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What is the value of $\alpha^{8}+\beta^{8}+\gamma^{8}$ if $\alpha$, $\beta$ and $\gamma$ are roots of the equation $x^3+x-1$? What is the value of $\alpha^{8}+\beta^{8}+\gamma^{8}$ if $\alpha$, $\beta$ and $\gamma$ are roots of the equation $x^3+x-1$? Is there a shorter way of finding the answer apart from finding the individual values of the roots.
|
$a,b,c$ be the roots of $x^3+x-1=0$, then $a^3=1-a \implies a^8=\frac{(1-a)^3}{a}=\frac{1-a^3-3a+3a^2}{a}=\frac{1-(1-a)-3a+3a^2}{a}$
$$\implies a^8=3a-2,\implies a^8+b^8+c^8=3(a+b+c)-6=-6$$
as the sum of the roots is zero.
|
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|
Hint to prove $\sin^4(x) + \cos^4(x) = \frac{3 + \cos(4x)}{4}$ Could go from LHS to RHS by adding zero but I need to know how to do this WITHOUT knowing the half-angle formula. So from RHS to LHS, you an expand $\cos4x$ twice. I get as close as
$$\frac{ \cos^4x + \sin^4x + 3(1 - 2\sin^2x\cos^2x)}{4}$$
|
Answer:
After Euler formula We know that:
$Cos(x) ^4 = \frac{cos(4x)}{8}+\frac{cos(2x)}{2}+\frac{3}{8}$
And
$sin(x) ^4 =\frac{cos(4x)}{8}-\frac{cos(2x)}{2}+\frac{3}{8}$
So,
$ Cos(x) ^4 +sin(x) ^4=\frac{cos(4x)}{4}+\frac{3}{4} =\frac{cos(4x)+3}{4}$
|
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|
Simplify $\frac{x^3+1}{x+\sqrt{x-1}}$
Simplify $$A=\dfrac{x^3+1}{x+\sqrt{x-1}}.$$
Firstly, $x-1\ge0$ and $x+\sqrt{x-1}\ne0:\begin{cases}x-1\ge0\\x+\sqrt{x-1}\ne0\end{cases}.$ The first inequality is equivalent to $x\ge1$. Can we use that in the second inequality? I mean can we say that $x+\sqrt{x-1}>0$ because $x>0$ and $\sqrt{x-1}\ge0.$ (I am asking if we can use in the inequalities after $x\ge1$ that $x$ is actually greater than or equal to $1$.)
After that we have $A=\dfrac{(x+1)(x^2-x+1)}{x+\sqrt{x-1}}.$ What from here? Thank you in advance!
|
We have:
$$x^2-x+1 = x^2-(x-1) = (x-\sqrt{x-1}) (x+\sqrt{x-1})$$
Can you do anything with that?
|
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|
How to solve this integral $I = \int\dfrac{\cos^3x}{\sin x + \cos x}dx$? $\displaystyle\int\dfrac{\cos^3x}{\sin x + \cos x}dx$
I added $J =\displaystyle \int\dfrac{\sin^3x}{\sin x + \cos x}dx$
then $I + J = \displaystyle\int\dfrac{\cos^3x + \sin^3x}{\sin x + \cos x}dx = x + \dfrac{1}{2}\cos2x + C$
but I can't find how to solve $I-J$
And is that the true way to solve it?
Please help!
|
Reducing the degree may be also a good idea...
$$
\begin{aligned}
I &=
\int\frac{\cos^3x}{(\cos x + \sin x)}\;dx =
\int\frac{\cos^3x(\cos x - \sin x)}{(\cos x + \sin x)(\cos x - \sin x)}\;dx
\\
&=
\int\frac{\cos^4x - \cos^3x\sin x}{\cos^2 x - \sin^2 x}\;dx
=
\int\frac{(\cos^2x)^2}{\cos 2x }\;dx
-
\int\frac{\cos^2x\cdot\sin x\cos x}{\cos2 x}\;dx
\\
&=
\frac 14\int\frac{(1+\cos2x)^2}{\cos 2x }\;dx
-
\frac 18\int\frac{(1+\cos2x)\cdot2\cdot 2\sin x\cos x}{\cos2 x}\;dx
\\
&=\frac 14
\int\left(\frac1{\cos 2x }+2+\cos 2x\right)\;dx
+
\frac 18\int\frac{(1+\cos2x)\cdot(\cos 2x)'}{\cos2 x}\;dx
\\
&=
\left(\frac 1{16}\log\frac{1+\sin 2x}{1-\sin 2x}+\frac x2 +\frac 18\sin 2x\right)
+
\left(\frac 18\cos 2x+\frac 18\log\cos 2x\right)+\text{constant .}
\end{aligned}
$$
(Hope there is no computational error, but the idea to proceed is at any rate clear.)
(It turns out from where i am looking to the integral that introducing the "counterpart" $J$ is in the same time introducing a more complicated expression.)
|
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|
Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$?
Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$.
Here's my progress. Let $x = \sqrt[4]{2}$. Then our expression can be written as $x^4/(x^4 - x)$, which simplifies to $x^3/(x^3 - 1)$. Multiply top and bottom by $(x^3 + 1)$ to get $x^3(x^3 + 1)/(x^6 - 1)$. Multiply top and bottom by $(x^6 + 1)$ to get$$x^3(x^3 + 1)(x^6 + 1)/(x^{12} - 1) = x^3(x^3 + 1)(x^6 + 1)/7 = {1\over7}(8 + 4\sqrt[4]{2} + 2 \sqrt{2} + 2^{3/4}).$$However, Wolfram Alpha also tells me that we can write this as$${1\over{14}}\Big(16 + 4\sqrt{2} + 7\sqrt{{{64}\over{49}} + {{72{\sqrt2}}\over{49}}}\Big)$$But how do I derive that? Seems impossible!
|
I would not use wolfram alpha. Rather rely on the back of the book for steps if possible. Or better yet, do it by hand first. Wolfram alpha can you give another result that's equivalent but looks different.
Typically when you see a radical in a denominator of a fraction we prefer to rationalize denominator. So in this case, multiply top and bottom by the conjugate of the denominator (same as denominator but it will have a plus instead of minus). From there distribute numerator and foil denominator (should be easy). From there simplify and if need be rationalize denominator again. Best of luck.
|
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|
Let $P(z) = az^3+bz^2+cz+d$ , where $a, b, c, d $ are complex numbers with $|a| = |b| = |c| = |d| = 1.$
Let $P(z) = az^3+bz^2+cz+d$
, where $a, b, c, d $ are complex numbers with $|a| = |b|
= |c| = |d| = 1.$ Show that $|P(z)| ≥ \sqrt{6}$ for at least one complex number z satisfying
$|z| = 1.$
Attempt
By triangle inequality $$|az^3+bz^2+cz+d|\ge ||az^3+bz^2|-|cz+d||\ge ||az+b|-|cz+d||$$
As you can see i am not utilising $|a|=|b|=|c|=|d|=1$
Then i tried using triangle inequality differently:$$|az^3+bz^2+cz+d|\ge ||az^3|-|bz^2+cz+d||$$
That's all i have tried please tell me how to start.
Thank you!
|
The following is inspired by Bound on a complex polynomial on AoPS.
For $|z| = 1$ we have $\overline z = 1/z$, so that expanding $|P(z)|^2 = P(z)\overline{P(z)}$ gives
$$
|P(z)|^2 = 4 + 2 \operatorname{Re} \left( a \overline b z + a \overline c z^2
+ a \overline d z^3 + b \overline c z + b \overline d z^2 + c \overline d z \right) \, .
$$
Now let $\omega = e^{2 \pi i /3}$ be a third root of unity, and note that
$1 + \omega + \omega^2 = 0$. It follows that
$$
|P(z)|^2 + |P(\omega z)|^2 + |P(\omega^2 z)|^2 =
12 + 6 \operatorname{Re}(a \overline d z^3)
$$
because all the terms with $z$ and $z^2$ cancel in the sum.
We can choose $z_0$ on the unit circle such that $a \overline d z_0^3 = 1$. Then
$$
|P(z_0)|^2 + |P(\omega z_0)|^2 + |P(\omega^2 z_0)|^2 = 18
$$
so that one term on the left must be at least $6$, and that implies the desired conclusion.
|
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|
$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $ Calculate below limit
$$\lim_{x \rightarrow 1} \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} } $$
Using L'Hôpital's rule might be too tedious. I wonder if there is a trick given the resemblence of numerator and denominator?
|
$$ \frac{ x^2-1 + \sqrt{x^3+1} - \sqrt{x^4+1} }{ x-1 + \sqrt{x+1} - \sqrt{x^2+1} }=\frac{x+1-\frac{x^3}{ \sqrt{x^3+1} +\sqrt{x^4+1}}}{1-\frac{x}{ \sqrt{x+1}+\sqrt{x^2+1}}}\rightarrow\frac{2-\frac{1}{2\sqrt2}}{1-\frac{1}{2\sqrt2}}=$$
$$=\frac{4\sqrt2-1}{2\sqrt2-1}=\frac{15+2\sqrt2}{7}.$$
|
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|
We have sides $a,b,c$ of triangle $ABC$ which are equal to $\sqrt{6}, \sqrt{3}, 1$, Prove that angles $A=90+B$ We have sides $a,b,c$ of triangle $ABC$ which are equal to $\sqrt{6}, \sqrt{3}, 1$, Prove that angles $A=90+B$
I just did the question above in the following way:
$t=\frac{\sqrt{6}+\sqrt{3}+1}{2}$
From Heron we have that:
$=\sqrt{\frac{\sqrt{6}+\sqrt{3}+1}{2}*\frac{\sqrt{3}+1-\sqrt{6}}{2}*\frac{\sqrt{6}-\sqrt{3}+1}{2}*\frac{\sqrt{6}+\sqrt{3}-1}{2}}$
$=\frac{1}{4}\sqrt{8}$
I state that AK is a height of the triangle.
We have that it is enough if $sin(90+B)=sin(A)$
However we also have that $sin(90+B)=sin(BAK)$ and $sin(BAK)=\frac{BK}{1}=BK$ (BAK is a right angled triangle)
Also:
$\frac{1}{2}*sin(A)*\sqrt{3}=\frac{1}{4}\sqrt{8}$
$sin(A)=\frac{\frac{1}{2}*\sqrt{8}}{\sqrt{3}}$
So we just have to prove that $BK=\frac{\frac{1}{2}*\sqrt{8}}{\sqrt{3}}$
We have that:
$\frac{AK*BC}{2}=\frac{1}{4}\sqrt{8}$
$AK=\frac{\sqrt{4}}{2\sqrt{3}}$
Hence from Pythagoras we have that:
$BK=\sqrt{\frac{8}{12}}=\frac{\frac{1}{2}*\sqrt{8}}{\sqrt{3}}=sin(A)$
Hence it is proved.
I realize that my method of proving it is complex. Could you please show some simple solutions to this problem?
|
Cosine law for $A$ and $B$: $$6=3+1-2\sqrt3\cos A$$ $$3=6+1-2\sqrt6\cos B$$ $$\therefore\quad\cos A=-\frac{1}{\sqrt3},\qquad \cos B=\sqrt{\frac{2}{3}}$$
$$\therefore\quad\cos(A-B)=-\sqrt\frac{1}{3}\sqrt{\frac{2}{3}}+\sqrt{\frac{2}{3}}\sqrt\frac{1}{3}=0$$
|
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|
Where am I missing the factor $\frac12$ in my fourier series expansion?
Question: A periodic function $f(t)$, with period $2\pi$ is defined as,
$$
f(t) = \begin{cases}
0 & \text{ if } -\pi<t<0, \\
\pi & \text{ if } 0<t<\pi.
\end{cases}
$$
Find the Fourier series expansion of $f$.
Below is my working:
Since $f$ is neither odd nor even, we need to calculate both $a_n$ and $b_n$, the coefficients of cosine and sine respectively.
Note that for $n\geq 1$, we have
\begin{align*}
a_n & = \frac{2}{\pi}\int_0^\pi f(t)\cos(nt)dt \\
& = \frac{2}{\pi}\int_0^\pi \pi \cos(nt)dt \\
& = 2 \int_0^\pi \cos(nt)dt \\
& = 0.
\end{align*}
Also,
\begin{align*}
a_0 & = \frac{1}{\pi}\int_0^\pi f(t)dt \\
& = \frac{1}{\pi}\int_0^\pi \pi dt \\
& = \int_0^\pi dt \\
& = \pi.
\end{align*}
On the other hand, note that
\begin{align*}
b_n & = \frac{2}{\pi}\int_0^\pi f(t)\sin(nt)dt \\
& = \frac{2}{\pi}\int_0^\pi \pi \sin(nt)dt \\
& = 2\int_0^\pi \sin(nt)dt \\
& = 2 \left[ \frac{1-(-1)^n}{n} \right] \\
& = \frac{2[1-(-1)^n]}{n}.
\end{align*}
Therefore, the Fourier series of $f$ is
$$
f(t) = \pi + \sum_{n=1}^\infty \frac{2[1-(-1)^n]}{n} \sin(nt).
$$
However, based on Wolfram alpha, it seems that I am missing the factor $\frac12$ throughout the fourier expansion.
Wolfram alpha gives
$$2 \sin(t) + \frac23 \sin(3t) + \frac25 \sin(5t) + \frac27 \sin(7t) + \frac{\pi}{2}.$$
I am not able to fathom why this is the case.
Any explanation would be greatly appreciated.
|
To see that the coefficients are wrong,
The Fourier series for $f(x)$ may be written
\begin{aligned}
f(x) &= a_0 + \sum_{n=1} a_n \cos n x + \sum_{n=1} b_n \sin n x
\end{aligned}
The coefficients may be calculated
\begin{aligned}
a_0 &= \frac{\int_{-\pi}^{\pi} f(x) dx}{\int_{-\pi}^{\pi} 1 dx} \\
a_n &= \frac{\int_{-\pi}^{\pi} f(x) \cos nx dx}{\int_{-\pi}^{\pi} \cos^2 nx dx}, \quad n \geqslant 1 \\
b_n &= \frac{\int_{-\pi}^{\pi} f(x) \sin nx dx}{\int_{-\pi}^{\pi} \sin^2 nx dx}, \quad n\geqslant 1 \\
\end{aligned}
giving
\begin{aligned}
a_0 = \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) dx, \quad a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos n x~dx, \quad b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin n x~dx.
\end{aligned}
You can easily verify the integrals of $\sin^2nx$ and $\cos^2nx$.
|
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|
Claculate limit $\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$ I have a problem to calculte this limit:
$$\lim_{x\to 0}\frac{1-(\cos(1-\sqrt{\frac{\sin(x)}{x}}))}{x^4}$$
I used Taylor expansion for $\sin(x), \cos(x)$ and considered also $1-\cos(\alpha)=2\sin^2(\frac{\alpha}{2})$ and $\alpha=2-2\sqrt{\frac{\sin(x)}{x}}$ (I have no clue, what to do next with it), but with Taylor and ended up with:
$$\lim_{x\to 0}\frac{\sqrt{1-\frac{x^2}{6}+o(x^2)}+o(\sqrt{1-\frac{x^2}{6}+o(x^2)})}{x^4} $$
which tends to infinity
|
\begin{align}
\lim_{x\to 0}\frac{1-\left\{\cos\left(1-\sqrt{\dfrac{\sin(x)}{x}}\right)\right\}}{x^4}
&=\lim_{x\to0}\underbrace{\dfrac{1-\left\{\cos\left(1-\sqrt{\dfrac{\sin(x)}{x}}\right)\right\}}{\left(1-\sqrt{\dfrac{\sin x}{x}}\right)^2}}_{=\frac12}\times\left(\dfrac{1-\sqrt{\dfrac{\sin x}{x}}}{x^2}\right)^2\\
&=\dfrac12\times\left(\lim_{x\to0}\dfrac{1-\sqrt{\dfrac{\sin x}{x}}}{x^2}\right)^2\\
&=\dfrac12\times\left(\lim_{x\to0}\dfrac{x-\sin x}{x^3}\times\dfrac{1}{1+\sqrt{\dfrac{\sin x}{x}}}\right)^2\\
&=\dfrac12\times\left(\dfrac16\times\dfrac12\right)^2\\
&=\boxed{\dfrac1{288}}
\end{align}
|
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|
Use mathematical induction to prove that (n+2)(n+3)(n+7) is divisible by 6. Use mathematical induction to prove that $q(n)=(n+2)(n+3)(n+7)$ is divisible by $6$.
I have already proved the base case at n=1. I need help on the second part to prove $n=k+1$.
What I did: $(n+2)(n+3)(n+7)=6P$
\begin{align*}
((k+2)+1)&((k+1)+3)((k+1)+7) = (k+3)(k+4)(k+8)\\
= &(k+3)[(k+2)+2][(k+7)+1]\\
= &[(k+3)(k+2)+(2)(k+3)][(k+7)+1]\\
= &(k+2)(k+3)(k+7)+2(k+3)(k+7)+(k+2)(k+3)+2(k+3)\\
= &6P+2k^2+20k+42+k^2+5k+6+2k+6\\
= &6P+3k^2+27k+54\\
= &6p+3(k^2+9k+18)
\end{align*}
I'm not sure what to do, my proof turned out to be divisible by 3 instead of 6. Please let me know how I can move forward with this. Thank you!
|
This is fine so far. You have divisibility by $3$ and only need to check that $k^2+9k+18$ is even to get divisibility by $6$. This is true because $k^2$ and $k$ are either both even or both odd.
|
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|
solve ;$\sqrt{\frac{1-4\cos^2 4x}{8\cos (2x-2\pi/3)}}=\cos (2x-\pi/6)$ This is one more of my unsolved trigonometry questions:
solve ;$$\sqrt{\frac{1-4\cos^2 4x}{8\cos (2x-2\pi/3)}}=\cos (2x-\pi/6)$$
My Try
provided that $\cos (2x-2\pi/3)\neq 0$ and squaring both sides $$1-4\cos^2 4x=8\cos (2x-2\pi/3)\cos^2(2x-\pi/6)....(1)$$
$$1-4\cos^2 4x=4\cos(2x-2\pi/3)(1+\cos(4x-\pi/3))...(2)$$
For convenience we take $2x=t$:$$1-4\cos^2 2t=4\cos(t-2\pi/3)(1+\cos(2t-\pi/3))....(3)$$
$$=4\cos(t-2\pi/3)-2\cos(3t)+2\cos (t+\pi/3)....(4)$$
What do i do next?
Source:A Panchishkin- Trigonometric functions
|
It's been a tour de force, mate. Even with the help of Wolfram Mathematica.
After expanding with standard formulae and subbed $2x=u$ I got this mess
$$-2 \sin ^4 u-\sqrt{3} \sin u-2 \cos ^4 u+2 \cos ^3 u+\cos u+12 \sin ^2 u \cos ^2 u-6 \sin ^2 u \cos u-1=0$$
I substituted $$\sin u=\frac{2t}{1+t^2};\;\cos u=\frac{1-t^2}{1+t^2};\;t=\tan u/2$$
After a while I got
$$t \left(3 t^7+\sqrt{3} t^6-39 t^5+3 \sqrt{3} t^4+73 t^3+3 \sqrt{3} t^2-13 t+\sqrt{3}\right)=0$$
And thanks to Mathematica this amazing factorization, using the function
Factor[t (Sqrt[3] - 13 t + 3 Sqrt[3] t^2 + 73 t^3 + 3 Sqrt[3] t^4 -
39 t^5 + Sqrt[3] t^6 + 3 t^7), Extension -> Automatic]
$$\left(\sqrt{3}-t\right) t \left(t+\sqrt{3}-2\right) \left(t+\sqrt{3}+2\right) \left(3 t+\sqrt{3}\right) \left(-3 t^3+3 \sqrt{3} t^2+9 t-\sqrt{3}\right)=0$$
The linear factors are trivial. The cubic is another mess. I got these three approximate solutions
$t_1=-1.19175, t_2=0.176327, t_3=2.74748$
A little consolation is that $t=\tan x$
In $[0,2\pi]$ solutions are $$x=0,\frac{5 \pi }{6},\pi ,\frac{4 \pi }{3},\frac{11 \pi }{6},2 \pi,0.174533,2.26893,5.41052,1.22173$$
|
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|
Converting $A8B34_{16}$ to octal. According to online calculators $A8B34_{16}=2505464_{\ 8}$, yet I keep getting $2005464_{\ 8}$. I just want to know where I'm going wrong.
$A8B34_{16}=10( 16^4)+8( 16^3) + 11( 16^2) + 3( 16) + 4$
$=10(2^48^4)+8(2^38^3)+11(2^28^2)+6(8)+4$
$=160(8^4)+8^5+44(8^2)+6(8)+4$
$=(8^2+7[8])(8^4)+8^5+(5[8]+4)(8^2)+6(8)+4$
$=8^6+7(8^5)+8^5+5(8^3)+4(8^2)+6(8)+4$
$=2(8^6)+5(8^3)+4(8^2)+6(8)+4$
$=2005464_{\ 8}$
|
We are lucky here that $8$ and $16$ are powers of two. It suffices to break down the numbers into bits and regroup:
$$A8B34_{16}=1010,1000,1011,0011,0100_2$$
$$=10,101,000,101,100,110,100_2=2505464_8$$
|
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For what value of c is the estimator consistent? Suppose $X_1,\ldots,X_n$ are i.i.d. normally distributed with unknown mean $\mu$ and unknown variance $\sigma^2$. Let $\bar{X}_n$ be the sample mean. Consider estimating $\sigma$. For any given constant $c$, define the estimator\footnote{followed this video for both parts. In (ii) I use his calculation to shorten my shown work. https://www.youtube.com/watch?v=MqeS2NWB3I4}
\begin{equation*}
\hat{\sigma}_{n,c} = \frac{c}{n} \sum_{i=1}^n |X_i - \bar{X}_n|
\end{equation*}
\textbf{(i)} Compute $E(\hat{\sigma}_{n,c})$
\begin{equation*}
\begin{aligned}
& E[\hat{\sigma}_{n,c}] = E \left[ \frac{c}{n} \sum_{i=1}^{n} |X_i - \bar{X}_n| \right] \\
& = \frac{c}{n} E \left[ \sum_{i=1}^{n} |X_i - \bar{X}_n| \right] \\
& = \frac{c}{n} E \left[ \sum_{i=1}^{n} \sqrt{ (X_i - \bar{X}_n)^2 } \right] \\
& = \frac{c}{n} E \left[ \sum_{i=1}^{n} \sqrt{\left[(X_i - \mu) (\bar{X}_n - \mu)\right]^2} \right] \\
& = \frac{c}{n} E \left[ \sqrt{\sum_{i=1}^{n}(X_i - \mu)^2 - 2n(\bar{X}_n - \mu)^2 + n(X_i - \mu)^2} \right] \\
& = \frac{c}{n} \left[ \sqrt{ n\sigma^2 - nE((X_i - \mu)^2)} \right] \\
& = \frac{c}{n} \left[\sqrt{ n\sigma^2 - n\frac{\sigma^2}{n}} \right] \\
& \frac{c\sigma\sqrt{n-1}}{n}
\end{aligned}
\end{equation*}
\textbf{(ii)} For what $c$ is $\hat{\sigma}_{n,c}$ a consistent estimator of $\sigma$. Explain.\
\
\indent We know that the estimator is consistent if $lim_{n \to \infty}Var(\hat{\sigma}_{n,c}) = 0$. So we show:
\begin{equation*}
\begin{aligned}
& \operatorname{Var}(\hat{\sigma}_{n,c}) = \operatorname{Var} \left( \frac{c}{n} \sum_{i=1}^{n} |X_i - \bar{X}_n| \right) \\
& = E\left[\left( \frac{c}{n} \sum_{i=1}^{n} |X_i - \bar{X}_n| \right)^2 \right] - E\left[ \frac{c}{n} \sum_{i=1}^{n} |X_i - \bar{X}_n| \right] ^2 \\
& = \frac{c}{n} E\left[ \left( \sqrt{ (X_i - \bar{X}_n)^2} \right)^2 \right] - \frac{c^2\sigma^2(n-1)}{n^2} = \\
& \text{saving space by citing the proof from the video here}\\
& = \frac{c\sigma^2}{n} - \frac{c^2\sigma^2(n-1)}{n^2} \\
& \sigma^2 \left[ \frac{c}{n} - \frac{c^2(n-1)}{n^2} \right]
\end{aligned}
\end{equation*}
|
Notice that $\operatorname E(X_i-\overline X) = 0$ and
\begin{align}
\operatorname{var}(X_i-\overline X) & = \operatorname{var}(X_i) + \operatorname{var}(\overline X) - 2\operatorname{cov}(X_i,\overline X) \\[8pt]
& = \sigma^2 + \frac{\sigma^2} n - \frac{2\sigma^2} n \\[8pt]
&= \frac{n-1} n \sigma^2.
\end{align}
So
\begin{align}
\operatorname E\big( |X_i - \overline X| \big) & = \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{+\infty} \sqrt{\frac{n-1} n}\cdot\sigma|z| e^{-z^2/2} \, dz \\[12pt]
& = \sigma\sqrt{\frac{n-1}{2\pi n}} \cdot 2 \int_0^{+\infty} e^{-z^2/2} (z \, dz) \\[10pt]
& = 2\sigma \sqrt{\frac{n-1}{2\pi n}} \int_0^{+\infty} e^{-u} \, du \\[8pt]
& = 2\sigma \sqrt{\frac{n-1}{2\pi n}}.
\end{align}
|
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|
Recurrence relation $a_n = 4a_{n-1} - 3a_{n-2} + 2^n + n + 3$ with $a_{0} = 1$ and $a_{1} = 4$ This is a nonhomogeneous recurrence relation, so there is a homogeneous and a particular solution.
Homogenous:
$a_n - 4a_{n-1} + 3a_{n-2} = 0$
$r^2 - 4r + 3 = 0$
$(r - 3)(r - 1)$
$a_n^h = \alpha(3^n) + \beta(1^n)$
This is where my solution stops because I don't know how to solve the particular solution since it would be $a_n - 4a_{n-1} + 3a_{n-2} = 2^n + n + 3$ and I'm not sure what form it should be. Would it be $A_0(r^n) + A_1(n) + A_2$ where $A_n$ is a constant or not?
I've tried solving it with that form and it ended like this:
$A_0(2^n) + A_1(n) + A_2 - 4(A_0(2^{n-1}) + A_1(n-1) + A_2) + 3(A_0(2^{n-2}) + A_1(n-2) + A_2) = 2^n + n + 3$
After simplifying and dividing $2^{n-2}$:
$A_0(2^n) - 4A_0(2^{n-1}) + 3A_0(2^{n-2}) - 4 = n + 3 + 2A_1(n) + 2A_2 - 2A_1$
And that's where I stop since I don't know what to do next.
Thanks for answering.
|
Here's an alternative approach. Let $A(z)=\sum_{n\ge 0} a_n z^n$ be the ordinary generating function for $a_n$. Then the recurrence relation implies that
\begin{align}
A(z) - a_0 - a_1 z
&= \sum_{n\ge 2}\left(4a_{n-1} - 3a_{n-2} + 2^n + n + 3\right)z^n \\
&= 4z \sum_{n\ge 2} a_{n-1} z^{n-1} - 3z^2 \sum_{n\ge 2} a_{n-2} z^{n-2} + \sum_{n\ge 2} (2z)^n + z \sum_{n\ge 2} n z^{n-1} + 3\sum_{n\ge 2}z^n \\
&= 4z (A(z)-a_0)- 3z^2 A(z) + \frac{(2z)^2}{1-2z} + z\left(\frac{1}{(1-z)^2}-1\right) + \frac{3z^2}{1-z},
\end{align}
so
\begin{align}
A(z)
&= \frac{a_0 + a_1 z -4 a_0 z + \frac{4z^2}{1-2z} + \frac{z}{(1-z)^2}-z + \frac{3z^2}{1-z}}{1-4z+3z^2}\\
&= \frac{1 - z + \frac{4z^2}{1-2z} + \frac{z}{(1-z)^2} + \frac{3z^2}{1-z}}{1-4z+3z^2}\\
&= \frac{1 - 4 z + 14 z^2 - 24 z^3 + 12 z^4}{(1 - 2 z) (1 - 3 z)(1 - z)^3 } \\
&= -\frac{4}{1-2 z} + \frac{39/8}{1-3 z} + \frac{19/8}{1-z} - \frac{7/4}{(1-z)^2} - \frac{1/2}{(1-z)^3} \\
&= \sum_{n\ge 0}\left(-4\cdot 2^n + \frac{39}{8}\cdot3^n + \frac{19}{8} - \frac{7}{4}\binom{n+1}{1} - \frac{1}{2}\binom{n+2}{2}\right)z^n,
\end{align}
which immediately implies that
\begin{align}
a_n &= -4\cdot 2^n + \frac{39}{8}\cdot3^n + \frac{19}{8} - \frac{7}{4}\binom{n+1}{1} - \frac{1}{2}\binom{n+2}{2} \\
&= \frac{- 2^{n + 5} + 13\cdot 3^{n + 1} -2 n^2 - 20 n + 1}{8}.
