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Compute integral $\int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx$ While solving a more general problem, I came across the following integral
$$I= \int_{-\infty}^{\infty} x^{4n} e^{-x^2/2} \, dx$$
To start off, I defined $t:=x^2/2$ to get
\begin{align*}
I= \int_{-\infty}^\infty x^{4n} e^{-x^2/2} \, dx &= 2\int_0^\infty x^{4n} e^{-x^2/2} \, dx \\
&= 2 \int_0^\infty (\sqrt{2t})^{4n} e^{-t} \frac{dt}{\sqrt{2t}} \\
&= 2 (\sqrt{2})^{4n - 1} \int_0^\infty t^{2n - 1/2} e^{-t} \, dt
\end{align*}
This seems to be a gamma function integral. I would start integrating by parts
\begin{align*}
\int_0^{+\infty} t^{2n-\frac{1}{2}}e^{-t} \, dt&=[ -t^{2n-\frac{1}{2}}e^{-t}]_0^{+\infty} - (-)(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t} \, dt \\
&=(2n-\frac{1}{2})\int_0^{+\infty}t^{2n-\frac{1}{2}-1}e^{-t} \, dt
\end{align*}
But how to proceed further?
I looked up in a table and the result should be
$$I=2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right)$$
PS: Alternative methods avoiding the gamma function are very welcomed! :)
EDIT: I think that what Yves suggested was the following
$$I= \int_{-\infty}^\infty x^{4n} e^{-x^2/2}= (2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2} \int_0^{+\infty}t^{-\frac{1}{2}}e^{-t}\,dt=(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2}\ \ 2\int_0^{+\infty}e^{-t} \, d(\sqrt{t})$$
Alrigtht! I see that $(2n-\frac{1}{2}) (2n-\frac{3}{2}) (2n-\frac{5}{2})\cdots\frac{3}{2}\frac{1}{2} = \Gamma\left(2n+ \frac 1 2 \right)$
NAIVE question though: how to see that $\int_0^{+\infty}e^{-t} \, d(\sqrt{t})=2 (\sqrt{2})^{4n - 1}$?
| You're really done after the first integration by parts: as $\displaystyle \Gamma(z) = \int_0^\infty x^{z-1} e^{-x}\,dx,$
$\displaystyle 2 (\sqrt{2})^{4n - 1} \int_{0}^{\infty} t^{2n -\frac{1}{2}} e^{-t} dt =2 (\sqrt{2})^{4n - 1} \int_{0}^{\infty} t^{2n + \frac{1}{2}-1} e^{-t} dt =2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right).$
In other words, your integral is the gamma function with $z = 2n+\frac{1}{2}$.
Regarding the other method, you have:
You've shown that if $\displaystyle I_n = \int_0^{\infty} t^{2n-\frac{1}{2}}\,{dt} $ then $I_n = \left(2n-\frac{1}{2}\right)I_{n-1}$ so that $$I_n =(2n-\frac{1}{2}) \cdot (2n-\frac{3}{2}) \cdot (2n-\frac{5}{2})\cdots\frac{3}{2} \cdot I_1 $$
Where $\displaystyle I_1 = \int_0^{+\infty}t^{\frac{1}{2}}e^{-t}\,dt$, then $t = x^2$ shows that
$$\displaystyle I_1 = 2 \int_0^{\infty} x^2 e^{-x^2}\,{dx}. $$
Which evaluates to $\displaystyle \frac{1}{2}\sqrt{\pi}$ (see here). Therefore
$$\begin{aligned} I &= \sqrt{\pi} \cdot (2n-\frac{1}{2}) \cdot (2n-\frac{3}{2}) \cdot (2n-\frac{5}{2})\cdots\frac{3}{2} \cdot \frac{1}{2}\\& =\frac{\sqrt{\pi}}{2^{2n}}(4n-1) (4n-3) (4n-5)\cdots 3 \cdot 1 \\& =\frac{\sqrt{\pi}}{2^{2n}}\frac{(4n-1)(4n-2) (4n-3)(4n-4) (4n-5)\cdots 1}{(4n-2) (4n-4) \cdots 2 } \\& =\frac{\sqrt{\pi}}{2^{2n}}\frac{(4n-1)(4n-2) (2n-3)(2n-4) (2n-5)\cdots 3 \cdot \cdot 2 \cdot 1}{2^{2n} (2n-1) (2n-2) \cdots 1 } \\& = \frac{(4n)!\sqrt{\pi}}{2^{4n}(2n)!}.\end{aligned} $$
Therefore your original integral is $\displaystyle I =2 (\sqrt{2})^{4n - 1} \frac{(4n)!\sqrt{\pi}}{2^{4n}(2n)!}.$
This agrees with $\displaystyle I = 2 (\sqrt{2})^{4n - 1} \Gamma\left(2n+ \frac 1 2 \right)$ as $\displaystyle \Gamma\left(2n+ \frac 1 2 \right) = \frac{(4n)!\sqrt{\pi}}{2^{4n}(2n)!}$
(see here).
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Question about the proof for "The rational line Q does not have the completeness property" Referring to the attached png file, I have questions about the inequality:
$$(n^2)/(2n+1) > (p^2)/(2q^2 - p^2)$$
How is this inequality derived? How did the n get into the inequality? Where did $n^2/(2n+1)$, $p^2/(2q^2 - p^2)$, and later in the proof, $2q^2/(p^2 - 2q^2)$ come from? There were no derivations for these 3 expressions. Could someone please provide the derivations?
| Yeah....
So we have $(\frac pq)^2 < 2$ and we want to show $\frac pq$ is not an upper bound so we need to show there is $S > \frac pq$ so that $(\frac pq)^2 < S^2 < 2$.
$S$ is rational, so there is a rational $r$ so that $S =r\frac pq$ and $r > 1$.
We need to show that $r^2\frac {p^2}{q^2} < 2$ or that $r^2 p^2 < 2q^2$ for some rational $r$.
Now we know $\frac {p^2}{q^2} < 2$ so $2q^2 > p^2$ and so we need to show that
$(r^2 -1) p^2 < 2q^2-p^2$ or that $r^2 -1 < \frac {2q^2-p^2}{p^2}$.
Now the RHS of that is a constant fixed positive (albeit it possibly quite small) so we need to show that we can have an $r^2-1$ that is smaller than that.
And it is sufficient to show we can have $r$ so that $r^2$ is arbitrarily close to $1$.
Now we usually can show rationals are arbitrarily close but showing that if $n$ is arbitrarily large then $\frac 1n$ is arbitrarily small.
So if we let $r = 1+ \frac 1n$ the want to show, we can choose and $n$ so that
$(1 + \frac 1n)^2 -1 = \frac 2n + \frac 1{n^2} < \frac {2p^2-p^2}{p^2}$ we will be done.
And that's a matter of making $\frac 2n + \frac 1{n^2} = \frac {2n+1}{n^2}$ arbitrarily small
or
making $\frac {n^2 }{2n+1}$ arbitrarily large. Or at least, $\frac {n^2}{n+1} > \frac {p^2}{2q^2 -p^2}$.
We can do that as $\frac {n^2}{n+1} \to \infty$
And that's it: If $\frac {n^2}{n+1} > \frac {p^2}{2q^2 -p^2}$ it's just algebraic manipluation that
For $r = 1+\frac 1n = \frac {n+1}n$ then $r\frac pq = \frac {p(n+1)}{qn}$ when squared will be less than $2$.
$(r\frac pq)^2 = (1+\frac 1n)^2\frac {p^2}{q^2}=$
$(1 + \frac {2n+1}{n^2})\frac {p^2}{q^2}=$
$\frac {p^2}{q^2} + \frac {2n+1}{n^2}\frac {p^2}{q^2} < $
$\frac {p^2}{q^2} + \frac {2q^2-p^2}{p^2}\frac {p^2}{q^2} =$
$\frac {p^2}{q^2} + \frac {2q^2 -p^2}{q^2}=$
$\frac {p^2 + 2q^2 -p^2}{q^2}=\frac {2q^2}{q^2}=2$.
We are done.
.....
But this was a bass-ackward proof in that we started with what we needed and worked our way back to show we can have it. That's not an invalid way of doing a proof because we never assumed what we wanted to prove, it's okay to say "If we need this, we can do that by this, and, hey, we can do that" but it's not an ideal argument. We like proofs to be direct and forward.
So we start with: "We can always find an $n$ so that $\frac {p^2}{2q^2 - q^2} < \frac {n^2}{2n+1}$".
To which the reader reacts with ".... well, sure, but where the #@%! did that come from"
And the proofer can say "It doesn't matter. We can do it, right?"
And the reader says "well, sure, but..."
And the proofer goes "ninh, ninh, ni..."
And the reader goes "...but..."
And the proofer goes "NINH! .... so if we do that then....." And lo and behold, it works.
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute close-form of $\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}$ I came across the improper trigonometric integral recently shared in a Chinese web forum
\begin{align}
\int_0^{\frac\pi2}\frac{dt}{\sin t+\cos t+\tan t+\cot t+\csc t+\sec t}\\
\end{align}
which amuses me because of its appearance. What is more amusing is the claim, without providing the proof, that is has the close-form result
\begin{align}
\frac{\sqrt{8\sqrt2+2\sqrt7}}{7^{3/4} }\tanh^{-1} \frac{\sqrt{2\sqrt7(4\sqrt2-5)}}{4\sqrt2-5 +\sqrt7}
+\frac{\sqrt{8\sqrt2-2\sqrt7}}{7^{3/4} }\tan^{-1}\frac{\sqrt{2\sqrt7(4\sqrt2-5)}}{4\sqrt2-5 - \sqrt7}
\end{align}
I was able to verify it numerically; yet rather curious in how it could ever be derived. Based on my knowledge and experience, I do not assume it would be easy.
I was able to find an unsolved post here from a long time ago, which only reexpresses the corresponding indefinite integral via the tangent half-angle substitution as
$$\int\frac{2t(1-t)}
{2 t^4-3 t^3+3 t^2+t+1}dt$$
but gives up due to the complexity in partial fractionalization. I am doubtful that this approach would lead to the explicit expression claimed above. I am interested in any suitable methods producing the close-form.
| This feels more like a comment to Travis Willse's excellent answer, which essentially solves the problem, but here's the rest of the calculation:
So far we have ($u = \frac{\tan \left(\frac{\pi}{8} - \frac{t}{2}\right)}{\sqrt{2}-1}$)
\begin{align}
I &\equiv \int \limits_0^{\pi/2} \frac{\mathrm{d} t}{\sin(t) + \cos(t) + \tan(t) + \cot(t) + \sec(t) + \csc(t)} \\
&= 2 \int \limits_0^{\pi/4} \frac{\mathrm{d} t}{\sin(t) + \cos(t) + \tan(t) + \cot(t) + \sec(t) + \csc(t)} \\
&= \frac{2 \sqrt{2} \beta^2}{\sqrt{7}} \int \limits_0^1 \frac{1-u^2}{u^4 + \beta^4} \, \mathrm{d} u = \frac{2 \sqrt{2} \beta^2}{\sqrt{7}} \int \limits_0^1 \frac{1-u^2}{(u^2 + \mathrm{i} \beta^2) (u^2 - \mathrm{i} \beta^2)} \, \mathrm{d} u
\end{align}
with $\beta = \sqrt{\frac{5 + 4 \sqrt{2}}{\sqrt{7}}}$. We now need $a_{1/2} \in \mathbb{C}$ such that $a_1 (u^2 + \mathrm{i} \beta^2) + a_2 (u^2 - \mathrm{i} \beta^2) = 1 - u^2$ holds. The solution is $a_{1/2} = - \frac{1}{2} \left(1 \pm \frac{\mathrm{i}}{\beta^2}\right)$, so we can write
\begin{align}
I &= - \frac{2 \sqrt{2} \beta^2}{\sqrt{7}} \operatorname{Re} \left[\left(1 + \frac{\mathrm{i}}{\beta^2}\right) \int \limits_0^1 \frac{\mathrm{d} u}{u^2 - \mathrm{i} \beta^2} \right] = - \frac{2 \sqrt{2} \beta^2}{\sqrt{7}} \operatorname{Re} \left[\left(1 + \frac{\mathrm{i}}{\beta^2}\right) \int \limits_0^1 \frac{\mathrm{d} u}{u^2 + \left(\frac{1 - \mathrm{i}}{\sqrt{2}} \beta\right)^2} \right] \\
&= - \frac{2 \sqrt{2} \beta^2}{\sqrt{7}} \operatorname{Re} \left[\left(1 + \frac{\mathrm{i}}{\beta^2}\right) \frac{1 + \mathrm{i}}{\sqrt{2} \beta} \arctan \left(\frac{1 + \mathrm{i}}{\sqrt{2} \beta}\right) \right] \\
&= - \frac{2}{\sqrt{7}} \operatorname{Re} \left[\left(\beta - \beta^{-1} + \mathrm{i} \left(\beta + \beta^{-1}\right)\right) \arctan \left(\frac{1 + \mathrm{i}}{\sqrt{2} \beta}\right) \right] \\
&= - \frac{1}{\sqrt{7}} \operatorname{Re} \left[\left(\beta - \beta^{-1} + \mathrm{i} \left(\beta + \beta^{-1}\right)\right) \left(\arctan \left(\frac{\sqrt{2}}{\beta - \beta^{-1}}\right) + \mathrm{i} \operatorname{artanh} \left(\frac{\sqrt{2}}{\beta + \beta^{-1}}\right)\right) \right] \\
&= \frac{1}{\sqrt{7}} \left[\left(\beta + \beta^{-1}\right) \operatorname{artanh} \left(\frac{\sqrt{2}}{\beta + \beta^{-1}}\right)- \left(\beta - \beta^{-1}\right) \arctan \left(\frac{\sqrt{2}}{\beta - \beta^{-1}}\right)\right] \, .
\end{align}
In the penultimate step we have used
$$ \arctan(a + \mathrm{i} b) = \frac{1}{2} \left[\arctan \left(\frac{2 a}{1 - a^2 - b^2}\right) + \mathrm{i} \operatorname{artanh} \left(\frac{2 b}{1 + a^2 + b^2}\right)\right] $$
from here. Since $\beta \pm \beta^{-1} = \sqrt{2} \sqrt{4 \sqrt{\frac{2}{7}} \pm 1}$, we obtain
$$ I = \sqrt{\frac{2}{7}} \left[\sqrt{4 \sqrt{\frac{2}{7}} + 1} \operatorname{arcoth} \left(\sqrt{4 \sqrt{\frac{2}{7}} + 1}\right) - \sqrt{4 \sqrt{\frac{2}{7}} - 1} \operatorname{arccot} \left(\sqrt{4 \sqrt{\frac{2}{7}} - 1}\right)\right] \, , $$
which is a simplified version of the final result given in the question.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving a Polynomial Equation by Factoring. $n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.
| Factor out a $64$ to get:
$$64 \left(\frac{n^3}{64} + \frac{12n^2}{64} + \frac{48n}{64} +1\right) = 64 \left(\frac{n^3}{64} + 3 \cdot\frac{n^2}{16} + 3 \cdot \frac{n}{4} +1\right)$$
$$= 64 \left( \left(\frac{n}{4}\right)^3 + 3\left(\frac{n}{4}\right)^2 + 3 \left(\frac{n}{4}\right) + 1 \right)$$
$$= 64(n/4 + 1)^3 = (4 \cdot n/4 + 4 \cdot1)^3 = (n+4)^3$$
where we have used the binomial theorem in the last line, and the fact that $4^3 = 64$.
This can be seen as the reverse of the AC method, where we instead factor out the largest perfect square of the constant term. This is $64 = 8^2$ in this question.
| {
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"url": "https://math.stackexchange.com/questions/4060667",
"timestamp": "2023-03-29T00:00:00",
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Locus of a point of intersecting lines The perpendicular from the centre of the ellipse $$b^2x^2+a^2y^2=a^2b^2$$ to the tangent at any point $P$ meets the line joining point $P$ to the focus $(ae,0)$ at $Q$. Prove that the locus of $Q$ is $$x^2+y^2-2aex=b^2$$
My attempt
Tangent to an ellipse at $P(\theta)$
$$bx\cos\theta +by\sin\theta=ab\tag{1}$$
slope $m_1$ of a normal line to $(1)$
$$m_1=\frac{a\sin\theta}{b\cos\theta}$$
Equation of the normal passing through origin
\begin{align*}
y=m_1x
&=\frac{a\sin\theta}{b\cos\theta}x\tag{2}
\end{align*}
Equation of a line through $P$ and $S$
$$\begin{align*}
y=m_2(x-ae)&=\frac{b\sin\theta}{\left(a\cos\theta-ae\right)}(x-ae)
\tag{3}
\end{align*}$$
Point $Q(h,k)$ satisfies $(2)$ and $(3)$.
How to eliminate $\sin\theta$ and $\cos\theta$ to obtain the required locus?
| You are almost there. Now equating both,
$\displaystyle \frac{a}{b\cos\theta} x = \frac{b}{\left(a\cos\theta-ae\right)} (x-ae)$
$(x-ae+ae) (a^2\cos\theta-a^2e) = b^2\cos\theta(x-ae)$
$(x-ae) (a^2 \cos \theta - b^2 \cos\theta - a^2e)= a^3e^2 - a^3e \cos\theta$
We will use the fact that $a^2 - b^2 = a^2e^2$ now and again later.
$x-ae = \displaystyle \frac{a\cos\theta-ae}{1 - e \cos\theta}$
$x = \displaystyle \frac{b^2\cos\theta}{a(1-e\cos\theta)}$
$y = \displaystyle \frac{b \sin\theta}{1-e\cos\theta}$
$x^2 + y^2 - 2aex = (x-ae)^2 + y^2 - a^2e^2$
$ = \displaystyle \frac{a^2 \cos^2\theta+a^2e^2-2a^2e\cos\theta + (a^2-a^2e^2)\sin^2\theta}{(1-e\cos\theta)^2} - a^2e^2$
$ = a^2 - a^2e^2 = b^2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Decomposition of a sum of cubes I have to decompose the following polynomial:
$$(x+2y)^3+(x-2y)^3$$
Since I have the sum of two cubes, I have thought:
$$(x+2y)^3+(x-2y)^3=(x+2y+x-2y)[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]=2x\color{red}{[(x+2y)^2+(x-2y)^2-(x+2y)(x-2y)]}$$
Now in my book the result is $2x(x^2+12y^2)$, but I can't understand how rewriting the red term as $x^2+12y^2$, without trivially executing the squares. Can you give me a hint?
| Your function is odd in $x$ and even in $y$, to wit
$[(-x)+2y]^3+[(-x)-2y]^3=-[(x-2y)^3+(x+2y)^3]$
$[x+2(-y)]^3+[x-2(-y)]^3=[(x-2y)^3+(x+2y)^3]$
So after factoring out $x$ your remaining quadratic factor must be even in both $x$ and $y$. Therefore your red expression can only contain three nonzero terms $ax^2+by^2+c$. Putting $x=0$ then forces $b=12,c=0$ and $y=0$ forces $a=1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Sequence queue - converges or diverges? I have the next sequence queue: $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$. Does the queue converges or diverge?
My attempt:
I have tried to show that $\frac{1}{n^2}>|\frac{n^2-n-1}{n^4+n^2+1}|$ for any $n>0$, and then by the comparation test we get that since $\sum_{n=1}^{\infty}{\frac{1}{n^2}}$ converges, we have that $\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$ converges too, and by a theorem we have that since $\sum_{n=1}^{\infty}{|\frac{n^2-n-1}{n^4+n^2+1}|}$ $\implies$ $\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$ converges. Is that right?
| We are trying to prove that the series, $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$
converges.
Let us use the Limit Comparison Test. That is, if $a_n>0$ and $b_n>0$, then if $$\lim_{n\to \infty}\frac{a_n}{b_n}=L$$ for some real number $L$. Then, if $b_n$ converges, by the limit comparison test, $a_n$ must also converge.
Let $a_n=\frac{n^2-n-1}{n^4+n^2+1}$ and we say $b_n=\frac{1}{n^2}$. Then, $$\lim_{n\to \infty} \frac{\frac{n^2-n-1}{n^4+n^2+1}}{\frac{1}{n^2}}\\
\lim_{n\to \infty}\left(\frac{(n^2-n-1)(n^2)}{n^4+n^2+1}\right)\\
\lim_{n\to \infty}\frac{n^4-n^3-n^2}{n^4+n^2+1}\cdot \frac{\frac{1}{n^4}}{\frac{1}{n^4}}\\
\lim_{n\to \infty} \left(\frac{{1-\frac{1}{n}}-\frac{1}{n^2}}{1+\frac{1}{n^2}+\frac{1}{n^4}}\right)\\
\lim_{n\to \infty}\frac{1-0-0}{1+0+0}=\frac{1}{1}=1$$
Hence, we have shown that the $\lim_{n\to \infty}\frac{a_n}{b_n}=L$ converges because it equals some constant; in this case, $1$.
Now, because $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges by the p-test, (since $p=2>1$), we can conclude that the series, $$\sum_{n=1}^{\infty}{\frac{n^2-n-1}{n^4+n^2+1}}$$ must also converge by the limit comparison test.
| {
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Expected Value of Binomial Distribution What is the expected value of the absolute value of the difference between the number of incoming tails and the number of incoming heads when a coin is tossed 5 times?
Here is what I think : Let $X$ is random variable of number of incoming heads then
$p=\frac{1}{2}$ and $n=5$ so $X\sim Binom(5,\frac{1}{2})$
$X$ and $Y=5-X$ ($Y$ is random variable of number of incoming tails) then $E(K)=E(\left|X-(5-X)\right|)=E(\left|2X-5\right|)=E(2X-5)=2E(X)-5=2.\frac{5}{2}-5=0$ where $K=\left|X-Y\right|=\left|2X-5\right|$ and $E(X)=n.p=5.\frac{1}{2}=\frac{5}{2}$
But the right answer is $\frac{15}{8}$. Where am I making a mistake? Any help will be appreciated.
| Since $X\sim Binom(5,\frac{1}{2})$, the range of $X$ is $\{0,1,2,3,4,5\}$, which implies that $2X - 5$ is not necessarily always positive. Therefore in your proof, it is not true that $E(\left|2X - 5\right|) = E(2X - 5)$.
| {
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Number of possible solutions to $2 \le |x-1||y+3| \le 5$, where $x$ and $y$ are negative integers
If $$2 \le |x-1||y+3| \le 5$$ and both $x$ and $y$ are negative integers, find the number of possible combinations of $x$ and $y$ .
Below is my solution approach :-
As $x$ is a negative integer, hence $|x-1|$ in the $2 \le |x-1||y+3| \le 5$ will be come $-(x-1)$ or $(1-x)$ and the main equation will transform into $2 \le (1-x)|y+3| \le 5$.
$1st$ case when $y+3 \ge 0 \Rightarrow y \ge -3 \Rightarrow y \in \{{-3,-2,-1}\} $ as $y$ is a negative integer :
For $y=-3$ we can see that there is no valid solution for $x$ as $|y+3|$ part will become $0$, hence this case is invalid.
For $y=-2$ we get the solution set for $x$ to be $x \in \{{-4,-3,-2,-1}\} $ and total number of solutions possible in this case is $4$.
For $y=-1$ we get the solution set for $x$ to be $x \in \{{-1}\} $ and total number of solutions possible in this case is $1$.
So for this $1st$ case when $y+3 \ge 0$ we have in total 5 solutions.
$2nd$ case when $y+3 \lt 0 \Rightarrow y \lt -3$ and in this case $y$ will have infinite values and I am not able to proceed from here.
The answer for the total number of solutions provided is $10$ and you can see that I've been able to find out the $5$ solutions in my $1st$ case. Can someone please guide or help me about how to proceed in the 2nd case?
Thanks in advance !
| Both $|x-1|$ and $|y+3|$ are non-negative integers. As the product must lie between $2$ and $5$, the only candidates are:
$$(1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (3,1), (4,1), (5,1)$$
However, we can't force $|x-1|=1$ with a negative $x$, so this leaves just five possible solutions:
$$(2,1), (2,2), (3,1), (4,1), (5,1)$$
We can make $|y+3|$ equal $1$ or $2$ twice though, with $|-2
+3|=1,|-4+3|=1$ and similarly with $y=-1,-5$, so each pair can be made two times, once with each $y$, and this gives $10$ solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent. Determine whether the series $\sum_{n=1}^\infty \ln\left(\cos \frac{1}{n}\right)$ is convergent.
Attempt:
Notice that the $p-$ series $\sum_{n=1}^\infty \frac{1}{n^2}$ with $p=2$ is convergent. Then, we have
\begin{align*}
\lim\limits_{n \to \infty} \frac{\ln\left(\cos \frac{1}{n} \right)}{\frac{1}{n^2}} &= -\frac{1}{2} \cdot \lim\limits_{n \to \infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} \cdot \frac{1}{\cos \frac{1}{n}} \\
&= -\frac{1}{2}.
\end{align*}
Hence, the limit is equal to $-\frac{1}{2} \ne 0$. Since the $p-$ series is convergent with $p=2$, by the limit comparison test, the series is convergent.
Am I true?
| Just for the fun.
In the same spirit as @Alexander Conrad, we have the infinite series representation
$$\log \left(\cos \left(\frac{1}{n}\right)\right)=\sum_{k=1}^\infty (-1)^k\frac{ 2^{2 k-3} (E_{2 k-1}(1)-E_{2 k-1}(0)) }{k (2 k-1)!}\,n^{-2 k}$$ where appear Euler polynomials.
Truncated to any order, we can transform the Taylor series into a $[2k,2]$ Padé approximant $P_k$.
For example
$$P_2=-\frac {3(10n^2-1) } {4n^2(15n^2-4) }=-\frac{3}{16 n^2}-\frac{75}{16 \left(15 n^2-4\right)}$$
$$\sum_{n=1}^\infty P_2=-\frac{1}{128} \left(75+4 \pi ^2-10 \sqrt{15} \pi \cot \left(\frac{2 \pi}{\sqrt{15}}\right)\right)\sim -0.943375$$ Repeating for various values of $k$, we have more and more accurate values as shown below
$$\left(
\begin{array}{cc}
1 & -0.9295780173 \\
2 & -0.9433749128 \\
3 & -0.9449845778 \\
4 & -0.9452801699 \\
5 & -0.9453463832 \\
6 & -0.9453628819 \\
7 & -0.9453672911 \\
8 & -0.9453685316 \\
9 & -0.9453688948 \\
10 & -0.9453690046
\end{array}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $a$ and $b$ have no greatest common divisor in $\mathbb{Z}[\sqrt-5]$
Show that in the ring $R=\mathbb{Z}[\sqrt{-5}]$, $a=3\cdot 7 \cdot(1+2\sqrt{-5})$ and $b=(1+2\sqrt{-5})\cdot 7\cdot(1+2\sqrt{-5})$ have no greatest common divisor.
Since $N(a)=3^3\cdot 7^3$ and $N(b)=3^2\cdot 7^4$ so it is clear that any common divisor of $a$ and $b$ must have norm dividing $3^2\cdot 7^3$. Now we will get many possibilities. Am I on right track or there is some other smart way to solve this problem?
| Note that
$$
\gcd\left(3\cdot7\left(1+2\sqrt{-5}\right),7\left(1+2\sqrt{-5}\right)^2\right)=7\left(1+2\sqrt{-5}\right)\gcd\left(3,\left(1+2\sqrt{-5}\right)\right)
$$
Since $N(3)=9$ and $N\!\left(1+2\sqrt{-5}\right)=21$, we must have
$$
\left.N\!\left(\gcd\left(3,\left(1+2\sqrt{-5}\right)\right)\right)\,\middle|\ 3\right.
$$
But the only element whose norm divides $3$ is $1$.
This would say that the $\gcd$ is $7\!\left(1+2\sqrt{-5}\right)$
Because $3\cdot7=\left(1-2\sqrt{-5}\right)\left(1+2\sqrt{-5}\right)$, the $\gcd$ is also
$$
\gcd\left(\left(1-2\sqrt{-5}\right)\left(1+2\sqrt{-5}\right)^2,7\left(1+2\sqrt{-5}\right)^2\right)=\left(1+2\sqrt{-5}\right)^2\gcd\left(1-2\sqrt{-5},7\right)
$$
Since $N\!\left(1-2\sqrt{-5}\right)=21$ and $N(7)=49$, we must have
$$
\left.N\!\left(\gcd\left(1-2\sqrt{-5},7\right)\right)\,\middle|\ 7\right.
$$
But the only element whose norm divides $7$ is $1$.
This would say that the $\gcd$ is $\left(1+2\sqrt{-5}\right)^2$
$N\!\left(7\!\left(1+2\sqrt{-5}\right)\right)=1029$ and $N\!\left(\left(1+2\sqrt{-5}\right)^2\right)=441$, so these numbers are not associates (the only units here are $\pm1$, so this can be seen simply).
This contradiction says that a $\gcd$ does not exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Find limits for uniformly distributed random variable given equations of expectations My question is this:
Let $\eta$ be uniformly distributed random variable over the interval $[a, b]$.
Let E$\eta^2$ = 1 and E$\eta$ = -E$\eta^3$. Find a and b.
This is my approach:
$
\eta\sim Unif[a,b]\Rightarrow
f_\eta(x)=
\dfrac{1}{b-a}, \qquad a\leq x\leq b\\~\\
%
E\eta\phantom{^2}=\int_{a}^{b} x\phantom{^2}f_\eta(x)dx= \int_{a}^{b} \dfrac{x}{b-a}dx=\left[ \dfrac{x^2}{2b-2a} \right]_{a}^{b}=\dfrac{b^2-a^2}{2b-2a}=\dfrac{b+a}{2}\\~\\~\\
E\eta^2=\int_{a}^{b} x^2f_\eta(x)dx= \int_{a}^{b} \dfrac{x^2}{b-a}dx=\left[ \dfrac{x^3}{3b-3a} \right]_{a}^{b}=\dfrac{b^3-a^3}{3b-3a}=\dfrac{b^2+ab+a^2}{3}\\~\\~\\
E\eta^3=\int_{a}^{b} x^3f_\eta(x)dx= \int_{a}^{b} \dfrac{x^3}{b-a}dx=\left[ \dfrac{x^4}{4b-4a} \right]_{a}^{b}=\dfrac{b^4-a^4}{4b-4a}=\dfrac{(b^2+a^2)(b+a)}{4}\\~\\~\\
\begin{cases}
\dfrac{b^2+ab+a^2}{3}=1\\~\\
\dfrac{b+a}{2}=-\dfrac{(b^2+a^2)(b+a)}{4}
\end{cases}\Rightarrow
\begin{cases}
b^2+ab+a^2=3\\~\\
4(b+a)=-2(b^2+a^2)(b+a)
\end{cases}\Rightarrow
\begin{cases}
b^2+ab+a^2=3\\~\\
2b+2a=-b^3-b^2a-a^2b-a^3
\end{cases}\Rightarrow
$
This is where I am stuck. I can't solve this system of equations, but graphically I can see that $\pm\sqrt{3}$ is my solution. Can someone help me get to that solution or tell me what is wrong with my solution or maybe there is another approach?
| We have $$4(b+a)=-2(b^2+a^2)(a+b)$$ If $b+a\neq0$ then $$b^2+a^2=-2$$ which is absurd. Therefore, $$a=-b$$ Substitute this into $$\frac{b^2+ab+a^2}3=1$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Representing all rational numbers between $\dfrac{1}{2}$ and $1$ How do I show that
$$\dfrac{2^{\left\lfloor\frac12 a_1\right\rfloor} + 2^{\left\lfloor\frac12 a_2\right\rfloor} + \ldots + 2^{\left\lfloor\frac12 a_n \right\rfloor}}{2^{\left\lceil\frac12 a_1\right\rceil} + 2^{\left\lceil\frac12 a_2\right\rceil} + \ldots + 2^{\left\lceil\frac12 a_n \right\rceil}}$$
represents all rationals between $\dfrac{1}{2}$ and $1$, where $a_1, a_2, \ldots, a_n \in \mathbb{N}$?
I have tried representing few rationals such as $\dfrac{7}{8}$ and $\dfrac{5}{8}$ and it seems like we can represent all of them. However I'm not completely sure why and don't have a rigorous proof.
