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Definite integral of the product of modified Bessel function of the first kind, exponential, and a power term Is there a solution to the following integral?
$$
\int_0^{\infty} t^{-0.5}e^{-at}I_{l}\left(kt\right)dt,\;\;\;a,k>0
$$
Here, $I_{l}$ is the modified Bessel function of the first kind, and $a,k$ are constants. I have found solutions of similar integrals without the $t^{-0.5}$ term, and with $l=0$ (e.g. here and here). The closest question and explanation I could find was this. However, the subscript $l$ is important for my question because the entire integral sits inside a summation over integer values of $l\in[-\infty,\infty]$.
Any kind of help/suggestions will be greatly appreciated. Thanks!
|
$$
\int_0^{\infty}{t^{-\frac{1}{2}}I_l\left( kt \right) e^{-at}dt}
\\
=\sum_{n=0}^{\infty}{\frac{1}{n!\Gamma \left( n+l+1 \right)}\left( \frac{k}{2} \right) ^{2n+l}}\int_0^{\infty}{t^{-\frac{1}{2}+2n+l}e^{-at}dt}
\\
=\sum_{n=0}^{\infty}{\frac{1}{n!\Gamma \left( n+l+1 \right)}\left( \frac{k}{2} \right) ^{2n+l}\frac{\Gamma \left( 2n+l+\frac{1}{2} \right)}{a^{2n+l+\frac{1}{2}}}}
\\
=\frac{1}{\sqrt{2a\pi}}\left( \frac{k}{a} \right) ^l\sum_{n=0}^{\infty}{\frac{\Gamma \left( n+\frac{l}{2}+\frac{1}{4} \right) \Gamma \left( n+\frac{l}{2}+\frac{3}{4} \right)}{n!\Gamma \left( n+l+1 \right)}\left( \frac{k^2}{a^2} \right) ^n}
\\
=\frac{1}{\sqrt{2a\pi}}\left( \frac{k}{a} \right) ^l\frac{\Gamma \left( \frac{l}{2}+\frac{1}{4} \right) \Gamma \left( \frac{l}{2}+\frac{3}{4} \right)}{\Gamma \left( l+1 \right)}\,\,_2F_1\left( \frac{l}{2}+\frac{1}{4},\frac{l}{2}+\frac{3}{4};l+1;\frac{k^2}{a^2} \right)
\\
Q_{\nu}^{\mu}(z)=\frac{e^{\mu \pi i}}{2^{\nu +1}}\frac{\Gamma (\nu +\mu +1)}{\Gamma \left( \nu +\frac{3}{2} \right)}\frac{\Gamma \left( \frac{1}{2} \right) \left( z^2-1 \right) ^{\frac{\mu}{2}}}{z^{\nu +\mu +1}}F\left( \frac{\nu +\mu}{2}+1,\frac{\nu +\mu +1}{2};\nu +\frac{3}{2};\frac{1}{z^2} \right)
\\
Q_{l-\frac{1}{2}}(z)=\frac{1}{2^{l+\frac{1}{2}}}\frac{\Gamma (l+\frac{1}{2})}{\Gamma \left( l+1 \right)}\frac{\Gamma \left( \frac{1}{2} \right)}{z^{l+\frac{1}{2}}}F\left( \frac{l}{2}+\frac{3}{4},\frac{l}{2}+\frac{1}{4};l+1;\frac{1}{z^2} \right)
\\
\int_0^{\infty}{t^{-\frac{1}{2}}I_l\left( kt \right) e^{-at}dt}
\\
=\frac{1}{\sqrt{2a\pi}}\left( \frac{k}{a} \right) ^l\frac{\Gamma \left( \frac{l}{2}+\frac{1}{4} \right) \Gamma \left( \frac{l}{2}+\frac{3}{4} \right)}{\Gamma \left( l+1 \right)}\,\,_2F_1\left( \frac{l}{2}+\frac{1}{4},\frac{l}{2}+\frac{3}{4};l+1;\frac{k^2}{a^2} \right)
\\
=\frac{2}{\sqrt{2k\pi}}Q_{l-\frac{1}{2}}(\frac{a}{k})
$$
|
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|
Euler's partition method. How does someone use it? I came across partitions recently and am not very much informed about it but I have a question regarding Euler's method for this. I came to know about this formula from a YouTube video so, it may not be the full equation.
Symbols:
P(n) -> Partition of n
$\pi$ -> Product
n -> Any number
$\sum_{}^\infty (P(n)*x^n) = \prod\limits_{k=1}^\infty (\frac{1}{1-x^k})$
I can't understand:
*
*What does 'x' mean in this expression
*How can someone calculate partition with this.
The video I learnt this from is this.
Note: I am not informed regarding this subject so, I may have the question all wrong. I only want to know how this works. Today, I don't think this formula is used today after Ramanujan's work. I just want to know how the people before used it.
|
This example supplements MAAvL's answer and is an expansion of dust05's comment. The following is a bit pedantic, with including exponents 1 and writing $(1+x^2+x^{2+2}+\cdots)$ rather than $(1+x^2+x^4+\cdots)$, for example, but the idea is to make the connection to partitions very concrete.
\begin{align*}
&\sum_{n \ge 0} P(n) x^n = \prod_{k \ge 1} \frac{1}{1-x^k} \\
& = \left(\frac{1}{1-x}\right) \left(\frac{1}{1-x^2}\right) \left(\frac{1}{1-x^3}\right) \left(\frac{1}{1-x^4}\right) \cdots \\
& = (1+x^1+x^{1+1}+\cdots)(1+x^2+x^{2+2}+\cdots)(1+x^3+x^{3+3}+\cdots)(1+x^4+x^{4+4}+\cdots)\cdots \\
& = 1 + x^1 + (x^2 + x^{1+1}) + (x^3 + x^2 x^1 + x^{1+1+1}) + (x^4 + x^3 x^1 + x^{2+2} + x^2 x^{1+1} + x^{1+1+1+1})\cdots\\
& = 1 + x^1 + (x^2 + x^{1+1}) + (x^3 + x^{2+1} + x^{1+1+1}) + (x^4 + x^{3+1} + x^{2+2} + x^{2+1+1} + x^{1+1+1+1})\cdots\\
& = 1 + x + 2x^2 + 3x^3 + 5x^4 + \cdots
\end{align*}
Also, this generating function for $P(n)$ is used all the time in contemporary research on integer partitions (where it is often written using the $q$-Pochhammer symbol as $1/(q;q)_\infty$ or just $1/(q)_\infty$). As per the comments and links, though, there are other ways to compute $P(n)$ exactly (Euler's recurrence using the pentagonal number theorem) and asymptotically (the Hardy-Ramanujan-Rademacher formula).
|
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|
Calculate $\int_A e^{x^2+y^2-z^2-w^2} \, dx\,dy\,dz\,dw$ Calculate $\int_A e^{(x^2+y^2-z^2-w^2)}\,dx\,dy\,dz\,dw$ where $A=\{x, y, z, w \in \mathbb{R} \mid x^2+y^2+z^2+w^2\leq1\}$
attempt:
$$\int_A e^{x^2+y^2-z^2-w^2} \,dx\,dy\,dz\,dw
= \int_0^1 \int_{\mathbb{S}^3_r} e^{x^2+y^2-z^2-w^2} \;\mathrm{d}S \;\mathrm{d}r
\\= \int_0^1 \int_{\mathbb{S}^3_r} e^{2x^2+2y^2-r} \;\mathrm{d}S \;\mathrm{d}r
= \int_0^1 \int_0^{2\pi} \int_0^{\pi} e^{2r^2 \sin^2\phi - r} r^2 \sin\phi \;\mathrm{d}\phi \;\mathrm{d}\theta \;\mathrm{d}r
$$
I got stuck here. Any help please?
|
Note that $$A=\{x^2+y^2\leq1, z^2+w^2\leq1-(x^2+y^2)\}$$
So the integral becomes
$$I=\int_{\{x^2+y^2\leq1\}}e^{x^2+y^2}\int_{\{z^2+w^2\leq1-(x^2+y^2)\}}e^{-(z^2+w^2)}$$
Where the inner integral is by Fubini again:
$$(*) = \int_{-\pi}^{\pi}\int_0^cre^{-r^2}drd\theta$$
where $c=\sqrt{1-(x^2+y^2)}$.
Now, this equals $\pi(1-e^{-c^2})$, so we got
$$I = \pi\int_{\{x^2+y^2\leq1\}}e^{x^2+y^2} \cdot(1-e^{-(1-(x^2+y^2))})$$
Where from here, it is easier to continue by polar coordinates and similar integral as in $(*)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the sum and asymptotic expansion of $\sum_{u=1}^{[N/2]} \sum_{v=1}^{[\sqrt{N}]} \left[{|u^2-2v| \le N}\right] \left[{|-2uv| \le N}\right]$ Where $[x]$ in the summation limits is the floor function. In the summation body this is the Iverson bracket
Find the sum and asymptotic expansion of $\sum_{u=1}^{\lfloor{N/2}\rfloor} \sum_{v=1}^{\lfloor{\sqrt{N}}\rfloor} \left[|u^2-2v| \le N\right] \left[|-2uv| \le N\right]$.
I suspect that there may be a closed form solution. I have not made any progress to the analytic form of this sum.
I have some table values
N
Sum
1
0
10
10
10^2
100
10^3
981
10^4
10000
10^5
99928
This indicates that the sum approach $N$ as $N \rightarrow \infty$.
The second part is to find this asymptotic expansion.
|
Lets look at a table between perfect squares. Say from N = 16 to 24. Then
N
Sum
Difference
16
16
0
17
17
0
18
17
1
19
18
1
20
18
2
21
19
2
22
19
3
23
20
3
24
20
4
This pattern repeats for each perfect square sequence.
So with ${S}_{1} \left({N}\right) = \sum_{u=1}^{\lfloor{N/2}\rfloor} \sum_{v=1}^{\lfloor{\sqrt{N}}\rfloor} \left[{|u^2-2v| \le N}\right] \left[{|-2u| v \le N}\right]$.
Then consider two consecutive perfect squares $\sqrt{N} \in \mathbb{Z}$ or $N = {n}^{2}$ to $N = \left({n + 1}\right)^{2}$. Then from $N = {n}^{2}$ to $\left({n +1}\right)^{2} - 1$ there are $2\, n$ values. Let $k \in \left\{{0, 1, \cdots, 2\, n}\right\}$. Then for the values ${S}_{1} \left({{n}^{2}}\right.$, ${n}^{2} + 1$, $\cdots$, $\left.{\left({n +1}\right)^{2} - 1}\right) = N - \lfloor{k/2}\rfloor$. The sequence of $k = 0$ to $2\, n$ is $\lfloor{k/2}\rfloor = \lfloor{(N-\lfloor{\sqrt{N}}\rfloor^{2})/2}\rfloor$. Thus, applying induction, we can write for any perfect square interval $N \ge 9$
$${S}_{1} \left({N}\right) = N - \lfloor{\frac{N-\lfloor{\sqrt{N}}\rfloor^{2}}{2}}\rfloor$$
We add the final corrections for the values of $N \in \left\{{1, 2, 3, 5}\right\}$ resulting in
$${S}_{1} \left({N}\right) = N - \lfloor{\frac{N-\lfloor{\sqrt{N}}\rfloor^{2}}{2}}\rfloor - {\delta}_{N \in \left\{{1, 2, 3}\right\}} - {\delta}_{N=5}$$
For the asymptotic values as $N \rightarrow \infty$ we write the average value between two perfect squares of
$$\left<{\lfloor{\frac{N-\lfloor{\sqrt{N}}\rfloor^{2}}{2}}\rfloor}\right> = \frac{\text{sum of interval}}{2 \times \text{ length of the interval}}= \frac{\lfloor{\sqrt{N}}\rfloor^{2}-1}{2(\lfloor{\sqrt{N}}\rfloor+2)}\sim\frac{N-1}{2(\sqrt{N}+2)}\sim\frac{1}{2} \sqrt{N}-1+O(\frac{1}{N})$$
with the final answer
$${S}_{1}\left({N}\right) \sim N - \frac{1}{2} \sqrt{N}+1+O(\frac{1}{N})$$
|
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|
Integration using mean value theorem of $\int_0^{2\pi} \frac{2+\cos\phi}{5+4\cos\phi}\,\mathrm{d}\phi=\pi$ I am going through some worksheets to study for an exam and one question is to use the mean value theorem:
$$u(a)=\frac{1}{V_r}\int_{B_r(a)}u$$
with a suitable harmonic function to show that:
$$
\int_0^{2\pi} \frac{2+\cos\phi}{5+4\cos\phi}\,\mathrm{d}\phi=\pi
$$
How do I begin?
|
We can try Poisson kernel.
$$
P_r(\theta)=\frac{1-r^2}{1-2r\cos\theta+r^2} = \Re \left(\frac{1+re^{i\theta}}{1-re^{i\theta}}\right).
$$
To have $\cos \theta$ on the numerator, we use $1+P_r(\theta)$. Then
$$
1+P_r(\theta) = \frac{2-2r\cos\theta}{1-2r\cos\theta+r^2} = 1+\Re \left(\frac{1+re^{i\theta}}{1-re^{i\theta}}\right).
$$
Put $r=1/2$, we have
$$
1+P_{1/2}(\theta) = \frac{8-4\cos\theta}{5-4\cos\theta}.$$
Then we use $\theta+\pi$ in place of $\theta$.
$$
1+P_{1/2}(\theta+\pi)=\frac{8+4\cos\theta}{5+4\cos\theta} = 1+\Re \left(\frac{1-\frac12 e^{i\theta}}{1+\frac 12e^{i\theta}}\right).
$$
Use $f(z) = 1+\frac{1-z/2}{1+z/2}$ on the neighborhood of the closed unit disk $\overline{\mathbb{D}}$, then we have by the Mean Value Theorem for $u(z)= \Re(f(z))$,
$$
\frac1{2\pi} \int_0^{2\pi} \frac{8+4\cos\theta}{5+4\cos\theta} d\theta
=u(0) = 2.$$
The result follows by dividing both sides by $4$.
|
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|
Trigonometric equality proof $ \cos^2(\omega t) + \cos^2(\omega t + \delta) = \sin^2\delta + 2\cos(\omega t + \delta)\cos(\omega t)\cos(\delta)$ Looking to prove
$x = A\cos(\omega t)\\
y = A\cos(\omega t + \delta)
\\
\\$
YIELDS
$x^2-2xy\cos(\delta)+y^2=A^2\sin^2(\delta)$
Specifically we're trying to express the equations without any reference to $t$. If it helps, we can look in terms of ${x^2 + y^2 \over {A^2}}$
and just prove
$$ \cos^2(\omega t) + \cos^2(\omega t + \delta) = \sin^2\delta + 2\cos(\omega t + \delta)\cos(\omega t)\cos(\delta)$$
I know the angle sum identity $$\cos(\omega t + \delta) = \cos(\omega t) \cos(\delta) - \sin(\omega t)\sin(\delta)$$
$$ \cos^2(\omega t) + \cos^2(\omega t + \delta) \\ =\cos^2(\omega t) + [\cos(\omega t )\cos(\delta) - \sin(\omega t)\sin(\delta)]^2\\
=\cos^2(\omega t) + \cos^2(\omega t )\cos^2(\delta)+ \sin^2(\omega t) \sin^2(\delta)-2\sin(\omega t)\sin(\delta)\cos(\omega t )\cos(\delta) $$
But I can't get anywhere after that. I suspect that there is some other trig identity I either don't know or am overlooking somewhere.
This is from a physics textbook talking about two dimensional oscillation with the same $\omega$ but offset by a $\delta$ phase angle. It's not a problem i'm just trying to follow along the text and fill in the author's gaps.
|
By setting $\omega t = x, \omega t + \delta = y$, it is equivalent to showing:
$\cos^2 x + \cos^2 y = \sin^2(y-x) + 2\cos y\cos x \cos (y-x) $
$\cos^2 x + \cos^2 y - \sin^2(y-x) \\= \cos^2 x + \cos^2 y - (\sin y\cos x - \cos y\sin x)^2 \\= \cos^2 x + \cos^2 y - \sin^2 y\cos^2 x - \cos^2 y\sin^2 x + 2\sin y\cos x \cos y\sin x \\= \cos^2 x (1- \sin^2 y) + \cos^2 y (1-\sin^2 x) + 2\sin y\cos x \cos y\sin x \\= \cos^2 x \cos^2 y + \cos^2 y \cos^2 x + 2\sin y\cos x \cos y\sin x \\=2\cos x\cos y(\cos x \cos y + \sin x\sin y) \\= 2\cos x\cos y\cos (y-x)$
from which the result immediately follows.
|
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|
Inequality with induction. I was wondering if my solution to this problem is correct:
Show that for every integer $n\ge1$ the following is true:
$\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\le\frac{1}{\sqrt{3n+1}}$
Base case:
$\frac{2\cdot1-1}{2}\le\frac{1}{\sqrt{3*1+1}}$
$\frac{1}{2}\le\frac{1}{2}$
$\Rightarrow \text{which is true.}$
Induction assumption:
$\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\le\frac{1}{\sqrt{3n+1}}$
Which means we want to prove the following:
$\prod\limits_{k=1}^{n+1}\frac{2k-1}{2k}\le\frac{1}{\sqrt{3(n+1)+1}}$
$\prod\limits_{k=1}^{n+1}\frac{2k-1}{2k}\le\frac{1}{\sqrt{3n+4}}$
Which is the same as:
$\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\cdot(\frac{2(n+1)-1}{2(n+1)})\le\frac{1}{\sqrt{3n+4}}$
$\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\cdot(\frac{2n+1}{2n+2)})\le\frac{1}{\sqrt{3n+4}}$
And then, we know the following is true:
$\frac{1}{\sqrt{3n+4}}\lt\frac{1}{\sqrt{3n+1}}$
Which lets us say the following:
$\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\cdot(\frac{2n+1}{2n+2)})\le\frac{1}{\sqrt{3n+4}}\le\frac{1}{\sqrt{3n+1}}$
The beforementioned statement is true because $2n+2\gt2n+1$ and $n\ge1$, which means that:
$0\lt\frac{2n+1}{2n+2}\lt1$
We can now that this is true according to the induction assumption. Which proves the original statement.
|
To conclude for the induction step we should use the induction hypothesis that is
$$\prod\limits_{k=1}^{n+1}\frac{2k-1}{2k}=\prod\limits_{k=1}^{n}\frac{2k-1}{2k}\cdot\left(\frac{2n+1}{2n+2}\right)\stackrel{Ind.Hyp.}\le\frac{1}{\sqrt{3n+1}}\cdot\left(\frac{2n+1}{2n+2}\right)\stackrel{?}\le \frac{1}{\sqrt{3(n+1)+1}}$$
and the latter is true indeed
$$\frac{1}{\sqrt{3n+1}}\cdot\left(\frac{2n+1}{2n+2}\right)\stackrel{?}\le \frac{1}{\sqrt{3(n+1)+1}}$$
$$\iff \left(\frac{2n+1}{2n+2}\right)^2\le \frac{3n+1}{3n+4}$$
$$\iff (4n^2+4n+1)(3n+4)\le (4n^2+8n+4)(3n+1)$$
$$\iff 12n^3+28n^2+19n+4\le 12n^2+28n^2+20n+4$$
$$\iff 19\le 20$$
|
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|
Prove that the limit $\lim_{(x,y)\to(0,0)} \frac{y^4}{\sqrt{3x^2+y^2}}\cos(\frac{1}{x})$ exists using $\epsilon \text{ and } \delta$ definition Prove that the limit $\lim_{(x,y)\to(0,0)} \frac{y^4}{\sqrt{3x^2+y^2}}\cos(\frac{1}{x})$ exists using $\epsilon \text{ and } \delta$ definition.
I have so far reduced it algebraically to the point $\frac{y^4}{\sqrt{x^2+y^2}}$ but I am not sure if I have to use polar coordinates or something to continue, kinda lost from here. I eliminated the cosine because $\cos(\frac{1}{x})\leq 1$ so
$$\frac{y^4}{\sqrt{3x^2+y^2}}\cos(\frac{1}{x}) \leq \frac{y^4}{\sqrt{3x^2+y^2}}$$
And likewise I eliminated the three from $3x^2$. Do I have to use polar coordinates like $r=x^2 + y^2$???
I did some more working and got $\delta=\epsilon^{\frac{1}{3}}$...
|
Hint:$\left|\frac{y^4}{\sqrt{3x^2+y^2}}\right|\leqslant |y^3|$
Addition:
Let's take $\forall \varepsilon >0$. As we have limit $(x,y)\to (0,0)$, then we should find $\delta>0$, such, that $\sqrt{x^2+y^2}< \delta$ implies $|f|<\varepsilon$. As we have $|y|\leqslant \sqrt{x^2+y^2}< \delta$, then it gives, that $\delta^3<\varepsilon$ is enough.
|
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|
Evaluating $\lim_{n \to \infty}\left(\frac{n}{n^3+1}+\frac{2n}{n^3+2}+ \dots + \frac{n^2}{n^3+n}\right)$
Compute the limit:
$$\lim_{n \to \infty}\left(\frac{n}{n^3+1}+\frac{2n}{n^3+2}+ \dots + \frac{n^2}{n^3+n}\right)$$
I tried applying the sandwich rule, constructing an upper and lower bound for the sequence, but I can't get the bounds to have the same limit.
Hints?
|
We have $$\frac{n^2(n+1)}{2(n^3+n)}=\frac{n}{n^3 +n } +...+\frac{n^2}{n^3 +n }\leq \frac{n}{n^3 +1 } +...+\frac{n^2}{n^3 +n }\leq \frac{n}{n^3 +1 } +...+\frac{n^2}{n^3 +1 }=\frac{n^2(n+1)}{2(n^3+1)}$$
|
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|
How to show $\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \Sigma_{k=1}^n \sqrt{k-1} = \frac{2}{3}$? How to show $$\lim_{n \to \infty} \frac{1}{\sqrt{n}} \frac{1}{n} \sum_{k=1}^n \sqrt{k-1} = \frac{2}{3}$$
In fact, this is the lower sum of the integral of $\sqrt{x}$ from 0 to 1. So the value of the above must be $\frac{2}{3}$. But how to show?
|
A solution based on Stolz-Cesàro theorem. Let $f(x)=x^{\frac{3}{2}}$
\begin{align*}
\lim_{n \to \infty} \frac{\sum_{i=1}^{n-1}\sqrt{i}}{n\sqrt{n}} = \lim_{n \to \infty} \frac{\sqrt{n}}{(n+1)\sqrt{n+1} - n\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{n}}{(1 + \frac{1}{n})^{\frac{3}{2}} - 1} = \frac{1}{f'(1)} = \frac{2}{3}
\end{align*}
|
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|
Determinant of an interesting Toeplitz matrix Let $ab=1$. Find
$$\begin{vmatrix} c & a & a^2 & ... & a^{n-1} \\ b & c & a & \dots & a^{n-2} \\ b^2 & b & c& \dots &a^{n-3} \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ b^{n-1} & b^{n-2} & b^{n-3} & \dots & c \end{vmatrix}$$
I tried to decompose by line however it does not lead to a solution
|
Let's add a third answer for fun. Since $ab=1$, we can write
$$ \det
\begin{pmatrix}
c&a&a^2&\cdots&a^{n-1} \\
b& c & a &\dots& a^{n-2}\\
b^2 & b & c &\cdots & a^{n-3}\\
\vdots &\vdots & \vdots & &\vdots\\
b^{n-1} & b^{n-2} &b^{n-3}& \cdots &c
\end{pmatrix}
=\frac{1}{b}\cdot\frac{1}{b^2}\cdots\frac{1}{b^{n-1}}\det
\begin{pmatrix}
c&1&1&\cdots&1\\
b &bc&b&\cdots &b\\
b^2 &b^2& b^2c&\cdots &b^2\\
\vdots&\vdots&\vdots& &\vdots\\
b^{n-1} & b^{n-1}& b^{n-1}&\cdots& b^{n-1}c
\end{pmatrix}\\
=\det
\begin{pmatrix}
c & 1 & 1& \cdots & 1\\
1&c&1&\cdots&1\\
1&1&c&\cdots&1\\
\vdots &\vdots&\vdots & & \vdots \\
1&1&1&\cdots&c
\end{pmatrix}=(c-1)^{n-1}(c+n-1).
$$
where you can use any of these answers: How do I proof that the determinant of this matrix has the form $(z-1)^{n-1} (z+(n-1))$?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving cubic equations with sine and cosine sums. I was playing with math, and then I tried to rewrite some cubic equation with sine power reduction formula
$$y^3 + my^2 + ny + d = 0.$$
Let
$$y = \sin(x).$$
Then
$$y^2 = \frac{1 - \cos(2x)}{2},$$
$$y^3 = \frac{3\sin(x) - \sin(3x)}{4}.$$
So the equation will be
$$\frac{3\sin(x)-\sin(3x)}{4} + m\frac{1 - \cos(2x)}{2} + n\cdot\sin(x) +d.$$
We can redefine constant multipliers and we will end with
$$a\cdot\sin(x) + b\cdot\cos(2x) + c\cdot\sin(3x) +L = 0.$$
What would be the best approach to solve this without needing to solve polynomial equations whose degree is greater or equal to 3?
|
This approach has been used to resolve the diminished cubic $x^3 = 3px - 2q$
$x = 2\sqrt p\cos\theta\\
8p\sqrt p\cos^3 \theta = 6p\sqrt p \cos\theta + 2q\\
2p\sqrt p(4\cos^3 \theta - 3\cos\theta) = 2q\\
\cos 3\theta = \frac {q}{p\sqrt p}$
We detour into the complex plane if the original cubic does not have 3 real roots.
The substitution $y = x - \frac {m}{3}$ will allow you to turn your initial cubic into a diminished cubic.
|
{
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"url": "https://math.stackexchange.com/questions/4233791",
"timestamp": "2023-03-29T00:00:00",
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|
Infinite product involving triangular numbers $\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}$ The question is about the value for the following infinite product involving triangular numbers $T_n=\frac{n(n+1)}2$:
$$\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}=\prod_{n=1}^{\infty} \frac{n(n+1)}{n(n+1)+2}=\prod_{n=1}^{\infty} \frac{1}{1+\frac 2{n(n+1)}}$$
Searching on MSE I've found solutions for other similar problems related to Weierstrass' product for the sine function:
$$\frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right)$$
but nothing specific on that particular case.
I've tried to use Weierstrass' product but without success.
Also expansion in logarithmic sum allow me to find good estimation with a small number of terms but nothing more than that.
According to wolfram the value for the infinite product should be: $\frac{2\pi}{\cosh\left(\frac{\sqrt 7}2\pi\right)} \approx 0.197$.
How can we determine the value for the partial and the infinite product in a closed form?
|
For complex $a, b, c, d$ with non-negative real part we have
$$
\prod_{n=1}^{N-1}\frac{(n+a)(n+b)}{(n+c)(n+d)} = \frac{\Gamma(c+1)\Gamma(d+1)}{\Gamma(a+1)\Gamma(b+1)} \cdot \frac{\Gamma(a+N)\Gamma(b+N)}{\Gamma(c+N)\Gamma(b+N)} \, .
$$
If additionally $a+b=c+d$ then Stirling's formula shows that the second factor converges to $1$ for $N \to \infty$, so that
$$
\prod_{n=1}^{\infty}\frac{(n+a)(n+b)}{(n+c)(n+d)} = \frac{\Gamma(c+1)\Gamma(d+1)}{\Gamma(a+1)\Gamma(b+1)} \, .
$$
In our case,
$$
\prod_{n=1}^{\infty}\frac{T_n}{T_n+1} = \prod_{n=1}^{\infty}\frac{n(n+1)}{\left(n+\frac 12 + \frac{\sqrt 7}{2}i\right)\left(n+\frac 12 - \frac{\sqrt 7}{2}i\right)} \\
= \Gamma\left(\frac 32 + \frac{\sqrt 7}{2i}\right)\Gamma\left(\frac 32 - \frac{\sqrt 7}{2i}\right) \\
= \left(\frac 12 + \frac{\sqrt 7}{2}i\right)\left(\frac 12 - \frac{\sqrt 7}{2}i\right)\Gamma\left(\frac 12 + \frac{\sqrt 7}{2}i\right)\Gamma\left(\frac 12 - \frac{\sqrt 7}{2}i\right) \\
= 2 \frac{\pi}{\sin\left(\pi\left(\frac 12 + \frac{\sqrt 7}{2}i\right)\right)}
= \frac{2\pi}{\cosh\left( \frac{\sqrt 7}{2}\pi\right)} \, ,
$$
using the functional equation of the Gamma function and Euler's reflection formula. For the partial product we then get
$$
\prod_{n=1}^{N-1}\frac{T_n}{T_n+1} = \frac{2\pi}{\cosh\left( \frac{\sqrt 7}{2}\pi\right)} \cdot \frac{(N-1)! N!}{\Gamma\left(N+\frac 12 + \frac{\sqrt 7}{2i}\right)\Gamma\left(N + \frac 12 - \frac{\sqrt 7}{2i}\right)} \, .
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
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.
|
By Jensen for the convex function $f(x)=4^{x^2}$ we obtain:
$$4^{x^2}+4^{(x-1)^2}\geq2\cdot4^{\left(\frac{x+1-x}{2}\right)^2}=2\sqrt2.$$
Can you end it now?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Algebraic operations between two expressions: how to include a constant outside a fraction into the numerator of that fraction I'm looking at a problem where the derivative of a function is defined as:
$$f'(x) = a-\frac{3x^2(1+x^2)-2x(x^3)}{(1+x^2)^2}$$
Then at the next line, the expression is expressed with the numerator as a quadratic function of $x^2$:
$$\frac{a+(2a-3)x^2+(a-1)x^4}{(1+x^2)^2}$$
But it is not clear what the steps are to get to this second expression.
