Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Why are the integral curves of this vector field easy to find? I'm reading a book of differential geometry that states that it's easy to get the integral curves of this vector field:
$$X(x,y)=(x^2-y^2,2xy)$$
But proceeding the way the book says, first I have to take that equation as
$$\frac{dx}{dt}=x^2-y^2$$
$$\frac{dy}{dt}=2xy$$
But from that I don't know what more to do to proceed. I know that the integral curves must be circles of center $(0,t)$, and the vector field seems as this
| Note that the system implies
$$y'(x)=\frac{2xy(x)}{x^2-y(x)^2}.$$
(This is the approach mentioned by @Sameh Shenawy in the comments.)
This can be solved by standard methods.
For example, setting $y(x)=xf(x)$ we arrive at the separable differential equation
$$f'(x) = \frac{1}{x} \frac{f(x)(1+f(x)^2)}{1-f(x)^2}.$$
The relevant integral may be done by a partial fraction expansion.
In my opinion the method put forward by @WimC is much preferable.
Addendum
The gory detail ...
\begin{align*}
f'(x) &= \frac{1}{x} \frac{f(x)(1+f(x)^2)}{1-f(x)^2} \\
\frac{1-f^2}{f(1+f^2)}df &= \frac{dx}{x} \\
\frac{(1+f^2)-2f^2}{f(1+f^2)} df
&= \frac{dx}{x} \\
\int\left(\frac{1}{f} - \frac{2f}{1+f^2}\right)df
&= \int\frac{dx}{x} \\
\ln f - \ln(1+f^2) &= \ln\frac{x}{2r} \\
\ln\frac{f}{1+f^2} &= \ln\frac{x}{2r} \\
\frac{f}{1+f^2} &= \frac{x}{2r} \\
\frac{y/x}{1+y^2/x^2} &= \frac{x}{2r} \\
\frac{y}{x^2+y^2} &= \frac{1}{2r} \\
x^2+y^2-2ry &= 0 \\
x^2 + y^2 - 2ry + r^2 &= r^2 \\
x^2 + (y-r)^2 &= r^2.
\end{align*}
(The integration constant was chosen to have the form $-\ln 2r$ for convenience.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Trigonometric functions I need some help with these tasks and would be very grateful if someone shows me the way of solving them.
If $\sin\left(a\right)+\cos\left(a\right)=b$ and $\left|b\right|\le \sqrt{2}$, represent the following expressions with $b$:
$A=\sin\left(a\right)\cdot \cos\left(a\right)$,
$B=\left|\sin\left(a\right)-\cos\left(a\right)\right|$,
$C=\left|\sin^2\left(a\right)-\cos^2\left(a\right)\right|$,
$D=\left|\sin^3\left(a\right)+\cos^3\left(a\right)\right|$,
$E=\sin^4\left(a\right)+\cos^4\left(a\right)$
Answer: $A\:=\:\frac{1}{2}\left(b^2-1\right),\:B\:=\sqrt{2-b^2},C=\left|b\right|\sqrt{2-b^2},\:D\:=\:\frac{\left|b\right|}{2}\left(3-b^2\right),\:E=\frac{1}{2}\left(1+2b^2-b^4\right)$
| Hints
$A)$
$$(\sin a +\cos a)^2=1+2\sin a \cos a$$
$B)$
$$(\sin a - \cos a)^2=1-2\sin a \cos a$$
$C)$
$$\sin^2a-\cos^2a=(\sin a+\cos a)(\sin a - \cos a)$$
$D)$
$$\sin^3a+\cos ^3a=(\sin a +\cos a)(1-\sin a \cos a)$$
$E)$
$$\sin^4 a+\cos^4a=(\sin^2a+\cos^2 a)^2-2(\sin a\cos a)^2=1-2(\sin a\cos a)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I find the slope of an angle bisector, given the equations of the two lines that form the angle? The equation for the first line is $y = \frac{1}{2}x - 2$, and the equation for the second line is $y = 2x + 1$. They intersect at $(-2, -3)$.
Someone told me I can just average the slopes of the two lines to find the slope of the bisector, but I'm not sure if it's right.
| Line 1 is $y = \frac 12 x - 2$. Line 2 is $y = 2x +1$. They intersect at $(-2-3)$.
If on line 1 you go over on $x$ $2$ units you will go up on $y$ by $1$ unit. This will put you at $(0, -2)$. The distance traveled is $\sqrt {1^2 + 2^2} = \sqrt{5}$.
If on line 2 you go over on $x$ $1$ units you will go up on $y$ by $2$ unit. This will put you at $(-1, -1)$. The distance traveled is $\sqrt {2^2 + 1^2} = \sqrt{5}$.
The angle bisector will go through the midpoint of $(0,-2)$ and $(-1,-1)$$*$. So the angle bisector will go through $(-\frac 12, -1\frac 12)$. So the angle bisector goes through the point $(-2,-3)$ and $(-\frac 12, -1\frac 12)$ so the slope is $\frac{-1\frac 12 - (-3)}{-1\frac 12 -(-2)} = \frac {1\frac 12}{1\frac 12} = 1$.
$*$ because... $A = (-2,-3); B= (0,-2); C=(-1,-1);$ and $AB$ = $AC= \sqrt{5}$ so $\triangle BAC$ is isoceles, and the angle bisector of $\angle BAC$ passes through the midpoint of $BC$.
==== details in general ====
You friend is not quite right. You can average the angles but slopes are not angles and there is not a linear conversion between them. (There is a trigonometric conversion between them. But not a linear conversion.)
Bear with me.
Suppose the two lines intersect at $(u,v)$ and line $1$ has slope $m$ and line $2$ has slope $n$.
Move along the line $1$ from $(u,v)$ a distance of $1$ unit. You will have move $\delta $ in the $x$ direction and $m*\delta $ in the $y$ direction so your total distance is $\sqrt{\delta^2 + m^2\delta^2} = 1$.
So $\delta\sqrt{1 + m^2} = 1$ so $\delta = \frac 1{\sqrt{1 + m^2}}$.
So the point on line $1$ that is one unit away form $(u, v)$ is the point $(x_1, y_1) = (u + \frac 1{\sqrt{1 + m^2}}, v + \frac m{\sqrt{1 + m^2}})$.
Likewise the point on line $2$ that is one unit away from $(u,v)$ will be the point $(x_2, y_2) = (u + \frac 1{\sqrt{1 + n^2}}, v + \frac n{\sqrt{1 + n^2}})$
The angle bisector will contain the midpoint of $(x_1, x_2)$ and $(x_2, y_2)$.
The midpoint is $(x_m, y_m) = (u + \frac{[\frac 1{\sqrt{1 + m^2}}]+[\frac 1{\sqrt{1 + n^2}}]}2, w + \frac{[\frac m{\sqrt{1 + m^2}}]+[\frac n{\sqrt{1 + n^2}}]}2)$.
So.... the slope of the angle bisector will be:
$\frac {y_m - v}{x_m - u}= \frac{\frac{[\frac m{\sqrt{1 + m^2}}]+[\frac n{\sqrt{1 + n^2}}]}2}{\frac{[\frac 1{\sqrt{1 + m^2}}]+[\frac 1{\sqrt{1 + n^2}}]}2}=$
$\frac{[\frac m{\sqrt{1 + m^2}}]+[\frac n{\sqrt{1 + n^2}}]}{[\frac 1{\sqrt{1 + m^2}}]+[\frac 1{\sqrt{1 + n^2}}]}=\frac{m\sqrt{1 + n^2}+n\sqrt{1 + m^2}}{\sqrt{1 + m^2}+\sqrt{1 + n^2}}$
$[\frac{m\sqrt{1 + n^2}+n\sqrt{1 + m^2}}{\sqrt{1 + m^2}+\sqrt{1 + n^2}}=\frac{\frac 12\sqrt{1 + 2^2}+2\sqrt{1 + \frac 12^2}}{\sqrt{1 + \frac 12^2}+\sqrt{1 + 2^2}}=\frac{\sqrt{5}/2+ \sqrt{5}}{\sqrt{5}+ \sqrt{5}/2}=1]$
Which... I must confess is a formula I never learned and would never expect anyone to memorize. I'd expect if one needs to find the slope of an angle bisector one would calculate it for the specific lines.
You might be able to simplify that equation further.
=====
That equation is sort of an "average"; just not a standard arithmetic average.
You can calculate the average of the angles of the lines. But slopes are not angles and do not have a linear conversion.
If you know trigonemetry:
Slope = $\frac{rise}{run} = \frac{\sin \theta}{\cos \theta} = \tan \theta$ so the angle of a line is $\theta = \arctan m$.
So the angle of the angle bisector is $\psi = \frac {\theta + \omega}2 = \frac {\arctan(m) + \arctan(n)}2$
And the slope of the angle bisector is $k = \tan(\psi) = \tan(\frac {\arctan(m) + \arctan(n)}2)=\frac{m\sqrt{1 + n^2}+n\sqrt{1 + m^2}}{\sqrt{1 + m^2}+\sqrt{1 + n^2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Accumulation of a series of payments A loan is repayable by eight annual payments, starting in one year's time with an interest rate $i$. Payments one to three are half as much as payments four to eight. What is the accumulated value of payments one year before the end of the eight annual payments.
What I thought it would be:
Let $X$ be the payment per year. From year 1 to 3 we pay $X$ Annually, and $2X$ From year 3 to 7, so the accumulated value is $$X + X(1+i) + X(1+i)^2 + 2X(1+i)^3 + 2X(1+i)^4 + 2X(1+i)^5 + 2X(1+i)^6$$, but the correct answer is in fact:
$$2X + 2X(1+i) + 2X(1+i)^2 + 2X(1+i)^3 + X(1+i)^4 + X(1+i)^5 + X(1+i)^6$$
Could someone explain the logic behind this?
| You have the correct value of the payments, but it looks like your interest factors are a bit off. The first payment of $X$ takes place in one year. It will sit in the account for six years (since the accumulation takes place one year before the loan is paid off). This gives you that the first payment will accumulate to
\begin{equation*}
X(1+i)^6
\end{equation*}
The next two payments will follow suit, but they will each have a reduced exponent. The accumulated values of those payments are
\begin{align*}
X(1+i)^5\\
X(1+i)^4
\end{align*}
respectively.
The remaining payments are all $2X$, and they grow $3$, $2$, $1$, and $0$ years. This gives us the values
\begin{align*}
2X&(1+i)^3\\
2X&(1+i)^2\\
2X&(1+i)\\
2X&
\end{align*}
Summing the seven values I listed gives you the desired response.
\begin{equation*}
2X+2X(1+i)+2X(1+i)^2+2X(1+i)^3+X(1+i)^4+X(1+i)^5+X(1+i)^6
\end{equation*}
Although, I prefer your order:
\begin{equation*}
X(1+i)^6+X(1+i)^5+X(1+i)^4+2X(1+i)^3+2X(1+i)^2+2X(1+i)+2X
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Irreducible factorization of quadratic function with complex roots? How can we get the irreducible factorization of function
$$f(x)=1+4x+8x^2$$
which is $$f(x)=(1+(2+2i)x)(1+(2-2i)x)?$$
If we find the discriminant, which is $\Delta=-16$, we get
$$\sqrt{\Delta}=4i,x_1=\frac{i-1}{4},x_2=\frac{-1-i}{4}$$
Now we have $$f(x)=8\left(x-\frac{i-1}{4}\right)\left(x+\frac{i+1}{4}\right)$$
My question is how can we get $f(x)=(1+(2+2i)x)(1+(2-2i)x)$ from $f(x)=8\left(x-\frac{i-1}{4}\right)\left(x+\frac{i+1}{4}\right)$?
| The factorization you got also is an irreducible factorization. To get to the other one, note that
$$ 8 = 2^2 + 2^2 = (2+2i)(2-2i), $$
hence
$$ 8\left(x + \frac{i-1}4\right)\left(x + \frac{i+1}4\right)
= \left((2-2i)x + \frac{(2-2i)(i+1)}4\right) \left((2+2i)x + \frac{(2+2i)(i-1)}4\right) = \bigl((2-2i)x + 1\bigr)\bigl((2+2i)x + 1\bigr) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\mathrm{d}\theta $ Evaluate
$$\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\,\mathrm{d}\theta $$
Several days ago,I found this interesting integral from a paper about generalized log-sine integrals,but I can't remember the title of it. The answer of the integral is
\begin{align*}
-\mathrm{Ls}_{7}^{\left ( 3 \right )}\left ( \pi \right)&=\frac{9}{35}\log^72+\frac{4}{5}\pi ^{2} \log^52+9\zeta \left ( 3 \right )\log^42-\frac{31}{30}\pi ^{4}\log^32\\
&-\left [ 72\mathrm{Li}_5\left ( \frac{1}{2} \right )-\frac{9}{8}\zeta \left ( 5 \right )-\frac{51}{4}\pi ^{2}\zeta \left ( 3 \right ) \right ]\log^22\\
&+\left [ 72\mathrm{Li}_{5,1}\left ( \frac{1}{2} \right )-216\mathrm{Li}_6\left ( \frac{1}{2} \right )+36\pi ^{2}\mathrm{Li}_4\left ( \frac{1}{2} \right ) \right ]\log2+72\mathrm{Li}_{6,1}\left ( \frac{1}{2} \right )\\
&-216\mathrm{Li}_7\left ( \frac{1}{2} \right )+36\pi ^{2}\mathrm{Li}_5\left ( \frac{1}{2} \right )-\frac{1161}{32}\zeta \left ( 7 \right )-\frac{375}{32}\pi ^{2}\zeta \left ( 5 \right )+\frac{1}{10}\pi ^{4}\zeta \left ( 3 \right )
\end{align*}
where
$$\mathrm{Ls}_n^{\left ( k \right )}\left ( \alpha \right ):=-\int_{0}^{\alpha }\theta ^{k}\log^{n-1-k}\left | 2\sin\frac{\theta }{2} \right |\mathrm{d}\theta $$
is the generalized log-sine integral and
$$\mathrm{Li}_{\lambda ,1}\left ( z \right )=\sum_{k=1}^{\infty }\frac{z^{k}}{k^{\lambda }}\sum_{j=1}^{k-1}\frac{1}{j}$$
is the multiple polylogarithm.
I found a beautiful way to solve the integrals below
$$\int_{0}^{\frac{\pi }{2}}t^{2n}\log^{m}\left ( 2\cos t \right )\mathrm{d}t $$
Let's consider
$$\mathcal{I}\left ( x,y \right )=\int_{0}^{\frac{\pi }{2}}\cos\left ( xt \right )\left ( 2\cos t \right )^{y}\mathrm{d}t$$
By using Gamma function,the integral become
$$\mathcal{I}\left ( x,y \right )=\frac{\pi \, \Gamma \left ( y+1 \right )}{2\Gamma \left ( \dfrac{x+y+2}{2} \right )\Gamma \left ( \dfrac{y-x+2}{2} \right )}$$
Then we can get
$$\mathcal{I}\left ( x,y \right )=\frac{\pi }{2}\exp\left ( \sum_{k=2}^{\infty }\frac{\left ( -1 \right )^{k}}{k\cdot 2^{k}}\zeta \left ( k \right )\left [ \left ( 2y \right )^{k}-\left ( y-x \right )^{k}-\left ( x+y \right )^{k} \right ] \right )$$
On the other hand,using taylor series
$$\mathcal{I}\left ( x,y \right )=\sum_{n=0}^{\infty }\frac{\left ( -1 \right )^{n}}{\left ( 2n \right )!}x^{2n}\sum_{m=0}^{\infty }\frac{y^{m}}{m!}\int_{0}^{\frac{\pi }{2}}t^{2n}\log^m\left ( 2\cos t \right )\mathrm{d}t$$
So,the comparison of coefficient shows the answer.For example
$$\int_{0}^{\frac{\pi }{2}}t^{2}\log^2\left ( 2\cos t \right )\mathrm{d}t=4\cdot \frac{\pi }{2}\left [ \frac{12}{4\cdot 16} \zeta \left ( 4 \right )+\frac{1}{2}\frac{8}{2^{2}\cdot 4^{2}}\zeta \left ( 2 \right )^{2}\right ]=\frac{11}{1440}\pi ^{5}$$
I wonder can we use the same way to prove the integral in the beginning,if not,is there another way to handle it?
| I think you can apply the method only partially for the integral
\begin{align}
\int_{0}^{\pi }\theta ^{3}\log^{3}\left ( 2\sin\frac{\theta }{2} \right )\,\mathrm{d}\theta &= 2^4\int_{0}^{\pi /2 }\left(\frac{\pi}{2}-\theta \right)^{3}\log^{3}\left ( 2\cos\theta\right )\,\mathrm{d}\theta\\
&=2π^3\int_{0}^{\pi /2 }\log^{3}\left ( 2\cos\theta\right )\,\mathrm{d}\theta - 12 π^2\int_{0}^{\pi /2 } θ\log^{3}\left ( 2\cos\theta\right )\,\mathrm{d}\theta \\&+ 24 π\int_{0}^{\pi /2 } θ^2\log^{3}\left ( 2\cos\theta\right )\,\mathrm{d}\theta - 16\int_{0}^{\pi /2 }θ^3\log^{3}\left ( 2\cos\theta\right )\,\mathrm{d}\theta
\end{align}
Where for even powers of $\theta$ you can use your formula. The other integrals are not trivial.
Note that the approach you suggested is driven by the fact that if
$$\mathcal{I}\left ( x,y \right )=\int_{0}^{\frac{\pi }{2}}\cos\left ( x \theta \right )\left ( 2\cos \theta \right )^{y}\mathrm{d}\theta$$
Then we can solve the integral by differentiation with respect to both $x$ and $y$ but since we cannot get rid of $\sin(x \theta)$ we can apply the derivative even number of times.
$$\frac{\partial^{2n}\partial ^m}{\partial x^{2n}\partial y^m} \mathcal{I}\left ( 0,0 \right )=(-1)^n\int_{0}^{\frac{\pi }{2}}\theta^{2n} \log^m\left ( 2\cos \theta \right )\mathrm{d}\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 4,
"answer_id": 1
} |
How to construct $A$ and $b$, given the set of all solutions to $Ax=b$?
I want to find a matrix $A \in \mathbb{C}^{2x4}$ and
$b \in\mathbb{C}^{2}$ the solution of
$Ax=b$ is: $$L = \left\{\pmatrix{1\\2\\0\\-1} + x_1\pmatrix{1\\-2\\2\\1} + x_2\pmatrix{2\\2\\-1\\1}\right\}$$
Therefore $\dim(A) = 4$, $\dim(\ker(A)) = 2$, $\dim(\operatorname{im}(A)) = 2$.
$A$ and $b$ have the following format:
$$
A= \begin{pmatrix}
a_{11} & a_{12} & a_{13} & a_{14}\\
a_{21} & a_{22} & a_{23} & a_{24} \\
\end{pmatrix}
b= \pmatrix{b_1\\b_2}
$$
My idea is to solve the following equation for A and b:
$$A\,\pmatrix{1 + x_1 + 2x_2\\2 - 2x_1+ 2x_2\\2x_1 - x_2\\-1+x_1 + x_2} = b$$
Is this the right way to start? I get then a linear equation with 12 unknowns and only 2 equations. I know that there must be many solutions. Do I simply define some of them as 1 or 0?
| Using
$$
v_0=\pmatrix{1\\2\\0\\-1} \quad
v_1=\pmatrix{1\\-2\\2\\1} \quad
v_2=\pmatrix{2\\2\\-1\\1}
$$
If you want
$$
b = A(v_0 + x_1 v_1 + x_2 v_2) \iff \\
\underbrace{b - A v_0}_d = x_1 Av_1 + x_2 Av_2
= \underbrace{(Av_1 Av_2)}_C \underbrace{(x_1, x_2)^T}_u \iff \\
C u = d \quad
$$
to hold for all $u = (x_1, x_2)^T \in \mathbb{R}^2$ it must include $0$, so $C0 = 0 = d$ and thus $b = A v_0$ for whatever $A$ we end up with.
Further $C u = 0$ for all $u$ means $C e_i = c_i = 0$ thus $C=(c_1 \, c_2) = (0 \, 0) = 0$.
This requires $A v_1 = A v_2 = 0$.
With $A^T = (r_1 \, r_2)$ we need $r_i^T v_j = r_i \cdot v_j = 0 \iff r_i \perp v_j \iff v_j^T r_i = 0$.
So $\ker A = \langle v_1, v_2 \rangle \perp
\langle r_1, r_2 \rangle = \DeclareMathOperator{img}{img}\img A^T$
$$
V^T r = 0 \iff
\begin{bmatrix}
v_1^T \\
v_2^T
\end{bmatrix}
=
\begin{bmatrix}
1 & -2 & 2 & 1 \\
2 & 2 & -1 & 1
\end{bmatrix}
\to
\begin{bmatrix}
1 & -2 & 2 & 1 \\
0 & 6 & -5 & -1
\end{bmatrix}
\to \\
\begin{bmatrix}
1 & -2 & 2 & 1 \\
0 & 1 & -5/6 & -1/6
\end{bmatrix}
\to
\begin{bmatrix}
1 & 0 & 1/3 & 2/3 \\
0 & 1 & -5/6 & -1/6
\end{bmatrix}
$$
We find $r_1 = (-1, 1, 1, 1)^T$, $r_2 = (0, 3, 4, -2)^T$ do the job.
So
$$
A =
\begin{pmatrix}
-1 & 1 & 1 & 1 \\
0 & 3 & 4 & -2
\end{pmatrix}
\quad
b =
\begin{pmatrix}
0 \\
8
\end{pmatrix}
$$
is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Primes such that $2p^2-3p-1$ is a perfect cube Find all primes $p$ such that $2p^2-3p-1$ is a perfect cube.
I've noticed that $p=2,3$ are solutions. Setting $2p^2-3p-1=x^3$ we can factorise this as $(4p-3)^2=8x^3+17$. However im unable to proceed further.
| Let $$2p^2-3p-1=x^3$$
$$p(2p-3)=(x+1)(x^2-x+1)$$
If $p=2$ then $x=1$
If $p=3$ then $2p^2-3p-1=8=2^3$
Let $p>3$. Then $\gcd(p,2p-3)=1$ and $p<2p-3$
Then
$$p=x+1$$
$$2p-3=x^2-x+1$$
Thus $$-3=x^2-3x-1$$
$$x^2-3x+2=0$$
Hence $x=1$ or $x=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that this lengthy integral is equal to $\int_{0}^{\infty}{1\over x^8+x^4+1}\cdot{1\over x^3+1}\mathrm dx={5\pi\over 12\sqrt{3}}$ $$\displaystyle\int_{0}^{\infty}{1\over x^8+x^4+1}\cdot{1\over x^3+1}\mathrm dx={5\pi\over 12\sqrt{3}}$$
$x^8+x^4+1=(x^2+x-1)(x^2-x-1)(x^2+x+1)(x^2-x+1)$
$x^3+1=(x+1)(x^2-x+1)$
${A\over x+1}+{Bx+C\over x^2-x+1}={1\over x^3+1}$
${Ax+B\over x^2+x-1}+{Cx+D\over x^2-x-1}+{Ex+F\over x^2+x+1}+{Gx+H\over x^2-x+1}={1\over x^8+x^4+1}$
it would be a nightmare trying to decomposition of fraction.
I would like some help please?
| We want to compute:
$$ I = \int_{0}^{1}\frac{x^4-1}{x^{12}-1}\cdot\frac{\mathrm dx}{x^3+1}+\int_{0}^{1}\frac{x^{8}-x^{12}}{1-x^{12}}\cdot\frac{x\,\mathrm dx}{x^3+1}=\int_{0}^{1}\frac{1-x^3+x^6}{1+x^4+x^8}\,\mathrm dx $$
that is:
$$ I = \int_{0}^{1}\left(1-x^3-x^4+x^6+x^7-x^{10}\right)\frac{\mathrm dx}{1-x^{12}} $$
or:
$$ I=\sum_{k\geq 0}\left(\frac{1}{12k+1}-\frac{1}{12k+4}-\frac{1}{12k+5}+\frac{1}{12k+7}+\frac{1}{12k+8}-\frac{1}{12k+11}\right). $$
By the reflection formula for the digamma function we have:
$$ \sum_{k\geq 0}\left(\frac{1}{12k+\tau}-\frac{1}{12k+(12-\tau)}\right)=\frac{\pi}{12}\,\cot\frac{\pi \tau}{12} $$
and the wanted result trivially follows from $\cot\left(\frac{\pi}{12}\right)=2+\sqrt{3}$.
| {
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"url": "https://math.stackexchange.com/questions/2089367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Help to prove that $\int_{0}^{\infty}{\sqrt{x^2+1}+x^2\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{1\over (1+x^2)^2}\mathrm dx={5\over 6}$ I am trying to prove that
$$\displaystyle \int_{0}^{\infty}{\sqrt{x^2+1}+x^2\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{1\over (1+x^2)^2}\mathrm dx={5\over 6}$$
$u=(x^2+1)^{1/2}$ then $du=(x^2+1)^{-1/2}dx$
$$\int_{1}^{\infty}{u^2+(u^2-1)\sqrt{u^2+1}\over u^2\sqrt{u^2+1}}du$$
$v=(u^2+1)^{1/2}$ then $dv=(u^2+1)^{-1/2}du$
$$\int_{1}^{\infty}{v^3+v^2-2v-1\over v^2-1}dv$$
$\int_{1}^{\infty}{v^2\over v^2-1}-{1\over v^2-1}dv$ -$\ln{(v^2-1)}|_{1}^{\infty}$
I am sure I when wrong somewhere, but I can figured it out.
