Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$
*
*I managed to get $ab=cd$. Don't know how to proceed further.
| Let $a = c + \delta$. Then as $a+b = c+\delta + b = c+d$ we have $b= d-\delta$.
So we ahve $c^2 + d^2 = (c+\delta)^2 + (d-\delta)^2$ so
$2\delta(c-d) + 2\delta^2= 0$.
Case 1: $\delta = 0$.
The we go back to near the beginning. Then $a =c+\delta =c$ and $b=d-\delta = d$.
Case 2: $\delta \ne 0$ then
$(c-d) + \delta = 0$
$c = d-\delta$.
But $b = d-\delta$ so $c=b$.
Going way back to the very beginning:
$a + b = c + d$ and $c + d = b+d$ we get $a=d$.
So case 1: $a=c;b=d$ or case 2: $a=d$ and $b=c$. (or both: $a=b=c=d$ is always an option.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving $3z^2 + (2+3i)z + (5i-5) = 0$ In a Precalculus book, Complex numbers chapter, I am given the following exercise:
Solve the equation $3z^2 + (2+3i)z + (5i-5) = 0$, giving your answers in the form $z = a + bi$
From the way that the exercise is shown, I thought about using the quadratic formula to solve it. Starting with the discriminant, I found
$$\Delta = (2+3i)^2 - 4 \cdot 3 \cdot (5i-5) = 55-48i$$
If I could find the root of that complex number, I could finish this exercise. But how could I do that? (Wolfram gives be $8-3i$ as an answer for that root)
(by the way, this exercise comes before the root finding method for complex numbers)
| Suppose $a+bi\ne 0$. Suppose we want to solve $(x+yi)^2 = a+bi$.
Then $(x+yi)^2 =x^2 + 2xyi -y^2 = (x^2 -y^2) + 2xyi = a+bi$. As $a,b,x^2-y^2, 2xy\in \mathbb R$ we must have
$x^2 - y^2 =a$ and $2xy =b$.
So solve that.
$x = \frac b{2y}$ so $\frac{b^2}{4y^2} - y^2 = a$ so $y^4 +y^2a -\frac 14b^2=0$ so $y^2 = \frac {-1\pm\sqrt {1+ab}}{2}$ ad so on.
So lets solve $(x+yi)^2 =(x^2-y^2) + 2xy = 55 -48i$ so $x^2 - y^2 = 55$ and $2xy =-48$.
$y =-\frac {24}x$ so $x^2 - 24^2 \frac 1{x^2} = 55$ so $x^4 -55x^2 -24^2 = 0$
$x^2 = \frac {55 \pm \sqrt{5^2\cdot 11^2 + 4\cdot 2^2\cdot 12^2}}2=$
yea gods... calculator time .... $x^2 = \frac {55 \pm 73}2$. But as $x^2 > 0$ we have $x^2 = \frac {128}2 = 64$. So $x =\pm 8$.
And $y = \mp \frac {24}8 = \mp 3$.
So the discriminate is $\pm 8 \mp 3i$.
Just like wolfram said.
| {
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Various ways to calculate $\int \sin(x) \cos(x) \, \mathrm{d}x$ Consider the integral
$$\mathcal{I} := \int \sin(x) \cos(x) \, \mathrm{d} x$$
$
\newcommand{\II}{\mathcal{I}}
\newcommand{\d}{\mathrm{d}}
$
This is one of my favorite basic integrals to think about as an instructor, because on the face of it, there are a lot of different ways to solve it, many are accessible to Calculus I students, and can give some insights into the nature of integration and to some trigonometry identities. For instance:
*
*Substitution with $u = \sin(x)$ gives $$\II = \frac{\sin^2(x)}{2} + C$$
*
*Substitution with $u = \cos(x)$ gives $$ \II = - \frac{\cos^2(x)}{2} + C$$
*
*(Noted in comments by Koro) Make the substitution
$$
u = \sec(x) \implies \d u = \sec(x) \tan(x) \, \d x= \frac{\sin(x)}{\cos^2(x)} \, \d x$$
Then
$$\begin{align*}
\II &= \int \sin(x) \cos(x) \frac{\cos^2(x)}{\sin(x)} \, \d u\\
&= \int u^{-3} \, \d u\\
&= - \frac{1}{2\sec^2(x)} + C\\
&= - \frac{\cos^2(x)}{2} + C
\end{align*}$$
A similarly motivated substitution:
$$
u = \csc(x) \implies \d u = -\cot(x) \csc(x) \, \d x = - \frac{\cos(x)}{\sin^2(x)} \, \d x
$$
yields
$$\begin{align*}
\II &= -\int \sin(x) \cos(x) \frac{\sin^2(x)}{\cos(x)} \, \d u\\
&= -\int u^{-3} \, \d u\\
&= \frac{1}{2\csc^2(x)} + C\\
&= \frac{\sin^2(x)}{2} + C
\end{align*}$$
*
*Using $\sin(2x) = 2 \sin(x) \cos(x)$ readily leads to $$\II = -\frac{\cos(2x)}{4} + C$$
*
*Integration by parts differentiating $\sin(x)$ yields $$\II = \sin^2(x) - \II$$ which will yield a previous solution on solving for $\II$.
*
*Integration by parts differentiating $\cos(x)$ yields $$ \II = -\cos^2(x)- \II$$ which, similarly, yields a previous solution once we solve for $\II$.
*
*Using the Weierstrass substitution $t = \tan(x/2)$ gives
$$
\II = \int \frac{2t}{1+t^2} \frac{1-t^2}{1+t^2} \frac{2}{1+t^2} \, \mathrm{d}t = \frac{2t^2}{(1+t^2)^2} + C=\frac{2 \tan^2(x/2)}{(1 + \tan^2(x/2))^2} + C
$$
*
*We can use the complex sine and cosine:
$$
\sin(x) = \frac{e^{ix} - e^{-ix}}{2i} \qquad
\cos(x) = \frac{e^{ix} + e^{-ix}}{2}
$$
Then
$$
\II = \int \frac{e^{2ix} - e^{-2ix}}{4i} \, \mathrm{d} x = - \frac 1 8 \left( e^{2ix} + e^{-2ix} \right) + C = -\frac 1 4 \cos(2x) + C
$$
(Warning for Novices: The $C$ constant in each expression may not be the same as in others. These answers are equivalent ones for the integral, but the solutions without the $+C$ terms are not equal expressions. These hint at certain trigonometry identities, which is why I find them interesting.)
This has given us a set of solutions to $\II$ via a few basic methods, and a few less-basic but accessible methods.
My question is, what other solutions can you come up with for $\II$?
I'm particularly interested in answers which:
*
*are obviously not functionally equivalent to those already given
*give answers other than those already expressed (perhaps a hint at other identities or concepts of note?)
*use methods that you don't see in a typical calculus class, or methods that are rarely used
*use unusual but slick and effective techniques
or any combination thereof! I have no real motivation for this but my own curiosity, but I'm curious to see what you guys can think of.
| $$\begin{align*}
\int\sin{x}\cos{x}dx
&= \frac{1}{4}\int\frac{4\tan{x}\sec^2{x}}{\sec^2{x}\sec^2{x}}dx\\
&= \frac{1}{4}\int\frac{4\tan{x}\sec^2{x}}{(1+\tan^2{x})^2}dx\\
&=\frac{1}{4}\int\frac{2\tan{x}\sec^2{x}((1+\tan^2{x})-(\tan^2{x}-1))}{(\tan^2{x}+1)^2}dx\\
&=\frac{1}{4}\int\frac{2\tan{x}\sec^2{x}(1+\tan^2{x})-(\tan^2{x}-1) \cdot 2\tan{x}\sec^2{x}}{(\tan^2{x}+1)^2}dx\\
&=\frac{1}{4}\int\frac{(1+\tan^2{x})\frac{d(\tan^2x-1)}{dx}-(\tan^2{x}-1)\frac{d(\tan^2{x}+1)}{dx}}{(1+\tan^2{x})^2}dx\\
&=\frac{1}{4}\int\frac{d}{dx}\frac{(\tan^2x-1)}{(\tan^2x+1)}dx\\
&=\frac{(\tan^2x-1)}{4(\tan^2x+1)}+ C_0
\end{align*}$$
*
*If same answer with different methods are allowed then this could be possible.
$\begin{align*} I
& =\int SC \\
&= S\int C - \int( {S' .\int C})\\
& = S^2 - \int CS = S^2-I
\end{align*}$
$$\begin{align*}
2I
& =S^2\\
&\implies I = \frac{\sin^2x}{2} +c_0\\
\end{align*}$$
Of course! When you will take $CS$ type you will get $$-\frac{\cos^2{x}}{2}+c_1$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "30",
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Are there infinitely many primes $q$, such that $\sigma(p^k) = 2q$, where $p$ is prime and $p \equiv k \equiv 1 \pmod 4$? Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of $x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Here is my question:
Are there infinitely many primes $q$, such that $\sigma(p^k) = 2q$, where $p$ is prime and $p \equiv k \equiv 1 \pmod 4$?
MY ATTEMPT
Since $p$ is prime, and $k \geq 1$ is an integer (satisfying $k \equiv 1 \pmod 4$), then we get
$$q=\frac{\sigma(p^k)}{2}=\frac{p^{k+1}-1}{2(p-1)}=\Bigg(\frac{p^{(k+1)/2} + 1}{2}\Bigg)\cdot\Bigg(\frac{p^{(k+1)/2} - 1}{p - 1}\Bigg)=\Bigg(\frac{p^{(k+1)/2} + 1}{2}\Bigg)\cdot\sigma(p^{(k-1)/2}),$$
whereupon we obtain, since $q$ is prime, that either
$$\text{Case (1): } \frac{p^{(k+1)/2} + 1}{2} = 1$$
XOR
$$\text{Case (2): } \sigma(p^{(k-1)/2}) = 1$$
holds.
We then get, from the first condition, that
$$\sigma(p^{(k-1)/2}) = q$$
which, together with $\sigma(p^k)=2q$, implies that
$$\sigma(p^k) = 2\sigma(p^{(k-1)/2})$$
Dividing both sides by $p^k$, and noting that $p^k = (p^{(k-1)/2})(p^{(k+1)/2})$, we obtain
$$I(p^k) = \Bigg(\frac{2}{p^{(k+1)/2}}\Bigg)I(p^{(k-1)/2})$$
But since $p \equiv k \equiv 1 \pmod 4$, then
$$1 < I(p^k) < \frac{5}{4}$$
and
$$1 \leq I(p^{(k-1)/2}) < \frac{5}{4}.$$
This implies that
$$\frac{4}{5} < \frac{2}{p^{(k+1)/2}} = \frac{I(p^k)}{I(p^{(k-1)/2})} < \frac{5}{4}$$
which means that
$$\frac{2}{5} < \frac{1}{p^{(k+1)/2}} < \frac{5}{8},$$
further giving
$$\frac{8}{5} < p^{(k+1)/2} < \frac{5}{2},$$
which implies that
$$\frac{64}{25} < p^{k+1} < \frac{25}{4}.$$
We get
$$\frac{39}{25} < p^{k+1} - 1 < \frac{21}{4}.$$
But since $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$, then
$$p - 1 \geq 4.$$
In particular, we obtain the upper bound
$$q = \frac{\sigma(p^k)}{2} = \frac{p^{k+1} - 1}{2(p - 1)} < \frac{21}{32},$$
which contradicts the requirement that $q$ must be prime.
We then get, from the second condition, that
$$k = 1,$$
and therefore that
$$\frac{p+1}{2}=q$$
where both $p$ and $q$ are primes.
Alas, this is where I get stuck!
| It seems that it is not known if there are infinitely many primes $q$ such that $2q−1$ is also prime according to OEIS/A005382 and Solve $2p=q+1$ where $p$ and $q$ are prime.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the probability of $P(Y=6)$ Anna writes down a random sequence created by the following process. She repeatedly rolls a fair
6-sided die. If the number she rolls is larger than all of the numbers she has previously rolled (if
any), then she writes the new number down and then continues rolling. Otherwise, she does not
write the new number down and the process ends.
Let X be the length of Anna’s sequence, and Y be the last number in her sequence.
For example, if Anna rolled 1 then 4 then 5 then 4, her sequence would be 1,4,5 and the random
variables X, Y would take values X = 3 and Y = 5.
Question:
what is $P(Y=6)$
My method
I know I can write down all the possible ways that the last number of the sequence could be 6, and then I add up all the probabilities.
*
*$1,2,3,4,5,6 = (\frac{1}{6})(\frac{1}{5})(\frac{1}{4})(\frac{1}{3})(\frac{1}{2})(1)$
*$2,3,4,5,6 = (\frac{1}{6})(\frac{1}{5})(\frac{1}{4})(\frac{1}{3})(\frac{1}{2})$
*$3,4,5,6 = (\frac{1}{6})(\frac{1}{5})(\frac{1}{4})(\frac{1}{3})$
*$1,2,4,6 = (\frac{1}{6})(\frac{1}{5})(\frac{1}{4})(\frac{1}{3})$
etc.
But this method is way too time-consuming as there are too many possible sequences.
I also tried to do it a different way by using $1 - $(The possible sequences that the last number is not 6), but there are even more sequences to take into account this way.
Can anyone show me a much more efficient way of finding the solution to this problem?
| $Y = 6$ means we have maximum of six rolls. If none of the six rolls is $6$, we would have at least one of the numbers between $1$-$5$ appear twice breaking the sequence and the process would stop.
Probability of getting $6$ in the first roll is $\frac{1}{6}$
Probability of getting $6$ in the second roll is $ \frac{5}{6} \cdot \frac{1}{6}$.
Now what is the probability of getting $6$ in the third roll? The first two rolls have to one of the remaining $5$ numbers and in the increasing order, which is simply ${5 \choose 2}$ out of $6^2$ possibilities as the order is pre-determined. The third roll has to be $6$.
Similarly for getting $6$ in the fourth, fifth or the sixth roll. So,
$ \displaystyle \small P(Y = 6) = \frac{1}{6} \cdot \left[1 + {5 \choose 1} / 6 + {5 \choose 2} / 6^2 + {5 \choose 3} / 6^3 + {5 \choose 4} / 6^4 + {5 \choose 5} / 6^5\right]$
| {
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"timestamp": "2023-03-29T00:00:00",
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What are the sine and cosine of dyadic angles? What are the values of sine and cosine of dyadic angles? We know
$$\cos\pi = -1 \qquad \cos\frac{\pi}{2} = 0 \qquad \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}\,,$$
and we can calculate sine by appealing to symmetry. But I don't think I've seen the sine and cosine values, presented in terms of radicals, of $\pi/8$, $3\pi/8$, $\pi/16$, $\pi/2^n$, etc. Do these have nice algebraic presentations? How can we calculate these?
| This answer has some approaches that were not the easiest ways to solve the problem. I'm recording them here in a separate answer so as not to lose the information (even if what this tells us is not to do it this way.)
For a complete table of sines and cosines of $\pi/16,$ we only need to compute values for $0,$ $\pi/16,$ $\pi/8,$ $3\pi/16,$ and $\pi/4.$
We already have cosines of all of these (in another answer) except $\cos(3\pi/16).$
Using the angle-sum formula for cosine,
\begin{align}
\cos \frac{3\pi}{16}
&= \cos\frac\pi8 \cos\frac\pi{16} - \sin\frac\pi8 \sin\frac\pi{16} \\
&= \frac14\sqrt{2+\sqrt2} \sqrt{2+\sqrt{2+\sqrt2}}
- \frac14\sqrt{2-\sqrt2} \sqrt{2-\sqrt{2+\sqrt2}} \\
\end{align}
It should be possible to simplify this further (see other answer) but the path is not obvious.
As an alternative to partitioning $m\pi/2^n$ into powers of two, there are the (somewhat) well-known formulas for the sine and cosine of arbitrary multiples of angles:
\begin{align}
\sin(k\theta) &= \sum_{j=0}^n {n \choose j} \cos^j\theta \, \sin^{n-j}\theta \, \sin\frac{(k-j)\pi}{2}, \\
\cos(k\theta) &= \sum_{j=0}^n {n \choose j} \cos^j\theta \, \sin^{n-j}\theta \, \cos\frac{(k-j)\pi}{2}.
\end{align}
There is a relatively simple proof using complex numbers. That may not be a good proof to expect trig students to be able to follow, but it might give some incentive to study complex numbers later.
\begin{align}
\cos \frac{3\pi}{16}
&= \cos^3 \frac{\pi}{16} - 3 \cos \frac{\pi}{16} \sin^2 \frac{\pi}{16} \\
&= \cos \frac{\pi}{16} \left(\cos^2 \frac{\pi}{16} - 3 \sin^2 \frac{\pi}{16} \right) \\
&= \cos \frac{\pi}{16} \left(\frac14\left(2+\sqrt{2+\sqrt{2}}\right)
- \frac34 \left(2-\sqrt{2+\sqrt{2}}\right) \right) \\
&= \cos \frac{\pi}{16} \left(-1 + \sqrt{2+\sqrt{2}}\right)\\
&= \frac12 \left(-1 + \sqrt{2+\sqrt{2}}\right) \sqrt{2+\sqrt{2+\sqrt{2}}}
\end{align}
Again it should be possible to simplify this further but the path is not obvious.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{1}{a_3+a_4}+\frac{1}{a_1+a_2}\ge \frac{4}{a_1+a_2+a_3+a_4} $ Prove that $$\frac{1}{a_3+a_4}+\frac{1}{a_1+a_2}\ge \frac{4}{a_1+a_2+a_3+a_4} $$
I am supposed to use AM-HM but I don't how I should. By AM-HM
We have $$\frac{1}{a_3+a_4}+\frac{1}{a_1+a_2}\ge \frac{2}{a_1+a_2+a_3+a_4} $$
Did I do something wrong?
| Hint : Let $a_1+a_2=x$ and $a_3+a_4=y$. It remains to prove,
$$\frac{1}{x}+\frac{1}{y}\ge \frac{4}{x+y}\;\; \Leftrightarrow\;\; (x+y)\left(\frac{1}{x}+\frac{1}{y}\right)\ge 4$$
The A.M-H.M inequality for two terms,
$$\frac{x+y}{2}\ge \dfrac{1}{\dfrac{\frac{1}{x}+\frac{1}{y}}{2}}$$
| {
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Can trigonometric substitution be used to solve this integral? The integral in question is $$\int \frac{x+1}{9x^2+6x+5}dx.$$ I first completed the square in the denominator giving $(3x+1)^2+4$ and proceeded to perform a $u$-substitution with $u = 3x+1$, $du=3~dx$, and $x=(u-1)/3$. After simplifying, I was left with $$\frac{1}{9}\int\frac{u+2}{u^2+4}du.$$ It is at this point I used trigonometric substitution with $u = 2\tan\theta$ and $du = 2\sec^2\theta~d\theta$ (I'm aware the integral can be written as $\frac 1 9\int\frac{u}{u^2+4}du+\frac 1 9\int\frac{2}{u^2+4}du$ and solved this way). After performing the trig substitution, I was left with $$\frac 1 9\int(\tan\theta + 1)~d\theta = \frac 1 9 \ln|\sec\theta|+\frac 1 9 \theta + C.$$ Rewriting everything in terms of $x$ gave me $$\frac 1 9\ln\left(\frac{(3x+1)^2+4}{2}\right)+\frac{1}{9}\tan^{-1}\left(\frac 1 2(3x+1)\right)+ C,$$ which is incorrect. The correct answer is $$\frac{1}{18}\ln\left(9x^2+6x+5\right)+\frac{1}{9}\tan^{-1}\left(\frac 1 2(3x+1)\right)+ C.$$ What went wrong with my trig substitution?
| Since $u=2\tan\theta$, you have$$u^2=4\tan^2\theta=4(\sec^2\theta-1).$$So,$$\sec\theta=\sqrt{\frac{u^2+4}4}$$and therefore$$\frac19\ln(\sec\theta)=\frac1{18}\ln\left(\frac{u^2+4}4\right).$$
| {
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Finding the maximum value of the following product.
Let $x,y,z$ be three reals which satisfies $x^2+y^2+z^2=1$. Find the maximum value of $$P = (x^2-yz)(y^2-zx)(z^2-xy).$$
I think expanding would be a bad idea. I tried to apply the Weirstrass product inequality but it may give the minimum not maximum value.
| We claim that $P \leq \dfrac{1}{8}.$
Case $1:$ All three terms are greater than or equal to zero.
By AM-GM,
\begin{align}
\sqrt[3]{P} & \leq \dfrac{x^2+y^2+z^2-xy-yz-xz}{3} \\
&= \dfrac{1-(xy+yz+xz)}{3}.
\end{align}
It thus suffices to prove $\dfrac{1-(xy+yz+xz)}{3} \leq \dfrac{1}{2} \iff xy+yz+xz \geq \dfrac{-1}{2}.$
But the last inequality is easy, since
\begin{align}
(x+y+z)^2 \geq 0 & \Rightarrow 1+2(xy+yz+xz) \geq 0 \\
& \Rightarrow xy+yz+xz \geq \dfrac{-1}{2}.
\end{align}
In particular, the maximum value of $P$ is attainable when we have, say, $x=0, y=\dfrac{\sqrt{2}}{2}, z=\dfrac{-\sqrt{2}}{2}.$
Case $2$ : At least one term is smaller than zero.
W.L.O.G. let $z^2-xy <0.$ Now, if all three terms are smaller than zero, then $P < 0 < \dfrac{1}{8}.$ And similarly if only $z^2-xy <0.$ Thus, there must be exactly two terms smaller than zero, and let $y^2-zx <0$ too. Hence, by AM-GM, this gives us
$$\sqrt[3]{P} \leq \dfrac{x^2-yz+zx-y^2+xy-z^2}{3},$$
so it suffices to prove
\begin{align}
\dfrac{x^2-yz+zx-y^2+xy-z^2}{3} \leq \dfrac{1}{2} &\iff x^2-(y^2+z^2)-yz+xz+xy \leq \dfrac{3}{2} \\
& \iff x^2-(y+z)^2+xz+xy+yz \leq \dfrac{3}{2} \\
& \Leftarrow (x+y)(x+z) \leq \dfrac{3}{2}
.
\end{align}
Again, by AM-GM, $(x+y)(x+z) \leq \dfrac{(x+y)^2+(x+z)^2}{2}.$ Hence we are done if we could prove
\begin{align}
(x+y)^2+(x+z)^2 \leq 3(x^2+y^2+z^2) & \iff 2xy+2xz \leq x^2+2y^2+2z^2 \\
& \iff x^2+y^2+z^2-2yz-2xz+y^2+z^2 \geq 0 \\
& \iff (x-y-z)^2 -2yz +y^2+z^2 \geq 0 \\
& \iff (x-y-z)^2+(y-z)^2 \geq 0,
\end{align}
Which is obvious.
| {
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Evaluate the partial sum $\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$ Evaluate the partial sum $$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$$
What I have tried:
Calculate the partial fractions which are (for sake of brevity) :
$$\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{n(n+2)}$$
So we get:
$$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)} = \sum_{n=1}^k \left(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{n(n+2)}\right)$$
Then calculating a few numbers for $n$ we get:
$$\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{6} \right) + \left(\frac{1}{4} - \frac{1}{3} + \frac{1}{8} \right) + \left(\frac{1}{6} - \frac{1}{4} + \frac{1}{10}\right) . . . \left(\frac{1}{2n} - \frac{1}{n+1} + \frac{1}{n+2}\right)$$
The first two fractions cancel out in the first bracket and we're left with $\frac{1}{6}$, as for the second bracket the first fraction is cancelled out by the second fraction in the third bracket.
I have noticed that the first fractrion so $\frac{1}{2n}$ cancel out by every even term in the denominator for $-\frac{1}{n+1}$ so the equation becomes:
$$\left(-\frac{1}{2n+1}+\frac{1}{n+2}\right) = \left(\frac{n-1}{(2n+1)(n+2)} \right)$$
Have I approached this correctly? I would greatly appreciate some assistance on any improvements!
| Calling $S_n = \sum_k^n \frac{1}{k}$ we have
$$
T_n = \frac 12 S_n - \left(S_n-1+\frac{1}{n+1}\right)+\frac 12\left(S_n-1-\frac 12+\frac{1}{n+1}+\frac{1}{n+2}\right)= \frac 14+\frac 12\left(\frac{1}{n+2}-\frac{1}{n+1}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4272062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For all $a, b$ in $\mathbb{Z}$, show $a +b$ is a factor of $a^{2n} - b^{2n}$. Given solution approach for first problem below, unable to approach the second one.
#1. For all $a, b$ in $\mathbb{Z}$, show $a - b$ is a factor of $a^n - b^n$.
By weak form of induction.
Base step: $n=1$.
Hypothesis step: Let, $n=k$; then:
$(a-b) \mid (a^k - b^k)$.
Inductive step: $n=k+1$,
$a^{k+1}-b^{k+1}=a^k(a-b)+(a^k-b^k)b$
Have two terms both divisible by $a-b$.
#2. For all $a, b$ in $\mathbb{Z}$
, show $a + b$ is a factor of $a^{2n} - b^{2n}$.
By weak form of induction.
Base step: $k=1$.
$(a+b) \mid (a^2 - b^2) = (a+b)\mid (a-b)(a+b)$
Hypothesis step: Let, $k=2n, n \in\mathbb{Z}$; then:
$(a+b) \mid (a^{2n} - b^{2n})$.
Inductive step: $k+1=2(n+1)$,
$a^{k+1}-b^{k+1}=a^{2(n+1)} - b^{2(n+1)} = a^{2n}.a^2 - b^{2n}.b^2$
| Not an answer, but too long for a comment:
You can get Heropup's result by applying the induction step you did in 1) twice. Note that $$\begin{aligned}
a^{2(n + 1)} - b^{2(n + 1)} &= a^{2n + 1}(a - b) + (a^{2n + 1} - b^{2n + 1})b \\
&= a^{2n + 1}(a - b) + (a^{2n}(a - b) + (a^{2n} - b^{2n})b)b
\end{aligned}
$$ Which after simplifying gives
$$
a^{2(n + 1)} - b^{2(n + 1)} = a^{2n + 1}(a - b) + a^{2n}(a - b)b + (a^{2n} - b^{2n})b^2
$$
The right term $(a^{2n} - b^{2n})b^2$ is divisible by $a + b$ by the induction hypothesis on $a^{2n} - b^{2n}$. For the remaining terms, note that $$
a^{2n + 1}(a - b) + a^{2n}(a-b)b = (a - b)(a^{2n + 1} + a^{2n}b) = (a + b)(a - b)a^{2n},
$$ hence they are also divisible by $a + b$, thus $a^{2(n + 1)} - b^{2(n + 1)}$ is divisible by $a + b$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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In triangle. Prove that: $2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9+\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$ Problem: Given a,b,c are length of triangle. Prove that: $$2(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9+\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$$
Happy Vietnamese Women's Day
Phan Ngoc Chau, Oct 20th 2021
My approach: Note that: $(b-c)^2-a^2=(b-c-a)(b-c+a)<0$ so $\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}<\sqrt{17}$
And as well- known result: $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9$
But it is not true to show: $3\sqrt{17}\le9$. It is quite ugly for me to get a proof. Anyone help me ? Thanks!
| I found a stronger one:
Problem: Given a,b,c are length of triangle. Prove that: $$2\sum_{cyc}{\frac{a+b-c}{c}}\sum_{cyc}{\frac{a+b-c}{2(a+b)-c}}+3\ge\sum_{cyc}{\sqrt{17+\frac{4(b+c)((b-c)^2-a^2)}{abc}}}$$
Note that: $\frac{c}{2(a+b)-c}+\frac{a}{2(b+c)-a}+\frac{b}{2(c+a)-b}\ge1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that: $2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge\sqrt{5ab+4ac}+\sqrt{5bc+4ba}+\sqrt{5ca+4cb}$ Problem: For $a,b,c\ge0: ab+bc+ca>0.$ Prove that: $$2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge\sqrt{5ab+4ac}+\sqrt{5bc+4ba}+\sqrt{5ca+4cb}$$
Recently, i have seen a post on AoPS link
My approach: After squaring both side, i get: $$2(a^2+b^2+c^2)+\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+2(a+b+c)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\ge\sum{\sqrt{(5ab+4ac)(5bc+4ba)}}$$
Since: $a+b+c\ge\sqrt{ab}+\sqrt{bc}+\sqrt{ca}$. It is desired to prove: $$2(a^2+b^2+c^2)+5\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})+2(ab+bc+ca)\ge\sum{\sqrt{(5ab+4ac)(5bc+4ba)}}$$
But it is more complicated. I hope we can find a good approach for problem. Thanks!
| Similarly: If $a,b,c\ge0$ then: $$\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge \sqrt{4bc+ca+ab}+\sqrt{4ca+ab+bc}+\sqrt{4ab+bc+ca}$$
$$2(a+b+c)+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge \sqrt{bc+4ca+4ab}+\sqrt{ca+4ab+4bc}+\sqrt{ab+4bc+4ca}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to get roots in the form of a quadratic from the quartic $x^4 - 4x^2 + 16$? So I need to factor this function into quadratics:
$$x^4 - 4x^2 + 16$$
I know that there are only complex solutions to this question, however, it is still possible to obtain quadratic factors without requiring the imaginary unit to be present.
I tried by simplifying it into a quadratic by replacing $x^2 =a$.
But I just end up getting (after completing the square) $$(x^2-2)^2+12$$
which is not in factored format. So could someone help me?
| We see that, the roots of the equation are not real numbers. So, if we want to factor polynomials whose coefficients are real numbers, and if that's possible, then we have
$$\begin{align}x^4-4x^2+16=(x^2+ax+b)(x^2-ax+b) \end{align} \tag 1$$
This factorization uses the following fact:
If $x_1,x_2$ are the root of the biquadratic, then $-x_1,-x_2$ are also roots of the biquadratic. So, you have
$$\begin{align}x^4-4x^2+16&=(x^2+ax+b)(x^2-ax+b)\\
&=(x^2+b)^2-a^2x^2\\
&=x^4+x^2(2b-a^2)+b^2 \end{align}$$
This implies,
$$\begin{align}\begin{cases}2b-a^2=-4\\b^2=16\end{cases}&\implies a^2=12,\thinspace b=4\\
&\implies a=±2\sqrt 3,\thinspace b=4.\end{align}$$
Explanation: $~(1)$
Since the roots of the polynomial are not real numbers, it is not possible to factor the coefficients with real numbers such that
$$x^4-4x^2+16=(x-a)(x^3+bx^2+cx+d)$$
That is, the polynomial $x^4-4x^2+16$ will be factored with real coefficients as follows:
$$x^4-4x^2+16=(x^2+ax+b)(x^2+mx+n)$$
Then, let $x_1, \thinspace x_2$ are the root of the polynomial $x^2+ax+b$. So, the roots of the second multiplier polynomial must be $-x_1,\thinspace -x_2$. Based on the Vieta's formulas we immediately obtain the coefficients of the second multiplier polynomial as follows:
$$\begin{align}m&=-\left(-x_1+(-x_2)\right)\\
&=x_1+x_2\\
&=-a\end{align}$$
and
$$\begin{align}n&=-x_1\times (-x_2)\\
&=x_1x_2\\
&=b\end{align}$$
Thus, our polynomial will be factored as
$$\begin{align}x^4-4x^2+16=(x^2+ax+b)(x^2-ax+b).\end{align}$$
| {
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How to solve this infinite system of equations? I'm trying to solve an infinite set of coupled equations. j is a real integer index where $-\infty<j<\infty$. {a,b,c,d,f} are variables, and K's are constants. I need to find closed form solutions for $\{a_{j},b_{j},c_{j}\}$ in terms of only the K's. I believe the respective solutions for $\{a_{j},b_{j},c_{j}\}$ should have the same form for all j. I don't need explicit forms for the d's or f's.
$$a_{j}+b_{j}+c_{j}=K_{j}$$
$$3a_{j}+2b_{j}+c_{j}=c_{j+1}$$
$$6a_{j}+3b_{j}+d_{j}+3f_{j}=0$$
$$3a_{j}+2b_{j}+d_{j}+2f_{j}=0$$
$$d_{j}+f_{j}-f_{j-1}=0$$
Can this be done?
| Given
$$
\left\{ \begin{array}{l}
a_j + b_j + c_j = K_j \\
3a_j + 2b_j + c_j = c_{j + 1} \\
6a_j + 3b_j + d_j + 3f_j = 0 \\
3a_j + 2b_j + d_j + 2f_j = 0 \\
d_j + f_j - f_{j - 1} = 0 \\
\end{array} \right.
$$
we can use 3rd and 4th rows to enucleate $d$ anf $f$ as a combination of $a$ and $b$
$$
\left\{ \begin{array}{l}
f_j = - 3a_j - b_j \\
d_j = 3a_j \\
\end{array} \right.
$$
Thereafter we can enucleate $a$ as
$$
\begin{array}{l}
\left\{ \begin{array}{l}
a_j + b_j + c_j = K_j \\
3a_j + 2b_j + c_j = c_{j + 1} \\
3a_{j - 1} - b_j + b_{j - 1} = 0 \\
\end{array} \right.\;\; \Rightarrow \;\left\{ \begin{array}{l}
a_j + b_j + c_j = K_j \\
3a_j + 2b_j + c_j = c_{j + 1} \\
3a_j + b_j = b_{j + 1} \\
\end{array} \right.\;\; \Rightarrow \\
\Rightarrow \;\left\{ \begin{array}{l}
a_j = K_j - b_j - c_j \\
\left\{ \begin{array}{l}
3K_j - b_j - 2c_j = c_{j + 1} \\
3K_j - 2b_j - 3c_j = b_{j + 1} \\
\end{array} \right. \\
\end{array} \right. \\
\end{array}
$$
where in the last row we have increased the index: we shall take that in due account, after solving the system, by accomodating the initial conditions.
