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how do you solve $a^2+b^2+c^2=d^3$ let $ a,b,c,d$ be 4 integers such that $\gcd(a,b,c,d)=1$. How do you find the integral solutions of the equation: $$a^2+b^2+c^2=d^3$$
|
It is a theorem that one can identically solve,
$$x_1^2+x_2^2+\dots+x_n^2 = (y_1^2+y_2^2+\dots+y_n^2)^k$$
for any positive integer n and k. Thus the kth power of n squares is itself the sum of n squares. For example, for $n =3$, we have,
$k=2:$
$$(a^2-b^2-c^2)^2+(2ab)^2+(2ac)^2 = (a^2 + b^2 + c^2)^2$$
$k=3:$
$$a^2(a^2 - 3b^2 - 3c^2)^2 + b^2(-3a^2 + b^2 + c^2)^2 + c^2(-3a^2 + b^2 + c^2)^2=(a^2 + b^2 + c^2)^3$$
and so on. See Theorem 1 at https://sites.google.com/site/tpiezas/004.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find z: $|\frac{z-1}{z-i}|=1$ and $|\frac{z-3i}{z+i}|=1$ Give a complex number z:
$|\frac{z-1}{z-i}|=1$ and $|\frac{z-3i}{z+i}|=1$
Find z ?
Could someone help me through this problem ?
|
${\left| {z - 1} \right|^2} = {\left| {z - i} \right|^2}$ and ${\left| {z - 3i} \right|^2} = {\left| {z + i} \right|^2}$, for $z = x + iy$ this gives you ${\left( {x - 1} \right)^2} + {y^2} = {x^2} + {\left( {y - 1} \right)^2}$ and ${x^2} + {\left( {y - 3} \right)^2} = {x^2} + {\left( {y + 1} \right)^2}$, since ${\left| {z - a - ib} \right|^2} = {\left( {x - a} \right)^2} + {\left( {y - b} \right)^2}$.
After simplifying, you get 2 linear equations with 2 unknowns
$\left\{ \begin{gathered}
2x + 1 = 2y + 1 \\
- 6x + 9 = 2y + 1 \\
\end{gathered} \right.$
, which you can solve.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Convergence of $\sum_{n=1}^{\infty}\frac{1}{n}\Big(\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}\Big)$ $\displaystyle\sum_{n=1}^{\infty}\frac{\sqrt[3]{(n+1)^{2}} - \sqrt[3]{n^{2}}}{n}$
Converging or Diverging? I guess I have to lower the fraction so that the roots will get away and I will have $\frac{1} {n}$ that diverges. But I have no idea how to do that.
Any ideas?
|
Recall that $(1+\varepsilon)^p\approx 1+p\varepsilon$; thus
$$ \frac{\sqrt[3]{(n+1)^2} - \sqrt[3]{n^2}}{n}
= \frac{n^{2/3} \sqrt[3]{(1+\frac1n)^2} - \sqrt[3]{n^2}}{n}
\approx \frac{n^{2/3}\left(1+\frac2{3n}\right) - n^{2/3}}{n}
= \frac{2}{3n^{4/3}}
$$
So it should converge.
To justify the conclusion more formally, we can replace $(1+\varepsilon)^p\approx 1+p\varepsilon$ with the more exact statement
$$ 1+p\varepsilon\le (1+\varepsilon)^p\le e^{p\varepsilon} $$
(The left-hand inequality is Bernoulli's inequality; the right-hand one is $1+x\le e^x$.) The informal work above showed that the series converges, so we'll take the upper bound; using it in place of $\approx$ above gives
$$ \frac{\sqrt[3]{(n+1)^2} - \sqrt[3]{n^2}}{n}
\le n^{2/3} \cdot \frac{e^{2/(3n)} - 1}{n}
= \frac2{3n^{4/3}} \cdot \frac{e^{2/(3n)} - 1}{\bigl(\frac2{3n}\bigr)}
$$
and now apply the standard limit $\displaystyle\lim_{h\to 0} \frac{e^h-1}{h}$.
|
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|
Prove inequality using AM-GM inequality. $$\sqrt{a}+\sqrt{b}\geq \sqrt{a+b}$$ I know it is trivial to prove straight forward but I need to prove it using AM-GM inequality.
|
\begin{align*}
\sqrt a + \sqrt b
&= \frac{(\sqrt a + \sqrt b + \sqrt2 \sqrt[4]{ab}) +
(\sqrt a + \sqrt b - \sqrt2 \sqrt[4]{ab})}{2} \\
&\ge \sqrt{\big(\sqrt a + \sqrt b + \sqrt2 \sqrt[4]{ab}\big)
\big(\sqrt a + \sqrt b - \sqrt2 \sqrt[4]{ab}\big)} \\
&= \sqrt{\big(\sqrt a + \sqrt b\big)^2 - 2 \sqrt{ab}} \\
&= \sqrt{a+b}
\end{align*}
(To apply AM/GM thus requires $\sqrt a + \sqrt b \ge \sqrt2\sqrt[4]{ab}$; happily, $\sqrt a + \sqrt b \ge 2\sqrt[4]{ab}$, by AM/GM again.)
|
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|
Exact closed form expression of $(2^0+...+2^n)+(2^1+...+2^{n+1})+...+(2^n+...+2^{2n})$ Exact closed form of this expression $(2^0+...+2^n)+(2^1+...+2^{n+1})+...+(2^n+...+2^{2n})$
I assume this means there is just one $2^0$ and one $2^{2n}$ and a double of all the terms in between?
|
Since all terms are divisible by $2^0+\cdots+2^n$, you can split off that common factor:
$$
(2^0+\cdots+2^n)\times(2^0+\cdots+2^n)=(2^{n+1}-1)^2=2^{2n+2}-2^{n+2}+1.
$$
In fact, the above misreads the question to be about $(2^0+\cdots+2^n)+(2^1+\cdots+2^{n+1})+\cdots+(2^n+\cdots+2^{2n})$ rather than about $(2^0+\cdots+2^n)+(2^1+\cdots+2^{n+1})+(2^n+\cdots+2^{2n})$ as it actually states (in title and body). The latter question is less interesting, so I'll assume that OP actually meant the former with the extra ellipses and leave the answer above. But just in case a sum of precisely $3$ parenthesised sum is meant; that gives
$$
\begin{align}
(2^0+\cdots+2^n)\times(1+2+2^n)
&=(2^{n+1}-1)\times(2^n+3)
\\&=2^{2n+1}+3\times2^{n+1}-2^n-3.
\\&=2^{2n+1}+5\times2^n-3.
\end{align}
$$
|
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|
Sum of Harmonic numbers $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$ Finding the closed form of:
$$\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$$
where, $\displaystyle H_n^{(2)} = \sum\limits_{k=1}^{n}\frac{1}{k^2}$
It appears when we try to determine the summation $\displaystyle \sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}$, using the generating function:
\begin{align}
\sum_{n=1}^{\infty} \frac{H_n}{n^3} \, x^{n} &= - \frac{1}{2} \, \sum_{n=1}^{\infty}\frac{1}{n^2} \, \sum_{k=1}^{n} \frac{(1-x)^k}{k^2} - \frac{\zeta(2)}{2} \, \operatorname{Li}_2(x) + \frac{7 \, \zeta(4)}{8} - \frac{1}{4} \, \operatorname{Li}_2^2(1-x) + \frac{\zeta^2(2)}{4} + \operatorname{Li}_4(x) \\
& \hspace{5mm} + \frac{1}{4} \, \log^2 x \, \log^2(1-x) + \frac{1}{2}\log x \, \log (1-x) \, \operatorname{Li}_2(1-x) + \zeta(3) \, \log x - \log x \,
\operatorname{Li}_2(1-x)
\end{align}
when we write, $\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{n^2}\sum\limits_{k=1}^{n}\frac{(1-x)^k}{k^2} = \zeta(2)\operatorname{Li}_2(1-x) + \operatorname{Li}_4(1-x) - \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{n^2}(1-x)^n$
Combined with Cleo's closed form here, I know what the closed form should be, but how do I derive the result ?
|
HINT: Consider $\displaystyle \sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}=\sum\limits_{n=1}^{\infty} \frac{H_{n-1}^{(2)}}{2^nn^2}+\operatorname{Li}_4\left(\frac{1}{2}\right)$ and then express the remaining sum as a double integral. After some work, you get
$$\int_0^1 \frac{\displaystyle\log(x)\operatorname{Li}_2\left(\frac{x}{2}\right)}{x-2} \ dx+\operatorname{Li}_4\left(\frac{1}{2}\right)$$
and after letting $x\mapsto 2x$ combined with the integration by parts, you arrive at some integrals
pretty easy to finish. I'm confident you can finish the rest of the job to do.
$$\frac{\pi^4}{1440}-\frac{\pi^2}{3}\log^2(2)+\frac{1}{24}\log^4(2)+\frac{7}{24}\pi^2\log^2(2)+\frac{1}{4}\log(2)\zeta(3)+\operatorname{Li}_4\left(\frac{1}{2}\right)$$
|
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|
Prove that $\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}=\frac{3-2\sqrt{2}}{4}$ How can I prove that
$$\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}=\frac{3-2\sqrt{2}}{4}$$
|
$$\sum_{n=0}^{\infty }\frac{(2n+1)!!}{(n+1)!}2^{-(2n+4)}= \sum_{n=0}^{\infty }\frac{(2n+1)!}{(n+1)!n!}2^{-(3n+4)} = \dfrac{1}{16}\sum_{n=0}^{\infty }{2n+ 1\choose n}x^{n}$$
with $x = \dfrac{1}{8}$.
We are going to compute $f(x) = \sum_{n=0}^{\infty }{2n+ 1\choose n}x^{n}$ using the integral presentation of binomial coefficients:
$${2n+ 1\choose n} = \dfrac{1}{2\pi i}\int_{|z| =1}(z+1)^{2n+1}z^{-n-1}dz$$
so we have
\begin{align}
f(x) = &\dfrac{1}{2\pi i}\int_{|z| =1}\sum_{n=0}^\infty \left(\dfrac{(z+1)^2}{z}x\right)^n \dfrac{1+z}{z}dz \\
= & \dfrac{1}{2\pi i}\int_{|z| =1} \dfrac{1}{1-\dfrac{(z+1)^2}{z}x} \dfrac{1+z}{z}dz \\
= & \dfrac{1}{2\pi i}\int_{|z| =1} \dfrac{1+z}{z-(z+1)^2x} dz \\
\end{align}
The last integral is not too hard to evaluate with $x = \dfrac{1}{8}$
|
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|
Problem involving cube roots of unity Given that $$\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2\omega^2\;\;\;\;\;(1)$$ $$\frac{1}{a+\omega^2}+\frac{1}{b+\omega^2}+\frac{1}{c+\omega^2}=2\omega\;\;\;\;\;(2)$$
Find $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=?\;\;\;\;\;\;(3)$$
I tried adding the given two equations, and simplified them. I'll show the working for one term here. $$\frac{1}{a+\omega}+\frac{1}{a+\omega^2}=\frac{2a-1}{a^2-a+1}=\frac{(2a-1)(a+1)}{(a^3+1)}=\frac{2a^2+a-1}{a^3+1}$$
$$\frac{1}{a+1}=\frac{a^2-a+1}{a^3+1}=\frac{1}{2}\left[\frac{(2a^2+a-1)-3(a-1)}{a^3+1}\right]=\frac12\left[\frac{2a^2+a-1}{a^3+1}\right]-\frac32\left[\frac{a-1}{a^3+1}\right]$$
Now, I cannot eliminate that extra term which I get at the end of the above expression. Is there hope beyond this? Or is there a better alternative?
|
Put $A(x)=(x+a)(x+b)(x+c)=x^3+ux^2+vx+w$. Then
$$\frac{A^{\prime}(x)}{A(x)}=\frac{1}{x+a}+\frac{1}{x+b}+\frac{1}{x+c}$$ Hence we get $A^{\prime}(\omega)=2\omega^2 A(\omega)$ and $A^{\prime}(\omega^2)-2\omega A(\omega^2)=0$, or if $Q(x)=xA^{\prime}(x)-2A(x)$, that $Q(\omega)=Q(\omega^2)=0$. Hence $Q(x)$ is divisible by $(x-\omega)(x-\omega^2)=x^2+x+1$. We have
$$Q(x)=x^3-vx-2w=(x-1)(x^2+x+1)+(-vx+1-2w)$$
Hence $v=0$, $w=1/2$, and $Q(x)=(x-1)(x^2+x+1)$, $A'(1)=2A(1)$, and we are done.
|
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|
How does the recursion relation work in the solution to this differential equation (using series)? Sorry for the vague title but it would not let me post the first step and last step of this equation (too many characters!).
How does $$\dfrac{a_0}{3n(3n-1)(3n-3)(3n-4)\cdots 9 \cdot 8 \cdot 6 \cdot 5 \cdot 3 \cdot 2} = \dfrac{a_0\Gamma(\frac{2}{3})}{3^nn!3^n\Gamma(n+\frac{2}{3})}?$$
This is a text book example I'm having trouble following. Solve the differential equation
$y'' = xy$ using series. This makes sense in general. But when we get down to solving for the coefficients of the series, it confuses me.
Based of the relations $$a_nn(n-1) = 0 \text{ for } n=0,1,2$$ and $$a_nn(n-1) = a_{n-3} \text{ for } n = 3,4,\ldots$$
we have
$$\begin{align}
a_{3n} &= \dfrac{a_0}{3n(3n-1)(3n-3)(3n-4)\cdots 9 \cdot 8 \cdot 6 \cdot 5 \cdot 3 \cdot 2} \\
&= \dfrac{a_0}{3^n n! 3^n(n-\frac{1}{3})(n-1-\frac{1}{3})(n-2-\frac{1}{3})\cdots (\frac{5}{3}) (\frac{2}{3})}\\
&= \dfrac{a_0\Gamma(\frac{2}{3})}{3^nn!3^n\Gamma(n+\frac{2}{3})}\\
\end{align}$$
Particularly, that middle step there is throwing me off.
The book also provided
$$\begin{align}
a_{3n+1} &= \dfrac{a_1}{(3n+1)(3n)(3n-2)(3n-3)\cdots 10 \cdot 9 \cdot 7 \cdot 6 \cdot 4 \cdot 3} \\
&= \dfrac{a_1}{3^n n! 3^n(n+\frac{1}{3})(n-1+\frac{1}{3})(n-2+\frac{1}{3})\cdots (\frac{7}{3}) (\frac{4}{3})}\\
&= \dfrac{a_1\Gamma(\frac{4}{3})}{3^nn!3^n\Gamma(n+\frac{4}{3})}\\
\end{align}$$
and
$$a_{3n+2} = 0,$$
but that last one makes sense.
As a side note, why do they put $3^n$ twice in the denominator? Why not $3^{2n}$?
|
From what I understand, the product I denote as $P$
$$(3n+1)(3n)(3n−2)(3n−3)⋯10⋅9⋅7⋅6⋅4⋅3$$
is really (by writing the product pairwise)
$$P = (3\cdot 4)(6\cdot 7)(9\cdot 10)\cdots(3n\cdot(3n+1))$$
the pattern is multiplying a pair, one of which is a multiple of three and the other, the increment of that. First, notice there are $n$ pairs, we can count from $3$ to $3n$, we can factor a 3 from each of these pairs to get $$P = 3^n(1\cdot4)(2\cdot 7)(3\cdot10)\cdots(n\cdot(3n+1))$$
as we can see, each pair now contains $1,2,3,...,n$. We can factor that as well to obtain $$P = 3^nn!(4)(7)(10)\cdots\left(3n+1\right)$$
Now I think they also factor out a 3 from each to obtain another $3^n$:
$$P = 3^nn!3^n(4/3)(7/3)(10/3)\cdots(n+1/3)$$
Also, they don't write $3^n3^n$ as $3^{2n}$ because they might want to make it clearer for students as to where each $3^n$ contribution came from.
|
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|
Help finding the residue of $1/(z^8+1)$ Help finding the residue of $1/(z^8+1)$
I'm integrating over $\{ Re^{it} | 0 \leq t \leq \pi \}$, and I found 4 simple poles at $z_0=e^{in\pi/8}$ where $n = 0,...,3$ and I'm trying to calculate $res(1/(z^8+1),z_0)$
calculating this: $$\lim_{z\to z_0} (z-z_0)f = \lim_{z\to z_0}\frac{z-z_0}{1+z^8},$$ now are there any algebra tricks I can do to simplify this?
|
$$z^8+1=(z-i^{\frac{1}{8}})(z+i^{\frac{1}{8}})(z-i^{\frac{3}{8}})(z+i^{\frac{3}{8}})(z-i^{\frac{5}{8}})(z+i^{\frac{5}{8}})(z-i^{\frac{7}{8}})(z+i^{\frac{7}{8}})$$
$$Res_{z=\sqrt[8]i}= \lim_{x\to\sqrt[8]i}\dfrac{(z-\sqrt[8]i)}{(z-i^{\frac{1}{8}})(z+i^{\frac{1}{8}})(z-i^{\frac{3}{8}})(z+i^{\frac{3}{8}})(z-i^{\frac{5}{8}})(z+i^{\frac{5}{8}})(z-i^{\frac{7}{8}})(z+i^{\frac{7}{8}})}
=\lim_{z\to\sqrt[8]i}\dfrac{(z-i^\frac{1}{8})}{(z-i^{\frac{1}{8}})(z+i^{\frac{1}{8}})(z-i^{\frac{3}{8}})(z+i^{\frac{3}{8}})(z-i^{\frac{5}{8}})(z+i^{\frac{5}{8}})(z-i^{\frac{7}{8}})(z+i^{\frac{7}{8}})}
=\lim_{z\to\sqrt[8]i}\dfrac{1}{(z+i^{\frac{1}{8}})(z-i^{\frac{3}{8}})(z+i^{\frac{3}{8}})(z-i^{\frac{5}{8}})(z+i^{\frac{5}{8}})(z-i^{\frac{7}{8}})(z+i^{\frac{7}{8}})}
=-\frac{i^\frac{1}{8}}{8}$$.
Now apply the same procedure to the other singularities.
|
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|
Finding the limit of a function with ArcTan I've found difficulties finding this limit ( without using Taylor series approximation, as it's intended for the secondary-school ):
$$
\lim_{x\ \to\ \infty}\left[\,
{x^{3} \over \left(\,x^{2} + 1\,\right)\arctan\left(\,x\,\right)} - {2x \over \pi}
\,\right]
$$
Thanks.
|
i think the answer is $4/\pi^2.$
i will make a change of variable $u = 1/x$ and will use the fact $\tan 1/u = \pi/2 - \tan u$ for $u > 0.$
$\begin{eqnarray}
lim_{x \to \infty} \frac{x^3}{(1+x^2) \tan x} - \frac{2x}{\pi} =
\lim_{u \to {0+}} \frac{1}{u(1+u^2)\tan(1/u)} - \frac{2}{\pi u} \\
= \frac{2}{u(1+u^2)(\pi-2\tan u)} - \frac{2}{\pi u} \\
= \frac{2\pi - 2 (1+u^2)(\pi - 2 \tan u)}{\pi u (1+u^2)(\pi - 2\tan u)}\\
= \frac{2 \pi - 2 (\pi - 2\tan u + \cdots)}{\pi u(1 + u^2)(\pi - 2 \tan u)}\\
= \frac{4 \tan u + \cdots}{\pi^2 u + \cdots} = \frac{4}{\pi^2}\\
\end{eqnarray}$
|
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|
$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$ How to find the sum of the following series:
$\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot$
Any hints.
|
According to Maple it's $1 - 5^{1/3}/2$. It looks like this comes from the
Maclaurin series for $(1-t)^{-1/3}$.
|
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Show that $2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$ If $a,b,c$ are positive real numbers, not all equal, then prove that
$$2(a^3+b^3+c^3)>a^2(b+c)+b^2(c+a)+c^2(a+b)>6abc$$
How can I show this?
|
Hint: $a^3 + b^3 > a^2b + ab^2$,
and use AM-GM inequality for $6$ terms for the other one.
|
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|
How to integrate $\frac{\sqrt{z+1}}{z}$ How to integrate $\frac{\sqrt{z+1}}{z}$
Anyone could help me? Thanks
|
$u=\sqrt{z+1}$, $z=u^2-1$. Then
$$\begin{gathered}
\int {\frac{{\sqrt {z + 1} }}
{z}dz} = \int {\frac{{2{u^2}}}
{{{u^2} - 1}}du} = 2\int {\left( {1 + \frac{1}
{{{u^2} - 1}}} \right)du} = 2\int {\left[ {1 + \frac{1}
{2}\left( {\frac{1}
{{u - 1}} - \frac{1}
{{u + 1}}} \right)} \right]du} \hfill \\
= 2\left( {u + \frac{1}
{2}\ln \left| {\frac{{u - 1}}
{{u + 1}}} \right| + C} \right) = 2\left( {\sqrt {z + 1} + \frac{1}
{2}\ln \left| {\frac{{\sqrt {z + 1} - 1}}
{{\sqrt {z + 1} + 1}}} \right| + C} \right) \hfill \\
\end{gathered} $$
|
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|
Feyman's Triangle? How do you find the area of the inner triangle if the outside triangle is equilateral If triangle $ABC$ is equilateral,$BD/BC=1/3, CE/CA=1/3,$ and $ AF/AB=1/3$. What is the ratio of the area of triangle? I have problems analyzing this triangle I tried to use phythagorean, heron's formula, the formula $A$=$\frac{ab\sin(\theta)}{2}$ but I still can't figure the inside area. I've also done some research they said it's $1:7$? But what I want is to understand how they got $ 1:7$.
|
There are many proofs of this result (see here for a generalized version), but perhaps the easiest is using the theorem of Menelaus, along with some area ratios.
