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How to factorize $2ab^2+2b^2c-ac^2-bc^2-a^2b-a^2c$? If $a,b,c$ are in AP $\implies \frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ are in AP $\implies
\frac{ab+bc+ca}{bc}, \frac{ab+bc+ca}{ca}, \frac{ab+bc+ca}{ab}$ are in AP $\implies \frac{bc+ca}{bc},\frac{ab+bc}{ca},\frac{bc+ca}{ab}$ are in AP $\implies \frac{bc}{bc+ca}, \frac{ca}{ab+bc}, \frac{ab}{bc+ca}$
are in HP.
If $P,Q,R$ are in HP $\implies Q=\frac{2PR}{P+R}.$ By this method proving that $\frac{bc}{bc+ca}, \frac{ca}{ab+bc}, \frac{ab}{bc+ca}$, if $a,b,c$ are in AP; is not straight forward.
Curiously, then we need to factorize $(2ab^2+2b^2c-ac^2-bc^2-a^2b-a^2c)$.
The question is: How do you factorize the last expression?
|
If rearranging terms as suggested in the previous answer is difficult to imagine here you have another approach but this has much simplification work.
Let the given polynomial be denoted by f(a,b,c)
Since this is homogeneous polynomial of a,b,c when you replace a by x root of x of f(x,b,c) =0 must be in the form of mb+nc. Therefore you can substitute mb+nc for x in f(x,b,c) = 0 and that gives m =2 , n = -1 when you equal the coefficients to zero .
Now x = 2b - c is a root and x - 2b +c is a factor of f(x,b,c) .
Now you can divide f(x,b,c) by x-2b+c to find the other factor and finally x must be changed back to a .
|
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|
Fourier series of $\sqrt[3]{\sin x}$ So I've done some experiments with how to add distortion to audio, and one of the methods I'm proposing is to take the cube root of an audio signal as a way to add overdrive. As the waveform that you're supposed to start with is a sine wave, I decided to tackle the problem of finding the magnitudes of the harmonics for $\sqrt[3]{\sin\left(2\pi fx\right)}$.
Based on what I've read about Fourier transforms, the coefficient of $\sin\left(nx\right)$ for $\sqrt[3]{\sin x}$is:
$\displaystyle a_n=\frac{1}{\pi}\int^{\pi}_{-\pi}\sin\left(nx\right)\sqrt[3]{\sin x}\,dx$
Since $\sqrt[3]{\sin x}$ contains only odd harmonics, I calculated the values of $a_n$ for odd n. Here's what I got out of the first seven terms:
$\displaystyle\left\{a_{1}, a_{3}, a_{5}, a_{7}, a_{9}, a_{11}, a_{13}\right\}=\frac{3\operatorname{\Gamma}\left(\frac{1}{3}\right)^{3}}{4\pi^{2}\sqrt[3]{2}}\left\{1, \frac{1}{5}, \frac{1}{10}, \frac{7}{110}, \frac{1}{22}, \frac{13}{374}, \frac{26}{935}\right\}$
which leads to the following:
$a_{2n+1}=\displaystyle\left(\frac{3\operatorname{\Gamma}\left(\frac{1}{3}\right)^{3}}{4\pi^{2}\sqrt[3]{2}}\right)\frac{\left(\frac{1}{3}\right)_{n}}{\left(\frac{5}{3}\right)_{n}}\qquad n\geqslant 0,\,n\in\mathbb{Z}$
where $\left(a\right)_{q}$ is the Pochhammer symbol.
Putting this all together, I get:
$\displaystyle\sqrt[3]{\sin\left(2\pi fx\right)}=\frac{2\pi^2\sqrt[3]{2}}{3\operatorname{\Gamma}\left(\frac{1}{3}\right)^3}\sum_{n\geqslant 0,\,n\in\mathbb{Z}}\left(\frac{\left(\frac{1}{3}\right)_{n}}{\left(\frac{5}{3}\right)_{n}}\,\sin\left(\left(4n+2\right)\pi fx\right)\right)$
Is this correct?
EDIT: By "cube root", I am referring to the real solution, where $\sqrt[3]{-x}=-\sqrt[3]{x}$ for $x\geqslant 0$.
|
The solution to the problem is straighforward, and I'm going to present it here with a minor generalisation.
Remember that the Fourier series to order $n$ of a function $f(x)$ is defined as
$$f_s(x,n) = \sum_{k=-n}^{n}c(k) e^{i k x}\tag{1a}$$
where the Fourier coefficients are defined as
$$c(k) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-i k x}\;dx\tag{1b}$$
It is more convenient to consider the related function
$$f(x) = |\sin(x)|^a\tag{2}$$
which is defined in the range $-\pi \lt x \lt \pi$.
Here we have introduced a paramter $a \gt -1$ which can be set to $\frac{1}{3}$ in the end.
The integral for the Fourier coefficient can be done in closed form:
$$c(k) = \frac{1}{2\pi} \int_{-\pi}^{\pi} |\sin(x)|^a e^{-i k x}\;dx = 2^{-a} \cos \left(\frac{\pi k}{2}\right) \binom{a}{\frac{k+a}{2}}\tag{3}$$
Notice that $c(k)$ vanishes for odd $k$.
The Fourier series then become
$$f_s(x,n) = c(0) + 2 \sum_{k=1}^{n} c(k) \cos(k x)\tag{4}$$
For the first few orders $n$ these are
$$
\left(
\begin{array}{cc}
0 & 2^{-a} \binom{a}{\frac{a}{2}} \\
2 & 2^{-a} \binom{a}{\frac{a}{2}}-2^{1-a} \binom{a}{\frac{a+2}{2}} \cos (2 x) \\
4 & 2^{-a} \binom{a}{\frac{a}{2}}-2^{1-a} \binom{a}{\frac{a+2}{2}} \cos (2 x)+2^{1-a} \binom{a}{\frac{a+4}{2}} \cos (4 x) \\
6 & 2^{-a} \binom{a}{\frac{a}{2}}-2^{1-a} \binom{a}{\frac{a+2}{2}} \cos (2 x)+2^{1-a} \binom{a}{\frac{a+4}{2}} \cos (4 x)\\
&-2^{1-a} \binom{a}{\frac{a+6}{2}} \cos (6 x) \\
\end{array}
\right)
\tag{5}$$
The approximations are shown in the graph
Notice that for odd $n \ge 1$ we have $f_s(x,n) = f_s(n-1,x)$
We can also let $n$ go to infinity in $(4)$ which results in a representation of our function in terms of a hypergeometric function:
$$\begin{align} |\sin(x)|^{a} =2^{-a} \binom{a} {\frac{a}{2}} \left(-1+\\2 \Re\left(\, _2F_1\left(1,-\frac{a}{2};\frac{a}{2}+1;e^{-2 i x}\right)\right)\right)
\end{align}\tag{5}$$
|
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|
Algebra of o-symbols In theorem $7.8$, Tom Apostol in his Calculus Vol. $1$ gave and proved the following basic rules for algebra of o-symbols
*
*$o(g(x)) \pm o(g(x)) = o(g(x)) $
*$o(cg(x)) = o(g(x)) = o(g(x)) $, if $c \ne 0$
*$f(x) \cdot o(g(x)) = o(f(x)g(x)) $
*$o(o(g(x))) = o(g(x)) $
*$\frac{1}{1 + g(x)} = 1 - g(x) + o(g(x))$
Now when proving that $\tan x = x + \frac{1}{3}x^3 + o(x^3)$ as $x \to 0$, he used the following equality, only mentioning that $g(x) = -\frac{1}{2}x^2 + o(x^3)$ and that he is applying the 5th case of the theorem above:
$$
Equation 1: \frac{1}{1 - \frac{1}{2}x^2 + o(x^3)} = 1 + \frac{1}{2}x^2 + o(x^2)
$$
By just applying that 5th case of the theorem, we get
$$
Equation 2: \frac{1}{1 - \frac{1}{2}x^2 + o(x^3)} = 1 + \frac{1}{2}x^2 - o(x^3) + o(-\frac{1}{2}x^2 + o(x^3))
$$
Now, I can prove that the right handside of Eq1 is equal to the right hand side of Eq2 (e.g. using this method), by proving that:
*
*$o(-\frac{1}{2}x^2 + o(x^3)) = o(x^2)$
*$o(x^2) + o(x^3) = o(x^2)$
But I wonder if there is an obvious reason why the above two equalities directly follow from the Theorem $7.8$?
Edit, to add a suggested proof for $o(x^2) + o(x^3) = o(x^2)$, which I hope was not even necessary, because it could be inferred trivially from the theorem $7.8$, potentially by the application of step $4$ $n-m$ times.
Let $m < n \land f(x) = o(x^m) \land g(x) = o(x^n)$.
Then we have:
$f(x) + g(x) = o(x^m) + o(x^n) = o(x^m) \cdot (1 + x^{n - m})$, because of T7.8:3.
\begin{equation}
\implies \frac{f(x) + g(x)}{1 + x^{n - m}} = o(x^m) \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \cdot (1 + 0) \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \cdot (\lim_{x \to 0} 1 + \lim_{x \to 0} x^{n-m})
\end{equation}
The last equality holds because $\lim_{x \to 0} x^{n-m} = 0$. Then further we have:
\begin{equation}
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \cdot (\lim_{x \to 0} \frac{x^m + x^n}{x^m}) \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m + x^n} \cdot \frac{x^m + x^n}{x^m} \\
\implies 0 = \lim_{x \to 0} \frac{f(x) + g(x)}{x^m} \\
\implies f(x) + g(x) = o(x^m)
\end{equation}
The last equality follows by definition of little-o. By just substituting $f$ and $g$, we get $o(x^m) + o(x^n) = o(x^m)$.
|
Please correct me if I'm wrong, but to prove $o(x^2) + o(x^3) = o(x^2)$, when $x \to 0$ we can take a simpler route, than what was explained in the answer (the edited part).
Namely, because of property $4$ of theorem $7.8$, we have the following:
\begin{equation}
\tag{By the definition of the $o$ notation}
x^3 = o(x^2)
\end{equation}
\begin{equation}
\tag{By substituting the $x^3$ from above}
\implies o(x^3) = o(o(x^2))
\end{equation}
\begin{equation}
\tag{By property 4 of theorem 7.8}
\implies o(x^3) = o(o(x^2)) = o(x^2)
\end{equation}
|
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|
Proving $ab^n + ba^n < a^{n+1} + b^{n+1}$ by Induction Given $0<a<b , ab \in \mathbb{R}$ , $n \geq 1$
By induction, Assuming $ab^n + ba^n < a^{n+1} + b^{n+1}$. - $(1)$
To prove $ab^{n+1} + ba^{n+1} < a^{n+2} + b^{n+2}$
Multiplying (1) by ab we get
$a^2b^{n+1} + b^2a^{n+1} < a^{n+2}b + b^{n+2}a$
How to proceed ?
Thanks
|
Your equation is equivalent to
$$0 < aa^n-a^nb-ab^n+bb^n$$
$$0 < (a-b)(a^n-b^n)$$
Since $0 < a < b,$ where $ a,b \in \mathbb{R}$,
$$0>(a-b)$$
that leaves
$$0 > (a^n-b^n)$$
Working on $n$, we have
$$(a^n-b^n) < 0$$
$$n\ln\frac{a}{b} < 0.$$
Again, $\ln\frac{a}{b}<0$ when $a<b$ so
$$n > 0.$$
|
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|
Convergence or divergence of $\int_2^\infty \frac{1-\cos(4-x^2)}{x-2} \, \mathrm{d}x$ Is the integral
$$ \int_2^\infty \frac{1-\cos(4-x^2)}{x-2} \, \mathrm{d}x $$
convergent or divergent? As the numerator goes to $0$ infinite times I can't apply the comparison test, and I can't seem to find any function to compare the limit. I thought of integrating by parts but I do not know how to prove the integral is divergent. Any hints?
|
Alternative solution:
Let
$$I_n := \int_3^{\sqrt{2n\pi + 4}} \frac{1 - \cos(4 - x^2)}{x - 2}\mathrm{d} x
\overset{x = \sqrt{y + 4}} = \int_5^{2n\pi} \frac{1 - \cos y}{\sqrt{y + 4} - 2}\, \frac{1}{2\sqrt{y + 4}}\mathrm{d} y.$$
We have
\begin{align*}
I_n &= \int_5^{2n\pi} \frac{1 - \cos y}{y}\, \frac{\sqrt{y + 4} + 2}{2\sqrt{y + 4}}\mathrm{d} y\\[6pt]
&\ge \int_5^{2n\pi} \frac{1 - \cos y}{y}\, \frac{\sqrt{y + 4}}{2\sqrt{y + 4}}\mathrm{d} y\\[6pt]
&= \int_5^{2n\pi} \frac{1 - \cos y}{2y}\mathrm{d} y\\[6pt]
&\ge \sum_{k=2}^n \int_{(2k-1)\pi - \pi/2}^{(2k-1)\pi + \pi/2} \frac{1 - \cos y}{2y}\mathrm{d} y\\[6pt]
&\ge \sum_{k=2}^n \int_{(2k-1)\pi - \pi/2}^{(2k-1)\pi + \pi/2} \frac{1}{2y}\mathrm{d} y\\[6pt]
&= \sum_{k=2}^n \frac12\ln\frac{(2k - 1)\pi + \pi/2}{(2k - 1)\pi - \pi/2}\\
&\ge \sum_{k=2}^n \frac12
\left(1 - \frac{(2k - 1)\pi - \pi/2}{(2k - 1)\pi + \pi/2}\right)\\
&= \sum_{k=2}^n \frac{1}{4k - 1}\\
&\ge \sum_{k=2}^n \frac{1}{4k}
\end{align*}
where we have used
$\ln u \ge 1 - u^{-1}$ for all $u\ge 1$ (easy to prove).
Thus, $\int_2^\infty \frac{1-\cos(4-x^2)}{x-2} \, \mathrm{d}x$ is divergent.
|
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|
Prove the independence of a certain segment in a special triangle Let $\triangle ABC$ be a triangle with obtuse angle $\angle A$ and $\overline{AB} = 1$. Also, let $\angle C = \gamma$ and $\angle B = 2\gamma$. If $E$ and $F$ are intersection points of perpendicular bisector of $\overline{BC}$ and circle $(A, \overline{AB})$, prove that $\overline{EF}$ is constant (not dependent of $\gamma$).
Note that circle $(A, r)$ means a circle with point $A$ as its center and $r$ as its radius.
It's not hard to see that $\angle FAE = 120^\circ$ but I don't know how to prove it either. Hope someone can help.
|
Note that if we fix the length $AB$, then the length of the chord $EF$ depends only on the distance from $A$ to $\overline{EF}$. Let $\overline{AD}$ be an altitude of $\triangle ABC$. Then
\begin{align*}
BD &= AB \cos(2\gamma) & BC &= AD \cot(2\gamma) + AD \cot(\gamma) = AB \sin(2\gamma) (\cot(2\gamma) + \cot(\gamma))
\end{align*}
Note that the distance from $A$ to $\overline{EF}$ is nothing but
$$\frac{BC}2-BD = AB \left( \frac12 \sin(2\gamma) (\cot(2\gamma)+\cot(\gamma)) - \cos(2\gamma) \right)$$
We need to show, then, that the above is a constant function of $\gamma$. Writing things out with sines and cosines, we get
\begin{align*}
\frac{BC}2-BD &= AB \left( \frac12 \sin(2\gamma) \left( \frac{\cos(2\gamma)}{\sin(2\gamma)} + \frac{\cos(\gamma)}{\sin(\gamma)} \right) - \cos(2\gamma) \right) \\
&= AB \left( \frac12 \cos (2\gamma) + \frac 12 \cdot \frac{\cos(\gamma)\sin(2\gamma)}{\sin(\gamma)} - \cos(2\gamma) \right) \\
&= \frac{AB}2 \left( \frac{\cos(\gamma)\sin(2\gamma)}{\sin(\gamma)} - \cos(2\gamma) \right) \\
&= \frac{AB}2 \left( \frac{\cos(\gamma) \cdot 2 \sin(\gamma) \cos(\gamma)}{\sin (\gamma)} - (2\cos^2(\gamma)-1) \right) \\
&= \frac{AB}2 \left( 2\cos^2(\gamma) - (2\cos^2(\gamma)-1) \right) \\
&= \frac{AB}2
\end{align*}
Indeed, we see that this is independent of $\gamma$, which implies that $EF$ is independent of $\gamma$ as well, as mentioned above.
There is almost certainly a faster, more purely geometric way to prove the result, but this gets the job done.
|
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One of the angles of a trapezoid $ABCD$ is $30^\circ$ and its diagonals are perpendicular to each other One of the angles of a trapezoid $ABCD (AB\parallel CD)$ is $30^\circ$ and its diagonals are perpendicular to each other. If the midsegment is $10$ and one of the bases is $8$, find the other base, the diagonals and the legs of $ABCD$.
Let $\measuredangle ABC=30^\circ$ and $MN$ be the midsegment $(M;N \in AD;BC)$. Let's $AB=a;DC=b;AD=c;BC=d.$ So $$MN=\dfrac{AB+CD}{2}\iff 10=\dfrac{a+8}{2}\iff20=a+8\iff a=12.$$ Clearly $a>b$.
I am not completely sure how we are supposed to interpret the fact that the diagonals of the trapezoid are perpendicular. I have tried the following: let $AC\cap BD=O;AO=x;OC=y;BO=z;OD=t.$ Then $$x^2+z^2=a^2=144\\y^2+z^2=d^2\\y^2+t^2=b^2=64\\x^2+t^2=c^2$$ From here we can derive $$a^2+b^2=x^2+y^2+z^2+t^2=c^2+d^2\\c^2+d^2=144+64=208$$
Now we are searching for another relationship between $c$ and $d$. I wasn't able to find such.
Yet let $CC_1\perp AB,DD_1\perp AB$. Since $\measuredangle C_1BC=30^\circ\Rightarrow CC_1=DD_1=h=\dfrac{d}{2}$. Now I decided to see what happens if $AD_1=x\Rightarrow C_1B=4-x$. Nothing helpful, though.
|
Hint
The triangles $\Delta OCD$ and $\Delta OAB$ are similars with raio $8/12 = 2/3$. It means we can write $OD=2x$, $OB=3x$, $OC=2y$ and $OA=3y$.
We can also write $$\tan (OBA) = \frac yx \quad (1)$$ and $$\tan (OBC)=\tan (30-OBA)=\frac{\tan (30)-\tan(OBA)}{1+\tan(30)\tan(OBA)}=\frac 23 \frac yx.\quad (2)$$
Solve the system with equations $(1)$ and $(2)$, and find $y/x$.
You also know that $$4x^2+4y^2=64.$$
Can you finish?
EDIT:
Setting $\tan(OBA)=y/x=k$ on equation $(2)$
$$\frac{\sqrt(3)/3-k}{1+\sqrt(3)/3 k}=\frac 23 k.$$
|
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Determine the geometric centre of a circle with a quarter missing The question
I have a circle of radius $a$, (where $a$ is a known positive constant), centred at Cartesian coordinates $(a,a)$. The bottom left quarter of the circle is missing.