Any help?
| Let $A=\sqrt{x^2+1}$ and $B=\sqrt{x^2+2}$ so that have $A^2-1=x^2$. With these you have
\begin{align}
{\sqrt{x^2+1}+x^2\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{1\over (1+x^2)^2}&=\frac{A+(A^2-1)B}{AB}\frac{1}{A^4}\\
&=\frac{1}{BA^4}-\frac{1}{A^3}-\frac{1}{A^5}\\
\end{align}
Note that the integral of all of these terms is very simple to evaluate, so have fun :-)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Different approach to tackle $\int_{0}^{\infty}{2\sqrt{x^2+1}+\gamma(x^2-x)\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{\mathrm dx\over (x^2+1)^2}=1$ $$\int_{0}^{\infty}{2\sqrt{x^2+1}+\gamma(x^2-x)\sqrt{x^2+2}\over \sqrt{(x^2+1)(x^2+2)}}\cdot{\mathrm dx\over (x^2+1)^2}=1\tag1$$
Where $\gamma=0.577...$ it is Euler's Constant
$$\int_{0}^{\infty}{2\over \sqrt{x^2+2}}\cdot{\mathrm dx\over (x^2+1)^2}+\gamma\int_{0}^{\infty}{x^2-x\over (x^2+1)^{5/2}}\mathrm dx\tag2$$
$$\gamma\int_{0}^{\infty}{x^2-x\over (x^2+1)^{5/2}}\mathrm dx=\gamma\int_{0}^{\infty}{x^2\over (x^2+1)^{5/2}}\mathrm dx-\gamma\int_{0}^{\infty}{x\over (x^2+1)^{5/2}}\mathrm dx\tag3$$
Enforcing $u=x^2$ then $du=2xdx$
Recalling from the beta function
$$\int_{0}^{\infty}{u^m\over (u+1)^{m+n+2}}du=B(m+1,n+1)\tag4$$
$${\gamma\over 2}\int_{0}^{\infty}{u\over (u+1)^{5/2}}\mathrm du-{\gamma\over 2}\int_{0}^{\infty}{1\over (u+1)^{5/2}}\mathrm du={B(3/2,1)\over2}-{B(1,3/2)\over2}=0\tag5$$
So this part of integration must be 1! It is ready done my @Marco on my previous post.
$$\int_{0}^{\infty}{2\over \sqrt{x^2+2}}\cdot{\mathrm dx\over (x^2+1)^2}\tag6$$
Can anyone prove this integral $(1)$ via another method? Thank you! Sorry for not addressing the question properly.
| In fact, letting $x=\tan t$ and $u=\sin t$, one has
\begin{eqnarray}
&&\int_{0}^{\infty}{2\over \sqrt{x^2+2}}\cdot{\mathrm dx\over (x^2+1)^2}\\
&=&2\int_0^{\pi/2}\frac{1}{\sqrt{1+\sec^2t}}\frac{1}{\sec^4t}\sec^2tdt\\
&=&2\int_0^{\pi/2}\frac{\cos^3 t}{\sqrt{1+\sec^2t}}dt\\
&=&2\int_0^{\pi/2}\frac{\cos^3 t}{\sqrt{2-\sin^2t}}dt\\
&=&2\int_0^{1}\frac{1-u^2}{\sqrt{2-u^2}}du\\
&=&u\sqrt{2-u^2}|_0^1\\
&=&1.\\
&&\int_{0}^{\infty}{x^2-x\over (x^2+1)^{5/2}}\mathrm dx\\
&=&\int_{0}^{\pi/2}{\tan^2t-\tan t\over (\sec^2t)^{5/2}}\sec^2t\mathrm dx\\
&=&\int_{0}^{\pi/2}(\tan^2t-\tan t)\cos t\mathrm dx\\
&=&\int_{0}^{\pi/2}\cos t\sin t(\sin t-\cos t)\mathrm dx\\
&=&\int_{0}^{\pi/2}\cos t\sin^2 t\mathrm dx-\int_{0}^{\pi/2}\cos^2 t\sin t\mathrm dx\\
&=&0.
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to simplify this fraction with different powers? I happen to be stuck trying to simplify this:
$$\left[\frac{(3x+2)(x+1)^\frac{3}{2}-(\frac{3}{2}x^2+2x)(\frac{3}{2})(x+1)^\frac{1}{2}}{(x+1)^3}\right]$$
here's the simplified solution that I'm trying to figure out how it was reached
| We have
$$\frac{(3x+2)(x+1)^\frac{3}{2}-(\frac{3}{2}x^2+2x)(\frac{3}{2})(x+1)^\frac{1}{2}}{(x+1)^3}$$
divide numerator and denominator by $(x+1)^{\frac12}$ giving
$$\frac{(3x+2)(x+1)-(\frac{3}{2}x^2+2x)(\frac{3}{2})}{(x+1)^\frac52}$$
multiply the $(3x+2)(x+1)$ to get $3x^2 + 5x + 2$ and the $(\frac32x^2 + 2x)(\frac32)$ to get $\frac94x^2 + 3x$ giving:
$$\frac{3x^2 + 5x + 2-\frac94x^2 - 3x}{(x+1)^\frac52}$$
then simplify the numerator:
$$3x^2 + 5x + 2-\frac94x^2 - 3x = \frac{12}{4}x^2 - \frac94x^2 + 5x - 3x + 2 = \frac34x^2 + 2x + 2$$
to get
$$\frac{\frac34x^2 + 2x + 2}{(x+1)^\frac52}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $abc=1$ prove that $\sum\limits_\text{cyc}\frac{1}{a+3}\geq\sum\limits_\text{cyc}\frac{a}{a^2+3}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}\geq\frac{a}{a^2+3}+\frac{b}{b^2+3}+\frac{c}{c^2+3}$$
I tried TL, BW, the Vasc's Theorems and more, but without success.
I proved this inequality!
I proved also the hardest version: $\sum\limits_{cyc}\frac{1}{a+4}\geq\sum\limits_{cyc}\frac{a}{a^2+4}$.
Thanks all!
| Partial Hint and too long for a comment :
Case where two varaibles are superior two one .
If we define :
$$f\left(x\right)=\frac{1}{e^{x}+3}-\frac{e^{x}}{e^{2x}+3},g\left(x\right)=\ln\left(1+e^{-x}\right)-\ln\left(2\right)$$
Show that for $x\in(-\infty,\infty)$ :
$$f''(x)+g''(x)>0$$
Now apply Jensen's inequality on two variable .
Remains to show :
$$-\left(g\left(a\right)+g\left(b\right)-2g\left(\frac{-c}{2}\right)\right)+f\left(c\right)\geq 0$$
Or :
$$-\left(g\left(a\right)+g\left(b\right)-2g\left(\frac{-c}{2}\right)\right)+\ln\left(f\left(c\right)+1\right)\geq 0$$
Wich is true for $a+b+c=0, a,b\in(0,2) \operatorname{or} a,b\in(2,\infty)$,
Last hint remark with the constraint $x,y\in [1,\infty)$:
$$\frac{\left(1+\frac{1}{xy+3}-\frac{xy}{\left(xy\right)^{2}+3}\right)\left(1+\sqrt{xy}\right)^{2}}{\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)}\ge \frac{\left(3+\frac{1}{xy+3}-\frac{xy}{\left(xy\right)^{2}+3}\right)\left(3+\sqrt{xy}\right)^{2}}{3\left(3+\frac{1}{x}\right)\left(3+\frac{1}{y}\right)}\ge \frac{4\left(3+\frac{1}{xy+3}-\frac{xy}{\left(xy\right)^{2}+3}\right)\left(3+\sqrt{xy}\right)}{3\left(3+\frac{1}{x}\right)\left(3+\frac{1}{y}\right)}\ge1$$
Case where two variable are less than one
Now consider the function :
$$p\left(x\right)=\frac{1}{x+3}-\frac{x}{x^{2}+3}$$
There exists a positive constant $C$ and $b\in(0,1),x>0$ such that :
$$p\left(Cx\right)-p\left(Cx^{b}\right)\geq 0$$
$$C=\left(3-2\sqrt{2}\right)^{\frac{1}{3}}+\left(3+2\sqrt{2}\right)^{\frac{1}{3}}$$
To be continued...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is $\sum\limits_{k=0}^{n} {{n}\choose{k}}\frac{(-1)^{k}}{2k+2}$ equal to $\frac{1}{2n+2}$? On p. 146 of Principles of Mathematical Analysis, Rudin claims that $\int_{0}^{1} x(1-x^2)^ndx=\frac{1}{2n+2}$. But evaluating the integral I found it equal to $$\int_{0}^{1}x(\sum\limits_{k=0}^{n}{{n}\choose{k}}(-x^2)^k)dx=\int_{0}^{1}\sum\limits_{k=0}^{n}{{n}\choose{k}}(-1)^{k}x^{2k+1}dx=\sum\limits_{k=0}^{n}{{n}\choose{k}}(-1)^{k}\frac{x^{2k+2}}{2k+2}$$ evaluated at $x=1$, which equals $\sum\limits_{k=0}^{n} {{n}\choose{k}}\frac{(-1)^{k}}{2k+2}$. So now I'm just wondering, why is this equal to $\frac{1}{2n+2}$?
| $$\begin{align*}
\sum_{k=0}^n\binom{n}k\frac{(-1)^k}{2k+2}&=\frac{1}2\sum_{k=0}^n\frac{1}{k+1}\binom{n}k(-1)^k\\
&=\frac{1}2\sum_{k=0}^n\frac{1}{n+1}\binom{n+1}{k+1}(-1)^k\\
&=\frac{1}{2n+2}\sum_{k=1}^{n+1}\binom{n+1}k(-1)^{k-1}\\
&=\frac{-1}{2n+2}\sum_{k=1}^{n+1}\binom{n+1}k(-1)^k1^{n+1-k}\\
&=\frac{-1}{2n+2}\left(\sum_{k=0}^{n+1}\binom{n+1}k(-1)^k1^{n+1-k}-\binom{n+1}0(-1)^01^{n+1}\right)\\
&=\frac{-1}{2n+2}\left((-1+1)^{n+1}-1\right)\\
&=\frac{1}{2n+2}
\end{align*}$$
by the binomial theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Question about inverse function $f(x)=\frac{1}{2}(x+\sqrt{x^2+4})$ This was from iranian university entrance exam .Suppose $f(x)=\frac{1}{2}(x+\sqrt{x^2+4})$ find $f^{-1}(x)+f^{-1}(\frac{1}{x}),x \neq 0$.
It is easy to find $f^{-1}$ and solve this like below ...
$$y=\frac{1}{2}(x+\sqrt{x^2+4}) \\(2y-x)^2=(\sqrt{x^2+4})^2\\4y^2+x^2-4xy=x^2+4\\4y^2-4=4xy\\x=\frac{y^2-1}{y}=y-\frac{1}{y} \\ \to f^{-1}(x)=x-\frac{1}{x}\\
f^{-1}(x)+f^{-1}(\frac{1}{x})=x-\frac{1}{x}+(\frac{1}{x}-\frac{1}{\frac{1}{x}})=0\\$$ And now my question is ...
is there an other method to solve this question ?
I was thinking about $f(x)f(-x)=1$ but I can't go anymore ...
Any hint ,or other Idea ?
Thanks in advanced.
| Perhaps your observation can be made to work (note $f(x) \neq 0$):
\begin{align*}
f(x)f(-x) & =1\\
f(-x) & = \frac{1}{f(x)}\\
f^{-1}(f(-x)) & = f^{-1}\left(\frac{1}{f(x)}\right)\\
-x & = f^{-1}\left(\frac{1}{f(x)}\right)\\
f^{-1}\left(\frac{1}{f(x)}\right)+x & = 0\\
f^{-1}\left(\frac{1}{f(x)}\right)+f^{-1}(f(x)) & = 0\\
f^{-1}\left(\frac{1}{y}\right)+f^{-1}(y) & = 0
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/2091552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the minimum value of $\dfrac {9x^2\sin^2 x+4}{x\sin x}$
Question: What is the minimum value of $$\dfrac {9x^2\sin^2x+4}{x\sin x}\tag1$$For $0<x<\pi$.
I solved this, but somewhere, I did something wrong. My work is as follows:
First, set $y=x\sin x$. Therefore,
$$\begin{align*}\dfrac {9y^2+4}{y}=\dfrac 4y+9y\tag2\end{align*}$$
And by the AM-GM Inequality, we have$$\begin{align*}\dfrac {9y+\dfrac 4y}2 & \geq\sqrt{\dfrac 4y\cdot 9y}\tag3\\9y+\dfrac 4y & \geq12\tag4\end{align*}$$
$(4)$ is minimized only when the RHS is equal to zero. So$$9y+\dfrac 4y=0\implies 9y^2=-4\implies y^2=-\dfrac 49\tag5$$
However, if I square root both sides, I get an imaginary value. Where did I go wrong? The book says the solution is $12$ (received by plugging in $x=\dfrac 23$).
| You already noted that
$$9y + \frac{4}{y} \geq 12$$
so if the LHS were exactly $12$, it would certainly be minimal!
If equality is to be attained, you have
$$9y + \frac{4}{y} - 12 = 0,$$
multiplying by $y$ gives
$$9y^2 - 12y + 4 = (3y - 2)^2 = 0$$
Hence $y = 2/3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the maximum value of the expression ${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$.
Find the maximum value of the expression :
$${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$$
where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$.
Taking $x=y=z=\frac {1}{3}$, I get the expression as $\frac {3x}{1+x^2}$, which is equal to $\frac {1}{1+{\frac{1}{9}}}$ or $\frac {9}{10}$.
How can I actually solve the problem without making unnecessary assumptions ?
| Let $x=\frac{a}{3}$, $y=\frac{b}{3}$ and $z=\frac{c}{3}$.
Hence, $a+b+c=3$ and we need to prove that $$\sum_{cyc}\frac{a}{9+a^2}\leq\frac{3}{10}$$ or
$$\sum_{cyc}\left(\frac{1}{10}-\frac{a}{9+a^2}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^2-10a+9}{9+a^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{a^2-10a+9}{9+a^2}+\frac{2}{3}\right)\geq2$$ or
$$\sum_{cyc}\frac{(a-3)^2}{9+a^2}\geq\frac{6}{5}.$$
Let $k=3\sqrt2-6$. Hence, by C-S
$$\sum_{cyc}\frac{(a-3)^2}{9+a^2}=\sum_{cyc}\frac{(a-3)^2(a+k)^2}{(a+k)^2(9+a^2)}\geq\frac{\left(\sum\limits_{cyc}(a-3)(a+k)\right)^2}{\sum\limits_{cyc}(a+k)^2(a^2+9)}=\frac{\left(\sum\limits_{cyc}(a^2-3-2k)\right)^2}{\sum\limits_{cyc}(a+k)^2(a^2+9)}.$$
Thus, it remains to prove that
$$5\left(\sum\limits_{cyc}(a^2-3-2k)\right)^2\geq6\sum\limits_{cyc}(a+k)^2(a^2+9),$$
which is fourth degree.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative and $abc=w^3$.
Hence, the last inequality is a linear inequality of $w^3$, which says that
it's enough to prove the last inequality for en extremal value of $w^3$,
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$. We obtain
$$(a-1)^2(\sqrt2a-3\sqrt2+3)^2\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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if $a+b=k\pi,$ then finding value of $k$ if the range of parameter $t$ in the interval $\left(0, 2\pi\right)$ satisfying
$\displaystyle \frac{(-2x^2+5x-10)}{(\sin t) x^2 + 2(1+ \sin t )x + 9\sin t +4} > 0$ for all real values of $x$
is $(a,b)$ and $a+ b=k\pi$ . then finding $k$
above $\displaystyle \displaystyle \frac{(2x^2-5x+10)}{(\sin t) x^2 + 2(1+ \sin t )x + 9\sin t +4} <0$
and $\displaystyle 2x^2-5x+10>0$ for all real values of $x$
so $(\sin t)x^2+2(1+\sin t)x+9\sin t+4<0$
wan,t be able to go further, could some help me with this
| $ax^2+bx+c<0$ for all real values of $x$ iff $a<0$ and $b^2<4ac$.
In our case, $a = \sin t, b=2(1+\sin t), c = 9\sin t +4$. So, $\sin t$ is negative.
Moreover, $b^2<4ac$ gives
$$4(1 + 2\sin t +\sin^2t)<4(9\sin^2 t + 4\sin t)$$
$$1 + 2\sin t +\sin^2t<9\sin^2 t + 4\sin t$$
$$0<8\sin^2 t + 2\sin t - 1$$
$$0<(4\sin t -1)(2\sin t +1)$$
As $\sin t$ is negative, the first term is always negative. So, we want the second term to be negative, which means $\sin t < -\frac12$, i.e. $t\in (\frac{4\pi}3, \frac{5\pi}3)$. So, $a=\frac43$ and $b=\frac53$. Thus, $k = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096024",
"timestamp": "2023-03-29T00:00:00",
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If $x,y>1$ and $(\ln x)^2+(\ln y)^2=\ln x^2+\ln y^2.$ then maximum value of $x^{\ln y}$ If $x,y>1$ and $(\ln x)^2+(\ln y)^2=\ln x^2+\ln y^2.$ then maximum value of $x^{\ln y}$
we can write $(\ln x)^2-\ln (x^2) = -((\ln y)^2-\ln (y^2))$ and assuming $x\geq y$
now let $f(x)=(\ln x)^2-\ln (x^2) $
$\displaystyle f'(x)=2 \ln x \cdot \frac{1}{x}-\frac{1}{x^2}\cdot 2x =\frac{2(\ln x-1)}{x}$
and $\displaystyle f''(x) = \frac{2(2-\ln x)}{x^2}$
wan,t be able to go further, could some help me with this, thanks
| Let $\ln x = a$ and $\ln y=b$. Then the equation is written as
$$a^2+b^2 = 2a + 2b.$$
We can now find maximum values of a and b by making up a quadratic in $a$
and $b$, i.e.
$$
(a-1)^2 + (b-1)^2 = 2.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the derivative of $7^{x^2-x}$ using the chain rule $7^{x^2-x}$
I know there is a formula ($(a^u)' = u' \cdot a^u \cdot \ln a$) for this but I wanted to understand the logic behind that formula so I tried using the chain rule to solve this:
$$(7^{x^2-x})' = (x^2-x)'\cdot7^{x^2-x-1}\cdot 7' = (2x-1)\cdot7^{x^2-x-1}\cdot0 = 0$$
Then I tried replacing $7^{x^2-x}$ with $(e^x)^{\ln7\cdot(x-1)}$:
$$((e^x)^{\ln 7 (x-1)})' = (\ln 7 (x-1))' \cdot (e^x)^{\ln 7 (x-1)-1} \cdot (e^x)' = (\ln7) \cdot (e^x)^{\ln 7 (x-1)-1} \cdot e^x = (\ln7) \cdot (e^x)^{\ln 7 (x-1)} = (\ln7) \cdot 7^{x(x-1)} =(\ln7) \cdot 7^{x(x-1)}$$
But the right answer is $(\ln7) \cdot 7^{x(x-1)} \cdot (2x-1)$. What did I do wrong?
| You can go straight to the chain rule.
Call $f(x)=7^x$ and $g(x)=x^2-x$, then
$$g'(x)=2x-1 \quad \text{and} \quad f'(x)=7^x(\ln 7)$$
so,
$$[f(g(x))]'=g'(x)f'(g(x))=(2x-1)7^{g(x)}(\ln7)=(2x-1)7^{x(x-1)}(\ln7)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Another way to express $\text{cis } 75^\circ + \text{cis } 83^\circ + \text{cis } 91^\circ + \dots + \text{cis } 147^\circ$ The number
$$\text{cis } 75^\circ + \text{cis } 83^\circ + \text{cis } 91^\circ + \dots + \text{cis } 147^\circ$$ is expressed in the form $r \, \text{cis } \theta$, where $0 \le \theta < 360^\circ$. Find $\theta$ in degrees.
Hint: $\text{cis}\ \theta=\cos \theta+i \sin \theta$
Edit:
I simplified the expression down to$$\frac{\text{cis} \ 75^\circ \sin 40^\circ \ \text{cis} \ 40^\circ }{\sin 4^\circ \ \text{cis} \ 4^\circ}.$$
What should I do now?
| Hint:
$e^{i(75\pi/180)}+e^{i(83\pi/180)}+e^{i(91\pi/180)}+\cdots +e^{i(147\pi/180)}$
G.P series with common ratio $r=e^{i\frac{8\pi}{180}}$
Sum of GP series $S_n = \frac{1-r^n}{1-r}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $19\mid 5^{2n+1}+3^{n+2} \cdot 2^{n-1}$ How can I prove that $$5^{2n+1}+3^{n+2} \cdot 2^{n-1} $$ can be divided by 19 for any nonnegative n? What modulo should I choose?
| Denote $\mathcal{P}(n)$ the statement that $5^{2n+1} + 3^{n+2}\cdot 2^{n-1}$ is divisible by $19$. You can check for yourself that $\mathcal{P}(1)$ is true. A setup for the proof of $\mathcal{P}(n+1)$:
\begin{align}
5^{2(n+1)+1} + 3^{(n+1)+2}\cdot 2^{(n+1)-1} & = \\
25 \times 5^{2n+1} + 6 \times 3^{n+2}\cdot 2^{n-1} & = \\
19 \times 5^{2n+1} + 6 \times 5^{2n+1} + 6 \times 3^{n+2}\cdot 2^{n-1} & = \cdots
\end{align} can you finish the proof from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluate $\lim\limits_{x\to \infty}x^2 \left( a^{1 \over x} - a^{1 \over x+1} \right)$ I am trying to evaluate
$$\lim\limits_{x\to \infty}x^2 \left( a^{1 \over x} - a^{1 \over x+1} \right)$$
for $a>0$.
The idea:
$$a^{1 \over x} = e^{\ln{a^{1 \over x}}}=e^{{1 \over x}\ln{a}} = 1 + {\ln{a} \over x} + {1 \over 2}{(\ln{a})^2 \over x^2}+{1 \over 6}{(\ln{a})^3 \over x^3}+\dots$$
$$a^{1 \over x}-a^{1 \over x+1} = e^{\ln{a^{1 \over x}}}-e^{\ln{a^{1 \over x+1}}}=e^{{1 \over x}\ln{a}}-e^{{1 \over x+1}\ln{a}} = \ln{a} \left({1\over x} - {1 \over x+1} \right)+ {1 \over 2}(\ln{a})^2\left({ {1\over x^2} - {1 \over (x+1)^2}}\right)+{1 \over 6}(\ln{a})^3\left({ {1\over x^3} - {1 \over (x+1)^3}}\right)+\dots$$
$$x^2 \left(a^{1 \over x}-a^{1 \over x+1}\right) =\ln{a} \left(x - {x^2 \over x+1} \right)+ {1 \over 2}(\ln{a})^2\left({ 1 - {x^2 \over (x+1)^2}}\right)+{1 \over 6}(\ln{a})^3\left({ {1\over x} - {x^2 \over (x+1)^3}}\right)+\dots$$
This is where I get stuck. I suppose the whole sum should converge to $\ln{a}$ but can't quite find the way to get there. I would appreciate any comments.
| You seem to have some issue with your Taylor expansion of $\exp$. Here, we will only need it to second or third order, but you forgot the linear term:
when $u\to 0$, we have $$e^u = 1+\boxed{\color{red} u} +\frac{u^2}{2} + o(u^2)$$
which is crucial here.
