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Given $\frac{dS}{dt}=-\frac{SI}{S + I + R}$ and $R = -(SI + S + I)/ \frac{dS}{dt}$ Determine $\frac{d^2S}{dt^2}$ I am given a system of equations:
\begin{align}
\frac{dS}{dt} &= -\frac{SI}{S + I + R}\\
\frac{dI}{dt} &= \frac{SI}{S + I + R} -I \\
\frac{dR}{dt} &= I - R\\
\end{align}
I am to subsitute $R = -\left(SI + S + I\right) / \frac{dS}{dt} $ to determine $\frac{d^2S}{dt^2}$
I chose to work only with the first equation.
When I substitute what is given for $R$ into the denominator I get:
\begin{align}
S + I + R &= S + I + R = -\left(SI + S + I\right) / \frac{dS}{dt}\\
&=\frac{\frac{dS}{dt}(S+I)-(SI+S+I)}{\frac{ds}{dt}}
\end{align}
I substitute this into the equation for $\frac{dS}{dt}$ and I get:
\begin{align}
\frac{dS}{dt} &= \frac{-SI \left(\frac{dS}{dt}\right)}{\frac{ds}{dt}\left(S+I\right)-(SI+S+I)}\\
\left(\frac{dS}{dt}\right)^2(S+I)-\frac{dS}{dt}\left(SI+S+I\right) &= -SI\left(\frac{dS}{dt}\right)\\
\left(\frac{dS}{dt}\right)^2(S+I) &= \left(\frac{dS}{dt}\right)\left(S+I\right)\\
\left(\frac{dS}{dt}\right)^2 &= \left(\frac{dS}{dt}\right)\\
\end{align}
How do I get the desired result of
$$
\frac{d^2S}{dt^2}= \frac{2}{S}\left(\frac{dS}{dt}\right)^2
$$
|
Try differentiating $\frac{dS}{dt}$ directly.
$\begin{align}\frac{d^2S}{dt^2} &= -\frac{(S+I+R)\frac{d(SI)}{dt}- SI\frac{d(S+I+R)}{dt}}{(S+I+R)^2}\\
\\&=-\frac{(S+I+R)\left[I\frac{d(S)}{dt}+S\frac{d(I)}{dt}\right]- SI\frac{d(S+I+R)}{dt}}{(S+I+R)^2}\\
\\&=-\frac{(S+I+R)\left[-I\frac{SI}{S + I + R}+S\frac{SI}{S + I + R} -SI\right]+SIR}{(S+I+R)^2}\\
\\&=-\frac{(S+I+R)\left[\frac{-SI(R+2I)}{S + I + R}\right]+SIR}{(S+I+R)^2}\\
\\&=\frac{2SI^2}{(S+I+R)^2} \\
\\& = \frac{2}{S}\left(\frac{SI}{S+I+R}\right)^2\\
\\&=\frac{2}{S}\left(\frac{dS}{dt}\right)^2\end{align}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find n where $(2n+1)2^{4n+5} = 3 \pmod{7}$ For $n$ normal number, the book solved it like this:
If $n$ can be divided by $3$ (which is $n = 3k$) then $n = 21L + 9$.
If $n$ can't be divided by 3(Which is either $n = 3k +1$ or $n = 3k + 2$) then $n = 21L + 1$ or $n = 21L + 2$ .
But I didn't solve it like this.
My logic is that since 3 is a prime number then $(2n+1)2^{4n+5} = 3 \pmod{7}$ means either $2^{4n+5} = 3\pmod{7}$ and $ 2n + 1 = 1\pmod{7} $ or the other way around.
But since there is no $n$ value that can make $2^{4n+5} = 3\pmod{7}$ then it means:
$ 2n + 1 = 3 \pmod{7}$ and by calculating we find in the end $n = 7k' + 1$
And $ 2^{4n+5} = 1\pmod{7}$ and by calculating we find that $n = 3k''$
So it means that $n = 7k' + 1$ AND $n = 3k''$
So, I know that my solution is faulty but can anyone explains to me the right solution or point out where I went wrong?
|
Problem: Find $n$ where $(2n+1)2^{4n+5} \equiv 3 \bmod{7}$
My approach:
Starting at $n=0,1,2...$, all $\bmod 7$, knowing that $2^3\equiv 1$, we can see that $2^{4n+5}\equiv 2^{n+2}$ cycles through $\{4, 1, 2\}$. Since $4^{-1}\equiv 2$, the implied $3$-cycle of $required$ values for $(2n+1)$ to satisfy is thus $\{2\cdot 3, 1\cdot 3, 4\cdot 3\} = \{\color{red}6,\color{green}3,\color{blue}5\}$, each of which can be found at one point in the $7$-cycle of its values $\{1, \color{green}3, \color{blue}5, 0, 2, 4, \color{red}6\}$.
So we need one of $n \underset{(3,7)}{\equiv} (0,6),(1,1),(2,2)$ which can be worked through the Chinese Remainder Theorem to give $n \equiv \{6,1,2\} \bmod 21$.
Your approach fails because of your assumption that only $1\times 3 = 3\times1 \equiv 3 \bmod 7$, whereas also $2\times 5 \equiv 4\times 6 \equiv 3 \bmod 7$; a prime modulus implies all coprime numbers have a multiplicative inverse.
The book's approach is incomprehensible, and $n \equiv \{\color{red}9,1,2\} \bmod 21$ is wrong.
|
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|
Find the derivative of $f(x)=x^{x^{\dots}{^{x}}}$.
Find the derivative of $f(x)$:
$$f(x)=x^{x^{\dots}{^{x}}}$$
Let $n$ be the number of overall $x's$ in $f(x)$. So for $n=1$, $f(x)=x$. I then tried to determine a pattern by solving for the derivative from $n=1$ to $n=5$. Here's what I got:
\begin{align}
n = 2 \Longrightarrow f(x) &= x^x \\
f'(x) &= \frac{d}{dx}\left(e^{x\ln \left(x\right)}\right) \\
&= e^{x\ln \left(x\right)}\frac{d}{dx}\left(x\ln \left(x\right)\right) \\
&= e^{x\ln \left(x\right)}\left(\ln \left(x\right)+1\right) \\
&= x^x\left(\ln \left(x\right)+1\right)
\end{align}
\begin{align}
n = 3 \Longrightarrow f(x) &= x^{x^{x}} \\
f'(x) &= \frac{d}{dx}\left(e^{x^x\ln \left(x\right)}\right) \\
&= e^{x^x\ln \left(x\right)}\frac{d}{dx}\left(x^x\ln \left(x\right)\right) \\
&= e^{x^x\ln \left(x\right)}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right) \\
&= x^{x^x}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)
\end{align}
\begin{align}
n = 5 \Longrightarrow f(x) &= x^{x^{x^{x^{x}}}} \\
f'(x) &= ... \\
&= x^{x^{x^{x^x}}}\left(x^{x^{x^x}}\ln \left(x\right)\left(x^{x^x}\ln \left(x\right)\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)+x^{x^x-1}\right)+x^{x^{x^x}-1}\right)
\end{align}
However, I am not sure if I see a pattern here that can help solve the question.
|
All the other answers are for the case where there are infinite x's. However, there are finite ($n$) x's in this problem!
Let's first denote the function as $f_n(x)=x^{x^{\cdots^x}}$ with $n$ x's in the exponent. For example, $f_1(x)=x^x, f_0(x)=x.$
Take the logarithm of both sides, we get $\ln f_n(x)=f_{n-1}(x)\ln x$. Taking the derivative gives $$\frac{f'_n(x)}{f_n(x)}=f'_{n-1}(x)\ln x+\frac{f_{n-1}(x)}{x}$$
$$f'_n(x)=f_n(x) f'_{n-1}(x)\ln x+\frac{f_n(x) f_{n-1}(x)}{x}$$
That's just a basic recurrence relation. We know that $$a_n=Ca_{n-1}+D\implies a_n=c C^{n-1} + \frac{D(C^n-1)}{C-1}$$ where $c$ is an arbitrary constant.
Similarly, we may get $$f'_n(x)=c(f_n(x)\ln x)^{n-1}+\frac{f_n(x)f_{n-1}(x)}{x(f_n(x)\ln x-1)} ((f_n(x)\ln x)^n-1)$$
To get the value of the parameter $c$, let $n=1$, then $f'_1(x)=c+x^x$. Hence, $c=x^x \ln x$. Substituting in, we get the final answer (in a rather cumbersome form) $$f'_n(x)=x^x \ln x(f_n(x)\ln x)^{n-1}+\frac{f_n(x)f_{n-1}(x)}{x(f_n(x)\ln x-1)} ((f_n(x)\ln x)^n-1)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3992667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Two Lebesgue's integrals I need to calculate these two integrals:
$$\lim_{n\rightarrow\infty}\int_\mathbb{R}\frac{nx^2+1}{nx^4+n^2}d\lambda(x)$$
$$\int_0^1\sum_{n=1}^\infty n\cdot\chi_{(\frac{1}{n+1},\frac{1}{n})}d\lambda$$
I did the second one like this:
$$\int_0^1\sum_{n=1}^\infty n\cdot\chi_{(\frac{1}{n+1},\frac{1}{n})}d\lambda=\int_0^11\cdot\chi_{(\frac{1}{2},1)}+2\cdot\chi_{(\frac{1}{3},\frac{1}{2})}+\dots d\lambda=$$
$$=\int_0^1\chi_{(\frac{1}{2},1)}d\lambda+2\cdot\int_0^1\chi_{(\frac{1}{3},\frac{1}{2})}d\lambda+3\cdot\int_0^1\chi_{(\frac{1}{4},\frac{1}{3})}d\lambda+\dots$$
First integral if equal to the area of rectangle $(\frac{1}{2},1)\times [0,1]$ which is $\frac{1}{2}$. Every next integral is the area of rectangle $(\frac{1}{n+1},\frac{1}{n})\times [0,1]$ times $n$. From that we get:
$$\frac{1}{2}+2\cdot \frac{1}{6}+3\cdot \frac{1}{12}+\dots=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots=\sum_{n=1}^\infty\frac{1}{n+1}=\infty.$$
For the first one I think I can use one of the theorem of Lebesgue which allow me to rewrite it as
$$\lim_{n\rightarrow\infty}\int_\mathbb{R}\frac{nx^2+1}{nx^4+n^2}d\lambda(x)=\int_\mathbb{R}\lim_{n\rightarrow\infty} \frac{nx^2+1}{nx^4+n^2}d\lambda(x)$$
I'm not sure how will that limit behave in that case though. I would be grateful for any hint on how to complete this.
|
You have for $\vert x \vert \ge 1$
$$\begin{aligned}
\left\vert \frac{nx^2+1}{nx^4+n^2} \right\vert &\le \frac{x^2+1}{x^4+n} \le \frac{x^2+1}{x^4+1}
\end{aligned}$$
and for $\vert x \vert \le 1$
$$\left\vert \frac{nx^2+1}{nx^4+n^2} \right\vert \le \frac{n+1}{n^2} \le 2.$$
Therefore $f_n(x) = \frac{nx^2+1}{nx^4+n^2} $ is integrable for all $n \ge 1$, and bounded by
$$g(x) = \begin{cases}
\frac{x^2+1}{x^4+1} & \vert x \vert \ge 1\\
2 & \vert x \vert \lt 1
\end{cases}$$ which is integrable. As $\{f_n\}$ converges pointwise to zero, you can apply Lebesgue dominated convergence theorem to conclude that
$$\lim_{n\rightarrow\infty}\int_\mathbb{R}\frac{nx^2+1}{nx^4+n^2}d\lambda(x) = 0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Failing a basic integration exercise; where did I go wrong? (This is a basic calculus exercise gone wrong where I need some feedback to get forward.)
I've attempted to calculate an integral by first integrating it by parts and then by substituting. The result I got is not correct though. Can I get a hint about where started to make mistakes?
Here are my steps:
$$\int \sqrt{x}\cdot\sin\sqrt{x}\,dx$$
*
*Applying integration by parts formula
$$\int uv' \, dx = uv - \int u'v \, dx$$
where $u=\sqrt{x}$, $v'=\sin\sqrt{x}$, $u'=\frac{1}{2\sqrt{x}}$ and $v=-\cos\sqrt{x}$, resulting in:
$$\int \sqrt{x}\cdot\sin\sqrt{x}\,dx = \sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{x}} \cdot -\cos\sqrt{x} \, dx$$
*Next, I tried substituting $x$ with $g(t)=t^2$ in the remaining integral, i. e. replacing each $x$ in the integral with $t^2$ and the ending $dx$ with $g'(t)=2t\,dt$. I would later bring back the $x$ by substituting $t=\sqrt{x}$ after integration. Continuing from where we left by substituting:
$$\sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{x}} \cdot -\cos\sqrt{x} \, dx$$
$$\Longrightarrow \sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{t^2}} \cdot -\cos\sqrt{t^2}\cdot2t \, dt$$
*Then I pulled out the constant multipliers from the integral:
$$\sqrt{x} \cdot -\cos\sqrt{x} - \frac{1}{2} \cdot -1 \cdot 2 \int \frac{1}{\sqrt{t^2}} \cdot \cos\sqrt{t^2}\cdot t \, dt$$
which turned out to eliminate each other (resulting in just $\cdot1$), so we end up with:
$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \int \frac{1}{\sqrt{t^2}} \cdot \cos\sqrt{t^2}\cdot t \, dt$$
*Reducing the integrand by reducing $\frac{1}{\sqrt{t^2}}\Rightarrow\frac{1}{t}$, which again is eliminated by multiplying with the integrand's $t$, and reducing $\cos\sqrt{t^2}\Rightarrow\cos t$. Therefore resulting in:
$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \int \cos t \, dt$$
where the integral can be solved as $\int \cos t \, dt = \sin t + C$. So now we are at:
$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \sin t + C$$
The correct answer however is
$$\int \sqrt{x} \sin\sqrt{x} \, dx = 4 \sqrt{x} \sin\sqrt{x} - 2 (x - 2) \cos\sqrt{x} + C$$
So something somewhere in my process went horribly wrong. What?
|
The mistake is
$$v'=\sin\sqrt{x}\,dx \to v=\int \sin\sqrt{x}\,dx\neq -\cos\sqrt{x}+k, \quad k\in \Bbb R$$
$$\int \sin\sqrt{x}\,dx=2\left(-\sqrt{x}\cos \left(\sqrt{x}\right)+\sin \left(\sqrt{x}\right)\right)+k',\quad k'\in \Bbb R$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral of a polynomial fraction: $\int \frac{4x^2-3x+2}{4x^2-4x+3}dx$ I have initially used long division to obtain: $$1+\frac{x-1}{4x^2-4x+3}$$
Now, I'm having difficulty solving $$\int \frac{x-1}{4x^2-4x+3}dx$$
Even if I do the $u$-substitution i.e. $u=4x^2-4x+3$ hence, $du=8x-4$ and $dx=\frac{1}{8x-4}du$, I can't seem to figure out how to deal with the $x-1$ in the numerator.
Any help regarding this will be appreciated. Thank you.
|
Hints:
$$2\frac{x-1}{4x^2-4x+3}=\frac{2x-2}{(2x-1)^2+2}=\frac{2x-1}{(2x-1)^2+2}-\frac1{(2x-1)^2+2}.$$
In the first term, the numerator is the derivative of the denominator (to a coefficient). In the second, we recognize the derivative of the arc tangent, modulo simple transformations.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve:$(1-y^2 +\frac{y^4}{x^2})p^2- (2\frac{y}{x})p + \frac{y^2}{x^2}=0$ I have the differential equation: $(1-y^2 +\frac{y^4}{x^2})p^2- (2\frac{y}{x})p + \frac{y^2}{x^2}=0$ where $p=\frac{dy}{dx}$
I have solved this up until $\frac{dy}{dx}$, but I am not able to reduce it to an exact differential equation.
Steps until here:
$$\frac{1}{x^2}\Bigl((x^2-x^2 y^2 +y^4)p^2- (2xy)p + y^2\Bigr)=0$$
As $\frac{1}{x^2}$ cannot equal zero,
$$(x^2-x^2 y^2 +y^4)p^2- (2xy)p + y^2=0$$
Applying the quadratic formula:
$$p=\frac{2xy \pm \sqrt{4x^2y^2-4x^2y^2+4x^2y^4-4y^6}}{2(x^2-x^2 y^2 +y^4)}$$
$$p=\frac{xy \pm y^2\sqrt{x^2-y^2}}{x^2-x^2 y^2 +y^4}$$
$$p=\frac{y\Bigl(x \pm y\sqrt{x^2-y^2}\Bigr)}{x^2-y^2(x^2-y^2)}$$
$$p=\frac{y\Bigl(x \pm y\sqrt{x^2-y^2}\Bigr)}{\Bigl(x-y\sqrt{x^2-y^2}\Bigr)\Bigl(x+y\sqrt{x^2-y^2}\Bigr)}$$
Thus,
$$p=\frac{dy}{dx}=\frac{y}{\Bigl(x-y\sqrt{x^2-y^2}\Bigr)}$$
Or,
$$p=\frac{dy}{dx}=\frac{y}{\Bigl(x+y\sqrt{x^2-y^2}\Bigr)}$$
From here, we can write the two equations as:
$$xdy-ydx=y\sqrt{x^2 - y^2}dy$$
$$xdy-ydx=-y\sqrt{x^2 - y^2}dy$$
I tried reducing this equation to an exact differential equation, but failed to do so. Kindly help me proceed from here. Thank you!
|
$$xdy-ydx=-y\sqrt{x^2 - y^2}dy$$
$$yx'-x=y\sqrt{x^2 - y^2}$$
$$\left(\dfrac xy \right)'=\sqrt{\dfrac {x^2}{y^2} -1}$$
$$w'=\sqrt{w^2 -1}$$
Where $w=\dfrac xy$.Then integrate.
|
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|
Help solving $\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}$ Reading an article online I came across this integral I've been trying to solve since yesterday morning and I literally tried every integration method I know and I still can't solve it. I am a beginner to integrals so help will be appreciated!
I tried substitution, integration by parts and I can't seem to find the correct way!
$$\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}$$
Does anyone know how to proceed from here? What to do? Because I am clueless. Thanks!
|
Try this.
\begin{align}
\int \:\frac{dx}{x+\sqrt{9x^2-9x+2}}
&={\displaystyle\int}\left(\dfrac{\sqrt{9x^2+9x+2}}{8x^2+9x+2}-\dfrac{x}{8x^2+9x+2}\right)\mathrm{d}x \\
{\displaystyle\int}\dfrac{\sqrt{9x^2+9x+2}}{8x^2+9x+2}\,\mathrm{d}x
&={\displaystyle\int}\dfrac{\sqrt{\left(3x+\frac{3}{2}\right)^2-\frac{1}{4}}}{8x^2+9x+2}\,\mathrm{d}x\tag{solve for 1st integral} \\
&=\frac12 {\displaystyle\int}\dfrac{\sqrt{9\left(2x+1\right)^2-1}}{8x^2+9x+2}\,\mathrm{d}x \\
&=\frac12{\displaystyle\int}\dfrac{\sqrt{9u^2-1}}{4u^2+u-1}\,\mathrm{d}u \tag{$u = 2x+1$, $\mathrm{d}x=\dfrac{1}{2}\,\mathrm{d}u$} \\
&=\frac12 {\displaystyle\int}\dfrac{\sqrt{\sec^2\left(v\right)-1}\sec\left(v\right)\tan\left(v\right)}{3\left(\frac{4\sec^2\left(v\right)}{9}+\frac{\sec\left(v\right)}{3}-1\right)}\,\mathrm{d}v \tag{$u=\dfrac{\sec\left(v\right)}{3}$, $v=\operatorname{arcsec}\left(3u\right), \mathrm{d}u=\dfrac{\sec\left(v\right)\tan\left(v\right)}{3}\,\mathrm{d}v$} \\
&=\frac32 {\displaystyle\int}\dfrac{\sec\left(v\right)\tan^2\left(v\right)}{4\sec^2\left(v\right)+3\sec\left(v\right)-9}\,\mathrm{d}v \\
&=\frac32 {\displaystyle\int}\dfrac{\frac{\left(\tan^2\left(\frac{v}{2}\right)+1\right) \,\cdot\, 4\tan^2\left(\frac{v}{2}\right)}{\left(1-\tan^2\left(\frac{v}{2}\right)\right)^3}}{\left(\frac{4\left(\tan^2\left(\frac{v}{2}\right)+1\right)^2}{\left(1-\tan^2\left(\frac{v}{2}\right)\right)^2}+\frac{3\left(\tan^2\left(\frac{v}{2}\right)+1\right)}{1-\tan^2\left(\frac{v}{2}\right)}-9\right)}\,\mathrm{d}v \\
&
\end{align}
|
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|
Prove this formula $\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}= \sum_{n=1}^{+\infty}r^{n}\sin\left(nx\right)$ I wish to use geometric power series of $\dfrac{1}{1-x}$ to prove this formula by Euler
$$\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}=r\sin(x)+r^2\sin(2x)+r^{3}\sin(3x)...=1+\sum_{n=1}^{+\infty}r^{n}\sin\left(nx\right)$$
In this thread, I have understood how to prove the same formula for $\cos(nx)$
Please see the answer of A-Level Student to see, his answer is:
Let
$$\begin{align}
C&=1+r\cos x+r^2\cos2x+r^3\cos3x+\cdots\\
S&=r\sin x+r^2\sin2x+r^3\sin3x+\cdots\\
\end{align}$$
Then
$$\begin{align}
C+iS&=1+r(\cos x+i\sin x)+r^2(\cos 2x+i\sin2x)+r^3(\cos3x+i\sin3x)+\cdots\\
&=1+re^{ix}+(re^{ix})^2+(re^{ix})^3+\cdots\\
&=\frac{1}{1-re^{ix}}=\frac{(1-re^{-ix})}{(1-re^{ix})(1-re^{-ix})}=\frac{1-r\cos x+ri\sin x}{1-2r\cos x+r^2}
\end{align}$$
Hence, equating real and imaginary parts we obtain
$$\begin{align}
C&=\frac{1-r\cos x}{1-2r\cos x+r^2}\\
S&=\frac{r\sin x}{1-2r\cos x+r^2}\\
\end{align}$$
I wish to use a similar method to prove the above mentioned series for $\sin(nx)$, but I am not sure value which series for of $C$ and $S$, should I choose.
I have chosen $C=1+r\sin(x)+r^2\sin(2x)+r^{3}\sin(3x)...$
$S=rcos(x)+r^2cos(2x)+r^3cos(3x)+...$
And I let $1 + S + iC$, with $i$ being imaginary number.
So I have
$1 +S + iC = 1+r(\cos(x)+i\sin(x))+r^2(\cos(2x)+i\sin(2x)+...$
= $1 + S +iC = 1+ re^{ix} + (re^{ix})^2+ (re^{ix})^3+ (re^{ix})^4+ (re^{ix})^5+...$
And the $1+ re^{ix} + (re^{ix})^2+ (re^{ix})^3+ (re^{ix})^4+ (re^{ix})^5+... = \dfrac{1}{1-e^{ix}}$
It is here that I am stuck, I don't think my choice of $S$ and $C$ is correct. I know that the Euler identity for $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$, but I don't know how to apply this formula.
|
Use infinite GP:
$$S=\sum_{n=1}^{\infty} r^n \sin nx=\Im \sum_{n=1}^{\infty} r^ne^{inx}=\Im \frac{re^{ix}}{1-re^{ix}}$$
$$\implies S=\Im \frac{re^{ix}(1-re^{-ix})}{(1-re^{ix})(1-re^{-ix})}=\frac{r\sin x}{1-2r\cos x+r^2}$$
This result has been check with Mathematica numerically. For instance for $r=1/2,x=\pi/3$, we get $S=1/\sqrt{3}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\left \lfloor \frac{1+\lfloor na+1/a\rfloor}{a} \right \rfloor=n$ If $a \geq \frac{1+\sqrt{5}}{2}$ and $n \in \Bbb W$, prove that
$$\left \lfloor \frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a} \right \rfloor=n.$$
I could prove only when $a$ is an integer, that is $a \geq 2$. If $a \in \Bbb Z$ we have: $$\left \lfloor \frac{1+na^2}{a} \right \rfloor=na+\left\lfloor \frac{1}{a} \right\rfloor=na$$ so we get: $$\left \lfloor \frac{1+\left\lfloor \frac{1+na^2}{a}\right\rfloor}{a} \right \rfloor=\left \lfloor \frac{1+na}{a} \right \rfloor=n+\left\lfloor \frac{1}{a} \right\rfloor=n.$$ But what if $a \notin \Bbb Z$?
|
We wish to show that $\exists\epsilon\in[0,1)$ such that $$1+\left\lfloor\frac1a+na\right\rfloor=a(n+\epsilon)=na+\epsilon a\implies\left\lfloor\frac1a+\{na\}\right\rfloor=\{na\}+b$$ where $b=\epsilon a-1\in[-1,a-1)$. As the LHS is either $0$ or $1$, either $b=-\{na\}$ or $1-\{na\}$ so we will always find an $\epsilon\in(0,1)$ except the case where $$b=1-\{na\}>a-1\implies\{na\}<2-\phi$$ as $a\ge\phi$. This means that $$\frac1a+\{na\}<\frac1a+2-\phi\le\frac1\phi+2-\phi=1\implies\left\lfloor\frac1a+\{na\}\right\rfloor=0$$ which is a contradiction as $b=1-\{na\}$.
|
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|
Prove that the algebraic expression is greater than zero Prove that $x^8 - x^5 + x^2 -x + 1 >0$ for $x \in \mathbb{R}$.
It's from the (junior) high-school competition and the idea is that not everyone there knows calculus, so that I'm looking for more "basic" justification.
My idea was to use AM-GM: $x^8 + x^2 \geq 2\sqrt{x^{10}} = 2x^5$. Thus,
$$x^8 - x^5 + x^2 -x + 1 \geq x^5-x + 1 $$
So that we can now focus on proving that $x^5 - x + 1 > 0 $ but I don't really see how to do so without calculus...
|
I considered parsing the polynomial as $ \ (x^8 - x^5) \ + \ (x^2 -x + 1) \ $ . Restricting techniques to ones that would likely appear in junior high school and high school curricula, we can show that $ \ x^2 - x + 1 \ $ is never equal to zero since its discriminant is $ \ -3 \ $ . So its value can never change sign; since it equals $ \ 1 \ $ for $ \ x \ = \ 0 \ $ , it is always positive. We can factor
$ \ (x^8 - x^5) \ = \ x^5 · (x^3 - 1) \ = \ x^5 · (x - 1) · (x^2 + x + 1) \ . \ \ \ $ [using "difference of two cubes"]
By an argument similar to the above, $ \ x^2 + x + 1 \ $ is also always positive. The product of the first two factors is positive for $ \ x \ > \ 1 $ and for $ \ x \ < \ 0 \ $ , hence the complete polynomial is positive in these intervals. It is equal to $ \ 1 \ $ at $ \ x \ = \ 0 \ $ and $ \ x \ = \ 1 \ $ , so it remains to establish the proposed inequality for $ \ 0 \ < \ x \ < \ 1 \ $ .
Using "completion of the square", we write our polynomial as $ \ (x \ - \ \frac{1}{2} )^2 \ + \ \frac{3}{4} \ - \ (x^5 \ - \ x^8) \ $ . From what we know about power-functions, we can say that $ \ x^5 \ - \ x^8 \ > \ 0 $ in the interval $ \ 0 \ < \ x \ < \ 1 \ $ , but I don't think there's a convincing argument using the available means to show that $ \ 0 \ < \ x^5 \ - \ x^8 \ < \ \frac{3}{4} \ $ . [We know that calculus will give us a maximum value for $ \ x^5 \ - \ x^8 \ $ of $ \ ( \frac{5}{8} )^{5/3} · \frac{3}{8} \ \approx \ 0.171 $ at $ \ x_{min} \ = \ (\frac{5}{8} )^{1/3} \ \approx \ 0.855 \ $ . ]
Instead we may exploit symmetry and separate the polynomial into "even" and "odd" components as $ \ (x^8 + x^2 + 1) \ - \ (x^5 + x ) \ . $ [This is a variation on what user143137 does.] For the first polynomial, $ \ x^8 + x^2 + 1 \ \geq \ 1 \ $ for all real numbers (the graphs of power functions should be familiar at this level). The second polynomial, $ \ x^5 + x \ $ , is negative for $ \ x \ < \ 0 \ $ , so we only need to be concerned about $ \ 0 \ < \ x \ < 1 \ $ . In that interval, we have $ 1 \ < \ x^8 + x^2 + 1 \ < \ 3 \ $ and $ \ 0 \ < \ x^5 + x \ < 2 \ $ . Each of these polynomials increases "monotonically" (although we might not use that word at this level), so we can conclude that $ \ (x^8 + x^2 + 1) \ - \ (x^5 + x ) \ > \ 0 \ . $
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
What's the distance between two circumcenters This problem was came to a Facebook post of mathigon ......the problem seems trickier than I expected...can you help me to find the distance??
|
Let $\angle CAB = \angle ABC = \angle BAD = \angle DBA = \theta$
Area of the $\bigtriangleup CAD = \frac{1}{2} 1^2 \sin 2 \theta = \frac{1}{2} \sin 2 \theta$
Area of the sector $CAD$ (consider the right side of the circle on the left) $ = \frac{1}{2} 1^2 2\theta = \frac{1}{2} (2 \theta)$
So the common area between the circles $ = 2 (\frac{1}{2} (2 \theta) - \frac{1}{2} \sin 2 \theta)$
However, it is given that the common area between the circles $= \frac{\pi}{2}$
Hence we need to solve the following equation for $\theta$
$$2 \left(\frac{1}{2} \left(2 \theta \right) - \frac{1}{2} \sin 2 \theta \right) = \frac{\pi}{2}$$
Solving, $\theta = 1.15494$
Finally length of $AB$ is $2 \cos \theta = 0.80794684217$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Is it possible to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ without using trigonometric substitution? The normal approach to evaluate $\int\frac{dx}{\sqrt{x^2+4}}$ is using the substitution $x=2\tan\theta$. But I wonder is is possible to do it without using trigonometric substitution? I tried this approach:
$$\int\frac{dx}{\sqrt{x^2+4}}=\int\frac{xdx}{\sqrt{x^2(x^2+4)}}=\frac12\int\frac{du}{\sqrt{u(u+4)}}\quad\text{where}\quad u=x^2$$
But I can't see a way to evaluate final integral without completing square and using the substitution $u+2=2\sec t$ (which is a trigonometric substitution!)
|
Knowing the answer, as i mentioned it in the comments, the substitution in
the answer of J.G. has best chances to work, and it works in a line.