\end{align}
|
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|
Prove that $ f(f(x)) \geq 0$ for all real x Let $f(x)= a x^2 + x +1 , x \in \mathbb{R} $. Find all values of parameter $a \in \mathbb{R} $ such that $f(f(x)) \geq 0 $ holds for all real $x$.
$f(x)> 0 $ iff $a> 0 $ and $ 1- 4a \leq 0$ which gives $a \geq \frac{1}{4} $ . But we have:
$f(f(x))= a ( a x^2 + x +1)^2 + a x^2 + x +1 +1 $ , now the degree is 4 and I am not sure what to do...Can anyone help? Thanks in advance.
|
Firstly, $1-4a\leq0$ is valid and since $a=0$ is not valid, it's enough to check $0<a<\frac{1}{4}$, which gives
$$ax^2+x+1=\frac{-1+\sqrt{1-4a}}{2a}$$ and
$$ax^2+x+1=\frac{-1-\sqrt{1-4a}}{2a}$$ have no real roots.
It's enough to work with the first equation, which gives:
$$1-4a\left(1-\frac{-1+\sqrt{1-4a}}{2a}\right)\leq0$$ or
$$2\sqrt{1-4a}\leq4a+1,$$ which gives $$a\geq\frac{2\sqrt3-3}{4}.$$
It's interesting that in the case $a=\frac{2\sqrt3-3}{4}$ we obtain:
$$f(f(x))=\frac{1}{64}(\sqrt3x+2\sqrt3+4)^2((26\sqrt3-45)x^2+4(7-4\sqrt3)x+44-24\sqrt3)\geq0.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Summation of $n$th partial products of the square of even numbers diverges, but for odd numbers they converge in this series I'm looking at. Why? So I have the two following series:
$$\sum_{n=1}^\infty \frac{\prod_{k=1}^n(2k)^2}{(2n+2)!}$$
$$\sum_{n=0}^\infty \frac{\prod_{k=0}^n(2k+1)^2}{(2n+3)!}$$
I figured out the $n$th partial products:
$$\prod_{k=1}^n(2k)^2=4^n(n!)^2$$
$$\prod_{k=0}^n (2k+1)^2=\frac{((2n+1)!)^2}{4^n(n!)^2}$$
So putting these back into my series they become the following:
$$\sum_{n=1}^\infty \frac{\prod_{k=1}^n(2k)^2}{(2n+2)!}=\sum_{n=1}^\infty\frac{4^n(n!)^2}{(2n+2)!}$$
Now this diverges as expected by the limit test test. However when I look at my other series:
$$\sum_{n=0}^\infty \frac{\prod_{k=0}^n(2k+1)^2}{(2n+3)!}=\sum_{n=0}^\infty\frac{(
(2n+1)!)^2}{4^n(n!)^2(2n+3)!}$$
By the limit test maybe diverges or maybe doesn't, and the ratio test is inconclusive. Since I wasn't sure what to use for the a comparison test I threw this into wolfram alpha and it told me it converges which is baffling to me since both series are very similar if we write them out:
$$\sum_{n=1}^\infty \frac{\prod_{k=1}^n(2k)^2}{(2n+2)!}=\frac{2^2}{4!}+\frac{2^24^2}{6!}+\frac{2^24^26^2}{8!}\cdot\cdot\cdot\cdot$$
$$\sum_{n=0}^\infty \frac{\prod_{k=0}^n(2k+1)^2}{(2n+3)!}=\frac{1^2}{3!}+\frac{1^23^2}{5!}+\frac{1^23^25^2}{7!}+\cdot\cdot\cdot$$
They both have the nth parial product of the even/odd integers squared in the numerator, and are over a factorial that is two greater than $n$, so I'm not sure why one is diverging and the other is converging. Is wolframalpha wrong, as it can be at times? Or is there someething here that I am missing?
|
Elaborating after @Erick Wong's comments.
You properly found that
$$a_n=\frac{4^n(n!)^2}{(2n+2)!}$$ Take logarithms
$$\log(a_n)=n \log(4)+2\log(n!)-\log((2n+2)!)$$ Use Stirling approximation twice and continue with Taylor series to find
$$\log(a_n)=\left(\frac{3}{2} \log \left(\frac{1}{n}\right)+\log \left(\frac{\sqrt{\pi
}}{4}\right)\right)-\frac{11}{8 n}+O\left(\frac{1}{n^2}\right)$$ that is to say
$$a_n \sim \frac{\sqrt \pi}{4 n^{\frac 32}}\exp\left(-\frac{11}{8 n}\right) <\frac{\sqrt \pi}{4 n^{\frac 32}}$$
$$\sum_{n=1}^\infty \frac{\sqrt \pi}{4 n^{\frac 32}}=\frac{\sqrt{\pi }}{4} \zeta \left(\frac{3}{2}\right)\approx 1.15758$$
Sooner or later, you will learn that
$$\sum_{n=1}^\infty \frac{4^n(n!)^2}{(2n+2)!}=\frac{\pi ^2-4}{8}\approx 0.73370$$
Doing the same with
$$b_n=\frac{(2n+1)!^2}{4^n(n!)^2(2n+3)!}$$
$$\log(b_n)=2\log((2n+1)!)-n \log(4)-2\log(n!)-\log((2n+3)!)$$
$$\log(b_n)=\left(\frac{3}{2} \log \left(\frac{1}{n}\right)+\log \left(\frac{1}{2 \sqrt{\pi
}}\right)\right)-\frac{17}{8 n}+O\left(\frac{1}{n^2}\right)$$ that is to say
$$b_n \sim \frac{1}{2 \sqrt \pi n^{\frac 32}}\exp\left(-\frac{17}{8 n}\right) < \frac{1}{2 \sqrt \pi n^{\frac 32}}$$
$$\sum_{n=1}^\infty \frac{1}{2 \sqrt \pi n^{\frac 32}}=\frac{1}{2 \sqrt \pi }\zeta \left(\frac{3}{2}\right)\approx 0.73694$$
Sooner or later, you will learn that
$$\sum_{n=0}^\infty \frac{(2n+1)!^2}{4^n(n!)^2(2n+3)!}=\frac{ \pi -2}{2} \approx 0.57080$$
Edit
Notice that
$$\sum_{n=1}^\infty a_n\,x^n=\frac{\sin ^{-1}(x)^2-x^2}{2 x^2}$$
$$\sum_{n=0}^\infty b_n\,x^n=\frac{\sin ^{-1}(x)-x}{x^3}$$
|
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|
Find the roots of the polynomial $x^3-2$. Find the roots of the polynomial $x^3-2$.
If $\alpha$ be the root of this polynomial i.e., $\alpha^3=2$,
then $(\zeta \alpha)^3=2$, where $\zeta$ is $3^{rd}$ root of unity.
Hence the solutions of this equations are
$$\zeta{2}^\frac{1}{3}$$, $\zeta$ a $p^{th}$ root of unity
And we know that $\zeta_n=e^{\frac{2\pi i}{n}}$
So the roots are $2^\frac{1}{3}, -(2^\frac{1}{3}) ~and~ 2^\frac{1}{3}\frac{1}{2}(\sqrt{3}i-1).$
Is this correct?
Note: I don't like factoring method. I studied in schools time. So don't help with that method.
Thanks!
|
Let's say $a > 0$ is a real number.
The roots of $x^3 = a$ are these
$x_1 = a^\frac{1}{3} ( \cos{2\pi/3} + i \cdot \sin{2\pi/3} )$
$x_2 = a^\frac{1}{3} ( \cos{4\pi/3} + i \cdot \sin{4\pi/3} )$
$x_3 = a^\frac{1}{3} ( \cos{6\pi/3} + i \cdot \sin{6\pi/3} )$
We can simplify these formulas further...
The root $x_3$ happens to be real.
The other two are complex conjugated to each other.
Seems your problem is that your knowledge of the roots of unity is a bit rusty.
You can refresh it here:
Roots of unity
Once you know how to find the roots of $x^3=1$,
what's left to do is just to add a factor of $a^{\frac{1}{3}}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$x+1$ is always invertible in $\Bbb Z_{x^3}$.
Claim: For any $x > 1$ and $x \in \Bbb N$, $x+1$ is always invertible in $\Bbb Z_{x^3}$.
Proof: We know that $x^3 +1 = (x+1)(x^2-x+1)$. Since $x > 1$ and $x \in \Bbb N$, we have $x^2-x+1 >0$ and $(x+1)(x^2-x+1) = x^3 +1 \equiv 1 \mod x^3$. Hence $x+1$ is always invertible in $\Bbb Z_{x^3}$.
Is the proof correct?
|
You proved the Bezout equation $\,1 = (\color{#c00}{x+1})f(x) - \color{#c00}{x^3}\,$ for $\,f(x) \in \Bbb Z[x]\,$
This implies that $\,\color{#c00}{x+1}\,$ and $\,\color{#c00}{x^3}\,$ are coprime in every ring, since any common divisor divides the RHS so also the LHS $= 1$, i.e. every common divisor is a unit (invertible).
$\!\bmod x^3\!:\ $ reducing Bezout $\, \Rightarrow 1\equiv (x+1)f(x)\,$ $\Rightarrow x+1\,$ is invertible
|
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|
Proving $\frac{7 + 2b}{1 + a} + \frac{7 + 2c}{1 + b} + \frac{7 + 2a}{1 + c} \geqslant \frac{69}{4}$. Here's the inequality
For positive variables, if $a+b+c=1$, prove that
$$
\frac{7 + 2b}{1 + a} +
\frac{7 + 2c}{1 + b} +
\frac{7 + 2a}{1 + c} \geqslant
\frac{69}{4}
$$
Here equality occurs for $a=b=c=\frac{1}{3}$ which is not-so-usual, so I decide to write the inequality as
$$
\frac{21 + 2q}{3 + p} +
\frac{21 + 2r}{3 + q} +
\frac{21 + 2p}{3 + r} \geqslant
\frac{69}{4}
$$Where the constraint now is $p + q + r = 3$ and the equality occurs for $p = q = r = 1$. Now we are left to proving that just
$$
2\sum_{cyc}{\frac{q}{3 + p}} +
21\sum_{cyc}{\frac{1}{3 + p}} \geqslant \frac{69}{4}
$$
Now it is sufficient to prove that
$$
\sum_{cyc}\frac{q}{3 + p}\geqslant\frac{3}{4} \quad \textrm{and} \quad \sum_{cyc}\frac{1}{3 + p} \geqslant \frac{3}{4}
$$ The second is true but I can't prove that the first is true.
|
Another way:flipping the inequality it suffices to prove $$\sum_{cyc}\frac{7a-2b}{1+a}\le 15/4...(3)$$.
Indeed by jensen on $f(x)=\frac{x}{1+x}$ $$\sum_{cyc} \frac{a}{1+a}\le 3\sum_{cyc}\frac{\frac{(a+b+c)}{3}}{1+\frac{(a+b+c)}{3}}=3/4...(1)$$
This can also be done like this :
as $$\frac{x}{1+x}\le \frac{9x+1}{16}$$ which is just ${(x-1/3)}^2\ge 0$
$$\sum \frac{a}{1+a}\le \sum \frac{9a+1}{16}=3/4$$
Also similar to my previous answer (by Titu's lemma) $$\sum_{cyc}\frac{b}{1+a}=\sum \frac{b^2}{b+ab}\ge 3/4...(2)$$
By $(1)$ and $(2)$ inequality $(3)$ is completed.
|
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|
find all $(x,y,z)$ such that $27^{3x^2 + 2y}+27^{3y^2 + 2z}+27^{3z^2 + 2x}=1$
Find all ($x,y,z$) such that
$$ 27^{3x^2 + 2y}+27^{3y^2 + 2z}+27^{3z^2 + 2x}=1$$
I am a high school student and would appreciate it if anyone could solve it using high school mathematics.
here is what I have tried till now:
$$ 3x^2 +2y < 0 $$ (this can also be said for the other 3 equation. it also means that x,y,z<0)
$$ -2y>3x^2 $$
$$ 4y^2>9x^2 $$ ( -2y and 3x^2 are both positive)
$$ 3y^2>27/4 x^2 $$
$$ -2z>3y^2>27/4 x^4 $$
$$ 4z^2 > 729/16 x^8 $$
$$ 3z^2> 2187/64 x^8 $$
$$ -128/2187<x^7 $$
$$ -2/3 < x $$
$$ -2/3< x,y,z < 0 $$
i tried this but as you can see didn't do much good
|
Hint:By am-gm $$1\ge3\sqrt[3]{{27}^{3(x^2+y^2+z^2)+2(x+y+z)}} = 3^{3(x^2+y^2+z^2)+2(x+y+z)+1}..(1)$$
But $$3(x^2+y^2+z^2)\ge {(x+y+z)}^2$$ when x=y=z
substituting in (1)
$$1\ge 3^{{(x+y+z+1)}^2}$$
which is possible only when $x+y+z=-1$ and $x=y=z$
|
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|
Difference between anti-derivative and indefinite integral My teacher gave me the following integral to evaluate:
$$\int \frac{x^2}{(x\sin(x)+\cos(x))^2}dx$$
After half an hour of uselessly fumbling around with trig identities I gave up and plugged it into an integral calculator: https://www.integral-calculator.com/. However I am confused: it displayed ANTIDERIVATIVE COMPUTED BY MAXIMA as $$-\dfrac{\left(2x^2-2\right)\sin\left(2x\right)+4x\cos\left(2x\right)}{\left(x^2+1\right)\sin^2\left(2x\right)+4x\sin\left(2x\right)+\left(x^2+1\right)\cos^2\left(2x\right)+\left(2-2x^2\right)\cos\left(2x\right)+x^2+1}+C$$ and I pressed the simplify button to obtain $$-\dfrac{\left(x^2-1\right)\cos\left(x\right)\sin\left(x\right)+2x\cos^2\left(x\right)-x}{\left(x^2-1\right)\sin^2\left(x\right)+2x\cos\left(x\right)\sin\left(x\right)+1}+C$$ However the "MANUALLY" COMPUTED ANTIDERIVATIVE displayed the following $$\dfrac{\sin\left(x\right)-x\cos\left(x\right)}{x\sin\left(x\right)+\cos\left(x\right)}+C$$ Which was computed by the following method:
$$\int \frac{x^2}{(x\sin(x)+\cos(x))^2}dx= \int \Bigg(\frac{x\sin(x)}{x\sin(x)+cos(x)}-\frac{x\cos(x)(\sin(x)-x\cos(x))}{(x\sin(x)+\cos(x))^2}\Bigg) dx$$
Using integration by parts
$$ \int \frac{x\cos(x)(\sin(x)-x\cos(x))}{(x\sin(x)+\cos(x))^2} dx= \dfrac{\sin\left(x\right)-x\cos\left(x\right)}{x\sin\left(x\right)+\cos\left(x\right)}+ \int\frac{x\sin(x)}{x\sin(x)+cos(x)}dx$$
$$\Rightarrow \int \frac{x^2}{(x\sin(x)+\cos(x))^2}dx=\dfrac{\sin\left(x\right)-x\cos\left(x\right)}{x\sin\left(x\right)+\cos\left(x\right)}+ \int\frac{x\sin(x)}{x\sin(x)+cos(x)}dx-\int\frac{x\sin(x)}{x\sin(x)+cos(x)}dx=\dfrac{\sin\left(x\right)-x\cos\left(x\right)}{x\sin\left(x\right)+\cos\left(x\right)}+C $$
My question is: why do I get different results from computing the anti-derivative and the indefinite integral? I simplified the anti-derivative so shouldn't it be simplified to the indefinite integral above? Are these two equations equal? Are functions for anti-derivatives and indefinite integrals vastly different? Any help will be appreciated
|
Computing the Indefinite Integral
$$
\begin{align}
&\int\frac{x^2}{(x\sin(x)+\cos(x))^2}\,\mathrm{d}x\\
&=\int\frac{x^2}{(x\sin(x)+\cos(x))^2}\frac{\mathrm{d}(x\sin(x)+\cos(x))}{x\cos(x)}\tag1\\
&=-\int\frac{x}{\cos(x)}\,\mathrm{d}\frac1{x\sin(x)+\cos(x)}\tag2\\
&=-\frac{x}{\cos(x)}\frac1{x\sin(x)+\cos(x)}+\int\frac1{x\sin(x)+\cos(x)}\,\mathrm{d}\frac{x}{\cos(x)}\tag3\\
&=-\frac{x}{\cos(x)}\frac1{x\sin(x)+\cos(x)}+\int\frac1{x\sin(x)+\cos(x)}\frac{\cos(x)+x\sin(x)}{\cos^2(x)}\,\mathrm{d}x\tag4\\[1pt]
&=\tan(x)-\frac{x}{\cos(x)}\frac1{x\sin(x)+\cos(x)}+C\tag5\\[2pt]
&=\frac{\sin(x)-x\cos(x)}{x\sin(x)+\cos(x)}+C\tag6
\end{align}
$$
Explanation:
$(1)$: $\mathrm{d}(x\sin(x)+\cos(x))=x\cos(x)\,\mathrm{d}x$
$(2)$: $\frac1{u^2}\mathrm{d}u=-\mathrm{d}\frac1u$
$(3)$: integrate by parts
$(4)$: $\mathrm{d}\frac{x}{\cos(x)}=\frac{\cos(x)+x\sin(x)}{\cos^2(x)}\,\mathrm{d}x$
$(5)$: $\sec^2(x)=\frac{\mathrm{d}}{\mathrm{d}x}\tan(x)$
$(6)$: $\tan(x)=\frac{\sin(x)}{\cos(x)}$ and simplify
The Integrals are the Same
$$
\begin{align}
&-\frac{\left(x^2-1\right)\cos(x)\sin(x)+2x\cos^2(x)-x}{\left(x^2-1\right)\sin^2(x)+2x\cos(x)\sin(x)+1}+C\\
&=-\frac{(x\sin(x)+\cos(x))(x\cos(x)-\sin(x))}{(x\sin(x)+\cos(x))^2}+C\tag7\\
&=\frac{\sin(x)-x\cos(x)}{x\sin(x)+\cos(x)}+C\tag8
\end{align}
$$
Explanation:
$(7)$: multiply and use $\cos^2(x)-\sin^2(x)=2\cos^2(x)-1$
$(8)$: cancel common factors
|
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|
(AIME 1994) $ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $
$($AIME $1994)$ Find the positive integer $n$ for which
$$ \lfloor \log_2 1 \rfloor + \lfloor \log_2 2 \rfloor + \lfloor \log_2 3 \rfloor + \ldots + \lfloor \log_2 n \rfloor = 1994 $$ where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.
The first few terms of this series shows me that summation $\lfloor \log_2 n \rfloor$ for $n=1$ to $n=10$ give $2^{n +1}$.
|
As noted, the sequence goes like
$$ 0,\underset{2}{\underbrace{1,1}},\underset{4}{\underbrace{2,2,2,2}},\underset{8}{\underbrace{3,3,3,3,3,3,3,3}},4,4,\ldots$$
i.e, every natural number $k$ occurs $2^k$ times.
So desired is $$ \sum k\cdot 2^k =1994$$
It's quick enough to attack directly:
$$ 1\cdot2 + 2\cdot4 + 3\cdot8 + 4\cdot16 + 5\cdot 32 + 6\cdot 64 + 7\cdot 128 = 1538$$
Next up is $8$ repeating $x$ times till $1994$ $$1538 + 8\cdot x = 1994$$
$$\Rightarrow x=57 $$
Last term of our sequence can be found by counting the number of repeating units: $$n = (1+2+4+\ldots+128) + 57 = \boxed{312}$$
|
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|
How to find the dot product using the law of cosines I'm working with the following problem:
We have a triangle with sides $AB=3$ and $BC=2$, the angle $ABC$ is 60 degrees. Find the dot product $AC \cdotp AB$
Since we don't actually know the side $AC$ my first step is to calculate this side via the law of cosines.
$$AC^2=AB^2 +BC^2 -2AB\cdot BC\cos x$$
$$\implies AC^2=3^2 +2^2 -2\cdot2\cdot3\cos 60^\circ$$
$$\implies AC^2=9 +4 -12\cdot0.5$$
$$\implies AC^2=13-6=7$$
$$\implies AC=\sqrt 7.$$
My next step is to calculate the angle $BAC$; we should be able to use the law of cosines here as well:
$$BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cos x$$
$$\implies4=9+(\sqrt 7)^2 -2\cdot3\sqrt 7\cos x$$
$$\implies4=9+7 -6\sqrt 7\cos x$$
$$\implies-2=-6\sqrt 7\cos x$$
$$\implies\frac{1}{3}=\sqrt 7\cos x$$
$$\implies\frac{1}{3\sqrt 7}=\cos x.$$
If we want to calculate our dot product using only the vector lengths, we would use the fact that $A\cdotp B=|A||B|\cos x$, which in this case would mean that:
$$AB \cdot AC=3\sqrt7\frac{1}{3\sqrt 7}=1,$$
which is quite wrong since the answer is supposed to be $12$.
Can someone please tell me where I've made a mistake?
In the proposed solution another method seem to be used. The textbook claims that $AB \cdotp AC = AB \cdotp (AB + BC)=AB \cdotp AB + AB \cdotp BC = |AB|^2 + |AB||BC|cos(60)=9+6*\frac{1}{2}=12$
|
$$\vec{AB}=\vec{AC}+\vec{CB}$$
and
$$\vec{AC}\cdot\vec{AB}=\vec{AB}^2-\vec{CB}\cdot\vec{AB}=9-3\cdot2\cdot\cos60°=6.$$
|
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|
Why is $\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$ $\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$ what theorem or algebra leads to this equality?
EDIT: The sum should have been infinite.
|
Let $S = \sum_{k=1}^\infty \left(\frac{1}{3}\right)^k$. Then,
$$ S = \frac{1}{3} + \left(\frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^3 + \ldots = \frac{1}{3} \left[ 1 + \underbrace{\frac{1}{3} + \left(\frac{1}{3}\right)^2 + \ldots}_{=S} \right] =
\frac{1}{3} \left[ 1 + S \right]$$
So, we have $S = \dfrac{1}{3} + \dfrac{1}{3}S \Rightarrow \dfrac{2}{3} S = \dfrac{1}{3} \Rightarrow S = \dfrac{1/3}{2/3}$.
|
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|
$\lim_{n\to\infty}\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})$ I need to find $\lim_{n\to\infty}{\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})}$ without using L'Hopital's rule, derivatives or integrals.
Empirically, I know such limit exists (I used a function Grapher and checked in wolfram) and it's equal to $-\frac{1}{4}$. I noticed that
$$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=\sqrt{n^3} \Big(\frac{1}{\sqrt{n+1}+\sqrt{n}}-\frac{1}{\sqrt{n+1}+\sqrt{n-1}}\Big) $$
That doesn't seem to lead to $-\frac{1}{4}$ when $n\to \infty$ . I tried another form of the original expression:
$$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=2\sqrt{n^3} \Bigg(\frac{\sqrt{n^2 - 1} - n} {\sqrt{n + 1} + \sqrt{n - 1} + 2\sqrt{n}}\Bigg)$$
If I multiply by the conjugate, we obtain
$$2\sqrt{n^3} \Bigg(\frac{\sqrt{n^2 - 1} - n} {\sqrt{n + 1} + \sqrt{n - 1} + 2\sqrt{n}}\Bigg)=-\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}$$
Now that doesn't seem to be of any use either. Any ideas?
|
Multiplying $\sqrt{n+1} - \sqrt n$ by $\dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+1}+\sqrt n}$ yields $\dfrac 1 {\sqrt{n+1}+\sqrt n}.$
Similarly $\sqrt n - \sqrt{n-1} = \dfrac 1 {\sqrt n + \sqrt{n-1}}.$
So then we have
\begin{align}
& \big(\sqrt{n+1} - \sqrt n\big) - \big(\sqrt n - \sqrt{n-1} \big) \\[8pt]
= {} & \frac 1 {\sqrt{n+1}+\sqrt n} - \dfrac 1 {\sqrt n + \sqrt{n-1}} \\[12pt]
= {} & \frac{\sqrt{n-1}- \sqrt{n+1}}{(\sqrt{n+1}+\sqrt n)( \sqrt n + \sqrt{n-1})} \\[12pt]
= {} & \frac{\sqrt{n-1}- \sqrt{n+1}}{(\sqrt{n+1}+\sqrt n)( \sqrt n + \sqrt{n-1})} \cdot \frac{\sqrt{n-1} + \sqrt{n+1}}{\sqrt{n-1} + \sqrt{n+1}} \\[12pt]
= {} & \frac{-2}{(\sqrt{n+1}+\sqrt n)( \sqrt n + \sqrt{n-1})(\sqrt{n-1} + \sqrt{n+1})}
\end{align}
If this is multiplied by $\sqrt{n^3}$ it becomes
$$
-2 \cdot \frac {\sqrt n} {\sqrt{n+1}+\sqrt n} \cdot \frac{\sqrt n}{ \sqrt n + \sqrt{n-1}} \cdot \frac {\sqrt n}{\sqrt{n-1} + \sqrt{n+1}} \longrightarrow \frac{-1} 4.
$$
|
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|
How to simplify $\frac {\sin 3A - \cos 3A}{\sin A + \cos A} + 1$? So I started by using $\sin 3A$ and $\cos 3A$ identities and then I added the lone $1$ to the trigonometric term. (Done in the picture below)
But after this I don't have any clue on how to proceed.
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-\left(4 \cos ^{3} \theta-\sin \theta\right)}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4\sin^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta}{\sin \theta+\cos \theta} + 1$$
$$=\frac{3 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+3 \cos \theta+\sin \theta+\cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta-4 \sin ^{3} \theta-4 \cos ^{3} \theta+4 \cos \theta}{\sin \theta+\cos \theta}$$
$$=\frac{4 \sin \theta+\cos \theta-\sin ^{3} \theta-\cos ^{3} \theta}
{\sin \theta+\cos \theta}
$$
Original image
|
Recall the cube sum identity: $x^3+y^3 = (x+y)(x^2-xy+y^2)$
Continuing from where you left off:
\begin{align}&\quad4\frac {\sin\theta+\cos\theta-\sin^3\theta-\cos^3\theta}{\sin\theta+\cos\theta}\\
&=4-4\frac {\sin^3\theta+\cos^3\theta}{\sin\theta+\cos\theta}\\
&=4-4(\sin^2\theta - \sin\theta\cos\theta+\cos^2\theta)\\
&=4-4(1-\frac12\sin2\theta)\\
&=2\sin2\theta
\end{align}
|
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|
Growth rate of $f(n) = \frac{1-\cos(\frac{n\pi}{n+1})}{1-\cos(\frac{\pi}{n+1})}$ I encountered this expression when studying the growth of condition numbers on a linear system. I was trying to verify that the order of the growth is $O(n^2)$. However I'm stuck upon proving this. Any hints would be appreciated. Thanks.
|
Hint
$$1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right)$$
Then $$\begin{aligned}
f(n) &= \frac{1-\cos(\frac{n\pi}{n+1})}{1-\cos(\frac{\pi}{n+1})}\\
&= \frac{\sin^2\left(\frac{n \pi }{2(n+1)}\right)}{\sin^2\left(\frac{\pi}{2(n+1)}\right)}\\
& \le \frac{1}{\sin^2\left(\frac{\pi}{2(n+1)}\right)}\\
&\simeq \frac{\pi^2}{4}n^2
\end{aligned}$$
|
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|
determine $\iiint zdv$ using spherical coordinates Let $D$ be the region enclosed by the surface $z = \sqrt{4-x^2-y^2}$ and $z = -\sqrt{x^2+y^2}$
Determine:
$$\iiint zdv$$
using Spherical coordinates.
