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Finding $\lim x_n$ when $\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$
Let $x_n$ be the unique solution of the equation $$\left( 1+\frac{1}{n}\right)^{n+x_n}=1+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{n!}$$
Find $\lim_{n \to \infty} x_n$
I think that the limit must be $\frac{1}{2}$, because $\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}$ is decreasing and convergent to $e$, while $1+\frac{1}{1!}+\dots+\frac{1}{n!}$ is increasing and convergent to $e$, so $$\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}>1+\frac{1}{1!}+\dots+\frac{1}{n!}$$
which means that $\frac{1}{2}>x_n$. I also know that for $a \in [0,\frac{1}{2}), \left(1+\frac{1}{n}\right)^{n+a}$ is eventually increasing, but I don't know how to get a lower bound for $x_n$ which goes to $\frac{1}{2}$
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The definition of $x_n$ can be directly used to evaluate the limit. To simplify typing let $s_n=\sum_{i=0}^{n}1/i!$. Then we have $$x_n=\frac{\log s_n-n\log(1+n^{-1})}{\log(1+n^{-1})}$$ Since $n\log(1+(1/n))\to 1$ the desired limit is equal to the limit of $$n\log s_n-n^2\log(1+(1/n))$$ and now we need to use Taylor series for $\log(1+x)$ to get $$\lim_{n\to\infty} n^2\log(1+(1/n))-n=-\frac{1}{2}$$ and thus our original limit is equal to the limit of $$\frac{1}{2}+n\log s_n-n$$ The next step needs some gymnastics with inequalities. We have $$e-s_n=\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\dots$$ which is less than $$\frac{1}{n!}\left(\frac{1}{n+1}+\frac{1}{(n+1)^2}+\dots\right)=\frac{1}{n!}\frac{1/(n+1)}{1-1/(n+1)}=\frac{1}{n\cdot n!} $$ and therefore we have $$1<\frac{e}{s_n}<1+\frac{1}{ns_n\cdot n!} $$ and applying logs we can see that $$0<n-n\log s_n<n\log\left(1+\frac{1}{ns_n\cdot n!} \right) \leq \frac{1}{s_n\cdot n!} $$ Since $s_n\to e$ by Squeeze Theorem we can see that $n-n\log s_n\to 0$ and therefore the desired limit is $1/2$.
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|
Proving that $\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 } $
Let $x,y,z\geq 1$ and $x+y+z=6$. Then $$\frac {1}{3x^2+1}+\frac {1}{3y^2+1}+\frac {1}{3z^2+1}\geq \frac {3}{16 }. $$
I tried to use Cauchy- Schwartz inequality but it doesn't work.
|
By Jensen's inequality with $\phi\left(t\right)=\frac{1}{3t^2+1}$ we have
$$\phi\left(\frac13\sum_{i=1}^{3}x_i\right)\le \frac13\sum_{i=1}^{3}\phi\left(x_i\right) $$
since $x+y+z= 6$ we get
$$\frac{1}{13}=\phi\left(2\right)\le \frac13\left(\frac{1}{3x^2+1}+\frac{1}{3y^2+1}+\frac{1}{3z^2+1}\right) $$
hence $$\left(\frac{1}{3x^2+1}+\frac{1}{3y^2+1}+\frac{1}{3z^2+1}\right)\ge \frac{3}{13}\ge \frac{3}{16} $$
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Limit of Gamma functions I need to prove:
$$\lim_{p \to\infty}\frac{\Gamma \left ({\dfrac{p+1}{2}} \right )}{\Gamma(p/2)\sqrt{p\pi}}= \frac{1}{\sqrt{2\pi}}.$$
Where $\Gamma (x)$ represents the gamma function.
To provide some context, I need to find the limiting distribution of a student's t distribution , which should converge to the normal distribution. The key part of the proof translates to solving the above limit.
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As $p\to +\infty$ the Gamma function has the asymptotic approximation:
$$\Gamma\left(\frac{p+1}{2}\right) \sim \large e^{\frac{p}{2}\left(1 - \ln(2) - \ln\left(\frac{1}{p}\right)\right)}\left(\sqrt{2\pi} - \frac{1}{6p}\sqrt{\frac{\pi}{2}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)$$
and
$$\Gamma\left(\frac{p}{2}\right) \sim \large e^{\frac{p}{2}\left(1 - \ln(2) - \ln\left(\frac{1}{p}\right)\right)}\left(2\sqrt{\pi}\frac{1}{\sqrt{p}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)$$
Hence substituting into the limit:
$$\frac{e^{\frac{p}{2}\left(1 - \ln(2) - \ln\left(\frac{1}{p}\right)\right)}\left(\sqrt{2\pi} - \frac{1}{6p}\sqrt{\frac{\pi}{2}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)}{e^{\frac{p}{2}\left(1 - \ln(2) - \ln\left(\frac{1}{p}\right)\right)}\left(2\sqrt{\pi}\frac{1}{\sqrt{p}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)\cdot \sqrt{p\pi}} = \frac{\left(\sqrt{2\pi} - \frac{1}{6p}\sqrt{\frac{\pi}{2}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)}{\left(2\sqrt{\pi}\frac{1}{\sqrt{p}} + \mathcal{O}\left(\frac{1}{p}\right)^{3/2}\right)\sqrt{p\pi}}$$
Which, as $p \to +\infty$ goes to
$$\frac{\sqrt{2\pi}}{2\pi} = \color{red}{\frac{1}{\sqrt{2\pi}}}$$
As wanted.
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|
Integration with multiple constants Question:
$$\int \frac{ax^2-b}{x\sqrt{{c^2x^2-(ax^2+b)^2}}}\ \text dx$$
My approach:
I can't understand whether I should integrate it normally or use a trigonometric function.
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Let $$I =\int\frac{(ax^2-b)}{x\sqrt{c^2x^2-(ax^2+b)^2}}dx = \int\frac{ax^2-b}{x\cdot x \sqrt{c^2-(ax+\frac{b}{x})^2}}dx$$
So $$I = \int\frac{a-\frac{b}{x^2}}{\sqrt{c^2-(ax+\frac{b}{x})^2}}dx$$
Put $\displaystyle \left(ax+\frac{b}{x}\right) = t\;,$ Then $\displaystyle \left(a-\frac{b}{x^2}\right)dx = dt$
$$I = \int\frac{1}{\sqrt{c^2-t^2}}dt = \arcsin\left(\frac{t}{c}\right)+\mathcal{C}$$
So $$I =\arcsin\bigg(\frac{ax^2+b}{cx}\bigg)+\mathcal{C}$$
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|
Prove that $f(x)=\sin^2 x-x^2\cos x, \forall x\in [0,\frac{\pi}{2}]$ is monotonic increasing I would like to show that $f(x)=\sin^2 x-x^2\cos x, \forall x\in [0,\frac{\pi}{2}]$ is monotonic increasing.
If we can show that $f'(x)>0$, $\forall x\in [0,\frac{\pi}{2}]$, then $f(x)$ is increasing there.
We have
$f'(x)=\sin 2x-2x\cos x+x^2\sin x$.
How can I show that $f'(x)>0, \forall x\in [0,\frac{\pi}{2}]$
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One has \begin{align}2\sin x\cos x-2x\cos x+x^2\sin x &=2\cos x(\sin x-x)+x^2\sin x \\ &\ge x^2\sin x-\frac{x^3\cos x}{3} \\&=x^2\cos x\left(\tan x-\frac{x}3\right)\\&\ge \frac23x^3\cos x\ge0,\end{align}where we have used the Taylor expansion of $\sin x$ and the fact that $\tan x\ge x$ on $\left[0,\dfrac\pi2\right]$.
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Repeated linear factors in partial fractions I have a question about the following partial fraction:
$$\frac{x^4+2x^3+6x^2+20x+6}{x^3+2x^2+x}$$
After long division you get:
$$x+\frac{5x^2+20x+6}{x^3+2x^2+x}$$
So the factored form of the denominator is
$$x(x+1)^2$$
So
$$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{(x+1)^2}$$
Why is the denominator under $C$ not simply $x+1$? It is $x$ times $(x+1)^2$ and not $(x+1)^3$
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You need an $(x+1)^2$ in the denominator of at least one of the partial fractions, or when you sum them they would not have an $(x+1)^2$ term in the denominator of the sum.
But you might find it easier to solve:
$\frac{5x^2+20x+6}{x(x+1)^2}=\frac{A}{x}+\frac{Bx + C}{(x+1)^2}$
And that is completely valid.
|
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Expected number of turns in dice throwing I generated a transition probability matrix for a scenario where I throw five dice and set aside those dice that are sixes. Then, I throw the remaining dice and again set aside the sixes - then I repeat this procedure until I get all the sixes. $X_n$ here represents the number of dices that are sixes after n rolls.
$$\begin{pmatrix}\frac{5^5}{6^5} & \frac{3125}{6^5} & \frac{1250}{6^5} & \frac{250}{6^5} & \frac{25}{6^5} & \frac{1}{6^5}\\\ 0 & \frac{625}{6^4} & \frac{500}{6^4} & \frac{150}{6^4} & \frac{20}{6^4} & \frac{1}{6^4} \\\ 0& 0 & \frac{125}{6^3}& \frac{75}{6^3}& \frac{15}{6^3} & \frac{1}{6^3} \\\ 0 & 0& 0& \frac{25}{6^2}& \frac{10}{6^2}& \frac{1}{6^2}& \\ 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6} \end{pmatrix}$$
I want to figure out how many turns it takes for me on average to get all sixes.
I'm not even sure where to start with this problem. Is it a right approach to write a program where I calculate $P^n$ and see when the 6th column all equals to 1?
Any pointers would be greatly appreciated.
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Just a quick correction to your transition matrix...
$$P =\begin{pmatrix}\frac{5^5}{6^5} & \frac{3125}{6^5} & \frac{1250}{6^5} & \frac{250}{6^5} & \frac{25}{6^5} & \frac{1}{6^5}\\\ 0 & \frac{625}{6^4} & \frac{500}{6^4} & \frac{150}{6^4} & \frac{20}{6^4} & \frac{1}{6^4} \\\ 0& 0 & \frac{125}{6^3}& \frac{75}{6^3}& \frac{15}{6^3} & \frac{1}{6^3} \\\ 0 & 0& 0& \frac{25}{6^2}& \frac{10}{6^2}& \frac{1}{6^2}& \\ 0 & 0 & 0 & 0 & \frac{5}{6} & \frac{1}{6}
\\ 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}
$$
and it only represents your transition probabilities on a single round of the game, not after n rolls.
One thing you can try doing is calculating
$$X_n = [1 \ 0 \ 0 \ 0 \ 0 \ 0] P^n = [p_{n,1} \ p_{n,2} \ p_{n,3} \ p_{n,4} \ p_{n,5} \ p_{n,6}]$$
The expected number of rounds played in this game would be given by
$$
S=1+\sum_{n=1}^\infty (p_{n,1}+p_{n,2}+p_{n,3}+p_{n,4}+p_{n,5})
$$
i.e. You're adding up all the elements in the first five columns for $n=1,2,3...$, since those represent the probability of still playing the game after n rounds. You can easily approximate $S$ using a spreadsheet. Calculating it analytically using this method might be tedious, but I would assume is definitely doable.
|
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Given odd positive integer, $n=2m-1$, $n \equiv 1 \pmod 4 \implies m \equiv 1 \pmod 2$ $n=2m-1\implies n+1=2m$. So, $n\equiv 1\pmod 4, 2m=n+1\equiv 2\pmod 4\implies 2m=n+1\equiv 0\pmod 2$ $\implies n\equiv -1\pmod 2\implies n\equiv 1\pmod 2$. But, how to find $m$ from this last line of equivalence relations is not clear.
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$$n = 2m-1\equiv 1\pmod{4}\quad\Rightarrow\quad
2m \equiv 2\pmod{4}.$$
Now, note that in general, if $k\ne 0$ and $ak\equiv bk\mod{ck}$, this means that $ak$ is $bk$ plus a multiple of $ck$, say $rck$:
$$ak = bk + rck.$$
Dividing through by $k$ gives $a = b+cr$, so that $a\equiv b\mod c$. Applying this to $2m\equiv 2\mod{(4 = 2\cdot 2)}$ gives
$$m\equiv 1\pmod{2}.$$
|
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|
Find the coefficient of $x^{25}$ in $(1 + x^3 + x^8)^{10}$
Find the coefficient of $x^{25}$ in $(1 + x^3 + x^8)^{10}$.
Here is my solution I am looking to see if it is correct or if there is another way to do it, thanks!
The only way to form an $x^{25}$ term is to gather two $x^8$ and three $x^3$ . Since there are ${{10}\choose{2}} =45$ ways to choose two $x^8$ from the $10$ multiplicands and $8$ ways to choose three ${{8}\choose{3}}= 56$ ways to choose $x^3$ from the remaining $8$multiplicands, the answer is using the product rule $45×56 = 2520$.
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A formal way:
$$(1 + x^3 + x^8)^{10}=\sum_{k=0}^{10}\binom{10}{k}x^{3k}(1+x^5)^k=\sum_{k=0}^{10}\sum_{j=0}^{k}\binom{10}{k}\binom{k}{j}x^{3k+5j}.$$
Now $25=3k+5j$ for $ 0\leq j\leq k\leq 10$ is solved iff $(k,j)=(5,2)$ and
$$[x^{25}](1 + x^3 + x^8)^{10}=\binom{10}{5}\binom{5}{2}=2520.$$
|
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|
Finding value of product of Cosines
Finding $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{9\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{27\pi}{20}\right)$$
My Try: $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}+\sin \frac{\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$
So we have $$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\sin\frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)$$
Could some help me to solve it, Thanks in Advanced
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You can end it by the following way:
$$\left(\frac{1}{2}+\cos \frac{\pi}{20}\right)\left(\frac{1}{2}+\sin\frac{\pi}{20}\right)\left(\frac{1}{2}+\cos \frac{3\pi}{20}\right)\left(\frac{1}{2}-\sin\frac{3\pi}{20}\right)=$$
$$=\left(\frac{1}{2}+\cos9^{\circ}\right)\left(\frac{1}{2}+\sin9^{\circ}\right)\left(\frac{1}{2}+\cos27^{\circ}\right)\left(\frac{1}{2}-\sin27^{\circ}\right)=$$
$$=\left(\frac{1}{4}+\frac{1}{2}(\cos9^{\circ}+\sin9^{\circ})+\frac{1}{2}\sin18^{\circ}\right)\left(\frac{1}{4}+\frac{1}{2}(\cos27^{\circ}-\sin27^{\circ})-\frac{1}{2}\sin54^{\circ}\right)=$$
$$=\left(\frac{1}{4}+\frac{1}{2}(\sin81^{\circ}+\sin9^{\circ})+\frac{1}{2}\sin18^{\circ}\right)\left(\frac{1}{4}+\frac{1}{2}(\sin63^{\circ}-\sin27^{\circ})-\frac{1}{2}\cos36^{\circ}\right)=$$
$$=\left(\frac{1}{4}+\sin45^{\circ}\cos36^{\circ}+\frac{1}{2}\sin18^{\circ}\right)\left(\frac{1}{4}+\sin18^{\circ}\cos45^{\circ}-\frac{1}{2}\cos36^{\circ}\right)=$$
$$=\left(\frac{1}{4}+\frac{\sqrt5+1}{4\sqrt2}+\frac{\sqrt5-1}{8}\right)\left(\frac{1}{4}+\frac{\sqrt5-1}{4\sqrt2}-\frac{\sqrt5+1}{8}\right)=$$
$$=\left(\frac{1}{8}+\frac{\sqrt5}{4\sqrt2}+\frac{1}{4\sqrt2}+\frac{\sqrt5}{8}\right)\left(\frac{1}{8}+\frac{\sqrt5}{4\sqrt2}-\frac{1}{4\sqrt2}-\frac{\sqrt5}{8}\right)=$$
$$=\left(\frac{1}{8}+\frac{\sqrt5}{4\sqrt2}\right)^2-\left(\frac{1}{4\sqrt2}+\frac{\sqrt5}{8}\right)^2=$$
$$=\frac{1}{64}+\frac{\sqrt5}{16\sqrt2}+\frac{5}{32}-\frac{5}{64}-\frac{\sqrt5}{16\sqrt2}-\frac{1}{32}=\frac{1}{8}-\frac{1}{16}=\frac{1}{16}.$$
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How to find $\lim_\limits{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$
How to find $$\lim_{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}$$
My attempt:
$$\lim _{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}=\lim _{n\to \infty }\left(1-\frac{5}{n^4}\right)^{\left(4\cdot 2018n+4\right)}=\lim _{n\to \infty }\left(\left(1-\frac{5}{n^4}\right)^{\left(4\cdot 2018n\right)}+\left(1-\frac{5}{n^4}\right)^4\right)=\lim _{n\to \infty }\left(\left(\left(1-\frac{5}{n^4}\right)^{n}\right)^{4*2018}+\left(1-\frac{5}{n^4}\right)^4\right)$$
so we get :
*
*$\lim _{n\to \infty }\left(1-\frac{5}{n^4}\right)^4=1$
let:
$b_n=\frac{-5}{n^3}$
$\lim _{n\to \infty }\frac{-5}{n^3}=0$
*
*$\lim _{n\to \infty }\left(\left(1-\frac{5}{n^4}\right)^n\right)^{4\cdot 2018}=\lim \:_{n\to \:\infty \:}\left(\left(1+\frac{b_n}{n}\right)^n\right)^{4\cdot \:2018}=\left(e^0\right)^{4\cdot 2018}=1$
so the answer is :
$$\lim _{n\to \infty }(1-\frac{5}{n^4})^{(2018n+1)^4}=1+1=2$$
Is this answer correct if not how can I find the limit ?
thanks.
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The standard way is to consider
$$(1-\frac{5}{n^4})^{(2018n+1)^4}=e^{(2018n+1)^4 \log{(1-\frac{5}{n^4}) }}\to e^{-5\cdot2018^4}$$
indeed by standard limits
$${(2018n+1)^4 \log{(1-\frac{5}{n^4}) }}=(2018n+1)^4(\frac{5}{n^4})
\frac{\log {(1-\frac{5}{n^4})}}{\frac{5}{n^4}}\to -5\cdot2018^4$$
Or as an alternative
$$(1-\frac{5}{n^4})^{(2018n+1)^4}=\left[(1-\frac{5}{n^4})^{\frac{n^4}5}\right]^{\frac5{n^4}(2018n+1)^4}\to e^{-5\cdot2018^4}$$
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|
Find all $z$ s.t $|z|=1$ and $|z^2 + \overline{z}^2|=1$ Here's my attempt. Let $z=x+i\ y$, then $$z^2=x^2-y^2+i\ 2xy$$ and $$\bar z^2=x^2-y^2-i\ 2xy$$
Then, $$z^2+\bar z^2=2x^2-2y^2$$ so $$1=|z^2+\bar z^2|=\sqrt{(2x^2)^2+(-2y^2)^2}$$
Simpliflying the expression above, we get $$1=4x^4+4y^4$$
which gives us $$\frac14=x^4+y^4$$. I am stuck here. Is it wrong approach? is there an easier one?
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Let $z=\cos t+i\sin t$. Then $|z^2+\bar{z}^2|=1$ becomes
$$2|\cos(2t)|=1 $$
or
$$ \cos(2t)=\pm\frac{1}{2}$$
So
$$ \cos t=\pm\sqrt{\frac{1+\cos(2t)}{2}}=\pm\frac{\sqrt3}{2}\text{ or }\pm\frac{1}{2}$$
and
$$ \sin t=\pm\sqrt{\frac{1-\cos(2t)}{2}}=\pm\frac12\text{ or }\pm\frac{\sqrt3}{2}. $$
Thus
$$ z=\cos t+i\sin t=\pm\frac{\sqrt3}{2}\pm\frac{1}{2}i,z=\pm\frac{1}{2}\pm\frac{\sqrt 3}{2}i. $$
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|
Find the limit of $\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$ I was trying to solve the following task but I stumbled across something I do not understand:
Calculate:
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
my attempt was to factorize n^2 out of the squareroot:
$$$$
$$\lim_{n \to \infty}\frac{1}{n^2-\sqrt{n^4+4n^2+n}}$$
\begin{align}
\\a_n & = \frac{1}{n^2-\sqrt{n^4+4n^2+n}} \\
& =\frac{1}{n^2-\sqrt{n^4\left(1+\frac{4}{n^2}+\frac{1}{n^3}\right)}} \\
& =\frac{1}{n^2-n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}} \\
\end{align}
Therefor, I thought that:
$$\lim_{n \to \infty}\frac{1}{n^2-n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}}$$
$$ = \lim_{n \to \infty}\frac{1}{\left(n^2-n^2\right)\sqrt{1}} = \infty$$
I also tried a different way where I got to the result of $-\dfrac{1}{2}$.
I am not going to show that method here but it starts with using the 3rd binomial formula. Then, having the squareroot at the top of the fraction, I factorized $n^2$ and it all worked.
Why does this method like shown above not work?
I am very happy for any help.
P.S. This is not the only example where this kind of getting to a solution does not work for me. Are there cases where I am not allowed to factorize something?
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You nearly had the answer yourself: In the line
$$\begin{align}
\\a_n & = \frac{1}{n^2-\sqrt{n^4+4n^2+n}} \\
& =\frac{1}{n^2-\sqrt{n^4\left(1+\frac{4}{n^2}+\frac{1}{n^3}\right)}} \\
& =\frac{1}{n^2-n^2\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}}} \\
\end{align}$$
For large $n$ $$\sqrt{1+\frac{4}{n^2}+\frac{1}{n^3}} \approx 1 + \frac{1}{2} (\frac{4}{n^2}+\frac{1}{n^3})$$ and $n^2$ this gives
$$\approx n^2 + \frac{1}{2} (4+\frac{1}{n})$$ so that the final result is
$$\frac{1}{n^2-n^2-2-\frac{1}{2 n}}$$ from which you can determine the desired result.
|
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|
Power Series of $\ \frac x3 \ln(1+x^2).$ I have a question regarding the power series representation of the function $$\ \frac x3 \ln(1+x^2).$$
I got the answer $$\sum_{n=1}^\infty \frac{(-1)^{n+1}{x^{2n+1}}}{3n} $$
by using the Maclaurin series expansion for $\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}{x^{n}}}{n}$ and replacing $x$ with $x^2$ and multiplying the series with $\frac{x}{3}$.
However, my teacher got the answer $$\frac 23\sum_{n=1}^\infty \frac{(-1)^{n+1}{x^{2n+3}}}{2n+1}. $$
My teacher mentioned differentiation and integration, as one of the first steps in his method was taking $\ln(1+x^2)$ and representing it as $\int \frac{2x dx}{1+x^2}.$ He then represented the original function as $\frac x3 \int \frac{2x dx}{1+x^2}.$ However, after this I could not follow the rest of the procedures. I am completely lost as to how my teacher obtained this answer, and any help would be appreciated in explaining how one may arrive to this conclusion.
|
So continuing from where your teacher left off, we have $\frac{2x}3 \int\frac{xdx}{1+x^2}$. We start with the series expansion for $\frac{1}{1+y}$:
$$ \frac1{1+y} = \sum_{n=0}^\infty (-1)^ny^n$$
We substitute $y = x^2$:
$$ \frac1{1+x^2} = \sum_{n=0}^\infty (-1)^nx^{2n}$$
We multiply by $x$:
$$ \frac x{1+x^2} = \sum_{n=0}^\infty (-1)^nx^{2n+1}$$
And now we integrate:
$$\int\frac{xdx}{1+x^2} = \sum_{n=0}^\infty \frac{(-1)^nx^{2n+2}}{2n+2} $$
Finally, we multiply by $\frac{2x}3$:
$$ \frac{2x}3 \int\frac{xdx}{1+x^2} = \frac23 \sum_{n=0}^\infty \frac{(-1)^nx^{2n+3}}{2n+2}$$
I imagine this was the intended methodology, but the answer I get doesn't match your teacher's. In fact, it matches yours by making the change of index $m=n+1$:
$$\frac{2x}3 \int\frac{xdx}{1+x^2} = \frac23 \sum_{m=1}^\infty \frac{(-1)^{m-1}x^{2m+1}}{2m}= \frac13 \sum_{m=1}^\infty \frac{(-1)^{m+1}x^{2m+1}}{m}$$
|
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|
What is the value of $\sin^2 (\frac{\pi}{10}) \sin^2 (\frac{3\pi}{10})$?
PROBLEM
$$ \prod_{i=0}^4 \left(1 + \cos \left(\frac{(2k+1)\pi}{10}\right)\right)$$
My Try
$$
\left(1 + \cos \frac{\pi}{10}\right)
\left(1 + \cos \frac{9\pi}{10}\right)
\left(1 + \cos \frac{7\pi}{10}\right)
\left(1 + \cos \frac{3\pi}{10}\right)
= \sin^2 \left(\frac{\pi}{10}\right)
\sin^2 \left(\frac{3\pi}{10}\right)
$$
I am not able to proceed further.
