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how to solve $x^{113}\equiv 2 \pmod{143}$ I need to solve $x^{113} \equiv 2 \pmod{143}$
$$143 = 13 \times 11$$
I know that it equals to $x^{113}\equiv 2 \pmod{13}$ and $x^{113}\equiv 2 \pmod{11}$
By Fermat I got
1) $x^{5} \equiv 2 \pmod{13}$
2) $x^{3} \equiv 2 \pmod{11}$
Now I'm stuck..
|
$x^5 \equiv 2 (\mod13)$
$x^{12} \equiv 1(\mod 13)$ (Femat's little theorem.)
$5\cdot5\equiv 1 (\mod 12)\\
(2^5)^5 \equiv 2 (\mod 13)\\
(2^5) \equiv 6 (\mod 13)\\
x\equiv 6 (\mod 13)$
Now use the same logic mod 11
$x^{10} \equiv 1(\mod 11)\\
3\cdot7 \equiv 1(\mod 10)\\
(2^7)^3 \equiv 2(\mod 11)\\
(2^7) \equiv 7(\mod 11)\\
7^3 \equiv 2(\mod 11)\\
x\equiv 7(\mod 11)$
Can you get home from here?
|
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|
Prove that $\Gamma(-k+\frac{1}{2})=\frac{(-1)^k 2^k}{1\cdot 3\cdot 5\cdots(2k-1)}\sqrt{\pi}$. I was able to prove that
$$
\Gamma\left (k+\frac{1}{2} \right )=\frac{1\cdot 3\cdot 5\cdots(2k-1)}{2^k}\sqrt{\pi}.\tag{$k\geq 1$}$$
using the Legendre's duplication formula. But I can't do the same to
$$\Gamma\left ( -k+\frac{1}{2} \right )=\frac{(-1)^k 2^k}{1\cdot 3\cdot 5\cdots(2k-1)}\sqrt{\pi}.\tag{$k\geq 1$}$$
If possible, I'd like you to give a hint. If it is not possible to use Legendre's duplication formula, then I tried this way, for $n\geq 1$,
\begin{align*}
\Gamma\left ( -n+\frac{1}{2} \right )&=\left ( -n-\frac{1}{2} \right )\Gamma\left ( -n-\frac{1}{2} \right )\\
&=\dots\\
&=\left ( \frac{1}{2} \right )\left ( \frac{3}{2} \right )\cdots\left ( \frac{-2n-3}{2} \right )\left ( \frac{-2n-1}{2} \right )\Gamma\left ( \frac{1}{2} \right )
\end{align*}
and not anymore. How do I count how many factors there are to right side beside $\Gamma(1/2)$?
|
I think that the reflection formula
$$
\Gamma(z) \Gamma(1 - z) = \frac{\pi}{\sin{(\pi z)}} \qquad z \not \in \mathbb{Z},
$$
should work better here, given that you already have computed $\displaystyle{\Gamma\left(k + \frac{1}{2} \right)}$.
|
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|
Binomial theorem question. Find the value of the constant $k$ $$\left[(k+x)\left(2-\frac{x}{2}\right)\right]^6$$
where the coefficient of $x^{2}$ is $84$.Find the value of the constant $k$.
I tried to expand the equation but got a equation of degree 6 for some reason.
|
We have $$\frac{1}{2^6}*(x+k)^6(x-4)^6\\(x+k)^6(x-4)^6=(P(x)+15x^2k^4+6xk^5+k^6)(Q(x)+15x^2*4^4-6x*4^5+4^6)\\15*4^4*k^6-36*4^5*k^5+4^6*15k^4$$ We have isolated the coefficient of $x^2$ so we have $$15*4k^6-36*4^2k^5+15*4^3k^4=84$$ hence
$$5k^6-48k^5+80k^4=7$$
This equation is irreducible and Wolfram gives $$\color{red}{k\approx 0.603231}$$
|
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|
Determine matrix of linear map Linear map is given through:
$\phi\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix} $
$\phi\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$
Determine matrix $A$ linear map.
Here I have solution but I dont understand how to get it.
$A=\begin{pmatrix} -3 & -3 \\ -2 & 4 \end{pmatrix} $
|
We note that when we are constructing a matrix for a linear map, we need to know the matrix related to which basis. Here, if you consider the vectors
$\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ and $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$
as the basis of your vector space, the given matrix by you is not the correct one. Actually, as we have
$$\phi\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix}=7\times \begin{pmatrix} 3 \\ -2 \end{pmatrix}-8\times \begin{pmatrix} 3 \\ 0 \end{pmatrix}$$
$$\phi\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}=3\times \begin{pmatrix} 3 \\ -2 \end{pmatrix}-6\times \begin{pmatrix} 3 \\ 0 \end{pmatrix}$$
The matrix should be (according to this basis!)
$$A=\begin{pmatrix} 7 & 3 \\ -8 & -6 \end{pmatrix}$$
Perhaps your given matrix is related to another basis.
|
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|
If $x= m-m^2-2$ then find $x^4+3x^3+2x^2-11x+6$ where m is a cube root of unity If $$x= m-m^2-2$$ then find $$x^4+3x^3+2x^2-11x+6$$ where $m$ is a cube root of unity.
My try:
Since $ m+ m^2+1=0$ the value of $x$ is $-1$.
Let $f(x)=x^4+3x^3+2x^2-11x+6$
then $ f(-1)=5$ So the answer is $5$.
Am I correct? In my book it says that answer is $1$. Please help me to identify my mistake.
|
Replacing $x=m-m^2-2$ in your given polynomial function, we have
$$f(x)=m^8-4 m^7+11 m^6-19 m^5+24 m^4-21 m^3+23 m^2-15 m+28$$ as a result. Now, noticing that $m$ is a cube root of $-1$, this is equal to
$$f(x)=m^2-4 m^1+11 m^0-19 m^2+24 m^1-21 m^0+23 m^2-15 m+28=$$
$$5m^2+5m+18=5(m^2+m+1)+13=\color{blue}{13}.$$
|
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|
Prove: if $|x-1|<\frac{1}{10}$ so $\frac{|x^2-1|}{|x+3|}<\frac{1}{13}$
Prove: $$|x-1|<\frac{1}{10} \rightarrow \frac{|x^2-1|}{|x+3|}<\frac{1}{13}$$
$$|x-1|<\frac{1}{10}$$
$$ -\frac{1}{10}<x-1<\frac{1}{10}$$
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
Adding 4 to both sides of $$ -\frac{1}{10}<x-1<\frac{1}{10}$$ gives:
$$\frac{39}{10}<x+3<\frac{41}{10}$$
$$|x+3|<\frac{39}{10}$$
Plugging those results in $$\frac{|x-1|*|x+1|}{|x+3|}<\frac{1}{13}$$
We get: $$\frac{\frac{1}{10}*\frac{19}{10}}{\frac{39}{10}}<\frac{1}{13}$$
$$\frac{19}{390}<\frac{1}{13}$$ Which is true, is this proof is valid as I took the smallest intervals, like $|x+3|<\frac{39}{10}$ and not $|x+3|<\frac{41}{10}$?
|
Your method (what you are trying to do) is correct, but you have a few errors.
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
This is wrong. It should be
$$|x+1|\lt \frac{21}{10}$$
$$\frac{39}{10}<x+3<\frac{41}{10}$$
$$|x+3|<\frac{39}{10}$$
This is wrong. It should be
$$|x+3|\lt\frac{41}{10}$$
but we use
$$|x+3|\color{red}{\gt} \frac{39}{10}\iff \frac{1}{|x+3|}\lt \frac{10}{39}$$
since $|x+3|$ is in the denominator.
Therefore, we get
$$\frac{|x^2-1|}{|x+3|}\lt\frac{1}{10}\cdot\frac{21}{10}\cdot\frac{10}{39}=\frac{7}{130}\lt \frac{10}{130}=\frac{1}{13}$$
|
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|
Solutions to the inequality $0>-x^2 +2x+3.$ I am trying to solve an inequality of
$$0>-x^2 +2x+3.$$
I am aware of two different ways of factorizing this.
$$(-x+3)(x+1)\quad\text{ and }\quad(x-3)(-x-1).$$
When I use $(-x+3)(x+1)$, I get the desired solution of $x>3$ and $x<-1$. However when I use $(x-3)(-x-1)$, this gives me $x<3$ and $x<-1$. How is this possible?
|
1) $0 > -x^2 + 2x + 3$ so
$0> (-x+3)(x+1)$
So $(-x + 3)(x+1)$ is negative. So one of the terms is positive and the other is negative.
So either $-x +3 > 0$ AND $x+1 < 0$
OR
$-x + 3 < 0$ AND $x+1 > 0$
If $-x +3 > 0$ AND $x+1 < 0$ then $x < 3$ and $x < -1$. Notice these are redundant statements. If $x < -1$ then OF COURSE $x < 3$. There's no need to state $x < 3$. So
So if $-x +3 > 0$ and $x+1 < 0$ then $x < -1$.
OR
If $-x +3 < 0$ AND $x+1 > 0$ then $x > 3$ and $x > -1$. Notice these are redundant statements. If $x > 3$ then OF COURSE $x > -1$. There's no need to state $x > - 1$. So
So if $-x +3 < 0$ and $x+1 > 0$ then $x > 3$.
So the solution is when EITHER $x > 3$ OR $x < -1$. It can not be when both as you stated as they are contradictory statements.
2) $0 > -x^2 + 2x + 3$ so
$0 > (x-3)(-x-1) $
So EITHER $x-3 < 0$ and $-x-1 > 0$
OR
$x-3 > 0$ and $-x-1 < 0$
So either both $x < 3$ and $x < -1$ so $x < -1$
OR
both $x > 3$ and $x > -1$ so $x > 3$.
1) and 2) give the exact same results.
|
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|
Let $a, b, c>0$, such that $a+b+c=1$, prove that $\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\ge\frac{9}{4}$ Let $a, b, c>0$, such that $a+b+c=1$, prove that:
$$\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\ge\frac{9}{4}$$
|
\begin{align}
& A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{{{a}^{2}}}{ab+ac}+\frac{{{b}^{2}}}{bc+ab}+\frac{{{c}^{2}}}{ac+bc} \\
\\
& [(ab+ac)+(bc+ab)+(ac+bc)]A\ge {{(a+b+c)}^{2}} \\
\\
& (2ab+2bc+2ca)A\ge 1 \\
\\
& A\ge \frac{(a+b+c)^2}{(2ab+2bc+2ca)}=\frac{1}{(2ab+2bc+2ca)}\ge \frac{3}{2} \\
\\
& (a+b+c)\left( \frac{a}{{{(b+c)}^{2}}}+\frac{b}{{{(c+a)}^{2}}}+\frac{c}{{{(a+b)}^{2}}} \right)\ge \left( \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \right)^2\ge \frac{9}{4} \\
\end{align}
|
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|
Express the function $ f $ without using absolute value signs $\left|\frac{x-2}{x+3}\right|e^{\left|x-2\right|}$? Good evening to everyone:
This is the equation $$ f(x) = \left|\frac{x-2}{x+3}\right|e^{\left|x-2\right|} $$ What I've tried is: $$ \frac{x-2}{x+3}\ge 0 => x-2 \ge 0 => x \ge 2$$ Then $$ \frac{-x+2}{-x-3} < 0 => -x+2 <0 => x > 2$$ These 2 combined give the result $$ \left|\frac{x-2}{x+3}\right| = \frac{-x+2}{-x-3} $$ for $x>2 $ and $$x-2 \ge 0 => x \ge 2$$ and $$ -x+2<0 => -x<-2 => x>2$$ therefore the result is $$ \left|x-2\right| =-x+2$$ for $x>2$. So the equation becomes $$ f(x) = \frac{-x+2}{-x-3}e^{-x+2} $$ for $ x>2 $ .But my teacher says it's not right. Can you clarify it for me please. Thanks for any possible response.
Edit: The real answers are: $$\frac{x-2}{x+3}e^{2-x}\:$$ for $ x<3$ $$ \frac{2-x}{x+3}e^{2-x}\: $$ for $-3 < x \le 2 $ and $$ \frac{x-2}{x+3}e^{x-2} $$ for $ x > 2 $
|
Hint:
1) $$\left|\frac{x-2}{x+3} \right|=\frac{x-2}{x+3} \Leftrightarrow x\in (-\infty;-3) \cup [2;+\infty)$$
2)$$\left|\frac{x-2}{x+3} \right|=-\frac{x-2}{x+3}\Leftrightarrow x\in (-3;2)$$
|
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|
Proof of AM-GM inequality for $n=3$: $\frac{a+b+c}{3}\geq\sqrt[3]{abc}$ Sorry for bad formatting, I couldn't mark the 3rd root on the right hand side...
I've figured this out into the point where
(and yeah, the problem is to prove that this applies to all non-negative real numbers)
$(a+b+c)/3\geq (abc)^{1/3}$
$$(a+b+c)^3\geq27abc\\
a^3+3a^2b+3a^2c+3ab^2+6abc+3ac^2+b^3+3b^2c+3bc^2+c^3\geq27abc$$
I'm not sure how to proceed. Any advice is appreciated - a full answer would of course be better. Thanks in advance!
|
The many online proofs for general $n$ and for $n=2$ leave little to do for $n=3$ except maybe to look for an algebraic identity proving the inequality. Let $(a,b,c) = (x^3,y^3,z^3)$, then:
$x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2 - xy - yz - zx) = (1/2) (x+y+z)(\sum (x-y)^2)$
The conclusion is slightly more precise than the inequality limited to non-negative variables:
the Arithmetic mean minus Geometric mean of $a,b,c$ has the same sign as $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c}$, except for $a=b=c$ where the difference is $0$.
The polynomial in the question, $(a+b+c)^3 - 27abc$, is not factorizable.
|
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|
What happens when you add $x$ to $\frac{1}{3}x$? I am dealing with an equation that requires me to add $x$ to $\frac{1}{3}x$:
$x + \frac{1}{3}x$ = ??
I know this might be simple to any of you on this site, because you are all asking questions with symbols I have never seen, but this is confusing to me.
I guess one way of thinking about this is - You are adding $x$ to $yx$, right?
Or just adding another $\frac{1}{3}$?
The complete equation that I am working on is [- don't laugh at its simplicity ;)]:
$\frac{2}{3}b + 5 = 20 - b$
So, when worked out... I got:
$\frac{2}{3}b + b = 15$
And this is where I get stuck.
|
$$
x + \frac{1}{3} x = 1 \cdot x + \frac{1}{3} x = \left( 1 + \frac{1}{3} \right) x
= \frac{4}{3} x
$$
Your full equation is slighty different, but the same rules (distributivity) apply:
$$
15 = \frac{2}{3} b + b = \left( \frac{2}{3} + 1 \right) b
= \frac{5}{3} b \iff \\
b = \frac{3}{5} \cdot 15 = 9
$$
|
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|
If $x^2+y^2 \equiv 0\pmod{p}$, then $p \equiv 1 \pmod{4}$
Prove that if $x^2+y^2 \equiv 0\pmod{p}$ where $p$ is a prime and $x,y$ are not both divisible by $p$, then $p \equiv 1 \pmod{4}$.
I tried using that $x^2 \equiv -y^2 \pmod{p}$ and conjectured that $-1$ must a quadratic residue modulo $p$, but I am not sure how that would help.
|
No. The way the question is worded, $p=2$ also works.
Anyway, $y \neq 0 \pmod p,$ this means $y$ has a multiplicative inverse $\pmod p,$ for no better reason than $\gcd(y,p)=1$ and we have integers with $ys+pt=1.$
$$x^2 + y^2 \equiv 0 \pmod p,$$
$$x^2 \equiv -y^2 \equiv 0 \pmod p,$$
$$ \frac{x^2}{y^2} \equiv -1 \pmod p,$$
$$ \left( \frac{x}{y} \right)^2 \equiv -1 \pmod p.$$
|
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|
A slightly problematic integral $\int{1/(x^4+1)^{1/4}} \, \mathrm{d}x$ Question. To find the integral of- $$\int {\frac{1}{(x^4+1)^\frac{1}{4}} \, \mathrm{d}x}$$
I have tried substituting $x^4+1$ as $t$, and as $t^4$, but it gives me an even more complex integral. Any help?
|
Let $$I = \int\frac{1}{(x^4+1)^{\frac{1}{4}}}dx$$
Put $x^2=\tan \theta,$ Then $2xdx = \sec^2 \theta d\theta$
So $$I = \int\frac{\sec^2 \theta}{\sqrt{\sec \theta}}\cdot \frac{1}{2\sqrt{\tan \theta}}d\theta = \frac{1}{2}\int\frac{1}{\cos \theta \sqrt{\sin \theta}}d\theta = \frac{1}{2}\int\frac{\cos \theta}{(1-\sin^2 \theta)\sqrt{\sin \theta}}d\theta$$
Now Put $\sin \theta = t^2\;,$ Then $\cos \theta d\theta = 2tdt$
So $$I = \int\frac{1}{1-t^4}dt = -\int\frac{1}{(t^2-1)(t^2+1)}dt = -\frac{1}{2}\int\left[\frac{1}{1-t^2}+\frac{1}{1+t^2}\right]dt$$
So $$I = \frac{1}{2}\ln \left|\frac{t-1}{t+1}\right|-\frac{1}{2}\tan^{-1}(t)+\mathcal{C}$$
|
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|
What are the differences between: $\sqrt{(-3)^2}$, $\sqrt{-3^2}$ and $(\sqrt{-3})^2$. First, is $\sqrt{-3}$ is equal to $-3$ or is it imaginary?
What is the difference between:
*
*$\sqrt{(-3)^2}$
*$\sqrt{-3^2}$
*$(\sqrt{-3})^2$
Can I write $(\sqrt{-3})^2 = -3$?
And, given the rule that $\sqrt{a^n}$ is equal to ($\sqrt{a})^n$, can I say that $\sqrt{-3^2}=-3$?
|
$\sqrt{(-3)^2}=\sqrt{9}=3$,
$\sqrt{-3^2}=\sqrt{-9}=\pm 3i$,
$(\sqrt{-3})^2=(\sqrt{-1}\sqrt{3})^2=(\pm i\sqrt{3})^2=-3$,
$\sqrt{a^n}=(\sqrt{a})^n $ iff $ a\geq 0$,
|
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|
Compute $\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\theta$ I''m stuck in a exercise in complex analysis concerning integration of rational trigonometric functions. I have the solution but I don't understand a specific part. Here it goes:
We want to find $\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\theta$.
Let $z=e^{i\theta}$ so that $d\theta=dz/iz$, $\cos \theta = \frac12 (z+z^{-1})$ and $\sin \theta = \frac {1}{2i}(z-z^{-1})$. We have
$$\begin{align}
I&=\int_0^{2\pi} \frac{\sin^2\theta}{5+3\cos \theta}d\theta\\\\
&=\oint_C \frac{-\frac14(z-z^{-1})^2}{5+3\frac12 (z+z^{-1})}\frac{dz}{iz}\\\\
&=\oint_C \frac{\frac i4 (z^2-2+z^{-2})}{5z+ \frac32 z^2 + \frac32}dz\\\\
&=\oint_C \frac{\frac i2 (z^2-2+z^{-2})}{3z^2 + 10z + 3}dz\\\\
&=\frac i2 \oint_C \frac{z^4-2z^2+1}{z^2(3z^2 + 10z + 3)}dz \quad (1)
\end{align}$$
where $C$ is the unit circle in the complex $z$-plane.
If I understand the last step correctly we multipliy numerator/denominator by $z^2$ in order to get rid of the negative power of $z$.
Now for the part that I don't understand in the solution. It is said that
$$(1) = \frac i2 \oint_C \frac{z^4-2z^2+1}{z^23(z+3)(z+\frac13)}dz = \frac i6 \oint_C \frac{z^4-2z^2+1}{z^2(z+3)(z+\frac13)}dz$$
Where does this $3$ come from in the denominator? If I factor out $3z^2 + 10z + 3$ I simply get $(z+3)(z+\frac13)$.
What is even stranger to me is that, after the computation of residues, the author is getting the correct solution of $\frac{2\pi}{9}$ (I verified with Wolfram) while I'm getting $\frac{2\pi}{3}$ so I'm missing a factor of $\frac13$ (i.e. I'm missing that $3$ in the denominator).
Can someone help me with this?
|
I think your error comes from the wrong use of the quadratic formula for factoring a second order polynomial. Factoring a polynomial and using the quadratic formula are two different things. Recall that for a quadratic polynomial
$$\color{blue}{ax^2+bx+c = a(x-x_1)(x-x_2)}$$
where you are missing the coefficient $a=3$. The quadratic formula will only give you the solutions of the quadratic equation.
|
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|
Evaluate $\int_0^\infty \frac{dx}{x^2+2ax+b}$ For $a^2<b$, is there an identity of evaluating the following integral?
$$\int_0^\infty \frac{dx}{x^2+2ax+b}$$
What about:
$$\int_0^\infty \frac{dx}{(x^2+2ax+b)^2}$$
My attempt is using partial fractions and completing the square, but I still failed to obtain a nice result.
|
Let the first integral be $I$ and the second one be $J$, then by putting $x=y-a$ and $y=z\sqrt{b-a^2}$ we have
\begin{align}
I(a,b)&=\int_0^{\infty}\frac{dx}{(x+a)^2+b-a^2}\\[10pt]
&= \int_a^{\infty}\frac{dy}{y^2+b-a^2}\\[10pt]
&=\frac{1}{\sqrt{b-a^2}}\int_{\large\frac{a}{\sqrt{b-a^2}}}^{\infty}\frac{dz}{z^2+1}\\[10pt]
&=\frac{1}{\sqrt{b-a^2}}\left(\frac{\pi}{2}-\arctan\left(\frac{a}{\sqrt{b-a^2}}\right)\!\right)\\[15pt]
\end{align}
and the 2nd integral is just a first derivative of $I$ with respect to $b$ times $-1$
\begin{equation}
\\[15pt]J(a,b)=-\frac{\partial I}{\partial b}=\frac{\partial }{\partial b}\left[\frac{1}{\sqrt{b-a^2}}\left(\arctan\left(\frac{a}{\sqrt{b-a^2}}\right)-\frac{\pi}{2}\right)\right]\\[10pt]
\end{equation}
Can you take it from here?
|
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|
Polynomial identity in positive terms, including AM-GM bound Consider $n$ nonnegative numbers $x_1 \cdots x_n$. An easy consequence of the AM-GM inequality
$$
\frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}
$$
is a lower bound on a polynomial
$$
(x_1 + x_2 + \cdots + x_n)^n \geq n^n (x_1 x_2 \cdots x_n)
$$
which holds with equality iff $x_1 = x_2 = \cdots = x_n$.
Question 1 (existence):
Can one write the LHS polynomial as an identity which is a sum of only nonnegative terms, including the RHS? These terms can again be composites (other than the considered difference LHS - RHS itself), if it can be guaranteed that they are nonnegative.
Question 2 ($n=4$):
What's a formula for $n=4$?
Question 3 (general $n$):
Is there a principle for composing a formula for general $n$?
First solutions / remarks:
Here are ways of doing that for $n=2$:
$$
(x + y)^2 = 4 x y + (x - y)^2
$$
and $n=3$:
$$
(x+y+z)^3 = 27 x y z + (x^3 + y^3 + z^3 - 3 x y z) + 3 (z-y)^2 x + 3 (x-z)^2 y+ 3 (y-x)^2 z
$$
where the second term (in brackets) is nonnegative again by AM-GM: $ x^3 + y^3 + z^3 \geq 3 \sqrt[3]{x^3 y^3 z^3} = 3 x y z $, or directly by the identity:
$x^3 + y^3 + z^3 - 3 xyz = \frac12 (x+y+z)((x-y)^2 + (y-z)^2 + (z-x)^2) \ge 0$.
