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Integrating a Rational Function with a Square Root : $ \int \frac{dx}{\sqrt{x^2-3x-10}} $
Integrate $$ \int \frac{dx}{\sqrt{x^2-3x-10}}. $$
I started off with the substitution $\sqrt{x^2-3x-10} = (x-5)t$. To which I got
$$ x = \frac{2+5t^2}{t^2-1} \implies \sqrt{x^2-3x-10} = \left(\frac{2+5t^2}{t^2-1} - 5\right)t = \frac{7t}{t^2-1} $$
and
$$ dx = \frac{-14t}{(t^2-1)^2} dt. $$
Substituting back in
\begin{align*}
\int \frac{t^2-1}{7t} \frac{-14t}{(t^2-1)^2}dt &= -2 \int \frac{dt}{t^2-1} \\
&= - \ln\left| \frac{t-1}{t+1} \right| +c \\
&= - \ln\left| \frac{\frac{\sqrt{x^2-3x-10}}{x-5} - 1}{\frac{\sqrt{x^2-3x-10}}{x-5} + 1} \right| +c \\
&= - \ln\left| \frac{\sqrt{x^2-3x-10} - x + 5}{\sqrt{x^2-3x-10} + x - 5} \right| + c.
\end{align*}
Is this a valid answer? Wolfram Alpha's answer is slightly different. I know this can come from a difference in method and/or the functions differ by a constant, but with an expression like this it's hard to check.
|
See
$$- \ln\left| \frac{\sqrt{x^2-3x-10} - x + 5}{\sqrt{x^2-3x-10} + x - 5} \right| = \ln\left| \frac{\sqrt{x^2-3x-10} + x - 5}{\sqrt{x^2-3x-10} - x + 5} \right| $$
and
\begin{align*}
\frac{\sqrt{x^2-3x-10} + x - 5}{\sqrt{x^2-3x-10} - x + 5}
=&
\frac{\sqrt{x^2-3x-10} + (x - 5)}{\sqrt{x^2-3x-10} - (x - 5)}
\times
\frac{\sqrt{x^2-3x-10} + (x - 5)}{\sqrt{x^2-3x-10} + (x - 5)}
\\
=&
\frac{2x^2-13x+15+2(x - 5)\sqrt{x^2-3x-10} }{7(x - 5)}
\\
=&
\frac{2x-3+2\sqrt{x^2-3x-10} }{7}
\end{align*}
so
$$- \ln\left| \frac{\sqrt{x^2-3x-10} - x + 5}{\sqrt{x^2-3x-10} + x - 5} \right| = \ln|2x-3+2\sqrt{x^2-3x-10} |-\ln7$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The limit of a fraction What is the value of
$$\lim_{x\to\infty}\frac{\sqrt{x^4-1}}{2x^2+3x-1}$$
I tried to factorise the denominator or the numerator but it takes me nowhere. My teacher said it equals to $\frac{1}{2}$. But I don't get it. Can anyone help me?
|
$$\lim_{x\rightarrow+\infty}\frac{\sqrt{x^4-1}}{2x^2+3x-1}=\lim_{x\rightarrow+\infty}\frac{\sqrt{1-\frac{1}{x^4}}}{2+\frac{3}{x}-\frac{1}{x^2}}=\frac{1}{2}$$
Also
$$\lim_{x\rightarrow-\infty}\frac{\sqrt{x^4-1}}{2x^2+3x-1}=\lim_{x\rightarrow-\infty}\frac{\sqrt{1-\frac{1}{x^4}}}{2+\frac{3}{x}-\frac{1}{x^2}}=\frac{1}{2}$$
Thus, indeed, the answer is $\frac{1}{2}$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Why can complex numbers be written in exponential form? $z=r(\cos \theta+i\sin \theta)$ is $z=re^{i\theta}$. Why can complex numbers be written in exponential form? $z=r(\cos \theta+i\sin \theta)$ is $z=re^{i\theta}$.
I have studied that the exponential form of a complex number $z=r(\cos \theta+i\sin \theta)$ is $z=re^{i\theta}$.
Can someone explain why?
|
Lets consider a function from $\mathbb R\to \mathbb C$
$z(\theta) = \cos \theta + i\sin \theta\\
z(\theta)z(\phi) = (\cos \theta + i\sin \theta)(\cos \phi + i\sin \phi) = \cos(\theta + \phi) + i\sin (\theta+\phi) = z(\theta + \phi)$
That is a property of an exponential function. We do not know the base.
For some base:
$\exp (iy) = z(y) =\cos y + i\sin y$
and:
$\exp (x + iy) = \exp(x)\exp(iy) =\exp(x) (\cos y + i\sin y)$
And then you can define $e$ to be the required base. In much of complex analysis, it does not matter that it is the same $e$ as you have learned to be Euler's constant.
However, if you have taken calculus, you should recognize these Taylor expansions.
$e^x = \sum_\limits{n=0}^{\infty} \frac {x^n}{n!}\\
\cos x = \sum_\limits{n=0}^{\infty} \frac {(-1)^nx^{2n}}{(2n)!}\\
\sin x = \sum_\limits{n=0}^{\infty} \frac {(-1)^nx^{2n+1}}{(2n+1)!}$
what is
$e^{ix}$ ?
$e^{ix} = \sum_\limits{n=0}^{\infty} \frac {{ix}^n}{n!}\\
1 + ix + \frac {(ix)^2}{2} + \frac {(ix)^3}{3!}+ \frac {(ix)^4}{4!} \cdots\\
1 + ix + \frac {-x^2}{2} + \frac {-ix^3}{3!} + \frac {x^4}{4!} \cdots$
collect the real terms and the imaginary terms
$(1 - \frac {x^2}{2} + \frac {x^4}{4!}\cdots )+ i( x - \frac {x^3}{3!} + \frac {x^5}{5!} \cdots)\\
e^{ix} = \cos x + i\sin x$
Without calculus.
we can define
$e = \lim_\limits{n\to\infty}(1+\frac {1}{n})^n\\
e^x =\lim_\limits{n\to\infty} (1+\frac {1}{n})^{nx} $
Make a substitution $m = nx$
$e^x =\lim_\limits{m\to\infty} (1+\frac {x}{m})^m $
Then look at what happens as $m = 1, 2,3, etc.$
We have already shown that multiplication of complex numbers multiplies the lengths and adds the angles.
As $m$ increases hopefully you can see how that sequence of line segments begins to lie on the curve of the circle.
and when $m$ is very large comes to rest on $\cos x + i\sin x$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Are the vectors linearly dependent or independent? Problem: Let $\vec{v}=(5,9), \ \vec{u}=(3,-2)$ and $\vec{w}=(2,1)$. Determine the nature of their linear dependency.
Attempt: So, we are looking for constants $k_1,k_2,k_3$ such that $k_1\vec{v}+k_2\vec{u}+k_3\vec{w}=0.$ We can write this as
$$k_1\left[\begin{matrix} 5 \\ 9 \end{matrix}\right]+k_2\left[\begin{matrix} 3 \\ -2 \end{matrix}\right]+k_3\left[\begin{matrix} 2 \\ 1 \end{matrix}\right]=\left[\begin{matrix} 0 \\ 0 \end{matrix}\right]$$
which in canonical form is a system of linear equations in terms of $k_1,k_2$ and $k_3$.
$$
M\vec{k}=\left[ {\begin{array}{cc}
5 & 3 & 1 \\
9 & -2 & 1 \\
\end{array} } \right]\cdot \left[\begin{matrix} k_1 \\ k_2 \\ k_3 \end{matrix}\right]=\left[\begin{matrix} 0 \\ 0 \end{matrix}\right].$$
Row echelon form on $M$ gives
$$
\left[ {\begin{array}{cc}
5 & 3 & 1 \\
0 & -\frac{37}{5} & -\frac{4}{5} \\
\end{array} } \right],$$
This means that there are infinite solutions, but in order for them to be linearly independant, there should only exist one unique solution. Thus we have shown that the vectors $\vec{v},\vec{u},\vec{w}$ are linearly dependant. This means that $\text{span}\{\vec{v},\vec{u},\vec{w}\}=\mathbb{R}^2,$ because the addition of one of the vectors does not add another dimension to the span of the other two. The third vector that is linearly dependant of the other two, lies in their span.
Have I understood the concept of linear dependancy and span correctly? Any constructive input is very welcome!
|
You have made a mistake, it should be:
$$M\vec{k}=\left[ {\begin{array}{cc}
5 & 3 & 2 \\
9 & -2 & 1 \\
\end{array} } \right]\cdot \left[\begin{matrix} k_1 \\ k_2 \\ k_3 \end{matrix}\right]=\left[\begin{matrix} 0 \\ 0 \end{matrix}\right]$$
The solutions are of the form $\alpha\cdot \left[\begin{matrix} -7 \\ -13 \\ 37 \end{matrix}\right]$ for $\alpha \in \mathbb{R}$.
Thus, $$-7\vec{v}-13\vec{u}+37\vec{w} = 0$$
so $\{\vec{v}, \vec{u}, \vec{w}\}$ is linearly dependent in $\mathbb{R}^2$.
In fact, any three distinct vectors in a two-dimensional space must be linearly dependent.
Edit:
I found the solutions by reducing the matrix to row echelon form:
$$\left[ {\begin{array}{cc}
5 & 3 & 2 \\
9 & -2 & 1 \\
\end{array} } \right] \sim \left[ {\begin{array}{cc}
5 & 3 & 2 \\
-1 & -8 & -3 \\
\end{array} } \right] \sim \left[ {\begin{array}{cc}
1 & 8 & 3 \\
5 & 3 & 2 \\
\end{array} } \right] \sim \left[ {\begin{array}{cc}
1 & 8 & 3 \\
0 & -37 & -13 \\
\end{array} } \right] \sim \left[ {\begin{array}{cc}
1 & 0 & \frac{7}{37} \\
0 & 1 & \frac{13}{37} \\
\end{array} } \right]$$
The first row gives $k_1 = -\frac{7}{37}k_3$, and the second row gives $k_2 = -\frac{13}{37}k_3$.
Thus, the solutions are given by $k_3\cdot \left[\begin{matrix} -\frac{7}{37} \\ -\frac{13}{37} \\ 1 \end{matrix}\right]$ for $k_3 \in \mathbb{R}$, which we can also write as $$\alpha\cdot \left[\begin{matrix} -7 \\ -13 \\ 37 \end{matrix}\right] \text{ for } \alpha\in\mathbb{R}$$ by setting $k_3 = 37\alpha$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$ The problem is the following: For each $a \in \Bbb{Z}$ work out $\gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
This is what I have at the moment:
Let's call $d = \gcd(3^{16} \cdot 2a + 10, 3^{17} \cdot a + 66)$
Then $d \ \vert \ 3^{16} \cdot 2a + 10 \ \land d \ \vert \ 3^{17} \cdot a + 66$
$\Rightarrow d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 3 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 2$
$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 30 \ \land \ d \ \vert \ 3^{16} \cdot 6a + 132$
$\Rightarrow d \ \vert \ 3^{16} \cdot 6a + 132 - (3^{16} \cdot 6a + 30) \ = \ 102 \ =
\ 2 \cdot 3 \cdot 17$
Also, $d \ \vert \ (3^{16} \cdot 2a + 10) \cdot 33 \ \land \ d \ \vert \ (3^{17} \cdot a + 66) \cdot 5$
$\Rightarrow d \ \vert \ 3^{17} \cdot 17a$ (almost with the same method as before)
So I get $d \ \vert \ 102 \ \land d \ \vert \ 3^{17} \cdot 17a$
After this, I can't see how to continue.
|
The prime factorization of $102$ is $2 \times 3 \times 17$. When do $2$, $3$ and $17$ divide your numbers?
|
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|
Proof that $lim_{x->0,y->0}{\frac {\sqrt {a+x^2y^2} -1} {x^2+y^2}} (a>0)$ doesn't exist while a $\ne 1$. How to prove that $\lim_{x\to 0,y\to 0}{\frac {\sqrt {a+x^2y^2} -1} {x^2+y^2}} (a>0)$ doesn't exist while a $\ne 1$?
I already calculated that when a = 1 by multiplying $\sqrt {a+x^2y^2} + 1$ on both denominator and numerator and using the fact $x^2y^2<(x^2+y^2)^2/4$.
Any help will be appreciated.
|
Evaluating on the path $x=y$:
$$\lim_{x,y\to0}{\frac{\sqrt{a+x^2y^2}-1}{x^2+y^2}}=\lim_{x\to0}{\frac{\sqrt{a+x^4}-1}{2x^2}}=\lim_{x\to0}{\frac{\sqrt a-1}{2x^2}}$$
If $a\ne1$ then $\sqrt a-1\ne0$ and a singularity exists at $x=y=0$, so the limit does not exist.
|
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|
Three dice are rolled simultaneously. In how many different ways can the sum of the numbers appearing on the top faces of the dice be $9$?
Three dice are rolled simultaneously. In how many different ways can the sum of the numbers appearing on the top faces of the dice be $9$?
What I did:
I know that the maximum value on the dice can be $6$. So, restricting the values of dice, $6-x+6-y+6-z=9$, $x+y+z=9$, $n=9$ and $r=3$.
Applying partition again:
$$\binom{9+3-1}{3-1}=\binom{11}2=55$$
But the answer is $25$. Please, someone explain this one.
This is a gmat exam question.
|
We must find the number of solutions of the equation
$$x_1 + x_2 + x_3 = 9 \tag{1}$$
in the positive integers subject to the restrictions that $x_1, x_2, x_3 \leq 6$.
A particular solution of equation 1 in the positive integers corresponds to the placement of $3 - 1 = 2$ addition signs in the $9 - 1 = 8$ spaces between successive ones in a row of nine ones.
$$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$
For instance, placing addition signs in the third and seventh spaces yields
$$1 1 1 + 1 1 1 1 + 1 1$$
which corresponds to the solution $x_1 = 3$, $x_2 = 4$, $x_3 = 2$.
The number of solutions of equation 1 in the positive integers is
$$\binom{9 - 1}{3 - 1} = \binom{8}{2}$$
since we must choose which two of the eight spaces between successive ones to fill with addition signs.
However, we have included solutions in which some $x_i > 6$. Since the summands are positive integers, this can only occur if one of the $x_i$'s is $7$ and the other two $x_i$'s are $1$. There are three choices for which $x_i$ is equal to $7$. Hence, the number of ways the sum of the numbers on the three dice could add up to $9$ is
$$\binom{8}{2} - \binom{3}{1}$$
We can formalize the argument about the restrictions as follows. Suppose $x_1 > 6$. Then $x_1' = x_1 - 6$ is a positive integer. Substituting $x_1' + 6$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 6 + x_2 + x_3 & = 9\\
x_1' + x_2 + x_2 & = 3 \tag{2}
\end{align*}
Equation 2 is an equation in the positive integers with
$$\binom{3 - 1}{3 - 1} = \binom{2}{2}$$
solutions. By symmetry, the number of solutions of equation 1 in which $x_1 > 6$ is equal to the number of solutions in which $x_2 > 6$ and to the number of solutions in which $x_3 > 6$. Hence,
$$\binom{3}{1}\binom{2}{2}$$
solutions of equation 1 violate the restrictions. Therefore, the number of admissible solutions is
$$\binom{8}{2} - \binom{3}{1}\binom{2}{2}$$
|
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|
How to approach the divergence of $\sum\frac{n!}{(x-1)(x-2)...(x-n)}$ If $x>0$, prove that the following series is divergent:
$\sum\frac{n!}{(x-1)(x-2)\ldots(x-n)}$
where $n=1,2,3\ldots$
I have proved that the absolute values series is divergent, but I cannot establish an inequality between both series. Help me, please!
|
In another way
$$
\eqalign{
& t_n = {{n!} \over {\left( {x - 1} \right)\left( {x - 2} \right) \cdots \left( {x - n} \right)}} = {{n!} \over {\left( {x - 1} \right)^{\,\underline {\,n\,} } }} \cr
& {{t_{n + 1} } \over {t_n }} = {{n + 1} \over {\left( {x - n - 1} \right)}} = {x \over {x - n - 1}} - 1 \cr
& \left| {{{t_{n + 1} } \over {t_n }}} \right| = \left| {{x \over {x - n - 1}} - 1} \right| = \left| {{x \over {n + 1 - x}} + 1} \right| \ge 1\quad \left| \matrix{
\;0 < x \hfill \cr
\;x - 1 < n \hfill \cr} \right. \cr}
$$
Now note that
$$
\eqalign{
& 0 < \left| {{x \over {n + 1 - x}} + 1} \right| < 1\quad \Rightarrow \cr
& \Rightarrow \quad - 1 < {x \over {n + 1 - x}} + 1 < 1 \cr
& \Rightarrow \quad - 2 < {x \over {n + 1 - x}} < 0 \cr
& \Rightarrow \quad \left[ \matrix{
\left\{ \matrix{
0 < n + 1 - x \hfill \cr
- 2n - 2 + 2x < x < 0 \hfill \cr} \right. \hfill \cr
\quad \vee \hfill \cr
\left\{ \matrix{
n + 1 - x < 0 \hfill \cr
0 < x < - 2n - 2 + 2x \hfill \cr} \right. \hfill \cr} \right.\quad \Rightarrow \quad \left[ \matrix{
\left\{ \matrix{
x < n + 1 \hfill \cr
x < 0 \hfill \cr} \right. \hfill \cr
\quad \vee \hfill \cr
\left\{ \matrix{
n + 1 < x \hfill \cr
2\left( {n + 1} \right) < x < 2x \hfill \cr} \right. \hfill \cr} \right.\quad \Rightarrow \quad \left[ \matrix{
x < 0\left( { \wedge \;0 \le n} \right) \hfill \cr
\quad \vee \hfill \cr
\left( {2 \le } \right)2\left( {n + 1} \right) < x \hfill \cr} \right. \cr}
$$
and the second condition is just implying a limited sum.
So we have that for negative $x$ the sum converge, while it doesn't for positive $x$.
For example, for $x=-1$ we have
$$
\left. {t_n } \right|_{\,x = - 1} = {{n!} \over {\left( { - 2} \right)^{\,\underline {\,n\,} } }} = \left( { - 1} \right)^{\,n} {{n!} \over {2^{\,\overline {\,n\,} } }} = \left( { - 1} \right)^{\,n} {{n!} \over {1^{\,\overline {\,n + 1\,} } }} = \left( { - 1} \right)^{\,n} {1 \over {n + 1}}
$$
By using the definition of the Gamma function as the limit of the partial Gamma, we can have a better
overlook on the situation
$$
\eqalign{
& t_n = {{n!} \over {\left( {x - 1} \right)^{\,\underline {\,n\,} } }} = \left( { - 1} \right)^{\,n} {{n!} \over {\left( {1 - x} \right)^{\,\overline {\,n\,} } }} = \left( { - 1} \right)^{\,n} {{\left( { - x} \right)} \over {n^{\, - x} }}{{n^{\, - x} n!} \over {\left( { - x} \right)^{\,\overline {\,n + 1\,} } }} \cr
& \mathop {\lim }\limits_{n \to \infty } t_n = \mathop {\lim }\limits_{n \to \infty } \left( { - 1} \right)^{\,n} {{\left( { - x} \right)} \over {n^{\, - x} }}\Gamma ( - x) = \mathop {\lim }\limits_{n \to \infty } \left( { - 1} \right)^{\,n} n^{\,x} \,\Gamma ( - x + 1) \cr}
$$
|
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|
Find limit of trigonometric function $$\lim_{x\rightarrow0} \frac{\tan^3(3x)-\sin^3(3x)}{x^5}$$
I think it should be decomposed with $$\lim_{x\to0} \frac{\sin x}{x}=1$$ but I'm always getting indefinity $\frac{0}{0}$.
|
$$\begin{align}&\lim_{x \to 0} \dfrac{\tan^3 3x - \sin^3 3x}{x^5} &\\= &\lim_{x \to 0} (\sin^3 3x)\dfrac{1 - \cos^3 3x}{x^5}&\\=\, &27 \lim_{x \to 0} \dfrac{(1 - \cos 3x)}{x^2}(1 + \cos 3x + \cos^2 3x) &\\=\, &81 \lim_{x\to 0} \dfrac{(1 - \cos 3x)}{x^2}\end{align}$$
Let $x = 2y$
$$\lim_{y \to 0} \dfrac{(1 - \cos 6y)}{(2y)^2} = \lim_{y \to 0} \dfrac{(2\sin^2 3y)}{(2y)^2} = \lim_{y \to 0} \dfrac{(\sin^2 3y)}{2(y)^2} = \dfrac92$$
So, $$\lim_{x \to 0} \dfrac{\tan^3 3x - \sin^3 3x}{x^5} = \dfrac{729}2$$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Let $G$ be a group and $a,b \in G$ show that $|b| = 5$
Show that if $|a|=2$, $ab^2a^{-1}=b^3$ and $b\ne e$, then $|b|=5$.
My attempt
$$a^2 = e \implies a * a = e \implies a = a^{-1}$$
$$ab^2a=b^3$$
$$(ab^2a)^2=(b^3)^2$$
I) $$ab^2aab^2a = b^5$$
$$ab^2a=b^3$$
$$ab^2ab^2=b^3b^2$$
II) $$ab^2ab^2 = b^5$$
I and II$\implies$$b^4 = b^5 \implies b = e \implies b^5 = e$
Of course it is wrong, I've been trying to solve this problem for over 2 hours but I can't find a way to show $b^5=e$ through the given properties.
|
One has
$ab^2 = b^3a$, then $(ab^2)^2 = b^3ab^3a = b^3a(ab^3a)ba = b^6aba$
$b^5 = b^3b^2 = ab^2a^{-1}b^2=ab^2ab^2=(ab^2)^2 = b^6aba$.
So, $baba = e$, then $b^{-1} = aba$, or $b^{-2} = ab^2a = b^3$, then $b^5=e$.
|
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|
Complex quintic equation Given the equation $x^5=i$, I need to show by both algebraic and trigonometrical approaches that
$$\cos18^{\circ}=\frac{\sqrt{5+2\sqrt5}}{\sqrt[5]{176+80\sqrt5}}$$
$$\sin18^{\circ}=\dfrac1{\sqrt[5]{176+80\sqrt5}}$$
Trying by trigonometric approach,
$x^5$ = i $\;\;\;\;$ -- eqn. (a)
=> x = $i\sin(\dfrac{\pi}{2} +2k\pi)$ => $i\sin(\pi\dfrac{4k + 1}{2}) $
Taking the value of k=0, for getting the principal root of 18$^{\circ}$, have x = $i\sin(\dfrac{\pi}{10}) $
Solving algebraically, the solution approach is : $(a+bi)^5$ = i $\;\;\;\;$ -- eqn. (b)
=> $a^5 + 5ia^4b -10a^3b^2 -10ia^2b^3 +5ab^4 +ib^5$
Separating the real & imaginary parts:
$a^5 -10a^3b^2 +5ab^4=0$$\;\;\;$ -- eqn. (c); $\;\;\;\;$$5a^4b -10ia^2b^3+b^5=1$$\;\;\;$ -- eqn. (d)
Solving (c), we have : $a(a^4 -10a^2b^2 +5b^4)=0$$\;\;\;$ -- eqn. (c);
Either $a$ = $0$, or $(a^4 -10a^2b^2 +5b^4)=0$$\;\;$ -- eqn. (c'),
dividing both sides by $b^4$, and having c = a/b, $(c^4 -10c^2 +5)=0$$\;\;$ -- eqn. (c''),
having d = $c^2$, get : $(d^2 -10d +5)=0$$\;\;$ -- eqn. (c'''), with factors as : d =$5\pm 2\sqrt5$
finding value of c for the two values, get square roots of the two values for d.
//Unable to proceed any further with (c''').
Only root of significance, from eqn. (c) is $a = 0$.