I can see that the $f'(x)$ can be written as:
$$a-\frac{3x^2+x^4}{(1+x^2)^2}$$
But then to include $a$ into the numerator and come to the second expression has got me stuck...
Anyone suggestions or a hint?
|
You have\begin{align}f'(x)&= a-\frac{3x^2(1+x^2)-2x(x^3)}{(1+x^2)^2}\\&=\frac{a(1+2x^2+x^4)-3x^2-3x^4+2x^4}{(1+x^2)^2}\\&=\frac{a+(2a-3)x^2+(a-1)x^2}{(1+x^2)^2}.\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4235808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How can I handle the integation which contains $~ \left( 1+ t ^{2} \right) ^{2} ~$ and $~ \left( 1- t ^{2} \right) ^{2} ~~$? $$ \left( a,d \in \mathbb R_{> 0} \right) ~~\wedge~~ \left( a < d \right) $$
I have to find out the value of the below integral .
$$ \alpha := \int_{0 }^{1 } \frac{ 2 a \left( 1- t ^{2} \right) ^{2} }{ \left( 1+ t ^{2} \right) \left\{ d ^{2} \left( 1+ t ^{2} \right) ^{2} - a ^{2} \left( 1- t ^{2} \right) ^{2} \right\} } \,dt $$
$$ = 2a \int_{0 }^{1 } \frac{ \left( 1- t ^{2} \right) ^{2} }{ \left( 1+ t ^{2} \right) \left\{ d ^{2} \left( 1+ t ^{2} \right) ^{2} - a ^{2} \left( 1- t ^{2} \right) ^{2} \right\} } \,dt $$
My tries so far are as below .
$$ t= \tan^{}\left( x \right) $$
$$ t : 0 ~\rightarrow~1 $$
$$ x : 0 ~\rightarrow~ \frac{\pi}{4} $$
$$ \frac{ dt }{ dx } = \sec^{2}\left( x \right) $$
$$ 1+ t ^{2} = \sec^{2}\left( x\right) $$
$$ \therefore ~~ \frac{ dt }{ dx } = \left( 1 + t ^{2} \right) $$
$$ \frac{ dt }{ \left( 1+ t ^{2} \right) }= dx $$
$$ 1+ t ^{2} = \sec^{2}\left( x \right) ~~ \leftarrow~~ \text{I will subtract each right and the left term by }~~ -2 t ^{2} $$
$$ 1+ t ^{2} -2 ^{2} = \sec^{2}\left( x \right) - 2 t ^{2} $$
$$ 1- t ^{2} = \sec^{2}\left( x \right) - 2 \left( \sec^{2}\left( x \right) -1 \right) $$
$$ = \sec^{2}\left( x \right) -2 \sec^{2}\left( x \right) + 2 $$
$$ = - \sec^{2}\left( x \right) +2 $$
$$ = 2- \sec^{2}\left( x \right) $$
$$ \therefore ~~ \alpha = 2a \int_{0 }^{\frac{\pi}{4} } \frac{ \left( 2- \sec^{2}\left( x \right) \right) ^{2} }{ \left\{ d ^{2} \left( \sec^{2}\left( x \right) \right) ^{2} -a ^{2} \left( 2 - \sec^{2}\left( x_{} \right) \right)^{2} \right\} } \,dt $$
I've been got stucked from here .
|
Substitute $t=\tan \frac x2$
\begin{align}
&\int_{0 }^{1 } \frac{ 2 a \left( 1- t ^{2} \right) ^{2} }{ \left( 1+ t ^{2} \right) \left\{ d ^{2} \left( 1+ t ^{2} \right) ^{2} - a ^{2} \left( 1- t ^{2} \right) ^{2} \right\} } \,dt \\
=& \int_0^{\frac\pi2}\frac{a \cos^2x}{d^2 - a^2\cos^2x}dx
= \frac1a\left(-\frac\pi2 + \int_0^{\frac\pi2}\frac{d^2}{d^2 - a^2\cos^2x}dx\right)\\
= &\frac1a\left(-\frac\pi2 + \int_0^{\frac\pi2}\frac{\>d(\tan x)}{\frac{d^2 - a^2 }{d^2}+ \tan^2x}\right)
=\frac\pi{2a}\left( \frac d{\sqrt{d^2-a^2}}-1\right)
\end{align}
|
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|
Let $a,b,c$ be non-negative real numbers .Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $ Let $a,b,c$ be non-negative real numbers
Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
My idea is to use the $(p,q,r)$ method:
$p=a+b+c$
$q=ab+bc+ca$
$r = abc $
$\Rightarrow a^2+b^2+c^2 = p^2-2q $
$ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
or
$ p^2-2q +\sqrt{2} r + 2\sqrt{2} +3 \geq (2+\sqrt {2} )p $
or
$ (p - (1+\sqrt{2} ))^2 -2q +\sqrt{2} r \geq 0 $
or
The problem I am facing is exactly what I want to prove : $\sqrt{2} r \geq 2q$ is completely wrong .
I hope to get help from everyone. Thanks very much !
|
$uvw$ helps. See here: https://math.stackexchange.com/edit-tag-wiki/5758
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, our inequality it's $$9u^2-6v^2+\sqrt2w^3+3+2\sqrt2\geq3(2+\sqrt2)u,$$ which is a linear inequality of $v^2$.
Thus, by $uvw$ it's enough to prove our inequality for equality case of two variables.
Let $b=a$.
Thus, we need to prove a quadratic inequality of $c$ and $\Delta\leq0$ it's an inequality of one variable $a$.
Can you end it now?
|
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|
$\alpha, \beta$ are roots of $x^2+px+q=0, f(x)=(\alpha+\beta)x-\left(\frac{\alpha^2+\beta^2}2\right)x^2+\left(\frac{\alpha^3+\beta^3}3\right)x^3-...$
If $\ln(1+x)=x-\dfrac {x^2}2+\dfrac {x^3}3-\dfrac{x^4}4+...$ and $\alpha$ & $\beta$ are roots of the equation $x^2+px+q=0$. If $f(x)=(\alpha+\beta)x-\left(\dfrac{\alpha^2+\beta^2}2\right)x^2+\left(\dfrac{\alpha^3+\beta^3}3\right)x^3-\left(\dfrac{\alpha^4+\beta^4}4\right)x^4+...$ then find $f'(x)$ in terms of $p,q$
By Vieta's formula
$\alpha+\beta=-p,\; \alpha\beta=q$
$\ln(1+(\alpha+\beta)x)=(\alpha+\beta)x-\dfrac {(\alpha+\beta)^2}2x^2+\dfrac {(\alpha+\beta)^3}3x^3-\dfrac{(\alpha+\beta)^4}4x^4+...$
$f'(x)=(\alpha+\beta)-(\alpha^2+\beta^2)x+(\alpha^3+\beta^3)x^2-(\alpha^4+\beta^4)x^3$
$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=p^2-2q$
$\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=-p^3+3pq$
Not able to work around and conclude.
|
$f(x)=(\alpha+\beta)x-\left(\dfrac{\alpha^2+\beta^2}2\right)x^2+\left(\dfrac{\alpha^3+\beta^3}3\right)x^3-\left(\dfrac{\alpha^4+\beta^4}4\right)x^4+...$
$f(x)=\ln(1+\alpha x)+\ln(1+\beta x)$
$f(x)=\ln(1+(\alpha+\beta)x+\alpha\beta x^2)$
$f(x)=\ln(1-px+qx^2)$
Can you continue??
|
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|
Evaluating $ \sum_{r=0}^\infty { 2r \choose r} x^r $ I recently came across this question during a test
Evaluate the following $$ \sum_{r=0}^ \infty \frac {1\cdot 3\cdot 5\cdots(2r-1)}{r!}x^r $$
I converted it to the follwing
into $\sum_{r=0}^\infty { 2r \choose r} x^r $
Then I tried using Beta function to further expand the binomial coefficient and tried to write it as sum of integrals but a pesky "r" term ends up being multiplied
How can we evaluate this with and without using calculus?
|
Use the Duplication formula for Gamma function
$$
\Gamma \left( {2\,z + 1} \right) = \Gamma \left( {2\,\left( {z + 1/2} \right)} \right)
= 2^{\,2\,z} \frac{{\Gamma \left( {z + 1} \right)\Gamma \left( {z + 1/2} \right)}}{{\Gamma \left( {1/2} \right)}}
$$
So
$$
\begin{array}{l}
\left( \begin{array}{c}
2r \\ r \\
\end{array} \right) = \frac{{2r!}}{{r!r!}} = \frac{{\Gamma \left( {2r + 1} \right)}}{{\Gamma \left( {r + 1} \right)\Gamma \left( {r + 1} \right)}} = \\
= 2^{\,2\,r} \frac{{\Gamma \left( {r + 1} \right)\Gamma \left( {r + 1/2} \right)}}{{\Gamma \left( {r + 1} \right)\Gamma \left( {r + 1} \right)}}
= 4^{\,\,r} \frac{{\Gamma \left( {r + 1/2} \right)}}{{\Gamma \left( {r + 1} \right)\Gamma \left( {1/2} \right)}} = \\
= 4^{\,\,r} \frac{{\left( {r - 1/2} \right)!}}{{r!\left( { - 1/2} \right)!}} = 4^{\,\,r} \left( \begin{array}{c}
r - 1/2 \\
r \\
\end{array} \right) = \\
\left| {\;0 \le r \in \Bbb{Z}} \right. \\
= \left( { - 1} \right)^{\,\,r} 4^{\,\,r} \left( \begin{array}{c}
- 1/2 \\ r \\
\end{array} \right) \\
\end{array}
$$
|
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|
Elementary solutions of the equation of a quadratic formula We know that when $A=0$, the quadratic equation $Ax^2+Bx+C=0$ has one solution, $-\frac{C}{B}$.
Since for every quadratic equation, a quadratic equation has at most two solutions: $\frac{-B + \sqrt{B^2 - 4AC}}{2A}$ and $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$, I think that in the case $A=0$ one of the results of the two formulas should match $-\frac{C}{B}$. Yet, if $A=0$, $\frac{-B + \sqrt{B^2 - 4AC}}{2\times0}$, we will get $\frac{0}{0}$, and with $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$ we will get $-\frac{B}{0}$, so none of the two possible solutions will match $-\frac{C}{B}$. Shouldn't there be at least one solution that has the same value as $-\frac{C}{B}$?
|
The solution $-\frac CB$ is the limit at $0$ of one of those two solutions. Suppose that $B\geqslant0$. Then $\sqrt{B^2}=B$ and\begin{align}\lim_{A\to0}\frac{-B+\sqrt{B^2-4AC}}{2A}&=\frac12\lim_{A\to0}\frac{\sqrt{B^2-4AC}-B}A\\&=\frac12\left.\frac{\mathrm d}{\mathrm dA}\sqrt{B^2-4AC}\right|_{A=0}\\&=-\frac C{\sqrt{B^2}}\\&=-\frac CB.\end{align}The case in which $B\leqslant0$ is similar.
|
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|
Proving a curve has no curve point I am trying to show that the curve $x^2 + y^2 - 3 = 0$ has no rational point, where a rational point is defined as a solution $(x,y) \in \mathbb{Q}^2$ to this equation.
My attempt was to proceed by contradiction. Assume there is such a point $(x,y) \in \mathbb{Q}^2$ and let $x = \frac{a}{b}$, $y = \frac{c}{d}$ for $a,b,c,d \in \mathbb{Z}$, $b,d \neq 0$ where we assume without loss of generality that these fractions are fully reduced. Then we substitute:
\begin{align*}
\frac{a^2}{b^2} + \frac{c^2}{d^2} - 3 = 0.
\end{align*}
Rearranging and finding a common denominator of $b^2 d^2$ on the left:
\begin{align*}
\frac{a^2 d^2 + b^2 c^2}{b^2 d^2} = 3.
\end{align*}
We multiply through by $b^2 d^2$:
\begin{align*}
a^2 d^2 + b^2 c^2 = 3b^2 d^2.
\end{align*}
At this point, I'm stuck. I could try to argue that $3$ divides the left-hand side and apply Euclid's lemma, but I can't figure out how to get a contradiction out of the assumption that $\frac{a}{b}$ and $\frac{c}{d}$ have no common factors.
|
You can rearrange the equality to get $a^2 d^2 = 3 b^2 d^2 - b^2 c^2 = b^2(3 d^2 - c^2)$ which means that $b^2 | a^2 d^2 \implies b | ad$, but $\gcd(a, b) = \gcd(c, d) = 1$ means that this implies $b | d$, and by a similar argument $d | b$, but that only happens if $b = d$.
From there, you can reduce your equation to $a^2 + c^2 = 3b^2$. Then you can prove that this requires both $a$ and $c$ to be multiples of 3 and you're just about home.
|
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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?
|
As noticed you went in the wrong direction, for the induction step it is better to start from the induction hypothesis, for example as follows
$$\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^n} < \frac{1}{2}$$
$$\implies 1+\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^n} <1+ \frac{1}{2}$$
$$\implies \frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^n} + \frac {1}{3^{n+1}} < \frac{1}{3}+ \frac{1}{6}= \frac{1}{2} $$
which complete the proof.
|
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|
Find $\int \frac{{\rm d}x}{1+e\cos x}$
Find $$\int \frac{{\rm d}x}{1+e\cos x}$$
*
*When $e$ lies between $0$ and $1$
*When $e$ is greater than $1$
I got $\int \frac{dx}{1+e\cos x}$ when $e$ lies between $0$ and $1$, by substituting $\tan\frac{x}{2} =t$.
$$\int \frac{dx}{1+e\cos x} = \frac{2}{\sqrt{1-e^2}} \tan^{-1}\left( \frac{\sqrt{1-e}}{\sqrt{1+e}}\tan \frac{x}{2} \right) +C,$$
which will not be valid for $e >1.$
What different attempt we can do for $e>1$?
|
\begin{align}
& \int \frac{dx}{1+e\cos x} = \int \frac{\frac{2\,dt}{1+t^2}}{1+e\cdot\frac{1-t^2}{1+t^2}} \\[8pt]
= {} & \int \frac{2\,dt}{(1+t^2) + e(1-t^2)} \\[8pt]
= {} & \int \frac{2\,dt}{(1+e) + (1-e)t^2} \\[8pt]
= {} & \frac 1 {1+e} \int \frac{dt}{1 + \frac{1-e}{1+e}\cdot t^2} \\[8pt]
& \text{Then if } e<1 \text{ we get this:} \\
= {} & \frac 1 {(1+e)\sqrt A} \int \frac{\sqrt A\,dt}{1+At^2}
\text{ where } A = \frac{1-e}{1+e} \\[8pt]
= {} & \frac 1 {(1+e)\sqrt A} \int \frac{du}{1+u^2} \text{ and so on.} \\[8pt]
& \text{If } e>1 \text{ then we have} \\[8pt]
& \frac 1 {1+e} \int \frac{dt}{1 - Bt^2} \text{ where } B=-A \\[8pt]
= {} & \frac 1 {1+e} \int \frac{dt}{(1 - t\sqrt B)(1 + t\sqrt B)} \\[8pt]
= {} & \frac 1 {1+e} \int \left( \frac C {1-t\sqrt B} + \frac D{1+t\sqrt B} \right) \, dt \\
& \text{and then you have to do} \\
& \text{some algebra to find $C$ and $D$.}
\end{align}
|
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|
PLANE TYPE COLUMN BASED QUESTION
My approach is as follow
$D = \left| {\begin{array}{*{20}{c}}
{2\left( {a + b} \right)}&{3\left( {b + c} \right)}&{4\left( {a + c} \right)}\\
{2\left( {b + c} \right)}&{3\left( {a + c} \right)}&{4\left( {a + b} \right)}\\
{2\left( {a + c} \right)}&{3\left( {a + b} \right)}&{4\left( {b + c} \right)}
\end{array}} \right|$
$\alpha = a + b;\beta = b + c,\gamma = a + c$
$D = \left| {\begin{array}{*{20}{c}}
{2\alpha }&{3\beta }&{4\gamma }\\
{2\beta }&{3\gamma }&{4\alpha }\\
{2\gamma }&{3\alpha }&{4\beta }
\end{array}} \right| \Rightarrow D = 2 \times 3 \times 4\left| {\begin{array}{*{20}{c}}
\alpha &\beta &\gamma \\
\beta &\gamma &\alpha \\
\gamma &\alpha &\beta
\end{array}} \right| \Rightarrow D = 24\left| {\begin{array}{*{20}{c}}
\alpha &\beta &\gamma \\
\beta &\gamma &\alpha \\
\gamma &\alpha &\beta
\end{array}} \right|$
$ \Rightarrow D = 24\left[ {\alpha \left( {\beta \gamma - {\alpha ^2}} \right) - \beta \left( {{\beta ^2} - \alpha \gamma } \right) + \gamma \left( {\alpha \beta - {\gamma ^2}} \right)} \right] = 24\left[ {3\alpha \beta \gamma - \left( {{\alpha ^3} + {\beta ^3} + {\gamma ^3}} \right)} \right] = 24T$
${\left( {\alpha + \beta + \gamma } \right)^3} = \left( {{\alpha ^3} + {\beta ^3} + {\gamma ^3}} \right) + 3\left[ {\left( {\alpha + \beta + \gamma } \right)\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right) - \alpha \beta \gamma } \right]$
$ \Rightarrow T = 3\alpha \beta \gamma - \left( {{\alpha ^3} + {\beta ^3} + {\gamma ^3}} \right) = 3\left( {\alpha + \beta + \gamma } \right)\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right) - {\left( {\alpha + \beta + \gamma } \right)^3}$
$ \Rightarrow T = \left( {\alpha + \beta + \gamma } \right)\left( {3\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right) - {{\left( {\alpha + \beta + \gamma } \right)}^2}} \right)$
$ \Rightarrow T = \left( {\alpha + \beta + \gamma } \right)\left( {3\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right) - {{\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2} + 2\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)} \right)}^2}} \right)$
$ \Rightarrow T = \left( {\alpha + \beta + \gamma } \right)\left( {\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right) - \left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2}} \right)} \right)$
$ \Rightarrow T = - \frac{{\left( {\alpha + \beta + \gamma } \right)}}{2}\left( {2\left( {{\alpha ^2} + {\beta ^2} + {\gamma ^2}} \right) - 2\left( {\alpha \beta + \beta \gamma + \alpha \gamma } \right)} \right)$
$ \Rightarrow T = - \frac{{\left( {\alpha + \beta + \gamma } \right)}}{2}\left( {{{\left( {\alpha - \beta } \right)}^2} + {{\left( {\beta - \gamma } \right)}^2} + {{\left( {\gamma - \alpha } \right)}^2}} \right)$
$ \Rightarrow T = - \left( {a + b + c} \right)\left( {{{\left( {a - c} \right)}^2} + {{\left( {b - a} \right)}^2} + {{\left( {c - b} \right)}^2}} \right)$
$ \Rightarrow T = - 2\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - \left( {ac + ba + cb} \right)} \right)$
$D \ne 0;a + b + c \ne 0;{a^2} + {b^2} + {c^2} \ne ab + bc + ca$, hence $P \to 3$
Using this how do we find the other options?
|
$\bullet$ If $a+b+c=0$ and $ a^2 + b^2 + c^2 \neq ac + ba + cb $ $\implies $ $T=0$ so the equations have infinitely many solutions
$\bullet$ If $a+b+c\neq0$ and $ a^2 + b^2 + c^2 = ac + ba + cb $ $\implies $ $T=0$ and $a=b=c\neq0$ so the equations represent identical planes
$\bullet$ If $a+b+c\neq0$ and $ a^2 + b^2 + c^2 \neq ac + ba + cb $ $\implies $ $T\neq0$ so the equations represent planes meeting only at a single point
$\bullet$ If $a+b+c=0$ and $ a^2 + b^2 + c^2 = ac + ba + cb $ $\implies $ $T=0$ and $a=b=c=0$ so the equations represent the whole of $\mathbb{R^3}$
|
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|
Sum of digits of $2^n - k$ in binary Denote the sum of digits of $n$ in binary as $s_2(n)$.
Trying to figure the relation $f(n,k) = s_2(2^n - k)$ for positive integers $n,k$ such that $2^n > k$. I see no pattern here and was thinking there should be at least a good lower bound as a function of $n,k,s_2(k)$.
It is clear from binary addition that $s_2(2^n + k) = s_2(k) + 1$.
Can we get some formula for $f(n,k)$ or at least lower bounds if not known?
E.g for $n=3$ we have:
If $k=1$ then $s_2(2^3-1)=s_2(7) = 3$
If $k=2$ then $s_2(2^3-2)=s_2(6) = 2 $
If $k=3$ then $s_2(2^3-3)=s_2(5) = 2$
If $k=4$ then $s_2(2^3-4)=s_2(4) = 1$
If $k=5$ then $s_2(2^3-5)=s_2(3) = 2$
If $k=6$ then $s_2(2^3-6)=s_2(2) = 1$
If $k=7$ then $s_2(2^3-7)=s_2(1) = 1$
|
Note that
$$
s_2((2^n-1)-k)=n - s_2(k),
$$
since subtracting from $11\cdots 1$ has the effect of complementing every bit of $k$. Therefore,
$$
s_2(2^n-k)=s_2((2^n-1)-(k-1))=\boxed{n-s_2(k-1).}
$$
|
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|
Find volume of a solid figure that lies between: $x^2+y^2+z^2\leq 4$, $2x^2+2y^2-2z^2\geq 1$, $2z^2\geq x^2+y^2$ Find volume of a solid figure that lies between:
$x^2+y^2+z^2\leq 4$,
$2x^2+2y^2-2z^2\geq 1$,
$2z^2\geq x^2+y^2$
I just really can't figure out the limits of integration... any hint would be great
-----edit----
Spherical coordinates:
$x=\rho \sin\varphi \cos\theta$
$y=\rho \sin\varphi \sin\theta$
$z = \rho \cos\varphi$
From sphere: $\rho ^2 \leq 4$ or $\rho \leq 2$.
From hyperboloid: $ 2\rho ^2( \sin^2\varphi - \cos^2 \varphi)\geq 1$
From cone: $\rho ^2(2 \cos^2\varphi - \sin^2 \varphi)\geq 0$
Intersection of sphere and hyperboloid, I got the ellipse:
$\frac{x^2}{\frac{9}{4}}+\frac{y^2}{\frac{9}{4}}=1$ or in spherical coordinates: $\rho^2 \sin^2\varphi=\frac{9}{4}$
|
Let's consider the region above $z = 0$. Due to symmetry the volume bound is same above and below $z = 0$.
The region is defined by,
a) $x^2 + y^2 + z^2 \leq 4$
In spherical coordinates, $\rho \leq 2$ (the sphere)
b) $2x^2+2y^2-2z^2\geq 1$
In spherical coordinates, $ - 2 \rho^2 \cos 2\phi\geq 1$
c) $2 z^2 \geq x^2 + y^2$
In spherical coordinates, $ \tan \phi \leq \sqrt2$
At intersection of hyperboloid and sphere,
$-8 \cos2\phi = 1$ (plugging in $\rho = 2$ in $(b)$)
So, $\cos 2\phi = - \frac{1}{8}$
Notice that this is the region which is outside the hyperboloid and inside the cone and the sphere.
So, $ \sqrt{- (\sec 2\phi) / 2} \leq \rho \leq 2$ (please note for the given limits of $\phi$, $\sec 2\phi$ is negative and so the value inside the square root is positive).
$\arccos \left(\frac{\sqrt7}{4}\right) \leq \phi \leq \arccos \left(\frac{1}{\sqrt3}\right)$ (the lower bound is same as $\cos 2 \phi \leq - 1/8$ and the upper bound is same as $\tan \phi \leq \sqrt2$. I have just rewritten them differently).
$0 \leq \theta \leq 2\pi$
Can you take it from here?
|
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|
Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\frac{7}{x+4}\\
-1(x+4)<7\\-x < 11\\
x>-11.$$
This results in the answer $x>-11, x>3,$ which doesn't make sense.
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If $k>0$, then $a<b\iff ak<bk$. However, if $k<0$, then $a<b\iff ak\color{red}{>}bk$.
So when multiplying an inequality by a constant, we should be certain whether that constant is positive or negative. In the case of $-1<\frac{7}{x+4}$, multiplying by $x+4$ is not a good idea because we don't know if $x+4<0$ or $x+4>0$. A better idea is to multiply by $(x+4)^2$, which is positive.* Then,
\begin{align}
-1<\frac{7}{x+4}&\iff-(x+4)^2<7(x+4) \\[5pt]
&\iff -x^2-8x-16<7x+28 \\[5pt]
&\iff 0<x^2+15x+44 \\[5pt]
&\iff0<(x+4)(x+11) \\[5pt]
&\iff x<-11\text{ or }x>-4 \, .
\end{align}
*Technically, $(x+4)^2$ could also be zero, if $x=-4$. However, this is a non-issue, as if $x=-4$, then $7/(x+4)$ is undefined, so it is not a solution to $-1<7/(x+4)$; and $x=-4$ is not a solution to $-(x+4)^2<7(x+4)$ either.
|
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|
Question about calculating a series involving zeta functions On this page it had shown that the sum of $\frac{1}{n^3(n+1)^3}=10-\pi^2$. I'm curious about, what is the value of $$\sum_{n=1}^\infty\frac1{n^3(n+k)^3}$$
For some positive integer $k$.
According to partial fraction expansion, we can show that $$\frac1{n^3(n+k)^3}= 6\bigg(\frac1{nk^5}-\frac1{(n+k)k^5}\bigg)-3\bigg(\frac1{k^4n^2}+\frac1{k^4(n+k)^2}\bigg)+\frac1{k^3n^3}-\frac1{k^3(n+k)^3}$$
It is obvious to show that the first part and the last part are telescoping series, and for the last part, we can see that $$\frac1{k^3n^3}-\frac1{k^3(n+k)^3}=\frac1{k^3}\bigg(\frac1{n^3}-\frac1{(n+k)^3}\bigg)=\frac1{k^3}\sum_{i=1}^{k}\frac1{i^3}=\zeta(6)+\sum_{i<j}\frac1{i^3j^3}=\sum_{n=1}^\infty\frac1{n^3(n+k)^3}$$
Which leads to the original question.
The particular values of the sum are
$k$
$$\sum_{n=1}^\infty\frac1{n^3(n+k)^3}$$
$1$
$10-\pi^2$
$2$
$\frac {21}{32}-\frac1{16}\pi^2$
$3$
$\frac {809}{5832}-\frac1{81}\pi^2$
We can easily know that the sum is in the form of $a+b\pi^2$ and $b=\frac1{k^4}$. So what about the value of $a$?
Edit: Some notes on $\zeta(3)$:
By squaring $\zeta(3)$,
$$(\zeta(3))^2=\zeta(6)+\sum_{i\ne j}\frac1{i^3j^3}$$.
Note that $i$ and $j$ are both integers and we can assume that $i$ is strictly larger than $j$, or we could say that $i=n$, $j=n+k$ for some positive integer $k$. Hence
$$(\zeta(3))^2=\zeta(6)+2\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac1{n^3(n+k)^3}$$
Assume $\sum_{n=1}^\infty\frac1{n^3(n+k)^3} = a_k-\frac{\pi^2}{k^4}$.
Thus we can know
$$\begin{align}(\zeta(3))^2&=\zeta(6)+2\sum_{k=1}^{\infty}\bigg(a_k-\frac{\pi^2}{k^4}\bigg)\\&=\frac{\pi^6}{945}+2\sum_{k=1}^{\infty}a_k-2\pi^2\zeta(4)\\&=2\sum_{k=1}^{\infty}a_k-\frac{4\pi^6}{189}\end{align}$$
For $\sum_{k=1}^{10}a_k$, we can calculate that
$$\begin{align}(\zeta(3))^2&\approx 2\sum_{k=1}^{8}a_k-\frac{4\pi^6}{189}\\&\approx 1.42163941214...\end{align}$$
And $(\zeta(3))^2\approx1.44494079841...$
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Here's a smooth and Elementary way of solving the problem .
