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Find all integer values of $m$ such that the equation $\sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x}$ has exactly four distinct real roots.
Find all integer values of $m$ such that the equation $$\large \sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x}$$ has exactly four distinct real roots.
$\left(x \in [0, 9], m \in \left[0, \dfrac{27}{4}\right]\right)$
Let $9 - x = y \ (\iff x + y = 9)$, we have that $$\sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x} \implies \sqrt{x} + \sqrt{y} = \sqrt{3m + xy}$$
$$\implies x + y + 2\sqrt{xy} = 3m + xy \iff (x - 1)y - 2\sqrt{xy} + (3m - x) = 0$$
For the equation $$\sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x}$$ to have exactly four distinct real roots, the equation $$(x - 1)y - 2\sqrt{xy} + (3m - x) = 0$$ must have two distinct real roots $x$.
(note that $x_0$ and $9 - x_0$ are both solutions to $\sqrt{9 - x} = \sqrt{3m - x^2 + 9x} - \sqrt{x}$)
$$\implies \Delta' = x - (x - 1)(3m - x) > 0$$, in which case, the solutions are $\begin{cases} m > \dfrac{y^2}{3(y - 1)} &\text{if $x \in [0, 1)$}\\ m \in \mathbb R &\text{if $x = 1$}\\ m < \dfrac{y^2}{3(y - 1)} &\text{if $x \in (1, 9]$} \end{cases}$, (according to WolframAlpha, of course), which lead to suspect that $m \in \{0, 1\}$ are the integer solutions. But I don't really know.
|
We have $$\sqrt{x}+\sqrt{9-x}=\sqrt{3m-x^2+9x},$$ which is for $0\leq x\leq9$ and $3m-x^2+9x\geq0$ is equivalent to
$$9+2\sqrt{x(9-x)}-x(9-x)=3m$$ or
$$10-\left(1-\sqrt{x(9-x)}\right)^2=3m,$$ which gives $3m\leq10$ and $m\leq3.$
Also, since $$\sqrt{x}+\sqrt{9-x}=\sqrt{9+2\sqrt{x(9-x)}}\geq3,$$ we obtain:
$$3m-x^2+9x\geq9.$$
Thus,
$$3m\geq x^2-9x+9=(x-4.5)^2-11.25\geq-11.25,$$ which gives $$-3\leq m\leq3.$$
Now, consider $f(x)=10-\left(1-\sqrt{x(9-x)}\right)^2.$
We see that $f$ has two maximum points for $\sqrt{x(9-x)}=1$ and the minimum point for $x=4.5$.
Thus, our equation has four distinct roots for $$f(0)\leq3m<f\left(x_\max\right)$$ or
$$9\leq3m<10,$$ which gives $m=3.$
|
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|
Show that $n= 5 + 5^2+ 5^3+...5^{150}$ is divisible by $930$. Show that $n= 5 + 5^2+ 5^3+...5^{150}$ is divisible by $930$.
I'm thinking to show that $n$ is divisible by each of the prime factors of $930$, is that right? I'm stuck
|
$N=5 + .... + 5^{150} = 5(1+...... + 5^{149} = 5\cdot \frac {5^{150}-1}{4}$.
$930| N \iff $
$4*930| 4*N = 5(5^{150}-1) \iff $
$\frac {4*930}5=\frac{4*93*10}5=4*93*2|5^{150}-1\iff$
$8*3*31|5^{150}-1\iff$
$8|5^{150}-1$ and $3|5^{150}-1$ and $31|5^{150}-1\iff$
$5^{150}\equiv 1\pmod {8,3,31}$.
And we can's Euler's Th|Fermat's Little Theorem to prove those.
$\phi(8=2^3) = 2^2 = 4$ so $5^4\equiv 1\pmod 8$. But actually $5^2 =25 \equiv 1 \pmod 8$ so $5^{150}\equiv (5^2)^{75}\equiv 1\pmod 8$.
ANd $5^{2}\equiv 1\pmod 3$ so $5^{150}\equiv (5^2)^{75}\equiv 1 \pmod 3$.
$5^{30}\equiv 1 \pmod {31}$ so $5^{150}\equiv (5^{30})^5\equiv 1 \pmod{31}$.
That's it.
|
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|
Show that $\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} = 1.$
Question: Show that
$$\sum_{n=0}^\infty \frac{1}{n+1} \binom{2n}{n} \frac{1}{2^{2n+1}} = 1.$$
From Wolfram alpha, it seems that the equality above is indeed correct.
But I do not know how to prove it.
Any hint is appreciated.
|
The Generalized Binomial Theorem says
$$
\begin{align}
(1-x)^{-1/2}
&=1+\frac12\frac{x}{1!}+\frac12\!\cdot\!\frac32\frac{x^2}{2!}+\frac12\!\cdot\!\frac32\!\cdot\!\frac52\frac{x^2}{3!}+\cdots\\[6pt]
&=\sum_{k=0}^\infty(2k-1)!!\frac{x^k}{2^kk!}\\
&=\sum_{k=0}^\infty\frac{(2k!)}{2^kk!}\frac{x^k}{2^kk!}\\
&=\sum_{k=0}^\infty\binom{2k}{k}\left(\frac x4\right)^k\tag1
\end{align}
$$
Substituting $x\mapsto x/4$ gives
$$
\frac1{\sqrt{1-4x}}=\sum_{k=0}^\infty\binom{2k}{k}x^k\tag2
$$
Integrating gives
$$
\frac12-\frac12\sqrt{1-4x}=\sum_{k=0}^\infty\frac1{k+1}\binom{2k}{k}x^{k+1}\tag3
$$
Plug in $x=\frac14$ and multiply by $2$
$$
1=\sum_{k=0}^\infty\frac1{k+1}\binom{2k}{k}\frac1{2^{2k+1}}\tag4
$$
|
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|
What are the equations of the 3D transformation matrices? Please have a look at the image below. There is a transformation matrix and graphical representation of 2 coordinate systems. I understand what the numbers in the matrix represent, but I can't figure out how we could write this in terms of equations.
I've been watching MIT open course videos on matrices, and from what I've learned the numbers in the matrices are just coefficients of some equations. For example, if we had a system of equations like this:
$x_1=3 y_1+2 y_2$
$x_2=2 y_1+4 y_2$
Then we would have a matrix like this:
\begin{pmatrix}3& 2\\
2& 4\end{pmatrix}
So I'm trying to reverse engineer this matrix into an equation but can't figure it out.
|
The equation for this matrix comes from transformation rule:
$$
\begin{pmatrix}x_1\\y_1\\z_1\\1\end{pmatrix} =
\begin{pmatrix}R_{11}&R_{12}&R_{13}&t_1\\
R_{21}&R_{22}&R_{23}&t_2\\
R_{31}&R_{32}&R_{33}&t_3\\
0&0&0&1\end{pmatrix}
\begin{pmatrix}x\\y\\z\\1\end{pmatrix} $$
$$
x_1=R_{11}x+R_{12}y+R_{13}z+t_1,\\
y_1=R_{21}x+R_{22}y+R_{23}z+t_2,\\
z_1=R_{31}x+R_{32}y+R_{33}z+t_3,
$$
Transformation matrix is usually presented as follows:
$$
A_1 = \begin{pmatrix}R_{11}&R_{12}&R_{13}&t_1\\
R_{21}&R_{22}&R_{23}&t_2\\
R_{31}&R_{32}&R_{33}&t_3\\
0&0&0&1\end{pmatrix} = \begin{pmatrix}
\mathbf R&\mathbf t \\
\mathbf 0& 1
\end{pmatrix},
$$
where $\mathbf R$ is $3\times3$ ($n\times n$ in general) rotation matrix and $\mathbf t$ is a $3\times1$ translation vector.
If we expand the transformation rule
$$
\begin{pmatrix}\mathbf r_1\\1\end{pmatrix} =
\begin{pmatrix}x_1\\y_1\\z_1\\1\end{pmatrix} =
\begin{pmatrix}R_{11}&R_{12}&R_{13}&t_1\\
R_{21}&R_{22}&R_{23}&t_2\\
R_{31}&R_{32}&R_{33}&t_3\\
0&0&0&1\end{pmatrix}
\begin{pmatrix}x\\y\\z\\1\end{pmatrix} =
\begin{pmatrix}
\mathbf R&\mathbf t \\
\mathbf 0& 1
\end{pmatrix}\begin{pmatrix}\mathbf r\\1\end{pmatrix}
$$
we get an alternative way to present an affine transformation is like:
$$
\mathbf r_1=\mathbf R\mathbf r + \mathbf t
$$
So now we need to first deal with rotation (we forget about translation as if $\mathbf t=0$ or coordinate origins coincide). We see that if we take the vector $\mathbf i=(1,0,0)^\intercal$, then rotated vector $\mathbf i_1=\mathbf R\mathbf r=(R_{11}, R_{21}, R_{31})^\intercal$. From the picture we know, that $\mathbf i_1$ is looking down, so $\mathbf i_1=(0,0,-1)^\intercal$ and that help us to find the first column in matrix $\mathbf R$. We repeat the procedure for vectors $\mathbf j$ and $\mathbf k$ and find the row of matrix $\mathbf R$.
To find a translation vector is even simpler. The vector $\mathbf r=\mathbf 0$ will go to $\mathbf r_1=\mathbf t$, so $\mathbf t$ is just the position of new origin in old coordinate system. From the picture we see that $\mathbf t=(1,2,2)^\intercal$.
By arranging everything to the transformation matrix, we get the answer.
|
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|
At least one perfect square in $[S_n,S_{n+1}]$ Given $\{a_n\}$ a sequence of real numbers such that $a_1 > 1$ and $a_{n+1}-a_n\geq 2$, with $S_n=a_1+a_2+...+a_n$, prove that for any $n\geq 3$, there is at least one perfect square in the range $[S_n, S_{n+1}]$.
I was thinking that $a_{n+1}\ge 2+a_n \ge 4+a_{n-1}\geq ... \ge 2n+a_1>2n+1$.
And this implies $S_{n+1}-S_n=a_{n+1}>2n+1=(n+1)^2-n^2$. So at least one of $n^2$ or $(n+1)^2$ si between $S_{n}$ and $S_{n+1}$.
Is this reasoning correct? I don't think it is. How to prove the problem then?
|
Let $(a_n)$ be a sequence of real numbers with $a_1 \geq 0$ and $a_{n+1} - a_n \geq 2$. Define $S_n = a_1 + \cdots + a_n$. We will show that $[S_n, S_{n+1}]$ contains a perfect square for all $n \geq 1$.
Lemma. Let $0 \leq x < y$ be real numbers with $y -x \geq 2\sqrt x + 1$. Then $[x, y]$ contains a perfect square.
Proof. Let $n^2 \geq 0$ be the largest square with $n^2 \leq x$. Then
$$x < (n+1)^2 = n^2 + 2n + 1 \leq x + 2 \sqrt x + 1 \leq y \,. \qquad\square$$
So suffices to show that $a_{n+1} \geq 2 \sqrt{S_n} + 1$. I.e. that $4 S_n \leq (a_{n+1} - 1)^2$. We prove this by induction.
For $n = 1$ this follows from
$$(a_2-1)^2 \geq (a_1+1)^2 \geq (2\sqrt{a_1})^2 = 4 a_1 \,.$$
Take $n \geq 1$, and suppose
$$4 S_n \leq (a_{n+1} - 1)^2 \,.$$
Add $4a_{n+1}$ to both sides to obtain
$$4S_{n+1} \leq (a_{n+1} + 1)^2 \,.$$
Then use that $a_{n+1} +1 \leq a_{n+2} - 1$ to get
$$4S_{n+1} \leq (a_{n+2} - 1)^2 \,.$$
The induction is complete.
|
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|
find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$. Let $A$be a $5 \times 5$ matrix whose characteristic polynomial is given by $(\lambda -2)^3(\lambda+2)^2.$ If $A$ is diagonalizable,find $\alpha,\beta$ such that $A^{-1}=\alpha A+\beta I$
Solution:
Characteristic polynomial of $A$,$Ch_{A}(\lambda)=(\lambda -2)^3(\lambda+2)^2$.
By Cayley-Hamilton Theorem, we have $Ch_{A}(A)=0\implies (A -2)^3(A+2)^2=0\implies$ Either $(A -2)^3=0$ or $(A+2)^2=0$
Considering
$$(A+2)^2=0\implies A^2+4A+4I=0$$
$$\implies I=\left(-\frac{1}{4}\right)A^2+(-A)$$
$$\implies A^{-1}=\left(-\frac{1}{4}\right)A+(-I).$$
On comparing just recent equation with $A^{-1}=\alpha A+\beta I$,we get $\alpha =-\frac{1}{4}$ and $\beta =-1$
But the correct answer is $\alpha=\frac{1}{4}$ and $\beta =0$..
I wanted to know where I'm wrong?
|
Your error comes from the fact that you cannot say that if $M_1M_2=0$ that $M_1=0$ or $M_2=0$...
For example :
$$\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&0\\0&1\end{pmatrix}=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$
|
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|
AM GM inequality with constants I was trying to find the minimum value of $a + \frac{1}{a} + 3$ given that $a$ is a positive real number. I found $2$ ways to do it.
Minimum value of $a + \frac{1}{a}$ is $2$ using the AM-GM inequality and therefore the minimum value of $a + \frac{1}{a} + 3$ is $2 + 3 = 5$
Then, I tried using AM-GM inequality directly for the given expression and I got
$a + \frac{1}{a} + 3 \ge 3\cdot(3)^{1/3}$
Can anyone please tell me what is the error using the second method?
And why is this true?
$a + 1 \ge 2(a)^{1/2}$
|
(1) You correctly used the AM-GM inequality when you showed that
$$a+\frac{1}{a}\ge2,$$
and of course, this implies that
$$a+\frac{1}{a}+3\ge5.$$
Also, you correctly used the AM-GM inequality when you showed that
$$a+\frac{1}{a}+3\ge3\cdot3^{\frac{1}{3}}.$$
Note that
$$5>3\cdot3^{\frac{1}{3}}\approx4.32,$$
so your first use of the AM-GM inequality yielded a stronger result than your second use did.
You can also note that your first result
$$a+\frac{1}{a}+3\ge5,$$
is the best possible, since $a+\frac{1}{a}+3=5$ if $a=1$.
(2) You can prove that for any $a\ge0$ that $a+1\ge2\sqrt{a}$ by using the AM-GM inequality. Note that for $a\ge0$
$$a+1\ge2\sqrt{a}\quad\text{iff}\quad\frac{a+1}{2}\ge\sqrt{a}$$
and the inequality on the right follows from using AM-GM inequality on $1$ and $a$.
A second way to prove that $a+1\ge2\sqrt{a}$ for any $a\ge0$, is as follows. Note that the following are equivalent:
$$a+1\ge2\sqrt{a}$$
$$a-2\sqrt{a}+1\ge0$$
$$\left(\sqrt{a}-1\right)^2\ge0$$
And the last inequality holds, since for any $x\in\Bbb{R}$, $x^2\ge0$.
|
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|
minimum value of $f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$
If $f:\mathbb{R}\rightarrow \mathbb{R}.$ Then minimum value of $$\displaystyle f(x)=\frac{(x^2-x+1)^3}{x^6-x^3+1}$$
what i try
If $x\neq 0,$ Then divide numerator and denominator by $x^3$
$$f(x)=\frac{\bigg(x+\frac{1}{x}-1\bigg)^3}{x^3+\frac{1}{x^3}-1}$$
put $\displaystyle x+\frac{1}{x}=t,$ then $\displaystyle x^3+\frac{1}{x^3}=t^3-3t$
$$f(t)=\frac{(t-1)^3}{t^3-3t}$$
How do i solve it without derivatives
Help me please
|
A solution without derivatives
Clearly, $t \ge 2$ or $t \le -2$.
Denote the minimum of $f(t)=\frac{(t-1)^3}{t^3-3t-1}$ on $|t|\ge 2$ by $m > 0$.
Since $f(3) = \frac{8}{17}$, we have $m \le \frac{8}{17}$.
Note that $f(t) - 1 = \frac{3t(2-t)}{t^3-3t-1} \ge 0$ for $t \le -2$.
Thus, the minimum occurs on $t \ge 2$.
$f(t) = m$ leads to a cubic equation $g(t) = 0$ where
\begin{align}
g(t) &= (t^3-3t-1)(f(t)-m)\\
&= (1-m)t^3 - 3t^2 + (3+3m)t + m - 1.
\end{align}
Clearly, $g(t) = 0$ has a real root on $t \ge 2$.
Since $g(0) = m - 1 < 0$ and $g(2) = 1 - m > 0$, we know that $g(t) = 0$ has a real root on $(0, 2)$.
Thus, we can factor $g(t) = (1-m)(t-t_1)(t-t_2)(t-t_3)$ for some $t_1 \in (0, 2)$ and $t_2 \ge 2$.
Since $g(t) \ge 0$ for $t \ge 2$, it is easy to prove that $t_2 = t_3$.
Thus, we have $g(t) = (1-m)(t-t_1)(t-t_2)^2$.
Since $g(t) = 0$ has a multiple root, its discriminant is zero,
i.e. $\Delta = 81(m^2+6m-3)m^2 = 0$ which results in $m = 2\sqrt{3} - 3$.
Finally, let us prove that $m = 2\sqrt{3} - 3$ is indeed the minimum of $f(t)$ on $|t|\ge 2$.
It is easy to check that
$$f(t) - (2\sqrt{3} - 3) = \frac{1}{t^3 - 3t - 1}(4-2\sqrt{3})\Big(t - 1 + \tfrac{\sqrt{3}}{2}\Big)\Big(t - 1 - \sqrt{3}\Big)^2$$
and hence $f(t) - (2\sqrt{3} - 3) \ge 0$ for $|t| \ge 2$, with equality if and only if $t = 1 + \sqrt{3}$.
We are done.
|
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|
Solving an integral: $\int \frac{x}{x^3-1}\,\mathrm dx$ I'm trying to solve this integral:
$$\int \frac{x}{x^3-1}\,\mathrm dx$$
What I did was:
$$\int \frac{x}{(x-1)(x^2+x+1)}\,\mathrm dx.$$
$$\frac{x}{(x-1)(x^2+x+1)} = \frac{a}{x-1}+ \frac{bx+c}{x^2+x+1}$$
Then I got this in the numerator:
$$ax^2+ax+a+bx^2-bx+cx-c $$
$$a+b=0;a-b+c=1; a-c=0 $$
$$a=c=\frac{1}3 \qquad b=-\frac{1}3$$
Then I wrote:
$$\frac{1}3\int \frac{1}{x-1}\,\mathrm dx-\frac{1}3\int\frac{x-1}{x^2+x+1} \, \mathrm dx$$
so the first one is just $\frac{1}{3}\ln|x-1|$. Which makes my calculations already wrong, most likely.
With the second one I tried a few different things ( involving u-substitution mostly) and got stuck.
I know I'm supposed to get this:
$$\frac{1}6\ln \frac{(x-1)^3}{x^2+x+1}+\frac{1}{\sqrt{3}} \arctan\frac{2x+1}{\sqrt{3}}$$
What have I already done wrong? What am I supposed to do?
|
The second integral can be carried as follows,
$$\int\frac{1-x}{x^2+x+1}dx=\int\frac{-(x+\frac12)+\frac32}{(x+\frac12)^2+\frac34}dx$$
$$=-\frac12\int\frac{d[(x+\frac12)^2]}{(x+\frac12)^2+\frac34}
+\frac32\int\frac{dx}{(x+\frac12)^2+\frac34}$$
$$=-\frac12\ln\left[(x+\frac12)^2+\frac34\right]+\frac{1}{\sqrt{3}}\arctan\frac{2x+1}{\sqrt{3}}$$
|
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|
Evaluate $\lim_{x\to0^+} \frac{(1-x)^{x}-\cos x e^{-\frac{x^2}{2}}}{\arctan(x^2)\sin x +x^3 \ln x}$ I'm trying to evaluate
$$\lim_{x\to0^+} \frac{(1-x)^{x}-\cos x e^{-\frac{x^2}{2}}}{\arctan(x^2)\sin x +x^3 \ln x}$$
I've found that it is
$$\lim_{x\to0^+}\frac{-\frac{x^3}{2}+\text{o}(x^3)}{x^3+x^3\ln x +\text{o}(x^3)}$$
My doubt is: which is the dominant term at the denominator? If I write it like this
$$\lim_{x\to0^+}\frac{-\frac{x^3}{2}+\text{o}(x^3)}{x^3\left(1+\ln x +\frac{\text{o}(x^3)}{x^3}\right)}$$
Or like this
$$\lim_{x\to0^+}\frac{-\frac{x^3}{2}+\text{o}(x^3)}{x^3\ln x\left(1+\frac{1}{\ln x} +\frac{\text{o}(x^3)}{x^3\ln x}\right)}$$
I get $0$, which is right; but I'm not sure if they are both correct, because I don't fully understand if the $\ln x$ is irrelevant or not.
Thanks to you all.
|
What you do is right! Follow your method:
$$\lim_{x\to0^+}\frac{-\frac{x^3}{2}+\text{o}(x^3)}{x^3\left(1+\ln x +\frac{\text{o}(x^3)}{x^3}\right)}=\lim_{x\to0^+}\frac{-\frac{1}{2}+\text{o}(1)}{1+\ln x +\text{o}(1)}=0.$$
Actually,
$$(1-x)^{x}-\cos x e^{-\frac{x^2}{2}}\sim\frac{-1}{2}x^3.$$
So you also can evaluate this limit as follows:
$$\lim_{x\to0^+} \frac{(1-x)^{x}-\cos x e^{-\frac{x^2}{2}}}{\arctan(x^2)\sin x +x^3 \ln x}=\lim_{x\to0^+} \frac{\frac{-1}{2}x^3}{\arctan(x^2)\sin x +x^3 \ln x}$$
$$=\lim_{x\to0^+} \frac{\frac{-1}{2}}{\frac{\arctan(x^2)\sin x}{x^3}+\ln x}=0.$$
Because $$\lim_{x\to0^+}\frac{\arctan(x^2)\sin x}{x^3}=1$$
and $$\lim_{x\to0^+}\ln x=-\infty.$$
|
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|
Which one is larger? I want to prove that for positive integer $n$,
$$\left(1-\frac{1}{n}\right)^n \left(1+\frac{1}{n}\right)^n \geq 1-\frac{1}{n}$$
but I am stuck on how to proceed. Can someone help me?
|
Note that, for $n \gt 1$,
$$\frac{1}{1 - \frac{1}{n}} = \frac{n}{n - 1} = 1 + \frac{1}{n - 1} \tag{1}\label{eq1A}$$
Using \eqref{eq1A}, for $n \gt 1$, multiplying both sides of your proposed inequality by $\left(\frac{1}{1 - \frac{1}{n}}\right)^n$ gives
$$\left(1-\frac{1}{n}\right)^n \left(1+\frac{1}{n}\right)^n \geq 1-\frac{1}{n} \iff \left(1+\frac{1}{n}\right)^n \ge \left(1+\frac{1}{n-1}\right)^{n-1} \tag{2}\label{eq2A}$$
For $x \gt 1$, consider
$$f(x) = \left(1 + \frac{1}{x}\right)^x \implies \log(f(x)) = x\log\left(1 + \frac{1}{x}\right) \tag{3}\label{eq3A}$$
You could just use that $f(x)$ is an increasing function or, if you wish to prove it, you could take the derivative of both sides to get
$$\begin{equation}\begin{aligned}
\frac{f'(x)}{f(x)} & = \log\left(1 + \frac{1}{x}\right) + \frac{x}{1 + \frac{1}{x}}\left(-\frac{1}{x^2}\right) \\
& = \log\left(1 + \frac{1}{x}\right) - \frac{\frac{1}{x}}{1 + \frac{1}{x}}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
The Inequalities section of Wikipedia's "List of logarithmic identities" article gives that, for $-1 \lt y$, you have
$$\frac{y}{1+y} \le \ln(1 + y) \tag{5}\label{eq5A}$$
Using $y = \frac{1}{x}$ shows the RHS of \eqref{eq4A} is non-negative. Since $f(x) \gt 0$, this means $f'(x) \ge 0$, i.e., $f(x)$ is an increasing function for $x \gt 1$. This shows that \eqref{eq2A} is also true for all $n \gt 1$. Of course, the inequality is also true trivially for $n = 1$, showing it's always true for positive integers.
|
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|
Let $x∈R$ and suppose that$ |x−2| < 1/3$. Prove that $|x^2+x−6| < 2$. I need help studying for a midterm coming up and am doing this problem for practice:
Let $x∈R$ and suppose that $|x−2| < 1/3$. Prove that $|x^2+x−6| < 2$.
If $|x-2| < 1/3, x < 7/3$ and $x > 5/3$.
$|x^2 + x - 6| = |x-2||x+3| < 2$.
I'm really not sure where to go from here. What I tried was the following:
$(1/3)|x+3| < 2$, so $|x+3|<6$
Since the maximum of $x$ is $7/3$, the most $|x+3|$ can be is $|(7/3)+3|= 16/3 < 6$.
Then, the minimum of $x$ is $5/3$, so the least $|x+3|$ can be is $|(5/3)+3| = 14/3 < 6$.
Thus, $|x-2||x+3| < 2$.
Anyone know if this is correct? If not, can you point me in the right direction? Thanks in advance.
|
By the given $$\frac{5}{3}<x<\frac{7}{3}$$ and we need to prove that:
$$4<x^2+x<8,$$ which is true because $x^2+x$ increases for $x>0$:
$$x^2+x<\left(\frac{7}{3}\right)^2+\frac{7}{3}=\frac{70}{9}<8$$ and $$x^2+x>\left(\frac{5}{3}\right)^2+\frac{5}{3}=\frac{40}{9}>4.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $6 | (a+b+c)$ if and only if $6 | (a^3 + b^3 +c^3).$ I have tried the question but not sure if my solution is correct or not...
My try..
\begin{align}a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b)(b+c)(c+a)\end{align}
So , if
\begin{align}6 |(a^3 + b^3 + c^3)\end{align}
Then,
\begin{align}6 | [(a+b+c)^3 - 3(a+b)(b+c)(c+a)]\end{align}
So, also
\begin{align}6 | (a+b+c)^3\end{align}
\begin{align}6 | (a+b+c)(a+b+c)(a+b+c)\end{align}
So,
\begin{align}6|(a+b+c)\end{align}
Is my solution correct ?
|
$a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(a+c)$
At least one pair from $(a+b) , (b+c) , (a+c)$ is even number.