Let the two-dimensional region $R$ be formed by this circle with a missing quarter. I'd like to compute the geometric centre of the region $R$ (or the centre of mass if we were to assume it has uniform density $\rho(x,y)=1)$.
My attempt
I know that for uniform density, the geometric centre of a region $R$, $(\bar x, \bar y)$ is
$$\bar x={1\over|R|}\iint_Rx\,\mathrm dA$$
$$\bar y={1\over|R|}\iint_Ry\,\mathrm dA$$
where $|R|$ is the area of the region. I got the area of the region by $|R|=\frac{3\pi a^2}{4}$.
I recognise that these integrals are better suited to be calculated in polar coordinates so I parametrise as such:
$$\begin{cases}x&=a\cos(\theta)+a\\y&=a\sin(\theta)+a\end{cases}$$
where $-\frac{\pi}{2}\leq\theta\leq\pi$.
I will then show a calculation for $\bar x$, which is erroneous (the calculation for $\bar y$ is the same):
$$\begin{align*}\bar x&={1\over |R|}\iint_R x\,\mathrm dA\\&={1\over |R|}\int_?^?\int_?^?(a\cos(\theta)+a)\,\mathrm da\,\mathrm d\theta\end{align*}$$
I think I may have made a mistake setting up the integral limits--this part confuses me:
$$\begin{align*}&={1\over |R|}\int_{-\pi\over 2}^\pi\int_0^a(a\cos(\theta)+a)\,\mathrm da\,\mathrm d\theta\\&=\frac{4}{3\pi a^2}\times\frac{3\pi a^2+2a}{4}\\&=\frac{2+3\pi}{4}\end{align*}$$
which does not even depend on $a$. I have clearly done something wrong.
|
Choose polar coordinates centered at the center of your circle so the change-of-variables looks like
\begin{cases}
x = a + r \cos \theta, \\
y = a + r \sin \theta.
\end{cases}
Notice that when $r=0$, we're at the center $(x, y) = (a, a)$. As you correctly observed, the region of interest is defined by the inequalities
$$
0 \leq r \leq a
\quad \text{and} \quad
-\tfrac{\pi}{2} \leq \theta \leq \pi.
$$
The Jacobian determinant for the change of variables to these shifted polar coordinates are the same as for standard polar coordinates:
$$
\frac{\partial(x, y)}{\partial(r, \theta)}
=
\left\lvert
\begin{matrix}
\partial x/ \partial r & \partial x/ \partial \theta \\
\partial y/ \partial r & \partial y/ \partial \theta
\end{matrix}
\right\rvert
=
\left\lvert
\begin{matrix}
\cos \theta & -r \sin \theta \\
\sin \theta & r \cos \theta
\end{matrix}
\right\rvert
= r \, (\cos^2 \theta + \sin^2 \theta)
= r
$$
Let's calculate $\bar{x}$ since, by symmetry, $\bar{y} = \bar{x}$.
\begin{align}
\bar{x} &= \frac{1}{\frac{3}{4} \pi a^2}
\int_{-\tfrac{\pi}{2}}^{\pi} \int_0^a (a + r\cos\theta)\, r \, \mathrm{d}r \, \mathrm{d}\theta \\
&= \frac{4}{3 \pi a^2}
\int_{-\tfrac{\pi}{2}}^{\pi} \int_0^a (ar + r^2\cos\theta) \, \mathrm{d}r \, \mathrm{d}\theta \\
&= \frac{4}{3 \pi a^2}
\int_{-\tfrac{\pi}{2}}^{\pi} \left.\biggl( \frac{ar^2}{2}
+ \frac{r^3\cos\theta}{3} \biggr)\right\rvert_0^a \, \mathrm{d}\theta \\
&= \frac{4}{3 \pi a^2} \cdot \frac{a^3}{6}
\int_{-\tfrac{\pi}{2}}^{\pi} \bigl( 3 + 2\cos\theta \bigr) \, \mathrm{d}\theta \\
&= \frac{2a}{9 \pi} \Bigl. \bigl( 3\theta + 2\sin\theta \bigr)
\Bigr\rvert_{-\tfrac{\pi}{2}}^{\pi} \\
&= \frac{2a}{9 \pi} \cdot \biggl( 3 \cdot \frac{3\pi}{2} + 2 \cdot 1 \biggr) \\
&= \frac{(9\pi + 4) \, a}{9 \pi}
\end{align}
Here's an interactive picture. You can drag the center of the circle and see the center of mass.
|
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Calculate: $\Delta=\left|\begin{array}{ccc} b c & c a & a b \\ a(b+c) & b(c+a) & c(a+b) \\ a^{2} & b^{2} & c^{2} \end{array}\right|$ Calculate:
$$\Delta=\left|\begin{array}{ccc}
b c & c a & a b \\
a(b+c) & b(c+a) & c(a+b) \\
a^{2} & b^{2} & c^{2}
\end{array}\right|$$
Does anyone know any easy way to calculate this determinant?
I tried the classic way but I was wondering if anyone knows an easier method.
|
If you add the first row to the second, the second row becomes three copies of $ab+bc+ca$, and you can write
$$\Delta=(ab+bc+ca)\begin{vmatrix}bc&ca&ab\\1&1&1\\a^2&b^2&c^2\end{vmatrix}.$$
Now, subtract the first column from the second and third to get
$$\Delta=(ab+bc+ca)\begin{vmatrix}bc&c(a-b)&b(a-c)\\1&0&0\\a^2&b^2-a^2&c^2-a^2\end{vmatrix};$$
you can then factor out $(a-b)(a-c)$ to get
$$\Delta=(ab+bc+ca)(a-b)(a-c)\begin{vmatrix}bc&c&b\\1&0&0\\a^2&-a-b&-a-c\end{vmatrix}.$$
Expanding by minors along the second row gives
\begin{align*}
\Delta&=(ab+bc+ca)(a-b)(a-c)(c(a+c)-b(a+b))\\
&=(ab+bc+ca)(a-b)(b-c)(c-a)(a+b+c).
\end{align*}
|
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|
Solving $\tan ^{-1}(\frac{1-x}{1+x})=\frac{1}{2}\tan ^{-1}(x)$ I was solving the following equation,$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$
But I missed a solution (don't know where's the mistake in my work).
Here's my work:
$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$
Putting $x = \tan(\theta)$
$$\begin{align} \tan ^{-1}\left(\frac{1-\tan\theta}{1+\tan\theta}\right)&=\frac{1}{2}\tan ^{-1}\left(\tan\theta\right)\\
\tan ^{-1}\left(\tan\left(\frac\pi 4 - \theta\right)\right)&=\frac{1}{2}\theta\\
\frac\pi 4 - \theta&=\frac{1}{2}\theta\\
\frac\pi 4&=\frac{1}{2}\theta + \theta\\
\frac\pi 4&=\frac{3}{2}\theta\\
\frac\pi 6&=\theta\\
\frac\pi 6&=\tan^{-1}(x)\\
\tan\frac\pi 6&=(x)\\
\frac1{\sqrt{3}}&=x\end{align}$$
I got only one solution, but the answer in my textbook is $\pm\frac{1}{\sqrt{3}}$. Where is the mistake?
|
Apply $tan$ to both sides
$ \dfrac{1 - x}{1 + x} = \tan \left( \dfrac{1}{2} \tan^{-1}(x) \right) \\
= \dfrac{ x }{ \sqrt{1 + x^2} + 1 } $
From this,
$ (1 - x) (\sqrt{1 + x^2} + 1 ) = x (1 + x) $
$ (1 - x) \sqrt{1 + x^2} + 1 - x = x + x^2 $
$ \sqrt{ 1 + x^2} = \dfrac{(x^2 + 2 x - 1)}{ ( 1 - x)} $
$ 1 + x^2 = \dfrac{ (x^2 + 2x - 1)^2 }{ (x - 1)^2 } $
$ (x^2 + 1) (x - 1)^2 = (x^2 + 2 x - 1)^2 $
$ (x^2 + 1)(x^2 - 2 x + 1) = x^4 + 4 x^2 + 1 + 4 x^3 - 4 x - 2 x^2 $
$ x^4 - 2 x^3 + x^2 + x^2 - 2 x + 1 = x^4 + 4 x^3 + 2 x^2 - 4 x + 1 $
Cancelling $x^4 $ and $1$ on both sides, and re-arranging,
$ 6 x^3 - 2 x = 0 $
Dividing by $(2 x)$ (The root $0$ is extraneous)
$ 3 x^2 - 1 = 0 $
Hence, $ x = \pm \dfrac{1}{\sqrt{3}} $
|
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|
How to solve $'''+ ''+ = 0$? $$'''+ ''+ = 0$$
What method can I use to further solve the equation?
|
We can write the first as
\begin{align}
(xy'')^{'} + \left(\dfrac{x^2}{2}\right)' = 0
\end{align}
so it becomes a problem of integration, since of course the integrating process can be applied linearly. As such
\begin{align}
xy'' + \dfrac{x^2}{2} = C_1 &\implies y'' = \dfrac{C_1}{x} - \dfrac{x}{2}\\
&\implies \int y''(x) dx = \int \left(\dfrac{C_1}{x} - \dfrac{x}{2}\right) dx\\
&\implies y' = C_1 \ln x- \dfrac{x^2}{4}+C_2\\
&\implies \int y'(x) dx = \int \left( C_1 \ln x- \dfrac{x^2}{4}+C_2 \right)dx\\
&\implies y(x) = C_1 \int \ln x dx - \dfrac{1}{4} \int x^2 dx + C_2 \int 1 dx\\
&\implies y(x) = C_1 (x \ln x - x ) - \dfrac{x^3}{12} + C_2 x + C_3.
\end{align}
|
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|
How do I prove this these inequalities, preferably without calculus? I'm trying to solve an inequality which stems from this problem:
If x, y, z are positive real numbers and $x^5+y^5+z^5=3$, prove $\frac{x^8}{y^3}+\frac{y^8}{z^3}+\frac{z^8}{x^3} \ge 3$
I thought that solving this directly would require an inequality I'm not familiar with, so I decided to set $y=ax$ and $z=bx$ (note $a>0$, $b>0$), reducing the inequality to be proven to:
$\frac{1}{a^3}+\frac{a^8}{b^3}+b^8≥1+a^5+b^5$ for all $a, b > 0$.
I tried proving it by proving the function $f$ with $f(a)=\frac{1}{a^3}+\frac{a^8}{b^3}+b^8≥1+a^5+b^5$ is always positive for $a>0$ and all $b>0$. I graphed it out, and noticed it only attained a value of zero when $a=1$ and $b=1$ (for values of b between 0 and 10). Afterwards, it seemed to only increase. However, the 11th degree polynomials are simply too much for me to handle.
So, my questions would be:
How do I prove the original inequality?
And how do I prove the inequality I have 'reduced' the original one to?
An answer without calculus and without advanced inequality identities would be preferred, but I would prefer an answer with them than no answer at all.
|
Note that by the rearrangement inequality for triples $(x^3,y^3,z^3)$ and $(x^2,y^2,z^2)$ we can conclude that $$y^3x^2+z^3y^2+x^3z^2 \le x^5+y^5+z^5$$
Now by Cauchy-Schwarz inequality
$$(\frac{x^8}{y^3}+\frac{y^8}{z^3}+\frac{z^8}{x^3})(y^3x^2+z^3y^2+x^3z^2) \ge (x^5+y^5+z^5)^2 \implies$$ $$\frac{x^8}{y^3}+\frac{y^8}{z^3}+\frac{z^8}{x^3} \ge \frac{(x^5+y^5+z^5)^2}{y^3x^2+z^3y^2+x^3z^2} \ge \frac{(x^5+y^5+z^5)^2}{x^5+y^5+z^5} =x^5+y^5+z^5 = 3$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find $\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$? By factorization:
$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}\tag{1}$$
$$=\lim_{x\to-\infty} \frac{x\sqrt{1+\frac{2}{x}}}{-x}$$
$$=\lim_{x\to-\infty}-\sqrt{1+\frac{2}{x}}$$
If I input $x=-\infty$, the limiting value seems to be $-1$. But according to desmos, the limiting value should be $1$.
By L'Hopital's rule:
$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$$
$$=\lim_{x\to-\infty} \frac{\dfrac{x+1}{\sqrt{x^2+2x}}}{-1}$$
$$=-\lim_{x\to-\infty} \dfrac{x+1}{\sqrt{x^2+2x}}$$
$$=-\lim_{x\to-\infty} \frac{1}{\dfrac{x+1}{\sqrt{x^2+2x}}}$$
$$=-\lim_{x\to-\infty} \dfrac{\sqrt{x^2+2x}}{x+1}$$
$$=-\lim_{x\to-\infty} \frac{\dfrac{x+1}{\sqrt{x^2+2x}}}{1}$$
$$=-\lim_{x\to-\infty} \dfrac{x+1}{\sqrt{x^2+2x}}$$
I can't get a determinate form.
My questions:
*
*How do I find $(1)$ using factorization?
*How do I find $(1)$ using L'Hopital's rule?
Related
|
Your approach is almost correct, you've just made a common mistake regarding square roots.
Writing out your manipulation of the numerator, you did $\sqrt{x^2 + 2x} = \sqrt{x^2(1 + \frac2x)} = \sqrt{x^2}\sqrt{1 + \frac2x} = x\sqrt{1 + \frac2x}.$ However, recall that because the principal square root is always positive (by definition) we actually have that $\sqrt{x^2} = |x|,$ and because we're looking at the limit as $x$ approaches $-\infty$ we consider negative $x,$ so $|x| = -x,$ explaining the sign discrepancy.
|
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|
Restoring third coordinate for triangle by its orthogonal projection and similar triangle Suppose we have triangle $\Delta OAB$ lying on plane $z=0$ with coordinates $O(0,0,0), A(x_a,y_a,0), B(x_b,y_b,0)$
Also there is triangle $\Delta EFG$, but we know only coordinates of its orthogonal projection on plane $z=0$. $E(x_e,y_e,0), F(x_f,y_f,z_f), G(x_g,y_g,z_g)$. So coordinates $z_f, z_g$ are unknown.
We know, that triangles are similar, so: $$\angle OAB = \angle EFG\\\angle ABO = \angle FGE\\ \angle BOA = \angle GEF$$
My hypothesis is that only finite number of pairs $z_f, z_g$ exists, that satisfies the above conditions (that may be wrong). If so - I need to find such pairs (prefer analytical answer).
I've tried the following approach and faced unsolvable system of equations. Some different approaches or hints for solving will help a lot.
*
*Rotate, scale and move triangle $\Delta EFG$, so it becomes $\Delta OCD$ with coordinates $O(0,0,0), C(x_c,y_c,z_c), D(x_b,y_b,z_d)$
*Use the law of sines to find relative sides:
Let $$CD = o $$
then $$OD = \frac{sin(\angle OCD)} {sin(\angle DOC)}\\OC = \frac{sin(\angle CDO)} {sin(\angle DOC)}$$
for simplicity $$OD = m*o\\OC = n*o$$
3) Use distance formula to get system of 3 equations (with unknown $o, z_c, z_b$) I was unable to solve:
$\left\{
\begin{array}{c}
o = \sqrt{(x_c-x_b)^2 + (y_c-y_b)^2 + (z_c-z_d)^2}\\
m*o = \sqrt{x_b^2 + z_d^2}\\
n*o = \sqrt{x_c^2 + y_c^2 + z_c^2}
\end{array}
\right.$
Thanks in advance!
Final result (based on @Robin's Premium Coffee answer):
$$K_1 := (x_f-x_g)^2+(y_f-y_g)^2\\
K_2 := (x_e-x_g)^2+(y_e-y_g )^2\\
K_3 := (x_e-x_f)^2+(y_e-y_f)^2\\
K_4 := \frac{b^2}{a^2}\\
K_5 := \frac{b^2 K_2}{a^2} -K_3\\
K_6 := K_4^2 a^4-2K_4 a^4+a^4-2K_4 a^2 o^2-2a^2 o^2+o^4\\
K_7 := 2K_1 a^4+2K_1 K_4 a^4-2K_5 a^4+2K_4 K_5 a^4-2K_1 a^2 o^2-2K_2 a^2 o^2-2K_2 K_4 a^2 o^2-2K_5 a^2 o^2+2K_2 o^4\\
K_8 := K_1^2 a^4+K_5^2 a^4+2K_1 K_5 a^4-2K_1 K_2 a^2 o^2-2K_2 K_5 a^2 o^2+K_2^2 o^4\\
z_g = \pm \sqrt{\frac{-K_7\pm\sqrt{K_7^2-4K_6K_8}}{2K_6}}\\
z_f = \pm \sqrt{K_4g^2+K_5}
$$
2 of 8 solutions are valid. It was enough for me to write a program to do all the calculations and check the results. Due to precision limitations equal triangles may cause discriminant to be negative (not a big problem).
|
For $\triangle OAB$ we have
$ o^2 = \overline{AB}^2 , a^2 =\overline{OB}^2 , b^2 = \overline{OA}^2 $
And for $\triangle EFG$ we have
$ e^2 = \overline{FG}^2 = (x_f - xg)^2 + (y_f - y_g)^2 + (z_f - z_g)^2 = K_1 + (z_f - z_g)^2 $
$ f^2 = \overline{EG}^2 = (x_e - x_g)^2 + (y_e - y_g)^2 + (0 - z_g)^2 = K_2 + z_g^2 $
$ g^2 = \overline{EF}^2 = (x_e - x_f)^2 + (y_e - y_f)^2 + (0 - z_f)^2 = K_3 + z_f^2 $
Since $\triangle OAB$ is similar to $\triangle EFG$ then
$ o^2 = \alpha e^2 = \alpha ( K_1 + (z_f - z_g)^2 )$
$ a^2 = \alpha f^2 = \alpha (K_2 + z_g^2 )$
$ b^2 = \alpha g^2 = \alpha (K_3 + z_f^2 )$
Eliminating $\alpha$ from the above three equations we end up with two equations, involving $z_f$ and $z_g$
$ o^2 (K_2 + z_g^2 ) = a^2 (K_1 + (z_f - z_g)^2 )$
$ a^2 (K_3 + z_f^2 ) = b^2 (K_2 + z_g^2 ) $
Which are two quadratic equations in the two unknowns $ z_f$ and $z_g$ and have at most $4$ solutions.
And these solutions (if they exist) come in pairs that are mirror images about the plane $z=0$, because if $(z_f, z_g)$ is a solution, then so is $(-z_f, -z_g)$.
|
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|
Find $\sum_{k=1}^\infty\frac{1}{x_k^2-1}$ where $x_1=2$ and $x_{n+1}=\frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2}$ for $n \ge 2$ Given $x_1=2$
and $x_{n+1}=\frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2}, n\geq 2$
Prove that $y_n=\sum_{k=1}^{n}\frac{1}{x_k^2-1}, n\geq 1$ converges and find its limit.