For example
$$\dfrac{5}{8} = \dfrac{2^{\left\lfloor\frac12 3\right\rfloor} + 2^{\left\lfloor\frac12 4\right\rfloor} + 2^{\left\lfloor\frac12 5 \right\rfloor}}{2^{\left\lceil\frac12 3\right\rceil} + 2^{\left\lceil\frac12 4\right\rceil} + 2^{\left\lceil\frac12 5 \right\rceil}}$$
$$\dfrac{7}{8} = \dfrac{2^{\left\lfloor\frac12 1\right\rfloor} + 2^{\left\lfloor\frac12 2\right\rfloor} + 2^{\left\lfloor\frac12 4 \right\rfloor}}{2^{\left\lceil\frac12 1\right\rceil} + 2^{\left\lceil\frac12 2\right\rceil} + 2^{\left\lceil\frac12 4 \right\rceil}}$$
Any help would be highly appreciated. Thanks
| We can observe that $2^{\lfloor a/2 \rfloor}$ and $2^{\lceil a/2 \rceil}$ are simply powers of $2$ which are either equal or consecutive. Hence:
$$2^{\lfloor a/2 \rfloor} \leqslant 2^{\lceil a/2 \rceil} \leqslant 2 \cdot 2^{\lfloor a/2 \rfloor}$$
Taking these bounds in our expression shows that any rational expressed in the given form must be between $\frac{1}{2}$ and $1$. Now, we will show that all rationals in $[\frac{1}{2},1]$ can be expressed in the given form by inducting over the size of the denominator.
The result holds good when the denominator is $1$ or $2$, by expressing $1$ and $\frac{1}{2}$ as they are. Now, assume that we are to show that $\frac{x}{y}$ can be expressed in the given form.
First, we can see that since $\frac{x}{y}<1$, it follows:
$$\frac{x-1}{y-1}<\frac{x}{y}<1$$
Moreover, if we assume $\frac{x-1}{y-1}<\frac{1}{2}$, then:
$$\frac{x-1}{y-1}<\frac{1}{2}<\frac{x}{y} \implies 2x-1<y<2x$$
which is clearly a contradiction. Hence, $\frac{x-1}{y-1}$ lies in $[\frac{1}{2},1]$ and can thus be expressed in the given form. Now, simply write $\frac{x}{y}$ in the given form by writing:
$$\frac{x}{y} = \frac{(x-1)+1}{(y-1)+1}$$
Note: Sometimes, the fraction $\frac{x-1}{y-1}$ might not be in simplified form. Simply consider the expression of this rational number in its simplest form, and repeat the terms in the numerator and denominator till the sum in the numerator is $x-1$ and in the denominator is $y-1$.
Example: Assume we need to express $\frac{16}{25}$ in the given form. We first write:
$$\frac{16}{25} = \frac{15+1}{24+1}$$
Now, we have:
$$\frac{15}{24}=\frac{5}{8}=\frac{2+2+1}{4+2+2}$$
Hence:
$$\frac{16}{25} = \frac{3(2+2+1)+1}{3(4+2+2)+1} = \frac{2+2+2+2+2+2+1+1+1+1}{4+4+4+2+2+2+2+2+2+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4080410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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} |
Show $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6+\sqrt{14})$ Sifting through my old Galois Theory notes, I found a proof that $\sqrt{5 + \sqrt{21}} = \frac{1}{2}(\sqrt 6 + \sqrt{14})$, but my proof was chicken-scratch so I can no longer decipher it.
Can somebody come up with a way of showing this?
| You're probably missing a $+$
The key is to convert it to the form $a^2+2ab+b^2$. Here you have $\sqrt{21}$ inside the outer square root, so first multiply and divide by $\sqrt 2$. Now $a$ and $b$ can be found easily as $7+3 = 10$ and $7\cdot 3 = 21$
$\begin{align}\sqrt{5+\sqrt{21}}& = \frac1{\sqrt2}\sqrt{10+2\sqrt{21}} = \frac1{\sqrt2}\sqrt{(7+3)+2\sqrt{7\cdot3}}\\& = \frac1{\sqrt2}\sqrt{(\sqrt7+\sqrt3)^2} = \frac1{\sqrt2}(\sqrt 7+\sqrt 3)\\& =\frac1{\sqrt2\cdot\sqrt2}(\sqrt {14}+\sqrt 6)\\&=\frac1{2}(\sqrt 6+\sqrt {14})\end{align}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find $(a, b, c)$ that satisfies the system of equations $2a + 3b -4c = 7, a-b+2c=6$ $\begin{align}
2a + 3b -4c = 7, \\
a - b + 2c = 6
\end{align}$
I haven't seen a three-variable system of equations with only two equations before. I try to make the problem into a two-variable system.
Solving for $a$ in the first equation
$\begin{align}
a = \frac{-3b +4c+7}{2}
\end{align}$.
Solving for $a$ in the second equation:
$\begin{align}
a = b-2c+6
\end{align}$.
Then substitute the equations into $a$ for the original equations.
$\begin{align}
2(b-2c+6) + 3b -4c = 7 \\
2b-4c+12+3b-4c=7 \\
5b-8c = -5
\end{align}$
$\begin{align}
\frac{-3b +4c+7}{2} - b + 2c = 6 \\
-3b+4c+7-2b+4c=12 \\
-5b+8c=5
\end{align}$
This create a new system of equations:
$\begin{align}
5b-8c =-5 \\
-5b+8c=5
\end{align}$
This leads to the solution $0=0$. So I assume this means I can select any $b$ and $c$. This however isn't the case, I can't find $a$ that satisfies both equations with this premise.
The problem additionally asks to solve for $b$ in terms of $a$ and solve for $c$ in terms of $a$. I did so using elimination.
$\begin{align}
b = -4a +19 \\
c = \frac{-5a+25}{2}
\end{align}$
So we have both $b$ and $c$ depending on $a$. There is an infinite number of choices for $a$ so there are infinitely many solutions to this system.
Could I not find one or more ordered triples that satisfy both equations if I did not (following the instruction) create this final system of equations? Until now I thought the number of equations in the system was simply related to the number of variables.
| Once you get the equations for b and c in terms of a, then you are done. Just say that all ordered pairs (a,b,c) that work are in the form ($k$, $-4k+19$, $\frac{-5k+25}{2}$) for any real number k. If you plug back into the two formulas, you will see that they will both simplify to c=c for some constant c, meaning any value of k will work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4083679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Has the equation $x^2-21 = 17y$ integer solutions? Has the equation $x^2-21 = 17y$ integer solutions?
Attempt:
I saw this: The equation $x ^ 2 + py + a = 0$ can be solved as an integer precisely, if $-a$ is a quadratic remainder modulo p.
I get: $x^2-17y-21=0$
Now i have to show $21$ is quadratic remainder modulo $-17$? I dont know if this is correct...
$(\frac{21}{-17}) = (\frac{3}{-17}) * (\frac{7}{-17})$
for $(\frac{3}{-17}) = (-1) (\frac{-17}{3})(\text{Quadratic reciprocity})= (\frac{17}{3}) = (\frac{2}{3}) = -1$
for $(\frac{7}{-17}) = (-1) (\frac{-17}{7})(\text{Quadratic reciprocity})= (\frac{17}{7}) = (\frac{3}{7}) = (-1)(\frac{7}{3})(\text{Quadratic reciprocity}) = (\frac{2}{3}) = -1$
insert, we get:
$(-1) * (-1) =1$ and we have integer solutions?
| $x^2-21=17y$ has integer solutions precisely when $x^2-4=17(y+1)$ has integer solutions, which occurs precisely when it's possible to solve $x^2 \equiv 4 \pmod{17}$. This has solutions, of course, when $x \equiv \pm 2 \pmod{17}$.
Checking: $x =2, x^2-21=-17=17(-1); x=15, x^2-21=204=17(12)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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An upper bound for $\{r_n\}$ which satisfies $r_n = r_{n-1} + \frac{K}{r_{n-1}}$ I have a sequence $\{r_n\}$ of positive numbers defined recursively by
$$\tag{1} r_n = r_{n-1} + \frac{K}{r_{n-1}}.$$
for some positive $K$. The sequence is clearly strictly increasing and tends to $+\infty$ as $n\to \infty$. Also, by induction one can prove
$$\tag{2} r_n^2 \ge 2(n-1)K +r_1^2$$
for all $n$.
Question: Does there exists positive $K_2$ depending on $r_1, K$ such that
$$\tag{3} r_n^2 \le 2nK + K_2$$
for all large $n$?
Squaring (1) on both sides give
$$ r_n^2 = r_{n-1}^2 + 2K + \frac{K^2}{r_{n-1}^2},$$
If I try to prove (3) using induction, (2) gives
$$ \frac{K^2}{r_{n-1}^2} \le \frac{K^2}{ 2(n-2)K +r_1^2},$$
which is not very helpful, since $\sum \frac{1}{n-2}$ is not summable.
| tl;dr: You cannot prove the upper bound you want, as the true growth of the sequence is "slightly more" than $2Kn$:
$$
r_n^2 = 2Kn + \frac{K}{2}\log n + O(1)
$$
First, for the sake of "elegance", set $a_n := \frac{r_n}{\sqrt{K}}$ for all $n\geq 0$, so that the recurrence relation becomes
$$
a_n = a_{n-1} + \frac{1}{a_{n-1}}, \qquad n\geq 1 \tag{1}
$$
Squaring both sides, we get $a_n^2 = a_{n-1}^2 + 2 + \frac{1}{a_{n-1}^2}$, or, equivalently,
$$
a_n^2 - a_{n-1}^2 = 2 + \frac{1}{a_{n-1}^2} \tag{2}
$$
Summing (2) from $1$ to $n$, we get
$$
a_n^2 - a_0^2 = 2n + \sum_{k=0}^{n-1}\frac{1}{a_k^2} \tag{3}
$$
and in particular $a_n^2 \geq 2n + a_0^2$ for all $n\geq 0$. We also have, analogously to (3), that
$$
a_n^2 - a_1^2 = 2(n-1) + \sum_{k=1}^{n-1}\frac{1}{a_k^2}
$$
and using our lower bound $a_k^2 \geq 2k$, we get
$$
a_n^2 \leq 2n-2+a_1^2 + \frac{1}{2}\sum_{k=1}^{n-1}\frac{1}{k}
\geq 2n + \frac{1}{2} \log n +a_1^2
$$
At this point, we have shown, for all $n\geq 1$,
$$
\boxed{2n + a_0^2 \leq a_n^2 \leq 2n + \frac{1}{2} \log n +a_1^2} \tag{4}
$$
This is great, but let us not stop there. Plugging (4) back into (3), we have
$$
a_n^2 - a_0^2 \geq 2n +\frac{1}{a_0^2} + \sum_{k=1}^{n-1}\frac{1}{2n + \frac{1}{2} \log n +a_1^2}
\geq 2n + \frac{1}{2}\log n + C \tag{5}
$$
for some constant $C\in\mathbb{R}$ which depends on $a_0,a_1$ only (and therefore on $a_0$ only, since $a_0$ determines $a_1$): where we used that
$$
\sum_{k=1}^{n-1}\frac{1}{2n + \frac{1}{2} \log n + a_1^2}
= \frac{1}{2}\sum_{k=1}^{n-1} \left(\frac{1}{n}\cdot \frac{1}{1+\frac{\log n + a_1^2}{n}}\right)
= \frac{1}{2}\sum_{k=1}^{n-1} \frac{1}{n} - \sum_{k=1}^{n-1}\left(\frac{\log n}{2n^2} + o\!\left(\frac{\log n}{n^2}\right)\right)
$$
and the second series is convergent.
Overall, we have therefore established that there exist a constant $C\in\mathbb{R}$ (depending on the initial term $a_0$ only) such that, for all $n\geq 1$,
$$
\boxed{2n + \frac{1}{2}\log n + C \leq a_n^2 \leq 2n + \frac{1}{2} \log n + a_1^2} \tag{6}
$$
that is, $a_n^2 = 2n + \frac{1}{2}\log n + O(1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Advanced complex numbers/roots of unity Let $a, b, c, d$ be real numbers, none of which are equal to $-1$, and let $\omega$ be a complex number such that $\omega^3 = 1$ and $\omega \neq 1.$ Given that,
$$
\frac{1}{a + \omega} + \frac{1}{b + \omega} + \frac{1}{c + \omega} + \frac{1}{d + \omega} = \frac{2}{\omega},
$$
how can I deduce
$$
\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c +1} + \frac{1}{d + 1}?
$$
I have tried clearing the denominators of the first equation, but that just results in a large mess. I don't know how to continue from there.
| I will write $u$ instead of $\omega$, so that $u$ is the primitive third root of unity, and the other one is $u^2=\bar u$.
Below, $v$ will be an element in the list $L=\{1,u,u^2\}$ of all third roots of unity.
The given relation can be rewritten:
$$
2=u\sum\frac 1{a+u}=\sum\frac u{a+u}=\sum\frac 1{1+au^2}
\ ,
$$
where the above sums have $4$ terms each, they are obtained by substituting instead of $a$ formally the values $a,b,c,d$.
It is natural to consider now the polynomial expressions which are the numerators of
$\displaystyle
2-\sum\frac 1{1+av}
$ for $v\in L$. They are
$$
\begin{aligned}
P_v
&:=2(1+av)(1+bv)(1+cv)(1+dv)
\\
&\qquad\qquad
-\sum \color{gray}{\underbrace{(1+av)}_{\text{omitted}}}(1+bv)(1+cv)(1+dv)
\\
&=2v\; abcd + (bcd+cda+dab+abc) - v(a+b+c+d) -2
\\
\\[3mm]
&\qquad\text{From here we get immediately:}\\[3mm]
vP_v &=
2v^2\; abcd + v(bcd+cda+dab+abc) - v^2(a+b+c+d) -2v
\\
\sum_{v\in L}v\;P_v
&=
2\left(\sum v^2\right)\; abcd + \left(\sum v\right)(bcd+cda+dab+abc)
\\
&\qquad\qquad - \left( \sum v^2\right)(a+b+c+d) -2\left(\sum v\right)
\\
&=0
\ .
\end{aligned}
$$
We have used
$\sum v=\sum v^2=1+u+u^2=0$ for $v$ running in $L$ in the above sums. From $P_u=0$, and its conjugated cousin $P_{u^2}=0$, we obtain $P_1=0$. So:
$$
2=\sum\frac 1{1+a}\ .
$$
$\square$
| {
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The equation ${4 \choose k}=6$ Find the solution of the equation $${4 \choose k}=6.$$
So $k$ must be a natural number $(n\in \mathbb{N})$ and we can find that when $k=1 \rightarrow {4\choose 1}=4$ and when $k=2\rightarrow {4\choose2}=6$, so the solution of the equation is actually $k=2$. Can we solve the problem without calculating the exact value of ${4\choose k}$? I tried: $$\dfrac{V_4^k}{k!}=6$$ but don't see how to solve this. Thank you in advance!
| For this case, you can use Pascal's triangle and binomial expansion.
$$
Pascal's\ \ triangle:
\\
\newcommand\pad[1]{\rlap{#1}\phantom{50}}
\begin{matrix}
n=0:&&&&&&&1\\
n=1:&&&&&&1&&1\\
n=2:&&&&&1&&2&&1\\
n=3:&&&&1&&3&&3&&1\\
n=4:&&&\pad{1}&&\pad{4}&&\pad{6}&&\pad{4}&&\pad{1}\\
n=5:&&\pad{1}&&\pad{5}&&\pad{10}&&\pad{10}&&\pad{5}&&\pad{1}
\end{matrix}
$$
$$
binomial\ \ expansion:
\\
\sum_{k=0}^n \binom{n}{k} = \binom{n}{0} + \binom{n}{1} + \ldots + \binom{n}{n-1} + \binom{n}{n} = 2^n = (1 + 1)^n $$
In your case,
$$ (1 + 1)^4 = 1 + 4 + 6 + 4 + 1 $$
So $ k = 2 $.
For example, for $ (1 + 1)^2 = 1 + 2 + 1 $:
$$ \binom{2}{0} = 1 , \binom{2}{1} = 2, \binom{2}{2} = 1 $$
More information about Pascal's triangle and binomial expansion: https://en.wikipedia.org/wiki/Pascal's_triangle
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solving $\tan^{-1}\left(\frac{2x+1}{x+1}\right)+\tan^{-1}\left(\frac{2x-1}{x-1}\right)=2\tan^{-1}\left(1+x\right)$
Find all $x$ that satisfy the equation $$\tan^{-1}\left(\frac{2x+1}{x+1}\right)+\tan^{-1}\left(\frac{2x-1}{x-1}\right)=2\tan^{-1}\left(1+x\right)$$
Attempt:
Taking tangent on both sides, and using the identities $\displaystyle \tan\left(\alpha+\beta\right)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ and $\displaystyle \tan\left(2\alpha\right)=\frac{2\tan\alpha}{1-\tan^{2}\alpha}$,
We obtain $$\frac{\frac{\left(2x+1\right)}{x+1}+\frac{\left(2x-1\right)}{x-1}}{1-\frac{\left(4x^{2}-1\right)}{x^{2}-1}}=\frac{2\left(x+1\right)}{1-\left(x+1\right)^{2}}$$ and after removing the root $x=0$, we are left with a simple equation $(x^2-2)(2x+1)=0$. So, the possible roots are $x=0,\pm \sqrt{2}, -\dfrac{1}{2}$ , now my question is, how do I verify which roots satisfy the original equation, because one cannot see directly after plugging, whether it's true or not because it involves arctangents.
To avoid this situation, I re-wrote the original equation as $$\tan^{-1}\left(\frac{2x+1}{x+1}\right)-\tan^{-1}\left(1+x\right)=\tan^{-1}\left(1+x\right)-\tan^{-1}\left(\frac{2x-1}{x-1}\right)$$ and because while using the identity $\tan^{-1}\left(\alpha\right)-\tan^{-1}\left(\beta\right)=\tan^{-1}\left(\frac{\alpha-\beta}{1+\alpha\beta}\right)$, there are no restrictions on $\alpha,\beta$ I went ahead with this, and got the equation $$\frac{\frac{\left(2x+1\right)}{x+1}-\left(x+1\right)}{2\left(x+1\right)}=\frac{\left(x+1\right)-\frac{\left(2x-1\right)}{x-1}}{2x}$$ which simplified to $\left(x+1\right)^{2}\left(x-2\right)+x^{2}\left(x-1\right)=0$ and surprisingly from here, I am not even getting those roots!
I used this method to avoid the extraneous roots, what is going wrong here?
Edit: The immediately above issue has been resolved, however, the first issue remains unresolved.
Note that all calculations have to be performed by hand, since these types of questions are asked in exams in which the allotted time per question averages out to 2-3 minutes.
| Too large for comment.
Like Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$\dfrac{(2x+1)(2x-1)}{(x+1)(x-1)}$ will be $\le1$
$\iff0\ge\dfrac{(2x+1)(2x-1)}{(x+1)(x-1)}-1=\dfrac{3x^2}{x^2-1}$
$\iff x^2-1\le0\iff-1\le x\le1$
Similarly, $(x+1)^2$ will be $\le1\iff-2\le x\le0$
For $x=-\sqrt2<-1,$ $$\tan^{-1}\dfrac{2x+1}{x+1}+\tan^{-1}\dfrac{2x-1}{x-1}=\pi+\tan^{-1}\dfrac{\dfrac{2x+1}{x+1}+\dfrac{2x-1}{x-1}}{1-\dfrac{2x+1}{x+1}\cdot\dfrac{2x-1}{x-1}}>\dfrac\pi2$$
But $$2\tan^{-1}(x+1)=2\tan^{-1}(-\sqrt2+1)=-2\tan^{-1}(\sqrt2-1)<0$$
Actually $\sqrt2-1=\csc\dfrac\pi4-\cot\dfrac\pi4=\tan\dfrac\pi8\implies\tan^{-1}(\sqrt2-1)=\dfrac\pi8$
| {
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"url": "https://math.stackexchange.com/questions/4087642",
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"source": "stackexchange",
"question_score": "8",
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} |
$a_1Give that $a_1<a_2<a_3<a_4$ are positive integers such that $$\sum_{i=1}^4\frac{1}{a_i}=\frac{11}{6}$$, find the value of $a_4-a_2$.
My try:
Since $11$ is prime, atleast one of $a_1,a_2,a_3,a_4$ should be divisible by $6$.
Now we should express the fraction $\frac{11}{6}$ as $\frac{A+B+C+D}{E}$ such that $E$ is divisible by $A,B,C,D,E$.
I tried $$\frac{11}{6}=1+\frac{1}{2}+\frac{1}{3}$$ But fourth fraction is not possible.
Any hint?
| Use the identity $\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$ with $n=3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$\angle DAB = 10^\circ$, $\angle CAD = 20^\circ$, $\angle CDB = 140^\circ$, $CD = BD$. Find $\angle ABD$ $\angle DAB = 10^\circ$
$\angle CAD = 20^\circ$
$\angle CDB = 140^\circ$
The triangle $\triangle BCD$ is isosceles. $CD = BD$
Find the angle $x$
The answer is $30º$.
Playing with GeoGebra, I found that $AC = AD$. But I don't know how to prove it.
| Set $\angle ACD= y, x+y=110^\circ$
For $\triangle ADC$ and $\triangle ADB$:
\begin{align}\frac{DB}{AD}=\frac{\sin 10^\circ}{\sin x}=\frac{CD}{AD}=\frac{\sin 20^\circ}{\sin (110^\circ -x)}\end{align}
\begin{align}\frac{\sin 10^\circ}{\sin 20^\circ}=\frac{sin x}{\cos(x-20^\circ)}\end{align}
\begin{align}\tan x=\frac{\cos 20^\circ}{2\cos 10^\circ -\sin 20^\circ}=\frac{\sqrt 3}{3}^*, x=30^\circ\end{align}
EDIT:
\begin{align}\frac{\cos 20^\circ}{2\cos 10^\circ -\sin 20^\circ}=\frac{\cos 20^\circ}{\cos 10^\circ+\cos 10^\circ-\sin 20^\circ}=\frac{\cos 20^\circ}{\cos 10^\circ+\sin 80^\circ-\sin 20^\circ}=\frac{\cos 20^\circ}{\cos 10^\circ+\cos 50^\circ}=\frac{\cos 20^\circ}{2\cos 20^\circ\cos 30^\circ}=\frac{\sqrt 3}{3}\end{align}
| {
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"source": "stackexchange",
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Given that $3\sin x + 2\sin y = 4$. Find the maximum value of $3\cos x + 2\cos y$.
Given that $3\sin x + 2\sin y = 4$. Find the maximum value of $3\cos x + 2\cos y$.
I wonder if calculus can be used to solve this problem? If not how to solve this? I am thinking about $(3\cos x + 2\cos y)^2=9\cos^2x+12\cos x\cos y + 4\cos^2y$ and work in this way but I don't quite see how.
| Although Koro's solution is my favourite one, you can also solve it just using AM-GM as follows:
*
*Set $s=\sin x, t = \sin y$
*$\Rightarrow$ to maximize is $3\sqrt{1-s^2}+2\sqrt{1-t^2}$ subject to $3s+2t=4$ for $(s,t)\in [0,1]^2$
Since the objective function is non-negative you can maximize the square of it:
\begin{eqnarray*} 9(1-s^2)+4(1-t^2) + 12 \sqrt{(1-s^2)(1-t^2)}
& \stackrel{AM-GM}{\leq} & 13 -(9s^2+4t^2) + 12 \frac{2-s^2-t^2}2 \\
& \stackrel{9s^2+4t^2 =16-12st}{\leq} & 9 + 12st - 6(s^2+t^2)\\
& \stackrel{AM-GM}{\leq} & 9 + 6(s^2+t^2) - 6(s^2+t^2) \\
& = & 9
\end{eqnarray*}
Equality is reached for $s=t = \frac 45$, hence, for $\sin x = \sin y =\frac 45$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4092309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Any pattern in the following sequence I am practicing finding patterns and sequences with numbers and found one that I cannot seem to find the logic.
$$
\frac{2}{3} ,\frac{4}{3} ,\frac{4}{5} ,\frac{6}{5} ,\frac{6}{7},\frac{6}{8}...
$$
The last term 6/8 breaks my understanding of a possible pattern.
If the last term didn't exist, I could suggest the following.
$$
\frac{2}{3} ,\frac{4}{3} ,\frac{4}{5} ,\frac{6}{5} ,\frac{6}{7},\frac{8}{7},\frac{8}{9},\frac{10}{9}...
$$
Is it possible that the original sequence does not have a pattern with which I can find the nth term?
| For
$$\{2,4,4,6,6,6\}$$ there are $27$ sequences in $OEIS$ and for
$$\{3,3,5,5,7,8\}$$ there are $30$ other.
Now, for
$$\frac{2}{3} ,\frac{4}{3} ,\frac{4}{5} ,\frac{6}{5} ,\frac{6}{7},\frac{6}{8}...$$ it is hard to say
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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recurrence relation - how to solve using generating functions How to solve this recurrence relation using generating functions (or in another way)?
$$\begin{cases} A(0)=0\\A(b)=1\\A(n) = pA(n+1)+qA(n-1)
\end{cases}\\ n=1,2,3,4,...,b-1.
$$
| Your recurrence relation can be written as
$$A(n+2) = \frac{A(n+1) - qA(n)}{p}, \qquad n=0,1,2,...,b-2.$$
which can be regarded as a generalization of the Fibonacci numbers. Let us set $A(1)=A$: notice that as far as the recurrence relation is concerned, the value of $A$ is arbitrary and the sequence proceeds indefinitely. We can solve the recurrence relation directly for the first few $n$:
*
*$\underline{n=0}$
$$A(2) = \frac{1}{p} A,$$
*$\underline{n=1}$
$$A(3) = \frac{1-pq}{p^2}A,$$
*$\underline{n=2}$
$$A(4) = \frac{1 - 2pq}{p^3} A,$$
*$\underline{n=3}$
$$A(5) = \frac{1 -3pq + p^2q^2}{p^4}A,$$
*$\underline{n=4}$
$$A(6) = \frac{1 - 4pq + 3p^2q^2}{p^5} A,$$
and so on. In fact, it is quite evident that by induction
$$A(n) = \frac{P_n(p,q)}{p^{n-1}}A,$$
where $P_n$ is a polynomial of degree $n-1$ in $p$ and $q$. It is clear that the value of $A$ can be adjusted, if we wish, as to satisfy $A(b)=1$:
$$A(b) = 1 \Longleftrightarrow A=\frac{p^{b-1}}{P_b(p,q)}.$$
In determining the polynomials $P_n(p,q)$ is where I think the generating function really comes in useful. By setting
$$G_{p,q}(x) = \sum_{n=0}^\infty A(n)\,x^n,$$
you have the equation
\begin{align}
G_{p,q}(x) &= A(1)x + \sum_{n=2}^\infty A(n)x^n = Ax + \sum_{n=2}^\infty\frac{A(n-1)-qA(n-2)}{p}x^n = \nonumber\\
&=Ax + \sum_{n=2}^\infty\frac{A(n-1)}{p}x^n - \sum_{n=2}^\infty\frac{qA(n-2)}{p}x^n = \nonumber\\
&=Ax + \sum_{n=1}^\infty\frac{A(n)}{p}x^{n+1} - \sum_{n=0}^\infty\frac{qA(n)}{p}x^{n+2} = \nonumber\\
&= Ax + \frac{xG_{p,q}(x)-qx^2G_{p,q}(x)}{p},
\end{align}
whose solution is simply
$$G_{p,q}(x) = \frac{Apx}{p-x(1-qx)}.$$
You can now use the generating function to represent the polynomials $P_n(p,q)$, obtaining a complete information on your sequence:
$$P_n(p,q) = \frac{p^{n-1}}{n!\,A} G^{(n)}_{p,q}(0) = \frac{p^n}{n!}\frac{\mathrm{d}^n}{\mathrm{d}x^n}\frac{x}{qx^2-x+p}\bigg|_{x=0}.$$
Edit: The polynomials can actually be written in closed form quite neatly, starting from the identity
$$\frac{\mathrm{d}^n}{\mathrm{d}x^n}\frac{x}{qx^2-x+p}\bigg|_{x=0} = n\frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}}\frac{1}{qx^2-x+p}\bigg|_{x=0}.$$
Indeed, by Leibniz's theorem
\begin{align}
\frac{\mathrm{d}^n}{\mathrm{d}x^n}\frac{x}{qx^2-x+p} &= \sum_{k=0}^n{n\choose k} \frac{\mathrm{d}^k}{\mathrm{d}x^k}x\frac{\mathrm{d}^{n-k}}{\mathrm{d}x^{n-k}}\frac{1}{qx^2-x+p} = \nonumber\\
&= x\frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}\frac{1}{qx^2-x+p} + n\frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}}\frac{1}{qx^2-x+p},
\end{align}
as the terms with higher derivatives of $x$ vanish. However when we evaluate the derivative in $x=0$, the first terms vanishes too. This leaves us with the task of calculating the derivatives of $1/(qx^2-x+p)$. We want to decompose this into simpler fractions: the roots of the denominator are
$$\phi_\pm(p,q)=\frac{1\pm\sqrt{1-4pq}}{2q},$$
which we can use to decompose our fraction as
$$\frac{1}{qx^2-x+p} = \frac{1}{\sqrt{1-4pq}}\bigg[\frac{1}{x-\phi_+(p,q)}-\frac{1}{x-\phi_-(p,q)}\bigg],$$
hence
\begin{align}
\frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}}\frac{1}{qx^2-x+p} = \frac{1}{\sqrt{1-4pq}}\bigg[\frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}}\frac{1}{x-\phi_+}-\frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}}\frac{1}{x-\phi_-}\bigg] = \frac{(-1)^{n-1}(n-1)!}{\sqrt{1-4pq}}\bigg[\frac{1}{(x-\phi_+)^n}-\frac{1}{(x-\phi_-)^n}\bigg],
\end{align}
so finally setting $x=0$, we obtain
\begin{align}
\frac{\mathrm{d}^{n-1}}{\mathrm{d}x^{n-1}}\frac{1}{qx^2-x+p}\bigg|_{x=0} &=\frac{(n-1)!}{\sqrt{1-4pq}}\bigg[\frac{1}{\phi_-(p,q)^n}-\frac{1}{\phi_+(p,q)^n}\bigg] = \nonumber\\
&=\frac{(n-1)!}{\sqrt{1-4pq}}(2q)^n\frac{(1+\sqrt{1-4pq})^n-(1-\sqrt{1-4pq})^n}{(4pq)^n} = \nonumber\\
&=\frac{(n-1)!}{(2p)^n}\frac{(1+\sqrt{1-4pq})^n-(1-\sqrt{1-4pq})^n}{\sqrt{1-4pq}}.
\end{align}
Therefore, the polynomials $P_n$ can be expressed in closed form as
\begin{equation}
P_n(p,q) = \frac{1}{2^n}\frac{(1+\sqrt{1-4pq})^n-(1-\sqrt{1-4pq})^n}{\sqrt{1-4pq}}.
\end{equation}
The implication that the right-hand is actually a polynomial is absolutely remarkable! In conclusion,
\begin{equation}
A(n) = 2^{b-n}p^{b-n}\frac{(1+\sqrt{1-4pq})^n-(1-\sqrt{1-4pq})^n}{(1+\sqrt{1-4pq})^b-(1-\sqrt{1-4pq})^b}.
\end{equation}
| {
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If $a$,$b$ and $y$ are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$ If a b and y are the roots of $3x^3+8x^2-1=0$ find $(b+1/y)(y+1/a)(a+1/b)$
This is what I have done so far, but apparently it is incorrect. I want to know why.
$(b+1/y)(y+1/a)(a+1/b)$
$(by+1/y)(ay+1/a)(ab+1/b)$
$(aby^2+1/ay)(ab+1/b)$
which is equal to
$a^2b^2y^2 + 1/aby$
Using Vieta's formula I get:
$aby = -d/a$
$aby = 1/3$
Subbing in original:
$(1/9 + 1)/1/3$
which is...
10/27
The answer is supposedly 2/3, want to know what I did wrong.
| You can also calculate the expression directly using the fact that
$p(x) = 3(x-a)(x-b)(x-y)\Rightarrow 3aby = 1$.