We use this figure. Consider collinear points $A, K, E$ on the sides of $\triangle DBG$. Menelaus's Theorem tells us:
$$\begin{align}
\frac{DA}{AB} \frac{BK}{KG} \frac{GE}{ED} &= 1 \\
\frac{3}{1} \frac{BK}{KG} \frac{2}{1} &= 1\\
\implies \frac{BK}{KG} &= \frac{1}{6}\\
\implies \frac{KG}{BG} &= \frac{6}{7}
\end{align}$$
Thus
$$\begin{align}
[\triangle KAG] &= \frac{KG}{BG} [\triangle BAG] \\
&= \frac{6}{7} \frac{BA}{DA}[\triangle DAG]\\
&= \frac{6}{7} \frac{1}{3} [\triangle DAG]\\
&= \frac{6}{21} [\triangle DAG]
\end{align}$$
By symmetry, we have that
$$ [\triangle KAG] = [\triangle LDA] = [\triangle MGD] = \frac{6}{21}[\triangle DAG] $$
Thus,
$$
\begin{align}
[\triangle LKM] &= [\triangle DAG] - ([\triangle KAG] + [\triangle LDA] + [\triangle MGD])\\
&= ( 1 - \frac{6}{21} - \frac{6}{21} - \frac{6}{21})[\triangle DAG])\\
&= \frac{1}{7}[\triangle DAG]
\end{align}$$
As desired.
|
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|
All solutions of the recurrence relation Find all solutions of the recurrence relation $$ a_n = 2a_{n-1}+15a_{n-219}-64a_{n-3}+k $$
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Plug $a_n=b^n$ into the above equation, ignoring the RHS, and get that $b$ satisfies
$$b^3-2 b^2-15 b+36=0 $$
Solutions are $b=3$ (double root) and $b=-4$. The general homogeneous solution is then
$$a_n = (A+B n) 3^n + C (-4)^n $$
For the particular solution, plug in $a_n = D \cdot 2^n$ and get that
$$D \left (1 - 1 - \frac{15}{4} + \frac{36}{8} \right ) 2^n = 2^n \implies D = \frac{4}{3} $$
Thus, the general solution is
$$a_n = (A+B n) 3^n + C (-4)^n + \frac{4}{3} 2^n$$
|
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|
Solving messy integral with modulus and trigonometry. If $$a\in \mathbb R,\int_{a-\pi}^{3\pi+a}|x-a-\pi|\sin(x/2)dx=-16$$ then a can be?
I tried substituting $x-a=u$ and then breaking into two integrals removing modulus then used $\int \sin x=-\cos x,\int x\sin x=\sin x-x\cos x$
|
The evaluation of the integral on the left hand side of the integral equation
\begin{equation*}
\int_{a-\pi }^{3\pi +a}\left\vert x-a-\pi \right\vert \sin \left( x/2\right)\tag{1}
\,dx=-16
\end{equation*}
can be carried out starting with the substitution suggested by GFauxPas $y=x-a-\pi $ in a comment, [edit] splitting the integral into two, one for $-2\pi <y<0$ and the other for $0\leq y<2\pi $, and proceeding with the substitution $z=\frac{y+a}{2}$ and integration by parts [edit end]:
\begin{eqnarray*}
-16 &=&\int_{a-\pi }^{3\pi +a}\left\vert x-a-\pi \right\vert \sin \left(
x/2\right) \,dx \\
&=&\int_{-2\pi }^{2\pi }\left\vert y\right\vert \sin \left( \frac{y+a}{2}+
\frac{\pi }{2}\right) \,dy,\qquad\qquad y=x-a-\pi \\
&=&\int_{-2\pi }^{2\pi }\left\vert y\right\vert \cos \left( \frac{y+a}{2}
\right) \,dy \\
&=&-\int_{-2\pi }^{0}y\cos \left( \frac{y+a}{2}\right) \,dy+\int_{0}^{2\pi
}y\cos \left( \frac{y+a}{2}\right) \,dy, \\
&=&-\left[ 2y\sin \frac{y+a}{2}+4\cos \frac{y+a}{2}\right] _{-2\pi }^{0} +\left[ 2y\sin \frac{y+a}{2}+4\cos \frac{y+a}{2}\right] _{0}^{2\pi }\tag{$\ast$} \\
&=&-8\cos \frac{a}{2}+4\pi \sin \frac{a}{2}-8\cos \frac{a}{2}-4\pi \sin
\frac{a}{2} \\
&=&-16\cos \frac{a}{2},
\end{eqnarray*}
because
\begin{eqnarray*}
I(y) &=&\int y\cos \left( \frac{y+a}{2}\right) \,dy \\
&=&\int 2\left( 2z-a\right) \cos z\,dz,\qquad \qquad z=\frac{y+a}{2} \\
&=&4\int z\cos z\,dz-2a\int \cos z\,dz
\end{eqnarray*}
and
\begin{eqnarray*}
I(y) &=&4\left( z\sin z-\int \sin z\,dz\right) -2a\sin z \\
&=&\left( 4z-2a\right) \sin z+4\cos z \\
&=&2y\sin \frac{y+a}{2}+4\cos \frac{y+a}{2}.\tag{$\ast$}
\end{eqnarray*}
Hence $(1)$ is equivalent to the simple trigonometric equation
\begin{equation*}
\cos \frac{a}{2}=1,\tag{2}
\end{equation*}
whose solution is
\begin{equation*}
a=4k\pi ,\text{ }k\in\mathbb{Z}.\tag{3}
\end{equation*}
|
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|
Closed form of $\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$ Is it possible to get a closed form of the following integral
$$\Phi(a,b)=\int_{0}^{1}\frac{dx}{(x^2+a^2)\sqrt{x^2+b^2}}$$
|
\begin{align}
\int_{0}^{1}\frac{\mathrm dx}{(x^2+a^2)\sqrt{x^2+b^2}}&=\int_{0}^{\large k_1}\frac{b\sec^2t}{(b^2\tan^2t+a^2)\sqrt{b^2\tan^2t+b^2}}\mathrm dt\tag1\\
&=\int_{0}^{\large k_1}\frac{\cos t}{b^2\sin^2t+a^2\cos^2t}\mathrm dt\tag2\\
&=\int_{0}^{\large k_1}\frac{\cos t}{\left(b^2-a^2\right)\sin^2t+a^2}\mathrm dt\tag3\\
&=\frac{1}{a\sqrt{b^2-a^2}}\int_{0}^{\large k_2}\frac{\mathrm dy}{y^2+1}\tag4\\
\Phi(a,b)&=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\frac{1}{a\sqrt{b^2-a^2}}\arctan\left(\frac{\sqrt{b^2-a^2}}{a\sqrt{1-b^2}}\right)}}
\end{align}
Explanation :
$(1)\;$ Use substitution $\;\displaystyle x=b\tan t\;$ and $\;\displaystyle k_1=\arctan \left(\frac{1}{b}\right)\quad\Rightarrow\quad\tan k_1=\frac{1}{b}$
$(2)\;$ Use identities $\;\displaystyle \sec^2t=1+\tan^2t\;$ and $\;\displaystyle \tan t=\frac{\sin t}{\cos t}$
$(3)\;$ Use identity $\;\displaystyle \cos^2t=1-\sin^2t$
$(4)\;$ Use substitution $\;\displaystyle \sin t=\frac{ay}{\sqrt{b^2-a^2}}\,$ and $\;\displaystyle k_2=\frac{\sqrt{b^2-a^2}}{a\sqrt{1-b^2}}\,$ since $\;\;\;\quad\displaystyle\tan k_1=\frac{1}{b}\quad\Rightarrow\quad\sin k_2=\frac{1}{\sqrt{1-b^2}}$
|
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|
Concise induction step for proving $\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$ I recently got a book on number theory and am working through some of the basic proofs. I was able to prove that $$\sum_{i=1}^n i^3 = \left( \sum_{i=1}^ni\right)^2$$ with the help of the identity $$\sum_{i=1}^ni = \frac{n(n+1)}{2}$$ My proof of the induction step goes as follows (supposing equality holds for all $k \in \{1,2,\dots n \})$: $$\sum_{i=1}^{n+1} i^3 = \sum_{i=1}^{n} i^3+(n+1)^3 \\ = \left( \sum_{i=1}^ni\right)^2+(n+1)^3 \\ = \left( \frac{n(n+1)}{2}\right)^2 + (n+1)^3 \\ = \frac{n^2(n+1)^2}{4}+\frac{4(n+1)^3}{4} \\ = \frac{n^4+2n^3+n^2}{4}+\frac{4n^3+12n^2+12n+4}{4} \\ = \frac{n^4+6n^3+13n^2+12n+4}{4} \\ = \frac{(n^2+3n+2)^2}{4} \\ = \frac{[(n+1)(n+2)]^2}{4} \\ = \left(\frac{(n+1)(n+2)}{2}\right)^2 \\ = \left(\sum_{i=1}^{n+1}i\right)^2$$ I was a little disappointed in my proof because the algebra got really hairy. It took me a long while to see that I could twice unfoil the polynomial $n^4+6n^3+13n^2+12n+4$ and all in all the solution seems pretty inelegant. Does anyone have a smoother way to prove the induction step or bypass the algebra? I feel like their must be some concise way to get the same result.
|
We could also start the induction step from the RHS. This way we need only to cope with a square of a binom and not with a cube of a binom.
Induction step $n \rightarrow n+1$:
\begin{align*}
\left(\sum_{i=1}^{n+1}i\right)^2&=\left(\sum_{i=1}^{n}i+(n+1)\right)^2\\
&=\left(\sum_{i=1}^{n}i\right)^2 + 2(n+1)\sum_{i=1}^{n}i+(n+1)^2\\
&=\sum_{i=1}^{n}i^3 + 2(n+1)\frac{n(n+1)}{2}+(n+1)^2\tag{1}\\
&=\sum_{i=1}^{n}i^3 + (n+1)^3\\
&=\sum_{i=1}^{n+1}i^3\\
\end{align*}
In (1) we do the induction step and use the formula $\sum_{i=1}^ni=\frac{n(n+1)}{2}$
|
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|
The minimum value of $\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$ The problem is to find the minimum of $A$, which I attempted and got a different answer than my book:
$$A=\frac{a(x+a)^2}{\sqrt{x^2-a^2}}$$ where $a$ is a constant
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)(x+a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)^2}{(x+a)(x-a)}$
$A'=\frac{(x^2-a^2)^{\frac{1}{2}}(2a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)}{(x-a)}$
$0=(x^2-a^2)^{\frac{1}{2}}(2a)-\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)$
$(x^2-a^2)^{\frac{1}{2}}(2a)=\frac{1}{2}(x^2-a^2)^{-\frac{1}{2}}(a)(x+a)$
$2a\sqrt{x^2-a^2}=\frac{a(x+a)}{2\sqrt{x^2-a^2}}$
$2a(x+a)(x-a)=a(x+a)$
$2(x-a)=a$
$2x-2a=a$
$x=\frac{3}{2}a$
My book says x=2a. I'm not sure where I went wrong. Thanks in advance
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I assume the constant $a$ is positive. This is because if $a\lt 0$ there is no minimum.
We equivalently want to minimize $\frac{(x+a)^4}{x^2-a^2}$ with $x\gt a$, or equivalently minimize
$$\frac{(x+a)^3}{x-a}$$
with $x\gt a$. The derivative is
$$\frac{3(x+a)^2(x-a)-(x+a)^3}{(x-a)^2},$$
difficult to get wrong. The top is
$$(x+a)^2[3(x-a)-(x+a)], \quad\text{which is}\quad (x+a)^2(2x-4a).$$
Remark: Under suitable conditions, maximizing/minimizing $f(x)$ is equivalent to maximizing $f^2(x)$. Replacing $f(x)$ by $f^2(x)$ is a frequently useful little trick.
|
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|
Evaluation of $2$ limit problems Evaluation of $(a)\;\;\lim_{x\rightarrow \infty}\left\{\sqrt{x^4+ax^3+3x^2+bx+2}-\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$
and $(b)\;\; \displaystyle \lim_{x\rightarrow 0^{-}}\frac{\lfloor x \rfloor +\lfloor x^2 \rfloor+...............+\lfloor x^{2n+1} \rfloor +n+1}{1+\lfloor x^2 \rfloor +|x|+2x}.$
$\bf{My\; Try::(a):}$ Given $\lim_{x\rightarrow \infty}\left\{\sqrt{x^4+ax^3+3x^2+bx+2}-\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$
Now Multiply both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\left\{\sqrt{x^4+ax^3+3x^2+bx+2}+\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$.
So $\displaystyle \lim_{x\rightarrow \infty}\frac{\left\{\sqrt{x^4+ax^3+3x^2+bx+2}-\sqrt{x^4+2x^3-cx^2+3x-d}\right\}}{\left\{\sqrt{x^4+ax^3+3x^2+bx+2}+\sqrt{x^4+2x^3-cx^2+3x-d}\right\}}\times \left\{\sqrt{x^4+ax^3+3x^2+bx+2}+\sqrt{x^4+2x^3-cx^2+3x-d}\right\}$
So we Get $\displaystyle \lim_{x\rightarrow \infty} \frac{(a-2)x^3+(3-c)x^2+(b-3)x+(2-d)}{x^2\cdot \left(\sqrt{1+\frac{a}{x}+\frac{3}{x^2}+\frac{b}{x^3}+\frac{2}{x^4}}+\sqrt{1+\frac{2}{x}-\frac{c}{x^2}+\frac{3}{x^3}-\frac{d}{x^4}}\right)}$
Now If limit is finite , Then $(a-2)=0\Rightarrow a=2$
can we solve the first question is any other way, If yes the plz explain me,
and how can i solve $(2)$ one.
Thanks
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Not a full answer, just expanding my comment about $(a)$.
We can rewrite the function (keeping in mind that $x \to \infty$) like this:
\begin{align}
f(x)&:=\sqrt{x^4 + ax^3 + 3x^2 + bx + 2} - \sqrt{x^4 + 2x^3 -cx^2 + 3x - d} =\\
&=x^2\left(1 + \frac a{2x} + \frac 3{2x^2} + \frac b{2x^3} + \frac 1{x^4} - 1 - \frac 1x + \frac c{x^2} - \frac 3{2x^3} + \frac d{2x^4} + \mathcal{o}\left(\frac 1{x^4}\right)\right)=\\
&=\left(\frac a2 - 1\right)x + \frac 32 + c + \mathcal{o}(1)
\end{align}
Now it's easy to see that
$$\lim_{x \to \infty} f(x) = \begin{cases}\frac 32 + c&\qquad a = 2\\
\infty&\qquad a \neq 2
\end{cases}$$
I've made use of the expansion:
$$\sqrt{1 + t} = 1 + t/2 + \mathcal{o}(t)\qquad\text{for}\ t \to 0$$
|
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How do I solve $\int\frac{x}{(x^2+x+1)^{\frac{1}{12}}}\,dx$? I'm trying to solve $$\int_0^1\frac{x}{(x^2+x+1)^{\frac{1}{12}}}\mathrm dx$$ To calculate it I first tried to calculate the primitive function. So let $$\int\frac{x}{(x^2+x+1)^{\frac{1}{12}}}\mathrm dx$$ first I added $+1-1$ to numerator then split it and tried to obtain the derivative of $x^2+x+1$ on numerator after this passages I obtain $$=\frac{6}{11}(x^2+x+1)^{\frac{11}{12}}-\frac12\int\frac{1}{(x^2+x+1)^{\frac{1}{12}}}\mathrm dx$$ My problem starts here: how can I deal with the last integral $$\int\frac{1}{(x^2+x+1)^{\frac{1}{12}}}\mathrm dx$$ I tried to substitute $(x^2+x+1)^{\frac{1}{2}}=x+t$ but it makes it more complex..Then wolphram says that we can't express it by elementary function. How can I proceed? Note that the book where I found the integral asks to calculate it from $0$ to $1$ and it reports a solution which is $$\frac12(\sqrt{3}-\frac72+\frac{3}{2(1+2\sqrt{3})}+\ln(1+2\sqrt{3}))$$.Thank you in advance
|
Mathematica 11.1 gives that this integral is equal to
$$
\frac{2}{11} \left(i 3^{17/12} \, _2F_1\left(1,\frac{11}{6};\frac{23}{12};\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)-i \sqrt{3} \, _2F_1\left(1,\frac{11}{6};\frac{23}{12};\frac{1}{6} \left(3+i
\sqrt{3}\right)\right)-3+ 3^{23/12}\right)
$$
the function of interest is then
$$
_2F_1\left(1,\frac{11}{6};\frac{23}{12};x\right) = \sum_{n=0}^\infty \frac{\Gamma(\frac{23}{12})\Gamma(\frac{11}{6}+n)}{\Gamma(\frac{11}{6})\Gamma(\frac{23}{12}+n)}x^n
$$
Removing the complex terms we can also the write this as
$$
\frac{6}{11} \left( 3^{11/12} -1 + 3^{5/12}\frac{\Gamma(\frac{23}{12})}{\Gamma(\frac{11}{6})}\underbrace{\sum_{n=0}^\infty \frac{\Gamma(\frac{11}{6}+n)}{\Gamma(\frac{23}{12}+n)}\left(3^{-\frac{11}{12}-\frac{n}{2}}\sin\left(\frac{n \pi}{6}\right)-\sin\left(\frac{n\pi}{3}\right)\right)}_{Q}\right)
$$
so if you want a closed form you need to work out the value of $Q$, which doesn't look super promising.
Edit: After playing with this we can write the sum as
$$
\int_0^1 \frac{x}{(x^2+x+1)^(1/12)}\;dx = \frac{1}{\Gamma(\frac{1}{12})}\sum_{k=0}^\infty \frac{(-1)^k}{k!}B\left(-1,2+2k,\frac{11}{12}-k\right)\Gamma\left(\frac{1}{12}-k\right)= 0.47017004410654...
$$
where $B$ is the incomplete beta function.
|
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show that the series converges. Using the definition, show that the series $\sum_{n=1}^\infty \frac{1}{n(n+3)}$ converges. Determine the limit.
This is what I've managed to do so far:
$\frac{1}{n(n+3)}$ = $\frac{1}{3n}$ -$\frac{1}{3(n+3)}$
$\sum_{n=1}^\infty \frac{1}{n(n+3)}$ = $\sum_{n=1}^\infty \frac{1}{3n}$ -$\frac{1}{3(n+3)}$
S$_{k}$= $\sum_{n=1}^{k} \frac{1}{n(n+3)}$ = $\sum_{n=1}^{k} \frac{1}{3n}$ -$\frac{1}{3(n+3)}$ = $\frac{1}{3}$-$\frac{1}{12}$+$\frac{1}{6}$-$\frac{1}{15}$+$\frac{1}{9}$-$\frac{1}{18}$+$\frac{1}{12}$-$\frac{1}{21}$ $\cdots$ +$\frac{1}{3k-9}$-$\frac{1}{3k}$+$\frac{1}{3k-6}$-$\frac{1}{3k+3}$+$\frac{1}{3k-3}$-$\frac{1}{3k+6}$+$\frac{1}{3k}$-$\frac{1}{3k+9}$=$\frac{1}{3}$+$\frac{1}{6}$+$\frac{1}{9}$-$\frac{1}{3k+3
}$-$\frac{1}{3k+6}$-$\frac{1}{3k+9}$
Hence
$\sum_{n=1}^\infty \frac{1}{n(n+3)}$ = $\lim_{k\to \infty}$S$_{k}$=$\lim_{k\to \infty}$($\frac{1}{3}$+$\frac{1}{6}$+$\frac{1}{9}$-$\frac{1}{3k+3
}$-$\frac{1}{3k+6}$-$\frac{1}{3k+9}$)=$\frac{1}{3}$+$\frac{1}{6}$+$\frac{1}{9}$= $\frac{11}{18}$
I did the long addition just to make things clearer for myself. I was just wondering if there was anything I have missed out?
|
prove that for the partial sum is hold
$$-1/3\, \left( m+1 \right) ^{-1}-1/3\, \left( m+2 \right) ^{-1}-1/3\,
\left( m+3 \right) ^{-1}+{\frac {11}{18}}
$$
and take then the Limit for $m$ tends to infinity
|
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|
Proving $1+2^n+3^n+4^n$ is divisible by $10$ How can I prove
$$1+2^n+3^n+4^n$$ is divisible by $10$ if $$n\neq 0,4,8,12,16.....$$
|
After checking that $S(n)=1+2^n+3^3+4^n\equiv0$ mod $10$ for $n=1$, $2$, and $3$ (i.e., computing $S(1)=10$, $S(2)=30$ and $S(3)=100$), induction does the rest, thanks to the difference
$$\begin{align}
S(n+4)-S(n)&=2^n(2^4-1)+3^n(3^4-1)+4^n(4^4-1)\\
&=15\cdot2^n+80\cdot3^n+255\cdot4^4\\
&\equiv5\cdot2^n+0\cdot3^n+5\cdot4^n\mod 10\\
&=5\cdot2(2^{n-1}+2^{2n-1})\\
&\equiv0\mod 10
\end{align}$$
Remark: This also shows that $S(4n)\equiv4$ mod $10$, since $S(0)=1+1+1+1=4$.
|
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|
In triangle ABC, find the range of sinAsinC. In $\triangle$ABC, $\angle$B=$\frac{\pi}{3}$. Find the range of sinAsinC.
I used C=120-A to simplify the equation in sinC and then applied $-1<sinC<1$ but answer did not matched.
|
From given, $A+C = \frac{2\pi}3$.
$$\sin A\sin C = \frac{\cos(A-C)-\cos(A+C)}{2} = \frac{\cos(\frac{2\pi}3-2C)-\cos\frac{2\pi}3}{2}$$
Find the maximum and the minimum of the first $\cos$ term.
For maximum, choose $C = \frac\pi3$, which gives $\cos(\frac{2\pi}3-2C) = 1$.
For minimum, choose $C = 0$ or $\frac{2\pi}{3}$, which gives $\cos(\frac{2\pi}3-2C) = -\frac12$.
$$0 =\frac{-1/2+1/2}{2} < \frac{\cos(\frac{2\pi}3-2C)-\cos\frac{2\pi}3}{2} \le \frac{1+1/2}2 = \frac34$$
|
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$4 $ balls selected at random from $ 15$ $15$ balls $ 4$ selected
$1$ blue, $2 $ green, $3$ red, $4$ white, $5$ yellow
What is the probability that $2$ are red and at least $1$ is white?