Using this, we get, since $\frac{\ln a}{x} \xrightarrow[x\to\infty]{}0$,
$$\begin{align}
a^{\frac{1}{x}} &= e^{\frac{\ln a}{x}}
= 1+\frac{\ln a}{x}+\frac{\ln^2 a}{x^2} + o\left(\frac{1}{x^2}\right) \tag{1}\\a^{\frac{1}{x+1}}&= e^{\frac{\ln a}{x+1}}
= 1+\frac{\ln a}{x+1}+\frac{\ln^2 a}{(x+1)^2} + o\left(\frac{1}{x^2}\right) \\
&= 1+\frac{\ln a}{x}\left(1-\frac{1}{x} + o\left(\frac{1}{x}\right) \right)+\frac{\ln^2 a}{x^2} + o\left(\frac{1}{x^2}\right) \\
&= 1+\frac{\ln a}{x}-\frac{\ln a}{x^2}+\frac{\ln^2 a}{x^2} + o\left(\frac{1}{x^2}\right) \tag{2}
\end{align}$$
using also that $\frac{1}{1+u} = 1-u+o(u)$ when $u\to 0$ (in the third line). Thus, the difference of the two will be
$$
a^{\frac{1}{x}} - a^{\frac{1}{x+1}} = \frac{\ln a}{x^2} + o\left(\frac{1}{x^2}\right)
$$
from (1) and (2) (the other terms cancel), and the limit will then be
$$
\lim_{x\to\infty}x^2\left( a^{\frac{1}{x}} - a^{\frac{1}{x+1}}\right) = \lim_{x\to\infty}x^2\left( \frac{\ln a}{x^2} + o\left(\frac{1}{x^2}\right) \right) = \boxed{\ln a}
$$
as desired.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $x^2+xy+y^2+ \sqrt{3}y + 1=0$ for $xy$. I have a problem on my textbook:
$x,y \in\mathbb{R}$ and $x^2+xy+y^2+ \sqrt{3}y + 1=0$ is given. Find the value of $xy$.
I couldn't find.
| Setting $t=xy\in\mathbb R$, the equation becomes (knowing that $y\ne0$):
$$\frac{t^2}{y^2}+t+y^2+\sqrt3y+1=0.$$
The discriminant of this quadratic equation is $$\Delta=b^2-4ac=\frac{-3y^2-4\sqrt3y-4}{y^2}=-3\left(1-\dfrac2{\sqrt3y}\right)^2$$ and for a real solution to exist, it must be zero.
Then
$$xy=t=-\frac b{2a}=-\frac{y^2}2=-\frac23.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the sum to n term of the series. $1+3x+5x^2+7x^3................, X\ne1$ Here, $a=1, d=2, b=1, r=x$
\begin{align}
S_n&= \frac{ab}{1-r}+\frac{bdr(1-r{^n}^{-1})}{(1-r)^2}-\frac{[a+(n-1)d]br^n}{1-r}\\
S_n&=\frac 1 {1-x}+\frac{ 2x(1-x{^n}^{-1})} {(1-x)^2}-\frac{[1+(n-1)(2)]x^n} {1- x}\\
&= \frac 1 {1-x}+\frac{2x}{(1-x)^2}-\frac {2x.x{^n}^{-1}}{(1-x)^2}-\frac{[1+2n-2]x^n}{(1-x)}\\
&= \frac 1 {1-x}+\frac{2x}{(1-x)^2}-\frac {2x^n}{(1-x)^2}-\frac{[2n-1]x^n}{(1-x)}\\
\end{align}
Is it correct. I have not got the answer, please show me how to move to this answer without skipping any line
$\frac {1-3x} {(1-x)^2}+\frac {2x^n}{(1-x)^2}-\frac{(2n-1)x^n}{(1-x)}$
| Here's a pretty simple approach:
$
S = 1 + 3x + 5x^2 + 7x^3 + \cdots + (2t + 1)x^t \\
\implies xS = x + 3x^2 + 5x^3 + \cdots + (2t - 1)x^t + (2t + 1)x^{t+1} \\
$
where $t = n - 1$. Subtracting $xS$ from $S$,
$$
(1 - x)S = 1 + 2x + 2x^2 + 2x^3 + \cdots + 2x^t + (2t+1)x^{t+1}.
$$
Replacing $t$ by $n - 1$,
$
(1-x)S = [-1 + (2n - 1)x^n] + 2(1 + x + x^2 + \cdots + x^{n-1})\\
\implies (1-x)S = [-1 + (2n - 1)x^n] + 2\frac{1 - x^n}{1 - x} \\
\implies S = \frac{(2n - 1)x^n}{1 - x} - \frac{1}{1-x} + 2\frac{1-x^n}{(1-x)^2}.
$
By the way, it's not important for the condition $0<x<1$ to hold, unless you're summing an infinite series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Arc Length of an Ellipse using integration I was thinking about what the arc length of an ellipse is, but throughout my calculations I got stuck. Here is how I approached the problem:$$$$We have an ellipse in the form:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow y=\pm \frac{b}{a}\sqrt{a^2-x^2}$$By applying the formula of the arc length of a function, we get:$$L=4\int_0^a\sqrt{1+\frac{b^2x^2}{a^2(a^2-x^2)}}dx=4\int_0^a\sqrt{\frac{a^4+(b^2-a^2)x^2}{a^2(a^2-x^2)}}dx$$Now I made a little subsitution recalling trigonometry: $$x=a\sin(u)\\dx=a\cos(u)du$$So the Integral now can be expressed as:$$L=4\int_0^{\frac{\pi}{2}}a\cos(u)\sqrt{\frac{a^4+(b^2-a^2)a^2\sin^2(u)}{a^2(a^2-a^2\sin^2(u))}}du=\\4\int_0^{\frac{\pi}{2}}a\cos(u)\sqrt{\frac{a^4+(b^2-a^2)a^2\sin^2(u)}{a^2(a^2\cos^2(u)+a^2\sin^2(u)-a^2\sin^2(u))}}du=\\4\int_0^{\frac{\pi}{2}}\sqrt{a^2+(b^2-a^2)\sin^2(u)}du$$So we have:$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+\frac{(b^2-a^2)}{a^2}\sin^2(u)}du$$
Letting $m=\frac{(b^2-a^2)}{a^2}$ we finally get:$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+m\sin^2(u)}du$$
However, at this point I do not know any way on how to integrate this function because the $m$ is 'in the way'. Does anyone have any hints?
| I suppose that you missed an $a$ somewhere and have in fact, for your last expression,$$L=4a\int_0^{\frac{\pi}{2}}\sqrt{1+m\sin^2(u)}\,du=4a\,E(-m)$$ where appears the complete elliptic integral.
This cannot be expressed in terms of elementary functions but some approximations are available by the great Ramanujan. In particular
$$L=\pi \left(3 (a+b)-\sqrt{(3 a+b) (a+3 b)}\right)\tag 1$$ and $$L=\pi (a+b) \left(1+\frac{3\frac{
(a-b)^2}{(a+b)^2}}{10 \sqrt{4-3\frac{
(a-b)^2}{(a+b)^2}}}\right)\tag 2$$
Appliead to your example $(a=4,b=8)$, $(1)$ would give $4\pi \left(9-\sqrt{35}\right)\approx 38.7537 $ and $(2)$ would give $\frac{2\pi}{55} \left(330+\sqrt{33}\right)\approx 38.3554$ while the exact value should be $16 E(-3)\approx 38.7538$.
You also could use an infinite sum formulation $$L=\pi(a+b)\sum_{n=0}^\infty \binom{\frac{1}{2}}{n}^2 h^n \qquad \qquad \text{where} \qquad h=\frac{
(a-b)^2}{(a+b)^2}$$ Limiting to $p$ terms, the convergence is quite fast
$$\left(
\begin{array}{cc}
p & L_p \\
0 & 37.69911184 \\
1 & 38.74630939 \\
2 & 38.75358160 \\
3 & 38.75378361 \\
4 & 38.75379237 \\
5 & 38.75379285 \\
6 & 38.75379288
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2108199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Minimum value of a function with two variables. Suppose I have a curve (for example $(x-2)^2 + (y-2)^2 = 1$. I want to find the point on the curve where $x+y$ (or any other expression say $x^2+y^2$ ) is minimum. How can I do it ?
P.S:- I know the method to find minimum and maximum for a two variable function but I don't understand the above case.
| A) Consider the extremal values of
$x^2 + y^2$ for points on the circle.
A bit Geometry:
$(x-2)^2 + (y-2)^2 = 1$ is a circle with center $(2,2)$ and radius $= 1$.
The straight line $y = x$ is a symmetry axis for this circle, it passes through $(0,0)$ and the centre $(2,2)$, and intersects the circle twice.
For any point $(x,y)$ on the circle, $(x^2 + y^2)$ is the squared distance from the origin.
The shortest and longest squared distances from $(0,0)$ to the circles are the points of intersection of the line $y = x$ with the circle:
$$(x-2)^2 + (y-2)^2 = 1\tag 1$$
$$y = x \tag 2$$ Combining $(2)$ and $(1)$ gives:
$$(x-2)^2 + (x-2)^2 = 1$$
$$2 \times (x-2)^2 = 1 ;$$
Two solutions:
$$x_1 = 2 - \frac {\sqrt 2} 2$$ $$x_2 = 2 + \frac {\sqrt 2} 2$$
$y_1 = x_1$ and $y_2 = x_2$ (cf. eq. 2)
Min squared distance :
$$(x_1)^2 + (y_1)^2 = 2 \times (x_1)^2 = 2 \times ( 4 + \frac 1 2 - 2 \sqrt 2) = 9 - 4 \sqrt 2$$
Max squared distance :
$$(x_2) ^2 + (y_2)^2 = 9 + 4 \sqrt 2$$
$$9 - 4\sqrt 2 \le x^2 + y^2 \le 9 + 4 \sqrt 2$$
B) Now to the extremal values for $x + y$ for points on the circle.
Consider the straight line
$$y = - x + C\tag; $$
where $C$ is the $y$-intercept, i.e. y-value for $x = 0$. This line is perpendicular to the axis of symmetry $y = x$.
Points of intersection of $y = - x + C$ with the circle, are
$(X,Y)$, inserting :
$Y+ X = C$
So finding the smallest and largest y- intercept $C$, with
$y = -x + C$ intersecting the circle, will solve the problem.
$C (min)$ : $y = -x +C$ passes through $(x_1,y_1)$ .
$C( max)$: $y= -x + C$ passes through $(x_2,y_2)$.
Inserted:
$$C( min) = 2\times( 2 - \frac{\sqrt2}2);$$
$$C( max) = 2\times(2 + \frac{\sqrt2}2).$$
Finally we get:
$4 - \sqrt2 \le x + y \le 4 + \sqrt2$
Comments welcome.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Factorise $\,a(a - 4)(a^2 - 4a - 1) - 20\,$ I am making an attempt to factorise the above equation.
To do this i am expanding the already factored part of the above equation to obtain ->
$\,f(a) = a^4 - 8a^3 + 15a^2 + 4a - 20\,$
Now i do some random guesses to find that f(2) = 0, hence i know that (a-2) will divide f(a) so i perform long division to get f(a) = (a-2)( Something ) and so on... until i get - > $\,f(a) = (a - 2)^2(a + 1)(a - 5)\,$
I feel that i am approaching the problem in an incorrect and long way. Please let me know if i can get the factorised form in a cleaner and faster way.
PS: Another linked question is to factorise $\,f(x) = (x + 1)(x + 3)(x + 5)(x + 7) + 15\,$
|
Another linked question is to factorise $\,f(x) = (x + 1)(x + 3)(x + 5)(x + 7) + 15\,$
Let $z=x+4$ then:
$$
\begin{align}
(z-3)(z-1)(z+1)(z+3)+15 & = (z^2-1)(z^2-9)+15=z^4-10z^2+24 \\
& = (z^2-4)(z^2-6) = (x^2+8x+12)(x^2+8x+10) \\
& = (x+2)(x+6)(x^2+8x+10)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2115282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit $\lim_{x\to 0} \frac{\tan ^3 x - \sin ^3 x}{x^5}$ without l'Hôpital's rule. I need to solve $$\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}$$
I did like this:
$\lim \limits_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5} = \lim \limits_{x\to 0} \dfrac{\tan ^3 x}{x^5} - \dfrac{\sin ^3 x}{x^5}$
$=\dfrac 1{x^2} - \dfrac 1{x^2} =0$
But it's wrong. Where I have gone wrong and how to do it?
| $$\lim_{x\to 0} \dfrac{\tan ^3 x - \sin ^3 x}{x^5}=\lim_{x\to 0} \dfrac{\frac{\sin ^3 x}{\cos^3x} - \sin ^3 x}{x^5}=\lim_{x\to 0} \dfrac{\sin ^3 x(1-\cos^3x)}{x^5\cos ^3 x}$$
with $\cos x\approx 1-\frac12x^2$ as $x\to0$
$$\lim_{x\to 0} \dfrac{\sin ^3 x}{x^3}\times\dfrac{1-(1-\frac12x^2)^3}{x^2\times1}=\lim_{x\to 0}\frac32-\frac34x^2+\frac18x^4=\frac32$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Can you explain why $(\sqrt{n})^3 = \sqrt{n^3}$? That this, why does, for example, the square root of $n$, cubed, give the same value as the square root of $n$ cubed?
| The reason that taking a square root fundamentally "makes sense" (i.e. is well-defined) as a function is as follows:
For any $x \geq 0$: if $a,b \geq 0$ are such that $a^2 = x$ and $b^2 = x$, then $a = b$
Now, let's look at those two values you have: that is, take $a = (\sqrt{n})^3$ and $b = \sqrt{n^3}$. We note that
$$
a^2 = (\sqrt{n})^3 \cdot (\sqrt{n})^3 = (\sqrt{n})^6 = ((\sqrt{n})^2)^3 = n^3\\
b^2 = \sqrt{n^3} \cdot \sqrt{n^3} = \sqrt{n^3 \cdot n^3} = \sqrt{(n^{3})^2} = n^3
$$
since $a,b$ are both non-negative and $a^2 = b^2$, it must be that $a = b$.
Here's an equivalent perspective: if $p$ and $q$ are integers, then $(n^{p})^q = (n^q)^p$. This is because, with the way exponentiation is defined, we have
$$
(n^q)^p = \left(\overbrace{n \cdots n}^{p \text{ times}}\right)^q =
\overbrace{\overbrace{n \cdots n}^{p \text{ times}} \cdots \overbrace{n \cdots n}^{p \text{ times}}}^{q \text{ times}} =
\overbrace{n \cdots n}^{pq \text{ times}} = n^{pq}\\
%
(n^q)^p = \left(\overbrace{n \cdots n}^{q \text{ times}}\right)^p =
\overbrace{\overbrace{n \cdots n}^{q \text{ times}} \cdots \overbrace{n \cdots n}^{q \text{ times}}}^{p \text{ times}} =
\overbrace{n \cdots n}^{qp \text{ times}} = n^{qp}
$$
and of course, $pq = qp$. When we extend the definition of exponentiation to rational numbers $p$ and $q$, the new exponents inherit the property $(n^p)^q = (n^q)^p = n^{pq}$. Thus, we have
$$
(\sqrt{n})^3 = (n^{1/2})^3 = n^{3/2} = (n^3)^{1/2} = \sqrt{n^3}
$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can both $n+3\; \text{and}\; n^2+3$ both be cubic numbers at same time?
Can both $n+3\; \text{and}\; n^2+3$ both be cubic number at same time? Where $n$ is an integer number. Not necessarily positive.
I tried writing $x^3 = n+3$ and expressing $n^2+3$ in terms of $x$. I found $x^6 -6x^3+12$ but this doesn't help. How do I prove this?
| Slightly overkill, but if $n + 3$ and $n^2 + 3$ are both cubes, then so is their product, and so
$$ (n + 3)(n^2 + 3) = n^3 + 3n^2 + 3n + 9 = (n + 1)^3 + 2^3 $$
would be a cube. But as is well known, the only possible solutions to this can occur when one of the cubes is $0$, and so we have that either $n = -3$ or $n = -1$, and we can verify that neither of these yield solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Right Triangle: Given hypotenuse and ratio of legs, find legs We are given a Right triangle where the Hypotenuse = $20$ cm.
The opposite side is $3$ times longer than the bottom side.
Is it possible to calculate the length of the opposite side?
(Tried substitution)
$$a^2 + b^2 = 400$$
$$a = 3b$$
$$(3b)^2 + b^2 = 400$$
$b = 10$ = not correct
| Suppose the length
of the hypotenuse is $c$
and the other two sides
have lengths $a$ and $b$.
We know that
$c^2 = a^2+b^2$.
If
$\dfrac{a}{b} = r$,
then
$a = br$
so that
$c^2
= a^2+b^2
= (br)^2+b^2
= b^2(r^2+1)
$.
Therefore,
if you know $c$ and $r$,
$b^2
=\dfrac{c^2}{r^2+1}
$
and
$a^2
= b^2r^2
=\dfrac{c^2r^2}{r^2+1}
$.
In your case,
$c=20$ and $r = 3$,
so
$b^2
=\dfrac{20^2}{10}
=40
$
and
$a^2
= b^2r^2
=40\cdot 9
=360
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\frac {1} {r-1} = \frac {1} {r+1} + \frac {2} {r^2+1} + \frac {4} {r^4+1} +\cdots$ Problem:
Show that
$$\frac {1} {r-1} = \frac {1} {r+1} + \frac {2} {r^2+1} + \frac {4} {r^4+1} +\cdots $$
for all $r > 1$, with a hint given that $\displaystyle\frac {1} {r-1} - \frac {1} {r+1} = \frac {2} {r^2-1}$.
Thoughts:
I am having difficulty seeing a connection from the hint to the problem. Any insights appreciated.
| First, replace $r$ by $r^{2^n}$ in your identity to get:
$$\frac{1}{r^{2^n}-1}-\frac{1}{r^{2^n}+1}=\frac{2}{r^{2^{n+1}}-1}$$
Then consider the partial sum: $s_n=\frac{1}{r+1}+\frac{2}{r^2+1}+\ldots+\frac{2^{n-1}}{r^{2^{n-1}}+1}$
You have:
$$\begin{align}\frac{1}{r-1}-s_1&=\frac{1}{r-1}-\frac{1}{r+1}=\frac{2}{r^2-1}\\\frac{1}{r-1}-s_2&=\frac{1}{r-1}-s_0-\frac{2}{r^2+1}\\
&=\frac{2}{r^2-1}-\frac{2}{r^2+1}=\frac{4}{r^4+1}\end{align}$$
And so on, in general (by induction):
$$\frac{1}{r-1}-s_n=\frac{2^n}{r^{2^n}+1}$$
Then take the limit:
$$\frac{1}{r-1}-s=\lim_{n\to\infty}\frac{2^n}{r^{2^n}+1}=\lim_{u\to\infty}\frac{u}{r^u+1}=0$$
(with substitution $u=2^n$, and using $r>1)$. So $s=\frac{1}{r-1}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $\frac{a^8+b^8+c^8}{a^3b^3c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ for $a,b,c>0$ As in the title. Prove the inequality $$\frac{a^8+b^8+c^8}{a^3b^3c^3}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ for $a,b,c>0$.
Thsi inequality can be proved in a pretty straightforward manner utilizing the Muirhead's inequality, yet I ought to prove it using the rearrangement inequality. I can't however figure out any suitable sequences and permutations of them.
| Without loss of generality, let $abc=1$. Then the inequality becomes
$$ a^8+b^8+c^8\ge ab+bc+ca.$$
Note
\begin{eqnarray}
2(a^8+b^8+c^8)&=&(a^8+b^8)+(b^8+c^8)+(c^8+a^8)\\
&\ge&2a^4b^4+2b^4c^4+2c^4a^4\\
&=&(a^4b^4+b^4c^4)+(b^4c^4+c^4a^4)+(c^4a^4+a^4b^4)\\
&\ge&2a^2b^4c^2+2a^2b^2c^4+2a^4b^2c^2\\
&=&2a^2+2b^2+2c^2\\
&=&(a^2+b^2)+(b^2+c^2)+(c^2+a^2)\\
&\ge&2ab+2bc+2ca
\end{eqnarray}
and hence
$$ a^8+b^8+c^8\ge ab+bc+ca.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Inequality in a non obtuse triangle $a$, $b$ and $c$ are the side of a triangle $\triangle ABC$ that is not obtuse and let $R$ be it the circumcenter.
Prove that
$$\frac{\left(\sum_{cyc} a^2\right)^2}{\prod_{cyc}(a^2+b^2)}<\frac{5}{2R^2}$$
My attempt is to write everything in terms of $R$, $r$ and $s$ and manage to use something like Blundon’s inequality. The problem is I didn't find a simple formula for $\prod_{cyc}(a^2+b^2)$.
| Let $a^2+b^2-c^2=z$, $a^2+c^2-b^2=y$ and $b^2+c^2-a^2=x$.
Hence, $a^2=\frac{y+z}{2}$, $b^2=\frac{x+z}{2}$, $c^2=\frac{x+y}{2}$ and we need to prove that:
$$\frac{8(x+y+z)^2}{\prod\limits_{cyc}(2x+y+z)}<\frac{5}{2\cdot\frac{a^2b^2c^2}{16S^2}}$$ or
$$\frac{8(x+y+z)^2}{\prod\limits_{cyc}(2x+y+z)}<\frac{5(xy+xz+yz)}{2\cdot\frac{\prod\limits_{cyc}(x+y)}{8}}$$ or
$$2\prod_{cyc}(x+y)(x+y+z)^2<5(xy+xz+yz)\prod_{cyc}(2x+y+z),$$
which is obvious and very very not strong.
By the way, the inequality $$\frac{(a^2+b^2+c^2)^2}{(a^2+b^2)(a^2+c^2)(b^2+c^2)}<\frac{1}{2R^2}$$
gives $$2\prod_{cyc}(x+y)(x+y+z)^2<(xy+xz+yz)\prod_{cyc}(2x+y+z),$$
which is stronger and also obvious because it's just
$$\sum_{sym}(x^3y^2+2x^3yz+5x^2y^2z)>0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of the series $\sum_{k=1}^{\infty} \frac {1}{(k)(k+2)(k+4)}$. Problem:
Find the sum of the series $\sum_{k=1}^{\infty} \frac {1}{(k)(k+2)(k+4)}$.
Thoughts
I first tried to write out the series to detect some kind of a pattern, I suspect I need a way of rewriting the expression $\frac {1}{(k)(k+2)(k+4)}$ , but not sure how to proceed. To help visualize it looks like:
$\frac {1} {(1)(3)(5)} + \frac {1} {(2)(4)(6)} + \frac {1} {(3)(5)(7)} + ...$
| $$\frac{1}{k(k+2)(k+4)} = \frac{1}{8} \left( \left( \frac{1}{k} - \frac{1}{k+2} \right) - \left( \frac{1}{k+2} - \frac{1}{k+4} \right) \right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the indicated derivative Find the indicated derivative
$\frac {d}{dt}$ $\frac {(6t-5)^6}{t+9}$
I'm stuck after getting to this part
$\frac {(6t-5)^6-36(6t-5)^5(t+9)}{(t+9)^2}$
How do they get to the answer
$\frac {(6t-5)^5(30t+329)}{(t+9)^2}$
| Can also use product and chain rules: $$6(6t-5)^5\cdot 6\cdot (t+9)^{-1}+(6t-5)^6\cdot -1\cdot (t+9)^{-2}\cdot 1$$
$$\frac{(t+9)\cdot 36\cdot (6t-5)^5}{t+9} -\frac{(6t-5)^6}{(t+9)^2}$$
$$\frac{(6t-5)^5[(36t+324)-(6t-5)]}{(t+9)^2}$$
$$\frac{(6t-5)^5(30t+329)}{(t+9)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Problem with $\int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}} \text{dx}$ I wrote
\begin{eqnarray}
I
&=& \int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}} \text{dx}\\
&=&
\int \frac{\sin^{3}x}{\sin{x}\cos{x}}\text{dx}+\int \frac{\cos^{3}x}{\sin{x}\cos{x}}\text{dx}\\
&=&
\int \frac{\sin^{2}x}{\cos^2{x}}\cos{x}\text{dx}+\int \frac{\cos^{2}x}{\sin^2{x}}\sin{x}\text{dx}\\
&=&
\int \frac{\sin^{2}x}{1-\sin^2{x}}\cos{x}\text{dx}+\int \frac{\cos^{2}x}{1-\cos^2{x}}\sin{x}\text{dx}\\
&=&
\int\frac{u^2}{1-u^2}du-\int\frac{m^2}{1-m^2}dm\\
&=&
\color{blue}{\int\frac{u^2}{1-u^2}du-\int\frac{u^2}{1-u^2}du}\\
&=&
0
\end{eqnarray}
and Other
Let $x=\dfrac{\pi}{2}-t$ so
$$I=\int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}} \text{dx}=-\int \frac{\sin^{3}t+\cos^{3}t}{\sin{t}\cos{t}} \text{dt}=-I$$
but correct answer is
$$
I=\ln\left|\dfrac{1+\tan\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}\right|+\ln\left|\tan\dfrac{x}{2}\right|-\sin x+\cos x+C
$$
Question.1 Where is wrong.?
