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Show $60 \mid (a^4+59)$ if $\gcd(a,30)=1$
If $\gcd(a,30)=1$ then $60 \mid (a^4+59)$.
If $\gcd(a,30)=1$ then we would be trying to show $a^4\equiv 1 \mod{60}$ or $(a^2+1)(a+1)(a-1)\equiv 0 \mod{60}$. We know $a$ must be odd and so $(a+1)$ and $(a-1)$ are even so we at least have a factor of $4$ in $a^4-1$. Was thinking I could maybe try to show that there is also a factor of $3$ and $5$ necessarily giving that $a^4-1\equiv0 \mod{60}$.
Other things I was thinking was that as $Ord_n(a) \mid \phi(n)=\phi(60)=16$ that we just need to show that $Ord_n(a) \in \{1,2,4 \}$.
Any hints? I have the exam soon =/
|
The question is answered successfully, I would like find the number for which $a^4\equiv1\pmod n$ for all $a$ such that $(a,n)=1$
Using Carmichael Function, If $\lambda(n)=2^m$ where integer $m\ge 1$
$n$ can not have any odd prime factor with exponent $>1$ and for each prime factor $p_i,p_i-1$ must be of the form $2^{2^{r_i}}$ where integer $r_i\ge0$
$$\text{Then if }n=2^s\cdot\prod p_i=2^s\cdot (2^{2^{r_i}}+1)$$
If $s\ge 3, \lambda(n)=$lcm $(2^{2^{r_i}},2^{s-2})$ as $\lambda(2^s)=2^{s-2}$ for $s\ge3$
$\implies\lambda(n)=$lcm $(2^{s-2},2^{2^{r_1}},2^{2^{r_2}},\cdots)=$lcm$(2^{s-2},2^{2^{\text{max}(r_i)}})$
If $\lambda(n)$ divides $4=2^2,$ $$2^{\text{max}(r_i)}\le 2 \text{ and }s-2\le 2 \implies 0\le max(r_i)\le1 \text{ and } 3\le s\le 4 $$
For the maximum value of $n,r_1=0,r_2=1$ and $s=4$
So, the maximum value of $n$ will be $3^1\cdot5^1\cdot2^4=240$
$$\implies x^4\equiv1\pmod {240}\implies x^4\equiv1\pmod {60}\equiv1-60=-59$$
|
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|
Can anyone show me, how to solve these system of Equations: $$\begin{align*}
x+y+z &= 2 \\
(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y) &= 1 \\
x^2(y+z)+y^2(z+x)+z^2(x+y) &= -6
\end{align*}$$
Can anyone explain me the solution. I asked it in mathoverflow but god knows why it was closed. Please help and detail the process in step by step. It is a RMO question (The Pre- Indian team selection exam for IMO)
|
$$x+y+z=2$$
$$x^2+y^2+z^2+2(xy+zx+yz)=4$$
The second equation is $x^2+y^2+z^2+3(xy+zx+zy)=1 \implies zx+yz+yx=-3$
You get $x^2+y^2+z^2=10$
Using the $x+y+z=2$ in the third equation.
$x^2(2-x)+y^2(2-y)+z^2(2-z)=-6 \implies 2(x^2+y^2+z^2)-x^3+z^3+z^3=-6$
A little re-arranging would give
$26 =x^3+y^3+z^3 $
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$26-3xyz=(2)(10-(-3)) \implies 26-26=3xyz \implies $ one of the $x,y,z$ is zero.
Consider $z=0$ (Generality is not lost) Now you have $x^2+y^2=10$ and $yx=-3$
You get solutions as $(x,y,z)=(3,-1,0),(0,-1,3)(0,3,-1),(-1,0,3),(3,0,-1)$ and $(-1,0,3)$.
|
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|
Solving $\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$ Where do I start to solve a equation for x like the one below?
$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$
After squaring it, it's too complicated; but there's nothing to factor or to expand?
Ideas?
|
Note that
$$ \sqrt{7x-4} - \sqrt{7x-5} = \frac{(7x-4)-(7x-5)}{\sqrt{7x-4} + \sqrt{7x-5}} = \frac{1}{\sqrt{7x-4} + \sqrt{7x-5}} $$
and likewise
$$ \sqrt{4x-1} - \sqrt{4x-2} = \frac{1}{\sqrt{4x-1} + \sqrt{4x-2}}. $$
Plugging these to our equation, we obtain
$$ \sqrt{7x-4} + \sqrt{7x-5} = \sqrt{4x-1} + \sqrt{4x-2}. $$
Thus we have
$$ \sqrt{7x-4} = \sqrt{4x-1} \quad \text{and} \quad \sqrt{7x-5} = \sqrt{4x-2}, $$
yielding $x = 1$.
|
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|
Find the limit without use of L'Hôpital or Taylor series: $\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$ Find the limit without the use of L'Hôpital's rule or Taylor series
$$\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$
|
Well, you have
$$\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right) = \left(\frac{1}{x} - \frac{1}{\sin x}\right)\left(\frac{1}{x} + \frac{1}{\sin x}\right)$$
$$=\left(\frac{1}{x^2} - \frac{1}{x\sin x}\right)\left(1+ \frac{x}{\sin x}\right)$$
Since the limit of $\left(1+ \frac{x}{\sin x}\right)$ is $2$, your answer will be
$$2\lim_{x \rightarrow 0}\left(\frac{1}{x^2} - \frac{1}{x\sin x}\right)$$
$$=2\lim_{x \rightarrow 0}\left(\frac{x}{\sin x}\right)\left(\frac{\sin x - x}{x^3}\right)$$
$$=2\lim_{x \rightarrow 0}\frac{\sin x - x}{x^3}$$
An elementary calculus way of showing this limit is $-{1 \over 6}$ can be found here.
So the overall answer is $-\frac{1}{3}$.
|
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|
What is the least nonnegative number $a$ congruent to $3^{340}\pmod{341}$? Find the least nonnegative number $a$ congruent to $3^{340} \pmod{341}$.
What steps should I take to get to the answer?
|
As $341=11\cdot31,$
Using Fermat's Little Theorem, $ 3^{10}\equiv1 \pmod {11}$
$\implies 3^{340}=(3^{10})^{34}\equiv1^{34}\equiv1\pmod{11}\ \ \ \ (1)$
Again applying Fermat's Little Theorem, $3^{30}\equiv1\pmod {31}$
$\implies 3^{340}=(3^{30})^{11}\cdot 3^{10}\equiv 1\cdot3^{10}\equiv 3^{10} \pmod {31}$
Now, $3^3\equiv-4\pmod {31}\implies 3^{10}=3\cdot (3^3)^3\equiv3\cdot (-4)^3\equiv 3(-2)\equiv-6$
$\implies 3^{340}\equiv-6\ \ \ \ (2)$
Now, use CRT in $(1),(2)$ to find $x\equiv56\pmod{341}$
|
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Proving that $\int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a²} \mathrm d\theta = 0$ for $|a|>1$
Let $|a|>1$. Show that $$\int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a²} \mathrm d\theta = 0$$
I'm also trying to solve this problem. I wasn't sure if I was supposed to start a new thread or contribute to previous asked questions about this integral.
I have some ideas that we could do something like this
$$ -i \oint _{|z|=1} \frac{1-az}{(z-2a) }dz = \int_{\theta=0}^{\theta=2\pi} \frac{1-a e^{i\theta}}{e^{i\theta}-2a } e^{i\theta}d\theta = \int_{\theta=0}^{\theta=2\pi} \frac{1-a e^{i\theta}}{1-2ae^{-i\theta} }d\theta \quad (2)$$
Then we have that the real part of $(2)$ is almost equal to the desired integral, except for the $a^2$ in the denominator.
An inequality to the right for the desired integral is then given from the real part of $(2)$. All we want to find is an inequality to the left for the desired integral. That hopefully and easily could be calculated with help of
$$ \oint _{|z-z_0|=r}(z-z_0)dz = \frac{0 \quad n \neq -1}{2\pi i \quad n = -1} $$
Any suggestions?
|
The standard complex analysis approach is to make the substitution $z = e^{i\theta}$. This will map $[0,2\pi]$ onto the unit circle traversed counter-clockwise once. Using Euler's formulas,
$$ \cos \theta = \frac{e^{i\theta}+e^{-i\theta}}2 = \frac{z+z^{-1}}{2}.$$
We also have $dz = ie^{i\theta}\,d\theta$, i.e.
$$d\theta = \frac1{iz}dz.$$
Doing all this we get
\begin{align}
\int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a^2} d\theta &=
\int_{|z| = 1} \frac{ 1-a(z+z^{-1})/2}{1-a(z+z^{-1})+ a^2} \cdot \frac{1}{iz} dz \\
&= \frac1{2i} \int_{|z| = 1} \frac{2z- a(z^2+1)}{z(z-a(z^2+1)+ a^2z)} dz.
\end{align}
Note that $z(z-a(z^2+1)+ a^2z) = z(1-az)(z-a)$, so the integrand has simple poles at $z = 0$, $z = a$ and $z = 1/a$.
If $|a| < 1$, we get
\begin{align}
\int_0^{2\pi}\frac{ 1-a\cos\theta}{1-2a\cos\theta + a^2} d\theta
&= \pi \left( \operatorname{Res}\limits_{z=0} \frac{2z- a(z^2+1)}{z(1-az)(z-a)} +
\operatorname{Res}\limits_{z=a} \frac{2z- a(z^2+1)}{z(1-az)(z-a)} \right) \\
&= \pi \left( \frac{-a}{-a} + \frac{2a- a(a^2+1)}{a(1-a^2)} \right) \\
&= \pi ( 1 + 1 ) = 2\pi.
\end{align}
I'll leave the case $|a| > 1$ to you.
|
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Prove that $5^n - 2^n$ is divisible by $3$ for all nonnegative integers $n$ using mathematical induction Using mathematical induction, prove for all integers n 1 that $5^n - 2^n$ is divisible by 3.
Can someone help me with this?
|
This one could be solved by induction but is it much neater if you use that
\[ a^n-b^n = (a-b) \cdot \sum_{i=0}^{n-1} a^i\cdot b^{n-1-i} \]
If you use this one in your problem you see that
\[ 5^n- 2^n = (5-2) \cdot \sum_{i=0}^{n-1} 5^i \cdot 2^{n-1-i}=3 \cdot \sum_{i=0}^{n-1} 5^i \cdot 2^{n-1-i}\]
Which can be dived by $3$.
If you really want to use Induction use that
\[ 5=3+2\]
(ok this hint seems ridicolous it shall be used like this)
\[ 5^{n+1} - 2^{n+1} = (3+2) \cdot 5^n - 2 \cdot 2^n= 3 \cdot 5^n+ 2(5^n-2^n)\]
|
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Finding the fraction $\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}$ when knowing the sums $a+b+c+d$ to $a^4+b^4+c^4+d^4$ How can I solve this question with out find a,b,c,d
$$a+b+c+d=2$$
$$a^2+b^2+c^2+d^2=30$$
$$a^3+b^3+c^3+d^3=44$$
$$a^4+b^4+c^4+d^4=354$$
so :$$\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}=?$$
If the qusetion impossible to solve withot find a,b,c,d then is there any simple way to find a,b,c,d
Is there any help?
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One way to approach such questions is to view $a, b, c, d$ as roots of a single quartic equation $x^4-s_1x^3+s_2x^2-s_3x+s_4=0$, when we have Vieta's relations
$$s_1=a+b+c+d$$$$s_2=ab+ac+ad+bc+bd+cd=(a+b)(c+d)+ab+cd$$$$s_3=abc+abd+acd+bcd$$$$s_4=abcd$$
We then let $p_k=a^k+b^k+c^k+d^k$ with: $$p_0=4$$$$p_1-s_1=0$$$$p_2-p_1s_1+2s_2=0$$$$p_3-p_2s_1+p_1s_2-3s_3=0$$
With the given values of $p_1, p_2, p_3$ these determine $s_1, s_2, s_3$.
For higher powers $(k\ge4)$ we have $$a^k-s_1a^{k-1}+s_2a^{k-2}-s_3a^{k-3}+s_4a^{k-4}=0$$ using the quartic equation. Adding together the corresponding equations for $a , b, c, d$ we obtain the recurrence for $p_k$$$p_k-s_1p_{k-1}+s_2p_{k-2}-s_3p_{k-3}+s_4p_{k-4}=0$$
Since we know $p_4$ we can use the recurrence to find $s_4$, and this is then sufficient to compute $p_5, p_6$ from the recurrence.
If this looks complicated, this is misleading. The relations between the elementary symmetric functions $s_r$ and the sums of powers $p_r$ are useful to know, and render this kind of problem essentially mechanical. You can see that you can solve the problem without solving explicitly for $a, b, c, d$.
Following this scheme we find successively $s_1=2, s_2=-13, s_3=-14, s_4=24$. Then the recurrence gives:$$p_5-2p_4-13p_3+14p_2+24p_1=0$$ and$$p_6-2p_5-13p_4+14p_3+24p_2=0$$
And this gives $p_5=812, p_6=4890$
|
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Simple divisibility problem
Suppose $a$ and $b$ are distinct integers and $n\mid a^n-b^n$. Prove that $n\mid\frac{a^n-b^n}{a-b}$.
Here is how I do it: Write $a=b+d$, then $a^n-b^n=(b+d)^n-b^n=\binom{n}{1}b^{n-1}d+\binom{n}{2}b^{n-2}d^2+ \cdots + \binom{n}{n-1}bd^{n-1}+d^n \equiv 0 \pmod{n}$. Since $n| \binom{n}{k}$ for $k=1,2,\cdots,n-1$, $n|d^n$, which implies $n|d^{n-1}$. Therefore $(a^n-b^n)/(a-b)=\binom{n}{1}b^{n-1}+\binom{n}{2}b^{n-2}d+ \cdots + \binom{n}{n-1}bd^{n-2}+d^{n-1} \equiv 0 \pmod{n}$ and we are done. I wanna ask why $n|d^n$ will implies $n|d^{n-1}$ and that if there are any other possible solutions, thanks in advance.
|
We will repeatedly use the trick that if $n=xy$, then $a^n$ is also an $x$th power.
We will approach this by first showing it is true for prime powers.
Base case: $p$ is a prime and $p | a^p - b^p$. Then, we have $a \equiv a^p \equiv b ^p \equiv b \pmod{p}$, and hence $\frac{a^p - b^p}{a-b} \equiv p a^{p-1} \equiv 0 \pmod{p}$.
If $p^{k} | a^{p^k} - b^{p^k}$, then $p \mid \frac{ a^{p^i} - b^{p^i}} { a^{p^{i-1}}-b^{p^{i-1}}}$ for all values of $i$, and hence if you take the product of all these terms (which telescope), $p^k | \frac{a^{p^k} - b^{p^k}}{a-b}$.
Now, we will show that if $p^k || n$, and $n| a^n-b^n$, then $p^k | \frac{a^n-b^n}{a-b}$.
Set $n = p^k \alpha$. Then, $p^k | (a^\alpha) ^{p^k} - (b^\alpha)^{p^k}$, and applying the above we get $p^k | \frac{ (a^\alpha) ^{p^k} - (b^\alpha)^{p^k}} { a^\alpha - b^\alpha}$. Now, multiply this by $\frac{ a^\alpha - b^\alpha} { a-b}$ which is an integer. Hence, $p^k | \frac{a^n-b^n}{a-b}$, and we are done.
|
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Evaluating the time average over energy For more info see the article equations 37
Edit: The $\varepsilon ^3 $ has vanished due to time average. But how to get the 4th order?
Let us define some function for scalar field
$$\phi= \sum_{k=1}^\infty\varepsilon^k \phi_k \tag{1}=\varepsilon^1 \phi_1+\varepsilon^2 \phi_2 +\varepsilon^3 \phi_3 ...$$
position scaling
\begin{equation}\tag{2}
\zeta^i=\varepsilon x^i
\end{equation}
time scaling
\begin{equation}\tag{3}
\tau=\omega(\varepsilon) t
\end{equation}
\begin{equation}
\Delta S -S +S^3=0\,,\quad S=p_1\sqrt{\lambda} \,.
\end{equation}
\begin{equation}
p_3=\frac{1}{\lambda^2\sqrt{\lambda}}\left[
\left(\frac{1}{24}\lambda^2-\frac{1}{6}\lambda g_2^2+\frac{5}{8}g_5
-\frac{7}{4}g_2g_4+\frac{35}{27}g_2^4\right)Z
-\frac{1}{54}\lambda g_2^2S(32+19S^2)
\right]
\end{equation}
the solutions for equation (1) can be written as,
\begin{eqnarray}
\phi_1&=&p_1\cos \tau \tag{4}\\
\phi_2&=&\frac16 g_2p_1^2\left(\cos(2\tau)-3\right) \tag{5}\\
\phi_3&=&p_3\cos \tau+\frac{1}{72}(4g_2^2-3\lambda)p_1^3\cos(3\tau)\tag{6}
\end{eqnarray}
My intention is to determine the time average energy density
The energy corresponding can be written as
\begin{equation}
E = \int d^{D}x\,{\cal E}\,, \quad
{\cal{E}} = \frac{1}{2} \left(\partial_t \phi\right)^2
+ \frac{1}{2} \left(\partial_i \phi\right)^2
+U(\phi)\,,
\end{equation}
where ${\cal{E}}$ denotes the energy density.
\begin{equation}
E = \frac{1}{\varepsilon^D}\int d^{D}\zeta\,{\cal E}\,, \quad{\rm where}\quad
{\cal{E}} = \frac{1}{2} (1-\varepsilon^2)\left(\partial_\tau \phi\right)^2
+ \varepsilon^2\frac{1}{2} \left(\partial_i \phi\right)^2
+U(\phi)\,.
\end{equation}
Because of the periodic time dependence we shall compute
the energy density averaged over a period,
\begin{equation}
\bar{E}=\frac{1}{2\pi}\int_0^{2\pi}d\tau E\,,
\end{equation}
The bar over a quantity will denote its time average.
Using the results of the $\varepsilon$ expansion,
Eqs. (4-6)
the time averaged energy density,
$\bar{{\cal E}}$, up to fourth order in $\varepsilon$ can be written as
(My problem is here that how will this term arise)
\begin{eqnarray}
\bar{{\cal E}}&=&\frac{\varepsilon^2}{2\lambda}S^2
-\frac{\varepsilon^4}{216\lambda^3}\biggl[
\lambda S^2(64g_2^2+27\lambda)(S^2+2)
-54\lambda^2(\nabla S)^2
\nonumber\\
&&-SZ(135g_5-378g_2g_4+280g_2^4-36\lambda g_2^2+9\lambda^2)
\biggl] \,.
\end{eqnarray}
For more info see the article equations 37
If you have problem to get the question then ask me please, I will
edit for you. Thanks advance,
|
Look,
$$
{\cal{E}} = \frac{1}{2} \left(\partial_t \phi\right)^2
+ \frac{1}{2} \left(\partial_i \phi\right)^2
+U(\phi)$$
Now fill in the expansion
$$\phi= \varepsilon^1 \phi_1+\varepsilon^2 \phi_2 +\varepsilon^3 \phi_3 + \ldots$$
this should give
$${\cal{E}} = \frac{1}{2} \left(\varepsilon^1 \partial_t\phi_1+\varepsilon^2 \partial_t\phi_2 +\varepsilon^3 \partial_t\phi_3 + \ldots \right)^2
+ \frac{1}{2} \left(\varepsilon^1 \partial_i\phi_1+\varepsilon^2 \partial_i\phi_2 +\varepsilon^3 \partial_i\phi_3 + \ldots\right)^2
+\frac{1}{8}\left(\varepsilon^1 \phi_1+\varepsilon^2 \phi_2 +\varepsilon^3 \phi_3 + \ldots\right)^2\left( \varepsilon^1 \phi_1+\varepsilon^2 \phi_2 +\varepsilon^3 \phi_3 + \ldots -2\right)^2$$
This is to second order in $\varepsilon$
$${\cal{E}}=\varepsilon^2 \left[\frac{1}{2}(\partial_t \phi_1)^2+\frac{1}{2}(\partial_i \phi_1)^2+\frac{1}{2}(\phi_1)^2\right] + O(\varepsilon^3)$$
Filling in $\phi_1 = p_1 \cos \tau$, you get
$${\cal{E}}=\varepsilon^2 \left[\frac{1}{2}(p_1 \omega(\varepsilon)\sin\tau)^2+\frac{1}{2}(\partial_ip_1 \cos\tau)^2+\frac{1}{2}(p_1 \cos\tau)^2\right] + O(\varepsilon^3)$$
or
$${\cal{E}}=\varepsilon^2 \left[\frac{1}{2}p_1^2+\frac{1}{2}(\partial_ip_1 \cos\tau)^2\right] + O(\varepsilon^3)$$
With the rescaling of the spatial coordinates introduced in the paper, the second term is proportional to $\varepsilon^2$ and thus as a whole to $\varepsilon^4$. Then using $S=p_1\sqrt{\lambda}$ we get
$${\cal{E}}=\varepsilon^2 \frac{1}{2\lambda}S^2 + O(\varepsilon^3)$$
So, if you continue working diligently in the same fashion, you should be able to find out what the fourth order correction is.
|
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A geometric reason why the square of the focal length of a hyperbola is equal to the sum of the squares of the axes? When I teach conics, I give a simple geometric argument for the Pythagorean relation for ellipses. (By "Pythagorean relation" I mean the fact that the square of the focal length $c$ is the difference of the squares of the semimajor and semiminor axes, i.e. $c^2=|a^2-b^2|$ for the ellipse defined by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.) The picture argument goes as follows: just draw the obvious isosceles triangle from the two foci to one of the two points at the end of the minor axis. We see two right triangles; in each, one leg is the focal length $c$ and the other leg is the semiminor axis, while the hypotenuse is the semimajor axis. The result follows from the Pythagorean theorem applied to this right triangle.
But when it comes to the Pythagorean relation for hyperbolas, I can't think of an equally convincing picture to draw. Instead I have to give a rather inelegant algebraic argument for the relation $c^2=a^2+b^2$ (for the hyperbola defined by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, with focal length $c$), directly from the definition of the hyperbola. The messy details are below.*
Now, obviously there is an asymmetry between the argument for the ellipse and the argument for the hyperbola: the ellipse argument is almost trivial and doesn't involve working straight from the locus definition, but the hyperbola argument is tedious and does require working straight from the definition. So, my question:
Is there an equally persuasive picture I could draw to convince high
school students of the Pythagorean relation for hyperbolas -- without having to get technical with points at infinity, say?
The natural triangle to draw is the right triangle with legs from the center to a vertex and from that vertex to an asymptote, which has legs of length $a$ and $b$; but then the problem becomes why the length along the asymptote should be equal to $c$. I doubt there is a purely (Euclidean) geometrical reason because this triangle lies on the asymptote, not on the hyperbola itself; I suspect some sort of limiting procedure will be required.
[*] The tedious algebraic details:
The foci lie at $(\pm c,0)$.
Say $a>b$. Then hyperbola is the locus of points $(x,y)$ such that
$$\left|\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}\right|=2a$$
Assume for simplicity the expression between the absolute value bars
is positive. (The algebra is even more tedious if we don't make such a
simplifying assumption.) Then moving the right root to the other side
and squaring gives
$$(x+c)^2+y^2=4a^2+(x-c)^2+y^2+4a\sqrt{(x-c)^2+y^2}$$
Simplifying gives
$$xc-a^2=a\sqrt{(x-c)^2+y^2}$$
and upon squaring again we have
$$x^2c^2-2xca^2+a^4=a^2(x^2-2xc+c^2+y^2)=a^2x^2-2xca^2+c^2a^2+a^2y^2$$
which reduces to
$$x^2(c^2-a^2)-y^2a^2=a^2(c^2-a^2)$$
Defining $b^2:=c^2-a^2$ brings the equation into the form
$x^2/a^2+y^2/b^2=1$, and the Pythagorean relation follows from the
definition of $b$.
|
Since the hyperbola is defined as the locus where the difference of the two distances is constant, I don't see any analogous argument. But certainly a limit argument does work: As $(x,y)\to \infty$ (with $x>0$), we have $\left|\frac yx\right|\to\frac ba$, and so
\begin{align*}
\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}&=\\\left(\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}\right)&\frac{\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}}{\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}}\\
&=\frac{4c}{\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}}\\&=\frac{4c}{\sqrt{(1+c/x)^2+(y/x)^2}+\sqrt{(1-c/x)^2+(y/x)^2}}\\
&\to \frac{2ac}{\sqrt{a^2+b^2}}\,.
