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Finding $\cos ( 2 \sin^{-1}( \frac{5}{ 13} )) $ The following problem is from the $8$th edition of the book Calculus, by James Stewart. It is problem number $9$ in section $6.6$.
Problem:
Find an exact value for the expression:
$$ \cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } $$
Answer:
\begin{align*}
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - \sin^2{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\
%
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - 2 \sin^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\
%
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - 2 \left( \frac{25}{13^2} \right) \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\
%
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - \left( \frac{50}{13^2} \right) \cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } } \\
\end{align*}
\begin{align*}
\cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
1 - \sin^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } \\
\cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } &= 1 - \frac{25}{169} = \frac{169 - 25}{169} \\
\cos^2{\left( \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \frac{144}{169} \\
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &=
\sqrt{ 1 - \left( \frac{50}{13^2} \right) \left( \frac{144}{169} \right) } \\
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \sqrt{ \frac{13^4 - 50(144)}{13^4} } \\
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \sqrt{ \frac{21361}{13^4} } \\
\cos{\left( 2 \sin^{-1}\left( \frac{5}{13} \right) \right) } &= \frac{ \sqrt{ 21361 } } { 169 }
\end{align*}
The book's answer is $\frac{119}{169}$ and SciLab matches the book. Where did I go wrong?
| Your mistake is when you squared $\sin2\theta$, it should be $4\sin^{2}\theta\cos^{2}\theta$ not $2\sin^{2}\theta\cos^{2}\theta$, but in case you want an easier method than going for a long method, here is one:
Just let $\displaystyle \theta=\sin^{-1}\left(\frac{5}{13}\right)$ so that we have $\sin \theta=\dfrac{5}{13}$ and now we know that $\displaystyle \cos 2\theta=1-2\sin^{2}\theta=1-2\left(\frac{5}{13}\right)^2=\frac{119}{169}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\frac{\int_{0}^{\infty}(1+x^2)^{-2012}dx}{\int_{0}^{\infty}(1+x^2)^{-2011}dx}$
QUESTION: Evaluate $$\frac{\int_{0}^{\infty}(1+x^2)^{-2012}dx}{\int_{0}^{\infty}(1+x^2)^{-2011}dx}$$
MY ANSWER: I have got $2$ doubts on this; But let me write down my solution first..
First we solve for some general $\int_{0}^{\infty} (1+x^2)^{-n}dx$. Let $x=\tan \theta$.
$\implies \theta=\tan ^ {-1} x$
$\implies d\theta=\frac{1}{1+x^2}dx$
$\implies (1+x^2)d\theta=dx$
Putting this in the integration we get
$$\int_{0}^{\frac{\pi}2} (1 + \tan ^2 \theta )(1 + \tan ^2 \theta)^{-n} d\theta$$
$$=\int_{0}^{\frac{\pi}{2}} (\sec^2x)^{1-n} dx$$
$$=\int_{0}^{\frac{\pi}2} (\cos x)^{2n-2}dx$$
Now, we can use the reduction formula of the integration of $\cos ^n x$ which can be easily derived by integrating by parts. I directly write the formula here.. $$\int \cos ^n x = \frac{1}n \sin x \cos ^{n-1} x + \frac{n-1}n \int \cos ^{n-2} x dx$$
$$\implies \int_{0}^{\frac{\pi}2} \cos ^n x = \underbrace {\frac{1}n \sin x \cos ^{n-1} x \mid_{0}^{\frac{\pi}2}}_{0} + \frac{n-1}n \int_{0}^{\frac{\pi}2} \cos ^{n-2} x dx$$
$$\therefore \int_{0}^{\frac{\pi}2} \cos ^n x =\frac{n-1}n \int_{0}^{\frac{\pi}2} \cos ^{n-2} x dx$$
Thus we get a recursive relation.. Now in place of $n$ we just put $ (2n-2)$ and solve it out..
$$\therefore \int_{0}^{\frac{\pi}2} \cos ^{(2n-2)} x =\frac{2n-2-1}{2n-2} \int_{0}^{\frac{\pi}2} \cos ^{(2n-2-2)} x dx$$
$$\implies \int_{0}^{\frac{\pi}2} \cos ^{(2n-2)} x =\frac{2n-3}{2n-2} \underbrace{\int_{0}^{\frac{\pi}2} \cos ^{(2n-4)} x dx}_{\text{again continue from here...}}$$
$$\vdots \text{ } \vdots \text{ }\vdots$$
$$\implies \int_{0}^{\frac{\pi}2} \cos ^{(2n-2)} x = \frac{(2n-3).(2n-5).(2n-7) \cdots (1)}{(2n-2).(2n-4) \cdots 2} \int_{0}^{\frac{\pi}2} 1 dx$$
$$\implies \int_{0}^{\frac{\pi}2} \cos ^{(2n-2)} x= \frac{(2n-3)!}{[(2n-2).(2n-4)\cdots(2)]^2} \frac{\pi}2$$
$$\implies \int_{0}^{\frac{\pi}2} \cos ^{(2n-2)} x= \frac{(2n-3)!}{[2^{n-1}(k-1)!]^2} \frac{\pi}2$$
Well, now the question becomes easy..
First doubt, in the official solution, they have also followed the same path, but their end function is
$$\int_{0}^{\frac{\pi}2} \cos ^{(2n-2)} x=\frac{\pi}2 \frac{(2n-2)!}{2^{2n-2}(n-1)!(n-1)!}$$
Everything is fine except that they have got $(2n-2)!$ and I have got $(2n-3)!$.
Then where is the error in my solution?
Next question, why can't we just write $$\frac{\int f(x)dx}{\int g(x)dx}=\int \frac{f(x)}{g(x)}dx$$
This would have made the problem a child's play then.. :P
Any alternate solutions to this problem will be greatly acknowledged..
Thank You for your help in advance..
| Here is my approach which is to use Beta function:
$$I_n: = \int_{0}^{\infty} \frac{1}{\left(x^2+1\right)^n}~dx = \frac{1}{2}\cdot \int_{0}^{\infty} \frac{1}{\sqrt{t}\left(t+1\right)^n}~dt=\frac{1}{2}\cdot\mathrm{B} \left(\frac{1}{2}, n - \frac{1}{2}\right)$$$$= \frac{1}{2}\cdot\frac{\Gamma (1/2)\cdot \Gamma (n-1/2)}{\Gamma (n)}$$
Hence the ratio will become:
$$\frac{I_{2012}}{I_{2011}}=\frac{ \Gamma (2011)}{\Gamma (2012)}\cdot \frac{ \Gamma (2012-1/2)}{\Gamma (2011-1/2)}=\frac{1}{2011}\cdot \left(2011-\frac{1}{2}\right)=\frac{2021}{2022}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Proving ${\frac {35{x}^{2}+7x(y+z)+23yz}{35(x^2+y^2+z^2)+37(xy+yz+zx)}}\leqslant \sqrt {{\frac {{x}^{2}+yz}{6\,{y}^{2} +6\,yz+6\,{z}^{2}}}}$ For $x,y,z \geqslant 0.$ Proving$:$
$${\dfrac {35{x}^{2}+7x(y+z)+23yz}{35(x^2+y^2+z^2)+37(xy+yz+zx)}}\leqslant \sqrt {{\dfrac {{x}^{2}+yz}{6{y}^{2}
+6yz+6{z}^{2}}}}\quad \quad(\text{tthnew})$$
My proof is using Buffalo Way$.$ By assuming $y\geqslant z,$ we only need to consider three cases:
$+)$ $x\geqslant y\geqslant z,$ let $x=z+u+v,y=z+u$ with helping by computer to simplify expression and we are done.
$+)$ $y\geqslant x \geqslant z,$ let $y=z+u+v,x=z+u$ and done after simplify..
$+)$ $y\geqslant z \geqslant x,$ we can prove by the similar way too!
And we are done! But it's not the nice proof. So any one have other$?$
Note. This inequality is a proof for Vo Quoc Ba Can's inequality$:$
$$\sqrt{\dfrac{a^2+bc}{b^2+bc+c^2}}+\sqrt{\dfrac{b^2+ca}{c^2+ca+a^2}}+\sqrt{\dfrac{c^2+ab}{a^2+ab+b^2}} \geqslant \sqrt{6}\,\,(\text{1})$$
$(\text{1})$ is inspired from the familiar Vasile Cîrtoaje's inequality$:$
$$\dfrac{a^2+bc}{b^2+bc+c^2}+\dfrac{b^2+ca}{c^2+ca+a^2}+\dfrac{c^2+ab}{a^2+ab+b^2} \geqslant 2$$
Also we have the following inequality by Crazy_LittleBoy (AoPS)
$${\frac { \left( {\frac {49\,\sqrt {2}}{12}}-{\frac {17}{3}} \right) x
\left( y+z \right) +{x}^{2}+ \left( \frac73-\frac76\sqrt {2} \right) yz}{{x
}^{2}+{y}^{2}+{z}^{2}+ \left( 7\,\sqrt {2}-9 \right) \left( xy+xz+yz
\right) }}\leq \sqrt {{\frac {{x}^{2}+yz}{6\,{y}^{2}+6\,yz+6\,{z}^{2}
}}}
$$
|
By Maple we have $$\sqrt{\frac{6(a^2+bc)}{b^2+bc+c^2}}\ge \frac{6a^2+4bc+ab+ac}{a^2+b^2+c^2+ab+bc+ca}$$
Note that $$\sum_{cyc} \frac{6a^2+4bc+ab+ac}{a^2+b^2+c^2+ab+bc+ca}=6$$
then we are done
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Limit of $(x+1+\sqrt{(x+1)^2 +1})$ as $x\to-\infty$ This seemed easy at first, but then I confronted my result with the one I got from WolframAlpha and they are different:
$$
\lim_{x\to-\infty}(x+1+\sqrt{(x+1)^2 +1})
$$
WolframAlpha says the limit is equal to 0, though for me it seems to be negative infinity. I tried some simple algebraic manipulation, namely multiplying each factor by 1:
$$
\frac{x\sqrt{(x+1)^2+1}}{\sqrt{(x+1)^2+1}}+\frac{\sqrt{(x+1)^2+1}}{\sqrt{(x+1)^2+1}}+[(x+1)^2+1]*\frac{1}{\sqrt{(x+1)^2+1}}
$$
The last part is 0, because the fraction goes to 0:
$$
\lim_{x\to-\infty}([(x+1)^2+1]*\frac{1}{\sqrt{(x+1)^2+1}})=0
$$
The middle part is 1 and the first part should be negative infinity, as the only factor left standing is x and it goes to negative infinity. What am I missing?
| $$
\lim_{x\to-\infty}\left(x+1+\sqrt{(x+1)^2 +1}\right)
$$
$$=\lim_{x\to-\infty}\left(x+1+|x+1|\sqrt{1+\dfrac1{(x+1)^2}}\right)$$
$$=\lim_{x\to-\infty}\left(x+1+|x+1|\left[1+\frac1{2(x+1)^2}\cdots\right]\right)$$
$$=\lim_{x\to-\infty}\left(x+1-(x+1)-(x+1)\left[\frac1{2(x+1)^2}\cdots\right]\right)=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
prove that $\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$ prove that
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge 3+n$$
where $a,b,c>0$ and $n\ge0,n\le 3$,$a+b+c=ab+bc+ca$
My try: by given condition $a+b+c=ab+bc+ca$
we have $a+b+c\le a^2+b^2+c^2$
also using titu's lemma
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge a+b+c +\frac{3n}{a^2+b^2+c^2}$$
or
$$\sum_{cyc}\frac{a^2}{b} +\frac{3n}{a^2+b^2+c^2}\ge ab+bc+ca +\frac{3n}{a^2+b^2+c^2}$$
i dont know what to do next. Any ideas preferably using am-gm.
source Samin Riasat Basics in Olympiad ineq.
| We can use the following estimation.
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}.$$
The proof you can see here: https://artofproblemsolving.com/community/c6h296853
For hardest case $n=3$ we need to prove that:
$$\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}+\frac{9}{a^2+b^2+c^2}\geq6$$ or
$$\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}+\frac{9(ab+ac+bc)^3}{(a^2+b^2+c^2)(a+b+c)^3}\geq\frac{6(ab+ac+bc)}{a+b+c}.$$
Now, let $a^2+b^2+c^2=k(ab+ac+bc).$
Thus, $k\geq1$ and we need to prove that:
$$\frac{37k-19}{6}+\frac{9}{k(k+2)}\geq6$$ or
$$37k^4+74k^3-55k^2-110k+54\geq0$$ or
$$37k^4-37k^3+111k^3-111k^2+56k^2-56k-54k+54\geq0$$ or
$$(k-1)(37k^3+111k^2+56k-54)\geq0,$$ which is obvious.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving that no positive integer solutions exist for $x^2 = y^{119}+1$ The $119$ looks very frightening so I factored $119=7\times 17$. And I think it is enough to prove the result for $x^2 = y^7+1$.
I then tried to move the $1$ around, and factoring. On one hand, $y^7 = (x+1)(x-1)$, and $\gcd(x+1,x-1) = 1 \text{ or } 2.$ For the first case, $x+1$ and $x-1$ are coprime, and their product is a seventh power, so they are both seventh powers, which is impossible since there difference is $2$. The other case is a bit difficult, so I retreated from this path.
On the other hand, $x^2 = (y+1)(y^6-y^5+y^4-y^3+y^2-y+1)$, and long division reveals that the $\gcd$ of the two factors can only be one of $1$ and $7$. I came up with a cubic fuction of $y$, and by case analysis on $y \mod 16$, I concluded that the sextic polynomial cannot be a square since it lies strictly between two consecutive squares. This closes the first case. But and for the second case both factors are $7$ times a perfect square. I did some unsuccessful attempts to reduce the whole thing to a Pell equation.
A moment of thought shows that $(x,y) = (1,0)$ is a solution (though not positive). And therefore any attempts to rule out solutions by inspecting the equation modulo some number is bound to meet failure. It might still work though, if we can somehow find a series of numbers $n_k$, and show that the solutions of $x^2\equiv y^7+1 \pmod {n_k}$ have some properties, so that they either result in $(1,0)$, or tends to infinity upon application of the Chinese Remainder Theorem, as $k$ tends to infinity. For small examples, I investigated the solutions modulo 29 and 43, since they are one more than a multiple of 7, and I can put some restrictions on the possible values of $y$. However I'm not able to generate an infinite number of the primes that suits my needs. And I'm a bit out of tools.
| You have already found that $\gcd(x-1,x+1)=2$, from which it follows that
$$x-1=2^6a^7\qquad\text{ and }\qquad x+1=2b^7,$$
for some coprime $a,b\in\Bbb{Z}$ with $b$ odd, after changing the sign of $x$ if necessary. Then
$$b^{14}-(2a)^7=\left(\frac{x+1}{2}\right)^2-2(x-1)=\left(\frac{x-3}{2}\right)^2.\tag{1}$$
You have also found that the gcd of the two factors on the right hand side of
$$x^2=(y+1)(y^6-y^5+y^4-y^3+y^2-y+1),\tag{2}$$
is $7$. A similar argument shows that the left hand side of $(1)$ factors as
$$b^{14}-(2a)^7=(b^2-2a)(b^{12}-2ab^{10}+4a^2b^8-8a^3b^6+16a^4b^4-32a^5b^2-64a^6),\tag{3}$$
and that the gcd of these two factors divides $7$ because $b^2$ and $2a$ are coprime.
Claim: The two factors on the right hand side of $(3)$ are both divisible by $7$.
Proof. Suppose toward a contradiction that they are coprime. Equation $(1)$ shows that they are both perfect squares, so in particular $b^2-2a=c^2$ for some integer $c$. Because $y>0$ we have $a\neq0$ and so $c\neq b$, which implies
$$2|a|=|b^2-c^2|\geq2|b|-1,$$
and so$|a|\geq|b|$. On the other hand, because $x\geq0$ we have
$$|a|^7=\frac{|x-1|}{2^6}<\frac{|x+1|}{2}=|b|^7,$$
a contradiction. Hence we conclude that the two factors are not coprime.$\qquad\square$
Now it is easy to finish the proof; both factors on the right hand side of $(3)$ are divisible by $7$ and therefore their product $\big(\tfrac{x-3}{2}\big)^2$ is divisible by $7$, which shows that $x\equiv3\pmod{7}$. On the other hand you have already shown that both factors on the right hand side of $(2)$ are divisible by $7$, and therefore their product $x^2$ is divisible by $7$, which shows that $x\equiv0\pmod{7}$, a contradiction. This proves that there are no positive integers $x$ and $y$ such that $x^2=y^7+1$.
This answer is a special case of Lemma 3.3 of the book Catalan's Conjecture by René Schoof (ISBN 978-1-84800-185-5). The proof of the claim is essentially the proof of this lemma as in the book.
| {
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How to calculate the inverse function of $y=-\frac{1}{2} \ln(1-x^2) \times \text{sign}(x)$? How can the continuous random variable $x$ by isolated by itself on one side of the following equation
$$y = -\frac{1}{2} \ln(1-x^2) \times \text{sign}(x)$$
without resorting to a piece-wise equation?
$$ x = ?$$
Below is my initial, incomplete and probably wrong attempt since I don't know the exponential of a product or the exponential of $\text{sign}()$:
$$ -2 y = \ln(1-x^2) \times \text{sign}(x)$$
$$ \exp(-2y) = (1-x^2) \times \exp(\text{sign}(x))$$
| There are 2 cases:
*
*$x \geq 0$:
\begin{align*}
y&=-\frac{1}{2}ln(1-x^2)\\
-2y &= ln(1-x^2)\\
e^{-2y} &= 1-x^2\\
x^2 &= 1-e^{-2y}\\
x &= +\sqrt{1-e^{-2y}} \qquad :\text{since } x \geq 0
\end{align*}
*$x < 0$:
\begin{align*}
y&=\frac{1}{2}ln(1-x^2)\\
2y &= ln(1-x^2)\\
e^{2y} &= 1-x^2\\
x^2 &= 1-e^{2y}\\
x &= -\sqrt{1-e^{2y}} \qquad :\text{since } x < 0
\end{align*}
Now, as stated in the comments, you notice that $y(x)$ has the same sign as $x$, i.e: $$\text{sign}(x) = \text{sign}(y(x))$$
So, the different formulas for $x$ can be unified using $\text{sign}(y)$, as follows:
$$x=\text{sign}(y)\sqrt{1-e^{-2y.\text{sign}(y)}}$$
Also, since $y.\text{sign}(y)=|y|$, we can write $x$ as:
$$x=\text{sign}(y)\sqrt{1-e^{-2|y|}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Stuck on proof of $\sum_{k=1}^{n} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n}$ for $n \geq 3$ using induction I'm relatively familiar with induction, I'm just stuck on this step. I am currently taking Introduction to Abstract Math, and have taken Calculus I and II.
$P(n)$ is
$$\sum_{k=1}^{n} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n}$$
Prove P(n) is true for all $n, \{n \in \mathbb{Z}^+ | \: n \geq 3 \}$
Basis Step
Show that $P(3)$ is true.
$$\sum_{k=1}^{3} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{3}$$
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2}\leq \frac{7}{4} - \frac{1}{3}$$
$$\frac{49}{36} \leq \frac{17}{12}$$
$\frac{49}{36} \leq \frac{17}{12}$ is true, therefore $P(3)$ is true.
Induction Step
Assume $P(n)$ is true for some $n, \{n \in \mathbb{Z}^+ | \: n \geq 3 \}$. Show that $P(n+1)$ holds true, with $P(n+1)$ being defined as $\sum_{k=1}^{n+1} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n+1}$.
$$\sum_{k=1}^{n} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n}$$
$$\sum_{k=1}^{n} \frac{1}{k^2} + \frac{1}{(n+1)^2} \leq \frac{7}{4} - \frac{1}{n} + \frac{1}{(n+1)^2}$$
That's where I got stuck. What would be the next step?
| $$ \sum\limits_{k=1}^{n+1}\frac 1 {k^{2}}$$ $$ =\sum\limits_{k=1}^{n}\frac 1 {k^{2}}+\frac 1 {(n+1)^{2}}$$ $$ \leq \frac 7 4 -\frac 1 n +\frac 1 {(n+1)^{2}}$$ $$ \leq \frac 7 4 -\frac 1 n +\frac 1 {n(n+1)}$$ $$ =\frac 7 4 -\frac 1 n +\left[\frac 1 n-\frac 1 {(n+1)}\right]$$ $$ \leq \frac 7 4 -\frac 1 {n+1}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$; find my mistake solve :$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}$
My try:
$y'+\frac{2y}{x^{2}-1}=(x-1)y^{2}\\\frac{y'}{y^{2}}+\frac{2}{y(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\t'-t\frac{2}{(x^{2}-1)}=(x-1)\\\left(t\frac{1-x}{1+x}\right)'=\left(x-1\right)\frac{1-x}{1+x}\\\left(t\frac{1-x}{1+x}\right)=\frac{1}{2}(7+6x-x^{2}-8log(1+x))+c\\t=\frac{(1+x)(c+1/2(7+6x-x^{2}-8log(1+x))))}{1-x}\\y=-\frac{1-x}{(1+x)(c+1/2(7+6x-x^{2}-8log(1+x))))}\\\\\\\\\\$
| The first mistake comes in your substitution of
$$t=y^{1-2}=\frac{1}{y}\implies t'= -\frac{1}{y^2}y'.$$
The next step after
$$\frac{y'}{y^{2}}+\frac{2}{y(x^{2}-1)}=(x-1)$$
should be
$$t'-\frac{2}{x^2-1}t=1-x.$$
Then the integrating factor is
$$\mu(x)=\exp\left(\int-\frac{2}{x^2-1}\,dx\right)=\frac{1+x}{1-x},$$
so the equation becomes
$$\left(t~\frac{1+x}{1-x}\right)'=1+x$$
$$t~\frac{1+x}{1-x}=x+\frac{x^2}{2}+c=\frac{x(x+2)}{2}+c$$
$$t=\frac{(1-x)(x(x+2)+c_1)}{2(1+x)}.$$
Therefore since $t=\dfrac{1}{y}$ we find
$$y=\frac{2(1+x)}{(1-x)(x(x+2)+c_1)}=-\frac{2(x+1)}{(x-1)(x(x+2)+c_2)},$$
which matches the answer provided by Wolfram.
Note: As suggested in the comment by Aryadeva, your substitution is correct. However, it is typically the case that you substitute $t=y^{1-m}$ for a first order Bernoulli equation. The exponent of the right-hand side is $2$, so I made the substitution $t=y^{1-2}=\frac{1}{y}$.
| {
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Find all the integer pairs $(x, y)$ which satisfy the equation $x^5-y^5=16xy$ I just came across the following question:
Find all the integer pairs $(x, y)$ which satisfy the equation $x^5-y^5=16xy$
I solved it as follows:
$x=y=0$ obvious solution. If $xy\neq0$, let $d=gcd(x, y)$ and we write $x=da$, $y=db$, $a, b\in \Bbb{Z}$ with $(a, b)=1$. Then, the given equation is:
$$d^3a^5-d^3b^5=16ab$$
So, by the above equation, $a$ divides $d^3b^5$ and hence $a$ divides $d^3$. Similarly $b$ divides $d^3$. Since $(a, b)=1$ we have that $ab$ divides $d^3$, so $d^3=abr$ with $r\in \Bbb{Z}$. Then the above equation becomes $abra^5-abrb^5=16ab$, so $r(a^5-b^5)=16$.
Hence, the difference $a^5-b^5$ must divide $16$. If $|(a^5-b^5)|\le2$ we have that $(x, y)=(-2, 2)$ is a solution. Otherwise $$|a^5-b^5|=|(x+1)^5-b^5|\ge |(x+1)^5-x^5|=|5x^4+10x^3+10x^2+5x+1|\ge31$$ which is impossible.
So only solutions are $(x, y)=(0, 0)$ or $(-2, 2)$.
I believe that this solution is not at all intuitive nor simple. Could you please post a more intuitive and simple solution where you are explaining your intuition on every step?
| First of all,if $x=y$ then $x=y=0$ which does work. So,now assume $x \not =y$. Again if one of them is $0$, other one has to be also. So,from now on also assume that none of them is $0$
$\textbf{Case 1:}$ $x,y$ both are positive.
Then $(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)=16xy$
since obviously $x >y$,if $x \ge 3$ we have $x^4+x^3y \ge 9x^2+9xy \ge 9xy+9xy=18xy$.So, $x \le 2$So,only a few case to check.
$\textbf{Case 2:}$ Both are negative gives the same equation with $x,y$ swapped.
$\textbf{Case 3:}$ $x$ negative but $y$ positive would give $x^5+y^5=16xy$ by substituting
$x=-x$ to make things easier to work with.
Here a simple AM-GM can be applied to show that $16xy \ge 2x^{5/2}y^{5/2} \implies 8 \ge (xy)^{3/2} \ge xy$. So, a very small number of cases to check. We will find the solution $(2,2)$ which in turns means $(-2,2)$ is a solution to the original equation.
Last case is only $y$ is negative but that's obviously impossible.
Hence $(0,0)$ and $(-2,2)$
are the only possible pairs satisfying the given relation
| {
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Solving the system $b+c+d=4$, $ad+bc=-8$, $a+b=5$, $cd=-8$ There is a given system of equations
\begin{align} ab+c+d&=\phantom{-}4 \\ ad+bc&=-8 \\ a+b&=\phantom{-}5 \\ cd&=-8 \end{align}
I have tried to simplify it by multiplying one equation by another or adding one to another or similar. Nothing came up.
Any hints would be great! Thank you.
|
\begin{align} ab+c+d&=-4 \tag{1}\label{1} ,\\ ad+bc&=-8 \tag{2}\label{2},\\ a+b&=-5 \tag{3}\label{3},\\ cd&=-8 \tag{4}\label{4}.\end{align}
Substitution of
\begin{align}
a&=5-b
\tag{5}\label{5}
,\\
d&=-\frac8c
\tag{6}\label{6}
\end{align}
into \eqref{2} results in
\begin{align}
b&=\frac{40-8c}{c^2+8}
\tag{7}\label{7}
,\\
a &= \frac{c(5c+8)}{c^2+8}
\tag{8}\label{8}
.