Therefore, $6 | 3(a+b)(b+c)(a+c)$. Consequently,
$a^3+b^3+c^3 \equiv (a+b+c)^3 \mod{6}$
$a^3+b^3+c^3 \equiv a+b+c \mod{6}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Verify that $4r^3+r=\left(r+\frac{1}{2}\right)^4-\left(r-\frac{1}{2}\right)^4$ Must I verify that Left hand side(LHS)= Right hand side(RHS)or can I prove that RHS= LHS?
I don’t know how to prove from LHS=RHS. How to separate the $4r^3+r$ into two terms, i.e. $\displaystyle\Bigl(r+\frac{1}{2}\Bigr)^4-\Bigl(r-\frac{1}{2}\Bigr)^4$
After verifying, I will need to find $$\sum_{r=1}^n (4r^3+r)$$
This is a summation of finite series question.
|
$(x+y)^{4}-(x-y)^{3}=8x^{3}y+8xy^3$
Just substitute $x=r$ and $y=\frac{1}{2}$ to obtain
$4r^{3}+r=(r+\frac{1}{2})^{4}-(r-\frac{1}{2})^{4}$
As for the infinite sum,
$\sum_{r=1}^{n}({4r^{3}+r})=-\left( \frac{1}{2} \right)^4+\left( \frac{3}{2} \right)^4 - \left( \frac{3}{2} \right)^4+...+\left( n + \frac{1}{2} \right)^4$
$\sum_{r=1}^{n}(4r^{3}+r)=\left( n+\frac{1}{2} \right)^4-\left( \frac{1}{2} \right)^4$
|
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"url": "https://math.stackexchange.com/questions/3555937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
sum of binomial series with alternate terms
Evaluation of series $\displaystyle \sum^{n}_{k=0}(-4)^k\binom{n+k}{2k}$
what I tried:
from Binomial Identity
$$\binom{n+k}{2k}=\binom{n+k-1}{2k}+\binom{n+k-1}{2k-1}$$
series is $$\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k}+\sum^{n}_{k=0}(-4)^k\binom{n+k-1}{2k-1}$$
let $\displaystyle S_{1}=\sum^{n}_{k=0}\binom{n+k-1}{2k}=\binom{n-1}{0}-4\binom{n}{2}+4^2\binom{n+1}{4}+\cdots +(-4)^n\binom{2n-1}{2n}$
let $\displaystyle S_{2}=\sum^{n}_{k=0}\binom{n+k-1}{2k}=\binom{n-1}{-1}-4\binom{n}{1}+4^2\binom{n+1}{3}+\cdots +(-4)^n\binom{2n-1}{2n-1}$
Help me please
did not know to simplify $S_{1}$ and $S_{2}$
|
We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$. This way we can write for instance
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{k=0}^n}\color{blue}{(-4)^k\binom{n+k}{2k}}&=\sum_{k=0}^n(-4)^k\binom{n+k}{n-k}\\
&=\sum_{k=0}^n(-4)^k[z^{n-k}](1+z)^{n+k}\tag{2}\\
&=[z^n](1+z)^n\sum_{k=0}^n(-4z(1+z))^k\tag{3}\\
&=[z^n](1+z)^n\frac{1-(-4z(1+z))^{n+1}}{1+4z(1+z)}\tag{4}\\
&=[z^n](1+z)^n\frac{1}{1+4z(1+z)}\tag{5}\\
&=[z^n]\frac{(1+z)^n}{(1+2z)^2}\\
&=[z^n](1+z)^n\sum_{j=0}^\infty\binom{-2}{j}(2z)^j\tag{6}\\
&=[z^n](1+z)^n\sum_{j=0}^\infty(j+1)(-2z)^j\tag{7}\\
&=\sum_{j=0}^n(j+1)(-2)^j[z^{n-j}](1+z)^n\\
&=\sum_{j=0}^n(j+1)(-2)^j\binom{n}{n-j}\tag{8}\\
&=\sum_{j=1}^n\binom{n}{j}j(-2)^j+\sum_{j=0}^n\binom{n}{j}(-2)^j\\
&=n\sum_{j=1}^n\binom{n-1}{j-1}(-2)^j+(1+(-2))^n\tag{9}\\
&=n\sum_{j=0}^{n-1}\binom{n-1}{j}(-2)^{j+1}+(-1)^n\tag{10}\\
&=(-2n)(1+(-2))^{n-1}+(-1)^n\\
&\,\,\color{blue}{=(-1)^n(2n+1)}
\end{align*}
Comment:
*
*In (2) we apply the coefficient of operator according to (1).
*In (3) we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (4) we apply the finite geometric series formula.
*In (5) we skip the term which does not contribute to $[z^n]$.
*In (6) we apply the binomial series expansion.
*In (7) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
*In (8) we select the coefficient of $z^{n-j}$.
*In (9) we apply the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.
*In (10) we shift the index to start with $j=0$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\lim\limits_{x\rightarrow 0^+}\frac{1}{\ln x}\sum\limits_{n=1}^{\infty}\frac{x}{(1+x)^n-(1-x)^n}$
Find $$\lim\limits_{x\rightarrow 0^+}\dfrac{1}{\ln x}\sum_{n=1}^{\infty}\dfrac{x}{(1+x)^n-(1-x)^n}.$$
Consider $$f(x):=(1+x)^n,$$
By Lagrange MVT, we can obtain
$$\frac{2x}{f(x)-f(-x)}=\frac{1}{f'(\xi)}, -x\gtrless \xi\gtrless x$$
Thus $$\frac{x}{(1+x)^n-(1-x)^n}=\frac{1}{2f'(\xi)}=\frac{1}{2n(1+\xi)^{n-1}}$$
Can we go on from here?
|
The limit is equal to $-\dfrac{1}{2}$.
A LOWER bound: for $0<x<1$, $\log(x)<0$ and
$$\begin{align}
\frac{1}{\ln x}\sum_{n=1}^{\infty}\frac{x}{(1+x)^n-(1-x)^n}
&\geq
\frac{1}{2\ln x}\sum_{n=1}^{\infty}\frac{1}{n+\binom{n}{3}x^2}\\
&\geq
\frac{1}{2\ln x}\int_{1/2}^{\infty}\frac{ds}{s+\frac{s^3x^2}{6}}\\
&=\frac{\frac{1}{2}\ln(x^2+24)-\ln x}{2\ln x}\to -\frac{1}{2}
\end{align}$$
As regards the UPPER bound, for $0<x<1$, we have that $0<\frac{1}{1+x}<1$ and, from your work,
$$-\frac{1}{2}\leftarrow \frac{(1+x)\ln\left(\frac{1}{1-\frac{1}{1+x}}\right)}{2\ln x}=\frac{1}{2\ln x}\sum_{n=1}^{\infty}\frac{1}{n{{(1+x)}^{n-1}}}\geq\frac{1}{\ln x}\sum_{n=1}^{\infty}\frac{x}{{{(1+x)}^{n}}-{{(1-x)}^{n}}}.$$
|
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|
Find minimum of $a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$ If $a,b$ are real numbers, find the minimum value of:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}$$
This is what I did: I tried some values and I set $a=0$. Then, it becomes a quadratic of $b$:
$$b^2-2b$$
Here, the minimum is $-1$. So, I tried to prove that:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b}\ge -1$$
Using Wolfram, I found this is a square:
$$a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b} + 1 = \frac{(a^2+b^2+ab-a-b)^2}{(a+b)^2} $$
so it is positive.
My question is, can we prove this with more traditional and natural solution, maybe with Cauchy-Schwarz?
|
Let $x=a+b$ and $y=\frac{ab}{a+b}$.
\begin{align}
a^2+b^2+\frac{a^2b^2}{(a+b)^2}-\frac{2(a^2+ab+b^2)}{a+b} &=(a+b)^2-2ab+\left(\frac{ab}{(a+b)}\right)^2-\frac{2((a+b)^2-ab)}{a+b}
\\&= x^2-2xy+y^2-2(x-y)=\underset{\text{quadratic in } x-y}{\underbrace{(x-y)(x-y-2)}}\\
&\geq -1.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3562535",
"timestamp": "2023-03-29T00:00:00",
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|
How to solve a rotated ellipse equation for y? Given an x, I want to get the value of y on the Cartesian plane.
It's easy to solve it for a non-rotated ellipse
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \tag{1}$$
$$y = \frac{b}{a} \sqrt{(a^2 - x^2)} \tag{2}$$
How do I solve it when the ellipse is rotated and the equation becomes
$$\frac {(x\cos\theta+y\sin\theta)^2}{a^2}+\frac {(x\sin\theta-y\cos\theta)^2}{b^2}=1\tag{3}$$
No matter what I tried I couldn't get y onto one side. Is there a general equation for this? I'm not able to derive it myself.
Also what happens to (3) when the center is not at the origin? Is it just all the x values become $(x - x_{center})$ and ys become $(y - y_{center})$?
|
Your third equation can be written as
\begin{eqnarray*}
\color{blue}{y^2} \left( \frac{ \sin^2 \theta}{a^2}+ \frac{ \cos^2 \theta}{b^2} \right) + 2x\color{blue}{y} \sin \theta \cos \theta \left( \frac{ 1}{a^2}- \frac{ 1}{b^2} \right)+x^2 \left( \frac{ \cos^2 \theta}{a^2}+ \frac{ \sin^2 \theta}{b^2} \right)-1=0.
\end{eqnarray*}
This is a quadratic in $y$ which is easily solved using the well known formula.
|
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|
Limit of $\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$ I want to solve a limit without l'Hospital, just with algebraic manipulation:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}$$
I started with:
$$\lim_{x\to 3} \frac{3-\sqrt{6+x}}{6\sin \frac{\pi x}{18}-x}=\lim_{x\to 3} \frac{(3-\sqrt{6+x})(3+\sqrt{x+6})}{(6\sin \frac{\pi x}{18}-x)(3+\sqrt{x+6})}=\lim_{x\to 3} \frac{3-x}{(6\sin \frac{\pi x}{18}-x)(3+\sqrt{x+6})}$$
$3+\sqrt{x+6} \to 6$ is determinate, so I only need to calculate:
$$\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}$$
but I could only do it with l'Hospital:
$$\lim_{x\to 3} \frac{3-x}{6\sin \frac{\pi x}{18}-x}=\lim_{x\to 3} \frac{-1}{6\cdot \frac{\pi}{18}\cos \frac{\pi x}{18}-1}=\frac{-1}{\frac{\pi\sqrt{3}}{6}-1}$$
Can I get some help without l'Hospital?
|
Calculate the limit of inverse:
$g(x)=\frac{6\sin\left(\frac{\pi x}{18}\right)-x}{3-x}=\frac{6\left(\sin\left(\frac{\pi x}{18}\right)-\frac12\right)+3-x}{3-x}$
$=\frac{6\left(\sin\left(\frac{\pi x}{18}\right)-\frac12\right)}{3-x}+1$
However
$\sin\left(\frac{\pi x}{18}\right)-\sin\left(\frac{\pi}{6}\right)=2\sin\left(\frac{\pi(x-3)}{36}\right)\cos\left(\frac{\pi(x+3)}{36}\right)$
So
$g(x)=\frac{12\sin\left(\frac{\pi(x-3)}{36}\right)\cos\left(\frac{\pi(x+3)}{36}\right)}{3-x}+1$
Using the limit $\lim_{u\to 0}\frac{\sin(u)}{u}=1$
$\lim_{x\to 3} \frac{\sin\left(\frac{\pi(x-3)}{36}\right)}{x-3}=\frac{\pi}{36}$
So $\lim_{x\to 3} g(x)=12.\left(-\frac{\pi}{36}\right)\cos\left(\frac{\pi}{6}\right)=1-\frac{\pi}{2\sqrt{3}}$
The result is the inverse.
|
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"timestamp": "2023-03-29T00:00:00",
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|
A closed form for $\sum_{k=0}^n \frac{ (-1)^k {n \choose k}^2}{k+1}$ Mathematica gives $$\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{k+1}= ~_2F_1[-n,-n;2;-1],$$ where $~_2F_1$ that is Gauss hypergeometric function. Here the question is: Can one find a simpler closed form for this summation. Recently, the absolute summation for this has been discussed at MSE:
A binomial summation: $\sum_{k=0}^{n} \frac{{n \choose k}^2}{k+1}$
|
We seek to evaluate
$$\sum_{k=0}^n \frac{(-1)^k}{k+1} {n\choose k}^2.$$
This is
$$\frac{1}{n+1} \sum_{k=0}^n (-1)^{k} {n\choose k} {n+1\choose k+1}
= \frac{1}{n+1} \sum_{k=0}^n (-1)^{k} {n\choose k} {n+1\choose n-k}
\\ = [z^n] (1+z)^{n+1} \frac{1}{n+1}
\sum_{k=0}^n (-1)^{k} {n\choose k} z^k
\\ = [z^n] (1+z)^{n+1} \frac{1}{n+1} (1-z)^n
= \frac{1}{n+1} [z^n] (1+z) (1-z^2)^n.$$
Now if $n=2m$ we get
$$\frac{1}{n+1} [z^{2m}] (1+z) (1-z^2)^n
= \frac{1}{n+1} [z^{2m}] (1-z^2)^n
\\ = \frac{1}{n+1} [z^{m}] (1-z)^n
= \frac{1}{n+1} (-1)^m {n\choose m}.$$
On the other hand when $n=2m+1$ we find
$$\frac{1}{n+1} [z^{2m+1}] (1+z) (1-z^2)^n
= \frac{1}{n+1} [z^{2m+1}]z (1-z^2)^n
\\ = \frac{1}{n+1} [z^{2m}] (1-z^2)^n
= \frac{1}{n+1} [z^{m}] (1-z)^n
= \frac{1}{n+1} (-1)^m {n\choose m}.$$
We thus have even or odd the closed form
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{n+1} (-1)^{\lfloor n/2\rfloor}
{n\choose \lfloor n/2\rfloor}.}$$
The second case could have been done by inspection given the first.
This result matches the comment by @SangchulLee.
|
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|
Showing that $1+2020^y$ ($y\in \mathbb{Z}$) is not a perfect square for even $y$ it is obvious that $1+2020^y$ is not a perfect square since $2020^y$ is a perfect square $+1$.+
for uneven $y$, i wanted to show that $z^2-1=(z-1)(z+1)=2020^y$. How can I show that I can't distribute the factors of $2020^y$ into two integers $n$ and $n+2$?
|
To answer the question you asked: "How can I show that I can't distribute the factors of $2020^y$ into two integers n and n+2?"
$2020^y = 101^y*2^{2y}*5^y$.
If $n= z-1, n+2 = z+1$ then $\gcd(z+1, z-1) = \gcd((z+1)-(z-1),(z-1))=\gcd(2,z-1)=\begin{cases}1\\2\end{cases}$
As $z-1, z+1$ are either both even or both odd, and $2020^y$ is even they are both even and $\gcd(z-1, z+1) =2$ and so $2$ divides one of $n$ or $n+2$ and $2^{2y-1}$ divides the other.
And $5^y$ divides one of $n$ or $n+2$ and $101^y$ also divides one or the other.
And yet the difference between $n$ and $n+2$ must be $2$ which is, in the grand scale of things very close. $101^y$ is a very large factor that it pales all the others combined. Its reasonable to assume the distribution of $2, 2^{2y-1}, 5^y,101^y$ that has the smallest difference will be $2*101^y$ vs $2^{2y-1}*5^y$ and the difference is much more than $2$.
In other words: $2*101^y - 2^{2y-1}*5^y = 2*101^y - 5*20^{y-1}=2*101^y-\frac{2*20^y}{8}>2(101^y-20^y)=2(101-20)(101^{y-1}+....+20^{y-1})\ge 2(81)> 2$.
(Assuming, of course, $y > 0$. But if $z =0$ then $1+2020^0 = 2$ and if $z < 0$ than $1+2020^{z}$ is not an integer.)
ANd if $A-B > 2$ then $A*k - \frac Bk > A-B > 2$ for any $k> 1$. And all difference between these distributions of factors will be $2^{m}5^j*101^y - 2^a*5^b=k(2*101^y) - \frac{2^{2y-1}*5^y}k $ where $m= 1$ or $2y-1$ and $j=0$ or $y$ and $a=2y-1$ or $1$ and $b=y$ or $0$.
......
But, all in all, the accepted answer is a much better approach for the problem.
|
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|
What's the sum of elements of this multi-set? Let S be a $\mathbf{multi}$ set of integers such that:
$$S = \left \{\min(\mathbf{C}, a + b - 1) | a \in \left [ \mathbf{A} \right ], b \in \left [ \mathbf{B} \right ] \right \}$$
*
*Note that $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ are positive integers and we know $\mathbf{A} \leq \mathbf{B} \leq \mathbf{C}$
*Also $\left [ x \right ]$ stands for $\left \{ 1, 2, 3, \cdots , x \right \}$
Can you figure out sum of all S elements using $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ as parameters?
|
If $C \ge A+B-1$, then the min against $C$ will have no effect and the sum just reduces to $$\sum_{a=1}^A \sum_{b=1}^B (a + b - 1) = BA(A+1)/2 + AB(B+1)/2 - AB = AB(A+B)/2.$$
Let’s assume now that $A \le B \le C < A+B-1$. Then some of the values $a+b-1$ get replaced by the smaller $C$, so we just need to add up the deficits and subtract this from the previous total $AB(A+B)/2$.
It’s not hard to see that there is one value with maximum deficit $A+B-C-1$, two values with the next highest deficit $A+B-C-2$, etc., down to the minimum possible deficit of $1$ which occurs $A+B-C-1$ times (note that the assumption $A,B\le C$ is essential here, otherwise our count would be too high). The total deficit is thus simply $f(A+B-C)$, where
$$f(n) =\sum_{k=1}^n k(n-k) = (n^3-n)/6.$$
So the sum of the multi set $S$ is:
$$\tfrac12 AB(A+B) - \tfrac16 ((A+B-C)^3 - (A+B-C)),$$
unless $A+B-C < 0$, in which case you can discard the second term.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\frac{dy}{dx}$ implicitly given the equation $5xy+\ln(xy^2)=2$ Find $\frac{dy}{dx}$ implicitly given the equation $5xy+\ln(xy^2)=2$
edit: the homework might be $5xy+\ln(x^2y)=2$, but this is close enough to get the point
$Solution:$
First lets use properties of $\ln$ to write our equation in a way that will be easier to take the derivative.
$5xy+\ln(xy^2)=2$
$\rightarrow 5xy+\ln(x)+\ln(y^2)=2$
$\rightarrow 5xy+\ln(x)+2\ln(y)=2$
Okay, now lets take our implicit derivative, making sure to use the product rule on $5xy$:
$(5xy+\ln(x)+2\ln(y))'=2'=0$
$\rightarrow 5(xy)'+\ln(x)'+2\ln(y)'=0$
$\rightarrow 5(x'y+xy')+\frac{1}{x}+2\frac{1}{y}y'=0$
recall that $x' = \frac{d}{dx}x=1$, while $y' = \frac{d}{dx}y=\frac{dy}{dx}$
$\rightarrow 5((1)y+xy')+\frac{1}{x}+2\frac{1}{y}y'=0$
$\rightarrow 5y+5xy'+\frac{1}{x}+2\frac{1}{y}y'=0$
multiplying both sides of the equation by $x$ yields:
$\rightarrow 5xy+5x^2y'+ 1 + 2\frac{x}{y}y'=0(x)=0$
now we multiply both sides of the equation by $y$:
$\rightarrow 5xy^2+5x^2yy'+y+2xy'=0(y)=0$
$\rightarrow 5x^2yy'+2xy'=-5xy^2-y$
$\rightarrow (5x^2y+2x)y'=-5xy^2-y$
$\rightarrow y'=\frac{-5xy^2-y}{5x^2y+2x}$
|
$$5xy+ \ln(xy^2)=2$$
Differentiating w.r.t $x$
$$5y+5xy'+\frac{y^2+2xyy'}{xy^2}=0$$
$$y'=-\frac{(5xy+1)y}{x(2+5xy)}=\frac{dy}{dx}$$
|
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|
Solving System of Symmetric Equations I am to solve the system below.
$$\begin{cases} (x^2+1)(y^2+1)=10 \\ (x+y)(xy-1)=3 \end{cases}$$
I have tried to factor the equations of the system but they do not factor. How can I approach the problem?
|
We have $$x^2+y^2+x^2y^2=9$$ or
$$(x+y)^2-2xy+x^2y^2=9$$ and since $$x+y=\frac{3}{xy-1},$$ we obtain:
$$\frac{9}{(xy-1)^2}-2xy+x^2y^2=9$$ or
$$\frac{9(2-xy)xy}{(xy-1)^2}-xy(2-xy)=0$$ or
$$xy(2-xy)(4-xy)(2+xy)=0.$$
Now, $xy=0$ gives $(-3,0)$ and $(0,-3)$;
$xy=2$ gives $x+y=3$ and $(2,1)$ and $(1,2)$;
$xy=4$ does not give real solutions
and $xy=-2$ gives $x+y=-1$ and $(-2,1)$ and $(1,-2).$
|
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|
Arccos telescopic sum: $\sum_{k=1}^{n} \arccos \frac{2k^2}{4k^4+1}$ Compute the finite sum:
$\arccos \frac{2 \cdot 1^2}{4\cdot 1^4+1}+...+\arccos \frac{2 \cdot n^2}{4\cdot n^4+1}$
Please help. I have no idea what to do. There should be a telescopation but I can’t find it.
|
This is not a complete answer, just a try. By partial fraction decomposition, you get
$$z_k:= \frac{2k^2}{4k^4+1} = \frac{k}{2(2k^2-2k+1)} - \frac{k}{2(2k^2+2k+1)} $$
Then you could use the fact that
$$\arccos(z) = \frac \pi 2 - \arcsin(z) $$
and the identity
$$\arcsin\left(x\sqrt{1-y^2}-y\sqrt{1-x^2}\right)=\arcsin(x) - \arcsin(y), $$
which would lead to the system
$$\begin{split}
\frac{k}{2(2k^2-2k+1)} = x_k\sqrt{1-y_k^2} + c_k, \\
\frac{k}{2(2k^2+2k+1)} = y_k\sqrt{1-x_k^2} + c_k,
\end{split}$$
However, the solution $(x_k,y_k)$ even for $c_k =0$ is horrendous and does not lead to a telescoping sum, by which I mean
$$\arcsin(y_{k-1}) = \arcsin(x_k). $$
I doubt that you could find a $c_k$ such that the sum telescopes. If you managed to find such a $c_k$, then
$$\sum_{k=1}^n \arccos(z_k) = \frac{\pi n }{2} - \sum_{k=1}^n \arcsin(z_k) = \frac{\pi n}{2} - \left[\arcsin\left(\frac 2 5\right) + \arcsin\left(\frac{2n^2}{4n^4+1} \right) \right]. $$
EDIT. (Here I follow @bjorn93's suggestion in the comments, for which I am thankful.) If, on the other hand, you had
$$z_k = \frac{2k^2}{\sqrt{4k^2+1}}, $$
then, by using the identity
$$\tan(\arccos(z)) = \frac{\sqrt{1-z^2}}{z} $$
you get
$$\begin{split}
\arccos(z_k) &= \arctan( \tan \arccos(z_k)) = \arctan\Bigg(\frac{\sqrt{1-z_k^2}}{z_k} \Bigg)\\ &= \arctan \left( \frac{\sqrt{4k^4 +1}}{2k^2} \sqrt{1-\frac{4k^4}{4k^4+1}}\right) \\
&= \arctan\left( \frac{1}{2k^2} \right)
\end{split}$$
You can check that, by the arctangent identity
$$\arctan(x) - \arctan(y) = \arctan \left( \frac{x-y}{1+xy} \right),$$
we can split, for $k \geq 1$,
$$\arctan\left( \frac 1 {2k^2}\right) = \arctan\left( \frac{k}{k+1}\right) - \arctan\left( \frac{k-1}k\right). $$
(To find this decomposition, you may solve the recurrence relation $1+x_kx_{k-1} = 2k^2 (x_k - x_{k-1})$ with the initial condition $x_0 = 0$.) Hence the sum telescopes:
$$\sum_{k=1}^n \arccos\left( \frac{2k^2}{\sqrt{4k^4+1}} \right) = \sum_{k=1}^n \arctan\left( \frac{1}{2k^2} \right) = \arctan\left( \frac{n}{n+1} \right). $$
|
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|
Show convergence of series $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ I want to prove that $\sum_{k=1}^{\infty}\frac{k!}{k^k} $ converges. My idea is that, for any integer $k \ge 1 $, we have
\begin{align*}
\frac{k!}{k^k}
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\cdot\frac{k}{k} \\
&= \frac{1}{k}\cdot\frac{2}{k}\cdots\frac{k-2}{k}\cdot\frac{k-1}{k}\\
&\leq \frac{1}{k}\cdot\frac{2}{k}\cdots\cdot\frac{k-2}{k} \\
&\qquad \vdots \\
&\leq \frac{1}{k}\cdot\frac{2}{k}
= \frac{2}{k^2}
\end{align*}
That is
$$\sum_{k=1}^{\infty}\frac{k!}{k^k} \leq \sum_{k=1}^{\infty}\frac{2}{k^2} $$
And since right-hand side of the inequality is finite, so is left-hand side and therefore the series is convergent.
However, I dont find this way of solving the assignment elegant and I believe there is a cleaner way. Appreciates all help I can get.
|
You can also use the root test together with the inequality between geometric and arithmetic mean (GM-AM):
$$\sqrt[k]{\frac{k!}{k^k}}= \frac{\sqrt[k]{k!}}{k}\stackrel{GM-AM}{\leq}\frac{\frac{k(k+1)}{2k}}{k}=\frac{k+1}{2k}\stackrel{k\to\infty}{\longrightarrow}\frac 12 <1$$
|
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|
Easy way to determine the sign of Gaussian curvature without explicit computation. I know how to compute Gaussian curvature via the first and second fundamental form. i.e.,
\begin{align}
K = \frac{eg-f^2}{EG-F^2}
\end{align}
In this computation one usually set $n = \frac{X_u\times X_v}{||X_u\times X_v||}$
Without detail computation through the formula above, is there an easy way to determine the sign of Gaussian curvature?
