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How to get the value of $a + b + c$? $(0 \leq a < b < c) \in Z$,
$a + b + c + ab + ac + bc + abc = 1622$
$a + b + c = ?$
I've assumed that $a = 0$, by that we can get rid of this part $ab + ac + abc$
Now $bc + b + c = 1622$.
But I found that was useless and got stuck.
|
Hint:
$$a+b+c+ab+ac+bc+abc=(1+a)(1+b)(1+c)-1$$
Therefore $(1+a)(1+b)(1+c)=?$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do I solve $\lim \limits_{x \to \frac{π}{3}} \frac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$ Alright, I know, there are easier ways to solve this, like L'hopitals Rule etc.
But I'm not solving it for the answer, just doing it for the fun so I tried using substitution method.
Put $t= x- \dfrac{π}{3}$
$\lim \limits_{x \to \frac{π}{3}} \dfrac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$
$= \lim\limits_{t \to 0} \dfrac{2 \sin \left(t+\frac{π}{3} \right) - \sqrt{3}}{\cos \left( \frac{3t}{2} + \frac{π}{2}\right)}$
$= \lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}$
Where do I go from here, I'm not able to eliminate the $t$ fully from the Nr and Dr, any help?
Or any other alternative way that uses only the fact that $\lim\limits_{x \to 0} \dfrac{ \sin x}{x} = 1$?
Thanks :)
|
So we already have, after putting $\;t:=x-\frac\pi3\;$ :
$$\dfrac{2 \sin \left(t+\frac{π}{3} \right) - \sqrt{3}}{\cos \left( \frac{3t}{2} + \frac{π}{2}\right)}=\frac{2\left(\sin t\cdot\frac12+\frac{\sqrt3}2\cos t\right)-\sqrt3}{-\sin\frac{3t}2}=-\frac{\sin t}{\sin\frac{3t}2}-\sqrt3\frac{\cos t -1}{\sin\frac{3t}2}=$$$${}$$
$$=-\frac23\cdot\frac{\frac{3t}2}{\sin\frac{3t}2}\cdot\frac{\sin t}t+\sqrt3\cdot\frac23\cdot\frac{\frac{3t}2}{\sin\frac{3t}2}\cdot\frac{1-\cos t}t\xrightarrow[t\to0]{}-\frac23\cdot1\cdot1+\frac2{\sqrt3}\cdot1\cdot0=-\frac23$$
|
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|
Proof that $x^{(n+4)} \bmod 10 = x^n \bmod 10\,$ for $\,n\ge 1$ While solving a programming challenge in which one should efficiently compute the last digit of $a^b$, I noticed that apparently the following holds (for $n > 0$)
$x^{(n+4)} \mod 10 = x^n \mod 10$
How can this be proven?
|
We want to show that $x^{n+4}-x^n=x^n(x^4-1)$ is a multiple of $10$.
We first notice that if $x^n$ is odd then $x^4-1$ will be even and vice versa. Consequently, the product will always be even.
If $x$ is divisible by $5$ then the product is divisible by $10$ because we have an even product which is divisible by $5$.
If $x$ is not divisible by $5$ then $x=5m+k$ for some $m$ and $k$ where $k\in\{1,2,3,4\}$.
Now, $x^4-1=(5m)^4+4(5m)^3k+6(5m)^2k^2+4(5m)k^3+k^4-1$ and every term is divisible by $5$ with the possible exception of $k^4-1$.
But there are only four values of $k$ and if you raise each to the fourth power and subtract $1$ you get a multiple of $5$. Once again, this gives us an even number divisible by $5$ which is a multiple of $10$.
|
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|
Solving $ \left(\frac{ ( x^{3}+1 )^{3}+8 }{16}\right) ^{3}+1=2x$
Solving :
$$\left(\frac{ ( x^{3}+1 )^{3}+8 }{16}\right) ^{3}+1=2x$$
My Try :
$$\left(\frac{ ( x^{3}+1 )^{3}+8 }{16}\right) ^{3}=\left(\dfrac{(x+1)^3+2^3}{2^4}\right)^3$$
$$ x^3+a^3=(x+a)^3-3ax(x+a)$$
$$\left(\frac{ (x^3+3)^3-6(x^3+1)(x^3+3)}{16}\right) ^{3}=\left(\dfrac{(x^3+3)(1-6(x^3+1))}{2^4}\right)+1=2x$$
Now what ?
|
Let $\frac{x^3+1}{2}=y$ and $\sqrt[3]{2x-1}=z.$
Thus, from the given we obtain $\frac{y^3+1}{2}=z.$
Now, let $x>y$.
Thus, $$x>y=\frac{x^3+1}{2}>\frac{y^3+1}{2}=z,$$ which says that
$$x>y>z.$$
But from $y^3+1=2z$ and $z^3+1=2x$ we obtain
$$0=y^3-z^3+2(x-z)>0,$$ which is a contradiction.
By the same way we'll get a contradiction for $x<y$.
Id est $x=y$, $$x^3-2x+1=0$$ or
$$x^3-x^2+x^2-x-x+1=0$$ or
$$(x-1)(x^2+x-1)=0,$$
which gives the answer:
$$\left\{1,\frac{-1+\sqrt5}{2},\frac{-1-\sqrt5}{2}\right\}$$
|
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|
Prove that $\sin a + \sin b + \sin(a+b) = 4 \sin\frac12(a+b) \cos \frac12a \cos\frac12b$ Helping my son with his trigonometry review. We know that
$$\sin(a+b) = \sin a \cos b + \cos a \sin b$$
We also know that
$$\sin a \cos b = \frac12 \left(\sin(a-b) + \sin(a+b)\right)$$
And we have
$$\sin a + \sin b = 2 \sin\frac12(a+b)\cos\frac12(a-b)$$
From there, we seem to be missing how to get to the right-hand side of the equation.
We first expand
$$\sin a + \sin b = 2 \sin\frac12(a+b) \cos\frac12(a-b)$$
Then we add $\sin(a+b)$, which is $\sin a \cos b + \cos a \sin b$.
We now have:
$$2 \sin\frac12(a+b) \cos\frac12(a-b) + \frac12 \left(\sin(a-b) + \sin (a+b)\right) + \frac12 \left( \sin(b-a) + \sin (a+b)\right)$$
From there, we can't see how to obtain the right-hand side of the equation which is
$$4 \sin\frac12(a+b) \cos\frac12a \cos\frac12b$$
|
It may help setting $A=a/2$ and $B=b/2$, so the left-hand side becomes
$$
\sin2A+\sin2B+\sin2(A+B)
$$
The right-hand side screams “sum-to-product”! OK, let's apply the formula
$$
\sin2A+\sin2B=2\sin(A+B)\cos(A-B)
$$
so the left-hand side becomes
\begin{align}
\sin2A+\sin2B+\sin2(A+B)
&=2\sin(A+B)\cos(A-B)+2\sin(A+B)\cos(A+B)\\
&=2\sin(A+B)\bigl(\cos(A-B)+\cos(A+B)\bigr) \\
&=2\sin(A+B)(\cos A\cos B+\sin A\sin B+\cos A\cos B-\sin A\sin B)\\
&=4\sin(A+B)\cos A\cos B
\end{align}
|
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|
A Difficult Definite Integral $\int_0^{2\pi}(t-\sin t)(1-\cos t)^2{\rm d}t.$ Problem
Evaluate $$\int_0^{2\pi}(t-\sin t)(1-\cos t)^2{\rm d}t.$$
Comment
It's very complicated to compute the integral applying normal method. I obtain the result resorting to the skillful formula
$$\int_0^{2\pi}xf(\cos x){\rm d}x=\pi\int_0^{2\pi}f(\sin x){\rm d}x,$$
where $f(x) \in C[-1,1].$
\begin{align*}
\require{begingroup}
\begingroup
\newcommand{\dd}{\;{\rm d}}\int_0^{2\pi} (t-\sin t)(1-\cos t)^2 \dd t
&= \int_0^{2\pi} t(1-\cos t)^2 \dd t - \int_0^{2\pi} \sin t(1-\cos t)^2 \dd t \\
&= \pi\int_0^{2\pi} (1-\sin t)^2 \dd t - \int_0^{2\pi} (1-\cos t)^2 \dd (1-\cos t) \\
&= \pi\int_0^{2\pi} \left(\frac32-\frac12\cos2t-2\sin t\right) \dd t - \left[\frac13(1-\cos t)^3\right]_0^{2\pi}\\
&= \pi\left[\frac32t-\frac14\sin2t+2\cos t\right]_0^{2\pi}\\
&= 3\pi^2
\endgroup
\end{align*}
But any other solution?
|
\begin{align}
&\int_0^{2\pi}(t-\sin t)(1-\cos t)^2{\rm d}t=\int_0^{2\pi}t+t{\cos }^2 t-2t\cos t-\sin t -\sin t{\cos }^2 t+\sin 2t{\rm d}t\\
&=\left[\frac{1}{2}t^2+\frac{1}{2}t^2+\frac{1}{4}t\sin 2t-\frac{1}{4}t^2+\frac{1}{8}\cos 2t-2t\sin t-2\cos t+\cos t+\frac{1}{3}{\cos }^3t-\frac{1}{2}\cos 2t\right]_0^{2\pi} \\
&=\left[\frac{3}{4}t^2+\frac{1}{4}t\sin 2t-\frac{3}{8}\cos 2t-2t\sin t-\cos t+\frac{1}{3}{\cos }^3 t\right]_0^{2\pi}\\
&=3\pi ^2
\end{align}
|
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|
If $a$ has order 3 $\pmod p$, $p$ prime, then $1+a+a^2 \equiv 0 \pmod p$ and $1+a$ has order $6$ There is a little proposition left to the readers in my number theory book, and I'm not sure how to answer this in a proper way:
If $a$ has order $3 \pmod p$, $p$ prime, then $1+a+a^2 \equiv 0 \pmod p$ and $1+a$ has order $6$.
Some hints?
|
How $a$ has order $3$, then $a\not\equiv 1$.
Now, note that $(a-1)(1+a+a^2) = a^3-1 \equiv 0$ (mod $p$). As $\mathbb{Z}_p$ is a field and $a-1\neq 0$, so $1+a+a^2=0$ (mod $p$).
To see why $1+a$ has order $6$, note that
$$(1+a)^6=(1+2a+a^2)^3=a^3=1$$
So, the order of $1+a$ divide $6$.
If $1+a$ has order $1$, then $1+a=1\Rightarrow a=0$. This contradicts the fact that $a$ has order $3$.
If $1+a$ has order $2$, then $(1+a)^2=1\Rightarrow 1+2a+a^2=1\Rightarrow a=1$. But $1$ has order $1$, not $3$.
If $1+a$ has order $3$, then $(1+a)^3=1\Rightarrow (1+a)(1+2a+a^2)=1\Rightarrow (1+a)a=1\Rightarrow a+a^2=1\Rightarrow 1+a+a^2=2\Rightarrow 0=2$, so $p=2$. But $\mathbb{Z}_2$ has no elements with order $3$.
So, the only possible order for $1+a$ is $6$.
|
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|
If $-1\leq x, y \leq 1$ and $x\sqrt{1-y^2} + y\sqrt{1-x^2} = 1$, find $x^2+y^2$ Let $x,\, y\in\mathbb
R,\ -1\leq x,\, y\leq 1$ such that $x\sqrt{1-y^2} + y\sqrt{1-x^2} = 1.$
Find the sum $S = x^2+y^2.$
|
I think Michael's solution is much easier, but it's not something most people would think of. Instead, let's try doing this algebraically. First, I am going to isolate one of the terms on the left, so subtract by $y\sqrt{1-x^2}$:
$$x\sqrt{1-y^2}=1-y\sqrt{1-x^2}$$
Now, when we square both sides, the square root on the left goes away:
$$x^2(1-y^2)=1+y^2(1-x^2)-2y\sqrt{1-x^2}$$
Simplify some terms:
$$x^2-x^2y^2=1+y^2-x^2y^2-2y\sqrt{1-x^2}$$
Add both sides by $x^2y^2$:
$$x^2=1+y^2-2y\sqrt{1-x^2}$$
Subtract both sides by $1+y^2$ to isolate the square root on the right:
$$x^2-y^2-1=-2y\sqrt{1-x^2}$$
Square both sides to get rid of the square root:
$$x^4+y^4+1-2x^2y^2-2x^2+2y^2=4y^2(1-x^2)$$
Simplify some terms:
$$x^4+y^4+1-2x^2y^2-2x^2+2y^2=4y^2-4x^2y^2$$
Add both sides by $4x^2y^2-4y^2$:
$$x^4+y^4+1+2x^2y^2-2x^2-2y^2=0$$
Now, this is kind of hard to notice, but the $x^4+y^4+2x^2y^2$ terms kind of go together because that's $(x^2+y^2)^2$ and the $-2x^2-2y^2$ go together because that's $-2(x^2+y^2)$. If we make these substitutions, we get:
$$(x^2+y^2)^2-2(x^2+y^2)+1=0$$
Substitute $S=x^2+y^2$:
$$S^2-2S+1=0$$
Factor and solve:
$$(S-1)^2=0\rightarrow S=1$$
|
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|
Find the interval of convergence of the series $ \sum^{\infty}\limits_{k=0} ((-1)^k+3)^k(x-1)^k $ I wish to find the interval of convergence of the following series
\begin{align} \sum^{\infty}_{k=0} ((-1)^k+3)^k(x-1)^k \end{align}
PROOF
Wittingly,
\begin{align} \left[(-1)^k+3\right]^k= \begin{cases} 0,&\text{if}\;j=0;\\4^j,&\text{if}\;j=2k;\\2^j,&\text{if}\;j=2k+1.\end{cases} \end{align}
Thus,
\begin{align} \limsup_{k\to\infty} \sqrt[k]{\left|((-1)^k+3)^k(x-1)^k\right|}&=|x-1|\limsup_{k\to\infty}\sqrt[k]{ 2^{2k+1}}\\&=|x-1|\limsup_{k\to\infty} 2^{(2k+1)\times \frac{1}{k}} \\&=4|x-1| \end{align}
The series converges absolutely when $\limsup_{k\to\infty} \sqrt[k]{\left|((-1)^k+3)^k(x-1)^k\right|}<1,$ i.e.,
\begin{align} |x-1|< \dfrac{1}{4}\iff \dfrac{3}{4}<x<\dfrac{5}{4} \end{align}
QUESTION:
Why must \begin{align} \limsup_{k\to\infty} \sqrt[k]{\left|((-1)^k+3)^k(x-1)^k\right|}= \limsup_{k\to\infty} \sqrt[k]{\left| 2^{2k+1}(x-1)^k\right|}\end{align}
as stated in the book before me and not
\begin{align} \limsup_{k\to\infty} \sqrt[k]{\left|((-1)^k+3)^k(x-1)^k\right|}= \limsup_{k\to\infty} \sqrt[k]{\left| 4^{2k}(x-1)^k\right|}\;?\end{align}
|
Hint
$$
\sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k =\lim_{n\to \infty}\left(\sum_{k=0}^n 4^{2k}(x-1)^{2k}+2(x-1)\sum_{k=0}^n 2^{2k}(x-1)^{2k}\right)
$$
and now
$$
\sum_{k=0}^n 4^{2k}(x-1)^{2k} = \frac{1-4^{2(n+1)}(x-1)^{2(n+1)} }{1-4^{2}(x-1)^{2}}
$$
which converges if $|4(x-1)| < 1$ etc. so the result is
$$
\sum^{n}_{k=0} ((-1)^k+3)^k(x-1)^k=\frac{2(x-1)}{1-2^2 (x-1)^2}+\frac{1}{1-4^2 (x-1)^2}
$$
for $\frac 34\lt x\lt \frac 54$
Attached the plot showing in red the sum approximation for $n = 10$ and in black the asymptotic result found
|
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|
$(x*(\log_{2}(x))^2)/2 = x^{3/2}$ how to solve it? Is there a manual solution for this equation? Or I should use Wolfram:
result from Wolfram.
|
There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:
The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $\mathbb{C}$ and there are two possible functions $W_0$ with $\textrm{Re}(W_0(x))\geq -1$ and $W_{-1}$ with $\textrm{Re}(W_{-1}) < -1$.
Then:
\begin{eqnarray}
\frac{1}{2}x\ \left(\log_2(x)\right)^2 &=& x^{3/2} \\
\frac{1}{2}x^{-1/2}\ \left(\log_2(x)\right)^2 &=& 1 \\
\frac{1}{\sqrt{2}}x^{-1/4}\log_2(x) &=& \pm 1 \\
x^{-1/4}\frac{\ln(x)}{\ln(2)} &=& \pm \sqrt{2} \\
x^{-1/4}\ln(x) &=& \pm \sqrt{2}\ln(2) \\
e^{(-1/4)\ln(x)}\ln(x) &=& \pm \sqrt{2}\ln(2) \\
e^{(-1/4)\ln(x)}\left(-\frac{1}{4}\ln(x)\right) &=& \pm \frac{1}{4}\sqrt{2}\ln(2) \\
\end{eqnarray}
Then with Lambert function:
\begin{eqnarray}
-\frac{1}{4}\ln(x) &=& W\left(\pm\frac{1}{4}\sqrt{2}\ln(2)\right) \\
x &=& \exp \left\lbrace-4W\left(\pm\frac{1}{4}\sqrt{2}\ln(2)\right)\right\rbrace \\
\end{eqnarray}
Then the solutions are:
\begin{eqnarray}
x_1 &=& \exp \left\lbrace-4W_0\left(\frac{1}{4}\sqrt{2}\ln(2)\right)\right\rbrace = 0.44836908960\ldots \\
x_2 &=& \exp \left\lbrace-4W_0\left(-\frac{1}{4}\sqrt{2}\ln(2)\right)\right\rbrace=4 \\
x_3 &=& \exp \left\lbrace-4W_{-1}\left(\frac{1}{4}\sqrt{2}\ln(2)\right)\right\rbrace=-153792.65205358\ldots-i101297.96245405\ldots \\
x_4 &=& \exp \left\lbrace-4W_{-1}\left(-\frac{1}{4}\sqrt{2}\ln(2) \right)\right\rbrace=6380.45994697086\ldots \\
\end{eqnarray}
The function is numeric too, but it is very elegant.
There is an article in wikipedia for this Lambert W function.
|
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|
Why is $\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ $15\sqrt{5}$ and not $15\sqrt[4]{25}$? I have an expression I am to simplify:
$$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$$
I arrived at $15\sqrt[4]{25}$. My textbook tells me that the answer is in fact $15\sqrt{5}$. Here is my thought process:
$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ = $\frac{15*\sqrt[4]{5}*\sqrt[4]{25}}{\sqrt[4]{5}}$ =
(cancel out $\sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15\sqrt[4]{25}$$
Where did I go wrong and how can I arrive at $15\sqrt{5}$?
|
Write $$\sqrt[4]{\frac{125}{5}}=\sqrt[4]{25}=\sqrt{5}$$
|
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|
Evaluating an improper integral $\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$ I tried to solve the integral: $$\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$$
using $ x = \sqrt{\tan(u)}$ and $dx = \frac{ \sec^2(u)}{2\sqrt{\tan(u)}} du,$
but I ended up with an even worse looking integral $$ \int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\tan(u)}}{\sec^2(u)}du.$$
Wolfram gave an answer of $ \dfrac{\pi}{8\sqrt{2}},$ but how would one get to that answer?
|
Write $$\frac{x^2}{(1+x^4)^2} = \frac{4x^3}{(1+x^4)^2} \cdot \frac{1}{4x}.$$ Then integration by parts with the choice $$u = \frac{1}{4x}, \quad du = -\frac{1}{4x^2} \, dx, \\ dv = \frac{4x^3}{(1+x^4)^2} \, dx, \quad v = -\frac{1}{1+x^4},$$ yields $$I_1(x) = \int \frac{x^2}{(1+x^4)^2} \, dx = -\frac{1}{4x(1+x^4)} - \int \frac{1}{4x^2(1+x^4)} \, dx.$$ Now write $$\frac{1}{x^2(1+x^4)} = \frac{1}{x^2} - \frac{x^2}{1+x^4},$$ thus $$I_1(x) = -\frac{1}{4x(1+x^4)} + \frac{1}{4x} + \frac{1}{4} \int \frac{x^2}{1+x^4} \, dx = \frac{x^3}{4(1+x^4)} + \frac{1}{4} I_2(x),$$ where we now seek to evaluate $I_2(x)$. This is accomplished in a number of ways; one is to perform the factorization $$1+x^4 = (1 + \sqrt{2} x + x^2)(1 - \sqrt{2} x + x^2),$$ and compute the partial fraction decomposition $$\frac{x^2}{1+x^4} = \frac{1}{2\sqrt{2}} \left( \frac{x}{1 - \sqrt{2} x + x^2} - \frac{x}{1 + \sqrt{2} x + x^2} \right).$$ Then each term is integrable in closed form. This is not the easiest or most elegant approach, but it is perhaps the most elementary, requiring no knowledge beyond AP Calculus BC.
|
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|
prove $a,b,c$ in A.P if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$
In $\Delta ABC$, if $\tan\dfrac{A}{2}=\dfrac{5}{6}$ and $\tan\dfrac{C}{2}=\dfrac{2}{5}$, then prove that the sides $a,b,c$ are in A.P.
My Attempt
$$
\sin A=\frac{2.5}{6}.\frac{36}{61}=\frac{60}{61}\\
\sin C=\frac{2.2}{5}.\frac{25}{29}=\frac{20}{29}\\
$$
it is solved in my reference some fomula involving $2s=a+b+c$, can I prove it using the basic known properties of triangles ?
|
Hint:
Like In $\Delta ABC$, find $\cot\dfrac{B}{2}.\cot\dfrac{C}{2}$ if $b+c=3a$,
$$2b=a+c$$ will hold true if $\tan\dfrac A2\tan\dfrac B2=\dfrac13$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Super hard system of equations
Solve the system of equation for real numbers
\begin{split}
(a+b) &(c+d) &= 1 & \qquad (1)\\
(a^2+b^2)&(c^2+d^2) &= 9 & \qquad (2)\\
(a^3+b^3)&(c^3+d^3) &= 7 & \qquad (3)\\
(a^4+b^4)&(c^4+d^4) &=25 & \qquad (4)\\
\end{split}
First I used the identity
$$(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(bc+ad)^2$$
Use this identity to (4) too
and simplify (3),
we obtain $$(a^2+b^2-ab)(c^2+d^2-cd)=7$$
And suppose $x=abcd$ use $ac=x /bd , bc=x/ad$
But got stuck...
|
This is not an answer but it is too long for a comment.
Looking at this system of equations, I had a very strange feeling (which I cannot explain).
Using a CAS, I solved equations $(1)$, $(2)$, $(3)$ for $a,b,c$ as functions of $d$ and obtained $8$ solutions which are listed below
$$\left\{a= \frac{2}{d},b= \frac{\sqrt{2}}{d},c=
-\frac{d}{\sqrt{2}}\right\},\left\{a= \frac{\sqrt{2}}{d},b= \frac{2}{d},c=
-\frac{d}{\sqrt{2}}\right\},\left\{a= \frac{2}{d},b=
-\frac{\sqrt{2}}{d},c= \frac{d}{\sqrt{2}}\right\},\left\{a=
-\frac{\sqrt{2}}{d},b= \frac{2}{d},c= \frac{d}{\sqrt{2}}\right\},\left\{a=
-\frac{1}{d},b= -\frac{\sqrt{2}}{d},c= -\sqrt{2} d\right\},\left\{a=
-\frac{\sqrt{2}}{d},b= -\frac{1}{d},c= -\sqrt{2} d\right\},\left\{a=
-\frac{1}{d},b= \frac{\sqrt{2}}{d},c= \sqrt{2} d\right\},\left\{a=
\frac{\sqrt{2}}{d},b= -\frac{1}{d},c= \sqrt{2} d\right\}$$
The problem is that, replacing in $(4)$ any of these solutions the resulting equation is $25=25$ !
|
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|
Global extremas of $f(x,y)=2x^2+2y^2+2xy+4x-y$ on the region $x\leq 0, y\geq 0, y\leq x+3$ I have the following two variable function $f(x,y)=2x^2+2y^2+2xy+4x-y$ and I need to find its global extremas on the region $x\leq 0, y\geq 0, y\leq x+3$.
As in the other questions I asked in the last days, I have no solution to this old exam question so I would like your feedback on wether my results are correct or not.
In the inside of our domain, we have :
$$f_x=4x+2y+4=0$$
$$f_y=4y+2x-1=0$$
So $y=1$ and $x=\frac{-3}{2}$ . This point lies in our domain so it is a candidate.
On the border, we have $$f(x, 0)=2x^2+4x, f'=4x+4 = 0 $$ so $x=-1$
We also have $$f(0,y)=2y^2-y, f'=4y-1=0$$ so $y=\frac{1}{4}$
And $$f(x, x+3)=2x^2+2(x+3)^2+2x(x+3)+4x-(x+3), f'=3(4x+7)=0$$ (WolframAlpha gets the same derivative for this $f(x,x+3)$). So we get $x=-\frac{7}{4}, y=x+3=\frac{5}{4}$
Now, we also have our endpoints $(-3,0), (0,3), (0,0)$
Now we plug each of our candidate into our function and look for the greatest/ smallest value which will be our global maximum/ minimum.
$(-\frac{3}{2},1) : -\frac{7}{2}$
$(-1,0):-2$
$(0,\frac{1}{4}) : - \frac{1}{8}$
$(- \frac{7}{4}, \frac{5}{4}) : - \frac{27}{8}$
$(-3,0) : 6$
$(0,0) : 0$
$(0,3) : 15$
So there seems to be a global maximum of $15$ at $(0,3)$ and a global minimum of $- \frac{7}{2}$ at $(- \frac{3}{2},1)$.
Are my results correct or did I miss/ forgot something ?
Also, this problem took me quite some time. Not enormously, but still. Is there a shortcut or a better way to proceed to see that the function actually reaches its global maximum at one of the endpoints and its minimum at $(- \frac{3}{2}, 1)$ ?
Thanks for your feedback !
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A solution without calculus.
Adopt new variables $x = u - v$, $y = u + v$. The objective function:
\begin{align*}
2x^2 + 2y^2 + 2xy + 4x - y &= 2(u^2 - 2uv + v^2) + 2(u^2 + 2uv + v^2) + 2(u^2 - v^2) + 4(u-v) - (u+v) \\
&= 6u^2 + 2v^2 + 3u - 5v \\
&= 6 \left( u + \frac{1}{4} \right)^2 + 2 \left(v - \frac{5}{4}\right)^2 - \frac{7}{2}.
\end{align*}
The constraints:
\begin{align*}
x \leq 0 &\Longrightarrow u \leq v \\
y \geq 0 &\Longrightarrow u \geq -v \\
y \leq x + 3 &\Longrightarrow v \leq \frac{3}{2}.