Please help me.
|
$$\sin^2 {\frac {\pi}{10}} \sin^2 \frac {3\pi}{10}=\left(\frac {4\sin {\frac {\pi}{5}}. \cos {\frac {2\pi}{10}}\cos {\frac {4\pi}{10}}}{4\sin \frac {\pi}{5}}\right)^2=\left(\frac {\sin \frac {4\pi}{5}}{4\sin \frac {\pi}{5}}\right)^2=\frac {1}{16}$$
|
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|
Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Compute the limit $\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}$
Here is what I have done so far:
\begin{align}
\lim_{x\to 0} \frac{3^x+2^x-2}{4^x+2^x-2}
&=
\lim_{x\to 0} \left(\frac{3^x-1}{4^x+2^x-2}+\frac{2^x-1}{4^x+2^x-2}\right)\\
&=\lim_{x\to 0} \left(\frac{3^x-1}{x}\frac{x}{4^x+2^x-2}+\frac{2^x-1}{x}\frac{x}{4^x+2^x-2}\right)\\
&=\ln 3\lim_{x\to 0} \frac{x}{4^x+2^x-2}+\ln 2\lim_{x\to 0} \frac{x}{4^x+2^x-2}
\end{align}
|
Note that by standard limits since
$$\frac{a^x-1}{x}\to \log a$$
we have that
$$\frac{3^x+2^x-2}{4^x+2^x-2}=\frac{\frac{3^x-1}x+\frac{2^x-1}x}{\frac{4^x-1}x+\frac{2^x-1}x}\to\frac{\log 3+\log2}{\log 4+\log 2}=\frac{\log 6}{\log 8}$$
|
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|
This formula gives $8^{-1}$ (mod $n$). Is there a deeper pattern lurking here? Pick $n\equiv 1$ (mod $4$) which is not a multiple of $3$ and such that $n>5$. Consider the sum
$$S(n):=2\cdot\frac{n-1}2+3\cdot\frac{n-3}2+\ldots+m(m+2)+m+1,$$
where $m:=\frac{n-1}4$. For example, for $n=17$ we are considering
$$2\cdot 8+3\cdot 7+4\cdot 6+5.$$
Then we can show, by expanding $$S(n)=\sum_{i=0}^{m-2}(2+i)\left(\frac{n-1}2-i\right)+m+1$$
and applying the formulas for triangular and pyramidal numbers, that in fact
$$S(n)\equiv 8^{-1}(\text{mod } n),$$
as the numerator of the resulting fraction reduces to 1 (with a denominator of $8$).
(We cannot do the same for multiples of 3 because of the factor of 6 in the denominator of the pyramidal formula).
The simplicity of the final expression is a happy finding. I wonder:
1) Is there a neater explanation for it? Should we in fact expect the numerator to be 1?
2) More in general, can we pinpoint by some method other similar sums with particularly simple reductions?
For example, for $n\equiv 3$ (mod $4$) we get $S'(n)\equiv -11\cdot 2^{-5}$ (mod $n$) with $S'(n)$ the sums of pairs from $2\cdot\frac{n-1}2$ until possible without repetitions. Could we predict this, or choose to avoid this formula in behalf of another?
3) Is there some simple way of getting prime factors other than $2,3$ for the inverse of the final result, apart from adding more factors to the products (i.e., taking sums of products of $k$ numbers)?
|
This is a partial answer.
Using that
$$\sum_{i=0}^{N}1=N+1,\qquad \sum_{i=0}^{N}i=\frac{N(N+1)}{2},\qquad\sum_{i=0}^{N}i^2=\frac{N(N+1)(2N+1)}{6}$$
we have
$$\small\begin{align}8S(n)&=8\sum_{i=0}^{m-2}(2+i)\left(\frac{n-1}{2}-i\right)+8m+8\\\\&=-8\left(\sum_{i=0}^{m-2}i^2\right)+(4n-20)\left(\sum_{i=0}^{m-2}i\right)+(8n-8)\left(\sum_{i=0}^{m-2}1\right)+8m+8
\\\\&=-8\cdot \frac{(m-2)(m-1)(2m-3)}{6}+(4n-20)\cdot\frac{(m-2)(m-1)}{2}+(8n-8)\cdot (m-2+1)+8m+8
\\\\&=-8\cdot \frac{(m-2)(m-1)(2m-3)}{6}+(16m-16)\cdot\frac{(m-2)(m-1)}{2}+32m (m-1)+8m+8
\\\\&=\frac 43m(4m^2+9m-1)
\\\\&=\frac 43\cdot\frac{n-1}{4}\left(4\left(\frac{n-1}{4}\right)^2+9\cdot\frac{n-1}{4}+1\right)
\\\\&=\frac{1}{12}(n-1)(n^2+7n-12)
\end{align}$$
Now, integers such that
$$n\equiv 1\pmod 4\quad\text{and}\quad n\not\equiv 0\pmod 3\quad \text{and}\quad n\gt 5$$
can be written as
$$n=12k+1\quad\text{or}\quad n=12k+5$$
where $k$ is a positive integer, and we have
*
*$8S(12k+1)=(12k+1)(12k^2+8k-1)+1\equiv 1\pmod{12k+1}$
*$8S(12k+5)=(12k+5)(12k^2+16k+3)+1\equiv 1\pmod{12k+5}$
|
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|
Eigenvalue and Eigenvector of $\small\pmatrix{0 & 0 \\ 0 & -7}$ I need help working out the eigenvectors for this matrix.
$ \begin {pmatrix} 0 & 0 \\ 0 & -7 \end{pmatrix} $
The original matrix is $ \begin {pmatrix} 5 & 0 \\ 0 & -2 \end{pmatrix} $ , eigenvalues are 5,-2,
but I am not sure how to about the eigenvectors,
as for 5
$ \begin {pmatrix} 0 & 0 \\ 0 & -7 \end{pmatrix} $ $ \begin{pmatrix} x \\ y \end{pmatrix}$ = $ \begin{pmatrix} 0 \\ 0 \end{pmatrix}$
from the first equation, $x$ and $y$ are both zero, but from the second equation $y = 0$, so what is the eigenvector?
|
From first equation you deduce whatever is x and y the equation holds
$$0x+0y=0$$
From second equation you deduce that $y=0$
$$0x-7y=0 \implies -7y=0 \implies y=0$$
So
$$(x,y)=(x,0)=x(1,0)$$
|
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|
If $A$, $B$ and $C$ are the angles of a triangle then find the value of $\Delta$ I'll state the question from my book below:
If $A$, $B$ and $C$ are the angles of a triangle, then find the determinant value of
$$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}.$$
Here's how I tried solving the problem:
$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}$
$R_2 \to R_2 - R_1$
$R_3 \to R_3 -R_1$
$= \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B-\sin^2A & \cot B-\cot A & 0 \\ \sin^2C-\sin^2A & \cot C-\cot A & 0\end{vmatrix}$
Expanding the determinant along $C_3$
\begin{align}
&= (\sin^2B-\sin^2A)(\cot C-\cot A)-(\cot B-\cot A)(\sin^2C-\sin^2A) \\
&= \sin(B+A) \sin(B-A) \left[\frac {\cos C}{\sin C} - \frac {\cos A}{\sin A}\right] - \left[\frac {\cos B}{\sin B} - \frac {\cos A}{\sin A}\right]\sin(C+A) \sin(C-A) \\
&= \frac {\sin(B+A) \sin(B-A) \sin(A-C)} {\cos A \cos C} - \frac {\sin(A-B) \sin(C+A) \sin(C-A)} {\cos A \cos C} \\
&= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin(A+B)} {\cos C} - \frac {\sin(A+C)} {\cos B}\right] \\
&= \frac {\sin(B-A) \sin (A-C)} {\cos A} \left[\frac {\sin C} {\cos C} - \frac {\sin B} {\cos B}\right] \\
&= \frac {\sin(B-A) \sin (A-C) \sin (C-B)} {\cos A \cos B \cos C}
\end{align}
I tried solving further but the expression just got complicated. I don't even know if the work I've done above is helpful. My textbook gives the answer as $0$. I don't have any clue about getting the answer. Any help would be appreciated.
|
A hint, as requested:
What you've done looks pretty good; I haven't checked every bit of the algebra, but the symmetry of the result makes me think that you've probably done it right.
What you have not done is use the important fact given at the start: that $A, B,$ and $C$ are the angles of a triangle (hence they sum to $\pi$). I don't see right away how to use that fact, but it's clearly important.
|
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|
Find the Cartesian equation of the locus described by $|z+2-7i| = 2|z-10+2i|$
Find the Cartesian equation of the locus described by $|z+2-7i| =
2|z-10+2i|$ Write your answer in the form $(x+a)^2+(y+b)^2=k$.
This was a question from my end of year exams just gone and I'm unsure as to where I have gone wrong :(. If anyone could point me in the right direction I'd very much appreciate that.
Here was my working:
Let $z = x+iy$
$|z|$ = $\sqrt{x^2+y^2}$
$$|z+2-7i| = 2|z-10+2i| \\
\implies |(x+iy)+2-7i| = 2|(x+iy)-10+2i| \\
\implies |(x+2)(y-7y)i| = 2|(x-10)+(y+2)i| \\
\implies \sqrt{(x+2)^2+(y-7)^2i^2} = 2\sqrt{(x-10)^2+(y+2)^2i^2} \\
\implies (x+2)^2-((y-7^2) = 2((x-10)^2-(y+2)^2)$$
$$(x^2+4x+4)-(y^2-14y+49) = 2((x^2-20x+100)-(y^2+2y+4)) \\
\implies x^2+4x+4-y^2+14y-49 = 2x^2-40x+200-2y^2-4y-8 \\
\implies -237 = x^2-44x-y^2-18y\\
\implies x^2-44x-y^2-18 = -237$$
|
Consider z= (x+iy),Now put the value of Z in the given question.
after applying rules of modulus
we get `x^2+Y^2-28x+10y+121=0
Now we can arrange in square form like
(x-14)^2+(y+5)^2=45
|
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|
Prove that $24|n^2-1$, if $(n,6)=1$ $n^2-1 = (n-1)(n+1)$
Then $24|(n-1)(n+1)$
$(n,6)=1$: $\exists a,b\in\mathbb{Z}$ that $n = 6\cdot a+b$
Investigate the residues, which arise when dividing the number n by two and three:
$\frac{6\cdot a+b}{3} = \frac{6\cdot a}{3}+\frac{b}{3} = 2\cdot a+\frac{b}{3}$
$\frac{6\cdot a+b}{2} = 3\cdot a+\frac{b}{2}$
Is this correct? How to prove it using resiudes?
Thank You.
|
Given $n-1, n, n+1$ and $n+2$ then $2$ divides two of them and $4$ divides one of them and $8$ divides the product $(n-1)n(n+1)(n+2)$. If $n$ is odd then we know the two that $2$ and $4$ divides must be $n-1$ and $n+1$ and that $8$ divides the product $(n-1)(n+1)$.
Given $n-1, n , n+1$ then $3$ divides exactly one of them. If $3\not \mid n$ then $3$ divide either $n-1$ or $n + 1$ so $3|(n+1)(n-1)$.
So if $n$ is odd and $3 \not \mid n$, then $3*8|(n+1)(n-1)$.
That will be the case if $\gcd(n,6) = 1$. So if $\gcd(n,6) =1$ then $n$ is odd and $3\not \mid 6$ and so $24 = 3*8|(n+1)(n-1) = n^2 - 1$.
......
Using residues:
$\gcd(n,6) = 1$ so $\gcd(2, n) = 1$ so $\gcd(8^n, 1) $ so $n \equiv 1, 3,5, 7 \equiv \pm 1, \pm 3\mod 8$. So $n^2 \equiv 1, 9\equiv 1 \mod 8$.
$\gcd(n,6) = 1$ so $\gcd(3, n) =1$ so $n \equiv \pm 1 \mod 3$. So $n^2 \equiv 1 \mod 8$.
So $n^2 -1 \equiv 0 \mod 3$ and $n^2 -1 \equiv 0 \mod 8$.
So $3|n^2 - 1$ and $8|n^2 -1$ and $24|n^2 - 1$.
|
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|
Monotonic recursive sequence with recursive term in denominator: $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ I am trying to show that $s_{n+1} = \frac{1}{2} \left(s_n + \frac{3}{s_n}\right)$ is a monotonic decreasing sequence for $n \ge 2$. Currently, my approach using induction is stuck because of the $\mathbf{s_n}$ term appearing in the denominator.
What I have so far:
Calculating a few terms: $s_1 = 1$, $s_2 = 2$, $s_3 = \frac{7}{4} < s_2$.
For this to be monotonic decreasing, we have to show $s_{k+2} < s_{k+1}$.
For the base case, we assume for $k$ we have $s_{k+1} < s_k.$
The inductive step: to show $s_{k+2} < s_{k+1}$, i.e. to show $$\color{grey}{s_{k+2} = }\frac{1}{2} \left(s_{k+1} + \frac{3}{s_{k+1}}\right) < \frac{1}{2} \left(s_{k} + \frac{3}{s_{k}}\right) \color{grey}{= {s_{k+1}}_.}$$
In order to show $\dfrac{1}{2} \left(s_{k+1} + \frac{3}{s_{k+1}}\right) < \dfrac{1}{2} \left(s_{k} + \frac{3}{s_{k}}\right)$, I am stuck:
Given, $s_{k+1} < s_k$, I cannot then say $s_{k+1} + \mathbf{\frac{3}{s_{k+1}}} < s_k + \mathbf{\frac{3}{s_{k}}}$, because the $s_k, s_{k+1}$ terms which appears in the denominator may reverse the inequality. Any pointers on how to proceed further?
Disclaimer: I am revising real analysis on my own from the Kenneth Ross book, this is not strictly homework.
|
Presuming $s_1 > 0$, by AM-GM inequality,
$$s_{n} = \frac{s_{n-1}+\frac{3}{s_{n-1}}}{2} \ge \sqrt{3}$$ for $n \ge 2$.
Also,
$$s_{n+1}-s_{n} = \frac{1}{2}\left(\frac{3}{s_n}-s_n\right)\le 0 \iff s_{n} \ge \sqrt{3}$$
for $n \ge 2$.
|
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|
A step in finding the slant asymptote I'm curious why this textbook shows an extra step when finding the slant asymptote in this function: $$f(x) = \frac {x^3}{x^2 + 1}.$$
So long division gives
$$f(x) = \frac{x^3}{x^2+1} = x - \frac{x}{x^2+1},$$
so
$$f(x) - x = - \frac{x}{x^2 + 1} = -\frac{\frac{1}{x}}{1 + \frac{1}{x^2}} \rightarrow 0$$ as $x \rightarrow \pm \infty$.
Why did they show that last step? Isn't it enough to have the $- \dfrac{x}{x^2 + 1}$?
|
Because $-\frac{x}{x^2+1}$ is in the indeterminate form $\frac{\infty}{\infty}$ and thus the additional step is to factor out an $x^2$ term before to take the limit in order to have a form $\frac{0}{1}$.
|
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|
What is the mistake in this? What is wrong in this proof?
In $\triangle ABC$ with right angle at $B$, $BD$ is drawn as an altitude on $AC$. Let $AB=a$ and $BC=b$. So by similarity, $AD=\frac{a^2}{\sqrt{a^2+b^2}}$ and similarly $CD=\frac{b^2}{\sqrt{a^2+b^2}}$. Now, $\triangle ABD,BDC$ are also right angled at $D$. So applying the Pythagorean theorem,
$$BD^2=a^2-\frac{a^2}{\sqrt{a^2+b^2}}=b^2-\frac{a^2}{\sqrt{a^2+b^2}}$$
Solve these two equations as:
$$a^2-b^2=a^2-\frac{b^2}{\sqrt{a^2+b^2}}$$
So $\sqrt{a^2+b^2}=1$. That means for all values of $a$ and $b$, $a^2+b^2=1$.
But this is absurd.
|
Your problem is actually not even really a misuse of a rule. It's a typo.
$$BD^2=a^2-\frac{a^2}{\sqrt{a^2+b^2}}=b^2-\frac{a^2}{\sqrt{a^2+b^2}}$$
This should be:
$$BD^2=a^2-\frac{a^2}{\sqrt{a^2+b^2}}=b^2-\frac{b^2}{\sqrt{a^2+b^2}}$$
Reducing that down, we get:
$$(a^2)\sqrt{a^2+b^2}-a^2=(b^2)\sqrt{a^2+b^2}-b^2$$
Factoring...
$$(a^2)\left(\sqrt{a^2+b^2}-1\right)=(b^2)\left(\sqrt{a^2+b^2}-1\right)\\
a^2=b^2$$
|
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|
Prove that $B$ is non singular and that $AB^{-1}A=A$
$$A_{n\times n}=\begin{bmatrix}a & b & b & b &. &.&.&&b\\b & a
&b&b&.&.&.&&b\\b & b &a&b&.&.&.&&b\\b & . &.&.&.&.&.&&b\\b & .
&.&.&.&.&.&&b\\b & b &b&b&.&.&.&&a\end{bmatrix}\text{ where }
a+(n-1)b =0$$
Define $l^t=\begin{bmatrix}1&1&1&1&1&....1\end{bmatrix}$ Where $l$ is
a $ n\times1$ vector, and:
$$B= A+ \frac{l\cdot l^t}{n}$$
Prove that $B$ is non singular and that $AB^{-1}A=A$
What i did:
$\text{A has a 0 eigenvalue , so A is a singular matrix}$
$\text{B has an eigenvalue of 1 with eigenvector} $$\,\, v^{t}= \begin{bmatrix}1&1&1&1&1&....1\end{bmatrix}$
Any idea about how to proceed?
Thanks.
|
Let $\mathbf{1}_n$ denote the $n\times n$ matrix with all entries equal to one and $E_n$ denote the $n\times n$ identity matrix. Then
$$
A_n = b\mathbf 1_n + (a-b)E_n = b\mathbf 1_n -nbE_n
$$
and
$$
B_n = A_n + \frac{1}{n}\mathbf 1_n = \left(b+\frac{1}{n}\right)\mathbf 1_n - nbE_n.
$$
In general, for a matrix $X_n=\alpha \mathbf 1_n + \beta E_n$, that is
$$
X =
\begin{pmatrix}
\alpha+\beta & \alpha & \alpha & \cdots & \alpha \\
\alpha & \alpha+\beta & \alpha & \cdots & \alpha \\
\alpha & \alpha & \alpha+\beta & \cdots & \alpha \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\alpha & \alpha & \alpha & \cdots & \alpha+\beta
\end{pmatrix},
$$
we can calculate the determinant by first subtracting the second row from the first to get
$$
\det X_n = \det
\begin{pmatrix}
\beta & -\beta & 0 & \cdots & 0 \\
\alpha & \alpha+\beta & \alpha & \cdots & \alpha \\
\alpha & \alpha & \alpha+\beta & \cdots & \alpha \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\alpha & \alpha & \alpha & \cdots & \alpha+\beta
\end{pmatrix},
$$
and then Laplace expand with respect to the first row to obtain
$$
\det X_n =
\beta
\det X_{n-1} + \beta
\det
\begin{pmatrix}
\alpha & \alpha & \cdots & \alpha \\
\alpha & \alpha+\beta & \cdots & \alpha \\
\vdots & \vdots & \ddots & \vdots \\
\alpha & \alpha & \cdots & \alpha+\beta
\end{pmatrix}.
$$
For the second matrix, subtract the first row from all others to get
$$
\det X_n =
\beta
\det X_{n-1} + \beta
\det
\begin{pmatrix}
\alpha & \alpha & \cdots & \alpha \\
0 & \beta & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \beta
\end{pmatrix}
=
\beta \det X_{n-1} + \alpha \beta^{n-1}
$$
Since $\det X_1=\alpha+\beta$, this recursion yields $$\det X_n=n\alpha\beta^{n-1} +\beta^n=\beta^{n-1}(n\alpha+\beta).$$
Thus, $X_n$ is invertible if and only if $\beta\neq 0$ and $n\alpha+\beta\neq 0$. In this case, we may guess that (or consider the adjugate to see that) $X^{-1}$ is of the form $X^{-1} = \gamma \mathbf 1 + \delta E_n$ as well and work out that
$$
X^{-1} = -\frac{\alpha}{\beta(n\alpha+\beta)} \mathbf 1_n + \frac{1}{\beta} E_n.
$$
In the example at hand,
$$
\det B_n = (-nb)^{n-1}(n\left(b+\frac{1}{n}\right)-nb)= (-nb)^{n-1}
$$
which is non-zero if and only if $b$ is non-zero. And
$$
B_n^{-1} = \frac{nb+1}{n^2b} \mathbf 1_n - \frac{1}{nb} E_n.
$$
However, to check that $AB^{-1}A=A$ is satisfied it is enough to show $A^2=AB$, since all matrices of the given form commute.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Showing $x^4-x^3+x^2-x+1>\frac{1}{2}$ for all $x \in \mathbb R$
Show that
$$x^4-x^3+x^2-x+1>\frac{1}{2}. \quad \forall x \in \mathbb{R}$$
Let $x \in \mathbb{R}$,
\begin{align*}
&\mathrel{\phantom{=}}x^4-x^3+x^2-x+1-\frac{1}{2}=x^4-x^3+x^2-x+\dfrac{1}{2}\\
&=x^2(x^2-x)+(x^2-x)+\dfrac{1}{2}=(x^2-x)(x^2+1)+\dfrac{1}{2}\\
&=x(x-1)(x^2+1)+\dfrac{1}{2}.
\end{align*}
Is there any way to solve this question?
|
Ok you already proved it if $x \in (-\infty ,0]\cup [1,+\infty) $ , it suffice to prove it for $x\in (0,1)$ to conclude : you can write it as ,
$(x^{2}-\frac{x}{2})^{2}+\frac{3x^{2}}{4}-x+\frac{1}{2}=(x^{2}-\frac{x}{2})^{2}+\frac{x^{2}}{4}+\frac{1}{2} (x-1)^{2}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\Delta=\left|\begin{smallmatrix} 3x&-x+y&-x+z\\ -y+x&3y&-y+z\\-z+x&-z+y&3z\end{smallmatrix}\right|=3(x+y+z)(xy+yz+zx)$
Prove that$$
\Delta=\begin{vmatrix}
3x&-x+y&-x+z\\
-y+x&3y&-y+z\\
-z+x&-z+y&3z
\end{vmatrix}=3(x+y+z)(xy+yz+zx)
$$
using factor theorem and polynomials.
My Attempt
$$
\begin{matrix}
\color{red}{3x}&\color{blue}{-x+y}&\color{red}{-x+z}&\color{blue}{3x}&\color{red}{-x+y}\\
-y+x&\color{red}{3y}&\color{blue}{-y+z}&\color{red}{-y+x}&3y\\
\color{red}{-z+x}&\color{blue}{-z+y}&\color{red}{3z}&\color{blue}{-z+x}&\color{red}{-z+y}
\end{matrix}
$$
$\Delta$ is a symmetric polynomial of degree $3$.
If we set $x+y+z=0$ we have $\Delta=0$, thus $(x+y+z)$ is a factor, which I can guess from the column operation $C_1\to C_1+C_2+C_3$.
But, how do I find that the remaining factor is $(xy+yz+zx)$ ?
Of course I'll get $\Delta=0$ if I substitute $(xy+yz+zx)=0$, but how do I guess the term in the first place from the fact that $x+y+z$ is one term and $\Delta$ is a symmetric polynomial ?
Similar Problem
Please check the attempted solution by @David Holden in Demonstrate using determinant properties that the determinant of A is equal to $2abc(a+b+c)^3$,
where he has made the substitution $\Delta(a,b,c)=abc(a+b+c)\left(\lambda(a^2+b^2+c^2)+\mu(ab+bc+ca)\right)$ after obtaining $abc(a+b+c)$ as a factor in order to obtain the remaining factor, which is confusing for me. Is there a better way to guess the remaining factor in my case?
|
As you have mentioned, $x+y+z$ is a factor. As the determinant is symmetric, the factor other than $x+y+z$ must also be symmetric. As it is quadratic, it must be in the form $a(x^2+y^2+x^2)+b(xy+yz+zx)$. (We don't have to guess. $a(x^2+y^2+x^2)+b(xy+yz+zx)$ is the general form of symmetric quadratic polynomial in $x$, $y$, $z$.)
Put $x=1$, $y=z=0$, we have $a=0$.
Put $x=y=z=1$, we have $b=3$.
|
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|
Verifying $\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ$ $$\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ.$$
Ηere is what I have so far
$$\sin 4θ = 2\sin 2θ \cos 2θ = 4\sin θ \cos θ \cos 2θ.$$
Not sure if this is the correct path I should take to solve this problem. I have been stuck hard for about an hour now.
|
$$\begin{align}\sin(n+1)x&=\sin nx\cos x+\cos nx\sin x\\
\sin(n-1)x&=\sin nx\cos x-\cos nx\sin x\\
\hline\sin(n+1)x+\sin(n-1)x&=2\sin nx\cos x\end{align}$$
Now that we know that
$$\sin(n+1)x=2\sin nx\cos x-\sin(n-1)x$$
We can go forward to
$$\begin{align}\sin 0x&=0\\
\sin1x&=\sin x\\
\sin2x&=2\sin x\cos x-\sin0x=2\sin x\cos x\\
\sin 3x&=2\sin 2x\cos x-\sin1x=4\sin x\cos^2x-\sin x\\
\sin4x&=2\sin3x\cos x-\sin2x=8\sin x\cos^3x-4\sin x\cos x\\
&=8\sin x\cos^3x-4\sin x\cos x(\cos^2x+\sin^2x)\\
&=4\sin x\cos^3x-4\sin^3x\cos x\end{align}$$
|
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|
number of distinct solution $x\in[0,\pi]$ of the equation satisfy $8\cos x\cos 4x\cos 5x=1$
The number of distinct solution $x\in[0,\pi]$ of the equation which satisfy $8\cos x\cos 4x\cos 5x=1$
Try: $$4\bigg[\cos(6x)+\cos(4x)\bigg]\cos 4x=1$$
$$2\bigg[\cos(10x)+\cos(2x)+1+\cos (8x)\bigg]=1$$
So $$\bigg[\cos(10x)+\cos(8x)+\cos(2x)\bigg]=-1$$
Could some help me to solve it, Thanks
|
Let $\cos{x}=t$.