Remark:
The obvious advantage of such a procedure would be that one could determine lower bounds of the LHS by any term on the RHS or weighted sum of terms on the RHS, with weights between 0 and 1. In particular, these lower bounds could be chosen according to prior knowledge: if all $x_i$ are known to be roughly equal, the AM-GM bound is a good one. If the $x_i$ are known to differ much, one would choose other terms on the RHS as lower bound.
|
Yes, we can.
We can do it by Bricks Throwing Method.
It's very ugly but it works.
For four variables we need to work with $(a+b+c+d)^4-256abcd$.
$3(a^4+b^4+c^4+d^4)\geq\sum\limits_{cyc}a^3(b+c+d)$ because we throw one brick:
$\sum\limits_{cyc}(3a^4-a^3(b+c+d))=\frac{1}{2}\sum\limits_{sym}(a^4-a^3b-ab^3+b^4)=\frac{1}{2}\sum\limits_{sym}(a-b)^2(a^2+ab+b^2)$.
More brick trowing:
$\sum\limits_{cyc}a^3(b+c+d)\geq\frac{1}{2}\sum\limits_{sym}a^2b^2$ gives
$\sum\limits_{cyc}a^3(b+c+d)-\frac{1}{2}\sum\limits_{sym}a^2b^2=\frac{1}{2}\sum\limits_{sym}(a^3b-2a^2b^2+ab^3)=\frac{1}{2}\sum\limits_{sym}ab(a-b)^2$ and so on...
I'll write a full proof.
More brick trowing:
$\frac{1}{2}\sum\limits_{sym}a^2b^2-\sum\limits_{cyc}a^2(bc+bd+cd)=\frac{1}{2}\sum\limits_{sym}(a-b)^2(c^2+d^2)$
The last brick trowing:
$\sum\limits_{cyc}a^2(bc+bd+cd)-12abcd=\frac{1}{4}\sum\limits_{sym}(a-b)^2cd$.
After using these bricks throwing we'll write $(a+b+c+d)^4-256abcd$ like a sum of squares.
|
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|
Find all the numbers $n$ such that $3\cdot 2^n+2\cdot 3^n\equiv 1 \pmod 7$
Find all the numbers $n$ such that $3\cdot 2^n+2\cdot 3^n\equiv 1 \pmod 7$
Attempt:
$\star$ denotes $3\cdot 2^n+2\cdot 3^n$
$$\text{for }n=1:\quad\star\equiv 5\not\equiv 1\\
\text{for }n=2:\quad\star\equiv 5\not\equiv 1\\
\text{for }n=3:\quad\star\equiv 3\not\equiv 1\\
\text{for }n=4:\quad\star\equiv 0\not\equiv 1\\
\text{for }n=5:\quad\star\equiv \color{red}1\\
\text{for }n=6:\quad\star\equiv 5\not\equiv 1\\
\vdots$$
this is how should I aprroach this? or maybe there is smarter way?
|
You have the right idea. Working mod $7,\,$ by little Fermat $\,2^6\equiv 1\equiv 3^6\pmod 7\,$ so the sequence $\,f_n \equiv 3\cdot 2^n + 2\cdot 3^n \,$ has period $\,6,\,$ i.e $\, f_{n+6}\equiv f_n.\,$ So we need only check the first cycle for $\,n = 0,\ldots, 5.\,$ Using $\,2^3\equiv 1,\ 3^3\equiv -1\,$ we compute the cycle as
$$\begin{eqnarray} 3\ (1&,&2&,&4&,&\ \ \,1&,&\,\ \ 2&,&\,\ \ 4)\\
+\ 2\ (1&,&3&,&9&,&{-}1&,&{-}3&,&{-}9)\\
\equiv \ \ (5&,& 5&,& 2&,&\,\ \ \color{#c00}1&,&\,\ \ 0 &,&\,\ \ \color{#0a0}1)\end{eqnarray}$$
Therefore, we conclude that $\ f_n \equiv 1\iff $ $\,\color{#c00}{n\equiv 3}\ $ or $\,\ \color{#0a0}{n \equiv 5}\, \pmod 6$
|
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|
Factorization of $x^n + y^n$, what sort of coefficients show up? We know that$$a^2 + b^2 = (a + bi)(a - bi).$$ What are the complete factorizations of $a^3 + b^3$, $a^4 + b^4$, $\ldots$ , $a^k + b^k$, etc.? What sort of coefficients show up?
|
The problem simplifies if you divide through by $a$ and lent $\frac{b}{a} = x$. Once you have the factors of $1+x^n$ it is easy to re-introduce $a$ and $b$.
The pattern in these is easy to see. For odd $n$,
$$a^3 + b^3 = (a+b) (a+b\mbox{ cis } (\frac{2\pi}{3})) (a+b\mbox{ cis }(\frac{4\pi}{3})) $$
$$a^5 + b^5 = (a+b) (a+b\mbox{ cis } (\frac{2\pi}{5})) (a+b\mbox{ cis }(\frac{4\pi}{5})) (a+b\mbox{ cis } (\frac{6\pi}{5})) (a+b\mbox{ cis }(\frac{8\pi}{5}))$$
and so forth. For even $n$,
$$a^4 + b^4 = (a+b\mbox{ cis } (\frac{1\pi}{4})) (a+b\mbox{ cis } (\frac{3\pi}{4})) (a+b\mbox{ cis } (\frac{5\pi}{4})) (a+b\mbox{ cis } (\frac{7\pi}{4}))$$
and so forth.
You can in fact express the full factorization in terms of easy to swallow radicals for $n=2,3,4,5,6,8$. Also for $n=17$ for example; see Fermat primes.
|
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|
Finding an Explicit Formula for $a_n$ from a Recurrence Relation I have the following recurrence relation
$$\beta ^n n(n+2)a_n = \sum_{k=0}^{n-1} \beta^k (\alpha+k+1) a_k , \quad a_0=1, \quad n \ge 1 \tag{1}$$
where $\beta$ and $\alpha$ are positive real numbers.
I know that one can easily find $a_n$ by back substitution. Here is my question
Is it possible to find an explicit formula for $a_n$?
My Work
I just computed the first three terms to see what I can guess. But I could not come up with anything!
$$\begin{align}
n&=1, \quad a_1=\beta^{-1} \left[ \frac{\alpha + 1}{1 \times 3} \right] \\
n&=2, \quad a_2=\beta^{-2}\left[ \frac{\alpha+1}{2 \times 4} + \frac{(\alpha+1)(\alpha+2)}{(1 \times 3)(2 \times 4)} \right] \\
n&=3, \quad a_3=\beta^{-3}\left[ \frac{\alpha+1}{3 \times 5}+\frac{(\alpha+1)(\alpha+2)}{(1 \times 3)(3 \times 5)}+\frac{(\alpha+1)(\alpha+3)}{(2 \times 4)(3 \times 5)}+\frac{(\alpha+1)(\alpha+2)(\alpha+3)}{(1 \times 3)(2 \times 4)(3 \times 5)} \right]
\end{align}$$
|
Inspired by the comment of Achille Hui, I noticed that
$$\begin{align}
\beta ^n n(n+2)a_n &= \sum_{k=0}^{n-1} \beta^k (\alpha+k+1) a_k \\
\beta ^{n-1} (n-1)(n+1)a_{n-1} &= \sum_{k=0}^{n-2} \beta^k (\alpha+k+1) a_k
\end{align}$$
and subtracting the above equations will lead to
$$\begin{align}
\beta ^n n(n+2)a_n - \beta ^{n-1} (n-1)(n+1)a_{n-1} &= \beta^{n-1}(\alpha+n)a_{n-1} \\
\beta n(n+2)a_n &= [n^2+n+(\alpha-1)] a_{n-1}
\end{align}$$
and finally
$$a_n=\frac{[n-(\frac{-1+\sqrt{5-4\alpha}}{2})][n-(\frac{-1-\sqrt{5-4\alpha}}{2})]}{\beta n(n+2)}a_{n-1}$$
one can easily solve the above recurrence relation to get
$$a_n = \frac{\left(\frac{3+\sqrt{5-4\alpha}}{2}\right)_n \left( \frac{3-\sqrt{5-4\alpha}}{2}\right)_n}{\beta^n n! (3)_n}$$
where $(x)_n$ is the Pochhammer symbol.
|
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|
Proving inequality using Lagrange multipliers I have this question. Prove that for all $ x,y\geq 0 $, $$ \dfrac{x^n+y^n}{2}\geq \bigg(\dfrac{x+y}{2}\bigg)^n $$ using the method of Lagrange Multipliers, via $$ \min \dfrac{x^n+y^n}{2}, \text{where $x+y=s$} $$ for some $s\geq 0$.
This is what I did. I consider $f(x,y)=\dfrac{x^n+y^n}{2}$ and the constraint $x+y=s$.
Then, I get $\begin{cases}
\dfrac{nx^{n-1}}{2}=\lambda\\
\dfrac{ny^{n-1}}{2}=\lambda
\end{cases}$. Using that system, I get that $x=y$. Going back to the constraint, I get $$2x=2y=s\ i.e.\ x=y=\dfrac{s}{2}.$$ I think I almost get the answer since $\bigg(\dfrac{s}{2}\bigg)^n=\bigg(\dfrac{x+y}{2}\bigg)^n$ but I get lost.
Need help. Thank you so much
|
Unless you restrict the problem to $n \geq 1$ the statement is false.
Consider $x=1, y=2, n=\frac12$:
$$
\sqrt{2} < \frac32 \implies 1+2+2\sqrt{2} < 6 \implies (1+\sqrt{2})^2 < \sqrt{6}^2\\ \implies 1+\sqrt{2}<\sqrt{6} \implies \frac{1+\sqrt{2}}{2} < \sqrt{\frac{3}{2}} \implies \frac{1^n+2^n}{2} < \left( \frac{1+2}{2}\right)^n
$$
|
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|
Maximize $k=x^2+y^2$ Subject to $x^2-4x+y^2+3=0$ Question
Let $x$ and $y$ be real numbers satisfying the equation $x^2-4x+y^2+3=0$. Find the maximum and minimum values of $x^2+y^2$.
My work
Let $k=x^2+y^2$
Therefore, $x^2-4x+y^2+3=0$ ---> $k-4x+3=0$ .
What do I do next? How do I find an expression in terms of $k$ that I can maximize?
|
Hint
The equation you have is that of a circle, $(x-2)^2+y^2= 1$, and you want the maximum of $x^2+y^2=4x-3$. Clearly from the latter we need the extreme possible $x$, which from the circle's equation will be when $x\in \{1,3\}$.
|
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|
Closed form of function $f(n) = \frac1n \sum\limits _{k=1}^{n-1} f(k)$? Could anyone help me get to the closed form of the function? Here $\mathbb{N}=\{1,2,3,\ldots\}$.
Find all functions $f:\mathbb{N}\to\mathbb{R}$ such that $f(1)=1$ and
$$f(n) = \frac 1 n \sum _{k = 1}^{n-1}f(k)$$
for every integer $n>1$.
So far, I see that
$$f(2)=\frac{1}{2}f(1)=\frac12,$$
$$f(3)=\frac{1}{3}\left(f(1)+f(2)\right)=\frac{1}{3}\left(1+\frac{1}{2}\right)=\frac12,$$
$$f(4)=\frac14\left(f(1)+f(2)+f(3)\right)=\frac14\left(1+\frac12+\frac12\right)=\frac12.$$
I guess that $f(n)=\frac12$ for every $n\in\mathbb{N}$. How to show this?
|
This is similar to the first answer, but without rewriting things. Note that
$$f(n+1) = \frac{1}{n+1}\sum_{x=1}^n f(x) = \frac{1}{n+1}\sum_{x=1}^{n-1}f(x) + \frac{1}{n+1}f(n).$$
Using the definition for $f(n)$, this becomes
$$f(n+1) = \frac{1}{n+1}\sum_{x=1}^{n-1} f(x) + \frac{1}{n(n+1)}\sum_{x=1}^{n-1}f(x) = \left(\frac{1}{n+1}+\frac{1}{n(n+1)}\right)\sum_{x=1}^{n-1}f(x).$$
By making a common denominator, we get that
$$\frac{1}{n+1}+\frac{1}{n(n+1)} = \frac{n}{n(n+1)}+\frac{1}{n(n+1)} = \frac{n+1}{n(n+1)} = \frac{1}{n},$$
thus
$$f(n+1) = \frac{1}{n}\sum_{x=1}^{n-1} f(x) = f(n).$$
|
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|
Maximize the product of the partitions of an integer
Let $n>0$ be an integer. Consider all partitions of $n$, i.e. all possible ways of writing $n$ as a finite sum of positive integers,
$$n=n_1+n_2+\cdots+n_k.$$
What partition maximizes the product $n_1n_2\cdots n_k$?
Examples:
*
*$2=2$
*$3=3$
*$4=2+2$
*$5=3+2$
*$6=3+3$
*$7=3+2+2$
*$8=3+3+2$
*$9=3+3+3$
*$10=3+2+2$
*$20=3+3+3+3+3+3+2$
*$30=3+3+3+3+3+3+3+3+3+3$
*$40=3+3+3+3+3+3+3+3+3+3+3+3+2+2$
*$50=3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+2$
Conjecture: It is the partition that has the maximum number of $3$'s among those consisting only of $3$'s and $2$'s.
|
Here is a rough idea. The problem can be divided into two parts. Firstly, given a $n$, for a fixed number of partitions $k$, it is not hard to find that all of the partitions should be equal. However, since we are dealing with integers, we might need to find the nearest integers.
Having the first part, we would like to know for which $k$ we get the optimum solution. we want to optimize the logarithm of the product instead. Please note that I have simplified things by letting numbers to be real.
$ln(n)=\sum_{i=1}^{k}ln(x_i)$
considering all $x_i$ are the same, we get
$ln(n)=k*ln(\frac{n}{k})$
taking the derivative of the function with respect to $k$ and putting it equal to zero, we get
$ln(\frac{n}{k})=1$
Therefore
$k\simeq\frac{n}{e}$
finally
$\frac{n}{k}=e\simeq2.7$
|
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|
How to prove: $\sum_{k=m+1}^{n} (-1)^{k} \binom{n}{k}\binom{k-1}{m}= (-1)^{m+1}$ Show that if $m$ and $n$ are integers with $0\leq m<n$ then $$\sum_{k=m+1}^{n} (-1)^{k} \binom{n}{k}\binom{k-1}{m}= (-1)^{m+1}$$
Attempts:
*
*$(-1)^{k}\binom{n}{k}$ is the coefficient of $x^{k}$ in the expansion of $(1-x)^{n}$
*And $\binom{k-1}{m}$ is the coefficient of $x^{m}$ in the expansion of $(1+x)^{k-1}$.
Thats all what I could come up with.
|
Following the hint of Arthur, note that $$\frac{1}{m!}\sum_{k=1}^{n}\dbinom{n}{k}\left(-1\right)^{k}x^{k-1}=\frac{\left(1-x\right)^{n}-1}{xm!}
$$ and if we differentiate the LHS $m$ times, we get $$\frac{1}{m!}\sum_{k=1}^{n}\dbinom{n}{k}\left(-1\right)^{k}\left(k-1\right)\left(k-2\right)\cdots\left(k-m\right)x^{k-m-1}=\sum_{k=m+1}^{n}\dbinom{n}{k}\left(-1\right)^{k}\frac{\left(k-1\right)!}{m!\left(k-m-1\right)!}x^{k-m-1}
$$ so we have $$\sum_{k=m+1}^{n}\dbinom{n}{k}\left(-1\right)^{k}\dbinom{k-1}{m}=\frac{1}{m!}\frac{d^{m}}{dx^{m}}\left(\frac{\left(1-x\right)^{n}-1}{x}\right)_{x=1}
$$ now note that $$\frac{d}{dx}\left(\frac{\left(1-x\right)^{n}}{x}-\frac{1}{x}\right)=-\frac{n\left(1-x\right)^{n-1}}{x}-\frac{\left(1-x\right)^{n}}{x^{2}}+\frac{1}{x^{2}}
$$ and $$\frac{d}{dx}\left(-\frac{n\left(1-x\right)^{n-1}}{x}-\frac{\left(1-x\right)^{n}}{x^{2}}+\frac{1}{x^{2}}\right)=\frac{2\left(1-x\right)^{n}}{x^{3}}+\frac{2n\left(1-x\right)^{n-1}}{x^{2}}+\frac{n\left(n-1\right)\left(1-x\right)^{n-2}}{x}-\frac{2}{x^{3}}
$$ and so on, so you can see that, since $m<n
$ we have only one term that doesn't vanish at $x=1$. Hence $$\frac{1}{m!}\frac{d^{m}}{dx^{m}}\left(\frac{\left(1-x\right)^{n}-1}{x}\right)_{x=1}=\frac{1}{m!}\left(-1\right)^{m+1}m!=\left(-1\right)^{m+1}
$$ so finally
$$\sum_{k=m+1}^{n}\dbinom{n}{k}\left(-1\right)^{k}\dbinom{k-1}{m}=\color{red}{\left(-1\right)^{m+1}}$$
as wanted.
|
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|
integrate $\int \frac{x^4-16}{x^3+4x^2+8x}dx$
$$\int \frac{x^4-16}{x^3+4x^2+8x}dx$$
So I first started with be dividing $p(x)$ with $q(x)$ and got:
$$\int x-4+\frac{8x^2+32x-16}{x^3+4x^2+8x}dx=\frac{x^2}{2}-4x+\int \frac{8x^2+32x-16}{x^3+4x^2+8x}dx$$
Using partial sum I have received:
$$\int \frac{8x^2+32x-16}{x^3+4x^2+8x}dx=8\int -\frac{1}{4x}+\frac{5x}{4(x^2+4x+8)}+\frac{5}{x^2+4x+8} dx=-2ln|x|+8(\frac{5}{4}\int\frac{x}{(x^2+4x+8)} +5\int \frac{1}{x^2+4x+8})=-2ln|x| +10\int\frac{x}{(x^2+4x+8)}dx +40\int \frac{1}{x^2+4x+8}dx$$
How do I continue from here?
|
You're almost there! Note that $$\frac{x}{x^2 + 4x + 8} =\frac{1}{2}\left( \frac{2x + 4 - 4}{x^2 + 4x + 8}\right) = \frac{1}{2}\left(\frac{2x + 4}{x^2 + 4x + 8}\right) - \frac{2}{x^2 + 4x + 8}$$
So that we have $$10\int\frac{x}{(x^2+4x+8)}dx +40\int \frac{1}{x^2+4x+8}dx = 5 \int \frac{2x + 4}{x^2 + 4x + 8} \, \mathrm{d}x + 20 \int \frac{\mathrm{d}x}{x^2 + 4x + 8}$$
which become a logarithmic and arctangent standard integral respectively.
|
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|
Need help with the proof of the following theorem of the equation $x^4+y^4=z^4$ Theorem 7-2. The equation $x^4+y^4=z^4$ is not solvable in nonzero integers.
Proof: It suffices to show that there is no primitive solution of the equation $$x^4+y^4=z^2$$ (why?)
Suppose that $x,y,z$ constitute such solution; with no loss in generality we may take $x>0$, $y>0$, $z>0$, and y even. Writing the supposed relation in the form
$$(x^2)^2+(y^2)^2=z^2$$
we see from Theorem 7-1 (the theorem of Pythagorean triples ) that $x^2=a^2-b^2$, $y^2=2ab$,$z=a^2+b^2$ where $(a,b)=1$ and exactly one of a an b is odd. If a were even, we would have $$1 \equiv x^2=a^2-b^2 \equiv -1(mod \text{ } 4)$$
(how is the latter true)so b is even. We apply theorem of Pythagorean triples again , this time to the equation $$x^2+b^2=a^2$$
(why do we do the latter?) and obtain $x=p^2-q^2$,$b=2pq$, $a=p^2+q^2$, where $(p,q)=1$, $p>q>0$, and not both p an q are odd. From $y^2=2ab$ we have $$y^2=4pq(p^2+q^2)$$
Here any two of p, q and $p^2+q^2$ are relatively prime and hence each must be a square (why are they relatively prime and why must each of them be a square?): $p=r^2$, $q=s^2$, $p^2+q^2=t^2$, whence $t>1$ and $$r^4+s^4=t^2$$
Now $x=r^4-s^4$, $y=2rst$, $z=a^2+b^2=r^8+6r^4s^4+s^8$, so that $$z>(r^4+s^4)^2=t^4$$
or $t<z^{1/4}$. It follows that if one nonzero solution of $x^4+y^4=z^2$ were known, another solution $r,s,t$ could be found for which $rst \neq 0$ and $1<t<z^{1/4}$. If we started from $r,s,t$ instead of $x,y,z$, a third solution $r',s',t'$ could then be found such that $1<t'<t^{1/4}$, and so on. But this would yield an infinite decreasing sequence of positive integers, $z,t,t',....,$ which is impossible. Thus there is no nonzero solution. (Can you explain this better?)
Thanks for your help
|
If $(x,y,z)$ is a solution to $x_1^4+x_2^4=x_3^4$, then $(x,y,z^2)$ is a solution to $x_1^4+x_2^4=x_3^2$. This is the same thing as saying if $x_1^4+x_2^4=x_3^2$ has no solution, then neither does $x_1^4+x_2^4=x_3^4$.
$1\equiv x^2 \equiv a^2-b^2 \equiv 0-1 \equiv -1 \mod 4$, because $a$ is even and $b$ is odd, which cannot be the case.
We know $x^2=a^2-b^2$, so naturally $x^2+b^2=a^2$, for which we can apply the same trick again.
$p$ and $q$ are coprime by construction. Whenever you have a primitive Pythagorean triple, the terms being added are coprime. $p=r^2, q=s^2$ and $p^2+q^2=t^2$ because they are all coprime factors of a perfect square $y^2$.
The last paragraph is just saying that we could consistently construct $z>t>t'>t''>t'''>\ldots$, all of which need to be positive whole numbers, which is not possible. That should be pretty intuitive. For example, consider the sequence $64, 32, 16, 8, \ldots$. If they are all positive whole numbers, then it should be pretty clear that this sequence needs to end. Positive ensures you can't go into negatives, and whole numbers ensures that the sequence can't start squishing together.
|
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|
inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ How can I prove the inequality $\sqrt{\cos x}>\cos(\sin x)$ for $x\in(0,\frac{\pi}{4})$ ?
The derivative of $f(x):=\sqrt{\cos x}-\cos(\sin x)$ is very unpleasant, so the standard method is probably not be the right choice...
|
\begin{align*}
\sqrt{\cos x} & > \cos \sin x \\
\cos x & > \cos^2 \sin x \\
1 - 2 \sin^2 \frac x2 & > 1 - \sin^2 \sin x \\
\sin^2 \sin x & > 2 \sin^2 \frac x2 \\
\sin \sin x & > \sqrt 2 \sin \frac x2 \\
\end{align*}
Note that for $0<x<\frac \pi 4$ we have
$$\frac \pi 2 > \frac 1{\sqrt 2} = \sin \frac \pi 4 > \sin x = 2\sin \frac x2 \cos \frac x2 > 2 \sin \frac x2 \cos \frac \pi 8 > 0.$$
Since sine is increasing on the interval $\left(0,\frac\pi2\right)$ we can write
$$\sin \sin x > \sin \left(2\sin \frac x2 \cos \frac \pi 8 \right).$$
Therefore it is enough to prove that
$$\sin \left(2\sin \frac x2 \cos \frac \pi 8 \right) > \sqrt 2 \sin \frac x2.$$
Recall that for $t>0$ we have $\sin t > t - \frac{t^3}6$. Putting $t=2\sin \frac x2 \cos \frac \pi 8$ and keeping in mind that $2\sin \frac x2 \cos \frac \pi 8 < \frac 1{\sqrt 2}$ we obtain
\begin{align*}
\sin \left(2\sin \frac x2 \cos \frac \pi 8 \right) &> 2\sin \frac x2 \cos \frac \pi 8 - \frac 16 \left(2\sin \frac x2 \cos \frac \pi 8 \right)^3 >\\
&>2\sin \frac x2 \cos \frac \pi 8 - \frac 16 \left(\frac 1{\sqrt 2}\right)^2\left(2\sin \frac x2 \cos \frac \pi 8 \right) = \\
& = \frac{11}6 \sin \frac x2 \cos \frac \pi 8.