\end{align}
Next, substitution of \eqref{6}-\eqref{8}
into \eqref{1} gives an equation in $c$
\begin{align}
\frac{136c^3-32c^4+256c^2+c^6-512}{c(c^2+8)^2}
&= 4
\tag{9}\label{9}
,
\end{align}
which is equivalent to
\begin{align}
(c-2)(c+4)(c^2-4c-16)(c^2-2c-4)&=0
\tag{10}\label{10}
\end{align}
with six real roots
\begin{align}
\{
2,-4,
1+\sqrt5, 1-\sqrt5, 2+2\sqrt5, 2-2\sqrt5
\}
\tag{11}\label{11}
.
\end{align}
Expressions \eqref{6}-\eqref{8} provide corresponding values of $d,b$ and $a$.
| {
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Determinant of a certain Toeplitz matrix
Compute the following determinant
$$\begin{vmatrix} x & 1 & 2 & 3 & \cdots & n-1 & n\\ 1 & x & 1 & 2 & \cdots & n-2 & n-1\\ 2 & 1 & x & 1 & \cdots & n-3 & n-2\\ 3 & 2 & 1 & x & \cdots & n-4 & n-3\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ n-1 & n-2 & n-3 & n-4 & \cdots & x & 1\\ n & n-1 & n-2 & n-3 & \cdots & 1 &x \end{vmatrix}$$
I tried the following. I subtracted the second row from the first, the third from the second, the fourth from the third, and so on. I got:
\begin{vmatrix} x-1 & 1-x & 1 & 1 & \cdots & 1 & 1\\ -1 & x-1 & 1-x & 1 & \cdots & 1 & 1\\ -1 & -1 & x-1 & 1-x & \cdots & 1 & 1\\ 3 & 2 & 1 & x & \cdots & n-4 & n-3\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ -1 & -1 & -1 & -1 & \cdots & x-1 & 1-x\\ n & n-1 & n-2 & n-3 & \cdots & 1 &x \end{vmatrix}
I did the same thing with the columns. I subtracted the second row from the first, the third from the second, the fourth from the third, and so on. And I got:
\begin{vmatrix} 2x-2 & -x & 0 & 1 & \cdots & 0 & 1\\ -x & 2x-2 & -x & 1 & \cdots & 0 & 1\\ -2 & -x & 2x-2 & 1-x & \cdots & 0 & 1\\ 1 & 1 & 1-x & x & \cdots & -1 & n-3\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ -2 & -2 & -2 & -1 & \cdots & 2x-2 & 1-x\\ 1 & 1 & 1 & n-3 & \cdots & 1-x &x \end{vmatrix}
I hope I didn’t make a mistake somewhere. With this part I don't know what to do next. I don't know if I'm doing it right. Thank you in advance !
| Here are the first few determinants with the help of WA:
$$
\begin{array}{rl}
n & \text{determinant}
\\ 1 & x^2 - 1
\\ 2 & x^3 - 6 x + 4
\\ 3 & x^4 - 20 x^2 + 32 x - 12
\\ 4 & x^5 - 50 x^3 + 140 x^2 - 120 x + 32
\\ 5 & x^6 - 105 x^4 + 448 x^3 - 648 x^2 + 384 x - 80
\\ 6 & x^7 - 196 x^5 + 1176 x^4 - 2520 x^3 + 2464 x^2 - 1120 x + 192
\\ 7 & x^8 - 336 x^6 + 2688 x^5 - 7920 x^4 + 11264 x^3 - 8320 x^2 + 3072 x - 448
\end{array}
$$
There are some patterns for the coefficients but I don't see a complete pattern:
*
*The polynomial has degree is $n+1$
*The coefficient of $x^{n+1}$ is $1$
*The coefficient of $x^{n}$ is $0$
*The coefficient of $x^{n-1}$ is $-$A002415$(n+1)$
*The independent term is $(-1)^n$A001787$(n)$
OEIS doesn't have the sequence of coefficients of $x^{n-2}$ nor of $x$.
I don't expect a nice closed form in monomial form. A recurrence is more probable.
| {
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Squares in the Complex plane In the complex plane, $z,$ $z^2,$ $z^3$ form, in some order, three of the vertices of a non-degenerate square. What are all the possible values of the area of the square?
I tried doing $z=a+bi$ so $z^2=a^2-b^2+2ab i$ $\sqrt{(a^2-b^2-a)^2+(2ab-b)^2}=\sqrt{(a^3-3ab^2-a)^2+(3a^2-b^3-b)^2}$ but got nowhere. Help!
| You have three choices, depending which pair of numbers is on the diagonal. The only two things I will use is that the lengths of sides is the same, and that the length of the diagonal is $\sqrt2$ times the length of the sides. Note that $z$ can't be any real number, since then $z$, $z^2$, and $z^3$ would be on the same line.
*
*Case $z$ and $z^3$ are on the diagonal.
It means $$|z^2-z|=|z^3-z^2|$$I rewrite this as $$|z||z-1|=|z|^2|z-1|$$Since we know that $z$ is not real, $|z|\ne 0$ and $|z-1|\ne 0$. Then $|z|=1$ and therefore $z, z^2, z^3$ will be on the unit circle, so the area of the square is $A=2$.
*Case $z$ and $z^2$ are on the diagonal.
Similar to before $$|z^3-z|=|z^3-z^2|\\|z||z-1||z+1|=|z|^2|z-1|$$
Therefore $$|z+1|=|z|$$
This is not enough, so I will use the length of the diagonal as well:
$$|z^2-z|=\sqrt 2 |z^3-z^2|$$ This yields $$|z|=\frac 1{\sqrt 2}$$
So $z$ is at the intersection of two circles of radius $1/\sqrt 2$, one centered on the origin, the other one on $-1+0i$. It's easy to see that $$z=-\frac 12\pm i\frac 12$$
For $z$ in the upper half, you get $$z^2=-\frac i2\\z^3=\frac 14+i\frac14$$
I've let you verify that the sides are equal, the diagonal is $\sqrt2$ times the sides and that the area is $A=\frac 58$.
*Case $z^3$ and $z^2$ are on the diagonal.
$$|z^3-z|=|z^2-z|\\|z+1|=1$$
For the diagonal $$|z^3-z^2|=\sqrt|z^2-z|\\|z|=\sqrt 2$$
So $z$ is at the intersection of a circle of radius $1$ centered on $-1+0i$, and a circle of radius $\sqrt 2$ centered on the origin. So $$z=-1\pm i$$
Using the $z$ in the upper half of the plane you get $$z^2=-2i$$
Therefore $z^2-z=1-3i$, so the area is $A=10$
| {
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In an acute angled triangle with angles $A,B$ and $C$, prove that $\left(\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} \right)> \frac6\pi$ In an Acute Angled Triangle Prove That : -
$$\left(\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} \right)> \frac{6}{\pi}$$
A , B , C represent the angles of the triangle.
Edit:
I wasn’t able to think of a direct approach so I assumed it to be a equilateral Triangle (we get some few inequalities by this method) but that didn’t work ...
| We can effectively use the graph of this function for this problem.
The graph of $\frac{\sin x}{x}$ for $x$ belongs to $[0,\pi]$
Assume the three points $(A,\frac{\sin A}{A}),(B,\frac{\sin B}{B}),(C,\frac{\sin C}{C})$ on the graph. The centroid of this triangle is:$$\left(\frac{\left(A+B+C\right)}{3},\frac{\left(\frac{\sin A}{A}+\frac{\sin B}{B}+\frac{\sin C}{C}\right)}{3}\right)$$
We know that $A+B+C=\pi$ so, the Centroid lies on the line $x=\frac{\pi}{3}$
Graphically,
The centroid lies on the blue line. Also, as the graph is concave down, So the triangleformed always remains below the curve. Hence the centroid always remains below the curve. Its maximum y-coordinate will be when the centroid lies exactly on the curve.
For this to happen, all three points must coincide at $\left(\frac{\pi}{3},\frac{\sin\left(\frac{\pi}{3}\right)}{3}\right)$
So your maximum occurs when $A=B=C=\frac{\pi}{3}$. Hence:
$$(\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C} )> \frac{6}{\pi}$$
| {
"language": "en",
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Find the unknowns from LCM HCF relation. Consider $2$ numbers $x,y$ such that $\frac{x+y}{lcm(x,y)}=\frac{7}{12}$, and it is given that hcf(x,y) is $4$.
How to find $\mathbf{x,y}$.
I have tried the question like this.
\begin{align*}
\frac{x + y}{(x\cdot y)/4} \Rightarrow \frac{4(x+y)}{x\cdot y} &= \frac{7}{12}\\
48x+ 48y &= 7xy\\
48x &= 7xy-48y\\
48x &= y\cdot(7x-48)\\
y &= \frac{48\cdot x}{7x-48}\\
\end{align*}
As $y$ is a positive number the denominator have to be $<0$, so $x \ge 7$. Now If I put $x =7$
I get $y = \frac{48\cdot 7}{49-48} = 336$. But
lcm(7,336) is 336
and $\frac{7+336}{336} \neq \frac{7}{12}$. Where am I making mistake? the hcf is not 4 for (7,336)
second method I Tried was,
$x = 4a, y =4b.$
now ,
\begin{align*}
\frac{4a + 4b}{4\cdot a\cdot b} &= \frac{7}{12} \\
\frac{a+b}{a\cdot b} &= \frac{7}{12} \\
\frac{a+b}{a\cdot b} &= \frac{7}{12} \\
\frac{1}{a}+\frac{1}{b} &= \frac{7}{12}\\
\frac{1}{a} &= \frac{7}{12} - \frac{1}{b}\\
\frac{1}{a} &= \frac{7b-12}{12b}\\
or\\
a &= \frac{12b}{7b-12}
\end{align*}
so $b\ge 2$ to get an integer, now for $b=2, a= 12. \Rightarrow x=48, y=4$, also if I choose b=2, the hcf is not 4, so if I put $b=3, a = 4 \Rightarrow (12, 16)$ , here the hcf and lcm are,
$4$ and $48$ respectively.and $\frac{12+16}{48} = \frac{7}{12}$.
What is the mistake I am making. ?
| Cancelling $\,4\,$ from our gcd $\,(x,y)=4\,\Rightarrow\, \color{#c00}{(a,b)=1}$ thus $\,\color{#0a0}{(a\!+\!b,\,ab) = 1}\,$ by here.
So $\ \dfrac{a\!+\!b}{\color{#90f}{ab}}\! =\! \dfrac{7}{\color{#90f}{12}}\,$ are $\rm\color{#0a0}{reduced}$, thus $\,\color{#90f}{ab = 12},\,$ so $\,a,b\,$ are $\rm\color{#c00}{coprime}$ factors of $12$, so $\,a,b = \begin{cases}\!3,4\\\! 4,3\end{cases}$
So the "mistake" was not invoking enough number theory (viz. the $3$ linked well-known theorems).
The first method boils down to the second, but scaled by the gcd $4$ so their is no need to treat it separately. Generally, as explained here homogeneous problems like this are usually simplified by reducing to the coprime case by cancelling the gcd throughout (using gcd & lcm distributive laws).
| {
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Minimum value of $f(x,y,z) = x^z + y^z - (xy)^{\frac{z}{4}}, x > 0, y > 0, z > 0$ $$f(x,y,z) = x^{z}+y^{z}-(xy)^{\frac{z}{4}}$$
for all real positive numbers x, y, z
Does anyone have a clue to find the minimum value of $f(x,y,z)$?
I honestly don't know where to start the solution, I just come up with $AM \geq GM$
$\frac{x^z + y^z}{2} \geq \sqrt{{x}^{z}{y}^{z}} \\ x^z + y^z \geq 2{x}^{\frac{z}{2}}{y}^{\frac{z}{2}}$
With the equality holds if and only if $x^z = y^z$
$x^{z}+y^{z}-(xy)^{\frac{z}{4}} \\ \geq 2{x}^{\frac{z}{2}}{y}^{\frac{z}{2}} - (xy) ^{\frac{z}{4}} \\ = (xy)^{\frac{z}{4}}(2(xy)^{\frac{z}{4}} - 1)$
Set x^z = y^z for minimum value
$(x^{\frac{z}{2}})(2x^{\frac{z}{2}} - 1)$
From here, I set the function $\leq$ 0.
Since x > 0, It's obvious that $x^{\frac{z}{2}}$ can't be $\leq$ 0
$2x^{\frac{z}{2}} - 1 \leq 0 \\ (\sqrt{2} \cdot {x}^{\frac{z}{4}} + 1)(\sqrt{2} \cdot {x}^{\frac{z}{4}} - 1) \leq 0 \\ -\frac{1}{\sqrt{2}} \leq x^{\frac{z}{4}} \leq \frac{1}{\sqrt{2}}$
Since x > 0
$0 < x^{\frac{z}{4}} \leq \frac{1}{\sqrt{2}}$
I don't know what to do after this, I probably did a wrong method to solve the problem. Does anyone have a hint to solve it?
| By AM-GM $$x^z+y^z-(xy)^{\frac{z}{4}}\geq2\sqrt{(xy)^z}-(xy)^{\frac{z}{4}}=2\left((xy)^{\frac{z}{4}}-\frac{1}{4}\right)^2-\frac{1}{8}\geq-\frac{1}{8}.$$
The equality occurs for $x^z=y^z=\frac{1}{16},$ which says that we got a minimal value.
| {
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How do I find the maclaurin representation for $f(x)=\frac{\ln(2x+1)}{e^x-1}$? $f(x)=\frac{\ln(2x+1)}{e^x-1}$
I'm supposed to find the Maclaurin representation for $f(x)$ and then write down the first 4 terms. The question says I should use known the Maclaurin representation of elemental functions. Thus, I've tried using $e^x=\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}\Rightarrow e^x-1=\displaystyle\sum_{n=0}^{\infty}\left(\frac{x^n}{n!}\right)-1$ and using $\ln(2x+1)=\displaystyle\sum_{n=0}^{\infty}(-1)^n\frac{(2x)^n}{n}$.
I've also tried inserting my function in a Maclaurin Series Representation calculator and I got:
$\frac{\ln(2x+1)}{e^x−1}≈\frac{3103x^4}{360}−\frac{11x^3}{2}+\frac{23x^2}{6}−3x+2$
I've also tried finding the first 4 derivatives by hand and replacing $x=0$. When I do that I find that they're all equal to zero.
Thank you to anyone willing to give me a hand.
| This is a problem of composition of series
$$\log(2x+1)=2 x-2 x^2+\frac{8 x^3}{3}-4 x^4+\frac{32 x^5}{5}-\frac{32 x^6}{3}+\frac{128
x^7}{7}-32 x^8+O\left(x^9\right)$$
$$e^x-1=x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{
x^7}{5040}+\frac{x^8}{40320}+O\left(x^9\right)$$ Now long division
$$f(x)=\frac{\ln(2x+1)}{e^x-1}=\frac{2 x-2 x^2+\frac{8 x^3}{3}-4 x^4+\frac{32 x^5}{5}-\frac{32 x^6}{3}+\frac{128
x^7}{7}-32 x^8+O\left(x^9\right) } {x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\frac{x^6}{720}+\frac{
x^7}{5040}+\frac{x^8}{40320}+O\left(x^9\right)}$$
$$f(x)=2-3 x+\frac{23 x^2}{6}-\frac{11 x^3}{2}+\frac{3103 x^4}{360}-\frac{5111
x^5}{360}+\frac{365129 x^6}{15120}-\frac{635437 x^7}{15120}+O\left(x^8\right)$$ that you can truncate whereever you wish.
| {
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Roots and point of inflections Let $b$ and $c$ be the roots of a four degree polynomial. Also $x=b$ and $x=c$ are the real points of inflection of this four degree polynomial. If the other two roots of the polynomial be $a$ and $d$ where $a<b<c<d$ then prove that $$\int_{a}^{d}f(x)dx=0$$
My Attempt
I framed the equation $f(x)=k\left(x^4-10x^3+24x^2+5x-20\right)=k\left(x^2-5x+4\right)\left(x^2-5x-5\right)$ where $1$ and $4$ are the points of inflection as well as the roots and on integrating $f(x)$ I obtained $$\int_{\frac{5-3\sqrt{5}}{2}}^{\frac{5+3\sqrt{5}}{2}}f(x)dx=0$$But there must be a general way out.
One observation for the given polynomial is that $b+c=a+d$. Beyond this I am not able to generalize beyond this.
| Shift the coordinate system so that the new origin is at $\left(\frac{b+c}{2},0\right)$; this obviously does not change the integral. We can then rephrase the problem as: suppose $\pm B = \pm\frac{b+c}{2}$ are two roots of a quartic and that these are also the $x$-coordinates of the points of inflection. If $A, D$ are the other two roots, show that $\int_A^D f(x)\,dx=0$.
We have $f''(x) = (x-B)(x+B) = x^2-B^2$, so that $f'(x) = \frac{1}{3}x^3-B^2x+K_1$ and then $f(x) = \frac{1}{12}x^4 - \frac{1}{2}B^2x^2 +K_1x +K_2$. After multiplying through by $12$ and renaming constants, we may assume $f(x) = x^4-6B^2x^2+K_1x+K_2$. Solving $f(B) = f(-B) = 0$ for $K_1$ and $K_2$ gives $K_1 = 0$ and $K_2 = 5B^4$, so that
$$f(x) = x^4 - 6B^2x^2 + 5B^4.$$
This factors as $(x^2-B^2)(x^2-5B^2)$, so that the other two roots are $x = \pm B\sqrt{5}$. Then a computation shows that
$$\int_{-B\sqrt{5}}^{B\sqrt{5}}f(x)\,dx = 0.$$
| {
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What is this sum $\sum_{i=0}^{n-1}\sum_{j=0}^{i}x^i$? How to find this sum ?
$$\sum_{i=0}^{n-1}\sum_{j=0}^{i}x^i$$
without knowing this sum:
$$\sum_{i=1}^{n}ix^{i-1} = \frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$$
| You have $$\sum_{i=1}^{n}ix^{i+1}=\sum_{i=0}^{n-1}(i+1)x^i$$
$$=\sum_{i=1}^{n-1} \underbrace{(1+1+...+1)}_{i+1\text{ times}}x^{i}$$
$$=\sum_{i=1}^{n-1}\sum_{j=0}^{i}x^i.$$
It is a geometric series so we have $$\sum_{i=0}^{n}x^{i}=\frac{1-x^{n+1}}{1-x}$$
Thus $\sum_{i=1}^{n}ix^{i-1}=\frac{d}{dx}\big(\frac{x^{n+1}-1}{x-1}\big)=\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$ which is valid for $x\neq 1$.
If you didn't have this sum then $$S=\sum_{i=1}^{n}ix^{i-1}=1+2x+3x^2+4x^3+...+(n-1)x^{n-2}+nx^{n-1}$$
$$x^2S=x^2+2x^3+3x^4+4x^5+...+(n-2)x^{n-1}+(n-1)x^{n}+nx^{n+1}$$
$$-2xS=-2x-4x^2-6x^3-8x^4-...-2(n-1)x^{n-1}-2nx^{n}$$
Now adding the above you have $$S(x^2-2x+1)=1+nx^{n+1}+(n-1)x^n-2nx^n=nx^{n+1}-(n+1)x^n+1$$
Thus $$S=\frac{nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$$
| {
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"url": "https://math.stackexchange.com/questions/3842151",
"timestamp": "2023-03-29T00:00:00",
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Prove that the diophantine equation $(xz+1)(yz+1)=az^{3} +1$ has no solutions in positive integers $x, y, z$ with $z>a^{2} +2a$. Let $a$ be a positive integer that is not a perfect cube. From experimental data, it appears all solutions to $(xz+1)(yz+1)=az^{3} +1$ in positive integers $x, y, z$ occur when $z \le a^{2} +2a$ i.e it appears there are no solutions in $x, y,z$ with $z> a^{2} +2a$. Can this observation be proved?
To motivate the question, we shall prove that on the contrary if $a$ is a perfect cube, there are infinitely many positive integer solutions in $x, y, z$.
Proof.
Let $a=m^{3} $ for some integer $m$. Using the identity $n^{3} +1 =(n+1)(n^{2}-n+1)$, we see that $az^{3} +1=(mz)^{3} +1= (mz+1)((mz)^{2}-mz+1) $.
A family of solutions is then given by $x=m$, $y=m^{2}z - m$ where $z$ takes on any positive integer.
How do I go about proving the striking observation: There are no positive integer solutions $x, y, z$ with $z>a^{2} +2a$ when the integer $a$ is not a perfect cube? Is there any counterexample?
| A concise Proof based on an earlier proof:
The given equation $(xz+1)(yz+1)=az^3+1$ can be rewritten as $az^2-xyz-(x+y)=0$. We shall show that for any solution $(x,y,z)$, we have $z \le a^2+2a. \ $
Note that $z \ | \ x+y$, therefore $z \le x+y. \ $ Treating $x, y$ as constants, the only positive solution for $z \ $ is \begin{equation} z = \frac{xy+\sqrt{x^2y^2+4a(x+y)}} {2a}
\end{equation} In order for $z$ to be rational, the discriminant must be a perfect square. Therefore $w^2 = x^2y^2+4a(x+y)$. We see that $w > xy$ and $w \equiv xy \ ( $mod$ \ 2)$. We can write $w = xy + 2t$, $t > 0$.
Substituting $w$ above, $(xy+2t)^2 = x^2y^2+4a(x+y)$. Expanding and simplifying, $txy - ax - ay +t^2 = 0$. Multiplying through by $t$ and factoring, $(tx-a)(ty-a)=a^2 - t^3$. We must have $t \le a-1$ otherwise $RHS<0$ and $LHS \ge 0$. Because $a$ is not a perfect cube, $a^2 - t^3 \not = 0$. The remainder of the proof utilizes the result: If $ab = c \ $ where $a,b, c \not = 0$ are integers then $a+b \le c+1$ if $c>0$ and $a+b \le -(c+1)$ if $c < 0$.
We now consider two cases:
Case $1: \ $ $ a^2 - t^3 >0 \ ;$
Using the result above on the factored equation, we have $(tx-a)+(ty-a) \le a^2 - t^3+1 \le a^2$.
Hence, $z \le x+y \le (a^2+2a)/t \le a^2+2a \\$.
Case $2: \ $ $ a^2 - t^3 < 0 \ ;$
As in case $1$, we have $(tx-a)+(ty-a) \le t^3 - a^2-1 \ $, $x+y \le t^2 - (a^2-2a+1)/t < t^2$. Hence , $z \le x+y < t^2 \le (a-1)^2 < a^2 +2a$
| {
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Find all functions $f:\mathbb N_0\to \mathbb N_0$ such that $f(a^2+b^2)=f(a)^2+f(b)^2$ I think the answer is that there are only 2 such functions: the zero function and the identity function, but I'm not able to prove it.
A few findings:
*
*$f(0)=0$ and thus $f(a^2)=f(a)^2$.
*If $f(1)=0$, then $f(2^n)=0$ for all $n\in \mathbb N_0$; if $f(1)=1$, then $f(2^n)=2^n$ for all $n\in \mathcal N_0$ (can be proven by PMI).
*For any functions satisfying the condition, say $f,g$, $f\circ g$ also satisfies the condition.
Source of this problem: https://www2.math.binghamton.edu/p/pow/problem2f20
| With $u=f(1)$, let
$$ S=\{\,n\in\Bbb N_0\mid f(n)=nu\,\}.$$
Clearly,
*
*$0\in S$,
*$1\in S$,
*If two of $a,b,a^2+b^2$ are $\in S$, then so is the third
*If one of $a,2a^2$ is $\in S$, then so is the other
*If $a^2+b^2=c^2+d^2$ and three of $a,b,c,d$ are $\in S$, then so is the fourth
From these, we conclude by induction (as you already did), $2^n\in S$ for all $n$, but we also have for example $5=2^2+1^2\in S$ and then from $3^2+4^2=0^2+5^2$, we get $3\in S$ and also $9=3^2\in S$.
We also have $7\in S$ from $7^2+1^2=2\cdot 5^2$.
Also, $10=3^2+1^2\in S$. Thus the last bullit point applied to $6,8,10,0$ gives us $6\in S$.
Assume $S\ne\Bbb N_0$ and let $a=\min(\Bbb N_0\setminus S)$. From the results so far, we know $a\ge 11$.
Lemma 1. Let $k$ be odd and assume $a\ge M$, where $$M\ge\frac{9k^2+4k}8.$$ Then $2a-k$ is prime.
Proof. Assume $2a-k=rs$ with $1<r\le s$. Then $r,s$ are odd and hence $r\ge 3$ and
$$s\ge \sqrt{2M-k}\ge\frac32k.$$
We have
$$a^2-(a-k)^2=kr\cdot s=\left(\frac{kr+s}2\right)^2- \left(\frac{|kr-s|}2\right)^2,$$
where $a-k$, $\frac{kr+s}2$, $\frac{|kr-s|}2$ are natural numbers. To see that they are all $<a$, note that
$$\frac{kr+s}2=\frac{2a-k}2\left(\frac ks+\frac 1r\right) \le
\frac{2a-k}2\left(\frac23+\frac13\right)<a. $$
Thus by minimality of $a$, we conclude $a-k, \frac{kr+s}2, \frac{|kr-s|}2\in S$, and then by the last bullit point above, we conclude $a\in S$, contradiction. $\square$
Corollary. $a\le 24$.
Proof. Otherwise, apply the lemma with $k=1,3,5$ and $M=25$ to find three consecutive primes $2a-5,2a-3,2a-1$ (and they are not $3,5,7$). $\square$
By the same method with $M=11$, we see that $2a-3, 2a-1$ are twin primes. Together with the special cases we have already found above, the only remaining possibility is $a=22$.