I'll use the fact that $$\frac{1}{a.b}=\frac{1}{(b-a)}{\left(\frac{1}{a}-\frac{1}{b}\right)}$$
So , $$\sum_{n=1}^{\infty}\left(\frac{1}{n(n+k)}\right)^3=\sum_{n=1}^{\infty}\frac{1}{k^3}\left(\frac{1}{n}-\frac{1}{(n+k)}\right)^3$$ $$\Rightarrow \frac{1}{k^3}\sum_{n=1}^{\infty}\underbrace{\frac{1}{n^3}-\frac{1}{(n+k)^3}}_{H^{(3)}_{k}}-\frac{3}{n(n+k)}\left(\frac{1}{n}-\frac{1}{(n+k)}\right)\tag{*}$$
Now we need to compute this sum .$$\sum_{n=1}^{\infty}\frac{3}{n(n+k)}\left(\frac{1}{n}-\frac{1}{(n+k)}\right)\Rightarrow3\color{red}{\sum_{n=1}^{\infty}\frac{k}{n^2(n+k)^2}}\tag{1}$$
Second sum which we have to compute is $$\sum_{n=1}^{\infty}\left(\frac{1}{n(n+k)}\right)^2=\sum_{n=1}^{\infty}\frac{1}{k^2}\left(\frac{1}{n}-\frac{1}{(n+k)}\right)^2$$ $$\Rightarrow\frac{1}{k^2}\sum_{n=1}^{\infty}\frac{1}{n^2}+\frac{1}{(n+k)^2}-\frac{2}{n(n+k)}$$ $$=\frac{1}{k^2}\left(\sum_{n=1}^{\infty}\frac{1}{n^2}+\color{red}{\underbrace{\sum_{n=1}^{\infty}\frac{1}{(n+k)^2}+\sum_{n=1}^{k}\frac{1}{n^2}}_{\zeta(2)}}-H_{k}^{(2)}-\color{green}{\sum_{n=1}^{\infty}\frac{2}{n(n+k)}}\right)\tag{2}$$ The third sum which we need to compute is $$\Rightarrow\sum_{n=1}^{\infty}\frac{1}{n(n+k)}=\sum_{n=1}^{\infty}\frac{1}{k}\left(\frac{1}{n}-\frac{1}{(n+k)}\right)=\frac{1}{k}\left(\sum_{n=1}^{\infty}\frac{1}{n}-\sum_{n=1}^{\infty}\frac{1}{(n+k)}\right)$$
$$\Rightarrow \frac{1}{k}\sum_{n=1}^{k}\frac{1}{n}=\color{green}{\frac{H_k^{(1)}}{k}}\tag{3}$$
Now finally assembling all the sums we get , $$\sum_{n=1}^{\infty}\left(\frac{1}{n(n+k)}\right)^2=\frac{1}{k^2}\left(2\zeta(2)-H_k^{(2)}-\frac{2H_k^{(1)}}{k}\right)\tag{4}$$
Using the value of sum from equation $(4)$ to equation $(*)$ , we'll get $$\frac{1}{k^3}\left(H_k^{(3)}-\frac{3}{k}\left(2\zeta(2)-H_k^{(2)}-\frac{2H_k^{(1)}}{k}\right)\right)$$ $$\Rightarrow\frac{1}{k^3}\left(H_k^{(3)}-\frac{6\zeta(2)}{k}+\frac{3H_k^{(2)}}{k}+\frac{6H_k^{(1)}}{k^2}\right)$$ $$\Rightarrow\color{red}{\underbrace{\frac{H_k^{(3)}}{k^3}+\frac{3H_k^{(2)}}{k^4}+\frac{6H_k^{(1)}}{k^5}}_{a}}-\frac{\pi^2}{k^4}$$
|
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|
Volume of regular tetrahedron in a cube
Given a cube with side length $a$, a regular tetrahedron is constructed such that two vertices of the tetrahedron lie on the cube’s body diagonal and the other two vertices lie on the diagonal of one of the faces of the cube. Determine the volume of the tetrahedron.
All I can do is to visualise such a tetrahedron, but unable to see how to get the side of it since it's vertices could be anywhere on the diagonals though at equal distances from the vertices of the cube. If somehow, I can get it's side $l$ , then it's volume would be $\dfrac{l^3}{6\sqrt{2}}$.
Another fact that I know is that volume of a tetrahedron is $1/6$ of the volume of the parallelopiped formed by it's three side vectors, but don't know if it's useful here.
Any help is appreciated!
Thanks.
|
The key insight is to first determine the minimum distance between the body diagonal and the face diagonal. Without loss of generality let the cube have unit side length and take the body diagonal to be the segment joining $(0,0,0)$ to $(1,1,1)$; then there are six face diagonals that do not intersect the body diagonal. These can be partitioned into two groups of three such that each group lies in a plane, and form the sides of equilateral triangle of side length $\sqrt{2}$ perpendicular to the body diagonal, which passes through the centers of these triangles.
Consequently, the minimum distance between the body diagonal and any such face diagonal is simply $$\frac{\sqrt{2}}{2\sqrt{3}} = \frac{1}{\sqrt{6}}.$$
Since the tetrahedron is regular, with two vertices on the body diagonal and two on the face diagonal, this means distance between two non-adjacent edges of the tetrahedron is $1/\sqrt{6}$. If the side length of the tetrahedron is $2s$, then this implies the distance between $(s, -\frac{1}{2\sqrt{6}}, 0)$ and $(0, \frac{1}{2 \sqrt{6}}, s)$ is $2s$; i.e., we require $$s^2 + \frac{1}{6} + s^2 = 4s^2,$$ or $$s = \frac{1}{2\sqrt{3}}.$$
Therefore, the tetrahedron's volume is $$V = \frac{(2s)^3}{6\sqrt{2}} = \frac{1}{18\sqrt{6}},$$ and the volume for the original cube of side length $a$ is $$V(a) = \frac{a^3}{18 \sqrt{6}}.$$
We could also have found this by noting that the circumscribed cube to the tetrahedron has edge length $1/\sqrt{6}$, thus the edge of the tetrahedron, being also the face diagonal of the circumscribed cube, has length $\sqrt{2}$ times this. And the circumscribed cube's volume is just $\frac{1}{6 \sqrt{6}}$, which is $3$ times the volume of the inscribed tetrahedron, since the four congruent tetrahedra inside the cube but outside the regular tetrahedron each has volume $1/6$ that of the cube. So the desired tetrahedron's volume is again $1/(18 \sqrt{6})$.
For your understanding and enjoyment, please see the animation below, which illustrates the six tetrahedra that can be thus formed with a single body diagonal. A representative set of coordinates for a single tetrahedron, up to symmetry, is $$\left\{ \left(\frac{6 - \sqrt{6}}{12}, \frac{6 + \sqrt{6}}{12}, 0 \right), \left(\frac{6 + \sqrt{6}}{12}, \frac{6 - \sqrt{6}}{12}, 0 \right), \left(\frac{1}{6}, \frac{1}{6}, \frac{1}{6}\right), \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) \right\}.$$
|
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|
Find number of arrangements of a cube if sum of numbers on each face must be same
Each vertex of a cube is to be labelled with an integer 1 through 8, without repetition, such that sum of numbers of the four vertices of a face is the same for each face. Arrangements that can be obtained through rotations of the cube are considered to be the same. How many different arrangements are possible?
My attempt: Fix 8 on E.
Then, D+C=F+G (Since H is common to both faces)
Similarly, C+H=A+F and D+A=H+G
We can extend this analogy to other sides, and we see that D+E=B+G, E+H=A+B, E+F=B+C.
This means that {1,2}, {1,3}, {7,8}, {8,6} cannot be on one line segment since there are no other distinct numbers which add up to give 3,4,15 or 14 respectively.
How can I get the other conditions?
Edit: As @John and @Alexander mentioned in the comments, the sum of numbers on each face should be 18. The possible sets of numbers on opposite faces will be {8,7,1,2}, {3,4,5,6}; {8,6,1,3}, {4,5,7,2}; {8,5,2,3}, {1,4,7,6} and {8,5,1,4}, {2,3,6,7}.
Case 2 will not have any possibilities since there are no 2 numbers except for 8,3 which add up to 11. In case 4, 4 must be opposite to 8 and in case 1, 7 must be opposite to 8.
Now we can try to count it on a case by case basis, but I can't think of a foolproof method without repetitions. Can someone help :)
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Note that the sum of the numbers on each face must be $18$ because $\frac{1+2+\cdots+8}{2}=18$.
So now consider the opposite edges (two edges that are parallel but not on the same face of the cube); they must have the same sum value too. Now think about the points $1$ and $8$. If they are not on the same edge, they must be endpoints of opposite edges, and we should have $1+X=8+Y$. However, this scenario would yield no solution for $[2,7]$, which is a contradiction.
The points $1$ and $8$ are therefore on the same side and all edges parallel must also sum to $9 .$
Now we have $4$ parallel sides $1-8,2-7,3-6,4-5$. Thinking about $4$ endpoints, we realize they need to sum to 18. It is easy to notice only $1-7-6-4$ and $8-2-3-5$ would work.
So if we fix one direction $1-8($ or $8-1)$ all other 3 parallel sides must lay in one particular direction. $(1-8,7-2,6-3,4-5)$ or $(8-1,2-7,3-6,5-4)$
Now, the problem is the same as arranging 4 points in a two-dimensional square, which is $\frac{4 !}{4}=6$.
It is also available here. There is also a video solution here.
|
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Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$
*
*I managed to get $ab=cd$. Don't know how to proceed further.
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Having gotten $ab=cd$, take a linear combination with $a^2+b^2=c^2+d^2$:
$\color{blue}{ab=cd}$
$\color{brown}{a^2+b^2=c^2+d^2}$
$\color{brown}{a^2}\color{blue}{-2ab}\color{brown}{+b^2}=\color{brown}{c^2}\color{blue}{-2cd}\color{brown}{+d^2}$
Those polynomials are squares:
$(a-b)^2=(c-d)^2$
$a-b=\pm(c-d)$
Then if $a-b=+(c-d)$ and $a+b=c+d$, we are forced to have $a=c,b=d$. Can you fill in the result for the case $a-b=-(c-d)$?
|
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|
Solve the following equation: $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$ I have tried many ways but not success
$\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$
$\Leftrightarrow \sqrt{x}(\sqrt{x+1}-\sqrt{x-1})=x+1$
or $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1 \Leftrightarrow \sqrt{\dfrac{x}{x+1}}-\dfrac{\sqrt{x(x-1)}}{x+1}=1 \Leftrightarrow \dfrac{x}{x+1}=1+\dfrac{2\sqrt{x(x-1)}}{x+1}+\dfrac{x(x-1)}{(x+1)^2} \Leftrightarrow \dfrac{(x+1)^2+2\sqrt{x(x-1)}(x+1)+x(x-1)-x(x+1)}{(x+1)^2}=0 \Leftrightarrow x^2+1+2\sqrt{x(x-1)}(x+1)=0 $
I need your help!
|
Using your first approach, square both sides:
$$x \left((x+1) - 2\sqrt{x^2-1} + (x-1)\right) = (x+1)^2.$$ Now rearrange and collect like terms to isolate the square root:
$$2x \sqrt{x^2 - 1} = 2x^2 - (x+1)^2 = x^2 - 2x - 1.$$
Next, square again:
$$4x^2 (x^2 - 1) = (x^2 - 2x - 1)^2.$$
The resulting quartic has a rather complicated closed form, but it does exist; the only thing left is to eliminate extraneous roots.
|
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|
How do I solve this probability Anna writes down a random sequence created by the following process. She repeatedly rolls a fair
6-sided die. If the number she rolls is larger than all of the numbers she has previously rolled (if
any), then she writes the new number down and then continues rolling. Otherwise, she does not
write the new number down and the process ends.
Let X be the length of Anna’s sequence, and Y be the last number in her sequence.
For example, if Anna rolled 1 then 4 then 5 then 4, her sequence would be 1,4,5 and the random
variables X, Y would take values X = 3 and Y = 5.
Question
What is $Pr(X=2|Y=3)$
My workings so far
$Pr(X=2|Y=3)=$ $\frac{Pr(X=2 \land Y=3)}{Pr(Y=3)}$
I know that $Pr(Y=3)= \frac{1}{6}(1+\frac{2}{6}+\frac{\binom{2}{2}}{6^2})$
Then, $Pr(Y=3)=\frac{49}{216}$
The problem I'm having trouble figuring out is calculating $Pr(X=2\land Y=3)$
Can anyone help me with this, I'm very confused. Thank you so much in advance.
|
What is $Pr(X=2|Y=3)$
Use Bayes Theorem.
In order for $Y$ to equal $3$, several things have to have happened:
*
*You rolled zero, one, or two rolls below a $3$.
*Then, you rolled a $3$.
*Then, you rolled a number below a $4$.
$$P(X = 2| Y = 3) = \frac{P(X = 2 \wedge Y = 3)}{P(Y = 3)}.$$
I will repeat some of your analysis in order to clarify how to calculate the numerator as well as the denominator.
$\underline{\text{Case 1:}}$
Roll a 1, then a 3, then a number below a 4.
$(1/6) \times (1/6) \times (1/2).$
Enumeration is $\frac{1}{72} = \frac{6}{432}$.
$\underline{\text{Case 2:}}$
Roll a 2, then a 3, then a number below a 4.
$(1/6) \times (1/6) \times (1/2).$
Enumeration is $\frac{1}{72} = \frac{6}{432}$
$\underline{\text{Case 3:}}$
Roll a 1, then a 2, then a 3, then a number below a 4.
$(1/6) \times (1/6) \times (1/6) \times (1/2).$
Enumeration is $\frac{1}{432}.$
$\underline{\text{Case 4:}}$
Roll a 3, then a number below a 4.
$(1/6) \times (1/2).$
Enumeration is $\frac{1}{12} = \frac{36}{432}.$
The denominator, which represents $p(Y = 3)$, is the sum of all 4 cases:
$\frac{49}{432}.$
The numerator, which represents $p(X = 2 \wedge Y = 3)$ is the sum of Case 1 and Case 2.
$\frac{12}{432}$.
Therefore, the desired probability is
$$\frac{\frac{12}{432}}{\frac{49}{432}} = \frac{12}{49}.$$
|
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|
Find the sum of the series $3x+8x^2+15x^3 + ....$ I'm trying to express the following series $3x+8x^2+15x^3 + ....$ as a sum and hope to find its sum for $|x| < 1$
Here is what I have so far:
To me the series looks to be the derivative of the following form:
$1 + \frac{3}{2}x^2+\frac{8}{3}x^3+\frac{15}{4}x^4 ...$
Given that the denominator progresses as $\frac{1}{n}$ I had thought of taking it out like so:
$\frac{1}{n}(1+3x^2+8x^3+15x^4 ....)$
Now I'm unsure of the sequence that the constants follow but lets denote this as $a$ and take it out of the sequence of $x$ values, then I get:
$\frac{1}{n}(1+a(x^2+x^3+x^4+x^5 ...))$
We know that the sum of the series of $x$ follows the following geometric series: $\frac{x^2}{(1-x)}$, plugging this into the equation:
$\frac{1}{n}(1+\frac{ax^2}{1-x})$
Then taking the first derivative I get:
$\frac{ax(2+x)}{n(1-x)^2}$
I'm unsure as to whether the $n$ can still be introduced or whether it's removed - How is this answer optimally derived?
Following the hint below:
$$\sum_{n=2}^{\infty}(n^2-1)x^n = \sum_{n=1}^{\infty}(n^2-1)x^n =\sum_{n=1}^{\infty}n^2x^n - \sum_{n=1}^{\infty}x^n = \frac{x(1+x)}{(1-x)^3}-\frac{x}{1-x}$$?
|
If you check your $\frac{x(1+x)}{(1-x)^3}-\frac{x}{1-x}$ you will find it is $3x^2+8x^3+15x^4 + \cdots$ so you need to remove a factor of $x$ to get $\frac{(1+x)}{(1-x)^3}-\frac{1}{1-x}$ though I would not write it that way
If you know the coefficient of $x^n$ is $n(n+2)$ then one approach could be to manipulate $\frac{1}{1-x}=1+x+x^2+x^3+\cdots$ and say
*
*$\frac{1}{1-x}=1+x+x^2+x^3+x^4+x^5+\cdots$
*$\frac d{dx} \frac{1}{1-x}=1+2x+3x^2+4x^3+5x^4+\cdots$
*$x \frac d{dx} \frac{1}{1-x}=x+2x^2+3x^3+4x^4+5x^5+\cdots$
*$\frac d{dx}\left(x \frac d{dx} \frac{1}{1-x}\right)=1+ 4x+9x^2+16x^3+25x^4+\cdots$
*$\frac d{dx}\left(x \frac d{dx} \frac{1}{1-x}\right) -\frac{1}{1-x} = 3x+8x^2+15x^3+24x^4+\cdots$
which seems to be what you want.
You can then find $$\frac d{dx}\left(x \frac d{dx} \frac{1}{1-x}\right) -\frac{1}{1-x} = \frac{2 x}{{{\left( 1-x\right) }^{3}}}+\frac{1}{{{\left( 1-x\right) }^{2}}}-\frac{1}{1-x}= \frac{x(3-x)}{(1-x)^3}$$
|
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|
Issue solving an integral I'm having issues solving the following integral:
$$ \int{\frac{e^x}{e^{2x}-e^x-2}}dx $$
From what I can tell, I should substitute $u = e^x$ and $du = e^xdx$ and end up with
$$ \int{\frac{1}{u^2 - u - 2}}du $$
where I can then use partial fraction decomposition to simplify the integrand as follows
$$ \frac{1}{u^2 - u - 2} = \frac{A}{u-2} + \frac{B}{u+1} $$
$$ A(u+1) + B(u-2) = 1 $$
$$ Au + A + Bu -2B = 1 $$
$$\begin{pmatrix}1 & 1 & 0\\1 & -2 & 1\end{pmatrix} \to \begin{pmatrix} 1 & 0 & \frac{1}{3} \\ 0 & 1 & -\frac{1}{3} \end{pmatrix} \to A=\frac{1}{3}, B=-\frac{1}{3} $$
So I have
$$ \int{ \left( \frac{ \frac{1}{3} }{u-2} - \frac{ \frac{1}{3} }{u + 1} \right) } du = \frac{1}{3} \int{\frac{1}{u - 2}du} - \frac{1}{3}\int{\frac{1}{u+1}}du $$
which gives me
$$ \frac{1}{3}\ln(u-2) - \frac{1}{3}\ln(u + 1) + C, \textrm{ where } u =e^x $$
so my final answer is
$$ \frac{1}{3}( \ln(e^x-2) - \ln(e^x + 1) ) + C $$
However, wolframalpha is telling me that the answer is
$$ \frac{2}{3}\tanh^{-1}\left(\frac{1}{3} - \frac{2e^x}{3}\right) + C $$
I really have no idea how the inverse tangent hyperbolic function would even get there. We haven't even covered any integrals that give $\tanh^{-1}$ in class.
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It turns out that$$\operatorname{arctanh}'(x)=\frac1{1-x^2}.$$But\begin{align}\frac1{u^2-u-2}&=\frac1{\left(u-\frac12\right)^2-\frac94}\\&=-\frac49\frac1{1-\left(\frac13-\frac{2u}3\right)^2},\end{align}and therefore\begin{align}\left(\frac23\operatorname{arctanh}\left(\frac13-\frac{2u}3\right)\right)'&=-\frac49\frac1{1-\left(\frac13-\frac{2u}3\right)^2}\\&=\frac1{u^2-u-2}.\end{align}
|
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|
Cycle notation and transpositions For example, consider the permutation
$$ \pi=\left(\begin{matrix}
1&2&3&4&5\\
4&3&2&5&1
\end{matrix}\right).$$
You can write it with two cycles as
$$ \pi =(145)(23).$$
Now I want to write $\pi$ as a product of transpositions. I know one way to do that is
$$ \pi =(14)(45)(23)$$
because $(145)=(14)(45) $. However, I don't understand the logic behind this notation. Transpositions are cycles as well, so wouldn't that notation imply $\pi(1)=4 $, and $\pi(4)=1\ne 5$ (according to the first transposition)? Or $\pi(4)=5 $ and $\pi(5)=4\ne 1$ (according to the second transposition)? So could someone elaborate this notation to me? I guess that I have misunderstood the cycle notation.
|
Multiplication of cycles can be written as composition of permutations (order right to left). This way we can see the commonality when representing $\pi$ as multiplication of transpositions as well as multiplication of other cycles.
We obtain
\begin{align*}
\color{blue}{\pi}&\color{blue}{=(1\,4\,5)(2\,3)}\\
&=(1\,4\,5)(2)(3)\circ(1)(2\,3)(4)(5)\\
&=\begin{pmatrix}
1&2&3&4&5\\
4&3&2&5&1\\
\end{pmatrix}
\\
\\
\color{blue}{\pi}&\color{blue}{=(1\,4)(4\,5)(2\,3)}\\
&=(1\,4)(2)(3)(5)\circ\left((1)(2)(3)(4\,5)\circ(1)(2\,3)(4)(5)\right)\\
&=(1\,4)(2)(3)(5)\circ(1)(2\,3)(4\,5)\\
&=\begin{pmatrix}
1&2&3&4&5\\
4&3&2&5&1\\
\end{pmatrix}
\\
\\
\color{blue}{\pi}&\color{blue}{=(1\,4)(4\,5)(2\,3)}\\
&=\left((1\,4)(2)(3)(5)\circ(1)(2)(3)(4\,5)\right)\circ(1)(2\,3)(4)(5)\\
&=(1\,4\,5)(2)(3)\circ(1)(2\,3)(4\,5)\\
&=\begin{pmatrix}
1&2&3&4&5\\
4&3&2&5&1\\
\end{pmatrix}
\end{align*}
|
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|
Find all complex numbers which make the following equations true: $ |z+1| =1 $ and $ |z^2+1| =1 $ Find all complex numbers which make the following equations true:
$$ |z+1| =1 $$
$$ |z^2+1| =1 $$
Solution:
If $ |z+1| =1 $ holds true, then
$$z+1 = 1.e^{i2n\pi}$$
$$z = 1.e^{i2n\pi}-1$$
$$z = 1.e^{i2n\pi}-1e^{i2m\pi}$$
If $ |z^2+1| =1 $ holds true, then
$$z^2+1 = 1.e^{i2p\pi}$$
$$z^2 = 1.e^{i2p\pi}-1$$
$$z^2 = 1.e^{i2p\pi}-1e^{i2q\pi}$$
*
*How to proceed after this?
*Am I supposed to do in this manner or break complex number z into real part x and imaginary part y and get two equations and thus solve for x and y?
|
If we interpret the absolute-value expressions as descriptions of loci of points at specified distances from particular locations, then
$ |z + 1| \ = \ 1 \ \ $ is the set of points at unit distance from $ \ z \ = \ -1 \ \ $ [the unit circle centered at that point] and
$ |z^2 + 1| \ = \ |(z + i)·(z - i)| \ = \ 1 \ \ $ is the set of points for which the product of the distances of each point from $ \ z \ = \ i \ $ and $ \ z \ = \ -i \ $ is a constant $ \ 1 \ $ [the "unit" lemniscate with foci at those points].
Solving for the intersections of these two curves using their Cartesian expressions becomes a bit complicated. We might instead convert these to "polar form":
$ (x + 1)^2 + y^2 \ = \ 1 \ \rightarrow \ x^2 + y^2 + 2x + 1 \ = \ 1 \ \rightarrow \ r \ = \ -2·\cos \theta \ \ ; $
$ [ \ x^2 + (y + 1)^2 \ ] · [ \ x^2 + (y - 1)^2 \ ] \ = \ 1 \ \rightarrow \ x^4 + 2x^2y^2 + y^4 \ = \ 2·(y^2 - x^2) $ $ \rightarrow \ r^2 \ = \ -2·\cos(2 \theta) \ \ . $
Equating these polar curve equations will give us information about intersections other than the origin (which must be checked separately):
$$ -2·\cos(2 \theta) \ \ = \ \ ( \ -2·\cos \theta \ )^2 \ \ \Rightarrow \ \ -2 \ · \ ( \ 2·\cos^2 \theta \ - \ 1 \ ) \ \ = \ \ 4 · \cos^2 \theta $$
$$ \Rightarrow \ \ 8 · \cos^2 \theta \ \ = \ \ 2 \ \ \Rightarrow \ \ \cos \theta \ \ = \ \ \pm \ \frac12 \ \ . $$
The solution $ \cos \theta \ = \ + \frac12 \ $ is "spurious", since it cannot be used in the equation $ \ r \ = \ -2·\cos \theta \ $ (alternatively, we can read the "negative radii" as duplicating the results for the positive-radius solutions). For $ \cos \theta \ = \ - \frac12 \ \ , $ we have $ \ \theta \ = \ \frac{2 \pi}{3} \ \ , \ \ \frac{4 \pi}{3} \ \ $ and $ \ r \ = \ 1 \ $ on the circle. For the lemniscate, we find $ \ 2\theta \ = \ \frac{4 \pi}{3} \ \ , \ \ \frac{8 \pi}{3} \ = \ \frac{2 \pi}{3} \ \ , $ so we again have $ \ r \ = \ 1 \ \ . $ The "off-origin" solutions are thus
$$ z_1 \ \ = \ \ e^{ \ i·2 \pi/3} \ \ = \ \ -\frac12 + i·\frac{\sqrt3}{2} \ \ \ , \ \ \ z_2 \ \ = \ \ e^{ \ i·4 \pi/3} \ \ = \ \ -\frac12 - i·\frac{\sqrt3}{2} \ \ = \ \ \overline{z_1} \ \ . $$ [It might be mentioned that these are the two complex cube-roots of unity.]
The origin is also an intersection point of these curves, since $ \ r \ = \ 0 \ $ and $ \ r^2 \ = \ 0 \ $ are possible values for the two curve functions, although they occur for different values of $ \ \theta \ $ for each curve. So the third (trivial?) solution for the pair of equations is $ \ z \ = \ 0 \ \ \ . $
|
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|
Prove that: $\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\sqrt{3(a^2+b^2+c^2-ab-bc-ca)}\ge \frac{3}{2}$ Problem:
Given non- negative real numbers such that: $ab+bc+ca>0: a+b+c=3.$ Prove that: $$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\sqrt{3(a^2+b^2+c^2-ab-bc-ca)}\ge \frac{3}{2}$$
My approach:
We have: $a^2+b^2+c^2\ge ab+bc+ca$. It is desired to show that: $$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}\ge\frac{3}{2}$$.
But it seems not work by calculation example. I hope we can get nice proof for problem. Thanks!
|
The following stronger inequality is also true.
Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$ and $ab+ac+bc\neq0$. Prove that:
$$\frac{1}{a^2+b^2}+\frac{1}{b^2+c^2}+\frac{1}{c^2+a^2}+\frac{7}{27}\sqrt{a^2+b^2+c^2-ab-bc-ca}\ge \frac{3}{2}.$$
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, the condition and $\sum\limits_{cyc}(a^2-ab)=9u^2-9v^2$ don't depend on $w^3$ and $f(w^3)=\sum\limits_{cyc}\frac{1}{a^2+b^2}$ increases, which by $uvw$ says that it's enough to prove our inequality in two cases:
*
*$w^3=0$;
*Two variables are equal, which easy to check.
Can you take it from here?
About $uvw$ see here:
https://artofproblemsolving.com/community/c6h278791
|
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|
Simplifying a derivative of a function involving roots. I am working on a question where I need to differentiate $$y=\{x+\sqrt{1+x²}\}^\frac{3}{2}\tag1$$
Using the chain rule I have found the first derivative $$\frac{dy}{dx} =\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{1}{2}\{1+\frac{x}{\sqrt{1+x²}}\}\tag2$$
Which can be written as
$$\frac{dy}{dx} = \frac{3y}{2\{x+\sqrt{1+x²}\}}\{1+\frac{x}{\sqrt{1+x²}}\}\tag3$$
This is as far as I have got in terms of simplifying. The solution says that $(2)$ is equal to $(4)$ $$\frac{\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{3}{2}}{\sqrt{1+x²}}=\frac{3y}{2\sqrt{1+x^2}}\tag4$$
How do you get from $(2)$ to $(4)$? I have included $(3)$ in order to show my workings.
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Since it is desired to have an expression for the derivative that includes $ \ y \ \ , $ we will want to keep in mind that
$$ y \ \ = \ \ [ \ x \ + \ \sqrt{1+x^2} \ ]^{3/2} \ \ \Rightarrow \ \ x \ + \ \sqrt{1+x^2} \ \ = \ \ y^{2/3} $$
and thus $ \ [ \ x \ + \ \sqrt{1+x^2} \ ]^{1/2} \ = \ y^{1/3} \ \ . $
The first derivative can then be written as
$$ \frac{dy}{dx} \ \ = \ \ \frac32 · [ \ x \ + \ \sqrt{1+x^2} \ ]^{1/2} \ · \ \left( \ 1 \ + \ \frac{x}{\sqrt{1+x^2}} \ \right) \ \ = \ \ \frac32 · y^{1/3} \ · \ \left( \frac{\sqrt{1+x^2} \ + \ x}{\sqrt{1+x^2}} \ \right) $$
$$ = \ \ \frac32 · y^{1/3} \ · \ \frac{y^{2/3} }{\sqrt{1+x^2}} \ \ = \ \ \frac{3 · y }{2 · \sqrt{1+x^2}} \ \ . $$
|
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|
Finding the complex integral as follows Let $C$ be a contour that is formed by an arc $y = x^2$ from $(0,0)$ to $(1,1)$, and a line segment from $(1,1)$ to $(0,1)$. Find the integral $\int_C f(z) dz$ where $f(z) = 2xy+ i(-x^2+y^2)$.
Attempt:
Let $C_1$ be the arc $y=x^2$ and $C_2$ be the line segment such that $C= C_1 \cup C_2$.
Parametrizing $C_1$ and $C_2$ gives
\begin{align*}
C_1 &: z = z(t) = t + it^2, \,0 \le t \le 1 \\
C_2 &: z = z(u) = (1-u) + i, \,0 \le u \le 1.
\end{align*}
and $z'(t) = 1 + i2t$ and $z'(u) = -1$.
Now, for $C_1$, we have
\begin{align*}
\int_{C_1} f(z) dz &= \int_{C_1} f(z(t))z'(t) \ dt \\
&= \int_0^1 (2t(t^2) + i \cdot (-t^2 + (t^2)^2)(1+i2t)) \ dt \\
&= \int_0^1 (2t^3 + i \cdot (t^4 - t^2)(1+i2t)) \ dt \\
&= \int_0^1 (-2t^5 + 4t^3) \ dt + i \cdot \int_0^1 (5t^4 - t^2 ) \ dt \\
&= \left[-\frac{t^6}{3} + t^4\right]_0^1 + i \cdot \left[t^5 - \frac{t^3}{3}\right]_0^1 \\
&= \left(-\frac{1}{3} + 1\right) + \left(1 - \frac{1}{3}\right) \\
&= \frac{2}{3} + i \cdot \frac{2}{3}.
\end{align*}
Now, for $C_2$, we have
\begin{align*}
\int_{C_2} f(z) dz &= \int_{C_2} f(z(u)) z'(u) du \\
&= \int_0^1 (2(1-u) + i \cdot (-(1-u)^2 + 1^2)(-1)) du \\
&= - \int_0^1 (2-2u + i \cdot (-u^2 + 2u) du \\
&= -([2u - u^2]_0^1 + i \cdot \left[-\frac{u^3}{3} + u^2\right]_0^1) \\
&= -((2 - 1) + i \cdot \left(-\frac{1}{3} + 1)\right) \\
&= -1 - i \cdot \frac{2}{3}.