\end{align*}
Therefore it is enough to show that
$$\frac{11}6 \sin \frac x2 \cos \frac \pi 8 > \sqrt 2 \sin \frac x2$$
or
$$\frac{11}6 \cos \frac \pi 8 > \sqrt 2.$$
This is true as
\begin{align*}
\frac{11}6 \cos \frac \pi 8 & > \frac 53 \cos \frac \pi 6 = \\
& = \frac 53 \cdot \frac{\sqrt 3}2 = \\
& = \sqrt{\frac{25}{12}} >\\
& > \sqrt 2.
\end{align*}
|
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|
Solve for $x,y,z$ from the linear equations. The main question is :
$$\begin{align}
(b+c)(y+z)-ax &= b-c \tag{1} \\
(c+a)(z+x)-by &= c-a \tag{2} \\
(a+b)(x+y)-cz &= a-b \tag{3}\\
\end{align}$$
Solve for $x,y,z$ if $a+b+c\ne0$
My method :
Opening all brackets, we get,
$$by+bz+cy+cz-ax=b-c$$
$$cz+cx+az+ax-by=c-a$$
$$ax+ay+bx+by-cz=a-b$$
Adding,
$$(a+b+c)(x+y+z)=0$$
Thus,
$$(x+y+z)=0$$
Also,
$$by+bz+cy+cz-b+c=by$$
Thus,
$$b(y+z-1)+c(y+z+1)=ax$$
Similarly,
$$c(z+x-1)+a(z+x+1)=by$$
$$a(x+y-1)+b(x+y+1)=cz$$
I can't go any further, nothing is clicking. Please help me.
|
You're almost there; nice work!
Plug $x+y+z = 0$, in the form $(y+z) = -x$, into your first equation to get $(-x)(a+b+c) = b-c,$ so $$x = \frac{c-b}{a+b+c}.$$
Do similar things to the other two equations.
|
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|
Proving that $x^{16} > 5$ when given a polynomial of degree $15$. I am unable to prove the following
If $x^{15}-x^{13}+x^{11}-x^9+x^7-x^5+x^3-x = 7$ prove that $x^{16} > 15$.
|
Note that your polynomial is a geometric series with first term $-x$, common ratio $-x^2$ so that it can be written as $$\frac{x(x^{16} - 1)}{x^2 + 1} = 7$$
So $$x^{16} = \frac{7(x^2 + 1)}{x} + 1$$
And now, all you need to do is show that $\frac{x^2 + 1}{x} = x + \frac{1}{x} > 2$, which is not hard.
|
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|
If $\sin x + \sin y = 1$ and $\cos x + \cos y = 0$, solve for $x$ and $y$
*
*$\sin x + \sin y = 1$
*$\cos x + \cos y = 0$
Any valid pair of $(x, y)$ is fine, as the restrictions on the board in the image below are obscured.
I got the question from chapter 26 of a comic called Yamada-kun.
How can I solve this equation?
|
I will only look for solutions on $[0, 2\pi)$
We need to make $2$ observations:
$1)$ The maximum value of $\sin \theta$ is $1$. Therefore, if any of $\sin x$, $\sin y$ is negative, $\sin x + \sin y < 1$. It follows that both $x$ and $y$ must be $\in [0, \pi]$.
$2)$ $\cos x= -\cos y \iff \cos x = \cos (\pi - y)$ . (You can prove this using the cosine subtraction formula). Since we are working only on the interval $[0, \pi]$, we must have $x=\pi-y$. This makes sense, because the cosines of symmetric angles on opposite sides of the $y$ axis will cancel out to $0$.
Substitute $x=\pi-y$ into the first equation:
$$\sin (\pi-y) + \sin y = 1$$
Using $\sin (a-b)= \sin a \cos b - \cos a \sin b$ we can deduce that $\sin (\pi-y)=\sin y$
$$\sin y + \sin y = 1$$
$$\sin y = \dfrac 12$$
$$y = \dfrac {\pi}{6}, y= \dfrac {5\pi}{6}$$
$$x=\pi-y$$
Therefore the solutions are $ \left( \dfrac {\pi}{6}, \dfrac {5\pi}{6} \right)$ and $ \left(\dfrac {5\pi}{6}, \dfrac {\pi}{6} \right)$. The symmetry of the solutions is expected, since both original equations and symmetric.
|
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|
Why does $\sum_{n=1}^{\infty} \frac{2^n+1}{5^n+1}$ converge?
Why does $\sum_{n=1}^{\infty} \frac{2^n+1}{5^n+1}$ converge?
I've tried by using the ratio test but I don't get so far, I'm a little lost with it. Any help will be really aprecciated.
|
We have $a_n=\frac{2^n+1}{5^n+1}$.
\begin{align*}
\frac{a_{n+1}}{a_{n}}=&\frac{2^{n+1}+1}{2^{n}+1}\cdot\frac{5^{n}+1}{5^{n+1}+1}\\
&=\frac{2+\frac{1}{2^n}}{1+\frac{1}{2^n}}\cdot\frac{1+\frac{1}{5^n}}{5+\frac{1}{5^n}}
\end{align*}
When $n$ tends to $+\infty$, $|\frac{a_{n+1}}{a_{n}}|$ approaches $2/5<1$. Hence the series converges.
|
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|
If $ax^2 +bxy+cy^2+5x-2y+3$ divided by $x-y+1$ has remainder $0$, determine $a$, $b$, and $c$.
If $ax^2 +bxy+cy^2+5x-2y+3$ divided by $x-y+1$ has remainder $0$, determine $a$, $b$, and $c$.
I do not know how to approach this problem and would appreciate advice how to proceed.
|
In $ax^2+bxy+cy^2+5x−2y+3 = (x−y+1)(dx+ey+f)$, put $x=0, y=1$ to get $c+1 = 0$ and hence $c=-1$. Put $y=0, x=-1$ to get $a-2=0$ and hence $a=2$. Putting $x=1, y = 2$, we get $a+2b+4c+5-4+3 = 0$ and hence $b=-1$.
|
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|
Solve $\sec (x) + \tan (x) = 4$ $$\sec{x}+\tan{x}=4$$
Find $x$ for $0<x<2\pi$.
Eventually I get $$\cos x=\frac{8}{17}$$
$$x=61.9^{\circ}$$
The answer I obtained is the only answer, another respective value of $x$ in $4$-th quadrant does not solve the equation, how does this happen? I have been facing the same problem every time I solved this kind of trigonometric equation.
|
Rewrite your equation as
$$
\frac{1}{\cos x}+\frac{\sin x}{\cos x}=4
$$
that becomes
$$
\frac{1+\sin x}{\cos x}=4
$$
This reminds the formula for the tangent of the half angle
$$
\tan\frac{\alpha}{2}=\frac{\sin\alpha}{1+\cos\alpha}
$$
but sine and cosine are mixed up. Not a problem: set $x=\pi/2-t$ and take the reciprocal:
$$
\frac{\sin t}{1+\cos t}=\frac{1}{4}
$$
that's
$$
\tan\frac{t}{2}=\frac{1}{4}
$$
Thus
$$
t=2\arctan\frac{1}{4}
$$
so
$$
x=\frac{\pi}{2}-2\arctan\frac{1}{4}\approx1.08084
$$
In degrees, $x=61.92757^\circ$.
A different strategy could be rewriting the equation as
$$
\sin x=4\cos x-1
$$
and squaring (but this may add spurious solutions):
$$
1-\cos^2x=16\cos^2x-8\cos x+1
$$
that gives
$$
\cos x(17\cos x-8)=0
$$
We know that $\cos x\ne0$ (because the original problem has $\sec x$) and so we remain with $\cos x=8/17$.
This leads to $x=\arccos(8/17)$ or $x=2\pi-\arccos(8/17)$, but the second solution must be discarded, because it leads to $\sin x<0$, whereas
$$
4\cdot\frac{8}{17}-1>0
$$
so this is incompatible with $\sin x=4\cos x-1$.
|
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|
Number theory problem, fractions and gcd The problem says:
If $a$ and $b$ are positive integers such that $\dfrac{a+1}{b}+\dfrac{b+1}{a}$ is an integer, then show that $\sqrt{a+b}\ge \gcd(a,b)$.
Adding $\frac{2ab}{ab}$ to $\frac{a+1}{b}+\frac{b+1}{a}$ yields that ab divides (a+b+1)(a+b), but I haven´t been able to continue from there.
|
If $\gcd(a,b)=1$, then clearly $\sqrt{a+b} \geq \gcd(a,b)$. Now assume $\gcd(a,b)=d>1$. Then if $\frac{a+1}{b}+\frac{b+1}{a}=n$, then by your simplification, we have $$(a+b+1)(a+b)=(n+2)ab$$Note that $d^2$ divides the righthand side, so it must also divide the lefthand side. Since $d$ divides $a$ and $d$ divides $b$, $d$ must divide $a+b$. Since $a+b+1$ and $a+b$ are coprime, and since $d$ divides $a+b$, we must also have $d^2$ dividing $a+b$. In particular, $a+b\geq d^2 \implies \sqrt{a+b}\geq d$.
|
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|
How to solve this using power series method $\left(x^2+2\right)y''\:+\:xy'\:-\:y=0$ $\left(x^2+2\right)y''\:+\:xy'\:-\:y=0$
What's next after this?
$\sum _{n=2}^{\infty }\:n\left(n-1\right)a_nx^n+2\:\sum _{n=0}^{\infty \:}\left(n+2\right)\left(n-1\right)a_{n+2}x^n+\sum _{n=2}^{\infty \:\:}na_n^{\:}x^n\:-\:\sum _{n=0}^{\infty \:}a_n^{\:}x^n = 0.$
|
Restarting from scratch, we assume that the DE $$ (x^2+2)y''+xy'-y=0 \tag{1}$$
has an analytic solution in a neighbourhood of zero, given by
$$ y(x) = \sum_{n\geq 0} a_n x^n \tag{2} $$
In terms of the coefficients, $(1)$ translates into:
$$ (x^2+2)\sum_{n\geq 2} n(n-1)a_n x^{n-2} + x\sum_{n\geq 1}na_n x^{n-1} - \sum_{n\geq 0}a_n x^n = 0 \tag{3}$$
that is equivalent to:
$$ \sum_{n\geq 2}n(n-1)a_n x^n +\sum_{n\geq 0}2(n+2)(n+1)a_{n+2}x^n + \sum_{n\geq 1} n a_n x^n -\sum_{n\geq 0}a_n x^n = 0 \tag{4} $$
Now we consider what is the coefficient of $x^n$ in the last expression and we get that, for every $n\geq 2$,
$$ n(n-1) a_n + 2(n+2)(n+1) a_{n+2} + (n-1)a_n = 0\tag{5} $$
hence:
$$ \frac{a_{n+2}}{a_n} = - \frac{(n-1)}{2(n+2)} \tag{6} $$
and by induction:
$$ a_{n+2k} = \frac{(-1)^k}{2^k}\cdot\frac{(n+2k-3)(n+2k-5)\cdot\ldots\cdot(n-1)}{(n+2k)(n+2k-2)\cdot\ldots\cdot(n+2)}\,a_n \tag{7}$$
so a solution of $(1)$ is given by a hypergeometric function.
|
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|
Which answer is correct? Finding the limit of a radical as $x$ approaches infinity. When I take $$\lim_{x \to -∞} \sqrt{x^2+7x}+x,$$ I multiply by the conjugate over the conjugate to get
$$\lim_{x \to -∞}\frac{7x}{\sqrt{x^2+7x}-x},$$ and multiply by either $\frac{\frac{1}{x}}{\frac{1}{x}}$ or $\frac{\frac{1}{-x}}{\frac{1}{-x}}$ to get an undefined answer or $\frac{-7}{2}.$
My teacher's solution involves multiplying by $\frac{\frac{1}{-x}}{\frac{1}{-x}}:$
$$=\lim_{x \to -∞}\frac{-7}{\sqrt{x^2/x^2+7x/x^2}+1}$$
$$=-\frac{7}{\sqrt{1+0}+1}$$
$$=\frac{-7}{2}$$
However, I multiplied
by $\frac{\frac{1}{x}}{\frac{1}{x}}$ and got the following:
$$\lim_{x \to -∞}\frac{7}{\sqrt{x^2/x^2+7x/x^2}-1}$$
$$\frac{7}{\sqrt{1+0}-1}$$
$$\frac{7}{0}$$
Which is undefined.
Why does multiplying by what is essentially $1$ cause different answers in general, and in particular for evaluating limits?
|
Your teacher has left out some justification, which is probably why you are confused.
In general, we have the following identity for all real $x$:
$$\sqrt{x^2}=|x|.\tag{1}$$ Also, for nonnegative real $a$ and positive real $b,$ we have $$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac ab}.\tag{2}$$ Putting $(1)$ and $(2)$ together, if $x$ a non-zero real number and $a$ is a nonnegative real number, then $$\frac1{|x|}\sqrt{a}=\sqrt{\frac a{x^2}}.\tag{$\star$}$$
Your teacher took advantage of $(\star),$ without being explicit about it. In particular, since we're taking the limit as $x\to-\infty,$ then we may as well assume that $x\le-7$ (since it has to be, eventually), so that $x^2+7x$ is nonnegative, $x$ is non-zero, and $-x=|x|.$ Thus, we have $$\frac{1}{-x}\sqrt{x^2+7x}=\frac1{|x|}\sqrt{x^2+7x}=\sqrt{\frac{x^2+7x}{x^2}}.$$
On the other hand, $$\frac{1}{x}\sqrt{x^2+7x}=\frac1{-|x|}\sqrt{x^2+7x}=-\frac1{|x|}\sqrt{x^2+7x}=-\sqrt{\frac{x^2+7x}{x^2}}.$$ From there, you should come up with the same answer.
|
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|
Minimum value of $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$ If $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$
Find the minimum value of the function
I tried using the AMGM inequality and differentiation but didn't know how to solve it any ideas?
This is from a math competition. ( I would like to see the most efficient way as I think differentiation in a math competition is not that efficient)
Using Mogjals comment and using the AM-GM inequality , setting $A = \sqrt{x^2 + (1-x)^2}$ and $B=\sqrt{(1-x)^2 +(1+x)^2}$
then
$A+B \geq 2\sqrt{AB}$
$$ \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2} \geq 2\sqrt{2}\sqrt{x^2+1}\sqrt{2x^2-2x+1}$$
With equality if and only $A=B$ so $x=-\frac{1}{2}$
|
Using Minkowski Inequality
$$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\geq \sqrt{(a+c)^2+(b+d)^2}$$
and equality hold when $\displaystyle \frac{a}{b} = \frac{c}{d}$
So $$\sqrt{x^2+(1-x)^2}+\sqrt{(1-x)^2+(1+x)^2}\geq \sqrt{[x+(1-x)]^2+[(1-x)+(1+x)]^2}=\sqrt{5}$$
and Equality hold when $$\frac{x}{1-x}=\frac{1-x}{1+x}\Rightarrow x^2-2x+1=x^2+x\Rightarrow x=\frac{1}{3}$$
|
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|
Is it possible to $\int \sqrt{\cot x}$ by hand $$\int \sqrt{\cot x}{dx}$$
$$\int \sqrt{\frac{\cos x}{\sin x}}{dx} $$
Using half angle formula
$$\int \sqrt{\frac{1-\tan^2 \frac{x}{2}}{2\tan \frac{x}{2}}}{dx}$$
But I am not getting any lead from here .I think it is not possible to integrate $\sqrt{\cot x}$ by hand .
I calculated the result with the help of integral calculator
$$\dfrac{\ln\left(\left|\tan\left(x\right)+\sqrt{2}\sqrt{\tan\left(x\right)}+1\right|\right)-\ln\left(\left|\tan\left(x\right)-\sqrt{2}\sqrt{\tan\left(x\right)}+1\right|\right)+2\arctan\left(\frac{2\sqrt{\tan\left(x\right)}+\sqrt{2}}{\sqrt{2}}\right)+2\arctan\left(\frac{2\sqrt{\tan\left(x\right)}-\sqrt{2}}{\sqrt{2}}\right)}{2^\frac{3}{2}}$$
|
Using Ted Shifrin Hint
$$I = \int \sqrt{\cot x}dx$$ Put $\cot x= t^2\;,$ Then $$\csc^2 xdx = -2tdt\Rightarrow dx = -\frac{2t}{1+t^4}dt$$
So $$I = -\int\frac{2t^2}{1+t^4}dt = -\int \frac{\left[(t^2+1)+(t^2-1)\right]}{t^4+1}dt$$
Now Let $$J = \int \frac{t^2+1}{t^4+1}dt = \int \frac{1+\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt = \int\frac{\left(t-\frac{1}{t}\right)'}{\left(t-\frac{1}{t}\right)^2+\left(\sqrt{2}\right)^2}dt$$
So $$J = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t^2-1}{t}\right)=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\cot x-1}{\sqrt{\cot x}}\right)+\mathcal{C_{1}}$$
Let $$K = \int \frac{t^2-1}{t^4+1}dt = \int \frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt = \int\frac{\left(t+\frac{1}{t}\right)'}{\left(t+\frac{1}{t}\right)^2-\left(\sqrt{2}\right)^2}dt$$
So we get $$K = \frac{1}{2\sqrt{2}}\ln \left|\frac{t^2-\sqrt{2}t+1}{t^2-\sqrt{2}t-1}\right| = \frac{1}{2\sqrt{2}}\ln \left|\frac{\cot x-\sqrt{2}\sqrt{\cot x}+1}{\cot x-\sqrt{2}\sqrt{\cot x}-1}\right|+\mathcal{C_{2}}$$
So $$I = \int \sqrt{\cot x}dx = -\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\cot x-1}{\sqrt{\cot x}}\right)-\frac{1}{2\sqrt{2}}\ln \left|\frac{\cot x-\sqrt{2}\sqrt{\cot x}+1}{\cot x-\sqrt{2}\sqrt{\cot x}-1}\right|+\mathcal{C}$$
|
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|
Integrating $\displaystyle\int \frac{1+x^2}{1+x^4}dx$ I am trying to integrate this function, which I got while solving $\int\frac{1}{\sin^4( x) + \cos^4 (x)}$:
$$\int \frac{1+x^2}{1+x^4}\mathrm dx$$
I think to factorise the denominator, and use partial fractions. But I cant seem to find roots of denominator. I also am unable to think substitution.
|
Let $$I = \int\frac{1}{\sin^4 x+\cos^4 x}dx = \int\frac{1}{\sin^2 x\cos^2 x\left(\tan^2 x+\cot^2 x\right)}dx$$
$$I =\int\frac{\sin^2 x+\cos^2 x}{\sin^2 x\cos^2 x\left(\tan^2 x+\cot^2 x\right)}dx= \int\frac{\sec^2 x+\csc^2 x}{(\tan x-\cot x)^2+\left(\sqrt{2}\right)^2}dx$$
Now Put $\tan x-\cot x = t\;,$ Then $(\sec^2 x+\csc^2 x)dx=dt$
$$I = \int\frac{1}{t^2+(\sqrt{2})^2}dt = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t}{\sqrt{2}}\right)+\mathcal{C}$$
So $$I = \int\frac{1}{\sin^4 x+\cos^4 x}dx = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan x-\cot x}{\sqrt{2}}\right)+\mathcal{C}.$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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|
$x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$ Determine all the possibilities for rational roots of the polynomial $x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$. Then determine how many of the real roots of the polynomial may be positive and how many may be negative. Factor the polynomial to confirm your results.
The answer is possible rational roots: $+-1$; number of possible real roots - positive: four or two or zero, negative: zero; actual roots: $x = 1, 1, 1, 1$ (a quadruple root).
Using the rational root theorem, you divide the factors of the constant, $1$, by the factors of the lead coefficient, also a 1. That step gives you only two different possibilities for rational roots: $1$ and $-1$.
The signs change four times in the original polynomial, indicating $4$ or $2$ or $0$ positive real roots. Replacing each $x$ with $-x$, you get $x^4 + 4x^3 + 6x^2 + 4x + 1 = 0$. The signs never change. The polynomial is the fourth power of the binomial $(x - 1)$, so it factors into $(x - 1)^4 = 0$, and the roots are $1, 1, 1, 1$. There are four positive roots (all the same number, of course).
Can someone explain, the factorization of the polynomial? I do not understand, how it factors into $(x - 1)^4$.
|
The rational root theorem tells you that any rational root $\frac{p}{q}$ will have $p\mid 1$ and $q \mid 1$ so $\frac{p}{q} = \pm 1$, you can then verify that only $x=1$ is a root and be done.
|
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|
Problem in the solution of a trigonometric equation $\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$
I needed to solve the following equation:
$$\tan\theta + \tan 2\theta+\tan 3\theta=\tan\theta\tan2\theta\tan3\theta$$
Now, the steps that I followed were as follows.
Transform the LHS first:
$$\begin{split}
\tan\theta + \tan 2\theta+\tan 3\theta
&= (\tan\theta + \tan 2\theta)
+ \dfrac{\tan\theta + \tan 2\theta}
{1-\tan\theta\tan2\theta} \\
&= \dfrac{(\tan\theta + \tan 2\theta)(2-\tan\theta\tan2\theta)}
{1-\tan\theta\tan2\theta}
\end{split}$$
And, RHS yields
$$\begin{split}
\tan\theta\tan2\theta\tan3\theta
&= (\tan\theta\tan2\theta)\dfrac{\tan\theta + \tan 2\theta}
{1-\tan\theta\tan2\theta}
\end{split}$$
Now, two terms can be cancelled out from LHS and RHS, yielding the equation:
$$
\begin{split}
2-\tan\theta\tan2\theta &= \tan\theta\tan2\theta\\
\tan\theta\tan2\theta &= 1,
\end{split}$$
which can be further reduced as:
$$\tan^2\theta=\frac{1}{3}\implies\tan\theta=\pm\frac{1}{\sqrt3}$$
Now, we can yield the general solution of this equation:
$\theta=n\pi\pm\dfrac{\pi}{6},n\in Z$. But, setting $\theta=\dfrac{\pi}{6}$ in the original equation is giving one term $\tan\dfrac{\pi}{2}$, which is not defined.
What is the problem in this computation?
|
When you cancel out the terms from LHS and RHS, you drop the solutions when these terms are 0 or do not exist (because the denominator is 0). A trivial example would the $\theta=0$, which certainly is a solution, but you did not find it because of the canceled terms.
|
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|
Determinant of a $4 \times 4$ matrix $A$ and $(\det(A))^5$ Calculate $\det(A)$ and $\det(A)^5$:
$$A= \begin{bmatrix}a&a&a&a\\a&b&b&b\\a&b&c&c\\a&b&c&d\end{bmatrix}$$
I found $\det A$ with Laplace expansion: $$a(-abc+b^2c+a c^2-b c^2+a b d-b^2 d-a c d+b c d) .$$
But how can I determine $\det A^5$ easily/fast? I know that there is the possibility to use $\det(A^5)=\det(A)^5)$, but this is too long for given time to resolve the problem.
|
@Michael Freimann
What follows is not an answer. It is just an aside comment, that might interest some readers.
The kind of matrices considered in this question can be described by the following formula for their coefficients
$$A_{ij}=a_{min(i,j)}$$ for a given sequence $a_1,a_2,...a_n$, this sequence being here {a,b,c,d}.