\end{align*}
|
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|
Finding another way of doing this integral $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$ Problem :
Integrate : $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$
I have the solution : We can substitute $\sqrt{x}= \cos^2t$ and proceeding further,
I got the the answer which is $-2\sqrt{1-x}+\cos^{-1}\sqrt{x}+\sqrt{x-x^2}+C$
Can we do this problem some other way as well.. Thanks..
|
Write $y = \sqrt{(1-\sqrt{x})/(1+\sqrt{x})}$. Then $y^2 = (1 - \sqrt{x})/(1+\sqrt{x}) = -1 + 2/(1+\sqrt{x}),$
so
$2/(1+y^2) = 1 + \sqrt{x},$
and so
$$x = \Bigl(\dfrac{2}{1+ y^2} - 1\Bigr)^2.$$
Thus $$y\, dx = y \, d\Bigl(\dfrac{2}{1+y^2} - 1\Bigr)^2
= -8 y^2 \Bigl(\dfrac{2}{1+y^2} - 1\Bigr)\dfrac{dy}{(1+y^2)^2}.$$
The indefinite integral of any rational function of $y$ can be
expressed in terms of elementary functions of $y$, and hence
our original integral, $\int y dx,$ can be expressed in terms of
elementary functions of $y$.
Working out the details will be painful, though, compared to the
trig-based substitutions.
|
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|
A Challenging Euler Sum $\sum\limits_{n=1}^\infty \frac{H_n}{\tbinom{2n}{n}}$ Recently, I encountered a strange series involving Harmonic Numbers and Binomial Coefficients both.
According to Mathematica:
$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}} = -\frac{2\sqrt{3} \pi}{27}(\log (3)-2)+\frac{2}{27} \left( \psi_1 \left( \frac{1}{3}\right)-\psi_1 \left(\frac{2}{3} \right)\right)$$
Here $\psi_n(z)$ denotes the Polygamma Function. Can anybody provide a nice proof of the above statement?
My Failed Attempt
Using the Beta-function identity,
$$\frac{1}{\binom{2n}{n}}=(2n+1)\int_0^1 y^n(1-y)^n \ dy$$
$$\displaystyle \begin{aligned} \sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}} &= \sum_{n=1}^\infty (2n+1)H_n \int_0^1 (y-y^2)^n dy \\ &= \int_0^1 \sum_{n=1}^\infty (2n+1)H_n (y-y^2)^n \ dy \end{aligned}$$
Here, I used the identity
$$\sum_{n=1}^\infty (2n+1)H_n t^n=\frac{2t-(1+t)\log(1-t)}{(t-1)^2}\quad |t|<1$$
and got
$$\sum_{n=1}^\infty \frac{H_n}{\binom{2n}{n}}=\int_0^1 \frac{2y-2y^2-(1+y-y^2)\log(y^2-y+1)}{(y^2-y+1)^2}dy$$
How should I continue from here? I tried making some substitutions but nothing worked. Am I going in the right direction?
Please help.
|
Note: there is a minus sign missing in front of the 1st term on the left side of your evaluation.
Put the integral into the form
$$I=\int_0^{1}\frac{\left(y^2-y+1\right)\ln\left(y^2-y+1\right)-2\ln\left(y^2-y+1\right)-2\left(y^2-y+1\right)+2}{\left(y^2-y+1\right)^2}dy.$$
Making the change of variable $y=\frac12+\frac{\sqrt{3}}{2}\tan\varphi$ (so that $y^2-y+1=\frac{3}{4\cos^2\varphi}$) and simplifying, this reduces to
$$I=\frac{8}{3\sqrt{3}}\int_{-\pi/6}^{\pi/6}\left\{\Bigl(\frac34-2\cos^2 \varphi\right)\left(\ln 3-2-2\ln (2\cos \varphi)\Bigr)-2\cos^2 \varphi\right\}d\varphi.\tag{1}$$
The only nontrivial integrals here are of the form
$$\int_{-\pi/6}^{\pi/6}\ln (2\cos \varphi)\,d\varphi,\qquad \int_{-\pi/6}^{\pi/6}\left(2\cos^2\varphi-1\right)\ln (2\cos \varphi)\,d\varphi.$$
The second integral can be easily done by parts - it is equal to
\begin{align}\int_{-\pi/6}^{\pi/6}\left(2\cos^2\varphi-1\right)\ln (2\cos \varphi)\,d\varphi&=\Bigl[\sin\varphi\cos\varphi\ln(2\cos \varphi)\Bigr]^{\pi/6}_{-\pi/6}+\int_{-\pi/6}^{\pi/6}\sin^2\varphi\,d\varphi=\\&=\frac{\pi}{6}+\frac{\sqrt{3}}{4}\ln3-\frac{\sqrt{3}}{4}.
\end{align}
Using this in (1), we reduce it to
$$I=\frac{8}{3\sqrt{3}}\left[\frac{\pi\left(2-\ln3\right)}{12}+\frac12\int_{-\pi/6}^{\pi/6}\ln(2\cos\varphi)\,d\varphi\right].$$
Therefore, the proof of your identity reduces to showing that
$$\int_{0}^{\pi/6}\ln(2\cos\varphi)\,d\varphi=\frac{\psi_1\left(\frac13\right)-\psi_1\left(\frac23\right)}{12\sqrt{3}},\tag{2}$$
However, the left side is clearly expressible in terms of polylogarithms, so (2) should follow from their known special values.
Added:
Indeed, as Raymond Manzoni pointed out, the difference of the formulas (5) and (7) here gives
$$\psi_1\left(\frac13\right)-\psi_1\left(\frac23\right)=6\sqrt{3}\,\mathrm{Cl}_2\left(\frac{2\pi}{3}\right)\tag{3}$$
Clausen function $\mathrm{Cl}_2\left(x\right)$ is basically the imaginary part of dilogarithm function, characterized by the integral representation
$$\mathrm{Cl}_2\left(x\right)=-\int_0^x\ln\left(2\sin\frac{t}{2}\right)dt.\tag{4}$$
Using (3), (4) and the fact that $\mathrm{Cl}_2(\pi)=0$, we deduce from (2) the necessary statement.
|
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|
Integrals using Arctangens We want to find $\displaystyle \int\dfrac{12}{16x^2 +1}$
I rewrote it to the form $ 3 \cdot \dfrac{1}{u^2 + 1} \cdot u' $ where $u=4x$.
I found out that the correction sheet does the same thing, but their next step leaves my puzzled:
$$ F(x) = 3 \arctan (4x) + C$$
Where did the $u' = 4$ go to?
|
As you noted, if $u = 4x$, then $du = 4dx$, yielding
$$\begin{split}
\int \frac{12dx}{16x^2+1}
&= 3 \int \frac{4dx}{(4x)^2+1} \\
&= 3 \int \frac{u'dx}{(u)^2+1} \\
&= 3 \int \frac{du}{u^2+1} \\
&= 3 \arctan(u) + C \\
&= 3 \arctan(4x)+C
\end{split}
$$
|
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Limit with roots I have to evaluate the following limit:
$$ \lim_{x\to 1}\dfrac{\sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2+1}-\sqrt{x^4+1}} . $$
I rationalized both the numerator and the denominator two times, and still got nowhere. Also I tried change of variable and it didn't work.
Any help is grateful. Thanks.
|
To ease typing, we temporarily manipulate the top and bottom separately.
The top is $\sqrt{(x-1)(x+1)}+\left(\sqrt{x+1}-\sqrt{x^3+1}\right)$ (we changed the order). "Rationalize" the part in parentheses. The top becomes
$$\sqrt{(x-1)(x+1)}-\frac{x(x-1)(x+1)}{\sqrt{x+1}+\sqrt{x^3+1}}.\tag{1}$$
Do the same rationalizing trick with the last two terms of the bottom. We get
$$\sqrt{x-1}-\frac{x^2(x+1)(x-1)}{\sqrt{x^2+1}+\sqrt{x^4+1}}.\tag{2}$$
Divide the expressions (1) and (2) by $\sqrt{x-1}$. We get that our original ratio is equal to
$$ \frac{\sqrt{x+1}-\frac{x(x+1)\sqrt{x-1}}{\sqrt{x+1}+\sqrt{x^3+1}}}{1-\frac{x^2(x+1)\sqrt{x-1}}{\sqrt{x^2+1}+\sqrt{x^4+1}}} .\tag{3}$$
Finally, let $x\to 1^+$. The messy terms in the top and bottom of (3) approach $0$ because of the surviving factor of $\sqrt{x-1}$. So the required limit as $x$ approaches $1$ from the right is $\sqrt{2}$.
|
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|
no. of solution of the equation $[x]^2+a[x]+b = 0$ is If $a$ and $b$ are odd integer. Then the no. of solution of the equation $[x]^2+a[x]+b = 0$ is
where $[x] = $ greatest Integer function
My Try:: Let $[x] = y$. Then equation become $y^2+ay+b = 0$
Now If given equation has real Roots, Then $\displaystyle y = \frac{-a\pm \sqrt{a^2-4b}}{2}$
Now $a^2-4b = k^2\Leftrightarrow a^2-k^2=4b^2$. where $k\in \mathbb{Z}$
Now How can I solve after that.
Help required
Thanks
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If $y := \lfloor x \rfloor$ is even, $y^2$ is even, $ay$ is even, $b$ is odd, hence $y^2 + ay + b$ is odd, so $y^2 + ay+b \ne 0$. If on the other hand, $y$ is odd, $y^2$ and $ay$ are odd also, as is $b$, so $y^2 + ay+ b$ is odd again. Hence $y^2 + ay + b$ is an odd integer for every integer $y$. So the equation does not have any solution.
|
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|
Solve this $\;2\sqrt[4]{\frac{x^4}{3} + 4} = 1 + \sqrt {\frac{3}{2}} |y| \ldots$ Solve
$\begin{cases}
2\sqrt[4]{\frac{x^4}{3} + 4} = 1 + \sqrt {\frac{3}{2}} |y|\\[8pt]
2\sqrt[4]{\frac{y^4}{3} + 4} = 1 + \sqrt {\frac{3}{2}} |x|
\end{cases}$
|
Following Ron Gordon's suggestion, and noting that the if $(x,y)$ is a solution, then $(\pm x, \pm y)$ are all solutions, we need to solve:
$$2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}}x$$
Both sides to the fourth power:
$$\frac{16x^4}{3}+64=1+2\sqrt6x+9x^2+3\sqrt6x^3+\frac{9x^4}{4}\\$$
It turns out that this expression factors:
$$(x-\sqrt6)^2(37x^2+38\sqrt6x+126)=0$$
So that $x=\sqrt6$. The quadratic has no real roots.
Answers are then $(\pm\sqrt6,\pm\sqrt6)$.
|
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Limit of $\lim\limits_{n\to \infty}\left(\frac{1^\frac{1}{3}+2^\frac{1}{3}+3^\frac{1}{3}+\dots+n^\frac{1}{3}}{n\cdot n^\frac{1}{3}} \right)$ I want to evaluate this limit.
$$\lim_{n\to \infty}\left(\frac{1^\frac{1}{3}+2^\frac{1}{3}+3^\frac{1}{3}+\dots+n^\frac{1}{3}}{n\cdot n^\frac{1}{3}} \right)$$
What I did is:
set $f(x)=x^\frac{1}{3}$
$$f(1)+\int^\infty_0 x^\frac{1}{3} \, dx<f(1)+f(2)+\dots+f(n)<f(n)+\int^\infty_0 x^\frac{1}{3}\,dx$$
how it can help me to evaluate this limit?
I can convert the right\left expressions to limits? or the whole inequality to expression of limit.
I need some advice here,
Thanks!
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Hint: An integral over $[0,\infty)$ does not look right to me, it seems better to observe that with your $f$ as above
\begin{align*}
\frac{1^{1/3} + \cdots + n^{1/3}}{n \cdot n^{1/3}} &= \frac{f(1/n) + f(2/n) + \cdots + f(1)}{n}\\
&= \sum_{i=1}^n f(i/n) \cdot \frac 1n
\end{align*}
|
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Find the product of the following determinants (involving logarithms with different bases) Find the product of the following determinants:
$$\begin{vmatrix}
\log_3512 & \log_43 \\
\log_38 & \log_49
\end{vmatrix} * \begin{vmatrix}
\log_23 & \log_83 \\
\log_34 & \log_34
\end{vmatrix}$$
I tried to solve it like this:
$$\begin{vmatrix}
9\log_2 3 & \log_43 \\
3\log_32 & 2\log_43
\end{vmatrix} * \begin{vmatrix}
\log_23 & \log_83 \\
2\log_32 & 2\log_32
\end{vmatrix}$$ $$(9\log_2 3*2\log_43-3\log_32*\log_43)(\log_23*2\log_32-2\log_32*\log_83)$$
How do I get the final answer?
Please offer your assistance
|
HINT:
Putting $\log_32=a,$
$\log_3512=9\log_32=9a, \log_38=3\log_32=3a,$
$\log_43=\frac{\log 3}{\log 4}=\frac1{\frac{\log 4}{\log 3}}=\frac1{\log_34}=\frac1{2\log_32}=\frac1{2a}$ and so on
$$\begin{vmatrix}
\log_3512 & \log_43 \\
\log_38 & \log_49
\end{vmatrix} =\begin{vmatrix}
9a & \frac1{2a} \\
3a & \frac1a
\end{vmatrix} =\frac{15}2\text{ as }a=\log_32\ne0$$
Alternatively using $\log_ba=\frac{\log a}{\log b}$, $$\begin{vmatrix}
\log_3512 & \log_43 \\
\log_38 & \log_49
\end{vmatrix} =\begin{vmatrix}
9\frac{\log 3}{\log 2} & \frac{\log3}{2\log 2} \\
3\frac{\log2}{\log 3} & \frac{2\log3}{2\log 2}
\end{vmatrix} =9-\frac32$$
|
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|
Finding the Equation of Parabola
Write the equation in the form $y=a(x-h)^{2}+k$ with zeros -4 and 8, and an optimal value of 18.
I'm not sure what "optimal value" means first of all- I think it means that the maximum value has a y-value of 18. What I've done so far:
$y=a(x+4)(x-8)$. Then to calculate the x-value of the vertex: $\frac{-4+8}{2}=2$ and then you substitute $x=2$ into the original equation to get the y-value of the vertex, which is:
$y=(2+4)(2-8)\implies y=-36$. Then $y=a(x-2)^{2}-36$, and then since we know a point on the line, I subbed in $(-4,0)$, which means that $0=a(-4-2)^{2}-36\implies 0=36a-36\implies a=1.$ So what I'm getting is $y=-(x-2)^{2}-36$
|
Use factored form: $y=a(x+s)(x+t)$
$y=a(x+4)(x-8)$
'Optimal Value'= $y$ value of $(x,y)$
$18$ = $y$ value of $(x,y)$
$x$ value of $(x,y)= (s+t)/2
= (-4+8)/2
= 4/2
= 2$
Therefore $(x,y) = (2,18)$
Substitute $(2,18)$ into $y=a(x+4)(x-8)$
$18=a(2+4)(2-8)$
$18=a(6)(-6)$
$18=a(-36)$
$-36a=18$
$a=18/-36$
$a=1/-2$
Therefore the equation of the parabola is $y= 1/-2(x+4)(x-8)$
|
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|
trouble expanding taylor series about a point other than zero using geometric series I'm trying to understand how to use a Taylor series expansion to correctly expand a population growth function about a point other than zero using the geometric series.
For expansion about $t=0$, I get:
$$
\begin{align}
p(t) &= 2 ln (t+2) \\
&= 2 \int \frac{1}{t+2} dt \\
&= 2 \int \frac{1}{ 2(\frac{t}{2}+1)} dt \\
&= \int \frac{1}{ \frac{t}{2}+1} dt \\
&= \int \frac{1}{ 1 - (-\frac{t}{2} )} dt \\
&= \int (1 + (\frac{-t}{2})+(\frac{-t}{2})^2+(\frac{-t}{2})^3+\cdots) dt \\
&= \int (1 - \frac{1}{2}t+\frac{1}{4}t^2-\frac{1}{8}t^3+\cdots) dt \\
&= t - \frac{1}{4}t^2+\frac{1}{12}t^3-\frac{1}{16}t^4+\cdots + C \\
p(0) &= 2\ln 2 = C\\
p(t) &= 2\ln 2+t - \frac{1}{4}t^2+\frac{1}{12}t^3-\frac{1}{16}t^4+\cdots \\
\end{align}
$$
I'm not sure how to go about expanding $p(t) = 2ln(t+2)$ about $t=2$ using the above method. In Wolfram Alpha, the answer it spits out is:
$$\ln 16 +\frac{x-2}{2}-\frac{1}{16}(x-2)^2+\frac{1}{96}(x-2)^3 + O((x-2)^4)$$
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You can expand $2\log(t+4)$ around $t=0$ and then plug in $t = x-2$.
|
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|
Find the range of $ f(x)=9^x - 3^x+1$
Problem:-Find range of function $ f(x)=9^x - 3^x+1$, here the domain of $f$ is $\mathbb R$.
Solution: $ f(x)=9^x - 3^x+1$. Let $f(x)=y$. Then
$$ \begin{split}y&=9^x - 3^x+1\\
y&=3^{2x} - 3^x+1
\end{split}$$
Let $3^x= u$. Then $ y=u^2 - u+1$, so
$$ u^2 - u+1-y=0.$$
Am I doing right?
|
Looks good. To finish it off, we complete the square and notice that:
$$
y = u^2 - u + 1 = \left(u^2-u+\dfrac{1}{4}\right)+\dfrac{3}{4} = \left(u-\dfrac{1}{2}\right)^2+\dfrac{3}{4} \ge 0+\dfrac{3}{4}=\dfrac{3}{4}
$$
since the square of any real number must be nonnegative. Hence, the range is:
$$
\{y\in \mathbb{R} \mid y \ge3/4\}
$$
|
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|
How to find a minimal polynomial (field theory) I was asked to find a minimal polynomial of $$\alpha = \frac{3\sqrt{5} - 2\sqrt{7} + \sqrt{35}}{1 - \sqrt{5} + \sqrt{7}}$$ over Q.
I'm not able to find it without the help of WolframAlpha, which says that the minimal polynomial of $\alpha$ is $$19x^4 - 156x^3 - 280x^2 + 2312x + 3596.$$ (Truely it is - $\alpha$ is a root of the above polynomial and the above polynomial is also irreducible over Q.)
Can anyone help me with this?
Thank you!
|
$$\alpha-\alpha\sqrt{5}+\alpha\sqrt{7}=3\sqrt{5}-2\sqrt{7}+\sqrt{35},$$
$$\alpha-\sqrt{35}=(\alpha+3)\sqrt{5}-(\alpha+2)\sqrt{7},$$
$$(\alpha-\sqrt{35})^2=[(\alpha+3)\sqrt{5}-(\alpha+2)\sqrt{7}]^2,$$
$$\alpha^2+35-2\alpha\sqrt{35}=5(\alpha+3)^2+7(\alpha+2)^2-2\sqrt{35}(\alpha+2)(\alpha+3),$$
$$2\sqrt{35}(\alpha^2+4\alpha+6)=11\alpha^2+58\alpha+38,$$
$$[2\sqrt{35}(\alpha^2+4\alpha+6)]^2=(11\alpha^2+58\alpha+38)^2.$$ From the last equality, we get that minimal polynomial is $19x^4-156x^3-280x^2+2312x+3596$.
|
{
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|
Induction to prove that $1^2+ 3^2 + 5^2+ \cdots +(2n + 1)^2=\frac{(n+1)(2n+1)(2n + 3)}{3}$ Have I started this right? I know I have to add $(k+1)$ but why?
Use mathematical induction to prove that $$1^2+ 3^2 + 5^2+ \cdots +(2n + 1)^2=\frac{(n+1)(2n+1)(2n + 3)}{3}.$$
BASIS STEP: $P(0)$ is true since $$\begin{align}(2(0)+1)^2 &= \frac{(0+1)(2(0)+1)(2(0)+3)}{3}\\1 &= \frac{3}{3}\\1 &= 1.\end{align}$$
INDUCTIVE STEP: Assume $P(k)$ holds for an arbitrary integer $k>0$. $$1^2+ 3^2 + 5^2+ \cdots +(2k + 1)^2=\frac{(k+1)(2k+1)(2k + 3)}{3}$$
|
You need to assume that it holds for some $k\ge0,$ not some $k>0$, for otherwise the induction doesn't go through (and you've only proved it for $k=0$).
Now, given the assumption that $$1^2+ 3^2 + 5^2+ \cdots +(2k + 1)^2=\frac{(k+1)(2k+1)(2k + 3)}{3},$$ you need to add $\bigl(2(k+1)+1\bigr)^2$ to both sides, and show that that makes the right hand side equal to $$\frac{\bigl((k+1)+1\bigr)\bigl(2(k+1)+1\bigr)\bigl(2(k+1)+3\bigr)}3.$$ That's how you show inductively that $P(k+1)$ holds whenever $P(k)$ does.
As an alternative, show that $$0^2+1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6},$$ so that $$0^2+2^2+4^2+\cdots+(2k)^2=4(0^2+1^2+2^2+\cdot+k^2)=\frac{2k(k+1)(2k+1)}{3}$$ and $$\begin{align}0^2+1^2+2^2+\cdots+(2k+1)^2 &= \frac{(2k+1)\bigl((2k+1)+1\bigr)\bigl(2(2k+1)+1\bigr)}{6}\\ &= \frac{(2k+1)(2k+2)(4k+3)}{6}\\ &= \frac{(k+1)(2k+1)(4k+3)}{3}.\end{align}$$ Can you see how to proceed from there?
|
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|
Indefinite integration : $\int \frac{1+x-x^2}{\sqrt{(1-x^2)^3}}$ Problem :
Solve : $\int \frac{1+x-x^2}{\sqrt{(1-x^2)^3}}$
I tried :
$\frac{1-x^2}{\sqrt{(1-x^2)^3}} + \frac{x}{\sqrt{(1-x^2)^3}}$
But it's not working....Please guide how to proceed
|
you had a good guees! see
$$\int \frac{1+x-x^2}{(1-x^2)^{\frac{3}{2}}}dx = \int \frac{1-x^2}{(1-x^2)^{\frac{3}{2}}}+\frac{x}{(1-x^2)^{\frac{3}{2}}}dx
=\int (1-x^2)^{\frac{-1}{2}}dx+\int\frac{x}{(1-x^2)^{\frac{3}{2}}}dx$$
note that $[(1-x^2)^{\frac{-1}{2}}] '= -\frac{1}{2}(1-x^2)^\frac{-3}{2}(-2x)= \frac{x}{(1-x)^{\frac{3}{2}}}$ . Now let's us solve changing the variable $x= \sin t$ then $dx=\cos tdt$ therefore
$$\int (1-x^2)^{\frac{-1}{2}}dx=\int (1-\sin^2t)^{\frac{-1}{2}}\cos t \;dt=
\int \frac{\cos t}{cost} dt=\int dt=t =\arcsin x$$
So we get
$$\int \frac{1+x-x^2}{(1-x^2)^{\frac{3}{2}}}dx=\int (1-x^2)^{\frac{-1}{2}}dx+\int\frac{x}{(1-x^2)^{\frac{3}{2}}}dx=\frac{x}{(1-x)^{\frac{3}{2}}}+\arcsin x$$
|
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|
What is the $8^{th}$ term of $\left(3x-\frac{y}{2}\right)^{10}$? What is the $8^{th}$ term of $\left(3x-\frac{y}{2}\right)^{10}$?