I'm trying to solve the above but I'm unsure how to go about it.
My attempt has been:
$$\int_0^{2\pi}\int_0^{2}\int_{-\sqrt{x^2+y^2}}^{\sqrt{4-x^2-y^2}} zdv$$
which I then integrate the first layer leaving me with:
$$\int_0^{2\pi}\int_0^{2} 2-x^2-y^2dv$$
Should I have subbed in the spherical coordinates right at the start and integrated it as:
$$\int_0^{2\pi}\int_0^{2}\int_{-\sqrt{x^2+y^2}}^{\sqrt{4-x^2-y^2}}p\cos\phi p^2\sin\phi dp d\theta d\phi$$
Based on a users input, would I have to first rewrite $z = \sqrt{4-x^2-y^2}$ and $z = -\sqrt{x^2+y^2}$
In terms of spherical coordinates?
|
Combine $z = \sqrt{4-x^2-y^2}$ and $z = -\sqrt{x^2+y^2}$ to get $z=-\sqrt2$, which corresponds to the boundary $\cos\theta = \frac zr =-\frac{\sqrt2}2$, or $\theta =\frac{3\pi}4$, in spherical coordinates.
Then, the integral is
$$\int_V zdv = 2\pi \int_{\frac{3\pi}4}^\pi\int_0^{2}(r\cos\theta )\>r^2\sin\theta drd\theta=- 2\pi
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction? I've been trying to solve the following problem:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$.
I had the following idea, we write:
$$1^2+2^2+\dots + (2n+1)^2=\frac{(2n+1)((2n+1)+1)(2(2n+1)+1)}{6}=k$$
Let's pretend the identity we want to prove is true, then:
$$1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}=j$$
We take then $k-x=j$ and solve for $x$. If the given identities are true, $x$ must be the sum of $2^2+4^2+\dots+(2n)^2$, and we have that
$$x= \frac{2n (n+1) (2 n+1)}{3} $$
We still don't know that $x=2^2+4^2+\dots+(2n)^2$ but that can be easily proved by induction. I'd like to know: Is there some "neater" way that doesn't involve induction?
Despite the tag, I'd like to see an induction-free demonstration. I chose that tag because I couldn't think of anything better to choose.
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You have a typo. Call the known sum $P_n$. The desired sum is$$\begin{align}P_{2n+1}-4P_n&=\frac{(2n+1)(2n+2)(4n+3)-4n(n+1)(2n+1)}{6}\\&=\frac{(n+1)(2n+1)}{3}(4n+3-2n)\\&=\frac{(n+1)(2n+1)(2n+3)}{3}.\end{align}$$
|
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|
Systems of Congruences \begin{cases} \overline{xyz138} \equiv 0 \mod7 \\ \overline{x1y3z8} \equiv 5 \mod11 \\ \overline{138xyz} \equiv 6 \mod13 \end{cases}
I worked my way up to this:
\begin{cases} 2000x+200y+20z \equiv 2 \mod7 \\ 100000x+1000y+10z \equiv 4 \mod11 \\ 2000x+200y+20z \equiv 7 \mod 13 \end{cases}
I tried to subtract one equation from the other, but got nothing. No idea what to do.
|
If your workings are correct, you can rewrite your equations as:
\begin{cases}
20(\overline{xyz}) &\equiv 2 \pmod 7\\
10(x+y+z) &\equiv 4 \pmod {11}\\
20(\overline{xyz}) &\equiv 7 \pmod {13}
\end{cases}
which reduces to
\begin{cases}
\overline{xyz} &\equiv 5 \pmod 7\\
x+y+z &\equiv 7 \pmod {11}\\
\overline{xyz} &\equiv 1 \pmod {13}
\end{cases}
By CRT, $\overline{xyz} \equiv 40 \pmod {91}$, so there are not many numbers to check.
|
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|
How do you prove that $\int \frac{1}{\sqrt{x^2+r^2}} \,dx = \ln{\left\lvert x + \sqrt{x^2+r^2} \right\rvert + C}$? I would like to prove that
$$\int \frac{1}{\sqrt{x^2+r^2}} \,dx = \ln{\left\lvert x + \sqrt{x^2+r^2} \right\rvert} + C$$
I tried with applying derivative with respect to $x$ to $\ln{\left\lvert x + \sqrt{x^2+r^2}\right\rvert}$ and show it is equal to $\frac{1}{\sqrt{x^2+r^2}}$, but I ended up with $$ \frac{d}{dx} \ln{\left\lvert x + \sqrt{x^2+r^2}\right\rvert} = \dots = \frac{1}{x + \sqrt{x^2+r^2}}\left(1+\frac{1}{\sqrt{x^2+r^2}}\right)$$ and I do not see how it can be simplified to $\frac{1}{\sqrt{x^2+r^2}}$.
Any help would be appreciated (even involving complex calculus, if needed).
|
HINT
What about the substitution $x = r\sinh(u)$? In such case, one has that
\begin{align*}
\int\frac{1}{\sqrt{x^{2}+r^{2}}}\mathrm{d}x & = \int\frac{r\cosh(u)}{\sqrt{r^{2}\sinh^{2}(u) + r^{2}}}\mathrm{d}u = \int1\mathrm{d}u = u + c
\end{align*}
Can you take it from here?
EDIT
Since the $\sinh$ function is bijective, there is always a solution to
\begin{align*}
z = \sinh(w) = \frac{e^{w} - e^{-w}}{2} & \Longleftrightarrow (e^{w})^{2} - 2ze^{w} - 1 = 0\\\\
& \Longleftrightarrow e^{w} = \frac{2z + 2\sqrt{z^{2}+1}}{2}\\\\
& \Longleftrightarrow w = \ln(z + \sqrt{z^{2}+1})
\end{align*}
At your case, we have that
\begin{align*}
\int\frac{\mathrm{d}x}{\sqrt{x^{2}+r^{2}}} = \sinh^{-1}\left(\frac{x}{r}\right) + c = \ln\left(\frac{x + \sqrt{x^{2} + r^{2}}}{r}\right) + c = \ln\left(x + \sqrt{x^{2} + r^{2}}\right) + C
\end{align*}
and we are done.
Hopefully this helps!
|
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|
Let $x,y \in R$ such that $|x+y| + |x-y| = 2$
Let $x,y \in R$ such that $|x+y| + |x-y| = 2$, then find maximum values of $x^2 - 6x + y^2$ and $x^2 + y^2 + 10y$.
How do I go about solving this question? Is it possible to find all real values of $x$ and $y$ from the first equation? Please help.
|
Guide:
Notice that $|x+y|+|x-y|=2$ describes the boundary of a square.
$x^2-6x+y^2=(x-3)^2+y^2-9$
The largest value of $y^2$ is $1$. Given that $-1 \le x \le 1$. Try to solve the $-1 \le x \le 1$, how would you optimize $(x-3)^2$.
|
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|
Finding the key matrix of a 2x2 Hill Cipher I'm trying to find the Hill Cipher key from the following given info:
(1,3)^T is encrypted as (-9, -2)^T, and (7, 2)^T is encrypted as (-2, 9)^T.
However I don't seem to end up with the correct answer as the encryption is invalid when I try to use the key. Below is my solution, did I make a mistake somewhere?
So
$$\begin{bmatrix} a & b\\ c & d\end{bmatrix}\begin{bmatrix}1\\3 \end{bmatrix}= \begin{bmatrix}-9\\-2 \end{bmatrix}$$
and
$$\begin{bmatrix} a & b\\ c & d\end{bmatrix}\begin{bmatrix}7\\2 \end{bmatrix}= \begin{bmatrix}-2\\9 \end{bmatrix}$$
or in one matrix notation:
$$\begin{bmatrix} a & b\\ c & d\end{bmatrix} \begin{bmatrix} 1 & 7\\ 3 & 2\end{bmatrix} = \begin{bmatrix} -9 & -2\\ -2 & 9\end{bmatrix}$$
which allows us to find the encryption matrix by
$$\begin{bmatrix} a & b\\ c & d\end{bmatrix} = \begin{bmatrix} -9 & -2\\ -2 & 9\end{bmatrix} {\begin{bmatrix} 1 & 7\\ 3 & 2\end{bmatrix}}^{-1}$$
The determinant of $\begin{bmatrix} 1 & 7\\ 3 & 2\end{bmatrix}$ is $1\cdot 2 - 3\cdot 7 = 7 \pmod{26}$, so the inverse exists and equals
$$\begin{bmatrix} 2 & -7\\ -3 & 1\end{bmatrix}$$
This allows us to compute the encryption matrix
$$\begin{bmatrix} a & b\\ c & d\end{bmatrix} = \begin{bmatrix} -9 & -2\\ -2 & 9\end{bmatrix} {\begin{bmatrix} 2 & -7\\ -3 & 1\end{bmatrix}}$$
$$
\begin{bmatrix}
-9*2+-2*-3=14 & -9*-7+-2*1=9\\
-2*2+9*-3=21 & -2*-7+9*1=23\end{bmatrix}
=
\begin{bmatrix} 14 & 9\\ 21 & 23\end{bmatrix}
$$
Then $$\begin{bmatrix} 14 & 9\\ 21 & 23\end{bmatrix}\begin{bmatrix}1\\3 \end{bmatrix}= \begin{bmatrix}15\\12 \end{bmatrix}$$ which does not match the given info as it should have been equal to
$$\begin{bmatrix}-9\\-2 \end{bmatrix}$$
|
I fully agree with
$$\begin{bmatrix} a & b\\ c & d\end{bmatrix} = \begin{bmatrix} -9 & -2\\ -2 & 9\end{bmatrix} {\begin{bmatrix} 1 & 7\\ 3 & 2\end{bmatrix}}^{-1}$$
and indeed the determinant of the right hand side matrix equals $$2- 21 \pmod{26} = -19 = 7 \pmod{26}$$ which is relatively prime to $26$ so has an inverse.
The general formula for an 2-by-2 inverse is:
$$A^{-1} = \begin{bmatrix}a & b\\c & d\end{bmatrix}^{-1} = \frac{1}{\det A}
\begin{bmatrix}d &-b\\-c & a\end{bmatrix}$$ so your inverse is wrong and all entries need to be multiplies by the invesre (modulo $26$) of the determinant $7$.
This inverse can be found by applying the extended Euclidean algorithm to $7$ and $26$ and we get the Bézout identity
$$1 = -11 \cdot 7 + 3\cdot 26$$
from which it follows that $$-11 \cdot 7 = 1 \pmod{26}$$
so that the inverse of $7$ is $-11 \equiv 15$.
So we multiply all elements of $$\begin{bmatrix} 2 & -7\\ -3 & 1\end{bmatrix}$$ by $15$ to get the inverse matrix we're looking for (of course all modulo $26$)
and we get $$\begin{bmatrix} 4 & 18\\ 7 & 15\end{bmatrix}$$
and now you can do the multiplication from the first equation modulo 26:
$$\begin{bmatrix} -9 & -2\\ -2 & 9\end{bmatrix} \begin{bmatrix} 4 & 18\\ 7 & 15\end{bmatrix}$$ to find the encryption matrix $E$. I leave that final bit to you.
takeaway: division is multiplying by the inverse. The inverse is found by the extended Euclidean algorithm. For $n=26$ you could also find the table of inverses once (program or trial and error) and use them all the time.
|
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|
How to solve $\int \sqrt{1+\sin x}\, dx$? It's easy to get this:
$$\int \sqrt{1+\sin x}\, dx \\= \int \sqrt{ \sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}} + 2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}\,\, dx \\ = \int \left | \sin{\frac{x}{2}} + \cos{\frac{x}{2}} \right |\, dx \\= \sqrt{2} \int \left | \sin{\left ( \frac{x}{2} + \frac{\pi}{4} \right )} \right |\, dx$$
So, in fact, my question is how to solve the integrate $\int \left | \sin x \right| dx $.
Or how to deal with the integrate when have absolute in it?
|
Hint: $|\sin x|=(-1)^n \sin x$ for $\pi n \leqslant x < \pi (n+1)$, where $n=0, \pm 1,\pm2, \cdots$.
To keep antiderivative continuous in $x=\pi (n+1)$ you need to solve
$$\left((-1)^{n+1}\cos x +C_n\right) \Big|_{x=\pi (n+1)}=\left((-1)^{n+2}\cos x +C_{n+1}\right)\Big|_{x=\pi (n+1)}$$
taking some $C=C_0$.
|
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"url": "https://math.stackexchange.com/questions/3916490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
IMO 2019 Problem N5 Solution 2 IMO 2019 Problem N5 Solution 2.
https://www.imo-official.org/problems/IMO2019SL.pdf
Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\binom{an}{b}-1$ is divisible by $an+1$ for all positive integers $n$ with $an\geqslant b$. Suppose $b$ is a positive integer such that $b$ is $a$-good, but $b+2$ is not $a$-good. Prove that $b+1$ is prime.Solution 2. $\quad$ We show only half of the claim of the previous solution: we show that if $b$ is $a$-good, then $p\mid a$ for all primes $p\leqslant b$. We do this with Lucas' theorem.$\quad$ Suppose that we have $p\leqslant b$ with $p\not\mid a$. Then consider the expansion of $b$ in base $p$; there will be some digit (not the final digit) which is nonzero, because $p\leqslant b$. Suppose it is the $p^t$ digit for $t\geqslant 1$.$\quad$ Now, as $n$ varies over the integers, $an+1$ runs over all residue classes modulo $p^{t+1}$; in particular, there is a choice of $n$ (with $an\gt b$) such that the $p^0$ digit of $an$ is $p-1$ (so $p\mid an+1$) and the $p^t$ digit of $an$ is $0$. Consequently, $p \mid an+1$ but $p\mid \binom{an}{b}$ (by Lucas’ theorem) so $p\not\mid\binom{an}{b}-1$. Thus $b$ is not $a$-good.$\quad$ Now we show directly that if $b$ is $a$-good but $b+2$ fails to be so, then there must be a prime dividing $an+1$ for some $n$, which also divides $(b+1)(b+2)$. Indeed, the ratio between $\binom{an}{b+2}$ and $\binom{an}{b}$ is $(b+1)(b+2)/(an-b)(an-b-1)$. We know that there must be a choice of $an+1$ such that the former binomial coefficient is $1$ modulo $an+1$ but the latter is not, which means that the given ratio must not be $1\ \text{mod}\ an+1$. If $b+1$ and $b+2$ are both coprime to $an+1$ then the ratio is $1$, so that must not be the case. In particular, as any prime less than $b$ divides $a$, it must be the case that either $b+1$ or $b+2$ is prime.$\quad$ However, we can observe that $b$ must be even by insisting that $an+1$ is prime (which is possible by Dirichlet’s theorem) and hence $\binom{an}{b}\equiv (-1)^b=1$. Thus $b+2$ cannot be prime, so $b+1$ must be prime.
I struggle with "$an+1$ is prime (which is possible by Dirichlet’s theorem)". For Dirichlet's theorem to be applicable here, there has to be an arithmetic progression of numbers $an+1$ for which $\binom{an}{b}\equiv1$ and $\binom{an}{b+2}\not\equiv1$ modulo $an+1$. Because $b$ is $a$-good and $b+2$ is not, we are guaranteed that at least one such $n$ exists. But I don't see how a single $n$ leads to a whole arithmetic progression of such numbers.
Any ideas? Thanks.
|
I think that if $n$ satisfies $an\ge b+2,\binom{an}{b}\equiv 1\pmod{an+1}$ and $\binom{an}{b+2}\not\equiv 1\pmod{an+1}$, then $an+1$ is a composite number.
To prove this, it is sufficient to prove that if $n$ is such that $an+1$ is prime satisfying $an\ge b+2$ and $\binom{an}{b}\equiv 1\pmod{an+1}$, then $\binom{an}{b+2}\equiv 1\pmod{an+1}$.
We can have
$$(an+1)\mid (b+1)(b+2)\bigg(\binom{an}{b+2}-1\bigg)\tag1$$
since we have
$$\begin{align}&(b+1)(b+2)\bigg(\binom{an}{b+2}-1\bigg)
\\\\&=(b+1)(b+2)\cdot\frac{an(an-1)\cdots (an-b+1)(an-b)(an-b-1)}{(b+2)(b+1)b!}-(b+1)(b+2)
\\\\&=(an-b)(an-b-1)\cdot\frac{an(an-1)\cdots (an-b+1)}{b!}-(b+1)(b+2)
\\\\&=(an-b)(an-b-1)\binom{an}{b}-(b+1)(b+2)
\\\\&\equiv (-1-b)(-b-2)\times 1-(b+1)(b+2)\pmod{an+1}
\\\\&\equiv (b+1)(b+2)-(b+1)(b+2)\pmod{an+1}
\\\\&\equiv 0\pmod{an+1}\end{align}$$
It follows from $an+1\ge b+3$ that $(an+1)\not\mid (b+1)(b+2)$.
So, from $(1)$, we get $(an+1)\mid (\binom{an}{b+2}-1)$, i.e.
$$\binom{an}{b+2}\equiv 1\pmod{an+1}.\quad\blacksquare$$
|
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"url": "https://math.stackexchange.com/questions/3917879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Circle $O$ is tangential to triangle $ABC$ at $B$ and passes $C$. Express $AO$ as $\frac{a√b}{c√d}$ Let $ABC$ be a triangle with side lengths $AB = 7, BC = 8, AC = 9$. Draw a circle tangent to $AB$ at $B$ and passing through $C$. Let the center of the circle be $O$. The length of $AO$ can be expressed as $\frac{a√b}{c√d}$ for positive integers a, b, c, d where gcd(a, c) = gcd(b, d) = 1 and b, d are not divisible by the square of any prime. Find a + b + c + d.
I drew everything and cant find any patterns
|
According to the cosine rule
$$\cos B = \frac{7^2+8^2-9^2}{2\cdot 7\cdot8}=\frac27,\>\>\>\>\>\sin B= \frac{3\sqrt5}7
$$
Then, the radius $OB$ is
$$OB = \frac{BC}{2\cos\angle OBC}=\frac4{\sin B}=\frac{28}{3\sqrt5}
$$
Per the Pythagoras' Theorem
$$AO=\sqrt{AB^2+BO^2}=\sqrt{7^2+ \frac{28^2}{45}}=\frac{7\sqrt{61}}{3\sqrt5}
$$
Thus, $a+b+c+d=76$.
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Determine conic section $x^2-4xy+4y^2-6x-8y+5=0$ and its center So I got a task to determine the conic section of the following:
$$x^2-4xy+4y^2-6x-8y+5=0$$
I started using matrices and got to the equation :
$$5x`^2 + \frac {10x`}{\sqrt5}-\frac {20y`}{\sqrt5}+5=0$$
I completed the square and got to the equation :
$$\frac {\sqrt5(x`+\frac {1}{\sqrt5})^2}{4} + \frac {1}{\sqrt5}=y`$$
So I can see it's most likely a parabola.
But I'm confused - How can I surely know ? How can I find the "center" (tasked asked for center but probably meant vertex).
I know that somehow I need to find $x`$ and $y`$ and then multiply it by my orthogonal $P$ matrix and this will be the center.
|
The given equation of the conic is
$ x^2 - 4 x y + 4 y^2 -6x - 8y + 5 = 0 $
If we define $r = [x, y]^T $ then the given conic equation can be written in matrix-vector form as
$ r^T A x + b^T x + c = 0 $
where
$ A = \begin{bmatrix} 1 && - 2 \\ -2 && 4 \end{bmatrix} $
$ b = [-6, -8] $
$ c = 5
First we need to check if $A$ is invertible. Calculate the determinant
$ | A | = (1)(4) - (-2)^2 = 0 $
So $A$ is singular, therefore the conic is either a parabola or a pair of lines.
Diagonalize $A$ and put it in the form $A = R D R^T $, using the following steps.
Step 1: Calculate the angle $\phi = \dfrac{1}{2} \tan^{-1}\bigg( \dfrac{2 A_{12}}{A_{11} - A_{22} } \bigg) = \dfrac{1}{2} \tan^{-1}\big(\dfrac{-4}{-3}\big) $
It follows that $ \tan(2 \phi) = \dfrac{4}{3} \Longrightarrow \cos(2 \phi) = \dfrac{3}{5} , \sin(2 \phi) = \dfrac{4}{5} $
Step 2: It follows from step 1. that $\cos(\phi) = \sqrt{ \dfrac{(1 + \cos(2 \phi) }{2} } = \dfrac{2}{\sqrt{5}} $ and $\sin(\phi) = \sqrt{ \dfrac{(1 - \cos(2\phi) }{2} } = \dfrac{1}{\sqrt{5}} $
Step 3:Define the rotation matrix as
$R = \begin{bmatrix} \cos(\phi) && - \sin(\phi) \\ \sin(\phi) && \cos(\phi) \end{bmatrix} = \dfrac{1}{\sqrt{5}} \begin{bmatrix} 2 && - 1 \\ 1 && 2 \end{bmatrix}$
Step 4: Compute the diagonal elements of matrix $D$
$D_{11} = \dfrac{1}{2}(A_{11} + A_{22}) + \dfrac{1}{2} (A_{11} - A_{22}) \cos(2 \phi) + A_{12} \sin(2 \phi) = \dfrac{5}{2} - \dfrac{9}{10} - \dfrac{8}{5} = 0 $
$D_{22} = \dfrac{1}{2}(A_{11}+A_{22}) - \dfrac{1}{2}(A_{11} - A_{22}) \cos(2 \phi) - A_{12} \sin(2 \phi) = \dfrac{5}{2} + \dfrac{9}{10} + \dfrac{8}{5} = 5 $
Step 5: Now the equation of the conic is
$ r^T R D R^T r + b^T r + c = 0 $
So define $w = [w_1, w_2] = R^T r \Longleftrightarrow r = R w $, then
$ w^T D w + b^T R w + c = 0$
where $b^T R = \dfrac{1}{\sqrt{5}} \begin{bmatrix} -20 , -10 \end{bmatrix} $
This equation written explicitly is
$ 5 w_2^2 - \dfrac{20}{\sqrt{5}} w_1 - \dfrac{10}{\sqrt{5}} w_2 + 5 = 0 $
Dividing by $5$, it becomes
$ w_2^2 - \dfrac{4}{\sqrt{5}} w_1 - \dfrac{2}{\sqrt{5}} w_2 + 1 = 0 $
Completing the square in $w_2$:
$ (w_2 - \dfrac{1}{\sqrt{5}})^2 - \dfrac{4}{\sqrt{5}} w_1 + \dfrac{4}{5} = 0 $
Thus, $w_1 = \dfrac{\sqrt{5}}{4} \bigg( (w_2 - \dfrac{1}{\sqrt{5}})^2 + \dfrac{4}{5} \bigg) $
Which is clearly a parabola that has a vertex in the $w$-plane equal to $w =(\dfrac{1}{\sqrt{5}}, \dfrac{1}{\sqrt{5}})$
The vertex in the $xy$ plane is found by the relation $ r = R w $. Using this, we get
$ \text{Vertex} = \frac{1}{5} \begin{bmatrix} 1 \\ 3 \end{bmatrix} = ( \dfrac{1}{5}, \dfrac{3}{5})$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Small angle approximation on cosine The problem is
Using the small angle approximation of cosine, show that $3-2\cos(x)+4\cos^2(x)\approx 5-kx^2$ where k is a positive constant
I did solve it by using $\cos^2(x)=1-\sin^2(x)$ on the $\cos^2(x)$, by plugging $\sin^2(x)\overset{x\to 0}{\approx}x^2$ and $\cos(x)\overset{x\to 0}{\approx}1-\frac{x^2}{2}$ to get $$3-2(1-\frac{x^2}{2})+4(1-x^2)=5-3x^2$$ hence $k=3$. But why does using $\cos^2(x)\overset{x\to 0}{\approx}(1-\frac{x^2}{2})^2$ doesn't work out? I originally tried plugging that into the $\cos^2(x)$ but got another complete thing. why?
|
Directly subbing $\cos^2x=(1-x^2/2+\cdots)^2$ should work out, provided you expand properly:
$$\cos^2x=1-2(x^2/2)+\dots=1-x^2+\cdots$$
$$1-\sin^2x=1-(x-\cdots)^2=1-x^2+\cdots$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3927646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Matrix Difference Equations Using Undetermined Coefficients?
$\begin{cases}x[n + 1] = -x[n] + 3 \\ y[n + 1] = -y[n] - e^{-n}\end{cases}$
Although I realize this system happens to be decoupled, I want to solve it using a general technique which doesn't depend on this contingency. I first solve the homogeneous system, which has coefficient matrix $\begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}$, which in turn has eigenvalue $-1$ with all of $\mathbb{R}^2$ as its eigenspace, so the homogeneous solution should be $\vec{x_h} = A(-1)^n\begin{bmatrix}1 \\ 0\end{bmatrix} + B(-1)^n\begin{bmatrix}0 \\ 1\end{bmatrix}$.
Based on how undetermined coefficients works for systems of ODEs, I guess that $\vec{x_p} = \vec a + \vec b e^{-n}$ will be the form of the particular solution. Then the system becomes $\vec a + \vec b e^{-(n + 1)} = \begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}(\vec a + \vec b e^{-n}) + \begin{bmatrix}3 \\ -e^{-n}\end{bmatrix}$. Expanding, $\begin{bmatrix}a_1 + \frac{b_1}{e} e^{-n} \\ a_2 + \frac{b_2}{e} e^{-n}\end{bmatrix}$ = $\begin{bmatrix}3 - a_1 - b_1 e^{-n} \\ -a_2 - b_2 e^{-n} - e^{-n}\end{bmatrix}$, leading to $\begin{cases} a_1 = 3 - a_1 \\ \frac{b_1}{e} = -b_1 \\ a_2 = -a_2 \\\frac{b_2}{e} = -b_2 - 1\end{cases}$ with solution $a_1 = \frac{3}{2}, b_1 = 0, a_2 = 0, b_2 = \frac{e}{e + 1}$, and finally the general solution $\vec x = A(-1)^n\begin{bmatrix}1 \\ 0\end{bmatrix} + B(-1)^n\begin{bmatrix}0 \\ 1\end{bmatrix} + \begin{bmatrix}\frac{3}{2} \\ \frac{1}{e^n + e^{n - 1}}\end{bmatrix}$.
I couldn't make this match Wolfram's answer exactly, and the graphs aren't identical. There isn't much information on this topic on Youtube, etc., so I just did what felt natural based on solving matrix ODEs.
Does this undetermined coefficients technique from matrix ODEs transfer over to matrix difference equations in the above manner, as one would expect? If so, did I make the correct particular solution guess (a sum of the forms one would guess for each individual equation in the system)? I wasn't sure if the $3$ forcing term would "resonate" with the $(-1)^n$ characteristic mode, requiring an extra term in the guess, but $(-1)^n$ technically isn't constant, and the coefficient system turned out consistent, so I suppose this should be unnecessary. Where did I go wrong?
|
$\def\A{{\bf A}}
\def\B{{\bf B}}
\def\C{{\bf C}}
\def\M{{\bf M}}
\def\I{{\bf I}}
\def\O{{\bf 0}}
\def\x{{\bf x}}
\def\y{{\bf y}}$There is a minor sign error in your work. Note that $b_2 = -e/(1+e)$. Taking this into account, you will arrive at the correct solution.
Below we consider the technique in a fairly general way.
Consider a vector of discrete functions $\y_n$ such that
\begin{align*}
\y_{n+1}-\y_n = \M\y_n,\tag{1}
\end{align*}
where $\M$ does not depend on $n$.
For example, if
$$\y_n = \left[\begin{array}{cccccc}
1&n&n^2&n^3&n^4&\cdots
\end{array}\right]^T$$
then such an $\M$ exists,
$$\M = \left[\begin{array}{cccccc}
0&0&0&0&0&\cdots \\
1&0&0&0&0&\cdots \\
1&2&0&0&0&\cdots \\
1&3&3&0&0&\cdots \\
1&4&6&4&0&\cdots \\
\vdots& & & & &\ddots
\end{array}\right]$$
If
$$\y'_n = \left[\begin{array}{cccccc}
r^n&nr^n&n^2r^n&n^3r^n&n^4r^n&\cdots
\end{array}\right]^T$$
then
$$\M' = \left[\begin{array}{cccccc}
r-1&0&0&0&0&\cdots \\
r&r-1&0&0&0&\cdots \\
r&2r&r-1&0&0&\cdots \\
r&3r&3r&r-1&0&\cdots \\
r&4r&6r&4r&r-1&\cdots \\
\vdots& & & & &\ddots
\end{array}\right]$$
We can consider combinations of such discrete functions such as
$$\y_n = \left[\begin{array}{cccc}
1&n&r^n&nr^n \\
\end{array}\right]^T$$
which results in a block diagonal form for $\M$,
$$\M = \left[\begin{array}{cccccc}
0&0&0&0 \\
1&0&0&0 \\
0&0&r-1&0 \\
0&0&r&r-1
\end{array}\right]$$
Suppose that we wish to solve
\begin{align*}
\x_{n+1} &= \A\x_{n} + \B\y_n\tag{2}
\end{align*}
where $\A,\B$ are constant matrices and where $\y_n$ is a discrete function satisfying (1).
