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Stuck solving $z^2$ + $zw^*$= $18$ and $2z^*$=$w^*(1−i)$ as a system of equations Let $z$ and $w$ be complex numbers that satisfy
$z^2$ + $z\overline{w}$= $18$
and $2\overline{z}$=$\overline{w}(1−i)$
with $\Re(z)>0$. Find $w$.
I tried to find $z$ by subbing $\overline{w}$ = $\frac{18-z^2}z$ into $2\overline{z}$=$\overline{w}(1−i)$ and letting $z\overline{z}=1$.
and got $z^2 = 17 + i$. This was where i got stuck. Using De Moivre's Theorem to get $z$ but the result was irrational.
|
Preliminaries
$z = |z|e^{i\theta}\\
\bar z = |z|e^{-i\theta}\\
w = |w|e^{i\phi}\\
\bar w = |w|e^{-i\phi}\\
(1-i) = \sqrt 2 e^{-\frac {\pi}{4} i}$
Let's use the second equation to express $w$ in terms of $z.$
$2\bar z = \bar w(1-i)\\
2|z|e^{-i\theta} = (|w| e^{-i\phi})\sqrt 2 e^{-\frac {\pi}{4} i}\\
2|z|e^{-i\theta} = (\sqrt 2 |w|) e^{(-\phi}-\frac {\pi}4) i)$
Both the modulus and the argument are equal.
$|w|=\sqrt 2 |z|\\
\phi = \theta - \frac {\pi}{4}$
And substitute into the first equation.
$z^2 + z\bar w = 18\\
|z|^2 e^{2\theta i} + \sqrt 2 |z|^2 e^{\theta i - (\theta -\frac {\pi}{4})i} = 18\\
|z|^2 (e^{2\theta i} + \sqrt 2 e^{\frac {\pi}{4})i}) = 18\\
|z|^2 (\cos 2\theta + i\sin 2\theta + 1+i) = 18$
$|z|^2(\cos 2\theta + 1) = 18\\
|z|^2(\sin 2\theta + 1) = 0\\
\theta = -\frac{\pi}{4}\\
|z|^2 = 18$
$z = 3- 3i\\
w = -6 i$
Alternatively,
$z = x + yi\\
w = a + bi\\
2(x+yi) = (a-bi)(1-i)\\
2x + 2yi = a-b - (a+b) i\\
2x = a-b\\
2y = -(a+b)$
$(x^2 -y^2) + (2xy)i + (ax + by) + (ay-bx) = 18\\
x^2 - y^2 + ax + by = 18\\
2xy + ay - bx = 0$
$-a^2 + b^2 - a^2 - ab + ab - b^2 = 0\\
a^2 = 0\\
b^2 = 36$
|
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|
Is my calculation of the integral $\int \tan^{-1} x \, dx$ correct? Compute $\int \tan^{-1}x \,dx$.
First, set $u = \arctan(x)$ and $dv = dx$. We want to find $du$ and we already have $v = x$.
We start by taking the tangent of both sides, leaving us with
$$\tan(u) = x.$$
Next, using implicit differentiation, we get $\frac{du}{dx}\sec^2(u) = 1$, or $$\frac{du}{dx} = \frac{1}{\sec^2(u)} = \frac{1}{1+ \tan^2(u)} = \frac{1}{1+ \tan^2(\arctan(x))} = \frac{1}{x^2+1}.$$
Therefore, $du = \frac{1}{x^2 +1}\, dx.$
We know that $\int u\, dv = uv - \int v\, du$ by Integration By Parts.
So, $\int \arctan(x)\, dx = x\arctan(x) - \int \frac{x}{x^2+1}\, dx.$
We can simplify $\int \frac{x}{x^2+1}\, dx$ by using u-substitution, making a different $u = x^2+1$ and $\frac{du}{2x} = dx.$
Subbing these in, we get
$$\int \frac{x}{u} \cdot \frac{du}{2x} = \frac{1}{2}\int \frac{1}{u}\, du = \frac{1}{2}|\log u| = \frac{1}{2}|\log(x^2+1)| + C.$$
Plugging this back into $x\arctan(x) - \int \frac{x}{x^2+1}\, dx,$ we get
$\boxed{x\arctan(x) - \frac{1}{2} |\log(x^2+1)| + C}$, where C is a constant.
|
Your explanation of the integration by parts, which you combine with a change of variable, is a little tedious.
Using $x=\tan(u)$ which implies $dx=(\tan^2(u)+1)\,du$,
$$\int\arctan(x)\,dx=\int u(\tan^2(u)+1)\,du=u\tan(u)-\int\tan(u)\,du$$
and back to the original variable with $u=\arctan(x)$,
$$\int\arctan(x)\,dx=x\arctan(x)-\int\frac x{x^2+1}\,dx.$$
It is more direct to assume a factor $1$ and integrate it,
$$\int\arctan(x)\,dx=\int1\arctan(x)\,dx=x\arctan(x)-\int\frac{dx}{x^2+1}.$$
You could as well have continued with the change of variable:
$$\int\tan(u)\,du=-\int\frac{d\cos(u)}{\cos(u)}=-\log(|\cos(u)|)$$
and
$$\cos(u)=\frac1{\sqrt{x^2+1}}.$$
|
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|
If $(a-b^2)b>0$, then $\sqrt[3]{a+\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}+\sqrt[3]{a-\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}$ is rational From Hardy´s "A course of pure mathematics" 10th edition, problem 31 miscellaneous problems of chapter I.
If $(a-b^2)b>0$, then
$$
\sqrt[3]{a+\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}+\sqrt[3]{a-\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}
$$
is rational.
"Note from the book":
Each of numbers under a cube root is of the form
$$
\left( \alpha + \beta\sqrt{\frac{a-b^3}{3b}} \right)^3
$$
where $\alpha$ and $\beta$ are rational.
|
Hint . Put $$a=y^3$$
We get :
$$b\Bigg(\sqrt[3]{x^3+\Big(3+\frac{x^3}{3}\Big)\sqrt{\frac{x^3-1}{3}}}+\sqrt[3]{x^3-\Big(3+\frac{x^3}{3}\Big)\sqrt{\frac{x^3-1}{3}}}\Bigg)\quad (1)$$
Where $x=\frac{y}{b}$
Now if $b$ is rational the expression in front of $b$ must be a rational but I don't see how ...
In fact if $x=1$ the expression $(1)$ becomes :
$$2y$$
Or :
$$2\sqrt[3]{a}$$
If we follow the comment of @Paramanand Singh there is a nice inverse function to one of the summand in the bracket of the expression $(1)$ see here
|
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|
Finding remainder of $123^{456}$ divided by 88 using Chinese Remainder Theorem I tried using Chinese remainder theorem but I kept getting 19 instead of 9.
Here are my steps
$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equiv (-2)^9 \equiv -512 \equiv -6 \equiv 5 \pmod{11}\\
x_2 &= 123^{456} \equiv 123^0 \equiv 1 \pmod{8}\\
y_2 &= 1^{-1} \equiv 1 \pmod{8} \\
123^{456}
&\equiv \sum_{i=1}^2 x_i\times\frac{M}{m_i} \times y_i
\equiv 9\times\frac{88}{11}\times5 + 1\times\frac{88}{8} \times1 \equiv 371
\equiv 19 \pmod{88}
\end{split}
$$
|
By Euler's theorem, we first get $123^{40}\cong1\pmod{88}$, since $\varphi(88)=40$. This results in $35^{16}\pmod{88}$, easily.
Now we use CRT: $\begin{cases}x\cong 35^{16}\pmod8\\x\cong35^{16}\pmod{11}\end{cases}$.
So, $x\cong3^{16}\pmod8\implies x\cong1\pmod8$, and $x\cong2^{16}\pmod{11}\implies x\cong5^4\pmod{11}\implies x\cong9\pmod{11}$, together yielding $x\cong9\pmod{88}$ by CCRT (constant case of the Chinese remainder theorem).
|
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|
In quadrilateral $ABCD$, $\angle BAC=\angle CAD=2\,\angle ACD=40^\circ$ and $\angle ACB=70^\circ$. Find $\angle ADB$.
Let quadrilateral $ABCD$ satisfy $\angle BAC = \angle CAD = 2\,\angle ACD = 40^\circ$ and $\angle ACB = 70^\circ$. Find $\angle ADB$.
What I tried
*
*Ceva’s Theorem (Trigonometry version)
*Try to construct some equilateral triangle.
Which both failed.
Any hints or solutions please. Thanks in advance!
|
Since $\angle ACB=\angle ABC=70^\circ$, the triangle $ABC$ is isosceles and $\;\overline{AB}=\overline{AC}$.
By applying the law of sines to the triangle $ACD$, we get that:
$\overline{AD}=\overline{AC}\cdot\cfrac{\sin\angle ACD}{\sin\angle ADC}=\overline{AC}\cdot\cfrac{\sin 20^\circ}{\sin 120^\circ}=\cfrac{2\overline{AC}\sin 20^\circ}{\sqrt{3}}\;.$
And, by applying the law of sines to the triangle $ABD$, we get that:
$\overline{AD}\sin\angle ADB=\overline{AB}\sin\angle ABD\;.\quad\color{blue}{(*)}$
Let $\;\alpha=\angle ADB\;.$
Since $\;\overline{AD}=\cfrac{2\overline{AC}\sin 20^\circ}{\sqrt{3}}\;$, $\;\overline{AB}=\overline{AC}\;$ and $\;\angle ABD=100^\circ-\alpha\;,\;$ the equality $(*)$ turns into:
$\cfrac{2\overline{AC}\sin 20^\circ\sin\alpha}{\sqrt{3}}=\overline{AC}\sin(100^\circ-\alpha)\;,$
$2\sin 20^\circ\sin\alpha=\sqrt{3}\sin(90^\circ+10^\circ-\alpha)\;,$
$4\sin 10^\circ\cos 10^\circ\sin\alpha=\sqrt{3}\cos(10^\circ-\alpha)\;,$
$4\sin 10^\circ\cos 10^\circ\sin\alpha=\sqrt{3}\left(\cos10^\circ\cos\alpha+\sin 10^\circ\sin\alpha\right)\;,$
$4\sin 10^\circ\sin\alpha=\sqrt{3}\left(\cos\alpha+\tan 10^\circ\sin\alpha\right)\;,$
$\left(4\sin 10^\circ-\sqrt{3}\tan 10^\circ\right)\sin\alpha=\sqrt{3}\cos\alpha\;,$
$\tan\alpha=\cfrac{\sqrt{3}}{4\sin 10^\circ-\sqrt{3}\tan 10^\circ}\;.$
Hence,
$\angle ADB=\alpha=\arctan\left(\cfrac{\sqrt{3}}{4\sin 10^\circ-\sqrt{3}\tan 10^\circ}\right)\simeq\\\simeq 77,3361794^\circ.$
|
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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?
|
From condition we get $ab+bc+ca=7.$ Using the Cauchy-Schwarz inequality, we have
$$11 \geqslant a^2 + \frac{(b+c)^2}{2} = a^2 + \frac{(5-a)^2}{2},$$
so
$$a^2 + \frac{(5-a)^2}{2} \leqslant 11 \Rightarrow \frac 13 \leqslant a \leqslant 3.$$
Similar we get $\frac 13 \leqslant b,\,c \leqslant 3.$ Therefore
$$(a-3)(b-3)(c-3) \leqslant 0,$$
equivalent to
$$abc \leqslant 27-9(a+b+c)+3(ab+bc+ca).$$
So $abc \leqslant 27-9 \cdot 5+3 \cdot 7 = 3.$
Equality occur when $a=b=1,\,c=3$ and any permution.
|
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|
If $a > b > 1$ and $\frac{1}{log_a(b)} + \frac{1}{log_b(a)} = \sqrt{1229}$ , find the value of :- $\frac{1}{log_{ab}(b)} - \frac{1}{log_{ab}(a)}$
If $a > b > 1$ and $\frac{1}{log_a(b)} + \frac{1}{log_b(a)} = \sqrt{1229}$ , find the value of :- $\frac{1}{log_{ab}(b)} - \frac{1}{log_{ab}(a)}$ .
What I Tried :- I tried the problem this way :-
As $log_a(b) = \frac{log_b(b)}{log_b(a)}$ , we have $\frac{1}{log_a(b)} = \frac{log_b(a)}{log_b(b)} = log_b(a).$ So :-
$$log_b(a) + \frac{1}{log_b(a)} = \sqrt{1229}$$
$$\rightarrow \frac{log(a)}{log(b)} + \frac{log(b)}{log(a)} = \sqrt{1229}$$
$$\rightarrow \frac{(log(a))^2+(log(b))^2}{log(a)log(b)} = \sqrt{1229}$$
Now :-
$$\frac{1}{log_{ab}(b)} - \frac{1}{log_{ab}(a)}$$
$$\rightarrow \frac{log(a) + log(b)}{log(b)} - \frac{log(a) + log(b)}{log(a)}$$
$$\rightarrow \frac{(log(a))^2 + log(a)log(b) - log(a)log(b) - (log(b))^2}{log(a)log(b)}$$
$$\rightarrow \frac{(log(a))^2 - (log(b))^2}{log(a)log(b)}$$
$$\rightarrow \sqrt{1229} - \frac{2(log(b))^2}{log(a)log(b)}$$
I could conclude only upto this, other than that I have no idea . Now can anyone help me?
|
Instead of trying to find the values of $\log b$ and $\log a$, just look at the equation $\log_b a + \frac 1{\log_b a} = 1229$.
Now, if $(ab)^x = a$, then $a^xb^x = a$ so $b^x = a^{1-x}$, then $a = b^{\frac x{1-x}}$, therefore $\frac x{1-x} = \log_b a$, therefore $x = \frac{\log_b a}{\log_b a + 1} = \log_{ab} a$.
Similarly, if $(ab)^y = b$, then $a^y = b^{1-y}$ so $a = b^{\frac{1-y}y}$ so $\frac {1-y}y = \log_b a$ so $y = \frac 1{\log_b a + 1} = \log_{ab} b$.
We have to find $\frac 1y - \frac 1x = {\log_b a+1} - \frac{(\log_b a + 1)}{\log_b a} = \log_b a - \frac 1{\log_b a}$.
Let $\log_b a = z$. Then $z + \frac 1z = \sqrt{1229}$, and we have to find $z- \frac 1z$. This is usual from $$
\left(z+\frac 1z\right)^2 - \left(z-\frac 1z\right)^2 = 4 \times z \times \frac 1z = 4
$$
and the fact that $a>b$ so $z=\log_b a >1$, hence $z-\frac 1z< 0$. (So the square root is the positive square root).
|
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|
Evaluating $\sum_{a=1}^6\sum_{b=1}^6\sum_{c=1}^6\frac{ab(3a+c)}{2^a2^b2^c(a+b)(b+c)(c+a)}$ without a calculator? Is there a way to get this value without calculator?
$$\sum_{a=1}^6\sum_{b=1}^6\sum_{c=1}^6\frac{ab(3a+c)}{2^a2^b2^c(a+b)(b+c)(c+a)}$$
I'm currently studying AIME.
|
Let's consider the sum
$$
S(n) = \sum_{a = 1}^{n} \sum_{b = 1}^{n} \sum_{c = 1}^{n} \frac{ab(3a + c)}{2^a 2^b 2^c (a + b)(b + c)(c + a)}.
$$
The key is to look at what happens if we exchange the roles of $a$, $b$, and $c$. The sum is, for example, equal to
$$
\sum_{a = 1}^{n} \sum_{c = 1}^{n} \sum_{b = 1}^{n} \frac{ac(3a + b)}{2^a 2^b 2^c (a + b)(b + c)(c + a)},
$$
which we obtain be replacing every $b$ with a $c$, and every $c$ with a $b$ in the original sum. By changing the order of summation, this implies that
$$
S(n) = \sum_{a = 1}^{n} \sum_{b = 1}^{n} \sum_{c = 1}^{n} \frac{ac(3a + b)}{2^a 2^b 2^c (a + b)(b + c)(c + a)}.
$$
We do the same thing with all of the possible permutations of $a$, $b$, and $c$, and add the resulting expressions together. We get that
$$
6S(n) = \sum_{a = 1}^{n} \sum_{b = 1}^{n} \sum_{c = 1}^{n} \frac{ab(3a + c) + ac(3a + b) + ab(3b + c) + bc(3b + a) + ac(3c + b) + bc(3c + a)}{2^a 2^b 2^c (a + b)(b + c)(c + a)}.
$$
At this point, a minor miracle occurs. It turns out that
$$
ab(3a + c) + ac(3a + b) + ab(3b + c) + bc(3b + a) + ac(3c + b) + bc(3c + a)
$$
is equal to
$$
3(a + b)(b + c)(c + a)
$$
and so we actually obtain the much simpler expression
$$
6S(n) = \sum_{a = 1}^{n} \sum_{b = 1}^{n} \sum_{c = 1}^{n} \frac{3}{2^a 2^b 2^c} = 3 \sum_{a = 1}^{n} \frac{1}{2^a} \sum_{b = 1}^{n} \frac{1}{2^b} \sum_{c = 1}^{n} \frac{1}{2^c}
$$
and so
$$
2S(n) = \left(\sum_{k = 1}^{n} \frac{1}{2^k} \right)^3 = \left(1 - \frac{1}{2^n} \right)^3
$$
and so finally
$$
S(n) = \frac{1}{2} \left( 1 - \frac{1}{2^n} \right)^3
$$
as noticed by Claude Leibovici
|
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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?
|
First Way
Break down the $\sec^3{x}$ and integrate by parts
$$ I = \int \sec^3{x} dx = \int \sec^2{x} \sec{x} dx $$
Recalling:
$$ \int u dv = uv - \int vdu$$
Useful tip: Always try to recognize famous derivatives. Note that $ \sec^2{x}$ seems familiar so lets
use $dv = \sec^2{x} dx$ and $ u = \sec{x}$ then $v = \tan{x}$ and $ du = \sec{x} \tan{x} dx$.
$$ I = \tan{x} \sec{x} - \int \tan^2{x} \sec{x} dx$$
break up the $ tan^2{x} $ as $ \sec^2{x} - 1$
$$I = \tan{x} \sec{x} - \int (\sec^2{x} - 1) \sec{x} dx = \tan{x} \sec{x} + \int \sec{x} dx - \int \sec^3{x} dx$$
Note that $ \int \sec^3{x} = I$ so we added up for both sides and divide by two.
$$ I = \frac{1}{2} (\tan{x} \sec{x} + \int \sec{x} dx ) =\frac{1}{2} (\tan{x} \sec{x} + \ln{|\sec{x} + \tan{x}} |) + C $$
Second way :
You can also directly apply integration by parts at the very beginning.
In fact, using integration by parts can also be applied. Let $ dv = \frac{x}{\sqrt{x^2 -1}} \,dx $ and $u = x$
then:
$$ \int dv = v = \int \frac{x}{\sqrt{x^2 -1}} dx $$
Use $z^2 = x^2 - 1$ and $2zdz = 2x dx$
$$ \int dv = v = \int \frac{z dx}{z} = \int dz = z = \sqrt{x^2 - 1} $$
Finally plugging v, u, dv, and du in the integration by parts formula.
$$ \int \frac{x^2}{\sqrt{x^2 -1}} dx = x \sqrt{x^2 - 1} - \int \sqrt{x^2 - 1}dx$$
$ x = \sec{\theta} $, $dx = \sec{\theta}\tan{\theta} d\theta$
$$ \int \sqrt{x^2 - 1} dx = \int \sqrt{\sec^2{\theta} - 1} \sec{\theta}\tan{\theta} d\theta = \int \tan^2{\theta} \sec{\theta} d\theta = J $$
And then we apply the same method used in the first way, for that.
Hence, if you find J you get that
$$ \int \frac{x^2}{\sqrt{x^2 -1}} dx = x \sqrt{x^2 - 1} - J $$
|
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|
Solving $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$
I have to solve this irrational equation on $\mathbb{R}$ :
$$ \sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$$
I tried to do a substitution with $u=1-x$ but the only things I manage to reach is the following equation by squaring and using $(a-b)(a+b)=a^2 -b^2$:
$$ (\sqrt{1-x}-2x\sqrt{1-x^2})^2 = (2x^2 -1)^2$$
$$\implies 1 - x + 4 x^2 - 4 x^4 - 4x \sqrt{1 - x} \sqrt{1 - x^2} = 4x^4 - 4x^2 +1$$
$$ \implies -4x\sqrt{(1-x)(1+x)(1-x)} = 8x^4-8x^2$$
$$ \implies 4(1-x)\sqrt{1+x} = 8x^4 -8x^2$$
$$ \implies (1-x)\sqrt{1+x} = 2x^2 (1-x^2)$$
I don't manage to go forward. The only thing I know is that the solution (if there is one) is in [-1;1].
Could you help me, please ?
|
Let $x=\cos\alpha,$ where $\alpha\in[0,\pi]$.
Thus, we need to solve
$$\sqrt{1-\cos\alpha}=2\cos^2\alpha-1+2\cos\alpha\sqrt{1-\cos^2\alpha}$$ or
$$\sqrt{2\sin^2\frac{\alpha}{2}}=\cos2\alpha+2\cos\alpha\sin\alpha$$ $$\sqrt2\sin\frac{\alpha}{2}=\cos2\alpha+\sin2\alpha$$ or
$$\sin\frac{\alpha}{2}=\sin\left(\frac{\pi}{4}+2\alpha\right),$$
which gives $$\frac{\alpha}{2}=\frac{\pi}{4}+2\alpha+2\pi k,$$ where $k\in\mathbb Z,$ which is impossible for $\alpha\in[0,\pi]$ or
$$\frac{\alpha}{2}=\pi-\left(\frac{\pi}{4}+2\alpha\right)+2\pi k,$$ where $k\in\mathbb Z,$ which is possible for $k=0$ only, which gives $$\alpha=\frac{3\pi}{10}$$ and the root
$$\cos54^{\circ}$$ or
$$\sin36^{\circ}$$ or
$$\frac{\sqrt{5-\sqrt5}}{2\sqrt2}.$$
|
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Is there a simpler expression for this piecewise-defined function? As a math-for-fun exercise, I challenged myself to find a globally-defined, everywhere-differentiable antiderivative of $\sqrt{1-\sin(x)}$. Because of the fundamental theorem of calculus, the problem boils down to evaluating the integral
$$f(x)=\int_{-\frac{\pi}{2}}^{x}\sqrt{1-\sin(t)}\text{ }dt$$
After lots of thinking, many laborious calculations and a series of roadblocks, I arrived at this piecewise-defined expression:
$$f(x)=\begin{cases}
4\sqrt{2} \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + 2\sqrt{1+\sin(x)} & \text{if } 2\pi \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor - \frac{\pi}{2}\leq x \leq 2\pi \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + \frac{\pi}{2}\\
4\sqrt{2} \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor +4\sqrt{2} - 2\sqrt{1+\sin(x)} & \text{if } 2\pi \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + \frac{\pi}{2}\leq x \leq 2\pi \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + \frac{3\pi}{2}
\end{cases}$$
Using knowledge obtained during the solution process, I obtained the (perhaps) simpler expression
$$f(x)=\begin{cases}
4\sqrt{2} \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + 2\sqrt{1+\sin(x)} & \text{if } \cos(x)\geq 0\\
4\sqrt{2} \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor +4\sqrt{2} - 2\sqrt{1+\sin(x)} & \text{if } \cos(x)\leq 0
\end{cases}$$
Naturally, I wondered whether this expression could be simplified further. Maybe there's a non-piecewise expression for $f(x)$? I originally thought about writing
$$4\sqrt{2}\left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + 4\sqrt{2}=4\sqrt{2} \left( \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor +1 \right)=4\sqrt{2}\left \lceil \frac{x}{2\pi}+\frac{1}{4} \right \rceil$$
but quickly recognized that this is not valid when $\frac{x}{2\pi}+\frac{1}{4}$ is an integer. I also tried writing
$$f(x)=\begin{cases}
4\sqrt{2} \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor + 2\text{sgn}(\cos x)\sqrt{1+\sin(x)} & \text{if } \cos(x)\geq 0\\
4\sqrt{2} \left \lfloor \frac{x}{2\pi}+\frac{1}{4} \right \rfloor +4\sqrt{2} + 2\text{sgn}(\cos x)\sqrt{1+\sin(x)} & \text{if } \cos(x)\leq 0
\end{cases}$$
but this fails when $\cos(x)=0$. I don't have any other tricks up my sleeve. Could I get some assistance?
|
$$\sqrt{1-\sin x}=\sqrt{\cos^2\frac x2-2\cos\frac x2\sin\frac x2+\sin^2\frac x2}=\left|\cos\frac x2-\sin\frac x2\right|
\\=\sqrt2\left|\sin\left(\frac x2-\frac\pi4\right)\right|$$ and we can focus on the antiderivative of $|\sin t|=\pm\sin t$, where the sign alternates with period $\pi$.
For the antiderivative, we can take $-\cos(t\bmod\pi)$ with a jump of $2$ units at every $k\pi$, to compensate the discontinuity.
$$2\left\lfloor\frac t\pi\right\rfloor-\cos(t\bmod\pi)$$
Don't forget that $t=\dfrac x2-\dfrac\pi4$. If you want, you can emulate the modulo and the floor by elementary transformations of $\tan(\arctan(x))$.
|
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|
Evaluating the limit $\lim_{h\to 0} \frac{\sqrt {1 - (x+h) ^2} - \sqrt {1-x^2}} {h}$ Find the limit
$\lim_{h\to 0} \frac{\sqrt {1 - (x+h) ^2} - \sqrt {1-x^2}} {h}$
I tried substituting "h" and also multiplying $\frac{\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2}}{\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2}}$
|
Let $f(x)=\sqrt{1-x^2}$
$$f'(x)=\lim_{h \to 0}\frac{\sqrt{1-({x+h})^2}-\sqrt{1-x^2}}{h}=\frac{-x}{\sqrt{1-x^2}}$$
|
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|
finding a relation in $p:p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$
if $$p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$$ and $$p^2+ap+c=0.$$ Find $a,c$ also $|c|=2$
My progress:The general term $$T_{m+1}=\frac{(1)(3)\cdots(2m+1)}{(3)(6)\cdots(3m+3)}$$ from here i tried to make it in terms of factorial but it does not help.
Is there any algorithm to solve these problems?
