x^2yi+(x^2y-y^3)i-2xy^2=(x^3-3xy^2)+(3x^2y-y^3)i$$
So we get
$$z^3=-10+5i \Rightarrow (x^3-3xy^2)+(3x^2y-y^3)i=-10+5i \\ \begin{cases}x^3-3xy^2=-10 \\ 3x^2y-y^3=5\end{cases} \Rightarrow \begin{cases}x(x^2-3y^2)=-10 \\ y(3x^2-y^2)=5\end{cases}$$
Is everything correct so far? How can we calculate $x$ and $y$ ? Or should we do that in an other way?
| The exact algebraic answer, which expressible in terms of real quantities by real-valued radicals, is:
$$\bbox[5px,border:2px solid #C0A000]{\begin{align}&x=-\alpha_k\sqrt [3]{\frac{5}{3\alpha_k^2-1}}\\
&y=-\sqrt [3]{\frac{5}{3\alpha_k^2-1}}\end{align}}$$
where,
$$\begin{align}\alpha_k=-2+2\sqrt{5}\cos\left(\frac{1}{3}\arccos\left(-\frac{2\sqrt 5}{5}\right)-\frac{2\pi k}{3}\right)\end{align}$$
for $k\in\{0,1,2\}\thinspace .$
$\rm Construction:$
In this answer, we find all pairs of $(x,y)\in\mathbb R^{2}$ in terms of real quantities by real-valued radicals.
After expanding the parentheses, we have:
$$\begin{align}&\begin{cases}x(x^2-3y^2)=-10 \\y(3x^2-y^2)=5\end{cases}\\
\implies &\frac xy\left(\frac {\frac {x^2}{y^2}-3}{\frac {3x^2}{y^2}-1}\right)=-2\end{align}$$
Substituting $\dfrac xy=u$, leads to:
$$\begin{align}&\frac{u(u^2-3)}{3u^2-1}=-2\\
\implies &u^3+6u^2-3u-2=0\end{align}$$
We have $3$ distinct real roots, however this is the case of Casus irreducibilis.
Using the key substitution $u=v-2$, yields:
$$v^3-15v+20=0$$
We know that, the depressed monic cubic equation $x^3+px+q=0$ is solved by
$$\begin{align}x_k=2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\right)-\frac{2\pi k}{3}\right) \quad \text{for} \quad k=0,1,2.\end{align}$$
Thus we obtain:
$$\begin{align}u_k=-2+2\sqrt{5}\cos\left(\frac{1}{3}\arccos\left(-\frac{2\sqrt 5}{5}\right)-\frac{2\pi k}{3}\right) \quad \text{for} \quad k=0,1,2.\end{align}$$
Finally, setting $\dfrac xy=\alpha\,$ we get:
$$\begin{align}&y(3y^2\alpha^2-y^2)+5=0\\
\implies &(3\alpha^2-1)y^3+5=0\\
\implies &y^3=-\frac{5}{3\alpha^2-1}\\
\implies &y=-\sqrt [3]{\frac{5}{3\alpha^2-1}}\\
\implies &x=-\alpha\sqrt [3]{\frac{5}{3\alpha^2-1}}\end{align}$$
where $\alpha\in\{u_0,u_1,u_2\}\thinspace.$
Note that, casus irreducibilis cannot be solved in radicals in terms of real quantities, it can be solved trigonometrically in terms of real quantities.
Therefore, we can no longer escape the general trigonometric or another non-algebraic representations.
| {
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$(a^n-1)(b^n-1)$ can't be a square for all $n$ unless $ab$ is a square Let $a,b$ be positive integers $>1$ such that $(a^n-1)(b^n-1)$ is a square for all $n\ge 1$. Prove that $ab$ is a perfect square.
I'm not asking for a solution to this problem because I already know one. What I'm asking is would the following approach work? or rather I want a particular limit to equal zero.
Assume $(a^n-1)(b^n-1)$ is a perfect square $\forall n\ge1$ and let $x_n=\sqrt{(a^n-1)(b^n-1)}$. Note that $x_n\in \mathbb N $ Now $$x_{n+2}=\sqrt{(a^{n+2}-1)(b^{n+2}-1)}=ab\sqrt{\left(a^{n}-\frac{1}{a^2}\right)\left(b^{n}-\frac{1}{b^2}\right)}$$
and $ab\cdot x_n=ab\sqrt{(a^n-1)(b^n-1)}$. Define $d_n=abx_n-x_{n+2}$ and let us assume for the moment $$\lim_{n\to \infty}d_n=\lim_{n\to \infty}abx_n-x_{n+2}=0$$
Therefore $(d_n)$ converges to $0$, but $d_n\in \mathbb N $ hence $d_n$ is eventually $0$. Meaning for big enough $N$, we have $\forall n\ge N$ $$d_n=abx_n-x_{n+2}=0$$
in other words $abx_n=x_{n+2}$. From this I think there is a lot of ways to get a contradiction but this is what I did,
$$\forall n\ge N' \quad ab\mid x_{n}\mid x_n^2=(a^n-1)(b^n-1)=(ab)^n-a^n-b^n+1$$
which means $a\mid ab\mid a^n+b^n-1\implies a\mid b^n-1$. In particular $$\forall p\ge N' \quad b^p\equiv 1\pmod a$$
Where $p$ is a prime. But this implies $\operatorname{ord}_ab\mid p$ hence $r=\operatorname{ord}_ab=1$ or $p$. Using the very same method we can see $s=\operatorname{ord}_ba=1$ or $p$ both gives a contradiction. If one of $r$ or $s$ is $p$ the contradiction is immediate since $p\mid \phi(a)$ or $\phi(b)$ but clearly this not okay for large enough $p$. Thus both of $r$ and $s$ is $1$. But this means $$a\mid b-1 \text { and } b\mid a-1$$ Hence $a+b\le a+b-2\implies 0\le -2$ a contradiction.
I think that the solution doesn't contain errors (If it did tell me). The only thing that I'm not sure of is $\lim d_n=0$ because I just counldn't evaluate $$\lim_{n\to \infty}\sqrt{(a^n-1)(b^n-1)}-\sqrt{\left(a^{n}-\frac{1}{a^2}\right)\left(b^{n}-\frac{1}{b^2}\right)}$$
You can clearly multiply by the conjugate but after expanding you get $-\infty /\infty$.
| The limit of $d_n$ is not zero except if $a=1$ or $b=1$. Assume that $b \ge a \ge 2$. Then,
\begin{align*}
\sqrt{\Big(a^n-\frac{1}{a^2}\Big)\Big(b^n-\frac{1}{b^2}\Big)} & \ge \sqrt{\Big(a^n-\frac{1}{4}\Big)\Big(b^n-1\Big)} \\
& \ge \sqrt{(a^n-1)(b^n-1) + \frac{3}{4}(b^n-1)} \\
& \ge \sqrt{(a^n-1)(b^n-1)} + \frac{3(b^n-1)}{8\sqrt{(a^n-1)(b^n-1) + \frac{3}{4}(b^n-1)}}\\
& \ge \sqrt{(a^n-1)(b^n-1)} + \frac{3(b^n-1)}{8\sqrt{\frac{7}{4}(b^n-1)(b^n-1)}} \mbox{ as $b \ge a$}\\
& \ge \sqrt{(a^n-1)(b^n-1)} + \frac{3}{4\sqrt{7}}
\end{align*}
where the third inequality comes from: $\sqrt{x+y} - \sqrt{x} = \frac{y}{\sqrt{x+y}+\sqrt{x}}$ so $\sqrt{x+y} \ge \sqrt{x} + \frac{y}{2\sqrt{x+y}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4602746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$?
If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$ ?
The quadratic formula tells me that $x=\dfrac{-13 \pm 20\sqrt{3}}{40}.$
How to proceed?
| If the question is asking you to factor you need not necessarily solve for the roots.
Use the FOIL (First, Out, In, Last) method to expand $(ax + b)(cx + d)$ and simultaneously solve for the coefficients $a, b, c, d$.
We would have $$\begin{cases} ac = 20 \\ ad + bc = 13 \\ bd = -15 \end{cases}$$
Furthermore, once you become more familiar and proficient with FOIL, you can try to guess factors and evaluate cases to see which $2$ factors of $20$ and $2$ factors of $-15$ add up to $13$ directly. This would be as such:
$20 \to \{\pm1, \pm2, \pm4, \pm5, \pm10, \pm20\}$
$-15 \to \{\pm1, \pm3, \mp 5, \mp15\}$
See that $5 \cdot 5 + 4 \cdot -3 = 13$; this is exactly in the form $ad + bc$. Then, $5 \cdot 4 = 20$ and $5 \cdot -3 = -15$, as required.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify fraction within a fraction (Precalculus) Simplify $$\frac{x-2}{x-2-\frac{x}{x-\frac{x-1}{x-2}}}$$
My attempt:
$$=\frac{x-2}{x-2-\frac{x}{\frac{x(x-2)-(x-1)}{x-2}}} \ \ = \ \ \frac{x-2}{x-2-\frac{x}{\frac{x^2-2x-x+1}{x-2}}} \ \ = \ \ \frac{x-2}{x-2-\frac{x^2-2x}{x^2-3x+1}}$$
$$ =\frac{x-2}{\frac{(x-2)(x^2-3x+1)-(x^2-2x)}{x^2-3x+1}} \ \ = \ \ \frac{x-2}{\frac{x^3-3x^2+x-2x^2+6x-2-x^2+2x}{x^2-3x+1}} $$
$$ =\frac{(x-2)(x^2-3x+1)}{x^3-6x^2+9x-2} \ \ = \ \ \frac{x^3-3x^2+x-2x^2+6x-2}{x^3-6x^2+9x-2} $$
$$ =\frac{x^3-5x^2+7x-2}{x^3-6x^2+9x-2} $$
But the answer is given as $ \ \frac{x^2-3x+1}{x^2-4x+1} \ $, so I went wrong somewhere but can't see it. Any help is appreciated thanks.
| Your answer is correct.... Sort of!
The answer you arrived at, can be simplified further to arrive at the answer you wish to achieve, since your attempt is completely valid. So, we have:
$$\frac{x^3-5x^2+7x-2}{x^3-6x^2+9x-2}$$
$$\frac{x^3-5x^2+7x-8+6}{x^3-6x^2+9x-8+6}$$
$$\frac{(x^3-8)-(5x^2-7x-6)}{(x^3-8)-(6x^2-9x-6)}$$
$$\frac{(x-2)(x^2+2x+4)-(5x^2-10x+3x-6)}{(x-2)(x^2+2x+4)-(6x^2-12x+3x-6)}$$
$$\frac{(x-2)(x^2+2x+4)-[5x(x-2)+3(x-2)]}{(x-2)(x^2+2x+4)-[6x(x-2)+3(x-2)]}$$
Factoring out $(x-2)$ from the numerator and denominator and canceling them out, we get:
$$\frac{x^2+2x+4-5x-3}{x^2+2x+4-6x-3}$$
$$\frac{x^2-3x+1}{x^2-4x+1}$$
And that's the desired answer.
| {
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"url": "https://math.stackexchange.com/questions/4604736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there another simpler method to evaluate the integral $\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x , \textrm{ where } \theta \in (0, \pi)?$ Using ‘rationalization’, we can split the integral into two manageable integrals as:
$\displaystyle \begin{aligned}\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x = & \int_0^{2 \pi} \frac{1-\cos \theta \cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & \int_0^{2 \pi} \frac{d x}{1-\cos ^2 \theta \cos ^2 x}-\cos \theta \int_0^{2 \pi} \frac{\cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & 4 \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x-\cos ^2 \theta} d x+\int_0^{2 \pi} \frac{d(\cos \theta \sin x)}{\sin ^2 \theta+\cos ^2 \theta \sin ^2x} \\= & 4 \int_0^{\frac{\pi}{2}} \frac{d\left(\tan x\right)}{\sin ^2 \theta+\tan ^2 x}+\frac{1}{\sin \theta}\left[\tan ^{-1}\left(\frac{\cos ^2 \theta \sin x}{\sin \theta}\right)\right]_0^{2 \pi} \\= & \frac{4}{\sin \theta}\left[\tan ^{-1}\left(\frac{\tan x}{\sin \theta}\right)\right]_0^{\frac{\pi}{2}} \\= & \frac{4}{\sin \theta} \cdot \frac{\pi}{2} \\= & \frac{2 \pi}{\sin \theta}\end{aligned}\tag*{} $
Is there another simpler method to evaluate the integral?
Your comments and alternative methods are highly appreciated.
| Also using complex analysis for $\theta\ne \frac{\pi}{2}$ we have
$$
\int_0^{2 \pi} \frac{dz}{1+\cos \theta \cos x}
{=
\frac{1}{i}\oint_{|z|=1} \frac{2dz}{2z+\cos \theta (z^2+1)}
\\=
\frac{1}{i}2\pi i\text{Res}\left\{\frac{2}{2z+\cos \theta (z^2+1)}\right\}\Bigg|_{z=\frac{\sin \theta-1}{\cos\theta}}
\\=
2\pi\text{Res}\left\{\frac{2}{2z+\cos \theta (z^2+1)}\right\}\Bigg|_{z=\frac{\sin \theta-1}{\cos\theta}}
\\=
2\pi\left\{\frac{1}{1+z\cos \theta}\right\}\Bigg|_{z=\frac{\sin \theta-1}{\cos\theta}}
\\=
\frac{2\pi}{\sin \theta}.
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Where is error in this solution to $ \sqrt{x} + \sqrt{-x} = 2 $? Given the equation:
$$ \sqrt{x} + \sqrt{-x} = 2 $$
The solutions are $x = \pm 2i$. This can be seen via Wolfram Alpha
$$
\left( \sqrt{x} + \sqrt{-x} \right)^2 = 2^2
$$
$$
\sqrt{x}^2 + 2\sqrt{x}\sqrt{-x} + \sqrt{-x}^2 = 4
$$
$$
x + 2\sqrt{x}\sqrt{-x} - x = 4
$$
$$
\sqrt{-x^2} = 2
$$
$$
-x^2 = 4
$$
$$
x = \pm 2i
$$
However, my approach to the problem only found the positive value to this equation.
$$ \sqrt{x} + \sqrt{-x} = 2 $$
$$ \sqrt{x} + i\sqrt{x} = 2 $$
$$ (1+i) \sqrt{x} = 2 $$
$$ (1-i)(1+i) \sqrt{x} = 2(1-i) $$
$$ 2 \sqrt{x} = 2 (1-i) $$
$$ x = (1-i)^2 $$
$$ x = 1^2 - 2i + (-i)^2 = -2i $$
Where did I make a mistake here that resulted in me only getting one of the two solutions to this equation?
| For the function $ \ f(x) \ = \ \sqrt{x} + \sqrt{-x} \ \ , \ $ if we are restricted to $ \ x \ \in \ \mathbb{R} \ \ , \ $ the first term is only defined for $ \ x \ \ge \ 0 \ \ $ and the second term, for $ \ x \ \le \ 0 \ \ , \ $ so the domain for $ \ f(x) \ $ is just the intersection of these intervals, $ \ x \ = \ 0 \ \ , \ $ and its only permissible value is $ \ f(0) \ = \ 0 \ \ . \ $ Consequently, to solve $ \ f(x) \ = \ r \ \ , \ $ with $ \ \ r \ $ being a non-zero real number, we are "forced onto the Argand plane" and must search for complex-valued roots of the equation.
