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Value of $\lim\limits_{n\rightarrow\infty}\prod\limits_{k=3}^n\left(1-\tan^4\frac{\pi}{2^k}\right)=\frac{\pi^3}{m}$ Find the value of m for the following $$\lim\limits_{n\rightarrow\infty}\prod_{k=3}^n\left(1-\tan^4\frac{\pi}{2^k}\right)=\frac{\pi^3}{m}$$
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It is possible to find a closed form for your product, if fact we have for any $ k\geq 3 $ :
\begin{aligned}1-\tan^{4}{\left(\frac{\pi}{2^{k}}\right)}&=\frac{1-\tan^{2}{\left(\frac{\pi}{2^{k}}\right)}}{\cos^{2}{\left(\frac{\pi}{2^{k}}\right)}}\\ &=\frac{\cos{\left(\frac{\pi}{2^{k-1}}\right)}}{\cos^{4}{\left(\frac{\pi}{2^{k}}\right)}}\\ 1-\tan^{4}{\left(\frac{\pi}{2^{k}}\right)}&=\frac{8\sin^{3}{\left(\frac{\pi}{2^{k}}\right)}}{\sin^{3}{\left(\frac{\pi}{2^{k-1}}\right)}}\times\frac{\cos{\left(\frac{\pi}{2^{k-1}}\right)}}{\cos{\left(\frac{\pi}{2^{k}}\right)}}\end{aligned}
Thus, if $ n\geq 3 $, then : \begin{aligned}\prod_{k=3}^{n}{\left(1-\tan^{4}{\left(\frac{\pi}{2^{k}}\right)}\right)}&=8^{n-2}\left(\prod_{k=3}^{n}{\frac{\sin^{3}{\left(\frac{\pi}{2^{k}}\right)}}{\sin^{3}{\left(\frac{\pi}{2^{k-1}}\right)}}}\right)^{3}\left(\prod_{k=3}^{n}{\frac{\cos{\left(\frac{\pi}{2^{k-1}}\right)}}{\cos{\left(\frac{\pi}{2^{k}}\right)}}}\right)\\ &=2^{3n-5}\frac{\sin^{3}{\left(\frac{\pi}{2^{n}}\right)}}{\cos{\left(\frac{\pi}{2^{n}}\right)}}\end{aligned}
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|
Find minimum and maximum value of $\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$
Blockquote
Find min,max of function $$T=\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$$
*
*Max
By squaring both side and AMGM:
$$T^2=7+2\sqrt{5\cos^2x+1}\cdot \sqrt{5\sin^2x+1}$$
$$\le 7+5\left(\cos^2x+\sin^2x\right)+2=14$$
Or $$T\le \sqrt {14}$$
*About minimal value:By $\sqrt x +\sqrt y \ge \sqrt{x+y}$
So $T\ge \sqrt{5+1+1}=\sqrt 7$
I think the minimal value has a little wrong. Pls help me.
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The quantity $T(x)\geq0$ is a ${\pi\over2}$-periodic function of $x$. It is extremal, i.e., maximal or minimal, iff $$T^2=7+2\sqrt{(5\cos^2 x+1)(5\sin^2 x+1)}$$ is extremal, and this is the case iff the radicand on the RHS
is extremal. Letting $\sin^2 x=:t$ this means that
$$\phi(t):=\bigl(5(1-t)+1\bigr)(5t+1)=6+25(t-t^2)\qquad(0\leq t\leq1)$$
should be extremal. This $\phi(t)$ takes its minimum $6$ at $t=0$ and $t=1$, and takes its maximum $6+{25\over4}={49\over4}$ at $t={1\over2}$.
From $T^2=7+2\sqrt{6}=\bigl(1+\sqrt{6}\bigr)^2$ it follows that the minimum of $T$ is $1+\sqrt{6}$. Similarly, from $$T^2=7+2\cdot\sqrt{49\over4}=14$$ it follows that the maximum of $T$ is $\sqrt{14}$.
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Absolute minimum and maximum of $f(x,y,z)=x^4+y^4+z^4-4xyz$ I have to find the absolute maximum and minimum of the function $f(x,y,z)=x^4+y^4+z^4-4xyz$ over $x^2+y^2+z^2\leq 9$, $x,y,z\geq 0$.
I'm having problems to find the constrained extremas in $x^2+y^2+z^2=9$. I have tried by using the Lagrange's Multipliers theorem, by parametrizing the sphere and by algebraic manipulating the function but I haven't been able to come up with the solution.
Could anybody help me?
Thank you in advance.
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Introduce the dummy variable $w$. Then we want to minimize
$$
F(x,y,z,w)=x^4+y^4+z^4+w^4-1-4xyzw
$$subject to $x^2+y^2+z^2\leq 9, w=1$. By AM-GM
$$
(x^4+y^4+z^4+w^4)\geq4 \sqrt[4]{x^4y^4z^4w^4}=4xyzw,
$$with equality iff $x=y=z=w$. Then since we must have $w=1$, the minimum occurs at $(x,y,z)=(1,1,1)$, giving $f(1,1,1)=-1$.
Now for the maximum. Note that
$$
f(x,y,z) = \langle x^2,y^2,z^2\rangle \cdot \langle x^2,y^2,z^2\rangle-4xyz
$$By Cauchy-Schwarz,
$$
|f(x,y,z)|\leq (9)^2-4xyz;
$$equality occurs if any of the variables are $0$. At $(3,0,0)$ we have $f(3,0,0)=81$ and at $(3/\sqrt{2},3/\sqrt{2},0)$, we have $f(3/\sqrt{2},3/\sqrt{2},0)=81/2$.
|
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|
Find minimal value of $\left(2-x\right)\left(2-y\right)\left(2-z\right)$
Let $x,y,z>0$ such that $x^2+y^2+z^2=3$. Find minimal value of $$\left(2-x\right)\left(2-y\right)\left(2-z\right)$$
I thought the equality occurs at $x = y = z = 1$ (then it is easy), but the fact is $x = y = \frac{1}{3}; z = \frac{5}{3}$. So I just thought of using $uvw$, but I am not allowed to use it during my exam. Because of the equality I cannot use AMGM, Cauchy-Schwarz, etc.
I tried to use Mixing-Variables, but I failed. Please help.
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From Inequality on AoPS:
Without loss of generality $x \ge y \ge z > 0$, so that $z \le 1$. We have
$$
2(2-x)(2-y) = (x+y-2)^2 +4 - x^2-y^2 \ge 4 - x^2 - y^2 = 1+z^2
$$
and therefore
$$
(2-x)(2-y)(2-z) \ge \frac 1 2 (1+z^2)(2-z) =: f(z) \, .
$$
An elementary calculation shows that the minimum of $f$ on $[0, 1]$ is $f(1/3) = 25/27$, i.e.
$$
(2-x)(2-y)(2-z) \ge \frac{25}{27} \, .
$$
Equality holds if $(x, y, z)$ is a permutation of $(5/3, 1/3, 1/3)$.
|
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Integrate $\int_{0}^{1} \frac{\ln(1+a^2x^2)}{\sqrt{1-x^2}} dx $ by differentiating parameter $a$ Can anyone help me calculate the integral using differentiation with respect to a parameter:
$$I(a) = \int_{0}^{1} \frac{\ln(1+a^2x^2)}{\sqrt{1-x^2}} dx, |a|<1$$
UPD: I try to solve like this:
\begin{align*}
I'(a)&=\int_{0}^{1} \frac{2ax^2}{\sqrt{1-x^2}(1+a^2x^2)} dx
\\&=2a\left(\frac{1}{a^2}\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}} - \frac{1}{a^2}\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}(a^2x^2+1)}\right)
\\&= 2a\left(\frac{1}{a^2}\arcsin(x)|_0^1-\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}(a^2x^2+1)}\right)
\end{align*}
Then I try to solve the second integral:
I am doing a variable change like this:
$$x=\sin(x) ,y=\arcsin(x),dx=\cos(y)dy$$
After that, I solve this integral and finally get:
$$\frac{2ax^2\arctan{\frac{\sqrt{1+a^2}x}{\sqrt{1-x^2}}}}{\sqrt{1+a^2}}$$
And the problem is when substituting 1 into the result:
$$\lim_{x\to 1}{\frac{2ax^2\arctan{\frac{\sqrt{1+a^2}x}{\sqrt{1-x^2}}}}{\sqrt{1+a^2}}}$$
Nevertheless, this is a difficult way. Is there an easy way?
|
Note
$$\lim_{x\to 1}{\frac{\arctan{\frac{\sqrt{1+a^2}x}{\sqrt{1-x^2}}}}{\sqrt{1+a^2}}}= \frac{\arctan^{-1}(\infty)}{\sqrt{1+a^2}}=
\frac{\frac\pi2}{\sqrt{1+a^2}} $$
Then, you have
$$I’(a)=\frac\pi a\left(1-\frac1{\sqrt{1+a^2}} \right)
$$
and
$$ \int_{0}^{1} \frac{\ln(1+a^2x^2)}{\sqrt{1-x^2}} dx
= \int_0^a I’(t) dt =\pi \int_0^a \frac1t\left(1-\frac1{\sqrt{1+t^2}} \right) dt \\
= \pi\ln \frac {\sqrt{1+a^2}+1}2
$$
|
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|
Pierre runs a game at a fair, where each player is guaranteed to win $10 - Die rolling question Pierre runs a game at a fair, where each player is guaranteed to win $10.
Players pay a certain amount each time they roll an unbiased die, and must keep rolling until a ‘6’ occurs.
When a ‘6’ occurs, Pierre gives the player $10 and the game concludes.
On average, Pierre wishes to make a profit of $2 per game. How much does he need to charge for each roll of the die?
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The probability of not rolling $6$ for $k$ times is $\left(\frac{5}{6}\right)^k$.
The probability of not rolling $6$ for $k-1$ first times and then rolling $6$ on the $k$th roll is $\frac16\left(\frac{5}{6}\right)^{k-1}$.
So to find the expected number of rolls we are just to find the sum
$\sum\limits_{k=1}^\infty k\frac16\left(\frac{5}{6}\right)^{k-1}$.
Let $S_n=\sum\limits_{k=0}^{n} (k+1)\left(\frac{5}{6}\right)^{k}$,
$$\begin{align*}
\frac{5}{6}S_n
&=\sum\limits_{k=0}^{n} (k+1)\left(\frac{5}{6}\right)^{k+1}\\
%&=\frac56+\sum\limits_{k=1}^{n} (k+1)\left(\frac{5}{6}\right)^{k+1}\\
&=\sum\limits_{k=1}^{n+1} k\left(\frac{5}{6}\right)^{k}\\
&=\sum\limits_{k=0}^{n+1} k\left(\frac{5}{6}\right)^{k}\\
&=\sum\limits_{k=0}^{n+1} (k+1)\left(\frac{5}{6}\right)^{k}-
\sum\limits_{k=0}^{n+1} \left(\frac{5}{6}\right)^{k}\\
&=\sum\limits_{k=0}^{n+1} (k+1)\left(\frac{5}{6}\right)^{k}-
\frac{(5/6)^{n+2} - 1}{5/6-1}\\
&=\sum\limits_{k=0}^{n} (k+1)\left(\frac{5}{6}\right)^{k}
+(n+2)\left(\frac{5}{6}\right)^{n+1}
-\frac{(5/6)^n - 1}{5/6-1}\\
&=S_n
+(n+2)\left(\frac{5}{6}\right)^{n+1}
+6\left((5/6)^n - 1\right),
\end{align*}$$
$$\begin{align*}
S_n&=-6\left((n+2)\left(\frac{5}{6}\right)^{n+1}
+6\left(\left(\frac{5}{6}\right)^{n} - 1\right)\right)
\end{align*}$$
$$\begin{align*}
\lim\limits_{n\to\infty}S_n&=-6\left(0
+6\left(0 - 1\right)\right)=36
\end{align*}$$
So the expected number of rolls is $\frac{1}{6}\cdot 36=6$ and a roll cost to have income $2$ per game on average is $\frac{10+2}{6}=2$
|
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|
Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$
Prove that
$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$
with $a>0, b>0 , c> 0$ and $d>0.$
My attempt:
$$\begin{align*}\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)& = \dfrac{abcd+b^2c^2+a^2d^2+abcd}{abcd}\\
& =\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\
&=\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\
&=\dfrac{(ad)^2+(bc)^2+2(ad)(bc)}{abcd}\\
&=\dfrac{(ad+bc)^2}{abcd}\end{align*}$$
I don't know how to continue from this.
Can someone help me?
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Now, $$\frac{(ad+bc)^2}{abcd}-4=\frac{a^2d^2-2abcd+b^2c^2}{abcd}=\frac{(ad-bc)^2}{abcd}\geq0.$$Also, by C-S
$$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq\left(\sqrt{\frac{b}{a}\cdot\frac{a}{b}}+\sqrt{\frac{d}{c}\cdot\frac{c}{d}}\right)^2`=4$$
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|
A solution of Diophantine equation: $\big(x+y+z\big)^{3}=27x y z$ with $(x,y)∈Z$ A solution is:
$x=(r+s)^{3}(r-3s)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$y=8r^{3}(2s-r)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$z=(r^{2}-2r s+3s^{2})^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
Are there any more?
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$\big(x+y+z\big)^{3}=27xyz\tag{1}$
Let $s=x+y+z$ and substitute $z=s-x-y$ to equation $(1).$
$s^3-27xys+27x^2y+27xy^2=0\tag{2}$
For $x$ to be rational solution, the discriminant of the equation $(2)$ must be square.
Hence $-27y(4s-3y)(s-3y)^2$ must be square, then
$v^2 = (9y-6s)^2-36s^2$
We solve simultaneous equations $[6s=m^2-n^2,9y-6s=m^2+n^2].$
We get $y = \frac{2m^2}{9}, s = \frac{m^2-n^2}{6}, x = \frac{-(m-n)^3}{36m}, z=\frac{-(m+n)^3}{36m}.$
Finally, $(x,y,z)=(-(m-n)^3, 8m^3, -(m+n)^3).$
$(m,n)=(2,1): [x,y,z]=[-1, 64, -27]$
$(m,n)=(3,2): [x,y,z]=[-1, 216, -125]$
$(m,n)=(4,1): [x,y,z]=[-27, 512, -125]$
$(m,n)=(4,3): [x,y,z]=[-1, 512, -343]$
Added new solutions:
We can get new solution using a known solution of equation $(1).$
$(x,y,z)=(-(m-n)^3, 8m^3, -(m+n)^3)$ is a known solution.
Substitute $x = t-(m-n)^3, y = t+8m^3, z = kt-(m+n)^3$ to equation $(1)$,
and let $k= \frac{-(5m^4+8m^3n+2m^2n^2+n^4)}{4(m^2-2mn+n^2)m^2},$ then $t = \frac{-72(m^4-2m^3n+2n^3m-n^4)m^3}{9m^4-18m^3n-6n^3m-n^4}.$
Hence we get new parametric solution below.
$(x,y,z)=(-(3m+n)^3(m-n)^3, -64n^3m^3, (m+n)^3(-n+3m)^3).$
Similarly, we get other new solution using above new solution below.
$(x,y,z)=(-(m-n)^3(3m+n)^3(-n^2+6mn+3m^2)^3, 512(3m^2-n^2)^3n^3m^3, (-n+3m)^3(m+n)^3(-n^2-6mn+3m^2)^3).$
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|
Complex number - arctan $$z=-2+2\sqrt{3}i\implies x=-2, y=2\sqrt3$$
$$r=\sqrt{x^2+y^2}=\sqrt{4+12}=4$$
$$\text{Angle}=\arctan\left(\frac{2\sqrt3}{-2}\right)+\pi=\frac{2\pi}{3}=120^\circ$$
1) May I know how $\arctan\left(\dfrac{2\sqrt3}{-2}\right)+\pi$ turns into $\dfrac{2\pi}{3}$?
2) Can I use calculator to do the calculation and how?
Thanks for the kindness!
|
Notice, $\tan(-\theta)=-\tan\theta$
$$\therefore \arctan\left(\frac{2\sqrt3}{-2}\right)+\pi=\arctan\left(-\sqrt{3}\right)+\pi=-\frac{\pi}{3}+\pi=\frac{2\pi}{3}=120^\circ$$
You can use calculator to find amplitude & argument. Do remember $\tan^{-1}\sqrt3=\frac{\pi}{3}$
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|
Is there an easier way to solve the given problem? If $x + 2 + \sqrt{3}i=0$ then find the value of $2x^4+3x^3-x^2-15x+36$
If you try to find the values of $x^2, x^3 $and $ x^4$, and then put the values in, we can find the solution to be 1. But is there a better way to solve this problem? Where you can probably break the second polynomial into a simple one and then solve it, saving a lot of time and making the calculations less prone to errors?
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Real coefficients in the quartic inspired me to multiply $x + 2 + \sqrt{3}i\;$ by the conjugate, which gives
$$(x + 2 + \sqrt{3}i)(x + 2 - \sqrt{3}i)=x^2+4x+7.$$
Dividing the given expression by $x^2+4x+7$ (just to see if we get a simple form) leads to $$2x^4+3x^3-x^2-15x+36=\underbrace{(x^2+4x+7)}_{0}\cdot(2x^2-5x+5)+1$$
Yes, if $x + 2 + \sqrt{3}i=0\;$ then $2x^4+3x^3-x^2-15x+36=1.$
|
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Spivak Calculus chapter 1 problem 13 proof critique Question paraphrased: Prove that the maximum between two numbers $x$ and $y$ is given by:
$$\max(x,y)=\frac{x+y+|y-x|}{2}$$
Proof: Let $x$ and $y$ be two arbitrary numbers. Then, one and only one holds true: $x \ge y$ or $x \le y$
If $x \ge y$, $x=x+(y-y)=(x-y)+y=|x-y|+y$
Also, $$x=\frac {x+x}{2}=\frac{|x-y|+y+x}{2}=\frac{|x-y|+x+y}{2}$$
If $y \ge x$, $y=y+(x-x)=(y-x)+y=|y-x|+y$
Also, $$y=\frac {y+y}{2}=\frac{|y-x|+x+y}{2}=\frac{|x-y|+x+y}{2}$$
Thus, $$\max(x,y)=\frac{x+y+|y-x|}{2}$$
How can I improve this proof? I am not very convinced with $y=\frac {y+y}{2}$ and $x=\frac {x+x}{2}$ technique because it feels as if I have done this to satisfy the answer.
|
Your proof is ok.
Rephrasing:
1) Let $y\ge x$:
Then $\max(x,y)=y$ and
$\dfrac{y+x+|y-x|}{2}=2y/2=y$, since $|y-x|=y-x$.
2) Let $y<x$ :
Then $\max(x,y)=x$ and
$\dfrac{y+x+|y-x|}{2}=x$ since $|y-x|=x-y$.
|
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If real $x$ and $y$ satisfy $x^2+y^2-4x+10y+20=0$, then prove that $y+7-3\sqrt{2}\le x\le y+7+3\sqrt{2}$
Let $x, y \in \mathbb{R}$ such that $x^2 + y^2 - 4x + 10y + 20 = 0$. Prove that
$$y + 7 - 3\sqrt{2} \le x \le y + 7 + 3\sqrt{2}$$
I'm struggling with that problem. I've recognized that the given equation is one of a circle, namely; $$(x-2)^2 + (y+5)^2 = 9$$
And what needs to be proven can be derived to the form: $$7-3\sqrt{2}\le x-y\le 7 + 3\sqrt{2} \quad\Leftrightarrow\quad (x-y)^2 - 14(x-y) + 31 \le 0$$
But I really can't find the correlation between what's given and what has to be proven! Also, I'm wondering if this problem can be generalized for any circle parameters, something like an universal interval of $x-y$ for any circle. Any help is apprecatiated! Thank you in advance!
|
Writing $u = x - 2$ and $v = y + 5$ shows that the problem is equivalent to proving that if $u^2 + v^2 = 9,$ then $|u - v| \leqslant 3\sqrt2.$ We have $$
(u - v)^2 + (u + v)^2 = (u^2 - 2uv + v^2) + (u^2 + 2uv + v^2) = 2(u^2 + v^2) = 18,
$$
therefore $(u - v)^2 \leqslant 18,$ and the result follows by taking square roots.
More generally, for the circle $(x - a)^2 + (y - b)^2 = r^2,$ if we write $u = x - a$ and $v = y - b,$ essentially the same argument gives $(u - v)^2 \leqslant 2r^2,$ whence $|u - v| \leqslant r\sqrt2,$ i.e.
$$
a - b -r\sqrt2 \leqslant x - y \leqslant a - b + r\sqrt2.
$$
|
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|
Uniform convergence of a sequence of functions which is integral of another sequence I was going through some questions on pointwise and uniform convergence. Got stuck in one of those which says:
Let $g_n(x) = \sin^2(x+\frac{1}{n})$ be defined on $[0,\infty).$
and $f_n(x) = \int_0^xg_n(t)\,dt.$
I am supposed to discuss about its uniform-convergence of $(f_n).$
The terms are really looking complicated to try it by the definition. Should I first show that $(g_n)$ is uniformly convergent? How am I supposed to do even that?
Help, please.
|
You have
$$\begin{aligned}g_n(x)&=\sin^2\left(x + \frac{1}{n}\right) = \frac{1}{2}\left(1- \cos\left(2(x + \frac{1}{n})\right)\right)\\
&=\frac{1}{2}\left(1 - \cos 2x \cos\frac{1}{n} + \sin 2x \sin \frac{1}{n}\right).
\end{aligned}$$
Therefore
$$f_n(x)= \frac{1}{2}\left(x - \frac{1}{2}\cos\frac{1}{n}\sin 2x-\frac{1}{2}\sin\frac{1}{n}\left(\cos 2x -1\right)\right).$$
From there, you can prove that $\{f_n\}$ converges uniformly to
$$f(x) = \frac{x}{2} - \frac{1}{4} \sin 2x$$
as
$$\begin{aligned}\left\vert f_n(x) - f(x) \right\vert &= \frac{1}{4}\left\vert \left(1 - \cos\frac{1}{n} \right)\sin 2x + \sin\frac{1}{n}\left(\cos 2x -1\right)\right\vert\\
&\le \frac{1}{4}\left(\left\vert \left(1 - \cos\frac{1}{n} \right)\sin 2x\right\vert + \left\vert\sin\frac{1}{n}\left(\cos 2x -1\right)\right\vert\right)\\
&\le \frac{1}{4}\left(\left\vert 1 - \cos\frac{1}{n} \right\vert + 2\left\vert\sin\frac{1}{n}\right\vert\right)\\
\end{aligned}$$
and the RHS of above inequality converges to zero independently of $x$.
|
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|
Calculating $\int_A \frac{z}{y^2}$ with $A:=\{z\ge0,x\ge1,y\ge 0, x^2+z^2<\min(2x,y^2)\}$ I want to calculate the following improper integral:
$$\int_A \frac{z}{y^2}\\
A:=\{z\ge0,x\ge1,y\ge 0, x^2+z^2<\min(2x,y^2)\}$$
First I noticed that the conditions imply $x^2<2x\rightarrow x<2;1<x^2<y^2\rightarrow y>1$, and thus $B=A\cap \{y>2\}=\{z\in(0,2),x\in(1,2), x^2+z^2<2x\}$. Thus this part of the integral is fairly easy:
$$\int_B\frac{z}{y^2}=\int_2^\infty\int_1^2\int_0^\sqrt{2x-x^2}\frac{z}{y^2}dzdxdy=\int_2^\infty\frac1{y^2}dy\int_1^2\frac{2x-x^2}{2}dx=\frac{1}{6}$$
We are now left with a "proper" integral (i.e. the region on which we are integration is finite, and the integrand is bounded):
$$\int_{A\cap\{y\in(1,2)\}}\frac{z}{y^2}$$
I tried to split the domain in two regions: $A'=\{x\in[1,2);y\in(x,\sqrt{2x});z\in [0,\sqrt{y^2-x^2})\}, A''=\{x\in[1,2);y\in[\sqrt{2x},2];z\in[0,\sqrt{2x-x^2})$
Is my approach correct (I'm not sure, since the computations that follow bring me to an uncorrect result)? Is there any other approach to computing this integral, perhaps less messy?
|
We can partition the domain $A$ into regions $A_1,A_2,A_3$ . . .
The diagram below shows the projections of $A_1,A_2,A_3$ onto the $xy$-plane.
Region $A_1$ is defined by
$$
\left\lbrace
\begin{align*}
0\le\,&z\le \sqrt{y^2-x^2}\\[4pt]
x\le\,&y\le\sqrt{2x}\\[4pt]
1\le\,&x\le 2\\[4pt]
\end{align*}
\right.
$$
so letting $a_1$ denote the integral for region $A_1$, we get
\begin{align*}
a_1&=\int_1^2\int_x^{\sqrt{2x}}\int_0^{\sqrt{y^2-x^2}}\!\frac{z}{y^2}\;dz\,dy\,dx\\[4pt]
&=\frac{19-13\sqrt{2}}{30}\\[4pt]
\end{align*}
Region $A_2$ is defined by
$$
\left\lbrace
\begin{align*}
0\le\,&z\le \sqrt{2x-x^2}\\[4pt]
1\le\,&x\le\frac{y^2}{2}\\[4pt]
\sqrt{2}\le\,&y\le 2\\[4pt]
\end{align*}
\right.
$$
so letting $a_2$ denote the integral for region $A_2$, we get
\begin{align*}
a_2&=\int_\sqrt{2}^2\int_1^{\frac{y^2}{2}}\int_0^{\sqrt{2x-x^2}}\!\frac{z}{y^2}\;dz\,dx\,dy\\[4pt]
&=\frac{11-7\sqrt{2}}{30}\\[4pt]
\end{align*}
Region $A_3$ is defined by
$$
\left\lbrace
\begin{align*}
0\le\,&z\le \sqrt{2x-x^2}\\[4pt]
1\le\,&x\le 2\\[4pt]
2\le\,&y < \infty\\[4pt]
\end{align*}
\right.
$$
so letting $a_3$ denote the integral for region $A_3$, we get
\begin{align*}
a_3&=\int_2^\infty \int_1^2\int_0^{\sqrt{2x-x^2}}\!\frac{z}{y^2}\;dz\,dx\,dy\\[4pt]
&=\int_2^\infty\!\frac{1}{3y^2}\;dy\\[4pt]
&=\frac{1}{6}
\end{align*}
Combining the results, we get
$$
\int_{\Large{A}}\,\frac{z}{y^2}=a_1+a_2+a_3=\frac{7-4\sqrt{2}}{6}
$$
|
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|
Prove there are infinitely many positive integers which cannot be represented as a sum of four non-zero squares. Prove there are infinitely many positive integers which cannot be represented as a sum of four non-zero squares. Every positive integer can be written as the sum of four squares. But not all necessarily non-zero. Any hints on this?
|
Assume there are $4$ such non-zero squares that add up to $2^{2n+1}$ for any $n \ge 1$, i.e., you have
$$2^{2n+1} = a^2 + b^2 + c^2 + d^2 \tag{1}\label{eq1A}$$
However, note all perfect squares are congruent to $0$, $1$ or $4$ modulo $8$. Since you just asked for a hint, the rest of the answer is in the spoiler below.
Any positive integer of the form $2^{k}$ where $k \ge 3$, such as where $k = 2n + 1$ for $n \ge 1$, is congruent to $0$ modulo $8$ and can only be the sum of $4$ squares if they are all even (since all $4$ odd gives a congruence of $4$ modulo $8$, $3$ odd gives $3$ or $7$, $2$ odd gives $2$ or $6$, and just $1$ odd gives $1$ or $5$). Thus, you have $a = 2a_1$, $b = 2b_1$, $c = 2c_1$ and $d = 2d_1$. Substituting this into \eqref{eq1A} and dividing both sides by $4$ gives
$$2^{2(n-1) + 1} = 2^{2n - 1} = a_1^2 + b_1^2 + c_1^2 + d_1^2 \tag{2}\label{eq2A}$$
This is an equation of the same form, so as long as the power of $2$ is $\ge 3$, you can repeat the procedure. Repeating this $n$ times gives
$$2^{2(n-n) + 1} = 2^{1} = a_n^2 + b_n^2 + c_n^2 + d_n^2 \tag{3}\label{eq3A}$$
This is not possible since the RHS is at least $4$ but the LHS is just $2$. This means at least one (actually, $2$) of the squares in \eqref{eq1A} must have been $0$. Since \eqref{eq1A} only required that $n \ge 1$, and \eqref{eq3A} shows it works for $n = 0$ also, you have an infinite # of positive integers of the form $2^{2n+1}$ which cannot be represented as the sum of $4$ non-zero squares.