Now the way this question is worded makes it seem as though you should account for the possibility of the case being $2$ red balls + $1$ white + $1$ other AND the possibility of $2$ red balls + $2$ white
So the answer is:
$$\frac{C_{3,2} C_{4,1}C_{8,1}}{C_{15,4}} + \frac{C_{3,2} C_{4,2}}{C_{15,4}}\,?$$
|
$\frac{\binom{3}{2}\binom{4}{1}\binom{8}{1}}{\binom{15}{4}}+\frac{\binom{3}{2}\binom{4}{2}}{\binom{15}{4}}$
So you are right
|
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|
To prove $(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 \ge 9$ I used the following way but got wrong answer
$$A.M. \ge G.M.$$
$$ \frac{\sin \theta + \csc \theta}{2} \ge \sqrt{\sin \theta \cdot \csc \theta}$$
Squaring both sides,
\begin{equation*}
(\sin\theta + \csc\theta )^2 \ge 4 \tag{1}
\end{equation*}
Similarly
\begin{equation*}
(\cos\theta + \sec\theta )^2 \ge 4 \tag{2}
\end{equation*}
Adding equation (1) and (2)
\begin{equation*}
(\sin\theta + \csc\theta )^2+(\cos\theta + \sec\theta )^2 \ge 8
\end{equation*}
What is wrong?
|
Since
$$\left(\sin{\theta}+\dfrac{1}{\sin{\theta}}\right)^2+\left(\cos{\theta}+\dfrac{1}{\cos{\theta}}\right)^2=5+\dfrac{1}{\sin^2{\theta}}+\dfrac{1}{\cos^2{\theta}}$$
Use Cauchy-Schwarz inequality we have
$$\dfrac{1}{\sin^2{\theta}}+\dfrac{1}{\cos^2{\theta}}\ge\dfrac{(1+1)^2}{\sin^2{\theta}+\cos^2{\theta}}=4$$
|
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|
Proving uniform continuity and uniform discontinuity Could someone please explain to me how to show uniform continuity and not uniformly continuous for the following:
$f(x) = \frac{1}{x^2}$ for $A = [1, \infty)$ show uniform continuity
$f(x) = \frac{1}{x^2}$ for $B = [0, \infty)$ show that $f$ is not uniformly continuous
|
*
*$|\frac{1}{x^2}-\frac{1}{y^2}|\leq |\frac{y^2-x^2}{x^2y^2}|=|\frac{(y-x)(y+x)}{x^2y^2}|\leq |\frac{x+y}{x^2y^2}||y-x|\leq (|\frac{1}{xy^2}|+|\frac{1}{yx^2}|)|x-y|\leq 2|x-y|$
2.$\frac{1}{x^2}$ is not uniformly continuous. $|\frac{1}{x^2}-\frac{1}{y^2}|\leq |\frac{y^2-x^2}{x^2y^2}|$ Near $0$ quantity is very large.
|
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|
How to find upper and lower bound without using formula? I am studying discrete math for tomorrow's exam and got stuck in the below question. I tried to google it and couldn't find anything usefull.
Prove the following sum is theta(n^2) (we have to find O(n^2) and omega (n^2))
1) $P(n)= 1+2+3+4+\cdots +n$
2) $P(n) =n+(n+1)+(n+2)+\cdots +2n$
Note: you cannot use any formula you have to do it my algebraic manipulation.
This might be simple but I am not getting any clue right now and I dont have solution of it.
|
For 1) if $n = 2k$, then $P(n) = P(2k) = 1+2+3+\cdots + 2k = (1+2k) +(2+ (2k-1)) + (3+(2k-2)) + \cdots + (k+(2k-(k-1))) = (2k+1) +(2k+1) + \cdots + (2k+1) = k(2k+1) = \dfrac{n}{2}\cdot (n+1) = \dfrac{n(n+1)}{2}$.
If $n = 2k+1$, then:
$1 + 2 + 3 +\cdots + (2k+1) = (1+2+3+\cdots + 2k) + (2k+1) = (1+2+\cdots + +(n-1)) + n = \dfrac{(n-1)n}{2} + n = \dfrac{n(n+1)}{2}$.
For 2) $n + (n+1) + (n+2) +\cdots + 2n = (n+n+n+\cdots + n) + (1+2+3+\cdots + n) = n\cdot n + \dfrac{n(n+1)}{2} = n^2 + \dfrac{n(n+1)}{2}$
|
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|
How can I understand solving the equation? $$\begin{align}
&\left[(\sqrt[4]{p}-\sqrt[4]{q})^{-2} + (\sqrt[4]{p}+\sqrt[4]{q})^{-2}\right] : \frac{\sqrt{p} + \sqrt{q}}{p-q} \\
&= \left(\frac{1}{(\sqrt[4]{p}-\sqrt[4]{q})^{2}}+\frac{1}{(\sqrt[4]{p}+\sqrt[4]{q})^{2}}\right) \times \frac{(\sqrt{p} - \sqrt{q})(\sqrt{p} + \sqrt{q})}{\sqrt{p} + \sqrt{q}} \\
&= \frac{(\sqrt[4]{p}+\sqrt[4]{q})^{2} + (\sqrt[4]{p} - \sqrt[4]{q})^{2}}{(\sqrt{p} - \sqrt{q})^{2}} \times \frac{(\sqrt{p} - \sqrt{q})(\sqrt{p} + \sqrt{q})}{(\sqrt{p} + \sqrt{q})}\end{align}$$
How can I get this expression? $$\frac{(\sqrt[4]{p}+\sqrt[4]{q})^{2} + (\sqrt[4]{p} - \sqrt[4]{q})^{2}}{(\sqrt{p} - \sqrt{q})^{2}}$$
Only this solving I can't understand.
|
$$(A+B)^2+(A-B)^2=(A^2+2AB+B^2)+(A^2-2AB+B^2)=2(A^2+B^2)$$ By using this identity, You can simplify the numerator. $$\dfrac{(\sqrt[4]{p}+\sqrt[4]{q})^{2} + (\sqrt[4]{p} - \sqrt[4]{q})^{2}}{(\sqrt{p} - \sqrt{q})^{2}}=\dfrac{2(\sqrt{p}+\sqrt{q})}{(\sqrt{p} - \sqrt{q})^{2}}$$ Rest will be obvious. Good luck.
|
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|
How did Sir Isaac Newton develop and formulate the famous binomial theorem? After completing combination, I have started to read Binomial Theorem. My book only mentioned about Pascal's Triangle. And the formula was then given straightforward. But how did Sir Issac Newton actually develop this theorem?? How did he formulate all these??? Please help.
|
About the Binomial Theorem.
When expanding the powers of a binomial by hand and grouping the terms by identical powers, it is not very hard to observe the pattern:
$$(x+y)^0=1$$
$$(x+y)^1=x+y$$
$$(x+y)^2=x^2+2xy+y^2$$
$$(x+y)^3=x^3+3x^2y+3xy^2+y^3$$
$$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4$$
$$\dots$$
*
*the expansion is the sum of all $n+1$ possible products of the form $x^iy^j$, with $i+j=n$.
*the shape of Pascal's triangle clearly appears as the number of terms goes linearly increasing.
*the basic recurrence relation between Pascal's numbers appears with the following expansion:
$$\begin{align}(x+y)(x+y)^3
&=x^4+3x^3y+3x^2y^2+xy^3\ +\\
&\ \ \ \ \ \underline{\ \ \ \ +\ \ x^3y+3x^3y^2+3xy^3+y^4}\\
&=x^4+4x^3y+6x^2y^2+4xy^3+y^4
\end{align}$$
You get a new line of coefficients by adding the last line to itself shifted by one position, so that $C_m^{n+1}=C_{m-1}^n+C_m^n$.
All of this is rather straightforward to establish. More interesting is the link with combinatorial analysis.
If you expand $(x+y)^n$ without regrouping the terms,
$$(x+y)^0=1$$
$$\begin{align}(x+y)^1&=x\\
&+y\end{align}$$
$$\begin{align}(x+y)^2&=xx\\
&+xy+yx\\
&+yy\end{align}$$
$$\begin{align}(x+y)^3&=xxx\\
&+xxy+xyx+yxx\\
&+xyy+yxy+yyx\\
&+yyy\end{align}$$
$$\begin{align}(x+y)^4&=xxxx\\
&+xxxy+xxyx+xyxx+yxxx\\
&+xxyy+xyxy+xyyx+yxxy+yxyx+yyxx\\
&+yyxy+yyyx+xyyy+yxyy\\
&+yyyy\end{align}$$
$$\dots$$
you see that there are $2^n$ terms, and you observe that all combinations of $i$ letters $x$ and $j$ letters $y$ (with $i+j=n$) can be grouped as $x^iy^j$, hence
$$C_m^n=\binom nm.$$
At the same time, you see a bell-shaped histogram appear, that prefigures the Gaussian curve.
|
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|
How can I prove $\sum_{n=1}^{\infty }\frac{1}{n^3(n+1)^3}=10-\pi ^2$ Can the residue theorem prove this?
$$\sum_{n=1}^{\infty }\frac{1}{n^3(n+1)^3}=10-\pi ^2$$
|
Hint
$$\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$$
and $$(a-b)^3=(a^3-b^3)-3ab(a-b)$$
so
\begin{align*}\dfrac{1}{n^3(n+1)^3}&=\left(\dfrac{1}{n^3}-\dfrac{1}{(n+1)^3}\right)-3\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\left(\dfrac{1}{n}-\dfrac{1}{(n+1)}\right)\\
&=\left(\dfrac{1}{n^3}-\dfrac{1}{(n+1)^3}\right)-\dfrac{3}{n^2}+6\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)-\dfrac{3}{(n+1)^2}
\end{align*}
and use well kown
$$\zeta{(2)}=\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$$ can solve it
|
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|
How to transform the product to sum? I just wonder that how to prove that
$$
\prod_{m=1}^{n}\Big(x-2\cos\frac{m\pi}{n+1}\Big)=\sum_{k=0}^{[n/2]}(-1)^{k}\binom{n-k}{k}x^{n-2k}.
$$
Similarly, how to transform the product
$$
\prod_{m=1}^{n}\Big(x+2\cos\frac{2m\pi}{n}\Big)\overset{?}{=}\sum_{k}b_k x^k?
$$
|
Let, $\displaystyle P(x) = \sum\limits_{k=0}^{[n/2]} (-1)^k\binom{n-k}{k}x^{n-2k}$
We show, $\displaystyle P(\omega + \omega^{-1}) = 0$ where, $\displaystyle\omega = \exp{\frac{2\pi i}{2n+2}}$ is a $(2n+2)^{th}$ roots of unity.
$\begin{align} P(\omega+ \omega^{-1}) &= \sum\limits_{k=0}^{[n/2]} (-1)^k\binom{n-k}{k}(\omega+ \omega^{-1})^{n-2k} \\& = \sum\limits_{k=0}^{[n/2]}\sum\limits_{l=0}^{n-2k} (-1)^k\binom{n-k}{k}\binom{n-2k}{l}\omega^{n-2k-2l} \end{align}$
If we collect the coefficients of $\omega^{n-2r}$, where, $r = k+l$ together,
$\displaystyle \begin{align} \sum\limits_{k=0}^{r}(-1)^k\binom{n-k}{k}\binom{n-2k}{r-k} = \sum\limits_{k=0}^{r}(-1)^k\binom{r}{k}\binom{n-k}{r} = 1\end{align} \tag{1}$
Where, the identity $(1)$ can be derived from the general form:
$\displaystyle \sum\limits_{k=0}^{r}(-1)^k\binom{r}{k}\binom{n-k}{s} = \begin{cases} 1 & \textrm{ if } r = s \\ 0 &\textrm{ otherwise } \end{cases}$
which can be proved by induction or otherwise.
Thus, $\displaystyle P(\omega+\omega^{-1}) = \sum\limits_{r=0}^{n} \omega^{n-2r} = 0$
Hence, $\displaystyle P(x) = \prod\limits_{r=1}^{n} \left(x - \omega^{r} - \omega^{-r}\right) = \prod\limits_{r=1}^{n} \left(x - 2\cos \frac{\pi r}{n+1}\right)$
As @Roger209 points out in the comments, $\displaystyle P(x) = U_n\left(\frac{x}{2}\right)$, where $U_n$ in the Chebyshev Polynomial of $2^{nd}$ Kind. So, $\displaystyle P(2\cos \tau) = \frac{\sin (n+1)\tau}{\sin \tau}$, which has roots at $\displaystyle \tau = \dfrac{m\pi}{n+1}$, for $m = 1(n)n$.
Second Part:
We take the second polynomial: $\displaystyle Q(x) = \prod\limits_{m=1}^{n}\left(x + 2\cos \frac{2m\pi}{n}\right)$
Keeping in mind that the Chebyshev Polynomial of $1^{st}$ Kind $T_n$ satisfies, $$T_n(\cos \theta) = \cos n\theta$$
Thus, $\displaystyle T_n\left(\cos \frac{2m\pi}{n}\right) = \cos 2 m\pi = 1$, for $m = 1,2,\cdots,n$.
Thus, $\displaystyle T_n(x) - 1 = - 1 + \frac{n}{2}\sum\limits_{r=0}^{[n/2]}\frac{(-1)^r}{n-r}\binom{n-r}{r}(2x)^{n-2r} = 2^{n-1}\prod\limits_{m=1}^{n}\left(x - \cos \frac{2m\pi}{n}\right)$
I.e., $\displaystyle Q(x) = (-1)^n2\left(T_n\left(-\frac{x}{2}\right) - 1\right)$.
|
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|
Integration without complex analysis on rational-improper integral Evaluate:
$$\int_{0}^{\infty} \frac{1}{x^6 + 1} \,\mathrm dx$$
Without the use of complex-analysis.
With complex analysis it is a very simple problem, how can this be done WITHOUT complex analysis?
|
Setting $x=\dfrac1y,$
$$I=\int_0^\infty\frac1{1+x^6}dx=\cdots=\int_0^\infty\frac{x^4}{1+x^6}dx$$
$$2I=\int_0^\infty\frac{1+x^4}{1+x^6}dx=\int_0^\infty\frac{(1+x^2)^2-2x^2}{1+x^6}dx$$
$$=\int_0^\infty\frac{1+x^2}{1-x^2+x^4}dx-\frac23\int_0^\infty\frac{3x^2}{1+x^6}dx$$
$$I_1=\int_0^\infty\frac{1+x^2}{1-x^2+x^4}dx=\int_0^\infty\frac{1/x^2+1}{1/x^2-1+x^2}dx$$
$$=\int_0^\infty\frac{1/x^2+1}{\left(x-\dfrac1x\right)^2-3}dx =\int_{-\infty}^\infty\frac{dz}{z^2-3}$$
$$=\frac1{2\sqrt3}\ln\frac{z-\sqrt3}{z+\sqrt3}|_{-\infty}^\infty=\frac1{2\sqrt3}\ln\frac{1-\sqrt3/z}{1+\sqrt3/z}|_{-\infty}^\infty=\ln1-\ln1=0$$
$$I_2=\int_0^\infty\frac{3x^2}{1+x^6}dx=\arctan(1+x^3)|_0^{\infty}=\frac\pi2-\frac\pi4=\frac\pi4$$
|
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|
Why doesn't squaring the radicand of a square root introduce a plus-minus sign here? The question I have concerns the following problem:
*
*$\sqrt{4x-1} = \sqrt{x+2}-3$
*$(\sqrt{4x-1})^2 = (\sqrt{x+2}-3)^2$
*$\sqrt{4x-1}\times\sqrt{4x-1} = (\sqrt{x+2}-3)\times(\sqrt{x+2}-3)$
*$\sqrt{(4x-1)\times(4x-1)} = (\sqrt{x+2})^2-3\sqrt{x+2}-3\sqrt{x+2}+9$
*$\sqrt{(4x-1)^2} = \sqrt{(x+2)\times(x+2)}-6\sqrt{x+2}+9$
*$\underline{4x-1} = \sqrt{(x+2)^2}-6\sqrt{x+2}+9$
*$4x-1 = \underline{x+2}-6\sqrt{x+2}+9$
How did the underlined expressions in steps 6 and 7 not include $\pm$? I thought that $\sqrt{x^2} = \pm x$.
|
By definition, if $r$ is a nonnegative real number, then $\sqrt{r}$ denotes the unique nonnegative real number whose square is $r$. Note that $\sqrt{r}$ and $-\sqrt{r}$ both have the property that their square is $r$, but there is only one nonnegative square root.
If $x$ is any real number, then $x^2$ is nonnegative, so it has a square root in the real numbers. By definition, $\sqrt{x^2}$ is the unique nonnegative real number whose square is $x^2$. If $x$ is nonnegative, then $x$ itself satisfies this property: $x = \sqrt{x^2}$. If $x$ is negative, then $x$ cannot be $\sqrt{x^2}$ because the square root must be nonnegative (by definition); however, note that $(-x)^2 = x^2$, so $-x = \sqrt{x^2}$.
Putting these together, we see that $\sqrt{x^2}$ is equal to $x$ if $x \geq 0$, and $-x$ if $x < 0$. But this is exactly the definition of $\lvert x \rvert$, so we can more concisely phrase it as: $\sqrt{x^2} = \lvert x \rvert$.
|
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|
For a Diophantine equation $x^2+py^2=z^2$ show that $z$ is necessarily odd. For a Diophantine equation $x^2+py^2=z^2$ where $p$ is a prime of the form $p\equiv 1(mod4)$ and $(x,y,z)=1$.
Show that $z$ is necessarily odd.
|
Suppose to the contrary that $z$ is even.
If $y$ is even, then $x^2$ is even, so $x$ is even, contradicting the fact that $\gcd(x,y,z)=1$. Similarly, if $x$ is even, then $py^2$ is even, and since $p$ is odd $y^2$ is even, and therefore $y$ is even, and again $\gcd(x,y,z)\gt 1$.
So $x$ and $y$ are both odd. Thus $x^2\equiv 1\pmod{4}$ and $y^2\equiv 1\pmod{4}$, and therefore $x^2+py^2\equiv 2\pmod{4}$. This is impossible, since any even square is congruent to $0$ modulo $4$.
|
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Why do we have $u_n=\frac{1}{\sqrt{n^2-1}}-\frac{1}{\sqrt{n^2+1}}=O(\frac{1}{n^3})$? Why do we have
*
*$u_n=\dfrac{1}{\sqrt{n^2-1}}-\dfrac{1}{\sqrt{n^2+1}}=O\left(\dfrac{1}{n^3}\right)$
*$u_n=e-\left(1+\frac{1}{n}\right)^n\sim \dfrac{e}{2n}$
any help would be appreciated
|
Hint: $\left(\dfrac{1}{n^2-1}\right)^{1/2} = \left(n^2-1\right)^{-1/2} = \dfrac{1}{n}\left(1-\frac{1}{n^2}\right)^{-1/2} = \dfrac{1}{n} + \mathcal{O}\left(\frac{1}{n^3}\right)$
|
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|
Stochastic matrix A^n=const. Let $A\in M(n,n)$ be a doubly stochastic matrix such that
$$
A^k=\begin{pmatrix}
1/n & \cdots & 1/n\\
\vdots & \ddots & \vdots\\
1/n & \cdots & 1/n\\
\end{pmatrix}
$$
for some $k>1$. Does it follow that
$$
A=\begin{pmatrix}
1/n & \cdots & 1/n\\
\vdots & \ddots & \vdots\\
1/n & \cdots & 1/n\\
\end{pmatrix}?
$$
|
No, it does not follow that
$$
A=\begin{pmatrix}
1/n & \cdots & 1/n\\
\vdots & \ddots & \vdots\\
1/n & \cdots & 1/n
\end{pmatrix}.
$$
Here is a (minimal) counterexample. Take
$$
J_3=\begin{pmatrix}
1/3 & 1/3 & 1/3\\
1/3 & 1/3 & 1/3\\
1/3 & 1/3 & 1/3
\end{pmatrix},\qquad N=\begin{pmatrix}
1 & 1 & -2\\
-1 & -1 & 2\\
0 & 0 & 0
\end{pmatrix}.
$$
Observe that $N^2=J_3N=NJ_3=0$. Consequently for all $\varepsilon$,
$$
(J_3+\varepsilon N)^2=J_3^2=J_3.
$$
When $\varepsilon\leq 1/6$, $J_3+\varepsilon N$ is doubly stochastic, producing a counterexample in the case $n=3,k=2$.
|
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|
How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$? How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$?
I got answer $\frac{1}{\sqrt{2}}$, but the real answer is $\frac{\sqrt{2}}{2}$. Anyway, calculator for both answers return same numbers.
|
The number $\frac{1}{\sqrt{2}}$ is defined to be the number such that, when you multiply it by $\sqrt{2}$, you get $1$. Symbolically, it is the solution to $x\cdot \sqrt{2}=1$. That's how division works - it's the inverse of multiplication. What is $\frac{\sqrt{2}}2\cdot \sqrt{2}$? Why, it's $\frac{2}2=1$ - so it must be $\frac{1}{\sqrt{2}}$, as it satisfies the definition of division for $\frac{1}{\sqrt{2}}$.
|
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|
Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$ $a,b,c \in \mathbb{R}$ and $a+b+c=0$.
Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$
I think that $2^{a}.2^{b}.2^{c}=1$, but i don't know what to do next
|
Let $x=2^a,y=2^b,z=2^c$. Then $x,y,z>0$ and
$$ xyz=2^{a+b+c}=1.$$
By Holder's inequality
$$(x^3+y^3+z^3)(1+1+1)(1+1+1)\ge (x+y+z)^3.$$
Therefore,
$$x^3+y^3+z^3\ge\dfrac{(x+y+z)^3}{9}\ge (x+y+z)$$
because use AM-GM inequality
$$(x+y+z)^2\ge (3\sqrt[3]{xyz})^2=9.$$
|
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|
Removing the root squares from this expression? I would like to understand how to remove the root squares from this expression:
$$x = \frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}$$
How to do it?
|
Here's a push
You can try rationalising it by multiplying the expression by its conjugate.
$$\frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}\times\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{\sqrt{2} + \sqrt{3} - \sqrt{5}}=\frac {\sqrt{2}+ \sqrt{3}- \sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}$$
and you can continue it.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality $\frac 1{\sqrt{1+xy}}+\frac 1{\sqrt{1+yz}}+\frac 1{\sqrt{1+zx}}\ge \frac 9{\sqrt {10}}$ Let $x,y,z$ be non-negative real numbers such that $x+y+z=1$ , then is the following true? $$\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}}\ge \dfrac 9{\sqrt {10}}$$
|
By AM-HM or C-S we have
$$\frac{\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}}}{3} \geq \frac{3}{\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}}$$
Therefore
$$\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}} \geq \frac{9}{\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}}$$
To prove your claim, you need to show
$$\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx} \leq \sqrt{10}$$
By C-S
$$\left(\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx} \right)^2 \leq 3 (1+xy+1+yz+1+zx )=9+3(xy+xz+yz)$$
Finally, again by C-S we have
$$xy+xz+yz \leq x^2+y^2+z^2 \Rightarrow 3\\(xy+xz+yz) \leq x^2+y^2+z^2+2(xy+xz+yz) =(x+y+z)^2=1$$
Combining the last two inequalities we get
$$\left(\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx} \right)^2 \leq 10$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $x^4+3x^3+6x+4=0$... easier way? So I was playing around with solving polynomials last night and realized that I had no idea how to solve a polynomial with no rational roots, such as $$x^4+3x^3+6x+4=0$$
Using the rational roots test, the possible roots are $\pm1, \pm2, \pm4$, but none of these work.
Because there were no rational linear factors, I had to assume that the quartic separated into two quadratic equations yielding either imaginary or irrational "pairs" of roots. My initial attempt was to "solve for the coefficients of these factors".
