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let $f(x)=(3(x+x^2))/14$ and $x$ between $0$ and $2$ , zero otherwise be the pdf for a random variable $X$ ,Find the median and the mode? let f(x)=(3(x+x^2))/14 and x between 0 and 2 ,
zero otherwise be the pdf for a random variable X
Find the median and the mode
`
Could you please help me Is it correct or not?
|
If $f(x)$ represents a legal probability density function (which you should check based on the support of the distribution), then for the median we find an $m$ such that
$$\int_0^m f(x) dx = \frac{1}{2}$$
$$\iff \int_0^m (3(x+x^2))/14 dx = \frac{1}{2}$$
$$\iff \frac{3}{14}\int_0^m x+x^2 dx = \frac{1}{2}$$
$$\iff \int_0^m x+x^2 dx = \int_0^m x+x^2 dx = \frac{7}{3}$$
$$\iff \frac{1}{2}x^2 + \frac{1}{3}x^3 \mid_0^m = \frac{7}{3}$$
$$\iff \frac{1}{2}m^2 + \frac{1}{3}m^3 = \frac{7}{3} \iff \frac{3m^2+2m^3}{6} = \frac{7}{3} \iff 3m^2 + 2m^3 = 14.$$
You can show through some algebraic manipulation that $m \approx 1.522.$
As you stated before, we are trying to find
$$\arg\max_x f(x)$$
over $0 < x < 2.$
If we set
$$f'(x) = 0 \implies \frac{3}{14}(1 + 2x) = 0 \implies 1 + 2x = 0 \implies x = \frac{-1}{2},$$
which would not be correct for $0 < x < 2.$ In since if $x \geq 0$, $\frac{3}{14}(x+x^2)$ is monotonically increasing, and thus the maximum value would be hit at $x = 2$. Hence, the mode is $x = 2$, if $2$ is within the support of the pdf. Otherwise, this function would not have a global maximum (and thus no mode).
|
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|
Prove $\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$
$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove
$$\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x} <4$$
A natural though is that from the condition $x^2+y^2+z^2+xyz=4$, I tried a trig substitutions $x=2\cdot \cos A$, $y=2\cdot \cos B$ and $z=2\cdot \cos C$, where $A,B,C$ are angles of an acute triangle.
Our inequality becomes
$$\sqrt[2]{2(\cos A+ \cos B)}+\sqrt[3]{2(\cos B+\cos C)}+\sqrt[4]{2(\cos C+\cos A)} <4$$
I used formula $\cos(A)+\cos(B) < 2$ and estimate that $LHS <5$ but not $4$.
I do not know how to proceed.
|
PARTIAL SOLUTION
$$f(x,y,z)=\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x}$$
$$g(x,y,z)= x^2+y^2+z^2+xyz-4=0$$
$$ \begin{align}\mathcal{L}(x,y,z,\lambda) =& \;f(x,y,z) + \lambda g(x,y,z)\\
=&\;\sqrt[2]{x+y}+\sqrt[3]{y+z}+\sqrt[4]{z+x}+\lambda(x^2+y^2+z^2+xyz-4) \end{align}$$
$$\nabla_{x,y,z,\lambda}\mathcal{L}(x,y,z,\lambda) =\left(\frac{\partial \mathcal{L}}{\partial x},\frac{\partial \mathcal{L}}{\partial y},\frac{\partial \mathcal{L}}{\partial z},\frac{\partial \mathcal{L}}{\partial \lambda}\right)$$
Now comes the work finding the partial derivatives
$$\begin{align}
&\frac{\partial \mathcal{L}}{\partial x} =\; \lambda(2x+yz) + \frac{1}{2(x+y)^{1/2}}+\frac{1}{4(x+z)^{3/4}}\\
&\frac{\partial \mathcal{L}}{\partial y} =\; \lambda(2y+xz) + \frac{1}{2(x+y)^{1/2}}+\frac{1}{3(y+z)^{2/3}}\\
&\frac{\partial \mathcal{L}}{\partial z} =\; \lambda(2z+xy) + \frac{1}{3(y+z)^{2/3}}+\frac{1}{4(x+z)^{3/4}}\\
&\frac{\partial \mathcal{L}}{\partial \lambda} =\; x^2+y^2+z^2+xyz-4
\end{align}$$
$$\nabla_{x,y,z,\lambda}\mathcal{L}(x,y,z,\lambda)=0 \iff \begin{cases}
\lambda(2x+yz) + \frac{1}{2(x+y)^{1/2}}+\frac{1}{4(x+z)^{3/4}} & = 0 \\
\lambda(2y+xz) + \frac{1}{2(x+y)^{1/2}}+\frac{1}{3(y+z)^{2/3}} & = 0 \\
\lambda(2z+xy) + \frac{1}{3(y+z)^{2/3}}+\frac{1}{4(x+z)^{3/4}} & = 0 \\
x^2+y^2+z^2+xyz-4, & = 0
\end{cases}$$
At this point I cannot seem to solve the resulting equations. The symmetry involved implies some simple trick should suffice, though I have yet to spot one.
|
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|
Inverse tangents in a cyclic order
If
$$\theta= \tan^{-1}\left(\frac{a(a+b+c)}{bc}\right)+\tan^{-1}\left(\frac{b(a+b+c)}{ac}\right)+\tan^{-1}\left(\frac{c(a+b+c)}{ab}\right)$$then find $\tan\theta$
I tried to use these as sides of a triangle and use their properties, but other than that I am clueless. I cannot think of a substitution either. The answer happens to be zero. Any help is appreciated. Thanks in advance!!
|
If
$$\theta= \tan^{-1}\left(\color{red}{\sqrt{\frac{a(a+b+c)}{bc}}}\right)+\tan^{-1}\left(\color{red}{\sqrt{\frac{b(a+b+c)}{ac}}}\right)+\tan^{-1}\left(\color{red}{\sqrt{\frac{c(a+b+c)}{ab}}}\right)$$then $$\tan\theta=\color{red}{0}$$
Proof:
$$\tan^{-1}\left( \sqrt {\frac {x (x+y+z)}{yz}} \right)+\tan^{-1}\left( \sqrt {\frac {y (x+y+z)}{zx}} \right)+\tan^{-1}\left( \sqrt {\frac {z (x+y+z)}{xy}} \right)$$
Let $ x $ - distance from the vertex $A $ of the triangle $ABC$ to the points of contact of the inscribed circle.
$y, z $ - similary
Then: $x+y+z=p, yz= r\cdot r_a $ , where $ r_a $- excircle radius.
$$\frac {x}{r}=\frac {p}{r_a}= ctg\left ( \frac {\alpha}{2} \right)$$
Then: $$\alpha+\beta+\gamma =\pi $$
$$\tan^{-1}\left (ctg \left (\frac {\alpha}{2}\right)\right )+\tan^{-1}\left (ctg \left (\frac {\beta}{2}\right)\right )+\tan^{-1}\left (ctg \left (\frac {\gamma}{2}\right)\right )=\pi$$
Addition:
If
$$\theta= \tan^{-1}\left(\frac{a(a+b+c)}{bc}\right)+\tan^{-1}\left(\frac{b(a+b+c)}{ac}\right)+\tan^{-1}\left(\frac{c(a+b+c)}{ab}\right)$$then find $\tan\theta$
Use: $$\tan(u+v+w)=\frac{\tan(u+v)+\tan w}{1-\tan(u+v)\cdot\tan w}=\frac{\tan u+\tan v+ \tan w-\tan u \cdot \tan v \cdot \tan w}{X}$$
Then $$\tan\theta=\frac{\left(\frac{a(a+b+c)}{bc}\right)+\left(\frac{b(a+b+c)}{ac}\right)+\left(\frac{c(a+b+c)}{ab}\right)-\left(\frac{a(a+b+c)}{bc}\right)\cdot\left(\frac{b(a+b+c)}{ac}\right)\cdot\left(\frac{c(a+b+c)}{ab}\right)}{X}$$
|
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|
How to integrate $\frac{\sin(z)}{(z^2 + 1/2)^2}$ over the unit circle? How can I integrate $\frac{\sin(z)}{(z^2 + 1/2)^2}$?
$(z^2 + 1/2)^2$ isn't in a form where I can use the Cauchy Integral Formula, and we haven't covered using residues yet in my class so I'm assuming that's not the way to go (it's also fairly ugly to try to do so in this case).
|
You can apply Cauchy's integral theorem by expanding the factor $\frac{1}{\left(z^2 + \frac{1}{2}\right)^2}$ in partial fractions. You can obtain the partial fraction expansion without having to solve any equations as follows. The singularities are at $z = \alpha^{\pm} = \pm \frac{i}{2}\sqrt{2}$. The partial fraction expansion of the function $\frac{1}{z^2 + \frac{1}{2}}$ is thus given by:
$$\frac{1}{z^2 + \frac{1}{2}} = \sum_{\pm}\left(\lim_{z\to\alpha^{\pm}}\frac{z-\alpha^{\pm}}{z^2+\frac{1}{2}}\right)\frac{1}{z-\alpha^{\pm}} = \frac{i}{2}\sqrt{2}\left(\frac{1}{z+\frac{i}{2}\sqrt{2}} - \frac{1}{z-\frac{i}{2}\sqrt{2}}\right) $$
Squaring both sides gives:
$$\frac{1}{\left(z^2 + \frac{1}{2}\right)^2} = -\frac{1}{2}\left(\frac{1}{\left(z+\frac{i}{2}\sqrt{2}\right)^2} + \frac{1}{\left(z-\frac{i}{2}\sqrt{2}\right)^2}\right)+\frac{1}{\left(z+\frac{i}{2}\sqrt{2}\right)\left(z-\frac{i}{2}\sqrt{2}\right)}$$
Around the pole at $z = -\frac{i}{2}\sqrt{2}$ we have the expansion:
$$\frac{1}{\left(z^2 + \frac{1}{2}\right)^2} =-\frac{1}{2}\frac{1}{\left(z+\frac{i}{2}\sqrt{2}\right)^2} +\frac{i}{2}\sqrt{2}\frac{1}{z+\frac{i}{2}\sqrt{2}}+\text{nonsingular terms}$$
Around the pole at $z = \frac{i}{2}\sqrt{2}$ we have the expansion:
$$\frac{1}{\left(z^2 + \frac{1}{2}\right)^2} =-\frac{1}{2}\frac{1}{\left(z-\frac{i}{2}\sqrt{2}\right)^2} -\frac{i}{2}\sqrt{2}\frac{1}{z-\frac{i}{2}\sqrt{2}}+\text{nonsingular terms}$$
If we subtract from $\frac{1}{\left(z^2 + \frac{1}{2}\right)^2}$ the four terms from the two expansions then this will cancel out all the singularities (we're left with only removable singularities), therefore this must yield a polynomial. But since this tends to zero at infinity, it must be zero. The function $\frac{1}{\left(z^2 + \frac{1}{2}\right)^2}$ is thus equal to the sum of the four singular terms in the two expansions.
|
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How to find $L = \int_0^1 \frac{dx}{1+{x^8}}$ Let $L = \displaystyle \int_0^1 \frac{dx}{1+{x^8}}$ . Then
*
*$L < 1$
*$L > 1$
*$L < \frac{\pi}{4}$
*$L > \frac{\pi}{4}$
I got some idea from this video link. But got stuck while evaluating the second integral.
Please help!!
Thanks in advance!
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There is a "closed-form" expression for this integral.
Using partial fractions,
$$ \dfrac{1}{1+x^8} = - \dfrac{1}{8} \sum_{\omega} \dfrac{\omega}{x - \omega}$$
where the sum is over the roots of the polynomial $x^8 + 1$. Then integrate:
$$ \int_0^1 \dfrac{dx}{1+x^8} = -\dfrac{1}{8} \sum_\omega \omega (\log(1-\omega) - \log(-\omega))$$
where we use the principal branch of the logarithm (note that the line segments from $-\omega$ to $1-\omega$ never cross the negative real axis, which is the branch cut of this function, so this works).
These roots can be expressed explicitly as $\left(\pm \sqrt {2+\sqrt {2}}\pm i\sqrt {2-\sqrt {2}}\right)/2$ and $\left(\pm \sqrt {2-\sqrt {2}}\pm i\sqrt {2+\sqrt {2}}\right)/2$. The final result is a rather messy, but explicit, expression in terms of $16$ complex logarithms. It can then be simplified to
$$\frac{1}{16} \left( \ln \left( \left( 4-2\,\sqrt {2} \right) \sqrt {2-
\sqrt {2}}-4\,\sqrt {2}+7 \right) +\pi \right) \sqrt {2-\sqrt {2}}+\frac{1}{16} \left( \ln \left( \left( 4+2\,\sqrt {2} \right) \sqrt {2+\sqrt
{2}}+4\,\sqrt {2}+7 \right) +\pi \right) \sqrt {2+\sqrt {2}}$$
The numerical value turns out to be approximately $.9246517057$.
Of course, this is not the best way to answer the original question.
|
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|
Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$ Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$
According to Fermat's theorem:
$$1^7+7^7+13^7+19^7+23^7\equiv{1+7+13+19+23}\pmod{7}\equiv{63}\pmod{7}\equiv{0}\pmod{7}$$
Now we need to show: $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{9}$ , but how??
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Just to be a bit different, note that a binomial expansion tells us
$$7^7=(1+6)^7=1+7\cdot6+(\text{stuff})\cdot6^2\equiv43\equiv-2\mod 9$$
since $6^2\equiv0$ mod $9$, and thus, since $13\equiv4$, $19\equiv1$, and $23\equiv-4$ mod $9$, we have
$$1^7+7^7+13^7+19^7+23^7\equiv1^7-2+4^7+1^7+(-4)^7=1-2+4^7+1-4^7=0\mod 9$$
|
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|
Prove $(a+b)^3(b+c)^3(c+d)^3(d+a)^3\ge 16a^2b^2c^2d^2$ Let $a,b,c,d>0$ and such $a+b+c+d=1$, show that
$$(a+b)^3(b+c)^3(c+d)^3(d+a)^3\ge 16a^2b^2c^2d^2$$
since
$$a+b\ge 2\sqrt{ab}$$
I think this will not hold.because we have $256abcd\le 1$
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We need to prove that $\prod\limits_{cyc}(a+b)^3\geq16(a+b+c+d)^4a^2b^2c^2d^2$.
Let $a+b+c+d=4u$, $ab+ac+ad+bc+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.
Since $a$, $b$, $c$ and $d$ are positive roots of the equation
$x^4-4ux^3+6v^2x^2-4w^3x+t^4=0$, we see that (by the Rolle's theorem) the equations
$v^2x^2-2w^3x+t^4=0$ and $ux^2-2v^2x+w^3=0$ have two positive roots,
which gives $w^6\geq v^2t^4$ and $v^4\geq uw^3$.
Since $(a+b)(b+c)(c+d)(d+a)-(a+b+c+d)(abc+abd+acd+bcd)=(ac-bd)^2\geq0$,
we obtain
$\prod\limits_{cyc}(a+b)^3\geq\left(\sum\limits_{cyc}abc\right)^3(a+b+c+d)^3=4^6w^9u^3=\frac{4^6w^{12}u^3}{w^3}\geq\frac{4^6v^4t^8u^3}{w^3}\geq\frac{4^6uw^3t^8u^3}{w^3}=4^6t^8u^4=$
$=16(a+b+c+d)^4a^2b^2c^2d^2$. Done!
|
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Method to find smallest value of $x$ for which $x^2-x+C$ is composite. Problem statement: Given the function $f(x)=x^2-x+C$, where $x$ is a positive integer $>1$ and $C$ is a positive integer ($C=0$ is also allowed), find some method and/or set of rules to always find the smallest value of $x$ that will give a non-prime number for the function $x^2-x+C$.
To show the research that has been formulated so far, I have denoted these set of rules for the problem:
*
*If $C$ is an even number, the smallest $x$ value for a non-prime is $2$ because the function will always give a non-prime when $C$ is even.
*$x^2-x+C=x(x-1)+C$, this shows $x^2-x$ is always divisible by at least $x$.
*If $x=C$, then $x$ will result in a non-prime but isn't always the smallest non-prime.
The last bullet point here provides an initial method for when C is odd - it allows one to narrow their search for smallest values down to $C \geq x\geq 2$. Lastly, here are a few that have been completed for example:
*
*When $C=3$ the smallest $x$ for $f(x)$ that gives a non-prime is $3$
*When $C=5$ the smallest $x$ for $f(x)$ that gives a non-prime is $5$
*When $C=7$ the smallest $x$ for $f(x)$ that gives a non-prime is $2$
*When $C=9$ the smallest $x$ for $f(x)$ that gives a non-prime is $3$
*When $C=11$ the smallest $x$ for $f(x)$ that gives a non-prime is $11$
*When $C=13$ the smallest $x$ for $f(x)$ that gives a non-prime is $2$
*When $C=15$ the smallest $x$ for $f(x)$ that gives a non-prime is $3$
*When $C=1$ the smallest $x$ for $f(x)$ that gives a non-prime is $1$
*When $C=0$ the smallest $x$ for $f(x)$ that gives a non-prime is $2$
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If $C$ is even then $f(2)$ is not prime, and if $C=1$ then $f(4)$ is not prime. So suppose $C>2$.
Let $p$ be the smallest prime dividing $C$. Then $p$ also divides $f(p)$, so $f(p)$ is not prime unless $f(p)=p$. But this is the case if and only if $C=p(2-p)$, which is impossible as $C$ is positive. Hence $2\leq x\leq p$.
Alternatively, as $f(x)>1$ and $f(x)$ is increasing for $C>2$, you could try to find the smallest composite $D>C$ for which
$$x^2-x+(C-D)=0,$$
has integer roots. This is the case if and only if the discriminant is an integer square, that is, if and only if $1+4(D-C)$ is a square. So look for the least composite $D>C$ for which $1+4(D-C)$ is a square.
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Showing that $\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$ Integrate
$$I=\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$$
Substitution $x=\sqrt{\tan(u)}\rightarrow dx={\sec^2(u)\over 2\sqrt{\tan(u)}}du$
$x=1\rightarrow u={\pi\over 4}$
$x=0\rightarrow u=0$
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{1-\tan^2(u)}\over 1+\tan^2(u)}\cdot {\sec^2(u)\over \sqrt{\tan(u)}}du$$
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{1-\tan^2(u)\over \tan(u)}}du$$
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{\cot(u)-\tan(u)}}du$$
Recall
$$\cot(u)-\tan(u)={\cos^2(u)-\sin^2(u)\over \sin(u)\cos(u)}=2\cot(2u)$$
Substitute back into I
$$I={1\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{2\cot(2u)}}du$$
$$I={\sqrt2\over 2}\int_{0}^{{\pi\over 4}}{\sqrt{\cot(2u)}}du$$
Well I know that $$\int{\cos(2u)\over\sin(2u)}du={1\over 2}\ln(\sin(2u))+C$$ but
$$\int\sqrt{{\cos(2u)\over\sin(2u)}}du$$ I have not idea, so can anyone please give a hand? Thank.
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Let $y = x^2$ and $z = \sqrt{y^{-1} - y}$, we have
$$\begin{align}
\int_0^1 \frac{\sqrt{1-x^4}}{1+x^4}dx
= & \int_0^1 \frac{x\sqrt{x^{-2} - x^2}}{x^2(x^{-2}+x^2)}dx
= \int_0^1 \frac{\sqrt{x^{-2} - x^2}}{(x^{-2}+x^2)}\frac{dx}{x}\\
= & \frac12 \int_0^1 \frac{\sqrt{y^{-1}-y}}{y^{-1}+y}\frac{dy}{y}
= -\frac12 \int_{y=0}^1 \frac{\sqrt{y^{-1}-y}}{y^{-1}+y}\frac{d(y^{-1} - y)}{y^{-1}+y}\\
= & \int_0^\infty \frac{z^2 dz}{z^4+4}
= \int_0^\infty \frac{z^2 dz}{(z^2+2)^2 - (2z)^2}
= \int_0^\infty \frac{z^2 dz}{(z^2-2z+2)(z^2+2z+2)}\\
= & \frac14 \int_0^\infty\left(\frac{z}{z^2-2z+2} - \frac{z}{z^2+2z+2}\right)dz\\
= & \frac14 \int_0^\infty \left[\frac12\log\left(\frac{z^2-2z+2}{z^2+2z+2}\right)'
+ \left(\frac{1}{z^2-2z+2} + \frac{1}{z^2+2z+2}\right)\right] dz
\end{align}
$$
The first piece contributes
$\displaystyle\;\frac18 \left[\log\left(\frac{z^2-2z+2}{z^2+2z+2}\right)\right]_0^\infty\;$ which clearly vanishes.
For the second piece, substitute $z$ by $-z$ in its second term, we get:
$$\int_0^1 \frac{\sqrt{1-x^4}}{1+x^4}dx
= \frac14 \int_{-\infty}^\infty\frac{dz}{z^2-2z+2}
= \frac14 \int_{-\infty}^\infty\frac{dz}{(z-1)^2+1}
= \frac{\pi}{4}$$
|
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Finding the degree of $\sqrt[3]{2} + \sqrt[3]{3}$ over $\mathbb Q$ I am practicing writing down random algebraic numbers and finding their degrees over $\mathbb Q$ and have fumbled when coming to $\sqrt[3]{2} + \sqrt[3]{3}$. Mathematica tells me the minimal polynomial is $x^9-27 x^6-135 x^3-729$, which is a polynomial in $x^3$. I figured by cubing the element I could show that the cube has degree 3, which is some partial progress, but I have failed to even do that. If anyone has any neat tricks up their sleeves about these kinds of results, I would be forever indebted to you.
|
First, Mathematica is wrong. The minimal polynomial is $-125-87X^3-15X^6+X^9$.
The irreducible equation for $\sqrt[3]2$ over $\Bbb Q$ is $X^3-2$, and the irreducible equation for $\sqrt[3]2+\sqrt[3]3$ over $\Bbb Q(\sqrt[3]3\,)$would apparently be $F(X)=(X-\sqrt[3]3\,)^3-2=X^3-3\sqrt[3]3X^2+3\sqrt[3]9X-5$. Now, to get a $\Bbb Q$-polunomial out of $F$, you need to multiply it by its two conjugates, gotten by (1) replacing $\sqrt[3]3$ by $\omega\sqrt[3]3$, and (2) replacing $\sqrt[3]3$ by $\omega^2\sqrt[3]3$, where $\omega=\frac12(-1+\sqrt{-3}\,)$ is a chosen primitive cube root of $1$. The conjugate polynomials are
\begin{align}
F_1(X)&=X^3-3\omega\sqrt[3]3X^2+3\omega^2\sqrt[3]9X-5\>,\>\text{and}\\
F_2(X)&=X^3-3\omega^2\sqrt[3]3X^2+3\omega\sqrt[3]9X-5\,.
\end{align}
I’ll leave it to you to do the multiplication $FF_1F_2$. It’s tedious by hand, but far easier than computing a nine-by-nine determinant!
|
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Integration methods: euler's substitution I've looked it up on the internet but I'm having trouble as to how to proceed using Euler's substitution.
For example, how does one solve the following integrals using Euler's substitution:
*
*$\int_1^2\dfrac{dx}{x+\sqrt{x^2+1}}$
and
*
*$\int_0^{\infty} \dfrac{xdx}{\sqrt{x^5+1}}$
For the first one I tried the following:
$\sqrt{x^2+1} = x+t \implies x= \dfrac{1-t}{2}$
Therefore, $x+\sqrt{x^2+1} = 2-t^2$
We have $dx=-tdt$
Therefore: $\int_1^2\dfrac{dx}{x+\sqrt{x^2+1}} = -\int_1^2 \dfrac{tdt}{2-t^2} = \dfrac{1}{2}[log(2-t^2)]^2_1 $ Which is obviously wrong becaus $log(-2)$ doesn't exist.
|
For the first problem:
$$\int_1^2\dfrac{dx}{x+\sqrt{x^2+1}}$$
$$x+\sqrt{x^2+1} = t \implies \sqrt{x^2+1} = -x+t \implies x= \dfrac{t^2-1}{2t}$$
Calculate the derivative $dx$:
$$dx=\dfrac{t^2+1}{2t^2}dt$$
Substitution of $dx$ and $\int_1^2\frac{dx}{x+\sqrt{x^2+1}}$:
$$\int_1^2\dfrac{dx}{x+\sqrt{x^2+1}} = \int_1^2 \dfrac{t^2+1}{2t^2}*\dfrac{1}{t}dt = [\dfrac{1}{2}log(t)-\dfrac{1}{4}t^{-2}]_1^2$$
This is equal to:
$$\left[\dfrac{1}{2}log(x+\sqrt{x^2+1} )-\dfrac{1}{4}\times\dfrac{1}{(x+\sqrt{x^2+1} )^{2}}\right]_1^2$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Problem solving a word problem using a generating function
How many ways are there to hand out 24 cookies to 3 children so that they each get an even number, and they each get at least 2 and no more than 10? Use generating functions.