To do so, note that
\begin{eqnarray*} P
& = & (b+1/y)(y+1/a)(a+1/b) \\
& = & \frac{(by+1)(ay+1)(ab+1)}{aby} \\
& \stackrel{aby=\frac 13}{=} & 3\left(\frac 1{3a}+1\right)\left(\frac 1{3b}+1\right)\left(\frac 1{3y}+1\right) \\
& \stackrel{aby=\frac 13}{=} & \frac 13(1+3a)(1+3b)(1+3y)
\end{eqnarray*}
Now, a standard trick uses the observation
\begin{eqnarray*} p\left(\frac{t-1}3\right)
& = & 3\left(\frac{t-1}3-a\right)\left(\frac{t-1}3-b\right)\left(\frac{t-1}3-y\right) \\
& = & \frac 19\left(t-(1+3a)\right)\left(t-(1+3b)\right)\left(t-(1+3y)\right)
\end{eqnarray*}
So, you only need the constant member $$c =-\frac 19(1+3a)(1+3b)(1+3y)=-\frac 13 P$$ of
$$p\left(\frac{t-1}3\right)= 3\left(\frac{t-1}3\right)^3 + 8\left(\frac{t-1}3\right)^2 -1 $$$$\Rightarrow c= -\frac 19+\frac 89-1 = -\frac 29$$
Hence,
$$P= -3c = \frac 23$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine all positive integers $n$ such that $n^2+1$ divides $2^{n!}-1$ For this question, I have tried to start from reading similar question (How many rationals of the form $\frac{2^n+1}{n^2}$ are integers? from this post How many rationals of the form $\large \frac{2^n+1}{n^2}$ are integers?)
I notice that $2^{2!}-1$ is always odd so I tried to rewrite it as $2k+1$ but I am stuck afterward. I am thinking of how do I relate this to modular arithmetic or gcd. I would appreciate if anyone could help.
| In the link in comment it is proved that $m^2-1\big|2^{m!}-1$ . Let $m=n^2$, then $m^2-1=n^4-1\big|2^{m!=n^2!}-1$ and $n^2+1\big|(n^4-1)$. For example :
$n=4 \rightarrow 4^2+1=17\big|4^4-1$ and we have:
$2^{4!}-1=2^{24}-1=[(2^4)^3-1][(2^4)^3+1]$
$(2^4)^3+1=(2^4+1)[(2^4)^2-(2^4)+1]=17k$
therefore:
$4^2+1\big|2^{4!}-1$
In fact $n=4$ is the smallest number and general form of n is $2t$ such that $(2t)^2+1$ is a prime. Another example:
$10^2+1=101\rightarrow 2^{101}\equiv 1 \bmod 101$, $100!=100(s)$ and $2^{100!}=(2^{100})^s\equiv 1 \bmod 101\Rightarrow 10^2+1\big|2^{100!}-1$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Help with $ \lim \frac{n\pi}{4} - \left( \frac{n^2}{n^2+1^2} +\frac{n^2}{n^2+2^2} + \cdots + \frac{n^2}{n^2+n^2} \right) $ I have proved that $$\lim_{n\to \infty} \left( \frac{n}{n^2+1^2} +\frac{n}{n^2+2^2} + \cdots + \frac{n}{n^2+n^2} \right) = \frac{\pi}{4},$$
by using the Riemann's sum of $\arctan$ on $[0,1]$. Now I'm interested in computing the next limit
$$ L=\lim_{n\to \infty} \left[ \frac{n\pi}{4} - \left( \frac{n^2}{n^2+1^2} +\frac{n^2}{n^2+2^2} + \cdots + \frac{n^2}{n^2+n^2} \right) \right] $$
Note that
$$ L= \lim n \left[ \frac{\pi}{4} - \left( \frac{n}{n^2+1^2} +\frac{n}{n^2+2^2} + \cdots + \frac{n}{n^2+n^2} \right) \right] =[\infty \cdot 0].$$
I can't manage to calculate $L$. How can this indeterminate form be solved? Also, by using computer, I suspect that $L$ exists and is equal to $1/4$. Any idea or hint is welcome! (Actually, I dont need a complete solution, just a good start point). Have a nice day you all!
| Let $f(x):=\frac{1}{1+x^2}$ so$$L=\lim_{n\to\infty}L_n,\,L_n:=n\int_0^1f(x)dx-\sum_{i=1}^nf(i/n).$$Taylor-expanding $f$,$$L_n=\sum_{k\ge1}\frac{f^{(k)}(0)}{k!}c_{nk},\,c_{nk}:=n\int_0^1x^kdx-\frac{1}{n^k}\sum_{i=1}^ni^k\sim-\tfrac12$$by Falhaber's formula. So$$L=-\tfrac12\sum_{k\ge1}\frac{f^{(k)}(0)}{k!}=-\tfrac12(f(1)-f(0))=\tfrac14.$$
| {
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Trouble with this one step involving arcsin So I'm solving a very basic trigonometric equation and after substituting the function for u and solving the quadratic equation, I ended up with $\sin\left(2x+\frac{\pi}{3}\right)=\frac{1}{4}$.
I need to take the arcsin of both sides if I'm not mistaken, but after inputting the original equation into a graph, it gives me a different solution than $2x+\frac{\pi}{3}=\arcsin\left(\frac{1}{4}\right)$
So 100% I'm missing a rule regarding cancelling out a function. I would appreciate an elaboration :)
|
For $\alpha \in (-1,1)$ the equation $\sin(x) = \alpha$ has solutions
$$ x = \arcsin(\alpha) + 2 k \pi \qquad \text{and}\qquad x = \pi - \arcsin(\alpha) + 2 k \pi $$
Hence, in your case, the solutions are
$$
2x + \frac{\pi}{3}= \arcsin\left(\frac{1}{4}\right)+ 2 k \pi \implies x = -\frac{\pi}{6} +\frac{1}{2}\arcsin\left(\frac{1}{4}\right) + k\pi\\
$$
and
$$
2x + \frac{\pi}{3} = \pi - \arcsin\left(\frac{1}{4}\right) + 2 k \pi \implies x = -\frac{2\pi}{3} -\frac{1}{2}\arcsin\left(\frac{1}{4}\right) + k\pi
$$
where $\arcsin\left(\frac{1}{4}\right) \approx 0.2526$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Prove that $a^ab^bc^cd^d\geq\frac{1}{16}(ab+c+d+1)^2$
Given $a, b, c, d\in\mathbb{R^+}$ such that $abcd = 1$. Prove that
$a^ab^bc^cd^d\geq\frac{1}{16}(ab+c+d+1)^2$
I've tried,
$a^ab^bc^cd^d=(abcd)^ab^{b-a}c^{c-a}d^{d-a}=b^{b-a}c^{c-a}d^{d-a}\geq\frac{1}{16}(ab+c+d+1)^2$
i'm stucked
| Since $x \mapsto x\ln x$ is convex on $x > 0$, and $a + b + c + d \ge 4\sqrt[4]{abcd} = 4$, we have
\begin{align*}
a\ln a + b\ln b + c\ln c + d\ln d
&\ge 4 \cdot \frac{a + b + c + d}{4} \ln \frac{a + b + c + d}{4}\\
&\ge 4 \ln \frac{a + b + c + d}{4}.
\end{align*}
Also, we have
$$ab + c + d + 1 \le \frac{1}{4}(a + b)^2 + c + d + 1.$$
It suffices to prove that
$$\left(\frac{a + b + c + d}{4}\right)^4 \ge \frac{1}{16}\left(\frac{1}{4}(a + b)^2 + c + d + 1 \right)^2$$
or
$$(a + b + c + d)^2 \ge (a + b)^2 + 4(c + d) + 4$$
or
$$(c + d - 2)^2 + 2(a + b)(c + d) - 8 \ge 0$$
which is true since $2(a + b)(c + d) \ge 2\cdot 2\sqrt{ab} \cdot 2\sqrt{cd} = 8$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4116450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
When is $E(X-\mu)^2 \ne E(X^2)-\mu^2$? Consider the probability density function
$$f(x)=\frac{x}{18}, \quad x \in [0,18]$$
Find $E(x)$ and $Var(x)$.
Solution:
a.)
\begin{align*}
E(x)
&=\int_{-\infty}^{\infty} x f(x) dx \\
&=\int_{0}^{18} x \left(\frac{x}{18}\right) dx \\
&=\frac{1}{18} \int_0^{18} x^2 dx \\
&=108
\end{align*}
b.)
Method 1:
\begin{align*}
Var(x)
&=E(x-\mu)^2 \\
&=\int_{-\infty}^{\infty} (x-\mu) f(x) dx \\
&=\int_{0}^{18} (x-108)^2 \left(\frac{x}{18}\right) dx \\
&=83106
\end{align*}
Method 2:
\begin{align*}
Var(x)
&=E(x^2)-\mu^2 \\
&=\int_{-\infty}^{\infty} x^2 f(x) dx - (108)^2 \\
&=\int_{0}^{18} x^2 \left(\frac{x}{18}\right) dx - (108)^2\\
&=\frac{1}{18}\int_{0}^{18} x^3 dx - (108)^2\\
&=-10206
\end{align*}
Clearly the second method is wrong, since a variance cannot be negative. But I cannot find the error. Can someone explain why the two methods yield different values? Is it not always the case that $E(X-\mu)^2 \ne E(X^2)-\mu^2$?
| The probability density function integrates to 9 in the interval given, that is why it does not work.
| {
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Calculating $\lim_{(x,y)\rightarrow (\infty,\infty)}(x^2+y^2)e^{-(x+y)}$ Please help me calculate the:
$$\lim_{(x,y)\rightarrow (\infty,\infty)}(x^2+y^2)e^{-(x+y)}$$
My attempt is: First method: for $t>0$ we have $e^t\geq \frac{t^3}{3}$ now i have:
$$(x^2+y^2)e^{-(x+y)}\leq\frac{3(x^2+y^2)}{(x+y)}\rightarrow 0, (x,y)\rightarrow(\infty,\infty)$$
but i dint know how to continue
second method:the polar coord
$$\lim_{(x,y)\rightarrow (\infty,\infty)}(x^2+y^2)e^{-(x+y)}=\lim_{r\rightarrow \infty}(r^2)e^{-r(cos\theta+sin\theta)}=\lim_{r\rightarrow \infty}\frac{r^2}{e^{r(cos\theta+sin\theta)}}=\frac{1}{cos\theta+\sin\theta}\lim_{r\rightarrow \infty}\frac{2r}{e^r}=\frac{2}{cos\theta+\sin\theta}\lim_{r\rightarrow \infty}\frac{1}{e^r}=\frac{2}{cos\theta+\sin\theta}\cdot0=0$$ i use the L'Hopital rule,but i dint know its correct
| Aside from what I take to be a typo (the denominator under the $3(x^2+y^2)$ should be $(x+y)^3$, not just $(x+y)$), your first method looks fine. It could be elaborated to
$$(x^2+y^2)e^{-(x+y)}\le{3(x^2+y^2)\over(x+y)^3}={3x^2\over(x+y)^3}+{3y^2\over(x+y)^3}\le{3x^2\over x^3}+{3y^2\over y^3}={3\over x}+{3\over y}\to0+0=0$$
| {
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Does $\int_1^{\infty}\frac{x^2\tan^{-1}(ax)}{x^4+x^2+1}dx$ have closed form? I have been trying to find the closed form for integral below $$\int_1^{\infty}\frac{x^2\tan^{-1}(ax)}{x^4+x^2+1}dx ,\; \; a>0 $$
My progress to this integral $$\cong\frac{\pi^2}{8\sqrt 3}+\frac{\pi}{8}\log(3)-\frac{\pi}{6a\sqrt 2}+\frac{1}{2a^3}\left(\frac{\log(3)}{4} -\frac{\pi}{12\sqrt 3}\right)-\frac{1}{5a^5}\left(\frac{1}{2}-\frac{\pi}{12\sqrt 2}-\frac{\log(3)}{4}\right)+\frac{1}{7a^7}\left(\frac{\pi}{6\sqrt 3}-\frac{1}{2}\right)+\frac{1}{108a^9} +\frac{1}{9a^9}\left(\frac{\log(3)}{9}-\frac{\pi}{12\sqrt 3}\right)-\frac{1}{24a^{11}}+\cdots $$ Using the series of $\tan(ax)$ the above form is obtained. However, I dont find closed form for it. If $a\to\infty+$,then it is equal to $\frac{\pi^2}{8\sqrt 3}+\frac{\pi}{8}\log(3)$.
| We have \begin{align}\int_1^{\infty}\frac{x^2\arctan ax}{x^4+x^2+1}\,dx&=\int_0^1\frac{\arctan a/u}{u^2(1/u^4+1/u^2+1)}\frac{du}{u^2}\\&=\int_0^1\frac{\pi/2-\arctan u/a}{u^4+u^2+1}\,dx\\&=\frac\pi2\left(\frac14\log3+\frac{\pi\sqrt3}{12}\right)-\int_0^1\frac{\arctan bu}{u^4+u^2+1}\,du\end{align} using $u^4+u^2+1=(u^2+u+1)(u^2-u+1)$ and $b=1/a$. The remaining integral can be represented in terms of dilogarithms, such as in this answer, but I doubt there is a clean result with an arbitrary $b>0$.
In short, this integral does have a closed form, but is unlikely to be "simple"; e.g. case $a=7$.
(As $a\to+\infty$ the term $\int_0^1\frac{\arctan bu}{u^4+u^2+1}\,du$ has contribution $\to0$ so your observation is true.)
| {
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"source": "stackexchange",
"question_score": "3",
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Maximize $x^2y$ where $x^2+y^2=100$. (Looking for other approaches)
For positive numbers $x,y$ we have $x^2+y^2=100$. For which ratio of $x$ to $y$, the value of $x^2y$ will be maximum?
$1)\ 2\qquad\qquad2)\ \sqrt3\qquad\qquad3)\ \frac32\qquad\qquad4)\ \sqrt2$
It is a problem from a timed exam so I'm looking for alternative approaches to solve this problem quickly. Here is my approach:
We have $x^2=100-y^2$, so $x^2y=y(100-y^2)$. By differentiating with respect to $y$ and putting it equal to zero, we have:
$$(100-y^2)-2y^2=0\qquad\qquad y^2=\frac{100}3$$
Hence $x^2=\frac{200}3$ and $\frac{x^2}{y^2}=2$ therefore $\frac{x}{y}=\sqrt2$.
Is there another quick approach to solve it? (like AM-GM inequality or others)
| Alternatively, by AM-GM inequality, $x^2y = 2\sqrt{\dfrac{x^2}{2}\cdot \dfrac{x^2}{2}\cdot y^2}\le 2\sqrt{\dfrac{\left(\dfrac{x^2}{2}+\dfrac{x^2}{2}+ y^2\right)^3}{27}}= 2\cdot \sqrt{\dfrac{100^3}{27}}= \dfrac{2,000}{3\sqrt{3}}=\dfrac{2,000\sqrt{3}}{9}\implies (x^2y)_{\text{max}} = \dfrac{2,000\sqrt{3}}{9}$ at $\dfrac{x^2}{2} = y^2\implies x^2=2y^2\implies 3y^2=100\implies y = \dfrac{10\sqrt{3}}{3}, x = y\sqrt{2}=\dfrac{10\sqrt{6}}{3}\implies \boxed{\dfrac{x}{y} = {\sqrt{2}}}$.
Note: I just saw the comment by @Albus above after posting this answer. Any way, just leave it here....
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Set of primes in the spectrum of a rational number. Let $x \in \mathbb{Q}^{+}$ (numerator and denominator are in the reduced form). Can we comment anything about the set $\left\{ \left\lfloor nx \right\rfloor \; : \; n \in \mathbb{N} \right\}$? Is there any structure in this set of integers especially the possible set of prime integers?
Also, can anyone point me toward the topics in number theory which deals with such kind of problems? Any hint will be much appreciated.
| Let $m = \lfloor x \rfloor $. Let $r = x-m$ so $0\le r < 1$. $q$ is rational so let $j,k \in \mathbb N$ so that $\gcd(j,k) = 1$ and $r = \frac jk$. In other words..... Let $x = m + \frac jk$ were $m$ is integers and $\frac jk$ is a non-negative rational in reduced terms less than $1$.
$ nx = nm + nj\cdot \frac 1k = nm + q_n + \frac {a_n}k$ where $q_n = \lfloor {nj}\rfloor$ and $a_n \equiv nj \pmod k; 0 \le a < k$
So $\lfloor nx\rfloor= nm + q_n$ where $q_n = \lfloor {nj}\rfloor$.
....
Anyhow,... As $j, k$ are relatively prime $nj; n=1,...,k$ form a complete residue system $\mod k$. And $b'_n = \lfloor \frac {nj}k\rfloor;n=1,...,k$ (note $b'_1= 0$ and $b'_k = j$ always) will be a sequence of $k$ integers. We can define $b_{n+k} = b_n$ to extend this sequence to be repeating sequence of period $k$.
So $nx = nm + j\lfloor \frac nk\rfloor + b_n + \frac {a_n}k$ and $\lfloor nx\rfloor = nm + j\lfloor \frac nk\rfloor + b_n$
I gave, in the comments, and example of $x = 7\frac 35 = \frac {38}5$.
In this case $m=\lfloor \frac {38}5\rfloor = 7; r= \frac 35; j= 3; k=5; a_n \equiv 3,6,9,12,15 \pmod 5$ and $a_n = 3,1,4,2,0$ and $b_n = 0,1,1,2,3$
And $\lfloor n\cdot 7\frac 35\rfloor= 7n + 3\lfloor \frac n5\rfloor + b_n$.
Example: $\lfloor 139\cdot 7\frac 35\rfloor =$
$\lfloor 139\cdot 7 + (27\cdot 5 + 4)\frac 35 \rfloor =$
$\lfloor 139\cdot 7 + 27\cdot 3 + \frac {4\times 3}5\rfloor =$
$\lfloor 139 \cdot 7 + 27\cdot 3 + \frac {2\cdot 5 + 2}5\rfloor = $
$\lfloor 139 \cdot 7 + 27\cdot 3 + 2 + \frac 25\rfloor =$
$139\cdot 7 + 27\cdot 3 + 2$
Where $7 = \lfloor x \rfloor = m$. and $3 =$ the numerator of $\frac 35 = $ numerator of $x-m = j$ and $2=b_{4}$ (as $139\equiv 4 \pmod 5$).
| {
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"source": "stackexchange",
"question_score": "1",
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Angle between tangents to the curve $x^2+3y^2=9$
Tangents drawn from the point $(\alpha,\alpha^2)$ to the curve $x^2+3y^2=9$ include an acute angle between them, then find $\alpha$.
My attempt is by using the equation for pair of tangents from an external point to the curve. If the curve is $S=0$ then the equation for pair of tangents is $SS_1=T^2$, where $S_1=0$ is obtained after putting the external point in $S=0$. And $T=0$ is obtained after changing $x^2$ to $xx_1$ and $y^2$ to $yy_1$ in $S=0$, where $(x_1,y_1)$ is the external point from which the tangents are being drawn.
So,
$$SS_1=T^2$$
$$\implies (x^2+3y^2-9)(\alpha^2+3\alpha^4-9)=(x\alpha+3y\alpha^2-9)^2$$
Now, if $\theta$ is the angle between the tangents then $\tan\theta=|\frac{2\sqrt{h^2-ab}}{a+b}|$, where, $2h=$ coefficient of $xy=6\alpha^3, a=$ coefficient of $x^2=3\alpha^4-9$ and $b=$ coefficient of $y^2=3\alpha^2-27$
Therefore, $\tan\theta=|\frac{2\sqrt{9\alpha^6-9(\alpha^4-3)(\alpha^2-9)}}{3\alpha^4+3\alpha^2-36}|=|\frac{2\sqrt{\alpha^6-(\alpha^6-9\alpha^4-3\alpha^2+27)}}{\alpha^4+\alpha^2-12}|=|\frac{2\sqrt{9\alpha^4+3\alpha^2-27}}{\alpha^4+\alpha^2-12}|$
Not able to proceed next.
| $$y=mx+c$$ will be tangent of $$\dfrac{x^2}{3^2}+\dfrac{y^2}3=1$$ if $c^2=9m^2+3$
$$(y-mx)^2=c^2=9m^2+3$$
$$ m^2(9-x^2)+2mxy+3-y^2=0$$
As these pass through $(\alpha,\alpha^2),$
$$ m^2(9-\alpha^2)+2m\alpha^3+3-\alpha^4=0$$
which is a quadratic equation in $m$
$$m_1+m_2=-\dfrac{2\alpha^3}{9-\alpha^2}, m_1m_2=\dfrac{3-\alpha^4}{9-\alpha^2}$$
If the angle between the tangents be $p,$
$$\tan p=\left|\dfrac{m_1+m_2}{1-m_1m_2}\right|$$
| {
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"url": "https://math.stackexchange.com/questions/4127113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Possible pattern involving $x$ in the continued fraction expansion of $\frac{1}{\sqrt[3]{x^3+1}-x}$ Consider the expression $$\frac{1}{\sqrt[3]{x^3+1}-x}$$
Plugging in $10$ for $x$ and using $W|A$, we find that the continued fraction expansion is $[300; \mathbf{10}, 450, 8, ...]$.
For $x=11$, it's $[363; \mathbf{11}, 544, ...]$. The pattern continues (I have only checked up to $21$, but I assume that it's true) that the second term in the continued fraction expansion is $x$.
I have two questions:
*
*Why does this pattern occur and how do we prove that it occurs?
*Is there a similar pattern with later terms?
I haven't found an answer to 2, and I've had no luck as of yet for 1. All I have done so far is simplify it to $x\sqrt[3]{x^3+1}+\sqrt[3]{(x^3+1)^2}+x^2$.
Thanks in advance.
| Let $$\sqrt[3]{x^3+1} - x = a$$ Then $$a^3 + 3a^2x + 3ax^2 = 1$$ This shows that $a < \frac{1}{3x^2}$. Now we have everything we need to work out the continued fraction:
$$\begin{align*} \frac{1}{a} &= 3x^2 + a(3x + a)\\
&=3x^2 + \frac{1}{\frac{1}{a}\frac{1}{3x+a}}\\
&=3x^2 + \frac{1}{\frac{3x^2 + 3ax + a^2}{3x+a}}\\
&=3x^2 + \frac{1}{x + a\frac{2x + a}{3x+a}}\\
\end{align*}$$
Using the upper bound $a<\frac{1}{3x^2}$, you can show that $a\frac{2x + a}{3x+a} < 1$, which proves the relationship you observed.
You can continue this process to find more continued fraction terms. $$a\frac{2x + a}{3x+a} = \frac{1}{\frac{1}{a}\frac{3x+a}{2x+a}}$$
Again, replace $\frac{1}{a} = 3x^2 + 3ax + a^2$ and separate into term independent of $a$ and term proportionate to $a$
| {
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Summation over $a+b+c=5$ Let $a,b,c$ be positive integers. Compute
$$\sum_{a+b+c=5} (6-a)(6-b)(6-c).$$
The first thing I notice is symmetry, so that I can permute $3!=6$ ways, but i'm not really sure how that works with the condition $a+b+c=10.$ The other method is to fix $a$, but that is reall time-consuming and unfeasible if say $a+b+c=20.$ Is there a clever method to evaluate this sum?
I would like to have a generalized method, please.
| This problem is somewhat specific and nice, since when using symmetry it reduces to the summation of two terms.
We obtain
\begin{align*}
&\color{blue}{\sum_{\substack{a+b+c=5\\a,b,c\geq 1}}(6-a)(6-b)(6-c)}\\
&\qquad=6\sum_{\substack{a+b+c=5\\1\leq a<b<c\leq 3}}(6-a)(6-b)(6-b)\tag{1}\\
&\qquad\quad +3\sum_{\substack{a+b+c=5\\1\leq a=b<c\leq 3}}(6-a)^2(6-c)\tag{2}\\
&\qquad\quad +3\sum_{\substack{a+b+c=5\\1\leq a<b=c\leq 3}}(6-a)(6-c)^2\tag{3}\\
&\qquad\quad +\sum_{\substack{a+b+c=5\\1\leq a=b=c\leq 3}}(6-a)^3\tag{4}\\
&\qquad=0+3\cdot5^2\cdot3+3\cdot5\cdot4^2+0\tag{5}\\
&\qquad=9\cdot25+15\cdot16\tag{6}\\
&\,\,\color{blue}{\qquad=465}
\end{align*}
Comment:
*
*In (1) we observe that the conditions $a+b+c=5$ and $a,b,c\geq 1$ imply that each element $a,b,c$ is less or equal $3$.
We sum up all triples $(a,b,c)$ whereby each two elements are pairwise different. This can be done in $3!=6$ ways. After appropriately renaming the variables we can consider therefore $6$ ordered triples $(a,b,c)$ with $1\leq a<b<c\leq 3$.
*In (2) we sum up all triples $(a,b,c)$ whereby $a$ is equal to $b$ and different to $c$. This can be done in $\frac{3!}{2!1!}=3$ different ways.
*In (3) we have a similiar situation as in (2). Here we consider the constellation $a<b=c$ which can be done again in $\frac{3!}{2!1!}=3$ different ways.
*In (4) we cover the final case $a=b=c$ which can be done in $\frac{3!}{3!}=1$ different way.
*In (5) we observe
*
*there is no admissible triple $(a,b,c)$ with $1\leq a<b<c\leq 3$ and $a+b+c=5$.
*there is just one admissible triple with $1\leq a=b<c\leq 3$ and $a+b+c=5$, namely $\color{blue}{(a,b,c)=(1,1,3)}$.
*there is just one admissible triple with $1\leq a<b\leq c\leq 3$ and $a+b+c=5$, namely $\color{blue}{(a,b,c)=(1,2,2)}$.
*there is no admissible triple $(a,b,c)$ with $1\leq a=b=c\leq 3$ and $a+b+c=5$.
Conclusion: In (6) we note that thanks to symmetry only two admissible triples $\color{blue}{(a,b,c)\in\{(1,1,3),(1,2,2)\}}$ provide a contribution to the result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Problem with integration by substitution.
Using the substitution $u=1-x$, compute the integral of
$\int{x(1-x)^2}dx.$
My Work: Let $u=1-x.$ Then $\mathrm dx=-\mathrm du$ and $x=1-u,$ so
$$\int{x(1-x)^2}\,\mathrm dx\\
=-\int{(1-u)u^2}\,\mathrm du\\
=\frac{u^4}{4}-\frac{u^3}{3}+C\\
=\frac{(1-x)^4}{4}-\frac{(1-x)^3}{3}+C\\
=\frac{3x^4-8x^3+6x^2-1}{12}+C.$$
However, the answer gotten by expanding the brackets first then integrating does not seem to match that of the answer above arrived by substitution:
$$\int{x(1-x)^2}\,\mathrm dx\\
=\int{x-2x^2+x^3}\,\mathrm dx\\
=\frac{x^2}{2}-\frac{2x^3}{3}+\frac{x^4}{4}+C.$$
I cannot account for the $-\frac{1}{12}$ from the method using substitution.
| Well
\begin{align}
\int{x(1-x)^2}dx &= \int x^3-2x^2+x \mathrm{d}x \\
&=\frac{1}{4}x^4-\frac{2}{3}x^3+\frac{x^2}{2} +C_1
\end{align}
Whereas with $u=1-x \, \implies \mathrm{d}u=-\mathrm{d}x$ and
\begin{align}
\int (1+u)^2u (- \mathrm{d}u) &= -\int u^3+2u^2+u \mathrm{d}u \\
&=-\frac{u^4}{4}-\frac{2}{3}u^3-\frac{u^2}{2}+C_2
\end{align}
Now substitute in $x=1-u$ to see equivalence.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given 2d vectors $A$, $B$, $C$, $D$, $P$, find $x$ such that the line through $Ax+B(1-x)$ and $Cx+D(1-x)$ passes through $P$ I'm trying to parametrize the quad with an x and y coordinate (from 0 to 1) so that I can properly apply a texture to it for a computer graphics application. I came up with the idea for this myself, but am having a hard time getting to a solution.
Given the 2d vectors $A$, $B$, $C$, $D$, $P$,
construct points
$R$ = $A$x+$B$(1-x) and $Q$ = $C$x+$D$(1-x)
then $P$ = $R$y+$Q$(1-y)
Check my explanation image
substitute R and Q so
$P$ = ($A$x+$B$(1-x))y + ($C$x+$D$(1-x))(1-y)
since coords are 2d, this is actually two equations (lowercase=x uppercase=y)
$p$ = ($a$x+$b$(1-x))y + ($c$x+$d$(1-x))(1-y)
$P$ = ($A$x+$B$(1-x))y + ($C$x+$D$(1-x))(1-y)
take first equation and solve for y
$y = \frac{-cx+d(x-1)+p}{x(a-c)+b(1-x)+d(x-1)}$
substitute y in second equation
$P = (Ax+B(1-x))\frac{-cx+d(x-1)+p}{(x(a-c)+b(1-x)+d(x-1)}+(Cx+D(1-x))(1-(\frac{-cx+d(x-1)+p}{x(a-c)+b(1-x)+d(x-1)}))$
solve for x? I think that would be it, I'm just stuck on how to do this. Its too large and unwieldy. Is there another way?
| Give all vectors a third coordinate of $0$ so that, as members of $\Bbb{R}^3$, they're compatible with cross products. Also, translate all points by $-P$ to move $P$ to the origin. Then, for $P$, $Q$, $R$ to be collinear, "all we have to do" is force
$$R\times Q = O \tag{0}$$
where $O:=(0,0,0)$. To simplify some calculations, let's re-parameterize relative to the midpoints of $\overline{AB}$ and $\overline{CD}$ by defining $s := 2x-1$. Then we have
$$R = \frac12(A+B)+ \frac12(A-B)s \qquad\qquad Q = \frac12(C+D)+\frac12(C-D)s$$
and $(0)$ becomes (after multiplying-through by $4$),
$$s^2 ((A-B)\times(C-D)) \;+\; 2 s (A\times C-B\times D) \;+\;((A+B)\times(C+D)) \;=\; O \tag1$$
Since $A$, $B$, $C$, $D$ are all in $\Bbb{R}^2$, the first two components of any cross product are $0$. Thus, $(1)$ amounts to a quadratic in $s$ whose coefficients are the third components of those cross-products. Solving via the Quadratic Formula, we find
$$s = -\frac{(A\times C -B\times D)_3\pm \sqrt{\Delta}}{((A-B)\times(C-D))_3} \qquad\qquad\left(\;x = \frac12(1+s)\;\right)\tag2$$
where
$$\begin{align}
\Delta &:=(A\times C - B\times D)_3^2 - ((A-B)\times(C-D))_3 ((A+B)\times(C+D))_3 \\[4pt]
&\;=(A\times D)_3^2 + (B\times C)_3^2 - 2 \left(\, (A\times B)_3\,(C\times D)_3 + (A\times C)_3\,(B\times D)_3 \,\right)
\end{align}$$
(The second form will show invariance under a substitution below.)
Now, to find $y$ first observe that
$$\begin{align}
Ry+Q(1-y) &= (Ax+B(1-x)) y + (C x+D(1-x))(1-y) \\
&=(Ay + C(1-y))x+ (B y+D(1-y))(1-x) \\
&=R' x + Q'(1-x)
\end{align}$$
where $R' := Ay+C(1-y)$ and $Q':=By+D(1-y)$. Thus, we can find $y$ by the process above, which gives a counterpart of $(2)$ with the substitutions $A\to A$, $B\leftrightarrow C$, $D\to D$ (leaving $\Delta$ unchanged, as promised):
$$t = -\frac{(A\times B -C\times D)_3\mp \sqrt{\Delta}}{((A-C)\times(B-D))_3} \qquad\qquad\left(\;y = \frac12(1+t)\;\right)\tag3$$
Note that the sign on $\sqrt{\Delta}$ flips from its state in $(2)$ to get the proper correspondence between the solutions for $s$ and $t$. $\square$
For an example, consider
$$A = (1,2,0) \qquad B = (3,4,0) \qquad C = (-1,-1,0) \qquad D = (-2,-3,0) \qquad P=(0,0,0)$$
Then, writing $\lambda$ for $\pm 1$, we (and by "we", I mean "Mathematica") can calculate
$$\Delta = 8 \qquad s = 1+\lambda\sqrt{2} \qquad t = -3-2\lambda\sqrt{2} $$
$$x = \frac12 (2 +\lambda \sqrt{2}) \qquad y = -1-\lambda\sqrt{2}$$
so that
$$\begin{align}
R &= Ax+B(1-x) = (1 - \lambda \sqrt{2}, 2 - \lambda \sqrt{2}, 0) \\
Q &= Cx+D(1-x) = \left(\frac12 (-2 + \lambda \sqrt{2}), -1 + \lambda \sqrt{2}, 0\right)
\end{align}$$
and
$$Ry+Q(1-y) = (3 (\lambda^2-1), 4 (\lambda^2-1), 0) = (0,0,0) = P$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4144982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $\left\lfloor \frac{1}{1^{\frac{2}{3}}}+\frac{1}{2^{\frac{2}{3}}}+\frac{1}{3^{\frac{2}{3}}}+...+\frac{1}{1000^{\frac{2}{3}}}\right\rfloor$ Let $$M=\left\lfloor \frac{1}{1^{\frac{2}{3}}}+\frac{1}{2^{\frac{2}{3}}}+\frac{1}{3^{\frac{2}{3}}}+...+\frac{1}{1000^{\frac{2}{3}}}\right\rfloor$$.