My solution: I'am not sure if I'am correct :)
$^{10}C_r (3x)^{10-r} \left(-\frac{y}{2}\right)^r$
where $r= 7$ since we start at $r=0$
$^{10}C_7 (3x)^3 \left(-\frac{y}{2}\right)^7$
$\left(-120\right)(27x^3)\left(\frac{y^7}{128}\right)$
$= -\left(\frac{405}{16}\right) x^3 y^7$
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$$(3x-(1/2)y)^{10}$$ general term is $$T_k=\binom{n}{k}a^{n-k}b^k$$ in our case we have
$$a=3x,b=-\frac{1}{2}y,n=10,k=7$$
so
$$T_7=\binom{10}{7}(3x)^3(-\frac{1}{2}y)^7=-\frac{405}{16}x^3y^7$$
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|
Differentiation and Rate of Change Im slaving over this one
using rate of change and Differentiation
I tried the chain rule but that is not the route as it got passed back
Question: find rate of change v if
V = x^3 * y^2
where
x=cos3t and y =sin3t
which gives
v = cos^3(3t) * sin^2(3t)
This is what i have so far
dv/dx
3-3Sin3t^2 x Sin3t^2
-9sin3t^2 x Sin3t^2
-9Sin3t^4
dv/dy
cos3t^3 x 2x3Cos3t
Cos3t^3 x 6Cos3t
6Cos3t^4
so
v = -9Sin3t^4 + 6Cos3t^4
but get the feeling that the "product rule" changes my calculations further.
im a bit stuck, can any one give some advice or confirm my calcs ?
Thanks
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Since $x=\cos 3t$ and $y = \sin 3t$, by chain rule we have:
$$
\dfrac{dx}{dt} = -3 \sin 3t \qquad \text{and} \qquad \dfrac{dy}{dt} = 3 \cos 3t
$$
Hence, by product rule, we have:
$$ \begin{align*}
\dfrac{dV}{dt}
&= \left(3x^2 \dfrac{dx}{dt}\right)(y^2) + (x^3)\left(2y \dfrac{dy}{dt}\right) \\
&= (3\cos^2 3t) (-3 \sin 3t)(\sin^2 3t) + (\cos^3 3t)(2 \sin 3t) (3 \cos 3t) \\
&= -9\cos^2 3t\sin^2 3t\sin 3t + 6\cos^3 3t\cos 3t\sin 3t \\
\end{align*} $$
|
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|
Explanation of an example of linear transformation This is an example from a text (Linear Algebra, Freidberg). I am trying to follow along, and I feel like I should know this from vector calc but I am missing something silly.
The example is:
Define the Linear transformation $T:P_2(R)\rightarrow M_{2x2}(R)$ by
$$T(\,f(x)) = \begin {pmatrix} f(1)-f(2) & 0 \\ 0 & f(0) \end{pmatrix} \DeclareMathOperator{\span}{span}$$
since $\beta = ${$1, x, x^2$} is a basis for $P_2(R)$ we have
\begin{align*}
R(T) &= \span(T(\beta)) \\
&= \span(\{T(1), T(x), T(x^2)\}) \\
&= \span\left(\left\{\begin {pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \begin {pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix}, \begin {pmatrix} -3 & 0 \\ 0 & 0 \end{pmatrix} \right\}\right) \\
&= \span\left(\left\{\begin {pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}, \begin {pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix} \right\}\right).
\end{align*}
I wasn't sure how they got from the matrix defined above to the ones below. Neither $f(1)$ nor $f(2)$ are really defined, and plugging an $x$ or $x^2$ in there doesn’t help.
This is frustrating, because I am trying to just get through an example so I can figure out this stuff. The book says “thus we have found a basis for $R(T)$”, but it is far from obvious to me. I understand they are using the theorem that says
$$R(T) = \span(T(\beta)) = \span(\{T(v_1), T(v_2)...T(v_n)\}),$$
where the $v_i$ are vectors that form a basis.
It’s getting through the matrix calculation above I am struggling with.
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Your transformation is from the space of polynomials of degree at most 2 to the space of $2\times 2$ matrices with real coefficients.
They tell you that the polynomials $1$, $x$, and $x^2$ are a basis for $P_2(\mathbb{R})$; the next three matrices are just the results of applying your transformation to these three polynomials:
If $f(x)=1$ for all $x$, then
$$
T(\,f(x))=\begin{pmatrix}f(1)-f(2) & 0\\0 & f(0)\end{pmatrix}=\begin{pmatrix}1-1 & 0\\ 0 & 1\end{pmatrix}=\begin{pmatrix}0&0\\0&1\end{pmatrix}.
$$
If $f(x)=x$, then
$$
T(\,f(x))=\begin{pmatrix}f(1)-f(2) & 0\\0 & f(0)\end{pmatrix}=\begin{pmatrix}1-2 & 0\\0 & 0\end{pmatrix}=\begin{pmatrix}-1 & 0\\0 & 0\end{pmatrix}.$$
Finally, if $f(x)=x^2$, then
$$
T(\,f(x))=\begin{pmatrix}f(1)-f(2) & 0\\0 & f(0)\end{pmatrix}=\begin{pmatrix}1-4 & 0\\0 & 0\end{pmatrix}=\begin{pmatrix}-3 & 0\\0 & 0\end{pmatrix}.
$$
|
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|
Solving for $a,b,c,d$ where $a^2 + b^2 + c^2 + d^2 = 630^2$ How could one solve for $a,b,c,d$ where:
$$a^2 + b^2 + c^2 + d^2 = 630^2,\ a>b>c>d$$
$a,b,c,d$ squared is equal to the square of $630$, and $a$ is larger than $b$, and so forth.
$a,b,c,d$ is also of such form that:
$$n = x + (x+1) + (x+2) + (x+3), \ x\in\mathbb{N}$$
where $n$ could be substituted for $ a, b, c \ $or $d$.
In what ways could one attack this problem?
|
Since $a, b, c, d$ are all of the form $4x+6$ for some $x \in \mathbb{N}$, we see that $a,b,c,d \equiv 2 \pmod 4$. It follows that $a^2+b^2+c^2+d^2 \equiv 4+4+4+4 \equiv 0 \pmod 8$. However, $630^2 \equiv 4 \pmod 8$. Therefore there is no solution.
|
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|
Factor $x^4 - 11x^2y^2 + y^4$ This is an exercise from Schaum's Outline of Precalculus. It doesn't give a worked solution, just the answer.
The question is:
Factor $x^4 - 11x^2y^2 + y^4$
The answer is:
$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2)$
My question is:
How did the textbook get this?
I tried the following methods (examples of my working below):
*
*U-Substitution.
*Guess and Check.
*Reversing the question (multiplying the answer out).
Here is my working for each case so far.
(1) U-Substitution.
I tried a simpler case via u-substitution.
Let $u = x^2$ and $v = y^2$. Then $x^4 - 11x^2y^2 + y^4 = u^2 -11uv + v^2$. Given the middle term is not even, then it doesn't factor into something resembling $(u + v)^2$ or $(u - v)^2$. Also, there doesn't appear to be any factors such that $ab = 1$ (the coefficient of the third term) and $a + b = -11$ (the coefficient of the second term) where $u^2 -11uv + v^2$ resembles $(u \pm a)(v \pm b)$ and the final answer.
(2) Guess and Check.
To obtain the first term in $x^4 - 11x^2y^2 + y^4$ it must be $x^2x^2 = x^4$. To obtain the third term would be $y^2y^2 = y^4$. So I'm left with something resembling:
$(x^2 \pm y^2)(x^2 \pm y^2)$.
But I'm at a loss as to how to get the final solution's middle term from guessing and checking.
(3) Reversing the question
Here I took the answer, multiplied it out, to see if the reverse of the factoring process would illuminate how the answer was generated.
The original answer:
$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2) = [(x^2 - 3xy) - y^2][(x^2 + 3xy) - y^2]$
$= (x^2 - 3xy)(x^2 + 3xy) + (x^2 - 3xy)(-y^2) + (x^2 + 3xy)(-y^2) + (-y^2)(-y^2)$
$= [(x^2)^2 - (3xy)^2] + [(-y^2)x^2 + 3y^3x] + [(-y^2)x^2 - 3y^3x] + [y^4]$
$= x^4 - 9x^2y^2 - y^2x^2 + 3y^3x - y^2x^2 - 3y^3x + y^4$
The $3y^3x$ terms cancel out, and we are left with:
$x^4 - 9x^2y^2 - 2x^2y^2 + y^4 = x^4 - 11x^2y^2 + y^4$, which is the original question.
The thing I don't understand about this reverse process is where the $3y^3x$ terms came from. Obviously $3y^3x - 3y^3x = 0$ by additive inverse, and $a + 0 = a$, but I'm wondering how you would know to add $3y^3x - 3y^3x$ to the original expression, and then make the further leap to factoring out like terms (by splitting the $11x^2y^2$ to $-9x^2y^2$, $-y^2x^2$, and $-y^2x^2$).
|
A start: Divide the original by $y^4$, and set $z=x/y$. We are trying to factor $z^4-11z^2+1$ as a product of two quadratics. Without loss of generality we may assume they both begin with $z^2$. So we are looking for a factorization of type $(z^2+az+b)(z^2+cz+d)$. By looking at the coefficient of $z^3$ in the product, we can see that $c=-a$, and you are on your way.
The next step is to see what $c=-a$ and the coefficient of $x$ in the product being $0$ tells us about the relationship between $b$ and $d$.
The idea works smoothly for, say $(11.3)x^2y^2$ replacing $11x^2y^2$, except of course the factorization involves square roots.
Another way: Divide the original polynomial by $x^2y^2$. We get
$$\frac{x^4-11x^2y^2+y^4}{x^2y^2}=\frac{x^2}{y^2}-11+\frac{y^2}{x^2}.\tag{1}$$
Make the substitution $w=\frac{x}{y}+\frac{y}{x}$. Then $\frac{x^2}{y^2}+\frac{y^2}{x^2}=w^2-2$. The right-hand side of Expression (1) becomes $w^2-9$, which factors as $(w-3)(w+3)$. Thus (1) can be expressed as the product
$$\left(\frac{x}{y}+\frac{y}{x}-3\right)\left(\frac{x}{y}+\frac{y}{x}+3\right).\tag{2}$$
Finally, multiply Expression (2) by $x^2y^2$, by multiplying each term by $xy$, and we get our factorization.
The idea generalizes immediately to $x^4-ax^2y^2+y^4$ where $a\ge 2$, and can be pushed well beyond.
|
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|
Express $\cos 6\theta $ in terms of $\cos \theta$ I think I'm supposed to use the chebyshev polynomials, as in $$ \cos n \theta = T_n(x) = \cos(n \arccos x)$$
But no idea what now?
|
Since $$\cos 2a =2\cos ^{2}a -1, \qquad\sin 2a
=2\sin a\cos a,$$ $$\cos (a+b)=\cos a\cos b-\sin a\sin b,$$
and
$$
\begin{eqnarray*}
\cos 3\theta &=&\cos (2\theta +\theta ) \\
&=&\cos 2\theta \cos \theta -\sin 2\theta \sin \theta \\
&=&( 2\cos ^{2}\theta -1) \cos \theta -2\sin ^{2}\theta \cos
\theta \\
&=&( 2\cos ^{2}\theta -1) \cos \theta -2( 1-\cos ^{2}\theta
) \cos \theta \\
&=&4\cos ^{3}\theta -3\cos \theta ,
\end{eqnarray*}
$$
we have
$$
\begin{eqnarray*}
\cos 6\theta &=&\cos \left( 2\times 3\theta \right) \\
&=&2\cos ^{2}3\theta -1 \\
&=&2( 4\cos ^{3}\theta -3\cos \theta ) ^{2}-1 \\
&=&32\cos ^{6}\theta -48\cos ^{4}\theta +18\cos ^{2}\theta -1.
\end{eqnarray*}
$$
ADDED: From the definition of the Chebyshev polynomials
$$
\begin{equation*}
T_{n}(x)=\cos (n\arccos x)\Leftrightarrow T_{n}(\cos \theta )=\cos n\theta
,\quad \theta =\arccos x,
\end{equation*}
$$
we get
$$
\begin{eqnarray*}
T_{1}(x) &=&\cos (\arccos x)=x \\
T_{2}(x) &=&\cos (2\arccos x)=2x^{2}-1.
\end{eqnarray*}
$$
Since they satisfy the recurrence
$$
\begin{equation*}
T_{n+1}(x)=2xT_{n}(x)-T_{n-1}(x),
\end{equation*}
$$
we have
$$
\begin{eqnarray*}
T_{3}(x) &=&2xT_{2}(x)-T_{1}(x) \\
&=&2x( 2x^{2}-1) -x \\
&=&4x^{3}-3x \\
&& \\
T_{4}(x) &=&2xT_{3}(x)-T_{2}(x) \\
&=&2x( 4x^{3}-3x) -( 2x^{2}-1) \\
&=&8x^{4}-8x^{2}+1 \\
&& \\
T_{5}(x) &=&2xT_{4}(x)-T_{3}(x) \\
&=&2x( 8x^{4}-8x^{2}+1) -( 4x^{3}-3x) \\
&=&16x^{5}-20x^{3}+5x \\
&& \\
T_{6}(x) &=&2xT_{5}(x)-T_{4}(x) \\
&=&2x( 16x^{5}-20x^{3}+5x) -( 8x^{4}-8x^{2}+1) \\
&=&32x^{6}-48x^{4}+18x^{2}-1.
\end{eqnarray*}
$$
Therefore
$$
\begin{eqnarray*}
\cos 6\theta &=&T_{6}(\cos \theta ) \\
&=&32\left( \cos \theta \right) ^{6}-48\left( \cos \theta \right)
^{4}+18\left( \cos \theta \right) ^{2}-1 \\
&=&32\cos ^{6}\theta -48\cos ^{4}\theta +18\cos ^{2}\theta -1,
\end{eqnarray*}
$$
as above.
|
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|
If $A = \tan6^{\circ} \tan42^{\circ},~~B = \cot 66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$ My trigonometric problem is:
If $A = \tan6^{\circ} \tan42^{\circ}$ B = cot$66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$.
Working :
$$B = \cot 66^{\circ} \cot78^{\circ} = 1- \frac{\tan24^{\circ}+\tan18^{\circ}}{\tan42^{\circ}}$$
$$A= \tan6^{\circ} \tan42^{\circ} = 1- \frac{\tan6^{\circ} +\tan42^{\circ}}{\tan48^{\circ}}$$
but it seems this is the wrong way of doing this...please suggest. Thanks!
|
$$A=Tan(6^\circ)\,Tan(42^\circ)=Tan(2.3^\circ)Tan(45^\circ-3^\circ)=\frac{2Tan(3^\circ)}{1-Tan^2(3^\circ)}\frac{1-Tan(3^\circ)}{1+Tan(3^\circ)}=\frac{2Tan(3^\circ)}{\left(1+Tan(3^\circ)\right)^2}$$
$\:$
$$ B=Tan(24^\circ)\,Tan(12^\circ)=Tan(2.12^\circ)Tan(12^\circ)=\frac{2Tan^2(12^\circ)}{1-Tan^2(12^\circ)} \implies$$
$\:$
$$Tan(12^\circ)=\sqrt{\frac{B}{B+2}}$$ But $$Tan(12^\circ)=Tan(15^\circ-3^\circ)=\frac{(2-\sqrt{3})-Tan(3^\circ)}{(2-\sqrt{3})+Tan(3^\circ)}\implies$$
$$Tan(3^\circ)=\left(\frac{\sqrt{B+2}-\sqrt{B}}{\sqrt{B+2}+\sqrt{B}}\right)(2-\sqrt{3})$$ Substitute $\,$$Tan(3^\circ)$ above in the Expression of $A$ to get the Required Relation
|
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|
How can one calculate this diferential equation $y\ dy = 4x(y^2+1)^2\ dx \text{ and } y(0) = 1$
I'm trying:
$\dfrac{y\ dy}{ (y^2+1)^2 }= 4x\ dx$
but I can't figure out what to do now.
|
Good start, you separated the variables. Now integrate.
For the integral of the left-hand side, make the substitution $u=y^2+1$. You should arrive at
$$\int \frac{1}{2} \cdot\frac{1}{u^2} \,du,$$
which is easy.
The integral of the right-hand side is $2x^2+C$. Use the initial condition to find $C$.
Added detail: Let $u=y^2+1$. Then $du =2y\,dy$ and therefore $y\,dy=\frac{1}{2} \,du$. Thus
$$\int \frac{y\,dy}{(1+y^2)^2}=\int \frac{1}{2}\cdot \frac{du}{u^2}=-\frac{1}{2}\cdot\frac{1}{u}.$$
(We deliberately and not quite correctly left out the constant of integration, since we will have one on the right, and one is enough.)
Thus an antiderivative of the expression on the left is $-\dfrac{1}{2(1+y^2)}$.
Integrating on the right, we end up with
$$ -\frac{1}{2(1+y^2)}=2x^2 +C.$$
Put $x=0$. Since $y(0)=1$, we get $C=-\frac{1}{4}$.
We have arrived at
$$-\frac{1}{2(1+y^2)}=2x^2-\frac{1}{4}=\frac{8x^2-1}{4}.$$
Change sign and flip over. We get
$$2(1+y^2)=\frac{4}{1-8x^2.}$$
Now a small amount of manipulation gets us $y$ explicitly in terms of $x$. At the end we have to take a square root. Take the positive one, since $y(0)=1\gt 0$.
|
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|
Is this function bounded? Next question about integral $\int_{\partial M} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac1{||y-x||} dS_y$.
Let $\partial M$ be $C^2$ closed surface in $\mathbb{R}^3$, $M$ is open. Show that
$$
f(x) = \frac{\int_{\partial M} \left| \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} \right| dS_y}{\left| \int_{\partial M} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y\right|}
$$
is bounded in $\overline{M}$
I already know few things about this kind of integral, I already had two questions about it here and here. But I still can't show that $f$ is bounded.
$f$ is continuous in $M$. So if I show that
$$
\lim_{x\rightarrow x_0} f(x) = 1
$$
for every $x_0 \in \partial M$. Than $f$ is continuous in $\overline{M}$ hence it is bounded.
My idea of proving it is:
Find $U(x)\subset \partial M$ that $\left( n_y \cdot \nabla_y \frac{1}{||y-x||} \right) > 0$ for all $y\in U(x)$ and that
$$
\frac{\int_{\partial M\setminus U(x)} \left| \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} \right| dS_y}{ \int_{U(x)} \frac{1}{||y-x||} n_y \cdot \nabla_y \frac{1}{||y-x||} dS_y} \rightarrow 0\qquad \text{as dist}(x,\partial M) \rightarrow 0
$$
If you show that map $U(x)$ exists than you can easily show the limit $\lim_{x\rightarrow x_0 \in \partial M} f(x) = 1 $
|
Why not thinking it the other way around? Integrals like this are typical of electromagnetic theory.
Let
\begin{multline}
\iint_M \mbox{div}\left(\frac{1}{\|y-x\|}\nabla_y \frac{1}{\|y - x\|}\right) dy = \\
\iint_M \left\{\frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} + \left\|\nabla_y \frac{1}{\|y-x\|}\right\|^2\right\}d y \\
= \iint_M \left\{\frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} + \frac{1}{\|y-x\|^4}\right\}d y
\end{multline}
This integral presents problems when $x = y$, hence we take $M = M_\epsilon \cup B_\epsilon$, where $B_\epsilon$ is the ball with center in $x$ and radius $\epsilon$, and $M_\epsilon = M \setminus B_\epsilon$.
Given that for $x \neq y$,
$$
\Delta_y \frac{1}{\|y - x\|} = 0
$$
we have, for the first integral,
$$
\iint_M \frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} dy = \iint_{B_\epsilon} \frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} dy.
$$
Using Green's first identity,
\begin{multline}
\iint_M \frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} dy = \int_{\partial B_\epsilon} \frac{1}{\|y-x\|} \frac{\partial}{\partial n_y} \left(\frac{1}{\|y-x\|}\right) d S_y \\
- \iint_{B_\epsilon} \left\|\nabla_y \frac{1}{\|y-x\|}\right\|^2 d y,
\end{multline}
and then
\begin{multline}
\iint_M \mbox{div}\left(\frac{1}{\|y-x\|}\nabla_y \frac{1}{\|y - x\|}\right) dy = \\
\int_{\partial B_\epsilon} \frac{1}{\|y-x\|} \frac{\partial}{\partial n_y} \left(\frac{1}{\|y-x\|}\right) d S_y + \iint_{M_\epsilon} \frac{1}{\|y-x\|^4} d y.
\end{multline}
Now,
$$
\frac{\partial}{\partial n_y} \left(\frac{1}{\|y-x\|}\right)_{\|y - x\| = \epsilon} = \frac{\partial}{\partial r} \left(\frac{1}{r}\right)_{r = \epsilon} = - \frac{1}{\epsilon^2}
$$
then
$$
\iint_{B_\epsilon} \frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} dy = -\frac{1}{\epsilon^2} \int_{\|y - x\| = \epsilon} \frac{1}{\|y-x\|} d S_y
$$
Taking the change of variables $y = x + \epsilon \eta$, where $\|\eta\| = 1$,
$$
\iint_{B_\epsilon} \frac{1}{\|y-x\|}\Delta_y \frac{1}{\|y-x\|} dy = - \int_{\|\eta\| = 1} \frac{1}{\epsilon} d S_\eta = -\frac{4 \pi}{\epsilon},
$$
while
$$
\iint_{B_{2\epsilon}\setminus B_\epsilon} \frac{1}{\|y-x\|^4} d y \le \iint_{M_\epsilon} \frac{1}{\|y-x\|^4} d y \le \iint_{B_R \setminus B_\epsilon} \frac{1}{\|y-x\|^4} d y,
$$
where $B_\epsilon \subseteq B_{2\epsilon} \subseteq M_\epsilon \subseteq B_R$.
Taking the change of variable $y = x + \xi$ on the first integral, and $y = x + \eta$ on the last, where $\epsilon \le \|\xi\| \le 2\epsilon$, and $\epsilon \le \|\eta\| \le R$
$$
\iint_{\epsilon \le \|\xi\| \le 2\epsilon} \frac{1}{\|\xi\|^4} d\xi \le \iint_{M_\epsilon} \frac{1}{\|y-x\|^4} d y \le \iint_{\epsilon \le \|\xi\| \le R} \frac{1}{\|\eta\|^4} d\eta
$$
and then
$$
\frac{2 \pi}{\epsilon} \le \iint_{M_\epsilon} \frac{1}{\|y-x\|^4} d y \le 4 \pi \left(\frac{1}{\epsilon} - \frac{1}{R}\right).
$$
Finally,
$$
\frac{2 \pi}{\epsilon} \le \left| \iint_M \mbox{div}\left(\frac{1}{\|y-x\|}\nabla_y \frac{1}{\|y - x\|}\right) dy \right|
$$
and
$$
\iint_M \left| \mbox{div}\left(\frac{1}{\|y-x\|}\nabla_y \frac{1}{\|y - x\|}\right) \right| dy \le 4 \pi \left(\frac{2}{\epsilon} - \frac{1}{R}\right).
$$
Using the Divergence theorem, we conclude that
$$
1 \le f(x) = \frac{\int_{\partial M} \left|\left(\frac{1}{\|y-x\|}\nabla_y \frac{1}{\|y - x\|}\right)\cdot n_y \right| dS_y}{\left|\int_{\partial M} \left(\frac{1}{\|y-x\|}\nabla_y \frac{1}{\|y - x\|}\right)\cdot n_y dS_y \right|} \le 4
$$
Of course, the argument can be refined to improve the upper bound.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Covergence of $\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \dotsb$ I am investigating the convergence of the following series: $$\frac{1}{4} + \frac{1\cdot 9}{4 \cdot 16} + \frac{1\cdot9\cdot25}{4\cdot16\cdot36} + \frac{1\cdot9\cdot25\cdot36}{4\cdot16\cdot36\cdot64} + \dotsb$$
This is similar to the series $$\frac{1}{4} + \frac{1\cdot 3}{4 \cdot 6} + \frac{1\cdot3\cdot5}{4\cdot6\cdot8} + \dotsb,$$which one can show is convergent by Raabe's test, as follows: $$\frac{a_{n+1}}{a_n}= \frac{2n-1}{2n+2}= \frac{2n + 2 - 3}{2n+2}=1 - \frac{3}{2(n+1)}.$$
However, in the series I am looking at, $\frac{a_{n+1}}{a_n}=\frac{(2n-1)^2}{(2n)^2}= 1 - \frac{4n-1}{4n^2}$, so Raabe's test doesn't work (this calculation also happens to show that the ratio and root tests will not work).