\end{align*}
We may combine the constraints as an easier-to-visualize $|u| \leq v \leq \frac{3}{2}$.
The objective function clearly has a global minimum over all of $\mathbb{R}^2$ at $(u, v) = (-\frac{1}{4}, \frac{5}{4})$, i.e. $(x, y) = (-\frac{3}{2}, 1)$, which is within the constraints. Furthermore, as the objective function is convex, its maximum on a convex polygon must be at a vertex, and it is easy to check all three possible vertices. The maximum is $(u, v) = (\frac{3}{2}, \frac{3}{2})$ or $(x, y) = (0, 3)$.
|
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|
Maximizing $f$ in $\mathbb{R}^3$
Find the domain and the maximum value that the function
$$f(x,y,z)=\frac{x+2y+3z}{\sqrt{x^2+y^2+z^2}}$$
may attain in its domain.
I have found the domain of the function to be $\mathbb{R^3\backslash\mathbf{0}}$. To maximize I differentiated in terms of $x,y,z$ having
$$f_x=\frac{-2 x y-3 x z+y^2+z^2}{\left(x^2+y^2+z^2\right)^{3/2}},\quad f_y=\frac{2 x^2-x y+z (2 z-3 y)}{\left(x^2+y^2+z^2\right)^{3/2}},\quad f_z=\frac{3 \left(x^2+y^2\right)-z (x+2 y)}{\left(x^2+y^2+z^2\right)^{3/2}}$$
But to solve the system $f_x=0,f_y=0$ and $f_z=0$ is rather hard. What are the plausible values of $x,y,z$?
|
Consider the line
$x = t, y = 2t, z = 3t$
Along this line.
$f(t,2t,3t) = \frac {14 t}{\sqrt{14 t^2}} = \sqrt {14}$
let's find some orthogonal vectors.
$\mathbf u = (1,2,3)\\
\mathbf v = (2,-1,0)\\
\mathbf w = (0,3,-2)$
Any point in $\mathbb R^3$ is some linear combination
$c_1\mathbf u + c_2\mathbf v+ c_3\mathbf w$
$f(c_1\mathbf u + c_2\mathbf v+ c_3\mathbf w) = \frac {14c_1}{\sqrt{14c_1^2 + 5c_2^2 + 13c_3^2}}\\
|f|\le \sqrt 14 \text{ sgn}(c_1)$
|
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|
Prove the identity for $\tan3\theta$
Prove the identity for $$\tan3\theta= \frac{3\tan\theta - \tan^3 \theta}{1-3\tan^2 \theta}$$
Using de Moivre's theorem I have found that:
$$\cos3\theta = 4\cos^3\theta - 3\cos \theta$$
$$\sin 3\theta = 3\sin \theta-4\sin^3\theta$$
therefore:
$$\tan 3\theta = \frac{\sin 3\theta}{\cos 3 \theta}=\frac{3\sin \theta-4\sin^3\theta}{4\cos^3\theta - 3\cos \theta}$$
To then try and get the whole expression in terms of $\tan\theta$ I multiplied top and bottom of the fraction by $(4\cos^3\theta)^{-1}$. This gave me the following but I'm not now sure how to finish it off
$$\tan3\theta =\frac{\frac{3\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos \theta}{4\cos^3 \theta}}$$
|
We have
$$
\tan3\theta =\frac{\frac{3\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos \theta}{4\cos^3 \theta}} =
\frac{\frac{3(\cos^2 \theta + \sin^2 \theta)\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3(\cos^2 \theta + \sin^2 \theta)\cos \theta}{4\cos^3 \theta}} = \\
\frac{\frac{3\cos^2 \theta\sin \theta + 3\sin^3 \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos^3 \theta + 3\sin^2 \theta\cos \theta}{4\cos^3 \theta}} = \\
\frac{\frac{3\cos^2 \theta\sin \theta}{4\cos^3\theta} + \frac{3\sin^3 \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos^3 \theta}{4\cos^3 \theta} -\frac{3\sin^2 \theta\cos \theta}{4\cos^3 \theta}}
$$
and from there, the simplification is straightforward.
|
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|
Prove all elements $x_n \geq \sqrt{2}.$ Given $x_1 = 2,$ and $$x_{n+1} = \frac{1}{2}\bigg(x_n + \frac{2}{x_n} \bigg),$$ show that for all $n \in \mathbb{N},$ $x_n \geq \sqrt{2}.$
I tried the following. Suppose, for contradiction, that $x_{n+1} < \sqrt{2}.$ Then, $$\frac{1}{2}\bigg(x_n + \frac{2}{x_n} \bigg) < \sqrt{2},$$ and $$x_n + \frac{2}{x_n} < 2\sqrt{2}.$$ Given that $x_n \geq \sqrt{2},$ it follows that $$\sqrt{2} + \frac{2}{x_n} \leq x_n + \frac{2}{x_n} < 2\sqrt{2},$$ and $$\sqrt{2} + \frac{2}{x_n} < 2\sqrt{2}.$$ So, $$\frac{2}{x_n} < \sqrt{2},$$ and $$\sqrt{2} < x_n.$$ No contradiction. I've tried proving it directly, by squaring both sides and by leaving the expression as is, and I consistently find that the $x_n$ and the $x_n^{-1}$ are together problematic.
I also tried something that allowed me to arrive at the following: $$x_{n+1}x_n \geq 2.$$ But, I don't think that's helpful.
Help?:)
|
$x_1\geq\sqrt{2}$, $x_2\geq\sqrt{2}$. Now suppose $x_{n-1}\geq\sqrt{2}$, then by AM-GM we have $x_n=\frac{1}{2}(x_{n-1}+\frac{2}{x_{n-1}})\geq\frac{1}{2}(2\sqrt{x_{n-1}\frac{2}{x_{n-1}}})=\sqrt{2}$
|
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|
Prove that triangle $XYZ$ is equilateral Let $ABC$ be an acute angled triangle whose inscribed circle touches $AB$ and $AC$ at $D$ and $E$ respectively. Let $X$ and $Y$ be the points of intersection of the bisectors of the angles $ACB$ and $ABC$ with the line $DE$ and let $Z$ be the midpoint of the side $BC$. Prove that the triangle $XYZ$ is equilateral if and only if $\angle A = 60^o$.
I dont know why, but it seems to me that $\Delta ADE$ and $\Delta XYZ$ are similar (or maybe congruent :\ ). Is it true? Or no? Please help.
|
Let
$$A=(0,\ a)\\
B=(-b,\ 0)\\
C=(b,\ 0)$$
and thus
$$Z=(0,\ 0)$$
then we get
$$\tan(\angle ABC)=\frac ab$$
and thus
$$\tan(\frac 12\ \angle ABC)=\frac{a/b}{1+\sqrt{1+a^2/b^2}}=\frac a{b+\sqrt{a^2+b^2}}$$
So we can deduce the center $M$ of the incircle to be
$$M=(0,\ \frac {ab}{b+\sqrt{a^2+b^2}})$$
Now define lines $g$ and $h$ by
$$g=\overline{AC}:\ y=-\frac ab\ x+a\\
h=\overline{ME}:\ y=\frac ba\ x+\frac {ab}{b+\sqrt{a^2+b^2}}$$
Equating those will then provide
$$\frac {a^2+b^2}{ab}\cdot x=\frac {ab+a\sqrt{a^2+b^2}-ab}{b+\sqrt{a^2+b^2}}$$
or
$$x=\frac{a^2b}{(b+\sqrt{a^2+b^2})\ \sqrt{a^2+b^2}}$$
Inserting $x$ into $h$ further provides
$$y=\frac{ab^2}{(b+\sqrt{a^2+b^2})\ \sqrt{a^2+b^2}}+\frac{ab}{b+\sqrt{a^2+b^2}}\cdot\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}=\frac{ab}{\sqrt{a^2+b^2}}$$
Thus we have
$$E=(\frac{a^2b}{(b+\sqrt{a^2+b^2})\ \sqrt{a^2+b^2}},\ \frac{ab}{\sqrt{a^2+b^2}})$$
Now define line $k$ to be
$$k=\overline{BM}:\ y=\frac a{b+\sqrt{a^2+b^2}}\ (x+b)$$
and intersecting that with $\overline{DE}$, i.e. equating it with the $y$ value of $E$, provides
$$\frac a{b+\sqrt{a^2+b^2}}\ (x+b)=\frac{ab}{\sqrt{a^2+b^2}}\\
x+b=\frac{b(b+\sqrt{a^2+b^2})}{\sqrt{a^2+b^2}}=\frac{b^2}{\sqrt{a^2+b^2}}+b\\
x=\frac{b^2}{\sqrt{a^2+b^2}}$$
Thus we have calculated $Y$ to be
$$Y=(\frac{b^2}{\sqrt{a^2+b^2}},\ \frac{ab}{\sqrt{a^2+b^2}})=\frac b{\sqrt{a^2+b^2}}\ (b,\ a)$$
and this finally proves your conjecture:
$$\overline{AB}\parallel\overline{ZY}$$
q.e. $ABC$ and $XYZ$ are indeed similar triangles, provided $ABC$ was an isoceles triangle, as asumed by the chosen coordinatisation.
--- rk
|
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|
Explicit calculation of $\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x$ Is it possible to confirm the value of this integral using the methods of complex analysis or similar?
$$
\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x=\frac{\pi^2-9}{12}
$$
Of course, one can reduce it to the definition of the polylogarithm and a $\zeta$-function, but I was looking for an explicit calculation.
|
This approach is unnecessarily complicated, but quite fun:
As explained in the comments and the other answers, we may write
$$ I \equiv \int \limits_0^\infty \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)} \, \mathrm{d}x = \int \limits_0^\infty \frac{2 x}{(1+x^2)(\mathrm{e}^{2\pi x}-1)} \, \mathrm{d} x \, . $$
We can then use the geometric series, integration by parts and this Laplace transform to obtain
\begin{align}
I &= \sum \limits_{n=1}^\infty \int \limits_0^\infty \frac{2 x}{(1+x^2)^2} \mathrm{e}^{-2 \pi n x} \, \mathrm{d} x = \sum \limits_{n=1}^\infty \left[1 - 2 \pi n \int \limits_0^\infty \frac{\mathrm{e}^{-2 \pi n x}}{1+x^2} \, \mathrm{d} x \right] \\
&= \sum \limits_{n=1}^\infty \left[1 - 2 \pi n \left(\frac{\pi}{2} - \operatorname{Si}(2 \pi n)\right)\right] \, ,
\end{align}
where $\operatorname{Si}$ is the sine integral . Plugging in its definition and integrating by parts twice more we arrive at
$$ I = \sum \limits_{n=1}^\infty \left[1 - 2 \pi n \int \limits_1^\infty \frac{\sin(2 \pi n t)}{t} \, \mathrm{d} t \right] = 2 \sum \limits_{n=1}^\infty \frac{1}{2 \pi n} \int \limits_1^\infty \frac{\sin(2 \pi n t)}{t^3} \, \mathrm{d} t \, .$$
Now the Fourier series
$$ 1 - 2 \{t\} = 4 \sum \limits_{n=1}^\infty \frac{\sin(2 \pi n t)}{2 \pi n} \, , \, t \in \mathbb{R} \, , $$
and the fractional part integral
$$ \int \limits_0^1 x \left\{\frac{1}{x}\right\} \, \mathrm{d} t = \int \limits_1^\infty \frac{\{t\}}{t^3} \, \mathrm{d} t = 1 - \frac{\zeta(2)}{2} = 1 - \frac{\pi^2}{12} $$
are sufficient to derive the final result
$$ I = \frac{1}{2} \int \limits_1^\infty \frac{1 - 2\{t\}}{t^3} \, \mathrm{d} t = \frac{\pi^2}{12} - \frac{3}{4} = \frac{\pi^2 - 9}{12} $$
and thus prove $\pi > 3$ in a rather convoluted manner.
|
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$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}$ evaluation using expansion series $$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}$$
This is the limit which I got from the book of Joseph Edwards' Differential Calculus for Beginners the second exercise of the First Chapter Page -10 question number 19. Whose Answer was given as $\frac 1 2$.
So my try was as follows:
$$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}$$
$$\Rightarrow \lim \limits _{x \rightarrow 0} [ \frac{x^3(1 + \frac{x^4}{4} + \frac{x^8}{32}...) - (x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!}...)^{\frac 3 2} }{x^7} ]$$
Now, this is the quite a big problem, as we can see that sine expansion series has the power $\frac 3 2$ which is nearly impossible to evaluate.
The expansion series given in the book are: $(1+x)^n$ and $(1-x)^{-n}$ which are not useful here as there are infinite series where those are 2-terms.
Please anyone help me , I've just started Calculus :P
|
We have $$\lim \limits _{x \rightarrow 0} \frac{x^3 e^{\frac {x^4}{4}} - {\sin^{\frac 3 2} {x^2}} }{x^7}{=\lim \limits _{x \rightarrow 0} \frac{e^{\frac {x^4}{4}} - \left({\sin x^2\over x^2}\right)^{3\over 2} }{x^4}\\=\lim \limits _{x \rightarrow 0} \frac{e^{\frac {x^4}{4}} -1+1- \left({\sin x^2\over x^2}\right)^{3\over 2} }{x^4}\\=\lim \limits _{x \rightarrow 0} \frac{e^{\frac {x^4}{4}} -1 }{x^4}+\lim \limits _{x \rightarrow 0} \frac{1 - \left({\sin x^2\over x^2}\right)^{3\over 2} }{x^4}}$$the latter equality is true since $\lim \limits _{x \rightarrow 0} \frac{e^{\frac {x^4}{4}} -1 }{x^4}={1\over 4}$ is bounded. It remains to show that $$\lim \limits _{x \rightarrow 0} \frac{1 - \left({\sin x^2\over x^2}\right)^{3\over 2} }{x^4}={1\over 4}$$This is straight forward since $$\lim \limits _{x \rightarrow 0} \frac{1 - \left({\sin x^2\over x^2}\right)^{3\over 2} }{x^4}{=\lim \limits _{u \rightarrow 0} \frac{1 - \left({\sin u\over u}\right)^{3\over 2} }{u^2}\\=\lim \limits _{u \rightarrow 0} \frac{1 - \left({\sin u\over u}\right)^{3\over 2} }{u^2}{1 + \left({\sin u\over u}\right)^{3\over 2} \over 1 + \left({\sin u\over u}\right)^{3\over 2} }\\={1\over 2}\lim \limits _{u \rightarrow 0} \frac{1 - \left({\sin u\over u}\right)^3 }{u^2}\\={1\over 2}\lim \limits _{u \rightarrow 0} \frac{u^3-\sin^3 u }{u^5}\\={1\over 2}\lim \limits _{u \rightarrow 0} \frac{u^3-\left(u-{u^3\over 6}\right)^3 }{u^5}\\={1\over 2}\lim \limits _{u \rightarrow 0} \frac{1-\left(1-{u^2\over 6}\right)^3 }{u^2}\\={1\over 2}\lim \limits _{u \rightarrow 0} \frac{{u^2\over 6}\left(1+\left(1-{u^2\over 6}\right)+\left(1-{u^2\over 6}\right)^2\right) }{u^2}\\={1\over 2}\lim_{u\to 0}{{3u^2\over 6}\over u^2}\\={1\over 4}}$$hence the result.
|
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Evaluate $\lim\limits_{x \to 1}\dfrac{x-x^x}{1-x+\ln x}$. Problem
Evaluate $$\lim\limits_{x \to 1}\frac{x-x^x}{1-x+\ln x}$$.
Solution
Consider using Taylor's formula. Expand $x^x $ and $\ln x$ at $x=1$. We obtain
$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$\ln x=(x-1)-\frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
\begin{align*}
\lim_{x \to 1}
\frac{x-x^x}{1-x+\ln x}&=\lim_{x \to 1}\frac{x-[1+(x-1)+(x-1)^2+o((x-1)^2)]}{1-x+[(x-1)-\frac{1}{2}(x-1)^2+o((x-1)^2)]}\\
&=\lim_{x \to 1}\frac{-(x-1)^2-o((x-1)^2)}{-\frac{1}{2}(x-1)^2+o((x-1)^2)}\\
&=2.
\end{align*}
Please correct me if I'm wrong! Can we solve it by L'Hospital's rule?
|
Another solution by L'Hospital's rule
Let $x=:1+h$. Then
\begin{align*}
\lim_{x \to 1}
\frac{x-x^x}{1-x+\ln x}&=\lim_{h \to 0}\frac{(1+h)[1-(1+h)^{h}]}{\ln(1+h)-h}\\
&=\lim_{h \to 0}\frac{e^{h\ln(1+h)}-1}{h-\ln(1+h)}\\
&=\lim_{h \to 0}\frac{h\ln(1+h)}{h-\ln(1+h)}\\
&=\lim_{h \to 0}\frac{h^2}{h-\ln(1+h)}\\
&=\lim_{h \to 0}\frac{2h}{1-\frac{1}{1+h}}\\
&=2\lim_{h \to 0}(1+h)\\
&=2.
\end{align*}
|
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Prove that $\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$ is an integer. Prove that $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$ is an integer using mathematical induction.
I tried using mathematical induction but using binomial formula also it becomes little bit complicated.
Please show me your proof.
Sorry if this question was already asked. Actually i did not found it. In that case only sharing the link will be enough.
|
Base case for $k=1$: $$\frac{1^7}{7}+\frac{1^5}{5}+\frac{2*1^3}{3}-\frac{1}{105}=1$$
Now, assume for some k that $\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$ is indeed in integer.
Then $$\frac{(k+1)^7}{7}+\frac{(k+1)^5}{5}+\frac{2(k+1)^3}{3}-\frac{k+1}{105}=\\\frac{\sum_{i=0}^7\binom{7}{i}k^i}{7}+\frac{\sum_{i=0}^5\binom{5}{i}k^i}{5}+2\frac{\sum_{i=0}^3\binom{3}{i}k^i}{3}-\frac{k+1}{105}=$$
Extracting the highest indexed term from each sum (and the $-\frac{k}{105}$ at the end): $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}+\frac{\sum_{i=0}^6\binom{7}{i}k^i}{7}+\frac{\sum_{i=0}^4\binom{5}{i}k^i}{5}+2\frac{\sum_{i=0}^2\binom{3}{i}k^i}{3}-\frac{1}{105}$$
By the induction hypothesis, the sum of the first four terms is an integer so, if we can show the rest of the above sum is an integer, we will be done. Use the fact that, for any prime, $p$, $p|\binom{p}{k}$ where $1\leq k\leq p-1$. This is because $$\binom{p}{k}=\frac{p(p-1)...(p-k+1)}{k(k-1)...1}$$
$p$ divides the numerator but not the denominator (as $1\leq k\leq p-1$) so $p|\binom{p}{k}$
So each term in the remaining sum with index $i$ $\geq1$ and $\leq p-1$ ($p$ being the respective prime in each sum) is divisible by the corresponding $p$ in the denominator and produces an integer. The only non-integer terms left will be the ones at $i=0$, i.e. $$\frac{\binom{7}{0}k^0}{7}+\frac{\binom{5}{0}k^0}{5}+\frac{2\binom{3}{0}k^0}{3}-\frac{1}{105}=\frac{1}{7}+\frac{1}{5}+\frac{2}{3}-\frac{1}{105}=1$$
So $\frac{(k+1)^7}{7}+\frac{(k+1)^5}{5}+\frac{2(k+1)^3}{3}-\frac{k+1}{105}$ is a sum of integers making it an integer.
|
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|
How do I solve for $a$ and $b$ in $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$? Given $\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$
I need to solve for $a$ and $b$, so here we go,
$\lim\limits_{x \to ∞} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right)$
$= \lim\limits_{x \to ∞} x \left(2 +(3+x) \left(\ln (1+\frac{a}{x}) - \ln(1+\frac{b}{x}) \right) \right)$
$=\lim\limits_{x \to ∞} x \left(2 +(3+x)\left( \dfrac{a-b}{x} \right) \right)$
$=\lim\limits_{x \to ∞} \left(2x +(3+x)\left( a-b \right) \right)$
$=\lim\limits_{x \to ∞} \left(2x + 3(a-b) + x(a-b) \right) $
Since the limit exists, $2x$ and $x(a-b)$ must cancel out so $a-b = 2$
But if I do this, I'm left with $3(a-b) = 2$ which gives me a contradiction since $a-b$ is supposed to be $2$.
What do I do now? And where exactly have I gone wrong?
Thank you!
|
Your mistake lies in the stp where you replace the difference between logs with the ratio $(a-b) /x$. You can not replace an expression $A$ by another expression $B$ unless $A=B$. While evaluating limit you may find replacements of $A$ by $B$ even when $A\neq B$ but such replacements always represent some hidden steps which the the author assumes to be obvious (an example of such a replacement is in my solution where the factor $x$ is replaced by $x+b$ and it is based on the fact that $x/(x+b) \to 1$ as $x\to\infty$). Also see this answer for details about such replacements.
The given limit condition implies that $(x+3)\log((x+a)/(x+b))\to - 2$ Then clearly $a\neq b$ and we have $$(a-b)\cdot\frac{x+3}{x+b}\cdot\dfrac{\log\left(1+\dfrac{a-b}{x+b}\right)}{\dfrac{a-b}{x+b}}\to - 2$$ Thus $a-b=-2$ or $a=b-2$. Using this value of $a$ in original limit we see that $$x\left(2+(x+3)\log\left(1-\frac{2}{x+b}\right)\right)\to 2$$ Replacing the first factor by $x+b$ and putting $x+b=2/t$ we see that $$\lim_{t\to 0^{+}}\frac{1}{t}\left(2+\left(\frac{2}{t}+3-b\right)\log(1-t)\right)=1$$ or $$\frac{2t+2\log(1-t)}{t^2}+(3-b)\frac{\log(1-t)}{t}\to 1$$ By L'Hospital's Rule or Taylor series the first fraction tends to $-1$ so that $-1+b-3=1$ or $b=5$ and $a=3$.
|
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|
Evaluating $\lim_{x\to0^+}\frac{2x(\sin x)^2+\frac{2x^7+x^8}{3x^2+x^4}-\arctan(2x^3)}{\ln(\frac{1+x^2}{1-x^2})-2x^2+xe^{-{1\over x}}}$ To start with, the term $xe^{-{1\over x}}$ can be ignored.Then splitting the term $\ln(\frac{1+x^2}{1-x^2})=\ln(1+x^2)-\ln(1-x^2)$ and expanding both of them up to order 3 we have in the denominator $2x^2+{2x^6\over 3}-2x^2={2x^6\over3}$
As for the numerator $\arctan(2x^3)\sim2x^3$ and ${2x^7+x^8\over 3x^2+x^4}\sim{2\over3}x^5$.So if we expand $\sin x$ up to order 2,then take the square (ignoring higher order terms) and multiply by $2x$ we have $2x(\sin x)^2\sim2x^3-{2\over3}x^5$ which cancels out with the other terms,giving us an indeterminate form.So I think we should have $x^6$ term in the numerator.
But if I expand $\sin x$ up to order 3 and take square, I fail to obtain $x^5$ term(so that when multiplied by $2x$ gives me $x^6$).How can this be done?
|
You made a mistake by considering only the first asymptotic term of the algebraic fraction. Be careful:
$$\frac{2x^5+x^6}{3+x^2}=\frac{2}{3}x^5 + \frac{1}{3}x^6 + o(x^6).$$
Then your limit gives $\frac{1}{2}$, as results from the comment by JimB.
|
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|
$a+b=c+d$ , $a^3+b^3=c^3+d^3$ , prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$
Let $a, b, c, d$ be four numbers such that $a + b = c + d$ and $a^3 + b^3 = c^3 + d^3$. Prove that $a^{2009}+b^{2009}=c^{2009}+d^{2009}$.
I've got $a^2+b^2-ab=c^2+d^2-cd$.
I tried squaring or cubing it repeatedly but I didn't get what I wanted.
Now how do I proceed?
|
$$(a+b)^3=(c+d)^3$$
$$\iff a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$
If $a+b\ne0, ab=cd\ \ \ \ (1)$
$\iff\dfrac ad=\dfrac cb=k$(say) $\ \ \ \ (2)$
$a+b=c+d\implies dk+b=bk+d\iff d(k-1)=b(k-1)$
Either $d=b\iff a=c$
or $k=1$ use this in $(2)$
Or Using $(1), ab=cda+b=c+d$ So if $a,b$ are roots of $t^2+bt+c=0,$
$c,d$ will be the roots of same equation $\implies\{a,b\}\equiv\{c,d\}$
In both bases we can prove $$a^n+b^n=c^n+d^n$$
|
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|
Understanding some proofs-without-words for sums of consecutive numbers, consecutive squares, consecutive odd numbers, and consecutive cubes
I understand how to derive the formulas for sum of squares, consecutive squares, consecutive cubes, and sum of consecutive odd numbers but I don't understand the visual proofs for them.
For the second and third images, I am completely lost.
For the first one I can see that there are $(n+1)$ columns and $n$ rows. I'm assuming that the grey are even and that the white are odd or vice versa? So in order to have an even amount of odds and evens you must divide by two?
How can I create an image for the sum of consecutive odd numbers ($1+3+5+...(2n-1)^2 = n^2$)
|
The second picture gives a visual proof for the formula
$$3(1^2+2^2+3^2+\dots +n^2)=\frac{n(n+1)}{2}\cdot (2n+1)$$
for $n=5$. The sum of the areas of the $3\cdot 5$ squares on the right
$$3(1^2+2^2+3^2+4^2+5^2)$$
is equal to the area of the rectangle on the left with height $1+2+3+4+5=\frac{6\cdot 5}{2}$ (see the first formula) and base $2\cdot 5+1$.