Alternatively...
Starting from the last expression in the OP, we may use the Euler substitution
$$
t = \sqrt{\frac{u+4}u}=\sqrt{1+ \frac 4u}\ .
$$
Then we have formally successively $t^2=1+\frac 4u$, $t^2-1=\frac 4u$,
$u=\frac 4{t^2-1}$,
$du=-\frac{8t}{(t^2-1)^2}\; dt$.
The substitution from $x$ to $t$ is thus $\color{blue}{t=\frac{\sqrt{x^2+4}}x}$. (Assuming $x>0$.)
Then with the used substitutions
$$
\begin{aligned}
\int\frac{dx}{\sqrt{x^2+4}}
&=\frac12\int\frac{du}{\sqrt{u(u+4)}}
=\frac12\int\frac{du}{u\sqrt{\frac{u+4}u}}\\
&=\frac12\int\frac{-\frac{8t}{(t^2-1)^2}\; dt}{\frac 4{t^2-1}\cdot t}
=-\int\frac{dt}{t^2-1}\\
&=
\frac 12\log\frac{t+1}{t-1}+C
=
\frac12
\log
\frac
{\color{blue}{\frac{\sqrt{x^2+4}}x}+1}
{\color{blue}{\frac{\sqrt{x^2+4}}x}-1}+C
\\
&=\log(\sqrt{x^2+4}+x)+C'\ .
\end{aligned}
$$
(At the last point we note that $(\sqrt{x^2+4}+x)(\sqrt{x^2+4}-x)$ is a constant.)
|
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|
inequality:$(a+c)(b+d)(ac+bd)+4\ge 3(a+b+c+d)$
If $a,b,c,d>0$ such that $abcd=1$ prove that $$(a+c)(b+d)(ac+bd)+4\ge 3(a+b+c+d)$$
My first idea was to find counter examples and also check the bahaviour of inequality in the edge cases.Easy to see the equality occurs when $a=b=c=d=1$ However there is a big problem when $a\to 0^+ b\to 0^+,c\to 0^+,d\to \infty$ In this case the RHS $\to \infty$ and the LHS although appears $\to 0$ we see that as $b+d\to \infty$ so we cannot say much.
Mixed variables doesnt help much as we cannot take WLOG $a\ge b\ge c\ge d$
I also tried Holder $$(a+c)(b+d)(bd+ac)\ge {(\sqrt[3]{ab^2d}+\sqrt[3]{ac^2d})}^3=ad{(b^{2/3}+c^{2/3})}^3$$ which has only complicated it further
Looking forward to solutions without lagranges multiplier.
|
WLOG assume that $bd\ge 1$ and $ac\le 1$. The inequality rewrites as
$$(a+c)((b+d)(ac+bd)-3)+4\ge 3(b+d).$$
AM-GM and $bd\ge 1$ yield
$$(b+d)(ac+bd)-3\ge 2\sqrt{bd}\cdot 2\sqrt{acbd}-3 \ge 2\cdot 2-3=1>0,$$
hence by AM-GM
$$(a+c)((b+d)(ac+bd)-3) \ge 2\sqrt{ac}((b+d)(ac+bd)-3) = 2(b+d)((ac)^{3/2}+(bd)^{1/2})-6(bd)^{-1/2}.$$
Hence it is enough to prove that
$$2(b+d)((ac)^{3/2}+(bd)^{1/2}-\frac 32) + 4 \ge 6(ac)^{1/2} .$$
Using AM-GM again we obtain $$(ac)^{3/2}+(bd)^{1/2} \ge 4\sqrt[4]{(ac)^{3/2}\cdot \left(\frac{(bd)^{1/2}}{3}\right)^3}=4\cdot 3^{-3/4}>\frac 32,$$
hence by AM-GM
$$2(b+d)((ac)^{3/2}+(bd)^{1/2}-\frac 32) \ge 4(bd)^{1/2}\left((ac)^{3/2}+(bd)^{1/2}-\frac 32\right) = 4ac+4bd-6(bd)^{1/2}$$
and we are left to prove that
$$4ac+4bd+4\ge 6(ac)^{1/2} + 6(bd)^{1/2}.$$
This follows from AM-GM:
$$4ac+4bd+4\ge 3ac+3bd+6 = 3ac+3 + 3bd+3 \ge 6\sqrt{ac}+6\sqrt{bd}$$
|
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|
Inequality with $abc=1$ Let $a;b;c$ be positive real numbers and $abc=1$
Find maximum value of: $P=\dfrac{1}{\sqrt{a^2+2b^2}}+\dfrac{1}{\sqrt{b^2+2c^2}}+\dfrac{1}{\sqrt{c^2+2a^2}}$
I tried to use Cauchy-Schwarz: $a^2+2b^2 \geq \dfrac{1}{3}(a+2b)^2$
Then: $P \leq \dfrac{\sqrt{3}}{a+2b}+\dfrac{\sqrt{3}}{b+2c}+\dfrac{\sqrt{3}}{c+2a}$
So, what should I do next? Thank you.
|
There is no maximum .
Note that when $a\to 0^+,b\to 0^+,c\to \infty$ we get $P\to \infty$ .This is because the $\sqrt{a^2+2b^2}\to 0$
Note that we chose $c\to \infty$ because $c=\frac{1}{ab}$
Now even if you are looking for the minimum there is no minimum .
Note again that $a\to 0^+,b\to \infty,c\to \infty$ we see that $P\to 0$.
|
{
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"url": "https://math.stackexchange.com/questions/4013715",
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|
Finding eigenvectors of matrix of matrices
Let $A$ be a matrix with $\lambda_1,...,\lambda_n$ eigenvalues, and $v_1,...,v_n$ the corresponding eigvenvectors. Let $B= \begin{pmatrix} 0& A \\ A& 0 \end{pmatrix}$. It's known that the eigenvalues of $B$ are $\{\pm \lambda_1,..., \pm \lambda_n\}$. Find the eigenvectors of $B$.
I've managed to find $n$ of the $2n$ eigenvectors of $B$, those who correspond to the eigenvalues: $\{ \lambda_1,..., \lambda_n\}$:
$u_1= \begin{pmatrix} v_1 \\ v_1 \end{pmatrix},..., u_n = \begin{pmatrix} v_n \\ v_n \end{pmatrix}$. I assumed that the new eigenvectors should be built of the original eigenvectors, thus I started playing with combinations of the original eigenvectors, such as: $\begin{pmatrix} v_i \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ v_i \end{pmatrix}$ however those vectors didn't help at all. Eventually, looking at how $B$ is defined, I understood that since $B$ simply works the same as $A$ works on vectors for the first half and then again for the second half, I found that $u_1= \begin{pmatrix} v_1 \\ v_1 \end{pmatrix},..., u_n = \begin{pmatrix} v_n \\ v_n \end{pmatrix}$ can work.
However, I couldn't manage to find the corresponding eigenvectors for the $\{-\lambda_1,...,-\lambda_n\}$ eigenvalues. I can't see how applying the original matrix $A$ in any kind of form will get us to the negative of its' eigenvalues.
Any hint/clue?
|
Let's look at $u_i =\begin{pmatrix} v_i \\ -v_i \end{pmatrix}$:
$B \begin{pmatrix} v_i \\ -v_i \end{pmatrix} = \begin{pmatrix} 0& A \\ A& 0 \end{pmatrix} \begin{pmatrix} v_i \\ -v_i \end{pmatrix}= \begin{pmatrix} A(-v_i) \\ Av_i \end{pmatrix} = \begin{pmatrix} -\lambda_iv_i \\ \lambda_iv_i \end{pmatrix} = -\lambda_i\begin{pmatrix} v_i \\ -v_i \end{pmatrix}$
Thus we get the other $n$ eigenvectors, ending with total of $2n$ eigenvectors.
|
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|
Show $\log\left(\frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}}<\log\left(\frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}$ if $x\geq 5$ and $1\leq y\leq x-2$ Assume that all logarithms are natural.
Let $x$ and $y$ be integers that satisfy $x \geq 5$ and $1 \leq y \leq x-2$. I am trying to show that $$\log\left( \frac{x-y}{x}\right)-2\sqrt{\frac{x-y}{x}} < \log\left( \frac{x+y}{x}\right)-2\sqrt{\frac{x+y}{x}}.$$
This is equivalent to showing that $$\log\left( \frac{x-y}{x+y}\right)<2\left(\sqrt{\frac{x-y}{x}}-\sqrt{\frac{x+y}{x}}\right).$$
My first inclination was to apply the well known inequality, $\log(z)\leq 2(\sqrt{z}-1)$ if $z>0$, to the left side by letting $z=\frac{x-y}{x+y}$, but this did not help me.
I also tried double induction on $(x,y)$ (since $x$ and $y$ are integers) but I could not finish it.
I appreciate any help, thank you.
|
Try working with the inequalities on $x$ and $y$. We are given $x\in[5,\infty)$ and $y\in[1,x-2]$. First, we find $x-y\in[2,x-1]$ and $x+y\in[6,2x-2]$. For a fraction, we can find the minimum by maximizing the denominator and minimizing the numerator, for instance $\frac{x-y}{x+y}\in[\frac{2}{2x-2},\frac{x-1}{6}]$.
Then, $\log(z)$ and $\sqrt(z)$ are strictly increasing functions so we can plug in the minimum and maximum values for their arguments. So it will suffice to show
$$\log\left(\frac{x-1}{6}\right)<2\left(\sqrt{\frac{x-1}{5}}-\sqrt{\frac{6}{x}}\right)$$
We can split up the inequality to find the inequality you mentioned:
$$\log\left(\frac{x-1}{5}\cdot\frac{5}{6}\right)<2\left(\sqrt{\frac{x-1}{5}}-1\right)+2\left(1-\sqrt{\frac{6}{x}}\right)$$
$$\log\left(\frac{x-1}{5}\right)+\log\left(\frac{5}{6}\right)<2\left(\sqrt{\frac{x-1}{5}}-1\right)+2\left(1-\sqrt{\frac{6}{x}}\right)$$
And extracting the $\log(z)\le 2(\sqrt(z)-1)$ identity we need to show:
$$\log\left(\frac{5}{6}\right)<2\left(1-\sqrt{\frac{6}{x}}\right)$$
Which is clearly true because the lhs is negative and the rhs is positive for the allowed values of $x$.
|
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How to proceed $\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $
$$I=\int \frac{dx}{(x-2) \left(1+\sqrt{7 x-10-x^2}\right)} $$
I have taken the second term of denominator (in square root) as $(x - 2)t$. But cannot go further. Please suggest how to proceed or any alternate method to solve it.
|
Note that $7x-10-x^2 = \frac94-(\frac72-x)^2$. So, substitute $\frac72-x=\frac32\sin t$ to integrate as follows
\begin{align}
& \int \frac{dx}{(x-2)(1+\sqrt{7x-10-x^2})}\\
= & - \int \frac{2\cos t}{(1-\sin t)(2+3\cos t)}dt\\
= & - \int \left ( \frac{\cos t}{1-\sin t}-\frac {3\sin t}{2+3\cos t} - \frac {3}{2+3\cos t} \right)dt\\
=& \>\ln(1-\sin t) -\ln(2+3\cos t)+\frac3{\sqrt5}\tanh^{-1}\frac{\tan\frac t2}{\sqrt5}+C
\end{align}
|
{
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|
Eliminating $x$, $y$, $z$ from $\frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$ and $ax+by+cz=0$ Here is a Math Tripos problem I cannot solve.
Eliminate $x$, $y$, $z$ from the equations
$$\frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$$ and
$$ax+by+cz=0$$
I dont know if there is a general (practical) method for such problem. I think you just manipulate the equations to eliminate $x,y,z$ but have been unable to find the path in this question.
The answer is $a^3+b^3+c^3=a^2(b+c)+b^2(a+c)+c^2(a+b)$ if that yields any insight.
|
Let's denote $t$ be equal to
$$t = \frac{x^2-xy-xz}{a}=\frac{y^2-yx-yz}{b}=\frac{z^2-zx-zy}{c}$$
We have:
\begin{align}
t &=\frac{(y^2-yx-yz)+(z^2-zx-zy)-(x^2-xy-xz)}{b+c-a} \\
&=\frac{(y^2-2yz+z^2)-x^2}{b+c-a} \\
&=\frac{(y-z)^2-x^2}{b+c-a} \\
&=-\frac{(z+x-y)(x+y-z)}{b+c-a} \\
\end{align}
Hence,
\begin{align}
t^2 &=-\frac{(z+x-y)(x+y-z)}{b+c-a} \times \frac{x^2-xy-xz}{a} \\
&=-\frac{(z+x-y)(x+y-z)}{b+c-a} \times \frac{-x(y+z-x)}{a} \\
&=(x+y-z)(y+z-x)(z+x-y) \times \frac{x}{a(b+c-a)} \\
\end{align}
We have then
$$\frac{a(b+c-a)}{x} = \frac{b(c+a-b)}{y} = \frac{c(a+b-c)}{z} = u $$
$$\left(\text{both equal to } u =\frac{(x+y-z)(y+z-x)(z+x-y)}{t^2} \right)$$
We can deduce that
$$u = \frac{a^2(b+c-a)}{ax} = \frac{b^2(c+a-b)}{by} = \frac{c^2(a+b-c)}{cz} $$
Hence:
$$a^2(b+c-a)+b^2(c+a-b)+c^2(a+b-c) = u(ax+by+cz) = 0$$
(because $ax +by +cz = 0$)
Finally, we can conclude that
$$a^2(b+c)+b^2(c+a)+c^2(a+b) = a^3 + b^3+c^3$$
|
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|
The center of the circumcircle lies on a side of a triangle Consider a triangle $ABC$. Let the angle bisector of angle $A$ be $AP,P\in BC$. $BP=16,CP=20$ and the center of the circumcircle of $\triangle ABP$ lies on the segment $AC$. Find $AB$.
$$AB=\dfrac{144\sqrt5}{5}$$
By Triangle-Angle-Bisector Theorem $$\dfrac{BP}{PC}=\dfrac{AB}{AC}=\dfrac{16}{20}=\dfrac{4}{5}\\ \Rightarrow AB=4x, AC=5x.$$ The cosine rule on $ABC$ gives $$BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cdot\cos\alpha \\ \iff 1296=41x^2-40x^2\cos\alpha,$$ where $\measuredangle A=\alpha.$ Is any of this helpful for the solution? Any help would be appreciated. Thank you in advance!
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Observe that $\angle OPA=\angle PAO=\frac\alpha2=\angle BAP\implies OP\parallel AB$. Thus $$\frac{CO}{OA}=\frac{CP}{PB}\iff \frac dR=\frac{20}{16}=\frac54$$ Also, Power of a point yields $$\begin{align*}\text{Pow}(C)_{(APB)}=\lvert d^2-R^2\rvert&=20\cdot 36\\\iff \left\lvert\left(\frac54R\right)^2-R^2\right\rvert&=720\\\iff \frac9{16}R^2&=720\\\iff R&= 16\sqrt{5} \end{align*}$$ Use $OP\parallel AB$ again in order to infer
$$AB=R\cdot \frac{36}{20}=16\sqrt5\cdot \frac95=\frac{144\sqrt5}{5}$$
|
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|
Estimations of some new recurrence sequences
Problem 1. Given the recursion $a_{n+ 1}= \sqrt{a_{n}^{2}+ a_{n}}$ with $a_{1}= 1.$ Prove that
$$a_{n}\sim\frac{n}{2}+ \frac{1}{2n}\,{\rm as}\,n\rightarrow\infty$$
For problem 1, I only can prove $a_{n}\sim\left ( n+ 1 \right )/2$ by using the Laurent series for the increasing $a_{n} :$
$$a_{n+ 1}= a_{n}+ \dfrac{1}{2}+ \mathcal{O}\left ( \dfrac{1}{a_{n}} \right )$$
Problem 2. Given the recursion $a_{n+ 1}= \sqrt[3]{a_{n}^{3}+ a_{n}}$ with $a_{1}= 1.$ Prove that
$$a_{n+ 1}\sim\sqrt{\dfrac{2n}{3}}\,{\rm as}\,n\rightarrow\infty$$
What if we changed the given one by $a_{n+ 1}= \sqrt[3]{a_{n}^{3}+ 3a_{n}^{2}+ a_{n}}$ ?
For problem 2, similarly, I saw that $a_{n+ 1}\fallingdotseq a_{n}+ \dfrac{1}{3a_{n}}$, then how can we find the relative equations like ${y}'= \dfrac{1}{3y}$ by $n$ times ?
|
We may use the approach in [1].
For Problem 1:
First, clearly $\lim_{n\to \infty} a_n = \infty$.
Second, we have
\begin{align}
a_{n+1} - a_n &= \sqrt{a_n^2 + a_n} - a_n\\
&= a_n\left(\sqrt{1 + \frac{1}{a_n}} - 1\right)\\
&= a_n\left(1 + \frac{1}{2} \frac{1}{a_n}
+ o\left(\frac{1}{a_n}\right) - 1\right)\\
&\to \frac{1}{2},\quad \mathrm{as}\ n\to \infty.
\end{align}
Thus, $a_n \sim \frac{n}{2}$ as $n\to \infty$.
Third, we have
\begin{align}
a_{n+1} - a_n &= \sqrt{a_n^2 + a_n} - a_n\\
&= a_n\left(\sqrt{1 + \frac{1}{a_n}} - 1\right)\\
&= a_n\left(1 + \frac{1}{2} \frac{1}{a_n} - \frac{1}{8}\frac{1}{a_n^2}
+ o\left(\frac{1}{a_n^2}\right) - 1\right)\\
&\sim \frac{1}{2} - \frac{1}{8}\frac{1}{a_n}, \quad \mathrm{as}\ n\to \infty\\
&\sim \frac{1}{2} - \frac{1}{4n}, \quad \mathrm{as}\ n\to \infty.
\end{align}
Thus, $a_n \sim \frac{n}{2} - \frac{1}{4}\ln n$ as $n\to \infty$.
$\phantom{2}$
For problem 2:
First, clearly $\lim_{n\to \infty} a_n = \infty$.
Second, we have
\begin{align}
a_{n+1}^2 - a_n^2 &= (\sqrt[3]{a_n^3 + a_n})^2 - a_n^2 \\
&= a_n^2 \left(\left(1 + \frac{1}{a_n^2}\right)^{2/3} - 1\right)\\
&= a_n^2 \left(1 + \frac{2}{3} \frac{1}{a_n^2} + o\left(\frac{1}{a_n^2}\right) - 1\right)\\
&\to \frac{2}{3}, \quad \mathrm{as}\ n\to \infty.
\end{align}
Thus, $a_n^2 \sim \frac{2}{3}n$ as $n\to \infty$.
Thus, $a_n \sim \sqrt{\frac{2}{3}n}$ as $n\to \infty$.
Reference
[1] "The second term in asymptotic expansion", by Moubinool OMARJEE, Paris
|
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|
proving that $89 \mid 2^{44}-1$ i tried to prove that $2^{44} \equiv 1\pmod{89}$.
I noticed that by Fermat's little theorem $2^{88} \equiv 2^{44}\cdot 2^{44} \equiv 1\pmod{89}$
which means that $2^{44}$ is the inverse of itself $\rightarrow 2^{44} \equiv 1 \pmod{89}$ or $2^{44}\equiv 88 \pmod{89}$.
how can I rule out the second option?
|
$2^{44} - 1 = (2^{22})^2 - 1 = (2^{22}+1)(2^{22} - 1) = (2^{22}+1)(2^{11} + 1)(2^{11} - 1) = (2^{22}+1)(2^{11} + 1)\times 2047 = (2^{22}+1)(2^{11} + 1)\times 23 \times 89$
|
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|
A simpler way to prove that $\frac 32-x-\frac{1}{x+1}$ is negative for $x\geq1$, instead of computing derivative? I have this function
$$f(x)=\frac 32-x-\frac{1}{x+1}$$
I want to prove that $f(x)$ is negative for $x\geq1$. We can easily prove this by calculating the first derivative. Is there a simpler way to prove that? Instead of taking derivative?
|
\begin{align} \left(-x-\frac{1}{x+1}\leq -\frac{3}{2}\right)\quad \text{ and }\quad x\geq1\\
\\
\iff\quad \left(x+\frac{1}{x+1}\geq \frac{3}{2}\right)\quad \text{ and }\quad x\geq1 \\
\\
\iff \quad \left(x(x+1) + 1 \geq \frac{3}{2}(x+1)\right)\quad \text{ and }\quad x\geq1\\
\\
\iff \quad \left(x^2 - \frac{1}{2}x - \frac{1}{2} \geq 0\right)\quad \text{ and }\quad x\geq1\\
\\
\iff\quad \left(\left(x-\frac{1}{4}\right)^2 \geq \frac{9}{16}\right)\quad \text{ and }\quad x\geq1\\
\\
\iff\quad \left(x - \frac{1}{4} \geq \frac{3}{4}\quad \text{ or }\quad x - \frac{1}{4} \leq -\frac{3}{4}\right)\quad \text{ and }\quad x\geq1\\
\\
\iff\quad \left(x\geq1\quad \text{ or }\quad x\leq -\frac{1}{2}\right)\quad \text{ and }\quad x\geq1\\
\\
\iff\quad x\geq1\\
\end{align}
|
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|
Prove that $ABCD$ is a trapezoid $ABCD$ is a cyclic quadrilateral inscribed of circle $O$. The tangent line of circle $O$ at $B$ intercepts $CD$ at $K$. The tangent line of circle $O$ at $C$ intercepts $AB$ at $M$.
if $AB = BM$ and $CD=CK$, prove that $ABCD$ is a trapezoid.
I was going down the route of:
$\frac{BM}{\sin \angle BCM} = \frac{BC}{\sin \angle BMC}$, so $BM = \frac{BC \sin \angle BCM}{\sin \angle BMC}$
$\frac{AB}{\sin \angle ACB} = \frac{BC}{\sin \angle BAC}$, so $AB = \frac{BC \sin \angle ACB }{\sin \angle BAC}$
and $\angle BCM = \angle BAC$
$\frac{BM}{AB} = \frac{\sin \angle BCM \sin \angle BAC}{\sin \angle BMC \sin \angle ACB} = 1$
but the relationship is awkward here.
|
It's enough to prove that $BD=AC$.
Observe that, $\triangle BCK\sim \triangle DBK$ and $\triangle CBM\sim \triangle ACM$.
Hence, $\frac{BK}{DK}=\frac{CK}{BK}$
$\Rightarrow BK=CK\sqrt{2}$
Thereafter, $\frac{BD}{BC}=\frac{BK}{CK}=\sqrt{2}$
Similarly yield $\frac{AC}{BC}=\sqrt{2}$ and thus $BD=BC\sqrt{2}=AC$.
|
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|
Computing the value of $\zeta(6)$ and $\zeta(8)$ I know the closed form formula for calculating the values of Zeta function at even integers. I was able to derive it from the coefficients of the power series of $-\dfrac{\pi x}{2}\cot(\pi x)$. Now, I am looking at the way with which Euler was able to derive the values of $\zeta(2n)$.
I have seen this "trick" in Brilliant:
So I have:
$$\dfrac{\sin(x)}{x}=1-\dfrac{x^2}{3!}+\dfrac{x^4}{5!}-\dfrac{x^6}{7!}+\dfrac{x^8}{9!}...\tag{1}$$
$$\dfrac{\sin(x)}{x}=\left(1-\dfrac{x^2}{\pi^2}\right)\left(1-\dfrac{x^2}{4\pi^2}\right)\left(1-\dfrac{x^2}{9\pi^2}\right)\left(1-\dfrac{x^2}{16\pi^2}\right)...\tag{2}$$
$$\dfrac{\sin(ix)}{ix}=1-\dfrac{(ix)^2}{3!}+\dfrac{(ix)^4}{5!}-\dfrac{(ix)^6}{7!}...\tag{3}$$
$$\dfrac{\sin(ix)}{ix}=1+\dfrac{x^2}{3!}+\dfrac{x^4}{5!}+\dfrac{x^6}{7!}...\tag{4}$$
$$\dfrac{\sin(ix)}{ix}=\left(1-\dfrac{(ix)^2}{\pi^2}\right)\left(1-\dfrac{(ix)^2}{4\pi^2}\right)\left(1-\dfrac{(ix)^2}{9\pi^2}\right)\left(1-\dfrac{(ix)^2}{16\pi^2}\right)...\tag{5}$$
$$\dfrac{\sin(ix)}{ix}=\left(1+\dfrac{x^2}{\pi^2}\right)\left(1+\dfrac{x^2}{4\pi^2}\right)\left(1+\dfrac{x^2}{9\pi^2}\right)\left(1+\dfrac{x^2}{16\pi^2}\right)...\tag{6}$$
Multiply $(6)$ and $(2)$, I have:
$$\dfrac{\sin(x)\sin(ix)}{ix^2}=1+\left(\dfrac{x^2}{3!}-\dfrac{x^2}{3!}\right)+\left(\dfrac{x^4}{5!}+\dfrac{x^4}{5!}-\dfrac{x^4}{(3!)^2}\right)+...\tag{7}$$
$$\dfrac{\sin(x)\sin(ix)}{ix^2}=\left(1+\dfrac{x^2}{\pi^2}\right)\left(1-\dfrac{x^2}{\pi^2}\right)\left(1+\dfrac{x^2}{4\pi^2}\right)\left(1-\dfrac{x^2}{4\pi^2}\right)\left(1+\dfrac{x^2}{9\pi^2}\right)\left(1-\dfrac{x^2}{9\pi^2}\right)...\tag{8}$$
$$\dfrac{\sin(x)\sin(ix)}{ix^2}=\left(1-\dfrac{x^4}{\pi^4}\right)\left(1-\dfrac{x^4}{16\pi^4}\right)\left(1-\dfrac{x^4}{81\pi^4}\right)...\tag{9}$$
Comparing the coeffcients of $x^4$ in the infinite Maclaurin series and the infinite product, we will arrive at $\zeta(4)=\dfrac{\pi^4}{90}$.
Now that I wish to continue this process to see if I can get $\zeta(6)$, $\zeta(8)$, $\zeta(10)$, I don't know how to continue. The trick above works fine for $\zeta(4)$, but what about higher $\zeta(2n)$.
I look around and find this interesting post of user17762:
In his answer, he wrote the function on the LHS as $\dfrac{i\sin(z)\sin(\dfrac{z}{i})}{z^2}$. A little bit of manipulation shows that it is exactly like $\dfrac{\sin(x)\sin(ix)}{ix^2}$ or $\dfrac{\sin(x)\sinh(x)}{x^2}$
I have tried to multiply again $\dfrac{\sin(x)\sinh(x)\sin(x/i^2)}{x^2(\frac{x}{i^2})}=\dfrac{\sin^2(x)\sinh(x)}{x^3}$, using the hint found in the answer of the member above. Then I use Wolfram to check the series expansion of this function but the third term of $x^6$ is $\dfrac{5}{3024}$. I thought that I should a find a function where its coefficient of $x^6$ should be $\dfrac{1}{945}$. I am not sure if my reasoning is problematic.
Is there an algorithm to find the function on the LHS?
|
The generalisation you are looking for is given by the functions
$$f_n \colon \mathbb{R} \to \mathbb{R}, \, f_n (x) = \prod \limits_{k=0}^{n-1} \operatorname{sinc}\left(\mathrm{e}^{\mathrm{i} \pi \frac{k}{n}} x\right) \, , $$
for $n \in \mathbb{N}$ ($\operatorname{sinc}(z) = \frac{\sin(z)}{z}$ with $\operatorname{sinc}(0) = 1$). Since the product representation of $\operatorname{sinc}$ converges absolutely, we can rearrange factors to obtain
$$ f_n(x) = \prod \limits_{k=0}^{n-1} \prod \limits_{l=1}^\infty \left(1 - \mathrm{e}^{2 \pi \mathrm{i} \frac{k}{n}} \frac{x^2}{\pi^2 l^2} \right) = \prod \limits_{l=1}^\infty \prod \limits_{k=0}^{n-1} \left(1 - \mathrm{e}^{2 \pi \mathrm{i} \frac{k}{n}} \frac{x^2}{\pi^2 l^2} \right) = \prod \limits_{l=1}^\infty \left(1 - \left(\frac{x}{\pi l}\right)^{2n} \right)$$
for $x \in \mathbb{R}$ and $n \in \mathbb{N}$ (see this question for a simple proof of the final step). This representation confirms that these functions are real at real arguments. Clearly, the coefficient in front of $x^{2n}$ in the series expansion of $f_n$ is $[x^{2n}] f_n (x) = - \frac{\zeta(2n)}{\pi^{2n}}$ for $n \in \mathbb{N}$. Now simply multiply the power series representations, i.e.
$$ f_n (x) = \prod \limits_{k=0}^{n-1} \sum \limits_{l_k = 0}^\infty \frac{(-1)^{l_k} \mathrm{e}^{2 \pi \frac{k l_k}{n} \mathrm{i}} x^{2l_k}}{(2l_k+1)!} \, ,$$
and compute the coefficient this way to obtain the zeta values by comparison. As you can easily check using CAS, this works for every $n \in \mathbb{N}$ ($[x^6]f_3(x) = - \frac{1}{945}$ or $[x^8]f_4(x) = - \frac{1}{9450}$, for example), but I have not yet managed to prove $[x^{2n}] f_n(x) = \frac{(-1)^{n} \mathrm{B}_{2n} 2^{2n-1}}{(2n)!}$ analytically.
|
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|
Prove that the set $a+b\sqrt{2}$ where a and b are rational without zero is a group under multiplication We are given $G=\{a+b\sqrt{2}: a,b\in\mathbb{Q}\}$ and asked to show that $G-\{0+0\sqrt{2}\}$ is a group under multiplication.