I assumed that $x^4+3x^3+6x+4=0$ factored into something that looked like this
$$\left(x^2+ax+b\right)\left(x^2+cx+d\right)=0$$
because the coefficient of the first term is one. Expanding this out I got
$$x^4+ax^3+cx^3+bx^2+acx^2+dx^2+adx+bcx+bd=0$$
$$x^4+\left(a+c\right)x^3+\left(b+ac+d\right)x^2+\left(ad+bc\right)x+bd=0$$
Equating the coefficients of both equations
$$a+c = 3$$
$$b+ac+d = 0$$
$$ad+bc = 6$$
$$bd = 4$$
I found these relationships between the various coefficients. Solving this system using the two middle equations:
$$\begin{cases} b+a\left(3-a\right)+\frac4b=0 \\ a\frac4b+b\left(3-a\right)=6 \end{cases}$$
From the first equation: $$a = \frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}$$
Substituting this into the second equation:
$$\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\cdot\frac4b+b\cdot\left(3-\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\right)=6$$
$$3\left(b-2\right)^2 = \left(b^2-4\right)\cdot\pm\sqrt{9+4b+\frac{16}b}$$
$$0 = \left(b-2\right)^2\cdot\left(\left(b+2\right)^2\left(9+4b+\frac{16}b\right)-9\left(b-2\right)^2\right)$$
So $b = 2$ because everything after $\left(b-2\right)^2$ did not really matter in this case. From there it was easy to get that $d = 2$, $a = -1$ and $c = 4$. This meant that
$$x^4+3x^3+6x+4=0 \to \left(x^2-x+2\right)\left(x^2+4x+2\right)=0$$
$$x = \frac12\pm\frac{\sqrt7}{2}i,\space x = -2\pm\sqrt2$$
These answers worked! I was pretty happy at the end that I had solved the equation which had taken a lot of work, but my question was if there was a better way to solve this?
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Reorganize the term like $$(x^4+4+4x^2)+(3x^3+6x)-4x^2=(x^2+2)^2+3x(x^2+2)-4x^2$$ than it is easy to come up with $$(x^2+4x+2)(x^2-x+2)$$
|
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|
Simple limit problem without L'Hospital's rule $$\lim_{x \to 1}\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1}$$
We are not supposed to use any derivatives yet, but I can't find any formula that helps here. It's a $\frac{0}{0}$ indeterminate form, and all I think of doing is
$$\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1} = \frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1} \cdot \frac{\sqrt{x^4 + 1} + \sqrt{2}}{\sqrt{x^4 + 1} + \sqrt{2}} = \frac{x^4-1}{(\sqrt[3]{x}-1)\cdot(\sqrt{x^4+1}+\sqrt{2})}$$
but I don't see if this leads anywhere.
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HINT:
$$\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1} =\frac{x^4+1-2}{x-1}\cdot\frac{\sqrt[3]{x^2}+\sqrt[3]x+1}{\sqrt{x^4 + 1} + \sqrt{2}}$$
Now as $x\to1,x\ne1,x-1\ne0,$ so we can safely cancel out $x-1$
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|
Let $a_n=2\sqrt n-\sum_{k=1}^n\frac{1}{\sqrt{k}}$. Show $a_n$ converges and $1<\lim_{n\to\infty}a_n<2$. I am able to show that $\lim_{n\to\infty}a_n\le 2$, but I can't think of a way to show strict inequality.
Here's my work so far:
By Riemann sums,
$$\int_1^{n+1}\frac{1}{\sqrt x}\,dx<\sum_{k=1}^n\frac{1}{\sqrt k}<\int_0^n\frac{1}{\sqrt x}\,dx.$$
$$2\sqrt{n+1}-2<\sum_{k=1}^n\frac{1}{\sqrt k}<2\sqrt n$$
So,
$$0<2\sqrt n-\sum_{k=1}^n\frac{1}{\sqrt k}<2\sqrt n+2-2\sqrt{n+1},$$
which implies,
$$0<2\sqrt n-\sum_{k=1}^n\frac{1}{\sqrt k}<2\quad\forall\,n\ge1.$$
$$\begin{align}
a_{n+1}-a_n & =\left(2\sqrt{n+1}-\sum_{k=1}^{n+1}\frac{1}{\sqrt k}\right)-\left(2\sqrt n-\sum_{k=1}^n\frac{1}{\sqrt k}\right)\\
& =2(\sqrt{n+1}-\sqrt n)-\frac{1}{\sqrt{n+1}}\\
&=\frac{2}{\sqrt{n+1}+\sqrt n}-\frac{1}{\sqrt{n+1}}\\
&=\frac{\sqrt{n+1}-\sqrt n}{\sqrt{n+1}(\sqrt{n+1}+\sqrt n)}>0.
\end{align}$$
Since, $a_1=1$, $\{a_n\}$ is a monotonically increasing sequence,
and
$$1<a_n<2,\quad\forall\,n\ge1.$$
So, $\{a_n\}$ converges, and
$$1<\lim_{n\to\infty}a_n\le2.$$
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You just have to make your initial estimate more quantitative:
\begin{eqnarray*}
2\sqrt{n+1}-2-\sum_{m=1}^{n}\frac{1}{\sqrt{m}} & = & \int_{1}^{n+1}\frac{1}{\sqrt{x}}\,{\rm d}x-\sum_{m=1}^{n}\frac{1}{\sqrt{m}}\\
& = & \sum_{m=1}^{n}\int_{m}^{m+1}\underbrace{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{m}}}_{\leq0}\,{\rm d}x\\
& \leq & \int_{1}^{2}\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{1}}\,{\rm d}x\\
& = & 2\sqrt{2}-3<0
\end{eqnarray*}
and hence
\begin{eqnarray*}
a_{n}-2 & = & 2\sqrt{n}-\sum_{m=1}^{n}\frac{1}{\sqrt{m}}-2\\
& = & 2\sqrt{n+1}-2-\sum_{m=1}^{n}\frac{1}{\sqrt{m}}+2\left(\sqrt{n}-\sqrt{n+1}\right)\\
& \leq & 2\sqrt{2}-3+2\left(\sqrt{n}-\sqrt{n+1}\right)\\
& \leq & 2\sqrt{2}-3.
\end{eqnarray*}
This implies
$$
\lim_{n\to\infty}a_{n}\leq2+2\sqrt{2}-3<2.
$$
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|
Finding the limit of $\lim_{n\rightarrow\infty}\frac{1}{n}[(a+\frac{1}{n})^{2}+(a+\frac{2}{n})^{2}+...+(a+\frac{n-1}{n})^{2}] $ Here's my try:
$\lim_{n\rightarrow\infty}\frac{1}{n}[(a+\frac{1}{n})^{2}+(a+\frac{2}{n})^{2}+...+(a+\frac{n-1}{n})^{2}] = $$\lim_{n\rightarrow\infty}[\frac{1}{n}(a^{2}+\frac{2a}{n}+\frac{1}{n^{2}})+\frac{1}{n}(a^{2}+\frac{4a}{n}+\frac{4}{n^{2}})+...+\frac{1}{n}(a^{2}+\frac{2a(n-1)}{n}+\frac{(n-1)^{2}}{n^{2}})]= $$\lim_{n\rightarrow\infty}[n\frac{a^{2}}{n}+\frac{2a+4a+...+2a(n-1)}{n^{2}}+\frac{1^2+2^2+...+(n-1)^{2}}{n^{3}}]= $$\lim_{n\rightarrow\infty}[a^{2}+\frac{2a(1+2+...+(n-1))}{n^{2}}+\frac{(n-1)n(2n-1)}{3n^{3}}]= a^{2}+\lim_{n\rightarrow\infty}[\frac{2a(n-1)n}{2n^2}+\frac{2n^{2}-3n+1}{3n^2}]= a^{2}+a+\frac{2}{3}$
but the last fraction should be $\frac{1}{3}$, not $\frac{2}{3}$, Anybody can point out where's my mistake?
2 mistakes pointed out by user kingW3, thank you
The CORRECT solution for other newcomers like me:
$\lim_{n\rightarrow\infty}\frac{1}{n}[(a+\frac{1}{n})^{2}+(a+\frac{2}{n})^{2}+...+(a+\frac{n-1}{n})^{2}] = $$\lim_{n\rightarrow\infty}[\frac{1}{n}(a^{2}+\frac{2a}{n}+\frac{1}{n^{2}})+\frac{1}{n}(a^{2}+\frac{4a}{n}+\frac{4}{n^{2}})+...+\frac{1}{n}(a^{2}+\frac{2a(n-1)}{n}+\frac{(n-1)^{2}}{n^{2}})]= $$\lim_{n\rightarrow\infty}[\mathbf{(n-1)}\frac{a^{2}}{n}+\frac{2a+4a+...+2a(n-1)}{n^{2}}+\frac{1^2+2^2+...+(n-1)^{2}}{n^{3}}]= $$\lim_{n\rightarrow\infty}[a^{2}+\frac{2a(1+2+...+(n-1))}{n^{2}}+\frac{(n-1)n(2n-1)}{\mathbf{6}n^{3}}]= a^{2}+\lim_{n\rightarrow\infty}[\frac{2a(n-1)n}{2n^2}+\frac{2n^{2}-3n+1}{\mathbf{6}n^2}]= a^{2}+a+\frac{1}{3}$
|
Did you use that
$$\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{\color{red}{6}}\quad?$$
Remark: You can simply find the result using the Riemann sum.
|
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|
Limit of $ \int_{0}^{\frac{\pi}{2}}\sin^n xdx$ and probability I begin with this problem:
Calculate the limit of $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx$ when $n\rightarrow \infty$.
It's natural to think of recurrence relation. Let $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx=I_n$. By integration by part: $$\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx=-\sin^{n-1}x\cos x|_{0}^{\pi/2}+(n-1)\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{n-2}\cos^2 x xdx$$
or $I_n=(n-1)(I_{n-2}-I_n)$. Thus $$I_n=\frac{n}{n-1}\cdot I_{n-2}$$
So we can find a formula for $I_n$ with odd $n$ : $$I_n=\frac{2\cdot 4\cdot 6\cdots (n-1)}{3\cdot 5\cdot 7\cdots n} $$
From the recurrence relation, we have $(I_n)$ decreases and is bounded below by $0$, then it has a limit $L\ge 0$. We will prove that $L=0$.
Suppose $L>0$. First, consider the series $$1+\frac{1}{2}+\frac{1}{3}+\cdots$$
This series diverges. Suppose the series $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots$ converges. Since $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots>\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots$, the series $\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots$ also converges. This yields that $1+\frac{1}{2}+\frac{1}{3}+\cdots$ converges, which is a contradiction. So $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\cdots$ diverges.
By Lagrange theorem, there exists $c_n\in [n-1,n]$ ($n\ge 3$) such that
$$\ln (n-1)-\ln n=-\frac{1}{c_n}$$
For all $n\ge 3$: $$\frac{1}{c_3}+\frac{1}{c_5}+\frac{1}{c_7}+\cdots>\frac{1}{3}+\frac{1}
{5}+\frac{1}{7}+\cdots$$
so $\frac{1}{c_3}+\frac{1}{c_5}+\frac{1}{c_7}+\cdots$ diverges by comparison test. Thus $\left[-\left(\frac{1}{c_3}+\frac{1}{c_5}+\frac{1}{c_7}+\cdots\right)\right]$ also diverges.
Back to our problem. Since $L>0$, then $\lim \ln I_n=\ln L$. This means $\ln(2/3)+\ln(5/4)+\cdots=\ln L$, or $-\frac{1}{c_3}-\frac{1}{c_5}-\cdots=\ln L$, contradiction. Thus $L=0$.
I have some questions for the above problem:
*
*Is there another way to compute the limit of $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx$? My solution is quite complicated
*Rewrite $I_n$ as $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots \left(1-\frac{1}{n}\right)$. Consider a probability problem: in a test of multiple choice questions, each question has exactly one correct answer. The first question has $3$ choices, the second has $5$ choices, etc. By the above result, if the tests has more and more questions, then the probability for the student to be false at all is smaller and smaller. This sounds interesting, because the questions have more and more choices. Is there any generalized result for this (fun) fact?
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Here is a possibly simpler way for proving your theorem. Since $\sin x$ is strictly increasing in the range $[0,\pi/2]$, for each $\epsilon > 0$ we can bound
$$
\begin{align*}
\int_0^{\pi/2} \sin^n x \, dx &= \int_0^{\pi/2-\epsilon} \sin^n x \, dx + \int_{\pi/2-\epsilon}^{\pi/2} \sin^n x \, dx \\ &\leq \frac{\pi}{2} \sin^n(\tfrac{\pi}{2}-\epsilon) + \epsilon.
\end{align*}
$$
Since $\sin(\tfrac{\pi}{2}-\epsilon) < 1$, we conclude that $\int_0^{\pi/2} \sin^n x \, dx \leq 2\epsilon$ for all large enough $n$. In particular, $\limsup_{n\to\infty} \int_0^{\pi/2} \sin^n x \, dx \leq 0$. Since $\int_0^{\pi/2} \sin^n x \, dx \geq 0$, we conclude that the limit exists and equals $0$.
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|
Arithmetic pattern $1 + 2 = 3$, $4 + 5 + 6 = 7 + 8$, and so on I am noticing this pattern:
\begin{align}
1+2&=3\\
4+5+6&=7+8\\
9+10+11+12&=13+14+15 \\
16+17+18+19+20&=21+22+23+24 \\
&\vdots
\end{align}
Is there a general formula that expresses a pattern here, such as for the $n$th term, or the closed form of the entire summation? I think the $n$th term starts with $$n^2+(n^2+1)+\cdots=\cdots+[(n+1)^2-1].$$
I am also presuming once this formula is discovered, we can prove it by induction for any $n$.
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Let's examine a simple identity:
$$9+10+11+12=13+14+15$$
Notice that there are $4$ terms on the left and $3$ on the right - so let's "use up" one of the terms on the left. We can split $9=3+3+3$ and distribute these additional $+3$ terms to the other $3$ summands on that side to get:
$$(10+\color{red} 3)+(11+\color{red} 3)+(12+\color{red} 3)=13+14+15$$
and, of course, the left hand is exactly $13+14+15$ when we compute. Well, that was easy!
Let's generalize. The sum is always of the form
$$n^2+(n^2+1)\ldots+(n^2+n)=(n^2+n+1)+\ldots + (n^2+2n)$$
and there are $n+1$ terms on the left hand and $n$ terms on the right hand side. So how do we get from left to right? Easy! We take the $n^2$ term and divide it into $\underbrace{n+n+\ldots+n}_{n\text{ times}}$. Now, there are $n$ other summands on that side, so we distribute a $+n$ term to each of them without changing the value. This yields
$$(n^2+1+\color{red} n)+\ldots+(n^2+n+\color{red}n)=(n^2+n+1)+\ldots + (n^2+2n)$$
but this is obvious, since simplifying both sides gives
$$(n^2+n+1)+\ldots+(n^2+2n)=(n^2+n+1)+\ldots + (n^2+2n)$$
where both sides are identical.
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|
Calculating $\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}$ I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$
$$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$
$$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}.$$
I know that $$\sum_{n=0}^\infty\frac{1}{n!}=e$$
so $$\sum_{n=1}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{n!}-1=e-1$$
But, what can I do for $$\sum_{n=1}^\infty\frac{1}{(n+1)!}$$ ?
Am I allowed to start a sum for $n=-1$ ? How can I bring to a something similar to $$\sum_{n=0}^\infty\frac{1}{n!}$$?
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**Hint: **$$\sum_{n=1}^\infty \frac{1}{(n+1)!} = \sum_{n=2}^\infty \frac{1}{n!} \\ = \sum_{n=0}^\infty \frac{1}{n!}-\frac{1}{1!}-\frac{1}{0!}$$
|
{
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"url": "https://math.stackexchange.com/questions/1098034",
"timestamp": "2023-03-29T00:00:00",
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|
series convergence. Dirichlet test $$\sum(-1)^{\lfloor\frac{n^3+n+1}{3n^2-1}\rfloor}\frac{\ln(n)}{n}$$
I thought about using Dirichlet test. $\ln(n)/n$ is a decreasing sequence that tends to 0 but I have problem with proving it. I also can
t tell anything about $(-1)^{\lfloor\frac{n^3+n+1}{3n^2-1}\rfloor}$.
|
To prove that $\frac{\log(n)}{n}$ is decreasing, notice that:
$$\frac{\log(n+1)}{n+1}=\frac{\log n}{n+1}+\frac{\log\left(1+\frac{1}{n}\right)}{n+1}\leq \frac{\log n}{n+1}+\frac{1}{n(n+1)}$$
hence if $n> e,$
$$\frac{\log(n+1)}{n+1}< \log n\left(\frac{1}{n+1}+\frac{1}{n(n+1)}\right)=\frac{\log n}{n}.\tag{1}$$
To prove that
$$ (-1)^{\left\lfloor\frac{n^3+n+1}{3n^2-1}\right\rfloor}\tag{2} $$
has bounded partial sums, notice that:
$$ \frac{n^3+n+1}{3n^2-1} = \frac{n}{3} + \frac{4n+3}{9n^2-3} $$
and:
$$ \forall n\geq 3,\qquad 0<\frac{4n+3}{9n^2-3}<\frac{1}{5}, $$
so for any $n\geq 3$ we have that $(2)$ behaves like the sequence:
$$ \ldots, -1,-1,-1,1,1,1,-1,-1,-1,1,1,1,\ldots \tag{3}$$
that obviously has bounded partial sums.
|
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|
How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$? How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$?
I get that you can take out the $x$ like so: $x^3+4x^2+x-6=x(x^2+4x+1)-6$ but how do you get the $2$ from here?
|
*
*Struggle to find a real root. By trying small positive and negative integers, you find that $P(-2)=0$. So $x+2$ is a factor.
*Apply the change of variable $y=x+2$:
$$P(x)=P(y-2)=(y-2)^3+4(y-2)^2+(y-2)+6=y^3-2y^2-3y.$$
Now you easily factor as
$$P(y-2)=y(y^2-2y-3),$$
and revert to the original
$$P(x)=(x+2)\left((x+2)^2-2(x+2)-3\right)=(x+2)(x^2+2x-3).$$
Similarly, $x=1$ is a root of $x^2+2x-3$; set $z=x-1$, and
$$x^2+2x-3=(z+1)^2+2(z+1)-3=z^2+4z=z(z+4).$$
Hence, $$x^2+2x-3=(x-1)((x-1)+4)=(x-1)(x+3).$$
Tedious, but "without" long division.
|
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|
How to decompose this matrix exponential? I would some help with the steps to decompose the below matrix exponential.
$\exp\left[ \zeta \left ( \begin{matrix}
-\cos(x) & i \sin(x) \\
-i \sin(x) & \cos(x)
\end{matrix} \right ) \right] = \left ( \begin{matrix}
\cosh(\zeta)-\sinh(\zeta)\cos(x) & i \sinh(\zeta)\sin(x) \\
-i \sinh(\zeta)\sin(x) & \cosh(\zeta)+\sinh(\zeta)\cos(x)
\end{matrix} \right )$
I think the first matrix could be factored as Pauli matrices but I have not had any luck trying this.
|
First, check your equality when $\sin(x)=0$.
Second, notice that $\zeta\begin{pmatrix}-\cos(x)& i\sin(x)\\-i\sin(x)& \cos(x)\end{pmatrix}=\zeta\begin{pmatrix}i& 0\\ 0& 1\end{pmatrix}\begin{pmatrix}-\cos(x)& \sin(x)\\\sin(x)& \cos(x)\end{pmatrix}\begin{pmatrix}-i& 0\\ 0& 1\end{pmatrix}$
and $\begin{pmatrix}i& 0\\ 0& 1\end{pmatrix}\begin{pmatrix}-i& 0\\ 0& 1\end{pmatrix}=\begin{pmatrix}1& 0\\ 0& 1\end{pmatrix}$.
Next, $\zeta\begin{pmatrix}-\cos(x)& \sin(x)\\\sin(x)& \cos(x)\end{pmatrix}=\begin{pmatrix}1+\cos(x)& 1-\cos(x)\\-\sin(x)& \sin(x)\end{pmatrix}\begin{pmatrix}-\zeta& 0\\ 0& \zeta\end{pmatrix}\begin{pmatrix}\dfrac{1}{2}& \dfrac{-1+\cos(x)}{2\sin(x)}\\\dfrac{1}{2}& \dfrac{1+\cos(x)}{2\sin(x)}\end{pmatrix}$.
Notice that $\begin{pmatrix}1+\cos(x)& 1-\cos(x)\\-\sin(x)& \sin(x)\end{pmatrix}\begin{pmatrix}\dfrac{1}{2}& \dfrac{-1+\cos(x)}{2\sin(x)}\\\dfrac{1}{2}& \dfrac{1+\cos(x)}{2\sin(x)}\end{pmatrix}=\begin{pmatrix}1& 0\\ 0& 1\end{pmatrix}$.
Thus, you need to compute $\exp(R\begin{pmatrix}-\zeta& 0\\ 0& \zeta\end{pmatrix}R^{-1})=R\begin{pmatrix} e^{-\zeta}& 0\\ 0 & e^{\zeta} \end{pmatrix}R^{-1}$, where $R=\begin{pmatrix}i& 0\\ 0& 1\end{pmatrix}\begin{pmatrix}1+\cos(x)& 1-\cos(x)\\-\sin(x)& \sin(x)\end{pmatrix}$ and $R^{-1}=\begin{pmatrix}\dfrac{1}{2}& \dfrac{-1+\cos(x)}{2\sin(x)}\\\dfrac{1}{2}& \dfrac{1+\cos(x)}{2\sin(x)}\end{pmatrix}\begin{pmatrix}-i& 0\\ 0& 1\end{pmatrix}$.
In order to finish you must use the formulas for the $\cosh(\zeta)$ and $\sinh(\zeta)$.
|
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|
Reverse engineering a Taylor expansion We have the sum: $$S(x) = \frac{x^4}{3(0!)} + \frac{x^5}{4(1!)} + \text{ }...$$ And we are told to sum the series to obtain a finite expression. My guess was to reverse engineer the expression in order to find a function which has $S(x)$ or some form of it as a taylor expansion. $$\frac{1}{x^4}S(x) = \frac{1}{3} +\frac{x}{4}+\frac{x^2}{5(2!)}$$ But I am not familiar with a function which increments the denominator by $1$ every time we take the derivative.
|
$$\begin{eqnarray*}S(x)&=&x^4\sum_{n\geq 0}\frac{x^j}{j!}\cdot\frac{1}{j+3}=x^4\sum_{n\geq 0}\frac{x^j}{j!}\int_{0}^{1}y^{j+2}\,dy=x^4\int_{0}^{1}y^2\sum_{n\geq 0}\frac{(xy)^j}{j!}\,dy\\&=&x^4\int_{0}^{1}y^2 e^{xy}\,dy=x\int_{0}^{x}z^2 e^z\,dz=x\cdot\left[\left(z^2-2z+2\right)e^{z}\right]_{0}^{x}\\&=&x\left(-2+(x^2-2x+2)e^x\right)=\color{red}{-2x+(x^3-2x^2+2x)e^x}.\end{eqnarray*}$$
|
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|
Finding Cartesian coordinates of remaining vertices of triangle, given a vertex and angle from y-axis I have an isosceles triangle $ABC$, where the height $h$ and angle at vertex $A$ are known. The Cartesian coordinates of vertex $A$ are also known to be $\left(x,y\right)$.