*
*To prove a convergence we can just estimate $x_n > n$, therefore $y_n<z_n$, where $z_n=\sum_{k=1}^{n}\frac{1}{k^2-1}$ and $z_n$ converges, then $y_n$ converges too.
*We can notice that $x_n^2+2x_n+5=(x_n+1)^2+4$. So $x_{n+1}$ is one of the roots of the equation:
$x_{n+1}^2-(x_n+1)x_{n+1}-1=0$
So $x_{n+1}^2-1=(x_n+1)x_{n+1}$ and therefore: $y_n=\sum_{k=1}^n \frac{1}{(x_{n-1}+1)x_{n}}$
I'm stuck here.
|
Using the relation $x_{n+1}^2 - 1 = x_{n+1}(x_n + 1)$, we find that
\begin{align*}
\frac{1}{x_n + 1} - \frac{1}{x_{n+1} + 1}
&= \frac{x_{n+1}}{x_{n+1}^2 - 1} - \frac{1}{x_{n+1} + 1} \\
&= \frac{1}{x_{n+1}^2 - 1}.
\end{align*}
So it follows that
\begin{align*}
y_n
&= \frac{1}{x_1^2 - 1} + \sum_{k=1}^{n-1} \left( \frac{1}{x_k + 1} - \frac{1}{x_{k+1} + 1} \right) \\
&= \frac{1}{x_1^2 - 1} + \frac{1}{x_1 + 1} - \frac{1}{x_n + 1} \\
&\xrightarrow[n\to\infty]{} \frac{x_1}{x_1^2 - 1} = \frac{2}{3}.
\end{align*}
|
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|
Under what conditions is $\vert x-y \vert = \vert x \vert -\vert y \vert$? This is a fairly basic question, however, I don't know what I am missing here. Solving the equation $\vert x- y\vert = \vert x \vert -\vert y \vert$ yields:
\begin{align}
\vert x- y\vert &= \vert x \vert -\vert y \vert \\
(x-y)^2&=x^2+y^2-2\vert xy\vert \\
xy & = \vert xy \vert
\end{align}
Which is obviously true if $xy\geq 0$. However, if I set $x=0$, this equality does not hold. For instance $x=0$ and $y=5$, then:
$$
\vert0-5\vert=\vert0\vert-\vert5\vert \\
5 = -5
$$
On the other hand, if $y=0$, it works out fine. So, I would conclude that we need $xy\geq 0$ and $x \neq 0$ for this equation to hold. I wonder whether there is an "analytical way" to see this rather than plugging in numbers.
|
You have that
$$
|x-y| = |x| - |y| \qquad\implies\qquad
(x-y)^2 = x^2 + y^2 - 2|xy|
$$
which is an implication but not an equivalence. The problem is that squaring adds solutions, for example $1\neq-1$ but $1^2=(-1)^2$.
If you'd take the square root of the right equation, you'd get $$|x-y| = \big|\,|x|-|y|\,\big|$$
The left side, however, must have $|x|\geqslant|y|$. Thus:
$$
|x- y| = |x| - |y| \qquad\Longleftrightarrow\qquad
(x-y)^2 = x^2 + y^2 - 2|xy| \ \text{ and }\ |x|\geqslant|y|
$$
One way to solve the equation is the rather tedious split into cases:
$$|x-y| = |x| - |y| \quad\Longleftrightarrow\quad
\begin{cases}
x-y = x-y; &x\geqslant y, x\geqslant 0, y\geqslant 0 \\
x-y = x+y; &x\geqslant y, x\geqslant 0, y < 0 \\
x-y = -x-y; &x\geqslant y, x < 0, y\geqslant 0 \\
x-y = -x+y; &x\geqslant y, x < 0, y < 0 \\
y-x = x-y; &x < y, x\geqslant 0, y\geqslant 0 \\
y-x = x+y; &x < y, x\geqslant 0, y < 0 \\
y-x = -x-y; &x < y, x < 0, y\geqslant 0 \\
y-x = -x+y; &x < y, x < 0, y < 0 \\
\end{cases}$$
Some cases vanish because they are always false, others reduce due to $x=0$ or $x=y$:
$$|x-y| = |x| - |y| \quad\Longleftrightarrow\quad
\begin{cases}
\text{true}; &x\geqslant y\geqslant 0 \\
x = y; &x\geqslant y, x < 0, y < 0 \\
y = 0; &x < y, x < 0, y\geqslant 0 \\
\text{true}; &x < y < 0 \\
\end{cases}$$
That is:
$$|x-y| = |x| - |y| \quad\Longleftrightarrow\quad
\begin{cases}
x\geqslant y\geqslant 0 \\
x = y < 0 \\
x < y \leqslant 0 \\
\end{cases}$$
This 3 cases can finally be rewritten as 2 cases:
$$|x-y| = |x| - |y| \quad\Longleftrightarrow\quad
\begin{cases}
x \geqslant y\geqslant 0 \\
x \leqslant y \leqslant 0 \\
\end{cases}$$
|
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|
To prove an Inequality: $ ( x^2 +2x)e^x + (x^2-2 x)e^{-x} \ge 0$
$ \left(x^2 +2x\right)e^x + \left(x^2-2 x\right)e^{-x} \ge 0$.
I used photomath to plot its graph: $y=(x^{2}+2x))e^{x} + \frac{{x}^{2}-2x}{{e}^{x}}$
But how do I prove it without an image? Should I take the derivative of it and reason, please tell me the solution.
|
Rearranging gives
$$2x^2\frac{(e^x+e^{-x})}{2}+4x\frac{e^x-e^{-x}}{2}=2f(x)+4g(x)$$
Now we will use Taylor expansion !!!
note $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$
$-e^{-x}=-1+x-\frac{x^2}{2!}+\frac{x^3}{3!}-....$
So,
$$\frac{e^x-e^{-x}}{2}=x+\frac{x^3}{3!}+...$$
So $$x\frac{(e^x-e^{-x})}{2}=x^2+\frac{x^4}{3!}+\frac{x^6}{5!}+...\geq 0$$
So, $$4g(x)\geq 0$$(#)
Now by AM-GM we have $$2f(x)\geq2x^2\geq 0$$ (@)
By adding (@) and (#) gives the result!!
|
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|
Sameness of Riemann surfaces I need to show that the equations $x^3+y^3=1$ and $y^2=4x^3-1$ are the same Riemann surface in $\mathbb{CP}^2$ and as a consequence to show that there two meromorphic functions $f,g$ such that $f^3+g^3=1$.
I tried using Able's theorem but didn't get much
Edit: Using compactification I managed to find the points that needed to be addaad to the surface of each of the equations but I dont know what to do next
|
I'm thinking if this proof is correct or not:
$x^3+y^3=1$ is the Riemann surface in $\Bbb{CP}^2$ define by:
$$
\{[X,Y,Z] \in \Bbb{C}^3| X^3+Y^3=Z^3, (X, Y, Z) \neq (0, 0, 0)\}
$$
Let
$$
\begin{aligned}
X &= U \\
Y &= \frac{\sqrt{3}}{6}V -\frac{1}{2}W \\
Z &= \frac{\sqrt{3}}{6}V +\frac{1}{2}W \\
\end{aligned}
$$
Then $(X, Y, Z) \mapsto (U, V, W)$ is a biholomorphic automorphism in $\Bbb{CP}^2$ and:
$$
\begin{aligned}
X^3+Y^3-Z^3&=U^3+(\frac{\sqrt{3}}{6}V-\frac{1}{2}W)^3-(\frac{\sqrt{3}}{6}V+\frac{1}{2}W)^3 \\
&=U^3-\frac{1}{4}W(V^2+W^2)
\end{aligned}
$$
Hence
$$
\{[X,Y,Z] \in \Bbb{C}^3| X^3+Y^3=Z^3, (X, Y, Z) \neq (0, 0, 0)\} \\
= \{[U, V, W] \in \Bbb{C}^3|U^3-\frac{1}{4}W(V^2+W^2)=0, (U, V, W) \neq (0, 0, 0)\}
$$
Setting $\frac{U}{W}=u, \frac{V}{W}=v$, in affine coordinate, it's $v^2=4u^3-1$.
|
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|
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \\
&= (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)
\end{align}$$
now,
$$\begin{align}
a^4 + b^4 &= (a+b)^4 - (4a^3b + 6a^2b^2 + 4ab^3) \\
&= (a+b)^4 - (4ab(a^2 + b^2) + 6(ab)^2) \\
&= (a+b)^4 - (4ab((a + b)^2 - 2ab) + 6(ab)^2) \\
&= 7^4 - (4(2)(7^2 - 2(2)) + 6(2)^2) \\
&= 2017
\end{align}$$
so
$$\begin{align}
&\phantom{=}\; (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)\\
&= 7^6 - (6\cdot2\cdot(2017) + 15(2)^2 (7^2 - 2(2)) + 20(2)^3) \\
&= 90585
\end{align}$$
correct?
|
What the hey, I'll toss in one more (there are plenty of ways to work with the binomials, and some of the other answers already given are ones I'd have tried first):
We need $$ \ a + b \ = \ 7 \ \ , \ \ a - b \ = \ \sqrt{\Delta} \ = \ \sqrt{41} \ \ , \ \ ab \ = \ 2 \ \ , $$ $$ a^2 + b^2 \ = \ (a + b)^2 - 2ab \ = \ 49 - 4 \ = \ 45 \ \ , $$
as already presented. We can write
$$ (a + b)^6 \ + \ (a - b)^6 \ \ = \ \ 2a^6 \ + \ 2b^6 \ + \ 30·(a^4b^2 \ + \ a^2b^4) $$ $$ = \ \ 2·(a^6 \ + \ b^6) \ + \ 30·(ab)^2·(a^2 \ + \ b^2) \ \ = \ \ 2·(a^6 \ + \ b^6) \ + \ 30·2^2·45 $$
$$ = \ \ 7^6 \ + \ 41^3 \ \ = \ \ 49^3 \ + \ 41^3 \ \ . $$
If we need to do this "by hand", we have $ \ 49^3 + 41^3 \ = \ (49 + 41)·(49^2 - 49·41 + 41^2) \ \ , $ for which we can use little "binomial tricks" to produce $ \ (50 - 1)^2 \ - \ (50 - 1)·(40 + 1) \ + \ (40 + 1)^2 $ $ = \ 2401 - 2009 + 1681 \ = \ 2073 \ \Rightarrow \ 49^3 + 41^3 \ = \ 90·2073 \ = \ 207,300 - 20,730 $ $ = \ 186,570 \ \ . $ We also have $ \ 30·4·45 \ = \ 6·4·15^2 \ = \ 6·(4·225) \ = \ 6·900 \ = \ 5400 \ \ . $
Thus,
$$ 2·(a^6 \ + \ b^6) \ \ = \ \ [ \ (a + b)^6 \ - \ (a - b)^6 \ ] \ - \ 30·(ab)^2·(a^2 \ + \ b^2) $$
$$ = \ \ 186,570 \ - \ 5400 \ \ = \ \ 181,170 \ \ \Rightarrow \ \ a^6 \ + \ b^6 \ \ = \ \ 90,585 \ \ . $$
[This is somewhere between what you started out doing and Dan's method.]
|
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|
Find $\int \sin^4x dx$ In my textbook, I came upon the following problem:
Problem: Find $\int \sin^4x dx$
Since there was only given a solution sketch and I came to a different conclusion in one step, I ask if my following solution is correct.
Solution: In my solution, I use the following identities:
$$\sin^2x = \frac{1-\cos2x}{2} \tag{1} \label{1}$$
$$\cos^2x = \frac{1+\cos2x}{2} \tag{2} \label{2}$$
Now we can write
$$\int \sin^4x dx = \int \left( \frac{1-\cos2x}{2} \right)^2 dx = \int\frac{1}{4}dx - \frac{1}{2}\int \cos2xdx + \frac{1}{4} \int cos^22xdx$$
My textbook gave the last term in this equation as $\int\frac{1}{4}dx - \int \cos2xdx + \frac{1}{4} \int cos^22xdx$. But, since $(1-\cos2x)^2 = 1 - 2\cos2x + \cos^22x$ the middle integral should simplify to $- \frac{1}{2}\int \cos2xdx$. Is this correct? If not, how do we come to the conclusion of $-\int \cos2xdx$?
To complete the integration, we have
$$\int \frac{1}{4} dx = \frac{x}{4}$$
$$ - \frac{1}{2} \int \cos 2xdx = - \frac{1}{4}\int \cos u du = - \frac{1}{4} \sin2x = - \frac{\sin2x}{4}$$
and
$$\frac{1}{4} \int cos^22xdx = \frac{1}{4} \int \frac{1 + cos4x}{2}dx = \frac{4x + \sin 4x}{32}$$
so
$$\int \sin^4x dx = \frac{x}{4} - \frac{\sin2x}{4} + \frac{4x + \sin 4x}{32}$$
Are these conclusions and computations correct? Thanks for any help!
EDIT: According to the correction of @Eevee Trainer I now use the correct identity $\cos^2x = \frac{1+\cos2x}{2}$. The rest of the computation should be correct. Thanks all for your help and comments!.
|
An alternative approach: A complex variable identity reduces the problem to an exercise in the Binomial Theorem.
$$\begin{align}
\sin^4 x &= \left( \frac{1}{2i} (e^{ix}-e^{-ix})\right)^4 \\
&= \frac{1}{16} \left( e^{4ix} -4 e^{2ix} +6 -4 e^{-2ix} + e^{-4ix} \right)
\end{align}$$
so
$$\begin{align}
\int \sin^4 x \;dx &= \frac{1}{16} \left( \frac{1}{4i} e^{4ix} - \frac{4}{2i} e^{2ix} + 6x + \frac{4}{2i} e^{2ix} - \frac{1}{4i} e^{4ix} \right) + C \\
&= \frac{1}{16} \left( \frac{1}{4i} (e^{4ix}-e^{-4ix}) -\frac{4}{2i}(e^{2ix}-e^{-2ix}) +6x \right) + C \\
&=\frac{1}{16} \left( \frac{1}{2} \sin 4x -4 \sin 2x +6x \right) + C \\
&= \frac{1}{32} \sin 4x -\frac{1}{4} \sin 2x +\frac{3}{8} x + C
\end{align}$$
|
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|
help me to evaluate these integrals Find the value of $$\int\frac{x}{x^2-x+1} dx$$ and $$\int\frac{1}{x^2-x+1} dx$$ This was not the original question. Original question was tougher and I have simplified that to these two integrals. I am having a hard time evaluating these two integrals but what I know is that, in these two integrals there is arctan occurring in a term of both integrals. Maybe we have to do some operations in the numerators?
Any help will be greatly appreciated.
EDIT: The original question was: $$\int\frac{3x}{x^3+1}dx$$
EDIT: $$\begin{align}\int\frac{3x}{x^3+1}dx&=3\int\frac{x}{x^3+1}dx\\&=3\int\frac{x+1}{3x^2-3x+3}dx - 3\int\frac{1}{3x+3}dx\\&=\int\frac{x+1}{x^2-x+1}dx - \int\frac{1}{x+1}dx\end{align}$$
The value of last integral is $\ln|x+1|$ so we will consider it later. We will only consider the first integral now and we can write that as:
$$\int\frac{x}{x^2-x+1}dx + \int\frac{1}{x^2-x+1}dx$$ after this I am stuck.
|
For the second one: A common method would be: complete the square in the denominator...
$$
x^2-x+1 = x^2 - x + \frac14+\frac34
=\left(x-\frac12\right)^2+\frac34
$$
Substitute $x-\frac12 = y$.
$$
\int\frac{1}{x^2-x+1}\;dx = \int\frac{1}{y^2+(\sqrt{3}/2)^2}\;dy
$$
recognize an arctangent integral
$$
=\frac{2}{\sqrt{3}}\arctan\left(\frac{2}{\sqrt{3}}y\right)+C
=\frac{2}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+C
$$
|
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|
Integral $\int_0^1 \frac{\ln(1-\sqrt{3}t+t^2)}{t} dt$ $$I=\int_0^1 \frac{\ln(1-\sqrt{3}t+t^2)}{t} dt=-\frac{13}{72}\pi^2$$
Here is what I tried: $1-\sqrt{3}t+t^2=(z_1-t)(z_2-t)$ where $z_1=e^{\frac{\pi}{6}i}, z_2=e^{-\frac{\pi}{6}i}$
$$I=\int_0^1 \frac{\ln(z_1)+\ln(1-\frac{t}{z_1})}{t}+\frac{\ln(z_2)+\ln(1-\frac{t}{z_2})}{t} dt=\int_0^1 \frac{\ln(1-\frac{t}{z_1})}{t}+\frac{\ln(1-\frac{t}{z_2})}{t} dt$$
where $\ln(z_1)+\ln(z_2)=0$ and $\frac{1}{z_1}=z_2, \frac{1}{z_2}=z_1$
$$I=-\sum_{n=1}^\infty \frac{1}{n} \int_0^1 \frac{1}{t}\left(\frac{t}{z_1}\right)^n+\frac{1}{t}\left(\frac{t}{z_2}\right)^n dt = -\sum_{n=1}^\infty \frac{1}{n}\left( \frac{z_1^n+z_2^n}{n} \right)$$
where $z_1^n+z_2^n=2\cos(\frac{n\pi}{6})$
$$I=-2\sum_{n=1}^\infty \frac{\cos(\frac{n\pi}{6})}{n^2}$$
How to proceed next?
|
Continue with the identity $\operatorname{Li}_2(z)+\operatorname{Li}_2(1/z)=-\frac{\pi^2}6-\frac12\ln^2(-z)$
\begin{align}
I= & \int_0^1 \frac{\ln(1-e^{-i\frac{\pi}6}{t})}{t}+\frac{\ln(1-e^{i\frac{\pi}6}t)}tdt\\
= & -\operatorname{Li}_2\left(e^{-i\frac{\pi}6}\right)-\operatorname{Li}_2\left(e^{i\frac{\pi}6}\right)\\
= & \frac{\pi^2}6+ \frac12\ln^2\left(e^{i\frac{5\pi}6}\right)
= \frac{\pi^2}6+\frac12\left(i\frac{5\pi}6\right)^2\\= & -\frac{13\pi^2}{72}
\end{align}
|
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|
Expand the function $f(z) = \frac{1+2z^2}{z^2+z^4} $ into power series of $z$ in all areas of convergence. I'm studying for my upcoming complex analysis qualifying exam by working through problems in past exams. For this problem, I'd like to know (1) if my answer is correct and complete (i.e. whether I've made any errors/omissions), and (2) if there are any better ways of doing this. I'm specifically curious about whether or not there are other areas of convergence that I need to address. Thanks!
Question:
Expand the function
$$
f(z) = \frac{1+2z^2}{z^2+z^4}
$$
into power series of $z$ in all areas of convergence.