Let us first specify that multiplication takes place as expected:
$$(a+b\sqrt{2})(c+d\sqrt{2})=a(c+d\sqrt{2})+b\sqrt{2}(c+d\sqrt{2})=ac+2bd+(ad+bc)\sqrt{2}$$
It is clear that $1+0\sqrt{2}$ is the multiplicative identity.
But I have hit a blocker on finding the inverse. Dummit and Foote's hint is to 'rationalize the denominator' but I'm not sure what this means in this case. If I try straight up equating coefficients in
$$(a+b\sqrt{2})(c+d\sqrt{2})=1+0\sqrt{2}$$ then that's just $$ac+2bd=1$$ and $$ad+bc=0$$. That's not a very nice set up at all. Any hints?
|
Bear in mind the multiplication in $G$ is the same as the multiplication of the reals so whatever is true about multiplication of $\mathbb R$ will still be true of the multiplication of $G$.
Multiplication is associative. We don't have to prove that.
We do have to prove that mulitplicaition is closed on $G$. That is: if $w \in G$ and $u \in G$ then $wu \in G$. (which you did).
The identity in $\mathbb R$ is $1$ so the identity for $G$ will be $1$. We don't have to prove that $1*w = w$ for all $w\in G$ and we don't have to solve that if $e*w = w$ then $e=1$. We know that must be true. The only thing we have to prove is that $1$ is actually a member of $G$. (Which as $1 = 1 + 0\sqrt 2$ it obviously is.
And as the invers of $x$ is $\frac 1x$, we do know that the inverse of $w \in G$ will HAVE to be $\frac 1w$. The issue though is we have to prove that if $w \in G$ then $\frac 1w \in G$ also.
SO let's prove that:
If $w = a+ b\sqrt 2\ne 0$ then $\frac 1w = \frac 1{a+b\sqrt 2}$. And that is in $G$???? We can write $\frac 1{a+b\sqrt 2}$ as $c +d \sqrt 2$ where $c$ and $d$ is rational? Is that actually true? How would we show that? How would we figure out what $c,d $ are?
Well, the hint "rationalize the denominator" is very apt.
$\frac 1{a+b\sqrt 2} = \frac 1{a+b\sqrt 2}\cdot \frac {a-b\sqrt 2}{a-b\sqrt 2}=$
$\frac {a-b\sqrt 2}{a^2 -2b^2 } =\frac a{a^2 -2 b^2}- \frac b{a^2 -2b^2}\sqrt 2$
And as $a,b \in \mathbb Q$ we have $\frac a{a^2 -2b^2},-\frac b{a^2 -2b^2}\in \mathbb Q$. So $\frac 1{a+b\sqrt 2} = \frac a{a^2 -2 b^2}- \frac b{a^2 -2b^2}\sqrt 2\in G$.
And that's that.
=====
Actually $ac + 2bd = 1$ and $ad + bc = 0$ is a perfectly fine set up.
$d = -\frac {bc}a$ (assuming $a \ne 0$)
$ac + 2b(-\frac {bc}a) = ac- \frac {2b^2c}a =2$
$c (a-\frac {2b^2}a) = 1$
$c = \frac 1{a-\frac{2b^2}a}=\frac {a}{a^2 -2b^2}$ (assuming $a-\frac{2b^2}a\ne 0$.... which it cant because $(a-\frac{2b^2}a)c = 1 \ne 0$.)
So $d =\frac {b \frac {1a}{a^2 -2b^2}}a= \frac {b}{a^2 - 2b^2}$
....
And if $a$ does equal $0$ we have
$ac + 2bd = 2bd =1$ and so $b\ne 0$ and $ad +bc = bc = 0$.
Now $b \ne 0$ as 1) $2bd =1 \ne 0$ and also because we are told $a + b\sqrt 2 \ne 0$ and if $a =0$ then....
SO $b\ne 0$ and $bc = 0$ so $c = 0$.
And $2bd= 1$ so $b = \frac 1{2b}$.
... in other words $b\sqrt 2\cdot x = 1 \implies x = \frac 1{2b}\sqrt 2$
|
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|
Obtain the coefficient of $x^2$ in the expansion of $1+\frac{6}{2x+1}+\frac{5}{2-3x}$ Hello so this is a 2 part question and I managed to express that praction as a partial fraction which was equaled to $$1+\frac{6}{2x+1}+\frac{5}{2-3x}$$ I will add my work below I tried lot to Obtain the coefficient of $x^2$ but cant get to the answer of 237/8 and help is much appreciated. Thank you!
|
This is how to do it the hard way.
Let's suppose that, for $x$ close enough to $0$,
$$1 + \frac{6}{2x+1} + \frac{5}{2-3x} = a_0 + a_1x + a_2x^2 + \dots$$
$$(2x+1)(2-3x)\left(1 + \frac{6}{2x+1} + \frac{5}{2-3x}\right) = 19 - 7x - 6x^2 + \dots$$
$$(2x+1)(2-3x)(a_0 + a_1x + a_2x^2)
= 2a_0 + (a_0+2a_1)x + (-6 a_0 + a_1 + 2 a_2)x^2 + \dots$$
Comparing $19 - 7x - 6x^2$ to $2a_0 + (a_0+2a_1)x + (-6 a_0 + a_1 + 2 a_2)x^2$
\begin{align}
2a_0 &= 19 \\
a_0 &= \frac{19}{2} \\
\hline
a_0 + 2a_1 &= -7 \\
\frac{19}{2} + 2a_1 &= -7 \\
a_1 &= -\frac{33}{4} \\
\hline
-6 a_0 + a_1 + 2 a_2 &= -6 \\
-57 - \frac{33}{4} + 2a_2 &= -6 \\
a_2 &= \frac{237}{8}
\end{align}
So the coefficient of x^2 is $\frac{237}{8}$.
|
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|
Does $\sum{\frac{\sin (n+1/n)}{ \ln n}} \sim \sum{\frac{\sin n}{\ln n}}$? As was mentioned in the title, can we say that both the series
$$
\sum{\frac{\sin (n+1/n)}{\ln n}} \quad \text{ and } \quad \sum{\frac{\sin n}{\ln n}}
$$
converges or diverges simultaneously? I mean, it looks very intuitive since
$$
\sin(n+1/n) \approx \sin(n)
$$
for sufficiently large $n$. But I didn't find any rigorous Test or Rule to show that.
|
Since $\sin(a+b) = \sin a \cos b + \cos a \sin b$, by a Taylor expansion around $0$ we get
$$\begin{align*}
\sin\!\left(n+\frac{1}{n}\right) &= \sin n \cos \frac{1}{n} + \cos n \sin \frac{1}{n} \\
&= \sin n\left(1-\frac{1}{n^2} + o\!\left(\frac{1}{n^2}\right) \right)
+ \cos n\left(\frac{1}{n} + o\!\left(\frac{1}{n^2}\right) \right)
\end{align*}$$
which means that
$$\begin{align*}
\frac{\sin\!\left(n+\frac{1}{n}\right)}{\ln n}
&= \frac{\sin n}{\ln n}
+ \frac{\cos n}{n\ln n}+ o\!\left(\frac{1}{n^2}\right)
\end{align*}$$
since $|\sin|,|\cos|\leq 1$. This shows that, if you take for granted that
$$
\sum_{n=2}^\infty \frac{\cos n}{n\ln n}
$$
converges, then indeed
$$
\sum_{n=2}^\infty \frac{\sin\!\left(n+\frac{1}{n}\right)}{\ln n},\quad\sum_{n=2}^\infty \frac{\sin n}{\ln n}
$$
have same nature (since the remainder, "$\sum_n o\!\left(\frac{1}{n^2}\right)$", is absolutely convergent by comparison). The question then becomes: can one easily show that
$$
\sum_{n=2}^\infty \frac{\cos n}{n\ln n}
$$
converges? (One way: Dirichlet's test.)
|
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|
Remainder When Divided By 70 : Remainder Theorem Problem If $x$ = $16^3$ + $17^3$ + $18^3$ + $19^3$ , then $x$ divided by $70$ leaves a remainder of?
I tried to solve this problem by using the remainder theorem which states that remainder of $$Rem[\frac{a+b+c+....}{x}] = Rem[\frac{a}{x}] + Rem[\frac{b}{x}] + Rem[\frac{c}{x}] + .... $$ where 'Rem' means 'Remainder Of'.
Using this same logic I can write $$Rem[\frac{16^3+17^3+18^3+19^3}{70}] = Rem[\frac{16^3}{70}] + Rem[\frac{17^3}{70}] + Rem[\frac{18^3}{70}] + Rem[\frac{19^3}{70}] = 18 +13+11+69 = 111 \Rightarrow Rem[\frac{111}{70}] = 41$$
So my answer comes up as 41. Can someone please explain to me what I might be doing wrong or what I have understood wrong?
|
Another way, where there is no need to compute the numeric values of the four cubes.
Note that $16+19=17+18=35=70/2$, hence
\begin{align*}
16^3+17^3+18^3+19^3
&=16^3+(35-18)^3 +18^3+(35-16)^3\\
&\equiv 16^3+35^3-18^3+18^3+35^3-16^3\\
&=2\cdot 35^3\equiv 0\pmod{70}.
\end{align*}
In particular, when we expand the cube $(35-2n)^3$ we find
$$35^3-3\cdot 35^2\cdot 2n+3\cdot 35\cdot (2n)^2-(2n)^3\equiv 35^3-(2n)^3\pmod{70}$$
because $3\cdot 35^2\cdot 2n$ and $3\cdot 35\cdot (2n)^2$ are divisible by $70$.
|
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|
Need help with De'Moivre's Theorem, or maybe that is the wrong approach to this problem.
Let ($a_{n}$) and ($b_{n}$) be the sequences of real numbers such that $(2+i)^n=a_{n}+b_{n}i$ for all integers $n\geq 0$, where $i=\sqrt{-1}$. What is $\sum_{n=0}^{\infty}\frac{a_{n}b_{n}}{7^n}$?
I need some help with the question above. I tried using De'Moivre's Theorem to simplify $(2+i)^n$, but got hung up with simplifying $\sqrt{5}^n\operatorname{cis}(n\theta)$. I was attempting to find $a_{n}$ and $b_{n}$ first, then evaluate the summation.
|
Let $\theta$ be the acute angle satisfying $\tan \theta = \frac{1}{2}$,
$$a_n = \sqrt{5^n} \cos(n \theta)$$
$$b_n = \sqrt{5^n}\sin (n\theta)$$
$$\frac{a_nb_n}{7^n} = \left( \frac57\right)^n\sin(n\theta)\cos(n\theta)=\frac12 \left( \frac57\right)^n \sin (2n\theta)= \frac12 \Im\left[\left(\frac57e^{2i\theta}\right)^n\right]$$
\begin{align}\sum_{n=0}^\infty \frac{a_nb_n}{7^n} &= \frac12 \Im \left[ \sum_{n=0}^\infty \left( \frac57 e^{2i\theta} \right)^n\right] \\
&= \frac12 \Im \left[ \frac{1}{1-\frac57e^{2i\theta}}\right]\end{align}
Try to simplify from here.
Edit:
\begin{align}
\frac12 \Im \left[ \frac1{1-\frac57 e^{2i\theta}}\right] &= \frac12 \Im \left[ \frac{1-\frac57e^{-2i\theta}}{(1-\frac57 e^{2i\theta})(1-\frac57 e^{-2i\theta})}\right] \\
&=\frac12\left[ \frac{\frac57\sin 2 \theta}{1+\frac{25}{49}-\frac{10}7 \cos2\theta}\right] \\
&=\frac12\left[ \frac{\frac{10}7\sin \theta \cos \theta}{1+\frac{25}{49}-\frac{10}7 (2\cos^2 \theta -1)}\right] \\
&=\frac12\left[ \frac{\frac{10}7\cdot \frac25}{1+\frac{25}{49}-\frac{10}7 (2\cdot \frac45 -1)}\right] \\
&=\frac12\left[ \frac{\frac{10}7\cdot \frac25}{1+\frac{25}{49}-\frac{10}7 \cdot \frac35}\right] \\
&=\frac12\left[ \frac{\frac47}{1+\frac{25}{49}-\frac{6}7}\right] \\
&=\frac12\left[ \frac{28}{49+25-42}\right] \\
&= \frac{14}{32}\\
&= \frac{7}{16}
\end{align}
|
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floor function inequality $n+ \left \lfloor \tfrac{n}{3} \right \rfloor \le x$ I have the following problem which has two parts:
Given any $x>0$, I define a function $A$ by
\begin{equation}
A(x)=\max \{ n \in \mathbb{N} : n+ \left \lfloor \tfrac{n}{3} \right \rfloor \le x\}
\end{equation}
I want to compute $A(x)$ and
\begin{equation}
\lim_{x \to \infty} \dfrac{A(x)}{x}.
\end{equation}
I tried to use the fact that $ x \le \lfloor x \rfloor \le x+1 $ but I got stuck.
Do I have to do a case by case analysis of if $n$ is divisible by $3$?
|
Since $n\in\mathbb{N}$, it follows by contradiction that
$$n+\left\lfloor{\frac{n}{3}}\right\rfloor\leq x \iff n+\left\lfloor{\frac{n}{3}}\right\rfloor\leq \lfloor x\rfloor$$
Suppose that it is possible to make the above equation an equality. Writing $n=3k+r,0\leq r\leq 2$ we note that
$$\lfloor x\rfloor=4k+r$$
Thus the only cases where such an equality is allowed is when $\lfloor x\rfloor\equiv 0,1,2 \mod 4$. Explicitly, in this case we can write the solution $n=\lfloor x\rfloor-\lfloor x/4\rfloor$.
Now when $x=3\mod 4$ we see that, since we cannot achieve equality with $\lfloor x\rfloor$ we can move to the closest integer to that, $\lfloor x \rfloor-1.$ We can certainly achieve equality here, with $n=\frac{3(\lfloor x \rfloor-1)+2}{4}=\lfloor x\rfloor-\lfloor x/4\rfloor-1$. Thus
$$A(x)=\begin{Bmatrix}\lfloor x\rfloor-\lfloor x/4\rfloor&x\neq 3\mod 4\\ \lfloor x\rfloor-\lfloor x/4\rfloor-1&x= 3\mod 4 \end{Bmatrix}$$
It should be clear that
$$\lim_{x\to\infty}\frac{A(x)}{x}=\frac{3}{4}$$
EDIT: It turns out a more elegant form exists:
$$A(x)=\lfloor x\rfloor-\left\lfloor \frac{x+1}{4}\right\rfloor$$
One can see this by noticing that the following Iverson bracket, the difference between the two branches above, is succintly given by
$$[x=3\mod 4]=\left\lfloor \frac{x}{4}+\frac{1}{4}\right\rfloor-\left\lfloor \frac{x}{4}\right\rfloor$$
|
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"url": "https://math.stackexchange.com/questions/4058205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solving a Polynomial Equation by Factoring. $n^3+12n^2+48n+64$
I know the sum of two cubes formula, $(a+b)(a^2-ab+b^2)$. I'm not sure how to apply it here? Any help would be appreciated.
|
Here it is an approach based on the Binomial Theorem:
\begin{align*}
n^{3} + 12n^{2} + 48n + 64 = {3\choose 0}n^{3}4^{0} + {3\choose 1}n^{2}4^{1} + {3\choose 2}n4^{2} + {3\choose 3}n^{0}4^{3} = (n+4)^{3}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Binomial coefficients identity : $\sum_{k=1}^{n-m+1} k\binom{n-k+1}{m}=\binom{n+2}{m+2}$ For any positive integer m&n.$n\ge m$ , let $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) = {}^n{C_m}$. Prove that $\left( {\begin{array}{*{20}{c}}
n\\
m
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
{n - 1}\\
m
\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}
{n - 2}\\
m
\end{array}} \right) + .. + \left( {n - m + 1} \right)\left( {\begin{array}{*{20}{c}}
m\\
m
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{n + 2}\\
{m + 2}
\end{array}} \right)$
My approach is as follow
$n=4, m=2$
$\left( {\begin{array}{*{20}{c}}
4\\
2
\end{array}} \right) + 2\left( {\begin{array}{*{20}{c}}
3\\
2
\end{array}} \right) + 3\left( {\begin{array}{*{20}{c}}
2\\
2
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
6\\
4
\end{array}} \right) = 15$
$6 + 6 + 3 = 15 \Rightarrow LHS = RHS$
But not able to solve it via property
|
Here is a combinatorial approach. Consider the set
$$\{ 1,2,3,4,\ldots,n-1,n,n+1,n+2\}$$
We will pick $m+2$ distinct numbers. This can be done randomly in $\binom{n+2}{m+2}$ ways.
Another way to do the same would be prioritizing the second smallest element.
If second smallest number is $2$, we can choose $1$ as the smallest element and select $m$ out of remaining $n$ numbers. $\binom{n}{m}$ ways.
If second smallest number is $3$, we can choose the smallest in two ways out of $1,2$ and select $m$ out of remaining $n-1$ in $\binom{n-1}{m}$ ways.
In general, if second smallest choice is $i+1$, the smallest can be picked in $i$ ways and rest in $\binom{(n+2)-(i+1)}{m}=\binom{n-i+1}{m}$ ways.
Summing over all valid values of $i$, LHS=RHS is proved.
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the intergal of $\int {\dfrac{1}{(x^2+9)^2}}\,dx$ with recursive formula? I used partial integration and I got this:
${\dfrac{x}{(x^2+9)^2}} + 4 $$\int {\dfrac{x^2}{(x^2+9)^3}}\,dx$
According to the recursice formula I should continue like this:
${\dfrac{x}{(x^2+9)^2}} + 4 $$\int {\dfrac{x^2 +9-9}{(x^2+9)^3}}\,dx$
${\dfrac{x}{(x^2+9)^2}} + 4 $$\int {\dfrac{x^2 +9-9}{(x^2+9)^3}}\,dx$ - 4$\cdot 9 $$\int {\dfrac{1}{(x^2+9)^3}}\,dx$
And then end up with something like that:
${\dfrac{1}{2 \cdot 9} \cdot \dfrac{1}{x^2+9} + \dfrac{1}{2 \cdot 27}} arctg{\dfrac{x}{3}}\,$ + C
But I'm sure that I'm doing this wrong or just I'm too confused?
|
Integrate as follows
\begin{align}
\int {\dfrac{1}{(x^2+9)^2}}\,dx
&= \frac19 \int {\dfrac{1}{x^2+9}}\,dx - \frac19\int {\dfrac{x^2}{(x^2+9)^2}}\,dx\\
&= \frac19 \int {\dfrac{1}{x^2+9}}\,dx + \frac1{18}\int x\>d({\dfrac{1}{x^2+9}})\\
&= \frac1{18} \frac x{x^2+9} + \frac1{18}\int {\dfrac{1}{x^2+9}}\,dx\\
&= \frac1{18} \frac x{x^2+9} + \frac1{54}\tan^{-1}\frac x3+C
\end{align}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4064701",
"timestamp": "2023-03-29T00:00:00",
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|
Find a value of $\;\lim\limits_{n\rightarrow\infty}\frac{a_{n}}{n}$
(If you're good at it, you can do it by heart.)
Let $a_n$ be the average value of lengths of all the diagonal lines of regular n-gon whose side is 1.
Find a value of
$$\lim_{n\rightarrow\infty}\frac{a_{n}}{n}$$
Source: Fujino_Yusui
After messing around with the cosine theorem and doing some calculations, I found the exact formula of $a_{n},$ so I did the rest of the calculations and the answer is
$$\lim_{n\rightarrow\infty}\frac{a_{n}}{n}= \lim_{n\rightarrow\infty}\frac{2\cos\frac{\pi}{n}- 1}{\left ( \left ( 1- \cos\frac{\pi}{n} \right )n \right )\left ( n- 3 \right )}= \frac{2}{\pi^{2}}$$
But I want to see a shortcut that leads $a_{n}\sim\frac{n}{2\pi}\cdot\frac{4}{\pi}\,{\rm as}\,n\rightarrow\infty$ without cosine theorem, I need to the help.
|
Inscribe the n-gon in a circle. When calculating the length of a diagonal, instead of using the cosine theorem, you can bisect the central angle subtended on the diagonal. The bisector, the radia leading to the end points of the diagonal, and the diagonal will create two right triangles, from which you can get that the length of the diagonal is
$$ L = 2R \sin\frac{\alpha}{2}$$
where $R$ is the radius of the circle cicrumscribed on the polygon and $\alpha$ is the central angle subtended on the diagonal. You can also use this formula to calculate the length of a side of the n-gon:
$$ a = 2R \sin\frac{\pi}{n}$$
Since $a=1$, we get
$$ R = \frac{1}{2 \sin\frac{\pi}{n}} $$
$$ L = \frac{\sin\frac{\alpha}{2}}{\sin\frac{\pi}{n}}$$
$$ a_n = \frac{1}{n \sin\frac{\pi}{n}} \sum_{k=2}^{n-2} \sin\frac{\pi k}{n}$$
\begin{align} \lim_{n\to\infty} \frac{a_n}{n} &= \lim_{n\to\infty} \frac{1}{n^2 \sin\frac{\pi}{n}} \sum_{k=2}^{n-2} \sin\frac{\pi k}{n} = \\
&= \lim_{n\to\infty} \frac{1}{n \sin\frac{\pi}{n}} \cdot \lim_{n\to\infty} \frac{1}{n} \sum_{k=2}^{n-2} \sin\frac{\pi k}{n} = \\
&= \frac{1}{\pi} \cdot \int_0^1 \sin(\pi x) dx = \\
&= \frac{1}{\pi} \cdot \frac{-1}{\pi}\cos(\pi x)\big|_{x=0}^{x=1} =\\
&= \frac{1}{\pi}\cdot\frac{2}{\pi} = \frac{2}{\pi^2} \end{align}
|
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|
How can I prove this for $a+b+c=0$ We wish to prove that
$$\left(\frac{a-b}{c}+\ \frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)=9,$$ for $a+b+c=0$ and $$abc \neq 0 , a\neq b , b\neq c, c \neq a.$$
I tried working with only $2$ fractions at a time, switching some of the variables (ex.$a=-b-c$) and then adding them up, and multiplying first but could not go further.
Thank you in advance.
|
You can expand directly, by noticing that the first parantheses is
$$\frac{ab(a-b) + bc(b-c) + ca(c-a)}{abc} = -\frac{(a-b)(b-c)(c-a)}{abc}$$
So the expression becomes
$$\frac{a^3+b^3+c^3 + 3abc - a^2b - a^2c - b^2a -b^2c - c^2a - c^2b}{abc}$$
Collecting $a^3, -a^2b, - a^2c $ and similarly this becomes
$$\frac{a^2(a - b -c) + b^2(b-c-a) + c^2(c - a -b)}{abc} + 3$$
Now $a-b-c = 2a$ by the condition so this becomes just
$$2\frac{a^3+b^3+c^3}{abc} + 3$$
But there is an identity $a^3+b^3+c^3 - 3abc = (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)$, so we get $a^3+b^3+c^3 = 3abc$.
Thus we get that the initial expression is $9$.
|
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|
Möbius transformation mapping $\mathbb{D}$ into itself. I want to show that a Möbius transformation $f(z) = \frac{az + b}{cz + d}$ for $a,b,c,d \in \mathbb{C}$ maps the unit disk $\mathbb{D}$ into itself if and only if the coefficients satisfie
\begin{equation*}\label{ineq}
|\overline{b}d - \overline{a}c| + |ad - bc| \leq |d|^2 - |c|^2.
\end{equation*}
So far I have shown that if a Möbius transformation maps $\mathbb{D}$ into itself it satisfies the inequality.
Now my idea for the implication $\Leftarrow$:
I consider the inequality above and want to show that for
$|z| \leq 1 \implies$ $|f(z)| \leq 1$.
Instead of showing $|f(z)| \leq 1 $ I defined for $|d| > |c|$
\begin{equation*}
v = \frac{b \overline{d} - a\overline{c}}{|d|^2 - |c|^2}, \; R = \frac{|ad - bc|}{|d|^2 - |c|^2}
\end{equation*}
and want to show $|f(z) - v|^2 \leq R^2$ since by the provided inequality
from $|f(z) - v|^2 \leq R^2$ follows $f(z) \in K_R(v) \subseteq \mathbb{D}$.
For the case $|d| = |c|$, $f$ wouldn't be a möbius transformation.
Probelem: Now I have trouble showing the inequality above. I tried using the provided inequality
show that
\begin{equation*}
|f(z) - v|^2 = |f(z)|^2 - 2 \Re(f(z)\overline{v}) + |v|^2 \leq |f(z)|^2 - 2 \Re(f(z)\overline{v}) + 1 - 2R + R^2
\end{equation*}
but from then on I stuck by estimating
\begin{align*}
|f(z)|^2 &= f(z) \overline{f(z)} = \frac{|a|^2 |z|^2 + 2\Re(az \overline{b}) + |b|^2}{|c|^2 |z|^2 + 2\Re(cz \overline{d}) + |d|^2} \\
|v|^2 &= \frac{b \overline{d} - a\overline{c}}{|d|^2 - |c|^2}\frac{\overline{b} d - \overline{a}c}{|d|^2 - |c|^2} = \frac{|b|^2 |d|^2 - 2 \Re(b \overline{d} \overline{a}c) + |a|^2|c|^2}{(|d|^2 - |c|^2)^2} \\
&2 \Re(f(z)v) = 2 \frac{1}{|d|^2 - |c|^2}\Re \left( \frac{ab\overline{d}z - a^2 \overline{c}z - ab\overline{c} + b^2 \overline{d}}{cz + d}\right)
\end{align*}
or showing that
\begin{align*}
|f(z)|^2 - 2 \Re(f(z)\overline{v}) + 1 - 2R &\leq 0 \\
\end{align*}
I made some estimates based on the fact that $|z| \leq 1$ but most of them lead nowhere.
I would appreciate it if someone can give me a hint or can tell that the idea doesn't work.
|
I thought to share my solution if someone is interested:
\begin{align*}
|f(z) - w|^2 &= \left|\frac{az + b}{cz + d} - \frac{b \overline{d} - a\overline{c}}{|d|^2 - |c|^2} \right|^2 \leq r^2 = \frac{|ad - bc|^2}{(|d|^2 - |c|^2)^2} \\
&\iff \left|\frac{az + b}{cz + d}(|d|^2 - |c|^2) - b \overline{d} - a\overline{c} \right|^2 \leq |ad - bc|^2 \\
&\iff \left|(az + b)(|d|^2 - |c|^2) - (cz + d)b \overline{d} - a\overline{c} \right|^2 \leq |ad - bc|^2|cz + d|^2 \\
&\iff \left| \overline{d}(azd + bd - czb -bd ) - \overline{c}(azc + bc - cza - da) \right|^2 \leq (|ad - bc| |cz + d|)^2 \\
&\iff \left| \overline{d}z(ad - cb) - \overline{c}(bc - da) \right|^2 \leq (|ad - bc| |cz + d|)^2 \\
&\iff \left| \overline{d}z + \overline{c} \right|^2 \leq |cz + d|^2 \\
&\iff \left| \overline{d \overline{z} + c } \right|^2 \leq |cz + d|^2 \\
&\iff \left| d \overline{z} + c \right|^2 \leq |cz + d|^2
\end{align*}
and from $|z| \leq 1$ and $|d| \geq |c|$ we get
\begin{align*}
|d \overline{z} + c|^2 &= |d|^2 |z|^2 + 2 \Re(d \overline{cz}) + |c|^2 \leq |cz + d|^2 = |c|^2|z|^2 + 2 \Re(cz \overline{d}) + |d|^2 \\
&\iff |d|^2 |z|^2 + |c|^2 \leq |c|^2 |z|^2 + |d|^2 \\
&\iff |d|^2(|z|^2 - 1) \leq |c|^2(|z|^2 - 1) \\
&\iff |d|^2 \geq |c|^2
\end{align*}
Therefore the inequality holds.
For $d = c$ $f$ is not a mobius transformation as I already stated above in my question.
|
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|
Fastest way to find derivative of the function at the given point
What is derivative of $f(x)=\left(\dfrac{\sqrt[3]{x^2+2x}}{x^2-x}\right)^3$ at the point $x=2$? $$1)-\frac34\qquad\qquad2)-\frac54\qquad\qquad3)-\frac52\qquad\qquad4)-\frac{15}4$$
This is a problem from an timed Exam, so I am looking for the fastest way to solve this. Here is my solution:
We have $f(x)=\dfrac{x^2+2x}{(x^2-x)^3}$ . by using Quotient rule, derivative at $x=2$ is:
$$\dfrac{6\times8-(3\times3\times4)\times8}{8^2}=\frac34-\frac92=-\frac{15}4$$
Although the way I putted the values of functions instead of writing the whole derivative of function seems to be fast , it is hard to avoid algebraic mistakes and find the correct answer in the exam condition (having time pressure, stress and so on).