(This question came in a maths magazine
|
Since $p=\tfrac13\sum_{m\ge0}\tfrac{\binom{2m+1}{m}}{6^m}$ and $\binom{n}{k}=\oint_{|z|=1}\frac{(1+z)^ndz}{2\pi iz^{k+1}}$, simplifying a geometric-series integrand eventually gives$$p=-2\oint_{|z|=1}\frac{1+z}{1-4z+z^2}\frac{dz}{2\pi i}.$$The denominator's roots are $2\pm\sqrt{3}$, but only $2-\sqrt{3}$ is enclosed. So$$p=-2\lim_{z\to2-\sqrt{3}}\frac{1+z}{z-2-\sqrt{3}}=\sqrt{3}-1.$$(More generally, the above technique shows $\sum_{m\ge0}\binom{2m+1}{m}x^m=\frac{(1-4x)^{-1/2}-1}{2x}$ for $|x|<1/4$.) If we assume $a,\,c\in\Bbb Z$, the sought quadratic has roots $-1\pm\sqrt{3}$ so that $a=2,\,c=-2$ because the quadratic is $(p+1)^2-3=0$.
|
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|
How does one go about solving $arg(\frac{z-2i}{z-6}) = \frac{1}{2}\pi$ This should give $$\frac{z-2i}{z-6} = bi$$
but solving that gives me $$z = \frac{-2b +6b^2-6bi +2i}{1+b^2}$$
and substituting $z$ for $x + yi$ gives me $x = \frac{-2b +6b^2}{1+b^2}$ and $y=\frac{-6b +2}{1+b^2}$
And I have no clue how to continue now. The answer is suppose to be $x = -by$, but I have no clue how to get there (have I made a calculation error??)
EDIT: Thanks everyone!!
|
You have not made a mistake. Observe that $x$ is a multiple of $y$:
$$x = \frac{-2b + 6b^2}{1 + b^2} = -b \cdot \frac{2 - 6b}{1 + b^2} = -by$$
which is what the answer states.
|
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|
If $a^2+b^2+c^2+d^2=4$ then $(a+2)(b+2)\geq cd$ Let $a,b,c,d$ be real numbers with $a^2+b^2+c^2+d^2=4$. Prove that $(a+2)(b+2)\geq cd$.
My approach: I have considered an expression $$\begin{aligned}(a+2)(b+2)-cd=&4+2(a+b)+(ab-cd)\\=&(a^2+b^2+c^2+d^2)+2a+2b+(ab-cd)\end{aligned}$$ I was trying to write it as the sum of squares but I failed.
Can anyone show how to solve this problem please?
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Because by AM-GM $$(a+2)(b+2)=ab+2(a+b)+4=$$
$$=\frac{1}{2}(2ab+4a+4b+4+a^2+b^2)+2-\frac{1}{2}(a^2+b^2)=$$
$$=\frac{1}{2}(a+b+2)^2+2-\frac{1}{2}(a^2+b^2)\geq$$
$$\geq2-\frac{1}{2}(a^2+b^2)=\frac{1}{2}(c^2+d^2)\geq |cd|\geq cd.$$
|
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|
What is the solution to $\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$ $$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$$
I've already tried this many times but still not get the answer, this was one of my solution which I can't go further anymore.
$\frac{{x}^{2}}{(x+1)({\sqrt{x+1}}-1)^{2}}< \frac{{x}^{2}+3x+18}{({x+1})^{2}}$
$\frac{{x}^{2}(x+1)}{({\sqrt{x+1}}-1)^{2}}< {x}^{2}+3x+18$
$\frac{{x}^{3}+{x}^{2}}{({\sqrt{x+1}}-1)^{2}}< \frac{({x}^{2}+3x+18)({\sqrt{x+1}-1})^{2}}{({\sqrt{x+1}-1})^{2}}$
$\frac{{x}^{3}+{x}^{2}}{({\sqrt{x+1}}-1)^{2}}< \frac{{x}^{3}+5{x}^{2}+24x+36-(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}-1})^{2}}$
$\frac{{x}^{3}+{x}^{2}-{x}^{3}-5{x}^{2}-24x-36+(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
$\frac{-4{x}^{2}-24x-36+(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
$\frac{-4(x+3)^{2}+2({x}^2+3x+18)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
Does anyone have a hint for the solution?
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Note that $$x=(x+1)-1=(\sqrt{x+1}-1)(\sqrt{x+1}+1)$$ Thus the left-hand side simplifies to $$(\sqrt{x+1}+1)^2<\frac{x^2+3x+18}{x+1}$$ This doesn't look promising, but let's see if we can get an approximate solution by pulling out the leading term on the right-hand side. Well, $x^2+3x+18=(x+1)(x+2)+16$. Thus partial fractions gives $$(\sqrt{x+1}+1)^2<x+2+\frac{16}{x+1}$$
Next, if we can isolate the square root, then we can square both sides and turn this into a polynomial problem. When we expand out the square on the left-hand side, we discover a miracle: $$(\sqrt{x+1}+1)^2=(x+1)+1+2\sqrt{x+1}=x+2+2\sqrt{x+1}$$ The $x+2$s cancel! So $$2\sqrt{x+1}<\frac{16}{x+1}$$
I trust you can solve it from here.
|
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|
How to integrate hyperbolic function $\frac{1-\sin(x)}{1+\sin(x)}$ I'm trying to find a primitive of $ \frac{1-\sin(x)}{1+\sin(x)} $. By changing the variable to $t=\tan(\frac{x}{2})$ and letting $\sin(x) = \frac{2t}{1+t^2}$ I get the following integral:
\begin{align}
& \int \frac{1-\sin(x)}{1+\sin(x)} \, dx \\[8pt] = {} & \int \frac{ 1-\frac{2t}{1+t^2} } { 1+ \frac{2t}{1+t^2} } \frac{2}{1+t^2} \, dt \\[8pt]
= {} & 2\int \frac{ t^2 -2t +1 }{ (t^2 +2t+1)(t^2 + 1) } \, dt \\[8pt]
= {} & 2\int \frac{t^2 -2t +1 }{(t+1)^2(t^2+1)} \, dt \end{align}
Now I know that I could do a partial fraction expansion. I would get 3 simpler fraction but I also know that the result contains only two fraction by calculating it in xcas: $$ \int \frac{1-\sin(x)}{1+\sin(x)} \, dx = 2\left(-\frac{2}{\tan(\frac{x}{2}) +1} - \frac{x}{2}\right) $$
Is there an easier way to calculate this primitive ?
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$$\begin{align}\int\frac{1-\sin x}{1+\sin x}dx &= \int\left(\frac2{1+\sin x}-1\right)\,dx
\\&= \int\frac2{1+\frac{2t}{1+t^2}}\cdot\frac2{1+t^2}\,dt-x
\\&= \int\frac4{1+2t+t^2}\,dt-x
\\&= \int\frac4{(t+1)^2}\,d(t+1)-x
\\&= -\frac4{t+1}-x
\\&= -\frac4{\tan\frac x2+1}-x
\end{align}$$
|
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|
Can a triangle ABC be made if $\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$
I would like to know the simplest approach to find out whether a triangle ABC will be made if $$\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$$
The counterpart questions for sine and tangent can be handled as follows:
*
*If $\dfrac{\sin A}{2}=\dfrac{\sin B}{3}=\dfrac{\sin C}{7}$, we can rule out triangle because by the Sine Rule $a=2k$, $b=3k$, $c=7k \implies a+b <c.$
*If $\dfrac{\tan A}{2}=\dfrac{\tan B}{3}=\dfrac{\tan C}{7}$, we can see that a triangle will be made as $\tan A=2k, \tan B =3k,\tan C=7k$, when inserted in the identity $\tan A+ \tan B+ \tan C= \tan A \tan B \tan C \implies k=\sqrt{2/7}$.
|
Show for yourself that if $A,B,C$ are the angles of a triangle then
$$
\cos^2A + \cos^2B + \cos^2C = 1-2\cos A \cos B \cos C
$$
This is not very difficult, use the fact that $A+B+C = 180$ along with double angle identities.
Therefore, if each of those ratios equaled $k$ , we get $62k^2 = 1-84k^3$, which can be solved using Cardano's formula (or you can use IVT to just assert the existence of a root) to get you a root like $k \approx 0.117928$. From here, you get that such a triangle in fact exists, and roughly has angles $76.36,69.28$ and $34.36$.
|
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|
Need help with this proof via induction
Let $x_1,...,x_n > 0$.
I'm having troubles proving this formula via induction:
$$
(x_1 + \ldots + x_n)\left(\frac1{x_1} + \ldots + \frac1{x_n}\right) \ge n^2
$$
So far, I've managed to rewrite it like this:
$$
\sum_{k=1}^n x_k \sum_{k=1}^n \frac{1}{x_k} \ge n^2
$$
Also the base case seems simple enough: $1 + 1/1 \ge 1^2$.
However, this is where I got stuck and I can't seem to be able to solve this any further.
|
$$\tag{*} \underbrace{(x_1 + \ldots + x_n)}_{A}~\underbrace{\left(\frac1{x_1} + \ldots + \frac1{x_n}\right)}_B \ge n^2$$
$$
\begin{split}
(x_1 + &\ldots + x_n + x_{n+1})
\left(\frac1{x_1} + \ldots + \frac1{x_n} + \frac{1}{x_{n+1}}\right) \\
&= \left(A+x_{n+1}\right)\left(B+\frac{1}{x_{n+1}}\right) \\
&= AB + \frac{A}{x_{n+1}} + Bx_{n+1}+ 1 \overset{(*)} \\ \\
&\ge n^2 + \color{red}{\frac{A}{x_{n+1}} + Bx_{n+1}}+ 1 \\
&\ge n^2 + \color{red}{2\sqrt{AB}}+1\overset{(*)} \\ \\
&\ge n^2+2n+1 \\
&= (n+1)^2 \ \ \ \ \square
\end{split}
$$
where the red part follows from the basic inequality $a+b\ge 2\sqrt{ab} \ \ (a,b \ge 0)$
|
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|
A nice way to remember trigonometric integrals? Is there a "nice" way to remember trigonometric integrals, beyond what is typically taught in a standard calculus class? I'm currently in Calculus II, and up to now I've found calculus rather accessible. I love that, at least in my classes, we learn the "how" and the "why". I'm struggling, however, to remember "trigonometric integrals" on exams, etc, where notes aren't allowed. We're effectively given an integration table, and tasked with memorizing maybe 15 or 20 results in a couple week's time (no notes, and no calculator are allowed on exams, in-class quizzes, and technically calculators are not allowed on homework).
So, is there a "nice" way to remember these, beyond something like a mnemonic, etc? Perhaps some line of reasoning or a simple proof, etc? I tend to more easily remember things when I understand their derivation/intuition, if nothing else because I'm able to recreate it on the spot without memorizong the details.
Also, to clarify, by "trigonometric integrals", I'm referring to the integrals of the trigonometric functions ($\sin$, $\cos$, $\tan$, and $\sec$), the inverse trigonometric integrals ($\sin^{-1}$, etc), and integrals like:
$\int \frac{1}{x^2+1}$
...which work out to be trigonometric functions, products of trigonometric functions, etc.
|
The pythagorean theorem, as applied to trigonometry says
$\sin^2 \theta + \cos^2 \theta = 1$
This is the key piece of knowledge for these integrals.
The implications are:
$\cos \theta = \pm \sqrt {1-\sin^2 \theta}\\
\sin \theta = \pm \sqrt {1-\cos^2 \theta}\\
\tan^2 \theta + 1 = \sec^2 \theta$
How does this relate to these integrals...
Whenever you see $x^2 + 1$ in someplace inconvenient, like under a radical or in the denominator, you should be thinking of the substitution $x = \tan \theta.$ With this substitution it will become $\tan^2\theta + 1 = \sec^2 \theta$
Similarly, when you see $1-x^2$ you should be thinking $x=\sin\theta$ or $x = \cos \theta$ and the expression becomes $1-\sin^2\theta = \cos^2\theta$
And when you see $x^2 - 1$ it is a bit of a toss up. Sometimes, $x = \sin \theta$ works and sometimes $x = \sec\theta$ works better. It really has to do whether you have reason to think $|x|<1$ (in which case use the sine substitution) or $|x| > 1$ in which case use the secant substitution.
Taking it up a level.
When you see $x^2 + a^2$ then you should be thinking $x = a\tan \theta$ and when you see $a^2 x^2 + b^2$ think $x = \frac {b}{a}\tan \theta$ Finally, when you see $(x+a)^2 + b^2,$ think $x+a = b\tan \theta.$ These will simplify nicely.
Some examples.
The area of a portion of a circle...
The equation of our circle is $x^2 + y^2 = 1$
We want $\int_a^1 \sqrt {1-x^2} \ dx$
start with:
$x = \cos \theta\\
dx = -\sin\theta\ d\theta$
What happens to our limits of integration?
$a = \cos \theta\\
\theta = \arccos a\\
1 = \cos \theta\\
\theta = 0$
$\int_{\arccos a}^{0} \sqrt {1-\cos^2\theta} (-\sin\theta \ d\theta)$
We can reverse the order of integration if we change the sign. $1-\cos^2 \theta = \sin^2\theta$
$\int_0^{\arccos a} \sqrt {\sin^2\theta} (\sin\theta) \ d\theta\\
\int_0^{\arccos a} \sin^2\theta \ d\theta$
Apply a half-angle identity:
$\sin^2\theta = \frac 12 (1-\cos 2\theta)$
$\int_0^{\arccos a} \frac 12 (1-\cos 2\theta) \ d\theta$
$\frac 12 (\theta-\frac 12 \sin 2\theta)|_0^{\arccos a}$
At this point I like to use the double angle identity
$\frac 12 (\theta-\sin \theta\cos \theta)|_0^{\arccos a}$
$\sin \arccos a = \sqrt {1-a^2}$
$\frac 12 (\arccos a - a\sqrt {1-a^2})$
What does this mean geometrically?
The area of the red plus the green is $\frac 12 \theta = \frac 12 \arccos a$
The height of the red triangle is $\sqrt {1-a^2}$ and the area is $\frac 12 a\sqrt {1-a^2}$
One more example
$\int \frac {1}{x^2+x+1} \ dx$
The denominator looks like a bit of a bear. It doesn't factor, if it did, I would suggest partial fractions. As it doesn't we use "completing the square."
$x^2 + x + 1 = (x+\frac 12)^2 + \frac 34$
$\int \frac {1}{(x+\frac 12)^2 + \frac 34} \ dx$
$x+\frac 12 = \sqrt {\frac 34} \tan \theta\\
dx = \sqrt {\frac 34} \sec^2 \theta\ d\theta$
Don't let those radicals scare you, they are just constants.
$\int \frac {\sqrt {\frac 34}\sec^2\theta}{\frac 34 \tan^2\theta + \frac 34} \ d\theta\\
\frac {1}{\sqrt {\frac 34}}\int \frac {\sec^2\theta}{\sec^2\theta} \ d\theta\\
\frac {2}{\sqrt 3} \theta
$
Now we need to reverse the substitution
$x + \frac 12 = \sqrt {\frac 34} \tan \theta\\
\frac {2}{\sqrt 3} (x+\frac 12) = \tan \theta\\
\theta = \arctan (\frac {2\sqrt 3}{3}x + \frac {\sqrt 3}{3})$
$\frac {2\sqrt 3}{3} \arctan (\frac {2\sqrt 3}{3}x + \frac {\sqrt 3}{3})$
I hope this helps.
|
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|
Sketch the graph of the function $f(x) =\frac{(x+1)^3} {x^2 - x+1}$ I tried to follow the rules of sketching a graph, I found the intersection with the ox and oy in (0,1)and (-1,0) and I think that the oblique asymptote is y=x+4. What about the vertical ones? The roots of the denominator are complex ones. Can we say these are asymptotes?.. The first derivative is pretty hard to find roots of. How can I sketch its graph, please?
|
You have $$f(x) =\frac{(x+1)^3} {x^2 - x+1} = x + 4 + \frac{6x - 3}{x^2-x+1}.$$
As $\lim\limits_{x \to \pm \infty} \frac{6x - 3}{x^2-x+1} = 0$, $y=x+4$ is indeed an asymptote of $f$. Also the sign of $\frac{6x - 3}{x^2-x+1}$ is the one of $6x-3$ as $x^2-x+1$ is always positive. So $f$ crosses its asymptote at $x = 1/2$, is above it for $x > 1/2$ and below it for $x < 1/2$.
You can also find where the maximal vertical distance between $f$ and its asymptote is by vanishing the derivative of $g(x)=\frac{6x - 3}{x^2-x+1}$. You'll find two symmetric values about $x=1/2$.
Some elements to start a decent plot...
|
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|
Find the area bounded by the curve $x^4+y^4=x^2+y^2$ I am stuck with this problem which deals with evaluating an Area
The problem reads :
Find the area bounded by the curve $x^4+y^4=x^2+y^2$.
I tried factorizing the expression and expressing $y$ in terms of $x$, not able to proceed with that idea. Someone please help me out.
|
Expressing the curve in polar coordinates, we have\begin{align}x^4+y^4=x^2+y^2&\iff\rho^4\cos^4\theta+\rho^4\sin^4\theta=\rho^2\\&\iff\rho=\frac1{\sqrt{\cos^4\theta+\sin^4\theta}}.\end{align}Hence, the area that you're after is\begin{align}\int_0^{2\pi}\int_0^{1/\sqrt{\cos^4\theta+\sin^4\theta}}\rho\,\mathrm d\rho\,\mathrm d\theta&=\int_0^{2\pi}\frac2{3+\cos(4\theta)}\,\mathrm d\theta\\&=\pi\sqrt2.\end{align}
|
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|
Messy Gaussian Integral I am trying to understand how to better perform the following integral.
$$\int^{\infty}_{0} x^4 e^{\frac{-x^2}{\beta^2}}\mathrm{d}x$$
I've done a little research and found that $e^{-x^2}$ doesn't integrate easily, for it is the Gaussian integral. Many sources are pointing me to use polar coordinates. Is that the best way to go about this? How can I solve this integral?
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I just wanted to expand on Claude's answer. First a transformation:
$$\int_0^\infty x^n\exp\left(\frac{-x^2}{a^2}\right)\mathrm{d}x=a^{n+1}\int_0^\infty x^n\exp(-x^2)\mathrm{d}x$$
Let
$$I_n=\int_0^\infty x^n\exp(-x^2)\mathrm{d}x$$
Integration by parts. Let $u=\exp(-x^2)$, $\mathrm{d}u=-2x\exp(-x^2)\mathrm{d}x$, $\mathrm{d}v=x^n\mathrm{d}x$, $v=\frac{x^{n+1}}{n+1}$.
$$I_n=\int_0^\infty u~\mathrm{d}v=(uv)\big|^\infty_0-\int_0^\infty v~\mathrm{d}u$$
$$=\left(\frac{x^{n+1}}{n+1}\exp(-x^2)\right)\bigg|^\infty_0-\int_0^\infty-2x\exp(-x^2)\frac{x^{n+1}}{n+1}\mathrm{d}x$$
$$I_n=\frac{2}{n+1}I_{n+2}\implies I_{n+2}=\frac{n+1}{2}I_n$$
Now we need to compute $I_0,I_1$. It's obvious that $I_0=\sqrt{\pi}/2$.
$$I_1=\int_0^\infty x\exp(-x^2)\mathrm{d}x$$
Via a substitution $t=x^2$, $\mathrm{d}t=2x\mathrm{d}x$,
$$I_1=\frac{1}{2}\int_0^\infty e^{-t}\mathrm{d}t=\frac{1}{2}\Gamma(1)=\frac{1}{2}.$$
So
$$I_2=\frac{\sqrt{\pi}}{4}~;~I_3=\frac{1}{2}~;~I_4=\frac{3\sqrt{\pi}}{8},...$$
Since the recurrence relation jumps by two, we can separate the even and odd cases. For odd $n$,
$$I_n=I_1\cdot\left(\frac{(1+1)}{2}\frac{(3+1)}{2}\frac{(5+1)}{2}...\frac{n-2+1}{2}\right)=\frac{1}{2}\left(1\cdot 2\cdot 3\cdot~~...~~\cdot \frac{n-1}{2}\right)=\frac{1}{2}\left(\frac{n-1}{2}\right)!$$
And, since for $n\in\Bbb{N},~n!=\Gamma(n+1)$,
$$I_n=\frac{1}{2}\Gamma\left(\frac{n+1}{2}\right)$$
For even $n$, it's a slightly trickier.
$$I_n=I_0\left(\frac{(0+1)}{2}\frac{(2+1)}{2}\frac{(4+1)}{2}...\frac{n-2+1}{2}\right)=\frac{\sqrt{\pi}}{2}\left(\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{2}\cdot~~...~~\cdot\frac{n-1}{2}\right)$$
However one might notice that $\sqrt{\pi}=\Gamma(1/2)$. Using the recursive properties of the Gamma,
$$I_2=\frac{1}{2}I_0=\frac{1}{2}\frac{\Gamma(1/2)}{2}=\frac{\Gamma(3/2)}{2}$$
$$I_4=\frac{3}{2}I_2=\frac{3}{2}\frac{\Gamma(3/2)}{2}=\frac{\Gamma(5/2)}{2}$$
So it's easy to see in general that this actually lines up with what we got with the odd case.
$$I_n=\frac{1}{2}\Gamma\left(\frac{n+1}{2}\right)$$
Finally, $$\int_0^\infty x^n\exp\left(\frac{-x^2}{a^2}\right)\mathrm{d}x=a^{n+1}I_n=\frac{a^{n+1}}{2}\Gamma\left(\frac{n+1}{2}\right)$$
So the integral in question is
$$\frac{a^{4+1}}{2}I_4=\frac{a^5}{2}\Gamma(5/2)=\frac{3a^5\sqrt{\pi}}{8}.$$
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Meaning of drawing any number of balls from an urn containing $n$ balls There is a question as follows:
From an urn containing $n$ balls any number of balls are drawn. Show that the probability of drawing an even number of balls is $\frac{2^{n-1}-1}{2^n-1}$
Firstly what does it mean to draw any number of balls from the urn and how does the sample space look like for this question?
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since we need to find the probability of choosing even number of balls it means the probability is
$$ \frac{\binom{n}{2}+\binom{n}{4} + \ .... \ }{\binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \ .... \ \binom{n}{n}}
$$
which equals $$ \frac{2^{n-1} \ - \ \binom{n}{0}}{2^n - \binom{n}{0}} $$
which immediatly yeilds your answer.
To find the sum $\binom{n}{2}+\binom{n}{4} + \ .... $ and $\binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \ .... \ \binom{n}{n}$
to find $\displaystyle \binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \ .... \ \binom{n}{n}$ put
$x=1$ in $(1+x)^n$
and to find the second one put $x = -1$and add with the first equation
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|
Let ABCD be a trapezoid; find point M on AB, such that $DM+MC$ is minimal possible Let ABCD be a trapezoid with $A=B=90$, $AD=a$ and $AB=BC=2a$.
Find point M on AB, such that $DM+MC$ is minimal possible.
I have been trying to do this question, but without success. I have succeeded in proving that $DA+AC<DB+BC$ something which I believe would help in finding this minimal value, however I have not succeeded in finishing it off. Moreover, I have also found that from Pythagoras we have that $DA+AC=a+2a\sqrt{2}$ and $DB+BC=2a+a\sqrt{5}$. Can someone please help me finish the question off?
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Another way to solve -
Choose a point $M$ on $AB$ such that $AM = x, \,$ then BM = $(2a-x)$
$DM = \sqrt{AD^2 + AM^2} = \sqrt{a^2+x^2}$
Similarly $MC = \sqrt{BC^2 + BM^2} = \sqrt{4a^2+(2a-x)^2}$
$L = DM + MC = \sqrt{a^2+x^2} + \sqrt{8a^2+ x^2 - 4ax}$
To find min value of $L$,
$ \displaystyle \frac{dL}{dx} = \frac {x} {\sqrt{a^2+x^2}} + \frac{x-2a}{\sqrt{8a^2+ x^2 - 4ax}} = 0$
$\displaystyle x \sqrt{8a^2+ x^2 - 4ax} + (x-2a) \sqrt{a^2+x^2} = 0$
$\displaystyle x \sqrt{8a^2+ x^2 - 4ax} = (2a-x) \sqrt{a^2+x^2}$
$\displaystyle x^2 (8a^2+ x^2 - 4ax) = (2a-x)^2 (a^2+x^2)$
Solving we get,
$ \displaystyle (x + \frac{2a}{3})^2 = \frac{16a^2}{9}$
So, $x = \frac{2a}{3}$ or $x = -2a$
Discarding $x = -2a$, $AM = x = \frac{2a}{3}$ gives you min.
EDIT: Just added a diagram for what Calvin Lin suggested which uses reflection -
$DM + MC$ will minimize when it is a straight line.
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Remainder of dividing polynomial of the $n$th degree
What is the remainder when dividing the polynomial
$$P(x)=x^n+x^{n-1}+\cdots+x+1$$ with the polynomial
$$x^3-x$$ if $n$ is a natural odd number?
So, what I know so far is:
$$P(x)=Q(x)D(x)+R(x)$$
In this case I'll call $Q(x) = x^3-x$
$$Q(x) = 0 \iff x=\pm1$$
So from here:
$\begin{array} {l}P(1):\qquad&1^n+1^{n-1}+\cdots+1^1+1=R(1)\\P(-1):&(-1)^n+(-1)^{n-1}+\cdots+(-1)^1+1=R(-1)\end{array}$
Here is where I assumed that $R(x)=ax^2+b$ since $Q(x)=x^3-x$
And from the equations (assuming $n = \{2k+1 \mid k\in\mathbb{N}\}$):
$\begin{array}{l}P(1):\\&R(1)=n\\P(-1):\\&R(-1)=0\end{array}$
And here I get to:
$$\left\{
\begin{array}{ll}
ax^2+by &=n \\
ax^2+by &=0 \\
\end{array}
\right.$$
I would like to know where did I go wrong? Is my assumption for $R(x)$ incorrect, my calculation of the $P(1)$ and $P(-1)$ wrong or is there something else I didn't think about?
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Hint:
Experimenting the first few values of $n$: $\:n= 3,5,7,9,11$, you may conjecture the remainder for the general polynomial $P_n(x)=x^n+x^{n-1}+\dots+x^3+x^2+x+1\:$ is
$$R_n(x)=kx^2+(k+1)x+1,\quad\text{ where } k=\left\lfloor \frac n2 \right\rfloor.$$
and try to prove it by induction on $n$ (odd).
You can prove the inductive step (case $n \implies$ case $n+2$) you can try rewriting
$$ P_{n+2}(x)= x^2 P_n(x)+ (x+1)=x^2\bigl[ Q_n(x)(x^3-x)+R_n(x)\bigr]+(x+1), $$
so the remainder for $P_{n+2}$ will be as well the remainder for
$$x^2 R_n(x)+(x+1)=x^2(kx^2+(k+1)x +1)+(x+1)=kx^4+(k+1)x^3+x^2+x+1.$$
You'll only have to prove that the coefficient of $x^2$ thus obtained is $\left\lfloor\dfrac{n+2}2\right\rfloor$ and the coefficient of $x$ is $1$ more.
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If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, what is $|a + b + c + d|$? From the Pre-Regional Mathematics Olympiad, 2019:
If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, what is $|a + b + c + d|$?