The third picture gives a visual proof for the formula
$$4(1^3+2^3+3^3+\dots +n^3)=(n(n+1))^2$$
for $n=6$. Starting from the center and evaluating the areas of each concentric frame, the area of the large square of side $7\cdot 6$ is
$$4\cdot 1^2+8\cdot 2^2+12\cdot 3^2+16\cdot 4^2+20\cdot 5^2+24\cdot 6^2\\
=4(1^3+2^3+3^3+4^3+5^3+6^3)$$
|
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|
$\int\limits_0^\infty {x^4 \over (x^4-x^2+1)^4}\ dx$ I want to calculate
$$\int\limits_0^\infty \frac{x^4}{(x^4-x^2+1)^4}dx$$
I have searched with keywords "\frac{x^4}{(x^4-x^2+1)^4}" and "x^4/(x^4-x^2+1)^4". But there are no results
|
For $b > 0$, define $I_b(a)$ by
$$ I_b(a)
= \int_{0}^{\infty} \frac{x^{a}}{((x - x^{-1})^2 + 1)^b} \, \mathrm{d}x
= \int_{0}^{\infty} \frac{x^{a+2b}}{(x^4 - x^2 + 1)^b} \, \mathrm{d}x. $$
This integral converges if $|a+1| < 2b$. We can also prove that $I_b(a) = I_b(-a-2)$ holds, by using the substitution $x \mapsto 1/x$. Then
\begin{align*}
\int_{0}^{\infty} \frac{x^4}{(x^4 - x^2 + 1)^4} \, \mathrm{d}x
&= I_4(-4)
= I_4(2) \\
&= I_4(2) - I_4(0) + I_4(-2) \\
&= \int_{0}^{\infty} \frac{1}{((x - x^{-1})^2 + 1)^3} \, \mathrm{d}x
\end{align*}
So, by the Glasser's master theorem,
\begin{align*}
\int_{0}^{\infty} \frac{x^4}{(x^4 - x^2 + 1)^4} \, \mathrm{d}x
&= \int_{0}^{\infty} \frac{1}{(u^2 + 1)^3} \, \mathrm{d}u \\
&= \int_{0}^{\frac{\pi}{2}} \cos^4\theta \, \mathrm{d}\theta \tag{$(u=\tan\theta)$} \\
&= \frac{3\pi}{16}
\approx 0.58904862254808623221 \cdots.
\end{align*}
|
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|
How can I find the real and imaginary parts of $\sqrt{i+\sqrt{i+\sqrt{i+\dots}}}$? By using $x = \sqrt{i+\sqrt{i+\sqrt{i+\dots}}}$ , I can find a quadratic equation, but how can I separate the real and imaginary parts?
Edit:
My question was how to find the real and imaginary parts of the complex number $\sqrt{i+\sqrt{i+\sqrt{i+\dots}}}$.
After solving the quadratic equation we'll get the solution - $x_{1,2} = \frac{1}{2}\pm \frac{\sqrt{1+4i}}{2}$. Now the prblem is to find the real and imaginary parts of $\sqrt{1+4i}$.
Sorry for the confusion.
|
x = $\sqrt{(i+\sqrt{(i+\dots)})}$ = $\sqrt{(i+x)}$
$x^2 - x - i = 0$
Solving for x we get,
$x_{1,2} = \frac{1}{2}\pm \frac{\sqrt{1+4i}}{2}$
Now,
$a + ib =\sqrt{1+4i}$
$\implies a^2 - b^2 + 2iab = 1+4i$
So we have, $a^2-b^2 = 1$ and 2ab = 4.
Solving for a and b we can get
$a = \sqrt(\frac{\sqrt{17}+1}{2})$ and $b = \sqrt(\frac{\sqrt{17}-1}{2})$.
So the final answer is-
$x = \frac{1}{2} + (\frac{\sqrt{\sqrt{17}+1}}{2\sqrt{2}}) + i (\frac{\sqrt{\sqrt{17}-1}}{2\sqrt{2}})$.
|
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|
$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$ $$\sum\limits_{n=1}^{\infty}\arctan{\frac{2}{n^2+n+4}}$$
We know that : $\arctan{x} - \arctan{y} = \arctan{\frac{x-y}{1+xy}}$ for every $ xy > 1 $
I need to find two numbers which satisfy: $ab = n^2+2n+3 $ and $ a-b =2$ in order to telescope.
Edit: I am very sorry on the denominator it should have been $n^2+n+4$ not $n^2+2n+4$
Similarly here :
$$\sum\limits_{n=1}^{\infty}\arctan{\frac{8n}{n^4-2n^2+5}}$$
The result should be $ \arctan 2 $ on the first one and $ \pi/2 + \arctan2 $ on the second one.
|
Hint:
$$\dfrac2{n^2+n+4}=\dfrac{\dfrac12}{1+\dfrac{n(n+1)}4}=\dfrac{\dfrac{n+1}2-\dfrac n2}{1+\dfrac n2\cdot\dfrac{n+1}2}$$
For the second, $$\dfrac{8n}{n^4-2n^2+5}=\dfrac{\dfrac{(n+1)^2}2-\dfrac{(n-1)^2}2}{\dfrac{(n+1)^2}2\cdot\dfrac{(n-1)^2}2+1}$$
|
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|
The greatest value of $|z|$ if $\Big|z+\frac{1}{z}\Big|=3$ where $z\in\mathbb{C}$
$\bigg|z+\dfrac{1}{z}\bigg|=3$ then the greatest value of $|z|$ is ___________
My Attempt
$$
\bigg|z+\frac{1}{z}\bigg|=\bigg|\dfrac{z^2+1}{z}\bigg|=\frac{|z^2+1|}{|z|}=3\\
\bigg|z+\frac{1}{z}\bigg|=3\leq|z|+\frac{1}{|z|}\implies |z|^2-3|z|+1\ge0\\
|z|=\frac{3\pm\sqrt{5}}{2}\implies\color{red}{|z|\in(-\infty,\frac{3-\sqrt{5}}{2}]\cup[\frac{3+\sqrt{5}}{2},+\infty)}
$$
$$
\bigg|z+\frac{1}{z}\bigg|=3\geq\bigg||z|-\frac{1}{|z|}\bigg|\\
|z|^2-3|z|-1\leq0\text{ or }|z|^2+3|z|-1\geq0\\
|z|=\frac{3\pm\sqrt{13}}{2}\text{ or }|z|=\frac{-3\pm\sqrt{13}}{2}\\
\color{red}{|z|\in[\frac{3-\sqrt{13}}{2},\frac{3+\sqrt{13}}{2}]}\text{ or }\color{red}{|z|\in(-\infty,\frac{-3-\sqrt{13}}{2}]\cup[\frac{-3+\sqrt{13}}{2},+\infty)}
$$
The solution given in my reference is $\dfrac{3+\sqrt{13}}{2}$, why am I not able to find it in my attempt ?
Note: A similar question has been asked before, If $∣z+\frac{1}{z}∣=a$ then find the minimum and maximum value of $|z|$, but it does not have any answer which address why we choose a specific version of the triangle inequality.
|
A possible approach is using polar representation
*
*$z = |z|e^{i\phi}$
*$\Big|z+\frac{1}{z}\Big|=3 \Leftrightarrow (z+\frac{1}{z})(\bar z + \frac{1}{\bar z}) = 9 \Leftrightarrow \boxed{\left(|z|e^{i\phi} + \frac{1}{|z|} e^{-i\phi}\right)\left(|z|e^{-i\phi} + \frac{1}{|z|} e^{i\phi}\right) = 9}$
Expanding gives
\begin{eqnarray*}
|z|^2 + e^{-2i\phi} + e^{2i\phi} + \frac{1}{|z|^2} & = & 9 \\
& \Leftrightarrow & \\
|z|^2 + 2\cos 2\phi + \frac{1}{|z|^2} & = & 9 \\
\end{eqnarray*}
So, the maximum $|z|$ is reached for $\cos 2\phi = -1$, since the maximum solution of $q+\frac{1}{q}= C$ is $q= \frac{1}{2}(C+\sqrt{C^2-4})$ . Hence, solve
\begin{eqnarray*}
|z|^2 + \frac{1}{|z|^2} & = & 11 \\
& \stackrel{|z|\: max.}{\Rightarrow} & \\
|z| & = & \sqrt{\frac{11+\sqrt{117}}{2}}\\
\end{eqnarray*}
Here I show that this is the same as the given solution:
\begin{eqnarray*}
|z| & = & \sqrt{\frac{11+\sqrt{117}}{2}}\\
& = & \sqrt{\frac{22+2\cdot 3 \sqrt{13}}{4}}\\
& = & \sqrt{\frac{9 + 13+2\cdot 3 \sqrt{13}}{4}}\\
& = & \sqrt{\frac{(3 +\sqrt{13})^2}{4}}\\
& = & \frac{3 +\sqrt{13}}{2}\\
\end{eqnarray*}
|
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|
How can I simplify this fraction problem? I have the problem $\frac{x^2}{x^2-4} - \frac{x+1}{x+2}$ which should simplify to $\frac{1}{x-2}$
I have simplified $x^2-4$, which becomes:
$\frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}$
However, if I combine the fractions I get, $x^2-x-1$ for the numerator, which can't be factored. That's where I get stuck.
How can I get $\frac{1}{x-2}$ out of this problem?
|
\begin{align}
\frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}&=\frac1{x+2}\left(\frac{x^2}{x-2} - (x+1)\right)\\
&=\frac1{x+2}\left(\frac{x^2-(x-2)(x+1)}{x-2}\right)\\
&=\frac1{x+2}\left(\frac{x^2-(x^2-x-2)}{x-2}\right)\\
&=\frac1{x+2}\left(\frac{x+2}{x-2}\right)\\
&=\frac1{x-2}
\end{align}
|
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|
Show $\frac{n+1}{2n^2+5} \in \Theta (\frac{1}{n})$ using definition Show $\frac{n+1}{2n^2+5} \in \Theta (\frac{1}{n})$ using definition
First showing the upper bound
$\frac{n+1}{2n^2+5} \in O (\frac{1}{n})$
$\frac{n+1}{2n^2+5} \leq \frac{2n}{2n^2} = \frac{1}{n}$
for $n \geq 1$
now showing the lowerbound
$\frac{n+1}{2n^2+5} \in \Omega (\frac{1}{n})$
$\frac{n+1}{2n^2+5} > \frac{n}{2n^2+5} \geq \frac{n}{2n^2 + 5n^2} = \frac{n}{7n^2} = \frac{1}{7n}$
for $n \geq 1$
hence, $\frac{n+1}{2n^2+5} \in \Theta (\frac{1}{n})$
is this right?
|
Yes, you have shown found constant $c_1$ and $c_2$ such that
$$ \frac{c_1}{n}\le\frac{n+1}{2n^2+5} \le \frac{c_2}{n}.$$
You have proven what you want.
|
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|
For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer? For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer?
For example, if x = 3, $y = 4(3^2) + 8(3) + 5$ = 65 which is divisible by 13
if x = 8, y = 325 which is divisible by 13
if x = 16, y = 1157 which is divisible by 13
if x = 21, y = 1937 which is divisible by 13
I am guessing that values of x = 13i + 3 or x = 13i + 8 where i is an integer will result in a value of y that is evenly divisible by 13.
How do you prove that x = 13i + 3 or x = 13i + 8 will result in a value of y that is evenly divisible by 13?
Is there a general proof to find values of x that will result in a value of y that is evenly divisible by an odd integer p?
|
You are looking for solutions to $4x^2+8x+5\equiv 0 \pmod {13}$. You can do the usual quadratic formula and find $x=\frac {-8 \pm \sqrt{64-80}}8$. Don't let the negative number under the square root bother you because $\pmod {13}$ we have $-16 \equiv 10$. You need to find whether $10$ is a square $\bmod 13$. One way is to just try them. Here we find $6^2 \equiv 7^2 \equiv 10 \pmod {13}$. Now noting that $8\cdot 5 \equiv 1 \pmod {13}$ we can say that $x=5(-8\pm 6)\equiv 3,8 \pmod {13}$
There is a whole theory of quadratic reciprocity that says when things are square roots in rings of integers. I have not studied it.
|
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|
Simplify $\frac{x^3-x}{x^2+xy+x+y}$ $$\frac{x^3-x}{x^2+xy+x+y}$$
What I did:
$$\frac{x}{xy+x+y}$$
through simplifying the $x$'s.
But it's not right.
What did I do wrong?
|
$$
\frac{x^3 - x}{x^2 +xy + x +y }
= \frac{x(x^2 - 1)}{x( x+y) + 1( x +y )}
= \frac{x(x - 1)(x+1)}{(x+1)( x+y)}
= \frac{x(x - 1)}{( x+y)}
$$
Here you had to remember that $x^2 -1 = (x+1)(x-1)$. The last step assumes $x\neq -1$.
|
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How to prove $\frac{(x+y)(y+z)(z+x)}{4xyz}≥\frac{x+z}{y+z}+\frac{y+z}{x+z}$ for $x,y,z>0$?
Prove that for $x,y,z$ positive numbers:
$$
\frac{(x+y)(y+z)(z+x)}{4xyz}≥\frac{x+z}{y+z}+\frac{y+z}{x+z}
$$
I tried to apply MA-MG inequality:
$x+y≥2\sqrt{xy}$ and the others and multiply them but it becomes
$\frac{(x+y)(y+z)(z+x)}{4xyz}≥2$ but $\frac{x+z}{y+z}+\frac{y+z}{x+z}≥2$ so it doesn't work like that.
I don't know what to apply here.
|
We have
$$
\frac1 {4yz}+\frac1 {4xz}\ge \frac1{(y+z)^2}+\frac1{(x+z)^2}.
$$ Therefore
$$
\frac{(x+y)}{4xyz}(y+z)(z+x)\ge\left(\frac1{(y+z)^2}+\frac1{(x+z)^2}\right)(y+z)(z+x),
$$ which leads to the desired inequality.
|
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|
Prove that $1+b+(1+c)+1/c+1+a \ge 3$ if $a, b,$ and $c$ are positive real numbers. Let $a, b, c$ be positive real numbers. prove that
$$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)}\ge\frac{3}{1+abc},
$$
and that equality occurs if and only if $a = b = c = 1$.
What I tried:
$1$st part: I tried a brute force approach where I make a common denominator for all $4$ fractions, but the number get real big real fast.
|
AM-GM helps:
$$\sum_{cyc}\frac{1}{a(1+b)}=\frac{1}{1+abc}\sum_{cyc}\frac{1+abc}{a(1+b)}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\left(\frac{1+abc}{a(1+b)}+1\right)-\frac{3}{1+abc}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}-\frac{3}{1+abc}=$$
$$=\frac{1}{1+abc}\sum_{cyc}\left(\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}\right)-\frac{3}{1+abc}\geq$$
$$\geq\frac{1}{1+abc}\cdot6\sqrt[6]{\prod_{cyc}\frac{(1+a)}{a(1+b)}\prod_{cyc}\frac{b(1+c)}{1+b}}-\frac{3}{1+abc}=\frac{3}{1+abc}.$$
|
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|
Show that $y^3=4x^2-1$ has no positive integer solutions Problem: Show that $y^3=4x^2-1$ has no positive integer solutions.
I tried to consider maybe that cubes are never equal to $3 \bmod 4$, but $3^3 \equiv 3\bmod 4$.
Next I tried letting $x$ be even , so that $y$ is odd, (and if $x$ is odd, then $y$ is still odd), but that doesn't reveal much, to me at least.
Any hints appreciated.
|
This equation is satisfied if and only if some $y^3$ exists that can be expressed as a product of two consecutive odd numbers, $2x - 1$ and $2x + 1$.
Consider $p$ a prime divisor of $y$. Note that $p > 2$. One of the two factors must be divisible by $p^2$. If it's not divisible by $p^3$, then other factor must be divisible by $p$. But then both $2x - 1$ and $2x + 1$ are divisible by $p$, which means their difference $2$ must be divisible by $p$, hence $p = 2$, a contradiction.
Thus, $p^3$ must divide $2x - 1$ or $2x + 1$. This is true for all prime divisors of $y$, hence $2x - 1$ and $2x + 1$ must both be perfect cubes. There are two perfect cubes that are $2$ apart: $-1$ and $1$, thus making $x = 0$ and $y = -1$ the only solution.
|
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|
Proving $\left(1+\frac{1}{m}\right)^m < \left(1+\frac{1}{n}\right)^n$ Let $m,n\in \mathbb{N}$. If $m > n$ show that
$$\left(1+\frac{1}{m}\right)^m > \left(1+\frac{1}{n}\right)^n$$
My works: I tried to show if $g(x)=\left(1+\frac{1}{x}\right)^x$ then $g'(x) > 0$.
\begin{align}
g'(x) &= \frac{d e^ { x \ln(1+\frac{1}{x}) }} {dx} \\
&= e^{x\ln(1+1/x)} \left(\ln(1+\frac{1}{x}) - \frac{1}{1+x} \right) >0
\end{align}
|
If $m<n$, then for all $j=0...m$ we have $1-\frac jm \le 1-\frac jn$ and so $\frac{m-j}{m} \le \frac{n-j}{n}$ (for $j=0$ we have equality) and therefore:
$$
\begin{align}
\left(1+\frac1m\right)^m & = \sum_{k=0}^m \binom{m}{k}\frac{1}{m^k} = \sum_{k=0}^m \frac {1}{k!}\prod_{j=0}^{k-1}\frac{m-j}{m}\\
& \le \sum_{k=0}^m \frac {1}{k!}\prod_{j=0}^{k-1}\frac{n-j}{n} = \sum_{k=0}^m \binom{n}{k}\frac{1}{n^k}\\
& < \sum_{k=0}^n \binom{n}{k}\frac{1}{n^k} = \left(1+\frac1n\right)^n\\
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How do we factor out $x^2 - x -2$ in this expression? Suppose we're given that
$$x^4 - 2x^3 +x-2$$
How do we factor out $x^2 - x -2$ in this expression?
$$(x^4 -x^3- 2x^2)-(x^3-x^2-2x)+(x^2-x-2) = x^2(x^2 -x-2)-x(x^2-x-2)+(x^2-x-2) = (x^2-x-2)(x^2-x+1)$$
This satisfies with what we want to get. However, I do not seem to have understood what is done there. Could I get your assistance in order to understand it? Perhaps there's better way of factoring.
Regards
|
$$x^4-2x^3+x-2=x^4-x^3-2x^2-x^3+x^2+2x+x^2-x-2=$$
$$=x^2(x^2-x-2)-x(x^2-x-2)+(x^2-x-2)=(x^2-x-2)(x^2-x+1).$$
|
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|
Studying the convergence of the series $\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots$ Studying the convergence of the series $$\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots$$
I saw this problem and I tried to do it my own way but I don't know what I'm doing wrong because I'm getting a divergent series while in the solution it says that it's convergent.
This is what I tried to do
I can write $\sqrt{2}=2\cos\frac{\pi}{4}=2\cos\frac{\pi}{2^2}$
$\sqrt{2+\sqrt{2}}=\sqrt{2+2\cos\frac{\pi}{4}}=2\sqrt{\frac{1+\cos2\frac{\pi}{8}}{2}}=2\cos\frac{\pi}{8}=2\cos\frac{\pi}{2^3}$
Continuing like this I get $$\underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}_{n\text{ times}}=2\cos\frac{\pi}{2^{n+1}}$$
So I can write the series $$\sqrt{2}+\sqrt{2+\sqrt{2}}+\sqrt{2+\sqrt{2+\sqrt{2}}}+ \cdots=2\sum_{n=1}^\infty\cos\frac{\pi}{2^{n+1}}$$
Since $$\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}\cos\frac{\pi}{2^{n+1}}=1\ne0$$ the series diverges
|
Your conclusion that $\sum_{n=1}^\infty a_n$ diverges is correct. You don't need the explicit formula, it would suffice to observe that
$$
a_n = \underbrace{\sqrt{2+\sqrt{2+\cdots+\sqrt{2}}}}_{n\text{ times}} \ge \sqrt 2
$$
or that
$$
0 < a_n < a_{n+1}
$$
for all $n \in \Bbb N$.
|
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|
Is $8n^3+12n^2-2n-3$ divisible by 5, with $n$ congruent with 1,2,3 mod 5? I make some test using python and I find that this is the case for $n = 5k +p$ with $k$ an integer and $p =$1,2,or 3.
I was able to prove for $p = 1$ and $p=2$ but not for $p = 3$.
What I'm doing wrong? This is my work:
Let's assume 5 is a divisor of $8n^3+12n^2-2n-3$. Let's prove it for $n + 1.$
We have $8(n+1)^3+12(n+1)^2-2(n+1)-3=8(n^3+3 n^2+3n+1)+12(n^2+2n+1)-2(n+1)-3=(8n^3+12n^2-2-3)+24n^2+48n+18$
For induction we have $8n^3+12n^2-2n-3$, we need only that $5$ divide $24n^2+48n+18$
Since $n = 5k + p$, $k$ an integer and $p$ with values $1$,$2$ or $3$.
We need to see that $5$ divide $24n^2+48n+18= 24$ with $(5k+p)^2+48(5k+p)+18$,
Using congruence we have $(5k+p)^2$ is congruent with $p^2$ mod 5 and $5k + p$ with $p$ modulo 5.
$f(p)= 24p^2+48p+18$ is divisible with 5 para $p = 1, p = 2$.
So $f(1) = 90$ is divisible for 5 , $f(2)= 210$ and also divisible for 5.
But $f(3)$ is not divisible for 5.
|
Alternative solution. Note that
$$8n^3+12n^2-2n-3=(2n+3)(2n+1)(2n-1)$$
which is divisible by $5$ iff $(2n+3)$ or $(2n+1)$ or $(2n-1)$ is divisible by $5$. Now
$$\begin{align}
&2n+3\equiv 0\leftrightarrow 2n\equiv -3\equiv 2\leftrightarrow n\equiv 1 \pmod{5}\\
&2n+1\equiv 0\leftrightarrow 2n\equiv -1\equiv 4\leftrightarrow n\equiv 2 \pmod{5}\\
&2n-1\equiv 0\leftrightarrow 2n\equiv 1\equiv 6\leftrightarrow n\equiv 3 \pmod{5}\end{align}$$
Hence $8n^3+12n^2-2n-3$ is divisible by $5$ iff $n\equiv 1,2,3$ modulo $5$.
|
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|
Proving that $\lim_{v\rightarrow \infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} +\cdots + \frac{v^2}{v^3+v}\right]= 1 $ I wonder if my solution that $\lim_{v\to\infty}\left[\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} + \cdots + \frac{v^2}{v^3+v}\right]= 1 $ is correct.
$$\frac{v^2}{v^3 + 1} + \frac{v^2}{v^3 + 2} + ... + \frac{v^2}{v^3+v} = \frac{\frac{1}{v}}{1 + \frac{1}{v^3}} + \frac{\frac{1}{v}}{1 + \frac{2}{v^3}} + ... + \frac{\frac{1}{v}}{1+\frac{1}{v^2}} \rightarrow 0 $$
So, the limit is not equal to 1.
Where is my mistake ?
|
Consider the sequence$$1,\frac12+\frac12,\frac13+\frac13+\frac13,\ldots,\overbrace{\frac1n+\cdots+\frac1n}^{n\text{ times}},\ldots$$Its limit is $1$, right?! However, by your argument, the limit should be $0$.
Yes, $\lim_{n\to\infty}\frac1n=0$, but, on the other hand, you have $n$ terms. So, the fact that the limit of each term is $0$ does not imply that the limit of the whole sequence is $0$.
|
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|
Find $K=a^2b+b^2c+c^2a$ for roots $a>b>c$ of a cubic. If $a>b>c$ are the roots of the polynomial $P(x)=x^3-2x^2-x+1$ find the value of $K=a^2b+b^2c+c^2a$.
Using Vièta's formulas:
$$a+b+c=2$$
$$ab+bc+ca=-1$$
$$abc=-1$$
Using those I found that
$$a^2+b^2+c^2=6$$
$$a^3+b^3+c^3=11$$
$$a^2b+b^2c+c^2a+ab^2+bc^2+ca^2=1$$
but I can't separate K from it.
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, $u=\frac{2}{3},$ $v^2=-\frac{1}{3},$ $w^3=-1$ and
$$\sum_{cyc}a^2b=\frac{1}{2}\sum_{cyc}(a^2b+a^2c+a^2b-a^2c)=$$
$$=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2}(a-b)(a-c)(b-c)=$$
$$=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2}\sqrt{(a-b)^2(a-c)^2(b-c)^2}=$$
$$=\frac{1}{2}(9uv^2-3w^3)+\frac{1}{2}\sqrt{27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)}=$$
$$=\frac{1}{2}\left(9\cdot\frac{2}{3}\left(-\frac{1}{3}\right)-3(-1)\right)+$$
$$+\frac{1}{2}\sqrt{27\left(3\left(\frac{2}{3}\right)^2\left(-\frac{1}{3}\right)^2-4\left(-\frac{1}{3}\right)^3-4\left(\frac{2}{3}\right)^3(-1)+6\cdot\frac{2}{3}\left(-\frac{1}{3}\right)(-1)-(-1)^2\right)}=4.$$
The hint for another way:
Prove that
$$(a,b,c)=\left(1+2\cos\frac{2\pi}{7},1+2\cos\frac{4\pi}{7},1+2\cos\frac{6\pi}{7}\right)$$ and end it!
|
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|
What is the number of pairs of $(a,b)$ of positive real numbers satisfying $a^4+b^4<1$ and $a^2+b^2>1$?
The number of pairs of $(a,b)$ of positive real numbers satisfying
$a^4+b^4<1$ and $a^2+b^2>1$ is -
$(i)$0
$(ii)$1
$(iii)$2
$(iv)$ more than 2
Solution:We have $a,b>0$,
According to the given situation,$0<a^4+b^4<1<a^2+b^2\implies a^4+b^4<a^2+b^2\implies 0<a^2(a^2-1)<b^2(1-b^2)\implies b^2(1-b)(1+b)<0 ;a^2(a-1)(a+1)>0\implies a\in(-\infty,-1)\cup (0,1);b\in(-1,0)\cup(1,\infty)$
But, in particular, if we choose $a=-1/2,b=1/2$,then $a^4+b^4=1/16+1/16=1/8<1$ and $a^2+b^2=1/4+1/4=1/2<1$.Which contradicts the given condition $a^2+b^2>1$
Where is the mistake in my approach? How should I approach this problem(means how to think ?). Can Triangle inequality be used here?
Please provide some hint...
|
There are infinitely many solutions:
In fact for every $-1<a<1$ , $a\not = 0$ there exist infinitely many $b$'s such that the pair $(a,b)$ satisfies the condition.
Indeed, the conditions can be written as $b^4<1-a^4$ and $b^2>1-a^2$.
Therefore if $b$ satisfies $b<\sqrt[4]{1-a^4}, b>\sqrt[2]{1-a^2}$ then the pair $(a,b)$ satisfy the conditions in the question. (note that here we use the fact that $-1<a<1$ which implies that $1-a^4$ and $1-a^2$ are positive).
It is left to show that $\sqrt[2]{1-a^2}< \sqrt[4]{1-a^4}$. As this would imply that there are infinitely many $b$'s in between those two values.
Since $-1<a<1$ the terms $1-a^2,1-a^4$ are positive so I can square twice the above inequality.
$$(1-a^2)^2 <(1-a^4) $$
Since $(1-a^2)^2 = 1-2a^2 + a^4$ it is enough to show that
$$ 1-2a^2 + a^4 < 1-a^4$$
but this is equivalent to $$a^4<a^2$$
Finally, since $-1<a<1$ and $a\not = 0$ the assertion that $a^2>a^4$ holds.
|
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|
Evaluate $\lim\limits_{x\to+\infty} \sqrt{x}\left ( \sqrt[3]{x+1}-\sqrt[3]{x-1}\right )$?