Taking eqn.(d), and substituting $a = 0$, we get:$\;\;\;b^5$=1 => $b =1$
//Unable to prove any of the two values for $\sin18^{\circ}$, or $\cos18^{\circ}$
|
I think you're going at this backwards. Quickly, $\arg \mathrm{i} = \pi/2 = 90^\circ$, so fifth roots of $\mathrm{i}$ are points on the unit circle at small multiples of $18^\circ$. If the given values of cosine and sine are to be believed, then \begin{align*}
\mathrm{i} &= \left( \cos 18^\circ + \mathrm{i} \sin 18^\circ \right)^5 \\
&= \left( \frac{\sqrt{5+2\sqrt5}}{\sqrt[5]{176+80\sqrt5}} + \mathrm{i} \frac{1}{\sqrt[5]{176+80\sqrt5}} \right)^5 \text{.}
\end{align*}
Let $x$ be the expression we are raising to the fifth power. Then, squaring twice, (and leaving the denominator largely unchanged) \begin{align*}
x^2 &= \frac{2(2 + \sqrt{5}) + 2\mathrm{i}\sqrt{5+2\sqrt{5}}}{(\sqrt[5]{176+80\sqrt{5}})^2} \text{,} \\
x^4 &= \frac{8(2 + \sqrt{5}) + 8\mathrm{i}((2+\sqrt{5})\sqrt{5+2\sqrt{5}})}{(\sqrt[5]{176+80\sqrt{5}})^4} \text{, and} \\
x^5 = x^4 \cdot x &= \frac{16 \mathrm{i}(11+5\sqrt{5})}{176+80\sqrt{5}} = \mathrm{i} \text{.}
\end{align*}
This confirms that the given values of cosine and sine are a fifth root of $\mathrm{i}$. The other roots also have unit magnitude and arguments multiples of $360^\circ/5 = 72^\circ$ greater than the given root, so have arguments $90^\circ$, $162^\circ$, $234^\circ$, and $306^\circ$. We see that $x$ is the only one in the first quadrant, so is the desired root.
For a trigonometry method:
$$ 0 = \cos(90^\circ) = \cos(5 \cdot 18^\circ) = \cos( 2 \cdot 2 \cdot 18^\circ + 18^\circ) \text{.} $$
Then use sum of angles and half angle, twice. Do the same thing to sine. You'll get a pair of algebraic equations in $\sin 18^\circ$ and $\cos 18^\circ$. Then check that the given sine and cosine values actually satisfy them.
|
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|
Seeking the maximal parameter value s.t. two-variable inequality still holds Consider the expression
$$\frac 12\left(\frac{a^2}b+\frac{b^2}a\right)$$
in two variables $\,a,b\,$ residing in $\,\mathbb R^{>0}$.
The arithmetic mean $\,\frac{a+b}2\,$ is a lower bound for it
$\big[$ one has $\,a^2b\le(2a^3+b^3)/3\,$ by AM-GM, do the same with $\,ab^2$, then add and divide by $2ab\,$$\big]$,
but $\,\max\{a,b\}\,$ is not
$\big[$ choose $\,(a,b)=(3,2)\,$ for instance, the expression then evaluates to $\,2\tfrac{11}{12}\,\big]$.
Both arithmetic mean ($x=1$) and the maximum ($x=\infty$) are instances of the Hölder mean
$$\left(\frac{a^x+b^x}2\right)^{\frac 1x}\quad\text{with }\; x\,\in\,\{-\infty\}\cup\mathbb R\cup\{\infty\}$$
aka Power mean, known to be strictly increasing with $\,x\,$ if $\,a\ne b$, and my question is:
What is the maximal value of $\,x\,$ such that
$$\left(\frac{a^x+b^x}2\right)^{\frac 1x}\;\leq\;\frac 12\left(\frac{a^2}b+\frac{b^2}a\right)$$
holds for all $\,a,b>0\,$?
Let's pick up the specific "max counter-example" $\,(a,b)=(3,2)\,$.
The following plot screen-shot displays the zero, by courtesy of WolframAlpha:
Returning to the general case we can at least state that $\,x_{max}\geqslant 5\,$:
After taking the fifth power of the corresponding expression and clearing denominators (to arrive at the LHS as of below) I could find a
Certificate of positivity
$$\begin{eqnarray}
\left(a^3 + b^3\right)^5 -16a^5b^5\left(a^5 +b^5\right)\; & =\;\left(a+b\right)\left(a-b\right)^2\big[a^{12} + a^{11}b +2a^{10}b^2 +4a^9b^3 +8a^8b^4 \\
& +8a^6b^6 +8a^4b^8 +4a^3b^9 + 2a^2b^{10} +ab^{11} +b^{12}\big] \\
& + 3a^3b^3\left(a^2+b^2\right)\left(a+b\right)^3\left(a-b\right)^4
\end{eqnarray}$$
It shows also that the equality case holds iff $\,a=b$.
|
Here is a proof that the maximal $x$ must be $\le 9$.
We consider $a = 1 + u$, $b = 1$ and compute the Taylor development of
$$
f(u) = \frac 12 \left( (u+1)^2 + \frac{1}{u+1}\right) -
\left( \frac{(1+u)^x + 1}{2}\right)^\frac 1x
$$
at $u = 0$.
From the geometric series
we get
$$
\frac 12 \left( (u+1)^2 + \frac{1}{u+1}\right) =
1 + \frac 12 u + u^2 + O(u^3)
$$
and from the binomial series we get
$$
(1+u)^x + 1 = 2 + xu + \frac{x(x-1)}{2} u^2 + O(u^3)
$$
and then
$$
\left( \frac{(1+u)^x + 1}{2}\right)^\frac 1x = 1 + \frac 12 u + \frac{x-1}{8}u^2 + O(u^3)
$$
It follows that
$$
f(u) = \frac{9-x}{8} u^2 + O(u^3) \text{ for } u \to 0 \, .
$$
If $x > 9$ then $f(0) =f'(0) = 0$ and $f''(0) < 0$, i.e.
$f$ has a strict local maximum at $u = 0$, so that the desired
inequality does not hold for $a = 1+u$, $b=1$ with $u \ne 0$
sufficiently small.
|
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|
How to evaluate the sum : $\sum_{k=1}^{n} \frac{k}{k^4+1/4}$ I have been trying to figure out how to evaluate the following sum:
$$S_n=\sum_{k=1}^{n} \frac{k}{k^4+1/4}$$
In the problem, the value of $S_{10}$ was given as $\frac{220}{221}$.
I have tried partial decomposition, no where I go. Series only seems like it telescopes, otherwise there isn't another way.
Any ideas are appreciated!
|
By Sophie Germain's identity
$$ 4k^4+1 = (2k^2+2k+1)(2k^2-2k+1) \tag{A}$$
hence
$$ \frac{1}{2k^2-2k+1}-\frac{1}{2k^2+2k+1} = \frac{4k}{4k^4+1} = \frac{k}{k^4+1/4}\tag{B} $$
and we may notice that by setting $p(x)=2x^2-2x+1$ we have $p(x+1)=2x^2+2x+1$.
In particular
$$ \sum_{k=1}^{n}\frac{k}{k^4+1/4}=\sum_{k=1}^{n}\left(\frac{1}{p(k)}-\frac{1}{p(k+1)}\right) = \frac{1}{p(1)}-\frac{1}{p(n+1)}=1-\frac{1}{2n^2+2n+1} $$
equals $\frac{2n^2+2n}{2n^2+2n+1}$ for any $n\geq 1$.
Telescoping is not strictly necessary to be able to compute the value of similar series. For instance
$$ \sum_{k\geq 0}\frac{1}{k^4+4} = \frac{\pi\cos\pi+\sinh\pi}{8\sinh\pi}, $$
but this is a different story, related with Weierstrass products, the Poisson summation formula or the (inverse) Laplace transform.
|
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|
Finding the Smallest Convex Hull of an Adjacency Matrix Let's say I have a an adjacency matrix of a directed graph, with at most one 1 in each row (self-loops are allowed but are 1, not 2). My goal is to rearrange the matrix using only row and column swaps. If I treat the positive entries as points, I can shrink the size of the convex hull by putting empty columns and rows at the ends. But, I don't know if that admits the smallest cluster. How do I know if the convex hull is the smallest possible without using brute force?
Here is an example:
We have the 9X9 adjacency matrix,
\begin{Vmatrix}
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
\end{Vmatrix}
which with dots is:
\begin{Vmatrix}
& & & & \cdot & & & & \\
& & & & & \cdot & & & \\
\cdot & & & & & & & & \\
& & & & & \cdot & & & \\
& & & & & \cdot & & & \\
& & & & & & & & &\\
\cdot & & & & & & & & \\
\cdot & & & & & & & & \\
& & & \cdot & & & & & \\
\end{Vmatrix}
then rearranged is:
\begin{Vmatrix}
& \cdot & & & & & & & \\
& & \cdot & & & & & & \\
\cdot & & & & & & & & \\
& & \cdot & & & & & & \\
& & \cdot & & & & & & \\
\cdot & & & & & & & & \\
\cdot & & & & & & & & \\
& & & \cdot & & & & & \\
& & & & & & & & \\
\end{Vmatrix}
with the convex hull of the latter clearly having the smaller area.
|
This isn't a proof. I noticed that we can treat each row as an integer since it has at most one 1 in a row. The column position of the one gives us the value, with the zero row being 0. We can then sort numbers based on how many duplicates there are for that number. So, if we have 1,2,4,4,4,5,5,6,6,6 we can sort it to,
4,4,4,6,6,6,5,5,2,1. We then make the first value 1's, the next new value 2's, and so on. This would give us 1,1,1,2,2,2,3,3,4,5. Then we can do our first step backwards. I am not sure if this always works though.
|
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|
How to compute symmetrical determinant I'm learning of determinants and am trying to find a trick to compute this one
\begin{pmatrix}
2 & 1 & 1 & 1 & 1\\
1 & 3 & 1 & 1 & 1\\
1 & 1 & 4 & 1 & 1\\
1 & 1 & 1 & 5 & 1\\
1 & 1 & 1 & 1 & 6
\end{pmatrix}
I expanded it out and got $349$ but I feel there must be some trick to easily compute it.
|
There is a general pattern for this kind of matrices. Define
$$
A_n:=\mathbf 1_n+\operatorname{diag}(1,2,\dots,n)=\pmatrix{2&1&1&\dots &1&1\\
1&3&1&\dots&1&1\\
\vdots&\ddots&\ddots& & \vdots& \vdots\\
1&1&1&\dots &n&1\\
1&1&1&\dots &1&n+1},
$$
where $\mathbf 1_n$ is the $n\times n$ matrix whose entries are one.
View the last column as $\pmatrix{1\\1\\\vdots\\1\\1}+n\pmatrix{0\\0\\\vdots\\0\\1}$.
Denoting by $D_n$ the determinant of $A_n$, we get
$$
D_n=\det \pmatrix{2&1&1&\dots &1&1\\
1&3&1&\dots&1&1\\
\vdots&\ddots&\ddots& & \vdots& \vdots\\
1&1&1&\dots &n&1\\
1&1&1&\dots &1&1}+n\det\pmatrix{2&1&1&\dots &1&0\\
1&3&1&\dots&1&0\\
\vdots&\ddots&\ddots& & \vdots& \vdots\\
1&1&1&\dots &n&0\\
1&1&1&\dots &1&1}.
$$
The first determinant can be computed in the following way: subtract to each column the last one to be reduced to compute the determinant of an upper diagonal matrix. The second determinant is $D_{n-1}$ so we end up with the recurrence relation
$$
D_n=nD_{n-1}+(n-1)!.
$$
This can be solved by letting $a_n:=D_n/n!$: we get $a_n=a_{n-1}+1/n$ hence
$$D_n=\left(1+\sum_{j=1}^n\frac 1j\right)n!.$$
|
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|
Determining whether a vector is an element of the row space. I've been having trouble with determining whether a vector is an element of the row space. Specifically, for the matrix:
$$\begin{bmatrix}-2 & 6 & 7 & -1 \\ 7 & -3 & 0 & -3 \\ 8 & 0 & 6 & 7\end{bmatrix}$$
Whether the vector:
$$\begin{bmatrix} 2 \\ 1 \\ 3 \\ -2 \end{bmatrix}$$ is an element.
My first attempt was to row reduce the original matrix to:
$$\begin{bmatrix}1 & 0 & 0 & \frac{-13}{4} \\ 0 & 1 & 0 & -\frac{79}{12} \\ 0 & 0 & 0 & \frac{32}{7}\end{bmatrix}$$.
Then, using the fact that the row-spaces of row-equivalent matrices are equivalent, I equated:
$$\begin{bmatrix} 2 \\ 1 \\ 3 \\ -2 \end{bmatrix} = \alpha _1\begin{bmatrix} 1 \\ 0 \\ 0 \\ -\frac{13}{4} \end{bmatrix} + \alpha _2 \begin{bmatrix} 0 \\ 1 \\ 0 \\ -\frac{79}{12}\end{bmatrix} + \alpha _3 \begin{bmatrix} 0 \\ 0 \\ 1 \\ \frac{32}{7}\end{bmatrix}$$
You thus get $$ \alpha _1 = 2, \alpha _2 = 1, \alpha _3 = 3$$ which gives $$2\frac{-13}{4} + \frac{-79}{12} + 3\frac{32}{7} = -2$$ by equating the final rows of the vector.
But in fact, $$2\frac{-13}{4} + \frac{-79}{12} + 3\frac{32}{7} = \frac{53}{84}$$, and so the vector is not an element of the row space.
However, by using a different method, consider the matrix $$\begin{bmatrix}8 & 7 & -2 & 2 \\ 0 & -3 & 6 & 1 \\ 6 & 0 & 7 & 3 \\ 7 & -3 & -1 & -2\end{bmatrix}$$, which is the augmented matrix of the transpose of the original and the vector, and noting that it row-reduces to the identity matrix, you can saw that the system of equations is consistent and hence the vector is an element of the row space.
But clearly these cannot both be true. Where is the error in my thinking?
|
If your reduced row echelon form (including the augmented part) reduces to the identity matrix, it is not consistent.
For example $$\left[\begin{array}{c|c} 1 & 0 \\ 0 & 1\end{array}\right]$$
implies that $x=0$ and $\color{blue}{0=1}$ which is not consistent.
|
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|
Polynomial $P(x)$ with $P(x)\in\mathbb Q$ iff $x\in \mathbb Q$
Find all the polynomials $P \in \mathbb{R}[X]$ such that: $P(x) \in \mathbb{Q}$ if and only if $x \in \mathbb{Q}$.
I think $P(x)=x+c$ or $P(x)=-x+c$ where $c$ is some rational constant. But I have no idea approach this.
|
Partial Answer: We can show that $P(x)$ can't be a quadratic with rational coefficients. In fact, $P(x)$ can't be a quadratic at all.
Let $P(x)=ax^2+bx+c$, for $a,b,c\in\mathbb{Q}$. Then $x=\frac{-b+\sqrt{m^2+1}}{2a}$, where $m\in\mathbb{Z}$, will be an irrational $x$ with a rational $P(x)$.
Constant coefficient, and Non-zero Coefficients
Clearly, any constant term, $c$, in the polynomial would need to be rational, since $P(0)=a\cdot0^2+b\cdot0+c=c$. We can set $c=0$ since it wouldn't affect whether a certain $P$ fits the criteria. Similarly, we can generate infinite families of polynomials by choosing different $c\in\mathbb{Q}$.
If $a=0$, then $P(x)=bx$, but we already know that linear equations fit our criteria. If $b=0$, then $P(x)=ax^2$ gives a rational value of $P(\sqrt{2})$ so $b\neq0$. Furthermore, if $a=b=0$, then $P(x)=0$ which is trivially rational for all irrational $x$. Hence, $a,b\neq0$ for all $P$ (not just quadratics).
Construction of $x=\frac{-b+\sqrt{m^2+1}}{2a}$
For any $a,b$, we can construct an irrational $x$ such that $P(x)$ is rational. The quadratic formula tells us that $x=\frac{-b\pm\sqrt{b^2-4a(-P(x))}}{2a}$. For $x$ to be irrational, $\sqrt{b^2-4a(-P(x))}$, must also be irrational. So we can choose an $m\in\mathbb{Z}$, $m\neq0$, such that $\sqrt{b^2-4a(-P(x))}=\sqrt{m^2+1}$, where the right-hand-side is irrational since $m^2+1$ can't be a square number. Rearranging this, we get $P(x)=\frac{m^2-b^2+1}{4a}$. We can also substitute $m$ into the expression for $x$ as $x=\frac{-b\pm\sqrt{m^2+1}}{2a}$.
Verification that $P(x)\in\mathbb{Q}$
We can verify this: when we compute $P(x)$, we get $$\begin{aligned}
P\left(\frac{-b+\sqrt{m^2+1}}{2a}\right)
&=a\left(\frac{-b+\sqrt{m^2+1}}{2a}\right)^2+b\left(\frac{-b+\sqrt{m^2+1}}{2a}\right)
\\&=\frac{1}{4a^2}\left(a(b^2-2b\sqrt{m^2+1}+c^2+1)+2ab(-b+\sqrt{m^2+1})\right)
\\&=\frac{1}{4a^2}\left(ab^2-2ab\sqrt{m^2+1}+a(m^2+1)-2ab^2+2ab\sqrt{m^2+1}\right)
\\&=\frac{a(m^2+1)-ab^2}{4a^2}
\\&=\frac{m^2-b^2+1}{4a}\end{aligned}$$.
Further Results
Combining this answer with @pisco125's Lemma 1, we see that $P(x)$ must have rational coefficients, meaning that it can't be a quadratic at all.
|
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|
Proof that $\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$. I was trying to count the number of equilateral triangles with vertices in an regular triangular array of points with n rows. After putting the first few rows into OEIS, I saw that this was described by A000332: $\binom{n}{4} = n(n-1)(n-2)(n-3)/24$.
A000332 has the comment:
Also the number of equilateral triangles with vertices in an equilateral triangular array of points with n rows (offset 1), with any orientation.
The linked solution is insightful, but the proof is very mechanical.
I tried to write an inductive proof, but I'm unable to come up with a nice way to prove the final identity (in particular for $n \geq 4$):
$$\binom{n + 3}{4} = n + 3 \binom{n + 2}{4} - 3 \binom{n + 1}{4} + \binom{n}{4}$$.
Wolfram|Alpha admits that these are indeed equal, and this could certainly be shown by writing everything as a polynomial.
However, I was hoping to find some nice combinatorial identities that give some intuition as to why this equality holds.
|
This is calculus of finite differences:
$$\binom{n+1}2-\binom{n}2=\binom{n}1=n,$$
$$\binom{n+1}3-\binom{n}3=\binom{n}2,$$
$$\binom{n+1}4-\binom{n}4=\binom{n}3,$$
etc. Then
$$n=\binom{n+1}2-\binom{n}2
=\left(\binom{n+2}3-\binom{n+1}3\right)
-\left(\binom{n+1}3-\binom{n}3\right)
=\binom{n+2}3-2\binom{n+1}3+\binom{n}3
=\left(\binom{n+3}4-\binom{n+2}4\right)
-2\left(\binom{n+2}4-\binom{n+1}4\right)
+\left(\binom{n+1}4-\binom{n}4\right)
=\binom{n+3}4-3\binom{n+2}4+3\binom{n+1}4-\binom{n}4
$$
etc.
|
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|
If we know range of a function , how can construct range of other function?
If we know $\frac{-1}{2}\leq \frac{x}{x^2+1}\leq \frac{1}{2}$ how can obtain the range of the function below?
$$y=\frac{4 x}{9x^2+25}$$
The problem is what's the range with respect to $\frac{-1}{2}\leq \frac{x}{x^2+1}\leq \frac{1}{2}$
|
Write
$$\frac{4x}{9x^2+25}=\frac{4}{9x+\frac{25}{x}}$$ and use AM-GM in two cases:
1) $x>0$:
$$y\leq\frac{4}{2\sqrt{9x\cdot\frac{25}{x}}}=\frac{2}{15}.$$
2) $x<0$: $$y\geq-\frac{4}{2\sqrt{(-9x)\cdot\left(-\frac{25}{x}\right)}}=-\frac{2}{15}.$$
The equality occurs for $9x=\frac{25}{x}$ and since we have a continuous function,
we got the answer:
$$\left[-\frac{2}{15},\frac{2}{15}\right]$$
|
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|
Proof that the square of an odd number is of the form $8m+1$ for some integer $m$
Show that $n$ is odd $\rightarrow \exists m \in \mathbb Z,n^2 = 8m + 1$
Let $k$ be an integer so that $n = 2k+1$. Then $n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2\left(2k^2+2k\right) + 1$.
But this isn't equal to $8m+1$, so can I change that into $4(2m)+1$?
|
$$n^2 = (2k+1)^2 = 4k^2+4k+1 = 4k(k+1)+1$$
At least one of $k$ and $k+1$ is divisible by $2$, so let $k(k+1)=2m$. Then, $n^2 = 4(2m)+1 = 8m+1$.
|
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|
Basis for Kernel and Image of a linear map I have to find the basis for the kernel and image of the following linear map.
$ \phi: R^3 → R^2, ϕ \begin{pmatrix} \begin{pmatrix} x \\y \\z \end{pmatrix}\end{pmatrix}= \begin{pmatrix} x -y \\z \end{pmatrix} $
For the range, I think we can express any arbitrary linear transformation as:
$ x\begin{pmatrix} 1 \\ 0 \end{pmatrix} - y\begin{pmatrix} 1 \\ 0 \end{pmatrix} + z\begin{pmatrix} 0 \\ 1 \end{pmatrix} $. So I think that a basis for the range would be $\left\{{{\begin{pmatrix} 1 \\ 0 \end{pmatrix}},{\begin{pmatrix} 0 \\ 1 \end{pmatrix}}}\right\}$
As for the kernel, we set
$\begin{pmatrix} x -y \\z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
We get $x=y$ and $z=0$. Therefore a basis for the kernel would be
$\left\{\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\right\}$
I am doing this right?
Thanks in advance
|
Your calculations are correct.
|
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|
Find all n for which $2^n + 3^n$ is divisible by $7$? I am trying to do this with mods. I know that:
$2^{3k+1} \equiv 1 \pmod 7$, $2^{3k+2} \equiv 4 \pmod 7$, $2^{3k} \equiv 6 \pmod 7$ and $3^{3k+1} \equiv 3 \pmod 7$, $3^{3k+2} \equiv 2 \pmod 7$, $3^{3k} \equiv 6 \pmod 7$, so I thought that the answer would be when $n$ is a multiple of $3$ since then $2^{3k} + 3^{3k} \equiv 1 + 6 \pmod 7$, but it doesn't work for $n = 6$
|
Since $(x-2)(x-3)=x^2-5x+6$, we get that
$$
a_n=5a_{n-1}-6a_{n-2}
$$
is satisfied by $a_n=2^n+3^n$. Since $a_0\equiv2\pmod7$ and $a_1\equiv5\pmod7$, we get the following sequence mod $7$:
$$
\color{#C00}{2},\color{#C00}{5},6,\color{#090}{0},6,2,\color{#C00}{2},\color{#C00}{5},
$$
Thus, the sequence mod $7$ has period $6$. Therefore,
$$
n\equiv3\pmod6\iff2^n+3^n\equiv0\pmod7
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Then find the value of ... Let $a, \: b, \: c$ be three variables which can take any value (Real or Complex). Given that $ab + bc + ca = \frac{1}{2}$; $a + b + c = 2$; $abc = 4$. Then find the value of $$\frac{1}{ab + c - 1} + \frac{1}{bc + a - 1} + \frac{1}{ac + b - 1}$$
I always get stuck in such problems. Please give a detailed solution and also give me an approach that I can use to solve other such kind of problems.
|
Hint:
Let $y=\dfrac1{ab+c-1}$
$\implies y=\dfrac 1{\dfrac12--bc-ca+c-1}=\dfrac2{2c(1-a-b)-1}=\dfrac2{2c(c-1)-1}=\dfrac2{2c^2-2c-1} $
$$\iff2y c^2-2y c-(y+2)=0\ \ \ \ (1)$$
Again $y=\dfrac1{ab+c-1}=\dfrac c{4+c^2-c}$
$$\iff c^2y-c(y+1)+4y=0\ \ \ \ (2)$$
Solve $(1),(2)$ for $c,c^2$ and use $c^2=(c)^2$ to form a cubic equation in $y$
Now apply Vieta's formula
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Binomial theorem relating proof There is this identity
$$1 -\frac{1}{2}\binom{n}{1}+\frac{1}{3} \binom{n}{2}- \frac{1}{4}\binom{n}{3}+....+(-1)^n \frac{1}{n+1}\binom{n}{n}$$
And we are supposed to prove it using these two identities
$$k\binom{n}{k} = n\binom{n-1}{k-1}$$
and
$$\binom{n}{0} + \binom{n}{1} + \binom{n}{2} +....+ \binom{n}{n} = 2^n$$
I have been working on this problem for a long time. Can you guys help me?
|
\begin{eqnarray*}
\frac{1}{i+1}= \int_0^1 x^i dx
\end{eqnarray*}
Sub this into the sum & interchange the order of the sum & the integral
\begin{eqnarray*}
\sum_{i=0}^{n} (-1)^i\binom{n}{i} \frac{1}{i+1}&=& \int_0^1 \sum_{i=0}^{n} (-1)^i\binom{n}{i}x^i dx \\
&=& \int_0^1 (1-x)^n dx \\
&=& \left[ \frac{-(1-x)^n}{n+1} \right]^1_0 \\
&=& \color{blue}{ \frac{1}{n+1}} \\
\end{eqnarray*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve equation with a conjugate How to solve this equation:
$$
|z|^2 - 9 - \bar z + 3i = 0
$$
I know that:
$$
|z|^2=z \bar z
$$
$$
|z|=\sqrt{a^2+b^2}
$$
$$
z = a + bi \Rightarrow \bar z = a - bi
$$
But I still don't know to calculate z from that.
|
Be $z=x+iy$ so $|z|^2=x^2+y^2$, $\bar z=x-iy$. Replacing in the equation:$$|z|^2-9-\bar z+3i=0$$ $$x^2+y^2-9-x+iy+3i=0$$
Reagrouping,$$(x^2+y^2-9-x)+i(y+3)=0$$
So $Re(z)=0$ and $Im(z)=0$,$$y+3=0$$$$x^2+y^2-9-x=0$$
Solving for $y$ gives $y=-3$. Replacing in the other equation:
$$x^2+(-3)^2-9-x=0$$
$$x^2-x=0$$
Gives solutions to $x$, $x=0$ $x=1$.