\end{align*}
Hence,
\begin{align*}
\int_C f(z) \ dz &= \int_{C_1} f(z) \ dz + \int_{C_2} f(z) \ dz \\
&= \frac{2}{3} + i \cdot \frac{2}{3} - 1 - i \cdot \frac{2}{3} \\
&= - \frac{1}{3}.
\end{align*}
*
*Am I correct?
*How about an approach by Cauchy's theorem (if $f$ is analytic then the complex integral equals to zero.)? I doubt about using it since the contour $C$ isn't closed.
Thanks in advanced.
|
Since $f(z)=-iz^2$, $f$ is indeed analytic. If we "add" another path $C_3$ to the contour, namely the one from $(0,1)$ to $(0,0)$. Then $C$ is closed. Now we get $$\int_C f(z)dz=0$$ by Cauchy. Furthermore $$0=\int_C f(z)dz=\int_{C_1}f(z)dz+\int_{C_2}f(z)dz+\int_{C_3}f(z)dz$$ and $$\int_{C_1}f(z)dz+\int_{C_2}f(z)dz=-\int_{C_3}f(z)dz.$$ Now it suffices to only calculate the much simpler integral along the line segment from $(0,0)$ to $(0,1)$.
Now since a parametrization of $C_3$ is $it$, we get $$\int_{C_3}f(z)dz=\int_0^1f(it)idt=\int_0^1-i(it)^2idt=\int_0^1-i(it)^2idt=\int_0^1-t^2dt=-\frac{1}{3}$$
and furthermore $$-\int_{C_3}f(z)dz=-(-\frac{1}{3})=\frac{1}{3}.$$
And yes, you are indeed correct! This is just a much simpler way of calculating it.
|
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|
The smallest value of the expression $4x^2y^2+x^2+y^2-2xy+x+y+1$ What is the smallest value that $4x^2y^2+x^2+y^2-2xy+x+y+1$ can take with real numbers $x$ and $y$?
I suspect the following transformation can be done: $(2xy-1/2)^2 + (x+1/2)^2 + (y+1/2)^2 + 1/4$.
|
After $(2xy-1/2)^2 + (x+1/2)^2 + (y+1/2)^2 + 1/4$, use a change of variables $x - 1/2 = u, y - 1/2 = v$, and $(2xy-1/2) = 2(u + 1/2)(v + 1/2) - 1/2$ $ = 2(uv + u/2 + v/2 + 1/4) - 1/2$ to get:
$$(2uv+u+v)^2 + u^2 + v^2 + 1/4$$
$$=(2uv+u+v)^2 + (u+v)^2 - 2uv + 1/4$$
$$= (p+q)^2 + q^2 - p + 1/4$$
where $p=2uv, q = u+v$.
For $u, v$ to be real numbers, $q^2 - p ≥ 0$ as $u^2+v^2 ≥ 0$ for all real numbers $u,v$. Thus $(p+q)^2 + q^2 - p + 1/4 ≥ (p + q)^2 + 1/4 ≥ 1/4$.
This minimum is attained in the original expression when $x = y = -1/2 $.
|
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|
An uncommon continued fraction of $\frac{\pi}{2}$ I'm currently stuck with the following infinite continued fraction:
$$\frac{\pi}{2}=1+\dfrac{1}{1+\dfrac{1\cdot2}{1+\dfrac{2\cdot3}{1+\dfrac{3\cdot 4}{1+\cdots}}}}$$
There is an obscure clue on this: as one can derive the familiar Lord Brouncker’s fraction below
$$
\frac{4}{\pi}=1+\dfrac{1^{2}}{2+\dfrac{3^{2}}{2+\dfrac{5^{2}}{2+\dfrac{7^{2}}{2+\cdots}}}}
$$
from the Wallis' Formula:
$$
\dfrac{2}{\pi}=\frac{1 \cdot 3}{2 \cdot 2} \cdot \frac{3 \cdot 5}{4 \cdot 4} \cdot \frac{5 \cdot 7}{6 \cdot 6} \cdot \frac{7 \cdot 9}{8 \cdot 8} \cdots
$$
the first fraction can be proved in the same manner.
However, I'm not getting any close to it using the Wallis' Formula. Really appreciated if anyone could point me the right direction or explain further how to systematically derive those continued fractions from any given convergent cumulative product.
|
We know that $$\sin^{-1}x=\int\underbrace{\color{red}{\frac{1}{\sqrt{1-x^2}}}}_{\text{apply binomial theorem}}dx=\int1+\sum_{n=0}^{\infty}\frac{2n-1}{2^{n+1}}x^{2n}dx$$and if you solve this further you'll get $$\sin^{-1}x=x+
\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}+......$$ $$= x+\frac{1}{2}\cdot\frac{x^3}{3}+\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{x^5}{5}+\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\frac{x^7}{7}+......$$ $$=x+x\left(\frac{x^2}{2\cdot3}\right)+x\left(\frac{x^2}{2\cdot3}\right)\left(\frac{(3x)^2}{4\cdot5}\right)+.........$$ OR $$\sin^{-1}x=x\left(1+\sum_{n=0}^{\infty}\prod_{i=0}^{n}\frac{(2i+1)^2x^2}{(2i+2)\cdot(2i+3)}\right)$$
Now according to Euler's formula for Continued Fraction , i.e. $$S=a\left(1+\sum_{i=1}^{\infty}\prod_{j=1}^{i}r_j\right)=\large\frac{a}{1-\frac{r_1}{1+r_1-\frac{r_2}{1+r_2-\frac{r_3}{......}}}}$$
So , $$\sin^{-1}x=x\left(1+\sum^{\infty}_{n=1}\prod_{i=1}^{n}\frac{(2i-1)^2x^2}{(2i)\cdot(2i+1)}\right)$$ $$=x\Large\left(\frac{1}{1-\frac{\frac{x^2}{2\cdot3}}{1+\frac{x^2}{2\cdot3}-\frac{\frac{3^2x^2}{4\cdot5}}{1+\frac{3^2x^2}{4\cdot5}-.......}}}\right)$$
Since , $$\sin^{-1}1=\frac{\pi}{2}$$
Now just put $x=1$ in the above continued fraction to get to your answer .
|
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|
How would you prove this by induction? I've been trying to solve this one using induction for quite a while but I don't get to the solution. Any tips would be apreciated.
Let $a_1 = 47$, $a_2=80$ and for $n \geq 3$, $a_n =4a_{n-1} - 4a_{n-2}+3(n-2)^2$. Prove that for any positive integer $$ a_n=2^n(3+n)+3n^2+12n+24$$
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By Induction
Prove basis (trivial).
Induction step:
$\begin{align}
a_{n+1}=&4a_{n}-4a_{n-1}+3(n-1)^2 \\
=&4.2^n(3+n)+12n^2+48n+96 \\
-&4.2^{n-1}(2+n)-12(n-1)^2-48(n-1)-96 \\
+&3n^2-6n+3
\end{align}$,
by induction hypothesis.
$\implies a_{n+1}=2^{n+1}(3+n+1)+3(n+1)^2+12(n+1)+24$, by simplifying
$\therefore a_n=2^n(3+n)+3n^2+12n+24$, $\forall{n} \in \mathbb{N}$.
By solving the recurrence relation
Also, by solving the homogeneous recurrence relation $a_n=4a_{n-1}-4a_{n-2}$,
we have the characteristic equation $r^2-4r+4=(r-2)^2=0$,
hence $a_h(n)=c_12^n+nc_22^n$, for some constants $c_1, c_2$.
Also, guessing the particular solution $a_p(n)=An^2+Bn+C$ and by substitution, we have
$\begin{align}
An^2+Bn+C&=4A(n-1)^2+4B(n-1)+4C \\
&-4A(n-2)^2-4B(n-2)-4C \\
&+3(n^2-4n+4)
\end{align}$,
solving, we have $A=3$, $B=12$, $C=24$,
s.t., $a_p(n)=3n^2+12n+24$ and solution to the non-homogeneous recurrence relation is
$a_n=a_h+a_p=c_12^n+nc_22^n+3n^2+12n+24$,
now using boundary conditions $a_1=47$ and $a_2=80$, we have $c_1=3$, $c_2=1$
$\implies a_n=2^n(3+n)+3n^2+12n+24$
|
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|
Max of $ax+by$ given that $x^2+y^2 \le 1$ If a and b a positive real numbers find the maximum value of ax+by in terms of a and b given that $x^2+y^2 \le 1$.
I started by letting $k = ax+by$ giving the line $y=\frac{-ax}{b} + \frac{k}{b}$. Now we want to find the largest value of k such that the line still intersects the unit circle. Graphically it can be seen that this occurs when the line is tangent to the circle. This results in a right triangle with $\frac{k}{b}$ as the hypotenuse and 1 as one of the side lengths. This is where I am stuck because in the solution they had the following triangle;
How do we know that the other side length of the triangle is the gradient of the line?
|
Applying Cauchy-Schwarz Inequality
$$(ax+by)^2\leq (a^2+b^2)(x^2+y^2)\leq (a^2+b^2) \times 1 =(a^2+b^2)$$
$$ax+by \leq \sqrt{a^2+b^2}$$
|
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|
Write with one radical $\sqrt[4]{2^6}\cdot\sqrt{3^3}$
Write with one root (radical) $$\sqrt[4]{2^6}\cdot\sqrt{3^3}$$
In this lesson we have learnt that when the roots exist then $$\sqrt[n]{a}=\sqrt[nk]{a^k}$$
Using that here, we have $\gcd(4,2)=2,$ so
$$\sqrt[4]{2^6}\cdot\sqrt[2\cdot2]{\left(3^{3}\right)^2}=\sqrt[4]{2^6}\cdot\sqrt[4]{3^6}=\sqrt[4]{2^6\cdot3^6}=\sqrt[4]{\left(2\cdot3\right)^6}=\sqrt[4]{6^6}$$
Their solution, though, goes as
$$\sqrt[4]{2^6}\cdot\sqrt{3^3}=\sqrt{2^3}\sqrt{3^3}=\sqrt{6^3}=6\sqrt6$$
They haven't calculated the $\gcd(4,2)$ of the indices $4$ and $2$ as in the previous examples (e.g. $\sqrt[4]{3}\cdot\sqrt[3]{2}=...=\sqrt[12]{432}$). What am I missing? Thank you!
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Both answers are mathematically equal. What is different is just observation.
We see that,
$$\begin{align}\sqrt[4]{2^6}\cdot\sqrt{3^3}&=2^{\frac 64}\times 3^{\frac 32}\\
&=2^{\frac 32}\times 3^{\frac 32}\\
&=(2\times 3)^{\frac 32}\\
&=6^{\frac 32}\\
&=\sqrt{6^3}\\
&=\sqrt{6^2\times 6}\\
&=6\sqrt{6}\end{align}$$
Also, you can write
$$\begin{align}\sqrt[4]{2^6}\cdot\sqrt{3^3}&=2^{\frac 64}\times 3^{\frac 32}\\
&=2^{\frac 64}\times 3^{\frac 64}\\
&=(2\times 3)^{\frac 64}\\
&=6^{\frac 64}\\
&=\sqrt[4]{6^6}\end{align}$$
You already know that,
$$\begin{align}\sqrt[4]{6^6}&=6^{\frac 64}\\
&=6^{\frac 32}\\
&=\sqrt{6^3}\\
&=6\sqrt 6.\end{align}$$
|
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|
Cone Section: Intersection between a plane ax + by + cz = d and two cones of height $h$, raidus $r$ whose apexes lie at the centre, along the $z$ axis A double cone can be described by circles that increase with $|z|$ and have radius $R$ at $h = |z_R|$. It holds:
\begin{align}
\frac{R^2}{h^2}\,z^2 = x^2 + y^2
\end{align}
The plane
\begin{align}
ax + by + cz = d
\end{align}
should now intersect this cone. To do this, we move the plane to $z$ and insert this into the double cone equation.
\begin{align}
&z = \frac{d - ax - by}{c} \\
&\Longrightarrow \hspace{10pt} \underbrace{\frac{R^2}{c^2h^2}}_{\color{red}{=\,\kappa^2}}\,(d-ax-by)^2 = x^2 + y^2 \\
&\Longrightarrow \hspace{10pt} \kappa^2\,(d^2+a^2x^2+b^2y^2-2adx-2bdy + 2abxy) = x^2 + y^2 \\
&\Longrightarrow \hspace{10pt} \left(\kappa^2a^2-1\right)x^2+\left(\kappa^2b^2-1\right)y^2 + \left(2\kappa^2 ab\right)xy + \left(-2\kappa^2ad\right)x + \left(-2bd\right)y + \kappa^2d^2 = 0
\end{align}
Unfortunately, this is not yet the right solution. If I take, for example, the plane
\begin{align}
-h \cdot x + 0 \cdot y + 2R \cdot z = Rh
\end{align}
then I get:
\begin{align}
&\kappa^2 = \frac{1}{4h^2} \\
&\Longrightarrow \hspace{10pt} \left(\frac{1}{4h^2}h^2-1\right)x^2-y^2 + \left(-2\,\frac{1}{4h^2}\,(-h)\,Rh\right)x + \frac{1}{4h^2} R^2h^2 = 0 \\
&\Longrightarrow \hspace{10pt} -\frac{3}{4}x^2-y^2 + \frac{R}{2}\,x + \frac{R^2}{4} = 0 \\
&\Longrightarrow \hspace{10pt} \frac{3}{R^2}x^2 + \frac{4}{R^2}y^2 - \frac{2}{R}\,x = 1 \\
\end{align}
And I also know where the problem is: I am supposed to look at the function in the sectional plane. So somehow I have to transform my coordinates from the original coordinate system into a coordinate system of the plane. Unfortunately, I don't know how. Maybe the task would even be easier if I had transformed my system right at the beginning. This way I would only have to consider my cone at $z=0$. My approach so far has been as follows:
My origin should be the point $(0,0,\frac{d}{c})$. My new $z$-axis is defined by the normal of the plane, so $\vec{z}_{new} = \frac{\vec{n}}{|\vec{n}|}$ with $\vec{n} = (a,b,c)$. My new $x$-axis shall continue to be perpendicular to the $y$-axis, so $\vec{x}_{new} = (\xi_1,0,\xi_2)$. My new $y$-axis is to be formed by the cross product of these two. Unfortunately, that's as far as I've got so far.
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First, you need to express the unit normal to the plane in spherical coordinates:
$n = \dfrac{1}{\sqrt{a^2 + b^2+c^2}} (a, b, c) = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) $
Comparing the expressions, we have
$ \theta = \cos^{-1} \dfrac{c}{\sqrt{a^2 + b^2+c^2} }$
$ \phi = \text{ATAN2}(a,b) $
It follows that the steepest descent direction that lies in the plane is
$ u_1 = (\cos \theta \cos \phi, \cos \theta \sin \phi, - \sin \theta ) $
We can take $u_1$ as the $x$ axis in the plane, while the $y$ axis is given by
$ u_2 = (-\sin \phi, \cos \phi , 0 ) $
Using the $u_1, u_2$ axes, any point on the plane can be expressed as
$ r = (x, y, z) = r_0 + u_1 x_1 + u_2 x_2 $
Where $r_0$ is any point on the plane, for example $(0, 0, d / c )$
Written compactly,
$r = r_0 + V u \hspace{48pt} (1)$
with $V = [u_1, u_2] $ and $ u = [x_1, x_2]^T $
Now the equation of the cone is (in terms of $r$)
$ r^T Q r = 0 \hspace{48pt} (2)$
where
$Q = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && - \tan^2 \theta_c \end{bmatrix} $
Where $\theta_c$ is the semi-vertical angle of the cone, i.e. $\tan \theta_c = \dfrac{R}{h} $
Substitute $(1)$ into $(2)$
$ (r_0 + V u)^T Q (r_0 + V u) = 0 $
Expand
$ r_0^T Q r_0 + 2 u^T V^T Q r_0 + u^T V^T Q V u = 0 $
Let's find the matrix V^T Q ,
$V^T Q = \begin{bmatrix} \cos \theta \cos \phi && \cos \theta \sin \phi && \tan^2 \theta_c \sin \theta \\ - \sin \phi && \cos \phi && 0 \end{bmatrix} $
Therefore,
$V^T Q V = \begin{bmatrix} \cos^2 \theta - \tan^2 \theta_c \sin^2 \theta && 0 \\ 0 && 1 \end{bmatrix}$
If $ \cos^2 \theta - \tan^2 \theta_c \sin^2 \theta = 0 $, then we have a parabola.
If $ \cos^2 \theta - \tan^2 \theta_c \sin^2 \theta \gt 0 $ then $V^T Q V$ is invertible, and the conic section is an ellipse.
If $ \cos^2 \theta - \tan^2 \theta_c \sin^2 \theta \lt 0 $ then $V^T Q V$ is invertible, and the conic section is an hyperbola.
Let's do the second case. We can find the center of this ellipse in $u_1u_2$ plane as follows:
$u_0 = - (V^T Q V)^{-1} V^T Q r_0 $
Explicitly evaluating $u_0$, we find
$u_0 = \begin{bmatrix} \dfrac{(-d/c) (\tan^2 \theta_c \sin \theta)}{\cos^2 \theta - \tan^2 \theta_c \sin^2 \theta} \\ 0 \end{bmatrix} $
Then by completing the square, we can write,
$(u - u_0)^T V^T Q V (u - u0) = u_0^T V^T Q V u_0 - r_0^T Q r_0 $
Now, $u_0^T V^T Q V u_0 = \dfrac{(d/c)^2 \tan^4 \theta_c \sin^2 \theta}{\cos^2 \theta - \tan^2 \theta_c \sin^2 \theta}$
and $r_0^T Q r_0 = -(d/c)^2 \tan^2 \theta_c $
Hence,
$u_0^T V^T Q V u_0 - r_0^T Q r_0 = \dfrac{(d/c)^2}{\cos^2 \theta - \tan^2 \theta_c \sin^2 \theta} ( \tan^4 \theta_c \sin^2 \theta + tan^2 \theta_c \cos^2 \theta - \tan^4 \theta_c \sin^2 \theta) $
And this simplifies to,
$ \text{Right Hand Side} = \dfrac{(d/c)^2}{\cos^2 \theta - \tan^2 \theta_c \sin^2 \theta} \cos^2 \theta \tan^2 \theta_c $
Therefore, the semi-major axis of the ellipse is along $u_1$ and given by,
$ a = (d/c) \dfrac{\cos \theta \tan \theta_c }{\cos^2 \theta - \tan^2 \theta_c \sin^2 \theta }$
And the semi-minor axis of the ellipse is along $u_2$ and given by,
$ b = (d/c) \dfrac{\cos \theta \tan \theta_c }{\sqrt{\cos^2 \theta - \tan^2 \theta_c \sin^2 \theta } } $
|
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Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now I need to factor the quadratic $y^2-12y+12$.
So, I calculated discriminant
$$D=12^2-4\times 12=96\implies \sqrt D=4\sqrt 6.$$
This means that the multipliers of quadratic are not rational. So I don't know how to proceed anymore.
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Since I saw it working, I wanted to write this method for readers as well.
We see that, the polynomial $4x^2-2xy-4x+3y-3$ contains the term of $y$, but not the term of $y^2$.
Thus, we can come to the following conclusion:
If factorization is possible, then only one multiplier polynomial contains the term of $«y»$, but the other multiplier polynomial does not.
Suppose that, the polynomial $4x^2-2xy-4x+3y-3$ is a factorable polynomial.
Let $ax+by+c=0,\thinspace b≠0$ be the multiplier polynomial, where $-\frac ab=m$ and $-\frac cb=n$.
Then there exist $m,n\in\mathbb R$ such that, if $y=mx+n$, then $4x^2-2xy-4x+3y-3=0$.
This implies,
$$\begin{align}x^2(4-2m)+x(3m-2n-4)+(3n-3)=0\end{align}$$
Hence,
$$\begin{align}&\begin{cases}4-2m=0\\ 3m-2n-4=0\\ 3n-3=0\end{cases}\\
\implies &(m,n)=(2,1).\end{align}$$
This immediately tells us, $y-(2x+1)$ or $2x-y+1$ is a factor.
This means,
$$\begin{align}\color{red}{4x^2}-2xy-4x+3y\color{blue}{-3}\\
=(\color{red}{2x}-y+\color{blue}{1})(\color{red}{2x}\color{blue}{-3}).\end{align}$$
|
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|
When is the sum of the products of any three distinct numbers less than $n$, divisible by $n$? Define $A$ as the set of all integer triples $(a,b,c)$, such that $0<a<b<c<n$.
$$A=\left\{(a,b,c)\in \mathbb{Z^3}:1\le a<b<c\le n-1 \right\}$$
Define $S$ as the sum of the product $abc$ for each triplet in $A$.
$$S=\displaystyle \sum_{(a,b,c)\in A}a\cdot b\cdot c$$
For what values of $n$ is $S$ divisible by $n$? Find all possible remainders when $S$ is divided by $n$.
The smallest possible value of $n$ is $4$. If $n=4$, we have, $A=\{(1,2,3)\}$ and $S=6.$ Obviously, $4\nmid 6$.
Claim: $S\equiv 0\pmod n$ for all odd $n$.
Proof: If $(a,b,c)\in A$, then $(n-c,n-b,n-a)\in A$. Since,
$$abc+(n-c)(n-b)(n-a)\equiv abc+(-c)(-b)(-a)\equiv 0 \pmod n$$
and $(a,b,c)\not\equiv (n-c,n-b,n-a)$. Therefore, $n$ divides $S$.
If $n$ is even, the same proof does not work because $(a,b,c)$ and $(n-c,n-b,n-a)$ might be identical.
For example, if we set $a=1,b=\frac{n}{2}$ and $c=n-1$,we have, $(a,b,c)\equiv (n-c,n-b,n-a)$.
Also, I checked upto $n=100$ using a simple program:
L=[]
for n in range (4,101):
S=0
for a in range (1,n-2):
for b in range (a+1,n-1):
for c in range(b+1,n):
S=S+a*b*c
print('n:',n,' ','Sum:',S)
if S%n==0:
print(True)
else:
print(False)
L.append(n)
print()
print(L)
This is the output for the first few numbers
n: 4 Sum: 6
False
n: 5 Sum: 50
True
n: 6 Sum: 225
False
n: 7 Sum: 735
True
n: 8 Sum: 1960
True
n: 9 Sum: 4536
True
n: 10 Sum: 9450
True
n: 11 Sum: 18150
True
n: 12 Sum: 32670
False
n: 13 Sum: 55770
True
n: 14 Sum: 91091
False
This is the list of numbers $n $ for which $S$ is not divisible by $n$.
[4, 6, 12, 14, 20, 22, 28, 30, 36, 38, 44, 46, 52, 54, 60, 62, 68, 70, 76, 78, 84, 86, 92, 94, 100]
Why is $S\equiv 0\pmod n\;, \;\forall n\not\equiv 4 \;\text{or}\; 6 \pmod 8$? Also, when $n$ is even and does not divide $S$, is the below claim true ? $$S\equiv \frac{n}{2} \pmod n$$
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Set
$$A_{3,n} := \{(a,b,c) \in \mathbb{Z}^3: 1 \leq a < b<c \leq n-1 \}$$
$$S_{3,n} := \sum_{(a,b,c)\in A_{3,n}}abc$$
Note that $$6S_{3,n} = \sum_{1 \leq a,b,c \leq n-1, a\neq b , b \neq c, a \neq c}abc$$
$$= (\sum_{j=1}^{n-1}j)^3 - 3(\sum_{j=1}^{n-1}j^2)(\sum_{j=1}^{n-1}j)+2\sum_{j=1}^{n-1}j^3$$
$$=\frac{(n-1)^3n^3}{8}-3(\frac{n(n-1)}{2})(\frac{n(n-1)(2n-1)}{6})+\frac{n^2(n-1)^2}{2}$$
$$=\frac{1}{8}(n^2)(n-1)^2(n-2)(n-3)$$
Hence
$$S_{3,n} = \frac{1}{48}(n^2)(n-1)^2(n-2)(n-3)$$
thus $S_{3,n}$ is divisible by $n$ precisely when $\frac{1}{48}(n)(n-1)^2(n-2)(n-3)$ is an integer, which is precisely when
$$1) \text{ }\text{ }\text{ }(n)(n-1)^2(n-2)(n-3) \equiv 0 \mod 16$$
$$2) \text{ }\text{ }\text{ }(n)(n-1)^2(n-2)(n-3) \equiv 0 \mod 3$$
$2)$ always holds as $n \equiv 0,1\text{ or }2\text{ }\mod(3)$. $1)$ holds when $n \equiv 0,1,2,3, 5, 7 \mod 8$. Someone with more stamina can fill the rest of the details.
|
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How to find the exact value of the integral $ \int_{0}^{\infty} \frac{d x}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}}$? $\textrm{I first reduce the power two to one by Integration by Parts.}$
$\begin{aligned}\displaystyle \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x &=\int_{0}^{\infty} \frac{x^{6}}{\left(x^{6}+1\right)^{2}} d x\\&=\displaystyle -\frac{1}{6} \int_{0}^{\infty} x d\left(\frac{1}{x^{6}+1}\right)\\&
=\displaystyle -\left[\frac{x}{6\left(x^{6}+1\right)}\right]_{0}^{\infty}+\frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x \quad \textrm{ (Via Integration by Parts})\\&=\displaystyle \frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x\end{aligned}$
$\textrm{Then I am planning to evaluate }\displaystyle I= \int_{0}^{\infty} \frac{1}{x^{6}+1}\text{ by resolving }\frac{1}{x^{6}+1} \text{ into partial fractions.}$
But after noticing that $$I=\int_{0}^{\infty} \frac{d x}{x^{6}+1}\stackrel{x\mapsto\frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{4}}{x^{6}+1} d x,$$
I changed my mind and started with $3I$ instead of $I$ as below:
$$
\begin{aligned}
3 I &=\int_{0}^{\infty} \frac{x^{4}+2}{\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)} d x \\
&=\int_{0}^{\infty}\left(\frac{1}{x^{2}+1}+\frac{1}{x^{4}-x^{2}+1}\right) d x \\
&=\left[\tan ^{-1} x\right]_{0}^{\infty}+\int_{0}^{\infty} \frac{\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x \\
&=\frac{\pi}{2}+\frac{1}{2} \int_{0}^{\infty} \frac{\left(1+\frac{1}{x^{2}}\right)-\left(1-\frac{1}{x^{2}}\right)}{x^{2}+\frac{1}{x^{2}}-1} d x\\
&=\frac{\pi}{2}+\frac{1}{2}\left[\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1}-\int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^{2}-3}\right] \\
&=\frac{\pi}{2}+\frac{1}{2}\left[\tan ^{-1}\left(x-\frac{1}{x}\right)\right]_{0}^{\infty}-0 \\
&=\frac{\pi}{2}+\frac{1}{2}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right] \\
&=\pi \\ \therefore I &=\frac{\pi}{3}
\end{aligned}
$$
Now I can conclude that
$$\boxed{\displaystyle \quad \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x=\frac{\pi}{18} }.$$
:|D Wish you enjoy the solution! Opinions and alternative methods are welcome.
|
It is a nice solution for sure.
As you showed the problem is to compute
$$I=\int \frac {dx}{x^6+1}$$
My working
Writing
$$x^6+1=(x^2+1)(x^4-x^2+1)=(x^2+1)(x^2-a)(x^2-b)$$ Using partial fraction decomposition
$$\frac 1 {(x^2+1)(x^2-a)(x^2-b)}=$$ $$\frac{1}{(a+1) (a-b) \left(x^2-a\right)}-\frac{1}{(b+1) (a-b)
\left(x^2-b\right)}+\frac{1}{(a+1) (b+1) \left(x^2+1\right)}$$ Integrating
$$I=\frac{\tan ^{-1}(x)}{(a+1) (b+1)}-\frac 1{a-b}\Bigg[\frac{\tanh ^{-1}\left(\frac{x}{\sqrt{a}}\right)}{(a+1)\sqrt{a} }-\frac{\tanh
^{-1}\left(\frac{x}{\sqrt{b}}\right)}{(b+1)\sqrt{b} } \Bigg]$$ Using the bounds, we then have
$$J=\int_0^\infty \frac {dx}{x^6+1}=\frac{\pi }{2 (a+1) (b+1)}+\frac \pi{2(a-b)}\Bigg[\frac{\sqrt{-\frac{1}{a}}}{a+1}-\frac{\sqrt{-\frac{1}{b}}}{b+1}\Bigg]$$
$$a=\frac{1+i\sqrt{3}}{2}= \sqrt{-\frac{1}{a}}\qquad \text{and} \qquad
b=\frac{1-i\sqrt{3}}{2}= \sqrt{-\frac{1}{b}} $$ So, finally
$$J=\frac{\pi }{1+(a+b)+a b}=\frac \pi 3$$
|
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|
Finding a pair of commutative stochastic $3 \times 3$ matrices Could you please give me an example of a pair $\left(A,B\right)$ of commutative stochastic $3 \times 3$ matrices with real entries given $A \neq B$ and excluding the identity matrix, symmetric matrices and doubly stochastic matrices?
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What it we take $B$ a convex combination of powers of $A$? In generic cases, if two matrices commute $A$, $B$ commute one $B$ will be a polynomial in the other $A$ ( if the eigenvalues of $A$ are distinct).
