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Find the sum of series $ \sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$ Let it be known that $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}
{6}.$$ Given suchβfind $$\sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$$
Attempt:
I have tried using the fact that $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and then expanding or using known sum types as $\displaystyle \sum_{k=1}^{n} k=\frac{n(n+1)}{2}$ or $\displaystyle \sum_{k=1}^{n} k^3=\frac{n^2(n+1)^2}{4}$
but nothing seems to lead to anything!
|
I'm basically just going to use the identity you noted in the comments $\frac 1n-\frac 1{n+1}=\frac 1{n(n+1)}$ repeatedly and then the closed form for $\zeta(2)$.
$$\sum_{n=1}^\infty \frac 1{n^3(n+1)^3}=\sum_{n=1}^\infty \left(\frac 1n-\frac 1{n+1}\right)^3=\sum_{n=1}^\infty \frac 1{n^3}-\frac{3}{n^2(n+1)}+\frac{3}{n(n+1)^2}-\frac 1{(n+1)^3}$$
The first and last terms telescope so you get
$$=1-3\sum_{n=1}^\infty \frac 1{n(n+1)}\left(\frac 1n-\frac 1{n+1}\right)=1-3\sum_{n=1}^\infty \left(\frac 1n-\frac 1{n+1}\right)^2$$
$$=1-3\sum_{n=1}^\infty \frac 1{n^2}-\frac{2}{n(n+1)}+\frac 1{(n+1)^2}$$
The first and last terms are $\zeta(2)$ and $\zeta(2)-1$ respectively
$$=4-6\zeta(2)+6\sum_{n=1}^\infty \frac 1{n(n+1)}=4-6\zeta(2)+6\sum_{n=1}^\infty\frac 1n-\frac 1{n+1}=10-6\zeta(2)=10-\pi^2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solving $\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $. Simply bringing it to a common denominator does not lead me to success How can I solve this equation?
$$\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}=4 $$
Simply bringing it to a common denominator does not lead me to success
What I tried
|
\begin{equation}
\frac{(1+x)(1+2x)(1+3x)}{(4+x)(4+2x)(4+3x)}
=4 \\
\text{We multiply both sides by teh common denominator}\\
(1+x)(1+2x)(1+3x\quad =\quad 4\big((4+x)(4+2x)(4+3x)\big) \\
\text{we expand both sides} \\
x^3 + 11 x^2 + 6 x + 1\quad =\quad 24 x^3 + 176 x^2 + 384 x + 256\\
\text{we collect terms} \\
18 x^3 + 165 x^2 + 378 x + 255=0\\
\text{we factor} \\
3 (3 x + 5) (2 x^2 + 15 x + 17)=0\\ \\
\text{we solve} \\
x = -5/3
\qquad\quad\space\space \space \approx -1.6667\\
x = \frac{-15 + \sqrt{89}}{4}
\quad\approx -1.3915\\
x = \frac{-15 - \sqrt{89}}{4}
\quad\approx -6.1085
\end{equation}
|
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|
Difficult limit proof Let $a_1 = 1$ and define a sequence recursively by $$a_{n+1} = \sqrt{a_1+a_2+\dots+a_n}.$$ Show that $\lim_{n \to \infty} \frac{a_n}{n} = \frac{1}{2}$.
So far, I've written $a_{n+1} = \sqrt{a_n^2 + a_n}$, and have shown that if $a_n/n < 1/2$, then so is $a_{n+1}/{(n+1)}$. I'm really not sure how to relate the recurrence relation to the limit we want, though.
|
The Stolz-Cesaro theorem states the following:
Let $a, b$ be sequences of real numbers such that $b$ is monotone increasing and $\lim\limits_{n \to \infty} b_n = \infty$. If $\lim\limits_{n \to \infty} \frac{a_{n + 1} - a_n}{b_{n + 1} - b_n} = \ell$ then $\lim\limits_{n \to \infty} \frac{a_n}{b_n} = \ell$.
It's basically the discrete version of L'Hopital's rule. For more details, see the Wikipedia page.
In this case, let $b_n = n$. Then we must compute $\lim\limits_{n \to \infty} \frac{a_{n + 1} - a_n}{b_{n + 1} - b_n} = \lim\limits_{n \to \infty} a_{n + 1} - a_n$.
We have
\begin{equation}
\begin{split}
a_{n + 1} - a_n &= \sqrt{a_n^2 + a_n} - a_n \\
&= (\sqrt{a_n^2 + a_n} - a_n) \frac{\sqrt{a_n^2 + a_n} + a_n}{\sqrt{a_n^2 + a_n} + a_n} \\
&= \frac{a_n}{\sqrt{a_n^2 + a_n} + a_n} \\
&= \frac{1}{\sqrt{1 + 1 / a_n} + 1}
\end{split}
\end{equation}
Therefore, we see that $\lim\limits_{n \to \infty} a_{n + 1} - a_n = \lim\limits_{n \to \infty} \frac{1}{\sqrt{1 + 1/a_n} + 1} = \frac{1}{2}$.
Note that this does require showing that $\lim\limits_{n \to \infty} a_n = \infty$. But this is trivial, since we know that $a$ is increasing, and therefore $a_{n + 1} - a_n = \frac{1}{\sqrt{1 + 1 / a_n} + 1} \geq \frac{1}{\sqrt{1 + 1 / 1} + 1} = \frac{1}{\sqrt{2} + 1}$. From this it follows that $a_n \geq n \frac{1}{\sqrt{2} + 1}$ for all $n$.
|
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|
Solve differential equation. $2xy'$ + $y^2 - 1 =0$ Solve differential equation.
$2xy' + y^2 - 1 =0$
My work.
$2x\dfrac{dy}{dx}+y^2-1=0$
$\dfrac{dx}{2x}=\dfrac{dy}{1-y^2}$
$\int\dfrac{dx}{2x}=\int\dfrac{dy}{1-y^2}$
getting
$\frac{1}{2}\ln|x|-\frac{1}{2}(\ln|1+y|-\ln|1-y|)=0$
but solution in the book says $y=\dfrac{cx-1}{cx+1}$ or $y=1$ or $y =-1$
EDIT
$\frac{1}{2}\ln|\frac{1+y}{1-y}| = \frac{1}{2}\ln|x| +c$
from here I still can't get the answer. And $y=1$, $y=-1$ are in fact solutions.
|
So $\dfrac{1}{2}\ln|\dfrac{1+y}{1-y}| = \dfrac{1}{2}\ln|x| +c \implies \ln|\dfrac{1+y}{1-y}| = \ln|x| +c_1 \implies \ln|\dfrac{1+y}{1-y}| = \ln|c_2x|$ as any number can be written as logarithm of some number. Thus, $\frac{1+y}{1-y} =cx \implies \frac{2}{1-y}-1=cx, cx+1=\frac{2}{1-y}, 1-y=\frac{2}{cx+1}, y=1-\frac{2}{cx+1}=\frac{cx-1}{cx+1}$
|
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|
Convergence of $1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } $ as $n \to \infty$ I stumbled upon this problem while reading about the bias of the sample standard deviation.
How to show that:
$$\bigg(1 - \sqrt{ \frac{2}{n-1} } \cdot \frac{ \Gamma(n/2) }{ \Gamma( \frac{n-1}{2} ) } \bigg) \sim \frac{1}{4 n}$$
as $n \to \infty$ and n is positive integer.
I thought I should use Stirling's formula, or Gautschi's inequality. The following equality might also lead in the right direction:
$$\Gamma(1/2+n)=\frac{(2n)!}{4^nn!}\sqrt{\pi}$$
Can you help?
|
Using twice Stirling formula, we have
$$\log \left(\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma
\left(\frac{n-1}{2}\right)}\right)=\frac{1}{2} \log \left(\frac{n}{2}\right)-\frac{3}{4 n}-\frac{1}{2 n^2}-\frac{3}{8
n^3}+O\left(\frac{1}{n^4}\right)\tag 1$$
Using the binomial expansion and taking the logarithm and ussing Taylor series
$$\log \left(\sqrt{\frac{2}{n-1}}\right)=-\frac{1}{2} \log \left(\frac{n}{2}\right)+\frac{1}{2 n}+\frac{1}{4 n^2}+\frac{1}{6
n^3}+O\left(\frac{1}{n^4}\right)\tag 2$$
Adding $(1)$ to $(2)$ gives
$$\log \left(\sqrt{\frac{2}{n-1}}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma
\left(\frac{n-1}{2}\right)}\right)=-\frac{1}{4 n}-\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)\tag 3$$
$$\sqrt{\frac{2}{n-1}}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma
\left(\frac{n-1}{2}\right)}=1-\frac{1}{4 n}-\frac{7}{32 n^2}+O\left(\frac{1}{n^3}\right)\tag 4$$
$$1-\sqrt{\frac{2}{n-1}}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma
\left(\frac{n-1}{2}\right)}=\frac{1}{4 n}+\frac{7}{32 n^2}+O\left(\frac{1}{n^3}\right)\tag 5$$
Try for $n=10$; the exact value is
$$1-\frac{128 }{105}\sqrt{\frac{2}{\pi }}=0.0273407\cdots$$ while the truncated series gives
$$\frac{87}{3200}=0.0271875\cdots$$
|
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|
The series representation of $\sum _{i=0}^{n-1} (-1)^{i+1} \left(\frac{i}{n}\right)^{r}$ Lets consider the:
$$\left(\frac{1}{n}\right)^r-\left(\frac{2}{n}\right)^r+\left(\frac{3}{n}\right)^r-...+\left(\frac{n-1}{n}\right)^r$$
Trying to find any formal serise representation for variable $n$. Have tried the EulerβMaclaurin summation, seems there is no simple way to go to series from there.
I have separated the odd and even parts:
$$\sum _{u=0}^{n-1} (-1)^{u+1}\left(\frac{u}{n}\right)^r=\sum _{u=0}^{\frac{n-2}{2}} \left(\frac{2 u+1}{n}\right)^r-\sum _{u=0}^{\frac{n-1}{2}} \left(\frac{2 u}{n}\right)^r$$
Then for each sum:
$$\sum _{u=0}^{\frac{n-2}{2}} \left(\frac{2 u+1}{n}\right)^r=\frac{1}{2} \left(\left(\frac{2 \left\lfloor \frac{n-2}{2}\right\rfloor +1}{n}\right)^r+\left(\frac{1}{n}\right)^r\right)+\int_0^{\frac{n-2}{2}} \left(\frac{2 u+1}{n}\right)^r \, du + \sum _{k=1}^{\infty } \frac{B_{2 k} \frac{\partial ^{2 k-1}}{\partial u^{2 k-1}}\left(\frac{2 u+1}{n}\right)^r}{(2 k)!} \biggr|_0^{\lfloor \frac{n-2}{2} \rfloor}$$
$$\sum _{u=0}^{\frac{n-1}{2}} \left(\frac{2 u}{n}\right)^r=\frac{1}{2} \left(\frac{2 \left\lfloor \frac{n-1}{2}\right\rfloor }{n}\right)^r+\int_0^{\frac{n-1}{2}} \left(\frac{2 u}{n}\right)^r \, du+\sum _{k=1}^{\infty } \frac{B_{2 k} \frac{\partial ^{2 k-1}}{\partial u^{2 k-1}}\left(\frac{2 u}{n}\right)^r}{(2 k)!} \biggr|_0^{\lfloor \frac{n-1}{2} \rfloor}$$
Then after extracting for first 2 terms for example for odd $n$ we get nothing simple in terms of series representation:
$$\frac{(n-2)^r-(n-1)^r+\frac{r}{r+1}}{2 n^r}+... +\sum _{k=1}^{\infty }... - \sum _{k=1}^{\infty }...$$
Any hint is appreciated.
|
First note that
$$
\sum\limits_{j = 0}^{n - 1} {( - 1)^{j + 1} \left( {\frac{j}{n}} \right)^r } = - \frac{1}{{n^r }}\sum\limits_{j = 0}^{n - 1} {( - 1)^j j^r } .
$$
Employing the EulerβBoole summation formula with $f(x) = x^r$, $h = 0$, $m = r + 1$, we deduce
\begin{align*}
\sum\limits_{j = 0}^{n - 1} {( - 1)^j j^r } & = - \frac{1}{2}E_r (0)\left( {( - 1)^n - 1} \right) - \frac{1}{2}\sum\limits_{k = 0}^{r - 1} {( - 1)^n \binom{r}{k}E_k (0)n^{r - k} }
\\ & = \frac{{2^{r + 1} - 1}}{{r + 1}}B_{r + 1} \left( {( - 1)^n - 1} \right) + \sum\limits_{k = 0}^{r - 1} {( - 1)^n \binom{r}{k}\frac{{2^{k + 1} - 1}}{{k + 1}}B_{k + 1} n^{r - k} } .
\end{align*}
An alternative form using Euler polynomials is 24.4.8 in the DLMF.
|
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|
Eliminating $\theta$ from $\cos^3\theta +a\cos\theta =b$ and $\sin^3\theta +a\sin\theta =c$
Eliminate $\theta$ from the equations.
$$\cos^3\theta +a\cos\theta =b$$
$$\sin^3\theta +a\sin\theta =c$$
Can anyone solve this question?
|
So, if I understood properly your question, we have
$$
\left\{ \begin{array}{l}
\theta ,a,b,c \in R \\
\cos ^3 \theta + a\cos \theta - b = 0 \\
\sin ^3 \theta + a\sin \theta - c = 0 \\
\end{array} \right.
$$
and we are looking for which domain of the four variables $ \theta ,a,b,c$
the system of the two equations has one or more (real) solutions.
Both equations are depressed cubics.
The discriminants are
$$
\begin{array}{l}
\Delta _1 = - 4a^3 - 27b^2 \\ \Delta _2 = - 4a^3 - 27c^2 \\
\end{array}
$$
If
*
*$\Delta >0$ the equation has three distinct real roots;
*$\Delta = 0$ all roots are still real , and if $a=0$ gives three coincident roots,
otherwise one simple root at $-3a/b; -3a/c$ and two coincident at $-3a/(2b) ; -3a/(2c)$;
*$\Delta < 0$ the equation has only one real root.
Then, denoting by $c_1, c_2, c_3$ the roots of the equation in $\cos$, eventually reduced to only two or one,
and by $s_1, s_2, s_3$ those of the equation in $\sin$, you shall impose to have at least a couple for which
$$
\left| {c_{\,k} } \right| \le 1\quad \wedge \quad \left| {s_{\,k} } \right| \le 1\quad
\wedge \quad c_{\,k} ^2 + s_{\,k} ^2 = 1
$$
That means to write each cubic as the product of the three monomials $(x-c_k)$, etc.
and patiently translate the above bounds and conditions onto $a,b,c$ by means of Vieta's formulas.
We get in fact
$$
\left\{ \begin{array}{l}
c_1 + c_2 + c_3 = 0 \\ c_1 c_2 + c_1 c_3 + c_2 c_3 = a \\ c_1 c_2 c_3 = b \\
\end{array} \right.\quad \Rightarrow \quad \left\{ \begin{array}{l}
c_1 = - \left( {c_2 + c_3 } \right) \\ c_2 ^2 + c_2 c_3 + c_3 ^2
= \frac{3}{4}\left( {c_2 + c_3 } \right)^2 + \frac{1}{4}\left( {c_2 - c_3 } \right)^2 = - a \\
\left( {c_2 + c_3 } \right)c_2 c_3 = - b \\
\end{array} \right.
$$
and similarly for the sine, changing $b$ with $c$.
So that, if the three solutions are real, the ellipse in the second line demand $a$ to be negative, resulting
in the two curves are sketched below.
Note that, the properties of the cubic ensure that there is a minimum and maximum value
for $b$ (in function of $a$) for which the two curves simultaneously touch each other,
and such values are easy to determine imposing the tangency along $c_2 = c_3$.
And we shall impose that that point at least be within $[-1,1] $, etc.
Finally note that the intersection points are six, since we have not ordered the roots.
|
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|
Maximum distance between mid-point of chord of ellipse Let $E$ be the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$. For any three distinct points $P,Q$ and $Qβ²$ on $E$, let $M(P, Q)$ be the mid-point of the line segment joining $P$ and $Q$, and $M(P, Qβ²)$ be the mid-point of the line segment joining $P$ and $Qβ²$. Then the maximum possible value of the distance between $M(P,Q)$ and $M(P,Qβ²)$, as $P,Q$ and $Q β²$ vary on $E$, is ___ .
My approach is as follow
Let $P(4cos\alpha,3sin\alpha)$,$Q(4cos\theta_1,3sin\theta_1)$,$Qβ²(4cos\theta_2,3sin\theta_2)$
$M(P, Q)=(\frac{4(cos\theta_1+cos\theta_2)}{2},)$
Let the mid point be represented as $M\left( {P,Q} \right) = \left( {\frac{{4\left( {\cos \alpha + \cos {\theta _1}} \right)}}{2},\frac{{3\left( {\sin \alpha + \sin {\theta _1}} \right)}}{2}} \right);M\left( {P,Q'} \right) = \left( {\frac{{4\left( {\cos \alpha + \cos {\theta _2}} \right)}}{2},\frac{{3\left( {\sin \alpha + \sin {\theta _2}} \right)}}{2}} \right)$
$\frac{1}{4}\sqrt {{{\left( {4\cos \alpha + 4\cos {\theta _1} - \left( {4\cos \alpha + 4\cos {\theta _2}} \right)} \right)}^2} + {{\left( {3\sin \alpha + 3\sin {\theta _1} - \left( {3\sin \alpha + 3\sin {\theta _2}} \right)} \right)}^2}} $
$\frac{1}{4}\sqrt {16\left( {{{\cos }^2}{\theta _1} + {{\cos }^2}{\theta _2} - 2\cos {\theta _1}\cos {\theta _2}} \right) + 9\left( {{{\sin }^2}{\theta _1} + {{\sin }^2}{\theta _2} - 2\sin {\theta _1}\sin {\theta _2}} \right)} $
$\frac{1}{4}\sqrt {16{{\cos }^2}{\theta _1} + 9{{\sin }^2}{\theta _1} + 16{{\cos }^2}{\theta _2} + 9{{\sin }^2}{\theta _2} - 32\cos {\theta _1}\cos {\theta _2} - 18\sin {\theta _1}\sin {\theta _2}} $
$\frac{1}{4}\sqrt {18 + 7\left( {{{\cos }^2}{\theta _1} + {{\cos }^2}{\theta _2}} \right) - 14\cos {\theta _1}\cos {\theta _2} - 18\left( {\cos \left( {{\theta _1} - {\theta _2}} \right)} \right)} $
${\cos ^2}{\theta _1} = \frac{{1 + \cos 2{\theta _1}}}{2};{\cos ^2}{\theta _2} = \frac{{1 + \cos 2{\theta _2}}}{2}$
${\cos ^2}{\theta _1} + {\cos ^2}{\theta _2} = 1 + \frac{{2\cos \left( {{\theta _1} + {\theta _2}} \right)\cos \left( {{\theta _1} - {\theta _2}} \right)}}{2} = 1 + \cos \left( {{\theta _1} + {\theta _2}} \right)\cos \left( {{\theta _1} - {\theta _2}} \right)$
$\frac{1}{4}\sqrt {18 + 7\left( {1 + \cos \left( {{\theta _1} + {\theta _2}} \right)\cos \left( {{\theta _1} - {\theta _2}} \right)} \right) - 7\left( {\cos \left( {{\theta _1} + {\theta _2}} \right) + \cos \left( {{\theta _1} - {\theta _2}} \right)} \right) - 18\left( {\cos \left( {{\theta _1} - {\theta _2}} \right)} \right)} $
$\frac{1}{4}\sqrt {25 + 7\cos \left( {{\theta _1} + {\theta _2}} \right)\left( {\cos \left( {{\theta _1} - {\theta _2}} \right) - 1} \right) - 25\left( {\cos \left( {{\theta _1} - {\theta _2}} \right)} \right)} $
How do we proceed from here , we get the official answer 4 when $\theta_1=0$ & $\theta_2=\pi$
|
Define $X\equiv M(P,Q)$ and $Y\equiv M(P,Q')$. Using the midpoint theorem $(\triangle PXY\sim \triangle PQQ')$,
$$|XY|=\frac{1}{2}\cdot|QQ'|$$
$|QQ'|$ is maximum when $Q$ and $Q'$ lie on the major axis of the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}$ (the two farthest points on an ellipse). Hence, $Q=(\pm 4,0)$ and $Q'=(\mp4,0)$.
Therefore,
$$\max(|XY|)=\frac{1}{2}\cdot8=\boxed{4}$$
|
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|
Showing $ 2\sqrt{\frac{x+3}{x}}+8\sqrt{\frac{x+1}{x}}-\ln\left(\frac{(x+1)^{3/2}(x+3)}{(x-1)^{5/2}}\right)\geq 10 $ for $x\geq7$
Suppose that $x\geq 7$. I would like to show that
$$ 2\sqrt{\frac{x+3}{x}} + 8\sqrt{\frac{x+1}{x}}-\ln\left(\frac{(x+1)^{3/2}(x+3)}{(x-1)^{5/2}}\right)\geq 10 $$
I rewrote the inequality as
$$ 2\sqrt{\frac{x+3}{x}} +8\sqrt{\frac{x+1}{x}}-\ln\left(\frac{\left(\frac{x+1}{x}\right)^{3/2}\left(\frac{x+3}{x}\right)}{\left(\frac{x-1}{x}\right)^{5/2}}\right)\geq 10, $$
which is equivalent to
$$ 2\sqrt{\frac{x+3}{x}} -2\ln\left( \sqrt{\frac{x+3}{x}}\right)+8\sqrt{\frac{x+1}{x}}-3\ln\left( \sqrt{\frac{x+1}{x}}\right) \geq 10-5\ln\left(\sqrt{\frac{x}{x-1}} \right). $$
I am not sure if rewriting it in the last way helps. Wolfram Alpha shows me that the inequality is true, but I cannot figure out why. Any help is greatly appreciated. Thank you.
|
We need to prove that, for all $x \ge 7$,
$$2\sqrt{\frac{x+3}{x}}-2\ln \sqrt{\frac{x+3}{x}} +8\sqrt{\frac{x+1}{x}}-3\ln \sqrt{\frac{x+1}{x}} \geq 10+5\ln\sqrt{\frac{x}{x-1}} .$$
With the substitution $x = \frac{1}{y}$,
it suffices to prove that, for all $y\in [0, 1/7]$,
$$2\sqrt{1 + 3y} - \ln(1 + 3y) + 8\sqrt{1 + y} - \frac32\ln(1 + y) \ge 10 - \frac52\ln(1 - y).$$
Let $F(y) = \mathrm{LHS} - \mathrm{RHS}$.
We have
$$F'(y) = \frac{3}{\sqrt{1 + 3y}}
- \frac{3}{1 + 3y} + \frac{4}{\sqrt{1 + y}}
- \frac{3}{2(1 + y)} - \frac{5}{2(1 - y)}$$
and
$$F''(y) = - \frac{9}{2(1 + 3y)^{3/2}} + \frac{9}{(1 + 3y)^2} - \frac{2}{(1 + y)^{3/2}} + \frac{3}{2(1 + y)^2} - \frac{5}{2(1 - y)^2}$$
and
$$F'''(y) = \frac{81}{4(1 + 3y)^{5/2}} - \frac{54}{(1 + 3y)^3} + \frac{3}{(1 + y)^{5/2}} - \frac{3}{(1 + y)^3}
- \frac{5}{(1 - y)^3}.$$
We have, for all $y\in [0, 1/7]$,
\begin{align*}
F'''(y)
&\le \left(\frac{81}{4(1 + 3y)^2} - \frac{54}{(1 + 3y)^3}\right) + \left(\frac{3}{(1 + y)^2} - \frac{3}{(1 + y)^3}
- \frac{5}{(1 + y)^3}\right)\\
&= \frac{27(9y - 5)}{4(1 + 3y)^3} + \frac{3y - 5}{(1 + y)^3}\\
& < 0.
\end{align*}
Note that $F''(0) > 0$ and $F''(1/7) < 0$.
Thus, there exists $y_0 \in (0, 1/7)$
such that $F''(y_0) = 0$,
$F''(y) > 0$ on $y\in [0, y_0)$,
and $F''(y) < 0$ on $y\in (y_0, 1/7]$.
Note that $F'(0) = 0$ and $F'(1/7) < 0$.
Thus, there exists $y_1 \in (y_0, 1/7)$ such that $F'(y_1) = 0$, $F'(y) > 0$ on $(0, y_1)$,
and $F'(y) < 0$ on $(y_1, 1/7)$.
Note that $F(0) = 0$ and $F(1/7) > 0$.
Thus, $F(y) \ge 0$ on $[0, 1/7]$.
We are done.
|
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|
Solve $\lim_{x\to \pi/4} \frac{\sin x - \cos x}{x-\pi/4}$ As the title suggests, we have to solve the limit:
$\lim_{x\to \frac\pi4} \frac{\sin x - \cos x}{x-\frac \pi4}$
I'm able to solve it by using L'Hospital's rule and got an answer $\sqrt2$ but the problem is that this rule is not allowed at school level.
So I tried another method:
$$\lim_{x\to \frac\pi4} \frac{\sin x - \cos x}{x-\frac \pi4}$$
$$\lim_{h\to 0} \frac{\sin(Ο/4+h) - \cos(Ο/4+h)}{h}$$
By using the identity of $\sin(a+b)$ and $\cos(a+b)$, we get:
$$\lim_{h\to 0} \frac{[\sin Ο/4+ \cos Ο/4][\cos h + \sin h]}{h}$$
If we here substitute $h=0$, we get $β2/0$. Can we solve it further?
Please help!
BTW sorry for the bad formatting.
|
Note that $\cos$ is continuous at $\pi/4$ and $\cos(\pi/4)>0$, so $\cos(x)$ is certainly nonzero near $\pi/4$. This justifies writing
\begin{align}
\frac{\sin(x)-\cos(x)}{x-\frac{\pi}{4}} &= \frac{\cos(x)\left(\frac{\sin(x)}{\cos(x)}-1\right)}{x-\frac{\pi}{4}}\\
&= \cos(x)\cdot\frac{\tan(x)-1}{x-\frac{\pi}{4}}\\
&= \cos(x)\cdot\frac{\tan(x)-\tan\left(\frac{\pi}{4}\right)}{x-\frac{\pi}{4}}
\end{align}
for every $x$ sufficiently close to $\pi/4$.
As $x\to\pi/4$, $\cos(x)\to\cos(\pi/4)=\sqrt{2}/2$ and
\begin{align}
\frac{\tan(x)-\tan\left(\frac{\pi}{4}\right)}{x-\frac{\pi}{4}} &\to (\tan)'\left(\frac{\pi}{4}\right)\\
&= \sec^2\left(\frac{\pi}{4}\right)\\
&= 2
\end{align}
It follows from the product rule for limits that
\begin{align}
\lim_{x\to\frac{\pi}{4}}\frac{\sin(x)-\cos(x)}{x-\frac{\pi}{4}} &= \lim_{x\to\frac{\pi}{4}}\left(\cos(x)\cdot\frac{\tan(x)-\tan\left(\frac{\pi}{4}\right)}{x-\frac{\pi}{4}}\right)\\
&= \frac{\sqrt{2}}{2}\cdot 2\\
&=\sqrt{2}
\end{align}
|
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|
Prove that $\sum_\circ\frac{a^3+3b^3}{5a+b}\ge\frac23(a^2+b^2+c^2)$ for $a,b,c>0$
Prove the following for a, b and c being positive real numbers :-
$\sf{{ \dfrac{ a^3 + 3b^3 }{5a+b} + \dfrac{ b^3 + 3c^3}{5b+c} + \dfrac{ c^3 + 3a^3}{5c+a} \geqslant \dfrac{2}{3} (a^2 + b^2 + c^2) }} $
So here we have a question in which we have to prove that the equation which we are provided with, has real values of a, b and c.
I know on this site we have to show our work while asking a question, but actually I'm unable to even start it. Please can you give me hint for solving this question? Or at least hint where to start.
I got this question while practicing some previous year examination papers for my future test which is next month.
|
Split our the fractions; use CS to grouped summands; use $ab+bc+ca \le a^2+b^2+c^2$:
$$\begin{align}
&\dfrac{ a^3 + 3b^3 }{5a+b} + \dfrac{ b^3 + 3c^3}{5b+c} + \dfrac{ c^3 + 3a^3}{5c+a} \\
&= \color{blue}{\dfrac{ a^4 }{5a+b}} + 3\color{fuchsia}{\frac{b^3}{5a+b}} \\
&+ \color{blue}{\dfrac{ b^3}{5b+c}} + 3\color{fuchsia}{\frac{c^3}{5b+c}} \\
&+ \color{blue}{\dfrac{ c^3}{5c+a}} + 3\color{fuchsia}{\frac{a^3}{5c + a}} \\
&= \color{blue}{\dfrac{ a^4 }{5a^2+ab} + \dfrac{ b^4}{5b^2+bc} + \dfrac{ c^4}{5c^2+ac}} \\
&+ 3\left(\color{fuchsia}{\frac{b^4}{5ab+b^2} + \frac{c^4}{5bc+c^2} + \frac{a^4}{5ca + a^2}}\right) \\
&\ge \frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2) + ab+bc+ca} + 3\frac{(a^2+b^2+c^2)^2}{a^2+b^2+c^2 + 5(ab+bc+ca)} \\
&\ge \frac{(a^2+b^2+c^2)^2}{5(a^2+b^2+c^2) + a^2+b^2+c^2} + 3\frac{(a^2+b^2+c^2)^2}{a^2+b^2+c^2 + 5(a^2+b^2+c^2)} \\
&= \left(\frac{1}{6} + \frac{3}{6}\right)(a^2+b^2+c^2)\\
&= \dfrac{2}{3} (a^2 + b^2 + c^2)\end{align}$$
|
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|
the value of $\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}$ I want to compute this limit
$$\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}.$$
I tried to apply Hopital rule, but I cannot compute it.
|
Just to suggest another way: $\ln(1+x^2)=x^2-\frac{x^4}2+O(x^6)$ and $\sin(x^2)=x^2+O(x^6)$, so the numerator behaves like $(x^2+1)(x^2-\frac{x^4}2)-x^2\sim\frac{x^4}2$. The denominator on the other hand behaves like $x^3$.
Proof that the limit is $0$.
*
*By simple study of functions, we check that $\lvert\ln(x^2+1)-x^2\rvert\le\frac{x^4}2$ and $\lvert\sin(x^2)-x^2\rvert\le\frac{x^6}6$.
*Then we can bound the numerator as
\begin{align*}
\Bigl|(x^2+1)\ln(x^2+1)-\sin(x^2)\Bigr|&\le x^2\ln(x^2+1)+\lvert\ln(x^2+1)-x^2\rvert+|x^2-\sin(x^2)|\\
&\le x^4+\frac{x^6}2+\frac{x^4}2+\frac{x^6}2\\
&=\frac32x^4+x^6.
\end{align*}
*Now
$$\left\lvert\frac{(x^2+1)\ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}\right\rvert
\le\frac{\frac32x^2+x^4}{|x|}\cdot\frac{x^2}{\lvert\sin(x^2)\rvert}
=\left(\frac32|x|+|x|^3\right)\left\lvert\frac{x^2}{\sin(x^2)}\right\rvert.$$
|
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|
A question about the maximal domain of a function So I have got the following equation:
$$h(x) = \sqrt{\frac{1}{x+1}+1}$$
I need to find the maximal domain of the function.
I have tried doing it algebraically:
As this is a square root, $\frac{1}{x+1}+1$, must be greater than $0$ and there is an asymptote at $x = -1$.
$\frac{1}{x+1}+1 > 0$
$\frac{1}{x+1} > -1$
$ 1 >-x-1$
$1+x>-1$
$\to x>-2$, provided that $x \not= -1$
But this is incorrect $:($
The answers show a different maximal domain. Moreover, I do not know how they got their answer. I need help. Thanks!!!
|
In what follows, I will assume that $h$ is a real-valued function of a real variable.
The expression
$$\frac{1}{x + 1}$$
is undefined when $x = -1$. Thus, we require that $x > -1$ or $x < -1$.
Since we cannot take the square root of a negative number, we require that the radicand (the term inside the square root) be nonnegative. Hence,
\begin{align*}
\frac{1}{x + 1} + 1 & \geq 0\\
\frac{1}{x + 1} & \geq -1
\end{align*}
At this point, you made a mistake. You have to consider two cases, depending on the sign of $x + 1$.
Case 1: $x > -1$.
If $x > -1$, then $x + 1 > 0$, so the inequality is preserved if we multiply both sides of the inequality
$$\frac{1}{x + 1} \geq -1$$
by $x + 1$, which yields
\begin{align*}
1 & \geq -x - 1\\
x + 1 & \geq -1\\
x & \geq -2
\end{align*}
which is automatically satisfied if $x > -1$.
Case 2: If $x < -1$, then $x + 1 < 0$, so the direction of the inequality is reversed if we multiply both sides of the inequality
$$\frac{1}{x + 1} \geq -1$$
by $x + 1$. Hence,
\begin{align*}
1 & \leq -x - 1\\
x & \leq -2
\end{align*}
Therefore, the domain of the function is $(-\infty, -2] \cup (1, \infty)$.
|
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|
Find $\delta > 0$ such that $|x-2| < \delta$ implies $|x^2+2x-18| < \frac{1}{4}$ Find $\delta > 0$ such that $|x-2| < \delta$ implies $|x^2+2x-18| < \frac{1}{4}$.