How do you evaluate $$\lim\limits_{x\to+\infty} \sqrt{x}\left (\sqrt[3]{x+1}-\sqrt[3]{x-1}\right ) ?$$
Thanks in advance for your support.
|
If it is ok for you to use only the fact that $\color{blue}{f(t)} =\sqrt[3]{1+t}$ and $\color{blue}{g(t)}=\sqrt[3]{1-t}$ are differentiable at $t= 0$, then without calculating any derivative you may proceed as follows:
*
*Set $x = \frac{1}{t}$ and consider $t \rightarrow 0^+$
\begin{eqnarray*} \sqrt{x}\left ( \sqrt[3]{x+1}-\sqrt[3]{x-1}\right )
& \stackrel{x=\frac{1}{t}}{=} & \frac{\sqrt[3]{1+t} - \sqrt[3]{1-t}}{\underbrace{\sqrt[3]{t}\cdot \sqrt{t}}_{=t^{\frac{5}{6}}}}\\
& = & \frac{\sqrt[3]{1+t} -1 + 1 - \sqrt[3]{1-t}}{t}\cdot t^{\frac{1}{6}}\\
& = & \left(\underbrace{\frac{\sqrt[3]{1+t} -1}{t}}_{\stackrel{t\to 0^+}{\longrightarrow}\color{blue}{f'(0)}} - \underbrace{\frac{\sqrt[3]{1-t} - 1}{t}}_{\stackrel{t\to 0^+}{\longrightarrow}\color{blue}{g'(0)}}\right) \cdot t^{\frac{1}{6}}\\
& \stackrel{t\to 0^+}{\longrightarrow} & (f'(0) - g'(0))\cdot 0 = 0
\end{eqnarray*}
|
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|
Find $A, B \in M_2(\mathbb{Q})$ Find $A, B \in M_2(\mathbb {Q}) $ so that $A^2+2B^2=\begin{pmatrix} 4& - 4\\
-2 & 2
\end{pmatrix}$ and $AB+BA=\begin{pmatrix}3 & - 3\\
-1 & 1
\end{pmatrix}$.
My work so far : $(A +\sqrt 2 B) ^2=A^2+2B^2+\sqrt 2 (AB+BA) $.
After taking determinants we get that $\det (A +\sqrt 2 B)=0$ and this implies that $Tr A\cdot Tr B=Tr(AB) $and $\det A=-2\det B$. Here I am stuck.
Edit: We can similarly get that $\det(A-\sqrt 2 B) =0$ and this implies that $\det A=\det B=0$ and I think now we can easily find the traces and then just substitute back into the equations.
|
If we denote $v=(1,1)$, then the given conditions imply that
$$
\ker (A+\sqrt 2B)^2 =\ker(A-\sqrt 2B)^2=\text{span}\{v\}.
$$ Thus $A\pm \sqrt{2}B$ are non-invertible matrices, which are not $O$, and it follows that
$$
\ker(A\pm \sqrt{2}B)=\text{span}\{v\}
$$ since
$$
(0)< \ker(A\pm \sqrt{2}B) \le \ker (A\pm \sqrt 2B)^2=\text{span}\{v\}.
$$This implies that $Av=Bv=0$, so both $A,B$ are non-invertible.
Denote $\alpha =\text{tr}(A)$ and $\beta=\text{tr}(B)$. Then Caley-Hamilton theorem implies that
$$
A^2=\alpha A, \ \ B^2=\beta B
$$ thus giving $A^2+2B^2=\alpha A+2\beta B\ (*)$. Take trace to $(*)$, then we get $\alpha^2+2\beta^2=6$. By squaring $(*)$, we obtain
$$
\alpha^3A+4\beta^3B+2\alpha\beta(AB+BA)=\left(\begin{array}{cc}24& -24\\-12&12\end{array}\right)
$$ and by taking trace again, it follows
$$
\alpha^4+4\beta^4+8\alpha\beta=36=(\alpha^2+2\beta^2)^2=\alpha^4+4\beta^4+4\alpha^2\beta^2.
$$ Note that $\alpha\beta=0$ or $\alpha\beta=2$. But $\alpha\beta=0$ would imply $\alpha^2=6$ or $\beta^2=3$, which is absurd since $\alpha,\beta\in \Bbb Q$. So, $\alpha\beta$ is $2$, and $(\alpha+\sqrt 2\beta)^2=6+4\sqrt 2=(2+\sqrt 2)^2$ shows that $(\alpha,\beta)=\pm(2,1)$. Note that if $(A,B)$ is a solution for $(\alpha,\beta)=(2,1)$, then $(-A,-B)$ is a solution for $(\alpha,\beta)=(-2,-1)$.
Now, assume $\alpha=2,\beta =1$ and observe that
$$
A^2+2B^2=2(A+B)=\left(\begin{array}{cc}4& -4\\-2&2\end{array}\right).
$$ Squaring $A+B$ gives
$$
A^2+B^2+AB+BA=2A+B+\left(\begin{array}{cc}3& -3\\-1&1\end{array}\right)=\left(\begin{array}{cc}6& -6\\-3&3\end{array}\right).
$$ By solving the equations simultaneously we finally obtain the solution for $(\alpha,\beta)=(2,1)$:
$$
A=\left(\begin{array}{cc}1& -1\\-1&1\end{array}\right),\quad B=\left(\begin{array}{cc}1& -1\\0&0\end{array}\right).
$$ For the case where $(\alpha,\beta)=(-2,-1)$, put $-$ signs to the above $A,B$. These are the only solutions to the given equations.
|
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|
Solving a tough equation involving integer functions I am stuck on solving the equation, given $k\lt\frac{n}{2},\ n,k\ge3$:
$$ m=\lceil 2k-\frac{2}{n}\displaystyle\left(\lfloor\frac{n-\lfloor\frac{n}{k+1}\rfloor}{2}\rfloor\right)(k+1)\rceil$$.
I think the value of $m$ would be $k$. But, the detailed solution is beyond my reach. What inequalities and analysis should we use to reach the solution? Also, what if I modify the equation to:
$$m=\lceil 2\frac{nk}{n-1}-\frac{2}{n-1}\displaystyle\left(\lfloor\frac{n-\lfloor\frac{n}{k+1}\rfloor}{2}\rfloor\right)(k+1)\rceil$$
Will the value of $m=k+1$ at any value of $n,k$? Any hints Thanks beforehand.
|
We calculate the first expression for integer values $k\geq 0, n\geq 1$.
Let $a,b$ be non-negative integer with
\begin{align*}
n=a(k+1)+b\qquad\quad a,b\geq 0,\ \ 0\leq b<k+1\tag{1}
\end{align*}
we obtain
\begin{align*}
\color{blue}{ a(n,k)}
&\color{blue}{=\left\lceil 2k-\frac{2}{n}\left(\left\lfloor\frac{n-\left\lfloor\frac{n}{k+1}\right\rfloor}{2}\right\rfloor\right)(k+1)\right\rceil}\\
&=\left\lceil2k-\frac{2}{n}\left(\left\lfloor\frac{n-a}{2}\right\rfloor\right)(k+1)\right\rceil\tag{2}\\
\end{align*}
Here we use (1) to get $\left\lfloor\frac{n}{k+1}\right\rfloor=\left\lfloor\frac{a(k+1)+b}{k+1}\right\rfloor=a+\left\lfloor\frac{b}{k+1}\right\rfloor=a$ noting that $0\leq b<k+1$.
In (2) we consider two cases: $n-a$ even or odd.
Case 1: $n-a$ even
We obtain from (2)
\begin{align*}
\color{blue}{a(n,k)}
&=\left\lceil2k-\frac{2}{n}\left(\frac{n-a}{2}\right)(k+1)\right\rceil\\
&=\left\lceil2k-\left(\frac{n-a}{n}\right)(k+1)\right\rceil\\
&=\left\lceil2k-\left(1-\frac{a}{n}\right)(k+1)\right\rceil\\
&=\left\lceil k-1+\frac{a(k+1)}{n}\right\rceil\\
&=\left\lceil k-1+\frac{n-b}{n}\right\rceil\tag{3}\\
&=\left\lceil k-\frac{b}{n}\right\rceil\\
&\color{blue}{=}\begin{cases}
\color{blue}{k}&\color{blue}{\qquad n>k}\\
\color{blue}{k-1}&\color{blue}{\qquad n\leq k}\tag{4}
\end{cases}
\end{align*}
Comment:
*
*In (3) we use again the representation (1).
*In (4) we observe from (1) that $n=b$ if $n\leq k$ so that $\left\lceil k-\frac{b}{n}\right\rceil=\left\lceil k-1\right\rceil=k-1$.
Case 2: $n-a$ odd
We obtain from (2)
\begin{align*}
\color{blue}{a(n,k)}
&=\left\lceil2k-\frac{2}{n}\left(\frac{n-a-1}{2}\right)(k+1)\right\rceil\\
&=\left\lceil2k-\left(\frac{n-a-1}{n}\right)(k+1)\right\rceil\\
&=\left\lceil2k-\left(1-\frac{a+1}{n}\right)(k+1)\right\rceil\\
&=\left\lceil k-1+\frac{a(k+1)}{n}+\frac{k+1}{n}\right\rceil\\
&=\left\lceil k-\frac{b}{n}+\frac{k+1}{n}\right\rceil\\
&=k+\left\lceil \frac{k+1-b}{n}\right\rceil\\
&\,\,\color{blue}{=k+1}\\
\end{align*}
The last line follows since from (1) we have $b<k+1$.
|
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"url": "https://math.stackexchange.com/questions/3125924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Stuck on a Geometry Problem
$ABCD$ is a square, $E$ is a midpoint of side $BC$, points $F$ and $G$ are on the diagonal $AC$ so that $|AF|=3\ \text{cm}$, $|GC|=4\ \text{cm}$ and $\angle{FEG}=45 ^{\circ}$. Determine the length of the segment $FG$.
How can I approach this problem, preferably without trigonometry?
|
Drop perpendiculars $FH, EI$ and $FJ$ and label $BH=y$, $EH=z$ and $\angle GEI=\alpha$:
$\hspace{2cm}$
Use the similarity of $\triangle FEH$ and $\triangle EIG$:
$$\frac{GI}{HE}=\frac{EI}{FH} \Rightarrow \\
\frac{4-CI}{z}=\frac{CI}{y+2z}\Rightarrow \\
\frac{4-\frac{y+z}{\sqrt{2}}}{z}=
\frac{\frac{y+z}{\sqrt{2}}}{y+2z}\Rightarrow \\
\frac{4\sqrt{2}-y-z}{z}=\frac{y+z}{y+2z}\Rightarrow\\
3z^2+(4y-8\sqrt{2})z+y^2-4\sqrt{2}y=0 \stackrel{y=\frac3{\sqrt{2}}}{\Rightarrow} \\
6z^2-4\sqrt{2}z-15=0 \Rightarrow \\
z=\frac3{\sqrt{2}}.$$
Hence:
$$x+7=2(y+z)\sqrt{2}=12 \Rightarrow x=5.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3126169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 12,
"answer_id": 3
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|
integrate sin(x)cos(x) using trig identity. Book tells me the answer is:
$$ \int \sin(x)\cos(x) dx = \frac{1}{2} \sin^{2}(x) + C $$
however, I get the result:
$$
\sin(A)\cos(B) = \frac{1}{2} \sin(A-B)+\frac{1}{2}\sin(A+B)
$$
$$
\begin{split}
\int \sin(x)\cos(x) dx
&= \int \left(\frac{1}{2}\sin(x-x) + \frac{1}{2}\sin(x+x)\right) dx \\
&= \int \left(\frac{1}{2}\sin(0) + \frac{1}{2}\sin(2x)\right) dx \\
&= \int \left(\frac{1}{2}\sin(2x)\right) dx \\
&= -\frac{1}{2} \frac{1}{2}\cos(2x) +C\\
&= -\frac{1}{4} \cos(2x) +C
\end{split}
$$
How did the book arrive at the answer $\frac{1}{2}\sin^2(x)$?
|
HINT
The answers are equivalent. Recall that
$$
\cos (2x) = \cos^2 x - \sin^2 x = \left(1-\sin^2 x\right) - \sin^2 x = 1 - 2 \sin^2x
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3126227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find $z$ such that $(2+i)(1+i)=2+zi $ I tried multiplying both of the complex numbers, but the answer is $1+3i $.
Sorry for the poor English; I'm not a native speaker.
|
Let $z = a + bi$ where $a$ and $b$ are real.
$(2+i)(1+i) = 2 +zi$
$1 + 3i = 2+ (a+bi)i$
$-1 + 3i = ai -b = -b + ai$
So $a =3$ and $-b = -1$ so $b = 1$.
So $z = 3 + i$.
=====
You should get used to the fact that Complex numbers are just like real numbers.
If you $m=a + dz$ then $z = \frac {m-a}d$ and this is no different.
$(2+i)(1+i) = 2 + zi$ means
$zi = (2+i)(1+i) - 2$
$z = \frac {(2+i)(1+i) - 2}i$.
You've multiplied out $(2+i)(1+i) = 1+3i$ so
$z = \frac {(1+3i)-2}i = \frac {-1 + 3i}i = \frac {-1}i + 3$.
Now it's a matter of figuring out what $\frac {-1}i$ is.
$\frac {-1}i = \frac {-1}i\cdot \frac ii = \frac {-i}{-1} = i$.
So $z = i + 3 = 3+i$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3126792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How to deduce the following complex number problem I am stuck with the following problem that says:
Using the result $$x^n-1=(x^2-1)\prod_{k=1}^{(n-2)/2}[x^2-2x\cos \frac{2k\pi}{n}+1],$$ if $n$ be an even positive integer, deduce that $$\sin \frac{\pi}{32}\sin \frac{2\pi}{32}\sin \frac{3\pi}{32}.........\sin \frac{15\pi}{32}=\frac{1}{2^{13}}$$
Can someone point me in the right direction with some explanation? Thanks in advance for your time.
|
This doesn't use the given information, but it does give some intuition of what you are trying to show.
Consider $x^8 - 1$
The roots of unity are equally spaced around the unit circle.
$x^8-1 = (x-1)(x-r_1)(x-r_2)\cdots(x-r_7) = (x-1)(x^7 + x^6 + \cdots + 1)$
with $r_k = (\cos \frac {2k\pi}{8}, i\sin \frac{2k\pi}{8})$
$|(x^7 + x^6 + \cdots + 1)| = |x-r_1||x-r_2|\cdots|x-r_7|$
letting $x = 1$
$8 = |1-r_1||1-r_2|\cdots|1-r_7|$
$|1 - r_k| = 2\sin\frac {k\pi}{8}$ (highlighted in the figure for the case $k=3$)
$8 = 2^2\sin\frac {\pi}{8}\sin\frac {2\pi}{8}\sin\frac {5\pi}{8}\cdots\sin\frac {7\pi}{8}\\
\sin\frac {\pi}{8}\sin\frac {2\pi}{8}\cdots\sin\frac {7\pi}{8} = 2^{-4}$
Not a whole lot changes when we consider $x^{32} - 1$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
How many vertices so that the number of pentagons equals the number of triangles? Question: A set of points is chosen on the circumference of a circle so that the
number of different triangles with all three vertices among the points is equal to the
number of pentagons with all five vertices in the set. How many points are there?
Solution
$$\frac{x!}{3!(x-3)!}=\frac{x!}{5!(x-5)!}\\
\frac{x(x-1)(x-2)(x-3)!}{3!(x-3)!}=\frac{x(x-1)(x-2)(x-3)(x-4)(x-5)!}{5!(x-5)!}\\
\frac{x(x-1)(x-2)}{6}=\frac{x(x-1)(x-2)(x-3)(x-4)}{120}\\
6x(x-1)(x-2)(x-3)(x-4)=120x(x-1)(x-2)$$
Hence, $x=1, x=1, x=2, x=-1, x=8$. Since a pentagon has $5$ vertices, the only plausible solution is 8. I couldn't find the solution online, just making sure.
|
Yes, this reasoning is correct. (Of course, $(n - 5)!$ is not defined for integers $\leq 4$, so we can discard those solutions outright for that reason, too.)
Alternatively, if there are $3 + 5 = 8$ vertices, then we can define a natural bijection $$\{\textrm{triangles}\} \leftrightarrow \{\textrm{pentagons}\}$$ taking a triangle, throwing out its vertices, and forming a pentagon from the remaining five, so, e.g., mapping the blue triangle to the red pentagon below.
This reflects the symmetry ${8 \choose 3} = {8 \choose 5}$, which is a special case of the general identity ${n \choose k} = {n \choose {n - k}}$.
|
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"url": "https://math.stackexchange.com/questions/3130991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
solve $\frac{\sqrt{5}(\cos \theta - \sin \theta)}{3\sqrt{2}}=\tan \theta$ I came across a question from another forum -
find the $x$ in the following diagram:
I managed to deduce an equation from the following diagram:
which is:
$\dfrac{\sqrt{5}(\cos \theta - \sin \theta)}{3\sqrt{2}}=\tan \theta$
and I know the answer (from WolframAlpha) is:
$\cos \theta= \dfrac{3}{\sqrt{10}}$
but I'm not able to deduce the answer myself, any ideas?
By the way, $x=2\sqrt{5}$, which can be easily deduced by:
$\dfrac{3\sqrt{2}}{x}=\cos\theta$
I also tried to solve the original question geometrically:
Somehow, I managed to figure out that $y=3$ in the above diagram, but I can't prove it either.
|
If we start with this diagram:
we can deduce that:
$$
\overline{CE}=\sqrt{(k+1)^2+1}
\\
\overline{DE}=\frac{1}{k+1}\sqrt{(k+1)^2+1}
\\
\frac{\overline{DE}}{\overline{BC}}
=\frac{\frac{1}{k+1}\sqrt{(k+1)^2+1}}{k}
=\frac{\sqrt{(k+1)^2+1}}{k(k+1)}
=\frac{\sqrt{5}}{3\sqrt{2}}
\\
5k^2(k+1)^2=18(k+1)^2+18
\\
5k^4+10k^3-13k^2-36k-36=0
\\
(k-2)(5k^3+20k^2+27k+18)=0
$$
hence we have only one positive solution $k=2$.
Now we can deduce any segment length on the diagram from this solution.
|
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"url": "https://math.stackexchange.com/questions/3132010",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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|
Calculate the sum of fractionals
Let $n \gt 1$ an integer. Calculate the sum: $$\sum_{1 \le p \lt q \le
n} \frac 1 {pq} $$ where $p, q$ are co-prime such that $p + q > n$.
Calculating the sum for several small $n$ value I found out that the sum is always $\frac 1 2$.
Now, I'm trying to prove the sum is $\frac 1 2$ using induction by $n$. Suppose it's true for all values less or equals to $n$, trying to prove it for $n + 1$.
$$\sum_{1 \le p \lt q \le n +1,p+q>n+1} \frac 1 {pq} = \sum_{1 \le p \lt q \le n} \frac 1 {pq} + \sum_{1 \le p \lt q = n +1} \frac 1 {pq} = \sum_{1 \le p \lt q \le n} \frac 1 {pq} + \frac 1 {n+1} \sum_{1 \le p \lt n +1} \frac 1 {p} \tag1$$
In the second sum, $p$ and $n+1$ are coprime.
$$\sum_{1 \le p \lt q \le n,p+q>n+1} \frac 1 {pq} = \sum_{1 \le p \lt q \le n, p+ q>n} \frac 1 {pq} - \sum_{1 \le p \lt q \le n, p+ q=n+1} \frac 1 {pq} = \frac 1 2 - \sum_{1 \le p \lt q \le n, p+ q=n+1} \frac 1 {pq} \tag 2$$
From (1) and (2) I have to prove that $$\frac 1 {n+1} \sum_{1 \le p \lt n +1} \frac 1 {p} = \sum_{1 \le p \lt q \le n, p+ q=n+1} \frac 1 {pq} \tag3$$
where $p,q$ are co-prime and I'm stuck here.
|
Let $$
s_n=\sum_{1\le p<q\le n\\(p,q)=1}\frac1{pq}-\sum_{p+q\le n\\p<q,(p,q)=1}\frac1{pq}=a_n-b_n.
$$ Then we have that
$$
a_{n+1}-a_n=\sum_{1\le p<q=n+1\\(p,n+1)=1}\frac{1}p\cdot\frac1{n+1}=\frac1{n+1}\sum_{1\le p<n+1\\(p,n+1)=1}\frac1{p}
$$ and $$\begin{align*}
b_{n+1}-b_n&=\sum_{p+q=n+1\\p<q, (p,q)=1}\frac1{pq}
\\&=\frac1{n+1}\sum_{p+q=n+1\\p<q, (p,q)=1}\frac1 p+\frac 1 q
\\&=\frac1{2(n+1)}\sum_{p+q=n+1\\(p,q)=1}\frac1 p+\frac 1 q
\\&=\frac1{n+1}\sum_{p+q=n+1\\(p,q)=1}\frac1 p\tag{*}
\\&=\frac1{n+1}\sum_{1\le p<n+1\\(p,n+1)=1}\frac1 p\tag{**}.
\end{align*}$$
$(*)$ : $\displaystyle \sum_{p+q=n+1, (p,q)=1}\frac1 p=\sum_{p+q=n+1, (p,q)=1}\frac1 q$ by symmetry.
$(**)$ : $(p,q)=(p,p+q)=(p,n+1)=1$ by Euclidean algorithm.
This gives $s_{n+1}-s_n = (a_{n+1}-a_n)-(b_{n+1}-b_n)=0$, hence $s_n =s_2=\frac 12$ for all $n\ge 2$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Find the number of ordered triplets satisfying $5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$ Find the number of ordered triplets $(x,y,z)$ of real numbers satisfying $$5\left(x+\frac{1}{x}\right)=12\left(y+\frac{1}{y}\right)=13\left(z+\frac{1}{z}\right)$$
and
$$xy+yz+zx=1$$
My try:
Letting:
$$\left(x+\frac{1}{x}\right)=\frac{k}{5}\tag{1}$$
$$\left(y+\frac{1}{y}\right)=\frac{k}{12}\tag{2}$$
$$\left(z+\frac{1}{z}\right)=\frac{k}{13}\tag{3}$$
By $AM-GM$
$$\left(z+\frac{1}{z}\right)^2 \ge 4$$
So $$k^2 \ge 676$$
Adding $(1),(2),(3)$ we get:
$$x+y+z+\frac{xy+yz+zx}{xyz}=\frac{281k}{780}$$
$$x+y+z+\frac{1}{xyz}=\frac{281k}{780}$$
Any clue from here?
|
Let $\left(x;y;z\right)\rightarrow \left(\tan \alpha ;\tan \beta ;\tan \gamma \right)\left(0<\alpha ,\beta ,\gamma <90^o\right)$
Then we have
\begin{align}
\begin{cases}
5\left(\tan \alpha +\frac{1}{\tan \alpha }\right)=12\left(\tan \beta +\frac{1}{\tan \beta }\right)=13\left(\tan \gamma +\frac{1}{\tan \gamma }\right) (1) \\
\tan \alpha \tan \beta +\tan \gamma \tan \beta +\tan \alpha \tan \gamma =1 (2)
\end{cases}
\end{align}
Note that: $$\left(1\right)\Leftrightarrow 5\left(\frac{\tan ^2\alpha +1}{\tan \alpha }\right)=12\left(\frac{\tan ^2\beta +1}{\tan \beta }\right)=13\left(\frac{\tan ^2\gamma +1}{\tan \gamma }\right)$$
$$\Leftrightarrow \frac{5}{\sin 2\alpha }=\frac{12}{\sin 2\beta }=\frac{13}{\sin 2\gamma }$$
And $\left(2\right)\Rightarrow \cot \gamma =\tan \left(\alpha +\beta \right)$
$$\Rightarrow \tan \left(\frac{\pi }{2}-\gamma \right)=\tan \left(\alpha +\beta \right)\Leftrightarrow \alpha +\beta +\gamma =\frac{\pi }{2}$$
Can you solve it now ? :)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Confusion in this algebraic limit approaching infinity
Question : Evaluate $\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}}$.
My working:
$$\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}}$$
Dividing and transforming each fraction by $x^2$in the numerator and denominator.
$$=\lim_{x \to \infty}{\frac{\frac{1}{x^2}}{\frac{1}{x^2} +1} +...+\frac{\frac{1}{x}}{\frac{1}{x} + 1}}$$
Using algebra of limits we get,
$$0+0+...+0 = 0$$
But when solving this using Sandwich Theorem I get,
$$\text{Let }\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}} = g(x)$$
$$g(x) < \frac{1}{1+x^2} + \frac{2}{2 + x^2}+...+\frac{x}{1+x^2}$$
$$g(x) < \frac{x(x+1)}{2(1 + x^2)}$$
$$g(x) > \frac{1}{x+x^2} + \frac{2}{x+x^2} +...+\frac{x}{x+x^2}$$
$$g(x) > \frac{x(x+1)}{2(x+x^2)} \to g(x) > \frac{1}{2} $$
$$\text{So, we get } \frac{1}{2}<g(x)<\frac{x(x+1)}{2(1 + x^2)}$$
Applying limits,
$$\lim_{x \to \infty}{\frac{x^2+x}{2(1+x^2)}} = 1/2$$
$$\text{Finally, we get } \frac{1}{2}<g(x)<\frac{1}{2}$$
By Sandwich theorem, $\lim_{x \to \infty}{\frac{1}{1+x^2} + \frac{2}{2+x^2} + \frac{3}{3+x^2}+...+\frac{x}{x + x^2}} =\frac{1}{2}$
So, my question is Why am I getting two different answers when worked out differently.
If there is any error or misconception in my working, please correct me.
|
I may be totally wrong but looking at the top formula, it seems that you are rather considering
$$S_n=\sum_{i=1}^n \frac i{i+n^2}=n \left(n H_{n^2}-n H_{n^2+n}+1\right)$$ where appear harmonic numbers.
Using the asymptotics
$$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
p^2}+O\left(\frac{1}{p^4}\right)$$ Replacing and continuing with Taylor expansions for large value of $n$, you get
$$S_n=\frac{1}{2}+\frac{1}{6 n}-\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Help finding the derivative of $f(x) = \cos{\left(\sqrt{e^{x^5} \sin{x}}\right)}$ I am trying to find the derivative of $f(x) = \cos(\sqrt{(e^{x^5} \sin(x)})$.