Find $M$
My Attempt
Let $f(x)=x^{-\frac{2}{3}}$
Since $f''(x)>0$
we have
$\sum_{r=1}^{999}r^{-\frac{2}{3}}>\int_{1}^{999}x^{-\frac{2}{3}}dx$
Also, $\sum_{r=2}^{999}r^{-\frac{2}{3}}<\int_{1}^{1000}x^{-\frac{2}{3}}dx$
I am stuck here. Don't know how to proceed
| Let $$S = \sum_{r=1}^{1000} r^{-2/3}.$$ Then because for each positive integer $r$, $$r^{-2/3} = \int_{x=r}^{r+1} \lfloor x \rfloor^{-2/3} \, dx > \int_{x=r}^{r+1} x^{-2/3} \, dx,$$ it follows that
$$S > \sum_{r=1}^{1000} \int_{x=r}^{r+1} x^{-2/3} \, dx = \int_{x=1}^{1001} x^{-2/3} \, dx = 3(1001^{1/3} - 1) > 3(1000^{1/3} - 1) = 27.$$
Conversely,
$$r^{-2/3} = \int_{x=r-1}^r \lceil x \rceil^{-2/3} \, dx < \int_{x=r-1}^r x^{-2/3} \, dx,$$ hence $$S = 1 + \sum_{r=2}^{1000} r^{-2/3} < 1 + \int_{x=1}^{1000} x^{-2/3} \, dx = 1 + 27 = 28.$$ Note that we must "peel off" the first term in the summation because the upper bound $$1 < \int_{x=0}^1 x^{-2/3} \, dx = 3$$ is too loose. This establishes $$27 < S < 28,$$ hence $M = \lfloor S \rfloor = 27.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Convolution of two independent uniform variables Question: Let's say $X$ and $Y$ are two independent random variables with $Uniform (0,3)$ distribution. What is the probability density function of $X+Y$?
In solution, it is told to use convolution method by many sources since the pdf of the sum of two independent random variables $(X+Y)$ is the convolution of the individual pdfs.
However, I couldn't solve this problem by looking at $Uniform (0,1)$ version solutions of this question. What I've done by now is;
$fx+y(a) = ∫_0^3 fx(a-y)*fy(y)dy$
$= 1/3 *∫_0^3 fx(a-y)dy$
If what I've found until here is true (I have doubts about upper and lower limits of the integral), how should I determine the limits of pdf of X+Y? Some sources assume $a=.5$ and $a=1.5$ and solve it that way; while other sources assume $0<a<1$ and $1<a<2$ and solve it that way. In both solutions, they don't explain how they find upper and lower limit of the integral. Fully solve the problem given above to clearly explain.
| Let $Z = X + Y$, then probability distributionn function of $Z$ is defined as
$$
\begin{aligned}
F_Z(z) &= \text{Pr}(Z \leq z) = \text{Pr}(X + Y \leq z) = \text{Pr}(X \leq z - Y) = \int\limits_{-\infty}^{+\infty}F_X(z-y)f_Y(y)dy = \\
&= \frac{1}{3}\int\limits_{0}^{3}F_X(z-y)dy =
\left|
\begin{array}{l}
u = z-y \Leftrightarrow du = -dy \\
y = 0 \Leftrightarrow u = z \\
y = 3 \Leftrightarrow u = z-3
\end{array}
\right| = \frac{1}{3}\int\limits_{z-3}^zF_X(u)du = \\
& = \left\{
\begin{array}{rl}
0, & z \leq 0 \\
\frac{1}{3}\int\limits_{0}^z\frac{u}{3}du = \frac{z^2}{18}, & 0 < z \leq 3 \\
\frac{1}{3}\int\limits_{z-3}^3\frac{u}{3}du + \frac{1}{3}\int\limits_{3}^zdu=
\frac{1}{2}-\frac{(z-3)^2}{18} + \frac{z-3}{3} = -\frac{z^2-12z+18}{18}, & 3 < z \leq 6 \\
1, & z > 6
\end{array}
\right.
\end{aligned}
$$
So, we have that
$$
F_Z(z)=\left\{
\begin{array}{rl}
0, & z \leq 0 \\
\frac{z^2}{18}, & 0 < z \leq 3 \\
-\frac{z^2-12z+18}{18}, & 3 < z \leq 6 \\
1, & z > 6
\end{array}
\right. \Rightarrow f_Z(z) = F_Z'(z) =
\left\{\begin{array}{rl}
0, & z \leq 0 \\
\frac{z}{9}, & 0 < z \leq 3 \\
-\frac{z-6}{9}, & 3 < z \leq 6 \\
0, & z > 6
\end{array}
\right.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4146514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What method is required to find out "for what values of k is $4x^2+kx+\frac14$ a perfect square?"
For what values of $k$ is this expression a perfect square?
$$4x^2+kx+\frac14$$
I'm thinking maybe the perfect square trinomial is supposed to be used, as I had to use that formula on the problem prior to this one. However, I don't know how to apply it, since I don't know which of the above terms would be $a^2$, $2ab$ and $b^2$. How can one deduce what term corresponds to what term in the general formula? Or is this not the formula to use at all?
EDIT:
User Exodd showed a general formula that when I followed, gave the same answer as user Jitendra Singh, who answered this question with a different formula.
The general formula is this:
$$(ax+b)^2 = a^2x^2+2abx+b^2$$
Which is a variation on the simpler perfect square trinomial
$$(a+b)^2=a^2+2ab+b^2$$
Now, if one assumes that $4x^2$ is $a^2x^2$, $kx$ is $2abx$ and $\frac14$ is $b^2$, then the calculation is quite simple. $k$ is then equal to $2ab$, which is equal to $\pm 2$, the sign depending on whether the $4$, or $a^2$, is the product of $2 \times 2$ or $-2 \times -2$, which creates:
$$2 \times 2 \times \frac12 = 2$$
or
$$2 \times -2 \times \frac12 = -2$$
However, this is assuming the aforementioned correspondence. It could be that $4x^2$ is $a^2x^2$, $kx$ is $b^2$ and $\frac14$ is $2abx$. In that case:
$$\frac14 = 2 \times 2 \times \sqrt{kx} \times x$$
or
$$\frac14 = 2 \times -2 \times \sqrt{kx} \times x$$
Furthermore, there could be a different correspondence. What if $4x^2$ is $b^2$, $kx$ is $a^2x^2$ and $\frac14$ is $2abx$? There are other correspondences as well. What says that the first one, that gave the answer of $\pm2$, is correct?
| We consider a polynomial in $x$
\begin{align*}
p(x)=4x^2+kx+\frac{1}{4}
\end{align*}
which has degree $2$, constant term $\frac{1}{4}$, linear term $kx$ and square term $4x^2$.
We are looking for a representation as perfect square
\begin{align*}
q(x)=(ax+b)^2
\end{align*}
We recall two polynomials are equal, i.e. $p=q$ if they have the same degree and the coefficients of like powers of $x$ are equal.
\begin{align*}
p(x)&=q(x)\\
4x^2+kx+\frac{1}{4}&=a^2x^2+2abx+b^2\tag{1}
\end{align*}
From (1) we see $\deg p = \deg q$ and we obtain by comparison of coefficients
\begin{align*}
4&=a^2\\
k&=2ab\\
\frac{1}{4}&=b^2
\end{align*}
Note: The equality of polynomials is based upon an identity theorem of polynomials which can be found for instance in this paper as theorem 4 and which is nicely answered by @IvesDaoust in this MSE post.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Solve the inequality $\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$ $$\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2}$$
The answer for this inequality is given as $x ∈ (- \infty , -1) \cup(0, 1) \cup (2, \infty)$ but when I solve it, I am only able to get $x ∈ ( - \infty, -1)$.
How should I solve it to get the complete answer?
| Solve the Inequality :$$\frac 1{x+1} - \frac 1{x} \le \frac 1{x-1} - \frac 1{x-2} \\$$
$$\frac{-1}{x(x+1)} \le \frac{-1}{(x-2)(x-1)} \\$$
Consider two scenarios :
scenario 1 : Less than
$$\frac{-1}{x(x+1)}~< \frac{-1}{(x-2)(x-1) }$$.
$$\frac{-1~(x^2-x-2x+2)}{x(x+1)}+1~<0$$
$$\frac{(4~x~-2)}{x~(~x+1)}~<0$$
$$x \in (-\infty , \frac{1}{2}) $$
Scenario 2 : equal to
$$\frac{-1}{x(x+1)}~=\frac{-1}{(x-2)(x-1) }$$.
$$\frac{-1~(x^2-x-2x+2)}{x(x+1)}+1~=0$$
$$\frac{(4~x~-2)}{x~(~x+1)}~=0$$
$$ x = \frac{1}{2}$$
Final step : $$x \in (-\infty , \frac{1}{2}] $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Given the following set of equations, find: $xy + 2yz + 3zx$.
The positive reals x, y, z satisfy the equations
$$x^2 + xy + \frac{y^2}{3}=25$$
$$\frac{y^2}{3}+z^2 = 9$$
$$z^2+ zx + x^2 = 16$$
Find $$xy + 2yz + 3zx$$
My understanding:
*
*What struck me first were the squares $9, 16, 25$. This is the “Egyptian triangle.” It is a hint to the theorem of Pythagoras, to geometry, and geometrical
interpretation.
*Instead of $x, y, z$ only $xy + 2yz + 3zx$ is required. This may be an area,
maybe even the area $6$ of the Egyptian triangle. It is also a hint that I
should not try to find $x, y, z$.
*$\frac{y^2}{3}$ occurs twice, so it may be helpful to set $a^2=\frac{y^2}{3}$.
The equations finally become:
$$x^2+ \sqrt{3}xt + t^2 = 25$$
$$t^2+z^2=9$$
$$z^2+ zx + x^2 = 16$$
Also, one more observation, $zx$ must be a perfect square.
I am not sure if I'm on the right track, but I don't know how to proceed further. Hints and help would be appreciated, thanks.
| Adding the second and third equation and subtracting first equation , we get
$2z^2+xz=yx$
Substitute the value for $y$ in the first equation, we get
$$3x^4+7x^2z^2+3x^3z+4z^4+4z^3x=75x^2$$
which easily gets factored into
$$(x^2+z^2+xz)(3x^2+4z^2)=75x^2$$
$$(16)(3x^2+4z^2)=75x^2$$
One possible way to avoid all the algebraic mess is that on solving the above equation we get,
$z^2=\frac{27}{64}x^2$ you could take $a^2=\frac{27}{64}$. This simplifies the calculations a lot since $z=ax$ and $y=2a^2x+ax$
On subsitituting in second equation we get $27=4x^2(a^2+a^3+a^4)$
And we have to find $4x^2(a+a^2+a^3)=\frac{27}{a}$.
And we are done. Not the cleanest approach but it is a way if nothing else seems obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Reverse the order of integration of $\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) \,dx\,dy$. Q: Reverse the order of integration of $\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) \, dx\,dy$.
I imagined that $x$ should varie between $\dfrac{1}{9}$ and $\sqrt{\dfrac{2}{3}}$ and $y$ varie between $x^2$ and $\sqrt x$, then we get:
$$\int_{\frac{1}{3}}^\frac{2}{3} \int_{y^2}^{\sqrt y} f(x,y) \,dx\,dy = \int_{\frac{1}{9}}^{\sqrt{\frac{2}{3}}} \int_{x^2}^{\sqrt x} f(x,y) \, dy \, dx$$
But that is not the answer cuz i tested it with some $f(x,y)$ and got different values on the integral.
Can somebody help me? Thanks!
| Three piece decomposition $\int\limits_\frac{1}{9}^\frac{4}{9}\int\limits_\frac{1}{3}^\sqrt{x}+\int\limits_\frac{4}{9}^\sqrt{\frac{1}{3}}\int\limits_\frac{1}{3}^\frac{2}{3}+\int\limits_\sqrt{\frac{1}{3}}^\sqrt{\frac{2}{3}}\int\limits_{x^2}^\frac{2}{3}f(x,y)dydx$
Draw a picture. You will see it immediately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given a unitary vector $(x_1,\ldots,x_n)^T$ and a vector $y=(x_2,\ldots, x_n)^T$, prove the columns of a created matrix form an orthonormal basis
Let $X=\begin{bmatrix}x_1\\x_2\\\vdots\\x_n\end{bmatrix}\in M_{n1}(\Bbb F),x_1\ne-1$ be a unit vector wrt the standard scalar product and let $Y=\begin{bmatrix}x_2\\x_3\\\vdots\\x_n\end{bmatrix}\in M_{n-1,1}(\Bbb F).$
Prove that the columns of the matrix $$M=\begin{bmatrix}x_1&y^*\\y&\frac1{1+x_1}yy^*-I\end{bmatrix}$$ form an orthonormal basis for $M_{n1}(\Bbb F)$
Note: $y^*$ denotes the conjugate transpose of $y$.
This is what I tried:
I would like to prove the given
matrix $M$ is unitary by showing $MM^*=I$.
First, I wanted to analyse the Hermitian matrix $$yy^*=\begin{bmatrix}x_2\bar x_2&x_2\bar x_3&\ldots&x_2\bar x_n\\x_3\bar x_2&x_3\bar x_3&\ldots &x_3\bar x_n\\\vdots&\vdots&\ddots&\vdots\\x_n\bar x_2&x_n\bar x_3&\ldots&x_n\bar x_n\end{bmatrix}$$
We can see that $\operatorname{rank}(yy^*)=1\implies\operatorname{Ker}(yy^*)=n-1$.
Furthermore, $$yy^*y=y(y^*y)=\|y\|^2y,$$
which means $y$ is an eigenvector of the matrix $yy^*$ corresponding to the eigenvalue $\|y\|^2=1-|x_1|^2=1-x_1\bar x_1$ since $\|x\|=\sqrt{\sum\limits_{i=1}^n|x_i|^2}=1$, and we conclude $\sigma(yy^*)=\{0,1-x_1\bar x_1\}$.
If $x_1\in\Bbb R, M=M^*,$ but it can happen that $x_1\not\in\Bbb R$.
I tried writing: $$\left(\frac1{1+x_1}yy^*-I\right)y=\begin{bmatrix}\left(\frac{1-x_1\bar x_1}{1+x_1}-1\right)x_2\\\left(\frac{1-x_1\bar x_1}{1+x_1}-1\right)x_3\\\vdots\\\left(\frac{1-x_1\bar x_1}{1+x_1}-1\right)x_n\end{bmatrix}=\begin{bmatrix}-\frac{1+\bar x_1}{1+x_1}x_1x_2\\-\frac{1+\bar x_1}{1+x_1}x_1x_3\\\vdots\\-\frac{1+\bar x_1}{1+x_1}x_1x_n\end{bmatrix}$$
but it wasn't quite helpful.
How should I proceede?
Thank you very much for your time and patience!
| As given, $\ M\ $ is not always unitary. Take
$$
X=\begin{bmatrix}\frac{i}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}
\end{bmatrix}\ .
$$
then
$$
M=\begin{bmatrix}\frac{i}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}&\frac{1}{2+i\sqrt{2}}-1
\end{bmatrix}=\begin{bmatrix}\frac{i}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}&-\frac{1+i\sqrt{2}}{2+i\sqrt{2}}
\end{bmatrix}
$$
\begin{align}
MM^*&=\begin{bmatrix}\frac{i}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}&-\frac{1+i\sqrt{2}}{2+i\sqrt{2}}
\end{bmatrix}\begin{bmatrix}-\frac{i}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\
\frac{1}{\sqrt{2}}&\frac{i\sqrt{2}-1}{2-i\sqrt{2}}
\end{bmatrix}\\
&=\begin{bmatrix}1&\frac{i}{2}+\frac{i\sqrt{2}-1}{2\sqrt{2}-2i}\\
\frac{-i}{2}-\frac{1+i\sqrt{2}}{2\sqrt{2}+2i}&1
\end{bmatrix}\\
&=\begin{bmatrix}
1&\frac{2\sqrt{2}i}{2\sqrt{2}-2i}\\
\frac{-2\sqrt{2}i}{2\sqrt{2}+2i}&1
\end{bmatrix}\\
&=\begin{bmatrix}
1&\frac{-\sqrt{2}+2i}{3}\\
\frac{-\sqrt{2}-2i}{3}&1
\end{bmatrix}\ .
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/4154990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Q is to prove that integer just above ($\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}$ for all n belongs to natural numbers. Q is to prove that integer just above($\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}$ for all n belongs to natural numbers.
In Q , by integer just above means that:
For an example , which is the integer just above 7.3 . It is 8. Then , Q wants you to prove that 8 is divisible by $2^{n+1}$.
\begin{equation}
\begin{array}{l}
(\sqrt{3}+1)^{2 n}=(4+2 \sqrt{3})^{n}=2^{n}(2+\sqrt{3})^{n}\\
=2^{n}(2+\sqrt{3})^{n}=2^{n}\left[^n C _{0}2^{n}+^n C_{1} 2^{n-1} \sqrt{3}+^n C_{{2}} 2^{n-2} \sqrt{3}^{2}+\right.\\
\begin{array}{l}
(\sqrt{3}-1)^{2 n}=(4-2 \sqrt{3})^{n}=2^{n}(2-\sqrt{3})^{n} \\
=2^{n}(2-\sqrt{3})^{n}=2^{n}\left[{ }^{n} c_{0} 2^{n}-n_{C}, 2^{n-1} \sqrt{3}+{ }^{n} c_{2} 2^{n-1} \sqrt{1}^{2} \ldots\right.
\end{array}\\
I+f+f)=2^{n}\left[2(\text { Integer) }]=2^{n+1}\right. \text { . Integer }\\
I+1=2^{n+1} \text { . Integer }
\end{array}
\end{equation}
In the image is the way this question is solved.
My Q from this method of solving is that if we notice at the end , we somehow got $ 2^{n+1}$. If the Q has taken some other value like $3^{n+3}$ or Something else. Then , it was not possible to prove this question.
What is another method to prove this Q or can you help me justify that the above method can be used for all kinds of Q.
Thank you.
| Note that $0\lt \sqrt 3 -1 \lt 1$
Now look at the numbers $a=1+\sqrt 3, b=1-\sqrt 3$ with $a+b=2, ab=-2$ which are roots of the quadratic $x^2-2x-2=0$
Then with $v_n=a^n+b^n$ we have $v_{n+2}-2v_{n+1}-2v_n=0$ (spot the coefficients) or $v_{n+2}=2(v_{n+1}+v_n)$ now $|b^n|\lt 1$ and indeed $0\lt b^{2n}\lt 1$ so the difference $v_{2m}-a^{2m}=b^{2m}$ is small and positive.
It will be found that $v_n$ is an integer close to $a^n$ for all $n$, and the sign of the difference for even indices is right, and you can run an induction to complete the proof.
If you understand this you will be able to do the same trick with $A=a^2$ and $B=b^2$ to get a rather simpler induction.
For these kinds of questions with powers of an irrational number "miraculously close" to integers, there is very often a recurrence to be found lurking in the background.
If $a$ and $b$ are the roots of $x^2-px+q$=0 and we put $u_n=Ca^n+Db^n$ we can reason as follows:
$$a^2-pa+q=0$$ because $a$ is a root. Multiply through bay $a^n$ to give $$a^{n+2}-pa^{n+1}+qa^n=0$$
Now multiply through by the constant $C$ $$Ca^{n+2}-Cpa^{n+1}+Cqa^n=0$$
Similarly with the other root $b$ and the constant $D$ $$Db^{n+2}-Dpb^{n+1}+Dqb^n=0 $$
Now add these last equations to obtain $$Ca^{n+2}+Db^{n+2}-p\left(Ca^{n+1}+Db^{n+1}\right)+q\left(Ca^n+Db^n\right) = 0 = u_{n+2}-pu_{n+1}+qu_n$$
And this works for any constants $C, D$ and hence for $C=D=1$.
We can also argue that $p=a+b$ and $q=ab$ so that
$$u_{n+2}-pu_{n+1}+qu_n=$$$$=Ca^{n+2}+Db^{n+2}-(a+b)Ca^{n+1}-(a+b)Db^{n+1}+abCa^n+abDb^n=0$$ because the terms all cancel.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rate of change of radius of spherical balloon
A spherical balloon is filled with gas at a rate of $4 \text{ cm}^3/\text{s}$. What rate is the radius $r$ changing with respect to the time when the vol $V=36π \text{ cm}^3$?
[ans:$\frac 19, \frac{\pi}{9} $or 9$\pi$]
I tried this one with implicit differentiation:
$\frac{dV}{dt} = \frac {4}{3} \pi \cdot \frac{d}{dt}[r^2]$
*
*$4 = \frac {4}{3} \pi \cdot 2r \frac{dr}{dt}$
Volume formula: $36 π = \frac43 π r^2$
*
*$r^2= 27$
*$r= \pm 3 \sqrt{3}$
Substituting r into the differentiation to find $\frac{dr}{dt}$:
$4 = \frac {4}{3} \pi \cdot 2r \frac{dr}{dt}$
$4 = \frac {4}{3} \pi \cdot 2(3 \sqrt{3}) \frac{dr}{dt}$
$\frac{dr}{dt} = \frac{1}{2\sqrt{3}\pi}$ , which seems wrong. Please can someone tell me where my knowledge gap is?
| Just as another way to approach the calculation, since the volume of the balloon is given, rather than its radius:
You had the result from implicit differentiation,
$$ \frac{dV}{dt} \ \ = \ \ \frac43 \pi \cdot \frac{d}{dt} \ [r^3] \ \ = \ \ \frac43 \pi \cdot 3r^2 · \frac{dr}{dt} \ \ = \ \ 4 \pi r^2 · \frac{dr}{dt} \ \ . $$
We might wish to have this "related-rates" equation expressed in terms of the current volume of the balloon. To do this, it will be helpful to find a constant $ \ c \ $ for which
$$ \ c·(4 \pi r^2)^{3/2} \ = \ V \ = \ \frac43 \pi r^3 \ \ \Rightarrow \ \ c · 4^{3/2} · \pi^{3/2} · r^3 \ \ = \ \ \frac43 \pi r^3 $$
$$ \Rightarrow \ \ c \ \ = \ \ \frac43 \pi · \frac{1}{4^{3/2} · \pi^{3/2}} \ \ = \ \ \frac{4 · \pi}{3 · 4^{3/2} · \pi^{3/2}} \ \ = \ \ \frac{1}{3 · 4^{1/2} · \pi^{1/2}} \ \ = \ \ \frac{1}{6 · \sqrt{\pi}} $$
$$ \Rightarrow \ \ \left(\frac{1}{6 · \sqrt{\pi}} \right)·(4 \pi r^2)^{3/2} \ = \ V \ \ \Rightarrow \ \ 4 \pi r^2 \ \ = \ \ (6 · \pi^{1/2})^{2/3}· V^{2/3} \ \ = \ \ 6^{2/3} · \pi^{1/3} · V^{2/3} \ \ . $$
The related-rates equation becomes
$$ \frac{dV}{dt} \ \ = \ \ 6^{2/3} · \pi^{1/3} · V^{2/3} · \frac{dr}{dt} \ \ , $$
and inserting the given quantities leads us to
$$ 4 \ \ = \ \ 6^{2/3} · \pi^{1/3} · (36 \pi)^{2/3} · \frac{dr}{dt} \ \ \Rightarrow \ \ 4 \ \ = \ \ 6^{2/3} · \pi^{1/3} · (6^2)^{2/3} · \pi^{2/3} · \frac{dr}{dt} \ \ $$
$$ \Rightarrow \ \ 4 \ \ = \ \ 6^{2/3} · (6)^{4/3} · \pi^{1/3} · \pi^{2/3} · \frac{dr}{dt} \ \ \Rightarrow \ \ 4 \ \ = \ \ 6^2 · \pi · \frac{dr}{dt} \ \ \Rightarrow \ \ \frac{dr}{dt} \ \ = \ \ \frac{4}{36·\pi} \ \ = \ \ \frac{1}{9·\pi} \ \ . $$
[In calculus classes, instructors are generally not concerned about units, but they matter very much to scientists and engineers. Going back to the earlier equations we used to find the "proportionality constant", we have
$$ \frac{dV}{dt} \ \ = \ \ 4 \pi r^2 · \frac{dr}{dt} \ \ \Rightarrow \ \ \frac{dV/dt}{dr/dt} \ \ = \ \ 4 \pi r^2 \ \frac{\text{cm}^3\text{/sec}}{\text{cm/sec}} \ \ = \ \ 4 \pi r^2 \ \text{cm}^2 $$
$$ \ c· (4 \pi r^2\ \text{cm}^2)^{3/2} \ \ = \ \ V \ \text{cm}^3 \ \ \Rightarrow \ \ c \ \ = \ \ \frac{1}{6 \sqrt{\pi}} \ \frac{\text{cm}^3}{\text{cm}^3} \ \text{ or no units} \ \ $$
$$ 4 \ \frac{\text{cm}^3}{\text{sec}} \ \ = \ \ (6 \sqrt{\pi} \ \text{no units})^{2/3} · \left(36 \pi \ \text{cm}^3 \right)^{2/3} · \frac{dr}{dt} \ \ \Rightarrow \ \ \frac{dr}{dt} \ \ = \ \ \frac{4 \ \frac{\text{cm}^3}{\text{sec}}}{36 \pi \ \text{cm}^2} \ \ = \ \ \frac{1}{9 \pi} \ \frac{\text{cm}}{\text{sec}} \ \ . $$
Since $ \ \frac{dr}{dt} \ $ could be thought of as "differential length divided by differential time", we would expect the units to be in cm./sec. ]
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"Find the volume of the ellipsoid $x^2+\frac{y^2}{100}+\frac{z^2}{4} = 1$" was the question and here's my attempt. So, I asked this question and got downvotes and someone saying I should present my attempt. Here lies my thought process:
I know the volume is $\iiint dxdydz$ on any W$\subset R^3$ and I "just" had to find the proper values for the W, namely x, y, z, their range.
I said $-1≤x≤1$, $-10≤y≤10$ and $-2≤z≤2$.
Expressing them on the integral, for every variable we set the next one to 0 and the previous as constant so it's:
$-1≤x≤1$
$-10 \sqrt{1-x^2} ≤y≤10 \sqrt{1-x^2}$
$-2 \sqrt{1-x^2-\frac{y^2}{100}} ≤ z ≤ 2 \sqrt{1-x^2-\frac{y^2}{100}}$
and it actually went from $dxdydz$ to $dzdydz$
Any thoughts? Thanks in advance.
| What you did is correct. Translating it into an integral in Cartesian coordinates, it becomes$$\int_{-1}^1\int_{-10\sqrt{1-x^2}}^{10\sqrt{1-x^2}}\int_{-2\sqrt{1-x^2-\frac{y^2}{100}}}^{2\sqrt{1-x^2-\frac{y^2}{100}}}1\,\mathrm dz\,\mathrm dy\,\mathrm dx.$$You can compute it as follows:\begin{align}\int_{-1}^1\int_{-10\sqrt{1-x^2}}^{10\sqrt{1-x^2}}\int_{-2\sqrt{1-x^2-\frac{y^2}{100}}}^{2\sqrt{1-x^2-\frac{y^2}{100}}}1\,\mathrm dz\,\mathrm dy\,\mathrm dx&=\int_{-1}^1\int_{-10\sqrt{1-x^2}}^{10\sqrt{1-x^2}}4\sqrt{1-x^2-\frac{y^2}{100}}\,\mathrm dy\,\mathrm dx\\&=\int_{-1}^120\pi(1-x^2)\,\mathrm dx\\&=\frac{80}3\pi.\end{align}However, the computation becomes much easier if you do the change of coordinates$$\left\{\begin{array}{l}x=X\\y=10Y\\z=2Z\end{array}\right.$$Then your integral becomes$$20\int_{-1}^1\int_{-\sqrt{1-^2}}^{\sqrt{1-X^2}}\int_{-\sqrt{1-X^2-Y^2}}^{\sqrt{1-X^2-Y^2}}1\,\mathrm dZ\,\mathrm dY\,\mathrm dX,\tag1$$since, other than the factor $20$, $(1)$ is the volume of the unit sphere, which is $\frac43\pi$. So you get, again, that the answer is $\frac{80}3\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170326",
"timestamp": "2023-03-29T00:00:00",
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How can you quickly use the graph of $3x-6y=4$ to produce the graph of $3x-6y=10$? To start with, I am asked to graph the lines $3x-2y=k$ for $k=1,2,3$ and $4$.
I find that the equations produce different parallel lines.
For each line where $k=2,3$ and $4$, the coordinates points are $0.5k$ below the coordinates points of the line $3x-2y=1$.
If we let $x=0$, we can find the $y$-intercept of the lines.
\begin{align*}
-2y&=k\\
y &=\frac{k}{-2}
\end{align*}
and if we increase $k$ by $1$ we decrease $y$ by $0.5$.
\begin{align*}
-2y&=k+1\\
y &=\frac{k}{-2}-\frac{1}{2}
\end{align*}
I'm not certain the above logic suffices in quickly using the graph of $3x -6y = 4$ to produce the graph of $3x-6y=10$.
\begin{align*}
-6y&=4+6\\
y&=-\frac{2}{3}-1 \\
y&=-\frac{5}{3}
\end{align*}
Can my answer be improved?
| You can consider the $x$-intercepts instead: for the first line, it is the point $\bigl(\frac23, 0\bigr)$, and for the second line the point$\bigl(\frac53,0\bigr)$. The difference is equal to $1$, so you obtain the second line through a horizontal translation with vector $(1,0)$ of the first.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate $\iint \sqrt{2-x^2-y^2} \, dx \, dy$ over the circle $x^2+y^2=x\sqrt{2}$ so as my question states I have to calculate
$$\iint_\limits A \sqrt{2-x^2-y^2}\,dx\,dy$$ where $A$ is the circle defined by
$$x^2+y^2=x\sqrt{2}$$
After introducing polar coordinates $(x = r\cos\theta, y= r\sin\theta)$, I get that:
The left side of the equation of the circle is always nonnegative, so the right side also has to be nonnegative. This means that $x \ge 0$, or
$$\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].$$
Also we have that
$$r=\sqrt{2} \cos\theta$$
Here's where I have a slight problem. If I think about it, $2-x^2-y^2$ has to be $\ge 0$ for the square root to be real. Now, that means that $r \le \sqrt{2}$.
How do I determine the boundaries of $r$? I think that the boundaries for $r$ are $r \in [\sqrt{2}\cos\theta, \sqrt{2}]$. If yes, could anyone explain why exactly? I'm confused because I don't have an inequality in $r=\sqrt{2}\cos\theta$!
| Your are almost there. You need to integrate over the circle $x^2 + y^2 \leq \sqrt2 x$.
