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Polynomial division in 3 variables I'm trying to compute $$\frac{x^3-2x^2-xyz+2xy+yz}{x-2}$$
I have written this in the form $$\begin{align}x^3-2x^2-xyz+2xy+yz&=(x-2)(x^2+bx+cy+dz)\\
x^3-2x^2-(yz+2y)x+yz&= x^3 + (b-2)x^2 + (cy+dz-2b)x+(-2cy-2dz)
\end{align}$$
This leaves me with the following equations
$$\begin{align}b-2&=-2\\
cy+dz-2b&=yz+2y \\
-2cy-2dz &= yz
\end{align}$$
From this we can see that $b=0$
Therefore we are left with $$\begin{align}cy+dz&=yz+2y \\
cy +dz&=-\frac{yz}{2}
\end{align}$$
I'm now stuck, how can I find values for $c$ and $d$ from this? If I have hit a dead-end, feel free to suggest a different method for solving this equation
| Let $F(x)={x}^{3}-xyz-2\,{x}^{2}+2\,xy+yz$. Then by Bézout's theorem we have
$F(x)=(x-2) Q(x)+F(2).$
Now $F(2)=4y-yz$.
$$
F(x)-F(2)={x}^{3}-xyz-2\,{x}^{2}+2\,xy+2\,yz-4\,y=\left( x-2 \right) \left( {x}^{2}-yz+2\,y \right).
$$
Thus
$$
\frac{{x}^{3}-xyz-2\,{x}^{2}+2\,xy+yz}{x-2}={x}^{2}-yz+2\,y+\frac{4y-yz}{x-2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2167213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Beta function-like integral $$\int\limits_0^1 \frac{x^{1-\alpha} (1-x)^\alpha}{(1+x)^3} \, dx $$
After the substitution $z=\frac{1}{x} - 1$,
I've got this:
$$\int\limits_0^\infty \left(\frac{1}{z}-1\right)^\alpha\left(\frac{1}{z}+1\right)^{-3} dz $$
But it is still far from the Beta function.
|
The key is the linear-fractional transformation,
$$x\mapsto\frac{1-t}{1+t}.$$
Then $1-x=\frac{2t}{1+t}$ and $1+x=\frac{2}{1+t}$, and we find
$$\begin{align}
\int_{0}^{1}\frac{x^{1-a}\left(1-x\right)^{a}}{\left(1+x\right)^{3}}\,\mathrm{d}x
&=\frac12\int_{1}^{0}\left(\frac{1-x}{x}\right)^{a}\frac{x}{1+x}\cdot\frac{\left(-2\right)}{\left(1+x\right)^{2}}\,\mathrm{d}x\\
&=\frac12\int_{0}^{1}\left(\frac{2t}{1-t}\right)^{a}\frac{1-t}{2}\,\mathrm{d}t;~~~\small{\left[x=\frac{1-t}{1+t}\right]}\\
&=\frac{1}{2^{2-a}}\int_{0}^{1}t^{a}\left(1-t\right)^{1-a}\,\mathrm{d}t\\
&=\frac{1}{2^{2-a}}\operatorname{B}{\left(1+a,2-a\right)}.\blacksquare\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2168581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Solving: $3^m-2=n^2$
Solve $3^m-2=n^2$ for positive pairs of integers $(n,m)$.
My try:
$3^m-6=n^2-4\implies 3(3^{m-1}-2)=(n+2)(n-2)\implies 3=n+2\ or\ 3=n-2$ so we got $n=\pm 5 , \pm 1$. Used $\pm$ as we have $n^2$.
Putting $n=\pm 5 , \pm 1$ gives $m=3,1$ respectively. We kick out negative ones as problem ask for positive ones only.
But how do I prove that $n+2=3$ not something like multiples of $3$ and similar for $n-2$. More precisely something like the following:
$n+2=3d$, $n-2=k$ and $dk=3^{m-1}-2$.
Please help.
| If $m=3k$, then the equation is $n^2 = (3^k)^3 -2$. I put the elliptic curve $y^2=x^3-2$ into SAGE. The only integral point is $(x,y) = (3,5)$ which corresponds to the the solution $m=3, n=5$.
If $m=3k+1$, then multiply through by $9$ to get $(3n)^2 = (3^{k+1})^3 - 18.$ I put the curve $y^2 = x^3-18$ into SAGE. The only integral point is $(x,y) = (3,3)$, which corresponds to the solution $m=1, n=1$.
If $m=3k+2$, then multiply through by $81$ to get $(9n)^2 = (3^{k+2})^3 - 182.$ I put the curve $y^2 = x^3 - 182$ into SAGE and there are no integral points.
So there are only the two solutions to the equation. I don't see a more elementary way to do this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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Variation of parameters (issue with the constants) I have to solve this ODE using the variation of parameters method:
$$4y''+y=\frac{2}{\cos \left( \frac{x}{2} \right)}$$
Solving the homogeneous problem yields
$$y_h(x)=
c_1 \cos \left( \frac{x}{2} \right)+
c_2 \sin \left( \frac{x}{2} \right)$$
Now, to solve the variation of parameters problem, you have to solve
$$
\begin{bmatrix}
\cos \left( \frac{x}{2} \right) &
\sin \left( \frac{x}{2} \right) \\
-\frac{1}{2} \sin \left( \frac{x}{2} \right) &
\frac{1}{2} \cos \left( \frac{x}{2} \right)
\end{bmatrix}
\begin{bmatrix} c_1' \\ c_2' \end{bmatrix}=
\begin{bmatrix}
0 \\
2\sec \left( \frac{x}{2} \right)
\end{bmatrix}$$
Solving the first equation $
c_1'\cos \left( \frac{x}{2} \right)+
c_2'\sin \left( \frac{x}{2} \right)=0$. This gives
$c_1'=-c_2'\tan \left( \frac{x}{2} \right)$
Solving the second equation gives $
\frac{c_2'}{2} \cos \left( \frac{x}{2} \right)-
\frac{c_1'}{2} \sin \left( \frac{x}{2} \right)=
\frac{2}{\cos \left( \frac{x}{2} \right)}$.
Substituting gives $
\frac{c_2'}{2} \cos \left( \frac{x}{2} \right)+
\frac{c_2'}{2}
\frac{\sin^2 \left( \frac{x}{2} \right)}
{\cos \left( \frac{x}{2} \right)}=
\frac{2}{\cos \left( \frac{x}{2} \right)}$.
Solving for $c_2'$ gives $\frac{c_2'}{2}=2$.
Finally, $c_2'=4$ and $c_2=4x$. This means that $
c_1'=-4\tan \left( \frac{x}{2} \right)$ and $
c_1=8\ln \left( \cos \left( \frac{x}{2} \right) \right)$.
However, according to Wolfram, $c_1=2\ln \left( \cos\left( \frac{x}{2} \right)\right)$ and $c_2=x$.
Indeed, I tried solving with my values for $c_1$ and $c_2$ and it doesn't work. What did I do wrong?
| Try with the equation writen in this form. The method needs the coefficient to be 1 for the highest derivative (standard form).
$$y''+\frac{y}{4}=\frac{1}{2\cos \left( \frac{x}{2} \right)}$$
With it, we get $c_2'/2=1/2$ and $c_1'=-\tan \left( \frac{x}{2} \right)$. With them, the expected solution follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2170476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Trigonometry - Obtuse angle An obtuse angle $\theta$ has a $\sin \theta = \frac{x+1}{x+2}$ and $\cos \theta= \frac{x}{x+2}$.
Which angle is it?
I've been frustrated with this for the last few hours now. My problem is that I can't seem to find the length of any of the sides. I've tried working it out with the $\sin \theta$ and $\cos \theta$ sentence, but I always get an answer which still contains something unknown - and that doesn't help much. Is it even possible to find the exact amount of degrees here? Or is it implied that I should only tell if it's A,B or C (Which seems weird to me).
| $$\sin \alpha = \frac{x+1}{x+2}, \,\cos \alpha=\frac{x}{x+2},\, \alpha\in \left(\frac{\pi}{2}, \pi\right), \, x\neq -2$$
Using $\sin^2\alpha+\cos^2\alpha=1$:
$$1=\left(\frac{x+1}{x+2}\right)^2+\left(\frac{x}{x+2}\right)^2$$
$$1=\frac{2x^2+2x+1}{x^2+4x+4}$$
$${2x^2+2x+1}=x^2+4x+4$$
$${x^2-2x-3}=0$$
$$(x+1)(x-3)=0$$
So $x\in \{-1, 3\}$
For $x=-1$:
$\cos \alpha=\frac{-1}{1}=-1$
So $\alpha = \pi+2k\pi$, $k\in \mathbb{Z}$, but then $\alpha \not \in \left(\frac{\pi}{2}, \pi\right)$. Thus $x\neq -1$
For $x=3$
$\sin \alpha=\frac{4}{5}$
Then,as we assume, that $\alpha$ is obtuse, we have
$$\alpha = \pi - \arcsin\frac{4}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Co-ordinates of the vertices an icosahedron relative to its centroid.
This is a picture of an icosahedron. I need to know the coordinates of the vertices of the icosahedron relative to it's centroid in order to programme a projection of one on a three dimensional plane. (By the way, it has twelve vertices, so it is going to be a very long process)
Unfortunately, I have not been able to find one vertices.
| With $20$ faces, each face has an area of $\frac\pi5$ steradians. That means that the spherical excess in each face is $\frac\pi5$ radians. Thus, each angle in each spherical triangular face has an angle of $\frac\pi3+\frac\pi{15}=\frac{2\pi}5$.
We can use the Spherical Law of Cosines for Angles to get the angle subtended by each side. Since $\cos(\mathrm{A})=\cos(\mathrm{B})=\cos(\Gamma)=\frac{-1+\sqrt5}4$, we get
$$
\begin{align}
\cos(\gamma)
&=\frac{\cos(\Gamma)+\cos(\mathrm{A})\cos(\mathrm{B})}{\sin(\mathrm{A})\sin(\mathrm{B})}\\
&=\frac1{\sqrt5}
\end{align}
$$
Since $\cos\left(\frac{2\pi}5\right)=\frac{-1+\sqrt5}4$ and $\sin\left(\frac{2\pi}5\right)=\sqrt{\frac{5+\sqrt5}8}$ and $\cos\left(\frac{4\pi}5\right)=\frac{-1-\sqrt5}4$ and $\sin\left(\frac{4\pi}5\right)=\sqrt{\frac{5-\sqrt5}8}$, the set of vertices would be
$$
\begin{align}
a_1&=(1,0,0)\\
a_2&=\textstyle\left(\frac1{\sqrt5},\frac2{\sqrt5},0\right)\\
a_3&=\textstyle\left(\frac1{\sqrt5},\frac{5-\sqrt5}{10},\sqrt{\frac{5+\sqrt5}{10}}\right)\\
a_4&=\textstyle\left(\frac1{\sqrt5},\frac{-5-\sqrt5}{10},\sqrt{\frac{5-\sqrt5}{10}}\right)\\
a_5&=\textstyle\left(\frac1{\sqrt5},\frac{-5-\sqrt5}{10},-\sqrt{\frac{5-\sqrt5}{10}}\right)\\
a_6&=\textstyle\left(\frac1{\sqrt5},\frac{5-\sqrt5}{10},-\sqrt{\frac{5+\sqrt5}{10}}\right)\\
a_7&=\textstyle(-1,0,0)\\
a_8&=\textstyle\left(-\frac1{\sqrt5},-\frac2{\sqrt5},0\right)\\
a_9&=\textstyle\left(-\frac1{\sqrt5},\frac{-5+\sqrt5}{10},-\sqrt{\frac{5+\sqrt5}{10}}\right)\\
a_{10}&=\textstyle\left(-\frac1{\sqrt5},\frac{5+\sqrt5}{10},-\sqrt{\frac{5-\sqrt5}{10}}\right)\\
a_{11}&=\textstyle\left(-\frac1{\sqrt5},\frac{5+\sqrt5}{10},\sqrt{\frac{5-\sqrt5}{10}}\right)\\
a_{12}&=\textstyle\left(-\frac1{\sqrt5},\frac{-5+\sqrt5}{10},\sqrt{\frac{5+\sqrt5}{10}}\right)
\end{align}
$$
$a_1$ shares edges with $a_k$ for $2\le k\le6$.
For $2\le k\le6$, $a_k$ shares edges with $a_1,a_{2+\text{mod}(k-1,5)},a_{2+\text{mod}(k+2,5)},a_{8+\text{mod}(k,5)},a_{8+\text{mod}(k+1,5)}$.
$a_7$ shares edges with $a_k$ for $8\le k\le12$.
For $8\le k\le12$, $a_k$ shares edges with $a_7,a_{8+\text{mod}(k-2,5)},a_{8+\text{mod}(k+1,5)},a_{2+\text{mod}(k-1,5)},a_{2+\text{mod}(k,5)}$.
Animation of the Icosahedron Generated above
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2174594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What does this series converge to, if anything? $$\arctan{1} + \arctan{\frac{1}{2}} + \arctan{\frac{1}{3}} + \arctan{\frac{1}{4}} ...= ?$$
The infinite series for arctan is
$$\arctan{x} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ...$$
So I want to sum up the $\arctan{1\over n}$ where $n$ starts at $1$ and goes to infinity.
I originally thought the resulting series can be written this way:
$$(1 + \frac{1}{2} + \frac{1}{3} ...) - \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^3}{3} + \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^5}{5} - \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^7}{7} ...$$
But that is wrong. The right way to "insert" the series is:
$$1 + \frac{1}{2} + \frac{1}{3} ... - \frac{1}{3} - \frac{(\frac{1}{2})^3}{3} - \frac{(\frac{1}{3})^3}{3} ... + \frac{1}{5} + \frac{(\frac{1}{2})^5}{5} + \frac{(\frac{1}{3})^5}{5} ... ...$$
So it looks like a bunch of harmonic serieses manipulated. Harmonic series diverges, but I remember that doesn't necessarily mean a similar series diverges. I remember from Calculus 2 that, for example, $\lim_{n\to\infty}$ of $\sin(n)\over n$ converges to zero even though $\sin(n)$ does not converge.
So how do I figure out what this series converges to, if anything?
| For $0 < x \leq 1$, the series
$$ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ... $$
is an alternating series with decreasing terms. This means the terms alternate between overshooting and undershooting the actual value of $\arctan(x)$. In particular,
$$ 0 < x \leq 1 \implies x - \frac{x^3}{3} < \arctan(x) < x $$
We can use this to get a good bound on the partial sums
$$ \sum_{k=1}^n \arctan\left( \frac{1}{k} \right)
< \sum_{k=1}^n \frac{1}{k} = H_n$$
$$ \sum_{k=1}^n \arctan\left( \frac{1}{k} \right)
> \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{3 k^3} \right)
= H_n -
\sum_{k=1}^n \left( \frac{1}{3 k^3} \right)
> H_n - \frac{1}{3} \zeta(3)$$
So not only does the infinite sum go to infinity, it does so in basically the same fashion as harmonic numbers $H_n$ do, and furthermore the error in this estimate is strictly less than $0.4007$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2175526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Factor $(x+y)^4+x^4+y^4$ Title says it all, I just want to know how to factor $(x+y)^4+x^4+y^4$. I only know that it's possible to factor, but got no idea how to do it. If it were a single-variable polynomial I could try to find rational roots or something, but I'm lost with this one.
| $$x^4 + y^4 + (x + y)^4
=2 x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + 2 y^4$$
$$=x (x (x (2 x + 4 y) + 6 y^2) + 4 y^3) + 2 y^4
=2 x^4 + y (4 x^3 + y (6 x^2 + y (4 x + 2 y)))$$
$$=4 y^2 (x^2 + x y) + 2 (x^2 + x y)^2 + 2 y^4
=2 (x^4 + 2 x^3 y + 3 x^2 y^2 + 2 x y^3 + y^4)$$
$$=2 (x^2 + x y + y^2)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2176542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $x^3+y^3 \le 2$. As in the title. If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $$x^3+y^3 \le 2.$$
This seems to be a very tricky one. I tried applying various inequalities like AM-GM, unfortunately, none of techniques I'm familiar with seem to work here.
I'd greatly appreciate any hints.
| Another way. Let $y=kx$.
Hence, the condition gives $x^4x^2-(k^3-1)x-1\leq0$ or
$$x\leq\frac{k^3-1+\sqrt{k^6+4k^4-2k^3+1}}{2k^4}.$$
Thus, we need to prove that
$$\left(\frac{k^3-1+\sqrt{k^6+4k^4-2k^3+1}}{2k^4}\right)^3(k^3+1)\leq2$$ or
$$3k^{12}-3k^{10}+2k^9+3k^4-2k^3+1\geq(k^9+k^7-k^6+k^4-k^3+1)\sqrt{k^6+4k^4-2k^3+1}$$ or
$$(k-1)^2(2k^{10}+4k^9-k^4-2k^3+3k^2+2k+1)\geq0.$$
Done!
Also we can make the following.
By the condition and AM-GM we obtain:
$$x^2+y^2\geq x^3+y^4+y^2-y^3\geq x^3+y^3.$$
Thus, $x\leq\frac{k^2+1}{k^3+1}$ and it's enough to prove that
$$(k^3+1)\left(\frac{k^2+1}{k^3+1}\right)^3\leq2$$ or
$$\left(\frac{k^3+1}{2}\right)^2\geq\left(\frac{k^2+1}{2}\right)^3,$$
which is just Power Means inequality.
Done!
By this way we can prove a stronger inequality: $x^5+y^5\leq2$.
Indeed, we need to prove that:
$$(k^5+1)\left(\frac{k^2+1}{k^3+1}\right)^5\leq2,$$
which is true and nice, but it's another problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2178764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Sketching the graph of $Y=\sin(3x)+\sin(x)$ How do I sketch the graph of
$Y = \sin(3x) + \sin (x)$?.
I realize the amplitude is $1$, but am not sure. I found the period to be $2\pi$. My problem is that I don't know if this graph will be similar to the graph of $Y = \sin (x)$. I want to know how the graph will be like.
| A simple method to sketch periodic functions like this is:
1) find all zeroes ($x$ for which $Y=0$) in the first period
$$ \sin(3x) + \sin(x) = 0 $$
$$ x = n\pi,\, x=n\pi - \frac{\pi}{2}, \quad n\in\mathbb{Z} $$
So $Y$ crosses the x-axis at $x=0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ for $x \in [0,2\pi]$
Edit How to find zeros:
$$ \begin{align*}
& \sin(x) = -\sin(3x) \\
& x = \pi + 3x + 2\pi n_1, \quad n_1 \in \mathbb{Z} \quad \# \sin^{-1}\text{ of LHS and RHS} \\
& x = 2\pi n_2 - 3x, \quad n_2 \in \mathbb{Z} \\
& \text{and solve for } x
\end{align*}$$
2) find all critical points ($x$ for which $Y'=0$ or is undefined) in the first period
$$ \frac{dY}{dx} = 3\cos(3x) + \cos(x) = 0 $$
$$ x = n\pi - \frac{\pi}{2}, 2n\pi \pm 2\tan^{-1}\left(\sqrt{5\pm 2\sqrt{6}} \right), \quad n \in \mathbb{Z}$$
Edit How to find zeros:
$$ \begin{align*}
&\cos(x) + 3\cos(3x) = 12\cos^3(x) - 8\cos(x) = \cos(x)(3\cos^2(x)-2)=0 \\
&\cos(x)=0 \implies x=\frac{\pi}{2}+\pi n, n\in\mathbb{Z} \text{ or} \\
&3\cos^2(x)=2 \\
&\cos(x)=\pm\sqrt{\frac{2}{3}} \\
&\text{ and take the in inverse cosine of each side} \\
&x=\pm\cos^{-1}\left(\sqrt{\frac{2}{3}}\right) + 2\pi n \text{ or}\\
&x= \pm\left(\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)-\pi\right) + 2\pi n, n \in \mathbb{Z} \\
\end{align*}
$$
Where the above inverse cosine is equivalent to the inverse tangent expression I showed in my answer.
so $Y$ has critical points at $(0.615, \,1.54),\, (\frac{\pi}{2}, \,0),\, (2.526,\,1.54),\, (\pi,\,0),\, (3.757,\,-1.54),\, (\frac{3\pi}{2},\, 0),\,\text{ and } (5.668,\, -1.54)$
3) find concavity by looking at the sign of the second derivative (the sign of $Y''$). This, however, is not really necessary for a periodic function because you can just look at the intervals between, which the first derivative changes sign.
and you should get a graph like below, which repeats every $2\pi$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Sixth degree polynomial problem If the graph of $$y = x^6 - 10x^5 + 29x^4 - 4x^3 + ax^2$$ always lies above the line $y = bx + c$, except for $3$ points where the curve intersects the line.
What is the largest value of $x$ for which the line intersects the curve?
*
*A) 4
*B) 5
*C) 6
*D) 7
Through general idea of graphs and with the help of a graphing calculator I have concluded the approximate look of such a curve and a line.
However, I would like to know a proper mathematical solution to this problem.
| I begin the computation by the same expression as @Ákos Somogyi
$$\tag{1}x^6-10 x^5+29 x^4-4 x^3+ax^2-bx-c=\underbrace{(x-\alpha)^2(x-\beta)^2(x-\gamma)^2}_{p(x)^2}$$
But I consider at once that this polynomial is equal to
$$\tag{2}p(x)^2=(x^3+ux^2+vx+w)^2$$
for certain coefficients $u,v,w.$ Expanding the square in (2):
$$\tag{3}x^6+2ux^5+(u^2+2v)x^4+(2uv+w)x^3+(2uw+v^2)x^2+2vwx+w^2,$$
we easily obtain by identification of coefficients in (1) and (3):
$$u=-5, v=2, w=8.$$
from which we deduce: $a=-76, b=-32, c=-64$ and
$$p(x)=x^3-5x^2+2x+8=(x+1)(x-2)(x-4)$$
Thus the rightmost root is: $x=4$, as can be seen on the picture below. This picture represents the curve with equation $y=x^6-10x^5+29x^4-4x^3-76x^2$ and the straight line with equation $y=-32x-64$, tangent to the curve at 3 differents points.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How did they simplify this sum? Im having trouble seeing how this simplification is justified:
$$\frac{2}{\pi}\sum_{k=-\infty}^\infty \frac{e^{i(2k+1)x}}{(2k+1)^{2}} =\frac{4}{\pi}\sum_{k=0}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}} $$
There are no steps on how they simplified, could anyone help me with it?
| Note that we can write
$$\frac{2}{\pi}\sum_{k=-\infty}^\infty \frac{e^{i(2k+1)x}}{(2k+1)^{2}} =\frac{2}{\pi}\sum_{k=-\infty}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}+i\frac{2}{\pi}\sum_{k=-\infty}^\infty \frac{\sin(2k+1)x}{(2k+1)^{2}} \tag1$$
Next, we split the series for the cosine into series over negative and non-negative values of the index to find
$$\begin{align}
\sum_{k=-\infty}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}&=\sum_{k=-\infty}^{-1} \frac{\cos(2k+1)x}{(2k+1)^{2}}+\sum_{k=0}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}\\\\
&=\sum_{k=1}^\infty \frac{\cos(2k-1)x}{(2k-1)^2}+\sum_{k=0}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}\\\\
&=\sum_{k=0}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}+\sum_{k=0}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}\\\\
&=2\sum_{k=0}^\infty \frac{\cos(2k+1)x}{(2k+1)^{2}}\tag 2
\end{align}$$
Similarly, we split the series for the sine into series over negative and non-negative values of the index to find
$$\begin{align}
\sum_{k=-\infty}^\infty \frac{\sin(2k+1)x}{(2k+1)^{2}}&=\sum_{k=-\infty}^{-1} \frac{\sin(2k+1)x}{(2k+1)^{2}}+\sum_{k=0}^\infty \frac{\sin(2k+1)x}{(2k+1)^{2}}\\\\
&=-\sum_{k=1}^\infty \frac{\sin(2k-1)x}{(2k-1)^2}+\sum_{k=0}^\infty \frac{\sin(2k+1)x}{(2k+1)^{2}}\\\\
&=-\sum_{k=0}^\infty \frac{\sin(2k+1)x}{(2k+1)^{2}}+\sum_{k=0}^\infty \frac{\sin(2k+1)x}{(2k+1)^{2}}\\\\
&=0\tag 3
\end{align}$$
Putting together $(1)-(3)$ yields the coveted result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
division of polynomials in $\mathbb{Z_2}$ I'm working on a problem involving division of polynomials in $\mathbb{Z_2}$. The question goes like this...
Find a greatest common divisor of the polynomials $x^6+x^4+x+1$ and $x^7+x^4+x^3+1$.