Question.2 what conditions guarantee that our changing variable in indefinite integrals doesn't change our final solutions!.
| $$\sin^3x+\cos^3x=(\sin x+\cos x)(1-\sin x\cos x)$$
$$\implies\dfrac{\sin^3x+\cos^3x}{\sin x\cos x}=\dfrac{\sin x+\cos x}{\sin x\cos x}-\sin x-\cos x$$
As $\int(\sin x+\cos x)dx=\sin x-\cos x$ and $(\sin x-\cos x)^2=1-2\sin x\cos x,$
set $\sin x-\cos x=u$ in $$\int\dfrac{\sin x+\cos x}{\sin x\cos x}dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}$ find the limit:
$$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}$$
my try :
$$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}\\=\frac{\tan(\pi x)}{x^2(1-\frac{1}{x\sqrt{x}})}\\=\lim_{x\to 1} \frac{\tan(\pi x)}{x^2}.\frac{1}{{1-\frac{1}{x\sqrt{x}}}}$$
now ?
| Note that
$$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}=\lim_{x \to 1}\frac{\tan (\pi x-\pi)}{\pi x-\pi} \times \frac{\pi (x-1)}{x^2-\sqrt{x}}$$
Now note that $$\lim_{x \to 1}\frac{\tan (\pi x-\pi)}{\pi x-\pi} \times \frac{\pi (x-1)}{x^2-\sqrt{x}}=\lim_{x \to 1 }\frac{\tan (\pi x-\pi)}{\pi x-\pi} \times \frac{ \pi (\sqrt{x}+1)}{\sqrt{x}(x+\sqrt{x}+1)}$$
So we have that $$\lim_{x\to 1} \frac{\tan(\pi x)}{x^2-\sqrt{x}}=\lim_{x \to 1 }\frac{\tan (\pi x-\pi)}{\pi x-\pi} \times \lim_{x \to 1}\frac{ \pi (\sqrt{x}+1)}{\sqrt{x}(x+\sqrt{x}+1)}=\frac{2 \pi }{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Quick question regarding $\int \frac{\cos x + \sin x}{\sin 2x}\,dx$ I already asked a question here regarding the discrepancy between the answers I get when I work integrals by hand and the result I can verify using $Mathematica$. I only know enough about $Mathematica$ to type in the commands I need to verify results, plot graphs of polynomials, etc.
Usually I get to the same answer that the program produces, which I always find a bit gratifying, but other times I get results that are equivalent but quite different and often tedious to transfer for one form to the other to verify the equivalence.
Anyway, since this happens to me quite often lately, I wonder how I could check whether I am getting correct results in a reliable way.
For example
$$\int \frac{\cos x + \sin x}{\sin 2x}\,dx$$
I use the identity $\sin 2x = 2 \sin x \cos x $ and get
$$\frac 12 \int \frac {\cos x + \sin x}{\sin x \cos x}\, dx$$
$$=\, \frac 12 \int \frac {\cos x}{\sin x \cos x}+\frac {\sin x}{\sin x \cos x}\, dx$$
$$=\, \frac 12 \int \csc x \, dx \; + \; \frac 12 \int \sec x \,dx$$
$$=\, -\frac 12 \ln {|\csc x + \cot x|} + \frac 12 \ln {|\sec x + \tan x|}+C$$
This is not the same as the result from $Mathematica$, which gives
$$-\frac 12 \ln \left(\cos \frac x2 \right) - \frac 12 \ln \left( \cos \frac x2 - \sin \frac x2 \right)+ \frac 12 \ln \left( \sin \frac x2 \right) + \ln \left( \cos \frac x2 + \sin\frac x2 \right)$$
is this equivalent to my result
$$=\, -\frac 12 \ln {|\csc x + \cot x|} + \frac 12 \ln {|\sec x + \tan x|}+C$$
and how could I reliably verify my answers with $Mathematica$?
| It's the same result, if you change the parentheses for modulus:
$$ -\ln \left| \cos \frac x2 - \sin \frac x2 \right|+ \ln \left| \sin \frac x2 + \cos \frac x2 \right| = \ln \frac{\left| \sin \frac x2 + \cos \frac x2 \right|}{\left| \cos \frac x2 - \sin \frac x2 \right|} =$$
$$\ln \left| \frac{\left( \sin \frac x2 + \cos \frac x2 \right)^2}{\left( \cos \frac x2 - \sin \frac x2 \right)\left( \sin \frac x2 + \cos \frac x2 \right)} \right | = \ln \left| \frac{1 + \sin x}{\cos x} \right | = \ln |\sec x + \tan x| $$
And
$$ -\ln {|\csc x + \cot x|} = \ln \left|\frac{\sin x}{1+\cos x} \right| = \ln \left|\frac{2\sin \frac x2\cos \frac x2}{2\cos^2 \frac x2} \right| = \ln \left |\tan \frac x2 \right| = \ln \left | \sin \frac x2 \right| - \ln \left |\cos \frac x2 \right| $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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The roots of the equation $x^2+3x-1=0$ are also the roots of $x^4+ax^2+bx+c=0$
The roots of the equation $x^2+3x-1=0$ are also the roots of quartic equation $x^4+ax^2+bx+c=0$. Find $a+b+4c$.
This problem is from yesterday's Bangladesh National Math Olympiad 2017. I tried this using Vieta Root Jumping but no luck. After the contest my friend laughed at me "One doesn't simply try a 10 point problem with Vieta Root Jumping".
How to solve this problem?
| For the equation $x^2+3x-1=0$, the sum of the roots is $-3$ and the product is $-1$. For the equation $x^4+ax^2+bx+c=0$, the sum of the roots is $0$ and the product is $c$. If the roots of the first equation are roots of the second, the remaining two roots have sum $3$ and product $-c$. Therefore, we have
$$x^4+ax^2+bx+c=(x^2+3x-1)(x^2-3x-c)$$
The coefficient of the $x^2$ term is $-c-9-1=-c-10$. The coefficient of $x$ is $-3c+3$. Therefore,
$$a+b+4c=-c-10-3c+3+4c=-7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $k(x^6+y^6+z^6)+xyz(x^3+y^3+z^3)\geq0$
Let $x$, $y$ and $z$ be real numbers and $k=\frac{(2+\sqrt7)\sqrt[3]{3-\sqrt7}}{6}$. Prove that:
$$k(x^6+y^6+z^6)+xyz(x^3+y^3+z^3)\geq0$$
The equality occurs here for $\frac{x}{\sqrt[3]{\sqrt7-3}}=y=z$.
I tried the following way.
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative and $xyz=w^3$.
Hence, the inequality
$$\sum_{cyc}(kx^6+x^4yz)\geq\frac{k+1}{3(1-m+n)^2}\left(\sum_{cyc}(x^3+m(x^2y+x^2z)+nxyz)\right)^2$$
is a linear inequality of $w^3$ and it's enough to prove the last inequality
for an extremal value of $w^3$, which happens for equality case of two variables.
It's obvious that we can assume $y=z=1$,
but I did not find values of $m$ and $n$, for which the last inequality would be true.
| It suffices to prove that $k(x^6+y^6+z^6) + xyz (x^3+y^3+z^3) \ge 0$ for $xyz = -1$.
Then, it suffices to prove that $k(x^2+y^2+z^2) - (x+y+z) \ge 0$ for $x, y > 0$ and $xyz = -1$.
Note that $x^2+y^2 \ge \frac{(x+y)^2}{2}$, $z^2 = \frac{1}{(xy)^2} \ge \frac{16}{(x+y)^4}$
and $-z = \frac{1}{xy} \ge \frac{4}{(x+y)^2}$. It suffices to prove that
$$k\Big(\frac{(x+y)^2}{2} + \frac{16}{(x+y)^4}\Big) - (x+y) + \frac{4}{(x+y)^2} \ge 0$$
for $x, y > 0$. It suffices to prove that
$k \ge \frac{2u^5 - 8u^2}{u^6 + 32}$ for $u > 0$.
Let $f(u) = \frac{2u^5 - 8u^2}{u^6 + 32}$.
We have $f'(u) = -\frac{2u(u+2)(u^2-2u+4)(u^6-24u^3+32)}{(u^6+32)^2}$.
$f'(u)=0$ has exactly two real solutions $u_1 = \sqrt[3]{12+4\sqrt{7}}$
and $u_2 = \sqrt[3]{12-4\sqrt{7}}$ on $(0, \infty)$.
On $[0,\infty)$, $f(u)$ attains its maximum at $u_1 = \sqrt[3]{12+4\sqrt{7}}$.
Note that $f(\sqrt[3]{12+4\sqrt{7}}) = \frac{\sqrt{7}-1}{24}\sqrt[3]{(12+4\sqrt{7})^2} = k$. We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Compute $\lim_\limits{x\to0^+}\frac{\pi/2- \arctan(1/x^2)-\sin(x^3)-1+\cos(x)}{x\tan(x)+e^{x^3}-1}$ I have to compute
$$
\lim_{x\to0^+}\frac{\pi/2- \arctan(1/x^2)-\sin(x^3)-1+\cos(x)}{x\tan(x)+e^{x^3}-1}
$$
I separated the numerator so I got that
$$\dfrac{-1+\cos(x)}{x\tan(x)+e^{x^3}-1} \longrightarrow -\dfrac{1}2;$$
I know that the limit is $\dfrac{1}2$ and I know, by checking on Wolfram Alpha, that
$$\dfrac{-\sin(x^3)}{x\tan(x)+e^{x^3}-1}\longrightarrow 0$$ so
$$\dfrac{π/2- \arctan(1/x^2)}{x\tan(x)+e^{x^3}-1} \longrightarrow 1.$$
I tried using L'Hopital but it gets even more complicated. How can I solve it?
| Remember that, for $t>0$,
$$
\arctan t+\arctan\frac{1}{t}=\frac{\pi}{2}
$$
so we can write
$$
\frac{\pi}{2}-\arctan\frac{1}{x^2}=\arctan(x^2)
$$
This simplifies things a bit.
Now prove that
$$
\lim_{x\to0^+}\frac{x\tan x+e^{x^3}-1}{x^2}=1
$$
which follows from
$$
\lim_{x\to0^+}\frac{x\tan x}{x^2}=1
$$
and from
$$
\lim_{x\to0^+}\frac{e^{x^3}-1}{x^2}=
\lim_{x\to0^+}x\frac{e^{x^3}-1}{x^3}=0\cdot 1=0
$$
Thus the limit you have to compute is the easier
$$
\lim_{x\to0^+}\frac{\arctan(x^2)-\sin(x^3)-1+\cos x}{x^2}=
\lim_{x\to0^+}
\left(\frac{\arctan(x^2)}{x^2}-\frac{\sin(x^3)}{x^2}-\frac{1-\cos x}{x^2}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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If $a+b+c=1$ then $\sum\limits_{cyc}\frac{a}{\sqrt[3]{a+b}}\leq\frac{31}{27}$
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$ and $a+b+c=1$.
Prove that:
$$\frac{a}{\sqrt[3]{a+b}}+\frac{b}{\sqrt[3]{b+c}}+\frac{c}{\sqrt[3]{c+a}}\leq\frac{31}{27}$$
The equality occurs for $(a,b,c)=\left(\frac{19}{27},\frac{8}{27},0\right)$.
This inequality is similar to the following inequality, which was proposed by Walther Janous.
For all non-negatives $x$, $y$ and $z$ such that $xy+xz+yz\neq0$ prove that:
$$\frac{x}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}+\frac{z}{\sqrt{z+x}}\leq\frac{5}{4}\sqrt{x+y+z}$$
My proof:
By Cauchy-Schwarz $$\left(\sum_{cyc}\frac {x}{\sqrt {x+y}}\right)^2\leq\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}.$$
Id est, it remains to prove that
$$\sum_{cyc}\frac{x(2x+4y+z)}{x+y}\sum_{cyc}\frac{x}{2x+4y+z}\leq\frac{25(x+y+z)}{16}$$ or
$$\sum_{cyc}(8x^6y+72x^6z-14x^5y^2+312x^5z^2-92x^4y^3+74x^4z^3+$$
$$+122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0$$ or
$$\sum_{cyc}2xy(4x+y)(x-3y)^2(x+2y)^2+$$
$$+\sum_{cyc}(122x^5yz+217x^4y^2z+143x^4z^2y+564x^3y^3z+1338x^3y^2z^2)\geq0,$$ which is obvious.
If we want to use a similar way for the starting inequality, we need to use Holder, which gives very big numbers.
Maybe there is another reasoning?
Thank you!
| I have an half proof :
We have :
Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then we have :
$$\frac{b}{(a+b)^{\frac{1}{3}}}+\frac{a}{(a+c)^{\frac{1}{3}}}+\frac{c}{(c+b)^{\frac{1}{3}}}\leq \Big(\frac{2}{3}\Big)^{\frac{-1}{3}}$$
We need a lemma :
Lemma
Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then we have :
$$\sum_{cyc}0.5(a+b)^{\frac{2}{3}}\leq \Big(\frac{2}{3}\Big)^{\frac{-1}{3}}$$
Proof : It's Jensen's inequality apply to the concave function $x^{\frac{2}{3}}$.
Now we prove :
Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then we have :
$$\frac{b}{(a+b)^{\frac{1}{3}}}+\frac{a}{(a+c)^{\frac{1}{3}}}+\frac{c}{(c+b)^{\frac{1}{3}}}\leq \frac{a+b}{2(a+b)^{\frac{1}{3}}}+\frac{b+c}{2(b+c)^{\frac{1}{3}}}+\frac{a+c}{2(a+c)^{\frac{1}{3}}}$$
Proof : The inequality is equivalent to :
$$\frac{b-a}{2(a+b)^{\frac{1}{3}}}+\frac{c-b}{2(c+b)^{\frac{1}{3}}}+\frac{a-c}{2(a+c)^{\frac{1}{3}}}\leq 0$$
Wich is true with SOS method or rearrangement inequality .
Finally we have :
Let $a,b,c>0$ such that $a+b+c=1$ and $a\geq b \geq c $ then we have :
$$\frac{b}{(a+b)^{\frac{1}{3}}}+\frac{a}{(a+c)^{\frac{1}{3}}}+\frac{c}{(c+b)^{\frac{1}{3}}}\leq \Big(\frac{2}{3}\Big)^{\frac{-1}{3}} <\frac{31}{27}$$
Conclusion :
So the real case to threat is when $b\geq a\geq c $.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Fibonacci numbers is defined by the recurrence relation $F_{n+2} = F_{n+1} + F_n$, prove by induction Question: $F_0 = 0, F_1 = 1$,Prove by induction on $n$ that, for $n\ge 0$, $F_n$ is even if $n$ is a multiple of $3$
Base Case: Let $n = 1$, Substituting the value into the equation
$F_{1+2} = F_{1+1} + F_{1} = F_{2} + F_{1} = F_{1} + F_{0} + F_{1} = 1 + 0 + 1 = 2$
Thus, the equation holds true for the first multiple of three.
Induction Hypothesis: Considering a value $k+2$ that is some arbitrary multiple of 3 and return even for the equation.And $n = k$
$F_{k+2} = F_{k+1} + F_{k}$,
Induction Step: To prove the equation holds true for $n = k+3$, that $k+5$ is all some odd number(Fact: Adding $2$ to any odd number gives odd number),
$F_{k+5} = F_{k+4} + F_{k+3}$,
$RHS = F_{k+4} + F_{k+2} + F_{k+1}$ (From Induction Hypothesis)
How do i get further to this proof...
Here is the question,
is my argument true so far?
How do i proceed further to this proof?
PS: My prof havn't agreed to my proof initiation. It will be good to give some detail explanation. So I can argue back, if needed
PSS:Please no answers, Just help me(If possible)
| Base Case: Let $n = 1$, Substituting the value into the equation
$F_{1+2} = F_{1+1} + F_{1} = F_{2} + F_{1} = F_{1} + F_{0} + F_{1} = 1 + 0 + 1 = 2$
Thus, the equation holds true for the first multiple of three.
Induction Hypothesis: Considering a value $k+2$ that is some arbitrary multiple of 3 and return even for the equation.And $n = k$
$F_{k+2} = F_{k+1} + F_{k}$,
Induction Step: To prove the equation holds true for $n = k+3$, that $k+5$ is all some odd number(Fact: Adding $2$ to any odd number gives odd number),
$F_{k+5} = F_{k+4} + F_{k+3}$,
$RHS = F_{k+4} + F_{k+3}$
$F_{k+4} = F_{k+3} + F_{k+2}$ By Recurrence relation of $F_{n+2}$
Applying the above relation to the RHS, we get:
$RHS = F_{k+3} + F_{k+2} + F_{k+3}$
$= 2.F_{k+3} + F_{k+2}$
Since, multiplying any number with 2 will result in even and $F_{k+2}$ is even based on Induction Hypothesis.
Hence, the equation holds true for all the odd $n$ will result even.
@stud_iisc Thanks for your help!
| {
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"timestamp": "2023-03-29T00:00:00",
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Minimize $\int\limits _{0}^{1} f^2(x) dx$ if $\int\limits_{0}^{1} f(x) dx = 1$, $\int\limits_{0}^{1} xf(x)dx = 1$ Any hints? Using integration by parts didn't produce any result. Interestingly, I cannot find any function $f(x)$ which such a property.
| Using the standard inner product on $L^2$, we have
$$
\langle a+bx,f(x)\rangle=\int_0^1(a+bx)f(x)\,dx=a+b
$$
Also
$$
\|a+bx\|^2=\int_0^1(a+bx)^2\,dx=a^2+ab+b^2/3
$$
and, by Cauchy-Schwarz,
$$
\int_0^1f(x)^2\,dx\ge\frac{(a+b)^2}{a^2+ab+b^2/3}
$$
Setting $t=a/b$, we want to find the maximum of
$$
g(t)=3\frac{(t+1)^2}{3t^2+3t+1}
$$
Since
$$
g'(t)=-3\frac{(t+1)(3t+1)}{(3t^2+3t+1)^2}
$$
Note that $g(t)\ge0$, so the maximum is at $t=-1/3$, with
$$
g(-1/3)=4
$$
This value is actually attained:
$$
\int_0^1 (a(-1+3x))^2\,dx=a^2\int_0^2(1-6x+9x^2)\,dx=a^2(1-3+3)=a^2
$$
so $a=\pm2$; also
$$
\int_0^1a(-1+3x)\,dx=a(-1+3/2)=a/2
$$
so we need $a=2$. Finally
$$
\int_0^12x(-1+3x)\,dx=1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solutions to $8^n = a^3+b^3+c^3-3abc$ Let $n$ be a positive integer. What is the number of solutions to the equation
$$8^n = a^3+b^3+c^3-3abc$$
with integers $a\geq b\geq c\geq 0$?
We have the factoring
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
meaning that both factors must be powers of $2$.
| Strangely enough, the solution is finite.
for the equation:
$$X^3+Y^3+Z^3-3XYZ=q=ab$$
If it is possible to decompose the coefficient as follows:
$4b=k^2+3t^2$
Then the solutions are of the form:
$$X=\frac{1}{6}(2a-3t\pm{k})$$
$$Y=\frac{1}{6}(2a+3t\pm{k})$$
$$Z=\frac{1}{3}(a\mp{k})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2152550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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the coefficient of $ x^{n-1}$ in $(x+3)^n+(x+3)^{n-1}(x+2)+(x+3)^{n-2}(x+2)^2\cdots(x+2)^n$
Find the coefficient of $ x^{n-1}$ in:
$\ \\ (x+3)^n+(x+3)^{n-1}(x+2)+(x+3)^{n-2}(x+2)^2\cdots+(x+2)^n$
My Approach:
$(x+3)^n+(x+3)^{n-1}(x+2)+(x+3)^{n-2}(x+2)^2\cdots+(x+2)^n=\sum(x+3)^{n-r}(x+2)^r$
For each term of this summation:
we have to calculate the number of ways of choosing one (x+a) for which we would multiply the 'a' and not the 'x' part. They can be chosen in $\binom{n-r}{1}+\binom{r}{1}$ ways. In case the chosen is (x-3) the coefficient of $x^{n-1}$ would be 3 else it would be 2. Hence the coefficient of $x^{n-1}$ for the general term would be $3\binom{n-r}{1}+2\binom{r}{1}$.
This simplifies to $3(n-r)+2r=3n-r$. The coefficient of $x^{n-1}$ in the summation is $\sum_0^n 3n-\sum_{r=0}^n{n} r=3n^2-\frac{n(n+1)}{2}=\frac{6n^2-n^2-n}{2}=\frac{5n^2-n}{2}$.
The answer given is $5\binom{n+1}{2}$ but I can't figure out what the error is in this logic. It would be great if someone could help me answer this question.
| It is easier to simplify the given expression by using formula $$a^{r-1}+a^{r-2}b+\cdots+ab^{r-2} +b^{r-1}=\frac{a^{r}-b^{r}}{a-b}$$ which is easily proved by multiplying both sides of the equation by $a-b$. Here we have $r=n+1,a=x+3,b=x+2$ and hence the expression simplifies to $(x+3)^{n+1}-(x+2)^{n+1}$ and the desired coefficient is $$\binom{n+1}{2}(3^{2}-2^{2})=5\binom{n+1}{2}$$
You had your approach correct but you made a calculation mistake while calculating the sum $\sum_{r=0}^{n}3n$ which should evaluate to $3n(n+1)$ but you wrote $3n^{2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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There exists a unique $f \in A$ such that $Tf(x) = f(x)$ Let $A$ be a subset of $C([0,1])$ such that $A = \{ f \in C([0,1]) \ | \ f(0)=0, \ f(1)=1\}$. \
Let
\begin{align*}
Tf(x)=
\begin{cases}
\frac{3}{4}f(3x), \ 0 \leq x \leq \frac{1}{3} \\
\frac{1}{2}f(2-3x) + \frac{1}{4}, \ \frac{1}{3} < x \leq \frac{2}{3} \\
\frac{3}{4}f(3x-2) + \frac{1}{4}, \ \frac{2}{3} < x \leq 1 \\
\end{cases}
\end{align*}
I want to find an $f \in A$ such that $Tf(x)=f(x)$ and show that its nowhere differentiable, but I cant seem to find a function $f$ that agrees with $Tf(x)$..
Thanks in advance!
| Here's a hint to get you started.
For $Tf=f$ to hold, we need to satisfy three functional equations. On the first interval we need to satisfy $\frac{3}{4}f(3x)=f(x)$, on the second interval we need to satisfy $\frac{1}{2}f(2-3x) + \frac{1}{4}$ and on the third we need to satisfy $f(x)=\frac{3}{4}f(3x-2) + \frac{1}{4}$. You can then compose these equations over and over again to get a recursive form that lets you evaluate the function at a wide variety of points. Have you dealt with functions that are nowhere differentiable in the past? This is similar to the construction of other continuous, nowhere-differentiable, fractal-like functions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Which integers $k$ give $b=ka$? For which integers $k$ is it possible to have $b=ka$, where $a$ is a positive integer and $b$ is obtained from $a$ by moving the initial digit of $a$ to the end.
| Write $a=10^n u + w$, where $u$ is a digit and $0 \le w < 10^n$. Then $b=10w+u$.
Now, $b=ka$ iff $k(10^n u + w) = 10 w +u $ iff $ (k10^n-1)u= (10-k)w$.
Since $w < 10^n$, we have $(k10^n-1)u < (10-k)10^n$ and so
$(ku+k-10)10^n < u < 10$. This can only happen if $ku+k-10 \le 0$.