So we are left with
$$
\left\{ \begin{array}{l}
3K_j - b_j - 2c_j = c_{j + 1} \\
3K_j - 2b_j - 3c_j = b_{j + 1} \\
\end{array} \right.\;\; \Rightarrow \left( {\begin{array}{*{20}c}
{b_{j + 1} } \\ {c_{j + 1} } \\
\end{array}} \right) = - \left( {\begin{array}{*{20}c}
2 & 3 \\ 1 & 2 \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
{b_j } \\ {c_j } \\
\end{array}} \right) + 3K_j \left( {\begin{array}{*{20}c}
1 \\ 1 \\
\end{array}} \right)
$$
The matrix diagonalizes as
$$
\left( {\begin{array}{*{20}c}
2 & 3 \\
1 & 2 \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
1 & {\sqrt 3 } \\
1 & { - \sqrt 3 } \\
\end{array}} \right)^{ - 1} \left( {\begin{array}{*{20}c}
{2 + \sqrt 3 } & 0 \\
0 & {2 - \sqrt 3 } \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
1 & {\sqrt 3 } \\
1 & { - \sqrt 3 } \\
\end{array}} \right)
$$
and thus finally we get the decoupled difference equations
$$
\left( {\begin{array}{*{20}c}
{u_{j + 1} } \\ {v_{j + 1} } \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
1 & {\sqrt 3 } \\ 1 & { - \sqrt 3 } \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
{b_{j + 1} } \\ {c_{j + 1} } \\
\end{array}} \right) = - \left( {\begin{array}{*{20}c}
{2 + \sqrt 3 } & 0 \\ 0 & {2 - \sqrt 3 } \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
{u_j } \\ {v_j } \\
\end{array}} \right) + 3K_j \left( {\begin{array}{*{20}c}
1 \\ 1 \\
\end{array}} \right)
$$
which can be easily solved .
| {
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"question_score": "2",
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Finding $\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx $ Is there a way to show $$\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx = 2C$$ where $C=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} $ is Catalan’s constant, preferably without using complex analysis?
The following is an attempt to expand it as as a series:
\begin{align*}
\int_0^{\frac{\pi}{2}} \frac{x}{\sin x} dx
&= \int_0^{\frac{\pi}{2}} \frac{x}{1-\cos^2 x}\sin x\ dx \\
&= \sum_{n=0}^{\infty} \int_0^{\frac{\pi}{2}} x\sin x \ \cos^{2n}x \ dx \\
&= \sum_{n=0}^{\infty}\frac{1}{2n+1} \int_0^{\frac{\pi}{2}} \cos^{2n+1}x \ dx \\
&= \sum_{n=0}^{\infty} \frac{4^n}{\binom{2n}{n}(2n+1)^2}
\end{align*}
which is close but not quite there.
| 1st Solution. Applying the Weierstrass substitution $t=\tan(x/2)$, it follows that
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} \, \mathrm{d}x
&= \int_{0}^{1} \frac{2\arctan(t)}{2t/(1+t^2)} \, \frac{2\mathrm{d}t}{1+t^2}
= 2 \int_{0}^{1} \frac{\arctan t}{t} \, \mathrm{d}t.
\end{align*}
Now note that we have
$$ \frac{\arctan(t)}{t} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} t^{2n} $$
for $-1 \leq t \leq 1$. So by the Abel's theorem, its integral from $0$ to $1$ can be computed by performing term-wise integration. (In other words, we can interchange the order of infinite summation and integral.) This yields
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} \, \mathrm{d}x
&= 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_{0}^{1} t^{2n} \, \mathrm{d}t
= 2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}
= 2C.
\end{align*}
2nd Solution. Using $\sin x = (e^{ix} - e^{-ix})/2i$, the integral is recast as
$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} \, \mathrm{d}x
= \frac{2}{i} \int_{0}^{\frac{\pi}{2}} \frac{xe^{ix}}{1 - e^{2ix}} \, \mathrm{d}x. $$
To expand the integrand, se adopt the following regularization:
$$ \int_{0}^{\frac{\pi}{2}} \frac{xe^{ix}}{1 - e^{2ix}} \, \mathrm{d}x
= \lim_{r \to 1^-} \int_{0}^{\frac{\pi}{2}} \frac{xe^{ix}}{1 - re^{2ix}} \, \mathrm{d}x $$
(This limit is easily verified by the dominated convergence theorem.) Now taking advantage of the fact taht $|re^{2ix}| < 1$, we can invoke the geometric series formula to expand
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{xe^{ix}}{1 - re^{2ix}} \, \mathrm{d}x
&= \int_{0}^{\frac{\pi}{2}} xe^{ix} \sum_{n=0}^{\infty} \bigl( re^{2ix} \bigr)^n \, \mathrm{d}x \\
&= \sum_{n=0}^{\infty} r^n \int_{0}^{\frac{\pi}{2}} xe^{(2n+1)ix} \, \mathrm{d}x \\
&= \sum_{n=0}^{\infty} r^n \left( \frac{\pi}{2} \cdot \frac{(-1)^n}{2n+1} - \frac{1}{(2n+1)^2} + \frac{(-1)^n}{(2n+1)^2}i \right)
\end{align*}
Letting $ r \to 1^-$, by the Abel's theorem, the limit can be computed term-wise to yield
\begin{align*}
\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} \, \mathrm{d}x
&= \frac{2}{i} \sum_{n=0}^{\infty} \left( \frac{\pi}{2} \cdot \frac{(-1)^n}{2n+1} - \frac{1}{(2n+1)^2} + \frac{(-1)^n}{(2n+1)^2}i \right) \\
&= \frac{2}{i} \biggl( \frac{\pi^2}{8} - \sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} \biggr) + 2C
\end{align*}
However, since the integral is real-valued, the imaginary part must vanish. Therefore we obtain the desired identity as well as a lucky by-product:
$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} \, \mathrm{d}x = 2C
\qquad\text{and}\qquad
\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} = \frac{\pi^2}{8}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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Why does $|\sin(\sqrt{4\pi^2n^2+x})| =|\sin(\sqrt{4\pi^2n^2+x})-\sin(2\pi n)|$ I need in a demonstration that $$|\sin(\sqrt{4\pi^2n^2+x})|
=|\sin(\sqrt{4\pi^2n^2+x})-\sin(2\pi n)|$$
but I don't understand why this equality is true. Can someone explain it to me?
| From the angle-sum formulas $$\sin (A\pm B)=\sin A \cos B \pm \cos A \sin B$$ $$ \cos (A\pm B)=\cos A \cos B\mp\sin A\sin B$$ we obtain $$\sin C-\sin D=2\sin A\cos B$$ where $A=(C-D)/2$ and $B=(C+D)/2.$
If $y>0$ and $4\pi^2y^2+x>0,$ let $C=\sqrt {4\pi^2y^2+x}$ and $D=2\pi y.$ We have $$|A|=\frac {1}{2}|C-D|=\frac {1}{2}\frac {|C^2-D^2|}{C+D}=$$ $$=\frac {|x|}{2(C+D)}\le\frac {|x|}{2D}=\frac {|x|}{4\pi y}.$$ Therefore if $y>0$ and $4\pi^2y^2+x>0$ then $$|\sin \sqrt {(4\pi^2y^2+x)} -\sin 2\pi y\,|=|\sin C-\sin D|=$$ $$=|2\sin A \cos B|\le|2\sin A|\le 2|A|\le \frac {|x|}{2\pi y}$$ which $\to 0$ as $y\to \infty.$
| {
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How to determine regions of positive and negative concavity of $h(g)= \frac{(g-a)(g-b)}{(g+a)(g+b)}$? We may assume that $a<0,b<0, |b|>|a|$
How to check where the concavity of the function $h(g)= \frac{(g-a)(g-b)}{(g+a)(g+b)}$ is positive or negative?
$h''(g)=\frac{4 \, {\left(a^{3} b + 2 \, a^{2} b^{2} + a b^{3} - {\left(a + b\right)} g^{3} + 3 \, {\left(a^{2} b + a b^{2}\right)} g\right)}}{a^{3} b^{3} + 3 \, {\left(a + b\right)} g^{5} + g^{6} + 3 \, {\left(a^{2} + 3 \, a b + b^{2}\right)} g^{4} + {\left(a^{3} + 9 \, a^{2} b + 9 \, a b^{2} + b^{3}\right)} g^{3} + 3 \, {\left(a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right)} g^{2} + 3 \, {\left(a^{3} b^{2} + a^{2} b^{3}\right)} g}$
The second derivative seems unhelpful.
Asymptote at $g=-a$, and $g=-b$
| I don't think there is anything simpler (at least not to come up with) than just checking the second derivative. Here is the second derivative fully factored (over $\mathbb{Q}$):
$$\frac{4 (a+b) \left(a^2 b+a b^2+3 a b g-g^3\right)}{(a+g)^3(b+g)^3}$$
The main issue is that $a^2 b+a b^2+3 a b g-g^3$ is irreducible (except possibly for special values of $a, b$). Thus solving $h''(g) \ge 0$ becomes complicated. But we know that $h$ is convex iff $h'' \ge 0$ so those roots are going to show up no matter what method we try to use.
The discriminant of that cubic is $-27 a^2 (a - b)^2 b^2$. So there is a single real root (multiplicity $1$) so long as $a, b, a - b \neq 0$.
| {
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Can the vertex of an elliptic paraboloid be recovered from its quadric matrix form? A circular paraboloid with unit axis $\mathbf{a}$, vertex $\mathbf{c}\in\mathbb{R}^3$, and quadratic coefficient $b\in\mathbb{R}$ is defined implicitly as the set of points $\mathbf{r}\in\mathbb{R}^3$ such that the following equality holds:
$$
\mathbf{a}^T(\mathbf{r} - \mathbf{c})\big(b^2 + \mathbf{a}^T(\mathbf{r} - \mathbf{c})\big) + ||\mathbf{r} - \mathbf{c}||^2 = 0
$$
I can identically write this as a quadric form in homogeneous coordinate $\mathbf{r}_h = \begin{bmatrix} \mathbf{r} \\ 1\end{bmatrix}$ as
$$
\mathbf{r}_h^T\mathbf{Q}\mathbf{r}_h = 0
$$
where $\mathbf{Q}$ is symmetric indefinite. $\mathbf{Q}$ can be composed of the original parameters $\mathbf{a}$, $\mathbf{c}$, and $b$ as
$$
\mathbf{Q} = \begin{bmatrix} \mathbf{Q}_r & \mathbf{q}_d \\ \mathbf{q}_d^T & q_0 \end{bmatrix}
$$
where $\mathbf{Q}_r = \mathbf{a}\mathbf{a}^T - \mathbf{I}$, $\mathbf{q}_d=\frac{b^2}{2}\mathbf{a} - \mathbf{Q}_r\mathbf{c}$, and $q_0 = -b^2\mathbf{a}^T\mathbf{c} + \mathbf{c}^T\mathbf{Q}_r\mathbf{c}$.
Now, let's say I'm given the matrix $\mathbf{Q}$. I want a procedure for recovering the original parameters $\mathbf{a}$, $\mathbf{c}$, and $b$ from $\mathbf{Q}$. Here are my steps:
*
*Let $\lambda_{max}$ be the maximum eigenvalue of $\mathbf{a}\mathbf{a}^T = \mathbf{Q}_r + \mathbf{I}$, and $\mathbf{v}_{max}$ be the corresponding eigenvector. I know that $\mathbf{v}_{max} = \mathbf{a}$.
*I know $\mathbf{a}\in\mathcal{N}(\mathbf{Q}_r)$ and $\mathbf{a}^T\mathbf{a}=1$ (and I recovered $\mathbf{a}$ in the last step), so it follows that $\mathbf{q}_d^T\mathbf{a} = \frac{b}{2} \implies b = 2\mathbf{q}_d^T\mathbf{a}$.
*Now I am stumped. I cannot find any tricks that will give me $\mathbf{c}$ given $\mathbf{Q}$, $\mathbf{a}$, and $b$.
I am wondering if it is possible to recover $\mathbf{c}$ at all from the matrix $\mathbf{Q}$.
I have tried using the normal vector to the surface: $\frac{\partial}{\partial \mathbf{r}_h} \mathbf{r}_h^T\mathbf{Q}\mathbf{r}_h\Bigg\vert_{\mathbf{r} = \mathbf{c}} = 2\mathbf{r}_h^T \mathbf{Q} \Bigg\vert_{\mathbf{r} = \mathbf{c}} = 2\mathbf{c}_h^T\mathbf{Q} = \mathbf{a}_h^T \implies \mathbf{c}_h=\frac{1}{2}\mathbf{Q}^{-1}\mathbf{a}_h$, which seems to be sensitive to the homogeneous coordinate I choose for $\mathbf{a}_h$.
I have read here that because $\det(\mathbf{Q}_r) = 0$ for all paraboloids, paraboloids are classed as "non-central" quadric surfaces, but I have not found any information about what that means for identifying $\mathbf{c}$ based on $\mathbf{Q}$.
Thank you for any suggestions!
Edit: I fixed a typo in definition of $q_0$ that didn't affect the overall problem.
| Thanks intelligenti-pauca for pointing me in the right direction! I'll add some details on my approach here.
Let $\mathbf{P} = \begin{bmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{a} \end{bmatrix}$ be the matrix of eigenvectors of $\mathbf{Q}_r$. I want to compose $\mathbf{c}$ in the eigenbasis of $\mathbf{Q}_r$ as $\mathbf{c} = \mathbf{P} \mathbf{d}$, so I need to find the elements $d_1$, $d_2$, $d_3$ of $\mathbf{d}$ as intelligenti-pauca suggested such that
$$\mathbf{c} = \mathbf{v}_1 d_1 + \mathbf{v}_2 d_2 + \mathbf{a} d_3 = \mathbf{P}\mathbf{d}$$
I find the first two elements $d_1$ and $d_2$ by an inner product with $\mathbf{q}_d$:
$$d_1 = \mathbf{v}_1^T\mathbf{q}_d = \mathbf{v}_1^T\mathbf{a}\frac{b^2}{2}
- \mathbf{v}_1^T\mathbf{Q}_r\mathbf{c} = - \mathbf{v}_1^T\mathbf{Q}_r\mathbf{c}$$
The same holds for $d_2 = \mathbf{v}_2^T\mathbf{q}_d$, recall $\mathbf{v}_1^T\mathbf{a} = \mathbf{v}_2^T\mathbf{a} = 0$.
Now, to find $d_3$ I substitute $\mathbf{c} = \mathbf{P}\mathbf{d}$ into the expression for $q_0$:
$$\begin{aligned}
q_0 &= -b^2\mathbf{a}^T\mathbf{c} + \mathbf{c}^T\mathbf{Q}\mathbf{c} \\
&= -b^2\mathbf{a}^T \mathbf{P} \mathbf{d} + \mathbf{d}^T\mathbf{P}^T\mathbf{Q}_r\mathbf{P}\mathbf{d} \\
&= -b^2\begin{bmatrix} 0 & 0 & 1 \end{bmatrix}\mathbf{d} + \mathbf{d}^T \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}\mathbf{d} \\
&= -b^2d_3 - d_1^2 - d_2^2 \\
&\implies d_3 = \frac{q_0 + d_1^2 + d_2^2}{-b^2}
\end{aligned}$$
Now that I have $d_1$, $d_2$, and $d_3$ I transform them back into the original space to get $\mathbf{c}$:
$$\mathbf{c} = \mathbf{P}\mathbf{d}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Fourth Degree Pell Equation I got stuck at the following of my research problem:
Prove that only solution to equation $4b^2-3a^4=1$ for odd positive integers $a$, $b$ is $(1,1)$.
I made factorization -->$3a^4 = (2b-1)(2b+1)$
from here since 2b-1 and 2b+1 are coprime,
$$2b-1 = 3x^4, 2b+1 = y^4$$
or
$$2b-1 = x^4, 2b+1 = 3y^4$$
obviously first one has no solution from modulo 3. Second one yields $x^4+2 = 3y^4$ which is more specific than original equation.
Another approach that might be useful is to rewrite the original equation as $(2b)^2 - 3(a^2)^2 = 1$ , we get a recursive equation for values of $a^2$ by pell and it is sufficient to prove that there is no odd perfect squares other than 1 in this recursive sequence (a similar method is used here https://arxiv.org/pdf/1705.03011.pdf )
I am seeking an elementary solution which doesn't include algebraic NT. Thanks.
| Let $a$ and $b$ be positive integers such that
$$4b^2-3a^4=1.\tag{1}$$
Then indeed, as you note
$$3a^4=4b^2-1=(2b+1)(2b-1),$$
where the greatest common divisor of the two factors divides their sum, which is $2$. Of course it is clear from $(1)$ that $3a^4$ is odd, and hence the two factors are coprime. This means that either
$$2b+1=3x^4\qquad\text{ and }\qquad 2b-1=y^4,$$
for some coprime integers $x$ and $y$, or that
$$2b+1=x^4\qquad\text{ and }\qquad 2b-1=3y^4,$$
for some coprime integers $x$ and $y$. In the latter case, reducing mod $3$ shows that $b\equiv2\pmod{4}$ and so $x^4\equiv2\pmod{4}$, a contradiction. In the former case we find that
$$3x^4-y^4=2.$$
Then in the ring $\Bbb{Z}[\sqrt{-2}]$, which is a UFD
$$3x^4=y^4+2=(y^2+\sqrt{-2})(y^2-\sqrt{-2}),$$
where again the two factors on the right hand side are coprime, and so
$$y^2+\sqrt{-2}=(1+\sqrt{-2})(a+b\sqrt{-2})^4,$$
for some coprime integers $a$ and $b$. Expanding the right hand side and comparing the coefficients of $\sqrt{-2}$ shows that
$$a^4-8a^3b-12a^2b^2+16ab^3+4b^4=1.$$
This is a Thue equation, for which efficient deterministic algorithms to find all solutions exist. PARI/GP tells me that they are the following pairs $(a,b)$:
$$(-1,-1),\qquad (-1,0),\qquad (0,1),\qquad (1,1).$$
Correspondingly either $y=0$ or $y=1$, and so the only valid solution is $(x,y)=(1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to get the value of t For each number t, consider the triangle T with corners at, A = (1, 2, 3), B = (1, 0, −2),
C = (t, t, 1).
For which value of t does T have a right angle at A?
I know how it has to do with AB*CD = something
Rest i don't know how to solve this??
| As I commented earlier, you can use the Pythagoras theorem: for a right angle we know that $a^2+b^2=c^2$. We can find the lengths first:
$AB = \sqrt{0^2+2^2+5^2} = \sqrt{29}$
$BC = \sqrt{(1-t)^2 + t^2 +3^2} = \sqrt{10-2t+2t^2}$
$AC = \sqrt{(1-t)^2 + (2-t)^2 + 2^2} = \sqrt{9-6t+2t^2}$
Now we can plug in these values for $a,b$ and $c$. We want a right angle at $A$ so we need to solve: $(AB)^2 + (AC)^2 = (BC)^2$.
This gives: $t=7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4297689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Non-linear 1st order differential equation of $~x\frac{dy}{dx}+y=x\sqrt{y}~~~$where$~\sqrt{y}~$exists I think this is the first time when I handle of non-linear differential equation.
$$ \underbrace{x \frac{dy}{dx} + y = x\sqrt{ y } }_{x > 0} \tag{1} $$
$$ \frac{dy}{dx} + \frac{1}{ x } y = \sqrt{ y } $$
$$ \underbrace{\frac{d}{dx}\left(yx\right)}_{\text{LHS of eqn1} } = x \sqrt{ y } ~~ \leftarrow~~ \text{What can I do for next?} $$
Which website is suitable for this problem?
ADD
I got the following after I have read the post of @Gerd
$$ \frac{d}{dx}\left(yx\right) = x \sqrt{ y } \tag{2} $$
$$ \frac{d}{dx}\left(xy\right) = x^{\frac{1}{2} } \cdot x^{\frac{1}{2} } \sqrt{ y } $$
$$ = \sqrt{ x } \sqrt{ xy } $$
$$ z:=xy \tag{3} $$
$$ \frac{ dz }{ dx } = \sqrt{ x } \sqrt{ z } \tag{4} $$
$$ \frac{ 1 }{ \sqrt{ z } } \frac{dz}{dx} = \sqrt{ x } $$
$$ \int_{ }^{ } \frac{1}{ \sqrt{ z } } \frac{dz}{dx} \,dx = \int_{ }^{ } \sqrt{ x } \,dx $$
$$ \int_{ }^{ } \frac{1}{ \sqrt{ z } } \,dz = \int_{ }^{ } x^{\frac{1}{ 2 } } \,dx $$
$$ \int_{ }^{ } \frac{ 1 }{ z^{\frac{1}{2} } } \,dz = \frac{ x^{\frac{ 3 }{ 2 } } }{ \frac{ 3 }{ 2 } } + \text{const}_{1} $$
$$ \int_{ }^{ } z^{-1/2} \,dz = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ \frac{ z^{\frac{1}{2}} }{ \frac{1}{2} } = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ 2z^{\frac{1}{2}} = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ 2 \sqrt{ z } = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ 2 \sqrt{ yx } = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ 2 \sqrt{ yx } - \frac{ 2 }{ 3 } x\sqrt{ x } = \text{const}_{1} $$
$$ \underbrace{\sqrt{ x } \left( 2 \sqrt{ y } - \frac{ 2 }{ 3 } x \right) = \text{const}_{1} }_{\text{general solution} } ~~ \leftarrow~~ \text{Is this also correct?} $$
| $$y'+\frac{y}{x} = \sqrt{y}$$
$$\frac{y'}{\sqrt{y}} +\frac{\sqrt{y}}{x} = 1$$
Using
$$z=\sqrt{y}$$
$$\frac{dz}{dx} = \frac{1}{2\sqrt{y}}\cdot y'$$
$$2\cdot\frac{dz}{dx} = \frac{y'}{\sqrt{y}}$$
Replace
$$2\cdot\frac{dz}{dx} +\frac{z}{x} = 1$$
And now you have built a linear system
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to find the probability that the number of boxes which must be selected to obtain all three blue boxes is greater than or equal to $5$. I am trying to solve this problem.
There are four identical red boxes and three identical blue boxes. If the boxes are randomly selected one at a time without replacement, find the probability that number of boxes which must be selected in order to obtain all three blue boxes is at least five.
I tried
*
*We select 3 boxes and all boxes are blue. In this case, we have one way.
*We select 4 boxes and there are three blue boxes, then we have three cases: RED BLUE BLUE BLUE, BLUE BLUE RED BLUE, BLUE RED BLUE BLUE.
*We select 5 boxes and there are three blue boxes. The end box must blue. The number of select is C(4, 2) = 6.
*We select 6 boxes and there are three blue boxes. The end box must blue. The number of select is C(5, 3) = 10.
*We select 7 boxes and there are three blue boxes. The end box must blue. The number of select is C(6, 4) = 15.
The probability to select 3 boxes such that the number of boxes is greater than or equal to 5 is
$$(6 + 10 + 15)/(1 + 3 + 6 + 10 + 15)=31/35.$$
Is this correct?
If the blue boxes and red boxed are different. How to find The probability to select 3 boxes such that the number of boxes is greater than or equal to 5?
| New Interpretation
There is disagreement about what the OP is asking. It has been suggested that my original interpretation is wrong. In hindsight, it does seem very plausible that I was mistaken. Therefore, I left my original interpretation at the very end of my answer, just in case it turns out to be what the OP intended.
The new interpretation is to compute the probability that in the first $4$ boxes selected (without replacement), less than $3$ blue boxes were selected. This would correspond to there needing to be at least $5$ boxes selected, before all of the blue boxes were selected.
The number of ways that $4$ boxes can be selected out of $7$ is $\displaystyle \binom{7}{4} = 35$.
The number of ways that $m$ blue boxes were among the $4$ selected boxes [with $m \in \{0,1,2,3\}$] is
$$\binom{3}{m} \times \binom{4}{4 - m}.$$
This is based on the idea that if there were $m$ blue boxes (out of $3$ total blue boxes) selected, then there were $(4 - m)$ red boxes (out of $4$ total red boxes) selected.
Therefore, the desired computation is
$$\frac{1}{\binom{7}{4}} \times $$
$$\left[\binom{3}{0} \times \binom{4}{4}\right] ~+~
\left[\binom{3}{1} \times \binom{4}{3}\right] ~+~
\left[\binom{3}{2} \times \binom{4}{2}\right].$$
This equals
$$\frac{1}{35} \times [1 + 12 + 18] = \frac{31}{35}.$$
Edit
A shortcut, based on my approach would be to compute $~1 - ~$ the probability that in $4$ selected boxes, exactly $3$ of them were blue.
This would be
$$1 - \frac{\binom{3}{3} \times \binom{4}{1}}{\binom{7}{4}} ~=~ 1 - \frac{4}{35} = \frac{31}{35}.$$
Old Interpretation
I am interpreting the OP (i.e. original poster's) intent to be that you are given that at least $5$ boxes were selected (without replacement), and wish to calculate the probability that all of the $3$ blue boxes were selected.
Let $k$ denote the number of boxes that were selected, where $k \in \{5,6,7\}$.
Based on this, let $p(k)$ denote the (relative) probability that $k$ boxes were selected.
Assuming that $k$ boxes were selected, let $f(k)$ denote the probability that all $3$ blue boxes were selected.
Then, the desired computation is
$$\sum_{k=5}^7 \left[p(k) \times f(k)\right].\tag1 $$
There are clearly $2^7$ equally likely possibilities of the (raw) selection of boxes, because (presumably) each box either was selected [probability $= (1/2)$] or was not selected [probability $= (1/2)$].
Therefore, the relative probabilites are $[\binom{7}{7} = 1]$ versus $[\binom{7}{6} = 7]$ versus $[\binom{7}{5} = 21]$ with respect to the relative probabilities of either $7, 6$, or $5$ boxes being selected.
Therefore,
$$p(7), p(6), p(5) ~=~
\frac{\binom{7}{7}}{29}, \frac{\binom{7}{6}}{29}, \frac{\binom{7}{5}}{29} ~~\text{respectively}.\tag2 $$
If $k$ boxes were selected [with $k \in \{5,6,7\}$], then the chance that all $3$ blue boxes were selected is
$$f(k) = \frac{\binom{4}{k-3}}{\binom{7}{k}}. \tag3 $$
The explanation for this is that in (3) above, the denominator represents the number of ways that the $k$ boxes could be selected. Further, if all $3$ blue boxes were selected, then $k-3$ red boxes were selected from the 4 red boxes. Therefore, the numerator reflects the number of ways of choosing $k-3$ red boxes from the $4$ red boxes.
Utilizing the expressions in (1), (2), and (3) above, the desired computation is
$$\sum_{k=5}^7 \left[p(k) \times f(k)\right]$$
$$ ~=~
\left[\frac{\binom{7}{7}}{29} \times
\frac{\binom{4}{4}}{\binom{7}{7}}\right] ~+~
\left[\frac{\binom{7}{6}}{29} \times
\frac{\binom{4}{3}}{\binom{7}{6}}\right] ~+~
\left[\frac{\binom{7}{5}}{29} \times
\frac{\binom{4}{2}}{\binom{7}{5}}\right].\tag4 $$
Simplifying the math in (4) above gives
$$\frac{1}{29} \times [1 + 4 + 6] = \frac{11}{29}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding all primes $p$ satisfying $p=a^2+b^2$ and $p$ divides $a^3+b^3-4$ .
Determine all primes $p$ such that there exist integers $a, b$ satisfying $p=a^2+b^2$ and $a^3+b^3-4$ is divisible by $p$.
So, $p | ab(a+b)+4$ after writing $a^3+b^3-4=(a+b)(p-ab)-4$ and $p=2$ is a solution if $p|4$. If $p$ doesn't divide $4$, $ab(a+b)\equiv -4 \pmod p$. Also, $p=4m+1$ for some integer $m$ by Fermat's 2-square theorem.It seems like $5$ and $2$ are the only primes that satisfy the conditions but I can't seem to prove it.
I've tried manipulating the stuff around but I'm not getting anything. Can anyone give a small hint? Thanks.
| Observe, $$\begin{align*} 2ab(a+b)+8&\equiv 0 &&\pmod p \\ 2ab&\equiv (a+b)^2&&\pmod p\\ \therefore\;\; (a+b)^3+8&\equiv 0 &&\pmod p \end{align*}$$
So, $$p|(a+b+2)\; \;\text{or}\; \;p|(a+b)^2-2(a+b)+4$$
Case 1: $$p|a+b+2$$
$$a^2+b^2|a+b+2\implies a^2+b^2\le a+b+2 \le \sqrt{2(a^2+b^2)}+2$$
$$\therefore\;\; p\le \sqrt{2p}+2 \;\Leftrightarrow\; p\in \{2,3,5\}$$
Case 2:
$$p|2ab-2(a+b)+4\implies p=2 \;\; \text{or}\;\; p|ab-a-b+4$$
$$a^2+b^2|ab-a-b+4\implies a^2+b^2\le ab-a-b+4\le ab+4\le \frac{a^2+b^2}{2}+4$$
$$\therefore\;\; p\le \frac{p}{2}+4 \;\Leftrightarrow\; p\in \{2,3,5,7\} $$
If $p=2$, we can set, $(a,b)=(1,1)$
If $p=5$, we can set, $(a,b)=(1,2)$
If $p=3$ or $p=7$,we have, $p=a^2+b^2$ is impossible by Fermat's theorem on sums of two squares.
| {
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"url": "https://math.stackexchange.com/questions/4299869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving $ a \cos x + b \sin x = u $, $ -a \sin x + b \cos x = v $ in a different way gives me a solution that does not work for $ x = n\pi $ I am solving the following equations to find $ a $ and $ b $:
\begin{align*}
a \cos x + b \sin x &= u \\
-a \sin x + b \cos x &= v
\end{align*}
Solution 1
Multiplying the first equation with $ \cos x $ and the second one with $ -\sin x $ and then adding them both I get
$$
a(\cos^2x + \sin^2 x) + b(\sin x \cos x - \sin x \cos x) = u \cos x - v \sin x.
$$
So
$$
a = u \cos x - v \sin x.
$$
Multiplying the first equation with $ \sin x $ and the second one with $ \cos x $ I get
$$
a(\sin x \cos x - \sin x \cos x) + b(\sin^2 x + \cos^2 x) = u \sin x + v \cos x
$$
So
$$
b = u \sin x + v \cos x.
$$
Solution 2
I obtain $ a = u \cos x - v \sin x $ as described in the previous solution.