We search for homogeneous and particular solutions, $\x_n^h,\x_n^p$, such that
\begin{align*}
\x_n &= \x_n^h + \x_n^p.
\end{align*}
Clearly $\x_n^h = \A^n \x_0^h$.
Assume that $\x_n^p = \C\y_n$ where $\C$ is independent of $n$.
This implies that
$$(\C\M+(\I-\A)\C-\B)\y_n = \O.$$
We assume that the elements of $\y_n$ are independent discrete functions of $n$.
Thus, we must solve
\begin{align*}
\C\M+(\I-\A)\C&=\B.\tag{3}
\end{align*}
This can be solved using the vec trick, for example.
A unique solution will exist if $\M$ and $\A-\I$ share no eigenvalues.
For the original question we have
\begin{align*}
\y_n &= \left[\begin{array}{c}
1 \\
e^{-n}
\end{array}\right] \\
\M &= \left[\begin{array}{cc}
0&0\\
0&\frac 1 e -1
\end{array}\right] \\
\A &= -\I \\
\B &= \left[\begin{array}{cc}
3&0\\
0&-1
\end{array}\right]
\end{align*}
Equation (3) becomes
\begin{align*}
\left[\begin{array}{cc}
2c_{11} & \frac{1+e}{e}c_{12} \\
2c_{21} & \frac{1+e}{e}c_{22}
\end{array}\right]
&= \left[\begin{array}{cc}
3&0\\
0&-1
\end{array}\right]
\end{align*}
which is easily solved, with the result
\begin{align*}
\C &=
\left[\begin{array}{cc}
\frac32 & 0 \\
0 & -\frac{e}{1+e}
\end{array}\right]
\end{align*}
Thus,
\begin{align*}
\x_n &= \left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right]^n
\left[\begin{array}{c}
x_0^h \\
y_0^h
\end{array}\right]
+ \left[\begin{array}{cc}
\frac32 & 0 \\
0 & -\frac{e}{1+e}
\end{array}\right]
\left[\begin{array}{c}
1 \\
e^{-n}
\end{array}\right] \\
&= \left[\begin{array}{c}
(-1)^n x_0^h + \frac32 \\
(-1)^n y_0^h - \frac{e}{1+e}e^{-n}
\end{array}\right]
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Bound on the coordinates of torsion points on the elliptic curve: $y^2 = x^3 - k^2x + k^3$ Consider the elliptic curve $y^2 = x^3 - k^2x + k^3$ where $k$ is a non-zero integer.
I've come across a question that asks to show that if $(x,y)$ is a rational torsion point on the elliptic curve, then:
$$\vert{y}\vert \le 5\vert{k}\vert^3$$ and $$\vert{x}\vert \le 3\vert{k}\vert^2$$
Now by the Nagell-Lutz Theorem, $x$ and $y$ must be integers, and if $y$ is non-zero, then $y^2$ divides the discriminant of the curve, which I believe is $-4k^6 + 27k^6 = 23k^6$.
I therefore believe this should mean $y^2$ divides $k^6$ and so $\vert{y}\vert \le \vert{k}\vert^3$.
I'm therefore wondering why a factor of 5 has been included in the result of the question - have I missed something?
Further I'm not sure how to find the bound on $x$. I'm assuming I need to use the bound on $y$ but this seems difficult.
Any help much appreciated.
|
For $y$, $y^2|-23k^6$, then $|y|^2\leq |23k^6|\leq|25k^6|$, so $|y|\leq 5|k^3|$.
For $x$, we know $|x^3 - k^2x + k^3|\leq 25|k|^6$, thus $|x^3 - k^2x| \leq|k^3|+25|k|^6$, which is also $|x||x ^2- k^2|\leq|k^3|+25|k|^6$.
If $|x|>3k^2$, then $|x||x ^2- k^2|>3k^2(9k^4-k^2)$, so $3k^2(9k^4-k^2)<|k^3|+25|k|^6$, we now get $2k^6-3k^4<|k^3|$.
Now there are two possibilities.
If $k>0$, then $2k^3<3k+1$, so $k=1$. Now $y^2=x^3-x+1$ where $\Delta=-23$ and $y=1$ or $-1$, hence $x^3-x+1=1$, then we know $x=0,1,-1$ which doesn't satisfy $|x|>3|k|^2$.
If $k<0$, then $2k^3>3k+1$, this is always impossible too since $k$ is an integer.
|
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|
To find the position of the ant after 2020 moves is $(p, q)$, An ant is moving on the coordinate plane. Initially it was at (6, 0). Each move
of the ant consists of a counter-clockwise rotation of 60◦ about the origin fol-
lowed by a translation of 7 units in the positive x-direction. If the position of
the ant after 2020 moves is $(p, q)$, find the value of $p^2$ + $q^2$
For this problem,I just drew some and tried to figure out how this is going to go... But I have absolutely no idea what is the trajectory of the ant after 2020 moves.
I need help.
|
In the complex plane, a rotation of $60^\circ$ is achieved by multiplying by $e^{i\frac{\pi}{3}} = \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} = \frac{1}{2} + i \frac{\sqrt{3}}{2}.$ For ease of notation, let $\alpha = e^{i\frac{\pi}{3}}$.
Let $z_0 = 6$ be the initial position of the ant and $z_k$ be the location of the ant after $k$ moves. Then
$$\begin{align}
z_1 &= z_0 \alpha + 7 \\
z_2 &= z_1 \alpha + 7 &&= z_0 \alpha^2 + 7 \alpha + 7 \\
z_3 &= z_2 \alpha + 7 &&= z_0 \alpha^3 + 7 \alpha^2 + 7 \alpha + 7 \\
& \;\; \vdots \\
z_n &= z_{n-1} \alpha + 7 &&= z_0 \alpha^n + 7 \alpha^{n-1} + 7 \alpha^{n-2} + 7 \alpha^{n-3} + \ldots + 7 \alpha + 7 \\
\\
& &&= z_0 \alpha + 7 \left[ \alpha^{n-1} + \alpha^{n-2} + \alpha^{n-3} + \ldots + \alpha + 1 \right] \\
\\
& &&= z_0 \alpha + 7 \left[ \frac{\alpha^n - 1}{\alpha - 1} \right]
\end{align}$$
In particular, noting that $\alpha^6 = 1$, so $\alpha^7 = \alpha$, we have that
$$\begin{align}
z_7 &= z_0 \alpha^7 + 7 \left[ \frac{\alpha^7 - 1}{\alpha - 1} \right] \\
\\
&= z_0 \alpha + 7 \left[ \frac{\alpha - 1}{\alpha - 1} \right] \\
\\
&= z_0 \alpha + 7 = z_1
\end{align}$$
This means that every six moves returns the ant to its starting position. Thus, since $2020 \equiv 4 \bmod{6}$, we have that $z_{2020} = z_4$.
We now use the fact that $\alpha^n = \cos \frac{n\pi}{3} + i \sin \frac{n\pi}{3}$, and in particular, $\alpha^3 = -1$. Then
$$\begin{align}
z_{2020} = z_4 &= z_0 \alpha^4 + 7 ( \alpha^3 + \alpha^2 + \alpha+ 1 ) \\
\\
&= 6 \alpha^4 + 7 ( \alpha^2 + \alpha ) \\
\\
&= 6 ( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} ) + 7 \left[ ( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} ) + ( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} ) \right] \\
\\
&= 6 ( \frac{-1}{2} - i \frac{\sqrt{3}}{2}) + 7 \left[ ( \frac{-1}{2} + i \frac{\sqrt{3}}{2} ) + ( \frac{1}{2} + i \frac{\sqrt{3}}{2} ) \right] \\
\\
&= -3 -3 \sqrt{3}i + 7 \sqrt{3}i \\
\\
&= -3 + 4 \sqrt{3}i
\end{align}$$
Thus $p = -3$ and $q = 4 \sqrt{3}$, so the answer is
$$ p^2 + q^2 = 9 + 48 = \boxed{57} $$
|
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Interesting solutions for cyclic infinite nested square roots of 2 I have derived cosine values for following cyclic infinite nested square roots of 2 ( Hereafter simply referred as $cin\sqrt2$)
$cin\sqrt2[1-]$ represents $\sqrt{2-\sqrt{2-...}}$
$cin\sqrt2[1-1+]$ represents $\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+...}}}}$
$cin\sqrt2[1-1+]$ = $2\cos\frac{2\pi}{5}$
$cin\sqrt2[2-2+]$ = $2\cos\frac{4\pi}{15}$
$cin\sqrt2[3-3+]$ = $2\cos\frac{24\pi}{65}$
$cin\sqrt2[4-4+]$ = $2\cos\frac{16\pi}{51}$
$cin\sqrt2[5-5+]$ = $2\cos\frac{352\pi}{1025}$
$cin\sqrt2[6-6+]$ = $2\cos\frac{64\pi}{195}$
$cin\sqrt2[7-7+]$ = $2\cos\frac{5504\pi}{16385}$
$cin\sqrt2[8-8+]$ = $2\cos\frac{256\pi}{771}$
$cin\sqrt2[9-9+]$ = $2\cos\frac{87552\pi}{262145}$
$cin\sqrt2[10-10+]$ = $2\cos\frac{1024\pi}{3075}$
.
.
.
We need pattern for $cin\sqrt2[n-n+]$ ($n\in N$)
At present I have figured out the pattern as follows
If n is even $$cin\sqrt2[n-n+] = 2\cos(\frac{2^n\pi}{3\times(2^n+1)})$$
If n is odd $$cin\sqrt2[n-n+] = 2\cos(\frac{\frac{2^n+1}{3}\times 2^n\times\pi}{2^{2n}+1})$$
In both the cases, it is quite obvious that
$$\lim_{n\to\infty} cin\sqrt2[n-n+]=2\cos\frac{\pi}{3}=1$$
My question is how to combine both the equations into a single one?
|
We can multiply the fraction in the even-$n$ case by $2^n-1$ on top and bottom, yielding
$$\operatorname{cin}\sqrt2[n-n+]=2\cos\frac{2^n\pi(2^n-(-1)^n)}{3(2^{2n}-(-1)^n)}$$
|
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|
Evaluate $\lim_{x\rightarrow 0} \frac{\left( \cosh x \right) ^{\sin x}-1}{\sinh x\cos \left( \sin \left( x \right) -1 \right)}$ Evaluate the limit
$$
\lim_{x\rightarrow 0} \frac{\left( \cosh x \right) ^{\sin x}-1}{\sinh x(\cos \left( \sin \left( x \right) \right)-1)}
$$
My Attempt: I tried to use L'Hôpital's rule to evalute it, however I found that the $1$st and $2$nd derivative of the numerator is $0$ at $x=0$, and the $3$rd derivative is very complicated. And the method of Taylor's Series is too complicated here. So, my question is, is there any easier way to evaluate this limit?
The desired answer is $-1$.
|
And the method of Taylor's Series is too complicated here.
Not really, actually. Using the Taylor series approximations of $\cos$, $\sin$, $\ln$, $\exp$, and $\cosh$ (all standard) and composing them as we go along:
$$\begin{align*}
\frac{\left( \cosh x \right) ^{\sin x}-1}{\sinh x(\cos \left( \sin \left( x \right) \right)-1)}
&= \frac{e^{\sin x \ln \cosh x}-1}{\sinh x(\cos \left( \sin \left( x \right) \right)-1)}
= \frac{e^{(x+o(x)) \ln \left(1+\frac{x^2}{2}+o(x^2\right)}-1}{(x+o(x))(\cos \left( \left( x +o(x) \right) \right)-1)}\\
&= \frac{e^{(x+o(x)) \left(\frac{x^2}{2}+o(x^2)\right)}-1}{(x+o(x))\left( -\frac{x^2}{2} +o(x^2) \right)}
= \frac{e^{\frac{x^3}{2}+o(x^3)}-1}{-\frac{x^3}{2} +o(x^3)}\\
&= \frac{\frac{x^3}{2}+o(x^3)}{-\frac{x^3}{2} +o(x^3)}
= \frac{1+o(1)}{-1+o(1)} \xrightarrow[x\to0]{} \boxed{-1}
\end{align*}$$
|
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|
What are the equations of the three straight lines represented by $x^3+bx^2y+cxy^2+y^3=0$ when $b+c=-2$? I am given that $$x^3 + bx^2y + cxy^2 + y^3 = 0$$ represents three straight lines if $b + c = -2$.
Is there a way to find the equations of the three lines separately?
I tried factorizing the equation but wasn't able to get anywhere. Is there a way to get the separate equations using partial differentiation?
|
Observe that all three straight lines pass through the origin $(0, 0)$
Let the three straight lines are
$y = m_1x$, $y = m_2x$ and $y = m_3x$
Then
$$ (y -m_1x)(y -m_2x)(y -m_3x) \equiv x^3 + bx^2y + cxy^2 + y^3$$
Comparing the coefficients, we have
$m_1 m_2 m_3 = -1$
$m_1 m_2 + m_1 m_3 + m_2 m_3 = b$
$m_1 + m_2 + m_3 = -c$
These are the three roots of the equation
$m^3 + cm^2 + bm +1 = 0$
If $b + c = -2$, then $m=1$ must be a root of the above equation. The other quadratic factor is $m^2 - bm - m - 1$ and has the roots $\frac{1}{2} (1 + b \pm \sqrt{b^2 + 2 b + 5})$
Finally the three straight lines are
$y = x$
$y = \frac{x}{2} (1 + b + \sqrt{b^2 + 2 b + 5})$
and
$y = \frac{x}{2} (1 + b - \sqrt{b^2 + 2 b + 5})$
|
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|
Find the derivative using the definition of derivative (limit).
Given $f(x)=\dfrac{5x+1}{2\sqrt{x}}$. Find $\dfrac{df(x)}{dx}=f'(x)$ using
the definition of derivative.
I have tried as below.
\begin{align*}
f'(x)&=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}\\
&= \lim\limits_{h\to 0}
\dfrac{\dfrac{5(x+h)+1}{2\sqrt{x+h}}-\dfrac{5x+1}{2\sqrt{x}}}{h}\\
&= \lim\limits_{h\to 0}
\dfrac{\dfrac{\left(5(x+h)+1\right)\sqrt{x}-(5x+1)\sqrt{x+h}}{2\sqrt{x+h}\sqrt{x}}}{h}\\
&= \lim\limits_{h\to 0}
\dfrac{\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-5x\sqrt{x+h}-\sqrt{x+h}}{2\sqrt{x+h}\sqrt{x}}}{h}\\
\end{align*}
Now I can't find the limit. I confused how to simplify the limit. Anyone can give me hint to solve it?
Note:
We were asked to find this derivative using
$$f'(x)=\lim\limits_{h\to 0} \dfrac{f(x+h)-f(x)}{h}$$
instead of
$$f(x)=\dfrac{u(x)}{v(x)}\iff f'(x)=\dfrac{u'(x)v(x)-u(x)v'(x)}{v(x)^2}.$$
|
First, note that we don't need to worry about the derivative of $f$ when $x=0$, because your function isn't defined when $x=0$.
Using the Binomial expansion, as $h \to 0,$ which garuntees that $|\frac{h}{x}| < 1,$
$ \sqrt{x+h} = \sqrt{x} \sqrt{1 + \frac{h}{x}} = \sqrt{x} \left(1 + \frac{h}{x}\right)^\frac12 = \sqrt{x}\left(1 + \frac{\left(\frac12\right)}{1!\ } \left(\frac{h}{x}\right) + \frac{\left(\frac12\right)\left(\frac{-1}{2}\right) }{2!\ } \left(\frac{h}{x}\right)^2 + ...\right),$
i.e., $\ \sqrt{x+h} = \sqrt{x} + \frac{h}{2\ \sqrt{x} } + O(h^2).$
Substitute this into the last line of working in the question (but not the denominator), we get:
\begin{align*}
f'(x)&= \lim\limits_{h\to 0}
\left(\dfrac{\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-(5x+1)\sqrt{x+h}}{2\sqrt{x+h}\sqrt{x}}}{h}\right)\\
&= \lim\limits_{h\to 0}
\left(\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-(5x+1)\left(\sqrt{x}+ \frac{h}{2\ \sqrt{x} } + O(h^2)\right)}{2h\sqrt{x+h}\sqrt{x}}\right)\\
&= \lim\limits_{h\to 0}
\left(\dfrac{5x\sqrt{x}+5h\sqrt{x}+\sqrt{x}-5x\sqrt{x}-\frac52 h \sqrt{x} - \sqrt{x} - \frac{h}{2\ \sqrt{x} } }{2h\sqrt{x+h}\sqrt{x}} + O(\sqrt{h})\right).\\
\end{align*}
... and after some cancellation you will arrive at the answer.
|
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|
Finding the indefinite integral of $\int_0^\pi\sqrt{\sin^3x - \sin^5x}dx$ It's not hard to find that $\int\sqrt{\sin^3x - \sin^5x}dx = \frac{2\sin^{5/2}x}{5} + C$,
But the indefinite integral $\int_0^\pi\sqrt{\sin^3x - \sin^5x}dx = \frac{4}{5}$ does not conform to $\cfrac{2\sin^{5/2} \pi}{5} - \cfrac{2\sin^{5/2} 0}{5} = 0 - 0 = 0$
Why is that we can't simply use Newton-Leibniz axiom here and what is the way to find the value $\frac{4}{5}$ here ?
|
I'm going to elaborate on Kavi's answer:
The problem here is that the antiderivative $\frac{2}{5}\sin^{\frac{5}{2}}(x)+C$ is only valid when $x\in\left[0,\frac{\pi}{2}\right]$.
Looking at the Desmos plot I provided, it can be seen that the derivative of $\frac{2}{5}\sin^{\frac{5}{2}}(x)$ does not agree with $\sqrt{\sin^3(x)-\sin^5(x)}$ when $x\in\left(\frac{\pi}{2},\pi\right)$. This is because
\begin{align*}
\sqrt{\sin^3(x)-\sin^5(x)} &= \sqrt{\sin^3(x)\left[1-\sin^2(x)\right]}\\
&= \sin^{\frac{3}{2}}(x)\sqrt{\cos^2(x)}\\
&= \sin^{\frac{3}{2}}(x)|\cos(x)|
\end{align*}
and $\cos(x)$ is negative when $x\in\left(\frac{\pi}{2},\pi\right)$.
Don't lose hope, though! We can still find the correct antiderivative and apply $\text{FTC}2$. To begin, note that
$$\sqrt{\sin^3(x)-\sin^5(x)}$$
is continuous over $\left[0,\pi\right]$, so $\text{FTC}1$ asserts that the function
$$f(x)=\int_{0}^{x}\sqrt{\sin^3(t)-\sin^5(t)}\text{ }dt$$
is an antiderivative of $\sqrt{\sin^3(x)-\sin^5(x)}$. We can now boil down the problem of finding an antiderivative to finding a nice expression for this integral.
Using the identity $\sqrt{\sin^3(x)-\sin^5(x)}=\sin^{\frac{3}{2}}(x)|\cos(x)|$, we can write
$$f(x)=\int_{0}^{x}\sin^{\frac{3}{2}}(t)|\cos(t)|\text{ }dt$$
so for $x\in\left[0,\frac{\pi}{2}\right]$,
\begin{align*}
\int_{0}^{x}\sin^{\frac{3}{2}}(t)|\cos(t)|\text{ }dt &= \int_{0}^{x}\sin^{\frac{3}{2}}(t)\cos(t)\text{ }dt\\
&= \int_{0}^{\sin(x)}u^{\frac{3}{2}}\text{ }du\\
&= \frac{2}{5}\sin^{\frac{5}{2}}(x)
\end{align*}
For $x\in\left[\frac{\pi}{2},\pi\right]$,
\begin{align*}
\int_{0}^{x}\sin^{\frac{3}{2}}(t)|\cos(t)|\text{ }dt &= \int_{0}^{\frac{\pi}{2}}\sin^{\frac{3}{2}}(t)|\cos(t)|\text{ }dt+\int_{\frac{\pi}{2}}^{x}\sin^{\frac{3}{2}}(t)|\cos(t)|\text{ }dt\\
&= \frac{2}{5}\sin^{\frac{5}{2}}\left(\frac{\pi}{2}\right)-\int_{\frac{\pi}{2}}^{x}\sin^{\frac{3}{2}}(t)\cos(t)\text{ }dt\\
&= \frac{2}{5}-\int_{1}^{\sin(x)}u^{\frac{3}{2}}\text{ }du\\
&= \frac{2}{5}-\left(\frac{2}{5}\sin^{\frac{5}{2}}(x)-\frac{2}{5}\right)\\
&= \frac{4}{5}-\frac{2}{5}\sin^{\frac{5}{2}}(x)
\end{align*}
Thus,
$$f(x)=\begin{cases}
\frac{2}{5}\sin^{\frac{5}{2}}(x) & \text{if }x\in\left[0,\frac{\pi}{2}\right]\\
\frac{4}{5}-\frac{2}{5}\sin^{\frac{5}{2}}(x) & \text{if }x\in\left[\frac{\pi}{2},\pi\right]
\end{cases}$$
We can now (legally) evaluate the original integral with $\text{FTC}2$.
\begin{align*}
\int_{0}^{\pi}\sqrt{\sin^3(x)-\sin^5(x)}\text{ }dx &= f(\pi)-f(0)\\
&= \frac{4}{5}-\frac{2}{5}\sin^{\frac{5}{2}}(\pi)-\frac{2}{5}\sin^{\frac{5}{2}}(0)\\
&= \frac{4}{5}
\end{align*}
|
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|
Evaluating: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$ Limit i want to solve: $\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
This is how i started solving this limit:
*
*$\lim_{x \to \infty} \left(\frac {3x+2}{4x+3}\right)^x$
*$\left(\frac {3x+x-x+2+1-1}{4x+3}\right)^x$
*$\left(\frac {4x+3}{4x+3}- \frac {x+1}{4x+3}\right)^x$
*$\left(1- \frac {x+1}{4x+3}\right)^x$
*$\left(1-\frac{1}{\frac{4x+3}{x+1}} \right)^{x*\frac{4x+3}{x+1}*\frac{x+1}{4x+3}}$
*$e^{\lim_{x \to \infty} \left(\frac {x^2+x}{4x+3}\right)}$
*$e^{\infty} = \infty$
answer i got is $\infty$ but if i write this limit into online calculator i get 0 as answer. So where did i go wrong?
Thanks!
|
We have:
$$\lim_{x\to\infty}\bigg(\frac{3x + 2}{4x + 3}\bigg)^{x} = \lim_{x\to\infty}\bigg(\frac{3 + \frac{2}{x}}{4 + \frac{3}{x}}\bigg)^{x} = \bigg(\frac{3}{4}\bigg)^{\infty} = \boxed{0}$$
Your mistake was in assuming that $\frac{4x+3}{x + 1}\to\infty$ as $x\to\infty$, which meant you could apply the definition of $e$. However, $\frac{4x + 3}{x+1}\to 4$. Remember that a rational function only diverges if the degree of the numerator is greater than the degree of the denominator.
|
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|
How can I prove this reduction formula for $\int^1_0{x(1-2x^4)^n}dx$ The exercise in my textbook states
You are given that $$I_n=\int^1_0{x(1-2x^4)^n}dx$$
Show that $$I_n=\frac{(-1)^n}{4n +2} + \frac{2n}{2n+1}I_{n-1}$$
I have started out by splitting the power
$$\int^1_0{x(1-2x^4)^n}dx=\int^1_0{x(1-2x^4)(1-2x^4)^{n-1}}dx$$
$$\int^1_0{x(1-2x^4)^{n-1}}dx + \int^1_0{-2x^4(1-2x^4)^{n-1}}dx$$
$$I_n=I_{n-1} + \int^1_0{-2x^5(1-2x^4)^{n-1}}dx$$
Then setting up the second integral for integration by parts
$$ I_n=I_{n-1} + \frac{1}{2}\int^1_0{x^2*-4x^3(1-2x^4)^{n-1}}dx $$
Differentiating $x^2$ and integrating $-4x^3(1-2x^4)^{n-1}$
$$I_n=I_{n-1} + \frac{1}{2}\left(\left[\frac{x^2(1-2x^4)}{n}\right]^1_0 -\frac{2}{n}\int^1_0{x(1-2x^4)^n}dx\right)$$
Evaluating $\left[\frac{x^2(1-2x^4)}{n}\right]^1_0$
$$I_n=I_{n-1} + \frac{1}{2}\left(\frac{(-1)^n}{n} -\frac{2}{n}I_n \right)$$
Collecting like terms
$$I_n=I_{n-1} +\frac{(-1)^n}{2n} -\frac{1}{n}I_n$$
$$I_n=\frac{1}{1+\frac{1}{n}}I_{n-1} + \frac{(-1)^n}{2n(1+\frac{1}{n})}$$
Simplifying the fractions
$$I_n=\frac{n}{1+n}I_{n-1} + \frac{(-1)^n}{2n+2}$$
However this is not equivelant to the expression in the question, have I messed up somewhere or missed something out?
|
$$I_n=\int_0^1x(1-2x^4)^ndx$$
$u=x^2\Rightarrow dx=\frac{du}{2x}$ and so:
$$I_n=\frac12\int_0^1(1-2u^2)^ndu$$
$$2I_n=\int_0^1(1+\sqrt{2}u)^n(1-\sqrt{2}u)^ndu$$
If we do IBP:
$$a'=(1+\sqrt{2}u)^n\Rightarrow a=\frac{1}{\sqrt{2}(n+1)}(1+\sqrt{2}u)^{n+1}$$
$$b=(1-\sqrt{2}u)^n\Rightarrow b'=-\sqrt{2}n(1-\sqrt{2}u)^{n-1}$$
so:
$$2I_n=\frac{1}{\sqrt{2}(n+1)}\left[(1-2u^2)^n(1+\sqrt{2}u)\right]_0^1+\frac{n}{n+1}\int_0^1(1-2u^2)^{n-1}(1+\sqrt{2}u)^2du$$
$$2I_n=\frac{(-1)^n(1+\sqrt{2})-1}{\sqrt{2}(n+1)}+\frac{n}{n+1}\left(I_{n-1}+2\sqrt{2}\int_0^1u(1-2n^2)^{n-1}du+2\int_0^1u^2(1-2u^2)^{n-1}du\right)$$
|
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|
How to integrate $\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$ The question given is to calculate
$$\int \frac{\cos(x) + \sqrt 3}{1+ 4 \sin(x+\frac{\pi}{3}) + 4\sin^2(x+\frac{\pi}{3})}{\rm d}x$$
My attempt
I managed to figure out that the denominator is given out as a perfect square of $$\{1+2\sin(x + \frac{\pi}{3})\}$$ and broke the $\sin(x+\frac{\pi}{3})$ so it looks like
$$\int \frac{\cos(x) + \sqrt 3}{(1+ \sin(x) +\sqrt 3 \cos(x))^2}{\rm d}x$$
I can't figure out how to approach further. Please guide me through this question.
|
Hint
Let $y=\dfrac{a\cos x+b\sin x +c}{1+\sin x+\sqrt3\cos x}$
Find $\dfrac{dy}{dx}$
Compare
$-(a\cos x+b\sin x+c)(\cos x-\sqrt3\sin x)+(1+\sin x+\sqrt3\cos x)(-a\sin x+b\cos x)$
with $\cos x+\sqrt3$
to find $a,b,c$
For example, by comparing the coefficient of $\cos x\sin x$ we get $b=\sqrt3a$
By $\sin x,a=-\sqrt3c$
So we reach at the expression below with $c$ as constant multiple
$$(-\sqrt3\cos x+3\sin x+1)(\cos x-\sqrt3\sin x)+(1+\sin x+\sqrt3\cos x)(\sqrt3\cos x+3\sin x)$$
Can you take it from here?
|
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|
Find the minimum value of the ratio $\frac{S_1}{S_2}$ using given data
3 points $O(0, 0) , P(a, a2 ) , Q(b, b2 )$ are on the parabola $y=x^2$. Let S1 be the area bounded by the line PQ and the parabola and let S2 be the area of the triangle OPQ, then find min of $\frac{S1}{S2}$
$$S_1 = \frac 12 (a+b)(a^2+b^2) -\int_{-b}^a x^2.dx$$
$$=\frac 16 (a^3+b^3) +\frac 12 (ab^2+a^2b)$$
And $$S_2=\frac 12 (a^2b+ab^2)$$
How do I solve from here?
|
$$
\dfrac{S_1}{S_2}=\frac{\frac{1}{2} \left(a^2 b+a b^2\right)+\frac{1}{6} \left(a^3+b^3\right)}{\frac{1}{2} \left(a^2 b+a b^2\right)}=\frac{(a+b)^2}{3 a b}\geq \dfrac{4ab}{3ab}=\dfrac{4}{3}
$$
|
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|
Prove that $\displaystyle\left|\int_L \frac{z^3}{z^2+1} dz\right| \le \frac{9\pi}{8}$ where $L=\big(z: |z|=3, Re(z) \ge 0\big)$ I tried solving this by using the fact that the length of L is half the circumference of a circle with radius 3, which would mean that
if
$$\left|\frac{z^3}{z^2+1}\right| \le \frac{3}{8}$$ this problem is solved.
however, It appears to me as though this cannot be true:
$$\left|\frac{z^3}{z^2+1}\right| = \frac{\left|z\right|^3}{|z^2+1|} \ge \frac{27}{10}$$
using the triangle inequallity.