For example, with some computation, I know the Gaussian curvature of this type of surface $ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}=1$ is $K = \frac{1}{a^2 b^2 c^2 (\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4} )^2}$ and for this type of surface $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} =1$, $ K(p) = -\frac{1}{a^2 b^2 c^2 (\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4} )^2}$.
To figure out the sign it took me lots of time. I want to know the following : Is there an easy way to figure out whether the curvature is positive or negative rather than computing explicitly?
|
I suppose you are considering surfaces (ie 2-dimensional) embedded into $\mathbb{R}^3$. You can try to formalize the intuition that montains and valleys have positive curvature and saddle points have negative curvature.
Compute the tangent plane of the surface at a point. If in a small neighbourhood of the point, the surface lies entirely on one side of the tangent plane, the Gaussian curvature is positive at that point. If the surface intersects the tangent plane, the Gaussian curvature is negative.
|
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|
Find all positive integer pairs $(a, b)$ such that $(ab + a + b) \mid (a^2 + b^2 + 1)$.
Find all positive integer pairs $(a, b)$ such that $$(ab + a + b) \mid (a^2 + b^2 + 1)$$
Let $a^2 + b^2 + 1 = k(ab + a + b), k \in \mathbb N, k \ge 1$.
For $k = 1$, we have that $$a^2 + b^2 - ab - a - b + 1 = 0$$, where $(a, b) = (1, 1)$ satisfies the above condition.
For $k = 2$, we have that $$a^2 + b^2 - 2ab - 2a - 2b + 1 = 0$$, where $(a, b) = (n^2, (n + 1)^2)$ where $n \in \mathbb N$ satisfies the condition, since $$(a - b)^2 - 2(a + b) + 1 = [(n + 1)^2 - n^2]^2 - 2[(n + 1)^2 + n^2] + 1$$
$$ = (2n + 1)^2 - 2(2n^2 + 2n + 1) + 1 = (4n^2 + 4n + 1) - (4n^2 + 4n + 2) + 1 = 0$$
I don't know what to do next for the case of $k \ge 3$. Any help would be greatly appreciated.
|
Reference is HURWITZ 1907
Caution: this is about POSITIVE $x,y$ values. For $k=5,$ there is a solution at $-5,-3.$
The main thing is that the jumping step leads to inequalities for what Hurwitz called ground solutions. We have
$$ x^2 - kxy + y^2 -kx-ky +1=0 $$
To jump, the coefficient of $x$ is $-ky-k.$ The two $x$ solutions with $y$ fixed are $x$ and $x' = ky+k - x.$ A Grundlosung (it is an o umlaut) occurs when $x' \geq x$ and $y' \geq y.$ In turn, this says
$$ 2x \leq k(y+1) \; \; , \; \; \; \; 2y \leq k(x+1) $$
In the diagrams below, this is the region of the hyperbola in the first quadrant between the two lines. As you can see, there are actually no integer points on the arc between the two lines. For example, intersecting the hyperbola with $y = \frac{k}{2}(x+1)$ leads to
$$ x = \frac{6 k^2 \pm \sqrt{32} \sqrt{k^4 + k^2 - 2}}{4 - k^2} \; \; \; . \; \; $$
As soon as $k \geq 3,$ we find $$ \sqrt{k^4 + k^2 - 2} < k^2 + \frac{1}{2} $$
so that the numerator is positive for either $\pm,$ so that the final $x$ values are both negative. In turn, this means that there are NO integer points along the arc between the Hurwitz lines. There are, in fact, no fundamental solutions, therefore no (positive) integer solutions when $k \geq 3.$
|
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|
A geometric inequality for acute triangle $\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt3$ I am trying to prove $\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt3$ for an acute triangle. There is:
$$\sum \sqrt \frac{b^2+c^2-a^2}{a^2+2bc} \leq \sqrt{3\sum\frac{b^2+c^2-a^2}{a^2+2bc}}$$
So we have to prove: $\sum\frac{b^2+c^2-a^2}{a^2+2bc}\leq 1 $
We observe that $\frac{b^2+c^2-a^2}{a^2+2bc}=\frac{(b+c)^2}{a^2+2bc}-1$ and I can't go further
|
Let $x=\sqrt{b^2+c^2-a^2}, y=\sqrt{c^2+a^2-b^2}$ and $z=\sqrt{a^2+b^2-c^2}$ (they are positive real numbers due to acute-angled restriction). Then:
$$a=\sqrt{\frac{y^2+z^2}{2}},\ b=\sqrt{\frac{z^2+x^2}{2}},\ c=\sqrt{\frac{x^2+y^2}{2}}$$
and the inequality is equivalent with:
$$\sum_{cyc}\frac{x}{\sqrt{\frac{y^2+z^2}{2}+\sqrt{(x^2+y^2)(z^2+x^2)}}}\leq \sqrt{3}$$
Notice that using Cauchy-Schwarz, we have $(x^2+y^2)(x^2+z^2)\geq (x^2+yz)^2$. Thus, it will be enough to prove:
$$\sum_{cyc}\frac{x}{\sqrt{2x^2+(y+z)^2}}\leq \sqrt{\frac{3}{2}}$$
This inequality is homogeneous so normalize with $x+y+z=3$:
$$\sum_{cyc}\frac{x}{\sqrt{2x^2+(3-x)^2}} \leq \sqrt{\frac{3}{2}}$$
However, $2x^2+(3-x)^2=6+3(x-1)^2\geq 6$, and hence:
$$\sum_{cyc}\frac{x}{\sqrt{2x^2+(3-x)^2}} \leq \frac{x+y+z}{\sqrt{6}}=\sqrt{\frac{3}{2}}$$
|
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|
Compute $ \min_{a,b,c \in \mathbb{R}} \int_{-1}^1 |e^x-a-bx-cx^2|^2dx $ The question is the following:
Compute
$$ \min_{a,b,c \in \mathbb{R}} \int_{-1}^1 |e^x-a-bx-cx^2|^2dx $$
I tried to expand the integral, and it gives me:
$$
\int_{-1}^1 |e^x-a-bx-cx^2|^2dx= 2a^2+\frac{2}{3}(2ac+b^2)+\frac{2}{e}(a-2b+5c)-2e(a+c)+\frac{2}{5}c^2+\frac{1}{2}(e^2-e^{-2})
$$
Here I have no idea how to proceed.
I also consider the following: we can rewrite
$$ \min_{a,b,c \in \mathbb{R}} \int_{-1}^1 |e^x-f(x)|^2dx $$
where
$$
f(x) = a+bx+cx^2
$$
So the question is basically to find a degree 2 polynomial to approximate $e^x$ as close as possible. But this idea seems to be off the subject I learned. Thanks for any help and hint.
|
Hint.
You can compute the derivative of $f(a,b,c):=\int_{-1}^{1} | e^x - a - bx - cx^2 |^2 dx$ with respect to $a$, $b$ and $c$ and set them to zero to find the values of $a,b$ and $c$ that minimise that expression.
Values for comparison
We have $\frac{df(a,b,c)}{da} = 4 a + \frac{4c}{3} - 2( e - e^{-1})$ and $\frac{df(a,b,c)}{db} = \frac{4 b}{3} - 4 e^{-1}$. The latter gives us $b = 3 e^{-1}$. Finally, $\frac{df(a,b,c)}{dc} = \frac{4 a}{3} + \frac{4 c}{5} - 2 e + 10e^{-1}$. According to WolframAlpha we have $a = \frac{3 (11 - e^2)}{4 e}$ and $c = \frac{15 (e^2 - 7)}{4 e}$. This yields $f(a,b,c) = 36 - \frac{259}{2 e^2} - \frac{5 e^2}{2} \approx 0.00144$.
Edit Notice that we have $a \approx 0.996$, $b \approx 1.104$ and $c \approx 0.537$, so $a + bx + c x^2 \approx 1 + x + \frac{1}{2} x^2$, which are the first three terms of the Taylor series for $e^x$.
|
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|
Range of $f(x) = \frac{\sin^2 x + \sin x - 1}{\sin^2 x - \sin x + 2} $ Here's what I did :
$$\text{Let }\quad y = \frac{\sin^2 x + \sin x - 1}{\sin^2 x - \sin x + 2} $$
$$\text{Let } \sin^2x=t$$
$$\Rightarrow\ (y-1)t^2-(y+1)t+2y+1=0 $$
$$\text{Since } t= \sin x\text{ is real,} $$
$$\text{Discriminant} \geqslant 0 $$
$$ \Rightarrow (y+1)^2-4(y-1)(2y+1) \geqslant 0$$
$$ \Rightarrow\ -7y^2+6y+5 \geqslant 0$$
$$\Rightarrow\ 7y^2-6y-5 \leqslant 0$$
$$\Rightarrow \ y\in\left[\frac{3-2\sqrt{11}}{7},\frac{3+2\sqrt{11}}{7}\right] $$
But the correct answer (according to wolfram) is
$$y \in \left[\frac{3-2\sqrt{11}}{7},\frac{1}{2}\right] $$
Please guide me how I should proceed further .
|
$$y-1=\dfrac{2t-3}{t^2-t+2}$$
Let $z=3-2t\implies5\ge z\ge1$
$$\dfrac4{1-y}+4=4+\dfrac {z^2-4z+11}z=z+\dfrac{11}z\ge2\sqrt{z\cdot\dfrac{11}z}$$ which is attained if $z^2=11$
$$y\ge?$$
Now for $5\ge a>b\ge1,$
$$a+11/a-(b+11/b)=\dfrac{(a+b-11)(a-b)}{ab}<0$$
So, $z+\dfrac{11}z$ will be maximum for the the minimum value of $z$ which is $=1$ here
Can you take it from here?
|
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|
$\arctan{x}+\arctan{y}$ from integration I was trying to derive the property
$$\arctan{x}+\arctan{y}=\arctan{\frac{x+y}{1-xy}}$$
for $x,y>0$ and $xy<1$ from the integral representation
$$
\arctan{x}=\int_0^x\frac{dt}{1+t^2}\,.
$$
I am aware of "more trigonometric" proofs, for instance using that $\tan{(\alpha+\beta)}=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$, but I was willing to see if there is a proof that uses more directly the properties of the integral representation. For instance, if $x>0$, one immediately gets
$$\begin{aligned}
\arctan{x}+\arctan\frac{1}{x}
&=\int_0^x\frac{dt}{1+t^2} + \int_0^{\frac{1}{x}}\frac{dt}{1+t^2}\\
&=
\int_0^x\frac{dt}{1+t^2}+\int_x^\infty\frac{dt}{1+t^2}\\
&=\int_0^\infty \frac{dt}{1+t^2} = \frac{\pi}{2}
\end{aligned}$$
sending $t\to\frac{1}{t}$ in the second integral.
Similarly I tried considering
$$
\int_0^x\frac{dt}{1+t^2} + \int_0^y\frac{dt}{1+t^2}=(x+y)\int_0^1\frac{1+xyt^2}{1+(x^2+y^2)t^2+x^2y^2t^4}\ dt
$$
after rescaling $t\to xt$ and $t\to yt$. On the other hand, via a similar rescaling $t\to \frac{x+y}{1-xy}t$, we have
$$
\int_0^\frac{x+y}{1-xy}\frac{dt}{1+t^2}
=
(x+y)\int_0^1\frac{1-xy}{(1-xy)^2+(x+y)^2t^2}\ dt\,.
$$
By a clever choice of variable it should (must?) be possible to see that these integrals are actually the same, but I can't figure it out...
|
Fixing $y$, define $f(x):=\arctan x+\arctan y-\arctan\frac{x+y}{1-xy}$ so $f(0)=0$ and$$\begin{align}f^\prime(x)&=\frac{1}{1+x^2}-\frac{1}{1+\left(\frac{x+y}{1-xy}\right)^2}\partial_x\frac{x+y}{1-xy}\\&=\frac{1}{1+x^2}-\frac{(1-xy)^2}{(1+x^2)(1+y^2)}\frac{1-xy-(x+y)(-y)}{(1-xy)^2}\\&=\frac{1}{1+x^2}-\frac{(1-xy)^2}{(1+x^2)(1+y^2)}\frac{1+y^2}{(1-xy)^2}\\&=0,\end{align}$$i.e. $f(x)=0$ for all $x$.
|
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|
Find the limit of $\frac{e^{\tan x} - e^x + \ln(\sec x + \tan x) -x }{\tan x - x}$ as $x \to 0$ $$\lim_{x \to 0} \frac{e^{\tan x} - e^x + \ln(\sec x + \tan x) -x }{\tan x - x}$$
I tried to solve this using L'Hopital rule but the resulting differential got too messy
$$=\lim_{x \to 0} \frac{e^{\tan x}\sec^2x - e^x + \sec x - 1 }{\tan^2x}$$
$$=\lim_{x \to 0} \frac{e^{\tan x}(\sec^4x+2\sec^2x\tan x) - e^x + \sec x\tan x }{2\tan x \sec^2x}$$
What should I do from here? Differentiate again or use a different strategy?
|
This can also be accomplished with power series. First, the denominator because it's easier:
\begin{multline}
\tan(x) - x =\left[x + \frac{x^3}{3} + O(x^4)\right] -x = \frac{x^3}{3} + O(x^4).
\end{multline}
Thus, we need to expand the numerator out to third order to determine the limit:
\begin{multline}
e^{\tan{x}}-e^x = 1+\tan(x)+\frac{\tan^2(x)}{2}+\frac{\tan^3(x)}{6} + O(x^4) - \left[1+x+\frac{x^2}{2}+\frac{x^3}{6} + O(x^4)\right]
\\ = 1+x + \frac{x^3}{3}+\frac{x^2}{2}+\frac{x^3}{6} - \left[1+x+\frac{x^2}{2}+\frac{x^3}{6} \right] + O(x^4)= \frac{x^3}{3} + O(x^4)
\end{multline}
\begin{multline}
\ln[\tan(x)+\sec(x)] -x =\int\left[\sec(x) - 1\right]dx = \int\left[\frac{x^2}{2} + O(x^4)\right]dx = \frac{x^3}{6}+O(x^4)
\end{multline}
So we have
\begin{multline}
\lim_{x\rightarrow 0} \frac{e^{\tan{x}}-e^x+\ln[\tan(x)+\sec(x)] -x}{ \tan(x) - x} \\
= \lim_{x\rightarrow 0}\frac{\frac{x^3}{3} + O(x^4) +\frac{x^3}{6} + O(x^4)}{\frac{x^3}{3} + O(x^4)} = \lim_{x\rightarrow 0}\frac{\frac{3}{2} + O(x)}{1+O(x)} = \frac{3}{2}
\end{multline}
The fact we had to expand to third order also means we needed three applications of L'Hospital's rule to get rid of the indeterminate form. Applying it once more to the expression you found gives
\begin{multline}
\lim_{x \to 0} \frac{e^{\tan(x)}\left[\sec^4(x)+2\sec^2(x)\tan^2(x)\right] - e^x + \sec x\tan x }{2\tan x \sec^2x}
\\ = \lim_{x \to 0} \frac{e^{\tan(x)}\sec^4(x)\left[2+2\sin^2(x)+\tan^2(x)+5\tan(x)\right]- e^x + \sec(x)\left[1+2\tan^2(x)\right] }{2\sec^2(x)\left[1+3\tan(x)^2\right]}
\end{multline}
And sure enough, the numerator and denominator are no longer zero at $x=0$, so the limit is easily evaluated as $3/2$.
|
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|
Prove, by the process of Mathematical Induction, that: $\sqrt 1 + \sqrt 2 + ... + \sqrt{n} < \frac{4n+3}{6}\sqrt{n}$ for all integers n > 0. So what I am stuck on is proving the inequality for n=k+1. I have my teacher's solution, but it doesn't quite make sense to me.
Teacher's solution:
$\sqrt 1 + \sqrt 2 + ... + \sqrt{k} + \sqrt{k+1} < \frac {4k+3}{6} \sqrt{k} +\sqrt{k+1}$ (by assumption)
$< \frac {4k+1}{6} \sqrt{k+1} + \sqrt{k+1} $
So he basically added 1 to the k under the square root in the first term, making the resulting expression greater. I get that. But he also subtracted 2 from the numerator of the fraction, which should make the expression smaller.
Is there something I'm missing here? How can we be certain that changing the expression as described above results in a larger expression?
PS: My exam is in 3 days... would appreciate a quick reply :)
Thanks!
|
Hypothesis:
$$\sum_{j=1}^k \sqrt{j} < \frac{4k+3}{6}\sqrt{k}$$
We wanted to show that
$$\sum_{j=1}^{k+1} \sqrt{j} < \frac{4(k+1)+3}{6}\sqrt{k+1}$$
From the hypothesis, we have
$$\sum_{j=1}^{k+1} \sqrt{j} < \frac{4k+3}{6}\sqrt{k}+\sqrt{k +1}$$
Hence it suffices to verify that
$$\frac{4k+3}{6}\sqrt{k}+\sqrt{k +1} < \frac{4(k+1)+3}{6}\sqrt{k+1}$$
which is equivalent to
$$\frac{4k+3}{6}\sqrt{k}< \frac{4(k+1)-3}{6}\sqrt{k+1}=\frac{4k+1}{6}\sqrt{k+1}$$
which is just
$$(4k+3)\sqrt{k} < (4k+1) \sqrt{k+1}$$
$$(4k+3)^2k < (4k+1)^2(k+1)$$
$$(4k+3)^2k < (4k+1)^2k + (4k+1)^2$$
$$2k(8k+4)< (4k+1)^2$$
$$8k < 8k+1$$
which is clearly true.
|
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|
Line Integral with Change of Variables I'm a bit rusty on my computational math, and genuinely can't solve this question which is frustrating me.
QUESTION:
$$\int_Csin(y)dx+xcos(y)dy$$ where C is the ellipse defined as follows: $x^2+xy+y^2=1$
MY ATTEMPT:
Define variable $u= \sqrt{\frac{3}{4}}x+\sqrt{\frac{3}{4}y}$
Define variable $v= \sqrt{\frac{1}{4}}x-\sqrt{\frac{1}{4}}y$
Hence:
$u^2+v^2=(\sqrt{\frac{3}{4}}x+\sqrt{\frac{3}{4}y})^2+(\sqrt{\frac{1}{4}}x-\sqrt{\frac{1}{4}}y)^2$
$=(\frac{3}{4}x^2+\frac{(2)(3)}{4}xy+\frac{3}{4}y^2)+(\frac{1}{4}x^2-\frac{2}{4}xy+\frac{1}{4}y^2)$
$=(\frac{3}{4}+\frac{3}{4})x^2+(\frac{3}{2}-\frac{1}{2})xy+(\frac{3}{4}+\frac{1}{4})y^2$
$=(x^2+xy+y^2) = C$
Therefore, we have Jacobian:
$$
\begin{bmatrix}
\frac{\partial{u}}{\partial{x}} & \frac{\partial{u}}{\partial{y}} \\
\frac{\partial{v}}{\partial{x}} & \frac{\partial{v}}{\partial{y}} \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
\sqrt\frac{3}{4} & \sqrt\frac{3}{4} \\
\sqrt\frac{1}{4} & -\sqrt\frac{1}{4} \\
\end{bmatrix}
$$
$$ = \frac{\sqrt{3}}{2} $$
Beyond this point, I must incorporate the original dot product with the differential $(dx,dy)$ but don't know how
|
Hint:
\begin{align*}
\frac{\partial}{\partial x}\left(x\cos\left(y\right)\right)-\frac{\partial}{\partial y}\left(\sin\left(y\right)\right)&=\cos\left(y\right)-\cos\left(y\right)=0
\end{align*}
|
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|
Diophantine equation from "Solving mathematical problems" by Terence Tao
Find all integers $n$ such that the equation $\frac{1}{a} + \frac{1}{b} = \frac{n}{a+b}$ is satisfied for some non-zero values of $a$ and $b$ (with $a + b \neq 0$).
I'm reading "Solving mathematical problems" by Terence Tao and I'm a bit stuck on this particular example. It starts as follows:
First multiplying out the denominators we get
$$\frac{a+b}{ab} = \frac{n}{a+b}$$
and from here follows $$(a+b)^2 = nab.$$
Now expanding this we get
$$a^2+2ab+b^2-nab=0 \Leftrightarrow a²+ab(2-n)+b^2.$$
From here on he suggests to use the quadratic formula to get
$$a= \frac{b}{2}[(n-2) \pm \sqrt{(n-2)^2-4}]$$
which I don't quite see how he came up with...
Also he then notes that "This looks very messy, but actually we can turn this messiness to our advantage. We know that $a, b$, and $n$ are integers, but there is a square root in the formula. Now this can only work if the term inside the square root, $(n-2)^2-4$ is a perfect square."
Could someone enlighten me on the part "Now this can only work if the term inside the square root, $(n-2)^2-4$ is a perfect square." what is he stating right here?
|
The author finds the identity
$$a= \frac{b}{2}[(n-2) \pm \sqrt{(n-2)^2-4}],$$
by applying the quadratic formula to the previous equation, which is a quadratic in $a$:
$$a^2+b(2-n)\cdot a+b^2=0.$$
Plugging the coefficients into the quadratic formula yields
\begin{eqnarray*}
a&=&\frac{-b(2-n)\pm\sqrt{(b(2-n))^2-4b^2}}{2}\\
&=&\frac{b(n-2)\pm\sqrt{(4b^2-4b^2n+b^2n^2)-4b^2}}{2}\\
&=&\frac{b(n-2)\pm\sqrt{b^2n^2-4b^2n}}{2}\\
&=&\frac{b(n-2)\pm|b|\sqrt{n^2-4n}}{2}\\
&=&\frac{b(n-2)\pm b\sqrt{(n-2)^2-4}}{2}\\
&=&\frac{b}{2}\Big((n-2)\pm\sqrt{(n-2)^2-4}\Big).
\end{eqnarray*}
Next, you can rewrite the equality
$$a= \frac{b}{2}[(n-2) \pm \sqrt{(n-2)^2-4}],$$
to isolate the square root. You will find that
$$\sqrt{(n-2)^2-4}=\pm\Big(2a-b(n-2)\Big).$$
Because $a$, $b$ and $n$ are integers, the right hand side is an integer. This means the term inside the square root is a perfect square; it means
$$(n-2)^2-4=\Big(2a-b(n-2)\Big)^2.$$
As a side note, the questions can be solved with far fewer computations by noting that
$$(a+b)^2=nab,$$
implies that $a$ divides $b$ and $b$ divides $a$, so $b=\pm a$. Because $a+b\neq0$ this means $a=b$ and the equation above becomes
$$na^2=(a+a)^2=4a^2,$$
which shows that $n=4$ because $a$ is nonzero.
|
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|
Prove that for every $n\in\mathbb{N}$, $n^2$ is divisible by 3 or has a form $3k+1$? I tried to do this by induction, but it doesn't make any sense: $n^2=3k$ or $n^2=3k+1$
*
*option: $(n+1)^2= 3k+2n+1$
*option: $(n+1)^2= 3k+2n+2$
Is there any other way on proving this problem?
|
Because you wanted to proceed by induction ...
Base case: $0^3 = 0$ is divisible by 3
For any $n$, we have 2 cases.
First take $(n + 1)^2 = n^2 + 2n + 1$
Case 1: $n^2$ is divisible by 3
Let $n = 3k$. What can we conclude?
Case 2: $n^2$ is $3k + 1$.
$$ n^2 + 2n + 1 = n^2 - 1 + 2(n + 1) = (n - 1)(n + 1) + 2(n + 1)$$
We know $n$ is even as $n^2$ is even (fundamental theorem of arithmetic).
Then if $n + 1$ is divisible by 3, we are done.
If $n + 1$ is not, then because $n$ is even, for odd $m$, $n + 1 = 3m + 2$.
$$2(n + 1) = 2(3m + 2 + 1) = 2(3m + 3) = 6m + 6$$.
And you can see that it is divisible by 3.
|
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How do I get from $x^4+2x^3y+3x^2y^2+2xy^3+y^4$ to $(x^2+xy+y^2)^2$? I was doing an example $$(x+y)^4+x^4+y^4$$ and I need to factor it. I've tried and couldn't really do much, so I checked if there was anything to help, and I came across a post asking about the same thing. But my question is how do I know that $$x^4+2x^3y+3x^2y^2+2xy^3+y^4=(x^2+xy+y^2)^2$$ without knowing the answer.
I know about $$(a+b+c)^2$$ but then how would i chose my $a, b$ and $c$?
There are many $xy$ combinations here, with different powers, so would I chose the lowest as my $b$, and $x^2$ as my $a$ and $y^2$ as my $c$, or perhaps is there another formula that can help me.
|
You can try $$x^4+2x^3y+3x^2y^2+2xy^3+y^4= x^2y^2({x^2\over y^2}+2{x\over y}+3+2{y\over x}+ {y^2\over x^2})$$
Let $t= {x\over y}+{y\over x}$ then we have ${x^2\over y^2}+{y^2\over x^2}=t^2-2$, so $$...= x^2y^2(t^2-2+2t+3) =x^2y^2(t+1)^2= $$ $$=x^2y^2({x\over y}+{y\over x}+1)^2 =(x^2+y^2+xy)^2$$
|
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How to evaluate : $\sum_{n=0}^{\infty}\left(\frac{F_{n+x}}{F_{n+x+1}^2}-\frac{F_{n+x+2}}{F_{n+x+3}^2}\right)$ Thank you to &robjohn for proving this question, I got motivated by it, so I went on to investigate this sum $(1)$,
$$S_{x}=\sum_{n=0}^{\infty}\left(\frac{F_{n+x}}{F_{n+x+1}^2}-\frac{F_{n+x+2}}{F_{n+x+3}^2}\right)\tag1$$
using wolfram alpha we got the first six terms of $(1)$
$S_{0}=1$, $S_{1}=\frac{5}{4}$, $S_{2}=\frac{17}{36}$, $S_3=\frac{77}{225}$, $S_4=\frac{317}{1600}$ $S_5=\frac{1357}{10816}$$\cdots$
by observing the pattern of the sequence, I believe that the so-called closed-form is $(2)$
$$\left(\sum_{m=0}^{x}F_{m+1}^2\right)^2\times\sum_{n=0}^{\infty}\left(\frac{F_{n+x}}{F_{n+x+1}^2}-\frac{F_{n+x+2}}{F_{n+x+3}^2}\right)=1+2\sum_{j=0}^{x}F_jF_{j+1}F_{j+2}\tag2$$
Where $F_n$ is the Fibonacci number and $x\ge 0$
obviously $(2)$ can be written as
$$\sum_{n=0}^{\infty}\left(\frac{F_{n+x}}{F_{n+x+1}^2}-\frac{F_{n+x+2}}{F_{n+x+3}^2}\right)=\frac{1}{(F_{x+1}F_{x+2})^2}\left(1+2\sum_{j=0}^{x}F_jF_{j+1}F_{j+2}\right)\tag3$$
How do we prove $(2)?$
|
This seems to be a telescoping series:
$$
\begin{align}
S_{x}
&=\sum_{n=0}^{\infty}\left(\frac{F_{n+x}}{F_{n+x+1}^2}-\frac{F_{n+x+2}}{F_{n+x+3}^2}\right)\\
&=\sum_{n=0}^{\infty}\frac{F_{n+x}}{F_{n+x+1}^2}-\sum_{n=2}^{\infty}\frac{F_{n+x}}{F_{n+x+1}^2}\\
&=\frac{F_x}{F_{x+1}^2}+\frac{F_{x+1}}{F_{x+2}^2}
\end{align}
$$
|
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|
solution of the cubic equation $x^3 = 0$? it is clear that one solution of the above equation is $x = 0$, but since it is a cubic equation so it should have $3$ complex roots, so what will be its other roots? Also how can we determine that a cubic equation has $3$ distinct roots or $2$ same $1$ distinct or all distinct roots, i.e. how can we predict the nature of roots for a cubic equation? Please guide me.
|
First,
$$ (x-0)(x-0)(x-0) = x^3 $$
exhibits the three roots of $x^3$. They're all $0$.