We take a prospective root of the equation to be $ \ z \ = \ \rho·(\cos \theta \ + \ i·\sin \theta) \ = \ \rho·e^{ \ i \ · \ \theta} \ \ , \ \rho \ $ being a positive real number (the modulus of $ \ z \ ) \ \ . $ As heropup points out, the equation $ \ z^{1/2} \ + \ (-z)^{1/2} \ = \ r \ \ $ has a symmetry in that if $ \ z \ $ is a root, $ \ (-z) \ $ is also. As complex numbers have two square-roots, we have
$$ z^{1/2} \ \ = \ \ \pm\sqrt{\rho} \ · \ [ \ \cos ( \theta / 2 ) \ + \ i·\sin (\theta / 2) \ ] \ \ . $$
The root $ \ (-z) \ $ lies in exactly the opposite direction from the origin that $ \ z \ $ does, so we have
$$ -z \ \ = \ \ \rho·[ \ \cos (\pi \ + \ \theta) \ + \ i·\sin (\pi \ + \ \theta) \ ] $$
$$ \Rightarrow \ \ (-z)^{1/2} \ \ = \ \ \pm\sqrt{\rho} \ · \ [ \ \cos (\pi / 2 \ + \ \theta / 2 ) \ + \ i·\sin (\pi / 2 \ + \ \theta / 2 ) \ ] \ \ , $$
which tells us that the square-roots of $ \ (-z) \ $ lie at equal distances from the origin and in directions perpendicular to the square-roots of $ \ z \ \ . \ $
Thus we have, $ \ z^{1/2} \ + \ (-z)^{1/2} $
$$ = \ \ \pm \sqrt{\rho} \ · \ [ \ \cos ( \theta / 2 ) \ + \ i·\sin (\theta / 2) \ ] \ + \ \pm\sqrt{\rho} \ · \ [ \ \cos (\pi / 2 \ + \ \theta / 2 ) \ + \ i·\sin (\pi / 2 \ + \ \theta / 2 ) \ ] $$
$$ = \ \ \pm \sqrt{\rho} \ · \ [ \ \{ \ \cos ( \theta / 2 ) + \cos (\pi / 2 \ + \ \theta / 2 ) \ \} \ + \ i·\{ \ \sin ( \theta / 2 ) + \sin (\pi / 2 \ + \ \theta / 2 ) \ \} \ ] $$
$$ = \ \ \pm \sqrt{\rho} \ · \ [ \ 2· \cos \left( \frac{\pi / 2 \ + \ \theta}{2} \right) · \cos (-\pi / 4 ) \ \ + \ i· 2· \sin \left( \frac{\pi / 2 \ + \ \theta}{2} \right) · \cos (-\pi / 4 ) \ ] $$
$$ = \ \ \pm \sqrt{\rho} · \sqrt2 \ · \ [ \ \cos ( \pi / 4 \ + \ \theta / 2 ) \ + \ i· \sin ( \pi / 4 \ + \ \theta / 2 ) \ ] $$
or
$$ \pm \sqrt{\rho} \ · \ [ \ \cos ( \theta / 2 ) \ + \ i·\sin (\theta / 2) \ ] \ + \ \mp\sqrt{\rho} \ · \ [ \ \cos (\pi / 2 \ + \ \theta / 2 ) \ + \ i·\sin (\pi / 2 \ + \ \theta / 2 ) \ ] $$
$$ = \ \ \pm \sqrt{\rho} \ · \ [ \ \{ \ \cos ( \theta / 2 ) - \cos (\pi / 2 \ + \ \theta / 2 ) \ \} \ + \ i·\{ \ \sin ( \theta / 2 ) - \sin (\pi / 2 \ + \ \theta / 2 ) \ \} \ ] $$
$$ = \ \ \pm \sqrt{\rho} \ · \ [ \ -2· \sin \left( \frac{\pi / 2 \ + \ \theta}{2} \right) · \sin (-\pi / 4 ) \ \ + \ i· 2· \cos \left( \frac{\pi / 2 \ + \ \theta}{2} \right) · \sin (-\pi / 4 ) \ ] \ \ = \ \ 0 $$
$$ = \ \ \pm \sqrt{\rho} · \sqrt2 \ · \ [ \ \sin ( \pi / 4 \ + \ \theta / 2 ) \ - \ i· \cos ( \pi / 4 \ + \ \theta / 2 ) \ ] $$
applying here the sum-to-product identities.
We find four possible implied sums of the square-roots. When we apply our resulting expressions to $ \ z^{1/2} \ + \ (-z)^{1/2} \ = \ r \ \ , \ $ we find for $ \ r + i·0 \ > \ 0 \ \ , \ $
$$ \pm \sqrt{2·\rho} \ \ = \ \ r \ \ \Rightarrow \ \ \rho \ \ = \ \ \frac{r^2}{2} \ \ , $$
$$ \cos ( \pi / 4 \ + \ \theta / 2 ) \ \ = \ \ 1 \ \ \ , \ \ \ \sin( \pi / 4 \ + \ \theta / 2 ) \ \ = \ \ 0 $$
$$ \Rightarrow \ \ \frac{\theta}{2} \ + \ \frac{\pi}{4} \ \ = \ \ 0 \ + \ 2k \pi \ \ \Rightarrow \ \ \frac{\theta}{2} \ \ = \ \ \frac{(8k - 1)· \pi }{4} \ \ \Rightarrow \ \ \theta \ \ = \ \ -\frac{ \pi}{2} \ + \ 4k \pi \ \ , $$
and for $ \ r + i·0 \ < \ 0 \ \ , \ $
$$ \cos ( \pi / 4 \ + \ \theta / 2 ) \ \ = \ \ -1 \ \ \ , \ \ \ \sin( \pi / 4 \ + \ \theta / 2 ) \ \ = \ \ 0 $$
$$ \Rightarrow \ \ \frac{\theta}{2} \ + \ \frac{\pi}{4} \ \ = \ \ \pi \ + \ 2k \pi \ \ \Rightarrow \ \ \frac{\theta}{2} \ \ = \ \ \frac{(8k + 3)· \pi }{4} \ \ \Rightarrow \ \ \theta \ \ = \ \ \frac{3 \pi}{2} \ + \ 4k \pi \ \ . $$
When $ \ r \ $ is real then, the set of roots numbers just two, $ \ z \ = \ -\frac{r^2}{2}·i \ \ $ and, due to the symmetry of $ \ f(x) \ \ , \ (-z) \ = \ +\frac{r^2}{2}·i \ \ . $ For the specific equation of this problem, the roots are $ \ \pm \ \frac{2^2}{2}·i \ = \ \pm \ 2i \ \ . \ $
If we write the square-roots of $ \ z \ $ as $ \ u_{\pm} \ $ and those of $ \ (-z) \ $ as $ \ v_{\pm} \ \ , \ $ the graph below shows that positive and negative values of $ \ r \ $ are obtained by choosing different sums of square-roots.
| {
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"url": "https://math.stackexchange.com/questions/4609145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is there other method to evaluate $\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x, \textrm{ where }n\in N?$ Letting $x\mapsto \frac{1}{x}$ transforms the integral into
$\displaystyle I=\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x=-\int_0^1 \frac{x^{2 n-1} \ln x}{x^n+1} d x \tag*{} $
Splitting the integrand into two pieces like
$\displaystyle I=- \underbrace{\int_0^1 x^{n-1} \ln x d x}_{J} + \underbrace{\int_0^1 \frac{x^{n-1} \ln x}{x^n+1} d x}_{K} \tag*{} $
For the integral $J,$ letting $z=-n\ln x$ transforms $J$ into
$\displaystyle J= -\frac{1}{n^2} \int_0^{\infty} z e^{-z} d z =-\frac{1}{n^2}\tag*{} $
For integral $K$, using the series for $|x|<1,$
$\displaystyle \ln (1+x)=\sum_{k=0}^{\infty} \frac{(-1)^k}{k+1} x^{k+1},\tag*{} $
we have
$\displaystyle \begin{aligned}K& =-\frac{1}{n} \sum_{k=0}^{\infty} \frac{(-1)^k}{k+1} \int_0^1 x^{n(k+1)-1} d x \\& =-\frac{1}{n^2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2} \\& =-\frac{1}{n^2}\left[\sum_{k=1}^{\infty} \frac{1}{k^2}-2 \sum_{k=1}^{\infty} \frac{1}{(2 k)^2}\right] \\& =-\frac{1}{2 n^2} \cdot \frac{\pi^2}{6} \\& =-\frac{\pi^2}{12 n^2}\end{aligned}\tag*{} $
Putting them back yields
$\displaystyle \boxed{ I=\frac{1}{12 n^2}\left(12-\pi^2\right)}\tag*{} $
Is there alternative method? Comments and alternative methods are highly appreciated.
| Substitute $t=\frac1{x^n}$
$$\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x
=-\frac1{n^2}\int_0^1 \frac{t\ln t}{1+t}dt
= \frac{1}{n^2}\left(1-\frac{\pi^2}{12}\right)
$$
| {
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"url": "https://math.stackexchange.com/questions/4609287",
"timestamp": "2023-03-29T00:00:00",
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I solved this question with a Arithmetic mean ≥ Harmonic mean inequality. If there is another solution, please show me. $x,y,z>0$. Prove that:$$\frac{2xy}{x+y} + \frac{2yz}{y+z} + \frac{2xz}{x+z} ≤ x+y+z $$ My solution:$$\frac{x+y}{2}≥\frac{2xy}{x+y}$$ $$\frac{y+z}{2}≥\frac{2yz}{y+z}$$ $$\frac{x+z}{2}≥\frac{2xz}{x+z}$$ Summing up these inequalities $$\frac{2xy}{x+y} + \frac{2yz}{y+z} + \frac{2xz}{x+z} ≤ x+y+z $$I solved this question with a Arithmetic mean ≥ Harmonic mean inequality. If there is another solution, please show me.
| Well I think using means inequalities $HM\le GM\le AM$ is pretty much the way to go, and it all boils down in the end to $(x-y)^2\ge 0$ if you decide to prove it directly.
Indeed the $AM\ge HM$ inequality is just
$AM-HM=\dfrac{x+y}2-\dfrac 2{\frac 1x+\frac 1y}=\dfrac{x+y}2-\dfrac{2xy}{x+y}=\dfrac{(x+y)^2-4xy}{x+y}=\dfrac{(x-y)^2}{x+y}\ge 0$
Even using $AM\ge GM$ inequality leads to the result:
$\dfrac{2xy}{x+y}+\dfrac{2yz}{y+z}+\dfrac{2xz}{x+z}\le \dfrac{2(\frac{x+y}2)^2}{x+y}+\dfrac{2(\frac{y+z}2)^2}{y+z}+\dfrac{2(\frac{x+z}2)^2}{x+z}=\dfrac{x+y}2+\dfrac{y+z}2+\dfrac{x+z}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612207",
"timestamp": "2023-03-29T00:00:00",
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Integrating $\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}$
I'm struggling with the integral, $$\int\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}dx.$$
I tried it as follows:
Substituting $x-2 = \frac1t \implies dx = \frac{-dt}{t^2}.$
$$\therefore \int\frac{dx}{(x-2)^4 \sqrt{x^2 + 6x + 2}} = \int \frac{- dt}{\frac{t^2}{t^4} \sqrt{(\frac1t + 2)^2 + 6 (\frac1t + 2) + 2}} = \int \frac{-t^3}{\sqrt{18t^2 + 10t + 1}}\ dt$$
How to continue from here?
| Continuing where you left off,
$$\begin{align*}
I &= - \int \frac{t^3}{\sqrt{18t^2+10t+1}} \, dt \\[1ex]
&= -16 \int \frac{(s-5)^3}{(s^2-18)^4} \, ds
\end{align*}$$
by employing the substitution,
$$s = \frac{\sqrt{18t^2+10t+1}-1}t \implies t = \frac{2s-10}{18-s^2} \implies dt = \frac{2(18-10s+s^2)}{(18-s^2)^2} \, ds$$
The rest can be done by expanding into partial fractions.
Alternatively, starting over,
$$\begin{align*}
I &= \int \frac{dx}{(x-2)^4 \sqrt{x^2 + 6x + 2}} \\[1ex]
&= \frac18 \int \frac{(1-y^2)^3}{\left(y^2+\sqrt2\,y-4\right)^4} \, dy \tag{1} \\[1ex]
&= -\frac1{4\sqrt2} \int \frac{(z^2-2)^3}{\left(z^2+2z-8\right)^4} \, dz \tag{2}
\end{align*}$$
and again with partial fractions.
*
*$(1)$ : substitute
$$y = \frac{\sqrt{x^2+6x+2} - \sqrt2}{x} \implies x=\frac{2\sqrt2\,y-6}{1-y^2} \implies dx=\frac{2\left(\sqrt2\,y^2-6y+\sqrt2\right)}{(1-y^2)^2} \, dy$$
*
*$(2)$ : substitute $z=\sqrt2\,y$
| {
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Given the curve $y=\frac{5x}{x-3}$. Find its asymptotes, if any. Given the curve $y=\frac{5x}{x-3}$. To examine its asymptotes if any.
We are only taking rectilinear asymptotes in our consideration
My solution goes like this:
We know that, a straight line $x=a$, parallel to $y$ axis can be a vertical asymptote of a branch of the curve $y=f(x)$, iff $f(x)\to\infty$ when $x\to a+0$, $x\to a-0$ or $x\to a$. Similarly, a straight line $y=a$, parallel to $x$ axis can be a horizontal asymptote of a branch of the curve $x=\phi(y)$, iff $x\to\infty$ when $y\to a+0$, $y\to a-0$ or $y\to a$. Using these lemmas, we obtain $x=3$ and $y=5$ as the rectilinear asymptotes. This is because, $y=\frac{5x}{x-3}=f(x)$, then as $x\to 3$, $f(x)\to \infty$. Also, since $y=\frac{5x}{x-3}$, thus,$xy-3y=5x$ or $xy-5x=3y$, hence $x=\frac{3y}{y-5}=\phi(y)$. Due to which $\phi(y)\to \infty$ as $y\to 5$.
Now, I tried checking, if there is any oblique asymptote. We know that, $y=mx+c$, is an oblique asymptote of $y=f(x)$, iff $\exists $ a finite $m=\lim_{|x|\to\infty}\frac{y}{x}$ and $c=\lim_{|x|\to\infty} y-mx$. Now, we have the function $y=\frac{5x}{x-3}$ and hence, $\frac{y}{x}=\frac{5}{x-3}$ and hence, $m=\lim_{|x|\to\infty}\frac{y}{x}=\lim_{|x|\to\infty}\frac{5}{x-3}=0$. Hence, $c=\lim_{|x|\to\infty} y-mx=\frac{5x}{x-3}=\frac{5}{1-\frac 3x}=5$. So, the oblique asymptote of the given curve is $y=mx+c=5$, which is same as the horizontal asymptote, parallel to $x-axis$ as shown above. Is the solution correct? If not, where is it going wrong?
| A homographic function is the quotient of two first-degree polynomial functions, through an expression in the form, with $c\ne 0$,
$$f \left( x \right)=\dfrac {ax+b} {cx+d}=\frac{5x+0}{x-3}$$
$a=5, b=0, c=1, d=-3$.
*
*If $c=0$ then $y =\frac{a \cdot x}{d}+\frac{b}{d}$, which is the equation of a line of angular coefficient $a \over d$, which intersects the $y$-axis at the point of ordinate $b \over d$.
If the mixed product between the coefficients $a\cdot d=b\cdot c$, then $d=\frac{b\cdot c}{a}$ can be substituted, and then, picking up by common factor, $y=\frac{a(ax+b)}{c(ax+b)}$, which simplified gives $y=\frac{a}{c}$, i.e., a line parallel to the $x$-axis representing the horizontal asymptote of the homographic function (The same result is reached by exploiting the definition of limit, i.e., $$y=\lim_{x \to +\infty} \frac{(ax+b)}{(cx+d)} =\lim_{x \to +\infty} \frac{x(a+\frac{b}{x})}{x(c+\frac{d}{x})} = \frac{a+0}{c+0} =\frac{a}{c}$$ which is the horizontal asymptote).
*
*If $c \neq 0$ and $a \cdot d \ne b \cdot c$, then the homographic function represents an equilateral hyperbola with asymptotes parallel to the coordinate axes. In particular, the asymptotes have equations $y=\frac{a}{c}$ and $x=-\frac{d}{c}$. The $x=-\frac{d}{c}$ is a vertical asymptote because must be $cx+d\ne 0$. In our case therefore long calculations are not there. In fact $ad-bc=5\cdot(-3)-0\ne 0$. Therefore the asymptotes have equation $y=5$ (horizontal) and $x=-\frac{-3}{1}=3$ (vertical).