Is there something wrong with the way I'm going about this or is the statement really false?
|
I believe the inequality you stated is not true. Firstly, if $|z|=3$, by the reverse triangle inequality we get that $|z^2+1|=|z^2-(-1)| \geq | |z^2|-1| = |3^2-1| = 8$. Taking reciprocals gives us that $\dfrac{1}{|z^2+1|} \leq \dfrac{1}{8}$.
If we multiply both sides of the inequality by $|z^3|$, we obtain for $|z|=3$ that
$$\left|\frac{z^3}{z^2+1}\right|\leq \dfrac{27}{8}$$
Now if we apply standard integral estimates, we get $\displaystyle\left|\int_L \dfrac{z^3}{z^2+1}dz\right| \leq \dfrac{27}{8}\cdot\operatorname{length}(L)$. As you said, the length of $L$ is half the circumference of a circle with radius $3$. That is, $\operatorname{length}(L) = \dfrac{6\pi}{2} = 3\pi$.
Thus, $$\displaystyle\left|\int_L \dfrac{z^3}{z^2+1}dz\right| \leq \dfrac{27}{8}\cdot3\pi= \dfrac{81\pi}{8} $$
|
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|
Series convergence or divergence $\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(3n)}{\sqrt{n^2 + 2}}$ Can you give me some hint for the series
\begin{align*}
\sum_{n=1}^{\infty}\frac{(-1)^{n}\cos(3n)}{\sqrt{n^2 + 2}}
\end{align*} I have tried Leibniz test. It didn't help me.
|
Hint: use Dirichlet's test. See more at Wikipedia
It's clear that $\left\{\dfrac 1 {\sqrt{n^2 + 2}}\right\}$ is monotonic and converges to 0
Now the task is to prove $-\cos 3 + \cos 6 - \cos 9 + \cos 12 + ...$ is bounded
Recognize that $\cos 6 - \cos 3 = -2\sin \frac 9 2 \sin \frac 3 2$
$\cos 12 - \cos 9 = -2\sin \frac {21} 2 \sin \frac 3 2$
$\cos 18 - \cos 15 = -2\sin \frac {33} 2 \sin \frac 3 2$
and $-2 \sin 3\left(\sin \frac 9 2 + \sin \frac {21} 2 + \sin \frac {33} 2 ... \right)= \cos \frac {15} 2 - \cos \frac 3 2 + \cos \frac {27} 2 - \cos \frac {15} 2 +...$
You can see that $\sum\limits_{i=1}^{2n} (-1)^k \cos {3k} = \dfrac{\sin \frac 3 2}{\sin 3} \left|\cos(2n + 1.5) - \cos 1.5 \right|$, which is always bounded
Therefore the series converges
|
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|
Prove $\lim\limits_{x\to0} xf(x) = 0$? I am confused about the following proof on a textbook I'm reading.
Suppose $f:(0,b] \to \Bbb R$ is continuous, positive, and integrable on $(0,b]$. Suppose further that as $x \to 0$ from the right, $f(x)$ increases monotonically to $+\infty$. Prove that $\lim\limits_{x\to0} xf(x) = 0$.
It would be nice if I could write $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(x)dt = f(x)x$. Then for all $\epsilon > 0$ there exists a $\delta$ such that $0<x<\delta$ implies $0<f(x)x \le \int_a^{a+x} f(t)dt <\epsilon$. But this is not true since $f$ increases monotonically as $x\to0$. Instead of writing $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(x)dt = f(x)x$, I should probably write $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(a+x)dt = f(a+x)x$. But that way my proof does not work. Can someone give me a hint as to how to proceed?
EDIT: Thanks for the comments and I realized that the monotonicity of $f$ should be one of the conditions.
|
As written, this is not true. However, I am not sure if meant to include that $f(x)$ is a decreasing function as you mention later in your proof that it is a monic function. If this condition is not included, then the theorem is false. Consider the function
$$f(x)=\begin{cases}
g_n(x) & \frac{1}{n^3}\leq x \leq \frac{1}{n^3+1}\\
\frac{1}{\sqrt{x}}& \text{otherwise}
\end{cases}$$
(for all $n\in\mathbb{N}$) where $g_n(x)$ is any continuous function such that
*
*$g_n\left(\frac{2n^3+1}{2n^3(n^3+1)}\right)=\frac{2n^3(n^3+1)}{2n^3+1}$
*$g_n(x)$ is maximized at $x=\frac{2n^3+1}{2n^3(n^3+1)}$
*$g_n\left(\frac{1}{n^3}\right)=\frac{1}{\sqrt{n^3}}$
*$g_n\left(\frac{1}{n^3+1}\right)=\frac{1}{\sqrt{n^3+1}}$
*$g_n(x)$ is minimized at $x=\frac{1}{\sqrt{n^3}}$
Note that these conditions imply $f(x)$ is continuous. Further, it is obvious that $f(x)>0$ and that $\lim_{x\to 0^+} f(x)=\infty$. For the final condition, note that
$$\int_{0}^b f(x)dx<\int_{0}^b\frac{1}{\sqrt{x}}dx+\sum_{n=1}^\infty \int_{\frac{1}{n^3}}^{\frac{1}{n^3+1}} g_n(x)dx$$
$$\leq \frac{2b^{3/2}}{3}+\sum_{n=1}^\infty\int_{\frac{1}{n^3}}^{\frac{1}{n^3+1}} \frac{2n^3(n^3+1)}{2n^3+1}dx=\frac{2b^{3/2}}{3}+\sum_{n=1}^\infty\frac{1}{n^6+n^3} \frac{2n^3(n^3+1)}{2n^3+1}<\infty$$
However, we know that
$$xf(x)\bigg|_{x=\frac{2n^3+1}{2n^3(n^3+1)}}=\frac{2n^3+1}{2n^3(n^3+1)}f\left(\frac{2n^3+1}{2n^3(n^3+1)}\right)$$
$$=\frac{2n^3+1}{2n^3(n^3+1)}g_n\left(\frac{2n^3+1}{2n^3(n^3+1)}\right)=\frac{2n^3+1}{2n^3(n^3+1)}\frac{2n^3(n^3+1)}{2n^3+1}=1$$
This implies $\lim_{x\to 0^{+}}xf(x)\neq 0$. However, I believe the original theorem would be true if you add the stipulation that $f(x)$ is decreasing.
|
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Is $x=2,y=13$ the unique solution? Problem:
Find all positive integers $x$ and $y$ satisfying:
$$12x^4-6x^2+1=y^2.$$
If $x=1, 12x^4-6x^2+1=12-6+1=7,$ which is not a perfect square.
If $x=2, 12x^4-6x^2+1=192-24+1=169=13^2$, which is a perfect square. Thus, $x=2,y=13$ is a solution to the given Diophantine equation.
However, after testing a few more small cases, it seems as if $12x^4-6x^2+1$ can never be a perfect square if $x>2$. I have tried to prove this, but to no avail. Here is the gist of what I considered:
$12x^4-6x^2+1=y^2 \iff 12x^4-6x^2 = (y+1)(y-1)$. Since the L.H.S. is a multiple of $2$, it follows that $2 \mid (y+1)(y-1) \Rightarrow y$ is odd $\Rightarrow y+1$ and $y-1$ are both even.
Hence, $4 \mid (y+1)(y-1) \Rightarrow 4 \mid 12x^4-6x^2 \Rightarrow 4 \mid 6x^2 \Rightarrow x $ is even. Let $x=2m$ and $y+1=2k$, so $12x^4-6x^2=192m^4-24m^2=4k(k-1) \Rightarrow 48m^4-6m^2=k(k-1)$. At this point, I am at a loss of how to continue. We could continue with divisibility arguments, but it seems to be a never-ending process?
Another method I tried was to let $y=x+k$, thus $12x^4-6x^2+1 = x^2 + 2xk+k^2 \iff k^2+2xk-(12x^4-7x^2+1)=0$, which is a quadratic in terms of $k$. However, stuff like the discriminant or sum and product of roots did not seem to yield any important information.
Any hints provided to point me in the right direction will be much appreciated.
|
This is an addition to my previous answer, which by pointed out by Mike doesn't sufficiently solve the problem.
We use the result, $$3(2x-1)^2(2x+1)^2 = (2y-1)(2y+1).$$
We have that $\gcd(2a-1,2a+1)=1$ for positive integer $a$ (by Euclidean Algorithm if you need convincing, however this is trivial enough to state).
It follows that $2y-1$ cannot be divisor to $2x-1$ or $2x+1$ without being divisor to the full square term (in other words, if $2y-1 \vert 2x-1,$ then $2y-1 \vert (2x-1)^2$ and similar). Since it also applies the other way around, it follows that one of the following are true:
*
*$2y-1 = 1$ or $2y-1 = 3$
*$2y-1 = (2x-1)^2$ or $2y-1 = 3(2x-1)^2$
*$2y-1 = (2x+1)^2$ (cannot be $3(2x+1)^2$ as we must have $2y+1 > 2y-1$)
For the first, we have $y = 1$ or $y=2.$ No $x$ makes this true, so we can simply ignore this case.
The second case (after making $y$ in terms of $x$ and rearranging)
$2y-1 = (2x-1)^2$ gives us the polynomial $4x^4 + 4x^3 - 7x^2 +2x = 0,$ which has roots $-2, 0, 1/2, 1/2.$ None of these are positive integers.
$2y-1 = 3(2x-1)^2$ gives a polynomial with rational roots $1/2,0$ and the rest are irrational.
$2y -1 = (2x+1)^2$ gives us rational roots (-1/2, 0, 2). 2 is thus an answer!
I may have some cases I overlooked but it seems that this can solve it, albeit in a depressingly bashy way.
|
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|
Fermat factorisation in one step? Apparently if $N$ has a factor within $\sqrt[4] N$ of $\sqrt N$ then Fermat factorisation works in one step.
Specifically this would mean that for $r = \lfloor \sqrt N \rfloor +1$ we have
$$r^2 - N = s^2$$
for some integer $s$.
I've tried bounding $r^2 -N$ in-between $s^2\pm1$ for some $s$ but haven't got anywhere, and I certainly don't know how to make use of that factor close to $\sqrt N$.
Can someone prove this?
|
A small correction: The algorithm starts with $r = \bigl\lceil \sqrt{N}\,\bigr\rceil$.
Usually, that's the same as $\lfloor \sqrt{N}\rfloor + 1$, but when $N$ is a square it's a difference that matters. A square $N$ of course has a factor within $\sqrt[4]{N}$ of $\sqrt{N}$, but $(\lfloor \sqrt{N}\rfloor + 1)^2 - N = 2\sqrt{N} + 1$ then rarely is a square too, so that we need to start with $r = \bigl\lceil \sqrt{N}\,\bigr\rceil$ to be sure to find a factor in the first step.
And of course, $N$ needs to be odd or divisible by $4$, since Fermat's method cannot be applied to $N \equiv 2 \pmod{4}$.
Having this out of the way, we can assume that $N$ is not a square and $N \not\equiv 2 \pmod{4}$.
Let $N = pq$ with $p < \sqrt{N} < q$ such that $q - \sqrt{N} < \sqrt[4]{N}$ or $\sqrt{N} - p < \sqrt[4]{N}$. If $q < \sqrt{N} + \sqrt[4]{N}$, then
$$p = \frac{N}{q} > \frac{N}{\sqrt{N} + \sqrt[4]{N}} = \frac{N - \sqrt{N}}{\sqrt{N} + \sqrt[4]{N}} + \frac{\sqrt{N}}{\sqrt{N} + \sqrt[4]{N}} > \sqrt{N} - \sqrt[4]{N}\,,$$
from which we obtain $q - p < 2\sqrt[4]{N}$ and hence
$$\biggl(\frac{q+p}{2}\biggr)^2 = N + \biggl(\frac{q-p}{2}\biggr)^2 < N + \sqrt{N}\,.$$
If $p > \sqrt{N} - \sqrt[4]{N}$, then $q$ may be slightly larger than $\sqrt{N} + \sqrt[4]{N}$:
$$q = \frac{N}{p} < \frac{N}{\sqrt{N} - \sqrt[4]{N}} = \sqrt{N} + \sqrt[4]{N} + 1 + \frac{1}{\sqrt[4]{N} - 1}\,.$$
For $N > 16$ we thus have $q < \sqrt{N} + \sqrt[4]{N} + 2$, whence $q - p < 2\sqrt[4]{N} + 2$ and
$$\biggl(\frac{q+p}{2}\biggr)^2 = N + \biggl(\frac{q-p}{2}\biggr)^2 < N + \sqrt{N} + 2\sqrt[4]{N} + 1\,.$$
However, since $N < (\lfloor \sqrt{N}\rfloor + 1)^2$ we have
$$(\lfloor \sqrt{N}\rfloor + 2)^2 = (\lfloor \sqrt{N}\rfloor + 1)^2 + 2(\lfloor \sqrt{N}\rfloor + 1) + 1 > N + 2\sqrt{N} + 1 > N + \sqrt{N} + 2\sqrt[4]{N} + 1$$
for $N > 16$, as then $2\sqrt[4]{N} < \sqrt{N}$.
This implies that for $N > 16$ not a square and not $\equiv 2 \pmod{4}$, if $N = pq$ with one of $p,q$ within $\sqrt[4]{N}$ of $\sqrt{N}$, then
$$\frac{q+p}{2} = \lfloor \sqrt{N}\rfloor + 1\,,$$
which shows that Fermat's method succeeds at the first step.
The cases $N = 3,5,7,8,11,12,13,15$ are dealt with by inspection. The equations
$$3 = 2^2 - 1^2, \quad 5 = 3^2 - 2^2, \quad 8 = 3^2 - 1^2, \quad 12 = 4^2 - 2^2, \quad 15 = 4^2 - 1^2$$
show that the method succeeds at the first step for $N \in \{3,5,8,12,15\}$, and the primes $7,11,13$ have no factor within $\sqrt[4]{N}$ of $\sqrt{N}$ (which is a close call for $N = 7$, as $\sqrt{7} - \sqrt[4]{7} - 1 \approx 0.01917475$).
Altogether, when Fermat's method is applicable (i.e. $N \not \equiv 2 \pmod{4}$) and $N$ has a factor within $\sqrt[4]{N}$ of $\sqrt{N}$, then the method succeeds at the first step, $\bigl\lceil \sqrt{N}\,\bigr\rceil^2 - N$ is then a perfect square.
|
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|
show this $\lceil \frac{n}{1-a_{n}}\rceil =n+1$ let $a_{n}$ be squence such $a_{1}=2-\dfrac{\pi}{2}$, and
$$a_{n+1}=2a_{n}+\dfrac{a_{n}-1}{n},n\in N^{+}$$
show that
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
I try:since
$$f(1)=\lceil \dfrac{1}{\dfrac{\pi}{2}-1}\rceil=2$$
and $$a_{2}=2a_{1}+a_{1}-1=3a_{1}-1=5-\dfrac{3}{2}\pi$$
so
$$f(2)=\lceil \dfrac{2}{\dfrac{3\pi}{2}-4}\rceil=3$$
and so on ,if we want prove
$$f(n)=\lceil \dfrac{n}{1-a_{n}}\rceil =n+1$$
or
$$n<\dfrac{n}{1-a_{n}}\le n+1$$
or
$$0<a_{n}\le\dfrac{1}{n+1}$$
if we use induction to prove it.then
$$a_{n+1}=(2+\dfrac{1}{n})a_{n}-\dfrac{1}{n}\le (2+\dfrac{1}{n})\cdot\dfrac{1}{n+1}-\dfrac{1}{n}=\dfrac{1}{n+1}$$
is wrong,becasue we want to prove $a_{n+1}\le\dfrac{1}{n+2}$
|
We write the recurrence as
$$
a_{n+1}=\frac{2n+1}n a_n - \frac1n.
$$
Then we divide both sides by $\prod_{j=1}^n \frac{2j+1}j$. We obtain
$$
\frac{a_{n+1}}{\prod_{j=1}^n \frac{2j+1}j} = \frac{a_n}{\prod_{j=1}^{n-1}\frac{2j+1}j}-\frac1{(2n+1)\prod_{j=1}^{n-1} \frac{2j+1}j}.
$$
Repeatedly applying this, we have
$$
\frac{a_{n+1}}{\prod_{j=1}^n\frac{2j+1}j}=\frac{a_2}3-\sum_{k=2}^n \frac1{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j}
$$
Note that $a_2=3a_1-1$ and
$$
\frac13+\sum_{k=2}^{\infty} \frac1{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j}=2-\frac{\pi}2,
$$
as commented on Tesla Daybreak's answer.
Then we need $a_1=2-\frac{\pi}2$ to have a bounded sequence $\{a_n\}$.
$$
\frac{a_{n+1}}{\prod_{j=1}^n\frac{2j+1}j}=2-\frac{\pi}2-(\frac13+\sum_{k=2}^n \frac1{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j})
$$
The RHS is
$$
\sum_{k=n+1}^{\infty} \frac1{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j}
$$
Thus,
$$
a_{n+1}=\sum_{k=n+1}^{\infty} \frac{\prod_{j=1}^n \frac{2j+1}j}{(2k+1)\prod_{j=1}^{k-1}\frac{2j+1}j}
$$
After some cancellations,
$$
a_{n+1}=\frac1{2n+3} + \frac1{(2n+5)\frac{2n+3}{n+1}}+\frac1{(2n+7)\frac{2n+5}{n+2}\frac{2n+3}{n+1}}+\cdots.
$$
$$
\leq \frac1{2n+3}+\frac1{(2n+5)2}+\frac1{(2n+7)2}\leq \frac1{n+2}.
$$
|
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On Lyapunov stability of a solution I have recently encountered the following exercise: prove or disprove that the solution of the ODE $\dot x = -\frac{2x}{t} + t^2$ that satisfies $x(1)=1$ is Lyapunov stable. I have proved that the solution is stable, however, the correct answer (as per my teacher) is the opposite.
My proof is as follows:
$$\begin{aligned}
\dot x &= -\frac{2x}{t} + t^2 \\
t^2\dot x + 2xt &= t^4 \\
t^2\dot x + x(t^2)' &= t^4 \\
(t^2 x)' &= t^4 \\
t^2 x &= \frac{t^5}{5} + C \\
x &= \frac{t^3}{5} + \frac{C}{t^2} \\
\end{aligned}$$
The solution that satisfies $x(1)=1$ is $x=\varphi(t)=\cfrac{t^3}{5} + \cfrac{4}{5t^2}$.
The definition of Lyapunov stability:
$$\forall \varepsilon > 0 \ \ \exists \delta > 0 \ \ |x(t_0) - \varphi(t_0)|< \delta \implies \forall t > t_0 \ \ |x(t) - \varphi(t)| < \varepsilon$$
$$|x(t_0) - \varphi(t_0)|< \delta \implies \left|\frac{1}{5} + \frac{C}{t^2} - \frac{1}{5} - \frac{4}{5}\right| < \delta \implies \left|C - \frac{4}{5}\right| < \delta$$
$$|x(t) - \varphi(t)| < \varepsilon \implies \left|\frac{t^3}{5} + \frac{C}{t^2} - \frac{t^3}{5} - \frac{4}{5t^2}\right| < \varepsilon \implies \left|\frac{1}{t^2}\left(C-\frac{4}{5}\right)\right| < \varepsilon$$
Let $\delta=\varepsilon$. Then
$$\left|\frac{1}{t^2}\left(C-\frac{4}{5}\right)\right| < \varepsilon \implies \varepsilon\left|\frac{1}{t^2}\right| < \varepsilon$$
, which holds.
|
You obtained the following equation
$$\frac{d}{dt}(t^2 x) = t^4 \quad (1) $$
then you found the following general solution:
$$t^2 x(t) = \frac{t^5}{5} + C $$
which is also true, but you can not treat $C$ as an initial condition of the original differential equation. To use the defintion of stability, you sould integrate both sides of Equaiton (1) form $t_0$ to $t\geq t_0$. Integrating (1) gives:
$$t^2x(t)-t_0^2x(t_0)=\frac{1}{5}t^5-\frac{1}{5}t_0^5$$
or equivilanetly:
$$x(t)=\frac{t_0^2}{t^2}x(t_0)+\frac{1}{5}t^3-\frac{1}{5t^2}t_0^5$$
To prove the stability, the following should hold for any $\epsilon$ .....
$$\|x(t)-x(t_0)||=\bigg\|(\frac{t_0^2}{t^2}-1)x(t_0)+\frac{1}{5}t^3-\frac{1}{5t^2}t_0^5\bigg\|\leq\epsilon \qquad (2)$$
But if there exists at least one $\epsilon$ such that Equaiton (2) is not satisfied for all initial conditions then the system is not stable. Slecet $\epsilon =1$, then one can see that $\bigg\|(\frac{t_0^2}{t^2}-1)x(t_0)+\frac{1}{5}t^3-\frac{1}{5t^2}t_0^5\bigg\|$ is not bounded because of $t^3$.
|
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|
$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$ Let $a,b,c>0$ and $a^2+b^2+c^2=3$, prove
$$\frac{a^2+3b^2}{a+3b}+\frac{b^2+3c^2}{b+3c}+\frac{c^2+3a^2}{c+3a}\geqslant 3$$
This inequality looks simple but I do not know how to solve it. The straightforward method is to bash the inequality with brutal computation method and assume $x = \min \{ x,y,z \}$, $y = x+a$, $z = x + b$, etc. I wonder if we can prove it using classical inequality.
|
By Cauchy-Schwarz Inequality we have
$\begin{align}
~~~~~~\sum_\text{cyc}\left(a+3b\right)\sum_\text{cyc}\dfrac{a^{2}+3b^{2}}{a+3b}\ge\left(\sum_\text{cyc}\sqrt{a^{2}+3b^{2}}\right)^{2}\cdot
\end{align}$
So it suffices to show that
$\begin{align}
~~~~~~&\left(\sum_\text{cyc}\sqrt{a^{2}+3b^{2}}\right)^{2}\ge3\sum_\text{cyc}\left(a+3b\right)\iff\\&\iff\sqrt{a^{2}+3b^{2}}+\sqrt{b^{2}+3c^{2}}+\sqrt{c^{2}+3a^{2}}\ge\sqrt{12\left(a+b+c\right)}\cdot
\end{align}$
With $a^{2}+b^{2}+c^{2}=3$, the last inequality is a "known unsolved inequality" (and it's true).
See:
$\sum\limits_{cyc} \sqrt{a^2+3b^2}\geq \sqrt{12(a+b+c)}$
Prove that $\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\geq6$ if $(a+b+c)^2(a^2+b^2+c^2)=27$
a^2+b^2+c^2=3
Inequality 88
|
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How to find the closed form of $\sum_{k=0}^n\frac{1}{k+1} {\binom n k}^2$ How to sum the following identity: $$\sum_{k=0}^n\frac{1}{k+1}{\binom n k}^2$$ I know the answer is $\frac{(2n+1)!}{{(n+1)!}^2}$ but I have tried a lot to find the closed form through any algebraic way or combinatorial but not able to find it out, any help?
|
consider this polynomial (or OGF).
$$
\sum_{k=0}^n \binom{n}{k}^2 \frac{1}{k+1}=[x^{n+1}]\sum_{k=0}^n \binom{n}{n-k}x^{n-k}\times \binom{n}{k}\frac{x^{k+1}}{k+1}
$$
It's the convolution(product) of $(1+x)^n$ and $F(x)$ where
$$
\begin{aligned}
F(x)&=\sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{k+1}\\
F'(x)&=\sum_{k=0}^n \binom{n}{k}x^k=(1+x)^n\\
F(x)&=\frac{(1+x)^{n+1}}{n+1}
\end{aligned}
$$
Thus the polynomial is $(1+x)^n\frac{(1+x)^{n+1}}{n+1}$ and the $(n+1)$ -th coefficient is $\frac{1}{n+1}\binom{2n+1}{n+1}=\frac{(2n+1)!}{(n+1)!(2n+1-n-1)!}\frac{1}{n+1}=\frac{(2n+1)!}{{\left((n+1)!\right)}^2}$
|
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|
$3^{1234}$ can be written as $abcdef...qr$. What is the value of $q+r$? It was possible to find $3^{15} ≡ 7\pmod{100}$. Knowing that $7^{4k}≡1\pmod{100}, (3^{15})^{80}≡3^{1200}≡(7)^{80}≡1\pmod{100}$.
$(3^{15})(3^{15})3^{1200}≡1\cdot 7\cdot 7\pmod{100}, 3^{1230}\cdot 3^{4}≡49\cdot 81\pmod{100}≡69\pmod{100}$ to get $6+9=15$. I felt that there must be a better way to solve this problem. What ideas should I keep in mind?
|
You can directly find the last $2$ digits of $3^{1234}$ by writing it as $3^{1232}\cdot 9=(81)^{308}\cdot 9$.
Last two digits of $(81)^{308}=(80+1)^{308}\equiv 41 \pmod{100} $. Hence $41\cdot 9 \equiv 69 \pmod{100}$.
|
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Given that $p$ is a prime, and the sum of all positive divisors of $p^4$ is a perfect square, find the possible number of primes $p$ .
Given that $p$ is a prime, and the sum of all positive divisors of $p^4$ is a perfect square, find the possible number of primes $p$ .
What I Tried: I know that the only divisors of $p^4$ will be $(1 , p , p^2 , p^3 , p^4)$ . From here I can conclude :-
$$\rightarrow 1 + p + p^2 + p^3 + p^4 = k^2$$ For some positive integer $k$ . Now this is where I get stuck, I can write it as :-
$$\rightarrow \frac{(1 - p^5)}{1 - p} = k^2$$
But I didn't understand how I can find all values of $p$. Next what I did is :-
$$\rightarrow p(1 + p + p^2 + p^3) = (k + 1)(k - 1)$$
And I am stuck at the same problem.
Can anyone help me here?
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Just some thoughts. Still editing...
Note that the sum $1+p+p^2+p^3+p^4$ is odd, so $k^2\equiv1\bmod 8$, and hence, $p(1+p+p^2+p^3)$ is a multiple of $8$.
If $p=2$, this isn't the case, so $p$ must be odd, and hence, $1+p+p^2+p^3$ must be a multiple of $8$.
mod $8$, $1+p+p^2+p^3=1+p+1+p$, so $2+2p$ is a multiple of $8$, so $1+p$ is a multiple of $4$. So, $p\equiv3\bmod 4$.
Through similar logic $\bmod{12}$, we have $p\equiv-1\bmod{12}$, or $p=3$.