$\bf{Added:}$ Here is a stochastic matrix $A$ with eigenvalues $1$, $\frac{1}{6}$, $\frac{1}{6}$
$$A=\left( \begin{matrix} \frac{11}{18} & \frac{1}{3} & \frac{1}{18} \\ \frac{4}{9} & \frac{1}{2} & \frac{1}{18} \\ \frac{4}{9} & \frac{1}{3} & \frac{2}{9} \end{matrix} \right)$$
$\bf{Added:}$ Some calculations show that the $3\times 3$ stochastic matrices with eigenvalues $1$, $\lambda$, $\lambda$ are of the form
$$A=\left( \begin{matrix} a & b & 1-a-b \\ p & q & 1-p-q \\ u & v & w=1-v-w\end{matrix} \right)$$
where $$u = p+ \frac{(t-1)^2}{4} (a+b-p-q)\\
v = b-\frac{(t+1)^2}{4}(a+b-p-q)\\
w = (1-b-p) + t\,(a+b-p-q)$$
Positivity of entrier is also required. To check the result, see WA. This follows from the factorization of the discriminant of the characteristic polynomial of a stochastic $3\times 3$ matrix
$$\Delta= (1 - a - b p - q + a q + u + b u - q u + v - a v + p v)^2\\ \cdot (a^2 + 4 b p - 2 a q + q^2 - 2 a u - 4 b u + 2 q u + u^2 + 2 a v - 4 p v - 2 q v + 2 u v + v^2)$$
The first factor in the square is $\ge 0$ for stochastic matrices ( it being $0$ is the condition that the matrix has two eigenvalues $1$).
The second factor equals:
$$(a-q-u+v)^2 + (b+p-u-v)^2 - (b-p+u-v)^2$$
It is the solutions to this expression being $0$ tnat are parametrized by the above (we may be missing some solutions corresponding to $t=\infty$).
|
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How to prove $n{n+9 \choose 8} \bmod 10 = 5 \Rightarrow n \bmod 16 \in \{1,3,5,15\}$ In reading this question I noticed that the values of $n$ there for which $s(n)$ there has the last digit equal to $5$ are a subset of OEIS A103127. Combining that observation with another comment in the above question, plus a little numerical test, led to the following conjectures:
$$n{n+9 \choose 8} \bmod 10 = 5 \Rightarrow n \bmod 16 \in \{1,3,5,15\}$$
and:
$$n \bmod 16 \in \{1,3,5,15\} \land n \bmod 10 \neq 9 \Rightarrow n{n+9 \choose 8} \bmod 10 = 5$$
Any idea for how to prove them?
|
Consider mod 2: $ n { n+ 9 \choose 8} \equiv 1 \pmod{2}$.
We must have both
*
*$ n \equiv 1 \pmod{2} $, and
*$ n + 9 \equiv 8, 9, 10, 11, 12, 13, 14, 15 \pmod{16} $ by Lucas theorem.
This simplifies to $ n \equiv 15, 1, 3, 5 \pmod{16}$.
This answers the first part of your question.
Consider mod 5: $ n { n + 9 \choose 8 } \equiv 0 \pmod{5}$, so either
*
*$ n \equiv 0 \pmod{5}$, or
*$ n+ 9 \not \equiv 8, 9, 13, 14, 18, 19, 23, 24 \pmod{25} $ by Lucas Theorem.
This simplifies to $ n \equiv 1, 2, 3, 6, 7, 8, 11, 12, 13, 16, 17, 18, 19, 20, 21, 22, 23 \pmod{25}$.
Hence, the full solution set could be described in mod $16\times25 = 400$, which I leave you to find the $4\times 17 = 68$ residue classes.
Note that $n = 19 $ is an example of a solution that you've rejected in your classification since $ n \equiv 9 \pmod{10}$. However, it's indeed a solution as verified by Wolfram in your numerical test.
|
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|
cubic polynomial for $L^2$ approximation.
Consider $f(x) = \sin \pi x$. Find the cubic polynomial giving $L^2$ approximation to $f$ on $[0,1]$.
I suspect I need to find a polynomial of the form $p(x) = \sum\limits_{i=0}^3 c_ix^i$ and calculate this (EDIT: below):
$$\int_0^1\left|f(x) - p(x)\right|^2 dx$$
(EDIT)Following along with this pdf.
I am using the standard monomial basis $\{x^i\}_{i=0}^n$ where $n=3$ in my case. Then I need to solve the Hilbert matrix below
$$
\left[\begin{matrix}
1 & \frac 12 & \frac 13 & \frac 14 \\
\frac 12 & \frac 13 & \frac 14 & \frac 15 \\
\frac 13 & \frac 14 & \frac 15 & \frac 16 \\
\frac 14 & \frac 15 & \frac 16 & \frac 17
\end{matrix}\right]\left[\begin{matrix}
c_0 \\ c_1 \\ c_2 \\ c_3
\end{matrix}\right] = \left[\begin{matrix}
\int_0^1f(x) dx \\
\int_0^1f(x)x dx \\
\int_0^1f(x)x^2 dx \\
\int_0^1f(x) x^3 dx
\end{matrix}\right].
$$
We then get the following augmented matrix system to solve for the $c_i$'s:
$$
\left[\begin{matrix}
1 & \frac 12 & \frac 13 & \frac 14 & \frac 2\pi\\
\frac 12 & \frac 13 & \frac 14 & \frac 15 & \frac 1\pi\\
\frac 13 & \frac 14 & \frac 15 & \frac 16 & \frac 1\pi - \frac 4{\pi^3}\\
\frac 14 & \frac 15 & \frac 16 & \frac 17 & \frac 1\pi - \frac 6{\pi^3}
\end{matrix}\right]
\sim
\left[\begin{matrix}
1 & \frac 12 & \frac 13 & \frac 14 & \frac 2\pi\\
0 & \frac 1{12} & \frac 1{12} \frac 3{40} & 0 \\
0 & 0 & \frac1{120} & \frac 9{700} & \frac{\pi^2-12}{2\pi^2} \\
0 & 0 & 0 & \frac{-1}{4200} & 0
\end{matrix}\right].
$$
Which then gives use the following where constants as
$$
\begin{cases}
c_0 = \frac 2\pi + \frac k6 \\
c_1 = -k\\
c_2 = k \\
c_3 = 0
\end{cases}
$$
where $k=\frac{\pi^2-12}{2\pi^2}120$.
Finally the cubic polynomial is
$$
p(x) = \left(\frac 2\pi + 120k\right) + (-kx) + kx^2 + 0.
$$
Is this what the question is intending?
Btw, I am not sure what mathematical tag this falls under so help with categorizing this type of problem would be appreciated too!
|
The question is quite badly worded, in my opinion. I suppose it is asking you to find the cubic polynomial that is the best approximation of $f(x)$ in the $L2$ norm on the interval $[0,1]$.
In other words, you need to find the cubic polynomial $p$ that makes the error
$$
\int_0^1(f(x)-p(x))^2\,dx
$$
as small as possible.
|
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|
Proof of identity related to sum of binomial coefficients and powers of $2$ Given two positive integers $a,b$, prove that $\left(\sum_{n=a}^{a+b-1}\binom{n-1}{a-1}2^{-n}\right)+\left(\sum_{n=b}^{a+b-1}\binom{n-1}{b-1}2^{-n}\right)=1$
The context comes from this MSE problem, where I indirectly showed that this is true with a combinatorial argument for the specific case of $a=3$ and $b=4$. This is because the first sum computes the probability of $A$ winning and the second sum computes the probability of $B$ winning. Since these are the only $2$ possible events, their sum must be $1$. This argument can be easily extended to a more general case.
While I understand the combinatorial argument, I would like to see an algebraic method. I'm not sure where to start for it.
Also, from the linked post, I (from the other answer) found that $\sum_{n=a}^{a+b-1} \binom{n-1}{a-1}2^{-n}=2^{-a-b+1}\sum_{n=a}^{a+b-1}\binom{a+b-1}{n}$. Proving this would readily prove the original identity, but I'm not sure how to prove this equation.
I'm looking for something like a generating function proof. I found that $\sum_{n=a}^{a+b-1} \binom{n-1}{a-1}2^{-n}$ is the coefficient of $x^{b-1}$ in $\frac{1}{(2-x)^a(1-x)}$ and $2^{-a-b+1}\sum_{n=a}^{a+b-1}\binom{a+b-1}{n}$ is the coefficient of $x^{b-1}$ in $\frac{2^{-a-b+1}(1+x)^{a+b-1}}{1-x}$
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The first term is
$$\sum_{n=a}^{a+b-1} {n-1\choose a-1} 2^{-n}
= 2^{-a} \sum_{n=0}^{b-1} {n+a-1\choose a-1} 2^{-n}
\\ = 2^{-a} [z^{b-1}] \frac{1}{1-z} \frac{1}{(1-z/2)^a}
= (-1)^a \mathrm{Res}_{z=0}
\frac{1}{z^b} \frac{1}{1-z} \frac{1}{(z-2)^a}.$$
Now residues sum to zero and the residue at infinity is zero by
inspection so to evaluate the residue at zero we require the residues at
one and at two. For the residue at one we get
$$- (-1)^a \times \frac{1}{(1-2)^a} = -1.$$
For the residue at two we write
$$(-1)^a \mathrm{Res}_{z=2}
\frac{1}{(2+(z-2))^b} \frac{1}{-1-(z-2)} \frac{1}{(z-2)^a}
\\ = - 2^{-b} (-1)^a \mathrm{Res}_{z=2}
\frac{1}{(1+(z-2)/2)^b} \frac{1}{1+(z-2)} \frac{1}{(z-2)^a}
\\ = - 2^{-b} (-1)^a
\sum_{n=0}^{a-1} {n+b-1\choose b-1} (-1)^n 2^{-n}
(-1)^{a-1-n}
\\ = 2^{-b}
\sum_{n=0}^{a-1} {n+b-1\choose b-1} 2^{-n}
= \sum_{n=b}^{a+b-1} {n-1\choose b-1} 2^{-n}.$$
We have shown that
$$\sum_{n=a}^{a+b-1} {n-1\choose a-1} 2^{-n}
- 1 + \sum_{n=b}^{a+b-1} {n-1\choose b-1} 2^{-n} = 0$$
which is the claim.
|
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|
What's the measure of the segment $BC$ in the rhombus below? For reference: Given a rhombus $ABCD$, on $BC$
mark the point $P$ such that : $BP= 3PC$ and $AP^2+ 3DP^2 = 38$.
Calculate $BC$.(answer: $2\sqrt2$)
My progress:
$BP = 3CP\\
AP^2+3DP^2 = 38\\
AB=BC=CD=AD$
Th. Stewart:
$\triangle ABC:\\
AC^2.BP+AB^2.CP=AP^2BC+BC.CP.BP\\
AC^2. 3CP+AB^2,CP = BC(AP^2+3CP^2)\\
\boxed{CP(3AC^2+AB^2) = BC(AP^2+3CP^2)}(I)\\
\triangle DBC:\\
CD^2.BP+BD^2CP=DP^2.BC+BC.BP.CP\\
CD^2.3CP+BD^2.CP=BC(DP^2+3CP^2)\\
\boxed{CP(3CD^2+BD^2) = BC(DP^2+3CP^2}(II)$
(I)+(II):
$\boxed{CP(3(AC^2+CD^2)+AB^2+BD^2) = BC(AP^2+DP^2+6CP^2)(III)}$
...??
|
I would have simply placed the figure on a coordinate plane such that $$\begin{align}
C &= (x,0), \\
B &= (0,y), \\
A &= (-x,0), \\
D &= (0,-y), \\
\end{align}
$$
hence $$P = (\tfrac{3}{4}x, \tfrac{1}{4}y),$$
and
$$AP^2 = \left(\tfrac{7}{4} x\right)^2 + \left(\tfrac{1}{4}y\right)^2 = \frac{49x^2 + y^2}{16}, \\
DP^2 = \left(\tfrac{3}{4}x\right)^2 + \left(\tfrac{5}{4}y\right)^2 = \frac{9x^2 + 25y^2}{16}.$$
Thus $$38 = AP^2 + 3DP^2 = \frac{76(x^2 + y^2)}{16} = \frac{19}{4}(x^2 + y^2) = \frac{19}{4}BC^2,$$ from which the result immediately follows.
|
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Determinant of a $4 \times 4$ matrix
If $$ A = \begin{pmatrix}a&b&c&d\\-b&a&-d&c\\-c&d&a&-b\\-d&-c&b&a\end{pmatrix} $$ calculate $\det(A)$.
If you calculate
$$AA^t=\begin{pmatrix}a^2+b^2+c^2+d^2&0&0&0\\0&a^2+b^2+c^2+d^2&0&0\\ 0&0&a^2+b^2+c^2+d^2&0\\0&0&0&a^2+b^2+c^2+d^2\end{pmatrix}$$
$$\det(AA^t)=(a^2+b^2+c^2+d^2)^4\Leftrightarrow (\det(A))^2=(a^2+b^2+c^2+d^2)^4$$
The answer is $\det(A)=(a^2+b^2+c^2+d^2)^2$ and not $\det(A)=-(a^2+b^2+c^2+d^2)^2$. Short of calculating it by hand, why is it not negative?
|
Because the coefficient of $a^4$ (which is the product of the main diagonal) is $+1$ not $-1$.
|
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|
Factoring by grouping: how to deal when the factors extracted are of the form $(x+k)$ and $ (x-k) $. EDITED : the original post contained a mistake regarding the factorization of the numerator.
The source of the exercice I'm trying to do is : Barton's College Practice Placement Test, Q. 17 https://www.barton.edu/pdf/math/practice-math-placement-test.pdf
Question : Simplify $\frac {4x^2 -1} { 2x^2 +5x -3}$ .
I first tried to factorize the denominator :
By grouping I get : $2x^2 +5x -3 = (2x^2+2x)+ (3x-3) = 2x (x+1) +3(x-1)$.
My question is : is there a classic way to deal with that kind of situation? I mean a way to recover an $(x+1)$ instead of an $(x-1)$ in the second term ( or reciprocally)?
I know how to get $-( b-a)$ from $(a-b) $ but that wouldn't help here.
|
One way is to try to get the factors of the numerator into the denominator:
$$4x^2 -1 = (2x-1)(2x+1)$$
and
$$2x^2 +5x -3 = (2x^2 -x) + (6x-3) = x(2x-1) + 3(2x-1) = (2x-1)(x+3).$$
Another possibility (if you do not wish to study the discriminant formula) is to complete the square:
\begin{align}
2x^2 + 5x -3 &= 2\left(x^2 + 2\cdot \frac{5}{4}x - \frac{3}{2}\right)\\
&= 2\left(x^2 + 2\cdot \frac{5}{4}x + \frac{25}{16} - \frac{25}{16}- \frac{3}{2}\right)\\
&=2\left(\left(x + \frac{5}{4}\right)^2 - \frac{49}{16}\right)\\
&=2\left(x+ \frac54 + \frac74\right)\left(x + \frac54 - \frac74\right)\\
&=2\left(x+3\right)\left(x-\frac12\right)
\end{align}
|
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Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=xyz$. Prove the following $xyz\geq27,xy+yz+zx\geq27,x+y+z\geq9$ Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=xyz$.
Prove $$xyz\geq27,\\xy+yz+zx\geq27,\\x+y+z\geq9.$$
Here's what I tried:
$$\begin{align} \frac{x^3+y^3+z^3}{3}&\ge\sqrt[3]{x^3y^3z^3}\\
x^3+y^3+z^3&\ge3\sqrt[3]{x^3y^3z^3}\\
x^3+y^3+z^3&\ge3xyz \\ x^3+y^3+z^3&\ge 3(x^2+y^2+z^2)\end{align} $$
Not sure where to go from here, I think we have to use the arithmetic and geometrical means and their relation. ($A\ge G$)
|
(1)
$xyz=x^2+y^2+z^2 \ge 3\sqrt[3]{x^2y^2z^2}$
So $\sqrt[3]{xyz}\ge3$
$xyz\ge27$
(2)
$xy+yz+zx\ge3\sqrt[3]{x^2y^2z^2}\ge3\sqrt[3]{27^2}=27$
(3)
$x^2+y^2\ge2xy$
$x^2+z^2\ge2xz$
$y^2+z^2\ge2yz$
add and divided by 2
$\Rightarrow x^2+y^2+z^2\ge xy+yz+zx$
$\Rightarrow (x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz\ge3(xy+yz+zx)\ge3*27$
use (2)
So $x+y+z\ge9$
|
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Proving that $x^2-2xy+6y^2-12x+2y+41\ge 0$ where $x,y \in\Bbb{R}$ We use the result that $AX^2+BX+C \ge 0, \forall X ~ \in \Bbb{R} $ if $A>0$ and $B^2-4AC \le 0$
for proving that $f(x,y)=x^2-2xy+6y^2-12x+2y+41\ge 0$, where $x,y \in\Bbb{ R}.$
Let us re-write the quadratic of $x$ and $y$ as a quadratic of $x$ as
$$f(x,y)=x^2+x(-2y-12)+6y^2+2y+41\ge 0, \forall x \in \Bbb{R}.$$
$$\implies B^2-4AC=(2y+12)^2-4(6y^2+2y+41)=-20(y-1)^2 \le 0,$$
which is true and the equality holds if $y=1$ this further means that $f(x)=(x-7)^2.$
So $f(x,y) \ge 0$, the equality holds if $x=7$ and $y=1$.
The question is: What could be other ways of proving this.
Edit: it will be interesting to note that this quadratic of $x$ and $y$ would represent just a point that is $(7,1)$. It is more clear by the solution of @Aqua given below.
|
In the original variables, your polynomial is $(x-y-6)^2 + 5(y-1)^2 $ and so is non-negative, equal to zero only when $y=1$ and then $x=7,$ so $(7,1)$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
- 1 & 1 & 0 \\
- 6 & - 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & - 1 & - 6 \\
0 & 1 & - 1 \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
1 & - 1 & - 6 \\
- 1 & 6 & 1 \\
- 6 & 1 & 41 \\
\end{array}
\right)
$$
|
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|
System of non-linear equations (ISNMO, 1991) Let $x,y,z$ be real numbers. Solve the following system of equations: $$\begin{cases}
\frac{3(x^2+1)}x=\frac{4(y^2+1)}y=\frac{5(z^2+1)}z; \\
xy+xz+yz=1.
\end{cases}$$
I tried to solve this system using the following method:
Since $$\frac{3(x^2+1)}x=\frac{4(y^2+1)}y; \Rightarrow 3y(x^2+1)=4x(y^2+1).$$ Now, from the second row, we have $$3y(x^2+xy+xz+yz)=4x(y^2+xy+xz+yz); \Rightarrow (x+y)(xy+4xz-3yz)=0.$$
Thus, $$x=-y; z=\frac {xy}{3y-4x}.$$
But if we replace $y$ with $-x$, we obtain a complex solution. The same happens when we replace $z$ with $\frac {xy}{3y-4x}$ in the second row because it takes us back to the first expression $x=-y$.
Finally, in case we try to deal with $$\frac{3(x^2+1)}x=\frac{5(z^2+1)}z, z=\frac {xy}{3y-4x},$$
then it complicates the problem with huge expressions.
Wolfram Alpha shows that the real solutions are $(x;y;z)=(\frac 13;\frac 12;1)\cup(-\frac 13;-\frac 12;-1)$, but I do not know how to get to them. I would appreciate any help.
Thank you in advance.
|
Expanding on OP's work and my comments:
From the first double equality it is obvious that $x,y,z\ne 0$, and $x, y$ and $z$ have the same sign (since the numerators are positive). So one also has $x+y,y+z$, and $z+x\ne 0$
Then as in OP's idea, using the given:
$$xy+yz+xz=1 \tag{1}$$
One gets:
$$x^2+1=x^2+xy+yz+xz=x(x+y)+z(x+y)=(x+y)(x+z)$$
So the first double equality can be rewritten as:
$$\frac{3(x+y)(x+z)}{x}=\frac{4(y+x)(y+z)}{y}=\frac{5(z+x)(z+y)}{z}$$
From the first and middle side of the double equality above, as $x+y\ne 0$ one gets:
$$\frac{3(x+z)}{x}=\frac{4(y+z)}{y}\text{ which can be rewritten as }\frac{3z}{x}=\frac{y+4z}{y}$$
Which gives
$$xy-3yz+4xz=0\tag{2}$$
Similarly, from the middle and right side of the double inequality one gets:
$$5xy+yz-4xz=0\tag{3}$$
From $(1)$, $(2)$ and $(3)$ by denoting $u=xy, v=yz, w=zx$, one gets the linear system:
$$\left\{
\begin{alignat}{4}
u&+&v&+&w&=1\\
u&-&3v&+&4w&=0\\
5u&+&v&-&4w&=0
\end{alignat}
\right.$$
By solving with one's favorite method, one gets $u=\frac{1}{6}, v=\frac{1}{2}, w=\frac{1}{3}$
Furthermore $(xyz)^2=uvw=\frac{1}{36}$, so $xyz=\pm\frac{1}{6}$
So $x=\frac{xyz}{v}=\pm\frac{1}{3}, y=\frac{xyz}{w}=\pm\frac{1}{2}, z=\frac{xyz}{u}=\pm 1$
As $x,y$ and $z$ have the same sign, one gets the solutions $(x,y,z)=\left(\frac{1}{3},\frac{1}{2},1\right)$ and $\left(-\frac{1}{3},-\frac{1}{2},-1\right)$
|
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|
Solve the equation $12x^5+16x^4-17x^3-19x^2+5x+3=0$ Solve the equation $$12x^5+16x^4-17x^3-19x^2+5x+3=0$$ The divisors of $3$ are $\pm1;\pm3$ and the divisors of $12$ are $\pm1;\pm2;\pm3;\pm4;\pm6;\pm12$, so the possible rational roots are $$\pm1;\pm\dfrac12;\pm\dfrac13;\pm\dfrac14;\pm\dfrac16;\pm\dfrac{1}{12};\pm3;\pm\dfrac32;\pm\dfrac34$$ which makes $22$ possible roots. We can use Horner, but how can I reduce the number of possible roots?
|
We first try the simplest $x= \pm 1$, then $12+16-17-19+5+3=0 \Rightarrow x-1 $is a factor and $-12+16+17-19-5+3=0 \Rightarrow x+1 $ is also a factor.
By synthetic division, we have
$$
\begin{aligned}12 x^{5}+16 x^{4}-17 x^{3}-19 x^{2}+5 x+3 =(x+1)(x-1)\left(12 x^{3}+16 x^{2}-5 x-3\right)
\end{aligned}
$$
Now we can try $x=\dfrac{1}{2},
12\left(\frac{1}{2}\right)^{3}+16\left(\frac{1}{2}\right)^{2}-5\left(\frac{1}{2}\right)-3=0 \Rightarrow x-\frac{1}{2}$ is also a factor.
By synthetic division again,
$$
\begin{aligned}
& 12 x^{5}+16 x^{4}-17 x^{3}-19 x^{2}+5 x+3\\
=&(x+1)(x-1)(2 x-1)\left(6 x^{2}+11 x+3\right) \\
=&(x+1)(x-1)(2 x-1)(3 x+1)(2 x+3)
\end{aligned}
$$
Therefore the roots of the equation are $\pm 1,\dfrac{1}{2} ,-\dfrac{1}{3} \text{ and }-\dfrac{3}{2}.$
|
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|
How can i evaluate this integral with or without using CAS $\int_0^{\infty}\frac{\mathrm dx}{(x^2+1)\cosh(\pi x)}$ $$\int_0^{\infty}\frac{\mathrm dx}{(x^2+1)\cosh(\pi x)}$$
Firstly i used substitution $\pi x=t; \mathrm dx=\frac{\mathrm dt}{\pi}$
$$\pi \int_0^{\infty}\frac{\mathrm dt}{(t^2+\pi^2)\cosh t}$$
writing $\cosh t =\frac{e^{t}+e^{-t}}{2}=\frac{1+e^{-2t}}{2e^{-t}}$
$$2\pi\int_0^{\infty}\frac{e^{-t}}{(t^2+\pi^2)(1+e^{-2t})}$$
substituting $e^{-t}=v;-e^{-t}\mathrm dt=\mathrm dv$
$$2\pi\int_0^{1}\frac{\mathrm dv}{(\ln^2 v+\pi^2)(1+v^2)}$$
|
Similar to the question already mentioned by @Laxmi Narayan Bhandari, considering the infinite series
$$\text{sech}(z)=\pi\sum_{k=0}^\infty (-1)^k\frac{ (2 k+1)}{\pi ^2 \left(k+\frac{1}{2}\right)^2+z^2}$$
Making the problem more general
$$I=\int_0^\infty \frac{\text{sech}(a x)}{b x^2+1}\,dx \quad \text{with} \quad a>0 \quad \text{and} \quad b>0$$ we have
$$\frac{\text{sech}(a x)}{b x^2+1}=4 \pi\sum_{k=0}^\infty(-1)^k\frac{ (2 k+1)}{\left(b x^2+1\right) \left(4 a^2 x^2+\pi ^2 (2
k+1)^2\right)}$$ Using partial fraction decomposition
$$\frac{ 1}{\left(b x^2+1\right) \left(4 a^2 x^2+\pi ^2 (2
k+1)^2\right)}=$$ $$\frac{b}{\left(b x^2+1\right) \left(\pi ^2 b (2 k+1)^2-4 a^2\right)}-\frac{4 a^2}{\left(\pi ^2 b (2 k+1)^2-4 a^2\right) \left(4 a^2 x^2+\pi ^2 (2
k+1)^2\right)}$$
$$\int_0^\infty\frac{ dx}{\left(b x^2+1\right) \left(4 a^2 x^2+\pi ^2 (2
k+1)^2\right)}=$$
$$\frac{\pi \sqrt{b}}{2 \left(\pi ^2 b (2 k+1)^2-4 a^2\right)}-\frac{a}{(2 k+1) \left(\pi ^2 b (2 k+1)^2-4 a^2\right)}$$
After summations,
$$\color{red}{I=\int_0^\infty \frac{\text{sech}(a x)}{b x^2+1}\,dx=\frac{\psi \left(\frac{a}{2 \pi\sqrt{b} }+\frac{3}{4}\right)-\psi
\left(\frac{a}{2 \pi\sqrt{b} }+\frac{1}{4}\right)}{2 \sqrt{b}}}$$
If, as in your case, $a=\pi$ and $b=1$, the result is
$$\int_0^\infty \frac{\text{sech}(\pi x)}{ x^2+1}\,dx=2-\frac \pi 2$$ and, if $a=\pi$ and $b=4$ as in the linked question
$$\int_0^\infty \frac{\text{sech}(\pi x)}{ 4x^2+1}\,dx=\frac{\log (2)}{2}$$
For $a=\pi$, these are the only cases where the result simplifies.
Edit
Update
After @Gary comment, let $z=\frac{a}{2 \pi \sqrt{b}}$ which makes
$$I \sqrt{b}=\frac 12 \Bigg[ {\psi \left(z+\frac{3}{4}\right)-\psi
\left(z +\frac{1}{4}\right)}\Bigg]\sim\sum_{k=0}^\infty \frac {E_{2 k} } {(4z)^{2k+1} }$$
Using this expansion to build $P_{m,m+1}$ Padé approximants,we obtain very good approximations of $I \sqrt{b}$. For example
$$P_{3,4}=\frac {52 z+64 z^3 } {9+224 z^2+256 z^4 }$$ shows an absolute relative error of $0.0015$% for $z=2$ and
$$\Phi=\int_2^\infty \Big[I \sqrt{b}-P_{3,4}\Big]^2\,dz=5.01\times 10^{-13}$$
|
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|
Finding a root of a degree 5 polynomial
There are positive integers $m$ and $n$ such that $m^2 −n = 32$ and $$\sqrt[5]{m+\sqrt{n}} + \sqrt[5]{m-\sqrt{n}}$$ is a real root of the
polynomial $$x^5 −10x^3 + 20x −40$$ Find $m+n$.
Okay I write the other $4$ roots as $x_{1}, x_{2}, x_{3}, x_{4}$ and get $\sqrt[5]{m+\sqrt{n}} + \sqrt[5]{m-\sqrt{n}} = - (x_{1} + x_{2} + x_{3} + x_{4})$. After doing stuff with Vieta's, I get
$(x_{1} + x_{2} + x_{3} + x_{4})x_{1} x_{2} x_{3} x_{4} = -40$
$\frac{1}{x_{1}} +\frac{1}{x_{2}}+ \frac{1}{x_{3}}+ \frac{1}{x_{4}}- \frac{1}{x_{1} + x_{2} + x_{3} + x_{4}} = \frac{1}{2}$
$\sum_{1\le{i}<j\le{4}}{x_{i}x_{j}} - (x_{1} + x_{2} + x_{3} + x_{4})^2 =-10$
$(x_{1} + x_{2} + x_{3} + x_{4})(\sum_{1\le{i}<j\le{4}}{x_{i}x_{j}}) + x_{1} x_{2} x_{3} x_{4}(\frac{1}{x_{1}} +\frac{1}{x_{2}}+ \frac{1}{x_{3}}+ \frac{1}{x_{4}}) = 0$.