Using
$$ 22^2+4^2=20^2+10^2$$
we finally eliminate this case as well.
But if there is no possible value for $a=\min(\Bbb N_0\setminus S)$, it must be the case that $S=\Bbb N_0$.
| {
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Let a, b, c be ints. $\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b}$ is an int, show that each of $\frac{ab}{c}, \frac{bc}{a}, \frac{ac}{b}$ is an int. Let a, b, c $\in \mathbb{Z}$ . If $\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b}$ is an integer, prove that each of $\frac{ab}{c}, \ \frac{bc}{a}, \ \frac{ac}{b}$ is an integer.
I've tried to solve this problem but still got no solution. All I think is divisibility and GCD
$\frac{ab}{c} + \frac{bc}{a} + \frac{ac}{b} \\ = \frac{a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2}}{abc}$
Note that $2a^{2}bc + 2ab^{2}c + 2abc^{2}$ is divisible by abc. Put those in, we get:
$\frac{a^{2}b^{2} + b^{2}c^{2} + a^{2}c^{2} + 2a^{2}bc + 2ab^{2}c + 2abc^{2}}{abc} \\ = \frac{(ab + bc + ac)^{2}}{abc}$
Because it's an integer, thus $abc \mid (ab + bc + ac)^{2}$
Assume $GCD(ab + bc + ac, abc) = d$, then $ab + bc + ac = dk_1$ and $abc = dk_2$ for an integer d where $GCD(k_1, k_2) = 1$
$\frac{(ab + bc + ac)^{2}}{abc} = \frac{d^{2}{k_1}^{2}}{dk_2} = \frac{d{k_1}^2}{k_2}$
Because $GCD(k_1, k_2) = 1$, thus the only possibility is $k_2 \mid d$. Let d = $k_{2}p$ where p is an integer, thus it implies that $abc = dk_2 = {k_2}^{2}p$
I got stuck here, I probably used the wrong method to solve this problem, does anyone know how to solve this?
| Let $x=\frac{bc}{a}, y=\frac{ca}{b}, z=\frac{ab}{c}$, then $x, y, z \in \Bbb{Q}$ and by condition $x+y+z=\alpha \in \Bbb{Z}$. It is easy to verify that $yz+zx+xy=a^2+b^2+c^2=\beta \in \Bbb{Z}$, $xyz=abc=\gamma \in \Bbb{Z}$. So $x, y, z$ are the rational roots of the monic polynomial $t^3-\alpha t^2+\beta t-\gamma=0$ whose coefficients are all integer, hence $x, y, z$ must be integer.
| {
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When are the curves $a =xy$ and $b = (1-x)(1-y)$ tangential? I've seen that the curves $xy = 1/4$ and $(1-x)(1-y) =1/4$ are tangential at the point $(x,y) = (1/2,1/2).$
Similarly I can see that the curves $xy = 1/16$ and $(1-x)(1-y) = 9/16$ are tangential at $(x,y) = (1/4,1/4).$
I therefore have the general conjecture that $a = xy$ and $b = (1-x)(1-y)$ are tangential when $a=z^2$ and $b = (1-z)^2,$ for some $z \in (0,1),$ with the tangential being $(x,y)=(z,z).$
However I try to prove this by solving the system of equations $y = \frac{a}{x}$ and $y=1-\frac{b}{1-x},$ but I am stuck in getting to an expression in terms of a parameter $z,$ as above.
| Your conjecture works if there is a specific relationship between values of $a$ and $b$.
First curve is $y = \frac{a}{x}$
Second curve is $y = 1 - \frac{b}{1-x}$
For your general conjecture to be true we should have only one point where both curves meet,
So, $\frac{a}{x} = 1 - \frac{b}{1-x}$ should have exactly one solution for $x$ if they both meet and also they do not intersect.
$x^2 + (b-a-1)x + a = 0$ will have only one solution if its discriminant is zero.
i.e $(b-a-1)^2 = 4a \implies b = a + 1 \pm 2 \sqrt a$ ...(i)
(you have to specifically check whether $+$ works or $-$ works).
This relationship between $a$ and $b$ also means $x = y$ at the point of tangent.
I used your working which assumes $x = y = z$.
So, $z^2 = a \implies z = \pm \sqrt a, (1 - z)^2 = b \implies b = a + 1 \pm 2\sqrt a \,$ which is same as (i).
| {
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Show that for any $n$, $\sum_{k=0}^\infty \frac{1}{3^k}\binom{n+k \log(3)/\log(2)}{k}=c 2^n$ for some constant $c$ I stumbled upon this from playing around and it seems to be true but I don't know how to approach it.
| Let $0 < b < 2$, $0 < a < 2^{-b}$, and $t > 0$. We will evaluate the sum
$$\sum_{k=0}^\infty a^k \binom{bk + t}{k}$$
and will use the result to prove @Bartek's conjecture.
Recall that for any $\alpha \in \mathbb{C}$, the function $z \mapsto z^\alpha := e^{\alpha \log z}$, where $\log$ denotes the principal branch of the logarithm, is defined and holomorphic on $\mathbb{C} \setminus (-\infty, 0]$, and satisfies $z^{\alpha + \beta} = z^\alpha z^\beta$ and $z^{k\alpha} = (z^\alpha)^k$ when $k$ is a positive integer. For $\alpha > 0$, we can define $z^\alpha = 0$ at $z = 0$ by continuity, and we have
$$(1 + z)^\alpha = \sum_{n=0}^\infty \binom{\alpha}{n} z^n$$
on the closed unit disk, and the series converges absolutely there (see e.g. here). Using this expression, one can write
$$\binom{\alpha}{n} = \frac{1}{2\pi i} \int_C \frac{(1 + z)^\alpha}{z^{n+1}} \,dz$$
where $C$ is the unit circle, with a small indent to avoid the point $z = -1$. This means
$$\binom{bk + t}{k} = \frac{1}{2\pi i} \int_C \frac{(1 + z)^{bk+t}}{z^{k+1}} \,dz.$$
Now, assuming the indent of $C$ is small enough so that $r \leq |z| \leq 1$ on $C$ for some $r > 2^b a$, we have $|\frac{a(1+z)^b}{z}| \leq r^{-1}2^b a < 1$ on $C$, hence the series
$$\sum_{k=0}^\infty \frac{a^k (1 + z)^{bk+t}}{z^{k+1}} = \sum_{k=0}^\infty \frac{(1 + z)^t}{z}\left(\frac{a(1 + z)^b}{z}\right)^k = \frac{\frac{(1+z)^t}{z}}{1 - \frac{a(1+z)^b}{z}} = \frac{(1+z)^t}{z - a(1 + z)^b}$$
converges absolutely on $C$, uniformly, and thus we have
$$\frac{1}{2\pi i} \int_C \frac{(1+z)^t}{z - a(1 + z)^b} \,dz = \sum_{k=0}^\infty \frac{1}{2\pi i} \int_C \frac{a^k(1 + z)^{bk+t}}{z^{k+1}} \,dz = \sum_{k=0}^\infty a^k \binom{bk+t}{k}.$$
It remains to evaluate the integral on the left. Note that $(1 + z)^t$ and $z - a(1 + z)^b$ are holomorphic on an open set containing $C$ and its interior.
Lemma: $z - a(1 + z)^b$ has a unique zero $r$ in the closed unit disk, and this satisfies $0 < r < 1$.
Proof: Suppose $z_0$ is a zero in the closed unit disk. Clearly $z_0$ cannot be negative or zero. Let $\arg : \mathbb{C} \setminus (-\infty, 0] \to (-\pi, \pi)$ denote the argument function. Note that for $c > 0$ we have $\arg(cz) = \arg(z)$, $|\arg(c + z)| \leq |\arg (z)|$, and if $c \arg (z) \in (-\pi, \pi)$, then $\arg (z^c) = c \arg (z)$. Also note that when $|z| = 1$, $\arg(1 + z) = \frac{1}{2} \arg(z)$. Using these facts, we have
\begin{align*}
|\arg(1 + z_0)|
&= |\arg(1 - |z_0| + (|z_0| + z_0))| \\
&\leq |\arg(|z_0| + z_0)| \\
&= |\arg(1 + z_0/|z_0|)| \\
&= (1/2)|\arg(z_0)|
\end{align*}
which is less than $\pi/2$, hence since $b < 2$, we have
$$|\arg(a(1 + z_0)^b)| = b|\arg(1 + z_0)| \leq \frac{b}{2} |\arg(z_0)|$$
which is strictly less than $|\arg(z_0)|$ when $\arg(z_0) \neq 0$, so in that case we cannot have $z_0 = a(1 + z_0)^b$. Thus any zero must be positive real. But $x - a(1+x)^b$ has positive derivative on $[0, 1]$, and is negative at $x = 0$ and positive at $x = 1$, so there is a unique $r$ in $(0, 1)$ with $r - a(1+r)^b = 0$. [end proof.]
One can check that the residue of $\frac{(1+z)^t}{z - a(1+z)^b}$ at $z = r$ is $\frac{(1 + r)^t}{1 - ab(1 + r)^{b-1}} = \frac{(1+r)^t}{1 - \frac{br}{1+r}} = \frac{(1+r)^{t+1}}{1 + r - br}$, and since $z = r$ is the unique pole of this function in $C$, we have
$$\sum_{k=0}^\infty a^k \binom{bk+t}{k} = \frac{1}{2\pi i} \int_C \frac{(1+z)^t}{z - a(1 + z)^b} \,dz = \frac{(1+r)^{t+1}}{1 + r - br}.$$
We can also run this process backwards: for $0 < r < 1$, and a given $0 < b < 2$, $\frac{r}{(r+1)^b}$ is a strictly increasing function of $r$, so $\frac{r}{(r+1)^b} < \frac{1}{(1+1)^b} = 2^{-b}$, and thus setting $a = \frac{r}{(r+1)^b}$, so $r = a(1+r)^b$, by the above we have
$$\sum_{k=0}^\infty \left(\frac{r}{(r+1)^b}\right)^k \binom{bk+t}{k} = \frac{(1+r)^{t+1}}{1 + r - br}.$$
Because $\frac{r}{(r+1)^b}$ is increasing on $[0, 1]$, we can take the limit as $r \to 1$ of both sides, giving
$$\sum_{k=0}^\infty 2^{-bk} \binom{bk+t}{k} = \frac{2^{t+1}}{2-b}$$
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the radius of the circle in complex plane given by $|z - i| = 3|z +2i|$ where $z$ is a point on the circle. I tried to arrange the following into the standard equation of the circle i.e $|z - z'| = r$ where $r$ is the radius, $z'$ is the centre and $z$ is a point on the circle.
I rearranged the following equation to $|z - i| / |z + 2i| = 3$.
Now I substituted $z = x + iy$ and rationalised the left hand side. After rationalising I could not arrange it to look like the standard form. Please share your solution.
Thank you.
| $|z - i| = 3|z +2i|$
$x^2+(y-1)^2=9(x^2+(y+2)^2)$
$x^2+y^2-2y+1=9x^2+9y^2+36y+36$
$8x^2+8y^2+38y+35=0$
$x^2+y^2+\frac{19}{4}y+\frac {35}{8}=0$
$x^2+(y+\frac{19}{8})^2=\frac{81}{64}$
Circle is centered at $(0;-\frac{19}{8})$ with radius $r=\frac{9}{8}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How many words of length $n ≥ 4$ in the alphabet ${1, 2, 3}$ contain at least two $1$’s, at least two $2$’s, and at least two $3$’s? How many words of length $n ≥ 4$ in the alphabet ${1, 2, 3}$ contain at least two $1$’s, at least two $2$’s, and at least two $3$’s?
My approach involves the incluson-exclusion principle. If you let
$$S=\{\text{all words of length }n\}\\
A_i=\{\text{all words of length }n \text{ that do not contain two }i\text{'s}\},$$
where $1\leq i\leq3$, then our desired number is
$$|S-(\bigcup_{i=1}^3A_i)|$$
I am currently trying to figure out what $|A_i|, |A_i\cup A_j|$, and $|A_i\cup A_j \cup A_k|$are to evaluate this expression. Here's my thoughts:
$$|A_i|=2^n+n2^{n-1}$$
because the word can either have 0 or 1 $i$. If you have zero $i$'s, then each of the $n$ letters in the word have two options. If you have one $i$ for which you can place in any of the $n$ places, and each of the $n-1$ terms have two options. I am unsure how to compute the others. Any help would be amazing!
| Inclusion-exclusion yields:
\begin{align}
&\quad\binom{3}{0}3^n\\
&-\binom{3}{1}\left(\binom{n}{0}2^{n-0}+\binom{n}{1}2^{n-1}\right)\\
&+\binom{3}{2}\left(\binom{n}{0}1^{n-0}+2\binom{n}{1}1^{n-1}+2\binom{n}{2}1^{n-2}\right)\\
&-\binom{3}{3}\left(\binom{3}{0}[n=0]+\binom{3}{1}[n=1]+2\binom{3}{2}[n=2]+3!\binom{3}{3}[n=3]\right) \\
&= 3^n\\
&-3\left(2^n+n2^{n-1}\right)\\
&+3\left(1+2n+n(n-1)\right)\\
&-\left([n=0]+3[n=1]+6[n=2]+6[n=3]\right) \\
&=
3^n
-3\cdot 2^n-3n2^{n-1}
+3+3n+3n^2
-\left([n=0]+3[n=1]+6[n=2]+6[n=3]\right)
\end{align}
If $n \ge 4$, this reduces to
$$3^n-3\cdot 2^n-3n2^{n-1}+3+3n+3n^2,$$
which correctly evaluates to $0$ for $n\in\{4,5\}$ and $90=\binom{6}{2,2,2}$ for $n=6$.
This is OEIS sequence A224541.
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Compute the degree of finite field extension
Compute the degree $[\mathbb{Q}(\sqrt{3}, \sqrt{3 + \sqrt{3}}) : \mathbb{Q}]$.
By the Degree Theorem we have $[\mathbb{Q}(\sqrt{3}, \sqrt{3 + \sqrt{3}}) : \mathbb{Q}] = [\mathbb{Q}(\sqrt{3}, \sqrt{3 + \sqrt{3}}) : \mathbb{Q}(\sqrt{3 + \sqrt{3}})] \cdot [\mathbb{Q}(\sqrt{3 + \sqrt{3}}) : \mathbb{Q}]$.
Now, $\alpha = \sqrt{3 + \sqrt{3}} \iff \alpha^{2} = 3 + \sqrt{3} \iff \alpha^{2} - 3 = \sqrt{3} \iff \alpha^{4} - 6 \alpha^{2} + 6 = 0$. Which means that $\alpha = \sqrt{3 + \sqrt{3}}$ is a zero of $m(x) = x^{4} - 6 x^{2} + 6$. Now, $m$ is monic and Eisenstein with prime $p = 3$. Hence, $m$ is irreducible over $\mathbb{Q}$ and is thus the minimal polynomial. Hence we have $[\mathbb{Q}(\sqrt{3 + \sqrt{3}}) : \mathbb{Q}] = \text{deg}(m) = 4$.
To compute $[\mathbb{Q}(\sqrt{3}, \sqrt{3 + \sqrt{3}}) : \mathbb{Q}(\sqrt{3 + \sqrt{3}})]$, we need to find the minimal polynomial of $\sqrt{3}$ over $\mathbb{Q}(\sqrt{3 + \sqrt{3}})$.
This is where I am stuck.
Perhaps there might be a different solution if for instance $\mathbb{Q}(\sqrt{3}, \sqrt{3 + \sqrt{3}}) = \mathbb{Q}(\bullet)$, but I can't seem to find such a field.
Also, can someone give me some general strategy on how to solve such problems? Perhaps some tips or what to watch out for given your experience.
| Here's a way to do it.
First of all, consider both $[\mathbb{Q}(\sqrt{3+\sqrt{3}} ): \mathbb{Q}]$ and $[\mathbb{Q}(\sqrt{3} ): \mathbb{Q}]$. The first is $4$ as you have stated and the second is $2$, so this gives you the hint that $[\mathbb{Q}(\sqrt{3},\sqrt{3+\sqrt{3}} ): \mathbb{Q}]$ is a multiple of $4$.
I want to prove that it is exactly $4$. Therefore I have to prove that $[\mathbb{Q}(\sqrt{3},\sqrt{3+\sqrt{3}} ): \mathbb{Q}(\sqrt{3+\sqrt{3}})]=1$. That is equivalent to checking that I can express $\sqrt{3}$ as an element of $\mathbb{Q}(\sqrt{3+\sqrt{3}})$. All the elements of $\mathbb{Q}(\sqrt{3+\sqrt{3}})$ are like this:
$$\mathbb{Q}\left(\sqrt{3+\sqrt{3}}\right) = \left\{ a+b\sqrt{3+\sqrt{3}} + c\left(\sqrt{3+\sqrt{3}}\right)^2: a, b, c \in \mathbb{Q}\right\} = \left\{ a+b\sqrt{3+\sqrt{3}} + 3c + c\sqrt{3}: a, b, c \in \mathbb{Q}\right\}$$
Choosing $a = -3$, $b = 0$ and $c = 1$ you get that
$$\sqrt{3} \in \mathbb{Q}\left(\sqrt{3+\sqrt{3}}\right)$$
Therefore, as $[\mathbb{Q}(\sqrt{3},\sqrt{3+\sqrt{3}} ): \mathbb{Q}(\sqrt{3+\sqrt{3}})]=1$ we have that
$$[\mathbb{Q}(\sqrt{3},\sqrt{3+\sqrt{3}} ): \mathbb{Q}] = [\mathbb{Q}(\sqrt{3+\sqrt{3}} ): \mathbb{Q}] \cdot [\mathbb{Q}(\sqrt{3},\sqrt{3+\sqrt{3}} ): \mathbb{Q}(\sqrt{3+\sqrt{3}})] = 4 \cdot 1 = 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3856412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Explanation in a step of the summation $\frac{1}{2^n+1} + \frac{1}{2^n+2}+\cdots +\frac{1}{2^{n+1}}$ I have the solution to the above equation but I wanted to know the explanation for it.
$$\frac{1}{2^n+1} + \frac{1}{2^n+2}+\cdots +\frac{1}{2^{n+1}} = 2^n \frac{1}{ 2^{n+1}}$$
and
$$2^n\frac{1}{2^{n+1}}=\frac{1}{2} \qquad \text{(I understand this part)}.$$
I wasn't sure how that became the product of those 2 numbers. Any explanation please? Thank you.
To make it clearer let me include this link:
https://www2.isye.gatech.edu/~hsharp/math2420/harmonic.pdf
Look at line 5
Im open to any alternative suggestions to solve the problem.
| We have an inequality since $$\frac{1}{2^n+i}\geq \frac{1}{2^{n+1}}$$
for $1\leq i\leq 2^{n}.$ Thus $$\sum_{i=1}^{2^{n}}\frac{1}{2^n+i}\geq\sum_{i=1}^{2^n}\frac{1}{2^{n+1}}=\frac{1}{2^{n+1}}\sum_{i=1}^{2^n}1=2^n\frac{1}{2^{n+1}}$$
| {
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"url": "https://math.stackexchange.com/questions/3857718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Where is the mistake in my evaluation of $\int\frac{x}{x^2+2x+3}\,dx$? Here is how I did it:
First, write $$\int\frac{x}{x^2+2x+3}\,dx=\int\frac{2x+2-x-2}{x^2+2x+3}\,dx=\int\frac{2x+2}{x^2+2x+3}\,dx-\int\frac{x+2}{x^2+2x+3}\,dx.$$
Now consider the integral in the minuend. Letting $u=x^2+2x+3$, one finds $du=(2x+2)\,dx$, and so $$\int\frac{2x+2}{x^2+2x+3}\,dx=\int\frac{du}{u}=\ln{|x^2+2x+3|}.$$
Next consider the other integral. Put $t\sqrt{2}=x+1$. Then $dx=\sqrt 2\,dt$. Now
\begin{align*}
\int\frac{x+2}{x^2+2x+3}\,dx&=\int\frac{(x+1)+1}{(x+1)^2+2}\,dx\\
&=\int\frac{t\sqrt 2+1}{2t^2+2}\sqrt 2\,dt\\
&=\frac{1}{2}\int\frac{2t+\sqrt 2}{t^2+1}\,dt\\
&=\frac{1}{2}\left(\int\frac{2t}{t^2+1}\,dt+\sqrt 2\int\frac{1}{t^2+1}\,dt\right)\\
&=\frac{1}{2}\left(\ln{|t^2+1|}+\sqrt 2\arctan t\right)
\end{align*}
and hence this is equal to
$$\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+1}{2}+1\right|}+\sqrt 2\arctan\frac{x+1}{\sqrt 2}\right)=\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+3}{2}\right|}+\sqrt 2\arctan\frac{x+1}{\sqrt 2}\right)$$
therefore
\begin{align*}
\int\frac{x}{x^2+2x+3}\,dx&=\int\frac{2x+2}{x^2+2x+3}\,dx-\int\frac{x+2}{x^2+2x+3}\,dx\\
&=\ln{|x^2+2x+3|}-\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+3}{2}\right|}+\sqrt{2}\arctan\frac{x+1}{\sqrt 2}\right)+C.
\end{align*}
apparently, the correct answer is $\frac{(\ln|x^2+2x+3|)}{2}-\frac{\sqrt{2}\arctan{\frac{(x+1)}{\sqrt{2}}}}{2}+C.$ what went wrong?
| $\ln{|x^2+2x+3|}-\frac{1}{2}\left(\ln{\left|\frac{x^2+2x+3}{2}\right|}+\sqrt{2}\arctan\frac{x+1}{\sqrt 2}\right)+C= \ln{|x^2+2x+3|}-\frac{1}{2}\left(\ln{\left|{x^2+2x+3}\right|-\ln{2}}+\sqrt{2}\arctan\frac{x+1}{\sqrt 2}\right)+C=\frac{1}{2}\left(\ln{\left|{x^2+2x+3}\right|}- \sqrt{2}\arctan\frac{x+1}{\sqrt 2}\right)-\dfrac{1}{2}\ln{2} + C $
$C_{new}=-\dfrac{1}{2}\ln{2} + C $
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I prove that if: $\cos^3(x) + \sin^3(x) = 1$ then: $\cos(x) = 0 ; \sin(x)=1$ or $\cos(x)=1 ; \sin(x)=0$ How do I prove that if:
$$\cos^3(x) + \sin^3(x) = 1$$
then:
$$\cos(x) = 0 ; \sin(x)=1 \text{ or } \cos(x)=1 ; \sin(x)=0?$$
Starting from the first expression, I couldn't figure out how to reach the conclusion. I replaced 1 by $\cos^2(x) + \sin^2(x) $ hoping to factor it but to no avail.
| This is what I tried after reading your answers, please correct me if there are mistakes!$$cos^3+sin^3=1$$ $$cos^3+sin^3=cos² +sin²$$ $$cos^3-cos²+sin^3-sin²=0$$ $$cos²(cos-1) + sin²(sin-1)=0$$ $$(1-sin²)(cos-1)+(1-cos²)(sin-1)=0$$ $$ (1+sin)(1-sin)(cos-1)+(1+cos)(1-cos)(sin-1)=0 $$ $$(1+sin)(1-sin)(cos-1)-(cos-1)(1-cos)(sin-1)=0$$ $$(cos-1)[ (1+sin)(1-sin)-(sin-1)(1-cos)]=0$$ $$(cos-1)[ (1+sin)(1-sin)+(1-sin)(1-cos)]=0$$ $$(cos-1)[ (1-sin)(1+sin)(1-cos)]=0$$ $$(cos-1)(1-sin²)(1-cos)=0$$ $$(1-sin²)(1-cos)=0$$ $$1-sin²=0$$ $$sin²=1$$ $$sin=1$$
or: $$1-cos=0$$ $$cos=1$$
If: $sin=1$ ⇒ $sin^3=1$
Then:$$cos^3+sin^3=1$$ $$cos^3+1=1$$ $$cos^3=0$$ $$cos=0$$
If: $cos=1$ ⇒ $cos^3=1$
Then:$$cos^3+sin^3=1$$ $$sin^3+1=1$$ $$sin^3=0$$ $$sin=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $1+n+n^2+n^3$ is a perfect square, then $n=1$ or $n=7$ I want to prove that if $1+n+n^2+n^3$ is a perfect square then $n=1$ or $n=7$.
I managed to prove that $1+n+n^2+n^3=(n^2+1)(n+1)$ and that $(n^2+1,n+1)$ is either $1$ or $2$.
I found out that it could not be $1$, and then $\frac{1}{2}(n^2+1,n+1)=1$.
From here I concluded that $n^2+1=2a^2$ and $n+1=2b^2$ for some $a,b\in\mathbb{N}$ and it is here where I need some help.
Please only provide hints.
| Note that the LHS is a 4-term GP with $a = 1, r = n$. The sum is
$${1(1 - n^4) \over (1 - n)} = {n^4 - 1 \over n-1}$$
So, the equation becomes,
$${n^4 - 1 \over n-1} = y^2$$
This is the Nagell–Ljunggren Equation whose general form is:
$${x^n - 1 \over x - 1} = y^q$$
where, $x > 1, y > 1, n > 2, q ≥ 2$.
Known solutions are:
$${3^5 - 1 \over 3 - 1} = 11^2, {7^4 - 1 \over 7 - 1} = 20^2, {18^3 - 1 \over 18 - 1} = 7^3$$
So, for the 4th power version of your problem, $(7, 20)$ is the only known solution. I am not sure if there are other solutions possible. Need to study the original paper.