Their inverses have a tridiagonal form with simple coefficients. For example, here:
$A^{-1}=\begin{pmatrix} \frac{1}{a} + \frac{1}{b-a}&\frac{1}{a - b}&0&0\\
\frac{1}{a - b}&\frac{1}{b-a} + \frac{1}{c-b}&\frac{1}{b - c}&0\\
0&\frac{1}{b - c}&\frac{1}{c-b} + \frac{1}{d-c }&\frac{1}{c - d}\\
0&0&\frac{1}{c - d}&\frac{1}{d-c}\end{pmatrix}$
In particular, when $a=b=c=d=1$,
$$\text{The inverse of} \ \ \begin{pmatrix}
1&1&1&1\\
1&2&2&2\\
1&2&3&3\\
1&2&3&4
\end{pmatrix} \ \ \text{is} \ \
\frac{1}{2}\begin{pmatrix} 2&-1&0&0\\
-1&2&-1&0\\
0&-1&2&-1\\
0&0&-1&1\end{pmatrix}
$$
One can recognized in the last matrix an approximation of the opposite of the second derivative, a very important matrix in numerical analysis: see this (slides 10, 29, 35, 37, 42...).
This is in clear connection with the decomposition given by @user1551, which, for $a=c=d=1$, gives:
$$A=\pmatrix{1&0&0&0\\ 1&1&0&0\\ 1&1&1&0\\ 1&1&1&1}
\pmatrix{1&1&1&1\\ 0&1&1&1\\ 0&0&1&1\\ 0&0&0&1}$$
which is equivalent to a double discrete integration operator.
Of course, this property extends naturally to any dimension. Moreover, it has a nice correspondence with the "continuous world": see this.
|
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|
Answer mismatch in definite integral question:$\int_{0}^{2\pi}{\frac{x^2\sin(x)}{8+\sin^2(x)}}dx$
I faced a problem while evaluating the following integral.
$$\int_{0}^{2\pi}{\frac{x^2\sin(x)}{8+\sin^2(x)}}dx$$
MY ATTEMPT:
Let $$I=\int_{0}^{2\pi}{\frac{x^2\sin(x)}{8+\sin^2(x)}}dx$$
By King's Rule $$I=-\int_{0}^{2\pi}{\frac{(2\pi-x)^2\sin(x)}{8+\sin^2(x)}}dx$$
On adding the above two integrals,$$2I=\int_{0}^{2\pi}{\frac{(-4{\pi}^2+4 \pi x)\sin(x)}{8+\sin^2(x)}}dx$$
or, $$2I=\int_{0}^{2\pi}{\frac{(-4{\pi}^2)\sin(x)}{8+\sin^2(x)}}dx+\int_{0}^{2\pi}{\frac{(4 \pi x)\sin(x)}{8+\sin^2(x)}}dx$$
The first integral follows the property $f(x)=-f(2\pi-x)$ so it vanishes.
Thus,
$$2I=\int_{0}^{2\pi}{\frac{(4 \pi x)\sin(x)}{8+\sin^2(x)}}dx$$
or,
$$I=2 \pi\int_{0}^{2\pi}{\frac{ x\sin(x)}{8+\sin^2(x)}}dx$$
Applying king's rule and adding,
$$2I=2 \pi\int_{0}^{2\pi}{\frac{ 2\pi\sin(x)}{8+\sin^2(x)}}dx$$
$$I=2{\pi}^2\int_{0}^{2\pi}{\frac{ \sin(x)}{8+\sin^2(x)}}dx$$
$$I=2{\pi}^2\int_{0}^{2\pi}{\frac{ \sin(x)}{9-\cos^2(x)}}dx$$
$$I=[2\pi^2(\frac{1}{6} \log(3-\cos(x))-\frac{1}{6} \log(\cos(x)+3))]_{0}^{2\pi}+C$$
Thus,$I=0$
But the answer is not 0.Where did I go wrong?
P.S King's Rule is $$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$$
|
Symmetry is a bit more evident if we set $x=z+\pi$. Since $\sin(z+\pi)=-\sin(z)$,
$$ I = -\int_{-\pi}^{\pi}\frac{(z^2+2\pi z+\pi^2)\sin(z)}{8+\sin(z)^2}\,dz=-2\pi\int_{-\pi}^{\pi}\frac{z\sin(z)}{8+\sin(z)^2}=-4\pi\int_{0}^{\pi}\frac{z\sin z}{8+\sin(z)^2}\,dz $$
and since $\sin(\pi-z)=\sin(z)$,
$$ I = -4\pi^2\int_{0}^{\frac{\pi}{2}}\frac{\sin z}{8+\sin(z)^2}\,dz=4\pi^2\int_{0}^{\frac{\pi}{2}}\frac{\cos' z}{9-\cos(z)^2}\,dz =-4\pi^2\int_{0}^{1}\frac{du}{9-u^2}$$
and by partial fraction decomposition:
$$ I = -\frac{2\pi^2}{3}\int_{0}^{1}\left(\frac{1}{3-u}+\frac{1}{3+u}\right)\,du=-\frac{2\pi^2}{3}\left(\log\frac{3}{2}+\log\frac{4}{3}\right)=\color{red}{-\frac{2\pi^2}{3}\log 2}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
Solve $x \equiv 32 \pmod{81}$ and $x \equiv 59 \pmod{64}$.
$32 + 81k = 59 + 64n \implies 81k - 64n = 27$
$17k \equiv 27 \pmod{64}$.
$64 = 3(17) + 13$
$17 = 1(13) + 4$
$13 = 3(4) + 1$
So $1 = 13 - 3(4) = 13 - 3[17 - 13] = 4(13) - 3(17) = 4(64 - 3*17) - 3*17 =
4(64) - 15(17)$. Thus $k \equiv 1 \pmod{64} \implies k = 1 + 64y$ so we have $n = \frac{54 + 5184y}{64}$
But this is not possible . Help?
EDIT $k\equiv 43 \pmod{64}$ thus,
$k = 43 + 64y$ thus $x = 32 + 81(43 + 64y) = 3515 + 5824y$ so then $x \equiv 3515 \pmod{5824}$.
|
So $$x = 81 \cdot a + 64 \cdot b$$ is what you need; where $81a \equiv 59\pmod{64}$ which you can do by finding $k$ such that $81k \equiv 1 \pmod{64}$ and then multiplying both sides by $59$.
Then you also need $64b \equiv 32\pmod{81}$ which you can do by finding $m$ such that $64m \equiv 1 \pmod{81}$ and multiplying both sides by $32$. Effectively you've reduced the problem to computing multiplicative inverses modulo $64$ and $81$, which is easy to do via trial and error or via the Euclidean algorithm.
|
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|
Prove the Inequality proves true, three variables Prove that for all positive real numbers $a,b,c$ we have:
$$\frac{1}{4a}+\frac{1}{4b}+\frac{1}{4c}+\frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\geq\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$$
So far I have solved for $a=b=c=1$ and $a=1$, $b=2$, $c=3$ to show that the inequality holds true but I am needing the answer written more like a proof, I'm just not sure what theorems to apply actually prove the inequality holds true.
|
Also we can use $uvw$ here.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $\frac{3v^2}{4w^3}+\frac{\sum\limits_{cyc}(3u+a)(3u+b)}{\prod\limits_{cyc}(3u+a)}\geq\frac{\sum\limits_{cyc}(a+b)(a+c)}{9uv^2-w^3}$ or
$\frac{v^2}{4w^3}+\frac{15u^2+v^2}{54u^3+9uv^2+w^3}\geq\frac{3u^2-v^2}{9uv^2-w^3}$, which is $f(w^3)\geq0$,
where $f$ is a concave function (it's obvious that the coefficient before $w^6$ is negative).
But a concave function gets a minimal value for an extremal value of $w^3$,
which happens in the following cases.
*
*$w^3\rightarrow0^+$. In this case our inequality is obviously true.
*$b=c=1$, which gives $(a-1)^2\geq0$. Done!
|
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|
About $(x^2+y^2)dy-y^2dx=0$ differential equation Let us show $(x^2+y^2)dy-y^2dx=0$ differantial equation is homogeneous then solve the equation.
Firstly I did show we can also write $\frac{y^2}{x^2+y^2}=\frac{dy}{dx}$ equation depending on the first equation. Let $v=y/x$. Then we get $\frac{v^2}{v^2+1}=\frac{dy}{dx}x+v$ equation. If we go on and integrate that equation, we will get $\int\frac{(v^2+1)dv}{v^3-v^2+v}+\int\frac{dx}{x}=C$ equation. I failed to integrate the left term, what to do about it?
By the way, I am a bit bad at English so if you don't understand something, tell me to explain it.
|
First of all you have $y=vx$ and $\frac{dy}{dx}=x\frac{dv}{dx}+v$, so we need to solve
$$\frac{v^2}{1+v^2}=x\frac{dv}{dx}+v$$
$$\frac{v^2-v(1+v^2)}{1+v^2}=x\frac{dv}{dx}$$
or
$$\frac{1+v^2}{-v^3+v^2-v}dv=\frac{1}{x}dx.$$
The right is easy to integrate; the left you should be able to tackle with partial fraction:
$$\frac{1+v^2}{v(-v^2+v-1)}=-\frac{1}{v^2-v+1}-\frac{1}{v}.$$
The only difficulty is the first term. Note that
$$\frac{1}{v^2-v+1}=\frac{1}{(v-1/2)^2+3/4}.$$
Use the fact that $\frac{d}{dx}\arctan\frac{x}{r}=\frac{r}{r^2+x^2}$. Then what is the derivative of $\arctan\frac{x-1/2}{\sqrt 3/2}$?
|
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|
Why do we find vertical asymptote of functions (in rational fractions) by turning the denominator into $0$? Like for $$f(x) = \frac{1}{x-2}$$ we can find its vertical asymptote by finding what $x$ can be in order to turn the denominator into zero.
e.g. $x=2$
But why is that? Why does the vertical asymptote correspond to the zero of the denominator?
|
A function $f$ has a vertical asymptote at $x = a$ if $|f(x)|$ increases without bound as $x$ approaches $a$, that is, $f$ has a vertical asymptote if one or more of the following conditions holds:
\begin{align*}
\lim_{x \to a^+} f(x) & = \infty\\
\lim_{x \to a^+} f(x) & = -\infty\\
\lim_{x \to a^-} f(x) & = \infty\\
\lim_{x \to a^-} f(x) & = -\infty
\end{align*}
Why does the function
$$f(x) = \frac{1}{x - 2}$$
have a vertical asymptote at $x = 2$?
Consider what happens as $x$ approaches $2$ from the right.
$$
\begin{array}{c c c}
x & x - 2 & f(x) = \dfrac{1}{x - 2}\\ \hline
2.1 & 0.1 & 10\\
2.01 & 0.01 & 100\\
2.001 & 0.001 & 1000\\
2.0001 & 0.0001 & 10000
\end{array}
$$
Observe that as $x$ approaches $2$ from the right, we can make $f(x)$ as large as we want by making $x - 2$ sufficiently close to $0$. Consequently, $f$ increases without bound as $x$ approaches $2$ from the right. We write
$$\lim_{x \to 2+} f(x) = \infty$$
read the limit as $x$ approaches $2$ from above (or from the right) is infinity, to indicate that the function increases without bound as $x$ approaches $2$ from the right. Since this condition holds, the function $f(x) = \frac{1}{x - 2}$ has a vertical asymptote at $x = 2$.
Consider what happens as $x$ approaches $2$ from the left.
$$
\begin{array}{c c c}
x & x - 2 & f(x) = \dfrac{1}{x - 2}\\ \hline
1.9 & -0.1 & -10\\
1.99 & -0.01 & -100\\
1.999 & -0.001 & -1000\\
1.9999 & -0.0001 & -10000
\end{array}
$$
Observe that $x$ approaches $2$ from the left, we can make $f(x)$ as large a negative number as we like by taking $x - 2$ sufficiently close to $0$. Consequently, $f$ decreases without bound as $x$ approaches $2$ from the left. We write
$$\lim_{x \to 2-} f(x) = -\infty$$
read the limit of $f(x)$ as $x$ approaches $2$ from below (or from the left) is negative infinity, to indicate that $f$ decreases without bound as $x$ approaches $2$ from the left. This condition also implies that the function $f(x) = \frac{1}{x - 2}$ has a vertical asymptote at $x = 2$.
The reason why the function $f(x) = \frac{1}{x - 2}$ has a vertical asymptote at $x = 2$ is that as $x$ approaches from the right or from the left, the denominator approaches $0$ while the numerator does not. This occurs near the zero of the denominator since we can make the denominator arbitrarily small without making the numerator arbitrarily small.
More generally, for a rational function in simplest form, the function has a vertical asymptote at points where the denominator is equal to zero and the numerator is not equal to zero. It is not sufficient that the denominator equal zero.
Consider the function
$$g(x) = \frac{x^2 - 4}{x - 2}$$
Notice that
\begin{align*}
g(x) & = \frac{x^2 - 4}{x - 2}\\
& = \frac{(x + 2)(x - 2)}{x - 2}\\
& = x + 2 && \text{provided that $x \neq 2$}
\end{align*}
While both $f(x) = \frac{1}{x - 2}$ and $g(x) = \frac{x^2 - 4}{x - 2}$ are undefined at $x = 2$ since the denominator is equal to $0$ when $x = 2$, the function $g$ does not have a vertical asymptote at $x = 2$. As $x$ approaches $2$, $g(x)$ approaches $2 + 2 = 4$, so $|g(x)|$ does not increase without bound as $x$ approaches $2$. Instead, the graph of $g$ is a punctured line.
|
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|
If y=y(x) and $\frac{2+sin(x)}{y+1}\frac{dy}{dx}=-cosx,y(0)=1$, then find $y(\frac{\pi}{2})$. Question:
If y=y(x) and $\frac{2+sin(x)}{y+1}\frac{dy}{dx}=-cosx,y(0)=1$, then find $y(\frac{\pi}{2})$.
My attempt:-
$$\frac{2+sin(x)}{y+1}\frac{dy}{dx}=-cosx$$
$$\frac{dy}{1+y}=-\frac{cos(x)}{2+sin(x)}dx$$
Integrating both sides
$$\int\frac{dy}{1+y}=-\int\frac{cos(x)}{2+sin(x)}dx$$
$$log(1+y)=-log(2+sin(x))+C$$
$$log(1+y)=log(\frac{1}{2+sin(x)})+C$$
Antilog on both sides
$$1+y=\frac{1}{2+sin(x)}+C$$
$$y=\frac{-1-sin(x)}{2+sin(x)}+C$$
Now it is given that $y(0)=1$
So $$y(0)=1=\frac{-1-sin(0)}{2+sin(0)}+C$$
$$1=\frac{-1}{2}+C$$
$$C=\frac{3}{2}$$
$$y(\frac{\pi}{2})=\frac{-1-sin(\frac{\pi}{2})}{2+sin(\frac{\pi}{2})}+C$$
$$y(\frac{\pi}{2})=\frac{-2}{3}+\frac{3}{2}$$
$$y(\frac{\pi}{2})=\frac{5}{6}$$
But the answer happens to be $\frac{1}{3}$.
|
You made a mistake by antiloging. The line $ 1+y=\frac{1}{2+\sin(x)}+C $ is wrong.
It would be easier if you would write $\log C$ instead $C$ in the line before. So in the next step you would get:
$$1+y=\frac{1}{2+\sin(x)} \cdot C$$
|
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|
Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$ Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge \frac{1}{2}$ where $a,b,c,d \ge 0$
My attempt:I used two ways but I get to a wrong answer.
My first way:We know that $\frac{a}{b}+\frac{b}{a} \ge 2$ where $a,b \ge 0$
Then:
$\frac{a}{b+c}+\frac{b+c}{a}+\frac{b}{c+d}+\frac{c+d}{b}+\frac{c}{d+a}+\frac{d+a}{c}+\frac{d}{a+b}+\frac{a+b}{d} \ge 8$
And:
$\frac{b+c}{a}+\frac{c+d}{b}+\frac{d+a}{c}+\frac{a+b}{d}=\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}+\frac{a}{c}+\frac{a}{d}+\frac{b}{d}=$
$\frac{a}{c}+\frac{c}{a}+\frac{b}{d}+\frac{d}{b} \ge 4$
$+$
$\frac{b}{a}+\frac{c}{b}+\frac{d}{c}+\frac{a}{d}\ge 4\sqrt{\frac{b}{a}*\frac{c}{b}*\frac{d}{c}*\frac{a}{d}}=4$
Then:
$\frac{a}{b+c}+\frac{b+c}{a}+\frac{b}{c+d}+\frac{c+d}{b}+\frac{c}{d+a}+\frac{d+a}{c}+\frac{d}{a+b}+\frac{a+b}{d} \ge 8$
And:
$\frac{b+c}{a}+\frac{c+d}{b}+\frac{d+a}{c}+\frac{a+b}{d}=\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{d}{b}+\frac{d}{c}+\frac{a}{c}+\frac{a}{d}+\frac{b}{d}\ge 8$
Then we will get:
$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge 0$
Which is not true.
My second way:I don't have enough time then I just explain it.
I used caushy-shuartz and I get:
$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b} \ge 2$
|
$$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac} \ge \frac{(a+b+c+d)^2}{ab+bc+cd+da+2ac+2bd}=\frac{(a+b+c+d)^2}{(a+c)(b+d)+2ac+2bd}$$
From Cauchy.
$$(a+b+c+d)^2=(a+c)^2+(b+d)^2+2(a+c)(b+d) \ge 4ac +4bd+2(a+c)(b+d)$$
Since $(x+y)^2 \ge 4xy$.
So $$\sum_{cyc}\frac{a}{b+c} \ge 2 >\frac{1}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1889950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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|
Area enclosed by the curve $5x^2+6xy+2y^2+7x+6y+6=0$ We have to find the area enclosed by the curve
$$5x^2+6xy+2y^2+7x+6y+6=0.$$
I tried and I got that it is an ellipse, and I know its area is $\pi ab$ where $a$ and $b$ are the semiaxis lengths of the ellipse.
But I am unable to find the value of $a$ and $b$.
|
The center:
find the minimum of $(5x^2 + 6xy + 2y^2 + 7x + 2y)$
$\frac {\partial}{\partial x}(5x^2 + 6xy + 2y^2 + 7x + 2y) = 10x + 6y + 7= 0\\
\frac {\partial}{\partial y}(5x^2 + 6y + 2y^2 + 7x + 2y) = 6x + 4y + 6= 0$
Solve this system of equations
$x = 2, y = -4.5$
$2(x-2)^2 + 6(x-2)(y+4.5)+5(y+4.5)^2 = 0.5$
$4(x-2)^2 + 12(x-2)(y+4.5)+10(y+4.5)^2 = 1$
$\mathbf x^T\begin{bmatrix}
4&6\\
6&10\end{bmatrix}\mathbf x =1$
Now we can find the eigenvalues of that matrix, but rather than solving for them I am just going to call them $\lambda_1,\lambda_2$
$x'^2 \lambda_1 + y'^2 \lambda_2 = 1$
$\lambda_1\lambda_2 = \frac 1{(ab)^2}$
and $\lambda_1\lambda_2$ = determinant of the matrix above = $4$.
$ab = \frac 12$
$A = \frac \pi2$
The transformed equation:
$x'^2 \lambda_1 + y'^2 \lambda_2 = 1$
$\lambda_1 = 7 + \sqrt{45}, \lambda_2 = 7 - \sqrt{45}\\
x' = (x-2) (\cos \theta) + (y+4.5)(\sin \theta)\\
y' = (x-2)(\sin \theta) - (y+4.5)(\cos \theta)\\
\theta = \frac 12 \tan^{-1} 2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1890139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
}
|
How to prove a fomula is real or not? The formula is:
$$f(u):=\sqrt{1+i}\arctan\left(\dfrac{\sqrt{2}}{\sqrt{(-1+i)u}}\right)+\sqrt{1-i}\arctan\left(\frac{(-1)^{3/4}\sqrt{(-1-i)u}}{u}\right)\tag{1}$$
I obtained it from the following definite integral:
$$f(u):=\int_u^{\infty } \frac{\sqrt{x}}{x^2-2 x+2} \, dx\quad {\text{ where }} u>0\tag{2}$$
Numercial estimation indicates that, when $u>0$, $f(u)\in \mathbb{R}$ because the imaginary part of the results are close to numerical epsilon.
How to prove or disprove it? If the conclusion is true, is it possible to eliminate the imaginary unit $i$ from the fomula of $f(u)$ in $(1)$ and obtain a relatively simpler form of it which does not contain or implies any imaginary number?
update
$$f^*(u):=\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{\sqrt{2}}} \left(\left(\sqrt{2}-1\right) \left(\ln \left(u-\sqrt{2 \left(1+\sqrt{2}\right)} \sqrt{u}+\sqrt{2}\right)-\ln \left(u+\sqrt{2 \left(1+\sqrt{2}\right)} \sqrt{u}+\sqrt{2}\right)\right)-2 \arctan\left(-\sqrt{2 \left(1+\sqrt{2}\right)} \sqrt{u}+\sqrt{2}+1\right)+2 \arctan\left(\sqrt{2 \left(1+\sqrt{2}\right)} \sqrt{u}+\sqrt{2}+1\right)-2 \pi \right)$$
|
Inspired by How to simplify $\Re\left[\sqrt2 \tan^{-1} {x\over \sqrt i}\right]$?
I manage to write a primitive of $\frac{\sqrt{x}}{x^2-2 x+2}$ using real functions. Let
$$F(x):=\frac{1}{\sqrt{2\sqrt{2}-2}}\cdot
\arctan\left(\frac{\sqrt{2\sqrt{2}-2}\cdot \sqrt{x}}{x-\sqrt{2}}\right)$$
$$G(x):=\frac{1}{\sqrt{2\sqrt{2}+2}}\cdot
\mbox{arctanh}\left(\frac{\sqrt{2\sqrt{2}+2}\cdot\sqrt{x}}{x+\sqrt{2}}\right),$$
then you can verify that
$$\frac{d}{dx}\left(-F(x)-G(x)\right)= \frac{\sqrt{x}}{x^2-2 x+2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1890943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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|
How to determine the matrix in the following case Say we have a vector $\textbf{b}$ and $\textbf u$ such that:
$$A \mathbf b= \mathbf u$$
Where $A$ is a square matrix.
If $\mathbf b$ and $\mathbf u$ are known and $A$ is the unknown, How to get the matrix $A$ (perhaps it is not unique but how can one proceed to get it)?
|
$$\begin{pmatrix}a_{11} &a_{12} \\a_{21} &a_{22}\end{pmatrix}\begin{pmatrix}b_1\\ b_2\end{pmatrix}=\begin{pmatrix}u_1\\ u_2\end{pmatrix}$$
Is equivalent to
$$\begin{pmatrix}a_{11}b_1+a_{12}b_2 \\a_{21}b_1+a_{22}b_2 \end{pmatrix}=\begin{pmatrix}u_1\\ u_2\end{pmatrix}$$
or
$$\begin{pmatrix}b_1 &b_2 &0 &0\\0 &0 &b_1 &b_2 \end{pmatrix}\begin{pmatrix}a_{11}\\ a_{12}\\a_{21}\\a_{22} \end{pmatrix}=\begin{pmatrix}u_1\\ u_2\end{pmatrix}.$$
You can then solve for $A$ using row reduction of the appropriate augmented matrix.
This can be easily generalized: Suppose $A$ is $m\times n$ and $b$ is $n\times 1$. Let $a^T$ be the $1\times mn$ vector formed by the concatenation of the rows of $A$, and $B$ be the $m\times mn$ matrix given by
$$B=\begin{pmatrix}b^T & 0 & \cdots & 0\\0 &b^T &\cdots &0\\\vdots&\vdots&\ddots &\vdots\\
0&0&\cdots & b^T\end{pmatrix} $$
where each $0$ represents a row vector of $n$ zeros. Then the system is
$$Ba=u.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1891041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Proof verification: $a_1=2$, $a_{n+1}=3+\frac{a_n}{2}$ is increasing and bounded For base case, $a_2=3+\frac{a_1}{2}=4>a_1$
For $n=k$, let $a_{k+1}>a_k$
Adding 3 on both sides after dividing by 2,
$$3+\frac{a_{k+1}}{2}>3+\frac{a_k}{2}$$
$$a_{k+2}>a_{k+1}$$
Hence the sequence is increasing.