So the first couple steps are easy.
The coefficient is $x^{24}$
$g(x) = x^6(1+x^2+x^4+x^6+x^8)^3$ or what I got was $x^6 (1 + (x^2)^1 +...+ (x^2)^4)^3$
now finding the closed formula is where I am having problems
My answer: using the fact that $\dfrac{1-x^{n+1}}{1- x}$
I get $x^6\left(\dfrac{1-x^9}{1-x}\right)$ which is wrong
The correct answer: $x^6\left(\dfrac{1-x^{10}}{1-x^2}\right)$
If someone could explain in some detail on how to get the correct formula would be much appreciated. Thanks!
|
When computing $1+x^2+(x^2)^2+(x^2)^3+(x^2)^4$, the series is in powers of $x^2$ not $x$. So the proper expression is $$1+x^2+(x^2)^2+(x^2)^3+(x^2)^4=\frac{1-(x^2)^5}{1-(x^2)}=\frac{1-x^{10}}{1-x^2}.$$ By contrast, $$\frac{1-x^{9}}{1-x}=1+x^1+x^2+\cdots +x^8$$ which differs from the above series in it has both odd and even powers of $x$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all the primitive roots of $13$
Find all the primitive roots of $13$
My attempt:
Since that $13$ is a prime I need to look for $g$ such that $g^{13-1}\equiv 1\pmod{13}$
There are $\phi(12)=4$ classes modulo $12$
how can I find the classes?
|
My quick & dirty method was to note that $13-1=12$ has prime factors $2,3$ & quickly find squares and cubes of a numbers, since those cannot be primitive roots, and not checking those which have already turned up. Any indication of a "short cycle" (like $5$) also gets discarded immediately.
$2: 2^2 \equiv 4, 2^3 \equiv 8 \\
3: 3^2 \equiv 9, 3^3 \equiv 1 \text{(discard all)}\\
5: 5^2 \equiv 12 \equiv -1 \text{(discard all)} \\
6: 6^2 \equiv 10, 6^3 \equiv 8 \\
7: 7^2 \equiv 10, 7^3 \equiv 5 \\
11: 11^2 \equiv 4, 11^3 \equiv 5
$
leaving $\{2,6,7,11\}$ as the primitive roots which is in accordance with the expectation of $\phi(12)=4$ roots (and also the knowledge that for each primitive root $g \bmod p$, $-g$ is a primitive root iff $p \equiv 1 \bmod 4$).
|
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|
Write the index table for the primitive root $3$ of $25$
Write the index table for the primitive root $3$ of $25$
My attempt:
$$
\begin{array}{c|lcr}
k & 0&1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\
\hline
3^k & 1&3&9&2&6&18&4&12&11&8&24&22&16&23&19&7&21&13&14&17 \\
\end{array}
$$
so the index table will be
$$
\begin{array}{c|lcr}
a & 1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19&20&21&22&23&24\\
\hline
\text{ind}_3 a & 0&3&1&6&\color{red}?&4&15&9&2&\color{red}?&8&7&17&18&\color{red}?&12&19&5&14&\color{red}?&16&11&13&10 \\
\end{array}
$$
As you can see I can't find $\text{ind}_3(5),\text{ind}_3(10),\text{ind}_3(15)$ and $\text{ind}_3(20)$
|
By saying $3$ is a primitive root of $25$ it only means that the positive integers less than 25 and relatively prime to it are all of the form $3^k$ for an appropriate exponent $k$ so your table is essentially right, only just exclude the $5,10,15,20$ because they are not coprime with 25.
|
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|
Calculation of area union of 3 overlapping circles Please help calculate the total un-overlapped area of union inside a boundary consisting of perimeters of 3 different circles as shown. The two larger circles are 4.25 mm in diameter and the center circle is 3.28 mm.
All the circles are centered on x-axis, the distance between the 2 largest circle centers is 1.34 mm. The middle smaller circle is centered along y-axis.
|
Let's start by finding the combined area of the two larger circles. The area of one of the larger circles, which I will call $A_O$, is:
$$
\begin{align*}
A_O &= \frac14\pi d^2\\
&= \frac14\pi(4.25)^2\\
&\approx14.186
\end{align*}
$$
Thus the area of both circles, ignoring their intersection for a moment, is $2A_O$. To find the area between them, we can use the formula for the area of a symmetric lens,
$$A_\text{lens} = R^2(\theta-\sin\theta), \tag{*}\label{*}$$
where $R$ is the radius of the circles forming the lens, and $\theta$ the central angle in radians. We know $R = 4.25$, but we need to find $\theta$. To do so, draw in a few lines on the diagram:
(Note that I left out the smaller circle for simplicity's sake.) In the diagram, $\theta = m\angle AOB = 2\alpha$, and $R = OA = OC = OB$. We can find $\theta$ through $\triangle AOD$. The length of line segment $OD$ is equal to $OC-CD$, or the radius minus half the distance between the two circles. We also know the length of line segment $OA$, since it is the radius, so we can write:
$$
\begin{align*}
\alpha &= \arccos\left(\frac{OD}{OA}\right)\\
&= \arccos\left( \frac{2.125-\frac12(1.34)}{2.125} \right)\\
&\approx .816596 \ \text{radians}
\end{align*}
$$
Since $\theta = 2\alpha$, $\theta=1.63319$. We can now use $\eqref{*}$:
$$
\begin{align*}
A_\text{lens} &= R^2(\theta-\sin\theta) \\
&= (2.125)^2(1.63319-\sin1.63319)\\
&\approx 2.868
\end{align*}
$$
Thus, the area between the two larger circles, $A_L$, is:
$$
\begin{align*}
A_L &= 2A_O-A_\text{lens}\\
&= 2(14.186)-2.868\\
&= 25.504
\end{align*}
$$
Now we have to deal with the two small regions created by the smaller circle. For these, I will be using the (rather lengthy) formula for the area of an asymmetric lens,
$$
\begin{align*}
A_\text{asy} = & \ r^2\arccos\left( \frac{d^2+r^2-R^2}{2dr} \right)
+ R^2\arccos\left( \frac{d^2+R^2-r^2}{2dR} \right) - \\
& \frac12 \sqrt{(-d+r+R)(d+r-R)(d-r+R)(d+r+R)}, \tag{**}\label{**}
\end{align*}
$$
where $r$ is the radius of the smaller circle, $R$ is the radius of the larger circle, and $d$ is the distance between the centers of the two circles. Here, $r=1.64$, $R = 2.125$, and $d=1.455$. Observe the diagram below.
The light blue region is the asymmetric lens we will be finding the area of, and we are looking for the area of the two green regions. Using \eqref{**}, we find the area of one asymmetric lens to be $A_\text{asy}\approx5.650$, and thus the area of two asymmetric lenses is $2A_\text{asy}=11.300$.
Notice that the intersection of the two asymmetric lenses in the diagram is the symmetric lens from before. Thus, the area of the light blue and dark blue region is:
$$
\begin{align*}
A_\text{blue} &= 2A_\text{asy}-A_\text{lens}\\
&=11.300-2.868\\
&=8.432
\end{align*}
$$
Now, the area of the green regions is equal to the area of the small circle minus the area of the blue regions. The area of the small circle is:
$$
\begin{align*}
A_o &= \pi r^2 \\
&= \pi(1.64)^2\\
&\approx 8.450
\end{align*}
$$
And so the area of the green regions is:
$$
\begin{align*}
A_\text{green} &= A_o-A_\text{blue}\\
&=8.450-8.432\\
&=0.018
\end{align*}
$$
Finally, the area of the entire shape is the area of the two large circles plus the area of the green regions, which is:
$$
\begin{align*}
A &= A_L + A_\text{green} \\
&= 25.504 + 0.018\\
&= \boxed{25.522 \text{ mm}^2}
\end{align*}
$$
|
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|
How to find the equation of a parabola when only given the x-intercepts and the Axis of symmetry? i have been given the following problem.
A tennis ball is lobbed from ground level and must cover a horizontal
distance of 22m if it is to land just inside the opposite end of the
court. If the opponent is standing 4m from the baseline and he can hit
any ball less than 3m high, what is the lowest maximum height the lob
must reach to win the point.
The answer is 5 and 1/24 Metres (back of the textbook)
I have no idea how to approach this problem but my attempt to
approach this problem is by finding the X-intercepts which are 0,
22. Thus, finding the Axis of symmetry that is 22/2 = 11. That is
all i've done so far.
Explanation will be extremely helpful :)
|
The general equation of a parabola is
$$
f(x) = a x^2 + bx + c
$$
it has three parameters.
We know $f(11) = 0$, so we have
$$
f(11) = 0
$$
we also know
$$
3 = f(11-4) = f(7)
$$
further we lobbed from ground level, such that we are 22m away, so this means
$$
0 = f(-11) = a 11^2 - 11 b + c
$$
This gives the linear system in the unknowns $a,b,c$:
$$
\left[
\begin{array}{rrr|r}
121 & 11 & 1 & 0 \\
49 & 7 & 1 & 3 \\
121 & -11 & 1 & 0
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
242 & 0 & 2 & 0 \\
49 & 7 & 1 & 3 \\
121 & -11 & 1 & 0
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 0 & 1/121 & 0 \\
49 & 7 & 1 & 3 \\
0 & -11 & 0 & 0
\end{array}
\right]
\to
\\
\left[
\begin{array}{rrr|r}
1 & 0 & 1/121 & 0 \\
0 & 1 & 0 & 0 \\
49 & 0 & 1 & 3 \\
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 0 & 1/121 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1-49/121 & 3 \\
\end{array}
\right]
\to
\left[
\begin{array}{rrr|r}
1 & 0 & 1/121 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 72/121 & 3 \\
\end{array}
\right]
\to
\\
\left[
\begin{array}{rrr|r}
1 & 0 & 0 & -1/24 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 121/24 \\
\end{array}
\right]
$$
The highest point is achieved in the middle, having height
$$
f(0) = c = 121/24
$$
|
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|
Equation involving the Jacobi symbol: $\left( \frac {-6} p \right) = 1$? I have to determine the values of $p \in \{0, \dots, 23 \}$ such that $\left( \frac {-6} p \right) = 1$.
I have that:
$$\left( \frac {-6} p \right) = \left( \frac 2 p \right) \left( \frac {-3} p \right)$$
and I know that $\left( \frac 2 p \right) = 1$ if $p \equiv 1,7 \pmod 8$ and $\left( \frac 2 p \right) = -1$ if $p \equiv 3,5 \pmod 8$, and also that $\left( \frac {-3} p \right) = 1$ if $p\equiv 1 \pmod 3$.
Using the Chinese Remainder Theorem, I find that $p\equiv 1,?,7,?\pmod{24}$.
Thanks in advance!
|
We have $(-6/p)=1$ if (i) $(2/p)=1$ and $(3/p)=1$ or (ii) $(2/p)=-1$ and $(3/p)=-1$.
You seem to have taken care of (i). We get the solutions $p\equiv 1\pmod{24}$ and $p\equiv 7\pmod{24}$.
For (ii) we want a) $p\equiv 3\pmod{8}$ and $p\equiv 2\pmod{3}$ or b) $p\equiv 5\pmod{8}$ and $p\equiv 2\pmod{3}$.
a) has the solution $p\equiv 11\pmod{24}$ and b) has the solution $p\equiv 5\pmod{24}$.
|
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|
How many ways can all six numbers in the set $\{4, 3, 2, 12, 1, 6\}$ be ordered Is there an easy way to solve the problem?
How many ways can all six numbers in the set $S = \{4, 3, 2, 12, 1, 6\}$ be ordered so
that $a$ comes before $b$ whenever $a$ is a divisor of $b$?
By analyzing each number in $S$, I get the answer, but I don't like the way I solved the problem.
$$1, 2, 3, 4, 6, 12$$ $$1, 2, 3, 6, 4, 12$$ $$1, 2, 4, 3, 6, 12$$ $$1, 3, 2, 4, 6, 12$$ $$1, 3, 2, 6, 4, 12$$
|
I would solve this problem by looking at the divisor lattice for $12$ and then doing casework. Here is the divisor lattice:
You're asking us to put any number that is below another on a tree before in the list. Therefore, $1$ is the first number since it's at the bottom. The next number is either $2$ or $3$, since those are the only nodes connected to $1$.
*
*After $1,2$, then we have access to $4$, but we could also go with $3$.
*
*
*After $1,2,3$, then we have access to $6$, but we could also go with $4$. If we go with $4$, we get $1,2,3,4,6,12$ and if we go with $6$, we get $1,2,3,6,4,12$.
*
*
*After $1,2,4$, we have to go with $3$, then $6$, then $12$, so $1,2,4,3,6,12$.
*After $1,3$, we have to go with $2$, but then we can go with either $4$ or $6$.
*
*
*After $1,3,2,4$, we must choose $6$ then $12$, so $1,3,2,4,6,12$.
*
*
*After $1,3,2,6$, we must choose $4$ then $12$, so $1,3,2,6,4,12$.
Thus, through this casework, we get the same $5$ solutions you did, but we went through all possible cases, so we can be sure there are no more.
|
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|
Finding certain coefficients in Taylor expansion of $ \log(1 +qx^2 + rx^3)$ This exercise is part of the STEP $3$ paper from $2014$.
At a certain point in the problem, we 're supposed find $a_n$ for $n = {2,5,7,9}$
where $a_n$ is the coefficient of $x^n$ in the series expansion around $0$ of $ \log(1+qx^2 + rx^3)$.
I knew at first sight that this can't be as direct as it seems, precisely because it'd be tedious to expand all the powers of $(qx^2+rx^3)$ to find the coefficients we're looking for. When I saw I couldn't get anywhere with this,
I looked at the provided 'hints and solutions' and , to my surprise, this is all they said, 'the coefficients may be found easily by expanding $ \log(1 +qx^2 + rx^3)$' . '
While using a computational resource such as Wolfram | Alpha may find the solution quickly, how does one do it by hand?
|
I think the suggested solution is straightforward enough not to be dismissed out-of-hand.
We know
$$
\log(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} \pm\dotsm
$$
Let $u = qx^2 + rx^3$. Then
$$
\log(1+qx^2 + rx^3) = (qx^2 + rx^3) - \frac{1}{2}(qx^2 + rx^3)^2 + \frac{1}{3}(qx^2 + rx^3)^3 + \dotsm
$$
Only the first power of $(qx^2 + rx^3)$ is going to produce a term of degree $2$. So $a_2 = q$. The expansion of $(qx^2 + rx^3)^2$ is going to produce terms of degree $4$, $5$, and $6$, and the expansion of $(qx^2 + rx^3)^3$ is going to produce terms of degree $6$, $7$, $8$, and $9$. The expansion of $(qx^2+rx^3)^4$ is going to produce terms of degree $8$, $9$, and higher. The expansion of $(qx^2 + rx^3)^5$ is going to produce terms of degree $10$ and higher.
So $a_5$ comes from the cross term in the second binomial power:
$$
a_5 = -\frac{1}{2} (2qr) = -qr
$$
The degree $7$ term comes from $(qx^2)^2(rx^3)^1$. From the binomial theorem there is a factor of $3$, but there is also the $\frac{1}{3}$ in front of the power to cancel those. Therefore
$$
a_7 = q^2 r
$$
The degree $9$ term comes from $(qx^2)^0(rx^3)^3$ in the third power and $(qx^2)^3(rx^3)^1$ in the fourth power. So
$$
a_9 = \frac{1}{3}r^3 - \frac{1}{4}\cdot 4 q^3r = \frac{1}{3}r^3 - q^3 r
$$
|
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|
If $x$ and $y$ are non-negative integers for which $(xy-7)^2=x^2+y^2$. Find the sum of all possible values of $x$. I am not able to reach to the answer. I have used discriminant as $x$ and $y$ are both integers but it didn't give any hint to reach to answer. I am not able to understand how should I deal with these type of question.
|
$$(xy-7)^2=x^2+y^2 $$
$$x^2y^2-14xy+49 = x^2+y^2 $$
$$x^2y^2-12xy+49 = x^2+2xy+y^2 $$
$$(xy-6)^2+13=(x+y)^2 $$
$$13 = (x+y)^2-(xy-6)^2 $$
$$13 = (x+y+xy-6)(x+y-xy+6) $$
The factors on the right must be factors of the prime $13$.
We conclude $x+y=7$ and easily enumerate all solutions.
|
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|
Prove that $\lim\limits_{x\to \infty} a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}=0$ if and only if $ a+b+c=0$ Prove that
$$ \displaystyle \lim_{x\to\infty } \left({a\sqrt{x+1}+b\sqrt{x+2}+c\sqrt{x+3}}\right)=0$$
$$\text{if and only if}$$
$$ a+b+c=0.$$. I tried to prove that if $a+b+c=0$, the limit is $0$ first, but after getting here i got stuck
$$\lim_{x\to\infty } \left({\sqrt{x+1}\left(a+b\sqrt{1+\frac{1}{x+1}}+c\sqrt{1+\frac{2}{x+1}}\right)}\right)$$
Got here by substituting $\sqrt{x+2}$ with $\sqrt{(x+1)(1+\dfrac{1}{x+1})}$
Edit: x tends to infinity, not to 0. I transcribed wrongly.
|
You can assume $x>0$ and substitute $x=1/t^2$, with $t>0$, so the limit becomes
$$
\lim_{t\to0^+}\frac{a\sqrt{1+t^2}+b\sqrt{1+2t^2}+c\sqrt{1+3t^2}}{t}
$$
The limit is zero if $a+b+c=0$ and infinite otherwise.
|
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|
Simplify $\frac{\cos(2x)}{\cot(x)-1}-\frac{\sin(2x)}{2}$ I am given this expression to simplify:
$\frac{\cos(2x)}{\cot(x)-1}-\frac{\sin(2x)}{2}$
and I know the correct answer is $\sin^2(x)$
I was able to reduce the second fraction to a bit nicer
$\frac{\sin(2x)}{2}=\frac{2\sin(x)\cos(x)}{2\tan(x)\cot(x)}=\frac{2\sin(x)\cos(x)}{\frac{2\cos(x)\sin(x)}{\cos(x)\sin(x)}}=\frac{2\sin(x)\cos(x)}{1}\cdot\frac{\cos(x)\sin(x)}{2\cos(x)\sin(x)}=\cos(x)\sin(x)$
which changes the original into this
$\frac{\cos(2x)}{\cot(x)-1}-\cos(x)\sin(x)$
which is the prettiest I've got. No matter what I do with the first fraction I get something too complicated. Does anyone have any idea? Thx
|
$$\frac { \cos 2x }{ \cot x-1 } -\frac { \sin 2x }{ 2 } =\frac { \cos ^{ 2 }{ x } -\sin ^{ 2 }{ x } }{ \frac { \cos { x } -\sin { x } }{ \sin { x } } } -\sin { x } \cos { x } =$$
$$=\frac { \sin { x } \left( \cos { x } +\sin { x } \right) \left( \cos { x } -\sin { x } \right) }{ \cos { x } -\sin { x } } -\sin { x\cos { x } = } $$
$$=\sin { x } \cos { x } +\sin ^{ 2 }{ x } -\cos { x } \sin { x } =\sin ^{ 2 }{ x } $$
|
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|
Convex integral inequality I cannot prove that if $f(x)$ is convex on $[a,b]$ then
$f\Big(\frac{a+b}2\Big) \le \frac1{b-a}\int_a^b f(x)\,dx \le \frac{f(a)+f(b)}2 .$
|
For convenience, we can take $b=1,a=-1$.
For the first inequality, use
$$f(0) \le \dfrac{f(x) + f(-x)}{2}$$
and integrate both sides from $x=-1$ to $1$.
For the second, use
$$ f(x) = f\left( \dfrac{1-x}{2} (-1) + \dfrac{1+x}{2}(1)\right) \le
\dfrac{1-x}{2} f(-1) + \dfrac{1+x}{2} f(1) $$
and integrate both sides from $x=-1$ to $1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let
$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$
Prove that $I=J={\pi \over 2\sqrt3}$
Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\rightarrow u=0$
Rewrite $(1)$ as
$$I=\int_{0}^{\infty}{1\over (1+x^2)^2-x^2}dx$$
then
$$\int_{0}^{\pi/2}{\sec^2{u}\over \sec^4{u}-\tan^2{u}}du\tag3$$
Simplified to
$$I=\int_{0}^{\pi/2}{1\over \sec^2{u}-\sin^2{u}}du\tag4$$
Then to
$$I=2\int_{0}^{\pi/2}{1+\cos{2u}\over (2+\sin{2u})(2-\sin{2u})}du\tag5$$
Any hints on what to do next?
Re-edit (Hint from Marco)
$${1\over x^8+x^4+1}={1\over 2}\left({x^2+1\over x^4+x^2+1}-{x^2-1\over x^4-x^2+1}\right)$$
$$M=\int_{0}^{\infty}{x^2+1\over x^4+x^2+1}dx=\int_{0}^{\infty}{x^2\over x^4+x^2+1}dx+\int_{0}^{\infty}{1\over x^4+x^2+1}dx={\pi\over \sqrt3}$$
$$N=\int_{0}^{\infty}{x^2-1\over x^4-x^2+1}dx=0$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx={1\over 2}\left({\pi\over \sqrt3}-0\right)={\pi\over 2\sqrt3}.$$
|
Note that $$\frac{1}{x^{4}+x^{2}+1}=\frac{1}{2}\left(\frac{x+1}{x^{2}+x+1}-\frac{x-1}{x^{2}-x+1}\right)
$$ and $$\begin{align}
\int\frac{x+1}{x^{2}+x+1}dx= & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\int\frac{1}{x^{2}+x+1}dx \\
= & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{1}{2}\int\frac{1}{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}dx \\
= & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{2}{3}\int\frac{1}{\left(\frac{2x+1}{\sqrt{3}}\right)^{2}+1}dx \\
= & \frac{\log\left(x^{2}+x+1\right)}{2}+\frac{\arctan\left(\frac{2x+1}{\sqrt{3}}\right)}{\sqrt{3}}
\end{align}
$$ and in a similar way we can compute the other integral, hence $$\begin{align}
\int_{0}^{\infty}\frac{1}{x^{4}+x^{2}+1}dx= & \frac{1}{2}\left(\log\left(\frac{\sqrt{x^{2}+x+1}}{\sqrt{x^{2}-x+1}}\right)+\frac{\arctan\left(\frac{2x+1}{\sqrt{3}}\right)+\arctan\left(\frac{2x-1}{\sqrt{3}}\right)}{\sqrt{3}}\right)_{0}^{\infty} \\ = & \frac{\pi}{2\sqrt{3}}.
\end{align}$$
For the second note that $$\frac{1}{x^{8}+x^{4}+1}=\frac{1}{4}\left(\frac{1}{x^{2}-x+1}+\frac{1}{x^{2}+x+1}\right)+\frac{1}{2}\left(\frac{x^{2}-1}{x^{4}-x^{2}+1}\right)
$$ the first two integral are part of the previous calculation, so we have only to analyze $$I=\int_{0}^{\infty}\frac{x^{2}-1}{x^{4}-x^{2}+1}dx
$$ and $$I=\int_{0}^{1}\frac{x^{2}-1}{x^{4}-x^{2}+1}dx+\int_{1}^{\infty}\frac{x^{2}-1}{x^{4}-x^{2}+1}dx \tag{1}
$$ and note that if we put $x=1/y$ in the second integral of $(1)$ we get $$\int_{1}^{\infty}\frac{x^{2}-1}{x^{4}-x^{2}+1}dx=-\int_{0}^{1}\frac{y^{2}-1}{y^{4}-y^{2}+1}dy
$$ then $$I=0.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The solution of equation $4+6+8+10+\cdots +x =270$ is 15. The solution of equation
$4+6+8+10+\cdots +x =270$
is $x=15$.
How can I prove it?