I keep getting the wrong answer, and I'm not sure what I'm doing wrong.
$$\frac{d}{dx} e^{x^5} = e^{x^5} \cdot 5x^4$$
$$\frac{d}{dx} \sin(x) = \cos(x)$$
$$\frac{d}{dx} (e^{x ^5} \cdot \sin(x)) = [(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]$$
$$\frac{d}{dx} \sqrt{x} = \frac{1}{2\sqrt{x}}$$
$$\frac{d}{dx} \sqrt{(e^{x^5} \sin(x))} = \frac{[(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]}{2 \sqrt{e^{x^5} \sin(x)}}$$
Therefore, since $\frac{d}{dx} \cos(x) = -\sin(x)$, I have
$$ f'(x) = -\sin(\sqrt{e^{x^5} \sin(x)}) \cdot \frac{[(e^{x^5} \cdot 5x^4) \cdot \sin(x)] + [\cos(x) \cdot e^{x^5}]}{2 \sqrt{e^{x^5} \sin(x)}}$$
However, the website I'm using, "WeBWorK", says this is incorrect.
|
This is one case where logarithmic differentiation can make life easier.
$$
\frac{d}{dx}\left[\cos\left(\sqrt{e^{x^5}\,\sin(x)}\right)\right]
=-\sin{\left(\sqrt{e^{x^5}\,\sin(x)}\right)}\,\,\frac{d}{dx}\left[\sqrt{e^{x^5}\,\sin(x)}\right]$$
Let
$$f=\sqrt{e^{x^5}\,\sin(x)}\implies \log(f)=\frac 12 x^5 +\frac 12 \log(\sin(x))$$
$$\frac {f'}f=\frac{1}{2} \left(5 x^4+\cot (x)\right)\implies f'=\frac{1}{2} \left(5 x^4+\cot (x)\right)\sqrt{e^{x^5}\,\sin(x)}$$
$$\frac{d}{dx}\left[\cos\left(\sqrt{e^{x^5}\,\sin(x)}\right)\right]=-\frac{1}{2}\sin{\left(\sqrt{e^{x^5}\,\sin(x)}\right)}\,\, \left(5 x^4+\cot (x)\right)\sqrt{e^{x^5}\,\sin(x)}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inverse of a function 2 What is the inverse of the function $f(x)=x+3[2x]+2[3x]$?
The function is one by one and has an inverse.
|
Given
$$ f(x)=x+3\lfloor2x\rfloor+2\lfloor3x\rfloor $$
Since
$$ x=\left(x-\lfloor x \rfloor\right)+ \lfloor x \rfloor$$
we have
\begin{eqnarray}
f(x)&=&x-\lfloor x \rfloor + \lfloor x \rfloor\\
&&+3\lfloor2\left(x-\lfloor x \rfloor+ \lfloor x \rfloor\right)\rfloor+2\lfloor3\left(x-\lfloor x \rfloor+ \lfloor x \rfloor\right)\rfloor\\
&=&x-\lfloor x \rfloor+3\lfloor2\left(x-\lfloor x \rfloor\right)\rfloor+2\lfloor3\left(x-\lfloor x \rfloor\right)\rfloor\\
&&+\lfloor x \rfloor+6\lfloor x \rfloor+6\lfloor x \rfloor\\
&=&f\left(x-\lfloor x \rfloor\right)+13\lfloor x \rfloor
\end{eqnarray}
On the interval $[0,1)$, the function $f$ can be defined piecewise as follows:
\begin{equation}
f(x)=\begin{cases}
x & 0\le x<\frac{1}{3}\\x+2&\frac{1}{3}\le x<\frac{1}{2}\\x+5&\frac{1}{2}\le x<\frac{2}{3}\\x+7&\frac{2}{3}\le x<1
\end{cases}
\end{equation}
Combining with equation (1) we get
\begin{eqnarray}
f(x)&=&f\left(x-\lfloor x \rfloor\right)+13\lfloor x \rfloor\\
&=&+13\lfloor x \rfloor+\begin{cases}
x-\lfloor x \rfloor & 0\le x-\lfloor x \rfloor<\frac{1}{3}\\x-\lfloor x \rfloor+2&\frac{1}{3}\le x-\lfloor x \rfloor<\frac{1}{2}\\x-\lfloor x \rfloor+5&\frac{1}{2}\le x-\lfloor x \rfloor<\frac{2}{3}\\x-\lfloor x \rfloor+7&\frac{2}{3}\le x-\lfloor x \rfloor<1\end{cases}\\
&=&\begin{cases}
x+12\lfloor x \rfloor & 0\le x-\lfloor x \rfloor<\frac{1}{3}\\x+12\lfloor x \rfloor+2&\frac{1}{3}\le x-\lfloor x \rfloor<\frac{1}{2}\\x+12\lfloor x \rfloor+5&\frac{1}{2}\le x-\lfloor x \rfloor<\frac{2}{3}\\x+12\lfloor x \rfloor+7&\frac{2}{3}\le x-\lfloor x \rfloor<1\end{cases}\\
\end{eqnarray}
So for $n\in\mathbb{Z}$ we may define $f$ on the interval $[n,n+1)$ as follows:
\begin{equation}
f(x)=
\begin{cases}
x+12n & n+0\le x<n+\frac{1}{3}\\
x+12n+2&n+\frac{1}{3}\le x<n+\frac{1}{2}\\
x+12n+5&n+\frac{1}{2}\le x<n+\frac{2}{3}\\
x+12n+7&n+\frac{2}{3}\le x<n+1\end{cases}
\end{equation}
So the graph of the function consists of collection of straight line segments of slope 1. Each segment can be easily inverted.
Let us invert the first segment:
\begin{eqnarray}
y=x+12n&\text{ for }&n\le x<n+\frac{1}{3}\\
x=y+12n&\text{ for }&n\le y<n+\frac{1}{3}\\
y=x-12n&\text{ for }&n\le y<n+\frac{1}{3}\\
y=x-12n&\text{ for }&13n\le x<13n+\frac{1}{3}\\
\end{eqnarray}
Applying the same process to each of the four parts give the inverse function.
For each integer $n$ and for each $x$ in the specified interval, $f^{-1}(x)$ is defined as follows:
\begin{equation}
f^{-1}(x)=\begin{cases}
x-12n&\text{ for }13n+0\le x<13n+\frac{1}{3}\\
x-12n-2&\text{ for }13n+\frac{7}{3}\le x<13n+\frac{5}{2}\\
x-12n-5&\text{ for }13n+\frac{11}{2}\le x<13n+\frac{17}{3}\\
x-12n-7&\text{ for }13n+\frac{23}{3}\le x<13n+8
\end{cases}
\end{equation}
Note that the domain of the inverse is a collection of disjoint intervals.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Developing a general expression for powers of matrices Is there a relationship for matrices that are squared?
I am trying to determine two possible matrices $P$ where
$$P^2=\begin{bmatrix}0.6&0.4\\0.4&0.6\end{bmatrix}$$
I know that $P$ has to be a $2×2$ matrix and that $P$ will also be symmetric.
Thanks
|
If you assume that $$P=\begin{pmatrix}a&b\\b&c\end{pmatrix}$$
then you compute $$P^2=\begin{pmatrix}a^2+b^2&ab+bc\\ab+bc&b^2+c^2\end{pmatrix}=\begin{pmatrix}0.6&0.4\\0.4&0.6\end{pmatrix}$$
From this, you get the system of equations $$a^2+b^2=0.6,~ b(a+c)=0.4,~ b^2+c^2=0.6$$
Subtracting the third from the first, we get $a^2-c^2=0$, or $a=\pm c$. $a=-c$ is inconsistent with the second equation, so we must have $c=a$, and now two equations: $$a^2+b^2=0.6, 2ab=0.4$$
Adding, we get $a^2+2ab+b^2=(a+b)^2=1$, so $a+b=\pm 1$. Also, subtracting we get $a^2-2ab+b^2=(a-b)^2=0.2$, so $a-b=\pm \sqrt{0.2}$. Hence, there are four cases, corresponding to $$a+b=1, a-b=\sqrt{0.2}$$ $$a+b=-1, a-b=\sqrt{0.2}$$ $$a+b=1, a-b=-\sqrt{0.2}$$ $$a+b=-1, a-b=-\sqrt{0.2}$$
The first gives $a\approx 0.72, b\approx 0.28$. The second gives $a\approx -0.28, b\approx -0.72$. The third gives $a\approx 0.28, b\approx 0.72$. The fourth gives $a\approx -0.72, b\approx -0.28$. Hence there are just four answers, two of which are the negatives of the other two.
|
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"timestamp": "2023-03-29T00:00:00",
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|
minimum value of $2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$ Let $\alpha,\beta$ be real numbers ; find the minimum value of
$2\cos \alpha\sin \beta+3\sin \alpha\sin \beta+4\cos \beta$
What I tried :
$\bigg|4\cos \beta+(2\cos \alpha+3\sin \alpha)\sin \beta\bigg|^2 \leq 4^2+(2\cos \alpha+3\sin \alpha)^2$
How do I solve it ? Help me please
|
By C-S twice we obtain:
$$2\cos\alpha\sin\beta+3\sin\alpha\sin\beta+4\cos\beta=$$
$$=\sin\beta(2\cos\alpha+3\sin\alpha)+4\cos\beta\geq$$
$$\geq-\sqrt{(\sin^2\beta+\cos^2\beta)((2\cos\alpha+3\sin\alpha)^2+16)}\geq$$
$$\geq-\sqrt{\left(\sqrt{2^2+3^2)(\cos^2\alpha+\sin^2\alpha}\right)^2+16}=-\sqrt{29}.$$
The equality occurs for $(\cos\alpha,\sin\alpha)||(2,3)$ and $(\sin\beta,\cos\beta)||(2\cos\alpha+3\sin\alpha,4),$
which says that we got a minimal value.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
A simpler method to prove $\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1?$ This is the way I think about it:
$
1 = \log_aa = \log_bb = \log_cc \\~\\
\textbf{Using the ‘change of base rule':} \\
\log_{a}{b} = \frac{\log_{b}{b}}{\log_{b}{a}}, \hspace{0.3cm} \log_{b}{c} = \frac{\log_{c}{c}}{\log_{c}{b}}, \hspace{0.3cm} \log_{c}{a} = \frac{\log_{a}{a}}{\log_{a}{c}}\\~\\
\rightarrow \log_{b}{a} = \frac{\log_{b}{b}}{\log_{a}{b}}, \hspace{0.3cm} \log_{c}{b} = \frac{\log_{c}{c}}{\log_{b}{c}}, \hspace{0.3cm} \log_{a}{c} = \frac{\log_{a}{a}}{\log_{c}{a}}\\~\\
\rightarrow \large\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}\rightarrow\\~\\
\rightarrow\frac{\log_{b}{b}}{\log_{a}{b}} \times \frac{\log_{c}{c}}{\log_{b}{c}} \times \frac{\log_{a}{a}}{\log_{c}{a}} \overset? = 1\\
\rightarrow \frac{1}{\log_{a}{b} \times \log_{b}{c} \times \log_{c}{a}} \overset? = 1\\
\rightarrow \log_{a}{b} \times \log_{b}{c} \times \log_{c}{a} = 1\\
\rightarrow \log_{a}{b} \times \log_{b}{c} \times \log_{c}{a} \overset? = \log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}\\
\rightarrow \frac{\log_{a}{b} \times \log_{b}{c} \times \log_{c}{a}}{\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c}} = 1 \\
\rightarrow \frac{\log_{a}{b}}{\log_{b}{a}} \times \frac{\log_{b}{c}}{\log_{c}{b}} \times \frac{\log_{c}{a}}{\log_{a}{c}} = 1\\~\\
\small \text{------ Using the ‘change of base rule' again ------} \\~\\
\Large \therefore \hspace{0.2cm}\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1
$
Any other methods?
Thanks in advance.
|
You may show it directly: let
$$x=\log_a(c)\log_c(b)\log_b(a),$$
hence
\begin{align}a^{\color{red}{x}}&=a^{\log_a(c)\log_c(b)\log_b(a)}\\
&=\left(\left(a^{\log_a(c)}\right)^{\log_c(b)}\right)^{\log_b(a)}\\
&=(c^{\log_c(b)})^{\log_b(a)}\\
&=b^{\log_b(a)}\\
&=a\\
&=a^{\color{red}{1}}.
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
How is $\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $ equivalent to $\frac{ y dx + xdy - zdz}{0}=\frac{ xdx - ydy -zdz}{0}$? In the book of PDE by Kumar, it is given that
However, I couldn't figure out how is
$$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $$
is equivalent to
$$\frac{ y dx + xdy - zdz}{0 } = \frac{ x dx - y dy -z dz}{ 0} .$$
|
$$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } $$
Use the well known basic property of the fractions :
$$\text{if}\quad \frac{A}{B}=\frac{C}{D} \quad \text{then}\quad \frac{A}{B}=\frac{C}{D}=\frac{c_1A+c_2C}{c_1B+c_2D}$$
$c_1,c_2$are arbitrary constants (not both nul}.
This property is valid for more fractions :
$$\frac{A}{B}=\frac{C}{D}=\frac{E}{F}=\frac{c_1A+c_2C+c_3E}{c_1B+c_2D+c_3F}$$
In the case of Eq.$(1)$ with $c_1=y\quad;\quad c_2=x \quad;\quad c_3=-z$ :
$$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } =\frac{ydx+xdy-zdz}{yz(x+y)+xz(x-y)-z(x^2+y^2)}=\frac{ydx+xdy-zdz}{0}$$
This implies $ydx+xdy-zdz=0$.
On the same way, with $c_1=x\quad;\quad c_2=-y \quad;\quad c_3=-z$ :
$$\frac{dx }{z(x+y) } = \frac{dy}{z(x-y) } = \frac{dz }{x^2 + y^2 } =\frac{xdx-ydy-zdz}{xz(x+y)-yz(x-y)-z(x^2+y^2)}=\frac{xdx-ydy-zdz}{0}$$
This implies $xdx-ydy-zdz=0$.
|
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|
What is wrong with this proof of $3\arcsin x$? We know that
\begin{align}
2\arcsin x&= \arcsin \left(2x\sqrt{1-x^2}\right) \tag{1}\\
\arcsin x + \arcsin y &= \arcsin \left[x\sqrt{1-y^2}+y\sqrt{1-x^2}\right] \tag{2}\\
3\arcsin x &= \arcsin x + 2\arcsin x \tag{3}
\end{align}
Thus $x=x, y=2x\sqrt{1-x^2}$
using ($1$), ($2$) and ($3$):
\begin{align}
3\arcsin x &= \arcsin \left[x\sqrt{1-2x\sqrt{1-x^2}^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{1-4x^2(1-x^2)}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{1-2(2x^2)+(2x^2)^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x\sqrt{(2x^2-1)^2}+ 2x(1-x^2)\right]\\
&= \arcsin \left[x|2x^2-1|+ 2x(1-x^2)\right]
\end{align}
If $2x^2-1$ is positive, then $|2x^2-1|$ is $2x^2 -1$.
If $2x^2-1$ is negative, then $|2x^2-1|$ is $-2x^2+1$.
Range of $x$ is $-1\leq x \leq 1 \implies 0\leq x^2 \leq 1 \implies 0\leq 2x^2 \leq 2$.
For $x\in\left(\frac{-1}{\sqrt2}, \frac{+1}{\sqrt2}\right)$, then $2x^2-1$ is negative.
For $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}\ , 1\right)$, then $2x^2-1$ is positive.
Thus for $x\in\left( \frac{-1}{\sqrt2}, \frac{+1}{\sqrt2}\right)$
\begin{align}
3\arcsin x &= \arcsin [x|2x^2-1|+ 2x(1-x^2)]\\
&= \arcsin \left[-2x^3 +x+ 2x(1-x^2)\right]\\
&= \arcsin [3x - 4x^3]
\end{align}
Thus for $x\in\left(-1, \frac{-1}{\sqrt2}\right) \cup \left(\frac{+1}{\sqrt2}, 1\right)$
\begin{align}
3\arcsin x &= \arcsin [x|2x^2-1|+ 2x(1-x^2)]\\
&= \arcsin [2x^3- x+2x-2x^3]\\
&= \arcsin[x]
\end{align}
But clearly, $3\arcsin x = \arcsin[3x-4x^3]$.
So what is wrong whith this proof?
|
For all positive $x$ (the case of negative $x$ is symmetric), $$\sin3x=3\sin x-4\sin^3x.$$
So with $x=\arcsin t$ we have
$$\sin(3\arcsin t)=3t-4t^3.$$
This allows us to write
$$3\arcsin t=\arcsin(3t-4t^3)\lor3\arcsin t=\pi-\arcsin(3t-4t^3).$$
As the range of the arc sine is $\left[0,\dfrac\pi2\right]$, the first identity holds up to $\arcsin t=\dfrac\pi6$, i.e. for $t\in\left[0,\dfrac12\right]$, then comes the second identity, for $t\in\left[\dfrac12,1\right]$.
$$3\arcsin t=\begin{cases}
t\le-\dfrac12&\to-\pi-\arcsin(3t-4t^3),
\\-\dfrac12\le t\le\dfrac12&\to\arcsin(3t-4t^3),
\\t\le\dfrac12&\to\pi-\arcsin(3t-4t^3).\end{cases}$$
|
{
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|
Find the polynomials $f(x)$ and $g(x)$ with integer coefficients such that the following equation is true.
(a) Find the polynomials $f(x)$ and $g(x)$ with integer coefficients such that
$$
\dfrac{f(\sqrt 3 + \sqrt 5)}{g(\sqrt 3 + \sqrt 5)} = \sqrt 3
$$
(b) Find $f$ and $g$ so that $$
\dfrac{f(\sqrt 3 + \sqrt 5)}{g(\sqrt 3 + \sqrt 5)} = \sqrt 5
$$
I can't think of a way to solve this problem.
|
Just take $f(x)=x^2-2$, $g(x)=2x$ for part (a) and $f(x)=x^2+2$, $g(x)=2x$ for part (b). It's not hard to check that $f, g$ are satysfying all conditions.
However, how we can find such polynomials? Firstly, you can try to find $f, g$ in form $f(x)=a_1x^2+b_1x+c_1$ and $g(x)=a_2x^2+b_2x+c_2$ by solving corresponding system of linear equations.
There is also another way. Note that the condition of part (a) is equivalent to $h(\sqrt{3}+\sqrt{5})=0$, where $h(x)=f(x)-\sqrt{3}g(x)$. Now, define new function $\varphi$ which maps $a+b\sqrt{3}+c\sqrt{5}+d\sqrt{15}$ to $a+b\sqrt{3}-c\sqrt{5}-d\sqrt{15}$, where $a,b,c,d\in\mathbb{Z}$ (in other words, $\varphi$ change $\sqrt{5}$ into $-\sqrt{5}$). Since $f$ and $g$ has integer coefficients we have
$$
f(\varphi(x))=\varphi(f(x)), g(\varphi(x))=\varphi(g(x))
$$
if $x$ has form $a+b\sqrt{3}+c\sqrt{5}+d\sqrt{15}$, where $a,b,c,d\in\mathbb{Z}$.
Hence,
$$
h(\sqrt{3}+\sqrt{5})=f(\sqrt{3}+\sqrt{5})-\sqrt{3}g(\sqrt{3}+\sqrt{5})=\varphi(f(\sqrt{3}-\sqrt{5}))-\sqrt{3}\varphi(g(\sqrt{3}-\sqrt{5})).
$$
Moreover,
$$
\varphi(f(\sqrt{3}-\sqrt{5}))-\sqrt{3}\varphi(g(\sqrt{3}-\sqrt{5}))=\varphi(f(\sqrt{3}-\sqrt{5})-\sqrt{3}g(\sqrt{3}-\sqrt{5}))=\varphi(h(\sqrt{3}-\sqrt{5})).
$$
Thus, $\varphi(h(\sqrt{3}-\sqrt{5}))=0$, so $h(\sqrt{3}-\sqrt{5})=0$. It means that $\sqrt{3}\pm\sqrt{5}$ are roots of $h(x)$, so $h(x)$ is divisible by $(x-\sqrt{3}+\sqrt{5})(x-\sqrt{3}-\sqrt{5})=(x-\sqrt{3})^2-5=x^2-2\sqrt{3}x-2$.
At this point we can simply take $h(x)=x^2-2\sqrt{3}x-2$. Since $h(x)=f(x)-\sqrt{3}g(x)$ we obtain $f(x)=x^2-2$ and $g(x)=2x$. Analgously, we can get example for part (b).
|
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|
Finding $\lim_{n \to \infty} \int_0^n \frac{dx}{1+n^2\cos^2x}$ Find $$\lim_{n \to \infty} \int_0^n \frac{dx}{1+n^2\cos^2x}$$
I tried:
*
*mean value theorem.
*variable change with $ \tan x = t $ but I need to avoid the points which are not in the domain of $\tan$ and it's complicated.
|
Result
Defining
$$f(n) = \int_0^n \frac{1}{1+n^2 \cos(x)^2} \,dx\tag{1}$$
we find
$$\lim_{n\to\infty}f(n)=1$$
Derivation
We split the integration region $(0,n)$ into the intervals $(0,\pi/2) ,( k \pi -\pi/2, k\pi +\pi/2)$ with $k = 1, 2, ...,k_{max}= n/\pi$.
Hence letting
$x=k \pi + \xi$ so that $\cos(x) = (-1)^k \cos(\xi)$ and $\cos(x)^2 = \cos(\xi)^2$
we obtain
$$f(n) = \int_0^{\frac{\pi}{2}} \frac{1}{1+n^2 \cos(x)^2} \,dx + \\
\sum_{k=1}^{k_{max}}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{1+n^2 \cos(\xi)^2}\,d\xi\tag{2}$$
Notice that the integrals (except possible for the last one) are independent of $k$.
Now
$$\left(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{1}{1+n^2 \cos(\xi)^2}\,d\xi\right)/.(\tan(\xi)\to t) \\
= \left(\int_{-\infty}^{\infty}\frac{1}{1+n^2+t^2}\,dt\right)/.(\frac{t}{\sqrt{1+n^2}}\to s)\\
=\frac{1}{\sqrt{1+n^2}}\int_{-\infty}^{+\infty}\frac{1}{1+s^2}\,ds
=\pi \sqrt{\frac{1}{n^2+1}}\tag{3}$$
Since $k_{max}=\frac{n}{\pi}$ we get
$$f(n) \simeq \pi \sqrt{\frac{1}{n^2+1}} k_{max} = \pi \sqrt{\frac{1}{n^2+1}}\frac{n}{\pi} $$
$$\lim_{n\to\infty}f(n) = \lim_{n\to\infty}n\sqrt{\frac{1}{n^2+1}}=1$$
Hence the limit is $1$.
Remark: Notice that the first integral in $(2)$ and possibly the last one which might not fit exactly into the equally spaced pattern both are $O\left(\frac{1}{n}\right)$ and hence vanish in the Limit
Discussion
With the same technique you can show that the limit of the similar integral
$$g(n) = \int_0^n \frac{1}{1+ n \cos(x)}\,dx\tag{4}$$
does not exist as $g(n)$ diverges logarithmically.
|
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|
How to prove that function $f(p)= \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$ is less than 0.3 for all $p$? How to prove that function $f(p)= \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$ is less than 0.3 for all $p$?
Let's suppose that we have a function defined as follows:
$f(p) = \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$
where $p$ is a prime number and $[...]$ are Iverson brackets.
Here are the first few values of $f(p)$:
$f(3) = \frac{1}{(3^2 - 1)} \times \frac {(3^2-3)}{3}$
$f(3) = \frac{1}{8} \times \frac{6}{3}$
$f(3) = 0.25$
$f(5) = \frac{1}{(5^2 - 1)} \times [ \frac{(3^2-3)}{3} +\frac{(5^2-3)}{5} ]$
$f(5) = \frac{1}{24} \times [ \frac{6}{3} +\frac{22}{5} ]$
$f(5) = 0.26667$
$f(7) = \frac{1}{(7^2 - 1)} \times [ \frac{(3^2-3)}{3} +\frac{(5^2-3)}{5} +\frac{(7^2-3)}{7}]$
$f(7) = \frac{1}{48} \times [ \frac{6}{3} +\frac{22}{5} +\frac{46}{7}]$
$f(7) = 0.273214286$
$f(11) = \frac{1}{(11^2 - 1)} \times [ \frac{(3^2-3)}{3} +\frac{(5^2-3)}{5} +\frac{(7^2-3)}{7}+\frac{(11^2-3)}{11}]$
$f(11) = \frac{1}{120} \times [ \frac{6}{3} +\frac{22}{5} +\frac{46}{7}+\frac{118}{11}]$
$f(11) = 0.198679654$
How to prove that $f(p)$ is always less than 0.3 for all prime numbers $p$?
I wrote a program to calculate $f(p)$ for the first $1000$ primes up to $p = 7,927$ and graphically it appears to approach $0$ with a maximum value of $0.2732$ at $p=7$. From the graph, it looks like it should be easy, but for some reason, I cannot figure it out.
Notice that $f(3) < f(5) < f(7)$, but $f(7) > f(11)$ so a proof by induction may not work.
|
The result is true for $p = 3$, $5$, and $7$, so assume that $p = 2n+1$ for $n \ge 4$.
Note that all the primes $q$ occurring in the sum are odd. Thus
$$
\begin{aligned}
f(p) = f(2n+1) = & \ \frac{1}{4n^2 + 4n} \sum_{q=3}^{p, \text{ with $q$ prime}} \frac{q^2 - 3}{q}\\
< & \ \frac{1}{4(n^2 + n)} \sum_{q=3}^{p, \text{ with $q$ prime}} q \\
< & \ \frac{1}{4(n^2 + n)} \sum_{q = 3}^{p, \text{with $q$ odd}}
q \\
= & \ \frac{1}{4(n^2 + n)} \sum_{j=1}^{n} (2j+1)\\
= & \ \frac{n^2 +2n}{4(n^2 + n)} \\
= & \ \frac{3}{10} - \frac{(n-4)}{20(n+1)} \\
\le & \ \frac{3}{10}. \end{aligned}$$
In reality, $f(p) \rightarrow 0$, since
$$f(p) \le \frac{1}{p^2} \sum_{q \le p} q
\le \frac{1}{p^2} \sum_{q \le p} p
= \frac{1}{p^2} \cdot p \pi(p) \sim 1/\log p.$$
|
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|
Finding a basis of the annihilator of a subspace Let $V\subset\Bbb R^4$ be the subspace spanned by $e_1+e_2+e_3+e_4$ and $e_1+2e_2+3e_3+4e_4$. Find a basis of the subspace $V^\circ$ in the dual space $(\Bbb R)^*$.
My attempt: Let $a = e_1+e_2+e_3+e_4$ and $b = e_1+2e_2+3e_3+4e_4$, then $a$ and $b$ are lin. independent by looking at the row reducing matrix $ \color{white}{a}^{\begin{pmatrix}
1 & 1 & 1 & 1\\
1 & 2 & 3 & 4
\end{pmatrix}= \begin{pmatrix}
1 & 1 & 1 & 1\\
0 & 1 & 2 & 3
\end{pmatrix}}$. Thus we can extend the set $\{a,b\}$ to form the basis of $\Bbb R^4$. Example: $\{a,b,e_3,e_4\}$ forms a basis for $\Bbb R^4$ as $e_3,e_4$ are lin. independent of $a,b$. Let $\{f_1,f_2,f_3,f_4\}$ be the dual basis and we can say that $\{f_3,f_4\}$ forms the basis of $V^{\circ}$ because $f_3(a) = f_3(b) = f_4(a) = f_4(b) = 0$? Appreciate a hint.