Okay and letting $\sqrt[5]{m+\sqrt{n}}$ be $a$ and $\sqrt[5]{m-\sqrt{n}}$ be $b$,
$a^5 + b^5 = 2m$
$ab =2$, from $m^2-n=32$
and ultimately I write $a^5 + b^5 = (a+b)((a+b)^4 -6(a+b)^2 +4)$. I guess I can just get $a+b$ from those Vieta's equations up there... but it's a bit tedious maybe and I think I might be missing something that makes this whole question easier. Can someone help please? Thanks.
|
Your second idea is better $$(a+b)^5 = a^5+b^5 + 5ab(a^3+b^3)+10a^2b^2(a+b)$$
So $$x^5 = 2m + 10(a+b)\Big((a+b)^2-6\Big)+40(a+b)$$ or $$x^5+2m+10x^3-60x+40x$$
so $$x^5-10x^3+20x -2m =0$$ and thus $m=20$ so ...
|
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|
What are the minimum and maximum of $f(x) = x^6 - 5x^4 + 5x^2 - 1$? I am stuck in obtaining the critical points:
$f'(x) = 6x^5 - 20x^3 + 10x$
$f''(x) = 30x^4 - 60x^2 + 10$
For critical points, we have: $f'(x) = 0$
$2x(3x^4 - 10x^2 + 5) = 0$
Now, how should i solve them to get the critical points. And what should the minima and maxima of the given function be at the end?
|
$2x(3x^4 - 10x^2 + 5) = 0$
$x_1=0$
$3x^4 - 10x^2 + 5 = 0$
Let $y=x^2$ then $3y^2 - 10y + 5 = 0$ and $y_{1,2}=\frac{5\pm\sqrt{10}}{3}$
and
$x_2=-\sqrt{\frac{5-\sqrt{10}}{3}}$, $x_3=\sqrt{\frac{5-\sqrt{10}}{3}}$, $x_4=-\sqrt{\frac{5+\sqrt{10}}{3}}$, $x_5=\sqrt{\frac{5+\sqrt{10}}{3}}$
So the maximum is $\infty$
Function is minimum when $x=-\sqrt{\frac{5+\sqrt{10}}{3}}$ or $x= \sqrt{\frac{5+\sqrt{10}}{3}}$
|
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|
What is $ \lim_{x \rightarrow 0}\left(\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}\right) $? The answer of the following limit:
$$
\lim_{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)
$$
is 1 by Wolfram Alpha.
But I tried to find it and I got $2/3$ :
My approach :
$1)$
$
\ln(\cos x)=\ln\left(1-\frac{x^{2}}{2}+o\left(x^{3}\right)\right)=-\frac{x^{2}}{2}+o\left(x^{3}\right)
$
$2)$
$
\sin ^{2}(x)=\left(x-\frac{x^{3}}{3!}+o\left(x^{3}\right)\right)^{2}=x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)
$
$3)$
$\begin{aligned} \frac{1}{-\frac{x^{2}}{2}+o\left(x^{3}\right)}+\frac{2}{x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)}=\frac{-x^{2}+x^{2}-\frac{x^{4}}{3}+o\left(x^{3}\right)}{-\frac{x^{4}}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)} \end{aligned}$
$4)$
$\lim _{x \rightarrow 0} \frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}}{-\frac{1}{2}}=\frac{2}{3}$
So where is the mistake in my approach?
Note: $o$ denotes the little-o notation
Edit : I've understood where's my mistake is, but another question popped up reading the answers which is : does $o(1/x)$ tends to zero as x tends to zero?
|
When you got$$\require{cancel}\frac{\cancel{-x^{2}}+\cancel{x^{2}}-\frac{x^{4}}{3}+o\left(x^{3}\right)}{-\frac{x^{4}}{2}+o\left(x^{5}\right)},$$you divided both the numerator and the denominator by $x^4$, getting$$\frac{-\frac13+o(x^3)}{-\frac12+o(x^5)}.$$This is not correct, because $\frac{o(x^3)}{x^4}$ is not $o(x^3)$.
Note that$$\log(\cos x)=-\frac{x^2}2-\frac{x^4}{12}+O(x^6)$$and that therefore$$\frac1{\log(\cos x)}=-\frac{2}{x^2}+\frac{1}{3}+\frac{x^2}{30}+O(x^4).\tag1$$Also, since$$\sin^2(x)=x^2-\frac{x^4}3+O(x^6),$$you have$$\frac2{\sin^2(x)}=\frac{2}{x^2}+\frac{2}{3}+\frac{2x^2}{15}+O(x^4).\tag2$$And it follows from $(1)$ and $(2)$ that$$\lim_{x\to0}\left(\frac1{\log(\cos x)}+\frac2{\sin^2(x)}\right)=1.$$
|
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|
What's the approach for this inequality question? The question:
$a,b,c > 0; ab+bc+ca =3$, Prove that
$\sum_{cyc}\frac{a}{\sqrt{a^{3}+5}}\leq\frac{\sqrt{6}}{2}$
The sum is cyclic over $a,b,c$
I've looked at the problem for a long time but still can't think of an approach for this, so how can I solve this?
|
We have
\begin{align*}
\sum_{\mathrm{cyc}} \frac{a}{\sqrt{a^3 + 5}}
&= \sum_{\mathrm{cyc}} \frac{a}{\sqrt{\frac{a^3}{2} + \frac{a^3}{2} + \frac12 + \frac92}}\\
&\le \sum_{\mathrm{cyc}} \frac{a}{\sqrt{3\sqrt[3]{\frac{a^3}{2} \cdot \frac{a^3}{2} \cdot \frac12} + \frac92}}\\
&= \sum_{\mathrm{cyc}} \frac{2a}{\sqrt{6a^2 + 18}}\\
&= \sum_{\mathrm{cyc}} \frac{2a}{\sqrt{6a^2 + 6(ab + bc + ca)}}\\
&= \frac{1}{\sqrt{6}}\sum_{\mathrm{cyc}} 2\sqrt{\frac{a}{a + b}}
\sqrt{\frac{a}{a + c}}\\
&\le \frac{1}{\sqrt{6}}\sum_{\mathrm{cyc}}
\left(\frac{a}{a + b} + \frac{a}{a + c}\right)\\
&= \frac{\sqrt{6}}{2}.
\end{align*}
We are done.
|
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|
How many ways can $720$ be decomposed into a product of two positive integers?
How many ways can $720$ be decomposed into a product of two positive integers?
My solution: There are $5 \cdot 3 \cdot 2 = 30$ ways to choose the exponents a, b, c, such that $2^a \cdot 3^b \cdot 5^c= 720$. Soon there are $30$ dividers. As there are $30$ dividers, there are $\frac{30}{2} = 15$ different products in the form $xy=720$
Question: We want $ab$ such that $a,b \in \{2^5 \cdot 3^3 \cdot 5^2\}$, that is, we want combinations of these possible values of $a$ and $b$, such that $ab=720$. How could I solve this problem following this line of reasoning? A different solution from mine (Using counting principles)
|
I believe this may be another approach:
We want $ab=720$. Since we have $2^5 \cdot 3^3 \cdot 5^2$ and we want this to form the product of 2 primes such that it results in $720$, then:
$$\binom{5}{2} \cdot \binom{3}{2} \cdot \binom{2}{2}= 30.$$ But since we ended up counting the possibilities twice (for example: if $a=2$, then $b=360$; if $b=2$, then $a=360$), then we have $\frac{30}{2}=15$
|
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|
Solve $(y+1)dx+(x+1)dy=0$ Solve $(y+1)dx+(x+1)dy=0$
$$\frac{dy}{y+1}=-\frac{dx}{x+1}$$
then we get $\ln|y+1|=-\ln|x+1|+c$
$$\ln(|(y+1)(x+1)|)=c$$
$$|(y+1)(x+1)|=e^c=c_1$$
but answer is $y+1=\frac{c}{x+1}$ can you help to find where is my mistake?
|
$(y+1)dx+(x+1)dy=0$
$M(x, y) =y+1$
$N(x, y) =x+1$
Then, $\frac {\partial M}{\partial y}=1=\frac {\partial N}{\partial x}$
Hence, the differential equation is exact.
Choose, $u(x, y) $ be such that
$u_x =M $ and $u_y =N$
Then, $u(x, y) =\int {M {dx}}=x(y+1)+h(y) $
$u_y (x, y)=x+h'(y) $
$x+h'(y)=x+1 $
$h(y) =y$
Hence, $u(x, y) =x(y+1)+y $
Solution : $u(x, y) =C$
$\implies x(y+1)+y =C$
$\implies x(y+1)+y+1 =C+1$
$ (y+1)(x+1 )=c \space \space [c=C+1]$
Hence, $(y+1)=\frac{c}{(x+1)}$
Note:
\begin{align} \frac {d}{dx}{u(x, y(x))} &=u_x + u_y \frac {dy}{dx}\\&=M+N \frac {dy}{dx} =0 \end{align}
Hence,$ u(x, y) =C$
|
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|
Ways to find $\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\cdots$
$$\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10}+\cdots$$
is equal to?
My approach:
We can see that the $n^{th}$ term is \begin{align}a_n&=\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)\cdot(2n+2)}\\&=\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)\cdot(2n+2)}\color{red}{[(2n+2)-(2n+1)}]\\&=\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)}-\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)\cdot(2n+1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)\cdot(2n+2)}\\
\end{align}
From here I just have a telescopic series to solve, which gave me $$\sum_{n=1}^{\infty}a_n=0.5$$
Another approach : note : $$\frac{(2n)!}{2^nn!}=(2n-1)!!$$
Which gives $$a_n=\frac{1}{2}\left(\frac{(2n)!\left(\frac{1}{4}\right)^n}{n!(n+1)!}\right)$$
So basically I need to compute
$$\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{(2n)!\left(\frac{1}{4}\right)^n}{n!(n+1)!}\right) \tag{*}$$
I'm not able to determine the binomial expression of $(*)$ (if it exists) or else you can just provide me the value of the sum
Any hints will be appreciated, and you can provide different approaches to the problem too
|
If you look at the Binomial expansion of
$$(1-x)^{-\frac{1}{2}}$$ you get :-
$$\sum_{r=0}^{\infty}\frac{\binom{2r}{r}x^{r}}{4^{r}}$$
So $$\int_{0}^{1}(1-x)^{-\frac{1}{2}}dx=\sum_{r=0}^{\infty}\frac{\binom{2r}{r}}{4^{r}(r+1)}$$
So you get $$\frac{1}{2}\sum_{r=0}^{\infty}\frac{\binom{2r}{r}}{4^{r}(r+1)}=\frac{1}{2}\sum_{r=0}^{\infty}\frac{(2r)!}{4^{r}r!(r+1)!}=\frac{1}{2}\int_{0}^{1}(1-x)^{-\frac{1}{2}}dx=1$$
So $$\frac{1}{2}\sum_{r=1}^{\infty}\frac{(2r)!}{4^{r}r!(r+1)!}=\frac{1}{2}\sum_{r=0}^{\infty}\frac{(2r)!}{4^{r}r!(r+1)!}-\frac{1}{2}(1)=1-\frac{1}{2}=\frac{1}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Why am I getting a different value for $\sin\left(2\tan^{-1}\frac{4}{3}\right)$ than my calculator? The expression:
$$\sin\left(2\tan^{-1}\left(\frac{4}{3}\right)\right)$$
Way 1:
If I punch the above expression in my calculator, I get $\frac{24}{25}$.
Way 2:
$$\sin\left(2\tan^{-1}\left(\frac{4}{3}\right)\right)$$
$$\sin\left(\tan^{-1}\left(\frac{2\times\frac{4}{3}}{1-(\frac{4}{3})^{2}}\right)\right)$$
$$[\text{Using the formula $2\arctan(x)=\arctan\left(\frac{2x}{1-x^2}\right)$}]$$
$$\sin\left(\tan^{-1}\left(\frac{-24}{7}\right)\right)$$
$$-\sin\left(\tan^{-1}\left(\frac{24}{7}\right)\right)$$
$$-\sin\left(\sin^{-1}\left(\frac{24}{25}\right)\right)$$
$$-\frac{24}{25}$$
Why am I getting a different answer than that of my calculator?
|
One has to keep in mind, which quadrant the angle lies in from the beginning, and not derive it from calculations. Your calculations will be prone to mistakes because trigonometric and inverse trigonometric functions are not one-to-one functions.
Notice that $\tan^{-1} 4/3 > \tan ^{-1} 1 = \pi/4$. So, $\theta = 2\tan^{-1} 4/3 > \pi/2$ and belongs to second quadrant. However $\tan^{-1} \text{(negative value)}$ lies in fourth quadrant. So the conversion
$$\sin\left( \underset{\text{II quadrant}}{2\tan^{-1}\left(\frac{4}{3}\right)}\right) \rightarrow \sin\left( \underset{\text{IV quadrant}} {\tan^{-1}\left(\frac{-24}{7}\right)} \right)$$
is incorrect. But it can be corrected as follows
$$\sin\left( \underset{\text{II quadrant}}{2\tan^{-1}\left(\frac{4}{3}\right)} \right) = \sin\left( \underset{\text{II quadrant}}{\pi + \tan^{-1}\left(\frac{-24}{7}\right)} \right)$$
Now calculation should give correct answer.
|
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|
Determinant formula for coordinates of circumcenter and orthocenter of a triangle I've come across the following formulas for coordinates of circumcenter $O=(x_O,y_O)$ and orthocenter $H=(x_H,y_H)$ of a triangle, in a formula book, stated without derivation. For a triangle with vertices $(x_i,y_i),$ $i \in \{1,2,3\}$,
$$x_O=\frac{\begin{vmatrix} x_1^2+y_1^2 & y_1 & 1 \\ x_2^2+y_2^2 & y_2 & 1 \\ x_3^2+y_3^2 & y_3 & 1 \end{vmatrix}}{2\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}} \quad , \quad y_O=\frac{\begin{vmatrix} x_1 & x_1^2+y_1^2 & 1 \\ x_2 & x_2^2+y_2^2 & 1 \\ x_3 & x_3^2+y_3^2 & 1 \end{vmatrix}}{2\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}} \tag{A}$$
$$x_H=\frac{\begin{vmatrix} y_1 & x_2x_3+y_1^2 & 1 \\ y_2 & x_3x_1+y_2^2 & 1 \\ y_3 & x_1x_2+y_3^2 & 1 \end{vmatrix}}{\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}} \quad , \quad y_H=\frac{\begin{vmatrix} x_1^2+y_2y_3 & x_1 & 1 \\ x_2^2+y_3y_1 & x_2 & 1 \\ x_3^2+y_1y_2 & x_3 & 1 \end{vmatrix}}{\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}} \tag{B}$$
The circumcenter case is pretty clear to me. Comparing two different equations of circumcircle
$$\begin{vmatrix} x^2+y^2 & x & y & 1 \\ x_1^2+y_1^2 & x_1 & y_1 & 1 \\ x_2^2+y_2^2 & x_2 & y_2 & 1 \\ x_3^2+y_3^2 & x_3 & y_3 & 1 \end{vmatrix} = 0 = \lambda(x^2+y^2 -2x_{O}x -2y_{O}y+c)$$
and equating the coefficients of $x,y$, one obtains $(A)$.
But I have not been able to make progress towards derivation of $(B)$. Can somebody provide a proof of it?
Please avoid an algebra-tedious proof. Thank you!
|
Each point verifies
\begin{eqnarray*}
(x_k-x_O)^2 + (y_k-y_O)^2 &=& r^2 \\
x_k^2+y_k^2+
x_O^2+y_O^2-
2(x_kx_O+y_ky_O) &=&
r^2 \\
2(x_kx_O+y_ky_O)
+ (r^2-x_O^2-y_O^2) &=& x_k^2+y_k^2
\end{eqnarray*}
This yields to the linear system
$$
\begin{pmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{pmatrix}
\begin{pmatrix}
2x_O \\
2y_O \\
a
\end{pmatrix}
=
\begin{pmatrix}
x_1^2+y_1^2 \\
x_2^2+y_2^2 \\
x_3^2+y_3^2
\end{pmatrix}
$$
It can be solved using Cramer's rule.
which yields to (A).
Note $D$ the denominator.
For the orthocenter, we use the relation
\begin{eqnarray*}
x_H-x_O
&=&
(x_1-x_O)+(x_2-x_O)+(x_3-x_O) \\
x_H
&=&
(x_1+x_2+x_3)-2 x_O \\
D x_H
&=&
(x_1+x_2+x_3)D+
\begin{vmatrix}
y_1 & x_1^2+y_1^2 & 1 \\
y_2 & x_2^2+y_2^2 & 1 \\
y_3 & x_3^2+y_3^2 & 1
\end{vmatrix} \\
D x_H
&=&
\begin{vmatrix}
y_1 & -(x_1+x_2+x_3)x_1 & 1 \\
y_2 & -(x_1+x_2+x_3)x_2 & 1 \\
y_3 & -(x_1+x_2+x_3)x_3 & 1
\end{vmatrix}
+
\begin{vmatrix}
y_1 & x_1^2+y_1^2 & 1 \\
y_2 & x_2^2+y_2^2 & 1 \\
y_3 & x_3^2+y_3^2 & 1
\end{vmatrix}
\end{eqnarray*}
Doing computations yield (B).
|
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|
how to calculate the cube root of this complex number Can anyone help me solve this exercise?
calculate the cube roots of $\frac{1}{(2-2i)}$ I started by rationalising by doing $\frac{1}{(2-2i)}$= $\frac{1}{(2-2i)}$ * $\frac{(2+2i)}{(2+2i)}$ --->
$\frac{(2(1+i)}{8}$ ----> $\frac{(1+i)}{4}$
then how can I continue?
thank you all in advance
|
suppose $z=\dfrac{1+i}{4}=\dfrac{1}{2\sqrt{2}}\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$. Let $z_1=r(\cos\alpha +\sin\alpha)$ be a root of $z_1^3=z$. Then,
$$r^3(\cos3\alpha+i\sin3\alpha)=\dfrac{1}{2\sqrt{2}}\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$$
Comparing moduli we get $r^3=\dfrac{1}{2\sqrt{2}}$, hence $r=\dfrac{1}{\sqrt{2}}$. Comparing the stuff inside brackets we get
$$3\alpha=\dfrac{\pi}{4}+2k\pi$$ where $k\in \mathbb{N}$. We get 3 unique values for $\alpha$ which are - $\pi/12$, $3\pi/4$ and $17\pi/12$. So, you have your answer for the three roots by substituting $\alpha$ in $z_1$.
|
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|
Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$ Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$.
So far,
\begin{align*}
x^2 + y^2 + 121 + 2xy + 22x + 22y &= x^2 + y^2 + 121\\
2xy + 22x + 22y &= 0\\
(2x+22)y &= -22x\\
(x+11)y &= -11x
\end{align*}
At least 1 of $x,y$ must be a multiple of 11?
Dont know where to progress after this. All help appreciated.
|
Hint for another way:Very useful the "Simon's trick" cited above by @John Omielan to solve this kind of diophantine equations.We want to give another kind of solution.
From $11=\dfrac{-xy}{x+y}$ we have $-xy=11K$ and $x+y=K$ so we can put $y=11M$ from which $$-xM=K\text { and }x+11M=K\Rightarrow \begin{cases}x=\dfrac{-11M}{M+1}\\y=11M\end{cases}$$ a parametrization of the variables from which immediately the solutions corresponding to $M=0$ and $M=10$ the other being less immediate but not hard to find.
|
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|
Find 2nd order recurrence relation with given solution Let $(a)_{n\geq0}$ with $a_n=a_{n-1}+6a_{n-2}$ and $a_0=1, a_1=4$.
The sequence $(b)_{n\geq0}$ shall satisfy $$b_n=\sum_{k=0}^{n}\binom{n}{k}a_k$$
I have to find a 2nd order recurrence relation which has $(b_n)_{n\geq0}$ as solution. How does one approach this task?
I first solved the recurrence realtion for $(a_n)$:
$$a_n=\frac{6\cdot3^n+(-1)^{n+1}2^n}{5}$$
But how do I proceed?
|
For every $n \geq 0$, one has
\begin{align*} b_n&=\sum_{k=0}^{n}\binom{n}{k}a_k\\
&=\sum_{k=0}^{n}\binom{n}{k}\frac{6\cdot3^k+(-1)^{k+1}2^k}{5}\\
&=\frac{6}{5}\sum_{k=0}^{n}\binom{n}{k}3^k - \frac{1}{5}\sum_{k=0}^{n}\binom{n}{k}(-2)^k\\
&=\frac{6}{5}(3+1)^n - \frac{1}{5}(1-2)^n\\
&=\frac{6 \cdot 4^n - (-1)^n}{5}\\
\end{align*}
which satisfies the recursion
$$\boxed{b_{n+2} = 3b_{n+1} + 4b_n}$$
|
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|
Conditions for positivity of this symmetric real matrix I have the following real symmetric matrix $M$ of size 3:
\begin{align}
M =
\begin{pmatrix}
a & b & c \\
b & d & e \\
c & e & f
\end{pmatrix},
\end{align}
for real parameters $a$, $b$, $c$, $d$, $e$, $f$. What are the necessary and sufficient conditions on these parameters for $M$ to be a positive matrix ($M \geq 0$)?
|
$D_k : k$-th order principal minor of $A$
$A_{n×n}$ is positive matrix if $D_k\ge 0\space \forall k=1, 2,...,n$
$D_1 : det[a]_{1×1}= a\ge 0$
$D_2 : det \begin{align}
\begin{pmatrix}
a & b \\
b & d \\
\end{pmatrix}
\end{align}=ad-b^2\ge 0$
$\begin{align} D_3&= det
\begin{pmatrix}
a & b & c \\
b & d & e \\
c & e & f\\
\end{pmatrix} \\ &= a d f-a e^2+b^2 (-f)+2 b c e-c^2 d \ge 0\end{align}$
|
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|
Let $a,b\in \mathbb{R}$ such that $ab=2$. Find the max value of $\frac{3}{2\left(a+b\right)^2}$ and $a, b$ where max is attained, without calculus. My thinking:
By Arithmetic and Geometric mean inequality (AGM):
$ab\le \left(\frac{a+b}{2}\right)^2$
We know $ab=2$
$\rightarrow$ $2\le \left(\frac{a+b}{2}\right)^2$
$\rightarrow$ $2\le \frac{a^2+b^2}{4}$
$\rightarrow$ $\frac{1}{2}\le \:\frac{1}{\frac{a^2+b^2}{4}}$
$\rightarrow$ $\frac{1}{2}\le \:\frac{4}{a^2+b^2}$
$\rightarrow$ $\frac{1}{2}\left(\frac{3}{a^2+b^2}\right)\le \:\frac{4}{a^2+b^2}\left(\frac{3}{a^2+b^2}\right)$
$\rightarrow$ $\frac{3}{2\left(a+b\right)^2}\le \frac{12}{\left(a+b\right)^4}$
Therefore the max value is $\frac{12}{\left(a+b\right)^4}$
Maximum value is attained when $a=b$:
So we can write:
$\rightarrow$ $\frac{3}{2\left(a\right)^2}=\frac{12}{\left(a\right)^4}$
$\rightarrow$ $3a^4=24a^2$
$\rightarrow$ $a=0, -2\sqrt{2},+2\sqrt{2}$
Therefore the max value of $a,b$ are $0,-2\sqrt{2},+2\sqrt{2}$
I'm not completely sure about my answer, if anyone could provide some feedback, that would be amazing! Thanks in advance.
EDIT:
$ab\le \frac{\left(a+b\right)^2}{2}\:\rightarrow \:2\le \frac{\left(a+b\right)^2}{4}\:\rightarrow \:\frac{1}{2}\le \frac{\left(a+b\right)^2}{16}\:\rightarrow \:\frac{3}{2\left(a+b\right)^2}\le \:\frac{3}{16}$
|
Why you found the maximal value of $a, b$? The question is asking $\frac3{2(a+b)^2}$!
Using AM-GM is, yes, the best approach. Here is my answer.
Since using AM-GM, $8=4ab\le(a+b)^2$, so $\frac1{(a+b)^2}\le\frac18$.
So the maxima of $\frac3{2(a+b)^2}$ is $\frac32\cdot\frac18=\frac3{16}$ when $a=b=\pm\sqrt2$.
|
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|
Compute the sum $ \sum_1^{\infty} \ln{ \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} }$ I need to prove that the sum $ \sum_{1}^{\infty} \ln \left( \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} \right) $ converges and equals $ \ln(\frac{4}{3})$. I tried expanding the fraction into four terms, which cancel each other out, but I'm left with $\ln(4)$. Here's my attempt:
$
\sum_{1}^{\infty} \ln \left( \frac{ \left( n + 1 \right) \left( 3n + 1 \right) }{n \left( 3n + 4 \right)} \right)
= \sum_{1}^{\infty} \ln{(n+1)} - \sum_{1}^{\infty} \ln{n} + \sum_{1}^{\infty} \ln{(3n+1)} -\sum_{1}^{\infty} \ln{(3n+4)} \\
= \sum_{1}^{\infty} \ln{(n+1)} - \sum_{2}^{\infty} \ln{n} + \ln(1) + \sum_2^{\infty} \ln{(3n+1)} + \ln(4) -\sum_{1}^{\infty} \ln{(3n+4)} = \ln(4)$
The last step is simply about playing correctly with the indices and the lower bounds of the series. These telescopic series almost completely cancel each other out, though I can't find the right answer.
Please tell me what I missed.
Thanks
|
Let's do it carefully:
$\ln(\sum_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$
$\ln(\prod_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$
$\ln(\prod_{n=1}^k \frac {n+1}{n}\prod_{n=1}^k\frac {3n+1}{3n+4})$.
Now $\prod_{n=1}^k\frac {n+1}n$ clearly equals $k+1$ as it's just the telescoping series $\require{cancel} \frac {\cancel 2}1\frac {\cancel 3}{\cancel 2}\frac {\cancel4}{\cancel3}.... \frac {\cancel{k-1}}{\cancel{k-2}}\frac {\cancel k}{\cancel{k-1}}\frac {k+1}{\cancel k}$.
[But note $\prod_{n=1}^\infty \frac {n+1}n = \lim_{k\to \infty}(k+1) = \infty$ diverges. As others have pointed out you can't do an $\lim M\cdot N= \lim N\cdot \lim N = \infty \cdot 0 = ?????$ split which is what you are in essence trying to do.]
$\prod_{n=1}^k \frac {3n+1}{3n+4}= \frac {3n+1}{3(n+1)+1}$ is also a telescopic series. And its product will be $\frac {4}{3(k+1)+1}$ as it is the telescopic seriers $\frac {4}{\cancel 7}\frac{\cancel 7}{\cancel {11}}......\frac {\cancel {3(k-1)+1}}{\cancel{3k+1}}\frac{\cancel{3k+1}}{3(k+1)+1}=\frac 4{3k+4}$.
[Also note $\prod_{n=1}^{\infty} \frac {3n+1}{3n+4} =\lim_{k\to \infty} \frac 4{3k+4} = 0$. So if we tried to do a $\lim \prod \frac {(n+1)(3n+1)}{n(3n+4)} = \lim (n+1)\frac 4{3k+4} = \lim (n+1)\cdot \lim \frac 4{3k+4}=\infty \cdot 0 = ?????$ it will not work. That is just illegal.]
So $\ln(\prod_{n=1}^k \frac {n+1}{n}\prod_{n=1}^k\frac {3n+1}{3n+4})=\ln((k+1)\frac 4{3k+4})=\ln \frac{4(k+1)}{3(k+1)+1}=\ln \frac 4{3+\frac 1{k+1}}$
[But this is acceptable. If you have an $\lim MN = \lim M \cdot \lim N = \infty \cdot 0$ paradox, it might be fixable if the product $MN$ have terms that cancel out as it does in this case. $\lim (k+1)\frac 4{3(k+1)+1}\ne \lim(k+1)\lim\frac 4{3(k+1)+1} = \infty \cdot 0=???$ is not okay but $\lim (k+1)\frac 4{3(k+1)+ 1} = \lim \cancel{(k+1)}\frac 4{3\cancel{(k+1)} + \frac 1{\color{purple}{k+1}}}=\frac 43$ is perfectly legal.]
And that's pretty much it:
$\ln(\sum_{n=1}^\infty \frac {(n+1)(3n+1)}{n(3n+4)})=$
$\ln(\lim_{k\to \infty}\sum_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$
$\lim_{k\to \infty}\ln(\sum_{n=1}^k \frac {(n+1)(3n+1)}{n(3n+4)})=$ (whoa! careful! is that actually true? [yes, it is..... so long as the sum converges... which it must else the logs of the partial sums would not converge which we know they do. ])
$\lim_{k\to \infty}\ln(\prod_{n=1}^k \frac {n+1}{n}\prod_{n=1}^k\frac {3n+1}{3n+4})=$
$\lim_{k\to \infty}\ln \frac {4}{3+\frac 1{k+1}}=$
$\ln (\lim_{k\to \infty} \frac 4{3+\frac 1{k+1}}=$
$\ln \frac 43$
|
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|
Trouble integrating square of a sine I have the following expression which I am trying to evaluate:
$$ \frac{2}{L} \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2 kx$$ where $k = \frac{\pi}{L}$.
According to my calculator, the answer should be $\frac{1}{6}$ but I can't seem to get this result by manual integration.
Here's what I've tried so far:
$$\begin{align} &\frac{2}{L} \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2 kx dx \\
&= \frac{1}{L} \int^{\frac{L}{3}}_{\frac{L}{6}} (1 - \cos 2kx) dx \\
&= \frac{1}{L} \left[ x - \frac{1}{2k} \sin2kx \right]^{\frac{L}{3}}_{\frac{L}{6}} \\
&= \frac{1}{3} + \frac{1}{L} \left[- \frac{L}{2 \pi} \sin \frac{2 \pi x}{L} \right]^{\frac{L}{3}}_{\frac{L}{6}} \\
&= \frac{1}{3} - \left[\frac{1}{2 \pi} \sin \frac{2 \pi}{x} \right]^{3}_{6} \\
&= \frac{1}{3} + \frac{\sqrt{3}}{2 \pi} \neq \frac{1}{6} \end{align}$$
|
There are two errors: First,
$$
\frac{1}{L} \bigl[ x \bigr]^{\frac{L}{3}}_{\frac{L}{6}} = \frac 16
$$
and not $1/3$. Second,
$$
\left[\frac{1}{2 \pi} \sin \frac{2 \pi}{x} \right]^{3}_{6}
= \frac{1}{2 \pi} \left( \sin \frac{2\pi}{3}- \sin \frac{ \pi}{3}\right) = 0 \, .
$$
It seems that you added the terms instead of subtracting them.