How can I show that the sequence is bounded?
|
Let $a_n=b_n+6$. Then
$$ b_{n+1}+6=3+\frac{b_n}{2}+3 $$
is equivalent to $b_{n+1}=\frac{b_n}{2}$ or to $b_n=\frac{b_1}{2^{n-1}}$. Since $a_1=2$ we have $b_1=-4$, hence
$ b_n = -\frac{8}{2^n} $ and
$$\boxed{ a_n = \color{red}{6-\frac{8}{2^n}} }$$
Now the claim is trivial.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1893200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Smallest positive integer that gives remainder 5 when divided by 6, remainder 2 when divided by 11, and remainder 31 when divided by 35? What is the smallest positive integer that gives the remainder 5 when divided by 6, gives the remainder 2 when divided by 11 and 31 when divided by 35?
Also, are there any standard methods to solve problems like these?
|
\begin{align}
n \equiv 5 &\pmod{6} \\
n \equiv 2 &\pmod{11} \\
n \equiv 31 &\pmod{35}
\end{align}
The idea is to find three integers, $A, B,$ and $C$ such that
\begin{array}{|l|l|l|}
A \equiv 1 \pmod{6} &
B \equiv 0 \pmod{6} &
C \equiv 0 \pmod{6}
\\
A \equiv 0 \pmod{11} &
B \equiv 1 \pmod{11} &
C \equiv 0 \pmod{11}
\\
A \equiv 0 \pmod{35} &
B \equiv 0 \pmod{35} &
C \equiv 1 \pmod{35}
\end{array}
Your answer will then be $5A + 2B + 31C \pmod{2310}$ where
$2310 = 6 \cdot 11 \cdot 35$. Since $6, 11,$ and $35$ are pairwise prime, the Chinese remainder theorem guarantees that $A, B,$ and $C$ exist.
There is a formula that describes how to find $A, B,$ and $C$, but it isn't that much trouble to figure them out from scratch.
The equivalences $A \equiv 0 \pmod{11}$ and $A \equiv 0 \pmod{35}$ imply that $A$ must be a multiple of $385 = 11 \cdot 35.$ So $A = 385x$ for some integer $x$. We solve
\begin{align}
A &\equiv 1 \pmod 6 \\
385x &\equiv 1 \pmod 6 \\
x &\equiv 1 \pmod 6 \\
\hline
A &= 385(1) = 385
\end{align}
The equivalences $B \equiv 0 \pmod{6}$ and $B \equiv 0 \pmod{35}$ imply that $B$ must be a multiple of $210 = 6 \cdot 35.$ So $B = 210x$ for some integer $x$. We solve
\begin{align}
B &\equiv 1 \pmod{11} \\
210x &\equiv 1 \pmod{11} \\
x &\equiv 1 \pmod{11} \\
\hline
B &= 210(1) = 210
\end{align}
The equivalences $C \equiv 0 \pmod{6}$ and $C \equiv 0 \pmod{11}$ imply that $C$ must be a multiple of $66 = 6 \cdot 11.$ So $C = 66x$ for some integer $x$. We solve
\begin{align}
C &\equiv 1 \pmod{35} \\
66x &\equiv 1 \pmod{35} \\
-4x &\equiv 1 \pmod{35} \\
x &\equiv -9 \pmod{35} \\
\hline
C &= 66(-9) = -594
\end{align}
We then compute
\begin{align}
n
& \equiv 5A + 2B + 35C \pmod{2310} \\
&\equiv 5(385) + 2(210) + 31(-594) \pmod{2310} \\
&\equiv 101\pmod{2310} \\
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1897521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 5
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|
Find $\frac{x^2}{y^2}$ + $\frac{y^2}{x^2}$ If $\frac{x}{y}$ + $\frac{y}{x}$ = 3 Find $\frac{x^2}{y^2}$ + $\frac{y^2}{x^2}$
Any Ideas on how to begin ?
|
Start by squaring both sides of $\frac{x}{y}+\frac{y}{x}=3$. What happens to the cross terms?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1898615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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|
Solving a system of linear equations with undefined number of equations I need to solve this system of linear equations:
$k_0=1+k_1$
$k_1=1+\frac{1}{3} k_1 +\frac{2}{3} k_2$
$k_2=1+\frac{1}{3} k_1 +\frac{2}{3} k_3$
$k_3=1+\frac{1}{3} k_1 +\frac{2}{3} k_4$
$...$
$k_{N-1}=1+\frac{1}{3} k_1 +\frac{2}{3} k_N$
$k_N=0$
I need to find a general solution for $k_0$ all $N$.
The system is equivalent to:
$k_0=1+k_1$
$k_i=1+\frac{1}{3} k_1 +\frac{2}{3} k_{i+1} \;\;\; i\in 1,2,3,...,N-1$
$k_N=0$
that is:
$k_0=1+k_1$
$k_i=\frac{3}{2} k_{i-1} -\frac{1}{2} k_1-\frac{3}{2} \;\;\; i\in 2,3,...,N \;\;\; [1]$
$k_N=0 \;\;\; [2]$
As far as I know, $[1]$ and $[2]$ is a linear unhomogenous recurrence relation with constant coefficients of order 1 with initial condition $k_N=0$ . The unhomogenous term is $f(n)=-\frac{1}{2} k_1-\frac{3}{2}$.
Solving this recurrence relation gives as general solution:
$k_i=A \left ( \frac{3}{2} \right )^i+3+k_1 \;\;\; [3]$
with $A=\left ( \frac{2}{3} \right )^N \left ( -3-k_1 \right )$
Now I don't know how to proceed. As I said, I need to find $k_0$, but the solution $[3]$ is just for $i\in 2,3,...,N$. How can I use $[3]$ to find a general solution for $k_0$ for all $N$?
Thank you.
|
You have done all the work already. Now simply use $[3]$ for $i=2$ together with the second equation in your question to - after reducing a bit - obtain
\begin{align}
k_1&=\frac{9}{2}\left(\frac{3}{2} \right)^{N-2}-3 \\
k_0&=1+k_1=\frac{9}{2}\left(\frac{3}{2} \right)^{N-2}-2
\end{align}
EDIT: Or use Johannes Kloos much faster method.
EDIT 2: Actually, as OP points out, Johannes Kloos' method actually doesn't work, since we get $$0=-\left(\frac{2}{3}\right)^N(3+k_1)\left(\frac{3}{2}\right)^N+3+k_1 \Rightarrow 0=0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1898744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
If $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1$, what can we say about $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$?
Suppose that $a,b,c$ are three real numbers such that $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1$. What are the possible values for $\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}$?
After clearing the denominators, we have $$a(c+a)(a+b)+b(b+c)(a+b)+c(b+c)(c+a)=(a+b)(b+c)(a+c)\,.$$
That is,
$$a^3+b^3+c^3+abc=0\,.$$
But then I'm stuck. This question is related, but a bit different.
Thank you for your help!
|
Since $\sum\limits_\text{cyc}\,\frac{a}{b+c}=1$, we have $$a+b+c=(a+b+c)\,\left(\sum_\text{cyc}\,\frac{a}{b+c}\right)=\sum_{\text{cyc}}\left(\frac{a^2}{b+c}+a\right)\,.$$
That is,
$$a+b+c=\sum_{\text{cyc}}\left(\frac{a^2}{b+c}\right)+(a+b+c)\,.$$
Hence,
$$\sum_{\text{cyc}}\left(\frac{a^2}{b+c}\right)=0\,.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1901062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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|
integral $\int_0^{\pi} \left( \frac{\pi}{2} - x \right) \frac{\tan x}{x} \, {\rm d}x$ Evaluate , if possible in a closed form, the integral:
$$\int_0^{\pi} \left( \frac{\pi}{2} - x \right) \frac{\tan x}{x} \, {\rm d}x$$
Basically, I have not done that much. I broke the integral
\begin{align*}
\int_{0}^{\pi} \left ( \frac{\pi}{2}-x \right ) \frac{\tan x}{x} \, {\rm d}x &= \int_{0}^{\pi/2} \left ( \frac{\pi}{2} - x \right ) \frac{\tan x}{x} \, {\rm d}x + \int_{\pi/2}^{\pi} \left ( \frac{\pi}{2} - x \right ) \frac{\tan x}{x} \, {\rm d}x\\
&\!\!\!\!\!\!\overset{u=\pi/2-x}{=\! =\! =\! =\! =\! =\!} \int_{0}^{\pi/2} \frac{u \cot u}{\frac{\pi}{2}-u} \, {\rm d}u + \int_{-\pi/2}^{0} \frac{u \cot u}{\frac{\pi}{2}-u} \, {\rm d}u\\
&= \int_{-\pi/2}^{\pi/2} \frac{u \cot u}{\frac{\pi}{2}-u} \, {\rm d}u\\
&\approx 2.13897
\end{align*}
I have no idea how to evaluate this. I was thinking of IBP and then some kind of Fourier , but I cannot get it to work. Any ideas?
|
Too long for a comment. We have the following representations:
\begin{align*}
\int_{0}^{\pi} \left(\frac{\pi}{2} - x\right) \frac{\tan x}{x} \, dx
&= \frac{\pi^2}{4} + \frac{\pi}{2}\int_{0}^{1} \psi\left(\frac{1+x}{2}\right) \sin(2\pi x) \, dx \\
&= \frac{\pi^2}{4} - \pi \int_{0}^{\infty} \frac{dt}{(1+t^2)(e^{2\pi t} + 1)}.
\end{align*}
At this moment I have no good idea of dealing with these, though I am posting this in hope of providing an alternative start for others as well as for a self record.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Decomposing $\frac{1 - 8z + 27z^2 - 35z^3 + 14z^4}{(1-z)^2(1-2z)^2(1-3z)}$ I got stuck at this expression and couldn't decompose it further. The book I am referring to has the answer $$\frac{1 - 8z + 27z^2 - 35z^3 + 14z^4}{(1-z)^2(1-2z)^2(1-3z)} = \frac5{4(1-z)} + \frac1{2(1-z)^2} - \frac3{1-2z} - \frac2{(1-2z)^2} + \frac{17}{4(1-3z)}$$ But I am not able to derive the same. Can anyone please help me out on how to resolve it?
|
Was able to decompose it using following method
$let \frac{1−8z+27z^2−35z^3+14z^4}{(1−z)^2(1−2z)^2(1−3z)}=\frac{A}{(1-z)}+\frac{B}{(1-z)^2}+\frac{C}{(1-2z)}+\frac{D}{(1-2z)^2}+\frac{E}{(1-3z)}
$
Multiply both sides by $(1−z)^2(1−2z)^2(1−3z)$
$1−8z+27z^2−35z^3+14z^4 = A(1-z)(1-2z)^2(1-3z)+
\\B(1-2z)^2(1-3z)+
\\C(1-z)^2(1-2z)(1-3z)+
\\D(1-z)^2(1-3z)+
\\E(1-z)^2(1-2z)^2\\$
$= A(1-8z+23z^2-28z^3+12z^4) + \\B(1-7z+16z^2-12z^3)+\\C(1-7z+17z^2-17z^3+6z^4)+\\D(1-5z+7z^2-3z^3)+\\E(1-6z+13z^2-2z^3+4z^4)\\$
$ =(A+B+C+D+E)-\\(8A+7A+7C+5D+6E)z+\\(23A+16B+17C+7D+13E)z^2-\\(28A+12B+17C+3D+12E)z^3+\\(12A+6C+4E)z^4\\$
Comparing LHS with RHS gives us 5 equations:
$A+B+C+D+E=1$
$8A+7B+7C+5D+6E=8$
$23A+16B+17C+7D+13E=27$
$28A+12B+17C+3D+12E=35$
$12A+6C+4E=14$
Solving these 5 equation in 5 variables gives { A = 5/4, B = 1/2, C = -3, D = -2, E = 17/4 }
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1902832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Value of $4\sin^2(\alpha)$ If the point on $y=x\tan(\alpha)-\frac{ax^2}{2u^2\cos^2(\alpha)}$ ,$\alpha>0$ where the tangent is parallel to $y=x$ has an ordinate $\frac{u^2}{4a}$ then value of $4\sin^2(\alpha)$ is?. If we substitute $y$ in the given equation we get a quadratic in $x$ whose roots are $-\tan(\alpha)\pm\sqrt{\tan^2(\alpha)-1}$ . Also $\frac{dy}{dx}=1$ but after doing this I don't get an equation in $\alpha$ as there is an $u$ making things difficult . I don't know how to proceed. Note that answer is an integer from $[0-9]$. Any hints ?
|
Let $P \equiv \left(x,\frac{u^2}{4a}\right) $
Then by $$y=x\tan(\alpha)-\frac{ax^2}{2u^2\cos^2(\alpha)}$$
we get $$\frac{u^2}{4a}=x\tan(\alpha)-\frac{ax^2}{2u^2\cos^2(\alpha)}$$
$$\frac{u^2}{4}=ax\tan(\alpha)-\frac{(ax)^2}{2u^2\cos^2(\alpha)}\rightarrow (1)$$
And $$\frac{dy}{dx}=\tan(\alpha)-\frac{ax}{u^2\cos^2(\alpha)}$$
$$1=\tan(\alpha)-\frac{ax}{u^2\cos^2(\alpha)}$$
$$(\tan\alpha-1)(u^2\cos^2\alpha)=ax \rightarrow (2)$$
Thus by $(1)$ & $(2)$ ,
$$\frac{u^2}{4}=(\tan\alpha-1)(u^2\cos^2\alpha)\tan\alpha-\frac{(\tan\alpha-1)^2(u^2\cos^2\alpha)^2}{2u^2\cos^2\alpha}$$
$$\frac{u^2}{4}=(\tan\alpha-1)(u^2\cos^2\alpha)\tan(\alpha)-\frac{(\tan\alpha-1)^2(u^2\cos^2\alpha)}{2}$$
$$\frac{\sec^2 \alpha}{4}=(\tan\alpha-1)\tan\alpha-\frac{(\tan\alpha-1)^2}{2}$$
$$\frac{\sec^2 \alpha}{4}=\tan^2\alpha-\tan\alpha-\frac{\tan^2\alpha}{2}+\tan\alpha -\frac{1}{2}$$
$$\frac{\sec^2 \alpha}{4}=\frac{\tan^2\alpha}{2}-\frac{1}{2}$$
$$\sec^2 \alpha=2\tan^2\alpha-2$$
$$1+\tan^2 \alpha=2\tan^2\alpha-2$$
$$\tan^2\alpha =3$$
Thus $$4\sin ^2 \alpha =3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1904579",
"timestamp": "2023-03-29T00:00:00",
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|
Help with this integral substitution This came up in my physics exercise:
$$\int_{\theta=0}^{\pi}\frac{\cos\left ( \theta \right )\sin\left ( \theta \right )}{\sqrt{R^{2}+r^{2}-2Rr\cos\left ( \theta \right )}}d\theta$$
I've tried the substitution $u=$ everything under the square root but it didn't worked.
I don't like this integral.
Any help is appreciated.
Thanks in advance.
|
Let $u=R^2+r^2-2Rr\cos\theta,\;$ so $du=2Rr\sin\theta d\theta$ and $\cos\theta=\frac{R^2+r^2-u}{2Rr}$.
Then $\displaystyle\int_{0}^{\pi}\frac{\cos\left ( \theta \right )\sin\left ( \theta \right )}{\sqrt{R^{2}+r^{2}-2Rr\cos\left ( \theta \right )}}d\theta=\frac{1}{2Rr}\int_{(R-r)^2}^{(R+r)^2}\frac{\frac{R^2+r^2-u}{2Rr}}{\sqrt{u}}du=\frac{1}{4R^2r^2}\int_{(R-r)^2}^{(R+r)^2}\left(\frac{R^2+r^2}{\sqrt{u}}-\sqrt{u}\right)du$
$\displaystyle=\frac{1}{4R^2r^2}\left[2(R^2+r^2)\sqrt{u}-\frac{2}{3}u^{3/2}\right]_{(R-r)^2}^{(R+r)^2}$
$\displaystyle=\frac{1}{4R^2r^2}\left[2(R^2+r^2)\big[(R+r)-(R-r)\big]-\frac{2}{3}\big[(R+r)^3-(R-r)^3\big]\right]$
$\displaystyle=\frac{1}{4R^2r^2}\left[2(R^2+r^2)(2r)-\frac{2}{3}(6R^2r+2r^3)\right]=\frac{1}{4R^2r^2}\left[4r-\frac{4}{3}r^3\right]=\color{blue}{\frac{2r}{3R^2}}$
|
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|
Probability of a 7 card sequence (same suit) in a 32 card game Deck of four suits of 8 cards each (32 cards total). Two Players with 12 cards each.
The probability of 8 sequential cards is supposably:
$$\binom{4}{1} \times \frac{\binom{8}{8}\binom{24}{4}}{\binom{32}{12}} $$
What is the probability of getting 7 sequential cards? Is it:
$$ 2 \times \binom{4}{1} \times \frac{\binom{8}{7}\binom{1}{0}\binom{23}{5}}{\binom{32}{12}} $$
|
There are four suits, with 8 cards each.
If one player has 12 cards out of those 32, then the probability is indeed
$\frac{\color{red}{4\cdot 1 } \cdot \color{green}{ \binom{24}{4}} }{\color{blue}{\binom{32}{12}}}$, where $\color{red}{4\cdot 1}$ is the number of ways to choose a sequence of 8 colors, $\color{green}{ \binom{24}{4}}$ the number of ways to choose the remaining cards and $\color{blue}{\binom{32}{12}}$ the total number of ways to draw 12 cards.
However, if we take into account that there two players, the probability becomes:
$$\large{\frac{\color{red}{4\cdot 1 } \cdot \color{green}{ \binom{24}{4}} \cdot \color{green}{ \binom{20}{12}} \, + \color{red}{4\cdot 1 } \cdot \color{green}{ \binom{24}{4}} \cdot \color{green}{ \binom{20}{12}} \, - \color{red}{12\cdot 1 } \cdot \color{green}{ \binom{16}{4}} \cdot \color{green}{ \binom{12}{4}} }{\color{blue}{\binom{32}{12}\binom{20}{12}}}}$$
where $\large{\color{red}{4\cdot 1}}$ is the number of ways to choose a sequence of 8 colors, which is $\large{\color{red}{12\cdot 1}}$ for the probability that both have a 8 card sequence (4&3 possibilities for the suits), $\color{green}{ \binom{24}{4}\binom{20}{12}} $ the number of ways to choose the remaining cards for one of the players and all cards for the other player, and $\color{blue}{\binom{32}{12}\binom{20}{12}}$ the total number of ways to draw 12 cards for two players. We substract the probability that both players have a sequence of length 8.
As Doug M pointed out, you made a small mistake in the seven card case. He explains the one player case, so here is the two player case:
$$\large{\frac{\color{red}{4\cdot 2 } \cdot \color{green}{ \binom{24}{5}} \cdot \color{green}{ \binom{21}{12}} \, + \color{red}{4\cdot 2 } \cdot \color{green}{ \binom{24}{5}} \cdot \color{green}{ \binom{21}{12}} \, - \color{red}{12\cdot 4 } \cdot \color{green}{ \binom{16}{5}} \cdot \color{green}{ \binom{11}{5}} }{\color{blue}{\binom{32}{12}\binom{20}{12}}}}$$
Here, the red, green an blue numbers play the same role.
So $\large{\color{red}{4\cdot 2}}$ is the number of ways to choose a sequence of 7 colors, which is $\large{\color{red}{12\cdot 4}}$ for the probability that both have a 7 card sequence (4&3 possibilities for the suits times 2 times 2 because they are sequential), $\color{green}{ \binom{24}{4}\binom{20}{12}} $ the number of ways to choose the remaining cards for one of the players and all cards for the other player, and $\color{blue}{\binom{32}{12}\binom{20}{12}}$ the total number of ways to draw 12 cards for two players. We substract the probability that both players have a sequence of length 7.
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|
An inequality concerning three positive numbers If $a, b, c$ belong to positive real numbers then prove that
$$\frac{2}{b+c} +\frac{2}{c+a}+\frac{2}{a+b} \le \frac 1a +\frac 1b +\frac 1c.$$
I applied AM-GM inequalities on both sides but I cant even figure out what to do on left side.
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Applying the inequality between the Arithmetic Mean and the Harmonic Mean we get:$$\frac{2}{(a+b)}\le\frac{\frac{1}{a}+\frac{1}{b}}{2}=\frac{a+b}{2ab}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\tag{1}$$
$$\frac{2}{(b+c)}\le\frac{\frac{1}{b}+\frac{1}{c}}{2}=\frac{b+c}{2bc}=\frac{1}{2}\left(\frac{1}{b}+\frac{1}{c}\right)\tag{2}$$
$$\frac{2}{(a+c)}\le\frac{\frac{1}{a}+\frac{1}{c}}{2}=\frac{a+c}{2ac}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{c}\right)\tag{3}$$
On adding (1), (2), and (3) we obtain the required inequality:
$$\frac{2}{(a+b)} +\frac{2}{(b+c)}+\frac{2}{(c+a)} \le \frac{1}{a} +\frac{1}{b }+\frac{1}{c}$$
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|
Using trig identity to solve a cubic equation I need to use the identity $\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta) $ to solve the cubic equation $t^3+pt+q=0$ when $27q^2+4p^3<0$.
I believe I need to let $t=Acos(\phi)$ when $A=2\sqrt{\frac{-p}{3}}$, so that my equation looks more like the identity and I have $A^3\cos(\theta)^3+Ap\cos(\theta)+q=0$. Next I multiply by $\frac{4}{A^3}$ to get $4\cos(\theta)^3+\frac{4p}{A^2}cos(\theta)+\frac{4q}{A^3}=0$.
So then $4\cos(\theta)^3-3\cos(\theta)-\frac{3q}{Ap}=0$.
Now I am stuck and not sure what my next step is.
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For a general cubic,
$$ax^{3}+bx^{2}+cx+d=0$$
use what is known as the Tschirnhaus-Vieta approach: let $x=t+B$, then the cubic becomes
$$t^3+\left(3B+\frac{b}{a}\right)t^2+\left(3B^2+\frac{c+2bB}{a}\right)t+\left(B^3+\frac{bB^2+cB+d}{a}\right)=0$$
By setting $B=-\frac{b}{3a}$, the coefficient of $t^2$ becomes zero and we obtain the depressed cubic
$$t^3+pt+q=0 \tag{1}$$
where
$$p=\frac{3ac-b^{2}}{3a^2},\qquad q=\frac{2b^3-9abc+27a^{2}d}{27a^3}$$
The quantity $\Delta=-4p^3-27q^2$ is called the discriminant of the cubic (1).
*
*If $\Delta<0$, one root is real and two complex conjugate roots;
*If $\Delta=0$ at least two roots coincide, and they are all real;
*If $\Delta>0$ there are three unequal real roots.
Here we have $-\Delta=4p^3+27q^2<0$ and so $\Delta>0$ and we have three unequal real roots, also since $4p^3<-27q^2$ this implies $p<0$, and $p^3<0$, which is an equivalent condition for the three real roots.