I have provided one answer below, and would be interested in alternative solutions.
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The only part missing in the existing answer is the fact that the minimal polynomial of $a=\sqrt{2}+\sqrt{3}+\sqrt{6}$ has degree $4$, i.e. the degree of the extension $|\Bbb{Q}[a]:\Bbb{Q}|$ is $4$. Since $a$ is a root of an integer polynomial of degree $4$, the degree of the extension is $1, 2$ or $4$. It cannot be $1$ since $a$ is not rational. If the degree is $2$ then for some rational $p,q$ you have $a^2+pa+q=0$ or $11+2(\sqrt{6}+2\sqrt{3}+3\sqrt{2})+pa+q=0$. But that is impossible. So, indeed, the degree is $4$.
Another way to prove it is to consider homomorphisms from $\Bbb{Q}[a]$ to $\Bbb{Q}[\sqrt{2},\sqrt{3}]$ (which has degree $4$). There are $4$ of those.
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What is $\tan \alpha$, if $(a+2)\sin\alpha +(2a - 1)\cos\alpha =2a + 1$? I tried the following:
$$\begin{aligned}a\sin\alpha +2\sin\alpha + 2a\cos\alpha - \cos\alpha &= 2a+1\\
a(\sin\alpha +2\cos\alpha)+(2\sin\alpha-\cos\alpha)&=2a+1\end{aligned}$$
Therefore,
$$\sin\alpha +2 \cos\alpha=2$$
$$2\sin\alpha - \cos\alpha=1$$
From these two equations, we get
$$\sin\alpha=\frac{4}{5},\cos\alpha=\frac{3}{5}$$
Therefore,
$$\tan\alpha = \frac{\sin\alpha} {\cos\alpha} = \frac{4} {3}$$
Is this a correct method to solve the question? Since $a$ is a constant, it does not seem necessary to me that its coefficients on the two sides of the equation be equal. Should I find $\sin\alpha$ and $\cos\alpha$ using some other method? Are there specific cases where this method of equating the coefficients will break?
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Hint:
Use Weierstrass substitution to form a quadratic equation $$\tan\dfrac\alpha2=t$$
$$(a+2)\cdot\dfrac{2t}{1+t^2}+(2a-1)\cdot\dfrac{1-t^2}{1+t^2}=2a+1$$
$$\iff-t^2(2a+1+2a-1)+2t(a+2) +2a-1-(2a+1)=0$$
$$\iff2at^2-t(a+2)+1=0$$
Now use $\tan2A=\dfrac{2\tan A}{1-\tan^2A}$
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Consider a function $f(x)= \arcsin (\frac {2x}{1+x^2}) + \arccos (\frac{1-x^2}{1+x^2}) +\arctan (\frac{2x}{1-x^2})-a\arctan x$
Consider a function $f(x)= \arcsin (\frac {2x}{1+x^2}) + \arccos (\frac{1-x^2}{1+x^2}) +\arctan (\frac{2x}{1-x^2})-a\arctan x$, where $a$ is any real constant. Find the value of $a$ if $f(x)=0$ for all x
Replacing $x$ with $\tan y$
$$\arcsin (\sin 2y) +\arccos (\cos 2y) +\arctan (\tan 2y)-a\arctan x=0$$
$$\implies 2y+2y+2y-ay=0$$
$$a=6$$
Alternatively, since $\cos$ is an even function
$$2y-2y+2y-ay=0$$
$$a=2$$
There is another value possible, according to the answer, which is $-2$. How do I obtain that?
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For any $\;x\in\left]-\infty,-1\right[\;,\;$ it results that
$\arcsin\left(\dfrac{2x}{1+x^2}\right)=-\pi-2\arctan x\;,$
$\arccos\left(\dfrac{1-x^2}{1+x^2}\right)=-2\arctan x\;,$
$\arctan\left(\dfrac{2x}{1-x^2}\right)=\pi+2\arctan x\;.$
Hence, for all $\;x\in\left]-\infty,-1\right[,\;$ it results that
$\arcsin\left(\dfrac{2x}{1+x^2}\right)+\arccos\left(\dfrac{1-x^2}{1+x^2}\right)+ \arctan\left(\dfrac{2x}{1-x^2}\right)=$
$=-\pi-2\arctan x-2\arctan x+\pi+2\arctan x=$
$=-2\arctan x$
Consequently, $\;a=-2\;,\;$ for all $\;x\in\left]-\infty,-1\right[.$
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How to solve this geometric+arithmetic recurrence I've been stuck on this recurrence closed form question for a while now:
$S(n)=9S(n-1)+4n, n > 1$
$S(1) = 4$
After expanding a couple iterations to find a pattern, I came up with this:
$4*9^{n-1}+4*(n*\sum_{k=0}^{n-2}9^k-\sum_{k=0}^{n-2}k*9^k)$
$=4*9^{n-1}+4*(n*\frac{9^{n-1}-1}{9-1}-\sum_{k=0}^{n-2}k*9^k)$
However, I can't seem to get further than than that in simplifying it in terms of n.
Can anyone point me in the right direction or show me how best to simplify this recurrence?
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The observed pattern is fine. We have
\begin{align*}
S(n)&=4\cdot 9^{n-1}+4\left(n\frac{9^{n-1}-1}{9-1}-\sum_{k=0}^{n-2}k\,9^k\right)\\
&=\frac{1}{2}9^{n-1}\left(n+8\right)-\frac{1}{2}n-4\sum_{k=1}^{n-2}k\,9^{k}\tag{1}
\end{align*}
In (1) we have collected terms and we start the index $k$ with $1$, since the term with $k=0$ is zero. In order to get a closed form of $\sum_{k=1}^{n-2}k\,9^{k}$ by elementary means we can use the trick to write $k=\sum_{j=1}^k 1$ and then rearrange the sums to obtain geometric sums which admit a closed form.
We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^{n-2}k\,9^k}&=\sum_{k=1}^{n-2}\left(\sum_{j=1}^k1\right)9^k=\sum_{k=1}^{n-2}\sum_{j=1}^k9^k\\
&=\sum_{1\leq j\leq k\leq n-2}9^k=\sum_{j=1}^{n-2}\sum_{k=j}^{n-2}9^k\tag{2}\\
&=\sum_{j=1}^{n-2}\frac{9^{n-1}-9^j}{9-1}\tag{3}\\
&=\frac{1}{8}9^{n-1}\sum_{j=1}^{n-2}1-\frac{1}{8}\sum_{j=1}^{n-2}9^j\tag{4}\\
&\,\,\color{blue}{=\frac{1}{64}9^{n-1}(8n-17)+\frac{9}{64}}\tag{5}\\
\end{align*}
Comment:
*
*In (2) we write the index region conveniently to better see how we to rearrange the sums.
*In (3) we apply the geometric summation formula to the inner sum.
*In (4) we split the sum and factor out terms.
*In (5) we apply the geometric summation formula again.
Combining (1) and (5) we obtain for $n\geq 1$:
\begin{align*}
\color{blue}{S(n)}&=\frac{1}{2}9^{n-1}\left(n+8\right)-\frac{1}{2}n-\frac{1}{16}9^{n-1}(8n-17)-\frac{9}{16}\\
&\,\,\color{blue}{=\frac{1}{16}\left(9^{n+1}-8n-9\right)}
\end{align*}
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|
Prove $\forall z\in\mathbb C-\{-1\},\ \left|(z-1)/(z+1)\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$ I'm trying to prove
$$\forall z\in\mathbb C-\{-1\},\ \left|\frac{z-1}{z+1}\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$$
thus showing that the solutions to $\left|(z-1)/(z+1)\right|=\sqrt2$ form the circle of center $-3$ and radius $\sqrt8$. But my memories of algebra in $\mathbb C$ fail me. The simplest I get is writing $z=x+i\,y$ with $(x,y)\in\mathbb R^2-\{(-1,0)\}$ and doing the rather inelegant
$$\begin{align}\left|\frac{z-1}{z+1}\right|=\sqrt2&\iff\left|z-1\right|=\sqrt2\,\left|z+1\right|\\
&\iff\left|z-1\right|^2=2\,\left|z+1\right|^2\\
&\iff(x-1)^2+y^2=2\,((x+1)^2+y^2)\\
&\iff0=x^2+6\,x+y^2+1\\
&\iff(x+3)^2+y^2=8\\
&\iff\left|z+3\right|^2=8\\
&\iff\left|z+3\right|=\sqrt8\\
\end{align}$$
How can I avoid the steps with $x$ and $y$ ?
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We have that
$$\frac{z-1}{z+1}=\sqrt 2 e^{i\theta} \implies z=\frac{1+\sqrt 2 e^{i\theta}}{1-\sqrt 2 e^{i\theta}}$$
and $$z+3= \frac{4-2\sqrt 2 e^{i\theta}}{1-\sqrt 2 e^{i\theta}} \implies |z+3|^2=\frac{4-2\sqrt 2 e^{i\theta}}{1-\sqrt 2 e^{i\theta}} \frac{4-2\sqrt 2 e^{-i\theta}}{1-\sqrt 2 e^{-i\theta}} =\frac{24-16\sqrt 2 \cos \theta}{3-2\sqrt 2 \cos \theta}=8$$
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|
Where did I go wrong in applying the remainder theorem?
When the expression $3x^3 + px^2 + qx + 8$ is divided by $x^2 - 3x + 2$, the remainder is $5x + 6$. Find the values of $p$ and $q$.
I tried to answer this question using the factor theorem (remainder theorem in this case), but got the answer wrong:
$$ \text{Let} f(x) = 3x^3 + px^2 + qx + 8 $$
$$ \text{Using remainder theorem,}$$
\begin{align}
f(x) &= (x^2 - 3x + 2)Q(x) + (5x + 6) \\
f(x) &= (x - 2)(x - 1)Q(x) + (5x + 6)
\end{align}
$$ \text{When } x = 2, $$
\begin{align}
f(2) &= (2 - 2)(2 - 1)Q(2) + (5(2) + 6) \\
3(2)^3 + p(2)^2 + q(2) + 8 &= 10 + 6 \color{red}{\leftarrow (1)}
\end{align}
$$ \text{When } x = 1, $$
\begin{align}
f(1) &= (1 - 2)(1 - 1)Q(x) + (5 + 6) \\
3(1)^3 + p(1)^2 + q(1) + 8 &= 5 + 6 \color{limegreen}{\leftarrow (2)}
\end{align}
$$ \text{From } \color{red}{(1)}: $$
\begin{align}
3(8) + p(4) + 2q + 8 &= 16 \\
24 + 4p + 2q + 8 &= 16 \\
12 + 2p + q + 4 &= 16 \\
q &= 16 - 16 - 2p \\
q &= -2p \color{blue}{\leftarrow (3)}
\end{align}
$$ \text{From } \color{limegreen}{(2)}: $$
\begin{align}
3 + p + q + 8 &= 11 \\
p + q + 11 &= 11 \\
p &= 11 - 11 - q \\
p &= -q \\
-q &= p \\
q &= -p \color{mediumpurple}{\leftarrow (4)}
\end{align}
$$ \color{mediumpurple}{(4)} + \color{blue}{(3)}: $$
\begin{align}
-p &= -2p \\
2p - p &= 0 \\
\therefore p &= 0
\end{align}
$$ \text{Substitute } p = 0 \text{ into } \color{mediumpurple}{(4)}: $$
\begin{align}
q &= -0 \\
\therefore q &= 0
\end{align}
The answers were $ p = -8, q = 8 $.
Where did I go wrong?
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It's difficult to keep track of all the lines, but you have the correct idea. In one case we have
$$f(2)=5\cdot 2+6\Rightarrow 24+4p+2q+8=16\Rightarrow2p+q=-8$$
and in the other case we have
$$f(1)=5\cdot(1)+6\Rightarrow 3+p+q+8=11\Rightarrow p+q=0$$
From the second equation we get $q=-p$ so by plugging that in into the first equation we get $2p-p=-8$ which gives $p=-8$ and therefore $q=8$.
Since you got $p=-q$, the error must be related to your equation (1), and here it is: after the line
$$24+4p+2q+8=16$$
you've divided by two to get $12+2p+q+4$ on the left, but didn't divide $16$ with $2$ to get $8$ on the right.
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|
Remainder Theorem: $f(x) = x^3+1$ divided by $\operatorname {d}(x)= x$ Take for example $\operatorname {f}(x) = x^3+1$ divided by $\operatorname {d}(x)= x-1$ $$\frac {x^3+1}{x-1}$$
The Remainder Theorem tells us that the remainder will be $$\operatorname {f}(zero-of-\operatorname {d}(x))$$
In this case the remainder should be $$\operatorname {f}(1) = 1^3+1 = 2$$
so the remainder should be 2 for any value of $x$
But for $x=3$
$$f(3) = 3^3+1$$
when this is divided by $x-1$ which is $3-1$ which equals to $2$, $$\frac {3^3+1}{3-1}$$ $$= \frac {27+1}{2}$$ $$= \frac {28}{2}$$ $$= 14, remainder = 0 ≠ 2$$
Please explain me why this happened
|
Leave fractions aside. The remainder theorem tells you that, if $f(x)$ is a polynomial with integer coefficients and $a$ is an integer, then
$$
f(x)=(x-a)q(x)+f(a)
$$
so $f(a)$ is the remainder of the division by $x-a$ in the ring of polynomials with integer coefficients.
What you're conjecturing is that $f(a)$ should also be the remainder of the division of $f(b)$ by $b-a$, whenever $b$ is an integer.
This is not possible, in general. For instance, if $b=a+1$, then would you expect that $f(a)$ is the remainder of the division of $f(a+1)$ by $(a+1)-a=1$? Well, no! The remainder in this case is zero, whatever $f(a+1)$ is.
And what if $f(a)<0$?
Example, similar to yours but with $f(x)=x^3-1$ and $a=-1$: we have $f(-1)=-2$ and indeed
$$
x^3-1=(x+1)(x^2-x+1)-2
$$
With the standard definitions, $-2$ can never be a remainder.
What you can say, and no more, is that
$$
f(b)=(b-a)q(b)+f(a)
$$
and therefore $f(a)\equiv f(b)\pmod{b-a}$
|
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|
Solving limit - $\lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}$ $\lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}$
Since $x$ approaches $0$ and $y$ also approaches $0$ we can suspect that $0<x^2 + y^2<1$. For every $x,y\in\Bbb R$, we have that $\frac{1}{4}(x^2 + y^2)^2\geq x^2y^2$.
Now, $1\geq (x^2+y^2)^{x^2y^2}\geq (x^2+y^2)^{\frac{1}{4}(x^2 + y^2)^2}$, then substitute $(x^2 + y^2)=t$
$1\geq \lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}\geq \lim_{t\to0}t^{\frac{1}{4}t^2}=\lim_{t\to0}e^{\frac{1}{4}t^2\ln t}=e^0=1$
This is how my professor solved this limit. What I don't understand is this part:
$\frac{1}{4}(x^2 + y^2)^2\geq x^2y^2$
How can I prove it? And this would never come to my mind, is there maybe some other way to solve the limit? Grateful in advance.
|
Here's another approach to this limit. Notice for any $(x,y)\neq (0,0)$ that $$
x^2y^2 \ln(x^2+y^2)=f(x,y)(x^2+y^2)\ln(x^2+y^2)$$ where $f(x,y)=\frac{x^2y^2}{x^2+y^2}$. Clearly $(x^2+y^2)\ln(x^2+y^2)\rightarrow 0$ as $(x,y)\rightarrow (0,0)$ while $f(x,y)$ is bounded on the punctured disc $x^2+y^2<1$, $(x,y)\neq (0,0)$. To see this, observe $f(x,0)=0$ for $x\in(-1,0)\cup(0,1)$ and for $x^2+y^2<1,y\neq 0$ we have $$\bigg|\frac{x^2y^2}{x^2+y^2}\Bigg|\leq \Bigg|\frac{x^2y^2}{0+y^2}\Bigg|=x^2\leq x^2+y^2<1$$ This shows $f$ is bounded above by $1$ on the punctured disc, making $$\lim_{(x,y)\rightarrow (0,0)} x^2y^2\ln(x^2+y^2)=0$$ Finally, $$\lim_{(x,y)\rightarrow (0,0)}(x^2+y^2)^{x^2y^2}=\lim_{(x,y)\rightarrow (0,0)}e^{x^2y^2\ln(x^2+y^2)}=e^0=1$$
|
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|
Solve $\cos\theta-3\cos2\theta+\cos3\theta=\sin\theta-3\sin2\theta+\sin3\theta$ My attempt:
\begin{align*}
\cos\theta-3\cos2\theta+\cos3\theta&=\sin\theta-3\sin2\theta+\sin3\theta\\
\cos\theta-3\cos2\theta+4\cos^3\theta-3\cos\theta&=\sin\theta-3\sin2\theta+3\sin\theta-4\sin^3\theta\\
-2\cos\theta-3\cos2\theta+4\cos^3\theta&=4\sin\theta-3\sin2\theta-4\sin^3\theta
\end{align*}
I have faced several symmetric trigonometry problems most of them need to use product to sum identities, but this one I can't continue.
|
$\cos\theta+\cos3\theta = 2\cos\theta\cos2\theta$
$\sin\theta+\sin3\theta = 2\sin2\theta\cos\theta$
So,
$\cos2\theta( 2\cos\theta-3) = \sin2\theta(2\cos\theta -3)$
|
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|
Application of the Cauchy-Schwarz Inequality Need to prove the following:
$(\frac{1}{2}a+\frac{1}{3}b+\frac{1}{6}c)^2\leq \frac{1}{2}a^2+\frac{1}{3}b^2+\frac{1}{6}c^2$
using the Cauchy-Schwarz inequality however direct application yields $(\frac{1}{2}a+\frac{1}{3}b+\frac{1}{6}c)^2\leq \frac{7}{18}(a^2+b^2+c^2)$ which is a dead end.
|
Hint: $\frac 12a+\frac 13b+\frac16c=\frac16a+\frac16a+\frac16a+\frac16b+\frac16b+\frac16c$
|
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|
Solve for closed form $a_j$ given $a_j = 1 + pa_{j+1} + (1-p)a_{j-1}$ where $a_1 = 0$.
Solve for a closed form expression for $a_j$ given constant $p \in (0,1)$ and $q=1-p$ and:
\begin{align*}
a_j &= 1 + pa_{j+1} + qa_{j-1} \\
\end{align*}
Given that $a_1 = 0$.
The text says to use the hint that $a_j = c(1-j)$, but where do we get that hint? If we plug in that hint, we get $a_j = \frac{1-j}{p-q} = \frac{1-j}{2p-1}$ which correctly satisfies the original equation. But how would I solve this without the given hint?
Is this matrix version of the original equation helpful in any way?
\begin{align*}
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ -\frac{1}{p} & \left(1 - \frac{1}{p}\right) & \frac{1}{p} \end{pmatrix} \begin{pmatrix} 1 \\ a_{j-1} \\ a_j \end{pmatrix} &= \begin{pmatrix} 1 \\ a_j \\ a_{j+1} \end{pmatrix} \\
\end{align*}
|
Let $b_j = a_{j+1} - a_j$, $b_j$ satisfies
$$-pb_{j} + qb_{j-1} = 1$$
Let $c_j = b_j - \frac{1}{q-p}$, $c_j$ satisfies
$$c_j = \frac{q}{p}c_{j-1}$$
So
$$c_{j} = c_1\left(\frac{q}{p}\right)^{j-1}$$
We deduce
$$a_{j} = a_1 + \frac{j-1}{q-p} + c_1\sum_{k=0}^{j-2}\left(\frac{q}{p}\right)^{k}$$
with the convention
$$\sum_{k=n}^m A_k = 0,\text{ if } m < n$$
|
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|
Follow-up 'Diophantine' question: extrapolating to the general case I came across a problem involving a certain Diophantine equation a few days ago. I learnt quite a few extremely helpful things about them on this thread here, which I started: A model that can be followed when solving Diophantine equations - ideas? -
Feel free to read it for context for what follows:
Whilst toiling with the concept of Diophantine equations I seemed to notice a surprising pattern.
Any equation of the form:
$$\frac{1}{x}+\frac{2}{y}=\frac{3}{p}$$
(Where x and y are positive integers and p is a prime greater than or equal to 5)
always seems to have exactly 3 solutions.
Is it possible to prove this initial hypothesis/conjecture? Or have I thought incorrectly and can you disprove me? Looking forward to seeing what the responses yield!
|
$$ 3xy - 2px - py = 0$$
$$ 9xy - 6px - 3py = 0$$
$$ 9xy - 6px - 3py + 2p^2 = 2p^2$$
$$ (3x-p)(3y-2p) = 2p^2$$
\begin{array}{|c|c|c|c|c|c|}
\hline
&\color{red}{(3x-p)}\color{blue}{(3y-2p)} & x & y\\ \hline
A&\color{red}{1}\cdot\color{blue}{2p^2} & (p+1)/3 & 2p(p+1)/3 & \color{green}{?}\\ \hline
B&\color{red}{2p^2}\cdot\color{blue}{1} & p(2p+1)/3 & (2p+1)/3 & \color{green}{?}\\ \hline
C&\color{red}{2}\cdot\color{blue}{p^2} & (p+2)/3 & p(p+2)/3 &\color{green}{?}\\ \hline
D&\color{red}{p^2}\cdot\color{blue}{2} & p(p+1)/3 & 2(p+1)/3 &\color{green}{?}\\ \hline
E&\color{red}{p}\cdot\color{blue}{2p} & 2p/3 & 4p/3 &\color{red}{✗} \\ \hline
F&\color{red}{2p}\cdot\color{blue}{p} & p & p &\color{green}{\checkmark} \\ \hline
\end{array}
From the table we come across two cases :
*
*When $p+1 \equiv 0 \pmod 3$, $A$, $D$ and $F$ each produce one solution.
E.g. for $p=11$, $$ {\dfrac{1}{4}+\dfrac{2}{88}=\dfrac{3}{11},\dfrac{1}{44}+\dfrac{2}{8}=\dfrac{3}{11},\dfrac{1}{11}+\dfrac{2}{11}=\dfrac{3}{11}}$$
*
*When $p+2 \equiv 0 \pmod 3$, $B$, $C$ and $F$ each produce one solution.
E.g. for $p=7$, $$ {\dfrac{1}{35} +\dfrac{2}{5}=\dfrac{3}{7},{\dfrac{1}{3}+\dfrac{2}{21}=\dfrac{3}{7}},\dfrac{1}{7}+\dfrac{2}{7}=\dfrac{3}{7}}$$
Note that $2p+1 \equiv 0 \equiv 2p+1+3 \equiv p+2 \pmod 3$.
Thus the conjecture that $1/x + 2/y = 3/p$ has exactly three solutions in positive integers $x$, $y$ for prime $p \ge 5$ is correct!
Thanks to @Servaes for pointing out the critical mistake! :)
|
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|
Finding expectation of minimum of $(X,Y)$ where $(X,Y)$ is bivariate normal distribution. Let $(X,Z)$ be bivariate normal with parameters
$\mu_X := E(X) = 1, \mu_Z := E(Z ) = 1, \sigma_X^2 := Var(X) = 1$,
$ \sigma_Z^2
:= Var(Z ) = 1$,
and the correlation coefficient of (X, Z ) is $\rho$ with density
$$f(x,z)=\dfrac{1}{2\pi\sqrt{1-\rho^2}}\exp(-\dfrac{1}{2(1-\rho^2)}((x-1)^2-2\rho (x-1)(z-1)+(z-1)^2)).$$
Let $Y=\min(X,Z)$. Then what is $E[Y]$
For the above what I have so far is
$\textbf{MY Attempt}$
we first transform the variables so that $X=1+X_1$ and $Z=1+X_2$. Then
\begin{align*}
\mathbb{P}\left\{\min(X_1,X_2) \leq y\right\} &= 1 - \mathbb{P}\left\{ X_1 > y, X_2 > y \right\} \\
&= 1 - \int_{y}^{\infty} \int_y^{\infty} f_{X_1,X_2}(s,t)dsdt \\
&= 1 -\int_y^{\infty} f_{X_2}(t)\int_y^{\infty} f_{X_1|X_2}(s,t) ds dt
\end{align*}
where
$$
f_{X_2} (t)=\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}, \qquad f_{X_1|X_2}(s,t)=\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(s-\rho t)^2}{2(1-\rho^2)}},
$$
and so
\begin{align*}
\mathbb{P}\left\{\min(X_1,X_2) \leq y\right\} &= 1 - \int_{y}^{\infty} \varphi(t) \left(1 - \Phi\left( \frac{y -\rho t}{\sqrt{1-\rho^2}}\right) \right) dt. \\
&= 1 - \int_{y}^{\infty} \varphi(t) \Phi\left( \frac{\rho t - y }{\sqrt{1-\rho^2}} \right) dt
\end{align*}
To get the density we differentiate with respect to $y$ giving
\begin{align*}
f_{Y}(y) &= -\frac{\partial}{\partial y}\int_{y}^{\infty}\varphi(t)\Phi\left(\frac{\rho t - y}{\sqrt{1-\rho^2}}\right) dt \\
&= \varphi(y)\Phi\left(\frac{\rho y - y}{\sqrt{1-\rho^2}}\right) + \int_{y}^{\infty}\varphi(t) \frac{1}{\sqrt{1-\rho^2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{(\rho t - y)^2}{2(1-\rho^2)}}dt \tag{1}
\end{align*}
Completing the square of the last term in $(1)$ we have
\begin{align*}
\varphi(t)\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(\rho t-y)^2}{2(1-\rho^2)}} &= \frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{1}{2(1-\rho^2)}\left((1-\rho^2)t^2 +(\rho t -y)^2 \right)} \\
&=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{1}{2(1-\rho^2)}\left(t^2 - 2 t \rho y + \rho^2 y^2 + (1-\rho^2)y^2 \right)} \\
&= \frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(t-\rho y)^2}{2(1-\rho^2)}}.
\end{align*}
So putting this back in to $(1)$ we get
\begin{align*}
f_{Y}(y) &= \varphi(y)\Phi\left(\frac{\rho y -y}{\sqrt{1-\rho^2}}\right) + \varphi(y)\int_y^{\infty} \frac{1}{\sqrt{2\pi(1-\rho^2)}}e^{-\frac{(t-\rho y)^2}{2(1-\rho^2)}}dt \\
&= \varphi(y)\Phi\left(\frac{\rho y -y}{\sqrt{1-\rho^2}}\right) + \varphi(y)\left( 1 - \Phi\left(\frac{y - \rho y}{\sqrt{1-\rho^2}}\right)\right) \\
&= 2\varphi(y)\Phi\left(\frac{\rho y -y}{\sqrt{1-\rho^2}}\right).