I ve tried the geometric sequence but I cannot figure out the pattern.
|
Since $$1+2+3+4+.....+(n-1)+n = \frac{n(n+1)}{2}$$
$$2+4+6+8+.....+2(n-1)+2n = n(n+1)$$
That is $$4+6+8+.....+2(n-1)+2n = n(n+1)-2$$
Thus $$n(n+1)-2 \leq 270$$
$$n^2+n-272 \leq 0$$
$$(n+17)(n-16) \leq 0$$
So $$n=16$$
Thus $$4+6+8+10+12+14+16+18+20+22+24+26+28+30+32=270$$
I think the question should be $$(4+6+8+...)\cdot x =270$$
Then we can argue that factors of $270$ are $1,2,3,5,6,9,10,15,18,27,30,45,54,90,135,270$
Then the only chances are
$$4+6 =10 $$ which gives $x=27$
$$4+6+8 = 18$$ which gives $x=15$
$$4+6+8+10+12+14 =54 $$ which gives $x=5$
$$4+6+8+10+12+14+16+18+20+22+24+26+28+30+32 = 270 $$ which gives $x=1$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
With the linear approx. of $f(x)= sin(x)$ around $0$ Calculate $\lim_{\theta\to 0} \frac{sin\theta}{\theta}$ With the linear approximation of $f(x)= sin(x)$ around $0$, calculate:
$$ \lim_{\theta\to 0} \frac{\sin\theta}{\theta}$$
Figured I have to use L'Hospital's Rule, but I think I don't get how to calculate the derivative of theta.
$$\lim_{\theta \to 0} \frac{\sin \theta}{\theta}=\lim_{\theta \to 0} \frac{\frac{d}{d\theta}\sin\theta}{\frac{d}{d\theta}\theta}=\lim_{\theta \to 0} \frac{\cos \theta}{1}=\frac{\cos 0}{1}=1$$
|
Two ways to go about doing this:
Taylor Series
We can use the Taylor expansion of $\sin(x)$
$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\ldots$$
$$\frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^3}{5!}-\frac{x^6}{7!}+\ldots$$
So $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1$
Squeeze Theorem
The area of the triangle in the image above is $\displaystyle \frac{\sin\theta}{2\cos\theta}$.
The area of the small circular sector is $\displaystyle \frac{\theta}{2}$.
The area of the big circular sector is $\displaystyle \frac{\theta}{2\cos^2\theta}$.
So, for $0<\theta<\displaystyle \frac{\pi}{2}$, we have
$$\displaystyle \frac{\theta}{2}<\displaystyle \frac{\sin\theta}{2\cos\theta}<\displaystyle \frac{\theta}{2\cos^2\theta}$$
Since $\displaystyle \frac{2\cos\theta}{\theta}$ is positive for $0<\theta<\displaystyle \frac{\pi}{2}$, we can multiply by $\displaystyle \frac{2\cos\theta}{\theta}$:
$$\cos\theta<\frac{\sin\theta}{\theta}<\frac{1}{\cos\theta}$$
Since $\displaystyle \lim_{\theta \to 0} \cos\theta=\lim_{\theta \to 0} \frac{1}{\cos\theta}=1$, we can use Squeeze Theorem to conclude $\displaystyle \lim_{\theta \to 0}\frac{\sin\theta}{\theta}=1$.
|
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|
If the sides of a triangle satisfy $(a-c)(a+c)^2+bc(a+c)=ab^2$, and if one angle is $48^\circ$, then find the other angles.
In triangle $ABC$ one angle of which is $48^{\circ}$, length of the sides satisfy the equality:
$$(a-c)(a+c)^2+bc(a+c)=ab^2$$
Find the value in degrees the other two angles of the triangle.
I have no idea how to solve this problem
|
$$(a-c)(a+c)^2+bc(a+c)=ab^2$$
$$\implies a^3-a b^2-a c^2+b c^2+a^2 c+a b c-c^3=0$$
$$\implies a(a^2-b^2)-c^2(a-b)+ca(a+b)-c^3=0$$
$$\implies a(a+b)(a-b)-c^2(a-b)+ca(a+b)-c^3=0$$
$$\implies (a-b)(a(a+b)-c^2)+c(a(a+b)-c^2)=0$$
$$\implies (a-b+c)(a(a+b)-c^2)=0$$
Dividing the both sides by $a-b+c\gt 0$ gives
$$c^2=a^2+ab$$
from which we have $c\gt a$.
Case 1 : $c^2=a^2+b^2$
$a=b\implies c=\sqrt 2\ a$, which contradicts that the triangle has $48^\circ$.
Case 2 : $c^2\gt a^2+b^2$.
$a\gt b\implies C\gt 90^\circ, A=48^\circ, B\lt 42^\circ$.
By the law of cosines,
$$a^2=b^2+a^2+ab-2b\sqrt{a^2+ab}\ \cos(48^\circ)\implies \frac ba=4\cos^2(48^\circ)-1$$
Also,
$$c^2=a^2+ab=a^2+b^2-2ab\cos C\implies \cos C=\frac{b}{2a}-\frac 12=2\cos^2(48^\circ)-1=\cos(96^\circ)$$
Therefore, $A=48^\circ,B=36^\circ,C=96^\circ$.
Case 3 : $c^2\lt a^2+b^2$.
$a\lt b$ and $C\lt 90^\circ$.
If $B=48^\circ$, then
$$\begin{align}&b^2=a^2+a^2+ab-2a\sqrt{a^2+ab}\ \cos(48^\circ)\\&\implies 2\cos(48^\circ)=\sqrt{2\cos C+2}\ (1-2\cos C)\\&\implies 2\cos(48^\circ)=2\cos\frac{C}{2}\left(1-2\left(2\cos^2\frac{C}{2}-1\right)\right)\\&\implies\cos(48^\circ)=3\cos\frac C2-4\cos^3\frac C2\\&\implies\cos(48^\circ)=-\cos\left(\frac{3C}{2}\right)\\&\implies C=32^\circ,88^\circ\end{align}$$
If $C=48^\circ$, then
$$c^2=a^2+ab=a^2+b^2-2ab\cos(48^\circ)\implies \cos(48^\circ)=\frac 12(4\cos^2A-1)-\frac 12=\cos(2A)$$$$\implies A=24^\circ$$
Therefore, the answer is
$$\color{red}{(A,B,C)=(48^\circ, 36^\circ, 96^\circ),(44^\circ,48^\circ, 88^\circ),(24^\circ,108^\circ,48^\circ)}$$
|
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|
What is the fastest method to find which of $\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ and $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $ is bigger manually? What is the fastest method to find which number is bigger manually?
$\frac {3\sqrt {3}-4}{7-2\sqrt {3}} $ or $\frac {3\sqrt {3}-8}{1-2\sqrt {3}} $
|
Using the approximation $\sqrt3\approx1.7$, one can do a bit of mental arithmetic and see that
$${3\sqrt3-4\over7-2\sqrt3}\approx{1.1\over3.6}\lt1\qquad\text{while}\qquad{3\sqrt3-8\over1-2\sqrt3}\approx{-2.9\over-2.4}={2.9\over2.4}\gt1$$
This only works, of course, because the two numbers are not at all close. It's also not a rigorous proof, although it could be turned into one with a little more care. For example, since $\sqrt3\lt2$, we have
$${3\sqrt3-4\over7-2\sqrt3}\lt{6-4\over7-4}={2\over3}\qquad\text{while}\qquad{3\sqrt3-8\over1-2\sqrt3}={8-3\sqrt3\over2\sqrt3-1}\gt{8-6\over4-1}={2\over3}$$
(Note the element of luck here, in that the two crude bounds happen to agree.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve definite integral: $\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$ I need to solve:
$$\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$$
Here is my steps, first of all consider just the indefinite integral:
$$\int \arctan(\sqrt{x+2})dx = \int \arctan(\sqrt{x+2}) \cdot 1\ dx$$
$$f(x) = \arctan(\sqrt{x+2})$$
$$f'(x) = \frac{1}{1+x+2} \cdot \frac{1}{2\sqrt{x+2}} = \frac{1}{(2x+6)\sqrt{x+2}}$$
$$g'(x) = 1$$
$$g(x) = x$$
So:
$$\bigg[\arctan(\sqrt{x+2}) \cdot x\bigg]_{-1}^{1} - \int_{-1}^{1} \frac{x}{(2x-6)\sqrt{x+2}}\ dx$$
How should I proceed with the new integral?
|
Integration by parts yields
\begin{equation}
\int_{-1}^{1}\arctan\sqrt{x+2}\ dx=x\arctan\sqrt{x+2}\ \bigg|_{-1}^{1}-\frac12\int_{-1}^{1}\frac{x}{(x-1)\sqrt{x+2}}\ dx
\end{equation}
Making substitution $u^2=x+2$, then
\begin{align}
\int_{-1}^{1}\arctan\sqrt{x+2}\ dx&=\frac{\pi}{3}+\frac{\pi}{4}-\int_{1}^{\sqrt3}\frac{u^2-2}{u^2-3}\ du\\[10pt]
&=\frac{7\pi}{12}-\int_{1}^{\sqrt3}\frac{u^2-3+1}{u^2-3}\ du\\[10pt]
&=\frac{7\pi}{12}-\int_{1}^{\sqrt3}\ du-\int_{1}^{\sqrt3}\frac{1}{u^2-3}\ du\\[10pt]
\end{align}
Can you take it from here?
|
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|
A System of Infinite Linear Equations Suppose that $\{a_{i}\}_{i=-\infty}^{\infty}$ with $\sum_{i=-\infty}^\infty a_{i} \lt \infty$ is known and that $\{b_i\}_{i=-\infty}^{\infty}$ is such that
$$\sum_{i=-\infty}^\infty a_{i}b_{-i} =1,$$
and that, for all $k \in \mathbb Z/\{0\}$,
$$\sum_{i=-\infty}^\infty a_{i}b_{-i+k} =0.$$
Is it possible to solve for $\{b_{i}\}_{i=-\infty}^{\infty}$ as a function of $\{a_{i}\}_{i=-\infty}^{\infty}$?
I just can't figure out where to start from.
|
Note: This did not help, but was fun:
Let us try this for a finite sum:
$$
1 =
\begin{pmatrix}
a_{-1} & a_0 & a_1
\end{pmatrix}
\begin{pmatrix}
b_1 \\
b_0 \\
b_{-1}
\end{pmatrix}
=
\begin{pmatrix}
a_{-1} & a_0 & a_1
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{pmatrix}
\begin{pmatrix}
b_{-1} \\
b_0 \\
b_1
\end{pmatrix}
$$
For $k=1$:
$$
0
=
\begin{pmatrix}
a_{-1} & a_0 & a_1
\end{pmatrix}
\begin{pmatrix}
0 \\
b_1 \\
b_0
\end{pmatrix}
=
\begin{pmatrix}
a_{-1} & a_0 & a_1
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}
\begin{pmatrix}
b_{-1} \\
b_0 \\
b_1
\end{pmatrix}
$$
This leads to the equations
$$
a^\top D^{(k)} b = \delta_{k0}
$$
for a matrix $D^{(k)}$ with components
$$
d_{ij}^{(k)} = \delta_{(k-i)j}
$$
where we also use negative indices for the matrix elements, thus having the center element at $d_{00}^{(k)}$.
This leads to a matrix equation
$$
A b = y
$$
with
$$
A =
\begin{pmatrix}
& \vdots & \vdots & \vdots \\
\dotsb & a_0 & a_{-1} & a_{-2} & \dotsb \\
\dotsb & a_1 & a_0 & a_{-1} & \dotsb \\
\dotsb & a_2 & a_1 & a_{0} & \dotsb \\
& \vdots & \vdots & \vdots \\
\end{pmatrix}
\quad
y=
\begin{pmatrix}
\vdots \\
0 \\
1 \\
0 \\
\vdots
\end{pmatrix}
$$
and the formal solution
$$
b = A^{-1} y
$$
which is the central column of $A^{-1}$.
|
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|
Solutions of $\sin^2\theta = \frac{x^2+y^2}{2xy} $ If $x$ and $y$ are real, then the equation
$$\sin^2\theta = \frac{x^2+y^2}{2xy}$$
has a solution:
*
*for all $x$ and $y$
*for no $x$ and $y$
*only when $x \neq y \neq 0$
*only when $x = y \neq 0$
|
$$\begin{align}
& {{\sin }^{2}}\theta =\frac{1}{2}\left( \frac{x}{y}+\frac{y}{x} \right)\ge 1\,\,\,\,\quad,\,\,\,\,\,\,\,\,\,\frac{x}{y}>0\,\,\,\,\,\,\Rightarrow \,\,\,\,\theta =k\pi +\frac{\pi }{2}\, \\
& {{\sin }^{2}}\theta =\frac{1}{2}\left( \frac{x}{y}+\frac{y}{x} \right)\le -1\,\,\,\,\,,\,\,\,\frac{x}{y}<0\,\,\,\,\,\, \\
\end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $2^x+4^x=2$ This is the equation, but the result is different from wolframalpha:
$$2^x+4^x=2$$
$$2^x+2^{2x}=2^1$$
$$x+2x=1$$
$$x=\frac{1}{3}$$
WolframAlpha: $x=0$
Where is the error?
|
$2^x + 4^x= 2$ $\Rightarrow$ $2^x (1 + 2^x ) = 2$ $\Rightarrow $ $1 + 2^x = 2 ^{1 - x}$ $\Rightarrow$ $1 + 2^x = 2 ^{- x} \times 2$
Now Set $ y = 2^x$; then we have
$1 + y = y^{-1} \times 2 $ $\Rightarrow$ $y^2 + y -2 = 0$. Which has solutions $y = 1$ and $y = -2$.
$y = -2 $ is unacceptable, because $y = 2^x$ is a positive function. So $y = 2^x = 0$ is acceptable and will give us $ x = 0$. And we are done!
|
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"url": "https://math.stackexchange.com/questions/1838980",
"timestamp": "2023-03-29T00:00:00",
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|
Find $\lim\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}$ The question arise in connection with this problem
Prove that
$$\lim_{n\rightarrow \infty}\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}=1$$
Thanks to answer @Vincenzo Oliva. I forgot the Stolz-Cesàro theorem
|
We have
\begin{align}
&\sqrt{\frac{1}{n}} - \sqrt{\frac{2}{n}} + \sqrt{\frac{3}{n}} - \cdots + \sqrt{\frac{4n-3}{n}} - \sqrt{\frac{4n-2}{n}} + \sqrt{\frac{4n-1}{n}} \\
=& \sqrt{\frac{1}{n}} +\sum_{k=1}^{2n-1}(\sqrt{\frac{2k+1}{n}}-\sqrt{\frac{2k}{n}}) \\
=& \sqrt{\frac{1}{n}} + \frac{1}{\sqrt{n}}\sum_{k=1}^{2n-1}\frac{1}{\sqrt{2k+1}+\sqrt{2k}} \tag{1}
\end{align}
Moreover,
\begin{align}
\sum_{k=1}^{2n-1}\frac{1}{\sqrt{2k+1}+\sqrt{2k}} \leq \sum_{k=1}^{2n-1}\frac{1}{2\sqrt{2k}} = \frac{1}{2\sqrt{2}}\sum_{k=1}^{2n-1}\frac{1}{\sqrt{k}} < \frac{1}{2\sqrt{2}}\int_0^{2n}x^{-1/2}dx = \sqrt{n} \tag{2}
\end{align}
and
\begin{align}
\sum_{k=1}^{2n-1}\frac{1}{\sqrt{2k+1}+\sqrt{2k}} \geq \sum_{k=1}^{2n-1}\frac{1}{2\sqrt{2k+2}} = \frac{1}{2\sqrt{2}}\sum_{k=2}^{2n}\frac{1}{\sqrt{k}}>\frac{1}{2\sqrt{2}}\int_2^{2n}x^{-1/2}dx = \sqrt{n}-1 \tag{3}
\end{align}
Provided (2) and (3), we conclude that
$$
\lim_{n\rightarrow \infty} \sqrt{\frac{1}{n}} + \frac{1}{\sqrt{n}}\sum_{k=1}^{2n-1}\frac{1}{\sqrt{2k+1}+\sqrt{2k}} = 1
$$
|
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|
Evaluating $\int_0^1 \frac{\arctan x \log x}{1+x}dx$ In order to compute, in an elementary way,
$\displaystyle \int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$
(see Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$)
i need to show, in a simple way, that:
$\displaystyle \int_0^1 \dfrac{\arctan x \log x}{1+x}dx=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}$
$G$ is the Catalan's constant.
|
I finally get a solution (i swear i didn't know it when i have posted the question)
Define for $x\in [0,1]$ the function $F$:
$\displaystyle F(x)=\int_0^x \dfrac{\ln t}{1+t}dt$
Notice that $F(1)=-\dfrac{\pi^2}{12}$
(use Taylor's development)
and, after performing the change of variable $y=\dfrac{t}{x}$,
$\displaystyle F(x)=\int_0^1 \dfrac{x\ln(xy)}{1+xy}dy$
Since that:
$\Big[F(x)\arctan x\Big]_0^1=-\dfrac{\pi^3}{48}$
then,
$\displaystyle -\dfrac{\pi^3}{48}=\int_0^1 \dfrac{F(x)}{1+x^2}dx+\int_0^1 \dfrac{\arctan x\ln x}{1+x}dx$
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\int_0^1 \dfrac{x\ln(xy)}{(1+xy)(1+x^2)}dxdy$
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\int_0^1 \dfrac{x\ln(x)}{(1+xy)(1+x^2)}dxdy+\int_0^1\int_0^1 \dfrac{x\ln(y)}{(1+xy)(1+x^2)}dxdy$
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=\int_0^1\left[\dfrac{\ln x\ln(1+xy)}{1+x^2}\right]_{y=0}^{y=1} dx+
\displaystyle \int_0^1 \left[-\dfrac{\ln y\ln(1+xy)}{1+y^2}+\dfrac{\ln y\ln(1+x^2)}{2(1+y^2)}+\dfrac{y\ln y\arctan x}{1+y^2}\right]_{x=0}^{x=1}dy$
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx= \int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx-\int_0^1\dfrac{\ln y\ln(1+y)}{1+y^2}dy+\dfrac{\ln 2}{2}\int_0^1 \dfrac{\ln y}{1+y^2}dy+
\dfrac{\pi}{4}\times \int_0^1 \dfrac{y\ln y}{1+y^2}dy$
Using Taylor's development,
$\displaystyle \int_0^1 \dfrac{y\ln y}{1+y^2}dy=-\dfrac{\pi^2}{48}$
And it's well known that, $\displaystyle -G=\int_0^1\dfrac{\ln y}{1+y^2}dy$
Therefore,
$\displaystyle\int_0^1 \dfrac{F(x)}{1+x^2}dx=-\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{192}$
And finally,
$\displaystyle \int_0^1 \dfrac{\arctan x \ln x}{1+x}dx=\dfrac{G\ln 2}{2}-\dfrac{\pi^3}{64}$
(I hope there is no mistake, this proof is too wonderful to be true )
NB:
Added, July 2, 2019.
The above computation is the result of "reverse engineering".
I was searching for a way to express $\pi^3$ as in integral.
If you introduce the function, for $x\in [0;1]$,
\begin{align}\displaystyle F(x)&=\int_0^x \dfrac{\ln t}{1+t}dt\\
&=\int_0^1 \dfrac{x\ln(tx)}{1+tx}dt
\end{align}
Observe that,
\begin{align}\frac{\partial F(x)}{\partial x}&=\dfrac{\ln x}{1+x}\\
F(1)&=-\frac{\pi^2}{12}
\end{align}
Then,
\begin{align}-\frac{\pi^3}{48}&=\Big[F(x)\arctan x\Big]_0^1\\
\end{align}
And,
\begin{align}\frac{\partial F(x)}{\partial x}\arctan x=\frac{\arctan x\ln x}{1+x}\end{align}
Thus, one can apply integration by parts,
\begin{align}\int_0^1 \frac{\arctan x\ln x}{1+x}\,dx&=\int_0^1 \frac{\partial F(x)}{\partial x}\arctan x\,dx\end{align}
and so on,
|
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"url": "https://math.stackexchange.com/questions/1842284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
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|
Inequality with square root $x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}}$ Good morning to everyone! The inequality is the following:$$ x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} $$. I don't know how to solve it. Here's what I tried: $$x+\sqrt{x^2-10x+9}\ge \sqrt{x+\sqrt{x^2-10x+9}} \rightarrow 2x^2-11x+9+\sqrt{x^2-10x+9}\left(2x+1\right)\ge \:\sqrt{x^2-10x+9}$$ Therefore $ 2x+1 \ge 0 => x\ge \frac{1}{2} $ so $x$ belongs to $(\frac{1}{2},\infty) $ and $ x^2-10x+9 \ge 0 => x$ belongs to $ (-\infty,1] $ and $[9,\infty) $. Therefore the statements are contradicting each other.
|
The suggestion about $u\ge\sqrt{u}$ helps in removing one level of square roots. Now you have
$$
x+\sqrt{x^2-10x+9}\ge1
$$
or
$$
\sqrt{x^2-10x+9}\ge1-x
$$
If you notice that the expression under the square root is $(x-1)(x-9)$, you can possibly use the clever trick shown in another answer. On the other hand, such an inequality can be analyzed in a systematic way, by transforming it into two systems of inequalities:
$$
\mathrm{(I)}
\begin{cases}
x^2-10x+9\ge0 \\[4px]
1-x<0
\end{cases}
\qquad\text{or}\qquad
\mathrm{(II)}
\begin{cases}
\bigl[\,x^2-10x+9\ge0\,\bigr] \\[4px]
x^2-10x+9\ge(1-x)^2 \\[4px]
1-x\ge0
\end{cases}
$$
The first system comes from the consideration that the square root is non-negative as soon as it exists, so it will be greater than a negative number; the second system comes from the consideration that an inequality between non-negative numbers is equivalent to the same inequality between their squares.
System (I) can be transformed into
$$
\mathrm{(I)}
\begin{cases}
x\le1 \quad\text{or}\quad x\ge9 \\[4px]
x>1
\end{cases}
$$
which has $x\ge9$ as solution set.
System (II), where the top inequality is implied by the middle one, becomes
$$
\mathrm{(II)}
\begin{cases}
x^2-10x+9\ge1-2x+x^2 \\[4px]
x\le1
\end{cases}
$$
and it is easily seen that the solution set is $x\le1$.
Summing up, the solution set of the original inequality is
$$
\boxed{\quad\vphantom{\Big|}x\le 1\quad\text{or}\quad x\ge9\quad}
$$
|
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"url": "https://math.stackexchange.com/questions/1842439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How prove this inequality: $\frac1{1-a}+\frac1{1-b}+\frac1{1-c}\ge \frac1{ab+bc+ac}+\frac1{2(a^2+b^2+c^2)}$ for $a+b+c=1$? Let $a,b,c>0$ and such $a+b+c=1$ show that
$$\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}\ge \dfrac{1}{ab+bc+ac}+\dfrac{1}{2(a^2+b^2+c^2)}$$
Let $p=a+b+c=1,ab+bc+ac=q,abc=r$
$$\Longleftrightarrow -4q^3+q^2-3qr+2r\ge 0$$
it seem hard to prove.
why I say it hard prove:
use Schur inequality
$$p^3-4pq+9r\ge 0\Longrightarrow r\ge\dfrac{4q-1}{9}$$
it remains to prove that
$$\dfrac{4q-1}{9}(2-3q)+q^2-4q^3\ge 0$$
In fact, this inequality can't hold (4q-1)
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, our inequality it's $\frac{9u^2+3v^2}{9uv^2-w^3}\geq3u\left(\frac{1}{3v^2}+\frac{1}{18u^2-12v^2}\right)$,
which is equivalent to $f(w^3)\geq0$, where $f$ is a linear function.
But the linear function gets a minimal value for an extremal value of $w^3$.
$a$, $b$ and $c$ are positive roots of the equation $(x-a)(x-b)(x-c)=0$ or
$w^3=x^3-3ux^2+3v^2x$ and we see that the line $y=w^3$ and graph of $y=x^3-3ux^2+3v^2x$ have three common points (draw it!).
Thus, an extremal value of $w^3$ we get for equality case of two variables
and we need to check also the case $w^3\rightarrow0^+$.
*
*$b=c$. After homogenization we can assume $b=c=1$, which gives $a(a-1)^2\geq0$;
*$w^3\rightarrow0^+$.