My textbook gave some examples but skipped a lot of the work and cannot follow what is going on. One example from the book said to find a greatest common divisor of $x^4+x^3+x+1$ and $x^5+x^2+x+1$ in $\mathbb{Z_2}$
. They got...
$x^5+x^2+x+1$ = $(x+1)(x^4+x^3+x+1)+(x^3+x)...$ and similarly
$x^4+x^3+x+1 = (x+1)(x^3+x)+(x^2+1)$.
How did they get this result? This should help clear things a bit before attempting the problem.
EDIT
I've attempted the original book problem and my gcd came out to $x^3+1$.
| I don't know how to typeset long division here, so I'll try to describe the procedure step-by-step instead. Say, we want to divide $x^5+x^2+x+1$ by $x^4+x^3+x+1$ with remainder over $\mathbb{Z}_2$. Dividing the leading terms gives us $x^5/x^4=\color{magenta}{x}$ as the first term of the quotient. Now,
$$\color{magenta}{x}\cdot(x^4+x^3+x+1)=x^5+x^4+x^2+x,$$
and
$$(x^5+x^2+x+1)-(x^5+x^4+x^2+x)=-x^4+1=x^4+1,$$
the last equality being true because we're in $\mathbb{Z}_2$ where $-1=1$. Now we continue in the same fashion to divide the current polynomial $x^4+1$ by $x^4+x^3+x+1$. Dividing the leading terms gives us $x^4/x^4=\color{magenta}{1}$ as the next term of the quotient. Then,
$$\color{magenta}{1}\cdot(x^4+x^3+x+1)=x^4+x^3+x+1,$$
and
$$(x^4+1)-(x^4+x^3+x+1)=-x^3-x=x^3+x,$$
for the same reason. So we ended up with the quotient of $\color{magenta}{x+1}$ and the remainder of $\color{blue}{x^3+x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Constant sequence limit Let us consider the sequence $(a_n)_{n \ge 1}$ of positive real numbers such that
$$\lim _ {n \to \infty} \left( \frac {a_1}{a_2} + \frac {a_2}{a_3} + \dots + \frac {a_{n-1}}{a_n} + \frac {a_n}{a_1} -n\right)=0.$$
Prove that the sequence is constant.
My try: I considered the $AM -GM$ inequality and I proved that the sequence in parantheses is always greater or equal than $0$. I can't go any further.
| We use induction to prove that $ a_1 = \cdots = a_m$ for all $m = 1, 2, \cdots$. Since the base case is trivial, we focus on the induction step.
Assume that $a_1 = \cdots = a_m$. Then for $n > m$, we have
\begin{align*}
\frac{a_1}{a_2} + \cdots + \frac{a_n}{a_1} - n
&= \overbrace{\left(\frac{a_1}{a_2} + \cdots + \frac{a_{m-1}}{a_m} \right)}^{=m-1} + \left( \frac{a_m}{a_{m+1}} + \cdots + \frac{a_n}{a_1} \right) - n \\
&= \frac{a_m}{a_{m+1}} + \cdots + \frac{a_n}{a_m} - (n-m+1)
\end{align*}
Applying the AM-GM, we find that
\begin{align*}
\frac{a_{m+1}}{a_{m+2}} + \cdots + \frac{a_n}{a_m}
&\geq (n-m)\left( \frac{a_{m+1}}{a_{m+2}} \cdots \frac{a_n}{a_m} \right)^{\frac{1}{n-m}} \\
&= (n-m)\left( \frac{a_{m+1}}{a_m} \right)^{\frac{1}{n-m}}
\end{align*}
So if we let $r = a_m/a_{m+1}$, then it follows from the AM-GM and the assumption that
$$ 0 \leq r + (n-m)r^{-\frac{1}{n-m}} - (n-m+1) \leq \frac{a_1}{a_2} + \cdots + \frac{a_n}{a_1} - n \xrightarrow[n\to\infty]{} 0 $$
and hence by the squeezing lemma,
$$ \lim_{n\to\infty} \left( r + (n-m)r^{-\frac{1}{n-m}} - (n-m+1) \right) = 0. $$
This limit can be explicitly computed, and the result is
$$ \lim_{n\to\infty} \left( r + (n-m)r^{-\frac{1}{n-m}} - (n-m+1) \right) = r - \log r - 1. $$
It is easy to check that the RHS, considered as function of $r$, achieves the unique global minimum $0$ at $r = 1$. This implies $a_{m+1} = a_m$ and hence the induction hypothesis is true for $m+1$ as well.
Therefore the claim follows from the mathematical induction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a+b=2$ so $2+\frac{ab(a-b)^2}{4}\leq a^{1+\sqrt{b}}+b^{1+\sqrt{a}}\leq2+\frac{(a-b)^2}{4}$
Let $a$ and $b$ be non-negative numbers such that $a+b=2$. Prove that:
$$2+\frac{ab(a-b)^2}{4}\leq a^{1+\sqrt{b}}+b^{1+\sqrt{a}}\leq2+\frac{(a-b)^2}{4}$$
In the left inequality the equality occurs also for $a=2$ and $b=0$, which makes additional problems.
For $b=2-a$ we get inequalities of one variable, but derivatives are very ugly and I think it can't help (I mean here $f'(x)=0$...).
Since the equality occurs for $a=b=1$, we can try the Taylor series, but I failed in this way.
Thank you!
| (Partial solution) For the right inequality:
WLOG, assume $a \le b$. Then $0 < a \le 1 \le b < 2$.
We have $\sqrt{b} = 1 + \frac{b - 1}{\sqrt{b} + 1} \ge 1 + \frac{b - 1}{\frac{b + 1}{2} + 1} = \frac{3b + 1}{b + 3}$
and $\sqrt{a} = 1 - \frac{1 - a}{\sqrt{a} + 1} \le 1 - \frac{1 - a}{(a + 1)/2 + 1} = \frac{3a + 1}{a + 3}$.
It suffices to prove that
$$a^{1 + \frac{3b + 1}{b + 3}} + b^{1 + \frac{3a + 1}{a + 3}}\leq 2+\frac{(a-b)^2}{4}. \tag{1}$$
Let $a = 1 - x, b = 1 + x$ for $0 \le x < 1$. (1) becomes
$$(1 - x)^2 \cdot (1 - x)^{\frac{2x}{4 + x}} + (1 + x)^2 \cdot (1 + x)^{\frac{-2x}{4 - x}} \le 2 + x^2.$$
Fact 1: $(1 + u)^{\frac{-2u}{4 - u}} \le 1 - \frac{1}{2}\,{u}^{2} + \frac{1}{8}\,{u}^{3}-{\frac {1}{96}}\,{u}^{4}+{\frac {11}{384}}
\,{u}^{5} \triangleq F(u)$ for all $u$ in $(-1, 1)$. (Hint: Take logarithm and then take derivatives.)
By Fact 1, it suffices to prove that
$$(1 - x)^2F(-x) + (1 + x)^2F(x) \le 2 + x^2$$
which is written as $\frac{1}{96}x^4(50 - 9x^2) \ge 0$. True.
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $f'(1)$ given $f(x)$ and $f(1)$ Question:
Let $$f(x)=\int\frac{x-1}{x+1}\frac{dx}{\sqrt{x^3+x^2+x}}$$ such that $f(1)=\frac{2\pi}{3}$
Then $f'(1)$ is equal to
A) $0$
B) $\cfrac \pi3$
C) $\cfrac\pi4$
D) $2$
Attempt:
\begin{align}f(x)&=\int\frac{x-1}{x+1}\frac{1}{\sqrt{x^3+x^2+x}}dx\\
f(1)&=\int0dx\\
f(1)&=0+c\\
&= \frac{2\pi}{3}\\
\implies c&= \frac{2\pi}{3}
\end{align}
I don't know what to do beyond this step. This is an indefinite integral so we cannot differentiate it. How else we bring $f'(x)$ into picture? Any hints?
| $f'(x)$ is simply the term under integration.$$f'(x)=\frac{x-1}{x+1}\frac{1}{\sqrt{x + x^2 + x^3}}$$
Take an example.
$$\text{Let} ~~~g(x) = \int xdx ; ~~\text{find g(1), given g(0)=0.}$$.
As you have done , it should be $$g(1) = \int 1dx= x+c$$ using $ Since g(0)=0 \implies c=0$
So, $g(1)$ should $x$, isn't it weird?
Yes, it is, because it is wrong.
By integrating , we will get $$g(x)=\frac{x^2}{2} \implies g(1)=\frac12$$
Therefore, $$f'(1)=\frac{1-1}{1+1}\frac{1}{\sqrt{1 + 1^2 + 1^3}}=0$$
Remember, if :
$$f(x)=\int g(x)dx \implies f'(x)=g(x) ~; ~~$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The remainder when $33333\ldots$ ($33$ times) is divided by $19$ $A= 33333\ldots$ ($33$ times). What is the remainder when $A$ is divided by $19$?
I don't know the divisibility rule of $19.$
What I did was
$32\times(33333\times100000)/19$ and my remainder is not zero and this is completely divisible by $19.$
This is a gmat exam question.
| Here's something crude but it takes less than the whole $33$ steps:
$$
\begin{array}{rccccccccccccccccc}
& & & 1 & 7 & 5 & 4 & \\[12pt]
& 19) & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & 3 & \ldots \\
& & 1 & 9 \\[12pt]
\text{remainder} = 14 \rightarrow & & 1 & 4 & 3 \\
& & 1 & 3 & 3 \\[12pt]
\text{remainder} = 10 \rightarrow & & & 1 & 0 & 3 \\
& & & & 9 & 5 \\[12pt]
\text{remainder} = 8 \rightarrow & & & & & 8 & 3 \\
& & & & & 7 & 6 \\[12pt]
\text{remainder} = 7 \rightarrow & & & & & & 7 & 3
\end{array}
$$
If we ever get $0$ as a remainder -- say after $n$ steps -- then we can discard the first $n$ digits, and the the next set of $n$ digits if that many remain, then the next set of $n$ digits after that if that many still remain, and so on. In that way we reduce the problem to one with fewer than $n$ iterations of the digit $3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all solutions in positive integers to $a^b -b^a = 3$
Find all solutions in positive integers to $a^b -b^a = 3$
It appears that the only solution is $a=4, b=1$.
Modular arithmetic is not of much help (apart from deriving that $a,b$ must have opposite parity).
We know that $a^x$ dominates the polynomial $x^a$ but this does not seem to yield the result.
Tried hard to estimate a bound for $a^b - b^a$ but with no success.
Any help is appreciated. This is a problem from a regional mathematics Olympiad.
Added on 26 March, 2017:
I was trying to find a solution that involves no Calculus (or a minimum of Calculus) since this is a problem from a Junior Olympiad in which the students do not have a knowledge of Calculus. The following argument seems to work. Please point out flaws/mistakes, if any.
Lemma 1 For any $n \geq 4$, $$n^{n+1} - (n+1)^n > (n-1)^n - n^{n-1}$$
Proof
\begin{align*}
&\qquad n^{n+1} - (n+1)^n > (n-1)^n - n^{n-1} \\
&\Leftrightarrow n^{n+1} + n^{n-1} > (n-1)^n + (n+1)^n \\
&\Leftrightarrow n+\frac{1}{n} > \left(1-\frac{1}{n}\right)^n + \left(1+\frac{1}{n}\right)^n
\end{align*}
Since $\left(1+\frac{1}{n}\right)^n < 3$ and $\left(1-\frac{1}{n}\right)^n < 1$, we have
$$\left(1+\frac{1}{n}\right)^n+\left(1-\frac{1}{n}\right)^n < 4 < n +\frac{1}{n}$$
Lemma 2 For any $k$ and $n \geq 4$,
$$n^{n+k} -(n+k)^n > n^{n+1} - (n+1)^n$$
Proof
\begin{align*}
&\qquad n^{n+k} -(n+k)^n > n^{n+1} - (n+1)^n \\
&\Leftrightarrow n^{n+k} -n^{n+1} > (n+k)^n - (n+1)^n \\
&\Leftrightarrow n^k - n > \left(1+\frac{k}{n}\right)^n - \left(1+\frac{1}{n}\right)^n
\end{align*}
Since $\left(1+\frac{k}{n}\right)^n < 3^k$ and $\left(1-\frac{1}{n}\right)^n < 1$, we have
$$\left(1+\frac{k}{n}\right)^n+\left(1-\frac{1}{n}\right)^n < 3^k +1 < n^k - n $$
as $n > 4$.
Now suppose that $a^b - b^a = 3$. Let $a \geq 4$. If $b = a+k > a$, then we have
\begin{align*}
a^{a+k} - (a+k)^a &> a^{a+1} - (a+1)^a \\
&> (a-1)^a - a^{a-1} \\
&> \cdots \\
&> 3^4 - 4^3 = 17
\end{align*}
Thus there are no solutions with $b > a \geq 4$.
If $a > b \geq 4$, then
$$a^b - b^a = -(b^a - a^b) \leq -17 $$
from what we have seen above. Thus there are no solutions if $a > b \geq 4$. Thus all solutions can be only in the range $1 \leq a,b \leq 4$. Clearly, in this range only $a=4, b=1$ satisfies the given equation.
| Consider $g(t) = a^t - t^a .$
Then $g'(t) = (\ln a) a^t - at^{a-1}$.
Note that
$$ \left(\frac{t}{\ln (t)} \right )' = \frac{(-1 + \ln(t))}{\ln^2(t)}$$
and
$$\left( \frac{t-1}{\ln(t)}\right)' = \frac{t(\ln (t) -1)+1}{t \ln^2(t)} $$
Thus both $t /\ln t$ and $(t-1)/\ln t$ are strictly increasing for $t > e$.
Therefore, for $a,t > e$ and $t >a$, $a^t > t^a$ and
$$ \frac{t-1}{\ln t} > \frac{a-1}{\ln a}$$
$$\Rightarrow a^{t-1} > t^{a-1} $$
$$\Rightarrow a^t > at^{a-1} $$
$$\Rightarrow (\ln a)a^t > at^{a-1}$$
That is, for $a,t > e$ and $t >a$, $g(t)$ is strictly increasing.
Thus for positive integers $ n, m \geq 3, n \neq m$,
$|n^m-m^n| = n^m - m^n $ if $m >n$ and $m^n - n^m$ if $n > m$. Since g is increasing, we have $|n^m-m^n| \geq |n^{n+1}-(n+1)^n|$ or $|n^m-m^n| \geq |m^{m+1}-(m+1)^m|$.
Now, note that $h(x) = x^{x+1} - (x+1)^x$ is increasing for $x \geq 3$.
Thus $|n^m-m^n| > |3^4-4^3| = 17 > 3$. Thus no positive integer solutions exist for $n,m \geq 3$. The cases $n$ or $m \in \{1,2\}$ are easily verified, and the solution you found is the only one.
| {
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"timestamp": "2023-03-29T00:00:00",
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if such $a^2+b^2+c^2+1=d+\sqrt{a+b+c-d}$ show $2(a+b+c)=3$ Let $a,b,c,d\in R$ and such
$$a^2+b^2+c^2+1=d+\sqrt{a+b+c-d}$$
show that $$2(a+b+c)=3$$
It seem can use inequality to solve it?
| Let $a+b+c-d=x^2$, where $x\geq0$.
Hence, $$a^2+b^2+c^2+1=a+b+c-x^2+x$$ or
$$x^2-x+a^2+b^2+c^2-a-b-c+1=0$$ or
$$\left(x-\frac{1}{2}\right)^2+\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\left(c-\frac{1}{2}\right)^2=0,$$
which gives $a=b=c=\frac{1}{2}$ and $2(a+b+c)=3$.
Done!
| {
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Prove that all odd derivate of $\tan(x)$ at $x=0$ is at least $1$. This was the original excercise:
Prove that $$\frac{\sin{x}+\tan{x}}{2} \geq x$$ where $x \in (0,\frac{\pi}{2})$
This is how I did it:
Knowing that
\begin{align}\sin{0}&=0\\
\cos{0}&=1\\
\frac{\mathrm{d} \sin{x}}{\mathrm{d} x}&=\cos{x}\\
\frac{\mathrm{d} \cos{x}}{\mathrm{d} x}&=- \sin{x}\end{align}
and the Taylor series of $f(x)$ is: $$\sum_{n=0}^{\infty} \frac{x^n\frac{\mathrm{d}^n f(0)}{\mathrm{d} x^n}}{n!}$$
we can easily see that the Taylor series of $\sin(x)$ is:
$$\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$
We know what the tan(x) function is odd, so the n-th derivate will be $0$ for all even $n$. For odd numbers, it will be $1,2,16,272,\cdots$ So if we sum the $2$ Taylor series, we will get this:
$$\sin{x}+\tan{x}=2x+\frac{x^3}{3!} (-1+1)+\frac{x^5}{5!}(16+1)+\frac{x^7}{7!}(272-1)+...$$
Where we can see that all part will be positive (or $0$) on the given interval. Back to the original inequality:
$$2x+\frac{x^3}{3!} (-1+1)+\frac{x^5}{5!}(16+1)+\frac{x^7}{7!}(272-1)+...\geq 2x$$
$$\frac{x^3}{3!} (-1+1)+\frac{x^5}{5!}(16+1)+\frac{x^7}{7!}(272-1)+... \geq 0$$
Which is true in the given interval.
But how could I prove that all odd derivate of $\tan(x)$ at $x=0$ is at least $1$?
| Nice exercise. We have that $f(x)=\tan(x)$ fulfills the differential equation
$$ f'(x) = 1 + f(x)^2 \tag{1}$$
with $f(0)=0$ and $f'(0)=1$. The even derivatives at the origin are obviously zero, since $\tan(x)$ is an odd function, and the odd derivatives can be computed by applying $\frac{d^{2k}}{dx^{2k}}$ to both sides of $(1)$.
For instance
$$ f'''(x) = 2\,f'(x)^2+2\,f(x)\,f''(x)\tag{2}$$
so, by induction, the odd derivatives of $\tan(x)$ at the origin are positive integers.
An alternative derivation of tangent numbers comes from
$$ \cos(x)=\prod_{m\geq 0}\left(1-\frac{4x^2}{(2m+1)^2\pi^2}\right). \tag{3}$$
By applying $-\frac{d}{dx}\log(\cdot)$ to both sides of $(3)$,
$$ \frac{\tan(x)}{x}=\sum_{m\geq 0}\frac{2}{\left(m+\frac{1}{2}\right)^2 \pi^2-x^2}=\sum_{k\geq 0}\frac{2x^{2k}}{\pi^{2k+2}}\sum_{m\geq 0}\frac{1}{\left(m+\frac{1}{2}\right)^{2k+2}}\tag{4}$$
hence:
$$ \frac{\tan(x)}{x}=\sum_{k\geq 0}\frac{2 x^{2k}(2^{2k+2}-1)\zeta(2k+2)}{\pi^{2k+2}}\tag{5}$$
and it is enough to show that the denominator of $\frac{\zeta(2k+2)}{\pi^{2k+2}}\in\mathbb{Q}$ is a divisor of $(2k+1)!$. That directly follows from the Fourier series of Bernoulli polynomials.
About the origin exercise, you may simply notice that
$$ \frac{\sin(x)+\tan(x)}{2} = \frac{1}{2}\int_{0}^{x}\left(\cos(t)+\frac{1}{\cos^2(t)}\right)\,dt> \frac{1}{2}\int_{0}^{x}2\,dt = x $$
since $u+\frac{1}{u^2}> 2$ for any $u\in(0,1)$. The similar inequality
$$ \forall x\in\left(0,\frac{\pi}{2}\right),\qquad 2\sin(x)+\tan(x) > 3x $$
was proved in 1631 by Snellius.
| {
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When, where and **how often** do you find polynomials of higher degrees than two in mathematical, pure/applied, research? A formula for solving a polynomial of degree three, see this link; $ax^3+bx^2+cx+d=0$, is
$$\begin{align}
x\quad&=\quad
\sqrt[3]{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right)
+
\sqrt{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right) ^2
+
\left(
\frac{c}{3a} - \frac{b^2}{9a^2}
\right) ^3
}
}\\
&+\quad
\sqrt[3]{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right)
-
\sqrt{
\left(
\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}
\right) ^2
+
\left(
\frac{c}{3a} - \frac{b^2}{9a^2}
\right) ^3
}
}
\;-\;\frac{b}{3a}
\end{align}$$
Unlike quadratic, cubic, and quartic polynomials, the general quintic
cannot be solved algebraically in terms of a finite number of
additions, subtractions, multiplications, divisions, and root
extractions, as rigorously demonstrated by Abel (Abel's impossibility
theorem) and Galois. However, certain classes of quintic equations can
be solved [...] Source: http://mathworld.wolfram.com/QuinticEquation.html
At levels of $5^{\text{th}}$ degree polynomials, things are starting to look really serious in my eyes. My question is:
If it is possible to not answer subjectively: When, where and how often do you find polynomials of higher degrees than two in mathematical, pure/applied, research?
| Gaussian quadrature is a class of methods of numerical integration based on polynomial approximation. Typically, polynomials of degree $5$ or $10$ are employed. For many of the functions that arise in applied mathematics, these give very accurate results. Polynomials of arbitrarily high degree can in principle be used, but the additional accuracy is seldom relevant or worth the extra complexity entailed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2197171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 13,
"answer_id": 11
} |
find the value of x at second derivative The quesion is :
\begin{align}
\ y= \sqrt{x} ~\text{ find the value of x for which}~ y+ \frac{d^2y}{dx^2} = 0
\end{align}
\begin{align}
\
\end{align}
The answer given is : x= 1/2
I am not sure how they got this answer so any help would be appreciated.
| $y=\sqrt { x } $ so plug this in the equation $$\sqrt { x } +\frac { d }{ dx } \left( \frac { d\left( \sqrt { x } \right) }{ dx } \right) =0\\$$ now here $\frac { d }{ dx } \left( \frac { d\left( \sqrt { x } \right) }{ dx } \right) $ means you should find derivative twice of $y=\sqrt { x } $
$$ \sqrt { x } +\frac { d }{ dx } \left( \frac { 1 }{ 2\sqrt { x } } \right) =0\\ \sqrt { x } -\frac { 1 }{ 4x\sqrt { x } } =0\\ 4{ x }^{ 2 }-1=0\\ x=1/2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2197330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
Evaluate $\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx$
$$\int_0^{\frac{\pi}{2}} \frac{\sin^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{\cos^3(x)}{\cos^3(x)+\sin^3(x)} dx=\int_0^{\frac{\pi}{2}} 1-\frac{1}{1+\tan^3(x)} dx$$
Then to evaluate $\int \frac{1}{1+\tan^3(x)}dx$, I let $\tan(x)=t$ and hence $dt = \sec^2(x) dx$, which implies $dx=\frac{1}{\sec^2(x)}dt$.
Then I don't know how to continue.
| Replacing $x$ by $\pi/2-x$, the integral becomes
$$ I = \int_0^{\pi/2} \frac{\cos^3{x}}{\cos^3{x}+\sin^3{x}} \, dx. $$
Adding together with the original form of $I$,
$$ 2I = \int_0^{\pi/2} \frac{\sin^3{x}+\cos^3{x}}{\cos^3{x}+\sin^3{x}} \, dx = \int_0^{\pi/2} 1 \, dx = \frac{\pi}{2}, $$
so $I = \pi/4$.
This trick works on any integral of the form
$$ \int_0^a \frac{f(x)}{f(x)+f(a-x)} \, dx: $$
the answer is always $a/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Multi-scale approximation from periodic standard form I solved the first order two-scale approximation of $\ddot y + y + \epsilon y^3=0$ ($y(0)=\alpha , \dot y(0)=0$) to be $\alpha \cos(t(1+\frac{3\alpha^2}{8}\epsilon))$. Then I put that equation into periodic standard form using $u=y$, $v=\dot y$, $u=a \cos t + b \sin t$, $v=-a\sin t +b\cos t$ yielding: $$\dot a=\epsilon\sin t(a\cos t + b\sin t)^3$$ $$\dot b=-\epsilon\cos t(a\cos t + b\sin t)^3$$
Then I averaged out one cycle to get: $$\frac1{2\pi}\int_0^{2\pi}\dot a \,dt = \epsilon\frac{ 3 b}{8}(a^2+b^2)$$ $$\frac1{2\pi}\int_0^{2\pi}\dot b \,dt = \epsilon\frac{- 3 a}{8}(a^2+b^2)$$
So if we take $a$ and $b$ to be $\sin$ and $\cos$ respectively, the $a^2 +b^2$ goes away and we can quickly solve to see $b=\alpha cos(\frac38\epsilon t)$ and $a=0$ because of the boundary conditions.
So my two results from the two sets of coordinates are $\alpha\cos(t(1+\frac{3\alpha^2}8\epsilon))$ and $\alpha\cos(\frac38\epsilon t)$.