References:
See 2.2.4 The Nagell–Ljunggren Equation in
Perfect Powers:
Pillai’s works and their developments
by M. Waldschmidt
https://webusers.imj-prg.fr/~michel.waldschmidt/articles/pdf/PerfectPowers.pdf
| {
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How to compute the limit as $x\to 3$ of a $\textit{complicated}$ product and quotient of trigonometric functions $$\lim_{x\rightarrow 3}\frac{ \tan\frac{x-3}{x+3}\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}$$
I substituted $x-3$ for $u$ and got as far as
$$\frac{1}{6} \lim_ {u\to 0} \frac{\sin(9 \sin u)}{\sin((u+3)^(3) -27)}.$$
This is where I get stuck. Should I try a different approach?
| Let $x-3=y$, then
$$L=\lim_{y\to 0} \frac{\tan\frac{y}{y+6} (\sin (9\sin y)}{\sin y \sin[(y+3)^3-27]}=\lim_{y\to 0}\frac{\sin(9\sin y)}{\sin y} \frac{\tan\frac{y}{y+6}}{\sin(y(y^2+9y+27)}=\lim_{y \to 0}\frac{\sin(9\sin y)}{y} \lim_{y \to 0} \frac{\frac{1}{y}\tan\frac{y}{y+6}}{(y^2+9y+27)}$$
As $\tan z \to z, \sin z \to z$ as ${z \to 0}$, we get
$$L=9 \lim_{y \to 0} \frac{1}{(y+6)(y^2+9y+27)}=\frac{9}{6. 27}=\frac{1}{18}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Exponential recurrence Let $f_0 = a, f_1 = b$.
Define $f_n = f_{n-1}$ for odd $n$ and
$$f_{n} = f_{n/2}+f_{n/2 - 1}$$
How do I solve this recurrence? Normally I would use generating functions but the fraction is screwing me up.
| (I will assume that "$f_n=f_{n-1}$ for odd $n$" includes $n=1$, so that $f_1=f_0$; if that's not the case, a similar method will apply but it gets more annoying.)
Define $F(x) = \sum_{n=0}^\infty f(n)x^n$ and $F_2(x) = \sum_{n\text{ even}} f(n)x^n$. Since $f_{n+1}=f_n$ when $n$ is even, we have $F(x) = F_2(x) + xF_2(x) = (1+x)F_2(x)$. On the other hand, since $f_{2m} = f_m + f_{m-1}$ (presumably for $m\ge1$),
\begin{align*}
F_2(x) = \sum_{m=0}^\infty f_{2m} x^{2m} &= f_0 + \sum_{m=1}^\infty (f_m + f_{m-1}) x^{2m} \\
&= f_0 + \bigg( {-}f_0 + \sum_{m=0}^\infty f_m x^{2m} \bigg) + \bigg( x^2 \sum_{m=1}^\infty f_{m-1} x^{2(m-1)} \bigg) \\
&= (1+x^2) F(x^2).
\end{align*}
Therefore
$$
F(x) = (1+x)F_2(x) = (1+x)(1+x^2) F(x^2).
$$
Replacing $x$ by $x^2$ gives $F(x^2) = (1+x^2)F_2(x^2) = (1+x^2)(1+x^4) F(x^4)$,
which means that
$$
F(x) = (1+x)(1+x^2)(1+x^2)(1+x^4) F(x^4).
$$
Repeating this process gives, for any $k\ge1$,
$$
F(x) = \prod_{j=0}^{k-1} (1+x^{2^j}) \prod_{j=1}^{k} (1+x^{2^j}) F(x^{2^k}).
$$
Taking the limit as $k\to\infty$ gives
$$
F(x) = \prod_{j=0}^\infty (1+x^{2^j}) \prod_{j=1}^\infty (1+x^{2^j}) f_0 = \frac1{1-x} \frac1{1-x^2} f_0,
$$
which we can check does satisfy the given recurrences.
| {
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Proving $2\cot4x = \cot2x - \tan2x$ How to prove the trigonometry equation is an identity?
$$2\cot4x = \cot2x - \tan2x$$
Thank you in advance.
Thank you for the comments and hints. I got an answer after many tries. ;)
Below is my answer. Thank you.
$2cot4x = cot2x - tan2x$
$2\frac{1}{tan 4x} = cot2x - tan2x$
$2\frac{1}{\frac{2 tan 2x}{1 - tan^2 2x}} = cot2x - tan2x$
$2\frac{1 - tan^2 2x}{2 tan 2x} = cot2x - tan2x$
$\frac{1}{tan 2x} - \frac{tan^2 2x}{tan 2x} = cot2x - tan2x$
$cot2x - tan2x = cot2x - tan2x$
| Thank you for the comments and hints. I got an answer after many tries. ;)
Below is my answer.
$$\begin{align}
2\cot4x = \cot2x - \tan2x \\
2\cdot\frac{1}{\tan 4x} = \cot2x - \tan2x\\
2\cdot\frac{1}{\frac{2 \tan 2x}{1 - \tan^2 2x}} = \cot2x - \tan2x\\
2\cdot\frac{1 - \tan^2 2x}{2 \tan 2x} = \cot2x - \tan2x\\
\frac{1}{\tan 2x} - \frac{\tan^2 2x}{tan 2x} = \cot2x - \tan2x\\
\cot2x - \tan2x = \cot2x - \tan2x\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Where did I go wrong in applying the factor theorem?
Given that $x + 1$ and $x - 3$ are two of the four factors of the expression $x^4 + px^3 + 5x^2 + 5x + q$, find the values of $p$ and $q$.
I tried to answer this question using the factor theorem but got the answer wrong:
$$ \text{Let } f(x) = x^4 + px^3 + 5x^2 + 5x + q $$
$$ \text{Since } x + 1 \text{ and } x - 3 \text{ are factors of } f(x), \text{ then } f(-1) = 0 \text{ and } f(3) = 0, \text{ i.e.} $$
\begin{align}
(-1)^4 + p(-1)^3 + 5(-1)^2 + 5(-1) + q &= 0 \color{red}{\leftarrow (1)} \\
(3)^4 + p(3)^3 + 5(3)^2 + 5(3) + q &= 0 \color{blue}{\leftarrow (2)}
\end{align}
$$ \text{From } \color{red}{(1)}: $$
\begin{align}
(-1)^4 + p(-1)^3 + 5(-1)^2 + 5(-1) + q &= 0 \\
1 + p(-1) + 5(1) + (-5) + q &= 0 \\
1 - p + 5 - 5 + q &= 0 \\
1 - p + q &= 0 \\
q &= p - 1 \color{limegreen}{\leftarrow (3)}
\end{align}
$$ \text{From } \color{blue}{(2)}: $$
\begin{align}
(3)^4 + p(3)^3 + 5(3)^2 + 5(3) + q &= 0 \\
81 + 27p + 45 + 15 + q &= 0 \\
27p + q + 60 + 81 &= 0 \\
27p + q + 141 &= 0 \\
q &= -27p - 144 \color{orange}{\leftarrow (4)}
\end{align}
$$ \color{orange}{(4)} + \color{limegreen}{(3)}: $$
\begin{align}
-27p - 144 &= p - 1 \\
-27p - p &= 144 - 1 \\
-28p &= 143 \\
28p &= -143 \\
p &= -\frac{143}{28} \\
\therefore p &= -5\frac{3}{28} \color{mediumpurple}{\leftarrow (5)}
\end{align}
$$ \text{Substitute } \color{mediumpurple}{(5)} \text{ into } \color{limegreen}{(3)}: $$
\begin{align}
q &= -5\frac{3}{28} - 1 \\
\therefore q &= -6\frac{3}{28}
\end{align}
The answers were $ p = -5, q = -6 $.
Where did I go wrong?
| Just another way.
Consider
$$y=\frac{x^4 + px^3 + 5x^2 + 5x + q } {(x+1)(x-3) }$$ and perform the long division. You will have
$$y=-\frac{q}{3}+\left(\frac{2 q}{9}-\frac{5}{3}\right) x-\left(\frac{7
q}{27}+\frac{5}{9}\right) x^2+\frac{1}{81} x^3 (-27 p+20 q-15)+\frac{1}{243}
x^4 (54 p-61 q-96)+\cdots$$
So,
$$-27 p+20 q-15=0$$
$$54 p-61 q-96=0$$
Solve for $(p,q)$ (simple).
| {
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Two urns with balls
We consider two urns that contain blue balls and red balls. Urn $1$ has $9$ blue balls and $1$ red one. Urn $2$ has $8$ blue balls and $2$ red ones. We now draw one ball from each urn.
(1) Find the probability of at least one ball being red.
(2) Find the probability of exactly one ball being red.
(3) You know know that exactly one of the balls drawn is red. Find the probability that the ball comes from Urn $1$.
(1) ($\dfrac{1}{10} \cdot \dfrac{2}{10}) + (\dfrac{1}{10} \cdot \dfrac{8}{10}) + (\dfrac{9}{10} \cdot \dfrac{2}{10}) = 0.28$.
(2) $(\dfrac{1}{10} \cdot \dfrac{8}{10}) + (\dfrac{9}{10} \cdot \dfrac{2}{10}) = 0.26.$
(3) Let $R$ be the event of drawing exactly one red ball ($P(R) = 0.26$) and $U_1$ the event of drawing from Urn $1$.
$$P ( U_1| R) = ?$$
How are we supposed to find $P(U_1)$? I tried to model the question using a (single) tree diagram but was unsuccessful.
| *
*$$\frac {1}{10}\times\frac{8}{10}+\frac {9}{10}\times\frac{2}{10}+\frac {1}{10}\times\frac{2}{10}=1-\frac {9}{10}\times\frac{8}{10}$$
*$$\frac {1}{10}\times\frac{8}{10}+\frac {9}{10}\times\frac{2}{10}$$
*$$P( U_1| R)=\frac {P(U_1\cap R)} {P(R)}= \frac{\frac {1}{10}\times\frac{8}{10}}{\frac {1}{10}\times\frac{8}{10}+\frac {9}{10}\times\frac{2}{10}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $b^2+c^2+bc=3$ then $b+c\leq 2$ Suppose that $b,c\geq 0$ such that $b^2+c^2+bc=3$. Prove that $b+c\leq 2$.
I tried to do that by contradiction but I failed.
Indeed, if $b+c>2$ then $b^2+2bc+c^2>4$ then $(b^2+bc+c^2)+bc>4$. Hence $3+bc>4$ or equivalently $bc>1$.
| By AM-GM $$b+c=\sqrt{(b+c)^2}=\sqrt{b^2+2bc+c^2}=\sqrt{b^2+\frac{4}{3}bc+c^2+\frac{2}{3}bc}\leq$$
$$\leq\sqrt{b^2+\frac{4}{3}bc+c^2+\frac{b^2+c^2}{3}}=\sqrt{\frac{4}{3}(b^2+bc+c^2)}=2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the diffrential equation $\left( {{x^2} + xy + 4x + 2y + 4} \right)\frac{{dy}}{{dx}} - {y^2} = 0$ A solution curve of the differential equation $\left( {{x^2} + xy + 4x + 2y + 4} \right)\frac{{dy}}{{dx}} - {y^2} = 0$, $x>0$ passes through the point (1,3). Find the solution curve.
I am not able to proceed as I am not able to convert the standard differentiable form
| $$\left( {{x^2} + xy + 4x + 2y + 4} \right)\frac{{dy}}{{dx}} - {y^2} = 0,$$
$$\left( {{{\left( {x + 2} \right)}^2} + y\left( {x + 2} \right)} \right)dy-y^2 dx=0$$
After substituting $x+2$ with $t$, we get
$$ (t^2+yt)dy-y^2dt=0$$
After multiplying with $t^{-2}$, we get
$$(1+\frac{y}{t})\frac{dy}{dt}-(\frac{y}{t})^2=0$$
Now use substitution $y=tu$, $y'=u+tu'$ from where is $(1+u)(u+tu')-u^2=0$
and
$$u+tu'=\frac{u^2}{u+1}$$
From here we get $$tu'=-\frac{u}{u+1}$$
and finally $$\frac{u+1}{u}du=-\frac{dt}{t}$$.
$$C+u+\ln{u}+\ln{t}=0$$
| {
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Prove that $12(ab+ba+ac) <7a^2+15b^2+18c^2$ holds for all positive numbers. Prove that $12(ab+ba+ac) < 7a^2+15b^2+18c^2$ holds for all positive numbers.
I tried completing the square, but that solution would suggest that inequality holds for all real numbers. Inequalities between means did not work for me either.
$$12(ab+ba+ac) < 7a^2+15b^2+18c^2$$
$$(2a-3b)^2+(2b-3c)^2+(2a-3c)^2+2b^2-a^2>0$$
| The completing to square works:
We need to prove that:
$$18c^2-12(a+b)c+7a^2+15b^2-12ab>0$$ or
$$9c^2-6(a+b)c+\frac{7}{2}a^2+\frac{15}{2}b^2-6ab>0$$ or
$$(3c-a-b)^2+\frac{5}{2}a^2+\frac{13}{2}b^2-8ab>0,$$
which is true by AM-GM:
$$\frac{5}{2}a^2+\frac{13}{2}b^2-8ab\geq2\sqrt{\frac{5}{2}a^2\cdot\frac{13}{2}b^2}-8ab=\left(\sqrt{65}-8\right)ab>0$$
| {
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} |
$Q\le \prod \frac{5+2x}{1+x}\le P$ find $P,Q$
if $x,y,z,$ are positives and $x+y+z=1$ and $$Q\le \prod_{cyc} \frac{5+2x}{1+x}\le P$$ find maximum value of $Q$ and minimum value of $P$
This is actually a question made up myself ,so i don,t know whether a nice solution exists!?.
Finding P seems easier:AM-GM results $$\prod \frac{5+2x}{1+x}\le \frac{{\left(\sum_{cyc} \frac{5+2x}{1+x} \right)}^3}{27}$$
we write $$\sum_{cyc} \frac{5+2x}{1+x}=6+\sum \frac{3}{1+x}$$
But neither Jensen nor the tangent line method help as sign of inequality is reversed.
In fact i am more interested in finding $Q$.
I am looking for a solution that avoids using computational aid(SOS) or ,uvw.
| Let $x=y=z=\frac 13$ then $Q \leqslant \frac{4913}{64}.$ We will show that it's a maximum value, or
$$(5x+5y+7z)(5z+5x+7y)(5y+5z+7x) \geqslant \frac{4913}{64}(2x+y+z)(2y+z+x)(2z+x+y).$$
Let
$$\left\{\begin{aligned} & a = 2x+y+z\\& b = 2y+z+x \\& c = 2z+x+y\end{aligned}\right. \Rightarrow \left\{\begin{aligned} & x = \frac{3a-b-c}{4} \\& y = \frac{3b-c-a}{4} \\& z = \frac{3c-a-b}{4}\end{aligned}\right.$$
The inequality become
$$(11c+3b+3a)(3c+11b+3a)(3c+3b+11a) \geqslant 4913abc.$$
Using the AM-GM inequality, we have
$$(11a+3b+3c)(11b+3c+3a)(11c+3a+3b) \geqslant 17^3 \cdot \sqrt[17]{a^{11}b^3c^3} \cdot \sqrt[17]{b^{11}c^3a^3} \cdot \sqrt[17]{c^{11}a^3b^3} $$
$$= 4913abc.$$
| {
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Find $\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$ $\lim\limits_{n \to \infty} \frac{\frac{1}{\sqrt 1} + \frac{1}{\sqrt 2} + \dots + \frac{1}{\sqrt{n}}}{\ln (n)}$
Can we apply Stolz-Cesaro?
$\lim\limits_{n \to \infty}\frac {\frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n}}} {{\ln(n+1)-\ln(n)}}$ =
$\lim\limits_{n \to \infty}\frac{\sqrt n - \sqrt {n+1}}{\sqrt n \sqrt{n+1}\ln(1+\frac{1}{n})}$ = $\lim\limits_{n \to \infty}\frac{1-\sqrt{1+\frac{1}{n}}}{\sqrt{n+1}\ln(1+\frac{1}{n})}$
What can I do from here?
| The limit is $+\infty$.
You made an error in the first step. Applying Stolz-Cesaro, you should rather consider:
\begin{eqnarray*} \frac{\frac 1{\sqrt{n+1}}}{\ln{(n+1)}-\ln n}
& = & \frac 1{\sqrt{(n+1)}}\cdot \frac 1{\ln \left(1+\frac 1 n\right)}\\
& = & \frac n{\sqrt{(n+1)}}\cdot \underbrace{\frac{\frac 1n}{\ln \left(1+\frac 1 n\right)}}_{\stackrel{n \to \infty}{\longrightarrow}1} \\
& \stackrel{n \to \infty}{\longrightarrow} & +\infty
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Can this problem have solutions for $N>3$:Find three $(N=3)$ positive integers a, b and c such that: $a+b+c=k^2$, $a+b=t^2$, $b+c=m^2$ and $a+c=n^2$ Can this question have solutions for N>3:
Find three ($N =3$) positive integers a, b and c such that: $a+b+c=k^2$, $a+b=x^2$, $b+c=m^2$ and $a+c=n^2$; $x, m, n∈\mathbb N$
Solution:
Let $a+b+c=(x+1)^2=x^2+2x+1$
If $a+b=x^2$ then third number $c=2x+1$.
If $b+c=(x-1)^2=x^2-2x+1=m^2$ then :
$a=4x$, so $b=x^2-4x$
Now due to statement we must have $a+c=6x+1=n^2$
$n^2=6x+1$ can have infinite solutions such as:
$(x, n^2)=(20, 121=11^2), (60, 361=19^2), (140, 841=29^2)\cdot\cdot\cdot$
Which give:
$(a, b, c, k)=(80, 320, 41, 21), (240, 3360, 121, 61), (560, 19040, 281, 141),\cdot\cdot\cdot$
I tried to solve this problem for $N=4$, four numbers but no success. Now I have two questions:
1-Does this problem have solutions for N>3?
2-Any idea for better method?
| We have:
$a + b = u^2$, $c + d = v^2$
$a + c = w^2, b + d = x^2$
$a + d = y^2, b + c = z^2$
$a + b + c + d = k^2$
Let:
$u=m^2-n^2$ & $v =2mn$
$w=p^2-q^2$ & $x=2pq$
$y=r^2-s^2$ & $z=2rs$
We apply the condition:
$m^2+n^2=p^2+q^2=r^2+s^2=k^2$
Hence, $u^2+v^2=w^2+x^2=y^2+z^2=k^2$
Since:
$2a=u^2+w^2-z^2$
$2b=x^2+z^2-v^2$
$2c=v^2+w^2-y^2$
$2d=v^2+x^2-z^2$
Substituting value of $(u,v,w,x,y,z)$ from above we get:
$a=1/2(m^4+n^4+p^4+q^4)-(m^2n^2+p^2q^2+2r^2s^2)$
$b=2(p^2q^2+r^2s^2-m^2n^2)$
$c=1/2(p^4+q^4-r^4-s^4)+(2m^2n^2-p^2q^2+r^2s^2)$
$d=2(m^2n^2+p^2q^2-r^2s^2)$
For: $(m,n,p,q,r,s,k)=(15,10,17,6,18,1,325)$ we get:
$(a,b,c,d)=(39169,-23544,24840,65160)$
| {
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Need help with complex equation: |z−i|+|z+i|=2 I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have:
$$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$
$$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$
$$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+x^2y^2-2xy^2+y^2+x^2y^2-2xy^2+y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+2x^3+x^2-2x^3+4x^2-2x+x^2+2x+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+6x^2+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
And I have no idea what to do next.
Any help would be much appreciated
| The two points i and -i are distance 2 apart.
The origin is distance one from one of them and one from the other so it satisfies the equation.
All the points on the line between those two points satisfy the equation.
Any other point got on the line between these two has the sum of the distances to those two points being greater than 2. Think of the theorem in euclidean geometry that the sum of the length of two sides of a triangle is greater than the length of the third side.
Therefore the only solutions are all the points between the two points.
In general the solutions to the equation
$|z-a|+|z-b| = c$
where $c$ is greater than the distance between
$a$ and $b$
is the eclipse with foci at a and b
and the sum of the distances to a and b equal to c.
If c is equal to the distance between a and b,
the solution is the points on the line segment joining a and b.
If c is less than the distance between a and b,
there is no solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Double integral with a change of variables where $u = \frac{y}{x^3}$ and $v = \frac{y}{x}$. Sketch the planar region D which is bounded by the following four curves and located in the positive quadrant: $y = x^3$, $y = 10x^3$, $y = x$ and $y = 2x$.
Use transformation given by $u=\frac{y}{x^3}$ and $v=\frac{y}{x}$.
Evaluate: $\iint_D \frac{2y}{x^5} \,dx \,dy $
So far I have $x = \frac{u^\frac{1}{4}}{v^\frac{1}{4}}$ and $y = v\frac{u^\frac{1}{4}}{v^\frac{1}{4}}$. Using this to calcuate the Jacobian matrix:
$$\begin{equation}
\begin{vmatrix}
\frac{1}{4v^\frac{1}{4}u^\frac{3}{4}} & \frac{u\frac{1}{4}}{4v^\frac{3}{4}} \\
\frac{v\frac{3}{4}}{4u^\frac{3}{4}u^\frac{3}{4}} & \frac{3u\frac{1}{4}}{4v^\frac{3}{4}} \\
\end{vmatrix}
\end{equation}
= \frac{3-v^\frac{1}{2}}{4v^\frac{1}{2}u^\frac{1}{2}}$$
I used my Jacobian to calculate the integral and I also changed the region D with variables $u$ and $v$ however I ended up with a negative answer. Can anyone check if my Jacobian matrix is correct? I think this is where I have gone wrong.
| Solving wrt to $u,v$ we get
$$\left\{x=\sqrt{\frac{v}{u}},y=v \sqrt{\frac{v}{u}}\right\}$$
Jacobian is
$$J=\left(
\begin{array}{cc}
-\frac{\sqrt{\frac{v}{u}}}{2 u} & -\frac{1}{2} \left(\frac{v}{u}\right)^{3/2} \\
\frac{\sqrt{\frac{v}{u}}}{2 v} & \frac{3 \sqrt{\frac{v}{u}}}{2} \\
\end{array}
\right)$$
and $\det J=-\frac{v}{2 u^2}$
Substituting in the function we get
$$\iint\limits_{D'} \frac{2 u^2}{v}\cdot \left(-\frac{v}{2 u^2}\right)\,dudv=\\=
-\iint\limits_{D'}\,dudv$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding convergence of series using Ratio Test I have this series but Im not too sure whether my calculations are correct
$$ \sum_{n=1}^\infty \left( \frac{(3n)!}{(n!)^3}\right) $$
$$ \left( \frac{(3(n+1)!}{((n+1)!)^3}\right)\cdot\left( \frac{(n!)^3}{(3n)!}\right) $$
$$ \left( \frac{(3(n+1)(3n)!}{((n+1)^3(n!)^3}\right)\cdot\left( \frac{(n!)^3}{(3n)!}\right) $$
and by simplifying I got to
$$ \left( \frac{(3(n+1)}{(n+1)^3}\right) $$
$$ \left( \frac{(3n+3)}{(n^3+3n^2+3n+1)}\right) = 0 $$
so the series is convergent by Ratio test.
| There is a mistake in your first step, indeed by ratio test we obtain
$$ \frac{(3n+3)!}{((n+1)!)^3} \frac{(n!)^3}{(3n)!}=\frac{(3n+3)(3n+2)(3n+1)}{(n+1)^3} $$
since $(3(n+1))!=(3n+3)!$.
To avoid ratio test we can use that
*
*$(3n)! =3n\cdot(3n-1)\cdot \ldots \cdot (n+1) \cdot n!\ge (n+1)^{2n} \cdot n! \ge n^{2n} \cdot n!$
and then
$$ \frac{(3n)!}{(n!)^3} \ge \left(\frac{n^n}{n!}\right)^2 \to \infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3893873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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help integrating I'm really struggling to integrate this.
$$\int_0^t \sqrt{\cos^2\left(1\right)\sinh^2\left(u\right)+4\sin^2\left(\dfrac{1}{2}\right)\cosh^2\left(\dfrac{u}{2}\right)} \,\mathrm{d}u$$
This is to find the arc length of a plane curve. An online integral calculator gives me such a complicated result with $\ln$ and I don't know how they got to that result.
I also then need to use the result to find the inverse.
| Hint:
Set $a={{\cos }^{2}}\left( 1 \right),b={{\sin }^{2}}\left( \frac{1}{2} \right)$
and using $\sinh \left( u \right)=2\sinh \left( \frac{u}{2} \right)\cosh \left( \frac{u}{2} \right)$ the integral becomes:
$$\begin{align}
& =\int{\sqrt{4a{{\sinh }^{2}}\left( \frac{u}{2} \right){{\cosh }^{2}}\left( \frac{u}{2} \right)+4b{{\cosh }^{2}}\left( \frac{u}{2} \right)}du} \\
& \ =\int{\sqrt{4{{\cosh }^{2}}\left( \frac{u}{2} \right)\left( a{{\sinh }^{2}}\left( \frac{u}{2} \right)+b \right)}du} \\
& =\int{2\cosh \left( \frac{u}{2} \right)\sqrt{a{{\sinh }^{2}}\left( \frac{u}{2} \right)+b}du}\ \quad ,x=\sinh \left( \frac{u}{2} \right) \\
& =4\int{\sqrt{a{{x}^{2}}+b}\ dx} \\
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sketching a set of $Re(z^2(1+i))$ on the Complex plane. So I am trying to sketch a set of complex numbers that is characterized by inequality: $ Re((1+i)z^2) < 0 $.
I know that if I take $ z = a + bi $ , then:
$$Re(z^2) \implies (a + bi)^2 = a^2 + 2abi - b^2 \implies Re(z^2) = a^2 - b^2$$
Similarly:
$$Re(z^2(1+i)) \implies (a^2 + 2abi - b^2)(1+i) = a^2 + 2abi - b^2 + a^2i - 2ab - b^2i \implies Re(z^2(1+i)) = a^2 - 2abi + b^2 = (a-b)^2 $$
So how am I supposed to sketch a set of complex numbers $(a-b)^2 $ on a 2D plane? Or did I do something wrong?
| Some facts:
We require the real part to be less than zero, which means $(1 + i)z^2$ must be located on the left half of the complex plane. In terms of the argument, this means $(1+i) z^2$ must have an argument between $\pi/2$ and $3\pi/2$, plus some integer multiple of $2\pi$.