With these corrections you'll get the expected result $1/6$.
|
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|
Diagonalisation of stochastic matrices Suppose that $(X_n)_{n≥0}$ is a Markov chain on a state space $I = {1, 2}$ and stochastic matrix
$$P = \begin{bmatrix} \frac{1}{4} & \frac{3}{4} \\ \frac{1}{3} & \frac{2}{3} \end{bmatrix}$$
(a) Find the eigenvalues
$$|P-\lambda I| = \begin{vmatrix} \frac{1}{4}-\lambda & \frac{3}{4} \\ \frac{1}{3} & \frac{2}{3}-\lambda \end{vmatrix} = \lambda^2 - \frac{11}{12}\lambda -\frac{1}{12}$$
Then $\lambda_1 = 1, \lambda_2 = -\frac{1}{12}$
(b) You know $P$ and $P^0 = I_2$. Use this to find $a(i,j), b(i,j)$ so that you have an explicit form for $P^n$
We know by diagonalizing $P$, every entry of $P^n$ can be written as $(P^n)_{i,j}=a(i,j)\lambda_1^n+b(i,j)\lambda_2^n$
Edit:
$P^n = \begin{bmatrix} \frac{1}{4} & \frac{3}{4} \\ \frac{1}{3} & \frac{2}{3} \end{bmatrix}^n = \begin{bmatrix} 1 & -9 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -\frac{1}{12} \end{bmatrix}^n \begin{bmatrix} \frac{4}{13} & \frac{9}{13} \\ -\frac{1}{13} & \frac{1}{13} \end{bmatrix}$
Edit (2):
$\begin{bmatrix} 1 & -9 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -\frac{1}{12} \end{bmatrix}^n \begin{bmatrix} \frac{4}{13} & \frac{9}{13} \\ -\frac{1}{13} & \frac{1}{13} \end{bmatrix} = \begin{bmatrix} 1 & -9 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}^n \begin{bmatrix} 0 & 0 \\ 0 & -\frac{1}{12} \end{bmatrix}^n\begin{bmatrix} \frac{4}{13} & \frac{9}{13} \\ -\frac{1}{13} & \frac{1}{13} \end{bmatrix}$?
|
You're almost there. We have
\begin{align}
P^n &=
\begin{bmatrix}
\frac{1}{4} & \frac{3}{4} \\ \frac{1}{3} & \frac{2}{3}
\end{bmatrix}^n =
\begin{bmatrix}
1 & -9 \\ 1 & 4
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\ 0 & -\frac{1}{12}
\end{bmatrix}^n
\begin{bmatrix}
\frac{4}{13} & \frac{9}{13} \\ -\frac{1}{13} & \frac{1}{13}
\end{bmatrix}
\\ & =
\frac 1{13}
\begin{bmatrix}
1 & -9 \\ 1 & 4
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\ 0 & \left(-\frac{1}{12}\right)^n
\end{bmatrix}
\begin{bmatrix}
4 & 9 \\ -1 & 1
\end{bmatrix}
\\ & =
\frac 1{13}
\begin{bmatrix}
1 & -9 \\ 1 & 4
\end{bmatrix}
\begin{bmatrix}
4 & 9 \\ -\left(-\frac{1}{12}\right)^n & \left(-\frac{1}{12}\right)^n
\end{bmatrix}
\\ & =
\frac 1{13}
\begin{bmatrix}
4 + 9\left(-\frac{1}{12}\right)^n & 9 - 9\left(-\frac{1}{12}\right)^n\\
4 - 4\left(-\frac{1}{12}\right)^n & 9 + 4\left(-\frac{1}{12}\right)^n
\end{bmatrix}.
\end{align}
|
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|
Two partial fraction approaches, one is wrong, the other is right, why? I want to do a partial fraction on
\begin{equation}
\frac{z}{(z-4)(z+\frac{1}{2})}
\end{equation}
Method one, which apparently is wrong:
\begin{equation}
\frac{z}{(z-4)(z+\frac{1}{2})}=\frac{A}{z-4}+\frac{B}{z+\frac{1}{2}}
\end{equation}
\begin{array}
f A(z+\frac{1}{2})+B(z-4)=Az+\frac{1}{2}A+Bz-4B ;\\
Az+Bz=z \rightarrow A=1-B\\
\frac{1}{2}A-4B=0 \rightarrow A-8B=0 \\
A=\frac{8}{9}, B=\frac{1}{9}
\end{array}
which gives
\begin{equation}
\frac{8}{9}\frac{1}{z-4}+\frac{1}{9}\frac{1}{z+\frac{1}{2}}
\end{equation}
Then , the second method which is a reference from an exam correction, which is deemed right, disregards for the existence of z in the numerator at first:
\begin{equation}
\frac{1}{(z-4)(z+\frac{1}{2})}=\frac{A}{z-4}+\frac{B}{z+\frac{1}{2}}
\end{equation}
\begin{array}
f A(z+\frac{1}{2})+B(z-4)=Az+\frac{1}{2}A+Bz-4B ;\\
Az+Bz=0 \rightarrow A=-B\\
\frac{1}{2}A-4B=1 \rightarrow B=-1/4 + A/8 \\
A=\frac{2}{9}, B=-\frac{2}{9}
\end{array}
Then at the end, the author inserts z back in the numerator, obtaining:
\begin{equation}
\frac{2}{9}\frac{z}{z-4}-\frac{2}{9}\frac{z}{z+\frac{1}{2}}
\end{equation}
So why was the former wrong, when it follows all the rules of partial fractions? And why is the second right, when z is not included in the partial fraction decomposition?
Thanks
|
Both methods are algebraically correct. That is, both methods yield new functions that are the same as the original function.
Your method is the correct way to carry out the partial fraction decomposition algorithm. Moreover, your method results in the outcome desired from the partial fraction algorithm—a sum of rational functions, each of which has a denominator with a single distinct irreducible factor, and a numerator whose degree is less than the degree of the irreducible factor in its denominator. The professor's method yields a result without this last property.
(The reason we want this last property, in a calculus context at least, is that there are standard and often immediate methods for integrating that type of rational function, whereas a function like $\frac z{z-4}$ still requires further algebraic preprocessing to be integrated.)
|
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|
Mixed Poisson distribution with Lindley mixing I want to calculate the unconditional distribution of $x$
$$p(x|p)
= \int_0^\infty p(x,\theta|p) d\theta = \int_0^\infty p(x|\theta)p(\theta|p) d\theta$$
where $x|\theta \sim Poisson(\lambda)$ and $\lambda$ is a random variable that follows the Lindley distribution $$p(\theta|p) = \frac{p^2}{p+1}(\theta+1)e^{(-\theta p)}$$
The proven result is $$p(x|p) = \frac{p^{2}(p+2+x)}{(p+1)^{(x+3)}}$$
Doing so I have :
\begin{align*}
p(x|\lambda)
&= \int_0^\infty p(x,\theta|\lambda) d\theta\\
&= \int_0^\infty p(x|\theta)p(\theta|\lambda) d\theta\\
&= \int_0^\infty \frac{e^{-\theta}\theta^x}{x!} \frac{p^2}{p+1}(\theta+1)e^{(-\theta p)}d\theta\\
&= \frac{p^2}{(p+1)x!}\int_0^\infty e^{-\theta(1+p)}\theta^{x}(\theta+1) d\theta\\
\end{align*}
So far so good. But I split the integral (am I allowed to do that?) into :
$$
\frac{p^2}{(p+1)x!}\int_0^\infty e^{-\theta(1+p)}\theta^{x}(\theta+1) d\theta
= \frac{p^2}{(p+1)x!}\int_0^\infty e^{-\theta(1+p)}\theta^{x+1} d\theta +\int_0^\infty e^{-\theta(1+p)}\theta^{x} d\theta $$
Edit:
Given that :
\begin{align*}
\int_0^\infty \theta^x e^{-\theta (p+1)} d\theta
&=\frac{\Gamma(x+1)}{(p + 1)^{(x+1)}} \int_0^\infty \frac{(p + 1)^{(x+1)}}{\Gamma(x+1)}\theta^x e^{-\theta (p+1)} d\theta\\
&=\frac{\Gamma(x+1)}{(p + 1)^{(x+1)}} \int_0^\infty \text{Gamma}(\theta; x+1, p+1) d\theta\\
&=\frac{\Gamma(x+1)}{(p + 1)^{(x+1)}}
\end{align*}
then becomes : $$\frac{p^2}{(p+1)\Gamma(x+1)} \left(\frac{\Gamma(x+1)}{(p+1)^{x+1}} + \frac{\Gamma(x)}{(p+1)^x} \right)$$
Am I right ?
|
Assuming your work up to the last line (before edit) is correct we have with a little algebraic manipulation
\begin{align}
f_X(x|p)
&=\frac{p^2}{(p+1)x!}\int_0^\infty e^{-\theta(1+p)}\theta^{x}(\theta+1)\,\mathrm d\theta\\
&=\frac{p^2}{(p+1)^{x+2}}\int_0^\infty (1+\theta)\frac{(p+1)^{x+1}}{\Gamma(x+1)}\theta^{(x+1)-1}e^{-\theta(p+1)}\,\mathrm d\theta,
\end{align}
which can be written as the expected value
$$
f_X(x|p)=\frac{p^2}{(p+1)^{x+2}}\mathsf E(1+X),\quad X\sim\operatorname{Gamma}(x+1,p+1).
$$
By linearity of the expected value
$$
\mathsf E(1+X)=1+\mathsf EX=1+\frac{x+1}{p+1};
$$
hence,
$$
f_X(x|p)=\frac{p^2}{(p+1)^{x+2}}\left(1+\frac{x+1}{p+1}\right).
$$
Note:
If you need to show the steps in evaluating $\mathsf EX$ then write
$$
\mathsf EX=\int_0^\infty \theta\frac{(p+1)^{x+1}}{\Gamma(x+1)}\theta^{(x+1)-1}e^{-\theta(p+1)}\,\mathrm d\theta.
$$
Substituting $u=(p+1)\theta$ and making use of the integral definition of the gamma function will get you the final result.
Edit:
As requested by the OP:
\begin{align}
f_X(x|p)
&=\frac{p^2}{(p+1)x!}\int_0^\infty e^{-\theta(1+p)}\theta^{x}(\theta+1)\,\mathrm d\theta\\
&=\frac{p^2}{(p+1)}\int_0^\infty(1+\theta) \frac{1}{\Gamma(x+1)}e^{-\theta(1+p)}\theta^{x}\,\mathrm d\theta\\
&=\frac{p^2}{(p+1)}\int_0^\infty(1+\theta) \frac{1}{\Gamma(x+1)}\theta^{(x+1)-1}e^{-(p+1)\theta}\,\mathrm d\theta\\
&=\frac{p^2(p+1)^{-(x+1)}}{(p+1)}\int_0^\infty(1+\theta) \frac{(p+1)^{x+1}}{\Gamma(x+1)}\theta^{(x+1)-1}e^{-(p+1)\theta}\,\mathrm d\theta\\
&=\frac{p^2}{(p+1)^{x+2}}\int_0^\infty(1+\theta) \frac{(p+1)^{x+1}}{\Gamma(x+1)}\theta^{(x+1)-1}e^{-(p+1)\theta}\,\mathrm d\theta\\
&=\frac{p^2}{(p+1)^{x+2}}\int_0^\infty(1+\theta) \operatorname{Gamma}(\theta|x+1,p+1)\,\mathrm d\theta\\
\end{align}
|
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|
How to find exact value of integral $\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}}dx$? When I first encountered the integral $\displaystyle \int_{0}^{\infty} \frac{1}{x^{4}-x^{2}+1} d x$, I am very reluctant to solve it by partial fractions and search for any easier methods. Then I learnt a very useful trick to evaluate the integral.
Noting that
$$I(1):= \int_{0}^{\infty} \frac{d x}{x^{4}-x^{2}+1} \stackrel{x \mapsto \frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{2}}{x^{4}-x^{2}+1}
$$
Combining them yields
\begin{aligned}
I(1)&=\frac{1}{2} \int_{0}^{\infty} \frac{x^{2}+1}{x^{4}-x^{2}+1} d x\\&= \frac{1}{2}\int_{0}^{\infty} \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x \\
&= \frac{1}{2}\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1} \\
&= \frac{1}{2}\tan ^{-1}\left(x-\frac{1}{x}\right)_{0}^{\infty} \\
&= \frac{\pi}{2}
\end{aligned}
Later, I started to investigate the integrands with higher powers.
Similarly,
$$
I(2):= \int_{0}^{\infty} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{2}} \stackrel{x \mapsto \frac{1}{x}}{=}\int_{0}^{\infty} \frac{x^{6}}{\left(x^{4}-x^{2}+1\right)^{2}} d x
$$
By division, we decomposed $x^6$ and obtain $$
\frac{x^{6}}{\left(x^{4}-x^{2}+1\right)^{2}}=\frac{x^{2}+1}{x^{4}-x^{2}+1}-\frac{1}{\left(x^{4}-x^{2}+1\right)^{2}}
$$$$
I(2)=\int_{0}^{\infty} \frac{x^{2}+1}{x^{4}-x^{2}+1} d x-\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{2}}dx
$$
We can now conclude that $$I(2)=I(1)=\frac{\pi}{2} $$
My Question:
How about the integral $$\displaystyle I_{n}=\int_{0}^{\infty} \frac{d x}{\left(x^{4}-x^{2}+1\right)^{n}}$$ for any integer $n\geq 3$?
|
With CAS and help of MellinTransfrom:
$$\int_0^{\infty } \frac{1}{\left(x^4-x^2+1\right)^n} \, dx=\\\mathcal{M}_s^{-1}\left[\int_0^{\infty } \mathcal{M}_a\left[\frac{1}{\left(x^4-a x^2+1\right)^n}\right](s) \,
dx\right](1)=\\\mathcal{M}_s^{-1}\left[\int_0^{\infty } \frac{(-1)^{-s} x^{-2 s} \left(1+x^4\right)^{-n+s} \Gamma (n-s) \Gamma (s)}{\Gamma (n)} \,
dx\right](1)=\\\mathcal{M}_s^{-1}\left[\frac{(-1)^{-s} \Gamma \left(\frac{1}{4}-\frac{s}{2}\right) \Gamma \left(-\frac{1}{4}+n-\frac{s}{2}\right) \Gamma (s)}{4 \Gamma
(n)}\right](1)=\\\frac{\Gamma \left(\frac{1}{4}\right) \Gamma \left(-\frac{1}{4}+n\right) \, _2F_1\left(\frac{1}{4},-\frac{1}{4}+n;\frac{1}{2};\frac{1}{4}\right)}{4 \Gamma
(n)}+\frac{\Gamma \left(\frac{3}{4}\right) \Gamma \left(\frac{1}{4}+n\right) \, _2F_1\left(\frac{3}{4},\frac{1}{4}+n;\frac{3}{2};\frac{1}{4}\right)}{4 \Gamma (n)}$$
|
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|
Find all positive integers s.t. $\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \frac{1}{d^2}$ Find all positive integers $a, b, c, d$ such that :
$$\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \frac{1}{d^2}$$
The original problem came from atomic electron transitions :
I would like to find out non-trivial positive integer solutions; since it is trivial if $a = b$ and $c = d$, or $a = c$ and $b = d$.
I found some of the solutions, and they look like :
There is some pattern in these integers, but it seems difficult to obtain a gerneralized form of the solution.
|
(Remark: The following is inspired by Equation of 1/x^2 on AoPS. However, a bit more work is needed to characterize all solutions of the given equation. It does not suffice to start with Pythagorean triples.)
Let $a, b, c, d$ be positive integers satisfying
$$ \tag{$1$}
\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \frac{1}{d^2} \, .
$$
By replacing $(a, b, c, d)$ with $(c, d, a, b)$ or $(a, c, b, d)$, if necessary, we can assume that $a \le b \le c \le d$, and actually $a < b \le c < d$ since we want to exclude the “trivial solutions.” By dividing all numbers by $\gcd(a, b, c, d)$ we can also assume that $a, b, c, d$ have no factor in common.
I'll call $(a, b, c, d)$ a “primitive solution” of $(1)$ if $a < b \le c < d$ and $\gcd(a, b, c, d) = 1$.
Now let $(a, b, c, d)$ be a primitive solution of $(1)$ and let $l = \operatorname{lcm}(a, b, c, d) $ be their least common multiple. Then
$$
(x, y, z, t) = \left( \frac la, \frac lb,\frac lc,\frac ld\right)
$$
is a quadruple of positive integers satisfying
$$ \tag{$2$}
x^2-y^2 = z^2-t^2 \, .
$$
It is also not difficult to see that $x > y \ge z > t$ and $\gcd(x, y, z, t) = 1$. I'll call that a “primitive solution” of $(2)$.
So a primitive solution of $(1)$ leads to a primitive solution of $(2)$.
But the converse is also true: If $(x, y, z, t)$ is a primitive solution of $(2)$ and $L = \operatorname{lcm}(x, y, z, t) $ their least common multiple then
$$
(a, b, c, d) = \left( \frac Lx, \frac Ly,\frac Lz,\frac Lt\right)
$$
is a primitive solution of $(1)$.
Therefore it suffices to determine all primitive solutions of $(2)$. Writing that equation in the form
$$ \tag{$*$}
x^2 + t^2 = y^2 + z^2
$$
shows that this task amounts to find positive integers which can be represented in two (or more) ways as the sum of two squares. The smallest integer with this property is $N=50$, compare https://oeis.org/A007692.
Example: $50 = 1^2+7^2 = 5^2+5^2$, so $(x, y, z, t) = (7, 5, 5, 1)$ is a primitive solution of $(2)$ with least common multiple $L=35$. This gives the solution
$$
\frac{1}{5^2} - \frac{1}{7^2} = \frac{1}{7^2} - \frac{1}{35^2} \, .
$$
So a recipe to compute all primitive solutions of $(1)$ could look like this:
*
*Enumerate the positive integers $N$ which can be represented as the sum of two squares. The Sum of two squares theorem can help to find these numbers efficiently.
*For each such number $N$, determine all pairs $(u_i, v_i)$ of non-zero integers such that $u_i^2+v_i^2 = N$.
*If there are at least two such pairs, try all combinations $(u_i, u_j, v_j, v_i)$ with $i \ne j$.
*If $u_i > u_j \ge v_j > v_i$ and $\gcd(u_i, u_j, v_j, v_i) = 1$ then
$$
(a, b, c, d) = \left( \frac L{u_i}, \frac L{u_j},\frac L{v_j},\frac L{u_i}\right)
$$
with $L = \operatorname{lcm}(u_i, u_j, v_j, v_i)$ is a primitive solution of $(1)$.
|
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|
$\sum \frac{a}{b+c+d}\le \frac{2\sum a^2}{\sum ab}$ if $\sum a =4$ Let $a, b, c, d$ positive real numbers such that $a+b+c+d=4$. Prove that
$$\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}\le
\frac{2(a^2+b^2+c^2+d^2)}{ab+ac+ad+bc+bd+cd}.$$
My idea is to cancel the denominators, expand the inequality and use Muirhead, but the calculations are terrible... So I'm looking for a smarter solution.
|
Fill in any missing gaps.
*
*Multiplying by $(ab+ac+ad+bc+bd+cd)$, show that the inequality is equivalent to $$\sum \frac{ abc+abd+acd}{b+c+d} \leq \sum a^2. $$
*Show that $$ \frac{ abc+abd+acd}{b+c+d} \leq \frac{ab+ac+ad}{3} \leq \frac{3a^2+b^2+c^2+d^2}{6}$$
*Hence, summing up the cyclic inequalities, the conclusion follows.
Note: This could be combined into 1 (non-obvious) step by showing that
$$ \frac{a}{b+c+d} \leq \frac{ 3a^2 + b^2+c^2+d^2} { 3(ab+ac+ad+bc+bd+cd)}.$$
|
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|
Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that $x^2\cdot f(x)+f(1-x)=2x-x^4$ $\forall\; x\in \mathbb R$
Find all functions $f:\mathbb R \rightarrow \mathbb R$ such that: $\;x^2\cdot f(x)+f(1-x)=2x-x^4,\;\forall\; x\in \mathbb R.$
My solution:
Replace $x$ by $(1-x)$ and by eliminating $f(1-x)$.
I obtained $f(x)=1-x^2$ as provided in my book.
My doubt:
How to check that there is no other function satisfying Above property?
Also, I think $f(x)=0,\;\forall\;x\in \mathbb R$ can be solution too.
Is my thinking correct?
|
Your solution is incomplete, because the solution is not required to be continuous. Your system of equations
$$
\begin{pmatrix}
x^2 & 1 \\
1 & (1-x)^2
\end{pmatrix}
\begin{pmatrix}
f(x) \\
f(1-x)
\end{pmatrix}
=
\begin{pmatrix}
2x-x^4 \\
2(1-x) - (1-x)^4
\end{pmatrix}
$$
has infinitely many solutions if $x=\varphi =\frac{1+\sqrt{5}}{2}$ or
$x=1-\varphi =\frac{1-\sqrt{5}}{2}$. In this case, the matrix is not invertible, and we get
$$
\begin{pmatrix}
f(\varphi) \\
f(1-\varphi)
\end{pmatrix}
=\begin{pmatrix}
-\varphi \\
\varphi-1
\end{pmatrix}
+\lambda
\begin{pmatrix}
-1 \\
1+\varphi
\end{pmatrix}
$$
So for each $\lambda \in \mathbb{R}$, the following is a valid solution:
$$
f(x)=
\begin{cases}
1-x^2 & \text{ for } x\in\mathbb{R}\setminus\{\varphi, 1-\varphi\} \\
-\varphi -\lambda & \text{ for } x=\varphi \\
\varphi-1 +\lambda(1+\varphi) & \text{ for } x=1-\varphi
\end{cases}
$$
|
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|
Find prime number $p$ such that $19p+1$ is a square number.
Find prime number $p$ such that $19p+1$ is a square number.
Now, I have found out, what I think is the correct answer using this method.
Square numbers can end with - $1, 4, 9, 6, 5, 0$.
So, $19p+1$ also ends with these digits.
Thus, $19p$ ends with - $0, 3, 8, 5, 4, 9$.
As $p$ is either odd or $2$. So, $19p$ is either odd or ends with $8$.
So, we can say that $19p$ ends with - $3, 8, 5, 9$.
So, $p$ ends with - $7, 2, 5, 1$.
As $19*7$ end with $3$, $19*2$ ends with $8$ etc.
Thus, possible one digit values of $p$ - $2, 7, 5$.
$$19*2+1=39$$
$$19*7+1=134$$
$$19*5+1=96$$
None of these are square numbers.
Possible two digit values of $p$ - $11, 17$.
$$19*11+1=210$$
$$19*17+1=324=18^2$$
Thus, $p=17$.
But, I am not satisfied with the solution as it is basically trial and error. Is there a better way to do this?
|
It's generally not a good idea to try to solve problems like this by looking at decimal digits. Pay more attention to properties of divisibility.
You want $19p+1 = x^2$ for a positive integer $x$, so
$$
19p = x^2 - 1 = (x+1)(x-1).
$$
Comparing the two factorizations $19p$ and $(x+1)(x-1)$ is the key point.
Since $x^2 = 19p + 1 > 19$ the numbers $x+1$ and $x-1$ are both greater than $1$, so by uniqueness of prime factorization either (i) $x+1 = 19$ and $x-1 = p$, or (ii) $x+1 = p$ and $x-1 = 19$. In the first case $x = 18$, so $p = x-1= 17$. In the second case $x = 20$ and $p = x+1 = 21$. But $21$ is not prime, so the second case doesn't actually occur. Hence $19p+1$ can be a perfect square for prime $p$ only when $p = 17$, and when $p = 17$ we do get a solution since $(19)(17)+1 = 324 = 18^2$.
|
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|
How we may express four squares whose difference is each a square in terms of (preferably solid) geometry? The problem of finding four squares whose difference is each a square is much more exhaustive as I thought. A quest up to $2^{34}$ yields nothing. The largest almost solution found in the range up to $2^{34}$ is $(w,x,y,z)=(17155833660,17162453700,17170527465,17177153175)$ and with Arty's tool we are still searching.
My Question:
Can we express this problem in terms of a solid geometry such as for example 4D-polyhedra, polychoron or 5-cell or are there connections to Coxeter Groups? I am looking for (preferably solid) geometric or geometric algebraic ways to describe the following system of diophantine equations:
\begin{array}{lllllll}
z^2-y^2&=&\square_1&\qquad\qquad&y^2-x^2&=&\square_4\\
z^2-x^2&=&\square_2&\qquad\qquad&y^2-w^2&=&\square_5\\
z^2-w^2&=&\square_3&\qquad\qquad&x^2-w^2&=&\square_6
\end{array}
What I tried so far:
I first tried my hand at the simpler version, the three squares variant. Here we are able to describe the problem of finding three squares whose difference is each a square using a cuboid having one face diagonal irrational with the following system of diophantine equations:
\begin{equation}
a^2+b^2=d_{bc}^2\qquad c^2+a^2=d_{ac}^2\qquad a^2+b^2+c^2=d_{abc}^2
\end{equation}
The notation above reuses MathWorld's variables/nomenclature that has been utilized for describing a "Perfect Cuboid". The squares of $d_{abc}$, $d_{ac}$, $c$ and of $d_{abc}$, $d_{bc}$, $b$ have their differences square and therefore provide solutions for three squares whose difference is each a square:
\begin{equation*}
{
\begin{array}{lllllllll}
d_{abc}^2&-d_{ac}^2&=&b^2&\hspace{2em}\qquad&d_{abc}^2&-d_{bc}^2&=&c^2\\
d_{abc}^2&-c^2&=&d_{bc}^2&\hspace{2em}\qquad&d_{abc}^2&-b^2&=&d_{ac}^2\\
d_{ac}^2&-c^2&=&a^2&\hspace{2em}\qquad&d_{bc}^2&-b^2&=&a^2
\end{array}}
\end{equation*}
|
Suppose you have positive integers satisfying
\begin{array}{lllllll}
z^2-y^2&=&a^2&\qquad\qquad&y^2-x^2&=&d^2\\
z^2-x^2&=&b^2&\qquad\qquad&y^2-w^2&=&e^2\\
z^2-w^2&=&c^2&\qquad\qquad&x^2-w^2&=&f^2
\end{array}
Then
\begin{eqnarray*}
b^2&=&z^2-x^2=(z^2-y^2)+(y^2-x^2)=a^2+d^2,\\
c^2&=&z^2-w^2=(z^2-y^2)+(y^2-w^2)=a^2+e^2,\\
e^2&=&y^2-w^2=(y^2-x^2)+(x^2-w^2)=d^2+f^2,
\end{eqnarray*}
and so the system above is equivalent to
\begin{array}{lllllll}
z^2&=&a^2+y^2&\qquad\qquad&y^2&=&d^2+x^2\\
b^2&=&a^2+d^2&\qquad\qquad&e^2&=&d^2+f^2\\
c^2&=&a^2+e^2&\qquad\qquad&x^2&=&f^2+w^2
\end{array}
We can reorder these equations a bit, and substitute to express them all in terms of $a$, $d$, $f$ and $w$, to get
\begin{eqnarray*}
b^2&=&a^2+d^2\\
e^2&=&d^2+f^2\\
x^2&=&f^2+w^2\\
c^2&=&a^2+d^2+f^2\\
y^2&=&d^2+f^2+w^2\\
z^2&=&a^2+d^2+f^2+w^2
\end{eqnarray*}
This can be interpreted as a $4$-dimensional cuboid with integer sides having three (particular!) integer face diagonals, two (particular!) integer solid diagonals, and integer 'long' diagonal.
|
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|
What are the possible values of the angles of triangle $ABC$?
In a triangle $ABC$, let $AP$ be the bisector of $\angle BAC$ with $P$ on the side $BC$, and let $BQ$ be the bisector of $\angle ABC$ with $Q$ on the side $CA$.
We know that $\angle BAC=60^\circ$ and that $AB + BP = AQ + QB$.
What are the possible values of the angles of triangle $ABC$?
(Answer: $\angle ABC = 80^\circ, \angle BCA=40^\circ, \angle BAC=60^\circ$)
My progress:
The relationships I found:
$AQ+QB = AB+BP\implies b+x=c+a$
$\angle C = 120^\circ -2\theta$
Considering angle bisector $BQ$ of $\triangle ABC$: $\dfrac{c}{b}=\dfrac{AC}{d}$
Similarly considering $AP$: $\dfrac{c}{a}=\dfrac{b+d}{y}$
$\angle 60^\circ +\angle B+\angle C \implies \angle B+\angle C = 120^\circ$
$x = \dfrac{(b+d)\cdot c}{AC+c}$
$x^2 =\dfrac{(AC)c}{bd}$
From sine rule:
$\displaystyle\frac{\sin60}{AC}=\frac{\sin C}{c}=\frac{\sin2\theta}{b+d}=\frac{\sqrt3}{2AC}$
From cosine rule:
$c^2 = AC^2+BC^2-2\cdot AC\cdot BC\cdot \cos\angle C$
$AC^2 = c^2+BC^2-2c\cdot BC\cdot\cos2\theta$
...???
|
Hints: As can be seen in figure there are two key points you have to show:
1- Q is on perpendicular bisector of BC.
2- Triangle BPI is isosceles.