Now I substitute $ a = u \cos x - v \sin x $ in the first equation to get
$$
(u \cos x - v \sin x) \cos x + b \sin x = u.
$$
Simplifying we get
$$
u \cos^2 x - v \sin x \cos x + b \sin x = u.
$$
So
$$
b = \frac{u - u \cos^2 x + v \sin x \cos x}{\sin x} = \frac{u \sin^2 x + v \sin x \cos x}{\sin x}
$$
When $ \sin x \ne 0 $, we divide both numerator and denominator by $ \sin x $ to get
$$
b = u \sin x + v \cos x.
$$
My Question
In solution 2, we have got the solution $ b = u \sin x + b \cos x $ but we got this solution only for the case when $ \sin x \ne 0 $, i.e. $ x \ne n\pi $ for integers $ n $. In solution 1, we got the same solution for all values of $ x $.
Why did I get an incomplete solution in solution 2? How do I ensure that I get a complete solution in solution 2?
| All you have to do to complete the second solution is work in cases:
Case 1: $\sin x\ne 0$
.
.
.
$$b = \frac{u - u \cos^2 x + v \sin x \cos x}{\sin x} = \frac{u \sin^2 x + v \sin x \cos x}{\sin x} = u \sin x + v \cos x$$
Case 2: $\sin x = 0$. Then $\cos^2 x =1$ and the original equations become
\begin{align*}
a \cos x &= u \\
b \cos x &= v
\end{align*}
Multipliying by $\cos x$ both equations we get
\begin{align*}
a &= u \cos x = u \cos x - v \sin x\\
b &= v \cos x = u \sin x + v \cos x
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that: $\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{21}{2}$ Given 3 positive real numbers $x, y, z$ satisfies $xy+yz+xz=1$. Prove that: $$\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{21}{2}$$
(Only use AM-GM, Cauchy-Schwarz inequalities)
My progress:
Till now, I have not made much of a progress besides finding out that $x^2+y^2+z^2\geq1$ and $xyz\leq\frac{\sqrt{3}}{9}$ and turn the LHS into $$\frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(x+y)(y+z)}}+\frac{z}{\sqrt{(x+z)(y+z)}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$$
| Let $x=\tan\frac{\alpha}{2},$ $y=\tan\frac{\beta}{2}$ and $z=\tan\frac{\gamma}{2},$ where $\{\alpha,\beta,\gamma\}\subset\left(0,\frac{\pi}{2}\right).$
Thus, $\alpha+\beta+\gamma=\pi$ and since $$\left(\sin\frac{\alpha}{2}+\cot^2\frac{\alpha}{2}\right)''=\frac{2\cos{\alpha}+4-\sin^5\frac{\alpha}{2}}{4\sin^4\frac{\alpha}{2}}>0,$$ by Jensen we obtain:
$$\sum_{cyc}\left(\frac{x}{\sqrt{x^2+1}}+\frac{1}{x^2}\right)=\sum_{cyc}\left(\sin\frac{\alpha}{2}+\cot^2\frac{\alpha}{2}\right)\geq3\left(\sin\frac{\pi}{6}+\cot^2\frac{\pi}{6}\right)=\frac{21}{2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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In the figure, determine $x$ as a function of $R$. For reference: In the figure, determine $x$ as a function of $R$.
My progress: I don't know and I can affirm that $\angle FGD$ is $90^o$ ...I think you would need to demonstrate ...but I can't
$\triangle FDB: (R+r)^2 = R^2+(2R-2r)^2\\R^2+2Rr+r^2=R^2+4R^2-4Rr+4r^2\\6Rr-3r^2 = 4Rr\\6R-3r=4R \implies \boxed{r = \frac{2R}{3}}\\\triangle FGD: (R+r)^2 = (x+r)^2+(x+R)^2\\R^2+\frac{4R^2}{3} + \frac{4R^2}{9} = x^2+\frac{4xR}{3}+\frac{4R^2}{9}+x^2+2xR+R^2\\\frac{4R^2}{3} =2x^2+\frac{10xR}{3}\\\frac{4R^2}{3} = \frac{6x^2+10xR}{3}\implies 4R^2 = 6x^2+10xR \implies\\ 3x^2+5xR - 2R^2 = 0\\\therefore \boxed{\color{red}x = \frac{R}{3}}$
| If you are not sure about the perpendicularity, let's follow the lengthy method.
Call the centres of small circle, small half circle and big half circle $C$, $E$ and $F$ respectively. Also draw the perpendicular lines to $AB$ and $BC$ to meet each side at $D$ and $G$. Thus we see, $CE=r+x$ and $CF=R+x$.
Let $CD=a$, $CG=b$.
We know that the line connecting centres of two half circles goes through the tangent point. Therefore $EF=R+r$.
Now, it's all Pythagoras.
$\triangle EBF\to R^2+(2R-r)^2=(R+r)^2$
$\triangle CBD\to a^2+b^2=(2R-x)^2\qquad(*)$
$\triangle CFD\to a^2+(R-b)^2=(R+x)^2$
$\triangle CEG\to b^2+(2R-r-a)^2=(r+x)^2$
Solving these equations might not be easy, yet possible.
From the first equation we get $r=\frac23R$ as you've already derived. Now the remaining equations can be written as, $$a^2+b^2-x^2=4R^2-4Rx$$ $$a^2+b^2-x^2=2Rb+2Rx$$ $$a^2+b^2-x^2=-\frac43(R^2-2Ra-Rx)$$ From the first and second, we get $$b=2R-3x$$ From the first and the third we get $$a=2R-2x$$ Now substituting these results in (*) we get $$(3x-R)(x-R)=0$$ which gives $x=\dfrac R3$ as the valid solution.
| {
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How to turn into a product of $2$ factors the expression : $ x^4 -3x^2+1$? Source : Lebossé & Hémery, Algèbre et Analyse ( Classe de seconde , 1965).
The exercice requires : turn into a product of $2$ factors the expression $ x^4 -3 x^2 +1$.
The only way I see is to use the property : $P(x)$ being a polynomial , if $P(a) =0$ then $P(x)$ is divisible by $(x-a)$.
Using the substitution $X = x^2$, the expression can be turned into : $X^2 -3X +1$
Setting $X^2 -3X +1= 0$ yields $ X = \frac {3+\sqrt5} {2}$ or $ X = \frac {3-\sqrt5} {2}$.
This implies that : $x= \pm\sqrt{ \frac {3+\sqrt5} {2}}$ or $x= \pm\sqrt{ \frac {3-\sqrt5} {2}}$.
Finally, using the property referred to above, one gets :
$ x^4 -3 x^2 +1 = ( x- \sqrt{\frac {3+\sqrt5} {2}}) ( x+ \sqrt{\frac {3+\sqrt5} {2} })( x - \sqrt{\frac {3-\sqrt5} {2}})(x+ \sqrt{\frac {3-\sqrt5} {2}})$.
But this method yields 4 factors instead of 2.
Is there a better way to meet the requirement of the exercise?
Is there a classical identity hidden below the original expression?
| You can stop earlier:
$$
x^4 - 3x^2 + 1
= X^2 - 3X + 1
= (X - \frac{3+\sqrt5}{2})(X - \frac{3-\sqrt5}{2})
= (x^2 - \frac{3+\sqrt5}{2})(x^2 - \frac{3-\sqrt5}{2}).
$$
| {
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Solving a problem in Coordinate Geometry Suppose, $2x \cos \alpha - 3y \sin \alpha = 6$ is equation of a variable straight line. From two point $A(\sqrt{5},0$) and $B(-\sqrt{5},0)$ foot of the altitude on that straight line is $P$ and $Q$ respectively. Show that the product of the length of two line segment $AP$ and $BQ$ is free of $\alpha$.
This question appeared in my exam today. The way I did it is first constructed the equation of two perpendicular line to $2x \cos \alpha - 3y \sin \alpha = 6$ which goes through the points $A(\sqrt{5},0$) and $B(-\sqrt{5},0)$. In this way, for $A$ and $B$, I got the following equations respectively-
$$3x \sin \alpha + 2y \cos \alpha -3 \sqrt{5}\sin \alpha=0 \qquad(1)$$$$3x \sin \alpha + 2y \cos \alpha +3 \sqrt{5}\sin \alpha=0 \qquad(2)$$
Then I found that the line $(1)$ intersects the line $2x \cos \alpha - 3y \sin \alpha = 6$ at point $$P\left ( \frac{9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}, \frac{-18 \sin \alpha+6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )$$
and the line $(2)$ intersects the line $2x \cos \alpha - 3y \sin \alpha = 6$ at point $$Q\left ( \frac{-9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}, \frac{-18 \sin \alpha-6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )$$
Now using distance formula $$AP = \sqrt{ \left ( \frac{9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}- \sqrt5 \right )^2 + \left ( \frac{-18 \sin \alpha+6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )^2}$$ and $$ BQ= \sqrt{ \left ( \frac{-9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} +\sqrt5 \right )^2 + \left ( \frac{-18 \sin \alpha-6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )^2}$$
Now the multiplication product $AP \cdot BQ$ indeed gives a constant value of $4$ which is free of the arbitrary variable $\alpha$ as you can see here is the simplified version of product of those two quantity. But this is tedious and I do not think this is the only way to do it and an appropriate way to follow in exam with limited time. So, I am looking for an alternative, time saving proof of it. I was wondering if using parametric form would help me, but I think it would get as difficult equally.
I created a visualisation for you on desmos to help my problem understand better.
| One missing factor is the locus of points $P$ and $Q$.
For point P we have the system of equations
$2\cos\alpha x - 3\sin\alpha y=6$
$3\sin\alpha x + 2\cos\alpha y=3\sqrt5\sin\alpha$
Square both equations and add. Cross-product terms cancel and we end with
$(4\cos^2\alpha+9\sin^2\alpha)(x^2+y^2)=36+45\sin^2\alpha$
Eliminate $\cos\alpha$ using $\cos^2\alpha+\sin^2\alpha=1$, then
$(4+5\sin^2\alpha)(x^2+y^2)=36+45\sin^2\alpha$
$\color{blue}{x^2+y^2=9, \text{ a circle of radius 3}}$
Similarly $Q$ has this same locus.
We then draw this circle which intersects the $x$-axis at $C=(3,0)$ and $D=(-3,0)$. With this in hand we extend $\overline{AP}$ through $A$ to point $Q'$. By symmetry $AQ'=BQ$ (the lines $\overline{AP},\overline{BQ}$ are parallel and equidistant from the center, so $Q$ and $Q'$ are antipodes on the circle). At the same time for intersecting chords of a circle $(AP)(AQ')=(AC)(AD)$. So:
$(AP)(BQ)=(AP)(AQ')=(AC)(AD)=(3-\sqrt5)(3+\sqrt5)=4.$
This exercise illustrates a theorem from the geometry of conic sections. Given an ellipse or a hyperbola, we can drop a perpendicular from both foci to any tangent of the conic. The product of the altitudes will then be the square of the semiminor axis for an ellipse or the square of the semiconjugate axis for a hyperbola. We may interpret the latter case as presenting a negative product (the square of a pure imaginary semiminor axis) because the altitudes are oppositely oriented as if one of them were negative. The family of lines in the problem at hand are the tangents to an ellipse whose vertices are $(\pm3,0)$ (matching a diameter of the circular locus) with foci at the same points as those chosen for $A$ and $B$.
| {
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Calculate $\lim_{x\rightarrow \infty} \frac{1}{\ln(x+1)-\ln(x)}-x$ I'm supposed to compute
$$\lim_{x\rightarrow \infty}\left( \frac{1}{\ln(x+1)-\ln(x)}-x\right).$$
However, I keep getting the wrong answer, so I'll present my solution for you, and I hope you can give me any tips on how to solve it.
Rewriting using logarithm laws, we have
$$\lim_{x\rightarrow \infty} \left(\dfrac{1}{\ln\frac{x+1}{x}}-x\right).$$
Simplyfing further, we have:
$$\lim_{x\rightarrow \infty} \dfrac{1-x\ln\frac{x+1}{x}}{\ln\dfrac{x+1}{x}}= \lim_{x\rightarrow \infty} \frac{1-\ln(1+\frac{1}{x})^x}{\ln(1+\frac{1}{x})} = \frac{1-e}{0} \rightarrow -\infty.$$
However, the answer sheet tells me that it's $1/2$, and I don't really see where I did something wrong in the solution. Thanks.
| I shall try to evaluate the limit by using L’Hospital Rule twice.
$$
\begin{array}{l}
\displaystyle \quad \lim _{x \rightarrow \infty}\left(\frac{1}{\ln (x+1)-\ln x}-x\right)\\
\begin{aligned}=& \lim _{x \rightarrow \infty} \frac{1-x \ln \left(\frac{x+1}{x}\right)}{\ln \left(\frac{x+1}{x}\right)} \quad\left(\frac{0}{0}\right) \\=& \lim _{x \rightarrow \infty} \frac{-x\left(\frac{x}{x+1}\right)\left(-\frac{1}{x^{2}}\right)-\ln \left(\frac{x+1}{x}\right)}{\frac{x}{x+1}\left(-\frac{1}{x^{2}}\right)} \\
=& -\lim _{x \rightarrow \infty} \frac{\frac{1}{x+1}-\ln \left(\frac{x+1}{x}\right)}{\frac{1}{x}-\frac{1}{x+1}} \quad\left(\frac{0}{0}\right) \\=& -\lim _{x \rightarrow \infty} \frac{-\frac{1}{(x+1)^{2}}-\frac{x}{x+1}\left(-\frac{1}{x^{2}}\right)}{-\frac{1}{x^{2}}+\frac{1}{(x+1)^{2}}}
\end{aligned}\end{array}\\ \begin{array}{l}
\end{array}
$$
Simplifying the quotient yields
$$
\begin{array}{l}
\displaystyle \quad \lim _{x \rightarrow \infty}\left(\frac{1}{\ln (x+1)-\ln x}-x\right) \\ \displaystyle =\lim _{x \rightarrow \infty} \frac{\frac{1}{(x+1)^{2}}-\frac{1}{x(x+1)}}{-\frac{1}{x^{2}}+\frac{1}{(x+1)^{2}}} \\
=\displaystyle \lim _{x \rightarrow \infty} \frac{x^{2}-x(x+1)}{-(x+1)^{2}+x^{2}} \\
=\displaystyle \lim _{x \rightarrow \infty} \frac{-x}{-2 x+1} \\
=\dfrac{-1}{-2+\frac{1}{x}} \\
=\dfrac{1}{2}
\end{array}
$$
| {
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Showing the formal identity $(1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)\cdots =\frac{1}{1+x+x^2}$
How to show this formal identity (or you can assume $|x|<1$)?
$$(1-x+x^2)(1-x^2+x^4)(1-x^4+x^8)\cdots =\frac{1}{1+x+x^2}$$
I can show that the latter is $$=1-x+x^3-x^4+x^6-x^7+\cdots$$ but how to show this is equal to the infinite product. I think it has something to do with residue of the exponent modulo $3$.
| Notice that $$(1+x^n+x^{2n})(1-x^n+x^{2n}) = 1+x^{2n}+x^{4n}$$
Now if $p_n$ is product of first $n$ brackets on LHS then $$p_n ={1+x^{2^n}+x^{2^{n+1}}\over 1+x+x^2}$$
Taking $\displaystyle \lim _{n\to \infty}p_n$ for $|x|<1$ we get the desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving a exponential floor flunction equation I am trying to solve this question but can't simplify this further.
Question:
$$ 2^{\lfloor log_2{(x)} + \frac{1}{2} \rfloor} = 2^{\lfloor log_2(x-2^{\lfloor{ log_2{(\frac{x}{2})} + \frac{1}{2}}\rfloor}) + \frac{1}{2}\rfloor} + 2^{\lfloor log_2{(\frac{x}{2})+\frac{1}{2}}\rfloor}$$
My first step
$$
= 2^{\lfloor{log_2{(x - 2^{\lfloor{log_2{(x)} + \frac{1}{2}-1}\rfloor})} + \frac{1}{2} }\rfloor} + 2^{\lfloor{log_2{(x) + \frac{1}{2} - 1}}\rfloor} $$
as : $log_2(x) + \frac{1}{2} = \lfloor{log_2(x) + \frac{1}{2}}\rfloor + r $
Then I rewrite $\lfloor{log_2(x) + \frac{1}{2}}\rfloor = log_2(x) + \frac{1}{2} - r $
So : $ 2^{\lfloor{log_2{(x)} + \frac{1}{2}}\rfloor} = 2^{log_2{(x)} + \frac{1}{2} -r} = x \cdot 2^{v}$ where $v = \frac{1}{2} - r$
So that
$$ 2^{\lfloor log_2{(x)} + \frac{1}{2} \rfloor} = 2^{\lfloor{log_2{(x - x\cdot2^{v-1} )} + \frac{1}{2}}\rfloor} + x \cdot 2^{v-1} $$
$$ = 2^{\lfloor{log_2{(x(1 - 2^{v-1}) )} + \frac{1}{2}}\rfloor} + x \cdot 2^{v-1}$$
$$ = 2^{\lfloor{log_2{(x)}+ \frac{1}{2} + log_2{(1-2^{v-1})} }\rfloor} + x \cdot 2^{v-1} $$
I'm stuck here. Can anyone help?
| ${first}... \\ $
$\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} {x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]} =\mathrm{2}.\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]} \\ $
$. \\ $
${so}... \\ $
$. \\ $
$\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} {x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]} −\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]} \\ $
$. \\ $
${then}... \\ $
$. \\ $
${k}=\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]} =\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} \left({x}−{k}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]} \\ $
$\frac{{x}}{\mathrm{2}}={x}−{k}\Rightarrow\frac{{x}}{\mathrm{2}}={k}=\mathrm{2}^{\left[\left(\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right]} \\ $
${for}:\:\left\{\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right\}\:\in\left[\mathrm{0},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $
$\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}=\left[\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right]\Rightarrow{x}=\mathrm{2}^{{n}} ,\:{n}\in{Z} \\ $
${for}:\:\left\{\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right\}\:\in\left[\frac{\mathrm{1}}{\mathrm{2}},\mathrm{1}\right) \\ $
$\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}=\left[\mathrm{log}_{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right]+\mathrm{1}\Rightarrow\left({no}\:{solution}\right) \\ $
$. \\ $
${so}\:{the}\:{answer}:\:{x}=\mathrm{2}^{{n}} ,\:{n}\in{Z} \\ $
| {
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Show that the set can describe all rational numbers in (0, 1). Let A be a subset of (0, 1) ∩ Q, where Q is the set of all rational numbers.
Given that $\frac{1}{2}\in A$ and that $\frac{x}{x+1},\frac{1}{x+1}\in A$ if $x\in A$, show that A = (0, 1) ∩ Q.
What I've tried to do:
For simplicity, let (1) be the operation $x\rightarrow\frac{x}{x+1}$, and (2) the other one $x\rightarrow\frac{1}{x+1}$. As we have only fractions in A, (1) can be rewritten as $\frac{x}{y}\rightarrow\frac{x}{x+y}$, and (2) as $\frac{x}{y}\rightarrow\frac{y}{x+y}$
Basically, we need to show that $\frac{a}{b}\in A $ for each a, b - natural numbers with a < b. For this, I figured out I need to show that I can make any $\frac{a}{b}$ using (1) and (2) on $\frac{1}{2}.$ The 'a < b' part is obvious as (1) and (2) only increases the denominator.
We can easily prove that we can make any $\frac{1}{n}$, where $n\geq2$ by applying (1) on $\frac{1}{2}$ (n-2) times.
After this, I proved that we can make any $\frac{F_{k+1}n+F_k}{F_{k+2}n+F_{k+1}}$ by applying (2) on $\frac{1}{n}$ (k) times, where I noted with $F_k$ - the kth number in the sequence {0, 0, 1, 1, 2, 3, 5, 8...} (the Fibonacci numbers).
After testing some examples, I arrived at the conclusion that I can write any $\frac{a}{b}$ by applying (1) on $\frac{F_{k+1}n+F_k}{F_{k+2}n+F_{k+1}}$ a specific number of times but I couldn’t prove this. I got this problem from a set of problems based on induction so I think it can be solved using it.
| You can do it by induction on the denominator. That is, we can show that, for all $0 < a < n$, we have $\frac{a}{n} \in A$, and we show this by induction on $n \ge 2$. Clearly the result holds for $n = 2$, by our assumption that $\frac{1}{2} \in A$.
Suppose that, for some $n \ge 3$, we have $\frac{a}{k} \in A$, whenever $0 < a < k < n$. Now, suppose $0 < a < n$. Let's solve the two equations:
\begin{align*}
\frac{1}{x + 1} = \frac{a}{n} &\iff x = \frac{n - a}{a} \\
\frac{x}{x + 1} = \frac{a}{n} &\iff x = \frac{a}{n - a}.
\end{align*}
Note that at one of the two solutions is a rational number in $(0, 1)$, and both have denominators strictly less than $n$. If $a > n - a \iff a > \frac{n}{2}$, then the first solution lies in $(0, 1)$ and hence in $A$ by the induction hypothesis. If $a < n - a \iff a < \frac{n}{2}$, then the second solution lies in $A$. If $a = n - a \iff a = \frac{n}{2}$, then $\frac{a}{n} = \frac{1}{2} \in A$ by assumption.
Either way, by strong induction, the result holds.
| {
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Find the distance $ID$ in the triangle below For reference: In the triangle $ABC. \angle A = 2\angle C, IG|| AC$, where $I$ is the incenter and $G$ the centroid. If $IG =x$, calculate the distance $ID$ if $BD$ is an interior angle bisector. (Answer: $3x\sqrt2$)
My progress:
I was able to see these relationships but I couldn't finish
$\frac{BI}{DI} = \frac{a+c}{b}\\
IG \parallel AC \implies b =\frac{a+c}{2} \\
ID = \frac{BD}{3} = 2BI$
Th. Angle Bissector
$\frac{AD}{c}=\frac{CD}{a}\\
BD^2 = a.c - AD.CD\\
\triangle BIG \sim \triangle BDM \implies\\
\frac{BI}{BD} = \frac{x}{DM}\rightarrow \frac{2}{3} = \frac{x}{DM} \therefore DM = \frac{3x}{2}$
Th. Median:
$a^2+c^2 = 2BM^2+\frac{b^2}{2}$
| Question says $IG = a$. To avoid confusion, I will instead use $IG = x$ and standard notation for the side lengths of the triangle.
$IG \parallel AC \implies r = \dfrac h 3$ where $h$ is altitude from $B$ to $AC$
So using $A = r \cdot s, $ where $r$ is inradius, $s$ is sub-perimeter and $A$ is area,
$\displaystyle \frac h 3 = \frac{2 A}{a + b + c} = \frac{2 \cdot b \cdot h / 2}{a+b+c} \implies 2b = a+c$
Now using $\triangle AIB \sim \triangle CDB$
$\displaystyle \frac{BI}{c} = \frac{BD}{a}$
Given $BI = \dfrac{2}{3} BD, c = \dfrac{2a}{3}$
So we have, $ \displaystyle a = \frac{6b}{5}, c = \frac{4b}{5}$
As $ \displaystyle DM = \frac{3x}{2}, \dfrac{a}{c} = \frac{3}{2} = \frac{CD}{AD} = \frac{b + 3x}{b - 3x} \implies b = 15x$
Using formula for the length of angle bisector,
$ \displaystyle BD^2 = ac \left(1 - \frac{b^2}{(a+c)^2}\right) = \frac{3ac}{4} = 162 x^2$
$ \displaystyle \implies BD = 9 x \sqrt2 ~ $ and $ ~ ID = \dfrac{BD}{3} = 3 x \sqrt2$
| {
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Calculating $\lim_{x \to 0}\frac{x \tan2x-2x\tan x}{(1-\cos2x)^2}$ I'm a bit confused regarding this question. I've been trying to solve it and have gotten to the same answer ($3/8$) thrice now. I have no idea where I'm going wrong and would really appreciate some help figuring it out. Here's my solution:
$$\lim_{x \to 0}\frac{x\tan2x-2x\tan x}{(1-\cos2x)^2}$$
Applying L'Hospital's rule,
$$\lim_{x \to 0}\frac{\tan2x+x(\sec^22x)(2)-2\tan x-2x\sec^2x}{2(1-\cos2x)(\sin2x)(2)}$$
Dividing numerator and denominator by $2x$,
$$\begin{align*}
\lim_{x \to 0}&\frac{(\frac{\tan2x}{2x})+(\frac{2x(\sec^22x)}{2x})-(\frac{2\tan x}{2x})-(\frac{2x\sec^2x}{2x})}{4(1-\cos2x)\frac{(\sin2x)}{2x}}\\
&=\lim_{x \to 0}\frac{1+\sec^22x-1-\sec^2x}{4(1-\cos2x)}\\
&=\lim_{x \to 0}\frac{\sec^22x-\sec^2x}{4(1-\cos2x)}
\end{align*}$$
Applying L'Hôpital's rule (again),
$$\lim_{x \to 0}\frac{(2)(\sec2x)(\sec2x)(\tan2x)(2)-(2)(\sec x)(\sec x\tan x)}{4(\sin2x)(2)}$$
$$=\lim_{x \to 0}\frac{(2)(\sec^22x)(\tan2x)-(\sec^2x)(\tan x)}{4(\sin2x)}$$
Dividing numerator and denominator by $2x$,
$$\begin{align*}
&=\lim_{x \to 0}\frac{(2)(\sec^22x)(\frac{\tan2x}{2x})-(\sec^2x)(\frac{\tan x}{2x})}{4(\frac{\sin2x}{2x})}\\
&=\lim_{x \to 0}\frac{(2)(\sec^22x)(\frac{\tan2x}{2x})-(\sec^2x)(\frac{\tan x}{x})(\frac{1}{2})}{4(\frac{\sin2x}{2x})}\\
&=\lim_{x \to 0}\frac{(2)(\sec^22x)(1)-(\sec^2x)(1)(\frac{1}{2})}{4(1)}\\
&=\frac{(2)(\sec^20)(1)-(\sec^20)(1)(\frac{1}{2})}{4(1)}\\
&=\frac{(2)(1)(1)-(1)(1)(\frac{1}{2})}{4(1)}\\
&=\frac{2-\frac{1}{2}}{4}\\
&=\frac{\frac{3}{2}}{4}\\
&=\frac{3}{8}.
\end{align*}$$
| Your problem arises here:
\begin{align*}
\lim_{x \to 0}&\frac{(\frac{\tan2x}{2x})+(\frac{2x(\sec^22x)}{2x})-(\frac{2\tan x}{2x})-(\frac{2x\sec^2x}{2x})}{4(1-\cos2x)\frac{(\sin2x)}{2x}}\\
&=\lim_{x \to 0}\frac{1+\sec^22x-1-\sec^2x}{4(1-\cos2x)}\\
&=\lim_{x \to 0}\frac{\sec^22x-\sec^2x}{4(1-\cos2x)}
\end{align*}
particularly where you evaluate $\lim_\limits{x\to 0} \frac {\tan 2x}{2x} = 1$ and $\lim_\limits{x\to 0} -\frac {2\tan x}{2x} = -1$ and cancel them.
$\tan 2x - 2\tan x\approx 2x^3$ and that is not trivial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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unbiased estimator for the negative binomial distribution Let be $P(X=x)= \binom{x-1}{r-1} \theta^r (1-\theta)^{x-r}$
Show that $\frac{r-1}{x-1}$ is an unbiased estimator for
My attempt;
E$[\frac{r-1}{x-1}]$ $=$ $\sum_{x=0}^{\infty}\frac{r-1}{x-1}\binom{x-1}{r-1} \theta^r (1-\theta)^{x-r} = \sum_{x=0}^{\infty} \binom{x-2}{r-2} \theta^{r-1} (1-\theta)^{x-r+1}$
$\sum_{x=0}^{\infty} \binom{x-1-1}{r-1-1} \theta^{r-1} (1-\theta)^{x-r+1}$
Such that $E(x)$ from a $BinNeg (r,p)$ is $E(x)= \frac{r(1-p)}{p}$ then;
E$[\frac{r-1}{x-1}]$ must be $\frac{(r-1)(1-p)}{p}$
So that means that $\frac{r-1}{x-1}$ is not an unbiased estimator for
Is that right?
I'm not sure about the value of r
| There are some errors in your calculation. Note that the lower index of summation should be $r$, not $0$. Then we have
$$\begin{align}
\frac{r-1}{x-1} \binom{x-1}{r-1} \theta^r (1-\theta)^{x-r}
&= \frac{r-1}{x-1} \frac{(x-1)!}{(r-1)! (x-r)!} \theta^r (1-\theta)^{x-r} \\
&= \theta \frac{(x-2)!}{(r-2)! ((x-2)-(r-2))!} \theta^{r-1} (1-\theta)^{(x-1)-(r-1)} \\
&= \theta \binom{(x-1)-1}{(r-1)-1} \theta^{r-1} (1-\theta)^{(x-1)-(r-1)}.\end{align}$$
Note this last expression is $\theta$ times the PMF of a negative binomial distribution with parameters $r^* = r-1$ and $\theta$. Hence
$$\operatorname{E}[X] = \theta \sum_{x=r}^\infty \binom{(x-1)-1}{(r-1)-1} \theta^{r-1} (1-\theta)^{(x-1)-(r-1)} = \theta \sum_{x=r^*}^\infty \binom{x-1}{r^*-1} \theta^{r^*} (1-\theta)^{x-r^*} = \theta.$$
| {
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"url": "https://math.stackexchange.com/questions/4321610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is $ \lim_{x \rightarrow 0}\left(\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}\right) $? The answer of the following limit:
$$
\lim_{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)
$$
is 1 by Wolfram Alpha.
But I tried to find it and I got $2/3$ :
My approach :
$1)$
$
\ln(\cos x)=\ln\left(1-\frac{x^{2}}{2}+o\left(x^{3}\right)\right)=-\frac{x^{2}}{2}+o\left(x^{3}\right)
$
$2)$
$
\sin ^{2}(x)=\left(x-\frac{x^{3}}{3!}+o\left(x^{3}\right)\right)^{2}=x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)
$
$3)$
$\begin{aligned} \frac{1}{-\frac{x^{2}}{2}+o\left(x^{3}\right)}+\frac{2}{x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)}=\frac{-x^{2}+x^{2}-\frac{x^{4}}{3}+o\left(x^{3}\right)}{-\frac{x^{4}}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)} \end{aligned}$
$4)$
$\lim _{x \rightarrow 0} \frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}}{-\frac{1}{2}}=\frac{2}{3}$
So where is the mistake in my approach?