Note you can also use induction to prove $2^{2n+1} \; \forall \; n \ge 0$ cannot be represented by a sum of $4$ non-zero squares by using \eqref{eq3A} as the base case, and then using the modulo $8$ congruences to show you can reduce the $n = k + 1$ case to the $n = k$ case in the inductive step.
|
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|
How to prove that $1-\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)} \geq \frac{2^{N-1}+1}{2^N}$? The question is in the title. Numerical computation suggests the result is true, but I don't know how to prove it rigorously.
|
First note that $\frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)}=\frac{\frac{1}{2}}{2^{n-1}+1}-\frac{1}{2^{n+1}}$ (an alternative form). Also the series $\sum_{n=1}^{\infty}\frac{1}{2^{n-1}+1}<\sum_{n=1}^{\infty}\frac{1}{2^{n-1}}=2<\infty$ and hence it converges by the comparison test. Since $\sum_{n=1}^{N}\frac{1}{2^{n+1}}=\frac{1}{2}-(\frac{1}{2^{N+1}})$ we have $\sum_{n=1}^N \frac{2^{n-1}-1}{2^{n+1}(2^{n-1}+1)}=\frac{1}{2}\sum_{n=1}^{N}\frac{1}{2^{n-1}+1}-\frac{1}{2}+\frac{1}{2^{N+1}}<\frac{1}{2}\sum_{n=1}^{N}\frac{1}{2^{n-1}}-\frac{1}{2}+\frac{1}{2^{N+1}}=\frac{1}{2}(2-\frac{2}{2^{N}})-\frac{1}{2}+\frac{1}{2^{N+1}}=\frac{1}{2}-\frac{1}{2^{N}}+\frac{1}{2^{N+1}}$
|
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|
Using residue theorem to calculate integral $\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}$ - where is my mistake? I am to calculate:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x}
$$
We can set $\gamma(t)=e^{it}$ for $t \in [0, 2\pi]$ and then $z = e^{it}$, $\dfrac{dz}{iz}=dt$, $\sin t =\dfrac{1}{2i}(z-\frac{1}{z})$ so that:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x} = \int\limits_\gamma \frac{dz}{\left(10+\frac{3}{i}(z-\frac{1}{z})\right)iz} = \int\limits_\gamma \frac{dz}{\left(10iz+3z^2-3\right)}
$$
Roots of denominator are $-3i$ and $\frac{-i}{3}$ but since the winding number of $-3i$ is equal to 0 we have:
$$
\int\limits_\gamma \frac{dz}{10iz+3z^2-3} = 2 \pi\, i\, Res\left(f,\frac{-i}{3}\right)\cdot1
$$
Calculating residue:
$$
Res\left(f,\frac{-i}{3}\right) = \lim_{\large z \to \frac{-i}{3}} \frac{1}{(z+3i)}=\frac{3}{8i}
$$
summing up:
$$
\int\limits_0^{2\pi} \frac{dx}{10+6\sin x} = 2\pi i \cdot \frac{3}{8i} = \frac{3\pi}{4}
$$
But wolfram says it is equal to $\dfrac{\pi}{4}$. Could you help me spot my mistake?
|
You made mistake while finding residue
$$
Res\left(f,\frac{-i}{3}\right) = \lim_{z \to \dfrac{-i}{3}} \frac{1}{3(z+3i)}=\frac{1}{3\left(-\frac i3+3i\right)}=\frac{1}{8i}
$$
$$
\therefore \int\limits_0^{2\pi} \frac{dx}{10+6\sin x} = 2\pi i \cdot \frac{1}{8i} = \frac{\pi}{4}
$$
|
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|
Does the limit $\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}$ exists? I was trying to find if the the following limit exists:
$$
\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}
$$
From this topic I tried to switch to polar coordinates:
$$
\frac{x^3y^2}{x^6+y^4}=\frac{r^3\cos^3\theta\cdot r^2\sin^2\theta}{r^6\cos^6\theta+r^4\sin^4\theta}=\frac{r\cos^3\theta\sin^2\theta}{r^2\cos^6\theta+\sin^4\theta}\to 0
$$
But apparently this limit does not exists (proved it in another way). Why this method worked for the limit in that topic and here it didn't work?
|
Consider the path along the x-axis, y-axis and $y=x^{\frac{3}{2}}$.
Along the x-axis:
$$\lim_{(x,0) \to (0,0)} \frac{x^3 \cdot 0}{x^6 + 0}=0
$$
Along the y-axis:$$\lim_{(0,x) \to (0,0)} \frac{0 \cdot y^2}{0 + y^4}=0
$$
Along the curve $y=x^{\frac{3}{2}}$:
$$ \lim_{(x,x^{\frac{3}{2}}) \to (0,0)} \frac{x^3{(x^{\frac{3}{2}})}^2}{x^6 + {(x^{\frac{3}{2}})}^4 }=\frac{x^6}{2x^6}=\frac{1}{2}$$
Therefore, the limit does not exist.
|
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|
If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$? If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$?
I tried substitution on the bottom (for example $\frac{b-1}{\frac{1}{a}+1}$), but I then a very similar term. What should I do?
I can not find a way to decompose of combine.
Also, how do you prove that $x^4+y^4+z^2 \ge xyz \sqrt{8}$?
I got $x^4+y^4+z^2\geq3\sqrt[3]{x^4*y^4*z^2}=3xy\sqrt[3]{xyz^2}.$
I do not know what to do from here.
|
It's wrong.
Try, $b>1$ and $bc\rightarrow-1^-$.
It's true for positive variables.
Indeed, let $a=\frac{y}{x}$ and $b=\frac{z}{y},$ where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{x}{z}$ and we need to prove that
$$\sum_{cyc}\frac{\frac{y}{x}-1}{\frac{y}{x}\cdot\frac{z}{y}+1}\geq0$$ or
$$\sum_{cyc}\frac{y-x}{x+z}\geq0$$ or
$$(y-x)(y+x)(y+z)\geq0$$ or
$$\sum_{cyc}(y-x)(y^2+xy+xz+yz)\geq0$$ or
$$\sum_{cyc}(x^3-x^2z)\geq0,$$ which is true by Rearrangement:
$$\sum_{cyc}x^3=\sum_{cyc}(x^2\cdot x)\geq\sum_{cyc}(x^2\cdot z)=\sum_{cyc}x^2z.$$
The second inequality.
By AM-GM
$$x^4+y^4+z^2=x^4+y^4+2\cdot\frac{1}{2}z^2\geq4\sqrt[4]{x^4y^4\left(\frac{1}{2}z^2\right)^2}=$$
$$=\sqrt8|xyz|\geq\sqrt8xyz.$$
|
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|
Solve for $a_{n}$ where $a_{n} = 4a_{n-1} + 2^{2n-1}$, $a_0=1$ and $a_1=6$ . $a_{n} = 4a_{n-1} + 2^{2n-1}$, $a_0=1$ and $a_1=6$.
So I am trying to find $a_n$ by using the generating function let's call it $A(x)$.
the equation then is written as (if I doing this correctly) :
$A(x)-ao = 4xA(x) + \sum_{k=1}2^{2n-1}*x^k$
now for $\sum_{k=0}2^{2n-1}*x^k = \sum_{k=1}\frac{1}{2}4^{k}x^k \mapsto \frac{1}{2}\frac{1}{1-4x} - \frac{1}{2}$
Then we have: $(1-4x)A(x)= 1+\frac{1}{2}\frac{1}{1-4x} -\frac{1}{2}\Rightarrow A(x)=\frac{1}{2}\frac{1}{1-4x}+\frac{1}{2}\frac{1}{(1-4x)^2}\\ \Rightarrow a_n= ? $
$ (1-4x)^{-2} \mapsto 1 + \sum_{k=1}\binom{2+k-1}{k}*4^kx^{k}\\ \Rightarrow \frac{1}{(1-4x)^2}= 1+ \sum_{k=1}\binom{1+k}{k}*4^kx^{k} = 1+ \sum_{k=1}(k+1)4^kx^{k}= \sum_{k=0}(k+1)4^kx^{k}$
$And, \frac{1}{1-4x} \mapsto 4^n$
So $an = \frac{1}{2}4^n + \frac{1}{2}(n+1)4^n$ agrees with a1 = 6
|
Your initial equation for $A(x)$ is incorrect. You need to work the boundary condition $a_0 = 1$ into the equation. (The condition $a_1=6$ is redundant, since it can be computed from $a_0$ and the recurrence relation.)
There is more than one way to do this, but one way is to write the recurrence equation so it is true for all $n$. Although you did not state the restriction, the equation $a_n = 4 a_{n-1}+2^{2n-1}$ only holds for $n \ge 1$. It is important to state such restrictions explicitly in order to avoid errors.
So here is how to revise the equation so it holds for all $n$. Notice that if we simply let $n=0$ in the original equation, we have $1 = 0 + 2^{-1}$, which is obviously false. (We adopt the convention that $a_n = 0$ for all $n < 0$.) To patch things up, we write
$$a_n = 4 a_{n-1}+2^{2n-1} + (1/2) \delta_{0n}$$
where $\delta$ is the Kronecker delta function. Now we have a relation that is true for all $n$, so
$$a_n x^n = 4 a_{n-1} x^n +2^{2n-1} x^n + (1/2) \delta_{0n} x^n$$
for all $n$. Summing over $n=0,1,2,\dots$, we have
$$\sum_{n=0}^{\infty} a_n x^n = 4 \sum_{n=0}^{\infty} a_{n-1} x^n + \sum_{n=0}^{\infty}2^{2n-1} x^n + 1/2 $$
so
$$A(x) = 4x A(x) + \frac{1}{2} \cdot \frac{1}{1-4x} + \frac{1}{2}$$
Maybe you can take it from here.
|
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|
Prove that $\frac{1}{2} \lt \sum_{r=1}^{n} \frac{1}{n+r} \lt \frac{3}{4} , n>1$ I have proven the left hand side inequality. Can someone give me a hint for the right hand side ?
Proof for the left hand side:
$\sum_{r=1}^{n} \frac{1}{n+r} >\sum_{r=1}^{n} \frac{1}{2n}$.
|
By C-S $$\sum_{r=1}^n\frac{1}{n+r}=1+\sum_{r=1}^n\left(\frac{1}{n+r}-\frac{1}{n}\right)=1-\frac{1}{n}\sum_{r=1}^n\frac{r}{n+r}=$$
$$=1-\frac{1}{n}\sum_{r=1}^n\frac{r^2}{nr+r^2}\leq1-\frac{\left(\sum\limits_{r=1}^nr\right)^2}{n\sum\limits_{r=1}^n(nr+r^2}=1-\frac{\frac{n^2(n+1)^2}{4}}{n\left(\frac{n^2(n+1)}{2}+\frac{n(n+1)(2n+1)}{6}\right)}=$$
$$=\frac{7n-1}{2(5n+1)}\leq\frac{3}{4}.$$
|
{
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|
Find intersection between line and ellipsoid I want to find points $\space P(x,y,z) \space$ where a line intersect an ellipsoid with $$P = P_{1}+t(P_{2}-P_{1})$$
Here is where I stuck:
The ellipsoid can be described as:
$$\frac{(x-x_{3})^{2}}{a^{2}} + \frac{(y-y_{3})^{2}}{b^{2}} + \frac{(z-z_{3})^{2}}{c^{2}} = 1$$
after substitution:
$$\frac{(x_{2} - x_{1})^{2}t^{2}-2t(x_{2}-x_{1})(x_{3}-x_{1})+(x_{3}-x_{1})^{2})}{a^{2}} + \frac{(y_{2} - y_{1})^{2}t^{2}-2t(y_{2}-y_{1})(y_{3}-y_{1})+(y_{3}-y_{1})^{2})}{b^{2}} + \frac{(z_{2} - z_{1})^{2}t^{2}-2t(z_{2}-z_{1})(z_{3}-z_{1})+(z_{3}-z_{1})^{2}}{c^{2}} - 1 = 0$$
Substitution gives a quadratic equation of the form:
$$kt^{2} + lt + m = 0$$
Any help would be appreciated.
|
Translate the ellipsoid to origin, by subtracting $(x_3, y_3, z_3)$ from $P1$ and $P2$, so the line is parametrised as
$$\vec{p} = \vec{v}_0 + t \vec{v}_1$$
where
$$\begin{aligned}
\vec{v}_0 = (\chi_0, \gamma_0, \zeta_0) &= (x_1 - x_3, y_1 - y_3, z_1 - z_3) \\
\vec{v}_1 = (\chi_1, \gamma_1, \zeta_1) &= (x_2 - x_1, y_2 - y_1, z_2 - z_1) \\
\end{aligned}$$
and the ellipsoid is
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{b^2} = 1$$
Substituting $\vec{p}$ into the ellipsoid yields
$$\left(\frac{\chi_0 + t \chi_1}{a}\right)^2 + \left(\frac{\gamma_0 + t \gamma_1}{b}\right)^2 + \left(\frac{\zeta_0 + t \zeta_1}{c}\right)^2 = 1$$
Expand the terms and collect the coefficients for powers of $t$ and you get
$$\begin{aligned}
\left( \displaystyle \frac{\chi_1^2}{a^2} + \frac{\gamma_1^2}{b^2} + \frac{\zeta_1^2}{c^2} \right) & t^2 ~ + \\
\left( \displaystyle \frac{2 \chi_0 \chi_1}{a^2} + \frac{2 \gamma_0 \gamma_1}{b^2} + \frac{2 \zeta_0 \zeta_1}{c^2} \right) & t ~ + \\
\left( \frac{\chi_0^2}{a^2} + \frac{\gamma_0^2}{b^2} + \frac{\zeta_0^2}{c^2} - 1 \right) & ~ = 0 \\
\end{aligned}$$
This is a simple quadratic equation in $t$, which can have 0, 1, or 2 real roots $t$.
Remember that we only translated the coordinate system so that they were relative to the center of the ellipsoid, but we didn't scale or rotate the coordinate system. So, after you have found $t$, you can find the point it refers to, in the original coordinate system, using your original parametrised line,
$$P = P1 + t ( P2 - P1 )$$
i.e.
$$\left\lbrace ~ \begin{aligned}
x &= x_1 + t \chi_1 \\
y &= y_1 + t \gamma_1 \\
z &= z_1 + t \zeta_1 \\
\end{aligned} \right.$$
for each intersection point $t$.
|
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|
What went wrong in the evaluation of $\int \frac{1}{3-2\sin(x)}dx$? I tried to evaluate the following integral:
$$\int \frac{1}{3-2\sin(x)}dx\,\, $$ with universal substitution, using the fact that
$t(x):=\tan\left(\frac{x}{2}\right)$
$\sin(x)=\frac{2t(x)}{1+t(x)^2}$ and $t'(x)=\frac{1+t(x)^2}{2}$
$\displaystyle\int \frac{1}{3-2\sin(x)}dx=$$\displaystyle\int \frac{1}{3-2(\frac{2t(x)}{1+t(x)^2})}dx=\displaystyle\int \frac{1}{3-\frac{4t(x)}{1+t(x)^2}}dx=\displaystyle\int \frac{1}{\frac{3+3t(x)^2-4t(x)}{1+t(x)^2}}dx=\displaystyle\int \frac{1+t(x)^2}{3+3t(x)^2-4t(x)}dx$
Now I substituted: $u:=t(x) \Longrightarrow dx= \frac{du}{t'(x)}=\frac{2du}{1+t(x)^2}$
So:
$\displaystyle\int \frac{1+u^2}{3+3u^2-4u}\frac{2}{1+u^2}du=\displaystyle\int \frac{2}{3+3u^2-4u}du=2\displaystyle\int \frac{1}{3u^2-4u+3}du=2\displaystyle\int \frac{1}{(\sqrt{3}u+\frac{2}{\sqrt{3}})^2+\frac{5}{3}}du$
Here I used the formular:
$\displaystyle\int \frac{1}{t^2+m^2}dt=\left[\frac{1}{m}\arctan\left(\frac{t}{m}\right)\right]$
So:
$2\displaystyle\int \frac{1}{(\sqrt{3}u+\frac{2}{\sqrt{3}})^2+\frac{5}{3}}du=2\left[\frac{\sqrt{3}}{\sqrt{5}}\arctan\left(\frac{(\sqrt{3}u-\frac{2}{\sqrt{3}})\sqrt{3}}{\sqrt{5}}\right)\right]=2\left[\frac{\sqrt{3}}{\sqrt{5}}\arctan\left(\frac{3u-2}{\sqrt{5}}\right)\right]$
Resubstituting:
$2\left[\frac{\sqrt{3}}{\sqrt{5}}\arctan\left(\frac{3\tan\left(\frac{x}{2}\right)-2}{\sqrt{5}}\right)\right]$
Here is the issue:
Wolfram tells:
$\displaystyle\int \frac{1}{3-2\sin(x)}dx=2\left[\frac{1}{\sqrt{5}}\arctan\left(\frac{3\tan\left(\frac{x}{2}\right)-2}{\sqrt{5}}\right)\right]$
I cannot figure out where I went wrong.. like.. its only one factor..
could someone maybe show me what went wrong? Thank you :)
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You made a mistake when applying
$$\displaystyle\int \frac{1}{t^2+m^2}dt=\left[\frac{1}{m}\arctan\left(\frac{t}{m}\right)\right].$$
Define $t := (\sqrt{3}u + \frac{2}{\sqrt{3}})$ as you did implicitly, but make sure to rework $du$ into $c \cdot dt$.
|
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|
Is there a nice way to represent $\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n+m+1}$? Here, $H_n$ denotes the harmonic number. More colloquially, is there any way to represent $$\int_0^1 x^{n-1}\log^2\left(1+x\right)\ \mathrm{d}x$$ in a nice way? The latter is corollary to the original question; namely,
$$\int_0^1 x^{n-1}\log^2\left(1+x\right)\ \mathrm{d}x = \frac{\log^2(2)}{n}-\frac{2}{n}\sum_{k=1}^\infty \frac{(-1)^{k+1}H_{\ell}}{n+k+1}...$$
which I do not know how to simplify. Thank you!
|
Mathematica choked when I gave it the general expression (at least the way I tried). Here's how the sum works out for some small values of $m$.
\begin{gather}
m = 0 : \tfrac{1}{2}\log^2(2) \\
m = 1 : -1 + 2 \log(2) - \tfrac{1}{2}\log^2(2) \\
m = 2 : \tfrac{5}{4} - 2 \log(2) + \tfrac{1}{2}\log^2(2) \\
m = 3 : - \tfrac{55}{36} + \tfrac{8}{3} \log(2) - \tfrac{1}{2}\log^2(2) \\
m = 4 : \tfrac{241}{144} -\tfrac{8}{3} \log(2)-\tfrac{1}{2}\log^2(2) \\
m = 5 : - \tfrac{6589}{3600} + \tfrac{46}{15} \log(2) + \tfrac{1}{2}\log^2(2)
\end{gather}
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|
Solving $\sin^2x + 3\sin x\cos x + 2\cos^2x=0$, for $0\leq x\leq 2\pi$
Solve
$$\sin^2x + 3\sin x\cos x + 2\cos^2x=0$$
for $0\leq x\leq 2\pi$.
My answers are
$$x=2.03, 5.18 \qquad\text{or}\qquad
x=\frac{3\pi}{4},\frac{7\pi}{4} \qquad\text{or}\qquad
x=\frac{\pi}{2}, \frac{3\pi}{2},$$
but the answer states $x=2.03, 5.18$ or $x=3\pi/4,7\pi/4$ only.
I got $x=\pi/2, 3\pi/2$ from $(\cos x)^2=0$,
where it is a factor in one of my steps:
$$\cos^2x\left(\tan^2x+3\tan x+2\right)=0.$$
|
Look at this $$x^2+3xy+2y^2=(x+y)(x+2y)$$ can you see??
|
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|
Finding the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$
Find the value of $\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$.
My attempt: $$\sin^{-1}\frac{12}{13}+\cos^{-1}\frac{4}{5}+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}\frac{12}{5}+\tan^{-1}\frac{3}{4}+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot\frac{3}{4}})+\tan^{-1}\frac{63}{16}$$
$$=\tan^{-1}\frac{63}{-16}+\tan^{-1}\frac{63}{16}$$
$$=-\tan^{-1}\frac{63}{16}+\tan^{-1}\frac{63}{16}$$
$$=0$$
But the answer is given as $\pi$. What is my mistake?
|
Now I am going to prove that:
$\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$
for all $x,y\in\left]0,+\infty\right[$ such that $xy>1$.
Proof:
For all $x,y\in\left]0,+\infty\right[$ such that $xy>1$, it results that:
$x>\frac{1}{y}=\cot\left(\tan^{-1}y\right)=\tan\left(\frac{\pi}{2}-\tan^{-1}y\right)$.
Since $\frac{\pi}{2}-\tan^{-1}y\in \left]0,\frac{\pi}{2}\right[$ and tangent is an invertible and monotone increasing function in $\left]-\frac{\pi}{2},\frac{\pi}{2}\right[$, it follows that:
$\tan^{-1}x>\frac{\pi}{2}-\tan^{-1}y$, therefore:
$\tan^{-1}x+\tan^{-1}y-\pi>-\frac{\pi}{2}$. (*)
On the other hand,
$\tan^{-1}x+\tan^{-1}y<\pi$, therefore:
$\tan^{-1}x+\tan^{-1}y-\pi<0$. (**)
From (*) and (**), it follows that:
$\tan^{-1}x+\tan^{-1}y-\pi \in \left]-\frac{\pi}{2},0\right[$.
$\tan\left(\tan^{-1}x+\tan^{-1}y-\pi\right)=\frac{\tan\left(\tan^{-1}x\right)+ \tan\left(\tan^{-1}y-\pi\right)}{1- \tan\left(\tan^{-1}x\right)\cdot\tan\left(\tan^{-1}y-\pi\right)}$
$\tan\left(\tan^{-1}x+\tan^{-1}y-\pi\right)=\frac{\tan\left(\tan^{-1}x\right)+ \tan\left(\tan^{-1}y\right)}{1- \tan\left(\tan^{-1}x\right)\cdot\tan\left(\tan^{-1}y\right)}$
$\tan\left(\tan^{-1}x+\tan^{-1}y-\pi\right)=\frac{x+y}{1-xy}$
Since $\tan^{-1}x+\tan^{-1}y-\pi \in \left]-\frac{\pi}{2},0\right[$ and tangent is invertible in $\left]-\frac{\pi}{2},\frac{\pi}{2}\right[$, it follows that:
$\tan^{-1}x+\tan^{-1}y-\pi=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, therefore:
$\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$.
|
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|
Finding a formula for a function using the 2nd fundamental theorem of calculus $$f(t) =\begin{cases} t &\text{if }0 \leq t \leq 1, \\
2-t &\text{if }1 < t \leq 2, \\
t^2 &\text{if }t > 2.
\end{cases}$$
Find a formula for the function $F: [0, 2] \rightarrow \mathbb{R}$ defined by $F(x) = \int_0^xf(t)\mathrm dt$ for x $\geq$ 0
Working:
For $0 \leq t \leq 1$:
\begin{align*}
F(x) &= \int_{0}^xf(t)\mathrm dt\\
&= \int_{0}^x t\mathrm dt\\
&= \left[\frac{1}{2} t^2\right]^x_{0}\\
&= \frac{1}{2}x^2
\end{align*}
I'm confused from that point onwards - ie. how to find the second bit. My teacher broke up the integrand (I understand that bit) but then I think she used $F(1)$ for the first part of the integrand. I don't really understand why she picked $1$ in particular.
Also, two queries:
*
*Does the way you apply the 2nd Fundamental Theorem change depending on if you are dealing with $<$ vs. $\leq$
*My teacher said, for $0 \leq t \leq 1$, let $g(t) = \frac{1}{2}t^2$ so $g'(t) = f(t)$ but I don't really see why the Fundamental Theorem helps much except by making things more confusing.
|
Your given function is given piecewise.
$$ f(t) = \begin{cases}
t ,& 0 \leq t \leq 1 \\
2-t ,& 1 < t \leq 2 \\
t^2 ,& 2 < t
\end{cases} $$
You should expect your accumulation function to also be given piecewise.
\begin{align*}
F(x) &= \int_0^x f(t) \,\mathrm{d}t \\
&= \begin{cases}
\int_0^x f(t) \,\mathrm{d}t &, 0 \leq x \leq 1 \\
\int_0^1 f(t) \,\mathrm{d}t + \int_1^x f(t) \,\mathrm{d}t &, 1 < x \leq 2 \\
\int_0^1 f(t) \,\mathrm{d}t + \int_1^2 f(t) \,\mathrm{d}t + \int_2^x f(t) \,\mathrm{d}t &, 2 < x \\
\end{cases}
\end{align*}
(When $x$ is in $[0,1]$, we only need to know the first piece of $f$ to get the accumulation from $0$ to $x$. When $x$ is in $(1,2]$, we first accumulate from $0$ to $1$ using the first piece of $f$, then from $1$ to $x$ using the second piece. When $x > 2$, we fully accumulate on $[0,1]$ and on $(1,2]$ then as much of the third piece as needed to reach $x$.)
\begin{align*}
&= \begin{cases}
\int_0^x t \,\mathrm{d}t &, 0 \leq x \leq 1 \\
\int_0^1 t \,\mathrm{d}t + \int_1^x (2-t) \,\mathrm{d}t &, 1 < x \leq 2 \\
\int_0^1 t \,\mathrm{d}t + \int_1^2 (2-t) \,\mathrm{d}t + \int_2^x t^2 \,\mathrm{d}t &, 2 < x \\
\end{cases} \\
&= \begin{cases}
\left.\frac{t^2}{2} \right|_{t=0}^x &, 0 \leq x \leq 1 \\
\left. \frac{t^2}{2} \right|_{t=0}^1 + \left. (2t-\frac{t^2}{2})\right|_{t=1}^{x} &, 1 < x \leq 2 \\
\left. \frac{t^2}{2} \right|_{t=0}^1 + \left. (2t-\frac{t^2}{2})\right|_{t=1}^{2} + \left. \frac{t^3}{3} \right|_{t=2}^x &, 2 < x \\
\end{cases} \\
&= \begin{cases}
\frac{x^2}{2} - 0 &, 0 \leq x \leq 1 \\
\frac{1}{2} - 0 + \left( 2x-\frac{x^2}{2} \right) - \left( 2-\frac{1}{2} \right) &, 1 < x \leq 2 \\
\frac{1}{2} - 0 + (4-2) - (2-\frac{1}{2}) + \frac{x^3}{3} - \frac{8}{3} &, 2 < x \\
\end{cases} \\
&= \begin{cases}
\frac{x^2}{2} &, 0 \leq x \leq 1 \\
-1 + 2x - \frac{x^2}{2} &, 1 < x \leq 2 \\
\frac{x^3 - 5}{3} &, 2 < x \\
\end{cases}
\end{align*}
(A check: This is an accumulation, so we expect the first two pieces to glue together continuously at $x = 1$. Both are continuous so we can evaluate the limit (from the left) of the first piece by specializing $x \mapsto 1$, obtaining $1/2$. Similarly specializing in the second piece, we obtain $-1+2- 1/2 = 1/2$. So the two pieces do match at $x = 1$. Checking again at $x = 2$, the second and third pieces meet at height $1$.)