Any ideas for this one? It seems the only thing left is comparison...
|
$$2\cdot 4\cdots (2n)=2^n n!,\quad 1\cdot 3\cdots (2n-1)=\frac{(2n)!}{2^nn!}$$
By Stirling:
$$\frac{1^2\cdot 3^2\cdots (2n-1)^2}{2^2\cdot 4^2\cdots (2n)^2}=\frac{1}{16^n}\left(\frac{(2n)!}{n!^2}\right)^2\sim \frac{1}{16^n}\left[\frac{\left(\frac{2n}{e}\right)^{2n}\sqrt{4\pi n}}{\left(\frac{n}{e}\right)^{2n}2 \pi n}\right]^2=\frac{1}{\pi n}$$
Hence it diverges.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate sum of squares of first $n$ odd numbers Is there an analytical expression for the summation $$1^2+3^2+5^2+\cdots+(2n-1)^2,$$ and how do you derive it?
|
One can give a combinatorial version of the argument below. But it takes some time to tell the right story. So for now we settle for the magic math approach. We have
$$\binom{2k+1}{3}-\binom{2k-1}{3}=\frac{(2k+1)(2k)(2k-1)-(2k-1)(2k-2)(2k-3)}{3!}.\tag{1}$$
Note that the above equation even holds when $k=1$, if we use the convention that $\binom{1}{3}=0$.
The right-hand side of (1) magically simplifies to $(2k-1)^2$. It follows that our sum $1^2+3^2+5^2+\cdots +(2n-1)^2$ is equal to
$$ \binom{3}{3}-\binom{1}{3} +
\binom{5}{3}-\binom{3}{3} + \binom{7}{3}-\binom{5}{3} + \cdots + \binom{2n+1}{3}-\binom{2n-1}{3} .$$
Note the wholesale cancellation. We get
$$1^2+3^2+5^2+\cdots +(2n-1)^2=\binom{2n+1}{3}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/437835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Volume using disk method $y = 2\sqrt{x} \quad y=x$ $y = 2\sqrt{x}$
$y=x$
about $x=-2$
I know that $y=x$ is on the outside and they meet at 4.
I need these in terms of y since I rotate about y.
$x = y$
$x = \frac{y^2}{4}$
$$\pi \int_0^4 (y - (-2))^2 - \left(\frac{y^2}{4} - (-2)\right)^2 dy$$
$$\pi \int_0^4 (y + 2)^2 - \left(\frac{y^2}{4} + 2\right)^2 dy$$
$$\pi \int_0^4 y^2 + 4y +4 - \left(\frac{y^4}{16} +4y+4\right) dy$$
$$\pi \int_0^4 y^2 - \frac{y^4}{16} dy$$
$$\pi \left(\frac{y^3}{3} - \frac{y^5}{16*5}\right)$$
Gives me $\dfrac{128\pi}{15}$
Where did I go wrong?
|
As pointed out in the other answer, your expansion of the subtracted term is off:
$$\pi \cdot \int_0^4 (y + 2)^2 -\left (\frac{y^2}{4} + 2\right)^2 dy$$
So far, so good. Your set up looks fantastic.
But note that expanding the subtracted square, we get:
$$\pi \cdot \int_0^4 y^2 + 4y +4 - \left(\frac{y^4}{16} +4\cdot\frac{y^2}{4} +4\right) dy = \pi \cdot \int_0^4 y^2 + 4y +4 - \frac{y^4}{16} - {y^2} -4 \,dy $$
$$\pi \cdot \int_0^4 4y - \frac{y^4}{16} dy$$
NOW integrating gives us $$\pi\left(\frac {4y^2}{2} - \frac{y^5}{5\cdot 16}\right)\Big|_0^4 = \pi\left(32 - \frac{64}{5}\right) = \left(\frac{96}{5}\right)\pi $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that, if $0 < x < 1$, then $(1+\frac{x}{n})^n < \frac1{1-x}$ More fully,
if $n\ge 2$ is an integer
and
$0 < x < 1$,
prove that
$(1+\frac{x}{n})^n < \frac1{1-x}$.
In addition,
if $c > 1$ and
$0 < x \le \frac{c-1}{c}$,
prove that
$(1+\frac{x}{n})^n < 1+cx$.
Proofs by elementary means
(no calculus or limits)
are particularly sought.
As an example of the utility of this result,
set $x = \frac12$.
Then
this shows that
$2 > (1+\frac1{2n})^n$
or
$2^{1/n} > 1+\frac1{2n}$
.
This is an example
of what I call a
contra-Bernoulli inequality
(CBI)
which gives an upper bound to
$(1+y)^n$
as opposed to Bernoulli's inequality,
which gives a lower bound to
$(1+y)^n$
of $1+ny$.
Note that
any CBI of the form
$(1+y)^n < 1+c y$
for $n \ge 2$
requires that $y$ is bounded,
since
$(1+y)^n > 1+y^n$
so
$1+cy > 1+y^n$
or
$cy > y^n$
or
$y < c^{1/(n-1)}$.
|
\begin{align}
\left ( 1+\frac{x}{n} \right )^n
&=1+n\cdot \frac{x}{n}+\frac{n(n-1)}{2}\frac{x^2}{n^2}+\cdots+\frac{x^n}{n^n}\\
&<1+x+x^2+\cdots\\
&=\frac{1}{1-x}
\end{align}
for $n\geq 2$, $0<x<1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/438927",
"timestamp": "2023-03-29T00:00:00",
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|
Number of terms in the expansion of $(1+x^3+x^5) ^n$? How can we find the number of terms in the expansion of $(1+x^3+x^5)^n$ ?
|
Note that
$$(1+x^3+x^5)^n = \sum_{k=0}^n \binom{n}{k} x^{3 k} (1+ x^2)^k$$
There are $n+1$ terms. For the $k$th term, there are $k+1$ terms. Therefore, there are
$$\sum_{k=0}^n (k+1) = \frac12 (n+1)(n+2)$$
terms before combining. However, there are many terms that will combine (i.e., like powers). To get an accurate count, then, we must figure out what those powers are.
Let's write, ignoring the coefficients, the powers by value of $k$:
$$(1+x^3+x^5)^n = \begin{array}\\ 1 \\ x^3 & x^5\\x^6 & x^8 & x^{10}\\ x^9 & x^{11} & x^{13} & x^{15}\\ x^{12} & x^{14} & x^{16} & x^{18} & x^{20}\\ x^{15} & x^{17} & x^{19} & x^{21} & x^{23} & x^{25} \end{array}$$
and so on. In general, we want to find all values of $3 k+2 \ell$ that overlap for different lattice points $(k,\ell)$. Note that we have a first combining when $k=5$ (which explains why my test for small $n$ worked); $(5,0)$ and $(3,3)$.
Unfortunately, I know of no closed for expression for the number, $h(m)$, of pairs of lattice points $(k,\ell)$ that have $3 k+2 \ell$ being equal to a given integer $m$. The best that I can see written is
$$\frac12 (n+1)(n+2) - \sum_{m=0, m=3 k+2 \ell}^{5 n} [h(m)-1]$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving the multiple integral $\int \int_D \sqrt{x^2+y^2+3}dx dy$, D:{$1\le x^2+y^2\le 4, y\le \sqrt3x, y\ge \frac x{\sqrt 3}$} $\iint_D \sqrt{x^2+y^2+3}\, dx dy$, $D=\left\{(x,y) \in \mathbb{R}^2 \mid 1 \leq x^2+y^2\le 4, y\le \sqrt3x, y\ge \frac x{\sqrt 3} \right\}$.
So I've started by drawing two circles, One with $R_1 =1$, Other with $R_2=2$
Then i drawed the lines $y_1=\sqrt3x$ and $y_2 = \frac x{\sqrt 3}$.
Using the polar system i'll need to calculate:
$\int^{\pi/3}_{\pi/6}dθ\int_1^2\sqrt{r^2+3}*r*dr$
Using substiution on: $\int_1^2\sqrt{r^2+3}*r*dr$ i get: $\frac {7^{3/2}}{3}-\frac {4^{3/2}}{3}$.
Then i need to multiply by $\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$
And the final answer is: $\frac {7^{3/2}-4^{3/2}}{18}\pi$.
The answer should be: $\frac {5(7\sqrt 7 -8)}{18}$, I'm pretty close but it seems like i'm missing something, I tried solving this one a couple of times and still got to the same answer.
I need help, What am i missing?
|
I think you've done almost everything right, but $\;1\le r\le\frac6{\sqrt{10}}=\frac{3\sqrt2}{\sqrt 5}\;$ ! , so
$$\int\limits_1^2r\sqrt{r^2+3}\,dr=\left.\frac12\frac23(r^2+3)^{3/2}\right|_1^2=\frac13\left(7^{3/2}-8\right)$$
and together with
$$\int\limits_{\pi/6}^{\pi/3}d\theta=\frac{\pi}6\;,\;\;\text{we get:}$$
$$\frac{\pi}{18}(7\sqrt 7-8)$$
I can't understand where does that five you wrote there come from...
|
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|
Solve the equation $z^3=z+\overline{z}$ I have been trying to solve an equation $z^3=z+\overline{z}$, where $\overline{z}=a-bi$ if $z=a+bi$. But I cant find any clues on how to move forward on that one. Please help.
|
Writing out real and imaginary parts of $z$ and separating real and imaginary parts yields
$$
x^3+3ix^2y-3xy^2-iy^3=2x
$$
therefore,
$$
x^3-3xy^2=2x\implies x=0\quad\text{or}\quad x^2-3y^2=2
$$
and
$$
3x^2y-y^3=0\implies y=0\quad\text{or}\quad3x^2=y^2
$$
If $y=0$, then $x^3=2x\implies x\in\{0,\sqrt2,-\sqrt2\}$.
If $x=0$, then $y=0$.
If $x^2-3y^2=2$ and $3x^2=y^2$, then $-8x^2=2$.
Thus, $(x,y)\in\{(0,0),(\sqrt2,0),(-\sqrt2,0)\}$; that is, $z\in\{0,\sqrt2,-\sqrt2\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/440000",
"timestamp": "2023-03-29T00:00:00",
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|
system of differential linear equations $y'=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}y$ find the solution to the problem $y'=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}y, y(0)=\begin{pmatrix}4\\0\end{pmatrix}$
I know i have to find the eigenvalues and eigenvectors of the matrix $A=\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$
there's only one eigenvalue which is $1$ and only "one" eigenvector and we can choose $\begin{pmatrix} 1\\0\end{pmatrix}$
but now I dont know what to do.
what comes next?
I should say that I know how to find the exponencial of a matrix and also I want to know how to solve this kind of problem in general, so techniques to solve this particular one that wouldnt work on a more general problem dont help me much.
|
Let
$$I_2 = \begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\quad\text{ and }\quad J_2 = \begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}$$
Notice $I_2 J_2 = J_2 I_2$, $I_2^2 = I_2$ and $J_2^2 = 0_2$, we have:
$$e^{At} = e^{I_2t + J_2t} = e^{I_2t}e^{J_2t} = (I_2 + I_2t + I_2^2\frac{t^2}{2!} + \cdots)( I_2 + J_2t+ J_2^2\frac{t^2}{2}+\cdots)\\= I_2( 1 + t + \frac{t^2}{2} + \cdots)(I_2+J_2t) = e^t (I_2+J_2t) = e^t\begin{pmatrix}1 & t\\0 & 1\end{pmatrix}$$
This gives us:
$$y(t) = e^{At}y(0) = e^t \begin{pmatrix}1 & t\\0 & 1\end{pmatrix}\begin{pmatrix}4\\0\end{pmatrix} = \begin{pmatrix}4 e^t\\0\end{pmatrix}$$
The same approach allow you to deal with other linear ODE: $y' = A y$ where $A$ is not diagonalizable. Let's say you have use a similarity transform and bring $A$ into a Jordan
normal block. For each Jordan block of size $n$:
$$B = \begin{pmatrix}
\lambda & 1 & 0 &\ldots & 0\\
0 & \lambda & 1 &\ldots & 0\\
& & \ddots & \ddots & &\\
0 & 0 & \ldots & \lambda & 1\\
0 & 0 & 0 &\ldots & \lambda
\end{pmatrix}$$
You can rewrite $B$ as $\lambda I_n + J_n$ where $J_n$ is a $n\times n$ matrix with entries on the superdiagonal all equal to 1. Once again, $I_n J_n = J_n I_n$ and $J_n^n = 0_n$ and one has in general:
$$e^{Bt} = e^{(\lambda I_n + J_n)t} = e^{\lambda t} \left(1 + J_n t + J_n^2\frac{t^2}{2!} + \cdots J_n^{n-1}\frac{t^{n-1}}{(n-1)!}\right)
= e^{\lambda t}
\begin{pmatrix}
1 & t & \frac{t^2}{2!} &\ldots & \frac{t^{n-1}}{(n-1)!}\\
0 & 1 & t &\ldots & \frac{t^{n-2}}{(n-2)!}\\
& & & \ddots & &\\
0 & 0 & \ldots & 1 & t\\
0 & 0 & 0 &\ldots & 1
\end{pmatrix}$$
|
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|
Find all values of a for which the equation $x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0$ possesses at least two distinct negative roots Find all values of a for which the equation $$x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0 $$ possesses at least two distinct negative roots.
I am able to prove that all roots would be negative .How to proceed after this.
|
I decided to transform my comment into an answer.
As $x=0$ is not a root, divide the equation by $x^2$ and solve for $x+\frac{1}{x}$ as
RGB has pointed out.
You will get two solutions $y_1$ and $y_2$ for $x+\frac{1}{x}$:
$$x+\frac{1}{x}=\frac{-(a-1)-\sqrt{(a-1)^2+4}}{2}=y_1$$
and
$$x+\frac{1}{x}=\frac{-(a-1)+\sqrt{(a-1)^2+4}}{2}=y_2$$
If we want two negative roots the line $y=y_1$ must intercept twice the function $f(x)=x+\frac{1}{x}$ in the third quadrant. That's only possible if $y_1<-2$. See the figure bellow:
So let's solve the inequality:
$$y_1<-2 \Rightarrow $$
$$\frac{-(a-1)-\sqrt{(a-1)^2+4}}{2}<-2 \Rightarrow $$
$$-(a-1)-\sqrt{(a-1)^2+4}<-4 \Rightarrow $$
$$(a-1)+\sqrt{(a-1)^2+4}>4 \Rightarrow $$
$$\sqrt{(a-1)^2+4}>5-a \Rightarrow $$
$$a^2-2a+5>25-10a+a^2 \Rightarrow $$
$$8a>20 \Rightarrow $$
$$a>\frac{5}{2}$$
Therefore for $a>\frac{5}{2}$ the original equation will have two negative roots.
|
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|
Integral $ \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx} $ I have this difficult integral to solve.
$$ \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx} $$
Now my approach is this: split $(\sin x)^{\sqrt 2 + 1}$ and $(\sin x)^{\sqrt 2 - 1}$ as $(\sin x)^{\sqrt 2}.(\sin x)$ and $(\sin x)^{\sqrt 2 - 2}.(\sin x)$ respectively, and then apply parts. But that doesn't seem to lead anywhere. Hints please!
Edit:
This is what I did (showing just for the numerator)
$$ \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx $$
$$ = \int_0^{\pi/2} (\sin x)^{\sqrt 2}.(\sin x) dx $$
$$ = (-\cos x)(\sin x)^{\sqrt 2}\Bigg|_0^{\pi/2} + \int_0^{\pi/2}(\sin x)^{ \sqrt 2 - 1 }(\cos^2 x) dx $$
(taking $ v = \sin x $ and $ u = (\sin x)^{\sqrt 2} $ in the $ \int uv $ formula)
$$ = \int_0^{\pi/2}\left( (\sin x)^{ \sqrt 2 - 1 } - (\sin x)^{ \sqrt 2 + 1 } \right) dx $$
Similarly for the denominator. This does give a reduction formula but then I don't see how to really use it for finding the answer.
|
Just like what @O.L pointed out, you can use Beta function to express the solution. In fact, you can change variables to get
\begin{eqnarray*}
&&\frac{\int_0^{\pi/2}\sin^{\sqrt{2}+1}xdx}{\int_0^{\pi/2}\sin^{\sqrt{2}-1}xdx}=\frac{\int_0^{1}u^{\sqrt{2}+1}(1-u^2)^{-1/2}du}{\int_0^{1}u^{\sqrt{2}-1}(1-u^2)^{-1/2}du}\\
&=&\frac{\int_0^{1}t^{\frac{1}{2}(\sqrt{2}+1)}(1-t)^{-1/2}\frac{1}{2t^{1/2}}dt}{\int_0^{1}t^{\frac{1}{2}(\sqrt{2}-1)}(1-t)^{-1/2}\frac{1}{2t^{1/2}}dt}=\frac{\int_0^{1}t^{\frac{1}{2}\sqrt{2}}(1-t)^{-1/2}dt}{\int_0^{1}t^{\frac{1}{2}(\sqrt{2}-2)}(1-t)^{-1/2}dt}\\
&=&\frac{B(\frac{\sqrt{2}+2}{2},\frac{1}{2})}{B(\frac{\sqrt{2}}{2},\frac{1}{2})}=\frac{\Gamma(\frac{\sqrt{2}+2}{2})\Gamma(\frac{1}{2})}{\Gamma(\frac{\sqrt{2}+3}{2})}\frac{\Gamma(\frac{\sqrt{2}+1}{2})}{\Gamma(\frac{\sqrt{2}}{2})\Gamma(\frac{1}{2})}=\frac{\Gamma(\frac{\sqrt{2}+2}{2})\Gamma(\frac{\sqrt{2}+1}{2})}{\Gamma(\frac{\sqrt{2}+3}{2})\Gamma(\frac{\sqrt{2}}{2})}\\
&=&\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}+1}{2}}=2-\sqrt{2}.
\end{eqnarray*}
|
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"url": "https://math.stackexchange.com/questions/445808",
"timestamp": "2023-03-29T00:00:00",
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|
If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$
If $\alpha +\beta = \dfrac{\pi}{4}$ prove that $(1 + \tan\alpha)(1 + \tan\beta) = 2$
I have had a few ideas about this:
If $\alpha +\beta = \dfrac{\pi}{4}$ then $\tan(\alpha +\beta) = \tan(\dfrac{\pi}{4}) = 1$
We also know that $\tan(\alpha +\beta) = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$
Then we can write $1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$
I have tried rearranging $1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}$ but it has not been helpful.
I also thought if we let $\alpha = \beta$ then I could write $\tan(\alpha+ \alpha) = 1$
(does this also mean $\tan(2\alpha) = 1$?)
then: $\tan(\alpha + \alpha) = \dfrac{\tan\alpha + \tan\alpha}{1- \tan\alpha\tan\alpha}$
which gives: $1 = \dfrac{2\tan\alpha}{1-\tan^2\alpha}$
Anyway these are my thoughts so far, any hints would be really appreciated.
|
You've reached here :
$$1 = \dfrac{\tan\alpha + \tan\beta}{1- \tan\alpha\tan\beta}
$$
Let's continue it in this way:
$$\large\begin{align}
\Rightarrow & \tan\alpha + \tan\beta+\tan\alpha\tan\beta = 1\\
\Rightarrow &\tan\alpha(1+\tan\beta) + \tan\beta = 1\\
\Rightarrow &\tan\alpha(1+\tan\beta) + 1+ \tan\beta = 2\\
\Rightarrow &(1 + \tan\alpha)(1 + \tan\beta) = 2\\
\end{align}
$$
|
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|
Find $F_{n}$ in : $F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} = 3^{n}$ I'm stuck with the question for a while : Find in $$F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} = 3^{n}$$
the element $F_{n}$ .
Placing $n-1$ instead on $n$ results in :
$$F_{n-1} +2F_{n-2} + ... + (n-1+1)\cdot F_{0} = 3^{n-1}$$
$$ F_{n-1} +2F_{n-2} + ... + n\cdot F_{0} = 3^{n-1} $$
subtracting both :
$$F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} -(F_{n-1} +2F_{n-2} + ... + n\cdot F_{0} )
=3^{n} - 3^{n-1}
$$
But that doesn't help much . Any ideas ?
Thanks
|
The condition
$$F_{n} +2F_{n-1} + ... + (n+1)\cdot F_{0} = 3^{n}$$
is equivalent to the equality of coefficients of $z^n$ in below expression:
$$( F_0 + F_1 z + F_2 z^2 + \cdots )(1 + 2z + 3z^2 + \cdots ) = 1 + 3z + 3^2 z^2 + \cdots$$
Notice $\frac{1}{(1-z)^2} = 1 + 2z + 3z + \cdots$ and $\frac{1}{1-3z} = 1 + 3z + 3^2 z^2 + \cdots$, we have:
$$\frac{\sum_{k=0}^{\infty} F_k z^k}{(1-z)^2} = \frac{1}{1-3z} \quad\implies\quad\sum_{k=0}^{\infty} F_k z^k = \frac{(1-z)^2}{1-3z} = \frac{1 - 2z + z^2}{1-3z}$$
By comparing the coefficients of $z^n$, we get:
$$\begin{align}
F_0 &= 1\\
F_1 &= 1\cdot 3^1 - 2 \cdot 1 = 1\\
&\;\vdots\\
F_n &= 1\cdot 3^n - 2 \cdot 3^{n-1} + 1 \cdot 3^{n-2} = 4\cdot 3^{n-2},\quad\text{ for }n \ge 2
\end{align}$$
|
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|
Show that $\frac {\sin(3x)}{ \sin x} + \frac {\cos(3x)}{ \cos x} = 4\cos(2x)$ Show that
$$\frac{\sin(3x)}{\sin x} + \frac{\cos(3x)}{\cos x} = 4\cos(2x).$$
|
$$\frac {\sin3x} {\sin x} = 3-4\sin^2x$$
as $\sin3x = 3\sin x-4\sin^3x$
$$\frac {\cos3x} {\cos x} = 4\cos^2x - 3$$
as $\cos3x = 4\cos^3x -3\cos x$
adding the two above we get $4(\cos^2x - \sin^2x)=4\cos2x.$
|
{
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|
Properties of Triangle - Trigo Problem : In $\triangle $ABC prove that $a\cos(C+\theta) +c\cos(A-\theta) = b\cos\theta$ Problem :
In $\triangle $ABC prove that $a\cos(C+\theta) +\cos(A-\theta) = b\cos\theta$
My approach :
Using $\cos(A+B) =\cos A\cos B -\sin A\sin B and \cos(A-B) = \cos A\cos B +\sin A\sin B$, we get:
\begin{equation}
a(\cos C\cos\theta -\sin C\sin\theta) +c(\cos A\cos\theta +\sin A\sin\theta)\quad\\=\cos\theta(a\cos C+c\cos A) +\sin\theta(c\sin A -a\sin C)\qquad(\text{i})
\end{equation}
By using projection formula which states: $b=a\cos C +c \cos A $
(i) will become :
$$b\cos\theta +\sin\theta (c\sin A -a\sin C)$$ How do I proceed from here?
|
Law of sines states that
$$\frac{\sin{A}}{a} = \frac{\sin{C}}{c}$$
which means that $c \sin{A} - a \sin{C} = 0$. The desired result follows.
|
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|
Initial Value, First Order Differential Equation: Weird natural log separation Solve the initial value first order differential equation problem:
$y' = \displaystyle\frac{y^5}{x(1+y^4)},\ y(1) = 1$
\begin{align}
\frac{1+y^4}{y^5}dy &= \frac 1x dx\\
\left(\frac 1{y^5} + \frac 1y\right)dy &= \frac 1x dx\\
-\frac 1 {4y^4} + \text{ln}|y| &= \text{ln}|x| + C_1
\end{align}
This is where I get stuck. How do I solve for $y$ at this point?