So you have to solutions for z.$z=-3i$ and $z=1-3i$
|
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|
Recurrence relation for the number of codewords of length $n$ from {0, 1, 2} with no two $0$’s appearing consecutively I'm supposed to use recurrence relations to find the number of codewords $b_n$ of length $n$ from the alphabet $\{0, 1, 2\}$ such that no two $0$’s appear consecutively.
After reading about this online, I feel that I should do this as follows:
Let $c_n$ be the number of codewords that start with $0$, $d_n$ be the number of codewords that start with $1$, and $e_n$ be the number of codewords that start with $2$ (all with length $n$). Then
$$b_n = c_n + d_n + e_n$$
where $c_n = d_{n-1}+e_{n-1}$, $\ d_n = c_{n-1}+d_{n-1}+e_{n-1}$, and $e_n = c_{n-1}+d_{n-1}+e_{n-1}$ with $c_1=d_1=e_1=1$.
Is this method correct? I am having quite a bit of trouble with recurrence relations.
|
The simple answer is that given a codeword of length $n-1$ we can add $10$ or $20$ to the end. Given a code word of length $n$ we can add $1$ or $2$ to the end. All code words of length $n+1$ can be gotten this way uniquely. So: $$b_{n+1}=2b_n+2b_{n-1}$$
There is a nice nerdy general approach to this sort of problem.
Let $u_n$ be the number of codewords of length $n$ than end in $0$ and $v_n$ the number of codewords of length $n$ that do not end in $0$.
Now, $v_{n+1}=2v_n+2u_n$ - we can add $1,2$ to the end of any sequence of length $n$ And $u_{n+1}=v_n$, because we can add $0$ to any sequence that did not end in $0$.
$$\begin{pmatrix}u_{n+1}\\v_{n+1}\end{pmatrix}=\begin{pmatrix}0&1\\2&2\end{pmatrix}\begin{pmatrix}u_n\\v_n\end{pmatrix}$$
Thus, inductively:
$$\begin{pmatrix}u_{n}\\v_{n}\end{pmatrix}=\begin{pmatrix}0&1\\2&2\end{pmatrix}^n\begin{pmatrix}0\\1\end{pmatrix}
$$
Now, $A=\begin{pmatrix}0&1\\2&2\end{pmatrix}$ has characteristic polynomial $x^2-2x-2$, which means $A^2=2A+2I$, and hence you get:
$$\begin{pmatrix}u_{n+1}\\v_{n+1}\end{pmatrix}=A^{n+1}\begin{pmatrix}0\\1\end{pmatrix}=(2A+2I)A^{n-1}\begin{pmatrix}0\\1\end{pmatrix}=2\begin{pmatrix}u_{n}\\v_{n}\end{pmatrix}+2\begin{pmatrix}u_{n-1}\\v_{n-1}\end{pmatrix}$$
and therefore, since we are looking for $b_n=u_n+v_n=\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}u_{n}\\v_{n}\end{pmatrix}$, we get:
$$b_{n+1}=2b_{n}+2b_{n-1}.$$
More generally, if $W_n = BA^nC$ for compatible matrices $B,A,C$, where $A$ is an $m\times m$ matrix, then $W_n$ satisfies:
$$W_{n+m}+a_{m-1}W_{n+m-1}+a_{m-2}W_{n+m-2}+\cdots +a_0W_{n}=0$$
where $x^m+a_{m-1}x^{m-1}+\cdots+a_0$ is the characteristic polynomial of $A$.
For example if you have an alphabet $\Sigma$ with $m$ letters and restrictions on which letters can follow which other letters, you get $A=(a_{ij})$ for $i,j\in \Sigma$ such that $a_{ij}=1$ if $j$ is allowed to follow $i$ and $0$ otherwise. Then:
$$b_n=\begin{pmatrix}1&\cdots&1\end{pmatrix}A^{n-1}\begin{pmatrix}1\\\vdots\\1\end{pmatrix}$$
Therefore, the given the characteristic polynomial for $A$ (or even the minimal polynomial for $A$) you can determine the linear recursion.
In your case $A=\begin{pmatrix}0&1&1\\1&1&1\\1&1&1\end{pmatrix}.$ This $A$ has characteristic polynomial $x^3-2x^2-2x$, but the factor of $x$ is irrelevant because you still get (for $n>0$) that $A^{n+2}=(2A^2+2A)A^{n-1}=(2A+2I)A^{n}$.
|
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|
Solve $2\log_bx + 2\log_b(1-x) = 4$ I need to solve $$2\log_bx + 2\log_b(1-x) = 4.$$
I have found two ways to solve the problem. The first (and easiest) way is to divide through by $2$:
$$\log_bx + \log_b(1-x) = 2.$$ Then, combine the left side: $$\log_b[x(1-x)] = 2,$$ and convert to the equivalent exponential form, $$b^2 = x(1-x) \;\;\implies\;\; x^2 - x + b^2 = 0,$$ which by the quadratic equation, we get $$x = {1\over2}\left(1\pm\sqrt{1 - 4b^2}\right).$$
My question is: How do I know that neither solution is extraneous? These are the solutions in the back of the textbook, but I am left questioning when these are the solutions.
The alternative solution involves some more clever thinking:
$$\begin{align}2\log_bx + 2\log_b(1-x) &= 4\\\log_bx^2(1-x)^2 &= 4\\ b^4 &= x^2(x^2 - 2x + 1)\\ 0&=x^4 - 2x^3 + x^2 - b^4\\ &=x^2(x - 1)^2 - b^4\\ & = [x(x-1)]^2 - (b^2)^2 \\ &= [x(x-1) + b^2][x(x-1) - b^2],\end{align}$$ which implies that $$x = {1\over2}\left(1 \pm\sqrt{1 + 4b^2}\right) \;\;\;\text{or} \;\;\; {1\over2}\left(1 \pm\sqrt{1 - 4b^2}\right),$$
which is even worse, because now there are $4$ solutions to check.
The most I know is that since $b$ is a logarithmic base, $b>0$ and $b\ne 1$. But what happens when $b$ is something like $2$? Then you end up with a complex number under the root (of the solutions from the easier way), and that doesn't necessarily make sense if I'm trying to solve the logarithmic equation over the reals.
|
You have to see that your equation $b^2=x(1-x)$ it has two unknowns actually so there are more than one solution. In fact you have the equation of the circle
$$(x-\frac12)^2+y^2=(\frac12)^2$$
The conclusion is that all the points $(x,b)$ in this semi-circle with $b$ positive are solutions of your problem.
|
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|
Proofs Involving Real Numbers We have the following rules that must be used in solving the question
*
*$a>0$ and $b>c$ $\implies$ $ab>ac$
*$a<0$ and $b>c$ $\implies$ $ab<ac$
*$a>b$ and $b>c$ $\implies$ $a>c$
Prove that the following is true: $\forall x,y \in \mathbb{R}$ is true that
$$ x>y \implies x^3>y^3$$
Distinguish the cases according to the signs of $x$ and $y$.
I know that there are 4 such cases.
*
*$+x$ and $+y$
*$+x$ and $-y$
*$-x$ and $+y$
*$-x$ and $-y$
I have already ruled out that (3) is not possible as this would contradict the whole proof. But after this, I do not know where to proceed.
|
Another way:
$\begin{array}\\
x^3-y^3
&=(x-y)(x^2+xy+y^2)\\
&=\frac12(x-y)(2x^2+2xy+2y^2)\\
&=\frac12(x-y)(x^2+2xy+y^2+x^2+y^2)\\
&=\frac12(x-y)((x+y)^2+x^2+y^2)\\
\end{array}
$
so
$x^3-y^3$
has the same sign as
$x-y$.
I prefer this
rewriting of
$x^2+xy+y^2$
as
$\frac12((x+y)^2+x^2+y^2)$
(rather than
$(x+y/2)^2+3y^2/4$)
because it is
symmetrical in
$x$ and $y$
like
$x^2+xy+y^2$.
|
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|
Finding the value of the definite integral $\int_0^2{x\int_x^2{\frac{dy}{\sqrt{1+y^3}}}}dx$ If $$f(x) = \int_x^2{\frac{dy}{\sqrt{1+y^3}}}$$
then find the value of $$\int_0^2{xf(x)}dx$$
I have no idea how to solve this question. Please help.
|
The integration domain can be equivalently written as
$$
\Omega = \{(x,y): x<y<2 ~~\mbox{and}~~~ 0 < x < 2 \}
$$
or
$$
\Omega = \{(x,y): 0<x<y ~~\mbox{and}~~~ 0 < y < 2 \}
$$
Such that
\begin{eqnarray}
\int_0^2{\rm d}x\int_{x}^2{\rm d}y ~\frac{x}{\sqrt{1 + y^3}} &=& \int_0^2{\rm d}y\int_{0}^y{\rm d}x ~\frac{x}{\sqrt{1 + y^3}} \\
&=& \int_0^2{\rm d}y ~\frac{y^2}{2}\frac{1}{\sqrt{1 + y^3}} \\
&=& \frac{1}{2} \int_0^2{\rm d}y \frac{y^2}{\sqrt{1 + y^3}} \\
&=&\frac{1}{2}\times\frac{4}{3} = \frac{2}{3}
\end{eqnarray}
|
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|
Finding integral solutions of $x+y=x^2-xy+y^2$
Find integral solutions of $$x+y=x^2-xy+y^2$$
I simplified the equation down to
$$(x+y)^2 = x^3 + y^3$$
And hence found out solutions $(0,1), (1,0), (1,2), (2,1), (2,2)$ but I dont think my approach is correct . Is further simplification required? Is there any other method to solve this? I am thankful to those who answer!
|
Write $\Delta\geq0$.
It must help!
$$x^2-(y+1)x+y^2-y=0,$$
which gives $$(y+1)^2-4(y^2-y)\geq0$$ or
$$3y^2-6y-1\leq0$$ or
$$1-\frac{2}{\sqrt3}\leq y\leq1+\frac{2}{\sqrt3},$$
which gives $$0\leq y\leq2,$$
which gives all solutions:
For $y=0$ we get $x^2-x=0$, which gives $(0,0)$ and $(1,0)$.
For $y=1$ we get $x^2-2x=0$, which gives $(0,1)$ and $(2,1)$.
For $y=2$ we get $x^2-3x+2=0$, which gives $(2,2)$ and $(1,2).$
|
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|
Verification of epsilon-delta proof I'd like to write an epsilon-delta proof that $$\lim_{x\to0} \frac{1}{x^2-1} = -1$$
My strategy in doing the proof was to choose $\delta < 1$ and then bound the $|x+1|$ and $x^2$ in $$ \left|\frac{1}{x^2-1} + 1\right| = \frac{x^2}{|x+1||x-1|}$$ If my math is correct, this yields $\frac{2x^2}{|x+1||x-1|} < 2\epsilon$. Below is the proof proper; I would be grateful if someone could verify if it's valid.
"Choose $\epsilon > 0$, and let $\delta = \min({1,2\epsilon})$. Then $0 < |x| < \delta$ implies that $|\frac{1}{x^2 - 1} + 1| < \epsilon$."
|
Let $\varepsilon>0$ and $x\in (-1,1)$
$$\left|\frac{1}{x^2-1} + 1\right| <\varepsilon\iff \left|\frac{x^2}{x^2-1} \right|<\varepsilon \iff \left|-\frac{x^2}{1-x^2} \right|<\varepsilon \iff \frac{x^2}{1-x^2}<\varepsilon $$
$\iff x^2<\varepsilon -x^2\varepsilon \iff x^2(1+\varepsilon)<\varepsilon\iff |x|<\sqrt{\dfrac{\varepsilon}{1+\varepsilon}}$
So $|x|<\sqrt{\dfrac{\varepsilon}{1+\varepsilon}} \implies \left|\dfrac{1}{x^2-1} + 1\right| <\varepsilon$
thus $\delta =\min\bigg(1,\sqrt{\dfrac{\varepsilon}{1+\varepsilon}} \bigg)=\sqrt{\dfrac{\varepsilon}{1+\varepsilon}}$
|
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|
How do you simplify an expression involving fourth and higher order trigonometric functions? The problem is as follows:
Which value of $K$ has to be in order that $R$ becomes independent from $\alpha$?.
$$R=\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha )$$
So far I've only come up with the idea that the solution may involve $R=0$, therefore
$$\sin^6\alpha +\cos^6\alpha +K(\sin^4\alpha +\cos^4\alpha)=0$$
as a result the expression becomes $0$ thus independent from $\alpha$, however the result is like this
$$-K=\frac{\sin^6\alpha +\cos^6\alpha}{\sin^4\alpha +\cos^4\alpha}$$
I am not sure if this is the right way.
Moreover, how can I simplify this expression, as it has order four and six?
|
Recall that
$$\sin^2\alpha = 1 - \cos^2\alpha$$
and express everything in terms of $\cos^2\alpha$:
$$\begin{align}
R &= \left(\;\sin^2\alpha\;\right)^3 + \left(\;\cos^2\alpha\;\right)^3+K\left(\;\left(\;\sin^2\alpha\;\right)^2+\left(\;\cos^2\alpha\;\right)^2\;\right) \\
&= \left(\;1-\cos^2\alpha\;\right)^3 + \left(\;\cos^2\alpha\;\right)^3+K\left(\;\left(\;1-\cos^2\alpha\;\right)^2+\left(\;\cos^2\alpha\;\right)^2\;\right) \\
&= \left(\;1-x\;\right)^3 + \left(\;x\;\right)^3+K\left(\;\left(\;1-x\;\right)^2+\left(\;x\;\right)^2\;\right) \qquad\text{(writing $x$ for $\cos^2\alpha$)}\\
&= 1 - 3 x + 3 x^2 + K \left(\; 1 - 2 x + 2 x^2 \;\right) \\
&= 1 +K -(3+2K) x + (3+2K) x^2
\end{align}$$
Independence from $\alpha$ translates to independence from $x$. We need a value of $K$ that causes the non-constant terms of the polynomial to vanish. Clearly, $K = -3/2$. $\square$
|
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|
Simplification of an expression with products $\prod_{1\leq iThe determinant of a matrix $C=(c_{ij})_{n\times n}$ whose entries have the form $c_{ij}=\frac{1}{a_i+b_j}$ is given by
$$\det C=\frac{\prod_{1\leq i<j\leq n}(a_i-a_j)(b_i-b_j)}{\prod_{1\leq i,j\leq n}(a_1+b_i)}.$$
In these notes (p. 145), this formula is applied to certain matrices $G$ and $G_m$. The result is
$$\det G=\frac{\prod_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)^2}{\prod_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)},\quad \det G_m=\frac{\prod^{\prime}_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)^2}{\prod^{\prime}_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}\tag{1}$$
"where $^\prime$ means that the index $m$ has been skipped in the product".
The aim is to compute $\frac{\det G_m}{\det G}$. A direct substitution gives
$$\frac{\det G_m}{\det G}=\frac{\prod^{\prime}_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)^2}{\prod^{\prime}_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}\cdot \frac{\prod_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}{\prod_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)^2}$$
which, according to the notes, should simplify to
$$\frac{\det G_m}{\det G}=2 m^2 \pi^2\underset{{1\leq k\leq n}}{{\prod}^{\prime}}\frac{(m^2+k^2)^2}{(m^2-k^2)^2}.\tag{2}$$
Question: How to manipulate $(1)$ properly in order to get $(2)$?
|
We have
$$\frac{\det G_m}{\det G}=\frac{\prod^{\prime}_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)^2}{\prod_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)^2}
\cdot
\frac{\prod_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}{\prod^{\prime}_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}.$$
And
\begin{align*} \frac{\prod^{\prime}_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)}{\prod_{1\leq i<j\leq n}(i^2\pi^2-j^2\pi^2)} & = \frac{1}{\prod_{m< k \le n}(m^2\pi^2-k^2\pi^2) \prod_{1 \le k<m}(k^2\pi^2-m^2\pi^2)}, \\
& = \frac{(-1)^{m-1}}{\prod_{1 \le k \le n}^{\prime}(m^2\pi^2-k^2\pi^2)}.
\end{align*}
Similarly,
$$\frac{\prod_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}{\prod^{\prime}_{1\leq i,j\leq n}(i^2\pi^2+j^2\pi^2)}=\prod_{1 \le k \le n}^{\prime}(m^2\pi^2+k^2\pi^2)^2 \cdot 2m^2\pi^2.$$
The product $2m^2\pi^2$ above appears for $i=j=m$.
Hence,
\begin{align*} \frac{\det G_m}{\det G} & =\left[ \frac{(-1)^{m-1}}{\prod_{1 \le k \le n}^{\prime}(m^2\pi^2-k^2\pi^2)} \right]^2 \cdot \prod_{1 \le k \le n}^{\prime}(m^2\pi^2+k^2\pi^2)^2 \cdot 2m^2\pi^2,\\
& =2m^2\pi^2\prod_{1 \le k \le n}^{\prime}\frac{(m^2+k^2)^2}{(m^2-k^2)^2}.
\end{align*}
|
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|
Quadrilateral in Square
$S$ is a unit square. Four points are taken randomly, one on each side of $S$. A quadrilateral is drawn. Let the sides of this quadrilateral be $a,b,c,d$. Prove that $2\leq{}a^2+b^2+c^2+d^2\leq{}4$.
My Efforts:
Let
$\begin{align}m^2+t^2&=a^2&\mathfrak{a}\\n^2+o^2&=b^2&\mathfrak{b}\\p^2+q^2&=c^2&\mathfrak{c}\\r^2+s^2&=d^2&\mathfrak{d}\end{align}$
$\mathfrak{a+b+c+d}\text{ gives}$
$m^2+n^2+o^2+p^2+q^2+r^2+s^2+t^2=a^2+b^2+c^2+d^2$
Since $m+n=o+p=q+r=s+t=1$,
This gives
$1-2mn+1-2op+1-2qr+1-2st=a^2+b^2+c^2+d^2$
Simplifying,
$4-2{(mn+op+qr+st)}=a^2+b^2+c^2+d^2$
Since the minimum value of $2{(mn+op+qr+st)}$ is $0$, I get
$a^2+b^2+c^2+d^2\leq4$.
Is there any other way to do this? How do I get $2\leq{}a^2+b^2+c^2+d^2$?
|
Since by AM-GM
$$mn+op+qr+ts\leq\left(\frac{m+n}{2}\right)^2+\left(\frac{o+p}{2}\right)^2+\left(\frac{q+r}{2}\right)^2+\left(\frac{t+s}{2}\right)^2=1,$$ we obtain:
$$a^2+b^2+c^2+d^2=4-2(mn+op+qr+ts)\geq4\cdot1-2=2.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Non linear recursive sequence bounds: $14\le R_{100} \le 18$. Let $R_{n+1}=R_n+\frac{1}{R_n}$, where $R_1=1$.
I need to prove that $14\le R_{100} \le 18$.
Can anyone help please?
|
Since $$R_n^2=R_{n-1}^2+\frac{1}{R_{n-1}^2}+2,$$ we obtain
$$R_{100}^2=R_1^2+\frac{1}{R_{99}^2}+...+\frac{1}{R_1^2}+2\cdot99>199>14^2.$$
Since, $R_{10}^2>2\cdot9+1=19,$ we obtain
$$R_{100}^2=\frac{1}{R_{99}^2}+...+\frac{1}{R_1^2}+199<\frac{89}{R_{10}^2}+\frac{10}{R_1^2}+199=\frac{89}{19}+10+199<324.$$
|
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|
How to find $\lim_{x \to 1^-} (x+1) \lfloor \frac{1}{x+1}\rfloor $? find the limits :
$$\lim_{x \to 1^-} (x+1) \lfloor \frac{1}{x+1}\rfloor =?$$
My try :
$$\lfloor \frac{1}{x+1}\rfloor=\frac{1}{x+1}-p_x \ \ \ : 0\leq p_x <1$$
So we have :
$$\lim_{x \to 1^-} (x+1) (\frac{1}{x+1}-p_x) \\= \lim_{x \to 1^-} -(x+1)p_x=!??$$
Now what ?
|
For $x$ near $1$ let say $$\frac12<x<1 \Longleftrightarrow -\frac12<x-1<0$$ we have have that
$$ \frac12<x<1 \Longleftrightarrow \frac32<x+1<2 \Longleftrightarrow \frac12 <\frac{1}{x+1} <\frac 23$$
Therefore
$$ \left\lfloor \frac{1}{x+1}\right\rfloor =0~~~\forall~ \frac12<x<1 $$
That is ,
$$ (x+1)\left\lfloor \frac{1}{x+1}\right\rfloor =0~~~\forall~ \frac12<x<1 $$
that is
$$ \color{red}{\lim_{x \to 1^-}(x+1)\left\lfloor \frac{1}{x+1}\right\rfloor =0}~$$
|
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|
Integrate $\arctan{\sqrt{\frac{1+x}{1-x}}}$ I use partial integration by letting $f(x)=1$ and $g(x)=\arctan{\sqrt{\frac{1+x}{1-x}}}.$ Using the formula:
$$\int f(x)g(x)dx=F(x)g(x)-\int F(x)g'(x)dx,$$
I get
$$\int1\cdot\arctan{\sqrt{\frac{1+x}{1-x}}}dx=x\arctan{\sqrt{\frac{1+x}{1-x}}}-\int\underbrace{x\left(\arctan{\sqrt{\frac{1+x}{1-x}}}\right)'}_{=D}dx.$$
So, the integrand $D$ remains to simplify:
$$D=x\cdot\frac{1}{1+\frac{1+x}{1-x}}\cdot\frac{1}{2\sqrt{\frac{1+x}{1-x}}}\cdot\frac{1}{(x-1)^2} \quad \quad (1).$$
Setting $a=\frac{1+x}{1-x}$ for notation's sake I get
$$D=x\cdot\frac{1}{1+a}\cdot\frac{1}{2\sqrt{a}}\cdot\frac{1}{(x-1)^2}=\frac{x}{(2\sqrt{a}+2a\sqrt{a})(x^2-2x+1)},$$
and I get nowhere. Any tips on how to move on from $(1)?$
NOTE: I don't want other suggestions to solutions, I need help to sort out the arithmetic to the above from equation (1).
|
Continuing from
$$D=x\cdot\frac{1}{1+\frac{1+x}{1-x}}\cdot\frac{1}{2\sqrt{\frac{1+x}{1-x}}}\cdot\frac{1}{(x-1)^2},$$
and observing that $(x-1)^2 = (1-x)^2$, you can take the two factors of $(1-x)$ from the final denominator and multiply them into the first two fractions. This gives
$$\begin{align}
D&=x\cdot\frac{1}{(1-x)+(1+x)}\cdot\frac{1}{2(1-x)\sqrt{\frac{1+x}{1-x}}}\\
&=x\cdot\frac12\cdot\frac1{2\sqrt{(1+x)(1-x)}}\\
&=\frac x{4\sqrt{1-x^2}}
\end{align}
$$
|
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|
In a triangle, $\Delta ABC$ we are given that $3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3\cos A = 1$. Then what is the measure of angle $C$?