Thus, we need to solve
$$8t(8t^4-8t^2+1)(16t^5-20t^3+5t)=1$$ or
$$(8t^3+4t^2-4t-1)(8t^3-4t^2-4t+1)(16t^4-16t^2+1)=0,$$
which gives $10$ roots on $[-1,1]$.
We can just solve this equation.
An interesting root: $t=\cos\frac{2\pi}{7}$ or $t=\cos15^{\circ}.$
Indeed, $$16t^4-16t^2+1=0$$ gives
$$1-16\cos^2x(1-\cos^2x)=0$$ or
$$1-4\sin^22x=0$$ or
$$1-2(1-\cos4x)=0$$ or
$$\cos4x=\frac{1}{2}$$ or
$$x=\pm15^{\circ}+90^{\circ}k,$$ where $k=\mathbb Z$ and we get here four roots: $$\{15^{\circ},105^{\circ},75^{\circ},165^{\circ}\}.$$
Now, $$8t^3+4t^2-4t-1=0$$ gives
$$8\cos^3x+4\cos^2x-4\cos{x}-1=0$$ or
$$8\cos^3x-6\cos{x}+4\cos^2x+2\cos{x}-1=0$$ or
$$2\cos3x+2(1+\cos2x)+2\cos{x}-1=0$$ or
$$2(\cos{x}+\cos2x+\cos3x)=-1$$ and since $\sin\frac{x}{2}\neq0$ we obtain:
$$2\sin\frac{x}{2}(\cos{x}+\cos2x+\cos3x)=-\sin\frac{x}{2}$$ or
$$\sin\frac{3x}{2}-\sin\frac{x}{2}+\sin\frac{5x}{2}-\sin\frac{3x}{2}+\sin\frac{7x}{2}-\sin\frac{5x}{2}=-\sin\frac{x}{2}$$ or
$$\sin\frac{7x}{2}=0$$ or
$$x=\frac{360^{\circ}k}{7},$$ where $k\in\mathbb Z$, which gives
$$\left\{\frac{360^{\circ}}{7},\frac{720^{\circ}}{7},\frac{1080^{\circ}}{7}\right\}.$$
The equation $8t^3-4t^2-4t+1=0$ also gives three roots.
Another way.
Since $\sin{x}\neq0,$ by your work we obtain
$$2\sin{x}(\cos10x+\cos8x+\cos2x)=-\sin{x}$$ or
$$\sin11x-\sin9x+\sin9x-\sin7x+\sin3x-\sin{x}=-\sin{x}$$ or
$$\sin11x+\sin3x-\sin7x=0$$ or
$$2\sin7x\cos4x-\sin7x=0,$$ which gives
$$\sin7x=0$$ or $$\cos4x=\frac{1}{2}$$ and we get the same $10$ roots.
|
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|
Definite integration that results in inverse trigonometric functions I try to evaluate this integrat
$$\int_1^{\sqrt{3}}\frac{1}{1+x^2}dx$$
It seems simple.
$$\int_1^{\sqrt{3}}\frac{1}{1+x^2}dx=\arctan(\sqrt{3})-\arctan(1)$$
My question is what exact number it is?
Should it be $\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$?
Or should it be $(k_1\pi+\frac{\pi}{3})-(k_2\pi+\frac{\pi}{4})=(k_1-k_2)\pi+\frac{\pi}{12}$?
$k_1,k_2=0,\pm 1,\pm2,... $
I think the definite integral should be A number, but there seems can be many numbers for the results.
I think I may miss some very basic concption here.
Thank you very much for help.
|
$$
\arctan\sqrt3 - \arctan 1 = \frac \pi 3 - \frac \pi 4 = \frac{4\pi}{12} - \frac{3\pi}{12} = \frac{(4-3)\pi}{12} = \frac \pi {12}.
$$
One of a number of ways to see that this need not involve any of the "nonprincipal" values of the arctangent is this:
$$
\text{If } 1 \le x \le \sqrt 3 \text{ then } \frac 1 2 \ge \frac 1 {1+x^2} \ge \frac 1 4,
$$
$$
\text{so } \frac{\sqrt 3-1}2 \ge \int_1^{\sqrt 3} \frac{dx}{1+x^2} \ge \frac {\sqrt3-1} 4.
$$
|
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|
Powers for the sum of three squares Say $n$ can be represented as a sum of three non-zero squares. (i.e. $n = a^2 +b^2+ c^2$, for some $n,a,b,c \in \mathbb{N}$)
Is it possible that every natural power of $n$ is also a sum of three non-zero
squares? (i.e. $n^k = x^2+y^2+z^2$ for $x,y,z,k \in \mathbb{N}$)
|
Yes, this is possible. Take $n=3$. Clearly $3=1^2+1^2+1^2$. Now any power of $3$ is also the sum of three squares by the Three-Squares Theorem, because $3^k$ is not of the form $4^n(8m+7)$. This may include zero summands, though. Otherwise one has to give a direct solution. If $n=3^k$ with $k=2m+1$, then $3^k=(3^m)^2+(3^m)^2+(3^m)^2$. If $n=3^k$ with $k=2m$, then
$$
3^k=3^{2m}=(3^{m-1})^2+(2\cdot 3^{m-1})^2+(2\cdot 3^{m-1})^2
$$
|
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|
Prove that $3^{2n-1} + 2^{n+1}$ is always a multiple of $7$. I'm trying to prove the following statement:
$P(n) = 3^{2n-1} + 2^{n+1}$ is always a multiple of $7$ $\forall n\geq1$.
I want to use induction, so the base case is $P(1) = 7$ so that's okay.
Now I need to prove that if $P(n)$ is true then $P(n+1)$ is true.
So there exists a $d \in \mathbb{N}$ such that
$$ 3^{2n-1} + 2^{n+1} = 7d $$
From this I need to say that there exists a $k \in \mathbb{N}$ such that:
$$ 3^{2n+1} + 2^{n+2} = 7k $$
With a little algebraic manipulation, I have managed to say:
$$ 2 \cdot 3^{2n+1} + 9 \cdot 2^{n+2} = 7\cdot(18d) $$
But now I am stuck. How should I keep going?
|
If
$$2 \cdot 3^{2n+1} + 9 \cdot 2^{n+2} = 7\cdot 18d$$
Then
$$2 \cdot 3^{2n+1} + 2 \cdot 2^{n+2} = 7\cdot 18d - 7 \cdot 2^{n+2}$$
And we conclude
$$3^{2n+1} + 2^{n+2} = \frac{7(18d - 2^{n+2})}{2} = 7(9d - 2^{n+1})$$
|
{
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|
How to prove that $ \frac{(99)!!}{(100)!!} < \frac{1}{10}$ How to prove that $ \dfrac{(99)!!}{(100)!!}=\dfrac{1\cdot3\cdot5\cdot7\cdot9 \cdots99}{2\cdot4\cdot6\cdot8\cdot10\cdots100} < \dfrac{1}{10}$
Any hint to prove it?
|
$$
(2k-1)!! = \frac{(2k-1)!}{2^{k-1}(k-1)!}
$$
$$
(2k)!! = 2^k k!
$$
$$
\frac{(2k-1)!!}{(2k)!!} = \frac{1}{2^{2k-1}}\frac{(2k-1)!}{k!(k-1)!}=\frac{1}{2^{2k-1}}\frac12\frac{2k(2k-1)!}{k!k(k-1)!}=\frac{1}{2^{2k}}\frac{(2k)!}{k!k!}
= \frac{1}{2^{2k}}\binom{2k}{k}
$$
We can express that central binomial coefficient in terms of Catalan numbers:
$$
\frac{(2k-1)!!}{(2k)!!} = \frac{1}{2^{2k}} (k+1)C_k
$$
The asymptotic form of $C_k$ for $k>2$ is
$$
\frac{4^k}{k^{3/2}\sqrt{\pi}} (1-\frac{9}{8k}) < C_k <\frac{4^k}{k^{3/2}\sqrt{\pi}}
$$
So
$$
\frac{(2k-1)!!}{(2k)!!} < \frac{1}{2^{2k}} (k+1) \frac{4^k}{k^{3/2}\sqrt{\pi}}
= (1+\frac1k)\frac1{\sqrt{k\pi}}
$$
If you plug in $k=33$ this says
$$\frac{65!!}{66!!} < \frac{34}{33\sqrt{33\pi}} \sim 0.0996 < 0.1$$
And of course $\frac{99!!}{100!!}$ is less than that.
|
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|
In this approximation of $\pi$, do you need to know $\pi$ make these calculations? $\pi = 2n\dfrac{\cos (x)}{\sin (x)+1}$
where $x = 90°\dfrac{n-2}{n}$
and $n \to \infty$
A high school student came up with the idea for this approximation of $\pi$, and I helped develop it. It is based on an inscribed polygon. Is this a circular definition? Does it require knowledge of the value of $\pi$ to work?
|
With some trigonometric identities we can rewrite this as $\pi=\lim_{n\to\infty}n\tan\frac{\pi}{n}$, which doesn't require knowledge of $\pi$ provided we consider values of $n$ that are powers of $2$. The insight is that $\tan 2x =\frac{2\tan x}{1-\tan^2 x}$ implies $$\tan 2x\tan^2 x +2\tan x - \tan 2x=0,\,\tan x =\frac{-1+\sqrt{1+\tan^2 2x}}{\tan 2x}$$for small $x>0$. (The sign used in the quadratic formula follows from $\tan 2x \approx 2x,\,\tan x \approx x.$) Now use $\tan\frac{\pi}{4}=1$ to compute $\tan\frac{\pi}{2^k}$ for $k\ge 3$. Whereas the case $k=2$ gives $\pi\approx 4\cdot 1 = 4$, $k=3$ gives $\pi\approx 8\cdot(\sqrt{2}-1)\approx 3.3$.
How good is this approximation? Writing $n=2^k$ we have $$n\tan\frac{\pi}{n}\approx n(\frac{\pi}{n}+\frac{1}{3}(\frac{\pi}{n})^3)=\pi+\frac{\pi^3}{3n^2}=\pi+\frac{\pi^3}{3\cdot 4^k},$$so to get $d$ decimal places right requires $k\approx d\dfrac{\log 10}{\log 4}$.
|
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|
Approximate $\sqrt{(1.02)^3+(1.97)^3}$ using differential So $$\sqrt{(1.02)^3+(1.97)^3}=\sqrt{(1+0.02)^3+(2-0.03)^3}$$
So the differential will be $$\sqrt{1^3+2^3}+\frac{3(1)^2}{2\sqrt{1^3+2^3}}\Delta x+\frac{3(2)^2}{2\sqrt{1^3+2^3}}\Delta y=3+\frac{1}{2}\Delta x+2\Delta y$$
is $\Delta x=0.02 \text{ and } \Delta y=-0.03$ or $\Delta x=(0.02)^3 \text{ and } \Delta y=(-0.03)^3$?
|
For $$\sqrt{(1.02)^3+(1.97)^3}=\sqrt{(1+0.02)^3+(2-0.03)^3}$$
The correct choice is $$\Delta x=0.02 \text{ and } \Delta y=-0.03$$
Note that you have changed your $x$ from $1$ to $1.02$, which means your change in $x$ is
$ \Delta x=0.02$
Similarly for change in $y$ from $2$ to $1.97$ we have $\Delta y=-0.03$
|
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|
Finding all positive integers $x,y,z$ that satisfy $2^x=3^y7^z+1$
Find all positive integers $x,y,z$ that satisfy $$2^x=3^y7^z+1$$.
I think that $(x,y,z)=(6,2,1)$ is the only solution, But how can I prove this?
|
The equation $2^x=m 3^y7^z+1$ can have infinitely many solutions;
We use Fermat little theorem; for primes p and q we may write:
$2^{p-1} ≡ 1 \ mod p$ ⇒ $2^{ k p}≡ 1 \mod p$
$2^{q-1} ≡ 1 \ mod q $⇒ $2^{k q}≡ 1 \mod q$
If $p=7 $ we have:
$2^{7-1}=2^6 ≡ 1 \ mod 7$ ⇒ $2^{6 k_1}≡ 1 \mod 7$
For $p=3$ we have:
$2^{3-1}=2^2 ≡ 1 \mod 3$ ⇒ $2^{2 k_2}≡ 1 \mod 3$
Therefore:
$2^{6k}≡ 1 \mod 3\times 7$
That is one set of solutions for equation $2^x=m 3^y7^z+1$is: $(x, y, z)=(6k, 1, 1); k ∈ N$
Also we can see that $2^{6k}≡ 1 \mod 3^2 \times 7$, that is another set of solutions of equation $2^x=m 3^y7^z+1$ is:$(x, y, z)=(6k, 2, 1); k ∈ N$
Here m=1, then equation is $2^x= 3^y7^z+1$, the minimum value of x is when $k=1$ so $x=6$ and we have:
$2^6 -1 =3^y 7^z$ which its only solution is:(x,y,z)=(6,2,1) wich is a member of second set of solutions.
|
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|
Solving $x^4-15x^2-10x+ 24 = 0$ using Ferrari’s method
Ferrari’s method for solving a quartic equation
$$x^4-15x^2-10x+ 24 = 0$$
begins by writing:$$x^4= 15x^2+ 10x-24$$and then adding a term of
the form:$$-2bx^2+b^2$$to both sides.
(a) Explain why this is good idea and what it accomplishes.
(b) Use $b= 7$ to find the two quadratic equations that yield the
solutions.
$(a)$
I have
$$\begin{align*}x^4-2bx^2+b^2&=15x^2+ 10x-24-2bx^2+b^2\\\\
&=(15-2b)x^2+ 10x+(b^2-24)
\end{align*}$$
I then notice that $$x^4-2bx^2+b^2=(x^2-b)^2$$
so we get
$$(x^2-b)^2=(15-2b)x^2+ 10x+(b^2-24)$$
Pluggin in $b=7$ I get
$$(x^2-7)^2=x^2+ 10x+25$$
but I fail to see why this helps with regards to solving for $x$.
Any hints to lean me in the direction would be much appreciated.
|
$$(x^2-7)^2 = (x+5)^2$$
$$(x^2-7)^2-(x+5)^2=0$$
$$(x^2-7-x-5)(x^2-7+x+5)=0$$
|
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|
Studying the extrema of $f(x,y) = x^4 + y^4 -2(x-y)^2$
Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ such that $f(x,y) = x^4 + y^4 - 2(x-y)^2$. Study its extrema.
So here was my approach.
We have $$\frac{\partial f}{\partial x}(x,y) = 4(x^3 -x + y),\frac{\partial f}{\partial y}(x,y)= 4(y^3 -y + x) $$
I have to find $(x_0,y_0) \in \mathbb{R}^2$ such that $ \frac{\partial f}{\partial x}(x_0,y_0)= \frac{\partial f}{\partial y}(x_0,y_0) = 0 $
So we have:
$$\left\{\begin{matrix}
x(x^2-1)+ y =0\\
y(y^2 -1) + x =0
\end{matrix}\right. $$
Thus we have $x-x^3 = y$, and by replacing $y$ with $x-x^3$ in the second line, we get:
$$ y(y^2 -1) + x =0 = (x-x^3)((x-x^3)^2 -1)+x =x^5(-x^4 -x^2 -3) = 0 $$ And the only solution for this is $x=0$. As $y = x^3 -x$ we immediately have $y=0$.
So the only extremum possible is at $(0,0)$. Now, I need to study its Hessian matrix.
We have: $$\frac{\partial ^2f}{\partial x^2}(x,y) = 12x^2 -4 , \frac{\partial ^2f}{\partial y^2}(x,y) = 12y^2 -4, \frac{\partial^2 f}{\partial x \partial y}(x,y) = \frac{\partial^2 f}{\partial y \partial x}(x,y) = 4$$
Thus $$H(x,y) \begin{bmatrix}
12x^2 -4&4 \\
4& 12y^2 -4
\end{bmatrix} $$
At $(x,y)=(0,0)$ we have
$$H(0,0) \begin{bmatrix}
-4&4 \\
4& -4
\end{bmatrix} $$
As we have $\text{det}(H(0,0)) = 0$ and $\text{Tr}(H(0,0)) = -8$ the eigenvalues are $0$ and $-8$. As it has $0$ as eigenvalue, I need to study the differential at a higher order. But here I lack understanding. What exactly should I do? Should I compute the third order partial differentials and then restudy their Hessian Matrix?
|
The matrix $H(0,0)$ is negative semidefinite since
$$\left\langle \begin{bmatrix}-4 & 4 \\ 4 & -4\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}, \begin{bmatrix}x \\ y\end{bmatrix}\right\rangle = -4(x-y)^2 \le 0$$
with equality iff $x = y$.
This is a necessary condition for a local maximum, but not sufficient. Therefore, the test is inconclusive.
Indeed, $(0,0)$ is a saddle point. If we approach $(0,0)$ along $x = y$ then we have
$$f(x,x) = 2x^4 > 0$$
However, if we approach along $x = -y$, then
$$f(x,-x) = 2x^4-8x^2 < 0$$
near $(0,0)$.
|
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"url": "https://math.stackexchange.com/questions/2697218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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|
How to evaluate this anti-derivative? How to evaluate
$$ \int \frac{1}{ \ln x} \ \mathrm{d} x, $$
where $\ln x$ denotes the natural logarithm of $x$?
My effort:
We note that
$$ \int \frac{1}{ \ln x} \ \mathrm{d} x = \int \frac{x}{x \ln x} \ \mathrm{d} x = \int x \frac{ \mathrm{d} }{ \mathrm{d} x } \left( \ln \ln x \right) \ \mathrm{d} x = x \ln \ln x - \int \ln \ln x \ \mathrm{d} x. $$
What next?
Another approach:
We can also write
$$ \int \frac{1}{ \ln x } \ \mathrm{d} x = \frac{x}{\ln x } + \int \frac{1}{ \left( \ln x \right)^2 } \ \mathrm{d} x = \frac{x}{\ln x } + \frac{x}{ \left( \ln x \right)^2 } + \int \frac{ 2 }{ \left( \ln x \right)^3 } \ \mathrm{d} x = \ldots = x \sum_{k=1}^n \left( \ln x \right)^{-k} + n \int \left( \ln x \right)^{-n-1} \ \mathrm{d} x. $$
What next?
Which one of the above two approaches, if any, is going to lead to a function consisting of finitely many terms comprised of elementary functions, that is, the kinds of solutions that we are used to in calculus courses?
Or, is there any other way that can lead us to a suitable enough answer?
|
I = $\large\int\frac{1}{ln(x)}dx$
let ln(x) = u
$\,e^u = x$
$\,dx = e^udu$
I = $\,\int \frac{e^u}{u}du$
expanding e$^u$,
I=$\,\int\frac{1+u+\frac{u^2}{2!}+\frac{u^3}{3!}+\frac{u^4}{4!}+\frac{u^5}{5!}..............}{u}du$
I = $\,\int\frac{1}{u}+1+\frac{u}{2!}+\frac{u^2}{3!}+\frac{u^3}{4!}+\frac{u^4}{5!}...........du$
I = $\,ln(u)+ u + \frac{u^2}{2.2!}+\frac{u^3}{3.3!}+\frac{u^4}{4.4!}+\frac{u^5}{5.5!}......+ C$
I = $\,ln(ln(x)) + ln(x) + \frac{(ln(x))^2}{2.2!} + \frac{(ln(x))^3}{3.3!} + \frac{(ln(x))^4}{4.4!} + \frac{(ln(x))^5}{5.5!} + ....... + C $
I = $\,ln(ln(x)) + ln(x) + \sum_{k=2}^\infty \frac{(ln(x))^k}{k.k!} + C $
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Algebra precalculus factorization I must factor this out:$(a+1)(a+3)(a+5)(a+7)+15$
I know that it might have an artifice or technique to transform those factors, but i cannot find any idea on how to do it.
PS.The majority of my questions are about factoring polynomials. Is there an app or site on internet where I can see step by step the resolution of these exercices so then I won't bother you guys anymore with a lot of these kind of silly questions?
|
take $$ b = a+4 $$
$$ (b-3)(b-1)(b+1)(b+3) + 15 $$
$$ (b-1)(b+1)(b-3)(b+3) + 15 $$
$$ (b^2-1)(b^2-9) + 15 $$
$$ b^4 - 10 b^2 + 9 + 15 $$
$$ b^4 - 10 b^2 + 24 $$
take $u = b^2$
$$ u^2 - 10 u + 24 $$
$$ (u - 4)(u - 6) $$
$$ (b^2 - 4)(b^2 - 6) $$
$$ (a^2 + 8 a + 16 - 4) (a^2 + 8 a + 16 - 6) $$
$$ (a^2 + 8 a + 12) (a^2 + 8 a + 10) $$
$$ (a + 6)(a+2) (a^2 + 8 a + 10) $$
The roots of $a^2 + 8 a + 10$ are real numbers but not rational, $-4 \pm \sqrt 6$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
To compute the sum using 2011 th roots of unity
Question:
Let $x$ be a complex number such that $x^{2011}=1$ and $x\neq1$ then, compute the sum
$$S=\dfrac{x^2}{x-1}+\dfrac{x^4}{x^2-1}+\dfrac{x^6}{x^3-1}+\dots+\dfrac{x^{4020}}{x^{2010}-1}$$
My attempt:
$$x=(1)^{\frac{1}{2011}}\implies x=1,e^{\frac{2\pi i}{2011}},e^{\frac{4\pi i}{2011}},\dots,e^{\frac{2010\pi i}{2011}}$$
let, $\alpha= e^{\frac{2\pi i}{2011}},$ then roots of given equation are $x=\alpha,\alpha^2,\alpha^3,\dots,\alpha^{2010}$
also
$$\alpha^{2010}=\dfrac{1}{\alpha};\ \alpha^{2009}=\dfrac{1}{\alpha^2};\ \dots;
\ \alpha^{1006}=\dfrac{1}{\alpha^{1005}}$$
now, we have to compute:
$$S= \dfrac{\alpha^2}{\alpha-1}+\dfrac{\alpha^4}{\alpha^2-1}+\dfrac{\alpha^6}{\alpha^3-1}+\dots+\dfrac{\alpha^{4020}}{\alpha^{2010}-1}$$
then, I combined first term and last term, similarly second term and second last term and so on in this way i ended up here
$$S=\dfrac{\alpha^{2}-\alpha^{-3}}{\alpha-1}+\dfrac{\alpha^{4}-\alpha^{-6}}{\alpha^2-1}+.........+\dfrac{\alpha^{2010}-\alpha^{-1006}}{\alpha^{1005}-1}$$
here i got stuck , somehow i just wanted to exploit the property that sum of all roots of unity (who itself are in G.P) is zero but I couldn't
please give me hint or provide right way to solve this question.
thank you!
edit:
I also tried taking logarithmic derivative of $f(x)=x^{2011}-1$ but no progress further
|
Supposing that $\zeta$ is a primitive $n$th root of unity we consider
$$f(z) = \frac{z^2}{z-1} \frac{n z^{n-1}}{z^n-1}
= \frac{1}{z-1} \frac{n z^{n+1}}{z^n-1}
= \frac{n}{z-1} \left(z + \frac{z}{z^n-1}\right)
\\ = n \left(1 + \frac{1}{z-1} \right)
\left(1 + \frac{1}{z^n-1}\right).$$
We seek $$S_n = \sum_{k=1}^{n-1} \frac{\zeta^{2k}}{\zeta^k-1}.$$
We have by inspection that
$$S_n = \sum_{k=1}^{n-1} \mathrm{Res}_{z=\zeta^k} f(z).$$
With residues summing to zero this implies
$$S_n = - \mathrm{Res}_{z=1} f(z)
- \mathrm{Res}_{z=\infty} f(z).$$
For the first residue we note that
$$\frac{1}{z^n-1} = \frac{1}{n} \frac{1}{z-1} + \cdots$$
and the constant term is
$$\left. \frac{1}{z^n-1}
- \frac{1}{n} \frac{1}{z-1} \right|_{z=1}
\\ = \left. \frac{1}{z-1}
\left(\frac{1}{1+z+\cdots+z^{n-1}} - \frac{1}{n}\right)\right|_{z=1}$$
which is by L'Hopital
$$\left. - \frac{1+2z+\cdots+(n-1)z^{n-2}}{(1+z+\cdots+z^{n-1})^2}
\right|_{z=1} = - \frac{1}{2} (n-1) n \frac{1}{n^2}
= - \frac{1}{2n} (n-1)$$
and we may write
$$f(z) = n \left(1 + \frac{1}{z-1} \right)
\left(1 + \frac{1}{n} \frac{1}{z-1} - \frac{1}{2n} (n-1) \cdots\right)$$
so that the residue is
$$n \left( 1 + \frac{1}{n} - \frac{1}{2n} (n-1) \right) =
n + 1 - \frac{1}{2} (n-1) = \frac{1}{2} n + \frac{3}{2}.$$
For the residue at infinity we find with $n\ge 2$
$$- \mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z)
= -n \mathrm{Res}_{z=0} \frac{1}{z^2}
\left(1 + \frac{1}{1/z-1} \right)
\left(1 + \frac{1}{1/z^n-1}\right)
\\ = -n \mathrm{Res}_{z=0} \frac{1}{z^2}
\left(1 + \frac{z}{1-z} \right)
\left(1 + \frac{z^n}{1-z^n}\right)
\\ = -n [z^1]
\left(1 + \frac{z}{1-z} \right)
\left(1 + \frac{z^n}{1-z^n}\right)
= - n.$$
Collecting everything yields the closed form
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2} n - \frac{3}{2}.}$$
This will produce $1004$ for $n=2011$, confirming the answer by
@JackDAurizio that appeared first.
|
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|
Finding integer values of $n$
Finding integer values of $n$ for which the equation
$$x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0$$
has at least one integer solution.