If the angle between the y-axis and the line represented by $h$ is $θ$, is it possible to find the Cartesian coordinates of points $B$ and $C$?
|
Absolutely.
The angle between the $y$-axis and $AB$ is $\theta - \tfrac{A}{2}$, and the angle between the $y$-axis and $AC$ is $\theta + \tfrac{A}{2}$.
So you will get
\begin{align*}
B &= (x - r \sin(\theta - \tfrac{A}{2}), y + r \cos(\theta - \tfrac{A}{2}) \\
C &= (x - r \sin(\theta + \tfrac{A}{2}), y + r \cos(\theta + \tfrac{A}{2})
\end{align*}
where $r = AB = AC = \frac{h}{\cos(\tfrac{A}{2})}$.
|
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|
Complex Solution I was looking at a simple complex problem and I came by this:
Solve: $z^2 = 3 - 4i$
Let
$$z= x + yi$$
Rewrite as
$$(x + yi)^2 = 3 - 4i$$
Expand
$$x^2 +2xyi + y^2i^2 = 3 - 4i$$
Simplify
$$x^2 - y^2 + 2xyi = 3 - 4i$$
real/imaginary parts
$$x^2 - y^2 = 3$$
and
$$2xy = -4$$
$$xy = -2$$
Therefore $y = -\frac{2}{x}$.
Substitute $y = -\frac 2x$ into $x^2 - y^2 = 3$, giving
$$x^2 - \frac{4}{x^2} = 3$$
Multiply by $x^2$:
$x^4 - 4 = 3x^2$
Factor
$$x^4 - 3x^2 - 4 = 0$$
$$(x^2 + 1) (x^2 - 4) = 0$$
$$(x^2+ 1) (x + 2) (x - 2) = 0$$
Given $xy = -2$
for $x = 2$ then $y = -1$ therefore $z = 2 - i$
for $x = -2$ then $y = 1$ therefore $z = -2 + i$
Now, I understand everything, but why do we ignore the root of $i$ and $-i$ for the root $x^2+1=0$? Is it because we are only looking for real roots? When plugging these two back into $(x+iy)=z$, I saw that it gave the same two solutions as the real roots.
|
When you write $z=x+yi$ you are looking for real $x$ and $y$.
There's really nothing bad in considering also the other two possibilities, but they will give just the same numbers.
Indeed, if you take $x=i$, you obtain, from $xy=-2$, that $y=-2/i=2i$ and so
$$
z=x+yi=i+2i^2=-2+i
$$
which has already been obtained. The other solution is obtained when you consider $x=-i$.
You know that the equation has at most two solutions, so when you find two of them you're done.
|
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|
$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$ Understand the representation $$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$$=$$\frac{1}{2} \left(-\frac{2}{n+1}+\frac{1}{n+2}+\frac{1}{n}\right)$$
$$s_n=\frac{1}{2} \left(\left(\frac{1}{n+2}+\frac{1}{n}-\frac{2}{n+1}\right)+\left(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}\right)+\ldots +\left(1 \frac{1}{1}-\frac{2}{2}+\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{2}+\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{2}{5}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{3}+\frac{1}{7}\right)+\left(\frac{1}{6}-\frac{2}{7}+\frac{1}{8}\right)\right)$$
Im sorry for not being able to get the term in right order, but as you can see the first two terms in the parenthesis should be the last for a accurate representation.
$$\left(
\begin{array}{cc}
1 & 1-1+\frac{1}{3} \\
2 & \frac{1}{2}-\frac{2}{3}+\frac{1}{4} \\
3 & \frac{1}{3}-\frac{1}{2}+\frac{1}{5} \\
4 & \frac{1}{4}-\frac{2}{5}+\frac{1}{6} \\
5 & \frac{1}{5}-\frac{1}{3}+\frac{1}{7} \\
6 & \frac{1}{6}-\frac{2}{7}+\frac{1}{8} \\
\end{array}
\right)$$
I think that my pure understanding of the representation is what makes me confused. From the table I see that a pattern in the denominators emerge, when n=1, then (1-2+3), when n=2 then, 2-3+4. However given the part of sn where $$\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}$$ I fail to see how the first part of expression right above is not undefined when n=1.
I am convinced that it is my lack of knowledge of what the representation actually means. And I am also struggling to see what cancels to give$$\frac{1}{2} \left(\frac{1}{n+2}-\frac{1}{n+1}+\frac{1}{2}\right)$$
However given that this is true I am able to understand that the answer is (1/4). But as I said it is the representation I do not get. Could someone help me as this would be very useful for my next chapter in my textbook!
|
Hint: We are finding
$$\sum_1^\infty \frac{1}{2}\left(\left(\frac{1}{n}-\frac{1}{n+1}\right)-\left(\frac{1}{n+1}-\frac{1}{n+2}\right) \right).$$
|
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|
Prove or disprove: $\sum a_n$ convergent, where $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$.
Let $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$. Show that $a_n>0\ \forall\ n\ge1$.
Prove or disprove: $\sum\limits_{n=1}^\infty a_n$ is convergent.
I can't show that $a_n > 0\ \forall n\ge1$. I tried using induction but it wouldn't work.
Attempt:
$$
\begin{align}
2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}&=(2\sqrt{n}-(\sqrt{n-1}+\sqrt{n+1}))\cdot {2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\\&={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\end{align}$$
(The calculations are true for sure. No check is desired.)
Denote
$$a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\text{ and }b_n={1\over n^{2}}.$$
Then
$$\begin{align*}
\lim_{n\to \infty}{a_n\over b_n} & =\lim_{n\to \infty}n^{2}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=\lim_{n\to \infty}n^{2}\lim_{n\to \infty}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=0
\end{align*}$$
By the comparison test for series convergence, since $\lim_{n\to \infty}{a_n\over b_n}=0$, then if $b_n$ converges, which it does, so does $a_n$.
|
I will address both parts of your question:
Part 1: Proving that $a_n>0$.
From your final form of $a_n$
$$a_n=\frac{2n-2\sqrt{(n-1)(n+1)}}{2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}}$$
we can notice that the denominator is always positive (when $n\ge1$), so it remains to show that $2n>2\sqrt{(n-1)(n+1)}=2\sqrt{n^2-1}$. This is quite easy (thanks to Steven Stadnicki for pointing this out):
$$n^2>n^2-1\implies n>\sqrt{n^2-1}\implies2n>2\sqrt{n^2-1}$$
as desired.
Part 2: Proving that $\sum\limits_{n=1}^\infty a_n$ converges by the limit comparison test.
Unfortunately your solution is not quite correct as there is an error in your limit computation, and you should get $\lim\limits_{n\rightarrow\infty}n^2a_n=\infty$ which doesn't work for us. However, we can instead use $n^{3/2}$, and find that $\lim\limits_{n\rightarrow\infty}n^{3/2}a_n=\frac{1}{4}$, which proves the sum converges because $\displaystyle\sum_{n=1}^\infty\frac{1}{n^{3/2}}$ converges.
Limit Computations:
First we note
$\begin{align}a_n&=\frac{2n-2\sqrt{(n-1)(n+1)}}{2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}}
\\&=\frac{2\left(n-\sqrt{n^2-1}\right)}{2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}}
\\&=\frac{2\left(n-\sqrt{n^2-1}\right)}{2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}}\times\frac{n+\sqrt{n^2-1}}{n+\sqrt{n^2-1}}
\\&=\frac{2}{\left(2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}\right)\left(n+\sqrt{n^2-1}\right)}
\end{align}$
Hence
$\begin{align}\lim\limits_{n\rightarrow\infty} n^2a_n&=\lim\limits_{n\rightarrow\infty}\frac{2n^2}{\left(2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}\right)\left(n+\sqrt{n^2-1}\right)}
\\&=\lim\limits_{n\rightarrow\infty}\frac{2\sqrt{n}}{\left(2+\sqrt{1-\frac{1}{n}}+\sqrt{1+\frac{1}{n}}\right)\left(1+\sqrt{1-\frac{1}{n^2}}\right)}
\\&=\infty
\end{align}$
while
$\begin{align}\lim\limits_{n\rightarrow\infty} n^{3/2}a_n&=\lim\limits_{n\rightarrow\infty}\frac{2n^{3/2}}{\left(2\sqrt{n}+\sqrt{n-1}+\sqrt{n+1}\right)\left(n+\sqrt{n^2-1}\right)}
\\&=\lim\limits_{n\rightarrow\infty}\frac{2}{\left(2+\sqrt{1-\frac{1}{n}}+\sqrt{1+\frac{1}{n}}\right)\left(1+\sqrt{1-\frac{1}{n^2}}\right)}
\\&=\frac{1}{4}
\end{align}$
|
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|
Given $x,y,z>0$: $\frac{2}{3x+2y+z+1}+\frac{2}{3x+2z+y+1}=(x+y)(x+z)$. Find Minimum Value Of: $P=\frac{2(x+3)^2+y^2+z^2-16}{2x^2+y^2+z^2}$
Given $x,y,z>0$: $\frac{2}{3x+2y+z+1}+\frac{2}{3x+2z+y+1}=(x+y)(x+z)$ $(1)$
Find Minimum Value Of:
$P=\frac{2(x+3)^2+y^2+z^2-16}{2x^2+y^2+z^2}$
I found $2x+y+z\geq 2$ from (1) but it not work :(
Could some one help me solve this ?
|
I don't know why I have this reslut:
since Use Cauchy-Schwarz inequality,we have
$$\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}$$
so
$$\dfrac{2}{3x+2y+z+1}+\dfrac{2}{3x+2y+z+1}\ge \dfrac{8}{6x+3y+3z+2}\ge \dfrac{8}{6x+3\sqrt{2(y^2+z)}+2}$$
other hand
$$(x+y)(x+z)\le\dfrac{(2x+y+z)^2}{4}\le\dfrac{(2x+\sqrt{2(y^2+z^2)})^2}{4}$$
let
$$2x+\sqrt{2(y^2+z^2)}=u>0$$
use this condition we have
$$\dfrac{u^2}{4}\ge\dfrac{8}{3u+2}\Longrightarrow u^2(3u+2)\ge 32$$
so
$$\Longrightarrow 3u^3-6u^2+8(u+2)(u-2)\ge 0\Longrightarrow (u-2)(3u^2+8u-16)\ge 0$$
$$\Longrightarrow (u-2)(3u-4)(u+4)\ge 0$$
so
$$0<u\le\dfrac{4}{3},u>2$$
so when $0<u\le\dfrac{4}{3}$,then we have
$$\sqrt{2(y^2+z^2)}\le\dfrac{4}{3}-2x\Longrightarrow y^2+z^2\le 2(\dfrac{2}{3}-x)^2$$
so
$$\dfrac{2(x+3)^2+y^2+z^2-16}{2x^2+y^2+z^2}=1+\dfrac{12x+2}{2x^2+y^2+z^2}\ge 1+\dfrac{12x+2}{2x^2+2(\dfrac{2}{3}-x)^2}$$
then It is to find it
But for $u>2$, it is clear is $0$?
|
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|
Analyze the continuity of the following function Here in my book I have such an exercise with the explanation given below, but still there is something the authors didn't add, but simply put "...after some operations...". Here is such an exercise:
Analyze the continuity of the following function:
$$f:\Bbb R^2\to\Bbb R $$ $$ f(x,y)= \begin{cases}
\frac{3x^2y}{\sqrt{x^4+y^2}}& \text{if } (x,y)\neq(0,0)\\[2ex]
0& \text{if } (x,y)=(0,0)
\end{cases}$$
In my book is writen, after some operations, we find that $f$ is continuous on $\Bbb R^2 \setminus\{(0,0)\}$. What are these operation ? Thanks.
|
if $|y|\leq 1$, $$\sqrt{x^4+y^2}\geq \sqrt{x^4+y^4}$$
and thus
$$|f(x,y)|=\left|\frac{3x^2y}{\sqrt{x^4+y^2}}\right|\leq\underbrace{\left|\frac{x^2}{\sqrt{x^4+y^4}}\right|}_{\leq 1}|3y|\leq |3y|\underset{(x,y)\to(0,0)}{\longrightarrow 0}$$
therefore $f$ is continuous on $\mathbb R^2$.
Justification for $\left|\frac{x^2}{\sqrt{x^4+y^4}}\right|\leq 1$:
$$x^4\leq x^4+y^4\implies x^2\leq\sqrt{x^4+y^4}\implies \underbrace{\frac{x^2}{\sqrt{x^4+y^4}}}_{=\left|\frac{x^2}{\sqrt{x^4+y^4}}\right|}\leq 1.$$
|
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|
Prove the inequality $a^3+2 \geq a^2+2 \sqrt{a}$ Prove the inequality $a^3+2 \geq a^2+2 \sqrt{a},a \geq 0.$
One way to do it is using the formula
$$
a^3+2 - a^2-2 \sqrt{a}=(\sqrt{a}-1)^2(1+(a+1)(\sqrt{a}+1)^2) \geq 0.
$$
But I hope there is a better way.
|
Substitute $a=b^2$, then $a^3-a^2-2\sqrt a+2=b^6-b^4-2b+2$, which has $1$ as a root, so we can factor it out:
$$b^6-b^4-2b+2=(b-1)(b^5+b^4-2)$$
The polynomial $b^5+b^4-2$ has still one as a root, so factor more:
$$b^5+b^4-2=(b-1)(b^4+2b^3+2b^2+2b+2)$$
So $a^3-a^2-2=(b-1)^2(b^4+2b^3+2b^2+2b+2)$, which is nonnegative for $b\ge 0$.
|
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|
Prove that $1\cdot f(1)+ 2\cdot f(2)+ ...+ n\cdot f(n) \leq n(n+1)(2n+1)/6$ where $f(n+1) = f(f(n))+1$ Consider checking function $\mathbb{N}\to \mathbb{N}$ relationship $f(n+1) = f(f(n))+1$, for any positive integer $n$.
Prove that $1\cdot f(1)+ 2\cdot f(2)+ ...+ n\cdot f(n) \leq n(n+1)(2n+1)/6$ for any positive integer $n$.
|
i am going to try. the defining relation for $f$ gives us the inequality $$nf(n) = n\{f(f(n-1)) + 1\} =n + nf(f(n-1)) \ge n $$ because the range of $f$ consists of positive integers.
we can use $jf(j) \ge j$ again and again to get
$$ nf(n) \ge n + nf(f(n-1)) \ge n + nf(n-1) \ge n + n(n-1) = n^2$$
therefore
$$1f(1) + 2f(2) + \cdots + nf(n) \ge 1^2 + 2^2 + \cdots + n^2 = n(n+1)(2n+1)/6 $$
${\bf edit}:$ the argument used to establish $f(n) \ge n$ has a flaw but not the conclusion and it can be fixed.
we already have $f(n) \ge 1 \text{ for all } n \ge 1$ by the range of $f$ being positive integers. now using that and $f(n) = f(f(n-1) + 1$ establishes $f(n) \ge 2 \text{ for } n \ge 2 $. now by induction $f(n) \ge n.$
|
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|
$\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = ?$ $S_1=\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = ?$
Attempt: $S_1=\sum_{n=1}^\infty \dfrac {2n+1}{n^2(n+1)^2} = \dfrac {2n+2-1}{n^2(n+1)^2}$
$S_1=2 \sum_{n=1}^\infty [\dfrac {n+1}{n^2(n+1)^2} ] - \sum_{n=1}^\infty \dfrac {1}{n^2(n+1)^2}$
$=2 \sum_{n=1}^\infty [\dfrac {1}{n^2(n+1) } ] - \sum_{n=1}^\infty \dfrac {1}{n^2(n+1)^2}$
I am stuck on how to move ahead. Please guide me.
Thank you for your help.
|
First note that $$\sum_{n=1}^{\infty}\frac{2n+1}{n^{2}(n+1)^{2}}=\sum_{n=1}^{\infty}\left[\frac{1}{n^{2}}-\frac{1}{(n+1)^{2}}\right]$$
By telescoping the series, we can see that
$$
\lim\limits_{z\to\infty}\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\dots +\frac{1}{z^2}-\frac{1}{(z+1)^2}\right]
$$
$$
= \lim\limits_{z\to\infty}\left[1-\frac{1}{(z+1)^2}\right]=1-0=1
$$
|
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|
Closed form as sum and combinatorial of Fibonacci numbers How can I prove that the Fibonacci numbers that are defined as $F_n=F_{n-1}+F_{n-2}, \; n \geq 2$ and $F_0=0,\ F_1=1,\ F_2=1$ have the form:
$$F_n=\sum_{k=0}^{n-1} \binom{n-1-k}{k}, \; n\ge 2 $$
I am aware of the gen. function that is:
$$\frac{x}{1-x-x^2} =\sum_{n=0}^{\infty}F_n x^n $$
but I cannot extract the other formula.
|
This one can also be done using complex variables.
Suppose we seek to show that
$$F_n = \sum_{k=0}^{n-1} {n-1-k \choose k}.$$
Introduce the integral repesentation
$${n-1-k\choose k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-1-k}}{z^{k+1}} \; dz.$$
This gives for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-1}}{z}
\sum_{k=0}^{n-1} \frac{1}{z^k (1+z)^k} \; dz$$
which is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-1}}{z}
\frac{1-1/z^n/(1+z)^n}{1-1/z/(1+z)} \; dz
\\ =\frac{1}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{n-1}
\frac{1-1/z^n/(1+z)^n}{z-1/(1+z)} \; dz
\\ =\frac{1}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{n-1}
\frac{1+z-1/z^n/(1+z)^{n-1}}{(1+z)z-1} \; dz
\\ =\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n-1/z^n}{(1+z)z-1} \; dz.$$
This has two components, the first is
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^n}{(1+z)z-1} \; dz$$
which is zero, and the second is
$$-\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1/z^n}{(1+z)z-1} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{z}{1-z-z^2} \; dz.$$
Using the generating function this evaluates to
$$F_n$$ by inspection.
Observation. Wilf / generatingfunctionology will produce this result as well.
|
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|
$\sum_1^n 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} $ converge or not? how to check if this converge? $$\sum_{n=1}^\infty a_n$$
$$a_n = 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}$$
what i did is to show that:
$$a_n =2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} > 2\sqrt{n} - \sqrt{n+1} - \sqrt{n+1} = 2\sqrt{n} - 2\sqrt{n+1} = -2({\sqrt{n+1} - \sqrt{n}}) = -2(({\sqrt{n+1} - \sqrt{n}})\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}) = -2\frac{1}{\sqrt{n+1} + \sqrt{n}} > -2\frac{1}{2\sqrt{n+1}} = \frac{-1}{\sqrt{n+1}} = b_n $$
and we know that:
$$\sum_{n=1}^\infty b_n$$
doesnt converge cause
$$\sum_{n=1}^\infty \frac{1}{\sqrt{n}}$$
doesnt converge
so from here my conclusion is that $\sum_{n=1}^\infty a_n$ doesnt converge
but i know the final answer is that it does converge
so what am i doing wrong?
|
$$\begin{eqnarray*}\sum_{n=1}^{N}\left(2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}\right)&=&\color{red}{\sum_{n=1}^{N}\left(\sqrt{n}-\sqrt{n-1}\right)}+\color{blue}{\sum_{n=1}^{N}\left(\sqrt{n}-\sqrt{n+1}\right)}\\&=&\color{red}{\sqrt{N}}+\color{blue}{1-\sqrt{N+1}}\\&=&1-\frac{1}{\sqrt{N}+\sqrt{N+1}},\tag{1}\end{eqnarray*}$$
so, by letting $N\to +\infty$, we get that the original series converges towards $\color{green}{1}$.
|
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|
Find the range of $ x-\sqrt{4-x^2}$ $Y=x-\sqrt{4-x^2}$. How to find these types of functions' range?
I just know that the answer is $R=\{y\in\mathbb{R}\mid-2\sqrt{2}\leq y\leq 2\}$, but I have no idea how to find it step by step.
|
Hint
Let
$x=2\cos{t}$,then
$$x-\sqrt{4-x^2}=2\cos{t}\pm 2\sin{t}$$
case 1
if $t\in[0,\pi]$,then
$$x-\sqrt{4-x^2}=2\cos{t}-2\sin{t}=2\sqrt{2}\cos{\left(t+\dfrac{\pi}{4}\right)}\in [-2\sqrt{2},2] $$
if case 2,$t\in[\pi,2\pi]\Longrightarrow \dfrac{5}{4}\pi\le t+\dfrac{\pi}{4}\le\dfrac{9\pi}{4}$
then
$$x-\sqrt{4-x^2}=2\cos{t}+2\sin{t}=2\sqrt{2}\sin{\left(t+\dfrac{\pi}{4}\right)}\in [-2\sqrt{2},2]$$
|
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|
Range of trigonometric functions I would like to know if there is a simple approach to find the range of functions in the form:
$$\sin x\sin2x$$
$$\cos x\cos3x$$
$$\sin 2x\cos 4x$$
For example, finding the range of a function in the form:
$$a\cos\theta + b\sin\theta$$ is simple (the minimum value is $-\sqrt{a^2 + b^2}$ while the maximum value is $\sqrt{a^2 + b^2}$.
|
We have always $ a^2\leq a^2+b^2\ \Rightarrow\ |a|\leq\sqrt{a^2+b^2} \Rightarrow \frac{|a|}{\sqrt{a^2+b^2}}\leq 1\Rightarrow-1\leq\frac{a}{\sqrt{a^2+b^2}}\leq1$ so there exists a $\alpha $ that $ \cos\alpha=\frac{a}{\sqrt{a^2+b^2}}$. Beside We know that $\sin\alpha=\pm\sqrt{1-\cos^2\alpha}=\pm\sqrt{1-(\frac{a}{\sqrt{a^2+b^2}})^2}=\pm\frac{b}{\sqrt{a^2+b^2}}$. Now let be $y=a cos \theta+b\sin \theta $ then $\frac{y}{\sqrt{a^2+b^2}}=\frac{a}{\sqrt{a^2+b^2}}\sin \theta+\frac{b}{\sqrt{a^2+b^2}}\cos \theta=\cos\alpha\sin \theta\pm\sin\alpha\cos \theta=\sin (\theta\pm\alpha)$. Because of $-1\leq\sin (\theta\pm\alpha)\leq 1$ then $-1\leq\frac{y}{\sqrt{a^2+b^2}}\leq 1$. Hence $-\sqrt{a^2+b^2}\leq a\cos \theta+b\sin \theta\leq \sqrt{a^2+b^2}$.
|
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|
Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$ The following identity is true for any given $x \in [-1,1]$:
$$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$
But I don't know how to explain it.