Now using the trigonometric identity:
$$4\cos^3(\theta)-3\cos(\theta)-\cos(3\theta)=0 \tag{2}$$
transform the reduced cubic (1) to match (2). To do this let $t=A\cos(\theta)$ and substitute in to get
$$A^3\cos^3(\theta)+Ap\cos(\theta)+q=0\tag{3}$$
Now multiply (3) by $\frac{4}{A^3}$ to give:
\begin{align*}
4\cos^3(\theta)&+\frac{4p}{A^2}\cos(\theta)+\frac{4q}{A^3}=0\tag{4}
\end{align*}
To match (4) with (2) we need $\frac{4p}{A^2}=-3$, and so $A=2\sqrt{\frac{-p}{3}}$, hence we need $p<0$ for $A$ to be real, which we already have. Thus
\begin{align*}
4\cos^3(\theta)&-3\cos(\theta)-\frac{3q}{Ap}=0
\end{align*}
And so using (2)
\begin{align*}
\cos(3\theta)=\frac{3q}{2p}\sqrt{\frac{-3}{p}}=\frac{3q}{Ap}\tag{5}
\end{align*}
Now solve the trigonometric equation (5) as
\begin{align*}
\phi&=\arccos\left(\frac{3q}{Ap}\right)
\end{align*}
where $\phi$ is one angle that satisfies (5). The solutions for $3\theta$ are then
$$3\theta=\pm \phi +2k\pi,\quad k\in\mathbb{Z},\,\,\, \phi\in[0,\pi]$$
from which the solutions for $\theta$ follow
$$\theta=\frac{2k\pi\pm\phi}{3},\quad k\in\mathbb{Z},\,\,\, \phi\in[0,\pi]\tag{6}$$
We now find three distinct values for $\theta$ which relate to the roots, and are due to periodicity of the cosine function: so putting $k=0$, $1$, and $2$ into (6), and appealing to the fact that $\cos$ is an even function gives the three roots for (1) given by $t=A\cos(\theta)$ as:
\begin{align*}
t_1&=A\cos\left(\frac{\phi}{3}\right)\\
t_1&=A\cos\left(\frac{\phi+2\pi}{3}\right)\\
t_1&=A\cos\left(\frac{\phi+4\pi}{3}\right)
\end{align*}
Remark 1: If $p>0$, then $\Delta=-4p^3-27q^2<0$, and there are two complex conjugate roots, with the real root given by:
\begin{align*}
\bar{A}&=2\sqrt{\frac{p}{3}}\\
\bar{\phi}&=\operatorname{arcsinh}\left(\frac{3q}{Ap}\right)\\
x_1&=-\bar{A}\sinh\left(\frac{\bar{\phi}}{3}\right)
\end{align*}
Remark 2: To transform back to $(\star)$ to find the solution for $x$, we initially set $x=t+B=A\cos(\theta)+B$, and so our roots for $(\star)$ are:
\begin{align*}
x_1&=A\cos(\theta_1)+B=A\cos\left(\frac{\phi}{3}\right)+B\\
x_2&=A\cos(\theta_2)+B=A\cos\left(\frac{\phi+2\pi}{3}\right)+B\\
x_3&=A\cos(\theta_3)+B=A\cos\left(\frac{\phi+4\pi}{3}\right)+B
\end{align*}
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|
Show that $\frac{1}{2}\le \sum\limits_{k=0}^n\frac{1}{n+k}\le1$ for $n\in \mathbb N^+$
$$\frac{1}{2n}\le \frac{\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{n+n}}{n}\le\frac{1}{n}$$
I tried math induction and I tried take integral but I want to solve this with most elementary methods please give me hint or just show that. Thanks....
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It holds when $k=1$
$$\frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 2n } \ge \overset { n }{ \overbrace { \frac { 1 }{ 2n } +\frac { 1 }{ 2n } +...\frac { 1 }{ 2n } } } =n\frac { 1 }{ 2n } =\frac { 1 }{ 2 } \\ \\ \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 2n } \le \overset { n }{ \overbrace { \frac { 1 }{ n+1 } +\frac { 1 }{ n+1 } +...+\frac { 1 }{ n+1 } } } =\frac { n }{ n+1 } \le 1\\ $$
|
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|
Show that $\sum\limits_{\theta=\alpha,\beta,\gamma}{\sin^{2}\theta}=\sum\limits_{\theta=\alpha,\beta,\gamma}{\cos^{2}\theta}=\frac{3}{2}$ If $\sin\alpha+\sin\beta+\sin\gamma=\cos\alpha+\cos\beta+\cos\gamma=0$, then show that $$\sum\limits_{\theta=\alpha,\beta,\gamma}{\sin^{2}\theta}=\sum\limits_{\theta=\alpha,\beta,\gamma}{\cos^{2}\theta}=\frac{3}{2}$$
I am confused how I should start. Hint will do.
thanks in advance!
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Let $z_1 = \cos\alpha + i\sin\alpha, z_2 = \cos \beta + i\sin \beta, z_3 = \cos\gamma + i\sin\gamma$. Given that $z_1+z_2+z_3 = 0$. Since the orthocenter of the triangle with vertices $z_1, z_2, z_3$ is $z_1+z_2+z_3$ it follows that for this triangle, orthocenter and circum center coincide and hence the triangle is equilateral. Thus the vertices are of the form $z_1, z_1\omega, z_1\omega^2$, where $\omega $ is a cube root of unity. Thus $\beta = \alpha + \frac{2\pi}{3}$ and $\gamma = \alpha + \frac{4\pi}{3}$. Thus
\begin{align*}
z_1^2+z_2^2+z_3^2 & = z_1^2(1+\omega^2 + \omega^4) = 0
\end{align*} and hence
\begin{align*}
\cos 2\alpha+ \cos 2\beta + \cos 2\gamma = 0
\end{align*} Now,
\begin{align*}
\cos^2\alpha + \cos^2\beta + \cos^2\gamma &= \frac{1+\cos2\alpha}{2}+\frac{1+\cos2\beta}{2}+\frac{1+\cos2\gamma}{2}\\
&= \frac{3}{2}
\end{align*}
Use $\sin^2 \alpha = 1-\cos^2\alpha$ to get the other result.
|
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|
Range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ with $A,B,C$ angles in a triangle
What is the range of $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$ if $A,B,C$ are angles in a triangle?
For $\tan^2\frac A2+\tan^2\frac B2+\tan^2 \frac C2$, I know we have to apply the AM–GM inequality. So
$$\begin{align}\frac12\left(\tan^2\frac A2+\tan^2\frac B2+\tan^2\frac C2\right)&\ge\sqrt{\tan^2\frac A2\tan^2\frac B2\tan^2\frac C2}\\
&\ge\tan\frac A2\tan\frac B2\tan\frac C2\\
\tan^2\frac A2+\tan^2\frac B2+\tan^2\frac C2&\ge2\tan\frac A2\tan\frac B2\tan\frac C2
\end{align}$$
After this, the LHS is of the required form, but the RHS is having half-angle terms. Then again, if $A+B+C=\pi$, $\tan\frac{A+B+C}2=\tan90^\circ=\infty$. Now I am stuck; how to solve further?
The given options are:
*
*$>1$
*$<1$
*$\ge1$
*$\le1$
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Note $\tan^2(x)$ is convex, so by Jensen's inequality
$$\sum_{cyc} \tan^2 \frac{A}2 \geqslant 3\tan^2 \frac{A+B+C}{2\cdot3} = 1$$
Further as $\tan(x)$ is unbounded, clearly we can have the sum unbounded by allowing one of the angles to approach $\pi$.
|
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|
proof using the mathematical induction Prove $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$ is an integer for every positive integer k.
In proving for (n+1) integer,the expression is integer,I found $(n+1)^7$ term.I use binomial theorem to expand but finally it won't work(some term become integer While some other remaining as rational)
Any hint to solve the problem is appreciated.Thanks.
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We have, for instance, $(k+1)^7/7 = (\sum_{i=0}^7 \binom{7}{i}k^i)/7$. Note that the only times the coefficient of $k^i$ in the numerator is not divisible by $7$ is when $i=0,7$. Therefore, $(k+1)^7/7 = k^7/7 + 1/7 + \text{integer}$. Applying the same logic to the other terms, we find
$$\begin{align}\frac{(k+1)^7}{7}+\frac{(k+1)^5}{5}+\frac{2(k+1)^3}{3}-\frac{k+1}{105}&=\left(\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}\right) \\&+\left(\frac{1}{7}+\frac{1}{5}+\frac{2}{3}-\frac{1}{105}\right) + \text{integer}\\ &= \text{integer} + 1 +\text{integer} \\&= \text{integer}.\end{align}$$
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Easy way to compute $\int_{-\pi}^\pi \cos^4(x)dx$. Is there an easy way to compute $$\int_{-\pi}^\pi \cos^4(x)dx\ \ ?$$
What I did is using $$\cos^4(x)+1-1=1+(\cos^2(x)-1)(\cos^2(x)+1)=-\sin^2(x)(1+\cos^2(x))$$
and then using formula $$\cos^2(x)=\frac{\cos(2x)+1}{2}$$
and $$\sin^2(x)=\frac{1-\cos(2x)}{2},$$
but at the end, I have to compute $$...+\int_{-\pi}^\pi\cos^2(2x)dx.$$
Using finally that $$\cos^2(2x)=\frac{1+\cos(4x)}{2},$$
I concluded (and found $\frac{3\pi}{4}$ which is correct). But is there an other (shorter) method ?
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The integral has four-fold symmetry, so it is equivalent to:
$$\huge 4\int_0^\frac{\pi}{2} \cos^{4}(x) \, \text{d}x = 2 \,\text{B}\left(\frac{5}{2},\frac{1}{2}\right)$$
$$\huge = 2 \, \frac{\Gamma{(\frac{5}{2})}\Gamma{(\frac{1}{2})}}{\Gamma{(3)}} = 2 \, \frac{ \frac{3}{2} \times \frac{1}{2} \times \pi}{2} = \frac{3\pi}{4}$$
|
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For what value $k$ is $f(x) = \begin{cases} \frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\ k & x = 2 \end{cases}$ continuous at $x=2$? For what value $k$ is the following function continuous at $x=2$?
$$f(x) = \begin{cases}
\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\
k & x = 2
\end{cases}$$
All those square roots are weighing me down! And $k$? My mind's not where it's supposed to be today. Thanks in advance for posting a solution!
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We must have $f(2) = \lim_{x \to 2}(\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2})$ for $f(x)$ to be continuous at $x=2$. Since plugging in $2$ gives us $\frac{0}{0}$, we can use L'hopital's Rule and differentiate the top and bottom:
$$\lim_{x \to 2}(\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2})=\lim_{x \to 2}(\frac{2\frac{1}{2}(2x+5)^{-\frac{1}{2}}-\frac{1}{2}(x+7)^{-\frac{1}{2}}}{1})$$
$$\lim_{x \to 2}(\frac{2\frac{1}{2}(2x+5)^{-\frac{1}{2}}-\frac{1}{2}(x+7)^{-\frac{1}{2}}}{1})=\frac{1}{6}$$
So we see that we need $f(2)=\frac{1}{6}$, so $f(2) = k = \frac{1}{6}$.
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|
${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c}+ \sqrt{c}\over b+a} \ge {9-3\sqrt{3}\over2\sqrt{a+b+c}}$
$${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge {9+3\sqrt{3}\over2\sqrt{a+b+c}}$$
I tried AM-GM, which only gives large terms without any answer.
$${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge 3\sqrt[3]{\left({\sqrt{a+b+c} + \sqrt{a}\over b+c}\right) \left({\sqrt{a+b+c} + \sqrt{b}\over a+c}\right)\left({\sqrt{a+b+c} + \sqrt{c}\over b+a}\right)}$$
Now this will never simplify to anything.
using a different approach i got,
$${\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge \sqrt{\left({\sqrt{a+b+c} + \sqrt{a}\over b+c}\right) \left({\sqrt{a+b+c} + \sqrt{b}\over a+c}\right)}+\sqrt{\left({\sqrt{a+b+c} + \sqrt{c}\over b+a}\right)\left({\sqrt{a+b+c} + \sqrt{b}\over a+c}\right)}+\sqrt{\left({\sqrt{a+b+c} + \sqrt{c}\over b+a}\right)\left({\sqrt{a+b+c} + \sqrt{a}\over b+c}\right)}$$
which also does not simplify further.
breaking the individual terms on the LHS also does not help nor does multiplying each term on LHS by its conjugate. :<
This was a introductory problem, so i guess there must exist a easy solution which i can't find.
Any hints will be helpful.
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It must be
$$\sum\limits_{cyc}\frac{\sqrt{a+b+c}+\sqrt{a}}{b+c}\geq\frac{9+3\sqrt3}{2\sqrt{a+b+c}}$$
Let $a+b+c=3$.
Hence,
$$\sum\limits_{cyc}\frac{\sqrt{a+b+c}+\sqrt{a}}{b+c}-\frac{9+3\sqrt3}{2\sqrt3}=\sum\limits_{cyc}\frac{\sqrt3+\sqrt{a}}{3-a}-\frac{3\sqrt3+3}{2}=$$
$$=\sum\limits_{cyc}\left(\frac{1}{\sqrt3-\sqrt{a}}-\frac{1}{\sqrt3-1}\right)=\frac{1}{\sqrt3-1}\sum\limits_{cyc}\frac{\sqrt{a}-1}{\sqrt3-\sqrt{a}}=$$
$$=\frac{1}{\sqrt3-1}\sum\limits_{cyc}\left(\frac{\sqrt{a}-1}{\sqrt3-\sqrt{a}}-\frac{1}{2(\sqrt3-1)}(a-1)\right)=$$
$$=\frac{1}{2(\sqrt3-1)^2}\sum\limits_{cyc}\frac{(\sqrt{a}-1)^2(\sqrt{a}+2-\sqrt3)}{\sqrt3-\sqrt{a}}\geq0$$
|
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|
Minimum value of $a^2 + b^2$ so that the quadratic $x^2 + ax + (b+2) = 0$ has real roots
The equation $ x^2 + ax + (b+2) = 0 $ has real roots, where $a$ and $b$ are real numbers.
How would I find the minimum value of $a^2 + b^2$ ?
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Using what you already did ( the discriminant is no-negative and thus $\;a^2-4b\ge8\;$) then
$$a^2+b^2\ge8+4b+b^2=(b+2)^2+4$$
So the above rightmost expression is at least...
|
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|
Is there a less trickier way to solve this differential equation? Solve the following differential equation:
$$ y^3 dy+(x + y^2)dx = 0$$
Solution:
The following solution uses the substitution $y^2=tx$.
$$y^3\frac{dy}{dx}+x+y^2=0\tag1$$
Differentiating $y^2 = tx$,
$$ 2y\frac {dy}{dx}= t + x\frac{dt}{dx}$$
Now substituting in (1),
$$tx\ (t + x\frac{dt}{dx})+2(x+tx)=0$$
$$t^2x+tx^2\frac{dt}{dx}+ 2x + 2tx=0 $$
$$2tx\ dx + t^2x\ dx + 2x\ dx+tx^2\ dt=0$$
$$x\ dx\ (t^2 + 2t + 2)=-tx^2\ dt$$
$$\int{\frac{1}{x}}dx= \int{\frac{-t}{t^2+2t+2}}dt$$
$$lnx=\int{\frac{-t-1}{t^2+2t+2}}dt + \int{\frac{1}{t^2+2t+2}dt}$$
$$lnx=\int{\frac{-t-1}{(t+1)^2+1}}dt+\int{\frac{1}{(t+1)^2+1}}dt$$
$$lnx=-\frac{1}{2}ln[(t+1)^2+1]+ \arctan(\frac{y^2}{x}+1)$$
$$lnx=-\frac{1}{2}ln[(\frac{y^2}{x}+1)^2+1)]+ \arctan(\frac{y^2}{x}+1)$$
I have the following questions:
1) How did the person who solved this problem get the idea of substituting
$y=tx^2$ ? This substitution doesn't seem to be obvious.
2) Is there any other method (without using the above substitution) to solve this problem, in a less trickier manner?
|
The substitution presented does seem somewhat unmotivated.
Given
\begin{equation}
(x+y^2)dx+y^3dy=0
\end{equation}
one would be inclined to first try the substitution $x=uy^2$ so that the $y^2$ might factor out of the first term.
This gives $dx=y^2du+2uy\,dy$ yielding after a bit of algebra the separable equation
\begin{equation}
y^4(u+1)du+y^3(2u^2+2u+1)dy=0
\end{equation}
Separation gives
\begin{equation}
\dfrac{u+1}{2u^2+2u+1}\,du+\dfrac{1}{y}\,dy=0
\end{equation}
which is straightforward, perhaps aided by the substitution $v=2u+1$ with the first integral partitioned into its logarithmic and inverse tangent parts.
|
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|
Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?
$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$$
|
This is straightforward when you have another identity in your pocket, namely
$$\sec\theta \cdot \csc\theta = \tan\theta + \cot\theta \tag{$\star$}$$
This implies
$$\sec^2\theta \csc^2\theta = ( \tan\theta + \cot\theta )^2 = \tan^2\theta + 2 \tan\theta\cot\theta + \cot^2\theta = \tan^2\theta + 2 + \cot^2\theta$$
which gets you the equivalent of the identity in question. $\square$
By the way, proof of $(\star)$ arises from this trigonograph:
Computing the area of the big triangle in two ways gives
$$\frac{1}{2}\;\sec\theta\cdot\csc\theta = \frac{1}{2}\cdot 1 \cdot\left(\;\tan\theta + \cot\theta\;\right)$$
and the identity follows. $\square$
|
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|
General term of recurrence relation Find the general term of the following recurrence relation:
$$a_{1} = 2$$
$$a_{n+1} = \frac{2a_{n} - 1}{3}$$
I've tried to find the first few terms:
$$a_{1} = 2$$
$$a_{2} = \frac{2 \cdot 2 - 1}{3} = 1$$
$$a_{3} = \frac{2 \cdot 1 - 1}{3} = \frac{1}{3}$$
$$a_{4} = -\frac{1}{9}$$
$$a_{5} = -\frac{1}{27}$$
but can't see any pattern, especially considering the first 2 terms.
|
Following your (@Jack D'Aurizio) suggestion, here is what I understood:
The given recurrence relation: $3a_{n} = 2a_{n-1} - 1$, with initial conditions: $a_{1} = 2$ is a $1^{st}$ degree linear non-homogeneous recurrence relation.
Solution:
We are searching for solution1 of the form: $a_{n} = b_{n} + h_{n}$ (1), where $h_{n}$ is the sequence satisfying the associated homogeneous recurrence relation and $b_{n}$ is a solution which is similar to $f(n)$.
1.The associated homogeneous recurrence relation is: $3h_{n} = 2h_{n-1}$ and its characteristic equation is: $3r - 2 = 0$ or $ r = \frac{2}{3}$ and the solution is:
$$h_{n} = \alpha_{0} (\frac{2}{3})^{n-1} $$
From the initial conditions: $a_{1} = \alpha_{0} (\frac{2}{3})^{1} = 2$ or $\alpha_{0} = 3$, so the solution of the homogeneous recurrence is:
$$h_{n} = 3 (\frac{2}{3})^{n-1} $$
*In our case $b_{n} = -1$ and from (1) we have:
$$a_{n} = 3 (\frac{2}{3})^{n-1} - 1$$
1. $a_{n}$ satisfies both the recurrence relation and the initial conditions.
2. $f(n)$ - function depending only on $n$.
|
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|
connection between two distinct prime numbers $p,q$ such that $p^n=q^2+q+1$ Let $p,q$ be two distinct prime number such that $p^n=q^2+q+1$, where $n \in \Bbb{N}$. I want to proof that, this forces $n=1$.
|
I borrowed two books on contest problems; this one is An Introduction to Diophantine Equations, by Titu Andreescu, Dorin Andrica, and Ion Cucurezeanu. Most of what is needed for your problem, probably all, is in section 4.2, The Ring of Integers of $\mathbb Q[\sqrt d].$ This is mentioned above by Jack, suggesting working in the Eisenstein integers $\mathbb Z[\sqrt {-3}],$ which is a UFD.
This is a pretty good start...
We have $x^2 + x + 1 = p^n;$ I am not sure yet whether the restriction that $x$ be prime matters. We have, page 168, the unit
$$ \omega = \frac{-1 + \sqrt {-3}}{2}, $$
which is a cube root of $1.$ The main detail is that $\sqrt {-3}$ is irreducible in $\mathbb Z[\sqrt {-3}].$ On page 173 we have some examples of finding gcd of two quantities. Also on page 177, problem 2 is similar to yours, solution takes pages 315-319. Note that
$$ \omega^2 = -1 - \omega,$$
$$ \omega - \omega^2 = \sqrt {-3}. $$
Alright, we have factorization
$$ (x - \omega)(x - \omega^2) = p^n. $$
We need to know
$$ \gcd(x - \omega, x - \omega^2). $$ If some $\delta$ in the ring divides both, then it divides $ \omega - \omega^2 = \sqrt {-3}. $ However, this is irreducible, so
$$ \gcd(x - \omega, x - \omega^2)= 1. $$
It follows that each factor is such a power, that is
$$ x - \omega = (a + b \omega)^n, $$
for ordinary $a,b \in \mathbb Z.$ NOTE: remind me to put in alternatives
$$ x - \omega = \omega(a + b \omega)^n, \; \; \; x - \omega = \omega^2(a + b \omega)^n. $$
What I have so far is the implication
$ b = \pm 1; $ let me do exponent $n=5.$
$$ x - \omega = (a + b \omega)^5, $$
$$ x - \omega = a^5 + 5 a^4 b \omega + 10 a^3 b^2 \omega^2 + 10 a^2 b^3 + 5 a b^4 \omega + b^5 \omega^2. $$
Remember $\omega^2 = -1 - \omega,$ so that
$$ x - \omega = (a^5 - 10 a^3 b^2 + 10 a^2 b^3 - b^5) + (5 a^4 b - 10 a^3 b^2 + 5 a b^4 - b^5 ) \omega. $$ The coefficient of $\omega$ needs to be $-1,$ that is
$$ (5 a^4 b - 10 a^3 b^2 + 5 a b^4 - b^5 ) = -1 $$
and
$$ b(5 a^4 - 10 a^3 b + 5 a b^3 - b^4 ) = -1. $$
Well, $b$ is an ordinary integer, and $b$ divides $-1.$ that is,
$$ b = \pm 1. $$
Next, we need either
$$ 5 a^4 - 10 a^3 + 5 a - 1 = -1, $$ or
$$ 5 a^4 + 10 a^3 - 5 a - 1 = -1. $$ Either way,
divide out by $5a,$
$$ a^3 + 2 a^2 - 1 = 0 $$ OR
$$ a^3 - 2 a^2 + 1 = 0. $$
The rational root theorem tells us that $a = \pm 1$ as well.
|
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|
show that $ab+bc+ca|(b^2-ac)^2$ Let $a,b,c$ be postive integers,and such $(a,b)=(b,c)=(c,a)=1$,if
$$ab+bc+ca|(a^2-bc)^2$$
show that
$$ab+bc+ca|(b^2-ac)^2$$
Own idea,maybe we look for some identities?
|
$$0\equiv (a^2-bc)^2 \equiv (a^2-(-ab-ac))^2 = a^2(a+b+c)^2 \bmod (ab+bc+ca).$$
However, $\gcd(a,bc)=1 \implies \gcd(a,ab+bc+ca)=1,$ so
$$ab+bc+ca | (a+b+c)^2.$$
$$ab+bc+ca | (b^2+ab+bc)^2.$$
$$ab+bc+ca | (b^2-ca)^2.$$
|
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|
$ax^3+8x^2+bx+6$ is exactly divisible by $x^2-2x-3$, find the values of $a$ and $b$ Find the values of $a$ and $b$ for which the polynomial $ax^3+8x^2+bx+6$ is divisible by $x^2-2x-3$.
|
$$ x^2 - 2 x + 3 = ( x + 1) ( x -3) $$
$ x= -1, x = 3 $ should satisfy the polynomial. So,
$$ -a + 8 -b + 6 =0 ;\, a \,3^3 + 8 \,3^2 - 3 \,3 + 6 =0 ;$$
Siimplify and solve for $a,b. $
|
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|
Lengthy partial fractions? $$\frac{1}{(x+3)(x+4)^2(x+5)^3}$$
I was told to integrate this, I see partial fractions as a way, but this absurd! Is there an easier way?
|
Lengthy but not impossible. We have for sure
$$\begin{eqnarray*} f(x)&=&\frac{1}{(x+3)(x+4)^2(x+5)^3}\\&=&\frac{A}{x+3}+\frac{B}{x+4}+\frac{C}{x+5}+\frac{D}{(x+4)^2}+\frac{E}{(x+5)^2}+\frac{F}{(x+5)^3}\end{eqnarray*}\tag{1}$$
where
$$ F=\lim_{x\to -5}(x+5)^3 f(x) = \lim_{x\to -5}\frac{1}{(x+3)(x+4)^2} = -\frac{1}{2},$$
$$ D=\lim_{x\to -4}(x+4)^2 f(x) = \lim_{x\to -4}\frac{1}{(x+3)(x+5)^3} = -1,$$
$$ A = \lim_{x\to -3}(x+3) f(x)=\lim_{x\to -3}\frac{1}{(x+4)^2 (x+5)^3} = \frac{1}{8}.\tag{2} $$
By setting $g(x)=f(x)-\frac{F}{(x+5)^3}-\frac{D}{(x+4)^2}-\frac{A}{(x+3)}$ it follows that:
$$ \begin{eqnarray*}g(x)&=&\frac{20-3x-x^2}{8(x+4)(x+5)^2}\\&=&\frac{B}{x+4}+\frac{C}{x+5}+\frac{E}{(x+5)^2}\end{eqnarray*}\tag{3}$$
where
$$ E = \lim_{x\to -5}g(x)(x+5)^2 = -\frac{5}{4}, $$
$$ B = \lim_{x\to -4}g(x)(x+4) = 2\tag{3} $$
and $C=-\frac{17}{8}$ follows from computing $g(x)-\frac{E}{(x+5)^2}-\frac{B}{x+4}$, or from a general property of the partial fraction decomposition of "rapidly decaying" meromorphic functions, that in this case grants $C=-(A+B)$, since $A,B,C$ are the residues of $f(x)$ at its poles. You may deduce the same by noticing that $\lim_{x\to +\infty} x\cdot f(x)=0$.