Try: let $\alpha,\beta,\gamma$ be the roots of the equation. Then
$\alpha+\beta+\gamma=-(n+1)$ and $\alpha\beta+\beta\gamma+\gamma\alpha=1-2n$ and $\alpha\beta\gamma=2n^2+n+4$.
Now i did not understand how to solve it. Could some help me to solve it , Thanks
|
Suppose $x,n$ are integers such that
$$x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0\tag{eq1}$$
Suppose first that $n$ is odd.
Then, reducing mod $2$, we can replace $n$ by $1$, so
\begin{align*}
&x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0\\[4pt]
\implies\;&x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0\;(\text{mod}\;2)\\[4pt]
\implies\;&x^3-2x^2-x -7\equiv 0\;(\text{mod}\;2)\\[4pt]
\implies\;&x^3-x\equiv 1\;(\text{mod}\;2)\\[4pt]
\implies\;&0\equiv 1\;(\text{mod}\;2)\\[4pt]
\end{align*}
contradiction.
Hence $n$ must be even.
Then, reducing mod $2$, we can replace $n$ by $0$, so
\begin{align*}
&x^3+(n+1)x^2-(2n-1)x-(2n^2+n+4)=0\\[4pt]
\implies\;&x^3+x^2+x-4=0\;(\text{mod}\;2)\\[4pt]
\implies\;&x^3+x^2+x\equiv 0\;(\text{mod}\;2)\\[4pt]
\implies\;&x\equiv 0\;(\text{mod}\;2)\\[4pt]
\end{align*}
so $x$ must be even.
Writing $x=2w$, and $n=2m$, $(\text{eq}1)$ reduces to
$$4m^2 - (4w^2-4w-1)m -(4w^3+2w^2+w-2) = 0\tag{eq2}$$
Then, regarding $(\text{eq}2)$ as a quadratic equation in $m$, it follows that the discriminant
$$D = 16w^4+32w^3+40w^2+24w-31$$
is a perfect square.
Noting that $D$ is odd, $D$ must be an odd square
Identically, we have
$$
\begin{cases}
D = (4w^2+4w+3)^2 - 40\\[4pt]
D = (4w^2+4w+1)^2 + 16(w+2)(w-1)\\
\end{cases}
$$
hence we must have $-2 \le w \le 1$, else $D$ would be trapped strictly between
$(4w^2+4w+1)^2$ and $(4w^2+4w+3)^2$, which are consecutive odd squares.
If $w=-1$ or $w=0$, then $D < 0$, contradiction.
For $w=-2$, $(\text{eq}2)$ reduces to $(4m-7)(m-4)=0$, which yields the solution $(w,m)=(-2,4)$ for $(\text{eq}2)$, and the corresponding solution $(x,n)=(-4,8)$ for $(\text{eq}1)$.
For $w=1$, $(\text{eq}2)$ reduces to $(4m+5)(m-1)=0$, which yields the solution $(w,m)=(1,1)$ for $(\text{eq}2)$, and the corresponding solution $(x,n)=(2,2)$ for $(\text{eq}1)$.
Thus, $n=2$ and $n=8$ are the only values of $n$ for which $(\text{eq}1)$ has at least one integer solution for $x$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Validity of proving identities by showing LHS-RHS =0 or using reversible steps? When proving $\mathrm{LHS}=\mathrm{RHS}$, the most common way of doing it is by manipulating it in such a way to show that $\mathrm{LHS}$ equals to some expression which equals to $\mathrm{RHS}$. But what about these methods:
Method 1:
Showing that $\mathrm{LHS}-\mathrm{RHS}=0$
For instance, if we are required to prove
$x^2\cos^2(x)+\sin^2(x) = x^2 - x^2\sin^2(x) + 1 - \cos^2(x)$
We instead show that $\mathrm{LHS}-\mathrm{RHS}=0$ as follows:
$\mathrm{LHS} - \mathrm{RHS} = x^2\cos^2(x)+\sin^2(x) - x^2 + x^2\sin^2(x) - 1 + \cos^2(x)$
$ = x^2(\cos^2(x) + \sin^2(x)) - x^2 + (\sin^2(x) + \cos^2(x)) - 1$
$= x^2 - x^2 + 1 - 1 = 0$
Method 2: Showing that $\mathrm{LHS}=\mathrm{RHS}$ is equivalent with another equation which is true (taking care that we can always reverse the steps and showing it by putting $\iff$):
$x^2\cos^2(x)+\sin^2(x) = x^2 - x^2\sin^2(x) + 1 - \cos^2(x)$
$ \iff x^2\cos^2(x)+x^2\sin^2(x) - x^2 = 1 - \sin^2(x) - \cos^2(x)$
$ \iff x^2 - x^2 = 1 - 1$
$ \iff 0 = 0$
Are these two methods of proving valid? Are there any cases where we can make fallacious argument by using these methods? Are any one of them better than the other?
|
This question works both from LHS $\rightarrow$ RHS and RHS $\rightarrow$ LHS to prove it.
|
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|
Differentiate $y=\sin^{-1}\left(2x\sqrt{1-x^2}\right),\quad\frac{-1}{\sqrt{2}}
Find $\dfrac{\mathrm dy}{\mathrm dx}$ if
$y=\sin^{-1}\left(2x\sqrt{1-x^2}\right),\quad\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}$
I can solve it as follows:
$$
\begin{align}
y'&=\frac{1}{\sqrt{1-4x^2(1-x^2)}}\frac{d}{dx}\Big(2x\sqrt{1-x^2}\Big)\\&=\frac{1}{\sqrt{1-4x^2+4x^4}}\bigg(2x\frac{-2x}{2\sqrt{1-x^2}}+2\sqrt{1-x^2}\bigg) \\
&=\frac{1}{\sqrt{(1-2x^2)^2}}\bigg(\frac{-2x^2}{\sqrt{1-x^2}}+2\sqrt{1-x^2}\bigg)\\
&=\frac{1}{|1-2x^2|}\frac{-2x^2+2-2x^2}{\sqrt{1-x^2}}\\
&=\frac{2(1-2x^2)}{|1-2x^2|\sqrt{1-x^2}}\\
\end{align}
$$
As $\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\implies|x|<\frac{1}{\sqrt{2}}\implies 0<x^2<\frac{1}{2}\implies 0<2x^2<1\\\implies-1<-2x^2<0\implies 0<1-2x^2<1$
Thus, $|1-2x^2|=1-2x^2$
$$
y'=\frac{2(1-2x^2)}{(1-2x^2)\sqrt{1-x^2}}=\frac{2}{\sqrt{1-x^2}}
$$
My Attempt
But if I try to solve it by substituting $x=\sin\alpha\implies\alpha=\sin^{-1}x$
$$
y=\sin^{-1}\bigg(2\sin\alpha\sqrt{\cos^2\alpha}\bigg)=\sin^{-1}\bigg(2\sin\alpha|\cos\alpha|\bigg)
$$
Here, $\frac{-1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}\implies \frac{-\pi}{4}<\sin^{-1}x=\alpha<\frac{\pi}{4}\implies \cos\alpha>0\implies|\cos\alpha|=\cos\alpha$
$$
\begin{align}
y&=\sin^{-1}\bigg(2\sin\alpha\cos\alpha\bigg)=\sin^{-1}\bigg(\sin2\alpha\bigg)\\&\implies\sin y=\sin2\alpha=\sin\Big(2\sin^{-1}x\Big)\\
&\implies y=n\pi+(-1)^n(2\sin^{-1}x)=\begin{cases}n\pi+2\sin^{-1}x,\quad \text{n even}\\
n\pi-2\sin^{-1}x,\quad \text{n odd}
\end{cases}
\end{align}
$$
Thus,
$$
y'=\begin{cases}\frac{d}{dx}\Big(n\pi+2\sin^{-1}x\Big)=\frac{2}{\sqrt{1-x^2}},\quad \text{n even}\\
\frac{d}{dx}\Big(n\pi-2\sin^{-1}x\Big)=\frac{-2}{\sqrt{1-x^2}},\quad \text{n odd}
\end{cases}
$$
How do I eliminate the case $y'=\frac{-2}{\sqrt{1-x^2}}$ ?
|
The idea is very good. However, the notation $\sin^{-1}t$ usually doesn't mean “the set of all angles $\varphi$ such that $\sin\varphi=t$”, but rather
$\sin^{-1}t$ denotes the unique angle $\phi\in[-\pi/2,\pi/2]$ such that $\sin\varphi=t$.
The fact that $\sin^{-1}$ is not the inverse function of the sine is the reason why many people, including myself, prefer to use $\arcsin t$ instead of the logically wrong $\sin^{-1}t$. It is the inverse function of a restriction of the sine function.
This way, for $\varphi\in[-\pi/2,\pi/2]$ and $t\in[-1,1]$ it holds that
$$
\sin(\sin^{-1}t)=t
\qquad
\sin^{-1}(\sin\varphi)=\varphi
$$
If $\alpha=\sin^{-1}x$, then indeed $-\pi/4<\alpha<\pi/4$, so $\cos\alpha>0$. Thus
$$
\sqrt{1-x^2}=\sqrt{1-\sin^2\alpha}=\cos\alpha
$$
Then $2x\sqrt{1-x^2}=2\sin\alpha\cos\alpha=\sin2\alpha$ and $-\pi/2<2\alpha<\pi/2$.
Therefore $\sin^{-1}(\sin2\alpha)=2\alpha=2\sin^{-1}x$ and so
$$
y'=\frac{2}{\sqrt{1-x^2}}
$$
|
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|
Minimum of product of three trigonometric function
If $x,y,z$ are real numbers with $x\geq y\geq z \geq 15^\circ$ and $x+y+z=90^\circ$, then find the range of $\cos x\cos y\cos z$.
I tried Jensen inequality and AM-GM inequality:
$$3(\cos x\cos y\cos z)^{\frac{1}{3}}\leq \cos x+\cos y+\cos z\leq 3 \cos\bigg(\frac{x+y+z}{3}\bigg)=\frac{3\sqrt{3}}{2},$$
so
$$ \cos x\cos y\cos z\leq \frac{3\sqrt{3}}{8}.$$
Equality holds when $x=y=z=30^\circ$.
Also,$$ \cos x\cos y\cos z=\frac{1}{4}\bigg[\sin(2x)+\sin(2y)+\sin(2x+2y)\bigg].$$
I am unable to find the minimum of $\cos x\cos y\cos z$.
|
Note that $x\geqslant y\geqslant z\geqslant15°$ and $x + y + z = 90°$ implies $2y + z\leqslant90°$ and $15°\leqslant z\leqslant30°$, then\begin{align*}
\cos x\cos y\cos z &= \frac12(\cos(x + y) + \cos(x - y))\cos z\\
&= \frac12(\cos(90° - z) + \cos(90° - 2y - z))\cos z\\
&= \frac12(\sin z + \sin(2y + z))\cos z \geqslant \frac12(\sin z + \sin 3z)\cos z\\
&= \sin 2z\cos z\cdot\cos z = 2\sin z\cos^3 z.
\end{align*}
Now define $f(z) = \sin z\cos^3 z$. For $15°\leqslant z\leqslant30°$,\begin{align*}
f'(z) &= \cos^4 z - 3\sin^2 z\cos^2 z = (\cos^2 z - 3\sin^2 z)\cos^2 z\\
&= (1 - 4\sin^2 z)\cos^2 z \geqslant 0,
\end{align*}
thus denoting $z_0 = 15°$, then\begin{align*}
&\mathrel{\phantom{=}}{} \cos x\cos y\cos z \geqslant 2\sin z\cos^3 z \geqslant 2\sin z_0\cos^3 z_0\\
&= \sin 2z_0 \cos^2 z_0 = \sin 2z_0 \cdot \frac12(\cos 2z_0 + 1)\\
&= \frac12 \cdot \frac12\left( \frac{\sqrt3}2 + 1 \right) = \frac{2 + \sqrt3}8.
\end{align*}
Because for $(x, y, z) = (60°, 15°, 15°)$, $\cos x\cos y\cos z = \dfrac{2 + \sqrt3}8$, then the minimum is indeed $\dfrac{2 + \sqrt3}8$.
|
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|
Graph of the equation The graph of the equation $$ x^2y^3=(2x+3y)^5$$
is same as $$x=-y$$ Why is this so?
I understand it satisfies the equation, but is there a way to derive it? Don't the other roots count?
The graph
Edit:
Something I just realized.
$$ dy/dx=y/x$$ for the equations $$x^my^n=(x+y)^{m+n}$$
we have the solution $$|y/x|=e^c$$ where c is a constant. Perhaps something similar is happening here?
Is there anything more general than this? Because it feels like a complex equation but turns out to be quite simple.
|
$$
\begin{aligned}
(2x + 3y)^5 - x^2y^3 &= (2(x+y)+y)^5 - x^2y^3\\
&= (x+y)(32(x+y)^4 + 80(x+y)^3y + 80(x+y)^2y^2 + 40(x+y)y^3 + 10y^4) \\
&\qquad - y^3(x^2 - y^2)\\
&= (x+y)(32(x+y)^4 + 80(x+y)^3y + 80(x+y)^2y^2 + 40(x+y)y^3 + 10y^4) \\
&\qquad - y^3(x+y)(x+y-2y)\\
&= (x+y)(32(x+y)^4 + 80(x+y)^3y + 80(x+y)^2y^2 + 39(x+y)y^3 + 12y^4)
\end{aligned}
$$
Substitute $s = \frac{x+y}{y}$ you get an equation $32s^4 + 80s^3 + 80s^2 + 39s + 12 = 0$. Wolfram alpha tells you this equation has only complex solutions, so the second parenthesis in the first equation is positive unless $x=y=0$.
|
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|
Partial fraction decomposition with a 3rd degree numerator I have the following function to decompose using PFD:
$H(x) = \frac{x^3 + 4x^2 - 11x - 48}{x^3 + 6x^2 + 3x - 10}$
The poles are $1$, $-2$ and $-5$ so I tried to do it this way: finding $a$, $b$ and $c$ such as:
$H(x) = \frac{a}{x-1} + \frac{b}{x+2} + \frac{c}{x+5}$
But it didn't give me a correct result. Doing it that way gives me $(a,b,c) = (-3, 2, -1) $
Is there a specific form of numerator to introduce here? Thanks.
|
Since the degree of the numerator is $\geq$ to the degree of the denominator you should first divide them. Hence
$$\frac{x^3 + 4x^2 - 11x - 48}{x^3 + 6x^2 + 3x - 10}=1-\frac{2x^2 + 14x + 38}{x^3 + 6x^2 + 3x - 10}=1+\frac{a}{x-1} + \frac{b}{x+2} + \frac{c}{x+5}$$
for some real constants $a$, $b$, $c$ (the same that you already found by using the residue method!).
|
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|
Let $a$ be the real root of the equation $x^3+x+1=0$ Let $a$ be the real root of the equation $x^3+x+1=0$
Calculate $$\sqrt[\leftroot{-2}\uproot{2}3]{{(3a^{2}-2a+2)(3a^{2}+2a)}}+a^{2}$$
The correct answer should be $ 1 $. I've tried to write $a^3$ as $-a-1$ but that didn't too much, I guess there is some trick here :s
|
The real root $a$ of $x^3+x+1$ belongs to the interval $\left(-1,-\frac{1}{2}\right)$ and in order to check that
$$ \sqrt[3]{(3a^2-2a+2)(3a^2+2a)}=1-a^2 \tag{1}$$
holds it is enough to check that
$$ (3a^2-2a+2)(3a^2+2a)=1-3a^2+3a^4+a^6 \tag{2}$$
holds, or that
$$ (1+a+a^3)(-1+5a+a^3) = 0 \tag{3} $$
holds, but that is trivial.
|
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|
Why $\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}=\ln\frac{1+\sin\frac{5π}{12}}{1-\sin\frac{5π}{12}}$? Why does an integral $$\int \frac{dx}{1-x^2}$$ with the limitless (undefined) interval equal to $$\frac 12\ln\frac{1+x}{1-x},$$ yet an integral $$\int\limits_{\sin(-5π/12)}^{\sin(5π/12)}\frac{dx}{1-x^2}$$ with an interval from $\sin\frac{-5π}{12}$ to $\sin\frac{5π}{12}$ has $$\ln\frac{1+\sin\frac{5π}{12}}{1-\sin\frac{5π}{12}}$$ without one half attached to ln?
|
$$\int \frac{dx}{1-x^2} = $$
$$\frac {1}{2} \int \bigg(\frac{1}{1+x} + \frac{1}{1-x} \bigg) dx =$$
$$\frac 12\ln\frac{1+x}{1-x}$$Upon Evaluation at the upper and lower limits, for $$a=\sin (5\pi /{12})$$
Note that $$-a=\sin (-5\pi /{12})$$
Thus $$\frac 12\ln\frac{1+x}{1-x}\bigg|_{-a}^a = $$
$$\frac 12\bigg(\ln\frac{1+a}{1-a} -\ln\frac{1-a}{1+a}\bigg)=$$
$$\frac 12\ln \bigg(\frac{1+a}{1-a}\bigg)^2=$$
$$\ln \bigg(\frac{1+a}{1-a}\bigg)$$
|
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|
How to evaluate $\int_0^1 \frac{1-x}{\ln x}(x+x^2+x^{2^2}+x^{2^3}+x^{2^4}+\ldots) \, dx$? Evaluate the definite integral:
$$\int_0^1 \frac{1-x}{\ln x}(x+x^2+x^{2^2}+x^{2^3}+x^{2^4}+\ldots) \, dx$$
I think the series involving $x$ converges because $x\in[0,1]$, but I cannot form an expression for the series. If I let
$$
u_n=x^{2^{n-1}} \\ \frac{\ln u_n}{\ln x}=2^{n-1}
$$
but then this series does not converge. Even WolframAlpha cannot evaluate a definite integral together with an infinite series, so I am stuck on this.
|
Claim 1: For $k\geq 1$, we have that
\begin{align}
\int^1_0 \frac{1-x}{\log x}x^{2^k}\ dx = -\log \frac{2^k+2}{2^k+1}.
\end{align}
Claim 2: We have
\begin{align}
\prod^\infty_{k=0}\left( 1+\frac{1}{2^k+1}\right) = 3
\end{align}
Using the claims, we have the series
\begin{align}
-\sum^\infty_{k=0} \log \left(\frac{2^k+2}{2^k+1}\right) =-\log\left(\prod^\infty_{k=0} \left(1+\frac{1}{2^k+1}\right) \right) =-\log 3.
\end{align}
|
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|
Find $\frac{dy}{dx}$ if $y=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0
Find derivative of $f(x)=\sin^{-1}[x\sqrt{1-x}-\sqrt{x}\sqrt{1-x^2}]$, $0<x<1$
Let $x=\sin a$ and $\sqrt{x}=\cos b$
Then I'll get:
$$
y=\sin^{-1}[\sin a\cos b-\cos a\sin b]=\sin^{-1}[\sin(a-b)]\\
\implies\sin y=\sin(a-b)\\
\implies y=n\pi+(-1)^n(a-b)=n\pi+(-1)^n(\sin^{-1}x-\sin^{-1}\sqrt{x})
$$
Thus,
$$
y'=\frac{d}{dx}\big[n\pi+(-1)^n(a-b)\big]=\begin{cases}\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}\text{ if }n\text{ is even}\\
-\bigg[\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}\bigg]\text{ if }n\text{ is odd}
\end{cases}
$$
Is it the right way to solve this problem and how do I check the solution is correct ?
Note: I think there got to be two cases for the derivative as the graph of the function is
|
Let $\sin a=x\implies a=\sin^{-1}x$ and $\sin b=\sqrt{x}\implies b=\sin^{-1}\sqrt{x}$
$$
y=\sin^{-1}\Big[ \sin a|\cos b|-\sin b.|\cos a| \Big]\\
$$
$0<x<1\implies 0<\sin^{-1}x=a<\frac{\pi}{2}\implies |\cos a|=\cos a$ and
$0<x<1\implies 0<\sqrt{x}<1\implies0<\sin^{-1}\sqrt{x}=b<\frac{\pi}{2}\implies|\cos b|=\cos b$
$$
\begin{align}
y&=\sin^{-1}\Big[\sin a\cos b-\cos a\sin b\Big]\\
&=\sin^{-1}\big[\sin(a-b)\big]
\end{align}
$$
We have $0<x<\frac{\pi}{2}$ and $0<b<\frac{\pi}{2}\implies \frac{-\pi}{2}<-b<0$, Hence $\frac{-\pi}{2}<a-b<\frac{\pi}{2}$
$$
y=\sin^{-1}\big[\sin(a-b)\big]=a-b=\sin^{-1}x-\sin^{-1}\sqrt{x}
$$
$$
\begin{align}
\color{blue}{\frac{dy}{dx}}&\color{blue}{=\frac{1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x}\sqrt{1-x}}}\\
&=\frac{2\sqrt{x}\sqrt{1-x}-\sqrt{1-x^2}}{2\sqrt{x}\sqrt{1-x}\sqrt{1-x^2}}
=\frac{\sqrt{1-x}(2\sqrt{x}-\sqrt{1+x})}{2\sqrt{x}\sqrt{1-x}\sqrt{1-x^2}}
\end{align}
$$
$y'<0$ when $2\sqrt{x}<\sqrt{1+x}\implies 4x<1+x\implies 0<x<\frac{1}{3}$
$y'<0$ when $\frac{1}{3}<x<1$
|
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|
Integration by Substitution of Fraction involving e
Find $\int\frac{2}{e^{2x}+4}$ using $u=e^{2x}+4$
The answer is $\frac{1}{2}x-\frac{1}{4}\ln(e^{2x}+4)+c$
I must have made a mistake somewhere as my answer is not the same. Apologies the question may be too specific, but I am teaching myself calculus.
$\int\frac{2}{e^{2x}+4}$
let $u = e^{2x} +4$
$\frac{dy}{dx}=2e^{2x}$
$dx=\frac{1}{2}e^{-2x}du$
$u = e^{2x} +4$
$e^{2x} = u-4$
$e^{-2x} = \frac{1}{u-4}$
$dx=\frac{1}{2}(\frac{1}{u-4})du$
Hence the integral is:
$$
\begin{aligned}
&\int\frac{2}{e^{2x}+4} \\
=& 2\int\frac{1}{u}\frac{1}{2}\left(\frac{1}{u-4}\right)du\\
=&\int\frac{1}{u}\left(\frac{1}{u-4}\right)du\\
=&\int\frac{1}{u^2-4u}du\\
=&\int u^{-2}-\frac{1}{4}u^{-1}du\\
=&\frac{u^{-1}}{-1}-\frac{1}{4}\ln(u)+c\\
=&-\frac{1}{e^{2x}+4}-\frac{1}{4}\ln(e^{2x}+4)+c\\
=&-e^{-2x}-\frac{1}{4}-\frac{1}{4}\ln(e^{2x}+4)+c\\
\ne& \frac{1}{2}x-\frac{1}{4}\ln(e^{2x}+4)+c\\
\end{aligned}
$$
???? help
|
You have made a mistake:
Note that $$\int\frac{1}{u^2-4u}du \ne \int u^{-2}-\frac{1}{4}u^{-1}du$$
By the way
$$\int\frac{2dx}{e^{2x}+4} = \int\frac{2e^{2x}dx}{e^{2x}(e^{2x}+4)} = $$
$$\int\frac{du}{u(u+4)}$$
Where $$u= e^{2x}$$
is a short way to go.
|
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|
Find the greatest common divisor of $f(x)=2x^3+2x^2+x+4$ and $g(x)=x^4+3x^3+4x^2+3x$ Find the greatest common divisor of $f(x)=2x^3+2x^2+x+4$ and $g(x)=x^4+3x^3+4x^2+3x$ in $\mathbb{Z}_{11}[x]$
|
We have to perform the Euclidean algorithm modulo $11$:
$$\begin{align}
h_1(x)&:=g(x)+(5x-1)f(x)\equiv 7x^2+7,\\
h_2(x)&:=f(x)+(6x+6)h_1(x)\equiv -x+2,\\
h_3(x)&:=h_1(x)+(7x+3)h_2(x)\equiv 2.\\
\end{align}$$ Hence, in $\mathbb{Z}_{11}[x]$,
$$\gcd(g(x),f(x))=\gcd(h_1(x),f(x))=\gcd(h_1(x),h_2(x))=\gcd(h_3(x),h_2(x))=1.$$
|
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|
Calculating the largest possible area of a rectangle inscribed in an ellipse So i got the equation $4x^2 + 9y^2 = 3600$
What i've done so far is:
$A= (2x)(2y) = 4xy$
Then I find the expression of $y$
$9y^2= 3600 -4x^2$
$y = \pm \sqrt{3600 -4x^2 / 9} = 2/3(\sqrt {900 - x^2} \quad 2/3(900 -x^2)^{1/2}$
Then i set
$A = 4x(2/3(900 -x^2)^1/2 = (8/3)x(900 -x^2)^1/2$
Taking the derivative
$A'(x) = 8/3(900 -x^2)^{1/2} + (8/3)x(1/2)(900 -x^2)^{-1/2}(-2x)
= (2400 - (16/3)x^2)/(\sqrt{900-x^2})$
Set the $A' = 0$
$2400 - (16/3)x^2 = 0$
$(16/3)x^2 = 2400$
$(16/3)x = \sqrt{2400} = (20\sqrt{6}) / 3$
$x = (5\sqrt{6})/12$
Then i put the value of x in the equation and get
$A = (8/3)((5 \sqrt 2)/12)(900 - ((5\sqrt 2)/12)^2)^{1/2} = 81.6...$
Is this right or?
|
Alt. hint - by AM-GM:
$$3600 = (2x)^2 + (3y)^2 \ge 2 \sqrt{(2x)^2 \cdot (3y)^2} = 12 \,|x|\,|y| = 3\,A$$
Equality holds iff $\,2x=3y\,$.
|
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|
Evaluate $\int_0^\infty \frac{x^2}{x^4 + 6x^2 + 13}dx$ In the context of the residue theorem, I have this integral to evaluate. The function is even, and $|\int_0^\pi\frac{R^2e^{2i\theta}iRe^{i\theta}}{R^4e^{4i\theta}+6R^2e^{2i\theta} + 13}d\theta| \leq \int_0^\pi2\frac{R^3}{R^4}d\theta \to 0$, so the problem is to find the residues in the upper halfplane.