So as you said, $0 \leq r \leq \sqrt2 \cos\theta, - \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$
The integrand is $\sqrt{2-x^2-y^2} = \sqrt{2-r^2}$
and as $r \leq \sqrt2 \cos\theta, \theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \ , \ $ we have $\ r \leq \sqrt2$
So the integrand is defined in the entire region and the integral in polar coordinates is,
$$ \int_{-\pi/2}^{\pi/2} \int_0^{\sqrt2 \cos\theta} r \ \sqrt{2-r^2} \ dr \ d\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Local maxima, Local minima and saddle points I am trying to find the local maxima, minima, saddle points and choose the global maxima and minima of this function;
$f(x,y) = e^{-x^2-3y^2}.(x+y^3)$
I tried to find the partial derivatives for x and y. I find the equations and then I use $f_x = 0$ and $f_y = 0$ to find the critical points. However, these equations do not give me points to make a suggestion. Am I doing something wrong?
| We have
$$\begin{align} f_x &= e^{-x^2-3 y^2}-2 x e^{-x^2-3 y^2} \left(x+y^3\right) \\ f_y &= 3 y^2 e^{-x^2-3 y^2}-6 y e^{-x^2-3 y^2} \left(x+y^3\right) \end{align}$$
If we simultaneously set $f_x = f_y = 0$, we can divide out $e^{-x^2-3 y^2}$ and we are left with
$$1 - 2x(x+y^3) = 0 \\ 3 y^2 - y(x + y^3) = 0$$
Simultaneously solving
$$(x, y) = \left(\pm \dfrac{1}{\sqrt{2}},0\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is a symmetric matrix a subspace of nxn matrices? Let $M$ be a vector space for all 2x2 matrices.
Show that the set of all the symmetric matrices $M2$={$B ∈ M2 : B=B^t$} is a subspace of M.
My solution:
*
*The null matrix is symmetric
$$B=\begin{pmatrix}0&0\\0&0\end{pmatrix}$$
*Let $B$, $C$ $∈$ $M2$, where $B = B^t$, $C=C^t$. Then, $B+C=B^t + C^t$.
$$B = \begin{pmatrix}a&c\\c&b\end{pmatrix}$$
$$C = \begin{pmatrix}d&f\\f&e\end{pmatrix}$$
$$B+C = \begin{pmatrix}a+d&c+f\\c+f&b+e\end{pmatrix}$$
$$B^t = \begin{pmatrix}a&c\\c&b\end{pmatrix}$$
$$C^t = \begin{pmatrix}d&f\\f&e\end{pmatrix}$$
$$B^t+C^t = \begin{pmatrix}a+d&c+f\\c+f&b+e\end{pmatrix}$$
Therefore, $B + C ∈ M2$.
*Let $B ∈ M2$, $θ ∈ R$. Then, $θB = θB^t$
$$B = \begin{pmatrix}a&c\\c&b\end{pmatrix}$$ $$θB = \begin{pmatrix}θa&θc\\θc&θb\end{pmatrix}$$
$$B^t = \begin{pmatrix}a&c\\c&b\end{pmatrix}$$ $$θB^t = \begin{pmatrix}θa&θc\\θc&θb\end{pmatrix}$$
Therefore, $θB ∈ M2$.
So, $M2$ is, indeed, a subspace of $M$.
I am just learning linear algebra, so I apologize if I've made any mistakes.
Also, I just got into this community, so I am still learning the formatting, so I apologize for any mistakes as well.
Appreciate all the answers.
| A simpler way to prove this is to note that if $A,B$ are symmetric matrices then since $(\lambda A)^T = \lambda A^T = \lambda A$ and $(A+B)^T = A^T+B^T = A +B$, so are $\lambda A$ and $A+B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integrate $\log_2(1+b(ax^2+c)^{-2})$ I am wondering how to calculate $\int \log_2(1+b(ax^2+c)^{-2})$, where $a,b,c>0$.
I have used wolframalpha to calculate it, and the answer is like
While it gives me a nice result, I am wondering how to get it. (Wolfram|Alpha does not provide Step-by-Step Solution for this problem)
Answer:
Most results are from the answer of @Claude Leibovici, and part is from the suggestion of @Robert Z. Thanks for their help.
$$
\int\log_2 \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\,dx = \frac{1}{\log\left(2\right)}\int\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\,dx
$$
We denote $I \triangleq \int\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\,dx$. Use integration by parts to get rid of the logarithm,
$$
u\triangleq\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\implies du=-\frac{4 a b x}{\left(a x^2+c\right) \left(\left(a x^2+c\right)^2+b\right)}
$$
$$I=x\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)+\int\frac{4 a b x^2}{\left(a x^2+c\right) \left(\left(a x^2+c\right)^2+b\right)}\,dx$$
Now, partial fraction decomposition
$$\frac{4 a b x^2}{\left(a x^2+c\right) \left(\left(a x^2+c\right)^2+b\right)}=\frac{4 \left(a c x^2+b+c^2\right)}{a^2 x^4+2 a c x^2+b+c^2}-\frac{4 c}{a x^2+c}$$
Now, factor the denominator
$$a^2 x^4+2 a c x^2+b+c^2=a^2\left(x^2+\frac{c+ \sqrt{-b}}{a}\right) \left(x^2+\frac{c- \sqrt{-b}}{a}\right)$$
$$\frac{4 \left(a c x^2+b+c^2\right)}{a^2 x^4+2 a c x^2+b+c^2} = \frac{2 c+2 \sqrt{-b}}{a x^2+c+ \sqrt{-b}} + \frac{2c-2 \sqrt{-b}}{a x^2+ c- \sqrt{-b}} = \frac{2}{\frac{a}{c+ \sqrt{-b}} x^2+1} + \frac{2}{\frac{a}{c- \sqrt{-b}} x^2+1}$$
Thus
\begin{equation}
\begin{aligned}
I = x\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)
&+ 2 \sqrt{ \frac{ c+\sqrt{-b} }{a} } \tan^{-1} \left(\sqrt{\frac{a}{c+\sqrt{-b}}} x \right)\\
&+ 2 \sqrt{ \frac{ c-\sqrt{-b} }{a} } \tan^{-1} \left(\sqrt{\frac{a}{c-\sqrt{-b}}} x \right)\\
&- 4 \sqrt{ \frac{c}{a} } \tan^{-1} \left(\sqrt{\frac{a}{c}} x \right)
\end{aligned}
\end{equation}
| $$I=\int\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\,dx$$ Use integration by parts to get rid of the logarithm
$$u=\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)\implies du=-\frac{4 a b x}{\left(a x^2+c\right) \left(\left(a x^2+c\right)^2+b\right)}$$
$$I=x\log \left(1+\frac{b}{\left(a x^2+c\right)^2}\right)+\int\frac{4 a b x^2}{\left(a x^2+c\right) \left(\left(a x^2+c\right)^2+b\right)}\,dx$$ Now, partial fraction decomposition
$$\frac{4 a b x^2}{\left(a x^2+c\right) \left(\left(a x^2+c\right)^2+b\right)}=\frac{4 \left(a c x^2+b+c^2\right)}{a^2 x^4+2 a c x^2+b+c^2}-\frac{4 c}{a x^2+c}$$
Now, factor the denominator
$$a^2 x^4+2 a c x^2+b+c^2=a^2\left(x^2+\frac{c+i \sqrt{b}}{a}\right) \left(x^2+\frac{c-i \sqrt{b}}{a}\right)$$ Partial fractions again .....
Edit
Considering the more general case of
$$I_d=\int\log \left(1+\frac{b}{\left(a x^2+c\right)^d}\right)\,dx$$ where $d$ is an integer (if not, no hope for any result). As you wrote in comment
$$I_d=x\log \left(1+\frac{b}{\left(a x^2+c\right)^d}\right)+2abd \int\frac{x^2}{\left(b + \left(ax^2+c\right)^d\right)\left(ax^2+c\right)}\,dx$$ The denominator is a polynomial of degree $(d+1)$ in $x^2$ which can be written as
$$\left(b + \left(ax^2+c\right)^d\right)\left(ax^2+c\right)=a^{d+1} \prod_{i=1}^{d+1} (x^2-r_i)$$ where the $r_i$'s are the roots of the polynomial. So, partial fraction decomposition will give
$$2abd \int\frac{x^2}{\left(b + \left(ax^2+c\right)^d\right)\left(ax^2+c\right)}\,dx=\frac {2bd}{a^d}\,\sum_{i=1}^{d+1}\alpha_i \int \frac {dx}{x^2-r_i}$$ which does not make much problem (as long as you found the $r_i$'s and the $\alpha_i$'s (the latest depending on the former).
| {
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"question_score": "1",
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Solve for $x: \dfrac{x+6}{x-6}\left(\dfrac{x-4}{x+4}\right)^2+\dfrac{x-6}{x+6}\left(\dfrac{x+9}{x-9}\right)^2\lt\dfrac{2x^2+72}{x^2-36}$
Solve for $x: \dfrac{x+6}{x-6}\left(\dfrac{x-4}{x+4}\right)^2+\dfrac{x-6}{x+6}\left(\dfrac{x+9}{x-9}\right)^2\lt\dfrac{2x^2+72}{x^2-36}$
I tried taking LCM on LHS, but it's getting quite unwieldy: $$\frac{(x+6)^2(x-4)^2(x-9)^2+(x-6)^2(x+9)^2(x+4)^2}{(x^2-36)(x+4)^2(x-9)^2}\lt\frac{2(x^2+36)}{x^2-36}$$
Maybe $x^2-36$ can be cancelled from the denominators, provided $x\lt-6$ or $x\gt6$.
For $x\in(-6,6)$, the sign of inequality would change. But still the inequality remains in an unmanageable form. The symmetry of the question suggests that there should be some neater approach to it. But not able to figure that out.
EDIT: @Parasseux Nguyen gave the hint that $2(x^2+36)=(x+6)^2+(x-6)^2$. Here is my try after that:
$$\frac{(x+6)^2(x-4)^2(x-9)^2+(x-6)^2(x+9)^2(x+4)^2}{(x^2-36)(x+4)^2(x-9)^2}-\frac{(x^2+6)^2+(x^2-6)^2}{x^2-36}\lt0$$
$$\frac{(x+6)^2(x-4)^2(x-9)^2+(x-6)^2(x+9)^2(x+4)^2-(x^2+6)^2(x+4)^2(x-9)^2-(x^2-6)^2(x+4)^2(x-9)^2}{(x^2-36)(x+4)^2(x-9)^2}\lt0$$
$$\frac{8x(x+6)^2(x-9)^2+18x(x-6)^2(x+4)^2}{(x^2-36)(x+4)^2(x-9)^2}\lt0$$
Taking $2x$ common in the numerator still gives quartic in the bracket, with some not so neat coefficients. I wonder if this is the only way to go about it.
| The inequality is satisfied if
$$\begin{align} x & \;<\; -6\tag{1a}\\
\text{or}\quad\frac 65\big(1-\sqrt{26}\,\big)\,=\, -4.92 \;<\; x & \;<\; 0\;\;\text{but }\; x\neq -4\tag{1b}\\
\text{or}\quad 6 \;<\; x & \;<\; 7.32\,=\,\frac 65\big(1+\sqrt{26}\,\big)\,.\tag{1c}
\end{align}$$
To arrive at this result, exploit Paresseux Nguyen's helpful spot in the comments concerning the RHS, rescale by substituting $s=\frac x6$, and regroup all terms on one side:
$$\frac{s+1}{s-1}\:\left(1-\left(\frac{\frac 32s-1}{\frac 32s+1}\right)^2\right)\ +\ \frac{s-1}{s+1}\:\left(1-\left(\frac{\frac 23s+1}{\frac 23 s-1}\right)^{2}\right)\tag{2}$$
We have to find out for which values of $s$ the preceding function is strictly positive.
Let $t=\frac{s+1}{s-1}$ (which is a Möbius transformation).
It defines a homeomorphism from $\,\mathbb R\backslash\{1\}$ to itself, and
it is its own inverse, thus $s=\frac{t+1}{t-1}$. By courtesy of desmos.com, here is its graph including the asymptotes in orange:
Changing the variable in $(2)$ from $s$ to $t$ yields
$$f(t) := t\left(1-\left(\frac{t+5}{5t+1}\right)^{2}\right)\ +\ \frac{1}{t}\left(1-\left(\frac{5t-1}{t-5}\right)^{2}\right)\tag{3}$$
The function $f$ has two properties which are our friends:
Firstly
$$f\left(-\frac 1t\right) \:=\: -f(t)\;\text{ for all }
t\in\mathbb R\backslash\{-\tfrac 15,0,5\}\tag{4}$$
This is checked in a straightforward way, and can be used to cut down the analysis to values $t>0$.
And secondly, using $\,p^2-q^2=(p+q)(p-q)$ repeatedly, $f(t)$ can be completely factorised:
$$\begin{align} f(t) & \;=\; t\cdot\frac{(5t+1+t+5)(5t+1-t-5)}{(5t+1)^2}
+\frac 1t\cdot\frac{(t-5+5t-1)(t-5-5t+1)}{(t-5)^2}\\[2ex]
& \;=\;\frac{6\cdot 4(t+1)}{t\,(5t+1)^2}(t-1)
\underbrace{\left(t+\frac{5t+1}{t-5}\right)\left(t-\frac{5t+1}{t-5}\right)}_{\frac{(t^2+1)\left((t-5)^2-26\right)}{(t-5)^2}}
\\[2ex]
& \;=\;\frac{24(t+1)(t^2+1)(t+\sqrt{26}-5)}{t\,(5t+1)^2}
\cdot\frac{(t-1)\big(t-5-\sqrt{26}\big)}{(t-5)^2}
\tag{5}\end{align}$$
Assume $t>0$, then the first factor is positive, and from the second one we read off that $\,0<f(t)\,$ if
$$\begin{align} 0 \;<\; & t\;<\; 1\tag{6a}\\
\text{or}\quad 5+\sqrt{26} \;<\; & t\,.\tag{6c}
\end{align}$$
If $\,1<t<5+\sqrt{26}\,$ but $\,t\neq 5$, then $\,f(t)<0$. By $(4)$ we conclude that $\,0<f(t)\,$ also holds if
$$\,-1<t<5-\sqrt{26}=-\frac 1{5+\sqrt{26}}\quad\text{and excluding }\:t=-\frac 15\,.\tag{6b}$$
Finally, rollback the variables from $t$ via $s$ to $x$ to obtain the result given at the outset. The letters in the tags indicate how the intervals w.r.to the different variables correspond to each other.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4181715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Draw the Ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ a tangent plane which cuts off equal segments on the coordinate axis. Draw the Ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ a tangent plane which cuts off equal segments on the coordinate axis.
First, I want to find the normal line of the plane which intersects the $x,y,z$ at some distance $h$ away from the origin.
$$<-h,h,0> \times<-h, 0, h> = <h^2, h^2, h^2> = h^2<1,1,1>$$
So I now need to find when there is a point where the gradient is pointing in that direction
$$\triangledown f = <\frac{2x}{a^2},\frac{2y}{b^2},\frac{2z}{c^2}>$$
and the point must also satisfy the ellipsoid equation. $x = ka^2/2, y = kb^2/2,z=kc^2/2$ plugging into the ellipsoid equation yields a $$k = \frac{2}{\sqrt{a^2+b^2+c^2}}$$ which makes the gradient $$\triangledown f = <\frac{a^2}{\sqrt{a^2+b^2+c^2}},\frac{b^2}{\sqrt{a^2+b^2+c^2}},\frac{c^2}{\sqrt{a^2+b^2+c^2}}>$$
knowing $k$ I can also find the $<x,y,z>$ position and use $\triangledown f \cdot<x-x_o,y-y_o,z-z_o> =0$. I get a horrific result of :
$$a^2x+b^2y+c^2z=\frac{a^4+b^4+c^4}{\sqrt{a^2+b^2+c^2}}$$
Which is not correct. Could someone aid in creating a solution?
| Ellipsoid is given by, $ \displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$
If a plane is tangent to ellipsoid at point $(x_1, y_1, z_1)$, its equation is given by,
$\displaystyle \frac{x x_1}{a^2} + \frac{y y_1}{b^2} + \frac{z z_1}{c^2} = 1$
If $d \gt 0$, we also know that the plane passes through three points with coordinates, $(\pm d, 0, 0), (0, \pm d, 0)$ and $(0, 0, \pm d)$.
Now plugging them in the equation of the plane, we get
$x_1 = \pm \dfrac{a^2}{d}, y_1 = \pm \dfrac{b^2}{d}, z_1 = \pm \dfrac{c^2}{d}$
As $(x_1, y_1, z_1)$ is on the ellipsoid, from equation of the ellipsoid we get,
$d = \sqrt{a^2+b^2+c^2}$
That leads to equations of tangents being,
$\pm x \pm y \pm z = \sqrt{a^2+b^2+c^2}$
These are $8$ such planes, one each in each octant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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What is $\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y}$? I want to understand this multivariate limit.
WolframAlpha says $\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y} = 0$
But what if I take a curve $y = -x$, then the limit doesn't exist, right? Wouldn't this make the limit inexistent?
I tried to prove using sandwich theorem and definition, but didn't get anything good.
Any help would be appreciated.
| Method $-1.$
$$\begin{align}\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y} &=2\lim_{(x,y) \to (0,0)} \dfrac{\sin^2\left(\frac x2\right)}{x+y}\end{align}$$
Let $y(x):=r\sin^2\left(\frac x2\right)-x$, as $x\to 0$ and $r\in\mathbb R\setminus \left\{0\right\}$ then we get
$$\begin{align}2\lim_{(x,y(x)) \to (0,0)} \dfrac{\sin^2(\frac x2)}{x+y}=2\lim_{x\to 0}\frac {\sin^2\left(\frac x2\right)}{r\sin^2\left(\frac x2\right)}=\frac 2r\end{align}$$
This means, the limit doesn't exist. Because, the evaluation of the original limit is dependent on $r.$
Method $-2.$
$$\begin{align}\lim_{(x,y) \to (0,0)} \dfrac{1 - \cos x}{x+y} &=2\lim_{(x,y) \to (0,0)} \dfrac{\sin^2\left(\frac x2\right)}{x+y}\end{align}$$
Let $y(x):=\left(\frac x2\right)^3-x$ as $x\to 0$, then we have
$$\begin{align}2\lim_{(x,y(x)) \to (0,0)} \dfrac{\sin^2\left(\frac x2\right)}{x+y}=2\lim_{x\to 0} \dfrac{\sin^2\left(\frac x2\right)}{\left(\frac x2\right)^3}\longrightarrow \text{limit does not exist.}\end{align}$$
Because,
$$\begin{align}\lim_{x\to 0^+} \frac{\sin^2\left(\frac x2\right)}{\left(\frac x2\right)^3}=+\infty\\ \lim_{x\to 0^-} \frac{\sin^2\left(\frac x2\right)}{\left(\frac x2\right)^3}=-\infty\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4184194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integrate using $t$ substitution integrate this $\int \frac{1}{\sqrt{x+1}+\sqrt{x}}$
I took $\sqrt{x} = t$
Differentiated it
So $\frac{1}{2\sqrt{x}}=\frac{dt}{dx}$
Then I don't know how to proceed
I did some steps and it got too complicated
I also took $\frac{1}{\sqrt{x+1}+\sqrt{x}} = t$
But that also got complicated
A hint would be perfect
| A COMPLICATED SOLUTION:
Let $$\sqrt{1+x}+\sqrt{x}=u, \quad \sqrt{1+x}-\sqrt{x}=1/u \implies 2\sqrt{x}=u-1/u~~~~(*)$$
So we get $dx=\frac{(u^4-1)}{4u^2} du$
So $$I=\int \frac{dx}{\sqrt{1+x}+\sqrt{x}}=\frac{1}{2}\int (1-\frac{1}{u^4}) dx$$
$$\implies I= \frac{1}{2} \left(u+\frac{1}{3u^3}\right) =\frac{1}{2}\left(\sqrt{1+x}+\sqrt{x}+\frac{1}{3}(\sqrt{1+x}-\sqrt{x})^3\right)$$
$$\implies I=\frac{2}{3}[(1+x)^{3/2}-x^{3/2}].$$
Such tranformation as (*) solve more interesting problems.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How to un-simplify a general term I have a general term of $\frac{3n-1}{5n+7}$ and I want to find out which term $\frac{7}{12}$ is. I have a number in mind that I got from just trial and error but how would I apply a formula to find this. I am treating it as an arithmetic sequence with a common difference of 3 on top and 5 on bottom. I then apply the nth term formula but can't seem to arrive at the right conclusion.
| You can't treat $\frac{3n - 1}{5n + 7}$ as an arithmetic progression. For reference, let $a_{i} = \frac{3i - 1}{5i + 7}$. Then,
\begin{align*}
a_{n} - a_{n - 1} &= \frac{3n - 1}{5n + 7} - \frac{3(n - 1) - 1}{5(n - 1) + 7} \\
a_{n} - a_{n - 1} &= \frac{3n - 1}{5n + 7} - \frac{3n - 3 - 1}{5n - 5 + 7} \\
a_{n} - a_{n - 1} &= \frac{3n - 1}{5n + 7} - \frac{3n - 4}{5n + 2} \\
a_{n} - a_{n - 1} &= \frac{26}{25n^{2} + 45n + 14}
\end{align*}
As you can see, the difference of consecutive terms is not constant.
However, if you are solving for $n$ where $\frac{3n - 1}{5n + 7} = \frac{7}{12}$, just proceed with rules of rational equations.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 1,
"answer_id": 0
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Show that $\int_a^b \sin\left(x+\frac{1}{x}\right) dx <3.$ Show that $$\int_a^b \sin\left(x+\frac{1}{x}\right) dx <3$$ for all $a,b \in \mathbb{R}.$
Here is my solution:
\begin{align*}
\int_a^b \sin\left(x+\frac{1}{x}\right) dx&=\int_a^b \left(\sin x\cos \frac{1}{x}+\cos x\sin \frac{1}{x}\right) dx\\
&\le \int_a^b( \sin x+\cos x )dx\\
&=\sqrt{2}\int_a^b \sin \left(x+\frac{\pi}{4}\right)d\left(x+\frac{\pi}{4}\right)\\
&\le2\sqrt{2}<3.
\end{align*}
But this seems to be incorrect, since we can not guarantee $\sin x,\cos x\ge 0$ in the first line.
| Partial answer :
We use the triangle inequality so we want to show :
$$f(a,b)=\left|\int_{b}^{a}\cos\left(x\right)\sin\left(\frac{1}{\left(x\right)}\right)dx\right|+\left|\int_{b}^{a}\sin\left(x\right)\cos\left(\frac{1}{\left(x\right)}\right)dx\right|<3$$
With the inequality due to Ostrowski (see my last comment) we have $a,b\geq 4$:
$$f(a,b)\leq 2\left(\left|\cos\left(\frac{1}{\left(a+\frac{\pi}{2}\right)}\right)-\cos\left(\frac{1}{\left(b+\frac{\pi}{2}\right)}\right)\right|+\left|\cos\left(\frac{1}{\left(b+\frac{\pi}{2}\right)}\right)\right|\right)+2\left(\left|\sin\left(\frac{1}{\left(a\right)}\right)-\sin\left(\frac{1}{\left(b\right)}\right)\right|+\left|\sin\left(\frac{1}{\left(b\right)}\right)\right|\right)$$
The rest is smooth !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 1
} |
Asymptotics for $\sum_{k=0}^{n-1}(-1)^k \cot(\frac{2k+1}{4n}\pi)\log{(2\sin(\frac{2k+1}{4n}\pi))} $ I have been able to establish the identity
$$
I(n):=\int_0^1 \frac{y^{n-1}(1+y-y^{n+1}) - 1}{(1-y)(1+y^{2n})}\ dy = \frac{1}{n} \sum_{k=0}^{n-1}(-1)^k\cot(\frac{2k+1}{4n}\pi)\log{(2\sin(\frac{2k+1}{4n}\pi))}
$$
However, neither side seems conducive to discovering an asymptotic expansion as $n \to \infty.$
Purely on numerical evidence I conjecture that $I(n) \sim -\log(n).$
$$
\begin{array}{c|lcr}
n & I(n) &-\log{n} & -\log{n}/I(n) \\
\hline
10^3 & -6.702 & -6.908 & 1.031 \\
10^4 & -9.004 & -9.2106 &1.023 \\
10^5 & -11.307 &-11.513 & 1.018 \\
\end{array}
$$
I seek the dominant and one subdominant asymptotic terms. (If the Lambert W function is involved, then maybe one term is sufficient.) Here are some ideas.
(1) On the LHS (left-hand side), letting $n \to \infty,$ the $y^n$ terms integrand go to zero and you are left with $-1/(1-y),$ and the integral over it will diverge. However, plotting the integrand for large $n$, it is seen that the integrand follows the $-1/(1-y)$ curve until it sharply drops to zero near $y=1.$ It is probable that the position of the drop, $y_d,$ is analytically tractable, and one can approximate the integral as
$-\int_{0}^{y_d}dy/(1-y).$ However, I have no idea on how to approach a subdominant term.
(2) The RHS looks sort of like a Riemann sum, except for the pesky alternating sign, and the fact that odd integers appear in the arguments instead of every integer. A long time ago I read that the Euler-Maclaurin formula has been extended to character sums, which this looks like. I also know that Euler-Maclaurin can be used for asymptotic analysis. A solution in this manner would be most edifying, since I know so little about it.
| Note that
$$
I(n) = \int_0^1 {\frac{{(t^n - 1)^2 }}{{(t - 1)(t^{2n} + 1)}}dt} + \int_0^1 {\frac{{t^{n - 1} }}{{t^{2n} + 1}}dt} .
$$
Here
$$
\int_0^1 {\frac{{t^{n - 1} }}{{t^{2n} + 1}}dt} \le \int_0^1 {t^{n - 1} dt} = \frac{1}{n}.
$$
Now
\begin{align*}
\int_0^1 {\frac{{(t^n - 1)^2 }}{{(t - 1)(t^{2n} + 1)}}dt} & =
- \int_0^1 {\frac{{1 - t^n }}{{1 - t}}dt} + \int_0^1 {\frac{{1 - t^{2n} }}{{1 + t^{2n} }}\frac{{t^n }}{{1 - t}}dt} \\ & = - \sum\limits_{k = 1}^n {\frac{1}{k}} +\int_0^1 {\frac{{1 - t^{2n} }}{{1 + t^{2n} }}\frac{{t^n }}{{1 - t}}dt}.
\end{align*}
We have
$$
- \sum\limits_{k = 1}^n {\frac{1}{k}} = - \log n -\gamma + \mathcal{O}\!\left(\frac{1}{n}\right).
$$
Furthermore,
\begin{align*}
&\int_0^1 {\frac{{1 - t^{2n} }}{{1 + t^{2n} }}\frac{{t^n }}{{1 - t}}dt} = \int_0^{ + \infty } {\frac{{\tanh (ns)}}{{e^s - 1}}e^{ - ns} ds}
\\ &
= \int_0^{ + \infty } {\frac{{t/n}}{{e^{t/n} - 1}}e^{ - t} \frac{{\tanh t}}{t}dt} \\&
= \int_0^{ + \infty } {e^{ - t} \frac{{\tanh t}}{t}dt} + \int_0^{ + \infty } {\left( {\frac{{t/n}}{{e^{t/n} - 1}} - 1} \right)e^{ - t} \frac{{\tanh t}}{t}dt} .
\end{align*}
Here
\begin{align*}
\int_0^{ + \infty } {\left| {\frac{{t/n}}{{e^{t/n} - 1}} - 1} \right|e^{ - t} \frac{{\tanh t}}{t}dt} & \le \int_0^{ + \infty } {\frac{t}{{2n}}e^{ - t} \frac{{\tanh t}}{t}dt} \\ & = \frac{1}{{2n}}\int_0^{ + \infty } {e^{ - t} \tanh tdt} = \mathcal{O}\!\left( {\frac{1}{n}} \right),
\end{align*}
since $\left| {\frac{x}{{e^x - 1}} - 1} \right| \le \frac{x}{2}
$ for all $x>0$. Thus, in summary,
$$I(n)=-\log n+C+\mathcal{O}\!\left( {\frac{1}{n}} \right)$$ as $n\to +\infty$, where
$$
C = - \gamma + \int_0^{ + \infty } {e^{ - t} \frac{{\tanh t}}{t}dt} = -\gamma +2\log \left( {\frac{{2\Gamma \left( {\frac{5}{4}} \right)}}{{\Gamma \left( {\frac{3}{4}} \right)}}} \right)
=0.20597 \ldots \, .
$$
It is possible to obtain a complete asymptotic expansion as follows. We observe that
$$
\int_0^1 \frac{{t^{n - 1} }}{{t^{2n} + 1}} = \frac{\pi }{{4n}}.
$$
It is known that
$$
- \sum\limits_{k = 1}^n {\frac{1}{k}} \sim - \log n - \gamma - \frac{1}{{2n}} + \sum\limits_{k = 1}^\infty {\frac{{B_{2k} }}{{2kn^{2k} }}} ,
$$
where $B_{2k}$ denotes the Bernoulli numbers. Also, using the generating function of the Bernoulli numbers, it is found that
\begin{align*}
& \int_0^{ + \infty } {\left( {\frac{{t/n}}{{e^{t/n} - 1}} - 1} \right)e^{ - t} \frac{{\tanh t}}{t}dt} \\ & \sim - \frac{1}{{2n}}\int_0^{ + \infty } {e^{ - t} \tanh tdt} + \sum\limits_{k = 1}^\infty {\left[ {\frac{{B_{2k} }}{{(2k)!}}\int_0^{ + \infty } {t^{2k-1} e^{ - t} \tanh tdt} } \right]\frac{1}{{n^{2k} }}} .
\\ &
= - \frac{{\pi - 2}}{{4n}} + \sum\limits_{k = 1}^\infty {\frac{B_{2k}}{2k}\frac{ \zeta \left(2k,\frac{1}{4}\right) -2 \zeta \left( {2k ,\frac{3}{4}} \right) + \zeta \left( {2k,\frac{5}{4}} \right) }{{16^k }}\frac{1}{{n^{2k} }}} .
\end{align*}
Here $\zeta(s,a)$ is the Hurwitz zeta function. Therefore,
\begin{align*}
I(n) & \sim - \log n + C + \sum\limits_{k = 1}^\infty {\frac{B_{2k}}{2k}\left( 1+\frac{ \zeta \left(2k,\frac{1}{4}\right) -2 \zeta \left( {2k ,\frac{3}{4}} \right) + \zeta \left( {2k,\frac{5}{4}} \right) }{{16^k }}\right)\frac{1}{{n^{2k} }}}
\\ & = - \log n + C + \sum\limits_{k = 1}^\infty
\frac{{B_{2k} }}{{16^k k}}\left( {\zeta \left( {2k,\tfrac{1}{4}} \right) - \zeta \left( {2k,\tfrac{3}{4}} \right)} \right)
\frac{1}{{n^{2k} }},
\end{align*}
as $n\to +\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 1,
"answer_id": 0
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Prove if number is rational or irrational I've been asked to prove if $\frac{\sqrt{3+\sqrt5}}{\sqrt{2} + \sqrt {10}}$ is a rational number. I've tried a proof as follows:
Suppose the number is rational, so it can be written as the quotient of 2 numbers $a$ and $b$
\begin{align*}
\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + \sqrt{10}} = \frac{a}{b} \\
\frac{3+\sqrt{5}}{12 + 4\sqrt{5}} = \frac{a^2}{b^2} \\
\frac{3+\sqrt{5}}{4(3+\sqrt{5})} = \frac{a^2}{b^2} \\
\frac{1}{4} = \frac{a^2}{b^2}
\end{align*}
And because we get $\frac{1}{4}$ which is rational, we can conclude that the proof is right and there aren't any contradictions. Hence $\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2} + \sqrt{5}}$ is a rational number.
I guess my proof lacks something and I don't feel it's complete yet. I would appreciate any recommendations on how to improve my answer. Thanks in advance.
| Alternative approach:
In a problem like this, my first step is to clear the denominator.
$\displaystyle \frac{\sqrt{3 + \sqrt{5}}}{\sqrt{10} + \sqrt{2}} \times \frac{\sqrt{10} - \sqrt{2}}{\sqrt{10} - \sqrt{2}} = \frac{1}{8} \times \left[\sqrt{3 + \sqrt{5}}\right] \times \left[\sqrt{10} - \sqrt{2}~\right].$
Therefore, the problem reduces to an examination of
$$ \left[\sqrt{3 + \sqrt{5}}\right] \times \left[\sqrt{10} - \sqrt{2}~\right].\tag1$$
Since both of the factors in (1) above are positive, the problem reduces to determining whether there exists positive rational numbers $a,b$ such that $a^2 \times b^2$ equals the square of the expression in (1) above.