These are very similar. But why aren't my two estimations equal?
| To be a bit more precise: you're using a particular form of multi-scale approximation, namely the Poincaré-Lindstedt method. The leading order term of the expansion is, as you calculated, equal to
\begin{equation}
\alpha\,\cos\left( \left(1+\epsilon \frac{3 \alpha^2}{8}\right)t\right).
\end{equation}
The second method you're using is known as (periodic) averaging. Denoting the slow variable as $\tau$, so $\tau = \epsilon t$, we get the dynamical system
\begin{align}
\frac{\text{d} a}{\text{d} \tau} &= \frac{3}{8} b (a^2+b^2),\\
\frac{\text{d} b}{\text{d} \tau} &= -\frac{3}{8} a (a^2+b^2).
\end{align}
So far, so good. Now, I don't understand what you mean by 'we take $a$ and $b$ to be sin and cos, so the $a^2+b^2$ goes away'. You might mean something like the following.
We see that the dynamical system contains terms of the form $a^2+b^2$. Therefore, it seems a good idea to try the coordinate transformation
\begin{align}
a(\tau) &= r(\tau)\,\cos(\phi(\tau)),\\
b(\tau) &= r(\tau)\,\sin(\phi(\tau)),
\end{align}
because then $a^2 + b^2 = r^2$. In these coordinates, the dynamical system indeed simplifies considerably, and we get
\begin{align}
\frac{\text{d} r}{\text{d} \tau} &= 0,\\
\frac{\text{d} \phi}{\text{d} \tau} &= - \frac{3}{8} r^2.
\end{align}
The original initial conditions $y(0) = \alpha, y'(0) = 0$ imply $a(0) = \alpha$ and $b(0) = 0$, which implies $r(0) = \alpha$ and $\phi(0) = 0$ (or $r(0) = -\alpha$ and $\phi(0) = \pi$, but you can show that it boils down to the same thing). Since $r(\tau)$ is constant, we have
\begin{equation}
r(\tau) = \alpha,
\end{equation}
which implies
\begin{equation}
\phi(\tau) = -\frac{3 \alpha^2}{8}\tau.
\end{equation}
Putting all this information together, we obtain the leading order expression
\begin{align}
y = u = a(\tau) \cos(t) + b(\tau) \sin(t) &= \alpha \cos(t) \cos\left(-\frac{3 \alpha^2}{8} \epsilon t\right) + \alpha \sin(t) \sin\left(-\frac{3 \alpha^2}{8} \epsilon t\right) \\
&= \alpha \cos \left(t + \frac{3 \alpha^2}{8} \epsilon t \right)
\end{align}
by the angle sum formulas for $\sin$ and $\cos$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Inverse of matrix of ones + nI Having a vector $\mathbf{1} \in \mathbb{R}^{n}$ containing only ones, following equality should be true according to a paper I am currently reading:
\begin{equation}
\left( nI+\mathbf{1}\mathbf{1}^T \right)^{-1}= \frac{1}{n}\left( I - \frac{1}{2n} \mathbf{1}\mathbf{1}^T \right)
\end{equation}
EDIT: what is the general rule for constructing an inverse of a matrix with $n$ on diagonal and $1$ elsewhere and how is this rule derived?
| Example
Example for $n=5$:
$$
\mathbf{A}(n) =
\left(
\begin{array}{ccccc}
6 & 1 & 1 & 1 & 1 \\
1 & 6 & 1 & 1 & 1 \\
1 & 1 & 6 & 1 & 1 \\
1 & 1 & 1 & 6 & 1 \\
1 & 1 & 1 & 1 & 6 \\
\end{array}
\right), \qquad
\mathbf{A}^{-1}(n) =
\frac{1}{50}
\left(
\begin{array}{rrrrr}
9 & -1 & -1 & -1 & -1 \\
-1 & 9 & -1 & -1 & -1 \\
-1 & -1 & 9 & -1 & -1 \\
-1 & -1 & -1 & 9 & -1 \\
-1 & -1 & -1 & -1 & 9 \\
\end{array}
\right)
$$
Result
In general,
$$
\mathbf{A}_{r,c}(n) =
\begin{cases}
1 & r\ne c \\
n+1 & r = c
\end{cases}
\qquad
\mathbf{A}^{-1}_{r,c}(n) =
\begin{cases}
-\frac{1}{2n^{2}} & r\ne c \\
\frac{2n-1}{2n^{2}} & r = c
\end{cases}
$$
Proof strategy: induction
To compute the inverse matrix,
$$
\mathbf{A}^{-1} = \frac{\text{adj } \mathbf{A}} {\det \mathbf{A}}
$$
where the adjugate matrix, $\text{adj } \mathbf{A}$, is the transpose of the matrix of cofactors.
*
*Establish $\det \mathbf{A}(n) = 2n^{-2}$.
*Establish the matrix of minors is an $n\times n$ matrix of the form
$$
\left[ \begin{array}{rrrr}
\alpha & \beta & -\beta & \dots \\
\beta & \alpha & \beta & \dots \\
-\beta & \beta & \alpha \\
\vdots & \vdots & & \ddots
\end{array} \right]
$$
with
$$
\alpha = \left(2n-1 \right) n^{n-2}, \qquad \beta = n^{n-2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Prove that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$
Let $n,x,y,z$ be positive integers such that $$(x-y)^2+(y-z)^2+(z-x)^2 = (2n+1)xyz.$$ Prove that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$.
The original equation doesn't have $x^3,y^3,z^3$, so how can we show that $(2n+1)(x+y+z)+6$ divides $x^3+y^3+z^3$? Expanding the given equation seems to make the question more complicated, and also how do we get the $6$?
| We're given that
$$(x-y)^2+(y-z)^2+(z-x)^2=(2n+1)xyz$$
Hence
$$\frac{(x-y)^2+(y-z)^2+(z-x)^2}{xyz}=(2n+1)$$
Substitute this into the expression given:
$$
\begin{aligned}
(2n+1)(x+y+z)+6
=
\frac{(x-y)^2+(y-z)^2+(z-x)^2}{xyz}(x+y+z)+\frac{6xyz}{xyz}
\end{aligned}
$$
After expanding and simplifying, you will get to
$$\frac{2(x^3+y^3+z^3)}{xyz}$$
Which certainly does divide $x^3+y^3+z^3$ if one of $x,y,z$ is even...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $c$ is equal to $\binom{99}{0}-\binom{99}{2}+\binom{99}{4}-\binom{99}{6}+\cdots+\binom{99}{96}-\binom{99}{98},$ then find $\log_2{(-c)}$. If $c$ is equal to $$\binom{99}{0}-\binom{99}{2}+\binom{99}{4}-\binom{99}{6}+\cdots+\binom{99}{96}-\binom{99}{98},$$then find $\log_2{(-c)}$.
I think the binomial theorem might help but the bottom numbers are skipping by 2's. How would I apply it now?
| $$\sum_{99 \geq k \geq 0, \text{even}} i^k {99 \choose k}$$
$$=\sum_{99 \geq k \geq 0} \frac{(-1)^k+1^k}{2}i ^k{99 \choose k}$$
$$=\frac{1}{2} \sum_{k=0}^{99} i^k {99 \choose k}+\frac{1}{2} \sum_{k=0}^{99} (-i)^k {99 \choose k}$$
$$=\frac{1}{2}(1+i)^{99}+\frac{1}{2}(1-i)^{99}$$
$$=\frac{1}{2}(\sqrt{2})^{99}e^{99 \frac{\pi}{4}i }+\frac{1}{2}(\sqrt{2})^{99} e^{-99 \frac{\pi}{4}i}$$
$$=(\sqrt{2})^{99} \frac{e^{99 \frac{\pi}{4}i} +e^{-99\frac{\pi}{4}i}}{2}$$
$$=\left((\sqrt{2})^{99}\right)\left(\cos (\frac{99}{4}\pi) \right)$$
$$=(2^{49})(\sqrt{2})(\frac{-1}{\sqrt{2}})$$
$$=-2^{49}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How should i approximate this? Find the taylor series for the given function about the given point.
$$f(x,y)= e^{x^2+y^2}$$
My solution I thought was $e^{x^2}\cdot e^{y^2}$.
How should I make a summation about this?.
| $e^{x^2} = 1 + x^2 + \frac 12 x^4 + \frac 16 x^6\cdots\\
(e^{x^2})(e^{y^2} = (1 + x^2 + \frac 12 x^4 + \frac 16 x^6\cdots)(1 + y^2 + \frac 12 y^4 + \frac 16 y^6\cdots)\\
(e^{x^2})(e^{y^2} = 1 + x^2 + y^2 + \frac 12 (x^4 + y^4 + 2x^2y^2)+\frac 16 (x^6 + y^6 + 3x^4y^2 + 3x^2y^4) \cdots$
I am just multiplying the terms from one expansion by the terms in the other expansion.
but more generally
$f(x,y) = f(x_0,y_0) + f_x (x_0,y_0)(x-x_0) + f_y (x_0,y_0)(x-x_0) + \frac 12 (f_{xx}(x_0,y_0)(x=x_0)^2 + 2f_{xy}(x_0,y_0)(x-x_0)(y-y_0)+f_{yy}(x_0,y_0)(y-y_0)^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove this $1+2abcxyz \geq a^2x^2 + b^2y^2+c^2z^2$ Let $a,b,c,x,y,z\in\mathbb{R}$ such that \begin{align}\{a,b,c,x,y,z\}&\subset[-1,1]\\
1 + 2abc &\geq a^2+b^2+c^2\\
1+2xyz&\geq x^2+y^2+z^2\end{align}
Prove that:
$$1+2abcxyz \geq a^2x^2 + b^2y^2+c^2z^2$$
| We'll rewrite our conditions in the following forms.
$(1-b^2)(1-c^2)=(a-bc)^2$ and $(1-y^2)(1-z^2)\geq(x-yz)^2$.
Let $b^2=\frac{1}{1+v^2}$, $c^2=\frac{1}{1+w^2}$, $y^2=\frac{1}{1+q^2}$, $z^2=\frac{1}{1+r^2}$, $a-bc=ubc$ and $x-yz=pyz$,
where $v$, $w$, $q$ and $r$ are non-negatives.
Hence, our conditions give $vw\geq|u|$, $qr\geq|p|$ and we need to prove that
$$(1-b^2y^2)(1-c^2z^2)\geq(ax-bcyz)^2$$ or
$$\left(1-\frac{1}{(1+v^2)(1+q^2)}\right)\left(1-\frac{1}{(1+w^2)(1+r^2)}\right)\geq((1+u)(1+p)bcyz-bcyz)^2$$ or
$$((1+v^2)(1+q^2)-1)((1+w^2)(1+r^2)-1)\geq((1+u)(1+p)-1)^2$$ or
$$(v^2q^2+v^2+q^2)(w^2r^2+w^2+r^2)\geq(up+u+p)^2,$$
which is true by C-S:
$$(v^2q^2+v^2+q^2)(w^2r^2+w^2+r^2)\geq(vwqr+vw+qr)^2\geq$$
$$\geq\left(|up|+|u|+|p|\right)^2\geq(up+u+p)^2.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $\displaystyle\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}$ equal to $-\dfrac 12$ or just $-\dfrac{\sin y}{2y}$ Question:
$$\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}=?$$
Here is my try:
\begin{align}\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}&=\lim_{x\rightarrow y}\dfrac{-2 \sin (\frac 12(x+y))\sin (\frac{1}{2}(x-y))}{(x+y)(x-y)}\\
&=-\dfrac{2 \sin y}{2y}\dfrac 12\\
&=-\dfrac{\sin y}{2y}\end{align}
My question: Is $-\dfrac{\sin y}{2y}$ the final answer or can it be calculated further as $-\dfrac12$?
I also try different route:
Let $p=x-y$ so $x=p+y$ and $p\rightarrow 0$ as $x \rightarrow y$. Hence,
\begin{align}\lim_{x\rightarrow y}\dfrac{\cos x-\cos y}{x^2-y^2}&=\lim_{p\rightarrow 0}\dfrac{\cos (p+y)-\cos y}{p^2+2py+y^2-y^2}\\\\
&=\lim_{p\rightarrow 0}\dfrac{-2 \sin (\frac{1}{2}(p+2y))\sin (\frac{1}{2}p)}{p(p+2y)}\\\\
&=\lim_{p\rightarrow 0}\dfrac{-2 \sin (\frac{1}{2}(p+2y))}{(p+2y)} \dfrac{\sin (\frac{1}{2}p)}{p}\\\\
&=-\dfrac{2\sin y}{2y} \dfrac{1}{2}\\\\
&=-\dfrac{\sin y}{2y}\end{align}
Okay, so that left me with the same result. What is the correct final answer, $-\dfrac{\sin y}{2y}$ or $-\dfrac 12$?
Thanks
| You could also have used Taylor series around $x=y$. This is $$\cos(x)=\cos (y)-(x-y) \sin (y)-\frac{1}{2} (x-y)^2 \cos (y)+O\left((x-y)^3\right)$$ Then $$\cos(x)-\cos (y)=-(x-y) \sin (y)-\frac{1}{2} (x-y)^2 \cos (y)+O\left((x-y)^3\right)$$ $$\frac{\cos(x)-\cos (y) }{x^2-y^2}=-\frac{\sin (y)}{2 y}+(x-y) \left(\frac{\sin (y)}{4 y^2}-\frac{\cos (y)}{4
y}\right)+O\left((x-y)^2\right)$$ which shows the limit and also how it is approached.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove: Im$\left(\frac{1+re^{i\theta}}{1-e^{i\theta}}\right) = \frac{2r\sin\theta}{1-2r\cos\theta + r^2}$ Prove the identity: $\operatorname{Im}\left(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\right) = \dfrac{2r\sin\theta}{1-2r\cos\theta + r^2}$
My only lead is to assume $r$ and $\theta$ are real.
Then,
$\operatorname{Im}\left(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\right) = \dfrac{r\sin\theta + \sin\theta}{\sin^2+(1-\cos\theta)^2}$
| RTP: $Im\bigg(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\bigg) = \dfrac{2r\sin\theta}{1-2r\cos\theta + r^2}$=$\frac{2r\sin\theta}{(r-\cos\theta)^2+1-\cos^2\theta}$
$$Im\bigg(\dfrac{1+re^{i\theta}}{1-e^{i\theta}}\bigg) = Im\bigg(\frac{(1+r\cos \theta+ri\sin\theta )(1-\cos\theta +i\sin\theta)}{(1-\cos\theta-i\sin\theta)(1-\cos\theta+i\sin\theta)}\bigg)$$
$$=Im\bigg(\frac{1-\cos\theta+i\sin\theta+r\cos\theta-r\cos^2\theta+r\sin\theta\cos\theta \cdot i +r\sin\theta \cdot i-r\sin\theta\cos\theta\cdot i-r\sin^2\theta}{1-\cos\theta +i\sin\theta -\cos\theta+\cos^2\theta-\cos\theta\sin\theta\cdot i -i\sin\theta +\sin\theta\cos\theta\cdot i +\sin^2\theta}\bigg)$$
$$= \frac{\sin\theta+r\sin\theta}{1-2\cos\theta+\cos^2\theta+\sin^2\theta}$$
$$= \frac{(r+1)\sin\theta}{2(1-\cos\theta)}$$
I don't see how to arrive at the identity from here, could someone comment if they see an error/ way to proceed?
| {
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"url": "https://math.stackexchange.com/questions/2209744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of $\frac{(\sqrt{x^2+12}-4)}{2-\sqrt{x^3-4}}$ as $x$ goes to 2
$$\lim_{x\to 2}\frac{\sqrt{x^2+12}-4}{2-\sqrt{x^3-4}}$$
I know it must be very trivial, but yet I'm stuck at this problem and other similar problems for quite a long time. I'd be glad if someone showed me a full solution without using L'Hospital and delighted if someone could give some hints about how to atack this kind of problem... what is the thinking behind? Is it just about intuition? Or is it about test all sort of algebraic manipulation?
| You need to get used to the standard limit $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ which solves these algebraic limit problems instantly.
Thus we have
\begin{align}
L &= \lim_{x \to 2}\frac{\sqrt{x^{2} + 12} - 4}{2 - \sqrt{x^{3} - 4}}\notag\\
&= \lim_{x \to 2}\frac{\sqrt{x^{2} + 12} - \sqrt{16}}{(x^{2} + 12) - 16}\cdot\frac{x^{2} - 4}{8 - x^{3}}\cdot\frac{4 - (x^{3} - 4)}{\sqrt{4} - \sqrt{x^{3} - 4}}\notag\\
&= -\lim_{t \to 16}\frac{t^{1/2} - 16^{1/2}}{t - 16}\cdot\lim_{x \to 2}\frac{x^{2} - 2^{2}}{x - 2}\cdot\frac{x - 2}{x^{3} - 2^{3}}\cdot\lim_{u \to 4}\frac{u - 4}{u^{1/2} - 4^{1/2}}\text{ (}u = x^{3} - 4, t = x^{2} + 12)\notag\\
&= -\frac{1}{2}\cdot 16^{-1/2}\cdot 2\cdot 2^{1}\cdot\frac{1}{3\cdot 2^{2}}\cdot\dfrac{1}{\dfrac{1}{2}\cdot 4^{-1/2}}\notag\\
&= -\frac{1}{3}\cdot\frac{2}{4}\notag\\
&= -\frac{1}{6}\notag
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2211029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How many different Strings can I create with the characters of the String *SPEISESTAERKE*? How many different Strings can I create with the characters of the String SPEISESTAERKE ?
According to that post, I have tried the following. Note, that I use all the given characters at once to create all possible combinations:
$2 \times$S,
$1 \times$P,
$4 \times$E,
$1 \times$I,
$1 \times$A,
$1 \times$R,
$1 \times$K
$\binom {13}{2} \cdot \binom {11}{1} \cdot \binom {10}{4} \cdot \binom {6}{1} \cdot \binom {5}{1} \cdot \binom {4}{1} \cdot \binom {3}{1} \cdot \binom {2}{1} \cdot \binom {1}{1} = 129 729 600$
Unfortunately this is the wrong answer. The correct solution is $432 432 00$.
Question: How can I achieve the correct result?
| alternative answer - with correctly counted letters:
$\binom {13}{3} \cdot \binom {10}{4} \cdot \binom {6}{1} \cdot \binom {5}{1} \cdot \binom {4}{1} \cdot \binom {3}{1} \cdot \binom {2}{1} \cdot \binom {1}{1} = 432 432 00$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that the sum of numbers written on the board is always equal to $\binom{20}{2}$ We write the number $20$ as the sum of two numbers $a,b$ and write $ab$ on the board where $a,b \ge 1$. We do this again for $a,b$ until we get only (multiple instances of) the number $1$. Prove that the sum of numbers written on the board is always equal to $\binom{20}{2}$.
I don't know how should I work to get a combination; it is more likely to get a sum?
| This process can be described as follows:
$F(n) = a \cdot (n-a) + F(a) + F(n-a)$ where $F(0) = F(1) = 0$.
(In other words, given some $n$, we're choosing a value for $a$ and letting $b=n-a$, so that $a+b=n$)
Try with $a=1$:
$F(n) = 1 \cdot (n - 1) + F(1) + F(n-1) = n - 1 + F(n-1) = \sum_{k=1}^{n} (k-1) = \frac{n(n-1)}{2}$
This at least gives us an easy way to compute $F(n)$ assuming we always choose $a=1$.
Let's try again using induction, with the hypothesis that $F(n) = \frac{n(n-1)}{2}$ is true regardless of $a$ by substituting:
$F(n) = a \cdot (n-a) + \frac{a(a-1)}{2} + \frac{(n-a)(n-a-1)}{2} = \frac{n(n-1)}{2}$
All the $a$-terms cancel out, so $F(n)$ does not depend on $a$. It will always be equal to $\frac{n(n-1)}{2} = \binom{n}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Limit question: $\lim_{x \rightarrow 2} \frac{x+1}{x+2} = \frac{3}{4}$
Show that $$\lim_{x \rightarrow 2} \frac{x+1}{x+2} = \frac{3}{4}$$ by the definition.
This is what I've done so far:
$$\left\vert\frac{x+1}{x+2} - \frac{3}{4} \right| = \left| \frac{x-2}{4(x+2)} \right|$$
and I see that
$$|x-a| = |x-2|$$ which is in our numerator. What choice of $\delta$ can I pick (and what is the motivation) so that it satisfies the following limit definition?
$$\lim_{x \rightarrow a} f(x) = b \Leftrightarrow \forall \epsilon >0, \exists \delta : |x-a| < \delta \Rightarrow |f(x) - b| < \epsilon$$
| Note that
$$
|\frac{x-2}{4(x+2)}| = \frac{|x-2|}{4|x+2|}
$$
for all $x \neq -2$.
We have to "bound away" the annoying term $|x+2|$ by a preliminary upper bound for $|x-2|$. If $|x-2| < 1$, then $-1 < x-2 < 1$, so $1 < x < 3$, and hence $3 < x+2 < 5$; hence $3 < |x+2| < 5$. Then
$$
\frac{|x-2|}{4|x+2|} < \frac{|x-2|}{4\cdot 3} = \frac{|x-2|}{12}.
$$
Given any $\varepsilon > 0$, we have $\frac{|x-2|}{12} < \varepsilon$ if $|x-2| < 12\varepsilon$. So combining the two upper bounds for $|x-2|$ we conclude that $\delta := \min \{ 1, 12\varepsilon \}$ suffices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2212984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Question on calculating $\int_{\partial B_{3/2}(1)} \frac{z^7 + 1}{z^2(z^4 + 1)}dz$ My task is to calculate $$\int_{\partial B_{3/2}(1)} \frac{z^7 + 1}{z^2(z^4 + 1)}dz$$ using the Cauchy integral formula. My question is: Is there a simple trick or do I have to perform a partial fraction decomposition?
| A partial fraction decomposition is not necessary.
The function
\begin{align*}
f(z)=\frac{z^7 + 1}{z^2(z^4 + 1)}
\end{align*}
has a double pole at $z=0$ and simple poles at
$z\in\left\{e^{\frac{\pi i}{4}},e^{-\frac{\pi i}{4}},e^{\frac{3\pi i}{4}},e^{-\frac{3\pi i}{4}}\right\}$. The contour of integration $\partial B_{3/2}(1)$ is a circle with center $(1,0)$ and radius $\frac{3}{2}$. It encloses precisely the poles at
$0,e^{\frac{\pi i}{4}},e^{-\frac{\pi i}{4}}$.
We obtain
\begin{align*}
\int_{\partial B_{3/2}(1)} \frac{z^7 + 1}{z^2(z^4 + 1)}dz&=\int_{\partial B_{3/2}(1)}f(z)dz\\
&=2\pi i\left(\operatorname{res}\left(f,0\right)
+\operatorname{res}\left(f,e^{\frac{3\pi i}{4}}\right)
+\operatorname{res}\left(f,e^{-\frac{\pi i}{4}}\right)\right)\\
&=2\pi i\left(0+\frac{1}{4}\left(e^{\frac{3\pi i}{4}}+i\right)+\frac{1}{4}\left(-e^{\frac{\pi i}{4}}-i\right)\right)\\
&=2\pi i\left(-\frac{1}{2\sqrt{2}}\right)\\
&=-\frac{\pi i}{\sqrt{2}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find shortest distance between lines in 3D Find shortest distance between lines given by
$$\frac{x-2}{3}=\frac{y-6}{4}=\frac{z+9}{-4}$$
and
$$\frac{x+1}{2}=\frac{y+2}{-6}=\frac{z-3}{1}$$
Is there any shortcut method for this problems?
| Hint:
write the equations of the two lines in the form $\vec x=\vec p+t\vec q$:
$$
r_1) \qquad \begin{pmatrix}
x\\y\\z
\end{pmatrix}=\begin{pmatrix}
2\\6\\-9
\end{pmatrix}+t\begin{pmatrix}
3\\4\\-4
\end{pmatrix}
$$
$$
r_2) \qquad \begin{pmatrix}
x\\y\\z
\end{pmatrix}=\begin{pmatrix}
-1\\-2\\3
\end{pmatrix}+t\begin{pmatrix}
2\\-6\\1
\end{pmatrix}
$$
than, noted the the two lines are not parallel nor intersecting, use the formula from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 2
} |
Solve the area on quadrant one given by the two curves $x^2+y^2=9$ and $ 4x^2+3y^2=36$
I'm having difficulties solving this, but I was told the answer is $\frac{3}{4}\pi$
However, when I equate the two and get $x = 3$ I try to solve them but can't get the correct answer after integration. I was thinking if these are like $2$ circle problem and then apply definite integrals but then again I can't get the correct answer.
| The required region (shaded portion) is given below:
The upper curve is given by $y=\sqrt{\frac{36-4x^2}{3}}$ and the lower curve is $y=\sqrt{9-x^2}$. Thus, the area is given by
$$A=\int_{0}^3\sqrt{\frac{36-4x^2}{3}}dx-\frac{\pi 3^2}{4}$$
Using Wolfram Alpha,
$$\int_{0}^3\sqrt{\frac{36-4x^2}{3}}dx=\frac{3\sqrt{3}\pi}{2}$$ but as you have noted the integral is $\frac{\text{area of the ellipse}}{4}=\frac{\pi(3)\sqrt{12}}{4}=\frac{3\sqrt{3}\pi}{2}$. Hence, the required area is
$$A=\frac{3\sqrt{3}\pi}{2}-\frac{9\pi}{4}=\frac{(6\sqrt{3}-9)\pi}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$1^2$+$2^2$-$2×3^2$+$4^2$+$5^2$-$2×6^2$+$...$+$9997^2$+$9998^2$-$2×9999^2$ ∵ For any positive integer $n$
$n^2-2(n+1)^2+(n+2)^2$
= $n^2-2n^2-4n-2+n^2+4n+4=2$
∵ $1^2$+$2^2$-$2×3^2$+$4^2$+$5^2$-$2×6^2$+$...$+$9997^2$+$9998^2$-$2×9999^2$
= $1^2+(2^2-2×3^2+4^2)+(5^2-2×6^2+7^2)+...+(9995^2-2×9996^2+9997^2)+9998^2-2×9999^2$
= $1+(2+2+...+2)+9998^2-2×9999^2$
I do not know how many lots of two will be inside the aforementioned parenthesis.
| Filling a gap for the benefit of other readers:
$$\begin{align}
(n-1)^2 -2n^2 + (n+1)^2 &= (n^2-2n+1) -2n^2+(n^2+2n+1) \\
&=2n^2-2n^2+2n-2n+2 \\
&=2
\end{align}$$
or:
$$\begin{align}
(n-1)^2 -2n^2 + (n+1)^2 &= ((n+1)^2-n^2) - (n^2-(n-1)^2) \\
&=(n+1+n)(n+1-n) - (n+n-1)(n-(n-1)) \\
&=(2n+1) - (2n-1) \\
&=2
\end{align}$$
or simply "the difference between squares increases by two each time"
And the accumulated number of $2$s from this will be one-third the number of terms involved in these equations, $9996/3$.