I understand that the derivative of the equation is a truth clause, but why would the following be true, intuitively?
$$\int^{x}_{C1}\frac{1\cdot dx}{\sqrt{1-x^{2}}} + \int^{x}_{C2}\frac{-1 \cdot dx}{\sqrt{1-x^{2}}} =\\
\arcsin(x) - \arcsin(C1) + \arccos(x) - \arccos(C2) = 0 \\
\text{while } \arcsin(C1) + \arccos(C2) = \frac{\pi}{2}$$
I can't find the right words to explain why this is true?
Edit #1 (25 Jan, 20:10 UTC):
The following is a truth clause:
$$
\begin{array}{ll}
\frac{d}{dx}(\arcsin(x) + \arccos(x)) = \frac{d}{dx}\frac{\pi}{2} \\
\\
\frac{1}{\sqrt{1-x^{2}}} + \frac{-1}{\sqrt{1-x^{2}}} = 0
\end{array}
$$
By integrating the last equation, using the limits $k$ (a constant) and $x$ (variable), I get the following:
$$
\begin{array}{ll}
\int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\
\\
\arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = m \text{ (m is a constant)}\\
\\
\arcsin(x) + \arccos(x) = m + \arcsin(k) + \arccos(k) \\
\\
\text{Assuming that } A = m + \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1]
\end{array}
$$
Using Calculus, why is that true for every $x \in [-1,1]$?
Edit #2:
A big mistake of mine was to think that $\int^x_k0 = m \text{ (m is const.)}$, but that isn't true for definite integrals.
Thus the equations from "Edit #1" should be as follows:
$$
\begin{array}{ll}
\int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\
\\
\arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = 0\\
\\
\arcsin(x) + \arccos(x) = \arcsin(k) + \arccos(k) \\
\\
A = \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1]
\end{array}
$$
|
There are a couple of ways to see this. Firstly, draw a right triangle, call it $ABC$ (with $C$ being the right angle), with side lengths $a$, $b$ and $c$ with the usual convention. Then $\arcsin(\frac{b}{c})$ is the measure of the angle $CBA$. Additionally, $\arccos(\frac{b}{c})$ is the angle of the angle of the opposite angle $CAB$, so $\arccos(\frac{b}{c}) = \frac{\pi}{2}-\arcsin(\frac{b}{c})$ since the opposite angles must sum to $\frac{\pi}{2}$. From here, you get the result.
We could also do some calculus to figure it out. Let's let $f(x) = \arcsin(x)+\arccos(x)$. Then $f'(x) = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0$. Thus $f$ is constant. What is $f(0)$ equal to?
|
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|
Study the convergence of $\int_1^\infty \frac{x\ln x}{x^4-1} dx$
Study the convergence of $\int_1^\infty \frac{x\ln x}{x^4-1} dx$
So first we have two potentially problematic points which are $1,\infty$
We split the integral to $$\int_1^2 \frac{x\ln x}{x^4-1} dx + \int_2^\infty \frac{x\ln x}{x^4-1} dx$$
Now first I tried to study $I_1 = \int_1^2 \frac{x\ln x}{x^4-1} dx$.
I evaluated $\lim_{x\to 1^+} \frac{x\ln x}{x^4-1} = \frac{1}{4}$ so basically the function is "nice" at $x=1$ and we can evaluate the integral, but how exactly?
I tried various of methods without luck (integration by parts , substituting $t=\ln x$ , $(x^4 - 1) = (x^2-1)(x^2+1)$ , $(x^4-1) = (x-1)(1+x+x^2+x^3)$)
EDIT
Following the answer I got, I thought about:
Since $\frac{x\ln(x)}{x^4-1}$ is monotonically decreasing to $0$ then we can use the inequality:
$$\int_1^\infty \frac{x\ln(x)}{x^4-1} \le \sum_{n=1}^\infty \frac{n\ln(n)}{n^4-1} \le \sum_{n=1}^\infty \frac{n^2}{n^4-1} \le \sum_{n=1}^\infty \frac{n^2}{n^4} = \sum_{n=1}^\infty \frac{1}{n^2} \lt \infty$$
|
For showing, that $I_{2}$ converges, you might use $\ln(x)\leq x$, which is valid for $x>0$.
|
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|
Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:
*
*$f(x) \geq 0$ on the interval $0\leq x\leq 1$;
*$f(0)=0$ and $f(1)=0$;
*the area bounded by the graph of $f$ and the $x$-axis between $x=0$ and $x=1$ is equal to $1$.
Compute the arc length, $L$, for the function $f$. The goal is to minimize $L$ given the three conditions above.
$\mathbf{\color{red}{\text{Contest results:}}}$
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{User} & {} & {} & \text{Arc length} \\ \hline
\text{1} & \text{robjohn $\blacklozenge$} & {} & {} & 2.78540 \\
\text{2} & \text{Glen O} & {} & {} & 2.78567 \\
\text{3} & \text{mickep} & {} & {} & 2.81108 \\
\text{4} & \text{mstrkrft} & {} & {} & 2.91946 \\
\text{5} & \text{MathNoob} & {} & {} & 3.00000 \\\hline
\text{-} & \text{xanthousphoenix} & {} & {} & 2.78540 \\
\text{-} & \text{Narasimham} & {} & {} & 2.78 \\
\end{array}$$
Original question after contest statement: The contest question was inspired by this paper. Can anyone come up with a different entry than those listed in the table below?
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{Function} & {} & {} & \text{Arc length} \\ \hline
\text{1} & 1.10278[\sin(\pi x)]^{0.153764} & {} & {} & 2.78946 \\
\text{2} & (8/\pi)\sqrt{x-x^2} & {} & {} & 2.91902 \\
\text{3} & 1.716209468\sqrt{x}\,\mathrm{arccos}(x) & {} & {} & 2.91913 \\
\text{4} & (8/\pi)x\,\mathrm{arccos}(x) & {} & {} & 3.15180 \\
\text{5} & (15/4)x\sqrt{1-x} & {} & {} & 3.17617 \\
\text{6} & -4x\ln x & {} & {} & 3.21360 \\
\text{7} & 10x(1-\sqrt{x}) & {} & {} & 3.22108 \\
\text{8} & -6x^2+6x & {} & {} & 3.24903 \\
\text{9} & 9.1440276(2^x-x^2-1) & {} & {} & 3.25382 \\
\text{10} & (-12/5)(x^3+x^2-2x) & {} & {} & 3.27402 \\
\end{array}$$
|
An easy approach is to simply construct an ellipse with its upper half satisfying the above conditions.
An ellipse is defined via $2$ numbers $a$ and $b$ which are each the half of the major and minor axis of the ellipse.
Then all points $(x,y)$ which suffice the following equation are on the ellipse:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
Or to get the upper half of the eclipse as a function:
$$ y = b \, \sqrt{\left(1 - \frac{x^2}{a^2}\right) } $$
The area $A$ of the complete ellipse is given via $A = \pi\,a\,b$ and therefore our first condition translates to:
$$ \frac{1}{2}\,\pi\,a\,b = 1$$
Also, as we want that $f(0) = 0 = f(1)$, we have:
$$ 2\,a = 1 $$
That already gives us
$$ a = \frac{1}{2} \\ b = \frac{4}{\pi} $$
and therefore an ellipse with the correct size. However, this results in an ellipse which intersects the $x$-Axis at $x_1=-0.5$ and $x_2=0.5$. To meet our conditions, we move the ellipse $0.5$ to the right and get:
$$ y = \frac{4}{\pi} \, \sqrt{1 - 4\,(x-0.5)^2 }\tag{$\dagger$} $$
Now we simply let Wolfram Alpha do the computation for the arc length. The result is
$$ 2.919463, $$
and the graph in $(\dagger)$ appears below:
|
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|
Use induction to prove that sum of the first $2n$ terms of the series $1^2-3^2+5^2-7^2+…$ is $-8n^2$. Use induction to prove that sum of the first $2n$ terms of the series $1^2-3^2+5^2-7^2+…$ is $-8n^2$.
I came up with the formula $\displaystyle\sum_{r=1}^{2n} (-1)^{r+1}(2r-1)^2=-8n^2$ but I got stuck proving it by induction. Should I use another formula or is it correct and I should continue trying to prove it by induction?
|
First, show that this is true for $n=1$:
$\sum\limits_{r=1}^{2}(-1)^{r+1}\cdot(2r-1)^2=-8$
Second, assume that this is true for $n$:
$\sum\limits_{r=1}^{2n}(-1)^{r+1}\cdot(2r-1)^2=-8n^2$
Third, prove that this is true for $n+1$:
$\sum\limits_{r=1}^{2(n+1)}(-1)^{r+1}\cdot(2r-1)^2=$
$\color{red}{\sum\limits_{r=1}^{2n}(-1)^{r+1}\cdot(2r-1)^2}+\color{green}{(-1)^{(2n+1)+1}\cdot(2(2n+1)-1)^2}+\color{blue}{(-1)^{(2n+2)+1}\cdot(2(2n+2)-1)^2}=$
$\color{red}{-8n^2}+\color{green}{(-1)^{(2n+1)+1}\cdot(2(2n+1)-1)^2}+\color{blue}{(-1)^{(2n+2)+1}\cdot(2(2n+2)-1)^2}=$
$-8n^2-8(2n+1)=$
$-8(n^2+2n+1)=$
$-8(n+1)^2$
Please note that the assumption is used only in the part marked red.
|
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|
Generating function for Pell numbers Problem: The Pell numbers $p_n$ are defined by the recurrence relation \begin{align*} p_{n+1} = 2p_n + p_{n-1} \end{align*} for $n \geq 1$. The initial conditions are $p_0 = 0$ and $p_1 = 1$.
a) Determine the generating function \begin{align*} P(x) = \sum_{n=0}^{\infty} p_n x_n \end{align*} for the Pell numbers. What is the radius of convergence?
b) Determine (on the basis of the found generating function) an explicit formula for $p_n$.
Solution: Together with the initial conditions we have \begin{align*} P(x) &= 0 + x + \sum_{n=2}^{\infty} p_n x_n \\ &= x + \sum_{n=1}^{\infty} p_{n+1} x^{n+1} \\ &= x + \sum_{n=1}^{\infty} (2p_n + p_{n-1}) x^{n+1} \\ &= x + \sum_{n=1}^{\infty} 2p_n x^{n+1} + \sum_{n=1}^{\infty} p_{n-1} x^{n+1} \\ &= x + 2x \sum_{n=1}^{\infty} p_n x^n + x \sum_{n=1}^{\infty} p_{n-1} x^n. \end{align*}
Now we look at each series separately to see what we've got. The first series on the left expands as $(x + p_2 x^2 + p_3 x^3 + ...)$. This is nothing but the original $P(x)$ (because we can ignore the constant term $0$ right?). So for the first series we've got $2x P(x)$.
The second series on the right expands as $(0 + x^2 + p_2 x^3 + ...)$. We can factorize $x$ out such that we get $P(x)$ again. So everything together we have:
\begin{align*} P(x) = x + 2xP(x) + x^2 P(x), \end{align*} which gives us \begin{align*} P(x) (1-2x-x^2) = x, \end{align*} or \begin{align*} P(x) = \frac{x}{(1-2x-x^2)}. \end{align*}
But then I don't know how to determine the radius of convergence, and how to do b).
Any help would be appreciated.
|
The radius of convergence is the distance to the nearest singularity
For $(b)$ you can advance as (based on your calculations) using partial fraction
$$P(x)= \frac{x}{(1-2x-x^2)}= \frac{A}{x-a} + \frac{B}{x-b} $$
where $a,b$ are the roots of $1-2x-x^2$ and $A$ and $B$ need to be determined. The calculations gives:
$$\frac{-1-\sqrt{2}}{2 \sqrt{2} (x+\sqrt{2}+1)}+\frac{1-\sqrt{2}}{2 \sqrt{2} (x-\sqrt{2}+1)}$$
Then use the geometric series expansion.
Note:
$$ \frac{1}{a-t} = \frac{1}{a}\sum_{n=0}^{\infty} \frac{t^n}{a^n} $$
|
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|
The equation $x^4+y^4=z^2$ has no integer solution The equation $$x^4+y^4=z^2$$ has no integer solution for $(x, y, z), x \cdot y \neq 0 , z >0$.
We suppose that there is a solution $(x, y, z)$.
We consider the set $$M=\{z \in \mathbb{N} | \exists x, y \in \mathbb{Z}: x^4+y^4=z^2, x \cdot y \neq 0 \} \subseteq \mathbb{N}$$
Without loss of generality, we suppose that $(x, y, z)=1$.
One of $x$ and $y$ must be even and the other one odd.
We suppose that $x=2k+1, k \in \mathbb{Z}$ and $y=2l, l \in \mathbb{Z}$,
$\Rightarrow z=2m+1>0, m \in \mathbb{Z}$.
$$x^4+y^4=z^2 \Rightarrow y^4=z^2-x^4=(z-x^2)(z+x^2)$$
It stands that $gcd(z-x^2, z+x^2)=2$.
To show that $gcd(z-x^2, z+x^2)=2$, is the following the only way??
Let $(z-x^2, z+x^2)=d>1$. Then it has a prime divisor, let $p$.
$p \mid d , d \mid z-x^2 \Rightarrow p \mid z-x^2$
$p \mid d , d \mid z+x^2 \Rightarrow p \mid z+x^2$
So $p \mid 2x^2 \Rightarrow p \mid 2 \text{ OR } p \mid x$
and $p \mid 2z \Rightarrow p \mid 2 \text{ OR } p \mid z$
When $p \mid x \text{ AND } p \mid z$, we have that $p \mid y^4 \Rightarrow p \mid y$, so $p \mid (x, y, z)=1$, a contradiction.
So, it should be $p \mid 2 \Rightarrow p=2$.
Is this correct so far?? How can we continue to show that $d=2$ ??
$$$$
We have the following two cases
$$(1): \begin{cases}
z-x^2=2a^4\\
z+x^2=8b^4
\end{cases} \text{ with } (a, b)=1, a \equiv 1 \pmod 2 , a>0$$
$$(2): \begin{cases}
z-x^2=8b^4\\
z+x^2=2a^4
\end{cases} \text{ with } (a, b)=1, a \equiv 1 \pmod 2 , a>0$$
Why do we have these two cases??
|
Since $x,z$ are odd, certainly $z\pm x^2$ are even, hence the gcd is at least $2$. Any common divisor of $z+x^2$ and $z-x^2$ also divides their sum $2z$ and their difference $2x^2$ and their product $y^4$, i.e., $\gcd(z-x^2,z+x^2)\mid \gcd(2x^2,y^4,2z)\mid 2\gcd(x^2,y^4,z)\mid 2\gcd(x^4,y^4,z^4)=2\gcd(x,y,z)^4=2$
|
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|
Prove $\lim_{x\to2} (x^4 - 2x^3 + x + 3) = 5$ using Epsilon Delta Prove $\lim_{x\to2} (x^4 - 2x^3 + x + 3) = 5$ using Epsilon Delta
I am having difficulty finding the $\delta$ value.
Here is what I have done so far:
What I want to show:
$$\forall \epsilon > 0, \exists \delta > 0 \; such \; that \; 0<\mid x -2 \mid < \delta \implies \mid x^4 - 2x^3 + x - 2\mid < \epsilon $$
After a lot of manipulation, I was able to turn the equation to the following:
$$\mid x^4 - 2x^3 + x - 2\mid \; = \; \mid x-2 \mid \mid x-2 \mid \mid x + 1 \mid \mid x + 1 \mid + 3\mid x + 1 \mid \mid x - 2 \mid$$
I already know that $ \mid x -2 \mid < \delta$. I must find an upper bound for $\mid x + 1 \mid$.
$$\mid x + 1 \mid = \mid x + 1 -3 + 3 \mid \; \le \; \mid x - 2 \mid + 3 $$
Choose $\delta = 1$, then
$$\mid x - 2 \mid < 1 \implies \mid x -2 \mid + 3 < 4$$
Therefore $\mid x + 1 \mid < 4$
This means:
$$\mid x^4 - 2x^3 + x - 2\mid < (\delta)(\delta)(4)(4) + 3(\delta)(4) = 16\delta^2 + 12\delta = \epsilon$$
This is where I get stuck. How do I transform $16\delta^2 + 12\delta = \epsilon$ into solving for $\delta$ in terms of $\epsilon$?
I am not sure if I am over thinking the whole question or if the above is even remotely correct. Please let me know any mistakes that I have made.
|
$$\text{Write the polynomial as a sum of powers of $x-2$.}$$
$$x^4-2x^3+x+3=(x-2)^4+6(x-2)^3+12(x-2)^2+9(x-2)+5$$
The most efficient way to do this is to apply Ruffini-Horner repeatedly with the polynomial with the value $x=2$.
Then $$|x^4-2x^3+x+3-5|=|(x-2)^4+6(x-2)^3+12(x-2)^2+9(x-2)|\leq(1+6+12+9)|x-2|$$
when $|x-2|<1$. So, you only need to put $\delta=\min(1,\epsilon/(1+6+12+9)).$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Help with $\lim_{x\rightarrow +\infty} (x^2 - \sqrt{x^4 - x^2 + 1})$ $\lim_{x\rightarrow +\infty} (x^2 - \sqrt{x^4 - x^2 + 1}) = ?$
I don't know how to solve the indetermination there... is it possible to rearrange the expression in brackets in order to use L'Hospital or Taylor Series?
|
Hint: $$x^2-\sqrt{x^4-x^2+1}=\left(x^2-\sqrt{x^4-x^2+1}\right)\cdot\frac{x^2+\sqrt{x^4-x^2+1}}{x^2+\sqrt{x^4-x^2+1}}=\frac{(x^2)^2-\left(\sqrt{x^4-x^2+1}\right)^2}{x^2+\sqrt{x^4-x^2+1}}$$
Once you have that simplified, multiply by $\dfrac{1/x^2}{1/x^2}$ and recognize that $x^2=\sqrt{x^4}$ to distribute the $1/x^2$ 'into' the radical.
|
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|
Simplification of Complex Number. I would appreciate any hints for the following problem:
Given that
$z=\dfrac{1-\cos4\theta+i\sin4\theta}{\sin2\theta+2i\cos^2\theta}$
show that $\vert z\vert=2\sin\theta$ and arg $z=\theta$
Update:
Using John's suggestion I now have
$\dfrac{1-\cos4\theta+i\sin4\theta}{\sin2\theta+2i\cos^2\theta}=\dfrac{1-(\cos\theta+i\sin\theta)^4}{2\sin\theta\cos\theta+2i\cos^2\theta}=\dfrac{1-(\cos\theta+i\sin\theta)^4}{2\cos\theta(\sin\theta+i\cos\theta)}$
|
I'm assuming $0 \le \theta < \pi/2$. Since \begin{align}(1 - \cos 4\theta) + i\sin 4\theta &= 2\sin^2 2\theta + i(2\sin 2\theta \cos 2\theta)\\
& = 2\sin 2\theta (\sin 2\theta + i \cos 2\theta)\\
& = 2\sin 2\theta e^{i(2\theta - \pi/2)}
\end{align}
and
$$\sin 2\theta + 2i\cos^2 \theta = 2\sin \theta \cos \theta + 2i\cos^2 \theta = 2\cos \theta(\sin \theta + i \cos \theta) = 2\cos \theta e^{i(\theta- \pi/2)}$$
we have $$z = \frac{2\sin 2\theta}{2\cos \theta}e^{i[(2\theta - \pi/2) - (\theta - \pi/2)]} = (2\sin \theta) e^{i\theta}$$ Therefore $|z| = 2\sin \theta$ and $\arg z = \theta$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
If $a^2 + p^2 = b^2$ then $2(a+p+1)$ is a perfect square We are given $$ a^2 + p^2 = b^2 $$ where $a,b\in\mathbb{Z}$ and $p$ is prime. We are to show that $$2(a+p+1)$$ is a perfect square. Is there any elegant ways to go about this problem? Struggling to find a proof myself. Thanks in advance.
|
We have $a^2 + p^2 = b^2$ so $p^2 = b^2-a^2 = (b+a)(b-a)$.
Therefore $b-a=1$ and $p^2 = b+a=2a+1$
Therefore $2(a+p+1) = p^2 + 2p + 1 = (p+1)^2$
|
{
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|
Inclusion and exclusion in combinatorics You have 15 identical balls and must divide them into 4 drawers stacked on top of each other with the following limitations:
*
*You have at least 2 balls in each drawer
*There will be no more than 5 balls in the top drawer and there will be no more than 5 balls in the bottom drawer
I'm not sure which strategy is correct;
1. Place 2 balls in each drawer, left with 7 balls. we have 10C3 ways to divide those into 4 drawers. from this we don't subtract because if there are more than 5 balls in the top or bottom drawer we cannot have more than 5 in the other.
Or,
2. subtract from 10C3 2*(6C3); where 6C3 stands for dividing the remaining 3 balls (after placing 4 more in either drawer) into 4 drawers.
Thanks,
|
Generating Function Approach
Inclusion-Exclusion is what the question asks for, and there are some good answers using inclusion-exclusion, but here is another approach using generating functions that gives the same answer as inclusion-exclusion.