This resolves to determining whether there exists positive integers $c,d,e,f$ such that
$$\frac{c^2 e^2}{d^2 f^2} = \left[3 + \sqrt{5}\right] \times \left[12 - 2 \sqrt{20}\right].$$
$$\left[3 + \sqrt{5}\right] \times \left[12 - 2 \sqrt{20}\right] = \left[3 + \sqrt{5}\right] \times \left[12 - 4 \sqrt{5}\right] = 16.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Line $y = mx$ through the origin that divides the area between the parabola $y = x-x^2$ and the x axis into two equal regions. There is a line $y = mx$ through the origin that divides the area between the parabola $y = x-x^2$ and the x axis into two equal regions. Find m.
My solution:
When I compute my answer, I get $1-\frac{1}{\sqrt[3]{2}}$ (as seen in Wolfram below)
Solution in Wolfram:
https://www.wolframalpha.com/input/?i=%281-x%29%5E2%28-2x%2B2%29%3D1
Solution in Geogebra:
https://www.geogebra.org/calculator/kshvr8mw
My question: is there a way to find this solution without a GDC/Geogebra? In other words, a method which would lead to a simpler equation in the end?
Thank you.
| This is my solution.
First, we recognize that the quadratic has roots at $(0,0)$ and $(1,0)$. So, to first solve for the area under the quadratic:
$\int_0^1 x-x^2.dx=\left[\frac{x}{2}-\frac{x^3}{3}\right]_0^1$
$=\frac{1}{6}$
So, half of this area is $\frac{1}{12}$
Let $x=a$ be the abscissa of the intersection point of line $y=mx$ and curve $y=x(1-x)$ different from $0$.
Then:
$ma=a(1-a)$
$m=(1-a)$
So, the equation of the straight line is $y=(1-a)x$
$\int_0^a(x-x^2)-(1-a)x.dx=\frac{1}{12}$
$\int_0^a-x^2+ax.dx=\frac{1}{12}$
$\left[-\frac{x^3}{3}+\frac{ax^2}{2}\right]_0^a=\frac{1}{12}$
$-\frac{a^3}{3}+\frac{a^3}{2}=\frac{1}{12}$
$\frac{a^3}{6}=\frac{1}{12}$
$a^3=\frac{1}{2}$
$a=\sqrt[3]{\frac{1}{2}}$
As earlier determined, the equation of the straight line is $y=(1-a)x$, so $y=(1-\sqrt[3]{\frac{1}{2}})x$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Find $\sin^5 x + \cos^5 x$
Given $$\cos \left[\sqrt{\left(\sin x + \cos x\right)\left(1 - \sin x \cos x \right)}\right] \\{}= \sqrt{\cos \left(\sin x + \cos x \right) \cos \left(1 - \sin x \cos x\right)}.$$
Find $\sin^5 x + \cos^5 x.$
My try:
\begin{align*}
\sin^5x+\cos^5x
&=(\sin x+\cos x)(\sin^4x-\sin^3x\cos x+\sin^2x\cos^2x-\sin x \cos^3x+\sin^4x)\\
&=(\sin x+\cos x)[(\sin^2x+\cos^2x)^2-(\sin^2x+\cos^2x)\sin x \cos x-\sin^2x\cos^2x]\\
&=(\sin x+\cos x)[1-\sin x\cos x-\sin^2x\cos^2x]\\
&=\frac{1}{4}(\sin x+\cos x)(4-2\sin 2x- \sin^2 2x)\\
\end{align*} I don't know what next?
| I wanted to develop the idea expressed in comments that LHS is bounded below by $\cos(1)$ while RHS is bounded above by $\cos(1)$ and that we get equality for $x=0\text{ or }\frac{\pi}2$ only.
*
*The LHS part and equality case is not too complicated to prove:
$\require{cancel}\begin{align}f(x)
&=\Big(\cos(x)+\sin(x)\Big)\Big(\overbrace{1}^{\cos(x)^2+\sin(x)^2}-\sin(x)\cos(x)\Big)\\
&=\cos(x)^3+\cancel{\cos(x)\sin(x)^2}-\cancel{\sin(x)\cos(x)^2}+\cancel{\sin(x)\cos(x)^2}+\sin(x)^3-\cancel{\sin(x)^2\cos(x)}\\
&=\cos(x)^3+\sin(x)^3\end{align}$
But notice that $\forall n\in\mathbb N$ since $|\sin(x)|\le 1\implies |\sin(x)|^n\le |\sin(x)|^2$ and same for cosine we get,
$\Big|f(x)\Big|=\Big|\sin(x)^3+\cos(x)^3\Big|\le \Big|\sin(x)\Big|^3+\Big|\cos(x)\Big|^3\le \sin(x)^2+\cos(x)^2=1$
Notice also that $|f(x)|$ has period $\pi$ and that $\cos\searrow$ on $[0,\pi]$
Therefore (and since RHS undefined where $f(x)$ is negative):
$$\cos\left(\sqrt{|f(x)|}\right)\ge \cos(1)\quad\text{and}\quad LHS\ge \cos(1)$$
The equality case is for extremal points.
So let search for critical points $f'(x)=3\sin(x)\cos(x)(\sin(x)-\cos(x))=0\iff x=0,\frac{\pi}2,\frac{\pi}4$ over the $[0,\pi]$ interval, among which only $0,\frac{\pi}2$ lead to $f(x)=1$.
*
*However, I faced a wall when I tried to prove the RHS part, so it seems basilik's approach despite looking more complicated at first sight, is eventually more effective.
Does anyone attempted this approach too ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $n$ can be expressed as $2x^2+3y^2$ with integers $x$ and $y$, then so can $7n$. The problem is:
Let us say that a positive integer $n$ is obtainable if there exist integers $x$ and $y$ such that $n=2x^2+3y^2$. If $n$ is obtainable, prove that $7n$ is also obtainable.
I tried doing that
$$2(x+a)^2+3(y+b)^2=14x^2+21y^2$$
which simplifies to
$$4ax+2a^2+6yb+3b^2=12x^2+18y^2$$
but then I seem to be caught in a loop.
Am I on the right direction and can someone give me a hint?
This is a homework question btw.
| I would modify your substitution to avoid linear term.
$$2(x+ay)^2 + 3(y+bx)^2 = 14x^2+21y^2$$
$$2(x^2+2axy+a^2y^2)+3(y^2+2bxy+b^2x^2)=14x^2+21y^2$$
$$(2+3b^2)x^2 +(4a+6b)xy + (2a^2+3)y^2=14x^2+21y^2$$
Can we find $(a,b)$ such that the equality holds for all $x,y$? I will leave the investigation to you.
| {
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"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Finding $x_1+x_2$, where the $x_i$, and $y$ and $z$, are positive integers satisfying $x^2+y^2+z^2=2(xy+1)$ and $x+y+z=2022$
Let $x$, $y$, and $z$ be positive integers satisfying
$$(x)^2+(y)^2+(z)^2=2(xy+1) \quad\text{and}\quad x+y+z=2022$$
If $x_1$ and $x_2$ are 2 distinct solutions for $x$, what is the value of $x_1+x_2$?
Attempted to express $(x)^2+(y)^2+(z)^2=2(xy+1)$ as $(x-y)^2+(z)^2=2$, then I'm stuck.
Kindly advise! Thanks!
| Considering achile hui comment we have:
$z=1$ and $(x-y)^2=1$ and:
case 1):$\begin {cases}x-y=1\\x+y+1=2022\end{cases}$
which gives $x=1011$ and $y=1010$
case 2):$\begin {cases}x-y=-1\\x+y+1=2022\end{cases}$
which gives $x=1010$ and $y=1011$
| {
"language": "en",
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"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving Diamond & Shurman Exercise 3.7.1, about conjugacy class of $\Gamma_0^{\pm}(N)$. I am reading chapter 3 of A First Course in Modular Forms but have troubles in Exercise 3.7.1 (c) and (d).
(c) Show that the $\Gamma_0^{\pm}(N)$-conjugacy class of $\gamma \in \Gamma_0(N)$ is the union of the $\Gamma_0(N)$ conjugacy classes of $\gamma$ and $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \gamma \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$. Show that if $\gamma$ has order $4$ or $6$ then this union is disjoint.
(d) Let $\gamma = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}^{-1} = \begin{bmatrix} 3 & -2 \\ 5 & -3 \end{bmatrix}$, an order-$4$ element of $\Gamma_0(5)$. Show that $\gamma$ is not conjugate to its inverse in $\Gamma_0(5)$.
Here let $\mathrm{GL}_2(\mathbb{Z})$ be the group of invertible $2 \times 2$ matrices with integer entries, $\mathrm{SL}_2(\mathbb{Z})$ the group of $2 \times 2$ matrices with integer entries and determinant $1$, and
\begin{align*}
\Gamma_0^{\pm}(N) &= \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in \mathrm{GL}_2(\mathbb{Z}) : \begin{bmatrix} a & b \\ c & d \end{bmatrix} \equiv \begin{bmatrix} * & * \\ 0 & * \end{bmatrix} \pmod{N} \right\} \\
\Gamma_0(N) &= \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \in \mathrm{SL}_2(\mathbb{Z}) : \begin{bmatrix} a & b \\ c & d \end{bmatrix} \equiv \begin{bmatrix} * & * \\ 0 & * \end{bmatrix} \pmod{N} \right\}.
\end{align*}
I have proved the first part of (c): the map
\begin{align*}
\left(\Gamma_0^{\pm}(N) \setminus \Gamma_0(N)\right) &\longrightarrow \Gamma_0(N) \\
\alpha &\longmapsto \alpha \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}
\end{align*}
is a bijection, so
\begin{align*}
\left\{ \beta \gamma \beta^{-1}: \beta \in \Gamma_0^{\pm}(N) \setminus \Gamma_0(N) \right\}
= \left\{ \alpha \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \gamma \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \alpha^{-1}: \alpha \in \Gamma_0(N) \right\}.
\end{align*}
How to show the remaining parts? For part (d), I think one cannot apply (c) directly because (d) serves as an example of the following statement in P.93
The extended conjugacy class of $\gamma$ under $\Gamma_0^{\pm}(N)$ is not in general the union of the conjugacy class of $\gamma$ and $\gamma^{-1}$ under $\Gamma_0(N)$.
| Let
$$\Gamma_0^-(N)=\left\{ \begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\in\Gamma_0^\pm(N):ad-bc=-1 \right\},$$
so the $\Gamma_0^\pm(N)$-conjugacy classes of $\gamma$ is the union of $\Gamma_0(N)$-conjugacy classes of $\gamma$ and $\Gamma_0^-(N)$-conjugacy classes of $\gamma$. Let $\alpha=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\in\Gamma_0^-(N)$, then
$$\alpha\gamma\alpha^{-1}=\begin{bmatrix}
a & -b\\
c & -d
\end{bmatrix}\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}\gamma\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}\begin{bmatrix}
-d & b\\
-c & a
\end{bmatrix},$$
where $\begin{bmatrix}
a & -b\\
c & -d
\end{bmatrix}\in\Gamma_0(N)$ and $\begin{bmatrix}
-d & b\\
-c & a
\end{bmatrix}=\begin{bmatrix}
a & -b\\
c & -d
\end{bmatrix}^{-1}$. Let $\alpha=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\in\Gamma_0(N)$ this time, then
$$\alpha\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}\gamma\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}\alpha^{-1}=\begin{bmatrix}
a & -b\\
c & -d
\end{bmatrix}\gamma\begin{bmatrix}
d & -b\\
c & -a
\end{bmatrix},$$
where $\begin{bmatrix}
a & -b\\
c & -d
\end{bmatrix}\in\Gamma_0^-(N)$ and $\begin{bmatrix}
d & -b\\
c & -a
\end{bmatrix}=\begin{bmatrix}
a & -b\\
c & -d
\end{bmatrix}^{-1}$.
Thus the $\Gamma_0^-(N)$-conjugacy classes of $\gamma$ is equal to the $\Gamma_0(N)$-conjugacy classes of $\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}\gamma\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}$. The first statement of (c) is true.
By Proposition 2.3.3, if $\gamma$ has order $4$ then $\gamma$ is conjugate to $\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}^{\pm1}=\pm\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}$ in $\mathrm{SL}_2(\mathbb Z)$. If this two conjugacy classes intersects, i.e. $\gamma$ is conjugate to $\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}\gamma\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}$ in $\Gamma_0(N)$, then $\gamma$ is conjugate to $\gamma$ in $\Gamma_0^-(N)$, then $\pm\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}$ is conjugate to $\pm\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}$ in $\mathrm{GL}_2^-(\mathbb Z)$, where
$$\mathrm{GL}_2^-(\mathbb Z)=\left\{ \begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\in\mathrm{GL}_2(\mathbb Z):ad-bc=-1 \right\}.$$
Assuming $\pm\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}=\pm\alpha\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}\alpha^{-1}$ and $\alpha=\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\in\mathrm{GL}_2^-(\mathbb Z)$, then
$$\pm\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}=\pm\begin{bmatrix}
-(ac+bd) & a^2+b^2\\
-(c^2+d^2) & ac+bd
\end{bmatrix}.$$
Thus $a^2+b^2=-1$, a contradiction.
If $\gamma$ has order $6$, we deduce that $\begin{bmatrix}
0 & -1\\
1 & 1
\end{bmatrix}$ and $\begin{bmatrix}
1 & 1\\
-1 & 0
\end{bmatrix}$ are conjugate to themselves in $\mathrm{GL}_2^-(\mathbb Z)$ by similar way, and which yields
$$\begin{bmatrix}
0 & -1\\
1 & 1
\end{bmatrix}=\begin{bmatrix}
bc-ac-bd & a^2+b^2-ab\\
cd-c^2-d^2 & ac+bd-ad
\end{bmatrix},$$
$$\begin{bmatrix}
1 & 1\\
-1 & 0
\end{bmatrix}=\begin{bmatrix}
ac+bd-ad & ab-a^2-b^2\\
c^2+d^2-cd & bc-ac-bd
\end{bmatrix}.$$
Thus we have $\left( a-\frac12b \right)^2+\frac34b^2=-1$ and $\left( c-\frac12d \right)^2+\frac34d^2=-1$, contradictions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4199234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Proving on the equation $\frac1{x+1}+\frac1{2(x+2)}+\frac1{3(x+3)}+\cdots+\frac1{n(x+n)}=1$ Consider the equation:
\begin{gather*}
\frac{1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + \cdots + \frac{1} {n(x+n)} = 1.\tag1
\end{gather*}
Prove that:
*
*For every $n \geq 2$, equation $(1)$ always has a unique positive solution $x_n$.
*The sequence $(x_n)$ has a finite limit with $n \to \infty+$. Find that limit.
Here's all I did:
$\frac {1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + .... + \frac {1} {n(x+n)} = 1.$
$\Leftrightarrow 2.3...n(x+1)(x+2)...(x+n) + 1.3...n(x+1)(x+3)...n + ... + 1.2...(n-1)(x+1)(x+2)...(x+n-1) = n! (x+1)(x+2)...(x+n) $
Consider the polynomial $P(x) = 2.3...n(x+1)(x+2)...(x+n) + 1.3...n(x+1)(x+3)...n + ... + 1.2...(n-1)(x+1)(x+2)...(x+n-1) - n! (x+1)(x+2)...(x+n) $
Consider the highest coefficient of the polynomial:$ -n! < 0 $
Consider the lowest coefficient of the polynomial:$ (2.3...n)^2+(1.3...n)^2+...(1.2...(n-1))^2 - n! = (1.3...n)^2+...(1.2...(n-1))^2 > 0$
Let $x_1,x_2,x_3....$ be the solutions of$ P(x) $
Assume that equation (1) has no positive solution.
According to Viette's theorem: $x_1 x_2 x_3....x_n = (-1)^n \frac{a_0}{a_n} $
$\frac {a_0}{a_n} < 0 \Rightarrow $if $n$ is an odd number , then $x_1 x_2 x_3....x_n < 0$ , but
$(-1)^n \frac{a_0}{a_n} >0 \Rightarrow $ (nonsense) . Same with $n $being an even number.
So the polynomial must have at least one integer root.
Suppose the polynomial has at least 2 positive roots $x$ and $y$ .
$\frac {1}{y+1} + \frac{1}{2(y+2)} + \frac{1}{3(y+3)} + .... + \frac {1} {n(y+n)} =\frac {1}{x+1} + \frac{1}{2(x+2)} + \frac{1}{3(x+3)} + .... + \frac {1} {n(x+n)} .$
$\Rightarrow (y-x)( \frac {1}{(x+1)(y+1)}+ \frac {1}{2(x+2)(y+2)} +...+\frac {1}{n(x+n)(y+n)} ) = 0 $
$\Rightarrow x=y \Rightarrow$ (no sense)
So for every $n \geq 2$ , equation $( 1 )$ always has a unique positive solution $x_n$
That's all I did , and I have no idea what to do next , I want to establish a relationship between $x_n$ but for $n \geq 3$ the polynomial's solution is a " bad " solution . " so it's very difficult to build. Hope to get help from everyone. Thanks very much .
| Since you already find a solution for the first question. I write an answer for the second one. Observe that
$$\sum_{k=1}^n\frac{1}{k\cdot (k+1)}=\sum_{k=1}^n\frac{1}{k}-\frac{1}{k+1}=1-\frac{1}{n+1}.$$ Hence
$$\sum_{k=1}^{\infty}\frac{1}{k\cdot (k+1)}=1.$$
This means for $x=1$ the left handside of your equation tends to 1. One can easily prove, that for $x<1$ and sufficient large $n$ the lefthandside is greater than 1 and for $x>1$ and sufficient large $n$ the lefthandside is less than 1.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Ambiguity in Integration done by substitution. I was solving this integral $$I= \int x\sqrt{1-x^2} \mathrm dx$$ My solution is:
Let $x=\sin y$, then $|\cos y|=\sqrt{1-x^2}$ I now use method of substitution for the cases $\cos y\geq 0$ and $\cos y<0$. The correct answer is $-\frac{(1-x^2)^{3/2}}{3} +C$. I am able to obtain this answer for $\cos y\geq 0$ but in the case where $\cos y <0$, we get it to be $\frac{(1-x^2)^{3/2}}{3} +C$. Where have I gone wrong?
| I suspect you forgot to include the sign when you substituted $\cos y$ back to get a function in terms of $x$ at the end. Here is the complete derivation.
I'm assuming that you did the substitution and integration correctly: if $x=\sin y$, then $dx = \cos y\,dy$, so we get
$$\int x\sqrt{1-x^2}\,dx = \int\sin y|\cos y|\cos y\,dy.$$
If we are in a situation where $\cos(y)\lt 0$ (e.g., if we are picking $\frac{\pi}{2}\leq x\leq \frac{3\pi}{2}$), then this would become
$$\int -\sin y\cos^2y\,dy.$$
This can be done by substitution (again), letting $u=\cos y$, $du = -\sin y\,dy$, so we get
$$\int \sin y\cos^2y\,dy = \int u^2\,du = \frac{1}{3}u^3+C.$$
Now we need to substitute back, and we get $\frac{1}{3}\cos^3y+C$.
Now, if $x=\sin y$, then $1-x^2 = \cos^2y$. That means that $|\cos y| = \sqrt{\cos^2y} = (1-x^2)^{1/2}$. Since we are assuming that $\cos y\lt 0$, this gives
$ - \cos y = (1-x^2)^{1/2}$; equivalently,
$$\cos y = -(1-x^2)^{1/2}.$$
Note the minus sign!
So substituting back we have
$$\begin{align*}
\int x\sqrt{1-x^2}\,dx &= \frac{1}{3}\cos^3y + C \\
&= \frac{1}{3}\left(-(1-x^2)^{1/2}\right)^3 + C \\
&= \frac{(-1)^3}{3}(1-x^2)^{3/2} + C\\
&= -\frac{1}{3}(1-x^2)^{3/2} + C,
\end{align*}$$
giving the same answer as you got when you assumed $\cos(y)\geq 0$.
| {
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"source": "stackexchange",
"question_score": "1",
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} |
Find the point on the curve $x^{2} - yz=1$ that is closest to the origin The problem statement is as follows:
Find the point on the curve $x^{2} - yz=1$ that is closest to the origin.
To do this we minimize $D^{2} = x^{2} + y^{2} + z^{2}$ with the above constraint.
Attempted solution:
Isolate $z$ in the constraint:
$$
x^{2} - yz = 1 \implies z = \frac{x^{2}-1}{y}
$$
Substitute into distance formula:
$$
x^{2} + y^{2} + \left(\frac{x^{2}-1}{y}\right)^{2}
$$
Differentiate in terms of x and y and set to zero
$$
\frac{\partial D^{2}}{\partial x} = 2x+4x\left(\frac{x^{2}-1}{y^{2}}\right)=0,\frac{\partial D^{2}}{\partial y} = 2y-2\left(\frac{(x^{2}-1)^{2}}{y^{3}}\right)=0
$$
which has real solutions $x=0, y=+1$, giving the critical point as $x=0, y=1, z=-1$.
The correct answer is $x=\pm 1, y=0, z=0$.
What am I doing wrong?
| In the case of a quadratic form compared with the standard (squared) distance $x^2 + y^2 + z^2,$ from Lagrange multipliers: the extremal directions are the eigenvectors of the Hessian matrix of the other form, here $x^2 - yz \; . \; \;$ In this case the eigenvalues are three distinct integers, one may take eigenvectors with all integer coordinates, find them explicitly, see what drops out...
$$
\left(
\begin{array}{rrr}
2&0&0 \\
0&0&-1 \\
0&-1&0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
0&0&1 \\
1&-1&0 \\
1&1&0 \\
\end{array}
\right) =
\left(
\begin{array}{rrr}
0&0&2 \\
-1&-1&0 \\
-1&1&0 \\
\end{array}
\right)
$$
The locus $x^2 -yz=1$ is a hyperboloid of one sheet, the eigenvector $(0,1,1)^T$ is the axis, multiples of it never lie on your surface.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Given $a,b,c$ are sides of a triangle, Prove that :- $\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$
Given $a,b,c$ are sides of a triangle, Prove that :-
$$\frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2} < \frac{1}{2}$$
What I Tried:- I was able to solve the left hand side inequality. From RMS-AM Inequality on $a,b,c$ :-
$$\sqrt{\frac{a^2+b^2+c^2}{3}} \geq \frac{a+b+c}{3}$$
$$\rightarrow \frac{a^2+b^2+c^2}{3} \geq \frac{(a+b+c)^2}{9}$$
$$\rightarrow \frac{1}{3} \leq \frac{a^2+b^2+c^2}{(a+b+c)^2}$$
I got no progress for the second part. I also have a clue, that $a,b,c$ are sides of a triangle, which I have not used yet. So maybe that should be used somehow, but I am not getting it.
Can anyone help me? Thank You.
| We need to prove that:$$2(a^2+b^2+c^2)<(a+b+c)^2,$$ which is true because
$$(a+b+c)^2-2(a^2+b^2+c^2)=\sum_{cyc}(2ab-a^2)=$$
$$=\sum_{cyc}(ab+ac-a^2)=\sum_{cyc}a(b+c-a)>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Alternative series expansion for the elliptic integral of the second kind I found the following alternative series expansion for the complete elliptic integral of the second kind (E):
$ E(k) = \frac{(1+k')\pi}{4} \left\{ 1+\frac{1}{2^2}\left(\frac{1-k'}{1+k'}\right)^2+ \frac{1^2}{2^2\cdot 4^2} \left(\frac{1-k'}{1+k'}\right)^4 +...+ \left( \frac{(2n-3)!!}{2^n n!}\right)^2 \left(\frac{1-k'}{1+k'}\right)^{2n} +...\right\} $
where $k'=\sqrt{1-k^2}$, $\; k$ is the modulus and $k'$ is the complementary modulus.
I have been trying to prove it without success. This expansion seems to be related to the Landen transformation:
$K\left(\frac{1-k'}{1+k'}\right) = \frac{1+k'}{2}K(k)$
$ E\left(\frac{1-k'}{1+k'}\right) = \frac{1}{1+k'}\left[E(k)+k'K(k) \right] $
where K(k) is the complete elliptic integral of the first kind.
Here is my try:
From the last relations we have:
$ E(k) = (1+k')E\left(\frac{1-k'}{1+k'}\right) -k'K(k) = (1+k')E\left(\frac{1-k'}{1+k'}\right) -\frac{2k'}{1+k'}K\left(\frac{1-k'}{1+k'}\right)$
Then I used the classic expansion of $E$ and $K$:
$K(k) = \frac{\pi}{2} \sum_{n=0}^{\infty} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2k^{2n} $
$E(k) = \frac{\pi}{2} \sum_{n=0}^{\infty} \frac{1}{1-2n} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2k^{2n} $
To get:
\begin{align*}
E(k) = & (1+k')\frac{\pi}{2} \sum_{n=0}^{\infty} \frac{1}{1-2n} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-k'}{1+k'}\right)^{2n} -\frac{2k'}{1+k'} \frac{\pi}{2} \sum_{n=0}^{\infty} \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-k'}{1+k'}\right)^{2n}\\
=& \frac{\pi(1+k')}{2} \sum_{n=0}^{\infty}\left\{ \left[\frac{(2n-1)!!}{(2n)!!}\right]^2\left(\frac{1-k'}{1+k'}\right)^{2n} \left(\frac{1}{1-2n} - \frac{2k'}{(1+k')^2}\right)\right\}
\end{align*}
but the path came to a dead end. I will really appreciate your help.
| Your path is not a dead end! We have $\frac{2k'}{(1+k')^2} = \frac{1}{2}\left[1 - \left(\frac{1-k'}{1+k'}\right)^2\right]$, so your last line implies
\begin{align}
\frac{4 \operatorname{E}(k)}{\pi (1+k')} &= \sum \limits_{n=0}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n}\left[\frac{2}{1-2n} - 1 + \left(\frac{1-k'}{1+k'}\right)^2\right] \\
&= 1 - \sum \limits_{n=1}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n}\frac{2n+1}{2n-1} + \sum \limits_{n=0}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2(n+1)} \\
&= 1 - \sum \limits_{n=1}^\infty \left[\frac{(2n-1)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n}\frac{2n+1}{2n-1} + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n-2)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n} \\
&= 1 + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n} \left[(2n)^2 - \frac{2n+1}{2n-1} (2n-1)^2\right] \\
&= 1 + \sum \limits_{n=1}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n} = \sum \limits_{n=0}^\infty \left[\frac{(2n-3)!!}{(2n)!!}\right]^2 \left(\frac{1-k'}{1+k'}\right)^{2n}\, .
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/4209421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\int^{\infty}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ diverges First of all, I noticed that $\sin^2(\pi(x+\frac{1}{x}))$ "approaches" $\sin^2(\pi x)$ as $x \to \infty$. In other words, it seems that the area under one period of $\sin^2(\pi(x+\frac{1}{x}))$ should converge to the value of the area under one period of $\sin^2(\pi x)$ as $x \to \infty$.
The following is my best attempt at formalizing this idea:
Let $f(x) = \sin^2(\pi(x+\frac{1}{x}))$ and $\xi_n = \frac{n+\sqrt{n^2-4}}{2}$, where $n \geq 2$. Notice that $f(\xi_n)=0$.
Now, consider the limit of $\xi_{n+1} - \xi_{n}$ as $n \to \infty$:
$$\begin{align} \lim_{n\to\infty} \xi_{n+1}-\xi_n &= \lim_{n\to\infty}\frac{(n+1)+\sqrt{(n+1)^2-4}}{2}-\frac{n+\sqrt{n^2-4}}{2} \\ &= \frac{1}{2}\lim_{n\to\infty}1+\sqrt{(n+1)^2-4}-\sqrt{n^2-4} \\ &= \frac{1}{2}\lim_{n\to\infty}\frac{2+\frac{2}{n}+2\sqrt{1+\frac{2}{n}-\frac{3}{n^2}}}{\frac{1}{n}+\sqrt{1+\frac{2}{n}-\frac{3}{n^2}}+\sqrt{1-\frac{4}{n^2}}} \\ &= 1 \end{align}$$ What the result above tells us is that as x
increases, the period of $\sin^2(\pi(x+\frac{1}{x}))$ approaches the
period of $\sin^2(\pi x)$. Furthermore, $x+\frac{1}{x} \approx x$ as
$x\to \infty$, which implies that $\sin^2(\pi(x+\frac{1}{x}))$
"approaches" $\sin^2(\pi x)$ as $x \to \infty$. Using both of these
facts, we can claim that $\int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx$ approaches $\int^{1}_{0} \sin^2(\pi x) dx$ as $n \to \infty$.
We can also compute the value of $\int^{1}_{0} \sin^2(\pi x) dx$:
$$\begin{align} \int^{1}_{0} \sin^2(\pi x) dx &= \frac{1}{2}\int^{1}_{0}1-\cos(2\pi x)dx \\ &=\frac{1}{2}\{(1-\frac{1}{2\pi}\sin(2\pi))-(0-\frac{1}{2\pi}\sin(0))\} \\ &= \frac{1}{2}\end{align}$$ Thus, we can state that:
$$\lim_{n\to\infty} \int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx = \frac{1}{2}$$
Notice that we can write $\int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx$ as... $$\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx - \int^{\xi_{n}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$$ If we consider $\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ as a sequence, then $\lim_{n\to\infty} \int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx = \frac{1}{2}$
tells us that $\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ is
not a Cauchy sequence, and hence $\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ is divergent as $n \to \infty$.
Thus, we can finally claim that $\int^{\infty}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ is divergent.
Now, the part of my argument which I find to be weak is:
What the result above tells us is that as $x$
increases, the period of $\sin^2(\pi(x+\frac{1}{x}))$ approaches the
period of $\sin^2(\pi x)$. Furthermore, $x+\frac{1}{x} \approx x$ as
$x\to \infty$, which implies that $\sin^2(\pi(x+\frac{1}{x}))$
"approaches" $\sin^2(\pi x)$ as $x \to \infty$. Using both of these
facts, we can claim that $\int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx$ approaches $\int^{1}_{0} \sin^2(\pi x) dx$ as $x \to \infty$.
How can I make this part more rigorous/stronger? I don't really know how to achieve any more rigor with my current level of mathematics.
| Here's another approach that uses your intuition that $x+1/x\sim x$ for large $x$ by doing a change of variable. Denoting the integrand by $f(x)$, we have
$$
\int_0^{\infty} f(x)\,dx = \int_0^1 f(x)\,dx +\int_1^{\infty}f(x)\,dx
$$Set $y=x+1/x$, or $x = \frac{y\mp \sqrt{y^2-4}}{2}$, with $-$ in the first region and $+$ in the second region. The limits of integration become $0\to\infty,$ $1\to 2,$ and $\infty\to\infty.$ Then we have $dx = \frac{1}{2}\left(1\mp \frac{y}{\sqrt{y^2-4}}\right)dy$, giving
$$
\frac{1}{2}\int_{\infty}^2\sin^2(\pi y) \left(1-\frac{y}{\sqrt{y^2-4}}\right)\,dy+\frac{1}{2}\int_2^{\infty}\sin^2(\pi y) \left(1+\frac{y}{\sqrt{y^2-4}}\right)\,dy
$$
$$
=\frac{1}{2}\int_2^{\infty}\sin^2(\pi y) \left(-1+\frac{y}{\sqrt{y^2-4}}\right)\,dy+\frac{1}{2}\int_2^{\infty}\sin^2(\pi y) \left(1+\frac{y}{\sqrt{y^2-4}}\right)\,dy
$$
$$
=\int_2^{\infty}\sin^2(\pi y)\frac{y}{\sqrt{y^2-4}}\,dy
$$This new integral is actually easier to work with, as it is clear $y/\sqrt{y^2-4}>1$ for $y>2$, and then your argument basically holds. One could also use
Cauchy-Bunyakovsky-Schwarz, or the First MVT for Integrals.
| {
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"url": "https://math.stackexchange.com/questions/4211726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Comparing $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the values
I want to compare $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the actual value of square roots.
I tried to apply square on both side but it still carries the root terms.