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there any other simpler method to evaluate $\int_0^{\infty} \frac{x^n-2 x+1}{x^{2 n}-1} d x,$ where $n\geq 2?$ Though $n\geq 2$ is a real number, which is not necessarily an integer, we can still resolve the integrand into two fractions,
$$\displaystyle I=\int_0^{\infty} \frac{x^n-2 x+1}{\left(1+x^n\right)\left(1-x^n\right)} d x=\int_0^{\infty}\left(\frac{x}{1+x^n}-\frac{1-x}{1-x^n}\right) d x=J-K\tag*{} $$
For the integral $J$, we are going to transforms it into a Beta function by letting $y=\frac{1}{1+x^n}$.
$$\displaystyle \begin{aligned}J & =\frac{1}{n} \int_0^1 y^{-\frac{2}{n}}(1-y)^{\frac{2}{n}-1} d y \\& =\frac{1}{n} B\left(-\frac{2}{n}+1, \frac{2}{n}\right) \\& =\frac{\pi}{n} \csc \left(\frac{2 \pi}{n}\right) \quad \textrm{ (By Euler Reflection Formula)}\end{aligned}\tag*{} $$
Next, we are going to evaluate the integral
$ \displaystyle K=\displaystyle \int_{0}^{\infty} \frac{1-x}{1-x^{n}} d x \tag*{} $
by the theorem
$ \displaystyle \displaystyle \sum_{k=-\infty}^{\infty} \frac{1}{k+z}=\pi \cot (\pi z), \textrm{ where } z\notin Z.\tag*{} $
We first split the integral into two integrals
$$ \displaystyle \displaystyle \int_{0}^{\infty} \frac{1-x}{1-x^{n}} d x=\int_{0}^{1} \frac{1-x}{1-x^{n}} d x+\int_{1}^{\infty} \frac{1-x}{1-x^{n}} d x \tag*{}$$
Transforming the latter integral by the inverse substitution $ x\mapsto \frac{1}{x}$ m, we have
$$\displaystyle \displaystyle \int_{1}^{\infty} \frac{1-x}{1-x^{n}} d x=\int_{0}^{1} \frac{x^{n-3}-x^{n-2}}{1-x^{n}} d x \tag*{} $$
Putting back yields
$ \begin{aligned}\displaystyle K&=\int_{0}^{1} \frac{1-x+x^{n-3}-x^{n-2}}{1-x^{n}} d x\\\displaystyle &=\int_{0}^{1}\left[\left(1-x+x^{n-3}-x^{n-2}\right) \sum_{k=0}^{\infty} x^{n k}\right] d x\\ \displaystyle & =\sum_{k=0}^{\infty} \int_{0}^{1}\left[x^{n k}-x^{n k+1}+x^{n(k+1)-3}-x^{n(k+1)-2}\right] d x\\ & =\sum_{k=0}^{\infty}\left(\frac{1}{n k+1}-\frac{1}{n k+2}+\frac{1}{n(k+1)-2}-\frac{1}{n(k+1)-1}\right)\\ & =\sum_{k=0}^{\infty}\left[\frac{1}{n k+1}-\frac{1}{n(k+1)-1}\right]+\sum_{k=0}^{\infty}\left[\frac{1}{n(k+1)-2}-\frac{1}{n k+2}\right] \end{aligned}\tag*{} $
Modifying yields
$$\displaystyle \begin{aligned} K&=\frac{1}{n}\left[\sum_{k=0}^{\infty} \frac{1}{k+\frac{1}{n}}+\sum_{k=-1}^{-\infty} \frac{1}{k+\frac{1}{n}}\right]+\frac{1}{n}\left(\sum_{k=1}^{\infty} \frac{1}{k-\frac{2}{n}}+\sum_{k=0}^{-\infty} \frac{1}{k-\frac{2}{n}}\right)\\& =\frac{1}{n}\left(\sum_{k=-\infty}^{\infty} \frac{1}{k+\frac{1}{n}}+\sum_{k=-\infty}^{\infty} \frac{1}{k-\frac{2}{n}}\right)\end{aligned} \tag*{} $$
By the Theorem,
$$ \displaystyle \displaystyle \sum_{k=-\infty}^{\infty} \frac{1}{k+z}=\pi \cot (\pi z), \tag*{} $$
where $ \displaystyle z\notin Z,$
we have
$$ \displaystyle \displaystyle K=\frac{1}{n}\left[\pi \cot \left(\frac{\pi}{n}\right)+\pi \cot \left(\frac{-2 \pi}{n}\right)\right]=\frac{\pi}{n}\left[\cot \left(\frac{\pi}{n}\right)-\cot \left(\frac{2 \pi}{n}\right)\right] =\frac{\pi}{n} \csc \frac{2 \pi}{n} =J\tag*{} $$
We can now conclude that
$\displaystyle \boxed{I=J-K=0 }\tag*{} $
Is there any other simpler method to evaluate $\int_0^{\infty} \frac{x^n-2 x+1}{x^{2 n}-1} d x,$ where $n\geq 2?$
| We have
\begin{align*}
&\int_{0}^{\infty} \frac{x^n - 2x + 1}{x^{2n} - 1} \, \mathrm{d}x \\
&= \int_{0}^{\infty} \left( \frac{1}{x^n - 1} - \frac{2x}{x^{2n} - 1} \right) \, \mathrm{d}x \\
&= \mathrm{PV}\!\int_{0}^{\infty} \frac{1}{x^{n} - 1} \, \mathrm{d}x - \mathrm{PV}\!\int_{0}^{\infty} \frac{2x}{x^{2n} - 1} \, \mathrm{d}x \\
&= \mathrm{PV}\!\int_{0}^{\infty} \frac{1}{x^{n} - 1} \, \mathrm{d}x - \mathrm{PV}\!\int_{0}^{\infty} \frac{1}{u^{n} - 1} \, \mathrm{d}u \tag{$u = x^2$} \\
&= 0.
\end{align*}
Addendum. Here, we rigorously justify the third step. Expanding the principal value using the definition,
\begin{align*}
&\mathrm{PV}\!\int_{0}^{\infty} \frac{2x}{x^{2n} - 1} \, \mathrm{d}x \\
&= \lim_{\varepsilon \to 0^+} \int_{[0, 1-\varepsilon]\cup[1+\varepsilon,\infty)} \frac{2x}{x^{2n} - 1} \, \mathrm{d}x \\
&= \lim_{\varepsilon \to 0^+} \int_{[0, (1-\varepsilon)^2]\cup[(1+\varepsilon)^2,\infty)} \frac{1}{u^n - 1} \, \mathrm{d}u .\tag{$u = x^2$}
\end{align*}
Then the last integral is recast as
\begin{align*}
&\int_{[0, (1-\varepsilon)^2]\cup[(1+\varepsilon)^2,\infty)} \frac{1}{u^n - 1} \, \mathrm{d}u \\
&= \int_{[0, 1-2\varepsilon]\cup[1+2\varepsilon,\infty)} \frac{1}{u^n - 1} \, \mathrm{d}u
+ \int_{1-2\varepsilon}^{(1-\varepsilon)^2} \frac{1}{u^n - 1} \, \mathrm{d}u
- \int_{1+2\varepsilon}^{(1+\varepsilon)^2} \frac{1}{u^n - 1} \, \mathrm{d}u
\end{align*}
The first integral converges to the desired term $\mathrm{PV}\!\int_{0}^{\infty} \frac{1}{u^n - 1} \, \mathrm{d}u$, hence it suffices to prove that the remaining two integrals vanish as $\varepsilon \to 0^+$. However,
\begin{align*}
\int_{1-2\varepsilon}^{(1-\varepsilon)^2} \frac{1}{\left| u^n - 1 \right|} \, \mathrm{d}u
\leq \int_{1-2\varepsilon}^{(1-\varepsilon)^2} \mathcal{O}(\varepsilon^{-1}) \, \mathrm{d}u
\leq \mathcal{O}(\varepsilon)
\end{align*}
and similarly,
\begin{align*}
\int_{1+2\varepsilon}^{(1+\varepsilon)^2} \frac{1}{\left| u^n - 1 \right|} \, \mathrm{d}u
\int_{1+2\varepsilon}^{(1+\varepsilon)^2} \mathcal{O}(\varepsilon^{-1}) \, \mathrm{d}u
\leq \mathcal{O}(\varepsilon).
\end{align*}
Therefore both integrals vanish as $\varepsilon \to 0^+$ and hence the desired claim follows.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Convert a boolean expression to just NOR operatons How do we convert this expression to just $\text{NOR}$ operation?
$$A \cdot B + B \cdot C + C \cdot D$$
My attempt:
$A \cdot B + B \cdot C + C \cdot D = \overline{\overline{A \cdot B + B \cdot C + C \cdot D}} $
Using De Morgan's law
$ \to \overline{\overline{(A \cdot B)} \cdot \overline{(B \cdot C)} \cdot \overline{(C \cdot D)}} $
Using $X \text { NOR } Y = \overline{X} \cdot \overline{Y}$
$\to \overline{{(A \cdot B)} \text{ NOR } {(B \cdot C)} \text{ NOR } {(C \cdot D)}}$
Using $X \text { NOR } Y = \overline{X} \cdot \overline{Y}$
$\to \overline{{(\overline{A} \text{ NOR } \overline{B})} \text{ NOR } {(\overline{B} \text{ NOR } \overline{C})} \text{ NOR } {(\overline{C} \text{ NOR } \overline{D})}}$
Using $X \text { NOR } X = \overline{X} $
$\to \overline{((A \text{ NOR } A) \text{ NOR } (B \text{ NOR } B)) \text{ NOR } ((B \text{ NOR } B) \text{ NOR } (C \text{ NOR } C)) \text{ NOR } ((C \text{ NOR } C) \text{ NOR } (D \text{ NOR } D))}$
Using $X \text { NOR } X = \overline{X} $
$\to (((A \text{ NOR } A) \text{ NOR } (B \text{ NOR } B)) \text{ NOR } ((B \text{ NOR } B) \text{ NOR } (C \text{ NOR } C)) \text{ NOR } ((C \text{ NOR } C) \text{ NOR } (D \text{ NOR } D))) \text{ NOR } (((A \text{ NOR } A) \text{ NOR } (B \text{ NOR } B)) \text{ NOR } ((B \text{ NOR } B) \text{ NOR } (C \text{ NOR } C)) \text{ NOR } ((C \text{ NOR } C) \text{ NOR } (D \text{ NOR } D)))$
I am wondering if is there a better solution that return something simpler.
| WolframAlpha returns
(a NOR c) NOR (b NOR c) NOR (b NOR d)
You can derive it by using the steps here.
| {
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Find the equation of the circle touching the line $(x-2)\cos\theta+(y-2)\sin\theta=1$ for all values of $\theta$
Find the equation of the circle touching the line $(x-2)\cos\theta+(y-2)\sin\theta=1$ for all values of $\theta$
The answer is given on toppr website.
It says, $(x-2)\cos\theta+(y-2)\sin\theta=\cos^2\theta+\sin^2\theta$, then comparing coefficients, it says,
$x-2=\cos\theta, y-2=\sin\theta$
I am not sure about this step. Do you think this step is valid?
Another answer exists on sarthaks website.
It says equation of tangent to the circle $(x-h)^2+(y-k)^2=a^2$ at $(x_1,y_2)$ is
$(x-h)x_1+(y-k)y_1=a^2$
Is this valid? I am not sure about this.
| Here's a simpler and correct (verified by GeoGebra) solution.
The circle center is $(m,n)$, then its distance to the line must be constant as $\theta$ varies. If we write the line's equation as
$$x\cos\theta+y\sin\theta-2\cos\theta-2\sin\theta-1=0,$$
apply the distance formula to have
$$d=\frac{|m\cos\theta+n\sin\theta-2\cos\theta-2\sin\theta-1|}{\sqrt{\cos^2\theta+\sin^2\theta}}.$$
The denominator is $\sqrt1=1$, and the numerator is can be written as
$$\left|\sqrt{(m-2)^2+(n-2)^2}\sin(\theta+\varphi)-1\right|.$$
For it to be constant, $\sqrt{(m-2)^2+(n-2)^2}=0$, so $m=n=2$, and so $d=1$.
Our circle is therefore $\odot((2,2),1)$, equation
$$(x-2)^2+(y-2)^2=1.$$
| {
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"url": "https://math.stackexchange.com/questions/4619501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do we find the value of $\tan\biggr(\sum_{1\leq i\leq 21}f(i)\biggr)$ if $f(x) = \arctan\left(\frac{1}{x^{2 }+ x + 1}\right)$? Let's assume that $f : \mathbb{R}\rightarrow (-\frac{\pi}{2}, \frac{\pi}{2})$ such that for every $x\in \mathbb{R}$, $f(x) = \arctan\biggr(\frac{1}{x^2+x+1}\biggr)$. How can we find $\tan\biggr(\sum_{1\leq i\leq 21}f(i)\biggr)$?
Below is a fact that I've derived:
$$\tan\biggr(f(x)+f(x+1)\biggr) = \frac{\frac{1}{(x+1)^2+(x+1)+1}+\frac{1}{x^2+x+1}}{1-\frac{1}{((x+1)^2+(x+1)+1)(x^2+x+1)}} = \frac{2}{(x+1)^2}$$
But I am not sure how this can be used to approach the problem.
| As @TheBestMagician has mentioned, notice that:
\begin{align*}
\arctan\left(\frac{1}{x^{2} + x + 1}\right) = \arctan\left(\frac{(x + 1) - x}{1 + (x + 1)x}\right) = \arctan(x + 1) - \arctan(x)
\end{align*}
Based on such identity, the proposed sum telescopes.
Hopefully this helps!
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the angle $\angle BDC$
Let's assume that $\angle DBC = 50^{\circ}$, $[DC]$ bisects $\angle ACB$, and that $|AC| = |BC|-|AD|$. How could we find the angle $\angle BDC$?
Applying angle bisector theorem:
$$\frac{|AD|}{|BD|} = \frac{|AC|}{|BC|}$$
Since $|AC| = |BC|-|AD|$,
$$\frac{|AD|}{|BD|} = \frac{|BC|-|AD|}{|BC|} = 1-\frac{|AD|}{|BC|}$$
$$|AD|\biggr(\frac{1}{|BD|}+\frac{1}{|BC|}\biggr) = 1$$
But this won't lead me anywhere, I believe. Could we take complex geometric approach to this problem?
| Let $\displaystyle x = \frac{∠A}{2} \;,\; y = \frac{∠C}{2}\;$
Given $∠B=50° \quad → 2x+2y = 130°$
Law of Tangent, on ΔACD, given $AD = (a-b)$
$\displaystyle \frac{b-(a-b)}{b+(a-b)}
= \frac{2b}{a} - 1
= \frac{\tan \left(\frac{(180-2x-y)-y}{2}\right)}{\tan \left(\frac{(180-2x-y)+y}{2}\right)}
= \frac{\tan 25°}{\tan (90°-x)}
= (\tan 25°)(\tan x)
$
Law of Sine, on on ΔABC
$\displaystyle \frac{b}{a} = \frac{\sin 50°}{\sin 2x}$
Let $t_1 = \tan 25° \;,\; t_2 = \tan x\;$, combine the two expressions:
$\displaystyle
2 \left(\frac{2t_1}{1+t_1^2} \right)
÷ \left(\frac{2t_2}{1+t_2^2} \right) - 1 = t_1\,t_2 $
$\displaystyle → t_2 = \frac{1}{t_1} \;\lor\; \frac{2\,t_1}{1-t_1^2} = \tan(65°) \;\lor\; \tan(50°)$
$x=65° ⇒ ∠C = 0°$, not a triangle, thus not a solution.
$x=50° ⇒ ∠C = 30° ⇒ ∠BDC = \left(180 - 50 - \frac{30}{2}\right)\!° = 115°\;$
Perhaps simpler way to solve for x:
$\displaystyle 2\left(\frac{\sin 50°}{\sin 2x}\right)
= 1 + (\tan25°)(\tan x)
= \frac{(\cos 25°)(\cos x) + (\sin 25°)(\sin x)}{(\cos 25°)(\cos x)}$
$\displaystyle \require{cancel} \frac{\sin 50°}{(\sin x)\cancel{(\cos x)}}
= \frac{\cos(x-25°)}{(\cos 25°){\cancel{(\cos x)}}}$
$(\sin 50°)(\cos 25°) = \sin(x)\cos (x-25°) \quad → x=50°$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of primes in the interval $(\delta n,n)$ Let $0<\delta<1$. What is an asymptotic formula for $$\sum_{\delta n\leqslant p\leqslant n}p,$$where $p$ is a prime number?
| Let $s(n) = \sum_{p\leqslant n, p \text{ is prime}}p$. Then, as detailed in this MathOverflow answer, $s(n) \sim \frac{n^2}{2\log{n}}$.
Thus,
$$
\begin{align}
2s(n) - 2s(\delta n) &= \frac{n^2}{\log{n}} - \frac{\delta^2n^2}{\log{\delta n}} + o\left(\frac{n^2}{\log{n}}\right)\\
&= \frac{n^2 \log{\delta n} - \delta^2n^2\log{n}}{\log{n}\log{\delta n}} + o\left(\frac{n^2}{\log{n}}\right)\\
&= \frac{n^2(1 - \delta^2)}{\log{\delta n}} + \frac{n^2 \log{\delta}}{\log{n}\log{\delta n}} + o\left(\frac{n^2}{\log{n}}\right)\\
&= \frac{n^2(1 - \delta^2)}{\log{n} + \log{\delta}} + o\left(\frac{n^2}{\log{n}}\right)\\
&\sim \frac{n^2(1 - \delta^2)}{\log{n}}
\end{align}
$$
(Here, we have used the characterisation $a \sim b \iff a = b + o(b)$.)
| {
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Can $7^n+1$ be a perfect cube when $n>1$? $7^n+1$ (when $n>1$) can in principle be a perfect cube, since it sometimes is divisible by $8$, and sometimes leaves $8$ as a remainder when dividied by $9$. But when I tried to find perfect cubes that is form of $7^n+1$, I did not suceed.
Can $7^n+1$ be a perfect cube?
| This solution is based on a solution to a similar problem about $2^n + 1$ found here.
No.
Suppose that $7^n + 1 = m^3$ for some positive integers $n$ and $m$. Rearranging, we get $$7^n = m^3 - 1 = (m-1)(m^2 + m + 1).$$
Because $7$ is prime, we know that both $m-1$ and $m^2 + m + 1$ must both be powers of $7$ (including $7^0 = 1$).
*
*If $m-1 = 7^0 = 1$, then $m = 2$ and hence $n = 1$.
*If $m-1 \neq 7^0$, then $7 \mid (m-1)$. This implies $m \equiv 1 \pmod 7$. However, we would get $$m^2 + m + 1 \equiv 1^2 + 1 + 1 \equiv 3 \pmod 7.$$ This contradicts the fact that $m^2 + m + 1$ must be a power of $7$.
So, excluding $n = 1$, there is no other positive integers $n$ such that $7^n + 1$ is a perfect cube.
| {
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Proving $n(n-2)(n-1)^2$ is divisible by 12 I want to show that $n^4 - 4n^3 + 5n^2 -2n$ is divisible by $12$ whenever $n>0$.
I reduced this to $n(n-2)(n-1)^2$. My approach has been to check divisibility by $3$ and $4$.