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|
Eliminate $\theta$ and prove $x^2+y^2=1$ We have:$${ \begin{cases}{2x=y\tan\theta+\sin\theta} \\ {2y=x\cot\theta+\cos\theta}\end{cases} }$$
And want to prove $x^2+y^2=1$
My works:
I multiplied first equation by $\cos\theta$ and second one by $\sin\theta$ and get:
$${ \begin{cases}{2x\cos\theta=y\sin\theta+\sin\theta\cos\theta} \\ {2y\sin\theta=x\cos\theta+\sin\theta\cos\theta}\end{cases} }$$
By extracting $\sin\theta\cos\theta$ we get: $$2x\cos\theta-y\sin\theta=2y\sin\theta-x\cos\theta$$
$$x\cos\theta=y\sin\theta$$
But I don't know whether this helps or not.
|
You are almost there. You already have $$2x\cos\theta=y\sin\theta+\sin\theta\cos\theta \tag1$$
and $$x\cos \theta = y \sin \theta \tag 2$$
Then $$x\cos \theta = y \sin \theta = 2x\cos \theta - y \sin \theta = \sin \theta \cos \theta$$
and you know $\sin \theta \ne 0, \cos \theta \ne 0$. Can you end it now?
|
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|
If $p$ and $q$ are solutions of the equation $x \tan x = 1$, show the integral of $\cos^2 px$ entirely in terms of $p$ I am working through a pure maths text book out of interest. I have finished the chapter on integration and differentiation of trigonometric functions and am doing the end of chapter questions. This is causing me a problem:
Given that $p$ and $q$ are solutions of the equation $x \tan x = 1$,
Find an expression for $\int^1_0 \cos^2 px \ dx$ entirely in terms of $p$, not involving any trigonometric functions.
This is my working so far:
if $p$ and $q$ are solutions of $x \tan x = 1$,
$$p\frac{\sin p}{\cos p} = 1$$
$$\cos^2px = \frac{1 + \cos 2px}{2}$$
$$\int \cos^2px ~dx = \frac{1}{2}(x + \frac{\sin 2px}{2p})$$
$$= \frac{1}{2}(\frac{2px + \sin 2px}{2p})$$
I need to find
$$\left[
\frac{1}{2}((\frac{2px + \sin 2px}{2p})
\right]_{0}^1$$
But I cannot arrive at the answer in the book which is:
$$\frac{2 + p^2}{2(1 + p^2)}$$
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$$I=\int_0^1 \cos ^2(p x) \, dx=\frac{\sin (2 p)}{4 p}+\frac{1}{2}$$
it is given that $\tan p=\frac{1}{p}$
From the formula
$$\sin \alpha=\frac{2t}{1+t^2};\;t=\tan\frac{\alpha}{2}$$
we get
$$\sin 2p=\frac{2\tan p}{1+\tan^2 p}=\frac{2\cdot\frac{1}{p}}{1+\left(\frac{1}{p}\right)^2}=\frac{2 p}{p^2+1}$$
and finally
$$I=\frac{\frac{2 p}{p^2+1}}{4p}+\frac{1}{2}=\frac{p^2+2}{2 p^2+2}$$
|
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|
Proving $n\leq3^{n/3}$ for $n\geq0$ via the Well-Ordering Principle [2] I know this question was already asked in here, but it was never marked as answered and all the solutions base themselves on the fact that $3(m-1)^3 < m$, what comes from assuming $3^m < m$ and it's not clear to me.
I tried multiple ways to understand why this assumption was made, but I can't figure it out. My first assumption was that since $m \in S$ it's true that:
$m > 3^{m/3}$
and by consequence:
$m^3 > 3^{m}$
but it still does not prove $3^m < m$. Any help is very appreciated.
|
I already accepted @fleablood's answer. I'd just like to share my own proof after some work on it as further reference for other people working on this problem.
Theorem: $n \leq 3^{n/3}\text{ for }n \geq 0$
Proof:
Let C be the set of counterexamples to the theorem, namely
$$C ::= \{n \in \mathbb{N} | n > 3^{\frac{n}{3}}\}$$
For the purpose of a contradiction, assume $C$ is not empty. By the Well-ordering principle, there's a least element, $c \in C$. The $c$ element can't be:
*
*1, because $1<3^\frac{1}{3}$;
*2, because $2<3^\frac{2}{3}$;
*3, because $3=3^\frac{3}{3}$;
*4, because $4<4^\frac{4}{3}$.
Therefore, $c>4$.
If $c$ is the least counterexample for the theorem, $c-1$ must hold true, once it's less than $c$ itself. So:
$$
c-1 \leq 3^\frac{c-1}{3}\\
c-1^3 \leq 3^{c-1}\\
c-1^3 \leq (3^c)(3^{-1})\\
c-1^3 \leq \frac{3^c}{3}\\
3(c-1)^3 \leq 3^c\\
$$
$c \in C$, therefore, $c > 3^{\frac{c}{3}}$:
$$
c > 3^{\frac{c}{3}}\\
c^3 > 3^c\\
3^c < c^3
$$
So, we can state that $3(c-1)^3 \leq 3^c < c^3$ and that $3(c-1)^3 < c^3$. Now, by manipulation:
$$
3(c-1)^3 < c^3\\
3(c^3-3c^2+3c-1) < c^3\\
3c^3-9c^2+9c-3 < c^3\\
2c^3-9c^2+9c-3 < 0\\
2c^3+9c < 9c^2+3\\
$$
Dividing both sides by $c^2$, we get:
$$
2c+\frac{9}{c}<9+\frac{3}{c^2}\\
$$
As stated before, $c>4$, so:
$$
c>4\\
c^2>16\\
\frac{1}{c^2}<\frac{1}{16}\\
\frac{3}{c^2}<\frac{3}{16}\\
9+\frac{3}{c^2}<9+\frac{3}{16}\\
9+\frac{3}{c^2}<9.1875<10\\
9+\frac{3}{c^2}<10
$$
Back to the main inequality:
$$
2c+\frac{9}{c}<9+\frac{3}{c^2}<10\\
$$
Since $c \in \mathbb{N}$, we can assume $2c < 2c+\frac{9}{c}$:
$$
2c<2c+\frac{9}{c}<9+\frac{3}{c^2}<10\\
2c<10\\
c<5
$$
$c \in \mathbb{N}$, $c>4$ and $c<5$ is a contradiction and therefore $C$ must be empty. The theorem is true.
QED
|
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|
What am I doing wrong on this question about "varies directly" and "varies inversely"? I don't get this problem:
$R$ varies directly as $S$ and inversely as $T$. When $R = \frac{4}{3}, T = \frac{9}{14}$, $S = \frac{3}{7}$. Find $S$ when $R = \sqrt{48}$ and $T = \sqrt{75}$.
My attempt:
We know
$$ R = c_1 S$$
and
$$ R = \frac{c_2}{T}$$
Plugging things in, we get $c_1 = \frac{28}{9}$ and $c_2 = \frac{6}{7}$. Now, we just solve for $S$.
$$ \sqrt{48} = \frac{6}{7} S$$
$$ 4 \sqrt{3} = \frac{6}{7} S$$
$$ 2 \sqrt{3} = \frac{3}{7} S$$
$$S = \frac{14}{3} \sqrt{3} $$
But, as far as I know, the correct answer appears to be $30$. What am I doing wrong? Am I getting wrong the definition of "directly" and "inversely"?
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You can’t treat the relationship between $R$ and $S$ and the relationship between $R$ and $T$ as if they were independent: the hypothesis actually means that
$$R=c\cdot\frac{S}T$$
for some constant $c$.
Thus, the relationship $R=c_1T$ holds only when $T$ is held constant, and the relationship $R=\frac{c_2}T$ holds only when $S$ is held constant.
You know that
$$\frac43=c\cdot\frac{3/7}{9/14}=c\cdot\frac37\cdot\frac{14}9=\frac23\,,$$
so
$$c=\frac{4/3}{2/3}=2\,.$$
When $R=\sqrt{48}$ and $T=\sqrt{75}$ we have
$$\sqrt{48}=\frac{2S}{\sqrt{75}}\,,$$
so
$$S=\frac12\sqrt{48\cdot75}=30\,.$$
|
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|
If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$. If $x, y, z \in (0, 1)$ and $x+y+z = 2$, prove that $8(1-x)(1-y)(1-z) \leq xyz$.
I found this in a Facebook group.
I start by doing the math in the LHS:
$8(1-x-y+xy-z+xz+yz-xyz) = 8(1-2+xy+xz+yz-xyz) = 8(xy+xz+yz-1-xyz)$.
Then we set $z=2-x-y$ and replace:
I get LHS = $x^2y+xy^2+2(x+y)-x^2-y^2-3xy-1$ but don't know how to proceed.
Thank you for your help.
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as $1-x=\frac{1}{2}(y+z-x)$ and similarly for others we have to prove $$(x+y-z)(y+z-x)(x+z-y)\le xyz$$ which is schur's inequality
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|
Analytic Proof of Wilson's Theorem I'm trying to find a reference for the following (analytic) proof of Wilson's Theorem. I would appreciate it immensely if some of you knew and told me the person who first came up with it. Here is the proof:
We start by considering the Maclaurin series of the function $f(x)=\ln \left( \frac{1}{1-x} \right)$, that is,
$\displaystyle \ln \left( \frac{1}{1-x} \right)= \sum_{n=1}^{\infty} \frac{x^n}{n} \tag*{}$
for $x \in [-1, 1).$
By simply using the definition of $\ln$ on the equation above, we get
$\displaystyle e^{x+\frac{x^2}{2}+\frac{x^3}{3}+ \cdots}=\frac{1}{1-x}=1+x+x^{2}+ \cdots + x^{p} + \cdots \tag{1}$
We can then rewrite the leftmost side of $(1)$ as
$\displaystyle \begin{align} e^{x+\frac{x^2}{2}+\frac{x^3}{3}+…} & = e^{x}e^{\frac{x^2}{2}}e^{\frac{x^3}{3}} \cdots \\ & = \prod_{n=1}^{\infty} e^{\frac{x^n}{n}} \\ & = \prod_{n=1}^{\infty} \sum_{k=0}^{\infty} \frac{\left( \frac{x^n}{n} \right)^k}{k!} \\ & = \left( 1+\frac{x}{1!}+\frac{x^2}{2!}+ \cdots \right) \cdot \left( 1+\frac{x^{2}/2}{1!}+\frac{(x^{2}/2)^2}{2!} + \cdots \right) \cdots \left(1+\frac{x^{p}/p}{1!}+\frac{(x^{p}/p)^2}{2!}+ \cdots \right) \cdots \\ & = 1+\frac{x}{1!}+x^{2} \left( \frac{1}{2!} + \frac{1}{2}\right)+x^{3} \left( \frac{1}{3!} + \frac{1}{1!} \frac{1/2}{1!} + \frac{1/3}{1!} \right) + \cdots + x^{p} \left( \frac{1}{p!} + \frac{1}{(p-2)!} \frac{1/2}{1!} + \cdots + \frac{1/p}{1!} \right)+ \cdots \end{align} \tag{2}$
Combining the rightmost sides of $(1)$ and $(2)$ together gives us
$\displaystyle 1+x+x^{2}+ x^{3}+ \cdots + x^{p} + \cdots= 1+\frac{x}{1!}+x^{2} \left( \frac{1}{2!} + \frac{1}{2}\right)+x^{3} \left( \frac{1}{3!} + \frac{1}{1!} \frac{1/2}{1!} + \frac{1/3}{1!} \right) + \cdots + x^{p} \left( \frac{1}{p!} + \frac{1}{(p-2)!} \frac{1/2}{1!} + \cdots + \frac{1/p}{1!} \right)+ \cdots , \tag{3}$
from which follows that the coefficient of $x^{p}$ on the RHS of $(3)$ is $1$. That is,
$\displaystyle \frac{1}{p!}+\frac{1}{(p-2)!} \frac{1/2}{1!}+ \cdots + \frac{1/p}{1!} = 1. \tag{4}$
Notice that such a coefficient is of the form $\frac{1}{p!}+\frac{r}{s}+\frac{1}{p}$, where $\frac{r}{s}$ is the sum of a finite number of rationals, none of which has the factor $p$ in its denominator. Therefore, if $\frac{r}{s}$ is irreducible, then $p$ doesn’t divide $s$. We then proceed by rewriting $(4)$ as
$\displaystyle \frac{1}{p!}+\frac{r}{s}+\frac{1}{p}=1. \tag*{}$
Equivalently,
$\displaystyle 1-\frac{r}{s}=\frac{1}{p!}+\frac{1}{p}=\frac{1+(p-1)!}{p!}, \tag*{}$
and
$\displaystyle (s-r) \cdot (p-1)! = \frac{s(1+(p-1)!)}{p}. \tag*{}$
Hence, $\frac{s(1+(p-1)!)}{p}$ is an integer. Moreover, since $p$ doesn’t divide $s$, $\frac{1+(p-1)!}{p}$ is also an integer. Equivalently, $(p-1)! \equiv -1 \text{ (mod } p)$, as desired.
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Here is a related argument with power series.
For $|x| < 1$ there is an identity
$$
e^x = \prod_{n \geq 1} (1 - x^n)^{-\mu(n)/n}.
$$
This can also be viewed as an identity of formal power series, and checked by verifying that the logarithmic derivatives ($f \leadsto f'/f$) of both sides are the same and that they start with the same constant term.
If we remove each factor on the right side where $p \mid n$, we get a different identity:
$$
e^{x + x^p/p + \cdots + x^{p^k}/{p^k} + \cdots} = \prod_{\substack{n \geq 1\\ p\nmid n}} (1 - x^n)^{-\mu(n)/n}.
$$
Look at the coefficient of $x^p$ on both sides. On the left side, it is the coefficient of $x^p$ in $e^xe^{x^p/p}$, which is $1/p! + 1/p = (1 + (p-1)!)/p!$. On the right side, the coefficients of $(1 - x^n)^{\pm 1/n}$ are rational with no $p$ in the denominator when $p \nmid n$, so all coefficients on the right are rational with no $p$ in the denominator. Thus $(1 + (p-1)!)/p!$ is rational with no $p$ in the denominator, which means $p \mid (1 + (p-1)!)$, so $(p-1)! \equiv -1 \bmod p$.
This argument can be found in books on $p$-adic analysis where $p$-adic integrality of all the coefficients of the Artin-Hasse exponential is discussed.
This is related to the identity
$$
\log\left(\frac{1}{1-x}\right) = \sum_{n \geq 1} \frac{x^n}{n}
$$
by extracting from each $n$ the highest power of $p$ dividing it:
$$
\log\left(\frac{1}{1-x}\right) = \sum_{\substack{n \geq 1\\ p \nmid n}} \frac{g(x^n)}{n}
$$
where $g(x) = \sum_{k \geq 0} x^{p^k}/p^k$, so
$$
g(x) = \sum_{\substack{n \geq 1}{p \nmid n}} \frac{\mu(n)}{n}\log\left(\frac{1}{1-x^n}\right).
$$
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|
Find the value of $k$ in $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+k$ It is also given that $abc = 1$.
I used AM-GM inequality to reach till
$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{\frac{bc}{a}} + \sqrt{\frac{ac}{b}} + \sqrt{\frac{ab}{c}} $
How to go further
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We will prove the homogeneous inequality (where we used $abc=1$) $$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[6]{abc}$$ Since equality is attainable for $a=b=c$, this would show that the desired $k$ is $3$. Observe that, in virtue of AM-GM $$\frac{\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{b}}+\frac{c}{\sqrt{a}}+\frac{a}{\sqrt{c}}}6\geqslant \sqrt[6]{abc}\iff\frac{\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{b}}+\frac{c}{\sqrt{a}}+\frac{a}{\sqrt{c}}}2\geqslant 3\cdot \sqrt[6]{abc} $$
Also, AM-GM yields $$\frac{2\cdot\frac{a}{\sqrt{b}}+2\cdot \frac{a}{\sqrt{c}}+\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{a}}}{6}\geqslant \sqrt[6]{a^3}=\sqrt{a}$$ Adding the first inequality and the cyclic variations of the latter one, we arrive at the desired inequality.
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Equivalence class for the following relation A relation of $\mathbb{R}$ is defined as $a\sim b : a^4-b^2=b^4-a^2$
Show that $\sim$ is equivalence relation (I have done this part)
Determine the equivalence class $[-1]_\sim$
Prove or disprove: Every equivalence class in $\mathbb{R}/\sim$ contains exactly 2 real numbers.
I am facing difficulty in second and third part of the question.
For the second one I think it is $[-1]_\sim=\{-1,1\}$ as $-1\sim x: (-1)^4-x^2 =x^4-(-1)^2 $ so both $-1$ and $1$ are proving the equality.
I think the third statement is true but how can I prove it?
Please help.
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Follow definitions.
$[-1] = \{b\in \mathbb R|(-1)^2 - b^2=b^4 -(-1)^2\}=$
$\{b\in \mathbb R| 1-b^2 = b^4 -1\}=$
$\{b\in \mathbb R| b^4+b^2-2 = 0\}=$
$\{b\in \mathbb R| (b^2+2)(b-1)(b+1)=0\}=$
$\{b\in \mathbb R| b^2 = -2\lor b=1\lor b=-1\}=$
$\{1,-1\}$
As for the second:
$[a] = \{b\in \mathbb R| a^4-b^2 = b^4 -a^2\}=$
$\{b \in \mathbb R|b^4+b^2 -(a^4+a^2) = 0\}$
Now potentially $b^4 + b^2-(a^4 + a^2)=0$ may have up to four solutions.
Indeed in the case of $[-1]$ we found $b \in [-1]$ if $(b^2 + 2)(b+1)(b-1)=0$ had two elements because $b^2 + 2=0$ is impossible.
We can use quadratic formula to solve $b^4 + b^2 -(a^4+a^2)$ for $b$ but it's probably easier to factor:
$b^4 + b^2-(a^4 + a^2)=0$
$(b^4 -a^4) + (b^2 - a^2) = 0$
$(b^2 +a^2)(b^2 - a^2)+(b^2 -a^2) =0$
$(b^2 + a^2 + 1)(b^2-a^2)=0$
$(b^2 +a^2 + 1)(b-a)(b+a)=0$.
Well, $b^2 + a^2 + 1 > 0$ so the only solutions are $b=\pm a$ and that will always be the case that $[a] = \{a,-a\}$ so.....
but what if $a = 0$. Then $[0] = \{0,-0\} =\{0\}$ has only one element.
And we can show that directly $0^4 -b^2 = b^4 - 0^2$
$b^4 + b^2 = 0$ so $b^2(b^2 + 1)=0$ so $b^2 = 0$ or $b^2 =-1$ so $b = 0$.
So the third is false but only because $[0]$ has one element. For all others it is true.
Not $\sim$ is exactly the same as the relationship $a = \pm b$.
Note: $a^4 -b^2 = b^4 - a^2 \iff a =\pm b$
In hindsight had we proven that from the very beginning we wouldn't have need to do so much work.
Live and learn.
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|
how to compute a series whose terms are a rational function times an exponential function? How can I compute the following series?
\begin{equation}
\sum_{n=1}^\infty\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n
\end{equation}
I manipulated the term and got
\begin{equation}
\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n = \left(\frac{11}{14n-7}-\frac{1}{n+3}\right)\left(-\frac{1}{3}\right)^n\text{.}
\end{equation}
But I don't know what to do next.
|
$$S=\sum_{n=1}^\infty\frac{n+10}{2n^2+5n-3}\left(-\frac{1}{3}\right)^n$$
$$S=\sum_{n=1}^{\infty} \left(\frac{3}{2n-1}-\frac{1}{n+3}\right)(-3^{-1})^{n}=S_1-S_2$$
$$S_2=-27\sum_{n=1}^{\infty} \frac{(-3^{-1})^{n+3}}{n+3}$$
Use $\ln(1+z)=\sum_{k=1}^{\infty} (-1)^{k-1} \frac{z^k}{k}$
$$=-27\sum_{m=4}^{\infty}\frac{(-1)^m(3^{-1})^m}{m}=\frac{-47}{6}+27\ln(4/3)$$
Next use $\tan^{-1} x=\sum_{k=1}^{\infty} (-1)^{k-1}\frac{z^{2k-1}}{2k-1}$
$$S_1=\sqrt{3}\sum_{n=1}^{\infty} (-1)^n \frac{(1/\sqrt{3})^{2n-1}}{2n-1}=-\sqrt{3} \tan^{-1}(1/\sqrt{3})=\frac{-\pi}{2 \sqrt{3}}$$
Finally $$S=\frac{-\pi}{2 \sqrt{3}}+\frac{47}{6}-27\ln(4/3)$$
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|
Calculate limit using Maclaurin series $$\lim_{x \to 0}\frac{1-(\cos{x})^{\sin{x}}}{x^3}$$
So what I have tried is defining three functions, $f(x)=\cos{x}, g(x) = \sin{x}, h(x) = x^3$
What I did next was do Maclaurin expansion for three functions up until the third order:
$f(x) = 1 - \frac{x^2}{2} + o(x^3)$
$g(x) = x - \frac{x^3}{3} + o(x^3)$
$h(x) = x^3 + o(x^3)$
So when I plug in these values I get:
$$\lim_{x \to 0}\frac{1-(1-\frac{x^2}{2}+o(x^3))^{x-\frac{x^3}{3} + o(x^3)}}{x^3}$$
So after plugging in $x=0$ I get $\frac{0}{0}$ and when I try to apply L'Hôpital's rule I find myself in a situation where I need to calculate the derivative of the denominator several times (which would take too long). I'm positive that I'm correct until this point (I checked on wolframalpha, the result is 1/2).
Does anyone know how to proceed from here? Do I need to calculate the derivative of the denominator several times or is there a smarter way?
|
We can derive the limit using series expansion as follows:
\begin{align*}
\color{blue}{\lim_{x\to 0}}&\color{blue}{\frac{1-\left(\cos x\right)^{\sin x}}{x^3}}\\
&=\lim_{x\to 0}\frac{1-\exp\left((\sin x)\ln\left(\cos x\right)\right)}{x^3}\\
&=\lim_{x\to 0}\frac{1-\exp\left(\left(x-\frac{x^3}{6}+\mathcal{O}\left(x^4\right)\right)
\ln\left(1-\frac{x^2}{2}+\mathcal{O}\left(x^4\right)\right)\right)}{x^3}\\
&=\lim_{x\to 0}\frac{1-\exp\left(\left(x-\frac{x^3}{6}+\mathcal{O}\left(x^4\right)\right)
\left(-\frac{x^2}{2}+\mathcal{O}\left(x^4\right)\right)\right)}{x^3}\\
&=\lim_{x\to 0}\frac{1-\exp\left(-\frac{x^3}{2}+\mathcal{O}\left(x^4\right)\right)}{x^3}\\
&=\lim_{x\to 0}\frac{1-\left(1-\frac{x^3}{2}+\mathcal{O}\left(x^4\right)\right)}{x^3}\\
&\,\,\color{blue}{=\frac{1}{2}}
\end{align*}
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|
Solving an equation that contain floor function Let suppose I have three positive integers $a, b,c$ and one unknown $x$ ($x$ is also a positive integer). Solve for the smallest $x$ that satisfies this equation
$$\left \lfloor\frac{x}{a} \right\rfloor + \left \lfloor \frac{x}{b} \right\rfloor \geq c$$
where $\lfloor x\rfloor$ is the floor function
For instance let $a = 1$, $b=1$ and $c=5$ in this case
$$\left \lfloor x \right\rfloor + \left \lfloor x \right\rfloor \geq 5$$
here the smallest $x$ is $3$.
For another case, if $a=3$ and $b=5$ and $c=4$ we obtain
$$\left \lfloor\frac{x}{3} \right\rfloor + \left \lfloor \frac{x}{5} \right\rfloor \geq 4$$
here the smallest $x$ is $9$
My question is, can we find the smallest $x$ exactly ?
|
Assume that $a,b,c$ are positive
.
Let $y=\frac{abc}{a+b}$. Then $$\frac{y}{a}+\frac{y}{b}=c.$$
Therefore $$\left \lfloor\frac{y}{a} \right\rfloor + \left \lfloor \frac{y}{b} \right\rfloor$$ will be either $c$ or $c-1$.
If it is $c$
Then $y$ is a solution. We will still have a solution when $y$ is increased, until $y$ becomes the next multiple of either $a$ or $b$.
If it is $c-1$
Then increase $y$ until it becomes the next multiple of either $a$ or $b$. This will be the solution $x$ unless it is a multiple of both $a$ and $b$ in which case there is no solution. Given that we have a solution $x$, then we can increase it as above until the next multiple of either $a$ or $b$ is reached.
Example
Solve $\left \lfloor\frac{x}{5} \right\rfloor + \left \lfloor \frac{x}{7} \right\rfloor=8.$
We obtain $y=70/3$ and then $\left \lfloor\frac{y}{5} \right\rfloor + \left \lfloor \frac{y}{7} \right\rfloor=7.$
As $y$ is increased, the next two multiples of $5$ or $7$ are $25$ and $28$. The solution is $$25\le x<28.$$
Is this the sort of practical answer you were looking for?
Summary
The solutions can be completely expressed in terms of how $\frac{abc}{a+b}$ fits between successive terms of the ordered sequence of elements of $aZ\bigcup bZ$.
A formula for the revised question
A formula for the minimal $x$ is obtained as in the above answer and can be expressed as follows.
If $\frac{abc}{a+b}$ is in $aZ\bigcup bZ$ then that is the minimal $x$.
Otherwise $x=$ min$ (a\left \lfloor\frac{bc}{a+b}\right\rfloor+a,b\left \lfloor\frac{ac}{a+b}\right\rfloor+b)$.
|
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|
I found this $\binom{2p^2}{k}\equiv 2\binom{p^2}{k}\pmod {p^2},1\le ktoday,when I deal this problem,I found this result:
let $p>3$ be prime number, show that
$$\binom{2p^2}{k}\equiv 2\binom{p^2}{k}\pmod {p^2},1\le k<p^2$$
I found this when $k=1,2,3$ it is easy to prove it,because
$k=1$
$$\binom{2p^2}{k}=\binom{2p^2}{1}=2p^2\equiv 0=2\binom{p^2}{1}\pmod {p^2}$$
$k=2$
$$\binom{2p^2}{k}=\binom{2p^2}{2}=p^2(2p^2-1)\equiv 0=2\binom{p^2}{2}\pmod {p^2}$$
$k=3$
$$\binom{2p^2}{k}=\binom{2p^2}{3}=\dfrac{2p^2(2p^2-1)(2p^2-2)}{6}\equiv 0=2\binom{p^2}{3}\pmod {p^2}$$
but for all postive integer $k$,I think is also right.But I can't prove it
|
Let $n$ be a positive integer and $1\le k<n$. By a straightforward counting argument we have the identity
$$\binom{2n}{k}=\sum_{a,b\ge 0, a+b=k}\binom{n}a\binom{n}b.$$
Setting $n=p^2$ and rewriting the equation gives
$$\binom{2p^2}k=2\binom{p^2}{k}+\sum_{a,b\ge 1, a+b=k}\binom{p^2}a\binom{p^2}b.$$
It therefore suffices to show that $p\mid\binom{p^2}a$ for all $1\le a<p^2$ and this follows immediately by expanding $x^{p^2}+1=(x+1)^{p^2}$ modulo $p$ using the binomial theorem.
|
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|
On the integral $\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$ I came across this integral
$$\mathcal{J} = \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}}$$
According to W|A it equals $\frac{1}{2}$. However, I cannot find a way to crack it. It smells like a Beta integral , but I do not see any obvious subs. One could start by setting $u=x^2$ but there is no clear path after that.
A promising way might be the following
\begin{align*}
\int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{3x^4+2x^2+3}} &= \int_{0}^{1} \frac{\mathrm{d}x}{\sqrt{\left ( 3x^2 +2 \right ) x^2 +3}}\\
&=\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{3\left ( x^2+ \frac{1}{3} \right )^2 + \frac{8}{3}}} \\
&= \cdots
\end{align*}
A clever trigonometric sub might clear things but I don't see something. On the other hand , I don't the theory of elliptic integrals is needed here nor complex analysis ( would be interesting to see a solution using contours, though )
So, any ideas how to evaluate it?