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Did it the long way here. To find the basis of $V^{\circ}$, we want linear functionals $f\in V^{*}$ s.t. $f(v)=0$, where $v = a_1v_1+a_2v_2$ ; $a_1,a_2\in \mathbb{R}$.
$$v_1 = e_1+e_2+e_3+e_4 = \begin{pmatrix}
1 \\
1 \\
1 \\
1 \\
\end{pmatrix} \& \text{ }v_2 = e_1+2e_2+3e_3+4e_4 = \begin{pmatrix}
1 \\
2 \\
3 \\
4 \\
\end{pmatrix}
$$
$$ f(v) = 0 \iff f(a_1v_1+a_2v_2)= a_1f(v_1)+a_2f(v_2) = 0 \iff f(v_1)=f(v_2)=0$$
For any $f\in V^{*}, \exists w\in R^{4}$, s.t. $f(v)= w.v$
$$f(v_1) = w.v_1 = 0\iff w_1+w_2+w_3+w_4 = 0$$ Similarly,
$$f(v_2) = w.v_2 = 0\iff w_1+2w_2+3w_3+4w_4 = 0 $$
$$ w = \begin{pmatrix}
w_1 \\
w_2 \\
w_3 \\
w_4 \\
\end{pmatrix} = \begin{pmatrix}
t+2p \\
-2t-3p \\
t \\
p \\
\end{pmatrix} = t\begin{pmatrix}
1 \\
-2 \\
1 \\
0 \\
\end{pmatrix} + p\begin{pmatrix}
2 \\
-3 \\
0 \\
1 \\
\end{pmatrix}$$
So $V^{\circ} = \{tf_1+pf_2; t,p\in \mathbb{R}\}$, where $f_1(x,y,z,w) = x-2y+z$ and $f_2(x,y,z,w) = 2x-3y+w$. Thus the basis of $V^{\circ}$ is $\{f_1,f_2\}$
|
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|
Does this method always produce the minimal polynomial over $\Bbb{Q}$? I have used it a few times and so far I have not seen an instance in which it did not give me the minimal polynomial over $\Bbb{Q}$.
The method is the following: For an element $a \notin \Bbb{Q}$ set $a=X$ and use arithmetic operations until we get an expression that looks like $p=0$ where $p \in \Bbb{Q}[X]$.
Example with $p=\sqrt2$:
$$X=\sqrt2 \iff X^2=2 \iff X^2-2=0$$
Example with $\sqrt3+\sqrt5$:
$$X=\sqrt3+\sqrt5 \iff X^2=8+2\sqrt{15}\iff (X^2-8)^2=60 \iff X^4 - 16
X^2 + 4=0$$
Example with $\sqrt3+ \sqrt[3]2$:
$$X=\sqrt3+ \sqrt[3]2 \iff (X-\sqrt3)^3=2 \iff X^3- 3 \sqrt3 X^2 + 9 X- 3 \sqrt3=2 \iff$$
$$(X^3 +9X -2)^2 = (3 \sqrt3 X^2+3 \sqrt3)^2 \iff$$
$$X^6 + 18 X^4 - 4 X^3 + 81 X^2 - 36 X + 4 = 27 X^4 + 54 X^2 + 27 \iff$$
$$X^6 - 9 X^4 - 4 X^3 + 27 X^2 - 36 X - 23=0$$
It is clear that the resulting polynomial will be normed and have $a$ as its root, but the last two examples are not obviously irreducible. Can I rely on this method to give me the minimal polynomials?
|
For example, suppose you set $X = \sqrt{3 + 2\sqrt{2}}$, and you try to find the minimal polynomial using $X^2 = 3 + 2\sqrt{2}$; $X^2 - 3 = 2\sqrt{2}$; $(X^2 - 3)^2 = 8 = X^4 - 6X^2 + 9$; $X^4 - 6X^2 + 1 = 0$. Your method would thus find $X^4 - 6X^2 + 1$ as the minimal polynomial.
However, secretly, $X$ is actually equal to $1 + \sqrt{2}$, which has minimal polynomial $X^2 - 2X - 1$. In fact, $X^4 - 6X^2 + 1$ is not irreducible over $\mathbb{Q}$, since $X^4 - 6X^2 + 1 = (X^2 - 1)^2 - 4X^2 = (X^2 - 2X - 1) (X^2 + 2X - 1)$.
|
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|
Using Cardano's method to find an algebraic equation whose root is $\sqrt{2} +\sqrt[3]{3}$ $\sqrt{2} +\sqrt[3]{3}$ solution of an algebraic equation $ a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{0}=0 $
How to find this equation?
I tried Cardano's method, noting that
$$\sqrt{2} +\sqrt[3]{3} = \sqrt[3]{\sqrt{8}} +\sqrt[3]{3}$$
It means
$$\begin{align}
-\frac{q}{2}+\sqrt{ \frac{q^{2}}{4}+\frac{p^{3}}{27}}&=3 \\[4pt]
-\frac{q}{2}-\sqrt{ \frac{q^{2}}{4}+\frac{p^{3}}{27}}&=\sqrt{8}
\end{align}$$
but this system doesn't have natural solutions.
Maybe it has rational $p$ and $q$?
|
In the same spirit as Awe Kumar Jha, let $x=\sqrt 2 + \sqrt[3]{3}$, that is to say
$$x-\sqrt 2 =\sqrt[3]{3}\implies (x-\sqrt 2) ^3=3$$ Expand and group terms to get
$$x^3-3 \sqrt{2} x^2+6 x-(2 \sqrt{2}+3)=0$$ Use Cardano method; it will show that this is the only real root of the equation.
|
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|
Sum of all values that satisfy $\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$.
What is the sum of all values of $x$ that satisfy the equation
$\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$?
I start off by cross multiplying.
$$x^2=x^{x-3+\frac{4}{x}}$$
Taking the square root of both sides gives me:
$$x=\pm x^{\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}}$$
I start with the positive side first:
$$1=\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}$$$$x^2-5x+4=0$$$$x=4, 1$$
Now, I start with the negative side:
$$x=-x^{\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}}$$$$x^\frac{x^2-5x+4}{2x}=-1$$
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
$$x^{x^2-5x+4}=1$$
Now, taking the log of both sides, I have
$$(x^2-5x+4)\cdot\operatorname{log}(x)=0$$
Dividing both sides by $\operatorname{log}(x)$, I get $x^2-5x+4=0$ again, which should give me $x=4, 1$ again. So thus, the answer should be $4+1=5$.
However, this is wrong, as the answer key says that the answer is $-1+1+4=\boxed{4}$. I have checked $-1$ as a solution on all of my equations, and all of them work. How can I derive the $-1$ out from my equations? Or, if my approach or any of my equations are wrong, how do I get the answer $x=-1$?
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The reasoning here is faulty:
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
For one thing, what if $x$ is not an integer, and $2x$ is not even?
Starting from $x^\frac{x^2-5x+4}{2x}=-1$, you could absolute value both sides: $$\lvert x\rvert^\frac{x^2-5x+4}{2x}=1$$
and then take the logarithm: $$\frac{x^2-5x+4}{2x}\ln\lvert x\rvert=0$$ from which you conclude either $x=4$, $x=1$, or $\lvert x\rvert=1$. Check that $4$, $1$, and $-1$ satisfy the original equation, and you have a complete solution.
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|
Help on $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{H_n^3}{n}$ From this post:
The sum: $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{H_n^3}{n+1}=-\frac{9}{8}\zeta(3)\ln(2)+\frac{\pi^4}{288}-\frac{\ln^4(2)}{4}+\frac{\pi^2}{8}\ln^2(2)$$
by moving the $n+1$ back to $n$
What is the sum of: $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{H_n^3}{n}?$$
I guess it could be easy, but I cannot see it.
Can anyone please with this sum? I needs it to solve one of my current problem I am working on.
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Here might be a start. Observe that$$\begin{align*}\sum\limits_{n\geq1}(-1)^{n+1}\frac {H_n^3}{n+1} & =\frac {H_1^3}2-\frac {H_2^3}3+\frac {H_3^3}4-\cdots\\ & =\sum\limits_{n\geq2}(-1)^n\frac {H_{n-1}^3}n\end{align*}$$
Now use the fact that $H_n=H_{n-1}+\tfrac 1n$. Therefore, the original sum becomes$$\begin{align*}\sum\limits_{n\geq1}(-1)^{n+1}\frac {H_n^3}{n+1} & =\sum\limits_{n\geq2}(-1)^n\frac {1}n\left[H_n-\frac 1n\right]^3\\ & =\sum\limits_{n\geq2}(-1)^n\frac 1n\left[H_n^3-\frac {3H_n^2}n+\frac {3H_n}{n^2}-\frac 1{n^3}\right]\\ & =\sum\limits_{n\geq2}(-1)^{n}\frac {H_n^3}{n}-3\sum\limits_{n\geq2}(-1)^n\frac {H_n^2}{n^2}+3\sum\limits_{n\geq2}(-1)^{n}\frac {H_n}{n^3}+\frac {7\pi^4}{720}\\ & =-\sum\limits_{n\geq2}(-1)^{n+1}\frac {H_n^3}n+3\sum\limits_{n\geq2}(-1)^{n+1}\frac {H_n^2}{n^2}+3\sum\limits_{n\geq2}(-1)^{n}\frac {H_n}{n^3}+\frac {7\pi^4}{720}\end{align*}$$
The second sum can be found with the help of this question. Reindex the identity to start the sum from two to get
$$\sum\limits_{n\geq2}(-1)^n\frac {H_n}{n^3}\color{red}{=1+2\operatorname{Li}_4\left(\frac 12\right)+\frac {7\log 2}4\zeta(3)-\frac {11\pi^4}{360}+\frac {\log^42}{12}-\frac {\pi^2\log^22}{12}}$$
In a similar fashion, some digging on MSE gives this question which might be a big help in tackling the second sum:$$\sum\limits_{n\geq2}(-1)^n\frac {H_n^2}{n^2}$$
|
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|
Compute $I = \int_{0}^{\infty} \frac{8}{\gamma^{4}} x^{3} e^{-\frac{2}{\gamma^{2}}x^{2}} e^{jxt} dx$ I want to evaluate the following integral:
$$I = \int_{0}^{\infty} \frac{8}{\gamma^{4}} x^{3} e^{-\frac{2}{\gamma^{2}}x^{2}} e^{jxt} dx $$
Where $\gamma \in \mathbb{R}$ and $j = \sqrt{-1}$.
The first thing I do is let:
$$ u = \frac{2}{\gamma} x \quad \Rightarrow \quad du = \frac{2}{\gamma} dx$$
$$ x = \frac{\gamma}{2} u \quad \Rightarrow \quad dx = \frac{\gamma}{2} du$$
After substituting and cancellation, we get:
$$ I = \frac{1}{2} \int_{0}^{\infty} u^{3} e^{-\frac{1}{2} u^{2} + j \frac{1}{2} \gamma tu} du$$
We complete the square, giving:
$$ I = \frac{1}{2} \int_{0}^{\infty} u^{3} e^{-\frac{1}{2} (u - j\frac{1}{2} \gamma t)^{2} - \frac{1}{8} \gamma^{2} t^{2}} du$$
$$ I = \frac{1}{2} e^{-\frac{1}{8} \gamma^{2} t^{2}} \int_{0}^{\infty} u^{3} e^{-\frac{1}{2} (u - j\frac{1}{2} \gamma t)^{2}} du$$
This is where I am unsure of what to do. My approach is as follows:
We want to solve:
$$ I = \frac{1}{2}e^{-\frac{1}{8} \gamma^{2} t^{2}} \underset{R \rightarrow \infty}{lim} \int_{0}^{R} u^{3} e^{-\frac{1}{2} (u - j\frac{1}{2} \gamma t)^{2}} du$$
We let:
$$ z = u - j \frac{1}{2} \gamma t \quad \Rightarrow \quad dy = du$$
$$ u = y + j\frac{1}{2} \gamma t \quad \Rightarrow \quad du = dy$$
$$ b = R - j\frac{1}{2} \gamma t \quad a = -j \frac{1}{2} \gamma t $$
Therefore:
$$ I = \frac{1}{2}e^{-\frac{1}{8} \gamma^{2} t^{2}} \underset{R \rightarrow \infty}{lim} \int_{ -j \frac{1}{2} \gamma t}^{R - j \frac{1}{2} \gamma t} (y- j \frac{1}{2} \gamma t)^{3} e^{-\frac{1}{2} y^{2}} dy$$
We create a contour $\Gamma = C_{1} + C_{2} + C_{3} + C_{4} = [-j \frac{1}{2} \gamma t, R-j \frac{1}{2} \gamma t] + [R-j \frac{1}{2} \gamma t, R] + [R, 0] + [0, -j \frac{1}{2} \gamma t]$
Since $ \underset{y \rightarrow \infty}{lim}\frac{(y - j \frac{1}{2} \gamma t)^{3}}{e^{\frac{1}{2}y^{2}}} = 0 $, the integral over the vertical contour $C_{3}$ is zero.
However I still need to evaluate the following integrals:
$$ I_{1} = \underset{R \rightarrow \infty}{lim} \int_{0}^{R} (y- j \frac{1}{2} \gamma t)^{3} e^{-\frac{1}{2} y^{2}} dy $$
$$ I_{2} = \int_{-j \frac{1}{2} \gamma t}^{0} (y- j \frac{1}{2} \gamma t)^{3} e^{-\frac{1}{2} y^{2}} dy $$
This is where I am getting stuck.
Is this the right approach? Am I on the right track? How can I solve those two integrals? How can I solve my original problem?
Thanks!
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Hint: find $$\int_0^\infty e^{-2 x^2/\gamma^2} e^{jxt}\; dx $$
and differentiate $3$ times with respect to $t$.
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|
Pólya Enumeration Theorem and Chemical Compounds How many different chemical compounds can be made by attaching H, CH3, or OH radicals to each of the carbon atoms in the benzene ring of Figure 1? (Assume that all of the C–C bonds in the ring are equivalent.)
Figure 1:
I understand the overall method of the Pólya-Burnside Method of Enumeration but I don't understand how to apply it to a chemical compound problem like this. Any help would be great, thank you in advance!
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Here is a solution via the Polya Enumeration Theorem. The problem is the same as determining the number of ways of coloring the vertices of a hexagon with three colors.
The symmetry group of the hexagon is the dihedral group $D_6$. The cycle index of the group is
$$Z=\frac{1}{12} (x_1^6 + 4 x_2^3 +2 x_3^2 +2 x_6 + 3x_1^2 x_2^2)$$
The figure inventory for three colors (or three radicals) is $x+y+z$. Substituting the figure inventory into the cycle index yields
$$\frac{1}{12}[ (x+y+z)^6 + 4(x^2+y^2+z^2)^3 +2(x^3+y^3+z^3)^2 + 2(x^6+y^6+z^6) + 3(x+y+z)^2(x^2+y^2+z^2)^2 ]$$
If we were to expand this polynomial we could find all the possible ways of coloring the vertices with specific combinations of colors, such as one red vertex, two blue vertices, and three green vertices. But all we really want to know for this problem is the total number of ways with all possible combinations of three colors, which is the sum of the coefficients, and we can find that sum by setting $x=y=z=1$, with result
$$\frac{1}{12}(3^6 + 4 \cdot 3^3 + 2 \cdot 3^2 + 2 \cdot 3 + 3 \cdot 3^2 \cdot 3^2) = \boxed{92}$$
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Linearizing the trigonometric functions or: Squaring the circle by Fourier transformation It's an easy exercise to approximate the cosine and the sine function by a piecewise linear function on the unit interval $[0,1]$. Let $\tau = 2\pi$.
Let
$$\boxed{\cos_\bigcirc(x) = \cos(\tau x)\\\sin_\bigcirc(x) = \sin(\tau x)}$$
and compare this to
$$\boxed{\cos_\square(x) = \begin{cases}
+1 & \text{ for } \frac{0}{8} \leq x \leq \frac{1}{8} \\
+2 - 8x & \text{ for } \frac{1}{8} \leq x \leq \frac{3}{8} \\
-1 & \text{ for } \frac{3}{8} \leq x \leq \frac{5}{8} \\
-6 + 8x & \text{ for } \frac{5}{8} \leq x \leq \frac{7}{8} \\
+1 & \text{ for } \frac{7}{8} \leq x \leq \frac{8}{8} \\
\end{cases}
\\ \\\sin_\square(x) = \begin{cases}
+0 + 8x & \text{ for } \frac{0}{8} \leq x \leq \frac{1}{8} \\
+1 & \text{ for } \frac{1}{8} \leq x \leq \frac{3}{8} \\
+4 - 8x & \text{ for } \frac{3}{8} \leq x \leq \frac{5}{8} \\
-1 & \text{ for } \frac{5}{8} \leq x \leq \frac{7}{8} \\
-8 + 8x & \text{ for } \frac{7}{8} \leq x \leq \frac{8}{8} \\
\end{cases}}$$
These are the plots:
Observations
*
*It may come as a surprise or not that while $\cos_\bigcirc(x)$ and $\sin_\bigcirc(x)$ yield the unit circle by $x_\bigcirc(x) = \cos_\bigcirc(x)$ and $y_\bigcirc(x) = \sin_\bigcirc(x)$, the functions $\cos_\square(x)$ and $\sin_\square(x)$ yield the unit square by $x_\square(x) = \cos_\square(x)$ and
$y_\square(x) = \sin_\square(x)$. By "unit square" I mean the square with "radius" $1$, not with side length $1$. The unit circle in the incircle of this square:
*
*While the circumference of the unit circle is just the "number of the circle" $\tau$, the circumference of the unit square is $8$ (the "number of the square"). Note how $8$ is used in the definition of $\cos_\square(x)$ and $\sin_\square(x)$, compared to $\tau$ in the definition of $\cos_\bigcirc(x)$ and $\sin_\bigcirc(x)$. Note further that not by accident $8 \approx \tau$ and that not by accident $8 = 3^2 - 1^2$:
*
*There are natural generalizations of piecewise linear approximations of the cosine and the sine for arbitrary regular $n$-polygons which will approximate the true functions better and better as the polygons will approximate the circle better and better.
*The tangens $\tan_\bigcirc(x) = \tan(\tau x)$ is very well approximated already by
$\tan_\square(x) = \frac{\sin_\square(x)}{\cos_\square(x)}$:
*
*The functions $\cos_\square(x)$ and $\sin_\square(x)$ can be used to parametrize the square spiral analoguous to how $\cos_\bigcirc(x)$ and $\sin_\bigcirc(x)$ can be used to parametrize the Archimedean spiral.
Questions
*
*Under which name and in which contexts have the functions
$\cos_\square(x)$ and $\sin_\square(x)$ been studied before?
*Is there an elegant and/or more compact way to write the equations for
$\cos_\square(x)$ and $\sin_\square(x)$ in one closed expression, e.g.
by using the Heaviside function?
*Is there a closed formula for the Fourier transform of $\cos_\square(x)$ and $\sin_\square(x)$ (the "Fourier transform of the quadrature of the circle")?
This is how the Fourier transforms of $\cos_\square(x)$ and $\sin_\square(x)$ look like:
Summary
Thanks to user J.M. we now know the Fourier coefficients $\widehat{\cos}_\square(k)$
$$\boxed{\widehat{\cos}_\square(k) = 8 \cdot \begin{cases}
\ \ \ \ \ 0 & \text{ for } k \equiv 0 \mod 2 \\
+(\pi k)^{-2} & \text{ for } k \equiv 1 \mod 8 \text{ or } k \equiv 7 \mod 8\\
-(\pi k)^{-2} & \text{ for } k \equiv 3 \mod 8 \text{ or } k \equiv 5 \mod 8\\
\end{cases}}$$
and $\widehat{\sin}_\square(k)$ accordingly.
Now can perform the squaring of the circle by these steps:
*
*Consider $\cos_\bigcirc(x)$ and $\sin_\bigcirc(x)$ which "draw" the unit circle (with diameter $2$).
*Consider the functions $\cos_\square(x)$ and $\sin_\square(x)$ defined by
$$\cos_\square(x) := \sum_{k=0}^{\infty} \widehat{\cos}_\square(k)\cos_\bigcirc(kx)\\
\sin_\square(x) := \sum_{k=0}^{\infty} \widehat{\sin}_\square(k)\sin_\bigcirc(kx)$$
*The functions $\cos_\square(x)$ and $\sin_\square(x)$ "draw" the unit square (with diameter $2$).
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The work has been done by user J.M. (who seems to be a mathematician of sorts), so let me just put it in order.
First let me rewrite the definition of $\cos_\square(x)$ a bit:
$$\cos_\square(x) = \begin{cases}
+1 & \text{ for } -1 \leq 8x \leq 1 \\
+2 - 8x & \text{ for }\ \ \ \ \ 1 \leq 8x \leq 3 \\
-1 & \text{ for }\ \ \ \ \ 3 \leq 8x \leq 5 \\
-6 + 8x & \text{ for }\ \ \ \ \ 5 \leq 8x \leq7 \\
\end{cases}$$
The unit square has diagonal (= "diameter") $2\sqrt{2}$, so let's normalize $\cos_\square(x)$ by a factor $\frac{1}{\sqrt{2}}$ so the unit square has diameter $2$ (just like the unit circle has diameter $2$):
$$\boxed{\cos_\square(x) = \frac{1}{\sqrt{2}}\cdot\begin{cases}
+1 & \text{ for } -1 \leq 8x \leq 1 \\
+2 - 8x & \text{ for }\ \ \ \ \ 1 \leq 8x \leq 3 \\
-1 & \text{ for }\ \ \ \ \ 3 \leq 8x \leq 5 \\
-6 + 8x & \text{ for }\ \ \ \ \ 5 \leq 8x \leq7 \\
\end{cases}}$$
User J.M.'s expression for the Fourier coefficients $\widehat{\cos}_\square(k)$ can then be written in this form:
$$\boxed{\widehat{\cos}_\square(k) = 8 \cdot \begin{cases}
\ \ \ \ \ 0 & \text{ for } k \equiv 0 \mod 2 \\
+(\pi k)^{-2} & \text{ for } k \equiv 1 \mod 8 \text{ or } k \equiv 7 \mod 8\\
-(\pi k)^{-2} & \text{ for } k \equiv 3 \mod 8 \text{ or } k \equiv 5 \mod 8\\
\end{cases}}$$
|
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|
induction proof: number of compositions of a positive number $n$ is $2^{n-1}$ So my problem was to find the formula for the number of composition for a positive number $n$. Then prove its validity. I found the formula which was $2^{n-1}$. Having trouble with the induction.
$$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$
When $n=5$ $$ 1+2+4+(2^{5-2}+1)=2^{5-1}$$
$$ 1+2+4+(2^{3}+1)=2^{4}$$
$$ 1+2+4+(8+1)=16$$
$$ 16=16$$
Assume that $n=k$
$$2^0+2^1+2^2+...+(2^{k-2}+1)=2^{k-1}$$
Now show for $k+1$
$$2^0+2^1+2^2+...+(2^{k-1}+1)=2^{k}$$
Don't know what to do after. Help is appreciated.
$$2^0+2^1+2^2+...+(2^k)+(2^{k-1}+1)=2^{k}$$
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You actually don't need induction to prove this. There's a telescoping effect.
$$Claim:2^0+2^1+2^2+...+(2^{n-2}+1)=2^{n-1}$$
Carry the one.
$$(1 + 2^0)+2^1+2^2 + ... + 2^{n-2}=2^{n-1}$$
$1 = 2^0$
$$(2^0 + 2^0)+2^1+2^2 + ... + 2^{n-2}=2^{n-1}$$
Collapse the parentheses
$$(2^1 + 2^1) + 2^2 + ... + 2^{n-2}=2^{n-1}$$
Collapse the parentheses.
$$(2^2 + 2^2) + ... + 2^{n-2}=2^{n-1}$$
Keep doing this until
$$2^{n-2} + 2^{n-2} = 2^{n-1}$$.
|
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|
show this inequality to $\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $ Let $a,b$ and $c$ be positive real numbers. Prove that
$$\sum_{cyc} \frac {a^3b}{(3a+2b)^3} \ge \sum_{cyc} \frac {a^2bc}{(2a+2b+c)^3} $$
This problem is from Iran 3rd round-2017-Algebra final exam-P3,Now I can't find this inequality have solve it,maybe it seem can use integral to solve it?
my attempts:
I took $p=3a+2b,$ $2a+2b+c$. $k=3$ and I wanted to use this integral:
$$\dfrac{1}{p^k}=\dfrac{1}{\Gamma(k)}\int_{0}^{+\infty}t^{k-1}e^{-pt}dt$$
but I don't see how it helps.
|
The proof of my inequality.
let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\tfrac{a^3b}{(2a+3b)^3}+\tfrac{b^3c}{(2b+2c)^3}+\tfrac{c^3a}{(2c+3a)^3}\geq\tfrac{a^2bc}{(2a+2b+c)^3}+\tfrac{b^2ca}{(2b+2c+a)^3}+\tfrac{c^2ab}{(2c+2a+b)^3}.$$
Indeed, by Holder and AM-GM we obtain:
$$\sum_{cyc}\tfrac{a^3b}{(2a+3b)^3}=\sum_{cyc}\tfrac{\left(4(2a+3b)+(2b+3c)+2(2c+3a)\right)^3\left(\tfrac{4a^3b}{(2a+3b)^3}+\tfrac{b^3c}{(2b+3c)^3}+\tfrac{2c^3a}{(2c+3a)^3}\right)}{2401(2a+2b+c)^3}\geq$$
$$\geq\sum_{cyc}\frac{\left(4\sqrt[4]{a^3b}+\sqrt[4]{b^3c}+2\sqrt[4]{c^3a}\right)^4}{2401(2a+2b+c)^3}\geq\sum_{cyc}\frac{\left(7\sqrt[28]{a^{12+2}b^{4+3}c^{1+6}}\right)^4}{2401(2a+2b+c)^3}=\sum_{cyc}\frac{a^2bc}{(2a+2b+c)^3}.$$
|
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|
Why does the Lagrange multiplier $\lambda$ change when the equality constraint is scaled? Consider the problem
$$\begin{array}{ll} \text{maximize} & x^2+y^2 \\ \text{subject to} & \dfrac{x^2}{25} + \dfrac{y^2}{9} = 1\end{array}$$
Solving this using the Lagrange multiplier method, I get
$$x = \pm5, \qquad y = 0, \qquad \lambda = 25$$
However, if I rewrite the constraint as $9x^2+25y^2=225$, I get a different value of $\lambda$, namely, $\lambda = \frac 19$. I am unclear about why this should happen: the constraint is exactly the same, only rewritten after cross multiplication — why should that affect the value of the multiplier? What am I missing?
|
Let $f(x,y)=x^2+y^2+\lambda(9x^2+25y^2-225).$
Thus, from $$\frac{\partial f}{\partial x}=2x+18\lambda x=0$$ and
$$\frac{\partial f}{\partial y}=2y+50\lambda y=0$$ we obtain two possibilities: $\lambda=-\frac{1}{9}$ or $\lambda=-\frac{1}{25}.$
The second gives a minimal value, wile the first gives a maximal value:
$$f(x,y)=x^2+y^2-\frac{1}{9}(9x^2+25y^2-225)=25-\frac{16}{9}y^2\leq25,$$ where the equality occurs for $y=0.$
If we consider $f(x,y)=x^2+y^2+\lambda\left(\frac{x^2}{25}+\frac{y^2}{9}-1\right)$ so we'll get $\lambda=-25$
and we'll get the same answer of course.