\end{align*}
$\textbf{Hence I think I have the density of random variable Y. From here its difficult for me to find the expectation of Y.}$
|
You can use the following
$$\min(X,Z) = \frac{(X+Z)}{2}-\frac{|X-Z|}{2} $$
the distribution of $W =(X -Z)$ is normal with mean $0$ and variance $2(1-\rho)$
Now this is a random variable and we can take out the distribution of $|W|$ as
$$\mathbb{P}(|W|\leq t) = \mathbb{P}(W\leq t)-\mathbb{P}( W \leq -t)=F_W(t) - F_W(-t)$$
Hence we will have $$f_{|W|}(t) = f_{W}(t)+f_{W}(-t) = 2f_{W}(t) $$
there we can get the $\mathbb{E}(|W|) = \frac{2}{\sigma\sqrt{2\pi}}\displaystyle\int_0^{\infty}z\exp{ \frac{-z^2}{2\sigma^2} }dz = \sigma\sqrt{\frac{2}{\pi}}$
Hence we will have the answer using $\mathbb{E}\Big[\min(X,Z)\Big] = .5\mathbb{E}\Big[X\Big]+.5\mathbb{E}\Big[Z\Big]-.5\mathbb{E}\Big[|W|\Big]$.
|
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|
General tribonacci explicit formula I have been thinking about finding an explicit formula for the tribonacci numbers, where, namely:
$$a_n = a_{n-1}+a_{n-2}+a_{n-3}$$
and $a_1 = 0, a_1 = 1, a_2 = 1.$ Obviously, these beginning terms can be shifted, but we'll leave them as such for now.
This has proven difficult, and I'm still not sure how it's done, but what about the general sequence:
$$a_n = xa_{n-1}+ya_{n-2}+za_{n-3}$$
With arbitrary $a_1, a_2,$ and $a_3.$ How is a tribonacci explicit formula calculated?
Cheers.
|
Please see https://www.fq.math.ca/Scanned/36-2/wolfram.pdf in The Fibonacci Quarterly, sub-section 6.2.2. That approach also considers having initial functions instead of any given initial values.
Changing the constant coefficients to $x$, $y$ and $z$ from 1 can be solved similarly.
Specifically, we are seeking a solution for $f$ where $f: \Re \rightarrow \Im$ which has the property
\begin{equation} \label{rec-eq}
f(x) = \sum_{1 \leq l \leq 3}~f(x - l)
\end{equation} where $f(0)$, $f(1)$ and $f(2)$ are given initial values.
The real root of the associated characteristic equation $x^3 - x^2 - x - 1 = 0$ is
$r_1 = \frac{1 + (19 - 3\sqrt{33})^{\frac{1}{3}} + (19 + 3\sqrt{33})^{\frac{1}{3}}}{3}$. The approximate values of the complex roots
are $-0.419643377607081 \pm 0.606290729207199 i$.
A general solution is
$f(x) = K_1 r_1^x + K_2 r_1^{\frac{-x}{2}}\cos (\theta x) +
r_1^{\frac{-x}{2}} \left( \frac{K_2\left(\frac{1 -r_1}{2}\right) +
K_3}{\sqrt{\frac{1}{r_1} - \left(\frac{r_1 - 1}{2}\right)^2}} \right)
\sin (\theta x)$ where $\theta =\arccos \left(\frac{1 - r_1}{2}
\sqrt{r_1}\right)$ and $K_1$, $K_2$ and $K_3$ are solutions to the system
$\left[ \begin{array}{ccc}
1 & 1 & 0\\
r_1 -1 & -r_1 & 1\\
\frac{1}{r_1} & 0 & -r_1
\end{array}\right]
\left[\begin{array}{c}
K_1\\ K_2\\ K_3
\end{array}\right] =
\left[\begin{array}{c}
f(0)\\ f(1) - f(0)\\ f(2) - f(1) - f(0)
\end{array}\right].
$
|
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|
Find the value of $x^5 + \frac{1}{x^5}$ - question about correctness of method The task is: if $x+ \frac{1}{x}= 1$ find $x^5 + \frac{1}{x^5} $.
I used the binomial formula and proved that $x^5 + \frac{1}{x^5} = 1$, but I have a question about following method, I am not sure if it's correct. If I take square of the the first equality, I get:
$x^2 +2 + \frac{1}{x^2} = 1$ so $x^2 +\frac{1}{x^2} = -1$. Now, the sum of 2 squares is nonnegative and the right side is negative, so when I come to this part does it mean that this method is wrong?
In general, when proving such equalities, when are we allowed to take square (and we don't know if one side of equality is positive as in this task)? Thanks in advance.
|
$x+\frac{1}{x} =1$
$\Rightarrow $ $ (x+\frac{1}{x}) ^5=1$
$\Rightarrow $ $x^5+(\frac{1}{x})^5 +5(x^3+(\frac{1}{x})^3)+10
(x+\frac{1}{x})=1$
$\Rightarrow $ $x^5+(\frac{1}{x})^5=-9-5(x^3+(\frac{1}{x})^3)(*)$
Let's find :$ x^3+(\frac{1}{x})^3$
$ (x+(\frac{1}{x}))^3 =1$
$\Rightarrow $ $x^3+(\frac{1}{x})^3+3(x+\frac{1}{x})=1$
$\Rightarrow $ $x^3+(\frac{1}{x})^3=-2(**)$
After$(*,**)$ we can see that :
$x^5+(\frac{1}{x})^5=-9+10=1$
|
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|
Maximising volume of a cylinder when surface area fixed I know how to start off this problem but get bogged down when it comes to differentiating at the end.
A right circular cylinder is of radius r cm. and height pr cm. The total surface area of the cylinder is $S cm^2$ and its volume is $V cm^3$.
Find an expression for V in terms of p and S. If the value of S is fixed, find the value of p for which V is a maximum.
I have said:
V = $\pi r^2pr = \pi r^3p$
$S = 2\pi r^2p + 2\pi r^2 = 2\pi r^2(p + 1)$ (I am assuming a closed cylinder)
$r = 2\pi r^2(1 + p)$
So $r = \sqrt(\frac{S}{2\pi (1 + p)})$
Substituting into the the formula for volume:
$V = \pi p(\frac{S}{2\pi (1 + p)})^{3/2}$
But when I try to differentiate this to find the maximum value I get confused. The book says the answer is 2.
My working out, as far as it goes is as follows:
$V = \pi p(\frac{S}{2\pi (1 + p)})^{3/2}$
$dV/dp = \pi p(\frac{3}{2\pi(1 + p)})^{1/2}.\frac{-S2\pi}{4\pi^2(1 + p)^2} + \pi(\frac{S}{2\pi(1 + p)})^{3/2}$
Which = 0 when:
$3\pi^2pS(\sqrt(\frac{S}{2\pi(1 + p)})) = \pi(\sqrt(\frac{S}{2\pi(1 + p)}))^3$
But after this I get confused.
|
The first term of your derivative is wrong.$$\begin{align*}&\frac{dV}{dp}=\frac d{dp}\left[\frac{S\sqrt S}{2\sqrt{2\pi}}\frac p{(1+p)^{3/2}}\right]=\frac{S\sqrt S}{2\sqrt{2\pi}}\left[\frac1{(1+p)^{3/2}}-\frac{3p/2}{(1+p)^{5/2}}\right]\\&\frac{dV}{dp}=\frac{S\sqrt S}{4\sqrt{2\pi}}\left[\frac{2-p}{(1+p)^{5/2}}\right]\end{align*}$$which is $0$ when $p=2$. Verify that the derivative is negative for $p>2$ and positive for $p<2$.
|
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|
Find positive definite $X$ such that $AXA^T=\alpha X$. Given an invertible real matrix $A\in\mathbb{R}^n$, is it possible to obtain a (real) positive definite matrix $X$ (and scalar $\alpha>0$) such that
$$
AXA^T=\alpha X?
$$
Attempt: The previous equation can be written as $AXA^T-X +Q=0$ with $Q=(1-\alpha)X$. So, if $(1-\alpha)>0$ this looks like the Lyapunov equation here. However, the analytic solution to the Lypunov equation has the form
$$
X = \sum_{k=0}^\infty A^kQ(A^T)^k = (1-\alpha)\sum_{k=0}^\infty A^kX(A^T)^k
$$
So, even if such equation have a solution, I don't have a clue on how to obtain it from here.
Do you have other thoughts/suggestions?
|
Consider $A = \pmatrix{2 & 1 \\ 0 & 1}$. Then you seek a matrix
$$
X = \pmatrix{a & b \\ c & d}
$$
and a constant $K$ such that
$$
\pmatrix{2 & 1 \\ 0 & 1}\pmatrix{a & b \\ c & d}\pmatrix{2 & 0 \\ 1 & 1} = K \pmatrix{a & b \\ c & d}
$$
i.e.,
$$
\pmatrix{2a+c & 2b+d \\ c & d}\pmatrix{2 & 0 \\ 1 & 1} = K \pmatrix{a & b \\ c & d}
$$
so
$$
\pmatrix{4a+2c+2b+d & 2b+d \\ 2c+d & d} = K \pmatrix{a & b \\ c & d}
$$
Case 1: $d \ne 0$. Then $K = 1$, and we have
$$
\pmatrix{4a+2c+2b+d & 2b+d \\ 2c+d & d} = \pmatrix{a & b \\ c & d}
$$
whence $2b+d = b$ so $b = -d$, and similarly $c = -d$; thus our equation becomes
$$
\pmatrix{4a+2(-d)+2(-d)+d & -d \\ -d & d} = \pmatrix{a & -d \\ -d & d}\\
\pmatrix{4a-3d & -d \\ -d & d} = \pmatrix{a & -d \\ -d & d}
$$
so $4a - 3d = a$, hence $a = d$, and we have
$$
X = \pmatrix{d & -d \\ -d & d}
$$
which is not positive definite, because its determinant is zero.
Case 2: $d = 0$. Then we want
$$
\pmatrix{4a+2c+2b & 2b \\ 2c & 0} = K \pmatrix{a & b \\ c & 0}
$$
For $X$ to have full rank, we need $c \ne 0$, so $K = 2$, and we can write
$$
\pmatrix{4a+2c+2b & 2b \\ 2c & 0} = \pmatrix{2a & 2b \\ 2c & 0}
$$
which reduces to $4a + 2c + 2b = 2a$, so $a + b + c = 0$. That gives
$$
X = \pmatrix{-(b+c) & b \\ c & 0}
$$
or, replacing $b$ and $c$ with their negatives,
$$
X = \pmatrix{(b+c) & -b \\ -c & 0}
$$
The determinant here is $bc$, so $b$ and $c$ must have the same sign for $X$ to be positive definite; the trace is $b+c$, so that sign must be "+". But then the quadratic formula gives the eigenvalues as
$$
x = \frac{(b+c) \pm \sqrt{ (b+c)^2 + 4bc}}{2}
$$
and because $(b+c)^2 + 4bc > (b+c)^2$, one of the two choices in the numerator leads to a negative eigenvalue, so $X$ is not positive definite.
In short: for the matrix $A$ given above, there is no solution $X$.
|
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|
An integer $n \geq 2$ is called square-positive- proof? An integer $n \geq 2$ is called square-positive if there are $n$ consecutive positive integers whose sum is a square. Determine the first four square-positive integers.
So I have found the first four square-positive numbers, but I need to prove that why it $4$ is not a square-positive number and I also need to write a general formula for determining whether a number is square-positive or not. I have tried to write the sum of consecutive positive integers like this $a + a +1 + a + 2 + a+3 \dots a - 1$ and I wrote it like this for all numbers, and part of the proof for why $4$ isn't a square-positive number is that $4a + 6$ is not divisible with $4$. But I haven't got so far.
Here is my answer:
2 : 4 + 5 = 9 which is 3^2
3 : 2 + 3 + 4 = 9 which is 3^2
5: 18 + 19 + 20 + 21 + 22 = 100 which is 10^2
6: 35 + 36 + 37 + 38 + 39 + 40 = 225 which is 15^2
Interesting fact is that for all odd numbers and some even numbers like 6 and 10, you can find out which number is the first (the one you start with and then go forward here like 3, 2, 18 and 35) using this formula :
(I show it in an example because I still can't write it algebraically):
For example: the sum of 95 subsequent numbers is 5n + 10
(10^2 - 10) /5 = 18
So your first number is 18
And if you keep adding, 18 + 19 + 20 + 21 + 22 you get 100 which is 10^2, the same number you squared.
|
We prove the following result:
If $n=2^b \cdot d, d \text{ is odd, then } n \text{ is square-positive if and only if } b=0 \text{ or } b \text{ is odd}$.
$\underline{\text{Case 1}}$: $b$ is even and $b>0$. Let $b=2c$. We prove by contradiction $n$ cannot be square-positive. If it were, then there exist $n$ consecutive positive integers,
$$
a, a+1, \ldots, a+n-1
$$
such that the sum of these numbers is a square, i.e.,
$$
S=a+(a+1)+\cdots + (a+n-1) = n\frac{n+2a-1}{2} =T^2
$$
for some $T\in \mathbb{N}$.
Note that $n=2^{2c}\cdot d$ is even, and
$$
T^2 = S = 2^{2c-1}\cdot d \cdot (n+2a-1),
$$
which cannot be a perfect square because 1) $d(n+2a-1)$ is odd and, 2) $2c-1$, the highest power of its factor $2$ is odd, a contradiction.
$\underline{\text{Case 2}}: b=0$. Let $n=d=2m+1$, then the sum of the following $n$ consecutive numbers is a perfect square:
$$
n-m, n-m+1, \ldots, n+m.
$$
Indeed,
$$
S = (n-m)+(n-m+1)+\cdots +(n+m) = n \cdot \frac{(n-m)+(n+m)}{2}=n^2.
$$
$\underline{\text{Case 3}}: b=2c+1 $ for some $c\geqslant 0$, then $n=2^{2c+1}d$. We show that the sum of the following $n$ consecutive numbers is a perfect square:
$$
a, a+1, \ldots, a+(n-1),
$$
where
$$
a=\frac{1+d(K^2-2^{2c+1})}{2}
$$
and $K$ is some odd positive integer such that
$$
K^2 > 2^{2c+1}.
$$
For example, if $c\geqslant 1$, we can pick $K=2^{c+1}-1$. If $c=0$, we can pick $K=3$. Also note that $a$, the first number in the sequence, is a positive integer by construction.
Finally,
$$
S=n\frac{n+2a-1}{2} = \frac{2^{2c+1}d}{2} \left[ 2^{2c+1}d + d(K^2-2^{2c+1}) \right] = 2^{2c} d [ d K^2 ] = (2^c d K)^2. \blacksquare
$$
Example: $n=8=2^3 \cdot 1, c=1, d=1, K=3 \Rightarrow a=1$, and the sum of the sequence is $1+2+\cdots+8=36=6^2$.
|
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|
Prove by induction that $\frac{1}{2n}\leq\frac{1\text{·}3\cdot5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}$ What would be the right way to solve this by induction proof?
$$\frac{1}{2n}\leq\frac{1\text{·}3\cdot5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}$$
This is what I've done (reference https://www.slader.com/discussion/question/prove-that-12n-1-3-5-2n-12-4-2n-whenever-n-is-a-positive-integer/#):
*
*Show that $S\left(n+1\right)$ by induction proof. This is
$$\frac{1}{2(n+1)}\leq\frac{1\text{·}3\text{·}5\text{·}\ldots\text{·}(2n+1)}{2+4+6+\ldots+2(n+2)}$$
Multiplying both sides of the equation $\frac{2n\text{·}(2(n+1)-1)}{2n\text{·}(2(n+1)-1)}=\frac{2n\text{·}(2n+1)}{2n\text{·}(2n+1)}$
$$\frac{1}{2n}\leq\frac{1\text{·}3\text{·}5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}\times\frac{2n\text{·}(2(n+1)-1}{2n\text{·}(2(n+1)-1}$$
$$\frac{1}{2n}\times\frac{2n\text{·}(2n+1)}{2n\text{·}(2n+1)}$$
Rewriting we have the following
\begin{array}{c}
\frac{2n+1}{2n(2n+1)}\\
\frac{1}{2n+1}+\frac{1}{2n(2n+1)}
\end{array}
$$\frac{1}{2n}\leq\frac{1}{2n+1}+\frac{1}{2n(2n+1)}$$
|
Let $\varphi, \psi:\mathbb N\to\mathbb N$ given by
\begin{aligned}
\varphi(n) &= \sum_{k=1}^n 2k = n(n+1)\\
\psi(n) &= \prod_{k=1}^n (2k-1)\\
\end{aligned}
and notice that
\begin{aligned}
\varphi(n+1) &= \frac{n+2}n\varphi(n)\\
\psi(n+1) &= (2n+1)\psi(n)\\
\end{aligned}
If you assume that, for a certain $n\in\mathbb N$, the following is true
$$\frac 1{2n}\le \frac{1\times 3\times\ldots\times (2n-1)}{2 + 4 + \ldots + 2n} = \frac{\psi(n)}{\varphi(n)},$$
then you may write
\begin{aligned}
\frac 1{2(n+1)}
& = \frac{2n}{2(n+1)}\frac1{2n}\\
& \le \frac{2n}{2(n+1)}\frac{\psi(n)}{\varphi(n)}\\
& = \frac{2n}{2(n+1)}\frac1{2n+1}\frac{n+2}{n}\frac{\psi(n+1)}{\varphi(n+1)}\\
& = \frac{n+2}{(n+1)(2n+1)}\frac{\psi(n+1)}{\varphi(n+1)}\\
& \le \frac{\psi(n+1)}{\varphi(n+1)},
\end{aligned}
where the last inequality follows from the fact that
$$
\frac{n+2}{(n+1)(2n+1)} = \frac{1}{2n+1}\left(1 + \frac1{n+1}\right) \le \frac 12\left(1 + \frac 12\right) = 1.
$$
|
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|
Why using fewer terms of Taylor series doesn't give $0/0$ but gives a wrong answer? I was reading a Calculus book and I saw this problem which looks easy:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} = ?$$
It's a 0/0 limit and it's using some of the Taylor series of $\sin$ and $\cos$ expressions to solve the problem.
I know that the First and Second way should be correct because it's using more expressions of the Taylor series around 0.
What I can't figure out is WHY using fewer expressions of the Taylor series in the Third way doesn't give 0/0 but gives a wrong answer?
First way:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x-2 \sin x \cos x}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x(x-\sin x)}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x\left(x-x+\frac{x^3}{6}\right)}{x^3}=\lim _{x \rightarrow 0} \frac{2 \cos x\left(\frac{x^3}{6}\right)}{x^3}=\frac{1}{3}$$
Second way:
$$\lim _{x \rightarrow 0} \frac{2x(1-\frac{x^2}{2})-(2x-\frac{8x^3}{6})}{x^3}=\lim _{x \rightarrow 0} \frac{2x-x^3-2x+\frac{8x^3}{6}}{x^3}=\lim _{x \rightarrow 0} \frac{\frac{x^3}{3}}{x^3}=\frac{1}{3}$$
Third way:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} =\lim _{x \rightarrow 0} \frac{2 x \cos x-2x}{x^3}=\lim _{x \rightarrow 0} \frac{2x(\cos x -1)}{x^3}=\lim _{x \rightarrow 0} \frac{2x(-\frac{x^2}{2})}{x^3}=-1$$
|
In all the cases we should use remainder to proceed properly as follows
$$\frac{2 x \cos x- \sin 2x}{x^3}=\frac{2 x \left(1-\frac12 x^2+O(x^3)\right)- \left(2x-\frac16 (2x)^3+O(x^4)\right)}{x^3}=$$
$$=\frac{2x-x^3-2x+\frac43x^3+O(x^4)}{x^3}=\frac13+O(x) \to \frac13$$
without remainder we can easily get wrong with the solution.
|
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|
Finding $\lim_{x \to 0} \left( \frac{\sin(x)-3\sinh(x)+2x}{x^2(\tanh(2x)+ \sin(x))} \right)$ I am trying to evaluate the following limit:
$$L=\lim_{x \to 0} \left( \frac{\sin(x)-3\sinh(x)+2x}{x^2(\tanh(2x)+\sin(x))} \right)$$
Begin by rewriting the limit as:
$$L=\frac{\lim\limits_{x \to 0}\left(\cfrac{\sin(x)-3\sinh(x)+2x}{x^2} \right)}{\lim\limits_{x \to 0}(\tanh(2x)+\sin(x))} \tag{1}$$
Applying L'Hospital's Rule to the numerator only:
$$L=\frac{\lim\limits_{x \to 0}\left(\cfrac{\cos(x)-3\cosh(x)+2}{2x} \right)}{\lim\limits_{x \to 0}(\tanh(2x)+\sin(x))} \tag{2}$$
The numerator is still in an indeterminate form, applying L'Hopital to the numerator again:
$$L=\frac{\lim\limits_{x \to 0}\left(\cfrac{-\sin(x)-3\sinh(x)}{2} \right)}{\lim\limits_{x \to 0}(\tanh(2x)+\sin(x))} \tag{3}$$
Rewriting as a single limit:
$$L=-\frac{1}{2}\lim_{x \to 0}\frac{\sin(x)+3\sinh(x)}{\tanh(2x)+\sin(x)} \tag{4}$$
And applying L'Hospital's Rule...
$$L=-\frac{1}{2}\lim_{x \to 0}\left(\frac{\cos(x)+3\cosh(x)}{2\operatorname{sech}^2(2x)+\cos(x)} \right)=-\frac{2}{3} \tag{5}$$
But according to Wolfram Alpha, $L=-\frac{2}{9}$
So something must be wrong in my calculation (I guess it's the limit of a product bit)?
|
$$\sinh x=x+\frac{x^3}{6}+O\left(x^4\right)$$
$$\tanh(2x)=2 x-\frac{8 x^3}{3}+O\left(x^4\right)$$
$$\sin x=x-\frac{x^3}{6}+O\left(x^4\right)$$
The limit can be rewritten as
$$\frac{x-\frac{x^3}{6}-3 \left(\frac{x^3}{6}+x\right)+2 x}{x^2 \left(2x-\frac{8 x^3}{3}+x-\frac{x^3}{6}\right)}\to -\frac{2}{9}\text{ as }x\to 0$$
|
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|
$(ab+bc+ca)^3=abc(a+b+c)^3$, prove that $a,b,c$ are in $G.P.$ Suppose $a,b,c$ are non-zero real numbers such that
$$(ab+bc+ca)^3=abc.(a+b+c)^3$$
Prove that $a,b,c$ must be terms of a $G.P.$
I simplified this equation too
$$(ab)^3+(bc)^3+(ca)^3=abc.(a^3+b^3+c^3)$$
I tried to subtract $3(abc)^2$ from both sides and it gave a factorised form.
$$(ab+bc+ca).((ab)^2+(bc)^2+(ca)^2-abc.(a+b+c))=abc.(a+b+c).(a^2+b^2+c^2-ab-bc-ca)$$
How can I proceed after that?
|
This one is a long-way of solving (With a very obvious result).
\begin{gather*}
\implies(ab+bc+ca)^3-abc(a+b+c)^3=0\\
\end{gather*}
\begin{multline*}
\implies (a^3b^3+b^3c^3+c^3a^3+3a^2b^3c+3a^2bc^3+3a^3b^2c+3ab^2c^3+3a^3bc^2+3ab^3c^2)-\\abc(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3ca^2)=0
\end{multline*}
After simplifying
\begin{equation}
\implies a^3b^3+b^3c^3+c^3a^3-a^4bc-ab^4c-abc^4=0
\end{equation}
Factorising this will give
\begin{equation}
\implies(ab-c^2)(bc-a^2)(ca-b^2)=0
\end{equation}
Therefore, a,b,c must be in G.P.
|
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|
Why is $\sqrt{n} + \sqrt{n + 1} > 2\sqrt{n}$? I am trying to prove an inequality, but that itself is not the question here. Looking around for ideas how to proceed I found solutions for the same problem.
But in several occasions I couldn't follow the answers because of this step.
In both of these excerpts (taken from answered questions on this site, the links are below) two different square roots are combined like this.
\begin{align}
\sqrt{n} + \sqrt{n + 1} > 2\sqrt{n}\tag{Now it's an inequality}
\end{align}
Why is this possible even though the square roots are not identical?
Here the excerpts:
\begin{align}
= & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{\sqrt{n + 2} + \sqrt{n + 1}}\right] \quad \text{multiply conjugate}\\
> & 2 \sqrt{n + 2} - 2 + \left[\frac{1}{\sqrt{n + 1}} - 2\frac{1}{2\sqrt{n + 1}}\right] \\
\end{align}
taken from here
or
\begin{align}
\sqrt{k+1}-\sqrt{k}&=\frac{1}{\sqrt{k+1}+\sqrt{k}}\\
&\le\frac{1}{2\sqrt{k}}
\end{align}
taken from here
I hope someone can explain this to me or point me in the right direction
|
We have that
$$\sqrt{n} + \sqrt{n + 1} >\sqrt{n}+\sqrt{n}= 2\sqrt{n}$$
since $\sqrt{n + 1}>\sqrt{n}$ or also dividing by $\sqrt n$
$$\sqrt{n} + \sqrt{n + 1}> 2\sqrt{n} \iff 1+\sqrt{1+\frac1n}>2$$
which is true since $\sqrt{1+\frac1n}>1$.
|
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|
Prove that $n\leq a_1+a_2+...+a_n \leq n+1$ If $a_1=2,a_{n+1}=\sqrt{a_n+8}-\sqrt{a_n+3}$. Prove that $n\leq a_1+a_2+...+a_n \leq n+1$ for every $n\ge1$ and $\lim a_n=1$.
I have showed that by squaring and inequality techniques:
*
*$a_i<\sqrt{3}$ for every $i>1$.
*If $a_i>1$ then $a_{i+1}<1$ for every $i\ge 1$
I think that $\sqrt{3}$ can be improved, but I am not sure if it's useful.
|
We have
$
a_{n+1}-1 = (a_n-1) \left( \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right)
$
and
$(a_{n+1}-1)+(a_n-1) = (a_n-1) \left( 1+ \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right)$,
a) $ \left| \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right| < \left| \cfrac{1}{3+\sqrt{8+a_n}} + \cfrac{1}{2+\sqrt{3+a_n}} \right| < \left| \cfrac{1}{3+\sqrt{8+0}} + \cfrac{1}{2+\sqrt{3+0}} \right| = 5-(\sqrt 3 + \sqrt8) < 1 $
This way we have $|a_{n+1}-1| < q |a_n-1| < q^n |a_1-1|$ by induction. This way, $|a_{n+1}-1|$ tends to 0.
b) We have to show some results:
*
*$ \left( \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right) < 0$, therefore $a_{n+1}-1$ and $a_n-1$ have opposite signals. But $a_1-1=1>0$, then the even guys are negative, the odd ones are positive: $a_{2k} < 1 < a_{2k+1}$.