It follows that:
$$ \int\frac{dx}{(x+3)(x+4)^2(x+5)^3}\\=C+\frac{1}{8} \left(\frac{8}{x+4}+\frac{2}{(x+5)^2}+\frac{10}{x+5}+\log(x+3)+16\log(x+4)-17\log(x+5)\right).$$
|
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|
Non-trivial solution to a homogeneous system of linear equations. I have equations:
\begin{cases}
2x + y - z = 0\\
x - 2y - 3z =0\\
-3x - y + 2z =0
\end{cases}
After I put this in matrix row-reduced echelon form I get solutions of $x=0, y=0, z=0$. But my book says it has a non-trivial solution. Could someone explain how that could be?
|
We can write the LHS of your system as
$$
\left[
\begin{array}{rrr}
2 & 1 & - 1 \\
1 & -2 & - 3 \\
-3 & -1 &2
\end{array}
\right]
$$
and apply Gauß elimination:
$$
\to
\left[
\begin{array}{rrr}
1 & -2 & - 3 \\
2 & 1 & - 1 \\
-3 & -1 &2
\end{array}
\right]
\to
\left[
\begin{array}{rrr}
1 & -2 & - 3 \\
0 & 5 & 5 \\
0 & -7 & 7
\end{array}
\right]
\to
\left[
\begin{array}{rrr}
1 & -2 & - 3 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{array}
\right]
\to
\left[
\begin{array}{rrr}
1 & 0 & -1 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{array}
\right]
$$
So the solution can be written e.g. as
$$
L
= \{ (z, -z, z) \mid z \in \mathbb{R} \}
= \{ z \, (1, -1, 1) \mid z \in \mathbb{R} \}
$$
|
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|
A different way to prove that $\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx = \frac{3 G}{4} + \frac{\pi}{16} \, \ln 2$ From the fact that $$\int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy $$ is an integral representation of Catalan's constant ($G$), I was able to deduce that $$\int_{1+\sqrt{2}}^{\infty} \frac{\ln(1+x)}{1+x^{2}} \, dx = \int_{0}^{\pi/8} \ln(1+ \cot u) \, du = \frac{3G}{4} + \frac{\pi}{16} \, \ln 2\tag{1}.$$
What is another way to prove $(1)$ that preferably doesn't involve the dilogarithm function?
EDIT:
In response to Dr. MV's comment, the following is how I deduced $(1)$ from that integral representation of Catalan's constant.
$$ \begin{align} G&= \int_{0}^{1} \int_{0}^{1} \frac{1}{2-x^{2}-y^{2}} \, dx \, dy \\ &= \int_{0}^{1} \frac{1}{2-y^{2}} \int_{0}^{1} \frac{1}{1-\frac{x^{2}}{2-y^{2}}} \, dx \, dy \\ &= \int_{0}^{1} \frac{1}{\sqrt{2-y^{2}}} \, \text{artanh} \left(\frac{1}{\sqrt{2-y^{2}}} \right) \, dy \\ &= \int_{0}^{\pi/4} \text{artanh} \left(\frac{1}{\sqrt{2} \cos \theta} \right) \, d \theta \\ &= \frac{1}{2} \int_{0}^{\pi/4} \ln \left(\frac{\sqrt{2} \cos \theta +1}{\sqrt{2} \cos \theta -1} \right) \, d \theta \\ &= \frac{1}{2} \int_{0}^{\pi/4} \ln \left(\frac{(\sqrt{2} \cos \theta+1)^{2}}{2 \cos^{2} \theta -1} \right) \, d \theta \\ &=\int_{0}^{\pi/4} \ln (\sqrt{2} \cos \theta +1) \, d \theta - \frac{1}{2} \int_{0}^{\pi/4} \ln(\cos 2 \theta) \, d \theta \\ &= \int_{0}^{\pi/4} \ln\left(\sqrt{2} \cos \left(\frac{\pi}{4} - \phi\right)+1\right) \, d \phi - \frac{1}{4} \int_{0}^{\pi/2} \ln( \cos \tau) \, d \tau \\ &= \int_{0}^{\pi/4} \ln \left(\sin(\phi) + \cos(\phi)+1\right) \, d \phi - \frac{1}{4} \left(- \frac{\pi}{2} \, \ln 2 \right) \\ &= \int_{0}^{\pi/4} \ln (\sin \phi) \, d \phi + \int_{0}^{\pi/4} \ln \left(1+ \frac{1+ \cos \phi}{\sin \phi} \right) \, d \phi + \frac{\pi}{8} \, \ln 2 \\ &= - \frac{G}{2} - \frac{\pi}{4} \, \ln 2 + \int_{0}^{\pi/4} \ln \left(1+ \cot \frac{\phi}{2} \right) \, d \phi + \frac{\pi}{8} \, \ln 2 \\ &= - \frac{G}{2} - \frac{\pi}{8} \, \ln 2 + 2 \int_{0}^{\pi/8} \ln (1 + \cot u) \, du \end{align}$$
|
Using the following trigonometric identitiy:
$$(1+\cot x)^2 =\frac{ 2 \tan\left(\frac{\pi}{4}+x\right)}{\tan(2 x) \, \tan(x)},$$
We have
\begin{align}2 \int_0^{\pi/8} \ln(1+ \cot x) \, dx=
& \ln2 \int_0^{\pi/8} dx-\int_0^{\pi/8} \ln \tan 2 x \, dx
\\ -
& \int_0^{\pi/8} \ln \tan x \, dx+\int_0^{\pi/8} \ln \tan\left(\frac{\pi}{4}+x\right)\, dx \tag{1}
\\ =
& \frac{\pi}{8} \ln 2-\frac12 \int_0^{\pi/4} \ln \tan x \, dx
\\ -
& \int_0^{\pi/8} \ln \tan x \, dx-\int_{\pi/8}^{\pi/4} \ln \tan x \, dx \tag{2}
\\ =
& \frac{\pi}{8} \ln 2-\frac32 \int_0^{\pi/4} \ln \tan x \, dx
\\=
& \frac{\pi}{8} \ln 2 + \frac32 G.
\end{align}
Where, to arrive at $(2)$ from $(1)$ we have substituted $x \mapsto x/2$ in the second integral, and in the last integral we
substituted $x \mapsto \pi/4-x$, in tandem with the fact that $\tan(\pi/2-x) = 1/\tan(x)$.
Also, the integral $\int_0^{\pi/4} \ln \tan x \, dx = - G$ is well known and can be easily reproduced.
|
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|
Do the Taylor series of $\sin x$ and $\cos x$ depend on the identity $\sin^2 x + \cos^2 x =1$? I had this crazy idea trying to prove the Pythagorean trigonometric identity;$$\sin^2x+\cos^2x=1$$by squaring the infinite Taylor series of $\sin x$ and $\cos x$. But it came out quite beautiful, involving also a combinatorics identitie.
The proof:
$$\sin x=\frac{x}{1}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}\\\\\sin^2x=x^2-x^4\left (\frac{1}{1!3!}+\frac{1}{3!1!}\right )+x^6\left (\frac{1}{1!5!}+\frac{1}{3!3!}+\frac{1}{5!1!}\right )-...\\\\\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}\\\\\cos^2x\!=\!1\!-\!x^2\left(\!\frac{1}{0!2!}\!+\!\frac{1}{2!0!}\!\right)\!+\!x^4\left(\!\frac{1}{0!4!}\!+\!\frac{1}{2!2!}\!+\!\frac{1}{4!0!}\!\right)\!-\!x^6\left(\!\frac{1}{0!6!}\!+\!\frac{1}{2!4!}\!+\!\frac{1}{4!2!}\!+\!\frac{1}{6!0!}\!\right)\!+...$$We should have shown that the series for both $\sin x$ and $\cos x$ converge absolutely (since we changed the arrangement), but it's obvious since the absolute value of all terms of $\sin x+\cos x$ add up to $e^x$.$$\sin^2x+\cos^2x=\\=1-x^2\left(\frac{1}{0!2!}-\frac{1}{1!1!}+\frac{1}{2!0!}\right)+x^4\left(\frac{1}{0!4!}-\frac{1}{1!3!}+\frac{1}{2!2!}-\frac{1}{3!1!}+\frac{1}{4!0!}\right)-x^6\left(\frac{1}{0!6!}-\frac{1}{1!5!}+\frac{1}{2!4!}-\frac{1}{3!3!}+\frac{1}{4!2!}-\frac{1}{5!1!}+\frac{1}{6!0!}\right)+...=\\\\=1+\sum_{n=1}^{\infty}(-1)^nx^{2n}\sum_{k=0}^{2n}\frac{(-1)^k\binom{2n}{k}}{(2n)!}$$
Since we can show easily that $\sum_{i=0}^n(-1)^i\binom{n}{i}=0$ by expanding $(1-1)^n$ using Binom's formula. So:$$\sin^2x+\cos^2x=1-0+0-0+...=1$$
I think it's beautiful. I just wanted to ask, do Taylor's series of those functions depend on this identity? Because if they do, the proof will be circular.
|
You are just showing that the Pythagorean theorem is a consequence of some property of the (complex) exponential function, there is no circularity in such argument. For instance, we may define, for any $z\in\mathbb{C}$,
$$ f(z)=e^{z}=\sum_{n\geq 0}\frac{z^n}{n!} \tag{1}$$
and prove through a combinatorial argument that such a function fulfills $e^{z}\cdot e^{w}=e^{z+w}$.
Since the Taylor coefficients at $0$ of such analytic function are real, we have $f(\bar{z})=\overline{f(z)}$, hence for any $\rho\in\mathbb{R}$
$$ \left\| e^{i\rho}\right\| = e^{i\rho}\cdot e^{-i\rho} = e^0 = 1. \tag{2}$$
If we define $\cos(\rho)$ and $\sin(\rho)$ as the real/imaginary part of $e^{i\rho}$, we get that
$$ \sin(\rho)=\sum_{n\geq 0}\frac{(-1)^n \rho^{2n+1}}{(2n+1)!},\qquad \cos(\rho)=\sum_{n\geq 0}\frac{(-1)^n \rho^{2n}}{(2n)!}\tag{3} $$
and $(2)$ can be read as:
$$ \sin^2(\rho)+\cos^2(\rho) = 1.\tag{4}$$
|
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|
$X\sim\mathcal N(0,1);\ Y = \sqrt{|X|}$; find $f_Y(y)$
Let $X\sim\mathcal N(0,1)$ be a normal distribution and $Y = \sqrt{|X|}$. Find $f_Y(y)$, the probability density function of $Y$.
I figure trying to find the expected value of $Y$ and the variance is a good start, but I have no idea how to integrate $\frac{2}{\sqrt{2 \pi}}\int^\infty_0 x^{\frac{1}{2}}e^{\frac{-x^2}{2}}$. Is there a good way to integrate this or am I missing an easier method to solving this?
|
\begin{align}
f_Y(y) & = \frac d {dy} \Pr(Y\le y) = \frac d{dy} \Pr( -y^2 \le X \le y^2) \\[10pt]
& = \frac d {dy}\int_{-y^2}^{y^2} \frac 1 {\sqrt{2\pi}} e^{-x^2/2} \,dx = 2 \frac d {dy} \int_0^{y^2} \frac 1 {\sqrt{2\pi}} e^{-x^2/2} \,dx \\[10pt]
& = 2 \frac d {dy} \int_0^u \cdots = 2 \frac{du}{dy} \cdot \frac d {du} \int_0^u \cdots =2 (2y) \cdot \frac d {du} \int_0^u \frac 1 {\sqrt{2\pi}} e^{-x^2/2} \, dx \\[10pt]
& = 4y \cdot \frac 1 {\sqrt{2\pi}} e^{-u^2/2} = 4y \cdot \frac 1 {\sqrt{2\pi}} e^{-y^4/2} \text{ for } y\ge 0.
\end{align}
|
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|
what is remainder when in the following scenario? What is the remainder when $x^{2016} - 1$ is divided by
$ x^{5}+x^{4}+x^{3}+x^{2}+x+1 ?$
how to solve this kind of problem.
|
$$x^{2016}-1=(x-1)(x^{2015}+x^{2014}+\cdots+x+1)$$
When you divide $x^{2015}+x^{2014}+\cdots+x+1$ by $x^5+x^4+x^3+x^2+x+1$ by long division, $6$ terms cancel out from the dividend at each step. For example, in the first step, the quotient term is $x^{2010}$, which on multiplication with $x^5$ cancels out $x^{2015}$, $x^4$ cancels out $x^{2014}$, and so on till $x^{2010}$. After this, the process repeats starting with $x^{2009}$. In all, the dividend has $2016$ terms, which is divisible by $6$. Hence, $x^{2015}+x^{2014}+\cdots+x+1$ is divisible by $x^5+x^4+x^3+x^2+x+1$, and as a result, $x^{2016}-1$ is also divisible by it.
Thus, the remainder is $0$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Finding a closed form for a recurrence relation $a_n=3a_{n-1}+4a_{n-2}$ Consider the sequence defined by
$$
\begin{cases}
a_0=1\\
a_1=2\\
a_n=3a_{n-1}+4a_{n-2} & \text{if }n\ge 2
\end{cases}
.$$
Find a closed form for $a_n$.
I tried listing out examples, but I don't see any common pattern between them. All solutions are greatly appreciated.
|
This recurrence relation can be described with the following matrix:
$\operatorname A = \left[\begin{array}{cc}
3 & 4 \\
1 & 0 \\
\end{array} \right]$
This means that the equation, $$ \operatorname A\left[\begin{array}{cc}
a_{n-1} \\
a_{n-2} \\
\end{array} \right] = \left[\begin{array}{cc}
3a_{n-1} + 4 a_{n-2} \\
a_{n-1} \\
\end{array} \right] = \left[\begin{array}{cc}
a_{n} \\
a_{n-1} \\
\end{array} \right] $$
holds. Why is this beautiful? First of all, by repeated multiplication we have that $$\operatorname A^k\left[\begin{array}{cc}
a_{1} \\
a_{0} \\
\end{array} \right] = \left[\begin{array}{cc}
a_{k+1} \\
a_{k} \\
\end{array} \right]$$
which gives you a way to calculate the $ \mathbb N \ni n$th term in $O(\log n)$ time, using binary (matrix) exponentiation. Since the size of the matrix is really small, it is a very fast way indeed.
Okay, that's good and well, but you've asked for a closed form. Well, you can get that from this as well. Let's say that you've diagonalized this matrix, and you've gotten $ \operatorname A = S \Lambda S^{-1}$. In this case, $$ \operatorname \Lambda = \left[\begin{array}{cc}
4 & 0 \\
0 & -1 \\
\end{array} \right]\quad \text{and} \quad S = \left[\begin{array}{cc}
4 & 1 \\
1 & -1 \\
\end{array} \right]. $$
Using $\operatorname A^k = S\Lambda^k S^{-1}$, we get that
$$ \left[\begin{array}{cc}
a_{k+1} \\
a_{k} \\
\end{array} \right] = \operatorname A^k\left[\begin{array}{cc}
a_{1} \\
a_{0} \\
\end{array} \right]= S\Lambda^k S^{-1} \left[\begin{array}{cc}
a_{1} \\
a_{0} \\
\end{array} \right] = \left[\begin{array}{cc}
4^{k+1} & (-1)^k \\
4^k & (-1)^{k-1} \\
\end{array} \right] \left[\begin{array}{cc}
\frac{3}{5} \\
-\frac{2}{5} \\
\end{array} \right] =\left[\begin{array}{cc}
\frac{3}{5} 4^{k+1} + \frac{2}{5} (-1)^{k+1} \\
\frac{3}{5} 4^{k} + \frac{2}{5} (-1)^{k} \\
\end{array} \right].$$ which implies $$a_k = \frac{3}{5} 4^{k} + \frac{2}{5} (-1)^{k}.$$
Okay, okay this is all very well, but can't I do it faster? Well, in this case, you can.
Start by trying to find $a_n$ in the form of $q^n$, that is
$$ a_n = 3a_{n-1} + 4a_{n-2} \quad \text{becomes}\quad q^n = 3q^{n-1} + 4q^{n-2}$$
divide by $q^{n-2}$ and you'll get the quadratic equation $ q^2 - 3q - 4 = 0$. The roots of this polynomial are $-1$ and $4$. Assume that $a_n$ must be a linear combination of the $n$th power of these, that is $a_n$ has the form of $a_n = a4^n + b(-1)^n$.
To find $a$ and $b$, we can use that $a_0$ and $a_1$: substituting $n=0,1$ we get \begin{align} a + b &= a_0 = 1 \\ 4a - b &= a_1 = 2 \end{align}
Solving this gives you $a = \frac{3}{5}, b= \frac{2}{5}$, which implies $$a_k = \frac{3}{5} 4^{k} + \frac{2}{5} (-1)^{k}.$$
If you are familiar with the meaning of the word characteristic polynomial, then you might notice that the "two" methods I've described above are closely connected, since $k_A(x) = x^2 - 3x - 4$ is the characteristic polynomial of $\operatorname A$. Moreover, calculating its roots is the same as calculating the eigenvalues of $\operatorname A$. The only big difference is that you don't have to know anything about matrices to do the second method, which is also faster, and the reason it works is hidden in the 1st method.
A word about calculating big powers of $\operatorname A$. This is more relevant if the matrix has a big size and a similarly nice characteristic polynomial. Let's say you wanted to calculate $A^{10^{10000}}$, and let's say that $A$ is an $m\times m$ matrix with a characteristic polynomial of $k_A$, where $\deg(k_A)$ is small, compared to $10^{10000}$.
By the Cayley–Hamilton theorem, we know that $k_A(A) = 0$.
So the beautiful trick is to calculate $h(x) = x^{10^{10000}} \mod k_A(x)$ first, and then $h(A)$.
Essentially, given any polynomial $f(x)$, in order to calculate $f(A)$, one can find the corresponding element $r(x)$ in the ring $R[x]/k_A(x)$, and then calculate $r(A)$ using again, binary exponentiation.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Prove by induction that $\left [ \left (3+\sqrt{5} \right )^{n} + \left (3-\sqrt{5} \right )^{n} \right ]$ is multiple of $2^{n}$ Prove by induction that :
$\left [ \left (3+\sqrt{5} \right )^{n} + \left (3-\sqrt{5} \right )^{n} \right ]$ is multiple of $ 2^{n}$
My proof is :
At $n=1$
$$\frac{\left ( 3+\sqrt{5} \right )+\left ( 3-\sqrt{5} \right )}{2} = \frac{6}{2} = 3 \in \mathbb{Z}$$
Assume $P(k)$ is true
$$\frac{\left ( 3+\sqrt{5} \right )^{k}+\left ( 3-\sqrt{5} \right )^{k}}{2^{k}} = m \in \mathbb{Z}$$
$$\Rightarrow \left ( 3+\sqrt{5} \right )^{k}+\left ( 3-\sqrt{5} \right)^{k} = 2^{k}.m \rightarrow (*)$$
At $n=k+1$
$$\left ( 3+\sqrt{5} \right )^{k+1}+\left ( 3-\sqrt{5} \right )^{k+1} = \left [ \left ( 3+\sqrt{5} \right )^{k}+\left ( 3-\sqrt{5} \right )^{k} \right ]\left [ \left ( 3+\sqrt{5} \right )+\left ( 3-\sqrt{5} \right ) \right ]-\left [ \left ( 3+\sqrt{5} \right )^{k}\left ( 3-\sqrt{5} \right )+\left ( 3-\sqrt{5} \right )^{k}\left ( 3+\sqrt{5} \right ) \right ]$$
from $(*) $:
$$\left ( 3+\sqrt{5} \right )^{k+1}+\left ( 3-\sqrt{5} \right )^{k+1} =2^{k}.m.6-\left ( 3+\sqrt{5} \right )\left ( 3-\sqrt{5} \right ) \left [ \left ( 3+\sqrt{5} \right )^{k-1}+\left ( 3-\sqrt{5} \right )^{k-1} \right ] $$
but i can't resume my proof .. i can't do any thing with $\left [ \left ( 3+\sqrt{5} \right )^{k-1}+\left ( 3-\sqrt{5} \right )^{k-1} \right ]$ .
|
Another way to solve this problem is to define $a_n = (3 + \sqrt{5})^n + (3-\sqrt{5})^n$ and note that the characteristic equation of this sequence is $x^2 - 6x + 4 = 0$. Therefore $a_{n+2} = 6a_{n+1} - 4a_n$
Obviously we have that $2 \mid a_1$ and $4 \mid a_2$. Now assume that $2^n \mid a_n$ for all $n \le k$ for some integer $k$. Then we have:
$$2^{k+1} \mid 6a_{k} - 4a_{k-1} = a_{k+1}$$
Hence the proof.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Squeeze theorem inequality: $\frac{n}{1+n} < \ln(1+\frac{1}{n})^n < 1$ show that for $n>1$, $$\frac{n}{1+n} < \ln(1+\frac{1}{n})^n < 1$$
by using $\frac{1}{n} < \ln n $
and $\ln(1+n) < n$ for $n>1$
I am unable to prove $\frac{n}{1+n} < \ln(1+\frac{1}{n})^n $
|
$$\frac{1}{\frac{n+1}{n}}=\frac{n}{n+1}<\ln(\frac{n+1}{n})=\ln(1+\frac{1}{n})<\frac{1}n$$
For $n>1$, we have $0<\ln(1+\frac{1}{n})<1, $ so
$$\ln(1+\frac{1}{n})^n<1$$
Also, $$1>\ln(1+\frac{1}{n})^n=n*\ln(1+\frac{1}{n})>n(\frac{n}{n+1})>\frac{n}{n+1}\text{ , for }n>1$$
finishes the proof
|
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"timestamp": "2023-03-29T00:00:00",
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|
Equation with double absolute value I have problem with solving the following equation.
$$2x - |5-|x-2|| = 1$$
How to handle absolute value in absolute value?