In a triangle, $\Delta ABC$ we are given that $3 \sin A + 4 \cos B = 6$ and $4 \sin B + 3\cos A = 1$. Then what is the measure of angle $C$?
On squaring and adding both the equations,we get
$$9 + 16 + 24 \sin (A + B) = 37\implies$$
$$\sin (A + B) =
\frac{1}{2}\implies A+B=\frac{\pi}{6}\implies C=5\pi/6$$
But,in answer key it is given that $C=\pi/6$
I wanted to know where i'm wrong.
I've also gone through Finding $C$ if $ 3\sin A + 4\cos B = 6 $ and $ 3\cos A + 4\sin B = 1 $ in a triangle $ABC$ ,but here also,i didn't get my answer.
|
Note that $\sin(A+B) = \frac{1}{2}$ has two solutions in $(0,\pi)$, that are $A+B \in \{\frac{\pi}{6},\frac{5\pi}{6}\}$.
Now check which of them satisfies original equations. Lets suppose $A+B = \frac{\pi}{6}$. In triangle, all angles are positive, so both $0 < A,B \lt \frac{\pi}{6}$. This implies that
*
*$0 < \sin A< \frac{1}{2}$
*$\frac{\sqrt{3}}{2} < \cos B < 1$
From first equation,
$$3\sin(A) +4\cos(B) = 6$$
Maximum value of LHS is $4+\frac{3}{2}$, that too not achievable, and it is less than $6$. So this equation is never satisfied for $A+B = \frac{\pi}{6}$.
Similarly analyse the equations for second solution, that is $A+B = \frac{5\pi}{6}$
|
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|
What is the sum of $E=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+...$ What is the right way to assess this problem?
$$E=\frac{1}{3}+\frac{2}{3^{2}}+\frac{3}{3^{3}}+\frac{4}{3^{4}}+...$$
To find the value of the sum I tried to use the fact that it could be something convergent like a geometric series. However it does not seem to be the case as there is no common ratio between the terms. Since the numerator is increasing from $1,2,3,...$ makes it impossible to find a ratio. What should I do?.
|
You can split it up into a sum of geometric progressions:
\begin{align} 3^{-1} + 2 \cdot 3^{-2} + 3 \cdot 3^{-3} + \dotsb = 3^{-1} + 3^{-2} + 3^{-3} + \dotsb \\
+ 3^{-2} + 3^{-3} + \dotsb \\
+ 3^{-3} + \dotsb \\
\ddots \\
= 1(3^{-1}+ 3^{-2} + 3^{-3} + \dotsb) \\
+3^{-1}(3^{-1}+ 3^{-2} + \dotsb) \\
+ 3^{-2}(3^{-1} + \dotsb) \\
\ddots \\
= (1+3^{-1}+3^{-2}+\dotsb)(3^{-1}+3^{-2}+3^{-3}+\dotsb)
\end{align}
|
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|
Find $\alpha + \beta$ given that: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ , ( $\alpha , \beta \in \Bbb R)$
Find $\alpha + \beta$ given that: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ , ( $\alpha , \beta \in \Bbb R)$
I have a solution involving a variable change ($\alpha=x+2$) which is not beautiful in my opinion! I'm looking for a more beautiful solution.
|
Lets create a polynomial for both these equations.
Let $A(x) = x^3 - 6x^2 +13x -1 $. and $B(x) = A(x) -18$. Both these polynomials have only one real root, which is easily verified from their derivative.
Now let $\alpha+\beta = a$. Then $\beta = a-\alpha$. So transforming $A(x)$ such that $x \to a-x$ should give a polynomial whose root is $\beta$, that is $pB(x)$, here we get $p = -1$ by comparing coefficient of $x^3$.
$$A(a-x) = p B(x) \\
(a-x)^3 - 6(a-x)^2 + 13(a-x) - 1 = -(x^3 -6x^2 +13x -19)$$
These polynomials are identically equal, so we can simply compare coefficients of equal powers of $x$.
*
*Comparing coefficient of $x^3$ we get $p=-1$.
*Comparing coefficient of $x^2$ : $3a -6 = 6$, we get $a = 4$
*Comparing coefficient of $x$: $-3a^2+12a-13 = -13$. Now, $a = 4$ satisfies this equation.
*Comparing constant term: $a^3-6a^2+13a-1 = 19$. Again $a = 4$ satisfies this.
We see $a = 4$ is the only value which satisfies all three conditions.
Hence $\alpha + \beta = a = 4$
|
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|
How to find the maximum and minimum values of $\frac{8x(x^2-1)}{(x^2+1)^2}$ algebraically? The function is $f(x) = \frac{8x(x^2-1)}{(x^2+1)^2}$.
I have tried using calculus, only to fail.
|
HINT: prove that $$-2\le \frac{8x(x^2-1)}{(x^2+1)^2}\le 2$$
we have $$2-\frac{8x(x^2-1)}{(x^2+1)^2}=2\,{\frac { \left( {x}^{2}-2\,x-1 \right) ^{2}}{ \left( {x}^{2}+1
\right) ^{2}}}
$$
and $$2+\frac{8x(x^2-1)}{(x^2+1)^2}=2\,{\frac { \left( {x}^{2}+2\,x-1 \right) ^{2}}{ \left( {x}^{2}+1
\right) ^{2}}}
$$
the Minimum will be attained by $$x=-1-\sqrt{2}$$ and the Maximum by $$1-\sqrt{2}$$
|
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|
Show that $a_n=\frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+\frac{2^3}{3}+..+\frac{2^n}{n}\right)$ converges and find its limit Sequence: $$a_n=\frac{n+1}{2^{n+1}}\left(\frac{2}{1}+\frac{2^2}{2}+\frac{2^3}{3}+..+\frac{2^n}{n}\right)$$
My Attempt: I showed the sequence is increasing by considering the difference $a_{n+1}-a_n>0.$ But I am having trouble in showing that the sequence is bounded. I have managed to show that $$\frac{(n+1)(2^n-1)}{n\times 2^{n+1}}\leq a_n\leq \frac{(n+1)(2^n-1)}{2^{n+1}}<\frac{n+1}{2}.$$
How do I proceed?
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$$ \lim_{n\to +\infty}\frac{\sum_{k=1}^{n}\frac{2^k}{k}}{\frac{2^{n+1}}{n+1}}\stackrel{\text{Cesàro-Stoltz}}{=}\lim_{n\to +\infty}\frac{\frac{2^{n+1}}{n+1}}{\frac{2^{n+2}}{n+2}-\frac{2^{n+1}}{n+1}}=\color{red}{1}, $$
no major mystery here.
|
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|
Proving the double differential of z = -z implies z= sinx In common usage, we know that $$\frac{d^2z}{dx^2}=-z\,$$ implies $z$ is of the form $a\sin x + b\cos x$. Is there a proof for the same. I was trying to arrive at the desired function but couldn't understand how to get these trigonometric functions in the equations by integration. Does it require the use of taylor polynomial expansion of $\sin x\, \text {or}\, \cos x$?
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We assume the solution is of the form
$$z=\sum_{n=0}^{\infty}a_nx^n=a_0+a_1x+a_2x^2+\cdots+a_nx^n+\cdots$$
then
$$
z''
=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}
=1\cdot2a_2+2\cdot3a_3x+3\cdot4a_4x^2+\cdots
$$
from $z''=-z$ then with arbitrary $a_0$ and $a_1$ we have
\begin{eqnarray*}
&&
a_2=-\frac{1}{1\times2}a_0 ~~~,~~~
a_4=-\frac{1}{3\times4}a_2=\dfrac{1}{4!}a_0 ~~,~~~
a_6=-\frac{1}{5\times6}a_4=-\dfrac{1}{6!}a_0 ~~,~~~ \cdots\\
&&
a_3=-\frac{1}{2\times3}a_1 ~~~,~~~
a_5=-\frac{1}{4\times5}a_3=\dfrac{1}{5!}a_1 ~~,~~~
a_7=-\frac{1}{6\times7}a_5=-\dfrac{1}{7!}a_0 ~~,~~~ \cdots
\end{eqnarray*}
then
$$z=a_0\left(1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\cdots\right) + a_1\left(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\cdots\right)=\color{blue}{a_0\cos x+a_1\sin x}$$
|
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|
max and min of $f(x,y)=y^8-y^4x^6+x^4$ The function is continuous in $\mathbb{R}^2$ and $f(x,y)=f(x,-y)=f(-x,y)=f(-x,-y)$.
If I consider $f(x,0)=x^4$ for $x\rightarrow +\infty$ $f$ is not limited up so $\sup f(x,y)=+\infty$.
But the origin is absolute min?
$f(x,x)=x^8-x^{10}+x^4\rightarrow -\infty$ if $x\rightarrow \infty$? so f is not limit down
|
$f'_x=-6x^5 y^4+4x^3;\;f'_y=8y^7-4y^3x^6$
They must be both zero
$
\left\{
\begin{array}{l}
-2 x^3 (3 x^2 y^4-2)=0 \\
-4 y^3 \left(x^6-2 y^4\right)=0 \\
\end{array}
\right.
$
$(0,0)$ is a solution and then
$
\left\{
\begin{array}{l}
3 x^2 y^4-2=0 \\
x^6-2 y^4=0 \\
\end{array}
\right.
$
Solve for $y$ the first equation
$y^4=\dfrac{2}{3x^2}$
and plug in the second
$x^6-\dfrac{4}{3x^2}=0\to x^8=\dfrac{4}{3}\to x=\pm\sqrt[8]{\dfrac{4}{3}}$
and $y^4=\dfrac{2}{3\sqrt[4]\frac{4}{3}}=\dfrac{2}{3}\sqrt[4]{\dfrac{3}{4}}=\sqrt[4]{\dfrac{4}{27}}$
$y=\pm\sqrt[16]{\dfrac{4}{27}}$
The singular points are
$P_0(0,0);\;P_1\left(\sqrt[8]{\dfrac{4}{3}},\sqrt[16]{\dfrac{4}{27}}\right);\;P_2\left(-\sqrt[8]{\dfrac{4}{3}},\sqrt[16]{\dfrac{4}{27}}\right)$
$P_3\left(\sqrt[8]{\dfrac{4}{3}},-\sqrt[16]{\dfrac{4}{27}}\right);\;P_4\left(-\sqrt[8]{\dfrac{4}{3}},-\sqrt[16]{\dfrac{4}{27}}\right)$
To understand what is what we need second partial derivative test. Hessian matrix is:
$H(x,y)=\left(
\begin{array}{ll}
12 x^2-30 x^4 y^4 & -24 x^5 y^3 \\
-24 x^5 y^3 & 56 y^6-12 x^6 y^2 \\
\end{array}
\right)$
and its determinant
$D(x,y)=-24 x^2 y^2 \left(9 x^8 y^4+6 x^6+70 x^2 y^8-28 y^4\right)$
As the determinant contains all even powers of the variables and a coefficient negative, all $P_i,\;i=1,2,3,4$
Around zero $f(x,y)=x^4+O(x^5)+O(y^8)$ therefore $f$ is positive
around the zero and the origin is a local minimum.
In the following graph we can see that $f(x,y)>0$ in a neighborhood of $(0,0)$
There is no global minimum as the function goes to $-\infty$ and no global maximum as the function goes to $+\infty$
Hope this helps
$$...$$
|
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|
Determing Fourier series and what it converges to I am trying to determine the fourier series of $f(x)$ on $[-2,2]$ which I believe I have done correctly. I will show some details below of my calculation. However my question is how do I determine what the Fourier series of $f(x)$ on $[-2,2]$ converges to. My guess is that you must take the limit of $f(x)$, but do I take the limit as $x$ goes to $\infty$? However there is parts if the equation that are $(-1)^n$which confuses me if I was to take the limit.
$$f(x) = \begin{cases}
2,&-2\leq x \leq 0\\
x, &0 < x \leq 2\\
\end{cases}$$
So I know that,
$$f(x) = \frac{A_0}{2} + \sum _{n=1}^{\infty} A_n \cos \Big(\frac{n\pi x}{L}\Big) + \sum _{n=1}^{\infty} B_n \sin \Big(\frac{n\pi x}{L}\Big)$$
Then calculating $A_0,$
$$A_0 = \frac{1}{L} \int_{-L}^{L} f(x) dx = \frac{1}{2} \int _{-2}^{2} f(x)dx = \frac{1}{2} \Bigg [\int_{-2}^02dx + \int_0^{2} x \Bigg]= \frac{1}{2}[4+2] = 3 $$
Then calculating $A_n$,
$$A_n = \frac{1}{L}\int_{-2}^{2} f(x) \cos\Big(\frac{n\pi x}{2} \Big) = \frac{1}{2}\Bigg[\int_{-2}^0 2\cos\Big(\frac{n\pi x}{2}\Big) + \int _0^2 x\cos \Big(\frac{n\pi x}{2}\Big) \Bigg]\\ = \frac{2}{\pi^2 n^2 }(-1)^{n} - \frac{2}{\pi^2 n^2 }$$
Then calculating $B_n, $
$$B_n = \frac{1}{L}\int_{-2}^{2} f(x) \sin\Big(\frac{n\pi x}{2} \Big) = \frac{1}{2}\Bigg[\int_{-2}^0 2\sin\Big(\frac{n\pi x}{2}\Big) + \int _0^2 x\sin \Big(\frac{n\pi x}{2}\Big) \Bigg]\\ = \frac{1}{2}\Bigg[\frac{-4}{n\pi}\cos \Big(\frac{n\pi x}{2}\Big)\Big|_{-2}^0 \Bigg] + \frac{1}{2}\Bigg[\frac{-2x}{\pi n}\cos \Big(\frac{n\pi x}{2 }\Big) + \frac{4}{n^2\pi ^2}\sin \Big(\frac{n\pi x}{2 }\Big)\Big|_0^2 \Bigg] \\ = \frac{1}{2}\Bigg[\frac{-4}{\pi n } + \frac{4}{\pi n } \cos(\pi n )\Bigg] + \frac{1}{2}\Bigg[\frac{-4}{\pi n }\cos(\pi n) + \frac{4}{\pi ^2 n ^2 } \sin (\pi n) \Bigg] \\ = \frac{1}{2} \Bigg[\frac{-4}{n\pi} + \frac{4}{n\pi}(-1)^n - \frac{4}{\pi n}(-1)^n \Bigg] = \frac{-2}{n\pi}$$
Thus,
$$f(x) = \frac{3}{2}+ \sum_{n=1}^{\infty}\Bigg[\frac{2}{\pi^2 n^2 }(-1)^{n} - \frac{2}{\pi^2 n^2 }\Bigg] \cos\frac{n\pi x}{2} + \sum_{n=1}^{\infty}\Bigg[\frac{-2}{n\pi }\Bigg] \sin \frac{n\pi x}{2}$$
So again my thought would be to take the limit of this function, and I have a feeling that it approaches $\frac{3}{2}$. But I am not sure if this is correct or how to show it, looking for some help with this, thanks!
|
Since $f (x) $ is piecewise smooth. But has a jump discontinuity at $x=0$ , the Fourier series will converge to $$1/2[ f (0-)+f (0+)]$$.
You don't need to calculate the series to find wheter it converges or not.
You can read more about it at http://tutorial.math.lamar.edu/Classes/DE/ConvergenceFourierSeries.aspx
|
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|
Calculus inequality involving sine and cosine community! I saw the following inequality in a calculus assignment, which I thought was harder to prove than I expected: For every $x\in\mathbb{R},$$$(\sin x + a \cos x)(\sin x + b \cos x)\leq 1+(\frac{a+b}{2})^2.$$
By means of a graphing device, I noticed that when $a=b$, $f(x)=(\sin x + a\cos x)(\sin x + b \cos x)$ has $1+(\frac{a+b}{2})^2$ as its maximum, but in other cases the maximum of $f$ is strictly less (and sometimes far less) than this number. I'd appreciate any kind of help :)
|
$$|\sin x+a\cos x|=\left|\sqrt{1+a^2}\left( \frac{1}{\sqrt{1+a^2}}\sin x+\frac{a}{\sqrt{1+a^2}}\cos x \right)\right|=|\sqrt{1+a^2}(\cos \theta \sin x+\sin \theta \cos x)|=|\sqrt{1+a^2}( \sin(x+\theta) )|\le \sqrt{1+a^2}.$$ So the maximum value of $\sin x+a\cos x=\sqrt{1+a^2}.$
\begin{align*}
|(\sin x+a\cos x)(\sin x+b\cos x)|& = |\sin x+a\cos x||\sin x+b\cos x|\\
& \le \sqrt{1+a^2}\sqrt{1+b^2}=\sqrt{(1+a^2)(1+b^2)}\\
& \le \frac{(1+a^2)+(1+b^2)}{2} \tag{1}\\
&=1+\frac{a^2+b^2}{2}
\end{align*}
In $(1)$, I have used AM-GM Inequality.
|
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|
In an Acute angled Triangle $\Delta ABC$ $\csc \left(\frac{A}{2}\right)+\csc \left(\frac{B}{2}\right)+\csc \left(\frac{C}{2}\right)=6$ In an Acute angled Triangle $\Delta ABC$ if $$\csc \left(\frac{A}{2}\right)+\csc \left(\frac{B}{2}\right)+\csc \left(\frac{C}{2}\right)=6$$ Prove that the Triangle is Equilateral.
I tried using $A.M \ge H.M$ So we have
$$\frac{\sin \left(\frac{A}{2}\right)+\sin\left(\frac{B}{2}\right)+\sin \left(\frac{C}{2}\right)}{3} \ge \frac{3}{\csc \left(\frac{A}{2}\right)+\csc \left(\frac{B}{2}\right)+\csc \left(\frac{C}{2}\right)}=\frac{1}{2}$$ $\implies$
$$\sin \left(\frac{A}{2}\right)+\sin\left(\frac{B}{2}\right)+\sin \left(\frac{C}{2}\right) \ge \frac{3}{2}$$
Then i started to analyze the equality case i.e.,
$$\sin \left(\frac{A}{2}\right)+\sin\left(\frac{B}{2}\right)+\sin \left(\frac{C}{2}\right)=\frac{3}{2} \tag{1}$$
Let $x=\sin \left(\frac{A}{2}\right)$ and $y=\sin \left(\frac{B}{2}\right)$
Then $$\sin \left(\frac{C}{2}\right)=\sin \left(\frac{180-A-B}{2}\right)=\cos \left(\frac{A+B}{2}\right)=\sqrt{1-x^2}\sqrt{1-y^2}-xy$$
Then we have from $(1)$
$$f(x,y)=x+y-xy+\sqrt{1-x^2}\sqrt{1-y^2}-\frac{3}{2}$$
I partially differentiated with respect to $x$ and got
$$y=1-2x^2$$ $\implies$
$$\sin \left(\frac{B}{2}\right)=\cos A$$
Can we proceed further?
|
Let $f(x)=\frac{1}{\sin\frac{x}{2}}$.
Thus, $f''(x)=\frac{3+\cos{x}}{8\sin^3\frac{x}{2}}>0$ for all $x\in\left(0,\frac{\pi}{2}\right)$ and by Jensen we obtain:
$$\sum_{cyc}\frac{1}{\sin\frac{\alpha}{2}}\geq\frac{3}{\sin\frac{\alpha+\beta+\gamma}{6}}=6.$$
The equality occurs for $\alpha=\beta=\gamma=60^{\circ}$ only and we are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Induction for $3^n\geq n^2$ I need to prove by induction that $3^n\geq n^2$
What I have so far:
*
*Basis: $n = 1 \rightarrow 3^1 \geq 1^2 \rightarrow 3 \geq 1$
*Condition: $3^n\geq n^2$
*Assumption: $3^{n+1}\geq (n+1)^2$
*Basis: $3^n\geq n^2$
$3^{n+1}\geq (n+1)^2
= 3(3^n) \geq n^2+2n+1 $
That's where I got stuck. I have no idea how to continue from there.
|
We verify that it holds for $n=2$ since $3^2\ge2^2$
We assume that it holds for $n:$ $3^n\ge n^2(*)$ and we want to show that $3^{n+1}\ge(n+1)^2$
From $(*)$ we have by multiplying with $3$:
$3^{n+1}\ge3n^2$
But
$3n^2\ge(n+1)^2$ since $3n^2-(n+1)^2=2n^2-2n-2=n^2-2+(n-1)^2\ge0$
Since $n^2-2\ge0, (n-1)^2\gt0$ $\forall n\ge2$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Showing that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$.
Problem: Show that there are no $3$-digit primes $\overline {abc}$ such that $b^2-4ac=9$.
Solution. Upon the given condition, the quadratic equation $ax^2+bx+c$ can be written as $(px+q)(rx+s)$ and we have:
$$\overline{abc}=100a+10b+c=(10p+q)(10r+s)$$
i.e. $\overline{abc}$ is not a prime which is a contradiction, so there is no such prime available.
I don't get the reasoning completely.
|
$c$ can take only $1, 3, 7, 9$ . [$abc$ is prime]
$b$ can take only $1, 2, 3, 4, 5, 6, 7, 8, 9$. [$b\neq 0$ as $-4ac$ can't be a square of a number]
$a$ can take $1, 2, 3, 4, 5, 6, 7, 8, 9$.
$b^2-k^2=4ac$.
Thus $b$ and $k$ both should be both odd or both even and $b>k$ as $c$ can't be $<0$.
$4ac=\text{Multiple of 4,12,28 or 36}$ as $c$ can take only $1, 3, 7, 9$.
$$(b,k)=(2,0),(4,0),(6,0),(8,0),(4,2),(6,2),(8,2),(6,4),(8,4),(8,6)(3,1),(5,1),(7,1),(9,1),(5,3),(7,3),(9,3),(7,5),(9,5),(9,7).$$
For all this values of $b$ and $k$, $a$ and $c$ can't assume such values such that $abc$ becomes a prime number. Thus, $b^2-4ac$ is never a perfect square when $abc$ is prime.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the limit of the complex function. $$ \lim_{x\to a} \left(2- \frac{x}{a}\right)^{\left(\tan \frac{\pi x}{2a}\right)}.$$
I have simplified this limit to this extent :
$$e^{ \lim_{x\to a} \left(\left(1- \frac{x}{a}\right){\left(\tan \frac{\pi x}{2a}\right)}\right)}$$
I don't know how to simplify the limit after that. please help.
|
Let $y=(2- \dfrac xa)^{\tan \frac{\pi x}{2a}}$ then $\ln y=\tan\dfrac{\pi x}{2a}\ln(2- \dfrac xa)$ and let $1-\dfrac{x}{a}=\dfrac{1}{t}$ then
\begin{align}
\lim_{x\to a} (2- \dfrac xa)^{\tan \frac{\pi x}{2a}}
&= \lim_{t\to \infty} \tan\dfrac{\pi}{2}\left(1-\dfrac1t\right)\ln\left(1+\dfrac1t\right) \\
&= \lim_{t\to \infty} \cot\dfrac{\pi}{2t}\ln\left(1+\dfrac1t\right) \\
&= \lim_{t\to \infty} \dfrac{\cos\dfrac{\pi}{2t}}{\sin\dfrac{\pi}{2t}}\ln\left(1+\dfrac1t\right) \\
&= \lim_{t\to \infty} \dfrac{\dfrac{\pi}{2t}\cos\dfrac{\pi}{2t}}{\sin\dfrac{\pi}{2t}} \cdot t\ln\left(1+\dfrac1t\right) \dfrac{1}{t\dfrac{\pi}{2t}}\\
&= \lim_{t\to \infty} \dfrac{\dfrac{\pi}{2t}}{\sin\dfrac{\pi}{2t}} \cdot \ln\left(1+\dfrac1t\right)^t ~~\dfrac{\cos\dfrac{\pi}{2t}}{\dfrac{\pi}{2}}\\
&= \color{blue}{\dfrac{2}{\pi}}
\end{align}
using $\lim_{t\to \infty}\left(1+\dfrac1t\right)^t=e$, then $y=e^{\frac{2}{\pi}}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Showing the range of a function The function $f:\mathbb{Z}\rightarrow\mathbb{Z}$ is defined by
$$f(n)=\begin{cases}3n+2\quad \text{if $n$ is even}\\ 2n+3\quad \text{if n is odd.}\end{cases}$$
Prove that the range of $f$ is $\{m\in\mathbb{Z}|m\equiv 1,2,5,8 \ \text{or} \ 9 \pmod {12}\}$
I wasn't sure of a nice way to show this, so I just considered when $n = 2k$,
$f(n) = 6k+2$.