*$ \left( 1+ \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right) > 0$, therefore $a_{n+1}+a_n-2$ and $a_n-1$ have the same signals: $a_{2k}+a_{2k+1} < 2 < a_{2k+1}+a_{2k+2}$
Now we have two cases to consider:
*
*$ (a_1+a_2)+\ldots+(a_{2k+1}+a_{2k+2}) > 2+\ldots+2=2k+2$ and $ a_1+(a_2+a_3)+\ldots+(a_{2k}+a_{2k+1})+a_{2k+2} < 2+\ldots+2+1=2k+3$
*$ (a_1+a_2)+\ldots+(a_{2k+1}) > 2+\ldots+1=2k+1$ and $ a_1+(a_2+a_3)+\ldots+(a_{2k}+a_{2k+1}) < 2+\ldots+2+2=2k+2$
|
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|
Which values of $a$ and $b$ will give $x^{2}+2$ as a factor of $x^{17}+ax+b$ over $\mathbb{Z}_{3}$? The question:
Let $x^{17}+ax+b$ be a polynomial over $\mathbb{Z}_{3}$. For which values of $a$ and $b$ will $x^{2}+2$ be a factor?
I know that I can find a solution by a brute force technique, but is there a more efficient way to solve this problem? Thanks.
|
If $x^2 + 2$ is a factor of
$p(x) = x^{17} + ax + b, \tag 0$
any zero of $x^2 + 2$ must also satisfy $p(x)$;. the zeroes of $x^2 + 2$ in $\Bbb Z_3$ are $1$ and $2$:
$1^2 + 2 = 3 \equiv 0 \mod 3, \tag 1$
$2^2+ 2 = 4 + 2 = 6 \equiv 0 \mod 3; \tag 2$
indeed
$(x - 1)(x - 2) = x^2 - 3x + 2 = x^2 + 2 \mod 3; \tag 3$
if $1$ and $2$ are roots of $p(x)$, then
$1 + a + b = 1^{17} + a \cdot 1 + b = p(1) = 0 \mod 3, \tag 4$
$2^{17} + a \cdot 2 + b = p(2) = 0 \mod 3; \tag 5$
now,
$2^{17} = 2(2^4)^4 = 2(16)^4 = 2(1)^4 = 2 \mod 3; \tag 6$
thus (5) becomes
$2 + 2a + b = p(1) = 0; \mod 3; \tag 7$
thus (4) and (7) form the system
$a + b = -1 = 2 \mod 3, \tag 8$
$2a + b = -2 = 1 \mod 3; \tag 9$
we subtract (8) from (9):
$a = -1 = 2 \mod 3, \tag{10}$
and then substitute this into (8):
$b + 2 = 2 \mod 3, \tag{11}$
whence
$b = 0 \mod 3; \tag{12}$
therefore
$p(x) = x^{17} + 2x. \tag{13}$
Note that
$p(1) = 1 + 2 \cdot 1 = 3 = 0 \mod 3 \checkmark \tag{14}$
and
$p(2) = 2^{17} + 2 \cdot 2 = 2 + 4 = 6 = 0 \mod 3 \checkmark \tag{15}$
|
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|
Volume of solid rotated about the x-axis I am to find the volume of the area $R$ bounded by the curve $x=y^2+2$, $y=x-4$ and $y=0$.
I have already found the points of intersection by first setting the lines equal to each other and used the quadratic formula:
\begin{align*}
y^2+2=y+4
\\-y^2+y+2=0
\\
\\y_{1,2}=\frac{1\pm3}{2}\\
\\y_1 = 2 \\y_2=-1
\end{align*}
With regard to y:
\begin{align*}
A=\int_0^2{(y+4)-(y^2+2)}dy\\
A=\left[(2\cdot2)+\frac{2^2}{2}-\frac{2^3}{3}\right]-\left[(2\cdot0)-\frac{0^2}{2}-\frac{0^3}{3}\right]\\
A=4+2-\frac{8}{3}\\
A=\frac{10}{3}
\end{align*}
Then I tried finding the volume of $R$:
We have:
$f(y)=2+y-y^2 dy $
\begin{align*}
V= 2\pi\int_a^byf(y) dy\\
=2\pi\left(\int_2^4 y^3+2ydy\right)+2\pi\left(\int_4^6(y^3+2y)-(y^2+4y)dy\right)\\
=4\pi\left(\left[\frac{y^4}{4}+y^2\right]_2^4 + \left[\left(\frac{y^4}{4}+y^2\right)-\left(\frac{y^3}{3}+2y^2\right)\right]_4^6\right)\\
=4\pi \cdot 290\\
= \frac{\pi}{2}(145)
\end{align*}
But the right answer should be $16\frac{\pi}{3}$. What am I doing wrong?
|
Assuming $R$ is the region in the first quadrant (i.e. with $x\ge0$ and $y\ge0$), then the volume - using cylindrical shells - is
$$\begin{align}
V&=2\pi\int_0^2y((y+4)-(y^2+2))\,\mathrm dy\\[1ex]
&=2\pi\int_0^2(2y+y^2-y^3)\,\mathrm dy\\[1ex]
&=2\pi\left[y^2+\frac{y^3}3-\frac{y^4}4\right]_0^2\\[1ex]
&=2\pi\left(2^2+\frac{2^3}3-\frac{2^4}4\right)=\boxed{\frac{16\pi}3}
\end{align}$$
Your error is in the limits of integration. You seem to be using the bounds for the variable $x\in[2,6]$, not $y\in[0,2]$.
If you did want to set up an integral using the bounds for $x$, you can use disks and washers:
$$\begin{align}
V&=\pi\left(\int_2^4\left(\sqrt{x-2}\right)^2\,\mathrm dx+\int_4^6\left(\left(\sqrt{x-2}\right)^2-(x-4)^2\right)\,\mathrm dx\right)
\end{align}$$
and you would find this to have the same value.
|
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|
How to solve this equation with matrices can you please give me some hints to solve the following? I really don't know how to start.
$$X^2= \begin{pmatrix}
6 & 2 \\ 3 & 7
\end{pmatrix}.$$
I tried to express this matrix as $4\cdot I + \begin{pmatrix}
2 & 2 \\ 3 & 3
\end{pmatrix}$ And somehow solve it, but I really have no clue. Please some help.
|
Let $$X=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \implies X^2=\begin{bmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc\end{bmatrix}=\begin{bmatrix} 6 & 2\\ 3 & 7 \end{bmatrix}.$$
$$\implies a^2+bc=6~~~(1), b(a+d)=2~~~(2), c(a+d)=3~~~(3), d^2+bc=7~~~(4)$$
From (2) and (3) $b/c=2/3$, let $b=2k, c=3k$. From (1) and (4), we get $a^2-d^2=-1 \implies (a-d)(a+d)=-1$. Again from (2) and (3) we get $a+d=1/k, a-d=-1/k \implies a=(1/k-k)/2, d=(1/k+k)/2$. Finally,inserting $a,b,c$ in (1), we get
$$25k^4-26k^2+1=0 \implies k=\pm 1,\frac{\pm 1
}{5}$$
For $k\pm 1$, we get
$$X=\pm \begin{bmatrix} 0 & 2 \\ 3 & 1 \end{bmatrix}~~~(1)$$
For $k=\pm\frac{1}{5}$, we get
$$X=\pm \frac{1}{5} \begin{bmatrix} 12 & 2 \\ 3 & 13 \end{bmatrix}~~~(2)$$
This sheer brute force method gives 4 matrices. one can check their correctness by squaring them.
|
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|
Point $M$ lies inside $\triangle ABC$, $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC)$
In $\Delta ABC, \angle CAB = 30^\circ$ and $\angle ABC = 80^\circ$. Point $M$ lies inside the triangle such that $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC)$.
What I Tried: Here is a picture to keep track of the angle-chasing .
The red angle is $30^\circ$ , the green one is $10^\circ$ , the purple one is $20^\circ$ . Now I extend $MC$ to $AB$ such that it meets at $K$. First I noticed that $AK = KC$ but that didn't seem to help. Also the $2$ yellow ones are equal to $40^\circ$ each, the brown one is equal to $80^\circ$ . The light green angle is $60^\circ$ . We have to find the blue angle.
After all these information, it seems like I am still missing something, because from here you actually cannot deduce the value of the blue angle. Can anyone help?
Thank You.
|
Let $\angle MBA = \theta$, $\angle MBC = 80 - \theta$
Using trig form of Ceva's theorem,
$$ \sin MAC \sin MCB \sin MBA = \sin MCA \sin MBC \sin MAB$$
$$ \sin 10 \sin 40 \sin \theta = \sin 30 \sin (80-\theta) \sin 20 $$
$$ 4\sin 10 \cos 20 =\dfrac{\sin (80-\theta)}{\sin \theta} $$
$$ 2(\sin 30 - \sin 10) =\sin 80 \cot \theta - \cos 80 $$
$$ 1 - \sin 10 =\sin 80 \cot \theta $$
$$ \cot \theta = \dfrac{1 - \sin 10}{\cos 10} $$
$$ \cot \theta = \dfrac{(\cos 5 - \sin 5)^2}{\cos^2 5 - \sin^2 5} $$
$$ \cot \theta = \dfrac{\cos 5 - \sin 5}{\cos 5 + \sin 5} $$
$$ \cot \theta = \dfrac{1 - \tan 5}{1 + \tan 5} = \tan 40$$
$$\therefore \angle MBA = 50$$
$$\Rightarrow \boxed{180 - \angle BMC = 70}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to use matrix language to describe the solution process? I see that the linear equations $\left\{\begin{array}{l}
2 x+y-t=-2 \\
3 y+z+2 t=3
\end{array}\right.$ can be solved in the following way:
I have uploaded a picture to give the details of the algorithm for the mathematical solution. It is obvious that this method is quite different from the calculation method commonly used in linear algebra textbooks. I want to explain the principle of this algorithm clearly, but I don't even understand the principle at present.
The above similar contents are from page 206 of this textbook:
$$\left(\begin{array}{cccc|c}
2 & 1 & 0 & -1 & 2 \\
0 & 3 & 1 & 2 & -3 \\
\hline 1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0
\end{array}\right) \rightarrow\left(\begin{array}{cccc|c}
1 & 0 & 0 & 0 & 2 \\
3 & -6 & 1 & 5 & -3 \\
\hline 0 & 1 & 0 & 0 & 0 \\
1 & -2 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0
\end{array}\right)\rightarrow\left(\begin{array}{cc|cc|c}
1 & 0 & 0 & 0 & 0 \\
3 & 1 & 0 & 0 & -9 \\
\hline 0 & 0 & 1 & 0 & 0 \\
1 & 0 & -2 & 1 & -2 \\
0 & 1 & 6 & -5 & 0 \\
0 & 0 & 0 & 1 & 0
\end{array}\right)\rightarrow\left(\begin{array}{cc|cc|c}
1 & 0 & 0 & 0 & 0 \\
3 & 1 & 0 & 0 & 0 \\
\hline 0 & 0 & 1 & 0 & 0 \\
1 & 0 & -2 & 1 & -2 \\
0 & 1 & 6 & -5 & -9 \\
0 & 0 & 0 & 1 & 0
\end{array}\right)$$
Then we can get the following relationship:
$$\left(\begin{array}{l}
x \\
y \\
z \\
t
\end{array}\right)=\left(\begin{array}{c}
0 \\
-2 \\
9 \\
0
\end{array}\right)+\left(\begin{array}{cc}
1 & 0 \\
-2 & 1 \\
6 & -5 \\
0 & 1
\end{array}\right)\left(\begin{array}{l}
c_{1} \\
c_{2}
\end{array}\right)$$
But I want to use matrix language to describe and prove the solution process:
$$\begin{array}{c}
A.X = B\\
M.A.X = M.B\\
M.A.X.N = M.B.N\\
\text{If }M.B.N = 0\\
\text{then }M.A.X.N = 0\\
\end{array}$$
How can I simply describe and prove the solution process?
|
I don't understand your approach in the picture for to solve this problem. However, this problem in the matricix language is easy. Note that if $$(S.E.L): \left\{\begin{aligned} 2x+y+0z-t=-2 \\ 0x+3y+z+2t=3 \end{aligned} \right.$$
so, we can write the S.E.L in the form $$AX=B \quad \text{and} \quad (A|B)$$
Taking the coefficients of the system of linear equations, we construct the matrix $A$, this's $$A=\begin{pmatrix} 2 & 1 & 0 & -1 \\ 0 & 2 & 1 & 2 \end{pmatrix}$$
and taking the unknows of the S.E.L, we have $$X=\begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix}$$
and finally, $$B=\begin{pmatrix} -2 \\ 3 \end{pmatrix}$$
Therefore, you have $$\left\{\begin{aligned} 2x+y+0z-t=-2 \\ 0x+3y+z+2t=3 \end{aligned} \right. \iff \underbrace{\begin{pmatrix} 2 & 1 & 0 & -1 \\ 0 & 2 & 1 & 2 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix}}_{X}=\underbrace{\begin{pmatrix} -2 \\ 3 \end{pmatrix}}_{B}$$
Now, you can use Gaussian elimination here $$(A|B) \iff \begin{pmatrix} 2 & 1 & 0 & -1 & | & -2 \\ 0 & 2 & 1 & 2 & | & 3\end{pmatrix}$$
|
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|
Perfect square involving the exponential law If $n$ is a natural number, and $2^{10} + 2^{13} + 2^n$ is a perfect square, what is the value of $n$?
I've attempted to factor out $2^{10}$ and got $2^{10}(1 + 2^3 + 2^{n-10})$. How can I move further?
|
We can factor out the $2^{10}$ first:
$$2^{10}(1+2^3+2^{n−10})$$
We can simplify $2^3 = 2\cdot{2}\cdot{2} = 8$. Also, because $2^{10}$ is a perfect square, we can remove it from the equation.
$$1+8+2^{n-10}=9+2^{n-10}$$
Now we can add a new variable, $m$, where $m$ can be any positive integer. We can rewrite our equation as:
$$9+2^{n-10}=m^2$$
We can now test values for $n-10$. If we do this, we'll find that $9 + 2^4 = 9+16=25=5^2$.
So $n-10=4$, and $n=14.$
|
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|
complex numbers exponential form I wish to show that $\cos^2(\frac{\pi}{5})+\cos^2(\frac{3\pi}{5})=\frac{3}{4}$
I know the solutions to $z^5+1=0$ are $-1$, $e^{i\frac{\pi}{5}}$, $e^{-i\frac{\pi}{5}}$, $e^{i\frac{3\pi}{5}}$, $e^{i\frac{-3\pi}{5}}$ and that
$z^5+1=(z+1)(z^4-z^3+z^2-z+1)$
which means that the solutions to $z^4-z^3+z^2-z+1=0$ are $e^{i\frac{\pi}{5}}$, $e^{-i\frac{\pi}{5}}$, $e^{i\frac{3\pi}{5}}$, $e^{i\frac{-3\pi}{5}}$
and so $z^4-z^3+z^2-z+1=(z-e^{i\frac{\pi}{5}})$$(z-e^{-i\frac{\pi}{5}})$ $(z-e^{i\frac{3\pi}{5}})$$(z-e^{i\frac{-3\pi}{5}})$
but I am not sure where to go from there.
|
Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$
Divide both sides by $z^2$ and replace $2u=z+\dfrac1z$
The roots of $$(2u)^2-2-2u+1=0$$ are $$\cos\frac\pi5,\cos\frac{3\pi}5$$
Now use https://mathworld.wolfram.com/VietasFormulas.html
|
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|
Prove that the plane through the point $(\alpha, \beta, \gamma)$ and the line $x = py + q = rz + s$ Prove that the plane through the point $(\alpha, \beta, \gamma)$ and the line $x = py + q = rz + s$ is
$${\begin{vmatrix} x& py+q& rz+s\\
\alpha& p\beta+q& r\gamma+s\\
1& 1& 1\\
\end{vmatrix}} = 0$$
My Attempt:
The equation of line is
$x = py + q = rz + s$
$$\implies x = p (y + \frac {q}{p}) = r ( z + \frac {s}{r}) $$
$$\implies \frac { x - 0}{1} = \frac {y + \frac {q}{p}}{\frac {1}{p}} = \frac {z + \frac {s}{r}}{\frac {1}{r}}$$
The equation of plane through this line is
$$a(x-0) + b(y+\frac {q}{p}) + c(z+\frac {s}{r}) = 0$$
where $a,b,c$ are the direction ratios of the line normal to the plane.
Now,
$$a\cdot 1 + b\cdot (y+\frac {q}{p}) + c\cdot (z + \frac {s}{r}) = 0$$
Also, the plane passes through the point $(\alpha, \beta, \gamma)$,
$$a(\alpha - 0) + b(\beta + \frac {q}{p}) + c(\gamma + \frac {s}{r}) = 0$$
How to proceed further from here?
|
Hint:
$\frac { x - 0}{1} = \frac {y + \frac {q}{p}}{\frac {1}{p}} = \frac {z + \frac {s}{r}}{\frac {1}{r}}$
One of the points on the line is $P \,$ $(0,-\frac{q}{p}, -\frac{s}{r}$).
Direction vector from this point to $Q \,$ $(\alpha, \beta, \gamma)$ is $\vec{PQ} \, (\alpha, \beta + \frac{q}{p}, \gamma + \frac{s}{r})$.
As the given line and $\vec{PQ}$ are in the plane we take a cross product of vectors $\, (\alpha, \beta + \frac{q}{p}, \gamma + \frac{s}{r})$ and $(1, \frac{1}{p}, \frac{1}{r}) \,$to find the normal vector to the plane say, $(a, b, c)$.
Then the equation of the plane is $a(x-\alpha) + b(y-\beta) + c(z-\gamma) = 0$
|
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|
Compute E(sin(X)|X+Y) Okay, let $X,Y$ be independent random variables with the same exponential distribution. Compute $E(X|X+Y)$ and $E(\sin(X)|X+Y)$.
Solution:
$X+Y=E(X+Y|X+Y)=E(X|X+Y)+E(Y|X+Y)=2\cdot E(X|X+Y)$
Hence $E(X|X+Y)=\frac{X+Y}{2}$.
$\sin(\frac{X+Y}{2})=E(\sin(\frac{X+Y}{2})|X+Y)=E(\sin \frac{X}{2}\cos\frac{Y}{2}+\cos\frac{X}{2}\sin\frac{Y}{2}|X+Y)=2E(\sin\frac{X}{2}\cos\frac{Y}{2}|X+Y)=E(\sin(X)|X+Y)$
Hence
$E(\sin(X)|X+Y)=\sin(\frac{X+Y}{2})$.
Is that correct?
|
Set $Z=X+Y$.
First we'll show $X|Z=z\sim \mathcal{U}(0,z)$ for $z>0$ fixed. For $x\in(0,z)$ we have
$$f_{X|Z=z}(x|z)=\frac{f_{XY}(x,z-x)\sqrt{2}}{\int_0^zf_{XY}(x,z-x)\sqrt{2}dx}=\frac{1}{z}$$ Here we're using the fact that $f_{XY}(x,y)=\lambda^2e^{-\lambda(x+y)}$ for $(x,y)\in (0,\infty)^2$ and $f_{XY}(x,y)=0$ elsewhere. Therefore, $$E(X|Z=z)=\int_0^zxf_{X|Z=z}(x|z)dx=\frac{z}{2}$$ Moreover, $$E(\sin(X)|Z=z)=\int_0^z\sin(x)f_{X|Z=z}(x|z)dx=\frac{2}{z}\sin^2\Big(\frac{z}{2}\Big)$$ So we get $E(X|X+Y)=\frac{X+Y}{2}$ while $E(\sin(X)|X+Y)=\frac{2}{X+Y}\sin^2\Big(\frac{X+Y}{2}\Big)$.
Note: The problem arises in your calculation when you claim that $$2E\Big(\sin(X/2)\cos(Y/2)\Big|X+Y=z\Big)=E(\sin(X)|X+Y=z)$$
|
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|
How to use the Fourier transform for show that $ \int_{0}^{\infty} \frac{sin^2(x)}{x^2} dx = \frac{\pi}{2} $? Suppose
$$ \int_{0}^{\infty} f(x)\cos(\alpha x)dx = F(\alpha) = \left\{ \begin{array}{lcc}
1-\alpha & if & 0\le \alpha \le 1 \\
\\ \hspace{0.6cm} 0 & if & \alpha>1 \\
\end{array}
\right.
$$
$$\mathcal{F}(F(\alpha) = \frac{2}{\pi} \int_{0}^{\infty} F(\alpha)\cos(\alpha x) d\alpha = \int_{0}^{1} (1-\alpha)\cos(\alpha x) d\alpha = \frac{1-\cos(x)}{x^2} $$
How to use this for show that $ \int_{0}^{\infty} \frac{\sin^2(x)}{x^2} dx = \frac{\pi}{2} $ ?
|
$$ \mathcal{F}(F(\alpha) = \frac{1-\cos(x)}{x^2} = \frac{4\sin^2(\frac{x}{2})}{\pi x^2} $$
Let $\alpha = 0$
$$ \int_{0}^{\infty} \frac{4\sin^2(\frac{x}{2})}{\pi x^2} \ cos(0) dx = 1 $$
$$ t=x/2 \Rightarrow x=2t \Rightarrow \frac{dx}{dt} = 2 $$
$$ 2 \int_{0}^{\infty} \frac{4\sin^2(t)}{4 \pi t^2} dt = 1 \iff \frac{2}{\pi} \int_{0}^{\infty} \frac{\sin^2(t)}{t^2}dt = 1 \iff \int_{0}^{\infty} \frac{\sin^2(t)}{t^2}dt = \frac{\pi}{2} $$
|
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|
Prove $\sum_1^n 1/i^2 \le 2 - 1/n$ for all natural $n$. I'm trying to do this by induction. It works for $n=1$ because we get $1 \le 2 - 1 = 1$. Now suppose for some natural $k \ge 1$, we have $\sum_{i=1}^k \le 2 - 1/k$. I must show $\sum_{i=1}^{k+1} \le 2 - 1/(k+1)$. I started out from the hypothesis and added $1/(k+1)^2$ to both sides of the inequality. I get
\begin{align*}
1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{k^2} + \frac{1}{(k+1)^2} \le 2 - \frac{1}{k} + \frac{1}{(k+1)^2}
\end{align*}
and it's not obvious what to do from there. Any hints?
|
Hint: prove that LHS "grows" slower than RHS
Prove that $$\frac{1}{n^2} < (2-\frac 1n) - (2-\frac{1}{n-1}), n>1$$
|
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|
Given $T_1 = 0$ and for $i > 1$: $T_i = 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} T_j$ Prove via induction that $T_i = \sum\limits_{j=1}^{i-1} \frac{1}{j}$
Given $T_1 = 0$ and for $i \in \mathbb{N}, i > 1$:
\begin{align*}
T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} T_j \\
\end{align*}
Manually computing terms $2,3,4$:
\begin{align*}
T_2 &= 1 \\
T_3 &= 1 + \frac{1}{2} \\
T_4 &= 1 + \frac{1}{3} \left(1 + 1 + \frac{1}{2} \right) = 1 + \frac{1}{2} + \frac{1}{3} \\
\end{align*}
Prove by induction that this is:
\begin{align*}
T_i &= \sum\limits_{j=1}^{i-1} \frac{1}{j} \\
\end{align*}
The given equation is clearly satisfied for terms $1,2,3,4$. We need to show the inductive step. Assuming it's true for all $j < i$, show that it is also true for $i$.
\begin{align*}
T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} \sum\limits_{k=1}^{j-1} \frac{1}{k} \\
T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-2} \frac{i-1-j}{j} \\
\end{align*}
|
Let's say it is true for $i$, and we want to prove for $i+1$ ($i>1$):
I think you can develop the left hand:
$$T_{i+1} = 1 + \frac{1}{i}\sum_{j=1}^{i}T_j = 1 + \frac{1}{i}\sum_{j=1}^{i-1}T_j + \frac{T_i}{i}$$
$$iT_{i+1} = i + \sum_{j=1}^{i-1}T_j + T_i$$
$$\frac{i}{i-1}T_{i+1} = \frac{1}{i-1} + 1 + \frac{1}{i-1}\sum_{j=1}^{i-1}T_j + \frac{T_i}{i-1}$$
$$\frac{i}{i-1}T_{i+1} = \frac{1}{i-1} + T_i + \frac{T_i}{i-1}$$
$$iT_{i+1} = 1 + (i-1)T_i + T_i$$
$$T_{i+1} = T_i + \frac{1}{i}$$
And that is the right hand:
$$T_{i+1} = \sum_{j=1}^{i}\frac{1}{j} = \sum_{j=1}^{i-1}\frac{1}{j} + \frac{1}{i} = T_i + \frac{1}{i}$$
|
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|
Prove inverse trigonometric equation $2\tan^{-1}2=\pi-\cos^{-1}\frac{3}{5}$ The question is:
Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$
I did the question without using the Hint, but I don't know how to do it using the hint.
Quick working out of what I've done:
\begin{aligned}
\text { If } \theta &=\tan ^{-1} 2 \\
\tan \theta &=2 \\
0 & < \theta < \frac{\pi}{2}
\end{aligned}
\begin{aligned}
\cos 2 \theta &=2 \cos ^{2} \theta-1 \\
&=2 \times\left(\frac{2}{\sqrt{5}}\right)^{2}-1 \\
&=\frac{3}{5} \\
2 \theta =& \cos ^{-1} \frac{3}{5}, \quad \text { since } 0 < 2\theta < \pi
\end{aligned}
\begin{array}{l}
2 \tan ^{-1} 2=\cos ^{-1} \frac{3}{5} \text { . } \\
\text { Note: } \cos ^{-1} x \text { has point symmetry } \\
\text { in }\left(0, \frac{\pi}{2}\right) \text { . }
\end{array}
$$
\begin{array}{l}
\cos ^{-1} x+\cos ^{-1}(-x)=\pi \\
\cos ^{-1} \frac{3}{5}=\pi-\cos ^{-1}\left(-\frac{3}{5}\right) \\
\therefore \quad 2 \tan ^{-1} 2=\pi-\cos ^{-1}\left(-\frac{3}{5}\right)
\end{array}
$$
But I didn't use the Hint given in the question for this working out. How do I use the hint? Thank you !
|
It isn't necessary to use the hint. What you have done is correct and valid! Nonetheless, here's the intended method using hint.
Let $\theta=2\tan^{-1}2$ and $\alpha=\pi-\cos^{-1}\frac 3 5$. Then,
$$\tan\theta=\frac{2\tan(\tan^{-1}2)}{1-\tan^2(\tan^{-1}2)}$$
$$\tan\theta=-\frac{4}{3}\quad (*)$$
and using the hint,
$$\tan\alpha=-\tan(\cos^{-1}\frac{3}{5})$$
Using $\cos^{-1}\frac{A}{\sqrt{A^2+B^2}}=\tan^{-1}\frac{B}{A}$, we get
$$\tan\alpha=-\tan(\tan^{-1}\frac{4}{3})$$
$$\tan\alpha=-\frac 4 3\quad (**)$$
From $(*)$ and $(**)$, we conclude
$$\theta=\alpha$$
Or $\text{LHS}=\text{RHS}$ as desired.