Note: $o$ denotes the little-o notation
Edit : I've understood where's my mistake is, but another question popped up reading the answers which is : does $o(1/x)$ tends to zero as x tends to zero?
| As an alternative by $\cos x= e^t$ with $t\to 0$ and then $\sin^2 x=1-e^{2t}$ we have
$$\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}=\\=\frac1t+\frac2{1-e^{2t}}=\frac{1-e^{2t}+2t}{t(1-e^{2t})}=\frac{1-1-2t-2t^2+2t+o(t^2)}{t(1-1-2t+o(t))}=\frac{-2t^2+o(t^2)}{-2t^2+o(t^2)}=\frac{1+o(1)}{1+o(1)}\to 1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not? Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not?
$$a^2+ab+b^2=(a+b)^2-ab=0$$ iff $$(a+b)^2=ab \tag{1}$$
but $(a+b)^2 = a^2+2ab+b^2 $ so equation 1 couldn't possibly be true.
Also, when $a=b\ne 0$, $(a^2+ab+b^2)(a-b) = a^3-b^3 =0$.
| Fix $b\neq0$ and try to solve in $a$. This a second order equations so the possible values of $a$ in terms of $b$ are
$$ a_{1,2}=\frac{-b\mp \sqrt{-3b^2}}{2}$$
So, for any non zero real value $b$ the possible values of $a$ are complex conjugated because $-3b^2<0$. Therefore you cannot find two non-zero reals such that $(a+b)^2=ab$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $z^3 =-1$ I need to find the solution of $z^{3} =-1$.
Let$$z=[ r,\theta ]$$
then using de Moivre's theorm$$ z^{3} =\left[ r^{3} ,3\theta \right] $$
write the number $-1$ in modulus argument from,
$$[ 1,( 2n-1) \pi ]$$
the we can write$$\left[ r^{3} ,3\theta \right] =[ 1,( 2n-1) \pi ]$$
Therefore,$$ r^{3} =1 \space \text{and} \space \theta =\frac{( 2n-1) \pi }{3}$$
Then$$z_{n} =1\left( \cos\frac{( 2n-1) \pi }{3} +i\sin\frac{( 2n-1) \pi }{3}\right)$$
\begin{align}
&n=0 \quad z_{0} =\frac{1}{2} -i\frac{\sqrt{3}}{2}\\
&n=1 \quad z_{1} =\frac{1}{2} +i\frac{\sqrt{3}}{2}\\
&n=2 \quad z_{1} =-1\\
\end{align}
| You can find the complex roots of unity as below. The equation you have solved is not the one that will give you complex roots of unity.
$$z^3=1 \implies z^3-1=0$$
Factor the LHS (using any of a variety of methods):
$$(z-1)(z^2+z+1)=0$$
Use the zero-product property:
$$z-1=0\implies z=1$$
$$z^2+z+1=0\implies z=\frac{-1\pm i\sqrt{3}}{2}$$
(where we used the quadratic formula above)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A limit with geometrical progression I got stuck trying to solve this limit.
$$ \lim_{n\to\infty} \frac{1 + \frac{1}{2} + \frac{1}{4} + ....+ \frac{1}{2^{n}}}{1 + \frac{1}{3} + \frac{1}{9}+...+\frac{1}{3^{n}}} $$
The first thing I did was calculating the sums of the geometrical progressions found in both numerator and denumerator. And after that I got
$$ \lim_{n\to\infty} \frac{\frac{(\frac{1}{2})^n - 1}{\frac{1}{2} - 1}}{\frac{(\frac{1}{3})^n - 1}{\frac{1}{3} - 1}} $$
And here I am stuck because I don't know how to continue. Any help?
| We see, that $\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+ \frac{1}{2^{n}}\right)$ and $\left(1 + \frac{1}{3} + \frac{1}{9}+...+\frac{1}{3^{n}}\right)$ are the geometric series.
Therefore,
$$
\lim_{n\to\infty}\left(1+\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots + \frac {1}{2^{n}}\right)=\sum_{n=0}^{\infty}\left(\frac{1}{2}\right)^n=\frac{1}{1-\frac{1}{2}}={2},
$$
$$
\lim_{n\to\infty}\left(1 + \frac{1}{3} + \frac{1}{9}+...+\frac{1}{3^{n}}\right)=\sum_{n=0}^{\infty}\left(\frac{1}{3}\right)^n=\frac{1}{1-\frac{1}{3}}=\frac{3}{2}
$$
We will have
$$
\lim_{n\to\infty}\frac{\left(1+\frac 1 2 + \frac 1 4 + \frac 1 {8} + \cdots + \frac {1}{2^{n}}\right)}{\left(1 + \frac{1}{3} + \frac{1}{9}+...+\frac{1}{3^{n}}\right)}=\frac{2}{\frac{3}{2}}=\frac{4}{3}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4332598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Solving $\sqrt3\;\cos(x)= \sin\;(\frac {x}{2})$ I'm currently stuck on this trigonometric equation:
$$\sqrt3 \;\cos(x)= \sin\;(\frac {x}{2})$$
Here's what I've tried so far (2 methods):
Method n.1
$$\sqrt3 \;\cos(x)= \sin\;(\frac {x}{2})$$
$$\sqrt3 \;(1-\sin^2x)= \sin\;(\frac {x}{2})$$
$$\sqrt3-\sqrt3\;\sin^2x= \sin\;(\frac {x}{2})$$
$$2\sqrt3-2\sqrt3\;\sin^2x- \sin\;x=0$$
$$// \; let\; \sin \;(x)=y \;//$$
$$2\sqrt3-2\sqrt3\;y^2- y=0$$
$$y=\frac {1\pm\sqrt{1^2-4(-2\sqrt3)(2\sqrt3)}}{2(-2\sqrt3)}$$
$$y=\frac {1\pm\sqrt{1^2(8\sqrt3)(2\sqrt3)}}{-4\sqrt3}$$
$$y=\frac {1\pm\sqrt{1^2+16\times3}}{-4\sqrt3}$$
$$y=\frac {1\pm\sqrt{49}}{-4\sqrt3}$$
$$y=\frac {1\pm7}{-4\sqrt3}$$
$$y_1=\frac {1+7}{-4\sqrt3}=\frac{-2}{\sqrt3}=\frac{-2\sqrt3}{3}$$
$$y_2=\frac {1-7}{-4\sqrt3}=\frac{-3}{-2\sqrt3}=\frac{\sqrt3}{2}$$
$$\arcsin(\frac{-2\sqrt3}{3})=\nexists$$
$$\arcsin(\frac{\sqrt3}{2})=60°$$
$$x_1= 60°+360°k$$
$$x_1= 120°+360°k$$
Method n.2
$$\sqrt3 \;\cos(x)= \sin\;(\frac {x}{2})$$
$$\sqrt3 \; \cos(x)=\pm \frac{\sqrt{ 1-cos (x)}}{2}$$
Any help is appreciated, thank you in advance! :D
| hint
If we make the substitution$$x=\pi+2y$$
the equation becomes
$$-\sqrt{3}\cos(2y)=\cos(y)=\sqrt{3}(1-2\cos^2(y))$$
or
$$2\sqrt{3}\cos^2(y)+\cos(y)-\sqrt{3}=0$$
with
$$\Delta=25$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4337047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding x which maximizes the volume of any cuboid missing one side So my grandpa and me where doing some maths for fun and we stumbled upon this problem:
Given a rectangle with unknown sidelenghts a and b, we need to cut off a square in each corner (call the side length of the square x) so that we get a net of a cuboid missing one side:
the problem:
The goal is to choose x such that we maximize the volume of this cuboid.
This was my idea:
$$V= (a-2x)(b-2x)*x $$ $$= 4x^3 - 2 ax^2 -2bx^2 + abx$$
now we differentiate in respect to x:
$$\frac{d}{dx} (4x^3 - 2 ax^2 -2bx^2 + abx)$$
$$=12x^2-4ax -4bx + ab$$
$$=12x^2-(4a+4b)x + ab$$
using the quadratic formula:
$$x_{1,2}=\frac{-(-(4a+4b) \pm \sqrt{(-(4a+4b))^2-4*12*ab}}{2ab}$$
$$x_{1,2}=\frac{(4a+4b) \pm \sqrt{16a^2+32ab+16b^2-48ab}}{2ab}$$
$$x_{1,2}=\frac{(4a+4b) \pm \sqrt{16a^2-16ab+16b^2}}{2ab}$$
$$x_{1,2}=\frac{4(a+b) \pm \sqrt{16(a^2-ab+b^2)}}{2ab}$$
$$x_{1,2}=\frac{4(a+b) \pm 4\sqrt{(a^2-ab+b^2)}}{2ab}$$
$$x_{1,2}=\frac{2(a+b) \pm 2\sqrt{(a^2-ab+b^2)}}{ab}$$
$$x_{1,2}=\frac{2(a + b \pm \sqrt{(a^2-ab+b^2)})}{ab}$$
Before calculating further I checked the solutions for x in Wolfram Alpha and it gave me this result (which is correct):
$$x_{1,2}= \frac{1}{6}(\pm\sqrt{a^2-ab+b^2}+a+b)$$
My solution is obviously incorrect, but I don't quite get what's wrong with my approach. I'm pretty sure that i calculated everything right. Am I not allowed to use the quadratic formula in that case, or what am I missing? Can somebody explain?
| Why $x_{1,2}=\frac{-(-(4a+4b) \pm \sqrt{(-(4a+4b))^2-4\cdot12\cdot ab}}{2ab}$? You made a serious mistake since using the root fomula of quadratic equation.
Since the equation is $12x^2-4(a+b)x+ab=0$, so...
$$
x_{1,2}=\frac{-(-4(a+b))\pm\sqrt{(-4(a+b))^2-4\cdot12\cdot ab}}{2\cdot12}
\\
=\frac{a+b\pm\sqrt{a^2-ab+b^2}}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4341517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Power series of $1/(1-x-x^2)$ I have a difficulty in finding the power series of: $$\frac{1}{1-x-x^2}$$ At first I did a polynomial division and I found out that: $$\frac{1}{1-x-x^2}=1+x+2x+3x^2+5x^3+8x^4+13x^5+...$$ where the coefficients of this Taylor Series are the numbers of the Fibonacci Sequence.
However, this is not a proof, so when I tried to prove it, I reached the following point:$$\frac{1}{1-x-x^2}=\sum_{n=0}^{\infty}(x+x^2)^n=\sum_{n=0}^{\infty}x^n(1+x)^n\\ =\sum_{n=0}^{\infty}x^n\sum_{k=0}^{n}\binom{n}{k}x^{n-k}=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{n}{k}x^{2n-k}$$
How can I proceed from this point, or I have to take another path?
| It is possible to continue from your approach. Let's fix a natural number $N$ and compute its coefficient. Given $n$, the inner sum can only contribute terms with powers $n, n+1, \ldots, 2n$. So to compute the coefficient of $x^{N}$, we should consider those terms with $n = N/2, N/2 + 1, \ldots N$. It now makes a difference whether $N$ is even or odd - let's assume it's even for simplicity.
The coefficient of $x^{N}$ is
$$\binom{N/2}{0} + \binom{N/2+1}{2} + \ldots + \binom{N}{N} = \sum_{i=0}^{N/2} \binom{N/2+i}{2i}$$
A similar argument for $N$ odd gives
$$\binom{(N+1)/2}{1} + \binom{(N+3)/2}{3} + \ldots \binom{N}{N}$$
Let's call these sums $S_{N}$. Sorry for the overloaded notation - assuming again that $N$ is even, so $N+1$ is odd, We can write $S_{N}+S_{N+1}$ as:
\begin{align*}
\binom{N/2}{0} + \binom{N/2+1}{2} + \ldots + \binom{N}{N} + \\
\binom{N/2 + 1}{1} + \binom{N/2+2}{3} + \ldots + \binom{N + 1}{N} \\
= \binom{N/2}{0} + \binom{N/2+2}{2} + \binom{N/2+3}{4} + \ldots \binom{N+1}{N+1}
\end{align*}
(applying Pascal's identity), and luckily $\binom{N/2}{0} = 1 = \binom{N/2 + 1}{0}$, so this final sum is indeed $S_{N+2}$. It remains to check the case $N$ odd, and that $S_{0} = S_{1} = 1$, but those checks are easy.
There is a more compelling approach: Assume we can write $\frac{1}{1-x-x^2} = \sum_{n=0}^{\infty} a_{n} x^{n}$.
Then, multiplying by $1-x-x^2$,
\begin{align*}
1 &= \sum_{n=0}^{\infty} a_{n} (1-x-x^2)x^{n}\\
&= a_{0} + a_{1} x - a_{0}x + \sum_{n=2}^{\infty} (a_{n} - a_{n-1} - a_{n-2}) x^n \\
\end{align*}
and we deduce, by equating coefficients, that
\begin{align*}
a_0 &= 1 \\
a_1 &= a_0 \\
a_n &= a_{n-1} + a_{n-2} \,\,\text{for $n \ge 2$}
\end{align*}
and hence, that $(a_n)$ are the Fibonacci numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
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Geometry problem to prove Euler line of a triangle $I$ is the incenter of $\triangle{ABC}$ and $D$ is the tangent point of the incircle with side $BC$. Points $E$ and $F$ are on side $BC$ so that $\angle{BAE}=\angle{ACB}, \angle{CAF}=\angle{ABC}$. $G$ and $H$ are incenters of $\triangle{ABE}$ and $\triangle{ACF}$ respectively. Prove that $AD$ is the Euler Line of $\triangle{GHI}$.
I tried to find related material to solve this problem but in vain. Any hints or helpful suggestion? Thanks!
With hint to use barycentric coordinates, here is what I have got:
Let $s=\dfrac{a+b+c}{2}, t=\sin A+\sin B+\sin C$, then
$A(1:0:0)\\
B(0:1:0)\\
C(0:0:1)\\
I(\dfrac{a}{2s}:\dfrac{b}{2s}:\dfrac{c}{2s})\\
D(0:\dfrac{s-c}{a}:\dfrac{s-b}{a})\\
E(0:1-\dfrac{c\sin C}{a\sin A}:\dfrac{c\sin C}{a\sin A})\\
F(0:\dfrac{b\sin B}{a\sin A}:1-\dfrac{b\sin B}{a\sin A})\\
G(\dfrac{\sin C}{t}:1-\dfrac{(a+c)\sin C}{at}:\dfrac{c\sin C}{at})\\
H(\dfrac{\sin B}{t}:\dfrac{b\sin B}{at}:1-\dfrac{(a+b)\sin B}{at})
$
| $\begin{array}{} ∠(ACB)=∠(BAE)=γ & ∠(ABC)=∠(CAF)=β & ∠(AEK)=β+γ \end{array}$
$\begin{array}{} A=\left( c\,cos(β),c\,sin(β) \right) & B=(0,0) & C=(a,0) \end{array}$
$\begin{array}{} ΔABC & cos(β)=\frac{a^2-b^2+c^2}{2ac} & cos(γ)=\frac{a^2+b^2-c^2}{2ab} \\ sin(β)=\frac{2S}{ac} & sin(β)=\frac{2S}{ab} & S=Δ(ABC)=\sqrt{s(s-a)(s-b)(s-c)} & s=\frac{a+b+c}{2}\end{array}$
$\begin{array}{} x_{A}=\frac{a^2-b^2+c^2}{2a} & y_{A}=\frac{2S}{a} \end{array}$
$\begin{array}{} ΔAEK & AK=c\,sin(β) & EK=\frac{AK}{tan(β+γ)} \\ tan(β+γ)=\frac{cos(β)sin(β)+cos(γ)sin(γ)}{cos(β)^2+cos(γ)^2-1}=\frac{4S}{a^2-b^2-c^2} & EK=\frac{a^2-b^2-c^2}{2a} & BE=BK-EK \\ \end{array}$
$\begin{array}{} BE=\frac{c^2}{a} & E=\left( \frac{c^2}{a},0 \right) & ∠(AEK)=∠(AFK) & AE=AF & AE=\frac{bc}{a} \end{array}$
$\begin{array}{} BF=BK+EK & BF=\frac{a^2-b^2}{a} & F=\left( \frac{a^2-b^2}{a},0 \right) \end{array}$
$\begin{array}{} \text{G is the incenter ΔABE } & x_{G}=\frac{x_{A}·BE+x_{B}·AE+x_{E}·c}{BE+AE+c} & x_{G}=\frac{c(a-b+c)}{2a} & y_{G}=\frac{2Sc}{a(a+b+c)} \end{array}$
$\begin{array}{} \text{H is the incenter ΔACF} & FC=a-BF & FC=\frac{b^2}{a} \\ x_{H}=\frac{2a^2-b(a+b-c)}{2a} & y_{H}=\frac{2Sb}{a(a+b+c)} & \, \\ \text{I is the incenter ΔABC} & x_{I}=\frac{a-b+c}{2} & y_{I}=\frac{2S}{a+b+c}\end{array}$
$l.x+m.y+n=0$ is the Euler line of $ΔGHI$. $l$ and $m$ are the coefficients of the Euler line in Cartesian coordinates. With CAS we simplify the matrix and get:
$\begin{array}{} l=\left| \begin{array}{} x_{G} & y_{G}^2+2y_{I}\,y_{H}+3x_{G}^2 & 1 \\ x_{I} & y_{I}^2+2y_{G}\,y_{H}+3x_{I}^2 & 1\\ x_{H} & y_{H}^2+2y_{G}\,y_{I}+3x_{H}^2 & 1 \\ \end{array} \right| & m=\left| \begin{array}{} y_{G} & x_{G}^2+2x_{I}\,x_{H}+3y_{G}^2 & 1 \\ y_{I} & x_{I}^2+2x_{G}\,x_{H}+3y_{I}^2 & 1\\ y_{H} & x_{H}^2+2x_{G}\,x_{I}+3y_{H}^2 & 1 \\ \end{array} \right| \end{array}$
$\begin{array}{} l=\frac{(a-c)(a-b)(a^2-b^2-c^2+2bc)(a^2-b^2-c^2+bc)}{a^2(a+b+c)} & \\ m=\frac{4S(a-b)(a-c)(b-c)(a^2-b^2-c^2+bc)}{a^2(a+b+c)^2} & \\ \end{array}$
$\begin{array}{} m_{Euler}=\frac{-l}{m} & m_{Euler}=\frac{-1}{4}\frac{(a+b+c)(a-b+c)(a+b-c)}{(b-c)S} \\ (a+b+c)(a-b+c)(a+b-c)=\frac{16S^2}{-a+b+c} & m_{Euler}=\frac{4S}{(a-b-c)(b-c)} \\ \end{array}$
$\begin{array}{} \text{J is the centroid of ΔGIH} & x_{J}=\frac{3a^2-2ab-b^2+2ac+c^2}{6a} & y_{J}=\frac{2S}{3a} \end{array}$
$\begin{array}{} \text{Euler line} & y-y_{J}=m_{Euler}(x-x_{J}) & \frac{y-y_{J}}{x-x_{J}}=m_{Euler} & \text{(x,y) is any point on the Euler line }\end{array}$
$\begin{array}{} \text{checking if the points belong to the line} & \frac{y_{A}-y_{J}}{x_{A}-x_{J}}-m_{Euler} =0& A=\left( \frac{a^2-b^2+c^2}{2a},\frac{2S}{a} \right) & \text{true} \\ \frac{y_{D}-y_{J}}{x_{D}-x_{J}}-m_{Euler}=0 & D=\left( \frac{a-b+c}{2},0 \right) & \text{true} \\ \end{array}$
| {
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} |
How to solve $\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$? I am stuck with this equation. All I could do is this:
$\lfloor x \rfloor$ = $\lfloor n + m \rfloor$ such that $n \in N$ and $m<1$.
We get:
$\lfloor x \rfloor + \lfloor \frac{1}{x} \rfloor = 1$
$\lfloor n + m \rfloor + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + \lfloor m \rfloor + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + 0 + \lfloor \frac{1}{n+m} \rfloor = 1$
$n + \lfloor \frac{1}{n+m} \rfloor = 1$
From here on I have no idea what to do!
Edit: It is easy to see that any value $1<x<2$ satisfies the equation, but can I find all the solutions?
| In general, to solve $\lfloor a \rfloor + \lfloor b \rfloor =c$ you can consider the pairs of values that $a$ and $b$ can take.
In this case your possibilities are very small: $a=0, b=1$ or $a=1, b=0$.
Can you proceed from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4348588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How many ways to deal with the integral $\int_{0}^{\frac{\pi}{4}} \ln (\tan x) d x$ and $\int_{0}^{\frac{\pi}{4}} \ln (\sin x) d x$? After finding that $\displaystyle \int_{0}^{\frac{\pi}{2}} \ln (\tan x) d x =0$ in my post, I was curious about the value of the integral with different upper limit $\dfrac{\pi}{4} $.
The answer is surprisingly simple and elegant i.e.
$$
\int_{0}^{\frac{\pi}{4}} \ln (\tan x) d x=-G \text {, }
$$
where $G$ is the Catalan’s constant.
We first let $y=\tan x$, then $d y=\sec ^{2} x d x=\left(1+y^{2}\right) d x$ and $I$ is converted to
$$
I=\int_{0}^{1} \frac{\ln y}{1+y^{2}} d y.
$$
Applying a power series yields $$
\begin{aligned}
I &=\sum_{n=0}^{\infty}(-1)^{n} \int_{0}^{1} y^{2 n} \ln y d y \\
&=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1} \int_{0}^{1} \ln y d\left(y^{2 n+1}\right) \\
&\left.=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1}\left(\left[y^{2 n+1} \ln y\right]_{0}^{1}-\int_{0}^{1} y^{2 n+1} \cdot \frac{1}{y} d y\right)\right) \\
&=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{2 n+1}\left(-\int_{0}^{1} y^{2 n} d y\right) \\
&=-\sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)^{2}} \\
&=-G.
\end{aligned}
$$
where $G$ is the Catalan’s constant.
For the second integral in the question, we use the identity $$
\ln (\sin x)=\ln (\tan x)+\ln (\cos x), $$
we have
$$\int_{0}^{\frac{\pi}{4}} \ln (\sin x) d x=\int_{0}^{\frac{\pi}{4}} \ln (\tan x) d x+ +\int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x$$
By my post in Quora, $$\int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x=\frac{G}{2}-\frac{\pi}{4} \ln 2.$$
Now we conclude that $$
\int_{0}^{\frac{\pi}{4}} \ln (\sin x) d x=-G+\left(\frac{G}{2}-\frac{\pi}{4} \ln 2\right)=-\frac{G}{2}-\frac{\pi}{4} \ln 2
$$
:|D Wish you enjoy the solution!
Is there any other simpler method to deal with the integral?
| We can also find the integrals using the Fourier series of $\ln (\sin x)$ on $(0, \pi)$,
$$
\ln (\sin x)=-\ln 2-\sum_{n=1}^{\infty} \frac{1}{n} \cos (2 n x) \quad \forall x \in(0, \pi).
$$
Integrating both sides from $0$ to $\dfrac{\pi}{4} $ yields
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{4}} \ln (\sin x) d x &=-\int_{0}^{\frac{\pi}{4}} \ln 2 d x-\sum_{x=1}^{\infty} \frac{1}{n} \int_{0}^{\frac{\pi}{4}} \cos (2 n x) d x \\
&=-\frac{\pi}{4} \ln 2-\sum_{n=1}^{\infty} \frac{1}{n}\left[\frac{\sin 2 n x}{2 n}\right]_{0}^{\frac{\pi}{4}} \\
&=-\frac{\pi}{4} \ln 2-\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2}} \sin \left(\frac{n \pi}{2}\right) \\
&=-\frac{\pi}{4} \ln 2-\frac{1}{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2 n+1)^{2}} \\
&=-\frac{\pi}{4} \ln 2-\frac{1}{2} G
\end{aligned}
$$
Similarly, using the result in my post in Quora, $$
\begin{aligned}
\int_{0}^{\frac{\pi}{4}} \ln (\tan x) d x &=\int_{0}^{\frac{\pi}{4}} \ln (\sin x) d x-\int_{0}^{\frac{\pi}{4}} \ln (\cos x) d x \\
&=\left(-\frac{\pi}{4} \ln 2-\frac{1}{2} G\right)-\left(\frac{G}{2}-\frac{\pi}{4} \ln 2\right) \\
&=-G
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4350260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Power series solution of $y'-y=x^2$ $y'-y=x^2$
I try to get a power series solution.
$y(x)=\sum_{n=0}^{\infty}a_{n}X^{n},y'(x)=\sum_{n=1}^{\infty}na_{n}X^{n-1}=\sum_{n=0}^{\infty}(n+1)a_{n+1}X^{n}$.
$\sum_{n=0}^{\infty}(n+1)a_{n+1}X^{n}-\sum_{n=0}^{\infty}a_{n}X^{n}=X^{2}$
Then $ \space a_{n+1}=\frac{a_n}{n+1} \space \forall n\neq 2$
$n=2 \implies 3a_3-a_2=1$
Then $a_1=a_0, a_2=\frac{a_0}{2}, a_3=\frac{2+a_0}{6}, a_4=\frac{2+a_0}{4!}, a_5=\frac{2+a_0}{24\cdot 4}$
Then the solution is $y=a_0+a_0x^1+\frac{a_0}{2}x^2+\frac{2+a_0}{6}x^3+\frac{2+a_0}{4!}x^4+\frac{2+a_0}{24\cdot 4}x^5 \cdots$
Is it correct? Am I supposed to find $a_0$ ?
Thanks!
| Maxima (Linux version of Macsyma) gives this solution to your
differential equation: $y(x) = C\exp(x)-(x^2+2x+2)$. You need a start value to get the integration constant $'C'$. Then you can proceed with the standard power series for $\exp(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4354247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{4b^2+b+1}{4\left|b\right|}\ge \sqrt{b+1}$, $b\ge -1,\:b\ne 0$ I've literally tried everything for this question. I think I'm supposed to use the Arithmetic-Geometric Mean Inequality (AGM) but I don't know how to apply AGM to this question so I tried algebraic manipulation and it feels like I'm getting nowhere.
$\frac{4b^2+b+1}{4\left|b\right|}\ge \sqrt{b+1}$
$\frac{\left(4b^2+b+1\right)^2}{16b^2}\ge b+1$
$b^2+\frac{b}{2}+\frac{9}{16}+\frac{1}{8b}+\frac{1}{16b^2}\ge b+1$
$b^2-\frac{b}{2}-\frac{7}{16}+\frac{1}{8b}+\frac{1}{16b^2}\ge 0$
$\frac{16b^4+2b-7b^2-8b^3+1}{16b^2}\ge 0$
$16b^4+2b-7b^2-8b^3+1\ge 0$
I don't know where to go from here.
I don't want a complete answer, just give me a hint, thanks.
EDIT: I think I figured it out:
AGM = $\frac{x+y}{\:2}\ge \sqrt{xy}$
= $\frac{x+y}{\:2}\ge \sqrt{x}\sqrt{y}$
Let x = $4b^2$ and y = $b+1$
Then we have, by AGM:
$\frac{\left(4b^2\right)+\left(b+1\right)}{2}\ge \sqrt{4b^2}\sqrt{b+1}$
$\frac{4b^2+b+1}{2}\ge 2\left|b\right|\sqrt{b+1}$
$\frac{\frac{4b^2+b+1}{2}}{2\left|b\right|}\ge \sqrt{b+1}$
$\frac{4b^2+b+1}{4\left|b\right|}\ge \sqrt{b+1}$, as needed. QED
| Hint: Remember that $x^2 =|x|^2$ and write LHS as $${4b^2\over 4|b|} +{b+1\over 4|b|}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the area inside the curve $r^2=2\cos(5\theta)$ and outside the unit circle. I found the area of one full rose-petal($A_1$) and the area enclosed by the petal and the unit circle($A_2$), subtracted these from one another to get the area enclosed by the curve outside of the circle, and multiplied the answer by the amount of rose pedals:
$f(\theta)=\pm\sqrt(2\cos(5\theta))$
$g(\theta)=1$
$f(\theta)= g(\theta) \Leftrightarrow \theta = {\pm\frac{\pi}{15}}$
$\cos(\theta)=0 \Leftrightarrow \theta = {\pm\frac{\pi}{10}}$
$$A_{1}=\frac{1}{2}\int_{-\frac{\pi}{15}}^{\frac{\pi}{15}}2\cos(5\theta)d\theta$$
$$A_{2}=\frac{1}{2}\int_{-\frac{\pi}{10}}^{-\frac{\pi}{15}}2\cos(5\theta)d\theta+\frac{1}{2}\int_{-\frac{\pi}{15}}^{\frac{\pi}{15}}1^2d\theta+\frac{1}{2}\int_{\frac{\pi}{15}}^{\frac{\pi}{10}}2\cos(5\theta)d\theta$$
$A = 6*(A_1-A_2)=\frac{-2}{5\pi} - \frac{1}{5}(-12\sqrt3 + 12)$, not the correct answer?
| The intersection points of one of the petals of the curve $r^2 = \sqrt{2 \cos(5\theta)}$ and unit circle $r = 1$ are indeed $r = 1, \theta = \pm \frac{\pi}{15}$.
But as you need area outside the unit circle, the integral should be
$A = 6 \cdot \displaystyle \frac 12 \int_{-\pi/15}^{\pi/15} \left(2 \cos(5\theta) - 1\right) ~ d\theta$
$ = \dfrac{6 \sqrt3 - 2 \pi}{5}$
Mistake in your work is in the calculation of $A_1$. It should be,
$ \displaystyle \frac{1}{2}\int_{-\frac{\pi}{10}}^{\frac{\pi}{10}}2 \cos(5\theta)~d\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A good explanation about this limit (Hierarchy of Infinities) I am searching for a good explanation (whether it be analytical or graphical) for why if I have this limit
$$\lim_{x\to+\infty} \sqrt{x^2 - 2x + 1} - x$$
I cannot use the hierarchy of infinities like
$$\sqrt{x^2\left(1 - \frac{2}{x} + \frac{1}{x^2}\right)} - x \longrightarrow \sqrt{x^2} - x \longrightarrow x - x = 0$$
I know the initial limit shows an indeterminate form $\infty - \infty$, but I also know hierarchy of infinities shall work too. In any case, I know the limit must be computed in such a way to remove the minus sign (and hence the indeterminate form) like
$$\lim_{x\to+\infty} \sqrt{x^2 - 2x + 1} - x \cdot \dfrac{\sqrt{x^2 - 2x + 1} + x}{\sqrt{x^2 - 2x + 1} + x} = \lim_{x\to+\infty} \dfrac{x^2 - 2x + 1 - x^2}{\sqrt{x^2 - 2x + 1} + x}$$
Which leads to
$$\lim_{x\to+\infty} \frac{1 - 2x}{\sqrt{x^2 - 2x + 1} +x} = \lim_{x\to+\infty} \frac{1-2x}{2x} = -1$$
The questions hence are:
$\bullet$ Why the initial method leads to a wrong answer?
$\bullet$ Is the "correct" method just a trick used to remove the minus sign, in order to have a sum and hence no indetermination problem?
Thank you!
| The other answers already explain why the initial method does not work, so I will only give an easiest way to derive this limit : just notice that for $x > 1$, one has
$$\sqrt{x^2 - 2x + 1} - x = \sqrt{(x-1)^2} - x = -1$$
So $$\boxed{\lim_{x\to+\infty} \sqrt{x^2 - 2x + 1} - x = -1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving an infinite alternating product Idea is to alternately multiply infinite many values from $1$ to $x$, with const distance $h$.