After all, is there a better approach (faster) to solve this problem ?
|
Perhaps slightly...but very slightly...easier: using that $\;(x^2-x)^3=x^3(x-1)^3\;$ and also $\;x=2\implies (x-1)^n=1\;$
$$f(x)=\frac{x^2+2x}{(x^2-x)^3}=\frac{x+2}{x^2(x-1)^3}=\frac1{x(x-1)^3}+\frac2{x^2(x-1)^3}\implies$$
$$f'(x)=\left(\frac1{x(x-1)^3}\right)'+2\left(\frac1{x^2(x-1)^3}\right)'=$$
$$=-\frac{(x-1)+3x}{x^2(x-1)^4}-2\frac{2(x-1)+3x(x-1)}{x^3(x-1)^4}\implies$$
$$f'(2)=-\frac{1+6}{4\cdot1}-2\frac{2+6}{8\cdot1}=-\frac74-2=-\frac{15}4$$
Another Option: Write your function multiplicatively and use the product rule, which surely will be easier than before:
$$f(x)=x^{-2}(x+2)(x-1)^{-3}\implies $$
$$f'(x)=-2x^{-3}(x+2)(x-1)^{-3}+x^{-2}(x-1)^{-3}-3x^{-2}(x+2)(x-1)^{-4}\implies$$
$$f'(2)=-2\cdot\frac18\cdot4+\frac14-\frac34\cdot4=-1+\frac14-3=-4+\frac14=-\frac{15}4$$
I'd go with this last option. As thumb rule, product rule nice than quotient rule
|
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|
Mathematical Induction: Can I assume that $P(k-1)$ is also true in the induction hypothesis? Suppose that $y \in \mathbb{R}$, $y \neq 0$ and $y + \frac{1}{y} \in \mathbb{Z}$. Prove using mathematical induction that $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.
Basis step: The statement is true for $n=1$ since: $y^{1} + \frac{1}{y^{1}} = y^{n} + \frac{1}{y^{n}} \in \mathbb{Z}$.
Induction hypothesis: Assume that $y^{k} + \frac{1}{y^{k}}$ and $y^{k-1} + \frac{1}{y^{k-1}}$ are integers for all $n \geq 1$.
Induction step: We show that $y^{k+1} + \frac{1}{y^{k+1}}$ is true.
$y^{k+1} + \frac{1}{y^{k+1}} = (y^{k} + \frac{1}{y^{k}})(y + \frac{1}{y}) - (\frac{y^{k}}{y} + \frac{y}{y^{k}})$
$y^{k+1} + \frac{1}{y^{k+1}} = (y^{k} + \frac{1}{y^{k}})(y + \frac{1}{y}) - (y^{k-1} + \frac{1}{y^{k-1}})$
Since it is given that $y + \frac{1}{y} \in \mathbb{Z}$ and by the induction hypothesis, $y^{k} + \frac{1}{y^{k}}$, $y^{k-1} + \frac{1}{y^{k-1}} \in \mathbb{Z}$:
$\Rightarrow (\mathbb{Z})(\mathbb{Z})-\mathbb{Z}=\mathbb{Z}$ (Closure property for integers)
Therefore, $y^{n} + \frac{1}{y^{n}}$ is an integer for all $n \geq 1$.
This is the only way I was able to do the proof. However, I want to know if I am allowed to assume that $P(k-1)$ is also true in the induction hypothesis since as far as I know we can only assume that $P(k)$ is true. Would also appreciate feedback and alternative solutions for my proof!
|
For $n=2$ and $n=1$ the statement is correct.
Because, $$\left(y^2+\frac{1}{y^2}\right)=\left(y+\frac 1y\right)^2-2$$
Then ,suppose that $n=k$ and $n=k-1$ are also correct.....
|
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|
Can we obtain a sufficient constraint over the parameters $A$, $B$, and $C$ in the given equation? I have this equation
$$ A\;\big(\cos (x-y)+\cos x+\cos y\big)+B\;\big(\sin (x-y)-\sin x+\sin y\big)+C=0, $$
where $-\pi\leq x,y \leq \pi$, and $A$, $B$, and $C$ are some real constants.
Then, according to the ranges of the trigonometric functions, is there a sufficient constraint over the parameters $A$, $B$, and $C$ to have the real solutions? If so, how can we obtain that condition?
|
Using Cauchy-Schwarz inequality repeatedly: $C^2 = (A\cos(x-y)+B\sin(x-y) +A\cos x-B\sin x + A\cos y + B\sin y)^2 \le 3(m^2+p^2+q^2), m^2 \le A^2+B^2, n^2 \le A^2+B^2, p^2 \le A^2+B^2\implies C^2 \le 3(A^2+B^2+A^2+B^2+A^2+B^2)= 9(A^2+B^2). $, Thus the necessary condition is: $C^2\le 9A^2+9B^2$.
|
{
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|
Floor Function Quotient Fact So I was perusing wikipedia for some floor function facts for a problem I am working on and I saw this:
Let $x,y\in \mathbb{R}$
$$\lfloor x\rfloor +\lfloor y\rfloor \leq \lfloor x+y\rfloor \leq \lfloor x\rfloor +\lfloor y\rfloor +1$$
However when I adapted my problem to this
Let $n,m,k\in \mathbb{Z},k\neq0$
$$\lfloor \frac{n}{k}\rfloor +\lfloor \frac{m}{k}\rfloor \leq \lfloor \frac{n}{k}+\frac{m}{k}\rfloor \leq \lfloor \frac{n}{k}\rfloor +\lfloor \frac{m}{k}\rfloor +1$$
I really didn't want to work with the inequalities, so I devised an equality that encompasses the idea from above as
$$\lfloor\frac{n+m}{k}\rfloor=\lfloor\frac{n}{k}\rfloor+\lfloor\frac{m}{k}\rfloor+\lfloor\lbrace\frac{n}{k}\rbrace+\lbrace\frac{m}{k}\rbrace\rfloor$$
$\lfloor num \rfloor$ is the floor of num and $\lbrace num \rbrace$ is the fractional component of num
However, I don't know if it this is true. So help on a proof of the last equation would be much appreciated.
Bonus points if we can also figure out if the equality is true or not for $n,m,k\in \mathbb{R},k\neq0$
Thanks in advanced!
Edit: I wanted a go at the proof after seeing the answer too, so check below for the alternate proof
Let $x,y\in \mathbb{R}$ and let $n\in \mathbb{Z}$
We will use the fact that $x = \lfloor x \rfloor + \lbrace x \rbrace$ and $\lfloor x + n \rfloor = \lfloor x \rfloor + n$.
So $\lfloor x + y \rfloor = \lfloor \lfloor x \rfloor + \lbrace x \rbrace + \lfloor y \rfloor + \lbrace y \rbrace \rfloor$
then commute
$\lfloor \lfloor x \rfloor + \lbrace x \rbrace + \lfloor y \rfloor + \lbrace y \rbrace \rfloor= \lfloor \lbrace x \rbrace + \lbrace y \rbrace+ \lfloor x \rfloor + \lfloor y \rfloor \rfloor$
and we know that $\lfloor x \rfloor,\lfloor y \rfloor \in \mathbb{Z}$
so by our second fact
$\lfloor \lbrace x \rbrace + \lbrace y \rbrace+ \lfloor x \rfloor + \lfloor y \rfloor \rfloor = \lfloor \lbrace x \rbrace + \lbrace y \rbrace+ \lfloor x \rfloor \rfloor + \lfloor y \rfloor$
$= \lfloor \lbrace x \rbrace + \lbrace y \rbrace \rfloor + \lfloor x \rfloor + \lfloor y \rfloor$
And rearrange to get
$\lfloor x + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor + \lfloor \lbrace x \rbrace + \lbrace y \rbrace \rfloor$
|
To prove : for $A,B \in \Bbb{R} $
$\lfloor(A + B)\rfloor = \lfloor A\rfloor
+ \lfloor B\rfloor + \lfloor \{A\} + \{B\}\rfloor.$
By definition:
$A = \lfloor A\rfloor + \{A\}.$
$B = \lfloor B\rfloor + \{B\}.$
Therefore,
$$(A + B) = \lfloor A\rfloor + \lfloor B\rfloor
+ \{A\} + \{B\}. \tag1$$
Also, since $0 \leq \{A\},\{B\} < 1$
$0 \leq \{A\} + \{B\} < 2.$
Also, if $P \in \Bbb{R}$ and $z \in \Bbb{Z}$ such that
$0 \leq (P - z) < 1$ then $\lfloor P\rfloor = z.$
Consider two cases separately.
$\underline{\text{Case 1:}~~0 \leq \{A\} + \{B\} < 1}$
Then
*
*$\lfloor \{A\} + \{B\}\rfloor = 0$
*Using equation (1) above,
$0 \leq (A + B) - (\lfloor A\rfloor + \lfloor B\rfloor) < 1. $
Therefore,
$\lfloor (A + B)\rfloor = \lfloor A\rfloor + \lfloor B\rfloor =
\lfloor A\rfloor + \lfloor B\rfloor + \lfloor \{A\} + \{B\}\rfloor.$
Therefore, the assertion holds in Case 1.
$\underline{\text{Case 2:}~~1 \leq \{A\} + \{B\} < 2}$
Then
*
*$\lfloor \{A\} + \{B\}\rfloor = 1$
*Using equation (1) above,
$0 \leq (A + B) - [(\lfloor A\rfloor + \lfloor B\rfloor) + 1]< 1. $
Therefore,
$\lfloor (A + B)\rfloor = \lfloor A\rfloor + \lfloor B\rfloor + 1 =
\lfloor A\rfloor + \lfloor B\rfloor + \lfloor \{A\} + \{B\}\rfloor.$
Therefore, the assertion holds in Case 2.
|
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|
Help with $\int \frac{1}{(1-t^2)t^2} \, dt$ $$\int \frac{1}{(1-t^2)t^2} \, dt$$
By using partial fractions I get:
$$\frac{1}{(1-t)(1+t)t^2} = \frac{A}{t} + \frac{B}{t^2} + \frac{C}{t+1} + \frac{D}{1-t}$$
$$1 = t^3 (-A-C+D) + t^2 (-B+C+D) + At+ B$$
$$-A-C+D = 0$$
$$-B+C+D=0$$
$$A=0$$
$$B=1$$
So, $A=0, B=1, C=\frac{1}{2}, D = \frac{1}{2}$
Then by replacing the variables with the results we get:
$$\int \frac{1}{(1-t^2)t^2} \, dt = \int \frac{1}{t^2} \, dt + \frac{1}{2} \int \frac{1}{t+1} \, dt + \frac{1}{2} \int \frac{1}{1-t} \, dt$$
$$- \frac{1}{t} + C + \frac{1}{2} \ln|t+1| + C + \frac{1}{2} \ln|1-t| + C$$
$$- \frac{1}{t} + \frac{1}{2} \ln|t+1| + \frac{1}{2} \ln|1-t| + C$$
Is this correct?
|
I agree that$$\int\frac{1}{t^2(1-t^2)}dt=\int\left(\frac{1}{t^2}+\frac{1}{1-t^2}\right)dt=-\frac1t+\frac12\ln\left|\frac{1+t}{1-t}\right|+K,$$where the locally constant function $K$ can change values either side of $t=\pm1$, so its values can be different for $t<-1,\,t\in(-1,\,1),\,t>1$. (I've called it $K$ rather than $C$ since $C$ is already a coefficient in your partial fractions calculation.) As @Vasya noted, the only issue is your $\ln|1-t|$ coefficient.
|
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|
Solve definite integral $\int_{\frac{\sqrt2}2}^{\sqrt{3}} \dfrac{dx}{x\sqrt{x^2+1}}$ by variable substitution I have this integral
$$\int_{\sqrt{2}/2}^{\sqrt{3}} \dfrac{dx}{x\sqrt{x^2+1}}$$
I have to solve it by substituting x.
I believe that the easiest way would be to substitute with $t = \sqrt{x^2+1}$. The other ways I tried didn't help me much.
so.. let $t = \sqrt{x^2+1}$. Suppose $\sqrt{6}/2 \leq t \leq 2$. In this case, we can say that $x = \varphi(t) = \sqrt{t^2-1}$. $\varphi(\sqrt{6}/2) = \sqrt{2}/2$ and $\varphi(2) = \sqrt{3}$.
Now we can substitute.
$$\int_{\sqrt{6}/2}^{2} {\dfrac{1}{\varphi(t)\sqrt{\varphi^2(t) + 1}}\varphi'(t)dt} =
\int_{\sqrt{6}/2}^{2} {\dfrac{dt}{t\sqrt{t^2-1}}}$$
But I'm stuck here. Actually the indefinite integral of the integrand is equal to $-\ln{\sqrt{t^2-1}}$, but then by using the fundamental theorem of calculus I get a wrong result. What am I doing wrong?
The right answer is $\ln{\dfrac{3+\sqrt{6}}{3}}$. Thanks in advance.
|
Let $$ \mathscr{I} = \int \dfrac{dx}{x\sqrt{x^2+1}}$$
Put $ x = \dfrac1t \implies dx = -\dfrac{dt}{t^2}$
$$ \begin{align} \mathscr{I} &\ = -\int t\dfrac{1}{\sqrt{\frac{1}{t^2}+1}} \dfrac{dt}{t^2} \\ &\ =- \int\dfrac{dt}{\sqrt{t^2+1}} \\ &\ = - \ln \left| t + \sqrt{t^2+1} \right| + C
\end{align} $$
Now, substitute the changed limits
$$ \begin{align} \int_{\frac{\sqrt{2}}{2}}^{\sqrt{3}} \dfrac{dx}{x\sqrt{x^2+1}} &\ = \ln \left| t + \sqrt{t^2+1} \right|_{\frac{1}{\sqrt3}} ^\sqrt2 \\ &\ = \ln \dfrac{\sqrt2 + \sqrt3}{\frac{1}{\sqrt3} + \frac{2}{\sqrt3} } \\ &\ = \ln{\dfrac{3+\sqrt{6}}{3}} \end{align} $$
|
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|
Integrating $\int \frac{r^3}{\sqrt{16 + r^2}} dr$ I need to calculate: $$\int \frac{r^3}{\sqrt{16 + r^2}} dr$$
I am currently in the midst of learning AP Calculus BC and the lesson from which this problem came from goes over integration by parts.
Basically, it utilizes that : $$\frac{d}{dx} \left[ f(x)g(x) \right] = f'(x)g(x) + f(x)g'(x)$$ in order to get: $$\int f'(x)g(x) dx = f(x)g(x) - \int f(x)g'(x) dx.$$
Sometimes, we have to apply this process multiple times.
However, for this question, I cannot seem to figure out what to set $f(x)$ and $g(x)$ to. So far, I have tried setting $f(x) = x \rightarrow f'(x) = 1$ and $g(x) = \frac{x^3}{\sqrt{16 + x^2}}$.
This left me needing to calculate $$\int x \cdot \frac{2x^2 (x^2 + 24)}{ (16 + x^2)^{\frac{3}{2}}}$$ which I have no idea how to do.
For my second attempt, I tried letting $f(x) = \frac{1}{4} x^4 \rightarrow f'(x) = x^3$ and $g(x) = \frac{1}{\sqrt{16 + x^2}}.$
This led to me needing to calculate: $$\int \frac{1}{4} x^4 \cdot \bigg(- \frac{x}{(16 + x^2)^{\frac{3}{2}}} \bigg)$$ which I have also been unable to compute.
What should I set $f(x)$ and $g(x)$ to? Thank you in advance for any help.
|
The fastest way to solve this integral (IMO), does not use IBP. We let
\begin{align*}
u &= \sqrt{16 + r^2} &&\implies \boxed{r^2 = u^2 - 16}\\
\frac{du}{dr} &= \frac{2r}{2\sqrt{16+r^2}}&&\implies\boxed{du = {\frac{r}{\sqrt{16+r^2}}}\,dr.}
\end{align*}
Making this substitution:
\begin{align*}\int \frac{r^{3}}{\sqrt{16+r^2}}\,dr &= \int r^2\,du\\
&= \int (u^2-16)\,du\\
&= \frac{u^{3}}{3} - 16u + C\\
&= \boxed{\frac{(16+r^2)^{3/2}}{3} - 16\sqrt{16+r^2} + C.}
\end{align*}
|
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|
Why is $\sqrt{6 + \sqrt{6 + \sqrt{6 + ...}}} = 3$ This is a problem from SASMO Grade 8 (Secondary 2) Sample Questions.
Solve for $x$
$\sqrt{x + \sqrt{x + \sqrt{x + ...}}} = 3$
Answer: $x=6$
I have tried this on a calculator: the more $x$ we add, the closer we are to $3$. But how can we prove it, and is there a way to figure this out by hand?
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Let $x= \sqrt{6+ \sqrt{6+\sqrt{6+ \sqrt{\cdot\cdot\cdot}}}}$
then $x^2= 6+ \sqrt{6+ \sqrt{6+ \sqrt{\cdot\cdot\cdot}}}= 6+ x$. Solve $x^2-x- 6= 0$.
Clearly this has to be positive. What is the positive root of $x^2- x- 6= 0$?
(There is only one positive root and this clearly has to be positive.)
|
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|
Calculating $\cos^{-1}{\frac{3}{\sqrt10}} + \cos^{-1}{\frac{2}{\sqrt5}}$ $$\cos^{-1}{\frac{3}{\sqrt{10}}} + \cos^{-1}{\frac{2}{\sqrt 5}}= ?$$
Let $\cos^{-1}{\frac{3}{\sqrt{10}}}=\alpha,
\cos^{-1}{\frac{2}{\sqrt 5}}=\beta$ then, $\cos\alpha=\frac{3}{\sqrt{10}}, \cos\beta=\frac{2}{\sqrt5}$
Therefore $$\cos\alpha=\frac{3\cdot2}{2\sqrt2\sqrt5}= \frac{3}{2\sqrt2}\cdot\cos\beta$$
This is all I did till now. Could you go further with this to answer?
|
Hint : apply this formula:
$$\cos^{-1} x +\cos^{-1}y=\cos^{-1}[xy-\sqrt{(1-x^2)(1-y^2)}]$$
Put $x=\frac2{\sqrt {10}}$ and $y=\frac 2{\sqrt 5}$
|
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|
How should I check that $6^{\log_{10} x} + 8^{\log_{10} x} = x $ does not have other solutions? The question is to solve the equation $6^{\log_{10} x} + 8^{\log_{10} x} = x $
I know one of the solutions is $x=100$ using Pythagorean triples but I can't show that this is the only solution.
I'm looking for an idea without using differentiation!
any hints would be appreciated. tnx
|
\begin{align*}
6^{\log_{10} x} + 8^{\log_{10} x}
&= 6^{\log_6(x)/ \log_6 10} + 8^{\log_8(x)/ \log_8 10} \\
&= x^{1/\log_6 10} + x^{1/\log_8 10} \text{.}
\end{align*}
We have
\begin{align*}
6^{\log_{10} x} + 8^{\log_{10} x}
&= x \qquad \text{when} \\
\frac{6^{\log_{10} x} + 8^{\log_{10} x}}{x} &= 1 \qquad \text{ so when} \\
\frac{x^{1/\log_6 10} + x^{1/\log_8 10}}{x} &= x^{-1 +1/\log_6 10} + x^{-1 + 1/\log_8 10} \\
&= 1 \text{.}
\end{align*}
We know this happens when $x = 100$. Also,
$$ \frac{\mathrm{d}}{\mathrm{d}x} \left( x^{-1 +1/\log_6 10} + x^{-1 + 1/\log_8 10} \right) = \left( -1 +1/\log_6 10 \right) x^{-2 +1/\log_6 10} + \left( -1 + 1/\log_8 10 \right) x^{-2 + 1/\log_8 10} \text{.} $$
Recall the original equation, in which the domain of $\log_{10} x$ is $x > 0$. For $x > 0$, the powers of $x$ in this derivative are positive. Also, $-1 +1/\log_6 10 = -0.22{\dots} < 0$ and $-1 +1/\log_8 10 = -0.09{\dots} < 0$, so the derivative is negative on the entire domain. Therefore, $\displaystyle \frac{6^{\log_{10} x} + 8^{\log_{10} x}}{x}$ is strictly monotonically decreasing, so can only take the value $1$ at most once.
We can even determine that $-1 +1/\log_6 10 < 0$ without a calculator. Since $10 > 6$, $\log_6 10 > 1$ and taking reciprocals, $\frac{1}{\log_6 10} < \frac{1}{1} = 1$, so we don't add enough to $-1$ to get a positive result. A parallel argument shows the same thing about $-1 +1/\log_8 10$.
|
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|
Identities similar to $\frac{a^3 + b^3 + c^3}{3} \cdot \frac{a^7 + b^7 + c^7}{7} = \left( \frac{a^5 + b^5 + c^5}{5} \right)^2$ This question was inspired by the post here, which asks for a proof of the following fact:
If $a+b+c = 0$ then show that
$$\frac{a^3 + b^3 + c^3}{3} \cdot \frac{a^7 + b^7 + c^7}{7} = \left( \frac{a^5 + b^5 + c^5}{5} \right)^2$$
I was curious about finding all exponents for which this works, arriving at the following problem:
For which distinct positive integers $m,n$ (with $m+n$ even) is it true that
$$\frac{a^m + b^m + c^m}{m} \cdot \frac{a^n + b^n + c^n}{n} = \left( \frac{a^\frac{m+n}{2} + b^\frac{m+n}{2} + c^\frac{m+n}{2}}{(m+n)/2} \right)^2$$
for all $a,b,c$ with $a+b+c = 0$?
I added the distinctness condition because $m=n$ always trivially works.
So far, the only solution I know of is $(3,7)$, as per the post quoted. A Mathematica search yielded that there are no other solutions with $a,b \leq 100$.
A solution for this problem would of course be really nice, but I don't really have high hopes for one! The shortest proof of the $(3,7)$ case uses Newton's formulas for power sums, which are defined recursively and don't have a nice closed form, as far as I know. So any kind of intuition/heuristic about why there should/shouldn't be any more solutions would also be very interesting.
Some progress: I decided to run a much bare-bone simulation in Mathematica, by checking if the identity holds at least for $(a,b,c) = (1,-\frac{1}{2},-\frac{1}{2})$, and indeed for $m,n \leq 2000$ it only holds for $(m,n) = (3,7)$. Plugging in, we have
$$\frac{1}{mn} \left(1 + \frac{(-1)^m}{2^{m-1}}\right)\left(1 + \frac{(-1)^n}{2^{n-1}}\right) = \frac{4}{(m+n)^2}\left( 1 + \frac{(-1)^\frac{m+n}{2}}{2^{\frac{m+n}{2} - 1}} \right)^2.$$
Thus we can turn the problem into a Diophantine equation, which might make things simpler?
|
The first step is to pin down the parity of $m$ and $n$. Since $m+n$ is even, we must have $m\equiv n\pmod{2}$.
Considering the case $(a,b,c)=(x,-x,0)$, we have $$x^{m+n}\cdot\frac{(1+(-1)^m)(1+(-1)^n)}{mn}=x^{m+n}\cdot\frac{4\left(1+(-1)^{\frac{m+n}{2}}\right)^2}{(m+n)^2}$$ Canceling $x$ and considering when these vanish, we find $m\equiv n\equiv\frac{m+n}{2}\pmod{2}$.
In the even case, we simplify to obtain $$\frac{1}{mn}=\frac{4}{(m+n)^2}\text{;}$$ thus $4mn=(m+n)^2$. Rearranging, $(m-n)^2=0$. (The original version of this paragraph had an error; HT to Tanny Sieben for pointing out the appropriate correction.)
In the odd case, things are more tricky. Following your idea, let's take $(a,b,c)=(2,-1,-1)$. (Sorry; fractions are annoying to type.)
Then $$\frac{(2^m-2)(2^n-2)}{mn}=\left(\frac{2^{\frac{m+n}{2}}-2}{\frac{m+n}{2}}\right)^2$$ Again expanding, we have $$(m+n)^2(2^{m+n}-2^{n+1}-2^{m+1}+4)=4mn(2^{m+n}-2^{\frac{m+n}{2}+2}+4)$$ Rearranging via the identity $(m+n)^2-4mn=(m-n)^2$, we have $$(m-n)^2(2^{m+n}+4)=(m+n)^2(2^{n+1}+2^{m+1})-mn2^{\frac{m+n}{2}+4}$$
At this point, a change of variables suggests itself: let $a=\frac{m+n}{2}$ and $d=\frac{m-n}{2}$. Note that $a>d>0$, $a$ is odd, and $d$ is even. In any case, $$4d^2(2^{2a}+4)=4a^2(2^{a+d+1}+2^{a-d+1})-(a^2-d^2)2^{a+4}$$ Canceling a factor of $2^{a+3}$, $$d^2(2^{a-1}+2^{1-a})=a^2(2^d+2^{-d})-2(a^2-d^2)$$ Rearranging, $$d^2(2^{a-1}+2^{1-a}-2)=a^2(2^d+2^{-d}-2)$$
But our equation is now a perfect square: $$d^2\left(2^{\frac{a-1}{2}}-2^{\frac{1-a}{2}}\right)^2=a^2\left(2^{\frac{d}{2}}-2^{-\frac{d}{2}}\right)^2$$ Taking square roots, and rearranging, $$\frac{2^{\frac{a-1}{2}}-2^{\frac{1-a}{2}}}{a}=\frac{2^{\frac{d}{2}}-2^{-\frac{d}{2}}}{d}\tag{1}$$
Call these $L(a)$ and $R(d)$, and consider them as functions over $\mathbb{R}^+$. What can we say?
Well, both $L$ and $R$ are increasing to infinity, with $\lim_{a\to0^+}{L(a)}=-\infty$ and $\lim_{d\to0^+}{R(d)}=\ln{(2)}$. Thus, for each $d$, there is a unique $a$ satisfying (1). Checking some small integers, we find (as expected) that there is no solution for $d\in\{1,3,4,5\}$, but there is, of course, $(d,a)=(2,5)$.
Moreover, for $d\geq5$, we have $L(d+2)>R(d)$. Thus $d<a<d+2$; that is to say, if there is another integer solution, it arises from $a=d+1$.
Are there any such solutions? Well, if $a=d+1$, then $n=1$ (w/oLoG) and $m=2d+1$. So $$\left(\frac{a^{d+1}+b^{d+1}+c^{d+1}}{d+1}\right)^2=\left(\frac{a^{2d+1}+b^{2d+1}+c^{2d+1}}{2d+1}\right)\left(\frac{a+b+c}{1}\right)$$ But the latter is $0$ by assumption (!). It is easy to see then that we must have $d=0$, the trivial case.
|
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Let $f(x)=(x+1)(x+2)..(x+100)$ and $g(x)=f(x)f’’(x)-(f’(x))^2$. Find number of roots $n$ of $g(x)=0$ After relevant simplification ie. Log on both sides and then differentiating, the expression received is
$$\frac{f’(x)}{f(x)} =\sum_{k=1}^{100} \frac{1}{x+k}$$
Differentiating again will give
$$\frac{g(x)}{(f(x))^2}=-\sum_{k=1}^{100} \frac{1}{(x+k)^2}$$
So in turn, the roots of $g(x)=0$ are simply the roots of $f(x)=0$, so ans should be 100, but that’s not correct.
Where am I going wrong?
|
$$\frac{f’(x)}{f(x)} =\sum_{k=1}^{100} \frac{1}{x+k}$$
D.w.r.t. $x$ both sides
$$\frac{f(x)f''(x)-f'^2(x)}{f^2(x)}=-\sum_{k=1}^{100} \frac{1}{(x+k)^2}<0 \ne 0.$$
So $$g(x)=f(x)f''(x)-f'^2(x)=\sum_{k=1}^{n}\frac{f^2(x)}{(x+k)^2}$$
Then $$-g(x)=(x+2)^2(x+3)^2(x+4)^2....(x+n)^2+ (x+1)^2(x+3)^2(x+4)^2....(x+n)^2+......>0 \ne 0,$$ has no real root.
|
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|
Integral involving the modified Bessel function of the second kind K_0 The following integral
$$
\int_0^{+\infty} K_0(\alpha\sqrt{x^2+z^2})\, dx, \alpha>0,
$$
can be computed according to Gradshteyn-Ryzhik 6.596 (3) taking $\nu=0$ and $\mu=-1/2$. Its value is $\frac{\pi}{2\sqrt{\alpha}} e^{-\alpha|z|}$.
My question is: what about the integral
$$
\int_0^{+\infty} \cosh(\beta x) K_0(\alpha\sqrt{x^2+z^2})dx, \alpha>0, \beta \geq 0?
$$
|
Assume that $\alpha>\beta$ and $|\arg z|<\frac{\pi}{4}$. Then using http://dlmf.nist.gov/10.32.E10 and interchanging the order of integrations,
\begin{align*}
& \int_0^{ + \infty } {\cosh (\beta x)K_0 (\alpha \sqrt {x^2 + z^2 } )dx}
\\ &
= \frac{1}{2}\int_0^{ + \infty } {\exp \left( { - t - \alpha ^2 \frac{{z^2 }}{{4t}}} \right)\int_0^{ + \infty } {\cosh (\beta x)\exp \left( { - \alpha ^2 \frac{{x^2 }}{{4t}}} \right)dx} \frac{{dt}}{t}}
\\ &
= \frac{{\sqrt \pi }}{{2\alpha }}\int_0^{ + \infty } {\exp \left( { - t - \alpha ^2 \frac{{z^2 }}{{4t}} + \left( {\frac{\beta }{\alpha }} \right)^2 t} \right)\frac{{dt}}{{t^{1/2} }}}
\\ &
= \frac{1}{2}\sqrt {\frac{\pi }{{\alpha ^2 - \beta ^2 }}} \int_0^{ + \infty } {\exp \left( { - s - \frac{{(\alpha ^2 - \beta ^2 )z^2 }}{{4s}}} \right)\frac{{ds}}{{s^{1/2} }}}
\\ &
= \sqrt {\frac{{\pi z}}{{2\sqrt {\alpha ^2 - \beta ^2 } }}} K_{ - 1/2} (\sqrt {\alpha ^2 - \beta ^2 } z)
\\ &
= \frac{\pi }{{2\sqrt {\alpha ^2 - \beta ^2 } }}e^{ - \sqrt {\alpha ^2 - \beta ^2 }z } .