I have been having trouble with this question from my Analysis class because $f(x)$ doesn't factor. Here is the first scratch work I did attempting to find a suitable $\delta$:
Let $\delta_1 =1$. Then $|x-2| < \delta \implies -1 < x-2<1 \implies 3 < x+2 < 5$. Hence $|x+2|<5$.
We want $|x^2+2x-18|< \frac{1}{4}$. This is the same as:
$$\frac{-1}{4} < x^2+2x-18 < \frac{1}{4}$$
$$-1 < 4x^2+8x-72 < 1$$
$$71 < 4x^2+8x < 73$$
$$71 < 4x(x+2) < 73$$
$$|4x(x+2)|=|4x||x+2| \leq |4x| \cdot5 < 73$$
$$|4x| < \frac{73}{5}$$
I got stuck here. In the other practice problems we did I always ended up with something like $|x-a|<...<...$ so I'm not sure what went wrong? I do not see how to factor $x^2+2x-18$ into something that utilizes $(x-2)$.
|
Let $f(x)=|x^2+2x-18|$. For all $\delta>0$, $|2-2|<\delta$ but $f(2)=10\geq\frac{1}{4}$. So no matter which value you pick for $\delta$ there is a $x$ ($2$) that wont satisfy $|x^2+2x-18|<\frac{1}{4}$.
|
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|
Prove that $17$ divides $3^{4n} + 4^{3n+2}$ for all natural numbers $n$ by induction. I would like to preface my very first question here by stating that this is not homework. Indeed, I am a mathematics major and computer science minor. However, I am doing extra work on my own to develop my understanding. I am unsure of the informal proof I have created, and my professor is not immediately available, so I am here asking for help.
My understanding goes about as far as multivariable calculus. After this semester, I will be heading into junior level classes. Anyway, that is enough preface.
Here is my informal proof by induction. I've only been writing proofs for about a month, so please be gentle.
$17$ divides $3^{4n} + 4^{3n+2} \ \forall n \in \Bbb{N}$
Proof: (PMI)
βcase 1: $n = 1$
ββ$3^{4(1)} + 4^{3(1)+2} = 1,105,$
β$\implies$ $17$ | $1,105$,
β$\implies$ $17m$ = $1,105, \exists m \in \Bbb{N},$
β$\implies m = 65,$
β$\implies$ $17$ | $3^4 + 4^5.$
βcase 2: $n = k$
ββ Assume: $17$ | $3^{4k} + 4^{3k+2}$ given $k \in \Bbb{N}.$
ββ WTS: $17$ | $3^{4(k+1)} + 4^{3(k+1)+2}$
β$\implies$ $\exists k_1 \in \Bbb{N}$ s.t. $17k_1 = 3^{4k} + 4^{3k+2}.$
β$\implies$ $3^{4(k+1)} + 4^{3(k+1)+2} = 3^{4k+4} + 4^{3k+3+2},$
ββββββββββββ$= 3^{4} \cdot 3^{4k} + 4^{2} \cdot 4^{3} \cdot 4^{3k},$
ββββββββββββ$= 3^{4}(3^{4k}) + 4^{3}(4^{3k+2}),$
ββββββββββββ$=81(3^{4k} + 4^{3k+2}) - 81(4^{3k+2}) + 64(3^{4k} + 4^{3k+2})-64(3^{4k})$,
ββββββββββββ$=81(3^{4k} + 4^{3k+2} - 4^{3k+2}) + 64((3^{4k} + 4^{3k+2} - 3^{4k})$,
ββββββββββββ$=(81-64)(3^{4k} - 4^{3k+2})$,
ββββββββββββ$=17(3^{4k} - 4^{3k+2})$,
ββββββββββββ$=17k_1$,
ββββββββββ$\implies 17$ | $3^{4k+4} + 4^{3k+5}$.
$\Box$
If I am way off base here, I would greatly appreciate your solutions (to study) or nudges in the right direction. Thank you.
|
Assume true for $n=k$. For $n=k+1$ we have (using $3^4=81$ and $4^3=64$) $81\times 3^{4k}+64\times 4^{3k+2}=17\times 3^{4k}+64\times(3^{4k}+4^{3k+2})$, which is divisible by $17$.
|
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|
$3 \sin x + 4 \cos y = 5$, $4 \sin y + 3 \cos x = 2$ How to find $\sin x$, $\sin y$, $\cos x$, $\cos y$, 2020 contest question https://www.cemc.uwaterloo.ca/contests/past_contests/2020/2020CSMC.pdf
https://www.cemc.uwaterloo.ca/contests/past_contests/2020/2020CSMCSolution.pdf
Question 5 from 2020 CSMC math contest:
$$3 \sin x + 4 \cos y = 5,$$ $$4 \sin y + 3 \cos x = 2.$$
Find $\sin(x+y)$.
The solution page added that it is also possible to find $\sin x$, $\sin y$, $\cos x$, $\cos y$ with a different approach but I could not figure it out.
|
Hint: starting with $\sin x \cos y + \cos x \sin y=\frac{1}{6}$ and using that $\cos y=\frac{5-3\sin x}{4}$, $\sin y=\frac{2-3\cos x}{4}$ we can obtain
$$ \sin x \cdot (5-3\sin x)+\cos x \cdot (2-3\cos x)=\frac{2}{3}$$
$$ 5\sin x+2\cos x =\frac{11}{3} \implies \cos x=\frac{11}{6}-2.5\sin x $$We also have that $\cos^2 x+\sin^2x=1$ Substitute $\cos x$ and solve quadratic to find $\sin x$.
|
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|
How to use the epsilon delta definition to prove that $\lim\limits_{x\to 1} \frac{x^3-1}{x-1} = 3$
Not sure if I am doing this right, however, this is what I have:
Let $\epsilon > 0$. We need to find a $\delta > 0$ such that $0<|x-1|<\delta$ leads to the conclusion $|f(x)-3|<\epsilon$.
$$|\frac{x^3-1}{x-1} - 3| < \epsilon$$
We know that $x^3-1 = (x-1)(x^2+x+1)$, so
$$|x^2+x+1-3|<\epsilon$$
$$|(x+2)(x-1)|<\epsilon$$
Do I need to use the triangle inequality? Not really sure where to go from here. Any help is appreciated.
|
We must find a $\delta > 0$ such that $|x-1|<\delta \implies \left|\frac {x^3 - 1}{x-1} - 3\right| < \epsilon$
$\left|\frac {x^3 - 1}{x-1} - 3\right|\\
\left|\frac {x^3 - 3x + 2}{x-1}\right|\\
\left|\frac {(x-1)^2(x+2)}{x-1}\right|$
$\left|\frac {(x-1)^2}{x-1}\right| = |x-1|<\delta$ when $x\ne 1$
Let $\delta \le 1$ then $|x+2| \le 4$
$\left|\frac {(x-1)^2(x+2)}{x-1}\right| < 4\delta$
$\delta = \min(1,\frac {\epsilon}{4})$
|
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|
finding the argument of $1-\cos 2\theta -i\sin 2 \theta$ "Let z = $1-\cos 2\theta -i\sin 2 \theta$, $0\leq \theta \leq \pi$. Find the modulus and argument of z in terms of $\theta$ in their simplest forms."
The modulus is pretty easy, I got $2\sin \theta$ which is correct.
The problem is, when finding the argument of z, I got $\tan \alpha = \frac{-1}{\tan \theta}$, where $\alpha$ is the argument.
the correct answer is $\alpha = \theta - \pi/2$, however I am curious why $\alpha = \theta + \pi/2$ doesn't work. This also fits the equation above, but is wrong. Can someone explain?
|
$$z = 1-\cos 2\theta -i\sin 2 \theta$$
Modulus
$$|z| = \sqrt{(1-\cos 2 \theta)^2 + (-\sin 2 \theta)^2} = \\
= \sqrt{1 + \cos^2 2 \theta - 2 \cos 2 \theta + \sin^2 2 \theta} = \\
= \sqrt{2 - 2 \cos 2 \theta} = \\
= \sqrt{2 - 2 \cos^2 \theta + 2 \sin^2 \theta} = \\
= \sqrt{2 - 2 \cos^2 \theta + 2 (1-\cos^2 \theta)} = \\
= \sqrt{2 - 2 \cos^2 \theta + 2 -2\cos^2 \theta} = \\
= \sqrt{4 - 4 \cos^2 \theta} = \\
= 2 \sqrt{1 - \cos^2 \theta} = \\
2 |\sin\theta|.$$
Argument
First of all, notice that
$$\eta = \arctan\frac{- \sin 2\theta}{1 - \cos 2\theta}
=\arctan\frac{- 2 \sin \theta \cos \theta}{1 - \cos^2 \theta + \sin^2 \theta}
=\arctan\frac{- 2 \sin \theta \cos \theta}{2\sin^2 \theta} = \\
= -\arctan\text{cotan} \theta = \theta - \frac{\pi}{2}.
$$
since:
$$\arctan(\text{cotan} \beta) = \frac{\pi}{2} - \beta.$$
Additionally, notice that:
$$\mathcal{Re}[z] = 1-\cos 2 \theta \geq 0 ~\forall \theta \in [0, \pi]$$
Then, in this case, under the assumption that $\arg{z} \in (-\pi, \pi]$:
$$\arg{z} = \eta = \theta - \frac{\pi}{2}.$$
Why does $\alpha = \theta + \frac{\pi}{2}$ not work?
First of all, notice that:
$$\mathcal{Im}[z] = -\sin 2 \theta \geq 0 ~\forall \theta \in \left[\frac{\pi}{2}, \pi\right]$$
Hence, for $\theta \in \left[\frac{\pi}{2}, \pi\right]$, $z$ represents a vector in the first quadrant of the complex plane. This means that its argument must be positive and less that $\frac{\pi}{2}$. Indeed:
$$\eta = \theta - \frac{\pi}{2} \in \left[0, \frac{\pi}{2}\right] ~\forall \theta \in \left[\frac{\pi}{2}, \pi\right],$$
while
$$\alpha = \theta + \frac{\pi}{2} \in \left[\pi, 3\frac{\pi}{2}\right] ~\forall \theta \in \left[\frac{\pi}{2}, \pi\right].$$
The last shows that $\alpha = \theta + \frac{\pi}{2}$ does not work.
|
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|
If $\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}$, prove that $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
Problem: Let $a, b, c, x, y, z$ be real numbers
such that $abc \ne 0$, $ ~ a + b + c \ne 0$, $~ (a^2 - bc)(b^2 - ca)(c^2 - ab)\ne 0$, $xyz\ne0$, $x+y+z\ne0$, $(x-y)^2+(y-z)^2+(z-x)^2\ne0$
and $$\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab} \ne 0.$$
Prove that $$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}.$$
The converse can be proved easily.
My Attempt:
$
\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}\\
\Rightarrow\frac{x^2-yz-y^2+zx}{a^2-bc-b^2+ca}=\frac{y^2-zx-z^2+xy}{b^2-ca-c^2+ab}=\frac{z^2-xy-x^2+yz}{c^2-ab-a^2+bc}\ [By\ Addendo]\\\Rightarrow\frac{x-y}{a-b}\cdot\frac{x+y+z}{a+b+c}=\frac{y-z}{b-c}\cdot\frac{x+y+z}{a+b+c}=\frac{z-x}{c-a}\cdot\frac{x+y+z}{a+b+c}\\\Rightarrow\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}\ \left[Considering\ (x+y+z)\ and\ (a+b+c)\ \neq0\right]
$
Known Exceptional Counter-examples:-
Macavity: $a=1, b=2, c=3, x=y=z$
River Li: $a = 1, b=2, c = -3, x = 7, y = 7, z = -14$
The conditions in the question have been edited to exclude exceptional counter-examples.
|
@Jean Marie and @orangeskid gave very nice proofs.
Their idea, in algebraic manner, leads to the following proof:
We have the following identity
$$(p^2 - qr)^2 - (q^2 - rp)(r^2 - pq)
= p(p + q + r)(p^2 + q^2 + r^2 - pq - qr - rp).$$
Using this identity, we have
$$\frac{(x^2 - yz)^2 - (y^2 - zx)(z^2 - xy)}{(a^2 - bc)^2 - (b^2 - ca)(c^2 - ab)}
= k\,\frac{x}{a}, \tag{1}$$
and
$$\frac{(y^2 - zx)^2 - (z^2 - xy)(x^2 - yz)}{(b^2 - ca)^2 - (c^2 - ab)(a^2 - bc)}
= k\, \frac{y}{b}, \tag{2}$$
and
$$\frac{(z^2 - xy)^2 - (x^2 - yz)(y^2 - zx)}{(c^2 - ab)^2 - (a^2 - bc)(b^2 - ca)}
= k \, \frac{z}{c}, \tag{3}$$
where
$$k = \frac{(x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)}{(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)}.$$
Note: One can prove $k \ne 0$.
See the remarks at the end.
We are done.
Remarks:
If $k = 0$, we have $x + y + z = 0$.
Using $z = -x - y$, we have
$$\frac{x^2 + xy + y^2}{a^2-bc}=\frac{x^2 + xy + y^2}{b^2-ca}=\frac{x^2 + xy + y^2}{c^2-ab} \ne 0$$
which results in
$a^2 - bc = b^2 - ca = c^2 - ab$
and thus $(a-b)(a + b + c) = 0$ and $(b - c)(a + b + c) = 0$.
Thus, $a = b = c$ which contradicts the given conditions.
|
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"url": "https://math.stackexchange.com/questions/4286648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Dealing more efficiently with fractional forms in system of equations As an example, suppose we have to solve the following system of two equations and two unknowns:
$$
\begin{cases}
-\frac{10}{x}-\frac{8}{y} &= \frac{8}{3} \\
-\frac{6}{x}+\frac{6}{y} &= -\frac{1}{3}
\end{cases}
$$
My approach and solution
I opted to solve it by combination, referring to the first equation as (1) and the second equation as (2), I started by eliminating the y's:
\begin{align}
\frac{6}{8}(1)+(2): -\frac{60}{8x}-\frac{6}{x} &= \frac{6}{3}-\frac{1}{3} \\
\frac{-60-48}{8x} &= \frac{5}{3} \tag{*}\\
40x &=3(-108) \tag{**} \\
x &=-\frac{81}{10} \tag{1'}
\end{align}
Substituting (1') into (2):
\begin{align}
\frac{100}{81}-\frac{8}{y}&=\frac{8}{3} \\
y &= -\frac{162}{29}
\end{align}
So I find the tuple of $(-\frac{81}{10};-\frac{162}{29})$ as solution.
Questions
*
*For such systems, should we not be concerned about the conditions of existence of the system? Namely, in this case both $x\neq 0$ and $y\neq 0,$ in the same way that we would do when solving an equation. If yes, how do we formally write the domain of existence for a system?
*Is my transition from step $(*)$ to $(**)$ allowed? My understanding is that, yes for all $x\neq 0.$
*I am really eager to learn whether there are simple ideas that simplify the system (and other similar systems) before we start solving it. As shown in my approach, it was slightly awkward dealing with the fractions throughout and the "large" numbers, which inherently may render the approach more prone to mistakes. Any alternative, quicker approach (I only know by combination and substitution) would be much appreciated.
|
$$
\begin{cases}
-\frac{10}{x}-\frac{8}{y} &= \frac{8}{3},\quad{|}\cdot{3}\\
-\frac{6}{x}+\frac{6}{y} &= -\frac{1}{3}.\quad{|}\cdot{(-5)}
\end{cases}
\iff
\begin{cases}
-\frac{30}{x}-\frac{24}{y} &= \frac{24}{3},\\
\frac{30}{x}-\frac{30}{y} &= \frac{5}{3}.
\end{cases}
\iff
$$
$$
\iff
\begin{cases}
-\frac{54}{y} &= \frac{29}{3},\\
\frac{30}{x}-\frac{30}{y} &= \frac{5}{3}.
\end{cases}
\iff
\begin{cases}
{y} = \frac{-162}{29},\\
\frac{30}{x}-\frac{30}{y} &= \frac{5}{3}.
\end{cases}
\iff
\begin{cases}
{y}=\frac{-162}{29},\\
{x}=\frac{-81}{10}.
\end{cases}
$$
Also the parameters $x\ne{0}$, $y\ne{0}$.
Good luck!
|
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"timestamp": "2023-03-29T00:00:00",
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|
General form for $\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$ I'm wondering if there is a general form for the following sum:
$$\sum_{n=1}^{\infty} (-1)^n \left(m n \, \text{arccoth} \, (m n) - 1\right)$$ for $m \in \mathbb{N}$
I have obtained the following closed-forms for these special cases:
Where $G$ is Catalan's constant and $\text{Cl}_2$ is the Clausen function of order 2.
$$\sum_{n=1}^{\infty}(-1)^n \left(2n \, \text{arccoth} \, (2n) - 1\right) = \frac{1}{2} - \frac{2G}{\pi}$$
$$\sum_{n=1}^{\infty} (-1)^{n}\left( 3n \, \text{arccoth} \, (3n)-1\right) = \frac{1}{2} - \frac{5}{2\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{3}\right) + \frac{1}{4} \ln (3)$$
$$\sum_{n=1}^{\infty} (-1)^n \left(4n \, \text{arccoth} \, (4n) - 1\right) = \frac{1}{2}+ \frac{G}{\pi} - \frac{4}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{1}{4} \ln \left(3- 2\sqrt{2}\right)$$
etc.
Given that $G = \text{Cl}_2 \left(\frac{\pi}{2}\right)$, I am curious to know if the general sum is expressible in terms of the Clausen function. These above sums were determined by using the Mittag-Leffler expansion of $\csc (z)$, i.e $\csc(z) = \frac{1}{z} + 2z \sum_{n=1}^{\infty} (-1)^n \frac{1}{z^2 - \left(\pi n\right)^2}$ and substituting it into the integral $\int_{0}^{\pi/m} x \csc (x) \, dx$
If one uses the following other method, we can determine the odd and even terms of the sums
$$\sum_{n=1}^{\infty} \left( 4n \, \text{arccoth} \, (4n)-1\right) = \frac{1}{2} - \frac{G}{\pi}- \frac{1}{4} \ln (2)$$
$$\sum_{n=1}^{\infty} \left( (4n-2) \, \text{arccoth} \, (4n-2) - 1\right) = \frac{G}{\pi} - \frac{1}{4} \ln (2)$$
$$\sum_{n=1}^{\infty} \left(6n \, \text{arccoth} \, (6n) - 1\right) = \frac{1}{2} - \frac{3}{2\pi} \, \text{Cl}_2 \left( \frac{\pi}{3}\right)$$
$$\sum_{n=1}^{\infty} \left( (6n-3) \, \text{arccoth} \, (6n-3) - 1\right) = \frac{1}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{3}\right) - \frac{1}{4} \ln(3)$$
$$\sum_{n=1}^{\infty} \left( 8n \, \text{arccoth} \, (8n) - 1\right) = \frac{1}{2} - \frac{2}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{1}{4} \ln (2-\sqrt{2})$$
$$\sum_{n=1}^{\infty} \left( (8n-4) \, \text{arccoth} \, (8n-4) - 1\right) = \frac{2}{\pi} \, \text{Cl}_2 \, \left(\frac{\pi}{4}\right) - \frac{G}{\pi} - \frac{1}{4} \ln(2+\sqrt{2})$$
EDIT
I am now interested in
$$\sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right)$$
for $|m|>1$ as asked here.
Here is how I originally proved it for the odd terms:
For the first odd term sum, begin with the known result $$2G = \int_{0}^{\frac{\pi}{2}} x \csc (x) \, dx$$
then integrate by parts to get:
$$2G = \int_{0}^{\frac{\pi}{2}} \ln \left(\cot (x) + \csc (x) \right) \, dx = \int_{0}^{\frac{\pi}{2}} \ln (\cos (x) + 1) \, dx - \int_{0}^{\frac{\pi}{2}} \ln (\sin (x)) \,dx$$
Then use the well-known result $\int_{0}^{\frac{\pi}{2}} \ln (\sin(x)) \,dx = - \frac{\pi}{2} \ln (2)$ and use the identity $\cos (x) + 1 = 2 \cos^2\left( \frac{x}{2}\right)$ and make the substitution $\frac{x}{2} = u$.
$$\implies 2G - \pi \ln (2) = 4 \int_{0}^{\frac{\pi}{4}} \ln (\cos (u)) \, du$$
Now use the Weierstrass product for $\cos (z)$, namely $\cos(z) = \prod_{n=1}^{\infty} \left(1-\frac{4z^2}{\pi^2 (2n-1)^2}\right)$ to obtain:
$$2G - \pi \ln (2) = 4 \sum_{n=1}^{\infty} \int_{0}^{\frac{\pi}{4}} \ln \left( 1-\frac{4u^2}{\pi^2 (2n-1)^2}\right) \, du$$
After integrating, obtain $\pi \sum_{n=1}^{\infty} \ln \left(1-\frac{1}{4(1-2n)^2}\right) = -\frac{\pi}{2} \ln (2)$ and the result quickly follows. The other odd sums are the same idea. The even sums just comes from combining the two results from the alternating sum and the odd term sum.
|
$\newcommand{\D}{\mathfrak{D}}\newcommand{\d}{\,\mathrm{d}}$Contour integration naturally leads us to the situation of Random Variableβs answer.
Fix a real $s>1$ and call this series $S$. Let $\log$ be the principal logarithm, so that $z\mapsto\log\frac{zs+1}{zs-1}$ is holomorphic on $\Bbb C\setminus[-s^{-1},s^{-1}]$. Note that this interval is strictly contained in the open unit disk since $s>1$.
For $0<\epsilon<s^{-1}$ let $\D_\epsilon$ be the contour which runs: $-s^{-1}+i\epsilon\to s^{-1}+i\epsilon$ in a straight line, round in a clockwise semicircle of radius $\epsilon$ to $s^{-1}-i\epsilon$, from there to $-s^{-1}-i\epsilon$ in a straight line and back in another clockwise semicircle to $-s^{-1}+i\epsilon$: a βdogbone contourβ.
Define $f:\Bbb C\to\hat{\Bbb C}$ by: $$z\mapsto\frac{zs}{2}\log\frac{zs+1}{zs-1}-1$$
Since $f$ satisfies the growth condition $zf(z)\to0$ uniformly as $|z|\to\infty$ we have, using the usual large box contour with corners $2\pi(N+1/2)(\pm1\pm i)$ for large $N$: $$4\pi i S=\oint_{\D_\epsilon}\pi\csc(\pi z)f(z)\d z$$For any such $\epsilon$.
Extracting the β$-1$β out of $f$, $\oint_{\D_\epsilon}\pi\csc(\pi z)(-1)\d z=2\pi i$ always holds. For $-1<x<1$ let $x^{\pm}$ denote a limit as $z\to x$ from above and below the real axis. Then the branch jump can be, with simple algebra, found to be: $$\log\frac{x^+s+1}{x^+s-1}-\log\frac{x^-s+1}{x^-s-1}=-2\pi i$$Since we are free to take $\epsilon\to0^+$, this information gives - after dividing by $4\pi i$ - the equation: $$S=\frac{1}{2}-\frac{\pi s}{4}\int_{-s^{-1}}^{s^{-1}}x\csc\pi x\d x$$
The exact same method gives: $$Sβ=\frac{1}{2}-\frac{\pi s}{4}\int_{-s^{-1}}^{s^{-1}}x\cot\pi x\d x$$If $Sβ$ is the non-alternating series.
Then Random Variableβs answer(s) expresses these last integral via the Clausen function.
I find it prettier to rewrite these as: $$\begin{align}2\sum_{n\ge1}(\pi ns\cdot\operatorname{arcoth}(\pi ns)-1)&=1-\int_0^{s^{-1}}\phi\cot\phi\d\phi\\2\sum_{n\ge1}(-1)^n(\pi ns\cdot\operatorname{arcoth}(\pi ns)-1)&=1-\int_0^{s^{-1}}\phi\cdot\csc\phi\d\phi\end{align}$$
|
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|
$3$ Bracket Knockout Tournament Probability $12$ people in $3$ brackets ($A,B$, and $C$) compete against each other in a knockout tournament. The final round has $4$ contestants, the $3$ winners from brackets $A,B$ and $C$, plus $1$ of the dropouts from brackets $A$ and $B$ brought back into the game at random. What is the probability of winning for a player in group $A$ or $B$ (assuming equal chance to advance to the next stage of the tournament and equal chance to be brought back in).
The way I though of this was that there are two ways that a person in Group $A$ or $B$ can win. They can win the tournament by simply winning all their matches, or by losing, being brought back and winning the final round. Adding these two probabilities should give the answer.
I calculated their chance of winning by simply winning all their matches as $\frac1{12}$, since each contestant has equal chance to advance and there are $12$.
For the other scenario, I said that there is a $\frac12$ chance they lose their first match (since all participants have equal chance of advancing) , multiplied by a $\frac16$ chance they get voted back in, (since of the original $8$ people from $A$ and $B$ only $2$ will advance and there will be $6$ dropouts and all dropouts have equal probability of being brought back). Then multiplied by $\frac14$ since there are $4$ participants in the final round and each has equal chance of winning, So $\frac12 \cdot \frac16 \cdot \frac14$.
Overall I got $\frac1{12} + \frac12 \cdot \frac16 \cdot \frac14=\frac5{48}$, which is not the answer.
The options for the answer are:
*
*$\frac3{16}$
*$\frac3{32}$
*$\frac1{16}$
*$\frac18$.
Thanks
|
*
*A random player's chance of winning four games in a row is $\left(\frac{1}{2}\right)^4 = \frac{1}{16} \neq \frac{1}{12}$.
*You haven't counted the players who win their first match and lose their second (and are awarded the final random spot).
Its hard to count all the possibilities, but here we go:
Win all games: $\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2} = \frac{1}{16}$
Lose, Lottery Spot, Win, Win: $\frac{1}{2}\cdot \frac{1}{6}\cdot \frac{1}{2}\cdot \frac{1}{2} = \frac{1}{48}$
Win, Lose, Lottery Spot, Win, Win: $\frac{1}{2}\cdot\frac{1}{2}\cdot \frac{1}{6}\cdot \frac{1}{2}\cdot \frac{1}{2} = \frac{1}{96}$
Add these up to get $\frac{9}{96} = \frac{3}{32}$.
Alternatively,
$$\underbrace{\frac{3}{8}}_{\text{get to the final from A/B}} \cdot \underbrace{\frac{1}{4}}_{\text{win the final}} = \frac{3}{32}$$
Edit (to address OP comment):
Checking our answer against the Law of Total Probability:
The lottery mechanism makes it more difficult to check each player's chance of winning. The C players must win all their games. Each of the four C players has a one-sixteenth chance. The A and B players have better chances. Eight players must first earn one of three finals spots and then win the finals:
$$\underbrace{4\cdot\frac{1}{16}}_{C} + \underbrace{8 \cdot \frac{3}{32}}_{A/B} = \frac{1}{4} + \frac{3}{4} = 1$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Understanding Heron's formula proof
I was trying to understand the proof of Heron's formula. I understood everything up to step $(d)$ except how they arrived at the expression ($a$+$b$+$c$)($b$+$c$-$a$)($a$+$b$-$c$)($a$-$b$+$c$) and the steps afterwards I still don't understand I would appreciate if someone gave me insight into how the expressions came about
|
Hint:
\begin{align}
h^2
&=b^2 - d^2\\
&=b^2 - \frac{(b^2+c^2-a^2)^2}{4c^2}\\
&=\frac{(2bc)^2-(b^2+c^2-a^2)^2}{4c^2}\\
&=\frac{(2bc+b^2+c^2-a^2)(2bc-b^2-c^2+a^2)}{4c^2}\\
&=\frac{\big[(b+c)^2-a^2\big]\big[-(b-c)^2+a^2\big]}{4c^2}\\
&=\frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{4c^2}
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $(x+1)^6=x^6$ then prove that $x=\frac{-1}2-i\cot (\frac{\theta}2)$ where $\theta=\frac{2k\pi}6$, $k=0,1,2,3,4,5$ QUESTION
If $(x+1)^6=x^6$ then prove that $x=\Large{\frac{-1}2}$$-i\cot (\large\frac\theta2)$ where $\theta=\Large\frac{2k\pi}6$, $k=0,1,2,3,4,5$
How can we prove this question using complex numbers. We need to prove it using roots of complex numbers.
If we consider$$\frac{(x+1)^{6}}{x^{6}} = 1$$
then $$\frac{(x+1)}{x}=(1)^\frac{1}{6} = \cos (\theta)^\frac{1}{6} + i \sin (\theta)^\frac{1}{6}$$
On further simplification I get the answer$$x=\frac{-1}{2}-\frac{i\cot(\theta/2)}{2}$$
But we need to prove $x=\frac{-1}{2}
-i\cot (\theta/2)$. What can be the error?
My solved example is here
|
First note that $k = 0$ does not give a solution, as $cot(0)$ is undefined. This should match with the intuition that your equation
$$(x+1)^6 = x^6 \Longrightarrow 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1 = 0$$
should only have up to 5 roots by the fundamental theorem of algebra.
Otherwise, your solution is sound. You can verify that the roots are
$$x_1 = -\frac{1}{2} - i \frac{\sqrt{3}}{2} = -\frac{1}{2} - \frac{i}{2} \cot(\frac{\pi}{6}),$$
$$x_2 = -\frac{1}{2} - i \frac{1}{2\sqrt{3}} = -\frac{1}{2} - \frac{i}{2} \cot(\frac{2\pi}{6}),$$
$$x_3 = -\frac{1}{2} = -\frac{1}{2} - \frac{i}{2} \cot(\frac{3\pi}{6}),$$
$$x_4 = -\frac{1}{2} + i\frac{1}{2\sqrt{3}} = -\frac{1}{2} - \frac{i}{2} \cot(\frac{4\pi}{6}),$$
$$x_5 = -\frac{1}{2} + i\frac{\sqrt{3}}{2} = -\frac{1}{2} - \frac{i}{2} \cot(\frac{5\pi}{6}).$$
|
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|
Number of real solutions of $\begin{array}{r} {\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]} =\frac{3 x-1}{2} \end{array}$ Solve for $x \in \mathbb{R}$ $$\begin{array}{r}
{\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]}
=\frac{3 x-1}{2}
\end{array}$$ where $[x]$ denotes greatest integer less than or equal to $x$.
My try:
Letting $a=\frac{2x+1}{3}$ we get
$$\left[a\right]+\left[a+\frac{1}{2}\right]=\frac{9 a-5}{4} \tag{1}$$
Now knowing that:
$$a-1<[a]\leq a$$ and $$a-\frac{1}{2}<\left[a+\frac{1}{2}\right]\leq a+\frac{1}{2}$$
Adding both the above inequalities and using $(1)$ we get
$$2a-\frac{3}{2}<\frac{9a-5}{4}\leq 2a+\frac{1}{2}$$
So we get $$a \in (-1, 7]$$
Any help here?
|
HINT: The RHS $\frac{3x-1}{2}$ of the original equation must be an integer, which implies that $3x$ has to be an odd integer [typo noted via comment below]. Your bounds for $a$ leave only a few values to check for $x$, in particular $-2<x<9$; this an $3x$ an odd integer leaves only to check each $x$ satisfying $x= \frac{k}{3}$; $k$ is an odd integer satisfying $-5 <k < 27$.
|
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|
Use Induction to prove recurrence
if the above screenshot is not visible, here is the text format:
Question:
Solve the following recurrence and prove your result is correct using induction:
$a_1 = 0$
$a_n = 3(a_{n-1}) + 4^{n}$ for $n>=2$
Use induction to prove this recursive sequence.
So my approach was that, I plug in the $a_1 = 0$ into $a_2 = 3(0) + 4^{2} = 4^{2}$
and then
$a_3 = 3(4^{2}) + 4^{3}$
$a_4 = 3(3(4^{2}) + 4^{3}) + 4^{4} = 3^{2} *4^{2} + 3 * 4^{3} +4^{4}$
$a_5 = 3(3^{2} *4^{2} + 3 * 4^{3} +4^{4}) + 4^{5} = 3^{3} *4^{2} + 3^{2} * 4^{3} +3*4^{4} + 4^{5}$
$a_6 = 3^{4} *4^{2} + 3^{3} * 4^{3} +3^{2}*4^{4} + 3*4^{5} + 4^{6}$
...
$a_n = 3^{n-2} *4^{2} + 3^{n-3} * 4^{3} + ... + 3*4^{n-1} + 4^{n} = \sum_{k=2}^{n} 3^{n-k}*4^{k} $
now through induction, I'm not sure how to get there. So far what I know is that:
$a_{n+1} = 3(a_{n}) + 4^{n+1} = \sum_{k=2}^{n+1} 3^{n+1-k}*4^{k}$
= $3(\sum_{k=2}^{n} 3^{n-k}*4^{k}) + 4^{n+1} = \sum_{k=2}^{n+1} 3^{n+1-k}*4^{k}$
= $(\sum_{k=2}^{n} 3*3^{n-k}*4^{k}) + 4^{n+1} = \sum_{k=2}^{n+1} 3^{n+1-k}*4^{k}$
= $(\sum_{k=2}^{n} 3^{n+1-k}*4^{k}) + 4^{n+1} = \sum_{k=2}^{n+1} 3^{n+1-k}*4^{k}$
but then I'm stuck, because I cannot simplify it to make them equal to each other, or I'm in the wrong direction?
|
Here's a useful "trick".
If
$a_n
=ua_{n-1}+v^n
$
then
(here comes the trick),
dividing by $u^n$,
$\dfrac{a_n}{u^n}
=\dfrac{ua_{n-1}}{u^n}+\dfrac{v^n}{u^n}
=\dfrac{a_{n-1}}{u^{n-1}}+(v/u)^n
$.
Let $b_n = \dfrac{a_n}{u^n}$.