Attempted Solution:
We begin by noting that this function has poles at $z=0$ and $z=\pm i$. Then we proceed by recalling the geometric series
$$
\sum_{k=0}^{\infty} z^k = \frac{1}{1-z},
$$
which converges for $|z|<1$. Using this form, we can rewrite $f(z)$ as follows
\begin{align*}
f(z) &= \frac{1+2z^2}{z^2+z^4} \\
&= \frac{1}{z^2+z^4} + \frac{2z^2}{z^2+z^4} \\
&= \frac{1}{z^2(1-(-z^2))} + \frac{2z^2}{z^2(1-(-z^2))} \\
&= \frac{1}{z^2}\left(\frac{1}{1-(-z^2)}\right) + 2\left(\frac{1}{1-(-z^2)}\right)\\
&= \left(\frac{1}{z^2}+2\right)\left(\frac{1}{1-(-z^2)}\right)\\
&= \left(\frac{1}{z^2}+2\right) \sum_{k=0}^{\infty} (-z^2)^k \\
&= \sum_{k=0}^{\infty} \left(z^{-2}+2\right)(-1)^k z^{2k}
\end{align*}
for $0<|z|<1$, since there is a pole at $z=0$.
|
Yes, that's perfect. Note that you can even simplifie more your expression:
$$\sum_{k=0}^{\infty} \left(z^{-2}+2\right)(-1)^k z^{2k}=\sum_{k=0}^{\infty}(-1)^kz^{2(k-1)}+2\sum_{k=0}^{\infty}(-1)^kz^{2k}=\\\sum_{k=-1}^{\infty}(-1)^{k+1}z^{2k}+2\sum_{k=0}^{\infty}(-1)^kz^{2k}\\=\frac{1}{z^2}+\sum_{k=0}^{\infty}(-1)^{k+1}z^{2k}+2\sum_{k=0}^{\infty}(-1)^kz^{2k}=\frac{1}{z^2}-\sum_{k=0}^{\infty}(-1)^kz^{2k}+2\sum_{k=0}^{\infty}(-1)^kz^{2k}\\=\frac{1}{z^2}+\sum_{k=0}^{\infty}(-1)^kx^{2k}=\frac{1}{z^2}+1-z^2+z^4+O(z^6)$$
Note that the residue of $f$ in $z=0$ since $a_{-1}=0$
|
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|
If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$. If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$.
Now what I thought is to manipulate given result somehow to get something in the form of $a + b$:
\begin{align*}
b^{3} - b^{2} = a^{3} - a^{2} & \Longleftrightarrow b^{3} - a^{3} = b^{2} - a^{2}\\\\
& \Longleftrightarrow (b-a)(a^{2} + ab + b^{2}) = (b+a)(b-a)\\\\
& \Longleftrightarrow a^{2} + ab + b^{2} = b + a
\end{align*}
but what next?
|
Refer to 'Inequality of arithmetic and geometric means' in wiki
$\frac{{a + b}}{2} \geqslant \sqrt {ab} \Rightarrow \frac{{{{(a + b)}^2}}}{4} \geqslant ab$
Next,
$\begin{gathered}
{a^2} + ab + {b^2} = a + b \hfill \\
{(a + b)^2} = a + b + ab < a + b + \frac{{{{(a + b)}^2}}}{4} \hfill \\
\frac{{3{{(a + b)}^2}}}{4} < a + b \hfill \\
a + b < \frac{4}{3} \hfill \\
\end{gathered} $
|
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|
$T(n) = 2T\left(\lfloor n/2\rfloor +1\right) + T(n-1) + 1$, $T(3) = 1$ I am looking for asymptotic bound on $T$, where $T$ satisfies the following recurrence:
$$T(n) = 2T\left(\lfloor{\frac{n}{2}\rfloor} +1\right) + T(n-1) + 1\text{ , }T(3) = 1.$$
I plotted the graph for $T$, and it seems like $T$ grows somewhere between a polynomial and an exponential. I want to know whether $T$ has exponential growth or slower growth.
|
Your prediction is correct, $T(n)$ has growth rate between polynomial and exponential. Let $F(x)$ be defined by $$
F(x) = \sum_{n=0}^\infty \frac{c_n (x-3)^n}{n!}
$$
where $c_n$ is defined by $c_0=1$ and $c_n = 3 \cdot 2^{-\binom{n-1}2}$ for $n\ge 1$. This series converges (quite rapidly, I would add) for all $x\in \mathbb{C}$.
Claim: For all $n\ge 3$ $$
F(n) \ge T(n) \ge C + (2C+1)F(n+3)
$$
where $C = \frac{1-F(6)}{1+2F(6)} \approx -0.477$.
We observe that $F(x)$ clearly has great-than-polynomial growth rate. It is an entire function with order $0$, which means that for any $\alpha >0$, $$
\lim\limits_{x\rightarrow\infty} \frac{F(x)}{\exp(x^\alpha)} = 0
$$
See 1.
Proof of upper bound: We observe that $c_{n+1} = 2^{1-n}c_n$ for $n\ge 1$, which implies \begin{eqnarray}
F'(x-1) &=& 2F(x/2 + 1) + 1
\end{eqnarray}
because \begin{eqnarray}
2F(x/2 + 1) + 1&=&
1+\sum_{n=0}^{\infty}\frac{2c_n(\frac x2 -2)^n}{n!}
\\&=& 1+\sum_{n=0}^\infty \frac{2^{1-n}c_n(x-4)^n}{n!}\\
&=&3 + \sum_{n=1}^\infty \frac{2^{1-n}c_n (x-4)^n}{n!}\\
&=&\sum_{n=0}^\infty\frac{c_{n+1}(x-4)^n}{n!} = F'(x-1).\end{eqnarray}
We claim that $T(n) \le F(n)$ for all $n\ge3$, which can be shown inductively. It is trivial for $n=3$. We observe \begin{eqnarray}
T(n) &=& 2T(\lfloor n/2 \rfloor + 1)+T(n-1)+1\le 2 F(\lfloor n/2\rfloor + 1) + F(n-1) + 1 \\&\le& 2F(n/2 + 1) + F(n-1) + 1 = F'(n-1) +F(n-1)\\
&=& \int_{n-1}^{n} F'(n-1) dx + F(n-1) \le \int_{n-1}^n F'(x)dx + F(n-1)\\
&=& F(n)
\end{eqnarray}
where we make use of the fact that $F(x)$ is convex for $x\ge 3$.
Proof of lower bound: For convenience, denote $G(x) = C + (2C+1)F(x+3)$. We observe \begin{eqnarray}
G'(x) &=& (2C+1)F'(x+3) = (2C+1)(2F(\frac {x+4}2 + 1) + 1)\\&=&2(2C+1)F(\frac x 2 + 3) + 2C+1=2G(x/2) + 1
\end{eqnarray} and \begin{eqnarray}
G(3) &=& C + (2C + 1)F(6) = \frac{1-F(6)}{1+2F(6)} + 2\frac{F(6)-F(6)^2}{1+2F(6)}+F(6)\\
&=&\frac{1 + F(6) - 2F(6)^2}{1+2F(6)}+F(6) = \frac{(1-F(6))(1+2F(6))}{1+2F(6)} + F(6) = 1
\end{eqnarray}
We use induction to show that $T(n) \ge G(n)$ for all $n\ge 3$, using the above as our base case, we proceed (again making use of convexity):\begin{eqnarray}
T(n)&=&2T(\lfloor n/2 \rfloor + 1) + T(n-1)+1 \ge 2G(\lfloor n/2 \rfloor + 1) + G(n-1) + 1\\
&\ge& 2G(n/2) + G(n-1)+1= G'(n) + G(n-1) \\&\ge& \int_{n-1}^{n}G'(x)dx + G(n-1) = G(n)
\end{eqnarray}
completing the proof.
Update - Numerical demonstration: After doing some numerical simulations, it looks like $$
\lim_{n\rightarrow\infty} \frac{T(n)}{F(n)} \approx 0.2656...
$$
Here's a plot of the first thousand terms of the ratio $T(n)/F(n)$, with the horizontal red line at $y=0.2656$:
I did the computation all the way out to $50000$. The ratio stays flat on that line. It also looks like $$
\log F(x) \sim (\log x)^\alpha$$
for some $\alpha \approx 1.81$.
|
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|
Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$ I saw the problem from one math competition:
$$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.
I tried to solve it this way:
\begin{align*}
& \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x\left(x^{3}-3x+1\right)+1=0\\
\Leftrightarrow\,\, & \left(x^{3}-3x+1\right)\left(\sqrt{x^{2}-1}+x\right)=-1
\end{align*}
Now I tried to use substitutions like $a=x^3-3x$ and $b=\sqrt{x^{2}-1}$ but I cannot get anything. I know that solutions are $x=1$ and $x=\pm\sqrt{2}$.
|
Of course, $x= \pm \sqrt{2}$ are two roots of this equation. Assume $x^2 \ne 2$, we have:
\begin{align*}
&(x^{3} - 3x + 1)(\sqrt{x^{2} - 1} + x) = -1 &\Longleftrightarrow & x^{3} - 3x + 1 = \sqrt{x^{2} - 1}-x \\\\
\Longleftrightarrow & x^{3} - 2x + 1 = \sqrt{x^{2} - 1} & \Longleftrightarrow & x^{3} - 2x = \sqrt{x^{2} - 1}-1\\\\
\Longleftrightarrow & (x^{2} - 2)x = \frac{x^2-2}{\sqrt{x^{2} - 1}+1} & \Longleftrightarrow & x = \frac{1}{\sqrt{x^{2} - 1}+1}
\end{align*}
Clearly, the only root for the last equation is $1$ because $LHS \ge 1 \ge RHS$.
So, the initial equation has exactly 3 roots $\{ 1, \sqrt{2},-\sqrt{2}\}$.
|
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|
Prove $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{7\left(a^3+b^3+c^3+a^2b+b^2c+c^2a+b^2a+a^2c+c^2b\right)-3}$ with $a^2+b^2+c^2=1$ and $a,b,c\ge 0$ With $a,b,c\ge 0$ and $a^2+b^2+c^2=1$, prove $\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\ge \sqrt{7\left(a^3+b^3+c^3+a^2b+b^2c+c^2a+b^2a+a^2c+c^2b\right)-3}$.
I don't know when equality holds for this problem, can you give me some hint? Thank!
|
It suffices to prove that
$$(\sqrt{a + b} + \sqrt{b + c} + \sqrt{c + a})^2 \ge 7(a + b + c) - 3$$
or
$$2a + 2b + 2c + \sum_{\mathrm{cyc}} 2\sqrt{(a + b)(b + c)} \ge 7(a + b + c) - 3.$$
Using Cauchy-Bunyakovsky-Schwarz inequality and GM-HM inequality, we have
$$\sqrt{(a + b)(b + c)} \ge b + \sqrt{ca} \ge b + \frac{2ca}{c + a}
\ge b + \frac{2ca}{a + b + c}.$$
It suffices to prove that
$$4a + 4b + 4c + \frac{4(ab + bc + ca)}{a + b + c} \ge 7(a + b + c) - 3$$
or
$$4a + 4b + 4c + \frac{2 \cdot [(a + b + c)^2 - 1]}{a + b + c} \ge 7(a + b + c) - 3$$
or
$$(a + b + c - 1)(2 - a - b - c) \ge 0$$
which is true (easy).
We are done.
|
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|
Analytical approach to this 2nd order ODE? Is there any analytical approach to solving an ODE of the form below? It could be more elegant than numerical approximation...
$$y''-\frac{a}{y^3}+b=0,\quad\textrm{where $a$ and $b$ are constants.}$$
If anyone has any pointers it would be greatly appreciated!
|
Another option is
Solve
\begin{gather*}
\boxed{y^{\prime \prime}-\frac{a}{y^{3}}+b=0}
\end{gather*}
Writing the ode as
\begin{align*}
y^{\prime \prime}&=\frac{a -y^{3} b}{y^{3}}
\end{align*}
Multiplying both sides by $y^{\prime}$ gives
\begin{align*}
y^{\prime} y^{\prime \prime}&=\frac{\left(a -y^{3} b \right) y^{\prime}}{y^{3}}
\end{align*}
Integrating both sides w.r.t. $x$ gives
\begin{align*}
\int{y^{\prime} y^{\prime \prime}\, \mathrm{d}x} &=\int{\frac{\left(a -y^{3} b \right) y^{\prime}}{y^{3}}\, \mathrm{d}x}\\
\int{y^{\prime} y^{\prime \prime}\, \mathrm{d}x} &=\int{\frac{a -y^{3} b}{y^{3}}\, \mathrm{d}y} \tag{1}
\end{align*}
But
$$
\int{y^{\prime} y^{\prime \prime}\, \mathrm{d}x} = \frac{1}{2} \left(y^{\prime}\right)^2
$$
And
$$
\int{\frac{a -y^{3} b}{y^{3}}\, \mathrm{d}y} = -b y-\frac{a}{2 y^{2}}
$$
Therefore equation (1) becomes
\begin{align*}
\frac{1}{2} \left(y^{\prime}\right)^2 &=-b y-\frac{a}{2 y^{2}} + c_2
\end{align*}
Where $c_2$ is an arbitrary constant of integration.
This is first order ODE separable ode which is now solved for $y$.
Solving for $y^{\prime}$ gives 2 ode's
\begin{align*}
y^{\prime}&=\frac{\sqrt{-2 y^{3} b +2 c_{2} y^{2}-a}}{y}\tag{1A} \\
y^{\prime}&=-\frac{\sqrt{-2 y^{3} b +2 c_{2} y^{2}-a}}{y}\tag{2A}
\end{align*}
Solving (1A) only (as 2A is similar)
\begin{align*}
\frac{y}{\sqrt{-2 y^{3} b +2 y^{2} c_{2}-a}}\mathop{\mathrm{d}y} &= \mathop{\mathrm{d}x}\\
\int \frac{y}{\sqrt{-2 y^{3} b +2 y^{2} c_{2}-a}}\mathop{\mathrm{d}y} &= x +c_{1}
\end{align*}
Using Fricas integrator, the above becomes
\begin{align*}
\frac{\sqrt{2}}{3 \, \sqrt{-b} b} \Delta &= \left(x +c_{1}\right)
\end{align*}
Where
\begin{equation}
\begin{aligned}
\Delta ={} & c_{2} {\operatorname{weierstrassPInverse}}\left(\frac{4 \, c_{2}^{2}}{3 \, b^{2}}, -\frac{2 \, {\left(27 \, a b^{2} - 4 \, c_{2}^{3}\right)}}{27 \, b^{3}}, \frac{3 \, b y - c_{2}}{3 \, b}\right) - \\
&\quad 3 \, b {\operatorname{weierstrassZeta}}\left(\frac{4 \, c_{2}^{2}}{3 \, b^{2}}, -\frac{2 \, {\left(27 \, a b^{2} - 4 \, c_{2}^{3}\right)}}{27 \, b^{3}}, {\operatorname{weierstrassPInverse}}\left(\frac{4 \, c_{2}^{2}}{3 \, b^{2}}, -\frac{2 \, {\left(27 \, a b^{2} - 4 \, c_{2}^{3}\right)}}{27 \, b^{3}}, \frac{3 \, b y - c_{2}}{3 \, b}\right)\right)
\end{aligned}
\end{equation}
Same for second ode.
Fricas code
r:=integrate(y/sqrt(-2*y^3*b+2*y^2*c-a),y);
(3)
"((-6)*b*weierstrassZeta((4*c^2)/(3*b^2),(8*c^3+(-54)*a*b^2)/(27*b^3),weierst
rassPInverse((4*c^2)/(3*b^2),(8*c^3+(-54)*a*b^2)/(27*b^3),(3*b*y+(-1)*c)/(3*b
)))+2*c*weierstrassPInverse((4*c^2)/(3*b^2),(8*c^3+(-54)*a*b^2)/(27*b^3),(3*b
*y+(-1)*c)/(3*b)))/(3*b*((-2)*b)^(1/2))"
|
{
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"url": "https://math.stackexchange.com/questions/4491652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Find $\int_{t}^1 \frac{\sin(\pi x)^2}{(\sin(\pi x/2)^2-a^2)^2} \ dx. $ Let $a \in (0,1)$, I would like to integrate
$$\int_{t}^1 \frac{\sin(\pi x)^2}{(\sin(\pi x/2)^2-a^2)^2} \ dx. $$
Now $\sin(\pi x/2)^2$ is a monotonically increasing function from $0$ to $1$, therefore there exists a unique $t^*$ such that $\sin(\pi t^*/2)^2 = a^2.$ Now $t^* <t$ such that the integral is well-defined.
Please let me know if you have any questions.
|
Use the result in this post:
Note that
\begin{align}
\frac{\frac14\sin^2\pi x}{\sin^2\frac{\pi x}2-a^2}
=-a^2 + \cos^2\frac{\pi x}2 + \frac{a^2(1-a^2)}{\sin^2\frac{\pi x}2-a^2}
\end{align}
Integrate respectively to obtain
\begin{align}
&\int_ 0^{x_{\downarrow}} \frac{\sin^2\pi x}{\sin^2\frac{\pi x}2-a^2}dx\\
=& \ 2(1-2a^2) x_{\downarrow} +\frac2\pi\sin\pi x_{\downarrow}
-\frac{8a\sqrt{1-a^2}}\pi\tanh^{-1}\frac{\sqrt{1-a^2} \tan\frac{\pi x_{\downarrow}}2}a
\\\\
&\int^1_{x_{\uparrow}} \frac{\sin^2\pi x}{\sin^2\frac{\pi x}2-a^2}dx\\
=& \ 2(1-2a^2)(1- x_{\uparrow})-\frac2\pi\sin\pi x_{\uparrow}
+\frac{8a\sqrt{1-a^2}}\pi\coth^{-1}\frac{\sqrt{1-a^2} \tan\frac{\pi x_{\uparrow}}2}a
\end{align}
Set $a^2=t$ and take derivative with respect to $t$, and replace $t=a^2$ back. It is done.
Further, by taking the $(n-1)^{\text{th}}$ derivative with respect to $t$, you can generalize it into
$$\int^1_{x_{\uparrow}} \frac{\sin^2(\pi x)}{\left(\sin^2\left(\frac{\pi x}{2}\right)-a^2\right)^n}dx$$
Moreover, by taking the integral with respect to $t$, you can get
$$\int^1_{x_{\uparrow}} \sin^2(\pi x)\cdot \ln\left(\sin^2\left(\frac{\pi x}{2}\right)-a^2\right) dx$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4494665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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|
Evaluating $\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$ $$
\begin{align*}
&\text { Let } \mathrm{x}=2 \mathrm{y} \quad \because x \rightarrow 0 \quad \therefore \mathrm{y} \rightarrow 0\\
&\therefore \lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-6(2 y)^{2}-12 \cos 2 y}{(2 y)^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-24 y^{3}-12 \cos 2 y}{16 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3(1-\cos 2 y)-6 y^{2}}{4 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3.2 \sin ^{2} y-6 y^{2}}{4 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{ 3\left\{y-\frac{y^{3}}{3 !}+\frac{y^{5}}{5 !}-\cdots \infty\right\}^{2}-3 y^{2}}{2 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3\left[y^{2}-\frac{2 y^{4}}{3 !}+\left(\frac{1}{(3 !)^{2}}+\frac{2}{3 !}\right) y^{4}+\cdots \infty\right)^{2}-3 y^{2}}{2 y^{4}}\\
&=\lim _{y \rightarrow 0} \frac{3\left\{y^{2}-\frac{2 y^{4}}{3 !}+\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}+y^{4}+\cdots \infty\right)-3 y^{2}\right.}{2 y^{4}}\\
&=\lim _{y \rightarrow 0}\left[\frac{-\frac{6}{3 !}+3\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}\right\} y^{2}+\cdots \infty}{2}\right]\\
&=\lim _{y \rightarrow 0}\left[\frac{-1+3\left\{\frac{1}{(3 !)^{2}}+\frac{2}{5 !}\right\} y^{2}+\cdots \infty}{2}\right]\\
&=-\frac{1}{2} \text { (Ans.) }
\end{align*}
$$
Doubt
Can anyone please explain the 5,6,7 equation line? Thank you
|
Use $\cos z=1-z^2/2!+z^4/4!+...$
$$L=\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$$$$=\displaystyle\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 (1-\frac{x^2}{2}+\frac{x^4}{24}+...)}{x^{4}}=-\frac{1}{2}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4496409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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|
How to calculate $\sum _{n=1}^{\infty }\:\left(\frac{2}{3}\right)^n\sin\left(\frac{\pi }{3}n\right)$ I believe that Complex numbers should be used in order to calculate this.