\end{align*}
In the last step, we employed http://dlmf.nist.gov/10.39.E2. Note that your simplification in terms of an exponential when $\beta=0$ is not correct.
|
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Find the least value of $\frac{1}{x}+\frac{3}{y}+\frac{5}{z}$ Given that $x$,$y$ and $z$ are positive numbers and $x+y+z=1$ , Find the least value of $\frac{1}{x}+\frac{3}{y}+\frac{5}{z}$.
My approach:
We write,
$$\frac{1}{x}+\frac{3}{y}+\frac{5}{z}=\frac{x+y+z}{x}+\frac{(3)(x+y+z)}{y}+\frac{(5)(x+y+z)}{z}$$
This further simplifies to,
$$ 9+\left(\frac{y}{x}+\frac{3x}{y}\right)+\left(\frac{z}{x}+\frac{5x}{z}\right)+\left(\frac{3z}{y}+\frac{5y}{z}\right)$$
Now we can use AM-GM inequalitie in each of the parentheses.
So the least value of above expression works out to
$$9+2\left(\sqrt{3}+\sqrt{5}+\sqrt{15}\right)$$
Is this correct?? If not could you please suggest some ways of approaching this problem?.
Thank you!
|
One can show that $\sum \limits_{I=1}^{n}\dfrac{a_i^2}{b_i}\geq \dfrac{(a_1+\dots+a_n)^2}{b_1+\dots+b_n}.$ using this inequality it follows that $$\frac{1}{x}+\frac{3}{y}+\frac{5}{z}\geq (1+\sqrt{3}+\sqrt{5})^2$$
|
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|
Extraneous solutions without performing squaring operation
Solve in $[0, 2\pi]$ $$\sec x+\tan x=2\cos x$$
My mind boggled while solving the trigonometric equation in two different ways:
Method $1.$
Assuming $x\ne \frac{\pi}{2},\frac{3\pi}{2}$
We have:
$$\sec x+\tan x=2\cos x$$
$\implies$
$$\sec x-\tan x=\frac{1}{2\cos x}$$
Adding both we get
$$2\sec x=2\cos x+\frac{1}{2\cos x}$$
$\implies$
$$\frac{3}{2\cos x}=2\cos x$$
$\implies$
$$\cos x=\frac{\pm \sqrt{3}}{2}$$
which gives four solutions in $[0,2\pi]$ which are $\frac{\pi}{6},\frac{11\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6}$ but only two of them which are $\frac{\pi}{6}, \frac{5\pi}{6}$ satisfying the original equation.
Method $2.$
Writing the given equation as
$$\frac{1}{\cos x}+\frac{\sin x}{\cos x}=2\cos x $$
Then we get
$$1+\sin x=2(1-\sin^2 x)$$
$\implies$
$$2\sin^2 x+\sin x-1=0$$
$\implies$
$$\sin x=-1, \sin x=\frac{1}{2}$$
Since $x\ne \frac{\pi}{2},\frac{3\pi}{2}$ we can ignore $\sin x=-1$. Thus we have
$\sin x=\frac{1}{2}$ which gives the two solutions $\frac{\pi}{6}, \frac{5\pi}{6}$
My question is why am i getting two extraneous solutions in Method $1$ even though they are in the domain of $\sec x$ and $\tan x$?
|
Note that this is not limited to trigonometric equations.
You obtained $\sec(x)-\tan(x)=\frac 1{2\cos(x)}$ by taking the inverse of both sides of the equation.
Let examine the very simple case $x=2\iff \dfrac 1x=\dfrac 12$
Notice that when summing these two equations $x+\dfrac 1x=\dfrac 52$ you now end up with two solutions $2$ and $\frac 12$.
This happens because you symmetrized the equation to $S(x,\frac 1x)=0$, therefore whenever $x$ is solution, $\frac 1x$ is solution too and the extraneous $\frac 12$ appears.
This is more or less the same issue here, but of course because we are dealing with trigonometric functions, the extraneous solutions are not just the inverses, but the principle is the same.
|
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|
Solving $7^x\bmod {29} = 23 $ I have $$7^x\bmod {29} = 23 $$
It is possible to get $x$ by trying out different numbers but that will not be possible if $x$ is actually big.
Are there any other solutions for this equation?
Kind regards
|
I don't think there is anything but trial and error.
But look. $7^2 \equiv 23 \equiv -6 \pmod {29}$
$7^{2x} \equiv 36 \equiv 7\pmod {29}$.
So $7^{2x-1} \equiv 1 \pmod {29}$.
We know by FLT the $7^{28}\equiv 1$ so for the least power to which $7$ is a multiple of $1$ must be a divisor of $28$.
So $2x-1$ is a multiple of a divisor of $28$. Immediately we have $2x-1$ is odd so it must be a multiple of an odd divisor of $28$.
The only nontrivial odd divisor of $28$ is $7$. So test to see if $7^7\equiv 1 \pmod {29}$. We have $7^7 \equiv 1\pmod {29}$. That was lucky. so we have $2x-1 = 7k$ and as there are only $7$ values of $7^w\pmod {29}$, we might as well assume $x \le 6$. so $x = 4$ is only option. Let's try that.
$7^4 \equiv 23\pmod {29}$.
Yep.... that worked.
|
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|
determine the coefficient of $x^2yz$ in the expansion of $(2x-y+z+1)^7$
determine the coefficient of $x^2yz$ in the expansion of $(2x-y+z+1)^7$
I found this question in my textbook , i seem very easy in the beginning but my answer is wrong according to answerkey.
I said that the coefficient is equal to $\frac{7!}{2! \times 1 \times 1 \times 3!} \times 2^2 \times (-1)$
However ,the answer is $-280$. What am i missing ?
|
A small crosscheck. We use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write for instance
\begin{align*}
[x^k](ax+z)^n=\binom{n}{k}a^kz^{n-k}\tag{1}
\end{align*}
We obtain by successively applying (1)
\begin{align*}
\color{blue}{[x^2yz]}&\color{blue}{(2x-y+z+1)^7}\\
&=[yz]\binom{7}{2}2^2(-y+z+1)^5\\
&=[z]\binom{7}{2}2^2\binom{5}{1}(-1)(z+1)^4\\
&=\binom{7}{2}2^2\binom{5}{1}(-1)\binom{4}{1}\\
&\,\,\color{blue}{=-1\,680}
\end{align*}
in accordance with OPs result.
|
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|
How to solve $x^4-2x^3-x^2+2x+1=0$? How to solve $x^4-2x^3-x^2+2x+1=0$?
Answer given is: $$\frac{1+\sqrt5}{2}$$
I tried solving it by taking common factors:
$$x^3(x-2)-x(x-2)+1=0 $$ $$x(x-2)(x^2-1)+1=0 $$ $$(x+1)(x)(x-1)(x-2)+1=0$$
But it's not leading me anywhere.
|
$x^4-2x^3-x^2+2x+1 = 0$
You should split the middle terms and try to look for a common factor.
$x^4-x^3-x^2-x^3+x^2+x-x^2+x+1 = 0$
$x^2(x^2-x-1)-x(x^2-x-1)-1(x^2-x-1) = 0$
$(x^2-x-1)^2 = 0$
Take the square root:
$x^2-x-1 = ±0$
$x^2-x-1 = 0 \lor x^2-x-1 = -0$
Add $\frac{5}{4}$:
$x^2-x+\frac{1}{4} = \frac{5}{4} \lor x^2-x+\frac{1}{4} = \frac{5}{4}$
We have two identical quadratic equations so all roots will coincide and there will be two repeated roots instead of four distinct roots.
$(x-\frac{1}{2})^2 = \frac{5}{4}$
Take the square root:
$x-\frac{1}{2} = ±\sqrt{\frac{5}{4}}$
$x-\frac{1}{2} = \sqrt{\frac{5}{4}} \lor x-\frac{1}{2} = -\sqrt{\frac{5}{4}}$
Add $\frac{1}{2}$:
$x = \frac{1}{2} + \frac{\sqrt{5}}{2} \lor x = \frac{1}{2} - \frac{\sqrt{5}}{2}$
|
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|
Evaluate $\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$
Evaluate: $$\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$$
This looks like an unusual hockey stick sum. Here are my attempts:
Method 1:
The sum is equivalent to
$$S=\sum_{r=0}^n 2^{n-r} \binom{n+r}{n}=\sum_{r=0}^n 2^{r} \binom{2n-r}{n-r}$$
and I could evaluate neither of these.
Method 2:
$$S=\text{coefficient of $x^n$ in }:$$
$$2^n(1+x)^n\Bigg( 1+\frac{1+x}{2}+\left(\frac{1+x}{2}\right)^2+\cdots+\left(\frac{1+x}{2}\right)^n \Bigg)$$
$$=(1+x)^n\left(\frac{(1+x)^{n+1}-2^{n+1}}{(x-1)}\right)$$
It looks like a heavy task to collect all the $x^n$ coefficients from this expression.
Im out of ideas. Any hint is appreciated.
From Putnam 2020, Q A2
|
In seeking to evaluate
$$S_n = \sum_{r=0}^n 2^{n-r} {n+r\choose r}$$
we find that it is
$$[z^n] \frac{1}{1-2z} \frac{1}{(1-z)^{n+1}}
= \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{1-2z} \frac{1}{(1-z)^{n+1}}.$$
We will use the fact that residues sum to zero, which requires the
residue at $z=1/2$ and the residue at $z=1$ as well as the residue at
infinity. The latter is zero by inspection, however . We get for the
residue at $z=1/2$
$$-\frac{1}{2} \mathrm{Res}_{z=1/2}
\frac{1}{z^{n+1}} \frac{1}{z-1/2} \frac{1}{(1-z)^{n+1}}$$
We obtain
$$- \frac{1}{2} 2^{n+1} 2^{n+1} = - 2 \times 4^n.$$
We also have for the residue at $z=1$
$$\mathrm{Res}_{z=1} \frac{1}{z^{n+1}}
\frac{1}{1-2z} \frac{1}{(1-z)^{n+1}}
\\ = \mathrm{Res}_{z=1} \frac{1}{(1+(z-1))^{n+1}}
\frac{1}{-1-2(z-1)} \frac{1}{(1-z)^{n+1}}
\\ = (-1)^n
\mathrm{Res}_{z=1} \frac{1}{(1+(z-1))^{n+1}}
\frac{1}{1+2(z-1)} \frac{1}{(z-1)^{n+1}}.$$
This is
$$(-1)^n \sum_{r=0}^n (-1)^r {n+r\choose r} (-1)^{n-r} 2^{n-r}
= \sum_{r=0}^n 2^{n-r} {n+r\choose r} = S_n.$$
We have shown that $S_n - 2 \times 4^n + S_n = 0$ or
$$\bbox[5px,border:2px solid #00A000]{
S_n = 4^n.}$$
For the residue at infinity we get
$$-\mathrm{Res}_{z=0} \frac{1}{z^2} z^{n+1} \frac{1}{1-2/z}
\frac{1}{(1-1/z)^{n+1}}
= - \mathrm{Res}_{z=0}
z^n \frac{1}{z-2} \frac{z^{n+1}}{(z-1)^{n+1}}
\\ = - \mathrm{Res}_{z=0} z^{2n+1} \frac{1}{z-2}
\frac{1}{(z-1)^{n+1}} = 0.$$
|
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|
Find the $\max$ and the $\min$ of $f(x,y)=xy$ subject to $g(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$, using Lagrange multipliers
Find the $\max$ and the $\min$ of $$f(x,y)=xy$$ subject to the constraint $$g(x,y)=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$$ using Lagrange multipliers.
My Attempt:
|
As you wrote, you want to solve the system$$\left\{\begin{array}{l}y+\frac{2\lambda}{a^2}x=0\\x+\frac{2\lambda}{b^2}y=0\\\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.\end{array}\right.$$So, $x$ is both equal to $-\frac{2\lambda}{b^2}y$ and to $-\frac{a^2}{2\lambda}y$. So, since there is no solution with $x=y=0$, you must have $-\frac{2\lambda}{b^2}=-\frac{a^2}{2\lambda}$, or $\lambda=\pm\frac{ab}2$.
If $\lambda=\frac{ab}2$, then $x=-\frac aby$. Replacing $x$ with $-\frac aby$ in the third equation, you get that $y=\pm\frac b{\sqrt2}$, and then $x=\mp\frac a{\sqrt2}$. The case in which $\lambda=-\frac{ab}2$ is similar.
|
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|
Compute $\lim_{n\to\infty}\sqrt[3]{8n^3+4n^2+n+11}-{\sqrt{4n^2+n+9}}$ Compute $\lim_{n\to\infty}\sqrt[3]{8n^3+4n^2+n+11}-\sqrt{4n^2+n+9}$.
My first thoughts are the that I can remove a factor of $2n$ from each root, and obtain
$$ \lim_{n\to\infty}2n\sqrt[3]{1+\frac{1}{2n}+\frac{1}{8n^2}+\frac{11}{8n^3}}-2n\sqrt{1+\frac{1}{2n}+\frac{9}{4n^2}} $$
But factoring out $2n$ won't help evaluate the limit because of the indeterminate form... So then I thought of using L'Hop, but thinking about the chain rule there makes me think that its more effort than it's worth and there's a missing something I can't get to. Any suggestions?
|
The expression seems designed for sheer cussedness if one attempts many of the common methods, since the indices of the roots differ so the "conjugate-factor" method is not much of an option. (And one should avoid applying LHR to expressions with sums or differences of radicals unless one enjoys suffering...)
An approximation of some sort is more helpful. Ross Millikan's approach makes use of the "binomial approximation". Another way that is accessible, since roots of suitable polynomials are being taken, is to "complete" the cubes and squares:
$$ \lim_{n \ \rightarrow \ +\infty} \ \sqrt[3]{8n^3+4n^2+n+11} \ - \ {\sqrt{4n^2+n+9}} $$
$$ = \ \lim_{n \ \rightarrow \ +\infty} \ \ 2 \ \sqrt[3]{\left( n^3 \ + \ \frac12 \ n^2 \ + \ \frac18 \ n \ + \ \frac{11}{8} \right)} \ - \ 2 \ {\sqrt{\left(n^2 \ + \ \frac14 \ n \ + \ \frac94 \right)}} $$
$$ = \ \lim_{n \ \rightarrow \ +\infty} \ \ 2 \ \sqrt[3]{ \ \left[ n^3 \ + \ 3·\frac16 \ n^2 \ + \ 3·\frac{1}{36} \ n \ + \ \frac{1}{216}
\right]+ \ \left(\frac18 - \frac{1}{12} \right) n \ + \ \left(\frac{11}{8} - \frac{1}{216} \right)} \ - \ 2 \ \sqrt{\left[n^2 \ + \ 2·\frac18 \ n \ + \ \frac{1}{64} \right] \ + \ \left( \frac94 - \frac{1}{64} \right)} $$
$$ = \ \lim_{n \ \rightarrow \ +\infty} \ \ 2 \ \sqrt[3]{ \ \left( n \ + \ \frac16 \right)^3 \ + \ \left[ \ \text{small terms} \ \right]} \ \ - \ \ 2 \ \sqrt{ \ \left(n \ + \ \frac18 \right)^2 \ + \ \left[ \ \text{small terms} \ \right]} $$
$$ \approx \ \lim_{n \ \rightarrow \ +\infty} \ \ 2· \left( n \ + \ \frac16 \right) \ \ - \ \ 2 · \left| \ n \ + \ \frac18 \ \right| \ \ , $$
which agrees with the "binomial approximation" result. (One has to magnify a graph a fair bit to see that the limit is not zero.)
Incidentally, this also shows that the horizontal asymptote for this function is "one-sided". We observe that the asymptote corresponding to the limit "at negative infinity" for this expression is a line of slope $ \ 4 \ \ , $ giving us $$ \lim_{n \ \rightarrow \ -\infty} \ \sqrt[3]{8n^3+4n^2+n+11} \ - \ {\sqrt{4n^2+n+9}} \ \ = \ \ -\infty \ \ . $$
(We can consider this situation since $ \ 4x^2+x+9 > 0 \ $ for all real numbers.)
|
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|
Evaluating $\int_0^B \frac{1}{A^2 + (B^2-x^2)^2}\,dx$ I am trying to solve the following integral:
$$\int_0^B \frac{1}{A^2 + (B^2-x^2)^2}\,dx$$
However the polynomial have imaginary roots:
$$-(-i A + B^2)^{1/2},\; (-i A + B^2)^{1/2},\; -(i A + B^2)^{1/2},\;(i A + B^2)^{1/2}$$
I am only interested on the solution from $0$ to $B$. How should I proceed?
|
Let $p=\sqrt{1+\frac{A^2}{B^4}}$ and substitute $x=Bt$
$$I=\int_0^B \dfrac{1}{A^2 + (B^2-x^2)^2} dx
= \frac1{B^3} \int_0^1 \dfrac{1}{t^4-2t^2+p^2} dt
$$
Decompose the integrand as
\begin{align}
\frac{1}{t^4-2t^2+p^2} = \frac1{2ps}\left( \frac{t+s}{t^2+st+p}- \frac{t-s}{t^2-st+p}\right)
\end{align}
with $s=\sqrt{2(p+1)}$. Then, integrate to obtain
$$I= \frac{1}{4B^3p}\left(\frac1{\sqrt{2(p+1)}}
\ln\frac {\sqrt{1+p}+\sqrt{2}}{\sqrt{1+p}-\sqrt{2}}
+ \sqrt{\frac2{p-1}}\tan^{-1}\sqrt{\frac2{p-1}}\right)
$$
|
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|
Global maxima and minima of $f(x,y)=x^2 + y^2 + \beta xy + x + 2y$ I am self-learning basic optimization theory and algorithms from An Introduction to Optimization by Chong and Zak. I would like someone to verify my solution to this problem, on finding the minimizer/maximizer of a function of two variables, or any tips/hint to proceed ahead.
For each value of the scalar $\beta$, find the set of all stationary points of the following two variables $x$ and $y$
$$f(x,y) = x^2 + y^2 + \beta xy + x + 2y$$
Which of those stationary points are local minima? Which are global minima and why? Does this function have a global maximum for some value of $\beta$.
Solution.
We have:
\begin{align*}
f(x,y) &= x^2 + y^2 + \beta xy + x +2y\\
f_x(x,y) &= 2x+\beta y + 1\\
f_y(x,y) &= \beta x + 2y + 2\\
f_{xx}(x,y) &= 2\\
f_{xy}(x,y) &= \beta \\
f_{yy}(x,y) &= 2
\end{align*}
By the first order necessary condition(FONC) for optimality, we know that if $\nabla f(\mathbf{x})=0$, then $\mathbf{x}$ is a critical point.
Thus,
\begin{align*}
f_x(x,y) &= 2x+\beta y + 1 = 0\\
f_y(x,y) &= \beta x + 2y + 2 = 0
\end{align*}
Solving for $x$ and $y$, we find that:
\begin{align*}
x = \frac{\begin{array}{|cc|}
-1 & \beta \\
-2 & 2
\end{array}}{\begin{array}{|cc|}
2 & \beta \\
\beta & 2
\end{array}}=\frac{-2+2\beta}{4-\beta^2}=\frac{2\beta-2}{4 -\beta^2}
\end{align*}
\begin{align*}
y = \frac{\begin{array}{|cc|}
2 & -1 \\
\beta & -2
\end{array}}{\begin{array}{|cc|}
2 & \beta \\
\beta & 2
\end{array}}=\frac{-4+\beta}{4-\beta^2}=\frac{\beta -4}{4 - \beta^2}
\end{align*}
The second order necessary and sufficient conditions for optimality are based on the sign of the quadratic form $Q(\mathbf{h})=\mathbf{h}^T \cdot Hf(\mathbf{a}) \cdot \mathbf{h}$.
The Hessian of $f$ is given by,
$$Hf(\mathbf{x})=\begin{array}{|c c|}
2 & \beta \\
\beta & 2
\end{array}$$
Thus, $d_1 = 2 > 0$ and $d_2 = 4 - \beta^2$. Thus, $f$ has a local minimizer if and only if $4 - \beta^2 > 0$. $g(\beta) = 4 - \beta^2$ is a downward facing parabola. So, the values of this expression positive, if and only if $-2 < \beta < 2$. The function $f$ has no global maximum.
Question. How do I find the actual global minima?
|
Using elementary algebra, we have
$$f(x,y)=x^2+x(\beta y+1)+2y+y^2$$
Applying well known method,
$$ax^2+bx+c=a(x-m)^2+n$$
$$m=-\frac{b}{2a}, n=-\frac{\Delta}{4a}$$
In this case, we have
$$\begin{align}m:=-\frac{\beta y+1}{2}, ~n:&=\frac 14 \left(y^2(4-\beta ^2) + 2y(4-\beta)-1\right)&\end{align}$$
which gives,
$$f(x,y)=\left(x+\frac{\beta y+1}{2}\right)^2+\frac 14 \left(y^2(4-\beta ^2) + 2y(4-\beta)-1\right)$$
Putting
$$x=-\frac{\beta y+1}{2}$$
We get,
$$f(x,y)≥y^2\left(1-\frac{\beta ^2}{4}\right)+y\left(2-\frac{\beta}{2}\right)-\frac 14$$
Let $\beta ≠±2$, then applying the same method, we have
$$y^2\left(1-\frac{\beta ^2}{4}\right)+y\left(2-\frac{\beta}{2}\right)-\frac 14=\left(1-\frac{\beta^2}{4}\right)\left(y+\frac{4-\beta}{4-\beta ^2}\right)^2+\frac{2 β - 5}{4 - β^2}$$
If $|\beta| >2$, then $\min \left\{f(x,y)\right\}$ doesn't exist.
Let, $|\beta|<2 \iff -2<\beta <2$, then putting
$$y=\frac{\beta-4}{4-\beta ^2}$$
We get,
$$\min \left\{f(x,y)\right\}=\frac{2\beta-5}{4-\beta^2}$$
I'll leave it for you to analyze the case of $\beta=±2.$
This case is equivalent to finding the minimum of the following polynomial:
$$y \left(2-\frac{\beta}{2}\right)- \frac 14.$$
It is possible to see clearly that there is no minimum.
|
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|
Find the area of the square with vertex at $C$ The figure below shows $6$ squares connected edge by edge. The top three are unit squares.If $$ar(\Delta ABD)=\frac{3}{2} \times ar(\Delta ABC)$$ Find the area of the square with vertex at $C$.
My try:
I actually attacked this using Analytical geometry:
I assumed the following:
$FB$ is along $X$ axis and $EH$ is along $Y$ axis.
Let $F(0,0),A(1,1),B(2,0),G(0,-a),H(0,-b)$.
Thus the side length of the square $CGHK$ is $b-a$.
Let the side length of the square $LKMD$ is $x$.
So the coordinates of $L$ is $L(b-a-x,-b)$
So the vertex $D$ is $D(b-a-x,x-b)$.
Also the vertex $C$ is $C(b-a,-a)$
Now using the formula for area of triangle using determinant as:
$$\Delta=\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|$$
We have $$ar(ABD)=\frac{1}{2}\left|\begin{array}{ccc}
b-a-x & x-b & 1 \\
1 & 1 & 1 \\
2 & 0 & 1
\end{array}\right|$$ and
$$ar(ABC)=\frac{1}{2}\left|\begin{array}{ccc}
b-a & -a & 1 \\
1 & 1 & 1 \\
2 & 0 & 1
\end{array}\right|$$
Using the fact that:
$$ar(\Delta ABD)=\frac{3}{2} \times ar(\Delta ABC)$$
We get
\begin{array}{l}
\frac{a+2}{2}=\frac{3}{2}\left(\frac{b-2 a-2}{2}\right) \\
\Rightarrow \quad 6 a-3 b+10=0 \tag{1}
\end{array}
I need help to find another equation so that we can solve the simultaneous equations to find $a,b$ and hence $b-a$.
|
On your solution, you have some mistakes in the sign. First note that $|b| = 3$. With $F$ as origin, and $|FG| = a$ and $|DM| = x$,
Coordinates of $C$ is $(-3 + a, -a)$ and of $D$ is $(-3 + a - x, -3 + x)$. Now using determinant for area and solving, you get $ \displaystyle a = \frac{1}{2} \ $ so side length of the square with vertex at $C$ is $\displaystyle \frac{5}{2}$.
Here is a solution using trigonometry -
Assume the side length of the square with vertex at $C$ is $x$ and with vertex at $D$ is $y$.
As $\triangle ABC$ and $\triangle ABD$ have the same base $AB$ and given the ratio of area of triangles,
$\displaystyle h_2 = \frac{3 h_1}{2} \ $ where $h_1$ is altitude from $C$ to base $AB$ and $h_2$ is altitidue from $D$ to base $AB$.
If $\angle CBF = \alpha, \angle DBF = \beta$,
$\displaystyle \sin \alpha = \frac{3-x}{BC}, \cos \alpha = \frac{2+x}{BC}$
$\displaystyle \sin \beta = \frac{3-y}{BD}, \cos \beta = \frac{2+x+y}{BD}$
Also note that $ \displaystyle \angle ABF = \frac{\pi}{4}$
$\displaystyle h_1 = BC \sin \big(\frac{\pi}{4} + \alpha) = \frac{5}{\sqrt2}$
$\displaystyle h_2 = BD \sin \big(\frac{\pi}{4} + \beta) = \frac{5+x}{\sqrt2}$
As $ \displaystyle h_2 = \frac{3 h_1}{2}, x = \frac{5}{2}$
So area of square with vertex at $C$ is $\displaystyle \frac{25}{4}$.
|
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|
Here why is it wrong to differentiate both sides and put $x=2$ to find $g'(2)$? Here why is it wrong to differentiate both sides and put $x=2$ to find $g'(2)$?
If $\displaystyle I = \int \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} \,\mathrm dx = g(x) + c$, then $\left\lfloor \dfrac{1}{g'(2)} \right\rfloor = \cdots$,
(where $\lfloor \cdot \rfloor$ represents the greatest integer function)
To me it seemed like it's an identity and so it should be safe to differentiate both sides.
|
$$
\int \frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} \,\mathrm dx = g(x) + c
$$
$$
\frac{x-1}{(x+1)\sqrt{x^3+x^2+x}} = g'(x)
$$
$$
\frac 1 {3\sqrt{14}} = g'(2)
$$
$$
\frac 1 {g'(2)} = 3\sqrt{14}, \text{ which is between 11 and 12}
$$
$$
\left\lfloor \frac 1 {g'(2)} \right\rfloor = 11
$$
Addendum :
$$\begin{align}
11=\sqrt{121}&<\sqrt{126}=\sqrt{2\cdot3^2\cdot7}=3\sqrt{14}=\\
&=\sqrt{126}<\sqrt{144}=12
\end{align}$$
$\text{hence ,}$
$$11<3\sqrt{14}<12\;.$$
|
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|
Prove that $\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$
Prove that $$\frac{a^3}{b} + \frac{b^3}{a} \geqslant a^2 + b^2$$
for $a,b \in \mathbb{R}^+$. Establish when the equality holds.
My approach was to use AM-GM however trying both sides individually assert to $\geqslant 2ab.$ RHS is baby AM-GM if considered.
Edit:
From L.H.S -
$\dfrac{\frac{a^3}{b} + \frac{b^3}{a}}{2} \geqslant ab \implies \dfrac{a^3}{b} + \dfrac{b^3}{a} \geqslant 2ab$
|
Here is a trick, using only AM-GM, twice.
$$a^2+b^2\geq 2ab \Rightarrow \\
\frac{a^3}{b}+\color{red}{a^2}+\frac{b^3}{a}+\color{red}{b^2} \geq
\frac{a^3}{b}+\color{red}{ab}+\frac{b^3}{a}+\color{red}{ab}\geq\\
2\sqrt{\frac{a^3}{b}\cdot ab} + 2\sqrt{\frac{b^3}{a}\cdot ab}=2a^2+2b^2$$
Or
$$\frac{a^3}{b}+\color{blue}{a^2}+\frac{b^3}{a}+\color{blue}{b^2} \geq 2(\color{blue}{a^2+b^2})$$
and the result follows.
|
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|
Does $\sum_{n \geq 1}f_n(x)$ converges uniformly on $[0, +\infty)$? Let $f_n(x) = \frac{x^2}{1 + nx^2}\sin(\frac{nx}{x + 1})$.
Is it true that $$ \sum_{n \geq 1} \frac{x^2}{1 + nx^2}\sin(\frac{nx}{x + 1})$$
converges uniformly on $[0, +\infty)$
I want to say 'Yes' and use Abel-Dirichlet test with:
$$a_n(x) = \sin(\frac{nx}{x + 1})$$ and $$b_n(x) = \frac{x^2}{1 + nx^2}$$ which converges to zero. But I have some troubles with proving than $a_n(x)$ and $b_n(x)$ satisfy the conditions we need.
Am I right in my attempts and what is the best way to prove that $f_n(x)$ converges uniformly in this case.
|
The series converges uniformly on $[0,\infty)$ by the Dirichlet test.
Factoring the terms of the series as
$$\sum_{n=1}^\infty \underbrace{\frac{x^2}{1+nx^2}}_{b_n(x)} \underbrace{\sin \frac{nx}{x+1}}_{a_n(x)}$$
is not helpful for applying the Dirichlet test. The requirement that $b_n(x) \to 0$ as $n \to \infty$ both monotonically and uniformly for all $x \in [0,\infty)$ is met. However, the partial sums $\sum_{n=1}^N a_n(x)$ are not uniformly bounded for all $N \in \mathbb{N}$ and $x \in [0,\infty)$. This is a consequence of the fact that $\sum_{n=1}^N \sin nz$ is not uniformly bounded for all $z$ on any set with $2m\pi, \, (m=0,1,2,\ldots)$ as a limit point.