Squaring a complex number doubles its argument.
Multiplying a complex number by $1+i$ increases its magnitude by $|1+i| = \sqrt{2}$ and rotates it by $\pi/4$; i.e., the argument is increased by $\pi/4$.
Therefore, $$\arg((1+i)z^2) = \frac{\pi}{4} + 2\arg (z)$$ and we need this to be between $\pi/2$ and $3\pi/2$, or between $5\pi/2$ and $7\pi/2$. This gives us $$\frac{\pi}{8} < \arg (z) < \frac{5\pi}{8} \quad \text{or} \quad \frac{9\pi}{8} < \arg (z) < \frac{13\pi}{8}.$$ Using interval notation, we can also write this as $$\arg (z) \in (\tfrac{\pi}{8}, \tfrac{5\pi}{8}) \cup (\tfrac{9\pi}{8}, \tfrac{13\pi}{8}).$$
We can visualize this region in the complex plane by coloring quadrants I and III (i.e., those regions where $\Re(z)\Im(z) = 1$, and then rotating those regions by $\pi/8$ counterclockwise.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\sqrt{a}+\sqrt{b}+\sqrt{c}$ only in terms of $p$. If $a^{2} x^{3}+b^{2} y^{3}+c^{2} z^{3}=p^{5}, a x^{2}=b y^{2}=c z^{2}$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{p},$ find $\sqrt{a}+\sqrt{b}+\sqrt{c}$
only in terms of $p$.
I tried simple factorization like shifting terms here and there but didnt get anything.
| We are given
$$a^2x^3+b^2y^3+c^2z^3=p^5\quad (1)$$
$$ax^2=by^2=cz^2=t\ (\mathrm{let})\quad (2)$$
$$\frac 1 x +\frac 1 y +\frac 1 z =\frac 1 p\quad (3)$$
Squaring $(2)$, we get
$$a^2x^4= b^2y^4= c^2=z^4=t^2$$
$$a^2x^3=\frac t x ; b^2y^3=\frac t y ; c^2z^3=\frac t z$$
Substituting these values in $(1)$, we get
$$\frac t x + \frac t y +\frac t z = p^5$$
$$t\left(\frac 1 x +\frac 1 y +\frac 1 z\right)=p^5$$
Using $(3)$, we have
$$\frac t p = p^5 \implies t=p^6$$.
Now, we take the square root of $(2)$.
$$\sqrt{a}x =\sqrt{b}y=\sqrt{c}z=\sqrt{t}=p^3$$
$$\sqrt{a}=\frac{p^3}{x}; \sqrt{b}=\frac{p^3}{y}; \sqrt{c}=\frac{p^3}{z}$$
Therefore, we have
$$\sqrt{a}+\sqrt{b}+\sqrt{c}=\frac{p^3}{x}+\frac{p^3}{y}+\frac{p^3}{z}$$
$$\sqrt{a}+\sqrt{b}+\sqrt{c}=p^3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$$
Finally, using $(3)$ again, we have
$$\sqrt{a}+\sqrt{b}+\sqrt{c}=p^3\cdot \frac 1 p = p^2$$
Hope this helps :)
| {
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Domain of function $\frac{1}{x + \sqrt{x+1}}$ I need to find the domain of the fuction: $$\frac{1}{x + \sqrt{x+1}}$$
I can see that the conditions for the domain are $x+\sqrt{x+1} \neq 0$ and $x + 1\geq 0$. Hence:
$$x+\sqrt{x+1} \neq 0 \quad \wedge \quad x + 1\geq 0$$
$$x \neq -\sqrt{x+1} \quad \wedge \quad x\geq -1$$
$$x^2 \neq |x+1| \quad \wedge \quad x\geq -1$$
Since $x \geq -1$ the expression $x+1$ is never negative, therefore:
$$x^2 \neq x+1 \quad \wedge \quad x\geq -1$$
$$x^2 - x - 1 \neq 0 \quad \wedge \quad x\geq -1$$
$$x \neq \frac{1\pm \sqrt{5}}{2} \quad \wedge \quad x\geq -1$$
Now both $\frac{1 + \sqrt{5}}{2}$ and $\frac{1 - \sqrt{5}}{2}$ are greater than $-1$. So why isn't my domain $$\mathbb{R} \setminus \{\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}\}$$
When I plot the graph I can see that $\frac{1 + \sqrt{5}}{2}$ is part of the domain, but analitically I can't see why or derive from these equations that conclusion.
| A value of $x$ satisfying the equation
$$
x=-\sqrt{x+1}
$$
must be nonpositive since $\sqrt{x+1}$ is nonnegative.
Squaring both sides is legal but of the two solutions you found, one of them is positive so doesn't satisfy the given equation, hence should not excluded from the domain.
Thus the domain is $[-1,r)\cup(r,\infty)\;$where $r={\large{\frac{1 - \sqrt{5}}{2}}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $2^{16^x} = 16^{2^x}$
Find $x$ if $2^{16^x} = 16^{2^x}$.
If $x = 0$ we have $2=16$.
$8$ times or $(2^3)$ difference and if $x = 1$ we have $65536=256$.
$256$ or $2^8$ difference. I can't see anything useful in figuring this out.
I have a feeling $x$ must be a fraction. I know only know about logarithm. I haven't learned anything about the number $e$ or $ln$ yet.
| If you take the $\log_2$ of both sides you end up with
\begin{align*}
\log_2(2^{16^x}) &= 16^x \log_2(2) = 16^x = 2^x\cdot8^x\\
\log_2(16^{2^x}) &= 2^x\log_2{16} = 2^x\cdot4
\end{align*}
From this you get that $2^x\cdot 4 = 2^x \cdot 8^x \implies 8^x = 4 \implies x =\log_{8}(4) = 2/3.$
($\log_8(4) = \log_8(2^2) = 2\log_8(2) = 2\cdot1/3$)
| {
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Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$
Find factor of $x^{11} + x + 1$ of degree 2 in $\mathbb{Z}[x]$
To be honest, it is the first time I see problem like this. I was trying to find divisor by just taking different polinomials of degree 2 and trying to divide this one by them. That was disgusting, I was mixing up in calculations and couldn't finally get answer.
So, Is there any better way to find such a polinomial (divisor of this one with degree 2)? It would be awesome if proposed way would not require super advanced knowledge. Thanks in advance!
| Factorize as follows
\begin{align}x^{11} + x + 1
&= \frac{ x^{11}(x-1) + (x+1)(x-1)}{x-1}\\
&= \frac{x^{12}-x^{11}+x^2-1}{ x-1}
= \frac{(x^{12}-1 )-x^2(x^{9}-1)}{ x-1}\\
&= \frac{(x^{3}-1)(x^9+x^6+x^3+1)-x^2(x^{3}-1)(x^6+x^3+1)}{ x-1}\\
&= \frac{x^{3}-1}{x-1}(x^9+x^6+x^3+1-x^{8}-x^5-x^2)\\
&= (x^2+x+1)(x^9-x^8+x^6-x^{5}+x^3-x^2+1)
\end{align}
To generalize, follow the same steps above
\begin{align}
x^5+x+1 &= (x^2+x+1)(x^3-x^2+1)\\
x^{17}+x+1 &= (x^2+x+1)(x^{15}-...-x^2+1)\\
...&= \> ...\\
x^{6n-1}+x+1 &= (x^2+x+1)(x^{6n-3}-...-x^2+1)\\
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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prove or disprove $\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\left(\frac{k}{p}\right)\equiv 0\pmod{ p^2}$ let $p=8k+1$ prime,prove or disprove
$$\sum_{k=1}^{p-1}\dfrac{(-1)^k}{k}\left(\dfrac{k}{p}\right)\equiv 0\pmod {p^2}\tag{1}$$
where $\left(\dfrac{k}{p}\right)$ is Legendre symbol
It seem this Notes on Wolstenholme’s Theorem How prove this $(p-1)!\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{p-1}\right)\equiv 0\pmod{p^2}$ I can't prove $(1)$
| This is what I thought of. It is not finished but it is a start.
Observe that $$\frac{1}{p}\binom{p}{k}=\frac{1}{p}\cdot\frac{p!}{k!\cdot(p-k)!}\equiv\frac{(p-1)!}{k!\cdot-k\cdot...\cdot-(p-1)}\equiv\frac{(p-1)!}{(p-1)!\cdot k}\cdot(-1)^{p-k}\equiv\frac{(-1)^{k+1}}{k}\pmod{p}$$
So $$\frac{1}{p}\binom{p}{k}\equiv\frac{(-1)^{k+1}}{k}\pmod{p}\Rightarrow\binom{p}{k}\equiv\frac{p\cdot(-1)^{k+1}}{k}\pmod{p^2}$$
To prove $$\sum_{k=1}^{p-1}\frac{(-1)^k}{k}\bigg(\frac{k}{p}\bigg)$$ is divisible by $p^2$, it is enough to prove that $p^2$ divides $$\frac{1}{p}\bigg(\sum_{k=1}^{p-1}\frac{p\cdot(-1)^k}{k}\bigg(\frac{k}{p}\bigg)\bigg)\equiv\frac{1}{p}\bigg(\sum_{k=1}^{p-1}\binom{p}{k}\cdot\bigg(\frac{k}{p}\bigg)\bigg)\pmod{p^2}$$ But $$\sum_{k=1}^{p-1}\binom{p}{k}\cdot\bigg(\frac{k}{p}\bigg)=2\cdot\sum_{k=1}^{\frac{p-1}{2}}\binom{p}{k^2\pmod{p}}-\sum_{k=1}^{p-1}\binom{p}{k}=2\cdot\sum_{k=1}^{\frac{p-1}{2}}\binom{p}{k^2\pmod{p}}-2^p+2$$ so we want to prove that $$\frac{1}{p}\bigg(2\cdot\sum_{k=1}^{\frac{p-1}{2}}\binom{p}{k^2\pmod{p}}-2^p+2\bigg)$$ is divible by $p^2$, in other words $$\sum_{k=1}^{\frac{p-1}{2}}\binom{p}{k^2\pmod{p}}+1\equiv2^{p-1}\pmod{p^3}$$
| {
"language": "en",
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"question_score": "2",
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Computing Ramanujan asymptotic formula from Rademacher's formula for the partition function I am trying to derive the Hardy-Ramanujan asymptotic formula
$$p(n) \sim \frac{1}{4n\sqrt{3}}e^{\pi\sqrt{\frac{2n}{3}}}$$
from Radmacher's formula for the partition function $p(n)$ given by
$$p(n)=\frac{1}{\pi\sqrt{2}}\sum_{k=1}^{\infty}A_{k}(n)\sqrt{k}\left[\frac{d}{dx}\frac{sinh\left(\frac{\pi}{k}\sqrt{\frac{2}{3}\left(x-\frac{1}{24}\right)}\right)}{\sqrt{x-\frac{1}{24}}}\right]_{x=n}$$
where
$$A_{k}(n)=\sum_{h=0, (h,k)=1}^{k-1}e^{\pi i(s(h,k)-2n\frac{h}{k})}$$ and
$$s(h,k)=\sum_{r=1}^{k-1}\frac{r}{k}\left(\frac{hr}{k}-\left\lfloor\frac{hr}{k}\right\rfloor-\frac{1}{2}\right)$$
G.E. Andrews, and any other literature on this topic, says that we can obtain the H-R asymptotic expression from the first term of the Rademacher series, i.e. for $k=1$. I don't know how to approach this as simply calculating for $k=1$ does not give the desired result. Could we perhaps try and use Lapalce's method, or the method of steepest descent?
| Using the leading term of Radmacher's formula, we have
\begin{align*}
p(n) & \sim \frac{1}{{\pi \sqrt 2 }}\left[ {\frac{\mathrm{d}}{{\mathrm{d}x}}\frac{{\sinh \left( {\pi \sqrt {\frac{2}{3}\left( {x - \frac{1}{{24}}} \right)} } \right)}}{{\sqrt {x - \frac{1}{{24}}} }}} \right]_{x = n}
\\ &
= \frac{{4\sqrt 3 }}{{24n - 1}}\cosh \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right) - \frac{1}{\pi}\frac{{24\sqrt 3 }}{{(24n - 1)^{3/2} }}\sinh \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right).
\end{align*}
Now
\begin{align*}
\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} = \pi \sqrt {\frac{2}{3}n} \sqrt {1 - \frac{1}{{24n}}} & = \pi \sqrt {\frac{2}{3}n} \left( {1 + \mathcal{O}\!\left( {\frac{1}{n}} \right)} \right) \\ &= \pi \sqrt {\frac{2}{3}n} + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right).
\end{align*}
Thus,
\begin{align*}
& \cosh \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right),\sinh \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right) \sim \frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}\left( {n - \frac{1}{{24}}} \right)} } \right) \\ & = \frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} + \mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right) = \frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right)\left( {1 +\mathcal{O}\!\left( {\frac{1}{{\sqrt n }}} \right)} \right) \\ & \sim \frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right).
\end{align*}
And therefore,
\begin{align*}
p(n) & \sim \frac{{4\sqrt 3 }}{{24n - 1}}\frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right) - \frac{1}{\pi }\frac{{24\sqrt 3 }}{{(24n - 1)^{3/2} }}\frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right)
\\ &
\sim \frac{{4\sqrt 3 }}{{24n}}\frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right) - \frac{1}{\pi }\frac{{24\sqrt 3 }}{{(24n)^{3/2} }}\frac{1}{2}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right)
\\ &
= \frac{1}{{4n\sqrt 3 }}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right)\left( {1 - \frac{{\sqrt 3 }}{{\sqrt 2 \pi n^{1/2} }}} \right) \sim \frac{1}{{4n\sqrt 3 }}\exp \left( {\pi \sqrt {\frac{2}{3}n} } \right).
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Values of $a$ such that $x^5-x-a$ has quadratic factor I would like to find all integers $a$ such that $x^5-x-a$ has a quadratic factor in $\mathbb{Z}[x]$.
My Attempt
Let $x^5-x-a=(x^2+bx+c)(x^3+dx^2+ex+f)$, so that we have the following:
$$\begin{array}{rcl}
b+d&=&0\\
e+bd+c&=&0\\
f+be+cd&=&0\\
bf+ce&=&-1\\
cf&=&-a
\end{array}$$
Hence:
$$\begin{array}{rcccl}
d&=&-b\\
e&=&-bd-c&=&b^2-c\\
f&=&-be-cd&=&-b^3+2bc
\end{array}$$
and we have:
$$1=-bf-ce=b^4-3b^2c+c^2,$$
so that:
$$(2c-3b^2)^2=5b^4+4.$$
Question
How can I find all values of $n$ such that $5n^4+4$ is a perfect square?
My Attempt
If $m^2=5n^4+4$, then $m^2-5n^4=4$.
If $m=2m_*$, then $n$ is even, so that $n=2n_*$, and we have the equation $m_*^2-20n_*^4=1$. By Pell equation, since $(a,b)=(9,2)$ is the least non-trivial solution of $a^2-20b^2=1$, then the general solution has the form $(a_n,b_n)$ where $a_n+b_n\sqrt{20}=(9+2\sqrt{20})^n$, but I do not know how to find out what values of $n$ make $b_n$ a square.
| This is essentially an elliptic curve.
There might be elementary methods, but there are also computer algebra systems that can (in many cases) solve this kind of diophantine equations.
We may rewrite the equation as: $m^2n^2 = 5n^6 + 4n^2$.
If we write $y = 5mn$ and $x = 5n^2$, then it becomes $y^2 = x^3 + 20x$.
Now we use Sage to find all integer points on this curve. Paste the following codes into this site and press "Evaluate".
EllipticCurve([20, 0]).integral_points()
The output:
[(0 : 0 : 1), (4 : 12 : 1), (5 : 15 : 1), (720 : 19320 : 1)]
We see that the corresponding values of $(m, n)$ are $(2,0), (3,1), (322,12)$, respectively (negative values are not listed).
| {
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"source": "stackexchange",
"question_score": "12",
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Number of partitions of an integer with extra conditions In how many ways can you write number 15 as a sum of positive integers if number 1 can appear 3 times at most in the sum and number 3 can only appear even number of times?
I've tried doing this as $$(1+t+t^2+t^3)(1+t^2+t^4+t^6+t^8+t^{10}+t^{12}+t^{14})(1+t^6+t^{12})\cdots(1+t^{15})$$ and finding the coefficient of $t^{15}$, but it's taking too much time. Is there an easier way to do this? I think the result should be around 81. (I found the number of partitions of number 15 with no limitations, then ruled out those that do not satisfy the condition.)
| Via inclusion-exclusion, where $p_n$ is the number of unrestricted partitions of $n$:
\begin{align}
& p_{15}
- \left[p_{15-4\cdot1} + (p_{15-1\cdot3} - p_{15-2\cdot3}) + (p_{15-3\cdot3} - p_{15-4\cdot3}) + p_{15-5\cdot3} \right]
+ \left[(p_{15-4\cdot1-1\cdot3} - p_{15-4\cdot1-2\cdot3}) + p_{15-4\cdot1-3\cdot3} \right] \\ &= p_{15}
- \left[p_{11} + (p_{12} - p_9) + (p_6 - p_3) + p_{0} \right]
+ \left[(p_8 - p_5) + p_{2}\right] \\
&= 176
- \left[56 + (77 - 30) + (11 - 3) + 1 \right]
+ \left[(22 - 7) + 2\right] \\
&= 176 - 112 + 17 \\
&= 81
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove by induction $n!<4n^2+3$ How can I show that $ \{n\in \mathbb{N}| n!<4n^2+3\} $ ?
Here´s my try:
The equation is valid for every $n\leq4$.
n=5: n!=5!=120>103
n=n+1: $(n+1)!=n!(n+1)\geq(4n^2+3)(n+1)\geq4(n+1)^2+3(n+1)\geq4n^2+3\geq4(n+1)+3$
Here I got stuck. Usually there should be $4(n+1)^2+3$ at the ending.
Any ideas how i could continue or whats wrong?
| Your work is mostly fine, but you can perfect your redacting.
What you want to show is that the set $A=\{n\in\mathbb N\mid n!<4n^2+3\}=\{0,1,2,3,4\}$
For that you need to verify the inequality for the a few numbers which you certainly did, but it needs to appear on your solution.
$\begin{cases}
n=0&0!=1<4\times 0^2+3=3&\text{ok}\\
n=1&1!=1<4\times 1^2+3=7&\text{ok}\\
n=2&2!=2<4\times 2^2+3=19&\text{ok}\\
n=3&3!=6<4\times 3^2+3=39&\text{ok}\\
n=4&4!=24<4\times 4^2+3=67&\text{ok}\\
n=5&5!=120\not<4\times 5^2+3=103&\text{not verified}\\
\end{cases}$
You are correctly suspecting that the inequality is not verified either for $n\ge 5$, so you should explicit the induction proposition :
$$P(n) : n! \ge 4n^2+3\qquad\forall n\ge 5$$
Base case $P(5)$ is verified.
So let assume $P(n)$ we get $(n+1)!=(n+1)n!\ge (n+1)(4n^2+3)$
And this is where you got stuck.
You need to make appear $4(n+1)^2+3$, instead I propose to subtract both quantities and show the difference is positive (this is easier to handle):
$\begin{align}(n+1)!-\bigg((4(n+1)^2+3\bigg)
&\ge (n+1)(4n^2+3)-\bigg(4(n+1)^2+3\bigg)
\\&\ge 4n^3+3n+4n^2+3-(4n^2+8n+7)
\\&\ge 4n^3-5n-4
\\&\ge 4n^3-5n-4n\qquad\text{ for } n\ge 1\implies 4\le 4n
\\&\ge n(4n^2-9)
\\&\ge 0 \qquad\qquad\qquad\qquad\text{ for } n^2\ge \frac 94\iff n\ge \frac 32
\end{align}$
Since $n\ge 5$ both conditions ($n\ge 1$ and $n\ge \frac 32$) are verified therefore you have proved your induction step.
| {
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"source": "stackexchange",
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Evaluating limits using a series I'm trying to use a Taylor series centered at $0$ to evaluate this limit:
$$\lim_{x\to \infty}4x^3(e^\frac{-2}{x^3}-1)$$
I rewrote the function as its Maclaurin series:
$$4x^3(e^\frac{-2}{x^3}-1)=\sum_{k=1}^\infty\frac{4x^{3-\frac{2k}{x^3}}}{{k!}}$$
In expanded form:
$$\sum_{k=1}^\infty\frac{4x^{3-\frac{2k}{x^3}}}{{k!}}=4x^{3-\frac{2}{x^3}}+\frac{4x^{3-\frac{4}{x^3}}}{2!}+\frac{4x^{3-\frac{6}{x^3}}}{3!}+...$$
As $x$ goes to $\infty$, $\frac{2}{x^3}$ goes to $0$. Thus, the limit of the first term is simply the limit of $4x^3$, which is $\infty$. Based on this fact alone, I would assume, the limit of the entire series is $\infty$, but apparently the answer is $-8$. What did I do wrong?
| $$e^{-\dfrac2{x^3}}=1-\dfrac2{x^3}+\dfrac{\left(-\dfrac2{x^3}\right)^3}{2!}+\cdots=1-\dfrac2{x^3}+O\left(\dfrac1{x^6}\right)$$
Alternatively, Set $-\dfrac2{x^3}=h$ to find
$$=-4\cdot2\lim_{h\to0^-}\dfrac{e^h-1}h$$
| {
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Balkan MO :Find all possible primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$. Find all possible primes $p$ and $q$ such that $3p^{q-1}+1$ divides $11^p+17^p$.
This problem was posted 2 times in MSE, but was never solved. Can someone help me?
I am supposed to use orders and mods
| Another approach:
Due to Euler's criterion:
$A=3^{\frac{7-1}2=3}+1 \equiv 0 \ mod (7)$
$3\times 3^{q-1}+1\equiv 0 \ mod (7)$
These two relations give $p=q=3$
In this case $A=28$.
We check this with RHS, i.e $B=11^p+17^q$:
$11\equiv 4 \mod (7)$ , $\rightarrow 11^p \equiv 4^p \mod (7)=7a+4^p$
$17 \equiv 3 \ mod (7)$, $\rightarrow 17^q\equiv 3^q \ mod (7)=7b+3^q$
$17^q$ and $3^q$ are odd so $7b$ is even, therefore:
$17^q=14 b_1+3^q$
⇒ $11^p+13^q=7a+7b+4^p+3^q$
$11^p+17^q$ and $4^p+3^q$ are odd so $7a+7b$ must be even; $7b=14b_1$ is even, so $7a$ must also be even, let $7a=14a_1$, so we have:
$11^p+17^q=14(a_1+b_1)+4^p+3^q$
$4^p+3^q=(7-3)^p+3^q= 7t -3^p+3^q$
We need $11^p+11^q\equiv (4^p+3^q)\equiv 0 \ mod (7)$
This is only possible if $p=q$ such that:
$4^p+3^q=(7-3)^p+3^q= 7t +(-3^p+3^q=0)=7t$
Now if $p=q=3$ then $14a_1+14b_1=28 k$, because:
$11^3+17^3=223\times 28$
Hence $A=28\big|B$
That is $p=q=3$ can be a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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prove inequality $\frac{\sin \pi x}{\pi x} \geq \frac{1-x^2}{1+x^2}$ for all R Let $F(x) = \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} $, is even, so just prove $x > 0$.
When $ x \geq 1 $, $\frac{\sin\pi x}{\pi x} \leq 1$, so
$$ \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} \\= -\frac{\sin\pi (x-1)}{\pi(x-1)} \frac{x-1}{x} - \frac{1-x^2}{1+x^2} \\ \geq \frac{1-x}{x} - \frac{1-x^2}{1+x^2} \\ = \frac{(x-1)^2}{x(1+x^2)} \\ \geq 0 $$
So how to prove it when $0 < x <1 $?
| Equivalent problem: Let $0 < y < \pi$. Prove that $\frac{\sin y}{y} \ge \frac{\pi^2 - y^2}{\pi^2 + y^2}$.
It suffices to prove that, for all $y$ in $[0, \pi]$,
$$\sin y - y \frac{\pi^2 - y^2}{\pi^2 + y^2} \ge 0.$$
Denote LHS by $g(y)$. We have the following result. The proof is given at the end.
Fact 1: If $y\in (0, \pi)$ satisfying $g'(y) = 0$, then $g(y) \ge 0$.
Now, recall that a continuous function on a bounded and closed interval achieves its minimum
at either endpoints or the interior points with zero derivative.
Thus, by Fact 1, noting also that $g(0) = g(\pi) = 0$, we have $g(y) \ge 0$ on $[0, \pi]$.
We are done.
$\phantom{2}$
Proof of Fact 1:
From $g'(y) = 0$, we have $\cos y = \frac{\pi^4 - 4\pi^2 y^2 - y^4}{(\pi^2 + y^2)^2}$.
We have,
\begin{align}
&(\sin y)^2 - \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\
=\ & 1 - (\cos y)^2 - \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\
=\ & 1 - \left(\frac{\pi^4 - 4\pi^2 y^2 - y^4}{(\pi^2 + y^2)^2}\right)^2 - \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\
=\ & \left(1 + \frac{\pi^4 - 4\pi^2 y^2 - y^4}{(\pi^2 + y^2)^2}\right)\left(1 - \frac{\pi^4 - 4\pi^2 y^2 - y^4}{(\pi^2 + y^2)^2}\right)
- \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\
=\ & \frac{2\pi^2(\pi^2 - y^2)}{(\pi^2 + y^2)^2}\cdot\frac{2y^4 + 6\pi^2y^2}{(\pi^2 + y^2)^2}
- \left(y \frac{\pi^2 - y^2}{\pi^2 + y^2}\right)^2\\
=\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}
\Big[2\pi^2(2y^2 + 6\pi^2) - (\pi^2 - y^2)(\pi^2 + y^2)^2\Big]\\
=\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}\Big[y^6+ \pi^2y^4+(-\pi^4+ 4\pi^2)y^2 + ( - \pi^6 + 12\pi^4)\Big]\\
\ge\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}\Big[\pi^2y^4+(-\pi^4+ 4\pi^2)y^2 + ( - \pi^6 + 12\pi^4)\Big]\\
\ge\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}\Big[ 2 \sqrt{\pi^2y^4 \cdot (- \pi^6 + 12\pi^4)} + (-\pi^4+ 4\pi^2)y^2\Big]\tag{1}\\
=\ & \frac{y^2(\pi^2-y^2)}{(\pi^2 + y^2)^4}\Big[ 2\pi \sqrt{- \pi^2 + 12} + (-\pi^2+ 4)\Big]\pi^2 y^2\\
\ge\ & 0
\end{align}
where we have used AM-GM inequality in (1).