Then I found this modulo $12$ by considering when $k\equiv 0,1,2,3,\ldots,11$ mod $12$.
This seemed to alternate from $2$ to $8$.
Then I considered when $n=2\ell +1$, and $\ell\equiv 0,1,2,\ldots,11$ mod $12$ and found the other results needed.
However, the solutions did the exact same thing, except they considered when $k\equiv 0,1,2$ and $\ell \equiv 0,1,2$ both modulo $3$... not $12$.
Why does this work?
|
Since they're asking for a result modulo $12$, I would've gone for a more direct approach, like so:
$(3)(4) = 12$, so for the even cases, work modulo $4$. An even number $n$ satisfies $n \equiv 0,2 \pmod 4$. So $3n+2 = 2,8 \pmod{12}$.
$(2)(6) = 12$, so for the odd cases, work modulo $6$. An odd number $m$ satisfies $m \equiv 1,3,5 \pmod 6$. So $2n+3 \equiv 5,9,13 \equiv 5,9,1 \pmod{12}$.
So the complete range is $1,2,5,8,9 \pmod{12}$ as required.
|
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|
Convergence of $n \sqrt{n^2+1}-n^2$ I have been asked to study the convergence of the the serie $a_n = n\sqrt{n^2+1}-n^2$. I have been told that the answer is $\frac{1}{2}$ but I am not being able to demonstrate it. I have tried to integrate it and i get that $\int^\infty_1 n\sqrt{n^2+1}-n^2 = \frac{1}{3}[(n^2+1)^{3/2}-n^3]|^\infty_1$. I don't know if any of this is correct or how to follow from here.
I have also read that $\sqrt{n^2+1} - n$ converges to 0 but I don't know how to apply that conclusion here.
|
Consider using
$$\sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^{2}}{8} + \mathcal{O}(x^{3})$$
for which
\begin{align}
f(n) &= n \, \sqrt{n^{2} + 1} - n^{2} = n^{2} \, \left(\sqrt{1 + \frac{1}{n^{2}}} - 1 \right) \\
&= n^{2} \, \left(\frac{1}{2 \, n^{2}} - \frac{1}{8 \, n^{4}} + \mathcal{O}\left(\frac{1}{n^{6}}\right) \right) \\
&= \frac{1}{2} - \frac{1}{8 \, n^{2}} + \mathcal{O}\left(\frac{1}{n^{4}}\right).
\end{align}
Considering that $n \to \infty$ the limit becomes
$$\lim_{n \to \infty} f(n) = \frac{1}{2}.$$
|
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|
Proof for an identity (from Ramanujan written) I saw an identity by Ramanujan
$$\forall n \in \mathbb{N} ,n>1 :\lfloor \sqrt n+\sqrt {n+2}+\sqrt{n+4} \rfloor=\lfloor \sqrt {9n
+17}\rfloor$$ I tried to prove it by limit definition . I post my trial below . If possible check my prove (right , wrong) ?
Then Is there more Idea to proof ?
|
Assume $(m-1)^2<9n+17<m^2$. (We never get equality because $9n+17\equiv -1\pmod{3}$.) So we get: $$(m-1)^2+1\leq 9n+17\leq m^2-1$$
or $$\frac{(m-1)^2-16}{9}\leq n\leq\frac{m^2-18}{9}.$$
So $$\frac{\sqrt{(m-1)^2-16}}{3}\leq\sqrt{n}\leq\frac{\sqrt{m^2-18}}{3}\\
\frac{\sqrt{(m-1)^2+2}}3\leq\sqrt{n+2}\leq\frac m3\\
\frac{\sqrt{(m-1)^2+20}}3\leq\sqrt{n+4}\leq\frac{\sqrt{m^2+18}}{3}$$
Now by AM/GM, we get, for any real $a\geq 4$ that:
$$\begin{align}\frac{\sqrt{a^2-16}+\sqrt{a^2+2}+\sqrt{a^2+20}}{3}&>\left((a^2-16)(a^2+2)(a^2+20)\right)^{1/6}\\&=\left(a^6 + 6 a^4 - 312 a^2 - 64\right)^{1/6}
\end{align}$$
For $a\geq 8$ this gives a lower bound of $a$, because then:
$$0<6 a^4 - 312 a^2 - 64$$
This means that for $a=m-1$ with $m\geq 9$ we get that $\sqrt{n}+\sqrt{n+2}+\sqrt{n+4}\geq m-1$.
You can hand-check the values when $m<9$.
Finally, we'd like to show that $\sqrt{m^2-18}+m+\sqrt{m^2+18}\leq 3m$. This follows from concavity of $\sqrt{\cdot}$ function - $\sqrt{m^2-18}+\sqrt{m^2+18}\leq 2\sqrt{m^2}=2m$.
This gives that $\sqrt{m^2-18}+m+\sqrt{m^2+18}<3m$.
So we have that $m-1<\sqrt{n}+\sqrt{n+2}+\sqrt{n+4}<m.$ and we are done.
The key step is the AM/GM step, which gives a good approximation of the left sum because when $a$ is large, the interval $\left(\sqrt{a^2-16},\sqrt{a^2+20}\right)$ is small, so the AM/GM inequality is close to equality.
The other key is that $\sqrt{9n+17}$ is never an integer.
It might well be a numeric "accident" that it works when $m\leq 8$ (that is, when $n\leq 5$.) After all, it doesn't work for $n=1$.
More generally, for any odd $d$ such that $-1$ is not a square modulo $d$ and any value $a$, you get that, for large enough $n$:
$$\left\lfloor \sum_{k=0}^{d-1} \sqrt{n+ka}\right\rfloor = \left\lfloor \sqrt{d^2n+\frac{d^2(d-1)a}{2}-1}\right\rfloor$$
(Experimentation seems to indicate this is true with no condition on $d$, but I don't have a proof.)
For example, when $d=7$ and $a=2$ then you get for $n>22$:
$$\left\lfloor \sqrt{n}+\sqrt{n+2}+\cdots+\sqrt{n+12}\right\rfloor = \left\lfloor\sqrt{49n+293}\right\rfloor$$
For $d=3,a=1$ you actually get for all $n\geq 1$ that:
$$\left\lfloor \sqrt n+\sqrt{n+1}+\sqrt{n+2}\right\rfloor=\left\lfloor\sqrt{9n+8}\right\rfloor$$
For $d=7,a=1$ you get equality for all $n\neq 3$.
|
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|
Find Probability that even numbered face occurs odd number of times A Die is tossed $2n+1$ times. Find Probability that even numbered face occurs odd number of times
First i assumed let $a,b,c,d,e,f$ be number of times $1$ occurred, $2$ occurred and so on $6$ occurred respectively. Then we have
$$a+b+c+d+e+f=2n+1$$ is a linear equation where variables are non negative integers.
So number of non negative integer solutions of above is $\binom{2n+6}{5}$
But $b$, $d$ and $f$ should be odd. So let $b=2x+1$, $d=2y+1$ and $f=2z+1$ where $x,y,z$ are non negative integers.
so we get
$$a+2x+c+2y+e+2z=2n-2$$ which is possible in the following two cases
Case $1.$ $a,c,e$ are all even so let $a=2m+1$, $c=2q+1$ and $e=2r+1$ we get
$$m+x+q+y+r+z=n-1$$ so number of solutions is $\binom{n+4}{5}$
Case $2.$ Exactly two of $a,c,e$ are odd and other is even which can be possible in 3 ways.
So let $a=2m+1$, $c=2q+1$, $e=2r$ we get
$$m+x+q+y+r+z=n-2$$
Number of solutions is $3 \times \binom{n+3}{5}$
Hence Probability is
$$P(A)=\frac{\binom{n+4}{5}+3\binom{n+3}{5}}{\binom{2n+6}{5}}$$
Is my approach fine?
|
First let us simplify the problem: the probability of an even face is 1/2, so is the probability of an odd face. The problem is therefore equivalent to a number of coin tosses of a fair coin, with even/odd faces corresponding to heads/tails.
Define the events $E$: the number of heads is even and $F$: the number of tails is even. On the one hand, heads and tails are perfectly interchangeable, therefore
$$
P(E)=P(F).
$$
Moreover, given an odd number of trials, either the number of heads is even or the number of tails is even, which implies
$$
P(E)+P(F)=1.
$$
You may then conclude that
$$
P(E)=P(F)=1/2.
$$
Therefore, the probability that even numbered face occurs odd number of times is 1/2.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Computing and proving existence of integral We're supposed to prove the existence of the following integral and compute it.
$\int^{10}_{y=0}\int^{\pi/3}_{x=0}{xy \cos x y^2}\,\mathrm{d}x\mathrm{d}y$
My two cents on this. I was trying to use Fubini's theorem and change the order of integration, but ran into trouble evaluating the actual integral itself.
|
Well, we know that:
$$\mathscr{I}:=\int_0^{10}\int_0^\frac{\pi}{3}x\cdot\text{y}\cdot\cos\left(x\cdot\text{y}^2\right)\space\text{d}x\space\text{d}\text{y}=$$
$$\int_0^{10}\text{y}\cdot\left\{\int_0^\frac{\pi}{3}x\cdot\cos\left(x\cdot\text{y}^2\right)\space\text{d}x\right\}\space\text{d}\text{y}\tag1$$
Using:
$$\cos\left(x\cdot\text{y}^2\right)=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot\left(x\cdot\text{y}^2\right)^{2\text{n}}=$$
$$\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot x^{2\text{n}}\cdot\text{y}^{4\text{n}}\tag2$$
So, we can write:
$$\mathscr{I}=\int_0^{10}\text{y}\cdot\left\{\int_0^\frac{\pi}{3}x\cdot\left(\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot x^{2\text{n}}\cdot\text{y}^{4\text{n}}\right)\space\text{d}x\right\}\space\text{d}\text{y}=$$
$$\mathscr{I}=\int_0^{10}\text{y}\cdot\left\{\int_0^\frac{\pi}{3}\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot x^{1+2\text{n}}\cdot\text{y}^{4\text{n}}\space\text{d}x\right\}\space\text{d}\text{y}=$$
$$\mathscr{I}=\int_0^{10}\text{y}\cdot\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot\text{y}^{4\text{n}}\cdot\left\{\int_0^\frac{\pi}{3}x^{1+2\text{n}}\space\text{d}x\right\}\space\text{d}\text{y}\tag3$$
Now, using:
$$\int_0^\frac{\pi}{3}x^{1+2\text{n}}\space\text{d}x=\left[\frac{x^{2+2\text{n}}}{2+2\text{n}}\right]_0^\frac{\pi}{3}=\frac{\left(\frac{\pi}{3}\right)^{2+2\text{n}}}{2+2\text{n}}\tag4$$
So:
$$\mathscr{I}=\int_0^{10}\text{y}\cdot\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot\text{y}^{4\text{n}}\cdot\left\{\frac{\left(\frac{\pi}{3}\right)^{2+2\text{n}}}{2+2\text{n}}\right\}\space\text{d}\text{y}=$$
$$\int_0^{10}\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot\text{y}^{1+4\text{n}}\cdot\left\{\frac{\left(\frac{\pi}{3}\right)^{2+2\text{n}}}{2+2\text{n}}\right\}\space\text{d}\text{y}=$$
$$\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot\frac{\left(\frac{\pi}{3}\right)^{2+2\text{n}}}{2+2\text{n}}\cdot\int_0^{10}\text{y}^{1+4\text{n}}\space\text{d}\text{y}\tag5$$
Now, using:
$$\int_0^{10}\text{y}^{1+4\text{n}}\space\text{d}\text{y}=\left[\frac{\text{y}^{2+4\text{n}}}{2+4\text{n}}\right]_0^{10}=\frac{10^{2+4\text{n}}}{2+4\text{n}}\tag6$$
So:
$$\mathscr{I}=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot\frac{\left(\frac{\pi}{3}\right)^{2+2\text{n}}}{2+2\text{n}}\cdot\frac{10^{2+4\text{n}}}{2+4\text{n}}=\frac{3}{400}\tag7$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$ then find $f(x)$ let $f(\frac{x}{3})+f(\frac{2}{x})=\frac{4}{x^2}+\frac{x^2}{9}-2$
then find $f(x)$
My Try :
$$f(\frac{x}{3})+f(\frac{2}{x})=(\frac{2}{x})^2-1+(\frac{x}{3})^2-1$$
So we have :
$$f(x)=x^2-1$$
it is right ?Is there another answer?
|
Suppose $f(x)$ can be given by the power series expansion $f(x) = a_{0} + a_{1} x + a_{2} x^{2} + \cdots$. Now, for the equation
$$f\left(\frac{x}{3}\right) + f\left(\frac{2}{x}\right) = \frac{x^{2}}{9} - 2 + \frac{4}{x^{2}}$$
it is seen that:
\begin{align}
\frac{x^{2}}{9} + \frac{4}{x^{2}} - 2 &= 2 a_{0} + a_{1} \, \left(\frac{x}{3} + \frac{2}{x}\right) + a_{2} \left(\frac{x^{2}}{9} + \frac{4}{x^{2}}\right) + \cdots.
\end{align}
It is easy to determine that $a_{m+3} = 0$, for $m \geq 0$, $a_{1} = 0$, $a_{2} = 1$, and $a_{0} = -1$ which leads to the result
$$f(x) = x^{2} - 1.$$
|
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|
Explicit sum of $\sum_{n=0}^{\infty} \frac{6+3^n}{6^{n+2}}$ I want to find the sum of the infinite series $$\sum_{n=0}^{\infty} \frac{6+3^n}{6^{n+2}}$$
So far I have managed to simplify the expression to $$\sum_{n=0}^{\infty} \left(\frac {1}{6^{n+1}} + \frac{1}{9\cdot2^{n+2}}\right)$$.
The sum clearly converges as both expressions are equal to $0$ as $\lim_{n\to\infty}$. Any tips as to how I may find an explicit expression for the sum of this series?
thanks in advance
|
for the finite sum we have $$\sum_{n=0}^k\frac{6+3^n}{6^{n+2}}=-4/5\,{\frac {6+{3}^{k+1}}{{6}^{k+3}}}-{\frac {72+12\,{3}^{k+2}}{5\,{6
}^{k+4}}}+{\frac{23}{90}}
$$ now compute the limit for $k$ tends to infinity
|
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|
Limit of a sequence by definition I need to prove by definition that the limit of sequence :
$$a_n = \frac{5n^3-3n^2+1}{4n^3+n+2}$$
is $\dfrac54$ which means I need to show that :
$$\left|\frac{5n^3-3n^2+1}{4n^3+n+2} - \frac54\right| < \varepsilon$$
I tried for hours to solve it but could not,
Can anyone help please?
Thanks
|
$|f(n)|:=|\dfrac{5n^3-3n^2+1}{4n^3+n+2} -5/4| =$
$|\dfrac{-12n^2 -5n-6}{4(4n^3+n+2)}| \lt$
$|\dfrac{12n^2 +5n+6}{16n^3}| \lt$
$|\dfrac{12n^2+5n^2+6n^2}{n^3}| =$
$|\dfrac{23n^2}{n^3}| = |\dfrac{23}{n}|.$
Let $\epsilon >0$ be given:
Choose $M \gt 23/\epsilon,$ with $M$, real.
There is a $n_0 \gt M$.(Archimedes)
Then for $n \ge n_0:$
$|f(n)| \lt 23/n \le 23/n_0 \lt 23/M $
$\lt \epsilon.$
|
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|
Can't solve integral of Landau Lifschitz Classical Mechanics I am reading right now a book about classical mechanics (Landau Lifschitz) which contains an integral which I do not understand
$\int_{0}^{\alpha}\left [ \frac{dx_{2}(U)}{dU}- \frac{dx_{1}(U)}{dU}\right]dU\int_{U}^{\alpha}\frac{dE}{\sqrt{(\alpha - E)(E-U)}}$
Then the book states the second integral is elementary and results in $\pi$.
And the result is written as:
$\pi\left[ x_{2}(U) - x_{1}(U) \right ]$.
I am not able to solve the second part of the integral nor understand it properly. Can anybody give me a hint on how to start. I was googling already some integral tables but was not able to find it.
|
This is a $\int\frac{du}{\sqrt{a^2 - u^2}}$ form, which you can find in integral tables and which most relatively thorough elementary calculus texts discuss. For a very complete treatment of the kinds of classical integration methods that would be familiar to students of the Landau/Lifschitz era, see A Treatise on the Integral Calculus, Volume I, by Joseph Edwards.
To put your integral into this form, complete the square of
$$(\alpha - E)(E-U) = -E^2 + (\alpha + U)E - \alpha U,$$
which gives you
$$ -\left(E \; - \; \frac{\alpha + U}{2} \right)^2 \;\; + \;\; \frac{(\alpha + U)^2}{4} \; - \; \alpha U $$
Therefore,
$$\;a \; = \; \sqrt{\frac{(\alpha + U)^2}{4} \; - \; \alpha U \;} \;\;\; \text{and} \;\;\; u = E \; - \; \frac{\alpha + U}{2}$$
or
$$\;a \; = \; \sqrt{\frac{(\alpha - U)^2}{4}\;} \;\;\; \text{and} \;\;\; u = E \; - \; \frac{\alpha + U}{2}$$
or
$$\;a \; = \; \frac{1}{2}\alpha \; - \; \frac{1}{2}U \;\;\; \text{and} \;\;\; u = E \; - \; \frac{1}{2}\alpha - \frac{1}{2}U$$
For the variable change $u = E - \frac{1}{2}\alpha - \frac{1}{2}U,$ we have $du = dE$ and the limits $E = U$ to $E = \alpha$ become $u = \frac{1}{2}U - \frac{1}{2}\alpha$ to $u = \frac{1}{2}\alpha - \frac{1}{2}U.$
Thus, we get
$$ \int_{E=U}^{E=\alpha}\frac{dE}{\sqrt{(\alpha - E)(E-U)}} \;\; = \;\; \int_{E=U}^{E=\alpha}\frac{du}{\sqrt{a^2 - u^2}} \;\; = \;\; \int_{u = \frac{1}{2}U - \frac{1}{2}\alpha}^{u = \frac{1}{2}\alpha - \frac{1}{2}U}\frac{du}{\sqrt{a^2 - u^2}}$$
$$= \;\; \left[ \sin^{-1}\left(\frac{u}{a}\right) \right]_{\frac{1}{2}U - \frac{1}{2}\alpha}^{\frac{1}{2}\alpha - \frac{1}{2}U} \;\; = \;\; \sin^{-1}\left(\frac{\frac{1}{2}\alpha - \frac{1}{2}U}{a}\right) \; - \; \sin^{-1}\left(\frac{\frac{1}{2}U - \frac{1}{2}\alpha}{a}\right) $$
$$ = \;\; \sin^{-1}\left(\frac{\frac{1}{2}\alpha - \frac{1}{2}U}{\frac{1}{2}\alpha - \frac{1}{2}U}\right) \; - \; \sin^{-1}\left(\frac{\frac{1}{2}U - \frac{1}{2}\alpha}{\frac{1}{2}\alpha - \frac{1}{2}U}\right)$$
$$= \;\; \sin^{-1}(1) - \sin^{-1}(-1) \;\; = \;\; \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) \;\; = \;\; \pi $$
|
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|
How to calculate the $\gcd(10^6 +1 , 10^2 +1)$? I can't really figure out how to approach this question..
I have tried to factor 10^2 +1 out of 10^6 +1 , however, the '+1' part makes it difficult.
$10^6 + 1 = (10^2 +1) \cdot 10^3 + 899000$,
$10^2 + 1 = 899000 \cdot ...? $
Here is where I get stuck, because $899000$ is way bigger than $10^2 + 1$..
Does anyone know how to approach this?
Thanks!
|
Well, the "throw rocks at it until it falls off a cliff approach":
$\gcd (10^6 + 1, 10^2 + 1) = \gcd([10^6 + 1] - 10^4(10^2 + 1), 10^2 + 1)$
$=\gcd (-10^4 + 1, 10^2 + 1) = \gcd(-10^4 + 1 + 10^2(10^2 + 1), 10^2 + 1)$
$= \gcd(10^2 + 1,10^2+1) = 10^2 + 1$.
....
There's also the insightful. $x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + .... + x + 1)$ and if $n$ is odd then $x^n + 1 = (x+1)(x^{n-1} - x^{n-2} + .... + x^2 - x + 1)$ and let $x = 10^2$ and $n = 3$ and... oh,... that's Mike Rozenberg's elegant answer... isn't it?
|
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|
An expansion of $\pi \cot(\pi/k)/(k-2)$. Mathematica suggests that for $k>=3$,
$$
\sum _{r=0}^{\infty } \frac{\left(\frac{2}{k}\right)_r}{(r+1) \left(1+\frac{1}{k}\right)_r}
=
\frac{\pi \cot \left(\frac{\pi }{k}\right)}{k-2}
,
$$
where $(x)_r=x(x+1)\dots(x+r-1)$.
A quick look of DLMF does not find any expansion of $\cot(x)$ in this form. Any idea where this identity comes from?
Update, I did a bit more digging. It turns out
\begin{align}
\frac{1}{1-\frac{1}{k}}
\, _3F_2\left(1,\frac{1}{k},1-\frac{1}{k};1+\frac{1}{k},2-\frac{1}{k};1\right)
=
\frac{\pi \cot \left(\frac{\pi }{k}\right)}{k-2}
.
\end{align}
The last equality id due to
$$
\, _3F_2(1,b,1-b;b+1,2-b;1)=\frac{(\pi b (1-b)) \cot (\pi b)}{1-2 b}.
$$
See here.
In other words, we need to show
$$
\sum _{r=0}^{\infty } \frac{\left(\frac{2}{k}\right)_r}{(r+1) \left(1+\frac{1}{k}\right)_r}
=
\frac{1}{1-\frac{1}{k}}
\, _3F_2\left(1,\frac{1}{k},1-\frac{1}{k};1+\frac{1}{k},2-\frac{1}{k};1\right)
.
$$
Update. I have solved this one.
$$
\sum _{r=0}^{\infty } \frac{\left(\frac{2}{k}\right)_r}{(r+1) \left(1+\frac{1}{k}\right)_r} =
\, _3F_2 (1,2/k,1;1+1/k,2;1).
$$
Then we can apply Watson's sum
$$
{{}_{3}F_{2}}\left({a,b,c\atop\frac{1}{2}(a+b+1),2c};1\right)=\frac{\Gamma%
\left(\frac{1}{2}\right)\Gamma\left(c+\frac{1}{2}\right)\Gamma\left(\frac{1}{2%
}(a+b+1)\right)\Gamma\left(c+\frac{1}{2}(1-a-b)\right)}{\Gamma\left(\frac{1}{2%
}(a+1)\right)\Gamma\left(\frac{1}{2}(b+1)\right)\Gamma\left(c+\frac{1}{2}(1-a)%
\right)\Gamma\left(c+\frac{1}{2}(1-b)\right)},
$$
See here.
|
With Euler’s integral transform we get :
$$\begin{eqnarray} \sum\limits_{n=0}^\infty \frac{(2x)_n}{(n+1)(1+x)_n} & = & {_3 F_2}(1,1,2x;2,x+1;1) \\ & = &\frac{\Gamma(x+1)}{\Gamma(2x)\Gamma(1-x)} \int\limits_0^1 t^{2x-1} (1-t)^{-x} {_2 F_1}(1,1;2;t)dt \\ & = & \frac{\Gamma(x+1)}{\Gamma(2x)\Gamma(1-x)} \int\limits_0^1 t^{2x-1} (1-t)^{-x} \frac{-\ln(1-t)}{t}dt \\ & = & \frac{\Gamma(x)}{\Gamma(2x)\Gamma(1-x)}\frac{d}{dz} \int\limits_0^1 t^{2x-2} (1-t)^{-xz}dt |_{z=1} \\ & = & \frac{\Gamma(x)}{\Gamma(2x)\Gamma(1-x)} \Gamma(2x-1)\frac{x\Gamma(1-x)(\psi(x)-\psi(1-x))}{\Gamma(x)} \\ & = & \frac{\pi x\cot(\pi x)}{1-2x} \end{eqnarray}$$
Where we used the reflection formula for the digamma function $\psi$ in the last step. With $\,\displaystyle x:=\frac{1}{k}\,$ your question is answered.
|
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|
Hi, I need help with the assignment, Find the point $P$ on the sphere $x^2 + y^2 + z^2 − 8 x − 4 y − 10 z = 4$ which is furthest from the plane Find the point $P$ on the sphere
$x^2 + y^2 + z^2 − 8x − 4y − 10z = 4$
which is furthest from the plane
$6x + 2y − 9z = −23$.
|
Let the point $P \equiv (p, q, r)$
The distance of $P$ from $6x+2y−9z=−23$ is given by $\frac{6p+2q−9r+23}{\sqrt{6^2+2^2+9^2}} = \frac{6p+2q−9r+23}{11}$
In other words, we are required to maximize $\frac{6p+2q−9r+23}{11}$ subject to $p^2+q^2+r^2−8p−4q−10r-4= 0$
This is equivalent to the following problem:
Maximize $6p+2q−9r$ subject to $p^2+q^2+r^2−8p−4q−10r-4= 0$
The solution is standard:
Let $f = (6p+2q−9r) + \lambda (p^2+q^2+r^2−8p−4q−10r-4)$
Set $\frac{\partial f}{\partial p} = \frac{\partial f}{\partial q} = \frac{\partial f}{\partial r} = \frac{\partial f}{\partial \lambda} = 0$ and solve.