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|
Evaluate hypergeometric $_6F_5\left(\{\frac12\}_6;1,\{\frac32\}_4;1\right)$ Background: I'm searching for $_pF_q$ representations for MZVs. In related article On the interplay between hypergeometric series, Fourier-Legendre expansions and Euler sums by M. Cantarini and J. D’Aurizio, the series $_5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)$ is transformed to and Euler sum i.e. $\sum _{n=1}^{\infty } \frac{(-1)^n \left(\sum _{k=1}^n \frac{1}{2 k+1}\right){}^3}{2 n+1}$ by using FL expansion, which I invoke MZV values to give a closed-form successfully:
*
*$ \pi \, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=64 \Im(\text{Li}_4(1+i))-\frac{2}{3} \pi \log ^3(2)-\pi ^3 \log (2)-\frac{1}{32} \left(\psi ^{(3)}\left(\frac{1}{4}\right)-\psi ^{(3)}\left(\frac{3}{4}\right)\right)$
Problem: I wonder if the higher-weight case can be evaluated by similar means:
*
*$ \pi \sum _{n=0}^{\infty } \left(\frac{\binom{2 n}{n}}{4^n}\right)^2\frac{1}{(2 n+1)^4}=\pi \, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)$
FL expansion of $\frac{\log ^3(x)}{\sqrt{x}}$ is needed here, but I'm not able to compute it.
Update: Using Jack's formula one may deduce
*
*$\small \pi \, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};1\right)=-40 \Im(\text{CMZV}(4,\{4,1\},\{1,0\}))+\frac{152}{3} \Im(\text{CMZV}(4,\{4,1\},\{1,2\}))-256 \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{16}{3} \beta(4) \log (2)+\frac{25 \pi ^5}{64}+\frac{1}{6} \pi \log ^4(2)+\frac{3}{4} \pi ^3 \log ^2(2)$
See here for detailed expanlation.
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All right, I am going to re-do my computations from scratch. This will probably take some time, so please do not downvote this answer in the meanwhile. For any $n\geq 3$ we have
$$ (-1)^{n+1}x^{n+1/2}\cdot \frac{d^n}{dx^n} \frac{\log^3(x)}{\sqrt{x}} = A_n+B_n\log(x)+C_n\log^2(x)+D_n\log^3(x)=S_n\tag{S}$$
with $A_n,B_n,C_n,D_n\in\mathbb{Q}$ related to each other by induction / recurrence relations. Once the explicit form of these constants is known we also have the explicit FL-expansion of $\frac{\log^3(x)}{\sqrt{x}}$ by Rodrigues formula, since
$$ \int_{0}^{1}\frac{\log^3(x)}{\sqrt{x}}P_n(2x-1)\,dx = \frac{1}{n!}\int_{0}^{1}x^n (1-x)^n \left[\frac{d^n}{dx^n}\frac{\log^3(x)}{\sqrt{x}}\right]\,dx \tag{B}$$
and the RHS is given by derivatives of the Beta function.
The coefficient which is the easiest to guess is $D_n$:
$$ D_n = -\frac{1\cdot 3\cdot\ldots\cdot(2n-1)}{2^n} = -\frac{(2n)!}{4^n n!}$$
then we may differentiate both sides of $(S)$ and write down the induced recurrence relations.
$$ \left(n+\tfrac{1}{2}\right)S_n - S_{n+1} = B_n + 2C_n \log(x)+3D_n\log^2(x)\tag{R}$$
Focusing on the coefficient of $\log^2(x)$ on both sides we get
$$ \left(n+\tfrac{1}{2}\right)C_n-C_{n+1} = 3D_n \tag{D_n}$$
and it is practical to introduce rescaled coefficients for simplifying the recursion. Letting $D_n=\frac{(2n)!}{4^n n!}d_n$ (and the same for $A_n,B_n,C_n$) we get
$$ \left(n+\tfrac{1}{2}\right)(c_n-c_{n+1}) = 3d_n = -3 $$
so
$$ c_{n+1} = c_n+\frac{6}{2n+1} $$
and
$$ C_n = \frac{(2n)!}{4^n n!}\cdot 6\sum_{k=0}^{n-1}\frac{1}{2k+1}.\tag{C_n}$$
[...] Continuing on this route, once we define $\mathscr{H}_n^{(k)}$ as $\sum_{h=0}^{n}\frac{1}{(2h+1)^k}$ we get
$$ \boxed{\small\frac{\log^3(x)}{\sqrt{x}}\stackrel{\mathcal{D}}{=}32\sum_{n\geq 0}(-1)^{n+1}P_n(2x-1)\left[4\mathscr{H}_n^3+2\mathscr{H}_n^{(3)}-\frac{6\mathscr{H}_n^2}{2n+1}+\frac{6\mathscr{H}_n}{(2n+1)^2}-\frac{3}{(2n+1)^3}\right]}.$$
Since $\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{1}{(2n+1)^4}$ can be represented in terms of $\frac{2}{\pi}\int_{0}^{1}\frac{\log^3(x)}{\sqrt{x}}K(x)\,dx $, the simple FL-expansion of $K(x)$ gives that the first hypergeometric series can be computed in terms of five Euler sums with weight five. The simplest of them are
$$ \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^5} = \frac{5\pi^5}{1536}$$
and
$$ \sum_{n\geq 0}\frac{(-1)^n\mathscr{H}_n}{(2n+1)^4}=-\frac{1}{96}\int_{0}^{1}\frac{\log^3(x)}{\sqrt{x}}\sum_{n\geq 0}(-1)^n\mathscr{H}_n x^n\,dx $$
which equals
$$ -\frac{1}{96}\int_{0}^{1}\frac{\log^3(x)\arctan(x)}{x(1+x)}\,dx=\frac{5\pi^5}{24576}-\frac{1}{96}\int_{0}^{1}\frac{\log^3(x)\arctan(x)}{x+1}\,dx. $$
The factor $\frac{dx}{x+1}$ is invariant with respect to the substitution $x\to\frac{1-x}{1+x}$, so the last integral can also be expressed in terms of $\int_{0}^{1}\text{arctanh}^3(x)\left(\frac{\pi}{4}-\arctan x\right)\frac{dx}{x+1}$, where
$$ \int_{0}^{1}\text{arctanh}^3(x)\frac{dx}{x+1}\,dx = \int_{0}^{+\infty}x^3(1-\tanh x)\,dx = \frac{7\pi^4}{960}$$
by the integral representations for the $\eta$ and $\zeta$ functions.
|
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|
Proving $a^4+2a^3 b+2ab^3+b^4 ≥ 6a^2b^2$ Prove that for any positive real numbers $a$ and $b$,$$a^4 + 2a^3b + 2ab^3 + b^4 ≥ 6a^2b^2.$$I tried using Vieta's formula to show the product of the LHS is greater than the RHS, but I don't think I am correct.
|
You can easily prove it using AM-GM Inequality or Lagrange method. Using AM-GM Inequality,
$$ \frac {a^4 + b^4}{2} \ge a^2.b^2 $$
$$ \frac {a^2 + b^2}{2} \ge a.b$$
$$ \text {Also, you can write }a^3.b+a.b^3 = ab.(a^2+b^2)$$
$$ \text {So, } a^4 + b^4 +2a^3.b+2a.b^3 \ge 6.a^2.b^2$$
|
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Let $\langle x_n\rangle$ be a recursive relation. Find $\lim_{n\to\infty}\frac {x_n}{n^2}.$ Let $\langle x_n\rangle$ be a recursive relation given by $$x_{n+1}=x_n+a+\sqrt {b^2+4ax_n}, n\geq0, x_0 =0$$ and $a$ and $b$ are fixed positive integers. Find $$\lim_{n\to\infty}\frac {x_n}{n^2}.$$
|
Clearly, $\lim_{n\to \infty} x_n = \infty$. We have
\begin{align}
\sqrt{x_{n+1}} - \sqrt{x_n} &= \sqrt{x_n + a + \sqrt{b^2 + 4ax_n}} - \sqrt{x_n}\\[6pt]
&= \frac{a + \sqrt{b^2 + 4ax_n}}{\sqrt{x_n + a + \sqrt{b^2 + 4ax_n}} + \sqrt{x_n}}\\[6pt]
&\to \sqrt{a} \quad \mathrm{as}\quad n \to \infty.
\end{align}
By the Stolz-Cesaro theorem, we have
$$\lim_{n\to \infty} \frac{\sqrt{x_n}}{n} = \sqrt{a}$$
and thus
$$\lim_{n\to \infty} \frac{x_n}{n^2} = a.$$
We are done.
|
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If $ 3a+2b+c=7$ then find minimum value of $ a^2+b^2+c^2$ Question:- If $ 3a+2b+c=7$ then find the minimum value of $ a^2+b^2+c^2 $.
I used vectors to solve this problem.
Let $$α=3\hat{i}+2\hat{j}+\hat{k}$$
$$β=a\hat{i}+b\hat{j}+c\hat{k}$$
Using Cauchy-Schwarz inequality
we have, $|α.β|\le |α| |β|$
$=|3a+2b+c|\le\sqrt{14}\sqrt{a^2+b^2+c^2}$
$= 7\le\sqrt{14}\sqrt{a^2+b^2+c^2}$
So, $a^2+b^2+c^2\ge \frac72$
Therefore, the minimum value of $a^2+b^2+c^2$ is $\frac72$
I want to know are there any other method to find the minimum value of
$a^2+b^2+c^2$ such as using inequalities and calculus by assuming function $f(x,y,z)=x^2+y^2+z^2$.
|
Also, $$a^2+b^2+c^2=\frac{1}{14}(3^2+2^2+1^2)(a^2+b^2+c^2)\geq\frac{1}{14}(3a+2b+c)^2=3.5.$$
The equality occurs for $$(3,2,1)||(a,b,c),$$ which says that we got a minimal value.
|
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If $x, y, z\in\mathbb R^+ $ and $x^3+y^3=z^3,$ then prove that $x^2+y^2-z^2>6(z-x) (z-y). $ I made several unsuccessful attempts. Still couldn't think of a proper way to prove the inequality. Please suggest how to approach this problem. Thanks in advance.
EDIT 1. My approach (that I was talking about):
Given: $z^3=x^3+y^3.$
We have to prove:
$x^2+y^2-z^2>6(z-x) (z-y)$ i.e., $\underbrace{(z^2+zx+x^2) (z^2+zy+y^2) (x^2+y^2-z^2)}_{=E\text{ (say)}}>6(z^3-x^3) (z^3-y^3)=6x^3y^3.$
(Here one thing which I noticed is that $(x^2+y^2-z^2)>0,$ since each of the terms on the LHS except this one is positive and $6x^3y^3$ is also positive for $x, y, z>0.$)
Using AM $\ge$ GM, we have:
$E\ge 3zx\cdot3zy\cdot(x^2+y^2-z^2)\ge 9xyz^2(2xy-z^2).$
From here I couldn't think of a proper way to prove $E>6x^3y^3.$ But I'm still working on it. At present I'm trying to manipulate the expression $9xyz^2(2xy-z^2)$ to get the job done. If I find something useful I'll update here.
|
Because each expression is homogenous, we may assume that $ z = 1$ (by using the substitution $ x' = \frac{x}{z}$). The question becomes:
If $x^3 + y^3 = 1$, show that $x^2 - 6xy + y^2 + 6x + 6 y - 7 > 0$.
Note: This 2-variable inequality type is common, and there are several ways of dealing with it by exploiting $x+y$.
Proof: Let $ w = x+y$. Since $ (x^3+y^3) < (x+y)^3 \leq 4(x^3 + y^3)$, thus $ 1 < w \leq \sqrt[3]{4} $.
$(x+y)(x^2 - 6xy + y^2 + 6x + 6 y - 7) \\
= x^3 +y^3 - 5xy (x+y) + 6(x+y)^2 - 7 (x+y) \\
= 1 - \frac{5}{3} ( (x+y)^3 - 1) + 6(x+y)^2 - 7 (x+y) \\
= \frac{1}{3}(-5 w^3 + 18w^2 - 21w + 8) \\
= \frac{1}{3}(w-1)^2(-5w+8)
$
Since $ 1 < w \leq \sqrt[3]{4} < \frac{8}{5}$, the final expression is positive, and we conclude that $x^2 - 6xy + y^2 + 6x + 6 y - 7 > 0$.
Of course, we didn't need to do the initial substitution, and could have shown that $$(x+y)\left[ x^2+y^2-z^2 - 6 ( z-y)(z-x)\right] = \frac{1}{3} ( x+y -z )^2 ( 8z - 5x - 5y )> 0.$$
However, it's hard to see how the equality holds by applying $(x+y)^3 = z^3 + xy(x+y)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Number of Real Solutions of $\frac{7^{1+\cos(\pi x)}}{3}+3^{x^2-2}+9^{\frac{1}{2}-|x|}=1$
Find the number of real solutions of the equation $$\frac{7^{1+\cos(\pi x)}}{3}+3^{x^2-2}+9^{\frac{1}{2}-|x|}=1\,.$$
By hit and trial i got the solution at $x=\pm 1$ but i am not able to solve it as it involves power of 7
|
Note that $\frac{7^{1+cos\pi x}}{3}+3^{x^2-2}+9^{\frac{1}{2}-|x|}$ is an even function. So it's enough to consider $x\ge 0$.
Let $$f(x) = \frac{7^{1+cos\pi x}}{3}+3^{x^2-2}+9^{\frac{1}{2}-x}$$
If $x \gt \sqrt{2} \ $ then $3^{x^2 - 2}\gt 1$ and also $\frac{7^{1+cos\pi x}}{3} \gt 0 \ \ , \ 9^{\frac{1}{2}-x} \gt 0$. Therefore $f(x) \gt 1$ and there is no root in this case. In the interval $0\le x \le 1$ we will show $\frac{7^{1+cos\pi x}}{3}$ is a decreasing function and also $3^{x^2-2}+9^{\frac{1}{2}-x}$. So there sum will be decreasing and the only root occurs when $x = 1$. When $1\lt x \le \sqrt{2}$ we have to compute the derivative $$f'(x) = -\frac{7\pi\ln 7}{3}\times7^{\cos \pi x} \sin{\pi x} + (2x)(\ln 3) 3^{x^2 -2} + (-2)(\ln 3)3^{1-2x}$$
In the mentioned interval $-\sin \pi x$ and $\cos \pi x$ are increasing so $-\frac{7\pi\ln 7}{3}\times7^{\cos \pi x} \sin{\pi x}$ is increasing. In the similar manner we can show two other functions are increasing, so $f'(x)$ is increasing when $1\lt x \le \sqrt{2}$. So we have $f'(x) \gt f'(1) = 0$. We can conclude there is no root in $1\lt x \le \sqrt{2}$.
Let $0\le x \le 1$ then $$g(x) = \frac{7^{1+cos\pi x}}{3} \implies g'(x) = -\frac{7\pi\ln 7}{3}\times7^{\cos \pi x} \sin{\pi x} \le 0$$
And $$h(x) = 3^{x^2-2}+9^{\frac{1}{2}-x} \implies h'(x) = (2x)(\ln 3) 3^{x^2 -2} + (-2)(\ln 3)3^{1-2x} = 2\ln 3(x\times 3^{x^2 - 2} - 3^{1 - 2x}) \le 0$$
Because $$x\times 3^{x^2 - 2} - 3^{1 - 2x} \le 0 \iff \log_{3} (x\times 3^{x^2 - 2}) \le \log_{3} (3^{1 - 2x}) \iff \log_{3} (x) + x^2 - 2 \le 1 -2x \iff \log_{3} (x) \le -x^2 -2x + 3 \le 3$$
Clearly this answer shows how much Batominovski's solution is elegant!
|
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|
Find whether the sequence is convergent . Find whether the sequence $(a_n)$ given by $a_{n+1}= \sqrt{a_n}+\sqrt{a_{n-1}}$, where $a_1=1$ and $a_2=2$, is convergent.
So , $a_{n+1}-a_{n}= \sqrt{a_n} + \sqrt{a_{n-1}} -a_n \implies \sqrt{a_n}(1-\sqrt{a_n})+ \sqrt{a_{n-1}}.$
Now I assumed the sequence is $>1$ and I showed it by induction then ,
$a_{n+1}-a_{n} < \sqrt{a_{n-1}}$.Any help from here ?
|
We show that the sequence is convergent by showing that it's bounded and monotone.
Claim 1. $2 \le a_n \le 4$ for all $n \ge 2.$
Proof. We prove this via induction. For $n = 2, 3$, it is manually verified.
Let $P(n)$ denote the statement "$2 \le a_n \le 4$".
Assume that $n \ge 4$ and that $P(k)$ is true for all $2 \le k \le n-1$. We prove that $P(n)$ is true.
By hypothesis, we have
\begin{align}
a_n &= \sqrt{a_{n-1}} + \sqrt{a_{n-2}}\\
&\ge \sqrt{2} + \sqrt{2} = 2\sqrt{2}\\
&\ge2.
\end{align}
Similarly, we have
\begin{align}
a_n &= \sqrt{a_{n-1}} + \sqrt{a_{n-2}}\\
&\le \sqrt{4} + \sqrt{4}\\
&=4.
\end{align}
This proves the statement.
Claim 2. $a_n \le a_{n+1}$ for all $n \ge 1$.
Proof. Let $P(n)$ denote the statement "$a_n \le a_{n+1}$".
$P(n)$ can be manually verified for $n = 1, 2, 3.$
Assume that $n \ge 4$ and that $P(k)$ is true for all $1 \le k \le n-1$. We prove that $P(n)$ is true.
Using the hypothesis, we see that $$a_n \ge a_{n-1} \ge a_{n-2}.$$
(Note that $n \ge 3$, so all these terms are defined.)
By the previous claim, we also see that all the terms are positive and thus, we can conclude
$$\sqrt{a_n} \ge \sqrt{a_{n-2}}. \quad (*)$$
Now, we have
\begin{align}
a_{n+1} - a_n &= \sqrt{a_n} + \sqrt{a_{n-1}} - \sqrt{a_{n-1}} - \sqrt{a_{n-2}}\\
&= \sqrt{a_n} - \sqrt{a_{n-2}}\\
&\ge 0.
\end{align}
The last inequality followed using $(*)$.
This proves this claim as well.
By Claim 1, the sequence is bounded and by Claim 2, the sequence is monotone. Thus, the sequence converges. (The value of the limit can be found to be $4$.)
|
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|
For which $k$ does $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt[3]{abc}}$ hold? By generalizing this (1) and this (2) questions and performing some research
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+k-3\ge \left(2+\frac k3\right)\cdot \frac{a+b+c}{\sqrt[3]{abc}},\hbox{ for }a,b,c>0$$
for all $0\le k<k_0\approx 11.108$.
The main goal was to prove the original inequality from (2), however, letting $a=x^3,\,b=y^3,\,c=z^3$ and clearing the denominator, the inequality becomes
$$3 k x^3 y^3 z^3 + 3 \sum\limits_{sym}x^6 y^3 z^0 - \left(3+\frac k2\right)\sum\limits_{sym} x^5 y^2 z^2\ge 0\tag{1}$$
and I'm failing to apply Muirhead's inequality.
The method from this answer works only for $k\le 3$, and even with calculus I don't think that solving system of $3$ equations like $\frac{\partial}{\partial x}$LHS(1)$=0$:
$$5 k x^3 y^2 z^2 - 9 k x y^3 z^3 + 2 k y^5 z^2 + 2 k y^2 z^5 - 18 x^4 y^3 - 18 x^4 z^3 + 30 x^3 y^2 z^2 - 9 x y^6 - 9 x z^6 + 12 y^5 z^2 + 12 y^2 z^5=0$$
may lead to something neat.)
Any help is appreciated. Thanks.
The question: what is $k_0$.
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Suppose $abc=1,$ and let $b=c=t,\,a=\frac{1}{t^2},$ the inequality become
$$k \leqslant \frac{6(t+1)(t^3+t^2+1)}{t(2t+1)} = F(t).$$
Easy to find
$$k \leqslant k_0 = F(t_0) = \frac{9\sqrt{665}}{8}\sin{\left(\frac{\pi}{6}+\frac{1}{3}\arccos{\frac{13117\sqrt{665}}{442225}}\right)}-\frac{141}{16} = 11.1086,$$
for $ \displaystyle t_0 = \frac{\sqrt 5}{2} \cos \left(\frac{\arctan(2 \sqrt{31})}{3}\right)-\frac 14 = 0 .7345.$
Finally, we will show that the inequality below is true for all $k \leqslant k_0$
$$f(a,b,c) = (a+b+c)(ab+bc+ca) + k - 3 - \left(\frac{k}{3}+2\right)(a+b+c) \geqslant 0. $$
Indeed, asumme $a = \max \{a,b,c\}$ and $t = \sqrt{bc},$ then $a \geqslant 1,$ we have
$$f(a,b,c) - f(a,t,t) = (\sqrt b - \sqrt c)^2 \left[3(a^2+ab+bc+ca)+ 6 at - k - 6\right].$$
According to the AM-GM inequality, we have
$$a^2+ab+bc+ca \geqslant 4a t,$$
so
$$3(a^2+ab+bc+ca)+ 6 at \geqslant 18at \geqslant 18 > k_0+6 \geqslant k + 6.$$
Thefore $f(a,b,c) \geqslant f(a,t,t),$ and
$$f(a,t,t) = f\left(\frac{1}{t^2},t,t\right) = \frac{(t-1)^2}{3t^3} \left[6(t+1)(t^3+t^2+1)-t(2t+1)k\right] \geqslant 0.$$
The proof is completed.
Note. This is Ji Chen inequality.
|
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|
Find matrix $B$ such that $B^2 = I - A$
$A$ is a squared matrix such that $A^4 = \mathbf{0}$. Find $B$ in form of $A$ such that $B^2 = I - A$.
I tried in this way:
\begin{gather*}
A = I - B^2 \\
A ^ 4 = ( I - B)^4 (I + B)^4
\end{gather*}
So $I = B$ or $- I = B$
But the question wants $B$ as a form of $A$.
any help?
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Assume $B$ can be written as a polynomial in $A$. That is,
\begin{align*}
B = c_0I+c_1A+c_2A^2+c_3A^3
\end{align*}
for unknown constants $c_0,c_1,c_2$, and $c_3$. Higher powers vanish since $A^4=0$. The relation $B^2=I-A$ gives
\begin{align*}
I-A = B^2 = c_0^2I+2c_0c_1A+(2c_0c_2+c_1^2)A^2+2(c_0c_3+c_1c_2)A^3.
\end{align*}
Comparing coefficients on either side yields $c_0=1$, $c_1=-\tfrac{1}{2}$, $c_2=-\tfrac{1}{8}$, and $c_3=-\tfrac{1}{16}$. So
\begin{align*}
B = I-\tfrac{1}{2}A-\tfrac{1}{8}A^2-\tfrac{1}{16}A^3
\end{align*}
solves $B^2=I-A$.
|
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|
Find the values of $\theta$ for which the tangent line to the given curve is parallel to $x$ ,$y$ axis Given the curve $$r(\theta):=\sec\left(\theta\right)+a\cos\left(\theta\right) \tag{$a \in \mathbb R$}$$
Find the values of $\theta$ for which the tangent line to the curve is parallel to the $x$ and $y$ axis.
*
*The points for which the tangent line to the curve is parallel to the $y$ axis is given by :
$$\frac{dx}{d\theta}=0$$$$\left(\sec\left(\theta\right)\tan\left(\theta\right)-a\sin\left(\theta\right)\right)\cos\left(\theta\right)-\sin\left(\theta\right)\left(\sec\left(\theta\right)+a\cos\left(\theta\right)\right)=0$$
$$\tan\left(\theta\right)-a\sin\left(\theta\right)\cos\left(\theta\right)-\tan\left(\theta\right)-a\sin\left(\theta\right)\cos\left(\theta\right)=0$$
Assuming $a\ne0$:
$$\sin\left(\theta\right)\cos\left(\theta\right)=0$$
$$\theta=\frac{k\pi}{2}\tag{$k \in \mathbb Z$}$$
On the other hand duo the existence of $\sec$ function we see that the acceptable $\theta$'s are :
$$\theta=\frac{2k\pi}{2}=k\pi\tag{$k \in \mathbb Z$}$$
Implies the points $\left(x,y\right)=\left(r\cos\left(\theta\right),r\sin\left(\theta\right)\right)$ are all in the form:
$$\left(\color{red}{\left(\sec\left(k\pi\right)+a\cos\left(k\pi\right)\right)\cos\left(k\pi\right)},\color{blue}{\left(\sec\left(k\pi\right)+a\cos\left(k\pi\right)\right)\sin\left(k\pi\right)}\right)$$
We see that the curves with $a\ne 0$ do have such tangent lines parallel to the $y$ axis.(Moreover for $a=0$ we have the line $x=1$ and the tangent line to the line (curve $r=\sec(\theta)$) parallel to the $y$ axis is the line itself.)
*
*The points for which the tangent line to the curve is parallel to the $x$ axis is given by :
$$\frac{dy}{d\theta}=0$$$$\left(\sec\left(\theta\right)\tan\left(\theta\right)-a\sin\left(\theta\right)\right)\sin\left(\theta\right)+\cos\left(\theta\right)\left(\sec\left(\theta\right)+a\cos\left(\theta\right)\right)=0$$
$$\frac{1}{\cos^{2}\left(\theta\right)}+2a\cos^{2}\left(\theta\right)-a=0$$
$$2a\cos^{4}\left(\theta\right)-a\cos^{2}\left(\theta\right)+1=0$$
$$\cos^{2}\left(\theta\right)=\frac{a\pm\sqrt{a^{2}-8a}}{4a}$$
Which is true whenever $$0\le\frac{a\pm\sqrt{a^{2}-8a}}{4a}\le1$$
Since $a^{2}-8a \ge 0$,we see that the curves with $0<a<8$ does not have such tangent lines parallel to the $x$ axis,moreover $\frac{a\pm\sqrt{a^{2}-8a}}{4a}$ is never between $0$ and $1$ and the inequality is not even sharp,so based on this information,such tangents lines parallel to the $x$ axis don't exist,but this is not true.
So where was I wrong?
|
So where was I wrong?
It is wrong that "$\frac{a\pm\sqrt{a^{2}-8a}}{4a}$ is never between $0$ and $1$".
This answer proves the following two claims :
Claim 1 : $$0\le \frac{a+\sqrt{a^2-8a}}{4a}\le 1\iff a\ge 8$$
Claim 2 : $$0\le \frac{a-\sqrt{a^2-8a}}{4a}\le 1\iff a\le -1\quad\text{or}\quad a\ge 8$$
Claim 1 : $$0\le \frac{a+\sqrt{a^2-8a}}{4a}\le 1\iff a\ge 8$$
Proof :
It follows from $a\not=0$ and $a^2-8a\ge 0$ that $a\lt 0$ or $a\ge 8$.
For $a\lt 0$, we have
$$\begin{align}0\le \frac{a+\sqrt{a^2-8a}}{4a}\le 1&\iff 0\ge a+\sqrt{a^2-8a}\ge 4a
\\\\&\iff \sqrt{a^2-8a}\le -a\quad\text{and}\quad \sqrt{a^2-8a}\ge 3a
\\\\&\iff a^2-8a\le (-a)^2
\\\\&\iff a\ge 0\end{align}$$
where note that $\sqrt{a^2-8a}\ge 3a$ holds for $a\lt 0$ since RHS is negative.
For $a\ge 8$, we have
$$\begin{align}0\le \frac{a+\sqrt{a^2-8a}}{4a}\le 1&\iff 0\le a+\sqrt{a^2-8a}\le 4a
\\\\&\iff -a\le \sqrt{a^2-8a}\quad\text{and}\quad \sqrt{a^2-8a}\le 3a
\\\\&\iff a^2-8a\le (3a)^2
\\\\&\iff a\ge -1\end{align}$$
where note that $-a\le \sqrt{a^2-8a}$ holds for $a\ge 8$ since LHS is negative.