Revisiting this, we could use another grouping. The alternative pattern repeat of interest is:
$\begin{align}(n-2)^2+(n-1)^2-2n^2 &= n^2-4n+4 +n^2-2n+1 -2n^2 \\
&=-6n+5
\end{align}$
And the $n$ values here have the form $3k$, giving the full sum as
$$\begin{align}
\sum_{i=1}^{3333}(5-6(3i)) &= 3333\cdot5-18 \frac{3333\cdot 3334}{2} \\
&=3333(5-30006) = 3333\times -30001 \\
&=-99993333
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Geometric transformation (symmetric point to line) I'm trying my own exercise:
Let $(ε):y=3x-4$ be a line in $\Bbb R^2$ given in cartesian coordinates. Find the geometric transformation that sends every point in $\Bbb R^2$ to it's symmetrical as to the line $(ε)$.
My attempt:
Let $$\begin{array}\\f_1: \Bbb R^2 \rightarrow \Bbb R^2 \\ P \rightarrow f_1(P)=P'(x',y'): \\ x'=x\\y'=y-4 \end{array}$$(creates a new coordinate system $4$ units down the $y$-axis).
Thus, $(ε)$ in $O'x'y'$ will be $y'=3x'$
Let $$\begin{array}\\ f_2: \Bbb R^2 \rightarrow \Bbb R^2 \\ P' \rightarrow f_2(P')=P''(x'',y'') \end{array}$$ (rotation of the coordinate system by angle $θ$ clockwise)
$$\begin{pmatrix}x''\\ y''\end{pmatrix}=\begin{pmatrix}cos(\frac{\pi}{2}-arctan(3)) & sin(\frac{\pi}{2}-arctan(3)) \\ -sin(\frac{\pi}{2}-arctan(3)) & cos(\frac{\pi}{2}-arctan(3))\end{pmatrix}\begin{pmatrix}x'\\ y'\end{pmatrix}$$
So $$\begin{array}\\ f: \Bbb R^2 \rightarrow \Bbb R^2 \\ P \rightarrow f(P)=P''(x'',y''): \\ x''=xcosθ+(y-4)sinθ \\ y''=-xsin+(y-4)cosθ \end{array}$$ (the final transformation, where $θ=\frac{\pi}{2}-arctan(3)$).
So I've found the symmetric of the point $(1,4)$ using GeoGebra, which is $(4,3)$.
And I used this to test my answer but it is giving me a wrong result. I can't seem to find where the mistake is. Thanks in advance!
| Let us look for a solution of the form:
$$\tag{1}\begin{pmatrix}x'\\ y'\end{pmatrix}=\begin{pmatrix}a & \ \ b \\b & -a\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}+\begin{pmatrix}e\\ f\end{pmatrix}$$
(see Edit 2 below)
(a symmetry matrix is orthogonal with determinant -1)
It suffices to write that points of intersection of the line with axes, i.e.,
$$\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ -4\end{pmatrix} \ \ \text{and} \ \ \begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}-\tfrac{4}{3}\\ \ \ 0\end{pmatrix}$$
are invariant through this transformation, which gives, by plugging these expressions into (1), four equations with the 4 unknowns $a,b,c,d$:
$$\begin{cases}-4b+c&=&0\\ \ \ 4a+d&=&-4\\ \ \ \tfrac{4}{3}a+c&=&\tfrac{4}{3}\\ \ \ \tfrac{4}{3}b+d&=&0\end{cases}$$
out of which we obtain:
$$a=-4/5, b=3/5, c=12/5, d=-4/5.$$
Edit 1: Another solution (closer to what you have done).
We could have considered the succession of operations
$$\tag{3}S = T^{-1} \circ R^{-1} \circ S' \circ R \circ T \ \ \ \text{where}$$
*
*T is the translation with vector $\binom{4}{0}$ bringing line with equation $y=3x-4$ onto line (D) with equation $y=3x$.
*R is the rotation with angle -atan$(3)$ mapping (D) onto $x$ axis.
(we will denote by c and s resp. the cosine and sine of atan$(3)$):
*S' is the (very simple) symmetry with respect to the $x$ axis.
Let us now give (3) under a matricial form:
$$\binom{x'}{y'}=\binom{0}{-4}+\underbrace{\begin{pmatrix}c&-s\\s&c\end{pmatrix}\begin{pmatrix}1& \ \ 0\\0&-1\end{pmatrix}\begin{pmatrix}c&s\\-s&c\end{pmatrix}}_{\begin{pmatrix}(c^2-s^2)&2cs\\2cs&-(c^2-s^2)\end{pmatrix}}(\binom{x}{y}+\binom{0}{4})$$
(the form we find here has exactly the same structure as the solution of Emilio Novati ; recall formulas $\cos^2a-\sin^2a=\cos(2a)$ and $2 \cos a \sin a = \sin(2a)$).
It is essential to understand that, in (3), you have to "follow the same way back" in the reverse order, of course.
Edit 2: There are 2 kinds of plane (vectorial) isometries : rotations, and symmetries with resp. expressions:
$$\tag{2}\begin{pmatrix}a & \ \ b \\-b & \ \ a\end{pmatrix} \ \ \ \text{and} \ \ \ \ \begin{pmatrix}a & \ \ b \\ \ \ b & -a\end{pmatrix} \ \ \ \text{with} \ \ \ a^2+b^2=1.$$
Why that ? Because an isometry has an orthogonal matrix, and an orthogonal matrix must be such that its column constitute an orthonormal basis.
Clearly, the first column is arbitrary. Let us denote by $\binom{a'}{b'}$ the second column of the matrix of an isometry. It must be orthogonal to the first one, i.e., dot product $\binom{a'}{b'} . \binom{a}{b}=0 \ \iff aa'+bb'=0$ with $a'^2+b'^2=1$. There are two solutions: $\binom{a'}{b'} = \binom{-b}{a}$ (direct orientation basis) or $\binom{a'}{b'} = \binom{b}{-a}$ (indirect orientation), whence formulas (2).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Setting up inequalities is proving difficult Given $$(a+b)^2\leq4$$
$$(a-b)^2\leq4$$
Is this true?
$$2(a+b)^2+2(a-b)^2-b^2\leq16$$
The problem I need to solve is to find the maximum value of $(4a^2+3b^2)$ Please help.
| $(a+b)^2 \le 4$ so $2(a+b)^2 \le 8$.
$(a-b)^2 \le 4$ so $2(a-b)^2 \le 8$.
So $2(a+b)^2 + 2(a+b)^2 \le 8+8 = 16$.
And $b^2 \ge 0$ so
$2(a+b)^2 + 2(a+b)^2-b^2 \le 2(a+b)^2 + 2(a+b)^2 \le 16$.
$16 \ge 2(a+b)^2 + 2(a+b)^2-b^2 = 2(a^2 + 2ab +b^2) + 2(a^2 -2ab + b^2) -b^2 = 4a^2 + 3b^2$. So $16$ is a potential maximum. But we don't know if it is an achievable maximum.
But if $(a+b)^2 = 4$ its maximum; and $(a -b)^2=4$ its maximum; and $b^2 = 0$; it's minimum then $4a^2 + 3b^2$ will meet its maximum of $16$.
$b^2 = 0 \implies b=0$ and so $(a+b)^2 = a^2 = (a-b)^2 =4$ which implies $a=\pm 2$ allow $4a^2 + 3b^2$ to equal a maximum possible value of $16$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
sum of two determinants If
\begin{vmatrix}a&c\\b&d\end{vmatrix} and \begin{vmatrix}a&e\\b&f\end{vmatrix}, then sum of these determinant can be written as in terms of another determinant given by
\begin{vmatrix}a&c+e\\b&d+f\end{vmatrix}
is it right?
| The determinant is a multilinear function of its columns and so
$$
\begin{vmatrix}a&c+e\\b&d+f\end{vmatrix}
=\begin{vmatrix}a&c\\b&d\end{vmatrix} + \begin{vmatrix}a&e\\b&f\end{vmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to know if equation adds more info to a set of equations? I'm looking for a formalized way to be able to know whether adding an equation to a set of linear equations adds new information or not, does anyone know of a way?
For a really simple case, let's say I have these equations:
$A = 3\\
B+C = 4\\
D = 8
$
Obviously, the equation below doesn't add any new information:
$-A = -3$
But it gets less obvious when adding an equation like this:
$B + D = 6$
Or:
$C = 2$
When the list of equations is longer, it gets a lot less obvious to me as well:
$
A=3\\
B+C=4\\
D=8\\
C=4\\
F=3\\
E=12\\
F+G=4\\
H=1
$
If I want to add a new equation:
$D+E=7$
I can see that $D$ and $E$ are already fully specified, so this doesn't really add any new information, but a more complex expression would be harder for me to rule out.
Is there a formalized way to see whether or not adding a new equation will add more information? Possibly by looking at the matrix form of the equations?
Here is the last group of equations:
$
\begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{bmatrix}
*
\begin{bmatrix}
A \\
B \\
C \\
D \\
E \\
F \\
G \\
H \\
\end{bmatrix}
=
\begin{bmatrix}
3 \\
4 \\
8 \\
4 \\
3 \\
12 \\
4 \\
1 \\
\end{bmatrix}
$
And here is the equation I wanted to add to the set:
$
\begin{bmatrix}
0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\
\end{bmatrix}
*
\begin{bmatrix}
A \\
B \\
C \\
D \\
E \\
F \\
G \\
H \\
\end{bmatrix}
=
7
$
Thanks for any help you can provide!
| Given a set of equations,
$$
\begin{cases}
A+B = 5\\
2D = 3\\
A + B + C + 3D = 10,
\end{cases}
$$
You can identify it with a matrix like this,
$$
\begin{bmatrix}
1 & 1 & 0 & 0 & 5 \\
0 & 0 & 0 & 2 & 3 \\
1 & 1 & 1 & 3 & 10 \\
\end{bmatrix}
$$
Notice that each column of the matrix corresponds to each variable A, B, C and D in order, the last column corresponds to the result of the equation, and each number in the rows (except for the results of the equations) correspond to the integer coefficient that is next to each variable. For example the first row corresponds to the equation $A+B = 5$ that is the same as $A + B + 0C + 0D = 5$ so we should write in the first row 1 1 0 0 5.
This association is very useful because the matrix has the same properties as the system of equations when you want to add equations is the same as adding rows, so you already know that when two equations have the same "information" then the can "cancel each other out". For example,
\begin{cases}
Z+Y = 1\\
-5Z-5Y=-5,\\
\end{cases}
Divide the second equation by 5,
\begin{cases}
Z+Y = 1\\
-Z-Y=-1\\
\end{cases}
Sum the first one to the last one,
\begin{cases}
Z+Y = 1\\
-Z-Y + Z+ Y=-1 + 1\\
\end{cases}
We get,
\begin{cases}
Z+Y = 1\\
0=0\\
\end{cases}
And we already know that so we end with,
\begin{cases}
Z+Y = 1\\
\end{cases}
But this method also works with matrices,
We associate a matrix to the system of equations and do the same "operations",
\begin{cases}
Z+Y = 1\\
-5Z-5Y=-5,\\
\end{cases}
$\hspace{87 mm} \equiv$
\begin{bmatrix}
1 & 1 & 1 \\
-5 & -5 & -5 \\
\end{bmatrix}
We divide the last row by 5,
\begin{bmatrix}
1 & 1 & 1 \\
-1 & -1 & -1 \\
\end{bmatrix}
the we sum the first row to the last row,
\begin{bmatrix}
1 & 1 & 1 \\
-1 + 1 & -1 + 1& -1 + 1\\
\end{bmatrix}
is the same as,
\begin{bmatrix}
1 & 1 & 1 \\
0 & 0& 0\\
\end{bmatrix}
then we just have that the associated system of equation is the 1x1 matrix:
\begin{bmatrix}
1 & 1 & 1 \\
\end{bmatrix}
We showed that the last equation wasn't adding "new information" because we ended up with
\begin{cases}
Z+Y = 1\\
\end{cases}
Finally, what we have done with the associated matrix is to check if its rows are linearly independent, then two equations have "different information" if and only if their "correspondent rows" are linearly independent in their associated matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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$c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$ so prove $e - c_n \le \frac{1}{n! * n}$ $c_n = 1 + \frac{1}{1!} + \frac{1}{2!} +...+\frac{1}{n!}$, so $e - c_n \le \frac{1}{n! * n}$
I absolutely have no idea how to solve it, could anyone tell me the approach?
| Hint. Using the infinite summation representation of $e$, we have
$$ e - c_n = \sum_{r=0}^\infty \frac1{r!} - \sum_{r=0}^n \frac1{r!} = \sum_{r=n+1}^\infty \frac1{r!} = \frac1{n!}\left(\frac1{n+1} + \frac1{(n+1)(n+2)} + \cdots \right). $$
All that's left to do now is to show that the sum in brackets $()$ is $\le 1/n$.
As the OP pointed out in the comment below, we simply use that $n+r > n+1$ for $r > 1$, and hence
$$ \left(\frac1{n+1} + \frac1{(n+1)(n+2)} + \cdots \right) < \left( \frac1{n+1} + \frac1{(n+1)^2} + \frac1{(n+1)^3} \cdots \right) $$
$$ = \frac{\frac1{n+1}}{1-\frac1{n+1}} = \frac1{n+1-1} = \frac1n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Area of a right triangle if $a/b$ is $1.05$ and the difference between the radii of the circumscribed and inscribed circles is $17$? What's the area of a right triangle if the quotient of its legs is $1.05$ and the difference between the radii of the inscribed and circumscribed circles is $17$?
I've been trying to solve this and I've got:
($R$ - radius of circumscribed circle, $r$ - radius of inscribed circle)
$1.$ $ \frac{a}{b}=1.05$
$2.$ $c^2=a^2+b^2$
$3.$ $a + b - 2r = c$
$4.$ $c-2r=34$
$5.$ $ab=(a+b+c)r$
Using the first four equations, I can substitute for one of the legs from $1.$ and for $r$ through $4.$ which leaves me with
$b(2.05)-2c=34$
$c=b\sqrt{1.05^2+1}$
However, solving this simply evades me, as I don't find myself getting rid of the square root which I don't know how to calculate.
I do know my equations give the right answer so I'm probably missing a simpler way to solve the system of equations.
Help much appreciated.
| The inradius of a right triangle in terms of its sides is given by the relationship $$r \frac{a+b+c}{2} = rs =|\triangle ABC| = \frac{ab}{2},$$ hence $$r = \frac{ab}{a+b+c}.$$ The circumradius is trivially $$R = \frac{c}{2}.$$ Thus the given conditions may be summarized as $$17 = R - r = \frac{c}{2} - \frac{ab}{a+b+c} = \frac{(a+b)c + c^2 - 2ab}{2(a+b+c)} = \frac{(a+b)c + (a-b)^2}{2(a+b+c)}, \\ \frac{a}{b} = \frac{21}{20}, \\ a^2 + b^2 = c^2.$$ We note that because $21^2 + 20^2 = 29^2$, we have $$a : b : c \equiv 21 : 20 : 29,$$ thus the computation is greatly simplified by letting $a = 21k$, $b = 20k$, $c = 29k$, to obtain from the first equation $$17(2)(70)k = (41)(29)k^2 + (21-20)^2 k^2,$$ or $$2380k = 1190k^2,$$ or $k = 2$. Thus the desired triangle is $(a,b,c) = (42, 40, 58)$ and its area is $$|\triangle ABC| = 840.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Calculate $\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{2n+1}\right)$. Calculate
$$\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{2n+1}\right)$$
Question: I want to verify that my next attempt is correct, I do it too exhausted and in that state I do not trust my abilities.
My attempt: Note that $$\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{2n+1}\right)=\mathbb{Re}\left(\sum_{k=1}^{n}e^{\frac{2\pi k i}{2n+1}}\right).$$
In this sense, we know that
$$\begin{array}{rcl}\sum_{k=1}^{n}e^{\frac{2\pi k i}{2n+1}}&=&{\displaystyle \sum_{k=1}^{n}\left(e^{\frac{2\pi i}{2n+1}}\right)^{k} } \\ &=& {\displaystyle\frac{e^{\frac{2\pi i}{2n+1}}-\left(e^{\frac{2\pi i}{2n+1}}\right)^{n+1}}{1-e^{\frac{2\pi i}{2n+1}}} } \\
&=& {\displaystyle\frac{ \left(e^{\frac{2\pi i}{2n+1}}-e^{\frac{2(n+1)\pi i}{2n+1}}\right)\left( 1-e^{\frac{-2\pi i}{2n+1}} \right) }{\left(1-e^{\frac{2\pi i}{2n+1}}\right)\left(1-e^{\frac{-2\pi i}{2n+1}}\right) } } \\
&=& {\displaystyle\frac{e^{\frac{2\pi }{2n+1}}-e^{\frac{2\pi (n+1) }{2n+1}} -1 +e^{\frac{2\pi n }{2n+1}} }{2-\cos\left( \frac{2\pi }{2n+1} \right)} }
\end{array}$$
Therefore, we have
$${\mathbb{Re}\left(\sum_{k=1}^{n}e^{\frac{2\pi k i}{2n+1}}\right)=\displaystyle\frac{\cos\left(\frac{2\pi }{2n+1}\right)-\cos\left(\frac{2\pi (n+1) }{2n+1}\right) -1 +\cos\left(\frac{2\pi n }{2n+1}\right) }{2-\cos\left( \frac{2\pi }{2n+1} \right)} }. $$
Hence, we can conclude
$$\lim_{n\rightarrow \infty}\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{2n+1}\right)=\lim_{n\rightarrow \infty}\mathbb{Re}\left(\sum_{k=1}^{n}e^{\frac{2\pi k i}{2n+1}}\right)=\displaystyle\frac{\cos\left(0\right)-\cos\left(\pi\right) -1 +\cos\left(0\right) }{2-\cos\left( 0 \right)}=0 .$$
| For all $n\in\Bbb{Z}^+ $,
$$
\sum_1^n \cos\frac{2\pi k}{2n+1} = -\frac12$$
Proof:
$$
\sum_1^{2n+1} \cos\frac{2\pi k}{2n+1} = 0$$
because it is the real part of
$$
S=\sum_1^{2n+1} e^{i\frac{2\pi k}{2n+1}}$$
and $S$ can be multiplied by any $e^{i\frac{2\pi r}{2n+1}}$ for integer $r$ with the effect of just rotating the entries in the sum, thus not changing the sum; so $S$ is a number which when multiplied by different complex quantities remains the same, thus $S=0$.
$$
\sum_1^{2n+1} \cos\frac{2\pi k}{2n+1} = \sum_1^{n} \cos\frac{2\pi k}{2n+1}
+ \sum_{n+1}^{2n} \cos\frac{2\pi k}{2n+1} + \cos\frac{2\pi (2n+1)}{2n+1}
$$
Now since $\cos(\pi-x)=\cos(\pi+x)$,
$$
0=\sum_1^{2n+1} \cos\frac{2\pi k}{2n+1} = 2\sum_1^{n} \cos\frac{2\pi k}{2n+1}
+ \cos(2\pi) = 2\sum_1^{n} \cos\frac{2\pi k}{2n+1}
+ 1\\
\sum_1^{n} \cos\frac{2\pi k}{2n+1} = -\frac12
$$
So your limit is in fact $-\frac12$ and as I commented, it approaches it very rapidly (immediately at $n=1$.
Son' task where @Don Antonio's reasoning went astray...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$
Integrate: $\int (x^2+a^2)^{-3/2} \cdot dx$
My Approach:
$\int (x^2+a^2)^{-3/2} \cdot dx$
$\int (x^2+a^2)^{-3/2} \cdot d(a^2+x^2)\cdot \frac{dx}{d(x^2+a^2)}$
But this doesn't give the right answer. I showed this to my friend and he said $d(x^2+a^2)$ is not possible which makes sense since you can't take a small element of the form $(x^2+a^2)$. How can I then solve this integration without using trigonometry?
| We use a trick to get a reduction formula:
$$ \frac{1}{(x^2+a^2)^{3/2}} = \frac{1}{a^2} \frac{x^2+a^2-x^2}{(x^2+a^2)^{3/2}} = \frac{1}{a^2} \frac{1}{\sqrt{x^2+a^2}} - \frac{1}{a^2}\frac{x^2}{(x^2+a^2)^{3/2}}. $$
The first term is lower-order, the second can be integrated by parts:
$$ \int \frac{x^2}{(x^2+a^2)^{3/2}} \, dx = -\frac{x}{\sqrt{x^2+a^2}} + \int \frac{dx}{\sqrt{x^2+a^2}}. $$
Oh, but now the new integral on the right here cancels with the integral of the first fraction above! Hence
$$ \int \frac{dx}{(x^2+a^2)^{3/2}} = \frac{1}{a^2} \frac{x}{\sqrt{x^2+a^2}} +C, $$
and it's easy to verify this by differentiating.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Values for $a$ and $b$ in $y=\cos(x)+a\cos(bx)$ such that every real value for $x$ has either a positive or $0$ value for $y$ If there is a function in the form $y=\cos(x)+a\cos(bx)$ do there exists real number values for $a$ and $b$ such that for every real number value for $x$ there is either a positive number value for $y$ or a $0$ value for $y$?
| Let $f(x) = \cos(x)+a\cos(bx)$, and suppose $f(x) \ge 0$ for all $x \in \mathbb{R}$.
First suppose $b=0$. Then $f(x) = \cos(x) + a,$ which has range $[a-1,\infty).$
Thus, for $b=0$, we get the solution pairs $(a,b) = (t,0),\,$ for any $t \ge 1.$
Next, suppose $b \ne 0$.
If $|a| > 1$, then when $a\cos(bx)$ realizes a value of $-a$, $f(x)$ will be negative, contradiction.
Similarly, if $|a| < 1$, then when $\cos(x)$ realizes a value of $-1$, $f(x)$ will be negative, contradiction.
It follows that $|a| = 1,$ hence either
\begin{align*}
f(x) &= \cos(x) + \cos(bx)\\[4pt]
&\,\text{or}\\[4pt]
f(x) &= \cos(x) - \cos(bx)\\[4pt]
\end{align*}
First suppose $f(x) = \cos(x) + \cos(bx)$.
Then when one of the summands realizes a value of $-1$, the other must realize a value of $+1$ to compensate. But that means each of the summands has a period which is a multiple of the other. It follows that periods are equal, hence $b = \pm 1$. But $\cos\,$ is an even function hence, $b = \pm 1$ implies $\cos(bx) = \cos(x)$. But then $f(x) = 2\cos(x)$, contradiction, since then $f(\pi) < 0$.