Reweriting it to math:
$$
h = \frac{x-1}{n}
$$
$$
f(x) = \lim_{n \to \infty} \left( \prod_{m=0}^{n} (x - mh)^{(-1)^{m}} \right)
$$
I have played with the idea in Python and it seems for me that:
$$
f(x) = \sqrt{x}
$$
$$
x \gt 0
$$
I see some similarities with function for calculating the length $L$ of some curve $f(x)$ but there are two main differences:
*
*$\prod$ instead of $\sum$
*alterantion $(-1)^m$
How I can prove it ?
| $$P_n=\prod_{m=0}^n\left(x-\frac{m }{n}(x-1)\right)^{(-1)^m}$$
$$P_n=\frac{n x \left(\frac{2(1- x)}{n}+x\right) \left(\frac{2(1-x)}{n}\right)^{\left\lceil
\frac{1-n}{2}\right\rceil +\left\lfloor \frac{n}{2}\right\rfloor -1}
\left(\frac{(n-4) x+4}{2(1- x)}\right)_{\left\lfloor \frac{n}{2}\right\rfloor
-1}}{((n-1) x+1) \left(\frac{(n-3) x+3}{2(1- x)}\right)_{\left\lfloor
\frac{n-1}{2}\right\rfloor }}$$ is not very pleasant to work with.
However
$$P_{2n+1}=\frac{\Gamma \left(\frac{n x}{2 (1-x)}+\frac{1}{2}\right) \Gamma \left(\frac{nx }{2
(1-x)}+n+1\right)}{\Gamma \left(\frac{n x}{2 (1-x)}\right) \Gamma \left(\frac{nx
}{2 (1-x)}+n+\frac{3}{2}\right)}$$
Taking logarithms, using four times Stirling approximation and continuing with Taylor series for large values of $n$
$$\log(P_{2n+1})=\frac{1}{2} \log \left(\frac{x}{2-x}\right)-\frac{1-x^2}{2 (2-x) x}\frac 1n+O\left(\frac{1}{n^2}\right)$$
So, for large values of $n$
$$\color{blue}{P_{2n+1}\sim \sqrt{\frac{x}{2-x}}}$$
On the other hand
$$P_{2n}=\frac{x ((n-2) x+2) \left(\frac{(n-4) x+4}{2-2 x}\right)_{n-1}}{((n-1) x+1)
\left(\frac{(n-3) x+3}{2-2 x}\right)_{n-1}}$$
$$\log(P_{2n})=\frac{1}{2} \log (x(2-x) )-\frac{(x-1)^2}{2 (2-x) x}\frac 1n+O\left(\frac{1}{n^3}\right)$$
So, for large values of $n$
$$\color{blue}{P_{2n}\sim \sqrt{(2-x)x}}$$
Combining the two series
$$\log(P_{2n}\,P_{2n+1})=\log(x)+\frac{x-1}{4 n x}-\frac{x^2-1}{16 n^2 x}+O\left(\frac{1}{n^3}\right)$$
$$\large\color{red}{\sqrt{P_{2n} \,P_{2n+1}}\sim \sqrt{x}\exp\Big[\frac{x-1}{8 n x} \Big] \quad \to \quad \sqrt{x}}$$
Trying for $x=\frac 12$ and $n=100$, the exact value is
$$\sqrt{P_{200} \,P_{201}}=\frac 12\sqrt{ \frac{401}{804}}=0.7062267$$ while
$$\frac{1}{\sqrt{2} \sqrt[800]{e}}=0.7062234$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding a linear combination of a linear combination of vectors Problem
Assume $\vec x, \ \vec y$ are linearly independent.
Let $\vec a = 3\vec x + 6\vec y$ and $\vec b = 9 \vec x + 21\vec y$.
Find an expression for $\vec x + \vec y$ as a linear combination of $\vec a, \vec b$.
I.e., find $c_1, \ c_2 \ \in \ \mathbb R$ such that $\vec x + \vec y = c_1\vec a + c_2\vec b$.
Progress
I have two different answers from two different methods, and I'm not sure which one is right (if any).
Method 1
(Assume all variables are vectors so I don't have to type \vec a bajillion times.)
We have the equations
$$\begin{align}
3x + 6y &= a \\
9x + 21y &= b
\end{align}$$
This gives us the coefficient matrix
$$\begin{bmatrix}
3 & 6 & a \\
9 & 21 & b
\end{bmatrix}$$
whose reduction I've found to be
$$\begin{bmatrix}
1 & 0 & -\frac53a + \frac23b \\
0 & 1 & -a + \frac13 b
\end{bmatrix}$$
Adding the two rows, we get the equation
$$x+y = -\frac83a + b$$
which is my first attempt.
Method 2
This time I used the initial equations
$$\begin{align}
\vec a &= 3x + 6y \\
\vec b &= 9x + 21y\end{align}$$
and found $3a = 9x + 18y$ so that $b-3a = 3y \ \Rightarrow \ y = -a + \frac13b$ by eliminating the $x$ term
and then doing the same with the $y$ term by adding $-\frac72$ of the first equation to the second, giving $x = \frac73 a - \frac32 b$.
Now
$$x+y = \frac43a - \frac76 b$$
which is different from the result of method 1.
Question
One (or both) of these must be wrong.
*
*Is there a way to verify a potential solution here like we do with regular systems of equations?
*Which of my methods went awry? I smell careless mistakes in both of them to be honest.
| Eliminate the $x$ term in row two by replacing row two with (row two minus three times row one).
$$
\boxed{\begin{align*}
3x + 6y &= a \\
(9x + 21y) - 3(3x + 6y) &= (b) - 3(a)
\end{align*}}
\implies
\boxed{\begin{align*}
3x + 6y &= a \\
3y &= -3a + b
\end{align*}}
$$
Conveniently, the coefficients of $x$ and $y$ will match if we just replace row one with (row one minus row two).
$$
\boxed{\begin{align*}
(3x + 6y) - (3y) &= (a) - (-3a + b) \\
3y &= a - 3b
\end{align*}}
\implies
\boxed{\begin{align*}
3x + 3y &= 4a - b \\
3y &= a - 3b
\end{align*}}
$$
Finally, scale row one by a factor of a third to obtain:
$$
\boxed{\begin{align*}
\frac{1}{3}(3x + 3y) &= \frac{1}{3}(4a - b) \\
3y &= a - 3b
\end{align*}}
\implies
\boxed{\begin{align*}
x + y &= \frac{4}{3}a - \frac{1}{3} b \\
3y &= a - 3b
\end{align*}}
$$
We can check by substitution:
\begin{align*}
\frac{4}{3}a - \frac{1}{3} b
&= \frac{4}{3}(3x + 6y) - \frac{1}{3} (9x + 21y) \\
&= (4x + 8y) + (-3x + -7y) \\
&= x + y
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ equals $(a+b)/2$ If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ is equal $(a+b)/2$
Multiplying by the conjugate, simplifying and clearing I arrive at the following expression
$$\frac{n}{\sqrt{(n+a)(n+b)}+n} < \frac{\epsilon+(a+b)-ab}{(a+b)}$$
| Another fun solution involves the HM-GM-AM-Inequality:
By the mentioned inequality, we obtain
$$
\frac{2}{\frac1{n+a} + \frac1{n+b}} \leq \sqrt{(n+a)(n+b)} \leq \frac{n+a+n+b}2
$$
Now note that
$$
\frac{2}{\frac1{n+a} + \frac1{n+b}} = \frac{2(n+a)(n+b)}{n+a+n+b} = \frac{n (2n + a + b) + n(a+b)+ab}{2n+a+b} \geq n + \frac{n(a+b)}{2n+a+b}.
$$
By L'Hospital's rule, we have $\lim_{n \rightarrow \infty}\frac{n(a+b)}{2n+a+b} = \frac{a+b}2$. Letting $n \rightarrow \infty$ and subtracting $n$
in the original inequality, this yields
$$
\frac{a+b}2 \leq \lim_{n \rightarrow \infty} \left( \sqrt{(n+a)(n+b)}-n \right) \leq \frac{a+b}2
$$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Are there any formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}$, where $m$ and $n$ are natural numbers and $a>0$? As mentioned in my post, I started to investigate the integral $$
I(m,n,a)=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}},
$$
where $m$ and $n$ are natural number and $a$ is positive.
First of all, let’s start with the ‘simple’ case, $$
I(1, n, 1)=\int_{0}^{\infty} \frac{d x}{x^{n}+1}.
$$
However, $\displaystyle \int_{0}^{\infty} \frac{d x}{x^{n}+1}$ is itself not simple. I was forced to use a ready made formula
$$
\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).
$$
which I can’t prove by elementary method yet. Please tell me if you have any.
Then $$I(m,n,a) =\int_{0}^{\infty} \frac{\sqrt[n]{a} d\left(\frac{x}{\sqrt[n]{a}}\right)}{a\left[\left(\frac{x}{\sqrt[n]{a}}\right)^{n}+1\right]}
=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right) a^{-\frac{n-1}{n}}$$
Now differentiating $I(1, n, a)$ w.r.t. $a$ by $(n-1)$ times yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{(-1)^{m-1}(m-1) ! d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left(-\frac{n-1}{n}\right)\left(-\frac{2 n-1}{n}\right)\left(-\frac{3 n-1}{n}\right)
\cdots\left(-\frac{m n-n-1}{n}\right) a^{-\frac{m n-1}{n}}
\end{aligned}
$$
Rearranging and simplifying yields
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} a^{\frac{m n-1}{n}} }\prod_{k=1}^{m-1}(k n-1)}
$$
Putting $a=1$ gives our formula
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}}=\frac{\pi \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} }\prod_{k=1}^{m-1}(k n-1)}
$$
For verification, $$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{3}+1\right)^{10}} =\frac{\pi \csc \left(\frac{\pi}{3}\right)}{9 ! 3^{10}} \cdot 2 \cdot 5 \cdot 8 \cdot 11 \cdot 14 \cdot 17 \cdot 20 \cdot 23 \cdot 26
=\frac{1118260 \pi}{4782969 \sqrt{3}},
\end{aligned}
$$
which is checked by Wolframalpha
$$\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{12}+1\right)^{5}}&=\frac{\pi \csc \left(\frac{\pi}{12}\right)}{4 ! 12^{5}} \cdot 11 \cdot 23 \cdot 35 \cdot 47\\
&=\frac{416185 \pi(\sqrt{6}+\sqrt{2})}{5971968} \\
&\doteq 0.845906950943631,
\end{aligned}
$$
which is checked by Wolframalpha
My question is whether we can prove the formula without using $\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).$
| Glad to see that there are several nice solutions to the problem, now I want to add one more by discovering a reduction formula using integration by parts.
In order to use integration by parts, I applied the inversion substitution to the original integral.
For any fixed natural number $n$, we now define
$$
J_{m}:=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}} \stackrel{x \rightarrow \frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{m n-2}}{\left(x^{n}+1\right)^{m}} d x
$$
Noting that $$
\frac{d}{d x}\left(\frac{1}{\left(x^{n}+1\right)^{m-1}}\right)=-(m-1) \cdot \frac{nx^{n-1}}{\left(x^{n}+1\right)^{m}},
$$
we have
$$J_m=-\frac{1}{n(m-1)} \int_{0}^{\infty} x^{m n-n-1} d\left(\frac{1}{\left(x^{n}+1\right)^{m-1}}\right)
$$
By integration by parts, we get
$$
\begin{aligned}
J_{m} &\left.=-\frac{m-1}{n}\left[\frac{x^{m n-n-1}}{\left(x^{n}+1\right)^{m-1}}\right]_{0}^{\infty}+\frac{1}{n(m-1)} \int_{0}^{\infty} \frac{(m n-n-1) x^{m n-1-2}}{\left(x^{n}+1\right)^{m-1}}\right] \\
&=\frac{m n-n-1}{n(m-1)} \int_{0}^{\infty} \frac{x^{m n-n+2}}{\left(x^{n}+1\right)^{m-1}} d x\\&= \frac{\left(m-1-\frac{1}{n}\right)}{m-1} J_{m-1}
\end{aligned}
$$
Applying the reduction formula $m$ times yields inductively
\begin{aligned}
J_m&=\frac{m-1-\frac{1}{n}}{m-1} \cdot \frac{m-2-\frac{1}{n}}{m-2} \cdot \frac{m-3-\frac{1}{n}}{m-3} \cdot \frac{1-\frac{1}{n}}{1} J_{0} \\
& = \frac{1}{(m-1) !} \prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) J_{0}
\end{aligned}
Using the well-known formula,
$$
J_0=\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right),
$$
we can conclude that
$$\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}} =\frac{\pi \csc \left(\frac{\pi}{n}\right)}{n(m-1) !} \prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
In $\frac{1}{a^2-x^2}$, how can $a\sin(\theta)$ be substituted for $x$ when finding the anti-derivative? $$\int{\frac{1}{a^2-x^2}}dx\tag{1}$$
When finding the above anti-derivative, $x$ is substituted with $a\sin(\theta)$. However, the range of $x$ is $\mathbb{R}-\{-a,a\}$ while the range of $a\sin(\theta)$ is $[-a,a]$.
Graph of $\frac{1}{a^2-x^2}[a=0.5]$
Needless to say,
$$\mathbb{R}-\{-a,a\}\ne[-a,a]$$
So, how can $a\sin\theta$ be substituted for $x$ in $(1)$ when the range of $x$ and $a\sin\theta$ are not the same?
Related
| When $x\in (a, +\infty)$, you can substitute $x = a \cosh \theta$, thus
\begin{equation}
\int \frac{d x}{a^2 - x^2} = \int\frac{1}{a^2}\frac{a\sinh \theta d\theta}{1 -\cosh^2 \theta}
= -\frac{1}{a}\int\frac{d\theta}{\sinh\theta}
\end{equation}
You can then use $u = e^\theta$ or $t = \tanh\frac{\theta}{2}$.
Of course, you could decompose directly from the beginning
\begin{equation}
\frac{1}{a^2-x^2} = \frac{1}{2a}\left(\frac{1}{a-x} + \frac{1}{a+x}\right)
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4367289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $\sqrt{3}a \cos x + 2b \sin x=c$ has 2 distinct real roots, then what is the range of $2c/(3a+2b)$? Let $a$,$b$,$c$ be three non-zero real numbers such that the equation $\sqrt{3}a\cos x + 2b \sin x = c$ ,($x$ belongs to $[-\pi/2,\pi/2]$) has two distinct real roots m and n with $m+n = \pi/3$. If range of values of $2c/(3a+2b)$ is $[q, r)$ then what is the value of $q+r$?
I tried to solve this question by putting $m$ and $n$ in the given equation and then subtracting the two equations. On solving I got the relation between $a$ and $b$ as $a=2b$.
On substituting the value of b in the given equation, and simplifying, I got
$\sin(x+\pi/3) = c/(2a)$
Also, $2c/(3a+2b) =c/(2a)$ (as $2b=a$)
For $x$ belongs to $[-\pi/2, \pi/2]$, $\sin(x+\pi/3)$ ranges from $-1/2$ to $1/2$. So, $q+r=0$.
But the answer is given as $1.5$. I'm unable to understand how? Someone please help me out with this question. Thank you
| Well... think it reversely. If we write the two roots in advance, can we determine $a,b,c$? Of course, we can.
Let the roots to be $m,n$ so we have
$$\sqrt{3}a\cos m + 2b \sin m = c, \sqrt{3}a\cos n + 2b \sin n = c.$$
So we have $$\sqrt{3}a(\cos m-\cos n) + 2b (\sin m-\sin n)=0.$$
And using sum-to-difference we have $$\sqrt{3}a(\sin \frac{m-n}2\sin\frac{m+n}2) + 2b (\sin \frac{m-n}2\cos\frac{m+n}2)=0.$$
Since $m\ne n$, and $m+n=\frac\pi 3$ we have $$\sqrt{3}a(\frac 12) + 2b (\frac {\sqrt 3}2)=0.$$
or we have $a=-2b$.
So we have $3a+2b=2a$ and $$c/a=\sqrt{3}\cos m - \sin m.$$
And thus $$2c/(3a+2b)=2c/2a=c/a=\sqrt{3}\cos m - \sin m.$$
Notice that $m+n=\frac\pi 3$ and $m,n\le\frac\pi 2$, WLOG $m> n$, we have $\frac\pi6<m\le\frac\pi2$, and $\sqrt{3}\cos m - \sin m$ is clearly decreasing. So the range is $[-1,0)$. So $q+r=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4369197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If you are given the values of m, $2^a$ mod m, and $2^b$ mod m, is there a way to find the value of $2^{ab}$ mod m? If you are given the values of m, $2^a$ mod m, and $2^b$ mod m, is there a way to find the value of $2^{ab}$ mod m?
| Yes.
Let us first assume that $n$ is odd. If you know $2^a \pmod n$, $2^b \pmod n$, and $n$ [but not the integers $a$ and $b$], then let $a'$ and $b'$ be any two integers that satisfy the equations $2^{a'} \pmod n =2^{a} \pmod n$ and $2^{b'} \pmod n = 2^b \pmod n$. [Now, finding such integers $a'$ and $b'$, can be done even if it means checking $2^{f}\pmod n$ for $\varphi(n)$ values of $f$. That suffices for here if I am understanding correctly; I'm not sure if there is a much faster way.] Then once $a'$ and $b'$ are found, then the integers $a$ and $b$, whatever they may be, must satisfy $a'=a + c$ and $b'=b+d$, for some integers $c$ and $d$ where $c$ and $d$ satisfy $2^c \pmod n =2^d \pmod n =1$.
Then $$2^{a'b'} \pmod n = 2^{(a+c)(b+d)}\pmod n$$
$$= 2^{ab}(2^c)^{b+d}(2^d)^{a} \pmod n= 2^{ab} \pmod n.$$
So find integers $a'$, $b'$ such that $2^{a'}\pmod n = 2^a \pmod n$, and $2^{b'}\pmod n = 2^b \pmod n$, any such $a'$, $b'$ will do. Then $2^{ab}\pmod n$ must be $2^{a'b'}\pmod n$.
For $n$ even write $n=2^gm$ where $m$ is odd and $g$ is integral.
Then given $2^a \pmod n$ and $2^b \pmod n$ one has $2^a \pmod {2^g}$ and $2^b \pmod {2^g}$; if either $2^a \pmod {2^g}$ or $2^b \pmod {2^g}$ is $1$ then either $a$ or $b$ must be $0$ so $ab$ must be $0$, so $2^{ab}\pmod{2^g}$ must also be $0$.
Otherwise $a$ and $b$ must both be positive integers [since here $2^a \pmod {2^g} \not =1$ and $2^b \pmod {2^g} \not =1$]. So if either $2^a \pmod {2^g}$ or $2^b \pmod {2^g}$ is $0$ then either $a$ or $b$ is at least $g$. And so as both $a$ and $b$ are positive integers, it follows that $ab \ge \max\{a,b\}$ and so $2^{ab} \pmod {2^g}$ must also be $0$. Otherwise the one remaining case is both $2^a \pmod {2^g}$, $2^b \pmod {2^g}$ $\not \in \{0,1\}$; from this we can recover $a$ and $b$ directly to get $ab$ and thus $2^{ab}\pmod {2^g}$.
Given $2^a \pmod n$ and $2^b \pmod n$ we can recover $2^a \pmod m$ and $2^b \pmod m$. As $m$ is odd, from the case when $n$ is odd, we can recover $2^{ab}\pmod m$.
Given $2^{ab}\pmod {2^g}$ and $2^{ab}\pmod m$ we get $2^{ab}\pmod n$ by the Chinese Remainder Thm.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4371001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why should I only pick the positive number for $C$ in this IVP (separable differential equation)? I'm practicing some problems for an upcoming DEs test. I tried the following initial value problem:
$\displaystyle\frac{1}{2}\displaystyle\frac{dy}{dx}=\sqrt{y+1}\cos x,y(\pi)=0$
Here's my work:
$\begin{align}
\displaystyle\frac{1}{2}\displaystyle\frac{dy}{dx}&=\sqrt{y+1}\cos x \\
\int\displaystyle\frac{dy}{\sqrt{y+1}}&=\int 2\cos x\,dx \\
2\sqrt{y+1}&=2\sin x+c_1 \\
\sqrt{y+1}&=\sin x + C \\
y+1&=(\sin x + C)^2
\end{align}$
Plugging in the initial value I get:
$\begin{align}
0+1&=(\sin\pi+C)^2\\
C^2&=1 \\
C&=\pm 1
\end{align}$
So the final answer I got was $y=(\sin x \pm 1)^2-1$, but the solution in the textbook was $y=(\sin x +1)^2-1$. Why did they only pick $+1$ as an answer for $C$?
| First, note that since $\dfrac{dy}{dx}=2\sqrt{y+1}\cos(x)$ is a function which is continuous on an open subset of the domain of $f(x)=2\sqrt{y+1}\cos(x)$ containing the point $(\pi,0)$ and so is $\partial f/\partial y$, we are guaranteed that there exists a unique solution containing the point $(\pi,0)$. So it cannot be the case that both $y=(\sin(x)+1)^2+1$ and $y=(\sin(x)-1)^2+1$ are solutions.
Both functions satisfy the initial condition in that the point $(\pi,0)$ lies on the graphs of both.
Consider the choice C=1.
Let $y=(\sin(x)+1)^2-1=\sin^2(x)+2\sin(x)$
\begin{eqnarray}\frac{dy}{dx}&=&2\sin(x)\cos(x)+2\cos(x)\\
\frac{1}{2}\frac{dy}{dx}&=&(\sin(x)+1)\cos(x)\\
&=& |\sin(x)+1|\cos(x)\\
&=&\sqrt{\sin^2(x)-2\sin(x)+1}\cos(x)\\
&=&\sqrt{y+1}\cos(x)\end{eqnarray}
Now, consider the choice $C=-1$.
Let $y=(\sin(x)-1)^2-1$
Then \begin{eqnarray}\frac{dy}{dx}&=&2\sin(x)\cos(x)-2\cos(x)\\
\frac{1}{2}\frac{dy}{dx}&=&(\sin(x)-1)\cos(x)\\
&=& -|\sin(x)-1|\cos(x)\\
&=&-\sqrt{\sin^2(x)-2\sin(x)+1}\cos(x)\\
&=&-\sqrt{y+1}\cos(x)\end{eqnarray}
Which does not satisfy the differential equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $f(x)=x^2-2$ with $x\in [-2,2]$. Show that the equation $f^{n}(x)=x$ has $2^{n}$ real roots. [Where $f^{n}(x)=f(f^{n-1}(x))$] Let $f(x)=x^2-2$ with $x\in [-2,2]$. Show that the equation $f^{n}(x)=x$ has $2^{n}$ real roots. [Where $f^{n}(x)=f(f^{n-1}(x))$]
My solution:
Let $x=2\cos(\theta)$ for $\theta\in [0,\pi]$
$\implies$
$f(x)=4\cos^2 (\theta)-2=2\cos(2\theta)$
$f(f(x))=(f(x))^2-2=(2\cos^22\theta)^2-2=2(2\cos^{2}2\theta-1)=2\cos(2^2 \theta)$
$\implies$$f(f(x))=2\cos(4\theta)$
Similarly $f(f(f(x)))=2\cos(2^3\theta)$
$\vdots$
$\underbrace {f\circ f\circ\cdots \circ f}_\text{n times}(x)=2\cos(2^n \theta)$
$\implies$
From Question i.e, $f^{n}(x)=x$
$\implies$
$2\cos(2^n\theta)=2\cos\theta$
$\implies$
$2^n\theta=2m\pi \pm \theta$
$\implies$
$\theta=\dfrac{2m\pi}{2^{n}-1}$ $\quad$ or $\quad$ $\theta=\dfrac{2m\pi}{2^{n}+1}$ $\forall \; \theta \in [0,\pi]$
Checking Result for $n=1$
$\theta=\dfrac{2m\pi}{1}\;$ or $\; \theta=\dfrac{2m\pi}{3}$
$\implies$ $\theta = \dfrac{2\pi}{3}, 0$ i.e., $2$ solution.
Checking Result for $n=2$
$\theta=\dfrac{2m\pi}{3}\;$ or $\; \theta=\dfrac{2m\pi}{5}$
$\implies$ $0, \dfrac{2\pi}{3},\dfrac{2\pi}{5},\dfrac{4\pi}{5}$ i.e. $4$ solution.
$\implies$ There are $2^{n}$ distinct root of $f^{n}(x)=x\;$ equation.
Is my Solution Correct?
| I think that your idea is very nice.
I would add proofs for the following claims :
Claim 1 : $f^{n}(2\cos\theta)=2\cos(2^n\theta)$ (your idea is very nice, but I would prove this claim by induction rigorously)
Claim 2 : The number of integers $m$ such that $0\leqslant\dfrac{2m\pi}{2^{n}-1}\leqslant\pi$ is $2^{n-1}$, and the number of integers $M$ such that $0\leqslant\dfrac{2M\pi}{2^{n}+1}\leqslant\pi$ is $2^{n-1}+1$.
Claim 3 : $\dfrac{2m\pi}{2^{n}-1}=\dfrac{2M\pi}{2^{n}+1}$ holds if and only if $m=M=0$.
(It follows from Claim 2 and 3 that $f^n(x)=x$ has $2^{n-1}+(2^{n-1}+1)-1=2^n$ real roots.)
Claim 1 : $f^{n}(2\cos\theta)=2\cos(2^n\theta)$.
Proof : If $n=1$, $f(2\cos\theta)=(2\cos\theta)^2-2=2(2\cos^2\theta-1)=2\cos(2\theta)$. Suppose that $f^{n}(2\cos\theta)=2\cos(2^n\theta)$. Then, $f^{n+1}(2\cos\theta)=(f^n(2\cos\theta))^2-2=2(2\cos^2(2^n\theta)-1)=2\cos(2^{n+1}\theta)$.$\ \square$
Claim 2 : The number of integers $m$ such that $0\leqslant\dfrac{2m\pi}{2^{n}-1}\leqslant\pi$ is $2^{n-1}$, and the number of integers $M$ such that $0\leqslant\dfrac{2M\pi}{2^{n}+1}\leqslant\pi$ is $2^{n-1}+1$.
Proof : Since $0\leqslant\dfrac{2m\pi}{2^{n}-1}\leqslant\pi\iff 0\leqslant m\leqslant 2^{n-1}-1$, the number of such integers $m$ is $2^{n-1}$. Since $0\leqslant\dfrac{2M\pi}{2^{n}+1}\leqslant\pi\iff 0\leqslant M\leqslant 2^{n-1}$, the number of such integers $M$ is $2^{n-1}+1$.$\ \square$
Claim 3 : $\dfrac{2m\pi}{2^{n}-1}=\dfrac{2M\pi}{2^{n}+1}$ holds if and only if $m=M=0$.
Proof : Since $m=\dfrac{(2^n-1)M}{2^n+1}$ with $\gcd(2^n-1,2^n+1)=\gcd(2^n-1,2)=1$, $M$ has to be a multiple of $2^n+1$. It follows from $0\leqslant M\leqslant 2^{n-1}$ that $M=m=0$.$\ \square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculate the limit: $ \lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}$
Calculate the limit:
$$\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}
$$
I tried to change to polar coordinates like that:
\begin{align*}
\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}} &=
\lim\limits_{(x,y)\to(0,0)}{\frac{e^{-\frac{1}{x^2+y^2}}}{(x^2+y^2)^2-2x^2y^2}}\\
&=\lim\limits_{r\to0}{\frac{e^{-\frac{1}{r^2}}}{r^4(1-2\cos^2\theta \sin^2\theta)}}\\
\end{align*}
and I'm not sure how to continue from this point.
Thank you!
| Hint. For now, suppose that the limit exists and is equal to $L$. Then,
$$L = \lim\limits_{r\to0}{\frac{e^{-\frac{1}{r^2}}}{r^4(1-2\cos^2\theta \sin^2\theta)}}$$
Take $\theta = \frac\pi 2$ first. Then, $$L = \lim\limits_{r\to0}{\frac{e^{-\frac{1}{r^2}}}{r^4}}$$
Now, take $\theta = \frac\pi 4$. Then,
$$L = \lim\limits_{r\to0}{\frac{2e^{-\frac{1}{r^2}}}{r^4}} = 2L \implies L = 0$$
So, if the limit exists, it must be equal to $0$.
Proof of Existence.
Observe that $$\frac{e^{-1/r^2}}{r^4(\sin^4\theta+\cos^4\theta)} \leq \frac{2}{r^4e^{1/r^2}}$$
Now, since $r^5e^{1/r^2}\to \infty$ as $r\to 0^+$, there is $\delta'>0$ such that $$0<r<\delta' \implies r^5e^{1/r^2} > 2 \implies \frac{2}{r^4e^{1/r^2}} < r$$
Finally, choose $\delta = \min\{\delta',\epsilon\}$ to conclude that $$0<\sqrt{x^2+y^2}<\delta \implies 0 < r < \delta \implies \left|{\frac{e^{-\frac{1}{x^2+y^2}}}{x^4+y^4}}\right|<\epsilon$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Suppose that $x, y, z$ are three positive numbers that satisfy the equation $xyz=1, x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$.
Suppose that $x, y, z$ are three positive numbers that satisfy the equation $xyz=1, x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$. Then $z+\frac{1}{y}=\frac{m}{n}$, where $m$ and $n$ are coprime. Find $m+n+1$.
I tried this, $$y=29-\frac{1}{x}$$ $$y=\frac{29x-1}{x}$$
And, $$x+\frac{1}{z}=5$$ $$z=\frac{1}{5-x}$$
What can I do next?
| Another solution: we have $\frac{1}{x}=yz$, $\frac{1}{y}=xz$, and $\frac{1}{z}=xy$. So
\begin{align*}
x+xy&=5\\
y+yz&=29\\
z+xz&=Q
\end{align*}
where Q is the quantity we're trying to determine.
Multiplying these three equations together gives $145Q=xyz(x+1)(y+1)(z+1)$ and hence $145Q=(x+1)(y+1)(z+1)$.
Expanding, we have
\begin{align*}
(x+1)(y+1)(z+1)&=xyz+xy+yz+xz+x+y+z+1\\
&=2+xy+yz+xz+x+y+z\\
&=2+5+29+Q\\
&=36+Q
\end{align*}
and so $145Q=36+Q$, from which it follows that $Q=\frac{1}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int {\frac{{dx}}{{{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^2}}}} $ My first solution was:
$$\begin{array}{l}
\int {\dfrac{{dx}}{{{{(\sin \dfrac{x}{2} + \cos \dfrac{x}{2})}^2}}}} \\
= \int {\dfrac{{dx}}{{{{\sin }^2}\dfrac{x}{2} + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2}}}} \\
= \int {\dfrac{{dx}}{{1 + \sin x}}} \\
= \int {\dfrac{{1 - \sin x}}{{1 - {{\sin }^2}x}}dx} \\
= \int {\dfrac{1}{{1 - {{\sin }^2}x}}dx - \int {\dfrac{{\sin x}}{{1 - {{\sin }^2}x}}dx} } \\
= \int {\dfrac{{dx}}{{{{\cos }^2}x}}} - \int {\dfrac{{d(\cos x)}}{{{{\cos }^2}x}}} \\
= \tan x - \dfrac{1}{{\cos x}} + C
\end{array}$$
However, the answer from Wolfram Alpha is: $\dfrac{{ - 2}}{{\tan \dfrac{x}{2} + 1}} + C$, it tells me that the step:
$$\int {\dfrac{{1 - \sin x}}{{1 - {{\sin }^2}x}}dx} \\
= \int {\dfrac{1}{{1 - {{\sin }^2}x}}dx - \int {\dfrac{{\sin x}}{{1 - {{\sin }^2}x}}dx} }$$ is false.
What is the problem with that step? If it is correct then where is the error in my solution?
| $$\tan x-\frac{1}{\cos x}=\frac{2t}{1-t^2}-\frac{1+t^2}{1-t^2}-\frac{(1-t)^2}{(1+t)(1-t)}--\frac{1-t}{1+t}=1-\frac{2}{1+t}$$
where $t=\tan\frac12x$
so the answers are the same, modulo a constant
| {
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"url": "https://math.stackexchange.com/questions/4380188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve derivative/limit of $f(x)=x\sqrt{4-x^2}$ I'm trying to differentiate $x\sqrt{4-x^2}$ using the definition of derivative.