P.S: Are there techniques available for these type of problems?
|
A trigonometric substitution does indeed help here: Consider $f \colon [0,1] \to \mathbb{R},$
\begin{align}
f (a) &= \int \limits_0^1 \frac{\mathrm{d} x}{\sqrt{1 + 2 a x^2 + x^4}} = \int \limits_0^1 \frac{\mathrm{d} x}{\sqrt{(1 + x^2)^2 - 2 (1-a) x^2}} \\
&\!\!\!\!\!\!\!\!\stackrel{x = \tan\left(\frac{t}{2}\right)}{=} \frac{1}{2} \int \limits_0^{\pi/2} \frac{\mathrm{d} t}{\sqrt{1 - 2 (1-a) \sin^2 \left(\frac{t}{2}\right) \cos^2 \left(\frac{t}{2}\right)}} = \frac{1}{2} \int \limits_0^{\pi/2} \frac{\mathrm{d} t}{\sqrt{1 - \frac{1-a}{2} \sin^2(t)}} \\
&= \frac{1}{2} \operatorname{K} \left(\sqrt{\frac{1-a}{2}}\right) \, .
\end{align}
Your integral is
$$ \mathcal{J} = \frac{1}{\sqrt{3}} f \left(\frac{1}{3}\right) = \frac{\operatorname{K} \left(\frac{1}{\sqrt{3}}\right)}{2 \sqrt{3}} > \frac{1}{2} \, .$$
|
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|
$\lim\limits_{n\to\infty}n\big(\sum_{k=1}^n\frac{k^2}{n^3+kn}-\frac{1}{3}\big)$? calculate $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right).$$
I got it $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}.$$
Use Squeeze theorem we have
$$\frac{1}{n+1}\sum\limits_{k=1}^n(\frac{k}{n})^2<\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}<\dfrac{1}{n}\sum\limits_{k=1}^n(\frac{k}{n})^2$$
So $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\int_0^1x^2\mathrm{d}x=\frac{1}{3}.$$
Use $$\lim\limits_{n\to\infty}n\left(\int_0^1f(x)\mathrm{d}x-\frac{1}{n}\sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)\right)=\frac{f(0)-f(1)}{2}.$$
Hence $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right)=\frac{1}{2}.$$
If our method is correct, is there any other way to solve this problem? Thank you
|
There is another using generalized haromonic numbers since
$$\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=n^3 \left(H_{n^2+n}-H_{n^2}\right)+\frac{1}{2} \left(-2 n^2+n+1\right)$$
$$S_n=-\dfrac{1}{3}+\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\frac{1}{6} \left(-6 n^2+3n+1\right)+n^3 \left(H_{n^2+n}-H_{n^2}\right)$$ Using asymptotics
$$S_n=\frac{1}{4 n}-\frac{2}{15 n^2}+\frac{1}{12 n^3}+O\left(\frac{1}{n^4}\right)$$
$$nS_n=\frac{1}{4 }-\frac{2}{15 n}+\frac{1}{12 n^2}+O\left(\frac{1}{n^3}\right)$$ and we do not agree !
Try with $n=10$; the exact value is
$$10 S_{10}=10 \times \frac{140325051799081}{5909102214621606}\sim 0.237473$$ while the approximation gives $\frac{19}{80} \sim 0.237500$.
|
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|
Integration of $\int_0^{\infty}\frac{e^{(\lambda +is)t}-e^{-(\lambda +is)t}}{e^{\pi t}-e^{-\pi t}}dt$ Integration of $$\int_0^{\infty}\frac{e^{(\lambda +is)t}-e^{-(\lambda +is)t}}{e^{\pi t}-e^{-\pi t}}dt$$
I am thinking about this integral from last half and hour but still can't figure it out how to solve this. what i was think is to make this in form
$$\int_{0}^{\infty}\frac{\sinh(\lambda+is)t}{\sinh (\pi t)} dt$$ but still this will get me nowhere. please just give me a hint how to proceed this.
Thank you.
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Considering $\displaystyle \mathcal{I}(\alpha, \beta) = \int_{0}^{\infty} \frac{\sinh\alpha z}{\sinh \beta z}\,\mathrm{d}z$, we show that $\displaystyle \mathcal{I}(\alpha, \beta) = \frac{\pi}{2\beta} \tan\left(\frac{\alpha \pi}{2\beta}\right)$.
Then your integral is $\displaystyle \mathcal I(\lambda+is, \pi)$. Starting with $z \mapsto \frac{1}{\beta} \log(z)$, we have:
\begin{aligned} \displaystyle \mathcal{I}(\alpha, \beta) & = \frac{1}{\beta} \int_0^1 \frac{z^{\frac{\alpha}{\beta}} -z^{-\frac{\alpha}{\beta}}}{z^2-1}\;{dz} \\& = \frac{1}{\beta} \int_0^{1} \left( z^{-\frac{\alpha}{\beta}}-z^{\frac{\alpha}{\beta}}\right) \sum_{k \ge 0}z^{2k} \;{dz}\\& = \frac{1}{\beta} \sum_{k \ge 0}\int_0^1 z^{-\frac{\alpha}{\beta}+2k} -z^{\frac{\alpha}{\beta}+2k}\;{dz} \\& = \frac{1}{\beta} \sum_{k \ge 0} \frac{1}{1-\frac{\alpha}{\beta}+2k}-\frac{1}{1+\frac{\alpha}{\beta}+2k} \\& = \frac{1}{\beta} \sum_{k \ge 0} \frac{2\alpha /\beta }{(1+2 k)^2-(\alpha/\beta)^2} \\& = \frac{1}{2\beta} \sum_{k ~ \text{odd} } \frac{4\alpha/\beta}{k^2-(\alpha/\beta)^2} \end{aligned}
Consider the series $\displaystyle \displaystyle \pi\cot(\pi z) = \frac{1}{z}+ \sum_{n \ge 1} \frac{2z}{z^2-n^2} $. Map $\displaystyle z \mapsto \frac{z}{2} $ and divide by two then we have:
$\displaystyle \displaystyle \frac{\pi}{2}\cot\left(\frac{\pi}{2} z\right) = \frac{1}{z}+\frac{1}{2}\sum_{n \ge 1} \frac{z}{(z/2)^2-n^2} $$\displaystyle \displaystyle = \frac{1}{z}+ \frac{1}{2}\sum_{n \ge 1} \frac{4z}{z^2-(2n)^2} = \frac{1}{z}+\frac{1}{2}\sum_{n ~ \text{even}} \frac{4z}{z^2-n^2}$
Thus $\displaystyle \displaystyle \frac{1}{z}+\sum_{n \ge 1} \frac{2z}{z^2-n^2} -\frac{1}{z}-\frac{1}{2} \sum_{n ~ \text{even}} \frac{4z}{z^2-n^2}$ $\displaystyle \displaystyle =\pi\cot(\pi z)-\frac{\pi}{2}\cot\left(\frac{\pi}{2} z\right) = -\frac{\pi}{2} \tan\left(\frac{z\pi}{2}\right)$
Hence $\displaystyle \displaystyle \frac{1}{2} \sum_{n \textrm{ odd}} \frac{4z}{z^2-n^2} = -\frac{\pi}{2}\tan\left(\frac{\pi z}{2}\right)$ and so $\displaystyle \displaystyle \mathcal I(\alpha, \beta)=\frac{1}{2\beta} \sum_{k ~ \text{odd} } \frac{4\alpha/\beta}{(\alpha/\beta)^2-k^2} = \frac{\pi}{2\beta} \tan\left(\frac{\alpha \pi}{2\beta}\right).$
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|
Proving $\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq\frac{3\sqrt{3}}2$ for $a$, $b$, $c$, $s$ the sides and semi-perimeter of a triangle
If $a, b, c$ are the lengths of the sides of a triangle and s is the semiperimeter, prove that:
$$\sum_{cyc}\frac{\sqrt{s(s-a)}}{a}\geq \frac{3\sqrt{3}}{2}$$
My attempt: $$\sum_{cyc} cos \,\frac{A}{2}=\sum _{cyc} \sqrt{\frac{s(s-a)}{bc}}\leq \frac{3 \sqrt{3}}{2}\leq \sum_{cyc}\frac{\sqrt{s(s-a)}}{a}$$. As the first inequality is obvious, it is enough to show that $$\sum _{cyc} \sqrt{\frac{s(s-a)}{bc}}\leq \sum_{cyc}\frac{\sqrt{s(s-a)}}{a}$$. We prove that $$\sqrt{\frac{s(s-a)}{bc}} \leq\frac{\sqrt{s(s-a)}}{a}$$ or equivalently $a^{2}\leq bc$. Similary $b^{2}\leq ac$ and $c^{2}\leq ab$ and adding the inequalities we get $(a-b)^{2}+(b-c)^{2}+ (c-a)^{2}\leq 0$.
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Let $a = y+z,\,b = z+x,\,c = x+y$ for $x,\,y,\,z>0,$ we have
$$\sum \frac{\sqrt{s(s-a)}}{a} = \sum \frac{\sqrt{\frac{a+b+c}{2}\left(\frac{a+b+c}{2}-a\right)}}{a} = \sum \frac{\sqrt{x(x+y+z)}}{y+z}.$$
We will show that
$$\sum \frac{\sqrt{x(x+y+z)}}{y+z} \geqslant \frac{3\sqrt 3}{2}.$$
Indeed, suppose $x+y+z=3$ then the inequality become
$$\sum \frac{\sqrt{x}}{3 - x} \geqslant \frac{3}{2}.$$
We have
$$\left(\frac{\sqrt{x}}{3 - x}\right)^2 - \left(\frac{x}{2}\right)^2 = \frac{x(4 - x)(x-1)^2}{4(3-x)^2} \geqslant 0.$$
So
$$ \frac{\sqrt{x}}{3 - x} \geqslant \frac{x}{2},$$
or
$$ \sum \frac{\sqrt{x}}{3 - x} \geqslant \frac{x+y+z}{2} = \frac32.$$
The proof is completed.
|
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Inequality $a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0$ Let $a,b,c$ be the lengths of the sides of a triangle. Prove that:
$$a^2b(a-b) + b^2c(b-c) + c^2a(c-a) \geq 0.$$
Now, I am supposed to solve this inequality by applying only the Rearrangement Inequality.
My Attempt:
W.L.O.G. let $a \geq b \geq c.$
Then, L.H.S. = $a \cdot ab(a-b) + b \cdot bc(b-c) + c \cdot ac(c-a)$.
There are $2$ cases to consider:
Case $1$:
$ab(a-b) \geq bc(b-c) \geq ac(c-a)$. Then, the L.H.S. is similarly sorted, so:
\begin{align}
\text{L.H.S.} \ & \geq c \cdot ab(a-b) + a \cdot bc(b-c) + b \cdot ac(c-a) \\
& = abc(a-b) + abc(b-c) + abc(c-a) \\
& = a^2bc - ab^2c + ab^2c - abc^2 + abc^2 - a^2bc \\
& =0.
\end{align}
Case $2$: $bc(b-c) \geq ab(a-b) \geq ac(c-a)$. But Rearrangement doesn't seem to work for this case. Any hints on how to proceed?
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Method 1:As Chrystomath used Ravi's substituition we are left to prove $$\frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}\ge x+y+z$$ Now
*
*$x\ge y\ge z\implies x^2\ge y^2\ge z^2,\frac{1}{x}\le \frac{1}{y} \le \frac{1}{z}$ $$x^2\left(\frac{1}{x}\right)+y^2\left(\frac{1}{y}\right)+z^2\left(\frac{1}{z}\right)\le \frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}$$ $$\iff x+y+z\le \frac{y^2}{z}+\frac{x^2}{y}+\frac{z^2}{x}$$
*$x\ge z\ge y$ is analogous can you finish it off...
Method 2: As I have already given an answer by rearrangement here is a far more reliable way
WLOG $c=max(a,b,c)$
so let $a=x,b=x+u,c=x+u+w$ where $x,u,w \ge 0$ also by triangle inequality $w< x$ we have to prove $$-ux^2(x+u)-w{(x+u)}^2(x+u+w)+{(x+u+w)}^2x(u+w)\ge 0$$ $$\iff u^3(x-w)+u^2(x^2-w^2)+w^3x+w^2x^2+u^2x^2+uwx^2\ge 0$$ which is obvious!!
If $a=x+u,b=x,c=x+u+w$ then again the inequality is after expanding $$u^3x+u^2wx+u^2x^2+2uw^2x+w^3z+w^2x^2\ge 0$$ which is true
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|
a random variable has the density $f\left(x\right)=a+bx^{2}$ . Determine and b so that its mean will be 2/3 A random variable has the density $f\left(x\right)=a+bx^{2}$ with $0<x<1$. Determine and b so that its mean will be 2/3.
I'm a little confued trying to get a and b. This is what I already tried:
$$P\left(0<X<1\right)=\int_{0}^{1}ax^{2}dx+\int_{0}^{1}bx^{4}dx$$
$$P\left(0<X<1\right)=a\int_{0}^{1}x^{2}dx+b\int_{0}^{1}x^{4}dx=a\left(\frac{x^{3}}{3}\right)_{0}^{1}+b\left(\frac{x^{5}}{5}\right)_{0}^{1}$$
$$P\left(0<X<1\right)=a\left(\frac{1}{3}\right)+b\left(\frac{1}{5}\right)=\frac{a}{3}+\frac{b}{5}=\frac{5a+3b}{15}$$
$$E\left[X\right]=a\left(\frac{1}{3}\right)+b\left(\frac{1}{5}\right)=\frac{a}{3}+\frac{b}{5}=\frac{5a+3b}{15}$$
$$E\left[X\right]=\frac{2}{3}=\frac{5a+3b}{15}$$
$$10=5a+3b$$
I'm assuming x as a continuous random variable
In this case shouldn't be a=0 b=1?
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You are mixing up the $P(X\in(0,1))=1$ and $E(X)=\frac 2 3$. The other post presumably showed you the expected value correctly, so $P(X\in(0,1))=\int_{-1}^1 f(x) dx=\int_0^1(a+bx^2)dx=\left[ax+\frac{bx^3}{3}\right]_{x=0}^1=a+\frac b 3=1$
Solving, you obtain $a=\frac 1 3, b=2$.
To check:
$E(X)=\int_{-\infty}^\infty xf(x)dx=\int_0^1x(\frac 1 3 + 2x^2)dx=\left[\frac{x^2}{6}+\frac{x^4}{2}\right]_{x=0}^1=\frac 2 3$
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|
Prove that $2<\int^2_1ln(x^2+1)+x<3$ To show that $2<\int^2_1ln(x^2+1)+x$ , I took the derivative of the integrand and got that it's positive always in $[1,2]$, so its strictly increasing and the minimum in $[1,2]$ is at $x=1$, so $ln(2)+1<ln(x^2+1)+x$, and $ln(2)<ln(e)=1$ so: $ln(2)+1<2< ln(x^2+1)+x$ in $(1,2]$.
EDIT: I just realized that the lower bound doesn't work too.
About the upper bound I tried to substitute $2$ but it becomes $ln(5)+2>3$, so it doesn't really help.
I would appreciate any help. Thanks in advance.
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I began to solve this before Bacha's solution appeared, so I did not delete.
Hope I'm correct.
$$\int\ln(x^2+1)dx=(*)\\let\:f=\ln(x^2+1),g=x\\then\:df=\frac{2x}{x^2+1}dx,dg=dx\\\text{using integrating by parts where }\int fdg=dg-\int gdf\text{ we get}\\(*)=x\ln(x^2+1)-\int\frac{2x^2}{x^2+1}dx=x\ln(x^2+1)-2\int \left(1-\frac{1}{x^2+1}\right)dx=\\=x\ln(x^2+1)-2x+2\arctan(x)+C$$
And now calculate
$$\int\limits_1^2(\ln(x^2+1)+x)dx=\\=2\ln(2^2+1)-2*2+2\arctan(2)-\ln(1^2+1)+2*1-2\arctan(1)+\frac{1}{2}2^2-\frac{1}{2}1^2=\\=\ln\frac{25}{2}-\frac{1}{2}-\frac{\pi}{2}+2\arctan(2)$$
If we use calculator then $\ln\frac{25}{2}-\frac{1}{2}-\frac{\pi}{2}+2\arctan(2)\approx2.6692$. But if you want a more "honest" solution then we can try to estimate this from left and from right but it would hard... Nevertheless let's see how estimating would look.
From left:
$$\ln\frac{25}{2}-\frac{1}{2}-\frac{\pi}{2}+2\arctan(2)>\ln(e^2)-\frac{1}{2}-\frac{\pi}{2}+2\arctan(\sqrt{3})=2-\frac{1}{2}-\frac{\pi}{2}+\frac{2\pi}{3}=\\=1.5+\frac{\pi}{6}> 2$$
Ok.
From right:
$$\ln\frac{25}{2}-\frac{1}{2}-\frac{\pi}{2}+2\arctan(2)<\\<\left(\ln(e^2)+(\frac{25}{2}-e^2)*\frac{1}{e^2}\right)^{[2]}-\frac{1}{2}-\frac{\pi}{2}+2*\left(\arctan(\sqrt 3)+(2-\sqrt 3)*\frac{1}{\sqrt 3^2+1}\right)^{[1]}<\\<\left(2+5.5*\frac{1}{7}\right)-\frac{1}{2}-\frac{\pi}{2}+\left(\frac{2\pi}{3}+(2-\sqrt 3)*\frac{1}{2}\right)<\\<2.8-\frac{1}{2}-\frac{\pi}{2}+\frac{2\pi}{3}+(2-\sqrt 3)*\frac{1}{2}=2.3+\frac{\pi}{6}+(2-\sqrt 3)*\frac{1}{2}<2.3+\frac{\pi}{6}+0.3*\frac{1}{2}=\\=2.45+\frac{\pi}{6}<3$$
[1] $\forall x\geq 0:\:\arctan(x) + (x+\delta x)*\arctan'(x)\geq\arctan(x+\delta x)$ because $\forall y>x\geq 0:\: \arctan'(x)>\arctan'(y)$.
[2] Same as [1].
Ok, but too much work.
|
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|
Evaluate $\int_{0}^\infty\frac{2}{({M-1})!}\sqrt{\frac{a}{x}}K_1\left(2\sqrt{\frac{a}{x}}\right)x^{M-1}\frac{e^{-\frac{x}{b}}}{b^M}\,dx$ How to evaluate this integral or at least rewrite it as special (non-elementary) function step by step ?
$$\int_{0}^\infty\frac{2}{({M-1})!}\sqrt{\frac{a}{x}}K_1\left(2\sqrt{\frac{a}{x}}\right)x^{M-1}\frac{e^{-\frac{x}{b}}}{b^M}\,dx$$
Where $M$ is a positive integer that is $M=1,2,3,4,...$ , $a,b$ are two positive number and ${K_1}\left( . \right)$ is the modified Bessel function of order 1.
If you could provide some reference, I would be pretty much graceful !
Thank you for your enthusiasm !
|
It seems useful to rearrange the integral:
\begin{align}
I&=\int_{0}^\infty\frac{2}{({M-1})!}\sqrt{\frac{a}{x}}K_1\left(2\sqrt{\frac{a}{x}}\right)x^{M-1}\frac{e^{-\frac{x}{b}}}{b^M}\,dx\\
&=\frac{2b^{-M}\sqrt{a}}{({M-1})!}\int_{0}^\infty K_1\left(2\sqrt{\frac{a}{x}}\right)x^{M-3/2}e^{-\frac{x}{b}}\,dx\\
&=\frac{2}{({M-1})!}\left( \frac{a}{b} \right)^M\int_{0}^\infty K_1\left(\frac{2}{\sqrt s}\right)s^{M-3/2}e^{-\frac{a}{b}s}\,ds
\end{align}
which expresses that the integral is a function of $a/b$.
To obtain the Meijer function expression given by @MariuszIwaniuk in the comment, one may use the Mellin-Barnes representation for the modified Bessel function
\begin{equation}
K_{\nu}\left(z\right)=\frac{(\frac{1}{2}z)^{\nu}}{4\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma\left(t\right)\Gamma\left(t-\nu\right)(\tfrac{1}{2}z)^{-2t}\,dt
\end{equation}
valid for $c>\max(\Re\nu,0),|\arg z|<\pi/2$. Here, with $\nu=1,z=2/\sqrt s,c>1$, we obtain, by changing the order of integration
\begin{align}
I&=\frac{2}{({M-1})!}\left( \frac{a}{b} \right)^M\int_{0}^\infty K_1\left(\frac{2}{\sqrt s}\right)s^{M-3/2}e^{-\frac{a}{b}s}\,ds\\
&=\frac{1}{({M-1})!}\left( \frac{a}{b} \right)^M\frac{1}{2\pi i}
\int_{c-i\infty}^{c+i\infty}\Gamma\left(t\right)\Gamma\left(t-1\right)\,dt\int_{0}^\infty s^{t+M-2}e^{-\frac{a}{b}s}\,ds
\end{align}
The integral over $s$ gives directly a Gamma function,
\begin{equation}
I=\frac{1}{({M-1})!}\left( \frac{a}{b} \right)\frac{1}{2\pi i}
\int_{c-i\infty}^{c+i\infty}\Gamma\left(t\right)\Gamma\left(t-1\right)\Gamma(t+M-1)\left(\frac ba\right)^t\,dt
\end{equation}
From the definition of the Meijer function
\begin{equation}
{G^{m,n}_{p,q}}\left(z;\begin{array}{c}
a_1,\ldots,a_p\\
b_1,\ldots,b_q
\end{array} \right)=\frac{1}{2\pi\mathrm{i}}\int_{L%
}{\frac{\prod\limits_{\ell=1}^{m}\Gamma\left(b_{\ell}-t\right
)\prod\limits_{\ell=1}^{n}\Gamma\left(1-a_{\ell}+t\right)}{\left(\prod\limits_%
{\ell=m}^{q-1}\Gamma\left(1-b_{\ell+1}+t\right)\prod\limits_{\ell=n}^{p-1}%
\Gamma\left(a_{\ell+1}-t\right)\right)}}z^{t}\,dt
\end{equation}
we can use $z=b/a$, $a_1=1,a_2=2,a_3=2-M$, with $m=q=0, n=p=3$ and an integral along the vertical axis from $c-i\infty$ to $c+i\infty$, as the real parts of the the poles of $\Gamma\left(1-a_{\ell}+t\right)$ are all less than $1$. We deduce
\begin{equation}
I=\frac{1}{({M-1})!}\left( \frac{a}{b} \right)G^{0,3}_{3,0}\left(\frac{b}{a};\begin{array}{c}
1,2,2-M\\
\textrm{---}
\end{array} \right)
\end{equation}
The proposed expression can be retrieved by combining the identity
\begin{equation}
I=\frac{1}{({M-1})!}\left( \frac{a}{b} \right)G^{3,0}_{0,3}\left(\frac{a}{b};\begin{array}{c}
\textrm{---}\\
0,-1,M-1
\end{array} \right)
\end{equation}
and the second identity
\begin{equation}
I=\frac{1}{({M-1})!}G^{3,0}_{0,3}\left(\frac{a}{b};\begin{array}{c}
\textrm{---}\\
0,1,M
\end{array} \right)
\end{equation}
|
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|
Famous or common mathematical identities that yield $1$ To me the most common identity that comes to mind that results in $1$ is the trigonometric sum of squared cosine and sine of an angle:
$$
\cos^2{\theta} + \sin^2{}\theta = 1 \tag{1}
$$
and maybe
$$
-e^{i\pi} = 1 \tag{2}
$$
Are there other famous (as in commonly used) identities that yield $1$ in particular?
|
Here are two well-known infinite series whose sum is $1$:
$$\sum_{n=1}^{\infty} \frac{1}{2^n} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots = 1$$
$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \frac{1}{4 \cdot 5} + \frac{1}{5 \cdot 6} + \cdots = 1$$
The following series is less known, interesting although:
$$\sum_{n=1}^{\infty} \frac{1}{s_n} = \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{43} + \frac{1}{1807} + \cdots = 1$$
where denominators form Sylvester sequence: every term is equal to the product of all previous terms, plus one. For example
$$\begin{array}{rcl}
2& & \\
3 & = & 2+1\\
7 & = & 2 \cdot 3 +1 \\
43 & = & 2 \cdot 3 \cdot 7 +1 \\
1807 & = & 2 \cdot 3 \cdot 7 \cdot 43 +1 \\
\end{array}$$
and so on.
|
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|
Question on finding maximum of AM GM inequality How can i solve for minimum value of function $f(x)=\frac{x^2+x+1}{x^2-x+1} $ using AM GM inequality.
I am aware about the AM GM inequality, but am not able to split the terms into the correct form in order to apply the inequality
|
we prove the answer is $\frac{1}{3}$ which happens at $x=-1$ for proving it we have :
$$\frac{x^2+x+1}{x^2-x+1} \ge \frac{1}{3}$$
if and only if
$$3x^2+3x+3 \ge x^2-x+1$$
if and only if
$$2x^2+4x+2=2(x+1)^2 \ge 0$$
[edit] if you think that it is not AM_GM here is a solution which is the same but the use of AM_GM is more clear on it.
$$\frac{x^2+x+1}{x^2-x+1}=1+\frac{2}{x+\frac{1}{x}-1} \ge 1 +\frac{2}{-2-1}= \frac{1}{3}$$
(we used $\frac{1}{x+\frac{1}{x}-1}\ge \frac{1}{-3}$ for possitive $x$ it is trivially true and for negative ones it is an AM_GM on $-x+\frac{-1}{x} \ge 2$)
|
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|
$x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$ $x,y \in \mathbb{R}\setminus\{0\}$
Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.
My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{x+y}{xy}=\frac{x^2y^2-3xy}{x^2-xy+y^2}$$
But i don’t know what to do now.
Please if you know the key to this type of problems ‘Find all possible values’ post it.
|
The following is equivalent to @NN2's solution, only written a bit differently.
First substitute $x=1/a, y=1/b$, so that the equation becomes
$$ \tag{*}
a^3+b^3+3ab - 1 = 0 \, .
$$
We are looking for the possible values of $a+b$, this suggests to introduce $S=a+b$ and $P = ab$. Then
$$
0 = a^3+b^3+3ab - 1 = S^3+3PS + 3P -1 = (S-1)(S^2 + S-3P+1) \, .
$$
So $S=1$ is one possible value for $a+b$. Conversely, all $(a, b)$ with $a+b=1$ satisfy the equation $(*)$.
In order to determine the possible zeros of the second factor we use that $P \le S^2/4$ from the inequality between geometric and arithmethic mean:
$$
0 = S^2 + S-3P+1 \ge S^2 + S-\frac 34 S^2+1 = \frac 14 (S+2)^2 \ge 0
$$
is only possible if $S=-2$ and $P=1$, i.e. if $(a, b) = (-1, -1)$.
Therefore $S=a+b=1/x+1/y$ can only take the values $1$ and $-2$, and both values do actually occur.
|
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|
Prove that $\angle AEF =90^\circ$ given a square $ABCD$ Let $ABCD$ be a square and $E\in CD$ such that $DE=EC$. Let $F\in BC$ such that $BC=4FC$. Prove that $\angle AEF =90^\circ$.
My attempt:
Proving that $\angle AEF =90^\circ$ is the same as proving that $\triangle AEF$ is a right triangle. In other words, we wish to prove that $AE^2 +EF^2 = AF^2$. I’ve tried a lot of methods to reach this point but none of them worked.
|
Ignoring the similar triangles for variety...
Let the side length of the square be $s$. We have $|AF|^2 = |AB|^2 +|BF|^2 = s^2 + \left(\frac 34 s\right)^2 = \frac{25}{16}s^2$
Then $|AE|^2 = |AD|^2 +|DE|^2 = s^2 + \left(\frac 12 s\right)^2 = \frac{5}{4}s^2 = \frac{20}{16}s^2$
and $|EF|^2 = |EB|^2 +|BF|^2 = \left(\frac 12 s\right)^2 + \left(\frac 14 s\right)^2 = \frac{5}{16}s^2$
giving $|AF|^2 = |AE|^2 +|EF|^2$ as required
|
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|
Finding the value of $ax^4+by^4$
If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.
Here is my attempt:
$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unknowns it seems that doesn't work.
|
Answer to OP's comment:
For solving the following system:
$$79(x+y)=23xy+217$$$$ 217(x+y)=79xy+691$$
We first eliminate $(x+y) $ from both equations:
$$ (x+y) = \frac{1}{79} (23 xy + 217)\tag{1}$$
And,
$$ x+y = \frac{1}{217} (79 xy + 691) \tag{2}$$
We get:
$$ \frac{1}{217} (79 xy + 691)= \frac{1}{79} (23 xy + 217)$$
We get the result
$$ y = \frac{-6}{x} \tag{3}$$
Now we can substitute (3) into (1), and then if we multiply the whole equation by $x$ , we can solve for the quadratic in x
|
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|
Establishing $\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha$ without the use of Werner or prostaferesis formulas I have this identity:
$$\bbox[5px,border:2px solid #138D75]{\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha} \tag 1$$
If I write this like as:
$$(\cos 4\alpha+\cos 2\alpha)\cdot (\cos 4\alpha-\cos 2\alpha)=-\sin 6\alpha \sin 2\alpha$$
I can use Werner and prostaferesis formulas and I find the identity $(1)$.