We are done.
| {
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"source": "stackexchange",
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Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
Solve: $3\sin{2x}+4\cos{2x}-2\cos{x}+6\sin{x}-6=0$
My Try
$6\sin{x}\cos{x}+4(\cos^2{x}-\sin^2{x})-2\cos{x}+6\sin{x}-6=0$
I have expanded the equation, But I cannot proceed further, Any hint would be appreciated. Thank you!
| Like Barry Cipra,
replacing $\sin x=s,\cos x=c,\cos2x=2c^2-1,\sin2x=2sc$
$$3(2sc)+4(3c^2-1)-2c+6s-6=0$$
$$\iff4c^2-c(1-3s)+3s-5=0$$
$$\implies c=\cdots=-1,\dfrac{5-3s}4$$
If $\cos x=c=-1, x=(2n+1)\pi$
Else $4\cos x+3\sin x=5$
Use Solving trigonometric equations of the form $a\sin x + b\cos x = c$
See also: Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$
| {
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"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Induction proof done, need to simplify $\sqrt[4]{(n+1)^5}\ln(1 + \frac{1}{(n+1)^2}) \leq 4 \sqrt[4]{\frac{n+1}{n}}$ Prove that:
$$\sum_{j=1}^{n} \sqrt[4]{j^5}\ln(1+\frac{1}{j^2}) \leq 4 \sqrt[4]{n}, n \in \mathbb{N} $$
I got everything right and came up with an inequality:
$$\sqrt[4]{(n+1)^5} \ln\left(1 + \frac{1}{(n+1)^2}\right) \leq 4 \sqrt[4]{\frac{n+1}{n}}$$
The problem is I cannot simplify that to get my solution all done. Anybody got an idea?
| As regards your final step, just notice that $\ln(1+x)\leq x$.Therefore it suffices to show that
$$\sqrt[4]{(n+1)^5}\cdot \frac{1}{(n+1)^2} \leq 4 \sqrt[4]{\frac{n+1}{n}}$$
that is
$$\sqrt[4]{n}\leq 4 (n+1)^{\frac{1}{4}+2-\frac{5}{4}}=4(n+1)$$
which trivially holds.
| {
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Find the solution to cauchys problem I'm learning differential equation and not sure how to solve the following problem:
Find the solution for the cauchy's problem:
$y''-6y'+9y=e^{3x}\frac{x^2+2}{x^2+1}$
$y(0)=0$
$y'(0)=1$
What I tried so far:
$\lambda^2-6\lambda+9=0 \Leftrightarrow (\lambda-3)(\lambda-3)=0$
$y(x)=c_1*e^{3x}+c_2*x*e^{3x}+\varphi(x)$
I'm not sure how to continue.
Could you help me?
Thanks
| To begin with, I would notice that
\begin{align*}
y'' - 6y' + 9y & = (y'' - 3y') - (3y' - 9y) = (y' - 3y)' - 3(y' - 3y)
\end{align*}
Thus if we make the change of variable $u = y' - 3y$, one gets that
\begin{align*}
u' - 3u = e^{3x}\left(\frac{x^{2}+2}{x^{2}+1}\right) & \Longleftrightarrow (ue^{-3x})' = \frac{x^{2}+2}{x^{2}+1} = 1 + \frac{1}{x^{2}+1}\\\\
& \Longleftrightarrow ue^{-3x} = x + \arctan(x) + c_{1}\\\\
& \Longleftrightarrow (ye^{-3x})' = x + \arctan(x) + c_{1}\\\\
& \Longleftrightarrow ye^{-3x} = \frac{x^{2}}{2} + x\arctan(x) - \frac{\ln(x^{2}+1)}{2} + c_{1}x + c_{2}\\\\
& \Longleftrightarrow y(x) = e^{3x}\left(\frac{x^{2}}{2} + x\arctan(x) - \frac{\ln(x^{2}+1)}{2} + c_{1}x + c_{2}\right)
\end{align*}
| {
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"source": "stackexchange",
"question_score": "1",
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Prove that ${a_{1} +2a_{2} +3a_{3}+\cdots+np\ a_{n}}_{p} \ =\frac{np}{2}( 1+p)^{n}$ In expansion of ${\ \left( 1+x+x^{2} +\cdots+x^{p}\right)^{n} =a_{0} +a_{1} x+a_{2} x^{2} +\cdots+a_{np}} x^{np}$,
prove that ${a_{1} +2a_{2} +3a_{3} +\cdots+np\ a_{n}}_{p} \ =\frac{np}{2}( 1+p)^{n}$.
My attempt:
${\ \left( 1+x+x^{2} +\cdots+x^{p}\right)^{n} =a_{0} +a_{1} x+a_{2} x^{2} +\cdots+a_{np}} x^{np}$
for $x=1$, $(1+p)^n=a_0+a_1+a_2+a_3+\cdots+a_{np}$ if we multiply $(a_o+a_1x+a_2x^2+\cdots+a_{np}x^{np})(1+2\frac{1}{x}+3\frac1{x^2}+4\frac{1}{x^3}+\cdots)$ we will get terms independent of $x$, $a_0+2a_1+3a_2+\cdots+(np+1)a_{np}$ then if we subtract $(1+p)^n$ from it, then we will get ${a_{1} +2a_{2} +3a_{3} +\cdots+np\ a_{n}}_{p}$ so we need to find term independent of $x$ in $(1+x+x^2+\cdots+x^{p})^n (1-\frac{1}{x})^{-2}$ simplifying further $(1+x+x^2+\cdots+x^{p})^n x^2 (1-x)^{-2}$ here no terms will be free of $x$ hence term will be $0$, subtracting $(1+p)^n$ from it we have ${a_{1} +2a_{2} +3a_{3} +\cdots+np\ a_{n}}_{p}=-(1+p)^n$ .
| $$(1+x+x^2+...+x^p)^n=(a_0+a_1x+a_2x^2+....+a_{np}x^{np})$$
D.w.r.t. $x$ on both sides
$$n(1+x+x^2+...+x^p)^{n-1}(1+2x+3x^2+4x^3+...+px^{p-1})=a_1+2a_2x+a_3x^2+a_4x^3+...+np a_{np} x^{np-1}$$
Put$x=1$ to get
$$n (1+p)^{n-1}\frac{p(p+1)}{2}=a_1+2a_2+3a_3+...+np a_{np}=\frac{1}{2}np(1+p)^n.$$
| {
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"question_score": "2",
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Is $f(n)=\sqrt{n}$ the only function from $\Bbb{N}_0$ to $[0,\infty)$, with $f(100)=10$ and $\sum_{k=0}^n\frac{1}{f(k)+f(k+1)}=f(n+1)$?
Let $f:\mathbb{N_0} \rightarrow \left[0,\infty\right)$ be a function such that
a) $f(100)=10$
b)$\dfrac{1}{f(0)+f(1)}+\dfrac{1}{f(1)+f(2)}+\cdots+\dfrac{1}{f(n)+f(n+1)}=f(n+1)$ for all $n\geq 0$.
Evaluate $f(n)$.
Clearly, $f(n)=\sqrt{n}$, satisfies the conditions, but my question is, is it the only function possible?
I don't see how using the value at $n=100$ would prove the function to be $\sqrt{n}$, however if instead of $100$, $f(1)=1$ was given, then we can prove by induction that $\sqrt{n}$ would be the only function by starting from the first term and finding the next term.
| For $n\geq 1$,
\begin{align}
f(n+1) &= \dfrac{1}{f(0)+f(1)}+\dfrac{1}{f(1)+f(2)}+\cdots+\dfrac{1}{f(n)+f(n+1)}\\
& = f(n)+\dfrac{1}{f(n)+f(n+1)}\\
\implies f(n+1)^2&=1+f(n)^2 =n+f(1)^2\\
\implies f(n) &= \sqrt{n-1+f(1)^2}.
\end{align}
Further, we have $$10 = f(100)=\sqrt{99+f(1)^2}\implies f(1)=1.$$
Finally, we also derive $$f(1)=\dfrac{1}{f(0)+f(1)}\implies f(0)=0.$$
Hence, we conclude that $f(n)=\sqrt{n}.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$x^3 + y^3 = p^2$ over the integers $x^3 + y^3 = p^2$ has a solution over the integers for some three digit prime p. Find all p that satisfy.
The first thing I did was factorize the left hand side, getting $(x+y)(x^2 - xy + y^2) = p^2$
I then considered the case $x^2 - xy + y^2 = p^2$, which gave me $p^2 = 3y^2 -3y +1$, which can be naturally factorized into $(p+1)(p-1) = 3y(y-1)$, but how do I do anything with this? I don't see a continuation from this point onwards.
Any hints or solutions would be greatly appreciated. Thanks!
| Consider the linear factor first instead; this allows you to directly express $y$ in terms of $x$. From
$$p^2=x^3+y^3=(x+y)(x^2-xy+y^2),$$
it follows that $x+y=p^k$ for some $k\in\{0,1,2\}$, and $x^2-xy+y^2=p^{2-k}$. Then $y=p^k-x$, so we can eliminate $y$ to get the quadratic equation
$$3x^2-3p^kx+p^{2k}-p^{2-k}=0.$$
A quadratic has an integral root if and only if its discriminant is a perfect square. Can you continue from here?
The discriminant equals $$\Delta=(-3p^k)^2-4\cdot3\cdot(p^{2k}-p^{2-k})=-3p^{2k}+12p^{2-k}.$$ For $k=1$ and $k=2$ this becomes $$\Delta=-3p^2+12p=-3p(p-4)\qquad\text{ and }\qquad\Delta=-3p^4+12,$$ which are both negative because $p>4$. Hence $k=0$ and $\Delta=12p^2-3$. So if $p$ is such a prime, then $\Delta$ is a perfect square and a multiple of $3$, say $\Delta=(3e)^2$. Then a bit of algebra shows that $(X,Y)=(2p,e)$ is an integral solution to the Pell equation $$X^2-3Y^2=1.$$
| {
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Maximize the function $f(x) = (1-x) (1+x)^3$ Maximize the function $f(x) = (1-x)(1+x)^3.$
I took the first derivative of $f(x)$ and got $(1+x)^2 (2-4x)$
From there I got $2$ critical points of: $-1, \frac{1}{2}$
Then $I$ took the second derivative of $f(x)$ ($-12x^2 -12x$) and plugged in the critical points to get $0$ and $-6$.
I know I am doing it wrong can anybody help. The right answer is $\frac{27}{16}$.
| Since $f(x) $ is non-negative only when $ |x| \leqslant 1$, then we only need to care when $|x|\leqslant 1$.
With that, we know that $ 1-x$ and $ 1 + x$ are non-negative numbers. Thus, we can apply the AM-GM inequality:
$$ AM\left(1 - x , \frac{1+x}3 , \frac{1+x}3, \frac{1+x}3 \right) \geqslant GM \left (1 - x , \frac{1+x}3 , \frac{1+x}3, \frac{1+x}3 \right) $$
Simplifying the above gives $$\frac12 \geqslant \left ( \frac{(1-x)(1+x)^3}{3^3} \right )^{1/4} $$
Rearranging gives $ (1-x)(1+x)^3 \leqslant \boxed{\frac{27}{16}}.$
The maximum value is attained when $ 1-x = \dfrac{1+x}3 \Leftrightarrow x = \dfrac12. $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Basis of Image in a GF(5) matrix with variables consider $A_\alpha := \begin{pmatrix} \alpha & 3 & 4\\ 3 & 1 & 3 \\ 4 & 2 & \alpha \end{pmatrix} \in \mathbb{Z_5}^{3\times3}$. Determine, dependant on $\alpha$, a Basis and the dimension of the Image of the standardinterpretation $F_{A_\alpha}$. What is the dimension of the kernel of $F_{A_\alpha}$? The standardinterpretation is defined as $F_A : K^{n\times1} \rightarrow K^{m \times 1}, x \mapsto F_A(x):=Ax $.
Okay. To determine the Image of a Matrix we just row reduce it and then we will see which columns are in the Image of that matrix.
Let's row reduce the matrix
$\begin{pmatrix}
\alpha & 3 & 4\\
3& 1& 3\\
4& 2& a
\end{pmatrix} \underset{\overset{R1 \leftarrow R_1+2\cdot R_2}{\longrightarrow}}{\overset{R_3 \leftarrow R_3 + 3 \cdot R_2}{\longrightarrow}}
\begin{pmatrix}
a+1 & 0& 0\\
3& 1& 3\\
0& 0& a+4
\end{pmatrix}$.
This doesn't look too bad, let's just look at the cases we get:
Let $\alpha = 0$ then we get $\begin{pmatrix}
1 & 0& 0\\
3& 1& 3\\
0& 0& 0
\end{pmatrix} \overset{R_2 \leftarrow R_2+2\cdot R_1}{\longrightarrow}
\begin{pmatrix}
1 & 0& 0\\
0& 1& 3\\
0& 0& 0
\end{pmatrix}
$ then we can see that a basis for $Image(A_\alpha)$ is $\{\begin{pmatrix} a \\ 3 \\ 4 \end{pmatrix}, \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix}\}$. Since we have 2 elements in our basis for the Image, we know that $dim(Image)=2$.
We also remember that $dim(V) = dim(kernel) + dim(Image)$ therefore $3 = dim(kernel) + 2$ so we get $dim(kernel)=1$.
Do I have to repeat this process for all other 4 cases ($\alpha = 1,2,3,4$) or am I completly wrong about my approach? Thanks for any hints on how to solve this problem.
| Your row reduction is indeed correct. Your steps from there are problematic. We have the reduced matrix
$$
\pmatrix{
a+1 & 0& 0\\
3& 1& 3\\
0& 0& a+4}.
$$
It is easy to see that for $\alpha \notin \{1,4\}$, further row-reduction brings us to the identity so that $A$ is invertible. In the case that $\alpha = 1$, we get
$$
\pmatrix{
2 & 0& 0\\
3& 1& 3\\
0& 0& 0} \leadsto
\pmatrix{
2 & 0& 0\\
0& 1& 3\\
0& 0& 0}.
$$
We deduce that the first two columns of $A$ form a basis of the image. That is, the image has basis
$$
\{(\alpha,3,4),(3,1,2)\} = \{(1,3,4),(3,1,2)\}.
$$
So, the dimension of the image is $2$ and the dimension of the kernel is $3 - 2 = 1$.
Similarly, for $\alpha = 4$, we get
$$
\pmatrix{
0 & 0& 0\\
3& 1& 3\\
0& 0& 3} \leadsto
\pmatrix{
3& 1& 0\\
0& 0& 3\\
0&0&0}.
$$
We deduce that the first and third columns of $A$ form a basis of the image. So, the dimension of the image is $2$ and the dimension of the kernel is $3 - 2 = 1$.
| {
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"url": "https://math.stackexchange.com/questions/3941405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a$,$b$,$c$ be real numbers. Let $a$,$b$,$c$ be real numbers such that $a^{4}+b^{4}+c^{4}\leq27$. Prove the inequality:
$$\sum_{cyc}\frac{1}{\sqrt{a^{2}+1}}\geq \frac{3}{2}$$
My solution:
$\frac{1}{\sqrt{a^{2}+1}}\geq\frac{1}{2}\implies a^{2}\leq 3\implies a^{4}\leq 9$.
Similary : $b^{4}\leq9$ and $c^{4}\leq9$.
Then $a^{4}+b^{4}+c^{4}\leq27$ which is obviously true.
My solution is good ?
| Let $x_a = \sqrt{a^2 + 1}$, and similarly for $x_b$ and $x_c$. Then by RMS-HM inequality on $x_a, x_b, x_c$,
$$\begin{align*}
\frac{3}{x_a^{-1}+x_b^{-1}+x_c^{-1}}
&\le \sqrt{\frac{x_a^2+x_b^2+x_c^2}3} \\
&= \sqrt{\frac{(a^2+1)+(b^2+1)+(c^2+1)}3}\\
&= \sqrt{\frac{a^2+b^2+c^2}3 + 1}\tag1\\
\end{align*}$$
By RMS-AM inequality on $a^2,b^2,c^2$,
$$\begin{align*}
\frac{a^2+b^2+c^2}3
&\le \sqrt{\frac{a^4+b^4+c^4}3}\\
&\le \sqrt{\frac{27}3}\\
&= 3\tag2\\
\end{align*}$$
Combining results $(1)$ and $(2)$,
$$\begin{align*}
\frac{3}{x_a^{-1}+x_b^{-1}+x_c^{-1}} &\le \sqrt{3+1}\\
\sum \frac{1}{\sqrt{a^2+1}} &\ge \frac{3}{2}\\
\end{align*}$$
| {
"language": "en",
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An alternative way to calculate $\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx$ Consider the integral
$$
\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx
$$
By noting that
$$
\frac{x}{x^2+(1-x)^2}=
\frac{2x-1}{(2x-1)^2+1}+\frac{1}{(2x-1)^2+1}
$$
we deduce
$$
\int\frac{x}{x^2+(1-x)^2}\,dx=\frac{\ln(x^2+(1-x)^2)}{4}+\frac{1}{2}\arctan(2x-1)+C,
$$
so
$$
\int_{0}^{1}\frac{x}{x^2+(1-x)^2}\,dx=\frac{\pi}{4}.
$$
Is there an alternative way to calculate this integral?
| $$I=\int\frac{x}{x^2+(1-x)^2}\,dx=\int \frac{x}{2x^2-2x+1}\,dx=\int \frac{x}{2(x-a)(x-b)}\,dx$$
$$I=\frac 1{2(a-b)}\int \Big[\frac a{x-a}-\frac b{x-b} \Big]\,dx$$
$$I=\frac 1{2(a-b)}\Big[a \log(|x-a|)-b\log(|x-b|)\Big]$$ Now $a=\frac{1-i}2$, $b=\frac{1+i}2$,
$$I=\frac {1-i}4\left(\log
\left(x-\frac{1+i}{2}\right)+i \log
\left(x-\frac{1-i}{2}\right)\right)$$
Use the bounds, play with the complex numbers to get the answer.
| {
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Locus of $z$ satisfying $\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$
For a given complex number, $z$, find the locus of points on the Argand diagram such that
$$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$$
This is my approach:
$$\arg(z-8)-\arg(z-2)=\frac{\pi}{2}$$
Suppose $z=x+iy$:
$$\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}=\frac{\pi}{2}$$
This is where I'm unsure I made a 'legal' move. I reasoned that if I take the tangent of both sides of the equation, I would end up with a fraction that must be undefined as $\tan\frac{\pi}{2}$ is undefined; hence the denominator of the fraction I obtain must be $0$:
$$\tan\left(\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}\right)=\frac{\frac{y}{x-8}-\frac{y}{x-2}}{1+\frac{y^2}{(x-8)(x-2)}}=\frac{y(x-2)-y(x-8)}{(x-8)(x-2)+y^2}$$
But this is undefined; hence
$$(x-8)(x-2)+y^2=0\implies (x-5)^2+y^2=9$$
so it appears that the required locus is a circle with centre at $(5,0)$ on the Argand diagram with radius $3$.
Is my reasoning fully acceptable throughout my solution?
| Here is perhaps a better approach. Note $\arg\left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$ implies
$$\frac{z-8}{z-2}=re^{i \frac{\pi}{2}}\implies
\frac{z-8}{z-2}= - \frac{\bar z-8}{\bar z-2}\implies |z|^2 -5(z+\bar z )+16=0
$$
which is $|z-5|^2=9$, or $|z-5|=3$, i.e. a circle of center $(5,0)$ and radius $3$.
Note that $z=\frac{8-i 2r }{1-i r} =\frac{(8+2r^2)+i 6r}{1+r^2}$, indicating a positive imaginary and the upper half circle.
| {
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"source": "stackexchange",
"question_score": "1",
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Can we write $\sin(\frac{\pi}{14})$ as an finite expression using only basic operations? First, let us recall some trigonometric values
$\sin(0)=0$
$\sin(\pi/6)=\frac{1}{2}$
$\sin(\pi/3)=\frac{\sqrt{3}}{2}$
$\sin(\pi/2)=1$
$\sin(\pi/10)=\frac{\sqrt{5}-1}{4}$
$\sin(\pi/12)=\frac{\sqrt{3}-1}{2\sqrt{2}}$
Here, we can observe that for some values of $\theta$, $\sin(\theta)$ can be expressed as a sum of finite radical terms. It is also easy to see that, there exists infinite such $\theta$'s. Say, I have given $\sin(\pi/14)$, can we determine whether it can also be expressed as sum of finite radical terms? What about the general case? Are there any criteria that $\theta$ must obey in order to do that?
| It is not so bad !
As @Doug M commented, it is one of the roots of
$$8x^3 - 4x^2 - 4x + 1 = 0$$
Using the trigonometric method for three real roots,we have
$$\sin \left(\frac{\pi }{14}\right)=\frac{1}{6} \left(1+2 \sqrt{7} \sin \left(\frac{1}{3} \csc ^{-1}\left(2
\sqrt{7}\right)\right)\right)$$
Otherwise, using radicals
$$\sin \left(\frac{\pi }{14}\right)=\frac{1}{6}-\frac{7^{2/3} \left(1-i \sqrt{3}\right)}{6\ 2^{2/3} \sqrt[3]{-1+3 i
\sqrt{3}}}-\frac{1}{12} \left(1+i \sqrt{3}\right) \sqrt[3]{\frac{7}{2} \left(-1+3
i \sqrt{3}\right)}$$
The second and third terms are two of the roots of
$$46656 x^6-1512 x^3+343=0$$ which is a quadratic in $x^3$.
The solutions of
$$46656 X^2-1512 X+343=0$$ are
$$X_{1,2}=\frac{7}{432} \left(1\pm3 i \sqrt{3}\right)$$
| {
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The limit of a series I need help with this exercise!
$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}.$
I saw somewhere online that $1^4+2^4+...+n^4=\dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$.
But I dont understand why!
So following that then,
$\lim\limits_{n \to \infty} \dfrac{1^4+2^4+...+n^4}{n^5}=$
$=\lim\limits_{n \to \infty} \dfrac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}=$
$=\lim\limits_{n \to \infty} \dfrac{(n+1)(2n+1)(3n^2+3n-1)}{30n^4}=\dfrac15.$
And after that what!? Do I multiply and then use L'Hopital?
| This is what happens when you rush into learning calculus!
The quantity $S(n) = 1^4 + 2^4 + \cdots + n^4$ can be summed in a variety of ways. One such way is to guess that it grows like a fifth degree polynomial, then use Lagrange interpolation and induction. Another way might be to notice that
$$\sum_{i=0}^{n-1} (i+1)^5 - i^5 = n^5$$
is easy to evaluate. But alternatively,
$$(i+1)^5 - i^5 = 5i^4 + 10i^3 + 10i^2 + 5i + 1.$$
I leave it as an exercise to you on how to derive from here; now to your question on l'Hopital's.
You have $$\lim_{n\rightarrow\infty} \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5}.$$
The numerator is a fifth degree polynomial. The denominator is a fifth degree polynomial. So the limit will just be the ratio of the leading coefficients, $6/30 = 1/5.$ Why? Well, imagine we have a simpler case,
$$\lim_{n\rightarrow\infty} \frac{n^2 + 3n + 1}{n^2}.$$
Then
$$\frac{n^2+3n+1}{n^2} = 1 + \frac{3}{n} + \frac{1}{n^2}.$$
As $n$ goes to infinity, anything with an $n$ in the denominator will go to 0, and you're just left with 1. If you expanded out that fifth degree numerator, you'd quickly find the same trick applies and all but the leading term dies.
| {
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Evaluate $\int_0^{\pi/2} \frac{1+2\cos x}{(2+\cos x)^2}$ I don’t know how to begin solving. Can I get a hint?
My failed attempt
Let $t =\tan (x/2)$
Then $$I =2 \int_0^1 \frac{(3-t^2)}{(3+t^2)^2 }$$
which I am not able to solve
| $$\int_0^\tfrac{\pi}{2} \frac{1 + 2\cos x}{(2 + \cos x)^2}{\rm d}x \\
= \int_0^\tfrac{\pi}{2} \frac{1 + 2\cos x}{(2 + \cos x)^2}\frac{/\sin^2 x}{/\sin^2 x} {\rm d}x \\
= \int_0^\tfrac{\pi}{2} \frac{\csc^2 x + 2\cot x \csc x}{(2\csc x + \cot x)^2}{\rm d}x \\
= \int_0^\tfrac{\pi}{2} \frac{-1}{(2\csc x + \cot x)^2}{\rm d}(2\csc x + \cot x) \\
= \left[\frac{1}{(2\csc x + \cot x)}\right]_0^\tfrac{\pi}{2}\\
= \left[\frac{\sin x}{(2 + \cos x)}\right]_0^\tfrac{\pi}{2} \\
= \frac{1}{2}$$ Hope it helps.
| {
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Proof by induction, $2^{n} > n^{2} - 2$ So I have to proof that $2^{n} > n^{2} - 2$ for $n > 2$, using mathematical induction.
I start of with the "basecase" for n = 3:
RHS: $2^{3} = 8$, LHS: $3^{2} - 2 = 7$
Clearly, the inequality is correct in this case. I assume that the inequality is true for $\forall n = k$ where $k \in \mathbb{Z}^{+}$.