In both cases, squares are always congruent to $0$ or $1$. So
\begin{align*}n^2 &\equiv_3 0, 1 \\ \implies (n-1)^2 &\equiv_3 n-2, n-1\end{align*}
Given $-1\equiv_3 2$, then $(n-1)^2\equiv_3 n-2$ or $(n-1)^2\equiv_3 n-1$. But now I'm stuck here.
I know the polynomial is made up of a product of 3 consecutive numbers, so either two of them are even and one odd, or two of them are odd and one even. When I made the assumption $n$ is even, then $n-1$ is odd and $n-2$ is even, eventually landing at $n(n-2)(n-1)^2 \equiv_4 n(n-2)$ but so far nothing helpful has come up.
What am I missing?
| Suppose $(n,12)=1.$ We use Euler and lil' Fermat.
$$n^4-4n^3+5n^2-2n\equiv 6-6n\equiv 2(n-1)\pmod {2^2},$$
and
$$n^4-4n^3+5n^2-2n\equiv 6-6n\equiv 0\pmod 3.$$
So by CRT, using Bezout, we have $1×4×2×0-1×3×2(n-1)\equiv6(n-1)\equiv0\pmod{12}.$
Suppose $(n,12)\neq1.$ Then either $2\mid n$ or $3\mid n.$
If $2\mid n,$ then if $3\not \mid n$, we are done by the previous calculation and CCRT.
If $3\mid n,$ then $n^4-4n^3+5n^2-2n\equiv 0\pmod3.$ So we're done by CCRT.
If $2\not\mid n$, then $3\mid n,$ so we're back to Bezout, since we have already done the modular arithmetic: $$1×4×0-1×3×2(n-1)\equiv 0\pmod {12}.$$
| {
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Turn a real number (with a complex closed form) into its trigonometric form This post outlines 'fake' complex numbers (real numbers with complex closed form that usually come from the roots of unfactorable cubics (the example I need right now), or they can come from things like $i^i = e^{-\frac{\pi}{2}}$), in his own answer he gives the example: $\sqrt[3]{1+i \sqrt{7}}+\sqrt[3]{1-i \sqrt{7}}$, which from first glance doesn't look like it could simplified any futher, then states:
Using Euler's formula we can find a trigonometric representation of this number. $2 \sqrt{2} \cos{\left(\frac{\tan^{-1}{\left(\sqrt{7}\right)}}{3}\right)}$
Which when put into WolframAlpha is confirmed, they are both approximately equal to $2.602$
How did he do that? Euler's formula, $e^{ix} = \cos(x) + i\sin(x)$ doesn't seem relevant, the example has no clear $a+ib$ form (and it shouldn't because it is a real number, $b$ will just be $0$, so a will just be the answer anyway) and there is an inverse tangent present in the expression, also covered by Euler's formula somehow? How does he just go from to $A$ to $B$ like it's common knowledge?
Question applies to the example but it would be nice to have further advice on how to turn other fake complex numbers into real forms like this without WolframAlpha.
| Recall DeMoivre's formula:
$$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$
$$\operatorname{cis}(\theta)^n = \operatorname{cis}(n\theta)$$
This can trivially be derived from Euler's formula.
Also note that in polar coordinates,
$$1 + i\sqrt{7} = \sqrt{8} \operatorname{cis}(\tan^{-1}\sqrt{7})$$
So by DeMoivre's formula:
$$(1 + i\sqrt{7})^{1/3}$$
$$ = \left(\sqrt{8} \operatorname{cis}(\tan^{-1}\sqrt{7})\right)^{1/3}$$
$$ = 8^{1/6} \operatorname{cis}(\frac{1}{3}\tan^{-1}\sqrt{7})$$
$$ = \sqrt{2} \operatorname{cis}(\frac{\tan^{-1}\sqrt{7}}{3})$$
$$ = \sqrt{2} \cos(\frac{\tan^{-1}\sqrt{7}}{3}) + i\sqrt{2}\sin(\frac{\tan^{-1}\sqrt{7}}{3}) $$
Similarly,
$$(1 - i\sqrt{7})^{1/3}$$
$$ = \left(\sqrt{8} \operatorname{cis}(-\tan^{-1}\sqrt{7})\right)^{1/3}$$
$$ = 8^{1/6} \operatorname{cis}(-\frac{1}{3}\tan^{-1}\sqrt{7})$$
$$ = \sqrt{2} \operatorname{cis}(-\frac{\tan^{-1}\sqrt{7}}{3})$$
$$ = \sqrt{2} \cos(-\frac{\tan^{-1}\sqrt{7}}{3}) + i\sqrt{2}\sin(-\frac{\tan^{-1}\sqrt{7}}{3}) $$
$$ = \sqrt{2} \cos(\frac{\tan^{-1}\sqrt{7}}{3}) - i\sqrt{2}\sin(\frac{\tan^{-1}\sqrt{7}}{3}) $$
So adding the two cube roots gives $2\sqrt{2} \cos(\frac{\tan^{-1}\sqrt{7}}{3})$, QED.
More generally, if $z = r\operatorname{cis}\theta$, then
$$\sqrt[n]{z} + \sqrt[n]{\overline{z}} = 2\sqrt[n]{r} \cos\left(\frac{\theta}{n}\right)$$
Or if $z = x + iy$, then:
$$\sqrt[n]{x + iy} + \sqrt[n]{x - iy} = 2\sqrt[2n]{x^2 + y^2} \cos\left(\frac{\arg(x + iy)}{n}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4630629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving that ${(1-\frac{2}{x^2})}^x < \frac{x-1}{x+1}$ for any $x > 2$. Proof that ${\left(1-\dfrac{2}{x^2}\right)}^x\!< \dfrac{x-1}{x+1}$ for any $x > 2$.
any ideas?
| Hi @max We can use Binomial series as we have :
$$\left(1-\frac{2}{x^{2}}\right)\leq \left(\frac{\left(x-1\right)}{x+1}\right)^{\frac{1}{x}}=\left(\frac{\left(x-1\right)}{x+1}\right)^{1+\frac{1}{x}-1}$$
So at order two we have :
$$\left(\frac{\left(x-1\right)}{x+1}\right)^{-1}\left(1+\left(\frac{\left(x-1\right)}{x+1}-1\right)\left(1+\frac{1}{x}\right)+\frac{1}{2}\left(\frac{\left(x-1\right)}{x+1}-1\right)^{2}\left(1+\frac{1}{x}\right)\left(\frac{1}{x}\right)\right)\leq \left(\frac{\left(x-1\right)}{x+1}\right)^{\frac{1}{x}}$$
But :
$$\left(\frac{\left(x-1\right)}{x+1}\right)^{-1}\left(1+\left(\frac{\left(x-1\right)}{x+1}-1\right)\left(1+\frac{1}{x}\right)+\frac{1}{2}\left(\frac{\left(x-1\right)}{x+1}-1\right)^{2}\left(1+\frac{1}{x}\right)\left(\frac{1}{x}\right)\right)=\left(1-\frac{2}{x^{2}}\right)$$
We are done !
Some idea fo someone who wants to show Angelo problem :
For $x\geq 2$ we have :
$$\cos\left(\frac{2}{x}\right)-\cos\left(\frac{2}{\sqrt{x}}\right)\geq 2\left|\frac{x-1}{x^{2}+\frac{2}{3}+\frac{1}{3}x}\right|\geq \left(\frac{x-1}{x+1}\right)^{\frac{1}{x}}-\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)^{\frac{1}{\sqrt{x}}}$$
Remains to show :
$$\lim_{n\to \infty}\cos\left(\frac{2}{x^{2n}}\right)-\left(\frac{x^{2n}-1}{x^{2n}+1}\right)^{\frac{1}{x^{2n}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How is $\frac{\cot10^\circ-\cot20^\circ}{\tan40^\circ(\cot10^\circ+\cot20^\circ)-2}=\frac{\sqrt3}{3}$ Can someone show me how $$\dfrac{\cot10^\circ-\cot20^\circ}{\tan40^\circ(\cot10^\circ+\cot20^\circ)-2}=\dfrac{\sqrt3}{3}$$
Is this an obvious simplification, as I really cannot see it. The LHS is equivalent to $$\dfrac{\frac{\cos10^\circ}{\sin10^\circ}-\frac{\cos20^\circ}{\sin20^\circ}}{\frac{\sin40^\circ}{\cos40^\circ}\left(\frac{\cos10^\circ}{\sin10^\circ}+\frac{\cos20^\circ}{\sin20^\circ}\right)-2}$$ which is $$\dfrac{\frac{1}{\sin20^\circ}}{\frac{\cos20^\circ}{\sin10^\circ\cos40^\circ}-2}$$ Is there a straight-forward simplification, or is this the approach?
| You need factorisation and defactorisation formulas to solve the problem.
\begin{align}...&=\dfrac{\sin10^\circ \cos40^\circ}{\sin 20^\circ(\cos20^\circ-2\sin10^\circ\cos40^\circ)}\\
&=\dfrac{ \cos40^\circ}{2\cos 10^\circ(\cos20^\circ-2\sin10^\circ\cos40^\circ)}\\
&=\dfrac{ \cos40^\circ}{2\cos 10^\circ\cos20^\circ-2\sin20^\circ\cos40^\circ}\\
&=\dfrac{ \cos40^\circ}{\cos 30^\circ+\cos10^\circ-\sin60^\circ +\sin20^\circ}\\
&=\dfrac{ \cos40^\circ}{\cos10^\circ+\cos70^\circ}\\
&=\dfrac{ \cos40^\circ}{2\cos40^\circ \cos30^\circ}\\
&={1\over \sqrt{3}}
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that if $x,y>0$, $\left(\frac{x^3+y^3}{2}\right)^2≥\left(\frac{x^2+y^2}{2}\right)^3$ Through some rearrangement of the inequality and expansion, I have been able to show that the inequality is equivalent to
$$x^6-3x^4y^2+4x^3y^3-3x^2y^4+y^6≥0$$
However, I am not sure how to prove the above or if expansion and rearrangement are even correct steps.
| Note that your inequality is a trivial equality for $x=y$ i.e. $x^6=x^6$ as the factors $2$ cancel out.
This indicates that you can factorize $(x-y)$ out of your final equation (i.e. it's value is $0$ when $x=y$)
You get:
$$x^6-3x^4y^2+4x^3y^3-3x^2y^4+y^6
=(x-y)(x^5+x^4y-2x^3y^2+2x^2y^3-xy^4-y^5)$$
The degree $5$ RHS expression again is zero when $x=y$, this can be seen because very symmetric in $x,y$ and signs alternate for quantities such as $+x^4y$ and $-xy^4$, so they cancel out.
So we get to factorize one more time:
$$\underbrace{(x-y)^2}_{\ge 0}\ \underbrace{(x^4+2x^3y+2xy^3+y^4)}_{>0}$$
Now the degree $4$ RHS has only $+$ signs so the expression is strictly positive considering $x,y>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that the Maclaurin expansion of exact ODE general solution has the same form as the power series solution So, given the ODE
$$
y''+2y'+y=0
$$
I have found a power series solution, coefficient recurrence relation and the general solution in terms of elementary functions:
$$
\begin{aligned}
a_{n+2} &= -\frac{2a_{n+1}}{n+2} - \frac{a_n}{(n+1)(n+2)}, \: \: n=0,1,2,... \\
\\
y(x) &= \sum_{n=0}^{\infty}a_nx^n\\
&= a_0(1 - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{8}x^4 +...)+ a_1(x - x^2 + \frac{1}{2}x^3 - \frac{1}{6}x^4 + ...)\\ \\
&= C_1e^{-x} + C_2xe^{-x}\\ \\
&= C_1\sum_{n=0}^{\infty}\frac{(-x)^n}{n!} + C_2x\sum_{n=0}^{\infty}\frac{(-x)^n}{n!}
\end{aligned}
$$
Now, I want to confirm my power series solution by showing that the Maclaurin expansion of the exact general solution has the same form as the power series, and find relations for $C_1$ and $C_2$ in terms of $a_0$ and $a_1$. I.e. find A, B, C, D such that
$$
\begin{align}
C_1 &= Aa_0 + Ba_1\\
C_2 &= Ca_0 + Da_1
\end{align}
$$
I'm really struggling with this.. Equating coefficients feels like it's leading me nowhere and I don't know what else to try.
Edit: my original ODE power series solution
$$
\begin{align}
y = \sum_{n = 0}^{\infty}a_nx^n, \:\: y' = \sum_{n = 0}^{\infty}na_nx^{n-1}, \:\: y'' = \sum_{n = 0}^{\infty}(n-1)na_nx^{n-2}
\end{align}
$$
So the ODE becomes
$$
\begin{align}
&\sum_{n = 0}^{\infty}(n-1)na_nx^{n-2} + 2\sum_{n = 0}^{\infty}na_nx^{n-1} + \sum_{n = 0}^{\infty}a_nx^n = 0 \\
\implies
& \sum_{n = 2}^{\infty}(n-1)na_nx^{n-2} + 2\sum_{n = 1}^{\infty}na_nx^{n-1} + \sum_{n = 0}^{\infty}a_nx^n = 0\\
\implies
& \sum_{i=0}^{\infty}(i+1)(i+2)a_{i+2}x^{i} + 2\sum_{k=0}^{\infty}(k+1)a_{k+1}x^{k} + \sum_{n = 0}^{\infty}a_nx^n = 0\\
\implies
& \sum_{n=0}^{\infty}[(n+1)(n+2)a_{n+2}+(n+1)2a_{n+1} + a_n]x^n = 0\\
\end{align}
$$
Comparing Coefficients leads to the recurrence relation:
$$
\begin{align}
(n+1)(n+2)a_{n+2}+(n+1)2a_{n+1} + a_n &= 0\\
\implies
a_{n+2} &= -\frac{2a_{n+1}}{n+2} - \frac{a_n}{(n+1)(n+2)}, \: \: n=0,1,2,...
\end{align}
$$
Which gives the solution:
$$
y = a_0(1 - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{8}x^4 +...)+ a_1(x - x^2 + \frac{1}{2}x^3 - \frac{1}{6}x^4 + ...)
$$
Apparently this is incorrect but I am really struggling to see a mistake after re-doing this
| Just expand $(C_1+C_2x)e^{-x}$ to get
$$y=C_1+(C_1+C_2)x+(C_1/2+C_2)x^2+\cdots$$
from which we get $a_0=C_1$ and $C_1+C_2=a_1$, i.e. $C_2=a_1-a_0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Expressing complex roots in a trigonometric form
Given that
$$(z+2)^{12}=z^{12}$$
Show that its roots may be expressed in the form
$$-1-i\cot\left(\frac{1}{12}k\pi\right)$$
where $k=\pm1,\pm2,\pm3,\pm4,\pm5.$
My attempt at solving this:
$$\frac{(z+2)^{12}}{z^{12}}=1 \rightarrow \left({\frac{z+2}{z}}\right)^{12}=1$$
$$\frac{z+2}{z}=e^{i\frac{1}{6}k\pi} \rightarrow\frac{2}{z}=-1+e^{i\frac{1}{6}k\pi} \rightarrow z=\frac{2}{-1+e^{i\frac{1}{6}k\pi}}$$
I'm stuck from this point onwards. Have I made a mistake in my working or am unable to proceed due to my deficiency in problem solving?
| Your calculations are correct so far, you can continue with
$$
\frac{2}{-1+e^{i\frac{1}{6}k\pi}} = -1 + \frac{e^{i\frac{1}{6}k\pi}+1}{e^{i\frac{1}{6}k\pi}-1}
= -1 + \frac{e^{i\frac{1}{12}k\pi}+e^{-i\frac{1}{12}k\pi}}{e^{i\frac{1}{12}k\pi}-e^{-i\frac{1}{12}k\pi}}
= -1 +\frac{2\cos(\frac{1}{12}k\pi)}{2i\sin(\frac{1}{12}k\pi)}
$$
Note that $k=6$ (corresponding to $z=-1$) must be included to get all $11$ solutions of the original equation.
| {
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"url": "https://math.stackexchange.com/questions/4642052",
"timestamp": "2023-03-29T00:00:00",
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Range of $f(x)=x \sqrt{1-x^2}$ I have to find the range of $f(x)=x\sqrt{1-x^2}$ on the interval $[-1,1]$. I have done so by setting $x=\sin\theta$ and thus finding it to be $[-0.5,0.5]$.
Let $x=\sinθ$. Then, for $x\in[-1,1]$ we get that $θ \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Thus, $f(x)$ becomes:
$f(\theta)=\sin\theta \sqrt{1-(\sin\theta)^2}= \sin\theta |\cos\theta|$. Since for $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ we have that $cos\theta \geq 0$:
$f(θ)=\sinθ\cosθ=\frac{1}{2}\sin2θ$. Which has maximum value $\frac{1}{2}$ when $\theta = \frac{\pi}{4}$ and minimum value $-\frac{1}{2}$ when $\theta = -\frac{π}{4}$. When $\theta = \frac{\pi}{4}$ we have $x=\frac{\sqrt{2}}{2}$ and when $\theta = -\frac{\pi}{4}$ we have $x=-\frac{\sqrt{2}}{2}$, which are the maximum and minimum positions respectively. So, $f(\frac{\sqrt{2}}{2})=\frac{1}{2}$ and $f(-\frac{\sqrt{2}}{2})=-\frac{1}{2}$. Thus the range of $f(x)$ is $[-\frac{1}{2},\frac{1}{2}]$.