Let $c\rightarrow0^+$. After homogenization we can assume $b=1$,
which gives $(a-1)^2\geq0$. Done!
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
What's the mistake on my answer for this inequality $ \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}} $ Good evening to everyone! I have the following inequality: $$ \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}} $$. I don't know what's wrong with my answer: $$ \frac{\left(x+1\right)}{\sqrt{x^2+1}}>\frac{\left(x+2\right)}{\sqrt{x^2+4}} \rightarrow \left(x+1\right)\sqrt{x^2+4}>\left(x+2\right)\sqrt{x^2+1} \rightarrow \left(x+1\right)^2\left(x^2+4\right)>\left(x+2\right)^2\left(x^2+1\right) \rightarrow \left(x^2+2x+1\right)\left(x^2+4\right)>\left(x^2+4x+4\right)\left(x^2+1\right) \rightarrow 2x^3-4x<0 \rightarrow x\left(2x^2-4\right)<0 \rightarrow $$ $x$ belongs to $(-\infty,0)$ and $2x^2-4<0 \rightarrow x_1 = 1-\sqrt{2}$ and $x_2=1+\sqrt{2}$ therefore the final result is $x$ belongs to $(1-\sqrt{2},0) $. But my answer sheet shows that $x$ belongs to $(0,\sqrt{2})$. Where am I wrong? Thanks for any responses!
|
The first part is correct. The inequality holds iff $$(x+1)\sqrt{x^2+4}>(x+2)\sqrt{x^2+1}\ \ (*)$$ Hence (squaring), provided $x+1>0$ (and hence $x+2>0$) we have $(x+1)^2(x^2+4)>(x+2)^2(x^2+1)$ which is equivalent to (expanding etc) $$2x(2-x^2)>0$$ which implies $0<x<\sqrt2$.
On the other hand if $-2<x<-1$ we have $x+1<0,x+2>0$ and so $(*)$ is false.
Whilst if $x<-2$, then $x+1<0,x+2<0$ and so $(*)$ is equivalent to $$|x+1|\sqrt{x^2+4}<|x+2|\sqrt{x^2+1}$$ Squaring as before this is equivalent to $$2x(2-x^2)<0$$ which this time is always false (for $x<-2$).
So, summarising, the given inequality is true iff $$0<x<\sqrt2$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1842720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How can I solve this nice rational equation I am trying solve this equation
$$\dfrac{3x^2 + 4x + 5}{\sqrt{5x^2 + 4x +3}}+\dfrac{8x^2 + 9x + 10}{\sqrt{10x^2 + 9x +8}} = 5.$$ Where $x \in \mathbb{R}$. I knew that $x=-1$ is a given solution. But I can't solve it. I tried
We rewrite the given equation in the form
$$\dfrac{3(x^2 + x + 1) + x + 2}{\sqrt{5(x^2 + x + 1) - (x+2)}}+\dfrac{8(x^2 + x + 1) + x + 2}{\sqrt{10(x^2 + x + 1)- (x+2)} } = 5.$$
Put $a = x^2 + x + 1$ and $b = x + 2$, we get
$$\dfrac{3a + b}{\sqrt{5a - b}}+\dfrac{8a + b}{\sqrt{10a- b} } = 5.$$
From here, I stoped.
|
The equation of tangent line of the grap of the funtion $ y = \dfrac{3 x^2+4 x+5}{\sqrt{5 x^2+4 x+3}}$ at the point $x=-1$ is $y=\dfrac{x}{2}+\dfrac{5}{2}$ and the equation of tangent line of the grap of the funtion $ y = \dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}}$ at the point $x=-1$ is $y=\dfrac{5}{2}-\dfrac{x}{2}$. We prove that
\begin{equation}
\label{eq4_30_06_2016}
\dfrac{3 x^2+4 x+5}{\sqrt{5 x^2+4 x+3}} \geqslant \dfrac{x}{2}+\dfrac{5}{2}
\end{equation}
and
\begin{equation}
\label{eq5_30_06_2016}
\dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}} \geqslant \dfrac{5}{2}-\dfrac{x}{2}.
\end{equation}
The first inequality equavalent to
\begin{equation}
\label{eq1_30_06_2016}
2(3 x^2+4 x+5) \geqslant (x+5)\sqrt{5 x^2+4 x+3}.
\end{equation}
If $x+5 \leqslant 0$, it is always true.
If $x+5 > 0$, squaring both sides , we get
$$31 x^4+42 x^3+16 x^2+30 x+25 \geqslant 0.$$
Equavalent to
\begin{equation}
\label{eq_2_30_06_2016}
(x+1)^2\cdot \left(31 x^2-20 x+25\right) \geqslant 0.
\end{equation}
Similarly, if $x \geqslant 5$ the second equality is true. If $x < 5$, squaring both sides, we get
$$(x+1)^2 \cdot\left(246 x^2+175 x+200\right) \geqslant 0.$$
Equality of two equalities hold iff $x = -1$.
Add two equalities, we have
$$\dfrac{3 x^2+4 x+5}{\sqrt{5 x^2+4 x+3}}+\dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}}\geqslant 5.$$
Therefore, the equation
$$\dfrac{3 x^2+4 x+5}{\sqrt{5 x^2+4 x+3}}+\dfrac{8 x^2+9x+10}{\sqrt{10 x^2+9x+8}}= 5$$
has only solution $x=-1.$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Questions regarding dot product (possible textbook mistake) I am given the following exercise:
Show that $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert
\overrightarrow{a} \Vert + \Vert \overrightarrow{b} \Vert $ if and
only if $\overrightarrow{a}$ and $\overrightarrow{b}$ are parallel and
point to the same direction. Also, show that $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert \overrightarrow{a} - \overrightarrow{b} \Vert$ if and only if $\overrightarrow{a} \cdot \overrightarrow{b} = 0$
Regarding the first question:
Firstly, i feel like the first question is wrong since $\Vert \overrightarrow{a} + \overrightarrow{b} \Vert = \Vert
\overrightarrow{a} \Vert + \Vert \overrightarrow{b} \Vert $ for $\theta = 0$ or $\pi$ (so they must not necessarily point on the same direction). Is that correct?
if so, I proceeded the following way:
$$
\Vert \overrightarrow{a} + \overrightarrow{b} \Vert^2 = \Vert
\overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 \pm 2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta)\\
\Vert
\overrightarrow{a} \Vert^2 + 2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert +\Vert \overrightarrow{b} \Vert^2 = \Vert
\overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 \pm 2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta)\\
\cos(\theta) = \pm 1\\
\theta = 0 \text{ or } \theta = \pi
$$
Regarding the second question:
$$
\Vert
\overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 +2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta) = \Vert
\overrightarrow{a} \Vert^2 + \Vert \overrightarrow{b} \Vert^2 -2 \Vert \overrightarrow{a} \Vert \Vert \overrightarrow{b} \Vert \cos (\theta)\\
- \cos(\theta) = \cos(\theta)\\
\cos(\theta) = 0\\
\theta = \pi/2
$$
Can someone help me with these?
|
Both of your questions can be answered using the fact that
$$
\lVert a\rVert^2 = a\cdot a
$$
So,
$$
\begin{align}
\lVert a + b\rVert^2 &= (a + b)\cdot (a + b) = \lVert a\rVert^2 + \lvert b\rVert^2 + 2a\cdot b.
\end{align}
$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Is the following Harmonic Number Identity true?
Is the following identity true?
$$ \sum_{n=1}^\infty \frac{H_nx^n}{n^3} = \frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right] + \operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}$$
In this accepted answer, @Tunk Fey proved the above. (See $(4)$). However, I have the following $3$ queries :
*
*Why can we add the integrals after the substitution $x \mapsto 1-x$ in the following step? I doubt it since $\int f(x) \ \mathrm{d}x \neq \int f(1-x) \ \mathrm{d}x$ in general.
*
*Why do we omit the constant of integration in the following step? We should add the constant since it will affect the summation.
$$\begin{align}
\color{red}{\int\frac{\ln x\ln^2(1-x)}{x}\ dx}&=-\int\frac{\ln (1-x)\ln^2 x}{1-x}\ dx\\
&=\int\sum_{n=1}^\infty H_n x^n\ln^2x\ dx\\
&=\sum_{n=1}^\infty H_n \int x^n\ln^2x\ dx\\
&=\sum_{n=1}^\infty H_n \frac{\partial^2}{\partial n^2}\left[\int x^n\ dx\right]\\
&=\sum_{n=1}^\infty H_n \frac{\partial^2}{\partial n^2}\left[\frac {x^{n+1}}{n+1}\right]\\
&=\sum_{n=1}^\infty H_n \left[\frac{x^{n+1}\ln^2x}{n+1}-2\frac{x^{n+1}\ln x}{(n+1)^2}+2\frac{x^{n+1}}{(n+1)^3}\right]\\
&=\ln^2x\sum_{n=1}^\infty\frac{H_n x^{n+1}}{n+1}-2\ln x\sum_{n=1}^\infty\frac{H_n x^{n+1}}{(n+1)^2}+2\sum_{n=1}^\infty\frac{H_n x^{n+1}}{(n+1)^3}\\
&=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n+1} x^{n+1}}{(n+1)^2}-\sum_{n=1}^\infty\frac{x^{n+1}}{(n+1)^3}\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n+1} x^{n+1}}{(n+1)^3}-\sum_{n=1}^\infty\frac{x^{n+1}}{(n+1)^4}\right]\\
&=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\sum_{n=1}^\infty\frac{x^{n}}{n^3}\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\sum_{n=1}^\infty\frac{x^{n}}{n^4}\right]\\
&=\frac12\ln^2x\ln^2(1-x)-2\ln x\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^2}-\operatorname{Li}_3(x)\right]\\&+2\left[\sum_{n=1}^\infty\frac{H_{n} x^{n}}{n^3}-\operatorname{Li}_4(x)\right].
\end{align}$$
*
*Is the identity even true, since putting $x=\dfrac{1}{2}$ gives a numerically different result than the correct result, as pointed out by the user @Super Abound in the comments of that answer.
Please help.
|
The sum in question has a closed form in terms of polylogarithms. The proof is complicated, and I don't intend to reproduce it as I derived it some 15 years ago, and polylogs are not a primary interest now. You can always differentiate both sides and use polylog IDs in Lewin.
$$\sum_{k=1}^\infty \frac{y^k}{k^3}H_k=\zeta(4)+2 Li_4(y)-Li_4(1-y)+Li_4(-y/(1-y))+\\
\frac{1}{2} \log(1-y) \Big( \zeta(3) – Li_3(y)+Li_3(1-y)+Li_3(-y/(1-y)) \Big)
+ \\\frac{1}{12}\log^3(1-y)\log(y) -\frac{1}{24}\log^4(1-y)$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1847204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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|
Factoring $x^4-11x^2y^2+y^4$ I am brushing up on my precalculus and was wondering how to factor the expression
$$
x^4-11x^2y^2+y^4
$$
Thanks for any help!
|
$$(x^2 + a y^2) (x^2 + b y^2) = x^4-11x^2y^2+y^4$$
Hence,
$$a + b = -11 \qquad \qquad \qquad a b = 1$$
In SymPy,
>>> a, b = symbols('a b')
>>> solve_poly_system([a + b + 11, a*b - 1], a, b)
____ ____ ____ ____
11 3*\/ 13 11 3*\/ 13 11 3*\/ 13 11 3*\/ 13
[(- -- - --------, - -- + --------), (- -- + --------, - -- - --------)]
2 2 2 2 2 2 2 2
Thus,
$$\left( x^2 + \left(\frac{-11 - 3 \sqrt{13}}{2}\right) y^2 \right) \left( x^2 + \left(\frac{-11 + 3 \sqrt{13}}{2}\right) y^2 \right) = x^4-11x^2y^2+y^4$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1847983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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|
Evaluation of $\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$
$$\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$
$\bf{My\; Try::}$ Let $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$
Put $\displaystyle x = \frac{1}{u},$ Then $\displaystyle dx = -\frac{1}{u^2}du$ and changing limits, We get
$$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{1}{u(1+u^2)}du$$
So $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{t}{1+t^2}dt = \frac{1}{2}\int_{\frac{1}{e}}^{e}\frac{2t}{1+t^2}dt$$
So $$f(x) = \frac{1}{2}\ln (1+t^2)|_{\frac{1}{e}}^{e} = \ln(e) = 1$$
My question is how can we solve it in some shorter way, Help required, Thanks
|
We have $$\int_{1/e}^{\tan\left(x\right)}\frac{t}{1+t^{2}}dt=\frac{1}{2}\int_{1/e}^{\tan\left(x\right)}\frac{2t}{1+t^{2}}dt
$$ $$=\frac{1}{2}\log\left(1+\tan^{2}\left(x\right)\right)-\frac{1}{2}\log\left(1+\frac{1}{e^{2}}\right)
$$ and $$\int_{1/e}^{\cot\left(x\right)}\frac{1}{t\left(1+t^{2}\right)}dt=\int_{1/e}^{\cot\left(x\right)}\frac{1}{t}dt-\int_{1/e}^{\cot\left(x\right)}\frac{t}{1+t^{2}}dt
$$ $$=\log\left(\cot\left(x\right)\right)+1-\frac{1}{2}\log\left(1+\cot^{2}\left(x\right)\right)+\frac{1}{2}\log\left(1+\frac{1}{e^{2}}\right)$$ so $$\int_{1/e}^{\tan\left(x\right)}\frac{t}{1+t^{2}}dt+\int_{1/e}^{\cot\left(x\right)}\frac{1}{t\left(1+t^{2}\right)}dt$$ $$=1+\frac{1}{2}\log\left(1+\tan^{2}\left(x\right)\right)-\log\left(\tan\left(x\right)\right)-\frac{1}{2}\log\left(1+\cot^{2}\left(x\right)\right)=\color{red}{1}$$ assuming $0<\tan(x)<\infty.$
|
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"timestamp": "2023-03-29T00:00:00",
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}
|
Prove by induction that $n^5-5n^3+4n$ is divisible by 120 for all n starting from 3 I've tried expanding $(n+1)^5-5(n+1)^3+4(n+1)$ but I end up with $120k+5(n^4+2n^3-n^2-2n)$ where k is any positive whole number, and I can't manipulate $5(n^4+2n^3-n^2-2n)$ to factor with 120.
|
${\binom{n+2}{5}= }$ ${\frac{(n+2)(n+1)(n)(n-1)(n-2)} {5!} }$
*
*${n^5-5n^3+4n = (n-2)(n-1)(n)(n+1)(n+2)= 5!\binom{n+2}{5}}$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Factoring out a $7$ from $3^{35}-5$? Please Note: My main concern now is how to factor $7$ from $3^{35}-5$ using Algebraic techniques, not how to solve the problem itself; the motivation is just for background.
Motivation:
I was trying to solve the following problem
What is the remainder when $10^{35}$ is divided by $7$?
I used the binomial formula: $\dfrac {(7+3)^{35}}{7}= \dfrac {7^{35} + \cdot \cdot \cdot 3^{35}}{7}= \dfrac {7^{35} + \cdot \cdot \cdot + 35 \cdot 3^{34} \cdot 7}{7} + \dfrac {3^{35}}{7}$
Therefore, $10^{35}$ will have the same remainder as $3^{35}$ when divided by $7$.
I was stuck here, and I wanted to try to reverse engineer the answer. I know the remainder is $5$. Therefore I should be able to write $3^{35} -5 + 5$ as $7k+5$. However, I have no idea how to factor out a $7$ from $3^{35}-5$, or from $10^{35}-5.$ How could I find this $k$ non-explicitly?
|
As a (possibly) different approach: start by remarking that $$3^3=4\times 7-1$$
It follows that $$3^{33}=(4\times 7-1)^{11}=7N-1$$
Where $N$ is a simple expression in powers of $4$ and $7$ (with some binomial coefficients thrown in for good measure). Specifically $$N=\sum_{i=1}^{11}\binom {11}i (-1)^{11-i}4^i7^{i-1}$$
We then remark that $$3^{35}-5=3^{33}3^2-5=(7N-1)\times 9-5 = 7\times 9N-9-5=7\times (9N-2)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to solve the equation $\log_{2x+3}(6x^2+23x+21)=4-\log_{3x+7}(4x^2+12x+9)$? How to solve the equation $\log_{2x+3}(6x^2+23x+21)=4-\log_{3x+7}(4x^2+12x+9)$ ?
Can someone please tell me a few steps as to how to approach these category of problems? I know $2x+3>0$ and $3x+7>0$ is a must.
|
HINT:
Using Laws of Logarithms,
$$\log_{2x+3}(6x^2+23x+21)=\log_{2x+3}(2x+3)+\log_{2x+3}(3x+7)=1+\log_{2x+3}(3x+7)$$
Let $\log_{2x+3}(3x+7)=a$
$$\log_{3x+7}(4x^2+12x+9)=\log_{3x+7}(2x+3)^2=2\log_{3x+7}(2x+3)=\dfrac2a$$ as $\log_{3x+7}(2x+3)\cdot\log_{2x+3}(3x+7)=1$
|
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|
Find the determinant using colum or row operations I find problem in simplification. When I tried to simplify I ended up doing the regular process of finding the determinant value. The matrix is $\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ ab & bc & ca \end{pmatrix}$.
|
Maybe it's unlikely that you would have learned of this method, but via Dodgson condensation, you can reduce this to a problem of computing the determinants of several $2\times2$ matrices, which you might find easier to compute strictly via row operations.
$$\begin{vmatrix}1&1&1\\a&b&c\\ab&bc&ca\end{vmatrix}=\begin{vmatrix}\dfrac{\begin{vmatrix}1&1\\a&b\end{vmatrix}}{b}&\dfrac{\begin{vmatrix}1&1\\b&c\end{vmatrix}}{b}\\[1ex]
\dfrac{\begin{vmatrix}a&b\\ab&bc\end{vmatrix}}{b}&\dfrac{\begin{vmatrix}b&c\\bc&ca\end{vmatrix}}{b}
\end{vmatrix}$$
The first determinant (in the $(1,1)$ position), for instance, would be
$$\begin{vmatrix}1&1\\a&b\end{vmatrix}=\begin{vmatrix}1&1\\0&b-a\end{vmatrix}=\begin{vmatrix}1&0\\0&b-a\end{vmatrix}=(b-a)\begin{vmatrix}1&0\\0&1\end{vmatrix}=b-a$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
number of non differentiable points in $g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right)\right)$
If $\displaystyle f(x) = \lim_{n\rightarrow \infty}\frac{x^2+2(x+1)^{2n}}{(x+1)^{2n+1}+x^2+1}\;,n\in \mathbb{N}$
and $\displaystyle g(x) = \tan \left(\frac{1}{2}\arcsin\left(\frac{2f(x)}{1+(f(x))^2}\right)\right)$
Then number of points where $g(x)$ is not differentiable.
$\bf{My\; Try::}$ We can write $$f(x) =\lim_{n\rightarrow \infty}\frac{x^2+2(x+1)^{2n}}{(x+1)^{2n+1}+x^2+1}= \left\{\begin{matrix}
\frac{2}{x+1}\;\;,&x+1<-1\Rightarrow x<-2 \\
\frac{x^2}{x^2+1}\;\;, & -1<x+1<1\Rightarrow -2<x<0\\
\frac{2}{x+1}\;\;, & x+1>1\rightarrow x>0 \\
1\;\;, & x+1=1\Rightarrow x=0\\
\frac{3}{2}\;\;,& x+1=-1\Rightarrow x=-2
\end{matrix}\right.$$
Now How can i solve after that help Required, Thanks
|
First note that, for every $y$, $2y/(1+y^2)\in[-1,1]$.
Consider $g(y)=\frac{1}{2}\arcsin(2y/(1+y^2))$, so
$$
g'(y)=\frac{1}{\sqrt{1-\dfrac{4y^2}{(1+y^2)^2}}}
\cdot\frac{1+y^2-2y^2}{(1+y^2)^2}=\frac{2}{1+y^2}\frac{1-y^2}{|1-y^2|}
$$
Therefore $g'(y)$ does not exist for $y=\pm1$ and
$$
g'(y)=\begin{cases}
\dfrac{1}{1+y^2} & \text{if $|y|<1$}
\\[4px]
-\dfrac{1}{1+y^2} & \text{if $|y|<1$}
\end{cases}
$$
Thus
$$
g(y)=\begin{cases}
a+\arctan y & \text{if $|y|<1$}
\\[4px]
b-\arctan y & \text{if $|y|<1$}
\\[4px]
-\pi/4 & \text{if $y=-1$}
\\[4px]
\pi/4 & \text{if $y=1$}
\end{cases}
$$
Since $g(0)=0$, we have $a=0$; since $\lim_{y\to\infty}g(y)=0$, we have $b=\pi/2$.
Thus
$$
\tan\left(\frac{1}{2}\arcsin\frac{2y}{1+y^2}\right)=\tan(g(y))=
\begin{cases}
y & \text{if $|y|<1$}
\\[4px]
\dfrac{1}{y} & \text{if $|y|>1$}
\\[4px]
-1 & \text{if $y=-1$}
\\[4px]
1 & \text{if $y=1$}
\end{cases}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $x^2+\frac{1}{2x}=\cos \theta$, evaluate $x^6+\frac{1}{2x^3}$. If $x^2+\frac{1}{2x}=\cos \theta$, then find the value of $x^6+\frac{1}{2x^3}$.
If we cube both sides, then we get $x^6+\frac{1}{8x^3}+\frac{3x}{2} \cdot \cos \theta=\cos ^3 \theta$ but how can we use it to deduce required value?
|
$$2x^3+1=2x\cos(a)\implies x^3-x\cos(a)+{1\over 2}=0$$Let $x=y+{b\over y}\implies y={1\over 2}(x+\sqrt{x^2-4b})$ Using back substitution$$\left(y+{b\over y}\right)^3-\left(y+{b\over y}\right)\cos(a)+{1\over 2}=0$$$$\implies {1\over 2}y^3+y^6+b^3+y^4(3b-\cos(a))+y^2(3b^2-b\cos(a))=0$$
Now substitute $b={\cos(a)\over 3}$ and $z=y^3$ $$z^2+{z\over 2}+{\cos^3(a)\over 27}=0\implies z={1\over 36}\left(\sqrt{3}\sqrt{27-16\cos^3(a)}-9\right)$$
Now substitute back $z=y^3$ and it will give the $3$ values of $y$.
Once you get the values of $y$ substitute back the values of $x$ and you will get $3$ quadratics of $x$
BINGO!
|
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|
Evaluate $\sum_{k=2}^\infty\frac{1}{k^3-1}$ I was considering a specialization of the Cauchy product
$$ \left(\sum_{n=1} ^\infty x^n \right) \left(\sum_{n=1}^\infty (-1)^n x^n \right)=\frac{-x^2}{1-x^2},$$
that converges for $0<x<1$.