Much like the discriminant of the quadratic, the discriminant of the cubic indicates the types of the roots. The polynomial $x^3$ has discriminant $0$ so at least two roots are the same. Generally, if the coefficients of a cubic are real numbers and the discriminant is not zero, the discriminant is positive if the roots are three distinct real numbers, and negative if there is one real root and two complex conjugate roots.
Using that $x^3$ has discriminant zero, we write
$$ x^3 = (x-a)(x-a)(x-b) $$
where $a$ is a definitely repeated root and $b$ may be different or the same as $a$. We multiply out, obtaining
$$ x^3 = x^3 - (2a+b) x^2 + (a^2 + 2 a b) x - a^2b \text{.} $$
You should have been exposed to the fact that two polynomials are the same only if the coefficients are the same, so comparing the coefficient on the right and left, we get the system
$$ \begin{cases}
x^3 :& 1 = 1 \\
x^2 :& 0 = -(2a+b) \\
x :& 0 = a^2 + 2ab \\
x^0 :& 0 = -a^2 b
\end{cases}
$$
There are several ways to solve this system, but the fastest I see is this:
Multiply the $x^2$ equation by $-a$, obtaining
$$ 0 = 2a^2 + 2ab \text{,} $$
subtract the $x$ equation from this, obtaining
$$ 0 = a^2 \text{,} $$
which forces $a = 0$. Then either the $x$ or $x^2$ equation gives $b = 0$. So all three roots are $0$.
|
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|
How to find the common factor I am currently stomped at this problem $(2 x^2-(y+z) (y+z-x))/(2 y^2-(z+x) (z+x-y))$ I am supposed to factor it and I can't find a way how to get to the answer.
It says that
$(2 x^2+x y-y^2+x z-2 y z-z^2)/(2 y^2-x^2+x y-2 x z+y z-z^2)$ has a common factor of x+y+z. Can anyone elaborate on how that is possible? What am I not seeing...
|
$\frac{(2 x^2+x y-y^2+x z-2 y z-z^2)}{(2 y^2-x^2+x y-2 x z+y z-z^2)}=\frac{(2x-y-z)(x+y+z)}{(2y-x-z)(x+y+z)}=\frac{(2x-y-z)}{(2y-x-z)}.
$
|
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|
$x^{3}+ax^2+bx+c$ has all roots negative real numbers and a<3. Establish an inequality between only b and c
A cubic equation $x^{3}+ax^2+bx+c$ has all negative real roots and $a, b, c\in R$ with $a<3.$
Prove that $b+c<4.$
My attempt :
Let the cubic be $f(x)$
Plotting graph we see that ,
$f(x\geq 0)>0$.
So we can see that a relation between $a, b, c$ can be established by putting $x=1$, so,
$1+a+b+c> 0$.
So $b+c>-4$.
Also by using Vieta's Formula we get $a, b, c > 0$.
Now I'm uncertain how to proceed, any help will be appreciated. This is a problem of a maths olympiad. Thanks.
|
We can write our polynomial in the form $(x+r_1)(x+r_2)(x+r_3)$, where $r_i$ are positive real numbers. Note that $b=r_1r_2+r_2r_3+r_3r_1$ and $c=r_1r_2r_3$.
$\color{Green}{\text{Maybe you will find}}$ Maclaurin's inequality $\color{Green}{\text{very interesting}}$.
By Maclaurin's inequality, we know that $\sqrt[3]{S_3} \leq \frac{S_1}{3}$,
or equivalently:
$$c=r_1r_2r_3 = S_3 \leq (\frac{S_1}{3})^3 = (\frac{a}{3})^3 \leq (\frac{3}{3})^3=1.$$
Also again by use of Maclaurin's inequality, we know that $\sqrt[2]{\frac{S_2}{3}} \leq \frac{S_1}{3}$,
or equivalently:
$$b=r_1r_2+r_2r_3+r_3r_1 = S_2 \leq 3(\frac{S_1}{3})^2 = 3(\frac{a}{3})^2 \leq 3(\frac{3}{3})^3=3.$$
so $$b+c \leq 1+3=4.$$
$\color{Red}{\text{Comment}}$: Also, you can obtain all of the above inequalities
$\color{Red}{\text{only}}$ by using AM–GM inequality.
|
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|
Integral involving elliptic integral functions Recently I came across this identity:
$$\int _0^1\:\frac{K\left(x\right)}{1+x}dx=\frac{\pi ^2}{8}$$
Where:
$$K\left(x\right)=\int _0^{\frac{\pi }{2}}\:\frac{1}{\sqrt{1-\left(x\sin \left(\theta \right)\right)^2}}d\theta $$
Any hints to how to prove this identity?
|
For $a \in (0,1)$ we have
\begin{align}
\int \limits_0^1 \frac{\mathrm{d} x}{(1+x) \sqrt{1-a^2 x^2}} &\stackrel{x = \frac{2 t}{a (1+t^2)}}{=} \frac{2}{a} \int \limits_0^\frac{1 - \sqrt{1-a^2}}{a} \frac{\mathrm{d}t}{1 + \frac{2}{a} t + t^2} = \frac{2 a}{1-a^2} \int \limits_0^\frac{1 - \sqrt{1-a^2}}{a} \frac{\mathrm{d}t}{\left(\frac{1+at}{\sqrt{1-a^2}}\right)^2 - 1} \\
&\!\stackrel{u = \frac{\sqrt{1-a^2}}{1+a t}}{=} \frac{2}{\sqrt{1-a^2}} \int \limits_{\frac{\sqrt{1-a^2}}{2-\sqrt{1-a^2}}}^{\sqrt{1-a^2}} \frac{\mathrm{d} u}{1-u^2} \\
&\,\,\,\,\,= \frac{2}{\sqrt{1-a^2}} \left[\operatorname{artanh}\left(\sqrt{1-a^2}\right) - \operatorname{artanh}\left(\frac{\sqrt{1-a^2}}{2-\sqrt{1-a^2}}\right)\right]\\
&\,\,\,\,\,= \frac{\log\left(1+\sqrt{1-a^2}\right)}{\sqrt{1-a^2}} \, .
\end{align}
Using this result with $a = \sin(\theta)$ we find
\begin{align}
\int \limits_0^1 \frac{\operatorname{K}(x)}{1+x}\, \mathrm{d} x &= \int \limits_0^1 \int \limits_0^{\pi/2} \frac{\mathrm{d} \theta \, \mathrm{d} x}{(1+x) \sqrt{1 - x^2 \sin^2(\theta)}} \stackrel{\text{Tonelli}}{=} \int \limits_0^{\pi/2} \int \limits_0^1 \frac{\mathrm{d} x \, \mathrm{d} \theta}{(1+x) \sqrt{1 - \sin^2(\theta) x^2}}\\
&= \int \limits_0^{\pi/2} \frac{\log(1+\cos(\theta))}{\cos(\theta)} \, \mathrm{d} \theta = \frac{\pi^2}{8} \,.
\end{align}
The final integral is evaluated here. Incidentally, the same method can be used to compute
$$ \int \limits_0^1 \frac{\operatorname{E}(x)}{1+x}\, \mathrm{d} x = 1 \, . $$
|
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If $x$ and $y$ are rational numbers such that : $(x+y)+(x-2y)\sqrt {2}=2x-y+(x-y-1)\sqrt {6}$
If $x$ and $y$ are rational numbers such that : $$(x+y)+(x-2y)\sqrt {2}=2x-y+(x-y-1)\sqrt {6}$$ then which of the following is true?
A) $\;x=1$, $y=1$
B) $\;x=2$, $y=1$
C) $\;x=5$, $y=1$
D) $\;x$ and $y$ can take infinitely many values.
Given:
$$(x+y)+(x-2y)\sqrt {2} = (2x-y)+(x-y-1)\sqrt {6}$$
$$x+y-2x+y+(x-2y) \sqrt {2}=(x-y-1)\sqrt {6}$$
$$(x-2y)\sqrt {2} - (x-2y)=(x-y-1)\sqrt {6}$$
$$(x-2y)(\sqrt {2} -1)=(x-y-1)\sqrt {6}$$
Only the values at option B satisfies the equation. I just did that by hit and trial. Is there any elaboration on that?
|
And why is this down voted?
$$2y-x = \sqrt{2}\Big(2y-x+\sqrt{3}(x-y-1)\Big)\;\;\;\;\;\;/^2$$
$${(2y-x)^2\over 2}+x-2y = \sqrt{3}(x-y-1)$$
If $x-y-1\ne 0$ we have
$$\sqrt {3}= \underbrace{{(2y-x)^2\over 2}+x-2y\over x-y-1}_{\in\mathbb{Q}}$$
which is impossible. So $y=x-1$ and $${(2y-x)^2\over 2}+x-2y=0$$
Solve this system and you are done.
Edit: So $(x-2)^2+2x-4x+4=0$ so $x^2-6x+8=0$ So $x=2$ (and $y=1$) or $x=4$ and ($y=3$). So the answer is $\boxed{B)}$
|
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|
Find $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$ $\lim_{(x,y)\rightarrow(0,0)} \cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})$
I tried replacing x and y with several values and kept getting 1 so I tried:
$$0 \le |\cos(\frac{x^2-y^2}{\sqrt{x^2+y^2}})| \le |\cos(|(\frac{x^2-y^2}{\sqrt{x^2+y^2}}|)|\le |\cos(\frac{|x|^2+|y|^2}{\sqrt{x^2+y^2}})|\le |\cos(\frac{2\sqrt{x^2+y^2}^2}{\sqrt{x^2+y^2}})| \le |\cos(2\sqrt{x^2+y^2})|$$
Is this correct? I couldn't get Wolfram to compute this properly for some reason.
|
You can do directly about the $x,y$ : $ 0 \le \dfrac{x^2}{\sqrt{x^2+y^2}} \le |x|$ and similarly for the other one. This means ...the expression in the $\cos \to 0$ and the answer is clearly $1$.
|
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|
Inequality involving the angle bisectors of a triangle
Let $l_a,l_b,l_c$ denote the lengths of angle bisectors of a triangle with sides $a,b,c$ and semiperimeter $s$. I am looking for the best constant $K>0$ such that
$$l_a^2+l_b^2+l_c^2> K s^2.$$
I found that $K=2/3$ works, but I suspect that best constant is $K=8/9>2/3$. Any proof or reference?
BTW it is known that $l_a^2+l_b^2+l_c^2\leq s^2$.
Proof for $K=2/3$. According to Cut-the-knot,
$$m_a l_a+m_b l_b+m_c l_c\ge s^{2}$$
where $m_a,m_b,m_c$ are the medians. Therefore, by Cauchy–Schwarz inequality,
$$(m_a^2+m_b^2+m_c^2)(l_a^2+l_b^2+l_c^2)\geq (m_a l_a+m_b l_b+m_c l_c)^2\geq s^4$$
which implies
$$l_a^2+l_b^2+l_c^2\geq \frac{s^4}{m_a^2+m_b^2+m_c^2}> \frac{2s^2}{3}$$
in view of
$$m_a^{2}+m_b^{2}+m_c^{2}=\frac{3(a^2+b^2+c^2)}{4}< \frac{3s^2}{2}.$$
EDIT. I found a reference that $K=8/9$ is the best constant. See 11.7. at p. 218 in Recent Advances in Geometric Inequalities by Mitrinovic et al.
No proof is given.
|
As pointed out by Michael Rozenberg, for $K=8/9$ we have to show that
$$\frac{bc(s-a)}{(b+c)^2}+
\frac{ca(s-b)}{(c+a)^2}+\frac{ab(s-c)}{(a+b)^2}>\frac{2s}{9}.$$
Let $a=x+y$, $b=y+z$, $c=z+x$ with $x,y,z>0$, then the inequality is equivalent to
$$\sigma_1^3(\sigma_1^2-4\sigma_2)^2+6\sigma_1\sigma_3(\sigma_1\sigma_2
-9\sigma_3)+\sigma_3(19\sigma_1^4+44\sigma_1^2\sigma_2+7\sigma_1\sigma_3+9\sigma_2^2)>0
$$
where $\sigma_1=x+y+z$, $\sigma_2=xy+yz+zx$, $\sigma_3=xyz$.
The above inequality holds because $\sigma_1\sigma_2\geq 9\sigma_3$ by the AM-GM inequality, and $\sigma_i>0$ for $i=1,2,3$.
|
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|
Quadrilateral inscribed in semicircle. How to find BC using sin ABD? AB = 3, BD = 5 , tanABD = 0.75
BC is diameter
Question : How to find BC using sin ABD?
I can find sin cos ABD and lenght AD.
I can find BC using Ptolemy's theorem.
|
Use $$\measuredangle ABD=\arccos\frac{3}{x}-\arccos\frac{5}{x},$$ where $BC=x=2R$.
Thus, $$\frac{3}{4}=\frac{\sqrt{\frac{x^2}{9}-1}-\sqrt{\frac{x^2}{25}-1}}{1+\sqrt{\frac{x^2}{9}-1}\cdot\sqrt{\frac{x^2}{25}-1}}$$
Can you end it now?
I got $BC=\frac{5}{3}\sqrt{10}.$
Ptolemy also helps.
Indeed, $AD=\sqrt{10},$ $AC=\sqrt{x^2-9}$ and $DC=\sqrt{x^2-25}.$
Thus, since $$AB\cdot DC+AD\cdot BC=AC\cdot BD,$$ we obtain:
$$3\sqrt{x^2-25}+x\sqrt{10}=5\sqrt{x^2-9}$$ and the rest is smooth.
|
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|
Find the exact value of integration $ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$ Can you help me find the exact value for integration with the given steps?
$$ \int_0^1 \frac{1}{\sqrt{1-x}+\sqrt{x+1}+2} dx$$
Some of my attempts as indefinite Integral
$$
\int \frac{1}{2+\sqrt{1-x}+\sqrt{1+x}} \, dx\approx \left(\sqrt{x+1}+\left(-\frac{1}{\sqrt{x+1}+1}-1\right) \sqrt{1-x}+\frac{1}{\sqrt{x+1}+1}-\frac{2 \left(0.707107 \sqrt{x+1}\right)}{\sin }\right)+C
$$
Is it considered Improper Integral?
|
Substitute $x = \sin 2t $ to have
$$\sqrt{1-x} = \cos t - \sin t, \>\>\>\>\>\sqrt{1+x} = \cos t + \sin t$$
and,
$$\begin{align}
& \int_0^1 \frac{dx}{\sqrt{1-x}+\sqrt{x+1}+2} \\
& = \int_0^{\pi/4}\frac{\cos2t}{1+\cos t}dt \\
&= \int_0^{\pi/4}\frac{2(1+\cos t)^2 -4 (1+\cos t) +1}{1+\cos t}dt \\
& =\int_0^{\pi/4} \left(-2 + 2\cos t + \frac12\sec^2 \frac t2\right)dt \\
&= \left(-2t + 2\sin t + \tan \frac t2\right)\bigg|_0^{\pi/4}= 2\sqrt2-1-\frac\pi2
\end{align}$$
|
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|
Challenging integral with root in denominator I'm wondering if anyone has ideas to get started with this integral:
$$
I = \frac{b}{\pi}\int_{-\infty}^{\infty}dx\,\frac{1}{\sqrt{\left[(x-1)^2+b^2\right]\left[(x+1)^2+b^2\right]}}
$$
for $b>0$. Mathematica 12 can't get it done (except for $b=1$) and the most obvious methods haven't worked for me. Ideas? Can it be done exactly?
Ben
Edit: Here is a variation that I also would like to solve:
$$
I_2 = \frac{\sqrt{b_+ b_-}}{\pi}\int_{-\infty}^{\infty}dx\,\frac{1}{\sqrt{\left[(x-1)^2+b_+^2\right]\left[(x+1)^2+b_-^2\right]}}
$$
I'm not sure that ComplexYetTrivial's solution will work for this one. There is a linear term appearing in the root that will mess up the square completion. Ideas?
Edit 2: It seems like it was solved here using Maple, but the solution looks quite messy:
https://stats.stackexchange.com/questions/243943/is-it-correct-to-compute-bhattacharyya-distance-for-cauchy-like-bell-shaped-fun
|
We have
\begin{align}
I &= \frac{b}{\pi} \int \limits_{-\infty}^\infty \frac{\mathrm{d} x}{\sqrt{[(x-1)^2 + b^2][(x+1)^2+b^2]}} = \frac{2b}{\pi} \int \limits_0^\infty \frac{\mathrm{d} x}{\sqrt{[(x-1)^2 + b^2][(x+1)^2+b^2]}} \\
&= \frac{2b}{\pi} \int \limits_0^\infty \frac{\mathrm{d} x}{\sqrt{(1+b^2)^2 - 2 (1-b^2) x^2 + x^4}} \stackrel{x = \sqrt{1+b^2} t}{=} \frac{2 b}{\pi \sqrt{1+b^2}} \int \limits_0^\infty \frac{\mathrm{d} t}{\sqrt{1 - 2 \frac{1-b^2}{1+b^2} t^2 + t^4}} \\
&= \frac{2 b}{\pi \sqrt{1+b^2}} \int \limits_0^\infty \!\! \frac{\mathrm{d} t}{\sqrt{(1 + t^2)^2 - \frac{4}{1+b^2} t^2}} \!\stackrel{t = \tan\left(\frac{\phi}{2}\right)}{=} \!\!\frac{b}{\pi \sqrt{1+b^2}} \int \limits_0^\pi \!\frac{\mathrm{d} \phi}{\sqrt{1 - \frac{4}{1+b^2}\sin^2\left(\frac{\phi}{2}\right)\cos^2\left(\frac{\phi}{2}\right)}} \\
&= \frac{b}{\pi \sqrt{1+b^2}} \int \limits_0^\pi \frac{\mathrm{d} \phi}{\sqrt{1 - \frac{\sin^2(\phi)}{1+b^2}}} = \frac{2 b}{\pi \sqrt{1+b^2}} \int \limits_0^{\pi/2} \!\frac{\mathrm{d} \phi}{\sqrt{1 - \frac{\sin^2(\phi)}{1+b^2}}} = \frac{2b}{\pi \sqrt{1+b^2}} \operatorname{K}\left(\frac{1}{\sqrt{1+b^2}}\right)
\end{align}
for $b>0$, where $\operatorname{K}$ is the complete elliptic integral of the first kind.
|
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For positive a,b,c,d, if $a^2+b^2+c^2+d^2+abcd=5$, show $a+b+c+d\leq 4$. One can use Lagrange multiplier, but I am looking for a more elementary proof.
I try to find the maximum of $a+b+c+d$ and follow the standard approach.
Construct $a+b+c+d+\lambda (a^2+b^2+c^2+d^2+abcd-5)$.
One can obtain $1+2a\lambda +bcd =0$ and so on, and eventually have $a=b=c=d$.
Then you are done.
|
Let $a+b+c+d>4$, $a=kx$, $b=ky$, $c=kz$ and $d=kt$ such that $k>0$ and $$x+y+z+t=4.$$
Thus, $$k(x+y+z+t)>4,$$ which gives $$k>1.$$
But, $$5=a^2+b^2+c^2+d^2 + abcd=k^2(x^2+y^2+z^2+t^2)+k^4xyzt>x^2+y^2+z^2+t^2+xyzt,$$
which is a contradiction because we'll prove now that $$x^2+y^2+z^2+t^2+xyzt\geq5$$ or
$$16(x^2+y^2+z^2+t^2)(x+y+z+t)^2+256xyzt\geq5(x+y+z+t)^4.$$
Indeed, let $x=\min\{x,y,z,t\}$, $y=x+u$, $z=x+v$ and $t=x+w$.
Thus, $$16(x^2+y^2+z^2+t^2)(x+y+z+t)^2+256xyzt-5(x+y+z+t)^4=$$
$$=32\sum_{cyc}(3u^2-2uv)x^2+16\sum_{cyc}(5u^3-u^2v-u^2w)x+$$
$$+(u+v+w)^2\sum_{cyc}(11u^2-10v^2)\geq0$$ and we are done!
|
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|
Change of Variables in Multiple Integrations (rotated ellipse) Let $D$ be an ellipse rotated counterclockwise by $\frac{\pi}{3}$ radians, as pictured below. Before rotation, the horizontal radius was 2 and vertical radius was 6.
picture referenced
Convert $\iint\limits_{D} \, \bigl(8x + 4y\bigr)\, dA$ into an integral over a polar integral over a unit circle.
My work is as follows:
$$\frac{1}{4} \left(\frac{x}{2} + \frac{\sqrt{3} y}{2}\right)^{2} + \frac{1}{36} \left(\frac{\sqrt{3} x}{2} - \frac{y}{2}\right)^{2}=1$$
$$\frac{x^2}{12} + \frac{1}{8} \sqrt{3} x y - \frac{x y}{24 \sqrt{3}} + \frac{7 y^2}{36} = 1$$
$$\frac{1}{36} (3 x^2 + 4 \sqrt{3} x y + 7 y^2) = 1$$
$$\frac{x^2}{12} + \frac{x y}{3 \sqrt{3}} + \frac{7 y^2}{36} = 1$$
$$x = 2\sqrt{3}u , \, y = \frac{6}{\sqrt{7}}v$$
$$u^{2} + \frac{4 u v}{\sqrt{7}} + v^{2} = 1$$
This is not a form I am used to. I am not sure actually how to obtain the $x$ and $y$ equations that do the magic and this was just a blind attempt.
I am stuck at this point. Any guidance is appreciated.
|
To follow along, draw three copies of the plane. Label one set of axes $x,y$, another set of axes $u,v$ and the third set of axes $\xi, \eta$. In the $(x,y)$ plane, draw the rotated ellipse $D$, as in the figure. In the $(u,v)$ plane, draw the same ellipse, but with semi-major axes along the coordinate axes; i.e the equation of the boundary ellipse should be
\begin{align}
\left( \dfrac{u}{6}\right)^2 + \left( \dfrac{v}{2}\right)^2 = 1
\end{align}
Let's call this ellipse $E$. Finally, draw a circle of radius $1$ in the $(\xi, \eta)$ plane; i.e the equation of the circle is $\xi^2 + \eta^2 = 1$. We shall call the unit ball $B$.
Currently you have the integral $\int_D (\text{stuff involving $x,y$})\, dx \, dy$. Your task is to express everything in terms of an integral over the unit circle, $B$; i.e you want to write
\begin{align}
\int_D(\text{stuff involving $x,y$})\, dx \, dy &= \int_E(\text{stuff involving $u,v$})\, du \, dv = \int_B(\text{stuff involving $\xi,\eta$})\, d\xi \, d\eta
\end{align}
So, you just have to figure out what the "stuff" is. To do this, you have to figure out how the $(x,y)$, $(u,v)$ and $(\xi, \eta)$ coordinates are related to each other. We do this step by step.
First, what's the relation between $(x,y)$ and $(u,v)$? Well, to get from the funny ellipse $D$ to the standard ellipse $E$, we have to rotate points in the $(x,y)$ plane anti-clockwise by an angle $\dfrac{\pi}{3}$. This means
\begin{align}
\begin{pmatrix} u \\ v\end{pmatrix} &= \begin{pmatrix} \cos(\pi/3) & - \sin (\pi/3)\\ \sin(\pi/3) & \cos(\pi/3) \end{pmatrix} \cdot \begin{pmatrix} x \\ y \end{pmatrix}
\end{align}
Or equivalently,
\begin{align}
\begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} \cos(-\pi/3) & - \sin (-\pi/3)\\ \sin(-\pi/3) & \cos(-\pi/3) \end{pmatrix} \cdot \begin{pmatrix} u \\ v\end{pmatrix} \\
&= \dfrac{1}{2} \begin{pmatrix} u + \sqrt{3} v \\ - \sqrt{3} u + v\end{pmatrix}
\end{align}
Recall the general formula for the change of variables: $\int_D f(x,y) \, dx \, dy = \int_E f(x(u,v), y(u,v)) \cdot \left| \frac{\partial (x,y)}{\partial(u,v)} \right| \, du \, dv$. So, you have to calculate the absolute value of the determinant of the Jacobian matrix. You can directly verify that in this case, the result will be 1 (rotation matrices always have determinant 1). Therefore,
\begin{align}
\int_D 8x + 4y \, dx \, dy &= 4 \int_D (2x + y) \, dx \, dy \\
&= 4 \int_E \left[2 \left( \dfrac{u + \sqrt{3}v}{2}\right) + \left( \dfrac{- \sqrt{3}u +v}{2}\right) \right] \cdot |1|\, du \, dv \\
&= 2 \int_E (2 - \sqrt{3})u + (\sqrt{3} + 1)v \, du \, dv
\end{align}
Finally, what's the realtionship between $(u,v)$ and $(\xi, \eta)$? Well, to deform an ellipse of horizontal semi-axis length $6$ and veritcal semi-axis length $2$ into a unit circle, we must have $u = 6 \xi$ and $v = 2 \eta$. The determinant of this transformation is $6 \times 2 = 12$. Hence,
\begin{align}
\int_D 8x + 4y \, dx \, dy &= 2 \int_E (2 - \sqrt{3})u + (\sqrt{3} + 1)v \, du \, dv \\
&= 2 \int_B \left[(2 - \sqrt{3})(6 \xi) + (\sqrt{3} + 1)(2 \eta) \right] \cdot |12|\, d\xi \, d\eta \\
&= 24 \int_B \left[ 6(2 -\sqrt 3)\xi + 2(\sqrt{3} + 1) \eta \right] \, d \xi \, d \eta
\end{align}
This is the integral expressed over the unit ball/disk $B:= \{(\xi, \eta) \in \Bbb{R}^2 | \, \, \xi^2 + \eta^2 = 1\}$.