By taking the products of the generating functions for the number of ways to have $n$ balls in each drawer,
$$
\begin{align}
{\small\text{first and fourth drawers:}}\quad&x^2\frac{1-x^4}{1-x}=x^2+x^3+x^4+x^5\\
{\small\text{second and third drawers:}}\quad&x^2\frac1{1-x}=x^2+x^3+x^4+\dots
\end{align}
$$
we get the generating function of the number of ways to have $n$ balls total in all drawers
$$
\overbrace{x^2\frac{1-x^4}{1-x}}^{\text{first drawer}}\quad
\overbrace{x^2\frac{1\vphantom{x^4}}{1-x}}^{\text{second drawer}}\quad
\overbrace{x^2\frac{1\vphantom{x^4}}{1-x}}^{\text{third drawer}}\quad
\overbrace{x^2\frac{1-x^4}{1-x}}^{\text{fourth drawer}}\quad
=\quad\overbrace{x^8\frac{(1-x^4)^2}{(1-x)^4}}^{\text{all drawers}}
$$
The series for all drawers is
$$
\begin{align}
x^8\left(1-x^4\right)^2\overbrace{\sum_{k=0}^\infty\binom{-4}{k}(-x)^k}^{(1-x)^{-4}}
&=\left(x^8-2x^{12}+x^{16}\right)\sum_{k=0}^\infty\binom{k+3}{k}x^k\\
&=\sum_{k=0}^\infty\left[\binom{k-5}{k-8}-2\binom{k-9}{k-12}+\binom{k-13}{k-16}\right]x^k
\end{align}
$$
The coefficient for $x^{15}$ is
$$
\binom{10}{7}-2\binom{6}{3}+\overbrace{\binom{2}{-1}}^0
$$
|
{
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"url": "https://math.stackexchange.com/questions/1127521",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluating $(1+\cos\frac{\pi}{8})(1+\cos\frac{3\pi}{8}) (1+\cos\frac{5\pi}{8})(1+\cos\frac{7\pi}{8})$
How to find the value of $$\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right) \left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)$$
I used: $ 1+ \cos \theta=2\sin^2(\theta/2)$ to get $$2^4 \cdot\prod_{n=0}^3 \sin\left(\frac{(2n+1)\pi}{16}\right)$$
After that I tried using the formula:
$$\sin\left(1 \frac{\pi}n\right)\sin\left(2 \frac{\pi}n\right) \cdots \sin\left((n-1) \frac{\pi}n\right)= \frac n {2^{n-1}}$$
to find the value of the expression in the parenthesis in the first equation, but currently I'm stuck. I don't think this method will lead me to the correct answer.
Any help would be appreciated!
|
If $\cos4x=0,4x=(2n+1)\dfrac\pi2$ where $n$ is any integer
$x=(2n+1)\dfrac\pi8$ where $n\equiv0,1,2,3\pmod4$
Now $\cos4x=2\cos^22x-1=\cdots=8\cos^4x-8\cos^2x+1$
So, the roots of $8c^4-8c^2+1=0$ are $\cos(2n+1)\dfrac\pi8$ where $n\equiv0,1,2,3\pmod4$
So, the equation whose roots are $1+\cos(2n+1)\dfrac\pi8=y\iff\cos(2n+1)\dfrac\pi8=y-1$ is
$8(y-1)^4-8(y-1)^2+1=0\iff8y^4+\cdots+1=0$
Using Vieta's formula, $$\prod_{n=0}^3\left[1+\cos(2n+1)\dfrac\pi8\right]=\dfrac18$$
|
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"question_score": "6",
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|
Legendre polynomial to show identity, can't spot mistake Using Legendre polynomial generating function
\begin{equation}
\sum_{n=0}^\infty P_n (x) t^n = \frac{1}{\sqrt{(1-2xt+t^2)}}
\end{equation}
Or $$ P_n(x)=\frac{1}{2^n n!} \frac{d^n}{dx^n} [(x^2-1)^n] $$
Show$$ P_{2n}(0)=\frac{(-1)^n (2n)!}{(4)^n (n!)^2} $$
And $$ P_{2n+1}(0)=0$$
I expressed $$(x^2 -1)^{2n}= \sum_{k=0}^{2n} {2n \choose k}x^{4n-2k}(-1)^k$$
And using second formula given, the only term that remains after differentiating 2n times and substituting x=0 is where $$4n-2k=2n$$ so $$2n=2k$$ k=n i.e $$(-1)^n \frac{(2n)!}{(n!)^2} $$ but multiplying this by$$ \frac{1}{2^{2n} (2n)!} $$ doesn't give desired solution. Where have I gone wrong?
|
\begin{equation}
\begin{split}
P_n(x) &= \frac{1}{2^nn!}[(x+1)^n\cdot(x-1)^n]^{(n)}
\\ &= \frac{1}{2^nn!}\sum_{k=0}^{n}\binom{n}{k}[(x+1)^n]^{(k)} \cdot [(x-1)^n]^{(n-k)}
\\ &= \frac{1}{2^nn!}\sum_{k=0}^{n}\binom{n}{k}\frac{n!}{(n-k)!}(x+1)^{n-k} \cdot \frac{n!}{k!}(x-1)^k
\\ &= \frac{1}{2^n}\sum_{k=0}^{n}\binom{n}{k}^2(x+1)^{n-k}(x-1)^k
\end{split}
\end{equation}
so
\begin{equation}
\begin{split}
P_{2n}(0) &=\frac{1}{2^{2n}}\sum_{k=0}^{2n}\binom{2n}{k}^2(-1)^k = \frac{1}{2^{2n}}\binom{2n}{n}(-1)^n = \frac{(-1)^n(2n)!}{2^{2n}n!^2}
\end{split}
\end{equation}
and
\begin{equation}
\begin{split}
P_{2n+1}(0) &=\frac{1}{2^{2n+1}}\sum_{k=0}^{2n+1}\binom{2n+1}{k}^2(-1)^k = \frac{1}{2^{2n}}\cdot 0 = 0
\end{split}
\end{equation}
See legendre polynomial problem (pls help!) for the derivation of the value of $\sum_{k=0}^m\binom{m}{k}^2(-1)^k$.
|
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|
limit of $ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $ Hello I am trying to find the limit of
$ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $
I've tried applying L'H rule but it ends up getting really messy.
The answer is $ \frac {1}{e} $ so I assume it must simplify into something which I can apply the standard limit laws.
|
let $y = \dfrac{x(x+1)^{x+1}}{(x+2)^{x+2}},$ then
$\begin{align} \ln y &= \ln x + (x+1)\ln(x+1)-(x+2)\ln(x+2)\\
&=\ln x +(x+1)[\ln x + \ln(1 + 1/x)] -(x+2)[\ln x+ \ln(1 + 2/x)]\\
&=\left(1+x+1-x-2 \right)\ln x +(x+1)\left(1/x + \cdots\right)-(x+2)\left(2/x +\cdots\right)\\
&=-1 + \cdots
\end{align}$
so $$\lim_{x \to \infty} \ln y= -1 \text{ and } \lim_{x \to \infty} y= \frac{1}{e}.$$
|
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|
The set $\{1,2,3,\ldots,n\}$, where $n \geq 5$, can be divided into two subsets so that the sum of the first is equal to the product of the second A peer of mine showed me earlier today this problem, taken from a 7th grade math contest :
Let $A=\{1,2,3,\ldots,n\}$; (where $n \geq 5$) prove that $A$ can be divided into two disjoint subsets such that the sum of the elements in the first subset is equal to the product of the elements in the second subset.
This has been puzzling me for 15 minutes already, but I'm sure there's a simple, straight-forward way to do it since it's a 7th grade problem, albeit I can't see it.
Can anyone shed some wisdom here ?
|
The sum of all the numbers is
$$
\sum_{k=1}^nk=\frac{n(n+1)}2\tag{1}
$$
We will find $a$ and $b$ so that
$$
\overbrace{\left(\sum_{k=1}^nk\right)-a-b-1}^{\text{sum of all but $a$, $b$, and $1$}}=\overbrace{\vphantom{\left(\sum_1^n\right)}1ab}^{\text{product}}\tag{2}
$$
That is, we need to find $a$ and $b$ so that
$$
\begin{align}
\frac{n(n+1)}2
&=\overbrace{ab}^{\text{product}}+\overbrace{a+b+1}^{\text{sum}}\\
&=(a+1)(b+1)\tag{3}
\end{align}
$$
and $(3)$ can be solved using
$$
(a,b)=\left\{\begin{array}{}
\left(\frac{n-1}2,n-1\right)&\text{if $n$ is odd}\\
\left(n,\frac{n-2}2\right)&\text{if $n$ is even}\\
\end{array}\right.\tag{4}
$$
Thus, the product of $1$, $a$, and $b$ is the sum of the rest of the numbers from $1$ to $n$.
If $n\ge5$, then $a,b\ge2$ so that they don't interfere with $1$.
Examples
With $n=5$, $a=\frac{5-1}2=2$ and $b=5-1=4$ so $1\cdot2\cdot4=3+5$
With $n=6$, $a=6$ and $b=\frac{6-2}2=2$ so $1\cdot2\cdot6=3+4+5$
With $n=7$, $a=\frac{7-1}2=3$ and $b=7-1=6$ so $1\cdot3\cdot6=2+4+5+7$
With $n=8$, $a=8$ and $b=\frac{8-2}2=3$ so $1\cdot3\cdot8=2+4+5+6+7$
|
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|
Problem with infinite series sum I'm trying to calculate the infinite sum of the series
$$\sum_{i=1}^{n}\frac{(i-1)^2}{n^3}. $$
Computing it with WolframAlpha results in $$\frac{2n^2-3n+1}{6n^2}.$$
I'm trying to solve it by considering it as an arithmetic series and using the formula for the nth $$\sum_{i=1}^{n}a_i=\frac{n(a_1+a_n)}{2},$$ but I can't reach the correct result.
Any clue, bibliography, anything?
EDITED
Following the instructions of avz2611:
$$\frac{1}{n^3}\sum_{i=1}^{n}(i-1)^2=\frac{1}{n^3}\cdot\frac{n(n-1)^2}{2}=\frac{n^2-2n+1}{2n^2}$$
with $a_1=0$ and $a_n=(n-1)^2$.
|
$$1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}\\0^2+1^2+2^2+...+(n-1)^2=\frac{(n-1)(n-1+1)(2(n-1)+1)}{6}=\\\frac{n(n-1)(2n-1)}{6}$$so $$ \sum_{i=1}^{n}\frac{(i-1)^2}{n^3}=\frac{1}{n^3}\sum_{i=1}^{n}(i-1)^2=\\\frac{\frac{n(n-1)(2n-1)}{6}}{n^3}=\\\frac{(n-1)(2n-1)}{6n^2}$$
|
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|
Finding $\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
Find the limit: $\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
My attempt:
$\begin {align}\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}} &=
\lim _{n\to \infty}\frac{\sqrt{n+n^4\sqrt{1+\frac{1}{n^2}}}}{n^2\sqrt{3+\frac{1}{n^2}}}\\ &=
\lim _{n\to \infty}\frac{n^2\sqrt{1+n^3\sqrt{1+\frac{1}{n^2}}}}{n^2\sqrt{3+\frac{1}{n^2}}}\\&=
\lim_{n\to \infty}\frac{\sqrt{1+n^3\sqrt{1+\frac{1}{n^2}}}}{\sqrt{3+\frac{1}{n^2}}}\end{align}$
Which looks like is equal to $\infty$ because of the $n^3\cdot 1$ but it's wrong.
What am I doing wrong here?
|
Hint:
$$
\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}=\lim _{n\to \infty}\sqrt{\dfrac{n+\sqrt{n^2+1}}{3n+1}}.
$$
So you only need to determine the limit of the term inside the square root.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\int \sqrt{\frac{x}{x+1}}dx$ In order to integrate
$$\int \sqrt{\frac{x}{x+1}}dx$$
I did:
$$x = \tan^2\theta $$
$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta = \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int \tan^3\theta d\theta = \int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta$$
$$p = \cos\theta \implies dp = -\sin\theta d\theta$$
$$\int\frac{\sin^3 \theta}{\cos^3 \theta}d\theta = -\int\frac{(1-p^2)(-\sin\theta)}{p^3 }d\theta = -\int \frac{1-p^2}{p^3}dp = -\int \frac{1}{p^3}dp +\int \frac{1}{p}dp = -\frac{p^{-2}}{-2}+\ln|p| = -\frac{(\cos\theta)^{-2}}{-2}+\ln|\cos\theta|$$
$$x = \tan^2\theta \implies \tan\theta= \sqrt{x}\implies \theta = \arctan\sqrt{x}$$
$$= -\frac{(\cos\arctan\sqrt{x})^{-2}}{-2}+\ln|\cos\arctan\sqrt{x}|$$
But the result seems a little bit different than wolfram alpha. I Know there may be easier ways to solve this integral but my question is about this method I choose, specifically.
Is the answer correct? Also, if it is, is there a way to reduce $\cos\arctan$ to something simpler?
|
$\displaystyle\int\sqrt{\frac{x}{x+1}}dx$
$\displaystyle x=y^{2}-1\Rightarrow dx=2ydy$
$\displaystyle\int\sqrt{\frac{x}{x+1}}dx=\int\frac{\sqrt{y^{2}-1}}{y}2ydy=2\int\sqrt{y^{2}-1}dy$
$\displaystyle y=ch(z)\Rightarrow dy=sh(z)dz$
$\displaystyle sh(z)=\sqrt{ch^{2}(z)-1}=\sqrt{y^{2}-1} $
$\displaystyle\int\sqrt{\frac{x}{x+1}}dx=2\int sh^{2}(z)dz=2\int(\frac{1+ch(2z)}{2})dz=z+\frac{sh(2z)}{2}+c$
$\displaystyle\int\sqrt{\frac{x}{x+1}}dx=\ ch^{-1}{\sqrt{x+1}}+\frac{1}{2}sh\left[ 2ch^{-1}\sqrt{x+1} \right]+c$
$\displaystyle ch^{-1}(x)=ln(x+\sqrt{x^{2}-1})$
$\displaystyle \int\sqrt{\frac{x}{x+1}}dx=ln(\sqrt{x+1}+\sqrt{x})+\frac{1}{2}sh\left[ 2ln(\sqrt{x+1}+\sqrt{x}) \right]+c$
|
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"timestamp": "2023-03-29T00:00:00",
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|
From $\frac{1-\cos x}{\sin x}$ to $\tan\frac{x}{2}$ How can I write $\frac{1-\cos x}{\sin x}$ as $\tan\frac{x}{2}$? I wrote $\sin x$ as $2\sin\frac{x}{2} \cos\frac{x}{2}$ also used the double angle identity for $\cos$ but wasn't able to make much progress
|
you can do $\dfrac{1-\cos x}{\sin x} = \dfrac{\cos 0 - \cos x}{\sin 0 + \sin x}=\dfrac{2\cos x/2 \cos x/2}{2 \sin x/2 \cos x/2} = \tan(x/2).$
we used $\cos a -\cos b = 2\cos (a-b)/2 \cos(a+b)/2$ and $\sin a + \sin b = 2 \sin(a+b)/2 \cos (a-b)/2$ with $a = 0$ and $b = x.$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How can I prove that $2/9$x$ and $y$ are real numbers.
Given that $1<x^2-xy+y^2<2$, how can I show that $\frac 29<x^4+y^4<8$ ?
Then can I use that to prove that for any natural number $n>3$
$$x^{2^n}+y^{2^n}>\frac 2{3^{2^n}} \text{?}$$
|
$x^2 - xy + y^2 = 1$ is an ellipse. to find out the minor and major axis, we change variables $$x = \xi \cos t - \eta \sin t, y = \xi \sin t + \eta \cos t$$ where $t$ will be determined later.
the ellipse in $\xi, \eta$ coordinates is $$\xi^2(\cos^2 t -\sin t \cos t+\sin^2 t) + \xi \eta(-2\cos t \sin t+\sin^2 t - \cos^2 t+2\sin t \cos t) + \eta^2(\sin ^2 t +\sin t \cos t +\cos ^2 t) =1 $$
set $t = \frac{\pi}{4},$ you get $$\frac12\xi ^2+\frac32\eta^2= 1. $$
so the major and minor axis of the ellipse $x^2 - xy + y^2 = 1$ are
$\sqrt 2$ and $ \sqrt { \frac23}.$
therefore by symmetry the minimum of$x^4 + y^4$ on the hyperbola $x^2 -xy + y^2 = 1$ is achieved on the minor axis $x = -\sqrt\frac13, y = \sqrt \frac 13$ and the minimum value is $\dfrac{2}{9}.$
therefore by symmetry the minimum of$x^4 + y^4$ on the hyperbola $x^2 -xy + y^2 = 2$ is achieved on the major axis $x = \sqrt 2, y = \sqrt 2$ and the maximum value is $8.$
we have established $$ \dfrac{2}{9} \le x^4 + y^4 \le 8 \text { on } 1 \le x^2 - xy + y^2 \le 2.$$
|
{
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|
Proving $\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right) = \frac35$ $$\lim _{x\to \infty }\left(\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}\right)$$
Can someone help me to solve it?
result of online calculator: 3/5
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Rewrite $$A=\frac{\sqrt{x+1}-\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-3}}=\frac{\sqrt{x}\sqrt{1+1/x}-\sqrt{x}\sqrt{1-2/x}}{\sqrt{x}\sqrt{1+2/x}-\sqrt{x}\sqrt{1-3/x}}=\frac{\sqrt{1+1/x}-\sqrt{1-2/x}}{\sqrt{1+2/x}-\sqrt{1-3/x}}$$ Replace $\frac 1x$ by $y$; so $$A=\frac{\sqrt{1+y}-\sqrt{1-2y}}{\sqrt{1+2y}-\sqrt{1-3y}}$$ Now, use the fact that, when $z$ is small compared to $1$,$\sqrt{1+z}\approx 1+\frac z2$. Replace $z$ by the appropriate value for each radical.
I am sure that you can take from here.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding $\lim_{(x,y)\rightarrow (0,0)} \frac{\tan(x^2+y^2)}{\arctan(\frac{1}{x^2+y^2})} $ I'm having trouble understanding how the $\displaystyle\lim_{(x,y)\rightarrow (0,0)} \frac{\tan(x^2+y^2)}{\arctan(\frac{1}{x^2+y^2})}$. I used the product law to set it up as $\displaystyle\frac{\lim_{(x,y)\rightarrow (0,0)} \tan(x^2+y^2)}{\lim_{(x,y)\rightarrow (0,0)}\arctan(\frac{1}{x^2+y^2})}$ then found $\lim_{(x,y)\rightarrow (0,0)} \tan(x^2+y^2)=0$ but got caught up on $\lim_{(x,y)\rightarrow (0,0)}\arctan(\frac{1}{x^2+y^2})$. I looked up the solution and $\lim_{(x,y)\rightarrow (0,0)}\arctan(\frac{1}{x^2+y^2})= \pi/2$. Can someone explain how to solve the second limit?
|
$$\lim\limits_{(x,y)\to (0,0)} \frac{\tan\left(x^2+y^2\right)}{\arctan\left(\frac{1}{x^2+y^2}\right)}$$
Using polar coordinates, we have
$$\lim\limits_{r\to 0^+} \frac{\tan\left(r^2\cos^2\phi+r^2\sin^2\phi\right)}{\arctan\left(\frac{1}{r^2\cos^2\phi+r^2\sin^2\phi}\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{\tan\left(r^2\left(\cos^2\phi+\sin^2\phi\right)\right)}{\arctan\left(\frac{1}{r^2\left(\cos^2\phi+\sin^2\phi\right)}\right)}$$
$$=\lim\limits_{r\to 0^+} \frac{\tan\left(r^2\right)}{\arctan\left(\frac{1}{r^2}\right)}= \frac{0}{\left(\frac{\pi}{2}\right)}=0$$
|
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|
Induction exercise check-up Prove by induction on $n$ that $13$ divides $2^{4n+2} + 3^{n+2}$ for all natural $n$.
For base case it is divisble by 13, and
$2^{4n+6} + 3^{n+3}$ must be divisble too.
$16 * 2^{4n+2}+ 3* 3^{n+2}$
If we denote $2^{4n+2} + 3^{n+2}$ as some $13y$,
we have $3*13y + 13*2^{4n+2}$
Didn't I make a mistake in assuming that $2^{4n+2} + 3^{n+2}$ is divisble by 13? In induction we have to prove that if something is divisible/whatever by something then n+1 is divisible too which I did
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First, show that this is true for $n=1$:
$2^{4+2}+3^{1+2}=13\cdot7$
Second, assume that this is true for $n$:
$2^{4n+2}+3^{n+2}=13k$
Third, prove that this is true for $n+1$:
$2^{4(n+1)+2}+3^{(n+1)+2}=$
$2^4\cdot2^{4n+2}+3^1\cdot3^{n+2}=$
$16\cdot2^{4n+2}+3\cdot3^{n+2}=$
$(13+3)\cdot2^{4n+2}+3\cdot3^{n+2}=$
$13\cdot2^{4n+2}+3\cdot2^{4n+2}+3\cdot3^{n+2}=$
$13\cdot2^{4n+2}+3\cdot(\color{red}{2^{4n+2}+3^{n+2}})=$
$13\cdot2^{4n+2}+3\cdot\color{red}{13k}=$
$13\cdot2^{4n+2}+13\cdot3k=$
$13\cdot(2^{4n+2}+3k)$
Please note that the assumption is used only in the part marked red.
|
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|
How to prove $\sum_{k=0}^{\infty}k^2x^{k} = \frac{x(1+x)}{(1-x)^3}\text{, }|x| < 1$? How do I prove that the summation $$\sum_{k=0}^{\infty}k^2x^{k} = \dfrac{x(1+x)}{(1-x)^3}\text{, }|x| < 1\text{?}$$
|
$$\sum_{k=0}^{\infty}x^k = \frac{1}{1-x}$$
Taking derivatives both sides with respect to $x$, the first and the second derivatives yield respectively:
$$\sum_{k=1}^{\infty}kx^{k-1} = \frac{1}{(1 - x)^2}$$
$$\sum_{k=2}^{\infty}k(k-1)x^{k-2} = \frac{2}{(1-x)^3}$$
Now:
$$\sum_{k=0}^{\infty}k^2x^k = \sum_{k=1}^{\infty}k^2x^k = \sum_{k=1}^{\infty}(k^2 - k + k)x^k = \sum_{k=1}^{\infty}k(k-1)x^k + \sum_{k=1}^{\infty}kx^k = \sum_{k=2}^{\infty}k(k-1)x^k + \sum_{k=1}^{\infty}kx^k = x^2\sum_{k=2}^{\infty}k(k-1)x^{k-2} + x\sum_{k=1}^{\infty}kx^{k-1} = \frac{2x^2}{(1-x)^3} + \frac{x}{(1 - x)^2} = \frac{2x^2 + x(1 - x)}{(1-x)^3} = \frac{x^2 + x}{(1-x)^3} = \frac{x(1 + x)}{(1-x)^3}$$
|
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|
Prove that $\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$
Prove that the following inequality
$$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$$
holds for arbitrary real numbers $a$, $b$ and $c.$
Someone says, "It's very easy problem. It can also be proved by AM–GM inequality." But I can't reason out the answer to this question.