This is only possible construction.Using bisectors theorem may help.
|
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|
Calculating arc length of $y=\frac{x^5}{10}+\frac{1}{6x^2}$, $x \in [1, 2]$ I want to calculate arc length of a function: $y=\frac{x^5}{10}+\frac{1}{6x^2}$, $x \in [1, 2]$. I calculate the derivative of $y$ and multiply it by itself to prepare for the formula:
$((\frac{x^5}{10}+\frac{1}{6x^2})')^2 = \frac{9x^{14}-12x^7+4}{36x^6}$
Now I applied that to the formula below:
$s = \int_1^2\sqrt{1+(f(x)')^2}$
But it produces an integral that I am not able to solve. Are there other ways to go about solving this or how could I possibly calculate that integral?
|
This exercise resembles a set of contrived examples commonly used in textbooks when students are studying arc length. For example:
Let $y=\dfrac{x^5}{10}+\dfrac{1}{6x^3}$ with $x^3$ rather than $x^2$ in the second term.
[In these contrived examples, the exponent in the denominator of second term must be two less than the exponent in the numerator of the first term.]
Then one gets $y^\prime=\dfrac{x^4}{2}-\dfrac{1}{2x^4}$.
So $(y^\prime)^2=\dfrac{x^8}{4}-\dfrac{1}{2}+\dfrac{1}{4x^8}$
and $1+(y^\prime)^2=\dfrac{x^8}{4}+\dfrac{1}{2}+\dfrac{1}{4x^8}=\left(\dfrac{x^4}{2}+\dfrac{1}{2x^4}\right)^2$
and one then obtains
$$ s=\int_1^2\dfrac{x^4}{2}+\dfrac{1}{2x^4}\,dx $$
So, is it possible there was a typographical error in the original exercise?
|
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|
Find the maclaurin series of 4x/(8+x^3) So I've done multiple attempts at this practice problem, but my result seems wrong. I'm new to this chapter, and I think I miss something fundamental, but I can't tell what. Can anybody see where I'm going wrong?
The practice problem
Find the maclaurin series of f(x) by using known series:
$f\left(x\right)=\frac{4x}{8+x^3}$
Known series (from my book):
$\left(1+x\right)^r\approx1+rx+\frac{r\left(r-1\right)}{2!}x^2+\frac{r\left(r-1\right)\left(r-2\right)x^3}{3!}$
My attempt at solving the practice problem:
$f\left(x\right)=\frac{4x}{8+x^3}=4x\cdot\left(8+x^3\right)^{-1}$
$8\cdot\left(1+\frac{x^3}{8}\right)^{-1}\approx8\cdot\left[1+\left(-1\right)\left(\frac{x^3}{8}\right)+\frac{\left(-1\right)\left(\left(-1\right)-1\right)}{2!}\left(\frac{x^3}{8}\right)^2+\frac{\left(-1\right)\left(\left(-1\right)-1\right)\left(\left(-1\right)-2\right)\left(\frac{x^3}{8}\right)^3}{3!}\right]$
$8\cdot\left(1+\frac{x^3}{8}\right)^{-1}\approx8-x^3+\frac{x^6}{8}-\frac{x^9}{8^2}$
$\frac{4x}{8+x^3}=4x\cdot8\cdot\left(1+\frac{x^3}{8}\right)^{-1}\approx32x-4x^4+\frac{x^7}{2}-\frac{x^{10}}{16}$
$f(x)=\frac{4x}{8+x^3}=\sum_{n=0}^{\infty}{\left(-1\right)^n\cdot2^{5-3n}\cdot x^{3n}}$
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I shall use the fact that if, on some interval $(-a,a)$ (with $a>0$) you have $f(x)=\sum_{n=0}^\infty a_nx^n$, then the Maclaurin series of $f$ is $\sum_{n=0}^\infty a_nx^n$. In this case, you have\begin{align}\frac1{8+x^3}&=\frac18\cdot\frac1{1+\left(\frac x2\right)^3}\\&=\frac18\left(1-\left(\frac x2\right)^3+\left(\frac x2\right)^6-\left(\frac x2\right)^9+\cdots\right)\end{align}if $|x|<2$. So, still assuming that $|x|<2$,$$\frac{4x}{8+x^3}=\frac12x-\frac1{2^4}x^4+\frac1{2^7}x^7-\frac1{2^{10}}x^{10}+\cdots$$
|
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|
About lineal dependence of sets of vectors in $\mathbb{R}^{4}$ Which of these sets of vectors in $\mathbb{R}^{4}$ are linearly dependent?
\begin{eqnarray}
&\text{(a)}& (1,2,1,-2), (0,-2,-2,0), (0,2,3,1),(3,0,-3,6)\\
&\text{(b)}& (4,-4,8,0),(2,2,4,0),(6,0,0,2),(6,3,-3,0)\\
&\text{(c)}& (4,4,0,0),(0,0,6,6),(-5,0,5,5)
\end{eqnarray}
For this in $(a)$ and $(b)$ I compute the determinants of the matrices and they are different of zero, then, $(a)$ and $(b)$ aren't linear dependence, for $(c)$ I reduce the matrix to an escalar matrix and I find that range is 3, so, I don't find anything set of vector with linear dependence. This is true? or Do I miss something?
|
Let $S:=\{v_{1},v_{2},\ldots,v_{n}\}\subseteq V$ with $V$ a vector space $n-$dimensional over a field $\mathbb{F}$ so $S$ is linearly independent iff $\det(v_{1},v_{2},\ldots,v_{n})\not=0$. For a) since $\det\begin{bmatrix} 1 & 0 & 0 &3 \\ 2 & -2 & 2 & 0\\ 1 & -2 & 3 & -3\\ -2 & 0 & 1 & 6\end{bmatrix}=-24\not=0$ so linearly independent. For b) since $\det\begin{bmatrix} 4 & 2 & 6 & 6\\ -4 & 2 & 0 & 3\\ 8 & 4 & 0 & -3\\ 0 & 0 & 2 & 0 \end{bmatrix}=480\not=0$ so linearly independent. For c) since $A:=\begin{bmatrix} 4 & 4 & 0 & 0\\ 0 & 0 & 6 & 6\\ 5 & 0 &-5 & 5 \end{bmatrix}\sim \cdots \sim \begin{bmatrix} 4 & 4 & 0 & 0\\ 0 & -5 & -5 & 5\\ 0 & 0 & 6 & 6 \end{bmatrix} \sim \begin{bmatrix} \color{red}{1} & 1 & 0 & 0\\ 0 & \color{red}{1} & 1 & -1\\ 0 & 0 & \color{red}{1} & 1 \end{bmatrix}$ so ${\rm rank}(A)=3$ and since $S:=\{(4,4,0,0),(0,0,6,6),(5,0,-5,5)\}$ has $\#S=3$ then $\#S={\rm rank}(A)$ and hence we have $S$ is linearly independent. Therefore the sets a), b) and c) are linearly independent. In the last part we are using the fact: since $S$ has $n$ vectors so the rank of $A$ to be $n$ in order for $S$ to be linearly independent set.
|
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How to prove $2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} = 1$
Prove $$2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} = 1$$
I got this question when I was going through some basic trigonometric identities as follows
$2(\cos\frac{\pi}{5}-\cos\frac{2\pi}{5}) =1\tag1$
very much straightforward to recognise
$2(\cos\frac{\pi}{7}-\cos\frac{2\pi}{7} + \cos\frac{3\pi}{7})=1\tag2$
Following steps proves the identity
$8\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{4\pi}{7} = -1$
$4\cos\frac{\pi}{7}[2\cos\frac{4\pi}{7}\cdot\cos\frac{2\pi}{7}] = -1$
$4\cos\frac{\pi}{7}[\cos\frac{6\pi}{7}+ \cos\frac{2\pi}{7}] =-1 $
$4\cos\frac{\pi}{7}[\cos\frac{2\pi}{7}-\cos\frac{\pi}{7}] =-1 $
$4\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7} - 4\cos^{2}\frac{\pi}{7} = -1$
Further simplification will lead to equality 2
$2(\cos\frac{\pi}{9}-\cos\frac{\pi}{9} + \cos\frac{3\pi}{9}- \cos\frac{4\pi}{9}) = 1$
by using transformation formula we can prove above one also
But when it comes to following equalities
$2(\cos\frac{\pi}{11}-\cos\frac{2\pi}{11} + \cos\frac{3\pi}{11}- \cos\frac{4\pi}{11}+ \cos\frac{5\pi}{11} )= 1$
$2(\cos\frac{\pi}{13}-\cos\frac{2\pi}{13} + \cos\frac{3\pi}{13}- \cos\frac{4\pi}{13}+ \cos\frac{5\pi}{13} -\cos\frac{6\pi}{13} )= 1$
I was able to check the results with brute force in Wolfram|Alpha for above equalities, but not able to get the steps properly even though I tried manually
My question is how to generalise the summation formula and is there any method other than trigonometric approach?
$$2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} = 1$$
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Let's use the identity $(-1)^a\cos x=\cos(a\pi-x)$. Therefore,
$$\begin{align}&2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} \\= &2\sum_{k=1}^n\cos\left((k-1)\pi-\frac{k\pi}{2n+1}\right)\\=&-2\sum_{k=1}^n\cos\frac{2nk\pi}{2n+1}\end{align}$$
Take a factor $\csc\left(\frac{n\pi}{2n+1}\right)$ and then use product to sum formula.
$$\begin{align}&-\csc\left(\frac{n\pi}{2n+1}\right)\sum_{k=1}^n2\cos\frac{2nk\pi}{2n+1}\sin\frac{n\pi}{2n+1}\\=&-\csc\frac{n\pi}{2n+1}\sum_{k=1}^n\left[\sin\left((2k+1)\frac{n\pi}{2n+1}\right)-\sin\left((2k-1)\frac{n\pi}{2n+1}\right)\right]\end{align}$$
This is a telescoping sum and we are left with, $$\csc\left(\frac{n\pi}{2n+1}\right)\left[\sin\left(\frac{n\pi}{2n+1}\right)-\sin(n\pi)\right]$$
which is clearly equal to $1$.
|
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|
Find the remainder when $f(x) = \sum_{i=0}^5 x^i$ is divided into $f(x^{12})$ I am looking for confirmation that this proof works. Feel free to be very pedantic.
We wish to find the remainder when $f(x) = \sum_{i=0}^5 x^i$ is divided into $f(x^{12})$. Notice that $f(x)(x - 1) = x^6 - 1$, so let $r$ be a root of $f$. Then $f(r)(r - 1) = 0$ which implies that $r^6 - 1 = 0$, and so $r^6 = 1$. We rewrite the polynomial $f(x^{12})$ into quotient + remainder form as $f(x^{12}) = q(x)f(x) + g(x)$, and so $f(r^{12}) = q(r)f(r) + g(r) = q(r)\cdot0 + g(r) = g(r)$. However, note that $r^{12} = (r^6)^2 = 1$, and so $f(r^{12}) = f(1)$. Now we evaluate $f(1) = \sum_{i=0}^5 1 = 6$, and so the remainder is $6$.
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You are almost there. Let us write $Q(x) := f(x^{12})$. Then indeed: $$Q(x) = f(x^{12})=f(x)q(x)+r(x),$$ where $r$ is a polynomial of degree at most $4$.
You noted that $Q(a)=6$ for all $a$ satisfying $a^6=1$, and that the roots of $f$ are $\{a; a^6=1$; $a \not = 1\}$, so that $Q(a)=6$ for each $a$ such that $f(a)=0$, or equivalently, $Q(a)=0$ for each of the roots $a$ of $f$. That is great, but you actually needed to explicitly note is that the equation $Q(a)=6$ holds for each of the $5 >4 =$deg$(r)$ roots $a$ of $f$. And thus, as $r(a)=6$ for $5 >$ deg$(r)$ values of $a$, it follows that $r$ has to be $6$ everywhere, as the only such polynomial $r$ of degree $4$ or less that satisfies $r(a)$ for $5$ or more values of $a$, is $r(x)=6 \ \forall x$ .
|
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How do you prove: $\dfrac{1}{\sin ^2 (\frac{2\pi}{9})} - \dfrac{1}{\sin ^2(\frac{4 \pi}{9})} = 8\sin(\frac{\pi}{18})$ I have to prove:
$\displaystyle \tag*{} \alpha={\dfrac{1}{\sin ^2 (\frac{2\pi}{9})} - \dfrac{1}{\sin ^2(\frac{4 \pi}{9})}}= 8\sin(\frac{\pi}{18})$
I tried to make a common denominator of $\alpha$ and use a few identities to arrive at:
$\displaystyle \tag*{} \alpha = \dfrac{\sin(\frac{6\pi}{9})}{\sin(\frac{2\pi}{9})\sin^2(\frac{4\pi}
{9})}$
I am not sure how to proceed from this. I even arrived at other 2 forms of $\alpha$, but they make even more complicated. Any hints would be greatly appreciated. Thanks.
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Let $c_n:=\cos\frac{n\pi}{9},\,s_n:=\sin\frac{n\pi}{9}$ so we want to prove$$\alpha:=\frac{s_6}{s_2s_4^2}=8c_4.$$Since $s_{9-n}=s_n$ and $s_3=\frac{\sqrt{3}}{2}$,$$\frac{\alpha}{8c_4}=\frac{\sqrt{3}}{8s_1s_2s_4}.$$Now we just need to prove$$s_1s_2s_4=\frac{\sqrt{3}}{8},$$which is a duplicate.
|
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Proving that $\prod_{n=0}^\infty (1 + x^{2^{n}}) = \frac{1}{1-x}$ I would like to verify the truth of my proof that $$\large\prod_{n=0}^\infty (1 + x^{2^{n}}) = \frac{1}{1-x}.$$
Proof. The proof is rather straightforward, just using some observations that two consecutive factors form a sum of a geometric sequence.
\begin{align*}\prod (1 + x^{2^{n}}) &= \prod (1 + x^{2^{2n}})(1+x^{2^{2n+1}}) \\
&= \prod (x^{3\cdot 2^{2n}} + x^{2\cdot 2^{2n}} + x^{1\cdot 2^{2n}} + x^{0\cdot 2^{2n}}) \\
&= \prod \frac{x^{4 \cdot 2^{2n}} - 1}{x^{2^{2n}} - 1}.
\end{align*} Next, we note that $$\large\prod_{n=0}^\infty \frac{x^{4 \cdot 2^{2n}} - 1}{x^{2^{2n}} - 1} = \frac{\prod_{n=0}^\infty x^{4 \cdot 2^{2n}} - 1}{\prod_{n=0}^\infty x^{2^{2n}} - 1} \qquad (*)$$
Step (*) worries me a little, even though I feel like it is valid. (Would I need to prove that if the limit of the LHS exists, then it's equal to the RHS?)
Then $$\Large \frac{\prod_{n=0}^\infty x^{4 \cdot 2^{2n}} - 1}{\prod_{n=0}^\infty x^{2^{2n}} - 1} = \frac{\prod_{n=0}^\infty x^{4 \cdot 2^{2n}} - 1}{(1-x)\prod_{n=0}^\infty x^{2^{2n+2}} - 1} = \frac{1}{1-x},$$ and we are supposedly done.
EDIT: For $|x| < 1$
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Hint:
You can make the step $$\prod_n\frac{a_n}{b_n}=\frac{\prod_n a_n}{\prod_n b_n}$$
if and only if the sums
$$A=\sum_n\log a_n,\qquad B=\sum_n\log b_n$$
converge absolutely. Set $a_n=x^{4\cdot2^{2n}}-1$, $b_n=x^{2^{2n}}-1$, and see for which values of $x$ we have absolute convergence of $A, B$.
Edit:
As was pointed out in the comments, we have the definition of an infinite product $\prod_na_n$:
$$\prod_n a_n=\exp\left(\sum_n \log a_n\right),$$
which is equivalent to $$\log\left(\prod_na_n\right)=\sum_n\log a_n.$$
So the product is said to converge iff the sum $\sum_n\log a_n$ converges.
If the sums $A$ and $B$ converge absolutely, then we are justified in saying that
$$A-B=\sum_n\log a_n-\sum_n\log b_n=\sum_n(\log a_n-\log b_n)=\sum_n\log\frac{a_n}{b_n}.\tag 1$$
Since $A=\log \prod_n a_n$ and $B=\log\prod_n b_n$ by definition, we see that
$$A-B=\log \left(\prod_na_n\right)-\log \left(\prod_n b_n\right)=\log\left(\frac{\prod_na_n}{\prod_nb_n}\right).$$
On the other hand, by definition,
$$\sum_n\log\frac{a_n}{b_n}=\log\left(\prod_n\frac{a_n}{b_n}\right).$$
Thus from $(1)$, assuming the products are real-valued,
$$\log\left(\frac{\prod_na_n}{\prod_nb_n}\right)=\log\left(\prod_n\frac{a_n}{b_n}\right)$$
$$\Rightarrow \frac{\prod_na_n}{\prod_nb_n}=\prod_n\frac{a_n}{b_n}.$$
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Finding the maximum of $ \sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x}$ for $0\le x \le 13$ using the Cauchy-Schwarz inequality gives two different answers I am trying to find the maximum of the expression $$\sqrt{x+27} + \sqrt{13-x} + \sqrt{x} \qquad \text{for } 0 \le x \le 13.$$
Clearly using Cauchy is the way to go here, so I tried $$((x + 27) + 2(13-x) + x)(1 + 0.5 + 1) = (x + 27 + 26 - 2x + x)\frac 52 = \frac{265}{2} \ge (\sqrt{x+27} + \sqrt{13-x} + \sqrt{x})^2.$$
But then, we also have that $$\left( 1 + \frac{1}{3} + \frac{1}{2} \right) ((x + 27) + 3(13 - x) + 2x) = 121 \ge (\sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x})^2.$$
Clearly, these are two different values. What gives? How could I “guess” the correct combination of scalars that lead to the correct answer? (as the first attempt is incorrect).
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How to "guess":
The form is making something nice, so I try to guess something to make $x+27,13-x,x$ are all squares, and indeed $x=9$, son the previous three numbers are $6^2,2^2,3^2$. So, this can have
$$(\sqrt{x+27} + \sqrt{13-x} + \sqrt{x})^2\le(6+2+3)(\frac{x+27}6+\frac{13-x}2+\frac{x}3)=121$$
So we can get $\sqrt{x+27} + \sqrt{13-x} + \sqrt{x}\le 11$, equality taken if $x=3$
The correct values will be yielded when the equality of Cauchy-Schwartz holds.
How to really calculate:
I am going to try this for Cauchy-Schwartz:
$$(\frac 1u+1+\frac 1{1-u})(u(x+27)+(13-x)+(1-u)x)\ge(\sqrt{x+27} + \sqrt{13-x} + \sqrt{x})^2$$
This holds if
$$(x+27)u^2=13-x=x(1-u)^2$$
Solving for $x$ we have
$$x=\frac{13}{1+(1-u)^2}$$
Using $1+\frac{27}{x}=\frac{x+27}{x}=\frac{(1-u)^2}{u^2}$ we have
$$1+27(\frac{1+(1-u)^2}{13})=\frac{(1-u)^2}{u^2}$$
Clearing the denominators, we have
$$27u^4-54u^3+54u^2+26u-13=0$$
Factoring (in fact, try the rational roots first) gives
$$(3u-1)(9u^3-15u^2+13u+13)=0$$
And the latter $9u^3-15u^2+13u+13$ has no positive real roots. So, you can set $u=1/3$ and yield the same result.
By the way, both Cauchy-Schwartz and the derivative will give you the following inequality:
$$\frac{1}{\sqrt x}+\frac{1}{\sqrt{27+x}}=\frac{1}{\sqrt{13-x}}$$
This is equivalent to
$$\frac{1}{x^2}+\frac{1}{{(27+x)^2}}+\frac{1}{{(13-x)^2}}=\frac{1}{x{(27+x)}}+\frac{1}{(13-x)x}+\frac{1}{(27+x)(13-x)}$$
And you can solve this equation, but I don't want to present it here.
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Prove that $x^3+y^3+z^3+(x+y+z-1)^2\ge1+3xyz$
Let $x,y,z\ge0$ satisfy $\max\left \{ x,y,z \right \}\ge 1$. Prove that $$x^3+y^3+z^3+(x+y+z-1)^2\ge1+3xyz$$
My attempts:
From the condition we can deduce $x+y+z\ge 1$
The inequality can be written as $$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)+(x+y+z)^2-2(x+y+z)\ge0$$ or $$x^2+y^2+z^2-xy-yz-zx+x+y+z-2\ge0$$
Let $x+y+z=p\ge1;xy+yz+zx=q$, the problem is: $$p^2-3q+p-2\ge0$$
I don't know how to find the relation between $p$ and $q$, because this inequality is not symmetric (The inequality hold iff $(x;y;z)=(1;0;0);(0;1;0);(0;0;1)$)
Please give me a hint in the comments, no need to give a full answer
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Remark: I rewrote the proof using substitution $p = y + z, q = yz$.
WLOG, assume that $x \ge 1$. Let $p = y + z, q = yz$. We have $p^2 \ge 4q$.
We have
\begin{align*}
&x^3 + y^3 + z^3 + (x + y + z - 1)^2 - 1 - 3xyz\\[5pt]
=\, & x^3 + p^3 - 3pq + (x + p - 1)^2 - 1 - 3xq\\[5pt]
=\,& x^3 + p^3 + (x + p - 1)^2 - 1 - 3(p + x)q \\[5pt]
\ge\,& x^3 + p^3 + (x + p - 1)^2 - 1 - 3(p + x)\cdot \frac{p^2}{4}\\[5pt]
=\,& (x + p)(x^2 - xp + p^2) + (x + p)^2 - 2(x + p) - \frac34(p + x)p^2\\[5pt]
=\,& \frac14(x + p)[4x^2 - 4xp + 4p^2 + 4(x + p) - 8 - 3p^2]\\[5pt]
=\,& \frac14(x + p)[p^2 - 4(x - 1)p + 4x^2 + 4x - 8]\\[5pt]
=\,&\frac14(x + p)[(p - 2x + 2)^2 + 12(x - 1)]\\[5pt]
\ge\,& 0.
\end{align*}
We are done.
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"timestamp": "2023-03-29T00:00:00",
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|
What´s the length of the segment AC in the triangle below? For reference: In the triangle $\angle A$ is right and $D$ is a point on the side $AC$ such that the segments $BD$ and $DC$ have length equal to $1 m$. Let $F$ be the point on the side $BC$ so that $AF$ is perpendicular to $BC$. If the segment $FC$ measures $1m$, determine the length of $AC$.(Answer$:\sqrt[3]{2})$
My progress:
$ AF = h\\
a=m+1\\
h^2 =m.FC = m.1 = m\\
\triangle AFC: AC^2 =b= h^2+FC^2 \implies b = \sqrt{m+1}\\
\triangle ABC: a^2 = b^2+ c^2 \implies (m+1)^2=c^2+(\sqrt{m+1})^2\\
\therefore c = \sqrt{m^2+m}\\
\triangle BAD: BD^2=AD^2+c^2\implies 1 = (\sqrt{m+1}-1)^2+m^2+m\\
\therefore: m^2+2m-2\sqrt{m+1}+1=0$
...????
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$\triangle ABF\sim \triangle AFC$
We have:
$$\frac {c+m}{c+AD}=\frac {c+AD}{c}$$
So:
$$(c+AD)^2=c^2+cm\Rightarrow AD=\sqrt{c^2+mc}-c$$
Therefore:
$$AC=c+AD=\sqrt{c^2+mc}$$
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"timestamp": "2023-03-29T00:00:00",
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|
Idea for recursively generating formulas for sums of powers of integers. Is it correct? I am not even a beginner mathematician, being unable to understand simple proofs others easily grasp. Still I noodle around and I think I stumbled on a fun way to determine the formula for sums of powers of integers, i.e. $f(n) = 1^p + 2^p + 3^p + \cdots +n^p$, based on knowing the formulas for the previous powers $1, \ldots, p-1$.
Let $\operatorname{Sum}(g(x))$ mean the sum from $1$ to $n$ of $g(x)$. Then, for example, one can derive the formula for $\operatorname{Sum}(x^4)$ by combining the formulas for $\operatorname{Sum}(x)$, $\operatorname{Sum}(x^2)$, $\operatorname{Sum}(x^3)$ in a special way.
How so? Consider
$$\begin{align}
\operatorname{Sum}((x+1)^{p+1} - (x^{p+1})) &= 2^{p+1} - 1^{p+1} + 3^{p+1} - 2^{p+1} + ..(n+1)^{p+1} - n^{p+1} \\
&= (n+1)^{p+1} - 1^{p+1}
\end{align}$$ because all the interior terms cancel out.
But also, using the binomial expansion,
$$\begin{align}
\operatorname{Sum}((x+1)^{p+1} - (x^{p+1})) &= \operatorname{Sum}(c_1x^p + c_2x^{p-1} + c_3x^{p-2} + \cdots + 1) \\
&= c_1\operatorname{Sum}(x^p) + c_2\operatorname{Sum}(x^{p-1}) + \cdots +\operatorname{Sum}(1) \\
&= (n+1)^{p+1} - 1
\end{align}$$
so that
$$c_1\operatorname{Sum}(x^p) = (n+1)^{p+1} - 1 - \left(c_2\operatorname{Sum}(x^{p-1}) + c_3\operatorname{Sum}(x^{p-2}) + \cdots + \operatorname{Sum}(1)\right) \tag{1}$$
Since we stipulated that we know the formulas for $\operatorname{Sum}(x^m)$ from $1, \ldots, p-1$, and since all the $c_k$ are known, as well as $(n+1)^{p+1}-1$, we can derive $\operatorname{Sum}(x^p)$ with a little algebra.
To see this in action, let's derive the formula for the sum of squares, i.e. $p=2$. From $(1)$ we have
$$\begin{align}
3\operatorname{Sum}(x^2) &= (n+1)^3 - 1 - \left(3\operatorname{Sum}(x^1) + \operatorname{Sum}(1)\right) \\
&= (n+1)^3 - 1 - 3n(n+1)/2 -n
\end{align}$$
since $\operatorname{Sum}(x) = n(n+1)/2$ and $\operatorname{Sum}(1) = n$
The above simplifies to $\operatorname{Sum}(x^2) = n(n+1)(2n+1)/6$, agreeing with the known formula.
To show this is not a fluke, let's derive the formula for $\operatorname{Sum}(x^3)$. By $(1)$ we have
$$\begin{align}
4\operatorname{Sum}(x^3) &= (n+1)^{4} - 1 - \left(6\operatorname{Sum}(x^2) + 4\operatorname{Sum}(x) + \operatorname{Sum}(1)\right) \\
&= (n+1)^4 - 1 - \left(n(n+1)(2n+1) + 2n(n+1) + n\right) \\
&= n^2(n+1)^2
\end{align}$$
Thus $\operatorname{Sum}(x^3) = n^2(n+1)^2/4$, which agrees with the known formula.
Undoubtedly this is a well known approach, and I claim absolutely zero originality.
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I've been informed by someone that this is a well-known result and not value added. I apologize to the community.
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|
Examples where $(a+b+\cdots)^2 = (a^2+b^2+\cdots)$ Consider the two infinite series
$$
\frac{\pi}{\sqrt{8}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots
$$
and
$$
\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \cdots
$$
(Notice that the first series has signs that go two-by-two rather than every-other.)
Squaring the first equality also gives $\pi^2/8$ and so these two, when put together, satisfy the 'highschooler's dream' for squaring a sum: just square each term and sum,
$$
(a + b + c + \cdots)^2 = (a^2 + b^2 + c^2 + \cdots)
$$
with nothing like $2ab + 2ac + 2bc + \cdots$ needed.
A trivial example of this would be
$$
(a + 0)^2 = a^2 + 2a0 + 0^2 = a^2 + 0^2
$$
but it only succeeds because one addend is zero.
My questions are
*
*Are there any other simple nontrivial examples? I believe any other nontrivial example must be an infinite sum. edit: John Omielan provides the simple finite example $(1+1-\frac{1}{2})^2 = 1^2 + 1^2 + \frac{1}{2^2}$.
*Is there an "obvious" demonstration that the above sum (other than the direct evaluation) satisfies the highschooler's dream? Put another way, is there a simple demonstration that the infinite sum of "cross terms" vanishes?
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one example would be, if $\omega$ is a complex cube root of unity, then
$$(1+\omega+\omega^2)^2=1+\omega^2+\omega^4$$
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|
Prove that $\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$
Prove that
$$J_n=\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$$
For $n=2$, it is OK. For general $n$, it seems impossible by integration by parts. Any other method?
When I calculate an integral $I_n=\int_0^\frac{\pi}{4}\frac{\cos nx}{\cos^nx}dx$, we find $I_n/2^{n-1}-I_{n-1}/2^{n-1}=-1/2^{n-1}J_n$. So we need to find the $J_n$, as the problem states.
The proof of $I_n/2^{n-1}-I_{n-1}/2^{n-1}=-1/2^{n-1}J_n$ is as follows.
\begin{align}
I_n/2^{n-1}-I_{n-1}/2^{n-1}
&=\frac{1}{2^{n-1}} \int_0^\frac{\pi}{4}\left(\frac{\cos nx}{\cos^nx}-\frac{2\cos(n-1)x}{\cos^{n-1}x}\right)dx\\
&=\frac{1}{2^{n-1}}\int_0^\frac{\pi}{4}\frac{\cos[(n-1)x+x]-2\cos(n-1)x\cos x}{\cos^nx}dx\\
&=-\frac{1}{2^{n-1}}\int_0^\frac{\pi}{4}\frac{\cos(n-2)x}{\cos^nx}dx
=-1/2^{n-1}J_n
\end{align}
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You can probably use the following formula :
$\cos((n-2)x)=\sum_{j=0}^{[(n-2)/2]}(-1)^j\binom{n-2}{2j}\cos^{(n-2)-2j}(x)\sin^{2j}(x)$, which can be obtained by using Newton binom with $e^{i(n-2)x}=(\cos x+i\sin x)^{n-2}$.
You take the sum out of the integral and use adapted changes of variables.
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