Let $z = \frac{2}{3}e^{\frac{i\pi }{3}}$,
So,
$$\sum _{n=1}^{\infty }\left(\frac{2}{3}\right)^n\sin\left(\frac{\pi }{3}n\right)=\sum _{n=1}^{\infty }\text{Im}\left(\frac{2}{3}e^{\frac{i\pi}{3}}\right)^n=\text{Im}\left[\sum _{n=1}^{\infty }\left(\frac{2}{3}e^{\frac{i\pi}{3}}\right)^n\right]$$
Is this correct?
How do you go about solving it further? I believe this might be a infinite geometric series where,
$a_1 = z, q = z$?
|
It is also possible to do it directly in the real domain.
Since the series is absolutely convergent, we can rearrange terms to group the same values of the sinus.
*
*$n\equiv 0,3\pmod 6\implies \sin(\frac{n\pi}3)=0$
*$n\equiv 1,2\pmod 6\implies \sin(\frac{n\pi}3)=\frac 12\sqrt{3}$
*$n\equiv 4,5\pmod 6\implies \sin(\frac{n\pi}3)=-\frac 12\sqrt{3}$
Let $a=\frac 23$ then :
$\displaystyle\begin{align}\frac{2S}{\sqrt{3}}
&=\sum\limits_{n=0}^\infty\left(\frac 23\right)^{6n+1}+\sum\limits_{n=0}^\infty\left(\frac 23\right)^{6n+2}-\sum\limits_{n=0}^\infty\left(\frac 23\right)^{6n+4}-\sum\limits_{n=0}^\infty\left(\frac 23\right)^{6n+5}\\\\
&=(a+a^2-a^4-a^5)\sum\limits_{n=0}^\infty {(a^6)}^n\\\\
&=(a+a^2-a^4-a^5)\,\frac{1}{1-a^6}\\\\
&=\frac{a}{a^2-a+1}=\frac 67
\end{align}$
Therefore $S=\dfrac{3\sqrt{3}}{7}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4498220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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|
$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$ Someone on Youtube posted a video solving this integral.
I can't find on Math.stack.exchange this integral using search engine https://approach0.xyz
It is related to $\displaystyle \int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx$
Following is a solution that is not requiring the use of series:
\begin{align}J&=\int_0^1 \frac{\arcsin x\arccos x}{x}dx\\
&\overset{\text{IBP}}=\underbrace{\Big[\arcsin x\arccos x\ln x\Big]_0^1}_{=0}-\underbrace{\int_0^1 \frac{\arccos x\ln x}{\sqrt{1-x^2}}dx}_{x=\cos t }+\underbrace{\int_0^1 \frac{\arcsin x\ln x}{\sqrt{1-x^2}}dx}_{x=\sin t}\\
&=\int_0^{\frac{\pi}{2}} t\ln(\tan t)dt\\
&\overset{u=\tan t}=\int_0^\infty \frac{\arctan u\ln u}{1+u^2}du\\
&\overset{\text{IBP}}=\underbrace{\left[\arctan u\left(\int_0^u \frac{\ln t}{1+t^2}dt\right)\right]_0^\infty}_{=0}-\int_0^\infty \frac{1}{1+u^2}\left(\underbrace{\int_0^u \frac{\ln t}{1+t^2}dt}_{y(t)=\frac{t}{u}}\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln(uy)}{(1+u^2)(1+u^2y^2)}dy\right)du\\
&=-\int_0^\infty \left(\int_0^1 \frac{u\ln u}{(1+u^2)(1+u^2y^2)}dy\right)du-\int_0^1 \left(\int_0^\infty \frac{u\ln y}{(1+u^2)(1+u^2y^2)}du\right)dy\\
&=-\int_0^\infty \left[\frac{\arctan(uy)}{1+u^2}\right]_{y=0}^{y=1}\ln udu-\frac{1}{2}\int_0^1 \left[\frac{\ln\left(\frac{1+u^2}{1+u^2y^2}\right)}{1-y^2}\right]_{u=0}^{u=\infty}\ln ydy\\
&=-J+\int_0^1 \frac{\ln^2 y}{1-y^2}dy\\
&=\frac{1}{2}\int_0^1 \frac{\ln^2 y}{1-y}dy-\frac{1}{2}\underbrace{\int_0^1 \frac{y\ln^2 y}{1-y^2}dy}_{z=y^2}\\
&=\frac{7}{16}\int_0^1 \frac{\ln^2 y}{1-y}dy\\
&=\frac{7}{16}\times 2\zeta(3)=\boxed{\frac{7}{8}\zeta(3)}
\end{align}
NB:I assume $\displaystyle \int_0^1 \frac{\ln^2 y}{1-y}dy=2\zeta(3)$
Feel free to post your solution.
|
Use the relation: $$\arcsin(x)+\arccos(x)=\frac{\pi}2$$
So,
$$J=\frac{\pi}2\int_0^1 \frac{\arcsin(x)}{x}dx-\int_0^1\frac{\arcsin^2(x)}{x} dx=\frac{\pi}2I_1-I_2$$
Let $t=\arcsin(x)$
$$I_1=\int_0^{\pi/2} \frac{t}{\tan(t)} dt,~~~~I_2=\int_0^{\pi/2} \frac{t^2}{\tan(t)}dt$$
For each integral use series:
$$\frac{1}{\tan(t)}=\frac{1}{t}+\sum_{n=1}^\infty \frac{2t}{t^2-n^2\pi^2}$$
Integrate term by term and it is done.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4500073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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|
$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive 4 real roots.
$a, b, c, d, e$ are distinct real numbers. Prove that this equation has distinctive four real roots.
$$(x-a)(x-b)(x-c)(x-d) \\ +(x-a)(x-b)(x-c)(x-e) \\ +(x-a)(x-b)(x-d)(x-e) \\ +(x-a)(x-c)(x-d)(x-e) \\ + (x-b)(x-c)(x-d)(x-e) =0.$$
My attempt:
\begin{align}
& f(x)=\sum_{cyc} (x-a)(x-b)(x-c)(x-d). \\
& f(a)=(a-b)(a-c)(a-d)(a-e). \\
& f(b)=(b-a)(b-c)(b-d)(b-e). \\
& f(c)=(c-a)(c-b)(c-d)(c-e). \\
& f(d)=(d-a)(d-b)(d-c)(d-e). \\
& f(e)=(e-a)(e-b)(e-c)(e-d). \\
\ \\
& f(0)=\sum_{cyc} abcd.\\
\end{align}
I put some arbitrary integers in $a, b, c, d, e$ and drew a graph of $\displaystyle f(x)=\sum_{cyc} (x-a)(x-b)(x-c)(x-d).$
It sure makes 4 distinct roots... How can we prove this?
Values in the graph above:
$a=-1.34, b=-4.67, c=-2.91, d=0.33, e=-6.09$
|
This can also be approached without theorems from calculus, though much of it uses ideas from pre-calculus that lead into limits and the Intermediate Value Theorem. For this argument, we will use a simpler function with just three terms, $ \ g(x) \ = \ (x - a)·(x - b) + (x - a)·(x - c) + (x - b)·(x - c) \ \ , $ from which we will form the rational function
$$ R(x) \ \ = \ \ \frac{g(x)}{(x - a)·(x - b)·(x - c)} \ \ = \ \ \frac{1}{x \ - \ a} \ + \ \frac{1}{x \ - \ b} \ + \ \frac{1}{x \ - \ c} \ \ . $$
We know that $ \ R(x) \ $ has vertical asymptotes at $ \ x = a \ \ , \ \ x = b \ \ , \ \ x = c \ \ $ (or the domain of $ \ R(x) \ $ is $ \ x \ \neq \ a \ , \ b \ , \ c \ ) \ . \ $ It also has the horizontal asymptote $ \ y \ = \ 0 \ \ $ (or $ \ \lim_{ \ x \ \rightarrow \ \pm \ \infty} R(x) \ = \ 0 \ ) \ . $
For $ \ x \ < \ a \ \ , \ R(x) \ < \ 0 \ \ , \ $ while for $ \ x \ > \ c \ \ , \ R(x) \ > \ 0 \ \ . $
We are particularly interested in the intervals between the vertical asymptotes. In $ \ (a \ , \ b) \ \ , \ $ we see that the $ \ \frac{1}{x \ - \ a} \ $ term becomes a larger and larger positive number as $ \ x \ $ gets closer to $ \ a \ $, while the sum of the other terms is "nearly constant", so $ \ R(x) \ $ becomes a larger and larger positive number (or $ \ \lim_{ \ x \ \rightarrow \ a^{+}} R(x) \ = \ +\infty \ ) \ . \ $ By the same token, the $ \ \frac{1}{x \ - \ b} \ $ term becomes a larger and larger negative number as $ \ x \ $ gets closer to $ \ b \ $, while the sum of the other terms is "nearly constant", so $ \ R(x) \ $ becomes a larger and larger negative number (or $ \ \lim_{ \ x \ \rightarrow \ b^{-}} R(x) \ = \ -\infty \ ) \ . \ $ Since $ \ R(x) \ $ is defined and continuous (however that is expressed in a particular pre-calculus course) "everywhere in the interval", there is (at least) one value $ \ x \ = \ p \ \ , \ \ a \ < \ p \ < \ b \ $ for which $ \ R(p) \ = \ 0 \ \ . $
For the overall purpose of this discussion, we need to show that there is only one such value of $ \ x \ $ in the interval. We have $ \ \frac{1}{p \ - \ a} + \frac{1}{p \ - \ b} + \frac{1}{p \ - \ c} \ = \ 0 \ \ . \ $ For $ \ a \ < \ X \ < \ p \ \ , $
$$ R(X) \ = \ \left[ \ \frac{1}{X \ - \ a} \ - \ \frac{1}{p \ - \ a} \ \right] \ + \ \left[ \ \frac{1}{X \ - \ b} \ - \ \frac{1}{p \ - \ b} \ \right] \ + \ \left[ \ \frac{1}{X \ - \ c} \ - \ \frac{1}{p \ - \ c} \ \right] \ \ $$
$$ = \ \frac{p \ - \ X}{(X \ - \ a)·(p \ - \ a)} \ + \ \frac{p \ - \ X}{(b \ - \ X)·(b \ - \ p)} \ + \ \frac{p \ - \ X}{(c \ - \ X)·(c \ - \ p)} \ \ > \ \ 0 \ \ , $$
since the numerators and denominators of each term are positive. By a similar argument, we can show that $ \ R(X) \ < \ 0 \ \ $ for $ \ p \ < \ X \ < \ b \ \ . \ $ So there is no other zero of $ \ R(x) \ $ in $ \ ( a \ , \ b) \ \ . \ $ [I was unable to find a convincing proof that $ \ R(x) \ $ is one-to-one in the interval; and while solving $ \ g(x) \ = \ 0 \ $ does show, with some effort, that one of the zeroes of this quadratic polynomial lies in $ \ (a \ , \ b) \ $ and the other does not, that method is not easily generalizable.]
Hence, as the denominator of $ \ R(x) \ $ is not zero at $ x \ = \ p \ $ or $ \ x \ = \ q \ \ , \ \ g(x) \ $ has exactly two zeroes, one each in $ \ (a \ , \ b) \ $ and $ \ (b \ , \ c ) \ \ . \ $ We can generalize this line of reasoning to show that for a function $ \ p(x) \ $ being the sum of every distinct product of $ \ (n - 1) \ $ factors from the set $ \ \large\{ \ (x - a_1) \ , \ (x - a_2) \ , \ldots , \ (x - a_n) \ \large\} \ \ , \ $ so $ \ \binom{n}{n-1} \ = \ n \ $ terms in all, the $ \ a_i \ $ being distinct, there are $ \ (n - 1) \ $ intervals, $ \ (a_1 \ , \ a_2) \ \ , \ \ (a_2 \ , \ a_3) \ \ , \ldots , \ \ (a_{n-1} \ , \ a_n) \ \ , \ $ each of which contains just one zero of $ \ p(x) \ \ . $
|
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|
Find all $(p,n)$ such that $p^n+1\mid n^p+1$ where $p$ is prime. Let $p$ be a prime number. Find all $(p,n)$ such that $$p^n+1\mid n^p+1.$$
Where $n$ is a positive integer.
Here is some obvious solutions $(p,p)$ for any prime $p$ and $(2,4)$.
If $p=2$ then we get $2^n+1\mid n^2+1$ therefore $2^n\le n^2$ but for $n>4$ we have $2^n>n^2$ so no solution. And the only solution for $n\le 4$ is $(2,2)$ and $(2,4)$.
Now assume $p$ is odd then $p+1\mid p^n+1$ therefore $$n^p\equiv -1\pmod{p+1} \implies n^{2p}\equiv 1\pmod{p+1}$$
So $a=\operatorname{ord}_{p+1}{n}\in \{1,2,2p\}$ if $a=1$ then $n\equiv 1\pmod{p+1}\implies n^p\equiv 1\pmod {p+1}$ which is a contradiction.
But I don't know what to do if $a=2$ or $2p$.
|
If $p = 2$, then $n^2 + 1 \geq 2^n + 1$ is required (since $2^n + 1 \mid n^2 + 1$ and both sides are positive), which implies $n \leq 4$, which we can see only gives solutions $(p, n) = (2, 2), (2, 4)$.
Now if $p \neq 2$, then $p$ is an odd prime, and so $n$ must be odd since $p^n + 1$ is even. Note that $n \geq 3$ since $n \neq 1$, and so we have $p^n + 1 \mid n^p + 1 \implies p^{\frac{1}{p}} \leq n^{\frac{1}{n}}$. But $x^{\frac{1}{x}}$ is decreasing for $x \geq e$, so we must have $n \leq p$.
Now we note $p + 1 \mid p^n + 1 \mid n^p + 1$, so $n^p \equiv -1 \pmod{p+1}.$ Since $p$ is odd, this means $(-n)^p\equiv 1\pmod{p+1}.$
Euler's theorem tells us that $(-n)^{\phi(p+1)} \equiv 1 \pmod{p+1}$, and so by the $\gcd$ trick we get $(-n )^{\gcd(p, \phi(p+1))} \equiv 1 \pmod{p+1}$.
But we note that $2\mid p + 1$ so $\phi(p+1) \leq \frac{p+1}{2}$, so $p$ cannot be a factor of it. Therefore, $\gcd(p, \phi(p+1)) = 1$, and so $-n\equiv 1 \pmod{p+1}$. We conclude that $n \equiv -1 \pmod{p+1}$, and combining with $n \leq p$ yields $n = p$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Summation of reciprocal products When studying summation of reciprocal products I found some interesting patterns.
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)}=\frac{1}{1\cdot1!}-\frac{1}{1\cdot(N+1)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)}=\frac{1}{2\cdot2!}-\frac{1}{2\cdot(N+1)(N+2)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)}=\frac{1}{3\cdot3!}-\frac{1}{3\cdot(N+1)(N+2)(N+3)}$$
The pattern is $$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)\cdot\cdot\cdot\cdot(k+i)}\space=\space\frac{1}{i\cdot i!}-\frac{1}{i\cdot(N+1)(N+2)(N+3)\cdot\cdot\cdot\cdot(N+i)}$$
which is easy to prove by induction.
As an easy consequence it follows that $$\lim_{N \to \infty} \sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)\cdot\cdot\cdot\cdot(k+i)}\space=\space\frac{1}{i\cdot i!}$$
I did not find this summation of reciprocal products in my mathbooks. Is this a known theorem?
Can anyone point me in the right direction for further study?
|
I think I've posted this before,
but what the heck.
The telescoping occurs not only
for the product of consecutive integers
but for its reciprocal.
Here's the proofs.
$p_m(x)
=\prod_{k=0}^{m-1} (x+k)
$.
$\begin{array}\\
p_m(x+1)-p_m(x)
&=\prod_{k=0}^{m-1} (x+1+k)-\prod_{k=0}^{m-1} (x+k)\\
&=\prod_{k=1}^{m} (x+k)-\prod_{k=0}^{m-1} (x+k)\\
&=(x+m-x)\prod_{k=1}^{m-1} (x+k)\\
&=m\prod_{k=0}^{m-2} (x+1+k)\\
&=mp_{m-1}(x+1)\\
p_m(x)-p_m(x-1)
&=mp_{m-1}(x)\\
\dfrac1{p_m(x)}-\dfrac1{p_m(x+1)}
&=\dfrac1{\prod_{k=0}^{m-1} (x+k)}-\dfrac1{\prod_{k=0}^{m-1} (x+1+k)}\\
&=\dfrac1{\prod_{k=0}^{m-1} (x+k)}-\dfrac1{\prod_{k=1}^{m} (x+k)}\\
&=\dfrac1{\prod_{k=1}^{m-1} (x+k)}(\dfrac1{x}-\dfrac1{x+m})\\
&=\dfrac1{\prod_{k=1}^{m-1} (x+k)}(\dfrac{x+m-x}{x(x+m)})\\
&=\dfrac{m}{\prod_{k=0}^{m} (x+k)}\\
&=\dfrac{m}{p_{m+1}(x)}\\
\end{array}
$
$\begin{array}\\
q_m(n)
&=\sum_{x=1}^{n} p_{m}(x)\\
&=\dfrac1{m+1}\sum_{x=1}^{n} (p_{m+1}(x)-p_{m+1}(x-1))\\
&=\dfrac{p_{m+1}(n)}{m+1}\\
r_m(n)
&=\sum_{x=n}^{\infty}\dfrac1{p_{m}(x)}\\
&=\dfrac1{m-1}\sum_{x=n}^{\infty}\left(\dfrac1{p_{m-1}(x)}-\dfrac1{p_{m-1}(x+1)}\right)\\
&=\dfrac1{(m-1)p_{m-1}(n)}\\
\end{array}
$
|
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"url": "https://math.stackexchange.com/questions/4506521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Evaluate $\lim_{n \to \infty} \frac{a_n}{2 ^ {n - 1}}$ if $a_n = a_{n - 1} + \sqrt{a_{n - 1}^2 + 1}$
Let $a_i (i \in \mathbb{N}_{0})$ be a sequence of real numbers such that $a_0 = 0$ and $$a_n = a_{n - 1} + \sqrt{a_{n - 1}^2 + 1} \text{ } \forall n \geq 1$$ Evaluate the limit $$\lim_{n \to \infty} \frac{a_n}{2 ^ {n - 1}}$$
Hello, I am trying to solve this problem. I honestly have no idea how to approach this, but I think the answer will be $1$ because as $n \to \infty$, $a_n$ gets bigger and $a_n \approx 2a_{n-1}$ and since $a_1 = 1 = 2^{0}$, the limit will approach $1$. (Of course, this is just a guess).