Any trick I was missing here?
| Use the Taylor expansions:
$$\sqrt{1+x}=\sqrt{x} + \frac1{2\sqrt x} - \frac1{8x\sqrt x}+\frac1{16x^2\sqrt x}-\frac5{128x^3\sqrt x}+O(x^{-4})\\
\sqrt{2+x}=\sqrt{x} + \frac1{\sqrt x} - \frac1{2x\sqrt x}+\frac1{2x^2\sqrt x}-\frac5{8x^3\sqrt x}+O(x^{-4})$$
Now we will plug in and calculate (manually possible, though tedious):
$$2.23606...=\sqrt{1+4}>2+\frac1{4}-\frac1{64}+\frac1{512}-\frac5{16384}\approx 2.23602;\\
2.4494...=\sqrt{2+4}>2+\frac12-\frac1{16}+\frac1{64}-\frac5{1024}\approx 2.4482;\\
3.3166...=\sqrt{2+9}>3+\frac13-\frac1{54}+\frac1{486}-\frac5{17496}\approx 3.3165.$$
Hence:
$$\sqrt5+\sqrt6+\sqrt{11}>8.00062>8$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating integral $\int_0^{2\pi}\frac{1}{2-\sin(x)}\,dx$ I come across a problem, which is to evaluate: $$\int_0^{2\pi}\frac{1}{2-\sin(x)}\,dx$$
My attempt so far is:
$$\begin{align}\int_0^{2\pi}\frac{1}{2-\sin(x)}\,dx
&= \int_0^{2\pi}\frac{1}{2-\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}}\,dx\\
&= \int_0^{2\pi}\frac{1+\tan^2(\frac{x}{2})}{2+2\tan^2(\frac{x}{2})-2\tan(\frac{x}{2})}\,dx\\
&= \int_0^{2\pi}\frac{\frac{1}{2} \sec^2(\frac{x}{2})}{1+\tan^2(\frac{x}{2})-\tan(\frac{x}{2})}\,dx\end{align}$$
I'm wondering whether I'm in the right direction and how can I proceed?
| This is the step-by-step process I mentioned in the link to deal with the discontinuity at $x=\pi$.
Using Weierstrass substitution $t = \tan\left(\frac{x}{2}\right)$, where $-\pi < x < \pi$ or $\pi < x < 3\pi$, then $\sin(x) = \frac{2t}{1+t^2}$, $dx = \frac{2}{1+t^2}dt$, there is a discontinuity at $x=\pi$ that needs to be taken into account, first you find the indefinite integral and then evaluate the limits of the endpoints of the improper integrals. First split the integral:
$$\int\limits_0^{2\pi}\frac{1}{2-\sin(x)}\:dx = \int\limits_0^{\pi}\frac{1}{2-\sin(x)}\:dx +\int\limits_\pi^{2\pi}\frac{1}{2-\sin(x)}\:dx.$$
Then evaluate the indefinite integral. Since $\frac{1}{2-\frac{2t}{1+t^2}}= \frac{1+t^2}{2(t^2-t+1)}$, $\frac{1}{2-\sin(x)}\:dx = \frac{1}{t^2-t+1}\:dt$, hence
$$\int\frac{1}{2-\sin(x)}\:dx = \int\frac{1}{t^2-t+1}\:dt,$$
substituting $t = s + \frac{1}{2}$ to get rid of the $-t$, then $dt=ds$, and $$\int\frac{1}{t^2-t+1}\:dt = \int\frac{1}{s^2 +\frac{3}{4}}\:ds = \frac{2}{\sqrt3}\int\frac{1}{\left(\frac{2s}{\sqrt3}\right)^2+1}\:\frac{2}{\sqrt3}ds =\\ \frac{2}{\sqrt3}\int\frac{1}{u^2+1}\:du = \frac{2}{\sqrt3}\arctan(u) $$
with $u =\frac{2s}{\sqrt3}$ and the answer of the indefinite integral is := $\frac{2}{\sqrt3}\arctan\left(\frac{2\tan\left(\frac{x}{2}\right)-1}{\sqrt3}\right)$
$$\int\limits_0^{\pi}\frac{1}{2-\sin(x)}\:dx +\int\limits_\pi^{2\pi}\frac{1}{2-\sin(x)}\:dx = \\\lim_{b\rightarrow\pi^{-}}\frac{2}{\sqrt3}\arctan\left(\frac{2\tan\left(\frac{x}{2}\right)-1}{\sqrt3}\right)\Biggr|_0^b + \lim_{b\rightarrow\pi^{+}}\frac{2}{\sqrt3}\arctan\left(\frac{2\tan\left(\frac{x}{2}\right)-1}{\sqrt3}\right)\Biggr|_b^{2\pi} = \\ \lim_{b\rightarrow\pi^{-}}\frac{2}{\sqrt3}\arctan\left(\frac{2\tan\left(\frac{b}{2}\right)-1}{\sqrt3}\right) - \lim_{b\rightarrow\pi^{+}}\frac{2}{\sqrt3}\arctan\left(\frac{2\tan\left(\frac{b}{2}\right)-1}{\sqrt3}\right) - \frac{2}{\sqrt3}\arctan\left(\frac{-1}{\sqrt3}\right) + \frac{2}{\sqrt3}\arctan\left(\frac{-1}{\sqrt3}\right) = \frac{2}{\sqrt3}\cdot\frac{\pi}{2} - \left(-\frac{2}{\sqrt3}\cdot\frac{\pi}{2}\right) + 0 = \frac{2\pi}{\sqrt3}$$
since you can't evaluate at $\pi$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Hard and non-trivial inequality with three variable reals For all reals $a$, $b$, $c$, show that $$a^2+b^2+c^2 \geq a\sqrt[\leftroot{-1}\uproot{1}4]{\frac{b^4+c^4}{2}} + b\sqrt[\leftroot{-1}\uproot{1}4]{\frac{c^4+a^4}{2}} + c\sqrt[\leftroot{-1}\uproot{1}4]{\frac{a^4+b^4}{2}}.$$
I tried to use Holder inequality:
$$\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)(a^2+b^2+c^2)(a^2+b^2+c^2)\Big(b^4+c^4+c^4+a^4+a^4+b^4\Big) \geq \text{RHS}^4$$
or
$$(a^2+b^2+c^2)^4 \geq (1+1+1)(a+b+c)^2 \Big(\sum_\text{cyc}{\frac{a^2(b^4+c^4)}{2}} \Big) \geq \text{RHS}^4$$
But got stuck after that (the $\geq$ is reversed).
| It's enough to prove our inequality for non-negatives $a$, $b$ and $c$.
Now, by C-S $$\sum_\text{cyc}a\sqrt[4]{\frac{b^4+c^4}{2}}\leq\sum_\text{cyc}a\sqrt{b^2-bc+c^2}\leq\sqrt{\sum_\text{cyc}a\sum_\text{cyc}a(b^2-bc+c^2)}\leq\sum_\text{cyc}a^2$$
because $$\sqrt[4]{\frac{b^4+c^4}{2}}\leq\sqrt{b^2-bc+c^2}$$ it's just $$(b-c)^4\geq0$$ and
$$\sqrt{\sum_\text{cyc}a\sum_\text{cyc}a(b^2-bc+c^2)}\leq\sum_\text{cyc}a^2$$ it's
$$\sum_\text{cyc}(a^4-a^3b-a^3c+a^2bc)\geq0,$$ which is Schur.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$x_1$ , $x_2$ are roots of $x=5-x^2$. Find the equation with roots $\frac1{(x_1+1)^3}$ and $\frac1{(x_2+1)^3}$.
Suppose $x_1$ , $x_2$ are the roots of the equation $x=5-x^2$.Then
$\dfrac1{(x_1+1)^3}$ and $\dfrac1{(x_2+1)^3}$ are roots of which
equation?
$1)125x^2+16x=1$
$2)125x^2=16x+1$
$3)125x^2=12x+1$
$4)125x^2+12x=1$
I solved this problem with the following approache,
I've denoted the roots of the original equation by $\alpha$ and $\beta$ rather than $x_1$, $x_2$ ,
$S=\alpha+\beta=-1$ and $P=\alpha\beta=-5$. We find $S'$ and $P'$ of the new equation,
$$P'=\dfrac1{[(\alpha+1)(\beta+1)]^3}=\dfrac1{(S+P+1)^3}=-\frac1{125}$$
$$S'=\dfrac{(\beta+1)^3+(\alpha+1)^3}{[(\alpha+1)(\beta+1)]^3}=-\frac1{125}\times\left((\alpha^3+\beta^3)+3(\alpha^2+\beta^2)+3(\alpha+\beta)+2\right)$$
$$=-\dfrac{(S^3-3PS)+3(S^2-2P)+3S+2}{125}=-\dfrac{-16+33-3+2}{125}=-\frac{16}{125}$$
Hence the new equation is $125x^2+16x-1=0$. And the answer is first choice.
This was a problem from a timed exam. So can you solve it with other approaches (preferably quicker one) ?
| As you can see the roots given for the new equation are of the format $\frac{1}{(x+1)^3}$ hence you can use root transformation, either you directly transform $$x \rightarrow \frac{1}{(x+1)^3}$$ or you can do $$x\rightarrow \frac{1}{1+x}$$ and then calculate a bit, I'm going to do using the second method as it reduces the problems and calculations involving irrationalities. Say $$y=\frac{1}{1+x}\implies x=\frac{1-y}{y}$$ Now putting $\bf x$ on the original equation
I get,$$\frac{1-y}{y}=5-\left(\frac{1-y}{y}\right)^2$$ $$\implies5y^2+y-1=0$$
Now, you can observe that the roots of this equation are $\frac{1}{1+x_1},\frac{1}{1+x_2}$ .We see that the roots we want are just these roots raised to the power $3$ Hence, sum of the roots we need is$-(\frac{1}{5})^3+\frac{3}{5}\times\frac {-1}{5}=\frac{-16}{125}$ similarly you can count the product of roots which will be $\frac{-1}{5}$ hence the equation we need is $$\bf 125x^2+16x-1=0$$
$\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the possible number of unique google meet codes? This is a sample google meet code: kgk-imsy-era
Any idea how many such unique codes can be there?
Given that a particular letter in the code may or may not repeat. If it repeats, we do not know how may times it'll repeat.
Is it something like
$$^{10}C_{x} \cdot 26^{x} \cdot ^{25}P_{(10-x)} + ^{26}P_{10}$$
where, x = number of times a particular letter is repeated
Few points to note:
*
*The codes are always composed of 10 letters (and no numbers)
*The codes are case-insensitive
*The hyphens in between are just used to separate the letters and are not a part of the code
Thanks!
| I do not know about the procedure of google meeting codes , but i will write my answer according to given information by you.
We have two option such that a particular letter repeat or not repeat.
If it the particular letter does not repeat :
Firstly ,select this letter by $C(26,1)=26$ ways and select the rest by $C(25,9)$.After that arrange them by $P(10,10)$ .Then the answer is $$C(26,1) \times C(25,9) \times P(10,10)$$
If it the particular letter does repeat :
Firstly ,select this letter by $C(26,1)=26$ ways .However , we do not know how many times it will repeat ,so we will use generating function to calculate the all possible arrangements of letters such that
Generating function for selected letter = $$ \frac{x^2}{2} + \frac{x^3}{3!} + ..$$
$\color{red}{NOTE=}$ We started by $\frac{x^2}{2!}$ to avoid overcounting when the letter occurs only once
Generating function for the rest of letters = $$1+ x + \frac{x^2}{2!} + \frac{x^3}{3!} + ..$$
So ,we should find the coefficient of $x^{10}$ and multiply it by $10!$ or find the coefficient of $\frac{x^{10}}{10!}$ in the expansion of $$\bigg( \frac{x^2}{2} + \frac{x^3}{3!} + ..\bigg) \times \bigg(1+x + \frac{x^2}{2} + \frac{x^3}{3!} + ..\bigg)^{25} $$
You can find the answer by wolfram - alpha , so the answer will be the sum of the two condition such that $$\bigg[C(26,1) \times C(25,9) \times P(10,10) \bigg] + C(26,1) \times [x^{10}] \bigg[\bigg( \frac{x^2}{2} + \frac{x^3}{3!} + ..\bigg) \times \bigg(1+x + \frac{x^2}{2} + \frac{x^3}{3!} + ..\bigg)^{25} \bigg] $$
NOTE $2 =$ In my solution , it is assumed that the particular letter must occur.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all primes $r$ and $s$ such that $ r^7+s=s^3+r $
Find all primes $r$ and $s$ such that $\;r^7+s=s^3+r\;.$
I have tried to factorise the expression since we are dealing with primes so:
$$ s(s^2-1)= r(r^6-1)=r(r^2-1)(r^4+r^2+1).$$
Since $ s^3>r^7$, then $\;s^2-1>r^4-1$.
That is all what I have found, thank you in advance for your precious help!
| This isn't a full-fledged answer yet but it is too long for a comment: On the one hand,
$$r|(s^2-1) \Rightarrow s=kr \pm 1$$ $$\Rightarrow s \equiv_r \pm 1.$$
On the other hand, $s \in \theta(r^{\frac{7}{3}})$ and $s$ divides one of $r^3-1$, $r^3+1$ gives $as = r^3\pm 1$ for some $a \in O(r^{3-\frac{7}{3}}) =O(r^{\frac{2}{3}})$.
But $as \equiv_r \pm a$ because $s \equiv_r \pm 1$, so either $a=1$ [which is impossible] or $a \ge r-1$. But both $a \in O(r^{\frac{2}{3}})$ and $a\ge r-1$ is impossible for $r$ sufficiently large. So the solution set is bounded.
ETA: You could also use the fact that both $r^3-1$ and $r^3+1$ factor further to polynomials no larger than $r^2+r+1$ [as noted in @John Omielan 's answer below, I had forgotten this], to conclude that $s \le r^2+r+1$, because $r,s$ prime $\Rightarrow$ $s$ must divide one of $r^2+r+1$, $r^2-r+1$, $r-1$, $r+1$. Then this gives $r^7-r \le (r^2+r+1)^3-(r^2+r+1)$ [because $y^3-y$ is strictly increasing in $y$ for $y \ge 3$]. All solutions for $r^7-r \le (r^2+r+1)^3-(r^2+r+1)$ require $r \le 13$. This leaves very few possibilities to check, check the value of $s$ that satisfies $s^3-s=r^7-r$ for each $r \in \{3,5,7,11,13\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$3\times 2^m + 4 = n^2$ I have been trying to solve this problem but I couldn't find any way to find the answers. Find all natural numbers m and n such that:
$3\times\ 2^m + 4 = n^2$
This is a question from the Moroccan Maths Olympiad (2019.)
I tried bringing $4$ to the other side and factorising to get: $3\times 2^m=(n+2)(n−2),$ and I tried solving for that $n+2$ is a multiple of three then $n-2$ is a multiple of three but I couldn't find anything.
Can anyone help me please?
| Assume that $m, n \in \mathbb{N}$. From the equation, you can deduce that $$2 \mid n^2 \implies 2 \mid n\implies n = 2k \implies 3\cdot 2^m + 4 = 4k^2\quad(1)$$ Observe that $m = 1$ yields no integer solution, and $m = 2 \implies k = 2 \implies n = 4$.Thus consider $m \ge 3$, divide both sides of $(1)$ by $4$, you have: $3\cdot 2^{m-2}+1=k^2\implies k$ is odd $$\implies k = 2q+1\implies 3\cdot2^{m-2}+1=(2q+1)^2 \implies 3\cdot2^{m-2} = 4q(q+1)\quad(2)$$ Observe also that $m = 3,4$ yields no integer solution in $q$. So $m \ge 5$, and $(2)$ gives $$3\cdot 2^{m-4} = q(q+1)$$ $$\implies q =2, q+1 = 3 \implies q =2^{m-4}$$ $$\implies q =3, q+1 = 4 =2^{m-4}$$ $$\implies m =5\implies k = 5\implies n = 10 \implies (m,n) = (5,10)$$ $$\implies m =6\implies k = 7\implies n = 14 \implies (m,n) = (6,14)$$ are the remaining solutions. So altogether, the solutions are: $(2,4), (5,10), (6,14)$ are the $3$ solutions and they are the only solutions to the equation.
| {
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Finding $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right)$
What is the limit of $\,\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right)$ ?
My thought was $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\left(\frac{1}{\sin^2 x} - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\left(\frac{1}{x^2} - \frac{1}{x^2} \right) = 0$. I used $\lim\limits_{x\to 0} \frac{\sin x}{x} = 1$.
But $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\frac{x^2-\sin^2x}{x^2\sin^2x}$ and if I apply L'Hospital's Rule four times at the second expression, I get $\lim\limits_{x\to 0}\,\left(\frac{8\cos 2x}{24\cos 2x - 32\sin 2x - 8x^2\cos 2x}\right) = \frac {1}{3}$.
What am I missing?
| Without L'Hopital, we can rewrite the limmand as
$$\frac{x-\sin x}{x^3}\cdot\frac{x+\sin x}{x}$$
The limit of the product is the product of the limits when both exist. The term on the right goes to $2$, so all there is left is to evaluate the term on the left. By using the identity
$$\sin 3x = 3\sin x - 4\sin^3x$$
notice that
$$L = \lim_{x\to0}\frac{x-\sin x}{x^3} = \lim_{u\to0}\frac{3u-\sin 3u}{(3u)^3} = \lim_{u\to 0}\frac{3u - 3\sin u +4\sin^3 u}{27u^3}$$
$$ = \lim_{u\to 0}\frac{u-\sin u}{9u^3}+\frac{4\sin^3u}{27u^3} = \frac{1}{9}L+\frac{4}{27} \implies L = \frac{1}{6}$$
Therefore the original limit is $\frac{1}{6}\times 2 = \boxed{\frac{1}{3}}$
| {
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Finding the sum of the series $1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$ The following is a question which has been bugging me for quite a while,
Find the sum of the series
$$1^2 + 2 × 2^2 + 3^2 + 2 × 4^2 + 5^2 + 2 × 6^2 + . . . + 2(n − 1)^2 + n^2 ,$$
Where $n$ is odd
I started by denoting this entire series with $S$, from there it is apparent that $S$ infact consists of two lesser series I shall call;
*
*$S_A$ The sum of the squares of odd natural numbers up till $n^2$
*$S_B$ Twice the sum of the squares of even natural numbers up till $(n-1)^2$
I resolved this would be easier to tackle by noting that
$$S_A + \frac{1}{2}S_B = \sum_{r=1}^{n} r^2$$
Which is just the sum of the squares of the first $n$ natural numbers
$$\therefore S_A +\frac{1}{2}S_B=\frac{n}{6}(n+1)(2n+1)$$
Leaving only the value of $\frac{1}{2}S_B$ to be found, this is where I am currently facing difficulty as I am unsure on whether my working is correct;
For the sum of the squares of the first n even natural numbers;
$$2^2 + 4^2 .... (2n)^2=2^2\sum_{r=1}^{n} r^2$$
$$\implies \frac{2}{3}n(n+1)(2n+1) $$
Hence the sum of the first $n-1$ even natural numbers should be
$$ \frac{2}{3}n(n-1)(2n-1)$$
And
$$S= \frac{n}{6}(n+1)(2n+1) + \frac{2}{3}n(n-1)(2n-1)$$
$$\therefore S= \frac{1}{6}n(10n^2 -9n + 5n) $$
However the correct answer is
$$\frac{1}{2}n^2(n+1)$$
Where has my working gone wrong and how would I arrive at the correct answer?
| Let f(n) be the sum of squares from 1 up to $n^2$. Then the answer to your question is f(n) + 4 f((n-1)/2). Find a formula for f(n) and substitute it. It should be close to 1.5 f(n) since you are adding about half of the terms twice.
| {
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$u_1 + \sqrt{u_1u_2} + \sqrt[3]{u_1u_2u_3} + \cdots \leq e(u_1+u_2+u_3 +\cdots)$. $\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\lrp}[1]{\left(#1\right)}$
IMO Longlisted Problem 28.
Let $u_1, \ldots, u_n$ be a positive real numbers.
For each $k$ in $\set{1, \ldots, n}$ define $v_k=\sqrt[k]{u_1 \cdots u_k}$.
Show that
$$
v_1 + \cdots + v_n \leq e(u_1 + \cdots + u_n)
$$
Attempt.
I am able to show this only for $n=1, 2, 3$.
For $n=1$ the statement is trivial.
For $n=2$, we have
$$
v_1+ v_2
= u_1 + \sqrt{u_1 u_2}
\leq u_1 + \frac{1}{2}(u_1+u_2)
\leq \lrp{1 + \frac{1}{2}} (u_1+u_2)
\leq e(u_1+u_2)
$$
For $n=3$ we have
$$
e(u_1+u_2 + u_3)
=
(u_1+u_2+u_3) + \frac{1}{2}(u_1+u_2+u_3) + \lrp{ \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \cdots } (u_1+u_2+u_3)
$$
which gives
$$
e(u_1+u_2+u_3)
\geq
u_1 + \frac{1}{2}(u_1 + u_2) + \lrp{ \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \cdots }(u_1+u_2+u_3)
$$
and hence
$$
e(u_1+u_2+u_3)
\geq
u_1 + \frac{1}{2}(u_1 + u_2) + 3\lrp{\lrp{1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \cdots} - (1 + 1/2) } \frac{u_1+u_2+u_3}{3}
$$
Therefore
$$
e(u_1+u_2+u_3)
\geq
u_1 + \frac{1}{2}(u_1 + u_2) + 3(e - 1.5) \frac{u_1+u_2+u_3}{3}
\geq
u_1 + \frac{1}{2}(u_1 + u_2) + \frac{1}{3}(u_1+u_2+u_3)
$$
which using AM-GM gives the desired.
| The Polya's proof in my writing.
Let $c_k=\frac{(k+1)^k}{k^{k-1}}.$
Thus, $c_1c_2...c_k=(k+1)^k$ and by AM-GM we obtain:
$$\sum_{k=1}^n\sqrt[k]{u_1 u_2 ... u_k}=\sum_{k=1}^n\frac{\sqrt[k]{u_1c_1 u_2c_2 ... u_kc_k}}{k+1}\leq\sum_{k=1}^n\frac{u_1c_1+u_2c_2+...+u_kc_k}{k(k+1)}=$$
$$=\sum_{k=1}^n u_kc_k\sum_{i=k}^n\frac{1}{i(i+1)}=\sum_{k=1}^n u_kc_k\left(\frac{1}{k}-\frac{1}{k+1}+\frac{1}{k+1}-\frac{1}{k+2}+...+\frac{1}{n}-\frac{1}{n+1}\right)<$$
$$<\sum_{k=1}^nu_kc_k\cdot\frac{1}{k}=\sum_{k=1}^nu_k\frac{(k+1)^k}{k^{k-1}}\cdot\frac{1}{k}=\sum_{k=1}^nu_k\left(1+\frac{1}{k}\right)^k<\left(1+\frac{1}{n}\right)^n\sum_{k=1}^nu_k<e\sum_{k=1}^nu_k.$$
| {
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Deducing an inequality from a Taylor series expansion Consider the inequality
$$
1-\frac{x}{2}-\frac{x^2}{2} \le \sqrt{1-x} < 1-\frac{x}{2}
$$
for $0 < x < 1$. The upper bound can be read off the Taylor expansion for $\sqrt{1-x}$ around $0$,
$$
\sqrt{1-x} = 1 - \frac{x}{2} - \frac{x^2}{8} - \frac{x^3}{16} - \dots
$$
by noting that all the non linear terms are negative. Can the left side inequality be read-off the expansion by a similar reasoning? Please do not try to prove the left side inequality by other means (such as minimizing $\sqrt{1-x} - 1 + \frac{x}{2} + \frac{x^2}{2}$ using derivatives).
| We need to prove that
$$(2-x-x^2)^2\leq4(1-x),$$ which is
$$x^2(3+x)(1-x)\geq0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
If $a^{2}+b^{2} \leq 4$, prove that $a+b \leq 4$ \begin{equation}
\text { If } a^{2}+b^{2} \leq 4 \text { prove that } a+b \leq 4 \text { }
\end{equation} What I have tried:
\begin{equation}
a^{2} \leq a^{2}+b^{2} \leq 4 \text { and } b^{2} \leq a^{2}+b^{2} \leq 4
\end{equation}
\begin{equation}
|a| \leq 2 \text { and }|b| \leq 2
\end{equation}
\begin{equation}\text { So } a+b
\text { can get the maximum value, then } a \geq 0 \text { and } b \geq 0 \text { }
\end{equation}\begin{equation}
\text { } 0 \leq a \leq 2 \text { and } 0 \leq b \leq 2 \text {}
\end{equation}
\begin{equation}
(a-b)^{2}=a^{2}+b^{2}-2 a b \geq 0
\end{equation}\begin{equation}
a^{2}+b^{2} \geq 2 a b \text { So at this step I am not certain what to do next. }
\end{equation}
| By rearrangement inequality we have
$$2ab\le a^2+b^2\le 4$$
and
$$(a+b)^2=a^2+b^2+2ab\le 8\implies |a+b|\le 2\sqrt 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Why does a boundedness proof of $\exp(x)$ depend on the sign of $x$? To prove $\lim\limits_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}$ exists, we prove that the sequence
$$f_n=\left(1+\frac{x}{n}\right)^n$$
is bounded and monotonically increasing toward that bound.
Proof Attempt:
We begin by showing $f_n=\left(1+\frac{x}{n}\right)^n$ is monotonically increasing by looking at the ratio of consecutive terms:
\begin{align*}
\frac{f_{n+1}}{f_n}
&=\frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}{n}\right)^{n}} \tag{Definition of $f_n$} \\
&=\frac{\left(1+\frac{x}{n+1}\right)^{n+1}\left(1+\frac{x}{n}\right)}{\left(1+\frac{x}{n}\right)^{n}\left(1+\frac{x}{n}\right)} \tag{Multiplication by $\frac{\left(1+\frac{x}{n}\right)}{\left(1+\frac{x}{n}\right)}$} \\
&=\frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}{n}\right)^{n+1}}\left(1+\frac{x}{n}\right) \tag{Simplify $a^n\cdot a = a^{n+1}$} \\
&=\left(\frac{1+\frac{x}{n+1}}{1+\frac{x}{n}}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Simplify $\frac{a^{n+1}}{b^{n+1}}=\left(\frac{a}{b}\right)^{n+1}$} \\
&=\left(\frac{\frac{n+1+x}{n+1}}{\frac{n+x}{n}}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Common denominators} \\
&=\left(\frac{n+1+x}{n+1}\cdot \frac{n}{n+x}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Simplify $\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{a}{b}\cdot \frac{d}{c}$} \\
&=\left(\frac{n^2+n+nx}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Distribute $(n+1+x)n$} \\
&=\left(\frac{n^2+n+nx+x-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Add and subtract $x$} \\
&=\left(\frac{(n+1)(n+x)-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Factor $n^2+n+nx+x$} \\
&=\left(1+\frac{-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}{n}\right) \tag{Simplify $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$} \\
&\ge\left(1+\frac{-x}{(n+x)}\right)\left(1+\frac{x}{n}\right) \tag{Bernoulli: $(1+x)^n \ge 1+nx$} \\
&=\left(\frac{n}{n+x}\right)\left(\frac{n+x}{n}\right) \tag{Common denominators} \\
&=1 \tag{Simplify $\frac{a}{b} \cdot \frac{b}{a}=1$}
\end{align*}
Since $\frac{f_{n+1}}{f_n}>1$, then $f_{n+1}>f_n$, which shows the sequence $f_n$ is monotonically increasing for all $n \in \mathbb{N}$.
Next, we show $f_n=\left(1+\frac{x}{n}\right)^n$ is bounded above. Note that
\begin{align*}
f_n
&=\left(1+\frac{x}{n}\right)^n \tag{Definition of $f_n$} \\
&=\sum_{k=0}^n \binom{n}{k} (1)^{n-k} \left(\frac{x}{n}\right)^{k} \tag{Binomial Theorem} \\
&=1+\frac{n}{1!}\left(\frac{x}{n}\right)+\frac{n(n-1)}{2!}\left(\frac{x}{n}\right)^2+\frac{n(n-1)(n-2)}{3!}\left(\frac{x}{n}\right)^3+\cdots+\frac{n!}{n!}\left(\frac{x}{n}\right)^n \\
&=1+\frac{\frac{n}{n}}{1!}x+\frac{\frac{n(n-1)}{n^2}}{2!}x^2+\frac{\frac{n(n-1)(n-2)}{n^3}}{3!}x^3+\cdots+\frac{\frac{n!}{n^n}}{n!}x^n \tag{Simplify}\\
&=1+\frac{1}{1!}x+\frac{\left(1-\frac{1}{n}\right)}{2!}x^2+\frac{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)}{3!}x^3+\cdots+\frac{\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{n-1}{n}\right)}{n!}x^n \\
& \le 1+\frac{1}{1!}x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots+\frac{1}{n!}x^n \tag{$1-\frac{k}{n}<1$} \\
& = \sum_{k=0}^n \frac{1}{k!} x^k \tag{Sigma notation}\\
& \to %\underset{n \to \infty}{\to}
\sum_{k=0}^\infty \frac{1}{k!}x^k \tag{as $n \to \infty$} \\
& = \sum_{k=0}^K \frac{1}{k!} x^k + \sum_{k=K+1}^\infty \frac{1}{k!}x^k \tag{$\exists K$, $k>K$ implies $k! \ge (2x)^k$}\\
& \le \sum_{k=0}^K \frac{1}{k!} x^k + \sum_{k=K+1}^\infty \frac{1}{(2x)^k} x^k \tag{$k! \ge (2x)^k$ implies $\frac{1}{k!} \le \frac{1}{(2x)^k}$}\\
& = \sum_{k=0}^K \frac{1}{k!} x^k + \sum_{k=K+1}^\infty \frac{1}{2^k} \tag{$\frac{1}{(2x)^k}x^k=\frac{1}{2^k x^k}x^k = \frac{1}{2^k}$}\\
&= \sum_{k=0}^K \frac{1}{k!} x^k + \frac{1}{2^K} \tag{Geometric series evaluation}
\end{align*}
which is finite. Thus, the sequence $f_n$ is bounded. Since it is both monotonically increasing and bounded, it is convergent by the Monotone convergence theorem.
Is my proof correct? I am suspicious of the step which says "$\rightarrow \sum_{k=0}^n \frac{1}{k!}x^k$", and would like to avoid taking another limit in the middle of the boundedness proof.
I also compared my proof to the following references and saw something worrisome:
*
*Reference 1 <- Assumes $x\ge 0$ (Why?)
*Reference 2 <- Assumes $x \ge -1$ (Why?)
*Reference 3 <- Considers $x=0$, $x>0$, and $x<0$ separately (Why?)
All of the above proofs either assumed $x>0$ or considered cases where $x>0$ and $x<0$ separately, but I do not know why. In fact, the third reference considers $\left(1-\frac{x}{n}\right)^{-n}$ for $x>0$ (I think this is a typo and should read $x<0$), but I am not sure why the negative exponent is needed (we are talking about a negative value of $x$, not negative $n$.)
I could only find one proof that did not consider different cases on the sign of $x$:
*
*Bonus reference 4 <- Uses absolute values, but I am not sure why these are necessary either.
I would like to verify my proof and ask 3 questions:
*
*Why is it necessary to consider cases $x>0$ and $x<0$ separately? Did any step in my proof implicitly assume that $x>0$? If so, which one?
*Is there any way to avoid taking a limit in the middle of the boundedness proof?
*Substituting $n=1$ in my boundedness proof shows $1+x \le \sum_{k=0}^n \frac{1}{k!}x^k$. Does this imply $1+x \le \lim\limits_{n\to \infty} \left(1+\frac{x}{n}\right)^n$, since $f_n$ is an increasing function of $n$? Can this be seen explicitly, or would that require a separate proof?
Thank you.
| *
*You application of Bernoulli's inequality is correct as long as $y_n=-\frac{x}{(n+x)(n+1)}\geq-1$. In reference (2) which you mention in your posting, Mark Viola assumes $x> -1$, in which case $y_n>-1$ for all $n\geq1$ and so, the sequence $f_n$ (in your notation) is monotone increasing for all $n\in\mathbb{N}$. However, he also points out that for any $x$, one can consider the first $n_0\in\mathbb{N}$ so that $n_0+x>0$. Then $y_n>-1$ for all $n\geq n_0$ and so, the sequence $f_n$ is monotone increasing for all for $n\geq n_0$. This shows that the limit exists for all $x\in\mathbb{R}$.