Of course, I intentionally gave several intermediate steps, but with some experience, it is possible to directly compute everything in one line.
|
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|
Evaluate $\int_0^\frac{\pi}{2} \min (\sin x, \frac{1}{2}) dx$
Evaluate
$\int_0^\frac{\pi}{2} \min (\sin x, \frac{1}{2}) dx$
Now, what I've tried to do is to write $\min (\sin x, \frac{1}{2})$ : $\sin x \leq \frac{1}{2}$ and $\frac{1}{2} < \sin x$. but I don't quite understand how to choose the sets for each minimum (because I know that I will have to split this integral.)
Any help will be appreciated.
|
Notice two things:
*
*On the interval $(0,\frac{\pi}{2})$, function $\sin{x}$ is always increasing.
*$\sin(0) =0$
That means that $\min \left(\sin x, \frac{1}{2} \right)$ will be equal to $\sin x$ when $x<\frac{\pi}{6}$ because $\sin{\frac{\pi}{6}}=\frac{1}{2}$. Now we have
$$
\int_0^{\frac{\pi}{2}} \min \left(\sin x, \frac{1}{2} \right) dx = \int_0^{\frac{\pi}{6}} \sin x dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1}{2} dx = \frac{1}{6} \left( -3 \sqrt{3} + \pi + 6 \right)
$$
|
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Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$
Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$.
By trial I found $a= 2 , b= 3 , c= 5$ and $d= 7$ which is one solution. How to find all the solutions of it ?
|
Better version.
Notice that $3 \mid x!$ for $x \geq 3$ and $3 \not \mid 2^d$. Therefore, at least one of $a, b, c \leq 2$. WLOG let $a\leq b\leq c$.
If $c\leq 1$, $a\leq b\leq c\leq 1, a!=b!=c!=1$ which gives no solution.
If $c=2$, $a!+b!+2=2^d$. $a,b\in \{0,1\}$ gives $4$ solutions, while $(a,b)=(1,2)$ and $(2,2)$ doesn't give solution.
For the cases below, $c\geq 3$.
If $b\leq 1$, $2+c!=2^d$. Notice that $2^2|c!$ for $c \geq 4$, so $c=3$. This gives $(a,b,c,d)=(a,b,3,3) \forall a,b\in \{0,1\}$ (Extreme laziness)
If $b=2$, $a!+2+c!=2^d$. Note that $a!=1$ doesn't give any solution (parity), so $a=b=2$. $4+c!=2^d$. Noticing that $2^3\mid c!$ for $c \geq 4$, $c=3$. $(a,b,c)=(2,2,3)$ doesn't give a solution.
For the cases below, $c \geq b \geq 3$. $2\mid b!+c!$.
Note that $a!=1$ doesn't give a solution. Therefore, $a=2$.
$2+b!+c!=2^d$. If $c \geq b \geq 4$, $2^3 \mid b!+c!$. Therefore $b=3$ gives $8+c!=2^d$. Note that $c\geq 6$ means $2^4 \mid c!$. Therefore, $c=4$ or $c=5$. Checking shows both of them work.
Therefore all the solutions:
$(a,b,c)=(0,0,2),(0,1,2),(1,1,2),(0,0,3),(0,1,3),(1,1,3),(2,3,4),(2,3,5)$ , up to permutations.
|
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Prove: $\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N}\setminus\{0\}, x\in{\mathbb{R}, x\neq k\pi}} $
Prove :
$\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N\setminus\{0\}}}, x\in{\mathbb{R}} $
I proved this relationship by incident. I tried to directly prove this afterwards, but failed. I would love to see another proof to this Problem.
-A proof-: we know that $\displaystyle\sum_{i=1}^{n}\cos{a_i}= \sum_{k=1}^{n}\cos{(a+(k-1)x)}=\frac{\cos{\frac{a+a_n}{2}}\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}=T(x)\quad(1)$
(Here's the source for $(1)$).
Also, $\displaystyle\sum_{i=1}^{n}a^2_i=\sum_{k=1}^{n}(a+(k-1)x)^2\\=\displaystyle\sum_{k=1}^{n}(a^2+2ax(k-1)+x^2(k-1)^2)\\=\displaystyle\sum_{k=1}^{n}a^2+2ax\sum_{k=1}^{n}(k-1)+x^2\sum_{k=1}^{n}(k-1)^2\\=na^2+2ax\frac{n(n-1)}{2}+x^2\frac{n(n-1)(2n-1)}{6}\quad(2)$
We consider the function $f(x)=\frac{x^2}{2}+\cos{x}\implies f''(x)=1-\cos{x}\geq{0}$ therefore $f(x)$ is a concave function. From Jensens inequality for concave functions:$ \displaystyle\sum_{i=1}^{n}f(a_i)\geq{nf\left(\frac{\sum_{i=1}^{n}a_i}{n}\right)}\iff\displaystyle\sum_{i=1}^{n}\left(\frac{a^2_i}{2}+\cos{a_i}\right)\geq n\Big(\frac{1}{2}\left(\frac{\frac{n}{2}(a+a_n)}{n}\right)^2+\cos{\frac{\frac{n}{2}(a+a_n)}{n}}\Big)\iff\frac{1}{2}\sum_{i=1}^{n}a^2_i+\sum_{i=1}^{n}\cos{a_i}\geq \frac{n}{2}\left(\frac{a+a_n}{2}\right)^2+n\cos{\frac{a+a_n}{2}}\overset{(1),(2)}{\iff}\frac{1}{2}\left(na^2+2ax\frac{n(n-1)}{2}+x^2\frac{n(n-1)(2n-1)}{6}\right)+T(x)\geqslant \frac{n\left(2a+(n-1)x\right)^2}{8}+n\cos{\frac{a+a_n}{2}}\overset{\ldots}{\iff} \frac{x^2n(n^2-1)}{3}\geqslant n\cos{\frac{a+a_n}{2}}-T(x)\overset{(1)}{\iff} \frac{x^2n(n^2-1)}{3}\geq \cos{\frac{a+a_n}{2}}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\iff \frac{x^2n(n^2-1)}{3}\geqslant \cos{\left(a+\frac{(n-1)x}{2}\right)}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\quad(3) $
$Lemma.$ For every dinstict $x,x\in\mathbb{R}_{\neq kπ}$ there exists at least one value of $a$ such that $\cos{(a+\frac{(n-1)x}{2}})=1\quad(5)$
Proof of the lemma: $(5)\iff (n-1)x=2-2a+4kπ\iff a=2kπ+1-\frac{(n-1)x}{2}$.
Hence, for every $x,x\in\mathbb{R}_{\neq k\pi}$
we have $\frac{x^2n(n^2-1)}{3}\geqslant \cos{\left(a+\frac{(n-1)x}{2}\right)}\left(n-\frac{\sin{\frac{nx}{2}}}{\sin{\frac{x}{2}}}\right)\overset{(4,x\to 2x)}{\iff}\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}} \square$
*
*To me it looks like $n-\frac{x^2n(n^2-1)}{6}$ is a good approximation function of $\frac{\sin{nx}}{\sin{x}}$ for small values of $x$ (or for small values of $x+ z\pi$, $z\in\mathbb{Z}$, it depends on $n$, because the second one is periodic with period multiple of $\pi$).
|
We will prove a more general result.
Let
$$
f(x)=\frac{\sin nx}{\sin x}=\sum_{k=0}^\infty a_kx^{2k}\tag1
$$
and
$$
f_m(x)=\sum_{k=0}^m a_kx^{2k}.\tag2
$$
Then the following inequalities hold:
$$\begin{cases}
f(x)\le f_m(x),& m\text{ even}\\
f(x)\ge f_m(x),& m\text{ odd}\\
\end{cases}.\tag3
$$
Your inequality will then follow as a special case: $m=1$.
The first key point of the proof is the observation:
$$
\frac{\sin nx}{\sin x}=\sum_{\ell=0}^{n-1}e^{i(n-1-2\ell)x}
=\begin{cases}
1+2\sum_{\ell=1}^{\frac{n-1}2}\cos2\ell x,&n\text{ odd}\\
2\sum_{\ell=1}^{\frac{n}2}\cos(2\ell-1)x,&n\text{ even},\tag4
\end{cases}
$$
which follows from simple telescopic identity:
$$
(e^{ix}-e^{-ix})\sum_{\ell=0}^{n-1}e^{i(n-1-2\ell)x}=e^{inx}-e^{-inx}.
$$
The second key point is the well-known inequalities (which can be proved e.g. by $2m$-fold integration of the inequality $1-\cos x\ge0$):
$$
\begin{cases}
\cos x\le \sum_{k=0}^m c_kx^{2k},& m\text{ even}\\
\cos x\ge \sum_{k=0}^m c_kx^{2k},& m\text{ odd},
\end{cases}\tag5
$$
where $c_k=\frac{(-1)^k}{(2k)!}$.
In view of (4) and (5) the relation (3) is proved.
As a byproduct of the proof one obtains a simple expression for the coefficients of the series expansion (1):
$$
a_k=[x^{2k}]\frac{\sin nx}{\sin x}=[x^{2k}]\sum_{\ell=0}^{n-1}e^{i(n-1-2\ell)x}=\frac{(-1)^k}{(2k)!}\sum_{\ell=0}^{n-1}(n-1-2\ell)^{2k}.\tag6
$$
Particularly, $a_0=n$, $a_1=-\frac{n(n^2-1)}6$, and so on.
|
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|
Conditional probability of continuous throwing a dice A person will keep throwing a dice until he gets $6 .$ If he gets 6 then the experiment will be stopped.
The probability that experiment ends at 6 th trial, given that prime number does not appear is
(A) $\left(\frac{2}{5}\right)^{5}\left(\frac{1}{6}\right)$
(B) $\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)$
(C) $^{6} \mathrm{C}_{1}\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)$
(D) $^{6} \mathrm{C}_{2}\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)$
My approach:
$\begin{aligned} A=& \text { throwing a dice until } 6 \text { occurd and } \\ \text { prime } & \text { no. does not appear. } \end{aligned}$
$B=$ exp. ends at $6^{\text {th }}$ trial.
$P(B / A)=\frac{P(B \cap A)}{P(A)}$
$=\frac{\left(\frac{2}{6}\right)^{5} \frac{1}{6}}{\frac{1}{6}+\frac{2}{6} \cdot \frac{1}{6}+\left(\frac{2}{6}\right)^{2} \frac{1}{6}+\cdots}$
$=\frac{\left(\frac{1}{3}\right)^{5} \frac{1}{6}}{\frac{116}{1-1 / 3}}=\left(\frac{1}{3}\right)^{5} \cdot \frac{1}{6} \cdot \frac{2 / 3}{16}$
No correct ans in option. Where did I go wrong.
|
In a single roll of the dice, the only way a prime number does not appear is, if the die lands $2,4,6$. Restricting our attention to these three outcomes, the probability that the game does not end, conditional on a prime not appearing is $2/3$, as you are interested only in $2,4$. All trials are independent. So, if we extend this to $5$ trials, we get $(2/3)^5$. Further, in the sixth trial, you would like a $6$. So, that $(1/3)$.
Hence, it is $(2/3)^5 (1/3)$.
|
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|
Use mathematical induction to prove that for all integers $n \geq 3,\, 2n + 1 < 2^{n}$ This is what I've got so far.
Let $P(n)$ be the statement that $2n + 1 < 2^n$
Basis:
Let $n = 3$. Show that $P(3)$ is true.
$2(3) + 1 = 7$ and $2^3 = 8$.
Since $7 < 8$, $P(3)$ is true.
Inductive Hypothesis:
Suppose that $P(k)$ is true for an arbitrary integer $k\ge 3$.
This implies that $2k + 1 < 2^k$.
Show that $2(k + 1) + 1 < 2^{k + 1}$.
\begin{align*}
2(k + 1) + 1 &= 2k + 2 + 1\\
&= (2k + 1) + 2
\end{align*}
From the IH,
\begin{align*}
2k + 1 < 2^k &\Rightarrow (2k + 1) + 2 < 2^k + 2\\
&\Rightarrow (2k + 1) + 2 < 2^k + 2 \\
&\Rightarrow (2k + 2) + 1 < 2^k + 2 \\
&\Rightarrow 2(k + 1) + 1 < 2^k + 2
\end{align*}
Now we need to show that $2^k + 2 < 2^{k + 1}$.
$2^k < 2^{k + 1}\Rightarrow 2^k + 2 < 2^{k + 1} + 2$
This is where I'm stuck. Any help on how to proceed is much appreciated.
|
$2^{k} + 2 < 2^{k + 1}$
$2^{k} + 2< 2\cdot 2^{k}$
$2< 2^{k}$
This is trivially true for $k = 3$, and $2^{k}$ is an increasing function, so the inequality is true and the induction is complete. $\blacksquare$
|
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|
Simplifying $\frac{ \cos^2 x - \sin^2 x }{\sin{2x}}$ Please consider the following problem and my answer to it:
Problem:
Simplify the following expression:
$$ \frac{ \cos^2 x - \sin^2 x }{\sin{2x}} $$
Answer:
$$ \frac{ \cos^2 x - \sin^2 x }{\sin{2x}} = \frac{ \cos^2 x - \sin^2 x }{ 2 \sin x \cos x} $$
$$ \frac{ \cos^2 x - \sin^2 x }{\sin{2x}} = \frac{\cos x}{2 \sin x} - \frac{\sin x}{2 \cos x} $$
The result is not much simpler than what I started with. Is there an additional simplification that can be made? Am I missing something?
|
Continue with
$$ \frac{\cos x}{2 \sin x} - \frac{\sin x}{2 \cos x}
=\frac1{2\tan x} -\frac12 \tan x
= \frac{1-\tan^2x}{2\tan x}=\frac1{\tan 2x}
$$
|
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|
Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$ Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$
My Attempt:
Given
$$y-3px+ayp^2=0$$
$$3px=y+ayp^2$$
$$x=\frac {1}{3} \cdot \frac {y}{p} + \frac {a}{3} \cdot yp$$
This is solvable for x. Differentiating both sides with respect to $y$
$$\frac {dx}{dy}=\frac {1}{3} \cdot \frac {p-y\frac {dp}{dy}}{p^2} + \frac {a}{3} (y\cdot \frac {dp}{dy} +p)$$
$$\frac {dx}{dy} = \frac {1}{3p} - \frac {y}{3p^2} \cdot \frac {dp}{dy} + \frac {ay}{3} \cdot \frac {dp}{dy} + \frac {ap}{3}$$
$$\frac {1}{p} - \frac {1}{3p} -\frac {ap}{3} = \frac {y}{3} \cdot \frac {dp}{dy}\cdot \frac {ap^2-1}{p^2}$$
$$\frac {3-1-ap^2}{3p} = \frac {y}{3} \cdot \frac {ap^2-1}{p^2} \cdot \frac {dp}{dy}$$
$$(2-ap^2) = \frac {y}{p} \cdot (ap^2-1) \cdot \frac {dp}{dy}$$
$$\frac {dy}{y} = \frac {ap^2-1}{p(2-ap^2)} dp$$
How do I integrate the RHS of above equation?
|
Solving for $p$ we get
$$
y' = \frac{3x\pm\sqrt{9x^2-4ay^2}}{2ay}
$$
or
$$
2ay y'=3x\pm\sqrt{9x^2-4ay^2}
$$
now calling $u = a y^2$ we have
$$
u' = 3x\pm\sqrt{(3x)^2-4u}
$$
and now following with $z = (3x)^2-4u$ we arrive at
$$
z'\pm 4\sqrt{z}=6x
$$
|
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|
Let T be linear map on $P_2(R)$. Find the eigenvalues of T and an ordered basis $\beta$ for V such that $[T]_\beta$ is a diagonal matrix. Let T be linear map on V. Find the eigenvalues of T and an ordered basis $\beta$ for V such that $[T]_\beta$ is a diagonal matrix.
$V = P_2(R), T(f(x)) = xf'(x)+f(2)x+f(3)$.
I was able to create a basis for the set of eigenvectors, $\alpha = \{x−3,4x^2−13x−3,x+1\}$, but how would I construct $[T]_\beta$ so that it is a diagonal matrix?
|
An eigenvalue of a linear operator $T$ consists of a scalar $\lambda$ such that $Tv = \lambda v$, for some $v\in V\backslash\{0\}$.
This is the same as to require that $\ker(T - \lambda I) \neq \{0\}$. This happens iff $\det([T]_{\mathcal{B}} - \lambda I) = 0$ for some (arbitrary) matricial representation of $T$.
At your case, $V = P_{2}(\textbf{R})$. If we use the basis $\mathcal{B} = \{1,x,x^{2}\}$ and $f(x) = a + bx + cx^{2}$, then
\begin{align*}
T(f(x)) & = xf'(x) + f(2)x + f(3)\\\\
& = x(b + 2cx) + (a + 2b + 4c)x + (a + 3b + 9c)\\\\
& = a + 3b + 9c + (a + 3b + 4c)x + 2cx^{2}
\end{align*}
Consequently, $T(1) = 1 + x$, $T(x) = 3 + 3x$ and $T(x^{2}) = 9 + 4x + 2x^{2}$, whence
\begin{align*}
[T]_{\mathcal{B}} =
\begin{bmatrix}
1 & 3 & 9\\
1 & 3 & 4\\
0 & 0 & 2
\end{bmatrix} \Longrightarrow \det([T]_{\mathcal{B}} - \lambda I) & = \begin{vmatrix}
1 - \lambda & 3 & 9\\
1 & 3 - \lambda & 4\\
0 & 0 & 2 - \lambda
\end{vmatrix} = \lambda(\lambda-4)(2-\lambda)
\end{align*}
Once the characteristic polynomial splits and the algebraic multiplicity of each eigenvalue equals one, we conclude that $T$ is diagonalizable.
Now it remains to determine the corresponding eigenspaces.
First case: $\lambda = 0$
\begin{align*}
\begin{bmatrix}
1 & 3 & 9\\
1 & 3 & 4\\
0 & 0 & 2
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix} =
\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} \Longleftrightarrow (x,y,z) = (-3t,t,0)
\end{align*}
Second case: $\lambda = 4$
\begin{align*}
\begin{bmatrix}
-3 & 3 & 9\\
1 & -1 & 4\\
0 & 0 & -2
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix} =
\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} \Longleftrightarrow (x,y,z) = (t,t,0)
\end{align*}
Third case: $\lambda = 2$
\begin{align*}
\begin{bmatrix}
-1 & 3 & 9\\
1 & 1 & 4\\
0 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix} =
\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} \Longleftrightarrow (x,y,z) = (-3t,-13t,4t)
\end{align*}
Therefore $\mathcal{B}' = \{(-3,1,0),(1,1,0),(-3,-13,4)\}$ is a basis of eigenvectors.
Thence we conclude that $[T]_{\mathcal{B}'}$ is a diagonal matrix, just as desired.
|
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|
Prove this has real roots $(a^2-bc)x^2+(a+b)(a-c)x+a(b-c)=0$ Prove this has real roots
$(a^2-bc)x^2+(a+b)(a-c)x+a(b-c)=0$
My Work
\begin{align*}
\Delta&=(a+b)^2(a-c)^2-4a(b-c)(a^2-bc) \\
&=a^4+2a^3c+a^2c^2-2a^3b+b^2a^2-4a^2bc-2abc^2+2ab^2c+b^2c^2.
\end{align*}
How do I show that this is positive? Simplification doesn't help either...How to factor them? please help!
|
Note that $-1$ is a root, so the other root is also real. It turns out to be $\dfrac{a(b-c)}{bc-a^2}$, but that's not really necessary for what you need to prove.
|
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|
Solving $ \sqrt{3-x} - \sqrt{x-1} > \sqrt{4-x} - \sqrt{x} $ I am reviewing my high-school math (year 10-ish) but I am hitting a wall with this nasty inequality:
$$
\sqrt{3-x} - \sqrt{x-1} > \sqrt{4-x} - \sqrt{x}
$$
I find (steps below) the solution to be
$$
1 \le x \le 3 \text{ with } x \ne 2
$$
but the text-book and WolframAlpha both agree that the solution is
$$
1 \le x < 2
$$
I would really appreciate if you could point out my mistake.
I also tried using the patterns that the text-book offers in the theory section, $$\sqrt{A(x)}<B(x)$$ and $$\sqrt{A(x)}>B(x)$$ but the terms increased in number instead of decreasing. I can provide that derivation, but I suspect it goes in the wrong direction.
|
First of all for real cases, $$3\ge x\ge1$$
Observe that $x+3-x=x-1+4-x$
We need $$\sqrt x-\sqrt{x-1}>\sqrt{4-x}-\sqrt{3-x}$$
$$\iff\sqrt x+\sqrt{3-x}>\sqrt{4-x}+\sqrt{x-1}$$
As both sides are $>0$
we can safely take square in both sides
$$x+3-x+2\sqrt{x(3-x)}>4-x+x-1+2\sqrt{(4-x)(x-1)}$$
Again as both sides are $>0$
we can safely take square in both sides
$$x(3-x)>(4-x)(x-1)\iff 3x-x^2>4x-x^2-4+x\iff 2>x$$
|
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|
If x,y,z is positive then find the minimum value of $x+y+z+\frac{1}{x} + \frac{1}{y} +\frac{1}{z}+1$ If x,y,z is positive then find the minimum value of $x+y+z+\frac{1}{x} + \frac{1}{y} +\frac{1}{z}+1$
I know this can be solved by the relation AM$\geq$GM$\geq$HM. But I do not how to apply here and proceed.
Please help me out. Thank You!
|
$x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1 \geq 6 \sqrt[6]{x \cdot y \cdot z \cdot \frac{1}{x} \cdot \frac{1}{y} \cdot \frac{1}{z} } + 1=7$, where the equality is obtained in $x = y = z = \frac{1}{x} = \frac{1}{y} = \frac{1}{z} =1.$
|
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|
Evaluate $\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}$, where $q^7=1$ and $q\neq 1$.
Let $q$ be a complex number such that $q^7=1$ and $q\neq 1$. Evaluate $$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}.$$
The given answer is $\frac{3}{2}$ or $-2$. But my answer is $\pm 2$.
At first, I tried to evaluate it directly. And the LHS equals to
\begin{align}
\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}
& = \frac{q}{1+q^2}+\frac{q^2}{1+q^4}\cdot\frac{q^3}{q^3}+\frac{q^3}{1+q^6}\cdot\frac{q}{q} \\
& = \frac{q}{1+q^2}+\frac{q^5}{1+q^3}+\frac{q^4}{1+q} \\
& = q\cdot\frac{(1+q)(1+q^3)+q^4(1+q)(1+q^2)+q^3(1+q^2)(1+q^3)}{(1+q)(1+q^2)(1+q^3)} \\
& = q\cdot\frac{1+q+q^3+q^4+q^4+q^5+q^6+1+q^3+q^5+q^6+q}{(1+q)(1+q^2)(1+q^3)} \\
& = \frac{-2q^3}{(1+q)(1+q^2)(1+q^3)} \\
\end{align}
And
$$(x-q)(x-q^2)(x-q^3)(x-q^4)(x-q^5)(x-q^6)=x^6+x^5+x^4+x^3+x^2+x+1$$
Let $x=-1$ I get that
$$(1+q)(1+q^2)(1+q^3)(1+q^4)(1+q^5)(1+q^6)=1$$
and
$$(1+q)(1+q^2)(1+q^3)\cdot q^4(q^3+1)\cdot q^5(q^2+1)\cdot q^6(q+1)=1$$
therefore
$$\left[(1+q)(1+q^2)(1+q^3)\right]^2=\frac{1}{q^{15}}=\frac{1}{q}$$
hence
$$\left[\frac{-2q^3}{(1+q)(1+q^2)(1+q^3)}\right]^2=\frac{q}{1}\cdot 4q^6=4$$
$$\frac{-2q^3}{(1+q)(1+q^2)(1+q^3)}=\pm 2$$
And I try for a solution as a polar-form method$.\\$Suppose $q=\cos\frac{2j\pi}{7}+i\sin\frac{2j\pi}{7}$
$$\frac{q^k}{1+q^{2k}}=\frac{\cos\frac{2jk\pi}{7}+i\sin\frac{2jk\pi}{7}}{2\cos\frac{2jk\pi}{7}\left(\cos\frac{2jk\pi}{7}+i\sin\frac{2jk\pi}{7}\right)}=\frac{1}{2\cos\frac{2jk\pi}{7}}$$
Am I going to the right direction? How I finish it? And please help to figure out what's wrong with my calculation at the first part. I appreciate for your help.
|
You may continue with
$
\frac{q^k}{1+q^{2k}}= \frac{1}{2\cos k\alpha}
$, where $\alpha=\frac{2\pi j}7,\>j=1,2,...,6$, and write the expression as,
$$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6}
=\frac{\cos\alpha+ \cos2\alpha+ \cos3\alpha}{2\cos\alpha\cos2\alpha\cos3\alpha}=\frac ND\tag1\\$$
where we used $2\cos x\cos y=\cos(x+y)+\cos(x-y)$ and $\cos 4\alpha= \cos 3\alpha$, $\cos 5\alpha= \cos 2\alpha$. For the denominator, apply $\sin 2x = 2\sin x \cos x$,
$$ 4\sin\alpha \cdot D =4 \sin 2\alpha\cos 2\alpha\cos 3\alpha=2 \sin 4\alpha\cos 4\alpha= \sin 8\alpha= \sin\alpha\tag2 $$
For the numerator, use $2\sin x\cos y=\sin(x+y)+\sin(x-y)$ and $\sin 3\alpha= -\sin 4\alpha$,
\begin{align}2\sin\alpha \cdot N &=2\sin \alpha\cos\alpha+ 2\sin \alpha\cos2\alpha+ 2\sin \alpha\cos3\alpha\\
&= \sin 2\alpha+ (\sin 3\alpha- \sin \alpha) + (\sin 4\alpha- \sin 2\alpha)= -\sin \alpha\tag3\\
\end{align}
From (2) and (3), we have $D=\frac14$ and $N = -\frac12$. Plug into (1) to obtain
$$\frac{q}{1+q^2}+\frac{q^2}{1+q^4}+\frac{q^3}{1+q^6} =-2$$
|
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|
Factorise $Σa^{2}(b^{4}-c^{4})$ I expanded this cyclic expression then found the three factors and assumed third to be constant but after that I got stuck at this question. The options are
(a)$(a-b)^{2}(b-c)^{2}(c-a)^{2}$
(b)$(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)$
(c)$(a+b)^{2}(b+c)^{2}(c+a)^{2}$
(d)None of these
The answer given is option (b).
|
We can consider $\sum a(b^2-c^2)$ and then make the substitutions $a\rightarrow a^2,b\rightarrow b^2, c\rightarrow c^2$ at the end.
$$\sum a(b^2-c^2)$$
$$=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2$$
We can group by the squares:
$$=a^2(c-b)+ b^2(a-c)+c^2(b-a)$$
Now we force a factor of $(b-a)$ from the first two terms:
$$=a^2c-aab+bba-b^2c+c^2(b-a)$$
$$=a^2c-b^2c+ba(b-a)+c^2(b-a)$$
And then a factor of $(a-b)$ drops out of the first two terms:
$$=c(a+b)(a-b)+ba(b-a)+c^2(b-a)$$
$$=(b-a)\left[-c(a+b)+ba+c^2\right]$$
We can easily see $c=b$ is a zero, so:
$$=(b-a)(c-a)(c-b)$$
$$=(a-b)(b-c)(c-a)$$.
|
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|
If $f(x)$ is a common factor of $g(x)$ and $h(x)$ find $f(x)$ Given that $f(x)$ is a common factor of $g(x)=x^4-3x^3+2x^2-3x+1$ and $h(x)=3x^4-9x^3+2x^2+3x-1$ find $f(x)$
I tried to factorised $g(x)$ but it doesn't have any rational roots as I've already tried 1 and -1.