The intended solution is too much magic.
Substitute $a_n = \cot \theta$.
Now $$a_{n + 1} = a_n + \sqrt{a_{n}^2 + 1} = \cot \theta + \csc \theta$$
$$=\frac{\cos \theta + 1}{\sin \theta}$$
$$=\frac{2 \cos^2{\frac{\theta}{2}}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}$$
$$=\cot \frac{\theta}{2}$$
And now, we can solve the limit and answer is $\frac{4}{\pi}$.
But, is there a more normal method to solving this (other than just thinking out of nowhere that substituting $a_n = \cot \theta$ is helpful)?
Please note that my question is about finding an alternative solution that's much more "thinkable". So, this is not a duplicate.
Thanks
|
This is a general method that you may be able to use.
Suppose given a function
$$f(x) := x + \sqrt{x^2+1}. $$
Given a number $\,a_0\ge 0,\,$ define the sequence
$$a_n := f(a_{n-1}). $$
Note the series expansion $$ f(x) = 2x + \frac1{2x} - \frac1{8x^3} + \frac1{16x^5} + \cdots. $$
Iterate to get
$$ f^2(x) := f(f(x)) = 4x\left( 1 +\frac5{16x^2}
-\frac{21}{256x^4} + \cdots\right). $$
In general,
$$ f^n(x) := f(f^{n-1}(x)) = 2^nx\left( 1
+ \frac{c_n}{x^2} + \frac{d_n}{x^4} + \cdots\right). $$
With a bit of work, take the limit to get
$$ g(x) := \lim_{n\to\infty} \frac{f^{n}(x)}{2^n} =
x\left(1 +\frac{1/3}{x^2} - \frac{4/45}{x^4} +
\frac{44/945}{x^6} + \cdots\right). $$
In this particular case,
$$ \frac1{g(x)} = \frac1{x} - \frac1{3x^3} +
\frac1{5x^5} - \frac1{7x^7} + \cdots = \tan^{-1}(1/x).$$
So define $\,b_n := a_{n+1}\,$ where $\,b_0 = 1.\,$ Then
$$ \lim_{n \to \infty} \frac{a_n}{2^{n-1}} =
\lim_{n \to \infty} \frac{a_{n+1}}{2^n} =
\lim_{n \to \infty} \frac{b_n}{2^n} = 1/\tan^{-1}(1/b_0) = 4/\pi.$$
The purpose of this section is to derive the equation
$$ \frac1{g(x)} = \tan^{-1}(1/x) $$
which appears near the end of the previous section.
The general idea is to study discrete dynamical systems and
their limiting behavior. Thus, given a subset $\,X\,$ of
real numbers, a self-map $\,\phi:X\to X,\,$ and a point
$\,a_0\,$ in $\,X\,$ define the sequence $\,(a_0,a_1,a_2,
\dots)\,$ by recursion with $\,a_{n+1} = \phi(a_n)\,$ for
$n>0.\,$
Another way to think of this situation is from the standpoint of
iterated functions.
That is, given a self-map $\,\phi:X\to X,\,$ define a sequence of
functions by recursion as $\,\phi^0: x \mapsto x,\,$ and
$\,\phi^{n+1}: x \mapsto \phi(\phi^n(x)).\,$ Thus, the
sequence $\,a_n\,$ is also defined by $\,a_n = \phi^n(a_0).\,$
We are interested in fixed points of the self-map and convergence to
fixed points.
In our particular case, $\,f(x) = x+\sqrt{x^2+1}\,$ and the sequence
$\,a_n\,$ is unbounded. Thus, it is more convenient to define
$$ \phi(x) := \frac1{f(1/x)} = \frac{x}{1+\sqrt{1+x^2}} =
\frac{x}2 - \frac{x^3}8 + \frac{x^5}{16} - \frac{5x^7}{128} + \cdots $$
This implies that iterates of $\,\phi\,$ converge to the fixed point $0$.
The convergence to this limit is linear. More precisely, define
$$ A_t(x) := x\,t\left(1 - x^2\frac{1-t^2}3\left(1
- x^2\frac{3-2t^2}5 + x^4\frac{45-53t^2+17t^4}{105} + \cdots\right)\right)$$
Verify that this function satisfies the identities
$$ A_u(A_t(x)) = A_{ut}(x),\qquad A_{1/2}(x) = \phi(x). $$
The relation of this function to the sequence $\,b_n\,$ is
$\, b_n = 1/A_{2^{-n}}(b_0).\,$
We seek another function, $\,T(x),\,$ which "conjugates" multiplication.
More precisely,
$$ A_t(x) = T(t\,T^{-1}(x)), \quad
T^{-1}(x) = x - \frac{x^3}3 + \frac{x^5}5 + \cdots $$
Thus, $\,A_t(x)\,$ is completely determined by the function $\,T(x).\,$
The special case of $\,t=2,\,$ implies
$\, T^{-1}(\phi^{-1}(x)) = 2\,T^{-1}(x).\,$
Differentiate both sides to get
$$ {T^{-1}}'(\phi^{-1}(x))\cdot \frac{2\,(1+x^2)}{(1-x^2)^2}
= 2\,{T^{-1}}'(x).$$
Define $\,U(x) := {T^{-1}}'(x)\,$ and simplify to get
$$ U(\phi^{-1}(x)) \frac{(1+x^2)^2}{(1-x^2)^2} = U(x)\,(1+x^2). $$
Substitute $\,\phi(x)\,$ for $\,x\,$ to get
$$ U(x)\frac{(1+\phi(x)^2)^2}{(1-\phi(x)^2)^2} =
U(\phi(x))\,(1+\phi(x)^2). $$
Use the definition of $\,\phi(x)\,$ in the left side of the equation to get
$$ U(x)\,(1+x^2) = U(\phi(x))\,(1+\phi(x))^2. $$
Because this holds for all $\,x,\,$
both sides are equal to the constant $\,U(0)=1.\,$
This implies
$$ U(x) = \frac1{1+x^2}, \qquad T^{-1}(x) = \tan^{-1}(x). $$
|
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"url": "https://math.stackexchange.com/questions/4507788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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|
How many methods to tackle the integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x ?$ $ \text{We are going to evaluate the integral}$
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \tag*{} \\$
by letting $ y=\frac{\pi}{4}-x. $ Then
$$\begin{aligned} \displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \displaystyle &=\int_{\frac{\pi}{4}}^{0} \frac{\sin \left(\frac{\pi}{4}-y\right)+\cos \left(\frac{\pi}{4}-y\right)}{9+16 \sin 2\left[(\frac{\pi}{4}-y)\right]}(-d y) \\
\displaystyle &=\int_{0}^{\frac{\pi}{4}} \frac{\frac{1}{\sqrt{2}}(\cos y-\sin y)+\frac{1}{\sqrt{2}}(\cos y+\sin y)}{9+16 \cos 2 y} d y \\
\displaystyle &=\sqrt{2} \int_{0}^{\frac{\pi}{4}} \frac{\cos y}{9+16\left(1-2 \sin ^{2} y\right)} d y \\
\displaystyle &=\sqrt{2} \int_{0}^{\frac{1}{\sqrt 2} } \frac{d z}{25-32 z^{2}} \text { , where } z=\sin y\\
\displaystyle &=\frac{\sqrt{2}}{10} \int_{0}^{\frac{1}{\sqrt 2} }\left(\frac{1}{5-4 \sqrt{2}z}+\frac{1}{5+4 \sqrt{2} z}\right) d z \\
\displaystyle &=\frac{\sqrt{2}}{10(4 \sqrt{2})}\left[\ln \left|\frac{5+4 \sqrt{2} z}{5-4 \sqrt{2} z}\right|\right]_{0}^{\frac{1}{\sqrt 2} } \\
\displaystyle &=\frac{1}{40}\ln 9 \end{aligned}$$
|
Using the famous tangent half-angle substitution,
$$I=\int \frac{\sin (x)+\cos (x)}{9+16 \sin (2 x)}\,dx=\int \frac{-2 t^2+4 t+2}{9 t^4-64 t^3+18 t^2+64 t+9}\,dt$$
$$\frac{-2 t^2+4 t+2}{9 t^4-64 t^3+18 t^2+64 t+9}=\frac{-2 t^2+4 t+2}{\left(t^2-8 t+9\right) \left(9 t^2+8 t+1\right)}=$$
$$\frac{4-t}{20 \left(t^2-8 t+9\right)}+\frac{9 t+4}{20 \left(9 t^2+8 t+1\right)}$$
$$I=\frac{1}{40} \log \left(\left|\frac{9 t^2+8 t+1}{t^2-8 t+9}\right|\right)=\frac{1}{40} \log \left(\left|\frac{5-4 \cos (x)+4 \sin (x)}{5+4 \cos (x)-4 \sin
(x)}\right|\right)$$ Plug your numbers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4508889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
If $\vec a,\vec b,\vec c$ be three vectors such that $|\vec a|=1,|\vec b|=2,|\vec c|=4$ and then find the value of $|2\vec a+3\vec b+4\vec c|$ If $\vec a,\vec b,\vec c$ be three vectors such that
$\vert \vec a\vert =1,\vert \vec b\vert =2,\vert \vec c\vert=4$
and
$\vec a \cdot \vec b+\vec b \cdot \vec c+\vec c \cdot\vec a=-10$
then find the value of $\vert 2\vec a+3\vec b+4\vec c \vert$
My Attempt
$\vert \vec a+\vec b+\vec c \vert^2=a^2+b^2+c^2+2(\vec a \cdot \vec b+\vec b \cdot \vec c+\vec c \cdot \vec a)=1+4+16-20=1$
So, $\vert \vec a+\vec b+\vec c \vert=1$
Further by hit and trial I could see that if $\vec a=\vec i,\vec b=2\vec i,\vec c=-4\vec i$ (where $\vec i$ is unit vector along x-axis) satisfies all conditions.
So, $\vert 2\vec a+3\vec b+4\vec c \vert =\vert 2\vec i+6\vec i-16\vec i\vert =8$
But can there be a better way to do this.
|
You have
$$2a + 3b +4c = 3(a+b+c) -a+c.$$ therefore
$$\begin{aligned}
\lVert 2a + 3b +4c \rVert^2 &= 9 \lVert a + b +c \rVert^2 + \lVert a \rVert^2 + \lVert c \rVert^2 - 6 \lVert a \rVert^2 - 6 a \cdot b - 6 a \cdot c + 6 \lVert c \rVert^2 + 6 b \cdot c + 6 a \cdot c\\
&=9 + 1 + 16 -6 + 96 - 6 a \cdot b + 6 a \cdot c\\
&=116 - 6 a \cdot b + 6 b \cdot c
\end{aligned}$$
... and the problem is not fully determined.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Implicit Differentiation - $x^my^n = (x+y)^{m+n}$
Use implicit differentiation to find $\frac{\mbox{dy}}{\mbox{dx}}$ if $$x^my^n = (x+y)^{m+n}$$
Differentiation both sides with respect to $x$:
$$mx^{m-1}y^n + x^mny^{n-1}y' =(m+n)(x+y)^{m+n-1}(1 + y')$$
$$y' = \frac{(m+n)(x+y)^{m+n-1} - mx^{m-1}y^n}{x^mny^{n-1}-(m+n)(x+y)^{m+n-1}}$$
$$y'= \frac{nxy-my^2}{nx^2-mxy}$$ after using given $x^my^n = (x+y)^{m+n}$. The answer given to me is $\frac{y}{x}$. So, it seems $y'$ can be simplified even more.
How do we do this?
Thanks
|
For any homogeneous relation of $x$ and $y$, we always have $$\frac{dy}{dx}=\frac{y}{x}.$$ Because the relationship is always satisfied by $y=vx$ meaning $$\frac{dy}{dx}=v=\frac{y}{x}.$$ The relation $$x^my^n=(x+y)^{m+n}$$ is a homogeneous function of degree $m+n$.
For $x^2+xy+y^2=0$ and $4x^2y+5y^3+6x^3-7xy^2=0.$
After finding you have to use back the relationship to get
$\frac{dy}{dx}=\frac{y}{x}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int_{0}^{\pi}\sqrt\frac{1+\cos2x}{x^2+1}dx$.
Evaluate$$\int_{0}^{\pi}\sqrt\frac{1+\cos2x}{x^2+1}\,\mathrm dx$$
We have $\sqrt{1+\cos2x}=\sqrt{2\cos^2x}=\sqrt{2} |\cos x|$. Then we need to solve $\int_{0}^{\pi}\frac{\sqrt{2} |\cos x|}{\sqrt{x^2+1}}dx$. Using symmetry of $\cos x$ from $0$ to $\pi$ the integral becomes $I=2\sqrt2\int_{0}^{\pi/2}\frac{ \cos x}{\sqrt{x^2+1}}dx$.
*
*I tried using "By Parts", it didn't helped much.
*Take the substitution $x=\tan \theta$ then we have $I=\int_{0}^{\tan^{-1}\pi/2}\cos (\tan \theta)\sec \theta d\theta$.
I am having difficulty in solving with the existing methods that I know.
Edit: After knowing that I can not assume the symmetry of the given integral the question will be to solve
$$\sqrt2\left(\int_{0}^{\pi/2} \frac{\cos x}{\sqrt {x^2+1}}\,\mathrm dx-\int_{\pi/2}^{\pi} \frac{\cos x}{\sqrt {x^2+1}}\,\mathrm dx\right) $$
Again I will face same problem to proceed ahead.
|
Just for the fun !
Being skeptical about a possible closed form, starting from
$$I=\sqrt2\left(\int_{0}^{\frac \pi 2} \frac{\cos (x)}{\sqrt {x^2+1}}\, dx-\int_{\frac \pi 2}^{\pi} \frac{\cos (x)}{\sqrt {x^2+1}}\, dx\right)$$ I rewrote it as
$$I=\sqrt2\left(\int_{0}^{\frac \pi 2} \frac{\cos (x)}{\sqrt {x^2+1}}\, dx+\int_0^{\frac \pi 2} \frac{\sin (x)}{\sqrt{\left(x+\frac{\pi }{2}\right)^2+1}}\, dx\right)$$ and used my favored $\large 1,400$ years old approximations
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$
$$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$
Even if messy, the antiderivatives exist. In particular, we have
$$\int_{0}^{\frac \pi 2} \frac{\cos (x)}{\sqrt {x^2+1}}\, dx=-4 \sinh ^{-1}\left(\frac{\pi }{2}\right)+\frac{5 \pi }{4 \sqrt{\pi ^2-1}}\log \left(\frac{3+2 \pi ^2+2 \sqrt{-4+3 \pi ^2+\pi ^4}}{3+2 \pi ^2-2 \sqrt{-4+3
\pi ^2+\pi ^4}}\right)$$ which, numerically, is $\color{blue}{0.856256}$ while numerical integration gives $\color{red}{0.856581}$.
The formula for the second integral is too long but fully explicit. Numerically, its value is $\color{blue}{0.368287}$ while numerical integration gives $\color{red}{0.368308}$.
So, as a total, these approximations lead to $\color{blue}{I=1.731765}$ while numerical integration gives $\color{red}{I=1.732254}$ corresponding to a relative error of $\color{red}{0.028\text{%}}$.
Edit
Cocerning the second integral, using $x+\frac \pi 2=t$,
$$J=\int_0^{\frac \pi 2}\frac{16 (\pi -x) x}{\left(5 \pi ^2-4 (\pi -x) x\right) \sqrt{\left(x+\frac{\pi }{2}\right)^2+1}}\,dx=-\int_{\frac \pi 2}^\pi\frac{(\pi -2 t) (3 \pi -2 t)}{ \left(t^2-2 \pi t+2 \pi ^2\right)\sqrt{t^2+1}}\,dt$$ Factoring and using partial fraction
$$J=-\int_{\frac \pi 2}^\pi\frac{4}{\sqrt{t^2+1}}\,dt-\int_{\frac \pi 2}^\pi\frac{5 i \pi }{2 (t-(1+i) \pi ) \sqrt{t^2+1}}\,dt+\int_{\frac \pi 2}^\pi\frac{5 i \pi }{2 (t-(1-i) \pi ) \sqrt{t^2+1}}\,dt$$ For the first integral
$$-\int_{\frac \pi 2}^\pi\frac{4}{\sqrt{t^2+1}}\,dt=4 \left(\sinh ^{-1}\left(\frac{\pi }{2}\right)-\sinh ^{-1}(\pi )\right)$$ For the other integrals
$$\int_{\frac \pi 2}^\pi \frac{dt}{ (t-a)\sqrt{t^2+1}}= \frac{1}{\sqrt{a^2+1}}\log \left(\frac{(a-\pi ) \left(\sqrt{4+\pi ^2} \sqrt{a^2+1}+\pi a+2\right)}{(2
a-\pi ) \left(\sqrt{1+\pi ^2} \sqrt{a^2+1}+\pi a+1\right)}\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the last two non-zero digits of $70!$ The question itself is quite straightforward, however, I am unable to get an exact answer to the problem. I have narrowed it down to four possibilities one from $\{18, 43, 68, 93\}$.
The approach
We begin by finding the number of zeros that are contained in $70!$. This is done as shown below:
$$
\sum_{n=1}^\infty \left\lfloor \frac{70}{5^n} \right\rfloor = 16
$$
And clearly
$$
\sum_{n=1}^\infty \left\lfloor \frac{70}{2^n} \right\rfloor > \sum_{n=1}^\infty \left\lfloor \frac{70}{5^n} \right\rfloor \qquad \qquad \text{(i)}
$$
So therefore we define some integer $N$ such that:
$$
N = \frac{70!}{5^{16}2^{16}}
$$
From here we will find the remainder of $N$ modulo $4$ and $25$ respectively so we can use the Chinese Remainder Theorem to finish the problem.