I know that it can be found with derivatives as well. I was wondering how can quadratic theory can be used to find the Range.
| Alternatively, observe that the function is odd over $[-1,1]$. Hence, if the max value occurs at $x_0 > 0$, then the min occurs at $-x_0$. But if $x > 0$, then by AM-GM inequality: $f(x) =x\sqrt{1-x^2} \le \dfrac{x^2+ (1-x^2)}{2}=\dfrac{1}{2}$ with $=$ occurs when $x=\sqrt{1-x^2}\implies x^2 = 1-x^2\implies x=\dfrac{1}{\sqrt{2}}$. So the max is $\dfrac{1}{2}$ and the min is $-\dfrac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\lim_{x \to \infty} (e^{\frac{1}{x}}(x^3 - x^2 + \frac{x}{2}) - \sqrt{x^6 +1})$ using Taylor $\lim\limits_{x \to \infty}\left[e^{\frac{1}{x}}\left(x^3 - x^2 + \dfrac{x}{2}\right) - \sqrt{x^6 +1}\right]$
(I assume that we need to use $t = \frac{1}{x}$ in order to get lim, where argument goes to $0$. But for now I'm not doing this since it won't help much at this moment).
$- \sqrt{x^6 + 1} = -\left( 1 + \dfrac{x^6}{2} + r_n (x)\right)$
Hence, we need somehow get $\frac{x^6}{2}$ from the $e^{\frac{1}{x}}(x^3 - x^2 + \frac{x}{2})$
Doesn't matter what I've tried, after expressing $e^{\frac{1}{x}}$ as Taylor polynomial max degree of any positive term is $3$.
Of course, I can multiply this term with $x^3$, but then I have infinity in nominator, which is not good. I don't know how to cancel $- \frac{x^6}{2}$ with something.
Also after multiplication of $e^{\frac{1}{x}}$ in polynomial form by $(x^3 - x^2 + \frac{x}{2})$ I get $x^3$ which, again, doesn't cancel with anything.
Multiply-divide by conjugate method helps a bit, but then there is "unsolvable" expression in denominator.
Am I doing something wrong?
If you've got another results, please, share
| We have that
$$e^{\frac{1}{x}} = 1+\frac1x+\frac1{2x^2}+\frac1{6x^3} +o\left(\frac1{x^3}\right)$$
$$\sqrt{x^6 +1} = x^3\left(1+\frac1{x^6}\right)^\frac12= x^3+\frac12\frac1{x^3}+o\left(\frac1{x^3}\right)$$
and then
$$e^{\frac{1}{x}}\left(x^3 - x^2 + \frac{x}{2}\right) - \sqrt{x^6 +1}=$$
$$=x^3 - x^2 + \frac{x}{2}+x^2 - x + \frac{1}{2}+\frac x2 - \frac12 + \frac16- x^3+o(1)=$$
$$=\frac 1 6+o(1)\to \frac16$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does $a c>b^2$ if and only if $(1-a) (1-c)<(1-b)^2$, where $0Does $$a c>b^2$$ if and only if $$(1-a) (1-c)<(1-b)^2,$$ where $0<a<1, 0<c<1$, and $0<b<1$ If it is correct, then how to prove it?
| I finally find a easy way to prove it.
First, $ac>b^2$ implies $b^2<(\frac{a+c}{2})^2$. And note that $$(1-a)(1-c)-(1-b)^2 \leq (1-\frac{a+c}{2})^2-(1-b)^2=(2-\frac{a+c}{2}-b)(b-\frac{a+c}{2})<0$$
The last inequality is from $b^2\leq(\frac{a+c}{2})^2$ combined with $a,b,c,\in (0,1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this series $f(n) = f(n/2) + n$? Can I solve this as:
$f(n) = f(n/2) + n$
let find, $$f(n/2) = f(n/2/2) + n/2\\
f(n/2) = f(n/4) + n/2$$
Now,
$$\begin{split}
f(n) &= f(n/4) + n/2 + n\\
f(n) &= f(n/8) + n/4 + n/2 + n
\end{split}$$
hence so on.
$$\vdots$$
Now, $n = 2^i$.
$$\begin{split}
f(2^i) &= f(2^i/2^i) + 2^i/2^{i-1} + 2^i/2^{i-2} + \cdots + 2^i\\
f(2^i) &= f(1) + 2^1 + 2^2 + \cdots + 2^i\\
f(2^i) &= 2^0 + 2^1 + 2^2 + \cdots + 2^i\\
f(2^i) &= 2^{i+1} - 1\\
f(2^i) &= 2^i\cdot 2^1 - 1\\
f(n) &= 2n - 1\\
f(2^k) &= f(2^{k-k}) + k\\
f(2^k) &= f(2^0) + k\\
\cdots &= f(1) + k\\
\cdots &= 1 + k
\end{split}$$
As we know
$$\begin{split}
n &= 2^k\\
\log (n) &= k \log(2)\\
k &= \log (n) / \log(2)\\
k &= \log_2 (n)\\
f(n) &= \log_2(n) + 1
\end{split}$$
| Given $$f(n) = f(n/2) + n$$ for all even integers $n\geq 2$, substitute the linear ansatz $f(n)= A n + B$ to yield
$$A n + B = A/2 n + B/2 + n.$$ Comparing coefficients of $n^1=n$ and of $n^0=1$ yields
the constraint equations $$A = A/2 + 1$$ and $$B=B/2$$ which are simultaneously solved only by $A=2$ and $B=0$. So
$$f(n)=2n$$ is the only affine function satisfying the recurrence relation.
Similarly, the more general ansatz
$$f(n)= \sum_{k=0}^\infty A_k n^k$$ yields that all higher $A_k=0$ for $k\geq 2$.
Alternatively, one can assume there is a second solution $g(n)\neq f(n)$ for the appropriate $n$ and arrive at a contradiction. So
$$f(n)=2n$$ is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
| For example,
$$X = 1+2+3+4+5+6$$
Then twice $X$ is
$$2X = (1+2+3+4+5+6) + (1+2+3+4+5+6)$$
which we can rearrange as
$$2X = (1+2+3+4+5+6) + (6+5+4+3+2+1)$$
and add term by term to get
$$2X = (1+6)+(2+5)+(3+4)+(4+3)+(5+2)+(6+1)$$
to get
$$2X = 7+7+7+7+7+7 = 6*7 = 42$$
| {
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"url": "https://math.stackexchange.com/questions/2260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "136",
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Perfect numbers, the pattern continues The well known formula for perfect numbers is
$$
P_n=2^{n-1}(2^{n}-1).
$$
This formula is obtained by observing some patterns on the sum of the perfect number's divisors. Take for example $496$:
$$
496=1+2+4+8+16+31+62+124+248
$$
one can see that the first pattern is a sequence of powers of $2$ that stops at $16$, the second pattern starts with a prime number, in this case $31$, the rest of them are multiples of $31$, e.g. $2\cdot 31, 4\cdot 31$ and $8\cdot 31$.
But I found that the pattern proceeds after $31$, for example $31=2^5-2^0$, $62=2^6-2^1$, $124=2^7-2^2$ and finally $248=2^8-2^3$, so the perfect number can be written as
$$
496=1+2+4+8+16+(2^5-2^0)+(2^6-2^1)+(2^7-2^2)+(2^8-2^3)
$$
or
$$
496=(1+2+4+8+16+32+64+128+256) -(1+2+4+8).
$$
So the formula follows very naturally from this.
I've searched but didn't find this formulation anywhere.
Well, is this something new? Has anyone seen this somewhere?
| The patterns you're after tell us nothing about the perfect(ness) of a number.
Since they hold even if the number is not perfect, for example, take n = 6. $2^5 (2^6-1) = 2016 = 2^{10} + 2^9 + ... + 2^5$. Which is valid for any n, since in binary, 2^6-1 is 6-1=5 1's from left, and multiplication by 2^5 is adding 5 zeros from right.
Moreover, the following also hold for any n,
$2^5 (2^6-1) = 2^0 + ... + 2^{6-1} + (2^6-2^0) + (2^{6+1} + 2^1) + ... + (2^{2*6-2} - 2^{6-2})$.
$= 1 + ... + 32 + (2^6 - 2^0) + ... + (2^10 - 2^4)$.
but 2016 is not perfect.
Here, $2^n-1$ must be prime so that $2^{n-1}(2^n-1)$ is the unique prime factorization of the number, and in which case, the terms in $\sum_{k=0}^{n-1} 2^k + \sum_{k=0}^{n-2} 2^k(2^n - 1)$ are precisely the divisors of $2^{n-1}(2^n-1)$.
Theorem: If P is an even perfect number, then P = $2^{k-1} (2^k - 1)$ for some k>1 with (2^k - 1) prime.
Proof:
Suppose P is even perfect. Then we can find k>1 such that $P = 2^{k-1}m$, for m odd.
Now, $2^{k-1} - 1 = 1 + ... + 2^{k-2}$ (as explained earlier, 2^{k-1} - 1 has k-2 ones).
i.e explaining why 31 = 1 + 2 + ... + 16.
Write $P = (2^{k-1}-1 + 1)m$. Then $P = (1 + ... + 2^{k-2})m + m$.
The case m is prime: The proper divisors of P are then $1,2, ..., 2^{k-1}, m, 2m, ..., 2^{k-2}m$.
Since P is assumed perfect, $P = 1 + 2 + ... + 2^{k-1} + m + 2m + ... + 2^{k-2}m$.
But $P = (1 + ... + 2^{k-2})m + m$ as shown above.
Therefore, $m = 1 + 2 + ... + 2^{k-1} = 2^{k}-1$
i.e $P = 2^{k-1} (2^k - 1)$ (note 2^{k}-1 prime).
The case for m is not prime: Assume without loss of generality $m = p_1 p_2$, neither a unit or even. Then the divisors (hopefully I didn't miss a divisor) of P are:
$1,2, ..., 2^{k-1}, p_1, 2p_1, ... 2^{k-1}p_1, p_2, 2p_2, ... 2^{k-1}p_2, 2m, ..., 2^{k-2}m$.
Taking the sum of the divisors and equating with $P = (1 + ... + 2^{k-2})m + m$. Then
$m = 1 + 2 + ... + 2^{k-1} + (1 + ... + 2^{k-1})p_1 + (1 + ... + 2^{k-1})p_2 = (2^{k} - 1) (1 + p_1 + p_2)$, but then $p_1 = (1 + p_1 + p_2)$, i.e p_2=-1 or $p_2 = (1 + p_1 + p_2)$, in either case a contradiction.
| {
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Exploring the quadratic equation $x^2 + \lvert x\rvert - 6 = 0$ This question and the described solution are copied from a test-paper :
For the equation $x^2$ + |x| - 6 = 0 analyze the four statements below for correctness.
*
*there is only one root
*sum of the roots is + 1
*sum of the roots is zero
*the product of the roots is +4
Answer : (3)
Answer Explanation :
If x > 0 |x| = x.
Given equation will be $x^2 + x - 6 = 0$⇒ x = 2,- 3 ⇒ x = 2
If x < 0 |x| = - x.
Given equation will b e $x^2$ - x - 6 = 0 ⇒ x = -2, 3 ⇒x = - 2
Sum of roots is 2 - 2 = 0
Now I have a doubt on the statment "If x < 0 |x| = - x." I think modulus means that |x| is always positive ?! Also I can see that (2) seems to be the correct option isn't ?!
Please post your views.
| The statement is saying:
$|x| = -x$ (for $x < 0$)
and divide both sides by $x$ to give us:
$\frac{|x|}{x} = -1$
Let us test a few values to make sure this holds.
$x = -3$ and $|x| = 3$:
$\frac{|-3|}{-3} = -1$
$\frac{3}{-3} = -1$
$-1 = -1$
True.
$x = -5$ and $|x| = 5$
$\frac{5}{-5} = -1$
$-1 = -1$
True.
We can verify this for all $x < 0$.
If you don't believe me, here is a plot of $y = \frac{|x|}{x}$.
Alternative:
If we define
-a is the number such that a + (-a) = 0
ie,
if a = 3, -a = -3, because $3 + (-3) = 0$
if a = -2, -a = 2, because $-2 + 2 = 0$
We can call 3 and -3 "opposites".
The opposite of -2 is 2.
The opposite of 5 is -5.
Then we can translate the original statement then as:
If x < 0, |x| is the opposite of x
Let's test this out.
$x = -4$
$|x| = 4$
$-4 + 4 = 0$
$|-4|$ (or, 4) is indeed the opposite of $-4$.
More generally, the statement says that, if $x < 0$,
$x + |x| = 0$
Which we can verify as true for all values $x < 0$
If you don't believe me, you can look at the plot of $y = x + |x|$
| {
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How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from
$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$
But how can this be proved (geometrically or trigonometrically)?
| for any $\theta \in \mathbb{R}$ the transformation $\psi: x \rightarrow 2x^2 -1$ sends $cos \theta$ to $cos 2\theta$ . hence $\psi^2 $ sends $cos \theta$ to $cos4\theta$
if $\alpha = \frac{2\pi}5$ then $cos 4\alpha = cos \alpha$ so that $cos \alpha$ is a fixed point for $\psi$ and if $c=cos \alpha$ we have
$$
\psi^2(c) = c
$$
i.e.
$$
2(2c^2-1)^2-1= c
$$
or
$$
(c-1)(8c^3-1) = 0
$$
i.e.
$$
(c-1)(2c-1)(4c^2+2c-1) = 0
$$
if we disregard the roots corresponding to angles with rational cosines we have:
$$
4c^2+2c-1 =0,
$$
giving
$$
c = \frac14\left(-1 \pm \sqrt{5}\right)
$$
the negative root corresponds to the angle $\frac{4\pi}5$
| {
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Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$ I wanna know how to solve this equation: $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
I have some trouble to do that and I'd glad with any help I may get.
| Once someone gives the answer, it becomes easier!
I presume you are only looking for real roots and that $\displaystyle (5x-x^3)^{1/3}$ is the unique real number whose cube is $\displaystyle 5x - x^3$.
First observe that
$\displaystyle (x-2)^3 + 5x - x^3 = -6x^2 + 17x -8$
So if $\displaystyle a = (5x-x^3)^{1/3}$ and $\displaystyle b = x-2$
then the equation becomes
$$\displaystyle -2x^2(x-2) - (-6x^2 - 17x - 8) = 2x^2(5x-x^3)^{1/3}$$
and thus
$$\displaystyle -2x^2 b - (a^3 + b^3) = 2x^2a$$
and so
$$\displaystyle (a+b)(a^2 - ab + b^2 + 2x^2) = 0$$
Now for any real numbers $\displaystyle a,b,x$ we have that
$\displaystyle a^2 -ab + b^2 \ge 0$ and $\displaystyle 2x^2 \ge 0$ and thus we must have that
$\displaystyle a + b = 0$ or $\displaystyle 2x^2 = 0$
$\displaystyle a = -b$ can be cubed to give $\displaystyle 6x^2 - 17x + 8 = 0$.
$\displaystyle 2x^2 = 0$ can be easily eliminated.
Note that the transformations we did were equivalent, and so both roots of $\displaystyle 6x^2 - 17x + 8 = 0$ are also roots of the original equation, given the definition of cuberoot at the top of the answer.
If you define $\displaystyle z^{1/3}$ using the principal branch of $\log z$, then the above assumption of $\displaystyle a$ being real is valid only if $\displaystyle 5x - x^3 \ge 0$, which eliminates $\displaystyle \dfrac{17 + \sqrt{97}}{12}$ as a root.
| {
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What type of triangle satisfies: $8R^2 = a^2 + b^2 + c^2 $? In a $\displaystyle\bigtriangleup$ ABC,R is circumradius and $\displaystyle 8R^2 = a^2 + b^2 + c^2 $ , then $\displaystyle\bigtriangleup$ ABC is of which type ?
| In general, in $\triangle{ABC}$
If $\sin^2 A + \sin^2 B + \sin^2 C \gt 2$ then $\triangle{ABC}$ is acute angled.
If $\sin^2 A + \sin^2 B + \sin^2 C = 2$ then $\triangle{ABC}$ is right angled.
If $\sin^2 A + \sin^2 B + \sin^2 C \lt 2$ then $\triangle{ABC}$ is obtuse angled.
Assume $A \le B \lt \pi/2$ and $ A \le B \le C$.
Basically, if $k = \sin^2 A + \sin^2 B + \sin^2 C$
then we have that
$3-2k = \cos 2A + \cos 2B + \cos (2A+2B)$
i.e
$3-2k = 2\cos(A+B)\cos(A-B) + 2\cos^2(A+B) -1 $
i.e
$4-2k = 4\cos(A+B)\cos A\cos B$
So if $k > 2$, then $\cos(A+B) \lt 0$ hence acute.
$k = 2$, then $\cos(A+B) = 0$ hence right triangled.
$k < 2$, then $\cos(A+B) \gt 0$, hence obtuse.
In fact, we can go further and show that the maximum possible value of $k$ is $k = \frac{9}{4}$ which corresponds to $\triangle{ABC}$ being equilateral, as follows:
$4\sin^2 A + 4\sin^2 B + 4\sin^2 C = 9 + \delta$
i.e.