In particular, since $u\ge 1$, this can only happen when $k\le 5$.
$k=1$ implies $w=(11\cdots1)u$.
$k=2$ implies $(2\cdot 10^n-1)u=8w$. Since $2\cdot 10^n-1$ is odd, $8$ must divide $u$ and so $u=8$. But $ku+k \le 10$ implies $u \le 4$, so no solution.
$k=3$ implies $(3\cdot 10^n-1)u=7w$ and so $7$ divides $3\cdot 10^n-1$. This implies $n \equiv 5 \bmod 6$. Also, $ku+k \le 10$ implies $u \le 2$.
A sample solution is $w=42857$, which gives $a=142857$ for $u=1$ and $a=285714$ for $u=2$.
$k=4$ implies $(4\cdot 10^n-1)u=6w$. Since $4\cdot 10^n-1$ is odd, $2$ must divide $u$. But $ku+k \le 10$ implies $u \le 1$, so no solution.
$k=5$ implies $(5\cdot 10^n-1)u=5w$. So $5$ must divide $u$. But $ku+k \le 10$ implies $u \le 1$, so no solution.
Bottom line: The only solutions are $a=(11\cdots1)u$ for $k=1$ and $a$ formed by repeating $142857$ or $285714$ for $k=3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find Minima and Maxima of $ y = \frac{x^2-3x+2}{x^2+2x+1}$ $$ y = \frac{x^2-3x+2}{x^2+2x+1}$$
I guess I made some mistakes cause after taking the first derivative and simlifying I have
$$y = \frac{2x^3-4x^2+5}{(x+1)^2}$$
but then numerator has complex roots. which should not be, IMO
| Let's derive
$$y'={(2x-3)(x^2+2x+1)-(2x+2)(x^2-3x+2)\over (x^2+2x+1)^2}={5x^2-2x-7\over (x+1)^4}={5x-7\over (x+1)^3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Equation of three lines forming a equilateral triangle.
Prove that if general equation $$\cos \left(3\alpha\right)\left(x^3 -3xy^2\right) + \sin3\alpha\left(y^3 - 3x^2y\right) + 3a\left(x^2 + y^2\right) - 4a^3 = 0$$ represents three lines then they form a equilateral triangle of area $3a^2\sqrt 3$.
This equation looks like an good candidate to solve in polar coordinates than in cartesian coordinates.
Converting it into polar coordinates by parts
\begin{align}
\left(x^3 -3xy^2\right) &= r^3 \left(\cos^3 \theta - 3\cos \theta \sin^2 \theta\right) =r^3 \left(4\cos^3 \theta - 3\cos \theta\right) = r^3\cos 3\theta \tag1\label1 \\
\left(y^3 - 3x^2y\right) &= r^3\left(4\sin^3 \theta - 3\sin\theta\right) = r^3\left(\cos\left(3\pi/2 - 3\theta\right)\right) = -r^3\sin 3\theta\tag2\label2
\end{align}
Using $\eqref1$, $\eqref2$
\begin{align}
r^3\left(\cos 3\alpha\cos 3\theta - \sin3\alpha\sin 3\theta\right) + 3ar^2 - 4a^3 &= 0 \\
r^3\cos \left(3\alpha+ 3\theta\right) + 3ar^2 - 4a^3 &= 0
\end{align}
On dividing by $a^3$ and substituting $z = r/a$
$$z^3\cos \left(3\alpha+ 3\theta\right) + 3z^2 - 4 = 0.$$
Now this happens very often to me, I am not able to solve this cubic for $z$.I did try to find the factor by putting $\,z = \sin\left(3\alpha + 3\theta\right),\,\tan\left(3\alpha + 3\theta\right),\,\ldots$
What should I do now ? how to solve this cubic ?
| Real and imaginary parts of $ (x+iy)^3$ can be recognized here.
EDIT1:
Factorization given below tallies fully with the first general equation in OP's post.
$$ (x \cos(\alpha) - y\sin (\alpha) -a)*
(x \cos(\alpha + 2 \pi/3) - y\sin (\alpha + 2 \pi/3) -a) *
(x \cos(\alpha + 4 \pi/3) - y\sin (\alpha + 4 \pi/3) -a)=0\, \tag1 $$
Area = $ \dfrac { \sqrt3 }{4} \cdot { { (2 \sqrt3 a})^2} = 3 \sqrt3 a^2$
The following is a plot of three sides for $ \alpha = 12.345, a=1. $
BTW, such factorizations are useful in obtaining torsion in non-circular shafts using St. Venant's theory.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Finding extremal points on $f(x,y)$ This is the equation:
$$f(x,y) = xye^\left(-\frac{1}{2}(x^2 + y^2)\right)$$
This is what I've done:
$$\nabla f(x,y) = \begin{bmatrix}
(1-x^2)ye^\left(-\frac{1}{2}(x^2 + y^2)\right) \\
(1-y^2)xe^\left(-\frac{1}{2}(x^2 + y^2)\right)
\end{bmatrix}$$
Here's the thing I'm worried about, to find when $\nabla f(x,y) = 0$, i set each equation $= 0$.
$$\begin{align}
(1-x^2)ye^\left(-\frac{1}{2}(x^2 + y^2)\right) &= 0 \\
y e^\left(-\frac{1}{2}(x^2 + y^2)\right) &= x^2ye^\left(-\frac{1}{2}(x^2 + y^2)\right) \\
1 &= x^2 \\
x &= \pm 1\end{align}
$$
Is this legal, or do i lose some solutions when I divide away everything?
| Dividing by $y$ is one of the problematic steps.
Just start with the partial derivative, and divide by the exponential term (which is never zero) to get $(1-x^2)y=0$. From here you see either $1-x^2=0$ or $y=0$, from which you get $x=\pm 1$ or $y=0$.
If you do this for the other partial derivative you get $y \pm 1$ or $x=0$.
Combine this with the above to gather all the critical points.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$5^{2^{n-3}}\not \equiv -1 \mod 2^n$ for any $n\geq 3$ I am in the middle of solving an abstract algebra problem and I have narrowed it down to showing that
$$5^{2^{n-3}}\not \equiv -1 \mod 2^n \quad \text{ for any } \quad n\geq 3. $$
I am not very good with congruences. I know this is equivalent to showing that $5^{2^{n-3}}+1\not \equiv 0\mod 2^n$, i.e., $5^{2^{n-3}}$ is not a multiple of $2^n$, but I am still not sure how to prove this.
Thank you for your help.
| First we check this for $n = 3$. We have $5 \not\equiv -1 \mod{8}$. Thus let $n \geq 4$ and suppose $5^{2^{n - 3}} \equiv -1 \mod{2^n}$ so $5^{2^{n - 3}} + 1 = a2^n$ for some $a \in \mathbb{Z}$. We then have $5^{2^{n - 3}} + 1 = 4a2^{n - 2}$ with $2^{n - 2} \in \mathbb{Z}$ since $n \geq 4$ so $5^{2^{n - 3}} \equiv -1 \mod{2^{n-2}}$. We have $\phi(2^{n - 2}) = 2^{n - 2} - 2^{n - 3} = 2^{n - 3}$ and $\gcd(5,2^{n-2}) = 1$. But then by Euler's Theorem
\begin{align}
-1 \equiv 5^{2^{n - 3}} \equiv 5^{\phi(2^{n - 2})} \equiv 1 \mod{2^{ n - 2}}
\end{align}
which is a contradiction since $-1 \equiv 1 \mod{q}$ iff $q = 2$, and since $n \geq 4$ we have $2^{n - 2} \neq 2$. Thus the supposition was false and $5^{2^{n - 3}} \not\equiv - 1 \mod{2^n}$ for $n \geq 4$. Combining this with our base case we have the result holding for $n \geq 3$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
Floor Function. Let x and y be rational numbers let x and y be rational numbers.
A. $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor
= \left \lfloor x+y \right \rfloor$
B. $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor
\le \left \lfloor x+y \right \rfloor$
C. $\left \lfloor x \right \rfloor + \left \lfloor y \right \rfloor
\ge \left \lfloor x+y \right \rfloor$
D. None of the above.
Can anyone please explain why the answer is B? Thank You!
| If $x$ and $y$ are rational (or real) numbers, then there must exist numbers $h$ and $k$ such that $0 \le h,k < 1$, $x = \left \lfloor x \right \rfloor + h$, and $y = \left \lfloor y \right \rfloor + k$.
So:
\begin{align}
\left \lfloor x + y \right \rfloor
&= \left \lfloor
\left \lfloor x \right \rfloor + h +
\left \lfloor y \right \rfloor + k
\right \rfloor \\
&= \left \lfloor x \right \rfloor +
\left \lfloor y \right \rfloor
+ \left \lfloor h+k \right \rfloor\\
&\ge \left \lfloor x \right \rfloor +
\left \lfloor y \right \rfloor \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $A=+\frac{3}{4×8}-\frac{3×5}{4×8×12}+\frac{3×5×7}{4×8×12×16}-···$ Find $A$:
$$A=+\frac{3}{4×8}-\frac{3×5}{4×8×12}+\frac{3×5×7}{4×8×12×16}-···$$
My Try :
$$a_1=\frac{3}{4×8}-\frac{3×5}{4×8×12}=\frac{3×12-3×5}{4×8×12}=\frac{3(12-5)}{4×8×12}=\frac{3(7)}{4×8×12}$$
$$a_2=\frac{3(7)}{4×8×12}-\frac{3×5×7}{4×8×12×16}=\frac{3×7×16-3×5×7}{4×8×12×16}=\frac{3×7(16-7)}{4×8×12×16}\\=\frac{3×7(8)}{4×8×12×16}$$
now?
| $$A=\sum_{n=2}^\infty (-1)^{n} \frac {(2n)!/(2^n n!)}{4^n n!}$$
$$=\sum_{n=2}^\infty (-1)^{n} \frac {(2n)!}{8^{n} (n!)^2}$$
$$=\sum_{n=2}^\infty \left(-\frac{1} {8}\right)^{n} \binom {2n} {n}$$
Now, note that $$\frac{1}{\sqrt{1-4x}} = \sum_{n=0}^\infty \binom {2n} {n}x^n$$ and take $x=-1/8$, separate 1st and 2nd term of the infinite sum to find your answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Double integral over area bounded by ellipse Evaluate the double integral of $\int \int(x+y)^{2} dxdy$ over the area bounded by the ellipse $(\frac{x}{a})^2+(\frac{y}{b})^2=1$
Please check answer = $\frac{ab\pi}{4}(a^{2}+b^{2})$
| If we are going to do this without using a transformation. Notice our regions boundary is,
$$(\frac{x}{a})^2+(\frac{y}{b})^2=1$$
So that $x \in [-a\sqrt{1-(\frac{y}{b})^2},a\sqrt{1-(\frac{y}{b})^2}]$ and of course $y \in [-b,b]$. So we are interested in computing,
$$\int_{-b}^{b} \int_{-a\sqrt{1-(\frac{y}{b})^2}}^{a\sqrt{1-(\frac{y}{b})^2}} (x^2+2xy+y^2) dx dy$$
We integrate with respect to $x$ first. Notice that $2xy$ is odd with respect to $x$ so we can drop that term. So then what we get is,
$$=\int_{-b}^{b} \left( \frac{2}{3}(a\sqrt{1-(\frac{y}{b})^2})^3+y^2\left(2a\sqrt{1-(\frac{y}{b})^2}\right) \right) dy$$
$$=\int_{-b}^{b} \left(\frac{2}{3}(a\sqrt{1-(\frac{y}{b})^2})a^2\left(1-(\frac{y}{b})^2\right)+y^2\left(2a\sqrt{1-(\frac{y}{b})^2}\right) \right) dy$$
First let $\frac{y}{b}=u$ so $dy=bdu$ and we have that what we want is the sum of,
$$I_1=\frac{2}{3}a^3b \int_{-1}^{1}(1-u^2)\sqrt{1-u^2} du$$
$$I_2=2ab^3\int_{-1}^{1} u^2\sqrt{1-u^2} du$$
The integrals in each of these are both standard, and related integrals. Find one, you find the other by geometry.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the limit without L'hopital rule $\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=$? Find the limit without L'hopital rule
$$\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=?$$
My Try:
$$1-\cot( \frac{π}{4}x)=1-\frac{1}{\tan( \frac{π}{4}x)}=\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}\\\sin (πx)=\sin (\pi-\pi x)=-\sin \pi(x-1)\\\lim_{ x \to 1}\frac{1-\cot(\frac{π}{4}x)}{\sin πx}=\lim_{ x \to 1}\frac{\frac{\tan( \frac{π}{4}x)-1}{\tan( \frac{π}{4}x)}}{-\sin \pi(x-1)}\\u=x-1⇒x=u+1\\\lim_{ u \to 0}\frac{\frac{\tan( \frac{π}{4}(u+1))-1)}{\tan( \frac{π}{4}(u+1))}}{-\sin \pi u}$$
now ?
| $$\lim_{ x \to 1}\frac{\tan(\frac{\pi}{4}x)-1}{\tan(\frac{\pi}{4}x)\sin πx}=$$
$$\lim_{ x \to 1}\frac{\tan(\frac{\pi}{4}x)-\tan(\frac{\pi}{4})}{\tan(\frac{\pi}{4}x)\sin πx}=$$
$$\lim_{ x \to 1}\frac{\sin(\frac{\pi}{4}(x-1))}{\cos(\frac{\pi}{4}x)\cos(\frac{\pi}{4})\tan(\frac{\pi}{4}x)\sin(\pi-\pi x)}=$$
$$\lim_{ x \to 1}\frac{\sin(\frac{\pi}{4}(x-1))}{\cos(\frac{\pi}{4})\sin(\frac{\pi}{4}x)\sin\pi(1- x)}=$$
$$\lim_{ x \to 1}\frac{(\frac{\pi}{4}(x-1))}{\cos(\frac{\pi}{4})\sin(\frac{\pi}{4}x)\pi(1- x)}=-\frac12$$
| {
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"url": "https://math.stackexchange.com/questions/2165038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $x^3-3x+\sqrt{3}=0$ (to evaluate $\sin20^\circ$) gives imaginary roots $$ x^3 - 3x + \sqrt{3} = 0 $$
How to solve the above cubic equation?
I used Cardano's Method but, I am getting imaginary roots but, that should not happen because I encountered the above cubic while trying to evaluate $\sin(20^\circ)$.
If not the cubic, please tell me how to evaluate $\sin(20^\circ)$?
| The equation
$$x^3-3x+\sqrt 3=0$$
can be written as
$$-4\left(\frac x2\right)^3+3\cdot\frac x2=\frac{\sqrt 3}{2}$$
Using that
$$\sin(3\theta)=-4\sin^3\theta+3\sin\theta$$
we see that
$$\frac{x}{2}=\sin 20^\circ,\sin 40^\circ,\sin(-80^\circ)$$
since
$$\sin(3\times 20^\circ)=\sin(3\times 40^\circ)=\sin(3\times(-80^\circ))=\frac{\sqrt 3}{2}$$
with
$$\sin(-80^\circ)\lt \sin 20^\circ\lt \sin 40^\circ$$
Therefore,
$$\color{red}{x=2\sin 20^\circ,\quad 2\sin 40^\circ,\quad -2\sin(80^\circ)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Perform this integration: $\int\frac{t}{t^3 + 1}$ I am solving the following question
$$\int\frac{\sin x}{\sin^{3}x + \cos^{3}x}dx.$$
I have been able to reduce it to the following form by diving numerator and denominator by $\cos^{3}x$ and then substituting $\tan x$ for $t$ and am getting the following equation. Should Iis there any other way use partial fraction to integrate it further or
$$\int\frac{t}{t^3 + 1}dt.$$
| Hint:
$\int \frac{t}{t^3+1} dx = \int \frac{t+t^2-t^2}{t^3+1} dx =\int \frac{t(t+1)}{t^3+1} dx-\frac{1}{3}\int \frac{3t^2}{t^3+1} dx=\int \frac{t(t+1)}{(t+1)(t^2-t+1)} dx-\frac{1}{3}\ln(t^3+1)$
$=\frac{1}{2}\int \frac{2t}{t^2-t+1} dx-\frac{1}{3}\ln(t^3+1)=\frac{1}{2}\int \frac{2t-1+1}{t^2-t+1} dx-\frac{1}{3}\ln(t^3+1)$
$=\frac{1}{2}\int \frac{2t-1}{t^2-t+1} dx+\frac{1}{2}\int \frac{1}{t^2-t+1} dx-\frac{1}{3}\ln(t^3+1)$
$=\frac{1}{2}\ln(t^2-t+1)+\frac{1}{2}\int \frac{1}{t^2-t+1} dx-\frac{1}{3}\ln(t^3+1)$
The integral can be evaluated by partial fractions with complex roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Did I calculate the Image of this matrix correctly?
Calculate the image of the matrix $A= \begin{pmatrix} 1 & 2 & -1\\
-2 & -6 & 1 \\ 1 & -2 & 0 \end{pmatrix}$
First I transposed the matrix:
$A^{T}= \begin{pmatrix}
1 & -2 & 1 \\
2 & -6 & -2 \\
-1 & 1 & 0
\end{pmatrix}$
Then I used Gauss to get zero lines. Multiply first line with $2$, add that to second line:
$A^{T}= \begin{pmatrix}
2 & -4 & 2\\
4 & -10 & 0 \\
-1 & 1 & 0
\end{pmatrix}$
Now multiply third line with $10$ and add second line to third line:
$A^{T}= \begin{pmatrix}
2 & -4 & 2\\
4 & -10 & 0 \\
-6 & 0 & 0
\end{pmatrix}$
We cannot create more zeroes, so Gauss ends here. We transpose back:
$A= \begin{pmatrix}
2 & 4 & -6 \\
-4 & -10 & 0 \\
2 & 0 & 0
\end{pmatrix}$
Thus we have $\text{Img(A)}= \left\{ \begin{pmatrix}
2\\
-4\\
2
\end{pmatrix}, \begin{pmatrix}
4\\
-10\\
0
\end{pmatrix}, \begin{pmatrix}
-6\\
0\\
0
\end{pmatrix} \right\}$
It's very important for me to know, did I do it correctly? I would do it like this in my exam.
| This is a highly unusual method. However, it is technically correct. You have indeed found a basis of the image. Remember to say that the image is the span of those vectors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2167180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Another polynomial equation Let $r$ be a root of the polynomial $p(x)=(\sqrt{3}-\sqrt{2})x^3 + \sqrt{2}x-\sqrt{3}+1$. Find another polynomial $q(x)$, with all integer coefficients, such that $q(r)=0$.
| $\sqrt{3}r^3-\sqrt{2}r^3+\sqrt{2}r-\sqrt{3}+1=0\\\sqrt{3}r^3-\sqrt{3}+1=\sqrt{2}r^3-\sqrt{2}r$
Square both sides
$3(r^3-1)^2+1+2\sqrt{3}(r^3-1)=2(r^3-r)^2$
$3(r^3-1)^2+1-2(r^3-r)^2=-2\sqrt{3}(r^3-1)$
Square both sides again
$(3(r^3-1)^2+1-2(r^3-r)^2)^2-12(r^3-1)^2=0$
Therefore $q(x)=(3(x^3-1)^2+1-2(x^3-x)^2)^2-12(x^3-1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2169700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of three variables given two equations Given $$x^2+y^2+z^2=121$$
$$x\sqrt{11} + 4y + z\sqrt{22}=77$$
Find $$ \frac{\sqrt{11} + 4 + \sqrt{22}}{x+y+z} $$
I tried to plug in something for z at first, since x and y should have unique values for every value of z, but that didn't seem to work.
The answer is 7/11, which is clearly the second equation divided by the first but I don't understand how or why that would lead the final expression.
| HINT:
WLOG $x=11\cos u\cos v,y=11\cos u\sin v,z=11\sin u$
$$77=\sqrt{11}\cdot11\cos u\cos v+4\cdot11\cos u\sin v+\sqrt{22}\cdot11\sin u$$
$$\iff7=S=\sqrt{11}\cos u\cos v+4\cos u\sin v+\sqrt{22}\sin u$$
$$=3\sqrt3\cos u\cos\left(v-\arcsin\dfrac4{3\sqrt3}\right)+\sqrt{22}\sin u$$
For $\cos u\ge0,$
$$S\le3\sqrt3\cos u+\sqrt{22}\sin u$$
Now $3\sqrt3\cos u+\sqrt{22}\sin u=7\cos\left(u-\arccos\dfrac{3\sqrt3}7\right)\le7$
So, we need $\cos\left(u-\arccos\dfrac{3\sqrt3}7\right)=\cos\left(v-\arcsin\dfrac4{3\sqrt3}\right)=1$
$\implies u\equiv\arccos\dfrac{3\sqrt3}7, v\equiv\arcsin\dfrac4{3\sqrt3}=\arccos\dfrac{\sqrt{11}}{3\sqrt3}\pmod{2\pi}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving convexity for multivariable function I am trying to prove the convexity for the following function:
$$f(x) = \sqrt{(a^Tx)^2 + (b^Tx)^2}$$ where $\{a,b,x \} \in \mathbb{R}^n$. I showed that the Hessian is positive semidefinite (which proves the convexity), but the equations are very complicated, so I'm trying to use the definition of convex functions to prove it (i.e., $$f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2); \ \forall t \in [0,1]$$
However, I cannot seem to be able to prove this. Any ideas on how to approach this? Your help is much appreciated!
| Starting out from $f(tx_1 + (1-t)x_2) \leq tf(x_1) + (1-t)f(x_2)$ lets square both sides to get rid of the square-root on the left to get
\begin{align*}
(ta^Tx_1 + (1-t)a^Tx_2)^2 + (tb^Tx_1 + (1-t)b^Tx_2)^2 \leq t^2 ((a^Tx_1)^2 + (b^Tx_1)^2) + (1-t)^2 ((a^Tx_2)^2 + (b^Tx_2)^2) + 2t(1-t)\sqrt{((a^Tx_1)^2 + (b^Tx_1)^2)((a^Tx_2)^2 + (b^Tx_2)^2)}.
\end{align*}
Now writing out the squares on the left and canceling with right hand side terms leaves
\begin{align*}
2ta^Tx_1(1-t)a^Tx_2 + 2tb^Tx_1(1-t)b^Tx_2 \leq 2t(1-t)\sqrt{((a^Tx_1)^2 + (b^Tx_1)^2)((a^Tx_2)^2 + (b^Tx_2)^2)},
\end{align*}
where we can also drop the $2t(1-t)$ factor.
We can square again both sides to get
\begin{align*}
(a^Tx_1 a^Tx_2 + b^Tx_1 b^Tx_2)^2 \leq ((a^Tx_1)^2 + (b^Tx_1)^2)((a^Tx_2)^2 + (b^Tx_2)^2).
\end{align*}
Now multiplying out both sides and canceling leaves
\begin{align*}
2 a^Tx_1 a^Tx_2 b^Tx_1 b^Tx_2 \leq (a^Tx_1)^2(b^Tx_2)^2 + (b^Tx_1)^2 (a^Tx_2)^2,
\end{align*}
which is equivalent to
\begin{align*}
0 \leq (a^Tx_1)^2(b^Tx_2)^2 + (b^Tx_1)^2 (a^Tx_2)^2 - 2 a^Tx_1 a^Tx_2 b^Tx_1 b^Tx_2 = \left(a^Tx_1 b^Tx_2 - b^Tx_1 a^Tx_2\right)^2,
\end{align*}
which now obviously is true and should show the convexity of your function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Is this circle always interior to ellipse? Let $r = a-c$, where $a$ is the length of semi major axis and $c$ is the distance between origin and one of the foci of an ellipse.
I'm wondering if there a nice way to see that the below circle is always interior to the ellipse.
$$(x-c)^2 + y^2 = r^2$$
I think this is equivalent to saying that the minimum distance between a focus and a point on the ellipse is $a-c$. If so, is there a way to prove this using geometry or some other means ? I feel calculus is messy here... Thank you!
https://www.desmos.com/calculator/il5rpye8on
| If the semi-minor axis is $b$,
then $c^2 = a^2-b^2$.
For the ellipse,
$(x/a)^2+(y/b)^2 = 1$
or
$y^2
=b^2(1-(x/a)^2)
=b^2-(bx/a)^2
$.