Hope this is clear :)
|
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|
Infinite sum of inverse of products of successive primes If we iteratively remove the multiples of the succesive prime numbers from the natural numbers, starting from 2, i.e.,
$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ... \rightarrow$
$1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, ... \rightarrow$
$1, 5, 7, 11, 13, 17, 19, 23, 25, ... \rightarrow$
we are consecutively removing fractions
$$\frac{1}{2}, \; \frac{1}{2·3}, \; \frac{1}{2·3·5}, \; \frac{1}{2·3·5·7},\; ... $$
from the total amount of natural numbers. This suggests that the infinite sum
$$\frac{1}{2}+ \frac{1}{2·3}+ \frac{1}{2·3·5}+ \frac{1}{2·3·5·7}+ ...$$
is equal to 1, but I couldn't figure out a proof for this result. Some references about this particular sum would be appreciated too.
Thank you in advance
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This is not the case:
$$\frac12+\frac1{2\cdot 3}+\frac1{2\cdot 3\cdot 5}+\frac1{2\cdot 3\cdot 5\cdot 7}+\frac1{2\cdot 3\cdot 5\cdot 7\cdot11}+\cdots \le \frac12\left(1+\frac1{3}+\frac1{3^2}+\frac1{3^3}+\frac1{3^4}+\cdots \right)=\frac34$$
It seems that your series converges to
$$0.70523017179180096514743168288824851374357763910915432819226791381\ldots$$
and does so way faster than the geometric series used in the above comparison.
After all, the $n$th summand is $$\le \frac12\cdot\frac{2^n\cdot n!}{(2n)!}\approx \frac{2^nn^ne^{-n}\sqrt{2\pi n}}{2\cdot(2n)^{2n}e^{-2n}\sqrt{4\pi n} }=\frac1{2\sqrt 2}\cdot \left(\frac e{2n}\right)^n.$$
|
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"url": "https://math.stackexchange.com/questions/3924934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluating $\lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x}$ Since we have $\frac00$,I Applied L'Hopital rule :
$$\lim_{x\to 0} (1+x)^{\tfrac1x}\times\left(\cfrac{-\ln(1+x)}{x^2}+\cfrac{1}{x(x+1)}\right)$$$$=\lim_{x\to 0}\cfrac{x^2(x+1)(1+x)^{\tfrac1x}-(x+1)\ln(1+x)+x}{x^2(x+1)}$$
But as you can see it is getting very ugly.
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$$y=\dfrac{(1+x)^{\tfrac1x}-e}{x}$$
$$z=(1+x)^{\tfrac1x}\implies \log(z)=\tfrac1x\log(1+x)$$ Now Taylor
$$\log(z)=\tfrac1x\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right) \right)=1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)$$
$$z=e^{\log(z)}=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)$$
$$y=-\frac{e}{2}+\frac{11 e x}{24}+O\left(x^2\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3928465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Series of characteristic polynomials Consider the sequence of symmetric matrices with diagonal 2 and second-diagonal s $-1$, e.g.
$$
M_4= \begin{pmatrix}
2 & -1 & 0 & 0 \\
-1 & 2 & -1 & 0\\
0 & -1 & 2 & -1\\
0 & 0 & -1 & 2\\
\end{pmatrix}
$$
I've found out that the characteristic polynomials are
$$
\begin{cases}
P_1(x)=2-x\\
P_2(x)=(2-x)^2-1\\
P_n(x) = (2-x)P_{n-1}(x)-P_{n-2}(x)
\end{cases}
$$
Or with a variable change
$$
\begin{cases}
Q_1(y)=y\\
Q_2(y)=y^2-1\\
Q_n(y) = y Q_{n-1}(y)-Q_{n-2}(y)
\end{cases}
$$
Looking at the first 8 $P_n$
I see that all eigenvalues are real (as for any symmetric matrix), they are between 0 and 4.
*
*How can I prove that all eigenvalues are between 0 and 4?
*Are these polynomials known (have a name)?
*How can I prove that the polynomial are sandwitched between
$$
\frac{1}{x}+\frac{1}{4-x}\quad\text{and}\quad
-\frac{1}{x}-\frac{1}{4-x}
$$
|
The Chebyshev polynomials of the second kind satisfy the recurrence relation
$$
\begin{cases}
U_0(y) = 1 \\
U_1(y)=x\\
U_n(y) = 2y U_{n-1}(y)-U_{n-2}(y)
\end{cases}
$$
so that $Q_n(y) = U_n(y/2)$ and $P_n(x) = U_n(1-x/2)$.
The zeros of $U_n$ are
$$
y_k = \cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n
$$
in the range $(-1, 1)$, so that
$$
x_k = 2 - 2\cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n
$$
are the zeros of $P_n$ in the range $(0, 4)$.
Also for $|x| < 1$
$$
U_n(x) = \frac{\sin((n+1)\arccos(x))}{\sqrt{1-x^2}}
$$
which implies
$$
|U_n(x)| \le \frac{1}{\sqrt{1-x^2}}
$$
and therefore
$$
| P_n(x)| \le \frac{2}{\sqrt{x(4-x)}} \le \frac 1x + \frac{1}{4-x}
$$
for $0 < x < 4$, the last estimate follows from the inequality between harmonic and geometric mean.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})<3$ I have tried to use induction, but after I assume that P(n) is true, I can't go further to prove that P(n+1) is true as well. I also have tried to find an intermediate inequality, but I can't figure out which inequality I should start from.
Something that seemed to be useful was taking P(n) and multiplying it by $(1+\frac{1}{(n+1)^3})$, therefore I have come to this
$(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})<3 | \times(1+\frac{1}{(n+1)^3})$
$(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})(1+\frac{1}{(n+1)^3})<3(1+\frac{1}{(n+1)^3})$
but, as anyone could imagine, I came to contradiction because I've tried to prove that $3(1+\frac{1}{(n+1)^3})<3$ which is false.
Any help it would be useful.
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Using the fact $1+x\le e^x$ for all real $x,$ we have $$\left(1+ \frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\cdots\left(1+\frac{1}{n^3}\right)\le \dfrac{9}{4}\exp\left(\sum_{k=3}^{n}\dfrac{1}{k^3}\right).$$ Now use the fact that $$\sum_{k=1}^{\infty}\dfrac{1}{k^3}\lt\dfrac{\pi^2}{7}.$$
|
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|
Integrating $\int{\frac{x\,dx}{x^2 + 3x -4}}$ I saw this "Beat the Integral" problem and wanted to be sure I was approaching it correctly.
The integral is $$\int{\frac{x}{x^2 + 3x -4}dx}$$
So I decide I want to decompose this, because it's a degree-2 polynomial in the denominator and a degree-1 in the numerator, so that'd be a go-to. Also, since the degree of th enumerator is one less than the denominator we can maybe treat this as $du/u$ which would imply a natural log, though we don't quite know that yet.
So we decompose: I know that $x^2+3x-4$ can be factored as $(x-1)(x+4)$.
That gets me
$$\int{\frac{x}{(x-1)(x+4)}dx}$$
so I can do this:
$$\frac{Ax}{x-1}+\frac{B}{x+4} = \frac{x}{(x-1)(x+4)}$$
Which implies
$$A(x+4)+B(x-1) = x$$
and since the roots are at $x=1$ and $x=-4$, I can set it up like this:
$$5A = 1;-5B=1 \text{ and } A = \frac{1}{5}, B=-\frac{1}{5}$$
Leading to:
$$\frac{x}{5(x-1)}-\frac{1}{5(x+4)}$$
Which I can set up the integral like so:
$$\int{\frac{x}{5(x-1)}-\frac{1}{5(x+4)}dx}=\int{\frac{x}{5(x-1)}dx-\int{\frac{1}{5(x+4)}dx}}$$
I can now integrate by addition here $$\int{\frac{x}{x-1}}dx=\int{\frac{x-1+1}{x-1}}dx=x+\int{\frac{1}{x-1}}dx=x-\ln(x-1)$$
And doing the same thing for the second term and bringing back my $\frac{1}{5}$
$$\frac{1}{5}(x-\ln(x-1)+\ln(x+4))$$
I suspect there is a further simplification I could do. On a problem like this I also saw it integrated as an arctangent, but that seemed needlessly complex? In any case I was curious if I did this correctly.
|
Since the denominator has two linear factors, you can set up your partial fraction decomposition as
$$\frac{x}{(x+4)(x-1)} = \frac{A}{x+4} + \frac{B}{x-1}$$
This implies
$$ x = A(x-1) + B(x+4) $$
So $A+B=1$ and $4B-A=0$ $\Rightarrow$ $(A,B) = (\frac{4}{5},\frac{1}{5})$.
This means that
$$\int \frac{x}{(x-4)(x+1)} dx = \frac{1}{5} \int \frac{4}{x+4} + \frac{1}{x-1}dx $$
Then you can continue from here.
This is my first response, so I would very much appreciate any feedback on formatting or other site etiquette I may be unaware of. I hope this helps!
|
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|
Show that $x^\frac{1}{n}$ is continuous at all $a \in [0,\infty)$ I have to show $x^\frac{1}{n}$ is continuous using:
$|x-a| = |(x^\frac{1}{n})^n - (a^\frac{1}{n})^n| = |x^\frac{1}{n} - a^\frac{1}{n}||\displaystyle \sum_{k=1}^{n-1} (x^\frac{1}{n})^{n-1-k} + (a^\frac{1}{n})^{k} |$
Here's what I did:
$a=0$:
$|x^\frac{1}{n}| < \epsilon => |x| < \epsilon^n$
Pick $\delta = \epsilon^n$, and we have $f(x)$ is continuous at $x=0$
$a>0$:
$|x-a| = |x^\frac{1}{n} - a^\frac{1}{n}||\displaystyle\sum_{k=1}^{n-1} (x^\frac{1}{n})^{n-1-k} + (a^\frac{1}{n})^{k}| < \epsilon \cdot |\displaystyle\sum_{k=1}^{n-1} (x^\frac{1}{n})^{n-1-k} + (a^\frac{1}{n})^{k}|$
Since this sum evaluates to a real number $R > 0$, we can pick $\delta = R\epsilon $ and we get:
$0<|x-a| < \delta => |x^\frac{1}{n} - a^\frac{1}{n}| < \epsilon$
Is this correct or is there anything else I need to show?
Also, this is my first time formatting with MathJax, so if there are any errors please let me know!
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First you need the correct factorization
$$x- a = (x^{\frac{1}{n}} - a^{\frac{1}{n}})\sum_{k=0}^{n-1}x^{\frac{n-1-k}{n}} a^{\frac{k}{n}}$$
Whence,
$$|x^{\frac{1}{n}} - a^{\frac{1}{n}}|= \frac{|x-a|}{\left|\sum_{k=0}^{n-1}x^{\frac{n-1-k}{n}} a^{\frac{k}{n}} \right|} $$
Note that by the reverse triangle inequality $|a|- |x| \leqslant |x-a|$ which implies that $|x| \geqslant |a| - |x-a|$.
If $a > 0$ and $|x-a| \leqslant \frac{a}{2}$, then we have $x = |x| \geqslant |a| - |x-a| \geqslant a - \frac{a}{2} = \frac{a}{2}.$
Hence, $\displaystyle x^{\frac{n-1-k}{n}} a^{\frac{k}{n}} \geqslant 2^{\frac{k}{n}}\left(\frac{a}{2}\right)^{1 - \frac{1}{n}},$ and
$$\left|\sum_{k=0}^{n-1}x^{\frac{n-1-k}{n}} a^{\frac{k}{n}} \right|\geqslant \left(\frac{a}{2}\right)^{1 - \frac{1}{n}}\sum_{k=0}^{n-1}2 ^{\frac{k}{n}} = \left(\frac{a}{2}\right)^{1 - \frac{1}{n}}\frac{1}{2^{\frac{1}{n}} -1} := C_n$$
It follows that if $|x-a| < \delta = \min\left(\epsilon C_n, \frac{a}{2}\right)$, then
$$|x^{\frac{1}{n}} - a^{\frac{1}{n}}|\leqslant \frac{|x-a|}{C_n} < \epsilon,$$
and $x \mapsto x^{\frac{1}{n}}$
is continuous at $a > 0$.
|
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|
Are functions: $1$; $\cos(x)$; $\cos^2(\frac{x}{2})$ linearly independent? I used the definition.
$1$; $\cos(x)$; $\cos^2(\frac{x}{2})$
$c_1\cdot1+c_2\cdot\cos(x)+c_3\cdot\cos^2(\frac{x}{2}) = 0$
I tried converting $\cos^2(\frac{x}{2})$ into something better: $\frac{1+\cos(x)}{2}$
$c_1+c_2\cdot\cos(x)+c_3\cdot\frac{1+\cos(x)}{2} = 0$
$c_1+\cos(x)\cdot\left(c_2+c_3\cdot\frac{1+\cos(x)}{2\cos(x)}\right) = 0+0\cdot\cos(x)$
$c_1=0$
$c_2+c_3\cdot\frac{1+\cos(x)}{2\cos(x)}=0 \to c_2=-c_3\cdot\frac{1+\cos(x)}{2\cos(x)} \to \frac{c_2}{c_3}=\frac{-1-\cos(x)}{2\cos(x)}$
$c_2=-1-\cos(x)$ and $c_3=2\cos(x)$
There exist $c_i$-s outside of $c_i=0$ which can be solutions. So these functions are linearly dependent.
Is this correct?
Thanks.
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From $\cos^2(\tfrac{x}{2}) = \tfrac12+\tfrac12\cos(x)$ we get $$\tfrac12 + \tfrac12\cos(x) - \cos^2(\tfrac{x}{2}) = 0$$ that is, $c_1\cdot1+c_2\cdot\cos(x)+c_3\cdot\cos^2(\frac{x}{2}) = 0$ where $c_1=c_2=\tfrac12$ and $c_3=-1$.
|
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|
standard deviation can not be $6$. For the frequency distribution :
Variate x : $x_1 , x_2 , x_3 , x_ 4 ........, x_{15}$
Frequency f : $f_1 , f_2 , f_3 , f_ 4 ........, f_{15}$ where $0 < x_1 < x_2 < x_3 < x_ 4 ........< x_{15} = 10$ and $\sum f_i >0$.
Then the standard deviation can not be $6$.
Can anyone please help me by giving some hints? I have no idea.
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There is a standard result that $$\sigma_x^2 \leq \frac{R_x^2}{4}$$ where $\sigma_x^2$ is the variance of $X$ and $R_x$ is the range of $X$
Note that here $R_x = 10 - 0 = 10$
Using this result, $$\sigma_x^2 \leq \frac{10^2}{4} = 25 $$
Or, $$\sigma_x \leq 5$$
Proof:
Let $\min_{1\leq i \leq n} x_i = a$ and $\max_{1\leq i \leq n} x_i = b$
Then $R_x = b - a$
$\sigma_x^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2$
Now $\sum_{i=1}^n (x_i - c)^2$ is least when $c = \bar{x}$
Hence $\sum_{i=1}^n (x_i - \bar{x})^2 \leq \sum_{i=1}^n (x_i - \frac{a+b}{2})^2 = \sum_1 (x_i - \frac{a+b}{2})^2 + \sum_2 (x_i - \frac{a+b}{2})^2$
where $\sum_1$ includes the values of $x$ less than or equal to $\frac{a+b}{2}$ and $\sum_2$ includes values of $x$ greater than $\frac{a+b}{2}$
Or, $\sum_{i=1}^n (x_i - \bar{x})^2 \leq \sum_1 (a - \frac{a+b}{2})^2 + \sum_2 (b - \frac{a+b}{2})^2 = \sum_1 \frac{R_x^2}{4} + \sum_2 \frac{R_x^2}{4} = n \frac{R_x^2}{4}$
Or, $\sigma_x^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 \leq \frac{R_x^2}{4}$
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|
Surface area of $z=4-x^2-y^2$ over a square region
Find the surface area of the paraboloid $z=4-x^2-y^2$ over the square region $-2\le x\le 2 $ and $-2\le y\le 2$.
I can parametrize this surface with $x=u,\ y=v,\text{and} \>z=4-u^2-v^2$,
where $-2\le u\le 2$ and $-2\le v\le 2$. Then I can parametrize the surface with a vector function as follows:
$$\mathbf r(u,v)=\langle u, v ,4-u^2-v^2\rangle $$
Then
$\mathbf r_u(u,v)=\langle 1,0,-2u\rangle,\>
\mathbf r_v(u,v)=\langle 0,1,-2v\rangle$ and
$$\mathbf r_u\times\mathbf r_v=\left|\matrix{\mathbf i&\mathbf j&\mathbf k\cr 1 & 0 & -2u\cr 0 & 1 & -2v}\right|=\langle 2u,2v,1\rangle$$
so $\|\mathbf r_u\times\mathbf r_v\|=\sqrt{4u^2+4v^2+1}$.
The surface area is then defined by
$$A=\int\int_D\|\mathbf r_u\times \mathbf r_v\|\,dA=\int_{-2}^2\int_{-2}^2\sqrt{4u^2+4v^2+1}\,dv\,du$$
But now, how do I perform this integral? Here's an image of this surface.
Update
Due to symmetry, we can calculate the surface area over the region $R=\{(u,v):\ 0\le u\le 2\text{ and }0\le v\le 2\}$, then multiply the result by 4. That is, the surface area is
$$A=4\int_{0}^2\int_{0}^2\sqrt{4u^2+4v^2+1}\,dv\,du$$
Now, it was suggested that I try polar coordinates. Here's an image of the region $R=\{(u,v):\ 0\le u\le 2\text{ and }0\le v\le 2\}$.
Note that the angle of the dashed segment from $(0,0)$ to $(2,2)$ has an angle of $\pi/4$ with the $u$-axis. Also, note that
\begin{align*}
\cos\theta&=\frac2r\\
r\cos\theta&=2\\
r&=\frac{2}{\cos\theta}\\
r&=2\sec\theta
\end{align*}
Again, because of the symmetry of our image, we can determine the surface area over the region $\{(r,\theta):\ 0\le \theta\le \pi/4\text{ and } 0\le r\le 2\sec\theta\}$ and multiply the result by 8. Thus, using
$$ u=r\cos\theta\qquad\text{and}\qquad v=r\sin\theta$$
the surface area is
\begin{align*}
A&=8\int_0^{\pi/4}\int_0^{2\sec\theta}\sqrt{4r^2\cos^2\theta+4r^2\sin^2\theta+1}\ r\,dr\,d\theta\\
A&=8\int_0^{\pi/4}\int_0^{2\sec\theta}\sqrt{4r^2+1}\ r\,dr\,d\theta\\
\end{align*}
Then I was able to integrate and manipulate a bit:
\begin{align*}
A&=8\int_0^{\pi/4}\left[\frac1{12}(4r^2+1)^{3/2}\right]_0^{2\sec\theta}\,d\theta\\
A&=\frac23\int_0^{\pi/4}\left[(16\sec^2\theta+1)^{3/2}-1\right]\,d\theta\\
A&=\frac23\int_0^{\pi/4}(16\sec^2\theta+1)^{3/2}\,d\theta-\frac23\int_0^{\pi/4}d\theta\\
A&=\frac23\int_0^{\pi/4}(16\sec^2\theta+1)^{3/2}\,d\theta-\frac{\pi}{6}
\end{align*}
But now I am unable to calculate this current integral.
Thanks.
Update
Click the link below to see the work I had to do to understand Quanto's nice suggestion.
Surface Area Solution
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Let $\sinh t= {\frac{4}{\sqrt{17}}}\tan\theta$ to evaluate
\begin{align}
I &= \int_0^{\pi/4}(16\sec^2\theta+1)^{3/2}\,d\theta\\
&=\frac{17^2}4\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}
\frac{\cosh^4 t}{1+\frac{17}{16}\sinh^2t }dt
=34\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}
\frac{(1+\cosh 2t)^2}{\frac{15}{17}+\cosh2t}dt\\
&=\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}
\left(34 \cosh2t + 38 + \frac{8}{{15}+17\cosh2t}\right)dt\\
&= 8\sqrt{33} + 38 \sinh^{-1}\frac{4}{\sqrt{17}}+\ 4\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}\frac{d(\tanh t)}{16+\tanh^2t}\\
&= 8\sqrt{33} + 38 \sinh^{-1}\frac{4}{\sqrt{17}}+\cot^{-1}\sqrt{33}
\end{align}
Thus, the surface area is
$$A=\frac23I-\frac{\pi}{6}=\frac23\left(
8\sqrt{33} + 38 \sinh^{-1}\frac{4}{\sqrt{17}}+\cot^{-1}\sqrt{33} -\frac\pi4\right)
$$
|
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|
$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$ I have not solved inequalities in a while, so I am a little rusty. Could you help me with this inequality I have found?
$$a^2 + b^2 + c^2 + 6 \ge 3(a + b + c),$$ where $a, b, c > 0$ and $abc = 1$
My initial idea was $a ^ 2 + 2 \ge 2\sqrt 2a$ and the inequalities with $b $ and $c$, then adding these 3, we get, $a ^ 2 + b ^ 2 + c ^ 2 + 6 \ge 2\sqrt 2(a + b+ c)$, but then we get to $2\sqrt2 > 3$, which is false.
Edited: I found some variants of the original problem.
Problem 1: Let $a, b, c > 0$. Prove that $a^2 + b^2 + c^2 + 6 + (abc - 1) \ge 3(a+b+c)$.
Problem 2: Let $a, b, c$ be reals with $abc \le 1$. Prove that $a^2 + b^2 + c^2 + 6 \ge 3(a+b+c)$.
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My second solution:
WLOG, assume $c = \min(a,b,c)$.
If $c \le \frac{1}{2}$, we have
$$a^2+b^2+c^2 + 6 - 3(a+b+c) = (a-\tfrac{3}{2})^2 + (b - \tfrac{3}{2})^2 + c^2 - 3c + \tfrac{3}{2}
\ge 0.$$
If $c > \frac{1}{2}$, noting that $x^2 + 2 - 3x + \ln x \ge 0$ for all $x > \frac{1}{2}$
(see the remark at the end),
we have
$$a^2+b^2+c^2 + 6 - 3(a+b+c) = \sum_{\mathrm{cyc}} (a^2 + 2 - 3a + \ln a) \ge 0.$$
We are done.
Remark: Let $f(x) = x^2 + 2 - 3x + \ln x$. We have $f'(x) = \frac{(2x-1)(x-1)}{x}$.
Thus, $f(x)$ is strictly decreasing on $(\frac{1}{2}, 1)$, and strictly increasing on $(1, \infty)$.
Also $f(1) = 0$. Thus, we have $f(x) \ge 0$ for all $x > \frac{1}{2}$.
|
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|
Find the minimum value of $7x-24y$ LMNAS $25^{th}$ UGM, Indonesian
Suppose that $x,y\in\mathbb{R}$, so that :
$x^2+y^2+Ax+By+C=0$
with $A,B,C>2014$. Find the minimum value of $7x-24y$
$x^2+y^2+Ax+By+C=0$
can be written $\rightarrow$ $(x+\frac{A}{2})^2+(y+\frac{B}{2})^2+C-(\frac{A}{2})^2-(\frac{B}{2})^2=0$
Stuck,:>
|
Using Cauchy-Schwarz inequality,
$$\left(x+\frac{A}{2}\right)^2+\left(y+\frac{B}{2}\right)^2+C-\left(\frac{A}{2}\right)^2-\left(\frac{B}{2}\right)^2=0\\
\implies 7x-24y=7\left(x+\frac A2\right)-24\left(y+\frac B2\right)+\frac{-7A+24B}{2}\\
\ge -\sqrt{7^2+(-24)^2} \cdot \sqrt{\left(x+\frac A2\right)^2+\left(y+\frac B2\right)^2}+ \frac{-7A+24B}{2}\\
=-25 \frac{1}{2} \sqrt{A^2+B^2-4C} + \frac{-7A+24B}{2}\\
=\frac 12 \left(-25\sqrt{A^2+B^2-4C}-7A+24B\right).\blacksquare$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Expressing solution of matrix equations in matrix form I want to find $x_{11}$, $x_{12}$, $x_{21}$, $x_{22}$, $a$, and $b$ that solve the following system of equations
\begin{align*}
\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}& =
\begin{bmatrix}\gamma_{11} & \gamma_{12}\\\gamma_{21} & \gamma_{22}\end{bmatrix}
\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\end{bmatrix}\begin{bmatrix}a\\0\end{bmatrix},
\\[2ex]
\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}& =
\begin{bmatrix}\gamma_{11} & \gamma_{12}\\\gamma_{21} & \gamma_{22}\end{bmatrix}
\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\end{bmatrix}\begin{bmatrix}0\\b\end{bmatrix},
\\[2ex]
\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\end{bmatrix}\begin{bmatrix}\theta_{2}\\\theta_{1}\end{bmatrix}& =
\begin{bmatrix}\alpha_{1}\\\alpha_{2}\end{bmatrix}.
\end{align*}
All parameters ($\gamma_{11}$, $\gamma_{12}$, $\gamma_{21}$, $\gamma_{22}$, $\theta_1$, $\theta_2$, $\alpha_1$, and $\alpha_2$) are non-zero and $\gamma_{11}\gamma_{22}-\gamma_{12}\gamma_{21}\neq0$.