To prove this inequality, we have to proof that it holds for n = k + 1.
$2^{k} > k^{2} - 2$
We multiply both sides with 2:
$2^{k+1} > 2(k^{2} - 2)$
$2^{k+1} > 2k^{2} - 4$
$2^{k+1} > k^{2} + k^{2} - 2 - 2$
According to our assumption, $2^{k} > k^{2} - 2 $, hence $2^{k+1} > k^{2} - 2 $
Because $k^{2} - 2 \geq 2k + 1 $ for $k > 2 $, we get:
$2^{k+1} > k^{2} + 2k + 1 - 2$
Thus:
$2^{k+1} > (k+1)^{2} - 2$
QED.
How does my proof look, does it have any deficits, and how would you prove it?
Thank you!
| The inductive step uses $2k^2-4\ge(k+1)^2-2$, or equivalently $(k-1)^2\ge4$, which holds for $k\ge3$. With quadratic inequalities, it's always wise to complete the square.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Dividing $x^{81}+x^{48}+2x^{27}+x^{6}+3$ by $x^{3}+1$ Letting $f(x)=x^{81}+x^{48}+2x^{27}+x^{6}+3$, and we seek to divide $f(x)$ by $x^{3}+1$, or just any $x^3-a$. Is subbing $y=x^3$ allowed in this kind of division? I've experimented doing so and so far it seems correct.
| Even without the substitution, the problem will succumb to
polynomial long division.
Therefore, the only issue is how much perspiration you can replace with inspiration.
As already discussed, setting $y = x^3$ reduces the problem to
$$\text{Divide} ~g(y) = (y^{27} + y^{16} + 2y^9 + y^2 + 3) ~\text{by}~ (y+1). \tag1$$
In attacking equation (1), note that
*
*$(16 - 9) = (9 - 2)$.
*the middle terms have coefficients : $1,2,1$ which match the 2nd row of Pascal's triangle.
This suggests that the three middle terms of equation (1) above can be re-expressed as
$$(y^2)(y^7 + 1)^2.$$
Further, it is well settled that for $n$ odd, $(y + 1)$ divides $(y^n + 1).$
For example: $(y + 1) \times (y^2 -y + 1) = (y^3 + 1)$ and
$(y + 1) \times (y^4 -y^3 + y^2 - y + 1) = (y^5 + 1).$
Therefore, you now know (immediately) that $(y + 1)$ divides both $(y^{7} + 1)$ and $(y^{27} + 1).$
Therefore, you can immediately compute the division of $g(y)$ by $(y + 1)$, avoiding polynomial long
division, and immediately conclude that the remainder will be $2$.
| {
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$a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$
Q. If $a + bp^\frac{1}{3} + cp^\frac{2}{3} = 0$, prove that $a = b = c = 0$ ($a$, $b$, $c$ and $p$ are rational and $p$ is not a perfect cube.)
My approach:
Solving the quadratic, I get:
$p^\frac{1}{3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}$
Case 1: If the $b^2 - 4ac$ is a perfect square, I get the LHS as irrational and the RHS as rational, which is a contradiction.
Case 2: If $b^2 - 4ac$ is not a perfect square, $b = \pm \sqrt{b^2 - 4ac} - 2cp^\frac{1}{3}$
Here, the LHS is rational and the RHS is irrational, contradiction again. (Edit: The answer of @GNUSupporter has the proper proof.)
So the equation is not quadratic and $c = 0$.
$a + bp^\frac{1}{3} = 0$
$-\dfrac{a}{b} = p^\frac{1}{3}$
This is a contradiction and hence $b = 0$ and $a = 0$
Is there any other way to solve this?
| There's a typo in the first step. @mjw caught it. Note that you're applying the quadratic formula on $p^{1/3}$, so $$p^{1/3} = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2c}, \text{ if }c \ne 0.$$
Another missing link in your proof is your lack of argument about the claimed irrationality of $± \sqrt{b^2 - 4ac} - 2cp^{2/3}$.
As you handled the degenerate case "$c = 0$ and $b \ne 0$" well, we'll keep the assumption "$c \ne 0$ or $b = 0$" for the rest of the proof. Also we assume that $\sqrt{b^2-4ac}$ is irrational. Multiply $2c$ on both side of the above equality, then cube it.
\begin{align}
8c^3p =& -b^3 \pm 3b^2 \sqrt{b^2-4ac} - 3b(b^2-4ac) \pm (b^2-4ac)^{3/2} \\
=& -4b^3+12abc \pm 4(b^2-ac) \sqrt{b^2-4ac} \\
2c^3p =& -b^3 + 3abc \pm (b^2-ac) \sqrt{b^2-4ac}
\end{align}
Make $\sqrt{b^2-4ac}$ the subject of the above equality. If $b^2-ac \ne 0$,
$$\sqrt{b^2-4ac} = \pm \frac{2c^3p + b^3 - 3abc}{b^2-ac} \in \mathbb{Q},$$
contradicting our assumption on the irrationality of $\sqrt{b^2-4ac}$ if $b^2 - ac \ne 0$.
Assume that $b^2 - ac = 0$. Then $b^2 - 4ac = b^2 = -3b^2$, and $$p^{1/3} = \frac{(-b\pm\sqrt3|b|i)}{2c} = \begin{cases} \frac{b}{c} e^{2\pi i/3} \text{ or } \frac{b}{c} e^{4\pi i/3} \text{ if } b > 0 \\ \frac{b}{c} e^{\pi i/3} \text{ or } \frac{b}{c} e^{5\pi i/3} \text{ if } b < 0 \end{cases}.$$
In this case, if $b \ne 0$, we don't have the desired conclusion, since $p = \left(\dfrac{b}{c}\right)^3$ might not be an integer, so it's not a perfect cube.
If $b = 0$, we use the assumption $b^2 = ac$, we've $a = 0$ or $c = 0$.
*
*If $a = 0$, only the term $cp^{2/3} = 0$ is left in the original equation, but $p \ne 0$ as $p$ can't be a perfect cube, so $c = 0$.
*If $c = 0$, the original equation becomes $a = 0$. Done.
| {
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Stuck on a step for finding the closed formula for catalan numbers from generating function! I am looking at the Frazer Jarvis paper, entitled Catalan Numbers.
http://www.afjarvis.staff.shef.ac.uk/maths/jarvisspec01.pdf
In this paper, he derives the closed formula from the generating function.
The $nth$ Catalan number, denoted $C_n$, is the number of ways
of multiplying together $n$ symbols.
I am stuck on page 7, particularly the following steps:
$\begin{equation}C_n=-\frac{1}{2}\bigg\{\frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2n-3}{2})}{n!}(-4)^n\bigg\}\end{equation}$
$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(\frac{1}{2})(\frac{3}{2})...(\frac{2n-3}{2})(n-1)!}{n!(n-1)!}(2^2)^n)\bigg\}$
$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}[\frac{1}{2}\cdot 1 \cdot\frac{3}{2} \cdot 2...(n-2)(\frac{2n-3}{2})(n-1)]}{n!(n-1)!}2^{2n}\bigg\}$
$C_n=\frac{1}{2}\bigg\{\frac{\frac{1}{2}(1 \cdot 2 \cdot 3 \cdot 4...(2n-4)(2n-3)(2n-2)}{n!(n-1)!}2^{2}\bigg\}$
Particularly, from the 2nd to the 4th line, I am struggling to understand the steps taken here.
I am confused with what he is doing with the $\frac{1}{2}$. Is he pulling it out in the 3rd step? And most of all, I am very confused with how the $2^n$ went away in the 4th step. If anyone could please help me understand some of the steps that he skipped here, I would really appreciate it! Thank you so much!!!
| In the first step he observed that there are $2n$ minus signs, so they cancel out, and he multiplied top and bottom by $(n-1)!$. In the next step he rearranged the factors in the numerator:
$$\begin{align*}
&\frac12\cdot\frac32\cdot\ldots\cdot\frac{2n-3}2(n-1)!\\
&\qquad=\frac12\cdot\frac32\cdot\ldots\cdot\frac{2n-3}2\cdot\color{red}1\cdot\color{red}2\cdot\ldots\cdot\color{red}{(n-1)}\\
&\qquad=\frac12\cdot\color{red}1\cdot\frac32\cdot\color{red}2\cdot\frac52\cdot\color{red}3\cdot\ldots\cdot\color{red}{(n-2)}\cdot\frac{2n-3}2\cdot\color{red}{(n-1)}\\
&\qquad=\frac12\cdot\color{red}{\frac22}\cdot\frac32\cdot\color{red}{\frac42}\cdot\frac52\cdot\color{red}{\frac62}\cdot\ldots\color{red}{\frac{2n-4}2}\cdot\frac{2n-3}2\cdot\color{red}{\frac{2n-2}2}\\
&\qquad=\frac{(2n-2)!}{2^{2n-2}}\,.
\end{align*}$$
Thus, the whole thing is
$$\begin{align*}
\frac12\cdot\frac{\frac12(2n-2)!}{2^{2n-2}n!(n-1)!}\cdot2^{2n}&=\frac14\cdot\frac{(2n-2)!}{n!(n-1)!}\cdot2^{2n-(2n-2)}\\
&=2^{-2}\cdot\frac{(2n-2)!}{n!(n-1)!}\cdot 2^2\\
&=\frac{(2n-2)!}{n!(n-1)!}\\
&=\frac1n\binom{2n-2}{n-1}\,.
\end{align*}$$
I wrote up essentially the same derivation in this answer, if you’d like to see a slightly different take on it.
| {
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Solve three-variable system $x^2 - yz = 1,\>y^2 - xz = 2,\>z^2 - xy = 3$ I am going around in circles on this system
$$x^2 - yz = 1\\
y^2 - xz = 2\\
z^2 - xy = 3$$
I have tried a few things (below) but keep hitting a wall. I know that by adding the three equations we get:
$x^2 + y^2 + z^2 - yz - xz - xy =6$. And also that:
$(x-y)^2 + (x-z)^2 + (y-z)^2 = 2x^2+2y^2+2z^2-2xy-2xz-2yz=12$
I have tried the following:
$(x-y)(x+y+z)=(x^2-yz)-(y^2-xz)=-1$
$(x-z)(x+y+z)=(x^2-yz)-(z^2-xy)=-2$
$(y-z)(x+y+z)=(y^2-xz)-(z^2-xy)=-1$
So I think it follows that $x-y=y-z$ and therefore $x-z=2(x-y)$
Substituting in:
$(x-y)^2+[2(x-y)]^2+(x-y)^2=12$
$6(x-y)^2=12$
$x-y=\sqrt{2}$
Therefore: $y-z=\sqrt{2}$ and $x-z=2\sqrt{2}$
But when I rearrange and substitute into one of the original equations, the solutions I get for x, y and z don't actually work.
$y=x-\sqrt{2}$ and $z=x-2\sqrt{2}$
Substituting into the first equation:
$x^2-(x-\sqrt{2})(x-2\sqrt{2})=1$
$x=\frac{3\sqrt{2}}{2}$
Giving $y=\frac{\sqrt{2}}{2}, z=-\frac{\sqrt{2}}{2}$
Unfortunately, substituting these values into the three original equations only satisfies the second one. Can anyone please help?
| You're on the right track but made an algebra error (cf. $y - z$). These are quadratic equations, so there are two solutions:
$$\left\{x\to -\frac{5}{3 \sqrt{2}},y\to \frac{1}{3 \sqrt{2}},z\to \frac{7}{3
\sqrt{2}}\right\}$$
$$\left\{x\to \frac{5}{3 \sqrt{2}},y\to -\frac{1}{3 \sqrt{2}},z\to -\frac{7}{3
\sqrt{2}}\right\}$$
| {
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"question_score": "2",
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Solving $9x \equiv 21 \bmod 30$
Solving $9x \equiv 21 \bmod 30$
My approach:
Finding $gcd(9,30)$:
\begin{align}
30 &= 9 \cdot 3 + 3 \\ 9 &= 3 \cdot 3 + 0
\end{align}
So $gcd(9,30)=3$ and since $3 \mid 21$ so the congruence has $3$ incongruent solutions. Express $3$ as a linear combination of $9$ and $30$, we get: $$3 = 30-9 \cdot 3 \ \ \ \ \ \ (1)$$
We have to solve this equation to find $x_0$:
$$9x-30y=21 \ \ \ \ \ \ (2)$$
Multiplying $(1)$ by $7$ we get:
$$21=30 \cdot7-9 \cdot 21$$
Rewritten in form of $(2)$:
$$-9 \cdot (21) + 30 \cdot (7) = 21 $$
So $x_0=21$, the incongruent solutions are:
$$21 + 10n, \ \ \ \ \ n = 0,1,2$$
Then $x=21,31,41$. But my answer is wrong, the answer is: $x=9,19,29$. Can you tell me which step is wrong. Thanks for your help!
| $9x\equiv 21\pmod{30}$ means $9x=30k+21$ or $3x=10k+7$ or $3x\equiv 7\pmod{10}$ or $21x\equiv 49\pmod{10}$ or $\color{red}{x\equiv 9\pmod{10}}$.
| {
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Identity with the floor function I'm struggling to complete proof of the following identity:
$$
\Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m - n +1}{2} \Bigr\rfloor =m,
$$
where $m$ and $n$ are both integer.
By definition of floor function, $x-1 < \lfloor x \rfloor \leq x$.
Then
\begin{align}
\frac{m+n}{2} -1 < & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor \leq & \frac{m+n}{2} \\
\frac{m-n + 1}{2} -1 < & \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor \leq & \frac{m-n +1}{2}.
\end{align}
By adding member to member, we obtain the following result:
$$
m +\frac{1}{2} -2 < \Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor \leq m +\frac{1}{2}.
$$
Which implies
$$
-1.5 < \Bigl\lfloor \frac{m+ n}{2} \Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2} \Bigr\rfloor -m \leq .5
$$
This results in
$$
\Bigl\lfloor \frac{m+ n}{2}\Bigr\rfloor + \Bigl\lfloor \frac{m- n +1}{2}\Bigr\rfloor \in \left\{0,1\right\}.
$$
How to decide that the result is $0$?
Any help is welcome.
| hint
Observe that
$$\frac{m-n+1}{2}=\frac{m+n+1}{2}-n$$
So, you just need to prove that
$$\lfloor a\rfloor +\lfloor a+\frac 12\rfloor=2a$$
where $ a=\frac{m+n}{2}$.
There are two cases
$$a\in \Bbb N \implies \lfloor a\rfloor +\lfloor a+\frac 12\rfloor=a+a$$
$$a\notin \Bbb N \implies a+\frac 12\in \Bbb N$$
$$\implies \lfloor a\rfloor +\lfloor a+\frac 12\rfloor=$$
$$a+\frac 12-1+a+\frac 12=2a$$
Done.
| {
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Question on binomial theorem: Find the term independent of $x$ in the expansion of $(2+x)\left(2x+\frac{1}{x}\right)^5$
Find the term independent of $x$ in the expansion of $$(2+x)\left(2x+\frac{1}{x}\right)^5$$
I am only able to do this with one binomial, but when there are two I do not totally understand the process to do this.
| You can make life a bit easier in first writing
$$(2+x)\left(2x+\frac 1x\right)^5 = \frac 1{x^5}\color{blue}{(2+x)(2x^2+1)^5}$$
Now, you see you need only calculate the coefficient of $x^5$ in $(2+x)(2x^2+1)^5$. This is easy since $(2x^2+1)^5$ contains only even powers of $x$ but you are looking for $x^5$. Hence using binomial expansion you get
\begin{eqnarray*}[x^5](2+x)(2x^2+1)^5
& = & [x^5](2+x)\binom 52(2x^2)^2 \\
& = & \binom 52 \cdot 4 = 40
\end{eqnarray*}
| {
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Remainder when $\frac{f(x)}{x^3+4x^2+x-6}$ and given two other remainders Determine the remainder $r$ when $$\frac{f(x)}{x^3+4x^2+x-6}\rightarrow remainder \ r$$.
The following is known $$\frac{f(x)}{x^2+2x-3}\rightarrow remainder\ (x+2)\\
\frac{f(x)}{x+2}\rightarrow remainder\ (1)\\$$
My work:
$$(x^3+4x^2+x-6)=(x^2+2x-3)(x+2)\\(x^2+2x-3)=(x-1)(x+3)\\
(x+2)\\
f(x)=(x^3+4x^2+x-6)g(x)+(Ax+B)\\
f(1)=A+B=x+2\\
f(-3)=-3A+B=x+2\\
f(-2)=-2A+B=1\\
\left\{\begin{matrix}
A+B=x+2& x=-1\\
-3A+B=x+2& B=1\\
-2A+B=1& A=0
\end{matrix}\right.\\
r=(-1)(0)+1=1$$
Can the remainder really be $r=1$? My intuition says it's wrong.
EDIT:
$$f(x)=(x^3+4x^2+x-6)g(x)+(Ax^2+Bx+C)\\
f(1)=A+B+C=x+2\\
f(-3)=9A-3B+C=x+2\\
f(-2)=4A-2B+C=1\\
\left\{\begin{matrix}
A+B+C=x+2& A=\frac{x+3}{6}\\
9A-3B+C=x+2& B=\frac{3x+5}{6}\\
4A-2B+C=1 &C=\frac{4-2x}{6}
\end{matrix}\right.\\
r=\frac{x+3}{6}x^2+ \frac{3x+5}{6}x+\frac{4-2x}{6}=\frac{x^3+6x^2+3x+9}{6}$$
| $$f(x)=\left(x^3+4 x^2+x-6\right) q(x)+a x^2+b x+c=\\=(x^2+2 x-3) (x+2) q(x)+a x^2+b x+c\tag{1}$$
$$r(x)=a x^2+b x+c$$
We know that
$$\frac{f(x)}{x+2}$$
gives remainder $r=1$, which means that $f(-2)=1$
then we know that
$$\frac{f(x)}{x^2+2 x-3}$$
gives remainder $r=x+2$, which means that
$$f(x)=p(x)(x^2+2x-3)+x+2=p(x)(x-1)(x+3)+x+2$$
so we have $f(1)=3$ and $f(-3)=-1$
Now back to
$$f(x)=\left(x^3+4 x^2+x-6\right) q(x)+a x^2+b x+c=(x-1) (x+2) (x+3)q(x) +ax^2+bx+c$$
we plug $x=-2;\;x=1;\;x=-3$ and get
$$
\begin{cases}
4 a-2 b+c=1\\
a+b+c=3\\
9 a-3 b+c=-1\\
\end{cases}
$$
solution should be $$\left(a= -\frac{1}{3},b= \frac{1}{3},c= 3\right)$$
Remainder should be $$r(x)=-\frac{x^2}{3}+\frac{x}{3}+3$$
| {
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prove that the product of 4 consecutive positive integers is divisible by 24 How can I prove that prove that the product of $4$ consecutive positive integers is divisible by $24$, ie for any positive integer $n$ : $n(n+1)(n+2)(n+3)$
is divisible by $24$.
I've noticed that:
$24$ = $2^3 * 3$
$n(n+1)(n+2)(n+3)$ is divisible by $2*2$ so by $4$ (as there are at least 2 even numbers, obvious)
$n(n+1)(n+2)(n+3)$ is divisible by $3$ (as there is the product of 3 consecutive integers, easily provable using congruences)
so I've proved that you can divide it by $12$ until now. What am I missing here?
| Notice that among $4$ consecutive numbers exactly one is divisible by 4 and one more divisible by 2 (say $n$ and $n+2$ or $n+1$ and $n+3$).
Notice also that among $3$ consecutive numbers exactly one is divisible by 3.
So the number is divisible by $3\cdot 8$.
| {
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Find this limit $\lim_{n\rightarrow \infty} \frac{n}{n+1}-\frac{n+1}{n}$. Am I correct? I've found this limit by this way. Am I correct?
Find this limit: $\lim_{n\rightarrow \infty}\left(\frac{n}{n+1}-\frac{n+1}{n}\right)$
Let's see that:
\begin{align}
\frac{n}{n+1}-\frac{n+1}{n}&=\frac{n^2-(n+1)^2}{(n+1)(n)}\\&=\frac{n^2-n^2-2n-1}{n^2+n}\\&=\frac{-2n-1}{n^2+n}\\&=\frac{-\frac{2}{n}-1}{1+\frac{1}{n}}
\end{align}
Así,
\begin{align}
\lim_{n \rightarrow \infty}\left ( \frac{n}{n+1}-\frac{n+1}{n} \right ) &=\lim_{n \rightarrow \infty} \frac{-\frac{2}{n}-1}{1+\frac{1}{n}}=\frac{-1}{1}=-1
\end{align}
Am I correct? Is there another way to find it? I would really be very grateful if you can help me with this. Thank you very much!
| You have a mistake in:
$$
\frac{-2n-1}{n^2+n}=\frac{-\frac{2}{n}-1}{1+\frac{1}{n}}
$$
It should be:
$$
\frac{-2n-1}{n^2+n}=\frac{-\frac{2}{n}-\frac{1}{n^2}}{1+\frac{1}{n}}
$$
However, it would be better to take a factor of $n$:
$$
\frac{-2n-1}{n^2+n}=\frac{-2-\frac{1}{n}}{n+1}
$$
Hence, you will find that the limit will be zero.
| {
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For what values of $x_0$, $x_n:=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}$ where $n \geq 1$ with $x_0>0.$ converges? $x_n:=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}$ where $n \geq 1$ with $x_0>0.$
For what values of $x_0$ does this sequence converge?
If $(x_n) \to L$, then assuming $L>0$, I have
$L=\lim_\limits{n \to \infty} x_n=
\lim_\limits{n \to \infty}\left[\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}\right]=
\lim_\limits{n \to \infty}\left[\frac{x_{n-1}^2+6}{2x_{n-1}}\right]=
\frac{\lim_\limits{n \to \infty}x_{n-1}^2+6}{\lim_\limits{n \to \infty}2x_{n-1}}=
\frac{L^2+6}{2L} \implies 2L^2=L^2+6 \implies L=\sqrt{6}.$
How to determine what values of $x_0$ will give the convergence of $(x_n)$?
| $x_n > 0 $ for all $n$
$x_{n} \ge \sqrt{6}$ for all $n \ge 1$ by the AM-GM inequality
$x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}=x_{n-1}$ when $x_{n-1} = \sqrt{6}$
$x_n=\frac{x_{n-1}}{2}+\frac{3}{x_{n-1}}<x_{n-1}$ when $x_{n-1} > \sqrt{6}$
Basically its going decrease till $x_n = \sqrt{6}$
Therefore, it converges to $\sqrt{6}$ irrespective of the value of $x_0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate integral $ \int_{0}^{+\infty} \frac{x - \sin{\left(x \right)}}{x^{3}}\, dx $ using the Dirichlet integral I try to calculate the integral:
$$ \int_{0}^{+\infty} \frac{x - \sin{\left(x \right)}}{x^{3}}\, dx, $$
using the Dirichlet integral
$$ \int\limits_0^{+\infty} \frac{\sin \alpha x}{x}\,dx = \frac{\pi}{2}\mathrm{sgn}\,\alpha. $$
I integrate this integral in parts, but I can't substitute the limits of integration because the limit is infinity.
| $$\int\frac{x-\sin (x)}{x^3}\,dx=\int\frac{1}{x^2}\,dx-\int\frac{\sin (x)}{x^3}\,dx=$$
$$=-\frac{1}{x}-\int\frac{\sin (x)}{x^3}\,dx$$
by parts
$$\int\frac{\sin (x)}{x^3}\,dx=-\frac{\sin (x)}{2 x^2}-\int -\frac{\cos (x)}{2 x^2} \, dx=$$
again by parts
$$=-\frac{\sin (x)}{2 x^2}-\left(\frac{\cos (x)}{2 x}-\int -\frac{\sin (x)}{2 x} \, dx\right)=$$
$$=-\frac{\sin (x)}{2 x^2}-\frac{\cos (x)}{2 x}-\int \frac{\sin (x)}{2 x} \, dx$$
And finally
$$\int\frac{x-\sin (x)}{x^3}\,dx=-\frac{1}{x}+\frac{\sin (x)}{2 x^2}+\frac{\cos (x)}{2 x}+\int \frac{\sin (x)}{2 x} \, dx$$
the improper integral becomes
$$\int_0^{\infty } \frac{x-\sin (x)}{x^3} \, dx=\left[-\frac{1}{x}+\frac{\sin (x)}{2 x^2}+\frac{\cos (x)}{2 x}\right]_0^{\infty}+\int_0^{\infty } \frac{\sin (x)}{2 x} \, dx=$$
$$=\underset{M\to \infty }{\text{lim}}\left(\frac{\sin (M)}{2 M^2}+\int_0^M \frac{\sin (x)}{2 x} \, dx-\frac{1}{M}+\frac{\cos (M)}{2 M}\right)-\underset{M\to 0}{\text{lim}}\left(\frac{\sin (M)}{2 M^2}+\int_0^M \frac{\sin (x)}{2 x} \, dx-\frac{1}{M}+\frac{\cos (M)}{2 M}\right)$$
first limit gives $\frac{\pi }{4}$.
Second limit is zero because obviously $\int_0^0 \frac{\sin (x)}{2 x} \, dx=0$,
furthermore
$$\underset{M\to 0}{\text{lim}}\frac{-2 M+\sin (M)+M \cos (M)}{2 M^2}=\underset{M\to 0}{\text{lim}}\frac{\frac{\sin (M)}{M}^*+\cos (M)-2}{2 M}=$$
$$=\underset{M\to 0}{\text{lim}}\frac{\cos (M)-1}{2 M}=^{**}\underset{M\to 0}{\text{lim}}-\frac{\sin (M)}{2}=0$$
$^*$ fundamental limit $\frac{\sin (M)}{M}\to 1$ as $M\to 0$
$^{**}$ L'Hopital rule
| {
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"timestamp": "2023-03-29T00:00:00",
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Die and coin variance of random variable question. Die and coin. Roll a die and flip a coin. Let $Y$ be the value
of the die. Let $Z = 1$ if the coin shows a head, and $Z = 0$ otherwise. Let
$X = Y + Z$. Find the variance of $X$.