For the circle,
$\begin{array}\\
y^2
&=r^2-(x-c)^2\\
&=(a-c)^2-(x-c)^2\\
&=a^2-2ac+c^2-(x^2-2xc+c^2)\\
&=a^2-2ac-x^2+2xc\\
\end{array}
$
So,
we want to show that
$a^2-2ac-x^2+2xc
\le b^2-(bx/a)^2
$
or
$x^2(1-(b/a)^2)-2xc
\ge a^2-2ac-b^2
$
or
$x^2(c/a)^2-2xc
\ge c^2-2ac
$
or
$x^2(c/a)^2-2xc+a^2
\ge c^2-2ac+a^2
$
or
$(cx/a-a)^2
\ge (c-a)^2
$.
Since
$a \ge c$
and
$x \le a$,
$cx/x \le a$,
so this last is equivalent to
$a-cx/a \ge a-c$
or
$cx/a \le c$
or
$x \le a$,
which is true.
Therefore the circle
is always
within the ellipse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2175196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differential equation with (bilateral) Laplace transformation So I have this:
$y'(t) - \int_0^tf(t-x)*y(x)dx=f(t), t\geq0$
where $y(0)=1$ and $f(t)=e^{-3t}$
I try solving it by multiplying both sides with the $H(t)$ (<-- Heaviside) and performing the Laplace transformation. But after some cleaning up I get $\mathcal{L}(y*H)=\frac{s+4}{s^2+3s-1}$ which seems wrong because the denominater has the roots $s= \frac{-3}2 \pm \frac{\sqrt13}2$.
Have I done something wrong along the way? Please help!!!
| In the general case consider
$$y'(t) - \int_{0}^{t} e^{-a (t-x)} \, y(x) \, dx = e^{-at} \hspace{5mm} t\geq 0.$$
Since $t\geq 0$ consider applying the one sided Laplace transform for which
\begin{align}
s \, \overline{y} - y(0) - \frac{1}{s+a} \, \overline{y} &= \frac{1}{s+a} \\
\frac{s^2 + as -1}{s+a} \, \overline{y} &= y(0) + \frac{1}{s+a} \\
\overline{y} &= \frac{1}{s^2 + as -1} + y(0) \, \frac{s+a}{s^2 + as -1}
\end{align}
Now, $s^2 + as -1 = (s+\alpha)(s + \beta)$ with $2 \alpha = a + \theta$, $2 \beta = a - \theta$, $\theta = \sqrt{a^2 + 4}$ such that
\begin{align}
\overline{y} &= \frac{1}{\alpha - \beta} \, \left[ ((\alpha - a) \, y(0) -1) \, \frac{1}{s+\alpha} - ((\beta - a) \, y(0) -1) \, \frac{1}{s+\beta} \right] \\
&= \frac{1}{\alpha - \beta} \, \left[ (\alpha \, y(0) +1) \, \frac{1}{s + \beta} - (\beta \, y(0) +1) \, \frac{1}{s+\alpha} \right]
\end{align}
and finally
\begin{align}
y(t) = \frac{e^{-a t/2}}{\sqrt{a^2+4}} \, \left[ (a \, y(0) + 2) \, \sinh\left(\frac{\sqrt{a^2 + 4} \, t}{2} \right) + y(0) \, \sqrt{a^2 + 4} \, \cosh\left(\frac{\sqrt{a^2 + 4} \, t}{2} \right) \right].
\end{align}
For this particular problem $a=3$ and $y(0) = 1$ and yields
$$y(t) = e^{-3 t/2} \, \left[ \frac{5}{\sqrt{13}} \, \sinh\left(\frac{\sqrt{13} \, t}{2}\right) + \cosh\left(\frac{\sqrt{13} \, t}{2}\right) \right].$$
Suppose the conditions were $a=4$ and $y(0) = 0$ then
$$y(t) = \frac{1}{\sqrt{5}} \, e^{-2 t} \, \sinh(\sqrt{5} \, t).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the seq $\{\frac{x^4-1}{x^4+x-6}\}$ converges and prove using $N-\epsilon$ proof Prove $$\{\dfrac{x^4-1}{x^4+x-6}\}$$ converges and prove using $N-\epsilon$ proof.
I see that this sequence approach $1$.
So,
I want to show:
$$\forall \epsilon >0, \exists N>0, s.t, n>N \to |\dfrac{x^4-1}{x^4+x-6}-1| < \epsilon$$
I started this proof and was trying to find an ideal $N$.
$$|\dfrac{x^4-1}{x^4+x-6}-1| < \epsilon$$
$$|\dfrac{x^4-1}{x^4+x-6}-\frac{(x^4+x-6)}{x^4+x-6}| < \epsilon$$
$$|\dfrac{-1-x+6}{x^4+x-6}| = |\dfrac{5-x}{x^4+x-6}| = |\dfrac{x-5}{x^4+x-6}| < \epsilon$$
Now I wanted to do some bounding so I can make the function larger:
$$|\dfrac{x-5}{x^4+x-6}| \leq |\dfrac{x-5}{x^4}| \leq |\dfrac{2x}{x^4}| \leq |\dfrac{2}{x^3}|$$
but this is not the case when I graph it. Where is my bounding wrong?
As you can see, the red graph is actually bigger then the blue graph. Why?
| Note that for $n\ge 2$, $n^4+n-6\ge n^4/2$. Hence we can write
$$\left|\frac{n^4-1}{n^4+n-6}-1\right|=\left|\frac{n-5}{n^4+n-6}\right|\le \frac{n}{n^4/2}=\frac{2}{n^3}<\epsilon$$
whenever $n>(2/\epsilon)^{1/3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2178312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum steps adding edges to form a complete graph Consider a graph $G$ with $t$ vertices and $0$ edges. Turn it into the complete graph $K_t$ by repeatedly applying the following move $M$:
$M$: Choose $n$ vertices in $G$ and add edges between each of them to make a complete subgraph $K_n$ within $G$. This gives the new $G$.
Question: Given $t$ and $n$, what is the least number $m$ of times $M$ has to be applied before $G$ is $K_t$?
Notes:
If $n=2$ then $m$ is ${t \choose 2}$.
If $n=t-1$ then $m$ is 3.
I'd prefer a comprehensive derivation of some estimation over precise candidates for computation.
| As mentioned in comments there is a trivial lower bound on $m$: $m \ge \left\lceil\frac{t(t - 1)}{n(n - 1)}\right\rceil$. This bounds is not tight even for $t = 4$ and $n = 3$ when $m = 3$ while the bound is $2$.
Let's think deeper: $\deg_{K_t} v = t - 1$ for any $v \in K_t$, and initially $\deg_G v = 0$. Each move $M$ increrases degree of $v$ by at most $n - 1$. Then each of $t$ vertices requires at least $\left\lceil\frac{t - 1}{n - 1}\right\rceil$ moves. And each move affects exactly $n$ vertices. Thus we get $m \ge \left\lceil\left\lceil\frac{t - 1}{n - 1}\right\rceil\cdot\frac{t}{n}\right\rceil$. It is exact up to $(t, n) = (7, 4)$ when $m = 5$ and the bound is $4$. (I mean my bound is exact for $t \le 6$ and for $t = 7$ and $n \le 3$.)
The most trivial upper bound for $n \ge 2$ is $\frac{t(t - 1)}2$ that is the number of edges. Let's try to get something better. To cover all edges incident to a single vertex it would be enought to make $\left\lceil\frac{t - 1}{n - 1}\right\rceil$ moves $M$. After that we can remove this vertex and get $K_{t - 1}$ on all other vertices. And so on until we left only $n$ vertices that need only one move:
$$m \le \left\lceil\frac{t - 1}{n - 1}\right\rceil + \left\lceil\frac{t - 2}{n - 1}\right\rceil + \cdots + \left\lceil\frac{n - 1}{n - 1}\right\rceil.$$
To simplify this bound let's use $k = \left\lfloor\frac{t - 2}{n - 1}\right\rfloor$ and $r = (t - 2) - k(n - 1)$. Noting that $\left\lceil\frac xy\right\rceil = \left\lfloor\frac{x + y - 1}{y}\right\rfloor$ for integers $x$ and $y$, we get:
$$m \le \left\lfloor\frac{t + n - 3}{n - 1}\right\rfloor + \left\lfloor\frac{t + n - 4}{n - 1}\right\rfloor + \cdots + \left\lfloor\frac{n + n - 3}{n - 1}\right\rfloor\\
= (k + 1) \cdot (r + 1) + k \cdot (n - 1) + (k - 1)\cdot(n - 1) + \cdots + 2 \cdot (n - 1) + 1\\
= (k + 1) \cdot (r + 1) + \left(\frac{k(k + 1)}{2} - 1\right)(n - 1) + 1\\
= \frac{k(k + 1)}{2}(n - 1) + kr + k + r - n + 3.$$
Here I combine these two bounds into one table:
$$\begin{array}{c|c|c|c|c|c|c|c}
n\backslash t & 2 & 3 & 4 & 5 & 6 & 7 & 8\\\hline
2 & 1 / 1 & 3 / 3 & 6 / 6 & 10 / 10 & 15 / 15 & 21 / 21 & 28 / 28\\\hline
3 & & 1 / 1 & 3 / 3 & \color{red}{5} / 4 & \color{red}{8} / 6 & \color{red}{11} / 7 & \color{red}{15} / 11\\\hline
4 & & & 1 / 1 & 3 / 3 & \color{red}{5} / 3 & \color{red}{7} / \color{red}{4} & \color{red}{10} / 6\\\hline
5 & & & & 1 / 1 & 3 / 3 & \color{red}{5} / 3 & \color{red}{7} / \color{red}{3}\\\hline
6 & & & & & 1 / 1 & 3 / 3 & \color{red}{5} / 3\\\hline
7 & & & & & & 1 / 1 & 3 /3\\\hline
8 & & & & & & & 1 / 1
\end{array}$$
Non-exact bounds are marked with red color. As far as you see both bounds agree on $n = 2$, $n = t$ and $n = t - 1$ and the lower bound is much more close to the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Geometric and arithmetic sequence Let a,b and c be consecutive terms in a geometric sequence and a, 2b and c be consecutive terms in an arithmetic sequence. Determine the quotient b/a.
$a,2b,c$ arithmetic sequence
$2b=a+d$ and $c=a+2d$
And, $a+2b+c=3(a+c)/2\leftrightarrow2a+4b+2c=3a+3c\leftrightarrow$
$4b=a+c$
$a,b,c$ geometric sequence
$b=ar$ and $c=ar^{2}$
And, $a+b+c=\frac{a(1-r^{3})}{1-r}$
Came to this first:
$\frac{b}{a}=\frac{c}{b}$
$\frac{b}{a}=\frac{4b-a}{b}$
$\frac{b}{a}=4-\frac{a}{b}$
$\frac{b}{a}+\frac{a}{b}=4$
| $2b-a = d\\
c - 2b = d\\
c - a = 2d\\
c+a = 4b$
That is about as much as we can say right now about the arithmetic sequence. Lets look at the geometric sequence.
$\frac {b}{a} = r\\
\frac {c}{b} = r^2\\
b = ar\\
c = ar^2$
We want to solve for $r$
Substitute: $b= ar, c=ar^2$ into $a+c = 4b$
$ar^2+a = 4ar\\
a(r^2 - 4r + 1) = 0$
Use the quadratic formula to solve for $r.$
| {
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"url": "https://math.stackexchange.com/questions/2179905",
"timestamp": "2023-03-29T00:00:00",
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Diophantine equation $p^{3}-q^{5}=(p+q)^{2}$ $p$,$q$ are prime numbers.
solve the equation $~p^{3}-q^{5}=(p+q)^{2}$.
I tried but got only that $p$,$q$ are both $1$ (mod $4$) or $3$ (mod $4$).
Could you help me?
| $$p^3-q^5=(p+q)^2=p^2+2pq+q^2$$
So, $p^3-q^5\ge 0$, so $p^3\ge q^5$, so $p>q$.
$$p^2(p-1)=p^3-p^2=q^5+2pq+q^2=q(q^4+2p+q)$$
Hence, $p^2$ divides $q^4+2p+q$ and so $q$ divides (p-1).
So $p^2\le q^4+2p+q$, so $p\le q^2+1$. We can then verify that $q=2$ has no solution for $p$.
$p$ is prime, and $q^2+1$ and $q^2$ are not:
$$q<p\le q^2-1$$
let $p=aq+b$ with $1\le a< q$ and $1\le b<q$.
We get $$q^5=p^3-(p+q)^2=(b^3-b^2)+q.(\dots)$$
Hence $q$ divides $b^2(b-1)$, so $b=1$
Hence $p=aq+1$ with $1\le a< q$
We get $$q^5=p^3-(p+q)^2=q.(3a-2(a+1))+q^2.(\dots)$$
Hence, $q$ divides $a-2$, so $a=2$.
So $p=2q+1$
$$q^5=q^2.(8q+12-9)$$
$$0=q^3-8q-3=(q-3).(q^2+3q+1)$$
Therefore the only positive integer solution is $q=3$ and so $p=7$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Factors, system and so on... $x+y=1$; $x^{2}+y^{2}=221$; what is $(x³)+(y³)$ ?
I already solved this problem, but I wanna know if there is an easier way to solve it, because I'm studying for a very hard exam and I must learn how to choose the best tactic to solve math problems.
Check out what I did:
$x^{3}+y^{3}$
$(x+y)[x^{2}-xy+y^{2}]$
$(x+y)[x^{2}+2xy+y^{2}-3xy]$
$(x+y)[(x+y)^{2}-3xy]$
$(x+y)^{3}-3xy(x+y)$
$1^{3}-3xy1$
$1-3xy$
Then:
$x=1-y$
$(1-y)^{2}+y^{2}=221$
$1-2y+y^{2}+y^{2}=221$
$2y^{2}-2y-220=0$
$y=\frac{-(-2)±\sqrt{(-2)^{2}-8(-220)}}{4}$
$y’=(-10)$; $y''=11$
$x+y=1$ $\therefore$ $x=11; y=(-10)$
Then we have:
$1-3*11(-10)=331$
And that's the answer!
My question is: There is an alternative way to solve this problem?
| The posted proof is essentially correct, but note that the first half is not really necessary since you calculate the actual values of $x,y$ in the second half.
Regarding x=11; y=(-10) you should make a note that the system is symmetric in $x,y\,$, so $x=-10, y=11$ is a solution as well, but it doesn't matter which pair you choose since the expression to calculate ($x^3+y^3$) is itself symmetric in $x,y$.
There is an alternative way to solve this problem?
*
*$(x+y)^2=x^2+y^2+2xy \implies xy = \frac{1}{2}\left((x+y)^2-(x^2+y^2)\right)=\frac{1}{2}(1-221)=-110$
*$(x+y)^3=x^3+y^3+3xy(x+y) \implies x^3+y^3 = (x+y)^3-3xy(x+y) = 1 + 3 \cdot 110$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2180907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Prove that : $ S(n) = \frac{2^{n-1}}{\pi}\int_0^{2\pi}\cos (t)\cos (2t) \dots \cos (nt) dt $ For a positive integer $n$, denote by $S(n)$ the number of choices of the signs "+" or
"−" such that $±1 ± 2±\dots±n = 0$. Prove that :
$$ S(n) = \frac{2^{n-1}}{\pi}\int_0^{2\pi}\cos (t)\cos (2t) \dots \cos (nt) dt $$
I have no clue on how to approach this problem, but the following hint was given:
$$ G_n(x) = \left(x + \frac{1}{x}\right) \left(x^2 + \frac{1}{x^2} \right) \dots \left(x^n + \frac{1}{x^n}\right) $$
| The hint pretty much gives away the answer.
$S(n)$ is the term not depending on $x$ in $G_n(x)$. If in the expression
$$\left(x + \frac{1}{x}\right) \left(x^2 + \frac{1}{x^2} \right) \dots \left(x^n + \frac{1}{x^n}\right) = S(n) + \sum_{k \neq 0 }c_kx^k$$ we set $x = e^{it}$ and integrate from $0$ to $2\pi$, we obtain
$$ \int_0^{2\pi} (2\cos t)(2\cos 2t)\dots(2\cos nt)dt = 2\pi S(n) + 0$$
which gives us the desired formula which is
$$ S(n) = \frac{2^{n-1}}{\pi}\int_0^{2\pi}\cos (t)\cos (2t) \dots \cos (nt) dt $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Product of Roots of a $4$th-degree Polynomial Problem:
Let $a$, $b$, $c$, and $d$ be distinct real numbers such that
\begin{align*}
a &= \sqrt{4 + \sqrt{5 + a}}, \\
b &= \sqrt{4 - \sqrt{5 + b}}, \\
c &= \sqrt{4 + \sqrt{5 - c}}, \\
d &= \sqrt{4 - \sqrt{5 - d}}.
\end{align*}Compute $abcd$.
I know we can find a polynomial that $a$ is a root of, then do the same for $b, c,$ and $d,$ but how would I continue to do this problem?
| Hint:
$a,b,c,d$ are the solutions of equation $$x=\sqrt{4\pm\sqrt{5\pm x}}$$
Repeatedly square both sides and you should get a $4$th-degree polynomial.
And the product $abcd$ would be the constant term of the polynomial.
(Why? Consider the equation $(x-a)(x-b)(x-c)(x-d)=0$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
Find the laurent series of $\frac{1}{z(z-2)^3} $ Find the laurent series of $$\frac{1}{z(z-2)^3} $$ about $$ z_0 = 0, z_0 = 2 $$
I am having trouble with this question.
I have worked out that with some manipulation I can get f(z) into the form:
$$ \frac{-1}{8z(1-\frac{z}{2})^3} $$
However the solution to the question uses the following geometric series to find residue at $$z_0 = 0 $$
$$\sum_{n=0}^\infty n(n-1)z^{n-2} = \frac{2}{(1-z)^3} $$
I have tried repeatedly to manipulate the standard geometric series for 1/(1-z) however I can't seem to replicate this geometric series. This is also a recurring problem I am having when trying to find the laurent series of a function f(z) when it has a pole of order n where n > 1.
Any help about how to solve this and similar questions of the same form in future would be much appreciated.
| Let $z=2w$.
$$\frac{1}{z(z-2)^3} = -\frac{1}{16}\frac{1}{w(1-w)^3} $$
First comes partial fraction decomposition
\begin{align}
\frac{1}{w(1-w)^3} &= \frac{1}{w}+\frac{1}{1-w}+\frac{1}{(1-w)^2}+\frac{1}{(1-w)^3}
\end{align}
If you're seeking the residue, the problem is almost over. The latter three terms can be expressed as a geometric series and its derivatives, which are all Taylor series. The residue must be coming from the first term $-\frac{1}{16w} = -\frac{1}{8z}$. If you want to actually write down the Laurent series, consider
\begin{align}
\frac{1}{1-w} &= \sum_{n=0}^\infty w^n
\\
\\
\frac{1}{(1-w)^2} &= \sum_{n=1}^\infty nw^{n-1} = \sum_{n=0}^\infty (n+1)w^{n}
\\
\\
\frac{1}{(1-w)^3} &= \frac{1}{2}\sum_{n=2}^\infty n(n-1)w^{n-2} = \frac{1}{2}\sum_{n=0}^\infty (n+2)(n+1)w^{n}
\\
\\
\frac{1}{(1-w)^{N+1}} &= \frac{1}{N!}\sum_{n=N}^\infty \frac{n!}{(n-N)!}w^{n-N}
= \frac{1}{N!}\sum_{n=0}^\infty \frac{(n+N)!}{n!}w^{n}
= \sum_{n=0}^\infty \binom{n+N}{N}w^{n}
\end{align}
Then
\begin{align}
-\frac{1}{16}\frac{1}{w(1-w)^3}
&=
-\frac{1}{16}\left\{ \frac{1}{w}+\sum_{n=0}^\infty\left(w^n+(n+1)w^n+\frac{(n+2)(n+1)}{2}w^n\right)\right\}
\\
&=
-\frac{1}{16}\left\{ \frac{1}{w}+\sum_{n=0}^\infty\left(1+(n+1)+\frac{(n+2)(n+1)}{2}\right)w^n\right\}
\\
\\
&=
-\frac{1}{16}\left\{ \frac{1}{w}+\sum_{n=0}^\infty\frac{(n+2)(n+3)}{2}w^n\right\}
\\
\\
\frac{1}{z(z-2)^3} &=
-\frac{1}{16}\left\{ \frac{2}{z}+\sum_{n=0}^\infty\frac{(n+2)(n+3)}{2^{n+1}}z^n\right\}
\end{align}
Finally, let's discuss the approach for the expansion about $z_0=2$. We already have the partial fraction decomposition.
\begin{align}
\frac{-1}{16}\left(\frac{2}{z(1-z/2)^3}\right) &= \frac{-1}{16}\left(\frac{2}{z}+\frac{1}{1-z/2}+\frac{1}{(1-z/2)^2}+\frac{1}{(1-z/2)^3}\right)
\\
&= \frac{2}{(z-2)^3}-\frac{1}{4(z-2)^2}+\frac{1}{8(z-2)}-\frac{1}{8z}
\end{align}
The first few terms are already terms in the Laurent series. This will come down to expanding $1/z$ about $z_0=2$. You should find that this expansion is given by a Taylor series.
Looking over this, the general series for $\frac{1}{w(1-w)^N}$ about $w_0=0$ and $w_0=1$ look fairly easily. From that you can transform over to $\frac{1}{z(z-a)^N}$. Anyway, the two solutions respectively appear to be
$$\frac{1}{w(1-w)^N} = \frac{1}{w}+\sum_{n=0}^\infty \sum_{k=1}^{N} \binom{n+1}{k}w^n$$
$$\frac{1}{w(1-w)^N} = \sum_{n=-N}^{\infty}(-1)^{2n+1}(w-1)^N$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Remainder of $3^n$ modulo $15$ Let $n \in \mathbb{N}_{>0}$. Then we have $$\begin{align*} &3 \equiv 3 \bmod 15\\&3^2 \equiv 9 \bmod 15\\&3^3 \equiv 12 \bmod 15\\&3^4 \equiv 6 \bmod 15\\&3^5 \equiv 3 \bmod 15\end{align*}$$ Hence all remainders modulo $15$ of $3^n$ should be $3,9,12$ and $6$. Why is this the case? Also, How could I calculate $$3^{1291}\equiv ? \bmod 15$$
| As to why: there isn't a much better reason than "because you've just demonstrated that's what happened". But here's a little: $3^n$ is always a multiple of $3$. Since $15$ is also a multiple of $3$, the remainder of $3^n$ when divided by $15$ must always be a multiple of $3$ - so $3^n$ must be $0,3,6,9$, or $12$ mod $15$. But $15$ is also a multiple of $5$, and $3^n$ is not, so $3^n$ cannot be a multiple of $15$; so $3^n$ can't be $0$ mod $15$.
As for $3^{1291}$: you've established more than just that $3^n$ is $3,9,12$, or $6$ mod $15$; you've established that it takes those values in order. So, in other words, you know that $3^1$, $3^5$, $3^9$, $3^{13}$, and so on all come to $3$ mod $15$. In general, you know that every fourth power of three after the first - so $3^{4n+1}$ for any $n$ - is $3$ modulo $15$. That will get you close to $3^{1291}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to take $\int^1_{-1} \frac{xdx}{x^2+x+1}$? Here's the integral:
$$\int^1_{-1} \frac{xdx}{x^2+x+1}$$
So I am basicly confused about $xdx$ if it were only $dx$ I would just compleate the square and go ahead with table formula, but as i can judge it doesn't work properly witx $xdx$
Can someone give me a hint, please?
| Let $I = \int^1_{-1} \frac{x}{x^2+x+1} dx$
$2I = \int^1_{-1} \frac{2x + 1 - 1}{x^2+x+1} dx$
$2I = [\ln|x^{2}+x+1|]_{-1}^{1} -\int_{-1}^{1}\frac{1}{x^{2}+x+1}dx$
$2I = \ln3 -\int_{-1}^{1}\frac{1}{(x+\frac{1}{2})^{2}+\frac{3}{4}}dx$
Can you solve it from there?