It happens because $-\frac{1}{9}\cdot225=-25$ and
$$x^2+y^2-\frac{1}{9}(9x^2+25y^2-225)=x^2+y^2-25\left(\frac{x^2}{25}+\frac{y^2}{9}-1\right).$$
|
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|
prove this inequality with $x^3+y^3+z^3$
Let $x,y,z>0$. Show that
$$x^3+y^3+z^3+2(xy^2+yz^2+zx^2)\ge 3(x^2y+y^2z+z^2x).$$
I have a solution by using $y=x+a,z=x+a+b$; my question is: can this be solved using simple methods (such as AM-GM, C-S, and so on)?
|
We need to prove that $$\sum_{cyc}(x^3-3x^2y+2x^2z)\geq0$$ or
$$\sum_{cyc}(2x^3-6x^2y+4x^2z)\geq0$$ or
$$\sum_{cyc}(2x^3-x^2y-x^2z)\geq5\sum_{cyc}(x^2y-x^2z)$$ or
$$\sum_{cyc}(x+y)(x-y)^2\geq5(x-y)(x-z)(y-z).$$
Since our inequality is cyclic, we can assume that $x=\max\{x,y,z\}$ and since for $x\geq z\geq y$ we have $$(x-y)(x-z)(y-z)\leq0,$$ it's enough to prove our inequality for $x\geq y\geq z$.
Now, easy to see that if we'll change $(x,y,z)$ on $(x-h,y-h,z-h)$ for all $0\leq h<z$
so $(x-y)^2,$ $(x-z)^2$, $(y-z)^2$ and $(x-y)(x-z)(y-z)$ are not changed and the left side of the inequality
$$\sum_{cyc}(x+y)(x-y)^2\geq5(x-y)(x-z)(y-z)$$ decreases.
Thus, it's enough to prove our inequality for $h\rightarrow z^-$,
which says that it's enough to prove our inequality for $z\rightarrow0^+$.
Id est, we need to prove that
$$x^3-3x^2y+2xy^2+y^3\geq0,$$ which we can prove by AM-GM:
$$x^3-3x^2y+2xy^2+y^3=\frac{1}{2}(8\cdot\frac{x^3}{4}+6\cdot\frac{2xy^2}{3}+2y^3)-3x^2y\geq$$
$$\geq\frac{1}{2}\cdot15\sqrt[15]{\left(\frac{x^3}{4}\right)^8\left(\frac{2xy^2}{3}\right)^6\cdot2y^3}-3x^2y=\left(\frac{15}{2\sqrt[5]{72}}-3\right)x^2y>0.$$
|
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|
How to solve this with mathematical induction? $$\sum_{i=0}^n \frac{i}{2^i} = 2 - \frac{n+2}{2^n} $$
Let's skip the check, since when n = 1, I have $\frac{1}{2} = \frac{1}{2}$
What i will next do ? What for expression i may receive ?
$$\sum_{i=0}^{n+1}\frac{i}{2^i} = 2 - \frac{n+2}{2^n} $$
My trying :
$$ \biggl(\sum_{i=0}^{n+1}\frac{i}{2^i} = 2 - \frac{n+2}{2^n}\biggl) = \frac{n+1}{2^{n+1}} $$
or
$$ \biggl(\sum_{i=0}^{n+1}\frac{i}{2^i} = 2 - \frac{n+2}{2^n}\biggl) = \frac{2^{n+2}-(n+1)+2}{2^{n+1}} $$
Which answer should I use? And how will I decide next?
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For the inductive step you have to show, that $\sum_{i=0}^{n+1} \frac{i}{2^i}=2-\frac{(n+1)+2}{2^{n+1}}$.
We have to use the inductive claim and go like this:
$\begin{align}\sum_{i=0}^{n+1} \frac{i}{2^i}\\&=\color{red}{\sum_{i=0}^{n} \frac{i}{2^i}}+\frac{n+1}{2^{n+1}}\\&=\color{red}{2-\frac{n+2}{2^n}}+\frac{n+1}{2^{n+1}}\\&=2+\frac{-2(n+2)+n+1}{2^{n+1}}\\&=2+\frac{-2n-4+n+1}{2^{n+1}}\\&=2+\frac{-n-3}{2^{n+1}}\\&=2-\frac{n+3}{2^{n+1}}\end{align}$
|
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|
Discontinuity of Hypergeometric function along the branch cut I am trying to evaluate an expression involving the hypergeometric function evaluated near its (principal) branch cut discontinuity, which is placed on the real line from $1$ to infinity.
For $x>1$, DLMF gives the following expression for the difference (or ''imaginary part'') of the regularized hypergeometric function $\mathbf F(a,b;c;z)={}_2F_1(a,b;c;z)/\Gamma(c)$ across the branch cut,
$$
\mathbf F(a,b;c;x+i0) - \mathbf F(a,b;c;x-i0)$$
$$ = \tfrac{i2\pi}{\Gamma(a)\Gamma(b)}(x-1)^{c-a-b}\mathbf F(c-a,c-b;c-a-b+1;1-x)\,.
$$
However, I am interested in the analogous expression for the sum (or, the ''real part''), again for $x>1$,
$$
\mathbf F(a,b;c;x+i0)+\mathbf F(a,b;c;x-i0)=\,?
$$
I tried plugging specific values of this into mathematica, in particular for integer $D>2$, which yields
$$
\mathbf F\left(\tfrac{D-1}{2},\tfrac{D}{2};\tfrac{D+1}{2};x+i0\right)
+
\mathbf F\left(\tfrac{D-1}{2},\tfrac{D}{2};\tfrac{D+1}{2};x-i0\right)$$
$$=
(1+(-1)^D)\mathbf F\left(\tfrac{D-1}{2},\tfrac{D}{2};\tfrac{D+1}{2};x\right)
+
\tfrac{2i^{D+1}\pi^{3/2}x^{-\tfrac{D-1}{2}}}{\cos\left(\tfrac{D\pi}{2}\right)\Gamma\left(\frac{D-3}{2}\right)\Gamma\left(\tfrac{D-1}{2}\right)\Gamma\left(\tfrac{D}{2}\right)}\,.
$$
This makes perfect sense when $D$ is odd, since then the right-hand side reads
$$
2\sqrt{\pi}(-1)^{\frac{D-1}{2}} \frac{x^{-\frac{D-1}{2}}}{\Gamma\left(\frac{D-1}{2}\right)\Gamma\left(\frac{D}{2}\right)}\,,
$$
however, it doesn't make much sense to me for even $D$, since then the right-hand side seems to still have a nonzero imaginary part.
Is there a general expression analogous to that provided for the difference by DLMF, but expressing the sum across the branch cut?
|
It turns out that to answer the question it is sufficient to employ in a suitable way some other formulas already present in the DLMF.
A possibility is to use the ''inversion formula''
$$
\frac{\sin(\pi(b-a))}{\pi}\,
\mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};z\right)
=
\frac{(-z)^{-a}}{\Gamma(b)\Gamma(c-a)}
\,\mathbf F\left(\begin{array}{} a,a-c+1\\ \ \ a-b+1\end{array};\frac{1}{z}\right)
-
\frac{(-z)^{-b}}{\Gamma(a)\Gamma(c-b)}
\,\mathbf F\left(\begin{array}{} b,b-c+1\\ \ \ b-a+1\end{array};\frac{1}{z}\right)\,.
$$
Substituting $z=x+i0$ with $x>1$ and noting that, if we place the branch cut of the logarithm along the negative real axis, $(-(x+i0))^{-a}=e^{i\pi a}\,x^{-a}$, we have
$$
\frac{\sin(\pi(b-a))}{\pi}\,
\mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};x+i0\right)
=
\frac{e^{i\pi a}\,x^{-a}}{\Gamma(b)\Gamma(c-a)}
\,\mathbf F\left(\begin{array}{} a,a-c+1\\ \ \ a-b+1\end{array};\frac{1}{x}\right)
-
\frac{e^{i\pi b}\,x^{-b}}{\Gamma(a)\Gamma(c-b)}
\,\mathbf F\left(\begin{array}{} b,b-c+1\\ \ \ b-a+1\end{array};\frac{1}{x}\right)\,.
$$
This gives both real and imaginary parts, provided $b-a$ is not an integer (otherwise, the left-hand side vanishes).
For the specific values of $a$, $b$ and $c$ chosen in the example, $b-a=1/2$, so this formula applies and
$$
\Re \mathbf F\left(\begin{array}{} \frac{D-1}{2},\frac{D}{2}\\\ \ \frac{D+1}{2}\end{array};x+i0\right)
=\begin{cases}
(-)^{\frac{D}{2}+1}\sqrt{\pi} x^{-\frac{D}{2}}/\Gamma(\frac{D-1}{2}) &\text{even }D\\
(-)^{\frac{D-1}{2}}\pi x^{\frac{1-D}{2}}/\Gamma(\frac{D}{2}) &\text{odd }D\,.
\end{cases}
$$
Another workaround is provided by the ''reflection formula''
$$
\frac{\sin(\pi(c-b-a))}{\pi}\,
\mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};z\right)
=
\frac{1}{\Gamma(c-a)\Gamma(c-b)}
\,\mathbf F\left(\begin{array}{} \ \ \ \ \ \ \ \ a,b\\a+b-c+1\end{array};1-z\right)
-
\frac{(1-z)^{c-b-a}}{\Gamma(a)\Gamma(b)}
\,\mathbf F\left(\begin{array}{} \ \ c-a,c-b\\c-a-b+1\end{array};1-z\right)\,,
$$
hence
$$
\frac{\sin(\pi(c-b-a))}{\pi}\,
\mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};x+i0\right)
=
\frac{1}{\Gamma(c-a)\Gamma(c-b)}
\,\mathbf F\left(\begin{array}{} \ \ \ \ \ \ \ \ a,b\\a+b-c+1\end{array};1-x\right)
-
\frac{e^{i\pi(a+b-c)}(x-1)^{c-b-a}}{\Gamma(a)\Gamma(b)}
\,\mathbf F\left(\begin{array}{} \ \ c-a,c-b\\c-a-b+1\end{array};1-x\right)\,,
$$
for $x>1$. As a byproduct, we get the formula for the imaginary part quoted in the question,
$$
\Im \mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};x+i0\right) = \frac{\pi(x-1)^{x-a-b}}{\Gamma(a)\Gamma(b)}\,\mathbf F\left(\begin{array}{} \ \ c-a,c-b\\c-a-b+1\end{array};1-x\right)\,,
$$
together with
$$
\frac{\sin(\pi(c-b-a))}{\pi}\,\Re \mathbf F\left(\begin{array}{} a,b\\\ \ c\end{array};x+i0\right)
=
\frac{1}{\Gamma(c-a)\Gamma(c-b)}
\,\mathbf F\left(\begin{array}{} \ \ \ \ \ \ \ \ a,b\\a+b-c+1\end{array};1-x\right)
-
\frac{\cos(\pi(a+b-c))(x-1)^{c-b-a}}{\Gamma(a)\Gamma(b)}
\,\mathbf F\left(\begin{array}{} \ \ c-a,c-b\\c-a-b+1\end{array};1-x\right)\,.
$$
|
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|
If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$. Find $\lim_{n \to \infty} nt_n$ If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$
Find $\lim_{n \to \infty} nt_n$
First attempt: $t_n$ is positive(grouping two terms and performing subtraction we will get it) so is $nt_n$. Now can we prove it is monotonically decreasing? If so then $\lim_{n \to \infty} nt_n=\lim_{n \to \infty}[(\frac{1}{2+1/n}-\frac{1}{2+2/n})+(\frac{1}{2+3/n}-\frac{1}{2+4/n})+\cdots +(\frac{1}{4-1/n}-\frac{1}{4})]$ and each of these terms will go to zero so is the limit.
Second attempt: I was trying to use Riemann summation $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}=
\frac{1}{2n+1}+\frac{1}{2n+2}+\frac{1}{2n+3}+\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}+\frac{1}{4n}-2[\frac{1}{2n+2}+\frac{1}{2n+4}+\cdots \frac{1}{4n}]\Rightarrow \lim \frac 1n [nt_n]=\int_0^2\frac{dx}{2+x}-\int_0^1\frac{dx}{1+x}=\ln4-\ln 2-\ln 2=0$
So $\lim t_n=0$
So what will happen with $\lim nt_n$
Edit: As I got the answer is not $0$ because of the flaw. So can we have different approaches even with Riemann Sum to have the answer?
|
Here is a
completely elementary proof
by algebraic manipulation that
$\dfrac{1}{8n}-\dfrac{1}{16n^2}
\lt t_n
\lt \dfrac1{8n}
$.
$\begin{array}\\
t_n
&=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}\\
&=\sum_{k=2n+1}^{4n} \dfrac{(-1)^{k+1}}{k}\\
&=\sum_{k=n}^{2n-1} (\dfrac1{2k+1}-\dfrac1{2k+2})\\
&=\sum_{k=n}^{2n-1} \dfrac{1}{(2k+1)(2k+2)}\\
&<\sum_{k=n}^{2n-1} \dfrac{1}{(2k)(2k+2)}\\
&=\dfrac14\sum_{k=n}^{2n-1} \dfrac{1}{k(k+1)}\\
&=\dfrac14\sum_{k=n}^{2n-1} (\dfrac1{k}-\dfrac1{k+1})\\
&=\dfrac14(\dfrac1{n}-\dfrac1{2n})\\
&=\dfrac1{8n}\\
\text{and}\\
t_n
&=\sum_{k=n}^{2n-1} \dfrac{1}{(2k+1)(2k+2)}\\
&=\dfrac14\sum_{k=n}^{2n-1} \dfrac{1}{(k+1/2)(k+1)}\\
&>\dfrac14\sum_{k=n}^{2n-1} \dfrac{1}{(k+1/2)(k+3/2)}\\
&=\dfrac14\sum_{k=n}^{2n-1} (\dfrac1{k+1/2}-\dfrac1{k+3/2})\\
&=\dfrac14(\dfrac1{n+1/2}-\dfrac1{2n-1/2})\\
&=\dfrac14\dfrac{(2n-1/2)-(n+1/2)}{(n+1/2)(2n-1/2)}\\
&=\dfrac14\dfrac{n-1}{(n+1/2)(2n-1/2)}\\
&=\dfrac18\dfrac{n-1}{(n+1/2)(n-1/4)}\\
&=\dfrac18\dfrac{n+1/2-1/2}{(n+1/2)(n-1/4)}\\
&=\dfrac18(\dfrac{n+1/2}{(n+1/2)(n-1/4)}-\dfrac{1/2}{(n+1/2)(n-1/4)})\\
&=\dfrac18(\dfrac{1}{n-1/4}-\dfrac{1}{2(n+1/2)(n-1/4)})\\
&>\dfrac18(\dfrac{1}{n}-\dfrac{1}{2(n^2+n/4-1/8)})\\
&>\dfrac{1}{8n}-\dfrac{1}{16n^2}\\
\end{array}
$
|
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|
Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$
$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3}
\text{then}\ a^5+b^5+c^5= \ ?$$
A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.
Like the video, the best I can do with this is relying on expansion formulas and substitution. As trivial a problem this is, the numerous trinomials and binomials with mixed terms makes it very, very tedious.
What is the quickest/shortest approach to this problem (meaning it doesn't need to be solved algebraically)? You don't have to type the entire solution out, I think if I'm given a good hint then I can take it from there.
|
I think to use an identity must be the shortest way:-
${ (a^5+b^5+c^5)=(a^4+b^4+c^4)(a+b+c)-(a^3+b^3+c^3)(ab+bc+ca)+abc(a^2+b^2+c^2)} $
If we square the 1st equation we get
$ {ab+bc+ca=\frac {-1}2} $
On squaring the above equation we get
$ {(ab)^2+(bc)^2+(ca)^2+abc(a+b+c)=\frac 14} $
We also know
${ (a^3+b^3+c^3)-3abc=(a+b+c)(a^2+b^2+c^2-ab+bc+ca) }$
from here we get :-
$abc=\frac 16$
Substituting the known values
We get :-
${ (ab)^2+(bc)^2+(ca)^2=\frac {-1}{12} } $
Now squaring the 2nd equation:-
${ (a^4+b^4+c^4)+2\big ((ab)^2+(bc)^2+(ca)^2)\big )=4 }$
From here we get
$(a^4+b^4+c^4)=\frac{25}6$
Now we have all the values to be substituted in the identity that i mentioned therefore you get
$ { (a^5+b^5+c^5)= (\frac{25}{6}\cdot 1) - (3 \cdot \frac {-1}2)+ (\frac 16 \cdot 2) }$
$\Rightarrow$$(a^5+b^5+c^5)=6$
|
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|
Distance Formula Problem If two vertices of an equilateral triangle are $(1, -1)$ and $(-\sqrt{3}, - \sqrt{3})$, find the coordinates of the third vertex.
Step by Step procedure to get the answer.
Take $A=(1, -1)$, $B=(-\sqrt{3}, - \sqrt{3})$.
Let the third vertex be $C=(x, y)$.
The distance d between the two points $A=(x_1, y_1)$ and $B=(x_2, y_2)$ is given by the formula
$$
d= \sqrt{ (x_2-x_1)^2 + (y_2 - y_1)^2 }
$$
So we get the distance $AB = 2 \sqrt{2}$
. Similarly we have to find the distance between BC & AC using distance formula where we get the equation in the form of x & y even though the distance is $2\sqrt{2}$ because it is equilateral triangle. But the correct solution is not able to get. Please help.
|
Equilateral triangle
$A=(1,-1)$
$B=(-\sqrt{3},-\sqrt{3})$
$C=(x,y)$
$|AB|=2\sqrt{2}$
height of an equilateral triangle:
$h=\frac{1}{2}\sqrt{3}a$
$a=|AB|$
$h=\frac{1}{2}\sqrt{3}.2\sqrt{2}=\sqrt{6}$
noting that
$|BO|=\sqrt{(0-(-\sqrt{3}))^2+(0-(-\sqrt{3}))^2}=\sqrt{6}$
O = (0,0) origin Cartesian Cordinate System
BO is foot of the altitude
O is midpoint de CA
$\frac{xC+1}{2}=0⇒xC=-1$
$\frac{yC+(-1)}{2}=0⇒yC=1$
C=(-1,1)
|
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}
|
$f$ dividing by $x + 1$ have remainder 4, when dividing with $x^2 + 1$ have remainder 2x+3. Find remainder dividing polynomial with($x+1$)($x^2+1$) Problem: The polynomial $f$ dividing by ($x + 1$) gives the remainder 4, and when dividing with ($x^2 + 1$) gives the remainder (2x+3). Determine the remainder when dividing the polynomial with ($x + 1$)($x^2 + 1$)?
My attempt:
By Polynomial remainder theorem we know that
$$f(x)=q_1(x)(x+1)+4$$
$$f(x)=q_2(x)(x^2+1)+(2x+3)$$
By putting $x=1$ we know that $f(-1)=4, f(i)=2i+3, f(-i)=-2i+3$
We want to find $r(x)$ such that: $$f(x)=(x+1)(x^2+1)q_3(x) + r(x) .$$
By applying the previous idea we know that $f(-1)=r(-1)=4$, but the same idea $r(i)=2i+3$ and $r(-i)=-2i+3$ but this is only three point and we need to determinant polynomial $r(x)$ of degree 3...
Please solve without modular arithmetic.
|
The hint.
Since $\deg((x+1)(x^2+1))=3,$ the degree of $r$ must be less than $3$.
Let $r(x)=ax^2+bx+c$ and take $x=-1$, $x=i$ and $x=-i$.
Now, solve the following system.
$$a-b+c=4,$$
$$-a+bi+c=2i+3$$ and
$$-a-bi+c=-2i+3.$$
I got $$r(x)=1.5x^2+2x+4.5.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What am I doing wrong when testing for concavity with $f(x)=5x^{2/3}-2x^{5/3}$? I can't figure out what I'm doing wrong when testing for concavity with $f(x)=5x^{2/3}-2x^{5/3}$. I can't find the right intervals for where the graph is concave-down and concave-up.
My Steps:
find $f'(x)$:
$$f'(x)=\frac{10}{3}x^{-1/3}-\frac{10}{3}x^{2/3}=\frac{10}{3\sqrt[3]{x}}-\frac{10\sqrt[3]{x^2}}{3}$$
find $f''(x)$:
$$\frac{d}{dx} 3\sqrt[3]{x} = 0 + \frac{1}{3\sqrt[3]{x^2}} \cdot 3 = \frac{1}{\sqrt[3]{x^2}}$$
$$\frac{d}{dx}\frac{10}{3\sqrt[3]{x}}=\frac{0-[\frac{1}{\sqrt[3]{x^2}} \cdot 10]}{(3\sqrt[3]{x})^2}=\frac{-10}{(3\sqrt[3]{x})^2(\sqrt[3]{x^2})}=\frac{-10}{9\sqrt[3]{x^4}}$$
$$\frac{d}{dx}10\sqrt[3]{x^2}= 0 +10 \cdot \frac{d}{dx}x^{2/3}=\frac{10}{3\sqrt[3]{x}}$$
$$\frac{d}{dx}\frac{10\sqrt[3]{x^2}}{3}=\frac{0 - [\frac{10}{3\sqrt[3]{x}} \cdot 3]}{9} = \frac{-30}{9(3\sqrt[3]{x})}=\frac{-10}{9\sqrt[3]{x}}$$
$$f''(x)=-\frac{10}{9\sqrt[3]{x^4}}-\frac{-10}{9\sqrt[3]{x}}= \frac{10}{9\sqrt[3]{x}}-\frac{10}{9\sqrt[3]{x^4}}$$
set $f''(x) = 0$ and solve for $x$.
$$\frac{10}{9\sqrt[3]{x}}-\frac{10}{9\sqrt[3]{x^4}}=0 \to x=1$$
find $f''(-1)$.
$$\frac{10}{9\sqrt[3]{(-1)}}-\frac{10}{9\sqrt[3]{(-1)^4}}=\frac{-20}{9}$$
find $f''(2)$
$$\frac{10}{9\sqrt[3]{(2)}}-\frac{10}{9\sqrt[3]{(2)^4}}=\frac{5\sqrt[3]{4}}{18}$$
So the graph should be concave-down on $(-\infty, 1)$ and concave-up on $(1, \infty)$.
However, this is wrong.
|
Keep going with fractional exponents, it's much easier:
$$
f'(x)=\frac{10}{3}x^{-1/3}-\frac{10}{3}x^{2/3}
$$
Therefore
$$
f''(x)=-\frac{10}{9}x^{-4/3}-\frac{20}{9}x^{-1/3}=-\frac{10}{9}x^{-1/3}(x^{-1}-2)
$$
This is undefined for $x=0$, and it should be easy study the sign elsewhere.
|
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|
Computing $K$-theory elements in a $C^*$ algebra $A$ Let $A$ be a unital $C^*$ algebra. Let $p,q$ be projections in $M_n(A)$. Then $[p]-[q]$ defines an element in $K_0(A)$.
Now consider the matrices, the projections,
$$
\left[ \begin{pmatrix}
1-p & 0 \\
0 & q
\end{pmatrix} \right] - \left[\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \right]
$$
does this define the same $K_0(A)$ element as $[p]-[q]$?
|
I will assume you mean
$$\left[\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \right]-\left[ \begin{pmatrix}
1-p & 0 \\
0 & q
\end{pmatrix} \right] ,
$$
since your first matrix is not a projection. Recall that
$$
\left[\begin{pmatrix} p & 0 \\ 0 & 0 \end{pmatrix} \right]=\left[\begin{pmatrix} 0 & 0 \\ 0 & p \end{pmatrix} \right]=[p].
$$
Recall also that if $r,s$ are projections with $rs=0$, then $[r]+[s]=[r+s]$. Then
$$
\left[\begin{pmatrix} 1-p & 0 \\ 0 & q \end{pmatrix} \right]
=\left[\begin{pmatrix} 1-p & 0 \\ 0 & 0 \end{pmatrix} \right]+\left[\begin{pmatrix} 0 & 0 \\ 0 & q \end{pmatrix} \right]=[1-p]+[q].
$$
So
\begin{align}
\left[\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \right]-\left[ \begin{pmatrix}
1-p & 0 \\
0 & q
\end{pmatrix} \right]
&=\left[\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \right]-\left[ \begin{pmatrix}
1-p & 0 \\
0 & 0
\end{pmatrix} \right] -\left[ \begin{pmatrix}
0 & 0 \\
0 & q
\end{pmatrix} \right] \\ \ \\
&=\left[\begin{pmatrix} p+(1-p) & 0 \\ 0 & 0 \end{pmatrix} \right]-\left[ \begin{pmatrix}
1-p & 0 \\
0 & 0
\end{pmatrix} \right] -\left[ \begin{pmatrix}
0 & 0 \\
0 & q
\end{pmatrix} \right] \\ \ \\
&=[p]+[1-p]-[1-p]-[q]\\ \ \\
&=[p]-[q].