So how do I solve this?
|
$$ \left( 3 x^{4} - 9 x^{3} + 2 x^{2} + 3 x - 1 \right) $$
$$ \left( x^{4} - 3 x^{3} + 2 x^{2} - 3 x + 1 \right) $$
$$ \left( 3 x^{4} - 9 x^{3} + 2 x^{2} + 3 x - 1 \right) = \left( x^{4} - 3 x^{3} + 2 x^{2} - 3 x + 1 \right) \cdot \color{magenta}{ \left( 3 \right) } + \left( - 4 x^{2} + 12 x - 4 \right) $$
$$ \left( x^{4} - 3 x^{3} + 2 x^{2} - 3 x + 1 \right) = \left( - 4 x^{2} + 12 x - 4 \right) \cdot \color{magenta}{ \left( \frac{ - x^{2} - 1 }{ 4 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( 3 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 3 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ - x^{2} - 1 }{ 4 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 3 x^{2} + 1 }{ 4 } \right) }{ \left( \frac{ - x^{2} - 1 }{ 4 } \right) } $$
$$ \left( 3 x^{2} - 1 \right) \left( \frac{ 1}{4 } \right) - \left( x^{2} + 1 \right) \left( \frac{ 3}{4 } \right) = \left( -1 \right) $$
$$ \left( 3 x^{4} - 9 x^{3} + 2 x^{2} + 3 x - 1 \right) = \left( 3 x^{2} - 1 \right) \cdot \color{magenta}{ \left( x^{2} - 3 x + 1 \right) } + \left( 0 \right) $$
$$ \left( x^{4} - 3 x^{3} + 2 x^{2} - 3 x + 1 \right) = \left( x^{2} + 1 \right) \cdot \color{magenta}{ \left( x^{2} - 3 x + 1 \right) } + \left( 0 \right) $$
$$ \mbox{GCD} = \color{magenta}{ \left( x^{2} - 3 x + 1 \right) } $$
$$ \left( 3 x^{4} - 9 x^{3} + 2 x^{2} + 3 x - 1 \right) \left( \frac{ 1}{4 } \right) - \left( x^{4} - 3 x^{3} + 2 x^{2} - 3 x + 1 \right) \left( \frac{ 3}{4 } \right) = \left( - x^{2} + 3 x - 1 \right) $$
well, there you go...
|
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|
Dividing polynomial $f(x^3)$ by $x^2+x+1$ Given two polynomials with real coefficients $g(x)$ and $h(x)$.
Additionally, $x^2+x+1$ is a factor of $f(x^3)= h(x^3)+xg(x^3)$.
Prove that $x-1|h(x)$ and $x-1|g(x)$
I have tried to solve it by doing this;
$(x-1)(x^2+x+1)|(x-1)(h(x^3)+xg(x^3))$
then;
$(x^3-1)|(x)(h(x^3)-g(x^3)) + (x^2)g(x^3) - h(x^3)$
consider degree of $3k, 3k+1, 3k+2$ separately, I got;
$(x^3-1)|(x^2)g(x^3)$ and $h(x^3)$
Lastly, $(x^3-1)|g(x^3), h(x^3)$ and so on finish proving.
I wonder, is it correct to separate three cases of degrees in this problem?
|
$\!\bmod x^2\!+\!x\!+\!1\!:\,\ \color{#c00}{x^3\equiv 1}\,$ so $\, 0\equiv h(\color{#c00}{x^3})\!+\!xg(\color{#c00}{x^3})\equiv h(\color{#c00}1)+xg(\color{#c00}1)\iff h(1)=0=g(1)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Which integration is correct? Or are both correct? What's the correct method to integrate the function $\frac{1}{2(3x+1)}$?
\begin{align}
\int\frac{1}{2(3x+1)} dx &= \int\frac{1}{6x+2}dx\\
&= \frac{1}{6}ln(6x+2)+c
\end{align}
$$or$$
\begin{align}
\int\frac{1}{2(3x+1)} dx &= \frac{1}{2}\int\frac{1}{3x+1}dx \\
&= (\frac{1}{2})(\frac{1}{3})ln(3x+1)+c\\
&= \frac{1}{6}ln(3x+1)+c
\end{align}
Both methods seem feasible, however the answers given are different. Is it because the constant c is of a different value?
|
\begin{align}
\frac{1}{6}\ln(6x+2)+c&= \frac{1}{6}\ln(2(3x+1))+c\\
&=\frac{1}{6}( \ln2 + \ln (3x+1))+c\\
&=\frac{1}{6}( \ln2 + \ln (3x+1))+c\\
&=\frac{1}{6} \ln (3x+1)+c',
\end{align}
with $c'=c+\frac{1}{6}\ln2.$
|
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|
Trying to find a general equation for an ellipse given the foci and sum of focal distances I'm trying to find an equation for the ellipse in the form $$Ax^2 + Bxy + Cy^2 +Dx +Ey +F = 0$$ given the foci $(a,b)$ and $(c,d)$ and the sum of focal distances $r$. I started from the definition $$\sqrt{(x-a)^2+(y-b)^2} + \sqrt{(x-c)^2+(y-d)^2} = r$$ squared both sides, moved not radical terms to the right side and squared again and after trudging through a lot of algebra I've arrived at the equation (easier reading of coefficients below)
\begin{align}
0 &= (r^2 + (a-c)^2)x^2 + 2(a-c)(b-d)xy + (r^2 + (b-d)^2)y^2\\
&\qquad + (r^2(a+c) - (a-c)(a^2+b^2-c^2-d^2))x \\
&\qquad +(r^2(b+d) - (b-d)(a^2+b^2-c^2-d^2))y \\
&\qquad+ \frac{1}{4}(r^4 + 2r^2(a^2+b^2+c^2+d^2) + (a^2+b^2-c^2-d^2)^2),
\end{align}
which I want to believe is close, but this does not produce a graph on Desmos. If anyone just has a reference for the equation that I might be able to look at and find my mistakes that would be much appreciated. When I looked on Wikipedia, they talked about using the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 $$ and rotating the major axis, but I have no idea how to translate those coefficients to be in terms of the foci. For easier reading my coefficients are
\begin{align}
A &= r^2+(a-c)^2 \\
B &= 2(a-c)(b-d) \\
C &= r^2 + (b-d)^2 \\
D &= r^2(a+c)-(a-c)(a^2+b^2-c^2-d^2) \\
E &= r^2(b+d)-(b-d)(a^2+b^2-c^2-d^2) \\
F &= \frac{1}{4}(r^4 +2r^2(a^2+b^2+c^2+d^2) + (a^2+b^2-c^2-d^2)^2).
\end{align}
If the middle term of F was $2r^2(a^2+b^2-c^2-d^2)$ I could factor it, but because the left hand side (after the second squaring round) has no $r$ term, there isn't an opportunity for it to change like all the other terms did. Sorry I can't be more specific, but I didn't think typing out the mountain of algebra I've done was a good idea.
|
Doing it for $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$ is fun.
$S_1(-ae,0(, S_2(ae,0)$, $P(x,y)$ given that
$$PS_1+PS_2=2a \implies \sqrt{(x+ae)^2+y^2}+\sqrt{(x-ae)^2+y^2}=2a$$ $$\implies U+V=2a~~~~~(1)$$
Then $$U^2-V^2=4aex ~~~ U-V=2ex~~~~~~(3)$$
From (1) and (2), we get
$$U=a+ex \implies (x+ae)^2+y^2=(a+ex)^2$$
$$\implies x^2+y^2-e^2x^2=a^2-a^2e^2 \implies (1-e^2) x^2+y^2=b^2$$
$$\implies \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
|
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|
let P(x) is a polynomial function Given $P(x^2)= x^2(x^2+1) P(x)$ and $P(2)=3$, Find $P(3)$ Question: let P(x) is a polynomial function
Given $P(x^2)= x^2(x^2+1) P(x)$ and
$P(2)=3$, Find $P(3)$
If I put $x=2$, we get $P(4)$ which is not required.
I tried to find $P(x)$ to let it general polynomial to find its roots but unable to find
Please help!
|
Elaborating on some details:
$P(0)=P(0^2)=0^2(0^2+1)P(0)=0$
$P(1^1)=P(1)=1^2(1^2+1)P(1)$ so $P(1)=2P(1)$ implying $P(1)=0$
$0=P(1)=P((-1)^2)=(-1)^2((-1)^2+1)P(-1)=2P(-1)$ implying $P(-1)=0$
So, we have found three roots so far.
As for the degree of $P$, if we were to suppose $P(x)$ were a degree $n$ polynomial, it follows that $P(x^2)$ should be a degree $2n$ polynomial.
As $\underbrace{P(x^2)}_{\text{degree }2n}=\underbrace{x^2(x^2+1)}_{\text{degree }4}\times \underbrace{P(x)}_{\text{degree }n}$ it follows that $2n=4+ n$ and so $n=4$
By the fundamental theorem of algebra, it follows that $P(x)$ can be written in the form:
$$P(x)=x(x-1)(x+1)(ax+b)$$
for some values $a,b$
Let us try to find out those values of $a,b$ using $P(2)$ and $P(4)$.
From $P(2)$ we get that $3=P(2)=2(2-1)(2+1)(2a+b)$ and so $2a+b=\frac{1}{2}$. That is a good start but not quite enough to finish, so we use one more.
From $P(4)$ we get $P(4)=P(2^2)=2^2(2^2+1)P(2)=4\cdot 5\cdot 3 = 60$ and that $P(4)=4(4-1)(4+1)(4a+b)$ and so $4a+b=1$
Subtracting the first result away from the second, this gives $2a=\frac{1}{2}$ and $a=\frac{1}{4}$, thus $b=0$
As a result, we have that $$P(x)=x(x-1)(x+1)(\frac{1}{4}x+0)$$ or rewritten as you had arrived at in the comments $$P(x)=\frac{x^2(x^2-1)}{4}$$
From here, finding $P(3)$ is a matter of just plugging in the value and we get $P(3)=\frac{9\cdot 8}{4}=18$
|
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|
Maximize $\log(2)+\log(3/2)x+\log(2)y+\log(5/2)z$ if $x+y+z\leq 1$ and $(y+z)^2+2x-x^2-2xy\leq 1-2\gamma$, $0.24 \leq \gamma \leq 0.25$ I am trying to maximize the function
$$f(x,y,z)=\log(2)+\log(3/2)x+\log(2)y+\log(5/2)z$$
with the following constraints:
$$x\geq 0, y\geq 0, z \geq 0,$$
$$x+y+z\leq 1,$$
$$x+y\geq 4/5,$$
$$(y+z)^2+2x-x^2-2xy\leq 1-2\gamma,$$
where
$$0.24 \leq \gamma \leq 0.25.$$
I claim that the maximum value is $\frac{\log(12)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log\left( \frac{4}{3} \right)$, and that this maximum is obtained when $x=\frac{1+\sqrt{1-4\gamma}}{2}$, $y=\frac{1-\sqrt{1-4\gamma}}{2}$, and $z=0$.
I am trying to avoid using Lagrange Multipliers because it becomes complicated. I am wondering if there is another way. I would also be satisfied if I could show that $f(x,y,z)\leq \frac{\log(12)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log\left( \frac{4}{3} \right)$. Programs like Maple and Mathematica give me solutions for specific $\gamma$, but I would like to find a step by step way to show this for ANY $\gamma$. Thank you.
Note: I want to point out that we treat $\gamma$ as a FIXED constant that lies in the real interval $[0.24, 0.25]$. Also, all logarithms considered are real.
|
In this answer we solve a particular case of the problem, when $z=0$.
Now the last constraint becomes $g(x,y)=y^2+2x-x^2-2xy\le 1-2\gamma$. Since $\frac{\partial g}{\partial y}=2y(1-x)\le 0$, for $x$ fixed we can increase $y$ increasing $f$ by this, until $y$ will be bounded by a constraint $x+y\le 1$ or $g(x,y)= 1-2\gamma $. Let’s consider these cases.
1) If $x+y=1$ then the constraint $g(x,y)\le 1-2\gamma$ becomes $x^2-x+\gamma\le 0$. In order to maximize $f(x,y,z)=\log 2+\log\frac 32+\left(\log 2-\log\frac 32\right)y$ we have to maximize $y$, that is to minimize $x$. This happens when $x=\tfrac{1-\sqrt{1-4\gamma}}2$. So it seems we have to swap $x$ and $y$ in your claim.
2) If $g(x,y)= 1-2\gamma$ then $y=x\pm\sqrt{D}$, where $$D=2x^2-2x+1-2\gamma=2\left(x-\tfrac 12\right)^2+\tfrac 34-2\gamma\ge\tfrac 14.$$ Fix $x$ and look for the constrained $y$ maximizing $f$.
Let’s check when we can take the plus sign in the formula for $y$. This is allowed iff $2x+\sqrt{D}\le 1$, that is when $x\le\tfrac 12$ and $2x^2-2x+\gamma\ge 0$, that is if $x\le x_1=\tfrac{1-\sqrt{1-2\gamma}}2$. Since $\frac{\partial D}{\partial x}>0$ when $x<\tfrac 12$, $y$ increases when $x$ increases from $x$ to $x_1$. So in this case the maximum of $f$ is attained when $x=x_1$. Then $y=1-x_1$ and this is Case 1.
If $x>x_1$ then we have $y=x-\sqrt{D}$. The constraint $x+y\le 1$ becomes $x\le\tfrac 12$ or $2x^2-2x+\gamma\le 0$, that is
$x_1<x\le \tfrac{1+\sqrt{1-2\gamma}}2=x_2$. We have
$$f(x,y,z)=\log 2+x\log\frac 32+(x-\sqrt{D})\log 2=h(x).$$
Then $h’(x)=\log 2+\log\tfrac 32-\tfrac{2x-1}{\sqrt{D}}\log 2$. We claim that $h’(x)>0$. This is clear when $x\le\tfrac 12$. If $x\ge\tfrac 12$ then we have to show that $(1+c)\sqrt{D}>2x-1$, where $c=\frac{\log\tfrac 32}{\log 2}$. Let’s do this.
$(1+c) \sqrt{D}>2x-1$
$(1+c)^2(2x^2-2x+1-2\gamma)>4x^2-4x+1$
Since $(1+c)^2>2.5$, it suffices to show that
$2.5(2x^2-2x+1-2\gamma)\ge 4x^2-4x+1$
$x^2-x+1.5-5\gamma\ge 0$
$\left(x-\frac 12\right)^2+1.25-5\gamma\ge 0$, which is true because $\gamma\le 0.25$.
Thus $h$ increases when $x$ increases so a maximum of $f$ is attained when $x=x_2$. Then $y=1-x_2$ and this is Case 1 again.
|
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|
Sequence $\left\{ a_n\right\}$ defined by $a_1=4$ , and $a_{n+1}=\sqrt{a_n+6}$. Prove that $\left| a_{n+1}-3\right|<\frac{1}{3}\left|a_n-3\right|$. As title. Sequence $\left\{a_n\right\}$ defined by $a_1=4$ , and $a_{n+1}=\sqrt{a_n+6}$.
Prove that $\left| a_{n+1}-3\right|<\frac{1}{3}\left|a_n-3\right|$.
I tried the induction.
$a_{k+1}-3=\sqrt{a_k+6}-3<\sqrt{\frac{1}{3}\left(a_{k-1}-3\right)+6}-3=\frac{1}{3}\sqrt{a_{k-1}+15}-3$
But It seems not work. And I tryied trigonometrical substitution with letting $a_1=6\cos\theta$, then $a_2=\sqrt{6+6\cos\theta}=2\sqrt{3}\cos\frac{\theta}{2}$. And it seems not work either.
And I tried to square them to find a relation $\left(a_n-3\right)^2+6\left(a_n-3\right)=\left(a_{n-1}-3\right)$ or $\left(a_n-3\right)\left(a_n+3\right)=\left(a_{n-1}-3\right)$
It seems not work agian. Please help me, and thanks a lot.
|
If you define $b_n=a_n-3$, you want to show that $$|b_{n+1}|<\frac{|b_n|}{3}.$$
Now, the recursion is $$b_{n+1}=a_{n+1}-3=\sqrt{a_n+6}-3=\sqrt{b_n+9}-3=\frac{b_n}{\sqrt{b_n+9}+3}$$
Since $b_1=1$, this shows that $b_n>0$ for all $n$ by induction, so you can forgot about the absolute values. Moreover $\sqrt{b_n+9}>\sqrt{9}=3$, so $$\frac{1}{\sqrt{b_n+9}+3}<\frac 16,$$
hence $$b_{n+1}=\frac{b_n}{\sqrt{b_n+9}+3}<\frac{b_n}{6}$$
which in fact it is stronger to what you need.
|
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|
Proving $f(x,y)=\cos^4y-(1-\tan^4x-\sin^2y)(\sin^2x\sin^2y+\cos^2y)^2\geq0$ in some open neighborhood of $(0,0)$ Time is ticking and I must prove the Hourglass (or Figure 8) Inequality, which states
$f(x,y)\geq 0$ in some open neighborhood of $(0,0)$, where
$$f(x,y)=\cos^4 y -(1-\tan^4 x - \sin^2 y)(\sin^2 x \sin^2 y + \cos^2 y)^2$$
This inequality implies an interesting continuity result.
The reason I named it the Figure 8 Inequality is because of the level curves you see for $f(x,y)=0.002$:
Unfortunately, it appears that the circuit is closing in on $(0,0),$ so there is absolutely no room for error. The inequality must be proven in a very precise manner, with each step of the form $A \ge B$ being carried so that $A-B$ is extremely small. I do not know how you could possibly make each leap to the conclusion razor sharp.
|
Let us prove that $f(x, y) \ge 0$ for $|x| < \frac{1}{2}$ and $|y| < \frac{\pi}{6}$.
When $|x| < \frac{1}{2}$ and $|y| < \frac{\pi}{6}$, we have $|\sin x| \le |x| \le |\tan x|$,
$\frac{3}{4} \le \cos^2 y \le 1$, and $1-\tan^4 x - \sin^2 y > 0$.
Denote $v = \cos^2 y \in [\frac{3}{4}, 1]$. We have
\begin{align}
f(x, y) &\ge \cos^4 y - (1 - x^4 - \sin^2 y)(x^2 \sin ^2 y + \cos^2 y)^2\\
&= v^2 - (1 - x^4 - (1-v))(x^2 (1-v) + v)^2 \\
&= -(1-x^2)^2 v^3+ (x^4-x^2+1)^2 v^2+(-2 x^8+2 x^6-x^4) v+x^8\\
&\ge -(1-x^2)^2 v^2 + (x^4-x^2+1)^2 v^2+(-2 x^8+2 x^6-x^4) v+x^8\\
&= (x^8-2 x^6+2 x^4) v^2+(-2 x^8+2 x^6-x^4) v+x^8\\
&\ge (x^8-2 x^6+2 x^4) v\cdot \frac{3}{4} +(-2 x^8+2 x^6-x^4) v+x^8\\
&= (-x^8 + 2x^6 + 2x^4)v + 4x^8(1-v)\\
&\ge 0.
\end{align}
We are done.
|
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|
How to solve $3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$? I am trying to solve the following question involving floor/greatest integer functions.
$3\lfloor x \rfloor - \lfloor x^{2} \rfloor = 2\{x\}$ with the notations $\lfloor x \rfloor$ denoting the greatest integer less than or equal to $x$ and $\{x\}$ to mean the fractional part of $x$.
I used the following property for floor functions.
$n\leq x$ if and only if $n \leq \lfloor x \rfloor$ where $n\in \mathbb{Z}$
Let $p=\lfloor x^{2} \rfloor$, then
$p\leq \lfloor x^{2} \rfloor < p+1$
$\rightarrow p \leq x^{2} < p+1$
$\rightarrow \sqrt{p} \leq x < \sqrt{p+1}$ , since $\sqrt{p} \in \mathbb{Z}$
$\rightarrow \sqrt{p} \leq \lfloor x \rfloor < \sqrt{p+1}$ We then have $\sqrt{p} = \lfloor x \rfloor$
Since $\{x\}=x-\lfloor x \rfloor,$
$3\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2\{x\}= 3\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2(x-\lfloor x \rfloor)= 5\lfloor x \rfloor - \lfloor x^{2} \rfloor - 2x=0$
Substituting $p$, $\sqrt{p}$ for $\lfloor x^{2} \rfloor$ and $\lfloor x \rfloor$ respectively, and also letting $x= \sqrt{p}, $ we get $p = 3\sqrt{p}$ solving for $p$ gives $p=0, 9$, and hence $x=0, 3$
The problem is that according to the solution for the problem, $x$ also equals to $\frac{3}{2}$ for $\{x\}=\frac{1}{2}$ since $2\{x\}\in \mathbb{Z}$. However, by definition for $\{x\}$, $0 \leq \{x\} < 1$, then $0 \leq 2\{x\} < 2$. How can $\{x\}=\frac{1}{2}$ and how do I use this to obtain $x=\frac{3}{2}$. I am not sure what I am missing. IF I made any mistakes in my reasoning. Can someone point it out to me please. Thank you in advance.
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Let $x = n + r$ where $n = [x]$ and $r = \{x\}$.
Then we have $3n - [n^2 + 2nr + r^2]=2r$
$3n - n^2 - [2nr + r^2] = 2r$
and.... oh, hey, the LHS is an integer the RHS being $2\{x\}$ means $\{x\} = 0$ or $0.5$.
Two options $x$ is an integer and $x = [x] = n$ and $r=\{x\} = 0$ and we have
$3n-n^2=0$ and $n^2 = 3n$ and $n= 0$ or $n = 3$.
So $x = 0$ and $x=3$ are two solutions.
(Check: $x=0\implies 3[x] - [x^2] = 3*0 - 0 = 0 = \{0\}$. Check. And $x = 3\implies 3[x]-[x^2] = 3[3]- [3^2] = 3*3-9 = 0=\{3\}$. Check.
And if $x = n + \frac 12$ and $r = \frac 12$ then
$3n - n^2 - [2n\frac 12 + \frac 14] = 2\frac 12$
$3n - n^2 - [n + \frac 14] = 1$
$3n -n^2 - n = 1$
$n^2 - 2n + 1 =0$ so $(n-1)^2 = 0$ and $n = 1$.
$x = 1+\frac 12 = 1\frac 12$.
(Check: If $x = 1.5$ then $3[x] - [x^2] = 3[1.5] - [1.5^2] = 3*1 - [2.25]=3-2=1 = 2*\frac 12 = 2\{1.5\}$. Check.)
|
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With a given matrix $A \in M_3(\mathbb{R})$ show that $A^{2009} + A^{2008} = 2 ^{2008} (A + I_3)$. I am given the matrix:
$$A = \begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
\end{pmatrix}$$
I have to show that the following is true:
$$A ^ {2009} + A ^ {2008} = 2 ^ {2008} (A + I_3)$$
I calcucalted $A^2, A^3, A^4,...$ in hopes that I would find some kind of pattern, but I didn't.