You'll get $x = \frac{8\lambda -6}{2\lambda}$, $y = \frac{4\lambda -2}{2\lambda}$ and $z = \frac{10\lambda +9}{2\lambda}$ from the first three equations.
Substitute these values into the fourth equation (which is same as the equation of the sphere) and solve for $\lambda$. The solution is $\lambda = \pm \frac{11}{14}$
Corresponding to the two values of $\lambda$, the two possible values of $P$ are $(-\frac{42}{11}, -\frac{14}{11}, \frac{63}{11})$ and $(\frac{42}{11}, \frac{14}{11}, -\frac{63}{11})$ respectively.
|
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|
Given that $a$ is an odd multiple of $1183$, find the greatest common divisor of $2a^2+29a+65$ and $a+13$. Given that $a$ is an odd multiple of $1183$, find the greatest common divisor of $2a^2+29a+65$ and $a+13$.
I know there exists some slick technique to simplify this problem. Any hints are greatly appreciated.
|
HINT
let $d=(2a^2+29a+65,a+13)$. We notice that $-13$ is a root of $2a^2+29a+39$. Another root is $-\frac32$
So we have $2a^2+29a+65=(a+13)(2a+3)+26$
Thus $d=((a+13)(2a+3)+26,a+13)\Rightarrow$ $d|(a+13)(2a+3)+26, d|a+13\Rightarrow d|(a+13)(2a+3)\Rightarrow d|26$
|
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|
Prove that $\cosh(x)=\sec(\theta )$ if $x=\ln(\sec \theta + \tan \theta)$ I'm trying to prove that $\cosh(x)=\sec(\theta )$ if $x=x=\ln(\sec \theta + \tan \theta)$. I've substituted the value of $x$ into $\cosh(x)$ to get $$\frac{e^{\ln(\sec \theta + \tan \theta)}+e^{-\ln(\sec \theta + \tan \theta)}}{2}$$ and simplified to get $$\frac{(\sec \theta +\tan \theta)+\frac{1}{(\sec \theta + \tan \theta)}}{2}$$. However, I do not know where to continue from here. Help would be greatly appreciated.
|
You're almost there!
$$\frac{\sec \theta + \tan \theta+\frac{1}{\sec \theta + \tan \theta}{}}{2}=\frac{1}{2}\left(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}+\frac{1}{\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}\right)$$ This simplifies to $$\frac{1}{2}\left(\frac{1+\sin\theta}{\cos \theta}+\frac{\cos \theta}{1+\sin\theta}\right)=\frac{1}{2}\left(\frac{1+2\sin \theta+\sin^2\theta+\cos^2\theta}{\cos \theta(1+\sin \theta)}\right)$$ Hence your expression is $$\frac{1}{2}\left(\frac{2(1+\sin\theta)}{\cos \theta(1+\sin \theta)}\right)=\frac{1}{\cos \theta}=\sec \theta$$ as required.
|
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|
Find the number of real roots for $x+\sqrt{a^2-x^2}=b$, $a>0$, $b>0$, as a function of $a$ and $b$
Given: (1) $x+\sqrt{a^2-x^2}=b$, $(a,b,x)\subset \mathbb R$, $a>0$, $b>0$.
Find: number of roots for (1), given possible values for $a$ and $b$.
This is a question from a book for the preparation for math contests.
It states as final answer: (a) 1 root if $b<a$; and (b) 2 roots if $a<b<a\sqrt{2}$.
I'm having difficulties on finding this answer. I don't know whether it is correct or perhaps I'm not finding the right approach.
I started moving $x$ in (1) to the left, to get
$$\sqrt{a^2-x^2}=b-x$$
Before proceeding with squaring both sides, I saved 2 needed conditions for checking the final solution (c1) $a^2-x^2\ge 0$ and (c2) $b-x\ge 0$. Then squaring both sides, we get:
$$a^2-x^2=b^2+x^2-2bx\Leftrightarrow 2x^2-2bx+(b^2-a^2)=0$$
with discriminant $\triangle$ defined by:
$$\triangle=4(2a^2-b^2)$$
From this it is easy to see that a condition for 2 roots is (c3) $\sqrt{2}a>b,$ and for 1 root is (c4) $\sqrt{2}a=b,$ as $a>0$ and $b>0$. Then I find the roots as $$x=\frac{2b\pm \sqrt{\triangle}}{4}=\frac{b\pm \sqrt{2a^2-b^2}}{2}$$
From this point, I can't see a way to reach the stated answer, if it is right.
Full solutions or helpful hints are welcome. Sorry if it is a duplicate.
|
You just forgot to check those conditions you mentioned. Indeed, you should check that for what values of $a$ and $b$ the conditions $-a\le x\le a$ and $b-x \ge 0$ are satisfied. The difficulty is to consider a few if then conditions.
If $b=\sqrt{2}a$ then $x=\frac{b}{2}$ is a potential solution. We should make sure that this solution satisfies our inequalities, that is, we should have $-a\le\frac{b}{2}\le a$ and $b-\frac{b}{2}\ge0$ which is equivalent to $b\le 2a$. But $b=\sqrt{2}a$ so this leads to $\sqrt{2}\le 2$ which is identically true. So if $b=\sqrt{2}a$ then the only answer is $x=\frac{b}{2}=\frac{\sqrt{2}}{2}a$.
If $b\gt\sqrt{2}a$ then $\Delta:=\sqrt{2a^2-b^2}<0$ and there is not real solution.
If $b\lt\sqrt{2}a$ then two potential answers will be $x_1=\frac{b}{2}+\frac{\sqrt{2a^2-b^2}}{2}$ and $x_2=\frac{b}{2}-\frac{\sqrt{2a^2-b^2}}{2}$. I guess you can see what to do from here. Just put $x_1$ and $x_2$ into the inqualities and see what restrictions will be obtained on $a$ and $b$. Then check these restrictions with the condition $b\lt\sqrt{2}a$ to see under what conditions $x_1$ and $x_2$ can be a solution.
Below, you can see a plot of the function $f(x)=x+\sqrt{a^2-x^2}$ with $a=1$. You can see easily that for what values of $b$ the equation $f(x)=b$ has a solution and indeed exactly how many solutions.
|
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|
Prove directly from the definition that $({1\over2}+\frac{1}{2^2}+...+\frac{1}{2^n})_n $ is Cauchy Prove directly from the definition that $({1\over2}+\frac{1}{2^2}+...+\frac{1}{2^n})_n$ is cauchy
I know from the definition of Cauchy that |$x_n$-$x_m$|< ϵ but how do you do this with |$\frac{1}{2^n}- \frac{1}{2^m}$|
what I've tried:
if $n\gt m$ then
$$ |\frac{1}{2^n}- \frac{1}{2^m}| \le |\frac{2^m-2^n}{2^n2^m}| \le |\frac{2^m +2^n}{2^{n+m}}| \le \frac{2^n+2^n}{2^{2n}}= \frac{1}{2^{n-1}} \le \frac{1}{2^{N-1}} \le \epsilon $$
and rearrange to get N $ \ge 1+ \frac{ln(\epsilon)}{ln(2)}$
is this correct?
|
Let $S_n=\sum_{k=1}^n\frac{1}{2^k}$. Then, we have for any $\epsilon>0$
$$\begin{align}
\left|S_n-S_m\right|&=\left|\sum_{k=\min(m,n)+1}^{\max(m.n)}\frac{1}{2^k}\right|\\\\
&=\left|\frac{1}{2^{\min(n,m)}}-\frac{1}{2^{\max(m,n)}}\right|\\\\
&\le \frac{1}{2^{\min(n,m)-1}}\\\\
&<\epsilon
\end{align}$$
whenever $\min(n,m)>1-\frac{\log(\epsilon)}{\log(2)}$
|
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|
What is $\gcd (x^{3}+6x^{2}+11x+6,x^{3}+1)$ When applying straight-forward Euclid's algorithm the result have fractional coefficients, but by factoring linear terms you get $x+1$. Which answer is right?
|
$(x^3+6x^2+11x+6,x^3+1)=((x+1)(x^2-x+1),(x+1)(x+2)(x+3)) = (x+1)((x^2-x+1),(x+2)(x+3)) = (x+1)$
|
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|
Polynomial remainder when divisor has double roots How to find the remainder of the division of $P(x) = x^{444} + x^{111} + x - 1$ by $Q(x) = (x^2 + 1)^2$?
By the little Bézout's theorem, we can write
$$P(x) = (x^2 + 1)^2 W(x) + ax^3+bx^2+cx+d.$$
In fact, I've found out that, since $(x^2 + 1)$ divides $P(x)$, we can write:
$$P(x) = (x^2 + 1)^2 W(x) + cx^3+dx^2+cx+d.$$
Next, I'd usually plug the roots of $(x^2 + 1)^2$, which are $\{-i,i\}$, into the preceding expression, to get a system of equations that would allow me to find the unknowns $a,b,c,d$ (or just $c,d$ in this case). But as each root has multiplicity 2, I end up with an indeterminate system. What am I missing?
|
We have
\begin{eqnarray}
P(x)&=&(x^2+1)^2W(x)+ax^3+bx^2+cx+d\\
P'(x)&=&4x(x^2+1)W(x)+(x^2+1)^2W'(x)+3ax^2+2bx+c
\end{eqnarray}
and
$$
P(i)=i^{444}+i^{111}+i-1=(i^4)^{111}+i^{108}i^3+i-1=1-i+i-1=0
$$
It follows that
$$
0=P(i)=ai^3+bi^2+ci+d=-ai-b+ci+d=(d-b)+(c-a)i
$$
Therefore
$$
a=c,\quad b=d
$$
Also
$$
P'(x)=444x^{443}-111x^{110}-1
$$
therefore
$$
P'(i)=444i^{443}-111i^{110}-1=444i^{440}i^3-111i^{108}i^2-1=110-444i
$$
Substituting $c=a$, we get
$$
110-444i=P'(i)=3ai^2+2bi+c=c-3a+2bi=-2a+2bi
$$
it follows that
$$
a-bi=-55+222i
$$
Therefore
$$
a=c=-55,\quad b=d=-222.
$$
Hence the remainder is
$$
R(x)=-55x^3-222x^2-55x-222=-(55x+222)(x^2+1).
$$
|
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|
Simplifying expressions. How do you simplify the following expression?
$$Q=(1-\tan^2(x)) \left(1-\tan^2 \left(\frac{x}{2}\right)\right)\cdots \left(1-\tan^2\left(\frac{x}{2^n}\right)\right)$$
I've tried the the following :
$\tan x = \frac{\sin x}{\cos x}$
$$\cos x = \frac{\sin 2x}{2\sin x}$$
$$\tan x = \frac{2\sin^2x}{\sin2x}$$
$$Q= \left( 1- \left(\frac{2\sin^2x}{\sin2x}\right)^2\right) \left(1-\left(\frac{2\sin^2\frac{x}{2}}{\sin x}\right)^2\right) \cdots \left(1-\left(\frac{2\sin^2\frac{x}{2^{n+1}}}{\sin(\frac{x}{2^n})}\right)^2\right)$$
But i have no idea how to continue.
Help appreciated!
|
A more straightforward hint comes from the tangent double angle formula:
$$
\tan(2x) = \frac{2\tan{x}}{1-\tan^2 x}.
$$
You'll then find you have a telescoping product.
|
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|
Prove or Disprove There Exist Only One Pair of Four Positive Integers whose sum equals their product. I have the following proof:
"We know that $$1+2+3=1\cdot 2\cdot 3.$$ Prove or Disprove: There exist four positive integers whose sum equals their product."
Now this seems pretty obvious at first that it can never exists. But the proposition also never states the the integers on the let hand side have to match the integers on the right hand side. With that said, how can that be used to help me solve this proof. I was thinking about using Minimum Counterexample but am unclear how to start.
|
To complete the task, we shall prove that the only solution is $1,1,2,4$.
First, we shall prove the following:
Assume that $x,y$ are real numbers such that $y\ge x\ge 2$. Then
$xy\ge x+y$ and if $xy=x+y$ then $x=y=2$.
Proof:
$xy\ge 2y=y+y\ge x+y$.
If $x>2$ then $xy>2y\ge x+y$.
If $y>x$ then $xy\ge 2y>x+y$.
Now assume that $abcd=a+b+c+d$ and rename the numbers so that $a\le b\le c\le d$.
Suppose that $a\ge 2$. Since $2,2,2,2$ is not a solution, $d>2$.
Now we have that $abcd>abc+d$. Since $bc>a\ge 2$, $abc>a+bc\ge a+b+c$. Then $abcd>a+b+c+d$.
We get a contradiction. Then, $a=1$.
Now we have $bcd=1+b+c+d$. Suppose now that $b\ge 2$. Since $1,2,2,2$ is not a solution, $d> 2$. Similarly, we see that $bcd>bc+d\ge b+c+d$. So $b=1$.
The equation is now:
$$cd=2+c+d$$
so
$$c=\frac{d+2}{d-1}=1+\frac3{d-1}$$
Since $c$ is integer, $d-1$ must be $1$ or $3$. For $d-1=1$ we get the solution $1,1,4,2$ and for $d-1=3$ we get $1,1,2,4$.
|
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Maximum possible value of $P(10)$ The real numbers $a$, $b$, $c$, and $d$ are each less than or equal to $12$. The polynomial $$P(x)=ax^3+bx^2+cx+d$$ satisfies $P(2)=2$, $P(4)=4$ and $P(6)=6$. Find the maximum possible value of $P(10)$.
What I did was first I used the given information to get $3$ equation in $a, b,c,d$. Then I obtained values of $a$, $b$, $c$ in terms of $d$. Then using these values, I found $P(10)$ in terms of $d$ and substituted $d=12$. But I couldn't arrive at the answer.
Thanks in advance
|
From the system
$
\left\{
\begin{array}{l}
8 a+4 b+2 c+d=2 \\
64 a+16 b+4 c+d=4 \\
216 a+36 b+6 c+d=6 \\
\end{array}
\right.
$
we get
$b = -12 a, c = 1 + 44 a, d = -48 a$
with the constraints
$-12 a\leq 12\land 44 a+1\leq 12\land -48 a\leq 12$
which give the limitation
$-\frac{1}{4}\leq a\leq \frac{1}{4}$
Therefore plugging in the polynomial we have
$P(x)=a x^3-12 a x^2+(44 a+1) x-48 a$
and $P(10)=2 (96 a+5)$
which is maximum for $a=\frac{1}{4}$
and $P(10)=58$
Hope this is useful
|
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What is the number of 5 digit numbers divisible by 3? What is the number of $5$ digit numbers divisible by $3$ using the digits
$0,1,2,3,4,6,7$ and repetition is not allowed?
|
I mostly agree with the previous answer. You can add up all seven numbers in that set which is equal $23$. For a five digit number to be divisible by $3$ the sum of its digits has to be a multiple of three.
So we have to find all possible combinations of $5$ digit numbers that when all digits are added equals a multiple of $3$.
The sum of the $7$ numbers equals $23$ and we have to subtract $23$ by two numbers from the set to get the sum of five numbers. $23 - a - b$
Where $a+b$ has to equal a multiple of $3 +2$. e.g. $3n+2$. So like $5$ or $8$. Because $23 - (3n+2) = 21 - 3n$ which is divisible by $3$ meaning the five digit number is divisible by $3$.
So we find all possible pairs that equals a multiple of $3 + 2$
$(0,2) (1,4) (1,7) (2,3) (2,6) (4,7)$
Which is similar to the previous answer except there are more pairs. There are $5!$ or 120 ways to order $5$ digits so the answer is $6×120=720$
This is true assuming that $02367$ is considered to be a $5$ digit number. It might not be considered a $5$ digit number since it starts with zero. Otherwise the answer would be $720 - 5×4! = 600$
|
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|
Proof $\lim\limits_{n \rightarrow \infty} \sqrt{n} (3^{\frac{1}{n}} - 2^{\frac{1}{n}}) = 0$ How would one go about proving the limit of sequence:
$$\lim_{n \rightarrow \infty}\ \sqrt{n} (3^{\frac{1}{n}} - 2^{\frac{1}{n}}) = 0\,. $$
Epsilon definiton of limit seems to be too complicated and/or unsolvable (variable both in exponent and base). $\lim\limits_{x \rightarrow \infty} \ 3^{\frac{1}{n}}$ is clearly $1$, but how does it combine with $\sqrt{n}$?
Is there an obvious upper bound that I'm missing?
|
From
$$1=3-2=\left(3^{\frac{1}{n}}\right)^n-\left(2^{\frac{1}{n}}\right)^n=\\
\left(3^{\frac{1}{n}}-2^{\frac{1}{n}}\right)\left( 3^{\frac{n-1}{n}}+3^{\frac{n-2}{n}} 2^{\frac{1}{n}}+3^{\frac{n-3}{n}} 2^{\frac{2}{n}}+...+3^{\frac{1}{n}} 2^{\frac{n-2}{n}} + 2^{\frac{n-1}{n}}\right)$$
we have
$$0<\sqrt{n}\left(3^{\frac{1}{n}}-2^{\frac{1}{n}}\right)=\frac{\sqrt{n}}{3^{\frac{n-1}{n}}+3^{\frac{n-2}{n}} 2^{\frac{1}{n}}+3^{\frac{n-3}{n}} 2^{\frac{2}{n}}+...+3^{\frac{1}{n}} 2^{\frac{n-2}{n}} + 2^{\frac{n-1}{n}}}<\\
\frac{\sqrt{n}}{2^{\frac{n-1}{n}}+2^{\frac{n-2}{n}} 2^{\frac{1}{n}}+2^{\frac{n-3}{n}} 2^{\frac{2}{n}}+...+2^{\frac{1}{n}} 2^{\frac{n-2}{n}} + 2^{\frac{n-1}{n}}}=
\frac{\sqrt{n}}{n2^{\frac{n-1}{n}}}<
\frac{1}{\sqrt{n}}$$
Or
$$0<\sqrt{n} \left(3^{\frac{1}{n}}-2^{\frac{1}{n}}\right)<\frac{1}{\sqrt{n}}$$
And use the squeeze theorem.
|
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GCD of two elements in $\mathbb Z \left[\frac{1 + \sqrt{-11}}{2}\right]$ I have to find $(3 + \sqrt{-11}, 2 + 4\sqrt{-11})$ in $\mathbb Z \left[\frac{1 + \sqrt{-11}}{2}\right]$.
If $\mathbb Z \left[\frac{1 + \sqrt{-11}}{2}\right]$ is an Euclidean domain, the euclidean algorithm should be acceptable for computing the GCD of any two elements of the ring.
And here is the trouble: trying to divide given elements at some step I get a remainder with norm bigger than the divider's one. I'm stuck and can't find a solution.
Now I'm not sure if i use the method correctly. Could you please explain me how to do it in the right way or suggest another method?
|
Well, let's see: $$\frac{3 + \sqrt{-11}}{2 + 4 \sqrt{-11}} = \frac{5 - 11 \sqrt{-11}}{18}.$$ Since $$\frac{5}{18} \approx \frac{1}{2}$$ and $$\frac{-11}{18} \approx -\frac{1}{2}$$ we have $$3 + \sqrt{-11} = (2 + 4 \sqrt{-11})\left(\frac{1}{2} - \frac{\sqrt{-11}}{2}\right) - 20.$$
Yeah, that is a problem. Try $$\frac{5 - 11 \sqrt{-11}}{18} \approx \sqrt{-11}$$ instead. Then $3 + \sqrt{-11} = (2 + 4 \sqrt{-11})(\sqrt{-11}) + 47.$ Hmm, even worse.
There is one third possibility to try: $$3 + \sqrt{-11} = (2 + 4 \sqrt{-11})\left(-\frac{1}{2} - \frac{\sqrt{-11}}{2}\right) + (-18 + 2 \sqrt{-11}).$$ That's the worst one so far. Maybe this is not a Euclidean domain after all.
Wait, did we compare norms for the two GCD function arguments before doing anything else? In $\mathbb Z$ we don't need to worry about that, the algorithm winds up flipping them as needed. But maybe in this domain, $\mathcal O_{\mathbb Q(\sqrt{-11})}$, it is essential. Obviously $N(2 + 4 \sqrt{-11}) > N(3 + \sqrt{-11})$. So then we try $$\frac{2 + 4 \sqrt{-11}}{3 + \sqrt{-11}} = \frac{5}{2} + \frac{\sqrt{-11}}{2},$$ which is an integer in this domain.
That means $2 + 4 \sqrt{-11}$ is divisible by $3 + \sqrt{-11}$, surprise, huh!? Indeed, $N(2 + 4 \sqrt{-11}) = 180$ and $N(3 + \sqrt{-11}) = 20$. So $$2 + 4 \sqrt{-11} = (3 + \sqrt{-11}) \left(\frac{5}{2} + \frac{\sqrt{-11}}{2}\right) + 0.$$ The algorithm works.
|
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Solving system of linear equations with rationals Problem:$$(x+11):(x+6)=(y+12):(y+7)\land(y+1):(x-1)=y:x$$
When I tried using $\frac{a1}{b1}=\frac{a2}{b2}=\frac{k1a1+k2a2}{k1b1+k2b2}$ for $\frac{x+11}{x+6}=\frac{y+12}{y+7}$ where k1 = 1 and k2 = -1, this is what I got:$$\frac{x+11-y-12}{x+6-y-7}=\frac{x-y-1}{x-y-1}=1$$
Which would mean $\frac{x+11}{x+6}=1$ and $\frac{y+12}{y+7}=1$, which are false.
If I instead transform $(x+11):(x+6)=(y+12):(y+7)$ into $\frac{x+11}{y+12}=\frac{x+6}{y+7}$ then $$\frac{x+11-x-6}{y+12-y-7}=\frac{5}{5}=1$$
Which is possible for $\frac{x+11}{y+12}$ and $\frac{x+6}{y+7}$. The equations are easy and could be solved simply by other means. I must be missing something fundamental.
Please help. Thanks.
|
Hint:
First absurd result came up as $x-y-1$ is actually $=0$ here.
For example $$\dfrac23=\dfrac46=\dfrac{2\cdot2-4}{2\cdot3-6}=?$$
Just cross multiply to form two simultaneous equations in $x,y$
|
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Evaluate line integral $\int_c x\,dx+ y\,dy + z\,dz$, where $C$ is the straight line from $(1,0,0)$ to $(0,1,\pi/2)$
$ \displaystyle \int_c x\,dx+ y\,dy + z\,dz$, where $C$ is the straight line from
$(1,0,0)$ to $(0,1,\pi/2)$
Parametric form of line will be:
$$x=1 - t, \quad y= t, \quad z= \frac{\pi t} 2$$
The integral becomes
$$\int_0^1 2t -1 + \frac{\pi^2t} 4 \, dt$$
but the final answer given in my book is $\pi^2/8$
which I am not getting by this.