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By Minkowski's inequality,
\begin{align}&\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - a)^2} \\
&= \|(a,1-b)\|_2 + \|(b,1-c)\|_2 + \|(c,1-a)\|_2\\
&= \frac{\|(a,1-b)\|_2 + \|(1 - c,b)\|_2}{2} + \frac{\|(b,1-c)\|_2 + \|(1-a,c)\|_2}{2}\\
& + \frac{\|(c,1-a)\|_2 + \|(1-b, a)\|_2}{2}\\
&\ge \frac{\|(a + 1 - c, 1)\|_2}{2} + \frac{\|(b + 1 - a, 1)\|_2}{2} + \frac{\|(c + 1 - a, 1)\|_2}{2}\\
&\ge \frac{\|(a + 1 - c) + (b + 1 - a) + (c + 1 - a),3)\|_2}{2}\\
&= \frac{\|(3,3)\|_2}{2}\\
&= \frac{3\sqrt{2}}{2}.
\end{align}
|
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|
Convolutions and the Gaussian distribution Suppose $X_1$ and $X_2$ are independent random variables each with the standard Gaussian distribution. Compute, using convolutions, the density of the distribution of $X_1 + X_2$ and show $X_1 + X_2 = \sqrt{2}X$ where X has standard Gaussian distribution.
I wrote down the density of the distribution as $f_{x_1+x_2}(z)= \int_{-\infty}^{\infty}f_{x_1}(x)f_{x_2}(z-x)dx$, and I don't know how to proceed to the next part of the question.
Any help is appreciated, thanks in advance.
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For $X, Y\sim N(0,1)$, we have
$${f_{X + Y}}\left( x \right) = \int_{ - \infty }^\infty {{f_X}\left( {x - y} \right){f_Y}\left( y \right)dy} = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{\operatorname{e} ^{ - \frac{1}{2}{{\left( {x - y} \right)}^2}}}{e^{ - \frac{1}
{2}{y^2}}}dy} = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{\operatorname{e} ^{ - \frac{1}{2}\left[ {{{\left( {x - y} \right)}^2} + {y^2}} \right]}}dy}.$$
We have
$${\left( {x - y} \right)^2} + {y^2} = {x^2} - 2xy + 2{y^2} = {\left( {\sqrt 2 y - \frac{1}{{\sqrt 2 }}x} \right)^2} + \frac{{{x^2}}}{2}.$$
So
$$\begin{gathered}
{f_{X + Y}}\left( x \right) = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{\operatorname{e} ^{ - \frac{1}{2}\left[ {{{\left( {\sqrt 2 y - \frac{1}
{{\sqrt 2 }}x} \right)}^2} + \frac{{{x^2}}}{2}} \right]}}dy} = \frac{1}
{{2\pi }}{e^{ - \frac{{{x^2}}}{4}}}\int_{ - \infty }^\infty {{\operatorname{e} ^{ - \frac{1}{2}{{\left( {\sqrt 2 y - \frac{1}{{\sqrt 2 }}x} \right)}^2}}}dy} \hfill \\
= \frac{1}{{2\pi }}{e^{ - \frac{{{x^2}}}{4}}}\int_{ - \infty }^\infty {{\operatorname{e} ^{ - {{\left( {\frac{{\sqrt 2 y - \frac{1}{{\sqrt 2 }}x}}
{{\sqrt 2 }}} \right)}^2}}}dy} = \frac{1}{{2\pi }}{e^{ - \frac{{{x^2}}}
{4}}}\int_{ - \infty }^\infty {{\operatorname{e} ^{ - {{\left( {y - \frac{1}
{2}x} \right)}^2}}}d\left( {y - \frac{1}{2}x} \right)} \hfill \\
= \frac{1}{{2\pi }}{e^{ - \frac{{{x^2}}}{4}}}\sqrt \pi = \frac{1}{{2\sqrt \pi }}{e^{ - \frac{{{x^2}}}{4}}}. \hfill \\
\end{gathered}$$
Now, for $Z\sim N(0,1)$, wet have
$${F_{\sqrt 2 Z}}\left( x \right) = P\left( {\sqrt 2 Z < x} \right) = P\left( {Z < \frac{x}
{{\sqrt 2 }}} \right) = {F_Z}\left( {\frac{x}
{{\sqrt 2 }}} \right)$$
This leads to
$$\begin{gathered}
{f_{\sqrt 2 Z}}\left( x \right) = \frac{d}
{{dx}}\left[ {{F_{\sqrt 2 Z}}\left( x \right)} \right] = \frac{d}
{{dx}}\left[ {{F_Z}\left( {\frac{x}
{{\sqrt 2 }}} \right)} \right] = \frac{1}
{{\sqrt 2 }}{{F'}_Z}\left( {\frac{x}
{{\sqrt 2 }}} \right) = \frac{1}
{{\sqrt 2 }}{f_Z}\left( {\frac{x}
{{\sqrt 2 }}} \right) \hfill \\
= \frac{1}
{{\sqrt 2 }}.\frac{1}
{{\sqrt {2\pi } }}{e^{ - \frac{1}
{2}{{\left( {\frac{x}
{{\sqrt 2 }}} \right)}^2}}} = \frac{1}
{{2\sqrt \pi }}{e^{ - \frac{{{x^2}}}
{4}}} \hfill \\
\end{gathered} $$
We deduce that ${f_{\sqrt 2 Z}}\left( x \right) = {f_{X + Y}}\left( x \right)$.
|
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|
Use proof by induction to prove $ \frac{1}{n!}<\frac{1}{2^n-1} $ for all $n\geq 4$ Use proof by induction to prove that that $ \frac{1}{n!}<\frac{1}{2^n-1} $ for all $n\geq 4$, .\Base case: $$\frac{1}{4}=\frac{1}{24}\leq \frac{1}{2^4-1}$$
Inductive hypothesis: Assume there exists $k\in \mathbb{N}$ s.t.
$$ \frac{1}{k!}\leq\frac{1}{2^k-1} $$
Inductive step: Show that:$$ \frac{1}{(k+1)!}\leq\frac{1}{2^{k+1}-1} $$
Now, $$\frac{1}{(k+1)!}=\frac{1}{k!}\cdot\frac{1}{k+1}$$
Using the hypothesis $$\frac{1}{(k+1)!}\leq\frac{1}{2^k-1}\cdot \frac{1}{k+1}$$
Because $n\geq4, \frac{1}{k+1}<\frac{1}{2}$
$$\frac{1}{(k+1)!}\leq\frac{1}{2^k-1}\cdot \frac{1}{2}=\frac{1}{2^{k+1}-2}\leq\frac{1}{2^{k+1}-1}$$
Hence by mathematical induction we have proved that $ \frac{1}{n!}<\frac{1}{2^n-1} $ for all $n\geq 4$
Firstly I need to know if the proof is correct, secondly it has to be as concise as possible hence I would like to know if there are any lines I can change/delete
And lastly can anyone explain to me why every sentence starts with "\" It looks perfectly fine in www.sharelatex.com ;(
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Rewrite as$$n!\ge2^n.$$
Then
$$4!\ge2^4$$
and
$$n!\ge2^n\land n+1\ge2\implies(n+1)!\ge2^{n+1}.$$
|
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|
Solving quadratic diophantine equations in two variables I've looked at the recommended questions, but none of them seem to match my question.
Consider the equation $2015 = \frac{(x+y)(x+y-1)}{2} - y + 1$.
This can trivially be simplified to $4030 = x^2 + 2xy - x + y^2 - 3y +2$.
According to Wolfram Alpha, the integer solutions of this equation can be represented as:
$x = -\frac{n^2}{2} + \frac{5n}{2} + 2012,\:y = \frac{n^2}{2} - \frac{3n}{2} - 2013,\:n \in \mathbb{Z}$.
Can someone explain how this set of solutions is derived?
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The equation is
$$x^2 + 2xy + y^2 - x -3y - 4028 = 0.$$
This can be modified to
$$(x+y)^2 - (x+y) - (2y+4028) = 0.$$
Let's denote $u = x+y$ so we have
$$u^2 - u = 2(y+2014)$$
Solving for $y$, we have
$$y = \frac{u^2-u}{2}-2014.$$
Then, remembering that $u=x+y$, we can solve for $x$:
$$x = u-y = u-\frac{u^2-u}{2}+2014 = \frac{-u^2+3u}{2}+2014.$$
Then we set $n=u-1$ (just a shift (to make the solution look the same)) and get the solution
$$x = \frac{-n^2+5n}{2}+2012$$
$$y = \frac{n^2-3n}{2}-2013$$
|
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|
Adding distances/weights to absorbing markov chain in presence of an absorbing state, I want to calculate mean/expected 'distance' from any state to that absorbing state. What I mean by distance is that I want to give different lengths from one particular state to another.
For example, below i wrote a simple coin toss, 2 levels deep. One player is in Start S, with a coin toss he goes A or B. If he reaches A, with %100 probability he reaches End after. But if he reaches B, he either reaches End or he turns back to Start with equal probability.
\begin{align*}
\begin{pmatrix}
& S & A & B & E \\
S & 0 & \frac{1}{2} & \frac{1}{2} & 0\\
A & 0 & 0 & 0 & 1 \\
B & \frac{1}{2} & 0 & 0 & \frac{1}{2} \\
E & 0 & 0 & 0 & 1
\end{pmatrix}
\end{align*}
Later I found total transition matrix between transient states
\begin{align*}
(I-Q)^{-1} = \left(\begin{array}{rr}
\frac{4}{3} & \frac{2}{3} & \frac{2}{3} \\
0 & 1 & 0 \\
\frac{2}{3} & \frac{1}{3} & \frac{4}{3}
\end{array}\right)
\end{align*}
So if I sum first row, I find 8/3 , that is expected number of steps from Start to End
Now where I am stuck is, probabilities stay same but if going from one state to another differs, I dont know how to manipulate these matrices to take distances.
For example, distance from A-> E may be 1 km, but B -> E may be 5 kms. S-> A may be 20 kms, rest can be 1 km. How can I manipulate these matrices to get this information. Thanks in advance
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One way to approach this is to create another Markov chain where the states are the transitions that have positive probability in the original Markov chain, plus one state for the initial state. We only need to include transitions from transient states since that's all we're interested in. So we exclude the transition $E$ to $E$. In your example we would have states: $S,SA,SB,AE,BS,BE$. Our transition matrix, $F$, for this new Markov chain is:
\begin{align*} F =
\begin{pmatrix}
& S & SA & SB & AE & BS & BE \\
S & 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\
SA & 0 & 0 & 0 & 1 & 0 & 0 \\
SB & 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \\
AE & 0 & 0 & 0 & 0 & 0 & 0 \\
BS & 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\
BE & 0 & 0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}
\end{align*}
Entry $F_{i,j}$ in matrix $(I-F)^{-1}$ represents the expected total number of visits to state $j$ beginning in state $i$ (states $i,j$ being states in the new Markov chain), using the same reasoning as for the $Q$ matrix in the original Markov chain. Here, we have,
\begin{align*} (I-F)^{-1} =
\begin{pmatrix}
& S & SA & SB & AE & BS & BE \\
S & 1 & 2/3 & 2/3 & 2/3 & 1/3 & 1/3 \\
SA & 0 & 1 & 0 & 1 & 0 & 0 \\
SB & 0 & 1/3 & 4/3 & 1/3 & 2/3 & 2/3 \\
AE & 0 & 0 & 0 & 1 & 0 & 0 \\
BS & 0 & 2/3 & 2/3 & 2/3 & 4/3 & 1/3 \\
BE & 0 & 0 & 0 & 0 & 0 & 1 \\
\end{pmatrix}
\end{align*}
For example, $F_{0,5} = 1/3$ is the expected number of times we move in one step from state $B$ to state $E$ (in the original Markov chain) given that we start from state $S$.
We also need a distance vector for the distances between states: $S\; (=0),SA,SB,AE,BS,BE$:
\begin{align*} D =
\begin{pmatrix}
0 \\
20 \\
1 \\
1 \\
1 \\
5 \\
\end{pmatrix}
\end{align*}
The expected total distances covered is then given by,
\begin{align*} (I-F)^{-1} \times D =
\begin{pmatrix}
50/3 \\
21 \\
37/3 \\
1 \\
53/3 \\
5 \\
\end{pmatrix}
\end{align*}
Since we start at state $S$, the value required is $\dfrac{50}{3}$.
|
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|
System of Recurrence Relations Solve the following System of Recurrence Relation:
$$a_n = 2a_{n-1} - b_{n-1} + 2, a_0 = 0$$
$$b_n = -a_{n-1} + 2b_{n-1} - 1, b_0 = 1$$
Workings:
$b_n - 2b_{n-1} = -a_{n-1} - 1$
$a_n = 2a_{n-1} - b_{n-1} + 2$
$a_{n+1} = 2a_n - b_n + 2$
$-2a_n = -4a_{n-1} + 2b_{n-1} - 4$
$a_{n+1} + 2a_n = (2a_n - b_n + 2) + (4a_{n-1} - 2b_{n-1} + 4) $
$a_{n+1} + 2a_n = 2a_n + 4a_{n+1} - b_n -2b_{n-1} + 6$
$a_{n+1} + 2a_n = 2a_n + 4a_{n-1} - (-a_{n-1} - 1) + 6$
$a_{n+1} + 2a_n = 2a_n + 4a_{n-1} + a_{n-1} + 7$
$a_{n+1} = 5a_{n-1} + 7 (*)$
Now I'm not sure what to next. Can I shift $(*)$ down to $a-n$.
Any help will be appreciated.
|
One way is to define generating functions $A(z) = \sum_{n \ge 0} a_n z^n$ and $B(z) = \sum_{n \ge 0} b_n z^n$, write your recurrences with indices shifted so that there are no subtractions in indices:
$\begin{align}
a_{n + 1} &= 2 a_n - b_n + 2 \\
b_{n + 1} &= - a_n + 2 b_n - 1
\end{align}$
Multiply both recurrences by $z^n$, sum over $n \ge 0$, and recognise some sums:
$\begin{align}
\sum_{n \ge 0} a_{n + 1} z^n
&= 2 \sum_{n \ge 0} a_n z^n
- \sum_{n \ge 0} b_n z^n
+ 2 \sum_{n \ge 0} z^n \\
\sum_{n \ge 0} b_{n + 1} z^n
&= - \sum_{n \ge 0} a_n z^n
+ 2 \sum_{n \ge 0} b_n z^n
- \sum_{n \ge 0} z^n
\end{align}$
$\begin{align}
\frac{A(z) - a_0}{z}
&= 2 A(z) - B(z) + \frac{2}{1 - z} \\
\frac{B(z) - b_0}{z}
&= - A(z) + 2 B(z) - \frac{1}{1 - z}
\end{align}$
The solution to this system of equations is:
$\begin{align}
A(z)
&= \frac{z - 2 z^2}{1 - 5 z + 7 z^2 - 3 z^3} \\
&= \frac{1}{4 (1 - 3 z)}
- \frac{3}{4 (1 - z)}
+ \frac{1}{2 (1 - z)^2} \\
B(z)
&= \frac{1 - 4 z + 2 z^2}{1 - 5 z + 7 z^2 - 3 z^3} \\
&= - \frac{1}{4 (1 - 3 z)}
+ \frac{3}{4 (1 - z)}
+ \frac{1}{2 (1 - z)^2}
\end{align}$
Note that:
$$
(1 - z)^{-r}
= \sum_{n \ge 0} (-1)^n \binom{-r}{n} z^n
= \sum_{n \ge 0} \binom{n + r - 1}{r - 1} z^n
$$
Also, $\binom{n + r - 1}{r - 1}$ is a polynomial of degree $r - 1$ in $n$.
In particular, $\binom{n + 1}{1} = n + 1$. Picking the coefficients of $z^n$ of the above:
$\begin{align}
a_n
&= \frac{3^n}{4}
- \frac{3}{4}
+ \frac{n + 1}{2} \\
&= \frac{3^n - 3}{4} + \frac{n + 1}{2} \\
b_n
&= - \frac{3^n}{4}
+ \frac{3}{4}
+ \frac{n + 1}{2} \\
&= - \frac{3^n - 3}{4} + \frac{n + 1}{2}
\end{align}$
|
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|
Proving the following set of real numbers is a field Show that the following set $A$ of real numbers under addition and multipication is a field:
$A = {a + b\sqrt{2} : a,b \ \text{rational}}$
I am not sure if I am right but here is what I have thus far:
Closure: Let $a_{1} + b_{1}\sqrt{2}$, $a_{2} + b_{2}\sqrt{2}\in \mathbb{R}$. Then $(a_{1} + b_{1}\sqrt{2})+ (a_{2} + b_{2}\sqrt{2}) = (a_{1} + a_{2}) + (b_{1} + b_{2})\sqrt{2}\in A$.
Now, $(a_{1} + b_{1}\sqrt{2})\times (a_{2} + b_{2}\sqrt{2}) = (a_{1}\times a_{2} + 2\times b_{1}\times b_{2}) + (a_{1}\times b_{2} + b_{1}\times a_{2})\sqrt{2}\in A$
So both addition and multipication are closed on $A$
Associativity: As addition and multipication are associative on $\mathbb{R}$ it follows from Restriction of Operation Associativity that they are also associative on the set $A$.
Commutativity: As addition and multipication are commutative on $\mathbb{R}$ it follows from Restriction of Operation Commutativity that they are also commutative on the set $A$.
Identity: We have $(a + b\sqrt{2}) + (0 + 0\sqrt{2}) = (a + 0) + (b + 0)\sqrt{2} = a + b\sqrt{2}$ and similarly for $(0 + 0\sqrt{2}) + (a + b\sqrt{2})$. So, $(0 + 0\sqrt{2})$ is the identity for addition on $A$. Now, for multipication we have $(a + b\sqrt{2})(1 + 0\sqrt{2}) = (a\times 1 + 2\times b\times 0) + (b\times 1 + a\times 0)\sqrt{2} = a + b\sqrt{2}$ and similarly for $(1 + 0\sqrt{2})(a + b\sqrt{2})$. So, $(1 + 0\sqrt{2})$ is the identity for multipication on $A$.
Inverses: $(a + b\sqrt{2}) + (-a + (-b)\sqrt{2}) = (a - a) + (b - b)\sqrt{2} = 0 + 0\sqrt{2}$ and similarly for $(-a + (-b)\sqrt{2}) + (a + b\sqrt{2})$. So, $(-a + (-b)\sqrt{2})$ is the inverse of $(a + b\sqrt{2})$ for addition on $A$.\
For product inverse consider the difference of squares: $(a + b\sqrt{2})(a - b\sqrt{2}) = a^{2} - 2b^{2}$ which leads to $(a + \sqrt{2})(\frac{a - b\sqrt{2}}{a^{2} - 2b^{2}}) = 1 = 1 + 0\sqrt{2}$. So, demonstrating that the product inverse of $(a + b\sqrt{2})$ is $\frac{a}{a^{2} - 2b^{2}} - \frac{b\sqrt{2}}{a^{2} - 2b^{2}}$ as $a,b$ are rational, it follows that so are $\frac{a}{a^{2} - 2b^{2}}$ and $\frac{b}{a^{2} - 2b^{2}}$. So, the product inverse $(a + b\sqrt{2})\in A$
Distributivity: We have the Real Multipication Distributes over addition, so by Restriction of Operation Distrbutibity, multipication is distributive over addition on $A$.
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Please can we use the fact that it is closed under addition, subtraction, multiplication and division to prove that it is a field.
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"url": "https://math.stackexchange.com/questions/1157407",
"timestamp": "2023-03-29T00:00:00",
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|
Permutation's decomposition into transpositions
*
*Transposition is a cycle with 2 elements.
*Any permutation can be decomposed into a product of transpositions.
For example, for permutation
$\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 3 & 1 & 4
\end{pmatrix}$
decomposition is
$\begin{pmatrix}
1 & 3
\end{pmatrix}
\begin{pmatrix}
1 & 2
\end{pmatrix}$
but why are $\begin{pmatrix}
1 & 3
\end{pmatrix}$ and $\begin{pmatrix}
1 & 2
\end{pmatrix}$ transpositions? Aren't they not even cycles?
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A cycle of length 2 is called a transposition, of course it switches (transposes) two elements.
$$
(1 3) =
\left(
\begin{matrix}
1 & 2 & 3 & 4 \\
3 & 2 & 1 & 4
\end{matrix}
\right)
$$
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"url": "https://math.stackexchange.com/questions/1161680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How can I prove that $\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1$? How can I prove that
$$\sum_{n=1}^\infty \frac{1}{n(n+1)} = 1 \,\,\, ?$$
I do know a way to prove this (see my answer) but I'm curious to know what other approaches could be taken in dealing with it.
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A simple proof by induction starts by noting that
$$ \tag 1 \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{2}{k(k+2)},$$
$$ \tag 2 \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} + \frac{1}{(k+2)(k+3)} = \frac{3}{k(k+3)}.$$
By induction we then see that
$$ \tag 3 \sum_{n=0}^N \frac{1}{(k+n)(k+n+1)} = \frac{N+1}{k(k+N+1)}. $$
The special case $k=1$ gives then
$$ \tag 4 \sum_{n=0}^N \frac{1}{(n+1)(n+2)} = \sum_{n=1}^{N+1} \frac{1}{n(n+1)} = \frac{N+1}{N+2} \to 1, \,\, \text{ for }N \to \infty. $$
Here is instead a different proof using the residue theorem:
Consider the function
$$ \tag 5 f(z) \equiv \frac{\pi \cot(\pi z)}{z(z+1)}. $$
This function is meromorphic with simple poles at $z \in \mathbb{Z} \setminus \{0,-1\} $, and double poles at $z=0,1$.
Moreover, it vanishes fast enough that the contour integral of $f$ over a circle $C_R$ of radius R vanished when $R \to \infty$:
$$ \tag 6 \lim_{R \to \infty} \oint_{C_R} \frac{dz}{2\pi i} f(z) = 0.$$
On the other hand, applying the residue theorem and remembering that for each $n \in \mathbb{Z}$ the Laurent series of $\cot(\pi z)$ at first orders is
$$ \tag 7 \cot(\pi z) = \frac{1}{\pi z} - \frac{\pi z}{3} - \frac{\pi^2 z^2}{45} + \mathcal O(z^3),$$
we see that
$$ \tag 8 \lim_{R \to \infty} \oint_{C_R} \frac{dz}{2\pi i} f(z)
= \sum_{n \neq 0,-1} \frac{1}{n(n+1)} - 2. $$
Putting together (6) and (8) we thus obtain
$$ \sum_{n \neq 0,-1} \frac{1}{n(n+1)} = 2 \frac{1}{n(n+1)} = 2.$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/1163665",
"timestamp": "2023-03-29T00:00:00",
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|
Find all integers $x$, $y$, and $z$ such that $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$ Characterize all positive integers $x$, $y$, and $z$ such that:
$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{z}$$
For example, $\dfrac{1}{x+1} + \dfrac{1}{x(x+1)} = \dfrac{1}{x}$.
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An idea : you have
$$\frac{1}{x}+\frac{1}{y} = \frac{x+y}{xy}$$
It imply that $x+y | xy$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1166999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
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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.