Verifying mod 4
From (i) it is clear that $N$ will enough spare twos to be divisible by $4$. This leads us to the conclusion that:
$$
N \equiv 0 \pmod 4
$$
Verifying mod 25
We can split this part of the problem into two parts. Define another integer $M$ such that
$$
M = \frac{70!}{5^{21}}
$$
This allows us to write:
$$
\frac{1}{2^{16}} \cdot M = N
$$
Finding 2^16 mod 25
Trivially, we can find that
$$
2^{16} = 65536 \equiv 11 \pmod {25}
$$
Finding M mod 25
This one is a little trickier. To do this, we will use the fact that:
$$
(5n+1)(5n+2)(5n+3)(5n+4) \equiv -1 \pmod {25}
$$
Notice that we can group $M$ as follows:
$$
\begin{align*}
(1\cdot2\cdot3\cdot4)(6\cdot7\cdot8\cdot9)\cdots(66\cdot67\cdot68\cdot69) \ &\times \\
(1\cdot2\cdot3\cdot4)(6\cdot7\cdot8\cdot9)(11\cdot12\cdot13\cdot14) \ &\times \\
(1\cdot2)
\end{align*}
$$
Now, we can easily see:
$$
M \equiv (-1)^{12} \cdot (-1)^{3} \cdot {2} \equiv 23 \pmod {25}
$$
Bringing it together
Now we have the following equation
$$
N \equiv \frac{M}{2^{16}} \equiv \frac{23}{11} \pmod {25}
$$
This is where I got stuck.
From here I tried to write
$$\frac{23 + 25k}{11} \pmod{25}$$
and this yields four possible solutions $\{18, 43, 68, 93\}$. Is there a way that I can find the correct solution and/or a better way to solve the problem?
Thanks!
|
From here we will find the remainder of $N$ modulo $4$ and $25$
respectively so we can use the Chinese Remainder Theorem to finish the
problem.
$$ N \equiv 0 \pmod 4 $$
$$ N \equiv \frac{M}{2^{16}} \equiv \frac{23}{11} \pmod {25} $$
$$\color{red}{\text{THIS IS WHERE I GOT STUCK!}}$$
To simplify $\frac{23}{11} \pmod {25}$, you can compute the inverse of 11 mod 25, by solving the Diophantine equation $11 x - 25 q = 1$. This will give you $11^{-1} = 16 \pmod {25}$. Thus $N \equiv 23 \times 16 \equiv 18 \pmod{25} $.
Now you can apply your last planned step, which was to use the Chinese remainder theorem to go from $N \equiv 0 \pmod{4}, N \equiv 18 \pmod{25}$ to $N \equiv ?? \pmod{100}$.
But actually, since $100 = 25 \times 4$, we don't even need the heavy artillery here. Just list the 4 possibilities $\{18, 43, 68, 93\}$ and cross out the 3 which are not $\equiv 0 \pmod{4}$.
We end up with $N \equiv 68 \pmod{100}$.
My computer confirms that result:
$$70! = 11978571669969891796072783721689098736458938142546425857555362864628009582789845319\color{red}{68}0000000000000000$$
|
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|
Find the sum of radicals without squaring, Is that impossible?
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$
My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
So I was wondering about a way to find this sum without squaring? It seems impossible, but I still want to ask.
|
Use Vieta's formulas:
Let, $$x_1=\sqrt {3-\sqrt 5},\,\,\,x_2=\sqrt {3+\sqrt 5}$$
$$x^2-px+2=0\\p=\frac{x^2+2}{x}$$
and we have,
$$p=\frac{5+\sqrt 5}{\sqrt {3+\sqrt 5}}=\frac{5-\sqrt 5}{\sqrt {3-\sqrt 5}}$$
Then, using the rule
$$p=\frac ab=\frac cd\implies p=\frac{a+c}{b+d}$$
We get
\begin{aligned}p&= \frac{5+\sqrt 5+5-\sqrt 5}{\sqrt {3+\sqrt 5}+\sqrt {3-\sqrt 5}}\\
&=\frac {10}{p}=p\\
&\implies p=\sqrt {10}. \end{aligned}
|
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|
Finding the equation of a plane given three points Below is a problem I did from a Calculus text book. My answer matches the
back of the book and I believe my answer is right. However, the method
I used is something I made up. That is, it is not the method described
in the text book.
Is my method correct?
Problem:
Find the plane through the points $(1,1,-1)$, $(2,0,2)$ and $(0,-2,1)$.
Answer:
The general form of a plane is:
$$ Ax + By + Cz = D$$
Sometimes the following constrain is added:
$$ A^2 + B^2 + C^2 = 1$$
By inspection, we can see this plane is not parallel to the x-axis, the y-axis or the z-axis. Hence,
we can assume that the plane is of the form:
$$ Ax + By + Cz = 1 $$
Now we setup the following system of linear equations.
\begin{align*}
A + B - C &= 1 \\
2A + 2C &= 1 \\
-2B + C &= 1 \\
\end{align*}
To solve this system of equations, we get rid of $A$ and $B$ in the first equation.
\begin{align*}
2A &= 1 - 2C \\
A &= \frac{ 1 - 2C }{2} \\
-2B &= 1 - C \\
B &= \frac{ C - 1 }{2} \\
\left( \frac{ 1 - 2C }{2} \right) + \left( \frac{ C - 1 }{2} \right) - C &= 1 \\
1 - 2C + C - 1 - 2C &= 2 \\
- 2C + C - 2C &= 2 \\
-3C &= 2 \\
C &= -\frac{2}{3} \\
B &= \frac{ -\frac{2}{3} - 1 }{2} = -\frac{2}{6} - \frac{1}{2} \\
B &= -\frac{5}{6} \\
A &= \frac{ 1 - 2\left( -\frac{2}{3} \right) }{2} = \dfrac{1 + \dfrac{4}{3} }{2} \\
A &= \dfrac{7}{6}
\end{align*}
Hence the equation is:
$$ \left( \dfrac{7}{6} \right) A + \left( -\frac{5}{6} \right) B + \left( -\frac{2}{3} \right) C = 1 $$
Clearing the fraction, we get the final answer of:
$$ 7A - 5B - 4C = 6 $$
As pointed out by Paul, the correct answer is:
$$ 7x - 5y - 4z = 6 $$
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Yes, your method is perfectly fine, and nicely laid out, though at the very end in the original $x,y,z$ were accidentally replaced by $A,B,C$. :)
In terms of numerical computation, this is a reasonably efficient algorithm.
In terms of formulaic or abstract presentation, (conceivably what the book did), we realize that to describe a plane in 3D, we need a "normal" (=perpendicular) vector $N$ and a point $P$ on the plane. Then the plane is the set of points $X=(x,y,z)$ such that $(X-P)\cdot N=0$, where dot denotes vector "dot product" (="inner product"="scalar product").
A formulaic/conceptual trick (that hides necessary computations) is that the vector cross product $v\times w$ is orthogonal to both $v$ and $w$. So, given three points $P,Q,R$, the vectors $P-Q$ and $P-R$ (for example) are in the plane, so their cross product is orthogonal to it. So an equation for points $X=(x,y,z)$ to be in the plane is $(X-P)\cdot ((P-Q)\times(P-R))=0$.
It should probably be noted that the computation needed for the more symbolic approach is really roughly the same as the "more direct" approach: a three-by-three system of linear equations is to be solved, and Cramer's rule, with expansion-by-minors, is in-effect carried out by those vector operations. :)
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How do we prove that the simple continued fraction for $e^{2/n}=[1;\frac{n-1}{2},6n,\frac{5n-1}{2},1,1,...]$? Motivation: remarkably, the simple continued fraction - which is unique - for $e^{1/n},e^{2/n}$ is known for every $n\in\Bbb N$. The expansions for $e^{3/n}$, or even $e^{p/q}$, are not known in general, and it is a strange miracle that we can figure out any of them at all... Euler first derived the continued fraction for $e$ and $e^{1/n}$. The expansion for $e^{2/n}$ is viewable on the web, but I have not been able to find a proof of this beautiful fact.
This original text from Euler proves that, for real $s$: $$\begin{align}\tag{1}e^{2/s}&=1+\frac{2}{s-1+}\frac{1}{3s+}\frac{1}{5s+}\cdots\\e^{1/s}&=1+\frac{1}{s-1+}\frac{1}{1+}\frac{1}{1+}\frac{1}{3s-1+}\cdots\end{align}$$It derives the expansion for $e^{1/s}$ from that for $e^{2/s}$ by means of his "interpolation" formulae: $$\begin{align}a+\frac{1}{m+}\frac{1}{n+}\frac{1}{b+}\frac{1}{m+}\frac{1}{n+}\frac{1}{c+}\cdots&=\frac{1}{mn+1}\left((mn+1)a+n+\frac{1}{(mn+1)b+m+n+}\frac{1}{(mn+1)c+m+n+}\cdots\right)\\A+\frac{1}{a+}\frac{1}{m+}\frac{1}{n+}\frac{1}{b+}\cdots&=A+\frac{mn+1}{(mn+1)a+n+}\frac{1}{(mn+1)b+m+n+}\cdots\\a+\frac{1}{b+}\frac{1}{c+}\cdots&=a-n+\frac{mn+1}{m+}\frac{1}{n+}\frac{1}{\frac{b-m-n}{mn+1}+}\frac{1}{m+}\frac{1}{n+}\frac{1}{\frac{c-m-n}{mn+1}+}\cdots\end{align}$$Which all have nice special cases for $m=n=1$.
Wikipedia claims that in fact: $$\tag{$\ast$}e^{2/s}=1+\frac{1}{\frac{s-1}{2}+}\frac{1}{6s+}\frac{1}{\frac{5s-1}{2}+}\frac{1}{1+}\frac{1}{1+}\frac{1}{\frac{7s-1}{2}+}\frac{1}{18s+}\frac{1}{\frac{11s-1}{2}+}\frac{1}{1+}\cdots$$
Which I'd like to understand. First note that, by equivalence transformation (multiply by $2,1/2,2,1/2,\cdots$) the a first try might be to convert $(\ast)$ into: $$e^{2/s}=1+\frac{2}{s-1+}\frac{1}{3s+}\frac{1}{5s-1+}\frac{1}{\frac{1}{2}+}\frac{1}{2+}\frac{1}{\frac{7s-1}{4}+}\cdots$$Which unfortunately doesn't work out. Also, although two $1s$ have clearly been interpolated somehow to get $(\ast)$, we can't directly employ any of the standard results shown above, since, instead of $a,1,1,b,1,1,...$ it is of the form $a,b,c,1,1,d,e,f,1,1,...$.
If we try to realise $(1)$ as an interpolation, using the middle formula, we need to solve $2a+1=s-1\implies a=\frac{s-1}{2}$, $2b+2=3s\implies b=\frac{3s-2}{2}$, etc., to get: $$e^{2/s}=1+\frac{1}{\frac{s-1}{2}+}\frac{1}{1+}\frac{1}{1+}\frac{1}{\frac{3s-2}{2}+}\frac{1}{1+}\frac{1}{1+}\frac{1}{\frac{5s-2}{2}+}\cdots$$Which seems better, but is still not quite right.
I'm convinced the answer is going to be a simple transformation: the denominators of $(\ast)$ are of the form $\frac{x-1}{2},2x$ for $x$ the denominator in $(1)$. The presence of a $2$-multiplier and a $1/2$-multiplier suggests some simple cancellation. However, the two interpolating $1s$ are causing me a headache.
Does anyone know how we get $(\ast)$?
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If you have prior knowledge of the answer, you can figure it out. I don’t know how you might derive it “from first principles” though! I’m a little surprised by the lack of publicly available information about continued fractions - you’d think ones concerning notable values, such as $e$, would have well-documented explanations.
If $\alpha_1:=\frac{1}{e^{2/s}-1}-\frac{s-1}{2}$ then $e^{2/s}=1+\frac{1}{\frac{s-1}{2}+\alpha_1}$. A simple series calculation finds: $$\alpha_1=\frac{\sum_{n\ge2}\frac{2^{n-1}\cdot(n-1)}{s^n\cdot(n+1)!}}{\sum_{n\ge1}\frac{2^n}{s^n\cdot n!}}$$If $\alpha_2:=\frac{1}{\alpha_1}-6s$, then $e^{2/s}=1+\frac{1}{\frac{s-1}{2}+}\frac{1}{6s+\alpha_2}$ and one calculates: $$\alpha_2=\frac{\sum_{n\ge3}\frac{2^n\cdot(n-1)(n-2)}{s^n\cdot(n+2)!}}{\sum_{n\ge2}\frac{2^{n-1}\cdot(n-1)}{s^n\cdot(n+1)!}}$$Continuing, let $\alpha_3:=\frac{1}{\alpha_2}-\frac{5s-1}{2}$: $$\alpha_3=\frac{\sum_{n\ge3}\frac{2^n\cdot n(n-1)(n-2)}{s^n\cdot(n+3)!}}{\sum_{n\ge3}\frac{2^n\cdot(n-1)(n-2)}{s^n\cdot(n+2)!}}$$Then, if (deliberately skipping $4$) $\alpha_5:=\frac{2\alpha_3-1}{1-\alpha_3}$, we have $e^{2/s}=1+\frac{1}{\frac{s-1}{2}+}\frac{1}{6s+}\frac{1}{\frac{5s-1}{2}+}\frac{1}{1+}\frac{1}{1+\alpha_5}$ and: $$\alpha_5=\frac{\sum_{n\ge4}\frac{2^n\cdot(n-1)(n-2)(n-3)}{s^n\cdot(n+3)!}}{3\sum_{n\ge3}\frac{2^n\cdot(n-1)(n-2)}{s^n\cdot(n+3)!}}$$Et cetera.
Inductively these values follow the progression (performing both $\frac{1}{1+}\frac{1}{1+}$ steps together, as done above when skipping from $\alpha_3\to\alpha_5$): $$\frac{\sum_{n\ge m+1}\frac{2^{n-1}\cdot\prod_{k=1}^m(n-k)}{s^n\cdot(n+m)!}}{\sum_{n\ge m}\frac{2^n\cdot\prod_{k=1}^{m-1}(n-k)}{s^n\cdot(n+m-1)!}},\frac{\sum_{n\ge m+2}\frac{2^n\cdot\prod_{k=1}^{m+1}(n-k)}{s^n\cdot(n+m+1)!}}{\sum_{n\ge m+1}\frac{2^{n-1}\cdot\prod_{k=1}^m(n-k)}{s^n\cdot(n+m)!}},\frac{\sum_{n\ge m+2}\frac{2^n\cdot n\prod_{k=1}^{m+1}(n-k)}{s^n\cdot(n+m+2)!}}{\sum_{n\ge m+2}\frac{2^n\cdot\prod_{k=1}^{m+1}(n-k)}{s^n\cdot(n+m+1)!}},\frac{\sum_{n\ge m+3}\frac{2^n\cdot\prod_{k=1}^{m+2}(n-k)}{s^n\cdot(n+m+2)!}}{(m+2)\sum_{n\ge m+2}\frac{2^n\cdot\prod_{k=1}^{m+1}(n-k)}{s^n\cdot(n+m+2)!}}$$With $m=1,4,7,10,...$.
Using some convergence results we can see that, since $\alpha_n\to0$ (empirically: I haven’t proven that yet, and will edit it in when I do) the continued fraction converges to $e^{2/s}$ for nonzero positive $s$. Certainly all terms are positive (eventually) when $s>0$ so convergence “to the right thing” is automatic (see the linked post).
Empirically it also converges for negative $s$ but that’s more annoying to deal with, due to the negative continued fraction coefficients. Anyway, if $s$ is negative you can just take a reciprocal as $e^{-2/s}=\frac{1}{e^{2/s}}$.
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Find the derivative of $y=\sqrt{\ln\left(4x-x^2\right)}$ Find the derivative of $$y=\sqrt{\ln{\left(4x-x^2\right)}}$$
So we can rewrite the function as $$y=\left[\ln\left(4x-x^2\right)\right]^\frac12$$ Let's try to break it down a bit. So let's set $$a(x)=x^\frac12$$ and $$b(x)=\ln(4x-x^2)$$ then $$y=a(b(x))$$ The chain rule then tells us that $$y'=a'(b(x))b'(x)$$ Now $a'$ we can easily find, as it is just $$a'(x)=\dfrac{1}{2\sqrt{x}},$$ but how do we find $b'$? Well, let's do the same thing again, namely notice that it's a composition and break it down. So now let $\alpha(x)=\ln x$ and $\beta(x)=4x-x^2$. These two functions we know how to differentiate! Indeed $\alpha'(x)=\dfrac{1}{x}$ and $\beta'(x)=4-2x$. Furthermore, the chain rule also tells us now that, as $b(x)=\alpha(\beta(x))$, $b'(x)=\alpha'(\beta(x))\beta'(x)$. Putting this all together we get that $$y'(x)=a'(b(x))\alpha'(\beta(x))\beta'(x)\\=\dfrac{1}{2\sqrt{\ln(4x-x^2)}}\cdot\dfrac{1}{4x-x^2}\cdot(4-2x)=\dfrac{1}{2x\sqrt{\ln(4x-x^2)}}.$$ The answer seems to differ...
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$y'(x) =\frac{(\ln{{(4x-x^2)}})'}{2\times\sqrt{\ln{{4(x-x^2)}}}}=\frac{\frac{(4x-x^2)'}{4x-x^2}}{2\times\sqrt{\ln{{(4x-x^2)}}}}=\frac{\frac{4-2x}{x(4-x)}}{2\times\sqrt{\ln{{(4x-x^2)}}}}=\frac{\frac{x-2}{x(x-4)}}{\sqrt{\ln{{(4x-x^2)}}}}$.
Now,
write $\frac{x-2}{x(x-4)}$ such that $\frac A{x}+\frac B{x-2}=\frac{A(x-4)+Bx}{x(x-4)}=\frac{(A+B)x-4A}{x(x-4)}$.
This way,
$A+B=1$ ans $-4A=-2$, i.e. $A=1/2$ and $B=1/2$.
So,
$y'(x)=\frac{\frac{x-2}{x(x-4)}}{\sqrt{\ln{{(4x-x^2)}}}}=\frac{\frac 1{2x}+\frac 1{2(x-4)}}{\sqrt{\ln{{(4x-x^2)}}}}=\frac{1}{2x\sqrt{\ln{{(4x-x^2)}}}}+\frac{1}{2(x-4)\sqrt{\ln{{(4x-x^2)}}}}$.
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If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\ {g'(x)} & {x< k} \end{cases} $ , $\max (k)=?$ if $f$ is a differentiable function
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\
{g'(x)} & {x< k} \end{cases} $. If $f(x)$ is a differentiable
function, what is the maximum value of $k$, provided that $b+c=a$?
$1)\frac34\qquad\qquad2)1\qquad\qquad3)3\qquad\qquad4)4$
In order to $f(x)$ be differentiable function, we should have $g(k)=g'(k)=g''(k)$,
$$ak^2+bk+c=2ak+b=2a$$ $$(b+c)k^2+bk+c=(2b+2c)k+b=2b+2c$$
Here for each equation I tried to equate the coefficients of $k^2 , k^1 , k^0$ but I get $a=b=c=0$ which doesn't make sense at all. I don't know how to continue form here.
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(A solution with no square roots needed.)