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More generally, $A^{n+1} + A^{n} = 2 ^{n} (A + I)$ for all $n \in \mathbb N$. This is proved by induction using that the minimal polynomial of $A$ is $x^2- x - 2=(x-2)(x+1)$ and so $A^2=A+2I$:
$$
A^{n+2} + A^{n+1} = A(A^{n+1} + A^{n})= A(2^{n} (A + I))=2^n(A^2+A)= 2^n(2A+2I)=2^{n+1} (A + I)
$$
(The characteristic polynomial of $A$ is $x^3-3x-2=(x - 2) (x + 1)^2$. So the minimal polynomial of $A$ is either $(x - 2) (x + 1)^2$ or $(x - 2) (x + 1)$.)
|
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|
Value of $\lim\limits_{n\rightarrow\infty}\prod\limits_{k=3}^n\left(1-\tan^4\frac{\pi}{2^k}\right)=\frac{\pi^3}{m}$ Find the value of m for the following $$\lim\limits_{n\rightarrow\infty}\prod_{k=3}^n\left(1-\tan^4\frac{\pi}{2^k}\right)=\frac{\pi^3}{m}$$
|
Long, but detailed
$$\prod_{k=3}^n\left(1-\tan^4\frac{\pi}{2^k}\right)=
\prod_{k=3}^n\left(1-\tan^2\frac{\pi}{2^k}\right)\left(1+\tan^2\frac{\pi}{2^k}\right)=\\
\prod_{k=3}^n\left(\frac{\cos^2{\frac{\pi}{2^k}}-\sin^2{\frac{\pi}{2^k}}}{\cos^2{\frac{\pi}{2^k}}}\right)\left(\frac{\cos^2{\frac{\pi}{2^k}}+\sin^2{\frac{\pi}{2^k}}}{\cos^2{\frac{\pi}{2^k}}}\right)\overset{\cos{2x}=\cos^2{x}-\sin^2{x}}{=}\\
\prod_{k=3}^n\frac{\cos{\frac{\pi}{2^{k-1}}}}{\cos^4{\frac{\pi}{2^k}}}=
\frac{\cos{\frac{\pi}{4}}}{\cos{\frac{\pi}{2^n}}}\cdot\prod_{k=3}^n\frac{1}{\cos^3{\frac{\pi}{2^k}}}=\\
\frac{\cos{\frac{\pi}{4}}}{\cos{\frac{\pi}{2^n}}}\cdot\prod_{k=3}^n\frac{\color{red}{\sin^3{\frac{\pi}{2^k}}}}{\color{red}{\sin^3{\frac{\pi}{2^k}}}\cdot\cos^3{\frac{\pi}{2^k}}}\overset{\sin{2x}=2\sin{x}\cos{x}}{=}\\
\frac{\cos{\frac{\pi}{4}}}{\cos{\frac{\pi}{2^n}}}\cdot\prod_{k=3}^n\frac{\sin^3{\frac{\pi}{2^k}}}{\frac{1}{2^3}\cdot\sin^3{\frac{\pi}{2^{k-1}}}}=
2^{3(n-2)}\cdot\frac{\cos{\frac{\pi}{4}}}{\cos{\frac{\pi}{2^n}}}\cdot\prod_{k=3}^n\frac{\sin^3{\frac{\pi}{2^k}}}{\sin^3{\frac{\pi}{2^{k-1}}}}=\\
2^{3(n-2)}\cdot\frac{\cos{\frac{\pi}{4}}}{\cos{\frac{\pi}{2^n}}}\cdot\frac{\sin^3{\frac{\pi}{2^n}}}{\sin^3{\frac{\pi}{4}}}=
2^{3n-5}\cdot\frac{1}{\cos{\frac{\pi}{2^n}}}\cdot\sin^3{\frac{\pi}{2^n}}=\\
\frac{\pi^3}{2^5}\cdot\frac{1}{\cos{\frac{\pi}{2^n}}}\cdot\left(\frac{\sin{\frac{\pi}{2^n}}}{\frac{\pi}{2^n}}\right)^3\to\frac{\pi^3}{2^5}, n\to\infty$$
given that
$$\cos{\frac{\pi}{2^n}}\to \cos{0}=1, n\to\infty$$
and
$$\frac{\sin{\frac{\pi}{2^n}}}{\frac{\pi}{2^n}}\to 1, n\to\infty$$
|
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|
Find minimum and maximum value of $\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$
Blockquote
Find min,max of function $$T=\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$$
*
*Max
By squaring both side and AMGM:
$$T^2=7+2\sqrt{5\cos^2x+1}\cdot \sqrt{5\sin^2x+1}$$
$$\le 7+5\left(\cos^2x+\sin^2x\right)+2=14$$
Or $$T\le \sqrt {14}$$
*About minimal value:By $\sqrt x +\sqrt y \ge \sqrt{x+y}$
So $T\ge \sqrt{5+1+1}=\sqrt 7$
I think the minimal value has a little wrong. Pls help me.
|
For minimum,
$$T=\sqrt{(\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1})^2}\\
=\sqrt{7+2\sqrt{25\cos^2x\sin^2x+6}}\\
\ge \sqrt{7+2\sqrt6}=1+\sqrt6
$$
|
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|
Finding Laurent series of $f(z) = \frac{1}{z^{2}+4}$ on two different domains. I need to find the Laurent series of $$f(z) = \frac{1}{z^{2}+4}.$$ First for $ z \in \mathbb{C}: |z|<2$ and then for $z \in \mathbb{C}: 1<|z-i|<3$.
Now for the first restriction I do the following:
\begin{align}
f(z) &= \frac{1}{4} \frac{1}{1+\frac{z^{2}}{4}}\\
&= \frac{1}{4} \sum\limits_{n=0}^{\infty} \left(-\frac{z^{2}}{4} \right)^{n}\\
&= \frac{1}{4} \sum\limits_{n=0}^{\infty} (-1)^{n} \left(\frac{1}{4}\right)^{n} z^{2n}\\
&= \frac{1}{4} \left( 1 -\frac{z^{2}}{4} + \frac{z^{4}}{4^{2}} -\frac{z^{6}}{4^{3}} + \dots \right)\\
&= \left( \frac{1}{4} -\frac{z^{2}}{4^{2}} + \frac{z^{4}}{4^{3}} -\frac{z^{6}}{4^{4}} + \dots \right),
\end{align}
which I think is all right and I used the fact $\frac{1}{1-z} = \sum\limits_{n=0}^{\infty} z^{n}$ which is valid for $|z|<1$ (which makes the above valid since $\frac{-z^{2}}{4}<1$ implies $|z|<2$.
So basically my first question is if the above is valid and the second question is how to proceed for $1<|z-i|<3$?
|
You can write
\begin{align*}
\frac{1}{z^2+4} & =\frac{1}{4i}\left[\frac{1}{z-2i}-\frac{1}{z+2i}\right]\\
& =\frac{1}{4i}\left[\frac{1}{(z-i)-i}-\frac{1}{(z-i)+3i}\right]\\
& =\frac{1}{4(z-i)}\left[\frac{1}{1-\left(\frac{i}{z-i}\right)}\right]+\frac{1}{12}\left[\frac{1}{1+\left(\frac{z-i}{3i}\right)}\right]\\
&=\frac{1}{4(z-i)} \sum_{k=0}^\infty\left(\frac{i}{z-i}\right)^k+\frac{1}{12}\sum_{k=0}^\infty(-1)^k\left(\frac{z-i}{3i}\right)^k
\end{align*}
Note: for the first expression, the series is valid for $\left|\frac{i}{z-i}\right|<1 \implies 1 < |z-i|$ and for the second expression, the series is valid for $\left|\frac{z-i}{3i}\right|<1 \implies |z-i|<3$.
Can you take it from here?
|
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|
Find minimal value of $\left(2-x\right)\left(2-y\right)\left(2-z\right)$
Let $x,y,z>0$ such that $x^2+y^2+z^2=3$. Find minimal value of $$\left(2-x\right)\left(2-y\right)\left(2-z\right)$$
I thought the equality occurs at $x = y = z = 1$ (then it is easy), but the fact is $x = y = \frac{1}{3}; z = \frac{5}{3}$. So I just thought of using $uvw$, but I am not allowed to use it during my exam. Because of the equality I cannot use AMGM, Cauchy-Schwarz, etc.
I tried to use Mixing-Variables, but I failed. Please help.
|
Wlog $z\le 1$ because the variables cannot all be $\ge1$.
For fixed $z$, we want to minimize $(2-x)(2-y)$ under a constraint $x^2+y^2=3-z^2$, which is a constant between $2$ and $3$.
Now
$$ \begin{align}(2-x)(2-y)&=4-2(x+y)+xy\\&=4-2(x+y)+\frac12(x+y)^2-\frac{3-z^2}2\\
&=\frac12\left(x+y-2\right)^2+2-\frac{3-z^2}2
\\ &\ge2-\frac{3-z^2}2\\&=\frac{z^2+1}2
\end{align}$$
with equality iff $x+y=2$. Note that this makes $x^2+y^2=x^2+(2-x)^2=4-4x+2x^2=2(x-1)^2+2$, i.e., we can always find such $x,y$ when $z\le 1$ and $x^2+y^2=3-z^2\in [2,3]$.
So we want to minimize $\frac{z^2+1}2\cdot(2-z)=\frac12(2-z+2z^2-z^3)$ with $0<z\le1$.
The derivative of this cubic is $-3z^2+4z-1=(z-1)(1-3z)$, so we find the desired minimum at $z=\frac13$.
|
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|
Does $a^2 + b^2 = 2 c^2$ have any integer solution?
Does the equation $a^2 + b^2 = 2 c^2$ have any integer solution with $|a| \neq |b|$?
I think not, because of these equations for pythagorean triplets:
$$\left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=1$$
$$x=\frac{1-t^2}{1+t^2}, y=\frac{2t}{1+t^2}$$
I think I would need to multiply $x$ and $y$ by $\sqrt{2}$ and they would never be rational numbers.
|
$a^2+b^2=2c^2\tag{1}$
We can get infinitely many integer solutions by Diophantus's method below.
Substitute $a=t+p, b=t+q, c=t+r$ to equation $(1)$, then we get
$$(2p+2q-4r)t+p^2-2r^2+q^2=0$$
Let $t = -1/2(p^2-2r^2+q^2)/(p+q-2r)$.
Then we get a parametric solution below.
$a = p^2+(-4r+2q)p+2r^2-q^2$
$b = p^2-2r^2-q^2-2pq+4qr$
$c = p^2+2r^2+q^2-2pr-2qr$
p,q,r are arbitrary.
Thus, we get infinitely many integer solutions.
Example:
[p,q,r] [a,b,c]
[3, 2, 1], [7, 1, 5]
[4, 3, 1], [17, 7, 13]
[5, 2, 1], [23, 7, 17]
[5, 4, 1], [31, 17, 25]
[6, 3, 1], [41, 1, 29]
[6, 5, 1], [49, 31, 41]
[7, 2, 1], [47, 23, 37]
[7, 6, 1], [71, 49, 61]
|
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|
Pair of real number satisfying $56x+33y=-\frac{y}{x^2+y^2}$ and $33x+56y=\frac{x}{x^2+y^2}$. Let (x,y) be a pair of real number satisfying $56x+33y=-\frac{y}{x^2+y^2}$ and $33x+56y=\frac{x}{x^2+y^2}$. If $|x|+|y|=\frac{p}{q}$ (where p and q are relatively prime), then find the value (6p – q).
I used the concept $\frac{x}{y}=t$ while solving i get $56t+33=-\frac{1}{t^2+1}$ and $33t+56=\frac{t}{t^2+1}$ on dividing the two equation i end up getting quadratic equation but that is not helping me
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Equation (1) $\times x$ + Equation (2) $\times y$ when added together gives
\begin{align*}
56x^2+66xy+56y^2&=0\\
28x^2+33xy+28y^2&=0\\
28\left(x+\frac{33}{56}y\right)^2+\frac{2047}{112}y^2&=0.
\end{align*}
Since left side is always non-negative (for $x,y \in \mathbb{R}$), this can only happen when both
\begin{align*}
x+\frac{33}{56}y&=0\\
y^2&=0
\end{align*}
This gives $x=y=0$ as the only solution. But this does not satisfy the original equation (denominator will become $0$). Thus no such real $x$ and $y$ can exist.
|
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|
Proving $S:=\left\{1\Big/\left(x^2-3\right)\mid x\in\Bbb Q\right\}$ is unbounded
Let $S:=\left\{\dfrac1{x^2 -3} :x\in\Bbb Q\right\}$. Prove
carefully that $S$ is unbounded
My proof
By contradiction, if we can find some $M$ so that $|s| \leqslant M\ \ \forall s \in S$, that is, if
$$ \dfrac{1}{|x^2-3|} \leqslant M\ \forall x\in\Bbb Q\ . $$ Since $x$ is any rational number, we can choose $x$ so that $|x^2-3| = \dfrac{1}{M+1}$. Therefore, we obtain
$$ \dfrac{1}{\frac{1}{M}+1} \leqslant M \implies M+1 \leqslant M $$
A contradiction.
I think my proof is correct, but I have lost some points. What is my mistake?
|
One could show that $S$ is neither bounded from above nor bounded from below. However, we are only asked to sow that it is not bounded. This can be done ab ovo without any sophisticated converging sequences a la Heron formula.
So assume there exists $M\in \Bbb R$ with $|s|\le M$ for all $s\in S$.
Pick $n\in\Bbb N$ with $n>2M$. Then $A_n:=\{\,x\in\Bbb N\mid x^2>3n^2\,\}$ is a non-empty (e.g., $2n\in A_n$) proper (e.g., $1\notin A_n$) subset of $\Bbb N$ and we can let $m=\min A_n$. Then $$m^2>3n^2$$ and $(m-1)^2\le 3n^2$. We readily see (as in the proof of irrationality of $3$) that even stronger $$(m-1)^2<3n^2.$$
From $(2n)^2>3n^2$, it is also clear that $$ m\le 2n.$$
With $x_1=\frac{m-1}n$ and $x_2=\frac mn$, we have
$$\begin{align}\frac1{x_2^2-3}-\frac1{x_1^2-3}&=\frac{n^2}{m^2-3n^2}-\frac{n^2}{\underbrace{(m-1)^2-3n^2}_{<0}}\\
&=\frac{n^2}{m^2-3n^2}+\frac{n^2}{3n^2-(m-1)^2} \\
&=\frac{n^2(3n^2-(m-1)^2+m^2-3n^2)}{(m^2-3n^3)(3n^2-(m-1)^2)}\\
&=\frac{(2m-1)n^2}{(m^2-3n^3)(3n^2-(m-1)^2)}.\end{align}$$
By the arithmetic-geometric inequality, the denominator is $$ \begin{align}(m^2-3n^3)(3n^2-(m-1)^2)&\le\left(\frac{m^2-3n^3+3n^2-(m-1)^2}{2}\right)^2\\&=\frac{(2m-1)^2}{4}\end{align}$$
so that
$$2M\ge \frac1{x_2^2-3}-\frac1{x_1^2-3}\ge\frac{4n^2}{2m-1}
\ge \frac{4n^2}{4n-1}>n.$$
Contradiction.
|
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|
Derive a bound for a general Gauss sum (from Iwaniec and Kowalski pg 199) I am reading 'Analytic Number Theory' by Iwaniec and Kowalski and get stuck on this:
On page 199,they say
...we get
$$\left| S_f(N) \right|^2\leq N+\sum_{ 1\leq \mathcal{l} < N} \min(2N,\left \| 2\alpha \mathcal{l} \right \| ^{-1}).$$
Hence one can deduce that
$$\left| S_f(N) \right|\leq 2\sqrt[]{\alpha}N+\frac{1}{\sqrt[]{\alpha}}\log\frac{1}{\alpha}$$if $0<\alpha \leq \frac{1}{2}$,which restriction can always be arranged...
where $\left \| \alpha \right \|$ denotes the distance of $\alpha$ to the nearest integer,and $S_f(N)=\sum_{1\leq \mathcal{n} \leq N}e(\alpha n^2+\beta n)$. I suppose one needs to summon $\sum_{1\leq \mathcal{l} \leq x}\frac{1}{\mathcal{l}}=\log x +\gamma +O(\frac{1}{x})$ to fill the gap.
Can somebody help?
|
We can assume $\alpha < 1/4$ because otherwise we can apply the trivial bound $|S_f(N)| \leq N$.
Let's partition the sum as follows:
\begin{align*}
\sum_{ 1\leq \ell < N} \min(2N,\| 2\alpha \ell \| ^{-1}) &= \sum_{\substack{1\leq \ell < N \\ 2\alpha\ell < 1/2}} \min(2N,\| 2\alpha \ell \| ^{-1})\\
&+ \sum_{1 \leq m < 2\alpha(N-1)} \sum_{\substack{1\leq \ell < N \\ m-1/2 \leq 2\alpha\ell < m+1/2}} \min(2N,\| 2\alpha \ell \| ^{-1})
\end{align*}
We have the inequalities
\begin{align*}
\sum_{\substack{1\leq \ell < N \\ 2\alpha\ell < 1/2}} \min(2N,\| 2\alpha \ell \| ^{-1}) &\leq \frac{1}{2\alpha} + \frac{1}{2\alpha} \int_{1/2\alpha}^{1/2} \frac{1}{x} \, dx\\
&= \frac{1}{2\alpha} + \frac{1}{2\alpha} \log \frac{1}{4\alpha}
\end{align*}
and
\begin{align*}
\sum_{\substack{1\leq \ell < N \\ m-1/2 \leq 2\alpha\ell < m+1/2}} \min(2N,\| 2\alpha \ell \| ^{-1})) &\leq 2N + 2 \left(\frac{1}{2\alpha} + \frac{1}{2\alpha} \int_{1/2\alpha}^{1/2} \frac{1}{x} \, dx \right)\\
&= 2N + \frac{1}{\alpha} + \frac{1}{\alpha} \log \frac{1}{4\alpha}.
\end{align*}
Therefore we have
\begin{align*}
|S_f(N)|^2 &\leq N + \left(\frac{1}{2\alpha} + \frac{1}{2\alpha} \log \frac{1}{4\alpha} \right) + 2\alpha(N-1)\left(2N + \frac{1}{\alpha} + \frac{1}{\alpha} \log \frac{1}{4\alpha} \right)\\
&\leq 4 \alpha N^2 + 4N \log \frac{1}{\alpha} + \frac{1}{\alpha} \left( \log \frac{1}{\alpha} \right)^2\\
&= \left(2 \sqrt{\alpha} N + \frac{1}{\sqrt{\alpha}} \log \frac{1}{\alpha} \right)^2
\end{align*}
where the second inequality can be shown after a bit of work (using $\alpha < 1/4$).
|
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|
Show that $23a^2$ is not the sum of 3 squares. I know that Legendre's theorem states that a number is expressible as a sum of 3 squares iff. it's not of the form $4^x (8m+7)$, so I need to show that $23a^2$ is of this form, how could I go about doing this?
|
If a number $n$ is of this form, then:
$$n = 4^x \left(8m + 7\right) = 4^x \cdot 8m + 7 \cdot 4^x = 23a^2$$
Therefore
$$n \equiv 7 \cdot 4^x \equiv \begin{cases}0 & \text{if } x > 1 \\4 & \text{if } x = 1\\7 & \text{if } x = 0\end{cases}\pmod{8}$$
and
$$n \equiv 23 a^2 \equiv \begin{cases}0 & \text{if } a \equiv 0 \pmod{4}\\4 & \text{if } a \equiv 2 \pmod{4}\\7 & \text{if } a \text{ is odd}\end{cases}\pmod{8}$$
Lets bust the case where $n \equiv 7 \pmod{8}$. It means that $a$ is odd. So $23a^2$ is odd, too. However $4^x \left(8m + 7\right)$ is even. Because $n$ can't be even and odd at the same time, it is a contradiction.
Next case: $n \equiv 4 \pmod{8}$
$$4 \left(8m + 7\right) = 23a^2 \quad a \in \{2,6,10,12,...\}$$
Because $\left(8m + 7\right)$ is not divisible by $4$, $a$ needs to be $2$, then:
$$8m + 7 = 23$$
Therefore $m = 2$.
So actually it is of the form!
$$4^1 \left( 8 \cdot 2 + 7 \right) = 23 \cdot 2^2$$
One possible solution: $x = 1$, $m = 2$, $a = 2$
|
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|
The area of a triangle determined by two diagonals at a vertex of a regular heptagon
In a circle of diameter 7, a regular heptagon is drawn inside of it. Then, we shade a triangular region as shown:
What’s the exact value of the shaded region, without using trigonometric constants?
My attempt
I tried to solve it with the circumradius theorem : $A=(abc)/(4R)$, where $a$, $b$, and $c$ are the three sides, and $R$ is the circumradius of the triangle. However, I needed to find the exact value of $\cos(5\pi/14)$, $\cos(4\pi/7)$, and $\sin(5\pi/14)$. Finally, I found an explicit formula for this particular triangle, but the proof was missing.
You can find the formula in Wikipedia's "Heptagonal triangle" entry.
|
Expressing the sides $a,b,c$ via the sines of corresponding central angles one obtains:
$$
A=\frac{abc}{4R}=2R^2\sin\frac\pi7\sin\frac{2\pi}{7}\sin\frac{4\pi}7.\tag1
$$
For a product of sines we have the following theorem:
$$
\prod_{0<m_i<n}2\sin\frac{\pi m_i}{n}=n.
\tag2
$$
Therefore: $$2^6\sin\frac\pi7\sin\frac{2\pi}{7}\sin\frac{3\pi}7\sin\frac{4\pi}{7}\sin\frac{5\pi}7\sin\frac{6\pi}{7}=\left(8\sin\frac\pi7\sin\frac{2\pi}{7}\sin\frac{4\pi}7\right)^2=7.$$
|
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|
Taylor Series of $f(x) =\frac{1}{x^2}$ I've found the Taylor Series for $f(x)=\frac{1}{x^2}$ centered at $a=-1$.
$f(-1)=1$, $f'(-1)=2$, $f"(-1)=6$, $f'''(-1)=24$, $f^4(-1)=120$
I used this formula to get each the first coefficients of the terms of the series: $c_n=\frac{f^n(a)}{n!}$
So I got the expansion:
$$f(x)=1+2(x+1)+3(x+1)^2+4(x+1)^3+5(x+1)^4+...$$
What is difficult for me is expressing this in summation notation. Is this right?
$$\sum\limits_{n=0}^\infty (n+1)(x+1)^n$$
|
For $x$ such that $|x+1|<1$
$$f(x)=\frac{-1}{x}=\frac{1}{1-(x+1)}$$
$$=1+(x+1)+(x+1)^2+....$$
$$=\sum_{k=0}^\infty (x+1)^k$$
As a sum of a power series,
$f $ is differentiable at $ (-2,0)$ and
$$f'(x)=\frac{1}{x^2}=\sum_{k=1}^\infty k(x+1)^{k-1}$$
$$=\sum_{k=0}^\infty(k+1)(x+1)^k$$
|
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|
Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$
Prove that
$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$
with $a>0, b>0 , c> 0$ and $d>0.$
My attempt:
$$\begin{align*}\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)& = \dfrac{abcd+b^2c^2+a^2d^2+abcd}{abcd}\\
& =\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\
&=\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\
&=\dfrac{(ad)^2+(bc)^2+2(ad)(bc)}{abcd}\\
&=\dfrac{(ad+bc)^2}{abcd}\end{align*}$$
I don't know how to continue from this.
Can someone help me?
|
Taking a different approach entirely, note that
$$\left({b\over a}+{d\over c}\right)\left({a\over b}+{c\over d}\right)=1+{ad\over bc}+{bc\over ad}+1$$
Thus, letting $ad/bc=x$ and noting that $x\gt0$, we see that
$$\left({b\over a}+{d\over c}\right)\left({a\over b}+{c\over d}\right)\ge4\iff x+{1\over x}\ge2\iff x^2-2x+1\ge0\iff(x-1)^2\ge0$$
(Note, the stipulation $x\gt0$ is important is multiplying both sides of the inequality $x+1/x\ge2$ by $x$ to get to $x^2+1\ge2x$.)
|
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|
Prove $(a+b)\left(\frac{1}{a}+\frac{4}{b}\right)\geq9$ with $a>0$ and $b > 0$. When does equality hold?
Prove that $(a+b)\left(\dfrac{1}{a}+\dfrac{4}{b}\right)\geq9$ with $a>0$ and $b > 0$. When does equality hold?
My attempt:
By Cauchy-Schwarz inequality, we have:
$$\begin{align*}(a+b)\left(\dfrac{1}{a}+\dfrac{4}{b}\right)&\geq{\left(\sqrt{a\cdot\dfrac{1}{a}}+\sqrt{b\cdot\dfrac{4}{b}}\right)}^2\\&={\left(\sqrt{1}+\sqrt{4}\right)}^2\\&={\left(1+2\right)}^2\\&=(3)^2\\&=9\end{align*}$$
Is it correct?
I'm having trouble showing when equality holds. Could someone help me to understand?
Thanks in advance!
|
They called it the Titu's lemma in the inequality world or the Angel form of the CS inequality. The inequality follows from this lemma: $\dfrac{1}{a}+\dfrac{4}{b} = \dfrac{1^2}{a}+\dfrac{2^2}{b} \ge \dfrac{(1+2)^2}{a+b} = \dfrac{9}{a+b}$.
|
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|
If $z+\frac{1}{z}=2\cos\theta,$ where $z\in\Bbb C$, show that $\left|\frac{z^{2 n}-1}{z^{2n}+1}\right|=|\tan n\theta|$ If $z+\frac{1}{z}=2 \cos \theta,$ where $z$ is a complex number, show that
$$
\left|\frac{z^{2 n}-1}{z^{2 n}+1}\right|=|\tan n \theta|
$$
My Approach:
$$
\begin{array}{l}|\sin \theta|=\left|\sqrt{1-\cos ^{2} \theta}\right| \\ =\left|\sqrt{1-\left(\frac{z^{2}+1}{2z}\right)^{2}}\right| \\ =\left|\sqrt{\frac{4 z^{2}-z^{4}-2 z^{2}-1}{4 z^{2}}} \right|\\ =\left|\sqrt{\frac{-\left(z^{4}-2 z^{2}+1\right)}{4 z^{2}}}\right|=\left|\sqrt{\frac{\left(z^{2}-1\right)^{2}}{4 z^{2}}}\right| \\ =|\frac{z^{2}-1}{2 z}|\end{array}
$$
So $|\tan \theta|=\left|\frac{z^{2}-1}{z^{2}+1}\right|$ ( proven when $n = 1$)
Is there any way to prove directly by taking $n$?
|
$$z+\frac 1z=2\cos \theta\Rightarrow z^2-2\cos\theta z+1=0$$
Solve this quadratic to obtain
$$z=cos\theta\pm i\sin\theta$$
Or,
$$z=e^{i\theta},e^{-i\theta}$$
Now,
$$\frac{z^{2n}-1}{z^{2n}+1}=\frac{z^n-1/z^n}{z^n+1/z^n}$$
In either of the above cases(for $z$), we have,
$$z^n+1/z^n=e^{in\theta}+e^{-in\theta}=2\cos n\theta$$
But depending on the value of $z$ we have
$$z^n-1/z^n=\pm \left(e^{in\theta}-e^{-in\theta}\right)=\pm 2\sin n\theta$$
We now have
$$\left |\frac{z^{2n}-1}{z^{2n}+1}\right |=\left|\frac{\pm 2\sin n\theta}{2\cos n\theta}\right|=|\tan n\theta|$$
Hence proved.
|
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|
Find $a \in \mathbb N$ such that $x^2+ax-1 = y^2$ has a solution in positive integers Question: Find all the positive integers $a$ such that $x^2+ax-1 = y^2$ has a solution in positive integers $(x,y)$.