Wolfram Alpha gave has this following step, which I do not understand at all:
I tried the raising everything as exponents of $e$, but that seems to be a dead end:
\begin{align}
-\frac 1 {4y^4} + \text{ln}|y| &= \text{ln}|x| + C_1\\
e^{\left(\text{ln} y - \frac 1 {4y^4}\right)} &= e^{\text{ln} x + C_1}\\
y\cdot e^{\left(- \frac 1 {4y^4}\right)} &= x\cdot C_2,\ \text{where $C_2 = e^{C_1}$}
\end{align}
P.S. Natural logarithms don't seem to be working: \ln |y| produces $\ln |y|$. I used \text{ln}|y| for $\text{ln}|y|$ instead. Is this a bug?
|
You got nearly the conclusion. Let's elevate to power $-4$ after your equation :
\begin{align}
e^{\ln y - \frac 1 {4y^4}} &= e^{\text{ln} x + C_1}\\
e^{-4\ln y + \frac 1 {y^4}} &= e^{-4\,\text{ln} x - 4\;C_1}\\
\frac 1{y^4}\;e^{\frac 1 {y^4}} &= \frac {C_2}{x^4}\\
\end{align}
So that from the definition of the LambertW function $\displaystyle W(z)\;e^{W(z)}=z\,$ we get for $\,z:=\dfrac {C_2}{x^4}$ :
$$W\left(\dfrac {C_2}{x^4}\right)=\frac 1{y^4}$$
or
$$y^4=\frac 1{W\left(\dfrac {C_2}{x^4}\right)}$$
and the four solutions proposed by Alpha :
\begin{align}
y&=\frac 1{\sqrt[4]{\left(W\left(\dfrac {C_2}{x^4}\right)\right)}}\\
y&=\frac {-1}{\sqrt[4]{\left(W\left(\dfrac {C_2}{x^4}\right)\right)}}\\
y&=\frac i{\sqrt[4]{\left(W\left(\dfrac {C_2}{x^4}\right)\right)}}\\
y&=\frac {-i}{\sqrt[4]{\left(W\left(\dfrac {C_2}{x^4}\right)\right)}}\\
\end{align}
|
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|
How to find the sum of the series $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$? How to find the sum of the following series?
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n}$$
This is a harmonic progression. So, is the following formula correct?
$\frac{(number ~of ~terms)^2}{sum~ of~ all ~the~ denominators}$
$\Rightarrow $ if $\frac{1}{A} + \frac{1}{B} +\frac{1}{C}$ are in H.P.
Therefore the sum of the series can be written as :
$\Rightarrow \frac{(3)^3}{(A+B+C)}$
Is this correct? Please suggest.
|
That formula is not correct to sum the first few terms of the harmonic series. Trying it with even the first three would mean that $$\frac{3^2}{1+2+3} = \frac{9}{6} = 1.5 \neq \frac{1}{1} + \frac{1}{2} + \frac{1}{3}$$.
The harmonic series actually diverges, so the sum of the series as we let $n$ get large doesn't exist... You can, however, get partial sums as the harmonic numbers, however this is somewhat outside the scope of the algebra/precalculus topic you have it listed under. You can find more information here.
|
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|
The first two moments of $\int_0^1 B_s^2 \, ds$ I was trying to solve the following problem from Continuous Martingales and Brownian Motion by Daniel Revuz and Marc Yor, but got my solution back as the answer for variance was wrong. I have already recomputed it several times and spelled out everything explicitly without any shortcuts, but still do not see what I am missing. I would appreciate any help.
Problem (Chapter I, Exercise 1.15)
Let $B_t$ be a standard linear Brownian motion. Compute the first two moments of
\begin{equation*}
X = \int_0^1 B_s^2 \, ds.
\end{equation*}
Solution
The expected value is
\begin{equation*}
E(X) = \int_0^1 E \left( B_s^2 \right) \,ds = \frac{1}{2},
\end{equation*}
and the variance is
\begin{align*}
& \text{Var}(X) = E\left(\left(\int_0^1 B_s^2 ds\right)^2\right) - \frac{1}{4} = E\left( \int_0^1 \int_0^1 (B_s B_t)^2 \,ds\, dt \right) - \frac{1}{4} \\
& = E\left( \int_0^1 \int_0^t (B_s B_t)^2 \,ds \,dt \right) + E\left( \int_0^1 \int_t^1 (B_s B_t)^2 \,ds\, dt \right) - \frac{1}{4} \\
& = E\left( \int_0^1 \int_0^t (B_s ((B_t - B_s) + B_s))^2 \,ds\, dt \right) \\
& \qquad {} + E\left( \int_0^1 \int_t^1 (B_t ((B_s - B_t) + B_t))^2 \,ds\, dt \right) - \frac{1}{4} \\
& = E\left( \int_0^1 \int_0^t \left[ B^2_s(B_t - B_s)^2 + B_s^4 + 2 B_s^3(B_t - B_s) \right] \,ds\, dt \right) \\
& \qquad {} + E\left( \int_0^1 \int_t^1 \left[ B^2_t(B_s - B_t)^2 + B_t^4 + 2 B_t^3(B_s - B_t) \right] \,ds \,dt \right) - \frac{1}{4} \\
& = \int_0^1 \int_0^t \left(s(t - s) + 3 s^2\right) \,ds\, dt + \int_0^1 \int_t^1 \left(t(s - t) + 3 t^2\right) \,ds\, dt - \frac{1}{4} \\
& = \int_0^1 \int_0^t \left(st + 2 s^2\right) \,ds\, dt + \int_0^1 \int_t^1 \left(ts + 2 t^2\right) \,ds\, dt - \frac{1}{4} \\
& = \int_0^1 t \int_0^t s \,ds\, dt + 2 \int_0^1 \int_0^t s^2 \,ds\, dt + \int_0^1 t \int_t^1 s \,ds\, dt + 2 \int_0^1 t^2 \int_t^1 \,ds\, dt - \frac{1}{4} \\
& = \frac{1}{2} \int_0^1 t^3 dt + \frac{2}{3} \int_0^1 t^3 dt + \frac{1}{2} \int_0^1 t (1 - t^2) dt + 2 \int_0^1 t^2 (1 - t) dt - \frac{1}{4} \\
& = \frac{1}{8} + \frac{1}{6} + \frac{1}{2} \left( \frac{1}{2} - \frac{1}{4} \right) + 2 \left( \frac{1}{3} - \frac{1}{4} \right) - \frac{1}{4} \\
& = \frac{7}{24} + \frac{7}{24} - \frac{6}{24} = \frac{1}{3}.
\end{align*}
Thank you!
Regards,
Ivan
|
It turned out that my solution was actually correct.
In addition, here is a little empirical proof in MATLAB:
m = 10000; % number of paths
n = 10000; % number of integration segments
dt = 1 / n; % length of one segment
B = cumsum(normrnd(0, sqrt(dt), m, n), 2); % draw m paths of B
X = dt * sum(B.^2, 2); % integrate
E = mean(X) % expectation
V = mean((X - E).^2) % variance
Running the code, we get
E =
0.4981
V =
0.3248
|
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|
Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha - \cos4\alpha\sin\alpha$
Show that $\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha = \sin4\alpha\cos\alpha - \cos4\alpha\sin\alpha$
I know that $\sin2\alpha = 2\sin\alpha\cos\alpha$
so
$$\sin2\alpha\cos\alpha=2\sin\alpha\cos^2\alpha$$
and $\cos2\alpha\sin\alpha$ can be expressed in three ways:
$$(\cos^2\alpha-\sin^2\alpha)\sin\alpha =\sin\alpha\cos^2\alpha-\sin^3\alpha$$
$$(2\cos^2\alpha -1)\sin\alpha = 2\cos^2\alpha\sin\alpha - \sin\alpha$$
$$(1-2\sin^2\alpha)\sin\alpha = \sin\alpha - 2\sin^3\alpha$$
I tried adding these, but nothing came close to the required answer.
So then I tried calculating $\sin4\alpha$ (from the required answer):
$$\sin4\alpha=2\sin(2\alpha)\cdot\cos(2\alpha)$$
$$\sin4\alpha=2\cdot2\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)$$
$$\sin4\alpha=4\sin\alpha\cos\alpha(\cos^2\alpha-\sin^2\alpha)$$
so $$\sin4\alpha\cos\alpha= 4\sin\alpha\cos^2\alpha(\cos^2\alpha-\sin^2\alpha)$$
Still looking at the answer, I calculated $\cos4\alpha$
$$\cos4\alpha = 1- \sin^2(2\alpha)$$
If $$\sin2\alpha = 2\sin\alpha\cos\alpha$$
then $$\sin^22\alpha= (2\sin\alpha\cos\alpha)^2$$
and $$\cos4\alpha = 1 - 4\sin^2\alpha\cos^2\alpha$$
and $$\cos4\alpha\sin\alpha=\sin\alpha-4\sin^3\alpha\cos^2\alpha$$
I have tried subtracting my values for $\sin4\alpha\cos\alpha$ and $\cos4\alpha\sin\alpha$, but I have not come close to a solution.
|
Just remember that $\sin(\alpha+\beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ and $\sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$ and apply the r.h.s. of each of these equalities to your example.
|
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|
show that $\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}$ show that:
$$\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}$$
where $n=0,1,2,3,\ldots$.
is there any help?
thanks for all
|
Let $I_k=\int_{-\infty}^\infty\frac{dx}{(x^2+1)^{k+1}}$. An easy integration shows that $I_1=\frac{\pi}{2}$.
Evaluate $I_k$ using integration by parts, letting $du=dx$ and $v=\frac{1}{(x^2+1)^{k+1}}$. We get $u=x$ and $dv=-(k+1)(2x)\frac{1}{(x^2+1)^{k+2}}$. Thus
$$I_k=\left.\frac{x}{(x^2+1)^{k+1}}\right|_{-\infty}^\infty+\int_{-\infty}^\infty\frac{2(k+1)x^2}{(x^2+1)^{k+2}}\,dx=\int_{-\infty}^\infty\frac{2(k+1)x^2}{(x^2+1)^{k+2}}\,dx.$$
Rewrite the top of the last integrand as $2(k+1)(x^2+1-1)$. Then we obtain
$$I_k=2(k+1)I_k -2(k+1)I_{k+1}.$$
We can rewrite this as
$$I_{k+1}=\frac{2k+1}{2(k+1)}I_k.$$
Now the only thing that remains is to show that if $A_k=\frac{(2k)!\pi}{2^{2k}(k!)^2}$ then $A_k$ satisfies the same initial condition as $I_k$ (easy) and the same recurrence relation (straightforward). Since $I_1=A_1$ and $I_k$ and $A_k$ satisfy the same recurrence, the two sequences must be the same.
|
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|
Quadratic Diophantine Equation in Four Variables Consider the equation: $$d^2 = 6 + a^2 - 3b^2 + 3c^2$$ where $a, b, c, d$ are integers.
Is it necessarily the case that $a$ and $b$ have the same parity and that $c$ and $d$ have the opposite parity to that of $a$ and $b$? If so, why?
Example 1: $a = 11, b = 5, c = 12, d = 22$
Example 2: $a = 14, b = 6, c = 3, d = 11$
|
Every square is congruent to either $0$ or $1$ $\pmod 4$, hence consider the equation modulo $4$ to get $$\{0,1\} \equiv 2 + \{0,1\} - \{0,3\} + \{0,3\} \pmod{4}$$ where $\{m,n\}$ denotes both possibilities.
Now if $a$ and $b$ are both even we get $a^2 \equiv 0 \pmod{4}$ and $3b^2 \equiv 0 \pmod{4}$, hence $$\{0,1\} \equiv 2 + \{0,3\} \pmod{4}$$ Easy casework shows that only $d^2 \equiv 1 \pmod{4}$ and $3c^2 \equiv 3 \pmod{4}$ works, so $c$ and $d$ are odd.
If $a$ and $b$ are both odd, then $a^2 \equiv 1 \pmod{4}$ and $3b^2 \equiv 3 \pmod{4}$, so $$\{0,1\} \equiv \{0,3\} \pmod{4}$$ Only possibility is $d^2 \equiv 0 \pmod{4}$ and $3c^2 \equiv 0 \pmod{4}$, i.e. both $c$ and $d$ are even.
|
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|
How to get value of $a^2b^2+b^2c^2+c^2a^2$ If $a+b+c=0$ and $a^2+b^2+c^2=36$, what is the value of $a^2b^2+b^2c^2+c^2a^2$?
|
$$(a+b+c)=0\implies (a+b+c)^2=0 $$
$$\implies a^2+b^2+c^2+2(ab+bc+ca)=0$$
$$\implies 36+2(ab+bc+ca)=0$$
$$\implies (ab+bc+ca)=-18$$
$$\implies (ab+bc+ca)^2=324$$
just solve this whole square you can find answer
$$\implies a^2b^2+b^2c^2+c^2a^2+2(ab^2c+bc^2a+a^2bc)=324$$
$$\implies a^2b^2+b^2c^2+c^2a^2+2abc(b+c+a)=324$$
since $a+b+c=0$
so
$$\implies a^2b^2+b^2c^2+c^2a^2=324$$
|
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|
How to solve that ode I'm trying solve the following differential equation:
$$
x\left(\frac{dx}{dy}\right)^2+y\frac{dx}{dy}=x
$$
I tried to rewrite it this way:
$$
y(x)=x\frac{dy}{dx}+f\left(\frac{dy}{dx}\right)
$$
|
Let $v = \frac{dx}{dy}$ then we face $xv^2+yv=x$ this gives $v^2+\frac{y}{x}v-1=0$ solve this quadratic equation for $v$ to obtain:
$$ v = \frac{-y/x \pm \sqrt{y^2/x^2+4}}{2} $$
Hence,
$$ \frac{dx}{dy} = \frac{-y/x \pm \sqrt{y^2/x^2+4}}{2} $$
Suppose $z = y/x$ then $xz=y$ and $z\frac{dx}{dy}+x\frac{dz}{dy}=1$,
$$ \frac{y}{z^2}\frac{dz}{dy} = \frac{-z \pm \sqrt{z^2+4}}{2} $$
Thus,
$$ \frac{-2dz}{z^3\pm z^2\sqrt{z^2+4}} = \frac{dy}{y} $$
Perhaps this helps.
|
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|
$\|A-B\|^2 = ?$ We know that if $x,y \in \mathbb{R}$
\begin{equation}
(x-y)^2 = x^2 -2xy + y^2
\end{equation}
If $x,y$ are vectors in $\mathbb{R}^n$ we have
\begin{equation}
|x-y|^2=|x|^2 - 2 \ x \cdot y +|y|^2.
\end{equation}
where $x\cdot y$ is the usual scalar product. We know that there are several ways to define $\|A\|$ when $A$ is a matrix, for example
\begin{equation}
\|A\| = \sup \{ |Ax|: |x|=1\}
\end{equation}
Is there a similar formula as above? This is a formula to $\|A-B\|^2?$
|
If the norm is induced by some inner product $\langle\cdot, \cdot\rangle$, it must obey the parallelogram law:
\begin{align*}
\|x+y\|^2 + \|x-y\|^2 &= \|x\|^2 + 2\langle x, y\rangle + \|y\|^2 + \|x\|^2 - 2\langle x, y\rangle + \|y\|^2\\
&= 2\|x\|^2 + 2\|y\|^2
\end{align*}
A first indication that something is wrong is that the two matrices $\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right]$ and $\left[\begin{array}{cc}1 & 0\\0 & 1\end{array}\right]$ have the same norm. Indeed,
$$\left\|\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right]+\left[\begin{array}{cc}0 & 0\\0 & 1\end{array}\right] \right\|^2+\left\|\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right]-\left[\begin{array}{cc}0 & 0\\0 & 1\end{array}\right] \right\|^2 = 2,$$
$$2\left\|\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right]\right\|^2 +2\left\|\left[\begin{array}{cc}0 & 0\\0 & 1\end{array}\right] \right\|^2 = 4,$$
violating the law.
|
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|
How many isosceles triangles with total side length $100$ are there? Let the sum of the three sides of a triangle be $100,$ and all the sides are positive integers length, how many possible isosceles triangles are there?
|
x=2*a+100-2*a if (100-2*a)
2*34+32=100
2*35+30=100
2*36+28=100
2*37+26=100
2*38+24=100
2*39+22=100
2*40+20=100
2*41+18=100
2*42+16=100
2*43+14=100
2*44+12=100
2*45+10=100
2*46+8=100
2*47+6=100
2*48+4=100
2*49+2=100
number of solution: 16
|
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|
Sum of $1/(x+1) +2x/(x+1)(x+2) +3x^2/(x+1)(x+2)(x+3)+\ldots$ till n terms Sum of $$\frac{1}{(x+1)}+\frac{2x}{(x+1)(x+2)}+\frac{3x^2}{(x+1)(x+2)(x+3)}+\ldots$$ till n terms ...
I have no clue how to solve this ... I tried multiplying quantities ... such as $(x+2)-(x+1)$
|
We have
$$ \frac{kx^{k-1}}{(x+1)(x+2)\cdots(x+k-1)(x+k)}\\= \frac{x^{k-1}}{(x+1)(x+2)\cdots(x+k-1)}- \frac{x^{k}}{(x+1)(x+2)\cdots(x+k)},\, k\geq2$$
so by telescoping
$$\sum_{k=1}^{n} \frac{kx^{k-1}}{(x+1)(x+2)\cdots(x+k-1)(x+k)}=\frac{1}{x+1}+\frac{x}{x+1}-\frac{x^{n}}{(x+1)(x+2)\cdots(x+n-1)(x+n)}\\=1-\frac{x^{n}}{(x+1)(x+2)\cdots(x+n-1)(x+n)}$$
|
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|
How to compute $\prod\limits^{\infty}_{n=2} \frac{n^3-1}{n^3+1}$
How to compute
$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}\ ?$$
My Working :
$$\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}= 1 - \prod^{\infty}_{n=2}\frac{2}{n^3+1}
= 1-0 = 1$$
Is it correct
|
HINT:
If $$t_n=\frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}$$
$$t_{n+1}=\frac{n\{(n+1)^2+n+1+1\}}{(n+2)\{(n+1)^2-(n+1)-1\}}=\frac{n\{(n+1)^2+n+1+1\}}{(n+2)(n^2+n+1)}$$
and $$t_{n-1}=\frac{(n-2)\{(n-1)^2+n-1+1\}}{n\{(n-1)^2-(n-1)-1\}}=\frac{(n-2)(n^2-n+1)}{n\{(n-1)^2-(n-1)-1\}}$$
Alternatively, let $\displaystyle u_n=\frac{n-1}{n+1},v_n=\frac{n^2+n+1}{n^2-n+1}$ so that $t_n=u_n\cdot v_n$
$$\implies\prod_{2\le n\le r}u_n=\frac{1\cdot2\cdot3\cdots(r-2)(r-1)}{3\cdot4\cdot5\cdots r(r+1)}\frac2{r(r+1)}$$
$$\implies\prod_{2\le n\le r}v_n=\frac{7\cdot13\cdot21\cdots(r^2-r+1)(r^2+r+1)}{3\cdot7\cdot13\cdots \{(r-1)^2-(r-1)+1\}(r^2-r+1)}=\frac{r^2+r+1}3$$
$$\implies\prod_{2\le n\le r} t_n=\left(\prod_{2\le n\le r}u_n\right)\left( \prod_{2\le n\le r}v_n\right)=\frac{2(r^2+r+1)}{3r(r+1)} $$
Setting $r\to\infty,$ $\displaystyle \frac{2(r^2+r+1)}{3r(r+1)}=\frac23\lim_{r\to\infty}\frac{1+\frac1r+\frac1{r^2}}{1+\frac1r}=?$
|
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|
Double Integral Question on unit square Hints on solving following double integral will be appreciated.
$$\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{d}y \ \text{d}x$$
|
Since the integrand has a non-integrable singularity at the origin, we must evaluate it by iterated integration. So let's consider the inner integral first:
$$\begin{align}
\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\, dy &= \frac{1}{x}\int_0^{1/x} \frac{1-t^2}{(1+t^2)^2}\, dt \qquad\qquad\qquad\qquad\qquad (y = x\cdot t)\\
&= \frac{1}{2x} \int_0^{2\arctan(1/x)} \frac{1 - \tan^2 (\varphi/2)}{(1+\tan^2(\varphi/2))^2}(1+\tan^2(\varphi/2))\, d\varphi\quad (t = \tan (\varphi/2))\\
&= \frac{1}{2x} \int_0^{2\arctan(1/x)} \frac{\cos^2(\varphi/2) - \sin^2(\varphi/2)}{\cos^2(\varphi/2)+\sin^2(\varphi/2)}\, d\varphi\\
&= \frac{1}{2x} \int_0^{2\arctan(1/x)} \cos\varphi\,d\varphi\\
&= \frac{1}{2x} \sin \left(2\arctan(1/x)\right)\\
&= \frac{1}{x} \sin \left(\arctan(1/x)\right)\cos\left(\arctan(1/x)\right)\\
&= \frac{1}{x} \frac{1/x}{\sqrt{1+1/x^2}}\frac{1}{\sqrt{1+1/x^2}}\\
&= \frac{1}{1+x^2}
\end{align}$$
From the on, it's very simple,
$$\int_0^1 \frac{dx}{1+x^2} = \arctan 1 = \frac{\pi}{4}.$$
Note that the other order of integration,
$$\int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\, dx\,dy$$
leads to the value $-\frac{\pi}{4}$.
|
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|
Closed-form Geometric series of both increasing and decreasing variables? This question comes from the formula
$$x^n - a^n = (x-a)(x^{n-1}a^0 + x^{n-2}a^1 + .... + x^1a^{n-2} + x^0a^{n-1})$$
which can be verified by summing the second factor as a geometric series. My question is, how do you express the second factor as a geometric series in closed form? If the $x$ factors were instead the constant $1$, we might have
$$s_n =1\cdot a^0 + 1 \cdot a^1 +.... +1\cdot a^{n-1}=\frac{1\cdot (1-a^n)}{1-a}$$
but that's as far as I've gotten.
|
Observe that the series has common ratio $r = \frac{a}{x}$. Hence:
\begin{align*}
(x-a)(x^{n-1}a^0 + x^{n-2}a^1 + \ldots + x^1a^{n-2} + x^0a^{n-1})
&= (x-a) \cdot \dfrac{x^{n-1}(1-(\frac{a}{x})^n)}{1-\frac{a}{x}} \\
&= (x-a) \cdot \dfrac{x}{x} \cdot \dfrac{x^{n-1}(1-\frac{a^n}{x^n})}{1-\frac{a}{x}} \\
&= (x-a) \cdot \dfrac{x^{n}(1-\frac{a^n}{x^n})}{x-a} \\
&= x^n - a^n
\end{align*}
|
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|
Series $\frac{k^2}{k!}$ with $ k=1$ to $\infty$ A practice Math Subject GRE asked me to compute
$\sum_{k=1}^\infty \frac{k^2}{k!}$. The sum is equal to $2e$, but I wasn't able to figure this out using Maclarin series or discrete PDFs. What's the most elementary way to solve this? Thank you.
|
$e^x=1+x+\dfrac{x^2}{2!}+\cdots+\dfrac{x^k}{k!}+\cdots$
Differentiating both sides,
$e^x=0+1+\dfrac{2x}{2!}+\cdots+\dfrac{kx^{k-1}}{k!}+\cdots$
Multiplying both sides by $x$,
$xe^x=x+\dfrac{2x^2}{2!}+\cdots+\dfrac{kx^k}{k!}+\cdots$
Again differentiating both sides,
$(x+1)e^x=1+\dfrac{2^2x}{2!}+\cdots+\dfrac{k^2x^{k-1}}{k!}+\cdots$
Now putting $x=1$ gives,
$2e=\dfrac{1^2}{1!}+\dfrac{2^2}{2!}+\cdots+\dfrac{k^2}{k!}+\cdots$
|
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|
Prove that $\lceil(\sqrt{3}+1)^{2n}\rceil$ is divisible by $2^{n+1}$. Let $n$ be a positive integer. Prove that $\lceil(\sqrt{3}+1)^{2n}\rceil$ is divisible by $2^{n+1}$.
I tried rewriting $\lceil(\sqrt{3}+1)^{2n}\rceil$ as $m*2^{n+1}$ for some m, but couldn't get anywhere.
|
Let $(1+\sqrt{3})^{2n}=a_n+b_n\sqrt{3}$. Then $(1-\sqrt{3})^{2n}=a_n-b_n\sqrt{3}$.