We can apply the Dirichlet test sucessfully by factoring the terms with
$$a_n(x) = \frac{x}{x+1}\sin \frac{nx}{x+1},\quad b_n(x) = \frac{x(x+1)}{1 + nx^2}$$
We have for all $x \in (0,\infty)$,
$$\left|\sum_{n=1}^N a_n(x)\right| = \left|\frac{x}{x+1}\sum_{n=1}^N \sin \frac{nx}{x+1}\right|= \frac{x}{x+1}\frac{\left|\sin\frac{Nx}{2(x+1)}\right|\, \left| \sin \frac{N+1)x}{2(x+1)}\right|}{\left|\sin \frac{x}{2(x+1)}\right|} \\ \leqslant \frac{\frac{x}{x+1}}{\sin \frac{x}{2(x+1)}} \leqslant \frac{1}{\sin 1/2}$$
The last inequality is a consequence of monotonicity and
$$\lim_{x \to 0}\frac{\frac{x}{x+1}}{\sin \frac{x}{2(x+1)}}= 2, \quad \lim_{x \to \infty }\frac{\frac{x}{x+1}}{\sin \frac{x}{2(x+1)}}= \frac{1}{\sin 1/2} \approx 2.0858$$
Hence, the partial sums $\sum_{n=1}^N a_n(x)$ are uniformly bounded.
We also have $\lim_{n \to \infty} b_n(x) = 0$ both montonically and uniformly for all $x \in [0,\infty)$. Clearly $b_n(x)$ decreases with $n$ for fixed $x$. For uniform convergence note that for all $x \in [0,1)$ we have
$$0 \leqslant b_n(x) = \frac{x(x+1)}{1 + nx^2}= (x+1) \frac{\sqrt{n}x}{1+ nx^2}\frac{1}{\sqrt{n}}\leqslant \frac{2}{\sqrt{n}},$$
and for all $x \in [1,\infty)$ we have
$$0 \leqslant b_n(x) = \frac{x(x+1)}{1 + nx^2}\leqslant \frac{2x^2}{1+nx^2} \leqslant \frac{2}{n}$$
|
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|
Prove convergence of MacLaurin expansion of$ (1+x)^{1/2}$ at $|x|=1$. I know that the Maclaurin expansion of $(1+x)^{1/2}$ has a domain of convergence of $|x|\leq1$. It is easy to prove using the ratio test that this expansion is convergent when $|x|<1$, however, I don't know how to prove convergence when $|x|=1$. Does anybody know any simple way to prove this convergence?
|
If you carefully write it out, you will get (for $c$ either $3/2$ or $1/2$):
$$f(\pm1)\,-\,c\,\,=\,\,-\frac{2^{-2}}{1(2)}\,\pm\,\frac{2^{-3}(3)}{1(2)(3)}\,-\frac{2^{-4}(3)(5)}{1(2)(3)(4)}\,\pm\,\frac{2^{-5}(3)(5)(7)}{1(2)(3)(4)(5)}\,-\pm\dots.$$
We want to show absolute convergence of this series for either choice $x=+1$ or $x=-1$, so make every term positive and bring each $2^{-k}$ factor into the denominator:
$$\sum_{k\,\geq\,2} \frac{(2k-3)!!}{(2k)!!} \,=\, \sum_{k\,\geq\,2} \frac{({\sqrt{2k-3}})^2!!}{(2k)!!}.$$
The general term of this series we can bound above by the term of the harmonic series of exponent $p=3/2$, i.e. over-harmonic series or $p$-series with $p>1$. This via the estimate $\sqrt{n+1}\sqrt{n-1}\leq n$:
$$\frac{\sqrt{1}}{4(2)}\,+\,\frac{\sqrt{3}\sqrt{3}\sqrt{1}}{6(4)(2)}\,+\,\frac{\sqrt{5}\sqrt{5}\sqrt{3}\sqrt{3}\sqrt{1}}{8(6)(4)(2)}\,+\,\frac{\sqrt{7}\sqrt{7}\sqrt{5}\sqrt{5}\sqrt{3}\sqrt{3}\sqrt{1}}{10(8)(6)(4)(2)}\,+\dots$$
$$=\frac{\sqrt{1}}{4(2)}\,+\,\frac{\sqrt{3}}{6(4)}\,\frac{\sqrt{3}\sqrt{1}}{2}\,+\,\frac{\sqrt{5}}{8(6)}\frac{\sqrt{5}\sqrt{3}}{4}\frac{\sqrt{3}\sqrt{1}}{2}\,+\dots$$
$$\leq\frac{\sqrt{1}}{4(2)}\,+\,\frac{\sqrt{3}}{6(4)} \,+\,\frac{\sqrt{5}}{8(6)} \,+\,\frac{\sqrt{7}}{10(8)}\,+\dots=\sum_{k\,\geq\,2} \frac{(2k-3)^{\frac{1}{2}}}{(2k)(2k-2)}.$$
|
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|
Find all integers n such that $(\frac{n^3-1}{5})$ is prime
Find all integers n such that $(\frac{n^3-1}{5})$ is prime?
My Approach:
I wrote all the prime which i will get after dividing $(n^3-1)$ by $5$.
$n^3-1=10,15,25,35,55,...,215$
which lead me to $n^3=11,16,26,...,216$, then I obtained $n=6$
My doubt is that how to check more value of $n$ without using modular arithmetic because the book I'm referring has not introduced it yet.
Second approach: $\frac{(n^3-1)}{5}=\frac{(n-1)(n^2+n+1)}{5}$
But my second approach too does not lead me anywhere.
This problem is from the book Pathfinder for Olympiad Mathematics
|
First, since you need the fraction to be a prime, you have $n^3-1 = 5p$ with $p$ a prime number. Now, $n^3-1=(n-1)(n^2+n+1)$. So this leads to the idea that one of those factors is $5$ and the other $p$.
*
*If $n-1=5$, $n=6$ so $n^2+n+1= 36+6+1= 43$.
*If $n-1\neq 5$, $n^2+n+1=5$, so $n(n+1)=4$, and you can check in multiple ways that this equation doesn't have solutions. (one way is noticing the parity of the lhs and forcing the odd factor to be $1$, so the other won't reach to be $4$)
EDIT: As how mentioned @lhf , I'm missing if $n^3 - 1 = -5p$ So considering this case, you have these cases:
*
*If $n-1 = -5$, $n = -4$, so $n^2+n+1 = 16 - 4 +1 = 13$
*If $n-1 \neq -5$, then $n^2+n+1 = -5$, and this leaves to $n^2+n = -6$, obtaining with general formula this doesn't give any possible solution.
Also,
*
*If $n-1 = -5p$, $n^2+n+1=1$, but this will mean $(5p)^2-5p=0$, implying $5p(5p-1)=0$, forcing $5p-1=0$, but that leaves $p$ not being a prime.
*If $n-1 = 5p$, then $n = 5p + 1$, so $(5p+1)^2 + 5p + 1 +1 = 0$, giving $25p^2+15p + 3 = 0$, but this equation doesn't have any integer solution, since this would mean $5\mid 3$, which is a nonsense.
Then the solutions are $(n,p)$ are $(6,43), (-4,-13)$
|
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|
Logarithmic integral $ \int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}\ \mathrm{d}x $ I found this integral weeks ago.
$$ \int_0^1 \dfrac{x\ln(x)\ln(1+x)}{1+x^2}\ \mathrm{d}x $$
I tried to solve this integral using various series representation and ended up with a complicated double series which I have asked here. How can I solve this integral?
|
Let $K= \int_0^1 \frac{\ln x\ln(1-x)}{1+x} dx$ and note
\begin{align}
I&= \int_0^1 \dfrac{x\ln x\ln(1-x^2)}{1+x^2} dx
\overset{x^2\to x}
=\frac14K \\
J &= \int_0^1 \dfrac{x\ln x\ln\frac{1-x}{1+x}}{1+x^2} dx
\overset{x\to \frac{1-x}{1+x}}=\int_0^1 \dfrac{\ln x\ln\frac{1-x}{1+x}}{1+x} dx -J\\
&= \frac12K-\frac12 \int_0^1 \dfrac{\ln x\ln(1+x)}{1+x} \overset{ibp}{dx }= \frac12K+\frac14 \int_0^1 \underset {= \frac14\zeta(3)}{\frac{\ln^2(1+x)}x }dx
\end{align}
Then
\begin{align}
&\int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}dx=\frac12 (I-J)
=-\frac1{32}\zeta(3)-\frac18K \tag1\\
\overset{ibp}=& -\frac1{32}\zeta(3)-\frac18\int_0^1 \frac{\ln x\ln(1+x)}{1-x}dx+\frac18\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x}dx
\end{align}
where
\begin{align}\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x}dx
=&\frac14\int_0^1 \frac{\ln^2(1-\overset{\to x}{x^2})}xdx -\frac14\int_0^1 \frac{\ln^2\frac{\overset{\to x}{1-x}}{1+x}}xdx\\
=&\frac18\int_0^1 \frac{\ln^2(1-x)}xdx -\frac12\int_0^1 \frac{\ln^2 x}{1-x^2}dx\\
=&\frac18\cdot 2\zeta(3) -\frac12\cdot \frac74\zeta(3)
=-\frac58\zeta(3)
\end{align}
\begin{align}\int_0^1 \frac{\ln x\ln(1+x)}{1-x}dx
=&\int_0^1 \ln(1+x)\>d\left( \int_0^x \frac{\ln t}{1-t}dt \right)_{t=xy}\\
=& \ln2\int_0^1 \frac{\ln t}{1-t}dt
-\int_0^1\int_0^1 \frac{x\ln x+\overset{y\leftrightarrow x}{x \ln y}}{(1+x)(1-xy)}
dxdy \\
=& -\frac{\pi^2}{6}\ln2
-\int_0^1\int_0^1 \frac{\ln x}{1-xy} -\frac{\ln x}{(1+x)(1+y)} dydx \\
= & -\frac{\pi^2}{6}\ln2
+\frac12 \int_0^1\frac{\ln^2x}{1-x} dx +\ln2\int_0^1 \frac{\ln x}{1+x}dx\\
= & -\frac{\pi^2}{6}\ln2 +\frac12\cdot 2\zeta(3)-\frac{\pi^2}{12}\ln2
= -\frac{\pi^2}4 \ln2 +\zeta(3)
\end{align}
Plug into (1) to obtain
$$ \int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}dx= -\frac{15}{64}\zeta(3) + \frac{\pi^2}{32}\ln2$$
|
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|
Solve system of equations $3(x+\frac{1}{x}) = 4(y + \frac{1}{y}) = 5(z+\frac{1}{z})$, $xy+yz+zx = 1$ Find all $x,y,z>0$ such that
$$3(x+\frac{1}{x}) = 4(y + \frac{1}{y}) = 5(z+\frac{1}{z})$$ $$xy+yz+zx = 1$$
The only solution should be $x=\frac{1}{3}$, $y = \frac{1}{2}$, $z=1$.
There is a way to do it with $x = \tan \alpha$, etc., but I would like to find an even more elementary way.
So far, I have written
$$x^2 - \frac{k}{3}x + 1 = y^2 - \frac{k}{4}y + 1 = z^2 - \frac{k}{5}z + 1 = 0$$
and noticed that at least two of $x$, $y$, $z$ must be smaller than $1$. (So in the bad cases I can express uniquely $x,y,z$.) But then?
Any help appreciated!
|
Hint
Note that $xy+yz+zx=1$, then we have:
$$\frac{4}{3}=\frac{x+\frac{1}{x}}{y+\frac{1}{y}}=\frac{x+\frac{xy+yz+zx}{x}}{y+\frac{xy+yz+zx}{y}} = \frac{\frac{(x+y)(x+z)}{x}}{\frac{(y+z)(y+x)}{y}} = \frac{y(x+z)}{x(z+y)}=\frac{yz+yx}{xy+xz}$$
Do similiarly with $\frac{x+\frac{1}{x}}{z+\frac{1}{z}}$, you can imply the result
|
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|
The sum of three positive integers is $20$. Find the probability that they form the sides of a triangle.
The sum of three positive integers is $20$. Find the probability that they form the sides of a triangle.
Let $a,b,c$ be there positive integers. So, $a+b+c=20$
Total number of solutions would be $^{19}C_2=171$
For $a,b,c$ to form a triangle $0\lt a,b,c\le9$
So, number of such solutions= coefficient of $x^{20}$ in $(x+x^2+x^3+...+x^9)^3=$ coefficient of $x^{17}$ in $(1+x+x^2+...+x^8)^3=$ coefficient of $x^{17}$ in $(1-x^9)^3(1-x)^{-3}=1\times^{19}C_2-3\times^{10}C_2=36$
So, the required probability is $\frac{36}{171}$ but the answer given is $\frac8{33}$
In the hint, they have written total combinations of $a,b,c=\frac{144}{6}+\frac{27}{3}=33$, and favorable combinations of $a,b,c=\frac{12}{3}+\frac{24}{6}=8$
I think they have split $171$ as $144+27$ and $36$ as $24+12$ but why? and why to divide them with $6$ or $3$?
|
It is a bad question as they have not described how the sides have been chosen, but their answer almost reveals what they intended:
*
*There are $171$ compositions of $20$ into three positive integer parts (any order). Of these,
*
*$144$ compositions have the three parts distinct
*$27$ compositions have two of the parts equal and the third distinct
*$0$ compositions have all three parts equal ($20$ is not divisible by $3$)
*So removing duplicates which are the same but in a different order, there are
*
*$\frac{144}{3!}=24$ partitions which have the three parts distinct
*$\frac{27}{3}=9$ partitions which have two of the parts equal and the third distinct
*$0$ partitions have all three parts equal ($20$ is not divisible by $3$)
*making $24+9+0=33$ for the number of partitions of $20$ into three positive integer parts
The $36 \to 8$ is the same but where the two smaller numbers add up to more than the largest number
|
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|
Unique ways to write $n$ as sum of three distinct nonnegative integers up to the order of the summands How many ways are there to express a natural number, $n$, as the sum of three whole numbers, $a,b,c$, where $a,b,c$ are allowed to be 0 but are unique?
For example: $n=9$ there are only seven ways: $1+2+6, 1+3+5, 2+3+4, 0+1+8, 0+2+7, 0+3+6, 0+4+5$.
PS: Looking at some previous [answers]Generating function for counting compositions of $n$, I learnt this "thing" is known as partitions of $n$. But I did not find an answer that includes 0 and does not allow repetitions. Number theory is not my area, and all I would like to know is if some analytical formula can determine this.
|
Suppose $S$ is the set of ordered triplets $\{x,y,z\}$ such that $x+y+z=n$. As you know the size of $S$ is $n+2 \choose 2$. There are three types of triplets in $S$.
*
*$x,y,z$ are all distinct.
*Exactly two of $x,y,z$ are the same.
*$x=y=z$.
Let $t_1,t_2,t_3$ be the number of triplets of type 1, 2 and 3 respectively. So, $t_1+t_2+t_3=|S|$. $t_3=1$ if $n$ is divisible by 3 otherwise $t_3$ = 0. The number of pairs $\{u,v\}$ such that $2u+v=n$ is $ \lfloor{\frac{n}{2}} \rfloor - t_3 + 1 $. Each such pair contributes exactly 3 ordered triplets $\{u,u,v\},\{u,v,u\},\{v,u,u\}$ in $S$. Hence, $t_2= 3(\lfloor{\frac{n}{2}} \rfloor - t_3 + 1)$. You are looking for the number of unordered triplets $\{a,b,c\}$ such that $a+b+c=n$. This count is exactly equal to $t_1/6$ as each such unordered triplet $\{a,b,c\}$ contributes 6 ordered triplets of type 1 in $S$. So, the count is $$\frac{|S|-t_2-t_3}{6} = \frac{\binom{n+2}{2}-3(\lfloor{\frac{n}{2}} \rfloor - t_3 + 1)-t_3}{6}= \frac{\binom{n+2}{2}-3(\lfloor{\frac{n}{2}} \rfloor + 1)+2t_3}{6}.$$ Taking your example of $n=9$, it is
$$\frac{\binom{9+2}{2} -3(\lfloor{\frac{9}{2}} \rfloor + 1)+2}{6} = \frac{55-3\cdot5+2}{6}=7.$$
|
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|
Fourier expansion of $4x(1-x)$ Im triyng expand $4x(1-x)$ in a Fourier sine series in the interval $(-1\leq x \leq 1)$
where:
$f(x) = 4x(1-x), (0\leq x \leq 1)$
$f(x) = 4x(1+x), (-1\leq x \leq 0)$
But on my calculations the coeficient $b_n$ is going to zero, where is supose to be $b_n=\frac{32}{n^3\pi^3}$ if n is odd and $b_n=0$ if n is even. I'm not seeing where am i wrong.
so far what i got is:
$b_n = \int_{-1}^{0} (4x^2+4x)(\sin(nx\pi))dx + \int_{0}^{1} (4x^2-4x)(\sin(nx\pi))dx = \frac {(4\pi\sin(n\pi)+8\cos(n\pi)-8}{n^3\pi^3} - \frac {4\pi\sin(n\pi)+8\cos(n\pi)-8}{n^3\pi^3}$
Which is zero
|
It would take forever to typeset all the IBP, but we have
\begin{align}
b_n &= \int_{-1}^0 (4x+4x^2)\sin(n\pi x)dx+\int_{0}^1 (4x-4x^2)\sin(n\pi x)dx\\
&= \int_{-1}^0 4x\sin(n\pi x)dx+\int_{-1}^0 4x^2\sin(n\pi x)dx+\int_{0}^1 4x\sin(n\pi x)dx-\int_{0}^1 4x^2\sin(n\pi x)dx\\
&= \int_{-1}^1 4x\sin(n\pi x)dx+\int_{-1}^0 4x^2\sin(n\pi x)dx-\int_{0}^1 4x^2\sin(n\pi x)dx\\
&= 2\int_{0}^1 4x\sin(n\pi x)dx+\int_{-1}^0 4x^2\sin(n\pi x)dx-\int_{0}^1 4x^2\sin(n\pi x)dx\\
&=\frac{8(\sin(\pi n)-\pi n cos(\pi n))}{\pi^2 n^2}+\frac{4((\pi^2 n^2-2)\cos(\pi n)-2\pi n \sin(\pi n)+2)}{\pi^3 n^3}+\frac{4(\pi^2 n^2-2)\cos(n \pi)-8 \pi n \sin(\pi n)+8}{\pi^3 n^3},
\end{align}
where we've used that $4x \sin(n \pi x)$ is even. You can check that each integral gives each of the terms listed in the last equality. This yields, as you should check,
$$b_n=
\begin{cases}
0 & \text{$n$ even} \\
\frac{32}{\pi^3 n^3} & \text{$n$ odd}
\end{cases}$$
|
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Find $ \int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx$
Find $\displaystyle \int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx$
I couldn't find my answer so I looked up the solution which is as follows
$$\displaystyle \begin{align}&\int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx\\&= \int \frac{2\cos x+1}{(2+\cos x)^2}\,dx\\&= \int \frac{2\cos x+\cos^2 x+\sin^2 x}{(2+\cos x)^2}\,dx\\&=\int \frac{\cos x}{2+\cos x}\,dx+\int \frac{\sin^2 x}{(2+\cos x)^2}\,dx\\&=\frac{1}{2+\cos x}\cdot(\sin x)-\int \sin x\cdot\frac{\sin x}{(2+\cos x)^2}\,dx+\int \frac{\sin^2 x}{(2+\cos x)^2}\,dx\\&=\frac{\sin x}{2+\cos x}+C\end{align}$$
I was stuck with $\displaystyle\int \frac{2\cos x+1}{(2+\cos x)^2}\,dx$ and didn't think of replacing $1$ with $\cos^2 x+\sin^2 x$. I want to know the motivation behind this because the reason of this substitution wasn't very obvious to me at first and it only became clear to me in the second last step. Other ways to solve this are also welcome.
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Using the "universal substitution":
$$\begin{align}
\int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}dx
&= \int\frac{(\tan^2(x/2) + 1)\left(2+ \frac{\tan^2(x/2) + 1}{1-\tan^2(x/2)}\right)}{(1-\tan^2(x/2))\left(1+\frac{\tan^2(x/2) + 1}{1-\tan^2(x/2)}\right)^2} \\
&= \begin{bmatrix}u = \tan(x/2) \\ du = \frac{1}{2}\sec^2(x/2)dx\end{bmatrix} \\
&= -2 \int \frac{u^2-3}{(u^2 + 3)^2}du,\end{align}$$
which can be integrated using the known algorithms.
However, the original solution is indeed more attractive.
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|
High school integration: $-\frac{1}{2} \int \sec \theta d\theta$ Original question was: $\int \frac{1}{(x+3)^2 -4}$
Substitution: $x=2\sin \theta -3$
This is a follow up on my last thread because I cannot seem to get the answer.
Anyway, I got to:
$$-\frac{1}{2} \int \sec \theta d\theta$$
And had used the substitution $x=2\sin \theta -3$. But the answer should be $$\frac {1}{4} \ln\left|\frac{x+1}{x+5}\right| +c $$
However I'm getting $$\frac {1}{4} \ln\left|\frac{-(x+1)}{x+5}\right| +c $$
My steps were:
$=\frac{-1}{2} \ln\left|\sec\theta +\tan\theta\right| +c $
$=\frac{-1}{2} \ln\left|\frac{\sin\theta+1}{\cos\theta}\right| +c $
$\sin\theta +1 = \frac{x+5}{2}$
$ \frac{-1}{4} ln\left|\frac{(x+5)^2}{4\cos^2\theta}\right| +c $
$ \cos^2\theta = 1- \frac{(x+3)^2}{4}$
$ \frac{-1}{4} ln\left|\frac{(x+5)^2}{4 - (x+3)^2}\right| +c $
$ \frac{-1}{4} ln\left|\frac{(x+5)^2}{(2-(x+3)(2+(x+3))}\right| +c $
$ \frac{-1}{4} \ln\left|\frac{(x+5)^2}{(-x-1)(x+5)}\right| +c $
$$\frac {1}{4} \ln\left|\frac{-(x+1)}{x+5}\right| +c $$
However, the -(x+1) should just be (x+1). Does anyone know where I went wrong?
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A much easier approach is to start by applying the difference of squares identity to the denominator, factorising and doing partial fractions. You'll get to the "intended" answer much quicker. Avoid trigonometric substitutions where possible.
So the integrand becomes $$\begin{align}\frac 1{(x+3)^2 - 2^2} &= \frac 1{(x+1)(x+5)}\\ &= \frac 14\left(\frac{1}{x+1} - \frac{1} {x+5}\right),\end{align}$$ from which you should be able to finish.
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|
Let $(m,n)$ be the pair of integers satisfying $2(8n^3 + m^3) + 6(m^2 - 6n^2) + 3(2m + 9n) = 437.$ Find the sum of all possible values of $mn$.
Let $(m,n)$ be the pair of integers satisfying $2(8n^3 + m^3) + 6(m^2 - 6n^2) + 3(2m + 9n) = 437.$ Find the sum of all possible values of $mn$.
What I Tried:- Given this expression I had no idea to claim something out of it, because it looks like we need to $437$ in terms of sums of $2,3$ (the $6$ does not matter much here). I could break $(8n^3 + m^3)$, but it didn't help much.
Instead I simplified the whole expression and got :-
$$2m^3 + 6m^2 + 6m + 16n^3 - 36n^2 + 27n = 437.$$
$$\rightarrow 2m^3 + 6m^2 + 6m + 16n^3 - 36n^2 = 437 - 27n.$$
From here, $2 \vert (437 - 27n)$, so $n$ is odd. Let $n = (2k + 1)$. Substitute this back to the equation to get :-
$$64k^3 + 24k^2 + 3k + m^3 + 3m^2 + 3m = 215.$$
Now I am stuck here, it is given $(k,m)$ are integers, so they can range to negative values also. So I was not able to make an upper bound on $k$, and I have no other idea what I can claim from here.
Can anyone help me? Thank You.
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First, substitute $x=m+1$ and move the constant term to the left side:
$$2x^3+16n^3-36n^2+27n=439$$
Then, note that $n$ must be odd. We can replace $n=2y+1$ and divide out by $2$ and get:
$$x^3+64y^3+24y^2+3y=216$$
Some algebraic manipulation (multiplying both sides by $8$, adding $1$ to both sides, factoring, and then replacing $z=8y+1$):
$$8x^3+512y^3+192y^2+24y=1728$$
$$8x^3+512y^3+192y^2+24y+1=1729$$
$$(2x)^3+(8y+1)^3=1729$$
$$(2x)^3+z^3=1729$$
Expressing $m$ and $n$ in terms of $x$ and $z$ gives $m=x-1$, $n=\frac{z-1}{4}+1$.
$1729$ is famously the Ramanujan-Hardy taxicab number, the smallest natural number that can be written as the sum of two cubes in two different ways. $1729=1^3+12^3=10^3+9^3$. This results in possible $(x,z)$ of $(6,1)$ or $(5,9)$. In terms of $(m,n)$, we have $(4,3)$ and $(5,1)$.
Therefore, our possible values for $mn$ are $5$ and $12$.
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|
$\alpha,\beta$ are roots of the equation $3x^2-(m-2)x+(m-5)=0$ such that $\alpha^5+\beta^5=33$. Find the value of $m$.
$\alpha,\beta$ are roots of the equation $3x^2-(m-2)x+(m-5)=0$ such that $\alpha^5+\beta^5=33$. Find the value of $m$.
$$\alpha+\beta=\frac{m-2}3$$
Squaring and cubing it one by one and then multiplying them, and putting $$\alpha^5+\beta^5=33\\\alpha\beta=\frac{m-5}3,$$ I got a quintic equation in $m$ which I couldn't solve. Any help?
Also, any other method to approach this question?
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Observe,
$$\alpha+\beta=\frac{m-2}{3}\;\;\;\text{and}\;\;\; \alpha\beta=\frac{m-5}{3}\implies \alpha+\beta=\alpha\beta+1\implies (\alpha-1)(\beta-1)=0$$
Therefore, either $\alpha=1$ or $\beta=1$. Since $\alpha^5+\beta^5=33,$ the roots of the quadratic must be $1$ and $2$. Plugging this into one of the previous equations, we must have $\boxed{m=11}$
Note: The quadratic equation is $3x^2-9x+6=3(x-1)(x-2)=0$
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Show that $u-v=\sqrt5$ given that $u-v>0$ given that $u=b+b^4$ and $v=b^2+b^3$ We're also given that $b$ is a root of $z^5-1=0$ $b^4+b^3+b^2+b+1=0$
If $u=b+b^4$ and $v=b^2+b^3$, show that
i) $u+v=uv=-1$
ii) $u-v=\sqrt5$ given that $u-v>0$
I managed to do part i):
Plugging in $u$ and $v$: $(b+b^4)+(b^2+b^3)=-1$ (using $b^4+b^3+b^2+b+1=0$)
$uv=(b+b^4)(b^2+b^3)$
$=(b^3+b^4+b^6+b^7)$
$=b^3(1+b+b3+b4)$
$=b^3(-b^2)$
$=-b^5$
$=-1^5$
$=-1$ (using $z^5=1$)
For ii)
$u-v=(b+b^4)-(b^2+b3)>0$
$(b+b^4)-(-b-b^4-1)>0$
$2b+2b^4+1>0$
$b+b^4>\frac{-1}{2}$
Although it's a dead-end after that.
What's the general idea for going about solving part ii)?
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One approach: the quadratic equation $(x-u)(x-v) = 0$ which has roots $u$ and $v$ expands to $x^2 - (u+v)x + uv = 0$. Once you've computed $u+v = uv = -1$, you know that this is the quadratic equation $x^2 + x - 1 = 0$, which you can just solve.
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|
Inconsistent solution for the factorized equation When I try to solve this equation as follows:
\begin{aligned}-k_1^2 \, \cos^2\theta \, - 2 \left( k_1k_5 + k_4 \right) \cos\theta+k_4^2-k_5^2+1 \geqslant 0 \end{aligned}
Noting that $k_1$, $k_4$, $k_5$ are real constants. I assumed:
$$
\left\{ \matrix{
x=\cos\theta \hfill \cr
a=-k_1^2 \hfill \cr
b=-2 \left( k_1k_5 + k_4 \right) \hfill \cr
c=k_4^2-k_5^2+1 \hfill \cr } \right.
$$
then it will be:
\begin{aligned}a \, x^2 +b\,x +c \geqslant 0 \end{aligned}
then finding the roots as follows:
\begin{aligned}x_1,_2 =\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{aligned}
led to:
\begin{aligned}\left(\cos\theta +\frac{k_1k_5 + k_4+\sqrt{k_1^2+k_4^2+k_1^2k_4^2+2k_1k_4k_5 }}{k_1^2} \right)\\\left(\cos\theta +\frac{k_1k_5 + k_4-\sqrt{k_1^2+k_4^2+k_1^2k_4^2+2k_1k_4k_5 }}{k_1^2} \right) \geqslant 0\end{aligned}
The problem I'm facing that when substituting real values, I cannot have the same result out of the new equation, for example susbstituting:
$$
\left\{ \matrix{
\theta=0.17453292 \hfill \cr
k_1=3.5 \hfill \cr
k_4=1.75\hfill \cr
k_5=-3.75 \hfill \cr } \right.
$$
I get in the main equation this:
\begin{aligned}0.52375908\geqslant 0\end{aligned}
and in the factored equation, I get this:
\begin{aligned}-0.04275584\geqslant 0\end{aligned}
Please advise me – what is wrong in my solution?