Next suppose $f(x) = \cos(x) - \cos(bx)$.
If $x_0$ is such that $\cos(x_0) = -1$, then $\cos(bx_0)$ must also be $-1$, else $f(x_0) < 0$.
It follows that the period of $\cos(bx)$ divides the period of $\cos(x)$.
Similarly, if $x_0$ is such that $\cos(bx_0) = 1$, $\cos(x_0)$ must also be $1$, else $f(x_0) < 0$.
It follows that the period of $\cos(x)$ divides the period of $\cos(bx)$.
Therefore the periods must be equal, hence $b = \pm 1$. But then, since $\cos$ is an even function, $b = \pm 1$ implies $\cos(bx) = \cos(x)$, hence $f(x)$ is identically $0$.
This gives the solution pairs $(a,b) = (-1,\pm 1).$
To summarize, we have the solution pairs
\begin{align*}
(a,b) &= (-1,1)\\[4pt]
(a,b) &= (-1,-1)\\[4pt]
(a,b) &= (t,0),\,\text{for any $t \ge 1$}\\[4pt]
\end{align*}
and those are the only solutions.
| {
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"url": "https://math.stackexchange.com/questions/2223465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
The value of $\prod_2^\infty \left(1-2/(n(n+1))\right)$ I'm trying to find the value of $\prod_2^\infty \left(1-2/(n(n+1))\right)$. So far I have the following.
\begin{align*}
\prod_2^\infty \left(1-2/(n(n+1))\right) &= (1-2/6)*(1-2/12)*(1-2/20)*(1-2/30)... \\&= (1-1/3)*(1-1/6)*(1-1/10)*(1-1/15)... \\&= (2/3)*(5/6)*(9/10)*(14/15)... \\&= (2/3) * ((1*5)/(2*3)) * ((3*3)/(5*2)) * ((2*7)/(3*5))...
\end{align*}
The partial product with just the first four terms equals $(1/3)*(7/5)$ because almost everything cancels out. I claim that the term which $(1/3)$ gets multiplied by goes to zero since it goes from $5/3$ to $3/2$ to $7/5$... (making the value $1/3$), but I don't know the explicit formula for each factor of each term so I can't prove that. Does anyone know the explicit formula for the factors?
| $$\prod _{ 2 }^{ \infty } \left( 1-\frac { 2 }{ n\left( n+1 \right) } \right) =\prod _{ 2 }^{ \infty } \left( \frac { { n }^{ 2 }+n-2 }{ n\left( n+1 \right) } \right) =\prod _{ 2 }^{ \infty } \left( \frac { \left( n+2 \right) \left( n-1 \right) }{ n\left( n+1 \right) } \right) =\\ =\prod _{ 2 }^{ \infty } \frac { n+2 }{ n+1 } \cdot \prod _{ 2 }^{ \infty } \frac { n-1 }{ n } =\\ =\left( \frac { 4 }{ 3 } \cdot \frac { 5 }{ 4 } \cdot \frac { 6 }{ 5 } .... \right) \left( \frac { 1 }{ 2 } \cdot \frac { 2 }{ 3 } \cdot \frac { 3 }{ 4 } ... \right) =\color{blue}{\frac { 1 }{ 3 }} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to solve $a_n = 2a_{n-1} + 1, a_0 = 0, a_1 = 1$? $a_n = 2a_{n-1} + 1, a_0 = 0, a_1 = 1$
So to get the closed form of this recurrence relation, I would usually try to get it into the form of $a_n = ra_{n-1}$ and then $a_n = r^na_0$. But what am I supposed to do with the $1$?
Thanks!
| $a_2=2\cdot 1+1=3=2^2-1,\quad a_3=2\cdot 3+1=7=2^3-1,\quad a_4=2\cdot7+1=15=2^4-1,\quad a_5=2\cdot 15+1=31=2^5-1$
So in general we guess that $a_{n}=2^n-1,\quad n\in\mathbb{N}$
So we know it's true for $n=0$
Assume true for $n$, now we must prove true for $n+1$:
$a_{n+1} = 2\cdot a_n+1=2\cdot(2^n-1)+1=2^{n+1}+1$ as required, hence true for $n+1$, hence true $\forall n\in\mathbb{N}$
Therefore we conclude that $a_n=2^n-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Find the equation of the tangent line to the circle $3x^2+3y^2+3z^2-2x-3y-4z=22;3x+4y+5z=26$ at the point $(1,2,3)$. I try take the centre of the sphere, then equation of the perpendicular line joining the centre of circle and the centre of sphere, then find the centre of circle.Tell me if I am going right way?
| Let $(x,y,z)=(1,2,3)+t(a,b,c)$ is an equation of out tangent line.
Thus, $3a+4b+5c=0$.
Rewrite an equation of our sphere in the following form:
$$\left(x-\frac{1}{3}\right)^2+\left(y-\frac{1}{2}\right)^2+\left(z-\frac{2}{3}\right)^2=\frac{293}{36}$$
Thus, $\left(1-\frac{1}{3}\right)a+\left(2-\frac{1}{2}\right)b+\left(3-\frac{2}{3}\right)c=0$ or $4a+9b+14c=0$,
which with $3a+4b+5c=0$ gives $b=-2c$ and $a=c$ and we get the answer:
$$(x,y,z)=(1,2,3)+t(1,-2,1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Generators of $GL_2(\mathbb{Q}_p)$ A well known fact is that the group $GL_2(\mathbb{Q}_p)$ is generated by the following matrices:
$1) \text{ }
w=
\left( {\begin{array}{cc}
0 & 1 \\
1 & 0 \\
\end{array} } \right)
$
$2) \text{ } \mathbb{Q}_p^{\star}
\left( {\begin{array}{cc}
1 & 0 \\
0 & 1 \\
\end{array} } \right)
$
$3) \text{ }
\left( {\begin{array}{cc}
\mathbb{Z}_p^{\star} & 0 \\
0 & 1 \\
\end{array} } \right)
$
$4) \text{ }
\left( {\begin{array}{cc}
p & 0 \\
0 & 1 \\
\end{array} } \right)
$
$5) \text{ }
\left( {\begin{array}{cc}
1 & p\mathbb{Z}_p \\
0 & 1 \\
\end{array} } \right)
$
I need a reference for the above fact. In particular, for $b \in \mathbb{Q}_p$, I am trying to write lower unipotent elements $
\left( {\begin{array}{cc}
1 & 0 \\
b & 1 \\
\end{array} } \right)
$ in terms of the above matrices. Any help is welcome. Thanks.
| We first note that multiplication on the left by $(1)$ switches the rows of a matrix and multiplication on the right by $(1)$ switches the columns call these observations $(*)$. Using both parts of $(*)$ we see conjugation by $(1)$ transforms $\begin{pmatrix} 1 & b \\ 0 & 1\end{pmatrix}$ to $\begin{pmatrix} 1 & 0 \\ b & 1\end{pmatrix}$ so it is sufficient to make the upper-triangular matrix $\begin{pmatrix} 1 & b \\ 0 & 1\end{pmatrix}$.
First the case $b\in p\Bbb Z_p$ let $b= up^k$ for $u\in\Bbb Z_p^*$ and $k>0$. Then as $\begin{pmatrix} 1 & up \\ 0 & 1\end{pmatrix}^n = \begin{pmatrix} 1 & nup \\ 0 & 1\end{pmatrix}$ we write $\begin{pmatrix} 1 & up^k \\ 0 & 1\end{pmatrix}=\begin{pmatrix} 1 & up \\ 0 & 1\end{pmatrix}^{p^{k-1}}$. So when the dust settles your expression is just
$$\begin{pmatrix} 1 & 0 \\ b & 1\end{pmatrix}=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix}1 & up \\ 0 & 1\end{pmatrix}^{p^{k-1}}\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$$
Now when $k\le 0$, we just need to scale a bit. Note that multiplication by a multiple of the identity just scales the matrix, so if $b=up^{k}$, with $k\le 0$ we can write $\begin{pmatrix} 1 & up \\ 0 & 1\end{pmatrix}\begin{pmatrix}p^{k-1} & 0 \\ 0 & p^{k-1}\end{pmatrix}= \begin{pmatrix}p^{k-1} & b \\0 & p^{k-1}\end{pmatrix}$ using operation $(2)$. Then to scale this back down, first use $(*)$ to switch the top and bottom rows, then multiply by $\begin{pmatrix} p^{1-k} & 0 \\ 0 & 1\end{pmatrix}$ with $(4)$ and use $(*)$ again to switch the rows back and we have $\begin{pmatrix} p^{k-1} & b \\ 0 & 1\end{pmatrix}$. Now conjugate by $(1)$ to give $\begin{pmatrix} p^{k-1} & 0 \\ b & 1\end{pmatrix}$ and multiply by $\begin{pmatrix} p^{1-k} & 0 \\0 & 1\end{pmatrix}$ to arrive at the finished product.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Limit of $S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$ when $n\to\infty$ $$S_n = \frac{1}{2n + 1} + \frac{1}{2n + 3} + ... + \frac{1}{4n - 1}$$
the task is to find $$\lim_{n \to \infty} S_n$$
I've tried different ways, but all I could do is to make an estimation that the limit is somewhere between 0.5 and 1, but that's not the exact answer.
Manually doing fist elements gives
$$\frac{1}{3}, \quad \frac{1}{5} + \frac{1}{7}, \quad \frac{1}{7} + \frac{1}{9} + \frac{1}{11}, \quad ...$$
I'm trying to restate it as
$$\lim_{n \to \infty} \bigg[ \sum_{i = 1}^{\infty} \frac{1}{2i - 1} - \sum_{i = 1}^{n} \frac{1}{2i - 1} - \sum_{i = 4n}^{\infty} \frac{1}{2i - 1} \bigg]$$
so that I cut off the beginning of the series and it's tail.
| The sum can be written as
$$S(n) = \sum_{k=n}^{2n} \frac{1}{2k-1}$$
If we replace the sum by an integral we find
$$I(n) = \int_n^{2 n} \frac{1}{2 k-1} \, dk = \frac{1}{2} \log (2+\frac{1}{2n-1})$$
And the limit $n\to\infty$ is
$$S(n\to\infty) = \frac{1}{2} \log (2)\tag{1}$$
The justification of the replacement of the sum by an integral is easily done considering the inequalities (lower "sum" < S < upper "sum")
$$\int_n^{2 n} \frac{1}{2 k-1} \, dk < S(n) < \int_n^{2 n} \frac{1}{2 (k-1)-1} \, dk $$
which are obvious graphically.
Hence we have
$$ \frac{1}{2} \log \left(2+\frac{1}{2n-1}\right)< S(n) < \frac{1}{2} \log \left(2 + \frac{3}{2 n-3}\right)$$
and the limit $n\to\infty$ gives back the result (1).
Generalization
It is tempting to consider the more general sum (where $0<a<b, c<0, d<a \;c \;n$)
$$S(a,b,c,d,n)=\sum _{k=a\; n}^{b\; n} \frac{1}{c\; k-d}$$
The limit is similarly calculated with the result
$$\lim_{n\to \infty } \, S(a,b,c,d,n)=\frac{1}{c}\;\log \left(\frac{b}{a}\right)$$
Notice the independence of $d$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Area of equilateral triangle inscribed in right triangle An Isosceles right triangle $ABC$ , $AB=BC= 4 cm$
Point $p$ is a midpoint of $BC$ , points $q$, $s$ lies on $AC$,$AB$ respectively , such that the triangle $pqs$ is an equilateral triangle ;
what is the area of triangle $pqs$
| Based on your description, let's define:
$$ \begin{align}
\mbox{Hypotenuse AC}:\; y &= -x +4 \\
p &= (2,0) \\
q &= (x, y=-x+4) \\
s & = (0,h)
\end{align}$$
The three sides are equal, hence:
$$ \begin{align}
\mbox{sp = pq} \Rightarrow
4+h^2 &= (x-2)^2 + (x-4)^2 \quad (1)\\
\mbox{sp = sq} \Rightarrow
4+h^2 &= x^2 + (h+x-4)^2, \quad(2) \\
\mbox{pq = sq} \Rightarrow
(x-2)^2 + (x-4)^2 &= x^2 + (h+x-4)^2. \quad(3)\\
\end{align}$$
There are 3 equations with only 2 variables $x$ and $h$ but only two equations are needed to solve for the variables. Let's say we solve $(2)$ and $(3)$ for $x$ and $h$. Then we plug the solution in to $(1)$ and if $(1)$ is not satisfied, then the problem has no solution.
From $(3)$:
$$ x = \frac{4 + 8h -h^2}{4 + 2h}. \quad(4) $$
Plugging $(4)$ into $(2)$ will result in an equation in $h$ alone. This equation can be solved numerically. Once $h$ is obtained, $x$ will be found by $(4)$.
The next step is to plug the results for $x$ and $h$ into $(1)$. If the equality fails, the problem has no solution.
If the problem has a solution, then the area of the triangle $spq$ is given by
$$ A = \frac{\sqrt3}{4}(sp)^2 = \frac{\sqrt3}{4}(4+h^2). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230684",
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"source": "stackexchange",
"question_score": "8",
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"answer_id": 1
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Find tan(C/2) in The triangle ABC . $ABC$ is a triangle.
$\tan\frac{A}{2} = 0.5$
$\tan\frac{B}{2} =\frac{1}{3}$.
Find $\tan\frac{C}{2}$.
I tried to find it :
when $A+B+C = 180^{\circ}$
So $\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = 90^{\circ}$.
How can I complete on these information?
| ∴ $\dfrac C2 = 90^0 –(\dfrac A2 + \dfrac B2)$
Then, $\tan (\dfrac {C}{2}) = \cot (\dfrac A2 + \dfrac B2) = \dfrac {1}{\tan (\dfrac A2 + \dfrac B2)} = \dfrac {1 -\tan (\dfrac A2) \tan (\dfrac B2)}{\tan (\dfrac A2) + \tan (\dfrac B2)} = ...$
Added:- It is the application of the compound angle formula $\tan (x + y) = \dfrac {\tan x + \tan y}{1 - \tan x \times \tan y}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit proof by definition I have the following limit:
$$\lim_{x \to 6} \frac{x+1}{x-5} = 7$$
How do I prove this equation by the famous definition of the limit? (the one includes delta and epsilon).
I know how to prove simple limits like this one: $$\\\lim_{x \to 6} x-5 = 1
\\ |x - x_0| < \delta \implies |x - 6| < \delta
\\ |f(x) -L| < \epsilon \implies |x-5 -1| < \epsilon \implies |x - 6| < \epsilon
\\ \implies \delta = \epsilon$$
But in the first example I asked about I have:
$$|x-6| < \delta$$ $$|\frac{x+1}{x-5} -7| < \epsilon \implies |\frac{36-6x}{x-5}| < \epsilon$$
don't really know how to proceed, thanks in advance :)
| We have $$\bigg|\frac{x+1}{x-5}-7\bigg|=\bigg|\frac{x+1-7x+35}{x-5}\bigg|=\bigg|\frac{-6x+36}{x-5}\bigg|=6\cdot\bigg|\frac{x-6}{x-5}\bigg|=6\cdot\bigg|\frac{1}{x-5}\bigg|\cdot|x-6|$$
Assume that $|x-6|<\frac{1}{2}$. Then we get
$$-\frac{1}{2}<x-6<\frac{1}{2}$$ and so we get
$$-\frac{1}{2}+1<x-6+1<\frac{1}{2}+1$$ that is, we get
$$0<\frac{1}{2}<x-5<\frac{3}{2}$$ that is, we get
$$0<\frac{1}{x-5}<2.$$
Let $\epsilon>0$.
Define $$\delta=\min\bigg\{\frac{1}{2},\frac{\epsilon}{12}\bigg\}.$$ Then $\delta\leq\frac{1}{2}$ and $\delta\leq\frac{\epsilon}{12}$. Hence, if $0<|x-6|<\delta$ then we get
$$\begin{align}\bigg|\frac{x+1}{x-5}-7\bigg|&=6\cdot\bigg|\frac{1}{x-5}\bigg|\cdot|x-6|\\
&=6\cdot\frac{1}{x-5}\cdot|x-6|\\
&<12\cdot\delta\leq \epsilon.
\end{align}$$
| {
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} |
A proof that $\frac{\log^k(1)+\log^k(2)+\dotsb +\log^k(n)}{1^k+2^k+\dotsb +n^k} \to 0$ Let $\left(a_n\right)$ be the following sequence:
$$a_n =\frac{\log^k\left(1\right)+\log^k\left(2\right)+\dotsb +\log^k\left(n\right)}{1^k+2^k+\dotsb +n^k},$$ for a fixed $k \in \mathbb{N}$. Prove that $a_n \to 0$. There are many proofs for this one, but I think this is an elegant one. I'd like you to check it out, first because I consider it to be a nice and concise and elementary, and secondly because I'd like to make sure that there are no mistakes.
Let's get started: First of all, $a_n\geq 0 \ \forall \,n \in \mathbb{N}$. We write $a_n$ as
$$\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+2^k\left(\frac{\log\left(2\right)}{2}\right)^k+\dotsb +n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}.$$
It is obvious that $\left(\frac{\log\left(n\right)}{n}\right)^k \to 0$. Let $\varepsilon >0$. Then $\exists\, n_0 \in \mathbb{N}$ such that $\forall \, n \geq n_0, \left(\frac{\log\left(n\right)}{n}\right)^k < \varepsilon$.
We have
\begin{align*}
a_n&=\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k+ n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb+n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}\\
&=\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+2^k+\dotsb +n^k} + \frac{n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb+n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+2^k+\dotsb +n^k}\\
&\leq\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+\dotsb +n^k}\\ & \qquad \qquad+ \frac{\varepsilon\left(1^k+\dotsb +\left(n_0-1\right)^k\right)+n_0^k\left(\frac{\log\left(n_0\right)}{n_0}\right)^k+\dotsb +n^k\left(\frac{\log\left(n\right)}{n}\right)^k}{1^k+\dotsb +n^k}\\
&\leq\frac{1^k\left(\frac{\log\left(1\right)}{1}\right)^k+\dotsb +\left(n_0-1\right)^k\left(\frac{\log\left(n_0-1\right)}{n_0-1}\right)^k }{1^k+\dotsb +n^k} + \frac{\varepsilon\left(1^k+2^k+\dotsb +n^k\right)}{1^k+2^k+\dotsb +n^k}\\
&=\frac{1^k+2^k+\dotsb +\left(n_0-1\right)^k}{1^k+2^k+\dotsb +n^k}+\varepsilon.
\end{align*}
Now, by taking the limsup and the liminf as $n \to\infty$, and since
$$\frac{1^k+2^k+\dotsb +\left(n_0-1\right)^k}{1^k+2^k+\dotsb +n^k}\to 0,$$ we have
$$0 \leq \limsup\left(a_n\right) \leq \varepsilon \quad\text{and}\quad 0\leq \liminf\left(a_n\right) \leq \varepsilon.$$ But $\varepsilon$ was arbitrarily small, so
$$\liminf\left(a_n\right)=\limsup\left(a_n\right)=0=\lim\left(a_n\right).$$
This is more of a discussion and not so much of a question :)
| I have perhaps a more elegant proof:
$$\begin{align}a_n&=\frac{\log^k(1)+\dots+\log^k(n)}{1^k+\dots+n^k}\\&<\frac{\log^k(n)+\dots+\log^k(n)}{n^k}\\&=\frac{n\log^k(n)}{n^k}\\&=\frac{\log^k(n)}{n^{k-1}}\\&\to0\end{align}$$
The first step follows from the fact that $\frac ab<\frac cd$ if $c>a$ and $d<b$ for positive numbers $a,b,c,d$.
The limit then follows by letting $n^{k-1}=u^k$, which gives
$$\frac{\log^k(n)}{n^{k-1}}=\left[\frac k{k-1}\frac{\log(u)}u\right]^k$$
and the limit is then taken as given.
| {
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"answer_count": 3,
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} |
Given a $2\times 2$ matrix $A$, compute $A^7$. Let $A =
\begin{pmatrix}
e^{2x} & -1 \\
0 & e^{2x}-1 \\
\end{pmatrix}
$. Compute $A^7$.
I've tried the obvious way of multiplying $A$ with $A$, then $A^2$ with $A^2$, but I arrived at a messy result in the top right member of the matrix. Is there a general form to be noticed here?
| For the 7th power, it pays already off to diagonalize the matrix. It is easy to see that the eigenvalues are $e^{2x}$ and $e^{2x}-1$ with the eigenvectors $(1,0)^T$ and $(1,1)^T$. Thus, we can write $A$ as
$A = T D T^{-1}$
with $$T= \begin{pmatrix}1 &1\\
0 &1 \end{pmatrix}, \quad D= \begin{pmatrix}e^{2x} &0\\
0 &e^{2x}-1 \end{pmatrix}, \quad T^{-1}=\begin{pmatrix}1 &-1\\
0 &1 \end{pmatrix}.$$
The 7-th power is thus given by
$$A^7 = T D^7 T^{-1} = \begin{pmatrix}1 &1\\
0 &1 \end{pmatrix}\begin{pmatrix}e^{14x} &0\\
0 &(e^{2x}-1)^7 \end{pmatrix}\begin{pmatrix}1 &-1\\
0 &1 \end{pmatrix}= \begin{pmatrix}e^{14 x} &(e^{2x}-1)^7-e^{14 x}\\
0 &(e^{2x}-1)^7 \end{pmatrix}.$$
| {
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Can you explain why this series converges? $$1 - \frac{1}{3 \cdot 2!} + \frac{1}{5 \cdot 3!} - \frac{1}{7 \cdot 4!}+\cdots$$
I am new to math over flow, and I do not know how to format the math, sorry! Also, what should this converge to?
| $$1 - \frac{1}{3 \cdot 2!} + \frac{1}{5 \cdot 3!} - \frac{1}{7 \cdot 4!}+\cdots$$
A standard theorem says this converges if the corresponding series of absolute values converges:
$$1 + \frac{1}{3 \cdot 2!} + \frac{1}{5 \cdot 3!} + \frac{1}{7 \cdot 4!}+\cdots$$
The "comparison test" says this converges if the following converges:
$$1 + \frac 1 {2!} + \frac 1 {3!} + \frac 1 {4!}+\cdots$$
And that converges by a ratio test:
$$
\frac{1/(n+1)!}{1/n!} = \frac 1 {n+1} \to \text{ as } n\to\infty.
$$
| {
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 2
} |
Prove that the sequence $\left\{\frac{4n^2}{2n^3-5}\right\}$ converges to $0$. Prove that the sequence $$\left\{\frac{4n^2}{2n^3-5}\right\}$$ converges to $0$.
The first thing I would like to do is further bound the sequence so I can get rid of the $-5$ in the bottom of the fraction. How can I do this? For negatives it's tricky since you want to make it smaller, but removing a $-5$ actually makes the denominator bigger, and not smaller.
If this cannot be done, I must do the following:
$$\left\{\frac{(4/n)}{2-5/n^3}\right\}$$ and find a bound for $2-5/n^3$, separately. If I assume $n\geq 10$, then:
$$\frac{1}{n}<\frac{1}{10}\color{red}{\to} \frac{5}{n}<\frac{1}{2}
\color{red}{\to} \frac{-5}{n}>\frac{-1}{2}\color{red}{\to} 2-\frac{5}{n}>\frac{3}{2}\color{red}{\to} \frac{1}{|2-5/n^3|}<\frac{2}{3}$$
Then I have that:
$$\left|\frac{(4/n)}{2-5/n^3}\right|<\frac{2}{3}\cdot\frac{4}{n}<\frac{8}{3N}<\frac{8}{3(8/3\epsilon)}=\epsilon$$
Pick $N=\max(10,8/3\epsilon)$
This excess bounding could have been avoided if I can somehow bound the fraction even more?
| For $n\ge 2$, we have for all $\epsilon>0$
$$\frac{4n^2}{2n^3-5}\le \frac{4n^2}{n^3}=\frac4n<\epsilon$$
whenever $n>N=\max\left(2,\frac{4}{\epsilon}\right)$
| {
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Solve $\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}=13$ for $x$
Solve $$\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}=13$$ for $x$
Any help is appreciated. I'm entering a challenge and can't reach the solution.
| First, we can rearrange the given equation as follows:
\begin{align}\frac{4^{2x}+4^x+1}{2^{2x}+2^x+1}&=13\\\\
\frac{(2^2)^{2x}+(2^2)^x+1}{2^{2x}+2^x+1}&=13\\\\
(2^2)^{2x}+(2^2)^x+1&=13\left(2^{2x}+2^x+1\right)\\
2^{4x}+2^{2x}+1&=13\left(2^{2x}+2^x+1\right)\\
(2^x)^4+(2^x)^2+1&=13\left((2^x)^2+2^x+1\right)\\
y^4+y^2+1&=13\left(y^2+y+1\right)\tag{substitute $y:=2^x$}\\
y^4+y^2+1&=13y^2+13y+13\\
y^4-12y^2-13y-12&=0\end{align}
We are then left with a quartic equation to solve, and then we can solve $y=2^x$ for each of the values of $y$ we find.