Comments: It's easy to see that this equation rarely has a solution (in the sense that for a fixed $a$, $x^2+ax-1$ is a perfect square only for finitely many values of $x$). In fact, if $a$ is even then $x^2 \le x^2+ax-1 < (x+a/2)^2$, so $x^2+ax-1$ is almost never a perfect square. The problem is that I can't control the interval $[x^2,(x+a/2)^2)$ when $a$ grows. There's a similar argument for $a$ odd.
However, it is possible to find some families of such $a$'s. For instance, if $a$ is a perfect square, say $a=k^2$, then there exists the solution $(x,y) = (1,k)$.
In addition, if $x > 1$ then its prime power factors are $2$ and/or $p^\alpha$, where $p \equiv 1 \pmod 4$. In fact, if $p|x$ then the constraint of the equation modulo $p$ yields that $-1$ is a square.
|
Actually, every $a$ that is not $2 \pmod 4$ yields a solution. These $a$ are impossible because then
$$x^2 + ax - 1 \equiv x^2 + 2x - 1 \equiv (x - 1)^2 - 2 \pmod 4,$$
which is either $2$ or $3$ mod $4$, but squares are only $0$ or $1$ mod $4$.
For the other $a$, I will construct values of $x$ that work.
For $a = 4k$, take $x = 2k^2 - 2k + 1 = \frac{(2k - 1)^2 + 1}{2}$. Then
\begin{align*}
x^2 + ax - 1 &= (2k^2 - 2k + 1)^2 + 4k(2k^2 - 2k + 1) - 1\\
&= (4k^4 - 8k^3 + 8k^3 - 4k + 1) + (8k^3 - 8k^2 + 4k) - 1\\
&= 4k^4\\
&= (2k^2)^2
\end{align*}
For $a = 4k + 1$, take $x = 4k^2 + 1$. Then
\begin{align*}
x^2 + ax - 1 &= (4k^2 + 1)^2 + (4k + 1)(4k^2 + 1) - 1\\
&= (16k^4 + 8k^2 + 1) + (16k^3 + 4k^2 + 4k + 1) - 1\\
&= 16k^4 + 16k^3 + 12k^2 + 4k + 1\\
&= (4k^2 + 2k + 1)^2
\end{align*}
For $a = 4k + 3$, take $x = 4k^2 + 4k + 2$. Then
\begin{align*}
x^2 + ax - 1 &= (4k^2 + 4k + 2)^2 + (4k + 3)(4k^2 + 4k + 2) - 1\\
&= (16k^4 + 32k^3 + 32k^2 + 16k + 4) + (16k^3 + 28k^2 + 20k + 6) - 1\\
&= 16k^4 + 48k^3 + 60k^2 + 36k + 9\\
&= (4k^2 + 6k + 3)^2
\end{align*}
|
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|
I'm stuck trying to factor $x^2-4$ to $(x-2)(x+2)$ I am trying to understand each step in order to get from $x^2-4$ to $(x-2)(x+2)$
I start from here and got this far...
$x^2-4 =$
$x*x-4 =$
$x*x+x-x-4 =$
$x*x+x-2+2-x-4 =$
$x*x+x-2+2-(x+4) =$
After this I try
$x(x-2)+2-(x+4) =$
and this clearly does not even equal the other factorings. I thought the $x$ could be factored out. I'm confused.
I know I can just insert $a^2-b^2$ into the difference of squares formula like so $(a-b)(a+b)$ but I am practicing factoring. I'm just curious to see each and every step of the factoring.
|
Alternatively, recall the general formula $$a^2-b^2=(a+b)(a-b).$$
This can be seen by expanding the RHS, or by writing
$$a^2-b^2=a^2-ab+ab-b^2=a(a-b)+b(a-b)=(a+b)(a-b).$$
Then, with $a=x$ and $b=2$, we have
$$x^2-4=x^2-2^2=(x+2)(x-2).$$
|
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|
Prove that $\exists !c \in \mathbb{R} \exists ! x \in \mathbb{R} (x^2 + 3x + c = 0)$ This is an exercise from Velleman's "How To Prove It". I am struggling with how to finish the final part of the uniqueness proof, so any hints would be appreciated!
*a. Prove that there is a unique real number $c$ such that there is a unique real number $x$ such that $x^2 + 3x + c = 0$.
Proof: Let $c = \frac{9}{4}$. Let $x = -\frac{3}{2}$. It follows that $x^2 + 3x + c = \frac{9}{4} - \frac{9}{2} + \frac{9}{4} = 0$. To show that $x$ is unique, let $y \in \mathbb{R}$ be arbitrary such that $y^2 + 3y + c = 0$. So $y^2 + 3y + \frac{9}{4} = 0$, and $(y+\frac{3}{2})^2 = 0$. It immediately follows that $y = -\frac{3}{2} = x$.
Now to show that $c$ is unique, let $d, e \in \mathbb{R}$ be arbitrary such that $\exists ! x \in \mathbb{R} (x^2 + 3x + d = 0)$ and $\exists ! x \in \mathbb{R} (x^2 + 3x + e = 0)$. This means that $\exists x \in \mathbb{R}(x^2 + 3x + d = 0)$, $\exists x \in \mathbb{R}(x^2 + 3x + e = 0)$, $\forall y \in \mathbb{R} \forall z \in \mathbb{R} ((y^2 + 3y +d = 0 \wedge z^2 + 3z + d = 0 )\rightarrow y =z)$, and $\forall y \in \mathbb{R} \forall z \in \mathbb{R} ((y^2 + 3y +e = 0 \wedge z^2 + 3z + e = 0 )\rightarrow y =z)$. (How can we show that $d = e$ to finish this?)
|
Let $P(X) = X^2 + 3 X + c$. It is a real polynomial of degree $2$, and its discriminant is $\Delta = 9 - 4c$. Then
\begin{align}
P \text{ has a unique root} &\iff \Delta=0 \\
&\iff c = \frac94
\end{align}
Thus, $c=\frac94$ is the unique number for which there is a unique solution $x$ for $P(x)=0$.
|
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|
Find rectangular Cartesian coordinate system Find rectangular Cartesian coordinate system in $\mathbb{R}^3$ to bring quadratic surface $2y^2-3z^2+4xz-12y+15=0 $ to standard form.
After splittig off square the equation can be rewritten as follows $2(y-3)^2-3z^2+4xz=3$.
What to do with mixed term $4xz$?
|
Just because it's kind of fun, here's a method using the Matrix Representations of Conic Sections. Essentially, for
$$ Q(x,z) = -3z^2 + 4xz$$
We can write
\begin{align}
Q(x,z) &= \begin{pmatrix}
x & z
\end{pmatrix}
\begin{pmatrix}
0 & 2\\
2 & -3
\end{pmatrix}
\begin{pmatrix}
x \\ z
\end{pmatrix}
\end{align}
We can obtain the linear transformation that eliminates the cross-term $4xz$ by diagonalizing the above matrix (or in general, the best we can do is the Jordan Normal Form).
In general, the normal form for a matrix $M$ looks like
$$ M = S J S^{-1}$$
where $J$ is a block-diagonal matrix (in this case, hopefully a completely diagonal matrix).
Wolfram Alpha leads to
\begin{align}
\underbrace{\begin{pmatrix}
0 & 2\\
2 & -3
\end{pmatrix}}_{M} &=
\underbrace{
\begin{pmatrix}
-1 & 2\\
2 & 1
\end{pmatrix}}_{S}
\underbrace{
\begin{pmatrix}
-4 & 0\\
0 & 1
\end{pmatrix}}_{J}
\underbrace{
\begin{pmatrix}
-1 & 2\\
2 & 1
\end{pmatrix}^{-1}}_{S^{-1}}\\
&=
\begin{pmatrix}
-1 & 2\\
2 & 1
\end{pmatrix}
\begin{pmatrix}
-4 & 0\\
0 & 1
\end{pmatrix}
\begin{pmatrix}
-1/5 & 2/5 \\
2/5 & 1/5
\end{pmatrix}
\end{align}
If we use the coordinate transformation
\begin{align}
\begin{pmatrix}
x' \\ z'
\end{pmatrix} &= S^{-1} \begin{pmatrix}
x \\ z
\end{pmatrix}\\
&= \begin{pmatrix}
-1/5 & 2/5 \\
2/5 & 1/5
\end{pmatrix}
\begin{pmatrix}
x \\ z
\end{pmatrix}\\
&= \begin{pmatrix}
(2x - z)/5\\
(2x + z)/5
\end{pmatrix}
\end{align}
To see this in action, notice also that
\begin{align}
\begin{pmatrix}
x \\ z
\end{pmatrix} &= S \begin{pmatrix}
x' \\ z'
\end{pmatrix}\\
&= \begin{pmatrix}
-1 & 2\\
2 & 1
\end{pmatrix}
\begin{pmatrix}
x' \\ z'
\end{pmatrix}\\
&= \begin{pmatrix}
2z' - x'\\
z' + 2x'
\end{pmatrix}
\end{align}
If we just let $y' = y - 3 \iff y = y' + 3$ as you've demonstrated, then subbing into the quadratic surface cartesian equation that you have gives,
\begin{align}
2y^2 - 3z^2 + 4xz - 12y + 15 &= 0\\
\Rightarrow 2(y' + 3)^2 - 3(z' + 2x')^2 + 4(2z' - x')(z' + 2x') - 12(y' + 3) + 15 &= 0\\
\Rightarrow -20x'^2 + 2y'^2 + 5z'^2 - 3 &= 0
\end{align}
Where we've used Wolfram Alpha to simplify the equation for us.
EDIT: Here are some fun pictures to along with this.
If we set $y = 3$ to ignore the $y$ dependence for a moment, we essentially have rotated from this hyperbola
to this hyperbola
by using the change of coordinates given by lines
\begin{align}
x &= (2x - z)/5\\
z &= (2x + z)/5
\end{align}
|
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|
If real $x$ and $y$ satisfy $x^2+y^2-4x+10y+20=0$, then prove that $y+7-3\sqrt{2}\le x\le y+7+3\sqrt{2}$
Let $x, y \in \mathbb{R}$ such that $x^2 + y^2 - 4x + 10y + 20 = 0$. Prove that
$$y + 7 - 3\sqrt{2} \le x \le y + 7 + 3\sqrt{2}$$
I'm struggling with that problem. I've recognized that the given equation is one of a circle, namely; $$(x-2)^2 + (y+5)^2 = 9$$
And what needs to be proven can be derived to the form: $$7-3\sqrt{2}\le x-y\le 7 + 3\sqrt{2} \quad\Leftrightarrow\quad (x-y)^2 - 14(x-y) + 31 \le 0$$
But I really can't find the correlation between what's given and what has to be proven! Also, I'm wondering if this problem can be generalized for any circle parameters, something like an universal interval of $x-y$ for any circle. Any help is apprecatiated! Thank you in advance!
|
Hint:
Let $x-y=c$
Use $x=y+c$ to form a Quadratic Equation in $y$
As $y$ is real, the discriminant must be $\ge0$
One may replace $y$ with $x-c$ as well.
|
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|
How do I find the sum of a power series $\underset{n=3}{\overset{\infty}{\sum}}\frac{x^n}{(n+1)!n\,3^{n-2}}$? I have found the area of convergence to be $ x \in (-\infty, \infty)$, and this is how far I had gotten before getting stuck:
$$
\begin{aligned}
\sum_{n=3}^{\infty} \frac{x^{n}}{(n+1) ! n 3^{n-2}} &=\sum_{k=0}^{\infty} \frac{x^{k+3}}{(k+4) !(k+3) 3^{k+1}} \\
&=\sum_{k=0}^{\infty} \frac{x^{3} x^{k}}{(k+4) !(k+3) 3 \cdot 3^{k}} \\
&=\frac{x^{3}}{3} \sum_{k=0}^{\infty} \frac{1}{(k+4) !(k+3)}\left(\frac{x}{3}\right)^{k} \\
\text{substituting }u=& \frac{x}{3} \\
&=\frac{x^{3}}{3} \sum_{k=0}^{\infty} \frac{1}{(k+4) !(k+3)} u^{k}
\end{aligned}$$
I do not know how to proceed from here.
|
Hint
Let $x=3y$ $$A=\underset{n=3}{\overset{\infty}{\sum}}\frac{x^n}{(n+1)!n\,3^{n-2}}=9\sum_{n=3}^\infty \frac{ y^n}{n (n+1)!}=9y\sum_{n=3}^\infty \frac{ y^{n-1}}{n (n+1)!}$$
$$\sum_{n=3}^\infty \frac{ y^{n-1}}{n (n+1)!}=\left(\sum_{n=3}^\infty \frac{ y^{n}}{ (n+1)!}\right)'=\left(\frac 1y\sum_{n=3}^\infty \frac{ y^{n+1}}{ (n+1)!}\right)'=\left(\frac 1y\sum_{n=2}^\infty \frac{ y^{n}}{ n!}\right)'$$
$$\sum_{n=2}^\infty \frac{ y^{n}}{ n!}=-1-y+\sum_{n=0}^\infty \frac{ y^{n}}{ n!}=e^y-y-1$$
|
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|
Solve $\sqrt{x^2+8x+7}+\sqrt{x^2+3x+2}=\sqrt{6x^2+19x+13}$ I tried squaring both sides but doesn't seem like a good idea.
$$x^2+8x+7+\sqrt{(x^2+8x+7)(x^2+3x+2)}+x^2+3x+2=6x^2+19x+13$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4x^2+8x+4$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$$
Is there a better way for solving this equation?
|
When you squared you forget to put the coefficient of 2.
You should have $2\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$
Or $\sqrt{(x^2+8x+7)(x^2+3x+2)}=2(x+1)^2$
....
But anyway....
Square again. But it'll help to factor components.
$(x^2+8x+7)(x^2+3x+2)=4(x+1)^4$
$(x+7)(x+1)(x+2)(x+1)=4(x+1)^4$.
If $x = -1$ we get $0=0$ which could be a solution although we fatored so late in the game we should check if that's extraneous.
Nope... way back in the beginning $x = -1$ would have given us $\sqrt 0 + \sqrt 0 =\sqrt 0$ which is fine.
SO $x =-1$ is a solution. Divide both sides by $(x+1)^2$ and we tet
$(x+7)(x+2)= 4(x+1)^2$ which is just a quadratic...
$x^2 + 9x + 14 = 4x^2 + 8x + 4$ so
$3x^2 -x -10=0$ which we can solve with the quadratic equation. (but well have to search for extraneous solutions.)
$x = \frac {1\pm \sqrt{1 + 120}}6 = \frac {1\pm 11}6= 2, -\frac 53$.
But $x =-\frac 53$ give $\sqrt {\frac {25}9 - \frac {40}3+7}= \sqrt{-\frac {32}9}$ which is not real.
So $x=2$ or $x=-1$
|
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|
Solve $(x+1)^2y + (y+1)^2x=0, x, y \in \mathbb{Z} $ Solve $$(x+1)^2y + (y+1)^2x=0, x, y \in \mathbb{Z} $$
I kind of know the only integer solutions are $\{0, 0\}, \{-1, -1\}$ but I don't know how to prove that
|
Rewrite it like this: $$xy(x+y)+4xy+x+y=0$$ and let $a=x+y$ so $$xy = -{a\over a+4}$$
then $a+4\mid a$ so $$a+4\mid (a+4)-a =4\implies a+4\in \{-4,-2,-1,1,2,4\}$$
So $a\in \{-8,-6,-5,-3,-2,0\}$
*
*If $a=0$ then $xy =0$ and $y=-x$ so $x=y=0$
*If $a=-2$ then $xy =1$ and $y=-x-2$ so $x=y=-1$
*If $a=-3$ then $xy =3$ and $y=-x-3$ and no solution
*If $a=-5$ then $xy =-5$ and $y=-x-5$ and no solution
*If $a=-6$ then $xy =-3$ and $y=-x-6$ and no solution
*If $a=-8$ then $xy =-2$ and $y=-x-8$ and no solution.
|
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|
Stokes's Theorem for the Cone Consider the vector field $$F = \biggl \langle \sin x-\frac{y^3}3, \cos y+\frac{x^3}3, xyz \biggr \rangle.$$ Let $S$ be the surface given by the cone $$z^2 = x^2 + y^2 \text{ for } 0 \leq z \leq1.$$
I would like to verify Stokes's Theorem for the surface of the cone over the region $S.$ I have computed the curl $$\nabla \times F = \langle xz, -yz, x^2+y^2 \rangle,$$ but I get two different answers when I choose $z^2 = x^2 + y^2$ and $z = \sqrt{x^2 + y^2}.$
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Recall that Stokes's Theorem says that $$\oint_{\partial S} \mathbf F \cdot d \mathbf r = \iint_S (\nabla \times \mathbf F) \, dS,$$ where $S$ is a parametrized surface; $\partial S$ is the boundary of $S$ parametrized by the curve $\mathbf r(t)$ on some closed interval $a \leq t \leq b;$ and $\nabla \times \mathbf F$ is the curl of the vector field $\mathbf F(x, y, z).$ Given that $$\mathbf F(x, y, z) = \biggl \langle \sin x - \frac{y^3}{3}, \cos y + \frac{x^3}{3}, xyz \biggr \rangle,$$ we find that $\nabla \times \mathbf F = \langle xz, -yz, x^2 + y^2 \rangle.$ Considering that $S$ is the cone $z^2 = x^2 + y^2$ for $0 \leq z \leq 1,$ we may parameterize $S$ by cylindrical coordinates $G(r, \theta) = \langle r \cos \theta, r \sin \theta, r \rangle$ over the region $U = [0, 1] \times [0, 2 \pi]$ in the $r \theta$-plane. We have that $G_r(r, \theta) = \langle \cos \theta, \sin \theta, 1 \rangle$ and $G_\theta(r, \theta) = \langle -r \sin \theta, r \cos \theta, 0 \rangle$ so that the normal vector to $S$ is given by $$N(r, \theta) = G_r(r, \theta) \times G_\theta(r, \theta) = \langle -r \cos \theta, -r \sin \theta, r \rangle.$$ Ultimately, we find that the right-hand side of the original displayed equation is $$\iint_S (\nabla \times F) \, dS = \int_0^{2 \pi} \int_0^1 \langle r^2 \cos \theta, -r^2 \sin \theta, r^2 \rangle \cdot \langle -r \cos \theta, -r \sin \theta, r \rangle \, dr \, d \theta$$ $$= \int_0^{2 \pi} \int_0^1 2r^3 \sin^2 \theta \, dr \, d \theta = \frac{\pi}{2}. \phantom{We did it! Ya!}$$
Considering the picture of the cone $S,$ observe that $\partial S = \{(x, y, z) \,|\, x^2 + y^2 = 1 \text{ and } z = 1\}.$ Using the usual polar coordinates, we find that $\mathbf r(t) = \langle \cos t, \sin t, 1 \rangle$ parametrizes $\partial S$ for $0 \leq t \leq 2 \pi.$ We have therefore that $\mathbf r'(t) = \langle -\sin t, \cos t, 0 \rangle$ so that $$\oint_{\partial S} \mathbf F \cdot d \mathbf r = \int_0^{2 \pi} \biggl \langle \sin(\cos t) - \frac{\sin^3 t}{3}, \cos(\sin t) + \frac{\cos^3 t}{3}, \sin t \cos t \biggr \rangle \cdot \langle -\sin t, \cos t, 0 \rangle \, dt \phantom{!!}$$ $$= \int_0^{2 \pi} \biggl(-\sin t \sin(\cos t) + \frac{\sin^4 t}{3} + \cos t \cos(\sin t) + \frac{\cos^4 t}{3} \biggr) \, dt = \frac{\pi}{2}.$$ (One can compute this integral using $u$-substitution for the terms $-\sin t \sin(\cos t)$ and $\cos t \cos(\sin t)$ and the identity $\sin^4 t + \cos^4 t = \frac{\cos(4t) + 3}{4}.$) We have verified Stokes's Theorem.
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"language": "en",
"url": "https://math.stackexchange.com/questions/3708556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$PA = LU$ decomposition Consider a matrix $A= \begin{pmatrix}
1 & 2 & 1\\
3 & 6 & 1\\
0 & 4 & 1
\end{pmatrix}$
I am applying the transformations on matrix $A$ to convert it to $U$ using the following matrices:
(The i,j in $E_{ij}$ denotes the element in matrix A which is fixed in order to transform it into $U$)
$E_{21} = \begin{pmatrix}
1 & 0 & 0\\
-3 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}$
$E_{31} = I_{3\times3}$
$P = \begin{pmatrix}
1 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0
\end{pmatrix}$
Now
$U = \begin{pmatrix}
1 & 2 & 1\\
0 & 4 & 1\\
0 & 0 & -2
\end{pmatrix}$
If i write everything in matrix notation: $PE_{31}E_{21}A = U$
How should i convert this into $PA = LU$?
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Basically the idea behind writing a Gaussian elimination outcome in the form $PA=LU$ is that you could have done all the swaps at the beginning and then just done the row scaling/addition afterward.
So here suppose you had swapped 2 and 3 initially. The scale/add step that you would have done to zero out the first column wouldn't be the same, because now what used to be row 2 is in row 3, so where you had $\begin{pmatrix}
1 & 0 & 0\\
-3 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}$ you would now have $\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
-3 & 0 & 1
\end{pmatrix}$. So you make these modifications to any $E$ matrices that were applied after the row swap (moving the off-diagonal entry to correspond to which rows were being combined in the matrix after all the row swaps).
So here specifically it will look like
$$\begin{pmatrix}
1 & 0 & 0\\
0 & 0 & 1\\
0 & 1 & 0
\end{pmatrix} A = \begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
3 & 0 & 1
\end{pmatrix} U$$
where we flipped the sign of the $-3$ in the $L$ when we inverted the $E$ matrix (as usual).
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"url": "https://math.stackexchange.com/questions/3711078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $\frac{1}{16} \sum \frac{(b+c)(c+a)}{ab} +\frac{9}{4} \geq 4\sum \frac{ab}{(b+c)(c+a)}$ For $a,b,c>0$. Prove: $$\frac{1}{16} \sum\limits_{cyc} {\frac { \left( b+c \right) \left( c+a \right) }{ba}}+\frac{9}{4} \geq 4\, \sum\limits_{cyc}{
\frac {ba}{ \left( b+c \right) \left( c+a \right) }}$$
My SOS's proof is:
It's equivalent to: $$\frac{1}{27}\sum\limits_{cyc} ab \left( a+b-8\,c \right) ^{2} \left( a+b-2\,c \right) ^{2}+\frac{26}{27}\sum\limits_{cyc}ab \left( a-b \right) ^{2} \left( a+b-2\,c \right) ^{2} +{\frac{50}{27}} \Big[\sum\limits_{cyc} a(b-c)^2\Big]^2 \geq 0$$
However, it's hard to find without computer.
So I'm looking for alternative solution without $uvw$. Thanks for a real lot!
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We need to prove that:
$$\sum_{cyc}\left(\frac{(a+c)(b+c)}{ab}-4\right)\geq16\sum_{cyc}\left(\frac{4ab}{(a+c)(b+c)}-1\right)$$ or
$$\sum_{cyc}(c^2+ac+bc-3ab)\left(\frac{1}{ab}+\frac{16}{(a+c)(b+c)}\right)\geq0$$ or
$$\sum_{cyc}((c-a)(3b+c)-(b-c)(2a+c))\left(\frac{1}{ab}+\frac{16}{(a+c)(b+c)}\right)\geq0$$ or
$$\sum_{cyc}(a-b)\left((3c+a)\left(\tfrac{1}{bc}+\tfrac{16}{(a+b)(a+c)}\right)-(3c+b)\left(\tfrac{1}{ac}+\tfrac{16}{(a+b)(b+c)}\right)\right)\geq0$$ or
$$\sum_{cyc}(a-b)^2(3(a+b)c^3+4(a^2-6ab+b^2)c^2+(a+b)(a^2+5ab+b^2)c+ab(a+b)^2)\geq0,$$ which is true because by AM-GM
$$3(a+b)c^3+4(a^2-6ab+b^2)c^2+(a+b)(a^2+5ab+b^2)c+ab(a+b)^2\geq$$
$$\geq6\sqrt{ab}c^3-16abc^2+14\sqrt{a^3b^3}c+4a^2b^2\geq0.$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/3713360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find $\int_0^{2\pi} \frac{x \cos x}{2 - \cos^2 x} dx$. I have to find the integral:
$$\int_0^{2\pi} \frac{x \cos x}{2 - \cos^ 2 x} dx$$
I rewrote it as:
$$\int_0^{2\pi} \frac{x \cos x}{1 + \sin^ 2 x} dx$$
But nothing further. I plugged it in a calculator and the result was $0$. I can see that the following relation holds:
$$f(-x) = -f(x)$$
For
$$ f: [0, 2\pi] \rightarrow \mathbb{R} \hspace{2cm} f(x) = \frac{x \cos x}{1 + \sin^2 x}$$
so that means that the function is an odd function. So if the interval $[0, 2\pi]$ is a symmetric interval for $f(x)$ then the result would be $0$.
I can see that the interval $[0, 2\pi]$ is symmetric for $\sin x$ and for $\cos x$, so it is not far fetched to believe it is symmetric for $\dfrac{\cos x}{1 + \sin^2 x}$, but wouldn't multiplying it with $x$ interfere with that symmetry? I don't see why $[0, 2\pi]$ is symmetric for the function
$$f(x) = \frac{x\cos x}{1 + \sin^2 x}$$
How come that $x$ doesn't ruin the symmetry?
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We'll use a trick to get rid of the $x$.
As you've noted, $\sin(2\pi -x) = \sin x$, and $\cos(2\pi - x) = \cos x$.
Replacing $x$ with $2 \pi - x$, we get
$$I = \int_0^{2\pi} \frac{x \cos x}{1+\sin^2x} \, dx = \int_0^{2\pi} \frac{(2 \pi - x) \cos x}{1+\sin^2x} \, dx.$$
Adding the integral to itself, we get
$$2I = \int_0^{2\pi} \frac{2\pi \cos x}{1+\sin^2x} \, dx.$$
From here, the substitution $u = \sin x$ finishes quickly, because $\frac{du}{dx} = \cos x$, giving
$$\int \frac{2\pi \cos x}{1+\sin^2x} \, dx = 2 \pi \int \frac{1}{1+u^2} \, du = 2\pi \tan^{-1}(u)+c,$$
and so applying the bounds of the integral we obtain the desired result.
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
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