The cited specialization is $x=\frac{1}{j^{3/2}}$ for integers $j\geq 2$. Then since $ \sum_{k=1}^{n} (-1)^k=\frac{(-1)^n-1}{2}$ the Cauchy product is computed (I don't know if these kind of calculations were in the literature, my intention was find an identity for $\zeta(3)$, where $\zeta(s)$ is the Riemann Zeta function, which was failed; on the other hand I believe that it is possible get some generalization of the following by the nature of the Cauchy's products, then if is such the case feel free to study this, if it is interesting) as
$$\frac{1}{1-j^3}=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{j^{\frac{3}{2}(n+1)}} \left( \frac{(-1)^n-1}{2} \right) .$$
Now I compute $$\frac{1}{j^3}=\frac{1}{1-j^3}+\text{something}=\text{RHS}+\text{something},$$
where $\text{something}=\frac{1-2j^3}{j^3(1-j^3)}$, and thus taking the sum from $j=2$ to infinite, one has
$$\zeta(3)=1+\sum_{j=2}^\infty\frac{1-2j^3}{j^3(1-j^3)}+\sum_{n=1}^\infty(-1)^{n+1} \frac{(-1)^n-1}{2}\sum_{j=2}^\infty\frac{1}{j^{\frac{3}{2}(n+1)}}. $$
Thus if there are no mistakes I can write first this simplification $$\sum_{j=2}^\infty\frac{1}{1-j^3}=\sum_{n=1}^\infty\frac{-1-(-1)^{n+1}}{2} \left( \zeta(\frac{3}{2}(n+1))-1 \right) ,$$ and after that $$\sum_{j=2}^\infty\frac{1}{j^3-1}=\sum_{m=1}^\infty(\zeta(3m+1)-1).$$
The online calculator of Wolfram Alpha knons how compute
sum 1/(k^3-1) from k=2 to infinite
and
Question. Can you explain how obtain $$\sum_{k=2}^\infty\frac{1}{k^3-1}?$$
What is the meaning of such calculations from the online calculator? Thanks in advance.
|
The key is to use the identity:
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}.\tag{1}$$
that leads to:
$$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)(n+c)} = \frac{1}{b-c}\left(\frac{\psi(a)-\psi(c)}{a-c}-\frac{\psi(a)-\psi(b)}{a-b}\right).\tag{2} $$
We have $k^3-1 = (k-1)(k-\omega)(k-\omega^2)$, hence by taking $a=1$, $b=2-\omega$, $c=2-\omega^2$:
$$ \sum_{k\geq 2}\frac{1}{k^3-1} = \sum_{n\geq 0}\frac{1}{(n+a)(n+b)(n+c)} = \frac{i}{\sqrt{3}}\left(\frac{\psi(1)-\psi(2-\omega^2)}{\omega^2-1}-\frac{\psi(1)-\psi(2-\omega)}{\omega-1}\right)\tag{3} $$
or:
$$ \sum_{k\geq 2}\frac{1}{k^3-1} = \color{red}{\frac{2}{\sqrt{3}}\,\text{Im}\left(\frac{H_{1-\omega}}{1-\omega}\right)}=0.2216893951\ldots\tag{4} $$
The same can be achieved by noticing that the LHS of $(4)$ equals
$$ \sum_{k\geq 2}\left(\frac{1}{k^3}+\frac{1}{k^6}+\frac{1}{k^9}+\ldots\right) = \sum_{n\geq 1}\left(\zeta(3n)-1\right) \tag{5} $$
and by applying a discrete Fourier transform to the generating function for $\zeta(n)-1$.
|
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|
Integrate $\int \frac{x^2-2}{(x^2+2)^3}dx$ $$\int \frac{x^2-2}{(x^2+2)^3}dx$$
What I did :
Method $(1) $ Re writing $x^2-2 = (x^2+2)-4 $ and partial fractions.
Method $(2) $ Substituting $x^2 = 2\tan^2 \theta $
Is there any other easy methods ?
Some substitution ?
|
Just to follow the answer above, "..for dealing with a numerator that is a power of positive definite quadratic polynomial".
Use integration by parts,
\begin{align*}
I_{n}:=\int\frac{dx}{\left(x^{2}+a^{2}\right)^{n}} & =\frac{x}{\left(x^{2}+a^{2}\right)^{n}}+2n\int\frac{x^{2}}{\left(x^{2}+a^{2}\right)^{n+1}}dx\\
& =\frac{x}{\left(x^{2}+a^{2}\right)^{n}}+2n\int\frac{1}{\left(x^{2}+a^{2}\right)^{n}}dx-2na^{2}\int\frac{1}{\left(x^{2}+a^{2}\right)^{n+1}}dx\\
& =\frac{x}{\left(x^{2}+a^{2}\right)^{n}}+2nI_{n}-2na^{2}I_{n+1}
\end{align*}
and so
\begin{align*}
I_{n+1} & =\frac{1}{2na^{2}}\left(\left(2n-1\right)I_{n}+\frac{x}{\left(x^{2}+a^{2}\right)^{n}}\right),\quad n=1,2,\cdots;\\
I_{1} & =\int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\arctan\left(\frac{x}{a}\right)+C.
\end{align*}
For $n=2$,
\begin{align*}
I_{3}=\int\frac{dx}{\left(x^{2}+a^{2}\right)^{3}} & =\frac{3}{4a^{2}}I_{2}+\frac{1}{4a^{2}}\frac{x}{\left(x^{2}+a^{2}\right)^{2}}\\
& =\frac{3}{8a^{4}}I_{1}+\frac{3}{8a^{4}}\frac{x}{\left(x^{2}+a^{2}\right)}+\frac{1}{4a^{2}}\frac{x}{\left(x^{2}+a^{2}\right)^{2}}\\
& =\frac{3}{8a^{5}}\arctan\left(\frac{x}{a}\right)+\frac{3}{8a^{4}}\frac{x}{\left(x^{2}+a^{2}\right)}+\frac{1}{4a^{2}}\frac{x}{\left(x^{2}+a^{2}\right)^{2}}+C.
\end{align*}
|
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|
Circular permutation probability
A circular table has $9$ chairs that $4$ people can sit down randomly. What is the probability for no two people sitting next to each other?
My current idea is to calculate the other probability, which is there are two people sitting next to each other, but I don't know enough conditions that I have to catch.
I only know that there would be $6!$ ways to arrange the other $7$ objects, the people-people pair can interchange places ($P_1P_2$ - $P_2P_1$) and I can choose which people that would form the pair ($_4C_2?$). Those conditions alone would give me $12 \times 6!$ arrangements, which is not likely (as I thought that two people sitting next to each other is more likely to happen than the other)
What did I miss?
|
This problem can be approached using the Polya Enumeration Theorem
(PET). Note that the four people seated on the table create four gaps
between them that consist of empty chairs and the empty chairs in all
four gaps must add up to five, where we have rotational symmetry
acting on the gaps.
Now the cycle index of $Z(C_4)$ is seen by enumeration to be (just
factor the four permutations)
$$Z(C_4) = \frac{1}{4} (a_1^4 + a_2^2 + 2a_4).$$
With the gaps possibly being empty we get
$$[z^5] Z(C_4)\left(\frac{1}{1-z}\right).$$
This works out to
$$\frac{1}{4} [z^5] \left(\frac{1}{(1-z)^4}
+ \frac{1}{(1-z^2)^2} + 2\frac{1}{1-z^4}
\right)$$
or
$$\frac{1}{4} {5+3\choose 3} = 14.$$
With the gaps containing at least one chair we get
$$[z^5] Z(C_4)\left(\frac{z}{1-z}\right)$$
which works out to
$$\frac{1}{4} [z^5] \left(\frac{z^4}{(1-z)^4}
+ \frac{z^4}{(1-z^2)^2} + 2\frac{z^4}{1-z^4}
\right)
\\ = \frac{1}{4} [z^1] \left(\frac{1}{(1-z)^4}
+ \frac{1}{(1-z^2)^2} + 2\frac{1}{1-z^4}
\right)$$
or $$ \frac{1}{4} \times {1+3\choose 3} = 1.$$
This finally yields for the desired probability
$$\frac{1}{14}.$$
Observe that by rotational symmetry if we must have at least one chair
in each gap there is only one chair left to distribute and these all
yield the same arrangement under rotational symmetry, so there is
indeed just one such arrangement.
Addendum. For the case of the people being distinguishable (call
them $A,B,C$ and $D$ and let $E$ represent gaps as in the word
empty) we obtain for the total possibilities
$$[ABCDE^5] Z(C_4)\left((A+B+C+D)
\times (1+E+E^2+E^4+E^5+\cdots)\right)
\\ = [ABCDE^5] Z(C_4)\left((A+B+C+D) \frac{1}{1-E}\right).$$
Here the only contribution comes from the first term and we get
$$\frac{1}{4} [ABCDE^5] (A+B+C+D)^4 \frac{1}{(1-E)^4}
\\ = \frac{1}{4} {5+3\choose 3} [ABCD] (A+B+C+D)^4.$$
The case of gaps not being empty yields
$$\frac{1}{4} [ABCDE^5] (A+B+C+D)^4 \frac{E^4}{(1-E)^4}
\\ = \frac{1}{4} [ABCDE] (A+B+C+D)^4 \frac{1}{(1-E)^4}
\\ = \frac{1}{4} {1+3\choose 3} [ABCD] (A+B+C+D)^4.$$
We get for the probability
$$\frac{4\choose 3}{8\choose 3} = \frac{1}{14}.$$
Here we have attached the gap next to a person in a clockwise
direction to the term for that person.
|
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|
Quadratic Functional equations. Suppose $f$ is a quadratic ploynomial, with leading cofficient $1$, such that $$f(f(x) +x) = f(x)(x^2+786x+439)$$ For all real number $x$. What is the value of $f(3)$?
|
Stupid method:
Let $f(x) = x^2 + ax + b$. Then the left hand side of the equation is
$$\begin{split}
f(f(x) + x) & =f( x^2 + ax + b + x) \\
&= (x^2+ (a+1)x +b)^2 + a(x^2+(a+1) x+ b) + b
\end{split}$$
If you compare the $x^3$ coefficient, this gives $2(a+1) = 786 +a$. Comparing the constant coefficient tells you $b$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Indefinite Integral - How to do questions with square roots?
$$\int \frac{dx}{x^4 \sqrt{a^2 + x^2}}$$
In the above question, my first step would be to try and get out of the square root, so I would take $ t^2 = a^2 + x^2 $. But that gets me nowhere. How would you solve this, and if you are going to take a substitution what is the logic behind that substitution?
|
substitute $x=a\tan { \theta } ,dx=\frac { a\,d\theta }{ \cos ^{ 2 }{ \theta } }$
so
$$\\ \\ \\ \int \frac { dx }{ x^{ 4 }\sqrt { a^{ 2 }+x^{ 2 } } } =\int \frac { a\,d\theta }{ \cos ^{ 2 }{ \theta } { \left( a\tan { \theta } \right) }^{ 4 }\sqrt { a^{ 2 }+{ a }^{ 2 }\tan ^{ 2 }{ \theta } } } =\frac { 1 }{ { a }^{ 4 } } \int { \frac { \cos ^{ 3 }{ \theta } }{ \sin ^{ 4 }{ \theta } } } \, d\theta =\\ \\ =\frac { 1 }{ { a }^{ 4 } } \int { \frac { 1-\sin ^{ 2 }{ \theta } }{ \sin ^{ 4 }{ \theta } } d\sin { \theta } } =\frac { 1 }{ { a }^{ 4 } } \left[ \int { \frac { d\sin { \theta } }{ \sin ^{ 4 }{ \theta } } } -\int { \frac { d\sin { \theta } }{ \sin ^{ 2 }{ \theta } } } \right] =\\ =\frac { 1 }{ { a }^{ 4 } } \left[ -\frac { 1 }{ 3\sin ^{ 3 }{ \theta } } +\frac { 1 }{ \sin { \theta } } \right] =\frac { 1 }{ { a }^{ 4 } } \left[ -\frac { 1 }{ 3\sin ^{ 3 }{ \left( \arctan { \frac { x }{ a } } \right) } } +\frac { 1 }{ \sin { \left( \arctan { \frac { x }{ a } } \right) } } \right] +C\\ $$
|
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|
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$
Prove: $x^3+y^3\geq \frac{1}{4}(x+y)^3$ for all $x,y$ positive.
Let's look at
$$\begin{split} &(x-y)^2(x+y)\geq 0 \\
\iff &(x-y)(x+y)(x-y)\geq 0\\
\iff& (x-y)(x^2-y^2)\geq 0 \\
\iff &x^3-xy^2-yx^2+y^3\geq 0\\
\iff & 3x^3+3y^3\geq +3xy^2+3yx^2\\
\iff &3x^3+3y^3\geq (x+y)^3 -x^3-y^3 \\
\iff & 4x^3+4y^3\geq (x+y)^3 \\
\iff & x^3+y^3\geq \frac{1}{4}(x+y)^3
\end{split}$$
is the proof valid? is there a shorter way?
|
In general
$$x^p + y^p \geq \dfrac{(x+y)^p}{2^{p-1}}$$
which follows from Holder's inequality, which states that
$$\Vert a \Vert_p \Vert b \Vert_q \geq \vert a\cdot b \vert$$
where $\dfrac1p + \dfrac1q = 1$. To obtain your result, take $a = (x,y)$ and $b=(1,1)$.
|
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|
Evaluate $\int \frac{\sqrt{64x^2-256}}{x}\,dx$ QUESTION
Evaluate $$\int \frac{\sqrt{64x^2-256}}{x}\,dx$$
I've tried this problem multiple times and cant seem to find where I made a mistake. If someone could please help explain where I went wrong I would really appreciate it
MY ATTEMPT
*
*Typed
$\newcommand{\dd}{\; \mathrm{d}}\int \frac{\sqrt{64x^2-256}}x \dd x \to
\int \frac{\sqrt{64(x^2-4)}}x \dd x \to
\int \frac{8\sqrt{x^2-4}}x \dd x$
Use $x=a\sec\theta$, $\dd x=a\sec\theta \tan\theta \dd \theta$.
$a=2$ $\to$ $x=2\sec\theta$, $\dd x=2\sec\theta \tan\theta \dd \theta$.
$=\int \frac{8\sqrt{4\sec^2\theta-4}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta \to
\int \frac{8\sqrt{4(\sec^2\theta-1)}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta $
$=\int \frac{8\sqrt{4\tan^2\theta}}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta \to
\int \frac{8(2\tan\theta)}{2\sec\theta}(2\sec\theta\tan\theta) \dd \theta$
$=\int 16\tan^2\theta \dd \theta \to
16\int\tan^2\theta \dd \theta \to
\underset{\text{trig. formula}}{\underbrace{16(\theta+\tan\theta)+C}}$
$\Rightarrow 16(\tan\theta-\theta)+C = 16\tan\theta-16\theta$
$x=2\sec\theta$, $\sec\theta= \frac x2$
$\boxed{16\tan\left(\frac{\sqrt{x^2-4}}2\right) -16\sec^{-1}\left(\frac x2\right)+C}$
*
*Handwritten
|
\begin{align}
&\int \frac{\sqrt{64x^2-256}}{x}\,dx\\
=& \int \bigg( \frac{8x}{\sqrt{{x^2}-4}}-\frac{32}{x^2\sqrt{1-\frac4{x^2}}}\bigg)\,dx
=\ 8 \sqrt{{x^2}-4}+16\sin^{-1}\frac2x
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1857798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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|
Solving equations system: $xy+yz=a^2,xz+xy=b^2,yz+zx=c^2$ Solve the following system of equations for $x,y,z$ as $a,b,c\in\Bbb{R}$
\begin{align*}xy+yz&=a^2\tag{1}\\xz+xy&=b^2\tag{2}\\yz+zx&=c^2\tag{3}\end{align*}
My try: Assume that $x,y,z\ne 0$ (it is easy to check the case where some are zero).
Subtract first equation from second equation we will get the system \begin{align*}xz-yz&=b^2-a^2\\xz+yz&=c^2\end{align*}
Adding and subtract both equations we get $$x=\frac{b^2+c^2-a^2}{2z},y=\frac{a^2+c^2-b^2}{2z}$$I tried to set the expressions into one of the equations and get value of $z$, but it got quiet messy.
Is there any easier way of solving the system?
Any help will be appreciated, thanks!
|
Adding we get $$2(xy+yz+zx)=a^2+b^2+c^2$$
$$2xy=a^2+b^2+c^2-2c^2=a^2+b^2-c^2$$ etc.
Muliplying we get $$8x^2y^2z^2=\prod_{\text{cyc}}(a^2+b^2-c^2)$$
$$xyz=\pm\sqrt{\dfrac{\prod_{\text{cyc}}(a^2+b^2-c^2)}8}$$
and $xy=\dfrac{a^2+b^2-c^2}2$
$$z=\dfrac{xyz}{xy}=?$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1857910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Convergence of the series $\sum \frac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$
To prove that nature of the following series : $$\sum \dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}$$
they use in solution manual :
My questions:
*
*I don't know how to achieve ( * ) could someone complete my attempts for ( * ) and is it correct if i use (**) to prove that the series is convergent :
$$\fbox{$\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}-\dfrac{(-1)^{n}}{n}+O\left(\dfrac{1}{n^{\frac{4}{3}}}\right)$}\quad (*)$$
My thoughts :
\begin{align}
\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} &=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right) \right)^{-1}
\end{align}
note that :
$$(1+x)^{\alpha}=1+\alpha x+O(x^{2})$$
\begin{align}
\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} &=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1-\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right)+O\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right)^{2} \right)\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} -\dfrac{(-1)^{n}}{n}+\dfrac{(-1)^{n}}{n^{\frac{4}{3}}} +\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}\times O\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right)^{2} \\
&=\ldots\ldots \\
&= \mbox{ I'm stuc here i hope someone complete my attempts }
\end{align}
Or i should use :
note that :
$$(1+x)^{\alpha}=1+O(x)$$
\begin{align}
\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} &=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+\left(\dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \right) \right)^{-1}\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}} \left( 1+O\left( \dfrac{1}{n^{\frac{1}{3}}}+\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}\right) \right)\\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}+O\left( \dfrac{1}{n}+\dfrac{(-1)^{n}}{n^{\frac{4}{3}}}\right) \\
&=\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}+O\left( \dfrac{(-1)^{n}}{n^{\frac{4}{3}}}\right) \\
\end{align}
$$\fbox{$\dfrac{(-1)^{n}}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}} =\dfrac{(-1)^{n}}{n^{\frac{2}{3}}}+O\left( \dfrac{(-1)^{n}}{n^{\frac{4}{3}}}\right)$}\quad (**) $$
|
$\dfrac{1}{n^{\frac{2}{3}}+n^{\frac{1}{3}}+(-1)^{n}}-\dfrac{1}{n^{\frac{2}{3}}}+\dfrac{1}{n} = \dfrac{-n(-1)^n+n + n^{2/3}(-1)^n}{n^{5/3}(n^{2/3}+n^{1/3}+(-1)^n}$
The right hand side has $n$ in the numerator and $n^{7/3}$ in the denominator and hence is $O\left(\frac{1}{n^{4/3}}\right)$
|
{
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"url": "https://math.stackexchange.com/questions/1858012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
$f(x)$ is a quadratic polynomial with $f(0)\neq 0$ and $f(f(x)+x)=f(x)(x^2+4x-7)$
$f(x)$ is a quadratic polynomial with $f(0) \neq 0$ and $$f(f(x)+x)=f(x)(x^2+4x-7)$$
It is given that the remainder when $f(x)$ is divided by $(x-1)$ is $3$.
Find the remainder when $f(x)$ is divided by$(x-3)$.
My Attempt:
Let $f(x)=ax^2+bx+c$ and $a,c \neq 0 $
I got $a+b+c=3$
And by the functional equation
$a[ax^2+(b+1)x+c]^2+b[ax^2+(b+1)x+c]+c= [ax^2+bx+c][x^2+4x-7]$
Then by putting $x=0$ , $ac^2 +bc+c=-7c$
Since $c \neq 0 $ , we have $ac+b+8=0$
Then comparing the coeffiecient of $x^4$ , we get $a^3=a$
Since $a \neq 0$ , $a=-1 $ or $a =1 $
Then how to proceed with two values of $a$ ?
or Is there a polynomial satisfying these conditions?
|
In general, if $F$ is a field, whose algebraic closure is $\bar{F}$, and $q(X)\in F[X]$ is a polynomial of degree $2$ with leading coefficient $k\neq 0$, then all solutions to $$f\big(\alpha\,f(X)+X\big)=f(X)\,q(X)\,,\tag{*}$$ where $\alpha\neq 0$ is a fixed element of $F$ and $f(X)\in \bar{F}[X]$, are given by $$f(X)=0\,,$$ $$f(X)=+\frac{1}{\alpha\sqrt{k}}\,q\left(X-\frac{1}{\sqrt{k}}\right)\,,$$ and $$f(X)=-\frac{1}{\alpha\sqrt{k}}\,q\left(X+\frac{1}{\sqrt{k}}\right)\,,$$ where $\sqrt{k}$ is one of the square roots of $k$. Thus, if $X^2-k$ is irreducible in $F[X]$, then the only solution to (*) in $F[X]$ is $f(X)=0$.
Hint: It is easy to see that nonconstant solutions to (*) must be of degree $2$. Suppose that $$q(X)=k\,(X-u)(X-v)\,,$$ where $u,v\in\bar{F}$. Then, it follows immediately that $$f(X)=\pm \frac{\sqrt{k}}{\alpha}\,(X-p)(X-q)$$ for some $p,q\in\bar{F}$. Compute $$\frac{f\big(\alpha\, f(X)+X\big)}{k\,f(X)}$$ in terms of $X,p,q,\sqrt{k},\alpha$. This should be the same as $$\frac{q(X)}{k}=(X-u)(X-v)\,.$$
In particular, if $F=\mathbb{C}$, $q(X)=X^2+4\,X-7$ (so that $k=1$), and $\alpha=1$, then the nonzero solutions to the functional equation (*) are $$f(X)=+q(X-1)=+X^2+2\,X-10$$ and $$f(X)=-q(X+1)=-X^2-6\,X+2\,.$$ Note that the extra condition that $f(1)=3$ is incompatible with all such polynomials, whence there are no solutions to the OP's conditions.
|
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"url": "https://math.stackexchange.com/questions/1859313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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|
Find the eigenvalues of the following matrix where $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$ as well as the associated eigenvector
Find the eigenvalues of the following matrix $\begin{bmatrix}-12&-6&0&0\\8&2&0&0\\0&0&-14&-9\\0&0&42&25\end{bmatrix}$
The eigenvalues are $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$ . Find the associated eigenvector for $\lambda_1, \lambda_2, \lambda_3, \lambda_4$
So should I start out by doing $A - I_m \lambda$ where $I_m$ is the identity matrix? So I would have $\begin{bmatrix}-12&-6&0&0\\8&2&0&0\\0&0&-14&-9\\0&0&42&25\end{bmatrix} - \begin{bmatrix}\lambda&0&0&0\\0&\lambda&0&0\\0&0&\lambda&0\\0&0&0&\lambda\end{bmatrix} = \begin{bmatrix}-12-\lambda&-6&0&0\\8&2-\lambda&0&0\\0&0&-14-\lambda&-9\\0&0&42&25-\lambda\end{bmatrix}$ then would it be correct to then put the matrix in either Upper Triangular Form or Lower Triangular Form so that I can then take the product of the diagonal which is the determinant and then solve for the eigenvalues $\lambda$?
|
Note that your matrix is the block diagonal matrix
$$
M =
\left[\begin{array}{rr}
A & 0\\ 0 & B
\end{array}\right]
$$
where
\begin{align*}
A &=
\left[\begin{array}{rr}
-12 & -6 \\
8 & 2
\end{array}\right]
&
B &=
\left[\begin{array}{rr}
-14 & -9 \\
42 & 25
\end{array}\right]
\end{align*}
Next, note that the eigenvalues and corresponding eigenvectors of $A$ are
\begin{align*}
\lambda_1 &= -4 &
v_1 &= (3,-4) &
\lambda_2 &= -6 &
v_2 &= (1,-1)
\end{align*}
and the eigenvalues and corresponding eigenvectors of $B$ are
\begin{align*}
\lambda_3 &= 7 &
v_3 &= (3,-7) &
\lambda_4 &= 4 &
v_4 &= (1,-2)
\end{align*}
The eigenvalues and corresponding eigenvectors of $M$ are thus
\begin{align*}
\lambda_1 &= -4 &
v_1 &= (3,-4,0,0) &
\lambda_2 &= -6 &
v_2 &= (1,-1,0,0) \\
\lambda_3 &= 7 &
v_3 &= (0,0,3,-7) &
\lambda_4 &= 4 &
v_4 &= (0,0,1,-2)
\end{align*}
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
Differentiate and simplify. $m(x) = \frac{x}{\sqrt{4x-3}}$ My work so far is:
\begin{align}
m'(x) &= \frac{(1)(\sqrt{4x-3})-(x)(1/2)(4x-3)^{-1/2}(4)}{(\sqrt{4x-3})^2} \\
&= \frac{\sqrt{4x-3} - 2x(4x-3)^{1/2}}{4x-3}
\end{align}
and now I'm stuck on how to simplify further
|
Another way which makes life easier when you face products, quotients, powers : logarithmic differentiation $$m=\frac{x}{\sqrt{4x-3}}\implies \log(m)=\log(x)-\frac 12 \log(4x-3)$$ Differentiate $$\frac{m'} m=\frac 1x-\frac 12 \times \frac 4{4x-3}=\frac{2x-3}{x(4x-3)}$$ $$m'=m\times \frac{2x-3}{x(4x-3)}=\frac{x}{\sqrt{4x-3}}\times \frac{2x-3}{x(4x-3)}=\frac{2x-3}{(4x-3)^{3/2}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1860814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
How to solve this inequality problem? Given that $a^2 + b^2 = 1$, $c^2 + d^2 = 1$, $p^2 + q^2 = 1$, where $a$, $b$, $c$, $d$, $p$, $q$ are all real numbers, prove that $ab + cd + pq\le \frac{3}{2}$.
|
HINT:
For real $a-b,c-d,p-q;$
$$(a-b)^2+(c-d)^2+(p-q)^2\ge0$$
More generally for real $a,b;$
$$a^2+b^2=(a-b)^2+2ab\ge2ab\iff2ab\le a^2+b^2=?$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1861002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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|
Solve the equation $x^{x+y}=y^{y-x}$ over natural numbers
Solve the equation $x^{x+y}=y^{y-x} \tag 1$ where $x,y \in \mathbb{N}$
$x = 1,y = 1$ is a solution, now suppose $x \ne 1, y \ne 1$.