Then
$b_n
=b_{n-1}+r^n
$
where $r = v/u$
or
$b_n-b_{n-1}
=r^n
$.
This becomes a telescoping sum,
so
$\begin{array}\\
b_m-b_0
&=\sum_{n=1}^m (b_n-b_{n-1})\\
&=\sum_{n=1}^m r^n\\
&=\dfrac{r-r^{m+1}}{1-r}
\qquad\text{(if } r \ne 1.
\text{ If }r=1, \text{the sum is }m.)\\
&=\dfrac{\frac{v}{u}-(\frac{v}{u})^{m+1}}{1-\frac{v}{u}}\\
&=\dfrac{v-\frac{v^{m+1}}{u^m}}{u-v}\\
\text{so}\\
\dfrac{a_m}{u^m}-a_0
&=\dfrac{v-\frac{v^{m+1}}{u^m}}{u-v}\\
\text{or}\\
a_m
&=u^ma_0+\dfrac{vu^m-v^{m+1}}{u-v}\\
&=u^ma_0+\dfrac{v(u^m-v^{m})}{u-v}\\
\end{array}
$
Note that,
if the $a$s start at $k$
instead of $0$,
we can do
$\begin{array}\\
b_m-b_k
&=\sum_{n=k+1}^m (b_n-b_{n-1})\\
&=\sum_{n=k+1}^m r^n\\
&=\dfrac{r^{k+1}-r^{m+1}}{1-r}
\qquad\text{(if } r \ne 1.
\text{ If }r=1, \text{the sum is }m-k.)\\
&=\dfrac{(\frac{v}{u})^{k+1}-(\frac{v}{u})^{m+1}}{1-\frac{v}{u}}\\
&=\dfrac{\frac{v^{k+1}}{u^k}-\frac{v^{m+1}}{u^m}}{u-v}\\
\text{so}\\
\dfrac{a_m}{u^m}-\dfrac{a_k}{u^k}
&=\dfrac{\frac{v^{k+1}}{u^k}-\frac{v^{m+1}}{u^m}}{u-v}\\
\text{or}\\
a_m
&=u^{m-k}a_k+\dfrac{v^{k+1}u^{m-k}-v^{m+1}}{u-v}\\
&=u^{m-k}a_k+\dfrac{v^{k+1}(u^{m-k}-v^{m-k})}{u-v}\\
\end{array}
$
If this is for
$k=1$,
this becomes
$a_m
=u^{m-1}a_1+\dfrac{v^{2}(u^{m-1}-v^{m-1})}{u-v}
$.
|
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|
Rationalize the denominator of $\frac{4}{9-3\sqrt[3]{3} + \sqrt[3]{9}}$ Rationalize the denominator of $\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}}$
I keep making a mess of this. I tried vewing the denominator as
$a +\sqrt[3]{9}$, where $a=9-3\sqrt[3]{3}$ and secondly as
$b -3\sqrt[3]{3}+\sqrt[3]{9}$, where $b=9$.
Then using the sum and differences in cubes fratorization but this keeps adding radicals to the denominator.
How should I approach this/where could I be going wrong?
|
To solve the general case, use the formula for the rationalization of 3 cube roots. $$\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}=\frac{\left(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}+\sqrt[3]{c^{2}}-\sqrt[3]{ab}-\sqrt[3]{ac}-\sqrt[3]{bc}\right)\left(\left(3\sqrt[3]{abc}+a+b+c\right)^{2}-3\left(a+b+c\right)\sqrt[3]{abc}\right)}{\left(a+b+c\right)^{3}-27abc}$$ Your particular problem appears to be rigged such that the answer simplifies greatly. $$\frac{4}{9-3\sqrt[3]{3}+\sqrt[3]{9}}=\frac{2\sqrt[3]{3}+6}{15}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the easier way to find the circle given three points? Given three points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$, if $$\frac{y_2-y_1}{x_2-x_1} \neq \frac{y_3-y_2}{x_3-x_2} \neq \frac{y_1-y_3}{x_1-x_3},$$ then there will be a circle passing through them. The general form of the circle is $$x^2 + y^2 + dx + ey + f = 0.$$ By substituting $x = x_i$ and $y = y_i$, there will be a system of equation in three variables, that is:
\begin{align*}
\begin{pmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end{pmatrix}
\begin{pmatrix}
d \\
e \\
f
\end{pmatrix} &=
\begin{pmatrix}
-\left(x_1^2+y_1^2\right) \\
-\left(x_2^2+y_2^2\right) \\
-\left(x_3^2+y_3^2\right)
\end{pmatrix}.
\end{align*}
Solving this system gives the solution
\begin{align*}
d &= \frac{(x_3^2 + y_3^2 -x_1^2+y_1^2)}{x_1 - x_3} - e\left(\frac{y_1 - y_3}{x_1 - x_3}\right) \\
e &= \frac{(x_3^2 + y_3^2 -x_2^2+y_2^2)(x_1-x_3) - (x_3^2 + y_3^2 -x_1^2+y_1^2)(x_2-x_3)}{(y_2-y_3)(x_1-x_3) - (y_1-y_3)(x_2-x_3)} \\
f &= \frac{-(x_3^2 + y_3^2)(x_1-x_3) - (y_1-y_3)x_3}{x_1 - x_3} - e\left(\frac{y_3(x_1 - x_3) - x_3(y_1 - y_3)}{x_1 - x_3}\right)
\end{align*}
As there are a lot of things going around, the solution is prone to errors. Maybe this solution also has an error.
Is there a better way to solve for the equation of the circle?
|
I prefer a different approach.
You know that the center of the circle is the intersection between any two of the three perpendicular bisectors.
Therefore,
$$(x - x_1)^2 + (y - y_1)^2 = (x - x_2)^2 + (y - y_2)^2 \tag1 $$
and
$$(x - x_1)^2 + (y - y_1)^2 = (x - x_3)^2 + (y - y_3)^2. \tag2 $$
Let $f_1(x) = (x - x_1)^2 - (x - x_2)^2. ~~: ~~$ 1st degree polynomial
Let $f_2(x) = (x - x_1)^2 - (x - x_3)^2. ~~: ~~$ 1st degree polynomial
Let $g_1(y) = (y - y_2)^2 - (y - y_1)^2. ~~: ~~$ 1st degree polynomial
Let $g_2(y) = (y - y_3)^2 - (y - y_1)^2. ~~: ~~$ 1st degree polynomial
You can express $g_2(y)$ as $Ag_1(y) + B$, where $A,B$ are fixed real numbers.
This implies that
$$f_2(x) = g_2(y) = Ag_1(y) + B = Af_1(x) + B. \tag3 $$
Equation (3) above gives you an equation solely in $x$, that you can solve to determine the $(x = x_0)$ coordinate of the center of the circle. Once this is determined, you can compute $y_0$ via $g_1(y_0) = f_1(x_0).$
|
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|
How to apply the AM-GM relation in this inequality? Here is an inequality that I came across recently.
$((a+b)/2)^m$ $\le$ $(a^m + b^m)/2$, if $m<0$ or $m>1.$
Equality happens when $m = 0$ or $m = 1.$ Inequality reverses if $m$ is between 0 and 1.
I was able to prove this for $m = 2$. Here is what I did:
We know that $(a+b)^2/2 = (a^2 + b^2)/2 + ab.$ ----- (1)
Also, AM of $a^2$ and $b^2$ is greater than or equal to GM of $a^2$ and $b^2.$ That means $(a^2 + b^2)/2$ $\ge$ $ab.$
=> (1) --> $(a+b)^2/2$ $\le$ $(a^2 + b^2)/2 + (a^2 + b^2)/2.$
=> $(a+b)^2/2$ $\le$ $a^2 + b^2.$ Or, $((a+b)/2)^2$ $\le$ $(a^2 + b^2)/2.$
Now, if I try to generalize by replacing '2' with 'm', I'm unable to use the AM-GM inequality to get the desired result.
How to apply the AM-GM inequality in the general case? Or is there any other way to prove the general result?
|
For $m>1$ (and $m\in\mathbb{Z}$), assume $0\le a\le b$ and let $c=\frac{a+b}{2}$ and $d=\frac{b-a}{2}$.
Then $a=c-d$ and $b=c+d$.
The LHS is $c^m$, and the RHS is $ \frac{(c-d)^m +(c+d)^m}{2}\ge c^m$ due to any negative terms being cancelled.
|
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|
Sum or bounds of $\sum\frac{1}{a^n+b^n}$ I will keep it short as I have made $0$ progress on the problem (both solving it and researching) and am just interested. Is there any closed form or non-trivial bounds (so tighter than $1 + \frac{1}{a - 1}$) on the sum $\sum\frac{1}{a^n + b^n}$ where $a > 1$? Clearly it converges by a comparison test.
Thank you for your time.
|
We can suppose $b<a$. IΒ΄ll assume too that the sum begins in $n=1$, so the obvious lower bound (with $b=0$) is $\sum_{n=1}^\infty\frac{1}{a^n}=\frac{1}{a-1}$. I donΒ΄t know if there is an exact way to find the sum, but one way I found to get upper/lower bounds as tight as you want uses this equality:
$$\frac{1}{a^n+b^n}=\frac{1}{a^n}-\frac{b^n}{a^n}\frac{1}{(a^n+b^n)}.$$
Using it, you can deduce:
$$\displaystyle\sum\frac{1}{a^n+b^n}=
\sum\frac{1}{a^n}-\sum\frac{b^n}{a^n}\frac{1}{(a^n+b^n)}=
\sum\frac{1}{a^n}-\sum\frac{b^n}{a^{2n}}+\sum\frac{b^{2n}}{a^{2n}}\frac{1}{a^n+b^n}
=$$
$$=\sum\frac{1}{a^n}-\sum\frac{b^n}{a^{2n}}+\sum\frac{b^{2n}}{a^{3n}}-\sum\frac{b^{3n}}{a^{3n}}\frac{1}{a^n+b^n}=\cdots$$
Calling $\displaystyle s_k=\sum\frac{b^{kn}}{a^{kn}}\frac{1}{(a^{n}+b^n)}$, itΒ΄s easy to see that $s_k$ tends to $0$ when $k$ goes to infinity (as the greater sequence $t_k=\sum\frac{b^{kn}}{a^{kn}}$ goes to $0$).
So, $\displaystyle\sum\frac{1}{a^n+b^n}=\sum\frac{1}{a^n}-\sum\frac{b^n}{a^{2n}}+\sum\frac{b^{2n}}{a^{3n}}+\dots=\sum_{k=0}^\infty(-1)^k\sum_{n=1}^\infty\frac{b^{kn}}{a^{(k+1)n}}=\sum_{k=0}^\infty(-1)^k\frac{\frac{b^{k}}{a^{(k+1)}}}{1-\frac{b^{k}}{a^{(k+1)}}}=$
$$=\sum_{k=0}^\infty(-1)^k\frac{b^k}{a^{k+1}-b^k}.$$
The difference between this sequence and the initial one is that this one is alternating, and its terms are decreasing in absolute value. This implies that, calling the partial sums $S_n=\sum_{k=0}^n(-1)^k\frac{b^k}{a^{k+1}-b^k}$, $S_n$ will be an upper bound for the limit when $n$ is even, and a lower bound when $n$ is odd. Thus for example you have the upper bound
$S_2=\frac{1}{a-1}-\frac{b}{a^2-b}+\frac{b^2}{a^3-b^2}$, it is smaller than the obvious upper bound $\frac{1}{a-1}$ (which is $S_0$).
In general taking $S_n$ with $n$ even and big enough will give you upper bounds as tight as you want, and the same for lower bounds taking $n$ odd.\
Edit: After thinking again this seems pretty useless because you will probably get similar bounds just using the obvious upper bounds $K_n:=\sum_{i=1}^n\frac{1}{a^i+b^i}+\frac{1}{a^{i+1}(a-1)}$.
|
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|
Decomposition of numbers: Show that there are positive integers $x, y$ and $z$, with $x$ and $y$ coprime, such that $a+b=x^2z\,$ and $\,c+d=y^2z$ The positive integers $a$, $b$, $c$ and $d$ satisfy the equation $$(ad-bc)^2=(a+b)(c+d).$$ Show that there are positive integers $x$, $y$ and $z$, with $x$ and $y$ coprime, such that $$a+b=x^2z~\text{ and }~c+d=y^2z.$$
I tried using the fact that $$(a+b)(c+d)=\text{lcm}(a+b,c+d)\gcd(a+b,c+d)$$ and by letting $m=\text{lcm}(a+b,c+d)$ and $z=\gcd(a+b,c+d)$, I got $$a+b=z\Big(\frac{a+b}{z}\Big)=z(a+b)\Big(\frac{m}{(ad-bc)^2}\Big)=z\Big(\frac{k(a+b)^2}{(ad-bc)^2}\Big),$$ since $m=k(a+b)$ for some integer $k$. However, I hit a road block in terms of simplifying further. Any help is appreciated! Thank you!
|
Write $a+b=vz,c+d=wz$. By the fundamental theirem of arithmatic, $v=p_1^{e_1}p_2^{e_2}..., w=q_1^{f_1}q_2^{f_2}...$. Since $v$ and $w$ are coprime, $p_i\neq q_j\forall i,j$. Finally,
$$(ad-bc)^2=(a+b)(c+d)=vwz^2$$
$$l^2:=\left(\frac{ad-bc}{z}\right)^2=vw$$
Now $l$ has its own prime factorization $r_1^{g_1}r_2^{g_2}...$, so we get
$$r_1^{2g_1}r_2^{2g_2}...=p_1^{e_1}p_2^{e_2}...q_1^{f_1}q_2^{f_2}...$$
Since each prime number appears at most once in $p_i, q_j$, we get that each $r_k$ is matched to exactly one $p_i$ or $q_i$. This means that each $e_i,f_j$ is matched to a $2g_k$, and is thus even. It follows that both $v$ and $w$ are square, that is, $v=x^2$ and $w=y^2$.
|
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|
How to calculate this improper integral? Calculate the improper integral
$$\displaystyle\int_0^{\infty}{\frac{1}{\theta}e^{\cos\theta}\sin(\sin\theta){d\theta}}$$
My try:
We know that for any $a\in\mathbb{C}$ the integral
$$\displaystyle\int_0^{\infty}e^{-ax^2}=\frac{1}{2}\sqrt{\frac{\pi}{a}}$$
Let $a=\cos\theta+i\sin\theta$ we know $$\displaystyle\int_0^{\infty}{e^{x^2\cos\theta}\sin(x^2\sin\theta){dx}}=\frac{\sqrt{\pi}}{2}\sin\frac{\theta}{2}$$
then $$\displaystyle\int_0^{\infty}{\frac{1}{\theta}e^{x^2\cos\theta}\sin(x^2\sin\theta){dx}}=\frac{\sqrt{\pi}}{2}\frac{\sin\frac{\theta}{2}}{\theta}$$
$$\displaystyle\int_0^{\infty}d\theta\displaystyle\int_0^{\infty}{\frac{1}{\theta}e^{x^2\cos\theta}\sin(x^2\sin\theta){dx}}=\displaystyle\int_0^{\infty}\frac{\sqrt{\pi}}{2}\frac{\sin\frac{\theta}{2}}{\theta}d\theta$$
Let $F(x)=\displaystyle\int_0^{\infty}{\frac{1}{\theta}e^{x^2\cos\theta}\sin(x^2\sin\theta){d\theta}}$, then the result equals to $F(1)$,But I don't know what to do next.
|
1st Solution. Define the sine integral by
$$ \operatorname{Si}(x) = \int_{0}^{x} \frac{\sin t}{t} \, \mathrm{d}t. $$
Using integartion by parts, it can be proved that
$$ \operatorname{Si}(x) = \frac{\pi}{2} + \mathcal{O}\left(\frac{1}{x}\right) \qquad\text{as } x \to \infty. $$
Now note that $e^{\cos\theta}\sin\sin\theta = \operatorname{Im}(e^{e^{i\theta}}-1) = \sum_{n=1}^{\infty} \frac{1}{n!}\sin(n\theta)$. Then by the Fubini's theorem, for $R > 0$,
\begin{align*}
\int_{0}^{R} \frac{e^{\cos\theta}\sin\sin\theta}{\theta} \, \mathrm{d}\theta
&= \int_{0}^{R} \frac{1}{\theta} \sum_{n=1}^{\infty} \frac{\sin(n\theta)}{n!} \, \mathrm{d}\theta \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \int_{0}^{R} \frac{\sin(n\theta)}{\theta} \, \mathrm{d}\theta \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \operatorname{Si}(nR) \\
&= \sum_{n=1}^{\infty} \frac{1}{n!} \left( \frac{\pi}{2} + \mathcal{O}\left( \frac{1}{nR} \right) \right) \\
&= \frac{\pi}{2}(e - 1) + \mathcal{O}\left(\frac{1}{R}\right).
\end{align*}
So by letting $R \to \infty$, the integral converges to
$$ \int_{0}^{\infty} \frac{e^{\cos\theta}\sin\sin\theta}{\theta} \, \mathrm{d}\theta = \frac{\pi}{2}(e-1). $$
2nd Solution. It is well-known that
$$ \lim_{N\to\infty} \sum_{k=-N}^{N} \frac{1}{z + 2\pi k} = \frac{1}{2}\cot\left(\frac{z}{2}\right). $$
Moreover, this convergence is locally uniform (in the sense that the difference between the limit and the $N$-th partial sum, when understood as a meromorphic function on $\mathbb{C}$, converges to $0$ uniformly on any compact subsets of $\mathbb{C}$).
Using this and noting that $e^{\cos\theta}\sin\sin\theta = \operatorname{Im}(e^{e^{i\theta}} - e)$, we find
\begin{align*}
\int_{-(2N+1)\pi}^{(2N+1)\pi} \frac{e^{\cos\theta}\sin\sin\theta}{\theta} \, \mathrm{d}\theta
&= \operatorname{Im}\biggl( \int_{-(2N+1)\pi}^{(2N+1)\pi} \frac{e^{e^{i\theta}} - e}{\theta} \, \mathrm{d}\theta \biggr) \\
&= \operatorname{Im}\biggl( \int_{-\pi}^{\pi} (e^{e^{i\theta}} - e) \sum_{k=-N}^{N} \frac{1}{\theta + 2\pi k} \, \mathrm{d}\theta \biggr) \\
&\to \operatorname{Im}\biggl( \int_{-\pi}^{\pi} (e^{e^{i\theta}} - e) \frac{1}{2}\cot\left(\frac{\theta}{2}\right) \, \mathrm{d}\theta \biggr)
\qquad\text{as } N \to \infty.
\end{align*}
Now we substitute $z = e^{i\theta}$. Then using the identity $\cot(\theta/2) = i \frac{e^{i\theta} + 1}{e^{i\theta} - 1}$ and the residue theorem,
\begin{align*}
\int_{-\infty}^{\infty} \frac{e^{\cos\theta}\sin\sin\theta}{\theta} \, \mathrm{d}\theta
&= \operatorname{Im}\biggl( \frac{1}{2} \int_{|z|=1} \frac{(e^{z} - e)(z+1)}{z(z-1)} \, \mathrm{d}z \biggr) \\
&= \operatorname{Im}\biggl( \pi i \, \underset{z=0}{\operatorname{Res}} \frac{(e^{z} - e)(z+1)}{z(z-1)} \biggr) \\
&= \pi(e - 1).
\end{align*}
Dividing both sides by $2$, we conclude that
$$ \int_{0}^{\infty} \frac{e^{\cos\theta}\sin\sin\theta}{\theta} \, \mathrm{d}\theta = \frac{\pi}{2}(e - 1). $$
|
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Given $(a_n)$ such that $a_1 \in (0,1)$ and $a_{n+1}=a_n+(\frac{a_n}{n})^2$. Prove that $(a_n)$ has a finite limit. Given $(a_n)$ such that $a_1 \in (0,1)$ and $a_{n+1}=a_n+(\frac{a_n}{n})^2$. Prove that $(a_n)$ has a finite limit.
Clearly $a_n$ are increasing. Also, $$\frac{1}{a_{n+1}}=\frac{1}{a_n\left(1+\frac{a_n}{n^2}\right)}=\frac{1}{a_n}-\frac{\frac{1}{n^2}}{1+\frac{a_n}{n^2}}$$
From this how to proceed further? Any idea will be appreciated
|
EDIT. I realized that this problem has been solved (even by myself) several times in the past; see the links above. A bound of the form
$$ \frac{x}{1-x} \leq \lim a_n \leq \frac{x}{1-\sqrt{x}} $$
is also proved there.
Let $x = a_1 \in (0, 1)$.
Claim 1. We have $a_n \leq nx$ for all $n \geq 1$.
Indeed, the base case ($n=1$) is trivial. Next, if $ a_n \leq nx$, then
$$ a_{n+1} \leq nx + \left( \frac{nx}{n} \right)^2 = nx + x^2 \leq (n+1) x. $$
So by the mathematical induction, the claim follows.
Claim 2. We have $a_n \leq Cn^x$ for some constant $C = C(x) \in (0, \infty)$.
Indeed,
$$ a_{n+1}
= a_n \left( 1 + \frac{a_n}{n^2} \right)
\leq a_n \left( 1 + \frac{x}{n} \right)
\leq a_n e^{\frac{x}{n}} $$
where we utilized the inequality $1 + t \leq e^t$ in the last step. From this, we get
$$ a_n \leq x e^{x H_{n-1}}, $$
where $H_n = \sum_{k=1}^{n} \frac{1}{k}$ is the harmonic number. Then using $H_n = \log n + \mathcal{O}(1)$, the desired claim follows. (In fact, using the inequality $H_{n-1} \leq 1+ \log n$, we may choose $C = x e^{x}$.)
Claim 3. $(a_n)$ converges.
Let $C$ be the constant as in the previous claim. Then
$$ a_{n+1}
= a_n \left( 1 + \frac{a_n}{n^2} \right)
\leq a_n \left( 1 + \frac{C}{n^{2-x}} \right)
\leq a_n \exp\left(\frac{C}{n^{2-x}}\right), $$
and so,
$$ a_n \leq x \exp\left( \sum_{k=1}^{n-1} \frac{C}{k^{2-x}} \right) \leq x e^{C \zeta(2-x)}. $$
This proves that $(a_n)$ is bounded. Since $(a_n)$ is increasing, we conclude that $(a_n)$ converges.
Addendum. A numerical simulation suggests that $a_{\infty}$ as a function of $x = a_1$ satisfies
$$ a_{\infty}(x) \leq \frac{2x}{1-x} $$
for all $x \in (0, 1)$, although I could not prove this. On the other hand, we can obtain a lower bound of the same order via relatively simple argument:
Argument 1. From the recurrence relation and Claim 1,
\begin{align*}
\frac{1}{x} - \frac{1}{a_{\infty}(x)}
= \sum_{n=1}^{\infty} \frac{1}{n^2 + a_n(x)}
\geq \sum_{n=1}^{\infty} \frac{1}{n^2 + nx}
= \sum_{n=1}^{\infty} \frac{1}{x} \left( \frac{1}{n} - \frac{1}{n+x} \right)
= \frac{H_x}{x},
\end{align*}
where $H_x = \int_{0}^{1} \frac{1-t^x}{1-t} \, \mathrm{d}t$ is the harmonic number. Rearranging,
$$ a_{\infty}(x) \geq \frac{x}{1 - H_x} \sim \frac{1}{1-x} \quad \text{as} \quad x \to 1^-. $$
Argument 2. By the principle of mathematical induction, it is easy to show that
$$ a_n \geq x + x^2 + \cdots + x^n = x \cdot \frac{1-x^n}{1-x}. $$
Indeed, the claim is trivial for $n = 1$, and if it is true for $n$, then by the AM-GM inequality, $a_n \geq n x^{(n+1)/2}$, and so,
\begin{align*}
a_{n+1}
&\geq (x + x^2 + \cdots + x^n) + \left( \frac{nx^{(n+1)/2}}{n} \right)^2 \\
&= x + x^2 + \cdots + x^n + x^{n+1}.
\end{align*}
This establishes the inductive step and hence proves the desired claim.
|
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Demonstration for the equal number of odd and unequal partitions of an integer I'm having some problems trying to resolve one exercise from The art and craft of problem solving by Paul Zeitz. What this problem asks you is to prove that $F(x)$ is equal to 1 for all $x$, where
$$F(x) = (1-x)(1+x)(1-x^3)(1+x^2)(1-x^5)(1+x^3)(1-x^7)(1+x^4)\cdots$$
This would demonstrate that the number of unequal partitions and odd partitions of an integer is the same since the generating function $U(x)$ for the number of unequal partitions of an integer $n$ is
$$U(x) = \sum_{n=0}^\infty u_nx^n = (1+x)(1+x^2)(1+x^3)\cdots$$
and the generating function $V(n)$ for the number of odd partitions of an integer $n$ is
$$V(x) = \sum_{n=0}^\infty v_nx^n = (1+x+x^2+x^3+\cdots)(1+x^3+x^6+x^9+\cdots)(1+x^5+x^{10}+x^{15}+\cdots)\cdots = \frac{1}{(1-x)(1-x^3)(1-x^5)\cdots}$$
and so $F(x)=\frac{U(x)}{V(x)}$
The way the exercise suggests to carry out the demonstration is to prove that $F(x)$ is invariant under the substitution $x\mapsto x^2$. This is easy to chech
$$F(x^2) = \prod_{k=1}^\infty (1-(x^2)^{2k-1})(1+(x^2)^k)=\prod_{k=1}^\infty (1-x^{2k-1})(1+x^{2k-1})(1+x^{2k}) = F(x)$$
Now the idea is to keep iterating the substitution and I guess that the solution is that for values of $x$ such that $|x| < 1$ all the $x$ terms go to $0$, but I see two problems with this:
*
*This wouldn't work for values of $x$ greater than $1$ or less than $-1$. We don't care about these values because the function does not converge when evaluated with them? Why?
*After iterating the substitution $n$ times we get
$$F(x) = F(x^{2^n}) = \prod_{k=1}^\infty (1-x^{2^n(2k-1)})(1+x^{2^nk})$$
and although the $x$ terms go to $0$ for the previously mentioned values of $x$ as $n$ goes to infinity, it is still an infinite product and thus an indeterminate form $1^\infty$ (I think) which I haven't been able to resolve.
|
We already know from OPs derivation
\begin{align*}
F(x)=F\left(x^2\right)=\cdots=F\left(x^{2^q}\right)\qquad\qquad q\in\mathbb{N}_0\tag{1}
\end{align*}
We set $F(x)=\sum_{n=0}^\infty a_nx^n$ and obtain from (1) by coefficient comparison
\begin{align*}
F(x)&=F\left(x^2\right)\\
\sum_{n=0}^{\infty}a_nx^n&=\sum_{n=0}^{\infty}a_nx^{2n}\qquad\Longrightarrow\qquad a_{2k+1}=0\qquad\forall k\geq 0\tag{2}
\end{align*}
We see the coefficients of odd powers of $x$ in $F(x)$ are zero. Let's have a look at
\begin{align*}
&a_0+a_1x+a_2x^2+\color{blue}{a_3x^3}+\cdots\\
&\qquad=a_0+a_1x^2+a_2x^4+\color{blue}{a_3x^6}+\cdots\\
&\qquad=a_0+a_1x^4+a_2x^8+\color{blue}{a_3x^{12}}+\cdots\\
&\qquad=a_0+a_1x^8+a_2x^{16}+\color{blue}{a_3x^{24}}+\cdots\\
&\qquad\ \ \vdots
\end{align*}
We observe by iteratively using $F(x)=F\left(x^2\right)=F\left(x^4\right)=\cdots$ that since for instance $a_3=0$, also all coefficients
\begin{align*}
a_6&=[x^6]F(x)=[x^6]F\left(x^2\right)=[x^3]F(x)=0\\
a_{12}&=[x^{12}]F(x)=[x^{12}]F\left(x^4\right)=[x^3]F(x)=0\\
a_{24}&=[x^{24}]F(x)=[x^{24}]F\left(x^8\right)=[x^3]F(x)=0\\
&\ \ \vdots
\end{align*}
and in general all coefficients of the form
\begin{align*}
a_{3\cdot2^q}=[x^{3\cdot 2^q}]F(x)=[x^{3\cdot 2^q}]F(x^{2^q})=[x^3]F(x)=0\qquad\qquad q\geq 0
\end{align*}
are zero. Here we use the coefficient of operator $[x^m]$ to denote the coefficient of $x^m$ of a series.
We conclude: Each natural number $m\geq 1$ has a unique representation
\begin{align*}
\color{blule}{m=2^q\cdot r}\qquad\qquad q\geq 0, r\ \text{odd}
\end{align*}
as product of a power of $2$ (if $q\geq 1$) with an odd natural number $r$. It follows from (1) and (2)
\begin{align*}
\color{blue}{[x^{m}]F\left(x^{2^q}\right)}=[x^{2^{q}r}]F\left(x^{2^q}\right)=[x^r]F(x)\color{blue}{=0}\qquad\qquad q\geq 0
\end{align*}
Since from OPs representation of
\begin{align*}
F(x)=\prod_{k=1}^{\infty}\left(\left(1-x^{2k-1}\right)\left(1+x^k\right)\right)
\end{align*}
we also know that $[x^0]F(x)=1$, the claim \begin{align*}
\color{blue}{F(x)=1}
\end{align*}
follows.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Taylor expansion of $\sin \pi z$ at $z = -1$. Taylor expansion of $\sin \pi z$ at $z = -1$ is
$$\sin\pi z = -\sin(\pi(z+1)) = -\sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}(z+1)^{2n+1}$$ so that
$$\sin\pi z = \sum_{n=0}^\infty \frac{(-1)^{n+1}\pi^{2n+1}}{(2n+1)!}(z+1)^{2n+1}. \tag{$\dagger$}$$
But if I try this
\begin{align}
\sin\pi z & = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}z^{2n+1} \\& = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}(z+1-1)^{2n+1} \\ & = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}\left(\sum_{k=0}^{2n+1}\binom{2n+1}{k}(z+1)^k(-1)^{2n-k+1}\right).\tag{$\dagger^*$}
\end{align}
In this case, how can I reduce $(\dagger^*)$ as the above form $(\dagger)$? Just direct calculation?
|
Start from
$$\sin\pi z = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}z^{2n+1}$$
and
$$-1 = \cos\pi = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n}}{(2n)!}$$
Consider
$$\sin\pi z \cos\pi = \sin(\pi (z+1)) = \sum_{n=0}^\infty \frac{(-1)^n\pi^{2n+1}}{(2n+1)!}z^{2n+1}\left(\sum_{n=0}^\infty \frac{(-1)^n\pi^{2n}}{(2n)!}\right)$$
and compare the coefficients of $z^k$ to
$$\sum_{n=0}^\infty \frac{(-1)^{n+1}\pi^{2n+1}}{(2n+1)!}(z+1)^{2n+1}$$
They are the same.
|
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|
Let $P$ be a point inside $\triangle{ABC}$ such that $\angle{ABP}=10^\circ, \angle{BAP}=\angle{BCP} = 20^\circ$. Show $\triangle{ABC}$ is an isosceles Let $P$ be a point inside $\triangle{ABC}$ such that $\angle{ABP}=10^\circ$, $\angle{BAP}=\angle{BCP} = 20^\circ$. Show $\triangle{ABC}$ is an isosceles and find $\angle{CPA}$.
extend $AP$, and tried law of sines; also googled and searched on stackexchange, no clues.
This seems to be one of the hard IMO problems https://davidaltizio.web.illinois.edu/CollectionOfGeometryProblems.pdf or Sharygin geometry problems https://www.isinj.com/mt-aime/Problems%20in%20Plane%20Geometry%20-%20Sharygin%20(MIR,1982).pdf and no good solution yet.
As shown by answers, the original problem doesn't give enough conditions to prove what's asked. The missing part is $\angle PBC=30^\circ$. Then one can make a mirror point $P'$ of $P$ over $BC$, connect $AP'$, $AB=PC$, $\Delta BPC$ is equilateral, and $BAP$ and $P'AP$ are congruent, $\Delta AP'C$ is equilateral, so $AB=PC=AP'=P'C=AC$.
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initial conditions: Let $P$ be a point inside $β³ABC$ such that $β ABP=10^\circ$, $β BAP=β BCP=20^\circ$
if $C\in R=Ξ»_{A}βͺΞ»_{B}βͺl$ then the triangle ABC is an isosceles.
$\begin{array}{} Ξ»_{A}(A,1) & Ξ»_{B}(B,1) & (l) \text{ perpendicular bisector of the line segment AB.} \end{array}$
If $C$ belongs to the intersection of $R$ with the circle $\lambda'$ then we have the solution satisfying the conditions.