\end{align}
|
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|
Number of real roots of $p_n(x)=1+2x+3x^2+....+(n+1)x^n$ if $n$ is an odd integer
If $n$ be an odd integer. Then find the number of real roots of the polynomial equation $p_n(x)=1+2x+3x^2+....+(n+1)x^n$
$$
p_n(x)=1+2x+3x^2+....+(n+1)x^n\\
x.p_n(x)=x+2x^2+....nx^n+(n+1)x^{n+1}\\
p(x)[1-x]=1+x+x^2+....+x^n-(n+1)x^{n+1}\\
p(x)=\frac{1+x+x^2+....+x^{n}}{1-x}-\frac{(n+1)x^{n+1}}{1-x}=\frac{x^{n+1}-1}{x-1}\frac{1}{1-x}-\frac{(n+1)x^{n+1}}{1-x}\\
=\frac{x^{n+1}-1-(n+1)(x-1)x^{n+1}}{-(x-1)^2}=\frac{x^{n+1}-1+(n+1)x^{n+1}-(n+1)x^{n+2}}{-(x-1)^2}\\
=\frac{(n+2)x^{n+1}-(n+1)x^{n+2}-1}{-(x-1)^2}=\frac{1-(n+2)x^{n+1}+(n+1)x^{n+2}}{(x-1)^2}=0\\
\implies \boxed{(n+2)x^{n+1}-(n+1)x^{n+2}=1}
$$
I think I am stuck with my attempt, how can I find the real solutions ?
|
You have found that $p(x)$ is a rational function. It is zero if and only if its numerator is zero. Its numerator is a polynomial of the form $Ax^{n+2}+Bx^{n+1}+C$, where $A,B,C$ are constants. The derivative of that numerator is of the form $Dx^{n+1}+Ex^n=x^n(Dx+E)$ for some constants $D,E$. It's easy to see how many real roots that derivative has. Then Rolle's Theorem tells you something about how many real roots the original numerator has. Can you fill in the details and draw a conclusion?
|
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|
Definite Integral of $\int_0^1\frac{dx}{\sqrt {x(1-x)}}$ We have to calculate value of the following integral :
$$\int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \qquad \qquad (2)$$
What i've done for (2) :
\begin{align}
& = \int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \\
& = \int_0^1\cfrac{dx}{\sqrt {x-x^2}} \\
& = \int_0^1\cfrac{dx}{\sqrt {(x^2-x+\frac 14)-\frac 14 }} \\
& = \int_0^1\cfrac{dx}{\sqrt {(x-\frac 12)^2-(\frac 12)^2 }} \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {(\frac 12\sec \theta)^2-(\frac 12)^2 }} I\ used\ trigonometric\ substitution \ u=a\sec \theta, by \ it's \ form \ u^2-a^2 \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {(\frac 14\sec^2 \theta)-\frac 14 }} \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {\frac 14(\sec ^2\theta-1)}} \ using \\tan^2\theta=\sec^2\theta-1 \\
& = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {\frac 12(\sqrt{\tan^2\theta) }}} \\
& = \int_0^1\sec\theta d\theta = \sec\theta \tan \theta |_0^1 \\
\end{align}
But i got problems calculating $\theta$ value, using trigonometric substitution, any help?
|
Are you sure that you got the correct antiderivative for the integral of the secant function? The correct indefinite integral of secant is $\int\sec{x}=\ln{|\tan{x}+\sec{x}|}+C$. Thus:
$$
\int_{0}^{1}\sec{\theta}\,d\theta=\ln{|\tan{\theta}+\sec{\theta}|}\bigg|_{0}^{1}.
$$
EDIT:
Also note that:
$$
x-x^2=-(x^2-x)=-\left(x^2-x+\frac{1}{4}-\frac{1}{4}\right)=\\
-\left(\left[x-\frac{1}{2}\right]^2-\frac{1}{4}\right)=
\frac{1}{4}-\left(x-\frac{1}{2}\right)^2.
$$
So, I think the substitution that you should be using would be:
$$
x=\frac{1}{2}\sin{\theta}+\frac{1}{2},\\
dx=\frac{1}{2}\cos{\theta}\,d\theta,\\
x=\frac{1}{2}\sin{\theta}+\frac{1}{2}\implies\theta=\arcsin{(2x-1)},\ -\frac{\pi}{2}\le\theta\le\frac{\pi}{2}.
$$
Putting it all together, you get the following:
$$
\int\frac{1}{\sqrt{x-x^2}}\,dx=\int\frac{1}{\sqrt{\frac{1}{4}-\left(\frac{1}{2}\sin{\theta}+\frac{1}{2}-\frac{1}{2}\right)^2}}\frac{1}{2}\cos{\theta}\,d\theta=\\
\int\frac{1}{\frac{1}{2}\sqrt{1-\sin^2{\theta}}}\frac{1}{2}\cos{\theta}\,d\theta=
\int\frac{\cos{\theta}}{|\cos{\theta}|}\,d\theta=
\int\frac{\cos{\theta}}{\cos{\theta}}\,d\theta=\\
\int\,d\theta=\theta+C=
\arcsin{(2x-1)}+C.\\
\int_0^1\frac{1}{\sqrt {x(1-x)}}\,dx=\arcsin{(2x-1)}\bigg|_0^1=\\
\arcsin{(1)}-\arcsin{(-1)}=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi.
$$
Wolfram Alpha check.
|
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|
Solution to $\sqrt{\sqrt{x + 5} + 5} = x$ There are natural numbers $a$, $b$, and $c$ such that the solution to the equation
\begin{equation*}
\sqrt{\sqrt{x + 5} + 5} = x
\end{equation*}
is $\displaystyle{\frac{a + \sqrt{b}}{c}}$. Evaluate $a + b + c$.
I am not sure where I saw this problem. My guess is that it was from a high school math competition. The solution to the equation is $\frac{1 + \sqrt{21}}{2}$. This suggests use of the quadratic formula.
The solution set to the given equation is a subset of the solution set to
\begin{equation*}
x^{2} - 5 = \sqrt{x + 5} ,
\end{equation*}
\begin{equation*}
x^{4} - 10x^{2} + 25 = x + 5
\end{equation*}
\begin{equation*}
x^{4} - 10x^{2} - x + 20 = 0 .
\end{equation*}
Using the quartic equation (or Wolfram), the solutions to this equation are computed to be
\begin{equation*}
\frac{1 \pm \sqrt{21}}{2} , \qquad \frac{-1 \pm \sqrt{17}}{2} .
\end{equation*}
|
use the following way
$$x=\sqrt{5+\sqrt{5 + x} }$$
$$x=\sqrt{5+\sqrt{5 + \sqrt{5+\sqrt{5 + \sqrt{5+\sqrt{5 + ....} }} }} } $$
or
$$x=\sqrt{5+x }$$
$$x^2-x-5=0$$
$$x=\frac{1}{2}\pm\frac{\sqrt{21}}{2}$$
now use long division to get the other roots and then check which which one satisfies the original equation
|
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"timestamp": "2023-03-29T00:00:00",
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|
Extracting coefficients from two-dimensional generating function We have the two-dimensional recurrent series $F(r+1,s+2) = F(r,s) + F(r,s+1) + F(r,s+2)$ and the boundary conditions $F(r,0)=1$, $F(0,s)=0$ for all $s>0$ and $F(0,0)=1$ and $F(r,1)=r$. This series is for all $r\geq0$ and $s\geq0$. How do we find the general expression for the coefficient?
I used the 2D generating function $G(x,y) = \sum_{r,s\geq0} F(r,s) x^r y^s$ and applied the boundary conditions to obtain the following expression for G:
$$G(x,y) = \frac{xy}{(x-1)(1-x-xy-xy^2)}$$
which I can split to obtain
$$G(x,y) = \frac{-1}{1-x} \frac{xy}{1-x-xy-xy^2}$$
and the term $\frac{-1}{1-x}$ can be expressed as $-(1+x+x^2+x^3+\dots)$. I am having problems with the other term $\frac{xy}{1-x-xy-xy^2}$.
I referred to this post which handled a bivariate generating function but unfortunately I couldn't recast $\frac{xy}{1-x-xy-xy^2}$ into a term similar to $\frac{1}{1-y(x+1)}$ which was presented there. The closest I could get was:
$$\frac{xy}{1-x-xy-xy^2} = \frac{xy}{1-x(y^2+y+1)} = \frac{xy}{1-x[(y+1)^2-y]}$$
which is again not similar to the expression $H(x,y) = \frac{xy}{1-x-y}$ in this post.
All suggestions welcome.
|
We use the coefficient of operator $[z^n]$ do denote the coefficient of $z^n$ in a series.
We obtain for $m,n\geq 1$:
\begin{align*}
\color{blue}{[x^my^n]}&\color{blue}{\frac{xy}{1-x(1+y+y^2)}}\\
&=[x^{m-1}y^{n-1}]\frac{1}{1-x(1+y+y^2)}\tag{1}\\
&=[x^{m-1}y^{n-1}]\sum_{j=0}^\infty x^j\left(1+y+y^2\right)^j\tag{2}\\
&=[y^{n-1}](1+y+y^2)^{m-1}\tag{3}\\
&=[y^{n-1}]\sum_{j=0}^{m-1}\binom{m-1}{j}y^j(1+y)^j\tag{4}\\
&=\sum_{j=0}^{\min\{m-1,n-1\}}\binom{m-1}{j}[y^{n-1-j}](1+y)^j\tag{5}\\
&\,\,\color{blue}{=\sum_{j=0}^{\min\{m-1,n-1\}}\binom{m-1}{j}\binom{j}{n-1-j}}\tag{6}\\
\end{align*}
Comment:
*
*In (1) we use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.
*In (2) we do a geometric series expansion.
*In (3) we select the coefficient of $x^{m-1}$.
*In (4) we apply the binomial theorem.
*In (5) we apply the rule from (1) again. We also set the upper limit of the sum accordingly, since the exponent of $y^{n-1-j}$ is non-negative.
*In (6) we select the coefficient of $y^{n-1-j}$.
|
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|
Prove polynomial map between two affine curves is bijection except origin I wanna to solve this problem dealing with a polynomial map between two affine curves.
First curve is $A \subset \mathbb{C}^2$ defined by equation $s(1+t^2)-1=0$
and the second curve is $B \subset \mathbb{C}^2$ defined by equation $(x^2+y^2)^2+3x^2y-y^3 = 0$
I have a polynomial map $f: A \rightarrow B$ defined as $f(s,t)= ( s^2t(1-3t^2), s^2(1-3t^2) )$
The problem is to prove that $f$ is surjective and also one-to-one (except the point $(0,0)$
I have no idea how to approach this. Can someone give me a little nudge what to look for? I only see that in the image of $f$ this holds $x=yt$, but I dont see how it can help. Maybe looking for inverse might be the way?
Thanks in advance
|
We can approach this by solving the polynomial system
$$
\left\{\begin{matrix}
x = ts^2(1-3t^2)\\
y = s^2(1-3t^2)
\end{matrix} \right.
$$
with $(x,y) \neq (0,0)$ satisfying $(x^2+y^2)^2+3x^2y-y^3=0$ and $(s,t)$ satisfying $s(1+t^2)=1$. First assume that $y\neq 0$, then $t = \frac{x}{y}$. From the equation of $A$ we see that $s= \frac{y^2}{y^2+x^2}$ (Note that the equation of $B$ implies that $x^2+y^2 \neq 0$) and you can verify that they solve the system above. Hence for each $(x,y)$ with $y\neq 0$ we have a unique point
$$
(s,t) = \left(\frac{y^2}{y^2+x^2},\frac{x}{y} \right)
$$
such that $f(s,t)=(x,y)$.
Now we turn to the case $y=0$. The equation of $B$ becomes $x^4 =0$ whence $x=0$ (we are at the origin). In this case our system becomes
$$
\left\{\begin{matrix}
0 = ts^2(1-3t^2)\\
0 = s^2(1-3t^2)
\end{matrix} \right. .
$$
The equation of $A$ says that $s \neq 0$ hence $1-3t^2 =0$. Therefore $ t = \pm \sqrt{\frac{1}{3}}$ and $s = \frac{3}{4}$ give the two possible solutions. This means that the inverse image of the origin by $f$ is composed by two points.
|
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|
Find $a$ and $b$ for which $\int_{0}^{1}( ax+b+\frac{1}{1+x^{2}} )^{2}\,dx$ takes its minimum possible value.
Calculate for which values $a$ and $b$ the integral
$$\int_{0}^{1} \left( ax+b+\frac{1}{1+x^{2}} \right)^{2}\,dx$$
takes its minimum possible value?
For being honest I'm not sure how to try this, but my idea is to calculate its derivative using fundamental calculus theorem as $\left(ax+b+\frac{1}{1+x^{2}} \right)^{2}$ is a continuous function over $[0,1]$. And then, evaluate the integral over $0,1$ and the values where the derivative we calculate is zero and find which $a$ and $b$ does the work. Sorry but this is the first problem of this type I'm trying. Thanks
|
Let \begin{align}f(a,b)=\int_{0}^{1} \left( ax+b+\frac{1}{1+x^{2}} \right)^{2}dx\implies \frac{\partial f(a,b)}{\partial a}&=\int_0^12x\left( ax+b+\frac{1}{1+x^{2}} \right)\,dx\\&=\left[\frac{2ax^3}3+bx^2+\ln(1+x^2)\right]_0^1\end{align} so $$\frac{\partial f(a,b)}{\partial a}=\frac23a+b+\ln2=0\tag1$$ for critical points. Similarly \begin{align}f(a,b)=\int_{0}^{1} \left( ax+b+\frac{1}{1+x^{2}} \right)^{2}dx\implies \frac{\partial f(a,b)}{\partial b}&=\int_0^12\left( ax+b+\frac{1}{1+x^{2}} \right)\,dx\\&=\left[ax^2+2bx+2\tan^{-1}x\right]_0^1\end{align} so $$\frac{\partial f(a,b)}{\partial b}=a+2b+\frac\pi2=0\implies\frac12a+b+\frac\pi4=0\tag2$$ for critical points. Performing $(1)-(2)$ gives $$\frac16a=\frac\pi4-\ln2\implies a=\boxed{\frac{3\pi}2-6\ln2}$$ and putting this into $(2)$ gives $$b=-\frac12a-\frac\pi4=\boxed{-\pi+3\ln2}.$$
|
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|
For game where first player to N points wins, find the distribution of win probability and total number of points between players
Two players, A and B, play a series of points in a game with player A winning each point with probability p and player B winning each point with probability
q = 1 - p. The first player to win N points wins the game. Assume that
N = 3. Let X be a random variable that has the value 1 if player A wins the
series and 0 otherwise. Let Y be a random variable with value the number
of points played in a game. Find the distribution of X and Y when p = 1/2.
Are X and Y independent in this case? Answer the same questions for the
case p = 2/3.
For the case of p = 1/2:
$P(Y=3)=(\frac{1}{2})^3+(\frac{1}{2})^3=0.25$
$P(Y=4)=3[(\frac{1}{2})^3(\frac{1}{2})+(\frac{1}{2})^3(\frac{1}{2})]=0.375$
$P(Y=5)=6[(\frac{1}{2})^3(\frac{1}{2})^2+(\frac{1}{2})^3(\frac{1}{2})^2]=0.375$
$P(X=1)=P(X=0)=0.5$
For the case of p = 2/3:
$P(Y=3)=(\frac{2}{3})^3+(\frac{1}{3})^3=0.333$
$P(Y=4)=3[(\frac{2}{3})^3(\frac{1}{3})+(\frac{1}{3})^3(\frac{2}{3})]=0.370$
$P(Y=5)=6[(\frac{2}{3})^3(\frac{1}{3})^2+(\frac{1}{3})^3(\frac{2}{3})^2]=0.296$
$P(X=1)=(\frac{2}{3})^3+3(\frac{2}{3})^3(\frac{1}{3})+6(\frac{2}{3})^3(\frac{1}{3})^2=0.79$
$P(X=0)=1-P(X=1)=0.21$
Is this correct?
I think for p = 1/2, X and Y are independent and for p = 2/3 X and Y are not?
I can't find anything wrong with it but I'm having trouble working through the logic of why it makes sense that $P(Y=3,4,5) = 1$.
|
Your calculation looks correct. As for why $\mathbb{P}(Y\in\{3,4,5\})=1,$ think in the following way.
$Y$ is the number of points at the end of the game.
Here $N=3,$ so $Y$ to be at least $3$, since a player can win at most $k$ points in $k$ trials. Now suppose we have $5$ points at the end, but neither player $A$ nor player $B$ has $3$ points. Is this possible? No. That means $Y$ can be at most $5.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
to prove $x^2 + y^2+1\ge xy + y + x$ $$x^2 + y^2+1\ge xy + y + x$$
$x$ and $ y$ belong to all real numbers
my attempt
$(u-2)^2\ge0\Rightarrow \frac{u^2}{4}+1\ge u $
let $u=x+y\Rightarrow \frac{(x+y)^2}{4}+1\ge x+y$
$\Rightarrow (x+y)^2+1\ge \frac{3}{4}(x+y)^2+(x+y)$
$but \frac{(x+y)^2}{4} \ge xy $ by AM-GM inequality
$\Rightarrow (x+y)^2+1\ge \frac{3}{4}(x+y)^2+(x+y)\ge3xy+(x+y)$
hence $\Rightarrow x^2 + y^2+2xy+1\ge 3xy+x+y$
are the steps correct and is there any other better way??
|
Let $c=x^2+y^2+1-(xy+x+y)$
$\iff x^2-x(1+y)+1-y+y^2-c=0$
As $x$ is real, the discriminant must be $\ge0$
i.e., $$(1+y)^2\ge4(1-y+y^2-c)\iff4c\ge3(1-y)^2$$ which is $\ge0$ for real $y$
Alternatively,
$$ x^2-x(1+y)+1-y+y^2=\left(x-\dfrac{1+y}2\right)^2+\dfrac{3(1-y)^2}4$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3197998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Advanced Complex numbers Let $\omega$ be a nonreal root of $z^3 = 1.$ Let $a_1,$ $a_2,$ $\dots,$ $a_n$ be real numbers such that
$$\frac{1}{a_1 + \omega} + \frac{1}{a_2 + \omega} + \dots + \frac{1}{a_n + \omega} = 2 + 5i.$$Compute
$$\frac{2a_1 - 1}{a_1^2 - a_1 + 1} + \frac{2a_2 - 1}{a_2^2 - a_2 + 1} + \dots + \frac{2a_n - 1}{a_n^2 - a_n + 1}.$$
I have no clue how to do this. Can someone help?
|
Rationalize the denominator for each term of the left hand side & the right hand side
For integer $n$ not divisible by$3$
Let $p=\dfrac{2n\pi}3,2\cos p=-1\ \ \ \ (1)$
$$\dfrac1{a_j+w}=\dfrac1{a_j+\cos p+i\sin p}=\dfrac{(a_j+\cos p)-i\sin p}{a_j^2+2a_j\cos p+1}$$
Use $(1)$
Equate the real parts
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3198831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $A =\left[ \begin{smallmatrix} 3 & -4 \\ 1 & -1\end{smallmatrix}\right]$, then find $A^n$
If $$ A = \begin{bmatrix}
3 & -4 \\
1 & -1
\end{bmatrix}$$
Then find $A^n$
I have tried solving it using diagonalization $PDP^{-1}$ but I am getting only one independent eigenvector i.e $$ \begin{bmatrix}
2\\
1
\end{bmatrix}$$
Please tell me the correct method to solve it.
|
Indeed, $A$ has one and only one eigenvalue: $1$. And $(2,1)$ is an eigenvector corresponding to that eigenvalue. Now, consider the vector $(1,0)$. Then$$A.(1,0)=(3,1)=(2,1)+(1,0).$$So, if $M=\left[\begin{smallmatrix}2&1\\1&0\end{smallmatrix}\right]$, then$$M^{-1}.A.M=\begin{bmatrix}1&1\\0&1\end{bmatrix},$$or$$A=M.\begin{bmatrix}1&1\\0&1\end{bmatrix}\cdot M^{-1}.$$So,\begin{align}A^n&=M.\begin{bmatrix}1&1\\0&1\end{bmatrix}^n.M^{-1}\\&=\begin{bmatrix}2&1\\1&0\end{bmatrix}.\begin{bmatrix}1&n\\0&1\end{bmatrix}.\begin{bmatrix}0&1\\1&-2\end{bmatrix}\\&=\begin{bmatrix}2n+1&-4n\\n&1-2n\end{bmatrix}.\end{align}
Note that the equality $\left[\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right]^n=\left[\begin{smallmatrix}1&n\\0&1\end{smallmatrix}\right]$ has to do with the fact that$$(\forall x,y\in\mathbb R):\begin{bmatrix}1&x\\0&1\end{bmatrix}.\begin{bmatrix}1&y\\0&1\end{bmatrix}=\begin{bmatrix}1&x+y\\0&1\end{bmatrix}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3201878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Jordan normal as transformation with respect to the basis of eigenvectors I have the following matrix
$$A = \begin{pmatrix}
2 & 0 & 1 & -3 \\
0 & 2 & 10 & 4 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 3 \\
\end{pmatrix}$$
and its Jordan normal form is
$$J = \begin{pmatrix}
2 & 0 & 0 & 0 \\
0 & 2 & 1 & 0 \\
0 & 0 & 2 & 0 \\
0 & 0 & 0 & 3 \\
\end{pmatrix}$$
with the linearly independent set of eigenvectors:
$$P = \begin{pmatrix}
0 & 1 & 0 & -3 \\
1 & 10 & 0 & 4 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix}$$
where $J = P^{-1} A P$. I am told that the following relations hold:
$$A\mathbf{v_{1}} = 2\mathbf{v_{1}}, \qquad A\mathbf{v_{2}} = 2\mathbf{v_{2}}, \qquad A\mathbf{v_{3}} = \mathbf{v_{2}} + 2\mathbf{v_{3}}, \qquad A\mathbf{v_{4}} = 3\mathbf{v_{4}}$$
In the case of $\mathbf{v_{1}}, \mathbf{v_{2}}, \mathbf{v_{4}}$, it makes total sense since this is due to the fundamental relation $A\mathbf{v} = \lambda \mathbf{v}$. Also, I see how $A\mathbf{v_{3}} = A_{2} +2A_{3}$, where $A_{2}$ and $A_{3}$ are the second and third column vectors of thre matrix $A$. This is simply multiplying $\mathbf{A}$ by $\mathbf{v_{3}}$. However, my intuition is failing me in seeing how
$$A\mathbf{v_{3}} = \mathbf{v_{2}} + 2\mathbf{v_{3}}$$
I am told that this is simply because $J$ represents the transformation corresponding to $A$ with respect to the basis $\big\{\mathbf{v_{1}}, \mathbf{v_{2}}, \mathbf{v_{3}}, \mathbf{v_{4}}\big\}$. Any pointers that can help me understand this?
|
One way of seeing where equations like $\ A\mathbf{v_{3}} = \mathbf{v_{2}} + 2\mathbf{v_{3}}\ $ come from, which I found helpful when first introduced to Jordan forms, is the definition of the invariant subspace corresponding to an eigenvalue $\ \lambda\ $ as that spanned by the non-zero vectors $\ \mathbf{v} $ for which $\ \left(A-\lambda I\right)^{\,k} \mathbf{v}=0\ $ for some positive integer $\ k\ $. If $\ k=1\ $, then $\ \mathbf{v}\ $ is an eigenvector, but if $\ k>1\ $ and $\ \left(A-\lambda I\right)^{\,k-1} \mathbf{v}\ne0\ $ then it isn't. However, if we put $\ \mathbf{v}_i=\left(A-\lambda I\right)^{\,i} \mathbf{v}\ $ for $\ i=0,1,\dots,k-1\ $, then we have $\ A\mathbf{v}_i=\lambda \mathbf{v}_i + \mathbf{v}_{i+1}\ \ $, with $\ \mathbf{v}_{k-1}\ $ being an eigenvector.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integer solutions to $x^2 - 15 y^2 = 15$ How would one show that $x^2 - 15 y^2 = 15$ has no integer solutions ?
I got that $x = \pm \sqrt{15 (y^2 +1) }$ and then WLOG I assume that $x \in \mathbb{N}$ and $x = \sqrt{15 (y^2 +1) }$. From there, I want to show that for every $y \in \mathbb{Z}$, $15 (y^2 +1)$ is not a square.
Clearly, $y^2 +1$ is not a square but how do I know that $15(y^2 +1)$ is not one ?
Any help would be appreciated !
|
If $15(y^2+1)$ is a square then $y^2+1$ is divisible by $15$, so in particular it is divisible by $3$, a contradiction.
Note that your remark
Clearly, $y^2+1$ is not a square...
is false because $0^2+1=1^2=(-1)^2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3206087",
"timestamp": "2023-03-29T00:00:00",
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|
How to prove an inequality using mathematical induction? I have to prove the following:
$$ \sqrt{x_1} + \sqrt{x_2} +...+\sqrt{x_n} \ge \sqrt{x_1 + x_2 + ... + x_n}$$
For every $n \ge 2$ and $x_1, x_2, ..., x_n \in \Bbb N$
Here's my attempt:
Consider $P(n): \sqrt{x_1} + \sqrt{x_2} +...+\sqrt{x_n} \ge \sqrt{x_1 + x_2 + ... + x_n}$
$$P(2): \sqrt{x_1} + \sqrt{x_2} \ge \sqrt{x_1 + x_2}$$
$$ x_1 + x_2 + 2\sqrt{x_1x_2} \ge x_1 + x_2$$
Which is true because $2\sqrt{x_1x_2} > 0$.
$$P(n + 1): \sqrt{x_1} + \sqrt{x_2} + ... + \sqrt{x_n} + \sqrt{x_{n+1}} \ge \sqrt{x_1 + x_2 +...+ x_n + x_{n+1}}$$
From the hypothesis we have:
$$\sqrt{x_1} + \sqrt{x_2} + ... + \sqrt{x_n} + \sqrt{x_{n+1}} \ge \sqrt{x_1 + x_2 + ... + x_n} + \sqrt{x_{n + 1}} \ge \sqrt{x_1 + x_2 +...+ x_n + x_{n+1}}$$
Squaring both sides of the right part:
$$ x_1 + x_2 + ... + x_n + x_{n + 1} + 2\sqrt{x_{n+1}(x_1 + x_2 +... + x_n)} \ge x_1 + x_2 +...+ x_n + x_{n+1} $$
Which is true, hence $P(n + 1)$ is true as well.
I'm not sure if I did it correctly?
|
I'm will just show it my way, since I find it simpler.
First, we will start off with the basis step where $n = 1$.
Basis step
$\sqrt{x_1} \geq \sqrt{x_1}$, which is of course true.
Inductive step
We let the inequality be true for any $k$, where $0 \leq k \leq n$:
$\sqrt{x_1} + \sqrt{x_2} + ... + \sqrt{x_k} \geq \sqrt{x_1 + x_2 + ... + x_k}$
If this inequality must hold, the inequalty must, by induction, hold for $k + 1$:
$\sqrt{x_1} + \sqrt{x_2} + ... + \sqrt{x_k} + \sqrt{x_{k + 1}} \geq \sqrt{x_1 + x_2 + ... + x_k + x_{k + 1}}$
We add $x_{k + 1}$ to both sides of the inequality:
$\sqrt{x_1} + \sqrt{x_2} + ... + \sqrt{x_k} + \sqrt{x_{k + 1}} \geq \sqrt{x_1 + x_2 + ... + x_k} + \sqrt{x_{k + 1}} \geq \sqrt{x_1 + x_2 + ... + x_k + x_{k + 1}}$
The inequality is hereby proved.
|
{
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.