Obviously $x, y \ne 0$ and $x, y$ have same prime divisors.
Because $x+y \gt y-x$ it follows that $x \mid y$ therefore $y = kx$ and (1) becomes: $x^{(k+1)x}=(kx)^{(k-1)x} \tag2$ and, dividing by $x^{k-1}$ we get
$x^{2x}=k^{(k-1)x} \tag 3$ therefore $x^2=k^{k-1} \tag 4$ and here I've got stuck. Any help is appreciated.
UPDATE
From (4) we have $k \ge 3$ therefore $2 \le k-1$ and $ k \mid x$. Let $x=km$ and (4) becomes $k^2m^2=k^{k-1} \tag5$ and $m^2=k^{k-3} \tag6$
UPDATE 2
$x=t^{t^2-1}, y=t^{t^2+1}$ is solution $\forall t \in \mathbb{N}-\{0\}$
|
$x^2 = k^{k-1}, k$ must be odd, or k is a perfect square.
k is odd.
$k = 2n+1\\
x,y = (2n+1)^n, (2n+1)^{n+1}$
$k$ is a pefect square. $k = n^2$
$x^2 = n^{2(2n-1)}\\
x,y = n^{2n-1}, n^{2n+1}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1863458",
"timestamp": "2023-03-29T00:00:00",
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|
Doubt in finding number of integral solutions Problem :
writing $5$ as a sum of at least $2$ positive integers.
Approach :
I am trying to find the coefficient of $x^5$ in the expansion of $(x+x^2+x^3\cdots)^2\cdot(1+x+x^2+x^3+\cdots)^3$ .
which reduces to coefficient of $x^3$ in expansion of $(1-x)^{-5}$ ,which is $${7\choose3}= 35$$
but we can count the cases and say that answer must be $6$ :
$4 + 1 $
$3 + 2 $
$3 + 1 + 1$
$2 + 2 + 1 $
$2 + 1 + 1 + 1$
$1 + 1 + 1 + 1 + 1$
At which stage am I making a mistake ? Thanks
|
According to your generating function approach let's assume we want to find the number of compositions of $5$ having at least two parts.
Using generating functions we have to look for the coefficient of $x^5$ in
\begin{align*}
&(x+x^2+x^3+\cdots)^2+(x+x^2+x^3+\cdots)^3+(x+x^2+x^3+\cdots)^4\\
&\qquad+(x+x^2+x^3+\cdots)^5\\
&\qquad=\frac{x^2}{(1-x)^2}+\frac{x^3}{(1-x)^3}+\frac{x^4}{(1-x)^4}+\frac{x^5}{(1-x)^5}\tag{1}\\
\end{align*}
The coefficient of $x^5$ in each of the four terms denotes for $j=2,3,4,5$ the number of possibilities to write the number $5$ as sum of $j$ numbers $\geq 1$.
It's convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$.
We obtain for $j=2,3,4,5$
\begin{align*}
[x^5]\frac{x^j}{(1-x)^j}&=[x^{5-j}]\frac{1}{(1-x)^j}\tag{2}\\
&=[x^{5-j}]\sum_{k=0}^\infty\binom{-j}{k}(-x)^k\tag{3}\\
&=[x^{5-j}]\sum_{k=0}^\infty\binom{k+j-1}{j-1}x^k\tag{4}\\
&=\binom{4}{j-1}\tag{5}
\end{align*}
Comment:
*
*In (2) we use the rule
$
[x^{p-q}]A(x)=[x^p]x^qA(x)
$
*In (3) we use the binomial series expansion
*In (4) we use the binomial identity
$
\binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q
$
*In (5) we select the coefficient from $x^j$
We conclude from (1) and (5): The number of solutions is
\begin{align*}
\binom{4}{1}+\binom{4}{2}+\binom{4}{3}+\binom{4}{4}&=4+6+4+1\\
&=15
\end{align*}
The corresponding $15$ tuples are
\begin{align*}
&14,23,32,41\\
&113,122,131,212,221,311\\
&1112,1121,1211,2111\\
&11111
\end{align*}
Hint: The generating function $(x+x^2+x^3+...)^2(1+x+x^2+...)^3$ is not appropriate since it counts e.g. $$1+2+2$$ more than once.
In fact $1+2+2$ is counted three times
\begin{align*}
(x+x^2+x^3+...)^2&(1+x+x^2+...)^3\\
1+2\quad\quad&\quad+2+0+0\\
1+2\quad\quad&\quad+0+2+0\\
1+2\quad\quad&\quad+0+0+2\\
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
integrate $\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx=\int \frac{\sin^2x \cos^2x \sin x }{1+\sin^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{1+1-\cos^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{2-\cos^2x}dx$$
$u=cosx$
$du=-sinxdx$
$$-\int \frac{(1-u^2 )u^2}{2-u^2}du=\int \frac{u^4-u^2 }{2-u^2}du$$
Polynomial division give us
$$\int-u^2-1+\frac{2}{u^2-2}du=-\frac{u^3}{3}-u=2\int \frac{1}{u^2-2}du$$
Using partial fraction we get to:
$$\frac{1}{u^2-2}=\frac{1}{(u+\sqrt{2})(u^2-\sqrt{2})}=\frac{A}{u^2+\sqrt{2}}+\frac{B}{u^2-\sqrt{2}}$$
So we get to:
$$1=(A+B)u^2+\sqrt{2}(A-B)$$
So $A=B$ but $0*\sqrt{2}\neq 1$, where did it go wrong?
|
Consider
$$
\frac{t^2-t}{2-t}=(-1-t)+\frac{2}{2-t}
$$
so your integral is
$$
\int\left(-\frac{2}{u^2-2}-u^2-1\right)\,du
$$
and your reduction is (almost) good, but a sign went wrong. Now your computation of partial fractions is really messed up:
$$
\frac{2}{u^2-2}=\frac{A}{u-\sqrt{2}}+\frac{B}{u+\sqrt{2}}
$$
so you get
$$
\begin{cases}
A+B=0\\[6px]
A\sqrt{2}-B\sqrt{2}=2
\end{cases}
$$
so $A=1/\sqrt{2}$ and $B=-1/\sqrt{2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$3(x-\sqrt{xy}+y)^2\geq x^2+xy+y^2$ Prove that $3(x-\sqrt{xy}+y)^2\geq x^2+xy+y^2$ for all $x,y\geq 0$.
Expanding, the inequality becomes
$$3x^2+3xy+3y^2-6x\sqrt{xy}-6y\sqrt{xy}+6xy\geq x^2+xy+y^2$$
which is
$$x^2+4xy+y^2\geq3\sqrt{xy}(x+y)$$
We can try using AM-GM:
$$x^2+xy+xy\geq 3\sqrt[3]{x^4y^2}$$
This is close to the right-hand side but still different.
|
$x^2+xy+y^2 = (x+\sqrt{xy}+y)(x-\sqrt{xy}+y)$, so divide through and get
$3(x-\sqrt{xy}+y) \ge x+\sqrt{xy}+y$
$2(x+y) \ge 4\sqrt{xy}$
$\frac{x+y}{2} \ge \sqrt{xy}$
which is AM-GM.
|
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|
Integer solutions to $5m^2-6mn+7n^2 = 1985$
Are there integers $m$ and $n$ such that $$5m^2-6mn+7n^2 = 1985?$$
Taking the equation modulo $3$ gives $n^2-m^2 \equiv 2 \pmod{3}$. Thus, $3 \mid n$ but $3 \nmid m$. How can I use this to find a contradiction?
|
The discriminant of the quadratic form is $6^2-4\cdot 5\cdot 7=-2^3\cdot \color{blue}{13}$ and the equation is equivalent to:
$$ (5m-3n)^2+26 n^2 = 9925 \tag{1}$$
that gives:
$$ 6\equiv 9925 \equiv (5m-3n)^2 \pmod{\color{blue}{13}}\tag{2} $$
but $6$ is not a quadratic residue $\!\!\pmod{13}$, since $13\equiv 5\pmod{8}$ and $13\equiv 1\pmod{3}$, so:
$$ \left(\frac{6}{13}\right)=\left(\frac{-1}{13}\right)\left(\frac{2}{13}\right)\left(\frac{-3}{13}\right)=(+1)\cdot(-1)\cdot(+1)=\color{red}{-1}\tag{3}$$
and there are no integer solutions to the original equation.
For similar problems, quadratic reciprocity and prime factors of the discriminant are the key.
|
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|
Evaluate $\int \frac{x}{x^4+5}dx$
$$\int \frac{x}{x^4+5}dx$$
$u=x^2$
$du=2xdx\Rightarrow \frac{du}{2}=xdx$
$$\int \frac{x}{x^4+5}=\frac{1}{2}\int \frac{du}{u^2+5}$$
I want to get to the expression in the form of $\frac{da}{a^2+1}$ so I factor out $5$ to get to:
$$\frac{1}{2}\int \frac{du}{5[(\frac{u}{\sqrt{5}})^2+1]}$$
$\frac{du}{5[(\frac{u}{\sqrt{5}})^2+1]}$ is in the form of $\frac{\frac{a}{b}}{\frac{5c}{d}}$ so if I factor out the $5$ should it be $$\int \frac{1}{2}*\frac{1}{5}\frac{du}{[(\frac{u}{\sqrt{5}})^2+1]}$$
or
$$\int \frac{1}{2}*5\frac{du}{[(\frac{u}{\sqrt{5}})^2+1]}$$
|
Your last line isn't correct, you can procced as follows:
$$\int \frac{1}{2}*\frac{1}{5}\frac{du}{[(\frac{u}{\sqrt{5}})^2+1]}=\frac 1 {10}\int\frac{du}{u^2/5+1}$$
Set $t=\frac{u}{\sqrt5}$ and $dt=\frac{1}{\sqrt 5}du$
$$=\frac{1}{2\sqrt 5}\int\frac{dt}{t^2+1}$$
$$=\frac{\arctan t}{2\sqrt 5}=\frac{\arctan \left(u/\sqrt 5\right)}{2\sqrt 5}=\frac{\arctan \left(x^2/\sqrt 5\right)}{2\sqrt 5}+C$$
|
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|
Prove that $a_n$ is a perfect square
*
*Let $\,\,\,\left(a_{n}\right)_{\ n\ \in\ \mathbb{N}\,\,\,}$ be the sequence of integers defined recursively by
$$
a_{1} = a_{2} = 1\,,\qquad\quad a_{n + 2} = 7a_{n + 1} -a_{n} - 2\quad
\mbox{for}\quad n \geq 1
$$
*Prove that $a_{n}$ is a perfect square for every $n$.
We have $a_{3} = 4, a_{4} = 25, a_{5} = 169,\ldots$
Is there a way we can simplify the recursion or get its closed form in order to get that it is a perfect square ?.
|
Define the sequence given by $b_1=1,b_2=1$ and $b_{n+2}=3b_{n+1}-b_{n}$ for $n>0$.
To prove that $a_n=b_n^2$ we just have to show:
$b_{n+2}^2=7b_{n+1}^2-b_{n}^2-2$.
Of course $b_{n+2}^2=9b_{n+1}^2-6b_{n+1}b_n+b_{n-1}^2$
So to finish we just need to prove $3b_{n+1}b_n=b_{n+1}^2+b_n^2+1$.
Notice $b_{n+1}^2+b_n^2+1=b_{n+1}(3b_n-b_{n-1})+b_n^2+1=3b_{n+1}b_n-b_{n+1}b_{n-1}+b_n^2+1$.
So to finish we just have to show $b_{n+1}b_{n-1}-b_n^2=1$
To do this we notice that
$$
\begin{pmatrix}
3 & 1\\
-1 & 0
\end{pmatrix}^n
\begin{pmatrix}
2 & 1\\
1 & 1
\end{pmatrix}
=
\begin{pmatrix}
b_{n+3} & b_{n+2}\\
b_{n+2} & b_{n+1}
\end{pmatrix}$$
Therefore the the determinant of the matrix on the right is always $1$, and that determinant is $b_{n+3}b_{n+1}-b_{n+2}^2$
|
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|
Triple Integration in vector calc NIn the following example I have been asked to find the volume V of the solid bounded by the sphere $ x^2 + y^2 +z^2 = 2 $ and the paraboloid $ x^2 + y^2 = z $ by using triple integration.
I am not quite sure how to set up this triple integration and what each of the integrands should be.
|
It is comfortable to describe your solid in cylindrical coordinates since both the sphere and the paraboloid have the $z$-axis as an axis of symmetry (both surfaces are surfaces of revolution around the $z$-axis).
In cylindrical coordinates, the sphere is described by $\rho^2 + z^2 = 2$ and the paraboloid is described by $\rho^2 = z$. The solid bounded between them is described by $\rho^2 \leq z \leq \sqrt{2 - \rho^2}$ and so
$$ V = \int_{0}^1 \int_{0}^{2\pi} \int_{\rho^2}^{\sqrt{2 - \rho^2}} \rho \, dz \, d\theta \, d\rho = 2\pi \int_0^{1} \rho \left( \sqrt{2 - \rho^2} - \rho^2 \right) \, d\rho =
2\pi \left( \frac{1}{2} \int_1^2 \sqrt{u} \, du - \int_0^{1} \rho^3 \right)
= 2\pi \left( \frac{2^{\frac{3}{2}} - 1}{3} - \frac{1}{4}\right).$$
where we used the substitution $u = 2 - \rho^2, \, du = -2\rho$ for the first integral.
|
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|
Integral $\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$ $$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$$
$$\tan x + \cot x + \csc x + \sec x=\frac{\sin x + 1}{\cos x} +\frac{\cos x + 1}{\sin x} $$
$$= \frac{\sin x +\cos x +1}{\sin x \cos x}$$
$$t= \tan {\frac{x}{2}}$$
On solving ,
$$\frac{1}{\tan x + \cot x + \csc x + \sec x}=\frac{t(1- t)}{1+ t^2}$$
$$\implies \int \frac{\tan {\frac{x}{2}}(1-\tan {\frac{x}{2}})}{1+\tan^2 {\frac{x}{2}}}{dx}$$
I think, I have made the things more difficult. How can I proceed further? Is there any better substitution for it?
|
$$\begin{aligned}\int \frac{1}{\frac{\sin(x)}{\cos(x)}\:+\:\frac{\cos(x)}{\sin(x)}\:+\:\frac{1}{\sin(x)}\:+\:\frac{1}{\cos(x)}}dx
& = \int \:\frac{\sin \:\left(2x\right)}{2\left(\cos \:\left(x\right)+\sin \:\left(x\right)+1\right)}dx
\\& =\frac{1}{2}\cdot \int \:\frac{\sin \left(2x\right)}{\cos \left(x\right)+\sin \left(x\right)+1}dx
\\& =\frac{1}{2}\cdot \frac{1}{2}\cdot \int \:\frac{\sin \left(t\right)}{\sin \left(\frac{t}{2}\right)+\cos \left(\frac{t}{2}\right)+1}dt
\\& =\frac{1}{2}\cdot \frac{1}{2}\cdot \int \:\sin \left(\frac{t}{2}\right)+\cos \left(\frac{t}{2}\right)-1dt
\\& =\color{red}{\frac{1}{4}\left(-2x+2\sin \left(x\right)-2\cos \left(x\right)\right)+C}
\end{aligned}$$
Applyied substitution: $$\color{blue}{t=2x,\quad \:dt=2dx}$$
|
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|
Quadratic Inequality in terms of variable $x$
Find the values of $a$ for which the inequality $x^2+ax+a^2+6a<0\;\forall x \in (1,2)$
$\bf{My\; Try::}$ We can Write Equation as $$x^2+ax+\frac{a^2}{4}+\frac{3a^2}{4}+6a<0$$
So $$\left(x+\frac{a}{2}\right)^2+\frac{3a^2+24a}{4}<0$$
Now how can i solve after that, Help required, Thanks
|
$$0>4x^2+4ax+4a^2+24a=(2x+a)^2+3a^2+24a$$
Now $1<x<2\iff2+a<2x+a<4+a$
$\implies(2x+a)^2<$min$\{(2+a)^2,(4+a)^2\}$
Case$\#1:$ If $(4+a)^2\ge(2+a)^2\iff a\ge-3,(2x+a)^2<(2+a)^2$
$$0>(2+a)^2+3a^2+24a=4(a^2+7a+1)$$
We need $a^2+7a+1<0$
As the roots of $a^2+7a+1=0$ are $\dfrac{-7\pm\sqrt{49-4}}2=\dfrac{-7\pm3\sqrt5}2$
$\implies$ either $a>\dfrac{-7+3\sqrt5}2$ or $a<\dfrac{-7-3\sqrt5}2$
But $a\ge-3$ and $\dfrac{-7-3\sqrt5}2<-3$
Case$\#2:$ What if $(4+a)^2<(2+a)^2\iff a<-3$
Left for you.
|
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|
Eliminating $x=-3,5$ when we solve the system $y=|3-x|, 4y-(x^2-9)=-24$ $$y=|3-x|$$
$$4y-(x^2-9)=-24$$
I use two methods to solve it,
Method 1: Squaring both sides and eventually get a equation like this$$(x-5)(x-7)(x+3)(x+9)=0$$, $x=-9,-3,5,7$.
Method 2:Using $$y=3-x$$ and $$y=-(3-x)$$, and finally also get $x=-9,-3,5,7$
However, the answers are $x=-9,x=7$ only.
Is there any method to eliminate $x=-3,x=5$ from the real answers? Or any other methods to avoid adding these 2 wrong solutions into the answers?
|
Is there any method to eliminate $x=-3,x=5$ from the real answers?
For method 1 :
You have to have
$$x^2-33=4y=4|3-y|\ge 0\implies x^2\ge 33$$
For method 2 :
As f'' comments, note that $|3-x|=3-x$ holds only when $3-x\ge 0,$ i.e. $x\le 3$ and that $|3-x|=-(3-x)$ holds only when $3-x\lt 0$, i.e. $x\gt 3$.
So, if you have $y=3-x$, then you get
$$4(3-x)-(x^2-9)=-24\implies x=5,-9$$
with $x\le 3$. Therefore, $x=-9$ is the only solution in this case.
|
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|
Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$ The value of $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$$is equal to
*
*$\frac{\pi}{6}$
*$\frac{\pi}{4}$
*$\frac{\pi}{3}$
*$\frac{\pi}{12} $
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{3} -\sqrt{2}}{1+ \sqrt{6}}\right)$$
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) -\tan^{-1}{\sqrt3} + \tan^{-1} {\sqrt2} $$
$$\implies\frac{\pi}{2} -\frac{\pi}{3}=\frac{\pi}{6}$$
Another possibility is
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) +\tan^{-1}{\sqrt3} - \tan^{-1} {\sqrt2} $$
How to solve this ?
|
By using identities
$$\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}$$ and
$$\tan^{-1}{\alpha}-\tan^{-1}{\beta}=\tan^{-1}(\frac{\alpha-\beta}{1+\alpha\beta}),$$
after a little calculation, we get $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)=\tan^{-1}(\frac{1}{\sqrt{3}})=\frac{\pi}{6}$$
|
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|
Compute Triple integral Compute $$\iiint\limits_{x^2+y^2+(z-2)^2\leq1}\frac{dx\,dy\,dz}{x^2+y^2+z^2}$$
I tried to convert it to spherical cordinates and I got $$\iiint \sin\theta \,dr\,d\theta \,d\varphi$$
However I am lost on how to find the boundaries of the new domain, can anyone help?
|
Change coordinates by letting $z'=z-2$. Then, we have
$$\begin{align}
\iiint_{x^2+y^2+(z-2)^2\le 1} \frac{1}{x^2+y^2+z^2}\,dx\,dy\,dz&=\iiint_{x^2+y^2+z^2\le 1} \frac{1}{x^2+y^2+(z+2)^2}\,dx\,dy\,dz\\\\
&=\int_0^{2\pi}\int_0^\pi\int_0^1 \frac{r^2\sin(\theta)}{r^2+4r\cos(\theta)+4}\,dr\,d\theta\,d\phi \\\\
&=2\pi\int_0^1 \int_0^\pi\frac{r^2\sin(\theta)}{r^2+4r\cos(\theta)+4}\,d\theta\,dr \tag 1\\\\
&=\frac{\pi}{2}\int_0^1 r \log\left(\frac{(r+2)^2}{(r-2)^2}\right)\,dr \tag 2
\end{align}$$
Can you finish now?
Note that in going from $(1)$ to $(2)$ we enforced the substitution $x=\cos(\theta)$. Then, $dx=-\sin(\theta)\,d\theta$ and we have
$$\begin{align}
\int_0^\pi \frac{r^2\sin(\theta)}{r^2+4r\cos(\theta)+4}\,d\theta&=r^2\int_{-1}^1 \frac{1}{r^2+4rx+4}\,dx\\\\
&=\frac{r}{4}\left.\left(\log(r^2+4rx+4)\right)\right|_{-1}^1\\\\
&=\frac{r}{4}\left(\log(r^2+4r+4)-\log(r^2-4r+4)\right)\\\\
&=\frac{r}{4}\log\left(\frac{(r+2)^2}{(r-2)^2}\right)
\end{align}$$
|
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|
Integrate $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$$
$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{2}{3}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx = \frac{3}{2}[\ln(\tan \ x)^{\frac{2}{3}}]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \frac{3}{2}[\ln(\sqrt{3})^\frac{2}{3}-\ln(1)^\frac{2}{3}]=\frac{3}{2}[\ln(\sqrt{3})^\frac{2}{3}] = \bf \color{red}{\ln(\sqrt{3})}$$
However, the answer given is not this value, but instead $= \frac{3\sqrt[3]{3}-3}{2} \approx 0.663... $ while $\ln(\sqrt{3}) \approx 0.549...$
|
You have $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sec^2 x}{\sqrt[3]{\tan x}}dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{1}{\sqrt[3]{\tan x}}d\tan x$$
Replacing $t = \tan x$, you have the integration equals
$$\int_{1}^{\sqrt{3}} \frac{1}{\sqrt[3]{t}}dt = \frac{3}{2}t^{\frac{2}{3}} \mid_{1}^{\sqrt{3}} = \frac{3}{2}(\sqrt[3]{3}-1)$$
|
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|
Prove that $\displaystyle \sum_{1\leq kProve that
$$\sum_{1\leq k < j\leq n}\tan^2\left(\frac{k\pi}{2n+1}\right)\tan^2\left(\frac{j\pi}{2n+1}\right)=\binom{2n+1}{4}$$
|
It is enough to recall Cauchy's proof of the Basel problem, relying on the identity:
$$ \binom{2n+1}{1}t^n-\binom{2n+1}{3}t^{n-1}+\ldots+(-1)^n\binom{2n+1}{2n+1}=\prod_{k=1}^{n}\left(t-\cot^2\frac{k\pi}{2n+1}\right)\tag{1} $$
If we consider the "reciprocal polynomial"
$$ \binom{2n+1}{2n+1}t^n-\binom{2n+1}{2n-1}t^{n-1}+\ldots+(-1)^n\binom{2n+1}{1}=\prod_{k=1}^{n}\left(t-\tan^2\frac{k\pi}{2n+1}\right)\tag{2} $$
the original sum is just the second elementary symmetric polynomial of the roots, $e_2$, that by Vieta's theorem is given by the coefficient of $t^{n-2}$ in the LHS of $(2)$, so:
$$ \sum_{1\leq j<k\leq n}\tan^2\left(\frac{j\pi}{2n+1}\right)\tan^2\left(\frac{k\pi}{2n+1}\right)=\binom{2n+1}{2n-3}=\color{red}{\binom{2n+1}{4}}\tag{3}$$
as wanted.
|
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|
If $x+y+z=1$ and $x,y,z>0\,$ find $\min xy(x+y)^2+yz(y+z)^2+zx(z+x)^2$
If $x+y+z=1$ and $x,y,z>0\,$ find the minimum of $xy(x+y)^2+yz(y+z)^2+zx(z+x)^2$
$\bf{My\; Try::}$ We can write expression as $xy(1-z)^2+yz(1-x)^2+zx(1-y)^2$
$ = xy(1+z^2-2z)+yz(1+x^2-2x)+zx(1+y^2-2y)$
$ = xyz(x+y+z)+(xy+yz+zx)-6xyz= xy+yz+zx-5xyz$
Now how can I solve it after that, help required, thanks
|
The minimum does not exist, but arbitrarily small values can be reached by letting $x,y\to0$, $z\to1$; i.e., the infimum is $0$.