Treating the system as 6 separete equations, I can express the solution as
\begin{align*}
x_{11} & =\frac{a\gamma_{12}}{1-a\gamma_{11}}x_{21},\\
x_{12} & =\frac{b\gamma_{12}}{1-b\gamma_{11}}x_{22},\\
x_{21} & =\frac{1}{\left( a-b\right) \theta_{2}}\left( \frac{\gamma
_{11}\alpha_{2}-\gamma_{21}\alpha_{1}}{\gamma_{11}\gamma_{22}-\gamma
_{12}\gamma_{21}}-b\alpha_{2}\right), \\
x_{22} & =\frac{1}{\left( b-a\right) \theta_{1}}\left( \frac{\gamma
_{11}\alpha_{2}-\gamma_{21}\alpha_{1}}{\gamma_{11}\gamma_{22}-\gamma
_{12}\gamma_{21}}-a\alpha_{2}\right),
\end{align*}
with $a$ and $b$ distinct roots of
$$1-\left(\gamma_{11}+\gamma_{22}\right) z+\left(\gamma_{11}\gamma_{22}-\gamma_{12}\gamma_{21}\right) z^{2}=0.$$
I would like to represent the solution for $\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\end{bmatrix}$ in matrix form. Is this even possible? Any suggestion would be highly appreciated.
|
Let $G=[\gamma_{ij}]$ and $X=[x_{ij}].$
As the OP already figured out themselves, $a$ and $b$ are the roots of $1-\mathrm{tr}(G)\,z+\det(G)\,z^2=0.$ If we substitute $z=\lambda^{-1}$ and multiply with $\lambda^2,$ we get $\lambda^2-\mathrm{tr}(G)\,\lambda+\det(G)=0.$ This means that $a^{-1}$ and $b^{-1}$ are the eigenvalues of $G,$ and it might make sense to diagonalize $G.$ So let $$G = P \begin{pmatrix} a^{-1} & 0 \\ 0 & b^{-1}\end{pmatrix}P^{-1} = PDP^{-1} $$ with an invertible matrix $P,$ and let us see what we get.
$$
X\begin{pmatrix} 1 \\ 0\end{pmatrix}
=PDP^{-1}X\begin{pmatrix} a \\ 0\end{pmatrix}
$$
or
$$
P^{-1}X\begin{pmatrix} 1 \\ 0\end{pmatrix}
=DP^{-1}Xa\begin{pmatrix} 1 \\ 0\end{pmatrix}
=aDP^{-1}X\begin{pmatrix} 1 \\ 0\end{pmatrix}
$$
which means
$$
\left(I-aD\right)P^{-1}X\begin{pmatrix} 1 \\ 0\end{pmatrix} = 0
$$
This can easily be fulfilled iff the lower left element of $P^{-1}X$ is $0.$
From $$
X\begin{pmatrix} 0 \\ 1\end{pmatrix}
=PDP^{-1}X\begin{pmatrix} 0 \\ b\end{pmatrix}
$$
we get
$$
\left(I-bD\right)P^{-1}X\begin{pmatrix} 0 \\ 1\end{pmatrix} = 0
$$
which means that the upper right element of $P^{-1}X$ must be $0.$ If we put all this together, we have $P^{-1}X=C$ with a diagonal matrix $C,$ which in turn means that $X=PC.$
Now we use the last of the three given equations to figure out the diagonal elements of $C.$ In other words, we figure out by which factor the initial choice of the eigenvectors of $G$ must be scaled to get the columns of $X.$
We want $X\theta = \alpha$ or $PC\theta = \alpha$ or $C\theta = P^{-1}\alpha$ with $\theta = (\theta_2,\theta_1)^T$ (sic!) and $\alpha=(\alpha_1,\alpha_2)^T.$
From this, we can conclude
$$
C= \mathrm{diag}\left(P^{-1}\alpha\right)
\,
\mathrm{diag}\left(\theta_2^{-1},\,\theta_1^{-1}\right)
$$
Therefore, we get
$$
X=
P
\,
\mathrm{diag}\left(P^{-1}\alpha\right)
\,
\mathrm{diag}\left(\theta_2^{-1},\,\theta_1^{-1}\right)
$$
|
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|
testing if the infinite series are converges or diverges here are two problems on testing if the infinite series are conv. or div.
*
*$\sum_{n=1}^\infty \frac{{{7\sqrt[6]{n^{13}}+2n}}}{{\sqrt[3]{27n^{9}-10n+16}}}$
*$\sum_{n=2}^\infty \frac{1}{\ln(n!)}$
the first one I couldn't do :(
but the second one here is what I did using the ratio test I got that
$$\frac{\ln(n!)}{\ln((n+1)!)}=\frac{\ln(n)+\cdots+ \ln(2)}{\ln(n+1)+\cdots +\ln(2)}<1$$
which I thought would be effective but it isn't :(
|
In the first one, you have
\begin{align}
& \frac{n^{13/6} + \text{comparatively negligible things}}{n^3 + \text{comparatively negligible things}} \\[12pt]
= {} & \frac 1 {n^{5/6} + \text{comparatively negligible terms}} \ge \frac {\text{some constant}} {n^{5/6}} \\[6pt]
& \text{(and “constant” means not changing as $n$ changes)}
\end{align}
Here I would work on trying to prove that the $n$ term of the series is greater than or equal to some constant multiple of $n^{-5/6}.$
\begin{align}
& \frac{7\sqrt[6]{n^{13}}+2n}{\sqrt[3]{27n^9-10n+16}} = \frac{7 + \frac 2 {n^{7/6}}}{\sqrt[3]{27n^{5/2} + \frac {10} {n^{11/2}} + \frac{16}{n^{13/2}}}} \\[12pt]
\ge {} & \frac 7 {\sqrt[3]{27n^{5/2} + 10 n^{5/2} + 16n^{5/2}}} = \frac 7 {n^{5/6} \cdot\sqrt[3]{27+10+16}}
\end{align}
And $\displaystyle \sum_{n=1}^\infty \frac 1 {n^{5/6}} = +\infty.$
|
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|
Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$
Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$
My approach:
Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$
so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+x}=x \implies 12+x=x^{2} \iff (x+3)(x-4)=0$$
So, the answer is $\boxed{4}$.
Is correct my solution?
Can you show other ways for to solve this problem?
Can you suggest me any textbooks with similar problems?
Thank you so much!
|
Method 1 - contraction mapping theorem.
Let $g : [0,\infty) \to [0,\infty)$ be the map $x \mapsto \sqrt{x+12}$. For any $x, y \in [0,\infty)$, we have
$$\begin{align} & g(x) - g(y) = \sqrt{x+12} - \sqrt{y+12} = \frac{x - y}{\sqrt{x+12} + \sqrt{y+12}}\\
\implies & |g(x)-g(y)| \le \frac{|x-y|}{2\sqrt{12}}
\end{align}
$$
This means $g$ is a contraction mapping over $[0,\infty)$.
By Contraction mapping theorem, $g(x)$ have a unique fixed point over $[0,\infty)$. Furthermore, if one pick any $z \in [0,\infty)$ and construct a sequence $z_n$ by $$z_n = \begin{cases}z, &n = 0\\g(z) = \sqrt{z+12}, & n > 0\end{cases}$$
$z_n$ will converges to that fixed point.
Since $g(4) = 4$, that unique fixed point is $4$. By setting $z$ to $0$, we find
$$\begin{array}{rcl}
z_1 &=& \sqrt{12},\\
z_2 &=& \sqrt{12+\sqrt{12}},\\
z_3 &=& \sqrt{12+\sqrt{12+\sqrt{12}}}\\
&\vdots&
\end{array}$$
converges to $4$.
Method 2 - explicit bound.
Define $z_n$ as in method $1$ and let $y_n = 4 - z_n$ for $n \ge 1$.
Notice
$$y_{n+1} = 4 - z_{n+1} = 4 - \sqrt{12 + z_n}
= 4 - \sqrt{16-y_n}
= \frac{y_n}{4 + \sqrt{16-y_n}}\tag{*1}
$$
$y_{n+1}$ has same sign as $y_n$. Since $y_1 = 4 - \sqrt{12}> 0$, all $y_n$ are positive.
Notice $\sqrt{16-y_n} = \sqrt{12+z_n} \ge 12$, $(*1)$ implies
$$0 < y_{n+1} < \frac{y_n}{4+\sqrt{12}}$$
Replace $n$ by $1, 2, \ldots, m-1$ and combine the inequalities, we find for $m \ge 1$,
$$0 < y_m \le \frac{y_1}{(4+\sqrt{12})^{m-1}}\quad
\implies\quad 4 - \frac{4}{(4+\sqrt{12})^m}
\le z_m < 4$$
By squeezing,
$$\lim_{m\to\infty} z_m = \lim_{m\to\infty}
\underbrace{\sqrt{12 + \sqrt{12 + \sqrt{12 + \cdots}}}}_{m\text{ times}} = 4$$
|
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|
What is the greatest number of points in the plane such that the distance between any two of them is an odd integer? Suppose the origin is one of the points, we can take $(3,0)$ as the second point. If $(x,y)$ is some other point in the set I think we can use the fact that $x^2+y^2$, for $x$ and $y$ odd, is never an integer. There is also a lot of lines that the points certainly cannot belong to. But I could not go beyond.
|
The answer is $3$.
Assume the contrary, let's say we can find four points $O,A,B,C$ such that following six distances are all odd integers:
$$(a,b,c,a_1,b_1,c_1) = (BC,CA,AB,OA,OB,OC)$$
Since $O, A, B, C$ lies in the plane, they form a degenerate tetrahedron with volume $V = 0$. Express $V$ in terms of corresponding Cayley Menger determinant,
the distances satisfy
$$\left|\begin{matrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & a_1^2 & b_1^2 & c_1^2\\
1 & a_1^2 & 0 & c^2 & b^2 \\
1 & b_1^2 & c^2 & 0 & a^2\\
1 & c_1^2 & b^2 & a^2 & 0 \\
\end{matrix}\right| = 288V^2 = 0\tag{*1}$$
Recall if $u$ is an odd integer, then $u^2 \equiv 1 \pmod 8$.
Taking modulo $8$ on both sides of $(*1)$, we arrive at a contradiction.
$$4 = \left|\begin{matrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & 1 & 1 & 1\\
1 & 1 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 & 1\\
1 & 1 & 1 & 1 & 0 \\
\end{matrix}\right| \equiv 0 \pmod 8$$
This means it is impossible to find four points in the plane whose pairwise distances are all odd integers.
Since it is trivial to find three points at odd integral distances from each other (eg. the vertices of an equilateral triangle of side $1$), the greatest number we seek is $3$.
|
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|
Proof that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}$ Kazakstan 2012
Suppose that $a, b \in\mathbb{R}$, and $a,b>0$. If $\frac{1}{a}+\frac{1}{b}=2$
prove that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}.$
My idea :
$a+b+\frac{1}{1+\sqrt{ab}}$ can be written as $ab(\frac{a+b}{ab}+\frac{1-\sqrt{ab}}{1-ab})=2ab+\frac{1-\sqrt{ab}}{ab-a^2b^2}$, but where can I use the fact that $\frac{1}{a}+\frac{1}{b}=2?$
|
From the hypothesis: $a+b=2ab$
So the claim can be written as: $2ab+\frac{1}{1+\sqrt{ab}}\geqslant 2.5$
Now to make everything based on $ab$, we analyze the hypothesis:
For which real values $x$, there exist $a,b$ such that $ab=x , \frac{1}{a}+\frac{1}{b}=2$?
For solving this put $a$ as our variable. We want to find $a$ such that $\frac{1}{a}+\frac{a}{x}=2$ or equivalently: $2ax=x+a^2$ now if we see this as a polynomial, $a$ exists iff $4x^2-4x\geqslant 0$ which happens iff $x\geqslant 1$ or $0\geqslant x$.
So now we can state the problem as follows:
For any real number $x$ such that $x$ isn't between $0$ and $1$, prove that $2x+\frac{1}{1+\sqrt x }\geqslant 2.5$
Now note that $x$ can't be negative because of the square root and this function is increasing for $x\geqslant 1$ so the minimum value is taken on $x=1$ which completes the proof.
|
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|
show that two rational functions do not have a root in common Let $$f(x)=8-\binom{4}{0}\frac{1}{x+5}-\binom{4}{1}\frac{1}{x+3}-\binom{4}{2}\frac{1}{x+1}-\binom{4}{3}\frac{1}{x-1}-\binom{4}{4}\frac{1}{x-3}$$ and $$g(x)=8+\binom{4}{0}\frac{1}{x-5}+\binom{4}{1}\frac{1}{x-3}+\binom{4}{2}\frac{1}{x-1}+\binom{4}{3}\frac{1}{x+1}+\binom{4}{4}\frac{1}{x+3}.$$
Show that there are no common roots for $f(x)=0$ and $g(x)=0$.
My approach is to use the intermediate value theorem. By looking at the graph of $f(x)$, we can find $5$ distinct intervals each of which contains exactly one root for $f(x)$. Similarly, we can find them for $g(x)$. Finally, I was able to finish the proof by showing that any two intervals among the $10$ distinct intervals do not overlap.
I wonder if there are any other ways to prove this.
|
Here's a mechanical way that doesn't require any cleverness, although it does require one tool that you might not have seen yet.
It suffices to show that the polynomials
\begin{align*}
p(x) &= (x-5) (x-3) (x-1) (x+1) (x+3) (x+5) f(x) \\
&= 8 x^6-16 x^5-264 x^4+480 x^3+1720 x^2-2384 x-1080 \\
q(x) &= (x-5) (x-3) (x-1) (x+1) (x+3) (x+5) g(x) \\
&= 8 x^6+16 x^5-264 x^4-480 x^3+1720 x^2+2384 x-1080
\end{align*}
have no common roots. One can then compute the resultant of the two polynomials, which turns out to be
$$
\mathop{\rm res}(p,q) = -65{,}355{,}707{,}448{,}508{,}518{,}802{,}391{,}040;
$$
since the resultant is nonzero, the two polynomials have no common roots.
|
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|
Summation of Cosine Series Find the summation of the series $\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)} $
My approach is as follow
$\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)} \Rightarrow \sum\limits_{k = 0}^n {\left( {1 - \cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} \right)} $
$ \Rightarrow \sum\limits_{k = 0}^n {\left( 1 \right)} - \sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} \Rightarrow \left( {n + 1} \right) - \sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} $
$\sum\limits_{k = 0}^n {\cos \left( {\frac{{2k + 2}}{{n + 2}}\pi } \right)} = \sum\limits_{k = 0}^n {\cos \left( {\frac{{2\pi }}{{n + 2}} + \frac{2}{{n + 2}}k} \right)} ;a = \frac{{2\pi }}{{n + 2}};d = \frac{2}{{n + 2}}$
From the website I got the following formula but the summation of the series is from $0$ to $n-1$.
$\sum\limits_{k = 0}^{n - 1} {\cos \left( {a + kd} \right)} = \frac{{\sin \left( {\frac{{nd}}{2}} \right)}}{{\sin \left( {\frac{d}{2}} \right)}} \times \cos \left( {a + \frac{{\left( {n - 1} \right)d}}{2}} \right)$
Where as in the question it is from $0$ to $n$, a total of $n+1$ terms. How do I proceed
|
$\dfrac{\sin\frac{(n+1)d}{2}}{\sin \frac{d}{2}} \cdot \cos(a + \dfrac{nd}{2})$
For summation from $0$ to $n$
That will work
A very basic proof if you want would be substitute n = t -1 and then using the summation formula for cosine.
Proof of summation formula of cosine and sine up to n-1 terms
|
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|
Find $\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}$ As stated in the title.
My attempt. Dividing through $(x-2)^{\frac{2}{3}}$.
$$L=\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}=\lim_{x \to \infty}\frac{(\frac{x+1}{x-2})^{\frac{2}{3}}-(\frac{x-1}{x-2})^{\frac{2}{3}}}{(\frac{x+2}{x-2})^{\frac{2}{3}}-1}$$
L'Hopital
$$L=\lim_{x \to \infty}\frac{\frac{2}{3}(\frac{x+1}{x_2})^{-\frac{1}{3}}(\frac{x-2-(x+1)}{(x-2)^2})-\frac{2}{3}(\frac{x-1}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x-1)}{(x-2)^2})}{\frac{2}{3}(\frac{x+2}{x-2})^{-\frac{1}{3}}(\frac{x-2-(x+2)}{(x-2)^2})}=\lim_{x\ \to \infty}{\frac{3(x+1)^{-\frac{1}{3}}-(-1)(x-1)^{-\frac{1}{3}}}{(x+2)^{-\frac{1}{3}}(-4)}}$$
$$=\lim_{x \to \infty}{\frac{3(1+\frac{1}{x})^{-\frac{1}{3}}-(1-\frac{1}{x})^{-\frac{1}{3}}}{4(1+\frac{2}{x})^{-\frac{1}{3}}}}=\frac{3-1}{(4)(1)}=\frac{1}{2}.$$
Is this correct and is there a more elegant way of doing it?
EDIT: strictly speaking L'Hopital is not applicable with $x \to \infty$ so just got lucky here...
|
Here is another way to do it. Dividing by $x^{2/3}$ yields $$ \frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}} = \frac{(1+\frac 1x)^{\frac{2}{3}}-(1-\frac 1x)^{\frac{2}{3}}}{(1+\frac 2 x)^{\frac{2}{3}}-(1-\frac 2x)^{\frac{2}{3}}}
$$
Substitute $t = 1/x$. Then $$\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}} = \lim_{t \to 0+}\frac{(1+ t)^{\frac{2}{3}}-(1-t)^{\frac{2}{3}}}{(1+ 2t)^{\frac{2}{3}}-(1- 2t)^{\frac{2}{3}}} $$
Putting $f(t) = (1+t)^{2/3}$ yields \begin{align} \lim_{t \to 0+}\frac{(1+ t)^{\frac{2}{3}}-(1-t)^{\frac{2}{3}}}{(1+ 2t)^{\frac{2}{3}}-(1- 2t)^{\frac{2}{3}}} &= \lim_{t \to 0
+} \frac{f(t)-f(-t)}{f(2t)-f(-2t)} \\ &= \frac 12 \cdot\lim_{t \to 0+} \frac{f(t)-f(-t)}{t} \cdot \lim_{t \to 0+} \frac{2t}{f(2t)-f(-2t)}\end{align}
Note that $$ \lim_{t \to 0+} \frac{f(t)-f(-t)}{t}= \lim_{t \to 0+} \frac{f(t)-f(0)}{t} + \lim_{t \to 0+} \frac{f(-t)-f(0)}{-t} = 2f'(0)=\frac 23$$
Thus $$\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}} = \frac 12 \cdot\lim_{t \to 0+} \frac{f(t)-f(-t)}{t} \cdot \lim_{t \to 0+} \frac{2t}{f(2t)-f(-2t)} = \frac 12 \cdot \frac 23 \cdot \frac 32 = \frac 12$$
|
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|
How to prove $\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) \, dx=\ln 2$ How to prove the integral
$$\begin{align}&\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) dx\\=&\int_0^\infty\left (\int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\right)dx\\=&\ln 2\end{align}$$
I try to use integration by parts
$$\begin{align}
&\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) dx\\
=&\int_0^\infty \int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\\
=&x\left.\int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\right|_0^\infty+\int_0^\infty \cos x \int_x^\infty\frac{\sin t}{t}dx+\int_0^\infty \sin x \int_x^\infty\frac{\cos t}{t}dx
\end{align}$$
But the last two integrals are divergent.
|
The Laplace transform is useful for providing integral representations in terms of non-oscillating functions:
$$\text{si}(x)=\int_{x}^{+\infty}\frac{\sin t}{t}\,dt=\int_{0}^{+\infty}\frac{e^{-sx}}{1+s^2}\left(\cos(x)+s\sin(x)\right)\,ds $$
$$\text{Ci}(x)=\int_{x}^{+\infty}\frac{\cos t}{t}\,dt=\int_{0}^{+\infty}\frac{e^{-sx}}{1+s^2}\left(s\cos(x)-\sin(x)\right)\,ds $$
This allows to apply Fubini's theorem:
$$\begin{eqnarray*} \int_{0}^{+\infty}\text{si}(x)\text{Ci}(x)\,dx &=& \iiint_{(0,+\infty)^3}\frac{e^{-(a+b)x}}{(1+a^2)(1+b^2)}(\cos x+a\sin x)(b\cos x-\sin x)\,dx\,da\,db\\&=&\iint_{(0,+\infty)^2}\frac{1}{(1+a^2)(1+b^2)}\left(1-\frac{a}{a+b}+\frac{ab-3}{4+(a+b)^2}\right)\,da\,db\end{eqnarray*} $$
and the last double integral is pretty simple to be computed:
$$ \iint_{(0,+\infty)^2}\frac{da\,db}{(1+a^2)(1+b^2)}=\left(\frac{\pi}{2}\right)^2 $$
$$ \iint_{(0,+\infty)^2}\frac{a}{(1+a^2)(1+b^2)(a+b)}\,da\,db=\iint_{(0,+\infty)^2}\frac{b}{(1+a^2)(1+b^2)(a+b)}\,da\,db =\frac{\pi^2}{8}$$
and the difficult part
$$ \iint_{(0,+\infty)^2}\frac{ab-3}{(1+a^2)(1+b^2)(4+(a+b)^2)}\,da\,db=\int_{0}^{+\infty}\frac{-\pi(a^2+1)+2(a^2-1)\arctan\frac{a}{2}+2a\log(a^2+4)}{4(a^2+1)^2}\,da$$
can be broken by integration by parts.
|
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|
The number of pairs of integers $(x,y)$ satisfying $x ≥ y ≥ -20$ and $2x + 5y = 99$ is I tried to solve this question but I was unable to think how to get the number of integer pairs satisfying these conditions. Till now I have broken down this into :-
$2x+5y = 99$
==>$y=(99-2x)/5$
$y\geq-20$
==> $(99-2x)/5 \geq20$
==> $x\leq 99.5$
$x\geq y$
==> $x \geq (99-2x)/5$
==> $x\geq 99/7$
From here I am not able to process further to think of the way that will give me the solution.
|
y must be of the form $y=-(2k+1)5\geq-20$ which gives:
$(2k+1)5= (-5, -15) \geq -20$ $\rightarrow$ $2k+1=1, 3$
So for any $k\geq 0$ there will be an integer solution for x. in fact number of solutions is 2 when y is negative. Examples:
$2k+1=1$ $\rightarrow$ $y=-5>-20$, gives : $2x=25+99$ $\rightarrow$ $x=62$
$2k+1=3$ $\rightarrow$ $y=-15>-20$ ,gives: $2x=99+75$ $\rightarrow$ $x=87$
Now suppose x is negative then we must have:
$2x=-4, -14, -24, -34, -44, -54, -64, -74, -84, -94$
And corresponding y is:
$y=19, 17, 15, 13, 11, 9, 7, 5, 3, 1$
But this is not possible because dute to statement $x>y\geq-20$.
Now suppose x and y are both positive we have:
$x=47, 42, 37, 32, 27, 22,17, 12, 7, 2$
$y=1, 3, 5, 7, 9, 11, 13, 15, 17, 19 $
In this case only following solutions safices the condition:
$(x, y)=(47,1),(42, 3), (37, 5), (32, 7), (27, 9), (22, 11), (17, 13)$
Hence number of solutions is 9.
|
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|
Interchange order of integral Can someone tell me which step in the process did wrong? My result seems wrong
$$D = \{(x, y) \mid \sqrt{2x - x^2} \leq y \leq \sqrt{2x}, 0 \leq x \leq 2 \}$$
$$\iint_D f(x,y) \, dy \, dx$$
Then I can draw the Graph
According to the graph, we can know that this is the area enclosed by $x = \frac{y^2}{2}$ and $x = 2$ minus the semicircle area
$$\iint_D f(x,y)\,dy\,dx = -\int_0^4 \int_{\frac{y^2}{2}}^2 f(x,y)\,dx\,dy - 2\int_0^1 \int_1^{1 + \sqrt{1 - y^2}} f(x,y)\,dx\,dy $$
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The natural way to write the iterated integral is of course $$\iint_D f \, dA = \int_{x=0}^2 \int_{y=\sqrt{2x-x^2}}^{\sqrt{2x}} f \, dy \, dx.$$ The first alternative where the order of integration is switched, is as you attempted, the area of the parabolic sector minus the semicircle. Note that the parabolic segment satisfies $$y = \sqrt{2x} \iff x = \frac{y^2}{2},$$ so when $x = 2$, $y = \sqrt{4} = 2$, not $y = 4$ as shown in your diagram. Then the semicircular arc is a portion of the circle $y^2 = 2x - x^2 = 1 - (x-1)^2$, hence $$x = 1 \pm \sqrt{1 - y^2}.$$ Thus the semicircular region is, for $y \in [0,1]$, bounded above by $1 + \sqrt{1 - y^2}$ and below by $1 - \sqrt{1 - y^2}$, and we have
$$\iint_D f \, dA = \int_{y=0}^2 \int_{x=y^2/2}^2 f \, dx \, dy - \int_{y=0}^1 \int_{x=1 - \sqrt{1 - y^2}}^{1 + \sqrt{1 - y^2}} f \, dx \, dy.$$ A second alternative is to compute the area from $y = 0$ to $1$ as the sum of two sub-areas, plus a third sub-area from $y = 1$ to $2$: $$\iint_D f \, dA = \int_{y=0}^1 \left( \int_{x=y^2/2}^{1-\sqrt{1-y^2}} f \, dx + \int_{x=1+\sqrt{1-y^2}}^2 f \, dx \right) dy + \int_{y=1}^2 \int_{x=y^2/2}^2 f \, dx \, dy.$$
|
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|
Solve for $x$ in $ \sin x + \sin x \cos x - (\frac{1}{\sqrt{2}} + \frac{1}{2})= 0 $ How can to solve for $x$ in $$ \sin x + \sin x \cos x - \left(\frac{1}{\sqrt{2}} + \frac{1}{2} \right)= 0.$$
My try:
$$ \sin x + \sin x \cos x = \left(\frac{1}{\sqrt{2}} + \frac{1}{2} \right) $$
$$ (\sin x + \sin x \cos x)^2 = (\frac{1}{\sqrt{2}} + \frac{1}{2})^2
$$
$$ \sin^2x + 2\sin^2 x \cos x + \sin^2x \cos^2x - (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
$$ \sin^2 x( 1 + 2 \cos x + \cos^2 x ) - (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
$$ (1 - \cos^2x)( 1 + 2 \cos x + \cos^2 x ) - (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
$$ 1 + 2 \cos x + \cos^2 x - \cos^2 x - 2 \cos^3 x - \cos^4 x- (\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
$$ 1 + 2 \cos x - 2 \cos^3 x - \cos^4x -(\frac{1}{\sqrt{2}} + \frac{1}{2})^2 = 0
$$
thx for all ur help
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Using the tangent half-angle substitution, you end with
$$\left(1+\sqrt{2}\right) t^4+2 \left(1+\sqrt{2}\right) t^2-8 t+(1+\sqrt{2})=0$$By inspection (with all these $1$'s and $\sqrt{2}$'s), $\color{red}{t=(\sqrt{2}-1)}$ is a root.
Factoring, we are left with
$$\left(1+\sqrt{2}\right) t^3+t^2+\left(1+3 \sqrt{2}\right) t-(3+2 \sqrt{2})=0$$ for which $$\Delta=-64 \left(63+44 \sqrt{2}\right) \quad <0$$ Using the hyperbolic method for only one real root which is (!!)
$$\color{red}{t=\frac{1}{3} \left(1-\sqrt{2}+4 \sqrt{3-\sqrt{2}} \sinh \left(\frac{1}{3}
\sinh ^{-1}\left(\frac{5}{14} \sqrt{\frac{1}{7} \left(173+81
\sqrt{2}\right)}\right)\right)\right)}$$ I shall not try to find what is this number which is not recognized by inverse symbolic calculators.
To the first root corresponds
$$x_1=2 \tan ^{-1}\left(\sqrt{2}-1\right)=\frac \pi 4+2k\pi$$
and the second is
$$x_2\sim 2\tan ^{-1}(0.778669) \sim 1.323197$$
|
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Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :- $\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :-
$\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
What I Tried: I checked similar questions and answers in the Art of Problem Solving here and here and tried to get some ideas.