So, we get
$$0\le \frac{a+\sqrt{a^2-8a}}{4a}\le 1\iff a\ge 8$$
Claim 2 : $$0\le \frac{a-\sqrt{a^2-8a}}{4a}\le 1\iff a\le -1\quad\text{or}\quad a\ge 8$$
Proof :
For $a\lt 0$, we have
$$\begin{align}0\le \frac{a-\sqrt{a^2-8a}}{4a}\le 1&\iff 0\ge a-\sqrt{a^2-8a}\ge 4a
\\\\&\iff \sqrt{a^2-8a}\ge a\quad\text{and}\quad \sqrt{a^2-8a}\le -3a
\\\\&\iff a^2-8a\le (-3a)^2
\\\\&\iff a\le -1\end{align}$$
where note that $\sqrt{a^2-8a}\ge a$ holds for $a\lt 0$ since RHS is negative.
For $a\ge 8$, we have
$$\begin{align}0\le \frac{a-\sqrt{a^2-8a}}{4a}\le 1&\iff 0\le a-\sqrt{a^2-8a}\le 4a
\\\\&\iff \sqrt{a^2-8a}\le a\quad\text{and}\quad \sqrt{a^2-8a}\ge -3a
\\\\&\iff a^2-8a\le a^2
\\\\&\iff a\ge 0\end{align}$$
where note that $\sqrt{a^2-8a}\ge -3a$ holds for $a\ge 8$ since RHS is negative.
So, we get
$$0\le \frac{a-\sqrt{a^2-8a}}{4a}\le 1\iff a\le -1\quad\text{or}\quad a\ge 8$$
|
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|
Evaluate $\lim_{x \to 1^+} \frac{\sin{(x^3-1)}\cos(\frac{1}{1-x})}{\sqrt{x-1}}$ Evaluate $$\lim_{x \to 1^+} \frac{\sin{(x^3-1)}\cos(\frac{1}{1-x})}{\sqrt{x-1}}$$
My attempt: $\cos(\frac{1}{1-x})=\cos(\frac{1}{x-1})$, so if we let $t=x-1$ we get
$$\lim_{x \to 1^+} \frac{\sin{(x^3-1)}\cos(\frac{1}{x-1})}{\sqrt{x-1}}=\lim_{x \to 0^+} \frac{\sin{((t+1)^3-1)}\cos(\frac{1}{t})}{\sqrt{t}}$$ That's where I got stuck because this doesn't look any better I guess. Any hint?
|
For $x\rightarrow1^+$ we obtain: $$\frac{\sin{(x^3-1)}\cos(\frac{1}{1-x})}{\sqrt{x-1}}=\frac{\sin(x^3-1)}{x^3-1}\cdot\sqrt{x-1}(x^2+x+1)\cos\frac{1}{x-1}\rightarrow0.$$
|
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|
Solve the inequality $|3x-5| - |2x+3| >0$. In order to solve the inequality $|3x-5| - |2x+3| >0$, I added $|2x+3|$ to both sides of the given inequality to get $$|3x-5| > |2x+3|$$ Then assuming that both $3x-5$ and $2x+3$ are positive for certain values of $x$, $$3x-5 > 2x+3$$ implies $$x>8$$ If $3x-5$ is positive and $2x-3$ is negative for certain values of $x$, then $$3x-5 > -2x-3$$ implies $$5x >2$$ implies $$x > \dfrac{2}{5}$$ I'm supposed to get that $x < \dfrac{2}{5}$ according to the solutions, but I'm not sure how to get that solution.
|
For any two functions $f$ and $g$, we have
$$\begin{align}
|f(x)|-|g(x)|\gt0
&\iff|f(x)|\gt|g(x)|\\
&\iff(f(x))^2\gt(g(x))^2\\
&\iff(f(x))^2-(g(x))^2\gt0\\
&\iff(f(x)-g(x))(f(x)+g(x))\gt0
\end{align}$$
For $f(x)=3x-5$ and $g(x)=2x+3$, we see that $f(x)-g(x)=x-8$ and $f(x)+g(x)=5x-2$, and so
$$|3x-5|-|2x+3|\gt0\iff(x-8)(5x-2)\gt0$$
Since $8\gt2/5$, the two factor are both positive if $x\gt8$ and both negative if $x\lt2/5$, hence
$$|3x-5|-|2x+3|\gt0\iff x\gt8\lor x\lt2/5$$
Remark: This approach is essentially the same as that in Bernard's answer (which appeared while I was composing), the main difference being that the step from $(f(x))^2-(g(x))^2\gt0$ to $(f(x)-g(x))(f(x)+g(x))\gt0$ gives the quadratic in factored form.
|
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|
Why does $\frac {1}{a}-\frac{1}{b}=\frac {b-a}{ab}$? I really don't understand the expression.
$$\frac {1}{a}-\frac{1}{b}=\frac {b-a}{ab}$$
I generally have a hard time understanding non-intuitive things in math and this is one of them. Normally when I don't understand something I use an app , photomath, to help explain expressions/equations I don't understand however, I still need help with this expression.
I'm told that to get to $\frac {b-a}{ab}$ you need to expand the fraction to the least common denominator:
$$\frac {1}{a}-\frac{1}{b} \to \frac {\pmb b\times 1}{\pmb b a} - \frac {\pmb a \times 1}{\pmb a b} \to \frac {b}{ab} - \frac {a}{ab} \to \frac {b-a}{ab}$$
What I don't understand is this
$$ \frac {\pmb b \times 1}{\pmb b a} - \frac{\pmb a \times 1}{\pmb a b}$$
I don't understand how exactly the $a$ and $b$ seemingly 'appear' in the expression.
|
$\frac{1}{a}-\frac{1}{b} = \frac{1}{a} * \frac{1}{1} - \frac{1}{b} * \frac{1}{1}$
= $\frac{1}{a} * \frac{b}{b} - \frac{1}{b} *\frac{a}{a}$ = $\frac{b} {ab} - \frac{a}{ab} = \frac{b-a}{ab}$
using $ab=ba$ and $\frac{x}{x} = 1$ for any $x \neq 0$
|
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|
Need help with Alternative Factoring method I was working on some factoring, as I have always been terrible at it, when I found 3B1B's video on an easier method. There's a TL:DR at the bottom if you're familiar. The basics are as follows:
Imagine the graph of a quadratic. $x^2 - 1$ for example. It's got 2 roots $r$ and $s$ the same distance $d$ apart from a midpoint $m$.
The method only works for equations that look like $x^2 +b-c=0$, so if you've got an $a$, scale everything else down by dividing everything by $a$.
So now you've got that:
$r+s=b$ and $r \cdot s = c$.
We also know that:
$r=m-d$ and $s=m+d$.
We can do some algebra to the above to realize the following:
$m=\frac{-b}{2}$
and
$d = \sqrt{m^2-c}$
So neat! we've got a simple way to factorize! Except something must turn out wrong, because I've gotten a wrong result.
$2x^2 -5x -3 = 0$
scales down to
$x^2 -\frac{5}{2}x -\frac{3}{2} = 0$
And after some crunching, we get that $r$ and $s$ are $1$ and $\frac{6}{4}$, respectively.
That's not accurate. The right answers are $-\frac{1}{2}$ and $3$. what gives?
TL;DR:
Using alternative quadratic method on
$x^2 -\frac{5}{2}x -\frac{3}{2} = 0$ gets me the x-intercepts at $x=-3$. Why? and how can I make sure this doesn't happen again?
|
Given
$$x^2+bx+c=x^2-\frac{5}{2}x-\frac{3}{2}$$
We get $m=\frac{-b}{2}=\frac{5}{4}$ and $$d=\sqrt{m^2-c}=\sqrt{\frac{25}{16}+\frac{3}{2}}=\sqrt{\frac{49}{16}}=\frac{7}{4}$$
Therefore our roots are $\frac{5}{4}\pm\frac{7}{4}=\{\frac{-1}{2},3\}$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3748157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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|
Help with the last step in solving $\lim_{x\to0}\frac{(1+\sin x +\sin^2 x)^{1/x}-(1+\sin x)^{1/x}}x$ I managed to solve part of this limit but can't get the final step right. Here's the limit:
$$
\lim_{x\to0} {
\frac
{
\left(
1+\sin{x}+\sin^2{x}
\right)
^{1/x}
-
\left(
1+\sin{x}
\right)
^{1/x}
}
{
x
}
}
$$
Let's begin then.
Using $f(x) = e^{\log{f(x)}} = \exp{\left[\log{f(x)}\right]}$ it gets to:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
x
}
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
x
}
\right]
}
{
x
}
}
$$
Multiplying and dividing by the right terms:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
\sin{x}+\sin^2{x}
}
\frac
{
\sin{x}+\sin^2{x}
}
{
x
}
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
\sin{x}
}
\frac
{
\sin{x}
}
{
x
}
\right]
}
{
x
}
}
$$
Grouping the $\sin{x}$ in the first exponential function:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
\sin{x}+\sin^2{x}
}
\frac
{
\sin{x}
}
{
x
}
\left(1+\sin{x}\right)
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
\sin{x}
}
\frac
{
\sin{x}
}
{
x
}
\right]
}
{
x
}
}
$$
Applying the know limits, specifically:
$$\lim \limits_{x\to0}{\frac{\log{\left(1+\sin{x}+\sin^2{x}\right)}}{\sin{x}+\sin^2{x}}=1}$$$$\lim \limits_{x\to0}{\frac{\log{\left(1+\sin{x}\right)}}{\sin{x}}=1}$$$$\lim \limits_{x\to0}{\frac{\sin{x}}{x}=1}$$$$\lim \limits_{x\to0}{\left(1+\sin{x}\right)}=1$$
We should end up with:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
\sin{x}+\sin^2{x}
}
\frac
{
\sin{x}
}
{
x
}
\left(1+\sin{x}\right)
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
\sin{x}
}
\frac
{
\sin{x}
}
{
x
}
\right]
}
{
x
}
}
=
\frac
{
\exp{\left[1(1)(1)\right]}
-
\exp{\left[1(1)\right]}
}
{
0
}
= \frac{e-e}{0}
= \frac{0}{0}
$$
Undetermined form, yay. Any hint is really appreciated, and I'm really really sorry for all the spaghetti rendering, it's hard to look at.
|
Evaluate $$L=\lim_{x\to0} {\frac{\left(1+\sin{x}+\sin^2{x}\right)^{1/x}-\left(1+\sin{x}\right)^{1/x}}{x}}$$
If there are more terms to apply the limit to, you should separate the terms of which limits are apparent to reduce the probability of such failure.
\begin{align}
L&=\lim_{x\to0} {\frac{\left(1+\sin{x}+\sin^2{x}\right)^{1/x}-\left(1+\sin{x}\right)^{1/x}}{x}}\\
&=\lim_{x\to0} \left(1+\sin{x}\right)^{1/x}{\frac{\left(1+\frac{\sin^2{x}}{1+\sin{x}}\right)^{1/x}-1}{x}}\tag{1}\\
&=e\lim_{x\to0} {\frac{\left(1+\frac{\sin^2{x}}{1+\sin{x}}\right)^{1/x}-1}{x}}\tag{2}\\
&=e\lim_{x\to0} {\frac{\exp\left[\frac 1x\ln\left(1+\frac{\sin^2{x}}{1+\sin{x}}\right)\right]-1}{x}}\tag{3}\\
&=e\lim_{x\to0} {\frac{\exp\left\{\displaystyle\frac 1x\left[\sum_{i=1}^{+\infty}(-1)^{i+1}\frac1i\left(\frac{\sin^2{x}}{1+\sin{x}}\right)^i\right]\right\}-1}{x}}\tag{4}\\
&=e\lim_{x\to0} {\frac{{\displaystyle\sum_{j=1}^{+\infty}\frac{\left\{\displaystyle\frac 1x\left[\sum_{i=1}^{+\infty}(-1)^{i+1}\frac1i\left(\frac{\sin^2{x}}{1+\sin{x}}\right)^i\right]\right\}^j}{j!}}}{x}}\tag{5}\\
&=e\lim_{x\to0} {\frac{\frac 1x\frac{\sin^2{x}}{1+\sin{x}}}{x}}=e\tag{6}\\
\end{align}
where in step $(6)$, I used the term $i=j=1$ only, because
*
*$\displaystyle\lim_{x\to 0}(1+\sin x)=1$.
*$\displaystyle\lim_{x\to 0}\frac{\sin^a x}{x^b}=1\Leftrightarrow a=b$, where $a,b\in\mathbb R$.
*$a\rightarrow 2ij=b\rightarrow j+1\Rightarrow j(2i-1)=1$, where $i,\ j\geq 1$
You did something like $\displaystyle\lim_{x\to 0}\frac{e-e}{0}$ in the place of my step $(4)$ because of overwhelmingly many terms on both sides of the minus sign.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3748637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
}
|
$\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4)(1+x^8)...$ How can I prove the identity $\frac{1}{1-x} = (1+x)(1+x^2)(1+x^4)(1+x^8)\ldots$ for $|x|<1$? I am preferably looking for a derivation rather than using the RHS. I have tried using binomial expansion, but it only seems to give the LHS back. I also tried taking the logarithm of $\frac1{1-x}$ on seeing a product and using the Taylor series of $\ln{(1+x)}$, but this appears to be a dead end.
|
We can repeatedly double the power in the denominator as follows: (similar to rationalizing the denominator, if you are familiar wtih that)
$$
\begin{align}
&\frac1{1-x}=\frac1{1-x}\cdot\frac{1+x}{1+x}\\
=&\frac{1+x}{1-x^2}=\frac{1+x}{1-x^2}\cdot\frac{1+x^2}{1+x^2}\\
=&\frac{(1+x)(1+x^2)}{1-x^4}=\frac{(1+x)(1+x^2)}{1-x^4}\cdot\frac{1+x^4}{1+x^4}\\
=&\frac{(1+x)(1+x^2)(1+x^4)}{1-x^8}=\cdots
\end{align}
$$
Hopefully it is clear how this pattern continues forever. Now, since
$$
\lim_{n\to\infty}x^n=0
$$
when $|x| < 1$, we can see that the denominator approaches $1$. Therefore, the numerator is all that remains, and the identity is shown.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3748821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
}
|
If $a_{n+1}=2a_n −n^2+n$ Define a sequence $a_n$ that satisfy the recurrence relation as described above, with $a_1 = 3$ If
$$a_{n+1}=2a_n −n^2+n$$
Define a sequence $a_n$
that satisfy the recurrence relation as described above, with $a_1 = 3$
Find the value of $$\dfrac{ |a_{20} - a_{15} | }{18133} $$
Attempt
First evaluate $a_{0}$
$$a_{1} = 2a_{0} \Rightarrow a_{0}= \frac{3}{2}$$
Then, use Z-transform: $$a_{n+1} - 2a_{n} + n^2 - n = 0$$
$$z(\mathbf{A}(z)-a_{0}) - 2\mathbf{A}(z) + \dfrac{z(z+1)}{(z-1)^3} - \dfrac{z}{(z-1)^2} = 0$$
$$\Rightarrow \mathbf{A}(z) = \dfrac{z(3z^3 -9z^2 + 9z - 7)}{2(z-2)(z-1)^3}$$
$$\Rightarrow \mathbf{A}(z) = \dfrac{2z}{z-1} + \dfrac{z}{(z-1)^2} + \dfrac{z(z+1)}{(z-1)^3} - \dfrac{z}{2(z-2)}$$
The inverse of the Z-transform will be: $$\boxed{a_{n} = 2 + n + n^2 - 2^{n-1}}$$
Now: $$a_{20} = 422-2^{19}$$ $$a_{15} = 242-2^{14}$$
Is it Correct??
Any other precise solution will be highly appreciated
|
Let $a_m=b_m+c_0+c_1m+c_2m^2+\cdots+c_rm^r$
$b_{m+1}+c_0+c_1(m+1)+c_2(m+1)^2+\cdots=2(b_n+c_0+c_1m+c_2n^2+\cdots)+n-n^2$
If $r\ge3,$ compare the coefficients of $m^r$
$$c_r=2c_r\iff c_r=0$$
$\implies a_m=b_m+c_0+c_1m+c_2m^2$
Compare the coefficients of $n^2, c_2=2c_1-1\iff c_2=1$
Similarly, comparing the coefficients of $n$ and the constants, $c_1=1, c_0=2$
so that $b_{m+1}=2b_m=\cdots=2^tb_{m-t}$ for integer $t\ge0$
Again $3=a_1=b_1+c_0+c_1+c_2\implies b_1=-1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3749804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
}
|
Differentiate $\left(x^6-2x^2\right) \ln\left(x\right) \sin\left(x\right)$ Differentiate
$$\left(x^6-2x^2\right)\ln\left(x\right)\sin\left(x\right)$$
with respect to $x$
My work so far
$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\left(x^6-2x^2\right)\ln\left(x\right)\sin\left(x\right)\right]}}$
$=\class{steps-node}{\cssId{steps-node-3}{\class{steps-node}{\cssId{steps-node-2}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[x^6-2x^2\right]}}\cdot\ln\left(x\right)\sin\left(x\right)}}+\class{steps-node}{\cssId{steps-node-5}{\left(x^6-2x^2\right)\cdot\class{steps-node}{\cssId{steps-node-4}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\ln\left(x\right)\right]}}\cdot\sin\left(x\right)}}+\class{steps-node}{\cssId{steps-node-7}{\left(x^6-2x^2\right)\ln\left(x\right)\cdot\class{steps-node}{\cssId{steps-node-6}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[\sin\left(x\right)\right]}}}}$
$=\class{steps-node}{\cssId{steps-node-8}{\left(\class{steps-node}{\cssId{steps-node-10}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[x^6\right]}}-2\cdot\class{steps-node}{\cssId{steps-node-9}{\tfrac{\mathrm{d}}{\mathrm{d}x}\left[x^2\right]}}\right)}}\ln\left(x\right)\sin\left(x\right)+\left(x^6-2x^2\right)\cdot\class{steps-node}{\cssId{steps-node-11}{\dfrac{1}{x}}}\sin\left(x\right)+\left(x^6-2x^2\right)\ln\left(x\right)\class{steps-node}{\cssId{steps-node-12}{\cos\left(x\right)}}$
$=\left(\class{steps-node}{\cssId{steps-node-13}{6}}\class{steps-node}{\cssId{steps-node-14}{x^5}}-2\cdot\class{steps-node}{\cssId{steps-node-15}{2}}\class{steps-node}{\cssId{steps-node-16}{x}}\right)\ln\left(x\right)\sin\left(x\right)+\dfrac{\left(x^6-2x^2\right)\sin\left(x\right)}{x}+\left(x^6-2x^2\right)\ln\left(x\right)\cos\left(x\right)$
$=\left(6x^5-4x\right)\ln\left(x\right)\sin\left(x\right)+\dfrac{\left(x^6-2x^2\right)\sin\left(x\right)}{x}+\left(x^6-2x^2\right)\cos\left(x\right)\ln\left(x\right)$
$=x\left(\left(\left(6x^4-4\right)\ln\left(x\right)+x^4-2\right)\sin\left(x\right)+\left(x^5-2x\right)\cos\left(x\right)\ln\left(x\right)\right)$
I've had great difficulty in solving this. Was my method correct?
Also, would there be any shortcuts in solving this, or is this method the best way of getting the solution?
|
Let $f(x)=\left(x^6-2x^2\right)\ln x\sin x$. Then,
$$\ln f(x)=\ln(x^6-2x^2) +\ln (\ln x ) + \ln(\sin x)
$$
and
$$f’(x)=f(x)\left(\frac{6x^5-4x}{x^6-2x^2}+\frac{1}{x\ln x}+ \cot x\right)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3751546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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|
Evaluating $ \lim_{x \to 0}\left(-\frac{1}{3 !}+\frac{x^{2}}{5 !}-\frac{x^{4}}{7 !}+\frac{x^{6}}{9!}+\cdots\right) $ This question comes to my mind immediately after asking this question.
I was earlier unknown that limit of sum equal sum of limits only when there are finite terms. Now the problem is then how do I evaluate the following limit which earlier I used to do by applying individual limits.
$$
\lim_{x \to 0}\left(-\frac{1}{3 !}+\frac{x^{2}}{5 !}-\frac{x^{4}}{7 !}+\frac{x^{6}}{9!}+\cdots\right)
$$
I’m high school student
|
Observe that, for $x\ne 0$,
$$ -\frac{1}{3 !}+\frac{x^{2}}{5 !}-\frac{x^{4}}{7 !}+\frac{x^{6}}{9!}+\cdots=\frac{1}{x^3} \left(-\frac{x^3}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9!}+\cdots\right)
=\frac{1}{x^3} \left(\frac{x}{1!}-\frac{x^3}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\frac{x^{9}}{9!}+\cdots-\frac{x}{1!}\right) = \frac{1}{x^3}\left(\sin x-x\right) \\
$$
Then you may apply L'Hôpital three times and obtain that
$$
\lim_{x\to 0}\frac{1}{x^3}\left(\sin x-x\right)=-\frac{1}{3!}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3751916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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|
Geometric sequence problem including sum of the numbers Numbers: $a,b,c,d$ generate geometric sequence and $a+b+c+d=-40. $ Find these numbers if $a^2+b^2+c^2+d^2=3280$
I tried this problem and I have system of equations which I can't solve. I think there should be different way to handle this problem.
|
$$-48=\dfrac p{d^3}+\dfrac pd+pd+pd^3=\dfrac p{d^3}(1+d^2+d^4+d^6)=\dfrac{p(1+d^2)(1+d^4)}{d^3}$$
Similarly,
$$3280=\dfrac{p^2(1+d^4)(1+d^8)}{d^6}$$
$$\implies\dfrac{(-48)^2}{3280}=\dfrac{(1+d^2)^2(1+d^4)}{1+d^8}=\dfrac{\left(d+\dfrac1d\right)^2\left(d^2+\dfrac1{d^2}\right)}{d^4+\dfrac1{d^4}}$$
Let $d^2+\dfrac1{d^2}=u$
$$d^4+\dfrac1{d^4}=u^2-2, \left(d+\dfrac1d\right)^2=u+2$$
So, we have a quadratic equation in $u$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3756260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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|
Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \theta$. Now I can change the form of the denominator to become easy enough to substitute as follows:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +x^2\bigg)\bigg) ^\frac{3}{2} $$
Which makes it clear that $x$ needs to be substituted as $x = \frac{3}{2} \tan \theta $, and $dx = \frac{3}{2} \sec^2 \theta $
for later use. At this point I can represent $(1)$ in terms of my substituted trignometric function. The only problem comes with the denominator where I get stuck on the power. Here is how I went about solving it:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\left(\frac{3}{2} \tan \theta\right)^2 \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\frac{9}{4} \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \frac{9}{4}\bigg (\ 1 + \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$ 9^\frac{3}{2}\bigg( \ 1 + \tan^2 \theta \bigg) ^\frac{3}{2} $$
$$ 27\ ( \sec^2 \theta ) ^\frac{3}{2} $$
Now I have no idea how to evaluate this power of $sec$. The author says that it changes into $sec^3 \theta$ but I just can't fathom how that would go about. If what I understand is correct, the power it is raised to would be added to it's own making it $ \sec^\frac{7}{2} \theta$. My question is that how exactly is my reasoning wrong here?
|
You can easily integrate with suitable substitution as follows
Let $2x=3\tan\theta\implies dx=\frac{3}{2}\sec^2\theta\ d\theta$
$$\int \frac{x^3}{(4x^2+9)^{3/2}}dx=\int \frac{(\frac32\tan\theta)^3}{(9\tan^2\theta+9)^{3/2}}\ \frac{3}{2}\sec^2\theta\ d\theta$$
$$=\left(\frac{3}{2}\right)^4\frac{1}{3^3}\int \frac{\tan^3\theta}{\sec^3\theta}\sec^2\theta\ d\theta$$
$$=\frac{3}{16}\int \frac{\tan\theta(\sec^2\theta-1)}{\sec\theta}\ d\theta$$
$$=\frac{3}{16}\int (\sec\theta\tan\theta-\sin\theta)\ d\theta$$
$$=\frac{3}{16}(\sec\theta+\cos\theta)+C$$
substitute back to $x$,
$$=\frac{3}{16}\left(\frac{\sqrt{4x^2+9}}{3}+\frac{3}{\sqrt{4x^2+9}}\right)+C$$
$$=\color{blue}{\frac{2x^2+9}{8\sqrt{4x^2+9}}}+C$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3758050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
}
|
Small-angle approximation of $ \frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} $ I need to show the following:
$$ \frac{\sin^2 x}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} \approx 1-\frac{x^2}{6} $$ when $ x $ is small.
I think this problem is trickier than most other questions like it because in the original source there is comment saying "if you got $ 1+\frac{x^2}{6} $ [what I got] then think again!". My attempt was:
When $ x $ is small, $ \sin x \approx x $ so
$$ \frac{\sin^2{x}}{x^2 \sqrt{1-\frac{\sin^2 x}{3}}} = \frac{x^2}{x^2 \sqrt{1-\frac{x^2}{3}}} = \left ( 1 - \frac{x^2}{3} \right )^{-\frac{1}{2}} $$
Then using the binomial series approximation,
$$ \left ( 1 - \frac{x^2}{3} \right )^{-\frac{1}{2}} \approx 1 - \frac{1}{2}\left ( -\frac{x^2}{3} \right ) + ... = 1 + \frac{x^2}{6} $$
...and so it looks like I've fallen into whatever trap the question set.
Where is my error?
|
$${\sin x\over x}\approx1-{1\over6}x^2$$
so
$${\sin^2x\over x^2}\approx\left(1-{1\over6}x^2\right)^2\approx1-{1\over3}x^2$$
not just $1$. We get
$${\sin^2x\over x^2\sqrt{1-{\sin^2x\over3}}}\approx\left(1-{1\over3}x^2\right)\left(1+{1\over6}x^2\right)\approx1-{1\over6}x^2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3760642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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|
How to solve $\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{dx} $? $$\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{dx}$$
I have tried $u$-substitution and multiplying by the conjugate and then apply $u$-substitution. For the $u$-substitution, I have set $u$ equal to each square root term, set $u$ equal to the entire denominator, and set $u$ equal to each expression in the radical.
However, all my attempts have just made the integral more complex without an obvious way to simplify. Can someone provide insight please? Thank you.
|
Multiplying by the conjugate and applying a couple of substitutions does work.
\begin{align*}
\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{d}x &=\int \underbrace{\frac{\sqrt{2x}}{x-4}}_{\sqrt{x} \to u} + \underbrace{\frac{\sqrt{x+4}}{x-4}}_{\sqrt{x+4} \to t} \; \mathrm{d}x\\
&=2\sqrt{2} \int \frac{u^2}{u^2-4} \; \mathrm{d}u+ 2\int \frac{t^2}{t^2-8} \; \mathrm{d}t \\
&=2\sqrt{2} \int \frac{u^2-4+4}{u^2-4} \; \mathrm{d}u+ 2\int \frac{t^2-8+8}{t^2-8} \; \mathrm{d}t \\
&=2\sqrt{2}u +2\sqrt{2} \ln{\bigg |\frac{u-2}{u+2}\bigg |} + 2t +2\sqrt{2}\ln{\bigg |\frac{t-2\sqrt{2}}{t+2\sqrt{2}}\bigg |}+\mathrm{C} \\
&=2\sqrt{2x} +2\sqrt{2} \ln{\bigg |\frac{\sqrt{x}-2}{\sqrt{x}+2}\bigg |} + 2\sqrt{x+4} +2\sqrt{2}\ln{\bigg |\frac{\sqrt{x+4}-2\sqrt{2}}{\sqrt{x+4}+2\sqrt{2}}\bigg |}+\mathrm{C} \\
\end{align*}
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Find all $x,y$ such that $x^3+5y^3=(a^3+5b^3)^3$. Let $a,b$ be coprime integers. I am trying to find all integers $x,y$ such that:
$$x^3+5y^3=(a^3+5b^3)^3$$
What I have tried:
$$5y^3=(a^3+5b^3-x)[(a^3+5b^3)^2+(a^2+5b^3)x+x^2 ]$$
There are 2 Cases depending which factor of $5y^3$ is divisible by $5$:
Case 1: $5$ divides the first factor:
$$a^3+5b^3-x=5y_0^{3-n}y_1^3$$
$$(a^3+5b^3)^2+(a^2+5b^3)x+x^2 =y_0^ny_2^3$$
where $$y=y_0y_1y_2$$ $$(y_1,y_2)=1$$ $$n=\{0,1,2,3\}$$
It follows $$x=a^3+5b^3-5y_0^{3-n}y_1^3$$
We substitute $x$
$$(a^3+5b^3)^2+(a^2+5b^3)(a^3+5b^3-5y_0^{3-n}y_1^3)+(a^3+5b^3-5y_0^{3-n}y_1^3)^2 =y_0^ny_2^3$$
For simplicity, we write $u=a^3+5b^3$:
$$u^2+u(u-5y_0^{3-n}y_1^3)+(u-5y_0^{3-n}y_1^3)^2 =y_0^ny_2^3$$
$$u^2+u^2-5y_0^{3-n}y_1^3u+u^2-10uy_0^{3-n}y_1^3+25y_0^{2(3-n)}y_1^6=y_0^ny_2^3$$
$$3u^2-15uy_0^{3-n}y_1^3+25y_0^{2(3-n)}y_1^6=y_0^ny_2^3$$
What do I do at this point? Any input will be appreciated. Thanks.
|
The only solutions are the ones with $y=0$.