$\int_0^\infty\frac{x^2}{x^4 + 6x^2 + 13}dx = \frac12\int_{-\infty}^\infty\frac{x^2}{x^4 + 6x^2 + 13}dx = \pi i\sum_{\{\Im z > 0\}}$res$(\frac{x^2}{x^4 + 6x^2 + 13})$
There are two residues to calculate:
*
*$z = \sqrt{-3 + 2i}$: $\frac{\sqrt{-3 + 2i}}{4(-3+ 2i) + 12} = -\frac i8\sqrt{-3 + 2i}$
*$z = \sqrt{-3 - 2i}: \frac i8\sqrt{-3 - 2i}$
(Wolfram Alpha if you don't want to trust me)
Giving me overall for the integral:
$\frac\pi8 (\sqrt{-3 + 2i} - \sqrt{-3 - 2i}) = $1.427346... i
But the answer is clearly not meant to be imaginary.
|
You need both square-roots in the upper half-plane. The difference between those two square-roots is real.
|
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|
Closed form for fixed $m$ to $\int\frac{dx}{x(x+1)(x+2)(x+3)...(x+m)}$ $I=\displaystyle\int\frac{dx}{x(x+1)(x+2)(x+3)...(x+m)}$
Attempt:
$\dfrac{ A_0 }{ x }+\dfrac{ A_1 }{ x +1 }+\dfrac{ A_2 }{ x + 2 }...+\dfrac{ A_m }{ x +m } =\dfrac{1}{x(x+1)(x+2)(x+3)...(x+m)}$
But things got very messy.
I also thought that 1)applying integral by parts, or 2) taking terms one from left head, one from right hand and use some kind of a symmetry, or 3) using trigonometric identites etc.
I cannot see the solution, any hint, help would be perfect. Thank you in advanced.
|
$f(x) = \frac {1}{x(x+1)(x+2)\cdots(x+m)} = \frac {A_0}{x} + \cdots + \frac {A_n}{x+m}\\
\lim_\limits{x\to -n} (x-n)f(x) = A_n\\
A_n = \prod_\limits {i\ne n} \frac 1{i-n}$
|
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|
The difference between the radii of the largest and smallest circles having centres on the circumference of $x^2+2x+y^2+4y=4$ Owing to restriction of 150 characters in the title section I include the latter part of the problem here below in bold and italics
Also given that both the circles(largest and smallest) pass through a point $(a,b)$ lying outside the given circle, $x^2+2x+y^2+4y=4$
My attempt:
$x^2+2x+y^2+4y=4$
$\implies\space (x+1)^2+(y+2)^2=(3)^2$
I have drawn a circle with center $(-1,-2)$
& Also the largest circle with a center at the circumference of $x^2+2x+y^2+4y=4$ is the circle with $radius=6$
Also this circle passes through $(a,b)$
Now the smallest circle with center on the circumference of $x^2+2x+y^2+4y=4$ and passing through $(a,b)$, let us assume, has its center $(\alpha,\beta)$ and radius $=r$.
Thus $(\alpha-a)^2+(\beta-b)^2=r^2\cdot\cdot\cdot(1)$
Again if the largest circle has radius $R=6$
Then it has its center(by observation) at $(2,-2)$
Thus Equation of the largest circle is $(x-2)^2+(y+2)^2=36$
Again this largest circle passes through $(a,b)$
$\therefore$ $(a-2)^2+(b+2)^2=36 \cdot\cdot\cdot(2)$
I am stuck here. Please throw some light.
|
Draw the line passing through $(a,b)$ and the centre of the circle, meeting the circle at $P$ and $Q$. $P$ and $Q$ are the nearest and the farthest points on the circle from $(a,b)$. So they are the centres of the smallest and the largest circles. If $d$ is the distance between $(a,b)$ and the centre of the given circle. Then the radii of the smallest and the largest circles are respectively $d-3$ and $d+3$. So their difference is $6$.
|
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|
Pre-calc optimization of a rational function A student I am tutoring is in a pre-calc class and they just had a test on rational functions. One of the questions my student said they got was to minimize the area of the outer rectangle, given that the area of the inner rectangle is a fixed 48, and the distance to the outer rectangle on either side is 1.5 horizontally and 1 vertically.
If we call the horizontal side length of the inner rectangle x, and the vertical y, then we can get the area for the outer rectangle (SB, surface area big) as
SB = 144/y + 2y + 56
Without needing much convincing, I explained (since 56 is const., and factor of 2 is constant as well) we can reduce it to optimizing 72/y+y. And then I am stuck.
After thinking about it for a bit, I tried plotting f(y)=72/y and f(y)=y together with f(y)=72/y+y, and the minimum appears at 72/y=y, which incidentally is the answer we find using calculus. But I got no idea if this is coincidental or if this is indeed a key to answering this question.
Anyways, the problem is to minimize the area of the outer rectangle without the use of calculus. Any hints / ideas?
|
Explanation for the surface area:
\begin{align}
\text{Surface area of outer rectangle} &= (x+2)(y+3) \\
&= \left(\frac{48}{y} + 2 \right)(y+3) \\
&= 48 + 2y + \frac{144}{y} + 6 \\
&= \color{red}{54} + 2y + \frac{144}{y}
\end{align}
Though OP wrote $56$ instead, it didn't affect the minimiser (the value of $y$ which minimises the surface area of the outer rectangle)
Idea of why optimising $144/y + 2y + 56$ is equivalent to optimising $72/y + y$:
\begin{align}
\frac{144}{y} + 2y + 56 &= 2 \Bigg(\color{blue}{\frac{72}{y} + y} - \underbrace{2 \sqrt{\frac{72}{y} \cdot y}\Bigg) + 56 + 2\sqrt{72}}_{\mbox{constant}} \\&= 2\underbrace{\left( \frac{6\sqrt2}{\sqrt{y}} - \sqrt{y} \right)^2}_{\mbox{minimized when } \frac{6\sqrt2}{\sqrt{y}}
= \sqrt{y}} + 56 + 24 \sqrt2
\end{align}
The square term on the right is minimised when $\dfrac{6\sqrt2}{\sqrt{y}} = \sqrt{y}$, i.e. $y = 6\sqrt2$.
Shorter solution using a.m.-g.m.-inequality
$$54 + 2y + \frac{144}{y} \ge 54 + 2 \sqrt{2y \cdot \frac{144}{y}} = 54 + 24 \sqrt2$$
and equality holds iff
$$2y = \frac{144}{y} \iff y = \sqrt{72} = 6\sqrt2$$
or even better
\begin{align}
A &= (x+2)(y+3) = xy + (3x + 2y) + 6 = (3x + 2y) + 54 \\
&\ge 2 \sqrt{6xy} + 54 = 2\sqrt{6 \cdot 48} + 54 = 24 \sqrt{12} + 54,
\end{align}
and equality holds iff $3x = 2y$. Solving with the constraint $xy = 48$ gives $y = \sqrt{72}$.
|
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|
Is there a closed form of this sequence? The sequence is $\frac{1}{1}$ , $\frac{1}{2}$ , $\frac{2}{1}$ , $\frac{1}{3}$ , $\frac{2}{2}$ , $\frac{3}{1}$ , $\frac{1}{4}$ , $\frac{2}{3}$ , $\frac{3}{2}$ , $\frac{4}{1}$ , ...
and I need to find n th..
Here's my approach..
I bound them with who has same sum of denominator and numerator.
So $\frac{1}{1}$ is set_1 .
$\frac{1}{2}$ , $\frac{2}{1}$ are set_2..
$\frac{1}{3}$ , $\frac{2}{2}$ , $\frac{3}{1}$ are set_3 .
$\frac{1}{4}$ , $\frac{2}{3}$ , $\frac{3}{2}$ , $\frac{4}{1}$ are set_4 and so on..
If I suppose that n is in set_K , I can conclude n th number is $$ \frac{1+( (n-1)-\sum_{i=1}^K (i-1) ) }{ K - ( (n-1)-\sum_{i=1}^K (i-1) ) } $$
( because set_K starts with $\frac{1}{K}$ and n is ( n - $\sum_{i=1}^K (i-1)$ ) th member of set_K.. )
As a result my conclusion is $$\frac{2n-(K-1)K}{K(K+1) - 2(n-1) }$$ but I don't know the relationship between n and K..
Is there anyone to help me? I really want to know.. :(
|
You can break the sequence in groups of $N$ terms with $N=1,2,3,\dots$ of the form
$$\frac{k}{N+1-k}\quad \text {for $k=1,2,\dots ,N$}.$$
So the first $N-1$ groups have $\sum_{i=1}^{N-1}i=\binom{N}{2}$ terms.
Now, given $n$, there is a unique $N$ such that $\binom{N}{2}<n\leq \binom{N+1}{2}$, which means that the $n$-term is in the $N$-th group and it is equal to
$\frac{k}{N+1-k}$ with $k=n-\binom{N}{2}$. For an explicit formula solve $\binom{N}{2}<n\leq \binom{N+1}{2}$ with respect to $N$:
$$N=\left\lfloor\frac{1+\sqrt{8n-7}}{2}\right\rfloor$$
|
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|
Find the power series representation and interval of convergence for the function Find the power series representation and interval of convergence for the function $ \ f(x)=\frac{3}{2+x} \ $
Answer:
$f(x)=\frac{3}{x+2}=\frac{3}{2} (1+\frac{x}{2})^{-1}=\frac{3}{2} (1-\frac{x}{2}+(\frac{x}{2})^2-(\frac{x}{2})^3+..........) \ =\sum_{n=0}^{\infty} \frac{3}{2} (-1)^n (\frac{x}{2})^n $
The series converges if $ \ |\frac{x}{2}|<1 \ \Rightarrow |x|<2 \ $
Thus the interval of convergence is $ -2<x<2 \ $
Am I right so far?
|
Rewrite the series as $\; \sum_{n\ge 0}(-1)^n 16^n x^{4n+5}= x^5\sum_{n\ge 0}(-1)^n 16^n(x^4)^n$.
The radius of convergence of the power series (setting $x^4=u$) is given by
$$\frac1R=\limsup_n\,\bigl(16^n\bigr)^{\tfrac1n}=16,$$
so the given power series converges if
$$ x^4<\frac1{16}\iff |x|<\frac12.$$
|
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|
If sides $a$, $b$, $c$ of $\triangle ABC$ are in arithmetic progression, then $3\tan\frac{A}{2}\tan\frac {C}{2}=1$
If sides $a$, $b$, $c$ of $\triangle ABC$ (with $a$ opposite $A$, etc) are in arithmetic progression, then prove that
$$3\tan\frac{A}{2}\tan\frac{C}{2}=1$$
My attempt:
$a$, $b$, $c$ are in arithmetic progression, so
$$\begin{align}
2b&=a+c \\[4pt]
2\sin B &= \sin A+ \sin C \\[4pt]
2\sin(A+C) &=2\sin\frac {A+C}{2}\;\cos\frac{A-C}{2} \\[4pt]
2\sin\frac{A+C}{2}\;\cos\frac{A+C}{2}&=\sin\frac{A+C}{2}\;\cos\frac{A-C}{2} \\[4pt]
2\cos\frac{A+C}{2}&=\cos\frac{A-C}{2}
\end{align}$$
|
Another way to look at it is just compute the tangents in terms of the sides of the triangle $\triangle ABC$.
From Law of Cosines we have that $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, then follows
$$2\sin^2\dfrac{A}{2}=1-\cos A=\dfrac{a^2-(b-c)^2}{2bc}=\dfrac{(a+b-c)(a-b+c)}{2bc}=\dfrac{2(p-c)(p-b)}{bc}$$
where $2p=a+b+c$. Similarly, $$2\cos^2\dfrac{A}{2}=1+\cos A=\dfrac{(b+c)^2-a^2}{2bc}=\dfrac{(-a+b+c)(a+b+c)}{2bc}=\dfrac{2p(p-a)}{bc}.$$
Then, $\tan\dfrac{A}{2}=\sqrt{\dfrac{(p-b)(p-c)}{p(p-a)}}$. Using symmetry we are able to show that $\tan\dfrac{C}{2}=\sqrt{\dfrac{(p-a)(p-b)}{p(p-c)}}$.
Finally, since the sides of $\triangle ABC$ are in $AP$, then $a=b-k$ and
$c=b+k$ for some $k\in\Bbb R$,
$$\tan\dfrac{A}{2}\tan\dfrac{C}{2}=\sqrt{\dfrac{(p-b)(p-c)}{p(p-a)}}\sqrt{\dfrac{(p-a)(p-b)}{p(p-c)}}=\dfrac{p-b}{p}=\dfrac{\dfrac{3b}{2}-b}{\dfrac{3b}{2}}=\dfrac13$$
|
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|
Asymptotic expansion of complete elliptic integral of third kind Is there a way to compute the expansion of the complete elliptic integral of third kind
$\Pi(n,k)=\int_0^{\pi/2} \frac{d\varphi}{(1-n\sin^2\varphi)\sqrt{1-k^2\sin^2\varphi}}$
for
$\Pi(1+\epsilon,1-\epsilon)\ , \qquad \epsilon\to 0$,
and if so, what is it?
|
Assuming $\epsilon > 0$, I get
\begin{equation}
\Pi (1 + \epsilon, 1- \epsilon ) = - \frac{1}{\epsilon \sqrt{3}} \left( \textrm{arcsinh} ( \tfrac{1}{\sqrt{2}} ) + \frac{i \pi}{2} \right) - \frac{1}{4} \log \epsilon + \frac{1}{36} \left( 6 - 8 \sqrt{3} \: \textrm{arcsinh} ( \tfrac{1}{\sqrt{2}} ) + 27 \log 2 - 4 i \pi \sqrt{3} \right) + O(\epsilon).
\end{equation}
To prove it express the elliptic integral in terms of hypergeometric functions. According to http://functions.wolfram.com/EllipticIntegrals/EllipticPi/introductions/CompleteEllipticIntegrals/05/ one has
\begin{equation}
\Pi (1 + \epsilon, 1- \epsilon ) = \Pi \left(1 + \epsilon | (1- \epsilon)^2 \right) = - \frac{i \pi}{2 \epsilon} \sqrt{\frac{1+\epsilon}{3 - \epsilon}} - \frac{\pi (1- \epsilon)^2}{4} \times \sum_{j=0}^\infty (- \epsilon)^j {}_2 F_1 \left( \frac{3}{2}, \frac{3}{2} + j; 2; (1- \epsilon)^2 \right).
\end{equation}
The Wolfram's definition of elliptic integrals differs from your more standard definition, i.e. $\Pi(n, k) = \Pi(n | k^2)$.
To deal with the hypergeometric terms we can use a linear transformation. Equation 15.3.12 of Abramowitz/Stegun gives us
\begin{equation}
{}_2 F_1 \left( \frac{3}{2}, \frac{3}{2} + j; 2; z \right) = \frac{2 \Gamma(1 + j)}{\sqrt{\pi} \Gamma (\tfrac{3}{2} + j)} (1 - z)^{-1-j} \sum_{n=0}^j \frac{ ( \tfrac{1}{2} - j)_n ( \tfrac{1}{2} )_n}{n! (-j)_n} (1 - z)^n - \frac{(-1)^{1+j}}{\sqrt{\pi} \Gamma( \tfrac{1}{2}-j )} \sum_{n=0}^{\infty} \frac{( \tfrac{3}{2} )_n ( \tfrac{3}{2} + j )_n}{n! (1 +j + n)!} (1-z)^n \left[ \log(1 - z) - \psi(1 + n) - \psi(n + j + 2) + \psi(\tfrac{3}{2} + n) + \psi( \tfrac{3}{2} + n + j) \right].
\end{equation}
Since each hypergeometric function is multiplied by $\epsilon^j$, we only need to extract divergences up to and including order $\epsilon^{-j}$, after the substitution $z = (1 - \epsilon)^2$. The second term contains at most the logarithmic divergence, while only elements with $n=0,1$ matter in the first term. For $j \geq 1$ we find
\begin{equation}
{}_2 F_1 \left( \frac{3}{2}, \frac{3}{2} + j; 2; (1 - \epsilon)^2 \right) = \frac{j!}{2^j \sqrt{\pi} \Gamma (\tfrac{3}{2} + j)} \epsilon^{-j} \left[ \frac{1}{\epsilon} + \frac{j^2 + 3j - 1}{2j} + O(\epsilon) \right], \qquad j = 1,2,3,\ldots
\end{equation}
while for $j = 0$ an additional logarithmic divergence appears,
\begin{equation}
{}_2 F_1 \left( \frac{3}{2}, \frac{3}{2}; 2; (1 - \epsilon)^2 \right) = \frac{2}{\pi \epsilon} + \frac{\log ( \epsilon / 8) + 4}{\pi} + O(\epsilon).
\end{equation}
Now we can resum the expansions of the hypergeometrics, which leads to the final result. You can check it numerically.
|
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On the series $\sum \limits_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )$ I managed to prove through complex analysis that
$$\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right ) = 1 -2 \log 2$$
However, I'm having a difficult time proving this result with real analysis methods. Partial summation of the series gets nasty pretty quickly since it involves harmonic numbers. Another interesting fact to note about this series is that this part
$$\sum_{n=1}^{\infty} \left ( \frac{1}{2n-1} + \frac{1}{2n+1} \right )$$
diverges. This came as a complete surprise to me since I expected this to telescope. What remains now to prove the series I want is by using generating function.
\begin{align*}
\sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{2n-1} - \frac{1}{2n+1} \right )x^n &=\sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{2n-1} - \sum_{n=1}^{\infty} \frac{x^n}{2n+1} \\
&= -\log \left ( 1-x \right ) - \sqrt{x} \;\mathrm{arctanh} \sqrt{x} - \frac{{\mathrm {arctanh} }\sqrt{x} - \sqrt{x}}{\sqrt{x}}
\end{align*}
I cannot , however, evaluate the limit as $x \rightarrow 1^-$ of the last expression. Can someone finish this up?
Of course alternatives are welcome.
|
A very simple way:
$$ \begin{eqnarray*}\sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{2n-1}-\frac{1}{2n+1}\right)&=&\sum_{n\geq 1}\int_{0}^{1}\left( 2x^{2n-1}-x^{2n-2}-x^{2n}\right)\,dx\\&=&\int_{0}^{1}\sum_{n\geq 1}x^{2n-2}(2x-1-x^2)\,dx\\&=&\int_{0}^{1}\frac{2x-1-x^2}{1-x^2}\,dx\\&=&\int_{0}^{1}\frac{x-1}{x+1}\,dx=\left[x-2\log(1+x)\right]_{0}^{1}=\color{red}{1-2\log 2}.\end{eqnarray*} $$
|
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|
Matrix in canonical form of an orthogonal transformation
Let:
$$A = \frac 12 \begin{pmatrix} 1 & -1 & -1 &-1 \\ 1 & 1 & 1 &-1 \\ 1 & -1 & 1 & 1\\ 1 &1 &-1 & 1\end{pmatrix} $$
Prove there exists an orthogonal transformation $\phi$ of the
Euclidean Vector Space $\mathbb{R^4}$ such that $A=M_{Bs}(\phi)$ Then
find an orthonormal basis ($B'$) such that $M_{B'}(\phi)$ will be the matrix in
canonical form of the orthogonal transformation.
Note $Bs$ is the standard basis of $\mathbb{R^4}$ and the canonical matrix is a matrix of the form:
$$ \begin{pmatrix} \pm 1 & 0 & 0 \\ 0 & \cos\theta& -\sin\theta\\ 0 & \sin\theta & \cos\theta \end{pmatrix},$$
We have that $AA^T= I_4 \Rightarrow A$ is an orthogonal transformation. I don't know if for this case I should find the characteristic polynomial or the minimal polynomal... Can you help me or give me some hint to find the canonical matrix, please?
|
Turns out you just need to get the eigenvalues, which will be complex and come in conjugate pairs. For each eigenvalue, find an eigenvector, make real vectors of the real and imaginary parts, if necessary use Gram-Schmidt to make that pair of vectors orthonormal. For this one, it was only necessary to adjust the lengths of the real vectors, dividing by either $\sqrt 6$ or $\sqrt 2$ once I multiplied through to make the entries integers.
The characteristic polynomial is $\left( x^2 - x + 1\right)^2,$ the minimal polynomial is $\left( x^2 - x + 1 \right)$. The eigenvectors I used for eigenvalue $\omega = (1 + i \sqrt 3)/2$ were
$$
\begin{pmatrix}
\omega & - \bar{\omega} \\
\bar{\omega} & \omega \\
1&0 \\
0&1
\end{pmatrix}
$$
Notice that $\omega = (1 + i \sqrt 3)/2$ fits with the observation by Doug M that $A^6 = I$, along with $A^3 = -I$.
$$
\begin{pmatrix}
0&0&\frac{\sqrt 6}{2}&0 \\
\frac{1}{\sqrt 2}&\frac{-1}{\sqrt 2}&0&\frac{1}{\sqrt 2} \\
0&0&0&\frac{\sqrt 6}{2} \\
\frac{1}{\sqrt 2}&\frac{1}{\sqrt 2}&\frac{-1}{\sqrt 2}&0
\end{pmatrix}
\begin{pmatrix}
\frac{1}{2}&\frac{-1}{2}&\frac{-1}{2}&\frac{-1}{2} \\
\frac{1}{2}&\frac{1}{2}&\frac{1}{2}&\frac{-1}{2} \\
\frac{1}{2}&\frac{-1}{2}&\frac{1}{2}&\frac{1}{2} \\
\frac{1}{2}&\frac{1}{2}&\frac{-1}{2}&\frac{1}{2} \\
\end{pmatrix}
\begin{pmatrix}
\frac{1}{\sqrt 6}&\frac{1}{\sqrt 2}&\frac{-1}{\sqrt 6}&\frac{1}{\sqrt 2} \\
\frac{1}{\sqrt 6}&\frac{-1}{\sqrt 2}&\frac{1}{\sqrt 6}&\frac{1}{\sqrt 2} \\
\frac{2}{\sqrt 6}&0&0&0 \\
0&0&\frac{2}{\sqrt 6}&0
\end{pmatrix}
=
\begin{pmatrix}
\frac{1}{2}&\frac{\sqrt 3}{2}&0&0 \\
\frac{-\sqrt 3}{2}&\frac{1}{2}&0&0 \\
0&0&\frac{1}{2}&\frac{\sqrt 3}{2} \\
0&0&\frac{-\sqrt 3}{2}&\frac{1}{2}
\end{pmatrix}
$$
$$
\left(
\begin{array}{cccc}
0&0&\frac{\sqrt 6}{2}&0 \\
\frac{1}{\sqrt 2}&\frac{-1}{\sqrt 2}&0&\frac{1}{\sqrt 2} \\
0&0&0&\frac{\sqrt 6}{2} \\
\frac{1}{\sqrt 2}&\frac{1}{\sqrt 2}&\frac{-1}{\sqrt 2}&0 \\
\end{array}
\right)
\left(
\begin{array}{cccc}
\frac{1}{\sqrt 6}&\frac{1}{\sqrt 2}&\frac{-1}{\sqrt 6}&\frac{1}{\sqrt 2} \\
\frac{1}{\sqrt 6}&\frac{-1}{\sqrt 2}&\frac{1}{\sqrt 6}&\frac{1}{\sqrt 2} \\
\frac{2}{\sqrt 6}&0&0&0 \\
0&0&\frac{2}{\sqrt 6}&0 \\
\end{array}
\right) =
\left(
\begin{array}{cccc}
1&0&0&0 \\
0&1&0&0 \\
0&0&1&0 \\
0&0&0&1 \\
\end{array}
\right)
$$
From notes by Mark F. Schumaker:
|
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|
Number of non-negative integral solutions This is a problem I've seen a couple times around here, but I couldn't find one quite like this.
Say we have ten variables, $a, b,$ and $c_1, c_2, c_3,\dots, c_8$. How many non-negative integral solutions are there to the following problem such that $a\leq 5$ and $b\geq 5$:
$$a+b+c_1 +c_2+c_3 +c_4 +c_5+c_6+c_7+c_8 = 100$$
I understand that the total number of solutions when $a$ and $b$ are unrestrained is ${10+100-1\choose 100}$, but I don't know any real way to formulate these restraints without something like
$$\sum_{a=0}^{5}\sum_{b=5}^{100-a}{107 - b-a\choose 7}$$
or something of the sort. Is there an easier way?
|
We wish to solve the equation
$$a + b + c_1 + c_2 + c_3 + c_4 + c_5 + c_6 + c_7 + c_8 = 100 \tag{1}$$
in the nonnegative integers subject to the constraints $a \leq 5$ and $b \geq 5$.