First thing which I did is thinking of pairing the values, I took for example, $f(-1)$ and $f(1)$.
We have :-
$$\rightarrow f(-1) = \frac{1}{\frac{1}{3} + \sqrt{3}} = \frac{3\sqrt{3} + 1}{3}$$
$$\rightarrow f(1) = \frac{1}{3 + \sqrt{3}}$$
Adding both gives $\frac{7 + 6\sqrt{3}}{12 + 10\sqrt{3}}$, which more or less looks like a random sum.
So my idea of pairing did not work, or at least I couldn't pair them nicely or missed a pattern. So how would I start solving it?
Can anyone help?
|
Hint: We have that
$$f(x)+f(1-x)=\frac{1}{3^{x} + \sqrt{3}}+\frac{1}{3^{1-x} + \sqrt{3}}=\frac{1}{3^{x} + \sqrt{3}}+\frac{3^{x}/\sqrt{3}}{\sqrt{3} + 3^x}=\frac{1}{\sqrt{3}}$$
|
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|
The limit $\lim_{x \to ∞} (3x^4-5x^2+1)(1-e^{1/x^2})^2$ $\lim_{x \to ∞} (3x^4-5x^2+1)(1-e^{1/x^2})^2$
I found +∞ but the limit is 3. How to compute this limit ?
|
$\displaystyle \lim_{ x \to \infty} (3x^4-5x^2+1)(1-e^{1/x^2})^2$
$=\displaystyle \lim_{x \to \infty} \left(\frac{e^{1/x^2}-1}{\frac{1}{x^2}}\right)^2\left(\frac{3x^4-5x^2+1}{x^4}\right)$
$=1\times 3$
You know that: $$\lim_{u \to 0^+} \frac{e^u-1}{u}=1$$
Putting $u=1/x^2$ we get:$$\lim_{x \to \infty} \frac{e^{\frac{1}{x^2}}-1}{\frac{1}{x^2}}=1$$
|
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|
Integral of a rational expression involving quadratics. I want to solve this integral but I have some problems...
$$\int_2^3 \frac{(x^2-2x+1)}{(x^2+2x+1)}$$
I transformed both in $(x-1)^2$ and $(x+1)^2$ respectively but didn't find any answer. I tried as well to transform the rational expression into $1 -\frac{4x}{(x^2+2x+1)}$, but I wasn't capable of finding any antiderivative.
Thanks for responding!
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$$ \frac{x^2-2x+1}{x^2+2x+1}=1-\frac {4x+4-4}{(x+1)^2}=1-4\frac {x+1}{(x+1)^2}+\frac 4{(x+1)^2}$$
|
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|
Evaluating $\lim_{x \to 1} \frac{1 - x^{1/\pi}}{1 - x^{1/e}}$ without de l'Hospital rule Is it possible to evaluate this limit without de l'Hopital's rule?
$$\lim_{x \to 1} \frac{1 - x^{1/\pi}}{1 - x^{1/e}}$$
|
Why not to use
$$1-x^a=-a (x-1)+\frac{1}{2} \left(a-a^2\right) (x-1)^2+O\left((x-1)^3\right)$$
$$\frac {1-x^a}{1-x^b}=\frac{-a (x-1)+\frac{1}{2} \left(a-a^2\right) (x-1)^2+O\left((x-1)^3\right)}{-b (x-1)+\frac{1}{2} \left(b-b^2\right) (x-1)^2+O\left((x-1)^3\right)}$$
$$\frac {1-x^a}{1-x^b}=\frac{a}{b}+\frac{a (a-b)}{2 b}(x-1)+O\left((x-1)^2\right)$$ which shows the limit and also how it is approached.
|
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|
Prove that: $ \sum_{i=0}^{\infty} \frac{\tan \frac{\theta}{2^i}}{2^i}= \frac{1}{\theta} - 2 \cot 2 \theta$ My attempt:
Consider the following series:
$$ S = \sum_{i=0}^n \ln( \sec \frac{x}{2^i} )$$
Notice that $ \lim_{n \to \infty} \frac{dS}{dx}$ is the required sum.
Simplfying S,
$$ S = - \ln \left( \cos x \cdot \cos \frac{x}{2} ... \cos \frac{x}{2^n} \right)$$
or,
$$ S = - \ln \left( \frac{ \sin(2x)}{2^{n+1} \sin (\frac{x}{2^n})} \right)= \ln(2^{n+1}) + \ln( \sin \frac{x}{2^n}) - \ln( \sin(2x) )$$
Now,
$$ \frac{dS}{dx} = \frac{1}{2^n} \cot \frac{x}{2^n} - 2 \cot(2x)$$
The problem I'm having is proving that
$$ \lim_{ n \to \infty} \frac{1}{2^n} \cot \frac{x}{2^n} = \frac{1}{x}$$
|
Thanks to @gary,
$ \frac{1}{2^n} = u$,
$$ \lim_{u \to 0 } u \cot ( ux)= \lim_{u \to 0} \cos(ux) \frac{ux}{x\sin (ux)} = \frac{1}{x}$$
|
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|
Coin betting problem. A coin bet is placed between two friends such that the person who wins four tosses first is the winner. What is total number of possible ways in which the bet can play out?
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Avoiding elegance, there will end up being $k$ tosses, where $k \in \{4,5,6,7\}.$ Therefore, you want $\sum_{k=4}^7 f(k)$, where $f(k)$ is the number of different coin toss sequences that result in the game ending after exactly $k$ coin tosses.
Clearly, $f(4) = \binom{4}{0} + \binom{4}{4} = 2$.
That is, there are $\binom{4}{0}$ ways of getting zero heads and
$\binom{4}{4}$ ways of getting 4 heads.
For a game to end after exactly 5 coin tosses, two things must happen:
*
*One of the two people must be ahead 3 to 1, after 4 coin tosses.
*That person must also win the 5th coin toss.
There are $\binom{4}{3}$ ways that exactly 3 of the 1st 4 coin tosses come up heads. If this happens, for the game to end after exactly 5 coin tosses, with heads the winner, the 5th coin toss must also be heads. Therefore, there are $4$ coin toss sequences that allow the game to end after exactly 5 coin tosses, with heads the winner.
By symmetry, there are exactly 4 coin toss sequences that allow the game to end after exactly 5 coin tosses, with tails the winner.
Therefore $f(5) = 2 \times \binom{4}{1} = 8.$
For a game to end after exactly 6 coin tosses, two things must happen.
*
*After 5 coin tosses, someone is ahead 3 to 2.
*On the 6th coin tosses, the person who was ahead also wins the 6th toss.
By arithmetic very similar to the computation of $f(5)$, you have that
$f(6) = 2 \times \binom{5}{3} = 20.$
For the game to end on exactly 7 tosses, the game must be tied after 6 tosses, 3 to 3. This can happen in $\binom{6}{3}$ ways. In that event, the 7th toss might be either heads or tails.
Therefore, $f(7) = 2 \times \binom{6}{3} = 40.$
Final answer:
$$\sum_{k=4}^7 f(k) = 2 + 8 + 20 + 40 = 70.$$
Addendum
The answer may be re-expressed as
$$2 \times \left[\binom{3}{3} + \binom{4}{3} + \binom{5}{3} + \binom{6}{3}\right].$$
Examining Pascal's triangle, and its rule, re
$\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1},$
$\binom{3}{3} + \binom{4}{3} = \binom {4}{4} + \binom{4}{3} = \binom{5}{4}.$
Then,
$\binom{5}{4} + \binom{5}{3} = \binom{6}{4}.$
Then,
$\binom{6}{4} + \binom{6}{3} = \binom{7}{4}.$
Addendum-1
As the other responses indicate, it is not a coincidence that the computation yields $2 \times \binom{7}{4}.$
Originally, I wasn't able to grasp the analysis. Then it hit me.
It is true that you can't simply assume that the contest will go 7 coin flips, since for example, you will never have one person getting 5 heads out of 7. That is, the contest ends before the person can get the 5th head.
However: assume that the person flipping heads wins. There is a bijection between each of the winning sequences (which will be of length 4,5,6, or 7) and all the possible ways that exactly 4 of the 7 coin tosses come up heads [i.e. $\binom{7}{4}$].
To see this, pretend that if the game finishes after $k$ coin tosses, with Heads the winner and $k < 7$, then the coin toss sequence is continued with exactly $(7 - k)$ trailing tails. Assuming that the total number of Heads remains constant at 4, there is only 1 such continuation possible, since the remaining $(7 - k)$ pretended coin tosses must all be tails.
Because of this bijection, the number of sequences with Heads the winner must equal $\binom{7}{4}.$
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How to divide two inequalities in the form $a - \epsilon < x < a + \epsilon$? I have the following inequalities
$$
a - \epsilon_x < x < a + \epsilon_x \\
b - \epsilon_y < y < b + \epsilon_y
$$
with $x,y,a,b,\epsilon_x,\epsilon_y\in {\rm I\!R}$, $|\epsilon_x|>0$ and $|\epsilon_y|>0$.
How to bound $\frac{x}{y}$?
I first tried to inverse the second inequality
$$
\frac{1}{b + \epsilon_y} < \frac{1}{y} < \frac{1}{b - \epsilon_y}
$$
I am not sure if it is correct, but I then used this answer to obtain
$$
\min\left( \frac{a-\epsilon_x}{b+\epsilon_y}, \frac{a-\epsilon_x}{b-\epsilon_y}, \frac{a+\epsilon_x}{b+\epsilon_y} \right) < \frac{x}{y} < \max\left( \frac{a+\epsilon_x}{b-\epsilon_y}, \frac{a+\epsilon_x}{b+\epsilon_y}, \frac{a-\epsilon_x}{b-\epsilon_y} \right)
$$
Can I conclude that
$$
\frac{a-\epsilon_x}{b+\epsilon_y} < \frac{x}{y} < \frac{a+\epsilon_x}{b-\epsilon_y}
$$
or should I detail every special cases depending on the sign of $a$, $b$... ?
I assume that $0\notin(b-\epsilon_y, b+\epsilon_y)$.
I tried to detail the four cases
*
*if $a > 0$ and $b > 0$ then $\frac{a-\epsilon_x}{b+\epsilon_y} < \frac{x}{y} < \frac{a+\epsilon_x}{b-\epsilon_y}$
*if $a > 0$ and $b < 0$ then $\frac{a+\epsilon_x}{b+\epsilon_y} < \frac{x}{y} < \frac{a-\epsilon_x}{b-\epsilon_y}$
*if $a < 0$ and $b > 0$ then $\frac{a-\epsilon_x}{b-\epsilon_y} < \frac{x}{y} < \frac{a+\epsilon_x}{b+\epsilon_y}$
*if $a < 0$ and $b < 0$ then $\frac{a+\epsilon_x}{b-\epsilon_y} < \frac{x}{y} < \frac{a-\epsilon_x}{b+\epsilon_y}$
But it seems different from @mathcounterexamples.net 's answer (see below).
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If $0 \in (b- \epsilon_y, b + \epsilon_y)$, there is no way to bound $\frac{x}{y}$.
Otherwise you have $$\vert y \vert \gt m = \begin{cases}
b-\epsilon_y & \text{ if } b - \epsilon_y \gt 0\\
-(b + \epsilon_y) & \text{ if } b + \epsilon_y\lt 0
\end{cases}$$As $\vert x \vert \lt \vert a \vert + \epsilon_x$, you get
$$\left\vert \frac{x}{y} \right\vert \lt \frac{\vert a \vert + \epsilon_x}{m}.$$
|
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|
Find n where $(2n+1)2^{4n+5} = 3 \pmod{7}$ For $n$ normal number, the book solved it like this:
If $n$ can be divided by $3$ (which is $n = 3k$) then $n = 21L + 9$.
If $n$ can't be divided by 3(Which is either $n = 3k +1$ or $n = 3k + 2$) then $n = 21L + 1$ or $n = 21L + 2$ .
But I didn't solve it like this.
My logic is that since 3 is a prime number then $(2n+1)2^{4n+5} = 3 \pmod{7}$ means either $2^{4n+5} = 3\pmod{7}$ and $ 2n + 1 = 1\pmod{7} $ or the other way around.
But since there is no $n$ value that can make $2^{4n+5} = 3\pmod{7}$ then it means:
$ 2n + 1 = 3 \pmod{7}$ and by calculating we find in the end $n = 7k' + 1$
And $ 2^{4n+5} = 1\pmod{7}$ and by calculating we find that $n = 3k''$
So it means that $n = 7k' + 1$ AND $n = 3k''$
So, I know that my solution is faulty but can anyone explains to me the right solution or point out where I went wrong?
|
Here is one way to solve it:
*
*$(2n+1)\pmod 7\equiv 1,3,5,0,2,4,6,\cdots\ $ and cycling.
*$2^{4n+5}\pmod 7\equiv 4,1,2,\cdots\ $ and cycling.
Both can be proved by induction on $n$.
*
*$1\times 3\equiv 2\times 5\equiv 4\times 6\equiv 3\pmod 7\ $ verify there are no others.
Finally can you find $n$ for which the conditions are reunited ? (hint, Chinese theorem).
|
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|
Find the supremum and infimum of a set X Let the set $X=\{\frac{s}{s^2+2}, s\in\mathbb{Z}\}$ I have to find the supremum and the infimum. I have done in this way:
I can observe that $X=\{\frac{s}{s^2+2}, s\in\mathbb{N}\}\cup \{\frac{-t}{t^2+2}, t\in\mathbb{N}\}$ and now I study separetely the infimum and supremum of the two sets that give me in the union the set X.
1)$\{x_s=\frac{s}{s^2+2}, s\in\mathbb{N}\}$:
since $\frac{s}{s^2+2}>\frac{s+1}{(s+1)^2+2}$ $\forall s\geq 1$ then my sequence is eventually decreasing and so $\exists \lim_{s\to\infty} x_s=0=inf$. The sup is instead given by $\frac{s}{s^2+2}$ evaluated at s=1, so $sup=\frac{1}{3}$.
2) $ \{x_t=\frac{-t}{t^2+2}, s\in\mathbb{N}\}$:
since $\frac{-t}{t^2+2}>-\frac{t+1}{(t+1)^2+2}$ $\forall t\geq 1$ then my sequence is eventually increasing and so $\exists \lim_{t\to\infty} x_t=0=sup$. The inf is instead given by $\frac{-t}{t^2+2}$ evaluated at t=1, so $inf=\frac{-1}{3}$.
So finally $supA=max\{\frac{1}{3},0\}=\frac{1}{3}$ and then $infA=max\{\frac{-1}{3},0\}=\frac{-1}{3}$.
TO DO: check my solving.
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Let $f(s) = s/(s^2 + 2)$. Then clearly for all $s \in \mathbb Z$, $$f(-s) = -s/((-s)^2 + 2) = -s/(s^2 + 2) = -f(s).$$ So if $f(n) = \sup X$, we must have $f(-n) = \inf X$.
Next, observe that $f(s) > 0$ if $s > 0$, so $f(s) < 0$ if $s < 0$, and $f(s) = 0$ if $s = 0$. So it suffices to consider $s \in \mathbb Z^+$ and $\sup X$ only.
Finally, consider the quotient $$\frac{f(s+1)}{f(s)} = \frac{(s+1)(s^2 + 2)}{s(s^2 + 2s + 3)} = \frac{s^3 + s^2 + 2s + 2}{s^3 + 2s^2 + 3s} = 1 - \frac{(s-1)(s+2)}{s(s^2 + 2s + 3)}.$$ When $s > 0$, the denominator of the second term is always positive. Since the numerator of the second term is zero if $s = 1$ (we ignore the negative root since $s > 0$), and is positive for $s > 1$, it follows that this ratio is equal to $1$ only when $s = 1$, and is strictly less than $1$ if $s > 1$. Therefore, for $s \in \mathbb Z^+$, we see $f(s)$ is greatest when $s = 1$, and $\sup X = f(1) = \frac{1}{3}$, from which it follows that $\inf X = -\frac{1}{3}$.
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|
Continued radical of powers of 4 equals 3 Can someone explain to me why
$$\sqrt{4 + \sqrt{4^2 + \sqrt{4^3 + \sqrt{4^4 + \dots}}}} = 3???$$
I need an answer
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Let $f_0(x)=0$ and recursively $$ f_{n+1}(x)=\sqrt{x+f_n(4x)}.$$
Next, let $$ f(x)=\lim_{n\to\infty}f_n(x).$$
Our goal is to find $f(4)$.
Claim 1. For $x\ge 0$ and $n\in\Bbb N$,
$$\tag1 f_{n}(x)\le f_{n+1}(x).$$
Proof.
This is trivial for $n=0$. Assume $(1)$ holds for $n$ and all $x\ge0$.
Then
$$f_{n+1}(x)=\sqrt{x+f_n(4x)}\le \sqrt{x+f_{n+1}(4x)} =f_{n+2}(x)$$
for all $x\ge 0$. Hence the claim follows by induction. $\square$
Claim 2. For all $n\in\Bbb N_0$ and all $x\ge 0$,
$$\tag2 f_n(x)\le\sqrt x+1.$$
Proof. The claim is certainly true for $n=0$. Assume $(2)$ holds for all $x\ge0$ and for some $n\in\Bbb N$. Then
$$f_{n+1}(x)=\sqrt{x+f_n(4x)}\le\sqrt{x+2\sqrt{x}+1} =\sqrt{(\sqrt x+1)^2}=\sqrt x+1,
$$
so $(2)$ holds for all $x\ge0$ also for $n+1$. By induction, the claim follows. $\square$
Combining claims 1 and 2, we see that $f_n(x)$ converges point-wise (so $f(x)$ is in fact defined for all $x\ge0$) and that
$$\tag3 \sqrt x\le f(x)\le \sqrt x+1. $$
In particular, $f(4)\le 3.$
We also note that
$$f(x)=\sqrt{x+f(4x)}. $$
Let $g(x)=\frac{f(x)}{\sqrt x+1}\in[0,1]$.
Then
$$ g(x)=\frac{\sqrt{x+g(4x)(\sqrt{4x}+1)}}{\sqrt x+1}=\sqrt{\frac{x+g(4x)(2\sqrt x+1)}{x+2\sqrt x+1}}$$
i.e.,
$ g(x)^2$ is a convex combination of $g(4x)$ and $1$ and hence is not further from $1$ than $g(4x)$ is. Then also $|g(x)-1|\le |g(4x)-1|$. By $(3)$, $\lim_{x\to\infty}g(x)=1$, so that we must have $g(x)=1$ for all $x>0$. We conclude
$$ f(x)=\begin{cases}0&x=0\\\sqrt x+1&x>0\end{cases}$$
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|
Primes of the form $p=X^2+3Y^2$ I'm trying to work in $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ to show that an odd prime $p\in\mathbb{Z}$, $p\neq 3$ is of the form $p=X^2+3Y^2$ if and only if $p\equiv1$ mod $3$. The hint is to first show that if $p=1$ mod $3$, then both $p=u^2+3v^2$ and $4p=u^2+3v^2$ where this time, $u$ and $v$ are odd.
Using quadratic reciprocity, I showed that $-3$ is a square mod $p$, which shows that $p|x^2+3$ and thus $p$ isn't prime in $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$, which in turn gives me that $p=X^2+3Y^2$ using the usual proof. But I dont understand where the $4p=u^2+3v^2$ comes from. Obviously this looks like elements of $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ after dividing by $4$ but maybe I overlooked something simple?
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HINT.- If $p\ne3M+1$ then $p=3M-1$. But then
$$3M-1=X^2+3Y^2\Rightarrow X^2\equiv-1\pmod3$$ which is impossible because $\left(\mathbb Z/3\mathbb Z\right)^2= \{0,1\}$.
COMMENT.-Maybe this could help @Saegusa in his inquiring on $\mathbb Q(\sqrt{-3})$ mainly in $4p=u^2+3v^2$. This problem was conjectured by Fermat and proved by Lagrange and Legendre by other method.
The discriminant of the quadratic form $ax^2+bxy+cy^2$ is the number $\Delta=b^2-4ac$
Note that the discriminant of the form $x^2+3y^2$ is $-12$.
In order to solve this problem, using quadratic reciprocity, prove first that this discriminant $-12$ is a square modulo the prime $p=3m+1$.
In fact $$\left(\frac{-12}{p}\right)=\left(\frac{4}{p}\right)\left(\frac{3}{p}\right)\left(\frac{-1}{p}\right)=\left(\frac{3}{p}\right)\left(\frac{-1}{p}\right)$$ As we have $$\left(\frac{p}{3}\right)\left(\frac{3}{p}\right)=(-1)^{\dfrac{(p-1)(3-1)}{4}}=(-1)^{\dfrac{p-1}{2}}$$ multiplingt both sides by $\left(\frac{p}{3}\right)$ we get
$$\left(\frac{-12}{p}\right)=\left(\frac{p}{3}\right)(-1)^{\dfrac{p-1}{2}}(-1)^{\dfrac{p-1}{2}}=1$$ Then the equation $x^2=-12$ has solution modulo the prime $p=3m+1$.
► If $x$ is even put $Y=x$ so $Y^2\equiv{-12}\pmod p$ and $4\space|\space Y^2$
►If $x$ is odd put $Y=p-x$ so $Y^2\equiv{-12}\pmod p$ and again $4\space|\space Y^2$
Since $Y^2+12\equiv 0\pmod p$ i.e. $p\space| Y^2+12$ so $4p\space| Y^2+12$ because $4\space|\space Y^2$.
Definition.-A quadratic form $ax^2+bxy+cy^2$ with discriminant $\Delta\lt0$ is called reduced if the two following conditions hold:
$(1)|b|\le a\le c\\(2)\text{if either } |b|=a\space \text{or }a=c \space\text{then } b\ge0$.
PROPOSITION:- The only primitive reduced quadratic form having discriminant $-12$ is the form $x^2+3Y^2$.
The end is given by considering the form $p=px^2+bxy+cy^2$ of an evident solution $(x,y)=(1,0)$ and using Lagrange's algorith of reduccion of forms choose $b$ and $c$ carefully to reduce to the form $p=x^2+3y^2$.
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Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $-\frac{7}{3}
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $$-\frac{7}{3}<a+b<-2$$
I have shown that $a+b<-2$. My approach: $-3=8-11=a^3+b^3-6ab+2^3=\frac{1}{2}(a+b+2)((a-b)^2+(a-2)^2+(b-2)^2)$. From this we must have that $a+b<-2$.
Please give some idea/hint for the other part.
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$$a^3-6 a b+b^3=-11$$
can be written as
$$(a+b)^3-3 a b (a+b)-6 a b=-11$$
setting $$a+b=s;\;ab=p$$
we get
$$ s^3-3ps-6p+11=0$$
The equation $$z^2-sz+p=0\tag{1}$$
Gives the values of $a,b$. In order to have real roots, discriminant must be positive
$$s^2-4p\ge 0$$
Solve
$$\begin{cases}
s^3-3ps-6p+11=0\\
s^2-4p\ge 0\\
\end{cases}
$$
Solving the first equation wrt $p$ we get
$$p=\frac{s^3+11}{3 (s+2)}$$
From the second equation we get $p\le \frac{s^2}{4}$.
Substituting in the first we get the inequality
$$\frac{s^3+11}{3 (s+2)}\le \frac{s^2}{4}$$
$$\frac{s^3+11}{3 (s+2)}= \frac{s^2}{4}\to s^3-6 s^2+44=0\to s=-2.30214$$
The solution of the inequality is
$$-2.30214 \le s<-2$$
that is
$$-2.30214 \le a+b<-2$$
edit
which is a more accurate result than
$$-\frac73< a+b <-2$$
indeed if $s=-2.32>-\frac73$ then $p=1.54913$ and the equation $(1)$ becomes
$$z^2+2.32 z+1.54913=0$$
In other words the claim is wrong because values of $a+b$ from $-\frac73$ to $-2.30214$ the numbers $a$ and $b$ can't be real.
which has no real solutions because discriminant is negative.
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|
How to find the minimum of $xy$ from the equation $2x+y+6=xy$ , $x,y>0$ I have tried doing factoring and completing the square and also AM-GM
$(2-y)(x-1)=-8$
$xy=2x+y+6≥\sqrt{12xy} $
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AM-GM works perfectly fine with a small trick.
$2x+y+6=xy \Rightarrow y(x-1)=2x+6 \Rightarrow y = \dfrac{2x+6}{x-1} \Rightarrow xy =\dfrac{2x^2+6x}{x-1}$.
As @Yves Daoust points out $\dfrac{2x^2+6x}{x-1} = 2x+8+ \dfrac{8}{x-1} = 2(x-1)+10+\dfrac{8}{x-1} $.
But by AM-GM, $ 2(x-1)+10+\dfrac{8}{x-1} \geq 10 + 2\left(\sqrt{2(x-1) \cdot \dfrac{8}{x-1} } \right) = 10+2\sqrt{16}=18$. This is indeed the minimum of $xy$.
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|
Find all positive integers $x$ such that $x(x+2021)$ is a perfect square I completed the square as to get $(x+1010.5)^2-102110.25 = k^2$ but I don't know where to go from here.
Please help, thank you
I then got $(2x+2021)^2-4084441=4k^2$ then $(2k-2x-2021)(2k+2x+2021)=43^2*47^2$
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If the product of two relatively prime numbers is a perfect square, then both factors must themselves be perfect squares. So, let's consider that case first.
How many ways can $2021$ be expressed as a difference of two squares? The answer to that is $2$. Namely, $2021=1011^2-1010^2=45^2-2^2$. Hence, $x=2^2=4$ and $x=1010^2=1020100$ are the two solutions coprime to $2021$.
Next, suppose that $x=43n$ is divisible only by $43$ (but not by $47$), so $x+2021=43(n+47)$ must then also be divisible by $43$. Then, for $x(x+2021)$ to be a perfect square, both $n$ and $n+47$ must themselves be perfect squares, and since $47$ is prime, the only solution for $n$ is $n=23^2=529$, which leads to the solution $x=43 \cdot 529=22747$.
Likewise, suppose that $x=47n$ is divisible only by $47$ (but not by $43$), so $x+2021=47(n+43)$ must then also be divisible by $47$. Then, for $x(x+2021)$ to be a perfect square, both $n$ and $n+43$ must themselves be perfect squares, and since $43$ is prime, the only solution for $n$ is $n=21^2=441$, which leads to the solution $x=47 \cdot 441=20727$.
Finally, suppose that $x=2021n$ is divisible by $2021$ (or equivalently, by both $43$ and $47$), so $x+2021=2021(n+1)$ must then also be divisible by $2021$. Then, for $x(x+2021)$ to be a perfect square, both $n$ and $n+1$ must themselves be perfect squares. The only solution for $n$ is $n=0$, but this leads to $x=0$, which is not a positive integer.
Hence, there are exactly $4$ solutions for $x$, namely $4, 20727, 22747,$ and $1020100$. Interestingly, the number of solutions for $x$ happens to itself be one of the solutions!
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