This is one of those problems where generalizing makes it easier, so I'm going to substitute the RHS with the cube of a general integer
$$x^3+5y^3=z^3$$
If $z=0$ you get the $(0,0,0)$ solution and otherwise $\frac x y =-5^{1/3}$ which is impossible. Now I'm going to let $x_1=x/z$ and $y_1=y/z$ so
$$x_1^3+5y_1^3=1$$
Now we just need to prove that $(1,0)$ is the only rational point on this curve. Let $t=x-1$ and $u=k(x+1)$. You can think of this as tracing a line of slope $k$ through the rational point we already know.
$$(t+1)^3+5(kt)^3=1$$
$$k^3=-\frac{t^2+3t+3}{5t^2}$$
Now let $t_1=1/t$ (note that from now on we're assuming $y \neq 0$)
$$3t_1^2+3t_1+1=-5k^3$$
$$3(2t_1+1)^2=-20k^3-1$$
Now to transform this into an elliptic curve, and abusing the notation by having a variable collision because that's what the software wants $x=60k$ and $y=180(2t_1+1)$ then
$$y^2=x^3-10800$$
Now we can get sage to tell us that this has no solutions:
sage: E = EllipticCurve([0,0,0,0,-10800])
sage: E
Elliptic Curve defined by y^2 = x^3 - 10800 over Rational Field
sage: E.rank()
0
sage: E.torsion_points()
[(0 : 1 : 0)]
Which means the only solution is the point at infinity of the projective plane, which does not concern us.
|
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Is it possible to justify these approximations about prime numbers? A recently closed question asked for a possible closed form of the infinite summation
$$f(a)=\sum _{i=1}^{\infty } a^{-p_i}$$ for which I already proposed a first simple but totally empirical approximation.
Since we quickly face very small numbers, I tried to find approximations of
$$g(a)=\Big[\sum _{i=1}^{\infty } a^{-p_i}\Big]^{-1} \qquad \text{and} \qquad h(a)=\Big[\sum _{i=1}^{\infty } (-1)^{i-1} a^{-p_i}\Big]^{-1}$$
All calculations where done with integer values of $a$ for the range $2 \leq a \leq 1000$.
What I obtained is
$$\color{blue}{g(a)\sim\frac{(a-1) (2a^3+2a-1)}{2 a^2}}\qquad \text{and} \qquad \color{blue}{h(a)\sim\frac{(a-1) \left(a^3+2 a^2+3 a+4\right)}{a^2}}$$
If the corresponding curve fits were done, in both cases we should have $R^2 > 0.999999999$.
For the investigated values of $a$,
$$\text{Round}\left[\frac{(a-1) (2a^3+2a-1)}{2 a^2}-{g(a)}\right]=0$$
$$\text{Round}\left[\frac{(a-1) \left(a^3+2 a^2+3 a+4\right)}{a^2}-{h(a)}\right]=0$$
Not being very used to work with prime numbers, is there any way to justify, even partly, these approximations ?
|
I found this estimate for $g$: $\ g(a)\sim \dfrac{a^2(a^2-1)}{a^2+a-1}$.
First inequality
$f(a) = \displaystyle \sum_{i=1}^{+\infty} \dfrac{1}{a^{p_i}} \leqslant \sum_{i=2}^{+\infty} \dfrac{1}{a^i} -\sum_{i=2}^{+\infty} \dfrac{1}{a^{2i}} = \dfrac{1}{a^2}\dfrac{1}{1-\dfrac{1}{a}} -\dfrac{1}{a^4}\dfrac{1}{1-\dfrac{1}{a^2}}$
$f(a)\leqslant \dfrac{1}{a(a-1)}-\dfrac{1}{a^2(a^2-1)}= \dfrac{a^2+a-1}{a^2(a^2-1)}$
$\fbox{$g(a)\geqslant \dfrac{a^2(a^2-1)}{a^2-a+1}$}$
Second inequality
$f(a) \geqslant \dfrac{1}{a^2}+\dfrac{1}{a^3}+\dfrac{1}{a^5}+\dfrac{1}{a^7}$
$\fbox{$g(a)\leqslant \dfrac{a^7}{a^5+a^4+a^2+1}$}$
Quality of the approximation
$0\leqslant g(a)-\dfrac{a^2(a^2-1)}{a^2+a-1}\leqslant \dfrac{a^7}{a^5+a^4+a^2+1} - \dfrac{a^2(a^2-1)}{a^2+a-1}$
And:
$\dfrac{a^7}{a^5+a^4+a^2+1} - \dfrac{a^2(a^2-1)}{a^2+a-1} = \dfrac{a^2}{(a^5+a^4+a^2+1)(a^2+a-1)}$
So
$\fbox{$0\leqslant g(a)-\dfrac{a^2(a^2-1)}{a^2+a-1}\leqslant \dfrac{1}{a^5}$}$
And:
$\forall a \in [2,+\infty[ \ , \ \text{Round} \left( g(a)-\dfrac{a^2(a^2-1)}{a^2+a-1}\right) = 0 $
|
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|
What is value of this integral? $\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^{2})}dx$
What is value of this integral $$I=\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^2)}dx$$
My work :
\begin{align*}I&=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)(10+10x^2)}{(9+x^2)(1+9x^2)}dx\\
&=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)}{1+9x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)}{9+x^2}dx\\
&=j_{1}+j_{2}+j_{3}\\
\end{align*}
$$j_{1}=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx=\sum_{n=0}^{\infty}\frac{(-1)^n(2^{2(n+1)})}{n+1}\int_{0}^{\infty}\frac{x^{2(n+1)}}{9+x^2}dx$$
$$j_{2}=\log(4)\int_{0}^{\infty}\frac{1}{1+(3x)^2}dx+\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(2^{2(n+1)})}\int_{0}^{\infty}\frac{x^{2(n+1)}}{1+9x^2}dx=\frac{\log(4)\pi}{6}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)2^{2(n+1)}}\int_{0}^{\infty}\frac{x^{2(n+1)}}{1+9x^2}dx$$
$$j_{3}=\int_{0}^{\infty}\frac{\log(4+x^2)}{9+x^2}dx=\frac{\log(4)\pi}{54}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)2^{2(n+1)}}\int_{0}^{\infty}\frac{x^{2(n+1)}}{9+x^2}dx$$
Wait for a review to find solutions to this
|
First, the integral:
$$
I := \int_0^{\infty} \frac{\ln\left(1+4\,x^2\right)\left(1+9\,x^2\right)\left(9+\,x^2\right)+\left(9+x^2\right)\ln\left(4+x^2\right)\left(10+10\,x^2\right)}{\left(9+x^2\right)^2\left(1+9\,x^2\right)}\,\text{d}x
$$
thanks to the linearity of the integrals it can be written as the algebraic sum of two integrals:
$$
I =
\int_0^{\infty} \frac{\ln\left(1+4\,x^2\right)}{9+x^2}\,\text{d}x +
\int_0^{\infty} \frac{10\left(1+x^2\right)\ln\left(4 + x^2\right)}{\left(9+\,x^2\right)\left(1+9\,x^2\right)}\,\text{d}x\,.
$$
At this point, to calculate the first, it's sufficient to consider the function $J : [0,\,+\infty) \to \mathbb{R}$ of the law:
$$
J(a) := \int_0^{\infty} \frac{\ln\left(1+a\,x^2\right)}{9+x^2}\,\text{d}x
$$
whose first derivative is equal to:
$$
J'(a) = \int_0^{\infty} \frac{x^2}{\left(9+x^2\right)\left(1+a\,x^2\right)}\,\text{d}x = \frac{\pi}{2\left(\sqrt{a}+3\,a\right)}
$$
and therefore, going back, we get:
$$
J(a) = \int \frac{\pi}{2\left(\sqrt{a}+3\,a\right)}\,\text{d}a = \frac{\pi}{3}\,\ln\left(1 + 3\sqrt{a}\right) + c
$$
where equaling the two expressions of $J(0)$ must be $c = 0$, from which:
$$
\int_0^{\infty} \frac{\ln\left(1+a\,x^2\right)}{9+x^2}\,\text{d}x = \frac{\pi}{3}\,\ln\left(1 + 3\sqrt{a}\right).
$$
Similarly, to calculate the second, it's sufficient to consider the function $K : [0,\,+\infty) \to \mathbb{R}$ of the law:
$$
K(b) := \int_0^{\infty} \frac{10\left(1+x^2\right)\ln\left(4 + b\,x^2\right)}{\left(9+\,x^2\right)\left(1+9\,x^2\right)}\,\text{d}x
$$
whose first derivative is equal to:
$$
K'(b) = \int_0^{\infty} \frac{10\,x^2\left(1+x^2\right)}{\left(9+x^2\right)\left(1+9\,x^2\right)\left(4+b\,x^2\right)}\,\text{d}x = \frac{\left(3+\frac{10}{\sqrt{b}}\right)\pi}{3\left(12+20\sqrt{b}+3\,b\right)}
$$
and therefore, going back, we get:
$$
K(b) = \int \frac{\left(3+\frac{10}{\sqrt{b}}\right)\pi}{3\left(12+20\sqrt{b}+3\,b\right)}\,\text{d}b = \frac{\pi}{3}\,\ln\left(12+20\sqrt{b}+3\,b\right) + c
$$
where equaling the two expressions of $K(0)$ must be $c = \frac{2\,\pi}{3}\,\ln(2) - \frac{\pi}{3}\,\ln(12)$, from which:
$$
\int_0^{\infty} \frac{10\left(1+x^2\right)\ln\left(4 + b\,x^2\right)}{\left(9+\,x^2\right)\left(1+9\,x^2\right)}\,\text{d}x = \frac{\pi}{3}\,\ln\left(12+20\sqrt{b}+3\,b\right) + \frac{2\,\pi}{3}\,\ln(2) - \frac{\pi}{3}\,\ln(12)\,.
$$
In conclusion, setting $a = 4$ and $b = 1$, we get:
$$
I = \frac{\pi}{3}\,\ln\left(1 + 3\sqrt{4}\right) + \frac{\pi}{3}\,\ln\left(12+20\sqrt{1}+3\cdot 1\right) + \frac{2\,\pi}{3}\,\ln(2) - \frac{\pi}{3}\,\ln(12) = \frac{\pi}{3}\,\ln\left(\frac{245}{3}\right),
$$
which is what is desired.
|
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|
Simplify the radical $\sqrt{x-\sqrt{x+\sqrt{x-...}}}$ I need help simplifying the radical $$y=\sqrt{x-\sqrt{x+\sqrt{x-...}}}$$
The above expression can be rewritten as $$y=\sqrt{x-\sqrt{x+y}}$$
Squaring on both sides, I get $$y^2=x-\sqrt{x+y}$$
Rearranging terms and squaring again yields $$x^2+y^4-2xy^2=x+y$$
At this point, deriving an expression for $y$, completely independent of $x$ does not seem possible. This is the only approach to solving radicals which I'm aware of. Any hints to simplify this expression further/simplify it with a different approach will be appreciated.
EDIT: Solving the above quartic expression for $y$ on Wolfram Alpha, I got 4 possible solutions
|
Perhaps it is more instructive to consider instead the following: let $$y = \sqrt{x - \sqrt{x + \sqrt{x - \sqrt{x + \cdots}}}}, \\
z = \sqrt{x + \sqrt{x - \sqrt{x + \sqrt{x - \cdots}}}},$$ so that if $y$ and $z$ exist, they satisfy the system $$y = \sqrt{x - z}, \\ z = \sqrt{x + y},$$ or $$y^2 = x - z, \\ z^2 = x + y.$$ Consequently $$0 = z^2 - y^2 - y - z = (z-y-1)(y+z).$$ It follows that either $z = -y$ or $z = 1 + y$. The first case is impossible for $x \in \mathbb R$ since by convention we take the positive square root, so both $y, z > 0$. In the second case, we can substitute back into the first equation to obtain $y^2 = x - (1+y)$, hence $$y = \frac{-1 + \sqrt{4x-3}}{2},$$ where again, we discard the negative root.
So far, what we have shown is that if such a nested radical for $y$ converges, it must converge to this value. It is not at all obvious from the above whether a given choice of $x$ results in a real-valued $y$, for any meaningful definition of $y$ must be as the limit of the sequence $$y = \lim_{n \to \infty} y_n, \\ y_n = \underbrace{\sqrt{x - \sqrt{x + \sqrt{x - \cdots \pm \sqrt{x}}}}}_{n \text{ radicals}},$$ and although the choice $x = 1$ appears at first glance permissible, we quickly run into problems; $y_3 = \sqrt{1 - \sqrt{1 + \sqrt{1}}} \ne \mathbb R$. In particular, we need $x$ to satisfy the relationship $$x \ge \sqrt{x + \sqrt{x}},$$ which leads to the cubic $x^3 - 2x^2 + x - 1$ with unique real root $$x = \frac{1}{3} \left(2+\sqrt[3]{\frac{25-3
\sqrt{69}}{2}}+\sqrt[3]{\frac{25+3 \sqrt{69}}{2}}\right) \approx 1.7548776662466927600\ldots.$$ However, any such $x$ meeting this condition will lead to a convergent sequence. The idea is to show that $|y_{n+2} - y| < |y_n - y|$ for all $n \ge 1$; then since $\lim y_n$ has at most one unique limiting value as established above, the result follows.
|
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|
Find limits of $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor Expansion. Find the limit $\displaystyle \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor expansion.
My Try
$\displaystyle =\lim _{x \to 0} \frac {\cos x - \frac{\sin x}{x}} {x \sin x}$
$=\frac {\displaystyle\lim _{x \to 0}\cos x - \lim _{x \to 0}\frac{\sin x}{x}} {\displaystyle\lim _{x \to 0}x \sin x}$
$=\frac{1-1}{0}$
But still I end up with $\frac00$
Any hint for me to proceed would be highly appreciated.
P.S: I did some background check on this question on mathstack and found they have solved this with l'Hopital's rule and the answer seems to be $\frac{-1}{3}$.
What is $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$?
|
Denote $L$ the existing limit. Then, express it as
\begin{align}
L=\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}
&= \lim _{x \to 0}\frac {x (2\cos^2\frac x2 -1) -2 \sin \frac x2 \cos\frac x2} {2x^2 \sin \frac x2 \cos\frac x2}\\
&= \lim _{x \to 0} \frac {x (\cos^2\frac x2-1) +2 \cos\frac x2(\frac x2\cos \frac x2- \sin\frac x2)} {2x^2 \sin \frac x2 \cos\frac x2}\\
&= - \lim _{x \to 0}\frac{\sin\frac x2}{\frac x2} \frac1{4\cos\frac x2}
+ \lim _{x \to 0} \frac{\frac x2\cos \frac x2- \sin\frac x2} {4(\frac x2)^2 \sin \frac x2} \\
&= -\frac14+\frac14L
\end{align}
Thus, $L= -\frac13$.
|
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|
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
Find $a$, $b$ such that $x^2 - x -1$ is a factor of $ax^9 + bx^8 + 1$
The second polynomial can be rewritten as
$$ax^9 + bx^8 + 1 = f(x)(x^2 - x - 1)$$
The roots of this polynomial are $\frac{1 \pm \sqrt 5}{2}$. Substituting one of these roots in this equation gives us:
$$a\left( \frac{1 + \sqrt5}{2}\right)^9 + b\left( \frac{1 + \sqrt 5}{2}\right)^8 + 1 = 0$$
I was able to solve this far, but I gave up because the calculation past this point gets too tedious. The textbook has gone ahead and simplified this to
$$2^9 a + 2^8b(\sqrt 5 - 1) + (\sqrt5 - 1)^9 = 0$$
after which it simplifies to (divide by $2^8$ and solve the binomial expression)
$$2a + b(\sqrt 5 -1) = 76 - 34\sqrt5$$
Is there a more elegant way to solve this problem? Preferably one that does not include the magical use of a calculator or the evaluation of that ugly binomial expansion?
|
Say $c$ and $d$ are zeroes of $x^2-x-1$ then they are zeros of $ax^9+bx^8+1$ too.
Since $c^2 = c+1$ we have $$c^4=c^2+2c+1=3c+2$$ and $$c^8 = 9c^2+12c+4 = 21c+13$$
and finnaly $c^9 = 34c+21$.
So we have $$a(34c+21)+b(21c+13)+1=0$$ or $$\boxed{(34a+21b)c+ (21a+13b+1)=0}$$
Simmilay we have for $d$: $$\boxed{(34a+21b)d+ (21a+13b+1)=0}$$
so if we substract equation and since $c\ne d$ we have $$(34a+21b)(c-d) = 0\implies 34a+21b=0$$
and thus $$21a+13b+1=0$$
Now solve this system and you are done.
|
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|
How do I find integers $x,y,z$ such that $x+y=1-z$ and $x^3+y^3=1-z^2$? This is INMO 2000 Problem 2.
Solve for integers $x,y,z$: \begin{align}x + y &= 1 - z \\ x^3 + y^3 &= 1 - z^2 . \end{align}
My Progress: A bit of calculation and we get $x^2-xy+y^2=1+z $
Also we have $x^2+2xy+y^2=(1-z)^2 \implies 3xy=(1-z)^2-(1+z)=z(z-3) \implies y=\frac{z(z-3)}{3x}$ and $x=\frac{z(z-3)}{3y} $.
Note that since $z$,$x$,$y$ is an integer, we must have $3\mid z$.
So, let $z=3k$.
So we have $y=\frac{3k(3k-3)}{3x}=\frac{k(3k-3)}{x}$ and $x=\frac{z(z-3)}{3y}=\frac{k(3k-3)}{y}$ .
Then I am not able to proceed.
Hope one can give me some hints and guide me.
Thanks in advance.
|
We have $(1-z)(x^2-xy+y^2)=1-z^2.$
If $z=1$, so $x+y=0$ and we obtain $(t,-t,1)$, where $t$ is an integer.
Let $z\neq1$.
Thus, $$x^2-xy+y^2=z+1$$ and $$x+y=1-z,$$ which gives $$(1-z)^2-3xy=z+1$$ or
$$3xy=z^2-3z.$$
Thus, $z$ is divisible by $3$ and $$(1-z)^2-\frac{4}{3}(z^2-3z)\geq0$$ or
$$z^2-6z-3\leq0$$ or
$$3-\sqrt{12}\leq z\leq 3+\sqrt{12},$$
which gives $$0\leq z\leq 6$$
Can you end it now?
|
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|
Contest math application for Wilson's theorem
$$1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{23} = \frac{a}{23!}$$ Find the remainder when $a$ is divided by $13.$
I found this online and got stuck a bit. I approached the problem as such:
From the expression we get $$a=\frac{23!}{1}+\frac{23!}{2}+\dots+\frac{23!}{23!}$$
so $$a \equiv \frac{23!}{13} \pmod{13}$$
from here we have that $$a\equiv(1\cdot 2\dots11\cdot12) \cdot(1\cdot2\dots 9\cdot10) \equiv(12!) \cdot(10!) \pmod{13}.$$
And now according to Wilson's theorem $$a \equiv(-1)(10!) \pmod{13}.$$
I cannot seem to get rid of the $10!$; what should I do here?
|
Since $12!=10!\times 11\times12\equiv-1\bmod13$, it follows that $10!\times-2\times-1\equiv-1\bmod13$,
so $10!\times2\equiv-1\equiv12\bmod13$, so $10!\equiv6\bmod 13$.
|
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|
Elementary Arithmetic Problem In 1988, in a French math competition for middle school grades, the following problem was given:
Complete this multiplication so that all the digits from 0 to 9 appear:
$... × .. = ....1$
I am stumped. Of course the last digits of the two numbers are 3 and 7 but this is as far as I can go!
I am curious if there is a logical way to solve this problem without too much trial and error.
For the record, after significant experimentation my 3rd grade daughter has managed to obtain one solution, but this was pure luck.
|
The rule of nines: $jkl \equiv j+ k + l \pmod 9$ so
So if we have $abc\times de = fghi1$ and $a,b,....,f,g,h,i,1$ are the digits from $0,....,9$ then $abc + de +fghi1 \equiv 0 \pmod 9$
And if $abc \equiv j\pmod 9$ and $de \equiv k \pmod 9$ we have $fghi1 \equiv jk \equiv -(j+k)$.
Or $(j+1)k \equiv -j$ and $(k+1)j \equiv -k$
Possible values for $j,k$.
$(0,0)$, $(1,4)$, $(3,6)$, $(4,1)$, $(6,3)$
Furthermore
$(a+b+c)(d+e) \equiv f+g+h+i+1 \equiv 0+1+2+3+4+5+6+7+8+9 - (a+b+c+d+e)\pmod 9$
$ad +bd +cd + ae +be +ce \equiv -a-b-c-d-e \pmod 9$
And as $\{c,e\} = \{7,3\}$ we have
$ad + bd +cd +ae +be + 21 \equiv -a-b-d -10\pmod 9$ so
$ad + bd + cd +ae + be +a+b+d \equiv 5\pmod 9$.
If $c=7;e=3$ we have
$ad + bd + 8d+3a+3b +a+b\equiv (a+b)(d+4)-d \equiv 5\pmod 9$
$(a+b)(d+4) \equiv 5+d\pmod 9$
Now $a+b+7\equiv a+b - 2\equiv j$ and $d+3\equiv k$ for $j,k$ above.
We can have
1)$a+b\equiv 2$ and $d= 6$
2)$a+b\equiv 3$ and $d= 1$ (impossible as $1$ is accounted for)
3)$a+b\equiv 5$ and $d= 3$ (ditto $3$)
4)$a+b\equiv 6$ and $d= 7$ (ditto $7$)
5)$a+b\equiv 8$ and $d= 0,9$.($d=0$ is impossible as $de$ is two digits)
Case 1: $a+b\equiv 2$ and $d=64
$(a+b) \equiv 2$. As $a,b\ne 1,3,7,6$ we have $a+b=11$ and $a,b=2,9$
$(297,927)\times 63 = 18711, 58401$.
$927\times 63 = 58401$ is a working solution.
Case 2: $a+b \equiv 8$ and $d=9$. As $a,b\ne 1,3,7,9;a\ne b$ we have $a+b=8$ and $a,b = 2,6$
$(267, 627)\times 93$ do not work.
If $c=3,e = 7$ we have
$ad + bd + cd +ae + be +a+b+d \equiv 5\pmod 9$.
$ad + bd + 3d + 7a + 7b + a+b+d\equiv 5\pmod 9$
$(a+b)d + 4d +8(a+b) \equiv (a+b)(d-1) -5d \equiv 5\pmod 9$ so
$(a+b)(d-1)\equiv 5(d+1)\pmod 9$.
$a+b+3 \equiv j$ and $d+7\equiv d-1\equiv k$ for $j,k=(0,0), (1,4), (3,6)$, (4,1), (6,3)$ above.
Which all lead to contradictions
*
*$a+b\equiv 6$ and $d=1$ but $6*0\not \equiv 5*2$
*$a+b\equiv 8$ and $d=5$ but $8*4\not \equiv 5*6$
*$a+b\equiv 0$ and $d=7$ but $0*6\not \equiv 5*8$
*$a+b\equiv 1$ and $d=2$ but $1*1 \not \equiv 5*3$
*$a+b\equiv 3$ and $d=4$ but $3*2\not \equiv 5*5$
|
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|
Series Expansion by differentiation We know that 1/(1-x) = 1+x+x^2+x^3+....x^n
Say we need to find a suitable function for the expansion x+x^2+x^3+x^4+....x^(n+1)
We would differentiate 1/(1-x) = 1+x+x^2+x^3+....x^n
This would yield 1/(1-x)^2 = 1 + 2x+3x^2+4x^3 +...+nx^(n-1)**
Multiplying it by x would yield x/(1-x)^2 = x + 2x^2 + 3x^3 + 4x^4 +...nx^n Hence the respective function would be x/(1 - x)^2 = x + 2x^2 + 3x^3 + 4x^4 +...nx^n
But how would we obtain the function g[x] for the expansion 1+ 2x^2 + 3x^3 + 4x^4 +...
|
Your work is fine, now we just have to subtract $x$ and add $1$ that is
$$g(x)=\frac{x}{(1-x)^2}-x+1$$
|
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|
A question based on quadratic forms in linear algebra ( rank, representation) This particular question was asked in masters of mathematics exam of a university and I am unable to solve it. So I am asking it here.
Consider the quadratic form $Q(v)=v^{t} A v$, where
$$A=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{bmatrix},\quad v=(x, y, z, w)$$
Then
*
*$Q$ has rank 3 .
*$x y+z^{2}=Q(P v)$ for some invertible $4 \times 4$ real matrix $P$
*$x y+y^{2}+z^{2}=Q(P v)$ for some invertible $4 \times 4$ real matrix $P$.
*$x^{2}+y^{2}-z w=Q(P v)$ for some invertible $4 \times 4$ real matrix $P$.
Attempt: Determinant of $D_{1} $ =1 , $D_{2}$ = 1 , $D_{3}$ =-1 for some matrix and $D_{4}$ = -1 . So matrix is neither positive definite nor negative definite.
Also, I have read everything about quadratic forms from wikipedia as quadratic forms were not covered in my linear algebra class.
So, can anyone please tell how to solve this question.
Can anyone please tell any textbook of linear algebra which covers quadratic forms in detail?
I shall be really thankful.
|
The key to solving these questions (i.e. the classification of quadratic forms up to a change of basis) amounts to using Sylvester's law of intertia.
In particular, we find that $A$ has $3$ positive eigenvalues and $1$ negative eigenvalue, so its indices of innertia are $n_+ = 3, n_- = 1, n_0 = 0$.
With that, we can eliminate both choices $2,3$ bescause the symmetric matrix corresponding to these bilinear forms fails to be invertible, which means that they have zero-inex $n_0 > 0$. Choice $4$ is correct because the corresponding symmetric matrix
$$
\pmatrix{1&0&0&0\\0&1&0&0\\0&0&0&-1\\0&0&-1&0}
$$
also has $3$ positive eigenvalues and $1$ negative eigenvalue. In fact, in this case it is possible to find a simple (diagonal!) change of basis matrix $P$.
The key to finding the matrix associated with a quadratic form is to note that for a $4 \times 4$ symmetric matrix $A$, we have
$$
\begin{array}{ccccccccc} v^TAv= && a_{1,1}x^2& + & a_{1,2}xy & + & a_{1,3}xz & + & a_{1,4}xw\\
&+&a_{2,1}xy& + & a_{2,2}y^2 & + & a_{2,3}yz & + & a_{2,4}yw\\
&+&a_{3,1}xz& + & a_{3,2}yz & + & a_{3,3}z^2 & + & a_{3,4}zw\\
&+&a_{4,1}xw& + & a_{4,2}yw & + & a_{4,3}zw & + & a_{4,4}w^2
\end{array}
\\
\begin{array}{ccccccccc} \phantom{v^TAv}= && a_{1,1}x^2& + & 2a_{1,2}xy & + & 2a_{1,3}xz & + & 2a_{1,4}xw\\
&&& + & a_{2,2}y^2 & + & 2a_{2,3}yz & + & 2a_{2,4}yw\\
&&&&& + & a_{3,3}z^2 & + & 2a_{3,4}zw\\
&&&&&&&+& a_{4,4}w^2.