$(2 - 2\cos2A) + (2-2\cos 2B) + 4(1- \cos^2 (A+B)) = 9 + \delta$
i.e.
$1 + 2\cos2A + 2\cos 2B + 4\cos^2(A+B) = -\delta$
i.e.
$1 + 4\cos(A+B)\cos(A-B) + 4\cos^2(A+B) = -\delta$
i.e.
$\sin^2(A-B) + \cos^2(A-B) + 4\cos(A+B)\cos(A-B) + 4\cos^2(A+B) = -\delta$
i.e.
$\sin^2(A-B) + (\cos (A-B) + 2\cos(A+B))^2 = -\delta$.
Hence $\delta \le 0$ and so $\sin^2 A + \sin^2 B + \sin^2 C \le \frac{9}{4}$
The case $\delta = 0$ gives us $\sin(A-B) = 0$ and $\cos(A+B) = \frac{-1}{2}$.
Hence $A=B=C$.
Thus the max value of $\sin^2 A + \sin^2 B + \sin^2 C$ is $\frac{9}{4}$ and is achieved when $A=B=C$.
| {
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A Vector Arithmetic Function, Special Integer Partitions and Residue Sequences Consider the following arithmetic function, $f(a,b) = (a+1, \lceil a- \frac{a}{b} \rceil , \lceil a- \frac{2a}{b} \rceil, \dots, \lceil \frac{a}{b} \rceil, 1)$, where all terms except the first and last are computed by the application of a ceiling function.
For any pair of positive integers $(a,b)$, the vector $f(a,b)$ consists of the terms of an integer partition of length $a+1$. Moreover, $f(a,b)$ and $f(b,a)$ represent dual partitions and have reflected Ferrers Diagrams. This can be proved by recalling the identity
\begin{eqnarray}
\sum_{j = 0}^{b-1} \left \lceil \tfrac{a(b-j)}{b} \right \rceil = \sum_{j = 0}^{a-1} \left \lceil \tfrac{b(a-j)}{a} \right \rceil.
\end{eqnarray}
(Going in the other direction, one can use the methods outlined here to give a neat combinatorial proof of this ceiling summation identity.)
In general, these integer partitions contain no $0$s, contain at least one $1$ and at most $a$ $1$s.
Question: Are these partitions a special (read named) subset of the standard partitions?
Problem 1: Determine all pairs $(a,b)$ such that $f(a,b)$ are the terms of an integer partition of $k$ for a fixed $k > 1$. Determine a simple way to find $k$ from $a$ and $b$ (and not by simply adding the terms of the defining relation).
Problem 2: Find other symmetries like $f(a,b) \leftrightarrow f(b,a)$.
(Motivation) For example, $f(2,3) = (4,2,1)$, $f(3,2) = (3,2,1,1)$, $f(1,5) = (6,1)$ and $f(5,1) = (2,1,1,1,1,1)$, which are dual integer partitions of $7$.
Problem 3: Give an interpretation to the sequence {$f(a,b)$} in terms of $a$ and $b$ for fixed $a$ or $b$.
(Motivation) The sequence {$f(1,b)$} consists of terms of the form $(2,1, \dots, 1)$ with $b + 1$ $1$s, which are integer partitions of $b+2$; {$f(2,b)$} consists of terms of the form $(3, 1), (3,1,1), (3,2,1)$ or $(3, 2, \dots, 2, 1, \dots, 1)$ which are each integer partitions of those integers greater than 3 not divisible by $3$; {$f(3,b)$} consists of special integer partitions of integers congruent to {$1,2,5$} mod $6$. Similarly, {$f(4,b)$} consists of integer partitions of integers congruent to {$1, 2, 3$} mod $5$ or {$1,2,6,8$} mod $10$; {$f(5,b)$} consists of integer partitions of integers congruent to {$1, 2, 4$} mod $6$, etc.
Here is some Mathematica code that might be useful:
f[a_, b_] :=
Flatten[{a + 1, Table[Ceiling[a - j a/b], {j, 1, b - 1}], 1}];
a = 2;
Table[Total@f[a, b], {b, 1, 100}]
Thanks!
| Here's an answer to the second part of Problem 1: $$k = \frac{(a+1)(b+1) - \gcd(a,b) + 3}{2}.$$
According to your definition,
$$k = 1 + (a+1) + \sum_{j=1}^{b-1} \left\lceil \frac{ja}{b} \right\rceil.$$
Now, $\left\lceil \frac{ja}{b} \right\rceil = \left\lfloor \frac{ja + b - 1}{b} \right\rfloor.$ (See, for example, Exercise 3.12, p.96, in Concrete Mathematics, 2nd edition.)
It is known that $$\sum_{j=0}^{b-1} \left\lfloor \frac{aj + x}{b} \right\rfloor = d \left\lfloor \frac{x}{d} \right\rfloor + \frac{(a-1)(b-1)}{2} + \frac{d-1}{2},$$
for any real number $x$, where $d = \gcd(a,b)$. (See, for example, Eq. 3.32, p. 94, of Concrete Mathematics, 2nd edition.) We have $x = b-1$.
Then $$d \left\lfloor \frac{b-1}{d} \right\rfloor = d\frac{b-1}{d} - d \left\{\frac{b-1}{d}\right\} = b-1 - d \left\{\frac{b}{d} - \frac{1}{d}\right\} = b - 1 - d \frac{d-1}{d} = b-d,$$
where $\{x\}$ denotes the fractional part of $x$. (The second-to-last step holds because $d | b$.)
Putting everything together yields $$k = a + 2 + b - d + \frac{(a-1)(b-1)}{2} + \frac{d-1}{2},$$
and then combining terms gets the rest.
The elements in $f(a,b)$ are closely related to Spec$(\frac{a}{b}) = \{\lfloor \frac{a}{b}\rfloor, \lfloor\frac{2a}{b}\rfloor, \lfloor\frac{3a}{b}\rfloor, \ldots \}$, although you want ceilings, and you want to stop when the multiplier is $b-1$. You may find the discussion of Spec$(\alpha)$ in Concrete Mathematics helpful. (In fact, when I looked at your problem I thought, "This reminds me of the discussion of spectra in Concrete Math," and then using results in there it wasn't too difficult to find the expression for $k$.)
| {
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Proof by induction $\frac1{1 \cdot 2} + \frac1{2 \cdot 3} + \frac1{3 \cdot 4} + \cdots + \frac1{n \cdot (n+1)} = \frac{n}{n+1}$ Need some help on following induction problem:
$$\dfrac1{1 \cdot 2} + \dfrac1{2 \cdot 3} + \dfrac1{3 \cdot 4} + \cdots + \dfrac1{n \cdot (n+1)} = \dfrac{n}{n+1}$$
| $\textbf{HINT:}$ Actually this is the answer in itself.
$\frac{n}{n+1} + \frac{1}{(n+1)(n+2)} = \frac{n+1}{n+2}$.
This is all that you will need when you do induction.
| {
"language": "en",
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Does the sum of reciprocals of primes converge? Is this series known to converge, and if so, what does it converge to (if known)?
Where $p_n$ is prime number $n$, and $p_1 = 2$,
$$\sum\limits_{n=1}^\infty \frac{1}{p_n}$$
| Let's start with three lemmas:
*
*Suppose $A\subseteq\{1,2,3,\ldots\}$ and $\sum\limits_{n\in A} \dfrac 1 n < \infty$. Then $\sum\limits_{n\in B} \dfrac 1 n <\infty$ where $B$ is the closure of $A$ under multiplication.
*The closure of the set of primes under multiplication is all of $\{1,2,3,\ldots\}$.
*$\sum\limits_{n=1}^\infty \dfrac 1 n = \infty$.
The second lemma is obvious. The third has a number of well known simple proofs. Here is one of those:
\begin{align}
& \frac 1 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \frac 1 6 + \cdots \tag 1 \\[10pt]
= {} &\left(\frac 1 1 + \frac 1 2\right) + \left(\frac 1 3 + \frac 1 4\right) + \left(\frac 1 5 + \frac 1 6\right) + \cdots \\[10pt]
\ge {} & \left(\frac 1 2 + \frac 1 2 \right) + \left( \frac 1 4 + \frac 1 4 \right) + \left( \frac 1 6 + \frac 1 6 \right) + \cdots \tag 2 \\[10pt]
= {} & \frac 1 1 + \frac 1 2 + \frac 1 3 + \cdots
\end{align}
The inequality on line $(2)$ is strict if the sum on line $(1)$ is finite, and that leads us to a contradiction. ${}\qquad\blacksquare$
The proof of lemma 1 is most of the work; here it is:
\begin{align}
& \sum_{n\in B} \frac 1 n \le \overbrace{\sum_{\begin{smallmatrix} C\subseteq A \\[2pt] C \text{ is finite} \end{smallmatrix}} \prod_{k\in C} \frac 1 k = \prod_{a\in A} \sum_{x=0}^\infty \frac 1 {a^x}}^\text{factoring -- see below} = \prod_{a\in A} \frac 1 {1-\frac 1 a} \\[10pt]
= {} & \exp \sum_{a\in A} - \log\left( 1 - \frac 1 a\right) \le \exp \sum_{a\in A} \frac 1 a < \infty.
\end{align}
(As "Pipicito" points out in a comment below, some members of the set $B$ may occur more than once in the sum below and that is why $\text{“}{\le}\text{''}$ rather than $\text{“}{=}\text{''}$ should appear in the first step above.)
Here's the factorization in more detail: Let $A=\{a_1,a_2,a_3,\ldots\}$. Then the product to the right of $\text{“}{=}\text{''}$ under the $\overbrace{\text{overbrace}}$ above is
\begin{align}
& \left( 1 + \frac 1 {a_1} + \frac 1 {a_1^2} + \frac 1 {a_1^3} + \cdots \right) \\
\times {} & \left( 1 + \frac 1 {a_2} + \frac 1 {a_2^2} + \frac 1 {a_2^3} + \cdots \right) \\
\times {} & \left( 1 + \frac 1 {a_3} + \frac 1 {a_3^2} + \frac 1 {a_3^3} + \cdots \right) \\
\times {} & \quad \cdots \cdots \\
\vdots~
\end{align}
When you expand the product, you multiply a term from the first factor, a term from the second factor, a term from the third factor, etc., but all except finitely many of those are $1$. The reason all but finitely many are $1$ is that if you multiply infinitely many non-$1$s, then the product is $0$, since its a product of infinitely many positive numbers less than $1/2$. Then you add up all possible such finite products, and that gives you the sum to the left of $\text{“}=\text{''}$ under the $\overbrace{\text{overbrace}}$ above.
| {
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Maximize the product of two integer variables with given sum Let's say we have a value $K$. I want to find the value where
$A+B = K$ (all $A,B,K$ are integers) where $AB$ is the highest possible value.
I've found out that it's:
*
*$a = K/2$; $b = K/2$; when $K$ is even
*$a = (K-1)/2$; $b = ((K-1)/2)+1$; when $K$ is odd.
But is there a way to prove this?
| In the even case, assuming $A\leq B$, the numbers will be of the form $A=\frac{K}{2}-m$ and $B=\frac{K}{2}+m$ in order to get $A+B=K$, with $m=0,1,\ldots,\frac{K}{2}$. Then multiplying gives $AB=\left(\frac{K}{2}\right)^2 - m^2 \leq \left(\frac{K}{2}\right)^2$.
In the odd case, you'll have $A=\frac{K}{2}-m-\frac{1}{2}$ and $B=\frac{K}{2}+m+\frac{1}{2}$, with $m=0,1,\ldots,\frac{K-1}{2}$. Multiplying gives
$\begin{align*}
AB &= \left(\frac{K}{2}\right)^2 - \left(m+\frac{1}{2}\right)^2 = \frac{K^2}{4} - m^2 - m - \frac{1}{4} \\
& = \frac{K-1}{2}\left(\frac{K-1}{2}+1\right) - m^2 - m \leq \frac{K-1}{2}\left(\frac{K-1}{2}+1\right).
\end{align*}$
Alternatively, you can think of optimizing a quadratic function by finding the vertex. If you write $AB$ as a quadratic function of $A$ or $B$ alone using the constraint $A+B=K$ (as indicated in PEV's answer), then the closest integer to the vertex will solve your problem because the parabola opens down.
| {
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Integration by parts: $\int e^{ax}\cos(bx)\,dx$ I need to evaluate the following function and then check my answer by taking the derivative:
$$\int e^{ax}\cos(bx)\,dx$$
where $a$ is any real number and $b$ is any positive real number.
I know that you set $u=\cos(bx)$ and $dv=e^{ax} dx$,
and the second time you need to integrate again you set $u=\sin(bx)$ and $dv=e^{ax}dx$ again.
It eventually simplifies down to
$$\int e^{ax}\cos(bx)dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right).$$
Now I know to move the integral on the left side to the right side so that I can just divide by the constant to solve.
Here is my problem:
I know I need to solve the right side to be:
$$\frac{e^{ax}\left(a\cos(bx) + b\sin(bx)\right)}{a^2+b^2} + C.$$
To divide by the constant, I multiplied everything on the right side by
$$\frac{a^2}{b^2+1}.$$
but this leads me to get $b^2 + 1$ on the bottom instead of $a^2 + b^2$.
I will show what I am doing in detail:
After setting the initial integral equal to:
$$\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right) + C$$
I simplify:
$$\int e^{ax}\cos(bx)\,dx = \frac{a^2}{b^2+1}\left(\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx)\right)\right) + C$$
If this is already wrong, can someone point be in the right direction?
If I have not gone wrong yet, I can edit to show the rest of my work.
| So you have
$$\int e^{ax}\cos(bx)dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right).$$
Multiplying out you get
$$\int e^{ax}\cos(bx)\,dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) - \frac{b^2}{a^2}\int e^{ax}\cos(bx)\,dx.$$
At this point, you should move that last integral on the right hand side to the left hand side and add in the constant of integration on the right.
Moving the last integral to the left hand side, you get
$$\left(1 + \frac{b^2}{a^2}\right)\int e^{ax}\cos(bx)\,dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) + C,$$
and I think this is where you made your mistake.
You tried to clear that $1 + \frac{b^2}{a^2}$ by multiplying through by $\frac{a^2}{1+b^2}$. But this is incorrect:
$$1 + \frac{b^2}{a^2} = \frac{a^2+b^2}{a^2} \neq \frac{1+b^2}{a^2}$$
so that what you multiplied through did not clear that factor. You need to multiply by $\frac{a^2}{a^2+b^2}$ (or, more horribly, by
$$\frac{1}{1+\frac{b^2}{a^2}}$$
which is too horrible for words) for things to cancel out.
If you do that, from
$$\frac{a^2+b^2}{a^2}\int e^{ax}\cos(bx)\,dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a^2}e^{ax}\sin(bx) + C,$$
multiplying both sides by $\frac{a^2}{a^2+b^2}$, we get:
$$\int e^{ax}\cos(bx)\,dx = \frac{a}{a^2+b^2}e^{ax}\cos(bx) + \frac{b}{a^2+b^2}e^{ax}\sin(bx) + C$$
the intended answer.
By the way: you don't need to have the "$+C$" on the right hand side until there are no more indefinite integrals there; the constant of integration is implicit in the indefinite integral, so for example, in your penultimate displayed equation, the "$+C$" is superfluous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/20952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Using congruences, show $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is integer for every $n$ Using congruences, show that the following is always an integer for every integer
value of $n$:
$$\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n.$$
| Taking the lcm we have $\displaystyle \frac{1}{15} \cdot \Bigl[ 3n^{5}+5n^{3} + 7n\Bigr]$ Now show that the quantity $3n^{5} + 5n^{3}+7n$ is always divisible by $15$. Induction may be useful.
Clearly for $n=1$, $3+5+7=15$ is divisible by $15$. Assume that it is true for $n=k$. That is assume that $3k^{5}+5k^{3}+7k$ is divisible by $15$. Use this to show that the quantity
\begin{align*}
3(k+1)^{5}+5(k+1)^{3}+7(k+1) &= 3 \Bigl[k^{5} + {5 \choose 1}k^{4} + \cdots +1\Bigr] + 5(k+1)^{3}+ 7(k+1)
\end{align*}
is divisible by $15$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/21548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What are the steps to solve this simple algebraic equation? This is the equation that I use to calculate a percentage margin between cost and sales prices, where $x$ = sales price and $y$ = cost price:
\begin{equation}
z=\frac{x-y}{x}*100
\end{equation}
This can be solved for $x$ to give the following equation, which calculates sales price based on cost price and margin percentage:
\begin{equation}
x=\frac{y}{1-(\frac{z}{100})}
\end{equation}
My question is, what are the steps involved in solving the first equation for $x$? It's been 11 years since I last did algebra at school and I can't seem to figure it out.