Thus $(1+\sqrt{3})^{2n}+(1-\sqrt{3})^{2n}$ is the integer $2a_n$. Since $|1-\sqrt{3}|\lt 1$, we have
$$2a_n=\left\lceil(\sqrt{3}+1)^{2n}\right\rceil.$$
Note that
$$(1+\sqrt{3})^{2n+2}=(1+\sqrt{3})^{2n}(4+2\sqrt{3})=(a_n+b_n\sqrt{3})(4+2\sqrt{3}).$$ Thus
$$a_{n+1}=4a_n+6b_n \quad\text{and}\quad b_{n+1}=2a_n+4b_n.$$
Thus the largest power of $2$ that divides $2a_k$ increases by at least $1$ when we increment $k$. But $2a_0=2$. This completes the proof.
Remark: Whenever $a+b\sqrt{c}$ has a problem, where $a$ and $b$ and $c$ are integers, and $c$ is not a perfect square, its conjugate $a-b\sqrt{c}$ is likely to be helpful.
|
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|
Solving $x^4-2 x^2-x+1 = 0$ for $x$ How could one solve the following, giving the answer as a closed form, not as an estimation:
$$\text{Solve }x^4-2 x^2-x+1 = 0\text{ for } x$$
Where $x$ is $\text{real.}$
I found this one particularly hard. Any help will be very much appreciated.
|
As others have mentioned, there are no rational solutions but you can find the exact solution through the formula for a quartic function (though why one would need it is another question). The fact that the coefficient of $x^3$ is zero makes things a bit easier, but not by much.
For: $$a x^4 + c x^2 + d x + e=0$$
The solutions are:
$$x_{1,2}=-S\pm \frac{1}{2}\sqrt{-4S^2+\frac{d-2cS}{Sa}}$$
$$x_{3,4}=+S\pm \frac{1}{2}\sqrt{-4S^2-\frac{d+2cS}{Sa}}$$
Where:
$$S=\frac{1}{2\sqrt{3a}}\sqrt{-2c+Q+\frac{\Delta_0}{Q}}$$
$$Q=\frac{1}{\sqrt[3]{2}}\sqrt[3]{\Delta_1+\sqrt{\Delta_1^2-4\Delta_0^3}}$$
$$\Delta_0=c^2+12ae,\ \ \Delta_1=2c^3+27ad^2-72ace$$
In your case, the two real solutions are $x_{3,4}$, and you can verify that:
$$Q = \sqrt[3]{\frac{1}{2} \left(155+3 \sqrt{849}\right)}$$
$$S=\frac{1}{2\sqrt{3}} \sqrt{4+Q+\frac{16}{Q}}$$
$$ x= S \pm \frac{1}{2\sqrt{3a}} \sqrt{4 + \frac{1}{S} - 4 S^2} \simeq 0.5249, 1.4902$$
|
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|
Find the area bounded by the curves $y=1-\cos(2x)$, $y = 0$, for $0\leq x \leq 2 \pi$ Find the area bounded by the curve $y=1-\cos(2x)$, the $x$ axis, and the lines $x=0$ and $x=2\pi$
Can someone please show me how this question is completed. This is causing much confusion.
Thanks everyone!
|
First, we plot the graph, and look at the region to be integrated. Below, compliments of Wolfram Alpha, we have the graph of $\color{blue}{y = 1 - \cos 2\pi}$, above the line $\color{purple}{y = 0}$, from $x = 0$ to $x = 2\pi$:
From here, it's relatively straight forward: our bounds of integration are from $x = 0$ to $x = 2\pi$, and the integrand = $(\underbrace{1 - \cos 2x}_{\text{upper curve}}\quad - \underbrace{0}_{\text{lower curve}}) \,dx = (1 - \cos 2x)\,dx$
Putting this together gives us $$\begin{align}\int_0^{2\pi} (1 - \cos 2x)\,dx \quad & = \quad \int_0^{2\pi} \,dx \quad - \quad \int_0^{2\pi} \cos 2x\, dx\\ \\
& = x\Big\vert_0^{2\pi} \quad -\quad \dfrac 12 \sin 2x\Big\vert_0^{2\pi}\\ \\
& = (2\pi - 0) - \frac 12 (\sin(4\pi) - \sin(0))\\ \\
& = (2\pi - 0) - \frac 12(0 - 0) \\ \\
& = 2\pi\end{align}$$
|
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|
How to show that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ indirectly? I found this amazingly beautiful identity here. How to prove that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ without directly multiplying the factors? (I've already verified it that way). Moreover, how could someone possibly find such a factorization using complex numbers? Is it possible to find such a factorization because $A^3+B^3+C^3 - 3ABC$ is a symmetric polynomial in $A,B,C$?
|
Expanding on Potato's answer:
First lets make $a^3 + b^3 + c^3 -3abc\ $ a polynomial in $a$ so we get $a^3 - a3bc + b^3 + c^3$. This has $3$ roots so lets express it as $(a+p)(a+q)(a+r)$. But also know: $$a^3 + b^3 + c^3 -3abc = (a+b+c)(a^2 + b^2 + c^2 -ab -bc - ac)$$
Therefore lets say $p=b+c$. So our polynomial is now: $(a+q)(a+r)(a+b+c)$. Expanding this expression and equating the co-efficients of $a$ we get the following equations:
$$r+q + b+c = 0 $$
$$qr+br+bq+cr+cq = -3bc$$
$$bqr + cqr = b^3 + c^3$$
Doing sum of roots on the third equation we can get:
$$qr = b^2 - bc +c^2$$
subbing the this equation into the second equation we get:
$$(b+c)^2 + br+bq+cr+cq = 0$$
$$(b+c)^2 = -(b+c)(q+r)$$
$$b+c = -(q+r)$$
Now, since $b+c$ is real and lets assume $q$ and $r$ are complex, this means $r$ is the conjugate of $q$. So let $q=x+yi$. Therefore we get:
$$-2x = b+c$$
$$qr = |q|^2 = x^2+y^2 = b^2+c^2-bc$$
So we have $x = -(b+c)/2$ and substituting this in the above equation we get $y = \frac{\sqrt{3}}{2}(b-c)$.
This means:
$$q= -\frac{b+c}{2} + \frac{\sqrt{3}}{2}(b-c)i$$
$$ = -\frac{b}{2} + \frac{\sqrt{3}}{2}bi + -\frac{c}{2} - \frac{\sqrt{3}}{2}ci$$
$$ = bw^2 + cw$$
where w is the complex cubic root of unity. Hence this means that:
$$ r = cw^2 + bw$$
Therefore:
$$a^3 + b^3 + c^3 -3abc = (a+b+c)(a+ bw + cw^2)(a+bw^2 + cw)$$
|
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How to prove this inequality for $a_{n}<5$ Let $a_{1}=1,a_{2}=2$ and be related by the recurrence
$$a_{n+2}=\dfrac{a^2_{n+1}(1+a_{n})+(1+2a_{n})a_{n+1}-a^2_{n}}{a_{n+1}+a^2_{n}+a_{n}+1}$$
for $n\in N.$
How can I show that
$a_{n}<5,n\in N$?
my idea: I think we have
$$a_{n+2}=pa_{n+1}+qa_{n}$$
because we have
$$a_{n+2}a_{n+1}+a_{n+2}a^2_{n}+a_{n+2}a_{n}+a_{n+2}=a^2_{n+1}+a^2_{n+1}a_{n}+a_{n+1}+2a_{n}a_{n+1}-a^2_{n}\cdots (1)$$
then
$$a_{n+1}a_{n}+a_{n+1}a^2_{n-1}+a_{n+1}a_{n-1}+a_{n+1}=a^2_{n}+a^2_{n}a_{n-1}+a_{n}+2a_{n-1}a_{n}-a^2_{n-1}\cdots (2)$$
(1)+(2)
|
Let us start with @Did's simplified form of the recurrence relation (with typos corrected): $$b_{n+2}=\dfrac{b_nb_{n+1}^2}{b_n^2-b_n+b_{n+1}}\quad (n=1,2,...)\quad\text{with}\quad b_1=2\text{ and }b_2=3.$$Write $c_n:=\dfrac {b_{n+1}+b_n^2}{b_nb_{n+1}}\;(n=1,2,...).$ Then substituting for $b_{n+2}$ in the corresponding expression for $c_{n+1}$ and simplifying gives$$c_{n+1}=c_n\quad (n=1,2,...).$$Therefore $c_n=c_1=\frac76$ forall $n$; that is $(b_{n+1}+b_n^2)/b_nb_{n+1}=\frac76,$ or$$b_{n+1}=\dfrac{6b_n^2}{7b_n-6}.$$Now the map$f:x\mapsto 6x^2/(7x-6)$ satisfies $f(1)=f(6)=6$ and $1<f(x)<6$ for $1<x<6$ (the minimum is $\frac{144}{49}$ at $x=\frac{12}{7}$). Since $b_1\;(=2)$ is in this range, so are all $b_n$. Thus $b_n<6$ for all $n.$
|
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Finding minima and maxima of $\sin^2x \cos^2x$ I am trying to find the turning points of $\sin^2x \cos^2x$ in the range $0<x<\pi$.
So far I have:
$\begin{align} f'(x) & =\cos^2x \cdot 2\sin x \cdot \cos x + \sin^2x \cdot 2\cos x \cdot -\sin x \\
& = 2 \cos^3x \sin x - 2\sin^3x \cos x \\
& = 2 \sin x \cos x (\cos^2x - \sin^2x) \\
& = 2 \sin x \cos x (\cos x + \sin x) (\cos x - \sin x)
\end{align}$
And, after lots of simplification, I think I've found that:
$f''(x) = 2 \left[\left( \cos^2x-\sin^2x \right)^2 - 4\sin^2x \cos^2 x\right]$
My questions are:
*
*How can I evaluate $0 = \cos x + \sin x$ and $0 = \cos x - \sin x$ without resorting to graph plotting?
*Are there trigonometric identities that I could have used to simplify either derivative?
|
if you use the Double angle formulae
$$\sin^2(x)\cos^2(x)=(1-\cos2x)/2 \cdot (1+\cos2x)/2 = (1-\cos^2(2x))/4 = \sin^2(2x)/4 = (1-\cos4x)/8$$
so minimum is $1/8$ at $x=45\circ$ and maximum is 1/4 at $x=90\circ$
remember $\sin^2(x)=(1-\cos2x)/2$ and $\cos^2(x)=(1+\cos2x)/2$
|
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How to calculate the coordinates of a triangle's orthocentre? How does one calculate the coordinates of a triangle's orthocentre?
I was surfing through the net and got this formula:
$$x-\rm{coordinate}= \frac{x_1\tan A+x_2\tan B+x_3\tan C}{\tan A+\tan B+\tan C}$$
$$y-\rm{coordinate}= \frac{y_1\tan A+y_2\tan B+y_3\tan C}{\tan A+\tan B+\tan C} $$
How do you prove this? Is there any method using just the coordinates?
|
Here $A\equiv (x_1,y_1)$,$B\equiv (x_2,y_2)$,$C\equiv (x_3,y_3)$. I'll use the usual notation for $a,b,c,R$ and $A,B,C$.
Using simple trigonometry, $BP=c\cos B$, $PC=b\cos C\implies\dfrac{BP}{PC}=\dfrac{c\cos B}{b\cos C}$ $$\implies P_x= \dfrac{x_2b\cos C+x_3c\cos B}{b\cos C+c\cos B}= \dfrac{x_2b\cos C+x_3c\cos B}{a}$$
Using some trig again, $AH=2R\cos A$ and $HP=2R\cos B \cos C\implies \dfrac{AH}{HP}=\dfrac{\cos A}{\cos B\cos C}$ $$\implies H_x= \dfrac{\cos A\left(\dfrac{x_2b\cos C+x_3c\cos B}{a}\right)+x_1\cos B\cos C}{\cos A+ \cos B\cos C}$$
$$H_x=\dfrac{x_1a\cos B\cos C+x_2b\cos C\cos A+x_3c\cos A\cos B}{a\cos A+a\cos B\cos C}$$
$$H_x=\dfrac{x_1a\cos B\cos C+x_2b\cos C\cos A+x_3c\cos A\cos B}{a\cos B\cos C+b\cos C\cos A+c\cos A\cos B}$$
Dividing by $\cos A\cos B\cos C$:
$$H_x=\frac{x_1a\sec A+x_2b\sec B+x_3c\sec C}{a\sec A+b\sec B+c\sec C}$$
Now using Sine Law do the substitution for $a,b,c$ and then cancel the $2R$ to get:
$$H_x=\dfrac{x_1\tan A+x_2\tan B+x_3\tan C}{\tan A+\tan B+\tan C}$$
|
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"url": "https://math.stackexchange.com/questions/478069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
|
The series can be rewritten as
$$
\begin{align}
&1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\dots\\
&8\left(\frac1{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\dots\right)\\
&=8\sum_{k=1}^\infty\frac{(2k-1)!!}{(2k+2)!!}\tag{1}
\end{align}
$$
This is reminiscent of the series (obtained by the binomial theorem)
$$
\begin{align}
(1-x)^{-1/2}
&=1+\frac12x+\frac12\frac32\frac{x^2}{2!}+\frac12\frac32\frac52\frac{x^3}{3!}+\dots\\
&=\sum_{k=0}^\infty\frac{(2k-1)!!}{(2k)!!}x^k\tag{2}
\end{align}
$$
Substitute $x\mapsto x^2$ and multiply by $x$ to get
$$
x(1-x^2)^{-1/2}=\sum_{k=0}^\infty\frac{(2k-1)!!}{(2k)!!}x^{2k+1}\tag{3}
$$
Integration yields
$$
1-\sqrt{1-x^2}=\sum_{k=0}^\infty\frac{(2k-1)!!}{(2k+2)!!}x^{2k+2}\tag{4}
$$
Plugging in $x=1$ and subtracting $\frac12=\frac{(-1)!!}{2!!}$ yields
$$
\frac12=\sum_{k=1}^\infty\frac{(2k-1)!!}{(2k+2)!!}\tag{5}
$$
Multiplying by $8$ and applying $(1)$, we get
$$
\begin{align}
4
&=8\sum_{k=1}^\infty\frac{(2k-1)!!}{(2k+2)!!}\\
&=1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\dots\tag{6}
\end{align}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/479610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
A problem on limit It seems
$\lim_{n \rightarrow \infty} \sum_{r=1}^{n-1}\binom{n}{r-1}
\sum_{j=0}^{n-r+1} (-1)^j \binom{n-r+1}{j} \frac{1}{r+j+1}$
is $\frac{1}{2}$. Here's a plot of the function for $n \leq 300$:
But I can not prove this. Any hints?
|
Note that $\int_{0}^{1}x^{r+j}dx=\frac{1}{r+j+1}$.
So the sum becomes:
$\sum_{r=1}^{n-1}\binom{n}{r-1}\sum_{j=0}^{n-r+1}\binom{n-r+1}{j}(-1)^{j}\int_{0}^{1}x^{r+j}dx$
$=\int_{0}^{1}\sum_{r=1}^{n-1}\binom{n}{r-1}\sum_{j=0}^{n-r+1}\binom{n-r+1}{j}(-1)^{j}x^{r+j}dx$
$=\int_{0}^{1}\sum_{r=1}^{n-1}\binom{n}{r-1}x^{r}\big(\sum_{j=0}^{n-r+1}\binom{n-r+1}{j}(-x)^{j}\big)dx$
$=\int_{0}^{1}\sum_{r=1}^{n-1}\binom{n}{r-1}x^{r}(1-x)^{n-r+1}dx$
$=\int_{0}^{1}\sum_{r=0}^{n-2}\binom{n}{r}x^{r+1}(1-x)^{n-r}dx$
$=\int_{0}^{1}x\sum_{r=0}^{n-2}\binom{n}{r}x^{r}(1-x)^{n-r}dx=\int_{0}^{1}x\big((x+(1-x))^{n}-nx^{n-1}(1-x)-x^{n}\big)dx$
$=\int_{0}^{1}x\big(1-nx^{n-1}+nx^{n}-x^{n}\big)dx=\int_{0}^{1}xdx-n\int_{0}^{1}x^{n}dx+n\int_{0}^{1}x^{n+1}dx-\int_{0}^{1}x^{n+1}dx$
$=\frac{1}{2}-\frac{n}{n+1}+\frac{n}{n+2}-\frac{1}{n+2}=\frac{1}{2}+\frac{-n}{(n+1)(n+2)}-\frac{1}{n+2}=\frac{1}{2}-\frac{2n+1}{(n+1)(n+2)}$
$=\frac{1}{2}-\frac{2+\frac{1}{n}}{1+\frac{1}{n}}\cdot\frac{1}{n+2}$.
So $\lim_{n\to\infty}\sum_{r=1}^{n-1}\binom{n}{r-1}\sum_{j=0}^{n-r+1}(-1)^{j}\binom{n-r+1}{j}\big(\frac{1}{r+j+1}\big)=\frac{1}{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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|
Showing Whether a Sequence is Bounded Above or Not I am trying to solve the following problem about a sequence:
Consider the sequence ${a_n}$ where $a_n = 1 + \frac{1}{1 \cdot 3} + \frac {1}{1 \cdot 3 \cdot 5} + \frac {1}{1 \cdot 3 \cdot 5 \cdot 7} + ... + \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)}.$ Decide whether ${a_n}$ is bounded above or not, and prove your answer is correct.
I started to solve it in the following manner:
Let $a_n \geq a_{n+1} \Rightarrow \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)} \geq \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2(n+1)-1} \Rightarrow \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1)} \geq \frac{1}{1 \cdot 3 \cdot ... \cdot (2n-1) \cdot 2n+1}.$ Multiplying by the reciprocal of $a_n$, we have $1 \geq \frac{1}{2n+1}$ which shows that $a_n$ is increasing.
This is where I am stuck since I do not know how to proceed in showing it is bounded from what I have so far. Any assistance or criticism is welcome.
I am using the following textbook: Introduction to Analysis by Arthur Mattuck.
|
This may be informal, but just showing that the sequence is monotone decreasing would mean the first term is the upper bound for the sequence.
We may rewrite the sequence in the following way $a_{n+1}=\frac{a_n}{2n+1}$. We can then see whether the sequence is infact monotone decreasing
$a_n-a_{n+1}=a_n-\frac{a_n}{2n+1}=\frac{2na_n}{2n+1}$
We know $ n>0, ~a_n>0,~ 2>0,~2n+1>0$ which means
$\frac{2na_n}{2n+1}>0 \implies a_n>a_{n+1}$
This sequence is monotone decreasing and therefore is bounded inclusively by the first term of the sequence itself, which is 1.
|
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|
Idea of the Proof : Existence of a & b so that (Any integer greater than 8) = 3a + 5b
Claim: Prove that for every integer $n \geq 8$, there exist nonnegative integers $a$ and $b$ such that $n = 3a + 5b.$
Proclaimed solution : Let $n ∈ \mathbb{Z}$ with $n ≥ 8.$ $\text{ Then } n = 3q, \text{ where } q ≥ 3, \quad \text{ or } n = 3q+1, \text{ where } q ≥ 3, \quad \text{ or } n = 3q+2, \text{ where } q ≥ 2.$
We consider these three cases. (Rest of proof omitted)
I do not apprehend the above parts of the proof BEFORE (Rest of proof omitted). However, I understand and completed the rest of the proof that I omitted. I recognise that it uses the result that division of any integer by $n$ must yield a remainder of $0, 1, 2, ..., n - 2,
\text{ or } n - 1$.
$1.$ However, what is the idea or strategy of the proof?
$2.$ What motivates or propounds the claim?
$3.$ Is there a more natural or easier proof?
I referenced 1. Source: P102, Problem 4.9 on P101 (related to P90, Result 4.8) of Mathematical Proofs, 2nd ed by Chartrand et al.
|
$3a=3\cdot a+0\cdot5$ which needs $a\ge0\implies 3a\ge 0$
$3a+1=3(a-3)+2\cdot5$ which needs $a\ge3\implies 3a+1\ge 10$ so disallows $1,4,7$
$3a+2=3(a-1)+5\cdot5$ which needs $a\ge1\implies 3a+2\ge 5$ so disallows $2$
Iterating with $5$
$5b=5\cdot b+0\cdot3$ which needs $b\ge0\implies 5b\ge 0$
$5b+1=5(b-1)+2\cdot3$ which needs $b\ge1\implies 5b+1\ge 6$
$5b+3=5\cdot b+1\cdot3$ which needs $b\ge0\implies 5b+3\ge 3$
$5b+2=5(b-2)+3\cdot4$ which needs $b\ge2\implies 5b+2\ge 12$
$5b+4=5(b-1)+3\cdot3$ which needs $b\ge1\implies 5b+4\ge 9$
So, the only numbers those are not representable are $1,2,4,7$
|
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|
$g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)$ for all $x,y$. Find all functions $g:\mathbb{R}\to\mathbb{R}$ with $g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)$ for all $x,y$.
I think the solutions are $0, 2, x$. If $g(x)$ is not identically $2$, then $g(0)=0$. I'm trying to show if $g$ is not constant, then $g(1)=1$. I have $g(x+1)=(2-g(1))g(x)+g(1)$. So if $g(1)=1$, we can show inductively that $g(n)=n$ for integer $n$. Maybe then extend to rationals and reals.
|
Here's a much less murky solution with strong influence from Calvin Lin's excellent answer.
$$
g(x + y) + g(x)g(y) = g(xy) + g(x) + g(y) \tag{0}
$$
First of all, plugging in $y = 1$ gives
$$ g(x + 1) + g(x)g(1) = 2g(x) + g(1)$$
For readability let $g(1) = k$ and we have
$$
g(x + 1) = (2-k)g(x) + k \tag{1}
$$
Plugging in $y + 1$ for $y$ in (0) and reducing with (1):
\begin{align*}
g(x + y + 1) + g(x)g(y+1)
&= g(xy + x) + g(x) + g(y+1) \\
(2-k)g(x + y) + k + g(x) \left[ (2-k)g(y) + k \right]
&= g(xy + x) + g(x) + (2-k)g(y) + k \\
(2-k)\left[g(x + y) + g(x) g(y) - g(x) - g(y)\right] + g(x)
&= g(xy + x)
\end{align*}
Using (0) again we conclude
$$
g(xy + x) = (2-k)g(xy) + g(x)
$$
So as long as $x$ is nonzero, letting $z = xy$ we have
$$
g(z + x) = (2 - g(1))g(z) + g(x) \tag{2}
$$
for all real $x$ and $z$, $x \ne 0$.
From here on out we just work with the much cleaner equation (2).
Setting $z = 0$ gives $(2 - g(1))g(0) = 0$, so either $g(1) = 2$ or $g(0) = 0$.
If $g(1) = 2$, then $g(z + x) = g(x)$ so $g(x) = 2$ everywhere. Otherwise, $g(0) = 0$, so plug in $z = -x$ to (2):
$$
0 = (2 - g(1))g(-x) + g(x) \implies g(x) = (g(1) - 2) g(-x)
$$
It follows by applying this to $g(--x)$ that $g(x) = (g(1) - 2)^2 g(x)$. If $g(x)$ is 0 everywhere, this is a solution to the original (0); otherwise, this implies $g(1) - 2 = \pm 1$, so $g(1) = 1$ or $g(1) = 3$.
In the case $g(1) = 1$, $g(z + x) = g(z) + g(x)$. This implies $g(x) = mx$ for some constant $m$, and $g(1) = 1$ so $g(x) = x$.
In the case $g(1) = 3$, $g(z + x) = -g(z) + g(x)$. But exchanging $x$ and $z$, $g(z + x) = -g(x) + g(z)$, so $g$ is uniformly 0 (which we already assumed was not the case).
|
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"timestamp": "2023-03-29T00:00:00",
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|
Compute $ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$ Compute $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx$$
Of course we have $$ \int_{-\infty}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \frac{x^2}{(1+x^2)^2} dx = 2 \int_{0}^{\infty} \left( \frac{x}{1+x^2} \right) ^2 dx = \lim_{ A \to \infty } \int_{0}^{A} \left( \frac{x}{1+x^2} \right) ^2 dx $$
I think that firstly I should compute $ \int \left( \frac{x}{1+x^2} \right) ^2 dx $ but I don't have idea.
|
This integral may be evaluated using the residue theorem. The integrand has poles at $x=\pm i$; if we close with a contour in the upper half plane, then we need only worry about the pole at $x=i$. The integral is therefore
$$i 2 \pi \left [\frac{d}{dx} \frac{x^2}{(x+i)^2} \right ]_{x=i} = i 2 \pi\left [\frac{i 2 x}{(x+i)^3} \right ]_{x=i} = i 2 \pi \frac{-i}{4} = \frac{\pi}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/483180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Rolling ellipses I'm struggling to prove the following.