So where am I making the mistake ?
|
Given $x = 1-t, y = t, z = \frac{\pi}{2} t$ then,
\begin{align*} x\, \textrm{dx} + y\, \textrm{dy} + z\, \textrm{dz} &=(1-t)(-1)\,dt + t(1)\, \rm{dt} + \frac{\pi}{2} t \left( \frac{\pi}{2}\right) \, \rm{dt} \\ \\ & = \left(-1+t+t+ \frac{\pi^2}{4}t\right) \, \rm{dt} \\ \\ & = \left( t \left( 2+\frac{\pi^2}{4} \right) - 1 \right) \, \rm{dt} \end{align*}
Okay so that's fine. Now integrate this on $[0,1]$.
\begin{align*} \int_0^1 t \left( 2+\frac{\pi^2}{4} \right) - 1 \ \rm{dt} & = \left. \frac{1}{2}\left( 2+\frac{\pi^2}{4} \right) t^2 \right|_{t = 0}^{t=1} - \left. t \vphantom{\frac11} \, \right|_{t = 0}^{t=1} \\ \\ & = \frac{1}{2}\left( 2+\frac{\pi^2}{4} \right) - 1 \\ \\ & = \frac{\pi^2}{8}\end{align*}
|
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writing Cosines using De Moivre's formula Given the question:
Use De Moivre’s formula to find a formula for $\cos(3x)$ and $\cos(4x)$ in terms of $\cos(x)$ and $\sin(x)$. Then use the identity $\cos^2(x) + \sin^2(x) = 1$ to express these formulas only in terms of $\cos(x)$.
I started out by rewriting $\cos(3x)$:
$\cos(3x)$+$i \sin(3x)$=($\cos(x)$+$i \sin(x)$)$^3$
This could then be written into
$\cos(3x) = \cos^3(3x)-3 \cos(x) \sin^2(x)$
or
$\sin(3x) = \cos^2(x) \sin(x)- \sin^3(x)$
Then to use the identity I would substitute $1-\cos^2(x)$ for the $\sin^2(x)$
and in the second I would substitute $\sin(x)$ for $\sqrt{1 - cos(x)}$ right and would need to separate the $\sin^3(x)$ into $\sin^2(x) * \sin(x)$ and substitute from there. I'm a lot less confident about the second equation substitution. Would this be the right way to go about doing this problem?
|
$${ \left( \cos { x } +i\sin { x } \right) }^{ 3 }=\cos { \left( 3x \right) +i\sin { \left( 3x \right) } } \\ \cos ^{ 3 }{ x } +3i\cos ^{ 2 }{ x\sin { x } -3\cos { x\sin ^{ 2 }{ x } -i\sin ^{ 3 }{ x } = } } \cos { \left( 3x \right) +i\sin { \left( 3x \right) } } \\ \\ \cos { \left( 3x \right) =\cos ^{ 3 }{ x } -3\cos { x } \sin ^{ 2 }{ x } =\cos ^{ 3 }{ x } -3\cos { x } \left( 1-\cos ^{ 2 }{ x } \right) =4\cos ^{ 3 }{ x-3\cos { x } } } \\ \sin { \left( 3x \right) =3\cos ^{ 2 }{ x } \sin { x } -\sin ^{ 3 }{ x } } =3\left( 1-\sin ^{ 2 }{ x } \right) \sin { x } -\sin ^{ 3 }{ x } =3\sin { x } -4\sin ^{ 3 }{ x } \\ $$
|
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Find all pairs of complex numbers (x,y) satisfying $xy=-1$ and $x^3-y^3=-1$ the problem asks me to find all pairs of complex numbers (x,y) satisfying $xy=-1$ and $x^3-y^3=-1$ simultaneously. I set the two equations to equal each other which yields
$x^3-xy-y^3=0$ but from this point on on im stuck on how i would solve for $x$ and $y$. any help would be appreciated thanks!
|
It seems that anon gave some explanation of the suggestion I made.
Also, you do have questions using complex numbers, including some with $e^{m \pi i / n}.$
So, your original variable $x$ satisfies $x^6 + x^3 + 1 = 0.$ Note that we have $x^3 \neq 1,$ otherwise we would have $1+1+1=0,$ which is false. Next,
$$ 0 = (x^3 - 1)(x^6 + x^3 + 1) = x^9 - 1. $$ Notice that the comment by anon points out that the solutions $x$ are precisely the primitive 9th roots of unity.
So far, $$ x^9 = 1, \; \; \mbox{but} \; \; \; x^3 \neq 1. $$
Define $$ \omega = e^{2 \pi i / 9} = \cos \frac{2 \pi}{9} + i \, \sin \frac{2 \pi}{9} \; \; . $$
The six answers to the original problem are $$ x = \omega, \omega^2, \omega^4, \omega^5, \omega^7, \omega^8. $$
In each case we take
$$ y = - \bar{x}. $$
To be specific, when $x = a + b i,$ we get $y = -a + bi.$
For example, when $x = \omega,$ we get
$$ y = - \cos \frac{2 \pi}{9} + i \, \sin \frac{2 \pi}{9} = e^{7 \pi i / 9} \; \; . $$
|
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Can we generalize $\sum_{n=0}^\infty \binom{4n}{n}\frac{1}{(3n+1)\,2^{4n+1}}=\frac1{T}$ for tribonacci constant $T$? (Inspired by this post.) Given the tribonacci constant $\Phi_3$, the tetranacci constant $\Phi_4$, etc. How do prove that,
$$\sum_{n=0}^\infty \binom{4n}{n}\frac{1}{(3n+1)\,2^{4n+1}}=\frac{1}{2}{\;}_3F_2 \left(\frac14,\frac24,\frac34; \color{blue}{\frac23,\frac43};\frac{4^4}{\color{red}{2^4}\,3^3} \right)=\frac1{\Phi_3}$$
$$\sum_{n=0}^\infty \binom{5n}{n}\frac{1}{(4n+1)\,2^{5n+1}}=\frac{1}{2}{\;}_4F_3 \left(\frac15,\frac25,\frac35,\frac45; \color{blue}{\frac24,\frac34,\frac54};\frac{5^5}{\color{red}{2^5}\,4^4} \right)=\frac1{\Phi_4}$$
$$\sum_{n=0}^\infty \binom{6n}{n}\frac{1}{(5n+1)\,2^{6n+1}}=\frac{1}{2}{\;}_5F_4 \left(\frac16,\frac26,\frac36,\frac46,\frac56; \color{blue}{\frac25,\frac35,\frac45,\frac65};\frac{6^6}{\color{red}{2^6}\,5^5} \right)=\frac1{\Phi_5}$$
and so on? And what is the corresponding hypergeometric formula for $\displaystyle \frac1{\Phi_2}$ with golden ratio $\Phi_2$?
(Courtesy of Jack D'Aurizio) By the Lagrange inversion theorem, a solution to,
$$x^k-x+1=0$$
is given by,
$$x=\sum_{n=0}^\infty \binom{kn}{n}\frac{1}{(k-1)n+1}$$
Similarly, given the general polynomial of the $k$-nacci constants as,
$$x^k(2-x)-1=0$$
how do we show that,
$$\frac1{x}=\frac1{\Phi_k}=\sum_{n=0}^\infty \binom{(k+1)n}{n}\frac{1}{(kn+1)\,2^{(k+1)n+1}}$$
|
Note that $\frac{1}{kn+1}\binom{(k+1)n}{n}$ is the number of $(k+1)$-ary trees on $n$ vertices. Let $T_{k+1}$ be the ordinary generating function for the class of $(k+1)$-ary trees. Then $T_{k+1}$ satisfies the equation
$$
T_{k+1}(z)=1+zT_{k+1}^{k+1}(z).
$$
Let $a=\frac{1}{2}T_{k+1}\left(\frac{1}{2^{k+1}}\right)$, then we need to prove that $\frac{1}{a}=\Phi_k$. From the functional equation, we have
$$
2a=1+\frac{1}{2^{k+1}}(2a)^{k+1}=1+a^{k+1},
$$
so $a$ satisfies the equation
$$
1-2x+x^{k+1}=0,
$$
and thus $\frac{1}{a}$ satisfies
$$
x^{k+1}-2x^k+1=(x-1)(x^k-x^{k-1}-\dots-x-1)=0.
$$
We will leave the argument that $0<a<1$ for $k\ge 2$ to the reader. Then $\frac{1}{a}$ is a root of the equation $\phi_k(x)=x^k-x^{k-1}-\dots-x-1=0$. From Proposition 1 on p. 6 of this paper, we see that $\phi_k(x)$ has a unique root greater than $1$, so $a$ is uniquely determined by this polynomial. However, $\Phi_k$ is also the unique root greater than $1$ of $\phi_k(x)$, so $\frac{1}{a}=\Phi_k$.
Perhaps, it would be interesting to interpret both sides probabilistically. A few years ago, I think I have actually heard of that problem for $k=1$, where the sum is $1$, but I'm not sure it's still open or not.
|
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Determine if the series 1/5 + 1/8 + 1/11 + 1/14 + 1/17 ... is convergent or divergent I am trying to determine if the series $\frac{1}{5} + \frac{1}{8} + \frac{1}{11} + \frac{1}{14} + \frac{1}{17} $ is convergent or divergent?
I can see that the terms in the denominator differ by 3 each time.
|
The pattern for the denominator is found by first adding another term so that the series starts $\frac{1}{2} + \frac{1}{5} + \frac{1}{8} + ...$ then for the series $2 + 5 + 8 + 11 ... $ the general nth term is found as $ T_n = a + (n-1)d$ where $a = \frac{1}{2}$ is the first term and $d = 3$ the difference. So $T_n = 3n -1$ and the original series is found to be $\sum _{n=1}^\infty \frac{1}{3n-1}$
If we let $f(x) = \frac{1}{3x-1}$ then $f'(x) = \frac{-1}{(3x-1)^2} $ which is negative for all $n \geq1$, which means it is decreasing on $[1,\infty)$. This function is also positive for all $n \geq1$ and so the conditions are met for the integral test.
$$\int_1^\infty f(x) =\lim _{b\to\infty}\int_1^b \frac{1}{(3x-1)} $$
$$= \lim _{b\to\infty}\frac{1}{3}\ln(3x-1)|_1^b$$
$$= \lim _{b\to\infty}\frac{1}{3}[\ln(3b-1) - \ln(2)]$$
$$=\infty$$
By the integral test the series $\sum _{n=1}^\infty \frac{1}{3n-1}$ is therefore divergent.
|
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Representing a given number as the sum of two squares.
There are four essentially different representations of $1885$ as the sum of squares of two positive integers. Find all of them.
I am guessing the must be some nice way of solving $z= x^2 + y^2 $ where x and y are integers in general that i don't know.
Is there some general way to know how many solution exists or bound it without computing it?
|
Since $\mathbb{Z}[i]$ is a UFD, by letting
$$ r_2(n)=\left|\left\{(a,b)\in\mathbb{Z}^2:a^2+b^2=n\right\}\right|$$
we have
$$ r_2(n) = 4(\chi_4*1)(n) = 4\sum_{d\mid n}\chi_4(d) $$
where $\chi_4(m)$ equals $1$ if $m\equiv 1\pmod{4}$, $-1$ if $m\equiv 3\pmod{4}$ and zero otherwise.
This is essentially due to the Brahmagupta-Fibonacci identity
$$ (a^2+b^2)(c^2+d^2) = (ad-bc)^2+(ac+bd)^2 $$
equivalent to the multiplicativity of the norm over $\mathbb{Z}[i]$.
Since $1885=5\cdot 13\cdot 29$ and $5=1^2+2^2$, $13=2^2+3^2$, $29=2^2+5^2$, from
$$(1+2i)(2+3i)(2+5i)=-43-6i$$
$$(1-2i)(2+3i)(2+5i)=21+38i$$
$$(1+2i)(2-3i)(2+5i)=11+42i$$
$$(1+2i)(2+3i)(2-5i)=27+34i$$
we get that the wanted representations are
$$ 1885=6^2+43^2 = 21^2+38^2 = 11^2+42^2=27^2+34^2.$$
|
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Trigonometric equation result differs from given i've got an equation:
$$\sin^6(x) + \cos^6(x) = 0.25$$
and i'm trying to solve it using the sum of cubes formula, like this:
$$ (\sin^2(x))^3 + (\cos^2(x))^3 = 0.25 $$
$$ (\sin^2(x) + \cos^2(x))^2 (\sin^4(x) - \sin^2(x)\cos^2(x) + \cos^4(x)) = 0.25 $$
$$ 1 - 3\sin^2(x)\cos^2(x) = 0.25 $$
$$ -3 \sin^2(x)\cos^2(x) = -\frac{3}{4} $$
$$ \sin^2(2x) = 1 $$
$$ \sin^2(2x) = \sin^2(2x) + \cos^2(2x) $$
$$ \cos^2(2x) = 0 $$
and here it must be $x = \frac{\pi n}{2} \pm \frac{\pi}{4} $
but solution is $ x = \pi n \pm \frac{\pi}{4}$
What's wrong?
|
Both the answers are correct. Since, both represent the same solution set.
This is so because each odd number can be either represented as $2n+1$ and $2m-1$, or represented as $4p+1$ or $4q-1$. And hence $$2x=(2n\pm 1)\frac{\pi}{2} \implies \frac{n\pi}{2} \pm \frac{\pi}{4} $$
$$ \text{also} \; 2x =(4n\pm 1)\frac{\pi}{2} \implies n\pi \pm \frac{\pi}{4} $$ So, both the solution set are basically the same.
It although doesn't sound to be convincing, so you can try to write a few elements of your set
$$\mathrm S_1= \frac{n \pi}{2} \pm \frac{\pi}{4} =\left \{\ldots -\frac{3\pi}{4}, -\frac{\pi}{4},\frac{\pi}{4},\frac{3\pi}{4} \ldots\right \} $$
$$\mathrm S_2= n \pi\pm \frac{\pi}{4} =\left \{\ldots -\frac{3\pi}{4}, -\frac{\pi}{4},\frac{\pi}{4},\frac{3\pi}{4} \ldots\right \} $$
|
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Calculate Riemann sums for $f(x)=4+3x^3$ for $x \in [-2,1]$ I have the following exercise:
Let $f(x)$ = $4+3x^3$, $x$ $\in$ $[-2; 1]$. Calculate the Riemann sums for $f$ on $[-2; 1]$ by dividing the interval into n equal sub-intervals and taking midpoints as sample points. Find the limit of Riemann sums as $n \to \infty$.
I have used the general formula
$\lim_{n \to \infty}$ $\frac{1}{n}$$\sum_{k=1}^{n}$$f(\frac{x_{k-1}+x_k}{2})$
Then I used $x_k$ formula($x_k=a+\frac{k(b-a)}{n}$) and got $\frac{x_{k-1}+x_k}{2}$ = $\frac{6k-3-4n}{2n}$
Then I plugged in my $x_k$ sample points and got the following:
\begin{align}
&\lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}4+(3\frac{6k-3-4n}{2n})^3\\&=
\lim_{n \to \infty}\frac{3}{2n^3}\sum_{k=1}^{n}(6k-(3+4n))^3\\&=
\lim_{n \to \infty}\frac{3}{2n^3}\sum_{k=1}^{n}(216k^3-108k^2(3+4n)+18k(3+4n)^2-(3+4n)^3)\\&=
\lim_{n \to \infty}\frac{3}{2n^3}(-n(3+4n)^3)\sum_{k=1}^{n}216k^3-\sum_{k=1}^{n}108k^2(3+4n)+\sum_{k=1}^{n}18k(3+4n)^2\\&=
\lim_{n \to \infty}\frac{3}{2n^3}(-n(3+4n)^3)216\sum_{k=1}^{n}k^3-108(3+4n)\sum_{k=1}^{n}k^2+18(3+4n)^2\sum_{k=1}^{n}k\\&=
\lim_{n \to \infty}\frac{3}{2n^3}(-n(3+4n)^3)216(\frac{n(n+1)}{2})^2-108(3+4n)(\frac{n(n+1)(2n+1)}{6})+18(3+4n)^2(\frac{n(n+1)}{2}) \end{align}
The limit of the last part is $-\infty$, however, I should get $\frac{3}{4}$ as it is another way to calculate the definite integral. Can someone please tell if there is something I have done wrong or there is a better way to solve this problem?
|
Mistake $1$:
$$\lim_{n \to \infty}\frac{1}{\color{blue}n}\sum_{k=1}^{n}\left[4+\color{red}3(\frac{6k-3-4n}{2n})^3\right]= \color{red}{4}+
\lim_{n \to \infty}\frac{3}{\color{red}{8n^4}}\sum_{k=1}^{n}(6k-(3+4n))^3$$
Mistake $2$:
\begin{align}&\lim_{n \to \infty}\frac{3}{2n^3}\sum_{k=1}^{n}(216k^3-108k^2(3+4n)+18k(3+4n)^2-(3+4n)^3)\\&=
\lim_{n \to \infty}\frac{3}{2n^3}\color{red}{(-n(3+4n)^3)}\sum_{k=1}^{n}216k^3-\sum_{k=1}^{n}108k^2(3+4n)+\sum_{k=1}^{n}18k(3+4n)^2 \end{align}
Correction:
\begin{align}&\color{blue}4+\lim_{n \to \infty}\frac{3}{\color{blue}{8n^4}}\sum_{k=1}^{n}(216k^3-108k^2(3+4n)+18k(3+4n)^2-(3+4n)^3)\\&=4+
\lim_{n \to \infty}\frac{3}{\color{blue}{8n^4}}\left[\color{red}{(-n(3+4n)^3)}\color{blue}+\sum_{k=1}^{n}216k^3-\sum_{k=1}^{n}108k^2(3+4n)+\sum_{k=1}^{n}18k(3+4n)^2\right] \end{align}
Mistake $3$:
The general formula should be
$$\lim_{n \to \infty}\frac{\color{red}{b-a}}{n}\sum_{k=1}^{n}f\left(\frac{x_{k-1}+x_k}{2}\right)$$
|
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|
If $p$ is prime, $p\ne3$ then $p^2+2$ is composite I'm trying to prove that if $p$ is prime,$p\ne3$ then $p^2+2$ is composite. Here's my attempt:
Every number $p$ can be put in the form $3k+r, 0\le r \lt 3$, with $k$ an integer. When $r=0$, the number is a multiple of 3, so that leaves us with the forms $3k+1$ and $3k+2$. The first one will be even when $k$ is odd, and the second one will be even when $k$ is even. So we will see what happens for each form in the case that $k$ is even (for the first form) and $k$ is odd (for the second form):
*
*$p=3k+1$, $k$ is even
Since $k$ is even, we can write it as $k=2q$ for some $q$. Then
$p^2+2=(3(2q)+1)^2+2=(6q+1)^2+2=6^2q^2+12q+1+2=3(12q^2+4q+1)$
So $p^2+2$ is composite.
*$p=3k+2$, $k$ is odd
Then, $k$ can be written as $k=2q+1$, for some $q$. Then
$p^2+2=(3(2q+1)+2)^2+2=(6q+5)^2+2=6^2q^2+60q+25+2=3(12q^2+10q+9)$
And again, $p^2+2$ is composite.
QED
Is that a correct proof? Is not the same that comes in the answer books.
|
Another observation is that primes $p\geq 5$ (the remaining $p=2$ can be checked manually) are of the following two forms $6k+1$ or $6k+5$ (simply because $6k+2, 6k+3, 6k+4$ are never primes). Then
*
*$p^2+2=36k^2+12k+3$ is divisible by 3
*$p^2+2=36k^2+60k+27$ is divisible by 3
|
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|
Is there a positive sequence $(a_n)_{n=1}^\infty$ where both $\sum_{n=1}^\infty \frac{1}{a_n}$ and $\sum_{n=1}^\infty \frac{a_n}{n^2}$ converge? Suppose $a_n$ is a positive sequence but not necessarily monotonic.
For the series $\sum_{n=1}^\infty \frac{1}{a_n}$ and $\sum_{n=1}^\infty \frac{a_n}{n^2}$ I can find examples where both diverge: $a_n = n$, and where one converges and the other diverges: $a_n = n^2$.
Can we find example where both converges?
|
Both series cannot converge.
Suppose $\sum a_n/n^2$ converges. Since the divergent harmonic series can be written as
$$\sum_{n=1}^\infty \frac{1}{n} = \sum_{\frac{1}{n} \leqslant \frac{a_n}{n^2}} \frac{1}{n} + \sum_{\frac{1}{n} > \frac{a_n}{n^2}} \frac{1}{n},$$
and the first series on the RHS converges, it follows that the second series diverges.
Therefore,
$$\sum_{n=1}^\infty \frac{1}{a_n} > \sum_{\frac{1}{a_n} > \frac{1}{n}} \frac{1}{a_n} > \sum_{\frac{1}{a_n} > \frac{1}{n}} \frac{1}{n} = \sum_{\frac{1}{n} > \frac{a_n}{n^2}} \frac{1}{n}= +\infty$$
|
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On the way of solving Integral $\int_{0}^{2\pi}\frac{x\sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$. Question
For $n > 0$, Find
$$\int_{0}^{2\pi}\frac{x\sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$$
My Approach
Let
$I_{n}$ =$$\int_{0}^{2\pi}\frac{x \sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$$
I$_{n}$=$\int_{0}^{2\pi}$$\frac{\left(2\pi-x\right)\sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$$\Longrightarrow$2I$_{n}=$$\int_{0}^{2\pi}$$\frac{2\pi \sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$
$\Longrightarrow$I$_{n}=\int_{0}^{2\pi}\frac{\pi \sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$
I am stuck here!
|
I will assume $n \in \mathbb{N}$. If in the integral
$$I_n = \int^{2\pi}_0 \frac{x \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx,$$
we set $x \mapsto 2\pi - x$ it can be readily seen that
$$I_n = \pi \int^{2\pi}_0 \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx.$$
Note the integrand
$$f(x) = \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x},$$
is both an even and periodic function with a fundamental period of $\pi$. If $f$ is a continuous bounded function with period $\mathfrak{a}$, then
$$\int^{b + \mathfrak{a}}_b f(x) \, dx = \int^\mathfrak{a}_0 f(x) \, dx,$$
where $b \in \mathbb{R}$.
Using this result repeatedly on our integral we have
\begin{align*}
I_n &= \pi \int^{-\pi + 2\pi}_{-\pi} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \qquad \text{(periodic with period $2\pi$)}\\
&= \pi \int^\pi_{-\pi} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx\\
&= 2\pi \int^\pi_0 \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \qquad \text{(since it is even)}\\
&= 2\pi \int^{-\pi/2 + \pi}_{-\pi/2} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \qquad \text{(periodic with period $\pi$)}\\
&= 4 \pi \int^{\pi/2}_0 \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \qquad \text{(since it is even)}.
\end{align*}
Now setting $x \mapsto \dfrac{\pi}{2} - x$ gives
$$I_n = 4 \pi \int^{\pi/2}_0 \frac{\cos^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx.$$
Thus
$$I_n + I_n = 4\pi \int^{\pi/2}_0 \frac{\sin^{2n} x + \cos^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx = 4\pi \int^{\pi/2}_{0} \, dx,$$
or
$$I_n = \int^{2\pi}_0 \frac{x \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx = \pi^2.$$
|
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|
Solving linear congruent system
Consider the following congruences system:
$$3x \equiv 1 \pmod{8} \\ x \equiv 7 \pmod {12} \\ x \equiv 4 \pmod
{15}$$
Find a minimal solution for the system
I was able to show using the Euclidean algorithm that $(-5)\cdot 3\equiv 1 \pmod 8$. so every solution for the first congruence must be of the form $$x = -5 +8k$$
So our system can be written as:
$$x = -5 +8k_1 \\ x = 7 + 12k_2 \\ x = 4 + 15k_3$$
I have found the relations between the $k$'s, but I'm not sure how to extract the general form of a solution for $x$.