*An alternative approach shows that it is enough to consider the existence of the limit $e=\lim_n(1+\tfrac1n)^n$ to show that $\lim_n(1+\tfrac{a}{n})^n$ exists and equals $e^a$ for all $a\in\mathbb{R}$.
Indeed, from the existence of $e=\lim_n(1+\tfrac1n)^n$, we have that
$\lim_{x\rightarrow\infty}(1+\tfrac{1}{x})^x$ ($x$ ranges along positive reals) exists and has limit $e$. To see this, notice that
$n_x:=\lfloor x\rfloor\leq x\leq \lfloor x\rfloor +1 = n_x+1$ hence
$$\big(1+\tfrac{1}{n_x+1}\big)^{n_x}\leq\big(1+\tfrac1x\big)^{n_x}\leq\big(1+\tfrac1x\big)^x\leq\big(1+\tfrac{1}{n_x}\big)^x\leq\big(1+\tfrac{1}{n_x}\big)^{n_x+1}
$$
As $n_x\xrightarrow{x\rightarrow\infty}\infty$, one gets that
$$
\lim_{n_x\rightarrow\infty}\big(1+\tfrac{1}{n_x+1}\big)^{n_x}=\lim_{n_x\rightarrow\infty}\frac{\big(1+\tfrac{1}{n_x+1}\big)^{n_x+1}}{1+\tfrac{1}{n_x+1}}=e
$$
and
$$
\lim_{n_x\rightarrow\infty}\big(1+\tfrac{1}{n_x}\big)^{n_x+1}=\lim_{n_x\rightarrow\infty}\big(1+\tfrac{1}{n_x}\big)^{n_x}\big(1+\tfrac{1}{n_x+1}\big)=e
$$
Therefore $\lim_{x\rightarrow\infty}\big(1+\tfrac1x\big)^x=e$.
As a consequence, for $a>0$, we have
$$\lim_{n\rightarrow\infty}\big(1+\tfrac{a}{n}\big)^n=\lim_{n\rightarrow\infty}\left(\big(1+\tfrac{a}{n}\big)^{\tfrac{n}{a}}\right)^a=e^a
$$
Similarly, for $x<0$, let $y=-x$ so that $y>0$.
$$\big(1+\tfrac{1}{x}\big)^x=\big(1-\tfrac1y\big)^{-y}=\left(\frac{1}{1-\frac1y}\right)^y=\left(1+\tfrac{\tfrac1y}{1-\tfrac1y}\right)^y=\left(1+\tfrac{1}{y-1}\right)^{y-1}\big(1+\tfrac{1}{y-1}\big)$$
Thus,
$$\lim_{x\rightarrow-\infty}\big(1+\tfrac1x\big)^x=\lim_{y\rightarrow\infty}\left(1+\tfrac{1}{y-1}\right)^{y-1}\big(1+\tfrac{1}{y-1}\big)=\frac1e$$
As a consequence, for $a>0$
$$\lim_{n\rightarrow\infty}\big(1-\tfrac{a}{n}\big)^n=\lim_{n\rightarrow\infty}\left(\big(1-\tfrac{a}{n}\big)^{-\tfrac{n}{a}}\right)^a=e^{-a}$$
*
*Notice that all these requires a definition of real powers that is a rigorous definition of $a^x$ for $a>0$ and $x\in\mathbb{R}$. This of course is done by starting with rational powers, show monotonicity and then use axioms such as that of the supremum or its equivalents. Other modern approaches are based on integration. One defines $\ln(x)=\int^x_1\frac{1}{t}\,dt$ for $x>0$ and shows that $\ln$ is a nice continuous, monotone increasing functions (furthermore, it is differentiable by the fundamental theorem of Calculus) taking values in all of $\mathbb{R}$, with the property that $\log(ab)=\log(a)+\log(b)$ for all $a,b>0$. Then, the exponential function is defined as the inverse of $\ln$. Theorems such as the inverse function theorem, would imply differentiability
and the property that $\exp(a+b)=\exp(a)\exp(b)$. The number $e=\exp(1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $f(x)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+ax+1$ and $f(x) = 0$ has two distinct negative roots and equal positive roots, find least integral value of a If $f(x)={{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+ax+1$ be a polynomial where a and b are real numbers and $f(x) = 0$ has two distinct negative roots and equal positive roots, find least integral value of a.
Since the function has equal positive roots, I expect $f'(\alpha) = 0$ and $f''(\alpha) > 0$ for the positive root $\alpha$ But I'm not sure how I should proceed in this question. What are the different methods of doing this problem?
| You have
$$\begin{aligned}
f(x)&=x^4+ax^3+bx^2+ax+1\\
&= x^2\left(x^2+ax+b+\frac{a}{x}+\frac{1}{x^2}\right)\\
&=x^2\left(\left(x + \frac{1}{x}\right)^2+ a\left(x + \frac{1}{x}\right)+b-2\right)
\end{aligned}$$
As $x=0$ is not a root of $f(x)$, $x$ is a root of $f(x)$ if and only if $X =x + \frac{1}{x}$ is a root of $g(X)= X^2+a X + b-2$. As all roots of $f$ are supposed to be real, so are the ones of $g$. Hence $\Delta = a^2 + 8-4b \gt0$. The roots of $g$ are
$$X_1 = \frac{-a - \sqrt{\Delta}}{2} \lt 0 \text{ and } X_2 = \frac{-a + \sqrt{\Delta}}{2} \gt 0$$
If $x_1 \lt x_2 \lt 0 \lt x_3=x_4$ are the roots of $f$, we have $X_2 = 2$ as $2$ is the only value for which the equation $x + \frac{1}{x}=X_2$ has a unique positive root. Which implies
$$a^2 + 8-4b = (4+a)^2 = 16+8a+a^2 \iff 2a+b+2=0, $$ hence $$\Delta=a^2+8-4b= a^2+8+8a+8=a^2+8a+16.$$ And as the negative roots of $f$ are distinct$$X_1 = \frac{-a - \sqrt{\Delta}}{2}\lt -2.$$ Which leads to
$$a^2+8a+16 \gt (4-a)^2$$ i.e.
$$16a \gt 0$$ i.e. $a \gt 0$. The minimum integral value for $a$ is therefore $1$. In that case, $$\Delta = 25, X_2=2, x_3=x_4=1$$ and $x_1,x_2$ are the values for which $$x + \frac{1}{x} = -3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 2
} |
Evaluate : $S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$ Evaluate:$$S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7} + \frac{1}{9\cdot10\cdot11}+\cdots$$to infinite terms
My Attempt:
The given series$$S=\sum_{i=0}^\infty \frac{1}{(4i+1)(4i+2)(4i+3)} =\sum_{i=0}^\infty \left(\frac{1}{2(4i+1)}-\frac{1}{4i+2}+\frac{1}{2(4i+3)}\right)=\frac{1}{2}\sum_{i=0}^\infty \int_0^1 \left(x^{4i}-2x^{4i+1}+x^{4i+2}\right) \, dx$$
So,$$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4} + \frac{x^2}{1-x^4}\right)dx=\frac{1}{2} \int_0^1 \left(\frac{1+x^2}{1-x^4}-\frac{2x}{1-x^4}\right)\,dx = \frac{1}{2} \int_0^1 \left(\frac{1}{1-x^2}-\frac{2x}{1-x^4}\right)\,dx$$
$$=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\int_{0}^{1}\frac{2x}{1-x^4}dx=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\frac{1}{2}\int_{0}^{1}\frac{1}{1-y^2}dy=0(y=x^2)$$
which is obviously absurd since all terms of $S$ are positive.
But if I do like this then I am able to get the answer, $$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4}+\frac{x^2}{1-x^4}\right)dx=\frac{1}{2}\int_{0}^{1}\frac{(1-x)^2}{1-x^4}dx=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1+x}-\frac{x}{1+x^2}\right)dx=\frac{\ln2}{4}$$
What is wrong with the previous approach
| Try this.
$$
F(x) := \sum_{i=0}^\infty\frac{x^{4i+3}}{(4i+1)(4i+2)(4i+3)},\quad |x| \le 1
$$
Problem: compute $F(1)$.
Differentiate, to get geometric series
$$
F'''(x) = \sum_{i=0}^\infty x^{4i} = \frac{1}{1-x^4},\quad |x|<1
$$
Now integrate this three times, using $F(0) = F'(0) = F''(0)=0$.
$$
F(x) = \frac{x^2-1}{4}\arctan x + \frac{x^2}{4}\operatorname{atanh} x
+\frac{1}{8}\log(1+x) - \frac{1}{8}\log(1-x)
+\frac{x}{4}\log(1-x^2)-\frac{x}{4}\log(1+x^2)
$$
and then take limit
$$
\lim_{x\to 1^-}F(x) = \frac{\log 2}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
} |
Finding the range of $f(x)=\log_2\left(4^{x^2}+4^{(x-1)^2}\right)$
If the range of the function $f(x)=\log_2\left(4^{x^2}+4^{(x-1)^2}\right)$ is $\left(\frac pq,\infty\right)$, where $p, q$ are in their lowest form then find $(p+q), (p-q)$
Looks like the domain is $\mathbb R$.
If $x^2=0$, $(x-1)^2=1\implies f(0)=\log_2\left(1+4\right)=\log_25$
If $(x-1)^2=0, x^2=1\implies f(1)=\log_25$
So, I think $p=\log_25, q=1$
So, $p+q=\log_25+1, p-q=\log_25-1$
But the options are $1/3/4/5/10$.
I tried converting the base to $10$ and writing $5$ as $10/2$ but in vain.
| You have
$$\begin{align}4^{x^2}+4^{{(x-1)}^2}
&≥2\sqrt{4^{x^2+(x-1)^2}}\\
&=2^{{x^2+(x-1)^2}+1}\\
&≥2^{2\left(x-\frac 12\right)^2+\frac 32}\\
&≥2^{\frac 32}\end{align}$$
$$\log_2\left(4^{x^2}+4^{(x-1)^2}\right)≥\log_2{2^{\frac 32}}=\frac 32.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Solve $x+3<\sqrt{x+33}$ : Solution Verification First of all the part inside the square root should be greater than or equal to $0$. So,
$x+33 \ge0$
$\Rightarrow x \ge -33$
Now I solved the inequality,
$x+3<\sqrt{x+33}$
$\Rightarrow(x+3)^2<{x+33}$
$\Rightarrow x^2+9+6x<{x+33}$
$\Rightarrow x^2+5x-24 < 0$
$\Rightarrow (x+8)(x-3) < 0$
$\Rightarrow -8<x<3$
Now combining the two results i.e. $ x \ge -33$ and $ -8<x<3$ , the answer should be $ -8<x<3$ as per my understanding but this is not the correct answer that has been provided. What am I doing wrong? Please help me on this !!!
Thanks in advance !!!
| HINT
You can only square both sides iff both sides and the argument of the square root are nonnegative.
Under such conditions, the proposed inequation is equivalent to
\begin{align*}
x + 3 < \sqrt{x + 33} & \Longleftrightarrow
\begin{cases}
(x+3)^{2} < x + 33\\\\
x+3 \geq 0\\\\
x + 33 \geq 0
\end{cases}\\\\
& \Longleftrightarrow
\begin{cases}
x^{2} + 6x + 9 < x + 33\\\\
x\geq -3
\end{cases}\\\\
& \Longleftrightarrow
\begin{cases}
x^{2} + 6x - 34 < 0\\\\
x \geq -3
\end{cases}
\end{align*}
Moreover, notice whenever $-33 \leq x \leq -3$, the LHS is negative and the RHS is positive.
Hence the interval $[-33,-3]$ is also a solution.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4235929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find all odd functions $f:\mathbb R\to\mathbb R$ that satisfy $f(x+1)=f(x)+1$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$
Find all functions: $f:\mathbb R\to\mathbb R$ that satisfy all the following three conditions:
*
*$f(-x)=-f(x)$
*$f(x+1)=f(x)+1$
*$f\left(\frac{1}{x}\right)=\frac{f(x)}{x^2}$
I assume $f(x)=x$ satisfies the conditions. I could prove $f(0)=0 $ and $f(n)=n$ for all $n\in{\mathbb{N}}$, then I didn't know how to go any further.
| Well, I just saw Dan Vellerman's proof (hint actually), which is very elegant, and which I did not think of.
I will nevertheless add the proof that $f(x)=x$ on $\mathbb{Q}$ because the induction techniques used might be useful for some.
Let's prove that $f(x)=x$ on $\mathbb{Q}$.
*
*Notice that $f(0) = -f(-0) = -f(0)$. Therefore $f(0)=0$.
*Let's prove $f(x+n)=f(x)+n$ for all $n\in\mathbb{N}$. For $n=0$ this is obiously true. By induction $f(x+n+1) = f(x+1+n) = f(x+1)+n = f(x)+1+n = f(x) + n+1$. Therefore it is true for all $n\in\mathbb{N}$.
*Let us now prove $f\left( \frac{a}{b} \right) = \frac{a}{b}$ for all $a\in \mathbb{Z}$ and $b\in\mathbb{N}^+$. Suppose that $a\in\mathbb{N}$ and continue by induction on $a$. If $|a|=0$ we reduce to the first case above. Otherwise:
$$
f\left( \frac{a}{b} \right) = \frac{f\left( \frac{b}{a}\right) }{\left(\frac{b}{a}\right)^2} = \frac{f\left( \frac{(b \mod a) + \lfloor \tfrac{b}{a} \rfloor a}{a} \right) }{\left(\frac{b}{a}\right)^2} = \frac{f\left( \frac{(b \mod a) }{a} + \left\lfloor \tfrac{b}{a} \right\rfloor \right) }{\left(\frac{b}{a}\right)^2} = \frac{f\left( \frac{(b \mod a) }{a}\right) + \left\lfloor \tfrac{b}{a} \right\rfloor }{\left(\frac{b}{a}\right)^2} = \frac{\frac{(b \mod a) }{a} + \left\lfloor \tfrac{b}{a} \right\rfloor }{\left(\frac{b}{a}\right)^2} = \frac{\frac{b}{a}}{\left(\frac{b}{a}\right)^2} = \frac{a}{b}
$$
Note that the induction step was used in saying that $f\left(\frac{b \mod a}{a}\right) = \frac{b \mod a}{a}$ since $(b \mod a) < a$.
Finally if $a<0$ then $f\left(\frac{a}{b}\right) = f\left(-\frac{|a|}{b}\right) = -f\left(\frac{|a|}{b}\right) = -\frac{|a|}{b}= \frac{a}{b}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4240890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Derivative of $\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ I'm a calculus beginner. I was asked to find the derivative of the function: $$\sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt 2}.$$
I'm able to solve it in the following way:
I first calculate the derivative of $\frac{x+\sqrt{1-x^2}}{\sqrt 2}$ and get $\frac{1}{\sqrt 2}(1-\frac{x}{\sqrt{1-x^2}})$. Then the derivative of the given function is $\frac{1}{\sqrt{1-(\frac{x+\sqrt{1-x^2}}{\sqrt 2}})^2}\cdot \frac{1}{\sqrt 2}(1-\frac{x}{\sqrt{1-x^2}})$. Simplifying this gives the final answer $\frac{1}{\sqrt{1-x^2}}$. But the simplication process is quite lengthy and involves some bizarre calculations. Is there tricks/ways to solve these kinds of derivatives that do not involve too much calculations like above?
| Let $f(x) = \sin^{-1}\frac{x+\sqrt{1-x^2}}{\sqrt{2}} $. We have, for $\theta = \sin^{-1} x \in[-{\pi\over 2},{\pi\over 2}]$:
$$ f(\sin \theta) = \sin^{-1}\frac{\sin\theta+\sqrt{1-\sin^2\theta}}{\sqrt{2}} = \sin^{-1}\frac{\sin\theta+\cos\theta}{\sqrt{2}} = \sin^{-1}\big(\sin(\theta+{\pi\over 4})\big)$$
That is
$$ f(\sin\theta) = \left\{\begin{array}{ll}\theta+ {\pi\over 4} & \text{for }\theta\in [-{\pi\over 2},{\pi\over 4}] \\ \pi-(\theta+ {\pi\over 4})={3\pi\over 4}-\theta & \text{for } \theta\in [{\pi\over 4},{\pi\over 2}]\end{array} \right. $$
$$ f(x) = \left\{\begin{array}{ll}{\pi\over 4} + \sin^{-1} x& \text{for } x \in[-1,{\sqrt{2}\over 2}] \\ {3\pi\over 4} - \sin^{-1} x & \text{for } x \in[{\sqrt{2}\over 2},1]\end{array} \right. $$
Taking the derivative we have
$$ f'(x) = \left\{\begin{array}{ll}{1\over\sqrt{1-x^2}} & \text{for }x\in (-1,{\sqrt{2}\over 2}) \\ -{1\over\sqrt{1-x^2}} & \text{for } x \in({\sqrt{2}\over 2},1)\end{array} \right. $$
In the point $x={\sqrt{2}\over 2}$ the derivative doesn't exist (though left-sided and right-sided derivates do).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $2\sin{ix} -3i\cos{ix}=3i$ I have got an answer for this, but I know that it is wrong because it contains $\cos^{-1}\left(-\frac{3}{\sqrt{5}}\right)$, which is undefined. So I was wondering if the issue is with the method I am using or if I have made a mistake with the numbers.
I rewrote the equation as
$$3i\cosh{x}-2i\sinh{x}=-3i$$
by using $\cos{ix}=\cosh{x}$ and $\sin{ix}=i\sinh{x}$. I then compared this to the compound angle formula for $\cos{a+bi}$, which is:
$$r\cos(a+bi)=r\cos{a}\cosh{b}-ir\sin{a}\sinh{b}$$
So, letting $r\cos{a}=3i$ and $r\sin{a}=2$ gives me $a=\tan^{-1}\frac{2}{3i}$ and $r=i\sqrt{5}$. This means that my original equation becomes:
$$\cos\left(\tan^{-1}\left(-\frac{2}{3}i\right)+ix\right)=-\frac{3}{\sqrt{5}}$$ which is impossible because the $\cos$ function will only produce values between $-1$ and $1$.
I can't see anything that is wrong with what I have done so I was wondering if I am missing something.
| $$2\sin{ix} -3i\cos{ix}=3i<=>$$
$$ <=>\sqrt{4+9i^2} \sin \left(ix -\arcsin {\frac{3i}{\sqrt{4+9i^2}}}\right)=3i<=>$$
$$ <=>i \sqrt{5} \sin \left(ix -\arcsin {\frac{3}{\sqrt{5}}}\right)=3i<=>$$
$$ <=>\sin \left(ix -\arcsin {\frac{3}{\sqrt{5}}}\right)=\frac{3} {\sqrt{5}}<=>$$
$$ <=>\left[\begin{array} {1} ix -\arcsin {\frac{3}{\sqrt{5}}}=\arcsin {\frac{3}{\sqrt{5}}} + 2\pi n\\ ix -\arcsin {\frac{3}{\sqrt{5}}}=\pi-\arcsin {\frac{3}{\sqrt{5}}}+ 2\pi n \end{array} \right. <=>$$
$$ <=>\left[\begin{array} {1} ix =2\arcsin {\frac{3}{\sqrt{5}}} + 2\pi n\\ ix = \pi+ 2\pi n \end{array} \right. <=>$$
$$ <=>\left[\begin{array} {1} ix = \pi - i \ln{5} + 2\pi n\\ ix = \pi+ 2\pi n \end{array} \right. <=>$$
$$ <=>\left[\begin{array} {1} x = \pi i + \ln{5} + 2\pi n i\\ x = \pi i+ 2\pi n i \end{array} \right. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $ab+bc+ca=1$ what is the minimum value of $10a^2+10b^2+c^2$?
$a,b,c$ are positive real numbers such that $ab+bc+ca=1$. What is the minimum value of $10a^2+10b^2+c^2$?
I want to solve this problem without using Lagrange multipliers or calculus. I tried the following with some basic inequalities:
From AM-GM inequality $$5a^2+5b^2\geq10ab\\ 5b^2+\frac 12c^2\geq\sqrt{10}bc\\ \frac 12c^2+5a^2\geq\sqrt{10}ca$$
Summing them gives $$10a^2+10b^2+c^2\geq\sqrt{10}(\sqrt{10}ab+bc+ca)$$
But this doesn't help.
| Noting that
$$20 a^{2}+20 b^{2}+2 c^{2}-8(a b+b c+c a)=(2 a-2 b)^{2}+(4 b-c)^{2}+(4 a-c)^{2} \geqslant 0,
$$
we have
$$
\begin{aligned}
\quad 2\left(10 a^{2}+10 b^{2}+c^{2}\right)\geqslant 8 \quad \Rightarrow \quad 10 a^{2}+10 b^{2}+c^{2} \geqslant 4
\end{aligned}
$$
The minimum value of $10 a^{2}+10 b^{2}+c^{2}$ is $4$, which is attained if and only if$$
\left\{\begin{array}{l}
a=b \\
4 b=c \\
c=4 a
\end{array}\right.
\Leftrightarrow
a=b=\frac{1}{3} \text { and } c=\frac{4}{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Finding equation of tangent line to $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{6}$ at the point $(0,\frac{1}{2})$
Find the equation of the tangent line to $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{6}$ at the point $(0,\frac{1}{2})$
This is in the context of learning implicit differentiation.
First, I apply $\frac{dy}{dx}$ operator to both sides of the equation yielding:
$-\sin^{-2}(x) - \sin^{-1}(y)\frac{dy}{dx} = 0$
Second, I want to solve for $\frac{dy}{dx}$.
$\frac{dy}{dx} = -\sin^{-2}(x)\sin(y)$.
Third, I substitute the point $(0,\frac{1}{2})$ into the above equation to find the slope of the tangent line.
$\frac{dy}{dx}\mid_{(0,\frac{1}{2})} = -\sin^{-2}(0)\sin(\frac{1}{2}) = -0.479$
Finally, I substitute the slope into the point-slope equation of the line to obtain
$y = -0.479x + 0.2395$
Is this correct?
| Notice that $\frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}}$. This can be seen because, by definition of an inverse function, $\sin(\arcsin(x)) =x$. So differentiating both sides gives
$$
1 = \frac{d}{dx}x =\frac{d}{dx}\sin(\arcsin(x)) \overset{\color{purple}{\text{Chain rule}}}{=} \cos(\arcsin(x))\frac{d}{dx}\arcsin(x) =\underbrace{\sqrt{1 - \color{blue}{\sin^2(\arcsin(x)}}}_{\sqrt{1 -\color{blue}{x^2}}}\frac{d}{dx}\arcsin(x)
$$
where on the last step we use the fact that $\sin^2(x) +\cos^2(x) = 1$ to write $\cos(x) = \sqrt{1 - \sin^2(x)}$.
With this established, implicitly differentiating $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{6}$ with respect to $x$ gives
$$
\frac{1}{\sqrt{1-x^2}} + \frac{1}{\sqrt{1-y^2}} \frac{dy}{dx} =0 \implies \frac{dy}{dx} = -\sqrt{\frac{1-y^2}{1-x^2}}
$$
So substituting your point $\left(\color{purple}{x}, \color{blue}{y}\right) =\left(\color{purple}{0}, \color{blue}{\frac{1}{2}}\right)$ gives
$$
\frac{dy}{dx} = -\sqrt{\frac{1-\left( \color{blue}{\frac{1}{2}}\right)^2}{1-\left( \color{purple}{0}\right)^2}} = \color{green}{-\frac{\sqrt{3}}{2}}
$$
And by the slope-intercept form of a line, the desired tangent line is
$$
\boxed{y = \color{green}{-\frac{\sqrt{3}}{2}}x + \color{blue}{\frac{1}{2}}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that for every positive integer $n$, $1/3 + 1/9 + \cdots + 1/{3^n} < 1/2$ Base case is $n=1$: $\frac {1}{3} < \frac{1}{2}$. So the base case holds.
Inductive hypothesis for $n = k$: $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} < \frac{1}{2}$
Inductive Step for $n = k + 1$:
$$ \left( \frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} \right) + \frac {1}{3^{k+1}}< \frac{1}{2}.$$
Multiplying the $n = k + 1$ step by $3$:
$$ \left( 1 + \frac{1}{3} + \cdots + \frac {1}{3^{k-1}} \right) + \frac {1}{3^{k}}< \frac{3}{2}$$
$$\implies \frac{1}{3} + \cdots + \frac {1}{3^{k-1}} + \frac {1}{3^{k}} < \frac{3}{2} - 1 = \frac {1}{2}.$$
We know this to be true from our inductive hypoethsis. Hence, $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^n} < \frac{1}{2}.$
Is this proof correct?
| Also, we can use a formula for the sum of a geometric progression.
It's $$\frac{1}{3}\cdot\frac{1-\left(\frac{1}{3}\right)^{n}}{1-\frac{1}{3}}<\frac{1}{2}.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
What is maximum value of $f\left(x\right)=x\sqrt{x}-\frac{1}{\sqrt{1-x^2}}$?
Find maximum value of $$f\left(x\right)=x\sqrt{x}-\frac{1}{\sqrt{1-x^2}}$$
Domain of the function is $[0,1)$. Equating the derivative by zero we get,
$$\frac32\sqrt x-\frac{x}{(1-x^2)^{\frac32}}=0$$
$$\frac94x=\frac{x^2}{(1-x^2)^3}$$
$$9(x^2-1)^3+4x=0$$
Not sure how to find the roots of the equation. But I can proceed as follow,
$$\frac9{4x}=\frac{1}{(1-x^2)^3}$$
$$\frac1{\sqrt{1-x^2}}=\sqrt[\large6]{\frac9{4x}}$$So we are looking for maximum of $x\sqrt x-\sqrt[\large6]{\frac{9}{4x}}$
| You end with the problem of solving for $x$ the sextic equation
$$9 x^6-27 x^4+27 x^2+4 x-9=0$$ and, by inspection (as @NickD already commented) the solution is close to $0.6$.
Make $x=y+\frac 35$ to make
$$9 y^6+\frac{162 y^5}{5}+\frac{108 y^4}{5}-\frac{648 y^3}{25}-\frac{1728
y^2}{125}+\frac{53972 y}{3125}+\frac{636}{15625}=0$$ Neglect the terms $(y^3,\cdots,y^6)$ and solve the quadratic
$$-\frac{1728
y^2}{125}+\frac{53972 y}{3125}+\frac{636}{15625}=0$$ The smallest root is
$$y=-\frac{318}{5 \left(13493+\sqrt{183434809}\right)}\implies x=0.597647652\cdots$$ while the exact solution is $x=0.597647629\cdots$
Using this approximation, the maximum of the function is then
$$-0.78523029529810916\cdots$$ while the exact solution is
$$-0.78523029529810836\cdots$$ which is not too bad since obtained at the price of a quadratic equation.
| {
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"url": "https://math.stackexchange.com/questions/4248651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $x\geq y\geq z>0$. Prove that $ \frac {x^{2}y}{z} + \frac {y^{2}z}{x} + \frac {z^{2}x}{y}\geq x^{2} + y^{2} + z^{2}.$
Let $x\geq y\geq z>0$. Prove that $$ \frac {x^{2}y}{z} + \frac {y^{2}z}{x} + \frac {z^{2}x}{y}\geq x^{2} + y^{2} + z^{2}.$$
The problem is from Vietnamese MO 1991 and has been posted here before. However I am having some issues on a particular solution (that is not on the given link). The solution is from the book "Inequalities from around the world (1990-2005)" which is as follows:
Since $x\geq y\geq z>0$ we have that $$\begin{align} \frac {x^2y}{z} + \frac {y^2z}{x} + \frac {z^2x}{y} &= \frac {\color\red{x^3y^2 + y^3z^2 + z^3x^2}}{xyz}\\ &\geq \frac {\color\red{(x^3 + y^3 + z^3)(x^2 + y^2 + z^2)}}{\color\red{3}(xyz)}\\ &\geq \frac {3xyz(x^2 + y^2 + z^2)}{3(xyz)}\\ &= (x^2 + y^2 + z^2)
\end{align}$$
by Chebyshev sum inequality.$\ \ \ \blacksquare$
The solution seem to use Chebyshev sum inequality on $(x^3,y^3,z^3)$ and $(y^2,z^2,x^2)$ [the part in red]. But the problem is that the first sequence is decreasing while the second is not. To use Chebyshev inequality, the second sequence must be $(x^2,y^2,z^2)$ and then the result will be $$\begin{align} x^5+y^5+z^5\geq \frac{(x^3 + y^3 + z^3)(x^2 + y^2 + z^2)}{3} \end{align}$$
which is not helpful in the problem.
So, is the solution wrong or am I missing something?
Edit: From the comments, I realized that the solution is wrong. So can this problem be solved with Chebyshev sum inequality or rearrangement inequality? I have some solutions but
it seems to me that the problem can be solved with those inequalities but I didn't find one.
| (Note:
This was done without looking at any of the links.
It probably duplicates some of them.)
Let
$y = (1+b)z,
x = (1+a)y,
a, b \ge 0$.
$x = (1+a)(1+b)z$.
$\begin{array}\\
RHS
&=z^2((1+a)^2(1+b)^2+(1+b)^2+1)\\
LHS
&=\dfrac {x^{2}y}{z} + \dfrac {y^{2}z}{x} + \dfrac {z^{2}x}{y}\\
&=\dfrac {((1+a)(1+b)z)^{2}(1+b)z}{z} + \dfrac {((1+b)z)^{2}z}{(1+a)(1+b)z} + \dfrac {z^{2}(1+a)(1+b)z}{(1+b)z}\\
&=(1+a)^2(1+b)^3z^2 + \dfrac {(1+b)z^2}{1+a} + z^{2}(1+a)\\
&=z^2\left((1+a)^2(1+b)^3 + \dfrac {1+b}{1+a} + (1+a)\right)\\
\end{array}
$
so we want
$(1+a)^2(1+b)^3 + \dfrac {1+b}{1+a} + (1+a)
\ge ((1+a)^2(1+b)^2+(1+b)^2+1)$.
Throwing this into
Wolfy
(this could be done by hand),
$(1+a)^2(1+b)^3 + \dfrac {1+b}{1+a} + (1+a)
-((1+a)^2(1+b)^2+(1+b)^2+1)\\
=(2 a^2 + 4 a + 1) b^2 + \dfrac{a (a^2 + 3 a + 1) b}{a + 1} + \dfrac{a^2}{a + 1} + (a + 1)^2 b^3\\
\ge 0
$.
There is equality
if and only if
$a=b=0$
or
$z = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4249459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Need help solving a exponent+multiplication problem The question goes as follows:
$$\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}$$
Of course the question tests for knowledge in logic and simplification as who would keep calculating this sum. I observed that each term here can be represented as $$x^4+324$$I thought for a minute and I realised that you can express $324$ as $$324=18^2=(2\times3^2)^2$$
So substituting that into the equation, I get $$x^4+4\times3^4$$ which reminds me of $x^4+4y^4$. My professor gave me a hint to add $$(2\times x^2\times2y^2)-(2\times x^2\times2y^2)$$ to $$x^4+4y^4$$So I did, and I got $$x^4+4y^4+(2\times x^2\times2y^2)-(2\times x^2\times2y^2)$$
$$=x^4+4y^4+(2\times x^2\times2y^2)-(2\times x^2\times2y^2)$$
$$=x^4+4y^4+(4x^2y^2)-(2\times x^2\times2y^2)$$
$$=(x^2+2y^2)^2-4x^2y^2$$
$$=(x^2+2y^2)^2-(2xy)^2$$
$$=(x^2+2y^2+2xy)(x^2+2y^2-2xy)$$
Now I don't know how this will be helpful because ultimately, I sill have to compute a large sum for every number. Any body got tips on how I could proceed further to solve this sum? Thanks in advance.
| You found:
$$(x^4+4y^4)=(x^2+2y^2+2xy)(x^2+y^2-2xy)$$
Just plug in:
$(x, y)=(10, 3), (4, 3)$
for fraction:
$\frac{10^4+324}{4^4+324}=\frac{178\times 58}{58\times 10}$
$(x, y)=(22, 3), (16, 3)\rightarrow \frac{22^4+324}{16^4+324}=\frac{634\times 370}{370\times178} $
continues this procedure you finally get:
$\frac{178}{10}\times\frac{634}{178}\times\frac{1378}{634}\times \frac{2410}{1378}\times\frac{3730}{2410}$
$\frac {3730}{10}=373$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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