Set one ellipse in contact with a congruent one so that the minor axis of one is aligned with the major axis of the other. Now roll one round the other. The locus of the centre of the rolling ellipse is a circle centre the centre of the other, radius a + b.
Is there an obvious line of attack?
|
The locus is not a circle.
Counterexample.
Consider ellipses with semi-axes $a=2,b=1$. Let equation of static
ellipse is
$$
\dfrac{x^2}{2^2} + \dfrac{y^2}{1^2} = 1.
$$
Here 3 steps are shown:
Suppose the locus is a circle (with radius $r = a+b =3$).
Then must be an instant/moment (see $2$nd image), when ellipses are co-directed
(semimajor axes are parallel).
And 2 conditions must be true:
$$KM = LM = 3/2;\tag{1}$$
(since symmetry); and
$$len(BM) = len(AM),\tag{2}$$
where $len(...)$ is arc length. Yes, $len(BM) =^{\mbox{symmetry}} len(CM) =^{\mbox{rolling}} len(AM)$, since ellipse is rolling without slip.
1).
Looking at condition $(1)$, let's find coordinates of point $M$.
$M$ belong to ellipse, $|KM|=3/2$.
$\left\{
\begin{array}{l}
x^2+4y^2 = 4; \quad (M \mbox{belong to ellipse});\\
x^2+y^2 = 9/4;\quad (|KM| = 3/2);
\end{array}
\right.
\quad \implies x = \sqrt{5/3},\; y = \sqrt{7/12}$.
2). Let's estimate $len(BM)$ and $len(AM)$.
Equation of arc $AB$ is $y = f(x) =\sqrt{1-x^2/4}$.
Note, that
$f'(x) = \dfrac{-x/2}{2\sqrt{1-x^2/4}} =
\dfrac{-x}{\sqrt{16-4x^2}}$.
$\sqrt{1 +(f'(x))^2} = \sqrt{\dfrac{16-4x^2}{16-4x^2} + \dfrac{x^2}{16-4x^2}} = \sqrt{\dfrac{16-3x^2}{16-4x^2}}$.
So,
$\displaystyle len(BM) = \int\limits_0^{\sqrt{5/3}} \sqrt{1+(f'(x))^2} dx =
\int\limits_0^{\sqrt{5/3}} \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx \approx 1.32081$ (wolfram alpha).
$\displaystyle len(AM) = \int\limits_{\sqrt{5/3}}^{2} \sqrt{1+(f'(x))^2} dx =
\int\limits_{\sqrt{5/3}}^2 \sqrt{\dfrac{16-3x^2}{16-4x^2}}dx \approx 1.10131$ (wolfram alpha).
When condition $(1)$ is true, condition $(2)$ isn't true.
So, the locus isn't circle.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to show $\sqrt{4+2\sqrt{3}}-\sqrt{3} = 1$ I start with $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, then
$\begin{align*}
x +\sqrt{3} &= \sqrt{4+2\sqrt{3}}\\
(x +\sqrt{3})^2 &= (\sqrt{4+2\sqrt{3}})^2\\
x^2 + (2\sqrt{3})x + 3 &= 4+ 2\sqrt{3}\\
x^2 + (2\sqrt{3})x - 1 - 2\sqrt{3} &= 0
\end{align*}$
So I have shown that there is some polynomial whose solution is $x=\sqrt{4+2\sqrt{3}}-\sqrt{3}$, but I have not shown it to be 1.
|
Here’s an approach that doesn’t require you to spot a not-terribly-obvious factorization. Multiplying by the conjugate to get rid of some of the square roots is a fairly natural thing to do, and
$$\left(\sqrt{4+2\sqrt3}-\sqrt3\right)\left(\sqrt{4+2\sqrt3}+\sqrt3\right)=1+2\sqrt3\;.$$
Thus, the desired result holds if and only if
$$1+2\sqrt3=\sqrt{4+2\sqrt3}+\sqrt3\;,$$
or
$$1+\sqrt3=\sqrt{4+2\sqrt3}\;,$$
which is easily verified.
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Solving a system of equations using modular arithmetic modulo 5 Give the solution to the following system of equations using modular arithmetic modulo 5:
$4x + 3y = 0 \pmod{5}$
$2x + y \equiv 3 \pmod{5}$
I multiplied $2x + y \equiv 3 \pmod 5$ by $-2$, getting $-4x - 2y \equiv -6 \pmod{5}$.
$-6 \pmod{5} \equiv 4 \pmod 5$
Then I added the two equations:
$4x + 3y \equiv 0 \pmod{5}$
$-4x - 2y \equiv 4 \pmod{5}$
This simplifies to $y \equiv 4 \pmod{5}$.
I then plug this into the first equation: $4x + 3(4) = 0 \pmod{5}$
Wrong work:
Thus, $x = 3$.
But when I plug the values into the first equation, I get $2(3) + 4 \not\equiv 3 \pmod{5}$.
What am I doing wrong?
EDIT:
Revised work:
$x = -3 \pmod{5} = 2 \pmod{5}$.
Now when I plug the values into the first equation, I get $2(2) + 4 \equiv 8 \pmod{5} \equiv 3 \pmod{5}$.
|
There's also a "cheats" method available here. There are only $25$ possible values of $(x,y) \in (\mathbb{Z}_5)^2$. We can just check them one-by-one, and see which ones work.
We could do this by hand, or on a computer. In GAP:
for x in [0..4] do
for y in [0..4] do
if((4*x+3*y) mod 5=0 and (2*x+y) mod 5=3) then
Print([x,y],"\n");
fi;
od;
od;
returns the single solution $(x,y)=(2,4)$.
|
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|
Prove that a group generated by two elements of order $2$, $x$ and $y$, is isomorphic to $D_{2n}$, where $n = |xy|.$ I am completely stuck at the question
Let $G$ be a finite group and let $x$ and $y$ be distinct elements of order 2 in $G$ that generate $G$. Prove that $G \cong D_{2n}$, where $n = |xy|.$
I have proved that
Let $x$ and $y$ be elements of order 2 in any group $G$. If $t = xy$ then $tx = xt^{-1}$.
Can I get some hints?
|
Personally, I like the following presentation for $D_{2n}$:
$$\langle x,y\mid x^n=y^2=(xy)^2=1\rangle$$ Now assume that the following presentation is given:
$$G=\langle x,y\mid x^2=y^2=(xy)^n=1\rangle$$ So we have:
$$
\begin{align*}
G=\langle x,y&\mid x^2=y^2=(xy)^n=1\rangle\\
\cong
\langle x,y,a&\mid x^2=y^2=(xy)^n=1,a=xy\rangle\\
\cong
\langle x,y,a&\mid x^2=y^2=(xy)^n=a^n=1,a=xy\rangle\\
\cong
\langle x,y,a&\mid x^2=y^2=a^n=1,a=xy\rangle\\
\cong
\langle x,y,a&\mid x^2=y^2=a^n=1,a=xy,(ay^{-1})^2=1\rangle\\
\cong
\langle x,y,a&\mid y^2=a^n=1,a=xy,(ay^{-1})^2=1\rangle\\
\cong
\langle y,a&\mid y^2=a^n=1,(ay^{-1})^2=1\rangle\\
\cong
\langle y,a,b&\mid y^2=a^n=1,(ay^{-1})^2=1,b=y^{-1}\rangle\\
\cong
\langle y,a,b&\mid y^2=a^n=1,(ay^{-1})^2=1,b=y^{-1},b^2=1\rangle\\
\cong
\langle y,a,b&\mid a^n=1,(ay^{-1})^2=1,b=y^{-1},b^2=1\rangle\\
\cong
\langle y,a,b&\mid a^n=b^2=(ab)^2=(ay^{-1})^2=1,b=y^{-1}\rangle\\
\cong
\langle y,a,b&\mid a^n=b^2=(ab)^2=1,b=y^{-1}\rangle\\
\cong
\langle a,b&\mid a^n=b^2=(ab)^2=1\rangle
\end{align*}$$
Which is $D_{2n}$.
|
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}
|
Factorizing $f(x) = 2x^4 +x^3 -x^2 +8x-4$ I tried plugging in values of $x$ until I found $f(x)=0$. I got that $(x+2)$ is a factor. Is this the only way to find factors of this polynomial?
Then I used polynomial division to get $f(x) = (x+2)(2x^3 -3x^2+5x-2)$. However I tried different values of x but still couldn't get the cubic to equal zero. Does this mean the cubic has imaginary roots?
Also what does it mean for a polynomial to have imaginary roots? When you sketch them on the Ragland diagram, how does this bare relevance to the sketch of the function on the Cartesian plane?
Thanks
|
You might want to take some of the guessing out of finding rational roots by making use of the rational root theorem. From Wikipedia:
In algebra, the rational root theorem (or rational root test) states a constraint on rational solutions (or roots) of the polynomial equation
$$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 = 0$$
with integer coefficients.
If $a_0$ and $a_n$ are nonzero, then each rational solution $x$, when written as a fraction $x = p/q$ in lowest terms (i.e., the greatest common divisor of p and q is 1), satisfies
*
*p is an integer factor of the constant term $a_0$, and
*q is an integer factor of the leading coefficient $a_n.$
Using the Rational Root Theorem, we can check and confirm first that $-2$ is a root and $\frac 12$ is a rational root, and use polynomial long division to find the remaining quadratic:
$$f(x) = 2x^4 +x^3 -x^2 +8x-4 = (x+2)(2x^3 -3x^2+5x-2)=(x+2)(2x-1)(x^2-x+2)$$
Now, with only a quadratic left, we can find the roots of $x^2 - x + 2$ using the quadratic formula, and since $\Delta = b^2 - 4ac = 1 - 8 = -7 < 0$, those roots will be complex roots: $$\dfrac {1 \pm \sqrt 7 i}{2}$$
So, $f(x) = (x+2)(2x-1)(x^2-x+2)$ has roots $$-2, \frac 12, \dfrac {1 \pm \sqrt 7 i}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
find this limit $\lim_{x\to0^{+}}\frac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$ find the limit.
$$\lim_{x\to0^{+}}\dfrac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$$
my try:
$$\tan{x}=x+\dfrac{1}{3}x^3+o(x^3),\sin{x}=x-\dfrac{1}{6}x^3+o(x^3)$$
so
$$\tan{(\tan{x})}=\tan{x}+\dfrac{1}{3}(\tan{x})^3+o(\tan^3{x})=x+\dfrac{1}{3}x^3+\dfrac{1}{3}(x+\dfrac{1}{3}x^3)^3+o(x^3)$$
$$\tan{(\sin{x})}=\sin{x}+\dfrac{1}{3}(\sin{x})^3+o(\sin^3{x})=x-\dfrac{1}{6}x^3+\dfrac{1}{3}(x-\dfrac{1}{6}x^3)^3+o(x^3)$$
so
$$\tan{(\tan{x})}-\tan{(\sin{x})}=\dfrac{1}{2}x^3+o(x^3)$$
$$\tan{x}-\sin{x}=\dfrac{1}{2}x^3+o(x^3)$$
Have other metods? Thank you
|
Yours is easily the simplest $direct$ method.
My first idea for the limit is that that really looks like a difference quotient for the derivative of $\tan$. Distilling the core idea, I make the following conjecture:
Let $f(x)$ be continuously differentiable at $a$. Then,
$$ \lim_{\substack{(x,y) \to (a,a) \\ x \neq y}} \frac{f(x) - f(y)}{x-y} = f'(a)$$
which lets us immediately see the limit should be
$$\tan'(0) = \sec^2(0) = 1$$
|
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|
Right triangle with equal permeter and height - how to find side lengths? The question:
Suppose there is a right triangle with sides $a$ and $b$ and
hypotenuse $c$. Its perimeter is the same as its area, and $b = 6$.
What are its side lengths?
I just cannot figure out how to do this! The second sentence isn't particularly helpful:
$$a + c + 6 = \frac{6a}{2}$$
$$a + c + 6 = 3a$$
$$???$$
And I can't get anywhere with the Pythagorean Theorem either:
$$a^2 + 36 = c^2$$
$$c^2 - 36 = a^2$$
$$(c + 6)(c - 6) = a^2$$
$$???$$
How do I solve this puzzle?
|
If $a=6$, then area is equal to $(6b)/2=3b$. If that is equal to the perimeter, then $a+b+c=3b$ so $a+c=2b$ so $6+c=2b$. You also know that $a^2+b^2=c^2$ so $36+b^2=c^2$.
From the $6+c=2b$ you get $c=2b-6$, and so $36+b^2=(2b-6)^2$. Expand and you get a quadratic on $b$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Fermat primes relation to $2^n+1$ If $2^n+1$ is prime how to show that $n$ must be a power of $2$
Being at elementary level I am at a loss what to begin with?
|
Suppose that $n$ is odd. Then $$x^n=(x+1-1)^n=(-1)^n= -1\mod x+1$$ so that $x+1\mid x^n+1$. Thus if $2^n+1$ is not composite, it must be the case $n=2^u$. If you want to be explicit $$\begin{align}
\frac{{{x^n} + 1}}{{x + 1}} &= \frac{{{{( - x)}^n} - 1}}{{( - x) - 1}} \cr
&= \frac{{{u^n} - 1}}{{u - 1}} \cr
&= 1 + u + {u^2} + \cdots + {u^{n - 1}} \cr
&= 1 - x + {x^2} - + \cdots + {\left( { - 1} \right)^{n - 1}}{x^{n - 1}} \end{align} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Pre calculus fraction simplify question Simplify:
$$\frac{\frac{16x^4}{81} - y^4}{\frac{2x}{3} + y}$$
Wolfram alpha confirms the answer from the answer sheet: Wolframalpha answer
|
$$\frac{\frac{16x^4}{81} - y^4}{\frac{2x}{3} + y}=\frac{16x^{4}-3^4y^4}{81}\cdot \frac{3}{2x+3y}=\frac{(4x^2-9y^2)(4x^2+9y^2)}{81}\cdot \frac{3}{2x+3y}=\frac{(2x-3y)(2x+3y)(4x^2+3y^2)}{81}\cdot \frac{3}{2x+3y}.$$
|
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|
Different Ways of Integrating $3\sin x\cos x$ I am asking this question for my son who is in (equivalent) twelfth
grade and I failed to answer his query.
When he tries to integrate $3\sin x\cos x$, he finds that
this can be done in at least following three ways.
And these three ways do not produce equivalent results.
ONE
Let us assume, $\sin x = z$.
This gives,
\begin{align*}
\cos x &= \frac{dz}{dx}\\
\cos x dx &= dz
\end{align*}
So, we can write,
\begin{align*}
\int 3\sin x\cos x dx &=3 \int zdz\\
&=3 \frac{z^2}{2}\\
&=\frac{3}{2} \sin^2 x\\
&=\frac{3}{4}\times 2\sin^2 x\\
&=\frac{3}{4} (1 -\cos 2x)\\
\end{align*}
TWO
Let us assume, $\cos x = z$.
This gives,
\begin{align*}
-\sin x &= \frac{dz}{dx}\\
\sin x dx &= -dz
\end{align*}
So, we can write,
\begin{align*}
\int 3\sin x\cos x dx &=-3 \int zdz\\
&=-3 \frac{z^2}{2}\\
&=-\frac{3}{2} \cos^2 x\\
&=-\frac{3}{4}\times 2\cos^2 x\\
&=-\frac{3}{4} (1 +\cos 2x)\\
\end{align*}
THREE
\begin{align*}
\int 3\sin x\cos x dx &=\frac{3}{2}\int 2\sin x\cos x dx\\
&=\frac{3}{2}\int \sin 2x dx\\
&=-\frac{3}{2}\times\frac{1}{2} \cos 2x\\
&=-\frac{3}{4} \cos 2x\\
\end{align*}
The results found in above three methods are not the same.
If we try a simple approach of evaluating the integration results at, $x = \frac{\pi}{6}$, we get as follows.
From the first one,
$\frac{3}{4} (1 -\cos 2x) = \frac{3}{4} (1 -\cos \frac{2\pi}{6})
= \frac{3}{4} (1 -\cos \frac{\pi}{3})
= \frac{3}{4} (1 - \frac{1}{2})
= \frac{3}{4}\times\frac{1}{2}
= \frac{3}{8}$
From the second one,
$-\frac{3}{4} (1 +\cos 2x) = -\frac{3}{4} (1 +\cos \frac{2\pi}{6})
= -\frac{3}{4} (1 +\cos \frac{\pi}{3})
= -\frac{3}{4} (1 + \frac{1}{2})
= -\frac{3}{4}\times\frac{3}{2}
= -\frac{9}{8}$
From the third one,
$-\frac{3}{4} \cos 2x=-\frac{3}{4} \cos \frac{2\pi}{6}
= -\frac{3}{4} \cos \frac{\pi}{3}
= -\frac{3}{4} \times \frac{1}{2}
= -\frac{3}{8} $
Clearly, we are getting some nonequivalent results. We have failed to find
the mistakes or explanations behind this. Your help will be appreciated.
|
You're forgetting that an indefinite integral must include a constant of integration; for any chosen constant $C$, we have that
$$\frac{d}{dx}\left(-\frac{3}{4}\cos(2x)+C\right)=3\sin(x)\cos(x),$$
and that is precisely the relationship captured by the statement that
$$\int 3\sin(x)\cos(x)\,dx=-\frac{3}{4}\cos(2x)+C.$$
|
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|
Partial Derivates Question How would I go about taking the partial derivative of:
$\frac{a+b^2}{a^2b}$ with respect to $a$ or $b$?
Thanks!
|
The partial derivative of $\frac{a+b^2}{a^2b}$ with respect to $a$ is computed by treating $b$ as a constant.
Thus, $\dfrac{\partial u}{\partial a}[\frac{a+b^2}{a^2b}]$ = $\frac{a^2b(1) - (a+b^2)(2ab)}{a^4b^2}$ = $\frac{a^2b - 2a^2b - 2ab^3}{a^4b^2}$ = $\frac{-a-2b^2}{a^3b}$.
The partial derivative of $\frac{a+b^2}{a^2b}$ with respect to $b$ is computed by treating $a$ as a constant.
Thus, $\dfrac{\partial u}{\partial b}[\frac{a+b^2}{a^2b}]$ = $\frac{a^2b(2b) - (a+b^2)(a^2)}{a^4b^2}$ = $\frac{2a^2b^2-a^3-a^2b^2}{a^4b^2}$ = $\frac{a^2b^2-a^3}{a^4b^2}$=$\frac{b^2-a}{a^2b^2}$.
|
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|
Determinant of matrix obtained by commuting matrices The Question is to prove that :
For Commuting $n\times n$ matrices $A,B,C,D$ over a field $F$,
Determinant of $\left(\begin{array}{cccc}
A & B \\
C & D \\
\end{array} \right)$ is given by $\det(AD-BC)$
I have no idea how to proceed for this except at the case of $n=1$ where $\left(\begin{array}{cccc}
A & B \\
C & D \\
\end{array} \right)$=$\begin{pmatrix}
a& b \\
c & d \\
\end{pmatrix}$ for some $a,b,c,d\in F$ and I know $\det\left(\begin{array}{cccc}
a& b \\
c & d \\
\end{array} \right)=ad-bc=\det(ad-bc)=\det(AD-BC)$
So, for $n=1$ we have $\det\left(\begin{array}{cccc}
A & B \\
C & D \\
\end{array} \right)=\det(AD-BC)$
I have no idea how to proceed for general $n$ not even when $n=2$
Do I need to proceed by induction?I doubt that it may not work..
please provide some hints to prove this case...
Thank You.
|
Here is a method that works if $F = \mathbb{C}$. Let $\mathcal{F} = \{A, B, C, D\}$ and let $P\in M_{n}(\mathbb{C})$ such that $P^{-1}XP$ is upper triangular for each $X\in \mathcal{F}$. Then
\begin{equation*}
\det\begin{pmatrix}
A & B\\
C & D
\end{pmatrix} = \det\begin{pmatrix}
P^{-1} & 0\\
0 & P^{-1}
\end{pmatrix}\begin{pmatrix}
A & B\\
C & D
\end{pmatrix}\begin{pmatrix}
P & 0\\
0 & P
\end{pmatrix} = \det\begin{pmatrix}
P^{-1}AP & P^{-1}BP\\
P^{-1}CP & P^{-1}DP
\end{pmatrix}.
\end{equation*}
Further, $\det(AD-BC) = \det(P^{-1}(AD-BC)P) = \det((P^{-1}AP)(P^{-1}DP)-(P^{-1}BP)(P^{-1}CP))$. Thus it suffices to assume that $A$, $B$, $C$ and $D$ are all upper triangular. We prove the result by induction. For $n = 1$ there is nothing to prove. Expanding the determinant about the first column we have
\begin{align*}
\det\left(\begin{array}{ccc|ccc}
a_{11} & & \ast & b_{11} & & \ast\\
& \ddots & & & \ddots & \\
& & a_{nn} & & & b_{nn}\\
\hline
c_{11} & & \ast & d_{11} & & \ast\\
& \ddots & & & \ddots & \\
& & c_{nn} & & & d_{nn}
\end{array}\right) =& (-1)^{1+1}a_{11}\det\left(\begin{array}{ccc|cccc}
a_{22} & & \ast & 0 & b_{22} & & \ast\\
& \ddots & & \vdots & & \ddots & \\
& & a_{nn} & 0 & & & b_{nn}\\
\hline
c_{12} & \dots & c_{1n} & d_{11} & & & \ast\\
c_{22} & & \ast & & \ddots & & \\
& \ddots & & & & \ddots & \\
& & c_{nn} & & & & d_{nn}
\end{array}\right)\\
&+(-1)^{1+n+1}c_{11}\det\left(\begin{array}{ccc|cccc}
a_{12} & \dots & a_{1n} & b_{11} & & & \ast\\
a_{22} & & \ast & & \ddots & & \\
& \ddots & & & & \ddots & \\
& & a_{nn} & & & & b_{nn}\\
\hline
c_{22} & & \ast & 0 & d_{22} & & \ast\\
& \ddots & & \vdots & & \ddots & \\
& & c_{nn} & 0 & & & d_{nn}
\end{array}\right).
\end{align*}
Expanding both terms about the $n^{\text{th}}$ column we get
\begin{equation*}
\det\left(\begin{array}{c|c}
A & B\\
\hline
C & D
\end{array}\right) = (a_{11}d_{11}-b_{11}c_{11})\det\left(\begin{array}{ccc|ccc}
a_{22} & & \ast & b_{22} & & \ast\\
& \ddots & & & \ddots & \\
& & a_{nn} & & & b_{nn}\\
\hline
c_{22} & & \ast & d_{22} & & \ast\\
& \ddots & & & \ddots & \\
& & c_{nn} & & & d_{nn}
\end{array}\right).