Edit: In case you get stuck:
Let $x+\frac{1}{2} = \frac{\sqrt{3}}{2}\tan u$
$\Rightarrow dx = \frac{\sqrt{3}}{2}\sec^{2}u$
Lower limit:
$-1+\frac{1}{2} = \frac{\sqrt{3}}{2}\tan u\Rightarrow \tan u = -\frac{1}{\sqrt{3}} \Rightarrow u = -\frac{\pi}{3}$
Upper limit:
$\frac{3}{2}=\frac{\sqrt{3}}{2}\tan u \Rightarrow u = \frac{\pi}{6}$
$2I = \ln3 - \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{\frac{3}{4}(\tan^{2}u+1)} \cdot \frac{\sqrt{3}}{2}\sec^{2}u\ du = \ln3 - \int_{-\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{2}{\sqrt{3}}\ du$
$2I = \ln 3 - (\frac{2\pi}{3\sqrt{3}} +\frac{2\pi}{6\sqrt{3}}) = \ln 3 -\frac{\pi}{\sqrt{3}}$
$I = \frac{1}{2}(\ln 3 - \frac{\pi}{\sqrt{3}})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How to find max integer value of $6\sin(x)-8\cos(x)$ without using derivative I have $6\sin(x)-8\cos(x)$ and want to find its maximum value. If we use derivative and assuming that it is equal to 0, we get $6\cos(x)+8\sin(x)=0$ which implies that $\tan(x)=-\cfrac{3}{4}$ and $x = 143^\circ$ and we can conclude that the maximum value is equal to $10$ . How can I find the maximum integer value of $6\sin(x)-8\cos(x)$ without using derivative?
| Let $\dfrac{b}{a}=\tan\alpha$ then
$$\cos\alpha=\dfrac{1}{\sqrt{1+\tan^2\frac{\alpha}{2}}}=\dfrac{a}{\sqrt{a^2+b^2}}$$
so
\begin{eqnarray}
a\sin x+b\cos x
&=&
a(\sin x+\dfrac{b}{a}\cos x)\\
&=&
a(\sin x+\tan\alpha\cos x)\\
&=&
a(\sin x+\dfrac{\sin\alpha}{\cos\alpha}\cos x)\\
&=&
\dfrac{a}{\cos\alpha}(\sin x\cos\alpha+\sin\alpha\cos x)\\
&=&
\dfrac{a}{\cos\alpha}\sin(x+\alpha)\\
&=&
\sqrt{a^2+b^2}\sin(x+\alpha)
\end{eqnarray}
from
$$-1\leq\sin(x+\alpha)\leq1$$
we have
$$-\sqrt{a^2+b^2}\leq a\sin x+b\cos x\leq \sqrt{a^2+b^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Factoring quadratics - why does this method work? Given a quadratic like this:
$2x^2 + 7x + 3$
I need to find two numbers $a$ and $b$ that have these properties:
$a + b = 7$
$a \cdot b = 2 \cdot 3$
Which in this case is 1 and 6.
So I can rewrite the equation as:
$ 2x^2 + 1x + 6x + 3$
$= x(2x + 1) + 3(2x + 1)$
$= (2x + 1) (x + 3)$
I wanted to know why this method works. I found a proof in a khan academy video that goes like this, based on the multiplication of two binomials:
$(fx + g) (hx + j)$
$= fhx^2 + ghx + jfx + gj$
$= fhx^2 + x(gh + jf) + gj$
Then let's assume that $gh = a$ and $jf = b$ this would be the proof for finding the middle coefficients. I understand this part.
The video then tells me that therefore $a \cdot b = gh \cdot jf$ which can be rewritten as $a \cdot b = fh \cdot gj$.
I don't understand this part - because for me it doesn't really prove that I need to find a number that multiplies the first coefficient with the constant (the last number) in the equation. Is there maybe another way to explain that?
| Let's assume $ax^2 + bx + c = (rx + s)(tx + u)$. Then
\begin{align*}
ax^2 + bx + c & = (rx + s)(tx + u)\\
& = rx(tx + u) + s(tx + u)\\
& = rtx^2 + rux + stx + su\\
& = rtx^2 + (ru + st)x + su
\end{align*}
is an algebraic identity that holds for every real number $x$. In particular, it holds if $x = 0$, $x = 1$, and $x = -1$. If $x = 0$, we obtain
$$c = su \tag{1}$$
If $x = 1$, we obtain
$$a + b + c = rt + ru + st + su \tag{2}$$
Since $c = su$, we can cancel $c$ from the left hand side and $su$ from the right hand side to obtain
$$a + b = rt + ru + st \tag{3}$$
If $x = -1$, we obtain
$$a - b + c = rt - ru - st + su \tag{4}$$
Since $c = su$, we can cancel $c$ from the left hand side and $su$ from the right hand side to obtain
$$a - b = rt - ru - st \tag{5}$$
Adding equations 3 and 5 yields
$$2a = 2rt \tag{6}$$
Dividing both sides of equation 6 by 2 yields
$$a = rt \tag{7}$$
Since $a = rt$, we can cancel $a$ from the left hand side of equation 3 and $rt$ from the right hand side of equation 3 to obtain
$$b = ru + st \tag{8}$$
Furthermore, observe that
$$ac = (rt)(su) = (ru)(st) \tag{9}$$
Hence, we can factor a polynomial with rational coefficients with respect to the rational numbers if there exist two numbers ($ru$ and $st$) with product $ac$ and sum $b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Generating function for a recurrence: I made a mistake
Use generating functions to solve the recurrence
$$a_n = 3a_{n-1} - 4a_{n-3} + n3^n$$
where $a_0 = a_1 = 1$, $a_2 = 0$.
I tried to solve this like any other generating functions problem, but I'm getting the wrong answer. Some, but not all, of the terms in the final partial fraction expansion are correct; the others are slightly off. (I tried using characteristic equations and I got the right answer, but I have to use generating functions.) Where am I going wrong?
My attempt:
$$
\begin{align*}
a_n &= 3a_{n-1} - 4a_{n-3} + n3^n \\
\implies \sum_{n=3}^{\infty} a_nx^n &= \sum_{n=3}^{\infty} 3a_{n-1}x^n - \sum_{n=3}^{\infty} 4a_{n-3}x^n + \sum_{n=3}^{\infty} n3^nx^n \\
\implies G(x) - a_0 - a_1x - a_2x^2 &= 3x\sum_{n=2}^{\infty} a_{n}x^n - 4x^3\sum_{n=0}^{\infty} a_{n}x^n + \sum_{n=3}^{\infty} n3^nx^n \\
&= 3x(G(x) - a_0 - a_1x) - 4x^3G(x) + \sum_{n=3}^{\infty} n3^nx^n \\
\implies G(x)(1 - 3x + 4x^3) &= (a_2 - 3a_1)x^2 + (a_1 - 3a_0)x + a_0 + \sum_{n=3}^{\infty} n3^nx^n \\
&= -3x^2 - 2x + 1 + \sum_{n=3}^{\infty} n3^nx^n
\end{align*}
$$
and as $\frac{3x}{(1-3x)^2} = \sum_{n=0}^{\infty} n3^nx^n$, we have $\sum_{n=3}^{\infty} n3^nx^n = \frac{3x}{(1-3x)^2} - 3x - 18x^2$ and
$$
\begin{align*}
G(x)(1 - 3x + 4x^3) &= -21x^2 - 5x + 1 + \frac{3x}{(1 - 3x)^2} \\
&= \frac{1 - 5x - 21x^2}{(2x - 1)^2(x + 1)} + \frac{3x}{(1 - 3x)^2(2x - 1)^2(x + 1)}
\end{align*}
$$
Decomposing that into partial fractions gives
$$-\frac{57}{2(2 x - 1)} + \frac{513}{16 (3 x - 1)} - \frac{1}{2 (2 x - 1)^2} + \frac{27}{4 (3 x - 1)^2} - \frac{27}{16 (x + 1)}$$
However, the given answer (which I verified to be true) is
$$-\frac{28}{2 x - 1} + \frac{405}{16 (3 x - 1)} - \frac{1}{2 (2 x - 1)^2} + \frac{27}{4 (3 x - 1)^2} - \frac{27}{16 (x + 1)}$$
Where did I go wrong?
| The answer you got is right; the given answer is wrong. To easily check this, you can substitute $x=0$ in the second answer to get
$$-\frac{28}{2\cdot0 - 1} + \frac{405}{16 (3\cdot0 - 1)} - \frac{1}{2 (2\cdot0 - 1)^2} + \frac{27}{4 (3\cdot0 - 1)^2} - \frac{27}{16 (0 + 1)} = \frac{29}{4}$$
which should be equal to $a_0 = 1$ but isn't.
It is true that the formula for $a_n$ is $$a_n = 28(2^n) - \frac{405}{16}(3^n) - \frac12 (n 2^n) + \frac{27}{4} (n 3^n) - \frac{27}{16}((-1)^n)$$ which appears to match the coefficients in the second answer. But the coefficients in the formula don't correspond to the coefficients in the partial fraction decomposition, because the terms which give you $n 2^n$ and $n 3^n$ are slightly more complicated:
$$\frac{1}{(2x-1)^2} = \sum_{n=0}^\infty (n+1)2^n x^n$$
and
$$\frac{1}{(3x-1)^2} = \sum_{n=0}^\infty (n+1)3^n x^n.$$
So from your (correct) partial fraction decomposition, you should actually obtain a coefficient of $\frac{57}{2} - \frac12 = 28$ for $2^n$ and a coefficient of $-\frac{513}{16} + \frac{27}{4} = -\frac{405}{16}$ for $3^n$, which matches the correct formula for $a_n$.
| {
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"url": "https://math.stackexchange.com/questions/2195860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
gaussian quadrature weight function x How do I find an orthogonal polynomial for the Gaussian quadrature when the weight function is $x$?
In particular, I need to find $w_0,w_1,x_0, x_1$ such that
$$\int_0^1xf(x)dx \approx w_0f(x_0)+w_1f(x_1) $$
is exact when $f \in \mathcal{P}_3$.
Which polynomials can I use to find the abscissae $x_0$ and $x_1$?
| A shorter way to construct monic orthogonal polynomial $p_2$ is orthogonal to 1 and x, so
$p_2(x)=x^2+ax+b$;
The orthogonality implies
$$
\int_0^1 x p_2(x)dx=\frac{1}{4}-\frac{1}{3}a+\frac{1}{2}b=0;
$$
$$
\int_0 x^2 p_2(x)dx=\frac{1}{5}+\frac{1}{4}a+\frac{1}{3}b=0.
$$
Solving the system with respect to $a$ and $b$ we obtain $a=-\frac{6}{5}$ and $b=\frac{3}{10}$. The nodes are the roots of $p_2$, i.e. $x_0=\frac{3}{5}-\frac{\sqrt{6}}{10}$ and $x_1=\frac{3}{5}+\frac{\sqrt{6}}{10}$.
The weights are the solution of the system
$$w_0+w_1=\int_0^1x dx=\frac{1}{2}$$
$$w_0x_0+w_1x_1=\int_0^1x^2 dx=\frac{1}{3},$$
that is, $w_0=\frac{1}{4}-\frac{\sqrt{6}}{36}$ and $w_1=\frac{1}{4}+\frac{\sqrt{6}}{36}$.
The error (remainder term)
$$R(f)=\frac{f^{(4)}(\xi)}{4!}\int_0^1 x\left[p_2(x) \right]^2 dx=
\frac{f^{\left(4\right)}\left(\xi \right)}{14400},$$
for some $\xi\in(0,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the exact value of $\cos \frac{11\pi}{12}$ and $\sin \frac{11\pi}{12}$ Find the exact values of those two numbers but by using the two following complex numbers complexs numbers :
$$z = \frac{-\frac{\sqrt3}{2} -i\frac{1}{2}}{\sqrt8}$$
and
$$w = \frac{\frac{\sqrt2}{2} +i\frac{\sqrt2}{2}}{\sqrt8}$$
I added them up and computed $z/w$ but it led me to nowhere I feel like I have to do some combined computation with those rwo complex numbers and then by identification I can know the exact values but I haven't found the right operations to do yet.
help me please.
| $$\cos\frac{\pi}{12}+i\sin\frac{\pi}{12} = e^{i\pi/12} = e^{i\pi/3}\cdot e^{-i\pi/4} \tag{1}$$
$$ = \left(\frac{1}{2}+i\frac{\sqrt{3}}{2}\right)\cdot\left(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right) \tag{2}$$
$$ = \left(\frac{\sqrt{2}}{4}+\frac{\sqrt{6}}{4}\right)+i\left(\frac{\sqrt{6}}{4}-\frac{\sqrt{2}}{4}\right)\tag{3} $$
hence
$$ \cos\frac{11\pi}{12} = -\frac{\sqrt{2}+\sqrt{6}}{4},\qquad \sin\frac{11\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4}.\tag{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that $a_{n+1} = 1 + \frac{a_n}{a_n +1}$ is monotone increasing with $a_1=1$ This question has been driving me crazy, and I can't find the answer anywhere.
I tried proving it using induction.
As for the base case:
$a_1=1<1.5=a_2$
Next, suppose $a_k \leq a_{k+1}$
Then,
$$a_k \leq a_{k+1}$$
$$a_k +1 \leq a_{k+1}+1$$
$$ \frac{1}{a_k+1}\geq \frac{1}{ a_{k+1}+1} $$
$$ \frac{a_k}{a_k+1}\geq \frac{a_k}{ a_{k+1}+1} $$
$$ \frac{a_k}{a_k+1} +1\geq \frac{a_k}{ a_{k+1}+1}+1 $$
My goal was to get
$$ \frac{a_k}{a_k+1} +1\leq \frac{a_{k+1}}{ a_{k+1}+1}+1 $$
$$a_{k+1} \leq a_{k+2}$$
But obviously, I was not able to find a way.
I also tried a different inductive approach where I noted
$$a_{k+2}=\frac{5 a_n +3}{3 a_n +2}$$
You can check that this is true by seeing you get the correct $a_3=1.6$ using the given $a_1=1$.
Assuming $a_k \leq a_{k+1}$ I had that
$$ a_k \leq 1 + \frac{a_n}{a_n +1}$$
$$ \frac{a_n}{a_n +1} +1 \leq \frac{{a_n}^2 + 4a_n +2}{{a_n}^2+2a_n+1}$$
From the assumption $a_k \leq 1 + \frac{a_n}{a_n +1}$, we have that ${a_n}^2 \leq a_n +1$, so
$$ \frac{a_n}{a_n +1} +1 \leq \frac{{a_n}^2 + 4a_n +2}{{a_n}^2+2a_n+1} \leq \frac{(a_n+1) +4a_n +2}{{a_n}^2+2a_n+1} =\frac{5a_n+3}{{a_n}^2+2a_n+1}$$
If only I could change the denominator the same way while keeping the direction of the inequality, I could substitute $a_n+1$ for ${a_n}^2$ and I would get the desired $\frac{5 a_n +3}{3 a_n +2}$, proving that $$a_{k+1} \leq a_{k+2}$$ since $$a_{k+2}=\frac{5 a_n +3}{3 a_n +2}$$.
How do I prove the sequence is monotonic increasing? Both of these routes did not get me to the answer.
| Hint: $\,a_{n+1} = 2 - \cfrac{1}{a_n +1}\,$, so $\,a_{n+1}-a_n=\cfrac{1}{a_{n-1} +1}-\cfrac{1}{a_n +1}=\cfrac{a_n-a_{n-1}}{(a_n +1)(a_{n-1} +1)}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $2\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx+\int_0^1 \frac{\ln^2(1+x)}{1+x^2}dx=\frac{3}{16}\pi\ln^22-2G\ln2+\frac{7}{64}\pi^3$ How prove that
$$2\int_0^1 \dfrac{(\ln(1-x))^2}{1+x^2}dx+\int_0^1 \dfrac{(\ln(1+x))^2}{1+x^2}dx=\dfrac{3}{16}\pi\left(\ln2\right)^2-2G\ln2+\dfrac{7}{64}\pi^3$$
Where G is the Catalan's Constant.
$$ \int_0^1 \dfrac{(\ln(1+x))^2}{1+x^2}dx=\Big[\arctan x\left(\ln\left(1+x\right)\right)^2\Big]_0^1-2\int_0^1 \dfrac{\arctan x\ln(1+x)}{1+x}dx=\dfrac{1}{4}\pi\left(\ln2\right)^2-2J$$ see users FDP about Evaluating $$\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$$
| Let,
$\displaystyle A=\int_0^1 \dfrac{\left(\ln(1-x)\right)^2}{1+x^2}dx$
$\displaystyle B=\int_0^1 \dfrac{\left(\ln(1+x)\right)^2}{1+x^2}dx$
In integral defining $A$ perform the change of variable $y=\dfrac{1-x}{1+x}$,
$\begin{align}
A&=\int_0^1 \dfrac{\left(\ln\left(\tfrac{2x}{1+x}\right)\right)^2}{1+x^2}dx\\
&=B-2\int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx-\dfrac{1}{4}\pi(\ln 2)^2+\dfrac{\pi^3}{16}-2G\ln 2+\dfrac{1}{4}\pi(\ln 2)^2\\
&=B+\dfrac{\pi^3}{16}-2G\ln 2-2\int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx
\end{align}$
From Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$
using (6),
$\boxed{\displaystyle \int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx=\dfrac{\pi^3}{256}-\dfrac{1}{2}G\ln 2+\dfrac{9}{64}\pi(\ln 2)^2-\dfrac{3}{2}J}$
using (4),
$\boxed{B=\dfrac{1}{4}\pi(\ln 2)^2-2J}$
Therefore,
$\boxed{A=J+\dfrac{7}{128}\pi^3-\dfrac{1}{32}\pi(\ln 2)^2-G\ln 2}$
Therefore,
$\begin{align}2A+B&=\left(2J+\dfrac{7}{64}\pi^3-\dfrac{1}{16}\pi(\ln 2)^2-2G\ln 2\right)+\dfrac{1}{4}\pi(\ln 2)^2-2J\\
&=\boxed{\dfrac{7}{64}\pi^3+\dfrac{3}{16}\pi(\ln 2)^2-2G\ln 2}
\end{align}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to analyze this recursion? How can I analyze this recursion for $k>0$?
$$T(n)=n+T\left(\frac{n}{2}\right)+T\left(\frac{n}{4}\right)+T\left(\frac{n}{8}\right)+\cdots+T\left(\frac{n}{2^k}\right)$$
I want to prove that $T(n)=\theta(n\log(n))$.
Is it true that $T(n)=n+\frac{n}{2}+\cdots+\frac{n}{2}=n+\frac{n}{2}\log(n)$?
I got it by iterations method of these:
\begin{align}T(n)&=n+T\left(\frac{n}{2}\right)+T\left(\frac{n}{4}\right)+T\left(\frac{n}{8}\right)+\cdots+T\left(\frac{n}{2^k}\right)\\
T\left(\frac{n}{2}\right)&=\frac{n}{2}+T\left(\frac{n}{4}\right)+T\left(\frac{n}{8}\right)+T\left(\frac{n}{16}\right)+\cdots+T\left(\frac{n}{2^{k+1}}\right)\\
T\left(\frac{n}{4}\right)&=\frac{n}{4}+T\left(\frac{n}{8}\right)+T\left(\frac{n}{16}\right)+T\left(\frac{n}{32}\right)+\cdots+T\left(\frac{n}{2^{k+2}}\right)\\
T\left(\frac{n}{8}\right)&=\frac{n}{8}+T\left(\frac{n}{16}\right)+T\left(\frac{n}{32}\right)+\cdots+T\left(\frac{n}{2^{k+3}}\right)\\
&\vdots\end{align}
| We have
$$
T\left(\frac{n}{2}\right) = \frac{n}{2} + T\left(\frac{n}{4}\right) + \cdots + T\left(\frac{n}{2^{k+1}}\right) \tag{$1$}
$$
and
$$
T\left(n\right) = n + T\left(\frac{n}{2}\right) + \cdots + T\left(\frac{n}{2^{k}}\right) \tag{$2$}
$$
By $(2) - (1)$, we have
$$
T(n) = 2T\left(\frac{n}{2}\right) + \frac{n}{2} - T\left(\frac{n}{2^{k+1}}\right) \tag{$3$}
$$
implying (if $T(n) \geq 0$ for all $n$)
$$
T(n) \leq 2T\left(\frac{n}{2}\right) + \frac{n}{2}
$$Thus by master theorem,
$$
T(n) = \mathcal{O}(n \log n)
$$
We can not conclude $T(n) = \Omega(n \log n)$ if $k$ is fixed. As a counterexample, when $k = 1$, we have
$$
T(n) = T\left(\frac{n}{2}\right) + n
$$
and in this case, $T(n) = \Theta(n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that for $n\geq3$, $n^3\geq3n+5$
Prove that for $n\geq3$,$$n^3\geq3n+5$$
My try:
Let $f(x)=x^3-3x-5$. Clearly $x=1$ and $x=-1$ are critical points where $x=1$ is the local minima so the function $f(x)$ is an increasing function for $x>1$ and the result follows.
Please check and if possible provide another way to solve like induction. Thank you.
| Using induction:
$1)$ For $n=3$ we get $3^3\ge 3\cdot 3+5=14$
$2)$ Hypothesis: $n^3\ge 3n+5$
$3)$ For $n+1$
$$(n+1)^3=n^3+(3n^2+3n+1)\ge 3n+5+(3n^2+3n+1)\\
(n+1)^3\ge [3(n+1)+5]+(3n^2+3n-2)\quad (1)$$
but
$$3n^2+3n-2=3n(n+1)-2\ge0\text{ for } n\ge 3$$
then from $(1)$ we get
$$(n+1)^3\ge 3(n+1)+5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $p(x)=1+x+x^2+x^3+x^4+x^5$ find the remainder when $p(x^{12})$ is divided by $p(x)$
Question:
If $p(x)=1+x+x^2+x^3+x^4+x^5$ find the remainder when $p(x^{12})$ is divided by $p(x)$
Options:
(A) $0$
(B) $6$
(C) $1+x$
(D) $1+x+x^2+x^3+x^4$
My attempt:
Since $p(x)=1+x+x^2+x^3+x^4+x^5$ is a geometric series assuming $x<1$ with sum
$$S_1=\frac{1-x^6}{1-x}$$
and $p\left(x^{12}\right)$ also is a geometric series with sum
$$S_2=\frac{1-\left(x^{12}\right)^6}{1-x}$$
then
$$\frac{S_2}{S_1}= \frac{1-\left(x^{12}\right)^6}{1-x^6}$$
Now I can't seem to find a way to proceed further.
EDIT
I just found a mistake I made.
$$S_2=\frac{1-\left(x^{12}\right)^6}{1-x^{12}}$$
And not
$$S_2=\frac{1-\left(x^{12}\right)^6}{1-x}$$
So my division was wrong and so was my solution thereafter.
Sorry !
| Hint: $\dfrac{S2}{S1}= \dfrac{1-(x^{12})^6}{1-x^6}= \dfrac{1-(x^6)^{12}}{1-x^6}= \dfrac{1-y^{12}}{1-y}$, for $y=x^6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $x,y$ and $z$ when: $x+y=\sqrt{4z-1}, \ y+z=\sqrt{4x-1}\ \mathrm{and} \ z+x=\sqrt{4y-1}$ How to Solve for $x,y$ and $z$ when: $x+y=\sqrt{4z-1}, \ y+z=\sqrt{4x-1}\ \mathrm{and} \ z+x=\sqrt{4y-1}$.
I've added the equations and get: $2(x+y+z)=\sqrt{4x-1}+\sqrt{4y-1}+\sqrt{4z-1}$ and after this how I proceed further?
Even if I subtract the first two equations then I'm getting:
\begin{align*}
x-z&=\sqrt{4z-1}-\sqrt{4x-1}\\
\Rightarrow\left(x+\sqrt{4x-1}\right)^2&=\left(z+\sqrt{4z-1}\right)^2\\
\Rightarrow\left(x^2-z^2\right)+4(x-z)&=2z\sqrt{4z-1}-2x\sqrt{4x-1}
\end{align*}
after this also I can't understand how to help with this.
| Suppose $x,y,z$ are real numbers such that
\begin{align*}
x + y &= \sqrt{4z-1}\\[4pt]
y + z &= \sqrt{4x-1}\\[4pt]
z + x &= \sqrt{4y-1}\\[4pt]
\end{align*}
Let $s = x + y + z$. Since $x + y \ge 0,\;\;y + z \ge 0,\;\;z + x \ge 0$, we have $s \ge 0$.
Then from the original system of equations, we get
\begin{align*}
(s-z)^2 &= 4z-1\\[4pt]
(s-x)^2 &= 4x-1\\[4pt]
(s-y)^2 &= 4y-1\\[4pt]
\end{align*}
Let $f(u) = (s - u)^2 - (4u - 1) = u^2 - (2s + 4)u + (s^2+1)$.