Explicitly, $x=y=\varepsilon$ and $z=1-2\varepsilon$ gives
$$xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\leq3\times\varepsilon\cdot1\cdot(1+1)^2=12\varepsilon.$$
|
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The integral $\int_0^1 x^n \log(2-x) \, dx$ In this question I had stuck on the integral:
$$\int_0^1 x^n \log(2-x) \, {\rm d}x$$
Claude Leibovici says that $\displaystyle \int_0^1 x^n \log(2-x) \, {\rm d}x=\frac{\, _2F_1\left(1,n+2;n+3;\frac{1}{2}\right)}{2 (n+1) (n+2)}=a_n+b_n \log 2$ where $a_n, b_n$ are sequences of rational numbers. The statement seems to be true since if we run the integral for some values of $n$ we get that:
$$\begin{array}{||c|c|c|c|c||}
\hline
\text{Integral}&n & a_n & b_n & \text{Value} \\
\hline
\int_0^1 x \log(2-x) \, {\rm d}x&1 & -\frac{5}{4} & 2 & -\frac{5}{4}+ 2 \log 2 \\\\
\int_0^1 x^2 \log(2-x) \, {\rm d}x&2 & -\frac{16}{9} & \frac{8}{3} & -\frac{16}{9} + \frac{8}{3} \log 2 \\\\
\int_0^1 x^3 \log(2-x) \, {\rm d}x&3 & -\frac{131}{48} & 4& -\frac{131}{48} + 4 \log 2 \\
\hline
\end{array}$$
It seems that $a_n$ is a sequence of negative rational numbers and $b_n$ a sequence of positive rational numbers. However I don't see a pattern of how to connect those numbers. Does anyone see?
Maybe it is worth it to take a look at the more general case:
$$\int_0^1 x^n \log(\alpha -x) \, {\rm d}x \; \quad \mathbb{N} \ni \alpha \geq 2$$
That would be more tedious. What I had come as a solution but I could not see how to proceed was something like this:
\begin{align*}
\int_{0}^{1}x^n \log(\alpha -x) \, {\rm d}x &= \int_{0}^{1}x^n \log \left [ \alpha \left ( 1 - \frac{x}{\alpha} \right ) \right ] \, {\rm d}x \\
&=\int_{0}^{1}x^n \left [ \log \alpha + \log \left ( 1- \frac{x}{\alpha} \right ) \right ] \, {\rm d}x \\
&=\log \alpha \int_{0}^{1} x^n \, {\rm d}x + \int_{0}^{1} x^n \log \left ( 1- \frac{x}{\alpha} \right ) \, {\rm d}x \\
&= \frac{\log \alpha}{n+1} - \int_{0}^{1} x^n \sum_{n=1}^{\infty} \frac{\left ( \frac{x}{\alpha} \right )^n}{n} \, {\rm d}x \\
&= \frac{\log \alpha}{n+1} - \sum_{n=1}^{\infty} \frac{1}{n \alpha^n} \int_{0}^{1} x^{2n} \, {\rm d}x\\
&= \frac{\log \alpha}{n+1} - \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n \alpha^n (2n+1)} \\
&= ??
\end{align*}
It seems that:
$$\sum_{n=1}^{\infty} \frac{1}{n \alpha^n (2n+1)} = -2\sqrt{\alpha} \; {\rm arctanh} \left ( \frac{1}{\sqrt{\alpha}} \right ) - \log \left ( \frac{\alpha -1}{\alpha} \right ) +2$$
I guess that won't be too difficult to prove taking into account that:
$${\rm arctanh}(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1} $$
|
By integration by parts
$$ I_n=\int_{0}^{1}x^n\log(2-x)\,dx = \frac{1}{n+1}\int_{0}^{1}\frac{x^{n+1}}{2-x}\,dx \tag{1}$$
so it is enough to write $x^{n+1}$ as $(x^{n+1}-2^{n+1})+2^{n+1}$ to get:
$$\begin{eqnarray*} I_n &=& \frac{1}{n+1}\left(2^{n+1}\log 2-\int_{0}^{1}\sum_{k=0}^{n}2^k x^{n-k}\,dx\right)\\&=&\color{blue}{\frac{2^{n+1}}{(n+1)}\log 2}-\color{red}{\frac{1}{n+1}\sum_{k=0}^{n}\frac{2^k}{n-k+1}}.\tag{2}\end{eqnarray*} $$
|
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|
generating function: how many possibilities are there to throw 10 different dice so that their sum is 25 My discrete math textbook has the following solution to the problem using generating functions (the solution has to use generating functions):
$$f(x)=(x+x^2+x^3+x^4+x^5+x^6)^{10}$$
Once we start developing the formal power series we get the following:
$$x^{10}(1-x^6)^{10}\frac{1}{(1-x)^{10}}$$
Because of $x^{10}$ we now need to find the coefficient of only $x^{15}$:
$$\binom{10}{0}\binom{10-1+15}{15}-\binom{10}{1}\binom{10-1+9}{9}+\binom{10}{2}\binom{10-1+3}{3}$$
I completely understand the first two expressions but why are we using the third one:
$$\binom{10}{2}\binom{10-1+3}{3}$$
Where does the $3$ come from and why this isn't enough:
$$\binom{10}{0}\binom{10-1+15}{15}-\binom{10}{1}\binom{10-1+9}{9}$$
?
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The coefficient of $x^{25}$ in $(x+x^2+\ldots+x^6)^{10}$ equals the coefficient of $x^{15}$ in
$$ (1+x+\ldots+x^5)^{10} = (1-x^6)^{10}\cdot\frac{1}{(1-x)^{10}} $$
so the answer is given by
$$ [x^{15}] \left(\sum_{k=0}^{10}\binom{10}{k}(-1)^k x^{6k}\right)\cdot\left(\sum_{j\geq 0}\binom{9+j}{j}x^j\right) $$
and it is enough to consider just the terms associated with $k=0,1,2$, since $x^{6k}$ has an exponent greater than $15$ for any $k>2$. By computing the Cauchy product between the above series it follows that the answer is given by:
$$ \underbrace{\binom{10}{0}\binom{24}{15}}_{k=0,\;j=15}-\underbrace{\binom{10}{1}\binom{18}{9}}_{k=1,\;j=9}+\underbrace{\binom{10}{2}\binom{12}{3}}_{k=2,\;j=3}.$$
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How to prove that no prime factor of $x^2-x+1$ is of the form $6k-1$ Consider sequence $x^2-x+1$ ($1,3,7,13,21,31,43,57,73,91,\dots$). Let's consider prime factorization of each term.
$$3=3$$
$$7=7$$
$$13=13$$
$$21=3\times7$$
It seems that the only prime factors we ever get are 3 and those of the form $6k+1$. In fact, prime factorization of the first 10 000 terms of the sequence gives 7233 distinct primes and all of them (except 3) are $6k+1$.
That no member of the sequence is ever divisible by a prime of the form $6k-1$ is a purely empirical conjecture. Is there a formal proof for it (or a counterexample)?
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First, some facts about numbers in your sequence. Let $n=x^2-x+1$. If $x=3k-1$, then
\begin{align*}
n=x^2-x+1 &= (3k-1)^2-(3k-1)+1\\
&= 9k^2-6k+1-3k+1+1\\
&= 9k^2-9k+3\\
\implies n/3 &= 3k^2-3k+1
\end{align*}
$k^2$ and $k$ have the same parity. Therefore
$n/3=1\mod 6$. If $x$ is any other integer then $n=1\mod 6$ as can be easily checked. Thus $n$ is odd; and, after a factor of 3 has been extracted, if there is one, the result is 1, modulo 6. In any case $n\ne 5\mod 6$.
Suppose your conjecture is wrong, and let $p$ be the lowest prime counterexample, i.e. lowest prime $p\ne3$, not of the form $6k+1$, which divides a number that is in your sequence.
Let prime $p$ divide $n=x^2-x+1$. Then $p\ne 2$. Because $p\ne3$ and $p\ne1\mod 6$, $p=5\mod 6$.
Let $x=sp+r$ where $s$ is the nearest integer to $x/p$, so $|r|<p/2$. Then
\begin{align*}
n=x^2-x+1 &= (sp+r)^2-(sp+r)+1\\
&= s^2p^2+2srp+r^2-sp-r+1\\
&= Ap+r^2-r+1
\end{align*}
where $A=s^2p+2sr-s$. Thus
$p\mid r^2-r+1$, so, for some integer $m'$,
\begin{align*}
m'p &= r^2-r+1\\
&\leqslant r^2\\
&<p^2/4\\
\implies m' &< p/4.
\end{align*}
If $3\mid m'$, let $m=m'/3$, otherwise let $m=m'$. Then $3\nmid m$, and $m\leqslant m'<p/4$.
Because $m<p$ and $p$ is the smallest counterexample, each prime factor $q$ of $m$ is 1, modulo 6 (because $q\leqslant m<p$ so $q$ is not a counterexample). So, modulo 6, $m=1$, $p=5$, so $mp=5$. But $mp = r^2-r+1$ is in your sequence, so $mp\ne 5\mod 6$. This is a contradiction. (In fact, for any integer $m$ where $m=5\mod 6$, at least one prime factor of $m$ is $q=5\mod 6$.)
Therefore your conjecture is correct.
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$\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square
Prove that there exist infinitely many positive integers $n$ such that $\frac{1^2+2^2+\cdots+n^2}{n}$ is a perfect square. Obviously, $1$ is the least integer having this property. Find the next two least integers with this property.
The given condition is equivalent to $(2n+2)(2n+1) = 12p^2$ where $p$ is a positive integer. Then since $\gcd(2n+2,2n+1) = 1$, we have that $2n+2 = 4k_1$ and $2n+1 = k_2$. We must have that $k_1$ is divisible by $3$ or that $k_2$ is divisible by $3$. If $k_1$ is divisible by $3$ and $k_2$ is not, then we must have that $k_1$ is divisible by $9$ and so $2n+2 = 36m$. Then we need $3mk_2$ to be a perfect square where $k_2+1 = 36m$. Thus if $3mk_2 = r^2$, we get $m = \dfrac{1}{72}\left(\sqrt{48r^2+1}+1\right)$.
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Using this:
$$
\frac{\sum_{k=1}^N k^2}{N}= \frac{1}{6} (N+1) (2 N+1)=K^2
$$
the condition can be written as (as you pointed out):
$$(N+1) (2 N+1)=6K^2$$
through numerical search, the last part of the problem is solved
$$N=1,337,65521$$
Not really a satisfying answer i know.
|
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Question on the inequality Question.
prove that if ${ a }_{ 1 },{ a }_{ 2 },...{ a }_{ n }>0$ then $$ \frac { { a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ n } }{ n } \ge \frac { n }{ \frac { 1 }{ { a }_{ 1 } } +\frac { 1 }{ { a }_{ 2 } } +...+\frac { 1 }{ { a }_{ n } } } $$
Proof
$$\\ \left( { a }_{ 1 }+{ a }_{ 2 }+...+{ a }_{ n } \right) \left( \frac { 1 }{ { a }_{ 1 } } +\frac { 1 }{ { a }_{ 2 } } +...+\frac { 1 }{ { a }_{ n } } \right) =\underset { n\left( n-1 \right) /2\quad terms }{ \underbrace { \left( \frac { { a }_{ 1 } }{ { a }_{ 2 } } +\frac { { a }_{ 2 } }{ { a }_{ 1 } } \right) +...+\left( \frac { { a }_{ n-1 } }{ { a }_{ n } } +\frac { { a }_{ n } }{ { a }_{ n-1 } } \right) + } \quad n\ge } \\ \ge n+2\cdot \frac { n\left( n-1 \right) }{ 2 } ={ n }^{ 2 }$$
in this proof i didn't understand this step "$n\left( n-1 \right) /2\quad ?
terms$" I mean how the number of terms can be $n\left( n-1 \right) /2\quad $ .Can anybody explain it.
Thanks in advance!
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You have $a_i\cdot \frac1{a_i}$ for each $i$; these sum up to $n$.
Apart from that, you have for each of the $n\choose 2$ choices of $i<j$ the summands $\frac{a_i}{a_j}+\frac{a_j}{a_i}$, which is $\ge 2$
|
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Am I correct when I do this in questions about induction? I have a problem which is:
If $n$ is a fixed positive integer and greater than 1, show by induction that for each positive integer $r$, where $2\le r\le n$,
$$1-\frac{r(r-1)}{n} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots (1-\frac{r-1}{n})$$
Denote this inequality(proposition) with variable r by $(\theta)_r$
I proved that when $r=2$, the above inequality is true, since
$$1-\frac{2(1-1)}{n}=1-\frac{2}{n} \le 1-\frac{1}{n}$$
Then I suppose for some $2\le k \le n-1$, $(\theta)_k$ is true, i.e.
$$1-\frac{k(k-1)}{n} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k-1}{n})$$Then I multiply both sides by $(1-\frac{k}{n})$, and I arrived at
$$1-\frac{k^2}{n}+\frac{k^2(k-1)}{n^2} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k}{n})$$
After that I was stuck and I changed my proposition $(\theta)_r$ as follows:
$$1-\frac{r(r-1)}{n} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{r-1}{n})$$ and$$1-\frac{r(r+1)}{n} \le 1-\frac{r^2}{n}+\frac{r^2(r-1)}{n^2}$$
Then new proposition $(\theta_0)_r$ with variable $r$ holds when $r=2$, since
$$1-\frac{2(2+1)}{n}-1+\frac{2^2}{n}-\frac{2^2(2-1)}{n^2}=\frac{-4-2n}{n^2} \le 0$$
Then I suppose for some $2\le k \le n-1$, $(\theta_0)_k$ is true and go back to
$$1-\frac{k^2}{n}+\frac{k^2(k-1)}{n^2} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k}{n})$$ $$ \Rightarrow1-\frac{k(k+1)}{n}\le 1-\frac{k^2}{n}+\frac{k^2(k-1)}{n^2} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{k}{n})$$
And I conclude, by induction, that $$1-\frac{r(r-1)}{n} \le (1-\frac{1}{n})(1-\frac{2}{n})\cdots(1-\frac{r-1}{n})$$ holds for all positive integer $r$, where $2\le r \le n$
Am I correct when I modify the proposition in this way?
I feel that it is true and I can't see any logical fallacies so far...
Please correct me if this is wrong
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Maybe you could prove the inequality,
$$1-\frac{r(r+1)}{n} \le 1-\frac{r^2}{n}+\frac{r^2(r-1)}{n^2}$$
directly (without induction). Infact by expanding you obtain
$$1-\frac{r^2}{n}-\frac{r}{n}\le 1-\frac{r^2}{n}+\frac{r^3}{n^2}-\frac{r^2}{n^2},$$
that is, $-1\le \frac{r^2}{n}-\frac{r}{n},$ or $-n\leq r(r-1)$ which trivially holds because $2\le r\le n$.
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A question about substitute equivalent form into limit: $\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$ When I had the calculus class about the limit, one of my classmate felt confused about this limit:
$$\lim_{x\to \infty} \sqrt{x^2+x+1}-\sqrt{x^2+3x+1} = -1$$
What he thought that since $x^2 > x$ and $x^2 > 3x$ when $x \to \infty$ so the first square root must be $x$ and same for the second. Hence, the limit must be $0$.
It is obviously problematic.
And what I thought is that make prefect square under the limit, though I know the right solution is to rationalize the numerator.
After perfect-squaring, $$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}$$ I assert that since there is a perfect square and a square root. As $x \to \infty$, the constant does not matter. So
$$\lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}-\sqrt{\left(x+\frac{3}{2}\right)^2-\frac{5}{4}}= \lim_{x\to \infty} \sqrt{\left(x+\frac{1}{2}\right)^2}-\sqrt{\left(x+\frac{3}{2}\right)^2}\\ = \frac{1}{2}-\frac{3}{2} = -1 $$
But when I conquer this limit: this post.
Here is my argument:
$\sqrt{ax^2+bx+c} =O(\sqrt a(x+\frac{b}{2a}))$ when $x \to \infty$
$$\lim_{x\to\infty}\frac{\sqrt{x^2-2x+3}+\sqrt{4x^2+5x-6}}{x+\sqrt{x^2-1}} =\frac{x-1+2(x+5/4)}{x+x} = \frac{3}{2}$$
Very concise get this answer but it gets downvoted.
I do not know what is wrong with my strategy. But in general case: for instance this problem about cubic root my strategy seems to work really efficient:
$$\lim_{x\to \infty} \sqrt[3]{x^3+6x^2+9x+1}-\sqrt[3]{x^3+5x^2+x+1}$$
My solution is:
$$\lim_{x\to \infty} \sqrt[3]{x^3+bx^2+cx+d} = \lim_{x\to \infty} \sqrt[3]{\left(x+\frac{b}{3}\right)^3}$$
So the limit becomes: $$\lim_{x\to \infty} \sqrt[3]{(x+2)^3+O(x)}-\sqrt[3]{\left(x+\frac{5}{3}\right)^3 + O(x)} =\lim_{x\to \infty} (x+2) -\left(x+\frac{5}{3}\right) = \frac{1}{3} $$
This result gets verified by wolframalpha.
To put all into a nutshell, what is wrong with my solution to these three problem. Is there any counterexample to this substitution. Any help, you will be appreciated.
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@ZackNi As David indicated, you are assuming "infinity minus infinity = 0,which works out well because both radicals are squareroots and the headcoefficients of the polynomials are both 1.You can also get"infinity minus infinity = 0 from a squareroot and cuberoot, but using your approach will then yield to a wrong answer to the limit.
Generally, $\infty-\infty$ and $\infty*0$ situations should be converted into situations like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ on which other techniques can be applied to get the desired limit. In your problem the conjugate approach is also a good way to go. (My comment post got messed up and I can't delete it...)
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Definite integral and limit $\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$ Given $I_{n} = \int_{0}^{1} x^{n} \arctan(x)dx $
Calculate:
$\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$
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By integration by parts you have that
$$I_{n} = \int_{0}^{1} x^{n} \arctan(x)dx=\frac{1}{n+1}\left[x^{n+1}\arctan(x)\right]_0^1-\frac{1}{n+1}\int_{0}^{1} \frac{x^{n+1}}{1+x^2} dx\\=\frac{1}{n+1}\left(\frac{\pi}{4}-\int_{0}^{1} \frac{x^{n+1}}{1+x^2} dx\right)$$
Hence
$$(n+1)I_{n}-\frac{\pi}{4}=-\int_{0}^{1} \frac{x^{n+1}}{1+x^2} dx.$$
Now by using the same trick, you find that
$$\int_{0}^{1} \frac{x^{n+1}}{1+x^2} dx =\frac{1}{2(n+2)}+\frac{2}{n+2}\int_{0}^{1} \frac{x^{n+3}}{(1+x^2)^2}dx.$$
Finally
$$\lim_{n\rightarrow\infty} n\left(\left(n+1\right)I_{n}-\frac{\pi}{4}\right)=-\frac{1}{2}+\lim_{n\rightarrow\infty}\int_0^1 x^nf(x)dx=-\frac{1}{2}$$
where $f$ is a continuous function and
$$\left|\int_0^1 x^nf(x)dx\right|\leq \max_{[0,1]}|f(x)|\int_0^1 x^ndx\leq \frac{\max_{x\in [0,1]}|f(x)|}{n+1}\to0.$$
P.S. By this procedure you can find more terms in the expansion of the infinitesimal sequence $I-n$. So far we have that
$$I_n=\frac{\pi}{4(n+1)}-\frac{1}{2n(n+1)}+o(1/n^2).$$
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Functional equation of type linear expression
If $f:\mathbb{R}\rightarrow \mathbb{R}$ and $f(xy+1) = f(x)f(y)-f(y)-x+2\;\forall x, y\in \mathbb{R}$ and $f(0) = 1\;,$ Then $f(x) $ is
$\bf{My\; Try::}$ Put $x=y=0\;,$ We get $f(1) = (f(0))^2-f(0)-0+2 = 1-1+2=2$
Similarly put $x=y=1\;,$ We get $f(2) = (f(1))^2-f(1)-1+2 = 4-2-1+2 = 3$
So from above values function must be in the form of $f(x) = x+1$
But i did not understand how can i calculate it.
Help required, Thanks
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$$f(xy+1) = f(x)f(y)-f(y)-x+2$$
Let $y=0$. Then $$f(1)=f(x)f(0)-x+2$$
Let $f(1)=a$. Then
$$f(x)=x+a-2$$
Check:
$xy+1+a-2=(x+a-2)(y+a-2)-(y+a-2)-x+2$
$xy+a-1=xy+ax-2x+ay+a^2-2a-2y-2a+4-y-a+2-x+2$
$(a-2-1)x+(a-2-1)y+(a^2-6a+9)=0$
$(a-3)x+(a-3)y+(a-3)^2=0$ $(\forall x,y \in \mathbb R)$
Then $a=3$
Answer: $$f(x)=x+1$$
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Given $\tan a = \frac{1}{7}$ and $\sin b = \frac{1}{\sqrt{10}}$, show $a+2b = \frac{\pi}{4}$.
Given
$$\tan a = \frac{1}{7} \qquad\text{and}\qquad \sin b = \frac{1}{\sqrt{10}}$$
$$a,b \in (0,\frac{\pi}{2})$$
Show that $$a+2b=\frac{\pi}{4}$$
Does exist any faster method of proving that, other than expanding $\sin{(a+2b)}$?
Thank You!
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So, $\cos b=+\sqrt{1-\left(\dfrac1{\sqrt{10}}\right)^2}=\dfrac3{\sqrt{10}}$
$\implies\tan b=\dfrac{\dfrac1{\sqrt{10}}}{\dfrac3{\sqrt{10}}}\implies b=\arctan \dfrac13$
Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$$2b=\arctan\dfrac{2\cdot\dfrac13}{1-\left(\dfrac13\right)^2}=\arctan\dfrac34$$
Again as $a=\arctan\dfrac17$
Again Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$a+2b=\arctan\dfrac17+\arctan\dfrac34=\arctan\dfrac{\dfrac17+\dfrac34}{1-\dfrac17\cdot\dfrac34}=\arctan1=\dfrac\pi4$
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If $\cos x + \cos y - \cos(x+y) = \frac{3}{2}$, then how are $x$ and $y$ related?
If $$\cos x + \cos y - \cos(x+y) = \frac{3}{2}$$ then
*
*(a) $x + y = 0$
*(b) $x = 2 y$
*(c) $x = y$
*(d) $2 x = y$
It is problem of trigonometry, and I have the solution of the problem. However, after seeing the solution, I don't quite understand how is one supposed to know how to approach this problem. My request is:
Don't just solve the problem, but also tell me from where do you come to know which approach would work. (Tell me from where do you come to know which trick or approach to be used to solution this problem.)
Thanks in advance.
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Using Prosthaphaeresis Formula,
$$\cos x+\cos y=2\cos\dfrac{x+y}2\cos\dfrac{x-y}2$$
Double angle formula says:
$$\cos(x+y)=2\cos^2\dfrac{x+y}2-1$$
So we have $$2\cos^2\dfrac{x+y}2-2\cos\dfrac{x+y}2\cos\dfrac{x-y}2+\dfrac12=0\ \ \ \ (1)$$ which is a Quadratic Equation in $\cos\dfrac{x+y}2$ which is real,
so the discriminant $\left(2\cos\dfrac{x-y}2\right)^2-4=-4\sin^2\dfrac{x-y}2$ must be $\ge0$
So we must have $\sin^2\dfrac{x-y}2=0\implies\dfrac{x-y}2\equiv0\pmod\pi\implies x\equiv y\pmod{2\pi}$
Consequently, $(1)$ becomes, $$2\cos^2\dfrac{x+y}2-2\cos\dfrac{x+y}2+\dfrac12=0\ \ \ \ (2)\implies\cos\dfrac{x+y}2=\dfrac12$$
$$\implies\dfrac{x+y}2=2m\pi\pm\dfrac\pi3\iff\cdots$$
where $m$ is any integer
Can you take it from here?