But if we suppose of not to use these formulas I have done a try writing:
$$\cos^2 4\alpha=\cos^2 2\alpha-\sin 6\alpha \sin 2\alpha \tag 1$$
$$[\cos(2(2\alpha))]^2=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 2$$
$$\cos^4 2\alpha -2\sin^2 2\alpha\cos^22\alpha+\sin^4 2\alpha=(2\cos^2\alpha-1)^2-\sin 6\alpha \sin 2\alpha \tag 3$$
Then I have abandoned because I should do long computations and I think that is not the right way.
Is there any trick without to use the prostapheresis or Werner's formulas?
|
OK, I think I have something a little more straightforward. If I am using some of the forbidden fruit, I apologize. First of all, as hamam already noted, I'm going to let $2\alpha = x$ to save my sanity. So I want to show that
$$\cos^2 2x = \cos^2 x - \sin 3x\sin x.$$
Aside from the double-angle formula for $\cos 2x$, I will need the triple angle formula
\begin{align*}
\cos 3x &= \cos 2x\cos x -\sin 2x\sin x \\ &= (2\cos^2 x-1)\cos x - 2\sin^2x\cos x \\ &= 2\cos^3 x-\cos x - 2(1-\cos^2 x)\cos x \\ &= 4\cos^3 x-3\cos x.
\end{align*}
Note that $\cos 2x=\cos (3x-x) = \cos 3x\cos x + \sin 3x\sin x$, so the original formula is equivalent to
\begin{equation*}
\cos^2x -\cos^2 2x + \cos3x\cos x = \cos 2x. \tag{$\star$}
\end{equation*}
Well, by the triple angle formula above,
\begin{align*}
\cos 2x+ \cos^2 2x - \cos^2 x &= (2\cos^2 x-1)+(2\cos^2 x -1)^2 - \cos^2 x \\ & = 4\cos^4x - 3\cos^2 x = \cos3x\cos x,
\end{align*}
and this establishes ($\star$).
Having looked carefully at A-level student's, he's using the triple angle formula for sin, so these solutions are almost surely equivalent. :)
|
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|
Prove $\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$ Proof of the integral
$$\int_0^\infty\frac{\ln x}{x^3-1}\mathrm{d}x=\frac{4\pi^2}{27}$$
I try to substitute $u = \ln x$. Then $x = e^u,\>\mathrm{d}x = e^u\mathrm{d}u$ and the limits $(0,\infty)\to (-\infty,\infty)$.
The integral becomes $$\int_{-\infty}^\infty \frac{ue^u}{e^{3u}-1}\mathrm{d}u.$$
|
Split the integral to simplify as follows
\begin{align}\int_0^\infty\frac{\ln x}{x^3-1}{d}x
= &\int_0^1\frac{\ln x}{x^3-1}{d}x + \int_1^\infty\overset{x\to \frac1x}{\frac{\ln x}{x^3-1}}{d}x
=\int_0^1\frac{(1+x)\ln x}{x^3-1}{d}x \\
= &\int_0^1\frac{(x^2+x+1)\ln x}{x^3-1}{d}x
- \int_0^1\overset{x^3\to x}{\frac{x^2\ln x}{x^3-1}}{d}x\\
=&\int_0^1\frac{\ln x}{x-1}{d}x-\frac19 \int_0^1\frac{\ln x}{x-1}{d}x
=\frac89 \int_0^1\frac{\ln x}{x-1}{d}x\\
=&\frac89\cdot \frac{\pi^2}6
=\frac{4\pi^2}{27}
\end{align}
Integrate $\int_0^1\frac{\ln x}{x-1}{d}x=\frac{\pi^2}6$
|
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|
Solving recurrence with substitution I am trying to solve this recurrence $T(n) = 4T(n − 2) + 2^{2n}$ by using substitution and knowing $T(1)=1,T(2)=2$ and here is my attempt:Expanding $T(n)=4T(n-2)+2^{2n}$ using $T(n-2)=4T(n-4)+2^{2(n-2)}$, we get $T(n)=4(4T(n-4)+2^{2(n-2)})+2^{2n}=4^2T(n-4)+2^{2n-2}+2^{2n}=4^2(4T(n-6)+2^{2(n-4)})+2^{2n-2}+2^{2n}=4^3T(n-6)+2^{2n-4}+2^{2n-2}+2^{2n}=...=4^k(T(n-2k)+2^{2n-2k+2}+2^{2n-2k+4}+...+2^{2n}$.Substituting $T(n-2k)=2, i.e. T(2)=2$, $n-2k=2$ then $k=\frac{n-2}{2}$, so our equation looks like $2^{2k}+2^{2n-2k+2}+2^{2n-2k+4}+...+2^{2n}$ with $k=\frac{n-2}{2}$, $2^{2k}+2^{2n-2k+2}+2^{2n-2k+4}+...+2^{2n}=2^{n-2}+2^{n+4}+...+2^{2n}=2^{n-2}+\frac{2^{n+4}(1-2^{\frac{n-4}{2}})}{1-2}=2^{n-2}+2^{\frac{3n+4}{2}}-2^{n+4}$. However I think this answer doesn't satisfy $T(2)=2$ and I really don't know where went wrong. Is there another way to do this problem?
|
Hint
$$T(n)=4T(n-2)+4^{n}$$
$$T(n)=4^2T(n-4)+4^{n}+4^{n-1}$$
by induction, find that
$$T(n)=4^kT(n-2k)+4^{n}+4^{n-1}+...+4^{n-(k-1)}.$$
Now, take $n=2k$ and $n=2k+1$ and find your general formula.
|
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|
Prove that $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$ Problem. (Nguyễn Quốc Hưng) Let $0\le a,b,c\le 3;ab+bc+ca=3.$ Prove that $$\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\leq 3+\sqrt{3}$$
I have one solution but ugly, so I 'd like to find another. I will post my solution in the answer.
|
Some thoughts:
WLOG, assume $a\ge b \ge c$.
By Cauchy-Bunyakovsky-Schwarz inequality, we have
$$\sqrt{a+b} + \sqrt{b + c} + \sqrt{c + a}
\le \sqrt{ \left( \frac{a + b}{2} + (b+c) + \frac{c+a}{\sqrt{3}}\right)(2 + 1 + \sqrt{3})}.$$
It suffices to prove that
$$\frac{a + b}{2} + (b+c) + \frac{c+a}{\sqrt{3}} \le 3 + \sqrt{3}.$$
This is true. Is there nice solutions?
|
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|
Coefficients of $p\circ p$ for polynomial $p(x)=\sum_{k=0}^d a_kx^k$ Let $p(x)=\sum_{k=0}^d a_kx^k$ be a polynomial of degree $d$. Then
\begin{align*}
(p\circ p)(x) &= \sum_{k=0}^da_k\left(\sum_{j=0}^da_j x^j\right)^k\\[3pt]
&=\sum_{k=0}^d \sum_{\substack{k_0+\cdots+k_d=k\\[3pt]k_0,\dots,k_d\geqslant0}}a_k\frac{k!}{k_0!k_1!\cdots k_d!} x^{k_1+2k_2+3k_3+\cdots+d k_d}\prod_{t=0}^da_t^{k_t}
\end{align*}
by the multinomial theorem.
Is there a nicer way I can express the coefficients of $p\circ p$? Where can I read more about this polynomial and study its properties?
|
Not quite the answer you're looking for, but for fun taking a cubic $p(z) = a_0 + a_1 z + a_2 z^2 + a_3 z^3$, $p(p(z)) = c_0 + \cdots + c_9 z^9$ where:
\begin{align}
c_0 &= a_{0}^{3} a_{3} + a_{0}^{2} a_{2} + a_{0} a_{1} + a_{0} \\
c_1 &= 3 \, a_{0}^{2} a_{1} a_{3} + 2 \, a_{0} a_{1} a_{2} + a_{1}^{2} \\
c_2 &= 3 \, a_{0} a_{1}^{2} a_{3} + 3 \, a_{0}^{2} a_{2} a_{3} + a_{1}^{2} a_{2} + 2 \, a_{0} a_{2}^{2} + a_{1} a_{2} \\
c_3 &= a_{1}^{3} a_{3} + 6 \, a_{0} a_{1} a_{2} a_{3} + 3 \, a_{0}^{2} a_{3}^{2} + 2 \, a_{1} a_{2}^{2} + 2 \, a_{0} a_{2} a_{3} + a_{1} a_{3} \\
c_4 &= 3 \, a_{1}^{2} a_{2} a_{3} + 3 \, a_{0} a_{2}^{2} a_{3} + 6 \, a_{0} a_{1} a_{3}^{2} + a_{2}^{3} + 2 \, a_{1} a_{2} a_{3} \\
c_5 &= 3 \, a_{1} a_{2}^{2} a_{3} + 3 \, a_{1}^{2} a_{3}^{2} + 6 \, a_{0} a_{2} a_{3}^{2} + 2 \, a_{2}^{2} a_{3} \\
c_6 &= a_{2}^{3} a_{3} + 6 \, a_{1} a_{2} a_{3}^{2} + 3 \, a_{0} a_{3}^{3} + a_{2} a_{3}^{2} \\
c_7 &= 3 \, a_{2}^{2} a_{3}^{2} + 3 \, a_{1} a_{3}^{3} \\
c_8 &= 3 \, a_{2} a_{3}^{3} \\
c_9 &= a_{3}^{4} \\
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4029832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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|
Finding $A$, $B$, $C$ such that $(2x-A)(2x+4)-Bx+25=(x-C)^2-Cx^2$ for all $x$
Find $A$, $B$, $C$ such that
$$(2x-A)(2x+4)-Bx+25=(x-C)^2-Cx^2$$
for all $x$.
This is what I've got so far.
$$\begin{align}
\text{L.H.S.}
&= (2x-A)(2x+4)-Bx+25 \tag1\\
&= 4x^2-2Ax+8x-4A-Bx+25 \tag2\\
&= 4x^2-(2A+8)x-4A-Bx+25 \tag3 \\ \\
\text{R.H.S.}
&= (x-C)^2-Cx^2 \tag4\\
&= x^2-2Cx+C^2+Cx^2 \tag5
\end{align}$$
I have no idea how I should find the constants by calculation. Please help.
|
Since it is true for all $x$, this is an identity and hence all you need to do is compare coefficients as mentioned in comments. So, we have $$\begin{aligned} 4 &=1-C \\ 2A+B-8&=2C \\ -4A+25&=C^2\end{aligned}$$
Which gives you $C=-3,A=4,B=-6$.
Note that you have done some mistakes in sign.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4031469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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|
How to prove $\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}$ for $x\geq 1$? Intuitively when $x$ gets bigger, $\sqrt{x+1}$ will get closer to $\sqrt{x}$, so their difference will get smaller.
However, I just cannot get a proper proof.
|
$$\sqrt{x} - \sqrt{x-1}>\sqrt{x+1} - \sqrt{x}\quad| +\sqrt{x}$$
$$2\sqrt{x} > \sqrt{x-1}+\sqrt{x+1} \quad|^2$$
$$4x>2x+ 2\sqrt{(x-1)(x+1)} \quad| -2x $$
$$2x>2\sqrt{(x-1)(x+1)} \quad| /2$$
$$x=\sqrt{x^2-1} \quad|^2$$
$$x^2>x^2-1 \quad|-x^2$$
$$0>-1$$
The transfoemations are valid in both directions, top down and bottom up
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4032599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Is there a known closed form solution to $\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx$? $\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}\def\Catalan{\mathsf{Catalan}}$
Related question
Is there a known closed form solution to
\begin{align}
I_n&=
\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} \,dx
=?
\tag{0}\label{0}
\end{align}
It checks out numerically, for $n=1,\dots,7$ that
\begin{align}
I_1=
\int_0^1\frac{\ln(1+x^2)}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln2-\Catalan
\tag{1}\label{1}
,\\
I_2=
\int_0^1\frac{\ln(1+x^{2\cdot2})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(2+\sqrt2)-2\Catalan
\tag{2}\label{2}
,\\
I_3=
\int_0^1\frac{\ln(1+x^{2\cdot3})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln6-3\Catalan
\tag{3}\label{3}
,\\
I_4=
\int_0^1\frac{\ln(1+x^{2\cdot4})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(4+\sqrt2+2\sqrt{4+2\sqrt2})-4\Catalan
\tag{4}\label{4}
,\\
I_5=
\int_0^1\frac{\ln(1+x^{2\cdot5})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(10+4\sqrt5)-5\Catalan
=
\tfrac\pi2\,\ln(10\cot^2\tfrac\pi5)-5\Catalan
\tag{5}\label{5}
,\\
I_6=
\int_0^1\frac{\ln(1+x^{2\cdot6})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln((5\sqrt2+2\sqrt{12})(1+\sqrt2))-6\Catalan
\tag{6}\label{6}
,\\
I_7=
\int_0^1\frac{\ln(1+x^{2\cdot7})}{1+x^2} \,dx
&=
\tfrac\pi2\,\ln(14\cot^2\tfrac\pi7)-7\Catalan
\tag{7}\label{7}
,
\end{align}
so \eqref{0} seems to follow the pattern
\begin{align}
I_n&=
\tfrac\pi2\,\ln(f(n))-n\Catalan
\tag{8}\label{8}
\end{align}
for some function $f$.
Items \eqref{5} and \eqref{7} look promising
as they agree to $f(n)=2n\cot^2(\tfrac\pi{n})$,
but the other fail on that.
Edit:
Also, it looks like
\begin{align}
\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\tfrac\pi2\,\ln(f(n))+n\Catalan
\tag{9}\label{9}
\end{align}
and
\begin{align}
\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} \,dx
&=\pi\,\ln(f(n))
\tag{10}\label{10}
\end{align}
with the same $f$.
Edit
Thanks to the great answer by @Quanto,
the function $f$ can be defined as
\begin{align}
f(n)&=
2^n\!\!\!\!\!\!\!\!\!\!
\prod_{k = 1}^{\tfrac{2n-1+(-1)^n}4}
\!\!\!\!\!\!\!\!\!
\cos^2\frac{(n+1-2k)\pi}{4n}
\tag{11}\label{11}
.
\end{align}
$\endgroup$
|
Just to put in more closed form what @Varu Vejalla has evaluated
$I=\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx$
$\int_0^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=\frac{1}{2}\int_{-\infty}^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=\int_0^1\frac{\ln(1+x^{2n})}{1+x^2} dx+\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx=I+\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx$
$\int_1^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx [t=\frac{1}{x}]=\int_0^1\frac{\ln(1+t^{2n})}{1+t^2} dt-\int_0^1\frac{\ln(t^{2n})}{1+t^2} dx=I+2nG$
$$I=\frac{1}{4}\int_{-\infty}^\infty\frac{\ln(1+x^{2n})}{1+x^2} dx-nG$$
Branch points of $\log$ (roots of $1+x^{2n}$) are $e^{\frac{i\pi}{2n}},e^{-\frac{i\pi}{2n}}, e^{\frac{3i\pi}{2n}}, e^{\frac{-3i\pi}{2n}},... e^{\frac{(2n-1)i\pi}{2n}}, e^{-\frac{(2n-1)i\pi}{2n}} $
Next we integrate in the complex plane, closing the contour in the upper half-plane for the roots $e^{-\frac{(2k-1)i\pi}{2n}}$ and in the lower half-plane for the roots $e^{+\frac{(2k-1)i\pi}{2n}}$ - to integrate the single-valued function.
Finally we get
$$I=\frac{2\pi{i}}{4*2i}\log\Bigl((-i-e^{\frac{i\pi}{2n}})(i-e^{-\frac{i\pi}{2n}})...(-i-e^{\frac{(2n-1)i\pi}{2n}})((-i-e^{-\frac{(2n-1)i\pi}{2n}})\Bigr)-nG=$$
$$=\frac{\pi}{4}\log\Bigl((2+2\sin\frac{\pi}{2n})...(2+2\sin\frac{(2n-1)\pi}{2n})\Bigr)-nG$$
$$I=\frac{\pi}{4}\log\Bigl((1+\sin\frac{\pi}{2n})...(1+\sin\frac{(2n-1)\pi}{2n})\Bigr)+\frac{\pi{n}}{4}\log2-nG$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4036975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
}
|
Solve for integer values of $x,y,z$: $\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$ Solve for integer values of $x,y,z$;
$$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3.$$
My attempt:
*
*Note that all $x, y, z$ are non-zero, otherwise a denominator would be zero.
*Mutliplying by $xyz$ gives:
$$
x^2y^2+x^2z^2+y^2z^2 = 3xyz
$$
*Since (from 1.) the left hand side (LHS) is strictly positive, then $xyz \gt 0$
*Exploring the LHS of the equation from 1.:
$$
2x^2y^2+2x^2z^2+2y^2z^2 = (xy-xz)^2+(xy-yz)^2+(xz-yz)^2+2xyz(x+y+z)
$$
therefore
$$
x^2y^2+x^2z^2+y^2z^2 \geq xyz(x+y+z)
$$
*Replacing LHS of the last inequation by $3xyz$ (according to 2.) gives:
$$3xyz \geq xyz(x+y+z)$$ and since (from 3.) $xyz$ is positive, we can divide both sides by $xyz$, obtaining: $$x+y+z \leq 3$$
*$xyz$ can be positive in two cases:
Case I. $x, y, z$ are all positive. In this case $x \geq 1$, $y \geq 1$, $z \geq 1$,
so
$$
x+y+z \geq 3
$$
but according to 5. this must be an equation. Therefore all inequations for $x$, $y$ and $z$ must turn into equations, so $x=y=z=1$.
Case II. One of $x, y, z$ is positive, two others are negative. Without loss of generality we can assume $x \gt 0$, $y \lt 0$, $z \lt 0$
...
|
If $(x,y,z)$ is a solution then so is $(|x|,|y|,|z|)$, so without loss of generality $x,y,z>0$, and after rearranging we may assume without loss of generality that $x\geq y\geq z$. Then $\tfrac xy,\tfrac xz,\tfrac yz\geq1$, and hence
$$3=\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}> x\frac yz\geq x,$$
and similarly $y,z<3$, so $x,y,z\in\{1,2\}$. This leaves just four triplets $(x,y,z)$ to check.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4041487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
}
|
Find the minimum of $\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$ Find the minimum of $$f(x,y)=\sqrt{4y^2-12y+10}+\sqrt{18x^2-18x+5}+\sqrt{18x^2+4y^2-12xy+6x-4y+1}$$ It seems that $f_x=f_y=0$ is very hard to compute. Is there any easier idea?
|
Complete the square in the first two terms and let $Y=2y-3$ and $X=3x-3/2$ to get $$f(X,Y)=\sqrt{Y^2+1}+\sqrt{2X^2+\frac12}+\sqrt{2\left(X+\frac12\right)^2+\left(Y+\frac12\right)^2-2XY+\frac74}$$ so that \begin{align}f_X&=\frac{2X}{\sqrt{2X^2+1/2}}+\frac{2X-Y+1}{\sqrt{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}}=0\\f_Y&=\frac Y{\sqrt{Y^2+1}}+\frac{Y-X+1/2}{\sqrt{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}}=0\end{align} which can be rewritten as \begin{align}\frac{4X^2}{2X^2+1/2}&=\frac{(2X-Y+1)^2}{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}\\\frac{Y^2}{Y^2+1}&=\frac{(Y-X+1/2)^2}{2(X+1/2)^2+(Y+1/2)^2-2XY+7/4}\end{align} and extensively simplified to give the symmetrical forms \begin{align}X^2(4Y^2+16Y+12)+4(Y-1)X-(Y-1)^2&=0\\Y^2(4X^2+12X+5)+4(2X-1)Y-(2X-1)^2&=0\end{align} on clearing denominators. Equating the two expressions on the left gives $$8X^2Y-6XY^2+8X^2-3Y^2-2XY-4X+3Y=0$$ which factorises to $(4X-3Y)(2XY+2X+Y-1)=0$, where the two interaction terms significantly aid this observation. Substituting $X=3Y/4$ gives the polynomial $$9Y^4+36Y^3+35Y^2-4Y-4=0$$ which has roots at $Y=-2,\pm1/3$ after using the rational root theorem. Hence $X=-3/2,\pm1/4$ and a minimum of $\sqrt{10}$ is obtained at $$(X,Y)=\left(-\frac14,-\frac13\right)\implies(x,y)=\left(\frac5{12},\frac43\right).$$ In fact, it is the global minimum as the expression $2XY+2X+Y-1=0$ is an anomaly due to squaring. The square root in the third term of $f(X,Y)$ means that the condition $f_X=0$ only holds if $X\le-3/2$ or $(X,Y)=(0,1)$ and $f_Y=0$ only holds if $-3/2\le X<0$ or $(X,Y)=(1/2,0)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4043163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
}
|
limit $\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}}$ How can I find the limit to infinity of this function? As this is a $0/0$ equation, I tried using the L'Hôpital's rule in this but ended up making it more complex. I've also tried rationalising the denominator but it didn't lead to anywhere.
$$\lim_{x \to \infty} \frac{\sqrt{x+2} - \sqrt{x+1}}{\sqrt{x+1} - \sqrt{x}} $$
|
\begin{gather*}
\lim _{x\rightarrow \infty }\frac{\sqrt{x+2} -\sqrt{x+1}}{\sqrt{x+1} -\sqrt{x}}\\
=\lim _{x\rightarrow \infty }\frac{\sqrt{1+\frac{2}{x}} -\sqrt{1+\frac{1}{x}}}{\sqrt{1+\frac{1}{x}} -1}\\
\end{gather*}
(Taking $\sqrt{x}$ common from both numerator and denominator)
Now, from the binomial series expansion, $(1+ay)^{b} =1+aby$, when $y\to 0$
\begin{gather*}
\sqrt{1+\frac{2}{x}} =1+\frac{1}{2} \times \frac{2}{x}\\
\sqrt{1+\frac{1}{x}} =1+\frac{1}{2} \times \frac{1}{x}\\
\lim _{x\rightarrow \infty }\frac{\sqrt{1+\frac{2}{x}} -\sqrt{1+\frac{1}{x}}}{\sqrt{1+\frac{1}{x}} -1} =\lim _{x\rightarrow \infty }\frac{\left( 1+\frac{1}{2} \times \frac{2}{x}\right) -\left( 1+\frac{1}{2} \times \frac{1}{x}\right)}{\left( 1+\frac{1}{2} \times \frac{1}{x}\right) -1} =1
\end{gather*}
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4043952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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|
How to find the nth pair from a generated list of pairs? I have a mathematical question applied on informatic (python). I would like to find the fastest way to get the pair from a given index.
As example, we have all pair combinations of values from 0 to 10:
impor itertools
combinations = list(itertools.combinations((i for i in range(0,10)),2)
like this the variable combinations is equals to
[(0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8), (0, 9), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (2, 3), (2, 4), (2, 5), (2, 6), (2, 7), (2, 8), (2, 9), (3, 4), (3, 5), (3, 6), (3, 7), (3, 8), (3, 9), (4, 5), (4, 6), (4, 7), (4, 8), (4, 9), (5, 6), (5, 7), (5, 8), (5, 9), (6, 7), (6, 8), (6, 9), (7, 8), (7, 9), (8, 9)]
Thus, I have this function to get the nth pair
def row_col_from_index(index: int, no_discrete_values: int):
row_num = 0
col_num = 0
seq_value = 0
level = 1
no_columns = 10
while seq_value <= index + 1 - no_columns:
no_columns -= 1
seq_value += (no_discrete_values - level)
level += 1
row_num = level - 1
col_num = level + (index - seq_value)
return row_num, col_num
Here we get the 24nth pair (indeed array is 0 based)
row_col_from_index(23,10)
-> (2, 9)
combinations[23]
-> (2, 9)
This work but that will be slow to find the corresponding pair with a huge number of values.
So is there a mathematical trick to speedup the computation like a dichotomy approach or other…
And the reverse question for a given pair how to get its corresponding index ?
thanks
|
Let $n$ play the role of the number $10$: the pairs are: $(0,1),(0,2),\ldots,(0,n-1),(1,2),(1,3),\ldots,(1,n-1),\ldots,(n-2, n-1)$ (all $n\choose 2$ of them).
Let's find a (zero-based) $k$th pair: say it is $(x,y)$. We have $n-1$ pairs starting with $0$, $n-2$ pairs starting with $1$, etc., ending with one pair starting with $n-2$. Thus, the number of pairs starting with something smaller than $x$ is $(n-1)+(n-2)+\ldots+(n-x)=xn-\frac{x(x+1)}{2}$. Obviously, this is $\le k$. Therefore, we have an inequality:
$$xn-\frac{x(x+1)}{2}\le k$$
and we are looking at the largest such $x$. The inequality above is quadratic:
$$x^2+(1-2n)x+2k\ge 0$$
and the solutions are $x_{1,2}=\frac{2n-1\pm\sqrt{4n^2-4n+1-8k}}{2}$. In addition, we are looking for the solutions that are smaller than the "smaller" of the two solutions, as the "larger" one is obviously greater than $\frac{2n-1}{2}$, which is outside the range of $x$ that we are after.
Thus, $x=\left\lfloor \frac{2n-1\pm\sqrt{4n^2-4n+1-8k}}{2}\right\rfloor$. From there, we find the number of pairs preceding the pair $(x, x+1)$ as $xn-\frac{x(x+1)}{2}$ and so $y$ can be found as: $y=k-\left(xn-\frac{x(x+1)}{2}\right)+x+1$.
Something like:
def pair(n, k):
x = math.floor((2*n-1-math.sqrt(4*n*n-4*n+1-8*k))/2)
y = k - x*n + x*(x+1)//2 + x + 1
return x, y
For the inverse, notice that $k=y-(x+1) + xn-\frac{x(x+1)}{2}$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4046051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solving system of equation using Macaulay2 As an algebraic curve, the Klein quartic can be viewed as a projective algebraic curve over the complex numbers $\mathbb{C}$, defined by the following quartic equation in homogeneous coordinates $[x:y:z]$ on $\mathbb{P}^2_{\mathbb{C}}$:
$$x^3 y + y^3 z + z^3 x = 0.$$
Now we want to find the eigenvectors of this curve, therefore consider the following matrix:
$$
\begin{bmatrix}
3x^2y+z^3 & x^3 + 3y^2z & y^3+3z^2 \\
x & y & z \\
\end{bmatrix},
$$
after computing the $2$-minors we got the following system:
\begin{equation*}
\left\{
\begin{alignedat}{3}
% R & L & R & L & R & L
3x^2y^2+z^3y & -{} & (3y^2zx + x^4) & = 0 \\
3z^2x^2+y^3x & -{} & (3x^2yz + z^4)& = 0 \\
3y^2z^2+x^3z & -{} & (3z^2xy + y^4)& = 0
\end{alignedat} \ .
\right.
\end{equation*}
I am new to Macaualy 2 and I am wondering if Macaulay2 can help me to solve the above system numerically with loading NumericalAlgebraicGeomety package, (I know that because of infinity many solution we need normalize the system on unit sphere).
|
I divided by $z^4$ all equations and set $$\frac{x}{z}=X;\;\frac{y}{z}=Y$$
so I got
$$
\begin{cases}
-X^4+3 X^2 Y^2-3 X Y^2+Y=0\\
-3 X^2 Y+3 X^2+X Y^3-1=0\\
X^3-3 X Y-Y^4+3 Y^2=0\\
\end{cases}
$$
solved by Mathematica I got these solutions
$$
\begin{array}{rr}
X & Y\\
\hline
1 & 1 \\
0.307979 & 1.55496 \\
0.643104 & 0.198062 \\
5.04892 & 3.24698 \\
-1.80194 & 0.801938 \\
-0.445042 & -0.554958 \\
1.24698 & -2.24698 \\
-0.610856 & 0.106042 \\
-0.173596 & -1.63705 \\
0.667631 & 1.26654 \\
0.789553 & 0.52713 \\
1.89706 & 1.49783 \\
9.4302 & -5.76459 \\
\end{array}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4046318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.