($Ξ»'$ is the reflection of $Ξ»(O,1)$ with respect to line $PB$). $\lambda'$ is locus of C such that $\angle{BCP}={20^\circ}$. Two solutions:$C$ and $C_{1}$. $O$ is the circumcenter of the triangle $ABP$. $O=(\frac{1}{2},-\frac{\sqrt{3}}{2})$.
straight line $AP$ through $A(0,0)$ with slope equal to $tan (\angle{20}^\circ)$ and $BP$ through $B(1,0)$ with slope equal to $-tan (\angle{10}^\circ)$. ($P=APβ©PB$) ($(l)$ $x=\frac{1}{2}$)
$h=tan\left( \frac{Ο}{18} \right)$
$\begin{array}{} x_{P}=\frac{h^2-1}{h^2-3} & y_{P}=\frac{2h}{3-h^2} \end{array}$
$\begin{array}{} x_{OΒ΄}=\frac{3h^2-5}{2(h^2-3)} & y_{O'}=\frac{\sqrt{3}(h^2-3)-4h}{2(h^2-3)} \end{array}$
$C=\left( \frac{1}{2},\frac{\sqrt{3}(h^2-3)-4(h+\sqrt{2-h^2})}{2(h^2-3)} \right)$
$b^2=\frac{[h(8-2\sqrt{3}h)+6\sqrt{3}]\sqrt{2-h^2}+h^3(h-2\sqrt{3})-6h(h-\sqrt{3})+17}{(h^2-3)^2}$
$t^2=\frac{h^2-5-2\sqrt{3}\sqrt{2-h^2}}{h^2-3}$
$d=\frac{h^2+1}{h^2-3}$
$cos(ΞΈ)=\frac{-b^2+d^2+t^2}{2dt}$
$ΞΈ=cos^{-1}\left( \frac{h^3(h+2\sqrt{3})-2h\left( 3\sqrt{3}+4\sqrt{2-h^2}-1 \right) }{2(h^2+1)\sqrt{15+h^2(h^2-8)-2(h^2-3)\sqrt{6-3h^2}}} \right)$
$\thetaβ
115.476252^\circ$
where: $b=AC$, $t=CP$, $d=AP$, $\theta=\angle{APC}$
$ΞΈ_{2}=β (APC_{1})=80Β°$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$1^2 - 2^2 + 3^2 - 4^2 + \dots + 1999^2$ using closed form formula for sum of squares of first $n$ natural numbers The question is simple, the solution, not so much
Q.Find the sum of the given expression $1^2- 2^2 + 3^2 - 4^2 + \dots + 1999^2$
My idea is we know $1^2 + 2^2 + 3^2 + 4^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$
So for $n=1999$ I get the sum as $2,66,46,67,000$
From this I need to subtract the squares of the even terms twice because subtracting once leaves with only the sum of the squares of the odd nos.
I observed something :
$2^2 + 4^2 + 6^2 + \dots + 1998^2 = (2 \cdot 1)^2 + (2 \cdot 2)^2 + \dots + (2 \cdot 999)^2$
Therefore to obtain the sum of the square of the even terms, I can take $4$ as common and use the aforementioned formula for $n=999$ and multiply it by $4$.
therefore sum of square of even terms = $1,33,13,34,000$
I need to subtract this sum twice to get the answer, because subtracting once simply leaves me with the sum of the squares of the odd numbers.
The answer is now $1999000$, which still doesn't match the answer key. Can someone explain where I am going wrong ?
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We can use the identity $$(x+1)^2-x^2=x+1+x$$
to tackle the sum in 2 simpler ways:
Way 1
$$ \begin{aligned}&1^2 - 2^2 + 3^2 - 4^2 + 5^2-... -1998^2+ 1999^2
\\=& 1+(2+3)+(4+5)+β¦+(1998+1999)
\\= &19990000\end{aligned}$$
Way 2
$$ \begin{aligned}&(1^2 - 2^2 + 3^2 - 4^2 + 5^2-6^2...-1998^2)+ 1999^2
\\=&-((1+2)+(3+4)+(5+6)+β¦+(1997+1998))+1999^2
\\= &19990000\end{aligned}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integration of $\int_0^{\pi/2}\frac{d\theta}{a^2+b^2\cos^2(\theta)}$ gives results such that $\tan(\pi/2)$ comes which is "undefined". How to proceed? Integrating the below equation:
$$\int_{0}^{\pi/2}\frac{d\theta}{a^2+b^2\cos^2(\theta)}$$
gives,
$$\frac{1}{\left| a \right|\sqrt{a^2+b^2}}\left[\arctan\left(\frac{ a\tan(\theta)}{\sqrt{a^2+b^2}}\right)\right]_{0}^{\pi/2}$$
This portion $\arctan\!\Big(\frac{a\tan(\pi/2)}{\sqrt{a^2+b^2}}\Big)$ gives $\tan(\pi/2)$ which is "undefined", so how can we proceed with the following resulted equation as below: [This is the answer mentioned in the book to be proven as the result]
$$\frac{\pi}{2\left| a \right|\sqrt{a^2+b^2}}$$
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Assume that $a>0$. You have\begin{align}\int_0^{\pi/2}\frac1{a^2+b^2\cos^2\theta}\,\mathrm d\theta&=\int_0^{\pi/2}\frac{\sec^2\theta}{a^2\sec^2\theta+b^2}\,\mathrm d\theta\\&=\frac1a\int_0^{\pi/2}\frac{a\sec^2\theta}{a^2\tan^2(\theta)+a^2+b^2}.\end{align}Now, if you do $x=a\tan\theta$ and $\mathrm dx=a\sec^2(\theta)\,\mathrm d\theta$, your integral becomes$$\frac1a\int_0^\infty\frac1{x^2+a^2+b^2}\,\mathrm dx=\frac\pi{2a\sqrt{a^2+b^2}}=\frac\pi{2|a|\sqrt{a^2+b^2}},$$since$$\int\frac1{x^2+a^2+b^2}\,\mathrm dx=\frac1{\sqrt{a^2+b^2}}\arctan\left(\frac x{\sqrt{a^2+b^2}}\right).$$And if $a<0$, then\begin{align}\int_0^{\pi/2}\frac1{a^2+b^2\cos^2\theta}\,\mathrm d\theta&=\int_0^{\pi/2}\frac1{(-a)^2+b^2\cos^2\theta}\,\mathrm d\theta\\&=\frac\pi{2(-a)\sqrt{a^2+b^2}}\\&=\frac\pi{2|a|\sqrt{a^2+b^2}}.\end{align}
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|
Evaluating $\lim\limits_{R\to +β}\iint_{x^2+y^2\leq R^2}\left(\frac{1+2x^2}{1+x^4+6x^2y^2+y^4}-\frac{1+y^2}{2+x^4+y^4}\right)\,\mathrm dx\mathrm dy$
Evaluate $\lim\limits_{R\to\infty} J(R)$, where
$$J(R) = \iint_{x^2 + y^2\leq R^2} \left(\frac{1+2x^2}{1+x^4+6x^2y^2 + y^4} - \frac{1+y^2}{2+x^4+y^4}\right)\,\mathrm dx\mathrm dy.$$
First note that interchanging $x$ and $y$ does not change the value of the integral, so
$$J(R) = \iint_{x^2 + y^2 \leq R^2}\left(\frac{1+2y^2}{1+x^4+6x^2y^2 + y^4} - \frac{1+x^2}{2+x^4+y^4}\right)\,\mathrm dx\mathrm dy$$
and hence average the two equal expressions for $J(R)$ gives that
$$J(R) = \iint_{x^2 + y^2\leq R^2} (f(x,y) - g(x,y))\,\mathrm dx\mathrm dy,$$
where
$$f(x,y) = \frac{1+x^2 + y^2}{1+x^4+6x^2y^2+y^4},\quad g(x,y) = \frac{1+(x^2+y^2)/2}{2+x^4+y^4}.$$
Note that $f(x,y) = 2g(x+y,x-y)$.
Now, why is it true that
\begin{gather*}\iint_{R^2 \leq x^2 + y^2\leq 2R^2}g(x,y)\,\mathrm dx\mathrm dy\ \text{?}\tag{1}\end{gather*}
I tried using substitutions, but I couldn't get how to prove (1). But assuming (1) does hold, we have by converting to polar coordinates that
\begin{align*}
J(R) &= \int_0^{2\pi} \int_R^{R\sqrt{2}} \frac{1+r^2/2}{2+r^4(\cos\theta)} r\,\mathrm dr\mathrm d\theta\\
&= \int_0^{2\pi}\int_R^{R\sqrt{2}} \frac{1+r^2/2}{2+r^4(1-(\sin^2(2\theta)/2))}r\,\mathrm dr\mathrm d\theta
\end{align*}
Now use the substitution $r\mapsto r/R$ to obtain the equivalent integral
$$\int_0^{2\pi}\int_1^{\sqrt{2}} \frac{1+(Rr)^2/2}{2+(Rr)^4(1-\sin^2(2\theta)/2)}(R^2r)\,\mathrm dr\mathrm d\theta.$$
Why is it true that since integral is uniformly bounded for $R \gg 0$, we can take the limit over $R$ through the integrals to obtain that
$$\lim_{R\to\infty} J(R) = \int_0^{2\pi}\int_1^{\sqrt{2}} \frac{1}{r(2-\sin^2(2\theta))}\,\mathrm dr\mathrm d\theta = \frac{1}2\ln (2)\int_0^{2\pi} \frac{2}{3+\cos(4\theta)}\,\mathrm d\theta\ \text{?}$$
Note that by symmetry,
$$\int_0^{2\pi} \frac{2}{3+\cos(4\theta)}\,\mathrm d\theta = 2\int_0^\pi \frac{2}{3+\cos\theta}\,\mathrm d\theta$$
Now using the half-angle substitution $t = \tan(\theta/2)$, we have
\begin{gather*}
2\int_0^{\pi} \frac{2}{3+\cos\theta}\,\mathrm d\theta = 2\int_{0}^{\infty} \frac{4}{3(1+t^2)+(1-t^2)}\,\mathrm dt\\
= 2\int_0^\infty \frac{2}{2+t^2}\,\mathrm dt = \sqrt{2}\arctan(t/\sqrt{2})\biggr|_0^\infty = \sqrt{2}\pi.
\end{gather*}
Hence we have $\lim\limits_{R\to\infty} J(R) = \dfrac{\sqrt{2}}2\ln(2)\pi$.
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$\def\d{\mathrm{d}}\def\R{\mathbb{R}}\def\vector#1#2{\begin{pmatrix}#1\\#2\end{pmatrix}}\def\abs#1{\left|#1\right|}\def\brace#1{\left\{#1\right\}}\def\paren#1{\left(#1\right)}$To prove (1), it suffices to prove\begin{gather*}
\iint\limits_{x^2 + y^2 \leqslant R^2} f(x, y) \,\d x\d y = \iint\limits_{x^2 + y^2 \leqslant 2R^2} g(x, y) \,\d x\d y. \tag{1$'$}
\end{gather*}
Define $Ο: \R^2 β \R^2$ by $Ο\vector{x}{y} = \vector{x + y}{x - y}$, then $f = 2g \circ Ο$. Since $Ο^{-1}\vector{u}{v} = \dfrac{1}{2} \vector{u + v}{u - v}$, then\begin{align*}
&\mathrel{\phantom=} Ο\paren{ \brace{\vector{x}{y} \in \R^2 \,\middle|\, x^2 + y^2 \leqslant R^2 } }\\
&= \brace{ \vector{u}{v} \in \R^2 \,\middle|\, \paren{ \frac{1}{2}(u + v) }^2 + \paren{ \frac{1}{2}(u - v) }^2 \leqslant R^2 }\\
&= \paren{ \brace{\vector{u}{v} \in \R^2 \,\middle|\, u^2 + v^2 \leqslant 2R^2 } },
\end{align*}
and$$
JΟ(u, v) = \begin{pmatrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.
$$
Therefore,\begin{gather*}
\iint\limits_{x^2 + y^2 \leqslant R^2} f(x, y) \,\d x\d y = \iint\limits_{u^2 + v^2 \leqslant 2R^2} f(Ο^{-1}(u, v)) |\det JΟ^{-1}(u, v)| \,\d u\d v\\
= \iint\limits_{u^2 + v^2 \leqslant 2R^2} \frac{2g(u, v)}{|\det JΟ(u, v)|} \,\d u\d v = \iint\limits_{u^2 + v^2 \leqslant 2R^2} g(u, v) \,\d u\d v = \iint\limits_{x^2 + y^2 \leqslant 2R^2} g(x, y) \,\d x\d y.
\end{gather*}
Now in order to prove that\begin{gather*}
\small \lim_{R β +β} \iint\limits_{\substack{1 \leqslant r \leqslant \sqrt{2} \\ 0 \leqslant \leqslant 2Ο}} \frac{R^2 r (R^2 r^2 + 2)}{R^4 r^4 (2 - \sin^2(2ΞΈ)) + 4} \,\d r\d ΞΈ = \iint\limits_{\substack{1 \leqslant r \leqslant \sqrt{2} \\ 0 \leqslant \leqslant 2Ο}} \lim_{R β +β} \frac{R^2 r (R^2 r^2 + 2)}{R^4 r^4 (2 - \sin^2(2ΞΈ)) + 4} \,\d r\d ΞΈ, \tag{2}
\end{gather*}
note that for $1 \leqslant r \leqslant \sqrt{2} \leqslant R$,$$
\abs{ \frac{R^2 r (R^2 r^2 + 2)}{R^4 r^4 (2 - \sin^2(2ΞΈ)) + 4} } < \frac{R^2 r (R^2 r^2 + 2)}{R^4 r^4 (2 - \sin^2(2ΞΈ))} \leqslant \frac{R^2 r Β· 2R^2 r^2}{R^4 r^4} = \frac{2}{r} \leqslant 2,
$$
thus (2) is implied by the dominated convergence theorem.
|
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|
$a,b>0$ then: $\frac{1}{a^2}+b^2\ge\sqrt{2(\frac{1}{a^2}+a^2)}(b-a+1)$
Let $a,b>0$. Prove that: $$\frac{1}{a^2}+b^2\ge\sqrt{2\left(\frac{1}{a^2}+a^2\right)}(b-a+1)$$
Anyone can help me get a nice solution for this tough question?
My approach works for 2 cases:
Case 1: $b-a+1>0$ then squaring both side, we get equivalent inequality: $$\frac{1}{a^4}+b^4+2\frac{b^2}{a^2}\ge2\left(\frac{1}{a^2}+a^2\right)(b^2+a^2+1-2ab-2a+2b)$$ Or: $$\frac{1}{a^4}+b^4\ge\frac{2}{a^2}(a^2+1-2ab-2a+2b)+2a^2(b^2+a^2+1-2ab-2a+2b)$$
The rest is so complicated. Is there nice idea etc: AM-GM, C-S to prove this inequality.
Case 2: $b-a+1<0$ which is obviously true.
I hope we can find a better approach for the inequality. Thank you very much!
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Since OP posted their own version, I'd post my solution.
WTS $ b^2 - \sqrt{2 ( \frac{1}{a^2 } + a^2 )} \cdot b + [\frac{1}{a^2} + \sqrt{2 ( \frac{1}{a^2 } + a^2 )} \cdot (a-1) ] \geq 0$.
We will prove this by viewing it as a quadratic in $b$, and showing that the discriminant is $ \leq 0$ for $ a > 0$, hence the value is always $ \geq 0 $.
$$D = [ \sqrt{2 ( \frac{1}{a^2 } + a^2 )}] ^2 - 4[\frac{1}{a^2} + \sqrt{2 ( \frac{1}{a^2 } + a^2 )} \cdot (a-1) ] \\
= - \frac{ 2 ( a-1) ( a^3+a^2+a+1 - 2\sqrt{2} \sqrt{ a^4 + 1 }) }{a^2} .$$
Observe that $ ( a^3 + a^2 + a + 1 ) ^2 - 8 (a^4 + 1) = (a-1) ( a^5 + 3a^4 - 2a^3 + 2a^2 + 5a + 7) $.
Clearly the second term is positive for $ a > 0$.
Hence
*
*If $ a > 1 $, then $ a^3 + a^2 + a + 1 > \sqrt{ 8 (a^4 + 1)}$, so $ D < 0 $
*If $ a = 1$, then $ D = 0 $
*If $ 0 < a < 1$, then $a^3 + a^2 + a + 1 < \sqrt{ 8 (a^4 + 1)},$ so $ D < 0 $.
Equality holds when $ D = 0 \Rightarrow a = 1 $ with corresponding root $b = 1$.
Modification of OP's solution.
Lemma: $$\frac{ \sqrt{2} x^2 } { \sqrt{ \frac{1}{a^2} + a^2 } } + \frac{ a^2 - a + 1 } { a } \geq 2 x . $$
Proof: We apply AM-GM directly. It remains to show that
$$ \sqrt{ \frac{ 2a^2} { a^4 +1 }} \times { \frac{ a^2 - a + 1 } { a } } \geq 1$$
or that $2 ( a^2 - a + 1 ) ^2 \geq (a^4 + 1) $, which is true since $2 ( a^2 - a + 1 ) ^2 - (a^4 + 1) = (a-1)^4$.
Equality holds when $ a = 1, x = 1 $.
Corollary: Apply the lemma to $ x = \frac{1}{a}$ and $x = b$, sum them up and shift terms around to get the desired inequality.
Equality holds when $ a = 1, b = 1 $.
Notes
*
*(I thought that) My contribution here is to simplify the AM-GM that OP did.
*
*Since $ \frac{\frac{1}{a} + a +\sqrt{2} } { 2 + \sqrt{2} } + \frac{ \frac{ 1}{a} + a - \sqrt{2} } { 2 - \sqrt{2} } = 2\frac{ a^2 - a + 1 } { a } $, arguably this made it more complicated.
*It's hard to motivate the lemma, which was obtained from OP's AM-GM.
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|
Proving $\frac{1}{abc}+\frac{12}{a^2b+b^2c+c^2a}\ge5$
Let $a,b,c>0$, $a+b+c=3$. Prove that$$\frac1{abc}+\frac{12}{a^2b+b^2c+c^2a}\ge5$$
My approach using a well-known result:$$a^2b+b^2c+c^2a+abc\le\frac4{27}(a+b+c)^3$$
We need to prove that $\frac1{abc}+\frac{12}{4-abc}\ge5$ but this inequality does not hold for all $a,b,c$.
Is there any better idea to help me solve the problem? Thanks for your help!
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Remark
$$\displaystyle\sum f(a,b,c)=f(a,b,c)+f(b,c,a)+f(c,a,b)$$
means cyclic sum.
Homogenize it, then we get
denote
$$
\begin{aligned}
f\left( a,b,c \right) :=&\sum{a^5b}+3\sum{a^4b^2}+3\sum{a^3b^3}+\sum{a^2b^4}\\
&+15\sum{a^4bc}-92\sum{a^3b^2c}+42\sum{a^2b^3c}+81a^2b^2c^2
\end{aligned}
$$
we gonna prove $f(a,b,c)\geq 0$
actually, we have
$$
\begin{aligned}
&\left( a+b+c \right) ^2f\left( a,b,c \right)\\
=&
28\sum{a^2b^2c}\sum{a\left( a-b \right) \left( a-c \right)}+\frac{13}{24}\sum{\left( a^2b-abc \right)^2}\sum{\left( a-b \right) ^2}\\
&+\frac{59}{12}\sum{\left( a^2b-abc \right)}\sum{\left( a-b \right) ^2b^2c}+\frac{65}{4}\sum{a^2bc\left( -a^2+ba+ca+b^2-2bc \right) ^2}\\
&+7\sum{a^3b^2\left( b-c \right) ^2c}+\frac{15}{2}\sum{a^3b\left( b^2-ac \right) ^2}+\frac{83}{12}\sum{a^4b^2\left( b-c \right) ^2}\\
&+\frac{121}{12}\sum{a\left( a-b \right) ^2\left( b-c \right) ^2c^3}+\frac{17}{4}\sum{ab^2c\left( -a^2+ba+ca+b^2-2bc \right) ^2}\\
&+\frac{1}{2}\sum{\left( a-b \right) ^4\left( ba^3-3b^2ca+b^2c^2+b^3c \right)}\\
&+\frac{253}{12}\sum{abc^2\left( -a^2+ba+ca+b^2-2bc \right) ^2}+\frac{1}{2}\sum{\left( a^2-2ba+b^2-bc \right) ^2\left( c^2-ab \right) ^2}\\
\geq & 0
\end{aligned}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4341189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Finding perfect two terms Egyptian fraction If I have two unity fractions, like $\frac{1}{12} + \frac{1}{180}$, for instance. These two fractions can be re-writen as $\frac{1}{15} + \frac{1}{45}$ or even $\frac{1}{18} + \frac{1}{30}$, which satisfies the Egyptian Fraction concept of maximising the value of the smallest fraction. Is there a formula or an algorithm that when followed, one can find out all the possible equal pairs of any unity fraction pair?
|
For a given rational number $\dfrac{c}{d}$, we wish to find all pairs of positive integers $(a,b)$ such that $$\dfrac{c}{d} = \dfrac{1}{a}+\dfrac{1}{b}.$$
We can manipulate the equation as follows: $$cab = da+db$$ $$cab - da-db = 0$$ $$c^2ab-cda-cdb = 0$$ $$c^2ab-cda-cdb+d^2 = d^2$$ $$(ca-d)(cb-d) = d^2.$$
Hence, $ca-d$ and $cb-d$ must be complementary factors of $d^2$. So the pairs of positive integers $(a,b)$ such that $\dfrac{c}{d} = \dfrac{1}{a}+\dfrac{1}{b}$ are all of the form $$(a,b) = \left(\dfrac{f_1+d}{c},\dfrac{f_2+d}{c}\right)$$ for all integers $f_1,f_2$ such that $d^2 = f_1f_2$ and both $\dfrac{f_1+d}{c}$ and $\dfrac{f_2+d}{c}$ are integers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4341333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to solve system of linear equations that came up in looking at a probability problem, any tricks? The following system of $k$ equations came up in a probability problem I was looking at:
$$(n+1)x_1 - x_2 = 1, \quad x_1 + nx_2 - x_3 = 1 , \quad x_1 + nx_3 - x_4 = 1, \quad x_1 + nx_4 - x_5 = 1, \quad \ldots \quad x_1 + nx_{k-1} - x_{k} = 1, \quad x_1 + nx_{k} = 2,$$which involves solving$$\begin{pmatrix}
n+1 & -1 & 0 & 0 & \cdots & 0 & 0 & 0\\
1 & n & -1 & 0 & \cdots & 0 & 0 & 0 \\
1 & 0 & n & - 1 & \cdots & 0 & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
1 & 0 & 0 & 0 & \cdots & 0 & n & -1\\
1 & 0 & 0 & 0 & \cdots & 0 & 0 & n\\
\end{pmatrix} \begin{pmatrix}
x_1\\
x_2 \\
x_3 \\
\vdots \\
x_{k - 1}\\
x_k\\
\end{pmatrix} = \begin{pmatrix}
1\\
1 \\
1 \\
\vdots \\
1\\
2\\
\end{pmatrix}$$Now, given that this matrix looks pretty "sparse", I'm wondering if there's any shortcuts that can get us the answer quickly without me having to painstakingly row reduce this augmented matrix by brute force. I've started in the direction of brute force but couldn't see any immediate pattern/simplification.
For the record, the answer in the back of my book is$$x_r = {{n^r - n^{r - 1} + n^{k} - 1}\over{n^{k+1} - 1}}.$$
|
"Smart" Brute force isn't that bad. We have the following pattern:
$S_1: (n+1) x_1 - x_2 = 1$
$S_2 + nS_1: (n^2+n+1) x_1 - x_3 = 1+n$
$S_3 + nS_2 + n^2 S_1: (n^3+n^2+n+1) x_1 - x_4 = 1+n+n^2$.
$\vdots$
$S_{k-1}+nS_{k-2} + \ldots + n^{k-2}S_1: \frac{n^k-1}{n-1} x_1 - x_k = \frac{n^{k-1} - 1 } { n-1}$
$S_k: x_1 + nx_k = 2$.
From the last 2 equations, we can get $x_1 = \frac{n^k+n-2}{n^{k+1} - 1}, x_k = \frac{2n^k-n^{k-1} - 1 } {n^{k+1} - 1}$.
Then from the previous equations, we can get $x_i =\frac{ (n^i-1) x_1 - (n^{i-1} -1 ) } { n-1} = \frac{n^i - n^{i-1} + n^k -1}{n^{k+1} - 1 } $.
Observe it has the given form.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is the series $\sum_{n=1}^\infty\frac{1}{n}\left(\sum_{k=1}^n\frac{1}{k}\left(\frac{1}{2}\right)^{n-k}\right)$ convergent? $$
\sum_{n = 1}^\infty\dfrac{1}{n}\left(\sum_{k = 1}^n\dfrac{1}{k}\left(\dfrac{1}{2}\right)^{n - k}\right)
$$
Does the series converge?
I calculate it using Matlab, and it seems that the sum converges to 2.4673. I also tried to use the ratio test to prove the convergent, but stopped at to calculate the sum $\sum_{k = 1}^n\dfrac{1}{k}(1/2)^{n - k}$.
|
\begin{aligned}\sum_{k=1}^{n}{\frac{2^{k-n}}{k}}&=\frac{1}{2^{n-1}}\sum_{k=1}^{n}{\int_{0}^{1}{\left(2x\right)^{k-1}\,\mathrm{d}x}}\\ &=\frac{1}{2^{n-1}}\int_{0}^{1}{\frac{1-2^{n}x^{n}}{1-2x}\,\mathrm{d}x}\\ &=\frac{1}{2^{n}}\int_{-\frac{1}{2}}^{\frac{1}{2}}{\frac{1-\left(1-2x\right)^{n}}{x}\,\mathrm{d}x}\end{aligned}
The integral in the second line make sense because $ x\mapsto\frac{1-\left(2x\right)^{n}}{1-2x} $ can be extended by continuity at $ \frac{1}{2} $, which means it is piecewise continuous on $ \left[0,1\right]$, and, hence, is integrable.
In the third line We've just used a substitution $ y=\frac{1}{2}-x $. Now using the fact that for $ \alpha\geq 0 $, we have that $ \left(\forall x\geq -1\right), \left(1+x\right)^{\alpha}\geq 1+\alpha x $. (Bernoulli's inequality), we have : \begin{aligned}\frac{1}{2^{n}}\int_{-\frac{1}{2}}^{\frac{1}{2}}{\frac{1-\left(1-2x\right)^{n}}{x}\,\mathrm{d}x}\leq\frac{1}{2^{n}}\int_{-\frac{1}{2}}^{\frac{1}{2}}{\frac{2nx}{x}\,\mathrm{d}x}=\frac{n}{2^{n-1}}\end{aligned}
Since $ \sum\limits_{n\geq 1}{\frac{1}{2^{n-1}}} $ converges, our series also converges.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
$\lim_{n\to+\infty} {n^{(4/3)}} \Big( \sqrt[3]{n^2 +1} + \sqrt[3]{3- n^2} \Big)$ $\lim_{n\to+\infty} {n^{(4/3)}} \Big( \sqrt[3]{n^2 +1} + \sqrt[3]{3- n^2} \Big)$
I used an extension of the appropriate 1 (A ^ 2 - AB + B ^ 2) / (A ^ 2 - AB + B ^ 2). I got here
$\lim_{n\to+\infty} {n^{(4/3)}} \Big(4/ (({n^2 +1} )^{(2/3)} + (n^4-2n^2-3)^{(1/3)}+(3-n^2)^{(2/3)})\Big)$
I'll blame the leading term in the denominator -> n ^ (4/3)
$\lim_{n\to+\infty} \Big(4/ (({1+0} )^{(2/3)} + (1+0+0)^{(1/3)}+(0-1)^{(2/3)})\Big)$
.
.
.
$(0-1)^{(2/3)}$
I care about this term in the denominator because it is equal to -1, but due to the exponentiation it is always positive -> How will it affect the final result ?
Final result is 4/3 op 4/1 ?
|
WLOG $\dfrac1{n^2}=h$ to find $\sqrt[3]{3-n^2}=\sqrt[3]{\dfrac{3h-1}h}=\dfrac{\sqrt[3]{3h-1}}{h^{1/3}}=-\dfrac{\sqrt[3]{1-3h}}{h^{1/3}}$
So, the limit in question reduces to
$$\lim_{h\to0^+}\dfrac{\sqrt[3]{1+h}-\sqrt[3]{1-3h}}h=\lim_{h\to0^+}\dfrac{\sqrt[3]{1+h}-1}h-\lim_{h\to0^+}\dfrac{\sqrt[3]{1-3h}-1}h$$
Now for set $\sqrt[3]{1+mh}-1=r\implies mh=3r+3r^3+r^3$ $$\lim_{h\to0^+}\dfrac{\sqrt[3]{1+mh}-1}h=\lim_{r\to0^+}\dfrac{rm}{3r+3r^3+r^3}=\frac m3$$
|
{
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
A 3d line intersecting 2 other 3d lines The equation of the line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and intersecting the lines $9x + y + z + 4 = 0 = 5x + y + 3z$ & $x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$
My solution is as follow
$\frac{{x - a}}{2} = \frac{{y - b}}{3} = \frac{{z - c}}{4}$ represent the line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and intersecting the line and intersecting the line ${L_1}:9x + y + z + 4 = 0 = 5x + y + 3z$ & ${L_2}:x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$
Let $\frac{{x - a}}{2} = \frac{{y - b}}{3} = \frac{{z - c}}{4} = {\lambda _i}$ where $\left( {x,y,z} \right) = \left( {a + 2{\lambda _i},b + 3{\lambda _i},c + 4{\lambda _i}} \right)$
$ \Rightarrow 9a + 18{\lambda _1} + b + 3{\lambda _1} + c + 4{\lambda _1} + 4 = 0 \Rightarrow \frac{{9a + b + c + 4}}{{ - 25}} = {\lambda _1}$ after putting the values in $L_1$
$5a + 10{\lambda _1} + b + 3{\lambda _1} + 3c + 12{\lambda _1} = 0 \Rightarrow \frac{{5a + b + 3c}}{{ - 25}} = {\lambda _1}$
$9a + b + c + 4 = 5a + b + 3c \Rightarrow 4a - 2c + 4 = 0$
${L_2}:x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$
$\left( {x,y,z} \right) = \left( {a + 2{\lambda _2},b + 3{\lambda _2},c + 4{\lambda _2}} \right)$
$a + 2{\lambda _2} + 2b + 6{\lambda _2} - 3c - 12{\lambda _2} - 3 = 0 = 2a + 4{\lambda _2} - 5b - 15{\lambda _2} + 3c + 12{\lambda _2} + 3$
$ \Rightarrow a + 2b - 3c - 3 - 4{\lambda _2} = 0 = 2a - 5b + 3c + 3 + {\lambda _2} \Rightarrow \frac{{a + 2b - 3c - 3}}{4} = {\lambda _2} = \frac{{2a - 5b + 3c + 3}}{{ - 1}}$
$ \Rightarrow a + 2b - 3c - 3 = - 8a + 20b - 12c - 12 \Rightarrow 9a - 18b + 9c + 9 = 0 \Rightarrow a - 2b + c + 1 = 0$
$ \Rightarrow a - 2b + c + 1 = 0\& 2a - c + 2 = 0$
Let $b = - 1 \Rightarrow a + c + 3 = 0\& 2a - c + 2 = 0 \Rightarrow c = - \frac{4}{3}\& a = - \frac{5}{3}$
$\frac{{x + \frac{5}{3}}}{2} = \frac{{y + 1}}{3} = \frac{{z + \frac{4}{3}}}{4} = \lambda \Rightarrow \left( {x,y,z} \right) = \left( { - \frac{5}{3}\hat i - \hat j - \frac{4}{3}\hat k} \right) + \lambda \left( {2\hat i + 3\hat j + 4\hat k} \right)$
But my answer is not matching. Can you tell me the error that I have made. Each steps is elaborated
|
Your way of squeezing more than one equation in a line is confusing me and I will do it the usuual way, one line for an equation.
A linear equation represent a plane. A line is represented as the intersection of two planes, that is a systeme of two equations. The coefficients of $x,y,$ and $z$ of a linear equation are equal to a normal vector of the plane. If two planes are parallel then they have parallel normal vectors. so we can get all parallel planes to a plane by changing the constant term of the plane equation.
The equations
$$\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$$
represent two planes
$$\frac{x}{2} = \frac{y}{3}\\
\frac{y}{3} = \frac{z}{4}$$
All other equations that can deduced from this notation are equations that are dependent from the two equations we already selected and therefore can be ignored.
These two equations can be transfomed to
$$3x-2y=0\\
4y-3z=0$$
The line we are looking for is parallel to these planes and therefor lies on planes parallel to these planes.
These parallel planes have the equations
$$3x-2y=a \tag{e1}\\
4y-3z=b$$
(Maybe your way to establish the equations of the parallel line is wrong)
we want to calculate $a$ and $b$. We have two other lines
$$9x + y + z + 4 = 0 \tag{e2} \\
5x + y + 3z=0$$
$$x + 2y - 3z - 3 = 0 \tag{e3}\\
2x - 5y + 3z + 3=0$$
Both lines intersect line($e1)$.
So we intersect line $(e1)$ with $(e2)$, this gives the systems
$$3x-2y=a \tag{e4}\\
4y-3z=b\\
9x + y + z + 4 = 0 \\
5x + y + 3z=0$$
and
$$3x-2y=a \tag{e5}\\
4y-3z=b\\
x + 2y - 3z - 3 = 0 \\
2x - 5y + 3z + 3=0$$
$(x,y,z)$ are the coordinates of the intersection point. So these are differents for system $(e4)$ and $(e5)$, because the intersection points are different. $a$ and $b$ are the same in both systems.