\end{array}
$$
Now, the quadratic form from option $2$ can be written as
$$
\begin{array}{cccccccc} v^TAv = & 0 x^2& + & 1\,xy & + & 0xz & + & 0xw\\
&& + & 0y^2 & + & 0yz & + & 0yw\\
&&&& + & 1\,z^2 & + & 0zw\\
&&&&&&+& 0w^2.
\end{array}
$$
|
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|
Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-
Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$
Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\over2}+{c^2\over2}\right)$$
$$ \geq\left(\sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4}\right)^3\geq8 $$
$$ \Rightarrow \sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4} \geq2 \Rightarrow \sqrt[3]{2(ab)^2}+\sqrt[3]{2(bc)^2}+\sqrt[3]{2(ca)^2}\geq4 $$
I reached till here but can't take it forward.
Approach 2:
$$ \prod_{cyc} {(1+a^2)}=\prod_{cyc} \sqrt{(1+a^2)(1+b^2)} \geq \prod_{cyc} {(1+ab)}\ge8 $$
but it failed as
$$ \sum_{cyc}{ab}=3 \Rightarrow \sum_{cyc}{(1+ab)}=6 \Rightarrow 8\ge \prod_{cyc} {(1+ab)} $$
This approach is surely weak, but I think that the first approach is unfinished.
Probably brute-force would help but other solutions are always welcome.
Thanks!
|
Another way.
By C-S and AM-GM we obtain: $$\prod_{cyc}(1+a^2)\geq(a+b)^2(1+c^2)=(a+b)^2+(ac+bc)^2\geq$$
$$\geq4ab+(3-ab)^2=a^2b^2-2ab+9=(ab-1)^2+8\geq8.$$
|
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|
How to prove that $S=\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)?$ How to prove that
$$S=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)$$
My attempt:
We have for $|x|\leq1$ $$\tanh^{-1}(x)=\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$$
and :
\begin{align*}
\displaystyle\int_0^{\sqrt{2}-1}\frac{\tanh^{-1}(x)}{x}\ \mathrm{d}x&=\displaystyle\int_0^{\sqrt{2}-1}\frac{1}{x}\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\mathrm{d}x\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{1}{2n+1}\displaystyle\int_0^{\sqrt{2}-1}x^{2n}\mathrm{d}x\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}\\
&=S\\
\end{align*}
So :
\begin{align*}
S&=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}\\
&=\displaystyle\int_0^{\sqrt{2}-1}\frac{\tanh^{-1}(x)}{x}\mathrm{d}x\\
&=\displaystyle\int_0^{\sqrt{2}-1}\frac{1}{2x}\left(\log(1+x)-\log(1-x)\right)\mathrm{d}x\\
&=\frac{1}{2}(J_1-J_2)
\end{align*}
Where:
\begin{align*}
J_1&=\displaystyle\int_0^{\sqrt{2}-1}\frac{\log(1+x)}{x}\mathrm{d}x\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\displaystyle\int_0^{\sqrt{2}-1}x^n\mathrm{d}x\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n(\sqrt{2}-1)^{n+1}}{(n+1)^2}\\
\end{align*}
And:
\begin{align*}
J_2&=\displaystyle\int_0^{\sqrt{2}-1}\frac{\log(1-x)}{x}\mathrm{d}x\\
&=\displaystyle\int_0^{\sqrt{2}-1}-\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n+1}dx\\
&=-\displaystyle\sum_{n=0}^{\infty}\frac{1}{n+1}\left[\frac{x^{n+1}}{n+1}\right]_0^{\sqrt{2}-1}\\
&=-\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{n+1}}{(n+1)^2}\\
\end{align*}
Finally we find :
$$S=\frac{1}{2}\left(\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{n+1}((-1)^n+1)}{(n+1)^2}\right)$$
But I could not find a way to calculate $S$.
Any help please? and thank's in advance.
|
I shall be continuing from the last two integrals of your work $$\int_0^{\sqrt 2-1} \frac{\tanh^{-1}x}{x} dx =\frac{1}{2}\int_0^{\sqrt 2-1} \left(\frac{\log(1+x)}{x}-\frac{\log(1-x)}{x}\right)dx$$ These two integrals posses non-elementary antiderivatives in terms of special function Dilogarithm function so by definition we have $$=\frac{1}{2}\left(\operatorname{Li}_2(x)-\operatorname{Li}_2(-x)\bigg|_0^{\sqrt 2-1}\right)=\frac{1}{2}\left(\operatorname{Li}_2\left(\sqrt2 -1\right)-\operatorname{Li}_2\left(1-\sqrt 2\right)\right)\cdots(1)$$ Using the last identity [3] we have
$$\operatorname{Li}_2(1-\sqrt 2)= -\frac{\pi^2}{6}-\frac{1}{2}\ln^2(\sqrt 2-1)+\operatorname{Li}_2\left(\frac{1}{1-\sqrt 2}\right)=\frac{\pi^2}{6}-\frac{1}{2}\ln^2(\sqrt 2-1)-\operatorname{Li}_2(-1-\sqrt 2)$$ plugging the obtained to $(1)$ we have then $$\frac{\pi^2}{12}+\frac{1}{4}\ln^2(\sqrt 2-1)+\frac{1}{2}\left(\operatorname{Li}_2(\sqrt 2-1)+\operatorname{Li}_2(-\sqrt 2-1)\right)\cdots(2)$$ let $u= \sqrt 2-1 $ and $v=-1-\sqrt 2$ also $uv= -(\sqrt 2-1)(\sqrt 2+1)=-1$.
Using the Abel identity $$\operatorname{Li}_2(u)+\operatorname{Li}_2(v)=\operatorname{Li}_2(uv)+\operatorname{Li}_2\left(\frac{u-uv}{1-uv}\right)+
\operatorname{Li}_2\left(\frac{v-uv}{1-uv}\right)+\ln\left(\frac{1-u}{1-uv}\right)\ln\left(\frac{1-v}{1-uv}\right)$$ plugging the assigned values of $u$ and $v$ we have $$\operatorname {Li}_(\sqrt 2-1)+\operatorname{Li_2}(-1-\sqrt 2)=-\operatorname{Li}_2(-1)+\color{red}{\operatorname{Li}_2\left(\frac{1}{\sqrt 2}\right)+\operatorname{Li}_2\left(-\frac{1}{\sqrt 2}\right)}+\ln\left(\frac{\sqrt 2-1}{\sqrt{2}}\right)\ln\left(\frac{\sqrt{2}+1}{\sqrt 2} \right) =-\frac{\pi^2}{12}+\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{2}\right)-\ln^2(\sqrt {2}-1)+\frac{1}{2}\ln^2(2)=-\frac{\pi^2}{12}+\color{red}{\frac{\pi^2}{24}-\frac{1}{2}\ln^2(2)}-\ln^2(\sqrt 2-1)+\frac{1}{2}\ln^2(2)=-\frac{\pi^2}{24}-\ln^2(\sqrt 2-1)\cdots(2)$$ plugging back to $(2)$ we have $$\frac{\pi^2}{12}+\frac{1}{4}\ln^2(\sqrt 2-1)+\frac{1}{2}\left(-\frac{\pi^2}{24}-\ln^2(\sqrt 2-1)\right)=\frac{\pi^2}{16}-\frac{1}{4}\ln^2\left(\sqrt 2-1\right)$$
We use identity
$$\color{red}{\operatorname{Li}_2(z)+\operatorname{Li}_2(-z)=\frac{1}{2}\operatorname{Li}_2(z^2)}$$ and for $z=\frac{1}{\sqrt 2}$ we have $$\color{red}{\frac{1}{2}\operatorname{Li}_2\left(\frac{1}{2}\right)=\frac{1}{2}\left(\frac{\pi^2}{12}-\ln^2(2)\right)}$$
|
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|
Proof that $ \frac{2^{-x}-1}{x} = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!} $
If
$$2^x = \sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!} \hspace{1cm} \forall x \in \mathbb{R},$$
Proof that:
$$ \frac{2^{-x}-1}{x} = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!} $$
I did the following:
\begin{align*}
&2^x = \sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!} \\
\Rightarrow \quad & 2^{-x} = \frac{1}{\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}} \\
\Rightarrow \quad & 2^{-x}-1 = \frac{1}{\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}} -1 \\
\Rightarrow \quad & \dfrac{2^{-x}-1}{x} = \frac{\frac{1}{\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}} -1}{x} \\
\Rightarrow \quad & \frac{2^{-x}-1}{x} = \frac{1-\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}}{x\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}}
\end{align*}
From this part I don't know how to continue
|
We have
$$2^x = \sum_{n=0}^{\infty} \dfrac{(x\ln(2))^n}{n!}$$
$$2^{-x} = \sum_{n=0}^{\infty} \dfrac{(-x)^n(\ln(2))^n}{n!}$$
$$2^{-x}-1 = -1+\sum_{n=0}^{\infty} \dfrac{(-1)^nx^n(\ln(2))^n}{n!}=\sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}x^{n+1}(\ln(2))^{n+1}}{(n+1)!}$$
$$\frac{2^{-x}-1}x =\sum_{n=0}^{\infty} \dfrac{(-1)^{n+1}x^{n}(\ln(2))^{n+1}}{(n+1)!}$$
|
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|
Find the minimum positive integer $n$ and the matrix $A^{2020}$ Let $A=a\begin{pmatrix} 1 & -1\\
1 & 1 \end{pmatrix}$ $(a>0)$ and $I=\begin{pmatrix} 1 & 0\\
0 & 1 \end{pmatrix}$ satisfy $A^4+I=\begin{pmatrix} 0 & 0\\
0 & 0 \end{pmatrix}$
*
*Find $a$ For this I can do. I saw $a=\frac{1}{\sqrt{2}}$
*Find the minimum positive integer $n$ such that $A^n\begin{pmatrix} 0\\ 1 \end{pmatrix}= \begin{pmatrix} 1\\ 0 \end{pmatrix}$
*Find $A^{2020}$
So please help to tell me about this 2 and 3. Give me some hints or ideas!!
|
For 2: $A^{2}\left(\begin{array}{c}
0\\
1
\end{array}\right)=\left(\begin{array}{c}
-1\\
0
\end{array}\right)$, so$$A^6\left(\begin{array}{c}
0\\
1
\end{array}\right)=-A^2\left(\begin{array}{c}
0\\
1
\end{array}\right)=\left(\begin{array}{c}
1\\
0
\end{array}\right).$$Hint for 3: $2020$ is an odd multiple of $4$.
|
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|
Evaluating $\int_0^1\frac{\arctan x\ln\left(\frac{2x^2}{1+x^2}\right)}{1-x}dx$ Here is a nice problem proposed by Cornel Valean
$$
I=\int_0^1\frac{\arctan\left(x\right)}{1-x}\,
\ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x =
-\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{192}\,\pi^{3} +
2\Im\left\{%
\text{Li}_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}
$$
My Trial: By subbing $x=\frac{1-t}{1+t}$ we have
$$I=\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)\ln\left(\frac{(1-x)^2}{1+x^2}\right)}{x(1+x)}dx$$
$$=2\underbrace{\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)\ln(1-x)}{x(1+x)}dx}_{x\to (1-x)/(1+x)}-\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)\ln(1+x^2)}{x(1+x)}dx$$
$$=2\int_0^1\frac{\arctan x\ln(\frac{2x}{1+x})}{1-x}dx-\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)\ln(1+x^2)}{x(1+x)}dx$$
and got stuck here. Any idea? thanks.
|
Update: the problem and solution will be part of a new paper soon.
A solution by Cornel Ioan Valean
Let's denote the main integral by $\mathcal{I}$, and then we have
$$\mathcal{I=}\int_0^1\frac{(\pi/4-\arctan((1-x)/(1+x)))\log\left(\frac{2x^2}{1+x^2}\right)}{1-x}\textrm{d}x$$
$$=\underbrace{\frac{\pi}{4}\int_0^1\frac{\log\left(\frac{2x^2}{1+x^2}\right)}{1-x}\textrm{d}x}_{\displaystyle J}- \underbrace{\int_0^1\frac{\arctan((1-x)/(1+x))\log\left(\frac{2x^2}{1+x^2}\right)}{1-x}\textrm{d}x}_{\displaystyle
K}. \tag1$$
The integral $J$ easily reduces to known integrals. If we integrate by parts, we get
$$J=\frac{\pi}{2}\underbrace{\int_0^1\frac{\log (1-x)}{x}\textrm{d}x}_{\displaystyle -\pi^2/6}-\frac{\pi}{2}\underbrace{\int_0^1\frac{x \log (1-x)}{1+x^2}\textrm{d}x}_{\displaystyle 1/8 (\log^2(2)-5\pi^2/12 )}=-\frac{\pi}{16}\log^2(2)-\frac{11}{192}\pi^3,\tag2$$
where the last integral also appears in the book, (Almost) Impossible Integrals, Sums, and Series, page $8$.
For the integral $K$, a bit of magic will be necessary. The first key observation is that
$$K=\Im \biggr\{\int_0^1\frac{\log^2(x (1 + x)/(1 + x^2) + i x (1 - x)/(1 + x^2))}{1-x}\textrm{d}x\biggr\}.$$
Now, we may consider the generalization
$$G(a)=\int_0^1\frac{\displaystyle\log^2\left(\frac{ (1+a) x}{1 + a x}\right)}{1- x}\textrm{d}x,$$
and make the variable change $\displaystyle x\mapsto \frac{1-x}{1+a x}$ that leads to
$$G(a)=\int_0^1 \frac{\log^2(1-x)}{x}\textrm{d}x-a\int_0^1\frac{\displaystyle\log^2(1-x)}{1+ a x}\textrm{d}x,$$
and letting the variable change $x\mapsto 1-x$ in both integrals, we finally get
$$G(a)=\int_0^1 \frac{\log^2(x)}{1-x}\textrm{d}x-\frac{a}{1+a}\int_0^1\frac{\displaystyle\log^2(x)}{1 -a/(1+a) x}\textrm{d}x=2 \zeta(3)-2\operatorname{Li}_3\left(\frac{a}{1+a}\right),$$
where in the calculations we also needed the integral, $\displaystyle \int_0^1 \frac{a \log^2(x)}{1-a x}\textrm{d}x=2\operatorname{Li}_3(a)$, that appears in a generalized form in the same book, (Almost) Impossible Integrals, Sums, and Series, page $4$.
A first note: The variable change $\displaystyle x\mapsto \frac{x}{1+a-ax}$ would work more directly, and no need for a second variable change.
Then, based on the previous result we make the second key observation,
$$K=\Im \{G(i)\}.$$
Thus,
$$\small K=\Im \biggr \{\int_0^1\frac{\log^2(x (1 + x)/(1 + x^2) + i x (1 - x)/(1 + x^2))}{1-x}\textrm{d}x \biggr \}=2 \Im \biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}. \tag3 $$
At last, combining $(1)$, $(2)$, and $(3)$, we conclude that
$$\mathcal{I}=-\frac{\pi}{16}\log^2(2)-\frac{11}{192}\pi^3+2 \Im \biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}.$$
End of story
A second note: no software needed for calculating such integrals, or far more advanced ones alike.
Another nice example of an integral calculated by similar means
$$\int_0^1 \frac{1}{x(1+x)}\left(12 \log \left(\frac{(1-x)^2}{1+x^2}\right) \arctan^2(x)-\log ^3\left(\frac{(1-x)^2}{1+x^2}\right)\right) \textrm{d}x$$
$$=\frac{2043 }{64}\zeta (4)+\frac{15}{8} \log ^2(2)\zeta (2)-\frac{1}{2} \log ^4(2)-15 \operatorname{Li}_4\left(\frac{1}{2}\right).$$
|
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How to find the determinant of this $6\times 6$ X-matrix? This question was asked in my quiz and i was unable to solve it, so I am asking it here.
Find the value of determinant of this particular matrix .
$$\begin{pmatrix}1&0&0&0&0&2\\0&1&0&0&2&0\\0&0&1&2&0&0\\0&0&2&1&0&0\\0&2&0&0&1&0\\2&0&0&0&0&1\end{pmatrix}$$
I have no clue on how this kind of matrices can be solved. Can anyone give a general strategy on how to solve matrices whose size are greater that $3\times 3$?
That would be really helpful.
|
Let ${\rm R}_3$ be the $3 \times 3$ reversal matrix. Hence,
$$\det \left[\begin{array}{ccc|ccc} 1&0&0&0&0&2\\ 0&1&0&0&2&0\\ 0&0&1&2&0&0\\ \hline 0&0&2&1&0&0\\0&2&0&0&1&0\\ 2&0&0&0&0&1\end{array}\right] = \det \begin{bmatrix} {\rm I}_3 & 2{\rm R}_3\\ 2{\rm R}_3 & {\rm I}_3\end{bmatrix} = \det \left( {\rm I}_3 - 4 {\rm R}_3^2 \right) = (-3)^3 = \color{blue}{-27}$$
because ${\rm R}_3^2 = {\rm I}_3$.
matrices block-matrices permutation-matrices determinant
|
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|
How can we take the derivative of this function: $y = \frac{x}{x^2+1}$ from first principles (using the limit definition of the derivative)? I was taking the derivative of the function: $y = \frac{x}{x^2+1}$. I know that we can solve it by the quotient rule. But I tried using the limit definition of differentiation. This is how I did it:
$$\lim_{h \to 0} \frac{(x+h)/\left((x+h)^2+1\right) - x/(x^2+1)}{h}$$
$$=\lim_{h \to 0}\frac{(x+h)/(x^2+h^2+2xh+1) - x/(x^2+1)}{h}$$
Then I multiplied both the denominator and numerator by the common factor $(x^2+h^2+2xh+1)(x^2+1)$ and expanded:
$$=\lim_{h \to 0} \ \Biggl(\frac{x^3+x+x^2h+h - x^3-xh^2-2x^2h-x}{x^4+x^2h^2+2x^3h+x^2+x^2+h^2+2xh+1}\Biggr) \cdot \frac{1}{h}$$
I then simplified the expression:
$$=\lim_{h \to 0} \ \frac{h-x^2h-xh^2}{x^4h + x^2h^3+2x^3h^2+2x^2h+h^3+2xh^2+h}$$
Now what is to be done? It seems that I am very far from the actual derivative.
|
You are almost there, check that
$$f'(x)=\lim_{h \to 0} \frac{h-x^2h-xh^2}{x^4h + x^2h^3+2x^3h^2+2x^2h+h^3+2xh^2+h}$$
Divide by $h$ up and down:
$$f'(x)=\lim_{h \to 0}\frac{1-x^2-xh}{(x^4+2x^2+1)+x^2h^2+2x^3h+h^2+2xh}$$ $$\implies f'(x)=\frac{1-x^2}{(1+x^2)^2}$$
|
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|
Correct result for this integral $\int \frac{\sqrt{\sqrt{\sqrt{2 \cos \left(5 \sqrt{x}+4\right)+2}+2}+2}}{\sqrt{x}}\, dx$ Wolfram|Alpha and its CAS, Wolfram Mathematica are, as far as I know, the only website and software that give the correct solution to this integral,
$$ f(x) = \frac{\sqrt{\sqrt{\sqrt{2 \cos \left(5 \sqrt{x}+4\right)+2}+2}+2}}{\sqrt{x}} $$
$$ F(x) = \int f(x)\, dx$$
because deriving the function given as result and using the FullSimplify function of Mathematica we get to the original function that we wanted to integrate.
This is the solution:
$$ F(x) = \frac{1}{5} (-8) \sqrt{2} \sqrt{\cos \left(5 \sqrt{x}+4\right)+1} \sqrt{\sqrt{2} \sqrt{\cos \left(5 \sqrt{x}+4\right)+1}+2} \left(\sqrt{\sqrt{2} \sqrt{\cos \left(5 \sqrt{x}+4\right)+1}+2}-2\right) \sqrt{\sqrt{\sqrt{2} \sqrt{\cos \left(5 \sqrt{x}+4\right)+1}+2}+2} \csc \left(5 \sqrt{x}+4\right) + C $$
Fricas finds another expression for the integral that I'm curious about:
$$ F(x) = \frac{1}{5} (-8) \sqrt{2} \sqrt{\sqrt[4]{2} \sqrt{\sqrt{\cos \left(5 \sqrt{x}+4\right)+1}+\sqrt{2}}+2} \left(\sqrt{2} \cos \left(5 \sqrt{x}+4\right)-2 \sqrt{\sqrt{2} \sqrt{\cos \left(5 \sqrt{x}+4\right)+1} \left(\sqrt{\cos \left(5 \sqrt{x}+4\right)+1}+\sqrt{2}\right)+2 \sqrt{\cos \left(5 \sqrt{x}+4\right)+1}+\sqrt{2}}\right) \csc \left(5 \sqrt{x}+4\right) + C$$
However, in this video, an incorrect result is given although the integration process seems correct. As above, you know that the result is incorrect since deriving the resulting function doesn't result in the original function we wanted to integrate.
Is the expression given by Fricas equivalent to the one given by Wolfram|Alpha and Mathematica?
Final update: According to Mathematica, the expression given by FriCAS is not equal to the correct solution given by Mathematica and Wolfram|Alpha. This means it is not the solution to this problem.
Even though I have a Pro Premium subscription, the step-by-step solution is not available for this input. What are the steps taken by Mathematica/Wolfram|Alpha to get to the given result?
|
This is a partial solution, equivalent to the process used in the video, which is only valid if
$$\cos\frac t2$$
is non-negative.
Start with this identity:
$$\sqrt{2+2\cos t} = \sqrt{4\cos^2\frac t2} = 2\cos\frac t2$$
For applying this to the integrand, first make the substitution $t = \sqrt x$, then successively apply this property.
$$\begin{gather}
I = \int\sqrt{2 + \sqrt{2 + \sqrt{2 + 2\cos(5t+4)}}}\cdot 2dt\\
= \int\sqrt{2 + \sqrt{2+2\cos\left(\frac{5t+4}{2}\right)}} \cdot 2 dt\\
= \int\sqrt{2 + 2\cos\left( \frac{5t+4}{4}\right)} \cdot 2dt \\
= \int 4\cos\left(\frac{5t+4}{8}\right) dt \\
= \frac{32}{5}\sin\left( \frac{5\sqrt x + 4}{8} \right) + C
\end{gather}$$
|
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|
Extraneous solution from substituting in equations I came across this example of how you could end up with an extraneous solution but I was wondering how it arose.
We have the equation: $$x^2+x+1=0 $$ Since x=0 does not satisfy the equation, you can divide by x on both sides which yields: $$x+1+\frac{1}{x}=0$$ which is equivalent to our first equation. From our first eqution we can conclude that: $$-x^2=x+1$$ We now substitute that into the second equation to get:$$x^2=\frac{1}{x}$$ which results in $$x^3=1$$ which is equivalent to our previous equation since x cannot be 0. However, one solution from our last equation is x=1, which is not a solution to our original equation.
I have a vague idea that it may have to do with the fact that you get a cubic equation and you began with a quadratic, and that steps imply the following and not vice versa, but can you provide a very detailed answer as to why it arises? Can you please provide more examples?
|
This substitution ($x+1=-x^2$) expands a set of roots of the equation
because $-x^2$ also depends on $x$.
You can substitute $x+1=y$, for example.
More example, when a similar substitution gives similar problems.
Let we need to solve $$\sqrt[3]{2x+1}+\sqrt[3]{x+1}=\sqrt[3]{x-1}.$$
We obtain:
$$\left(\sqrt[3]{2x+1}+\sqrt[3]{x+1}\right)^3=x-1$$ or
$$2x+1+x+1+3\sqrt[3]{2x+1}\cdot\sqrt[3]{x+1}\left(\sqrt[3]{2x+1}+\sqrt[3]{x+1}\right)=x-1.$$
Now, since $$\sqrt[3]{2x+1}+\sqrt[3]{x+1}=\sqrt[3]{x-1},$$ which can get something bad, we obtain:
$$3\sqrt[3]{(2x+1)(x+1)(x-1)}=-3-2x$$ or
$$x(440x^2+630x+189)=0$$ and we got as one of options $x=0$.
Easy to see that $0$ is not a root of the starting equation and it happened
because we used a not correct substitution $\sqrt[3]{2x+1}+\sqrt[3]{x+1}=\sqrt[3]{x-1}.$
Now, we need to check that all roots of the equation $440x^2+630x+189=0$ are roots of the starting equation, which is not so easy.
If we want to avoid from these problems, so we need to use the following identity. $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$
|
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|
Closed form of hypergeometric $\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},1,1;z\right)$ How can we prove $$\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},1,1;z\right)=\frac{4 }{\pi ^2}K\left(\frac{1}{2}-\frac{\sqrt{k(z)}}{2}\right) K\left(\frac{1}{2}-\frac{1}{2 \sqrt{k(z)}}\right), |z|\le1$$
Where $k(z)=-2 \sqrt{z^2-z}-2 z+1$ (thanks to @maxim for the correct form)?
Update: A proof is given by verifying that both LHS and RHS of the following are solutions of certain order $4$ ODE, satisfying same initial conditions (Mathematica command DifferentialRootReduce should do it well). The rest are trivial by analytic continuation. Thus
$${_4 F_3} {\left( \frac 1 4, \frac 1 4, \frac 3 4, \frac 3 4; \frac 1 2, 1, 1; -\left( \frac w 2 - \frac 1 {2 w} \right)^{\! 2} \right)} = \frac 4 {\pi^2} K {\left( \frac 1 2 - \frac w 2 \right)} K {\left( \frac 1 2 - \frac 1 {2 w} \right)}$$
|
This is not an answer.
For small values of $z$, we have
$$\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},1,1;z\right)=1+\frac{9 z}{128}+\frac{3675 z^2}{131072}+\frac{266805
z^3}{16777216}+O\left(z^4\right)$$ while, with $ k(z)=-2 \sqrt{z^2-z}-2 z+1$,$$\frac{4 }{\pi ^2}K\left(\frac{1}{2}-\frac{k(z)}{2}\right) K\left(\frac{1}{2}-\frac{1}{2k(z)}\right)=1+\frac{9 z}{32}+\frac{1371 z^2}{8192}+\frac{31605
z^3}{262144}+O\left(z^{4}\right)$$
To be different, they are !
Edit
After @Maxim comment
$$w_\pm = \sqrt {1 - 2 z \pm 2 \sqrt {z^2 - z}}\implies -\left( \frac {w_\pm} 2 - \frac 1 {2 w_\pm} \right)^2=z$$
$$\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{3}{4},\frac{3}{4};\frac{1}{2},1,1;z\right)=\frac 4 {\pi^2} K {\left( \frac 1 2 - \frac {w_\pm} 2 \right)} K {\left( \frac 1 2 - \frac 1 {2 w_\pm} \right)}$$ seems to hold for positive and negative values of $z$.
|
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|
Given positive $x,y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $, find minimum $(x+y)$ I am given positive numbers $x, y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $. I need to find the minimum value of $(x+y)$. Here is my try. Using AM-GM inequality for nonnegative numbers, I have
$$
\frac{(x+y)}{2} \geqslant \sqrt{x} \sqrt{y}
$$
$$
\sqrt{x} \sqrt{y}(x-y) \geqslant 2 \sqrt{x} \sqrt{y}
\\ \therefore (x-y) \geqslant 2
$$
So, I have been able to arrive at this conclusion. But I am stuck here. Any help ?