Since $b \geq 5$, $b' = b - 5$ is a nonnegative integer. Substituting $b' + 5$ for $b$ in equation 1 yields
\begin{align*}
a + b' + 5 + c_1 + c_2 + c_3 + c_4 + c_5 + c_6 + c_7 + c_8 & = 100\\
a + b' + c_1 + c_2 + c_3 + c_4 + c_5 + c_6 + c_7 + c_8 & = 95 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers. A particular solution of equation 2 corresponds to the placement of nine addition signs in a row of $95$ ones. The number of solutions of equation 2 is
$$\binom{95 + 9}{9} = \binom{104}{9}$$
since we must choose which $9$ of the $104$ positions needed for $95$ ones and $9$ addition signs will be filled with addition signs.
We have addressed the constraint $b \geq 5$, but we still have the constraint that $a \leq 5$. If this constraint is violated, then $a \geq 6$. Therefore, we must subtract the number of nonnegative integer solutions of equation 2 in which $a \geq 6$ from the number of solutions of equation 2.
Suppose $a \geq 6$. Then $a' = a - 6$ is a nonnegative integer. Substituting $a' + 6$ for $a$ in equation 2 yields
\begin{align*}
a' + 6 + b' + c_1 + c_2 + c_3 + c_4 + c_5 + c_6 + c_7 + c_8 & = 95\\
a' + b' + c_1 + c_2 + c_3 + c_4 + c_5 + c_6 + c_7 + c_8 & = 89 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers with
$$\binom{89 + 9}{9} = \binom{98}{9}$$
solutions.
Hence, the number of solutions of equation 1 that satisfy the constraints $a \leq 5$ and $b \geq 5$ is
$$\binom{104}{9} - \binom{98}{9}$$
|
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|
Evaluate $\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdots(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}$ Evaluate $$\frac{(5+6)(5^2+6^2)(5^4+6^4)\cdot\dots\cdot(5^{1024}+6^{1024})+5^{2048}}{3^{1024}}.$$
I can't figure out where to start. I tried using logarithms but I couldn't get a pattern going. Any advice will be helpful, thanks in advanced.
|
Hint - telescope:
$$
\begin{align}
&\quad \frac{\big(\color{red}{(6-5)\cdot}(6+5)\big)\;(6^2+5^2)(6^4+5^4)\cdot\dots\cdot(6^{1024}+5^{1024})+5^{2048}}{\color{red}{(6-5)\cdot}3^{1024}} \\
&= \frac{\big((6^2-5^2)(6^2+5^2)\big)\;(6^4+5^4)\cdot\dots\cdot(6^{1024}+5^{1024})+5^{2048}}{3^{1024}} \\
&= \frac{\big((6^4-5^4)(6^4+5^4)\big)\;(6^8+5^8)\cdot\dots\cdot(6^{1024}+5^{1024})+5^{2048}}{3^{1024}} \\
&= \;\ldots
\end{align}
$$
|
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|
Given $P(x) = x^4 + ax^3 + bx^2 + cx + d.$ Which of the following is the smallest? The graph below shows a portion of the curve defined by the quartic polynomial $P(x) = x^4 + ax^3 + bx^2 + cx + d.$ Which of the following is the smallest?
$(A)$ $P(-1)$
$(B)$ The product of the zeros of $P$
$(C)$ The product of the non-real zeros of $P$
$(D)$ The sum of the coefficients of $P$
$(E)$ The sum of the real zeros of $P$
I know that
$P(-1) = 1-a+b-c+d $ also the product of zeroes is $d$. From the graph the real zeroes are around $1.7$ and $3.85$, so product of non-reals is $\frac{d}{1.7*3.85}$. The sum of the coefficients is $1+a+b+c+d$ and the sum of the zeros is $-a$. Now with this information I don't see immediately how to tell which is the smallest.
|
We have, since the non-real root are conjugates $$P(x)=(x-(m+ni))(x-(m-ni))(x-a)(x-b)$$ or
$$P(x)=x^4-(a+b-2m)x^3+(ab+2m(a+b)+m^2+n^2)x^2-(2abm+(a+b)(m^2+n^2))x+ab(m^2+n^2)$$ It is not hard to compare $m^2+n^2$ which corresponds to $(C)$ above with the other quantities $(A),(B),(D),(E)$.
For example, calculate a quartic polynomial having a graphic resemblance (this is not immediate!). I found
$$P(x)=x^4-5x^3+\dfrac{153}{64}x^2+\dfrac{181}{32}x+\dfrac{25}{8}$$ so I get the "analogues"$$\begin{cases}(A)=5.859375\\(B)=3.125\\(C)=0.390625\\(D)=7.171875\\(E)=6\end{cases}$$ from which the smallest is $(C)=\dfrac{25}{64}$, i.e. the product of non-real zeros $\dfrac{4\pm3i}{8}$of $P$.
|
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|
Finding the coefficients $p_0,p_1,p_2,q_1,q_2,q_3$ of Padé approximation Determine the Padé approximation of degree $5$ with $ n =2 $ and $ m= 3$ for $f(x) = e^{-x}$.
Suppose $r$ is a rational function of degree $N$.$$
r(x) = \frac{p(x)}{q(x)} = \frac{p_0 +p_1x + \cdots + p_nx^n}{q_0 +q_1x+ \cdots + q_mx^m}.
$$
The Padé approximation technique is$$
f(x)-r(x) = f(x)-\frac{p(x)}{q(x)} = \frac{f(x)q(x)-p(x)}{q(x)} = \frac{\left(\sum\limits_{i=0}^{\infty} a_ix^i \right)\left(\sum\limits^m_{i=0}q_{i}x^i \right)-\sum\limits^n_{i=0} p_ix^i}{q(x)}.
$$
To find the Padé approximation we need to choose $p_0,p_1,p_2,q_1$ and $q_2,q_3$. The Maclaurin expansion of $e^{-x}$ is $\sum\limits_{i = 0}^{\infty} \dfrac{(-1)^i}{i!} x^i$. Since $q_0 \neq 0$, thus $q_0= 1$ if $x= 0$, and $p_0 =1$. How does one find $p_1,p_2,q_1,q_2,q_3$ ao that$$
r_{2,3}(x) = \frac{p_0+p_1x+p_2x^2}{q_0+q_1x+q_2x^2+q_{3}x^3}?
$$
Does one have to transform this into a linear equation to solve?$$
\begin{bmatrix}
a_0 & & & &\\
a_1 & a_0 & & &\\
a_2 & a_1 & a_0&& \\
a_3 & a_2 & a_1& a_0&\\
a_4 & a_3 & a_2 & a_1&a_0\\
\end{bmatrix} \begin{bmatrix}
q_1\\
q_2\\
q_3\\
0\\
0
\end{bmatrix} - \begin{bmatrix}
p_1\\
p_2\\
0\\
0\\
0
\end{bmatrix} = -\begin{bmatrix}
a_1\\
a_2\\
a_3\\
a_4\\
a_5
\end{bmatrix}
$$
How do I solve this linear system so I may obtain the solution? Can someone demonstrate a concise method of Gaussian elimination to find $r_{2,3}$ and that can also work for $r_{1,4}$ or any $r_{n,m}$ for Padé approximation?
|
$\newenvironment{gmatrix}{\left\lgroup\begin{matrix}}{\end{matrix}\right\rgroup}$In general, to find the Padé approximation $r(x) = \dfrac{\sum\limits_{k = 0}^n p_k x^k}{\sum\limits_{k = 0}^m q_k x^k}$ for $f(x) = \sum\limits_{k = 1}^\infty a_k x^k$, it reduces to find the solution to$$
\begin{gmatrix}
a_0 &&&\\
a_1 & a_0 &&\\
\vdots & \ddots & \ddots &\\
a_{m + n} & \cdots & a_1 & a_0
\end{gmatrix} \begin{gmatrix}
q_0 \\ q_1 \\ \vdots \\ q_{m + n}
\end{gmatrix} = \begin{gmatrix}
p_0 \\ p_1 \\ \vdots \\ p_{m + n}
\end{gmatrix}, \tag{1}
$$
where $p_{n + 1} = \cdots = p_{m + n} = 0$, $q_{m + 1} = \cdots = q_{m + n} = 0$. In order to solve this system of equations, usually it is assumed that $q_0 = 1$ and (1) becomes$$
\begin{gmatrix}
a_0 &&\\
\vdots & \ddots &\\
a_{m + n - 1} & \cdots & a_0
\end{gmatrix} \begin{gmatrix}
q_1 \\ \vdots \\ q_{m + n}
\end{gmatrix} - \begin{gmatrix}
p_1 \\ \vdots \\ p_{m + n}
\end{gmatrix} = \begin{gmatrix}
a_1 \\ \vdots \\ a_{m + n}
\end{gmatrix},\ p_0 = a_0.
$$
Since\begin{align*}
&\mathrel{\phantom{=}}{} \begin{gmatrix}
a_0 &&\\
\vdots & \ddots &\\
a_{m + n - 1} & \cdots & a_0
\end{gmatrix} \begin{gmatrix}
q_1 \\ \vdots \\ q_{m + n}
\end{gmatrix} = \begin{gmatrix}
a_0 &&\\
\vdots & \ddots &\\
a_{m + n - 1} & \cdots & a_0
\end{gmatrix} \begin{gmatrix}
q_1 \\ \vdots \\ q_m \\ 0 \\ \vdots \\ 0
\end{gmatrix}\\
&= \begin{gmatrix}
a_0\\
\vdots & \ddots\\
a_{m - 1} & \cdots & a_0\\
a_m & \cdots & a_1 & 0\\
\vdots & \ddots & \vdots & \vdots & \ddots\\
a_{m + n - 1} & \cdots & a_n & 0 & \cdots & 0
\end{gmatrix} \begin{gmatrix}
q_1 \\ \vdots \\ q_m \\ 0 \\ \vdots \\ 0
\end{gmatrix} = \begin{gmatrix}
A & B\\
C & O
\end{gmatrix} \begin{gmatrix}
q_1 \\ \vdots \\ q_m \\ 0 \\ \vdots \\ 0
\end{gmatrix},
\end{align*}
where$$
A = \begin{gmatrix}
a_0\\
\vdots & \ddots\\
a_{m - 1} & \cdots & a_0
\end{gmatrix},\ B = -I_n,\ C = \begin{gmatrix}
a_m & \cdots & a_1\\
\vdots & \ddots & \vdots\\
a_{m + n - 1} & \cdots & a_n
\end{gmatrix},
$$
and $O$ is an $m × n$ zero matrix, and$$
-\begin{gmatrix}
p_1 \\ \vdots \\ p_{m + n}
\end{gmatrix} = -\begin{gmatrix}
p_1 \\ \vdots \\ p_n \\ 0 \\ \vdots \\ 0
\end{gmatrix} = \begin{gmatrix}
A & B\\
C & O
\end{gmatrix} \begin{gmatrix}
0 \\ \vdots \\ 0 \\ p_1 \\ \vdots \\ p_n
\end{gmatrix},
$$
then (1) is equivalent to$$
\begin{gmatrix}
A & B\\
C & O
\end{gmatrix} \begin{gmatrix}
q_1 \\ \vdots \\ q_m \\ p_1 \\ \vdots \\ p_n
\end{gmatrix} = \begin{gmatrix}
a_1 \\ \vdots \\ a_{m + n}
\end{gmatrix},\ p_0 = a_0. \tag{2}
$$
Now, to find the Padé approximation $r(x) = \dfrac{\sum\limits_{k = 0}^2 p_k x^k}{\sum\limits_{k = 0}^3 q_k x^k}$ for $f(x) = \mathrm{e}^{-x} = \sum\limits_{k = 1}^\infty \dfrac{(-1)^k}{k!} x^k$, from (2) there is $q_0 = 1$, $p_0 = 1$, and$$
\begin{gmatrix}
1 &&& -1\\
-1 & 1 &&& -1\\
\dfrac{1}{2} & -1 & 1\\
-\dfrac{1}{6} & \dfrac{1}{2} & -1\\
\dfrac{1}{24} & -\dfrac{1}{6} & \dfrac{1}{2}
\end{gmatrix} \begin{gmatrix}
q_1 \\ q_2 \\q_3 \\ p_1 \\ p_2
\end{gmatrix} = \begin{gmatrix}
-1 \\ \dfrac{1}{2} \\ -\dfrac{1}{6} \\ \dfrac{1}{24} \\ -\dfrac{1}{120}
\end{gmatrix},
$$
thus $q_1 = -\dfrac{3}{5}$, $q_2 = -\dfrac{3}{20}$, $q_3 = -\dfrac{1}{60}$, $p_1 = \dfrac{2}{5}$, $p_2 = -\dfrac{1}{20}$. Therefore,$$
r_{2, 3}(x) = \frac{1 + \dfrac{2}{5}x - \dfrac{1}{20} x^2}{1 - \dfrac{3}{5} x - \dfrac{3}{20} x^2 - \dfrac{1}{60} x^3}.
$$
|
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|
An inequality involving three consecutive primes Can you provide a proof or a counterexample to the following claim :
Let $p,q,r$ be three consecutive prime numbers such that $p\ge 11 $ and $p<q<r$ , then $\frac{1}{p^2}< \frac{1}{q^2} + \frac{1}{r^2}$ .
I have tested this claim up to $10^{10}$ .
For $p>5$ we get $\pi(2p)-\pi(p) \ge 2$ , a result by Ramanujan .
This means that $q<2p$ and $r<2p$ , so $\frac{1}{2p}<\frac{1}{q}$ and $\frac{1}{2p}<\frac{1}{r}$ which implies $\frac{1}{p} < \frac{1}{q} + \frac{1}{r}$ . If we square both sides of inequality we get $\frac{1}{p^2} < \frac{1}{q^2} + \frac{2}{qr} + \frac{1}{r^2}$ . Now , I don't know how to rule out term $\frac{2}{qr}$ .
|
Here is a stronger form of the above inequality. The prime number theorem implies that for every $\epsilon$, there exists a prime $p_{\epsilon}$ such that for all $p > p_{\epsilon}$, we have $1 - \epsilon < \frac{p}{q}, \frac{q}{r} < 1$. Hence for all positive real $a$ and all primes greater than some $p_{\epsilon_a}$
$$
\frac{2 - \epsilon_a}{p^a} < \frac{1}{q^a} + \frac{1}{r^a} < \frac{2}{p^a}
$$
|
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|
Geometric intuition of composition of hyperbolic and inverse hyperbolic trig functions We can evaluate $\cos(\arcsin x)$ with simple geometric intuition. Write $y = \cos(\arcsin x)$. Letting $\theta = \arcsin x$, $y = \cos \theta$ corresponds to a right triangle with angle $\theta$, adjacent side length $y$, and hypotenuse $1$. Pythagoras gives that the opposite side has length $\sqrt{1-y^2}$. Then $\sin \theta = \sqrt{1-y^2}/1 = \sqrt{1-y^2}$. Now $\sin \theta = \sin(\arcsin x) = x$ with appropriate bounds, so $x = \sqrt{1 - y^2}$ with appropriate bounds, so $y = \cos (\arcsin x) = \sqrt{1 - x^2}$ with appropriate bounds.
It can be shown that $\cosh (\mathrm{arcsinh} \ x) = \sqrt{1+x^2}$. Can we derive this using a similar intuitive geometric construction, perhaps on a hyperbola? If so, how?
|
Solved it myself. We can mirror the argument above, but using the hyperbolic identity $\cosh^2 - \sinh^2 = 1$ in place of Pythagoras.
Write $y = \cosh(\mathrm{arcsinh}\ x) = \cosh(\theta)$. Then $y = \cosh \theta, x = \sinh \theta$, so $y^2 - x^2 = 1 \implies y = \sqrt{1 + x^2}$.
|
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|
Find variance of $X$ given its cumulative distribution function A rv $X$ has the cumulative distribution function
$$ F(x) = \begin{cases} 0 \; \; \; \; x<1 \\ \frac{x^2-2x+2}{2} \; \; \; \; 1 \leq x < 2 \\ 1 \; \; \; \; x \geq 2 \end{cases} $$
Calculate the variance of $X$
attempt
First since $F'(x) = f(x)$, then
$$ f(x) = x-1 $$
on $[1,2)$ and $0$ otherwise. Now,
$$ Var(X) = E(X^2) - E(X)^2 = \int\limits_1^2 (x^3 - x^2) dx - \left( \int\limits_1^2 (x^2-x )\right)^2 $$
After solving this easy integral I get $0.73$ whereas my answer key says the answer is $\boxed{0.139}$. What is my mistake here? Am I applying the formulas wrong?
|
The cdf 'jumps' at $x=1$, so $P(X=1)=0.5$.
$$\begin{align} Var(X) &= E(X^2) - E(X)^2 = \int\limits_1^2 (x^3 - x^2) dx + 0.5 - \left( \int\limits_1^2 (x^2-x )dx+0.5\right)^2 \\
& \approx 1.4167+0.5-(0.83333+0.5)^2 \\
& \approx 0,1389 \end{align}$$
|
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|
Finding joint prob mass function
Suppose two balls are chosen from a box containing 3 white, 2 red and
5 blue balls. Let X = the number of white balls chosen and Y = the
number of blue balls chosen. Find the joint pmf of X and Y.
Attempt
We want $p_{XY}(x,y) = P(X = x \cap Y =y ) $
Let's count the number of ways to select $X$ white balls from $10$ given balls. ${3 \choose x}$ gives all possible ways to pick $x$ white balls. Now, to pick the remaining we can do that in ${7 \choose 7-x}$ ways. Similarly for the number of blue balls chosen, same reasoning gives ${5 \choose y } \times {5 \choose 5-x}$. Now, by multiplication princliple one has
$$ p_{XY}(x,y) = \frac{ {3 \choose x}{7 \choose 7-x}{5 \choose y } {5 \choose 5-x} }{{10 \choose 2 } }$$
Is this a correct joint pmf?
|
Numerator should be white's, blue's and red's:
$$ p_{XY}(x,y) = \frac{ {3 \choose x}{5 \choose y } {10-5-3 \choose 2-x-y}}{{10 \choose 2 } }$$
Our answers are the same if
$${7 \choose 7-x} {5 \choose 5-y} = {10-5-3 \choose 2-x-y}$$
which is not the case (Note: I guess you meant ${5 \choose 5-y}$ and not ${5 \choose 5-x}$).
Yours is wrong for 2 reasons.
*
*You seem to assume independence. We don't compute $P(X=x)$ and $P(Y=y)$ separately and then multiply them.
*The individual computations are wrong. Observe the additions below:
$$P(X=x) = \frac{{3 \choose x}{7 \choose 2-x}}{{10 \choose 2}}$$
Here, we see that $3+7=10$ and $x+2-x=2$
So, ${3 \choose x}{7 \choose 7-x}$ should instead be ${3 \choose x}{7 \choose 2-x}$
Next:
$$P(Y=y) = \frac{{5 \choose y}{5 \choose 2-y}}{{10 \choose 2}}$$
Here, we see that $5+5=10$ and $y+2-y=2$
So, ${5 \choose y } \times {5 \choose 5-y}$ should instead be ${5 \choose y } \times {5 \choose 2-y}$
Verify that the following is wrong as well:
$${7 \choose 2-x} {5 \choose 2-y} = {10-5-3 \choose 2-x-y}$$
Okay, so how did I get ${3 \choose x}{5 \choose y } {10-5-3 \choose 2-x-y}$? I first chose among the 3 x's. Instead of choosing among the 7 non-x's, I then chose among the 5 y's (the order doesn't matter for choosing x's and y's). Now instead of choosing among the 7 non-x's or the 5 non-y's (note that 7 non-x's and 5 non-y's overlap), I choose among the remaining 2 balls which are not x and not y.
|
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|
Find the remainder when $(x+1)^n$ is divided by $(x-1)^3$. I know this question has been answered before, but I have a slightly different different question.
I saw the solution of this question in my book and the author has solved it by substituting $x-1=y$ and then equating the coefficients of $y^2$, $y^1$ and $y^0$ to $A(y+1)^2$, $B(y+1)^1$ and $C$.
My question is why do we have to substitute $x-1=y$ and why can't equate coefficients of $x^2$, $x^1$ and $x^0$ to $A$, $B$ and $C$ without substituting? Thanks in advance.
|
Write $x:=u+1$. Then
$$(x+1)^n=(u+2)^n=\sum_{k=0}^n{n\choose k}2^{n-k}u^k\ .$$
Dividing by $(x-1)^3=u^3$ we get the remainder
$$r={n\choose0}2^n+{n\choose 1}2^{n-1}u+{n\choose 2}2^{n-2}u^2=2^n+n2^{n-1}(x-1)+n(n-1)2^{n-3}(x-1)^2\ .$$
This can be written in the form
$$r=2^{n-3}\bigl((n^2-n)x^2+(-2n^2+6n)x+(n^2-5n+8)\bigr)\ .$$
|
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|
$f(x) = \frac{x^3}{6}+\frac{1}{2x}$, $\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx = ?$ $$f(x) = \frac{x^3}{6}+\frac{1}{2x}$$
$$\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx = ?$$
Let's start by deriving the function, we have
$$f'(x) = \dfrac{x^4-1}{2x^2}$$
Hence we get
$$\int^{3}_{1} \sqrt{1 +\Big[\dfrac{x^4-1}{2x^2}\Big]^2}\, dx = ?$$
Am I right?
UPDATE:
If we have a definite integral:
$$\int^{3}_{1} \sqrt{1 +\Big[\dfrac{x^4-1}{2x^2}\Big]^2}\, dx = \dfrac{x^4-3}{6x}+C$$
Then
$$\dfrac{x^4-3}{6x}+C = \boxed{\frac{14}{3}}$$
|
Note that
$$f(x) = \frac{x^3}{6}+\frac{1}{2x}\implies f'(x)=\frac {x^2}{2}-\frac1{2x^2}$$
$$ \implies I=\int^{3}_{1} \sqrt{1 +[f'(x)]^2}\, dx =\int^{3}_{1} \sqrt{(\frac {x^2}{2}+\frac 1{2x^2})^2}\, dx $$
$$\implies I=\left . \frac {x^3}6-\frac 1 {2x}\right |^3_1 = \frac {14}3$$
And that you can't write this
$$\int^{3}_{1} \sqrt{1 +\Big[\dfrac{x^4-1}{2x^2}\Big]^2}\, dx = \dfrac{x^4-3}{6x}+C$$
since the integral is definite it's a number it can't be equal to a function...
|
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|
Negative variance.
I've an urn with 5 red and 3 white balls. I'm randomly drawing a ball
from the urn each round (without replacement). Let $X$ be the number
of balls taken from the urn until the first white ball is chosen. I'm
asked for the pmf, expected value and variance of $X.$
Since we can at most take 5 red balls before there is a white ball with 100 % certainty, we have that $k=1,2,3,4,5.$ We can ignore all the previous balls that we don't look at, the probability will therefore remain the same for all the draws, that is the pdf is $$P(X=k)=3/8, \ \ k=1,2,...,5.$$
For the expected value we then have that
$$E(X)=\sum_{k=1}^{5}k\cdot\frac{3}{8}=\frac{45}{5}\approx 5.6.$$
For the vairance, we also need $E(X^2),$ so
$$E(X^2)=\sum_{k=1}^{5}k^2\cdot\frac{3}{8} = \frac{165}{8}\approx 20.6.$$
So $$Var(X)= E(X^2)-E(X)^2=20.6-5.6^2=-10.67.$$
But negative variance, can't be correct! Where am I wrong?
|
Continuing from comments:
Letting $X$ be the random variable counting the number of balls pulled until pulling the first white ball, we try to find the probability distribution.
To do this, temporarily label each of the balls uniquely. For the $k$'th ball to be the first white ball, that implies that the first $k-1$ balls are all red. These $k-1$ red balls can be pulled in $5\frac{k-1}{~}=P(5,k-1)=\frac{5!}{(5-(k-1))!}$ ways. (For example, pulling two red balls can be done in $5\cdot 4$ ways and pulling three red balls can be done in $5\cdot 4\cdot 3$ ways, etc...). Then, pulling a white ball can be done in $3$ ways, so multiplying we get a total of $5\frac{k-1}{~}\cdot 3$ ways in which we can pull $k$ balls where only the $k$'th ball is white. Then, recognize that the number of ways of pulling $k$ balls without regards to color will be $8\frac{k}{~}$.
Taking the ratio then yields the probability we are after: $Pr(X=k)=\dfrac{5\frac{k-1}{~}\cdot 3}{8\frac{k}{~}}$ (for values of $k\geq 1$. For $k=0$ the probability is zero as pulling a white ball requires pulling at least one ball)
Alternatively, one could see the same result by using conditional probability arguments. For the $k$'th ball to be the first white ball to occur, that implies that the first ball is red which occurs with probability $\frac{5}{8}$, the second ball is red which occurs with probability $\frac{4}{7}$ given that the first ball was red, the third ball was red... on up until the $k-1$'st ball is red, followed by the final ball being white.