Each system has $4$ equations and $5$ variables. We can eliminate $3$ variables $x$,$y$,$z$ and then an equation with two variables $a$,$b$ is left. We get the two equations
$$2a+b+6=0 \tag{e6}\\
a-b+3=0$$
And if we solve this we get $$a=-3\\b=0\tag{e7}$$
And so the line we are looking for is
$$3x-2y=-3 \tag{e8}\\
4y-3z=0$$
The intersection of $(e2)$ and $(e8)$ is the point
$$x=-\frac{3}{5},y=\frac{3}{5},z=\frac{4}{5}$$
the intersection of $e3$ and $e8$ is the point
$$x=-3,y=-3,z=-4$$
You can use them to check the solution.
|
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|
Hopf fibration from $SO(3)$ Lie algebra generators? One can use the Pauli matrices $\sigma_i$ to generate $Cl_3(\mathbb{R})$ and taking commutators of these matrices gives the $SU(2)$ Lie algebra $\mathfrak{su}(2)=\biggl(\begin{matrix}
ia&-z\\
z&-ia\\
\end{matrix}\biggr)$
However, one can also generate $Cl_3(\mathbb{R})$ using the $4\times4$ quaternion Cayley Matrices:$R_i=\begin{pmatrix}
0&1&0&0\\
-1&0&0&0\\
0&0&0&-1\\
0&0&1&0\\
\end{pmatrix}, R_j =\begin{pmatrix}
0&0&1&0\\
0&0&0&1\\
-1&0&0&0\\
0&-1&0&0\\
\end{pmatrix}, R_k = \begin{pmatrix}
0&0&0&1\\
0&0&-1&0\\
0&1&0&0\\
-1&0&0&0\\
\end{pmatrix}$
These matrices act on 4-collumn unit spinors which from what I understand are elements of Spin(3), yet the associated Lie algebra of Spin(3) is generated by the following $3\times3$ matrices:
$\pi_1=\begin{pmatrix}
0&0&0\\
0&0&-1\\
0&1&0\\
\end{pmatrix}, \pi_2=\begin{pmatrix}
0&0&1\\
0&0&0\\
-1&0&0\\
\end{pmatrix}, \pi_3=\begin{pmatrix}
0&-1&0\\
1&0&0\\
0&0&0\\
\end{pmatrix}$
I'm stuck on how one derives these matrices from the $Cl_3(\mathbb{R})$ generators.
Moreover, in this paper: https://arxiv.org/pdf/1601.02569.pdf The author writes on page 8 that the Hopf fibration is given by the map: $\Psi \pi_i \Psi^T$ where $\Psi \in\mathbb{H}$ but how can a $3\times 3$ matrix act on a quaternion, which is a 4-column, or can be represented by a $2\times2$ or $4\times4$ matrix? This makes perfect sense to me if one replaces $\pi_i$ with $\sigma_i$ then the Hopf fibration arises if we turn the quaternions into $2\times 2$ matrices but I have no idea how $\pi_i$ act on quaternions.
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The $3\times 3$ Lie algebra matrices, you have quoted, would act on the quaternion expressed as a $3\times 3$ matrix, i.e. the SO(3) group.
https://en.wikipedia.org/wiki/Quaternions_and_spatial_rotation
Rather than on a quaternion represented as a 4-column, a $2\times2$ complex matrix [SU(2)], or a $4\times4$ real matrix, as is my understanding.
A quaternion $q=a+bi+cj+dk$ is written in SO(3):
\begin{equation}
\Psi\;=\;
\begin{pmatrix}
a^2+b^2-c^2-d^2 & 2(bc-ad) & 2(bd+ac) \\
2(bc+ad) & a^2-b^2+c^2-d^2 & 2(cd-ab) \\
2(bd-ac) & 2(cd+ab) & a^2-b^2-c^2+d^2
\end{pmatrix}
\end{equation}
The Lie algebra matrices can be used in matrix exponentiation, for example to define a $3\times 3$ quaternion (a rotation matrix) using the roll-pitch-yaw angles:
https://en.wikipedia.org/wiki/Conversion_between_quaternions_and_Euler_angles
\begin{align}
\exp(\theta \pi_1)\;&=\;
\begin{pmatrix}
1 & 0 & 0 \\
0 & \cos(\theta) & -\sin(\theta) \\
0 & \sin(\theta) & \cos(\theta)
\end{pmatrix}\\
\exp(\theta \pi_2)\;&=\;
\begin{pmatrix}
\cos(\theta) & 0 & \sin(\theta) \\
0 & 1 & 0 \\
-\sin(\theta) & 0 & \cos(\theta)
\end{pmatrix}\\
\exp(\theta \pi_3)\;&=\;
\begin{pmatrix}
\cos(\theta) & -\sin(\theta) & 0\\
\sin(\theta) & \cos(\theta) & 0 \\
0 & 0 & 1
\end{pmatrix}
\end{align}
|
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|
Let $x,y,z$ be real numbers in the interval $[-1,2]$ such that $x+y+z=0$. Prove that $\sum_{cyc}\sqrt{\frac{(2-x)(2-y)}{(2+x)(2+y)}}\ge 3$ Let $x,y,z$ be real numbers in the interval $[-1,2]$ such that $x+y+z=0$. Prove that
$ \sqrt{\frac{(2-x)(2-y)}{(2+x)(2+y)}}+\sqrt{\frac{(2-y)(2-z)}{(2+y)(2+z)}}+\sqrt{\frac{(2-z)(2-x)}{(2+z)(2+x)}}\ge 3$
My working: let $\frac{2-x}{2+x}=a^2,\frac{2-y}{2+y}=b^2,\frac{2-z}{2+z}=c^2$
$\implies a,b,c\in[0,\sqrt 3], \sum\frac{1-a^2}{1+a^2}=0\text{ or } 3+a^2+b^2+c^2-a^2b^2-b^2c^2-c^2a^2-3a^2b^2c^2=0$ and to prove $ab+bc+ca\ge3$
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As you suggested, let $p=a+b+c$, $q=ab+bc+ca$, $r=abc$. We use contradiction: if $q<3$, then $f(p,q,r)=3+a^2+b^2+c^2-a^2b^2-b^2c^2-c^2a^2-3a^2b^2c^2=3+p^2-2q-q^2+2pr-3r^2>0$ and thus leads to a contradiction. If $p,q$ is fixed, we must have $f$ takes its minimum when $r$ is at its extreme. Consider $F(X)=x^3-p^2X+qX$. This is a cubic curve, and the three roots is it intersect with $y=r$. So if $r$ takes the extreme values, we must have: either $a,b,c$ gets to the left, right walls ($0$ and $\sqrt{3}$), or gets to upper, lower walls (two of $a,b,c$ are equal). So we only need to prove the case with one of $a,b,c$ is zero, or $\sqrt{3}$, or $a=b$. That is, for this inequality (and this TYPE of inequalities) we only need to consider the special cases.
Case 1: $a=\sqrt{3}$.
Then $p=(\sqrt{3}+b+c)$, $q=\sqrt{3}(b+c)+bc$, $r=\sqrt{3}bc$. Let $s=b+c$ and $t=bc$, so $f(p,q,r)=6-2(b^2+c^2)-4b^2c^2=6-2s^2+4t-4t^2$. $q<3$ means $\sqrt{3} s+t< 3$ . To minimize $f$, we need $s$ to be maximized, so $s<\sqrt{3}-\frac{1}{\sqrt{3}}t$. So $f=6-2s^2+4t-4t^2>6-2(\sqrt{3}-\frac{1}{\sqrt{3}}t)^2+4t-4t^2=8t-\frac{14}{3}t^2$. Notice that $s^2>4t$ so we have $2\sqrt{t}\le s<\sqrt{3}-\frac{1}{\sqrt{3}}t$ and thus we can conclude that $t<1$. So $8t-\frac{14}{3}t^2\ge 0$ and thus $f>0$.
Case 2: $a=0$.
Then $p=b+c$, $q=bc<3$, $r=0$, so we have $p^2\ge 4q$, and thus $f=3+p^2-2q-q^2\ge 3+4q-2q-q^2=3+2q-q^2=(3-q)(1+q)>0$
Case 3: $a=b$
We have $f=3+2a^2-a^4+(1-2a^2-3a^4)c^2$. The condition $q<3$ transformed to $a(a+2c)<3$. Notice that this $f$ takes the minimum value, the $c^2$ (thus $c$) must take extreme values, so $c=0,c=\sqrt{3}$ or $c<\frac{3}{2a}-\frac{a}{2}$. Take $c=\frac{3}{2a}-\frac{a}{2}$, substitute it in, we have
$$f>3+2a^2-a^4+(1-2a^2-3a^4)(\frac{3}{2a}-\frac{a}{2})^2 = \frac{3(3-a^2)(a-1)^2(a+1)^2(a^2+1)}{4a^2}\ge 0$$
Thus $f>0$.
|
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|
Find $x$ in the figure
Find $x$ in the figure.
(Answer: $20^\circ$)
My progress:
Let $P$ such that $PDHC$ is cyclic.
$\angle ACH = 180^\circ-20^\circ-90^\circ = 70^\circ\implies AHC = 100^\circ$
$\therefore β PDC=β AHC=100^\circ$ and $β PDA=180^\circ-60^\circ-100^\circ=20^\circ$
$\therefore β BDH+x=60^\circ$
If I find out $\angle DCP$, ends...??
|
Leveraging colleague Vasily's perpendicular idea:
Let $P$ be a point on $AH$ such that the quadrilateral $PDHC$ is cyclic.
$\angle ACH = 180^\circ-20^\circ-90^\circ = 70^\circ\implies AHC = 100^\circ.$
$\therefore β PDC=β AHC=100^\circ$ and $β PDA=180^\circ-60^\circ-100^\circ=20^\circ.$
Trace $DE \perp AH ~ (E \in AC).$
$G=DC \cap AH$ and $F =DE \cap AH.$
$\angle DPG = 30^\circ$
$\therefore \angle DGP = 180^\circ - 100^\circ-30^\circ = 50^\circ \implies \angle FDG = 40^\circ$
$\therefore \boxed{\color{red}x = 60^\circ-40^\circ = 20^\circ}$
|
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|
Finding integer solutions to $(x-y)^2+(y-z)^2+(z-x)^2=2022$ This was from round A of the Awesomemath Summer Program application. The deadline was yesterday and we can discuss the problems now.
Find all integer triples $(x,y,z)$ which satisfy
$$(x-y)^2+(y-z)^2+(z-x)^2=2022$$
I tried a couple of things. First I noticed $x-y+y-z+z-x=0,$ so if we let $a=x-y, b=y-z,c=z-x$ than we have
$$\begin{align}
a^2+b^2+c^2&=2022 \tag1\\
a+b+c &=0 \tag2\\
(a+b+c)^2-2ab-2bc-2ca &=2022 \tag3
\end{align}$$ So,
$$-2ab-2bc-2ca=2022 \tag4$$
so
$$ab+bc+ca=-1011 \tag5$$
However, it is tricky to proceed from here.
I also recalled the factorization
$$x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}((x-y)^2+(y-z)^2+(z-x)^2)=1011 \tag6$$ This also equals $$x(x-y)+y(y-z)+z(z-x) \tag7$$
but this doesn't seem promising.
I also tried to factorize $2022=2 \cdot 3 \cdot 337$, but I am not sure what to do after prime factorizing.
So far, I feel like my first approach is the most promising, but I am not sure how to finish with it.
|
I am going from here: instead of dividing two, I am going to multiple two.
$$4044=2((x-y)^2+(y-z)^2+(z-x)^2)=(2x-y-z)^2+3(y-z)^2$$
WLOG $x\ge y\ge z$ Therefore set $2x-y-z=a$ and $y-z=b$ we are finding the solution of $a^2+3b^2=4044$. After trying, we have three solutions: $(63,5),(39,29),(24,34)$. $(63,5)$ yields $x=34+z,y=5+z$, $(39,29)$ yields $x=34+z,y=29+z$, and $(24,34)$ does not yield $x\ge y\ge z$ solution. These two solutions are "somewhat equivalent." So, the solutions $x=34+z,y=5+z$ and $x=34+z,y=29+z$ if ordered $x\ge y\ge z$, or, all the solutions are $x,y,z$, when they ordered as $a\ge b\ge c$ and $a-b,b-c$ is a permutation of $5$ and $29$.
|
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|
determinant of $a_{ij}=2$ where if $i=j$ and $1$ if $|i-j|=1$, otherwise zero I want to find the general formula for $n\times n$ matrix $A$ whose entries $a_{ij}$ is $2$ if $i=j$ and $1$ for $|i-j|=1$ and $0$ otherwise.
Followings are my explicit trials;
For $n=3$ case I have
\begin{align}
\begin{vmatrix}
2 & 1 & 0 \\
1 & 2 & 1 \\
0 & 1 & 2
\end{vmatrix} = 2\times (4-1) -1\times 2 =4
\end{align}
and for $n=4$
\begin{align}
\begin{vmatrix}
2 & 1 & 0 & 0 \\
1 & 2 & 1 & 0 \\
0 & 1 & 2 & 1 \\
0 & 0 & 1 & 2
\end{vmatrix}= 2 \begin{vmatrix}
2 & 1 & 0 \\
1 & 2 & 1 \\
0 & 1 & 2
\end{vmatrix} - \begin{vmatrix}
1 & 1 & 0 \\
0 & 2 & 1 \\
0 & 1 & 2
\end{vmatrix} = 5
\end{align}
How one can construct the general form of determinant?
Denoting $a_n$ be the determinant of $A$ for $n$, I see $a_n = 2 a_{n-1} - x $ (I am not sure the correct structure of $x$ at this moment)
Any ideas? Is there some nice trick or algorithm
|
You're on the right track. Call the matrix you've defined in dimension $n$ $A_n$. Using expansion by minors on the first row, you get
$$\det(A_n)=2\det(A_{n-1})-\det(B_{n-1}),$$
where $B_{n-1}$ is $A_n$ but with the first row and second column removed. Now, we wish to determine the determinant of $B_n$. The only nonzero entry of the first column of $B_n$ is the one in its top left (corresponding to the one in column $1$ of the original matrix), and so expanding $B_n$ by minors along the first column gives that $\det(B_{n-1})=\det(A_{n-2})$. So, you get the recurrence $a_n=2a_{n-1}-a_{n-2}$.
|
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|
Integrate using integration by parts $\int e^{-\frac{x}{2}}\frac{\sqrt{1-\sin{x}}}{1+\cos{x}}dx$ Integrate using integration by parts $\int e^{-\frac{x}{2}}\frac{\sqrt{1-\sin{x}}}{1+\cos{x}}dx$
My Attempt
I tried taking $u=e^{-\frac{x}{2}}$ to evaluate using the general formula $\int udv=uv-\int vdu$, but I ended up with a complex trigonometric expression which I couldn't simplify further ( I can show this complex result of mine if needed)
I'm wondering if there's a suitable substitution which would simplify the integration by part process. It would be great if anyone can give me a Hint to work this integral. Thank you in advance!
|
The answer is not-that correct: we assume $-\pi/2\le x\le\pi/2$. Look at this, we first consider the half-angle transform
$$\frac{\sqrt{1-\sin{x}}}{1+\cos{x}}=\frac{\cos \frac x2-\sin \frac x2}{2\cos^2 \frac x2}$$
Notice that $$\frac{{\rm d}}{{\rm d}x}\frac{1}{\cos \frac x2}=\frac{\sin \frac x2}{2\cos^2 \frac x2}$$
So we have
\begin{align}
&\int e^{-\frac{x}{2}}\frac{\sqrt{1-\sin{x}}}{1+\cos{x}}{\rm d}x=\int e^{-\frac{x}{2}}\frac{\cos \frac x2-\sin \frac x2}{2\cos^2 \frac x2}{\rm d}x\\
=&\int e^{-\frac{x}{2}}\frac{\cos \frac x2}{2\cos^2 \frac x2}{\rm d}x-\int e^{-\frac{x}{2}}\frac{\sin \frac x2}{2\cos^2 \frac x2}{\rm d}x\\
=&\int e^{-\frac{x}{2}}\frac{1}{2\cos \frac x2}{\rm d}x-e^{-\frac{x}{2}}\frac{1}{\cos\frac x2}-\int \frac 12 e^{-\frac{x}2}\frac{1}{\cos\frac x2}{\rm d}x=-e^{-\frac{x}{2}}\frac{1}{\cos\frac x2}
\end{align}
|
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|
Describe the locus of $w$ if $w=\frac{1-z}{1+z}$ and $z=1+iy$ (i.e $z$ is a complex number that moves along the line $x=1$) So I'm trying to solve the following problem: If $z=x+iy$, express $w=\frac{1-z}{1+z}$ in the form $a+bi$ and hence find the equation of the locus of $w$ if $z$ moves along the line $x=1$.
My attempt is:
$w=\frac{1-z}{1+z}\times\frac{1+\bar{z}}{1+\bar{z}}$
$=\frac{1-2i\Im{(z)}-|z|^2}{1+2\Re{(z)}+|z|^2}$
$=\frac{1-2yi-x^2-y^2}{1+2x+x^2+y^2}$
$\therefore a+bi=\frac{1-x^2-y^2}{(x+1)^2+y^2}-\frac{2y}{(x+1)^2+y^2}i$
Substituting $x=1$ gives the parametric equations $a=\frac{-y^2}{4+y^2}$ and $b=\frac{-2y}{4+y^2}$
My problem is that I can't reduce these two to only an equation in terms of $a$ and $b$ (best I can do is $b=2ya$, when I try to find $y$ in terms of $b$ it goes $4+y^2=-\frac{2y}{b}$ which is $y^2-\frac{2y}{b}+\left(\frac{y}{b}\right)^2=-4+\left(\frac{y}{b}\right)^2$ and so $\left(y-\frac{y}{b}\right)^2=-4+\left(\frac{y}{b}\right)^2$ but clearly this is not working) so I'm not sure if my steps are right into deriving the parametric equations or if the question just requires me to be more capable of reducing parametric equations. If they can be reduced into equations of just $a$ and $b$, could someone show me how?
|
A more systematic approach to get the result would be as follows:
First rearrange the relationship $w=\frac{1-z}{1+z}$ to make $z$ the subject, then apply the condition that the real part of $z$ is $1$.
Therefore, writing $w=u+iv$, we have $$z=\frac{1-w}{1+w}=\frac{1-u-iv}{1+u+iv}\cdot\frac{1+u-iv}{1+u-iv}$$
The real part of this is $$\frac{1-u^2-v^2}{(1+u)^2+v^2}=1$$
This simplifies to become $$u^2+v^2+u=0$$
i.e. the locus is the circle $$\left(u+\frac12\right)^2+v^2=\left(\frac12\right)^2$$
|
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|
Prove that $\sum \frac{a^3}{a^2+b^2}\le \frac12 \sum \frac{b^2}{a}$
Let $a,b,c>0$. Prove that
$$ \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\le \frac12 \left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right).\tag{1}$$
A idea is to cancel the denominators, but in this case Muirhead don't work because the inequality is only cyclic, not symmetric.
Another idea would be to apply Cauchy reverse technique:
$$(1)\iff\sum \left(a-\frac{a^3}{a^2+b^2}\right)\ge \frac12 \sum (2a-b^2/a)\iff \sum\frac{ab^2}{a^2+b^2}\ge \frac12\sum\frac{2a^2-b^2}{a}$$
$$\iff \sum \frac{(ab)^2}{a^3+ab^2}\ge \frac12\sum \frac{2a^2-b^2}a.$$
Now we can apply Cauchy-Schwarz, and the problem reduces to
$$\frac{(\sum ab)^2}{\sum a^3+\sum ab^2}\ge \frac12\sum \frac{2a^2-b^2}{a},$$
and at this point I am stuck. Here the only idea is to cancel the denominators, but as I say above it can't work.
|
(Just AM-GM is sufficient.)
From OP's work / River Li's solution, the stated inequality is equivalent to $\sum a - \frac{a^3}{a^2+b^2} \geq \sum a - \frac{1}{2} \sum \frac{b^2}{a}$, which is:
$$ \sum \frac{ab^2 } { a^2 + b^2 } \geq \sum a - \frac{1}{2} \sum \frac{ b^2}{a}$$
Let $ X = \sum \frac{ab^2 } { a^2 + b^2 } , Y = \sum a, Z = \sum \frac{ b^2}{a}$ for simplicity.
We want to show that $ X \geq Y - \frac{1}{2} Z$.
Notice that by AM-GM,
$$ \frac{ 4ab^2 }{ a^2 + b^2 } + \frac{ a^2 + b^2 } { a} \geq 4b,
\quad \frac{ b^2}{a} + a \geq 2b. $$
Summing up the cyclic versions gives us
*
*$4X + Y + Z \geq 4Y$
*$ Z + Y \geq 2Y \Rightarrow Z \geq Y$.
*Hence $4 X \geq 3Y - Z \geq 4Y - 2Z$ as desired.
Note: From $ 4X \geq 4Y - (Y+Z)$, the inequality could be strengthened to
$$ \sum \frac{a^3}{a^2+b^2} \leq \frac{1}{4} \sum \frac{ a^2+b^2}{a} \leq \sum \frac{1}{2} \frac{ b^2}{a} $$
|
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|
How to solve in integers the equation $2x^2+x+2x^2y-y^2+y=1$? Any hints? I could've reduced it to Pell's equation and solved it, if there wasn't $2x^2y$ part.
This is a part of a bigger problem, and that's all I have left to solve the main one.
|
\begin{align*}
2x^2 &+x+2x^2y-y^2+y=1
\\
\\
2(y+1)x^2 &+(1)x-(y^2-y+1)=0\\
x &= \dfrac{\sqrt{8 y^3 + 9} - 1}{4 (y + 1)}
\\ \\
y^2& - (2x^2+1)y -(2x^2+x-1)=0
\\
y &= \dfrac{2 x^2 + 1+ \sqrt{4 x^4 + 12 x^2 + 4 x - 3}}{2}
\end{align*}
Solving for $\space x\space$ suggests that
$\space y\ne-1,\space$
but solving for $\space y\space$ and viewing a graph of this function shows that this is not necessarily so. The graph of this function suggests that solutions are not far from the origin.
A spreadsheet may not show everything but, for $\space -100 \le x \le 100,\space$ the integer solutions are
$\quad (-1,0),\space (-1,3),\space (3,-1),
\space (3,20).$
|
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|
Six people (half are female, half are male) for seven chairs. Problem:
Suppose there are $7$ chairs in a row. There are $6$ people that are going to randomly
sit in the chairs. There are $3$ females and $3$ males. What is the probability that
the first and last chairs have females sitting in them?
Answer:
Let $p$ be the probability we seek. Out of $3$ females, only $2$ can be sitting at the end of the row. I consider the first and last chairs to be at the end of the row.
\begin{align*}
p &= \dfrac{ {3 \choose 2 } 3(2) (4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ 3(3)(2) (4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ 3(4)(3)(2) } { 7(5)(4)(3)(2) } = \dfrac{ 3(3)(2) } { 7(5)(3)(2) } \\
p &= \dfrac{ 18 } { 35(3)(2) } \\
p &= \dfrac{ 3 } { 35 }
\end{align*}
Am I right?
Here is an updated solution.
Let $p$ be the probability we seek. Out of $3$ females, only $2$ can be sitting at the end of the row. I consider the first and last chairs to be at the end of the row.
\begin{align*}
p &= \dfrac{ 3(2) (5)(4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ (5)(4)(3)(2) } { 7(5)(4)(3)(2) } \\
p &= \dfrac{1}{7}
\end{align*}
Now is my answer right?
|
Here is how I would think about it. There are seven of them - three male, three female and an empty chair (say, a ghost). In other words, for any given chair, there are seven equally likely possibilities.
So, $ \displaystyle P = {3 \choose 2} / {7 \choose 2} = \frac 17$
|
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|
$A+B+C+D=\pi$, and $0\leq A,B,C,D \leq \frac{\pi}{2}$. Prove that $\sin^4(A)+\sin^4(B)+\sin^4(C)+\sin^4(D)\leq 2$ I tried to simplify it from some ways.
(1).$\sin^4(A)+\sin^4(B)+\sin^4(C)+\sin^4(D)$
$=\left(\frac{1-\cos(2A)}{2}\right)^2+\left(\frac{1-\cos(2B)}{2}\right)^2+\left(\frac{1-\cos(2C)}{2}\right)^2+\left(\frac{1-\cos(2D)}{2}\right)^2$
$=1-\frac{1}{2}\left(\cos(2A)+\cos(2B)+\cos(2C)+\cos(2D)\right)+\frac{1}{4}\cdot\left(2+\frac{\cos(4A)+\cos(4B)+\cos(4C)+\cos(4D)}{2}\right)$
$=\frac{3}{2}-\frac{1}{2}\left(\cos(2A)+\cos(2B)+\cos(2C)+\cos(2D)\right)+\frac{1}{8}\left(\cos(4A)+\cos(4B)+\cos(4C)+\cos(4D)\right)$
(2).$\cos(2A)+\cos(2B)+\cos(2C)+\cos(2D)$
$=2\cos(A+B)\cos(A-B)+2\cos(C+D)\cos(C-D)$
since $A+B=\pi-(C+D)$, thus $\cos(A+B)=-\cos(C+D)$. It follows that
$=2\cos(A+B)\left(\cos(A-B)-\cos(C-D)\right)$
$=2\cos(A+B)\cdot\left(-2\sin(\frac{A-B+C-D}{2})\sin(\frac{A-B-C+D}{2})\right)$
$=2\cos(A+B)\cdot\left(-2\sin\left(\frac{\pi}{2}-(B+D)\right)\sin\left(\frac{\pi}{2}-(B+C)\right)\right)$
$=2\cos(A+B)\cdot\left(-2\cos(B+D)\cos(B+C)\right)$
$=-4\cos(A+B)\cos(A+C)\cos(A+D)$
(3).Similarly,$\cos(4A)+\cos(4B)+\cos(4C)+\cos(4D)$
$=2\cos(2A+2B)\cos(2A-2B)+2\cos(2C+2D)\cos(2C-2D)$
$=2\cos(2A+2B)\cos(2A+2C)\cos(2A+2D)$
Am I on the right track? Maybe it is a Jensen's-inequality problem but it seems not always concave up or down in the interval $[0,\frac{\pi}{2}]$. I am stuck here. Please help, and thank you.
|
Remark i): The desired inequality is still true without the condition $A, B, C, D \le \pi/2$.
Remark ii): According to my proof, for $A + B + C + D = \pi$, we have the following identity:
\begin{align*}
&2 - \sin^4 A - \sin^4 B - \sin^4 C - \sin^4 D \\
=\, & \sin^2 (A - B) \cos^2 (A + B)
+ \frac12(\cos (A + B) + \cos (A - B))^2\\
&\quad + 2\cos(C + D)\sin C \sin D
+ \sin^2 C\cos^2 C + \sin^2 D \cos^2 D.
\end{align*}
$\phantom{2}$
Problem 1: Let $A, B, C, D \ge 0$ with $A + B + C + D = \pi$. Prove that
$$\sin^4 A + \sin^4 B + \sin^4 C + \sin^4 D \le 2.$$
Proof:
WLOG, assume that $A \ge B \ge C \ge D$.
Clearly, $C + D \le \pi/2$. We have (See Remark 1 at the end for details)
$$\sin^2 (C + D) - \sin^2 C - \sin^2 D
= 2\cos (C + D) \sin C \sin D \ge 0. \tag{1}$$
Thus, we have
$$\sin^4 C + \sin^4 D
\le \sin^2 C + \sin^2 D \le \sin^2 (C + D).$$
It suffices to prove that
$$\sin^4 A + \sin^4 B + \sin^2(C + D) \le 2$$
or
$$\sin^4 A + \sin^4 B + \sin^2(A + B) \le 2$$
which is written as (see Remark 2 at the end for details)
$$\sin^2 (A - B) \cos^2 (A + B)
+ \frac12(\cos (A + B) + \cos (A - B))^2 \ge 0. \tag{2}$$
We are done.
$\phantom{2}$
Remark 1: We have
\begin{align*}
&\sin^2(C + D) - \sin^2 C - \sin^2 D\\
=\,& \frac{1 - \cos (2C + 2D)}{2} - \frac{1 - \cos 2C}{2} - \sin^2 D\\
=\,& \frac{\cos 2C - \cos (2C + 2D)}{2} - \sin^2 D\\
=\,& \sin (2C + D) \sin D - \sin^2 D\\
=\,& (\sin (2C + D) - \sin D)\sin D\\
=\,& 2\cos (C + D) \sin C \sin D.
\end{align*}
$\phantom{2}$
Remark 2: Let $u = A + B, v = A - B$ (correspondingly, $A = \frac{u + v}{2}, B = \frac{u - v}{2}$). We have
\begin{align*}
&2 - \sin^4 A - \sin^4 B - \sin^2(A + B)\\
=\,& 2 - \left(\frac{1 - \cos (u + v)}{2}\right)^2 - \left(\frac{1 - \cos (u - v)}{2}\right)^2 - \sin^2 u\\
=\,& \frac{3}{2} - \frac{\cos^2(u + v) + \cos^2 (u - v)}{4} + \frac{\cos(u + v) + \cos (u - v)}{2} - \sin^2 u \\
=\,& \frac{3}{2} - \frac{\cos^2 u \cos^2 v + \sin^2 u \sin^2 v}{2} + \cos u \cos v - \sin^2 u \\
=\,& - \cos^2 u \cos^2 v + \frac{1}{2}\cos^2 v + \frac{3}{2}\cos^2 u + \cos u \cos v \\
=\,& \cos^2 u \sin^2 v + \frac12(\cos u + \cos v)^2.
\end{align*}
|
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|
To find the number of ways to put $14$ identical balls into $4$ bins with the condition that no bin can hold more than $7$ balls. To find the number of ways to put $14$ identical balls into $4$ bins with the condition that no bin can hold more than $7$ balls.
I have tried the following:
The total no of ways to distribute $14$ identical balls into $4$ bins without any restriction is $$\binom{14+4-1}{4-1}= \binom{17}{3}.$$
Note that there can't be two bins with more than $7$ balls since we have only $14$ identical balls only.
Now, we count the no. of ways so that one bin has more than $7$ balls. So, it has at least $8$ balls and the remaining $6$ can be distributed in $\binom{6+4-1}{4-1}= \binom{9}{3}$ ways. We can choose one bin out of $4$ in $4$ ways.
Hence the reqd number of ways = $$\binom{17}{3} - 4 \times \binom{9}{3}.$$
|
Your required sum is the coefficient of $x^{14}$ in
$$(1+x+\dots+x^7)^4$$
or
$$[x^{14}]:\left(\frac{1-x^8}{1-x}\right)^4$$
We can expand $(1-x^8)^4$ into $1-4x^8$, and the other term is $\sum_{i=0}^\infty \binom{i+3}{3}x^i$.
We want the values only when $i=6$ or $i=14$, which gives your equation.
|
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Integer solutions of $ x^4+34x^2 y^2+y^4=z^2$ Now I'm trying to solve one problem, and already twice an expression of the form $a x^4+b x^2 y^2+a y^4 \,\,(x > y > 1)$ appears under the square root.
First was: $9 x^4 - 14 x^2 y^2 + 9 y^4$
Second: $x^4 + 34 x^2 y^2 + y^4$
I suppose that there will be other equations of this type.
Question: Are there integer solutions or a parametrization of equations$ a x^4+b x^2 y^2+a y^4=z^2$ in general, or of the variants proposed above in particular? Or maybe can the absence of solutions be proven?
Upd: No solutions
Second case: according to https://www.cambridge.org/core/services/aop-cambridge-core/content/view/0B14DEDE386776126F3B5A36CA325ECF/S0017089500007862a.pdf/x4-dx2y2-y2-z2-some-cases-with-only-trivial-solutions-and-a-solution-euler-missed.pdf (Thanks Will Jagy for link)
First case: $9 x^4 - 14 x^2 y^2 + 9 y^4$ can be written as $(3(x^2-y^2))^2 +(2xy)^2=z^2$ Pythagorean triple. So $2xy=2ab, 3(x^2-y^2)=a^2-b^2$
From this: $3y^4+(a^2-b^2)y^2-3a^2b^2=0$ can be obtain.
Discriminant $a^4+34a^2b^2+b^4$ (Second equation of my question. Strange coincidence). So there is no rational roots.
|
At first, it looks like the equation can be rewritten as a Pythagorean triple
$\space A^2+B^2=C^2\quad$ here shown by Euclid's formula:
$\quad A=(m^2-k^2) \quad B=2mk\quad C=(m^2+k)^2\space $ and a triple may be found given any legitimate value of $A,\space B, \space \text{or}\space C.$
$\text{Here we have}\quad(x^2-y^2)^2 + (6xy)^2 =z^2\space$ and $\space B= 6xy\ne 2mk\space$ so it is not a Pythagorean triple because it cannot quite be a right triangle. The best approach may be experimentation and observation.
\begin{align*}
x=0,\space y=0 &\implies z\in\mathbb{Z}^2
\\
x=y,\space\space \space x,y>0 &\implies z=6n^2\space ,n\in\mathbb{N}\\
\not\exists \space x>y>1
\\
x=0,\space y=4 &\implies z=\pm8
\\
x=4,\space y=0 &\implies z=\pm8
\end{align*}
|
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|
Using Lagrange Mult. to find maximum of $2x + y^2$ with constraints: $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ I was trying to maximize the function $2x + y^2$ with the following constraints: $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ using Lagrange multipliers. I already did all the work of proving that the maximum does exist and showing that we can apply the technique of Lagrange multiplies. After that I arrived at the following systems of equations:
$$\begin{cases}2 = 2\alpha x + \beta \\
2y = 2\alpha y + 2 \beta y \\
0 = 2\alpha z + \beta \\
x^2 +y^2 + z^2 =2 \\
x + y^2 + z = 0\end{cases}$$
Where $\alpha$ and $\beta$ are the Lagrange multipliers. The problem is that I'm having a lot of trouble solving this system for $(x,y,z)$.
My plan was trying to write $x,y,z$ in terms of $\alpha,\beta$ and then using the equations $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ find the values for $\alpha $ and $\beta$, but I wasn't able to find an expression for $y$ in terms of $\alpha$ and $\beta$.