I have tried multiple times to solve it but I get no solution.
|
No "easy way out". Solve the first absolute value $$|x-2|=\begin{cases}-x+2, & x<2\\\phantom{-}x-2, & x\ge 2 \end{cases}$$ to take cases
*
*Case 1: $x<2$. Then $$2x-|5-|x-2||=2x-|5-(-x+2)|=2x-|x+3|$$ And now take subcases (but do not forget that you are in $x<2$) depending on the second absolute value
$$|x+3|=\begin{cases}-x-3, & \hspace{26.5pt}x<-3\\ \phantom{-}x+3, & -3\le x <2 \\\end{cases}$$ So, $$2x-|x+3|=\begin{cases}2x-(-x-3)=3x+3, & \hspace{26.5pt}x<-3 \\2x-(x+3)=x-3, & -3\le x < 2 \end{cases}$$ And now treat each subcase separately: $$3x+3=1\iff x=-\frac23\not <-3$$ so, no solution here. Second subcase: $$x-3=1\iff x=4 \notin-3\le x<2$$ so, nothing here either.
*Case 2: $x\ge 2$. Then $$2x-|5-|x-2||=2x-|5-(x-2)|=2x-|7-x|$$ And now take subcases (but do not forget that you are in $x\ge 2$) depending on the second absolute value
$$|7-x|=\begin{cases} \phantom{-}7-x, & 2\le x<7\\-7+x, & 7\le x \end{cases}$$ So, $$2x-|7-x|=\begin{cases}2x-(7-x)=3x-7, & 2\le x<7 \\2x-(-7+x)=x+7, & 7\le x \end{cases}$$ And now treat each subcase separately: $$3x-7=1\iff x=\frac83\approx2.67\in [2,7)$$ so, bingo! first solution here. Second subcase: $$x+7=1\iff x=-6 \not \ge 7$$ so, nothing here.
The only solution is $x_0=\frac83=2\frac23$.
|
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|
Schrodinger equation to find general wave function I am trying to answer this question:
I have tried solving the Schrodinger equation using separation of variables.
However in the later time wave function i can not seem to get the exp(-3i...)
My current workings are:
|
Expanding
\begin{align*}
\Psi(x,0) &=
\frac{1}{\sqrt{a}} \sin \frac{\pi x}{a}+
\frac{2}{\sqrt{a}} \sin \frac{\pi x}{a} \cos \frac{\pi x}{a} \\
&=
\frac{1}{\sqrt{a}} \sin \frac{\pi x}{a}+
\frac{1}{\sqrt{a}} \sin \frac{2\pi x}{a} \\
\omega_{n} &= \frac{n^2 \pi^2 \hbar}{2ma^2} \\
\psi_{n} (x,t) &= \sqrt{\frac{2}{a}} e^{-\omega_{n} t} \sin \frac{n\pi x}{a} \\
\Psi(x,t) &=
\frac{1}{\sqrt{a}} \exp \left( -\frac{n^2 \pi^2 \hbar}{2ma^2} \right)
\sin \left( \frac{\pi x}{a} \right)+
\frac{1}{\sqrt{a}} \exp \left( -\frac{4n^2 \pi^2 \hbar}{2ma^2} \right)
\sin \left( \frac{2\pi x}{a} \right) \\
&= \frac{\psi_{1}(x,t)}{\sqrt{2}}+\frac{\psi_{2}(x,t)}{\sqrt{2}}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
There exists real numbers $x$ and $y$, such that $x$ and $y$ are irrational, and $x+y$ is also irrational. Using the fact that $\sqrt{2}$ is irrational, prove the following:
There exists real numbers $x$ and $y$, such that $x$ and $y$ are irrational, and $x+y$ is also irrational.
My attempt:
Let $x= \sqrt{2}$ and $y= \sqrt{2} + 1$.
Suppose $y= \sqrt{2} + 1$ is rational. Than $y= \sqrt{2} + 1 = \frac mn$, where $m,n$ are integers. So, $y= \sqrt{2} = \frac mn - 1$ which is rational. This contradicts the original statement that $\sqrt{2}$ is irrational. Thus, $y= \sqrt{2} + 1$ is irrational.
Now, $x+y= \sqrt{2}+ \sqrt{2} + 1 $ which equals $2\sqrt{2} + 1$.
Suppose $2\sqrt{2} + 1$ is rational. Then $2\sqrt{2} + 1 = \frac pq$ where $p,q$ are integers. So, $2\sqrt{2} = \frac pq - 1$ which equals $\sqrt{2}= \frac{p}{2q} + \frac 12 $ which is rational. This contradicts the original statement that $\sqrt{2}$ is irrational. Thus, $2\sqrt{2} + 1$ is irrational.
So, $x+y$ is irrational if $x$ and $y$ are irrational.
Does this proof make sense? Is there a simpler way?
|
Your argumentation is fine.
However, one has to be careful that the statement "So, $x+y$ is irrational if $x$ and $y$ are irrational." does not hold for all $x,y \in \mathbb{R}$ (but, of course, for your choice).
An easier choice should be $x = y = \sqrt{2}$.
|
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|
Derivative of $f(x)=\int^{x^2}_0 \frac{\sin(t)}{t}dt$
Let $f(x)=\int^{x^2}_0 \frac{\sin(t)}{t}dt$. Find $f'(x)$.
Is that integral undefined/nonexistant, or just impossible to integrate?
In this case does $f'(x)$ exist and can be solved normally?
Is it correct that $f'(x)=\frac{\sin(x^2)}{x^2}2x$?
|
The answer provided by @Clarinetist is the most efficient way to solve this problem. Here is another way.
Using the series definition of the sine function we have
\begin{equation}
\sin(z) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!} x^{2n-1}
\end{equation}
and
\begin{align}
f(x) &= \int\limits_{0}^{x^{2}} \frac{\sin(z)}{z} \mathrm{d}z \\
&= \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!} \int\limits_{0}^{x^{2}} z^{2n-2} \mathrm{d}z
\end{align}
The integral on the right hand side above is
\begin{equation}
\int\limits_{0}^{x^{2}} z^{2n-2} \mathrm{d}z = \frac{1}{2n-1} (x^{2})^{2n-1}
\end{equation}
and now we have
\begin{equation}
f(x) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!(2n-1)} (x^{2})^{2n-1}
\end{equation}
Let
\begin{equation}
g(x) = (x^{2})^{2n-1}
\end{equation}
then
\begin{equation}
\frac{\mathrm{d}g}{\mathrm{d}x} = \frac{2}{x}(2n-1)(x^{2})^{2n-1}
\end{equation}
Now we have
\begin{align}
\frac{\mathrm{d}f}{\mathrm{d}x} &= \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!(2n-1)} \frac{\mathrm{d}g}{\mathrm{d}x} \\
&= \frac{2}{x} \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n-1)!} (x^{2})^{2n-1} \\
&= \frac{2}{x} \sin(x^{2})
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Sum of the series $\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$
The sum of the series $$\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........$$
$\bf{My\; Try::}$ We can write it as $\displaystyle \binom{n}{0} = $ Coefficient of $x^0$ in $(1+x)^n$
Similarly $\displaystyle \binom{n-1}{1} = $ Coefficient of $x^1$ in $(1+x)^{n-1}$
Similarly $\displaystyle \binom{n-2}{2} = $ Coefficient of $x^2$ in $(1+x)^{n-2}$
Now, how can I solve it after that, Help Required, Thanks
|
Generating Functions
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^n(-1)^k\binom{n-k}{k}x^n
&=\sum_{k=0}^\infty\sum_{n=k}^\infty(-1)^k\binom{n-k}{k}x^n\\
&=\sum_{k=0}^\infty\sum_{n=0}^\infty(-1)^k\binom{n}{k}x^{n+k}\\
&=\sum_{n=0}^\infty x^n(1-x)^n\\
&=\frac1{1-x(1-x)}\\
&=\frac1{1-x+x^2}\\
&=\frac{1+x}{1+x^3}\\[6pt]
&=(1+x)(1-x^3+x^6-x^9+\dots)\\[11pt]
&=1+x-x^3-x^4+x^6+x^7-x^9-x^{10}+\dots
\end{align}
$$
Therefore,
$$
\sum_{k=0}^n(-1)^k\binom{n-k}{k}=\left\{\begin{array}{r}
1&\text{if }n\equiv0,1\pmod6\\
0&\text{if }n\equiv2,5\pmod6\\
-1&\text{if }n\equiv3,4\pmod6
\end{array}\right.
$$
|
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|
How to integrate this function over this surface? I'm solving a physics problem, and at some point I need to solve this integral
$$\iint\limits_S \frac{1}{x+2a} dxdy$$
where
$$S=\{(x,y,z):x^2+y^2\leq a^2, \ x\geq0, \ z=0\}$$
So $S$ is the right semicircle corresponding to the circle centered at the origin with radius $a$. This seemed a bit difficult so I decided to calculate the integral over the quarter circle in the first quadrant (let's call this surface $S'$) and then multiply that by two to get the value of the whole surface integral. Doing this, I got
$$2\iint\limits_{S'} \frac{1}{x+2a} dxdy=2\int\limits_0^a dy\int\limits_0^{\sqrt{a^2-y^2}}\frac{1}{2a+x}dx=2\int\limits_0^a\ln\left(\sqrt{a^2-y^2}+2a\right)-\ln(2a)\ dy=\int\limits_0^a 2\ln\left(\sqrt{a^2-y^2}+2a\right)\ dy \ - a\ln(2a)$$
And this function doesn't seem to have a primitive, or at least it is very hard to find it (I looked up in some tables and found nothing). The problem shouldn't be so hard so I expect that this integral can be calculated in a much easier way. Any ideas?
|
Integrating $y$ first and using the trig substitution $x = a\sin\theta$ leads to
$$
\int_{0}^{a} \frac{2\sqrt{a^{2} - x^{2}}\, dx}{2a + x}
= \int_{0}^{\pi/2} \frac{2a\cos^{2}\theta\, d\theta}{2 + \sin\theta}.
\tag{1}
$$
The tangent half-angle substitution $t = \tan\frac{\theta}{2}$ gives
$$
\cos\theta = \frac{1 - t^{2}}{1 + t^{2}},\qquad
\sin\theta = \frac{2t}{1 + t^{2}},\qquad
d\theta = \frac{2\, dt}{1 + t^{2}},
$$
upon which (1) becomes
$$
2a\int_{0}^{1} \frac{\left(\dfrac{1 - t^{2}}{1 + t^{2}}\right)^{2}}{2 + \dfrac{2t}{1 + t^{2}}} \cdot \frac{2\, dt}{1 + t^{2}}
= 2a\int \frac{(1 - t^{2})^{2}}{(1 + t^{2})^{2}(1 + t + t^{2})}\, dt.
\tag{2}
$$
Partial fractions is a bit laborious (six equations, six unknowns); for the record, the decomposition is
$$
\frac{(1 - t^{2})^{2}}{(1 + t^{2})^{2}(1 + t + t^{2})}
= \frac{A_{1}t + B_{1}}{(t^{2} + 1)^{2}} + \frac{A_{2}t + B_{2}}{t^{2} + 1} + \frac{A_{3}t + B_{3}}{t^{2} + t + 1},
$$
and the augmented matrix of the resulting linear system for $(A_{1}, B_{1}, A_{2}, B_{2}, A_{3}, B_{3})$ is
$$
\left[\begin{array}{@{}rrrrrr|r@{}}
0 & 0 & 1 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 & 0 & 1 & 1 \\
0 & 1 & 0 & 1 & 0 & 1 & 1 \\
1 & 0 & 2 & 1 & 2 & 0 & 0 \\
1 & 1 & 1 & 2 & 0 & 2 & -2 \\
1 & 1 & 1 & 1 & 1 & 0 & 0 \\
\end{array}\right].
$$
The end result is easily checked to be
$$
\frac{(1 - t^{2})^{2}}{(1 + t^{2})^{2}(1 + t + t^{2})}
= -\frac{4t}{(t^{2} + 1)^{2}} + \frac{4}{t^{2} + 1} - \frac{3}{(t + \frac{1}{2})^{2} + \frac{3}{4}}.
$$
Consequently, (2) becomes
$$
2a\left[\frac{2}{t^{2} + 1} + 4\arctan t - 2\sqrt{3} \arctan\bigl(\tfrac{2}{\sqrt{3}}(t + \tfrac{1}{2})\bigr)\right]\bigg|_{0}^{1}
= 2a\left[-1 + \pi + \frac{\pi}{\sqrt{3}}\right].
$$
|
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|
Find $a,b,c$ where $(a-1)(b-1)(c-1)$ is a divisor of $(abc-1)$ Find integers $a,b,c$ such that $1<a<b<c$ and $(a-1)(b-1)(c-1)$ is a divisor of $(abc-1)$ .I tried it solve using elementary number theory but I can't proceed . Somebody help me.
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Not completely different from Oleg517's. write $x=a-1$, $y=b-1$, $z=c-1$, we have $1\le x < y <z$ and $$\frac{abc-1}{(a-1)(b-1)(z-1)} = 1 + \frac{x+y+z + xy + yz + zx}{xyz}$$ is an integer.
On the other hand, $$0< x+y+z + xy + yz + zx \le xyz + (xy+yz+zx) = xyz + x(y+z) + yz < xyz + xyz + yz < 3xyz,$$ so $x+y+z + xy + yz + xz = xyz$ or $2xyz$.
First, solve $x+y+z + xy + yz + zx = xyz$: $x>1$ is easy to see; on the other hand, if $x\ge 3$, then the left side is bounded by $$xyz(1/20+1/15+1/12+1/3+1/4+1/5) <xyz$$. So $x=2$, and we get $y=4$, $z=14$ by solving $(y-3)(z-3)=11$.
Second, solve $x+y+z+xy+yz+zx=2xyz$: it is easy to see $x=1$ by arguments similar to the above, then $y=3$, $z=7$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/1963717",
"timestamp": "2023-03-29T00:00:00",
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|
Find Min of function $x+\frac{2}{x}$ for x>0 with following solution? I want to find Minimum of function $x+\frac{2}{x}$ for $x>0$ with following solution. My problem is how can we say we found the answer with following solution.
Solution:
Let Minimum be something like c. We have:
$x+\frac{2}{x} \ge c \to \frac{x^2+2}{x} \ge c \to x^2 + 2 \ge cx \to x^2 -cx +2 \ge 0 $
So $\Delta \le 0$ because the quadratic must not have two solutions so it will be only positive or zero. So:
$\Delta = c^2 - 8 \le 0 \to c^2 \le 8 \to -2\sqrt{2} \le c \le +2\sqrt{2}$
My question is how can we say with following solution that the minimum of $x+\frac{x}{2}$ must be $+2\sqrt{2}$? We find the maximum of $c$, How can we say that maximum is minimum of function? Is it a valid solution?
Thanks and sorry for my English.
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Let's look at the original inequality $$x + \frac{2}{x} \ge c$$ after you have concluded that $$-2\sqrt{2} \le c \le 2\sqrt{2}.$$ If we suppose $c = 0$, then is the original inequality true? Yes: $x + 2/x \ge 0$ is a true statement for $x > 0$. Indeed, any value of $c$ in the interval you obtained, when substituted into the first inequality, results in a true statement, by construction. But all of the inequalities resulting from choosing $c < 2 \sqrt{2}$ are looser than the inequality resulting from choosing $c = 2 \sqrt{2}$, in the sense that for any $c' < 2 \sqrt{2}$, we trivially have $$x + \frac{2}{x} \ge 2 \sqrt{2} > c',$$ hence $c = 2 \sqrt{2}$ results in the tightest inequality of the set of choices we can make.
The only missing part is to show that equality can be attained for this choice of $c$ thus guaranteeing that the inequality is strict, and of course, this occurs when $x = 2/x = \sqrt{2}.$
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"url": "https://math.stackexchange.com/questions/1964011",
"timestamp": "2023-03-29T00:00:00",
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|
If $abc=1$ so $\sum\limits_{cyc}\frac{a}{\sqrt{a+b^2}}\geq\frac{3}{\sqrt2}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{a}{\sqrt{a+b^2}}+\frac{b}{\sqrt{b+c^2}}+\frac{c}{\sqrt{c+a^2}}\geq\frac{3}{\sqrt2}$$
After substitution $a=\frac{y}{x}$... I tried C-S, but without success.
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I proved this inequality!!!
Let $a=\frac{x}{z}$, $b=\frac{y}{x}$ and $c=\frac{z}{y}$, where $x$, $y$ and $z$ are positives.
Hence, we need to prove that $\sum\limits_{cyc}\frac{x^2}{\sqrt{z(x^3+y^2z)}}\geq\frac{3}{\sqrt2}$.
Now by AM-GM $\sum\limits_{cyc}\frac{x^2}{\sqrt{z(x^3+y^2z)}}=\sum\limits_{cyc}\frac{2\sqrt2x^3}{2\sqrt{2x^2z(x^3+y^2z)}}\geq\sum\limits_{cyc}\frac{2\sqrt2x^3}{x^3+y^2z+2zx^2}$.
Thus, it remains to prove that $\sum\limits_{cyc}\frac{x^3}{x^3+y^2z+2zx^2}\geq\frac{3}{4}$, which is true, but my proof is still very ugly:
Let $x=\min{x,y,z\}$, $y=x+u$ and $z=x+v$.
Thus, we need to prove that:
$$8(u^2-uv+v^2)x^7+(2u^3+45u^2v+5uv^2+2v^3)x^6+$$
$$+(3u^4+40u^3v+153u^2v^2-40uv^3+3v^4)x^5+$$
$$+(5u^5+37u^4v+146u^3v^2+114u^2v^3-47uv^4+5v^5)x^4+$$
$$+(3u^6+20u^5v+130u^3v^3+21u^2v^4-18uv^5+3v^6)x^3+$$
$$+uv(7u^5+30u^4v+82u^3v^2+38u^2v^3-12uv^4+2v^5)x^2+$$
$$+u^2v^2(6u^4+20u^3v+27u^2v^2-6uv^2+v^4)x+u^5v^3(2u+5v)\geq0,$$ which is smooth.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1964115",
"timestamp": "2023-03-29T00:00:00",
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How to show $1+3x$ is a unit in $\mathbb{Z_6}[[x]]$ I am new to the subject of rings of formal power series. The problem is to show that $1+3x$ is a unit in $\mathbb{Z_6}[[x]]$, where $\mathbb{Z_6}[[x]]$ is the ring of formal power series with coefficients in $\mathbb{Z}_6$, also known as $ \mathbb{Z }/6\mathbb{Z}$.
My work is as follows:
$ \begin{align} (1+3x)^{-1} &= \frac{1}{1+3x}
\\&= \frac{1}{1-(-3x)}
\\ &= 1 +(-3x) + (-3x)^2 + (-3x)^3 ...
\\ &= 1 - 3x + 3^2x^2 -3^3x^3 ...
\\ &=_6 1 + 3x +3x^2 + 3x^3 + ... \end{align}$
This is supposed to show that the inverse of $1 + 3x$ is $ 1 + 3x + 3x^2 + ...$ in the ring $\mod 6$. I am confused what this actually means. When I plug in $x =3$ into $1+3x$, this produces $4$ in mod 6 and there is no modular inverse of $4$. So why is $1+3x$ called the inverse?
I notice that we can check the calculation manually.
$\begin{align}( 1 + 3x) ( 1 + 3x + 3x^2 + 3x^3 + ...) &= 1 + 3x + 3x + 3^2x^2 + 3x^2 +...
\\ &=_6 1 + 3x + 3x + 3x^2 + 3x^2 +...
\\ &=_6 1 + 6x + 6x^2 + ...
\\ &=_6 1 + 0 + 0 + ...
\\ &=_6 1
\end{align}$
What bothers me is that some values you cannot plug into x, the modulo inverse is not defined.
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First at all $1+3x$ is an element of $\mathbb{Z_6}[[x]]$in which all the coefficients except $a_0=1$ and $a_1=3$ are nuls. Your expand of $\frac{1}{1+3x}$ modulo $6$ is correct. It follows, changing $3$ by its equivalent modulo $6$, which is $-3$.
$$(1+3x)(1-3x+3x^2-3x^3+....\pm3x^n\mp3x^{n+1}\pm.......)$$ This multiplication gives $$1+(3x+3x^2+3x^3+.....)+(-3x-3x^2-3x^3-.....)=1$$ Thus it is verified that
$1-3x+3x^2-3x^3+....\pm3x^n\mp3x^{n+1}\pm.......$ is in fact the inverse of $1+3x$ in $\mathbb{Z_6}[[x]]$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1971415",
"timestamp": "2023-03-29T00:00:00",
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|
How should I calculate the determinant? $\left|\begin{array}{cccc}1&a&b&c+d\\1&b&c&a+d\\1&c&d&a+b\\1&d&a&b+c\end{array}\right|=
\left|\begin{array}{cccc}1&a&b&c\\1&b&c&a\\1&c&d&a\\1&d&a&b\end{array}\right|+
\left|\begin{array}{cccc}1&a&b&d\\1&b&c&d\\1&c&d&b\\1&d&a&c\end{array}\right|$
I tried to calculate the determinant but I couldn't do it after separating the determinant by the property. How should I calculate it?
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$${\begin{vmatrix}1&a & b &c\\1 &b &c &d \\1 &c &d &a\\1&d &a &b &\end{vmatrix}}=-{\begin{vmatrix}1&a & b &d\\1 &b &c &a \\1 &c &d &b\\1&d &a &c &\end{vmatrix}}$$
because the second determinant is the same as the first by cyclically permuting the columns $2,3$ and $4$, an cyclically permuting the four rows.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1971668",
"timestamp": "2023-03-29T00:00:00",
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|
A circle inside an ellipse Consider an ellipse with semi-axes $a$ and $b$, taller than it is wide with a small circle of radius $r$ inside. Assume the circle falls to the lowest point possible while staying inside the ellipse.
If $2r\le a-c$ then the circle and ellipse will meet at a single point at the bottom. If $2r>a-c$ the circle and ellipse will intersect at two points on the opposite side, leaving a space between the bottom of the circle and the bottom of the ellipse. For this case, given the radius of the circle and the dimensions of the ellipse how do I calculate the distance $d$ between the bottom of the circle and the bottom of the ellipse?
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I got $$w =b - \sqrt{\frac{(b^2-a^2)(a^2-r^2)}{a^2} }-r $$
Start with polar coordinates for the ellipse $$ \begin{pmatrix}x \\ y \end{pmatrix} = \tfrac{a b}{\sqrt{b^2-(b^2-a^2) \cos^2 \varphi}} \begin{pmatrix} \sin \varphi \\ -\cos\varphi \end{pmatrix}$$
where $(x,y)$ are the contact point coordinates relative to the center of the ellipse
The contact angle $\eta$ (see below) is found from the tangent vector of the ellipse as $$ \eta =
\varphi - \tan^{-1} \left( \frac{ (a^2-b^2) \sin\varphi \cos\varphi}{b^2 - (b^2-a^2)\cos^2 \varphi} \right) $$
the location of the circle is found from the relationships $$\begin{aligned} r \sin \eta = s \sin \varphi \\ z+r \cos \eta = s \cos \varphi \end{aligned}$$
This is solved for $$\varphi = \frac{\pi}{2} - \tan^{-1} \left( \frac{b^2}{a} \sqrt{ \frac{a^2-r^2}{b^2 r^2-a^4} } \right) $$
and $$ z = s \cos\varphi - r \cos \eta $$
Finally, the gap is $$\boxed{w = b -z - r}$$ and with a lot of simplifications (thank you CAS) I got the answer above.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1972994",
"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.