Can you continue from here?
| {
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Random walk with 3 cases. Let the walking start be at $x=0$. With probability $p_1$ new $x=x+1$, with probability $p_2$: $x=x-1$ and with probability $1-p_1-p_2 \geq 0$ walking ends.
The question is what is the probability of ending on point $n$.
I've computed it numerically and getting something close to normal distribution with peak at zero, but with different dispersion on sides of $x=0$. How to get it analytically?
The provided answer seems ok, but I hope I can find some smooth function for it. Maybe with another type of solution.
Upd: my idea of solution.
$$P_{n+1} (x) = p_1 P_{n} (x-\delta x) + p_2 P_{n} (x + \delta x)$$
$$P_{n+1} (x) - P_{n} (x) = p_1 [P_n(x-\delta x) - P_n(x)] + p_2 [P_n(x+\delta x) - P_n(x)] - q\cdot P_{n}(x)$$
Dividing by $\delta n$ and $\delta x$ getting something like:
$$\frac{\partial P(n,x)}{\partial n}= \frac{\delta x}{\delta n}\left[ (p_2 + p_1) \frac{\partial P(n,x)}{\partial x} - (1 - p_2 - p_1)\cdot P(n,x)\right]$$
Solving it leads to something like this:
$$P(n,x) = Ae^{\frac{-q x}{p_1 + p_2}}(n + \frac{x}{p_1 + p_2})$$
I'm not sure yet if it means something.
|
We derive a closed formula and show the following is valid for $n\geq 0$
\begin{align*}
\sum_{k=0}^\infty\binom{n+2k}{k}x^k&=
\frac{1}{\sqrt{1-4x}}
\left(\frac{1-\sqrt{1-4x}}{2x}\right)^n\tag{1}\\
\end{align*}
Note that
\begin{align*}
c(x)=\frac{1-\sqrt{1-4x}}{2x}=1+x+2x^2+5x^3+14x^4+\cdots
\end{align*}
is the generating function of the ubiquitous Catalan numbers.
We use the coefficient of operator $[w^k]$ to denote the coefficient of $w^k$ in a series. This way we can write e.g.
\begin{align*}
[w^k](1+w)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^{\infty}\binom{n+2k}{k}x^k}
&=\sum_{k=0}^\infty[w^{k}](1+w)^{n+2k}x^k\tag{2}\\
&=[w^0](1+w)^n\sum_{k=0}^\infty\left(\frac{x(1+w)^2}{w}\right)^k\tag{3}\\
&=[w^0]\frac{(1+w)^n}{1-\frac{x(1+w)^2}{w}}\tag{4}\\
&=-[w^{-1}]\frac{(1+w)^n}{x+(2x-1)w+xw^2}\\
&=-[w^{-1}]\frac{\left(1+w\right)^n}{x(w-w_0(x))(w-w_1(x))}\tag{5}\\
&=\frac{\left(1+w_0(x)\right)^n}{x(w_1(x)-w_0(x))}\tag{6}\\
&\color{blue}{=\frac{1}{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^n}\tag{7}
\end{align*}
and the claim (1) follows.
Comment:
*
*In (2) we apply the coefficient of operator.
*In (3) we use the linearity of the coefficient of operator and apply the rule
\begin{align*}
[w^{p+q}]A(w)=[w^p]w^{-q}A(w)
\end{align*}
*In (4) we apply the geometric series expansion.
*In (5) we represent the denominator with poles $w_0(x)$ and $w_1(x)$.
\begin{align*}
w_0(x)&=\frac{1-\sqrt{1-4x}}{2x}-1\tag{8}\\
w_1(x)&=\frac{1+\sqrt{1-4x}}{2x}-1\\
\end{align*}
*In (6) we calculate the residual of the simple pole at $w=w_0(x)$ using
\begin{align*}
\operatorname{res}_{w=a}f(w)=\lim_{w\to a}(w-a)f(w)
\end{align*}
*In (7) we use the representation from (8).
| {
"language": "en",
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"question_score": "8",
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Test the series $\sum_{k=0}^{\infty}\frac{(-1)^k 1 \cdot3\cdot5 \cdots(2k+1)}{1\cdot4\cdot7\cdots(3k+1)}$ Use the ratio test for absolute convergence to determine
whether the series absolutely or
diverge.
$$\sum_{k=0}^{\infty}\frac{(-1)^k 1 \cdot3\cdot5 \cdots(2k+1)}{1\cdot4\cdot7\cdots(3k+1)}$$
I don't understand how the general term becomes this $$\lim_{k\to\infty} \frac{a_{k+1}}{a_k} = \lim_{k\to\infty} \frac{2k+3}{3k+4}$$
Obviously this Absolutely converges, I just dont get the 2nd step.
| Let us write
$$a_k=\frac{ 1 \cdot3\cdot5 \cdots(2k+1)}{1\cdot4\cdot7\cdots(3k+1)}.$$
Then, we have
$$\begin{align}
\frac{a_{k+1}}{a_k}&=\frac{1\cdot3\cdot5\cdots [2(k+1)+1]}{1\cdot 4\cdot 7\cdots [3(k+1)+1]}\quad\cdot\quad\frac{1\cdot 4\cdot 7\cdots (3k+1)}{1\cdot3\cdot5\cdots(2k+1)}\\
&=\frac{1\cdot3\cdot5\cdots (2k+3)}{1\cdot 4\cdot 7\cdots (3k+4)}\quad\cdot\quad\frac{1\cdot 4\cdot 7\cdots (3k+1)}{1\cdot3\cdot5\cdots(2k+1)}\\
&=\frac{1\cdot3\cdot5\cdots(2k+1) (2k+3)}{1\cdot 4\cdot 7\cdots (3k+1)(3k+4)}\quad\cdot\quad\frac{1\cdot 4\cdot 7\cdots (3k+1)}{1\cdot3\cdot5\cdots(2k+1)}\\
&=\frac{1\cdot3\cdot5\cdots(2k+1)}{1\cdot 4\cdot 7\cdots (3k+1)}\quad\cdot\quad\frac{2k+3}{3k+4}\quad\cdot\quad\frac{1\cdot 4\cdot 7\cdots (3k+1)}{1\cdot3\cdot5\cdots(2k+1)}\\
&=\frac{2k+3}{3k+4}.
\end{align}$$
This proves the 2nd step you asked.
| {
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Ramanujan's Surd : Proof required. I came across a problem on the web, it is as follows:
Question:If $ x $, $ y $ and $ z $ are rational numbers such that
$ \sqrt[3]{\sqrt[3]{2}-1} = \sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}$ then find $ x,y,z $.
From Wikipedia|Nested Radicals|Some identities by Ramanujan, it is clear that the values of $x,y,\text{and} \ z$ is $\sqrt[3]{\frac{1}{9}}$,$-\sqrt[3]{\frac{2}{9}}$ and $\sqrt[3]{\frac{4}{9}}$ respectively.
However, even after going through the references from the wikipedia and from the web [2], [3], [4]; I still fail at understanding it.
Can someone please show me the proof of this result in an intuitive way?
My approach in attacking this problem is as follows,
Let, $y = \sqrt[3]{\sqrt[3]{2}-1}$
Cubing both sides,
$$
y^{3} = \sqrt[3]{2}-1 \\
\implies y^{3}+1=\sqrt[3]{2} \\
\implies (y^{3}+1)^{3}= 2
$$
Now, using the identity : $(a+b)^3=a^3 + 3a^{2}b + 3ab^{2} + b^{3}$
$$
y^{9} + 3y^{6}+3y^{3}+1 = 2 \\
\implies y^{9} + 3y^{6}+3y^{3} - 1 = 0
$$
Now, I get the cubic equation using the substitution, $y^{3} = m$
$$
m^{3} + 3m^{2}+3m - 1 = 0
$$
Seems like I get lost in my own web.
Can someone help me out?
| You want a proof of the identity $$\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{\frac{1}{9}}-\sqrt[3]{\frac{2}{9}}+\sqrt[3]{\frac{4}{9}}$$ which can be written as $$\sqrt[3]{x-1}=\frac{1-x+x^{2}}{\sqrt[3]{9}}$$ where $x=\sqrt[3]{2}$. Thus we let the RHS be denoted by $y$ and show that $y^{3}=x-1$. Since $x, y$ are real this proves the desired identity.
Now we can see that $$y^{3}=\frac{(1-x+x^{2})^{3}}{9}=\frac{(x^{3}+1)^{3}}{9(x+1)^{3}}=\frac{3}{x^{3}+3x^{2}+3x+1}=\frac{1}{1+x+x^{2}}=\frac{x-1}{x^{3}-1}=x-1$$ In the above algebraic manipulation $x^{3}$ has been replaced with $2$.
Ramanujan was expert in algebraic manipulation of numbers as well as symbolic expression so much so as to make MAPLE and MACSYMA almost unnecessary and he obtained most of such denesting of radicals as special cases of general algebraic identities. It was not exactly like solving equations to find value of $x, y,z$ as you ask in the question.
| {
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Solve : $x+y=2010$, $(\sqrt[3]{x}+\sqrt[3]{y})(\sqrt[3]{x}+11)(\sqrt[3]{y}+11)=2942$
Given$$x+y=2010,\\(\sqrt[3]{x}+\sqrt[3]{y})(\sqrt[3]{x}+11)(\sqrt[3]{y}+11)=2942,$$
find the value of $\sqrt[3]{x}+\sqrt[3]{y}$.
My attempts:
Let $u=\sqrt[3]{x} , v=\sqrt[3]{y}$.
$(u+v)(u^2-uv+v^2)=2010 \tag{1}$
$(u+v)(uv+11u+11v+121)=2942
\tag{2}$
$(1)+(2)$:
$$(u+v)(u^2+v^2+11u+11v+121)=4952$$
How to proceed?
| u and v look quite similar, so I will just use a and b instead:
$$3(a+b)(ab+11a+11b+121)+(a+b)(a^2-ab+b^2)=3\times2942+2010=10836$$
$$(a+b)(a^2+2ab+b^2+33a+33b+363)=10836$$
$$(a+b)((a+b)^2+33(a+b)+363)=10836$$
$$(a+b)^3+33(a+b)^2+363(a+b)-10836=0$$
Solving for $a+b$, the real root to this is $a+b=12$, so the answer is 12. I'm not sure if this is the desired approach, but it works.
| {
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Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$
$$\left|\frac{3x+1}{x-2}\right|<1$$
$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , }x \neq 2$$
$${-x+1}<{3x} \text{ and } 3x<{x-3} \text{ , }x \neq 2$$
$${1}<{4x} \text{ and } 2x<{-3} \text{ , }x \neq 2$$
$${\frac{1}{4}}<{x} \text{ and } x<{\frac{-3}{2}} \text{ , }x \neq 2$$
While the answer is
$${\frac{1}{4}}>{x} \text{ and } x>{\frac{-3}{2}}$$
| You can solve it more easily by splitting the modulus and squaring both sides:
Assuming $x\neq2$,
$$|3x+1|<|x-2|\implies(3x+1)^2-(x-2)^2<0$$
$$\implies(3x+1+x-2)(3x+1-x+2)<0$$
So $$(4x-1)(2x+3)<0\implies-\frac 32<x<\frac 14$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simple inequality $\frac{4x+1}{2x-3}>2$
$$\frac{4x+1}{2x-3}>2$$
I have started with looking at positive/negative situations and there are 2 or that both expressions are positive or both are negative.
If both a positive we can solve for $$\frac{4x+1}{2x-3}>2$$
$${4x+1}>2(2x-3)$$
$${4x+1}>4x-6$$
$${0x}>-7$$
Or both are negative and then $${4x+1}<2(2x-3)$$
And again $$0x<-7$$
So there is not answer?
| HINT
Notice that
$$
\frac{4x+1}{2x-3}
= \frac{4x-6}{2x-3} + \frac{7}{2x-3}
= 2 + \frac{7}{2x-3}
$$
which is more than $2$ when the fraction is positive. Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove expression is not prime $$
(n + 4)^4 + 4
$$
If n is natural number, how to prove that above expression is not prime?
I am stuck here
$$
(n+4)^2 \cdot (n+4)^2 + 2 . 2
$$
$$
\left(\left(n^2+4^2\right) \cdot 2\right)\left(\left(n^2+4^2\right) \cdot 2\right)
$$
| From the context of the question, I assume $n ≥ 0$.
Clearly for even $n$, $(n+4)^4+4$ is even. Hence suppose $n$ is odd, so $n = 2k+1$. Then $(n+4)^4+4 = (2k+5)^4+4 = 16 k^4 + 160 k^3 + 600 k^2 + 1000 k + 629 = (4 k^2 + 16 k + 17) (4 k^2 + 24 k + 37)$.
Clearly this is composite, as for $n≥0$, this is a product of two natural numbers greater than $1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is $1+x+x^2+x^3...$? What is the difference between these two series?
$$
\begin{align}
1+x+x^2+x^3+...+x^n+\mathcal O(x^{n+1})&=\frac{1}{1-x}\\
\\
1+r+r^2+r^3+...+r^{n-1}&=\frac{r^n-1}{r-1}\\
\end{align}
$$
I can't wrap my head around it; they both start with $1+x+x^2+x^3...$? Of course the first is the maclaurin series. But other than that, why don't they both yield the same result?
Thanks!
| The two series actually do produce the same result. You just have to compare them fairly.
To begin with, you've listed $n$ terms in the first series
(not including the "big-O" notation) and only $n-1$ terms in the second series. Let's make both series have $n$ terms, as follows.
(I'm also putting the "closed expression" side of the equation on the left
in order to have the big-O on the right, as is the standard practice.)
\begin{align}
\frac{1}{1-x} &= 1+x+x^2+x^3+...+x^n+\mathcal O\left(x^{n+1}\right),\\
\frac{r^{n+1}-1}{r-1} &= 1+r+r^2+r^3+...+r^n.
\end{align}
Notice that $\frac{(-y)}{(-y)} = \frac yy,$ so if we reverse the subtraction
on the top and bottom of $\frac{r^{n+1}-1}{r-1}$ at the same time,
we get the same result:
$$
\frac{1 - r^{n+1}}{1 - r} = 1+r+r^2+r^3+...+r^n.
$$
Now distribute the numerator of $\frac{1 - r^{n+1}}{1 - r}$ over the
denominator:
$$
\frac{1}{1 - r} - \frac{r^{n+1}}{1 - r} = 1+r+r^2+r^3+...+r^n.
$$
Add $\frac{r^{n+1}}{1 - r}$ to both sides:
$$
\frac{1}{1 - r} = 1+r+r^2+r^3+...+r^n + \frac{r^{n+1}}{1 - r}.
$$
The first equation occurs in a context where $x$ is very small
(close to zero), so that the $\mathcal O\left(x^{n+1}\right)$ term
represents just an error correction to the previous terms.
If we likewise suppose that $\lvert r\rvert \ll 1,$ so that we are
comparing apples to apples (both series dealing with the same kind of numbers), then
$\frac{r^{n+1}}{1 - r} = \mathcal O\left(r^{n+1}\right),$
and we can write
$$
\frac{1}{1 - r} = 1+r+r^2+r^3+...+r^n + \mathcal O\left(r^{n+1}\right).
$$
So this is the same as the first equation, but with the variable $r$
substituted for $x.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to deal with a Pell's equation type problem with two primes I have some understanding of Pell's equation $$x^2-dy^2=1$$ where $d$ is a prime. I know that you can take the continued fraction of $\sqrt{d}$ and use information about the period and convergents to find things out. Now I need to show that if $p,q$ are primes $\equiv 3 \pmod{4}$ then at least one of the equations $$px^2-qy^2=\pm 1$$ is soluble in integers $x,y$. I have not learned what to do when there is a prime in front of the $x^2$. How should I begin doing this?
| You have a primitive solution to $x^2-pqy^2=1$ ($x$, $y$ positive, $y$ minimal). Then $x$ is odd and $y$ is even (think modulo $4$). Also
$$x^2-1=(x+1)(x-1)=pqy^2$$
so that
$$\frac{x+1}2\frac{x-1}2=pq\left(\frac{y}{2}\right)^2.$$
As $(x\pm 1)/2$ are coprime integers then we have one of the following
*
*$(x+1)/2=u^2$, $(x-1)/2=pqv^2$
*$(x+1)/2=pu^2$, $(x-1)/2=qv^2$
*$(x+1)/2=qu^2$, $(x-1)/2=pv^2$
*$(x+1)/2=pqu^2$, $(x-1)/2=v^2$
We succeed with (2) and (3) (write $1=(x+1)/2-(x-1)/2$). We eliminate
(1) by minimality to solution of Pell and (4) gives $v^2-pqu^2=-1$
which is impossible modulo $p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2 $$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$
But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.
| Hint:
Expand $(3x-1)^2+(3y-1)^2+(3z-1)^2\geqslant 0$ And simplify.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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In a triangle $ABC$, if its angles are such that $A=2B=3C$ then find $\angle C$? In a triangle $ABC$, if its angles are such that $A=2B=3C$ then find $\angle C$?
$1$. $18°$
$2$. $54°$
$3$. $60°$
$4$. $30°$
My Attempt:
$$A=2B=3C$$
Let $\angle A=x$ then $\angle B=\dfrac {x}{2}$ and $\angle C=\dfrac {x}{3}$
Now,
$$x+\dfrac {x}{2} +\dfrac {x}{3}=180°$$
$$\dfrac {6x+3x+2x}{6}=180°$$
$$\dfrac {11x}{6}=180°$$
$$x=\dfrac {180\times 6}{11}$$
So, which is the correct option?
| Add:
$$180^\circ=A+B+C=2B+B+\frac23B=\frac{11}3B\implies B=\frac{540}{11},\ldots etc.$$
Or also
$$180^\circ=A+B+C=3C+\frac32C+C=\frac{11}2C\implies C=\frac{360}{11}$$
By the way, none of the options given in your question matches the above...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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There exist $3$ positive numbers $x,y,z$ such that $x\cdot y\cdot z=100$ but $x^2+y^2+z^2<65$. I have to show this using a form of calculus.
I know the answer is the cube root of $100$. I did this using intuition and upper and lower bounds but I have no idea how to show it from a calculus standpoint. Maybe showing the intersection of the surfaces?
| Hint: by AM-GM $\,
x^2+y^2+z^2 \ge 3 \sqrt[3]{x^2 y^2 z^2} = 30 \sqrt[3]{10} \simeq 64.63\,$, with equality iff $x=y=z\,$.
[ EDIT ] Equivalent proof using Jensen's inequality for the concave function $\ln(x)$ on $\mathbb{R}^+\,$:
$$
\ln \left(\frac{1}{3}x^2+\frac{1}{3}y^2+\frac{1}{3}z^2\right) \ge \frac{1}{3}\left(\ln(x^2)+\ln(y^2)+\ln(z^2)\right) = \ln(\sqrt[3]{x^2y^2z^2})
$$
By monotonicity of $\ln(x)$ on $\mathbb{R}^+$ it follows that $\cfrac{x^2+y^2+z^2}{3} \ge \sqrt[3]{x^2y^2z^2} = 10 \sqrt[3]{10}\,$, and because $\ln(x)$ is strictly concave the equality holds iff $x=y=z\,$.
[ EDIT #2 ] Yet another way to find the minimum of $x^2+y^2+z^2$ using "more" calculus.
Starting from $x^2+y^2 \ge 2xy\,$ with equality iff $x=y\,$, it follows that $x^2+y^2+z^2 \ge 2xy + z^2 = \cfrac{200}{z} + z^2\,$ with equality iff $x=y=\sqrt{\cfrac{100}{z}}\,$.
Let $f(z)=z^2+\cfrac{200}{z}\,$ then $f'(z)=2z - \cfrac{200}{z^2} = 0 \iff z^3 = 100\,$, so the minimum is attained for $z= \sqrt[3]{100}\,$, and $x = y = \sqrt{\cfrac{100}{\sqrt[3]{100}}}=\sqrt[3]{100}\,$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Express the value of $s\left(m\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}$ in terms of $m$. The previous question was: Find the range such that the equation $|x^2 -3x +2|=mx $ has 4 distinct real solutions: $a,b,c,d$, and that turned out to be $0<m<3-2\sqrt{2}$. The book says that the solution is $\frac{m^2+5}{2}$. I have been trying to express $a$ and $b$ as the two possible results from the quadratic equation resulting from $x^2−3x+2=mx$, and $c$ and $d$ as the ones from the equation $x^2−3x+2=-mx$, and plugging that into $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2}$, yet I do not see how a complicated expression can end up being the answer given by the book. Therefore I am sure I must be doing something wrong.
| Hint: $a+b=m+3$ and $ab=2$ by Vieta's relations for $x^2-(m+3)x+2=0\,$, so:
$$\frac{1}{a^2}+\frac{1}{b^2}=\frac{a^2+b^2}{a^2b^2} = \frac{(a+b)^2-2\cdot ab}{(ab)^2}=\frac{(m+3)^2-2 \cdot 2}{2^2}=\frac{m^2+6m+5}{4}$$
Repeat the similar calculation for $\,\cfrac{1}{c^2}+\cfrac{1}{d^2}\,$, then add them up.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve recurrence relation: $a_n=3a_{n−1}+5a_{n−2}−7a_{n−3}$ I hope can you help me, I want to solve this recurrence relation:
$a_n=3a_{n−1}+5a_{n−2}−7a_{n−3}$ with initial values $a_1=4, a_2=8, a_3=2$
Using synthetic division I get the roots: $1, 1-2\sqrt{2}$ and $1-2\sqrt{2}$
What are the characteristic equation for three different roots? I guess that:
$ a_n = c_1(r_1)^n + c_2(r_2)^n + c_3(r_3)^n$ where $r_1, r_2, r_3$ are the roots.
Can you help me solve this problem?
| Recurrence $a_n=3a_{n−1}+5a_{n−2}−7a_{n−3}$ has characteristic polynomial $x^3 - 3x^2 - 5x + 7 = 0$ which has roots $x=1$, $x=1 - 2\sqrt{2}$, and $x=1 +2\sqrt{2}$.
That means your recurrence has form $a_n = \alpha \cdot 1^n + \beta \cdot (1-2\sqrt{2})^n + \gamma \cdot (1+2\sqrt{2})^n$.
Substitute in the values you know:
$4 = \alpha \cdot 1^1 + \beta \cdot (1-2\sqrt{2})^1 + \gamma \cdot (1+2\sqrt{2})^1$
$8 = \alpha \cdot 1^2 + \beta \cdot (1-2\sqrt{2})^2 + \gamma \cdot (1+2\sqrt{2})^2$
$2 = \alpha \cdot 1^3 + \beta \cdot (1-2\sqrt{2})^3 + \gamma \cdot (1+2\sqrt{2})^3$
This leads to solution $\alpha = \frac{21}{4}$, $\beta = \frac{3}{8} + \frac{1}{2\sqrt{2}}$, and $\gamma = \frac{3}{8} - \frac{1}{2\sqrt{2}}$.