How can this system be solved?
|
To simplify things, solve for $z$ from the second constraint, and plug it into the first constraint, as follows
From $x + y^2 + z = 0 $ you get $ z = - x - y^2 $
Plug this into $ x^2 + y^2 + z^2 = 2$ you get $ x^2 + y^2 +(x + y^2)^2 = 2 $
which simplifies to $ 2 x^2 + y^2 (1 + 2 x) + y^4 = 2 $
The Lagrange function is
$ g = 2x + y^2 + \lambda ( 2 x^2 + y^2 (1 + 2 x) + y^4 - 2 ) $
$g_x = 2 + \lambda ( 4 x + 2y^2 ) $
$ g_y = 2 y + \lambda ( 2 y (1 + 2 x) + 4 y^3 )$
$g_\lambda = 2 x^2 + y^2 (1 + 2 x) + y^4 - 2$
Setting these to zero, implies
$ \lambda = \dfrac{-2}{ 4 x + 2 y^2} = \dfrac{ - 2 y}{ 2 y(1+ 2x) + 4y^3 } $
Cross multiplying,
$2 y (1 + 2 x) + 4 y^3 = y ( 4 x + 2 y^2 ) $
Dividing through by $2$,
$y (1 + 2 x) + 2 y^3 = y (2 x + y^2 ) $
$ y + 2 x y + 2 y^3 = 2 x y + y ^ 3 $
$ y + y^3 = 0 $
$ y( 1 + y^2) = 0 $
Hence, $ y = 0 $
Using the third constraint,
$ 2 x^2 - 2 = 0 $
So $x = \pm 1 $
Clearly the maximum occurs at $x = 1$, hence $z = -x - y^2 = -1$
Therefore, the maximum of $2$ is achieved at $(1, 0, -1)$
|
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|
Why $\sin(x) = x - \frac{x^3}{6} + o(x^5)$ is false? I'm wondering why the following use of little o is incorrect.
$$
\sin(x) = x - \frac{x^3}{6} + o(x^5)
$$
I know that te definition of little o is the following:
$$
f(x) = o(g(x)) \quad \text{when } x \to x_0 \qquad \text{if } f(x)=g(x) \, w(x) \quad \text{with } \lim_{x \to x_0} w(x) = 0
$$
In my case should be $x_0=0$, $f(x) = x - \frac{x^3}{6}$, and $g(x) = x^3$. It seems that the limit tends to infinity, but the same situation happens if I consider (the following is correct):
$$
\sin(x) = x - \frac{x^3}{6} + o(x^4)
$$
Answer suggested by the guys below: in the first case we have:
$$
w(x) = \frac{\sin(x) - \left( x - \frac{x^3}{3!} \right)}{x^5} =
\frac{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \left( x - \frac{x^3}{3!} \right)}{x^5} = \frac{1}{5!}
$$
so:
$$
\lim_{x \to 0} w(x) = \frac{1}{5!} \neq 0
$$
Instead in the second case:
$$
w(x) = \frac{\sin(x) - \left( x - \frac{x^3}{3!} \right)}{x^4} = \frac{x - \frac{x^3}{3!} - \left( x - \frac{x^3}{3!} \right)}{x^4} = 0
$$
so:
$$
\lim_{x \to 0} w(x) = 0
$$
|
Your problem is that the sentence $\sin(x) = x - \frac{x^3}{6} + o(x^5)$ is NOT saying that $f(x) = o(x^5)$ where $f(x) = x - \frac{x^3}{6}$. It is saying that $f(x) = o(x^5)$ where $f(x) = \sin(x) - x + \frac{x^3}{6}$. Now use the Taylor expansion for sine.
|
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|
Functional analysis problem. Find the smallest value of $f(x) = \frac{x^2}{x-9}$ on the interval $(9, +\infty)$.
We should basically find the biggest $a$ such that $\frac{x^2}{x-9} \geq a$. We can multiply both sides by $x-9$ since it's positive and than we get $x^2-ax+9a \geq 0.$ I don't know how to proceed.
|
Here is a purely algebraic way using the inequality between arithmetic and geometric mean (AM-GM):
For $x>9$ we have
\begin{eqnarray*} \frac{x^2}{x-9}
& = & \frac{x^2-81+81}{x-9} \\
& = & x+9 + \frac{81}{x-9} \\
& = & x-9 + \frac{81}{x-9} + 18 \\
& \stackrel{AM-GM}{\geq} & 2\sqrt{81} + 18\\
& = & 36
\end{eqnarray*}
Equality holds if and only if
$$x-9 = \frac{81}{x-9} \stackrel{x>9}{\Leftrightarrow} x=18$$
|
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|
What is the area of the triangle here?
We are given that the angle of $BAD$ is $2\alpha$ and the angle of $DAC$ is $\alpha$. $|AC| = 10$, $|BD| = 6$, $|DC| = 5$ units. Find the area of the triangle $ABC$.
The answer would be $33$. We need to show that if we drop an altitude from $A$ to $BC$ at point $E$, $|AE| = 6, |EC| = 8$. Somehow $\triangle AEC$ becomes $6$-$8$-$10$ triangle. Since $|AE| = 6$ and $|BC|=11$, the area becomes $33$.
But how can we prove that?
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Let area of $\triangle BAD = A_1$ and area of $\triangle CAD = A_2$. Let $AD = m$ and $AB = x$.
As they share a common height, the ratio $\frac{A_1}{A_2} = \frac 65$
Also, using an alternative formula for triangular area, $\frac{A_1}{A_2} = \frac{\frac 12xm\sin 2\alpha}{\frac 12(m)(10)\sin \alpha} = \frac{x\cos\alpha}{5}$ (where a double angle identity was used).
Equating the two expressions and rearranging, $\cos \alpha = \frac 6x$.
Now apply cosine rule on $\triangle ABC$, to give $11^2 = x^2 + 10^2 - 2(x)(10)\cos 3\alpha$.
Using a triple angle identity ($\cos 3\alpha = 4\cos^3 \alpha - 3\cos\alpha$),
$x^2 - 20x(4(\frac 6x)^3 - 3(\frac 6x)) - 21 = 0$
After a bit of algebra and setting $y = x^2$ to reduce the quartic to a quadratic, we get:
$y^2 + 339y - 17280 = 0$.
Solving for the positive root, $y = 45 \implies x =\sqrt{45}$.
At this point, we can proceed to find the area of $\triangle ABC$ in a couple of ways. One way is to apply Heron's formula using the three known sides. I instead opted to apply the previously used formula, so that the area $A = \frac 12(x)(10)\sin 3\alpha$.
From $\cos \alpha = \frac 6x$ we get $\sin \alpha = \frac 3{\sqrt{45}}$. Applying a triple angle formula ($\sin 3\alpha = 3\sin\alpha - 4\sin^3\alpha$),
$A = \frac 12(x)(10)(\frac 9{\sqrt{45}} - 4(\frac{27}{45\sqrt{45}})) = 5\sqrt{45}(\frac 9{\sqrt{45}} - \frac{12}{5\sqrt{45}}) = 45- 12 = 33$.
So the answer is $33$.
|
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|
$y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati) $y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati)
(a) Find the solutions.
(b) $y(x_0)=y_0$, prove two cases:
$$0<y_0<x_0 \implies \text{solution's domain is} [x_0,\infty) $$
$$0<x_0<y_0 \implies \text{solution's domain is} [x_0,x_0+\alpha) , \alpha \in \mathbb {R}.$$
I will be grateful for help in $(b)$.
My solution for (a):
$y_1(x)=x$
Denote $y=x+z$
$$(x+z)'=2-\frac{3}{x}(x+z)+\frac{2}{x^2}(x+z)^2$$
$$1+z'=-1-\frac{3z}{x}+\frac{2}{x^2}(x^2+2xz+z^2)$$
$$z'=\frac{z}{x}+\frac{2z^2}{x^2}$$
This is a Bernoulli differential Equation.
Denote $u=z^{-1} \implies -\frac{z'}{z^2} \implies z'=-u'z^2$
Then,
$$u'=\frac{u}{x}-\frac{2}{x^2}$$
Integration factor is $\mu=x$
$$\int(xu)'=\int-\frac{2}{x} \implies xu=-2\ln|x|+c \implies u=\frac{-2\ln|x|+c}{x} \implies z=\frac{x}{-2\ln|x|+c}$$
The solution is $y=x+\frac{x}{-2\ln|x|+c}$
Is my solution correct ?
Thanks !
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An alternative approach parametrizes solutions as $y=2\frac{u}{u'}$ to get
$$
y''=2-2\frac{u''u}{u'^2}=2-\frac6{x}\frac{u}{u'}+\frac{8}{x^2}\frac{u^2}{u'^2}
$$
This simplifies to
$$
0=u''-\frac{3}{x}u'+\frac{4}{x^2}u
$$
This now is an Euler-Cauchy equation, with correspondingly simple solutions.
|
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|
Show that it is impossible to find $a,b \in \mathbb{Q}$ such that $ \frac{\sqrt{5}}{10}=a+b \frac{\sqrt{3}}{7} $ Show that it is impossible to find $a,b \in \mathbb{Q}$ such that
$$
\frac{\sqrt{5}}{10}=a+b \frac{\sqrt{3}}{7}
$$
We have $\frac{1}{20}=a^{2}+\frac{2 \sqrt{3}}{7} a b+b^{2} \frac{3}{49}=\left(a^{2}+b^{2} \frac{3}{49}\right)+\frac{2 \sqrt{3}}{7} a b$ with $a \neq 0, b \neq 0$.
Suppose $a=0$, then
$$
\frac{1}{20}=b^{2} \frac{3}{49} \Leftrightarrow 3 b^{2}= \frac{49}{20}
$$
(here I don't know how to conclude)
If $b=0$, then
$$
\sqrt{5}=10 a
$$
which is impossible because $\sqrt{5} \neq \mathbb{Q}$.
|
Show that it is impossible to find $a,b \in \mathbb{Q}$ such that
$$
\frac{\sqrt{5}}{10}=a+b \frac{\sqrt{3}}{7}
$$
Alternative approach.
Lemma 1
$\sqrt{15}$ is not a rational number.
$\underline{\text{Proof}}$
Suppose that $~\displaystyle \sqrt{15} = \frac{p}{q} ~:$
$~p,q~$ relatively prime integers, and $~q > 0.$
Then $q\sqrt{15} = p \implies 15q^2 = p^2.$
However, since $p,q$ relatively prime, this implies that $15$ divides $p^2$. However, by the fundamental theorem of arithmetic, this implies that the prime factorization of $p$ contains both $3$ and $5$.
This implies that $15$ divides $p$.
This implies that $(15)^2$ divides $p^2.$
This implies that $(15)$ divides $q^2,~$ since $15q^2 = p^2.$
This is a contradiction, because $p,q$ are relatively prime.
Lemma 2
If $~k \in \Bbb{Z_{\neq 0}},~$ then $k\sqrt{15}$ is not an integer.
$\underline{\text{Proof}}$
Suppose $k\sqrt{15} = n \in \Bbb{Z}.$
Then $~\displaystyle ~\sqrt{15} = \frac{n}{k}.$
This contradicts Lemma 1.
Lemma 3
If $R,S \in \Bbb{Z_{\neq 0}}$,
then $R\sqrt{3} + S\sqrt{5}$ is not an integer.
$\underline{\text{Proof}}$
Suppose $~R\sqrt{3} + S\sqrt{5} = n \in \Bbb{Z}.$
Then $3R^2 + 5S^2 + 2RS\sqrt{15} = n^2,$ which is also an integer.
This implies that $2RS\sqrt{15}$ is an integer.
This contradicts Lemma 2.
Suppose that $~\displaystyle a,b = \frac{p_a}{q_a}, \frac{p_b}{q_b}~$ respectively,
where $p_a,q_a$ are relatively prime integers,
where $p_b,q_b$ are relatively prime integers,
and $q_a, q_b$ are both positive.
Then
$$\frac{p_a}{q_a} + \left(\frac{p_b}{q_b}\times \frac{\sqrt{3}}{7}\right) - \frac{\sqrt{5}}{10} = 0.\tag1 $$
In (1) above, you can ensure that each term has the common denominator of $(q_a \times q_b \times 7 \times 10).$
This implies that
$$\frac{70p_aq_b + 10q_ap_b\sqrt{3} - 7q_aq_b\sqrt{5}}{q_a \times q_b \times 7 \times 10} = 0. $$
This implies that
$$70p_aq_b + 10q_ap_b\sqrt{3} - 7q_aq_b\sqrt{5} = 0. $$
This implies that
$$\left(10q_ap_b\sqrt{3} - 7q_aq_b\sqrt{5}\right) ~~\text{is an integer}. $$
This is a contradiction, by Lemma 3.
Edit
I overlooked that you could have $p_b = 0$.
However, this would result in $7q_aq_b\sqrt{5}$ being an integer.
The exact same analysis used in Lemmas 1 through 3 may be repeated on $\sqrt{5}$ to show that that would be impossible.
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|
Determinant identity for symmetric 4x4 matrix with zero diagonal Does anybody know a reference or attribution for this identity?
$$
\det
\begin{pmatrix}
0 & {a_{12}}^2 & {a_{13}}^2 & {a_{14}}^2 \\
{a_{12}}^2 & 0 & {a_{23}}^2 & {a_{24}}^2 \\
{a_{13}}^2 & {a_{23}}^2 & 0 & {a_{34}}^2 \\
{a_{14}}^2 & {a_{24}}^2 & {a_{34}}^2 & 0
\end{pmatrix}
=\\
(a_{12} a_{34} + a_{13}a_{24} + a_{14} a_{23})
(a_{12} a_{34} - a_{13}a_{24} - a_{14} a_{23})
(-a_{12} a_{34} + a_{13}a_{24} - a_{14} a_{23})
(-a_{12} a_{34} - a_{13}a_{24} + a_{14} a_{23})
$$
I found the related identity
$$
\det
\begin{pmatrix}
0 & {a_{12}} & {a_{13}}^2 & {a_{14}}^2 \\
{a_{12}}^2 & 0 & {a_{23}}^2 & {a_{24}}^2 \\
{a_{13}}^2 & {a_{23}}^2 & 0 & {a_{34}}^2 \\
{a_{14}}^2 & {a_{24}}^2 & {a_{34}}^2 & 0
\end{pmatrix}
=
\det
\begin{pmatrix}
0 & {a_{12}a_{34}} & {a_{13}a_{24}} & {a_{14}a_{23}} \\
{a_{12}a_{34}} & 0 & {a_{14}a_{23}} & {a_{13}a_{24}} \\
{a_{13}a_{24}} & {a_{14}a_{23}} & 0 & {a_{12}a_{34}} \\
{a_{14}a_{23}} & {a_{13}a_{24}} & {a_{12}a_{34}} & 0
\end{pmatrix}
$$
in Muir's Treatise on the Theory of Determinants (1882), p. 41, but I'm interested in the factorization.
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The dehomogenized identity with $\,a_{j4}=1\,$ is Heron's formula written in terms of a CayleyβMenger determinant.
$$
\frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}
\;=\; \frac{1}{4}\sqrt{\,-
\begin{vmatrix}
0 & a^2 & b^2 & 1 \\
a^2 & 0 & c^2 & 1 \\
b^2 & c^2 & 0 & 1 \\
1 & 1 & 1 & 0
\end{vmatrix}}
$$
|
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Is the average of primes up to $n$ smaller than $\frac{n}{2}$, if $n > 19$? For $n = 19$, the average $\frac{(2 + 3 + 5 + 7 + 11 + 13 + 17 + 19)}{ 8} = \frac{77}{8}$ is greater than $\frac{19}{2}$ but for $n > 19$, it seems that the average is smaller than $n/2$.
I confirmed it holds for all $n < 10^7$ but have not been able to prove it for all n. How can I prove this?
|
The answer is yes for sufficiently large $n$. Sketch of proof:
*
*Using summation by parts (as in either one of these two existing answers), the average in question is
$$
\frac1{\pi(n)}\sum_{p\le n} p = \frac1{\pi(n)} \biggl( n\pi(n) - \sum_{k<n} \pi(k) \biggr) = n - \frac1{\pi(n)} \sum_{k<n} \pi(k).
$$
*Instead of just using $\pi(x) = \dfrac x{\log x} + O\biggl( \dfrac x{\log^2 x} \biggr)$, we use the more precise
$$
\pi(x) = \dfrac x{\log x} + \dfrac x{\log^2 x} +O\biggl( \dfrac x{\log^3 x} \biggr)
$$
that follows from the asymptotic expansion of the logarithmic integral.
*If we write
$\displaystyle
\pi(n) = \dfrac n{\log n} \biggl( 1 + \dfrac 1{\log n} + O\biggl( \dfrac1{\log^2 n} \biggr) \biggr)
$
and use $(1+\varepsilon)^{-1} = 1-\varepsilon+O(\varepsilon^2)$, we see that
$$
\frac1{\pi(n)} = \dfrac{\log n}n \biggl( 1 - \dfrac 1{\log n} + O\biggl( \dfrac1{\log^2 n} \biggr) \biggr).
$$
*Since the function $\dfrac x{\log^\alpha x}$ is increasing for $x>e^\alpha$, it's easy to see that $\displaystyle \sum_{k<n} \frac k{\log^\alpha k} = \int_2^n \frac t{\log^\alpha t}\,dt + O(n)$.
*Integration by parts yields $\displaystyle \int_2^n \frac t{\log^\alpha t}\,dt = \frac{n^2}{2\log^\alpha n} + O(1) + \frac{\alpha}2 \int_2^n \frac t{\log^{\alpha+1} t}\,dt$. It follows that
$\displaystyle
\int_2^n \frac t{\log^2 t}\,dt = \frac{n^2}{2\log^2 n} + O\biggl( \frac{n^2}{\log^3 n} \biggr)
$ and
$\displaystyle\int_2^n \frac t{\log t}\,dt = \frac{n^2}{2\log n} + \frac{n^2}{4\log^2 n} + O\biggl( \frac{n^2}{\log^3 n} \biggr)
$.
Putting these all together:
\begin{align*}
\frac1{\pi(n)} \sum_{p\le n} p &= n - \frac1{\pi(n)} \sum_{k<n} \biggl( \dfrac k{\log k} + \dfrac k{\log^2 k} + O\biggl( \dfrac k{\log^3 k} \biggr) \biggr) \\
&= n - \frac1{\pi(n)} \biggl( \int_2^n \frac t{\log t}\,dt + \int_2^n \frac t{\log^2 t}\,dt + O\biggl( \int_2^n \frac t{\log^3 t}\,dt + n\biggr) \biggr) \\
&= n - \dfrac{\log n}n \biggl( 1 - \dfrac 1{\log n} + O\biggl( \dfrac1{\log^2 n} \biggr) \biggr) \\
&\qquad{}\times\biggl( \biggl( \frac{n^2}{2\log n} + \frac{n^2}{4\log^2 n} \biggr) + \frac{n^2}{2\log^2 n} + O\biggl( \frac{n^2}{\log^3 n} \biggr) \biggr) \\
&= n - \biggl( \frac n2 + \frac n{4\log n} + O\biggl( \frac n{\log^2n} \biggr) \biggr) = \frac n2 - \frac n{4\log n} + O\biggl( \frac n{\log^2n} \biggr),
\end{align*}
which finishes the proof.
|
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|
Reflection across an Ellipsoid A reflective ellipsoid is given by
$$ \dfrac{x^2}{16} + \dfrac{y^2}{9} + \dfrac{(z - 2)^2}{4} = 1 $$
A light source, emitting rays in all directions, is placed at $A=(10,4,3)$. Find the point $C=(x,y,z)$ on the surface of the ellipsoid that minimizes the sum of distances $AC + CB$, where $B = (8, 12, 7) $
My attempt:
Parameterize the equation of the ellipsoid as follows
$ x = 4 \sin \theta \cos \phi $
$ y = 3 \sin \theta \sin \phi $
$ z = 2 + 2 \cos \theta $
Now the sum of the two distances is
$S = AC + CB = \sqrt{ (4 \sin \theta \cos \phi - 10)^2 + (3 \sin \theta \sin \phi - 4)^2 + (2 + 2 \cos \theta - 3)^2 } \\
+\sqrt{ (4 \sin \theta \cos \phi - 8)^2 + (3 \sin \theta \sin \phi - 12)^2 + (2 + 2 \cos \theta - 7)^2 }$
How can I minimize $S$ ?
|
Take a generic point on the ellipsoid, $(a,b,c)$ (first condition on $a,b,c$). To mimimize your sum, you need the ray of light going from $A$ to $B$ bouncing at the ellipsoid to be reflected at $C$. The normal to the ellipsoid at $(a,b,c)$ is the gradient $(a/8, 2b/9, (c-2)/2)$. You need two additional conditions: the angles formed between $CA$ and $CB$ and the normal vector should be equal and the three vectors should be coplanar (their triple product equal to zero). Solve the system of three equations.
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|
Finding the volume with triple integrals I want to find the volume of a function described by:
$$ G= \{(x,y,z)|\sqrt{x^2+y^2} \le z \le 1, (x-1)^2+y^2 \le 1\}$$
This question can be best solved in cylindrical coordinates. So if I follow that process, I get the following limits:
$$ r \le z \le 1$$
$$ 0 \le r \le 2\cos(\theta)$$
$$ -\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$$
I am fairly certain that these limits are correct. So continuing, to find the volume of G:
$$\begin{align}
\iiint r \ dz\ dr\ d\theta
&= \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \int^{2\cos(\theta)}_{0} \int^{1}_{r} r \ dz \ dr \ d\theta\\
&= \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \int^{2\cos(\theta)}_{0} r-r^2 \ dr \ d\theta\\
&= \int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \frac{(2\cos(\theta))^2}{2}-\frac{(2\cos(\theta))^3}{3} \ d\theta\\
&= \pi - \frac{32}{9}
\end{align}$$
Solving the above integral I get a negative answer which does not make sense considering the physical quantity is volume, which has to be positive. Where am I going wrong?
|
Your limits are not completely correct. Note that it should be
$$0 \le r \le 2\cos(\theta)\quad\textbf{and}\quad 0 \le r \leq 1$$
that is the upper limit is $1$ for $|\theta| \le \frac{\pi}{3}$ and it is $2\cos(\theta)$ for $\frac{\pi}{3}\leq|\theta| \le \frac{\pi}{2}$.
Therefore, by symmetry,
$$\begin{align}
V&=2\int^{\frac{\pi}{3}}_{0} \int^{1}_{0} (r-r^2) \ dr \ d\theta+2\int^{\frac{\pi}{2}}_{\frac{\pi}{3}} \int^{2\cos(\theta)}_{0} (r-r^2) \ dr \ d\theta\\
&=\frac{4\pi}{9}+\frac{3\sqrt{3}}{2}-\frac{32}{9}\approx 0.439.
\end{align}$$
|
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|
Minimize $xy$ over $x^2+y^2+z^2=7$ and $xy+xz+yz=4$.
Let $x,$ $y,$ and $z$ be real numbers such that $x^2 + y^2 + z^2 = 7$ and $xy + xz + yz = 4.$ Find the smallest possible value of $xy.$
I used Cauchy to get $$(x^2+y^2+z^2)(1^2+1^2+1^2)\geq (x+y+z)^2$$ and $$(x^2+y^2+z^2)(y^2+z^2+x^2)\geq(xy+xz+yz)^2,$$ but this doesn't do much as both of the inequalities already work. :( Any guidance would be appreciated!!
Thanks in advance!!!
P.S. I've been given that this can be solved by Lagrange multipliers, but I haven't learned it yet. Hopefully, someone can give me hints on a method that doesn't involve Lagrange multipliers. :)
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$x^2+y^2+z^2 = 7$ and $xy +xz + yz = 4$
implies
$x+y+z = \pm\sqrt{15}$
Now using the second equality that is given in the question, we get
$xy =4 -(x+y)z = 4 - (\pm\sqrt{15} - z)z$
*
*Case 1: When $x+y+z=\sqrt{15}$, minimising the right hand side with respect to $z$ gives $z = \frac{\sqrt{15}}{2}$. So, minimum value of $xy$ is $4 - (\sqrt{15} - z)z= 4-\frac{15}{4}=\frac{1}{4}$.
*Case 2: When $x+y+z=-\sqrt{15}$, minimising the right hand side with respect to $z$ gives $z = -\frac{\sqrt{15}}{2}$. So, minimum value of $xy$ is $4 - (-\sqrt{15} - z)z= 4-\frac{15}{4}=\frac{1}{4}$.
So the minimum value of $xy$ is $\frac{1}{4}$.
|
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|
How can I show that $\underset{\left(x,y\right)\rightarrow\left(0,0\right)}{\lim}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}$ exists, using limit def? I am trying to solve an exercise to show that this function $$f(x,y)=\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}$$ has a limit as $(x,y)$ approaches $(0,0)$:
$$\underset{\left(x,y\right)\rightarrow\left(0,0\right)}{\lim}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}$$
To do so, I have to use the limit precise definition:
$$\forall\epsilon>0\ \exists\delta>0βΆ\ \left|\left(x,y\right)-\left(a,b\right)\right|<\delta\ {\Rightarrow}\left|f\left(x,y\right)-L\right|<\ \epsilon$$
I've found out that the the potential limit $L=2$ by evaluating the limits across the x-axis, y-axis, arbitrary line and x^2 parabola and y^2 parabola.
I have plugged all the necessary values and ended up with this:
$$\forall\epsilon>0\ \exists\delta>0βΆ\ \sqrt{x^2+y^2}<\delta\ {\Rightarrow}\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|<\ \epsilon$$
Now, I am in trouble to construct inequalities that lead to prove that the following is true:
$$\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|<\sqrt{x^2+y^2}\ $$
Can someone help me out here?
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You have\begin{align}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}&=\frac{(x^2+y^2)\left(\sqrt{x^2+y^2+1}+1\right)}{\left(\sqrt{x^2+y^2+1}-1\right)\left(\sqrt{x^2+y^2+1}+1\right)}\\&=\sqrt{x^2+y^2+1}+1\end{align}and therefore$$\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|=\left|\sqrt{x^2+y^2+1}-1\right|.$$But$$\require{cancel}\left|\sqrt{x^2+y^2+1}-1\right|\leqslant\sqrt{x^2+y^2}.$$In fact, this inequality is equivalent to$$\cancel{x^2+y^2}+2-2\sqrt{x^2+y^2+1}\leqslant\cancel{x^2+y^2}$$and it is clear that we always have $2\leqslant2\sqrt{x^2+y^2+1}$.
Putting all together, we have$$\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|\leqslant\sqrt{x^2+y^2}$$and so, in order to prove that$$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}=2,$$for each $\varepsilon>0$, we only have to take $\delta=\varepsilon$.
|
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|
How do I determine the value of $d$ from the tangent graph of the form $y = a\times \tan (bx + c) + d$?
I understand that $d$ is the vertical shift, and I have already worked out the values of $a$, $b$ and $c$ to be the following (please tell me if I am wrong):
$\pmb{a}\quad$ Given point is $(-1, -7)$ so I substituted it in as the values of $x$ and $y$ ; $-7= a\times \tan\left(\dfrac{\pi}{4}\left(-1\right)+5\right)$ which I solved for $a$ and that gave me$\space$ -3.8040.
$\pmb{b}\quad$ Period so $\dfrac{\pi}{4}$
$\pmb{c}\quad$ $y$-intercept so $5$
But how do I find $d$ for this type of graph with only vertical asymptotes?
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The graph of a transformed $\tan$ function is
$ f(x) = A \tan( B (x - C) ) + D $
The first thing is that $ B = \dfrac{\pi}{T} $ where $T $ is the period.
Here the period is $ 4 $ , so $B = \dfrac{\pi}{4} $
$C $ is the horizontal shift to the right. Here the graph is shifted to the left by $1$, therefore, C = -1 $
Now we have two points on the graph: $ (-1, -7)$ and $(0, -5) $. Hence
$ -7 = 0 + D $ from which $D = -7 $
and
$ -5 = A \tan( \dfrac{\pi}{4} ( 0 + 1 ) ) - 7 $
From which $ A = 2 $
Therefore, the function is
$ f(x) = 2 \tan( \dfrac{\pi}{4} x + \dfrac{\pi}{4} ) - 7 $
|
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|
If a convex quadrilateral is inscribed and circumscribed and has perpendicular diagonals, then one diagonal divides it into congruent right triangles
A convex quadrilateral $ABCD$ is inscribed and circumscribed. If the diagonals $AC$ and $BD$ are perpendicular, show that one of them divides the quadrilateral into two congruent right triangles.
The first question I asked myself is how we determine which of the diagonals divides $ABCD$ in the desired way.
If $AB=a,BC=b,CD=c,AD=d$, then $a+c=b+d$ on one hand. Also we have $A+C=B+D=180^\circ$. I tried to do something with the areas. $$S_{ABC}=\dfrac12ab\sin B\\S_{ACD}=\dfrac12cd\sin D\\S_{ABCD}=\dfrac12ab\sin B+\dfrac12cd\sin D=\dfrac12\sin B(ab+cd)\\S_{ABCD}=\dfrac{AC\cdot BD}{2}$$ I don't really think anything of that helps. How to approach the problem?
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Let the diagonals intersect at $O$ and set: $a=AO$, $b=BO$, $c=CO$, $d=DO$.
The sums of opposite sides must be the same, giving:
$$
\strut\sqrt{a^2+b^2}+\sqrt{c^2+d^2}=
\sqrt{a^2+d^2}+\sqrt{b^2+c^2}.
$$
Squaring and simplifying we get:
$$
(a^2+b^2)(c^2+d^2)=
(a^2+d^2)(b^2+c^2),
$$
which reduces to
$$
(a^2-c^2)(b^2-d^2)=0.
$$
Hence, $a=c$ or $b=d$, that is one of the diagonals bisects the other one and is thus a diameter of the circumscribed circle.
|
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|
Finding max{f(x)} without derivative Consider the function $f(x)=\frac{\sqrt{x^3+x}}{x^2+x+1}$
the question is about: to find max{f} without using derivative.
I can find max with derivation and it is not hard to find. it is $f'(x)=0 \to x=1 $ so $max\{f\}=\frac{\sqrt 2}{3}$
but I am looking for an idea to do as the question said.
I am thankful for any help.(because I got stuck on this problem)
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Maybe a bit late answer but i thought it might be worth mentioning it.
Here is a direct elementary calculation of the maximum.
Let $x\geq 0$:
We only have to find the maximum of
$$\frac{x^3+x}{(x^2+x+1)^2} = \frac{x(x^2+1)}{(x^2+x+1)^2}$$
Simple rearrangements give
\begin{eqnarray*} \frac{x(x^2+1)}{(x^2+x+1)^2}
& = & \frac{x(x^2+x+1) -x^2}{(x^2+x+1)^2} \\
& = & \frac{x}{x^2+x+1} -\left(\frac{x}{x^2+x+1}\right)^2 \\
\end{eqnarray*}
Now, we have to maximize $t-t^2$ with $$0\leq t= \frac{x}{x^2+x+1} \stackrel{AM-GM}{\leq} \frac 13$$
with equality on the RHS for $x=1$.
The maximum of $t-t^2$ (a downward parabola with vertex at $t=\frac 12$) on $[0,\frac 13]$ is attained for $t=\frac 13$.
Hence,
$$\max_{x\geq 0}\frac{x^3+x}{(x^2+x+1)^2} = \frac 13 - \frac 19 = \frac 29 \Rightarrow \max_{x\geq 0}f(x) = \frac{\sqrt 2}{3}$$
|
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|
Solving $\frac{\ 3x}{x^2+x+1}+\frac{2x}{x^2-x+1}=3$ with other approaches
$$\frac{\ 3x}{x^2+x+1}+\frac{2x}{x^2-x+1}=3$$
$$x=?$$
I solved this problem as follow:
$x=0$ is not a root, we can divide numerator and denominator of each fraction by $x$:
$$\frac{3}{t+1}+\frac{2}{t-1}=3\quad\quad\text{where $t=x+\frac1x$}$$
$$5t-1=3t^2-3\Rightarrow t=2 , \frac{-1}6$$
Only $x+\frac1x=2$ is acceptable and $x=1$.
I'm looking for other elegant methods to solve this equation.
|
I don't know if you consider this elegant:
\begin{align}
0&=3-\frac{3x}{x^2+x+1}-\frac{2x}{x^2-x+1}\\
&=\left(1-\frac{3x}{x^2+x+1}\right)+2\left(1-\frac{x}{x^2-x+1}\right)\\
&=\frac{(x-1)^2}{x^2+x+1}+2\frac{(x-1)^2}{x^2-x+1}\\
&=(x-1)^2\left(\frac{1}{x^2+x+1}+\frac{2}{x^2-x+1}\right)
\end{align}
Since $x^2+x+1$ and $x^2-x+1$ are always positive, we can cancel the second part to get $(x-1)^2=0$
|
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|
Is there any closed form for $\int_{0}^{1} \frac{\ln ^nx}{\sqrt{1-x^{2}}}\,d x$? In my post, I found that $$\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^{2}}} d x = -\frac{\pi}{2} \ln 2 .$$
Then I try to generalize the result to the integral by the same technique. $$
J_{n}:=\int_{0}^{1} \frac{\ln ^nx}{\sqrt{1-x^{2}}} d x
$$
Letting $x=\cos \theta $ converting $I_n$ into
\begin{aligned}
J_{n} &=\int_{0}^{\frac{\pi}{2} } \frac{\ln^n (\cos \theta)}{\sin \theta} \sin \theta d \theta \\
&=\int_{0}^{\frac{\pi}{2}} \ln ^{n}(\cos \theta) d \theta \end{aligned}
By my post,
$$\begin{aligned}
(-2)^{n}J_n&= 2 \ln 2(-2)^{n-1} J_{n-1} + (n-1) !\sum_{k=0}^{n-2} \frac{2^{n-k}-2}{k_ !} \zeta(n-k) (-2)^{k} J_k \\J_n&= -\ln 2 J_{n-1} + (n-1) !\sum_{k=0}^{n-2} \frac{(-1)^{n-k}}{k !} \left(1-\frac{1}{2^{n-k-1}} \right)\zeta(n-k) J_k
\end{aligned}
$$
which is a reduction formula for $J_n.$
For example, $$
\begin{aligned}
\int_{0}^{1} \frac{\ln ^2x}{\sqrt{1-x^{2}}} d x&=-\ln 2 J_{1}+\frac{1}{2} \zeta(2) J_{0} \\
&=-\ln 2\left(-\frac{\pi}{2} \ln 2\right)+\frac{1}{2} \zeta(2) J_{0} \\
&=\frac{\pi \ln ^{2} 2}{2}+\frac{\pi^{3}}{24},
\end{aligned}
$$
which is checked by Wolframalpha.