So it would be something like
$$\underset{h\to 0}{\text{lim}}\frac{(h+x) \sqrt{4-\left(h^2+2 h x+x^2\right)}-x \sqrt{4-x^2}}{h}$$
I was trying to solve and I just can end up with something like
$$\underset{h\to 0}{\text{lim}}\frac{(x+h)\sqrt{4-x^2-2xh-h^2}-x\sqrt{4-x^2}}h \cdot \frac{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}{\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$
$$\underset{h\to 0}{\text{lim}}\frac{-3x^2h-3xh^2+4h-h^3+\sqrt{4-x^2}-\sqrt{4-x^2-2xh+h^2}}{h\sqrt{4-x^2-2xh-h^2}+\sqrt{4-x^2}}$$
Now if I group on h, I will have some tricky 3 instead of 2.
The idea is I should have something like $h(2x^2+4)$ that would cancel up.
I'm quite stuck can I ask a little of help? I know wolframalpha exists but it refuses to create the step by step solution with the error "Ops we don't have a step by step solution for this query".
The final result shall be
$$-\frac{2 \left(x^2-2\right)}{\sqrt{4-x^2}}$$
| I think your algebra could look more like:
$$\begin{align}
&\frac{(x+h)\sqrt{4-(x+h)^2}-x\sqrt{4-x^2}}{h}\cdot\frac{(x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}}{(x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}}\\
&=\frac{(x+h)^2(4-(x+h)^2)-x^2(4-x^2)}{h\left((x+h)\sqrt{4-(x+h)^2}+x\sqrt{4-x^2}\right)}\\
\end{align}$$
This leaves no radicals in the numerator.
In the numerator, once this is multiplied out, all $h$-free terms will have canceled out through adding terms to their negatives.
Then you can factor $h$ from the top and cancel the $h$ in the denominator. Then it will be OK to just let $h\to0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Given $a+b=8$, $ax+by=9$, $ax^2+by^2=33$, $ax^3+by^3=60$, find $ax^4+by^4$
Given $$\begin{align}a+b&=8\\ax+by&=9\\ax^2+by^2&=33\\ax^3+by^3&=60\end{align}$$
Find $ax^4+by^4$
I really have no idea here.
Without coefficients $a$ and $b$, we can simply figure out $xy$ and find what we want by linear recurrences, but that doesn't seem to work here.
| Following the hint in comments, letting $x+y=s$ and $xy=p$
we can write the following linear system:
$$60=33s-9p\\33=9s-8p$$
Solving gives $s=1,p=-3$ and
$$ax^4+by^4=60s-33p=159$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Decomposing $\sum_{i = 0}^{2n} x^i$ as a simple sum of squares As we have $\sum_{i = 0}^{2n} x^i = (x^{2n + 1} - 1) / (x - 1)$, the polynomial is positive.
So we know that there is a decomposition as a sum of squares. Is there a closed simple form for such a decomposition? For small value of $n$ we have
$$ x^2 + x + 1 = 1/4 ((2x+1)^2+ 3)$$
$$ x^4 + x^3 + x^2 + x + 1 = 1/16((4 x^2 + 2 x + 1)^2 + (2 x + 3)^2 + 6)$$
$$ x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = 1/144 ((12x^3+6x^2+4x+3)^2+ 3(2x^2+2x+5)^2 + 5(2x+3)^2 +15)$$
| Note that
$$\begin{split}
\sum_{k=0}^{2n}x^k
&= \frac{x^{2n+1}-1}{x-1} \\
&= (2n+1)\int_0^1 (1-t + tx)^{2n}\mathrm{d}t \\
&= \frac{2n +1}{2^{2n+1}}
\int_{-1}^1(x+1 + (x-1)t)^{2n}\mathrm{d}t \\
&= \sum_{k=0}^{2n}\frac{_{2n}C_k(2n +1)}{2^{2n+1}}
\int_{-1}^1(x+1)^{2n-k}(x-1)^{k}t^k\mathrm{d}t \\
&= \sum_{k=0}^{n}\frac{_{2n}C_{2k}(2n+1)}{2^{2n}(2k+1)}
(x+1)^{2n-2k}(x-1)^{2k}\text{;}
\end{split}$$
whence
$$\boxed{\sum_{k=0}^{2n}x^k
=\sum_{k=0}^{n}\frac{_{2n}C_{2k}(2n+1)}{2^{2n}(2k+1)}
((x+1)^{n-k}(x-1)^{k})^2}\text{.}$$
Consequently, if $\frac{_{2n}C_{2k}(2n+1)}{2^{2n}(2k+1)}$ is a sum of squares (all nonnegative rationals are), then $\sum_{k=0}^{2n}x^k$ is a sum of squares with rational coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Area of shaded region in a square
My approach
The area of shaded region = 8×the area of
Given that the points of the shaded region are closer to the center than the boundary of the square.
Let's talk about the boundary of the shaded region
The boundary of the shaded region therefore must be the locus of all the points whose distance from the center of the square = distance from the boundary.
Let's find the locus of the boundary of the shaded region
From the second figure
${\sqrt {h^2+k^2} } = {\sqrt {(h-h)^2 + (k-{a\over 2})^2 }}$
This simplifies to be:
$k = {a^2 - 4h^2\over 4a}$
$y = {a^2 - 4x^2\over 4a}$
Also the curve intersects the line( the hypotenuse of the triangle) y= x
For point of intersection :
$ {a^2 - 4x^2\over 4a} = x $
$4x^2 +4ax - a^2 = 0$
$x = {-4a ± \sqrt{16a^2 - 4×4×(-a^2)}\over 8}$
$x = a{(\sqrt{2} ± 1)\over 2}$
Solution 1: $x = a{(\sqrt{2} + 1)\over 2}$
Solution 2: $x = a{(\sqrt{2} - 1)\over 2}$
Solution 1 can be discarded as $ x = a{(\sqrt{2} + 1)\over 2} ≈ 1.207106 a > {a\over 2} $
Solution 2: $ a{(\sqrt{2} - 1)\over 2} ≈ 0.2071 a < {a\over 2} $
from 0 to $a{(\sqrt{2} - 1)\over 2}$ :
The curve (boundary of shaded region) lies above the line $ y=x$
so $dA = ({a^2\over 4a} - {4x^2\over 4a} - x)dx $
$dA = ({a\over 4} - {x^2\over a} - x)dx $
$\int_{0}^{A} dA = \int_{0}^{a{(\sqrt{2} - 1)\over 2}} ({a^2\over 4a} - {4x^2\over 4a} - x)dx $
A = $({a\over 4}x - {x^3\over 3a} - {x^2\over 2})]_{0}^{{a{(\sqrt{2} - 1)\over 2}}}$
$ A = {a^2\over 8}(\sqrt{2}-1) - {a^2\over 24}(\sqrt{2}-1)^3 - {a^2\over 8}
(\sqrt{2}-1)^2 $
$ = {a^2\over 8}(\sqrt{2}-1) \Biggl( 1 - {(\sqrt{2}-1)^2\over 3} - (\sqrt{2} -1)\Biggr)$
This simplifies to be equal to
${a^2\over 8}(\sqrt{2}-1)({3+5\sqrt{2}\over 3})$
$= {(7- 2\sqrt{2})\over 8×3}a^2$
Area of shaded figure = 8A
$A_{total}$ = ${(7- 2\sqrt{2})\over 3}a^2$
But the answer is :
${4\sqrt{2}-5\over 3}a^2 $
I don't know where I got it wrong, and also I have re calculated this and the result is same.
Did I miss something important or calculated wrongly
Any help of hint or suggestion or worked out solution would be appreciated.
| $$\int_0^{\frac a2(\sqrt 2 -1)}(\frac a4-\frac{x^2}a-x)dx=\frac{ax}4-\frac{x^3}{3a}-\frac {x^2}2\bigg|_0^{\frac a2(\sqrt 2 -1)}\\=a^2\left(\frac{\sqrt 2-1}{8}-\frac{(\sqrt 2-1)^3}{24}-\frac{(\sqrt 2-1)^2}8\right)\\=\frac{a^2}{24}(3\sqrt2-3-2\sqrt2+6-3\sqrt 2+1-3(2+1-2\sqrt 2))\\=\frac{a^2}{24}(4\sqrt 2-5)$$
Your error is when you did the simplification and obtained $3+5\sqrt 2$. That should be $3-\sqrt 2$. So $$\frac{a^2}8(\sqrt 2-1)\frac{3-\sqrt 2}3=\frac{a^2}{24}(\sqrt 2(3+1)-3-2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Reduce precision of fraction Say I have a reduced fraction where the numerator and denominator can only be integers:
$$ \frac{1071283}{28187739} $$
and I want to reduce it more, accepting the lose of precision.
I could just remove an equal numbers of integers from the right:
$$ \frac{107}{2818} $$
However playing around with it shows me that I can easily find a fraction that contains the same number of integers but has lost less precision compared to the original fraction:
$$ \frac{108}{2842} $$
How can I reduce a fraction of integers and lose the minimal amount of precision?
| As suggested in comments, use the continued fraction decomposition.
$$\frac{1071283}{28187739} = \frac{1}{26+\frac{1}{3+\frac{1}{4+\frac{1}{1+\frac{1}{9+
\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{1+\frac{1}{1+
\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{14+\frac{1}{2+
\frac{1}{3}}}}}}}}}}}}}}}}$$
Halting the decomposition sooner will give you a good approximation, among the fractions with denominators less than the one of the fraction.This gives the approximations
$$\left(\frac{1}{26} , \frac{3}{79}
, \frac{13}{342} , \frac{16}{421}
, \frac{157}{4131} , \frac{173}{4552}
, \frac{676}{17787} , \frac{849}{22339}
, \frac{1525}{40126},...\right)$$
So for example,
$$\frac{157}{4131} = \frac{1}{26+\frac{1}{3+\frac{1}{4+\frac{1}{1+\frac{1}{9}}}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4388232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Weird Problem on Polynomial Roots The polynomial $f(x)=x^3-3x^2-4x+4$ has three real roots $r_1$, $r_2$, and $r_3$. Let $g(x)=x^3+ax^2+bx+c$ be the polynomial which has roots $s_1$, $s_2$, and $s_3$, where
\begin{align*}
s_1 &= r_1+r_2z+r_3z^2, \\
s_2 &= r_1z+r_2z^2+r_3, \\
s_3 &= r_1z^2+r_2+r_3z,
\end{align*}and $z=\dfrac{-1+i\sqrt3}2$. Find the real part of the sum of the coefficients of $g(x)$.
I know the sum of the coefficients is $g(1)$, $g(x)=(x-s_1)(x-s_2)(x-s_3)$, and $z^3=1$. This means $s_1z=s_2$, and $s_2z=s_3$. Since $s_1^3=s_2^3=s_3^3$, I have $g(x)=x^3-s_1^3$. Since the answer is $g(1)$, I need to calculate $$1-s_1^3.$$ I expanded $s_1^3$ to get $$s_1^3=r_1^3+r_1^2r_2z+3r_1^2r_3z+3r_1r_2^2z^2+6r_1r_2r_3+3r_1r_3^2z+r_2^3+3r_2^2r_3z+3r_2^2r_3z+3r_2r_3^2z^2+r_3^3.$$ I'm pretty sure using Vieta's can finish this, but I'm not sure where else to apply Vieta's other than $r_1r_2r_3$. I also tried substituting $z^2=-z-1$, but it didn't do much. I also tried using $(r_1+r_2+r_3)^2$, but this also failed. Could someone give me some guidance?
Thanks in advance!
| You know the symmetric polynomials in the roots:
$\sigma_1 = \displaystyle\sum_{i=1}^{3} r_i$,
$\sigma_2 = \displaystyle\sum_{1\leq i < j\leq 3} r_i r_j$,
and $\sigma_3 = \displaystyle\prod_{i=1}^{3} r_i$.
Think of combining these into products of degree 3, namely:
$\sigma_1^3, \sigma_1 \sigma_2, \sigma_3$.
Some linear combination of these will give your expression for the real part of $s_1^3$.
This needs the fact that $z$ and $z^2$ both have the same real part.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4392446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to calculate optimal strategy for bimatrix game Consider the bimatrix game given by
$$A = \begin{bmatrix} 3 &2 &-2 \\
-1 &3 &0 \\
1 &-2 &0 \\ \end{bmatrix} $$
$$ B = \begin{bmatrix} -2 &3 &-1 \\
-1 &-2 &2 \\
-1 &-1 &-3 \\ \end{bmatrix} $$
Taking $y^*$ to be the optimal strategy for $P_2$, I get $Ay^{*T} = u_1$ and solving the system gives $u_1 = \frac{1}{16}$ and $y^* = (\frac 5{16}, \frac 18, \frac 9{16})$
Similarly let $x^*$ be the optimal strategy for $P_1$, $B^Tx^{*T} = u_2 $ gives $u_2 = -\frac{15}{14} $ and $x^* = (\frac 1{14}, \frac{5}{14}, \frac{4}{7} ) $
This is indeed an optimal strategy according to sage.
Now consider the bimatrix game
$$A = \begin{bmatrix}2&1&2\\
-2&-1&3\\
-3&1&-2\\ \end{bmatrix} $$
$$B = \begin{bmatrix}2&1&-2\\
1&-1&2\\
2&1&2\\ \end{bmatrix}$$
Taking $y^*$ as optimal strategy for $P_2$ and taking $Ay^{*T} = u_1$ gives $u_1 = \frac{25}{23}$ and $y* = (-\frac 8{23}, \frac{21}{23}, \frac {10}{23})$ which is absurd because probabilities must be positive.
What am I doing wrong? How to calculate the optimal mixed strategy for such games?
In $2 \times 2$ bimatrix games, I have seen this work every time but it is failing in $3 \times 3$ games. What is the method which can be used for larger bimatrix games also?
| What you appear to be doing is trying to find a strategy for each player that makes his or her opponent's payoff independent of the pure strategy chosen by that opponent. While a pair of such strategies will constitute a Nash equilibrium if they exist, it's unfortunately very possible (and very likely, in fact, if the matrices of the game have any more than a few rows and columns) that no such pair of strategies will exist.
This online bimatrix game solver returns $3$ Nash equilibria for your second bimatrix game:
\begin{array}{l}
1.& \text{Strategy for player }1:\ (0,1,0)\\
&\text{Strategy for player }2:\ (0,0,1)\\
&\text{Payoff to player }1:\pmatrix{0&1&0}\pmatrix{2&1&2\\
-2&-1&3\\
-3&1&-2}\pmatrix{0\\0\\1}=3\\
&\text{Payoff to player }2:\pmatrix{0&1&0}\pmatrix{2&1&-2\\
1&-1&2\\
2&1&2}\pmatrix{0\\0\\1}=2\\
2.& \text{Strategy for player }1:\ \left(\frac{1}{5},\frac{4}{5},0\right)\\
&\text{Strategy for player }2:\ \left(\frac{1}{5},0,\frac{4}{5}\right)\\
&\text{Payoff to player }1:\pmatrix{\frac{1}{5}&\frac{4}{5}&0}\pmatrix{2&1&2\\
-2&-1&3\\
-3&1&-2}\pmatrix{\frac{1}{5}\\0\\\frac{4}{5}}=2\\
&\text{Payoff to player }2:\pmatrix{\frac{1}{5}&\frac{4}{5}&0}\pmatrix{2&1&-2\\
1&-1&2\\
2&1&2}\pmatrix{\frac{1}{5}\\0\\\frac{4}{5}}=\frac{6}{5}\\
3.& \text{Strategy for player }1:\ (1,0,0)\\
&\text{Strategy for player }2:\ (1,0,0)\\
&\text{Payoff to player }1:\pmatrix{1&0&0}\pmatrix{2&1&2\\
-2&-1&3\\
-3&1&-2}\pmatrix{1\\0\\0}=2\\
&\text{Payoff to player }2:\pmatrix{1&0&0}\pmatrix{2&1&-2\\
1&-1&2\\
2&1&2}\pmatrix{1\\0\\0}=2
\end{array}
Note that none of these equilibrium strategies makes the payoff to the opponent of the strategy's user independent of that opponent's strategy.
One sure way of finding a Nash equilibrium for any bimatrix game is the Lemke-Howson algorithm.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What is the minimum value of $\frac{2a}{a+2b+2c} + \frac{2b}{b+2c+2a} +\frac{2c}{c+2a+2b}$ given $ a+b+c=2$? What is the minimum value of $\frac{2a}{a+2b+2c} + \frac{2b}{b+2c+2a} +\frac{2c}{c+2a+2b}$ given $ a+b+c=2$?
I know the answer is going to be $\frac{6}{5}$ because it will occur when there is equality between the three pronumerals.
The issue is this question was given to a group of students who have only learned the AM/GM inequality for two pronumerals.
I can do the question by using the AM/GM inequality with three pronumerals (along with a bit of not immediately obvious algebraic manipulation) however I'm struggling to see how I can solve this using the AM/GM inequality with only two values.
I've tried making 3 separate equations and adding them together, however so far I haven't been able to find how to do it.
| This is the most basic solution I can think of. This method is useful even if you only know AM-GM. Also, when I starts to learn the inequalities, I do them like this.
To prove
$$\frac{2a}{a+2b+2c} + \frac{2b}{b+2c+2a} +\frac{2c}{c+2a+2b}\ge 6/5$$
Cancelling out the denominators
$$5(2a(b+2c+2a)(c+2a+2b)+2b(a+2b+2c)(c+2a+2b)+2c(a+2b+2c)(b+2c+2a))\ge 6(a+2b+2c)(b+2c+2a)(c+2a+2b)$$
Expand them all, we have
$$40(a^3+b^3+c^3)+80(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)+150abc\ge 24(a^3+b^3+c^3)+84(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)+174abc$$
Or,
$$16(a^3+b^3+c^3)\ge 4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)+24abc$$
By AM-GM, we have
$$a^3+ab^2\ge 2\sqrt{a^3\times ab^2}=2a^2b$$
And similarly $b^3+ba^2\ge 2\sqrt{b^3\times ba^2}=2b^2a$. So add them together, we have
$$a^3+ab^2+b^3+ba^2\ge 2a^2b+2b^2a$$
And therefore,
$$a^3+b^3\ge a^2b+b^2a$$
So we have
$$8(a^3+b^3+c^3)=4(a^3+b^3)+4(c^3+b^3)+4(a^3+c^3)\ge 4(a^2b+b^2a)+4(c^2b+b^2c)+4(a^2c+c^2a)=4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)$$
So what remains is $$8(a^3+b^3+c^3)\ge 24abc$$
However, by above, we know that
$$8(a^3+b^3+c^3)\ge 4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)$$
So we need to prove that
$$4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)\ge 24abc$$
Which is also AM-GM:
$$4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)=4((a^2b+c^2b)+(b^2c+a^2c)+(c^2a+b^2a))\ge 4(2\sqrt{a^2b\times c^2b}+2\sqrt{b^2c\times a^2c}+2\sqrt{c^2a\times b^2a})=4(2abc+2abc+2abc)=4\times6abc=24abc$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Find $\sum_{n=0}^\infty \big( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\big)$ without power series or integration? In a calculus, I have to find a direct way to compute the sum
$$\sum_{n=0}^\infty \left( \frac{1}{3n+4}-\frac{3}{3n+2}+\frac{2}{3n+1}\right).$$
I tried to reform a partial sum:
$$\sum_{n=0}^N a_n = \frac12+3\sum_{n=1}^N \left(\frac{1}{3n+1}-\frac{1}{3n+2}\right).$$
A classic way then is to introduce the missing terms to use Euler's formula. Problem is, I don't know how to deal with the alternating signs.
What I can't do: use power series (the sum comes from one), neither integration (same problem). I need to find a direct way to sum this, that is, a way which only involves classic techniques with series.
Anyone to help me? Thanks.
| First note that
$$\sum_{n=0}^\infty z^n = \frac{1}{1-z}$$
implies
$$\sum_{n=1}^\infty \frac{z^n}{n} = \log\frac{1}{1-z} = -\log(1-z),$$
and so
\begin{align}
\sum_{n=0}^\infty \frac{z^n}{n+1} &= \frac{-\log(1-z)}{z} \tag1\\
\sum_{n=0}^\infty \frac{z^n}{n+2} &= \frac{-\log(1-z)-z}{z^2} \tag2\\
\sum_{n=0}^\infty \frac{z^n}{n+4} &= \frac{-6\log(1-z)-6z-3z^2-2z^3}{6z^4} \tag3
\end{align}
Now let
$$a_n = \frac{1}{n+4}−\frac{3}{n+2}+\frac{2}{n+1}$$
and apply $(1)$ through $(3)$ to yield
$$\sum_{n=0}^\infty a_n z^n = \sum_{n=0}^\infty \frac{z^n}{n+4} − 3\sum_{n=0}^\infty \frac{z^n}{n+2} + 2\sum_{n=0}^\infty \frac{z^n}{n+1}
= \frac{(-6+18z^2-12z^3)\log(1-z) - 6z - 3z^2 + 16z^3}{6 z^4}.$$
Finally apply
$$\sum_{n=0}^\infty a_{3n} = \sum_{n=0}^\infty a_n\frac{1+\omega^n+\omega^{2n}}{3} = \frac{1}{3}\sum_{n=0}^\infty a_n + \frac{1}{3}\sum_{n=0}^\infty a_n\omega^n + \frac{1}{3}\sum_{n=0}^\infty a_n(\omega^2)^n,$$
where $\omega=\exp(2\pi i/3)$,
to obtain
$$\frac{1}{3}\cdot\frac{7}{6} + \frac{1}{3}\left(\frac{-25-19 i\sqrt3}{12} + 3 i \sqrt3 \log\left(\frac{3 - i \sqrt3}{2}\right)\right) + \frac{1}{3}\left(\frac{-25 + 19 i\sqrt3}{12} - 3 i \sqrt3 \log\left(\frac{3 + i \sqrt3}{2}\right)\right) = \frac{\pi}{\sqrt3} - 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How to evaluate this integral with square trinomial and root? I have to evaluate this integral:
$$\int{\frac{1-x+x^2}{x\sqrt{1+x-x^2}}dx}$$
I have experience with methods such as:
*
*Using substitute for x;
for example, $${\frac{1}{x}=t;dx = -\frac{dt}{t^2}}$$
*Method of adding a new variable;
*Substitution method;
*Integration by the parts.
In addition, I just need to mention, that we have learned topic with square trinomial.
Just remake this expression:
$${1+x-x^2}={\frac{5}{4}-(x-\frac{1}{2})^2}$$
And, of course, then we can use some formulas for evaluation, when have expression in required view. Thanks!
| $\int{\frac{1-x+x^2}{x\sqrt{1+x-x^2}}dx}$
The rooted is a trinomial ($a x^{2}+2b x+c$), which has two real roots $\alpha$ and $\beta$, decomposing it we have:
$\sqrt{a x^{2}+2b x+c}=\sqrt{a(x-\alpha)(x-\beta)}=(x-\beta)\sqrt{\frac{a(x-\alpha)}{x-\beta}}$.
Let's say
$\frac{a(x-\alpha)}{x-\beta}=t^{2}$, and we get :
$x=\frac{a\alpha-\beta t^{2}}{a-t^{2}}$
$dx=\frac{2 a t(\alpha-\beta)}{(a-t^{2})^{2}}dt$.
Substituting these values in the integral, we have:
$\int{\frac{1-x+x^2}{x\sqrt{1+x-x^2}}dx}=$,
$=-\int{\frac{4(2t^{4}-t^{2}+2)}{(t^{2}+1)^{2}(t^{2}(\sqrt{5}+1)-\sqrt{5}+1)}dt}=$,
expanding
$=\int{\frac{2\sqrt{5}}{(t^{2}+1)^{2}}dt}$
+$\int{\frac{1-\sqrt{5}}{t^{2}+1}dt}$
-$\int{\frac{4}{ t^{2}(\sqrt{5}+1)-\sqrt{5}+1)}dt}$.
Calculating the individual integrals:
$I_{1}=\sqrt{5} .tan^{-1}(t)+\frac{\sqrt{5}t}{t^{2}+1}$,
$I_{2}=(1-\sqrt{5}). tan^{-1}(t)$,
$I_{3}=-ln(\frac{2t-\sqrt{5}+1}{2t+\sqrt{5}-1})$.
Ultimately you have:
$I=I_{1}+I_{2}+I_{3}=tan^{-1}(t)+\frac{\sqrt{5}t}{t^{2}+1}-ln(\frac{2t-\sqrt{5}+1}{2t+\sqrt{5}-1})$.
All that remains is to replace the value of $t$ with $x$:
$t=\frac{-2x-\sqrt{5}+1}{2x-\sqrt{5}-1}$,
getting the result.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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why substitution works in general Consider the equations $x^2 +y^2=1$ and $y=x^2-1$. If $y=x^2-1$ is substituted into $x^2 +y^2=1$ to obtain $x^2 +(x^2-1)^2=1$, then the $x$ values that result from solving this equation will be the $x$ values that produce equal $y$ values in the original equations $y=x^2-1$ and $x^2 +y^2=1$. I don't understand why this is true on conceptual a level, but rather just accept it as a process. I understand that if I am trying to solve the system $y=x+1$ and $y=2x-1$, I can set these equations equal to each other and $2x-1=x+1 \implies$ that the $y$ values are the same when $x=2$. This works because $x+1=2x-1 \iff x+2=2x \iff x=2$. I don't understand, however, why this sort of substitution works in general. Is there a proof that this technique works for higher order polynomials and more complicated equations? Please not only answer the questions posed above but also help me identify additional things that I may not understand.
| Let $(a,b)$ be a solution of the system which consists of the equations$$x^2+y^2=1\tag1$$and$$y=x^2-1.\tag2$$Asserting that $(a,b)$ is a solution of both $(1)$ and $(2)$ means that $a^2+b^2=1$ and that $b=a^2-1$. But then $a^2+(a^2-1)^2=1$, and therefore $a$ is indeed a solution of the equation $x^2+(x^2-1)^2=1$. And $b=a^2-1$ means that if, in the equality $y=x^2-1$, if you replace $x$ by $a$, then $y$ becomes $b$.
| {
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Can we prove that the equation $2^n+1=5^m$ has no integral solution other than $(2, 1)$? $\textrm{Assume, on the contrary, that }n\geq 4,\text{then }$
\begin{aligned} 2^{n} &=5^{m}-1 \\ &=(5-1)\left(5^{m-1}+5^{m-2}+\cdots+1\right) \\ &=4\left(5^{m-1}+5^{m-2}+\cdots+1\right) \\ 2^{n-2} &=5^{m-1}+5^{m-2}+\cdots+1 \\ 2^{n-2} & \equiv \underbrace{1+1+\cdots+1}_{m \text { ‘1’s }}(\bmod 4) \\ 0 & \equiv m \quad(\bmod 4) \\ \therefore \quad m &=4 k \text { for some integer } k . \end{aligned}
$\text {Again, }$
$\begin{aligned}2^{n} &=5^{4 k}-1 \\ \displaystyle \quad &=\left(5^{4}\right)^{k}-1 \\ \displaystyle \quad &=\left(5^{4}-1\right)\left(625^{k-1}+625^{k-2}+\cdots+1\right) \\ \displaystyle \quad &=624\left(625^{k- 1}+625^{k-2}+\cdots+1\right) \\&\Rightarrow 624 | 2^{n},\text {which leads to a contradiction. } \\\therefore \quad n=0,1,2 \text{ or }3.\end{aligned}$
After checking one by one, we can conclude that $(n, m)=(2,1) $
is the unique integer solution.
Your opinions and alternative proofs are highly appreciated.
| We suspect the largest solution is $4+1 = 5.$ Therefore write $2^n - 4 = 5^m - 5.$ Make new letters $x +2 = n$ and $y+1 = m. $ This becomes
$$ 4 (2^x - 1) = 5 (5^y-1) $$
ASSUME that $x \geq 1, y \geq 1.$
Since $2^x \equiv 1 \pmod 5,$ we see that $4 | x.$ Furthermore $2^4 - 1 | 2^x - 1.$ And $15 = 3 \cdot 5.$ So far, $3 | 2^x - 1.$ This means $3 | 5^y - 1$
Alright, $5^y \equiv 1 \pmod 3.$ This requires $2 | y.$ $5^2 - 1 = 24 = 3 \cdot 8. $ Thus $8 | 5^y - 1$ and $ 8 | 4(2^x - 1)$
The final $ 8 | 4(2^x - 1)$ is a contradiction: if $x \geq 1$ then $2^x - 1$ is odd.
Therefore $x=0$ and $y = 0.$
| {
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How do I find the centralizer for this matrix?
Let $B=\begin{pmatrix}
2 & 2 \\
0& 1 \\
\end{pmatrix} \in GL_2(\mathbb{R})$. Determine the centralizer for $B$.
Here's what I've done so far:
Let $D=\begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} \in GL_2(\mathbb{R})$.
$\begin{align}
BD&=DB\\
\begin{pmatrix}
2 & 2 \\
0& 1 \\
\end{pmatrix} \begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} &= \begin{pmatrix}
a & b \\
c & d \\
\end{pmatrix} \begin{pmatrix}
2 & 2 \\
0& 1 \\
\end{pmatrix}\\
\begin{pmatrix}
2(a+c) & 2(b+d) \\
c& d \\
\end{pmatrix}&= \begin{pmatrix}
2a & 2a+b \\
2c& 2c+d \\
\end{pmatrix}
\end{align} $
From the above, we have:
$\begin{aligned}
2a+2c&=2a\\
2b+2d&=2a+b\\
c&=2c\\
d&=2c+d
\end{aligned}
$
From which we derive:
$\begin{aligned}
c&=0\\
b+2d-2a&=0\\
d&=d \in \mathbb{R} \backslash \{0\}
\end{aligned}$
As you can see, I've managed to determine $c$ and $d$, but I can't seem to figure out $a$ and $b$. What am I missing?
| This means that an invertible $2 \times 2$ matrix $D$
commutes with $B$ if and only if it is of the form
$$D=\begin{pmatrix}
a & 2a-2d \\
0 & d \\
\end{pmatrix}$$
for $a,d \in \mathbb{R}\setminus\{0\}$.
| {
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Proving by Induction that $x_{2n+1} = 1+1/2+1/2^3+\cdots+1/(2^{2n-1})$ Let $x_1 = 1$, $x_2 = 2$ and $x_n = \frac{x_{n-1}+x_{n-2}}{2}$. The sequence $(x_n)$ is Cauchy, which I can easily prove. To find its limit I must first show (by Induction) the assertation
$$
P(n): x_{2n+1} = 1+\frac{1}{2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{2n-1}},\quad\forall n\in\mathbb{N}.
$$
Base case.
For $n=1$, $x_{2n+1} = x_{3} = \frac{1+2}{2} = 1+ \frac{1}{2},$ thus $P(1)$ holds.
Inductive step. Let $P(k)$ be true. Then for $P(k+1)$ we have
$$
x_{2(k+1)+1} = \frac{1}{2}(x_{2(k+1)}+x_{2(k+1)-1}) = \frac{1}{2}(x_{2k+2}+x_{2k+1}).
$$
But then I'm stuck. I am missing something here but, at present, I cannot think of a different approach.
| As you already noted, $$x_{2n+1} = 1 + \sum_{i=1}^{n-1}\frac1{2^{2i-1}} = \frac13\left(5-\frac1{2^{2n-1}}\right).$$
You might also note that $$x_{2n+2} = 3 - \sum_{i=0}^n\frac1{2^{2i}} = \frac13\left(5+\frac1{2^{2n}}\right).$$
Combine these two, you have
$$x_n = \frac13\left(5 + 4\left(-\frac12\right)^n\right) \qquad(1).$$
Now, let's prove $(1)$ using induction. First,
$n = 1$, $x_1 = 1$, so $(1)$ is true. Assume $(1)$ is true for all integers from 1 to $k$, then
$$x_{k+1} = \frac12\left(x_k+x_{k-1}\right) = \frac13\left(5+4\left(-\frac12\right)^{k+1}\right).$$
So $(1)$ is true for $k+1$, so $(1)$ is true for all $k > 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\int_{0}^{1} \sqrt{\frac{x^{m}}{1-x^{n}}} d x$? Stuck by the integral
$$\int_{0}^{1} \sqrt{\frac{x}{1-x^{3}}} d x,$$
I finally solve the the problem using Beta Function.