\end{equation*}
Conclude using the inductive hypothesis.
|
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|
Improving bound on $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}}$ An old challenge problem I saw asked to prove that $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < 3$. A simple calculation shows the actual value seems to be around $2.8$, which is pretty close to $3$ but leaves a gap. Can someone find $C < 3$ and prove $\sqrt{2 \sqrt{3 \sqrt{4 \ldots}}} < C$?
|
Iif $a(n)=\sqrt{ 2\sqrt {3\sqrt {4...\sqrt{n}}}}$,
$$a(n)^{2}= 2\sqrt {3\sqrt {4...\sqrt{n}}} \tag{1}$$
$$\dfrac{1}{2^{2}}.a(n)^{2}= 3\sqrt {4...\sqrt{n}} \tag{2}$$
$$\dfrac{1}{3^{2}}.\dfrac{1}{2^{2^{2}}}.a(n)^{2^{2}}= 4\sqrt{...\sqrt{n}} \tag{3}$$
$$\vdots$$
$$\prod_{i=0}^{n-2}\Big(\dfrac{1}{(n-i)^{2^{i}}}\Big).a(n)^{2^{n-1}}=1 \tag{n}$$
Then $$a(n)=(\prod _{i=0}^{n-2}{(n-i)^{2^{i}})}^{\dfrac{1}{2^{n-1}}} \tag{1}$$
and
$$a(n+1)=(\prod _{i=0}^{n-1}{(n+1-i)^{2^{i}})}^{\dfrac{1}{2^{n}}} \tag{2}$$
From $(1)$ and $2$,
$$\frac{a(n+1)}{a(n)}=(n+1)^{\frac{1}{2^{n}}}$$
Mathematica could not find the limit sequence for $n\longrightarrow \infty$.
|
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|
Increasing and bounded sequence proof Prove that the sequence $a_n= 1+ \frac 12+ \frac 13+\cdots+ \frac 1n-\ln(n)$ is increasing and bounded above. Conclude that it’s convergent.
This what I got so far
Proof:
Part 1: Proving $a_n$ is increasing by induction.
Base:
$a_1=1$
$a_2=1+\frac 12= \frac 32$
$a_1≤a_2$
So the base case is established.
Induction step: We assume that $a_{n-1}≤a_n$. We will show that $a_n≤a_{n+1}$.
Since
$a_{n-1}≤a_n$
$$1+ \frac 12+ \frac 13+\cdots+ \frac{1}{(n-1)}-\ln(n-1) \leq 1+ \frac 12+ \frac 13+\cdots+ \frac 1n-\ln n$$
How should I continue?
|
@Patrick's answer shows that the sequence is decreasing. However, we can show boundedness and covergence in a single step since the terms of the sequence are non-negative. I only provide a skeleton solution, which will need fleshing out with a limit argument.
We can begin by rewriting the sequence as:
\begin{equation*} \sum_{k=1}^n \frac{1}{k} -\log{n} = \sum_{k=1}^n \frac{1}{k} -\int_{1}^n \frac{1}{x} d{x}
= \overbrace{\sum_{k=1}^{n-1} \int_{k}^{k+1} \Bigl( \frac{1}{k} -\frac{1}{x} \Bigr) d{x} +\frac{1}{n}}^\circledast.
\end{equation*}
The integral's limits provide the inequalities $k\le x \le {k+1} $, with which we obtain the following bounds on the integrand:
$$0 \le \frac{1}{k} - \frac{1}{x} \le \frac{1}{k} -\frac{1}{k+1} = \frac{1}{k(k+1)} < \frac{1}{k^2}.$$ Then, after substituting our new integrand $\tfrac{1}{k^2}$, we obtain:
\begin{equation*} 0 \le \sum_{k=1}^n \frac{1}{k} -\log{n} =\overbrace{\sum_{k=1}^{n-1} \int_{k}^{k+1} \Bigl( \frac{1}{k} -\frac{1}{x} \Bigr) d{x} +\frac{1}{n}}^\circledast \le \overbrace{\sum_{k=1}^{n-1} \frac{1}{k^2}}^{\spadesuit} +\frac{1}{n}.
\end{equation*}
By comparison with the convergent series $\spadesuit$, we find that the original sequence converges.
|
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|
How many ways to form $7$-digit numbers from $\{1,2,\ldots,9\}$ in which $4$ is not the next of $5$? I have tried in several cases:
*
*the digits don't contain five only, so the number of possible ways is $8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2$
*the digits don't contain four only, so the number of possible ways is $8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2$
*the digits don't contain both four and five, so the number of possible ways is $7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$
*the digits contain both four and five, but I am stuck on this case.
The result is that I add up the numbers in the four cases.
Is that process true?
|
The correct results for the first three cases you enumerated are:
*
*The number contains $4$ but not $5$: we obtain $7\cdot\binom{7}{6}\cdot 6!$ different numbers (the $7$ accounts for the position of the $4$, the binomial for choosing $6$ other digits out of the $7$ possible and the $6!$ for ordering them).
*The number contains $5$ but not $4$: we obtain again $7\cdot\binom{7}{6}\cdot 6!$ different numbers.
*The number contains neither $4$ nor $5$: we obtain $\binom{7}{7}\cdot 7! = 7!$ different numbers.
For the fourth case, if the number contains both $4$ and $5$, we must consider different cases. The trick is to consider the placement of $4$ as the first number:
*
*If $4$ is placed as the first or last digit, $5$ can be placed in $5$ different places, this gives us $2\cdot 5\cdot \binom{7}{5}\cdot 5!$ different numbers (the $2$ is for the two positions considered for the $4$, the $5$ for the position of the $5$, the binomial for choosing $5$ other numbers and the $5!$ for ordering them).
*If $4$ is not the first or last digit, $5$ can be placed only in $4$ places, thus we get $5\cdot4\cdot\binom{7}{5}\cdot 5!$ different numbers.
|
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|
How many weak compositions of $n$ with $k$ parts up to reflection? The number of weak compositions of $n$ with $k$ parts is given by:
$$ \binom{n+k -1}{k-1}. $$
Consider the following identification: two compositions $(s_1 + \cdots + s_k), (t_1 + \cdots + t_k)$ are the same if $t_i = s_{k-i}$ for $i = 1,\ldots,k$. In other words, we consider them equivalent if they are reflections of each other.
Under this identification, how many weak compositions of $n$ with $k$ parts are there?
Equivalently, how many palindromic weak compositions are there (i.e. weak compositions that read the same forward and backward)?
|
This is a simple application of Polya Counting. Let us recall for a moment that the generating function of weak compositions of $n$ into $k$ parts is
$$Z(E_k)\left(\frac{1}{1-z}\right)$$
where $$Z(E_k) = a_1^k$$
is the cycle index of the identity group so that we get the OGF
$$\left(\frac{1}{1-z}\right)^k.$$
Now when we do coefficient extraction using the Newton binomial we obtain the count
$$[z^n] \left(\frac{1}{1-z}\right)^k = {n+ k-1\choose n},$$
as observed in the introduction.
Now to answer the question we need the cycle index of the group $F_k$ containing the identity and a single flip of the row of slots into which we distribute the elements of the composition. We have that when $k$ is even,
$$Z(F_k) = \frac{1}{2} a_1^k + \frac{1}{2} a_2^{k/2}$$
and when $k$ is odd,
$$Z(F_k) = \frac{1}{2} a_1^k + \frac{1}{2} a_1 a_2^{(k-1)/2}.$$
Therefore the count $q_n$ when $k$ is even is given by
$$[z^n]Z(F_k)\left(\frac{1}{1-z}\right)
= \frac{1}{2} [z^n] \left(\frac{1}{1-z}\right)^k
+ \frac{1}{2} [z^n] \left(\frac{1}{1-z^2}\right)^{k/2}.$$
This gives
$$q_n = \frac{1}{2} {n+k-1\choose n}
+ \frac{1}{2}\begin{cases} 0 & \text{if} \quad n \quad \text{is odd,}\\
{n/2 + k/2-1\choose n/2} & \text{if} \quad n \quad \text{is even.}
\end{cases}.$$
The count $q_n$ for odd $k$ is given by
$$[z^n]Z(F_k) \left(\frac{1}{1-z}\right)
= \frac{1}{2} [z^n] \left(\frac{1}{1-z}\right)^k
+ \frac{1}{2} [z^n] \frac{1}{1-z}\left(\frac{1}{1-z^2}\right)^{(k-1)/2}.$$
The second term in the sum simplifies to
$$(1+z)\left(\frac{1}{1-z^2}\right)^{(k-1)/2+1}.$$
This gives
$$q_n = \frac{1}{2} {n+k-1\choose n}
+ \frac{1}{2}\begin{cases} {(n-1)/2 + (k-1)/2\choose (n-1)/2} &
\text{if} \quad n \quad \text{is odd,}\\
{n/2 + (k-1)/2\choose n/2} &
\text{if} \quad n \quad \text{is even.}
\end{cases}.$$
Here is a MSE Link to a chain of Polya Counting applications.
|
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|
Is there an equation that will graph a line segment? Is there an equation that, given the two points of a line segment, will result, when graphed for x on a real graph, in a line segment?
|
A very simple way to do this would be to add and then subtract the square roots of an expression to define the domain. For example:
Given $(3,34)$ and $ (5, 52)$, first find the equation for the line that goes between the two points:
$(52-34)/(5-3)x + b = y$
$y = 9x + b$
$34 = 9(3) + b$
$b = 7$
Therefore, the equation of the line is: $y = 9x + 7$
To restrict the domain to $[3, 5]$ start by adding and subtracting $\sqrt(x-3)$. The equation becomes:
$y = 9x + 7 + \sqrt{x-3}- \sqrt{x-3}$
To restrict the upper part of the domain, add and subtract √(-x+5). The final equation becomes:
$y = 9x + 7 + \sqrt{x-3} - \sqrt{x-3} + \sqrt{-x+5} - \sqrt{-x+5}$
Hope this helps!
|
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|
Prove $\sqrt{s_n+1} = \frac{1}{2}(1+\sqrt{5})$ This is to prove how the limit of $s_n$ converges to $\frac{1}{2}(1+\sqrt{5})$.
Assume: $s_1 = 1$; for $n \geq 1$, $s_{n+1} = \sqrt{s_n + 1}$.
How to prove this converges to $\frac{1}{2}(1+\sqrt{5})$?
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Consider applying the Contraction Mapping Theorem. Just check that the conditions for the theorem are satisfied (I'll leave those details to you).
We have $s_{n+1} = \sqrt{s_n + 1}$. Take $F: x \to \sqrt{x +1}$. By the Contraction Mapping Theorem we can conclude that $F$ has a unique fixed point $s$.
This gives us:
\begin{align*}
\lim s_{n+1} =& \lim\sqrt{s_n +1} \\
s =& \sqrt{s+1} \\
s^2 - s - 1 =& 0
\end{align*}
Then by the quadratic formula we have that our unique fixed point $s = \dfrac{1+\sqrt5}{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Division by fractions Why is it that to divide by a fraction you need to multiply by the reciprocal of that fraction?
For instance:
$\eqalign{
& 1 \div \frac{1}{3} \cr
& = 1 \times 3 \cr
& = 3 \cr} $
Thank you.
|
Note that
$$1 \div \frac{1}{3} = \frac{1}{\frac{1}{3}} = \frac{1}{\frac{1}{3}}\times 1 = \frac{1}{\frac{1}{3}}\times \frac{3}{3} = \frac{1\times 3}{\frac{1}{3}\times 3} = \frac{3}{1} = 3.$$
More generally, you can use the above manipulations to show $\dfrac{a}{b}\div\dfrac{c}{d} = \dfrac{ad}{bc}$.
|
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|
Solving a system of three linear equations with three unknowns Is my working correct or am I completely wrong? Have I missed anything out? Any feedback is appreciated.
Question: Consider the following system of equations
$2x + 2y + z = 2$
$−x + 2y − z = −5$
$x − 3y + 2z = 8$
Form an augmented matrix, then reduce this matrix to reduced row echelon form and solve the system.
My answer/working:
Given:
$2x + 2y + z = 2$
$-x + 2y - z = -5$
$x - 3y + 2z = 8$
Matrix form:
$\begin{pmatrix} 2 & 2 & 1 & 2\\ -1 & 2 & -1 & -5 \\ 1& -3& 2 & 8 \end{pmatrix}$
$\begin{pmatrix}2 & 0 & 0 & 2\\ 0 & 3 & 0 & -3\\ 0 & 0 & \frac56 & \frac53\end{pmatrix}$
Solution: $x = 1; y = -1; z = 2;$
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You can reduce your matrix further. remember that you can multiply and/or divide each row so you end up obtaining
$$
\begin{pmatrix}
1 & 0 & 0 & 1\\
0 & 1 & 0 & -1\\
0 & 0 & 1 & 2\\
\end{pmatrix}
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
fourier expansion of $\coth$ and justifying an identity The problem:
Justify the following equalities:
$$\cot x = i\coth (ix) = i \sum^\infty_{n=-\infty} \frac{ix}{(ix)^2+(n\pi)^2}=\sum^\infty_{n=-\infty}\frac{x}{x^2+(n\pi)^2}$$
I am trying to figure out how to start this. When I insert the Euler identity of $
\coth$ (using the formula for complex Fourier series) I end up with: $$c_n = \frac{1}{2\pi}\int^{\pi}_{-\pi}\frac{e^{x} + e^{-x}}{e^{x} - e^{-x}}e^{inx} \ dx = \frac{1}{2\pi} \int^{\pi}_{-\pi}\frac{e^{(1+in)x} + e^{-(1+in)x}}{e^{x} - e^{-x}} \ dx$$
which is one ugly integral. So my question is a) did I make a mistake in the starting point and b) can this integral be simplified in some way that's better? Or is there some stupidly silly pattern I should be recognizing here? (I considered treating the integral as $\frac{1}{2\pi} \int^{\pi}_{-\pi} \frac{u}{du}$ or something like it).
I suspect I am missing something obvious.
Thanks folks. :-)
|
Adding a bit of diversity to this discussion we suppose that we seek to prove that
$$\coth x = \sum_{n=-\infty}^\infty \frac{x}{x^2+\pi^2 n^2}.$$
Rather than use the standard technique (absolutely nothing wrong with it) of applying a circular contour to
$$ f(z) = \pi \cot(\pi z) \frac{x}{x^2+\pi^2 z^2}$$
we use Mellin transforms, which are admittedly a bit more complicated in this case (than summing the residues at the two poles of $f(z)$ on the imaginary axis at $\pm ix/\pi$), but perhaps have some complex variable didactic value.
Rewrite this as
$$\coth x = \frac{1}{x} + 2 \sum_{n=1}^\infty \frac{x}{x^2+\pi^2n^2}
= \frac{1}{x} + 2x \sum_{n=1}^\infty \frac{1}{x^2+\pi^2n^2}
= \frac{1}{x} + 2x \sum_{n=1}^\infty \frac{1}{n^2} \frac{1}{(x/n)^2+\pi^2}.$$
Now the inner sum, call it $S(x),$ is harmonic and may be evaluated by inverting its Mellin transform.
Recall the Mellin transform identity for harmonic sums with base function $g(x)$, which is
$$\mathfrak{M}\left(\sum_{k\ge 1}\lambda_k g(\mu_k x); s\right) =
\left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s}\right) g^*(s)$$
where $g^*(s)$ is the Mellin transform of $g(x).$
In the present case we have
$$\lambda_k = \frac{1}{k^2}, \quad
\mu_k = \frac{1}{k} \quad \text{and} \quad
g(x) = \frac{1}{x^2+\pi^2}.$$
The Mellin transform of $g(x)$ is
$$\int_0^\infty \frac{1}{x^2+\pi^2} x^{s-1} dx,$$
which we evaluate using a semicircular contour in the upper half plane (no convergence issues as is easily verified), getting
$$g^*(s) (1 + e^{\pi i(s-1)})= 2\pi i
\operatorname{Res}\left(\frac{1}{z^2+\pi^2} z^{s-1}; z = +i\pi\right).$$
Using the branch of the logarithm with arguments from zero to $2\pi$, this becomes
$$g^*(s) = \frac{2\pi i}{1- e^{\pi i s}}\frac{1}{2\pi i} (i\pi)^{s-1}
= \pi^{s-1} \frac{e^{i(\pi/2)(s-1)}}{1- e^{i\pi s}}
= - \pi^{s-1} \frac{i e^{i(\pi/2)s}}{1- e^{i\pi s}}\\
= - \pi^{s-1} \frac{i}{e^{-i(\pi/2)s}- e^{i(\pi/2)s}}
= \frac{1}{2} \pi^{s-1} \frac{2i}{e^{i(\pi/2)s}- e^{-i(\pi/2)s}}
= \frac{1}{2} \frac{\pi^{s-1}}{\sin((\pi/2) s)}. $$
Now in the present case we have
$$\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} =
\sum_{k\ge 1} \frac{1}{k^2} \frac{1}{k^{-s}} = \zeta(2-s).$$
It follows that the Mellin transform $Q(s)$ of $S(x)$ is given by
$$ Q(s) = \frac{1}{2} \frac{\pi^{s-1}}{\sin((\pi/2) s)} \zeta(2-s).$$
The Mellin inversion integral for an expansion about zero is
$$\int_{1/2-i\infty}^{1/2+i\infty} Q(s)/x^s ds,$$
shifted to the left. The poles from the sine term are at the even integers, with those in the right half plane being canceled by the trivial zeros of the zeta function.
We have
$$\sum_{q\ge 0} \operatorname{Res}(Q(s)/x^s; s=-2q)
= \sum_{q\ge 0} \frac{(-1)^q}{\pi^{2q+2}} \zeta(2+2q) x^{2q}\\
= \sum_{q\ge 0} \frac{(-1)^q}{\pi^{2q+2}}
\frac{(-1)^q B_{2q+2} (2\pi)^{2q+2}}{2(2q+2)!} x^{2q}
= \sum_{q\ge 0} \frac{2^{2q+2} B_{2q+2}}{2(2q+2)!} x^{2q}.$$
We thus have for the initial sum
$$\frac{1}{x} + 2x S(x) =
\frac{1}{x} + 2x \sum_{q\ge 0} \frac{2^{2q+2} B_{2q+2}}{2(2q+2)!} x^{2q}
= \frac{1}{x} + \frac{1}{x} \sum_{q\ge 0} \frac{2^{2q+2} B_{2q+2}}{(2q+2)!} x^{2q+2}.$$
Recall that the generating function of the Bernoulli numbers is precisely
$$\frac{x}{e^x-1}.$$
Adapting the expression from the harmonic sum calculation to match this generating function (note that the odd index $B_m$ are zero when $m> 1$) we get
$$\frac{1}{x} - \frac{1}{x} + 1 + \frac{1}{x} \sum_{m\ge 0} \frac{B_m}{m!} (2x)^m.$$
Simplify one last time, getting
$$ 1 + \frac{1}{x}\frac{2x}{e^{2x}-1}
= \frac{e^{2x}+1}{e^{2x}-1}
= \frac{e^x+e^{-x}}{e^x-e^{-x}} = \coth x,$$
QED.
Just to recapitulate, the direction of this computation was as follows. We started by conjecturing that a certain harmonic sum represents the hyperbolic cotangent in a neighborhood of zero. Then we computed the Mellin transform of this sum. We inverted that transform for an expansion about zero. The expansion that we obtained was precisely the generating function of the Bernoulli numbers with two exceptions. Substituting that generating function into the sum and correcting for the exceptions we obtained the defining equation of the hyperbolic cotangent in terms of exponentials.
|
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|
Common linear and Quadratic factors (i) If $ax^5+bx^2+c$ has a factor of the form $x^2+px+1$ prove that:
$(a^2-c^2)(a^2-c^2+bc)=(ab)^{2}$
(ii)In this case prove that:
$ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor
|
Proving part (ii) independently is less tedious than part (i). Here it is:-
Let $f(x) = ax^5 + bx^2 + c$
Also $f(x) = (x^2 + px + 1)Q_3 (x)$; where $Q_3 (x)$ is a quotient polynomial of degree 3…….(1)
Let $g(x) = x^2 + px + 1$ such that g(x) = 0 has m and n as roots.
From product of roots, $n = 1/m$.
$f(n) = f(1/m) = … = (1/m)^5 . [a + b(m)^3 + c(m)^5]…....(2)$
From (1), $f(n) = f(1/m) = … = (1/m)^2 . [1 + p(m) + (m)^2]Q_3 (1/m)……(3)$
Equating (2) and (3), we have $a + bm^3 + cm^5 = (1 + pm + m^2).[m^3.Q_3 (1/m)]$
This means $a + bm^3 + cm^5$ has $(1 + pm + m^2)$ as factor.
i.e. $cx^5 + bx^3 + a$ has $(x^2 + px +1)$ as factor.
Thus, $ax^5+bx^2+c$ and $cx^5+bx^3+a$ have a common quadratic factor.
For part(i), I still need some time to workout.
|
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|
integrating $\int \frac{dt}{(t+2)^2(t+1)}$ I'm practicing to solve a whole, and I am not able to solve this one, could you help me? $$\int \frac{dt}{(t+2)^2(t+1)}$$I tried $$\frac{1}{(t+2)^2(t+1)}=\frac{A}{(t+2)^2}+\frac{B}{(t+2)}+\frac{C}{(t+1)}\\1=A(t+1)+B(t+2)(t+1)+C(t+2)^2\\t=-2\Longrightarrow 1=-A\Longrightarrow \fbox{$A=-1$}\\t=-1\Longrightarrow1=C(-1)^2\Longrightarrow\fbox{$C=1$}$$Making $t = 0$ and substituting $A$ and $C$ we have $$1=A+2B+4C=-1+2B+4=2B+3\\\fbox{$B=-1$}$$THEN$$\int \frac{dt}{(t+2)^2(t+1)}=\int -\frac{1}{(t+2)^2}-\frac{1}{(t+2)}+\frac{1}{(t+1)}\;dt\\=-\int \frac{1}{(t+2)^2}-\int\frac{1}{(t+2)}+\int\frac{1}{(t+1)}\;dt=-\int u^{-2\;}du-\ln|t+2|+\ln|t+1|\\=-\frac{u^{-1}}{-1}-\ln|t+2|+\ln|t+1|=\\\fbox{$\frac{1}{t+2}-\ln|t+2|+\ln|t+1|+c$}$$Only I could not do the derivative to "take the test", can you help me? Or contains an error in my resolution?
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$$\left(\frac1{t+2}+\log\frac{t+1}{t+2}\right)'=-\frac1{(t+2)^2}+\frac{t+2}{t+1}\frac1{(t+2)^2}=$$
$$=-\frac1{(t+2)^2}+\frac1{(t+1)(t+2)}=\frac{-t-1+t+2}{(t+1)(t+2)^2}=\frac1{(t+1)(t+2)^2}\;\color\red\checkmark$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find asymptotic of recurrence sequence Given a sequence $x_1=\frac{1}{2}$, $x_{n+1}=x_n-x_n^2$. It's easy to see that it limits to $0$.
The question is: is there exists an $\alpha$ such, that $\lim\limits_{n\to\infty}n^\alpha x_n\neq0$.
I tried to find explicit formula for $x_n$, but did not succeed. I stucked at that point: if $a_1=(\frac{1}{4})^2$, $a_n=(a_{n-1}+\frac{1}{4})^2$, then $x_n=-a_{n-2}+\frac{1}{4}$ for $n\ge3$.
Thanks for any ideas and help.
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Let $y_n = \frac{1}{x_n} - n$, we have
$$y_{n+1} = \frac{1}{x_{n+1}}-(n+1) = \frac{1}{x_n} + \frac{1}{1-x_n} - (n+1) = y_n + \frac{x_n}{1-x_n}$$
This implies for $n > 1$,
$$y_n = y_1 + \sum_{k=1}^{n-1} \frac{x_k}{1-x_k}
\quad\iff\quad
\frac{1}{x_n} = n + 1 + \sum_{k=1}^{n-1}\frac{x_k}{1-x_k}
\tag{*1}$$
Since $0 \le x_k < 1$, this immediately give us an upper bound of $x_n$:
$$\frac{1}{x_n} \ge n+1 \quad\iff\quad x_n \le \frac{1}{n+1}\tag{*2}$$
Substitute $(*2)$ in $(*1)$ and notice the map $x \mapsto \frac{x}{1-x}$ is an increasing function for $x \in [0,1)$, we obtain a lower bound for $x_n$:
$$\begin{align}
& \frac{1}{x_n} \le n+1 + \sum_{k=1}^{n-1}\frac{\frac{1}{k+1}}{1 - \frac{1}{k+1}}
= n + 1 + \sum_{k=1}^{n-1}\frac{1}{k} \le n + 1 + \log n + \gamma\\
\implies & x_n \ge \frac{1}{n + 1 + \log n + \gamma}\tag{*3}
\end{align}$$
where $\gamma$ is the
Euler–Mascheroni constant.
Combine these two bounds, it is clear we can take $\alpha = 1$ and $\displaystyle\lim_{n\to\infty} n^\alpha x_n = 1$.
|
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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.