My Question
Is there any closed form for the integral? Your comments and methods are warmly welcome.
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Under the substitution $x^2 \to x$ we get that
$$I_{n,m} =\int_{0}^{1} \frac{\ln^n(x)}{\left(1-x^2\right)^m}\mathrm{d}x = \frac{1}{2^{n+1}}\int_{0}^{1}\ln^n(x) x^{-\frac{1}{2}}(1-x)^{-m}\mathrm{d}x $$ Recalling the definition of the Beta function we see that
\begin{align*}
B\left(\frac{1}{2}+t, 1-m\right) =& \int_0^{1} x^{-\frac{1}{2}+t}(1-x)^{-m}\mathrm{d}x \\
\mathbin{\color{blue}{\implies}}\frac{\mathrm{d}^n}{\mathrm{d}t^n}B\left(\frac{1}{2}+t, 1-m\right)\Bigg\vert_{t=0} =&\int_{0}^{1}\ln^n(x) x^{-\frac{1}{2}}(1-x)^{-m}\mathrm{d}x =2^{n+1}I_{n,m}
\end{align*}
So we get the generalization
$$\int_{0}^{1} \frac{\ln^n(x)}{\left(1-x^2\right)^m}\mathrm{d}x =\frac{1}{2^{n+1}}\frac{\mathrm{d}^n}{\mathrm{d}t^n}B\left(\frac{1}{2}+t, 1-m\right)\Bigg\vert_{t=0} \qquad \text{for }\ \ m<1,\ n \in \mathbb{N}$$
This result could technically be expanded in terms of polygamma functions using the General Leibniz rule, but I expect this will just result in a way messier expression.
|
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|
Prove that the sequence $t_n$ defined by $\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$ converges determine if the following sequence converges
$t_n=\dfrac{1}{n}(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{n}})$
my solution:
$\text{ using AM-GM inequality, }\space\space t_n\ge (1\cdot2\cdot 3\cdots n)^{\dfrac{-1}{2n}}$
$\implies (1\cdot2\cdot 3\cdots n)^{\dfrac{-1}{2n}}\le t_n\le \dfrac{1}{n}(1+\cdots+1)=\dfrac{n}{n}=1$
$(1\cdot2\cdot 3\cdots n)^{\dfrac{1}{n}}=(1)^{\frac{1}{n}}\cdot (2)^{\frac{1}{n}}\cdot (3)^{\frac{1}{n}}\cdots\cdot (n)^{\frac{1}{n}}\stackrel{n\to\infty}{\to}1$
$\implies (1\cdot2\cdot 3\cdots n)^{\dfrac{-1}{2n}}\to1$
Hence, by sandwich theorem $t_n\to1$
Is this correct?
|
You can calculate the limit in an elementary way using
$$\sqrt{k+1}-\sqrt k = \frac 1{\sqrt{k+1}+\sqrt k} > \frac 1{2\sqrt{k+1}}$$
Hence,
\begin{eqnarray*} \frac 1n\sum_{k=1}^n \frac 1{\sqrt k}
& = & \frac 1n\left(1+ \sum_{k=1}^{n-1} \frac 1{\sqrt{k+1}}\right) \\
& < & \frac 1n\left(1+ 2\sum_{k=1}^{n-1} (\sqrt{k+1}-\sqrt k)\right) \\
& = & \frac 1n\left(1+ 2(\sqrt{n}-1)\right) \\
& = & -\frac 1n + \frac 2{\sqrt n} \stackrel{n\to\infty}{\longrightarrow} 0
\end{eqnarray*}
|
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|
Why is $\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2} \equiv \frac{(1+\cos x)^2}{\sin^2x+1+2\cos x+\cos^2x}$? Currently I have a problem to understand why $$\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2} \equiv \frac{(1+\cos x)^2}{\sin^2x+1+2\cos x+\cos^2x}\,?$$
My calculations get me:
$$\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2}\equiv \frac{(1+\cos x)^2}{1+\sin^2x}.$$
Why is this not correct? We have a rational number in the denumerator. Thus the numerator of this rational number should move above. The other formula did this too. But why is there $2\cos x+\cos^2x$?
Thanks in advance! Stay healthy!
|
If you multiply the numerator and the denominator by $(1+\cos x)^2$, you get:
$$\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2}=\frac{(1+\cos x)^2}{(1+\cos x)^2}\times\frac{1}{1+(\frac{-\sin x}{1+\cos x})^2}=\frac{(1+\cos x)^2}{(1+\cos x)^2\left(1+(\frac{-\sin x}{1+\cos x})^2\right)}\\=\frac{(1+\cos x)^2}{(1+\cos x)^2+\sin^2x}$$
|
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|
Finding the equation of the Parabola The parabola $y = x^2 + bx + c$ has the following properties:
*
*The point on the parabola closest to $(12,3)$ is the intersection with the $y$ axis of the parabola.
*The parabola passes through $(-5,0).$
How can I find $(b, c)$?
Here is my attempt:
The point $(0, c)$ is the intersection with the $y$ axis of the parabola. The distance from $(12, 3)$ is $\sqrt{144 + (c-3)^2}$, and we have the equation $-5b +25+c =0$.
But if we don't know vertex of the parabola how do we find $b$ & $c$?
|
$ y = x^2 + bx + c $
$ y' = 2 x + b $
At $(0,c)$ we have $y' = b $, and thus
$ b \bigg( \dfrac{c - 3}{0 - 12} \bigg) = -1 $
which simplifies to $ b (c - 3) = 12 $
We also have $(-5, 0)$ on the parabola, then $0 = 25 - 5 b + c $
Solving the above two equations yields two solutions
Case I: $b = -0.4, c = -27 $
Case II: $ b = 5 , c = 6 $
We need to check if the distance at the $y-intercept$ is indeed the shortest distance. At the critical points, we have
$ (2 x + b ) \bigg( \dfrac{x^2 + bx + c - 3 }{ x - 12 } \bigg) = -1 $
So that
$ 2 x^3 + 3 b x^2 + x ( 1 + 2 (c - 3) + b^2) + b ( c - 3) - 12 = 0 $
In case I, this gives three critical points, listed with their respective distances from $(12, 3)$
$A (-5.13231, 1.393538) , 17.20746 $
$B (0, -27) , 32.31099 $
$ C (5.732311, 3.566462), 6.293235 $
Clearly the $y$-intercept, which is point $B$ is not the closest point to $(12,3)$. Hence Case I is extraneous.
Case II leads to one critical point only which is $(0, 5)$ with a distance of $12.16553$.
Hence our parabola is
$ y =x^2 + 6 x + 5 $
|
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|
Why does $\lim_{n \to \infty}(1+\frac{1}{4^n})^{2^n} $ not equal $\sqrt{e}$? I tried to evaluate $\lim_{n \to \infty}(1+\frac{1}{4^n})^{2^n}$ with
$$\lim_{n \to \infty}(1+\frac{1}{4^n})^{2^n} = \lim_{n \to \infty}(1+\frac{1}{4^n})^{(4^\frac{1}{2})^n} = \lim_{n \to \infty}(1+\frac{1}{4^n})^{4^\frac{n}{2}} = \lim_{n \to \infty}((1+\frac{1}{4^n})^{4^n})^\frac{1}{2} = ({e})^\frac{1}{2} = \sqrt{e}.$$
However, I know that the correct answer is $\lim_{n \to \infty}(1+\frac{1}{4^n})^{2^n} = 1$. So at what point did I make a mistake? Which move was not allowed?
|
Note that in general $\sqrt{A^x} = A^{\frac{x}{2}} \not = A^{\sqrt{x}}$ [as in general $\frac{x}{2} \not = \sqrt{x}$]. Try $A=2$ and $x=9$ to see for yourself.
In particular here: $$\left(1+\frac{1}{4^n}\right)^{2^n} = \left(1+\frac{1}{4^n}\right)^{\sqrt{4^n}} \not = \sqrt{\left(1+\frac{1}{4^n}\right)^{4^n}}.$$
Instead:
$$\left(1+\frac{1}{4^n}\right)^{2^n} = \left(\left(1+\frac{1}{4^n}\right)^{4^n}\right)^{2^{-n}},$$ and as the inequality $4 \ge \left(1+\frac{1}{4^n}\right)^{4^n} \ge 1$ holds for each $n$, it follows that, on the one hand:
$$\lim_{n \rightarrow \infty} \left(1+\frac{1}{4^n}\right)^{2^n} = \lim_{n \rightarrow \infty}\left(\left(1+\frac{1}{4^n}\right)^{4^n}\right)^{2^{-n}}$$
$$\le \lim_{n \rightarrow \infty} 4^{2^{-n}} = 1.$$
And on the other hand:
$$\lim_{n \rightarrow \infty} \left(1+\frac{1}{4^n}\right)^{2^n} = \lim_{n \rightarrow \infty}\left(\left(1+\frac{1}{4^n}\right)^{4^n}\right)^{2^{-n}}$$
$$\ge \lim_{n \rightarrow \infty} 1^{2^{-n}} = 1.$$
Thus indeed,
$$\lim_{n \rightarrow \infty} \left(1+\frac{1}{4^n}\right)^{2^n} = 1.$$
|
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|
Find an orthonormal basis and the signature of the quadratic form
Consider the quadratic form given by the matrix below (in the canonical basis)
\begin{pmatrix}
1 & 1 & -1\\
1 & 1 & 3\\
-1 & 3 & 1
\end{pmatrix}
Find an orthonormal basis of it and find its signature.
First I calculated the eigenvalues, which are $4, \frac{-1+ \sqrt{17}}{2}, \frac{-1-\sqrt{17}}{2}$. Then I calculated the eigenvectors associated to $4$ and $\frac{-1+ \sqrt{17}}{2}$ and normalized them, which gave me
\begin{align}
e_1 &= \frac{1}{\sqrt{2}}\begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix}, \quad
e_2 = \frac{\sqrt{2}}{\sqrt{17+3\sqrt{17}}}\begin{bmatrix}
-\frac{3+\sqrt{17}}{2} \\
-1 \\
1
\end{bmatrix}
\end{align}
And the third vector of the basis I want to be orthogonal to $e_1$ and $e_2$, so
$$e_3 = \frac{1}{\sqrt{17+3\sqrt{17}}} e_1 \wedge e_2 =\frac{1}{\sqrt{17+3\sqrt{17}}} \begin{bmatrix}
2 \\
-\frac{3+\sqrt{17}}{2} \\
\frac{3+\sqrt{17}}{2}
\end{bmatrix} $$
I can't detail the calculations because they are very big. Perhaps someone can confirm the results. For the signature I know that the two possibilities are $(0,3)$ and $(2,1)$ but I don't know how to find the right one.
|
This business with eigenvalues and eigenvectors is not how you diagonalize a quadratic form. It will give a correct result if done correctly, but is way too long and computationally painful. Simply use a "complete the square" method. In the canonical coordinates, your quadratic form is
$$x^2 + y^2+z^2+2xy-2xz+6yz.$$
Now let us complete the squares:
$$\begin{align}
& x^2 + y^2+z^2+2xy-2xz+6yz \\
&= (x^2+2xy +y^2) + z^2-2xz+6yz \\
&= (x+y)^2 + z^2 -2xz+6yz \\
&= (x+y)^2 + z^2 -2z(x-3y) \\
&= (x+y)^2 + (z-(x-3y))^2 - (x-3y)^2.
\end{align}$$
This means that the quadratic form is simply $(x')^2+(y')^2-(z')^2$ with the change of coordinates $$x'=x+y,\quad y'=-x+3y+z,\quad z'=x-3y.$$
This already clearly shows that the signature is $(2,1)$.
So to find your orthogonal basis, you just have to invert the matrix
$$\begin{pmatrix}
1 & 1 & 0 \\
-1 & 3 & 1 \\
1 & -3 & 0
\end{pmatrix}$$
and the basis will be given by the columns of the inverse. No need to compute eigenvalues or eigenvectors with complicated expressions. This gives
$$\frac{1}{4}\begin{pmatrix}
3 & 0 & 1 \\
1 & 0 & -1 \\
0 & 4 & 4
\end{pmatrix},$$
so an orthonormal basis is $e_1 = \begin{bmatrix} 3/4 \\ 1/4 \\ 0\end{bmatrix}$, $e_2 = \begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$ and $e_3 = \begin{bmatrix} 1/4 \\ -1/4 \\ 1\end{bmatrix}$.
|
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|
Integral $\int( 3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}dx$ I tried to integrate
$$\int( 3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}dx$$
by multipling by $\sqrt{2x^2 + 2x + 1}$ in the numerator and the denominator to break it into $5$ fractions
The answer is possible but it's too long
Another better solution
|
A systematic approach after $t=2x+1$
\begin{align}
I=&\int (3x^2 +5x + 1 )\sqrt{2x^2 + 2x + 1}\>dx\\
= &\frac1{8\sqrt2}\int (4t+3)(t^2+1)^{3/2}-6 (t^2+1)^{1/2}\>dt\tag1
\end{align}
Integrate below by parts
\begin{align}
&\int (t^2+1)^{3/2}dt=\frac14t(t^2+1)^{3/2}+\frac34\int(t^2+1)^{1/2}dt\\
&\int (t^2+1)^{1/2}dt=\frac12 t(t^2+1)^{1/2}+\frac12\sinh^{-1}t
\end{align}
Plug into (1) to arrive at
$$I= \frac1{8\sqrt2}\left[\bigg(\frac34t + \frac43 \bigg)(t^2+1)^{3/2}-\frac{15}8 t(t^2+1)^{1/2}-\frac{15}8\sinh^{-1}t\right]
$$
|
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|
Derivative of the nth order and evaluation in $x=0$ I want to solve the following exercise; be $f(x)=\frac{1}{x}e^x-\frac{1}{x^2}e^x+\frac{1}{x^2}$ with $x\neq 0$ we have that the function has derivatives in all orders for $x=0$
I want to determine $f^{(n)}(0)$
It seems that the above is equivalent to the expression $\frac{1}{a(n)}$ where $a(n)$ represents an expression in terms of n, now, how can I determine this derivative? any suggestions? thanks.
|
We know $f$ admits a Taylor expansion at $x=0$:
\begin{align*}
f(x)=\sum_{k=0}^\infty\frac{f^{k}(0)}{k!}x^k
\end{align*}
Denoting with $[x^n]$ the coefficient of $x^n$ of a series $f(x)$ we want to calculate
\begin{align*}
f^{(n)}(0)=n![x^n]f(x)
\end{align*}
Recalling the series expansion
\begin{align*}
e^x=\sum_{j=0}^{\infty}\frac{x^j}{j!}\tag{1}
\end{align*}
we obtain for $n\geq 0$:
\begin{align*}
\color{blue}{f^{(n)}(0)}&=n![x^n]f(x)\\
&=n![x^n]\left(\frac{1}{x}e^x-\frac{1}{x^2}e^x+\frac{1}{x^2}\right)\\
&=n!\left([x^{n+1}]e^x-[x^{n+2}]e^x\right)\tag{2}\\
&=\frac{1}{n+1}-\frac{1}{(n+2)(n+1)}\tag{3}\\
&\,\,\color{blue}{=\frac{1}{n+2}}
\end{align*}
Comment:
*
*In (2) we apply the rule $[x^p]x^qf(x)=[x^{p-q}]f(x)$ and note $[x^n]\frac{1}{x^2}=0$.
*In (3) we select the coefficient of $x^{n+1}$ and $x^{n+2}$ from the exponential function (1) and multiply with $n!$.
|
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|
Find area of $x^2+axy+y^2=1$, $|a|\leq1$ I was wondering how to find the area of
$$x^2+axy+y^2=1,\>\>\>\>\>|a|\leq1$$
I have solved
$$\rho^{2}(\theta)=\frac{1}{1+\frac{a}{2}\sin 2\theta}$$
However, integrating this equation using trigonometric substitution is very cumbersome.
|
The equation of the ellipse can be written compactly using the position vector $r=[x,y]^T$ as
$ r^T Q r = 1 $
where $Q = \begin{bmatrix} 1 && \frac{1}{2} a \\ \frac{1}{2} a && 1 \end{bmatrix} $
To find the area you need the product of the semi-minor and semi-major axes.
If you diagonalize $Q$ and write it as $ Q = R D R^T $ , then the original equation becomes
$ r^T R D R^T r = 1 $
The eigenvalues of $D$ are the square of the reciprocals of the semi-major and semi-minor axes. Therefore its determinant
$|D| = \dfrac{1}{a^2 b^2} $
Since $R$ is a rotation matrix $ | Q | = | R D R^T | = | D | $
Therefore, $ a b = \dfrac{1}{\sqrt{|D|}} = \dfrac{1}{\sqrt{|Q|}} $
From $Q$, we get $|Q| = 1 - \dfrac{a^2}{4} $
Therefore, the area is given by
$\text{Area} = \pi a b = \dfrac{\pi}{ \sqrt{ 1 - \dfrac{a^2}{4} }} = \dfrac{2 \pi}{\sqrt{4 - a^2}} $
|
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|
why not $G \cong \frac{\mathbb{Z}}{3\mathbb{Z}} \times \frac{\mathbb{Z}}{3\mathbb {Z}} \times \frac{\mathbb{Z}}{3\mathbb{Z}} ?$
Determine the Galois group of $(x^2-2) (x^2-3)(x^2-5)$. Determine all the subfields of the splitting field of this polynomial.
My attempt: I found the answer here
$G= Gal(K/\mathbb{Q})$ where$ K= \mathbb{Q}(\sqrt 2, \sqrt 3, \sqrt 5)$
It is written that
$\sigma_2 \begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}$
$\sigma_3: \begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}$
$\sigma_5: \begin{cases}
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}$
then obviously $G= \langle \sigma_2, \sigma_3, \sigma_4 \rangle \cong \frac{\mathbb{Z}}{2\mathbb{Z}} \times \frac{\mathbb{Z}}{2\mathbb{Z}} \times \frac{\mathbb{Z}}{2\mathbb{Z}}$
My question: why $G=\langle \sigma_2, \sigma_3, \sigma_4 \rangle \cong \frac{\mathbb{Z}}{2\mathbb{Z}} \times \frac{\mathbb{Z}}{2\mathbb{Z}} \times \frac{\mathbb{Z}}{2\mathbb{Z}} ?$
Why not $\langle \sigma_2, \sigma_3, \sigma_5 \rangle \cong \frac{\mathbb{Z}}{3\mathbb{Z}} \times \frac{\mathbb{Z}}{3\mathbb {Z}} \times \frac{\mathbb{Z}}{3\mathbb{Z}} ?$
My thinking: $\sigma_2$ has $3$ choice similarly $\sigma_3 $ and $\sigma_5$ have $3$ choice so i think
$\langle \sigma_2, \sigma_3, \sigma_5 \rangle \cong \frac{\mathbb{Z}}{3\mathbb{Z}} \times \frac{\mathbb{Z}}{3\mathbb {Z}} \times \frac{\mathbb{Z}}{3\mathbb{Z}} $
|
$\sigma_2$ doesn't have "three choices" it has 8:
$$
\begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto \sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}\\
\begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto -\sqrt{3}\\
\sqrt{5}\mapsto -\sqrt{5}\\
\end{cases}
$$
Of course, to those of us in the know, only one of these deserve the name $\sigma_2$.
No, more important than the number of possibilities is its order. Choose
$$
\sigma_2:\begin{cases}
\sqrt{2}\mapsto -\sqrt{2}\\
\sqrt{3}\mapsto \sqrt{3}\\
\sqrt{5}\mapsto \sqrt{5}\\
\end{cases}$$
Then $\sigma_2^2$ is the identity. This means that $\langle \sigma_2\rangle\cong \Bbb Z/2\Bbb Z$. Same for the other two $\sigma$'s. The way the three chosen $\sigma$'s interact (they commute, their generated subgroups intersect trivially, and together they generate the whole group) is what yields
$$
G=\langle \sigma_2, \sigma_3, \sigma_5 \rangle \cong {\mathbb{Z}}/{2\mathbb{Z}} \times {\mathbb{Z}}/{2\mathbb{Z}} \times {\mathbb{Z}}/{2\mathbb{Z}}
$$
|
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|
Range of $\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - d)^2} + \sqrt{d^2 + (1 - a)^2}$ on $0 \le a,$ $b,$ $c,$ $d \le 1.$ Let $0 \le a,$ $b,$ $c,$ $d \le 1.$ Find the possible values of the expression
$$\sqrt{a^2 + (1 - b)^2} + \sqrt{b^2 + (1 - c)^2} + \sqrt{c^2 + (1 - d)^2} + \sqrt{d^2 + (1 - a)^2}.$$
I tried to use some inequalities to find the bounds of the expression, but it didn't really work. Also, I don't know calculus yet, so please keep the responses and hints non-calc.
Thanks in advance!!
|
Denote the expression by $P(a, b, c, d)$.
First, using Minkowski inequality, we have
\begin{align*}
P &\ge \sqrt{(a + b + c + d)^2 + (1 - b + 1 - c + 1 - d + 1 - a)^2}\\
&= \sqrt{x^2 + (4 - x)^2}\\
&= \sqrt{2(x - 2)^2 + 8}\\
&\ge 2\sqrt 2
\end{align*}
where $x = a + b + c + d$.
Also, $P(1/2, 1/2, 1/2, 1/2) = 2\sqrt 2$. Thus, the minimum of $P$ is $2\sqrt 2$.
Second, using $x + y + z + t \le \sqrt{4(x^2 + y^2 + z^2 + t^2)}$ for all $x, y, z, t \ge 0$ (well-known, the so-called AM-QM), we have
\begin{align*}
P &\le \sqrt{4[a^2 + (1 - b)^2 + b^2 + (1 - c)^2 + c^2 + (1 - d)^2 + d^2 + (1 - a)^2]}\\
&= \sqrt{16 - 8a - 8b - 8c - 8d + 8a^2 + 8b^2 + 8c^2 + 8d^2}\\
&\le 4
\end{align*}
where we have used $a \ge a^2$ etc.
Also, $P(1, 1, 1, 1) = 4$.
Thus, the maximum of $P$ is $4$.
Thus, the range of $P$ is $[2\sqrt 2, 4]$.
|
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|
Find $k\in\mathbb{R}$ given $w = k+i$ and $z=-4+5ki$ and $\arg(w+z)$ I am working on problem 2B-11 from the book Core Pure Mathematics, Edexcel 1/AS. The question is:
The complex numbers $w$ and $z$ are given by $w = k + i$ and $z = -4 + 5ki$ where $k$ is a real constant. Given at $\arg(w + z) = \frac{2\pi}{3}$, find the exact value of $k$.
I have attempted this question, but by answer is different from the given one and I can't figure out why. My answer:
Let $w + z = a + bi$.
$a = k - 4 \\
b = 1 + 5k$
The angle of the triangle formed by the vector $w + z$ (on Argand diagram) and the x-axis is $\pi -\frac{2\pi}{3} = \frac{\pi}{3}$
$\tan\frac{\pi}{3} = \frac{b}{a} \\
\tan\frac{\pi}{3} = \frac{1 + 5k}{k - 4} \\
k\tan\frac{\pi}{3} - 4\tan\frac{\pi}{3} = 1 + 5k \\
k\tan\frac{\pi}{3} - 5k = 1 + 4\tan\frac{\pi}{3} \\
k = \frac{1 + 4\tan\frac{\pi}{3}}{\tan\frac{\pi}{3} - 5} = -2.42$
However, the given answer is:
$\frac{4\sqrt{3} - 1}{5 + \sqrt{3}} = 0.88$ where $\tan\frac{\pi}{3} = \sqrt{3}$
I have just noticed that assuming the angle is $-\tan\frac{\pi}{3}$, my answer is correct. How can you tell the quadrant of $w + z$ when the constant $k$ appears in both $a$ and $b$?
|
$w = k + i$ and $z = -4 + 5ki$
Let $w + z = a + bi$.
$a = k - 4 \\b =i + 5ki$
Typo: $b=1+5k.$
The angle of the triangle formed by the complex number $w + z$ is $$\pi -\frac{2\pi}{3} = \frac{\pi}{3}.$$
"The triangle formed by $w + z$" is ambiguous.
In any case, your critical error is here, where you assume that $a$ and $b$ have the same sign:
$$\tan\frac{\pi}{3} = \frac{b}{a}.$$
Edit
The angle of the triangle formed by the vector w+z (on Argand diagram)
This is no less ambiguous than the original statement!
I still do not understand however, although I made that bad assumption, how do you know the sign of $a$ and $b,$ to be able to answer the question?
Since the argument of $a+bi$ is $\dfrac23\pi,$ then $a$ is negative and $b$ positive.
So, $$\operatorname{Arg}(a+bi)=\pi-\arctan\left(\frac b{-a}\right).$$
In general, $$\tan(\arg z)=\frac{\Im (z)}{\Re (z)}\quad\text{if }\Re (z)\ne0,$$ and $$\cot(\arg z)=\frac{\Re (z)}{\Im (z)}\quad\text{if }\Im (z)\ne0.$$
|
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|
Maclaurin series of $\ln \left( 1+\frac{\ln(1+x)}{1+x} \right)$ So the first thing I done was
$$\begin{align}\ln(1+x)&=x-\frac{1}{2}x^2+\frac{1}{3}x^3+o(x^3)\\&=(1+x)\left(x-\frac{3}{2}x^2+\frac{11}{6}x^3\right)+o(x^3)\end{align}$$
I've never seen this done but I'm pretty sure I can do this.
Now I want to divide:
$$(1+x)\left(x-\frac{3}{2}x^2+\frac{11}{6}x^3\right)+o(x^3)$$By $(1+x)$ but I don't know what
$\frac{o(x^3)}{1+x}$ will be.
Also is there another way to do this?
|
Let $u=1+x$ and $w=u+\log(u$). Rewrite
$$
f(x)=\log(1+\frac{\log(1+x)}{1+x})=\log(w) -\log(u)
$$
We seek the coefficients in the Taylor series expansion
$$
f(x) = \sum_{n=0}^{\infty}f^{\{n\}}(0)\frac{x^n}{n!}
$$
where the coefficient $f^{\{n\}}(0)$ is the nth derivative of the function evaluated at $x=0$. We find
$$
f^{'}(x)=\frac{w^{'}}{w}-\frac{1}{u}
$$
$$
f^{"}(x)=\frac{w^{"}}{w}-\frac{w^{'}}{w^2} +\frac{1}{u^2}
$$
$$
f^{\{3\}}(x)=\frac{w^{\{3\}}}{w}-\frac{2w^{"}}{w^2} +\frac{2w^{'}}{w^3}-\frac{2}{u^3}
$$
$$
f^{\{4\}}(x)=\frac{w^{\{4\}}}{w}- \frac{3w^{\{3\}}}{w^2}+\frac{6w^{"}}{w^3} -\frac{6w^{'}}{w^4}+\frac{6}{u^4}
$$
$$
f^{\{5\}}(x)=\frac{w^{\{5\}}}{w}- \frac{4w^{\{4\}}}{w^2}+\frac{12w^{\{3\}}}{w^3} -\frac{24w^{"}}{w^4}+\frac{24w^{'}}{w^5}-\frac{24}{u^5}
$$
We now observe
$$
w(0)=1\\
w^{'}(0)=2\\
w^{"}(0)=-1\\
w^{\{3\}}(0)=2\\
w^{\{4\}}(0)=-6=-3!\\
w^{\{5\}}(0)=24=4!
$$
Plugging these values into our derivative equations we find by induction,
$$
f(0)=0\\
f^{'}(0)=1\\
f^{"}(0)=-2\\
f^{\{3\}}(0)=6=3!\\
f^{\{4\}}(0)=-24=-4!\\
f^{\{5\}}(0)=5!\\
.\\
.\\
.\\
f^{\{n\}}(0)= (-1)^{n+1}n!\\
$$
Thus, interestingly enough,
$$
\log(1+\frac{\log(1+x)}{1+x})=\sum_{n=1}^{\infty}(-1)^{n+1}x^n
$$
|
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|
If $f(x)=\frac {x^ 2 -2x +4}{ x^ 2 +2x+4}$ for $x \in \mathbb{R}$, prove that the range of $f(x)$ is $[1/3, 3]$ One method to solve it would by putting $y = f(x)$ then multiplying the denominator with $y$ hence making a quadratic equation in x then we can just use the inequalities for $x$ being real to prove it.
For an alternative what I did is
$y = \frac{x^2 + 2x + 4}{x^2 - 2x + 4}$
Then$ \frac{1+y}{1-y} =\frac{x^2+4}{2x} $
Therefore$ \frac{1+y}{1-y} = {x/2+2/x}$
If we go further putting rhs range to lhs the answer gets altered . Is this method wrong if so why if not where did the process go wrong.
|
Function range definition : The set value of the dependent variable for which a function is defined.
Rewrite the equation given above as :
$$\frac{x^2-2x+4}{x^2+2x+4}=y$$
Multiply both sides by $x^2+2x+4$ :
$$\frac{x^2-2x+4}{x^2+2x+4}\left(x^2+2x+4\right)=y\left(x^2+2x+4\right)$$
Simplify :
$$x^2-2x+4=y\left(x^2+2x+4\right)$$
The range is a set of y for which the discriminant is greater or equal to zero :
$$x^2-2x+4=y\left(x^2+2x+4\right)$$
Expand the $y\left(x^2+2x+4\right)$ :
$$x^2-2x+4=x^2y+2xy+4y$$
Simplify to polynomial of $x$ :
$$\left(1-y\right)x^2-\left(2+2y\right)x+4-4y=0$$
For a quadratic equation of the form $ax^2+bx+c=0$, the discriminant is $b^2-4ac$. For :
$$a=1-y$$
$$b=-2-2y$$
$$c=4-4y$$
This becomes :
$$b^2-4ac$$
$$\left(-2-2y\right)^2-4\left(1-y\right)\left(4-4y\right)$$
Expand and simplify :
$$-12y^2+40y-12$$
Then :
$$-12y^2+40y-12\ge \:0$$
$$-3y^2+10y-3\ge \:0$$
$$-\left(3y-1\right)\left(y-3\right)\ge \:0$$
Multiply both side by $-1$ (reverse the inequality) :
$$\left(-\left(3y-1\right)\left(y-3\right)\right)\left(-1\right)\le \:0\cdot \left(-1\right)$$
Simplify :
$$\left(3y-1\right)\left(y-3\right)\le \:0$$
Identify the intervals and we will get :
$$\frac{1}{3}\le \:y\le \:3$$
$$ $$
Check if the range interval endpoints are included or not.
Take the point of $y=\frac{1}{3}$ and plug it into $\frac{x^2-2x+4}{x^2+2x+4}=y$ and we will get $x=2$.
Thus the solution exists for $y=\frac{1}{3}$. Therefore $y=\frac{1}{3}$ is included in the range.
Take the point of $y=3$ and plug it into $\frac{x^2-2x+4}{x^2+2x+4}=y$ and we will get $x=-2$.
Thus the solution exists for $y=3$. Therefore $y=3$ is included in the range.
$$ $$
Therefore, the range is :
$$\frac{1}{3}\le \:f\left(x\right)\le \:3$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Is $3 \times 2 = 3+3$ or $2+2+2$? $3\times 2$ is $3+3$ or $2+2+2$. We know both are correct as multiplication is commutative for whole numbers. But which one is mathematically accurate?
|
In every ring and even in every semiring, $3 \times 2$ is by definition $(1+1+1)\times (1+1)$ . Since multiplication distributes over addition, you get on the one hand
$$
(1+1+1) \times (1+1) = (1+1+1) + (1+1+1) = 3 +3
$$
and on the other hand
$$
(1+1+1) \times (1+1) = (1+1) + (1+1) + (1+1) = 2 + 2 + 2
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A basis that makes a matrix triangular. Find a basis for $\mathbb C^3$ so that the following matrix is in triangular form:
\begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{pmatrix}
What are the eigenvalues?
I found the eigenvalues to be $\lambda_1 = 1, \lambda_2 = \frac{1}{2}(-1 - i \sqrt{3} ), \lambda_3 = \frac{1}{2}(-1 + i \sqrt{3} ).$ And the eigenvectors to be $$\begin{pmatrix}1\\
1\\
1
\end{pmatrix}, \begin{pmatrix}\frac{ -(\sqrt{3} + i)^2}{4}\\
\frac{ i(\sqrt{3} + i)}{2}\\
1
\end{pmatrix},\begin{pmatrix} \frac{ 2i}{\sqrt{3} - i}\\
\frac{ -2i}{\sqrt{3} + i}\\ 1
\end{pmatrix}$$
But still I do not know how this will help me to find a basis for $\mathbb C^3$ so that the given matrix is in triangular form. I was told that I may find a dual basis, but still I do not know what exactly the steps I should do. Could anyone help me in solving this problem please? (not necessarily using dual basis and probably using an elegant and clean general way of doing this)
|
My calculations showed the following:
(Kindly try yourself and verify these results)
Eigenvalue: $\lambda_1 = 1$
Eigenvector for $\lambda_1 = 1$:
We solve the equation
$$
(A - \lambda_1 I) \mathbf{x} = \mathbf{0}
$$
The RREF of $A - \lambda_1 I$ is obtained as
$$
R_1 = \left[ \begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & -1 \\
0 & 0 & 0 \\
\end{array} \right]
$$
Thus, we solve the equations
$$
x_1 - x_3 = 0, \ \ x_2 - x_3 = 0
$$
Easy to see that we get an eigenvector by taking $x_3$
as a free variable and setting $x_3 = 1$.