Thanks
|
Given $\sqrt{xy}\left(x-y\right)=x+y$
let $yx=c$ , where $c>0$.
$$\sqrt{c}\left(x^{2}-c\right)=x^{2}+c$$
$$x^{2}-\frac{\left(c+1+2\sqrt{c}\right)}{\left(c-1\right)}c=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -[1]$$
Let a function $$F(x,c)=x^{2}-\frac{\left(c+1+2\sqrt{c}\right)}{\left(c-1\right)}c$$ be defined. Then $$\frac{\partial F(x,c)}{\partial c}=0$$ at a constant $x$ gives us $c \approx 2.618 \implies x \approx 3.33 $ ( using $[1]$ ) . So, $$x+\frac{c}{x}\geqslant 4$$
$$min(x+y)=4$$
when $x=2+\sqrt{2} \text{ and } y=2-\sqrt2$ as pointed out.
|
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|
If $V_n(a)$ counts sign changes in the sequence $\cos a, \cos2a,\cos3a,\ldots,\cos na,$ show that $\lim_{n\to\infty}\frac{V_n(a)}n=\frac{a}\pi$
Let $0\leq\alpha\leq \pi $. $V_n (\alpha) $ denote the number of sign changes in the sequence $\cos\alpha,\cos2\alpha,\cos3\alpha,\ldots,\cos n\alpha $. Then prove that $$\lim\limits_{n\to\infty}\dfrac{V_n (\alpha)}{n}=\dfrac{\alpha}{\pi}.$$
I saw a hint where $\dfrac{V_n (\alpha)}{n}$ is considered as the probability. I mean how this expression is a probability of something. If it is, how can I progress further in this way?
Update: I have a solution to this problem
In $n\alpha$ rotation the number of times full circle rotation occures $=\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor$
In one full circle rotation sign change occures 2 times. Hence in $\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor$ full rotation sign change occures $=2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor$
Now the rest angle is $n\alpha-\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor\times2\pi$
If we consider 0 as a change of sign in case of $\cos\left( \dfrac{\pi}{2}\right)$ and $\cos\left(\dfrac{3\pi}{2}\right)$ then:-
(1) If $0\leq n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi<\dfrac{\pi}{2 }$ sign changes 0 times
(2) If $\dfrac{\pi}{2 }\leq n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi<\dfrac{3\pi}{2 }$ sign changes 1 times
(3) If $\dfrac{3\pi}{2 }\leq n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi<2\pi$ sign changes 2 times
Let $f$ be a function such that $$f\left(\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor\right)=\begin{cases}0,\text{ when }\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=0\\ 1,\text{ when }\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=1\\ 1,\text{ when }\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=2\\ 2,\text{ when } \left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=3\end{cases}$$
Therefore $\dfrac{V_n(\alpha)}{n}=\dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor+ f\left(\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor\right)}{n}$
Hence $$\dfrac{V_n(\alpha)}{n}\geq \dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor}{n}$$ and $$\dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor+ 2}{n}\leq \dfrac{V_n(\alpha)}{n}$$
$\lim\limits_{n\to \infty}\dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor}{n}=\dfrac{\alpha}{\pi}$ and $\lim\limits_{n\to\infty} \dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor+ 2}{n}=\dfrac{\alpha}{\pi}$
Hence by Sandwich Theorem We get $\lim\limits_{n\to \infty}\dfrac{V_n(\alpha)}{n}=\dfrac{\alpha}{\pi}$ [Proved]
Is this correct?
|
*
*I will assume that the case where $\alpha\in \pi \mathbb{Q}$ is easy, because the sequence $\big(\cos(k\alpha)\big)_{k\ge1}$ is periodic in this case, and if we consider $0$ as positive number then $V_{2q}(p\pi/q)=2p\pm 1$ and the result holds in this case.
*Now we assume that $\alpha\notin \pi\mathbb{Q}$. This implies that the sequence $\big(k\alpha \mod(2\pi)\big)_{k\geq 1}$ is equidistributed in $[0,2\pi]$. See Equidistributed sequences.
Now, let $f$ be the $2\pi$ periodic function defined by
$$f(\theta)=\cases{0, & if $\cos\theta \cos(\theta+\alpha)\geq0$,\\
1,& if $\cos\theta \cos(\theta+\alpha)<0$.}$$
With this definition,
$$V_n(\alpha)=\text{card}\left\{k\in\{1,\ldots,n\}:f(k\alpha)=1\right\}$$
But if we define
$$\mathcal{I}=\cases{\left(\frac{\pi}{2}-\alpha,\frac{\pi}2\right)\cup
\left(\frac{3\pi}{2}-\alpha,\frac{3\pi}2\right)
,&if $0<\alpha<\pi/2$,\cr
\left[0,\frac{\pi}{2}\right)\cup
\left(\frac{3\pi}{2}-\alpha,\frac{3\pi}2\right)\cup\left(\frac{5\pi}{2}-\alpha,2\pi\right]
,&if $\pi/2<\alpha<\pi$.}$$
Then for $\theta\in[0,2\pi]$ we have
$$f(\theta)=1\iff \theta\in\mathcal{I}$$
So, equidistribution of the sequence implies that
$$\lim_{n\to\infty}\frac{V_n(\alpha)}{n}=\frac{\text{length}(\mathcal{I})}{2\pi}=\frac{\alpha}{\pi}$$
Done.$\qquad\square$
|
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|
Apporoaches to solve the given algebraic expression If $\displaystyle \ \ x^{4} \ +\ x^{2} \ =\ \frac{11}{5}$ then what is the value of the given expression
$$\displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} +\ \left(\frac{x-1}{x+1}\right)^{\frac{1}{3}} \ =\ \ ?$$
My Try :
As I can find the value of $\displaystyle x$, from the given equation but it will be tedious I think !.
$$\displaystyle x^{4} \ +\ x^{2} \ =\ \frac{11}{5}$$
$$\displaystyle \Longrightarrow \ x^{2} +1/2 \ =\ \frac{7}{2\sqrt{5}}$$
$$\displaystyle \Longrightarrow \ x^2 \ =\ \ \frac{7-\sqrt{5}}{2\sqrt{5}}$$
Which is getting too much complicated to solve the expression by putting the value of $\displaystyle x$.
What could be the other way to solve the given expression?
|
Let
$$ a = \displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} +\ \left(\frac{x-1}{x+1}\right)^{\frac{1}{3}}. $$
Notice that the first term is the multiplicative inverse of the second term. Thus
$$ a^3 = \displaystyle \frac{x+1}{x-1} + \frac{x-1}{x+1} + 3 \left(\displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} +\ \left(\frac{x-1}{x+1}\right)^{\frac{1}{3}} \right). $$
So
$$ a^3 - 3 a = \displaystyle \frac{x+1}{x-1} + \frac{x-1}{x+1}. $$
You can calculate the right hand side and then solve for $a$.
|
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|
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .
If $x + y = 5xy$ , $y + z = 6yz$ , $z + x = 7zx$ . Find $x + y + z$ .
What I Tried :
I used some clever ways to get $x + y + z = 26xyz$ , but I suppose we have some solution as a number .
All all $3$ to get :-
$$2(x + y + z) = 5xy + 6yz + 7zx$$
Or,
$$ 2(x + y + z) = (xy + xy + xy + xy + xy) + (yz + yz + yz + yz + yz + yz) + (zx + zx + zx + zx + zx + zx + zx)$$
That is,
$$ 2(x + y + z) = (xy + zx) + (xy + zx) + (xy + zx) + (yz + zx) + (yz + zx) + (yz + zx) + (yz + zx) + (xy + yz) + (xy + yz)$$
Now see that $(xy + zx) = x(y + z) = 6xyz$ , similarly $(yz + zx) = 5xyz$ and $(xy + yz) = 7xyz$
So
$$2(x + y + z) = 3(6xyz) + 4(5xyz) + 2(7xyz)$$
$$\Rightarrow (x + y + z) = \frac{52xyz}{2} = 26xyz$$
I tried till this , then I have no idea . Can anyone help?
|
Taking @user's hint a little further (and also assuming $x,y,z,\neq 0$), writing
\begin{align}
u &=\frac{1}{x}\\
v&=\frac{1}{y}\\
w&=\frac{1}{z}
\end{align}
Yields
\begin{align}
u+v &= 5\\
v+w &= 6\\
z+u&=7
\end{align}
thus
$$\begin{pmatrix}
1 & 1 & 0\\
0 & 1 & 1\\
1 & 0 & 1\\
\end{pmatrix}
\begin{pmatrix}
u\\
v\\
w\\
\end{pmatrix}=
\begin{pmatrix}
5\\
6\\
7
\end{pmatrix}
$$
Thus
$$
\begin{pmatrix}
u\\
v\\
w\\
\end{pmatrix} = \frac{1}{2}
\begin{pmatrix}
1 & -1 & 1\\
1 & 1 & -1\\
-1 & 1 & 1
\end{pmatrix}
\begin{pmatrix}
5\\
6\\
7
\end{pmatrix}$$
Whence
$$\begin{pmatrix}
u\\
v\\
w\\
\end{pmatrix}=\frac{1}{2}\begin{pmatrix}
6\\
4\\
8\\
\end{pmatrix}$$
And returning to the representation of these $u, v, w$ in terms of $x, y, z$ gives the desired result.
|
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|
What is a quick way (without calculator) to determine that $(2^9 + 1)^2 + 2^9 + 2$ is not prime? I came across the following expression $(2^9 + 1)^2 + 2^9 + 2$, which is divisible by 7 and thus not prime. Without this information and a calculator, how could I easily determine that this number is not prime?
Here is a way I thought of how to determine whether 7 is a factor, but I don't really like this approach because if I got that expression above, I'm not going to know to check for 7 right away, and this approach also requires me to know modulo arithmetic, which I'm not too familiar with.
$$
(2^9 + 1)^2 = (513*513) \\
(513*513) \% 7 = (513 \% 7 * 513 \% 7) \% 7 = (2 * 2) \% 7 = 4 \\
(2^9 + 2) \% 7 = 3 \\
(3 + 4) \% 7 = 0
$$
|
$$A=(2^9 + 1)^2 + 2^9 + 2$$
$$\begin{align}\color{red}{A-7}=\left(2^9-1+2\right)^2+2^9+2-7=(a+2)^2+2^9-5=a^2+4a+2^9-1=a^2+4a+a=\color{red}{a^2+5a}\end{align}$$
,where $a=2^9-1=\left(2^3\right)^3-1=\left(2^3-1\right)\left(2^6+2^3+1\right)=7×\left(2^6+2^3+1\right).$
So, $(A-7) \mod 7=0 \Longrightarrow A\mod 7=0.$
The shorter way can be done as follows: (based on the result we get)
$$\begin{align} (2^9 + 1)^2 + 2^9 + 2= 49×\left(2^6+2^3+1\right)^2+35×\left(2^6+2^3+1\right)+7.\end{align}$$
|
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|
Proving $\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$ For $a,b,c\ge 0$ Prove that $$\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$$
My Attempt WLOG $b=\text{mid} \{a,b,c\},$
$$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\dfrac{9}{16}(a-b)^2(b-c)^2(c-a)^2$$
\begin{align*} &=\frac{1}{9}(a+b+c)^2(a-2b+c)^4\\ &+\frac{2}{3}(a+b+c)^2(a-2b+c)^2(b-c)(a-b)\\ &+\frac{1}{16}(a-b)^2(b-c)^2(a+4b+7c)(7a+4b+c)\\&\geqslant 0\end{align*}
However, this solution is too hard for me to find without computer. Could you help me with figuring out a better soltuion?
Thank you very much
|
Without loss of generality suppose that $a\geq b\geq c$. Then,
$$
6\cdot\left(\frac{a^3+b^3+c^3}{3}-abc\right)=2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)
$$
Note that
$$
a+b+c\geq (a-c)+(b-c)=2(a-b)+(b-c)
$$
and
$$
(a-b)^2+(b-c)^2+(c-a)^2=2(a-b)^2+2(b-c)^2+2(a-b)(b-c).
$$
Denote $x=a-b$ and $y=b-c$, due to our assumption $x$ and $y$ are nonegative. Then, we need to prove that ($\sqrt{(a-b)^2(b-c)^2(c-a)^2}=xy(x+y)$)
$$
(2x+y)(2x^2+2y^2+2xy)\geq 6\cdot\frac{3}{4}xy(x+y),
$$
or
$$
4(2x+y)(x^2+xy+y^2)\geq 9xy(x+y)
$$
or
$$
4(2x^3+3x^2y+3xy^2+y^3)\geq 9xy(x+y)
$$
or
$$
8x^3+3x^2y+3xy^2+4y^3\geq 0,
$$
which is obvious.
|
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|
Finding $a$ such that the complex solutions of $z^4 - 6z^3 + 11az^2 - 3(2a^2 + 3a - 3) z + 1 = 0$ form a parallelogram in the complex plane Find all values of the real number $a$ so that the four complex roots of
$$z^4 - 6z^3 + 11az^2 - 3(2a^2 + 3a - 3) z + 1 = 0$$ form the vertices of a parallelogram in the complex plane.
I set $z = x + yi$ due to the given knowledge that $z$ is complex. I then substituted it in, which gave me $$(x+yi)^4 - 6(x+yi)^3 + 11a(x+yi)^2 - 3(2a^2 + 3a - 3)(x+yi) + 1 = 0.$$ However, I'm afraid it's a bit too bashy for it to be viable, so are there any other ways I can get started on this question?
|
Let $x$ be the parallelogram center. Then the roots are $x+y,\,x-y,\,x+z,\,x-z$ for some $y,z$. Then, by Vieta's formulas:
$$\begin{cases}
6=4x\\
6x^2-y^2-z^2=11a\\
4 x^3 - 2 x y^2 - 2 x z^2 = 3(2a^2+3a−3)\\
(x^2-y^2)(x^2-z^2)=1
\end{cases}$$
Although it is doable by hand, too lazy to perform it here, feed it to WA to get
$$a=1,\quad a=3.$$
|
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|
How to solve $\int {\cos ^4 x} \ dx$ using $\cos 2x$? How would I go about solving $$\int \cos{^4} x \ dx $$ given that $$\cos 2x = \cos^2 x − \sin^2 x = 1 − 2 \sin ^2 x = 2 \cos^2 x − 1?$$ I know that I could break it down to $$(\cos{^2}x){^2}$$ but how do I proceed from here?
|
Another way:
$$ \cos^4 x = \left( \frac{e^{ix}+e^{-ix}}{2} \right)^4=\frac{1}{16}
\left( e^{4ix} + 4 e^{2ix} + 6 + 4 e^{-2ix} + e^{-4ix} \right)$$
$$\begin{aligned}
\int \cos^4 x \, dx & = \frac{1}{16} \left(-\frac{i}{4} e^{4ix} -2i e^{2ix} + 6 x+ 2i e^{-2ix} +\frac{i}{4} e^{-4ix} \right)+\text{const.}\\
&= \frac{3}{8}x + \frac{1}{4} \sin 2 x +\frac{1}{32} \sin 4x + \text{const.}
\end{aligned}$$
|
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|
Find all pairs of integers $(x, y)$ such that $x^3+y^3=(x+y)^2.$
Find all pairs of integers $(x, y)$ such that $$x^3+y^3=(x+y)^2.$$
Since $x^3+y^3 = (x+y)(x^2-xy+y^2)$ we get that $$x^2-xy+y^2=x+y$$
this can be expressed as $$x^2-(y-1)x+y^2-y=0.$$
Since we want integers we should probably look at when the discriminant is positive?
$$\Delta = (y-1)^2-4(y^2-y)=-3y^2+6y+1$$
so for $\Delta \geqslant 0$
$$-\frac{2\sqrt3}{3}+1 \leqslant y \leqslant \frac{2\sqrt3}{3}+1$$
only possible solutions are $y=0,1,2.$ However I don't see how this is helpful at all here. What should I do?
|
You are almost there. Substitute $y = 0, 1, 2$ and solve for $x$ in each case.
When $y=0$, the equation is $x^3 = x^2$. The two solutions for $x$ are $0, 1$.
When $y = 1$, the equation is $x^3+1 = (x+1)^2$. Expanding and rearranging gets $x^3-x^2-2x=0$, and the solutions are $x = -1, 0, 2$.
When $y = 2$, the equation is $x^3+8 = (x+2)^2$. Expanding and rearranging gets $x^3-x^2-4x+4 = 0$, and the solutions are $-2, 1, 2$. (You could use RRT to get the solutions.)
So far, we have eight pairs, namely $$(0, 0), (1, 0), (-1, 1), (0, 1), (2, 1), (-2, 2), (1, 2), (2, 2).$$
However, also note that when $x = -y$, the equation is satisfied, since $$(-y)^3+y^3 = ((-y)+y)^2 \rightarrow 0 = 0$$
Therefore, all possible solutions are $$(0, 1), (1, 0), (1, 2), (2, 1), (2, 2), \text{ and } (x, -x).$$
|
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|
Maximum of the function $f_n(x) = (1 - (1 - x^2)^n) / x$
Consider the following function $f_n$ defined on the interval $(0,1]$:
$$f_n(x) = \frac{1 - (1 - x^2)^n}{x}$$
Determine the maximum of this function for any natural number $n$.
I have computed the derivative of this function: $f_n(x)' = \frac{2n x^2(1 - x^2)^{n - 1} + (1 - x^2)^n - 1 }{x^2}$, and from $f_n(x)' = 0$ and by denoting $t := 1 - x^2 \in [0, 1]$, I obtained: $(1 - 2n)t^n + 2nt^{n - 1} -1= 0$.
Now to determine the roots of this polynomial in $t$ I denoted the function $g(t) = (1 - 2n)t^n + 2nt^{n - 1} -1$, and computing the derivative of $g$, we get that $g$ is increasing when $t \in [0, 1 - \frac{1}{2n - 1}]$ and is decreasing when $t \in [1 - \frac{1}{2n - 1}, 1]$. Then $g(0) = -1$, $g(1) = 0$, and $g(1 - \frac{1}{2n - 1}) > 0$. Therefore from this we deduce that the unique maximum $t$ belongs to $[0, 1 - \frac{1}{2n-1}]$, or in other words the unique $x$ that maximizes $f$ lies in $[\frac{1}{\sqrt{2n - 1}}, 1]$.
But I couldnt go beyond this point and if anyone has any ideas I would really appreciate it.
|
I will add on to dezdichado's answer to find the exact constant on the rate of growth of the maximum. Let $x = \frac{c}{\sqrt{n}} + O\left(n^{-3/2}\right)$. Since it must be true that $$\lim_{n \to \infty}(1-2n)\left( 1-x^2 \right)^{n} + 2n\left(1-x^2 \right)^{n-1}-1=0$$
$x = \frac{c}{\sqrt{n}}$ can be plugged into this to get $$\lim_{n \to \infty}(1-2n)\left( 1-\left(\frac{c}{\sqrt{n}}\right)^2 \right)^{n} + 2n\left(1-\left(\frac{c}{\sqrt{n}}\right)^2 \right)^{n-1}-1=0$$
Equivalently, this is $$\lim_{n \to \infty}\left( 2c^2\left( 1-\frac{c^2}{n} \right)^{n-1}+\left( 1-\frac{c^2}{n} \right)^{n}-1 \right) = 0$$
Since $\lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = e^x$, this simplifies to $$(2c^2 +1)e^{-c^2}-1=0$$
$c$ is then the positive solution of $\left(2c^2+1\right)e^{-c^2}=1$. The maximal value is then given by $$\frac{2c}{2c^2+1}\sqrt{n} + O\left(n^{-1/2}\right) \approx 0.638 \sqrt{n} + O\left( n^{-1/2} \right)$$
Here are some numerical results concerning the accuracy of this approximation.
$$\left(
\begin{array}{cccc}
n & \text{actual maximum} & \text{approximation} & \text{relative error} \\
10 & 2.084592 & 2.018079 & 3.1907 \cdot 10^{-2} \\
100 & 6.401867 & 6.381727 & 3.1459 \cdot 10^{-3} \\
1000 & 20.18713 & 20.18079 & 3.1416 \cdot 10^{-4} \\
10000 & 63.81927 & 63.81727 & 3.1411 \cdot 10^{-5} \\
100000 & 201.8086 & 201.8079 & 3.1411 \cdot 10^{-6} \\
\end{array}
\right)$$
|
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|
Finding the smallest power of $A$ such that $A^n = I$ Let $A=\begin{bmatrix}0 & 1\\-1 & 1\end{bmatrix}$ then the smallest positive positive integer $n$ such that $A^n = I$ is :
(a) $1$ (b) $2$ (c) $4$ (d) $6$
proof: option (d) 6.
The characteristic polynomial of $A$ is $\lambda^2 - \lambda + 1$. So the eigenvalues of $A$ are $\omega, \omega^2$ where $\omega$ is a cube root of $-1$ and not equal to $-1$. So the eigenvalues of $A^n$ are $\omega^n$ and $\omega^{2n}$ and the eigenvalue of $I$ is $1$, so $\omega^{2n} = 1$ and $\omega^n = 1$ which implies $n$ as a multiple of $3$ and $2$ which means $6$ is the only option we have.
Is my reasoning correct?? And can we solve this without using eigenvalues?
|
Since $\lambda^2-\lambda +1$ is characteristic, it follows that $A^2 - A + I = 0$, i.e. \begin{align}
A^2 &= A - I\\
A^3 &= A^2 - A = -I\\
A^4 &= -A \\
A^5 &= A^2\cdot A^3 = I - A\\
A^6 &= A^3\cdot A^3 = I.
\end{align}
|
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|
Integrate $ \int \frac{1}{\sin^{4}x+\cos^{4}x}dx $ Show that$$ \int \frac{1}{\sin^{4}(x)+\cos^{4}(x)}dx \ = \frac{1}{\sqrt{2}}\arctan\left(\frac{\tan2x}{\sqrt{2}}\right)+C$$
I have tried using Weierstrass substitution but I can't seem to get to the answer... Should I be using the said method or is there another way I can approach the question? Since the integrand evaluates into an arctangent function I am assuming there is some trickery in the manipulation that can get me there. But I just can't seem to see it...
|
$$\sin ^4(x)+\cos ^4(x)=\left(\sin ^2(x)+\cos ^2(x)\right)^2-2\sin^2x\cos^2x=1-\frac{1}{2}\sin^2(2x)=\\=1-\frac{1}{2}\cdot\frac{1-\cos(4x)}{2}=\frac{1}{4}\left(3+\cos(4x)\right)$$
$$\cos(4x)=\frac{1-\tan^2(2x)}{1+\tan^2(2x)}=\frac{1-t^2}{1+t^2}$$
where $\tan(2x)=t$, $x=\frac{1}{2}\arctan t$, $dx=\frac{dt}{2(1+t^2)}$
$$\int \frac{1}{\sin ^4(x)+\cos ^4(x)} \, dx=\int \frac{4}{3+\cos(4x)} \, dx=$$
$$=4\int \frac{1}{3+4\frac{1-t^2}{1+t^2}}\cdot \frac{dt}{2(1+t^2)}=\int \frac{dt}{2+t^2}$$
$t=u\sqrt{2}$, $dt=du\sqrt 2$
$$\int \frac{dt}{2+t^2}=\int \frac{du\sqrt 2}{2+2u^2}=\frac{\sqrt 2}{2}\int\frac{du}{1+u^2} = \frac{\sqrt 2}{2}\arctan u+C=\frac{\sqrt 2}{2}\arctan \frac{t}{\sqrt 2}+C=$$
$$=\frac{\sqrt 2}{2}\arctan \frac{\tan(2x)}{\sqrt 2}+C$$
|
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|
If $\tan(\theta$) = $\frac{x}{7}$ for -$\frac{\pi}{2} < \theta < \frac{\pi}{2}$, find an expression for $\sin(2\theta$) in terms of $x$. I'm trying to solve this and am getting a very different answer from the book. Can someone please help.
I know that $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. So I have to take $\tan(\theta) = \frac{x}{7}$ and find values for sin and cos in terms of $x$. So here goes:
$\frac{\sin(\theta)}{\cos(\theta)} = \frac{x}{7}$
Then $\cos(\theta) = \frac{7\sin(\theta)}{x}$ and $\sin(\theta) = \frac{x\cos(\theta)}{7}$
Now to find cos and sin in terms of $x$ I plug in values into $\sin^2(\theta) + \cos^2(\theta) = 1$
$\left(\frac{x\cos(\theta)}{7}\right)^2 + \cos^2(\theta) = 1$ then $\cos^2(\theta) = \frac{49}{x^2+49}$
$\sin^2(\theta) + \left(\frac{7\sin(\theta)}{x}\right)^2 = 1$ then $\sin^2(\theta) = \frac{x^2}{x^2+49}$
Now plug these values into $2\sin(\theta)\cos(\theta)$ and I get $2\left(\frac{x}{\sqrt{x^2+49}}\right)\left(\frac{7}{\sqrt{x^2+49}}\right)$
This is very different than the answer in the book and the graphs are different too. Can someone please tell me what I'm doing wrong? The book answer is $1-\frac{x^2}{x}$
|
Your solution is not full.
For example, you can not say immediately that $\sin\theta=\frac{x}{\sqrt{49+x^2}}$ because $|\sin\theta|=\frac{|x|}{\sqrt{49+x^2}}.$
We can make the following.
$$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{\frac{2x}{7}}{1+\frac{x^2}{49}}=\frac{14x}{49+x^2}.$$
|
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|
Does the inequality $ \sqrt{a+b}\geq \sqrt{a/2}+\sqrt{b/2}$ have a name? This is somewhat embarrassing but if $a,b$ are nonnegative real numbers the following seems to hold
$$
\sqrt{a+b}\geq \sqrt{\frac{a}{2}}+\sqrt{\frac{b}{2}}.
$$
Does this inequality have a name?
|
From AM-GM
$$\frac{a+b}{2}\geq \sqrt{ab} \iff\\
a+b\geq \frac{a+b}{2}+2\sqrt{\frac{a}{2}\cdot\frac{b}{2}} \iff\\
a+b\geq \left(\sqrt{\frac{a}{2}}+\sqrt{\frac{b}{2}}\right)^2 \overset{a,b\geq0}{\iff}\\
\sqrt{a+b}\geq \sqrt{\frac{a}{2}}+\sqrt{\frac{b}{2}}$$
Which makes it a special case or corollary of it.
|
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|
Maximum value of $abc$ for $a+b+c=5$ and $a^2+b^2+c^2=11$
$a,b,c$ are three real numbers such that $a+b+c=5$ and $a^2+b^2+c^2=11$,
what's the maximum value of $abc$?
I thought of a way, $ab+bc+ca$ is not hard to find, $a,b,c$ satisfy the cubic equation $x^3 - 5 x^2 + 7 x - abc = 0$ , then use the discriminant of the cubic equation non-negative.
The discriminant of $x^3 + A x^2 + B x + C=0$ is
$A^2 B^2 - 4 B^3 - 4 A^3 C + 18 A B C - 27 C^2$
Is there an easier way?
|
As you did not exactly clarify what you mean by abc (their product?) I can only give you halve an answer. I assume you do not want a integer solution rather than a continous one and also real solutions.
With some calculations you will find:
$$b_{1/2}=\frac{1}{2}(+/- \sqrt{-3a^2+10a-3}-a+5) $$
and
$$c_{1/2}=\frac{1}{2}(+/- \sqrt{-3a^2+10a-3}-a+5) $$
This means you can define $f_{1/1}(x\in \mathbb{R}):=a*b_1*c_1$, $f_{1/2}(x\in \mathbb{R}):=a*b_1*c_2$ and so on (all combinations) and from them just calculate the maximum by using standard analysis methods.
|
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|
Continuity of piecewise $f$ and its partials, $f(x,y)= \frac{xy(x^2-y^2)}{x^2+y^2}$ for $x,y\neq 0$
Let $f: \mathbb{R}^2 \to \mathbb{R}$ and $$f(x,y) = \left\{
\begin{array}{ll}
0, & (x,y)=0 \\
\frac{xy(x^2-y^2)}{x^2+y^2}, & (x,y) \neq0 \\
\end{array}
\right.$$ Determine if $f, \partial_xf$ and $\partial_yf$ are continuous.