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Problem solving recurrence equation for $a_n = a_{n-1} + 2a_{n-2}, \,\,\, a_0=a_1=1$ I want to try the generating function $A(x)$ and a formula for $a_n$ for the following sequence:
$$a_n = a_{n-1} + 2a_{n-2}, \,\,\, a_0 = a_1 = 1$$
So I multiplied the above equation on both sides by $x^n$ and summed them for each $n \geq 2$. So:
$$\sum_{n=2}^{\infty} a_nx^n = \sum_{n=2}^{\infty} a_{n-1}x^n + 2\sum_{n=2}^{\infty} a_{n-2}x^n$$
$$\Longrightarrow \sum_{n=2}^{\infty} a_nx^n= \frac{1}{x}\sum_{n=1}^{\infty} a_{n}x^n + 2\frac{1}{x^2}\sum_{n=0}^{\infty} a_{n}x^n$$
$$\Longrightarrow A(x) - x - 1 = \frac{A(x)-1}{x} + 2\frac{A(x)}{x^2}$$
$$\Longrightarrow x^2A(x) - x^3 - x^2 = xA(x)-x + 2A(x)$$
$$\Longrightarrow A(x)(x^2-x-2) = x^3 + x^2 + x$$
$$\Longrightarrow A(x) = \frac{x(x^2 + x + 1)}{(x-2)(x+1)} = \frac{1}{3} \cdot \frac{1}{1+x} + \frac{14}{3} \cdot \frac{1}{2-x}$$
Now if I search for the series expansion, I find:
$$A(x) = \sum_{n=0}^{\infty} (\frac{1}{3} \cdot (-1)^n - \frac{14}{3} \cdot \frac{1}{2^{n+1}})x^n$$
$$\Longrightarrow a_n = \frac{1}{3} \cdot (-1)^n - \frac{14}{3} \cdot \frac{1}{2^{n+1}}$$
However, this is obviously wrong. I checked on Wolfram, and the correct answer should be:
$$a_n = \frac{1}{3} ((-1)^n+2^{n+1})$$
I redid the whole thing twice, but I got the same result, so I guess that there is some flaw in my logic, rather than the calculations (I could be wrong though).
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As the recurrence relation is linear with constant coefficients, let us look for a solution of the form $a_n = z^n$. We obtain
$$z^n (z^2 - z - 2) = 0$$
Given the nonzero initial conditions, $z^n \neq 0$. Hence, $z^2 - z - 2 = 0$. The solutions are $-1$ and $2$.
$$a_n = \beta_1 (-1)^n + \beta_2 2^n$$
From the initial conditions $a_0 = a_1 = 1$, we obtain $\beta_1 = \frac 13$ and $\beta_2 = \frac 23$. Thus,
$$a_n = \frac 13 \left( (-1)^n + 2^{n+1} \right)$$
|
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|
Given a matrix $B$, what is $det(B^4)$? My task is this:
Compute det$(B^4)$, where
$$B =\begin{pmatrix}
1 & 0 & 1\\
1 & 1 & 2\\
1 & 2 & 1
\end{pmatrix}$$
My work so far:
It can be shown that
$B =
\begin{pmatrix}
1 & 0 & 1\\
1 & 1 & 2\\
1 & 2 & 1
\end{pmatrix} \sim \begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 2\\
0 & 0 & -4
\end{pmatrix}\implies B$ is invertable.
Applying theorem for determinants to triangular matrices we get det$(B) = 1*1*(-4)=-4$.
Appling the multiplicative property theorem, we should get that det$(B^4)=$det$(B)$det$(B)$det$(B)$det$(B)=[$det$(B)]^4=(-4)^4= 256.$
Now I need help from the community to verify if the reasoning is correct, because I've already had a fellow student telling me that the answer is 16, and indeed the textbook I'm using states the same, which makes me abit confused. So please point out my errors or verify that each step holds true.
Thanks in advance.
|
I don't know how you have obtained
$$\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 2\\
0 & 0 & -4
\end{pmatrix}$$
But making elementary operations ($R2\leftarrow R2-R1$, $R3\leftarrow R3-R1$) I obtain
$$\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 1\\
0 & 2 & 0
\end{pmatrix}$$
and then ($R3\leftarrow R3-2\cdot R2$):
$$\begin{pmatrix}
1 & 0 & 1\\
0 & 1 & 1\\
0 & 0 & -2
\end{pmatrix}$$
|
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|
Proof for $\forall n\in \mathbb{N}: 7\mid(1 + 2^{2^n} + 2^{2^{n+1}})$ I am stuck at the following exercise:
Prove that
$\forall n\in \mathbb{N}: 7\mid(1 + 2^{2^n} + 2^{2^{n+1}})$.
I tried to prove the Expression by induction but I cannot find a way to prove the implication
$$7\mid(1 + 2^{2^n} + 2^{2^{n+1}}) \Rightarrow 7\mid(1 + 2^{2^{n+1}} + 2^{2^{(n+1)+1}}).$$
Any help would be appreciated very much.
|
Induction will work here. Assume $1+2^{2^n} + 2^{2^{n+1}}$ is divisible by 7.
Note that $2^{2^{n+2}} = 2^{2^n4} = 16^{2^n} = (14+2)^{2^n} = 14M+2^{2^n}$ for some integer $M$. Then $1+2^{2^{n+1}} + 2^{2^{n+2}} = 1+2^{2^{n+1}}+(14+2)^{2^n} = 1+2^{2^{n+1}} +2^{2^n}+14M$, which must be a multiple of 7, since the first three terms are from the induction hypothesis.
|
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|
Prove that $ABC$ is right-angled Prove that if $\cos^2{A} + \cos^2{B} + \cos^2{C} = 1$, then $ABC$ is right-angled.
I only found that $\sin^2{A} + \sin^2{B} + \sin^2{C} = 2$, but I have no idea what to do next.
Thank you in advance for your answers!
|
HINT:
$$F=\cos^2A+\cos^2B+\cos^2C=\cos^2A-\sin^2B+\cos^2C+1$$
Now using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$F=\cos(A+B)\cos(A-B)+\cos^2C+1$$
Now as $\cos(A+B)=\cos(\pi-C)=-\cos C,$
$$F=-\cos C\cos(A-B)+\cos C\{-\cos(A+B)\}=1-\cos C(2\cos A\cos B)$$
|
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|
Find $\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}, x \gt 4$
Find
$$\sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}}, x \gt 4$$
My attempt :-
$$3(x-1)+2\sqrt{2x^2 - 7x - 4}$$
$$\implies 3x - 3 + 2\sqrt{2x^2 - 7x - 4} + 2x^2 - 2x^2 -7x + 7x - 1 +1 $$
$$\implies 2x^2 -7x - 4 + 2\sqrt{2x^2 - 7x - 4} + (- 2x^2 + 10x + 1) $$
let $y = \sqrt{2x^2 - 7x - 4}$
$$\therefore y^2 + 2y + (- 2x^2 + 10x + 1)$$
solving for $y$ :-
$$y = {-2\pm \sqrt{4 - 4\times(- 2x^2 + 10x + 1)} \over 2} $$
$$\implies y = {-1\pm \sqrt{2x^2 - 5x}} $$
$$\therefore y^2 + 2y + (- 2x^2 + 10x + 1) = (y + 1 + \sqrt{2x^2 - 5x})(y + 1 - \sqrt{2x^2 - 5x})$$
$$\therefore \sqrt{{3\over2}(x-1)+\sqrt{2x^2 - 7x - 4}} = \sqrt{(y + 1 + \sqrt{2x^2 - 5x})(y + 1 - \sqrt{2x^2 - 5x})\over 2}$$
Sadly this does not remove the square root.
So the question is how can i factorize $3(x-1)+2\sqrt{2x^2 - 7x - 4}$ to remove that square root ?
just some hints are fine with me, thanks.
|
Hint: $2 x^2 - 7 x - 4 = (2x+1)(x-4)$, so you may guess that the desired result is of the form $a \sqrt{2x+1} + b \sqrt{x-4}$. Now square that, and see what $a$ and $b$ would give you $\dfrac{3}{2}(x-1) + \sqrt{2x^2-7x-4}$.
|
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|
If $7$ Divides $\binom{2^n}{2}+1$ then $n =3k+2$ for some positive integer $k$ It is a straightforward question.
If $$ 7 \text{ }\Bigg | \binom{2^n}{2}+1$$ then $n=3k+2$ for some
positive integer $k$.
This is just curiosity no motivation just rummaging through some old question in a notebook. A simple counter example would work.
Note if $\mathfrak a(n) =\binom{2^n}{2}+1$ then $7$ divides $\mathfrak a(n)$ for $n=2,5,8,11,14,17,\ldots$ and from this I am guessing that $n=3k+2$
|
The binomial coefficient is $2^n (2^n -1)/2 = 2^{n-1} (2\cdot 2^{n-1} -1)$.
For the divisibility to hold you need this to be $-1 \pmod{7}$.
Now the sequence nonnegative powers of $2$ modulo $7$ have a period of at most $6$, by Fermat's little theorem, but actually it is just has period $3$, and it starts
$2^0 = 1, 2^1 = 2, 2^2 =4, 2^3 =8\equiv 1, 2, 4, \dots$.
So $2^{3k}\equiv 1$, $2^{3k+1}\equiv 2$, and $2^{3k+2}\equiv 4$ modulo $3$.
It remains to check if you have the desire congruence
$$2^{n-1} (2\cdot 2^{n-1} -1) \equiv -1 \pmod{7}$$
somewhere.
Since
$$2 (2\cdot 2 -1) \equiv -1 \pmod{7}$$
Indeed you have it for $2^{n-1} \equiv 2 \pmod{7}$, and only for those,
that is $n-1 \equiv 1 \pmod{3}$, which is just what you claim.
|
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|
How to prove each element of the following sequence is a perfect square? Sequence $\{a_n\}$ satisfies the following formula:
$a_{n+2}=14a_{n+1}-a_n+12$,
and $a_1=1, a_2=1$.
It is easy to check that $a_3=25$ and $a_4=361$.
The question is how to prove each element of the sequence $\{a_n\}$ is a perfect square?
|
The above proofs are elegant. The following is what I was thinking:
Like what Steven did, we can use generation functions and write out $$a_n = c(2+\sqrt{3})^{2n} + d(2-\sqrt{3})^{2n} - 1$$ for some constants $c$ and $d$. Note that $c>0$ since $a_n$ increases as $n$ increases. If $d>0$, then $a_n = (\sqrt{c}((2+\sqrt{3})^n - \sqrt{d}(2-\sqrt{3})^n)^2 + (2\sqrt{cd} - 1 )$, and we want to show $ 2\sqrt{cd} - 1=0$, or $cd=1/4$.
Write $\sigma = (2+\sqrt{3})$, from $a_1=a_2$ we have $c\sigma^2 + d\sigma^{-2} = c\sigma^4 + d\sigma^{-4}$, which simplifies to $d=c\sigma^6$; then $d>c>0$. We also have $2=a_1+1= c\sigma^2 + d\sigma^{-2} = c\sigma^2(1+\sigma^2) = c\sigma^2*4\sigma$, so $c\sigma^3= 1/2$, which yields $2\sqrt{cd} - 1= 2c\sigma^3-1= 0$. We then have $a_n=b_n^2$, where $b_n=\sqrt{c}(2+\sqrt{3})^n - \sqrt{d}(2-\sqrt{3})^n$. The sequence $\{b_n\}$ are integers since $b_1=b_2=1$ and $b_{n+2} = 4b_{n+1}-b_n$.
|
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|
Evaluate $x+y=xy=3$ how would you evaluate $x^4+y^4$? I know how to evaluate $x^3 +y^3$ when $x+y=xy=3$, but how would you evaluate for?
$$x^4+y^4$$
Any help would be appreciated.Thanks in advance!
|
You can express $x^4+y^4$ in terms of $s=x+y$ and $p=xy$:
$$
x^4+y^4=as^4+bs^2p+cp^2
$$
(the degree of $s$ is $1$ and the degree of $p$ is $2$). We can determine $a,b,c$ by using particular values for $x$ and $y$.
*
*If $x=1$ and $y=0$, then $x^4+y^4=1$, $s=1$ and $p=0$
*If $x=1$ and $y=1$, then $x^4+y^4=2$, $s=2$ and $p=1$
*If $x=1$ and $y=-1$, then $x^4+y^4=2$, $s=0$ and $p=-1$
\begin{cases}
1=a \\[4px]
2=16a+4b+c \\[4px]
2=c
\end{cases}
Thus $a=1$, $b=-4$ and $c=2$:
$$
x^4+y^4=(x+y)^4-4xy(x+y)^2+2(xy)^2
$$
(you can check the correctness by expanding the right-hand side).
In the case $x+y=xy=3$, we have
$$
x^4+y^4=3^4-4\cdot 3\cdot3^2+2\cdot3^2=-9
$$
A different approach is to compute $x$ and $y$, which are the roots of $z^2-3z+3=0$, so we can use
$$
x=\frac{3+i\sqrt{3}}{2},\qquad y=\bar{x}=\frac{3-i\sqrt{3}}{2}
$$
The modulus of $x$ and $y$ is $\sqrt{x\bar{x}}=\sqrt{3}$, so
$$
x=\sqrt{3}\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)
=\sqrt{3}e^{i\pi/6},
\qquad
y=\sqrt{3}\left(\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)=
\sqrt{3}e^{-i\pi/6}
$$
Therefore
$$
x^4+y^4=9e^{2i\pi/3}+9e^{-2i\pi/3}=18\cos\frac{2\pi}{3}=-\frac{1}{2}\cdot18=-9
$$
|
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|
Positive number whose square exceeds its cube by the greatest amount
Find the positive number whose square exceeds its cube by the greatest amount.
Could you please help me with this question? It seems pretty easy at first glance, but I struggled to produce any answer. Thank you in advance!
UPD:
Sorry, I didn't mention, but the exercise is to solve it without any differentiation.
|
Here is a method that does not require calculus:
Lemma: If $x+y+z = k$ (where $k$ is a constant), the product $xyz$ is maximized when $x=y=z= \dfrac k3 $
Proof:
By $AM-GM$, we have
$$\dfrac {x+y+z}{3} \ge \sqrt[3] {xyz}$$
$$\dfrac {k}{3} \ge \sqrt[3] {xyz}$$
$$\left(\dfrac {k}{3}\right)^3 \ge xyz$$
We can see that the maximum of $xyz$ is $\left(\dfrac {k}{3}\right)^3$, with equality occurring $\iff x=y=z$
$$\left(\dfrac {k}{3}\right)^3 = x\cdot x \cdot x$$
$\implies x = \dfrac k3 $
$\implies x = y = z = \dfrac k3$
Remark: This lemma can be generalized for any $n$.
Now we want to find the maximum of $x^2 - x^3$, or equivalently
$$ f(x) = x\cdot x \cdot (1-x)$$
The maximum of $f$ occurs at the same $x$ value as the maximum of
$$g(x) = 2\cdot f(x) = x \cdot x \cdot (2-2x)$$
But notice that the factors in $g$ have a constant sum:
$$(x) + (x) + (2-2x) = 2$$
By the lemma, the product $x \cdot x \cdot (2-2x)$ reaches its maximum when $x = x = (2-2x) = \dfrac 23$
We can see that $x = \dfrac 23$, and the maximum is $$\left(\dfrac 23 \right )^2 - \left( \dfrac 23 \right)^3 = \dfrac {4}{27}$$
|
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|
Why is solution to inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ equal to interval $[0, \frac{3 - \sqrt{5}}{6})$? Given inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ we can easily determine, that it's domain is $D = [0, 1]$. Because each term is real, we can take square of the inequality, which yields:
$$\frac{1}{3} > \sqrt{x}\sqrt{1 -x}.$$
Squaring it again we get inequality:
$$9x^2 - 9x + 1 > 0,$$
solution to which is a domain $[0, \frac{3 - \sqrt{5}}{6}) \cup (\frac{3 + \sqrt{5}}{6}, 1]$. But, solution to original inequality is just $[0, \frac{3 - \sqrt{5}}{6})$. What am I omitting/where I'm doing mistakes?
|
Observe first that if must be $\;0\le x\le 1\;$ , so the given interval is within this constraint. Second, square:
$$\sqrt{1-x}-\sqrt x\ge\frac1{\sqrt3}\implies1-x-2\sqrt{x(1-x)}+x\ge\frac13\implies$$
$$2\sqrt{x(1-x)}\le\frac23\implies x(1-x)\le\frac19\implies 9x^2-9x+1\ge0\implies$$
The last quadratic's roots are given by
$$x_{1,2}=\frac{9\pm\sqrt{45}}{18}=\frac{3\pm\sqrt5}6\implies$$
$$ 9x^2-9x+1\ge0\iff \left(x-\frac{3-\sqrt5}6\right)\left(x-\frac{3+\sqrt5}6\right)\ge0\iff$$
$$x\le\frac{3-\sqrt5}6\;\;\text{or}\;\;x\ge\frac{3+\sqrt5}6\;\;\;\color{red}{(***)}$$
and we get what we want from the leftmost inequality.
Added on request: If we take
$$f(x)=\sqrt{1-x}-\sqrt x\implies f'(x)=-\frac1{2\sqrt{1-x}}-\frac1{2\sqrt x}=-\frac12\left(\frac1{\sqrt{1-x}}+\frac1{\sqrt x}\right)<0\implies$$
$\;f\;$ is monotone descending, and since
$$f\left(\frac{3-\sqrt5}6\right)=\sqrt{1-\frac{3-\sqrt5}6}-\sqrt\frac{3-\sqrt5}6=\frac1{\sqrt6}\left(\sqrt{3+\sqrt5}-\sqrt{3-\sqrt5}\right)\le\frac1{\sqrt3}\;(*)$$
since
$$(*)\iff \left(\sqrt{3+\sqrt5}-\sqrt{3-\sqrt5}\right)^2\le\left(\sqrt\frac63\right)^2\iff6-4\le2\;\;\color{green}\checkmark$$
so in $\;\color{red}{(***)}\;$ only the left inequality applies.
|
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Constructing a base $b$ repunit that is $a$ $\pmod p$ Suppose I want to construct a number of the form ($b^n-1$)/($b-1$) $=$ $a$ $\pmod p$ with $n$ and $p$ prime. What are the limitations on the base $b$ $>$ $1$ and prime $n$?
For example, if ($b^n-1$)/($b-1$) $=$ $4$ $\pmod 5$, what are possible choices for $b$ and (prime) $n$ in modular form?
I was able to find that $b$ nor prime $n$ $≠$ $2, 3, 5, 7$ or $11$. As an example that there exists such a number with $b > 1$ and $n$ prime: ($6^{19}-1$)/$5$ $=$ $4$ $\pmod 5$.
|
$(b^n-1) = (b-1)(b^{n-1} + b^{n-2} + ... + b^2 + b + 1)$
So (assuming $b \neq 1$!) it might be easier to directly consider
$$(b^{n-1} + b^{n-2} + ... + b^2 + b + 1) = a \mod p$$
If $p | b$, the above is true only if $a = 1$; and in that case it is true for all $n$, prime or not. So let's assume that $p$ is not a factor of $b$.
In the example you gave with $b=6, p=5$, we get that $b = 1 \mod p$, so your equation becomes $n = a \mod p$.
We only care about the modulo result for $b$; i.e, if it works for some $b$, it will also work for $(b+p)$, and similarly for $n$ (assuming it's still a prime!).
So, e.g., $(11^{19}-1)/10 = 4 \mod 5$ as well - what matters is that $b = 1 \mod 5$; then every $n$ (prime or not) satisfying $n = 4 \mod 5$ also works (e.g., $(31^{24}-1)/30 = 4 \mod 5$)
What else can be said? Let's say $0 < b < p$. Then $b^{p-1}=1 \mod p$; and the terms in your sum begin repeating every $p-1$ powers. So given $b$ and $p$, let $$S_k = \sum_{i=0}^{k-1}b^i \mod p$$
(with $S_0 = 0$, the 'empty sum'). Suppose $n = d(p-1) + m$, where $0 <= m < p-1$ (so $m = n \mod (p-1)$). Then
$$(b^{n-1} + b^{n-2} + ... + b^2 + b + 1) = S_n = d \cdot S_{p-1} + S_m \mod p$$
For $p = 5$, there are the following possibilities (in each case, $S_{p-1} = S_4$)
$$b=1 \rightarrow S_n = n $$
$$b=2 \rightarrow S_1=1, S_2= 3, S_3=2, S_4=0 $$
$$b=3 \rightarrow S_1=1, S_2= 4, S_3=3, S_4=0 $$
$$b=4 \rightarrow S_1=1, S_2= 0, S_3=1, S_4=0 $$
From this we see that for $p=5, a = 4$, the only solutions are:
$(b=1, n=4 \mod 5)$ and $(b=3, n=2 \mod 4)$. Note than when $b$ is such that $S_{p-1} = 0$,
$$S_n = d \cdot S_{p-1} + S_m = S_m = S_{n \mod p-1}$$
so we just look for those $b$ with some $S_m = a$. Of course, the only prime $n = 2 \mod 4$ is 2! So for example $(13^2 -1)/12 = 4 \mod 5$.
So we can also find the conditions for other values of $a$ from the above; for example for $a = 3$, we have: $$(b=1, n=3 \mod 5)$$ $$(b=2, n= 2 \mod 4)$$ $$(b=3 , n = 3 \mod 4)$$ and no solutions for $b = 4$.
For larger values of prime $p$, you can use the same approach. Some observations:
If $b = 1 \mod p$, then you can always look for primes of the form $n = a \mod p$.
If $b \neq 1 \mod p$, then $S_{p-1} = 0$ (proof omitted); in which case you always get:
$$S_n = d \cdot S_{p-1} + S_m = S_m$$
so you need only look for those $b$ such that $S_m = a \mod p$, where $n = m \mod p-1$.
|
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|
$4$ digit integers such that none of the digits appear more than twice
How many $4$ digit integers can be chosen such that none of the digits
appear more than twice?
$4$-digit integers with distinct digits= $9\times9\times8\times7 = 4536$
$4$-digit integers with two alike digits $(XXYY) = 9\times9\times2\times1\times \cfrac{4!}{2!\times2!} = 972$
Total = $5508$
Is the solution correct?
|
Direct approach. There are three possibilities. The first is ABCD, where all four digits are distinct. There are $9 \times 9 \times 8 \times 7 = 4536$ of these.
The second is AABC/ABAC/ABCA/BAAC/BACA/BCAA. Each of these has $9$ choices for the first position, and then $9 \times 8$ choices for the other two distinct digits. There are thus $6 \times 9 \times 9 \times 8 = 3888$ of these.
Finally, we have AABB/ABAB/ABBA. Each of these has $9$ choices for A and $9$ choices for B, so there are $3 \times 9 \times 9 = 243$ of these.
The total is $4536+3888+243 = 8667$.
If we allow leading zeros, the totals are $5040+4320+270 = 9630$.
Alternative approach. Count up the number of numbers that don't qualify, and then subtract from $9000$ (the number of numbers from $1000$ through $9999$, inclusive).
There are just $9$ numbers that have four of a kind: $1111, 2222, \ldots, 9999$.
There are also $9$ numbers that have three of a kind with three $0$'s: $1000, 2000, \ldots, 9000$.
For numbers that have three of a kind with three of a digit other than $0$: There are $9$ choices. If the other digit is in the first position, it can be anything other than the chosen digit or $0$, so there are $8$ choices. Otherwise, there are $9$ choices, for a total of $8+9+9+9 = 35$ possibilities for the other digit and its placement. This therefore accounts for $9 \times 35 = 315$ further disallowed numbers.
The total number of disallowed numbers is $9+9+315 = 333$, so the number of qualifying numbers must be $8667$.
If we allow leading zeros, the total number of possibilities increases to $10000$, and the number of disallowed numbers is $10+10 \times 4 \times 9 = 370$, so the number of permitted numbers is $9630$.
|
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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.