Rewritten and evaluated, this yields the following table:
$\begin{array}{|c|c|c|c|}\hline k&Pr(X=k)&\text{with fractions}&\text{simplified}\\
\hline
1&\dfrac{5\frac{0}{~}\cdot 3}{8\frac{1}{~}}&\dfrac{3}{8}&\dfrac{3}{8}\\\hline
2&\dfrac{5\frac{1}{~}\cdot 3}{8\frac{2}{~}}&\dfrac{5}{8}\cdot\dfrac{3}{7}&\dfrac{15}{56}\\\hline
3&\dfrac{5\frac{2}{~}\cdot 3}{8\frac{3}{~}}&\dfrac{5}{8}\cdot\dfrac{4}{7}\cdot\dfrac{3}{6}&\dfrac{5}{28}\\\hline
4&\dfrac{5\frac{3}{~}\cdot 3}{8\frac{4}{~}}&\dfrac{5}{8}\cdot\dfrac{4}{7}\cdot\dfrac{3}{6}\cdot\dfrac{3}{5}&\dfrac{3}{28}\\\hline
5&\dfrac{5\frac{4}{~}\cdot 3}{8\frac{5}{~}}&\dfrac{5}{8}\cdot\dfrac{4}{7}\cdot\dfrac{3}{6}\cdot\dfrac{2}{5}\cdot\dfrac{3}{4}&\dfrac{3}{56}\\
\hline6&\dfrac{5\frac{5}{~}\cdot 3}{8\frac{6}{~}}&\dfrac{5}{8}\cdot\dfrac{4}{7}\cdot\dfrac{3}{6}\cdot\dfrac{2}{5}\cdot\dfrac{1}{4}\cdot\dfrac{3}{3}&\dfrac{1}{56}\\\hline\end{array}$
As a sanity check, the values in the right column all add up to equal $1$, as expected.
Continue.
As an aside, yet another way you could arrive at the correct probability distribution is through the use of binomial coefficients. For the $k$'th ball to be the first white one, we need to select $k-1$ red balls and one white ball and further need the white ball selected to occur last.
This occurs with probability $Pr(X=k)=\dfrac{\binom{5}{k-1}\binom{3}{1}}{\binom{8}{k}}\cdot\frac{1}{k}$ which one can check equals the same expression as discussed earlier.
It seems you are insistent on writing this with factorials. The expression will be
$$Pr(X=k) = \dfrac{ \frac{5!}{(5-k+1)!}\cdot 3}{\frac{8!}{(8-k)!}}$$
Which doesn't simplify much. If you insisted you could move things around as
$$Pr(X=k)= \dfrac{ (8-k)!}{112\cdot (5-k+1)!}$$
|
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|
Computing $\cos^4 20^{\circ}+\cos^4 40^{\circ}+\cos^4 60^{\circ}+\cos^4 80^{\circ}$ Compute $$\cos^4 20^{\circ}+\cos^4 40^{\circ}+\cos^4 60^{\circ}+\cos^4 80^{\circ}$$
Suppose that this is a scenario where calculator isn't allowed.
I want to say that this expression has something to do with this equation $\cos^2 20^{\circ}+\cos^2 40^{\circ}+\cos^2 60^{\circ}+\cos^2 80^{\circ}=\frac 74$ but I can't seem to find a method to solve this without relying on brute forcing every term to $\cos20^{\circ}$
|
We can use the "inverse" rule : just like how we compute $\cos n \theta$ from powers of $\cos \theta$, we may compute powers of $\cos \theta$ using values of $\cos n\theta$, which is what we should desire here, since the sequence $20n$ has some regularity(remainder on division by $90$, values for which we know $\sin \backslash\cos$ like $60,120$ etc.) which we can exploit.
Indeed, we have the following situation : $2\cos^2 \theta = \cos 2 \theta - 1$, so squaring and using the $\cos 2\theta$ formula again gives:
$$
\cos^4 \theta = \frac{3 + 4\cos 2 \theta + \cos 4 \theta}{8}
$$
Let's use this formula:
$$
8\cos^4 20 = 3 + 4 \cos 40 + \cos 80 \\
8 \cos^4 40 = 3 + 4 \cos 80 + \cos 160 \\
8 \cos^4 80 = 3 + 4 \cos 160 + \cos 320
$$
Note that $\cos 40 + \cos 80 + \cos 160 = 0$, and $\cos 80 + \cos 160 + \cos 320 = 0$
! These can be checked by combining two of the three angles into the sum of cosines type formula.
Hence, adding all the above gives $\cos^4 20 + \cos^4 40 + \cos^4 80 = \frac 9{8}$. Adding $\cos^4 60 = \frac 1{16}$ gives the answer $\frac{19}{16}$.
Moral : the inverse formulas are really helpful in problems involving multiples of numbers with "regularity" as I described above.
|
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throwing a two-dice 7 with in total 6 dice We have been stuck on the following questions regarding the game Qwixx:
What is the probability that there is a combination of two dice which add up to exactly 7 when throwing with a total of 6 dice once?
Thank you in advance!
|
Note: what follows has several cases each of which is somewhat error prone. The calculation should be checked carefully.
The pairs that add to $7$ are $(1,6), (2,5), (3,4)$. We'll say that a pair is "hit" if at least one member of the pair comes up in your six tosses. We'll work from the complement...that is we will consider the probability that no two values sum to $7$ and then subtract from $1$.
Case I: only one pair is hit. In this case only one value can be hit, so the probability is $\frac 1{6^6}$. As there are $6$ possible values, the probability in this case is $$\frac 1{6^5}$$
Case II: exactly two pairs are hit. There are $3$ ways to choose the two pairs. Then $4$ ways to choose the values in those two pairs which are hit. Given those choices, we need every toss to be one of those two values (but not all the same), the probability of which is $\frac 1{3^6}-2\times \frac 1{6^6}=\frac {31}{23328}$ Combining we see that the probability is $$3\times 4 \times \frac {31}{23328}=\frac {84}{15552}=\frac {31}{1944}$$
Case III: all three pairs are hit. Then there are $8$ ways to choose the three values that occur. Given those choices we need the probability that all the values we get are contained in those three, and that we hit each of those three at least once. The probability of that is $\frac 1{2^6}-3\times \frac 1{3^6}+3\times \frac 1{6^6}=\frac {5}{432}$. Combining we get $$8\times \frac {5}{432}=\frac {5}{54}$$
It follows that the desired result is $$1-\frac 1{6^5}-\frac {31}{1944}-\frac {5}{54}=\boxed {\frac {6931}{7776}}=0.8913323\cdots$$
|
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Showing that a constraint is not convex
Is the following problem convex?
$$\begin{array}{ll} \text{minimize} & x+y^2\\ \text{subject to} & x + y \leq 2\\ & \frac{x+y^2}{x^2+2} \leq 3\\ & x^2 = 1\end{array}$$
The problem is, in fact, not convex, but I'm having problems showing that the constraint $$\frac{x+y^2}{x^2+2} \leq3$$ is not convex. Is there a way to do so without calculating the Hessian?
|
$$\frac{x+y^2}{x^2+2}\le 3$$
$$y^2 \le 3x^2-x+6$$
Hence if $y$ is fixed, and $x$ is large enough in magnitude, then this construct will be satisfied.
Let $y=3$, then $(2,3)$ and $(-2,3)$ both satisfies the inequality, but $(0,3)$ doesn't.
Remark:
For this question, we know that $x=1$ or $x=-1$.
If $x=1$, $y \le 1$ and $y^2 \le 8$, hence $-\sqrt{8} \le y \le 1.$
If $x=-1$, $y \le 3$ and $y^2 \le 10$, hence $-\sqrt{10} \le y \le 3.$
In particular $(1,0)$ and $(-1,0)$ are both feasible but $(0,0)$ is not feasible.
|
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Almost done solving IVP using Laplace transform .. Need advice/guidance I have the following IVP:
$$\begin{cases}y''-2y'+5y = -8e^{-t},\\ \ \\ y(0) = 2\\ \ \\ y'(0) = 12\end{cases}$$
I will show my steps so far.
After taking the transform of each term I get:
$$s^2Y(s) - 2sY(s) + 5Y(s) - 2s - 8 = \frac {-8}{s+1}$$
I simplified the L.H.S and got:
$$ Y(s)(s^2-2s+5) = \frac {-8}{s+1} + 2s+ 8 $$
Then the R.H.S becomes:
$$ Y(s)(s^2-2s+5) = \frac {2s^2+10s}{s+1}$$
I then get:
$$Y(s) = \frac {2s^2+10s}{(s+1)(s^2-2s+5)}$$
I then used partial fraction decomposition and got $A = -1, B = 3$, and $C = 5$
I then have the following:
$$Y(s) = \frac {-1}{s+1} + \frac {3s+5}{s^2-2s+5}$$
I am now stuck finding the inverse Laplace transform of the second term.
Please can someone check my work and let me know if I did this right and, if so, what is that inverse Laplace transform I mentioned at the last step?
Help is greatly appreciated. Thank you.
|
You have $s^2-2s+5=(s-1)^2+4$. So the first term looks like
$$
\frac {3s+5}{(s-1)^2+4}
=\frac {3(s-1)+8}{(s-1)^2+4}
=\frac {3(s-1)}{(s-1)^2+4}
+\frac {8}{(s-1)^2+4}.
$$
Since $\frac{s}{s^2+4}$ is the transform of $\cos 2t$ and $\frac2{s^2+4}$ of $\sin 2t$, with the shifting rule you get
$$
\mathcal L^{-1}\left[\frac {3s+5}{s^2-2s+5}\right]
=\mathcal L^{-1}\left[\frac {3s+5}{(s-1)^2+4}\right]
=3e^t\cos2t+4e^t\sin 2t
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Could someone elaborate on this algebraic transformation? I have a solution for an exercise and one part of it is not clear to me:
$
\frac{\frac{1}{\sqrt{x - 1}} - 1}{x - 2} = \frac{1 - \sqrt{x - 1}}{(x - 2)\sqrt{x - 1}}
$
Could anyone explain, please, how the result was obtained?
The full solution is this:
$
\frac{\frac{1}{\sqrt{x - 1}} - 1}{x - 2} = \frac{1 - \sqrt{x - 1}}{(x - 2)\sqrt{x - 1}} * \frac{1 + \sqrt{x - 1}}{1 + \sqrt{x - 1}} = \frac{2 - x}{(x - 2)\sqrt{x - 1}(1 + \sqrt{x - 1})} = \frac{-1}{\sqrt{x - 1}(1 + \sqrt{x - 1})}, x \ne 2
$
|
You simply multiply the top and bottom by $\sqrt{x-1}$:
$$\frac{\frac{1}{\sqrt{x-1}}-1}{x-2}=\frac{\Bigl(\frac{1}{\sqrt{x-1}}-1\Bigr)\sqrt{x-1}}{(x-2)\sqrt{x-1}}=\frac{1-\sqrt{x-1}}{(x-2)\sqrt{x-1}}$$
Edit: This is valid as long as the thing you are multiplying top and bottom by is not $0$. In this case, $\sqrt{x-1} \neq 0$ for or else the thing you were given containing a $\frac{1}{\sqrt{x-1}}$ would have been undefined in the first place.
|
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|
Minimum of the maximum Let $a, b,c,d,e,f,g$ seven nonnegative real numbers with the sum equal to 1. Let considered the sums $ a+b+c$, $b+c+d$, $c+d+e$, $d+e+f$, $e+f+g$. What is the minimum of the maximum of those sums when we consider all the possibilities?I have no idea. One hint, please?
|
The language is obfuscating but:
Suppose $a=b=c=0$ then $d+e+g = 1$. $d+e+g$ and that is the maximum of the sums. But that's pretty high maximum. What if we don't have three elements equal to $0$. Then any sum of three numbers will not include a fourth possible none zero term. So the sum of any three can not be as high as $1$. So in that case the maximum will be less than $1$.
So what's the minimum the maximum can be. That's what they are asking.
Let me explain WOLOG (without loss of generality). The letters we use don't matter. One of those numbers must be the least number of the seven. Let's call that one $a$. Without loss of generality we might as well assume that $a$ is the least element because ... some number must be and we can just relabel. Likewise we can assume $b$ is the second least and so on.
So WOLOG $a \le b \le c \le d\le e\le g$. And so the maximum of the sums is $d+e+g$. What is the least that $d+e+g$ can be if $a \le b \le c \le d\le e\le g$?
That is what is really being asked.
Now $1 = a+b+c+d+e+f+g \ge (a+b+c) + (a+b+c) = 2(a+b+c)$ so $a+b+c \le \frac 12$.
So $d+e+g = 1 - (a+b+c) \ge \frac 12$. So $d+e+g$ will be at least $\frac 12$ so $\frac 12$ is the absolute minimum $d+e+g$ can be.
But can $d + e+g = \frac 12$? Well, it can if $a+b+c= \frac 12$ but can $a+b+c = \frac 12$? Well it can if $a = b = c = \frac 16$. And then $d+e+g$ can equal $\frac 12$ if $d=e=g=\frac 16$.
So the minimum of the maximum is $\frac 12$.
It'll take a bit more work to show that the only way $d+e +g = \frac 12$ is if all terms are equal to $\frac 16$ but we weren't asked to show that.
But here's how:
If $g > \frac 16$ and $d+e+g = \frac 12$ then $d+e < \frac 12 - \frac 16 = \frac 13$. So $d+e < \frac 13$ and $2d = d + d \le d+e <\frac 13$ so $d < \frac 16$.
So $a\le b \le c \le d < \frac 16$ so $a + b + c < \frac 16 + \frac 16 + \frac 16 = \frac 16$. But then $\frac 12 = d+e+g = 1-(a+b+c) > \frac 12$ which is a contradiction.
So $d+e+g =\frac 12 \iff g = \frac 16$. $1 = a+b+c+d+e+g \ge a+b+c+d+g+g \ge a+b+c+g+g+g \ge a+b+g+g+g+g \ge a+g+g+g+g+g \ge g+g+g+g+g+g = 1$. Thus $ a+b+c+d+e+g = a+b+c+d+g+g = a+b+c+g+g+g = a+b+g+g+g+g = a+g+g+g+g+g = g+g+g+g+g+g$ so $a=b=c=d=e=g=\frac 16\iff d+e+g =\frac 12$.
|
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|
Calculate the following convergent series: $\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)}$ I need to tell if a following series convergent and if so, find it's value:
$$
\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)}
$$
I've noticed that
$$
\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)} = \frac{1}{3}\sum _{n=1}^{\infty }\:\frac{1}{n}-\frac{1}{n+3} = \frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{2}-\frac{1}{5}+\frac{1}{3}-\frac{1}{6}+\frac{1}{4}-\frac{1}{7}+\cdots\right)
$$
Wich means some values are zeroed, does that mean it's a telescoping sum?
I also know that $\sum _{n=1}^{\infty }\:\frac{1}{n\left(n+3\right)} \le \sum _{n=1}^{\infty }\:\frac{1}{n^2}\:$ so the series converges by the harmonic p-series.
|
Yes, you are correct that the sum telescopes quite nicely. The partial sum formula is$$\sum\limits_{n=1}^m\frac 1{n(n+3)}=\frac 13\left[\left(1+\frac 12+\cdots+\frac 1m\right)-\left(\frac 14+\frac 15+\cdots+\frac 1{m+3}\right)\right]$$Notice how anything past $\frac 14$ in the first sum is automatically canceled from the second sum. Hence, the partial sum formula is given as$$\sum\limits_{n=1}^m\frac 1{n(n+3)}=\frac 13\left(1+\frac 12+\frac 13-\frac 1{m+1}-\frac 1{m+2}-\frac 1{m-3}\right)$$As $m\to\infty$, the fractions containing $m$ vanish, leaving$$\sum\limits_{n\geq1}\frac 1{n(n+3)}\color{blue}{=\frac {11}{18}}$$
|
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|
Residue of $\frac{1}{z(e^{z}-1)}$ I was trying to find the residue of $\dfrac {1}{z (e^z - 1)}$. I have written the Taylor series for $e^z$ which is $1 + z + \dfrac{z^2}{2!} + \dfrac{z^3}{3!}..$ Thus, for $e^z - 1$ I have series of the form $z+\dfrac{z^2}{2!}+\dfrac{z^3}{3!}..$. But now I am stuck as I have a problem dividing 1 with my series.
|
We obtain
\begin{align*}
\color{blue}{\frac{1}{z\left(e^z-1\right)}}&=\frac{1}{z(z+\frac{1}{2}z^2+O(z^3))}\tag{1}\\
&=\frac{1}{z^2\left(1+\frac{1}{2}z+O(z^2)\right)}\tag{2}\\
&=\frac{1}{z^2}\left(1-\frac{1}{2}z+O(z^2)\right)\tag{3}\\
&\,\,\color{blue}{=\frac{1}{z^2}-\frac{1}{2z}+O(1)}
\end{align*}
and we conclude the residue is $\color{blue}{-\frac{1}{2}}$.
Comment:
*
*In (1) we expand $e^z-1$ up to $z^3$.
*In (2) we factor out $z$.
*In (3) we do a geometric series expansion
\begin{align*}
\frac{1}{z^2\left(1+\frac{1}{2}z+O(z^2)\right)}&=\frac{1}{z^2}\left(1-\left(\frac{1}{2}z+O(z^2)\right)-\left(\frac{1}{2}z+O(z^2)\right)^2+O(z^2)\right)\\
&=\frac{1}{z^2}\left(1-\frac{1}{2}z+O(z^2)\right)
\end{align*}
|
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|
Minimum value of $\dfrac{a+b+c}{b-a}$
$f(x)= ax^2 +bx +c ~ ~~(a<b)$ and $f(x)\ge 0~ \forall x \in \mathbb R$ .
Find the minimum value of $\dfrac{a+b+c}{b-a}$
Attempt:
$b^2 \le 4ac$
$f(1) = a+b+c$
$f(0) = c$
$f(-1) = a-b+c$
$a>0$
and $c>0$
I am unable to utilize these things to find the minimum value of the expression $\equiv \dfrac{a+b+\frac {b^2}{4a}}{b-a}$
The answer given is $3$.
|
Actually, you are on the right track! Now let's complete it with only elementary calculus.
$$\frac{a+b+\frac{b^2}{4a}}{b-a}=\frac{4a^2+4ab+b^2}{4a(b-a)}=\frac{(2a+b)^2}{4a(b-a)}$$
Knowing that $b>a$, let's set $b=ka, k>1$.
$$\frac{(2a+b)^2}{4a(b-a)}=\frac{(2+k)^2a^2}{4(k-1)a^2}=\frac{(2+k)^2}{4(k-1)}$$
Now you just need to find the minimum of this function for $k>1$. You can differentiate it and check that $k=4$ is a critical point, and you get $\frac{a+b+c}{b-a}=3$ there as a minimum.
|
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|
How do you set up this Tricky u-sub? Tricky u-sub. Can you point me in the right direction?
$$\int{{x^3}\sqrt{5-2x^2}}dx$$
$u = 5-2x^2$
$ du = -4x dx$
Obviously, this does not match fully.
I tried breaking up the $x^3 = x*x^2$
and I continued with:
$u=5-2x^2$
$2x^2 = 5-u$
$x^2=\frac{5-u}{2}$
But, I can't see how to make it all fit.
Am I on the right track?
|
Another approach:
We can write the integrand as
\begin{align*}
{x^3}\sqrt{5-2x^2}&=-\tfrac12x(5-2x^2)\sqrt{5-2x^2}+\tfrac52x\sqrt{5-2x^2}&&\text{then}\\[5pt]
\int{x^3}\sqrt{5-2x^2}dx&=-\frac12\int x\left(5-2x^2\right)^{3/2}dx+\frac52\int x\left(5-2x^2\right)^{1/2}dx\\[5pt]
&=\color{red}{\frac1{20}\left(5-2x^2\right)-\frac5{12}\left(5-2x^2\right)^{3/2}+C}
\end{align*}
|
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|
Calculate $\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately Is it possible to calculate $\sum_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately?
The original question was find the number of solutions to $2x+y+z=20$ which I calculated to be the coefficient of $x^{20}$ in $(1+x^2+x^4\dots)(1+x+x^2\dots)^2$ which simplified to the term above.
I know $\sum_{k=0}^{20}\binom{k+2}{2}=\binom{23}3$ but the $(-1)^k$ is ruining things.
|
$$\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2 = \binom{2}{2} \underbrace{-\binom{3}{2} + \binom{4}{2}}_{\binom{3}{1}} \underbrace{-\binom{5}{2} + \binom{6}{2}}_{\binom{5}{1}}- \ldots \underbrace{-\binom{21}{2} + \binom{22}{2}}_{\binom{21}{1}}$$
$$ = 1 + 3 + 5 + \ldots + 21 = 121$$
|
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|
How to prove that $(1 + \frac{1}{n})^{\sqrt{n(n+1)}} < e$? I've been trying to solve the following problem:
Show that $\ln{(k+1)} - \ln{k} = \ln{(1 + \frac{1}{k})} \leq \frac{1}{\sqrt{k(k+1)}}$
EDIT: the title was inaccurate, my bad.
So what we have to prove is that the upper limit of $(1 + \frac{1}{n})^{\sqrt{n(n+1)}}$ equals $e$.
|
More information...
As $n \to +\infty$, we have
\begin{align*}
\sqrt{n(n+1)} &= n + \frac{1}{2} - \frac{1}{8n} + O(n^{-2})
\\
\log\left(1+\frac{1}{n}\right) &= \frac{1}{n} - \frac{1}{2n^2} +
\frac{1}{3n^3} + O(n^{-4})
\\
\sqrt{n(n+1)}\log\left(1+\frac{1}{n}\right) &=
1-\frac{1}{24n^2}+O(n^{-3})
\end{align*}
Therefore, for large enough $n$,
\begin{align*}
\sqrt{n(n+1)}\log\left(1+\frac{1}{n}\right) & <
1
\\
\left(1+\frac{1}{n}\right)^{\sqrt{n(n+1)}} &< e
\end{align*}
|
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|
finding solution to simultaneous equations I have three equations
$$
\begin{align*}
c_1 & = \frac{a}{1+a+b} + \frac{a}{a+b}+\frac{a}{1+a}\\
c_2 & = \frac{b}{1+a+b} + \frac{b}{a+b}+\frac{b}{1+b}\\
c_3 & = \frac{1}{1+a+b} + \frac{1}{1+a}+\frac{1}{1+b}\\
\end{align*}$$
where $c_1,c_2,c_3$ are known constant.
I want to know if I can find $a$ and $b$ or not..
Thanks,
|
Before undertaking calculations which are guaranteed to be tedious, sometimes it helps to graph the equations involved. In this case if we turn these into equations in $x$ and $y$ by letting $x=a$ and $y=b$. Using the graphing site desmos.com and letting $c_1,\,c_2,\,c_3$ be 'sliders' we see that the places where the three graphs contain the same point cannot actually be solutions since they would lead to division by $0$.
For example, $a=x=-1$ and $b=y=0$ involve division by zero, as does $a=x=0$ and $b=y=-1$. We can get a solution $a=x=b=y=0$ but only for the case $c_1=c_2=0$ and $c_3=3$.
There are cases where a common intersection can be obtained but they require specific values of $c_1,\,c_2,\,c_3$. For example, if we let $c_1=c_2=c_3=\frac{4}{3}$ then $a=b=1$ is a solution.
So in general, no, one cannot solve for $a$ and $b$.
Here is the link to the Desmos graphs.
|
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|
Limit of $\frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n$ Compute the limit
$$
\lim_{n \to \infty} \frac{\sqrt[n]{(n+1)(n+2)\cdots(2n)}}n
$$
How can this be done? The best I could do was rewrite the limit as
$$
\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n}
$$
Following that log suggestion in the comments below:
\begin{align}
&\ln \left(\lim_{n \to \infty} \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\ &= \lim_{n \to \infty} \ln \left( \left(\frac{n+1}n \right)^{\frac 1n}\left(\frac{n+2}n \right)^{\frac 1n}\cdots\left(\frac{2n}n \right)^{\frac 1n} \right) \\
&= \lim_{n \to \infty} \left(\ln \left(\frac{n+1}n \right)^{\frac 1n} + \ln\left(\frac{n+2}n \right)^{\frac 1n} + \cdots + \ln\left(\frac{n+n}n \right)^{\frac 1n} \right) \\
&= \lim_{n \to \infty} \sum_{i=1}^n \ln \left(\frac{n+i}n \right)^{\frac 1n} \\
&= \lim_{n \to \infty} \frac 1n \sum_{i=1}^n \ln \left(1 + \frac in \right) \\
&= \int_1^2 \ln x \, dx \\
&= (x \ln x - x)\vert_1^2 \\
&= (2 \ln 2 - 2)-(1 \ln 1-1) \\
&= (\ln 4-2)-(0-1) \\
&= \ln 4-1 \\
&= \ln 4 - \ln e \\
&= \ln \left( \frac 4e \right)
\end{align}
but I read from somewhere that the answer should be $\frac 4e$.
|
Now take a logarithm and realize you have just gotten a Riemann sum. (Interval: $[1,2]$, widths: $1/n$, heights $\ln(1+k/n)$.)
|
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|
Minimum value of $\frac{pr}{q^2}$ in quadratic equation If $y^4-2y^2+4+3\cos(py^2+qy+r)=0$ has $2$ solutions and $p,q,r\in(2,5)$. Then minimum of $\displaystyle \frac{pr}{q^2}$ is.
solution I try
$$-3\cos(py^2+qy+r) =(y^2-1)^2+3\geq 3$$
$$\cos(py^2+qy+r)\leq -1\implies \cos(py^2+qy+r)=-1$$
$$py^2+qy+r=(4n+1)\pi\implies py^2+qy+r-(4n+1)\pi=0$$
for real and distinct roots
$$q^2-4p[r-(4n+1)\pi ]\geq 0$$
How do I find minimum of $\displaystyle \frac{p r}{q^2}$? Help me.
|
The condition for
$$
y^4-2y^2+4+3\sigma(y)=0
$$
with $-1 \le \sigma(y) \le 1$
to have real roots is that $\sigma(y) = -1$ or $p y^2+q y+ r = \pi$ but then $y = y_0 = \pm 1$ hence
$$
\min \frac{pr}{q^2}\\
\mbox{subjected to}\\
p y^2_0+q y_0 + r = \pi \\
(p,q,r)\in (2,5)^3
$$
and the minimum value is $\frac{1}{2(2-\pi)}$ for both $y_0 = \pm 1$
As example of solution
$$
p = 2\\
q = 2(\pi-2)\\
r = 2-\pi
$$
|
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|
Integration using "method of judicious guessing" My calculus-book gives an example of integration using the method of judicious guessing. But I do not intuit the method very well.