So in $\mathbb{R^2}_{\ne 0}$ we have that $f$ is a rational function hence continuous. For the case $(x,y) = 0$ I managed to get the following:
Let $\varepsilon>0$ $$|f(x,y)-f(0)|=\left|\frac{xy(x^2-y^2)}{x^2+y^2}\right| \leqslant \frac{|xy||(x^2+y^2)|}{x^2+y^2} = |xy| = (xy)^2 < \varepsilon$$ when $xy < \sqrt{\varepsilon.}$
However for $ \partial_xf = \frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2}$ I cannot seem to be able to get anything. Also I'm not sure does the fact this is also a rational function imply continuity at $\mathbb{R^2}_{\ne 0}$?
|
Hint:
Transform the partial derivatives via polar coordinates. Also there's an error in your partial derivative w.r.t. $x$:
$$\frac{\partial f}{\partial x}=\frac{y(x^4+4x^2y^2-y^{\color{red}4})}{(x^2+y^2)^2}=\frac{r^5(\cos^4\theta+4\cos^2\theta\sin^2\theta-\sin^4\theta)}{r^4}=r(\cos2\theta+\sin^22\theta),$$
therefor it clearly tends to $0$, and similarly for the other partial derivative.
|
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|
Probability of HTHT in n coin flips What is the probability of finding the sequence HTHT in $n$ flips of a coin? By brute force I get
*
*$n=4: \frac{1}{2^4}$
*$n=5: \frac{2(2)}{2^5}$
*$n=6: \frac{2^2(3)-1}{2^6}$
*$n=7: \frac{2^3(4)-2}{2^7}$
*$n=8: \frac{2^4(5)-6}{2^8}$
I hoped to extract a pattern but counting gets messy.
EDIT: My original counts for $n=7$ and $n=8$ are, as noted in the comments, incorrect. Recounting I get
\begin{align}
\frac{2^3(4)-4}{2^7}\ \text{for}\ n=7\ \text{and}\ \frac{2^4(5)-12}{2^8}\ \text{for}\ n=8
\end{align}
which agrees with the comment below.
|
Let $A_n$ be the number of sequences that don't include the pattern $HTHT$. Then
$$A_n = \pmatrix{1 & 1 & 1 &1} T^n {\pmatrix{1 & 0 & 0 &0}}^{t} \tag1$$
where
$$T= \pmatrix{
1 & 0 & 1 & 0 \\
1 & 1 & 0 & 1 \\
0 & 1 & 0 &0 \\
0 & 0 & 1 &0
} \tag2$$
Eq $(1)$ is obtained by counting recursively the "allowed" sequences that have $(0,1,2,3)$ trailing matching characters.
Some values for $A_n$, from $n=0$ to $n=10$: $(1, 2,4,8,15,28,53,100,188,354,667)$
This is OEIS sequence A118870.
Then the desired probability is
$$ p_n = 1 - \frac{A_n}{2^{n}}$$
I don't think there is a simple formula to express $(1)$ for arbitrary $n$
I'd expect that $p_n \le 1 - (15/16)^{n-3}$
|
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|
Complete $\int \frac{x^2}{\sqrt{9x^2-1}}dx$ I am trying to solve the integral $$\int \frac{x^2}{\sqrt{9x^2-1}}dx,$$ but I am not sure how to solve it.
I have thought substitute $x$ by $\frac{\sec(u)}{3}$. Then $dx = \frac{1}{3} \tan(u)\sec(u)$,
$\sqrt{9x^2 -1} = \sqrt{\sec^2(u) - 1} = \tan(u)$ and $u = \sec^{-1}$(3x)
$$\int \frac{x^2}{\sqrt{9x^2-1}}dx = \int \frac{\sec^3(u)}{9}du$$
I am just not sure how to finish that? Can someone help me with that?
|
There is a specific formula that helps in your case that helps "remove" the powers on a secant integral.
$$\int \sec ^n\left(x\right)dx=\frac{\sec ^{n-1}\left(x\right)\sin \left(x\right)}{n-1}+\frac{n-2}{n-1}\int \sec ^{n-2}\left(x\right)dx$$
Also I think you missed a 3 somewhere, during the substitution the dx should give you 1 three but the x squared should give you an additional 2.
Using this formula at $n=3$ we have:
$$\frac{1}{27}\int \sec ^3\left(u\right)du=\frac{1}{27}\left(\frac{\sec ^{2}\left(u\right)\sin \left(u\right)}{2}+\frac{1}{2}\int \sec\left(u\right)du\right)$$
$$=\frac{1}{27}\left(\frac{\sec ^{2}\left(u\right)\sin \left(u\right)}{2}+\frac{1}{2}\ln \left|\tan \left(u\right)+\sec \left(u\right)\right|\right)$$
$$=\frac{1}{27}\left(\frac{1}{2}\sec \left(u\right)\tan \left(u\right)+\frac{1}{2}\ln \left|\tan \left(u\right)+\sec \left(u\right)\right|\right)$$
Finally, $u=sec^{−1}(3x)$ and the $+C$:
$$=\frac{1}{27}\left(\frac{3}{2}x\sqrt{9x^2-1}+\frac{1}{2}\ln \left|\sqrt{9x^2-1}+3x\right|\right) + C$$
|
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|
Diophantine equation:$x^5+x^4+1=p^y$
find all triplets $(x,y,p)$ satisfying $x^5+x^4+1=p^y$ where x, y are positive integers and p is a prime.
My attempt:I didn't know how to start. So I tried finding some triplets. Interestingly, $(1,1,3)$ satisfies the given equation, but I am not able to find any more.
Next, I tried factorising the equation i.e. $x^5+x^4+1=(x^2+x+1)(x^3-x+1)=p^y$
Now I am stuck. I tried considering the gcd of the common factors but it does not help.
Any ideas??
|
Continuing your approach note that from the equality
$$
(x^3-x+1)(x^2+x+1)=p^y
$$
we obtain that $x^3-x+1=p^n$ and $x^2+x+1=p^m$ for some nonnegative integers $m$ and $n$.
For $x=1$ and $x=2$ we have solutions $(x,y,p)\in\{(1,1,3),(2,2,7)\}$.
Now note that for $x\geq 3$ we have
$$
x^3-x+1=(x^3-1)-(x-2)=(x-1)(x^2+x+1)-(x-2)>x^2+x+1,
$$
so $p^n>p^m$ or $n>m$. Hence, $p^m\mid p^n$, so $x^2+x+1\mid x^3-x+1$. Since
$$
x^3-x+1=(x-1)(x^2+x+1)-(x-2)
$$
we have $x^2+x+1\mid x-2$. However, for $x\geq 3$ we have $0<x-2<x^2+x+1$, so there are no solutions in case $x\geq 3$.
Therefore, all solutions to this equation are $(x,y,p)\in\{(1,1,3),(2,2,7)\}$.
|
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|
How to decide whether to use Bayes' theorem or not?
A bag contains 12 red balls and 6 white balls. Six balls are drawn one by one without replacement of which at least 4 balls are white. Find the probability that in next two drawn , exactly one white ball is drawn. ( Leave answer in $ \binom{n}{r}$)
What I did:
$$ P = \frac{\binom{6}{4} \binom{12}{2}}{\binom{18}{6}} \frac{\binom{2}{1} \binom{10}{1}}{\binom{12}{2}}+ \frac{\binom{6}{5} \binom{12}{1}}{\binom{18} {6}} \frac{\binom{1}{1} \binom{11}{1}}{\binom{12}{2}} + \frac{\binom{6}{6} \binom{12}{0}}{\binom{18}{6}} \cdot 0$$
The idea behind this: so the first factor in each term of sum, is the probability for a number of white ball- red ball combination to be drawn and the factor multiplied with it is the probability for drawing one white ball from the remaining pool.
However the solution key says,
$$ P= \frac{\binom{12}{2}\binom{6}{4} \binom{10}{1} \binom{2}{1} + \binom{12}{1} \binom{6}{5} \binom{11}{1}\binom{1}{1} }{\binom{12}{2} ( \binom{12}{2} \binom{6}{4} + \binom{12}{1} \binom{6}{5} + \binom{12}{0} \binom{6}{6})}$$
This seems to be some sort of Bayes' theorem application but I can't really see why we would need Bayseian thinking here.
|
First a formula: Suppose we have $R$ red balls & $W$ white balls, then the probability of selecting a specific sequence of $r $ red balls & $w $ white balls (with $r \le R, w \le W$) is
$ { (R+W - (r+w))! \over (R+W)!} { R! \over (R-r)! } {W! \over (W-w)! }$. Since these can
be rearranged in $\binom{w+r}{w}$ ways, we see that the probability of selecting $r $ red balls & $w $ white balls (in any order) is
$p((R,W),(r,w)) = \binom{w+r}{w} { (R+W - (r+w))! \over (R+W)!} { R! \over (R-r)! } {W! \over (W-w)! } = {1 \over \binom{W+R}{w+r} }\binom{W}{w} \binom{R}{r}$.
Let $B$ be the event that the $7$th and $8$th draw have exactly one white ball.
Let $A_k$ be the event that exactly $k$ white balls are drawn in the first $6$ draws.
Let $A= A_4 \cup A_5 \cup A_6$ (disjoint union).
We want to compute $P[B|A] = {P[B \cap A] \over P[A] } = {P[B \cap A_4]+P[B \cap A_5]+p[B \cap A_6] \over P[A_4]+P[A_5]+P[A_6]} = {P[B|A_4]P[A_4]+P[B|A_5]P[A_5]+p[B|A_6]P[A_6] \over P[A_4]+P[A_5]+P[A_6]} $.
Note that $P[B|A_k] = p((10+(k-4),2-(k-4)),(1,1))$.
Substituting we get
$P[B|A]= { p((10,2),(1,1)) p((12,6),(2,4))+p((11,1),(1,1)) p((12,6),(1,5))+p((12,0),(1,1)) p((12,6),(0,6)) \over p((12,6),(2,4))+p((12,6),(1,5))+p((12,6),(0,6))} $.
Substituting for $p$ gives the solution key answer above.
|
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|
How to find $\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$? I tried to solve it, but the answer I got was different from the answer given.
Answer given:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)} = \frac{n(2n+1)}{4(2n-1)(2n+3)}$$
My working:
$$\sum_{r=1}^{n}\frac{r}{(2r-1)(2r+1)(2r+3)}$$
$$= \sum_{r=1}^{n} \Biggl[\frac{1}{16(2r-1)}+\frac{1}{8(2r+1)}-\frac{3}{16(2r+3)}\Biggl]$$
$$= \frac{1}{16} + \frac{1}{24} - \require{cancel} \cancel{\frac{3}{80}}$$
$$+ \frac{1}{48} + \require{cancel} \cancel{\frac{1}{40}} - \require{cancel} \cancel{\frac{3}{112}}$$
$$+ \require{cancel} \cancel{\frac{1}{80}} + \require{cancel} \cancel{\frac{1}{56}} - \require{cancel} \cancel{\frac{3}{144}}$$
. . .
. . .
. . .
$$+ \require{cancel} \cancel{\frac{1}{16(2n-3)}} + \require{cancel} \cancel{\frac{1}{8(2n-1)}} - \frac{3}{16(2n+1)}$$
$$+ \require{cancel} \cancel{\frac{1}{16(2n-1)}} + \frac{1}{8(2n+1)} - \frac{3}{16(2n+3)}$$
$$= \frac{1}{16} + \frac{1}{24} + \frac{1}{48} - \frac{3}{16(2n+1)} + \frac{1}{8(2n+1)} - \frac{3}{16(2n+3)}$$
$$= \frac{2n^{2}+6n+3}{4(2n+1)(2n+3)}$$
|
Using partial fraction decmposition
$$\frac{r}{(2r-1)(2r+1)(2r+3)}=\frac{1}{8 (2 r+1)}-\frac{3}{16 (2 r+3)}+\frac{1}{16 (2 r-1)}$$ that you can rewrite
$$\frac 18 \left(\frac{1}{2 r+1}-\frac{1}{2 r+3} \right)+\frac 1{16} \left(\frac{1}{2 r-1}-\frac{1}{2 r+3} \right)$$ There would be a lot of telescoping
|
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|
Solve this integral $\int \frac{dx} {\sin^5{x}}$ Let us denote
$$I_{m,n} = \int \sin^m{x} \cos^n{x}\ dx$$
where $m,n$ are integers (possibly negative or zeros).
There are some well-known recurrent formulas for $I_{m,n}$
So... as an example I was trying to solve this particular integral
$$\int \frac{dx} {\sin^5{x}}$$
using these recurrent formulas and I finally got this answer
(the computations were not very long, just 1 or 2 A4 sheets):
$$F(x) = \frac{5}{8} \ln {|\tan \frac{x}{2}|} - \frac{5}{24} \cdot \frac{\cos{x}}{\sin^2{x}} - \frac{1}{6} \cdot \frac{\cos{x}}{\sin^4{x}}$$
But WA is not giving me a simple expression when I differentiate $F(x)$
Is my answer incorrect? How do I check it with some tool other than WA?
Any ideas how to verify?
EDIT: Now I fixed the issues in my calculations and I am getting this answer.
$$\frac{3}{8}\ln|\tan(\frac{x}{2})|-\frac{3}{8}\frac{\cos(x)}{\sin^{2}(x)}-\frac{1}{4}\frac{\cos(x)}{\sin^{4}(x)}$$
But it still does not match with the WA answer...
WA answer
Which one is correct here?
|
You made a mistake somewhere. From WA and after simplifications you should have (what I get):
$$\frac{3}{8}\ln|\tan(\frac{x}{2})|-\frac{3}{8}\frac{\cos(x)}{\sin^{2}(x)}-\frac{1}{4}\frac{\cos(x)}{\sin^{4}(x)}$$
Let me know!
I think the $24\ln|\tan(\frac{x}{2})|$ is clear and we multiplied it by $\frac{1}{64}$ to give the correct term.
Then re-write the remaining terms (from WA) as:
$$\frac{1}{64}\big[-\frac{1}{\sin^{4}(\frac{x}{2})}-\frac{6}{\sin^{2}({\frac{x}{2})}}+\frac{1}{\cos^{4}(\frac{x}{2})}+\frac{6}{\cos^{2}({\frac{x}{2})}}\big]$$
$$=\frac{6}{64}\big[\frac{\sin^{2}(\frac{x}{2})-\cos^{2}(\frac{x}{2})}{\sin^2({\frac{x}{2})\cos^{2}(\frac{x}{2})}}\big]+\frac{1}{64}\big[\frac{\sin^{4}(\frac{x}{2})-\cos^{4}(\frac{x}{2})}{\sin^4({\frac{x}{2})\cos^{4}(\frac{x}{2})}}\big].$$
Then using $\cos^2(x)-\sin^2(x)=\cos(2x)$ and $\sin(2x)=2\sin(x)\cos(x)$ the first term becomes:
$$\frac{6}{64}\big[\frac{\sin^{2}(\frac{x}{2})-\cos^{2}(\frac{x}{2})}{\sin^2({\frac{x}{2})\cos^{2}(\frac{x}{2})}}\big]=-\frac{3}{32}\big[\frac{\cos(x)}{\frac{\sin^{2}(x)}{4}}\big]=-\frac{3}{8}\frac{\cos(x)}{\sin^{2}(x)}.$$
For the other term we have (numerator is of the form $(A^2-B^2)=(A-B)(A+B)$):
$$-\frac{1}{64}\big[\frac{\cos^{2}(\frac{x}{2})-\sin^2({\frac{x}{2})}}{\frac{\sin^{4}(x)}{2^4}}\big]=-\frac{1}{4}\frac{\cos(x)}{\sin^{4}(x)}.$$
|
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|
Solve this integral $\int \frac{dx}{\sin^6x \cdot \cos^6x}$ I was solving this integral manually
$$\int \frac{dx}{\sin^6x \cdot \cos^6x}$$
It's not easy but it wasn't that hard too having all good formulas/recurrences in front of me and not rushing my calculations.
So I got this answer by hand:
$$(1/5) \cdot \sin(x)^{-5} \cdot \cos(x)^{-5} + \\ (2/3) \cdot \sin(x)^{-5} \cdot \cos(x)^{-3} + \\ (16/3) \cdot \sin(x)^{-5} \cdot \cos(x)^{-1} \\ - (32/5) \cdot \sin(x)^{-5} \cdot \cos(x) \\ - (32 \cdot 4/15) \cdot \sin(x)^{-3} \cdot \cos(x) \\ - (32 \cdot 8/15) \cdot \sin(x)^{-1} \cdot \cos(x) $$
Now... my problem is Wolfram Alpha does not even want to compute the derivative of this. Is it too long for the free WA version? I then tried using SymPy Live but that does not help either. It is not able to simplify the derivative to $\sin^6x \cdot \cos^6x$, it seems.
So I finally found this free derivative calculator (not sure what it's based on),
and it seems to be saying my answer is OK.
www.derivative-calculator.net
Is my answer OK indeed?
What is the best free online tool I can use for this task?
|
Since your result has been confirmed.
It would have been very fast to use $x=\tan^{-1}(t)$ which makes
$$I=\int \frac{dx}{\sin^6(x) \, \cos^6(x)}=\int\frac{\left(t^2+1\right)^5}{t^6}\,dt $$ that is to say
$$I=\frac{t^5}{5}-\frac{1}{5 t^5}+\frac{5 t^3}{3}-\frac{5}{3 t^3}+10 t-\frac{10}{t}+C$$ Back to $x$
$$I=\frac{\tan ^5(x)}{5}+\frac{5 \tan ^3(x)}{3}+10 \tan (x)-\frac{1}{5} \cot
^5(x)-\frac{5 \cot ^3(x)}{3}-10 \cot (x)+C$$ Simplifying
$$I=-\frac{1}{30} (10 \cos (2 x)-5 \cos (6 x)+\cos (10 x)) \csc ^5(x) \sec ^5(x)+C$$
|
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|
Help Finding Closed Form for Function of Decreasing Half Steps I would like to know if there is a closed form for this function:
$
f(x) =
\begin{cases}
\frac{1}{2}, & \text{if}\ 0 \leq x < \frac{1}{2} \\
\frac{1}{4}, & \text{if}\ \frac{1}{2} \leq x < \frac{1}{2} + \frac{1}{4} \\
\frac{1}{8}, & \text{if}\ \frac{1}{2} + \frac{1}{4} \leq x < \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \\
\frac{1}{16}, & \text{if}\ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \leq x < \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} \\
...
\end{cases}
$
Image of function
|
Fig. 1: Graphical representation of function $f$ (formula (0)). The red vertical segments are a plotting artefact.
Here is a solution:
$$y=f(x)=2^{\lfloor\log_2(1-x)\rfloor}\tag{0}$$
where $\lfloor \cdots \rfloor$ is the floor function and $\log_2$ is the logarithm function with base $2$.
Remark: Consideration of the extreme cases of the "integer part of $a$", which can be $a$ or $a-1$, one gets
*
*
*
*$ \ \ y=2^{\log_2(1-x)}=1-x$ or
*
*$ \ \ y=2^{\log_2(1-x)-1}=\tfrac12 (1-x)$
which are the equations of the lines containing resp. the "upper" and "lower" points of discontinuity of the graphical representation (dotted blue lines in the figures).
Proof of formula (0):
Fig. 2: Graphical representation of function $g$, given by formula (1), related to function $f$ by $f(x)=g(1-x)$.
In fact, it will be equivalent and simpler to work on function $g$ with graphical representation given in Fig. 2 and establish that its equation is
$$y=g(x)=2^{\lfloor\log_2(x)\rfloor}\tag{1}$$
Function $g$ is defined as taking values $$y=\tfrac{1}{2^{n+1}}\tag{2}$$ for
$$\tfrac{1}{2^{n+1}} < x \le \tfrac{1}{2^{n}}\tag{3}$$
Remark: $g$ is in fact a rather well known function giving the highest power of 2 smaller than or equal to a given number $x$.
Let us attempt to express $n$ as a function of $x$.
Applying the increasing function $\log_2$ to this double inequation, we get:
$$-(n+1) < \log_2(x) \le -n$$ or, reversing the sign and the direction of inequalities:
$$n \le -\log_2(x) < n+1$$
Otherwise said: $$n=\lfloor -\log_2(x) \rfloor.$$
Plugging this expression of $n$ into relationship (2) gives:
$$y=2^{-(n+1)}=y=2^{-\lceil -\log_2(x) \rceil}$$
which is equivalent to expression (1) (where $\lceil...\rceil$ is the "ceiling" function).
|
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|
How can I prove that $y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}>0$ when $x>0$ and $1I would like to prove that
$$y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}>0$$
for all real numbers $x > 0$ and $1 < y < 1.5$. This seems true when plotted on WolframAlpha, but I don't know how to prove it. I tried replacing some of the terms using the given inequalities to obtain a simpler function, but any perturbation I make seems to render the inequality untrue. How would you approach this problem?
|
We have that
$$g(y)=\frac{\partial }{\partial y}\left(y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}\right) =1-\frac{2xy^3}{(1+x^2)^2}+\frac{2x^3y}{(1+y^2)^2}$$
with
$$g'(y)=\frac{\partial }{\partial y}\left(y-x+x^{5}-\frac{xy^{4}}{2(1+x^{2})^{2}}-\frac{x^{3}}{1+y^{2}}\right) =-\frac{6xy^2}{(1+x^2)^2}-\frac{2x^3(-1+3y^2)}{(1+y^2)^3}<0$$
therefore the minimum is reached either for $y=1$ or $y=\frac32$, therefore we reduce to check and
$$x^5-\frac1{2}x^3-\frac{x}{2(x^2+1)^2}-x+1>0 \tag 1$$
$$x^5-\frac4{13}x^3-\frac{81}{32}\frac{x}{(x^2+1)^2}-x+\frac32>0 \tag 2$$
|
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|
how to find $\frac{a}{b}$ given that $2a^2 + 2007a + 3 = 0$ and $3b^2 + 2007b + 2 = 0$ Given that
$$\begin{cases}
2a^2 + 2007a + 3 = 0 \\
3b^2 + 2007b + 2 = 0
\end{cases}
$$
and $ab \ne 1$, how to solve for $\frac{a}{b}$?
My try:
\begin{align}
2007a = -3 - 2a^2 \\
2007b = -2 - 3b^2 \\
\frac{a}{b} = \frac{-3 - 2a^2}{-2 - 3b^2}
\end{align}
Also,
\begin{align}
2a^2 = -2007a - 3 \\
3b^2 = -2007b - 2 \\
\frac{a^2}{b^2} = \frac{-2007a - 3}{-2007b - 2}\frac{3}{2} \\
\frac{a}{b} = \pm\sqrt{\frac{-6021a - 9}{-4014b - 4}}
\end{align}
Well, it doesn't look very promising. So I want to seek help here. How to find $\frac{a}{b}$ without actually solving for the roots? Hints are welcomed.
|
$\mathcal{Hint}$
Consider $2x^2+2007x+3=0$
Note that $a$ is one the roots.
Now, replacing $x$ by $\frac{1}{x}$
How does that affect the equation and its roots?
|
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|
Best method for proving that $7\times11^{2n+1}-3^{4n-1}$ is divisible by $10$ I am asked to prove by induction that $7\times11^{2n+1}-3^{4n-1}$ is divisible by $10$.
I wonder whether there is a more direct method, for example factorizing by $10$.
If an expression is divisible by $10$, does this mean that I can factorize it by $10$?
Thanks in advance
|
Best method? Depends on what you know.
If you know modular arithmetic and Eulers Theorem then
$7\cdot 11^{2n+1} - 3^{4n-1}\equiv 7\cdot 1^{2n+1} - (3^{-1})\cdot 3^{4n}\equiv 7-(3^{-1})\pmod{10}$ and as $3\cdot 7 \equiv 1 \pmod {10}$ then $3^{-1}\equiv 7 \pmod {10}$ and $7-7\equiv 0\pmod {10}$ so $10|7\cdot 11^{2n+1} - 3^{4n-1}$.
But that assumes you are comfortable with many concepts
If you don't know any modular arithmetic:
$11^k = (10+1)^k = 10^k + k10^{k-1} + ...... + k\cdot 10 + 1 = 10 M + 1$.
And $3^{4n-1} = 3\times 3^{4n-2}= 3\times 9^{2n-1}$. And if $k=2n-1$ is odd the $9^k = (10-1)^k = 10^k - k10^{k-1} + ....... + k\cdot 10 - 1= 10N-1$.
So $7\cdot 11^{something} - 3^{1+2\times something\ odd} = 7(10M + 1)- 3(10N-1)= 70M -30N +10$.
|
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|
Evaluate $\int_0^2 (1+x^3)^{1/2} + (x^2+2x)^{1/3}dx$
$$\int_0^2 (1+x^3)^{1/2} + (x^2+2x)^{1/3}dx$$
I figured out that the functions are inverses of each other (kinda). If find the inverse of 1st function, we get
$y=(1+x^3)^{1/2}$
$y^2=1+x^3$
$(y^2-1)^{1/3} = x$
But the function given is, $((x+1)^2-1)^{1/3}$
Substituting $x+1$ would also change the limits. So that is not possible. I don't know what to do here now.
Any help would be appreciated.
|
By Integral of Inverse Functions we have:
$$\int_c^d f^{-1}(y)\,dy+\int_a^b f(x)\,dx=bd-ac$$
given that $f(a)=c, f(b)=d$.
Let $f(x)= (1+x^3)^{1/2}$. As you have observed, $f^{-1}(y) = (y^2-1)^{1/3}$.
Also we have $f(0)=1$ and $f(2)=3$. Thus:
$$\int_1^3 (y^2-1)^{1/3}\,dy + \int_0^2(1+x^3)^{1/2}\,dx = 3\times2-1\times0=6$$
Do you see how $\displaystyle\int_1^3 (y^2-1)^{1/3}\,dy = \int_0^2 ((x+1)^2-1)^{1/3}\, dx$?
|
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|
Is $(x^2+3)^3-4(x-1)^3$ an irreducible polynomial in $Q[x]$? I was trying to solve the problem below
Let $K \subset \mathbb{C}$ be a splitting field of $f(x) = x^3 − 2$ over $\mathbb{Q}$. Find a complex number $z$, such that $K = \mathbb{Q}(z)$.
All the roots of the polynomial $f(x)$ are $2^{1/3}$, $2^{1/3}\omega$ and $2^{1/3}\omega^2$, where $\omega = (-1+i\sqrt{3})/2$ is a cube root of unity. Thus $\mathbb{Q}(2^{1/3},\omega)$ or $\mathbb{Q}(2^{1/3},i\sqrt{3})$ is a splitting field of $f(x)$. $[\mathbb{Q}(2^{1/3}):\mathbb{Q}]=3$ as irreducble polynomial of $2^{1/3}$ is $x^3-2$(Eisenstien's criteria for $p=2$) and $[\mathbb{Q}(\omega):\mathbb{Q}]=2$, as irreducible polynomial of $\omega$ is $g(x) = x^2+x+1$( $g(x+1)$ is irreducble by eisenstien's criteria for $p=3$). Therefore, if we have a splitting field $\mathbb{Q}(z)$, then degree of irreducible polynomial of $z$ is $6$ by compositum of fields. Now we need to find $z$.
Me and my friend got the idea of taking $z$ as $2^{1/3} + i\sqrt{3}$.
$$\begin{eqnarray*} x & = & 2^{1/3} + i\sqrt{3} \\ \left( x - 2^{1/3} \right)^2 & = & -3
\\ -2^{2/3}x+2^{2/3} & = & -x^2 -3 \\ 4(x-1)^3 & = & (x^3+3)^3 \end{eqnarray*}$$
Thus $h(x) = (x^2+3)^3-4(x-1)^3$ is a degree six polynomial having root $2^{1/3} + i\sqrt{3}$. How do i show from here that it is irreducible in $Q[x]$? It's expanded form is $x^6+9x^4-4x^3+39x^2-12x+31$.
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Actually, the polynomial is irreducible over $\Bbb F_{19}$ and hence irreducible over $\Bbb Z$ and $\Bbb Q$. It has no root over $\Bbb F_{19}$ and no quadratic or cubic factor.
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"language": "en",
"url": "https://math.stackexchange.com/questions/3833776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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