I'd be glad for help!
|
HINT reduce the system to $\pmod {p_i}$ and apply CRT
You can expand in this way:
$\begin{cases} 3x \equiv 1 \pmod{2^3} \\ x \equiv 7 \pmod {2^2\cdot3} \\ x \equiv 4 \pmod{3\cdot5} \end{cases}$
$\begin{cases} x \equiv 3 \pmod{2^3} \\ x \equiv 3 \pmod {2^2} \\x \equiv 1 \pmod {3} \\ x \equiv 4 \pmod{5} \end{cases}$
$\begin{cases} x \equiv 3 \pmod{2^3} \\x \equiv 1 \pmod {3} \\ x \equiv 4 \pmod{5} \end{cases}$
CRT guarantees that solution exist unique $\pmod{120}$
https://en.wikipedia.org/wiki/Chinese_remainder_theorem
|
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Prove that $2^{3n+1} + 5$ is a multiple of 7 for all n ≥ 0. As the title states I need to prove that $(2^{3n+1}+5)$ is a multiple of 7 for all $n \geq 0$.
I can do this using induction but I also want to prove it using modular arithmetic. So here's what I've got so far.
Start with Fermat's little theorem:
\begin{align}2^6 &\equiv 1 \ \ \pmod{7} \\ 2^3 \cdot 2^3 &\equiv 1 \ \ \pmod{7} \\8 \cdot 2^3 &\equiv 1 \ \ \pmod{7} \\ 2^3 &\equiv 1 \ \ \pmod{7} \\ 2^{3n} &\equiv 1^n \pmod{7} \end{align}
I feel up to this point everything is correct.
How would I add the $1$ to the exponent?
\begin{align} 2^{3n+1} &\equiv 1^n (5) \pmod{7} \\ 2^{3n+1} &\equiv 5 \ \ \ \ \ \ \ \pmod{7}
\end{align}
Now adding the $5$, I am confused as to how to do that as well. I would just subtract the $5$ remainder correct? Such that:
$2^{3n+1} -5 \equiv 0 \pmod{7}$ but this is not what I intend to do.
I need, $2^{3n+1} +5 \equiv 0 \pmod{7}$
|
Here's a much quicker way to do it:
Notice that $2^{3n+1} = 2^{3n} \cdot 2 = 8^n \cdot 2$. Since $8 \equiv 1 \pmod 7$, we have $1 * 2 + 5 \equiv 0 \pmod 7$, which is clearly true.
|
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|
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$ (using induction)
Prove $5^n + 2 \cdot 3^n - 3$ is divisible by 8 $\forall n\in \mathbb{N}$
Base case $n = 1\to 5 + 6 - 3 = 8 \to 8 \mid 8 $
Assume that for some $n \in \mathbb{N}\to 8 \mid 5^n + 2 \cdot 3^n - 3$
Showing $8 \mid 5^{n+1} + 2 \cdot 3^{n+1} - 3$
$$5^{n+1} + 2 \cdot 3^{n+1} - 3$$ $$5\cdot 5^n + 2\cdot 3\cdot 3^n - 3$$ $$ (5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n $$ $$ 5\cdot(5^n + 2\cdot 3^n - 3) + 4\cdot 5^n + 2\cdot 2\cdot3^n - 4\cdot(5^n + 2\cdot 3^n - 3)$$
$$ [5\cdot(5^n + 2\cdot 3^n - 3)] - [4\cdot 3^n - 12]$$
The first term divides by 8 but I am not sure how to get the second term to divide by 8.
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$$4 \cdot 5^n+4 \cdot 3^n$$ is clearly divisible by 8 cause once you take the 4 out it is the sum of two odd numbers.
|
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|
How to calculate this hard integral $\int_0^{\infty} \frac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx$? How to prove that $\displaystyle \int_0^{\infty} \dfrac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx= \frac{1}{4}\pi^2\sqrt{2}-\sqrt{2}\ln^2\left(1+\sqrt{2}\right)$ ?
It's very difficult and I have no idea.
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$\begin{align}I&=\int_0^{\infty} \dfrac{\arctan(x)\sqrt{\sqrt{1+x^2}-1}}{x\sqrt{1+x^2}}\,dx
\end{align}$
Perform the change of variable $x=\dfrac{2y}{1-y^2}$,
$\begin{align}I&=\sqrt{2}\int_0^{1} \dfrac{\arctan\left(\dfrac{2y}{1-y^2}\right)}{\sqrt{1-y^2}}\,dy\\
&=2\sqrt{2}\int_0^{1} \dfrac{\arctan y}{\sqrt{1-y^2}}\,dy\\
\end{align}$
If one assume that,
$\displaystyle \int_0^{1} \dfrac{\arctan y}{\sqrt{1-y^2}}\,dy=\frac{1}{8}\pi^2-\frac{1}{2}\ln^2(\sqrt{2}+1)$
therefore,
$\displaystyle \boxed{I=\frac{1}{4}\pi^2\sqrt{2}-\sqrt{2}\ln^2(\sqrt{2}+1)}$
(for a proof of this, see: Evaluating a sum involving binomial coefficient in denominator )
|
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|
How many possibilities are there for two full houses to be dealt to two players in one game? Imagine dealing cards from a classic 52 card deck to two poker players.
How many possibilities are there for both of them to be dealt a full house(three cards in same rank and two cards of another rank) in same round?
As I know totally ${13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2}$ possibilities for one player (AAABB). According to that, I tried to solve my problem as shown below: $${13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2}{11 \choose 1}{4 \choose 2} + {13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 3}{11 \choose 1}{4 \choose 2}$$ for two players ((AAABB)(AAACC) or (AAABB)(CCCBB)). Is my solution correct or am I missing something?
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*
*choose the two three-card ranks. The number of ways is $\binom{13}{2}\binom{4}{3}\binom{4}{3}$.
*choose a two-card rank. The number of ways is $\binom{11}{1}\binom{4}{2}$.
*the next two-card rank may be the same or different as the first one. The number of ways is $\binom{2}{2}+\binom{10}{1}\binom{4}{2}$
*Take the two players into account, the number of ways is
$$2\binom{13}{2}\binom{4}{3}\binom{4}{3}\binom{11}{1}\binom{4}{2}\left[\binom{2}{2}+\binom{10}{1}\binom{4}{2}\right]$$
|
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Prove through induction that if $f(x) := \frac{2}{1-x^2}$ then $f^{(n)}(x) = n!(\frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+x)^{n+1}})$ The original formula is $f(x) := \frac{2}{1-x^2}$, for $f\colon\mathbb{R}- ({-1,1}) \to \mathbb{R}$
Through Induction prove that $f^{(n)}(x)=n!(\frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+x)^{n+1}})$.
So I start of with $n=0$ as the base case and see that:
$$f^{(0)}(x)=0!(\frac{1}{(1-x)^{0+1}}+\frac{(-1)^0}{(1+x)^{0+1}})$$
$$f^{(0)}(x)=(\frac{1}{(1-x)}+\frac{1}{(1+x)})$$
$$f^{(0)}(x)=\frac{1(1+x)+1(1-x)}{(1-x)(1+x)}$$
$$f^{(0)}(x)=\frac{2}{(1-x)(1+x)}$$
$$f^{(0)}(x)=\frac{2}{1-x^2}$$
Then I presumed that because it works for $n=0$ then it must also work for all $n∈\mathbb N$. Then I proceeded to state that it must also work for $n=n+1$.
$$f^{(n+1)}(x)=(n+1)!(\frac{1}{(1-x)^{n+1+1}}+\frac{(-1)^{n+1}}{(1+x)^{n+1+1}})$$
$$f^{(n+1)}(x) = f^{(n)'}(x)\\$$
$$=[(n!)((\frac{1}{(1-x)^{n+1}}+\frac{(-1)^n}{(1+x)^{n+1}}))]´$$
$$=(n!)[(\frac{1}{(1-x)^{n+1}}]´ + [\frac{(-1)^n}{(1+x)^{n+1}}]´$$
$$=n!(-(-n-1)(1-x)^{-n-2})+((-1)^n(-n-1)(1+x)^{-n-2})$$
But I'm kinda stuck on how to continue on from there ...
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you did all right until the last line
$$....=(n!)\left[(\frac{1}{(1-x)^{n+1}}\right]' + \left[\frac{(-1)^n}{(1+x)^{n+1}}\right]' =(n!)(\frac{n+1}{(1-x)^{n+2}} - \frac{(n+1)(-1)^n}{(1+x)^{n+2}} \\= (n+1)!\left(\frac{1}{(1-x)^{n+2}} + \frac{((-1)^{n+1}}{(1+x)^{n+2}}\right)$$
|
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|
Time and Work. How much work does C do per hour?
A, B and C need a certain unique time to do a certain work. C needs 1
hours less than A to complete the work. Working together, they require
30 minutes to complete 50% of the job. The work also gets completed if
A and B start working together and A leaves after 1 hour and B works
for a further 3 hours. How much work does C do per hour?
(A)16.66%
(B)33.33%
(C)50%
(D)66.66%
My attempt:
Let the total work be 100 units.
Let the work done by A,B and C be a units/hour,b units/hour,c units/hour respectively.
Let the time taken by A alone to complete the work be t hours.
ATQ:
\begin{align*}
(a+b+c) \cdot \frac{1}{2} & =50 \tag{1}\\
(a+b) \cdot 1+b \cdot 3 & =100 \tag{2}\\
c \cdot (t-1)& =100 \tag{3}\\
at & =100 \tag{4}
\end{align*}
Please help me solve these equations. When I am solving it is getting cumbersome.
Also if someone tells us some other way of solving, that would be helpful as well. Thanks.
|
Let us assume, they take $a, b,$ and $c$ hours to finish the job individually, respectively. Note the following points:
*
*They complete half the job in half an hour $\implies$ full job is completed in a hour. Hence, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$
*Also, A takes 1 hour more time than C. So, $a= c+1\implies \frac{1}{a}=\frac{1}{c+1}$
*If A works for a hour, and B works for four hours, then $\frac{1}{a}+\frac{4}{b}=1 \implies \frac{1}{b}=\frac{1-\frac{1}{a}}{4}=\frac{1-\frac{1}{c+1}}{4}=\frac{c}{4+4c}$.
Now, solving for $c$ gives us the quadratic: $3c^2-4c-4=0 \implies c=2, c= -\frac{2}{3}$. As, $c >0$, we have, the rate of work C does $=0.5$.
Solving by your method: $a =\frac{100}{t}, b = \frac{100-a}{4}=\frac{25t-25}{t}$. Then, we get, $$a +b + c =100 \implies 100+25t-25 + ct = 100t \implies t = \frac{75}{75-c}$$
Putting this in Eq.(3), we get, $$\frac{100}{c}=\frac{c}{75-c}\equiv 7500-100c = c^2 \equiv c^2+100c-7500=0 \implies c = \frac{1}{2}, -\frac{3}{2}$$
Rejecting the negative solution, we have $c =\frac12$, the same answer as obtained earlier.
No mistake in your approach!
|
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"url": "https://math.stackexchange.com/questions/2566277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Probability Problem with $10$ players being put on two teams Could somebody please check my answer to this problem?
Thanks
Bob
Problem:
$10$ kids are randomly grouped into an A team with five kids and a B team with five kids. Each
grouping is equally likely. There are three kids in the group, Alex and his two best friends Jose
and Carl. What is the probability that Alex ends up on the same team with at least one of his two
best friends?
Answer:
Let $p_{a}$, $p_j$, and $p_c$ be the probabilities that Alex is on team A, Jose is on team A, and Carl is on team A. Let $p$ by the probability that we seek.
\begin{eqnarray*}
p &=& 2 ( p_a p_j + p_a * p_c - p_a p_j p_c ) \\
p_a &=& \frac{1}{2} \\
p_j &=& \frac{1}{2} \\
p_c &=& \frac{1}{2} \\
p &=& =
2 ( \frac{1}{2} \Big( \frac{1}{2} \Big) + \frac{1}{2} \Big( \frac{1}{2} \Big) -
\frac{1}{2} \Big( \frac{1}{2} \Big) \Big( \frac{1}{2} \Big) \\
p &=& 2 ( \frac{1}{4} + \frac{1}{4} - \frac{1}{8} ) = 2 ( \frac{1}{2} - \frac{1}{8} ) \\
p &=& \frac{3}{4} \\
\end{eqnarray*}
|
The probability that Alex ends up on the same team as at least one of his two best friends can be found by subtracting the probability that neither of his friends is on the same team from $1$. If Alex is on a team, four of the other nine kids must be his teammates. They can be selected in
$$\binom{9}{4}$$
ways. If both Jose and Carl are not on his team, then Alex's teammates must be selected from the other seven available students. The number of such selections is
$$\binom{7}{4}$$
Hence, the probability that neither Jose nor Carl is on the same team as Alex is
$$\frac{\dbinom{7}{4}}{\dbinom{9}{4}} = \frac{35}{126} = \frac{5}{18}$$
Therefore, the probability that at least one of his two best friends are on the same team as Alex is
$$1 - \frac{\dbinom{7}{4}}{\dbinom{9}{4}} = \frac{13}{18}$$
As for why our results differ, notice that once Alex has been placed on a team, the probability that Jose will be placed on the same team is $4/9$ since there are only four positions left to be filled and nine students available. The same calculation applies to Jose. The probability that both his friends are on the same team as Alex is $\frac{4}{9} \cdot \frac{3}{8}$. To see this, observe that if Jose is selected to be on the same team as Alex, there are three positions left for Alex and eight people left to fill those positions. Hence, the probability that at least one of Alex's friends is on the same team is
$$P(A \cap J) + P(A \cap C) - P(A \cap J \cap C) = \frac{4}{9} + \frac{4}{9} - \frac{4}{9} \cdot \frac{3}{8} = \frac{13}{18}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac{AQ* RB}{QR}$ is constant, while point $P$ moves along the ray $AB$. Let $AB$ and $FD$ be chords in circle, which does not intersect and $P$ point on arc $AB$ which does not contain chord $FD$. Lines $PF$ and $PD$ intersect chord $AB$ in $Q$ and $R$. Prove that $\frac{AQ* RB}{QR}$ is constant, while point $P$ moves along the arc $AB$.
I have only a projective solution to this problem and no sythetic (and natural!) solution. Any idea? Please don't just draw some circle which will do the job with no logical explanation what was your motive to draw it.
|
Let $|AT|=a$, $|QT|=b$, $|QB|=c$,
$|AP|=u$, $|BP|=v$,
$|TP|=p$, $|QP|=q$,
$|OA|=|OB|=|OP|=|OD|=R$ (circumradius of
$\triangle ABP$,
$\triangle ADP$,
$\triangle DFP$,
$\triangle FBP$).
Prove that
\begin{align}
\frac{(a+b)(b+c)}{b}
&=\mathrm{const},
\tag{1}\label{1}
\quad\text{ while point $P$ moves along the arc $AB$}
.
\end{align}
\begin{align}
\frac{(a+b)(b+c)}{b}
&=a+b+c+\frac{a\,c}{b}=|AB|+\frac{a\,c}{b}
,\\
\end{align}
\begin{align}
\text{hence, \eqref{1} is equivalent to}\quad
\frac{a\,c}{b}&=\mathrm{const}
\tag{2}\label{2}
.\\
\triangle ATP:\quad
\frac{a}{\sin\phi}
&=
\frac{p}{\sin\alpha}
=\frac{u}{\sin(\alpha+\phi)}=\frac{2R\sin\beta}{\sin(\alpha+\phi)}
,\\
\triangle TQP:\quad
\frac{b}{\sin{\psi}}
&=
\frac{q}{\sin(\alpha+\phi)}
=\frac{p}{\sin(\beta+\theta)}
,\\
\triangle QBP:\quad
\frac{c}{\sin{\theta}}
&=
\frac{q}{\sin\beta}
=
\frac{v}{\sin(\beta+\theta)}=\frac{2R\sin\alpha}{\sin(\beta+\theta)}
,
\end{align}
\begin{align}
\frac{a^2}{\sin^2\phi}
&=
\frac{p\,2R\sin\beta}{\sin\alpha\sin(\alpha+\phi)}
,\qquad
\frac{c^2}{\sin^2\theta}
=
\frac{q2R\sin\alpha}{\sin\beta\sin(\beta+\theta)}
,\\
\frac{b^2}{\sin^2\psi}
&=
\frac{p\,q}{\sin(\alpha+\phi)\sin(\beta+\theta)}
,\\
\frac{a^2c^2}{b^2}\cdot
\frac{\sin^2\psi}{\sin^2\phi\sin^2\theta}
&=4R^2
,\\
\frac{a\,c}b&=2\,R\cdot\frac{\sin\phi\sin\theta}{\sin\psi}
=2\,R\cdot\frac{|AD|\cdot|BF|\,2\,R}{4\,R^2\,|DF|}
=\frac{|AD|\cdot|BF|}{|DF|}
=\mathrm{const}
.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Transform $\iint_D \sin(x^2+y^2)~dA$ to polar coordinates and evaluate the polar integral.
Transform the given integral in Cartesian coordinates to polar coordinates and evaluate the polar integral:
(b) $\iint_D \sin(x^2+y^2)~dA$, where $D$ is the region in the first quadrant bounded by the lines $x=0$ and $y=\sqrt{3}\cdot x$ and the circles $x^2+y^2=2\pi$ and $x^2+y^2=3\pi$.
In Cartesian coordinates I know $0\le y\le \sqrt{3 \pi}, \space0\le x \le \frac{\sqrt{3\pi}}{4}$
I also know $x=r\cos\theta$, $y=r\sin\theta$
How should I proceed from here?
$r$ remains the same $r= \sqrt{3\pi}$?
So $0\le r\le\sqrt{3\pi}$, $0\le\theta\le \frac{\pi}{2}$. Is this correct?
|
Given the above integral with the given bounds converted to polar coordinates, $$\iint_D \sin(x^2+y^2)~dA=\int_{\pi/3}^{\pi/2} \int_{\sqrt{2\pi}}^{\sqrt{3\pi}} \sin(r^2)\cdot r~dr~d\theta$$ evaluated gives ${\pi/6}$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Number of teams There are $8$ women and $7$ men, from which we must create a team of $4$ women and $3$ men. Two men doesn't like each other, so they don't want to be in one team.
I tried number of teams (without the specific two deviant men): $\binom{8}{4}\times \binom{7}{3}$ but it's not correct.
The given solution is: $\binom{6}{2}\binom{8}{4}\cdot2+\binom{5}{3}\cdot \binom{8}{4}=2800$, which is more than $\binom{8}{4}\times \binom{7}{3}=2450$.
|
Some thoughts:
1) No matter what, the number of women on the team can be computed as $\binom{8}{4}$
2) $\binom{8}{4}\times \binom{7}{3} = 2450$. $\binom{8}{4}\times \binom{6}{2}+\binom{8}{4}\times \binom{5}{3} = 1750$, not $2800$.
3) Count the teams containing neither of the special men. Then add the count of teams containing the first special man. Next add the count of teams containing the second special man. Finally, multiply that result by the (independent) count of the women. Once you see how to break up the count, the numbers work out easily.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Finding smallest possible value of expression with x and y I'm not supposed to use calculus here. I'm trying to find the smallest possible value of the expression $x^2+4xy+5y^2-4x-6y+7$ for real numbers $x$ and $y$.
Here's my attempt:
$x^2+4xy+5y^2-4x-6y+7=[(x-2)^2+3)+5((y-3/5)^2-9/25)]+4xy$
$3\le (x-2)^2+3$ for all real $x$
$-9/5\le 5((y-3/5)^2-9/25)$ for all real $y$.
Thus $6/5=3-9/5\le x^2+5y^2-4x-6y+7$ for all real $x$ and $y$. $(A)$
Therefore $x^2+4xy+5y^2-4x-6y+7 \ge 6/5+4xy$ for all real $x$ and $y$. $(B)$
The equality in $B$ does only hold when equality in $A$ holds. This happens when $x=2$ and $y=3/5$.
The smallest value of $x^2+4xy+5y^2-4x-6y+7$ is then $6/5+4*2*(3/5)=6$
Is this a legitimate procedure or am I doing something wrong? I'm a bit shaky on the statement $B$ and the following implications. Let me know.
|
minimal trickery, I see $u = x + 2y,$ so $4u = 4x+8y,$ your expression becomes
$$ u^2 + y^2 - 4u + 2y + 7, $$
$$ u^2 - 4 u + y^2 + 2 y + 7, $$
$$ (u-2)^2 - 4 + (y+1)^2 - 1 + 7, $$
$$ (u-2)^2 + (y+1)^2 + 2 $$
least value, $2,$ occurs when $y=-1$ and $x+2y = x - 2 = 2,$ so $x=4$ and $y=-1$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Slope of The Tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$ Find slope of tangent to $r=7\sin \theta$ , $\theta = \frac{\pi}{6}$
using $$\frac{dy}{dx} = \frac{\frac{dr}{dθ}\sin \theta + r \cos \theta}{\frac{dr}{dθ}\cos \theta-r \sin \theta}$$
I got
$$\frac {14\cos \theta \sin \theta}{7 \cos^2 \theta-7\sin \theta cos \theta}$$
|
When you have the equation in rectangular coordinates
$$
x^2+y^2=7y
$$
you can use implicit differentiation:
$$
2x+2yy'=7y'
$$
so
$$
y'=\frac{2x}{7-2y}
$$
For $\theta=\pi/6$, we have $r=7/2$, so $x=7\sqrt{3}/4$ and $y=7/4$; therefore the derivative at the point is
$$
y'(7\sqrt{3}/4)=\frac{2\dfrac{7\sqrt{3}}{4}}{7-2\dfrac{7}{4}}=\sqrt{3}
$$
If you want to do it directly with polar coordinates,
$$
dx=\cos\theta\,dr-r\sin\theta\,d\theta
\qquad
dy=\sin\theta\,dr+r\cos\theta\,d\theta
$$
which gives
$$
\frac{dy}{dx}
=\frac{\sin\theta\,dr+r\cos\theta\,d\theta}{\cos\theta\,dr-r\sin\theta\,d\theta}
=\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}
=\tan2\theta
$$
Your derivation is almost correct: you have $\sin\theta\cos\theta$ at the bottom right, but it should be $7\sin^2\theta$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Why is function domain of fractions inside radicals not defined for lower values than those found by searching for domain of denominator in fraction? Consider function $y = \sqrt{\frac{1-2x}{2x+3}}$. To find the domain of this function we first find the domain of denominator in fraction:
$2x+3 \neq 0$
$2x \neq -3$
$x \neq - \frac{3}{2}$
So, the domain of $x$ (for fraction to be valid) is $x \in \left(- \infty, - \frac{3}{2}\right) \cup \left(- \frac{3}{2}, + \infty\right)$.
Then we find the domain for whole fraction:
$\frac{1-2x}{2x+3} \ge 0$
$1-2x \ge 0$
$-2x \ge -1$
$x \le \frac{1}{2}$
My textbook says that the (real) domain of the whole $y$ function is $x \in \left(- \frac{3}{2}, \frac{1}{2}\right]$.
I understand why the function is not defined in values larger than $\frac{1}{2}$ (because condition is $x \le \frac{1}{2}$), but I don't understand why it can't be have values less than $- \frac{3}{2}$ (because condition says only $x \neq - \frac{3}{2}$).
I checked the domain of this function and the domain given in the textbook is correct. Function has imaginary values for $x$ values less than $- \frac{3}{2}$ or bigger than $\frac{1}{2}$. It is undefined in $- \frac{3}{2}$.
Real values only in $\left(- \frac{3}{2}, \frac{1}{2}\right]$ domain.
|
$\frac ab > 0$ means either 1) !!BOTH!! $a > 0; b>0$ OR 2) !!BOTH!! $a < 0; b< 0$.
And as $\frac ab$ existing means $b \ne 0$ then $\frac ab > 0$ means either 1) both $a \ge 0; b < 0$ or both $a \le 0; b < 0$.
So $\frac{1-2x}{2x+3} \ge 0$ means
1) $1 - 2x \ge 0;2x + 3 > 0$
2) $1-2x \le 0; 2x + 3 < 0$.
Number 1) yeilds:
$x \le \frac 12$ !!AND!! $x > -\frac 32$ or $-\frac 32 < x \le \frac 12$.
Number 2) yeilds the inconsistent
$x \ge \frac 12$ and $ x < -\frac 32$.
So we have 1) and domain of $f$ is $(-\frac 32, \frac 12]$.
Note if we had a consistent
$g(x) = \sqrt{\frac{2x-1}{2x +3}}$ you would have gotten the domain as $[\frac 12, \infty)$ when it should be
$(-\infty, -\frac 32) \cup [\frac 12, \infty)$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.