Thus, we get the eigenvector
$$\mathbf{v}_1 = \left[ \begin{array}{c}
1 \\
1 \\
1 \\
\end{array} \right] $$
Eigenvalue: $\lambda_2 = -{1 \over 2} - i {\sqrt{3} \over 2} $
Eigenvector for the eigenvalue $\lambda_2$:
We solve the equation
$$
(A - \lambda_2 I) \mathbf{x} = \mathbf{0}
$$
It is easy to verify that RREF of $A-\lambda_2 I$ is obtained as
$$
R_2 = \left[ \begin{array}{ccc}
1 & 0 & {1 \over 2} + i {\sqrt{3} \over 2} \\[2mm]
0 & 1 & {1 \over 2} - i {\sqrt{3} \over 2} \\[2mm]
0 & 0 & 0 \\[2mm]
\end{array} \right]
$$
Thus, we get an eigenvector as:
$$\mathbf{v}_2 = \left[ \begin{array}{c}
-{1 \over 2} - i {\sqrt{3} \over 2} \\
-{1 \over 2} + i {\sqrt{3} \over 2} \\
1 \\
\end{array} \right] $$
Eigenvalue: $\lambda_3 = \bar{\lambda}_2 = -{1 \over 2} + i {\sqrt{3} \over 2} $
Eigenvector for $\lambda_3$:
We solve the equation
$$
(A - \lambda_3 I) \mathbf{x} = \mathbf{0}
$$
It is easy to see that RREF of $A - \lambda_3$ is obtained as
$$
R_3 = \left[ \begin{array}{ccc}
1 & 0 & {1 \over 2} - i {\sqrt{3} \over 2} \\[2mm]
0 & 1 & {1 \over 2} + i {\sqrt{3} \over 2} \\[2mm]
0 & 0 & 0 \\[2mm]
\end{array} \right]
$$
Hence, we get an eigenvector as:
$$\mathbf{v}_3 = \bar{\mathbf{v}}_2 = \left[ \begin{array}{c}
-{1 \over 2} + i {\sqrt{3} \over 2} \\
-{1 \over 2} - i {\sqrt{3} \over 2} \\
1 \\
\end{array} \right] $$
Form the modal matrix:
$$
P = \left[ \matrix{
\mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 \cr} \right]
$$
The eigenvectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are
linearly independent, since they correspond to different eigenvalues.
It is easy to verify that
$$
P^{-1} A P = D = \left[ \begin{array}{ccc}
\lambda_1 & 0 & 0 \\
0 & \lambda_2 & 0 \\
0 & 0 & \lambda_3 \\
\end{array} \right]
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Equation of bisector between two straight lines given in symmetrical form Please help in solving the attached question.
I know in 2D, it is solved as
$\dfrac{ax+by+c}{\sqrt{a^2+b^2}}=\pm \dfrac{px+qy+s}{\sqrt{p^2+q^2}}$
Not sure if we use the same formula for 3D? And if yes, whatβs the value of the constant c and s here?
|
In $3D$ the equation of the two lines is given as follows
$ \dfrac{ x - x_1}{a} = \dfrac{y - y_1}{b} = \dfrac{z - z_1}{c} $
for the first line, and
$ \dfrac{ x - x_1}{d} = \dfrac{y - y_1}{e} = \dfrac{z - z_1}{f} $
and these two symmetrical form equations correspond to the parametric equations
$P_1(t) = (x_1, y_1, z_1 ) + t (a,b,c) $
$P_2(t) = (x_1, y_1, z_1) + t (d, e, f) $
From here find the unit vectors
$u_1 = \dfrac{ (a,b,c) }{\sqrt{a^2 + b^2 + c^2} } = (a_1, b_1, c_1)$
$u_2 = \dfrac{(d,e,f) }{\sqrt{d^2 + e^2 + f^2}} = (d_1, e_1, f_1) $
The direction vectors along the two bisectors are
$ v_1 = u_1 + u_2 $ and $ v_2 = u_1 - u_2 $
And therefore, the symmetrical form equations of the two bisectors are
$ \dfrac{ x - x_1}{a_1 + d_1} = \dfrac{y - y_1}{b_1 + e_1} = \dfrac{z - z_1}{c_1 + f_1} $
$ \dfrac{ x - x_1}{a_1 - d_1} = \dfrac{y - y_1}{b_1 - e_1} = \dfrac{z - z_1}{c_1 - f_1} $
For the question quoted, $(x_1,y_1, z_1) = (3,-4, 5) $, $d_1 = (2,-1,-2) $, $d_2 = (4, -12, 3)$
From here $u_1 = \dfrac{ (2, -1, -2) }{\sqrt{2^2 + (-1)^2 + (-2)^2} } = (\dfrac{2}{3}, -\dfrac{1}{3}, - \dfrac{2}{3} ) $
and
$u_2 = \dfrac{ (4, -12, 3) }{\sqrt{4^2 + (-12)^2 + 3^2}} = (\dfrac{4}{13}, -\dfrac{12}{13}, \dfrac{3}{13} ) $
Hence,
$v_1 = u_1 + u_2 = ( \dfrac{38}{39} , -\dfrac{49}{39}, - \dfrac{17}{39} )$
$v_2 = u_1 - u_2 = ( \dfrac{14}{39} , \dfrac{23}{39} , -\dfrac{35}{39} )$
Normalizing $v_1, v_2$ by multiplying through by $39$, the equations of the bisectors are
$ \dfrac{ x - 3}{38} = \dfrac{y +4}{-49} = \dfrac{z - 5}{-17} $
and
$ \dfrac{ x - 3}{14} = \dfrac{y +4}{23} = \dfrac{z - 5}{-35} $
|
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|
An inequality $2(ab+ac+ad+bc+bd+cd)\le (a+b+c+d) +2(abc+abd+acd+bcd)$
Let $a, b, c, d \ge 0$ such that $a+b, a+c, a+d, b+c, b+d, c+d \le 1$. Show that
$$2(ab+ac+ad+bc+bd+cd)\le (a+b+c+d) +2(abc+abd+acd+bcd).$$
I am trying to maximise
$$f(a,b,c,d)=(a+b+c+d) /(1-(1-a)(1-b)(1-c)(1-d) +abcd) $$ over the set of nonnegative real numbers $a, b, c, d$ subject to $a+b, a+c, a+d, b+c, b+d, c+d \le 1$.
When $a=b=c=d=0$, set $f(0,0,0,0)=1$. Some massage with Wolfram Alpha gives $2$ as a local maximum, which, if it were the global maximum, would then be equivalent to the above inequality.
It seems one standard way to solve inequalities of this class is the pqrs, uvwt method.
|
The form of the constraints on the variables suggest a transform based around their differences from each other and from $1/2$. Let $a = (1-x+y)/2$, $b = (1 - x - y)/2$, $c = (1-z+w)/2$, $d = (1-z-w)/2$. The positivity constraints are $x + |y| \le 1$ and $z + |w| \le 1$. The sum constraints are $x,z\in[0,1]$ and $|y|+|w|\le x+z$.
With these substitutions, the inequality can be reduced to
$$
(x+z)(1-xz) + zy^2 + xw^2 \ge 0,
$$
which is clearly true whenever $x,z\in [0,1]$.
This substitution also works with your function method. The expression for the function becomes
$$
\frac{4(2-x-z)}{4 - (x+z)(xz+1)+xw^2 + zy^2}.
$$
This function's value can only be increased by decreasing $|y|$ and $|w|$. Additionally, decreasing $|y|$ or $|w|$ can never cause the constraints to become unsatisfied. Thus its maximum value is will be the same as its maximum value assuming $y = w = 0$. That is, $a = b$ and $c = d$. The constraints then require $0\le a,c\le 1/2$. It's not hard to show the maximum value of $f(a,c,a,c)$ subject to these constraints is $2$, occurring at $a = c = 1/2$.
|
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|
Is there a mistake in solving this limit? I want to solve this:
\begin{equation}
L=\lim_{x\rightarrow 0} \frac{\sum\limits_{m=1}^{M}a_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}{(m^2+2x^2)^{-3/2}} } {\sum\limits_{m=1}^{M} c_{m}\exp\left\{\frac{bm^2}{ (m^2+2x^2)x^2}\right\}{(m^2+2x^2)^{-3/2}}}
\end{equation}
where
\begin{equation}
\begin{aligned}
m=1,2,\cdots,M\\
a_m,b,c_m \neq 0 \quad \text{and are all constants}
\end{aligned}
\end{equation}
Here is my approach:
Since $x\rightarrow 0$, I ignore the term $2x^4$ in the denominator of $\exp\left\{\frac{b_m}{ (m^2+2x^2)x^2}\right\}$ and I get:
\begin{equation}
\begin{aligned}
L&=\lim_{x\rightarrow 0} \frac{\exp\left\{\frac{b}{x^2}\right\}\sum\limits_{m=1}^{M}a_{m}{(m^2+2x^2)^{-3/2}} } {\exp\left\{\frac{b}{x^2}\right\}\sum\limits_{m=1}^{M} c_{m}{(m^2+2x^2)^{-3/2}}}\\
&=\lim_{x\rightarrow 0} \frac{\sum\limits_{m=1}^{M}a_{m}{(m^2+2x^2)^{-3/2}} } {\sum\limits_{m=1}^{M} c_{m}{(m^2+2x^2)^{-3/2}}}\\
&=\frac{\sum\limits_{m=1}^{M}\frac{a_{m}}{m^3} } {\sum\limits_{m=1}^{M} \frac{c_{m}}{m^3}}
\end{aligned}
\end{equation}
My question is:
*
*Is there a mistake in my derivation?
*If I made mistakes in the derivation, then, what is the correct derivation?
Thanks for helpful comments and answers!
|
Since @justt gave the answer, just try.
Take an example : $a_m=m$, $c_m=m^2$, $b=1$, $M=4$.
Without any simplification at all, the limit is
$$\frac{144+36 e^{3/2}+16 e^{16/9}+9 e^{15/8}}{144+72 e^{3/2}+48 e^{16/9}+36
e^{15/8}}=0.465476$$
Ignoring the term $2x^4$, the limit becomes $\frac{41}{60}$
|
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|
How do I get to this formula for the area of a triangle I am new here and new student to geometry.
In my geometry skript there is a task:
Show that the formula for the area $F$ of a triangle with sidelengths $a,b,c $ is given by
$$F^2 = - \frac{1}{16} \det \begin{pmatrix} 0 & c^2 & b^2 & 1 \\ c^2 & 0 & a^2 & 1 \\ b^2 & a^2 & 0 & 1 \\ 1& 1& 1& 0 \end{pmatrix}$$
This drives me crazy because I cant see how to get there.
I do know how to get to
$F= \frac12 \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} $
and $ F = \frac12 \det \begin{pmatrix} 1 & 1 & 1 \\ a & b & c \end{pmatrix} $
I also thought about using Heron's formula for the square of a triangle:
$ F^2 = s(s-a)(s-b)(s-c) $ where $s= \frac{a+b+c}{2} $
I did multiplied it all out and can't seem to find a pattern..
Hope this question is appropriate.
Maybe here is someone who has an idea :)
|
Start with
\begin{pmatrix} 0 & c^2 & b^2 & 1 \\ c^2 & 0 & a^2 & 1 \\ b^2 & a^2 & 0 & 1 \\ 1& 1& 1& 0 \end{pmatrix}
Perform the column transformations $C_2 \to C_2 - C_1$ and $C_3 \to C_3- C_1$ which don't change the determinant.
\begin{pmatrix} 0 & c^2 & b^2 & 1 \\ c^2 & -c^2 & a^2-c^2 & 1 \\ b^2 & a^2-b^2 & -b^2 & 1 \\ 1& 0& 0& 0 \end{pmatrix}
Perform the row transformations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ which also leave the determinant unchanged.
\begin{pmatrix} 0 & c^2 & b^2 & 1 \\ c^2 & -2c^2 & a^2-c^2-b^2 & 0 \\ b^2 & a^2-b^2-c^2 & -2b^2 & 0 \\ 1& 0& 0& 0 \end{pmatrix}
A simple Laplace expansion, first along the fourth row then along the third column of the resulting submatrix, will tell you that the determinant of this matrix is in fact the determinant of the central matrix
\begin{pmatrix} -2c^2 & a^2-c^2-b^2\\ a^2-b^2-c^2 & -2b^2 \end{pmatrix}
which is just equal to $$
4b^2c^2 - (a^2-b^2-c^2)^2 = (2bc)^2 - (a^2-b^2-c^2)^2\\=(a^2-b^2+2bc-c^2)(a^2-b^2-2bc-c^2) \\= (a^2 - (b+c)^2) (a^2-(b-c)^2) \\
= (a-b-c)(a+b+c)(a+b-c)(a+c-b)
$$
Recognizing the semi-perimeter $2S = a+b+c$, we just have $$
a-b-c = 2(a-S) \\
a+b+c = 2S \\
(a+b-c) = 2(S-c)\\
(a+c-b) = 2(S-b)
$$
which leads to $$
(a-b-c)(a+b+c)(a+b-c)(a+c-b) = 16S(S-c)(S-b)(a-S) = -16F^2
$$
where $F$ is the area of the triangle with sides $a,b,c$ and $F^2 = S(S-a)(S-b)(S-c)$ by Heron's formula. That is, the determinant at the start equals $-16F^2$, as desired.
|
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|
Prove that, for any positive real numbers a, b, and c, $a^3b^2 + a^2b^3 + a^3c^2 + a^2c^3 + b^3c^2 + b^2c^3 \geq \frac{a^5 + b^5 + c^5}{5}$. For this question I assume you would start with a rewritten form so that $a^3b^2 + a^2b^3 + a^3c^2 + a^2c^3 + b^3c^2 + b^2c^3$ = $(a^3 + b^3 + c^3)(a^2 + b^2 + c^2) - (a^5 + b^5 + c^5)$, but how would I progress from there to prove the inequality? I have tried algebraic manipulation, but it has led me no where. I have also tried the AM-GM-RMS inequality, but no luck! I might have made an algebraic error somewhere. Any help is appreciated, thanks!
|
$(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$ .for easy simplification note that($(a+b)^5=\sum_{k=0}^{5}{5 \choose k}a^{5-k}b^k$)
$(a+c)^5=a^5+5a^4c+10a^3c^2+10a^2c^3+5ac^4+c^5$
$(c+b)^5=c^5+5c^4b+10c^3b^2+10c^2b^3+5cb^4+b^5$
then $(a+b)^5+(a+c)^5+(c+b)^5=2a^5+2b^5+2c^5+5ab^4+5ac^4+5cb^4+10a^2b^3+10a^2c^3+10c^2b^3+10a^3b^2+10a^3c^2+10c^3b^2+5a^4b++5a^4c+5c^4b$
and this implie that:$\frac{(a+b)^5+(a+c)^5+(c+b)^5}{10}-\frac{2a^5+2b^5+2c^5}{10}\geq a^2b^3+a^2c^3+c^2b^3+a^3b^2+a^3c^2+c^3b^2$.(beacuse if $x=y+w$ then$ x \geq y,w$ for any positive numbers $x,y,x$)
so :$-\frac{2a^5+2b^5+2c^5}{10}\geq a^2b^3+a^2c^3+c^2b^3+a^3b^2+a^3c^2+c^3b^2-\frac{(a+b)^5+(a+c)^5+(c+b)^5}{10}$
so finally ;$\frac{a^5+b^5+c^5}{5}\leq -(a^2b^3+a^2c^3+c^2b^3+a^3b^2+a^3c^2+c^3b^2)+\frac{(a+b)^5+(a+c)^5+(c+b)^5}{10}$ (**)
this inequality holds for any positive numbers, but your inequality isn't always true, and you can wonder if the left hand side of your inequality will be smaller or equal the right hand side of (**),so you will see it is not always true.
|
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|
Proving a combinatorial identity related to convolution of central binomial coefficients I'm trying to calculate the coefficient values for $f(x) = \sum_{k=0}^n {n \choose k}^2 (1+x)^{2n-2k}(1-x)^{2k}$. TL;DR I don't know where to begin in order to prove:
$$\sum_{k=0}^r (-4)^k{n \choose k}{2n-2k \choose n}{n-2k \choose r-k} = {2r \choose r}{2n-2r \choose n-r}$$
Explanation: writing $f(x)$ as the coefficient of $y^n$ in:
\begin{align*}
g(x, y) & = [y + (1 + x)^2]^n[y - (1 + x)^2]^n \\\\
& = [(y + 1 + x^2)^2 - 4x^2]^n
\end{align*}
shows that this is an even function in $x$, so any nonzero coefficient of $x^l$ requires $l = 2r$ for some integer $r$. Making this substitution and calculating the coefficient of $y^n$ using the Multinomial Theorem gives
\begin{align*}
& [x^{2r}][y^n]\left((y+1+x^2)^2 - 4x^2 \right)^n \\
= & [x^{2r}]\sum_{k=0}^n {2n-2k \choose k; n-2k} (-4)^{k}x^{2k}(1+x^2)^{n-2k} \;\text{(Multinomial Expansion)} \\
= & [x^{r}]\sum_{k=0}^n (-4)^k {n \choose k}{2n-2k \choose n}x^k(1+x)^{n-2k} \;\text{($x^2 \to x$ substitution)} \\
= & \sum_{k=0}^r (-4)^k{n \choose k}{2n-2k \choose n}{n-2k \choose r-k}
\end{align*}
At this point, I can tell by numerical calculation of the first few values of $n$ and $r$ that this is equal to ${2r \choose r}{2n-2r \choose n-r}$, but despite many attempts at resolving the sum using generating functions and combinatorial arguments I haven't managed to prove it. I would appreciate any help with this or insight into my methods, the sum, or how to prove the value of the coefficient.
|
Suppose we seek to evaluate the coefficients
$$[x^q] f_n(x) = [x^q] \sum_{k=0}^n {n\choose k}^2
(1+x)^{2n-2k} (1-x)^{2k}.$$
Recall the generating function of the Legendre polynomials
$$\sum_{n\ge 0} P_n(y) t^n = \frac{1}{\sqrt{1-2yt+t^2}}$$
and the known fact
$$P_n(y) = \left[\frac{y-1}{2}\right]^n
\sum_{k=0}^n {n\choose k}^2
\left[\frac{y+1}{y-1}\right]^k.$$
Now put $y= -\frac{x^2+1}{2x}$ to get
$$P_n\left(-\frac{x^2+1}{2x}\right)
= (-1)^n \frac{(1+x)^{2n}}{4^n x^n}
\sum_{k=0}^n {n\choose k}^2
\left[\frac{(1-x)^2}{(1+x)^2}\right]^k.$$
It follows that
$$f_n(x) = (-1)^n 4^n x^n
P_n\left(-\frac{x^2+1}{2x}\right).$$
Using the OGF we have
$$(-1)^n 4^n x^n
[t^n] \frac{1}{\sqrt{1+((x^2+1)/x)t+t^2}}
\\ = [t^n] \frac{1}{\sqrt{1-4(x^2+1)t+16x^2t^2}}
\\ = [t^n] \frac{1}{\sqrt{1-4t}} \frac{1}{\sqrt{1-4x^2t}}.$$
We observe at this point that we must have $q=2r$ and continue with
$$[x^{2r}] [t^n] \frac{1}{\sqrt{1-4t}} \frac{1}{\sqrt{1-4x^2t}}
\\ = [x^r] [t^n] \frac{1}{\sqrt{1-4xt}} \frac{1}{\sqrt{1-4t}}
= [x^r] \sum_{k=0}^n {2k\choose k} x^k {2n-2k\choose n-k}
\\ = {2r\choose r} {2n-2r\choose n-r}.$$
This is the claim. One reference for Legendre polynomials is Gould's
Combinatorial Identities (page 38).
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Show that $p^k \mid \mid (x-y)$ iff $p^k \mid \mid (x^6-y^6)$. Definition. Let $n>1$ be an integer and $p$ be a prime. We say that $p^k$ fully divides $n$ and write $p^k \mid \mid n$ if $k$ is the greatest positive integer such that $p^k \mid n$.
Let $p>6$ be a prime. Let $x$ and $y$ be two distinct integers such that $p\nmid x, p\nmid y,$ and $p\mid (x-y)$. Show that $p^k \mid \mid (x-y)$ iff $p^k \mid \mid (x^6-y^6)$.
Attempt:
$(\implies)$ Let $p^k \mid \mid (x-y)$. Then $k$ is the greatest positive integer such that $p^k \mid (x-y)$. Write $x-y=p^km$ for some $m \in \Bbb Z$. Notice that
$$x^6-y^6 = (x-y)(x^5+x^4y+\cdots+y^5) = p^km(x^5+x^4y+\cdots+y^5)=p^k(m(x^5+x^4y+\cdots+y^5)).$$
Hence, $p^k \mid (x^6-y^6)$ and then $p^k \mid \mid (x^6-y^6)$.
Does this approach correct? If not, how to approach it? And how to approach the reverse direction?
Any ideas? Thanks in advanced.
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Your attempt so far looks good (as far as I can tell). What you really need to show is that $p\nmid (x^5+\dots+y^5)$ because by now you have only shown that $p^k \mid (x^6-y^6)$. Let's focus on this point, then the reverse direction gets easy. We have $p|(x-y)$ thus $x-y \equiv 0 \mod p$. From this we conclude $x\equiv y \mod p$, so $x^5 \equiv x^4y\equiv x^3y^2\equiv x^2y^3 \equiv xy^4 \equiv y^5 \mod p$. Now $(x^5+\dots + y^5) \equiv 6x^5 \mod p$ and since $p\nmid x$ we have $p\nmid x^5$ and we can conclude $ 6x^5\not \equiv 0 \mod p$. This completes your argument, for the reverse direction consider: $p^k\mid\mid (x^6-y^6)$ so $(x^6-y^6)=p^km=(x-y)(x^5+\dots +y^5)$. Now since $p\nmid (x^5+\dots y^5)$ we have $p^k \mid (x-y)$ and of course $p^k\mid\mid (x-y)$ because otherwhise we would have $p\mid m$ which is not possible.
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Find non-zeroes $a$ and $b$ such that $\lim[a^n(u_n - 1)] = b$ where $u_1 = 0$ and $4u_{n + 1} = u_n + \sqrt{6u_n + 3}, \forall n \in \mathbb Z^+$.
Consider sequence $(u_n)$, defined as $\left\{ \begin{aligned} u_1 &= 0\\ u_{n + 1} &= \dfrac{u_n + \sqrt{6u_n + 3}}{4}, \forall n \in \mathbb Z^+ \end{aligned} \right.$. Knowing that $a$ and $b$ are two real numbers not equal to zero such that $\lim[a^n(u_n - 1)] = b$, calculate the value of $a + b$.
[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
Why was this question there? WHY WAS THIS IN MY MID-SEMESTER EXAM? (>_<)
I don't need answers, I need to be angered. (This is a joke, please take it with a grain of salt. I do actually need answers to this problem.)
Here are my observations.
According to the WolframAlpha, the general term of the sequence above is $$u_n = \left(1 - \dfrac{1}{2^{n - 1}}\right)\left(1 - \dfrac{2 - \sqrt 3}{2^{n - 1}}\right), \forall n \in \mathbb Z^+$$
If we define sequences $(v_n)$ and $(w_n)$ as $v_n = 1 - \dfrac{1}{2^{n - 1}}, \forall n \in \mathbb Z^+$ and $w_n = 1 - \dfrac{2 - \sqrt 3}{2^{n - 1}}, \forall n \in \mathbb Z^+$ respectively, then it can be obtained that $u_n = v_nw_n, \forall n \in \mathbb Z^+$ and $$\left\{ \begin{aligned} v_1 &= 0\\ v_{n + 1} &= \dfrac{v_n + 1}{2}, \forall n \in \mathbb Z^+ \end{aligned} \right. \text{ and } \left\{ \begin{aligned} w_1 &= -1 + \sqrt 3\\ w_{n + 1} &= \dfrac{w_n + 1}{2}, \forall n \in \mathbb Z^+ \end{aligned} \right.$$
That means $$\begin{aligned} v_{n + 1}w_{n + 1} = \dfrac{v_nw_n + \sqrt{6v_nw_n + 3}}{4} &\iff \dfrac{(v_n + 1)(w_n + 1)}{4} = \dfrac{v_nw_n + \sqrt{6v_nw_n + 3}}{4}\\ &\iff v_n + w_n + 1 = \sqrt{6v_nw_n + 3}\\ &\iff w_n = (2v_n - 1) + \sqrt 3(1 - v_n), \forall n \in \mathbb Z^+ \end{aligned}$$
Anyhow, about the sequence $(u_n)$ itself, it is strictly increasing bounded. More specifically, we have that $u_n \in [0; 1), \forall n \in \mathbb Z^+$. The same goes for sequences $(v_n)$ and $(w_n)$.
Actually, $v_{n + 1} = \dfrac{v_n + 1}{2}$ and $w_{n + 1} = \dfrac{w_n + 1}{2}$, those look familiar, hmmm~
Of course, I forgot. If we let $v_n = \cos a_n, \forall n \in \mathbb Z^+$ and $w_n = \cos b_n, \forall n \in \mathbb Z^+$, then it can be obtained that $$\left\{ \begin{aligned} a_1 &= \dfrac{\pi}{2}\\ \cos(a_{n + 1}) &= \cos^2\dfrac{a_n}{2}, \forall n \in \mathbb Z^+ \end{aligned} \right. \text{ and } \left\{ \begin{aligned} b_1 &= \arccos(-1 + \sqrt 3)\\ \cos(b_{n + 1}) &= \cos^2\dfrac{b_n}{2}, \forall n \in \mathbb Z^+ \end{aligned} \right.$$
Nevermind, that didn't work as well as I had thought.
What was I doing this entire time? You might be wondering. Well, I'm trying to find the general term of sequence $(u_n)$ without the need of a laptop, since you can't take that into the testing room.
Anyhow, for the second part of the problem, first of all, let $\lim u_n = m$, then we have that $m = \dfrac{m + \sqrt{6m + 3}}{4} \iff m = 1$. Again, the same goes for sequences $(v_n)$ and $(w_n)$. Futhermore, $$\begin{aligned} \left\{ \begin{aligned} 2^n(v_n - 1) &= -2\\ 2^n(w_n - 1) &= 2\sqrt 3 - 4 \end{aligned} \right. &\iff \left\{ \begin{aligned} 2^n(v_n + w_n - 2) &= 2\sqrt{3} - 6\\ 4^n(v_n - 1)(w_n - 1) &= 8 - 4\sqrt{3} \end{aligned} \right.\\ &\implies 4^n\left[u_n - \left(\dfrac{2\sqrt{3} - 6}{2^n} + 2\right) + 1\right] = 8 - 4\sqrt{3}\\ &\iff 4^n(u_n - 1) = (8 - 4\sqrt{3}) - 2^n(6 - 2\sqrt{3})\\ &\iff 2^n(u_n - 1) = \dfrac{(8 - 4\sqrt{3})}{2^n} - (2\sqrt{3} - 6), \forall n \in \mathbb Z^+\\ &\implies \lim[2^n(u_n - 1)] = 2\sqrt{3} - 6 \end{aligned}$$
In conclusion, $a + b = 2 + (2\sqrt{3} - 6) = 2\sqrt{3} - 4$.
My question is more focused on the first part of the problem, on how the general term of sequence $(u_n)$. As always, thanks for reading (and even more if you could help~) By the way, the options were $-1, \sqrt 3 - 1, 2\sqrt 3 - 4$ and $4 - 2\sqrt 2$.
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Well, there may be something "lost in translation", indeed: if all those questions were multiple choice, which were the choices for $a+b$?
I'm asking because $u_n$ is converging to $1$ quite rapidly, and since
$$a=\lim_{n\to\infty}\frac{u_n-1}{u_{n+1}-1},$$ computing that ratio for a few $n$ (up to $n=10$, say) with a pocket calculator would suggest $a=2$, and calculating $b_n=2^n\,(u_n-1)$ for $n=10$ might be sufficient to identify $a+b$ among the given choices.
If you really want to discover the closed form, it's hardly a good idea to start from WolframAlpha's answer. Instead, let's make that square root rational. And since the limit of $u_n$ is $1$, and thus, the limit of $6\,u_n+3$ is $9$, let
$$6\,u_n+3=9\,v^2_n.$$ This gives
$$u_n=\frac{3\,v^2_n-1}2,$$ and
$$\frac{3\,v^2_{n+1}-1}2=\frac{\frac{3\,v^2_n-1}2+3\,v_n}4$$ simplifies to
$$v^2_{n+1}=\frac{v^2_n+2\,v_n+1}4=\left(\frac{v_n+1}2\right)^2,$$
i.e.
$$v_{n+1}=\frac{v_n+1}2.$$ So
$$v_{n+1}-1=\frac{v_n-1}2,$$ meaning
$$v_n-1=(1/2)^{n-1}(v_1-1)=(1/2)^{n-1}\left(\sqrt{1/3}-1\right).$$ This gives another closed form for $u_n$, but proving this to be identical with WolframAlpha's answer may be left as an exercise to the reader. ;-)
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Is there any closed form for $\frac{d^{2 n}( \cot z)}{d z^{2 n}}\big|_{z=\frac{\pi}{4}}$? Latest Edit
Thanks to Mr Ali Shadhar who gave a beautiful closed form of the derivative which finish the problem as
$$\boxed{S_n = \frac { \pi ^ { 2 n + 1 } } { 4 ^ { n + 1 } ( 2 n ) ! } | E _ { 2 n } | }, $$
where $E_{2n}$ is an even Euler Number.
In the post, I had found the sum
$$
\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{3}}= \frac{\pi^{3}}{32},
$$
and want to investigate it in a more general manner, $$
S_{n}=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2 n+1}}
$$
where $n\in N.$
$$
\begin{aligned}
S_{n}&= \lim _{N \rightarrow \infty} \sum_{k=0}^{N} \frac{1}{(4k+1)^{2 n+1}}-\sum_{k=0}^{N} \frac{1}{(4 k+3)^{2 n+1}} \\
&=\frac{1}{4^{2 n+1}} \lim _{N \rightarrow \infty} \left[\sum_{k=0}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}-\sum_{k=0}^{N} \frac{1}{\left(k+\frac{3}{4}\right)^{2 n+1}}\right] \\
&=\frac{1}{4^{2 n+1}} \lim _{N \rightarrow \infty} \left[\sum_{k=0}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}+\sum_{k=0}^{N} \frac{1}{\left(-k-\frac{3}{4}\right)^{2 n+1}}\right] \\
&=\frac{1}{4^{2 n+1}} \lim _{N \rightarrow \infty} \left[\sum_{k=0}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}+\sum_{k=-N}^{-1} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}\right] \\
&=\frac{1}{4^{2 n+1}}\left[\lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}\right]
\end{aligned}
$$
Using the Theorem:
$$(*):\pi \cot (\pi z)=\lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{1}{k+z} ,\quad \forall z \not \in Z.$$
Differentiating (*) w.r.t. $z$ by $2 n$ times yields
$$
\begin{aligned}
& \lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{(-1)^{2 n}(2 n) !}{(k+z)^{2 n+1}}=\frac{d^{2 n}}{d z^{2 n}}[\pi \cot (\pi z)] \\
\Rightarrow & \lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{1}{(k+z)^{2 n+1}}=\frac{\pi}{(2 n) !} \frac{d^{2 n}}{d z^{2 n}}[\cot (\pi z)]
\end{aligned}
$$
Now we can conclude that
$$\boxed{\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2n+1}}=\left.\frac{\pi^{2n+1}}{4^{2 n+1}(2 n) !} \frac{d^{2 n}}{d z^{2 n}}[\cot z]\right|_{z=\frac{\pi}{4}}}$$
My Question:
Is there any closed form for $\displaystyle \left.\frac{d^{2 n} (\cot z)}{d z^{2 n}}\right|_{z=\frac{\pi}{4}}$?
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Using $\csc(z)=\cot(z/2)-\cot(z)$, we have
$$\lim_{z\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\csc(z)=\underbrace{\lim_{z\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\cot(z/2)}_{z=2x\to\, dz=2 dx}-\underbrace{\lim_{z\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\cot(z)}_{0}$$
$$=2^{-2n}\lim_{x\to \frac{\pi}{4}}\frac{d^{2n}}{dx^{2n}}\cot(x)$$
$$\Longrightarrow \lim_{x\to \frac{\pi}{4}}\frac{d^{2n}}{dx^{2n}}\cot(x)=2^{2n}\lim_{z\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\csc(z).$$
Substitute $\,\,\,\,\displaystyle\lim_{z\to \frac{\pi}{2}}\frac{d^{2n}}{dz^{2n}}\csc(z)=|E_{2n}|$, (check the edit section at the end)
$$\boxed{\lim_{x\to \frac{\pi}{4}}\frac{d^{2n}}{dx^{2n}}\cot(x)=2^{2n}|E_{2n}|}$$
which matches @Mariusz Iwaniuk's answer in the comments since $(-1)^n E_{2n}=|E_{2n}$|.
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Fractions in Questions and Answers
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