$a_n = \frac{21}{4}+ (\frac{3}{8} + \frac{1}{2\sqrt{2}})\cdot (1-2\sqrt{2})^n + (\frac{3}{8} - \frac{1}{2\sqrt{2}})\cdot (1+2\sqrt{2})^n$
This can be flattened down a little:
$a_n = \frac{1}{8} ((3 - 2\sqrt{2}) (1 + 2 \sqrt{2})^n + (3 + 2 \sqrt{2}) (1 - 2 \sqrt{2})^n + 42)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving a problem on limits
$$\lim_{x\to 1}\left(\frac{p}{1-x^p}-\frac{q}{1-x^q}\right) , p,q\in N$$
I tried fitting the formula $\lim_{x\to a}\frac{x^n - a^n}{x-a} = na^{n-1}$ but could not proceed further. Also I don't think that we should directly apply L'Hôpital's rule as it will be too lengthy.
| Using Taylor expansions around $0$, namely:
*
*$(1+u)^\alpha = 1+u+\frac{\alpha(\alpha-1)}{2}u^2 + o(u^2)$
*$\frac{1}{1+u} = 1-u+o(u)$ (a specific case of the first, for $\alpha=-1$)
You have, setting $h\stackrel{\rm def}{=} 1-x \xrightarrow[x\to1]{}0$ (to center things around $0$, with which I am much more familiar than around $1$):
$$\begin{align}
\frac{p}{1-x^p}-\frac{q}{1-x^q}
&= \frac{p}{1-(1-h)^p}-\frac{q}{1-(1-h)^q}
\\
&= \frac{p}{1-(1-ph+\frac{p(p-1)}{2}h^2+o(h^2))}-\frac{q}{1-(1-qh+\frac{q(q-1)}{2}h^2+o(h^2))}\\
&= \frac{p}{ph-\frac{p(p-1)}{2}h^2+o(h^2)}-\frac{q}{qh-\frac{q(q-1)}{2}h^2+o(h^2)}\\
&= \frac{1}{h-\frac{p-1}{2}h^2+o(h^2)}-\frac{1}{h-\frac{q-1}{2}h^2+o(h^2)}\\
&= \frac{1}{h}\left(\frac{1}{1-\frac{p-1}{2}h+o(h)}-\frac{1}{1-\frac{q-1}{2}h+o(h)}\right)\\
&= \frac{1}{h}\left(1+\frac{p-1}{2}h+o(h)-(1+\frac{q-1}{2}h+o(h))\right)\\
&= \frac{1}{h}\left(\frac{p-1-(q-1)}{2}h+o(h)\right)\\
&= \frac{p-q}{2}+o(1)
\xrightarrow[h\to0]{}\boxed{\frac{p-q}{2}}
\end{align}$$
| {
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If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value of $\alpha^6+\beta^6$ is
If $\alpha$ and $\beta$ are the roots of $x^2-2x+4=0$ then the value
of $\alpha^6+\beta^6$ is
I know that, here, $\alpha\beta=4$ and $\alpha + \beta = 2$ and use that result to find $\alpha^2 + \beta^2$ using the expansion of $(a+b)^2$
But how to find $\alpha^6+\beta^6$ ?
| Use this one:
$$[(a+b)^2-2ab)][((a+b)^2-2ab)^2-3a^2b^2]$$
| {
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Solve the equation in integers $a,b$: $20a^3-b^3=1$
Solve the equation in integers $a,b$: $$20a^3-b^3=1.$$
Assume that $a \neq 0$. Then simplifying and rearranging the equation gives $$20a^3 = 2^2 \cdot 5 \cdot a^3 = b^3+1 = (b+1)(b^2-b+1).$$ Note that neither factors can be divisible by $3$, since both of them must be divisible by $3$ but $b^2-b+1 \not \equiv 0 \pmod{9}$ and $b^2-b+1 = 3$ gives a contradiction. Also since $b$ is odd, then $b+1$ is even and $b^2-b+1$ is odd. Also, $b^2-b+1 \not \equiv 0 \pmod{5}$, so $b+1 \equiv 0 \pmod{20}$.
Then since $$\gcd(b+1,b^2-b+1) = \gcd(b+1,-2b+1) = \gcd(b+1,3),$$ it follows that $(b+1)$ and $(b^2-b+1)$ are relatively prime. Thus $b+1 = 20n^3$ and $b^2-b+1 = m^3$ where $m,n$ are relatively prime. I then thought about bounding $b^2-b+1$ between two perfect cubes, but I didn't see how to do that.
Then since two solutions for $(a,b)$ are $(0,-1)$ and $(7,19)$ I wanted to show that $(b+1) \mid 20$, or equivalently that $a$ and $b+1$ are relatively prime. So suppose that for some prime $p$ we have $a \equiv 0 \pmod{p}$ and $b+1 \equiv 0 \pmod{p}$. It then follows that $b+1 \equiv 0 \pmod{p^3}$, so that $b = p^3k-1$ for some integer $k$. Then we have $$(p^3k-1)^2-(p^3k-1)+1 = p^3k(p^3k-3)+3 = m^3.$$ How can we get a contradiction from here?
| With the equation $20a^3 - b^3 = 1$, if $b$ = $55860$, $a \approx 20578.999951$. In fact, the prime factorisation of $55860$ is $2^2\times3\times5\times7^2\times19$.
$\frac{55860}{19} = 2940$ and funnily enough, $\frac{20579}{7} \approx 2939.857136$ which is close to $2940$. The prime factorisation of $20579$ is $13\times1583$.
If $a$ is below $20579$ and $b$ is below $55860$, you will come across other "near solutions", but this was as close as I could come. I hope this would help you one way or another to find another solution for this equation where $a$ is not $0,7$ and $b$ is not $-1,19$. I am probably too late now anyway :)
| {
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Solve $2\cos^2{x}=\sqrt{3}\sin{2x}$.
Problem:
Solve $2\cos^2{x}=\sqrt{3}\sin{2x}$ and give the sum of all the solutions in the interval $0\leq x\leq2\pi.$
Attempt: Using the fact that $\sin{2\theta}= 2\cos{\theta}\sin{\theta}$ on the RHS I get $$2\cos^2{x}=2\sqrt{3}\cos{x}\sin{x}.$$
Dividing by $2\cos{x}$ I get $$\cos{x}=\sqrt{3}\sin{x}.$$
Dividing by $\cos{x}$ again I get $$\tan{x}=\frac{1}{\sqrt{3}} \ \Longleftrightarrow \ x=\pi k+\frac{\pi}{6}, \ \ \forall \in \mathbb{Z.}$$
But it's not correct. Why?
| The right way to solve is:
$$2\cos^2{x}=2\sqrt{3}\cos{x}\sin{x} \to \cos x(\cos x-\sqrt{3}\sin x)=0$$
what give you:
$$\cos x=0 \text{ or } \cos x-\sqrt{3}\sin x=0$$
For
$$\cos x=0\to x=\frac{\pi}{2}+k\pi$$
and for,
$$\cos x-\sqrt{3}\sin x=0\to \tan x=\frac{\sqrt{3}}{3}\to x=\frac{\pi}{6}+k\pi$$
| {
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"url": "https://math.stackexchange.com/questions/2256618",
"timestamp": "2023-03-29T00:00:00",
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Three real roots of equation is the length of three sides of a right triangle, find the parameter Three real roots of the equation $x^3-2p(p+1)x^2+(p^4+4p^3-1)x-3p^3=0$ is the length of three sides of a right triangle, what is the value of $p$?
| Hint: let $a,b,c$ be the roots, then by Vieta's relations:
$$\require{cancel}
\begin{align}
a^2+b^2+c^2 &= (a+b+c)^2 - 2(ab+bc+ca) \\
&= \left(2p(p+1)\right)^2 - 2(p^4+4p^3-1) \\
&= 4p^2(p^2+\bcancel{2p}+1) - 2p^4-\bcancel{8p^3}+2 \\
&= 2(p^4 + 2p^2+1) \\
&= 2(p^2+1)^2
\end{align}
$$
For the roots to be the sides of a right triangle, one of them, say $a\,$, must satisfy $a^2=b^2+c^2$.
| {
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Solve the inequality $\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}$ Solve for $x$ real the inequality $$\sqrt { x + 2} - \sqrt { x + 3} < \sqrt { 2} - \sqrt { 3}.$$ Obviously $x\ge-2$. After that I tried to square the whole inequality, which led me to $x < - \frac { 18} { 4\sqrt { 6} - 5}$. Now, the answer is $[-2;0) $. Should there be a different approach?
| First lets multiply the inequality by $-1$ so we can square. $$\sqrt{x+2}-\sqrt{x+3}<\sqrt2-\sqrt3\\\sqrt{x+3}-\sqrt{x+2}>\sqrt{3}-\sqrt{2}$$Now lets square$$2x+5-2\sqrt{(x+3)(x+2)}>5-2\sqrt{6}\\x+\sqrt6>\sqrt{(x+3)(x+2)}$$
Now since $x+\sqrt{6}\geq-2+\sqrt 6>0$ we can square the inequality again
$$x^2+2\sqrt{6}x+6>x^2+5x+6\\(2\sqrt6-5)x>0$$
This happens when $x<0$ since $2\sqrt6=\sqrt{24}<\sqrt{25}=5$ is negative so $x$ must be as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
To prove $\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$ if $a,b,c$ are non zero positive reals prove $$\sum_{cyc}\frac{1}{a^3+b^3+abc} \le \frac{1}{abc}$$ I have used A.M G.M inequality as follows:
$$a^3+b^3+c^3 \ge 3abc$$ adding $abc$ both sides we get
$$a^3+b^3+abc \ge 4abc-c^3=c(4ab-c^2)$$ so
$$\frac{1}{a^3+b^3+abc} \le \frac{1}{c(4ab-c^2)}$$ But since $a,b,c$ are positive reals
$$4abc-c^3 < 4abc$$ so
$$\frac{1}{4abc-c^3} \gt \frac{1}{4abc}$$ but i am unable to proceed here
| Notice that $$\sum_{cyc}\frac{abc}{a^3+b^3+abc} = \sum_{cyc}\frac{1}{\frac{a^2}{bc}+\frac{b^2}{ac}+1} \leq \sum_{cyc}\frac{1}{\frac{a+b}{c}+1} = \sum_{cyc}\frac{c}{a+b+c} = 1 $$ Then divide both sides by $abc$.
Note: In the inequality part we used Cauchy-Schwartz to get $$\left(\frac{a^2}{b}+\frac{b^2}{a}\right)(b+a)\geq(a+b)^2\implies \frac{a^2}{b}+\frac{b^2}{a}\geq a+b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Finding the nth element of a sequence Suppose we have sequence given by
$$ (1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,...) $$
I need to prove that $a_n = \lfloor{ \sqrt{2n} + \frac{1}{2} \rfloor} $.
Try:
I notice the following. The last occurrence of 1 occurs at $a_1$, the last occurrence of $2$ occurs at $a_3$, the last occurrence of $3$ occurs at $a_6$. So, if we define $F(n)$ to be the last occurrence of digit $n$, then we guess
$$ P(n) = \frac{n(n+1)}{2} $$
Who show by induction. base case is clear. Suppose the formula holds true for some $k$, that is suppose $P(k) = \frac{ k(k+1) }{2}$.
Note $P(k+1) = P(k) + k+1$
since the (k+1)th last entry is k+1 positions after the kth entry. Thus,
$$ P(k+1) = \frac{k(k+1)}{2} + k+1 = \frac{(k+1)(k+2)}{2} $$
and guess is true by induction.
Now, let $a_n = m$. Since $a_n$ must lie between $P(m-1)$ and $P(m)$ we have
$$ P(m-1) < n \leq P(m) \implies \frac{ (m-1)m}{2} < n \leq \frac{m(m+1)}{2} \implies m^2 - m < 2n \leq m^2 + m $$
now, since $m^2 + m + 1/4 > m^2 + m$ and since adding $1/4$ to $m^2-m$ would still be less than $2n$, then
$$ m^2 - m + \frac{1}{4} < 2n < m^2 +m + \frac{1}{4} \iff (m-1/2)^2 <2n < (m+1/2)^2 $$
Thus,
$$ m - 1/2 < \sqrt{2n} < m + 1/2 \implies m < \sqrt{2n} < m + 1$$
Hence, $m = a_n = \lfloor{ \sqrt{2n} + \frac{1}{2} \rfloor}$.
Is this a correct argument? Any feedback would be really appreciated. IS there a better approach to solve this problem?
| In your sequence the first occurrence of $m\in\{1,2,\ldots\}$ appears at the position $1+2+\ldots+(m-1)+1 = \frac{m^2-m}{2}+1$ and the last one at the position $1+2+\ldots+m=\frac{m^2+m}{2}$. To prove the claim it is enough to show that the function $f:x\mapsto\left\lfloor\sqrt{2x}+\frac{1}{2}\right\rfloor$ maps every element of the interval $\left[\frac{m^2-m}{2}+1,\frac{m^2+m}{2}\right]$ into $m$.
Since $f$ is a weakly increasing function, it is enough to show that both the endpoints of the previous interval are mapped into $m$. We have
$$\left(m-\frac{1}{2}\right)^2 = m^2-m+\frac{1}{4} < m^2-m+2 < m^2+m+2 \leq (m+1)^2 $$
and the claim easily follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Matrix determinant as Dickson polynomial $\frac{x^{n+1}-y^{n+1}}{x-y}$
Given matrix $$
A=\begin{bmatrix}
x+y&xy&0& .&.&. &0\\
1&x+y&xy&0& .&.&0 \\
0&1&x+y&xy&.&.&. \\
.&.&.&.&.&.&. \\
.&.&.&.&.&.&0 \\
.&.&.&.&.&.&xy \\
0&.&.&.&0&1&x+y
\end{bmatrix}
$$
prove by induction that $$|A|=\frac{x^{n+1}-y^{n+1}}{x-y}$$ $x \neq y$, $A_{n \times n}$.
The determinant expression appears to be Dickson polynomial of second kind.
Let $D_n$ be the determinant of $A_n$. We can see that the appropriate recurrence relation is $$D_n=(x+y)D_{n-1}-xyD_{n-2}$$
Base cases:
$$D_1=x+y=\frac{x^2-y^2}{x-y}$$
$$
D_2=(x+y)^2-xy=x^2+xy+y^2=\frac{x^3-y^3}{x-y}
$$
Suppose that $$D_n=(x+y)D_{n-1}-xyD_{n-2}$$
Then we need to prove that $$D_{n+1}=(x+y)D_{n}-xyD_{n-1}$$
Which can be developed as:
$$
D_{n+1}=(x+y)((x+y)D_{n-1}-xyD_{n-2})-xyD_{n-1}=
$$
$$
=(x+y)^2D_{n-1}-xy(x+y)D_{n-2}-xyD_{n-1}=
$$
$$
=(x^2+xy+y^2)D_{n-1}-xy(x+y)D_{n-2}=
$$
$$
=\frac{x^3-y^3}{x-y}D_{n-1}-xy(x+y)D_{n-2}
$$
I tried doing this up to $D_{n-6}$ in order to get any insights into possible simplification but I'm pretty stuck.
| A linear algebra approach to the recurrence relation is to write it in matrix form: $$\mathcal{D}_n\equiv \begin{pmatrix} D_{n} \\ D_{n-1}\end{pmatrix}=\begin{pmatrix} x+y & -xy\\1 & 0 \end{pmatrix}\begin{pmatrix} D_{n-1} \\ D_{n-2}\end{pmatrix}\equiv M \mathcal{D}_{n-1},\hspace{4mm} \mathcal{D}_0 = \binom{1}{0}.$$ (I choose $D_0=1,D_{-1}=0$ so that the recurrence for $D_n$ in terms of $D_{n-1},D_{n-2}$ valid for $n\geq 1$.) Then $D_n=(\mathcal{D}_n)_1=(M^n \mathcal{D}_0)_1=(M^n)_{11},$ so computing $D_n$ amounts to matrix multiplication. But the eigenvalues and eigenvectors are simple enough in this case, yielding the spectral decomposition $$M=\Lambda D \Lambda^{-1}=\begin{pmatrix}x & y \\ 1 & 1\end{pmatrix}\begin{pmatrix}x & 0 \\ 0 & y\end{pmatrix}\begin{pmatrix}x & y \\ 1 & 1\end{pmatrix}^{-1}$$
and therefore
\begin{align}
M^n
&=(\Lambda D \Lambda^{-1})^n \\
&= \Lambda D^n \Lambda^{-1} \\
&= \begin{pmatrix}x & y \\ 1 & 1\end{pmatrix}\begin{pmatrix}x^n & 0 \\ 0 & y^n\end{pmatrix}\begin{pmatrix}x & y \\ 1 & 1\end{pmatrix}^{-1} \\
&= \begin{pmatrix}x^{n+1} & y^{n+1} \\ x^n & y^n\end{pmatrix}\cdot \frac{1}{x-y}\begin{pmatrix}1 & -y \\ -1 & x\end{pmatrix}\\
&=\frac{1}{x-y}\begin{pmatrix}x^{n+1}-y^{n+1} & x y^{n+1}-x^{n+1}y \\ x^{n}-y^{n}& xy^{n}-yx^{n}\end{pmatrix}.
\end{align}
From the upper-left corner we read off the anticipated result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
How to tell two ideals belong to the same ideal class group Potentially dumb question here: how can you tell if two ideals belong to the same ideal class group?
Let's say we're looking at $\textbf{Z}[\sqrt{10}]$. It has infinitely many ideals, but by the Minkowski bound we need only concern ourselves with ideals with a norm of $2$ or $3$.
My gut tells me $\langle 3, 1 - \sqrt{10} \rangle$ and $\langle 3, 1 + \sqrt{10} \rangle$ are in the same ideal class, but that could be altogether wrong.
| Let's show that the ideal $(3, 1+ \sqrt{10})\cdot (3, 1-\sqrt{10})^{-1}$ is principal. We have
$$\frac{(3, 1+ \sqrt{10})}{(3, 1-\sqrt{10})} = \frac{(3, 1+ \sqrt{10})\cdot(3, 1+ \sqrt{10})}{(3, 1-\sqrt{10})\cdot (3, 1+\sqrt{10})}$$
Now, for fractional ideals of number fields we have
$$(\alpha_1, \alpha_2)\cdot( \beta_1, \beta_2) = (\alpha_1 \beta_1, \alpha_1 \beta_2 + \alpha_2 \cdot \beta_1, \alpha_2 \cdot \beta_2)$$
( Gauss lemma for Dedekind rings, see Herman Weyl- Algebraic Number Theory)
Therefore, the denominator in the fraction $(*)$ is
$$(9, 6, 1-\sqrt{10}^2) = (9, 6, -9)= (3)$$
while the numerator is
$$(9, 6(1+ \sqrt{10}), (1+\sqrt{10})^2)=(9, 6(1+ \sqrt{10}), 11 + 2 \sqrt{10})=\\ =(9, 6(1+ \sqrt{10}), 2 + 2\sqrt{10}) =((1-\sqrt{10})(1+\sqrt{10}), 2(1+\sqrt{10})) \\=(1+\sqrt{10})\cdot (1-\sqrt{10}, 2)$$
Now $(1-\sqrt{10}, 2) = (1)$, since $((1-\sqrt{10})(1+\sqrt{10}), 2) = (9,2) = (1)$
We conclude:
$$\frac{(3, 1+ \sqrt{10})}{(3, 1-\sqrt{10})} = (\frac{1+\sqrt{10}}{3})$$
or $\frac{1+\sqrt{10}}{3}\cdot (3, 1-\sqrt{10})= (1+\sqrt{10}, 3)$.
This is obvious, once we know the factor...
$\bf{Added:}$ If $\alpha \cdot \beta = k^2$ then
$$\frac{\alpha}{k} = \frac{k}{\beta} \ \textrm{so}\ \frac{\alpha}{k} (k, \beta)= (\alpha, k)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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"answer_id": 3
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The sides $a$, $b$, $c$ of a triangle with area $D$ are such that $ab+bc+ca\ge 4\sqrt{3}D$
The sides $a$, $b$, $c$ of a triangle with area $D$ are such that $ab+bc+ca\ge 4\sqrt{3}D$.
I think I need to figure out a way to apply the A.M.-G.M. inequality here, but couldn't find a proper way of doing so.
| I realise this isn't answering your question, but it was interesting so I have decided to post it anyway. tl;dr I showed that $ab+ac+bc \geq 6D$, a weaker inequality, using AM-GM.
The area of triangle (in terms of its $3$ sides) is given by:
$D=\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$. This is known as Heron's formula.
This means we want to show that $ab+bc+ac \geq \sqrt{3(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$
First note that \begin{align*}(a+b+c)(-a+b+c)+(a-b+c)(a+b-c) &= (-a^2+b^2+c^2+2bc)+(a^2-b^2-c^2+2bc)\\ &= 4bc\end{align*}
Hence by AM-GM, $\frac{4bc}{2} = 2bc \geq \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} = 4D$
By symmetry we also have $2ab \geq 4D$, $2ac \geq 4D$.
Now by adding the three equations, we have $2(ab+ac+bc) \geq 12D$, so $ab+ac+bc \geq 6D$. Unfortunately $4\sqrt{3} \approx 6.9$, so proved a weaker result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Inequality $\frac{b}{ab+b+1} + \frac{c}{bc+c+1} + \frac{a}{ac+a+1} \ge \frac{3m}{m^2+m+1}$ Let $m=(abc)^{\frac{1}{3}}$, where $a,b,c \in \mathbb{R^{+}}$. Then prove that
$\frac{b}{ab+b+1} + \frac{c}{bc+c+1} + \frac{a}{ac+a+1} \ge \frac{3m}{m^2+m+1}$
In this inequality I first applied Titu's lemma ; then Rhs will come 9/(some terms) ; now to maximise the rhs I tried to minimise the denominator by applying AM-GM inequality.But then the reverse inequality is coming
Please help.
| By Holder:
$$\sum_{cyc}\frac{b}{ab+b+1}=1-\frac{(abc-1)^2}{\prod\limits_{cyc}(ab+b+1)}\geq1-\frac{(m^3-1)^2}{(\sqrt[3]{a^2b^2c^2}+\sqrt[3]{abc}+1)^3}=$$
$$=1-\frac{(m-1)^2(m^2+m+1)^2}{(m^2+m+1)^3}=1-\frac{(m-1)^2}{m^2+m+1}=\frac{3m}{m^2+m+1}.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Am I doing it wrong or this derivative is really lengthy? We have:
$$f(x)=\begin{vmatrix} (\alpha x+1)\cos^2 x & x & 1-x\\\beta\sin x& x^2 & 2x \\(\gamma x^2+1)\tan x & x & 1-x^2\end{vmatrix}$$
And we are supposed to calculate $$\lim_{x\rightarrow 0}\frac{1}{x^6}\int_{x^2}^{x^3}f(x).dx=A$$
and tell if $[A]$ (where $[x]$ is greatest integer less than or equal to $x$) is equal to $f''(0)$ or not.
Yes, i know how to differentiate a determinant But To be honest , this question looks barabaric in cases where I need to solve it under $5$ minutes or $8$ max.Does there exist any simpler way than
$$f'(x)=\begin{vmatrix} \alpha \cos^2 x+(\alpha x+1)(-2\sin x \cos x) & 1 & -1\\\beta\sin x& x^2 & 2x \\(\gamma x^2+1)\tan x & x & 1-x^2\end{vmatrix}+\begin{vmatrix} (\alpha x+1)\cos^2 x & x & 1-x\\\beta\cos x& 2x & 2 \\(\gamma x^2+1)\tan x & x & 1-x^2\end{vmatrix}+\begin{vmatrix} (\alpha x+1)\cos^2 x & x & 1-x\\\beta\sin x& x^2 & 2x \\2\gamma x \tan x+ \sec^2 (\gamma x^2+1) & 1 & -2x\end{vmatrix}$$
and proceeding?
Please guide.
| Let $(u,v,w)$ be the entries in $1^{st}$ column of the determinant $f(x)$, we have
$$f(x)
= \begin{vmatrix} u & x & 1-x\\ v & x^2 & 2x \\ w & x & 1-x^2\end{vmatrix}
= x\begin{vmatrix} u & 1 & 1-x\\ v & x & 2x \\ w & 1 & 1-x^2\end{vmatrix}
= x\begin{vmatrix} u & 1 & -x\\ v & x & x \\ w & 1 & -x^2\end{vmatrix}
= x^2\begin{vmatrix} u & 1 & -1\\ v & x & 1 \\ w & 1 & -x\end{vmatrix}
$$
Let $g(x)$ be the rightmost determinant. Expand it as a power series in $x$,
$$g(x) = \begin{vmatrix} u & 1 & -1\\ v & x & 1 \\ w & 1 & -x\end{vmatrix} = g_0 + g_1 x + g_2 x^2 + \cdots$$
and substitute it in the integral, we get
$$\begin{align}
\frac{1}{x^6}\int_{x^2}^{x^3} f(t) dt
&= \frac{1}{x^6}\int_{x^2}^{x^3} t^2 g(t) dt
= \frac{1}{x^6}\int_{x^2}^{x^3} ( g_0 t^2 + g_1 t^3 + \cdots ) dt\\
&= \frac{1}{x^6}\left( \frac{g_0}{3} [t^3]_{x^2}^{x^3} + \frac{g_1}{4} [t^4]_{x^2}^{x^3} + \cdots\right)
= \frac{1}{x^6}\left( \frac{g_0}{3} (x^9 - x^6) + \frac{g_1}{4} (x^{12}-x^8) + \cdots\right)\\
&= -\frac13 g_0 + O(x^2)
\end{align}
$$
Notice when $x = 0$, $(u,v,w) = (1,0,0)$, we find
$$A = \lim_{x\to0} \left(-\frac{g_0}{3} + O(x^2)\right)
= -\frac{g_0}{3} =
-\frac13
\begin{vmatrix} 1 & 1 & -1\\ 0 & 0 & 1 \\ 0 & 1 & 0\end{vmatrix}
= \frac13
$$
This leads to
$$\left\lfloor A \right\rfloor = 0\quad\ne\quad f''(0) = 2g(0) = 2g_0 = -2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2266649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Direct/transverse tangents of two circles If equations of two circles in a plane are given as
$$ f(x,y)=0,\, g(x,y)=0,\, $$
*
*Find equations of direct tangents in terms of $f,g$
and
*Find equations of transverse tangents in terms of $f,g$
We could calculate using coincident roots/point etc. of a transverse line but how can it be found using pole/polar properties ? or other methods?
| If $f(x,y)=0$ is the equation of a circle with center $(h,k)$ and radius $r,$ then the most general form of $f$ is
$$f(x,y) = a((x-h)^2 + (y-k)^2 - r^2)$$
where $a\neq0.$ Consider the following values of $f$:
\begin{align}
f(0,0) &= a(h^2 + k^2 + r^2), \\
f(1,0) &= a(h^2 - 2h + 1 + k^2 + r^2), \\
f(-1,0) &= a(h^2 + 2h + 1 + k^2 + r^2).