I'm guessing the first step is to divide both sides by $100$ like so:
\begin{equation}
\frac{z}{100}=\frac{x-y}{x}
\end{equation}
Then what? Do I multiply both sides by $x$? If so how to I reduce the equation down to a single $x$?
| First, clear the denominator by multiplying both sides by $x$:
\begin{align*}
z &= \frac{100(x-y)}{x}\\
zx &= 100(x-y)
\end{align*}
Then move all the terms that have an $x$ in it to one side of the equation, all other terms to the other side, and factor out the $x$:
\begin{align*}
zx &= 100x - 100y\\
zx - 100x &= -100y\\
x(z-100) &= -100y
\end{align*}
Now divide through by $z-100$ to solve for $x$; you have to worry about dividing by $0$, but in order for $z-100$ to be $0$, you need $z=100$; the only way for $z$ to be equal to $100$ is if $\frac{x-y}{x}=1$, that is, if $x-y=x$, that is, if $y=0$. Since, presumably, you don't get the things for free, you can assume that $y\neq 0$ so this division is valid. You get:
$$x = \frac{-100y}{z-100}.$$
Now, to get it into nicer form, use the minus sign in the numerator to change the denominator from $z-100$ to $100-z$. Then divide both the numerator and the denominator by $100$ to get it into the form you have:
\begin{align*}
x & = \frac{-100y}{z-100}\\
x &= \frac{100y}{100-z}\\
x &= \frac{\frac{1}{100}\left(100 y\right)}{\frac{1}{100}(100-z)}\\
x &= \frac{y}{1 - \frac{z}{100}}.
\end{align*}
Added: Alternatively, following Myself's very good point, you can go "unsimplify" $\frac{x-y}{x}$ to $1 - \frac{y}{x}$, to go from
$$\frac{z}{100} = \frac{x-y}{x} = 1 - \frac{y}{x}$$
to
$$\frac{y}{x} = 1 - \frac{z}{100}.$$
Taking reciprocals and multiplying through by $y$ gives
\begin{align*}
\frac{x}{y} = \frac{1}{1 - \frac{z}{100}}\\
x = \frac{y}{1-\frac{z}{100}}
\end{align*}
which is probably how the particular expression you had (as opposed to $\frac{100y}{100-z}$) arose in the first place.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that if $a^{k} \equiv b^{k} \pmod m $ and $a^{k+1} \equiv b^{k+1} \pmod m $ and $\gcd( a, m ) = 1$ then $a \equiv b \pmod m $ My attempt:
Since $a^{k} \equiv b^{k}( \text{mod}\ \ m ) \implies m|( a^{k} - b^{k} )$ and $a^{k+1} \equiv b^{k+1}( \text{mod}\ \ m ) \implies m|( a^{k+1} - b^{k+1} ) $
Using binomial identity, we have:
$$a^{k} - b^{k} = ( a - b )( a^{k - 1} + a^{k - 2}b + a^{k - 3}b + ... ab^{k - 2} + b^{k - 1} )$$
$$a^{k + 1} - b^{k + 1} = ( a - b )( a^{k} + a^{k - 1}b + a^{k - 2}b + ... ab^{k - 1} + b^{k} )$$
Now there are two cases:
1. If $m|(a - b)$, we're done.
2. Else $m|( a^{k - 1} + a^{k - 2}b + a^{k - 3}b + ... ab^{k - 2} + b^{k - 1} )$ and $m|( a^{k} + a^{k - 1}b + a^{k - 2}b + ... ab^{k - 1} + b^{k} )$
And I was stuck from here, since I could not deduce anything from these two observations. I still have $(a, m) = 1$, and I guess this condition is used to prevent $m$ divides by the two right hand side above.
A hint would be greatly appreciated.
Thanks,
Chan
| This is related to what Bill Dubuque said but is more elementary:
Since $a^k = b^k$ mod $m$, one has $m | (a^k - b^k)$, and similarly $m | (a^{k+1} - b^{k+1})$. Thus by the first equation $m$ divides $b(a^k - b^k) = (a^kb - b^{k+1})$. Subtracting, we have that $m | (a^{k+1} - b^{k+1}) - (a^kb - b^{k+1}) = a^{k+1} - a^kb = a^k(a - b)$. But since $(a,m) = 1$, the fact that $m | a^k(a - b)$ means $m | a - b$, or equivalently $a = b$ mod $m$ which is what you needed to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/23461",
"timestamp": "2023-03-29T00:00:00",
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Resolution of a non-homogeneous heat equation I'm looking for solution to this non-homogeneus problem.
$\frac{\partial u}{\partial t}-\frac{\partial^2 u}{\partial x^2}=F(x,t)$ for $0<x<\pi$, $t>0$
$u(x,0)=0$
$u(0,t)=\frac{\partial u}{\partial x}(x=\pi)=0$
Does anyone know how to proceed?
| Let $u(x,t)=\sum\limits_{s=0}^\infty C(s,t)\sin\dfrac{(2s+1)x}{2}$ so that it automatically satisfies $u(0,t)=\dfrac{\partial u}{\partial x}(x=\pi)=0$ ,
Then $\sum\limits_{s=0}^\infty\dfrac{\partial C(s,t)}{\partial t}\sin\dfrac{(2s+1)x}{2}+\sum\limits_{s=0}^\infty\dfrac{(2s+1)^2}{4}C(s,t)\sin\dfrac{(2s+1)x}{2}=F(x,t)$
$\sum\limits_{s=0}^\infty\biggl(\dfrac{\partial C(s,t)}{\partial t}+\dfrac{(2s+1)^2}{4}C(s,t)\biggr)\sin\dfrac{(2s+1)x}{2}=F(x,t)$
For $0<x<\pi$ , with reference to Wave equation with initial and boundary conditions - is this function right? ,
$\sum\limits_{s=0}^\infty\biggl(\dfrac{\partial C(s,t)}{\partial t}+\dfrac{(2s+1)^2}{4}C(s,t)\biggr)\sin\dfrac{(2s+1)x}{2}=\sum\limits_{s=0}^\infty\dfrac{2}{\pi}\int_0^\pi F(x,t)\sin\dfrac{(2s+1)x}{2}dx~\sin\dfrac{(2s+1)x}{2}$
$\therefore\dfrac{\partial C(s,t)}{\partial t}+\dfrac{(2s+1)^2}{4}C(s,t)=\dfrac{2}{\pi}\int_0^\pi F(x,t)\sin\dfrac{(2s+1)x}{2}dx$
$\dfrac{\partial}{\partial t}\left(e^{\frac{(2s+1)^2t}{4}}C(s,t)\right)=\dfrac{2e^{\frac{(2s+1)^2t}{4}}}{\pi}\int_0^\pi F(x,t)\sin\dfrac{(2s+1)x}{2}dx$
$e^{\frac{(2s+1)^2t}{4}}C(s,t)=\dfrac{2}{\pi}\int e^{\frac{(2s+1)^2t}{4}}\int_0^\pi F(x,t)\sin\dfrac{(2s+1)x}{2}dx~dt$
$e^{\frac{(2s+1)^2t}{4}}C(s,t)=A(s)+\dfrac{2}{\pi}\int_0^t e^{\frac{(2s+1)^2t}{4}}\int_0^\pi F(x,t)\sin\dfrac{(2s+1)x}{2}dx~dt$
$C(s,t)=A(s)e^{-\frac{(2s+1)^2t}{4}}+\dfrac{2e^{-\frac{(2s+1)^2t}{4}}}{\pi}\int_0^t e^{\frac{(2s+1)^2t}{4}}\int_0^\pi F(x,t)\sin\dfrac{(2s+1)x}{2}dx~dt$
$\therefore u(x,t)=\sum\limits_{s=0}^\infty\biggl(A(s)e^{-\frac{(2s+1)^2t}{4}}+\dfrac{2e^{-\frac{(2s+1)^2t}{4}}}{\pi}\int_0^t e^{\frac{(2s+1)^2t}{4}}\int_0^\pi F(x,t)\sin\dfrac{(2s+1)x}{2}dx~dt\biggr)\sin\dfrac{(2s+1)x}{2}$
$u(x,0)=0$ :
$\sum\limits_{s=0}^\infty A(s)\sin\dfrac{(2s+1)x}{2}=0$
$A(s)=0$
$\therefore u(x,t)=\dfrac{2}{\pi}\sum\limits_{s=0}^\infty\int_0^t e^{\frac{(2s+1)^2t}{4}}\int_0^\pi F(x,t)\sin\dfrac{(2s+1)x}{2}dx~dt~e^{-\frac{(2s+1)^2t}{4}}\sin\dfrac{(2s+1)x}{2}$
$u(x,t)=\dfrac{2}{\pi}\sum\limits_{s=0}^\infty\int_0^t e^{\frac{(2s+1)^2\tau}{4}}\int_0^\pi F(\xi,\tau)\sin\dfrac{(2s+1)\xi}{2}d\xi~d\tau~e^{-\frac{(2s+1)^2t}{4}}\sin\dfrac{(2s+1)x}{2}$
$u(x,t)=\dfrac{2}{\pi}\int_0^t\int_0^\pi\sum\limits_{s=0}^\infty e^{-\frac{(2s+1)^2(t-\tau)}{4}}\sin\dfrac{(2s+1)x}{2}\sin\dfrac{(2s+1)\xi}{2}F(\xi,\tau)~d\xi~d\tau$
| {
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"url": "https://math.stackexchange.com/questions/24480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Equation of the complex locus: $|z-1|=2|z +1|$ This question requires finding the Cartesian equation for the locus:
$|z-1| = 2|z+1|$
that is, where the modulus of $z -1$ is twice the modulus of $z+1$
I've solved this problem algebraically (by letting $z=x+iy$) as follows:
$\sqrt{(x-1)^2 + y^2} = 2\sqrt{(x+1)^2 + y^2}$
$(x-1)^2 + y^2 = 4\big((x+1)^2 + y^2\big)$
$x^2 - 2x + 1 + y^2 = 4x^2 + 8x + 4 + 4y^2$
$3x^2 + 10x + 3y^2 = -3$
$x^2 + \frac{10}{3}x + y^2 = -1$
$(x + \frac{5}{3})^2 +y^2 = -1 + \frac{25}{9}$
therefore, $(x+\frac{5}{3})^2 + y^2 = \frac{16}{9}$, which is a circle.
However, I was wondering if there is a method, simply by inspection, of immediately concluding that the locus is a circle, based on some relation between the distance from $z$ to $(1,0)$ on the plane being twice the distance from $z$ to $(-1,0)$?
| A perhaps easier way to see that it is a circle is to rewrite as
$\displaystyle |z'| = 2|z' + 2|, \text{where } z' = z -1$
And so
$\displaystyle 1/4 = |1/2 + 1/z'|$
Thus $\displaystyle \frac{1}{z'}$ gives a circle centred at $(-1/2,0)$ and radius $1/4$.
As we can see from the answers and comments to this recent question: If $0$, $z_1$, $z_2$ and $z_3$ are concyclic, then $\frac{1}{z_1}$,$\frac{1}{z_2}$,$\frac{1}{z_3}$ are collinear that, the map $\displaystyle z \to \frac{1}{z}$ maps circles to circles, if the circle does not pass through the origin.
In this case, it does not pass through the origin, and so the original curve you seek would be a circle too.
Of course the fact that $\displaystyle z \to \frac{1}{\bar{z}}$ is inversion with respect to the unit circle probably uses the Appollonius theorem, but it looks like a useful fact to keep in mind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/27199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Integer multiplication using FFT I've some trouble in understanding integer multiplication using FFT.
I'm using the algorithm described on wikipedia.
Here is an example of how I understand this algorithm:
$$a=173$$
$$b=95$$
Lets take $w=4$, then we have
$$a=13*2^{4\cdot 0}+10*2^{4\cdot1}$$
$$b=15*2^{4\cdot 0}+5*2^{4\cdot1}$$
In the vector notation it looks like this:
$$a = \left[ \begin{array}{cc} 13 \\ 10 \end{array} \right]$$
$$b = \left[ \begin{array}{cc} 15 \\ 5 \end{array} \right]$$
The FFT of those vectors are:
$$f(a) = \left[ \begin{array}{cc} \frac{23}{\sqrt{2}} \\ \frac{3}{\sqrt{2}} \end{array} \right]$$
$$f(b) = \left[ \begin{array}{cc} \frac{20}{\sqrt{2}} \\ \frac{10}{\sqrt{2}} \end{array} \right]$$
The product of those results entry by entry is:
$$c = \left[ \begin{array}{cc} 230 \\ 15 \end{array} \right]$$
The inverse FFT of $c$ is:
$$f^{-1}(c)= \left[ \begin{array}{cc} \frac{245}{\sqrt{2}} \\ \frac{215}{\sqrt{2}} \end{array} \right]$$
And now what should be done?
Edit
As Myself noticed those vectors should be 4dimensional instead of two dimensional. Here are correct calculations (in case anyone has similar problem):
$$a = \left[ \begin{array}{cc} 13 \\ 10 \\ 0 \\ 0 \end{array} \right]$$
$$b = \left[ \begin{array}{cc} 15 \\ 5 \\ 0 \\ 0 \end{array} \right]$$
The FFT of those vectors are:
$$f(a) = \left[ \begin{array}{cc} 11.5 \\ 6.5+5i \\ 1.5 \\ 6.5-5i \end{array} \right]$$
$$f(b) = \left[ \begin{array}{cc} 10 \\ 7.5+2.5i \\ 5 \\ 7.5-2.5i \end{array} \right]$$
The product of those results entry by entry is:
$$c = \left[ \begin{array}{cc} 115 \\ 36.25+53.75i \\ 7.5 \\ 36.25-53.75i \end{array} \right]$$
The inverse FFT of $c$ is:
$$f^{-1}(c)= \left[ \begin{array}{cc} 195 \\ 215 \\ 50 \\ 0 \end{array} \right]$$
So the final result is $ab=195\cdot 2^{0} + 215 \cdot 2^{4} + 50 \cdot 2^{8} = 16435$
| -- answer first posted as comment --
At this point I think you're supposed to reinterpret this result as a natural number that's the product of the numbers you started with. Wait, $\sqrt{2}$? Have you double-checked the calculations? Also I think your vectors should have 4 positions because now the result will be cropped (i.e. the circular convolution gives a different result than the real convolution).
| {
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"timestamp": "2023-03-29T00:00:00",
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Simple 4-cycle permutation I call a 4-cycle permutation simple if I can write it as $(i,i+1,i+2,i+3)$ so $(2,3,4,5)$ is a simple 4-cycle but $(1,3,4,5)$ is not. I want to write $(1,2,3,5)$ as a product of simple 4-cycles. So this is what I did:
$$
(1,2,3,5)=(1,2)(1,3)(1,5)
$$
but
$$\begin{align}
(1,3)&=(2,3)(1,2)(2,3)\\
(1,5)&=(4,5)(3,4)(2,3)(1,2)(2,3)(3,4)(4,5)
\end{align}$$
So now
$$(1,2,3,5)=(1,2)(2,3)(1,2)(2,3)(4,5)(3,4)(2,3)(1,2)(2,3)(3,4)(4,5)$$
Can you please give me a hint on how I can express
$$(1,2)(2,3)(1,2)(2,3)(4,5)(3,4)(2,3)(1,2)(2,3)(3,4)(4,5)$$
as a product of simple 4-cycles.
Note: We do permutation multiplication from left to right.
| (2,3,4,5) is (1,2,3,5) conjugated by (1,2,3,4) so (1,2,3,5) = (1,2,3,4)(2,3,4,5)(1,2,3,4)−1.
While this may appear to be coincidence, I think you'll find it works quite well in general as suggested by Carl Brannen.
| {
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"timestamp": "2023-03-29T00:00:00",
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Self-Contained Proof that $\sum\limits_{n=1}^{\infty} \frac1{n^p}$ Converges for $p > 1$ To prove the convergence of the p-series
$$\sum_{n=1}^{\infty} \frac1{n^p}$$
for $p > 1$, one typically appeals to either the Integral Test or the Cauchy Condensation Test.
I am wondering if there is a self-contained proof that this series converges which does not rely on either test.
I suspect that any proof would have to use the ideas behind one of these two tests.
| This is the most direct and elementary way I know how to prove the result, although it only works for powers in the range $[0,1] \cup [2,\infty)$ which is exactly the uninteresting set, and Joriki's answer is much better regardless. I had already written this, and perhaps somebody finds it useful.
First we consider $p=1$. Set $x_n$ equal to the greatest power of 2 less than $\frac{1}{n}$. That is, $$(x_n) = (\frac{1}{2}, \frac{1}{4},\frac{1}{4}, \frac{1}{8}, \frac{1}{8},\frac{1}{8},\frac{1}{8},\dots).$$
Note that
$$\sum_{i=1}^{\infty} x_n
= \sum_{i=0}^{\infty}\Big(\sum_{j=2^i}^{2^{i+1}-1}x_j\Big)
= \sum_{i=0}^{\infty}\Big(\sum_{j=2^i}^{2^{i+1}-1} \frac{1}{2^{i+1}}\Big)
= \sum_{i=1}^{\infty} \frac{1}{2},
$$
which diverges, and that $x_n < \frac{1}{n}$, which proves $\sum_{n=1}^{\infty} \frac{1}{n}$ also diverges.
If $p = 2$, let $x_n = \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$ if $n>1$ and 1 if $n=1$.
Observe that
$$\sum_{n=1}^{\infty} x_n = 1 + \sum_{n=2}^{\infty}\Big(\frac{1}{n-1} - \frac{1}{n}\Big) = 1 + 1 - \lim_{n\rightarrow\infty}(1/n) = 2,$$
and so in particular this series converges.