QUESTION: How does the derivative of $f_{mn}(x)$ "suggest that we try" $I=Px^4\left(\log {x}\right)^2 +Qx^4\log{x}+Rx^4+C$? Where does this trial formula come from?
The example goes as follows:
Find the derivative of $f_{mn}(x)=x^m\left(\log {x}\right)^n$ and use the result to suggest a trial formula for $I=\int x^3\left(\log {x}\right)^2dx$. Thus evaluate this integral.
Solution: We have
$$f'_{mn}(x)=mx^{m-1}\left(\log {x}\right)^n+nx^{m-1}\left(\log {x}\right)^{n-1}.$$
This suggests that we try
$$I=Px^4\left(\log {x}\right)^2+Qx^4\log{x}+Rx^4+C$$
for constants $P$, $Q$, $R$ and $C$. Differentiating we get
$$\frac{dI}{dx} = 4Px^3\left(\log {x}\right)^2 + 2Px^3\log{x} + 4Qx^3\log{x} + Qx^3 + 4Rx^3 = x^3\left(\log {x}\right)^2,$$
solving for $P$, $Q$ and $R$ we arrive at the right answer:
$$\int x^3\left(\log {x}\right)^2dx=\frac{1}{4}x^4\left(\log {x}\right)^2-\frac{1}{8}x^4\log{x}+\frac{1}{32}x^4+C.$$
Please note my level of mathematics is still "in development": I am learning without a teacher.
BACKGROUND: In my efforts I did notice the following, which also results in the right answer:
$$\frac{d}{dx}x^m\left(\log {x}\right)^n=mx^{m-1}\left(\log {x}\right)^n+nx^{m-1}\left(\log {x}\right)^{n-1}.$$
Integrating both sides we get:
$$x^m\left(\log {x}\right)^n=m\int x^{m-1}\left(\log {x}\right)^n dx+n\int x^{m-1}\left(\log {x}\right)^{n-1}dx.$$
Now we can define $g_{mn}(x)$ as follows:
$$g_{mn}\left(x\right)=\int x^{m-1}\left(\log {x}\right)^n dx=\frac{1}{m}x^m\left(\log {x}\right)^n-\frac{n}{m}\int x^{m-1}\left(\log {x}\right)^{n-1}dx.$$
Taking $m=4$ and $n=2$ we get:
\begin{align}
I&=\int x^3\left(\log {x}\right)^2dx=g_{42}(x)=\frac{1}{4}x^4\left(\log {x}\right)^2-\frac{1}{2}\int x^3\log{x}\,dx\\
&=\frac{1}{4}x^4\left(\log {x}\right)^2-\frac{1}{2}g_{41}(x)=\frac{1}{4}x^4\left(\log {x}\right)^2-\frac{1}{8}x^4\log{x}+\frac{1}{32}x^4+C.
\end{align}
|
Note that$$
f_{m, n}'(x) = mx^{m - 1} (\ln x)^n + nx^{m - 1} (\ln x)^{n - 1} = P_{m, n}(x, \ln x),
$$
where $P_{m, n}(x, y) = mx^{m - 1} y^n + nx^{m - 1} y^{n - 1}$ is a polynomial of $x$ and $y$, and $\deg_x P_{m, n} = m - 1$, $\deg_y P_{m, n} = n$.
First, the primitive function of $f_{m_0, n_0}(x) = x^{m_0} (\ln x)^{n_0}$ can be guessed as a linear combination of some $f_{m, n}$'s plus a constant $C$. Next, note that $P_{m, n}$ is a monomial of degree $m - 1$ with respect to $x$, i.e. $P_{m, n}(x, y) = x^{m - 1} Q_{m, n}(y)$, thus the combination should be of $f_{m, n}$'s all with $m = m_0 + 1$. Now, since $\deg_y P_{m, n} = n$, then the $f_{m_0 + 1, n}$'s in the combination should satisfy $0 \leqslant n \leqslant n_0$. Therefore, the guessed form is$$
F_{m_0, n_0}(x) = \sum_{k = 0}^{n_0} a_k f_{m_0 + 1, n}(x) + C.
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving by definition that $\lim_{(x,y) \to (1,2)}\frac{3x-4y}{x+y}=-\frac{5}{3}$
Proving by definition that $\lim_{(x,y) \to (1,2)}\frac{3x-4y}{x+y}=-\frac{5}{3}$
Take $\epsilon>0$, I want to find $\delta>0$ such that:
$$\lVert (x-1,y-2)\rVert <\delta \Rightarrow \left\lvert \frac{3x-4y}{x+y}+\frac{5}{3}\right\rvert<\epsilon$$
So I started by adding both fractions and obtained:
$$\left\lvert \frac{3x-4y}{x+y}+\frac{5}{3}\right\rvert=\frac{7}{3}\left\lvert\frac{2x-y}{x+y}\right\rvert=\frac{7}{3}\left\lvert\frac{2x-y-2+2}{x+y}\right\rvert=\frac{7}{3}\left\lvert\frac{2(x-1)-(y-2)}{x+y}\right\rvert$$
Now, I have $\lvert x-1\rvert\leq\sqrt{(x-1)^2+(y-2)^2}<\delta$ and $\lvert y-2\rvert\leq\sqrt{(x-1)^2+(y-2)^2}<\delta$
However, im not being able to bound $\frac{1}{\lvert x+y\rvert}$
Am I on the correct track? Any suggestions? Thank you very much.
|
On bounding $\frac{1}{|x+y|}$. If you take $\delta<1/2$, then
$$
|x-1|<\delta
$$
and
$$
|y-2|<\delta
$$
yields
$$
1/2<x
$$
and
$$
3/2<y
$$
and so
$$
|x+y|=x+y>2
$$
which is the part you are worried about.
The intuition behind this: Note that you are worried about $|x+y|$ getting very small, but of course it wont since in the limit, $x\approx 1$ and $y\approx 2$.
|
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|
Why $\frac{a^{2m}-1}{a+1}=\frac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+\cdots+a^2+1)?$ I know that $\dfrac{a^2-b^2}{a+b} = a-b$, because$$
a^2-b^2 = aa -ab+ab- bb = a(a-b)+(a-b)b = (a-b)(a+b).
$$
Also, I know that$$
\frac{a^n-b^n}{a+b} = a^{n-1}-a^{n-2}b+\cdots-ab^{n-2}+ b^{n-1}.$$
But I do not understand this equality below:
$$\frac{a^{2m}-1}{a+1}=\frac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+\cdots+a^2+1).$$
|
$(a^2-1)(a^{2m-2}+a^{2m-4}+\cdots+a^2+1)$
$=a^{2m}+a^{2m-2}+a^{2m-4}+\cdots+a^4+a^2-(a^{2m-2}+a^{2m-4}+\cdots+a^2+1)$
$=a^{2m}+a^{2m-2}-a^{2m-2}+a^{2m-4}-a^{2m-4}+\cdots+a^4-a^4+a^2-a^2-1$
$=a^{2m}-1$
$\Rightarrow \dfrac{a^{2m}-1}{a+1}=\dfrac{a^2-1}{a+1}(a^{2m-2}+a^{2m-4}+\cdots+a^2+1)$
|
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|
Find the second derivative of y when y is given in terms of x. (Solved) y = $(2+1/x)^3$
Find y''.
Explanation for help:
The correct answer is y'' = $6/x^3(2+1/x)(2+2/x)$.
So far, I'm at y' = $-3/x^2(2+1/x)^2$.
I'm not sure how to get from y' to y'', though.
Could someone please show how to solve for y'' starting from y'?
This problem has been solved.
|
Starting with $y' = −3/x^2(2+1/x)^2$, we solve for $y''$,
$$
y'' = \frac{\mathrm{d}}{\mathrm{d}x}\Big[-3/x^2\Big](2+1/x)^2 + \frac{\mathrm{d}}{\mathrm{d}x}\Big[(2+1/x)^2\Big](-3/x^2).
$$
We now simplify and then factor.
\begin{array}{rl}
y'' = & 6/x^3\cdot(2+1/x)^2 + (-2/x^2)\cdot(2+1/x)\cdot(-3/x^2) \\
= & 6/x^3\cdot(2+1/x)^2 + 6/x^4\cdot(2+1/x) \\
= & 6/x^3\cdot\Big((2+1/x)^2+1/x\cdot(2+1/x)\Big) \hspace{10pt}\text{($6/x^3$ was factored out here)} \\
= & 6/x^3\cdot(4+4/x+1/x^2+2/x+1/x^2) \\
= & 6/x^3\cdot(2/x^2+6/x+4) \\
= & 12/x^3\cdot(1/x^2+3/x+2) \\
= & 12/x^3\cdot(1/x+2)\cdot(1/x+1) \\
\end{array}
Hence we have
$$ y'' = 6/x^3\cdot(2+1/x)\cdot(2+2/x). $$
|
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|
Prove that $\sqrt{11}-1$ is irrational by contradiction I am working on an assignment in discrete structures and I am blocked trying to prove that $\sqrt{11}-1$ is an irrational number using proof by contradiction and prime factorization.
I am perfectly fine doing it with only $\sqrt{11}$, but I am completely thrown off by the $-1$ when it comes to the prime factorization part.
My current solution looks like this :
$$ \sqrt {11} -1 = \frac {a}{b}$$
$$ \sqrt {11} = \frac {a}{b} + 1$$
$$ \sqrt {11} = \frac {a+b}{b}$$
$$ 11 = \left(\frac {a+b}{b}\right)^2$$
$$ 11 = \frac {(a+b)^2}{b^2}$$
$$ 11 = \frac {a^2 + 2ab + b^2}{b^2}$$
$$ 11 b^2 = a^2 + 2ab +b ^2$$
$$ 10b^2 = a^2 + 2ab $$
At that point, is it acceptable to conclude that a² is a multiple of 11 even though we have a trailing $2ab$?
The required method is then to conclude using prime factorization that $a = 11k$ and replace all that in the formula above to also prove $b$, however, I am again stuck with the ending $2ab$.
Would it instead be correct to prove that $\sqrt{11}$ is rational using the usual method and that, by extension, $\sqrt{11} - 1$ is also rational?
Thank you
|
we prove $\sqrt11$ is irrational by contradiction.
let $\sqrt11$ is rational,$\exists a,b$ such that $(a,b)=1$ and
$\sqrt{11}=\frac{a}{b}$
$\implies 11=\frac{a^2}{b^2}$
$a^2=11.b^2 \implies 11\mid a^2 $ and $11$ is prime
therefore, $11\mid a, \exists a_1$ such that $a=11a_1$
$$a^2=(11a_1)^2=121a_1^2=11b^2$$
$$\implies b^2=11a_1^2 $$
$\implies 11\mid b$
it violated $(a,b)=1$.
by contradiction, $\sqrt{11}$ is irrational.
now I go to this problem,
to prove $\sqrt{11}-1$ is irrational by contradiction,
assume that $\sqrt{11}-1$ is rational.
We know that $1$ is rational number.
We also know that sum of two rational numbers is always rational number.
$[\sqrt{11}-1]+1$ is rational number.
but, $\sqrt{11}$ is irrational
by contradiction, $\sqrt{11}-1$ is irrational
|
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|
Solving Recurrence Relation for Series Solution of an ODE I am trying to solve the below problem:
Assume $y = \sum_{n=0}^{\infty}a_nx^n$ is a solution to $(x-1)y''-(x-3)y'-y=0$. Find $a_n$.
I took both derivatives of $y$, plug them into the equation, modify the indices until each series has the same $x^n$, and take terms out of the series such that they have the same index to bring all the terms under the same summation. I arrive at
$\sum_{n=0}^{\infty}[(n+1)(n)a_n-(n+1)(n+2)a_{n+2} - na_n+3(n+1)a_{n+1}-a_n]x^n + (-2a_2+3a_1-a_0)=0$
Therefore $$a_2 = \frac{3a_1}{2}-\frac{a_0}{2}$$
and
$$a_{n+2} = \frac{(n+3)a_{n+1}-a_n}{(n+2)}, n = 1,2,3...$$
I also find some terms as
$$a_3 = \frac{4a_2}{3}-a_1 = \left(\frac{4\cdot 3}{3\cdot 2} - 1 \right) a_1-\frac{4}{3\cdot 2}a_0$$
$$a_4 = \left(\frac{5\cdot 4\cdot 3}{4\cdot 3\cdot 2} - \frac{13}{8} \right) a_1-\left(\frac{5\cdot 4}{4\cdot 3\cdot 2} - \frac{1}{8} \right)a_0$$
$$a_5 = \left(\frac{6\cdot 5\cdot 4\cdot 3}{5\cdot 4\cdot 3\cdot 2} - \frac{53}{40} \right) a_1-\left(\frac{6\cdot 5\cdot 4}{5\cdot 4\cdot 3\cdot 2} - \frac{17}{60} \right)a_0$$
I can see a pattern for the first set of coefficients, but not the next. For example the first part for $a_1$ is $a_n = \frac{n+1}{2}$
Any help would be greatly appreciated. Thanks!
|
Following along up until the difference equation for the coefficients leads one to
$$a_{n+2} = \frac{(n+3) \, a_{n+1} - a_{n}}{n+2} \hspace{5mm} n \geq 0.$$
Now, for a few values of $n$ it is discovered that
\begin{align}
a_{2} &= \frac{3 a_{1} - a_{0}}{2!} \\
a_{3} &= \frac{4 a_{2} - a_{1}}{3} = \frac{10 a_{1} - 4 a_{0}}{3!} \\
a_{4} &= \frac{5 a_{3} - a_{2}}{4} = \frac{41 a_{1} - 17 a_{0}}{4!}.
\end{align}
The general form follows as
$$a_{n} = \frac{\alpha_{n} \, a_{1} - \beta_{n} \, a_{0}}{n!} \hspace{5mm} n \geq 0.$$
Further values of $n$ will lead to the discovery of $\alpha_{n} \in \{ 0, 1, 3, 10, 41, 206, 1237, \cdots \}$ which can be seen as $\lfloor (e-1) \, n! \rfloor$, or sequence A002627 of the Oeis. Similarly $\beta_{n} = \lfloor (e-2) \, n! \rfloor$ and $\beta_{n} \in \{ 1, 0, 1, 4, 17, 86, 517, \cdots \}$ and is sequence A056542.
From this the solution can then be seen as
$$f(x) = a_{1} \, \sum_{n=0}^{\infty} \lfloor (e-1) \, n! \rfloor \, \frac{x^{n}}{n!} - a_{0} \, \sum_{n=0}^{\infty} \lfloor (e-2) \, n! \rfloor \, \frac{x^{n}}{n!}.$$
This could also be seen in the form:
$$f(x) = a_{0} - a_{0} \, \sum_{n=2}^{\infty} \lfloor (e-2) \, n! \rfloor \, \frac{x^{n}}{n!} + a_{1} \, \sum_{n=1}^{\infty} \lfloor (e-1) \, n! \rfloor \, \frac{x^{n}}{n!}.$$
|
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|
Indefinite integral of $\sqrt{x^2-x}$ i was trying to compute the indefinite integral:
$$
\int\sqrt{x^2-x}dx
$$
but i got stuck: after a few (unsuccessful) attempts for some $u$-substitution, i tried integration by parts:
$$
\int\sqrt{x^2-x} \ dx=\int(x)'\sqrt{x^2-x} \ dx= \\ x\sqrt{x^2-x}-\int x(\sqrt{x^2-x})'dx= \\
=x\sqrt{x^2-x}-\frac{1}{2}\int x\frac{2x-1}{\sqrt{x^2-x}}dx= \\
=x\sqrt{x^2-x}-\int \frac{x^2}{\sqrt{x^2-x}}dx+\frac{1}{2}\int \frac{x}{\sqrt{x^2-x}}dx=...
$$
and now what? Can anybody help?
|
We can eliminate the square root with an Euler substitution of
$$t = \sqrt{x^2-x} - x \implies x = -\frac{t^2}{1+2t} \implies dx = -\frac{2t(1+t)}{(1+2t)^2} \, dt$$
Then
$$\begin{align*}
I &= \int \sqrt{x^2-x} \, dx \\[1ex]
&= -2 \int \left(t-\frac{t^2}{1+2t}\right) \frac{t(1+t)}{(1+2t)^2} \, dt \\[1ex]
&= -2 \int \frac{t^2(1+t)^2}{(1+2t)^3} \, dt \\[1ex]
&= - \frac18 \int \left(1 + 2t - \frac2{1+2t} + \frac1{(1+2t)^3}\right) \, dt
\end{align*}$$
|
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|
Evaluation of $\lim _{ x\rightarrow 0 }{ \frac { \sin ^{ -1 }{ x } -\tan ^{ -1 }{ x } }{ { x }^{ 3 } } } \quad $ without using L'Hospital rule My aim is to evaluate the following limit
$\displaystyle\lim _{ x\rightarrow 0 }{ \frac { \sin ^{ -1 }{ x } -\tan ^{ -1 }{ x } }{ { x }^{ 3 } } } \quad$
which evaluates to $\frac{1}{2}$ after using L'Hospital rule for rather $5$ steps, which of course is a tedious task. Could anyone suggest some shorter and elegant methods for evaluating this limit?
Thanks.
|
We have
$$
\sin(\arcsin(x)-\arctan(x))=\frac{x(1-\sqrt{1-x^2})}{\sqrt{1+x^2}}
$$
and
$$
\lim_{x\to 0}\left(\frac{\arcsin(x)-\arctan(x)}{x^3}\right)\equiv \lim_{x\to 0}\left(\frac{\sin(\arcsin(x)-\arctan(x))}{\sin(x^3)}\right)
$$
then
$$
\frac{\sin(\arcsin(x)-\arctan(x))}{\sin(x^3)} = \frac{x^3(1-\sqrt{1-x^2})}{x^2\sqrt{1+x^2}\sin(x^3)} = \left(\frac{x^3}{\sin(x^3)}\right)\frac{x^2}{x^2(1+\sqrt{1-x^2})\sqrt{1+x^2}}
$$
then
$$
\lim_{x\to 0}\left(\frac{\sin(\arcsin(x)-\arctan(x))}{x^3}\right)\equiv\lim_{x\to 0}\left(\frac{x^3}{\sin(x^3)}\right)\frac{1}{(1+\sqrt{1-x^2})\sqrt{1+x^2}} = \frac{1}{2}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}.$ Problem
Evaluate $$\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}.$$
My solution
Notice that $$\lim_{n \to \infty}\frac{n+n^2+n^3+\cdots +n^n}{n^n}=\lim_{n \to \infty}\frac{n(n^n-1)}{(n-1)n^n}=\lim_{n \to \infty}\frac{1-\dfrac{1}{n^n}}{1-\dfrac{1}{n}}=1,$$and
$$\lim_{n \to \infty}\frac{1+2^n+3^n+\cdots+n^n}{n^n}=\frac{e}{e-1}.$$
Hence,\begin{align*}\lim\limits_{n\rightarrow \infty}\frac{n+n^2+n^3+\cdots +n^n}{1^n+2^n+3^n+\cdots +n^n}&=\lim_{n \to \infty}\frac{\dfrac{n+n^2+n^3+\cdots +n^n}{n^n}}{\dfrac{1+2^n+3^n+\cdots +n^n}{n^n}}\\&=\frac{\lim\limits_{n \to \infty}\dfrac{n+n^2+n^3+\cdots +n^n}{n^n}}{\lim\limits_{n \to \infty}\dfrac{1+2^n+3^n+\cdots +n^n}{n^n}}\\&=1-\frac{1}{e}.\end{align*}
The solution posted above need to quote an uncommon limit. Is there another more simple and more direct solution?
|
We shall prove that
$$
\frac{1^n+2^n+\cdots+n^n}{n^n}\to \frac{\mathrm{e}}{\mathrm{e}-1}\tag{$\star$}
$$
First of all, $\log (1-x)<-x$, for all $x\in(0,1)$ and hence
$$
\log\left(1-\frac{k}{n}\right)<-\frac{k}{n}\quad\Longrightarrow\quad
\left(1-\frac{k}{n}\right)^n<\mathrm{e}^{-k}, \quad \text{for all $n>k$}
$$
and thus
$$
\frac{1^n+2^n+\cdots+n^n}{n^n}=\sum_{k=0}^{n-1}\left(1-\frac{k}{n}\right)^n
<\sum_{k=0}^{n-1}\mathrm{e}^{-k}<\sum_{k=0}^{\infty}\mathrm{e}^{-k}=\frac{1}{1-\frac{1}{\mathrm{e}}}=\frac{\mathrm{e}}{\mathrm{e}-1}.
$$
Hence
$$
\limsup_{n\to\infty}\frac{1^n+2^n+\cdots+n^n}{n^n}\le \frac{\mathrm{e}}{\mathrm{e}-1}. \tag{1}
$$
Meanwhile, for all $k\in\mathbb N$,
$$
\frac{(n-k)^n}{n^n}=\left(1-\frac{k}{n}\right)^n\to\mathrm{e}^{-k},
$$
and hence, for every $k\in\mathbb N$ fixed,
$$
\frac{1^n+2^n+\cdots+n^n}{n^n}\ge \frac{(n-k)^n+(n-k+1)^n+\cdots+n^n}{n^n}\\=\left(1-\frac{k}{n}\right)^n+\left(1-\frac{k-1}{n}\right)^n+\cdots+\left(1-\frac{1}{n}\right)^n+1\to \mathrm{e}^{-k}
+\mathrm{e}^{-k+1}+\cdots+1=\frac{\mathrm{e}-\mathrm{e}^{-k}}{\mathrm{e-1}}.
$$
Hence, for all $k\in\mathbb N$,
$$
\liminf_{n\to\infty}\frac{1^n+2^n+\cdots+n^n}{n^n}\ge \frac{\mathrm{e}-\mathrm{e}^{-k}}{\mathrm{e-1}}
$$
and thus
$$
\liminf_{n\to\infty}\frac{1^n+2^n+\cdots+n^n}{n^n}\ge \sup_{k\in\mathbb N}\frac{\mathrm{e}-\mathrm{e}^{-k}}{\mathrm{e-1}}=\frac{\mathrm{e}}{\mathrm{e-1}} \tag{2}
$$
Combining $(1)$ & $(2)$, we obtain $(\star)$.
|
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|
There is a triangle having two sides $ax^2 +2hxy +by^2 = 0$ and orthocenter $(p, q)$ Find the other side. There is a triangle having two sides $ax^2 +2hxy +by^2 = 0$ and orthocenter $(p, q)$. Find the equation of
other side.
I first let the sides are $OM$: $y = m_1x$ and $ON:$ $y = m_2x$. Where $m_1m_2= \frac{-2h}{b}$ and $m_1 +
m_2 = \frac{a}{b}$
Then I found the foot of the perpendicular of $(p, q)$ on $OM$. It was
$$Q\left(\frac{p-m_1q}{m_1^2 + 1},m_1\frac{p-m_1q}{m_1^2 + 1}\right) $$
Now $Q$ is the midpoint of
$O$ and $M$. Thus I got $M$ is
$$\left(2\frac{p-m_1q}{m_1^2 + 1},2m_1\frac{p-m_1q}{m_1^2 + 1} \right)$$
Similarly, N is
$$\left(2\frac{p-m_2q}{m_2^2 + 1} ,2m_2\frac{p-m_2q}{m_2^2 + 1} \right)$$
Using two point formula and making some manipulation, the equation of line $MN $
i.e. the third side became
$$\frac{(m_1^2 + 1)y- 2m_1p+2m_1^2q}{(m_1^2 + 1)x- 2p+2m_1q} =\frac{m_1m_2p}{(m_1+m_2)p+q}$$
From now where to go ?
|
Put $A(0,0),B(x,0),C(?,?)$. The coordinates of the vertex $C$ are intersection of the perpendiculars to the altitudes passing by $A$ and $B$, so we have after easy calculation $C(p,\frac{p(x-p)}{q})$.
Now one has $$(AC)^2=p^2+\left(\frac{p(x-p)}{q}\right)^2\\(BC)^2=(p-x)^2+\left(\frac{p(x-p)}{q}\right)^2$$ The quotient $\dfrac{AC}{BC}$ is root of the given condition
$ax^2+2hxy+by^2=0\iff a(\frac xy)^2+2h(\frac xy)+b=0$.
Consequently one has, putting $R=\dfrac{-2h\pm\sqrt{h^2-4ab}}{2a}$ (as data of problem)
$$\frac{(p-x)^2+\left(\frac{p(x-p)}{q}\right)^2}{p^2+\left(\frac{p(x-p)}{q}\right)^2}=R^2$$
This quadratic equation gives the value of $x$ from which we have calculated the three vertices of the triangle determining this way the three sides, $x$ itself being the required side.
|
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|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.