Then I generalize the result to
$$
I(m,n)=\int_{0}^{1} \sqrt{\frac{x^{m}}{1-x^{n}}} d x
$$
can be tackled by the Beta function by letting $$
y=x^{n} \text {, then } x=y^{\frac{1}{n}} \textrm{ followed by} \quad d x=\frac{1}{n} y^{\frac{1}{n}-1} d y
$$
$$
\begin{aligned}
I &=\int_{0}^{1} y^{\frac{m}{2 n}}(1-y)^{-\frac{1}{2}} \cdot\frac{1}{n} y^{\frac{1}{n}-1} d y \\
&=\frac{1}{n} \int_{0}^{1} y^{\frac{m+2}{2 n}-1}{(1-y)^{\frac{1}{2}-1}} d y \\
&=\frac{1}{n} B\left(\frac{m+2}{2 n} , \frac{1}{2}\right) \\
&=\frac{\Gamma\left(\frac{m+2}{2 n}\right) \Gamma\left(\frac{1}{2}\right)}{n\left(\frac{m+2+n}{2 n}\right)}\\&=\frac{\sqrt{\pi} \Gamma\left(\frac{m+2}{2 n}\right)}{n \Gamma\left(\frac{m+2}{2 n}+\frac{1}{2} \right)}
\end{aligned}
$$
For example, $$
\begin{aligned}
I(1,3) &=\frac{1}{3} B\left(\frac{1}{2}, \frac{1}{2}\right) \\
&=\frac{1}{3} \Gamma^{2}\left(\frac{1}{2}\right) \\
&=\frac{\pi}{3}
\end{aligned}
$$
My Question:
Can we simplify the last answer? Your suggestion or help are highly appreciated.
| Your first integral has an elementary antiderivative.
$$\int_0^1\sqrt{\frac{x}{1-x^3}}=\int_0^1\frac{x^\frac{1}{2}}{\sqrt{1-(x^\frac{3}{2})^2}}dx=\frac{2}{3}\int_0^1\frac{d(x^\frac{3}{2})}{\sqrt{1-(x^\frac{3}{2})^2}}=\frac{2}{3}\arcsin(x^\frac{3}{2})\Bigg|_0^1=\frac{2}{3}\frac{\pi}{2}=\frac{\pi}{3}$$
I doubt there is any simplification from that ratio of gammas, but we could convert it into a cool product form if m and n are integers using the definition of the gamma function
| {
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Finding roots of the equation $8x^3+4x^2-4x-1=0$ which are in form of cosine of angles. So, I start this by reducing it to a cubic with 0 coefficient of the 2nd degree term which gives me :
$$\left(x + \frac{1}{6}\right)^3 - \frac{7}{12}\left(x + \frac{1}{6}\right)-\frac{7}{216}=0$$
Replacing $\left(x + \frac{1}{6}\right)$ by $y$:
$$y^3-\frac{7}{12}y -\frac{7}{216}=0$$
Then replacing $y$ by $r \cos(\alpha)$ and comparing the equation with :
$$\cos^3 (\alpha) - \frac{3}{4}\cos(\alpha)-\frac{1}{4}\cos(3\alpha)=0$$
I get:
$$r = \sqrt{\frac{7}{3}}, \alpha = \frac{1}{3}\cos^{-1}\left(\frac{1}{2\sqrt{7}}\right)$$
which finally give:
$$\theta = \cos^{-1}\left(\frac{\sqrt{7}}{9}\cos^{-1} \left(\frac{1}{2\sqrt{7}}\right)-\frac{1}{6}\right)$$
which is approximately 1.329 on putting it in the calculator.
But if I consider the common problem:
$8x^3-4x^2-4x+1=0$ having roots $\cos(\pi/7), \cos(3\pi/7), \cos(5\pi/7)$.
If I replace $x$ by $-x$ in this equation, we get:
$$8x^3+4x^2-4x-1=0$$
which is same the expression in the actual problem. So its roots must be $-\cos(\pi/7), -\cos(3\pi/7), -\cos(5\pi/7)$. This clearly doesn't matches with the answer I got.
Where I go wrong? Also can someone confirm while comparing coefficients in $cos(\alpha)$ variable cubic equation, we treat $\cos(3\alpha)$ as a constant. My book give a special section in which it shows this method of comparing in this way. So I guess its probably right? But I am not sure how is it right, isn't $\cos(3\alpha)$ also dependent on the main variable $\cos(\alpha)$.
| From the book in the chat, we can spot that the wrong thing is $r$, it should be $\sqrt{7}/3$ not $\sqrt{7/3}$.
So, your $\theta$ should be (I think you have a wrong calculation there!)
$$\theta = \cos^{-1}\left(\frac{\sqrt{7}}{{\color{red} 3}}{\color{red}\cos}\left( {\color{red}{\frac13}} \cos^{-1}\left(\frac{1}{2\sqrt{7}}\right)\right)-\frac{1}{6}\right)$$
Which is actually $2\pi/7$, fit $-\cos(5\pi/7)$ root.
As for equation $$\cos^3\theta+\frac{3H}{r^2}\cos\theta+\frac{G}{r^3}=0$$
We notice that there are some spaces to solve $\theta$ by adjusting $r$. Notice that no matter how we adjust, the linear term can't be cancelled, but we need a way to make the equation to solve easily. The triple-angle formula for cosine has nonzero linear term. That is why we put $3H/r^2=-3/4$ (and hence $G/r^3=-\cos(3\theta)/4$) since we can use the triple-angle formula there to solve.
| {
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"source": "stackexchange",
"question_score": "3",
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Proving that $-\ln(x-\sqrt{x^2-1})=\ln(x+\sqrt{x^2-1})$ Show $-\ln(x-\sqrt{x^2-1})=ln(x+\sqrt{x^2-1})$
Can somebody verify that my solution is correct (or incorrect!) and also show me alternative methods to do this? Thanks!
Let $r=-\ln(x-\sqrt{x^2-1})$ and $s=ln(x+\sqrt{x^2-1})$
$\rightarrow e^r=(x-\sqrt{x^2-1})^{-1}$ and $e^s = x+\sqrt{x^2-1}$
$\rightarrow \frac{e^s}{e^r} = (x+\sqrt{x^2-1})(x-\sqrt{x^2-1}) = x^2 - (x^2 - 1) = 1$
Thus $e^r=e^s$ and so $r=s$
| Looks good.
But note that it's really just a matter of observing that $x \pm \sqrt{x^2-1}$ are reciprocals, since
$$\frac{1}{x + \sqrt{x^2-1}} = \frac{1}{x + \sqrt{x^2-1}}\cdot\frac{x - \sqrt{x^2-1}}{x - \sqrt{x^2-1}}$$
$$=\frac{x - \sqrt{x^2-1}}{x^2-(x^2-1)}$$
$$= \frac{x - \sqrt{x^2-1}}{1}$$
$$=x - \sqrt{x^2-1}$$
This means that their logarithms are opposites, since $\log u^{-1} = -\log u$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Examples where $(a+b+\cdots)^2 = (a^2+b^2+\cdots)$ Consider the two infinite series
$$
\frac{\pi}{\sqrt{8}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots
$$
and
$$
\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \cdots
$$
(Notice that the first series has signs that go two-by-two rather than every-other.)
Squaring the first equality also gives $\pi^2/8$ and so these two, when put together, satisfy the 'highschooler's dream' for squaring a sum: just square each term and sum,
$$
(a + b + c + \cdots)^2 = (a^2 + b^2 + c^2 + \cdots)
$$
with nothing like $2ab + 2ac + 2bc + \cdots$ needed.
A trivial example of this would be
$$
(a + 0)^2 = a^2 + 2a0 + 0^2 = a^2 + 0^2
$$
but it only succeeds because one addend is zero.
My questions are
*
*Are there any other simple nontrivial examples? I believe any other nontrivial example must be an infinite sum. edit: John Omielan provides the simple finite example $(1+1-\frac{1}{2})^2 = 1^2 + 1^2 + \frac{1}{2^2}$.
*Is there an "obvious" demonstration that the above sum (other than the direct evaluation) satisfies the highschooler's dream? Put another way, is there a simple demonstration that the infinite sum of "cross terms" vanishes?
| The first term
$$
\eqalign{
& \left( {a_1 + a_2 + \cdots + a_n } \right)^2 = R^2 \quad \Rightarrow \cr
& \Rightarrow \quad \left( {{{a_1 } \over R} + {{a_2 } \over R} + \cdots
+ {{a_n } \over R}} \right)^2 - 1 = 0\quad \Rightarrow \cr
& \left( {\left( {b_1 + b_2 + \cdots + b_n } \right) + 1} \right) \cdot
\left( {\left( {b_1 + b_2 + \cdots + b_n } \right) - 1} \right) = 0 \cr}
$$
is the equation (in $b_k$) of two diagonal planes, symmetric wrt the origin, with normal vector $(1,1, \ldots , 1)$,
through the point
$$
\pm \left( {{1 \over n},{1 \over n}, \ldots ,{1 \over n}} \right)
$$
and thus each at a distance from the origin of
$$
{1 \over {\sqrt n }}
$$
The second term
$$
\eqalign{
& a_1 ^2 + a_2 ^2 + \cdots + a_n ^2 = R^2 \quad \Rightarrow \cr
& \Rightarrow \quad b_1 ^2 + b_2 ^2 + \cdots + b_n ^2 = 1 \cr}
$$
is a unitary sphere centered at the origin.
Therefore the equality
$$
\eqalign{
& \left( {a_1 + a_2 + \cdots + a_n } \right)^2 = a_1 ^2 + a_2 ^2 + \cdots + a_n ^2 = R^2
\quad \Rightarrow \cr
& \Rightarrow \quad \left( {b_1 + b_2 + \cdots + b_n } \right)^2
= b_1 ^2 + b_2 ^2 + \cdots + b_n ^2 = 1 \cr}
$$
is satisfied whenever the points $b_k$ lie on one of the two circles resulting from the intersection,
and the points $a_k$ on any dilation of those circles, i.e. on the conic surface with vertex at the origin, axis $(1,1, \ldots , 1)$, cross-section defined by the above circle on the unitary sphere.
| {
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"source": "stackexchange",
"question_score": "27",
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$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$
For $a,b,c\ge0: ab+bc+ca+abc=4$ then: $$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$$
I used the condition and get: $a+\sqrt{bc+abc}=a+\sqrt{4-a(b+c)}\le a+2$
So we need to prove that: $$\frac{2-a}{a+2}+\frac{2-b}{b+2}+\frac{2-c}{c+2}\ge1$$
I tried to full expand but the rest seems complicated for me.
Can anyone help me full my idea? Every thinking is welcomed, thanks!
| Now I find a proof by AM-GM: $$4=2\sqrt{\frac{a(b+c)+bc+abc}{a+\sqrt{bc+abc}}(a+\sqrt{bc+abc})}\le a+\sqrt{bc+abc}+\frac{a(b+c)+bc+abc}{a+\sqrt{bc+abc}}=\frac{a(a+b+c)}{a+\sqrt{bc+abc}}+2\sqrt{bc+abc}$$
$$\implies \frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}\ge\frac{2(b+c)}{a+b+c}$$
Or: $$\frac{2-a}{a+\sqrt{bc+abc}}\ge\frac{b+c-a}{a+b+c}$$
Sum up similar inequalities, we get desired result!
Equality holds iff $(a,b,c)=(0,2,2)$ and pers
| {
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$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}}\text{ for }n\ge 2$ I'm completely lost on this question:
$$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}} \text{ for }n\ge 2$$
alternatively,
$$\prod_{a=2}^{n} {\left(1-\frac{1}{\sqrt{a}}\right)} \lt \frac{2}{n^2} \text{ for } n\geq2$$
I tried to do it by induction but ended up with $\frac{2}{k^2}(1-\frac{1}{\sqrt{k+1}})$ in the RHS which I have no idea what to do with. I also tried to do it with AM/GM inequality but also had little luck. The question did not specifically specify what type of proof to use so I'm assuming you can use any type of proof. The first that came to mind is induction obviously, but I'm really lost on this one. Any help or hints would be appreciated!
With induction, the thing that I'll need to prove is:
$$\frac{2}{k^2}(1-\frac{1}{\sqrt{k+1}}) \lt \frac{2}{(1+k)^{2}}$$
But I'm not sure how to approach it.
SOLVED
$$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}} \text{ for }n\ge 2$$
Prove by induction:
A. When $n=2$,
$$\begin{aligned}LHS &= 1-\frac{1}{\sqrt{2}}\newline
&\approx 0.293\newline
RHS&=\frac{2}{2^2}\newline
&=0.5\newline
&\gt LHS\newline
\therefore \text{true for }n&=2\end{aligned}$$
B. Assume the statement is true for the positive integer $n=k$, where $k\in\mathbb{N}$.
That is, assume that
$$\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{k}} \right)\lt \frac{2}{k^{2}}$$
Now prove the statement for $n=k+1$. That is, prove that
$$\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{k+1}} \right)\lt \frac{2}{(k+1)^{2}}$$
By the induction hypothesis:
$$\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{k}} \right)\left( 1-\frac{1}{\sqrt{k+1}} \right)\lt \frac{2}{k^{2}}\left( 1-\frac{1}{\sqrt{k+1}} \right)$$
Hence, we are required to prove for $k\geq2$ that $$\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\leq\frac{2}{(k+1)^2}.$$
For $k\geq2,$ \begin{aligned}&\text{LHS-RHS}\\
=&\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)-\frac{2}{(k+1)^2}\\
=&\frac{2\sqrt{k+1}}{1+\sqrt{k+1}}\left(\frac{1}{k^2}\left(1-\frac{1}{\sqrt{k+1}}\right)\left(1+\frac{1}{\sqrt{k+1}}\right)-\frac{1}{(k+1)^2}\left(1+\frac{1}{\sqrt{k+1}}\right)\right)\\
=&\cdots\\ =&\frac{2\sqrt{k+1}}{1+\sqrt{k+1}}\left(\frac{1}{k}-\frac{1}{\sqrt{k+1}}\right)\\
=&\frac{2(\sqrt{k+1}-k)}{k(1+\sqrt{k+1})} \\ \lt&0,\end{aligned} as required.
$\therefore$ It follows from parts A and B by mathematical induction that the result is true for all integers $n\gt2$
| For $k\geq2,$ \begin{align} & \frac{2}{k^2}\left(1-\frac{1}{\sqrt{k+1}}\right)\leq \frac{2}{(k+1)^2}\\
\iff & 1-\frac{k^2}{(k+1)^2}\leq \frac{1}{\sqrt{k+1}}\\
\iff & \frac{2k+1}{(k+1)^2}\leq \frac{1}{\sqrt{k+1}}\\
\iff & (2k+1)^2\leq (k+1)^3\\
\iff & k(k^2-k-1)\geq0\\
\iff & k^2-k-1\geq0\\
\iff & k\in\left(-\infty,\frac12-\frac{\sqrt5}2\right]\cup\left[\frac12-\frac{\sqrt5}2,\infty\right)\\
\iff & k\geq2
\end{align}
Therefore, $$ k\geq2\implies \frac{2}{k^2}\left(1-\frac{1}{\sqrt{k+1}}\right)\leq \frac{2}{(k+1)^2},$$ as required for your induction step.
Addendum
OP:
Hence, we are required to prove
$$\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\lt\frac{2}{(k+1)^2},$$ that is, $LHS-RHS<0.$
\begin{aligned}&LHS-RHS\\
=&\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)-\frac{2}{(k+1)^2}\\
=&\frac{1}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\Big(1+\frac{1}{\sqrt{k+1}}\Big)-\frac{1}{(k+1)^2}\Big(1+\frac{1}{\sqrt{k+1}}\Big)\\
=&\cdots\\ =&\frac{1}{k}-\frac{1}{\sqrt{k+1}}\newline \lt&0,\end{aligned} as required.
@ryang I thought multiplying the expression
by $\left(1-\frac{1}{k+1}\right)$ would change the expression
entirely so it wouldn't be 'equal'?
Yes, so the above attempt is erroneous.
But, writing "$⟺$" with "$<0$", isn't that the
conclusion that I'm trying to reach, and I would think that I can't > just assume that the expression is $<0\,?$
I explained here that to prove an inequality $P(x),$ it is valid and legitimate to rewrite it as a simpler but equivalent inequality $Q(x)$ then explain how $Q(x)$ is true. In effect, the flow of reasoning is this: since $P(x){\iff}Q(x)$ and $Q(x)$ is true, thus $P(x)$ must be true.
Is there potentially a better way to write this step?
All the $<$ ought to be changed to to $\leq$ since the requirement is to prove merely the latter.
Suggestion 1 (this presentation parallels my answer above):
Hence, we are required to prove for $k\geq2$ that $$\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\leq\frac{2}{(k+1)^2}.$$
Now, for $k\geq2,$
\begin{aligned}&\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\leq\frac{2}{(k+1)^2}\\
\iff &
\frac{1}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\Big(1+\frac{1}{\sqrt{k+1}}\Big)-\frac{1}{(k+1)^2}\Big(1+\frac{1}{\sqrt{k+1}}\Big)\leq0\\
\iff &\cdots\\ \iff &\frac{1}{k}-\frac{1}{\sqrt{k+1}}\leq0\\ \iff &
k\geq\sqrt{k+1}\\\text{and}\quad &k\geq\sqrt{k+1}.\end{aligned} Hence, $$k\geq2\implies
\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\leq\frac{2}{(k+1)^2},$$ as required.
Suggestion 2 (this presentation corrects your attempt):
Hence, we are required to prove for $k\geq2$ that $$\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\leq\frac{2}{(k+1)^2}.$$
For $k\geq2,$ \begin{aligned}&\text{LHS-RHS}\\
=&\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)-\frac{2}{(k+1)^2}\\
=&\frac{2\sqrt{k+1}}{1+\sqrt{k+1}}\left(\frac{1}{k^2}\left(1-\frac{1}{\sqrt{k+1}}\right)\left(1+\frac{1}{\sqrt{k+1}}\right)-\frac{1}{(k+1)^2}\left(1+\frac{1}{\sqrt{k+1}}\right)\right)\\
=&\cdots\\ =&\frac{2\sqrt{k+1}}{1+\sqrt{k+1}}\left(\frac{1}{k}-\frac{1}{\sqrt{k+1}}\right)\\
=&\frac{2(\sqrt{k+1}-k)}{k(1+\sqrt{k+1})} \\ \leq&0,\end{aligned} as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Writing a polynomial as polynomial powers I want to write the polynomial
$$p(x)=x^4+3x^3+4x^2-7x+6$$ as a polynomial in powers of $x-1$.
What I am trying to do is to determine the taylor series around $x=1$.To do so, consider that $p(1)=7, p'(1)=14, p''(1)=38, p'''(1)=42, p''''(1)=24$
From this it follows that
$p(x)=7+\frac{14}{1!}(x-1)+\frac{38}{2!}(x-1)^2+\frac{42}{3!}(x-1)^3+\frac{24}{4!}(x-1)^4$
Honestly I am not convinced with my answer, since comparing with what is proposed by the book the coefficient of $(x-1)$ is $-1$ and the coefficient of $(x-1)^2$ is $4$ but I don't understand why, any suggestion?
| Your expansion is correct.
Here is an unconventional way to do this.
$P(x) = x^4 + 3x^3 + 4x^2 - 7x + 6\\
P(x+1) = (x^4 + 4x^3 + 6x^2 + 4x + 1) + 3(x^3+3x^2+3x+1) + 4(x^2 + 2x+1) - 7(x+1) + 6\\
P(x+1) = x^4 + 7x^3 + 19x^2 + 14x + 7\\
P((x-1)+1) = P(x) = (x-1)^4+7(x-1)^3 + 19(x-1)^2 + 14(x-1)+7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4415773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How would you find the inverse of a complex function? Let's say you had a function $f \colon \mathbb{C} \longrightarrow \mathbb{C}$ defined by
$$ f(x+ \iota y) = (3+6x)+ \iota (2−3y). $$
How would you go about finding its inverse? This provoked my interest and curiosity, as it seems you can't simply manipulate the equation to get $x+ \iota y$ on the right hand side, and then swap $f(x+ \iota y)$ with $x+ \iota y$ and solve for $f(x+\iota y)$, as that yields a different result to what my textbook gives. So, how do you tackle a problem like this?
| Note that, for any complex numbers $x+\iota y$ and $u + \iota v$, we have $f ( x+ \iota y) = u + \iota v$ if and only if $x+ \iota y = f^{-1}(u+\iota v)$, or $f^{-1}(u+\iota v) = x+\iota y$.
Suppose that $f(x +\iota y) = u + \iota v$, then we have
$$
(3+6x)+\iota (2−3y) = u + \iota v,
$$
which implies
$$
\begin{align}
3 + 6x &= u \\
2 - 3y &= v,
\end{align}
$$
and these last two equations in turn imply
$$
\begin{align}
x &= \frac{ u-3 }{ 6 } \\
y &= \frac{ 2-v }{ 3 }.
\end{align}
$$
Therefore we have
$$
f^{-1}(u+\iota v) = x+\iota y = \frac{ u-3 }{ 6 } + \iota \frac{ 2-v }{ 3 },
$$
that is,
$$
f^{-1}(u+\iota v) = \frac{ u-3 }{ 6 } + \iota \frac{ 2-v }{ 3 }.
$$
Hence the function $f^{-1} \colon \mathbb{C} \longrightarrow \mathbb{C}$ is defined by the formula
$$
f^{-1} (x + \iota y) = \frac{ x-3 }{ 6 } + \iota \frac{ 2-y }{ 3 }.
$$
PS:
There is one more piece of calculation / verification that is needed to be carried out.
We find that, for any complex $u + \iota v$, we have
$$
\begin{align}
\left( f \circ f^{-1} \right) ( u + \iota v) &= f \left( f^{-1} (u + \iota v) \right) \\
&= f \left( \frac{ u-3 }{ 6 } + \iota \frac{ 2-v }{ 3 } \right) \\
&= 3 + 6 \left( \frac{ u-3 }{ 6 } \right) + \iota \left[ 2 - 3 \left( \frac{ 2-v }{ 3 } \right) \right] \\
&= u + \iota v \\
&= i_{\mathbb{C}} ( u+\iota v),
\end{align}
$$
which implies that
$$
f \circ f^{-1} = i_{\mathbb{C}},
$$
where $i_{\mathbb{C}}$ denotes the identity function on the set $\mathbb{C}$ of complex numbers.
Similarly, for any complex number $x + \iota y$, we have
$$
\begin{align}
\left( f^{-1} \circ f \right) ( x+ \iota y) &= f^{-1} \big( f( x+ \iota y) \big) \\
&= f^{-1} \big( (3+6x) + \iota (2-3y) \big) \\
&= \frac{ ( 3+6x ) - 3 }{ 6 } + \iota \frac{ 2- (2-3y) }{ 3 } \\
&= x + \iota y \\
&= i_{\mathbb{C}} (x + \iota y),
\end{align}
$$
which implies
$$
f^{-1} \circ f = i_{\mathbb{C}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
When should I use partial fractions in generating functions I am currently studying generating functions and I don't understand why should I use partial fraction decomposition when solving $x^n$ coefficient of a question.
For example, in this function
$$\frac{1}{(1-x)(1-2x)^2} $$
Why should I use partial fractions to divide it to this expression
$$\frac{1}{1-x} - \frac{2}{1-2x} + \frac{2}{(1-2x)^2} = \sum_{i=0}^\infty \left(x^i - 2(2x)^i + 2 \binom{i+1}{1}(2x)^i\right) $$
to get the coefficient of $x^n$ and not transfer it to sums from the start like
$\begin{array}{cc}\frac{1}{1-x} = \sum_{i=0}^\infty x^i & \frac{1}{(1-2x)^2} = \sum_{i=0}^\infty \binom{i+1}{1}x^i \end{array}$
Therefore,
$$\frac{1}{(1-x)(1-2x)^2} = \sum_{i=0}^\infty x^i * \sum_{i=0}^\infty \binom{i+1}{1}x^i$$
and then find the coefficient from convolution.
Or this function instead of doing partial fraction I can do this
$$\frac{1 - x^2}{(1-x)^3} = (1-x^2) * \sum_{i=0}^\infty \binom{i+2}{2}x^i$$
and use convolution again
| First example:
The partial fraction decomposition in the first case is a convenient representation since we can directly derive a closed formula (without sums) from it. Denoting with $[x^n]$ the coefficient of $x^n$ of a series we obtain at a glance
\begin{align*}
[x^n]\left(\frac{1}{1-x} - \frac{2}{1-2x} + \frac{2}{(1-2x)^2}\right)
&=1-2\cdot2^{n}+2(n+1)2^n\\
&\,\,\color{blue}{=1+n2^{n+1}}\tag{1}
\end{align*}
On the other hand, if we consider instead
\begin{align*}
\frac{1}{(1-x)}\frac{1}{(1-2x)^2}
\end{align*}
we know the right-hand term admits the binomial expansion
\begin{align*}
\frac{1}{(1-2x)^2}&=\sum_{n=0}^\infty\binom{-2}{n}(-2x)^n\\
&=\sum_{n=0}^\infty(n+1)(2x)^n\tag{2}
\end{align*}
Multiplication with the term $\frac{1}{1-x}$ has the nice property
\begin{align*}
\frac{1}{1-x}\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty\left(\sum_{k=0}^na_k\right)x^n
\end{align*}
to transform coeffcients $a_n$ to the sum $\sum_{k=0}^n a_k$. It follows from (2) the wanted coefficient
\begin{align*}
[x^n]\frac{1}{(1-x)}\frac{1}{(1-2x)^2}=\color{blue}{\sum_{k=0}^n(k+1)2^k}\tag{3}
\end{align*}
Comparison of (1) and (3) gives
\begin{align*}
\color{blue}{\sum_{k=0}^n(k+1)2^k=1+n2^{n+1}}
\end{align*}
Note, without comparison we need some additional steps in order to simplify (3) into the closed form (1).
Second example:
Here we can easily derive the coefficient without partial fraction decomposition. We obtain for $n\geq 0$:
\begin{align*}
[x^n]\frac{1-x^2}{(1-x)^3}&=[x^n]\frac{1+x}{(1-x)^2}\\
&=[x^n](1+x)\sum_{k=0}^\infty (k+1)x^n\\
&=\left([x^n]+[x^{n-1}]\right)\sum_{k=0}^\infty (k+1)x^n\\
&\,\,\color{blue}{=2n+1}
\end{align*}
Conclusion: It depends on the specific situation if partial fraction decomposition is the convenient approach or some alternative techniques are more appropriate instead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I prove algebraically that $\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$? I found out that $\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}$, by reverse engineering:
$$\sqrt{6}-\sqrt{2} =
\sqrt{ (\sqrt{6}-\sqrt{2} )^{2} } = \sqrt{ (\sqrt{6}-\sqrt{2} ) (\sqrt{6}-\sqrt{2} )}= \sqrt{ 6-\sqrt{6×2}-\sqrt{6×2}+2}=\sqrt{ 8-2\sqrt{12}}=\sqrt{ 8-4\sqrt{3}}$$
But how can I do it the other way, also for other similar problems?
I tried multiple problem solving sites, but none could simplify $\sqrt{8-4\sqrt{3}}$ further then $2\sqrt{2-\sqrt{3}}$
| If you didn't know the answer already for $\sqrt{8-4\sqrt{3}}$, you could start with $$8-4\sqrt{3}=(\sqrt{a}-\sqrt{b})^2$$
$$\implies8-4\sqrt{3}=a-2\sqrt{ab}+b$$
$$\implies a+b=8$$ and $$\sqrt{ab}=2\sqrt{3}\implies ab=12$$
and solve for $a$ and $b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A $17$-digit number and the number formed by reversing its digits are added together. Show that the sum has a even digit. A $17$ digit number is chosen, and it's digits are reversed, forming a new number. These two numbers are added together. Show that there sum has at least one even digit.
The solution given in the book is as follows:
Suppose there were a $17$-digit integer whose reversed sum contained no even digit. For convenience, we number the columns of digits from right to left and consider the usual addition algorithm. The ninth digit of our number will be added to itself. This would produce an even digit in the answer, unless there is a "carry" from $8$th column. But if there is such a carry, then there must be one also from the $10$th column to the $11$th column (the $10$th column is identical is identical to the $8$th except for the order of the digits). Hence the $7$th column has digits of the same parity, and requires a carry from the sixth column. Proceeding similarly we find that there is a carry in each odd numbered column . But there cannot be a carry into the 1st column , so we have a contradiction.
However, I am not getting the part of the solution where it says, "Hence the $7$th column has digits of the same parity, and requires a carry from the sixth column." I am not getting this part ...
| $E$ indicates even; $O$ indicates odd; $d_{n}$ indicates digit $n$; column $k$ indicates the column number (with $1$ being the rightmost column and $17$ being the leftmost column); $\checkmark$ indicates that the condition of at least $1$ even digit in the sum has been satisfied.
$$\begin{array}{|c|c|}
\hline
&d_1& d_2& d_3& d_4& d_5& d_6& d_7& d_8& d_9&d_{10}& d_{11}& d_{12}& d_{13}& d_{14}& d_{15}& d_{16}& d_{17}\\ \hline
&d_{17}& d_{16}& d_{15}& d_{14}& d_{13}& d_{12}& d_{11}& d_{10}& d_9&d_8& d_7& d_6& d_5& d_4& d_3& d_2& d_1\\ \hline
A& & & & & & & & &E& &\\ \hline
B& & & & & & & & & E& \\ \hline
C& & & & & & & E& & O& \\ \hline
D& & & & & & & O& & O& &E\\ \hline
X& & & & & & & O& & O& & E&\\ \hline
F& & & & & E& & O& & O& & O\\ \hline
G& & & & & O& & O& & O& & O& &E\\ \hline
H& & & & & O& & O& & O& & O& &E\\ \hline
I& & & E& & O& & O& & O& & O& &O\\ \hline
...& & & & & & & & & & & & &\\ \hline
N& O& & O& & O& & O& & O& & O& & O& & O& &\color{red}E\\ \hline
\end{array}$$
$A = $ no carry $\rightarrow \checkmark$
$B = $ even carry from column $8 \rightarrow \checkmark$
$C = $ odd carry from column $8 \rightarrow$ odd carry from column $10$ and $d_7$ and $d_{11}$ have different parity $\rightarrow \checkmark$
$D = $ odd carry from column $8 \rightarrow$ odd carry from column $10$ and $d_7$ and $d_{11}$ have same parity $\rightarrow d_7 + d_{11}$ is even and no carry from column $6 \rightarrow \checkmark$
$X = $ odd carry from column $8 \rightarrow$ odd carry from column $10$ and $d_7$ and $d_{11}$ have same parity $\rightarrow d_7 + d_{11}$ is even and even carry from column $6 \rightarrow \checkmark$
$F = $ odd carry from column $8 \rightarrow$ odd carry from column $10$ and $d_7$ and $d_{11}$ have same parity $\rightarrow d_7 + d_{11}$ is even and odd carry from column $6 \rightarrow$ odd carry from column $12$ and $d_5$ and $d_{13}$ have different parity $\rightarrow \checkmark$
$G = $ odd carry from column $8 \rightarrow$ odd carry from column $10$ and $d_7$ and $d_{11}$ have same parity $\rightarrow d_7 + d_{11}$ is even and odd carry from column $6 \rightarrow$ odd carry from column $12$ and $d_5$ and $d_{13}$ have same parity $\rightarrow d_5 + d_{13}$ is even and no carry from column $4 \rightarrow \checkmark$
$H = $ odd carry from column $8 \rightarrow$ odd carry from column $10$ and $d_7$ and $d_{11}$ have same parity $\rightarrow d_7 + d_{11}$ is even and odd carry from column $6 \rightarrow$ odd carry from column $12$ and $d_5$ and $d_{13}$ have same parity $\rightarrow d_5 + d_{13}$ is even and even carry from column $4 \rightarrow \checkmark$
$I = $ odd carry from column $8 \rightarrow$ odd carry from column $10$ and $d_7$ and $d_{11}$ have same parity $\rightarrow d_7 + d_{11}$ is even and odd carry from column $6 \rightarrow$ odd carry from column $12$ and $d_5$ and $d_{13}$ have same parity $\rightarrow d_5 + d_{13}$ is even and odd carry from column $4 \rightarrow$ odd carry from column $14$ and $d_{3}$ and $d_{15}$ have different parity $\rightarrow \checkmark$
Do you see the pattern?