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[A Quadratic equation][1]
The values of x that satisfy the equation are called solutions of the equation, and roots or zeros of the expression on its left-hand side. A quadratic equation has at most two solutions. If there is no real solution, there are two complex solutions. If there is only one solution, one says that it is a double root. A quadratic equation always has two roots, if complex roots are included and a double root is counted for two. A quadratic equation can be factored into an equivalent equation
\begin{aligned}a \, x^2 +b\,x +c = a\left( x-x_1\right)\left( x-x_2\right) \end{aligned}
So, the factored equation should have been:
\begin{aligned}-k_1^2\left(\cos\theta +\frac{k_1k_5 + k_4+\sqrt{k_1^2+k_4^2+k_1^2k_4^2+2k_1k_4k_5 }}{k_1^2} \right)\\\left(\cos\theta +\frac{k_1k_5 + k_4-\sqrt{k_1^2+k_4^2+k_1^2k_4^2+2k_1k_4k_5 }}{k_1^2} \right) \geqslant 0\end{aligned}
[1]: https://en.wikipedia.org/wiki/Quadratic_equation
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|
If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$? If $2\sin\theta+\cos\theta=\sqrt3$, what is the value of $\tan^2\theta+4\tan\theta$ ?
$1)1\qquad\qquad2)2\qquad\qquad3)3\qquad\qquad4)5$
First I tried plugging in some values for $\theta$ like $0,\frac{\pi}4,\frac{\pi}3,...$ but neither of these known angles worked. But by doing it I realized that for $\theta=\frac{\pi}4+k\pi\quad$, $\tan^2\theta+4\tan\theta=5\quad$ and $2\sin\theta+\cos\theta\neq\sqrt3$ Hence the fourth choice is wrong.
Also tried to expanding,
$$\tan^2\theta+4\tan\theta=\dfrac{\sin^2\theta}{\cos^2\theta}+\dfrac{4\sin\theta}{\cos\theta}=\dfrac{\sin^2\theta+4\sin\theta\cos\theta}{\cos^2\theta}$$But can't continue even writing $4\sin\theta\cos\theta=2\sin2\theta$ doesn't help.
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Alternative:
We have $$2\sin\theta=\sqrt3 -\cos \theta$$
Squaring, $$4-4\cos^2 \theta=3+\cos^2\theta-2\sqrt 3 \cos\theta$$
So, $$5\cos^2 \theta-2\sqrt 3 \cos\theta-1=0$$
Or $$\cos \theta=\frac {2\sqrt 3 \pm 4\sqrt 2}{10}$$
Since further conditions haven't been given, both values would be valid, and from here $\tan \theta$ can be calculated.
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|
Finding the remainder of a polynomial divided by $x^4+x^2+1$ if remainders when dividing by $x^2+x+1$, $x^2-x+1$ are $-x+1$, $3x+5$.
Find the remainder of $f$ divided by $g(x)=x^4+x^2+1$ if the remainder of $f$ divided by $h_1 (x)=x^2+x+1$ is $-x+1$ and the remainder of $f$ divided by $h_2(x)=x^2-x+1$ is $3x+5$.
My attempt was to write $f(x)=(x^4+x^2+1)q(x)+Ax^3+Bx^2+Cx+D=(x^2+x+1)(x^2-x+1)q(x)+Ax^3+Bx^2+Cx+D$ and then factor out $x^2+x+1$ to get $f(x)=(x^2+x+1)[(x^2-x+1)q(x)+B]+Ax^3+(C-B)x+(D-B)$, then do the same for $x^2-x+1$ and then use that along with the known remainder of $f$ divided by $x^2+x+1$ and $x^2-x+1$ to obtain $A$, $B$, $C$, $D$. However, $Ax^3$ is in the way so I don't know how to proceed nor do I have any other ideas to start with.
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Let $u=x^2+x+1,$ $v=x^2-x+1$ so that $uv=x^4+x^2+1.$ Let $P$ be the given polynomial.
$$P=uq_1-x+1\tag1$$
$$P=vq_2+3x+5\tag2$$
$$P=uvq_3+R\tag3$$
There are polynomials $\lambda$ and $\mu$ such that
$$\lambda u+\mu v=1\tag4$$
Suitable values are $\lambda={-x+1\over2}$ and $\mu= {x+1\over2} .$
By (1), (2) and (4),
$$P=\lambda uP+\mu vP=\lambda uvq_2+(3x+5)\lambda u+\mu uvq_1+(-x+1)\mu v$$
Therefore, modulo $uv,$ $R$ is
$$ (3x+5)\lambda u+(-x+1)\mu v\tag5 $$
Of course, we must reduce the degree in (5).
$$(3x+5)\lambda=(3x+5){-x+1\over2}={1\over2}(-5x+8-3v)\tag6$$
$$(-x+1)\mu=(-x+1){x+1\over2}={1\over2}(x+2-u)\tag7$$
By (5), (6) and (7),
$$R={1\over2}\big((-5x+8)(x^2+x+1)+(x+2)(x^2-x+1) \big)$$
$$= {1\over2}\big((-5x^3+3x^2+3x+8)+(x^3+x^2-x+2) \big)$$
$$\text{Thus,}\ \ \ R=-2x^3+2x^2+x+5. $$
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|
How many real solutions does this equation have ? $4^x\times9^\frac{1}{x} + 9^x\times4^\frac{1}{x} + 6^{x+\frac{1}{x}} = 108$ How many real solutions does this equation have ?
$4^x\times9^\frac{1}{x} + 9^x\times4^\frac{1}{x} + 6^{x+\frac{1}{x}} = 108$
I am pretty sure it has something to do with prime factorization of 6, so I wrote
$(2^x\times3^\frac{1}{x})^2 + (3^x\times2^\frac{1}{x})^2 + (2^x\times3^\frac{1}{x})\times(3^x\times2^\frac{1}{x}) = 2^2\times3^3$
If i let $a=(2^x\times3^\frac{1}{x})$ and $b=(3^x\times2^\frac{1}{x})$ the equation will look like this
$a^2+b^2+ab=2^2\times3^3$ which reminds me of $a^3-b^3=(a-b)(a^2+b^2+ab)$.
Unfortunately, I don't know what to do after this, maybe this is not even the right path to solving this?
Any tips ?
Thanks in advance !
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You obtained the form $$a^2+b^2+ab=108$$
Dividing both sides with $ab$, we obtain the following equation:
$$(\frac 23)^{x-\frac 1x}+\frac {1}{(\frac 23)^{x-\frac 1x}}+1=\frac {108}{6^{x+\frac 1x}}$$
Note that, from the original equation, it can be easily shown that there can never be any negative solutions, since all three terms are $<1$. I shall leave this to you. For $x>0$, we have:
$$x+\frac 1x\geq 2$$
This means that $$\frac {108}{6^{x+\frac 1x}}\leq \frac {108}{36}\leq3$$
However, from AM-GM on LHS, we get that LHS$\geq 3$. Hence only solution is $x=1$, when both sides equal $3$.
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|
An elementary way to show that the determinant is non zero
Show that the determinant of the matrix
\begin{pmatrix} a && -c && -b \\
b && a - 2c && -c -2b \\ c && b && a -2c
\end{pmatrix} is non zero for all integers $a,b,c$ where $abc \ne 0$
There is an interesting way to do this by using integral domains. It is easy to see that the polynomial $t^3 + 2t + 1$ is prime in $\mathbb{Z}[t]$ so the ring $\mathbb{Z}[t]/\langle t^3+2t+1\rangle$ is an integral domain and isomorphic to the ring $\{a + bu + cu^2 \vert \text{$a,b,c$ are integers}\}$ where $u$ is a root of the equation $t^3 +2t + 1 =0$.
Now consider integers $a,b,c$ and $x,y,z$ such that $abc \ne 0$ and $(a +bu + cu^2)(x + yu + zu^2) = 0$. These are element of an integral domain so we must have that $x = y = z = 0$ since at least one of $a,b$ and $c$ are nonzero. But expanding the above equation we get a system of linear equations in terms of $x,y$ and $z$.
$$
\begin{aligned}
ax- cy -bz &= 0 \\
bx+ (a-2c)y + (-c-2b)z &= 0\\
c x+b y+ (a-2c)z &= 0
\end{aligned}
$$
If the determinant is $0$ then this equation will have a nontrivial solution which is not possible.
This strategy requires the use of integral domains which is an abstract tool.
Are there any elementary ways to solve this? It gets messy when we try to do row operations or try to expand the determinant.
Update : After Carl Schildkraut's and JimmyK4542's answer.
A curious observation:
We can also prove that the element $t^3 + rt + 1$ is prime in $\mathbb{Z}[t]$ for $r \ne 0, -2$. Therefore by similar arguments
we can say that the determinant of the matrix
\begin{pmatrix} a && -c && -b \\
b && a - rc && -c -rb \\ c && b && a -rc
\end{pmatrix} is non zero for all integers $a,b,c,r$ where $abc \ne 0$ and $r \ne 0,-2$.
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This might be equivalent to what you did, but I'll go ahead and write this as an answer anyway.
Let $X = \begin{bmatrix}0 & 0 & -1 \\ 1 & 0 & -2 \\ 0 & 1 & 0\end{bmatrix}$. Then, $X^2 = \begin{bmatrix}0 & -1 & 0 \\ 0 & -2 & -1 \\ 1 & 0 & -2\end{bmatrix}$, and so, $$Y := \begin{bmatrix}a & -c & -b \\ b & a - 2c & -c -2b \\ c & b & a -2c\end{bmatrix} = aI+bX+cX^2.$$
The eigenvalues of $X$ are the three roots $\lambda_1,\lambda_2,\lambda_3$ of $\lambda^3+2\lambda+1 = 0$.
Since $Y = aI+bX+cX^2$, the eigenvalues of $Y$ are $a+b\lambda_k+c\lambda_k^2$ for $k = 1,2,3$.
Now, suppose $\det(Y) = 0$ for some integers $a,b,c$ with $abc \neq 0$. Then, $0$ must be an eigenvalue of $Y$, and so, $a+b\lambda_k+c\lambda_k^2 = 0$ for some $k \in \{1,2,3\}$. But by using the quadratic formula, we have $\lambda_k = \dfrac{-b \pm \sqrt{b^2-4ca}}{2c}$.
So now you just need to prove that none of the three roots of $\lambda^3+2\lambda+1 = 0$ can be expressed in the form $\dfrac{-b \pm \sqrt{b^2-4ca}}{2c}$ for non-zero integers $a,b,c$. This just requires showing that $\lambda^3+2\lambda+1$ is irreducible in $\mathbb{Z}[\lambda]$, which only requires the rational root theorem and checking that $\pm 1$ are not roots.
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|
$\min\left(\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\right)$ Let $k>0~$ be fixed. Find
$$\min\left(\sqrt{ka+1}+\sqrt{kb+1}+\sqrt{kc+1}\right)$$
over all $c, b, a\geq0~\text{satisfying}~ab+bc+ac=a+b+c>0~.$
==================
In case we have 2 positive numbers $a+b=ab$ results that $a, b > 1$ and $a= \dfrac{b}{b-1}$,
so the minimum of the expression $ab$ is 4 because $\dfrac{b}{b-1} \cdot b \geqslant 4$ --- equality when $a=b=2$
Then $(\sqrt{ka+1} + \sqrt{kb+1})^2 = kab+2+ 2\sqrt{k^2ab + kab +1} ~~~$ (I've used $a+b = ab$)
so the minimum is reached when $ab$ is minimal, that is 4 so we obtain $4k+2 + 2\sqrt{4k^2 + 4k + 1} = 4(2k+1)$
That is the minimum searched is $2\sqrt{2k+1}$.
In case of 3 numbers I wonder if an idea is to raise to power 2 and proceed in some similar way...
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Solution for $k\ge 24$:
Let $k\ge 24$ be a real number (fixed).
Let $f(a, b, c) = \sqrt{ka + 1} + \sqrt{kb + 1} + \sqrt{kc + 1}$.
Let us prove that $f(a, b, c) \ge f(2, 2, 0) = 2\sqrt{2k + 1} + 1$
for all $a, b, c\ge 0$ with $ab + bc + ca = a + b + c > 0$.
WLOG, assume that $c = \min(a, b, c)$.
From $ab + bc + ca = a + b + c$, we have
$(1 - c)(a + b) = ab - c$. Thus, $c \le 1$.
Thus, we have $(1 - c)\cdot 2\sqrt{ab} \le ab - c$
or $(\sqrt{ab} - 1 + c)^2 \ge c^2 - c + 1$ which results
in $\sqrt{ab} \ge \sqrt{c^2 - c + 1} + 1 - c$.
We have
\begin{align*}
f(a, b, c) &= \sqrt{\left( \sqrt{ka + 1} + \sqrt{kb + 1}\right)^2} + \sqrt{kc + 1}\\
&= \sqrt{k(a + b) + 2 + 2\sqrt{k^2ab + k(a + b) + 1}} + \sqrt{kc + 1}\\
&\ge \sqrt{k\cdot 2\sqrt{ab} + 2 + 2\sqrt{k^2ab + k\cdot 2\sqrt{ab} + 1}} + \sqrt{kc + 1}\\
&= \sqrt{k\cdot 2\sqrt{ab} + 2 + 2\cdot (k\sqrt{ab} + 1)} + \sqrt{kc + 1}\\
&= 2\sqrt{k\sqrt{ab} + 1} + \sqrt{kc + 1}\\
&\ge 2\sqrt{k(\sqrt{c^2-c+1} + 1-c) + 1} + \sqrt{kc + 1}.
\end{align*}
It suffices to prove that, for all $c$ in $[0, 1]$,
$$2\sqrt{k(\sqrt{c^2-c+1} + 1-c) + 1} + \sqrt{kc + 1}\ge 2\sqrt{2k + 1} + 1.$$
Let $\sqrt{c^2 - c + 1} - c = u$.
Then $0 \le u \le 1$ and $c = \frac{1 - u^2}{2u + 1}$.
It suffices to prove that, for all $u$ in $[0, 1]$,
$$2\sqrt{ku + k + 1} + \sqrt{k \cdot \frac{1 - u^2}{2u + 1} + 1} \ge 2\sqrt{2k + 1} + 1$$
or (squaring both sides)
\begin{align*}
&4\sqrt{ku + k + 1}\sqrt{k \cdot \frac{1 - u^2}{2u + 1} + 1}\\
\ge\ & (2\sqrt{2k + 1} + 1)^2 - 4(ku + k + 1)
- k \cdot \frac{1 - u^2}{2u + 1} - 1
\end{align*}
or (squarding both sides again)
\begin{align*}
&16(ku + k + 1)\left(k \cdot \frac{1 - u^2}{2u + 1} + 1\right)\\
\ge\ & \left[(2\sqrt{2k + 1} + 1)^2 - 4(ku + k + 1)
- k \cdot \frac{1 - u^2}{2u + 1} - 1\right]^2
\end{align*}
that is
$$\frac{(1 - u)k}{(2u + 1)^2}(Au^3 + Bu^2 + Cu + D) \ge 0$$
where
\begin{align*}
A &= 81k, \\
B &= -112\,\sqrt {2\,k+1}+73\,k-32, \\
C &= -104\,\sqrt {2\,k+1}+31\,k-16, \\
D &= -24\,\sqrt {2\,k+1}+7\,k.
\end{align*}
It is easy to prove that $A, B, C, D \ge 0$ for all $k \ge 24$.
We are done.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4185351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
$x^4+6x^2+1$ is reducible modulo $p$ for every prime $p$. Let $f(x)=x^4+6x^2+1 \in \mathbb{Z}[X]$. We have to prove that $f$ is reducible modulo $p$ for all prime $p$.
I am not sure how to proceed. Please give me some hints so that I can try solving the problem.
Edit I:
I have tried with the following primes.
*
*For p=2; $f(x)$ reduces to $x^4+1 = (x^2+1)^2$ since $\bar{6}=\bar{0}=\bar{2}$. Hence $f(x)$ is reducible mod 2.
*For p=3; $f(x)$ also reduces to $x^4+1=(x^2+x+2)(x^2+2x+2)$ in $\mathbb{Z}_3[X]$
Edit II: Following JyrkiLahtonen's comment I am adding a few more things:
A. Taking $y=x^2;\ f(x)$ becomes $y^2-6y+1$. Solving the equation $y^2-6y+1=0$ we get
$$
y=-3 \pm 2 \sqrt{2}
$$
Now replacing $y$ by $x^2$ we get the following
$$
x=\pm\sqrt{-3+2\sqrt{2}}=\pm(i-i\sqrt{2}) \ \ \ \text{and} \ \ \ x=\pm\sqrt{-3-2\sqrt{2}}=\pm(i+i\sqrt{2})
$$
Hence we have obtained all the roots of the polynomial $f(x)$; say $x_1,x_2,x_3,x_4$.
B. The three ways of expanding this quartic are as follows:
*
*$\{(x-x_1)(x-x_2)\}\{(x-x_3)(x-x_4)\}=(x^2+1-2ix)(x^2+1+2ix)\\
= (x^2+1)^2-(2ix)^2 \\
= (x^2+1)^2-2^2(-1)x^2$
*$\{(x-x_1)(x-x_4)\}\{(x-x_2)(x-x_3)\}=(x^2-1-2\sqrt{2}ix)(x^2-1+2\sqrt{2}ix)\\
= (x^2-1)^2-(2\sqrt{2}ix)^2\\
=(x^2-1)^2-2^2(-2)x^2$
*$\{(x-x_1)(x-x_3)\}\{(x-x_2)(x-x_4)\}=(x^2+3+2\sqrt{2})(x^2+3-2\sqrt{2})\\
=(x^2+3)^2-(2\sqrt{2})^2\\
=(x^2+3)^2-2^2(2)$
C. Now the final piece.
$f(x)$ can be factored in any of the above 3 ways. Consider any prime $p$.
*
*If -1 is square element in $\mathbb{F}_p$; there exist $a \in \mathbb{F}_p$ such that $-1=a^2$. So factoring $f(x)$ as in form B.1. Hence considering $f(x)$ in $\mathbb{F}_p[X]$ we have
$$
f(x)=(x^2+1)^2-2^2(-1)x^2=(x^2+1)^2-(2ax)^2=(x^2+1-2ax)(x^2+1+2ax)
$$
Hence $f(x)$ is reducible modulo $p$ if $-1$ is a square element.
*If 2 is square element in $\mathbb{F}_p$, where $p$ is an odd prime; there exist $a \in \mathbb{F}_p$ such that $2=a^2$. So we factor $f(x)$ as in form B.3. Now considering $f(x)$ in $\mathbb{F}_p[X]$ we have
$$
f(x) =(x^2+3)^2-2^2(2)=(x^2+3)^2-(2a)^2=(x^2+3+2a)(x^2+3-2a)
$$
Hence we get $f(x)$ is reducible modulo $p$ if $2$ is a square element.
*Now if both $-1$ and $2$ are non square elements in $\mathbb{F}_p$ then $-2$ is a square element in $\mathbb{F}_p$ (follows by using Legendre symbol as mentioned in the comment by Jyrki Lahtonen.) Hence factoring $f(x)$ as in the form B.2 we get that $f(x)$ is reducible modulo $p$.
Hence we have $f(x)$ is reducible modulo $p$ for all $p$.
This is a special case of a more general infinite class of polynomials with similar property. I have written the general case here.
|
Here is a bit of a different approach.
For $p=2$, we can factor
$$
\begin{align}
x^4+6x^2+1
&\equiv(x-1)^4&\pmod2
\end{align}
$$
For the odd primes, where $p\in\{1,3,5,7\}\pmod8$, we will use the results from this answer, to show that $-1$ is a quadratic residue mod $p$ iff $p\equiv1\pmod4$, and this answer, to show that $2$ is a quadratic residue mod $p$ iff $p\equiv\pm1\pmod8$. Together, those answers show that $-2$ is a quadratic residue mod $p$ iff $p\equiv1,3\pmod8$.
Since $p\equiv1,5\pmod8$ implies there is a $q$ so that $q^2\equiv-1\pmod p$, we can factor
$$
\begin{align}
x^4+6x^2+1
&=\left(x^2+1\right)^2-(-4)x^2\\
&\equiv\left(x^2-2qx+1\right)(x^2+2qx+1)&\pmod p\\
\end{align}
$$
Since $p\equiv1,3\pmod8$ implies there is a $q$ so that $q^2\equiv-2\pmod p$, we can factor
$$
\begin{align}
x^4+6x^2+1
&=\left(x^2-1\right)^2-(-8)x^2\\
&\equiv\left(x^2-2qx-1\right)(x^2+2qx-1)&\pmod p\\
\end{align}
$$
Since $p\equiv1,7\pmod8$ implies there is a $q$ so that $q^2\equiv2\pmod p$, we can factor
$$
\begin{align}
x^4+6x^2+1
&=\left(x^2+3\right)^2-8\\
&\equiv\left(x^2+3-2q\right)(x^2+3+2q)&\pmod p\\
\end{align}
$$
|
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"url": "https://math.stackexchange.com/questions/4186629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
Alternative proofs of convergence of geometric series The usual proof for the convergence of a geometric series of ratio $C: |C|\in [0,1)$ makes use of the formula $$\sum_{0\leq k \leq n} C^k = \frac{1-C^{n+1}}{1-C}.$$
I'm looking for alternative ways to prove it. The motivation for this is that, if someone who never saw this formula tried to prove the geometric series converges might have a hard time, unless maybe there are other, perhaps more insightful ways to prove it.
|
If $c = \dfrac1{1+b}$ where $b > 0$
then,
since,
for $k \ge 2$,
$(1+b)^k
\ge 1+bk+b^2k(k-1)/2
\gt b^2k(k-1)/2$
(readily proved by induction),
so
$c^k
= \dfrac1{(1+b)^k}
\le \dfrac{2}{b^2k(k-1)}
$
so
$\begin{array}\\
\sum_{k=2}^n c^k
&\le \sum_{k=2}^n\dfrac{2}{b^2k(k-1)}\\
&= \dfrac{2}{b^2}\sum_{k=2}^n\dfrac1{k(k-1)}\\
&= \dfrac{2}{b^2}\sum_{k=2}^n(\dfrac1{k-1}-\dfrac1{k})\\
&= \dfrac{2}{b^2}(1-\dfrac1{n})\\
&< \dfrac{2}{b^2}\\
&= \dfrac{2}{(\frac1{c}-1)^2}\\
&= \dfrac{2c^2}{(1-c)^2}\\
&\text{so}\\
\sum_{k=0}^n c^k
&< 1+c+\dfrac{2c^2}{(1-c)^2}\\
&= \dfrac{c^3 + c^2 - c + 1}{(1-c)^2}\\
\end{array}
$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4188906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
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|
Integer solution of $abc=a+b+c+2$. Let $a,b,c$ be integers greater than $1$. I am trying to prove that $$abc\geq a+b+c+2$$
with equality if and only if $a=b=c=2$.
I can prove the inequality by using the fact that $ab\geq a+b$. Since $ab\geq 4$ and $c\geq 2$, it follows that $abc\geq 4c$ and $abc\geq 2ab$. Therefore $abc\geq ab+2c\geq ab+c+c\geq a+b+c+c\geq a+b+c+2$.
The main problem I face is to justify that $$abc=a+b+c+2\implies a=b=c=2.$$
My idea is to assume that $a>2$ and try to get a contradiction.
|
I think your answer is correct. Assume that $a>2$, since $bc\geq b+c$, you get $$abc\geq ab+ac\geq a+b + a +c>a+b+c+2.$$
All $a,b,c$ must be less than or equal to $2$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4189755",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Find $a^2 + b^2 + c^2 + d^2$.
$$\begin{align}\dfrac{a^2}{2^2 - 1^2} + \dfrac{b^2}{4^2 - 1^2} + \dfrac{c^2}{6^2 - 1^2} + \dfrac{d^2}{8^2 - 1^2} &= \dfrac{a^2}{2^2 - 3^2} + \dfrac{b^2}{4^2 - 3^2} + \dfrac{c^2}{6^2 - 3^2} + \dfrac{d^2}{8^2 - 3^2}
\\&= \dfrac{a^2}{2^2 - 5^2} + \dfrac{b^2}{4^2 - 5^2} + \dfrac{c^2}{6^2 - 5^2} + \dfrac{d^2}{8^2 - 5^2} \\ &= \dfrac{a^2}{2^2 - 7^2} + \dfrac{b^2}{4^2 - 7^2} + \dfrac{c^2}{6^2 - 7^2} + \dfrac{d^2}{8^2 - 7^2}
\\ &= 1\end{align}$$
Find $a^2 + b^2 + c^2 + d^2$.
My Approach:
I tried forming Linear equations by taking $a^2=x, b^2=y, c^2=z, d^2=w$
Then I obtained
$\dfrac{x}{3}+\dfrac{y}{15}+\dfrac{z}{35}+\dfrac{w}{63}=1$
$\dfrac{x}{-5}+\dfrac{y}{7}+\dfrac{z}{27}+\dfrac{w}{55}=1$
$\dfrac{x}{-21}+\dfrac{y}{-9}+\dfrac{z}{11}+\dfrac{w}{39}=1$
$\dfrac{x}{-45}+\dfrac{y}{-33}+\dfrac{z}{-13}+\dfrac{w}{15}=1$
After solving with www.wolframalpha.com I obtained the result
$w=\frac{6435}{256}$, $x=\frac{315}{512}$, $y= \frac{693}{256}$, $z=\frac{3861}{512}$
Which lead me $x+y+z+w=36$ that is $a^2+b^2+c^2+z^2=36$
But how to solve this question without any calculator or online tools because it was asked to a $11th$ class student
Wolfram Alpha
|
We write down the following polynomial:
\begin{eqnarray}
f(x) &=& (x - 2^2)(x - 4^2)(x - 6^2)(x - 8^2)\\
&+& a^2(x - 4^2)(x - 6^2)(x - 8^2)\\
&+& b^2(x - 2^2)(x - 6^2)(x - 8^2)\\
&+& c^2(x - 2^2)(x - 4^2)(x - 8^2)\\
&+& d^2(x - 2^2)(x - 4^2)(x - 6^2).
\end{eqnarray}
The given conditions then imply that $f(1^2) = f(3^2) = f(5^2) = f(7^2) = 0$.
However, $f(x)$ is a monic polynomial of degree $4$. Therefore we have $$f(x) = (x - 1^2)(x - 3^2)(x - 5^2)(x - 7^2).$$
It only remains to evaluate $f(x)$ at $x = 2^2, 4^2, 6^2, 8^2$ respectively to get values of $a^2, b^2, c^2, d^2$.
As suggested in the comments, one can look at the coefficient before $x^3$ to get the value of $a^2 + b^2 + c^2 + d^2$ without doing much calculation.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Constructing a polynomial with integer coefficient's sharing roots with another two polynomials
Let $i$ be a root of the equation $ x
^2 + 1 = 0 $and let $ω$ be a root of the
equation $x^2 + x + 1 = 0$. Construct a polynomial $f(x) = \sum_k a_k x^k$ with $a_k \in \mathbb{Z}$ such that $f(i+w)=0$. Source
After some long thought, I got a 'partial' construction for $f$: let $q(x) =x^2 +1$ and $p(x) = x^2 + x+1$, then:
$$ f(x) = q(x- \omega) + p(x- i)$$
However, the above doesn't necessarily satisfy the 'integer' coefficients criteria and may contain an imaginary part, so I modified the construction a bit to fix the second problem and got:
$$ f(x) = q(x- \omega) + p(x-i) + \overline{ p(x-i) + q( x - \omega)}$$
Problem: The above has no guarantee to have integer coefficients and the expression also has terms of $\overline{x}$ i.e: complex conjugate of $x$.
|
Multiplying $\,\alpha=\omega+i\,$ by $\,i\,$ and using that $\,i^2=-1\,$:
$$
\begin{align}
\begin{cases}
\alpha &= \omega+i
\\ i \,\alpha &= i \omega - 1
\end{cases}
\end{align}
$$
Eliminating $\,i\,$ between the two equations, for example by taking $\,i=\alpha-\omega\,$ from the first equation and substituting into the second one:
$$
\alpha^2 - 2\omega\alpha + \omega^2+1=0
$$
Using that $\,\omega^2=-\omega-1\,$:
$$
\alpha^2 - 2\alpha\omega - \omega=0
$$
Multiplying by $\,\omega\,$ and using that $\,\omega^2=-\omega-1\,$:
$$
\require{cancel}
\begin{align}
\begin{cases}
\alpha^2 - \omega (2\alpha + 1) &= 0
\\ \omega\,\alpha^2 + (\omega+1)(2\alpha+1) &= 0
\end{cases}
\;\;\iff\;\; \begin{cases}
- \omega\,(2\alpha+1) + \alpha^2 &= 0
\\ \omega\,(\alpha^2+2\alpha+1) + 2\alpha+1&=0
\end{cases}
\end{align}
$$
Eliminating $\,\omega\,$ between the two equations:
$$
(\alpha^2+2\alpha+1)\alpha^2+(2\alpha+1)^2 = 0
\;\;\;\;\iff\;\;\;\; \alpha^4 + 2 \alpha^3 + 5 \alpha^2 + 4 \alpha + 1 = 0
$$
With $\,P(x)=x^2+1\,$ and $\,Q(x)=x^2+x+1\,$, the above procedure is equivalent to calculating the polynomial resultant of $\,P(x)\,$ and $\,Q(\alpha-x)\,$, as verified in WA. A similar procedure can be used for higher degrees to produce a polynomial having as roots the sums of roots of $\,P\,$ and $\,Q\,$ with coefficients in the same ring as those of $P,Q$, in this case $\mathbb Z\,$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4190723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
For $|x|\gt1,\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac2{x^{2^k}+x^{-2^k}}\right)=f(x)$
For $|x|\gt1,$ $$\lim_{n \to \infty}\prod_{k=0}^{n} \left(1+\frac2{x^{2^k}+x^{-2^k}}\right)=f(x)$$
Show that
*
*A) $\int_2^5f(x)dx=3+\ln16$
*B) $\lim_{x\to\infty}f(x)=1$
*C) $f(x)=0$ has one solution
*D) $f(x)$ is decreasing function in the domain of function.
$$f(x)=\lim_{n \to \infty}\prod_{k=0}^{n} \left(\frac{x^{2^k}+x^{-2^k}+2}{x^{2^k}+x^{-2^k}}\right)\\=\lim_{n \to \infty}\prod_{k=0}^{n}\frac{(x^{2^{k-1}}+x^{-2^{k-1}})^2}{x^{2^k}+x^{-2^k}}\\=\lim_{n\to\infty}\frac{(x^{2^{-1}}+x^{-2^{-1}})^2}{x^{2^0}+x^{-2^0}}\frac{(x^{2^{0}}+x^{-2^{0}})^2}{x^{2^1}+x^{-2^1}}\frac{(x^{2^{1}}+x^{-2^{1}})^2}{x^{2^2}+x^{-2^2}}...\frac{(x^{2^{n-1}}+x^{-2^{n-1}})^2}{x^{2^n}+x^{-2^n}}\\=\lim_{n\to\infty}(x^{2^{-1}}+x^{-2^{-1}})^2(x^{2^{0}}+x^{-2^{0}})(x^{2^{1}}+x^{-2^{1}})...\frac{(x^{2^{n-1}}+x^{-2^{n-1}})}{x^{2^n}+x^{-2^n}}$$
$x+\frac1x\gt2\implies x^2+\frac1{x^2}\gt2\implies x^4+\dfrac1{x^4}\gt2...\implies x^{2^n}+\dfrac1{x^{2^n}}\gt2$
Not able to proceed next.
|
Let $P$ denote the product, without the limit.