Then, $x,y,z$ are roots of $f$, hence, since $f$ is quadratic in $u$, at least two of $x,y,z$ must be equal.
Without loss of generality, assume $z = y$.
Suppose $x \ne y$. Then since $x,y$ are roots of $f$, Vieta's formulas yield
\begin{align*}
&x + y = 2s + 4\\[4pt]
&xy = s^2 +1\\[4pt]
\end{align*}
hence, since $s \ge 0$, we get $x,y > 0$. Then also $z > 0$, since $z=y$. But then
\begin{align*}
&x,y,z > 0\\[4pt]
\implies\; &x + y < s\\[4pt]
\implies\; &2s + 4 < s\\[4pt]
\implies\; &s < -4\\[4pt]
\end{align*}
contradiction.
It follows that $x = y$, hence $x = y = z$, so
\begin{align*}
&(s - z)^2 = 4x-1\\[4pt]
\implies\; &(2x)^2 = 4x -1\\[4pt]
\implies\; &(2x-1)^2 = 0\\[4pt]
\implies\ &x = {\small{\frac{1}{2}}}\\[4pt]
\implies\; &x = y = z = {\small{\frac{1}{2}}}\\[4pt]
\end{align*}
It's easily verfied that the triple
$$
(x,y,z) =
{\small{
\left(
\frac{1}{2},
\frac{1}{2},
\frac{1}{2}
\right)
}}
$$
satisfies the original system of equations, hence it's the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can we prove that $\displaystyle\sum_{i\in\mathbb N^+}1/i^2$ is convergence.(Elementary or advanced) I've tried to apply compare test with $\displaystyle\sum_{i\in\mathbb N^+}1/2^i$ like this:
$f(x)=\dfrac{\ln x}{x}$, and if we derivative this we would find that its extrem maximum point is $x=e$, therefore, for very big numbers we can say this;
$$n>>l\\ \Rightarrow \\ l^n>n^l$$ if we choose $l=2$
$$2^n>n^2\quad\rightarrow\quad \dfrac1{2^n}<\dfrac1{n^2}$$
But it is not what we wanted.
How can we prove this, and can we put a upperbound better than 2?
| Besides the comparison with the geometric series, you can also use a telescoping one:
\begin{align*}
\sum_{n=1}^\infty\frac{1}{n^2}=1+\sum_{n=2}^\infty\frac{1}{n\cdot n}\leq 1+\sum_{n=2}^\infty\frac{1}{(n-1)n}=1+\sum_{n=2}^\infty\left(\frac{1}{n-1}-\frac{1}{n}\right)=1+1=2.
\end{align*}
In order to get a better upper bound than 2, just start the telescoping series a little later. For instance:
\begin{align*}
\sum_{n=1}^\infty\frac{1}{n^2}=1+\frac{1}{4}+\sum_{n=3}^\infty\frac{1}{n\cdot n}\leq \frac{5}{4}+\sum_{n=3}^\infty\frac{1}{(n-1)n}=\frac{5}{4}+\sum_{n=3}^\infty\left(\frac{1}{n-1}-\frac{1}{n}\right)=\frac{5}{4}+\frac{1}{2}=\frac{7}{4}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $x$ in $\frac{x^2+12x+36}{x^2+4x-12}=2$ $\frac{x^2+12x+36}{x^2+4x-12}=2$
I factored top and bottom to $\frac{(x+6)(x+6)}{(x+6)(x-2)}=2$
and eliminated the common factor (x+6) for $\frac{x+6}{x-2}=2$
then $x+6=2(x-2)$ and $x+6=2x-4$ and $6+4=2x-x$ and finally $x = 10$
Plugging this back into the original fraction proves true
My instructor however in her exam prep test says
$\frac{x^2+12x+36}{x^2+4x-12}=2$
$$x^2+12x+36 = 2(x^2+4x-12)$$
$$x^2+12x+36 = 2x^2+8x-24$$
$$0 =2x^2-x^2 +8x -12x -24 -36$$
$$0 =x^2-4x -60$$
which she then factors to $(x-10)(x+6)$ saying the answer to the problem is $x=10$ or $x=-6$
But if you plug -6 back into the original fraction you get $0=2$. Where is the problem here? Thanks
| Other people have done an excellent job of pointing out why $-6$ cannot be a solution, but let me add exactly where your professor's approach breaks down. In the first step of cross-multiplying by $x^2+4x-12$, information is being "destroyed"; because you're multiplying by something that can be zero, you might be creating a $0=0$ situation and thus creating solutions that weren't present in the initial problem. Likewise, $\frac{2x}{x} = 1$ has no solution (because the left-hand side evaluates to $2$ whenever it is defined) but $2x = x$ has the new solution $x = 0$, corresponding to the manufactured $0=0$.
In fairness, though, this is an easy mistake to make, and one I wouldn't be surprised to see made by a perfectly competent teacher. If the numbers involved were slightly different, it would have turned out fine. I would argue that the real failure on the professor's part here was failing to check; a simple examination, as you noted, shows that $-6$ can't be a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the polynomial $a_nx^n+\cdots+a_1x+a_0$ has no rational roots
Let $\overline{a_n \ldots a_1a_0}$ be the decimal representation of $65^k$ for some $k \geq 2$. Prove that the polynomial $a_nx^n+\cdots+a_1x+a_0$ has no rational roots.
I thought about using the rational root theorem. We know that $a_0 = 5$ since $65$ ends in a $5$, so any rational root must be of the form $-\dfrac{5}{a}$ or $-\dfrac{1}{a}$ where $a$ is an integer factor of $a_n$. How can we continue?
| Let $f_n$ be the polynomial obtained from the digits of $65^n$. The other answer shows that if $p/q$ is a root of this polynomial in lowest terms, then $p-10q|65^n$; along with the bounds on $f_n$, this implies that we must have $p=-5$ and either $q=2$ or $q=6$. That is, the only possible rational roots are $-\frac{5}{6}$ and $-\frac{5}{2}$.
Note that $(6x+5)^n$ is also a polynomial which evaluates to $65^n$ at $x=10$. So we can write
$$
f_n(x)=(6x+5)^n+(x-10)R(x)
$$
for some polynomial $R$.
Setting $x=0$ in this expression, we have
$
5=5^n-10R(0)
$,
and so $R(0)=\frac{5^{n-1}-1}{2}$, which is not a multiple of $5$.
Now, suppose $n \geq 3$ and $x$ is either $-\frac{5}{6}$ or $-\frac{5}{2}$. Then $(6x+5)^n$ has $5$-adic valuation at least $3$ (that is, its numerator is a multiple of $5^3$ when it is written in lowest terms). In contrast, the $5$-adic valuation of $R(x)$ is $0$ (because the $5$-adic valuation of $R(0)$ is $0$) and $(x-10)$ is either $-\frac{5}{6}-10=-\frac{65}{6}$ or $-\frac{5}{2}-10=-\frac{25}{2}$, neither of which has $5$-adic valuation greater than $2$. So $(6x+5)^n$ and $(x-10)R(x)$ have unequal $5$-adic valuations, which means that $f_n(x)=(6x+5)^n+(x-10)R(x)$ cannot possibly be $0$.
It only remains to check that $f_2$ has no rational roots, which is an easy calculation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Value of n for which f(n) = $\,n^2 + 9n + 30\,$ is a perfect square. I attempted this by setting $f(n) = \,m^2.\,$
So $\,n^2 + 9n + 30 = m ^2\,$.
Then $\,9(n + 10/3) = (m + n)(m - n)\,$.
So $m = 10/3$ and $n = -17/3$ which is incorrect.
| Hint:
Let $n^2+9n+30=(n+a)^2$ where $a$ is any integer
$\iff n=\dfrac{a^2-30}{9-2a}$ which has to be an integer
Now if integer $d(>0)$ divides both $a^2-30,9-2a;$
$d$ must divide $2(a^2-30)+a(9-2a)=9a-60$
$d$ must divide $-9(9-2a)-2(9a-60)=39$
So, a necessary condition for $(9-2a)|(a^2-30)$ is $9-2a$ must divide $39$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213155",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Determine $s_{10}$ for $\sum_{n=1}^{\infty}\frac{1}{n^2}$ Consider the convergent series $$\sum_{n=1}^{\infty}\frac{1}{n^2}$$
To determine $s_{10}$ which is the sum of the first ten terms, the easiest way of course is to add them up.
However is there another way to figure out $s_{10}$?
| By creative telescoping
$$\begin{eqnarray*} \sum_{n\geq m}\frac{1}{n^2}&=&\sum_{n\geq m}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\sum_{n\geq m}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\\&+&\frac{1}{6}\sum_{n\geq m}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)-\frac{1}{6}\sum_{n\geq m}\frac{1}{n^3(n+1)^3}\tag{1}\end{eqnarray*} $$
hence by recalling that $\zeta(2)=\frac{\pi^2}{6}$ and plugging in $m=11$ in $(1)$ we get:
$$ H_{10}^{(2)} = \frac{\pi^2}{6}-\frac{1}{11}-\frac{1}{2\cdot 11^2}-\frac{1}{6\cdot 11^3}+\frac{1}{6}\sum_{n\geq 11}\frac{1}{n^3(n+1)^3}\tag{2} $$
hence $\frac{\pi^2}{6}-\frac{1}{11}-\frac{1}{2\cdot 11^2}-\frac{1}{6\cdot 11^3}$ is an approximation of $H_{10}^{(2)}$ with an error $\leq 10^{-6}$.
By Wolstenholme's theorem we know that $11$ is a divisor of the numerator of $H_{10}^{(2)}$ and the denominator of $H_{10}^{(2)}$ is clearly a divisor of $2^6\cdot 3^4\cdot 5^2\cdot 7^2$. These facts allow to turn the previous approximation into an exact evaluation:
$$ H_{10}^{(2)}=\frac{1968329}{1270080}\tag{3}$$
but I wonder why a reasonable person should follow this approach, instead of just adding ten terms of the form $\frac{1}{n^2}$.
| {
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Proving that $2^{744}-1$ is a multiple of $2^{248}+2^{124}+1$ and $2^{93}+2^{47}+1$ we have to prove that $2^{744}-1$ is multiple of $2^{248}+2^{124}+1$ and $2^{93}+2^{47}+1$
I could prove first part as follows:
$$2^{744}-1=\left(2^{372}\right)^2 -1=(2^{372}-1) (2^{372}+1)$$
But $$2^{372}-1= \left(2^{124}\right)^3 -1$$ Using $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ we get another factor as $2^{248}+2^{124}+1$
But can i have any hint for second part
| As I wrote in my comment, it is known that$$4a^4+1=4a^4+4a^2+1-4a^2=(2a^2+1)^2-(2a)^2=(2a^2+2a+1)(2a^2-2a+1)$$
You can try substituting $a$ with some number to show that $A=4a^4+1$ is multiple of $2^{93}+2^{47}+1$, and it won't be hard to prove that $2^{744}-1$ is multiple of $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215292",
"timestamp": "2023-03-29T00:00:00",
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evaluation of parametric integral I do not know how to compute integral $\int_0^1\frac{x^a-x^b}{\ln\;x}\cos(\ln\;x)\;\mbox{d}x$ where $a,b>0$. I only know that I should use $\frac{x^a-x^b}{\ln\;x}=\int_b^a x^y \;\mbox{d}y$ and Fubini's formula.
Thanks for any help.
| Note that
$$\cos(\log(x))+i\sin(\log(x)) = \,e^{i\log(x)} = x^i$$
Think about the extension such that $a,b \in \mathbb{C}$
$$\int^1_0 \frac{x^a-x^b}{\log(x)}\,dx = \log \left( \frac{b+1}{a+1}\right)$$
Let $a=c+i$ and $b = d+i$
$$\int^1_0 \frac{(x^c-x^d)(\cos(\log(x))+i\sin(\log(x)))}{\log(x)}\,dx = \log \left( \frac{c+i+1}{d+i+1}\right)$$
By computing the real part
\begin{align}\int^1_0 \frac{(x^c-x^d)\cos(\log(x))}{\log(x)}\,dx &= \log \left|\frac{c+i+1}{d+i+1}\right|\\
&=\frac{1}{2}\log\left[ \frac{(cd+1)^2}{(d^2+1)^2}+\frac{(d-c)^2}{(d^2+1)^2}\right]\end{align}
Note we used that $\Re \log|z| = \log\sqrt{x^2+y^2}$, hence we reach to
\begin{align}\int^1_0
\frac{(x^{c-1}-x^{d-1})\cos(\log(x))}{\log(x)}\,dx
&=\frac{1}{2}\log\left[\frac{c^2+1}{d^2+1}\right]\end{align}
Another approach
Consider
$$f(a) = \int^1_0 \frac{x^{c-1}-x^{d-1}}{\log (x)} \cos(a\log(x))\,dx$$
$$f'(a) = -\int^1_0 (x^{c-1}-x^{d-1}) \sin(a\log(x))\,dx$$
Let $t = -\log(x)$
$$f'(a) = \int^\infty_0 (e^{-ct}-e^{-dt}) \sin(at)\,dt = \frac{a}{a^2+c^2}-\frac{a}{d^2+a^2}$$
By integration
$$f(a)= \frac{1}{2}\log \left(\frac{a^2+c^2}{a^2+d^2} \right)$$
Third approach
$$\int^1_0 \frac{(x^a-x^b)\cos(\log(x)}{\log(x)}dx = \int^1_0\int_b^a x^y \cos(\log(x))\,dy dx = \int^a_b \int^1_0x^y \cos(\log(x))\, dx dy $$
Note that
$$ \int_b^a\int^1_0x^y \cos(\log(x))\, dx \,dy= \int_b^a\frac{(y+1)}{(y+1)^2+1}\,dy = \frac{1}{2}\log \left[ \frac{(a+1)^2+1}{(b+1)^2+1}\right]$$
| {
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"source": "stackexchange",
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Find the number of zeroes in ${999\cdots99}^m$
Find the number of $\text{zeroes}$ in $\mathrm{999 \cdots 99}^{\mathrm{m}}$ for some given $m \in \mathbb{Z^+}$ provided that there are $\mathrm{n~many~9's}$.
I observed that when $m=3$, the zeroes increase by $1$ starting from $1$.
For $m=4$, starting from $2$, it increases by $2$ and for $m=5$, it increases by $2$ again but for $m=6$, it increases by $3$.
I can only prove why ${(\underbrace{\mathrm{999 \cdots 99}}_{\text{n many}})}^3$ has $\text{n-1 zeroes}$. I am not stating the proof as it's done after expanding $(10^n-1)^3$.
How can it he generalized? (The question itself is the generalized version)
| Here are a few data points ($1 \le m,n \le 15$). It doesn't look very regular for low $n$.
\begin{array}{}
m \setminus n & 1 & 2 & 3 & 4 \\
1 & 0 \\
2 & n-1 \\
3 & n-1 \\
4 & 2(n-1) \\
5 & 2n-1 \\
6 & 0 & 3(n-2)+2 \\
7 & 0 & 3(n-2)+2 \\
8 & 1 & 4(n-2)+1 \\
9 & 1 & 1 & 4(n-2) \\
10 & 1 & 7 & 5(n-2) \\
11 & 2 & 1 & 5(n-3)+4 \\
12 & 0 & 6(n-2)+2 \\
13 & 0 & 1 & 2 & 6(n-3) \\
14 & 0 & 3 & 15 & 7(n-3)+15 \\
15 & 2 & 2 & 6 & 7(n-4)+9
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2217670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the maximum value $\frac{11a+4b}{4a^2-ab+2b^2}+\frac{11b+4c}{4b^2-bc+2c^2}+\frac{11c+4a}{4c^2-ca+2a^2}$ For the positive real numbers $a,b,c$ satisfy $ab+bc+ca=3abc$. Find the maximum value $$P=\frac{11a+4b}{4a^2-ab+2b^2}+\frac{11b+4c}{4b^2-bc+2c^2}+\frac{11c+4a}{4c^2-ca+2a^2}$$
i tried all methods such as: AM-GM; C-S,... but unsuccess. Help me give a hint
| If $a=b=c=1$ we get the value $9$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{11a+4b}{4a^2-ab+2b^2}\leq\frac{3(ab+ac+bc)}{abc}$$ or
$$\sum_{cyc}\left(\frac{3}{a}-\frac{11a+4b}{4a^2-ab+2b^2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)(a-6b)}{a(4a^2-ab+2b^2)}\geq0$$ or
$$\sum_{cyc}\left(\frac{(a-b)(a-6b)}{a(4a^2-ab+2b^2)}+\frac{1}{b}-\frac{1}{a}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2(a+b)}{ab(4a^2-ab+2b^2)}\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Working with functions
Suppose $f(3-x)=2x^2-5x+4$ and $f(x)=ax^2+bx+c$. What is $a+b+c$?
I don't know how to approach this. I thought of maybe doing $a(3-x)^2+b(3-x)+c=2x^2-5x+4$ and solving for $a+b+c$ but it got messy.
| We know that $f(x)=ax^2+bx+c$ and so $f(1)=a+b+c$. Also, $f(3-x)=2x^2-5x+4$ and hence we get
$$a+b+c=f(1)=f(3−2)=2(2^2)−5(2)+4=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit $\frac{e^\frac{-x^2}{2}-\cos(x)}{x^3\sin(x)}$ as $x\to 0$ We have to find the limit:
$$\lim_{x\to 0}\dfrac{e^\frac{-x^2}{2}-\cos(x)}{x^3\sin(x)}$$
I was stuck, but was able to find the limit using series expansion as $\dfrac{1}{4}$.
How can we calculate the limit with standard limits like
$$\lim_{x\to 0}\dfrac{e^x-1}{x}=1\\\lim_{x\to 0}\dfrac{\sin(x)}{x}=1$$
etc.
Also I didn't try L'hospital as that would be too complicated.
| We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{e^{-x^{2}/2} - \cos x}{x^{3}\sin x}\notag\\
&= \lim_{x \to 0}\frac{e^{-x^{2}/2} - 1 + 1 - \cos x}{x^{4}}\cdot\frac{x}{\sin x}\notag\\
&= \lim_{x \to 0}\frac{e^{-x^{2}/2} - 1 + 2\sin^{2}(x/2)}{x^{4}}\notag\\
&= \lim_{t \to 0}\frac{e^{-2t^{2}} - 1 + 2\sin^{2}t}{16t^{4}}\text{ (putting }x = 2t)\notag\\
&= \frac{1}{16}\lim_{t \to 0}\frac{e^{-2t^{2}} - 1 + 2t^{2} + 2\sin^{2}t - 2t^{2}}{t^{4}}\notag\\
&= \frac{1}{16}\lim_{t \to 0}\frac{e^{-2t^{2}} - 1 + 2t^{2}}{t^{4}} + \frac{1}{8}\lim_{t \to 0}\frac{\sin^{2}t - t^{2}}{t^{4}}\notag\\
&= \frac{1}{4}\lim_{u \to 0}\frac{e^{u} - 1 - u}{u^{2}} + \frac{1}{8}\lim_{t \to 0}\frac{\sin t + t}{t}\cdot\frac{\sin t - t}{t^{3}}\text{ (putting }u = -2t^{2})\notag\\
&= \frac{1}{4}\cdot\frac{1}{2} - \frac{1}{8}\cdot 2\cdot\frac{1}{6}\notag\\
&= \frac{1}{12}\notag
\end{align}
The limits $$\lim_{u \to 0}\frac{e^{u} - 1 - u}{u^{2}} = \frac{1}{2},\lim_{t \to 0}\frac{\sin t - t}{t^{3}} = -\frac{1}{6}$$ are easily evaluated via Taylor's series or L'Hospital's Rule.
| {
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"url": "https://math.stackexchange.com/questions/2221910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all functions $f:\mathbb R \rightarrow \mathbb R$, such that: $(x^2 − y^2)\cdot f(xy) = x\cdot f(x^2y) − y\cdot f(xy^2)$ Find all functions $f:\mathbb R \rightarrow \mathbb R$, such that:
$$(x^2 − y^2)\cdot f(xy) = x\cdot f(x^2y) − y\cdot f(xy^2)$$
for all $x,y \in \mathbb R$
My work so far:
1) $f(0)=0$
2) $y=1; y=\frac1x; y=-\frac1x; y=kx$
| Since my first answer contained a mistake so I am posting this revised version which only gives a partial result to the question.
Claim: Assume $f\colon\mathbb R\to\mathbb R$ satisfies $(x^2-y^2)f(xy) = xf(x^2y)-yf(xy^2)$ for all real $x,y$, and is continuous in $1$. Then $f(x) = f(1)x$ for all $x$; i.e. $f$ is linear.
Proof: Plugging in $y=-x$ yields
$$ 0 = xf(-x^3) +xf(x^3) \implies f(x) = -f(-x) \qquad\text{for } x\neq 0 $$
Moreover since $f(0)=0$, as seen by plugging in $(x,y)=(-1,0)$, we conclude that $f$ is an odd function. Now if we plug in $y=1$ instead we find that for all $x\neq 0$:
$$ (x^2 -1)f(x) = xf(x^2) -f(x) \implies x^2 f(x) = xf(x^2) \implies
\boxed{\frac{f(x)}{x} = \frac{f(x^2)}{x^2}}\qquad (I) $$
For $x>0$ we can subsitute $x\to\sqrt{x}$ to find $\frac{f(x)}{x} = \frac{f(\sqrt{x})}{\sqrt{x}} = \frac{f(\sqrt{\sqrt{x}})}{\sqrt{\sqrt{x}}} = \ldots = \frac{f(\sqrt[2^k]{x})}{\sqrt[2^k]{x}}$ for all $k\ge0$ by repeatedly applying the equation to itself. Then we can use the well known fact that $\lim\limits_{n\to\infty}\sqrt[n]{x} =1$ for all $x>0$, which together with the continuity of $f$ in $1$ implies that for $x>0$:
$$ f(1) =\lim_{k\to\infty}\frac{f(\sqrt[2^k]{x})}{\sqrt[2^k]{x}} = \lim_{k\to\infty} \frac{f(x)}{x} = \frac{f(x)}{x} $$
Thus by symmetry we have $f(x)=f(1)x$ for all $x$.
Remark: I wasn't able to deduce continuity of $f$ in $1$, but one possible approach might be to set $y = \frac{1}{x}$ and use $xf(x) = f(x^2)$, a variation of $(I)$, to deduce:
\begin{gather}
(x^2 - \frac{1}{x^2})f(1) = xf(x) - \frac{1}{x}f(\frac{1}{x}) = f(x^2) - f(\frac{1}{x^2})
\\
\implies \boxed{f(1) = \frac{f(x) - f(\frac{1}{x})}{x-\frac{1}{x}}}
\end{gather}
Which holds not only for positive $x$, but by symmetry for all $x\neq 0$, and then somehow use that $x\approx 1 \iff \frac{1}{x} \approx 1$ to show continuity. Alternatively one might try to find a pathological counter-example that is not continuous, similarly as is the case with Cauchy's functional equation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is it possible to simplify the product $\prod_{i=1}^n (9i^2 - 1)$? Is is possible to reduce this product to something that only relates (max.) on the calculation of "$\Gamma(x)$ constants (e.g. $\Gamma(\frac{1}{3})$)" and (single) factorials.
I've started with Mathematicas calculated identity:
$$\prod_{i=1}^n (9i^2 - 1) = \frac{\Gamma(\frac{2}{3} + n) \Gamma(\frac{4}{3} + n)}{2 π} * 3^{\frac{3}{2} + 2 n}$$
I can simplify $\Gamma(\frac{4}{3} + n) = \Gamma(\frac{1}{3} + (n+1))$ to $$ = \Gamma(\frac{1}{3}) * \frac{(3(n+1) - 2)!!!}{3^{n+1}} = \Gamma(\frac{1}{3}) * \frac{(3n + 1)!!!}{3^{n+1}}$$
Is it now possible to reduce $\Gamma(\frac{1}{3}) * \frac{(3n + 1)!!!}{3^{n+1}}$ and $\Gamma(\frac{2}{3} + n)$ to a 'representation' that is only with n-independent Gamma function-values and single factorials (not third)?
Thanks in advance! :)
| Unless I'm being dense, we have
\begin{align}
\prod_{i=1}^n (9i^2-1) &= \prod_{i=1}^n [(3i-1)(3i+1)] = [2\cdot 4]\cdot[5\cdot 7] \cdots [(3n-1)(3n+1)] \\
&=\frac{(3n+1)!}{3\cdot 6 \cdots 3n} = \frac{(3n+1)!}{3^n n!}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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