If we continue as above, we will eventually reach a point where if there is no carry to column $1$, there will be an even digit in column $1$. However, there can't be a carry to column $1 \implies$ the digit in column $1$ has to be even. Hence, we will always have at least $1$ even digit in the sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is $d(\mathbb{Q}(\sqrt{2},\sqrt{3}))$? $K=\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$ and $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}]=4$. We also know the conjugates of $\sqrt{2}+\sqrt{3}$ are,
$$x_1=\sqrt{2}+\sqrt{3}$$
$$x_2=\sqrt{2}-\sqrt{3}$$
$$x_3=-\sqrt{2}+\sqrt{3}$$
$$x_4=-\sqrt{2}-\sqrt{3}$$
$\mathbb{Q}(\sqrt{2}+\sqrt{3})=\{a+b(\sqrt{2}+\sqrt{3})+c(\sqrt{2}+\sqrt{3})^2+d(\sqrt{2}+\sqrt{3})^3|a,b,c,d \in \mathbb{Q}\}$. Thus an integral basis of $K$ is,
$$S=\{1,\sqrt{2}+\sqrt{3},(\sqrt{2}+\sqrt{3})^2,(\sqrt{2}+\sqrt{3})^3\}$$
$$S=\{1,\sqrt{2}+\sqrt{3},5+2\sqrt{6},(\sqrt{2}+\sqrt{3})^3\}$$
and hence,
$$d(K)=d(\mathbb{Q}(\sqrt{2}+\sqrt{3}))=D(S)$$
$$D(S)=\begin{vmatrix}
1 & \sqrt{2}+\sqrt{3} & (\sqrt{2}+\sqrt{3})^2 & (\sqrt{2}+\sqrt{3})^3\\
1 & \sqrt{2}-\sqrt{3} & (\sqrt{2}-\sqrt{3})^2 & (\sqrt{2}-\sqrt{3})^3\\
1 & -\sqrt{2}+\sqrt{3} & (-\sqrt{2}+\sqrt{3})^2 & (-\sqrt{2}+\sqrt{3})^3\\
1 & -\sqrt{2}-\sqrt{3} & (-\sqrt{2}-\sqrt{3})^2 & (-\sqrt{2}-\sqrt{3})^3
\end{vmatrix}^2$$
$$D(S)=\begin{vmatrix}
1 & \sqrt{2}+\sqrt{3} & 5+2\sqrt{6} & 11\sqrt{2}+9\sqrt{3}\\
1 & \sqrt{2}-\sqrt{3} & 5-2\sqrt{6} & 11\sqrt{2}-9\sqrt{3}\\
1 & -\sqrt{2}+\sqrt{3} & 5-2\sqrt{6} & -11\sqrt{2}+9\sqrt{3}\\
1 & -\sqrt{2}-\sqrt{3} & 5+2\sqrt{6} & -11\sqrt{2}-9\sqrt{3}
\end{vmatrix}^2$$
Using Mathematica,
$$D(S)=2677248 + 1078272\sqrt{6}$$
Why am I not getting an integer answer?
Appararently I don't know how to type, $D(S)= 147456$
| Well, my favorite method of defining (and calculating!) the discriminant of a free algebra is as the determinant of the trace pairing. Before I launch into this, however, in case your ring is of the form $\Bbb Q[a]$, so that a basis is the powers $\{a^m\}_{0\le m<n}$, where the rank is $n$, there’s a quick and dirty way of calculating the discriminant. This is just $\text{Norm}\bigl(f'(a)\bigr)$, where $f$ is the minimal polynomial of $a$. Using $a=\sqrt2+\sqrt3$ and its minimal polynomial, you do indeed get $2^{14}\cdot3^2$. But we know that this method is not apposite to your problem.
If it’s true that an integral basis of your field is $\{1,\sqrt2,\sqrt3, (\sqrt3-1)/\sqrt2\}$, we can label these $b_1$ through $b_4$ and write down the matrix whose $(i,j)$-th entry is $\text{Trace}(b_ib_j)$, and calculate its determinant. I’ll let you check my calculation that the matrix in question is
$$
\begin{pmatrix}
4&0&0&0\\0&8&0&-4\\0&0&12&0\\0&-4&0&8
\end{pmatrix}\,,
$$
whose determinant is $2304=2^8\cdot3^2$. Since this agrees with Daniel’s result, it seems my guess of the basis was correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4420178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Given that $ a,b,c$ are positive numbers and $(a+b)(b+c)(c+a)=1$, find the maximum value of $P=ab+bc+ca$ I am stuck with this problem:
Given that $ a,b,c$ are positive numbers and $(a+b)(b+c)(c+a)=1$, find the maximum value of $P=ab+bc+ca$
I tried to use: $(a+b)(a+c)(b+c)=(a+b+c)(ab+bc+ca)-abc \implies P=\frac{1+abc}{a+b+c}$ but I am stuck. Can anyone help me? Thank you!
I'm just 14 years old, so don't use derivative or something like that (my brother uses it but I cant understand).
| We have $(a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) - abc$ which results in
$$P = \frac{1 + abc}{a + b + c}.$$
Using AM-GM, we have
$$(a + b)(b + c)(c + a)
\ge 2\sqrt{ab}\cdot 2\sqrt{bc}\cdot 2\sqrt{ca} = 8abc$$
and
$$(a + b)(b + c)(c + a)\le \left(\frac{a + b + b + c + c + a}{3}\right)^3 = \frac{8}{27}(a + b + c)^3.$$
Thus, we have
$abc \le 1/8$ and $a + b + c\ge 3/2$.
Thus, we have
$$P = \frac{1 + abc}{a + b + c} \le \frac{1 + 1/8}{3/2} = 3/4.$$
Also, when $a = b = c = 1/2$,
we have $(a + b)(b + c)(c + a) = 1$ and $P = 3/4$.
Thus, the maximum of $P$ is $3/4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving an exponential diophantine equation I came across the following problem:
Find all positive integers $a,b$ such that
The following equation is satisfied by the two integers:
$5^b+2=7^c$
My Attempt: I tried equating both sides modulo 6. We know $7^c\cong 1\mod 6$. Hence $5^b\cong 5\mod 6$ which means $b=2k+1$. Thus the equation reduces to $7^c=2+5^{2k+1}$. Now I tried subtracting 7 from both sides to factor the terms which gives:
$7(7^c-1)=5(5^{2k}-1)$.
Now since 5 divides rhs it must divide lhs but since 7 is not divisible by 5 so 5 must divide the other factor
$7^(c-1)\cong 1\mod 5$ and thus $c=4p+1$. But all these efforts don't seem to help. Any help would be appreciated.
| Better use a modulo where you can really look at the specifics of the case $b \ge 2$ or $c \ge 2$, because without either of those, you won't manage to prove much looking only at possible remainders (since the equality can stand with $b=c=1$).
For instance, modulo $25$, the LHS is always $2 \pmod{25}$, but you can check that \begin{align*} 7^{4k} & \equiv 1 \pmod{25} \\ 7^{4k+1} & \equiv 7 \pmod{25} \\ 7^{4k+2} & \equiv -1 \pmod{25} \\ 7^{4k+3} & \equiv -7 \pmod{25} \end{align*}
so no solution for $b \ge 2$. For $b=1$, it is obvious that we need $c=1$, and that is indeed a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Express $2\cos^2x - 3\sin^2x$ in terms of $\cos(2x)$ I need to express $2\cos^2x - 3\sin^2x$ in terms of $\cos(2x)$.
I have substituted the $\sin^2x$ using trig identities and have reached:
$5\cos^2x - 3$.
But now I'm really stuck.
| So $2\cos^{2}x-3\sin^{2}x=5\cos^{2}x -3$ but also $\cos 2x=2\cos^{2}x-1$ substitution give \begin{align*} 2\cos^{2}x-3\sin^{2}x&=5\cos^{2}x -3,\\&=5\left(\frac{1}{2}+\frac{1}{2}\cos2x\right)-3,\\&=-\frac{1}{2}+\frac{5}{2}\cos 2x.\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4431956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Relation between roots and coefficients of equation If the roots of the equation $x^4 - x^3 +2x^2+x+1 = 0 $ are given by $a,b,c,d$ then find the value of $(1+a^3)(1+b^3)(1+c^3)(1+d^3)$
I found out that:
$$ (1+a^3)(1+b^3)(1+c^3)(1+d^3) = (abcd)^3+\sum(abc)^3 +\sum(ab)^3+\sum(a)^3 +1$$
But how do I calculate $\sum(abc)^3$ and $\sum(ab)^3$?
| Something seemed a bit suspicious about that polynomial: poking around a little, I found that
$$ ( x^4 \ - \ x^3 \ + \ 2x^2 \ + \ x \ + \ 1)·(x^2 \ + \ x \ - \ 1 ) \ \ = \ \ x^6 \ + \ 4x^3 \ - \ 1 \ \ . $$
Treating this as a "quadratic in $ \ x^3 \ \ , $ " we find that there are two "triplets" of zeroes given by $ \ x^3 \ = \ -2 \ \pm \ \sqrt5 \ \ . $ The factor $ \ (x^2 \ + \ x \ - \ 1 ) \ $ has the "familiar" real zeroes $ \ -\frac12 \ + \ \frac{\sqrt{5}}{2} \ = \ \frac{1}{\phi} \ \ $ and $ \ -\frac12 \ - \ \frac{\sqrt{5}}{2} \ = \ -\phi \ \ $ (the Golden Ratio emerges from the shadows once more!). The cubes of these numbers are indeed $ \ \frac{1}{\phi^3} \ = \ -2 + \sqrt5 \ $ and $ \ (-\phi)^3 \ = \ -2 - \sqrt5 \ \ , $ so the other four zeros of $ \ x^6 \ + \ 4x^3 \ - \ 1 \ \ , $ and thus the four zeroes of $ \ x^4 \ - \ x^3 \ + \ 2x^2 \ + \ x \ + \ 1 \ \ , $ are the four complex(-conjugate) cube-roots of $ \ -2 + \sqrt5 \ $ and $ \ -2 - \sqrt5 \ \ . $
Consequently, we may designate $ \ a^3 \ , \ b^3 \ = \ -2 \ + \ \sqrt5 \ \ , $ with $ \ b \ = \ \overline{a} \ \ , $ and $ \ c^3 \ , \ d^3 \ = \ -2 \ - \ \sqrt5 \ \ , $ with $ \ d \ = \ \overline{c} \ \ . $ We then seek the value of
$$ (1 + a^3)·(1 + b^3)·(1 + c^3)·(1 + d^3) \ \ = \ \ (1 + a^3)·(1 + \overline{a}^3)·(1 + c^3)·(1 + \overline{c}^3) $$
$$ = \ \ ( \ 1 \ + \ a^3 \ + \ \overline{a}^3 \ + \ [a·\overline{a}]^3 \ )·( \ 1 \ + \ c^3 \ + \ \overline{c}^3 \ + \ [c·\overline{c}]^3 \ ) \ \ . $$
It remains to determine the last terms in each of the factors above. Since $ \ a \ , \ \overline{a} \ , $ and $ \ \alpha \ = \ -\frac12 \ + \ \frac{\sqrt{5}}{2} \ $ all have the cube $ \ -2 + \sqrt5 \ , $ it follows that $ \ (a·\overline{a}·\alpha)^3 \ = \ (-2 + \sqrt5)^3 \ \ , $ and hence $ \ (a·\overline{a})^3 \ = \ (-2 + \sqrt5)^2 $ $ \ = \ 9 \ - \ 4\sqrt5 \ \ ; \ $ similarly, $ \ (c·\overline{c})^3 \ = \ (-2 - \sqrt5)^2 $ $ \ = \ 9 \ + \ 4\sqrt5 \ \ . $
So at last we obtain
$$ (1 + a^3)·(1 + b^3)·(1 + c^3)·(1 + d^3) $$ $$ = \ \ ( \ 1 \ + \ [-2 + \sqrt5] \ + \ [-2 + \sqrt5] \ + \ [9 - 4\sqrt5] \ )$$ $$· \ ( \ 1 \ + \ [-2 - \sqrt5] \ + \ [-2 - \sqrt5] \ + \ [9 + 4\sqrt5] \ ) $$
$$ = \ \ ( \ 6 \ - \ 2\sqrt5 \ ) \ · \ ( \ 6 \ + \ 2\sqrt5 \ ) \ \ = \ \ 36 \ - \ 20 \ \ = \ \ 16 \ \ . $$
[Evidently, this approach has connections to the more sophisticated methods employed by dxiv and Snacc.]
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $x^4+y^4+z^4=1$ Find the minimum value of $\sum_{cyc}\frac{x^3}{1-x^8}$. Let $x,y,z$ be positive reals such that $x^4+y^4+z^4=1$. Find the minimum value of
$$\sum_{cyc}\frac{x^3}{1-x^8}$$
First I tried Jensen. Let $$f(x)=\frac{x^3}{1-x^8}$$
Then $f$ is convex on $\mathbb R^+$, using Jensen $$\frac{1}{3}\sum_{cyc}f(x)\ge f\left(\frac{x+y+z}{3}\right)$$
but we don't have enough information about $x+y+z$.
By the way my guess for the minimum value is at $x=1/\sqrt[4]{3}$
| We have
\begin{align*}
\frac{1}{1 - x^8}
&= \frac{1}{8/9 + (1/9 - x^8)}\\
&= \frac{9}{8}\cdot \frac{1}{1+ (9/8)(1/9 - x^8)}\\
&\ge \frac98 \left(1 - \frac{9}{8}(1/9 - x^8)\right) \tag{1}\\
&= \frac{9}{64}(7 + 9x^8)\\
&\ge \frac{9}{64}\cdot 8\sqrt[8]{1^7 \cdot 9x^8} \tag{2}\\
&= \frac98 \sqrt[8]{9} \, x
\end{align*}
where we have used $\frac{1}{1 + u} \ge 1 - u$ for all $-1 < u < 1$ in (1) and AM-GM in (2).
Thus, we have
$$\sum_{\mathrm{cyc}} \frac{x^3}{1 - x^8} \ge \frac98 \sqrt[8]{9}\, (x^4 + y^4 + z^4) = \frac98 \sqrt[8]{9}.$$
Also, when $x = y = z = 1/\sqrt[4]{3}$, we have $x^4 + y^4 + z^4 = 1$ and $\sum_{\mathrm{cyc}} \frac{x^3}{1 - x^8} = \frac98 \sqrt[8]{9}$.
Thus, the minimum of $\sum_{\mathrm{cyc}} \frac{x^3}{1 - x^8}$ is $\frac98 \sqrt[8]{9}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Generating Function Approach giving wrong combination count while normal brute force casework approach giving correct answer I am solving a problem: How many words are less than four letters long and contain only the letters A, B, C, D, and E? Here, 'word' refers to any string of letters.
My Solution 1 uses Generating functions gives wrong answer
$(1+x+x^2+x^3)(1+x+x^2+x^3)(1+x+x^2+x^3)(1+x+x^2+x^3)(1+x+x^2+x^3)$
where $(1+x+x^2+x^3)$ is Generating function for each letter. This approach gave a wrong answer $(35+15+5+1)$ which is the sum of coefficients of $(x^3, x^2, x, 1)$ in $$x^{15} + 5 x^{14} + 15 x^{13} + 35 x^{12} + 65 x^{11} + 101 x^{10} + 135 x^9 + 155 x^8 + 155 x^7 + 135 x^6$$ $$+ 101 x^5 + 65 x^4 + 35 x^3 + 15 x^2 + 5 x + 1$$
Correct Approach using Case Work gives the correct answer
Case 1: The word is one letter long. Clearly, there are $5$ of these words.
Case 2: The word is two letters long. Constructing the set of these words, there are $5$ options for the first letter and $5$ options for the second letter, so there are $5^2 = 25$ of these words.
Case 3: The word is three letters long. By similar logic as above, we have $5$ options for the first letter, $5$ options for the second, and $5$ options for the third. Then there are $5^3 = 125$ of these letters.
Adding all our cases up, there are $5 + 25 + 125 = 155$ words that are less than four letters long and contain only the letters A, B, C, D, and E.
Could Someone help me in explaining why the Generating function approach failed here?
| Ordinary Generating functions does not calculate the permutations. It gives the combinations and thus order is not taken into account. So, second appraoch gives right answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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$A^2 - B^2$ is an invertible matrix and $A^5 = B^5$ and $A^3B^2 = A^2B^3$ . Then what is the determinant of the matrix $A^3 + B^3$? $A,B$ are two $3 \times 3$ real-matrices with following three properties.
*
*$A^2 - B^2$ is an invertible matrix
*$A^5 = B^5$
*$A^3B^2 = A^2B^3$
Then what is the determinant of the matrix $A^3 + B^3$ ?
| Let $V=\mathbb R^3$. The condition $A^5=B^5$ implies that $A^5V=B^5V$. Since $A$ is $3\times 3$, the descending chain $V\supseteq AV\supseteq A^2V\supseteq\cdots$ must have been stabilised at $A^3V$ and the similar is true for $B^3V$. Therefore from $A^5V=B^5V$ we get $W:=A^3V=B^3V$.
Furthermore, as the descending chains have been stabilised, both $A$ and $B$ are automorphisms on the invariant subspace $W$. The condition $A^3B^2=A^2B^3$ thus implies that $A=B$ on $W$. As $A^2-B^2$ is invertible, $W$ must be zero. Hence $A^3V=B^3V=0$, i.e., $A^3=B^3=0$. Consequently $\det(A^3+B^3)=0$.
Remark. Actually, for any number $c$, “$\det(A^3+B^3)=c$” is a correct answer, because no matrices $A$ and $B$ can satisfy all three given conditions in the first place. Suppose the contrary. By our previous argument, $A$ and $B$ must be nilpotent. Since the matrices are $3\times3$, the ranks of $A^2$ and $B^2$ are at most one. Therefore the rank of $A^2-B^2$ is at most two. Hence $A^2-B^2$ is singular, but this is a violation of the first condition.
However, if we change the first condition to “$A-B$ is invertible”, there are feasible pairs of $A$ and $B$, such as
$$
A=\pmatrix{0&2&0\\ 0&0&1\\ 0&0&0}
\text{ and }B=\pmatrix{0&1&0\\ 0&0&0\\ 1&0&0}.
$$
In this case, our previous argument still applies and we may conclude that $A^3=B^3=0$ and $\det(A^3+B^3)=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find $x$ such that $\frac{1}{2^{p-1}} -\frac{1}{a^p}>x$? Let $a, p\in\mathbb{R}$ be such that $a\ge 2$ and $p>1$. Consider the quantity
$$\frac{1}{2^{p-1}} -\frac{1}{a^p}.$$
Since $a\ge 2$, it seems clear to me that
$$\frac{1}{2^{p-1}} -\frac{1}{a^p}>0.$$
My question is: it is possible to find a positive real number $x$ (not depending on $a$) such that
$$\frac{1}{2^{p-1}} -\frac{1}{a^p}>x?$$
Could someone please help me in finding that?
Thank you in advance!
| $$
f(a) = \frac{1}{2^{p-1}} -\frac{1}{a^p}
$$
is increasing in $a$, therefore is
$$
f(a) \ge f(2) = \frac{1}{2^{p-1}} -\frac{1}{2^p}= \frac{1}{2^p}
$$
for $a \ge 2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Calculate the minimum value of $T = 3x + 2y$ with positives $x$ and $y$ satisfying $\log_3\frac{xy + 2}{x^2 + 1} = \frac{3^x + 3x^2 + 1}{3^x + xy}$.
Consider two positives $x$ and $y$ such that $\log_3\dfrac{xy + 2}{x^2 + 1} = \dfrac{3^x + 3x^2 + 1}{3^x + xy}$. Calculate the minimum value of $T = 3x + 2y$.
We have that $$\begin{aligned} \log_3\dfrac{xy + 2}{x^2 + 1} = \dfrac{3^x + 3x^2 + 1}{3^x + xy} \iff \log_3(xy + 2) - \log_3(3x^2 + 3) = \dfrac{(3x^2 + 3) - (xy + 2)}{3^x + xy} \end{aligned}$$
I don't know how to notate this, but let $z$ be a variable and $f(x), g(x)$ be functions such that $g(f(x), z) = f(x) + (3^z + yz)\log_3f(x)$, which means that $g(xy + 2, x) = g(3x^2 + 3, x)$ and $$\dfrac{\mathrm dg(f(x), x)}{\mathrm dx} = f'(x) + \dfrac{(3^x\ln 3 + y)\ln f(x)}{\ln 3} + \dfrac{(3^x + xy)f'(x)}{\ln 3f(x)}$$
Entering the equation $g'(x) = 0$ into WolframAlpha, function $f(x)$ can be written in the form of $$f(x) = \dfrac{3^x + xy}{\ln 3}W\left(\dfrac{\ln 3}{(3^x + xy)e^{(3^x + xy)/c}}\right)$$ with $c$ being a constant and $W(z)$ being the product log function/Lambert $W$ function. Now, I don't know about this function, so I'd consider this a dead end.
I do have a hunch that $xy + 2 = 3x^2 + 3$, I just don't know how to prove it.
How about continuing from this point on? It can be implied that $y = \dfrac{3x^2 + 1}{x}$, which makes $T$ equal to $3x + \dfrac{2(3x^2 + 1)}{x}$. From there, $T' = 0 \iff 9 - \dfrac{2}{x^2} = 0 \iff x = \dfrac{\sqrt 2}{3}$
Drawing the table of variations for $T$, it can be seen that $\min T = 9 \cdot \dfrac{\sqrt 2}{3} + \dfrac{2}{\dfrac{\sqrt 2}{3}} = 6\sqrt{2}$.
Now, if I could just prove that $xy + 2 = 3x^2 + 3$, it would be fantastic. So I turn to your support.
As always, thanks for reading, (and even more if you could help~)
| Build the auxiliary Lagrangian function
$$\mathscr L=3x+2y-\lambda\left(\log_3\frac{xy+2}{x^2+1}-\frac{3^x+3x^2+1}{3^x+xy}\right)+\mu_1x+\mu_2y.$$
The first order conditions for a minimum (see KKT multipliers) are
\begin{align}
\frac{\partial\mathscr L}{\partial x}&=0\\
\frac{\partial\mathscr L}{\partial y}&=0\\
\frac{\partial\mathscr L}{\partial \lambda}&=0\\
\mu_1x&=0\\
\mu_2y&=0
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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provided $f( x+1) =\ \lim _{n\rightarrow \infty }\left(\frac{n+x}{n-2}\right)^{n}$, what is f(x)? $$
f( x+1) =\ \lim _{n\rightarrow \infty }\left(\frac{n+x}{n-2}\right)^{n}
$$
Here is one of the solution from my workbook:
$$
\begin{array}{l}
f( x+1) \ =\ \lim _{n\rightarrow \infty }\left[\left( 1+\frac{2+x}{n-2}\right)^{\frac{n-2}{2+x}}\right]^{n\left(\frac{2+x}{n-2}\right)} =\ e^{2+x} =e^{1+( x+1)}\\
f( x) \ =\ e^{1+x}
\end{array}
$$
but is a bit preplexing for me :(
| First, $$\frac{n+x}{n-2} = 1 + \frac{1}{\frac{n-2}{2+x}}$$
from rearranging.
Then they use the following fact $$\lim_{k \to \infty} \left(1 + \frac{1}{k}\right)^k = e$$
which implies
$$\lim_{n \to \infty}\left(1 + \frac{1}{\frac{n-2}{2+x}}\right)^{\frac{n-2}{2+x}} = e.$$
By continuity, we can raise both sides to the $2+x$ power.
$$\lim_{n \to \infty}\left(1 + \frac{1}{\frac{n-2}{2+x}}\right)^{\frac{n-2}{2+x} \cdot (2+x)} = e^{2+x}.$$
Finally, note that
$$\begin{align}
\lim_{n \to \infty}\left(1 + \frac{1}{\frac{n-2}{2+x}}\right)^{\frac{n-2}{2+x} \cdot \frac{n}{n-2}(2+x)}
&=
\lim_{n \to \infty}\left(1 + \frac{1}{\frac{n-2}{2+x}}\right)^{\frac{n-2}{2+x} \cdot (2+x)}
\cdot
\underbrace{\lim_{n \to \infty}\left(1 + \frac{1}{\frac{n-2}{2+x}}\right)^{2}}_{=1}
\\
&=
\lim_{n \to \infty}\left(1 + \frac{1}{\frac{n-2}{2+x}}\right)^{\frac{n-2}{2+x} \cdot (2+x)}
\\
&= e^{2+x}.\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a any matrix $C$ satisfying the below conditions. Let the $A,B$ and $C$ are $3\times 3$ matrices.
For the vectors $v_1= \begin{bmatrix} 1 \\ 1 \\ -1 \\ \end{bmatrix}$,$v_2= \begin{bmatrix} 0 \\ 1 \\ -1 \\ \end{bmatrix}$ and $v_3= \begin{bmatrix} 1 \\ 1 \\ -2 \\ \end{bmatrix}$, $Av_1=v_1
Av_2=v_2, Av_3=-2v_3, Bv_1 = -v_1, Bv_2 = 2v_2$ and $Bv_3 = -v_3$. Find a any $C$ whose eigenvalues are $0,1,-1$ with $AC=CA$ and $BC=CB$.
I take the case $v_1, v_2$ and $v_3$ are eigenvectors of the $0,1$ and $-1$ respectively.
So Putting $P= [v_1 \vert v_2 \vert v_3 ]= \begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ -1 & -1& -2 \\\end{bmatrix}$ and $D = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0& -1 \\\end{bmatrix}$.
From those we get $P^{-1} = \begin{bmatrix} 1 & 1 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & -1 \\\end{bmatrix}$. If we take the $C = P^{-1}DP$ the conditions of the $AC=CA$ and $BC=CB$ would satisfy because $A, B$ are diagonalizable considering the $P$ (e.g $PD_1P^{-1} =A$)
Consequently my final answer is $C = P^{-1}DP = \begin{bmatrix} -1 & 1 & 1 \\ -2 & 2 & 1 \\ 3 & -3 & -2 \\\end{bmatrix}$. But the eigenvalues of the $C$ which I made are $0,1$ and $2$. I can't find any error in my solution. Please let me know my mistake.
| $$C = P^{-1}DP=\left(\begin{bmatrix} 1 & 1 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & -1 \\\end{bmatrix}\begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0& -1 \\\end{bmatrix}\right)\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ -1 & -1& -2 \\\end{bmatrix}$$
$$=\left(\begin{bmatrix} 0 & 1 & -1 \\ 0 & 1 & 0\\ 0& -1& 1 \\\end{bmatrix}\right)\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ -1 & -1& -2 \\\end{bmatrix}$$
$$= \begin{bmatrix} 2 & 2 & 3 \\ 1 & 1 & 1 \\ -2 & -2& -3 \\\end{bmatrix}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\log(x) -\frac{1}{2}\log(x-\frac{1}{2}) = \log(x+\frac{1}{2}) - \frac{1}{2}\log(x+\frac{1}{8})$
Find $x$ in the equation $\log(x) -\frac{1}{2}\log(x-\frac{1}{2}) = \log(x+\frac{1}{2}) - \frac{1}{2}\log(x+\frac{1}{8})$.
My attempt:
$$\log\left(x\right)\ -\ \frac{1}{2}\ \log\left(x-\frac{1}{2}\right)\ =\ \log\left(x+\frac{1}{2}\right)\ -\ \frac{1}{2}\log\left(x\ +\ \frac{1}{8}\right)$$
$$\log\left(x\right)\ -\ \frac{1}{2}\log\left(\frac{2x-1}{2}\right)\ =\ \log\left(\frac{2x+1}{2}\right)\ -\ \frac{1}{2}\log\left(\frac{8x+1}{8}\right)$$
$$\log\left(x\right)\ -\ \frac{1}{2}\log\left(2x-1\right)\ +\ \frac{1}{2}\log\left(2\right)\ =\ \log\left(2x+1\right)\ -\ \log\left(2\right)\ -\ \frac{1}{2}\log\left(8x+1\right)\ +\ \frac{1}{2}\log\left(8\right)$$
$$\log\left(x\right)\ -\ \frac{1}{2}\log\left(2x-1\right)\ =\ \log\left(2x+1\right)\ -\ \frac{1}{2}\log\left(8x+1\right)$$
$$\log\left(x\right)\ -\ \frac{1}{2}\log\left(2x-1\right)\ -\ \log\left(2x+1\right)\ +\ \frac{1}{2}\log\left(8x+1\right)=0$$
$$2\log\left(x\right)\ -\ \log\left(2x-1\right)\ -\ 2\log\left(2x+1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$
$$\log\left(x^2\right)\ -\ \log\left(2x-1\right)\ -\ \log\left(4x^2 + 4x +1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$
$$\log\left(x^2(8x+1)\right)\ -\ \log\left(2x-1\right)\ -\ \log\left(4x^2 + 4x +1\right) = 0$$
$$\log\left(\frac{x^{2}\left(8x+1\right)}{(2x-1)(4x^2 + 4x + 1)}\right)=\ 0\ \tag{1}$$
I'm not getting any idea what to do with this further. Also, the original equation has only $x = 1$ as a solution but the equation $(1.)$ has $2$ solutions $x = 1; x= -1/3$.
I think there might be a simpler solution to this equation rather than my cumbersome one.
Not sure what's the mistake.
| Simplifying $$\log(x) -\frac{1}{2}\log(x-\frac{1}{2}) = \log(x+\frac{1}{2}) - \frac{1}{2}\log(x+\frac{1}{8})$$
into
$$\log\left(\frac{x}{(x-\frac{1}{2})^\frac{1}{2}}\right) = \log\left(\frac{x+\frac{1}{2}}{(x+\frac{1}{8})^\frac{1}{2}}\right)$$
And taking the exponential,
$$\frac{x}{(x-\frac{1}{2})^\frac{1}{2}} = \frac{x+\frac{1}{2}}{(x+\frac{1}{8})^\frac{1}{2}}$$
Taking the square of both sides,
$$\frac{x^2}{x-\frac{1}{2}} = \frac{x^2+x+\frac{1}{4}}{x+\frac{1}{8}}$$
Simplifying the equation, we have
$$-3x^2 + 2x +1 = 0$$
Use quadratic equation to find two solutions $$x=1, x=-1/3$$
that you were looking for. I hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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.