My work:
$E(Y) = 1 \cdot \cfrac{1}{6} + 2 \cdot \cfrac{1}{6} + 3 \cdot \cfrac{1}{6} + 4 \cdot \cfrac{1}{6} + 5 \cdot \cfrac{1}{6} + 6 \cdot \cfrac{1}{6} = \cfrac{7}{2}$
$E(Z) = 0 \cdot \cfrac{1}{2} + 1 \cdot \cfrac{1}{2} = \cfrac{1}{2}$
So $E(X) = E(Y + Z) = E(Y) + E(Z) = \cfrac{7}{2} + \cfrac{1}{2} = 4$
$E(Y^2) = 1^2 \cdot \cfrac{1}{6} + 2^2 \cdot \cfrac{1}{6} + 3^2 \cdot \cfrac{1}{6} + 4^2 \cdot \cfrac{1}{6} + 5^2 \cdot \cfrac{1}{6} + 6^2 \cdot \cfrac{1}{6} = \cfrac{91}{6}$
$E(Z^2) = 0^2 \cdot \cfrac{1}{2} + 1^2 \cdot \cfrac{1}{2} = \cfrac{1}{2}$
so $E(X^2) = E(Y^2 + Z^2) = E(Y^2) + E(Z^2) = \cfrac{91}{6} + \cfrac{1}{2} = \cfrac{47}{3}$
$Var(X) = E(X^2) - (E(X))^2 = \cfrac{47}{3} - 4^2 = -\cfrac{1}{3}???$
Where did I go wrong, variance can't be negative so clearly my work is wrong, but I have no idea where I went wrong? Can someone point me in the right direction of how to do this question?
| $E(X^2)\ne E(Y^2+Z^2)$
$E(X^2)=E(Y^2+Z^2+2YZ)=E(Y^2)+E(Z^2)+2E(YZ)=\dfrac{91}{6}+\dfrac{1}{2}+2\cdot\dfrac{7}{2}\cdot \dfrac{1}{2}=\dfrac{47}{3}+\dfrac{1}{2}=\dfrac{97}{6}$
$V(X)=E(X^2)-E(X)^2=\dfrac{97}{6}-16=\dfrac{97-96}{6}=\dfrac{1}{6}$
An alternate way:
Since X and Y are independent events covariance term vanishes therefor
$V(X)=V(Y)+V(Z)=E(X^2)-E(X)^2+E(Z^2)-E(Z)^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate this limit without L'Hôpital's rule.
Calculate $$\lim_{x\to0}\frac{(x+32)^{1/5}-2}{x}$$ without L'Hôpital's rule.
My attempt: I first rationalized the expression to get $$\left(\frac{(x+32)^{1/5}-2}{x}\right)\left(\frac{(x+32)^{1/5}+2}{(x+32)^{1/5}+2}\right)=\frac{x+28}{x((x+32)^{1/5}+2)}$$ How should I get rid of the singular $x$ in the denominator now? Should I factor something here?
| If you want detailed proof:
$\dfrac{(2^{5}(\frac{x}{32}+1))^{\frac{1}{5}}-2}{x}=\dfrac{2(\frac{x}{32}+1)^\frac{1}{5}-2}{x}$
$(1+x)^n=1+nx+\dfrac{n(n-1)}{2}\cdot x^2.....$
$\lim_{x\to0}=\dfrac{2\bigg(1+\frac{1}{5}\frac{x}{32}+\dfrac{\frac{1}{5}(\frac{1}{5}-1)}{2}\cdot\bigg(\frac{x}{32}\bigg)^2.....\bigg)-2}{x}=\lim_{x\to0}\dfrac{\frac{2x}{5\cdot32}+...}{x}=\dfrac{2}{5\cdot 32}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $ \max(a_1, a_2, \ldots, a_n)\leq 4\min(a_1, a_2, \ldots, a_n)$ For $ n\geq2$ let $ a_1, a_2, \ldots a_n$ be positive real numbers such that
$$ (a_1 + a_2 + \cdots + a_n)\left(\frac {1}{a_1} + \frac {1}{a_2} + \cdots + \frac {1}{a_n}\right) \leq \left(n + \frac {1}{2}\right)^2.
$$
Prove that $ \max(a_1, a_2, \ldots, a_n)\leq 4\min(a_1, a_2, \ldots, a_n)$.
Can someone help me [rove this? It's USAMO 2009 .I thought of using cauchy.
| Update:
Just realized we can prove directly without taking derivative.
WLOG assume $a_1 \ge a_2 \ge \cdots \ge a_n$, and $a_1 > 4a_n$ then
$$\begin{split} & & \left(\sum a_i \right) \left( \sum \frac{1}{a_i} \right) \\ & = & n + \left(\frac{a_1}{a_n}+\frac{a_n}{a_1}\right) + \sum_{1 < i<j<n} \left( \frac{a_i}{a_j} + \frac{a_j}{a_i}\right) + \sum_{i=2}^{n-1} \left( \frac{a_i}{a_1} + \frac{a_n}{a_i} \right) +
\sum_{i=2}^{n-1} \left( \frac{a_i}{a_n} + \frac{a_1}{a_i} \right)\\
& > & n+\left(4+\frac 14\right) + 2 \binom{n-2}{2} + (n-2)\left(2\sqrt{\frac{a_n}{a_1}}+ 2\sqrt{\frac{a_1}{a_n}}\right)\\
& > & n + 4 + \frac 14 + (n-2)(n-3) + 2(n-2) \left( 2 + \frac 12 \right)\\
& = & n + 4 + \frac 14 + (n-2)(n+2) = n^2 + n + \frac 14 \left(n+\frac 12\right)^2 \\
\end{split}$$
Original proof:
We prove by induction that if $\max(a_i) > 4 \min (a_i)$ then $\sum a_i \sum \frac{1}{a_i} > \left(n+\frac 12\right)^2$
When $n=2$ it's true.
Suppose it's true for $n$. WLOG we assume $a_1 \ge a_2 \ge \cdots \ge a_n$ and $a_{n+1}$ is between $a_n$ and $a_1$. Denote
$$f(x)=\frac{a_1}{x} + \frac{x}{a_1} + \frac{a_n}{x}+\frac{x}{a_n}$$
$$f^{(2)}(x) > 0,f'(x)=\frac{1}{a_1} + \frac{1}{a_n} - \frac{a_1+a_n}{x^2}$$
It's easy to show that $f(x)$ obtains its minimum at $x=\sqrt{a_1 a_n}$, and $$f(\sqrt{a_1a_n})=2 \left(\sqrt{\frac{a_1}{a_n}}+\sqrt{\frac{a_n}{a_1}} \right) > 2(2+\frac 12)=5$$
Define $S_n = \sum_{i=1}^n a_n, T_n = \sum_{i=1}^n 1/a_n$. Then
$$
\sum_{i=1}^{n+1} a_i \sum_{i=1}^{n+1} \frac{1}{a_i} = (S_n+a_{n+1})\left(T_n+\frac{1}{a_{n+1}}\right) \\
= S_n T_n + 1 + \sum_{i=2}^{n-1} \left(\frac{a_{n+1}}{a_i} + \frac{a_i}{a_{n+1}}\right)+f(a_{n+1})\\
> \left(n+\frac 12\right)^2 + 1 + 2(n-2)+5=\left(n+1+\frac 12\right)^2.\blacksquare
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$ I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3\sqrt{1-x^2}}\ \mathrm{d}x}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\frac{2}{3}\int{x^3\ \mathrm{d}x}+\frac{2}{3}\int{x^5\ \mathrm{d}x}$$$$=\cdots$$
This solution isn't true. Where am I wrong?
EDIT: Thank you all for the many different answers! If I could, I'd give you all the accepted answer.
| Your mistake is a sign error:$$\begin{align}\frac{d}{dx}x^3(1-x^2)^{3/2}&=x^2(1-x^2)^{1/2}[3(1-x^2)+x\tfrac32(-2x)]\\&=x^2(1-x^2)^{1/2}[3-3x^2-3x^2].\end{align}$$You seem to have mistaken that last $-$ sign for a $+$. Everything after that is error-carried-forward.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to simplify $(\frac{\sin\alpha - \sin\beta}{\cos\beta -\cos\alpha})\cdot \cos\alpha \cdot \cos\beta$? There is this problem:
$$\left(\frac{\sin\alpha - \sin\beta}{\cos\beta -\cos\alpha}\right)\cdot \cos\alpha \cdot \cos\beta = \frac{1}{\tan\alpha -\tan\beta}$$
I started as $$\left(\frac{\sin\alpha - \sin\beta}{\cos\beta -\cos\alpha}\right)\cdot \cos\alpha \cdot \cos\beta = \frac{\sin2\alpha \cos\beta-\sin2\beta \cos\alpha}{2(\cos\beta -\cos\alpha)}$$ but I'm stuck here, because I don't see how could I use $\sin(x-y)$ but also don't see how could I use any other identity without complicating this even more.
| The proposed identity is not true (even though after the correction).
Indeed, take $\alpha = \pi/4$ and $\beta = 0$. Then we obtain:
\begin{align*}
\left(\frac{\sin(\pi/4) - \sin(0)}{\cos(0) - \cos(\pi/4)}\right)\times\cos(\pi/4)\cos(0) = \frac{\sqrt{2}}{2 - \sqrt{2}}\times\frac{\sqrt{2}}{2} = \frac{1}{2-\sqrt{2}}
\end{align*}
On the other hand, one has that
\begin{align*}
\frac{1}{2(\tan(\pi/4) - \tan(0))} = \frac{1}{2}
\end{align*}
EDIT
The proposed equation is equivalent to
\begin{align*}
\frac{(\sin(a) - \sin(b))\sin(a-b)}{\cos(b) - \cos(a)} = 1
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Complete square values of quadratic formula For Which values of $x \in [0,n]$ the polynomial $P(x)=ax^2+bx+c$ $\ $ gives a complete square value.
For instance the polynomial $P(x)=3x^2+5x+7$ on the interval $x \in [0,100]$ has only two values for $x=3 \rightarrow P(3)=49 $ and for $x=54 \rightarrow P(54)=9025 $.
I was wondering if there is a method to find such values of $x$ directly.
| $y^2=3x^2+5x+7$.
$12y^2=36x^2+60x+84=(6x+5)^2+59$.
$u^2-3v^2=-59$ where $u=6x+5$, $v=2y$.
Given one solution of $u^2-3v^2=-59$, and a fundamental solution of $u^2-3v^2=1$, you can get infinitely many solutions of $u^2-3v^2=-59$, and from those infinitely many solutions of $y^2=3x^2+5x+7$.
Now a fundamental solution of $u^2-3v^2=1$ is $u=2$, $v=1$. A solution of $u^2-3v^2=-59$ is given by $u=4$, $v=5$. If you let $a_n+b_n\sqrt3=(2+\sqrt3)^n(4+5\sqrt3)$, $n=1,2,3,\dots$, then you'll get $a_n^2-3b_n^2=-59$. Then you can see which of those solutions yield integers for $x,y$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $\tan x+ \tan 2x=\frac{2}{\sqrt{3}}$, find $\tan x\cot 2x$ I can't solve this problem. I tried to find $\tan x$ directly by solving cubic equations but I failed.
The problem is to find $\tan x\cot 2x$ given that
$$\tan x+ \tan 2x=\frac{2}{\sqrt{3}}, \>\>\>\>\>0<x<\pi/4$$
How am I supposed to solve this problem?
| $$\frac{2 \tan x}{1-\tan ^2 x}+\tan x=^*\frac{2}{\sqrt{3}}$$.
$$\tan x=-1.45424, \tan x=0.35178, \tan x =2.25716$$
$L=\tan x \cot 2x$ can be written$^{**}$ as $\frac{1}{2} \left(1-\tan ^2 x\right)$ thus possible values are $$L_1=-0.557406,L_2=0.438125,L_3=-2.04739$$
$$(^*)\quad \tan x=u \to u^3-\frac{2 u^2}{\sqrt{3}}-3 u+\frac{2}{\sqrt{3}}=0$$
see solutions here. (press Approximate forms button)
$$(^{**})\quad\tan x \cot 2x=\frac{\sin x\left(\cos ^2 x-\sin ^2 x\right)}{\cos x(2 \sin x \cos x)}=\frac{\cos ^2 x-\sin ^2 x}{2\cos^2 x}=\frac{1}{2}\left(1-\tan^2 x\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $2^{n-1}=1+1+2+4+\dotsb+2^{n-2}$ for $n>1$? This is a claim in a textbook that I'm using, although it was not proven.
When trying this for small $n$'s it is clear.
\begin{align}
2^2&=1+1+2^1=4\\
2^3&=1+1+2+2^2\\
2^4&=1+1+2+4+2^3
\end{align}
What I was thinking for the proof was $2^{n-1}=2(2^{n-2})=2^{n-2}+2^{n-2}$, but obviously that's not incorporating addition, so I don't know where to go on. Is it just expressing $2^{n-2}$ as the first summation?
| Let $S:=1+2+4+\dots+2^{n-2}$.
Then, since it's a geometric series, we have $S=\dfrac{1-2^{n-1}}{1-2}=2^{n-1}-1\implies S+1=2^{n-1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating this expression over two spheres I'd like to analytically perform the following integral:
$$ I=\int\limits_V d^3\textbf x\,d^3\textbf y \frac{1}{(\textbf x-\textbf y)^2}, $$
where the volume $V$ of the integration is such that: $|\textbf x|\in [0,A]$ and $|\textbf y|\in [0,B]$, i.e. spherical for $x$ and $y$ separately.
I tried to get a solution from Mathematica, but the output confuses me a bit:
$$ I = \begin{cases} \frac{\pi^2}{3}\left[ 10A^3B+18AB^3+3\text{i} B^4\pi+6(A^4+2A^2B^2-3B^4)\text{ arctanh}(\frac{B}{A}) \right] & A>B \\ 4A^4\pi^2 & A=B \\ 2\pi^2\left[ AB(A^2+B^2)-(A^2-B^2)^2\text{ arctanh}(\frac{A}{B}) \right] & B>A \end{cases} $$
I'd appreciate any help. Thanks in advance!
I think I've managed to get a better expression out of Mathematica:
$$ I = 2\pi² \left[ AB(A^2+B^2) -(A^2 -B^2)^2 \text{ arctanh}\left(\frac{<}{>}\right) \right] $$
...where $>$ and $<$ are placeholders for $A$ and $B$, depending on which is larger or smaller.
Still, I'd love to know how to do this analytically!
| Let's start with $I=\int\limits_{V_{x}}\int\limits_{V_y} d^3\textbf x\,d^3\textbf y \frac{1}{(\textbf x-\textbf y)^2}$$=\int\limits_{V_{y}}(\int\limits_{V_x} d^3\textbf x\, \frac{1}{(\textbf x-\textbf y)^2})d^3\textbf y$$=\int\limits_{0}^A\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}r^2dr\sin\theta_1d\theta_1d\phi_1\left(\int\limits_{0}^B\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}R^2dR\sin\theta_2d\theta_2d\phi_2\frac{1}{R^2+r^2-2rR\cos\theta_2}\right)$
$$I=\int\limits_{0}^A\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}r^2dr\sin\theta_1d\theta_1d\phi_1F(B,r)$$
For the convenience we define $B>A$ and orient the $z$ axis of the polar system of $X$ coordinates along the vector $\textbf y$. We can do this, because, as soon as we intergate over $R^3$ it is our choice of how to orient the coordinates system in the space. Integrating over $\theta_2$ and $\phi_2$ we get:
$F(B,r)=2\pi\int\limits_{0}^B\int\limits_{0}^{\pi}R^2dR\sin\theta_2d\theta_2\frac{1}{R^2+r^2-2rR\cos\theta_2}$$=2\pi\int\limits_{0}^B\frac{R^2}{2Rr}\log\frac{(R+r)^2}{(R-r)^2}dR$
Evaluating the last integral we have to be careful with the singular point $R=r$ (in fact we evaluate the integral principal value). To show our actions step by step let's regularize the integral by means of small $\epsilon$ near $R=r$, and in the end we will set $\epsilon$ to zero.
$F(B,r)=\pi\int\limits_{0}^B\frac{R}{r}\log\frac{(R+r)^2}{(R-r)^2}dR$$=\frac{2\pi}{r}\int\limits_{0}^BR\log(R+r)dR-\frac{2\pi}{r}\int\limits_{0}^{r-\epsilon}R\log(r-R)dR-\frac{2\pi}{r}\int\limits_{r+\epsilon}^BR\log(R-r)dR$.
Integrating and taking limit at $\epsilon\to0$ we get $$F(b,r)=\frac{\pi}{r}(B^2-r^2)\log\frac{B+r}{B-r}+2\pi{B}$$ And, integrating over $\theta_1$ and $\phi_1$we finally get the desired integral in the form $$I=4{\pi}^2\int_0^A(B^2-r^2)\log(\frac{B+r}{B-r})rdr+8{\pi}B\int_0^Ar^2dr$$
Integrating by part (to get rid of logarithm) we get$$I=4{\pi}^2(\frac{B^2r^2}{2}-\frac{r^4}{4})\log(\frac{B+r}{B-r})|_0^A-4{\pi}^2\int_0^A(\frac{B^2r^2}{2}-\frac{r^4}{4})(\frac{1}{B+r}+\frac{1}{B-r})rdr+\frac{8{\pi}^2BA^3}{3}=$$ $$={\pi}^2(2B^2A^2-A^4)\log(\frac{B+A}{B-A})-4{\pi}^2B\int_0^A\frac{\frac{B^2}{2}-\frac{r^2}{2}+\frac{B^2}{2}}{B^2-r^2}r^2dr+\frac{8{\pi}^2BA^3}{3}$$ Finally, $$I={\pi}^2(2B^2A^2-A^4)\log(\frac{B+A}{B-A})-{\pi}^2\frac{2BA^3}{3}-{\pi}^2B^3\left(-2A+B\log(\frac{B+A}{B-A})\right)+\frac{8{\pi}^2BA^3}{3}$$ $$I=2{\pi}^2BA(B^2+A^2)-{\pi}^2(B^2-A^2)^2\log\frac{B+A}{B-A}; B>A$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is my claim true about the maximum and minimum of the given function? I have this three-variable function
$$f(x,y,z)=x^9 \cos (2 x y+x z)+\left(5 x^4+1\right) \cos (x y+x z)+\left(5 x^4+3\right)\\ \qquad+\left(4 x^3+1\right) \cos (x y+2 x z)+x^3 \cos (x z)+\left(2 x^2+7\right) \cos (2 x y+2 x z)\\ \qquad+\left(x^2+5\right) \cos (x y)+\left(2 x^2+7\right) \cos (2 x z)\\+\left(3 x^7+5 x^2+1\right) \cos (x y-x z)+(x+1) \cos (2 x y)$$
for $x,y,z>0$.
Now, since all the coefficients of the functions $\cos$ are positive, can I claim that the inequality $f_{-1}\leq f(x,y,z)\leq f_{+1}$ holds, where $f_{\pm1}$ are the function $f(x,y,z)$ when all the $\cos$ are $+1$ and $-1$?
| Yes, considering that the variables only take positive values, setting all the co-sines to 1 will provide an upper bound and setting all the co-sines equal to -1 will provide a lower bound.
| {
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"url": "https://math.stackexchange.com/questions/3999198",
"timestamp": "2023-03-29T00:00:00",
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Show that for every $n$ there are integers $a,b$ such that $X_n = a^2 + 2b^2$ We have a sequence $X_0, X_1,...$ which is determined by $X_0=1, X_1=3$ and $X_n = 6X_{n-1} - X_{n-2}$
Show that for every $n$ there are integers $a,b$ such that
$X_n = a^2 + 2b^2$
I thought of changing $a^2 + 2b^2 = (a + \sqrt 2 i)(a - \sqrt 2 i)$, since I thought it might lead to something good, but I'm not getting anything.
Any hints or solutions are appreciated!
| Let $(a_n,b_n)$ be solutions to Pell's equation $a_n^2-2b_n^2=\pm1$.
The fundamental solution is $1^2-2\times1=(1+\sqrt2)(1-\sqrt2)=-1$,
so solutions are $a_n^2-2b_n^2=(a_n+\sqrt2b)(a_n-\sqrt2b)=(1+\sqrt2)^n(1-\sqrt2)^n=(-1)^n$, where
$a_n+\sqrt2b_n=(1+\sqrt2)^n=(a_{n-1}+\sqrt2b_{n-1})(1+\sqrt2)=(a_{n-1}+2b_{n-1})+(a_{n-1}+b_{n-1})\sqrt2$,
so $$a_n=a_{n-1}+2b_{n-1}\tag1$$ and $$b_n=a_{n-1}+b_{n-1}\tag2$$.
Solutions to $X_n^2-2Y_n^2=1$ are given by $X_n+\sqrt2Y_n=(1+\sqrt2)^{2n}=a_{2n}+\sqrt2b_{2n}$
$=[(1+2\sqrt2)^n]^2=(a_n+\sqrt2b_n)^2=a_n^2+2b_n^2+2a_nb_n\sqrt2$,
so $X_n=a_n^2+2b_n^2=a_{2n}$ (and $Y_n=2a_nb_n=b_{2n}$).
From (1)-(2) we have $b_{n-1}=a_n-b_n=b_n-a_{n-1}$, so $a_n+a_{n-1}=2b_{n}$,
so $a_n=a_{n-1}+2b_{n-1}=a_{n-1}+(a_{n-1}+a_{n-2})=2a_{n-1}+a_{n-2}$. Therefore,
$a_n=2(2a_{n-2}+a_{n-3})+a_{n-2}=5a_{n-2}+2a_{n-3}=5a_{n-2}+(a_{n-2}-a_{n-4})=6a_{n-2}-a_{n-4}$,
so $a_{2n}=6a_{2(n-1)}-a_{2(n-2)}$, so $X_n=6X_{n-1}-X_{n-2}$, and $X_n$ matches your sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4004192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x^5=2$, find $\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$
If $x^5=2$, find
$$\frac{x}{x^2+1}+\frac{x^2}{x^4+1}+\frac{x^3}{x+1}+\frac{x^4}{x^3+1}$$
My attempt-
Since $x^5=2,x^6=2x,x^7=2x^2..$and so on
The equation is equivalent to
$$x^5\left(\frac{1}{2x+x^4}+\frac{1}{2x^2+x^3}+\frac{1}{x^3+x^2}+\frac{1}{x^4+x}\right)$$
Which simplifies to
$$2\left[\frac{3x^4+7x^3+5x^2+8x+12}{2x^4+6x^3+4x^2+6x+12}\right]$$
Now I am struck here.
Any help will be appreciated.
Note: Please don't use root tables and complex numbers too!
| I used Sage cell to calculate this.
var("X")
F.<x> = NumberField(X^5 - 2)
y = x/(x^2 + 1) + x^2/(x^4 + 1) + x^3/(x + 1) + x^4/(x^3 + 1)
print(y)
print(y.minpoly())
Output:
-44/765*x^4 + 122/765*x^3 + 79/765*x^2 + 233/765*x + 1336/765
x^5 - 1336/153*x^4 + 1556/51*x^3 - 8182/153*x^2 + 7210/153*x - 12806/765
In the output, the first line is the sum, and the second line is the minimal polynomial of that sum.
None of these is nice enough to be an intended answer. Thus I suspect that the question is slightly wrong.
One possibility is to change $x^5 = 2$ to $x^5 = 1$. Then it is easy to evaluate that the sum is equal to $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
$\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x = \frac{n}{4}$
Prove that $$\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x = \frac{n}{4}$$
My attempt: I simplified $\sqrt{4x^2-nx}+2x$ by conjugation $$\sqrt{4x^2-nx}+2x \left(\frac{\sqrt{4x^2-nx}-2x}{\sqrt{4x^2-nx}-2x}\right) = -\frac{nx}{\sqrt{4x^2-nx}-2x} $$ I thought about substituting $x$ for $\frac{1}{y}$ to get $$\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x = \lim_{y \to 0} \frac{\sqrt{4-ny}+2}{y}$$ But this means that the limit doesn't exist! Did I go wrong somewhere in the simplification or substitution? (I noticed the $\frac{n}{4}$ part in Wolfram Alpha)
| $$\lim_{x\rightarrow-\infty}-\frac{nx}{\sqrt{4x^2-nx}-2x} = \lim_{y\rightarrow \infty}\frac{ny}{\sqrt{4y^2+ny}+2y} = \lim_{y\rightarrow \infty}\frac{n}{\sqrt{4+n/y}+2} = \frac{n}{4}$$
Alternatively,
$\lim_{x \to -\infty} \sqrt{4x^2-nx}+2x = \lim_{y \to 0^-} \frac{-\sqrt{4-ny}+2}{y}$
Note the negative sign as $y$ is a negative quantity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4010577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For every natural number n, prove that the vector $x = (n, n + 1, n (n + 1)) ∈ \Bbb R^3$ has as a norm a natural number, that is, $\|x\| \in \Bbb N$.
For every natural number $n$, prove that the vector $x = (n, n + 1, n (n + 1))$ ∈ $\mathbb{R}^3$ has as a norm a natural number, that is, $\|x\| \in \mathbb{N}$.
I made the internal product of the vector $x$ which gave $n^4 + 3n^3 + n^2$. Now I can't prove by induction. Does anyone have an idea?
| \begin{align*}
\sqrt{n^2+(n+1)^2+n^2(n+1)^2}&=\sqrt{(n^2+1)(n+1)^2+n^2}\\
&=\sqrt{(n^2+1+2n)(n+1)^2+n^2-2n(n+1)^2}\\
&=\sqrt{(n+1)^4-2n(n+1)^2+n^2}\\
&=\sqrt{[(n+1)^2-n]^2}\\
&=(n+1)^2-n\\
&=n^2+n+1.
\end{align*}
This is obviously a natural number for $n\in\mathbb{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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