Now
$x_n > \frac{1}{n \cdot n} = \frac{1}{n^2},$
so the series $$\sum_{n=1}^{\infty} \frac{1}{n^2}$$ converges as well.
Finally if $p>2$ then since $\frac{1}{n^p} < \frac{1}{n^2}$, the series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges, and
if $0<p<1$ we have $\frac{1}{n^p} > \frac{1}{n}$, and so $\sum_{n=1}^{\infty} \frac{1}{n^p}$ diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ This is being asked in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions, and here: List of abstract duplicates.
What methods can be used to evaluate the limit $$\lim_{x\rightarrow\infty} \sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x.$$
In other words, if I am given a polynomial $P(x)=x^n + a_{n-1}x^{n-1} +\cdots +a_1 x+ a_0$, how would I find $$\lim_{x\rightarrow\infty} P(x)^{1/n}-x.$$
For example, how would I evaluate limits such as $$\lim_{x\rightarrow\infty} \sqrt{x^2 +x+1}-x$$ or $$\lim_{x\rightarrow\infty} \sqrt[5]{x^5 +x^3 +99x+101}-x.$$
| Alternatively, rewrite this limit as
$$\lim_{x\rightarrow\infty}x\left(\sqrt[n]{1+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}}-1\right).$$
Consider the Taylor expansion around $0$ of $\sqrt[n]{1+z}$. We have
$$\sqrt[n]{1+z}=1+\frac{1}{n}z+O\left(z^{2}\right).$$ Setting $z=\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}$ we see that $z=O\left(\frac{1}{x}\right)$, and hence
$$\sqrt[n]{1+z}=1+\frac{1}{n}\left(\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}\right)+O\left(\frac{1}{x^{2}}\right)=1+\frac{a_{n-1}}{n}\frac{1}{x}+O\left(\frac{1}{x^{2}}\right).$$ Thus we have
$$x\left(\sqrt[n]{1+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}}-1\right)=\frac{a_{n-1}}{n}+O\left(\frac{1}{x}\right)$$
and we conclude
$$\lim_{x\rightarrow\infty}x\left(\sqrt[n]{1+\frac{a_{n-1}}{x}+\cdots+\frac{a_{0}}{x^{n}}}-1\right)=\frac{a_{n-1}}{n}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to calculate Jacobi Symbol $\left(\dfrac{27}{101}\right)$? How to calculate Jacobi Symbol $\left(\dfrac{27}{101}\right)$?
The book solution
$$\left(\dfrac{27}{101}\right) = \left(\dfrac{3}{101}\right)^3 = \left(\dfrac{101}{3}\right)^3 = (-1)^3 = -1$$
My solution
$$\left(\dfrac{27}{101}\right) = \left(\dfrac{101}{27}\right) = \left(\dfrac{20}{27}\right) = \left(\dfrac{2^2}{27}\right) \cdot \left(\dfrac{5}{27}\right)$$
$$= (-1) \cdot \left(\dfrac{27}{5}\right) = (-1) \cdot \left(\dfrac{2}{5}\right) = (-1) \cdot (-1) = 1.$$
Whenever I encounter $\left(\dfrac{2^b}{p}\right)$, I use the formula
$$(-1)^{\frac{p^2 - 1}{8}}$$
I guess mine was wrong, but I couldn't figure out where? Any idea?
Thank you,
| Your mistake is calculating $\left(\dfrac{2^2}{27}\right)\:.\:$ But since you wish to use only $2$'s it's much simpler, viz.
$$ \left(\dfrac{27}{101}\right)\ =\ \left(\dfrac{128}{101}\right)\ =\ \left(\dfrac{2}{101}\right)^{7}\ =\ -1 $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find equation of quadratic when given tangents? I know the equations of 4 lines which are tangents to a quadratic:
$y=2x-10$
$y=x-4$
$y=-x-4$
$y=-2x-10$
If I know that all of these equations are tangents, how do I find the equation of the quadratic?
Normally I would be told where the tangents touch the curve, but that info isn't given.
Thanks!
| Since the two pairs of tangents are symmetric with respect to the $y$-axe,
the quadratic function $f(x)=ax^{2}+bx+c$ must be even ($f(x)=f(-x)$),
which implies that $b=0$. The equations of the tangents to the graph of $f(x)=ax^{2}+c$ at points $%
\left( x_{1},f(x_{1})\right) $ and $\left( x_{2},f(x_{2})\right) $ are
$$\begin{eqnarray*}
y &=&f^{\prime }(x_{i})x-f^{\prime }(x_{i})x_{i}+f(x_{i})\qquad i=1,2 \\
&=&2ax_{i}x+c-ax_{i}^{2}.
\end{eqnarray*}$$
These equations must be equivalent to two of the given tangents, one from each pair, e.g. $y=2x-10$ and $y=x-4$:
$$\left\{
\begin{array}{c}
2ax_{1}x+c-ax_{1}^{2}=2x-10 \\
2ax_{2}x+c-ax_{2}^{2}=x-4%
\end{array}%
\right. $$
Finally we compare coefficients and solve the resulting system of $4$
equations:
$$\left\{
\begin{array}{c}
2ax_{1}=2 \\
c-ax_{1}^{2}=-10 \\
2ax_{2}=1 \\
c-ax_{2}^{2}=-4%
\end{array}%
\right. \Leftrightarrow \left\{
\begin{array}{c}
x_{1}=8 \\
x_{2}=4 \\
a=\frac{1}{8} \\
c=-2%
\end{array}%
\right. $$
Thus the quadratic is $f(x)=\frac{1}{8}x^{2}-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/31321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that if $p$ is an odd prime that divides a number of the form $n^4 + 1$ then $p \equiv 1 \pmod{8}$ Problem
Prove that if $p$ is an odd prime that divides a number of the form $n^4 + 1$ then $p \equiv 1 \pmod{8}$
My attempt was,
Since $p$ divides $n^4 + 1 \implies n^4 + 1 \equiv 0 \pmod{p} \Leftrightarrow n^4 \equiv -1 \pmod{p}$.
It follows that $(n^2)^2 \equiv -1 \pmod{p}$, which implies $-1$ is quadratic residue modulo $p$. Hence $p \equiv 1 \pmod{4} \Leftrightarrow p \equiv 1 \pmod{8}$.
Am I in the right track?
Thanks,
| As you have noticed, $p \equiv 1 \bmod 4$. Then $1 \equiv n^{p-1} = (n^4)^{(p-1)/4} \equiv (-1)^{(p-1)/4} \bmod p$. This means that $(p-1)/4$ is even, i.e., $p\equiv 1 \bmod 8$.
By induction, this argument generalizes to: if an odd prime $p$ divides a number of the form $n^k+1$, where $k$ is a power of $2$, then $p \equiv 1 \bmod {2k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/33392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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How can I sum the infinite series $\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad$ How can I find the sum of the infinite series
$$\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad ?$$
My attempt at a solution - I saw that I could rewrite it as
$$\frac{1}{5}\left(1 - \frac{4}{10} \left( 1 - \frac{7}{15} \left(\cdots \left(1 - \frac{3k - 2}{5k}\left( \cdots \right)\right)\right.\right.\right.$$
and that $\frac{3k - 2}{5k} \to \frac{3}{5}$ as $k$ grows larger. Using this I thought it might converge to $\frac{1}{8}$, but I was wrong, the initial terms deviate significantly from $\frac{3}{5}$.
According to Wolfram Alpha it converges to $1-\frac{\sqrt[3]{5}}{2}$. How can I get that ?
| You use this formula $$(1+x)^{-p/q} = 1 - \frac{\displaystyle\frac{p}{q}}{1!}\cdot x + \frac{ \displaystyle\frac{p}{q} \cdot \Bigl(\frac{p}{q}+1\Bigr)}{2!} x^{2} - \cdots$$
So your series can be rewritten as $$ \frac{1}{3} \times \frac{3}{5} - \frac{\displaystyle\frac{1}{3}\cdot\Bigl(\frac{1}{3} + 1\Bigr)}{2!} \times \Bigl(\frac{3}{5}\Bigr)^{2} + \frac{\displaystyle\frac{1}{3} \cdot \Bigl(\frac{1}{3}+1\Bigr) \cdot \Bigl(\frac{1}{3}+2\Bigr)}{3!} \times \Bigl(\frac{3}{5}\Bigr)^{3} - \cdots$$
So this is nothing but $$\Bigl(1+\frac{3}{5}\Bigr)^{-1/3} -1 = \frac{1}{3} \times \frac{3}{5} - \frac{\displaystyle\frac{1}{3}\cdot\Bigl(\frac{1}{3} + 1\Bigr)}{2!} \times \Bigl(\frac{3}{5}\Bigr)^{2} + \cdots$$
So our answer will be $$1 - \Bigl(\frac{8}{5}\Bigr)^{-1/3} = 1 -\frac{5^{1/3}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/34671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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"answer_id": 2
} |
Writing a percent as a decimal and a fraction I am having a problem understanding some manipulations with recurring decimals. The exercise is
Write each of the following as a
decimal and a fraction:
(iii) $66\frac{2}{3}$%
(iv) $16\frac{2}{3}$%
For (iii), I write a decimal $66\frac{2}{3}\% = 66.\bar{6}\% = 0.66\bar{6} = 0.\bar{6}$ and a fraction $66\frac{2}{3}\% = \frac{66*3 + 2}{3*100}\% = \frac{200}{3}\times\frac{1}{100} =\frac{2}{3}$.
For (iv), I follow a similar path to establish that the fraction is $\frac{1}{6}$ and the decimal is $0.1\bar{6}$.
What I don't understand is a part of the model answer for this exercise. They say
$66\frac{2}{3}\% = 66.\bar{6}\% = 0.\bar{6} = \frac{6}{9} = \frac{2}{3}$
and
$16\frac{2}{3}\% = 16.\bar{6}\% = 0.1\bar{6} = \frac{16-1}{90} = \frac{15}{90} = \frac{1}{6}$
I did not do much work with recurring decimals before and I don't know how to justify the 9 and 90 in the denominator.
Could you please explain?
| It happens to be well known that a number over 9 (other than 0 or 9) has a repeating decimal. That is: $\dfrac{1}{9} = 0.11 \bar{1}$, $\dfrac{4}{9} = 0.44 \bar{4} $, and so on. Why is this true?
It comes from the fact that if we have the equation $10x = 1.\bar{1}$, then we have $x = 0.\bar{1}$ by division. Subtracting, we get that $9x = 1$, or that $x = \dfrac{1}{9}$.
As happens - I typed this at the same time as Ross.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/35631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
} |
Residue of $f(z) = \frac{z-2}{z^2}\sin(\frac{1}{1-z})$ at $z = 1$ Given the complex function $f(z) = \frac{z-2}{z^2}\sin(\frac{1}{1-z})$, how can we calculate the residue at the essential singularity at $z = 1$?
| Let us define $\xi=z-1$ (such that we can calculate the residue at $\xi=0$). The function is given by
$$\frac{1-\xi}{(1+\xi)^2} \sin(\xi^{-1}).$$ For the residue, we need to obtain the coefficient in front of $\xi^{-1}$ in the Laurent expansion around $\xi=0$. We have the Laurent expansions
$$\sin(\xi^{-1}) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!\xi^{2k+1}}$$ and
$$\frac{1-\xi}{(1+\xi)^2} = \sum_{k=0}^\infty (-1)^k (2k+1) \xi^k$$ valid for $0<|\xi|<1$.
As the original function is the product of these functions, the residue (= coefficient in front of $\xi^{-1}$) is given by
$$
\begin{align}
\text{Res}_{z=1}\left[\frac{z-2}{z^2}\sin\left(\frac{1}{1-z}\right) \right]
&= \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} (-1)^{2k} (4k+1) \\
&=\sum_{k=0}^\infty \frac{(-1)^k [2 (2k +1 ) -1]}{(2k+1)!} \\
&= 2\cos(1) - \sin(1) \approx 0.24
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/37612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proving that a repeating decimal equals a fraction I'm having some trouble with this question:
Prove that 0.1636363636...=9/55, using infinite series. I'd appreciate any help you can give me. Thanks!
| You can do this: \begin{align*} 0.1636363\cdots &= \frac{1}{10} + \biggl[ 63 \times 10^{-3} + 63\times 10^{-5} + 63 \times 10^{-7} + \cdots\biggr] \\ &= \frac{1}{10} + 63 \cdot \Bigl[ 10^{-3} + 10^{-5} + 10^{-7} + \cdots \Bigr] \\ &= \frac{1}{10} + 63 \cdot \frac{10^{-3}}{1-10^{-2}} \\ &= \frac{1}{10} + \frac{7}{110}=\frac{9}{55}\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/38838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Using Horner's Method I'm trying to evaluate a polynomial recursively using Horner's method.
It's rather simple when I have every value of $x$ (like: $x+x^2+x^3...$), but what if I'm missing some of those? Example: $-6+20x-10x^2+2x^4-7x^5+6x^7$.
I would also appreciate it if someone could explain the method in more detail, I've used the description listed here but would like some more explanation.
| Horner's method or form is also sometimes called nested form. You can think of it as starting with the whole polynomial $$6x^7-7x^5+2x^4-10x^2+20x-6,$$ setting aside the constant term (if it's zero, you can set aside a zero here) and factoring out an $x$ from the remaining terms $$(6x^6-7x^4+2x^3-10x+20)x-6,$$ and then repeating the process for the innermost polynomial (the part that isn't yet fully in nested form:
$$((6x^5-7x^3+2x^2-10)x+20)x-6$$
$$(((6x^4-7x^2+2x)x-10)x+20)x-6$$
$$((((6x^3-7x+2)x+0)x-10)x+20)x-6$$
$$(((((6x^2-7)x+2)x+0)x-10)x+20)x-6$$
$$((((((6x)x-7)x+2)x+0)x-10)x+20)x-6$$
$$(((((((6)x+0)x-7)x+2)x+0)x-10)x+20)x-6$$
Alternately, if your polynomial is $$p(x)=\sum_{k=0}^{n}a_kx^k=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,$$ you can think of $p(x)$ in nested form as $p_0$ where $p_n=a_n$ and $p_{k-1}=p_kx+a_{k-1}$. That is, start with the leading coefficient ($a_n$). Multiply by $x$ and add the next coefficient (adding zero to fill in for "missing" powers of $x$), repeating until you've added the constant term ($a_0$). Back to the same example:
$$\begin{align}
p_7&=6\\
p_6&=(6)x+0\\
p_5&=((6)x+0)x-7\\
p_4&=(((6)x+0)x-7)+2\\
p_3&=((((6)x+0)x-7)+2)x+0\\
p_2&=(((((6)x+0)x-7)+2)x+0)-10\\
p_1&=((((((6)x+0)x-7)+2)x+0)-10)+20\\
p_0&=(((((((6)x+0)x-7)+2)x+0)-10)+20)-6
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/49051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Is it possible to solve a separable equation a different way and still arrive at the same answer? I have the following equation
$$(xy^2 + x)dx + (yx^2 + y)dy=0$$ and I am told it is separable, but not knowing how that is, I went ahead and solved it using the Exact method.
Let $M = xy^2 + x $ and $N = yx^2 + y$
$$My = 2xy \text{ and } Nx = 2xy $$
$$ \int M.dx ==> \int xy^2 + x = x^2y^2 + (x^2)/2 + g(y)$$
$$ \text{Partial of } (x^2y^2 + (x^2)/2 + g(y)) => xy^2 + g(y)'$$
$$g(y)' = y$$
$$g(y) = y^2/2$$
the general solution then is
$$C = x^2y^2/2 + x^2/2 + y^2/2$$
Is this solution the same I would get if I had taken the Separate Equations route?
| Taking the other route you get $y'y(x^2+1)=-xy^2-x$. Denote $z=y^2$, hence $z'=2yy'$. So you get $z'(x^2+1)=-2xz-x$. Hence you have $z'+\frac{2x}{x^2+1}z=-\frac{x}{x^2+1}$. This gives you $z=\frac{k(x)}{x^2+1}$. Substituting, you have $k'(x)=-x$, which implies $k(x)=-\frac{x^2}{2}+C$. Hence $z=-\frac{x^2}{2(x^2+1)}+\frac{C}{x^2+1}$. Finally, $y^2=z$.
From here you get $2y^2(x^2+1)+x^2=C$, which is the answer you got.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/50887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to add compound fractions? How to add two compound fractions with fractions in numerator like this one:
$$\frac{\ \frac{1}{x}\ }{2} + \frac{\ \frac{2}{3x}\ }{x}$$
or fractions with fractions in denominator like this one:
$$\frac{x}{\ \frac{2}{x}\ } + \frac{\ \frac{1}{x}\ }{x}$$
| Here is a start for the first one:
$$\frac{\frac{1}{x}}{2} + \frac{\frac{2}{3x}}{x} = \frac{x\frac{1}{x}}{2x} + \frac{2\frac{2}{3x}}{2x}
= \frac{1}{2x} + \frac{\frac{4}{3x}}{2x} = \frac{1}{2x} + \frac{4}{3x}\frac{1}{2x} = \frac{1}{2x} + \frac{4}{6x^2} = \frac{1}{2x} + \frac{2}{3x^2}$$
Now try to derive $\displaystyle\frac{3x + 4}{6x^2}$ from this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/51410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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