Note that:
$$\left(1+\frac {2}{x^{2^k}+\frac {1}{x^{2^k}}}\right)=\frac {\left(x^{2^{k-1}}+\frac {1}{x^{2^{k-1}}}\right)^2}{x^{2^k}+\frac {1}{x^{2^k}}}$$
Using this leads to a partial telescopic product. I shall skip the cancellation steps, and write directly the result as:
$$P=\frac {\left(\sqrt x+\frac {1}{\sqrt x}\right)^2}{x^{2^n}+\frac {1}{x^{2^n}}}\prod_{k=0}^{n-1} \left(x^{2^k}+\frac {1}{x^{2^k}}\right)$$
Let the remaining product be called $Q$, then multiplying and dividing $Q$ by $x-\frac 1x$, we get (again, simplification has been left for the reader):
$$Q=\frac {x^{2^n}-\frac {1}{x^{2^n}}}{x-\frac 1x}$$
Thus, we have:
$$f(x)=\frac {(\sqrt x+\frac {1}{\sqrt x})^2}{x-\frac 1x} \lim_{n\to \infty} \frac {x^{2^n}-\frac {1}{x^{2^n}}}{x^{2^n}+\frac {1}{x^{2^n}}}$$
The limit part evaluates to $1$, hence, after simplification we obtain:
$$f(x)=\frac {x+1}{x-1}$$
The rest is quite simple, all options shall be correct.
|
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|
Find the longest side of a triangle when the sides are in a sequence with a common difference as 2 The Question:
Sides of a triangle form an arithmetic sequence with a common difference of 2. The area of triangle is $24 cm^2$. If the length of the greatest side is $P$, find the length of $P$.
My thoughts on this statement arithmetic sequence with a common difference of 2, is that the sides of a triangles differ by $2cm$.
So let the sides of the triangle be $x,x+2,x+4$.
According to heron's formula for finding the area of a triangle:
$A=\sqrt{s(s-a)(s-b)(s-c)}$, where $a,b,c$ are the 3 sides of the triangle and $s$ is the semi perimeter of the triangle: $\frac{a+b+c}{2}=s$
For this: $\frac{a+b+c}{2}=s$, I substituted $x,x+2,x+4$,
So it becomes:
$$\frac{x+(x+2)+(x+4)}{2}=s$$
$$\text{After substitution, this becomes}$$
$$\frac{3x+6}{2}=S$$
Substituting that into heron's formula:
$$24=\sqrt{\frac{3x+6}{2}(\frac{3x+6}{2}-x)(\frac{3x+6}{2}-(x+2))(\frac{3x+6}{2}-(x+4))}$$
$$24=\sqrt{\frac{3x+6}{2}(\frac{3x+6}{2}-\frac{2x}{2})(\frac{3x+6}{2}-\frac{2x+4}{2})(\frac{3x+6}{2}-\frac{2x+8}{2})}$$
$$24=\sqrt{\frac{3x+6}{2}(\frac{x+6}{2})(\frac{x+2}{2})(\frac{x-2}{2})}$$
The positive root of the expression is $6$
So, the longest side I got is $$x+4=10$$
So, is my answer and logic correct?
|
Building on Jaap and Math Lover's comments
Your solution is perfectly correct but it is advisable not to try this in olympiads as you have limited time over here and you can't devote this much time to one comment otherwise you won't be able to do the other questions. Instead let $x-2,x,x+2$ be the sides of the triangle and then solve it and yay you got a expression much easier to solve. Just solve that qurtic equation and get the answer. You will get $x$ as $8$.
|
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|
If $\alpha\ne1,\alpha^6=1$ and $\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x,$ then find the value of $|x|$. The following question is taken from the practice set of JEE exam.
If $\alpha\ne1,\alpha^6=1$ and $\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x,$ then find the value of $|x|$.
$$\sum_{r=1}^6 {^6}C_r\alpha^{r-1}=x\\\implies {^6}C_1\alpha^0+{^6}C_2\alpha^1+{^6}C_3\alpha^2+{^6}C_4\alpha^3+{^6}C_5\alpha^4+{^6}C_6\alpha^5=x\\\implies {^6}C_1\alpha^1+{^6}C_2\alpha^2+{^6}C_3\alpha^3+{^6}C_4\alpha^4+{^6}C_5\alpha^5+{^6}C_6\alpha^6=\alpha x\\\implies {^6}C_0\alpha^0+{^6}C_1\alpha^1+{^6}C_2\alpha^2+{^6}C_3\alpha^3+{^6}C_4\alpha^4+{^6}C_5\alpha^5+{^6}C_6\alpha^6=\alpha x+{^6}C_0\\\implies (1+\alpha)^6=1+\alpha x$$
How shall I proceed next? I understand $\alpha$ is $6^{th}$ root of unity. How to use that here? Maybe $\alpha^6=1$ can be written as $$\alpha^6-1=0\\\implies (\alpha-1)(\alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1)=0$$
Also, $\alpha\ne1\implies \alpha^5+\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$. Is this of some help here? Thanks.
|
I'll do the case where $\alpha$ is a primitive $6$-th root of unity.
Then $\alpha$ satisfies $\alpha^2-\alpha+1=0$,$\;$hence
\begin{align*}
x&=\sum_{i=1}^6 {\small{\binom{6}{i}}}\alpha^{i-1}
\\[4pt]
&=
\alpha^5+6\alpha^4+15\alpha^3+20\alpha^2+15\alpha+6
\\[4pt]
&=
(\alpha^3+7\alpha^2+21\alpha+34)(\alpha^2-\alpha+1)+(28\alpha-28)
\\[4pt]
&=
28\alpha-28
&&
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!
\bigl(\text{since $\alpha^2-\alpha+1=0$}\bigr)
\\[4pt]
&=
28(\alpha-1)
\\[4pt]
&=
28\alpha^2
&&
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!
\bigl(\text{$\alpha^2-\alpha+1=0$ implies $\alpha-1=\alpha^2$}\bigr)
\\[10pt]
\implies\;|x|&=
|28\alpha^2|
\\[4pt]
&=
28|\alpha|^2
\\[4pt]
&=
28
&&
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!
\bigl(\text{since $|\alpha|=1$}\bigr)
\end{align*}
|
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|
Upper limit for $(1+1/2)(1+1/3)\dots(1+1/n)$ I am trying to find an upper limit for the product $(1+1/2)(1+1/3)\dots(1+1/n)$. I tried using AM-GM inequality as follows:
\begin{align}
(1+1/2)(1+1/3)\dots(1+1/n) \leq \left(\frac{1}{n-1}\left( -1 + n + 1/2 + 1/3 + 1/4 + \dots + 1/n\right)\right)^{n-1}
\end{align}
From Wikipedia,
$1/2 + 1/3 + 1/4 + \dots + 1/n \leq \log n$.
Therefore, I get
\begin{align}
(1+1/2)(1+1/3)\dots(1+1/n) \leq \left(\frac{1}{n-1}\left( -1 + n + 1/2 + 1/3 + 1/4 + \dots + 1/n\right)\right)^{n-1} \leq \left( 1+ \frac{\log n}{n-1} \right)^{n-1}
\end{align}
As $n \to \infty$, I find an upper limit is $n$.
I was wondering if there are more tighter upper limits. Any comments/thoughts are welcome.
|
The product is equal to
$$\prod_{i=2}^n \left(1+\frac{1}{i}\right)=\prod_{i=2}^n \frac{i+1}{i}=\frac{n+1}{2}$$
:D
|
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|
Given $3$ points on a unit circle, figure out something about them. Question: Given three points $(a, b), (c, d)$ and $(x, y)$ on the unit circle in a rectangular coordinate plane, find the maximum possible value of the expression $(ax + by - c)^2 + (bx - ay + d)^2 + (cx + dy + a)^2 + (dx - cy - b)^2. $
Answer: We will prove that the only value for this expression is $4$.
Without loss of generality, assume that the unit circle is the graph $x^2 + y^2 = 1$. This means that $b^2 = 1 - a^2, d^2 = 1 - c^2,$ and $y^2 = 1 - x^2$.
Expanding the expression, we get
$a^2x^2 + b^2y^2 + c^2 + 2axby - 2acx - 2byc + b^2x^2 + a^2y^2 + d^2 - 2bxay - 2ayd + 2bxd + c^2x^2 + d^2y^2 + a^2 + 2cxdy + 2ady + 2acx + d^2x^2 + c^2y^2 + b^2 - 2dxcy + 2bcy - 2dxb$.
This was a long expression! Fortunately, we see that most of the terms cancel out and we are left with $a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 + c^2x^2 + c^2y^2 + d^2x^2 + d^2y^2 + a^2 + b^2 + c^2 + d^2$.
This can be factored to $(a^2 + b^2 + c^2 + d^2)(x^2 + y^2 + 1)$.
Using the fact from above that $b^2 = 1 - a^2, d^2 = 1 - c^2,$ and $y^2 = 1 - x^2$, this expression simplifies to $(a^2 + 1 - a^2 + c^2 + 1 - c^2)(x^2 + 1 - x^2 + 1) = 2(2) = 4$.
So, the answer to our original question is $\boxed{4}$.
Please verify my proof to see if there are any flaws or mistakes with the proof. Thanks in advance!
|
Here is another way I would approach it. Given all the three points are on the unit circle, their coordinates can be written as,
$(a, b) \mapsto (\cos \theta_1, \sin \theta_1); \ (c, d) \mapsto (\cos\theta_2, \sin\theta_2)$;
$(x, y) \mapsto (\cos\theta_3, \sin\theta_3)$
So, $ \ (ax + by - c)^2 + (bx - ay + d)^2 + (cx + dy +a)^2$
$ + (dx - cy - b)^2 \ $ can be written as,
$(\cos(\theta_1 - \theta_3) - \cos \theta_2)^2 + (\sin(\theta_1 - \theta_3) + \sin \theta_2)^2 + (\cos(\theta_2 - \theta_3) + \cos\theta_1)^2 + (\sin(\theta_2 - \theta_3) - \sin\theta_1)^2$
$ = 2 - 2 \cos (\theta_1 + \theta_2 - \theta_3) + 2 + 2 \cos (\theta_1 + \theta_2 - \theta_3) = 4$
|
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|
Calculate surface of a triangle It's quite a simple question solvable with many methods but the $2$ of them I picked don't agree.
First of I have chosen the triangle with corners: $(3,0,0) \quad (0,3,0) \quad (0,0,6)$
Thus, this one here:
First Method is by evaluating the surface Integral: $A = \int_{S}1\,\mathrm{dS} = \int_{(u,v)}1\,|n|\,\mathrm{du\,dv}$
As parametrization I came up with: $\psi(u,v) = \left(\begin{array}{cc}3\,u\\ 3\,v \\ 6-6\,u-6\,v\end{array}\right) \quad u \in[0,1] \quad v \in [0,1] \quad u+v<1$
Hence $n = \partial_u\psi \times \partial_v\psi = \left(\begin{array}{cc}18\\ 18 \\ 9\end{array}\right)\quad$ $\Rightarrow \quad \displaystyle{A = \int_0^1 \int_0^{1-u}} \sqrt{2\cdot18^2+9^2}\,\mathrm{dv\,du} = 13,5$
Next Method is by using straight geometry: $A= \frac{1}{2}\,g\,h$ (base times hight)
It should be: $A = \frac{1}{2}\,\sqrt{3^2+3^2}\,h = \frac{1}{2}\,\sqrt{18}\,\sqrt{6^2+\left(\sqrt{3^2-\frac{1}{2}\sqrt{18}}\right)^2} \approx 13,9$
Now again, where does this difference come from? Probably one could check by calculation $\textbf{det}$
|
Given $V=6$ and $S=3$
The length of the triangle base is base is
$B=\sqrt{S^2+S^2}=\sqrt{3^2+3^2}=3\sqrt{2}. \quad$
The distance from the back corner to the base is
$Z=\sqrt{S^2-\bigg(\dfrac{B}{2}\bigg)^2}=\sqrt{3^2-\bigg(\dfrac{3\sqrt{2}}{2}\bigg)^2}
=\dfrac{3\sqrt{2}}{2}.\quad$
The altitude of the triangle is
$H=\sqrt{V^2+Z^2}
=\sqrt{6^2+\bigg(\dfrac{3\sqrt{2}}{2}\bigg)^2}
=\dfrac{9\sqrt{2}}{2}.\quad$
The area of the triangle is
$A=\dfrac{B}{2}H
=\dfrac{3\sqrt{2}}{2}\cdot\dfrac{9\sqrt{2}}{2}
=\dfrac{27}{2}=13.5
$
|
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|
A non-inductive proof regarding a property of natural numbers Here is the question:
For each natural number $n$, prove that
$$n(n + 1)(n +2) (n+3)$$
is never a perfect square.
My approach: Here,we notice that two of the factors given in the product are bound to be odd while
the remaining two are bound to be even. WLOG,let $n$ be even. If $n=2k$ for some integer $k$ then
$$n+2=2k+2=2(k+1).....(1)$$. Thus,we get that $n(n+2)=4(k+1)k$ where $k$ and $k+1$ can have no common factor.
If two consecutive odds $n+1$ and $n+3$ have a common factor $x$ then,
$$n+3-(n+1)=2=x(k_1-k_2)$$where $n+3=k_1x$ and $n+1=k_2x$. But,then $x=1$ is the only possible value. Thus, we get that their gcd is 1.Since,we now see that there can be no double repetition of the prime factors except that of $4$ in:
$$n(n + 1)(n +2) (n+3)=4(k+1)k(n+3)(n+1)......(2)$$ since the consecutives cannot contribute a common factor and
the numbers that have the same kind of divisibility by $2$ cannot have common factors other than $2$ (for the even case).So,we get that the product is never a perfect square.(We see that steps (2) and (1) are invertible if $n$ is odd.
It is a bit rough and I am not sure whether this is correct.Please help me verify this and correct the proof. I want to see the defects of my proof.
|
$$(n^2+3n)^2<n(n+1)(n+2)(n+3)=n(n+3)(n+1)(n+2)=(n^2+3n)(n^2+3n+2)=(n^2+3n)^2+2(n^2+3n)<(n^2+3n)^2+2(n^2+3n)+1=(n^2+3n+1)^2$$
$n(n+1)(n+2)(n+3)$ is between two consecutive perfect squares, it is not a perfect square
|
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|
Find segment ratio when three points form three lines dividing area of equilateral triangle evenly $\triangle{ABC}$ is equilateral. Points $E,F,G$ are inside it and form four triangles so that $S_1=S_2=S_3=S_4$. Find $\dfrac{AF}{FG}$.
I saw this math problem on social media but did not see a strict solution. I had some results after applying barycentric coordinates to the points but did not get the final result: if mark the area of $\triangle{ACE}, \triangle{BFA}, \triangle{CGB}, \triangle{ABC}$ as $S_4, S_5, S_6, S$, then
$(S-(6+2\sqrt{5})S_4)(S-(6-2\sqrt{5})S_5)+(S-(6+2\sqrt{5})S_5)(S-(6-2\sqrt{5})S_6)+(S-(6+2\sqrt{5})S_6)(S-(6-2\sqrt{5})S_1)=0$
Maybe I need some strict proof to show the symmetry for these triangles.
Thanks.
|
Join G to C , we have:
$\frac{A_{GFC}}{A_{AFC}}=\frac{FG}{FA}=\frac{EG}{GB}$
$A_{GFC}=A_{GFE}+A_{GEC}$
Putting this in first relation we get:
$\frac{FG}{FA}=\frac{EG}{GB}=1+\frac{EG}{EB=EG+GB}=1+\frac{FG}{GA=FA+FG}$
Rearranging we get:
$\frac{FG}{FA}-1=\frac{FG}{FA+FG}$
Or: $FG^2-FA^2=FA\times FG$
Let $FG=x$ and $FA=y$, we have:
$x^2-y^2-xy=0$
Solving for x we find:
$x=\frac{y\pm\sqrt{y^2+4y^2}}{2}$
$\frac xy=\frac {FG}{FA}=\frac{1+\sqrt 5}2$
|
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|
Comparing $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the values
I want to compare $\sqrt{5} + \sqrt{6} + \sqrt{11}$ and $8$ without calculating the actual value of square roots.
I tried to apply square on both side but it still carries the root terms.
Any trick I was missing here?
|
Note that $8>\sqrt{11}$. Therefore\begin{align}\sqrt5+\sqrt6+\sqrt{11}>8&\iff\sqrt5+\sqrt6>8-\sqrt{11}\text{ ($3$ square roots)}\\&\iff11+2\sqrt{30}>75-16\sqrt{11}\text{ (only $2$ square roots)}\\&\iff\sqrt{30}>32-8\sqrt{11}\\&\iff30>1728-512\sqrt{11}\text{ (a single square root)}\\&\iff512\sqrt{11}>1698\\&\iff256\sqrt{11}>849,\end{align}which is true, since $256^2\times11=720\,896$ and $849^2=720\,801$.
|
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|
Integral of exponential within a region Are there methods to compute the following integral for $a \leq b$? Here $x\in\mathbb{R}$
$$
\int\limits_{a \leq -\frac{x^2}{2} \leq b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\,dx
$$
Substitution
The error function is
$$
\text{erf}(z) = \frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2} dt
$$
Using the substitution $t^2=\frac{x^2}{2}$ we have $2dt = dx$ and $\sqrt{-b}\leq t \leq \sqrt{-a}$
$$
\frac{1}{\sqrt{2}}\cdot \frac{2}{\sqrt{\pi}}\int_{\sqrt{-b}}^{\sqrt{-a}} e^{-t^2}dt = \frac{1}{\sqrt{2}}\cdot \left[\frac{2}{\sqrt{\pi}}\int_{0}^{\sqrt{-a}} e^{-t^2}dt + \frac{2}{\sqrt{\pi}}\int^0_{\sqrt{-b}} e^{-t^2} dt\right]
$$
Using the definition of error function and substituting $t' = -t$
$$
\frac{1}{\sqrt{2}}\left[\text{erf}(\sqrt{-a}) - \frac{2}{\sqrt{\pi}}\int_0^{-\sqrt{-b}} e^{-t^2} dt\right] = \frac{1}{\sqrt{2}}\left[\text{erf}(\sqrt{-a}) - \text{erf}(-\sqrt{-b})\right]
$$
|
We can use $\operatorname{erf}$ in place of $\Phi$. By definition,
$$
\frac{d}{dx}\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}e^{-x^2}
\tag1$$
Now start with
$$
V = \int\limits_{a \leq -\frac{x^2}{2} \leq b} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\,dx
$$
where $a < b \le 0$.
Divide into two equal parts, $x>0$ and $x<0$, to get
$$
V = \frac{\sqrt2}{\sqrt{\pi}}\int_{\sqrt{-2b}}^{\sqrt{-2a}}e^{-x^2/2}dx
$$
In order to match $(1)$ substitute $x=\sqrt{2}\,y$, $dx = \sqrt{2}\,dy$ to get
$$
V =\frac{2}{\sqrt{2\pi}}\int_{\sqrt{-b}}^{\sqrt{-a}} e^{-y^2} dy
=\frac{2}{\sqrt{\pi}}\left(
\frac{\sqrt{\pi}}{2}\operatorname{erf}(\sqrt{-a})-
\frac{\sqrt{\pi}}{2}\operatorname{erf}(\sqrt{-b})
\right)
=\operatorname{erf}(\sqrt{-a})-\operatorname{erf}(\sqrt{-b})
$$
|
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|
How to calculate $\mathbb{E}\left(\varPhi^{2}\left(\frac{W_{s}}{\sqrt{1-s}}\right)\right)$? How can we calculate $$\mathbb{E}\left(\varPhi^{2}\left(\frac{W_{s}}{\sqrt{1-s}}\right)\right)$$ where $W$ denotes a Wiener process, and $\varPhi$ is the CDF of the standard normal random variable... If $\varphi$ denotes the PDF of the standard normal variable, then$$\mathbb{E}\left(\varPhi^{2}\left(\frac{W_{s}}{\sqrt{1-s}}\right)\right)=\int_{\mathbb{R}}\varPhi^{2}\left(\frac{x}{\sqrt{1-s}}\right)\frac{1}{\sqrt{2\pi s}}e^{-\frac{x^{2}}{2s}}dx=\int_{\mathbb{R}}\varPhi^{2}\left(\frac{x}{\sqrt{1-s}}\right)\frac{1}{\sqrt{s}}\varphi\left(\frac{x}{\sqrt{s}}\right)dx=...?$$ I can't continue...
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Let $$M_t = \Phi \left( \frac{W_t}{\sqrt{1-t}} \right) = g(t, W_t)$$ where $g(t,x) =\Phi \left( \frac{x}{\sqrt{1-t}} \right)$. We claim that $M_t$ is a martingale. First, observe that $\Phi^\prime (s) =\frac{1}{\sqrt{2 \pi}} \exp \left( \frac{s^2}{2} \right)$ and $\Phi^{\prime \prime}(s) = - \frac{s}{\sqrt{2 \pi}}\exp \left( \frac{s^2}{2} \right)$. Furthermore,
$$\begin{align*}
g_x(t,x) &= \frac{1}{\sqrt{1 - t}} \Phi^\prime \left( \frac{x}{\sqrt{1 - t}} \right) \\
g_t(t,x) &= \frac{x}{2\sqrt{1 - t}(1-t)} \Phi^\prime \left( \frac{x}{\sqrt{1 - t}} \right) \\
g_{xx}(t,x) &= \frac{1}{1-t} \Phi^{\prime \prime}\left( \frac{x}{\sqrt{1 - t}} \right)
\end{align*}$$
From here, you may check that $g_t(t,x) + \frac{1}{2}g_{xx}(t,x) = 0$, so that $M_t$ is driftless and hence a local martingale. It's in fact a true martingale by observing that $$dM_t = \frac{1}{\sqrt{2\pi}\sqrt{1-t}} \exp \left( - \frac{W_t^2}{2(1 - t)} \right) dW_t$$ and noting that the integrand (being bounded) belongs to $L^2 (\Omega \times [0,T])$ for $T < 1$.
The dynamics of $M_t^2$ are given easily by Itô's lemma: $$d(M^2)_t = (dM_t)^2 + 2M_t(dM_t) = \frac{1}{2\pi (1-t)} \exp \left( - \frac{W_t^2}{1-t} \right) dt + \text{martingale}$$
So that $$E(M_t^2) = M_0^2 + \int_0^t \frac{1}{2\pi (1-s)} E\exp\left( -\frac{W_s^2}{1-s}\right) ds$$
The inner expectation evaluates to $$E\exp\left( -\frac{W_s^2}{1-s}\right) = E\exp\left( -\frac{s}{1-s}W_1^2\right) = \left( \frac{1+s}{1-s} \right)^{-1/2} = \sqrt{ \frac{1-s}{1+s} }$$
Where in the second equality we have evaluated the MGF of $W_1^2 \sim \chi^2(1)$ at the point $- \frac{s}{1-s}$.
From this point on, the expectation is a standard calculus exercise which I leave to you.
|
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|
$f(z)$ is holomorphic on $|z| < 1$, with $|f(z)| ≤ 3$ and $f(\frac{1}{2}) = 2$. Show that $f(z) \neq 0$ when $|z| < \frac{1}{8}.$ This is an old Schwarz lemma problem from the August 2020 UMD qualifying exam for analysis, which is posted here. The precise wording from the test is:
Suppose $f(z)$ is a holomorphic function on the unit disk with $|f(z)| \le 3$ for all $|z| < 1$, and
$f(1/2) = 2$. Show that $f(z)$ has no zeros in the disk $|z| < 1/8$. (Hint: first show $f(0) \neq 0$).
There is another kind of standard problem that assumes $|f(0)| \ge r$ and asks to prove $|f(z)| \ge \frac{r - |z|}{1 - r|z|}$ on the disk $|z| < r$; see problem 5 from chapter IX.1 of Gamelin. This problem instead gives you a value for $f(a)$, $a \neq 0$.
|
We know that the hyperbolic unit disk metric, $d(z, w) := |\frac{z - w}{1 - \bar{w}z}|$, satisfies $d(g(z), g(w)) \le d(z, w)$ for analytic maps $g : \mathbb{D} \to \mathbb{D}$. Note that $|\frac{f(z)}{3}| \le 1$ when $|z| < 1$, and the maximum principle implies we can assume $|\frac{f(z)}{3}| < 1$, otherwise $|\frac{f(z)}{3}|$ is constantly $1$, contradicting $f(\frac{1}{2}) = 2$. Hence
\begin{align*}
\left|\frac{\frac{2}{3} - \frac{f(0)}{3}}{1 - \overline{\frac{f(0)}{3}}\cdot \frac{2}{3}}\right| = d(\frac{2}{3}, \frac{f(0)}{3}) = d(\frac{f(\frac{1}{2})}{3}, \frac{f(0)}{3}) \le d(\frac{1}{2}, 0) = \frac{1}{2}.
\end{align*}
If we assume that $|\frac{f(0)}{3}| \le \frac{2}{3}$, then the triangle inequality applied twice implies
\begin{align*}
\frac{2}{3} - \frac{|f(0)|}{3} \le \frac{1}{2} + \frac{|f(0)|}{9}.
\end{align*}
Simplifying this implies $|\frac{f(0)}{3}| \ge \frac{1}{8}$. This inequality also holds with the negation of our assumption, and so it holds generally.
Recall that $\phi_a (z) := \frac{z - a}{1 - \overline{a}z}$ is a conformal map $\mathbb{D} \to \mathbb{D}$ which swaps $a$ and $0$. So $\phi_{\frac{f(0)}{3}} \circ (\frac{f}{3})$ maps $0$ to $0$. The Schwarz lemma implies
\begin{align*}
\frac{|\frac{f(0)}{3}| - |\frac{f(z)}{3}|}{1 + |\frac{f(0)}{3}||\frac{f(z)}{3}|} \le \frac{\left||\frac{f(0)}{3}| - |\frac{f(z)}{3}|\right|}{1 + |\frac{f(0)}{3}||\frac{f(z)}{3}|} \le \left| \frac{\frac{f(z)}{3} - \frac{f(0)}{3}}{1 - \overline{\frac{f(0)}{3}} \cdot \frac{f(z)}{3}} \right|= |\phi_{\frac{f(0)}{3}}(\frac{f(z)}{3})| \le |z|.
\end{align*}
Rearranging this inequality implies
\begin{align*}
\frac{|\frac{f(0)}{3}| - |z|}{1 + |\frac{f(0)}{3}||z|} \le |\frac{f(z)}{3}|.
\end{align*}
If $|z| < \frac{1}{8}$, then $|\frac{f(0)}{3}| \ge \frac{1}{8}$ implies the first term in the above inequality is nonzero. And so we finally conclude that $f(z) \neq 0$ if $|z| < \frac{1}{8}$.
|
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|
Proving that the reduction formula for $I_{m,n}=\int\frac{x^m}{(a^2-x^2)^n}dx$ is $I_{m,n}=a^2I_{m-2,n}-I_{m-2,n-1}$
So I'm trying to prove the reduction formula for
$$I_{m,n}=\int\frac{x^m}{(a^2-x^2)^n}dx$$
which is listed as
$$I_{m,n}=a^2I_{m-2,n}-I_{m-2,n-1}$$
I tried integrating by parts by taking $u=(a^2-x^2)^{-n}\implies du=2nx(a^2-x^2)^{-n-1}dx$ and $dv=x^m\implies v=\frac{1}{m+1}x^{m+1}$ which gives
$$I_{m,n}=\frac{x^{m+1}}{(m+1)(a^2-x^2)^n}-\frac{2n}{(m+1)}\int\frac{x^{m+2}}{(a^2-x^2)^{n+1}}dx\\\implies I_{m,n}=\frac{x^{m+1}}{(m+1)(a^2-x^2)^n}-\frac{2n}{(m+1)}I_{m+2,n+1}\\\implies I_{m+2,n+1}=\frac{x^{m+1}}{2n(a^2-x^2)^n}-\frac{(m+1)}{2n}I_{m,n}$$
Shifting parameters $m$ and $n$ gives
$$I_{m,n}=\frac{x^{m-1}}{2(n-1)(a^2-x^2)^{n-1}}-\frac{(m-1)}{2(n-1)}I_{m-2,n-1}$$
which doesn't look anywhere close to the form above. Any help?
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Integrate both sides of the equation below
\begin{align}
\frac{x^m}{(a^2-x^2)^n}
= \frac{x^{m-2}(a^2+x^2-a^2)}{(a^2-x^2)^n}
= \frac{a^2 x^{m-2}}{(a^2-x^2)^{n}}- \frac{x^{m-2}}{(a^2-x^2)^{n-1}}
\end{align}
to arrive at
$$I_{m,n}=a^2I_{m-2,n}\> \>-I_{m-2,n-1}$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/4219283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
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