\end{align}
We can combine these as follows:
$$
f(1,0) + f(-1,0) - 2f(0,0) = 2a,
$$
from which we find $a$ in terms of $f,$ namely,
$a = \tfrac12(f(1,0) + f(-1,0)) - f(0,0).$
We also have
$$
f(-1,0) - f(1,0) = 4ah,
$$
and since we can already express $a$ in terms of $f,$ this lets us express $h$ in terms of $f.$
We can also plug the expression for $a$ into the equation
$f(0,0) - f(0,1) = a(2k - 1)$ in order to find $k$ in terms of $f,$
and we can then plug these expressions for $a,$ $h,$ and $k$
into the equation $f(0,0) = a(h^2 + k^2 + r^2)$ to find $r$ in terms of $f.$
Since this establishes that $a,$ $h,$ $k,$ and $r$ can all be expressed in terms of $f,$ from now on let's just see what we can express in terms of
$a,$ $h,$ $k,$ and $r,$ with the knowledge that whatever we can express in those terms can be expressed in terms of $f$ as well.
Likewise, if $g(x,y)=0$ is the equation of a circle with center $(H,K)$ and radius $R,$ then the most general form of $g$ is
$$g(x,y) = A((x-H)^2 + (y-K)^2 - R^2),$$
where $A,$ $H,$ $K,$ and $R$ can all be expressed in terms of $g.$
Let $\ell$ be the line satisfying the equation
$$ L(x,y) = nx + \left(\sqrt{1-n^2}\right)y - m = 0 $$
where $m$ and $n$ are real constants, $-1 \leq n \leq 1.$
The vector $\left\langle n, \sqrt{1-n^2}\right\rangle$ is a unit normal vector of $\ell.$ (Another way to express this is that $n=\cos\theta$ and
$\sqrt{1-n^2} = \sin\theta,$ where $\theta$ is the angle of the vector.)
Given any line in the plane, we can select values of $m$ and $n$ such that $\ell$ is the given line.
The distance from a point $(x,y)$ to the line $\ell$ is
$$
d((x,y),\ell) = \left\lvert nx + \left(\sqrt{1-n^2}\right)y - m\right\rvert.
$$
In order to find common tangents to the two circles whose equations are
$f(x,y)=0$ and $g(x,y)=0,$ therefore, we need to find $m$ and $n$ such that
the distance from $(h,k)$ to $\ell$ is $r$ and the distance from $(H,K)$ to $\ell$ is $R,$ that is,
\begin{align}
nh + \left(\sqrt{1-n^2}\right)k - m &= \pm r, \\
nH + \left(\sqrt{1-n^2}\right)K - m &= \pm R.
\end{align}
If we select the signs of $\pm r$ and $\pm R$ in these two equations,
and solve for $m$ and $n,$ we derive the equation of a line tangent to
the two circles.
There are four possible choices of signs, leading to four tangent
lines whose equations we can derive:
for $+r,+R$ or $-r,-R$ we get a direct tangent,
but for $+r,-R$ or $-r,+R$ we get a transverse tangent.
| {
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Show that $\frac{\tan \alpha + \sqrt{5}\sin \alpha}{2}(\sqrt{5}\cos \alpha -1)$ is equivalent to $\frac{5}{4}\sin(2\alpha)-\frac{1}{2}\tan \alpha$
Show that $\frac{\tan \alpha + \sqrt{5}\sin \alpha}{2}(\sqrt{5}\cos
\alpha -1)$ is equivalent to $\frac{5}{4}\sin(2\alpha)-\frac{1}{2}\tan
\alpha$.
I tried:
$$\frac{\frac{\sin \alpha}{\cos \alpha}+\sin \alpha}{2}(\sqrt{5}\cos \alpha-1) = \\
\frac{\frac{\sin \alpha+\sqrt{5}\sin \alpha \cos \alpha}{\cos \alpha}}{2}(\sqrt{5}\cos \alpha-1) = \\
\frac{\frac{\sin \alpha}{\cos \alpha}(\sqrt{5}\cos \alpha+1)}{2}(\sqrt{5}\cos \alpha-1) = \\
\frac{\tan \alpha (5\cos^2\alpha-1)}{2} = \\
???$$
What do I do next?
| Now simply $$\frac{5\sin\alpha \cos\alpha-\tan\alpha}{2}$$
$$\frac{5\times2.\sin\alpha \cos\alpha}{4}-\frac{\tan\alpha}{2}$$
$$\frac{5\times\sin 2\alpha}{4}-\frac{\tan\alpha}{2}\tag{sin2x=2 sinx cox}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using generating functions in combinatorics I know these questions get asked a lot but I cannot figure it out. It requires the use of generating functions to find the number of solutions (coefficient) to the equation:
$u_i+u_2+u_3+u_4 = 20$, where $1 \leq u_i \leq 5, i = 1,...,4$
How do I solve this question step-by-step?
Thanks.
|
The constraint $1\leq u_i \leq 5$ can be encoded using the finite geometric series formula as
\begin{align*}
z^1+z^2+z^3+z^4+z^5=\frac{z\left(1-z^5\right)}{1-z}\tag{1}
\end{align*}
Since (1) holds for each $u_i, 1\leq i\leq 4$ all possible configurations can be encoded as
\begin{align*}
\left(\frac{z\left(1-z^5\right)}{1-z}\right)^4\tag{2}
\end{align*}
We want to find the number of non-negative integer solutions of
\begin{align*}
u_1+u_2+u_3+u_4=20
\end{align*}
with the constraints given above.
In the following we denote with $[z^n]$ the coefficient of $z^n$. According to (2) we are looking for
\begin{align*}
[z^{20}]&z^4\frac{\left(1-z^5\right)^4}{(1-z)^4}\tag{3}\\
&=[z^{16}]\frac{\left(1-z^5\right)^4}{(1-z)^4}\tag{4}\\
&=[z^{16}]\left(1-4z^5+6z^{10}-4z^{15}\right)\sum_{k=0}^\infty\binom{-4}{k}(-z)^k\tag{5}\\
&=\left([z^{16}]-4[z^{11}]+6[z^{6}]-4[z]\right)\sum_{k=0}^\infty\binom{k+3}{3}z^k\tag{6}\\
&=\binom{19}{3}-4\binom{14}{3}+6\binom{9}{3}-4\binom{4}{3}\tag{7}\\
&=969-4\cdot364+6\cdot84-4\cdot 4\\
&=1
\end{align*}
in accordance with the obvious single solution $u_1=u_2=u_3=u_4=5$.
Comment:
*
*In (3) we select the coefficient of $[z^{20}]$ of the product of the generating function (2) which correspond to the valid ranges specified for $u_i$ with $1\leq i \leq 4$.
*In (4) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (5) we multiply out the numerator and skip terms with powers greater than $16$ since they do not contribute to $[z^{16}]$. We also apply the binomial series expansion.
*In (6) we use the linearity of the coefficient of operator, apply the same rule as in (4) four times and use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q$.
*In (7) we select the coefficients accordingly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$ For real numbers $a, b, c, d >0$ I have to prove that $$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$$ always holds. I tried using the mean inequalities for multiple numbers but I always get in trouble with the different orders of the two roots. And in case I make an estimate that can handle the orders of the roots, it is too strong. Can anyone help?
| Since
$$\frac{a^2+b^2+c^2+d^2}{4}\geq\frac{ab+ac+bc+ad+bd+cd}{6}$$
it's just
$$(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2\geq0,$$ it's enough to prove that
$$\sqrt{\frac{ab+ac+bc+ad+bd+cd}{6}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}},$$
which I proved here:
How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269645",
"timestamp": "2023-03-29T00:00:00",
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Prove the sqrt of 4 is irrational, where did I go wrong? So I know that $\sqrt{4}=2$, and therefore is rational, but, it seems like I proved it to be irrational? Could I have some help as to where I went wrong? Thank you!
Prove $\sqrt4$ is irrational.
Suppose per contradiction $\sqrt4$ is rational, that is, $\sqrt4 = \frac ab$ where $a$ and $b$ are integers. We assume $\frac ab$ is in reduced form.
$$\begin{align}\sqrt4^2 &= \left(\frac{a}{b}\right)^2\\
4 &= \frac{a^2}{b^2}\\
4b^2 &= a^2
\end{align}$$
This implies $4$ divides $a^2$, and thus $4|a$, and thus $2|a$. That is, $a = 2n$ for some integer $n$. The definition of an even integer is $2k$ for some integer $k$, so $a$ is even. Applying this new logic to our old statement we get:
$$\begin{align}
4b^2 &= (4n)^2\\
4b^2 &= 16n^2\\
b^2 &= 4n^2
\end{align}$$
This implies $4|b^2$, and thus $4|b$, and thus $2|b$. Since we know $b = 2n$ for some integer $n$, and the definition of an even integer, $b$ is therefore an even integer. Since we know that both $a$ and $b$ cannot be even, as that would contradict the fact that $\frac ab$ is in reduced form, then we know that there is no $a$ and $b$ such that $\sqrt4 = \frac ab$. Therefore, $\sqrt4$ is irrational.
| So $4|a^2$ does not imply that $4|a$. But your real error comes from
$$4b^2 = (4n)^2 4b^2 = 16n^2 b^2 = 4n^2$$ and you conclude that $4|b^2$. You should divide by $4$ to get
$$b^2=n^2$$ from which little can be concluded about divisibility by $4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2270591",
"timestamp": "2023-03-29T00:00:00",
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Help solving the differential equation $y''+y'+y=0$ with initial conditions $y(0)=4,y'(0)=-3$ using Laplace transform. UPDATED WTH ANSWER:
Please help me solve the following differential equation using Laplace transform:
$$y''+y'+y=0;y(0)=4,y'(0)=-3$$
My answer so far:
$$\mathcal{L}\{y''\} + \mathcal{L}\{y'\} + \mathcal{L}\{y\}=0$$
We know that $\mathcal{L}\{y'\} = s\mathcal{L}\{y\}-y(o)$ and $\mathcal{L}\{y''\}=s\mathcal{L}\{y'\}-y'(0)$. Upon substituting this to the equation, it becomes
$$s\mathcal{L}\{y'\}-y'(0)+s\mathcal{L}\{y\}-y(0)+\mathcal{L}\{y\}=0$$
$$s(s\mathcal{L}\{y\}-y(0))-y'(0)+s\mathcal{L}\{y\}-y(0)+\mathcal{L}\{y\}=0$$
$$s^2\mathcal{L}\{y\}-sy(0)-y'(0)+s\mathcal{L}\{y\}-y(0)+\mathcal{L}\{y\}=0$$
Upon substituting the values of initial conditions the equation becomes
$$s^2\mathcal{L}\{y\}-4s+3+s\mathcal{L}\{y\}-4+\mathcal{L}\{y\}=0$$
Solving for $\mathcal{L}\{y\}$:
$$\mathcal{L}\{y\}(s^2+s+1)-4s+3-4=0$$
$$\mathcal{L}\{y\}(s^2+s+1)=4s+1$$
$$\mathcal{L}\{y\}=\frac{4s+1}{s^2+s+1}$$
Modifying the numerator and denominator to obtain the form similar to one of the them in the Laplace transform table.
First applying the method of completing the square to the denominator:
$$\frac{4s+1}{s^2+s+1} = \frac{4s+1}{(s^2+s+\frac{1}{4})+(1-\frac{1}{4})}=\frac{4s+1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}$$
Then modifying the numerator:
$$\frac{4s+1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=\frac{4s+2-1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=\frac{4(s+\frac{1}{2})-1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}$$
Divide the fractions:
$$\mathcal{L}\{y\}=\frac{4(s+\frac{1}{2})}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}-\frac{1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}$$
Taking the inverse of Laplace transform of $\mathcal{L}\{y\}$ and let it be $y$. Then the equation becomes
$$y=4\mathcal{L}^{-1}[\frac{(s+\frac{1}{2})}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2)}] - \mathcal{L}^{-1}[\frac{1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}]$$
The inverse of the first fraction is:
$$4e^{\frac{-1x}{2}}\cos(\frac{\sqrt3x}{2})$$
To match the second fraction to one of the forms from the Laplace Transform table, we need to multiply the numerator and denominator by $\frac{\sqrt3}{2}$. Then it becomes:
$$\frac{2}{\sqrt3}\mathcal{L}^{-1}[\frac{\frac{\sqrt3}{2}}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}]$$
Therefore the inverse of the second fraction is:
$$\frac{2}{\sqrt3}e^{\frac{-1x}{2}}\sin(\frac{\sqrt3x}{2})$$
FINAL ANSWER:
$$y=4e^{\frac{-1x}{2}}\cos(\frac{\sqrt3x}{2})+\frac{2}{\sqrt3}e^{\frac{-1x}{2}}\sin(\frac{\sqrt3x}{2})$$
//
| $$\mathcal{L}(s)=\frac{4\left(s+\frac{1}{2}\right)-1}{(s+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}$$
therefore
$$ f(x)=e^{-\frac{1}{2}x}\left(4\cos \left(\frac{\sqrt{3}}{2}x\right)-\frac{2\sqrt{3}}{3}\sin \left(\frac{\sqrt{3}}{2}x\right)\right)$$
In case there remains some mystery concerning how to know to replace the $4s+1$ with $4\left(s+\frac{1}{2}\right)-1$, since after completing the square in the denominator we see that it contains $\left(s+\frac{1}{2}\right)^2$ we must then replace the $4s$ with $4\left(s+\frac{1}{2}\right)$. But seeing that this changes the numerator to $4s+2$ we subtract the $1$ to keep the numerator the same after the adjustment.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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The value of infinite product At a first look I observed that every term of this product is less than 1 so the limit will also be less than 1 but clueless about how to find limit .
$$P=\frac{7}{9}\cdot\frac{26}{28}\cdot\frac{63}{65}\cdots\frac{n^3-1}{n^3+1}\cdots.$$
| To restate your question:
$$
\prod_{n=2}^\infty \frac{n^3-1}{n^3+1}
$$
First we begin by expanding the terms:
$$
n^3-1 = (n-1)(n^2+n+1)
$$
$$
n^3+1 = (n+1)(n^2-n+1)
$$
By treating each term as a separate product,and shifting the index of the product, a telescope-like method can be used to simplify the terms into the desired answer.
\begin{align*}
\prod_{n\ge2}\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}
\\
= \frac{\left(\prod_{n \ge 0} n+1\right)\left(\prod_{n \ge 2} n^2+n+1\right)}{\left(\prod_{n \ge 2} n+1\right)\left(\prod_{n \ge 2} n^2-n+1\right)}
\\
= \frac{2\left(\prod_{n \ge 2} n^2+n+1\right)}{\left(\prod_{n \ge 1} (n+1)^2-(n+1)+1\right)}
\\
= \frac{2\left(\prod_{n \ge 2} n^2+n+1\right)}{\left(\prod_{n \ge 1} n^2+(2-1)n+(1-1+1)\right)}
\\
= \frac{2\left(\prod_{n \ge 2} n^2+n+1\right)}{\left(\prod_{n \ge 1} n^2+n+1\right)}
\\
= \frac{2}{3}
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/2272601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Coefficient of $x^9$ Find the coefficient of $x^9$ in $(1+x)(1+x^2)(1+x^3)\cdots(1+x^{100})$.
Approach:-
By inspection, coefficient of $x^9$ will not be there in terms after $(1+x^{9})$
Now checking coefficient in $(1+x)(1+x^2)...(1+x^9)$.
Now by hit and trial, I checked what are the possible combinations of getting $x^9$.
I got the answer as $8$. Is there any other and better approach?
| This question is equivalent to: How many sums of natural numbers equal $9$, subject to the condition that the terms be increasing? The terms in the sum are just exponents coming from all of the factors you mention.
For example, the sum $2+7=9$ corresponds to the term in your product where we take $x^2$, $x^7$, and $1$ from every other factor.
Thus, we list:
$9, 1+8, 2+7, 3+6, 4+5, 1+2+6, 1+3+5, 2+3+4$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as... If $t=\tan{\frac{x}{2}}$, then $\cos{x}$ can be expressed as
a) $\frac{1+t^2}{1-t^2}$
b) $\frac{2t}{1+t^2}$
c) $\frac{1-t^2}{1+t^2}$
d) $\frac{2t}{1-t^2}$
Attempt: I tried using the half angle formula but it just leaves me with an expression in terms of $\tan{x}$'s and I don't know how to go to $t$, let alone express $\cos{x}$ in $t$.
| Following your idea you can do:
$$\tan x=\frac{2\tan (x/2)}{1-\tan^2(x/2)}=\frac{2t}{1-t^2}$$
so,
$$\frac{\sin x}{\cos x}=\frac{2t}{1-t^2}\to \frac{\sin^2 x}{\cos^2 x}=\frac{4t^2}{(1-t^2)^2}$$
using proportion properties we get:
$$\frac{\cos^2 x}{\sin^2 x+\cos^2 x}=\frac{(1-t^2)^2}{4t^2+(1-t^2)^2}\to \cos^2x=\frac{(1-t^2)^2}{(1+t^2)^2}\to \cos x=\frac{1-t^2}{1+t^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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$x=(a+b\omega)^3$ with $\omega$ a primitive third root of unity. Then $x=(c+d\sqrt{-3})^3$. ($a,b,c,d \in \mathbb{Z}$) Let $x=(a+b\omega)^3$ with $\omega$ a primitive third root of unity. Then $x=(c+d\sqrt{-3})^3$. ($a,b,c,d \in \mathbb{Z}$). Can someone give me a hint?
Approach:
$x= a^3 + 3a^2 b\omega + 3ab^2\omega^2 + b^3$ and $\omega^2 = -\omega -1$. But then, I'm stuck.
Thanks in advance.
| Hint:
$$ (a + b \omega)^3 = (a \omega + b \omega^2)^3 = (a \omega^2 + b)^3.$$
Now write $$\omega = - \frac 1 2 + \frac{\sqrt{-3}}2, \ \ \ \ \ \ \ \ \omega^2 = - \frac 1 2 - \frac{\sqrt{-3}}2,$$
and separate into cases depending on whether $a$ and $b$ are odd or even...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2273646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I find the generating function of this sequence with recurrence relation? I looked up related documents and saw a similar problem solving.
But, I don't understand well and still can not solve the problem.
How can I find the sequence of generating function ???
$${ G }_{ n }(x)=\sum _{ k=0 }^{ n }{ f(n,k){ x }^{ k } } \quad
\text{with} \quad f(n,k)=f(n,k-1)-2f(n-1,k-1)$$
$$( { G }_{ 0 }=1,\quad f(n,0)=0 )$$
| Try writing this as a series in two indeterminates $x$ and $y$ such that
$$H(x,y)=\sum_{n\ge 0}G_n(x)y^n=\sum_{k,n\ge 0}f(n,k)y^nx^k$$
$$\implies H(x,y)=\sum_{n\ge 0}f(n,0)y^n + \sum_{k\ge 1}f(0,k)x^k + \sum_{k,n\ge 1}f(n,k)y^nx^k$$
From the i.c's we get
$$H(x,y)=1+\sum_{k,n\ge 1}f(n,k)y^nx^k$$
and using $f(n,k)=f(n,k-1)-2f(n-1,k-1)$ we can write
$$\begin{align}H(x,y)&=1+\sum_{k,n\ge 1}\left(f(n,k-1)-2f(n-1,k-1)\right)y^nx^k\\
&=1+x\sum_{k,n\ge 1}f(n,k-1)y^nx^{k-1}-2xy\sum_{k,n\ge 1}f(n-1,k-1)y^{n-1}x^{k-1}\\
&=1+x\sum_{\begin{subarray}{c}k&\ge 0\\n&\ge 1\end{subarray}}f(n,k)y^nx^k-2xy\sum_{k,n\ge 0}f(n,k)y^nx^k\end{align}$$
Using our i.c $f(0,0)=1$ again gives
$$\begin{align}H(x,y)&=1+x\left(\sum_{k,n\ge 0}f(n,k)y^nx^k - 1\right) - 2xy\sum_{k,n\ge 0}f(n,k)y^nx^k\\
&=1-x+xH(x,y)-2xyH(x,y)\\
&=\frac{1-x}{1-x(1-2y)}=\left(1-\left(\frac{-2x}{1-x}\right)y\right)^{-1}\\
H(x,y)&=\sum_{n\ge 0}\left(\frac{-2x}{1-x}\right)^ny^n\\
H(x,y)&=\sum_{n\ge 0}G_n(x)y^n\end{align}$$
equating coefficients of $y^n$
$$G_n(x)=\frac{(-2)^nx^n}{(1-x)^n}$$
this can be written by expanding the negative binomial
$$\begin{align}G_n(x)&=(-2)^nx^n\sum_{k\ge 0}\binom{n+k-1}{n-1}x^k=\sum_{k\ge 0}(-2)^n\binom{n+k-1}{n-1}x^{n+k}\\ &= \sum_{k\ge n}(-2)^{n}\binom{k-1}{n-1}x^k\end{align}$$
$$\implies f(n,k)=(-2)^n\binom{k-1}{n-1}\tag{Answer}$$
check the recurrence
$$f(n,k-1)-2f(n-1,k-1)=(-2)^n\binom{k-2}{n-1}+(-2)^n\binom{k-2}{n-2}=(-2)^n\left(\binom{k-2}{n-1}+\binom{k-2}{n-2}\right)=(-2)^n\binom{k-1}{n-1}=f(n,k)$$
which (due to Pascal's identity) checks out.
$$\begin{array}{c}
\text{Table of values for $f(n,k)$}\\
\begin{array}{c|rrrrrrrr}
n \backslash k & 0 & 1 & 2 & 3 & 4 & 5 & 6&\cdots\\\hline
0&1 &0 &0 &0 &0 &0 &0&\cdots\\
1&0 &-2 &-2 &-2 &-2 &-2 &-2&\cdots\\
2&0 &0 &4 &8 &12 &16 &20&\cdots\\
3&0 &0 &0 &-8 &\bbox[aqua]{-24} &-48 &-80&\cdots\\
4&0 &0 &0 &0 &\bbox[yellow]{16} &\bbox[lime]{64} &160&\cdots\\
5&0 &0 &0 &0 &0 &-32 &-160&\cdots\\
6&0 &0 &0 &0 &0 &0 &64&\cdots\\
\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\ddots\\
\end{array}\\
\end{array}$$
e.g. For $n=4$ and $k=5$
$$\begin{split}f(n,k)&=f(n,&k-1)&\:-\:&2f(n-&1,k-1)\\
\bbox[lime]{\phantom{A}}\,&=\,&\bbox[yellow]{\phantom{A}}&\:-\:&2\,&\bbox[aqua]{\phantom{A}}\end{split}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve for $x$ in $\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$ Solve for $x$ in $$\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$$
i have done in a lengthy way:
By inspection we observe that $x=41$ and $x=-24$ are the solutions
we have
$$\sqrt[4]{57-x}=5-\sqrt[4]{x+40}$$ squaring both sides we get
$$\sqrt{57-x}=25+\sqrt{x+40}-10 \sqrt[4]{x+40}$$ that is
$$\sqrt{57-x}-25=\sqrt{x+40}-10 \sqrt[4]{x+40}$$ again squaring both sides we get
$$682-x-50\sqrt{57-x}=x+40+100\sqrt{x+40}-20(x+40)^{\frac{3}{4}}$$
i got messed up here any better way and just a hint please
| $$\sqrt[4]{57-x}+\sqrt[4]{x+40}=5$$
Let $y=\sqrt[4]{57-x}$, and $z=\sqrt[4]{x+40}$
Then
$$y\ge 0, z \ge 0, y+z=5$$
$$x=57- y^4=z^4-40$$
Thus
$$y^4+ (5-y)^4-97$$
$$=2(y-3)(y-2)(y^2-5y+44)=0$$
Thus $y=2$ or $y=3$, which means $x=41$ or $x=-24$
We are done.
| {
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.