$g'(k) = g''(k)$ gives
$$ 2ak+b = 2a $$
If $a=0$ then this forces $b=0$, $0 = g'(k) = 2a$ so $a=0$, and finally $g(k)=c=0$. In the degenerate case where $f$ and $g$ are zero everywhere, $k$ can be any number at all. The problem should have eliminated this case, to have an answer.
So for the rest, assume $a \neq 0$.
$$ k = 1 - \frac{b}{2a} $$
Then $g(k) = g''(k)$ gives
$$ a\left(1 - \frac{b}{a} + \frac{b^2}{4a^2}\right) + b\left(1-\frac{b}{2a}\right) + c = 2a $$
$$ a + c - \frac{b^2}{4a} = 2a $$
Substituting $c=a-b$,
$$ \frac{b^2}{4a} + b = 0 $$
So either $b=0$ or $\frac{b}{a} = -4$. If $b=0$, $k=1$. If $\frac{b}{a} = -4$, $k=3$. The largest possible value of $k$ (again excluding the $a=b=c=0$ case) is $k=3$.
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How did people come up with the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$? Every resource that I've read proves the formula
$$ a^3 + b^3 = (a+b)(a^2-ab+b^2) \tag1$$
by just multiplying $(a+b)$ and $(a^2 - ab + b^2)$.
But how did people come up with that formula? Did they think like, "Oh, let's just multiply these polynomials, I don't know why, let's just do it." I don't think that people just pointed a finger at the sky and came up with that formula.
So, how to prove $(1)$?
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$«$Who first factored the expression $a^3+b^3$ and what was the method used?$»$ - I don't know the exact answer to this question, but the similar question I'm trying to answer is:
$«$How can we factor $a^3+b^3$ using the most basic algebraic techniques?$»$
It seems to me that, this formula basically comes from the Binomial theorem:
$$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$
But, without the binomial theorem you can obtain this result as follows:
$$\begin{aligned}(a+b)^3&=(a+b) \left((a+b)\times (a+b)\right)\\
&=(a+b)(a^2+2ab+b^2)\\
&=a^3+3a^2b+3ab^2+b^3\end{aligned}$$
Then observe that,
$$\begin{aligned}a^3+3a^2b+3ab^2+b^3&=(a^3+b^3)-3ab(a+b)\end{aligned}$$
This leads,
$$
\begin{aligned}a^3+b^3&=(a+b)^3-3ab(a+b)\\
&=(a+b)\left((a+b)^2-3ab\right)\\
&=(a+b)(a^2-ab+b^2).\end{aligned}
$$
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Calculate $\sin^5\alpha-\cos^5\alpha$ if $\sin\alpha-\cos\alpha=\frac12$ Calculate $$\sin^5\alpha-\cos^5\alpha$$ if $\sin\alpha-\cos\alpha=\dfrac12$.
The main idea in problems like this is to write the expression that we need to calculate in terms of the given one (in this case we know $\sin\alpha-\cos\alpha=\frac12$).
I don't see how to even start to work on the given expression as we cannot use $a^2-b^2=(a-b)(a+b)$ or $a^3-b^3=(a-b)(a^2+ab+b^2)$. So in other words, I can't figure out how to factor the expression (even a little).
The given answer is $\dfrac{79}{128}$.
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Another approach: Solve $ \sin{z}-\cos{z}=\frac{1}{2}$
This results in $e^{iz}=\frac{\sqrt{7}-i}{2(i-1)}$
Plugging this into the complex sin and cos definition results in:
$\sin{z}=\frac{1-1\sqrt{7}}{4}$ and $\cos{z}= \frac{-1-1\sqrt{7}}{4}$
When applying this in $sin^5z-cos^5z$ all uneven powers of $\sqrt(7)$ cancel out.
What remains is: $sin^5z-cos^5z=\frac{1+70+5\cdot 49}{2^9}=\frac{79}{128}$
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Line of best fit for $\{(n,n+\sin n) : n \in \mathbb{Z}\}$ It seems intuitive that the line of best fit for $\{(n,n+\sin n) : n\in \mathbb{Z}\}$ should be $y=x$.
More concretely, it seems like a reasonable conjecture would be:
If $y = m_k x + b_k$ is the line of best fit for the set of points $$\{ (n,n+\sin n) : n\in \mathbb{Z}, |n| \leq k \},$$ then $\lim_{k\to \infty} m_k = 1$ and $\lim_{k\to\infty} b_k = 0$.
Is this conjecture true? And if so, how would one go about proving it? And moreover still, if $\{a_n\}$ is a sequence in $\mathbb{R}$ which is uniformly distributed in some compact interval $[A,B]$, then how does the line of best fit change when considering the set $\{(n,n+a_n)\}$?
EDIT: Just to clarify, by "line of best fit" I mean using the method of least-squares.
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To start with, we're going to need a few identities:
$$\begin{eqnarray} \sum_{j = 1}^n j & = & \frac{n(n+1)}{2} \\
\sum_{j = 1}^n j^2 & = & \frac{n(n+1)(2n+1)}{6} \\
\sum_{j = 1}^n \sin j & = & \frac{\sin n - \sin (n+1) + \sin 1}{2(1 - \cos 1)} \\
\sum_{j = 1}^n j \sin j & = & \frac{(n + 1) \sin n - n \sin (n + 1)}{2(1 - \cos 1)} \end{eqnarray}$$
The first two are the triangular and square pyramidal numbers respectively, the third is the sum of sines formula as expressed in this answer, and the last is courtesy of Wolfram Alpha, although you can derive it using a similar approach to the sum of sines formula, but applying a formula for the sum of $k x^k$ (that in turn comes from differentiating the geometric series formula with respect to $x$).
Then, we're going to use this formula for the coefficients of a simple linear regression:
$$\begin{eqnarray} y & = & \alpha + \beta x \\
\beta & = & \frac{n \sum_j x_j y_j - \sum_j x_j \sum_j y_j}{n \sum_j x_j^2 - (\sum_j x_j)^2} \\
\alpha & = & \bar{y} - \beta \bar{x} \\
& = & \frac{1}{n}\left(\sum_j y_j - \beta \sum_j x_j \right) \end{eqnarray}$$
Now, we get to start substituting $x_j = j$ and $y_j = j + \sin j$ everywhere. Starting with the denominator of $\beta$:
$$\begin{eqnarray} n \sum_j j^2 - (\sum_j j)^2 & = & n \frac{n(n+1)(2n+1)}{6} - \left(\frac{n(n+1)}{2}\right)^2 \\
& = & \frac{n^2 (n+1)}{2}\left(\frac{2n+1}{3} - \frac{n+1}{2} \right) \\
& = & \frac{n^2 (n+1)(n-1)}{12}\end{eqnarray}$$
Next, the numerator:
$$\begin{eqnarray} n \sum_j j(j + \sin j) - \sum_j j \sum_j (j + \sin j) & = & n(\sum_j j^2 + \sum_j j \sin j) - \left( (\sum_j j)^2 + \sum_j j \sum_j \sin_j \right) \\
& = & \frac{n^2 (n+1)(n-1)}{12} + n \frac{(n+1) \sin n - n \sin(n+1)}{2(1 - \cos 1)} \\ && - \frac{n(n+1)}{2} \frac{\sin n - \sin(n+1) + \sin 1}{2(1 - \cos 1)} \\
& = & \frac{n^2 (n+1)(n-1)}{12}
\\ && + \frac{n \left((n+1) \sin n - (n - 1)\sin(n+1) - (n+1) \sin 1 \right)}{4(1 - \cos 1)}
\end{eqnarray}$$
Putting those together, we get:
$$\begin{eqnarray} \beta & = & 1 - \frac{n \left((n+1) \sin n - (n - 1)\sin(n+1) - (n+1) \sin 1 \right)}{4(1 - \cos 1)}\frac{12}{n^2 (n+1)(n-1)} \\
& = & 1 - \frac{3 \left((n+1) \sin n - (n-1) \sin(n+1) - (n+1) \sin 1 \right)}{n(n-1)(n+1)(1 - \cos 1)} \end{eqnarray}$$
And:
$$\begin{eqnarray} \alpha & = & \frac{1}{n}(\sum_j (j + \sin j) - \beta \sum_j j) \\
& = & \frac{1}{n}\left(\frac{\sin n - \sin(n+1) + \sin 1}{2(1 - \cos 1)} \right.
\\ && \left. + \frac{n(n+1)}{2} \frac{3 \left((n+1) \sin n - (n-1) \sin(n+1) - (n+1) \sin 1 \right)}{n(n-1)(n+1)(1 - \cos 1)} \right) \\
& = & \frac{(2n+1)\sin n - (2n-2)\sin(n+1) - (n+2)\sin 1}{n(n-1)(1 - \cos 1)}
\end{eqnarray}$$
Or, at least, that's probably close to being right, but the probability of an algebraic error creeping in there is pretty high. However, assuming that that's all roughly accurate, we can see that $\beta = 1 + O(n^{-2})$ and $\alpha = O(n^{-1})$, so in the limit as $n \rightarrow \infty$ we do indeed get that $y = x$ is the least squares regression line.
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Limit for $(x,y) \to (0,0)$ of $f(x,y)=\frac{x^2+y^4}{x}$ I tried to solve this limit, but I'm not sure whether it's correct. The candidate limit is $l=0$.
From $ | \frac{x^2+y^4}{x} | $, since $\sqrt{x^2-y^2} \leq \sqrt{x^2} = |x| $, we then have:
$$| \frac{x^2+y^4}{x} | \leq | \frac{x^2+y^4}{\sqrt{x^2-y^2}} | \leq ...$$
$$ ...\leq | \frac{r^2 cos^2 {\theta}+r^4 sin^4 {\theta}}{\sqrt{r^2 (cos^2 {\theta} - sin^2 {\theta})}} | \leq | \frac{r cos^2 {\theta}+r^3 sin^4 {\theta}}{\sqrt{(cos^2 {\theta} - sin^2 {\theta})}} \leq ... $$
$$ ... \leq | \frac{r+r^3}{\sqrt{cos(2\theta)}} | $$
The only case where the denominator is 0 is when $\theta = \frac{\pi}{2}$. However, this restriction is not a subset of the domain because we have $x \neq 0$ by hypothesis.
So the limit is 0.
Is this correct? I am not entirely sure about the last part!
|
I suggest consider curve $y=\sqrt[8]{x}=x^{\frac{1}{8}}$. Then we have on this curve
$$\frac{x^2+y^4}{x}=x + \frac{1}{\sqrt{x}}$$
limit of first summand exists, while second tends to infinity.
On another hand (accordingly comments) along the curve $y=x$ limit is obviously zero. Thus, we can postulate, that limit does not exist.
|
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|
$\int_{0}^{\infty}\tan^{-1}\left(\frac{2x}{1+x^2}\right)\frac{x}{x^2+4}dx$ $$\int_{0}^{\infty}\tan^{-1}\left(\frac{2x}{1+x^2}\right)\frac{x}{x^2+4}dx$$
I am given solution for this definite integral is $\frac{\pi}{2}\left(\ln\frac{\sqrt2+3}{\sqrt2+1}\right)$. Any idea or approach you would use to solve this?
|
Using integration by parts, we find
$$\int_0^\infty \tan^{-1}\left(\frac{2 x} {x^2 +1}\right)\frac{x}{x^2 +4}dx=\int_0^{\infty}\frac{(x^2-1)\ln(x^2+4)}{x^4+6x^2+1} dx.$$
Factoring the denominator as
$$x^4+6x^2+1=(x^2+3+2\sqrt{2})(x^2+3-2\sqrt{2})\\
=(x^2+(\sqrt{2}+1)^2)(x^2+(\sqrt{2}-1)^2)$$ and then applying the partial fraction decomposition, we obtain
$$\frac{\sqrt{2}+1}{2}\int_0^{\infty} \frac{\ln(x^2+4)}{x^2+(\sqrt{2}+1)^2} dx - \frac{\sqrt{2}-1}{2}\int_0^{\infty} \frac{\ln(x^2+4)}{x^2+(\sqrt{2}-1)^2} dx.$$
Finally we use An improper integral : $\int_{0}^\infty {\ln(a^2+x^2)\over{b^2+x^2}}dx$? $=\frac{\pi}{b} \, \ln(a+b) $, getting
$$\frac{\pi}{2}\ln(\sqrt{2}+3)-\frac{\pi}{2}\ln(\sqrt{2}+1)=\frac{\pi}{2}\ln\left(\frac{\sqrt2+3}{\sqrt2+1}\right).$$
|
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Showing $\int_1^\infty\left(\sqrt{\sqrt{x}-\sqrt{x-1}}-\sqrt{\sqrt{x+1}-\sqrt{x}}\right)dx=\frac4{15}\left(\sqrt{26\sqrt2-14}-2\right)$ A Putnam problem asked to show that some improper integral is convergent, but I was curious to see if it can be computed in closed form and Mathematica came up with this:
$$\int_1^{\infty} \left(\sqrt{\sqrt{x}-\sqrt{x-1}} -\sqrt{\sqrt{x+1}-\sqrt{x} } \right)dx=
\frac{4}{15} \left(\sqrt{26\sqrt{2}-14 }-2\right) \approx 0.739132$$
I did a few substitutions but it didn't turn out as an easy calculation. I remember that eliminating square roots required some Euler type substitutions. Any ideas of how one can arrive at such a surprising result?
|
This is a partial answer.
\begin{align}
a_n&=\int_n^{n+1} \left(\sqrt{\sqrt{x}-\sqrt{x-1}} -\sqrt{\sqrt{x+1}-\sqrt{x}} \right)\mathbb dx\\
&=\int_n^{n+1} \sqrt{\sqrt{x}-\sqrt{x-1}}\,\mathbb dx-\int_{n+1}^{n+2} \sqrt{\sqrt{x}-\sqrt{x-1}}\mathbb\,dx\\
\sum_{n=1}^{\infty}a_n&=\int_1^{2} \sqrt{\sqrt{x}-\sqrt{x-1}}\,\mathbb dx
=\int_1^{2} \frac1{\sqrt{\sqrt{x}+\sqrt{x-1}}}\,\mathbb dx
\end{align}
The WolframAlpha shows that the indefinite integral is
$$\int \frac1{\sqrt{\sqrt{x}+\sqrt{x-1}}}\,\mathbb dx=\frac{5\left(\sqrt{x-1}+\sqrt x\right)^4+3}{15\left(\sqrt{x-1}+\sqrt x\right)^{5/2}}+C$$
, which I didn't figure out how to get there.
I completed the answer below.
\begin{align}
\sqrt{x}+\sqrt{x-1}&\to t\\
dt&=\frac12\left(\frac1{\sqrt{x}}+\frac1{\sqrt{x-1}}\right)dx=\frac12\frac{t}{\sqrt{x}\sqrt{x-1}}dx\\
t^2&=2x-1+2\sqrt x\sqrt{x-1}\\
4x(x-1)&=(t^2-2x+1)^2=t^4-2(2x-1)t^2+(2x-1)^2\\
0&=t^4-4xt^2+2t^2+1\\
x&=\frac{(t^2+1)^2}{4t^2}\\
x-1&=\frac{(t^2-1)^2}{4t^2}\\
\int \frac1{\sqrt{t}}\,\mathbb dx&=\int \frac{2\sqrt{x}\sqrt{x-1}}{t^{3/2}}\,\mathbb dt=\int \dfrac{2\dfrac{t^4-1}{4t^2}}{t^{3/2}}\mathbb dt\\
&=\frac12 \int \left(t^{1/2}-t^{-7/2}\right)\mathbb dt\\
&=\frac13t^{3/2}+\frac15t^{-5/2}+C\\
&=\frac13\left(\sqrt{x}+\sqrt{x-1}\right)^{3/2}+\frac15\left(\sqrt{x}+\sqrt{x-1}\right)^{-5/2}+C
\end{align}
|
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|
FOPDE: $2x z_x + 3y z_y = x + y$ Solving the First Order PDE:$$2x z_x + 3y z_y = x + y$$
for $c_1$ I get: $$ c_1 = \frac{x^3}{y^2} $$
I do not know how to go about solving for $c_2$
The correct general solution should be:
$$z = \frac{x}{2} + \frac{y}{3} + f\bigg(\frac{x^3}{y^2}\bigg)$$
How would one come to this calculation?
I would really appreciate your help
|
$$2x z_x + 3y z_y = x + y$$
Charpit-Lagrange characteristic ODEs :
$$\frac{dx}{2x}=\frac{dy}{3y}=\frac{dz}{x+y}$$
A first characteristic equation from solving $\frac{dx}{2x}=\frac{dy}{3y}$ that you found correctly :
$$\frac{x^3}{y^2}=c_1$$
A second characteristic equation from
$$\frac{dx}{2x}=\frac{dy}{3y}=\frac{dz}{x+y}=\frac{3dx+2dy-6dz}{3(2x)+2(3y)-6(x+y)}=\frac{3dx+2dy-6dz}{0}$$
See explanation below.
$$\implies\quad 3dx+2dy-6dz=0$$
$$z-\frac12 x-\frac13 y=c_2$$
General solution of the PDE on the inplicit form $c_2=f(c_1)$
$$z-\frac12 x-\frac13 y=f\left(\frac{x^3}{y^2}\right)$$
$$\boxed{z(x,y)=\frac12 x+\frac13 y+f\left(\frac{x^3}{y^2}\right)}$$
Explanation about combining fractions :
Use the well known basic property of the fractions :
$$\text{if}\quad \frac{A}{B}=\frac{C}{D} \quad \text{then}\quad \frac{A}{B}=\frac{C}{D}=\frac{c_1A+c_2C}{c_1B+c_2D}$$
$c_1$ , $c_2$ are arbitrary constants (not both nul}.
This property is valid for more fractions :
$$\frac{A}{B}=\frac{C}{D}=\frac{E}{F}=\frac{c_1A+c_2C+c_3E}{c_1B+c_2D+c_3F}$$
In above case $c_1=3\:;\:c_2=2 \:;\: c_3=-6 $ .
|
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Is there any closed form for the integral $\int_0^{\frac{\pi}{2}} \frac{\ln ^n(\sin x) \ln ^m(\cos x)}{\tan x} d x?$ Inspired by the question in the post, I started to generalise the integral
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{\ln^n (\sin x) \ln (\cos x)}{\tan x} d x =& \frac{1}{n+1} \int_0^{\frac{\pi}{2}} \ln (\cos x) d\left(\ln ^{n+1}(\sin x)\right) \\
=& \frac{1}{n+1}\left[\ln (\cos x) \ln ^{n+1}(\sin x)\right]_0^{\frac{\pi}{2}}+\frac{1}{n+1} \int_0^{\frac{\pi}{2}} \frac{\sin x \ln ^{n+1}\left(\sin x\right)}{\cos x} d x\\\stackrel{IBP}{=} &\frac{1}{2^{n+2}(n+1)} \int_0^{\frac{\pi}{2}} \frac{\ln ^{n+1}\left(\sin ^2 x\right)}{\cos ^2 x} d\left(\sin ^2 x\right)\\=& \frac{1}{2^{n+2}(n+1)} \int_0^{1} \frac{\ln ^{n+1} y}{1-y} d y \quad \textrm{ where }y=\sin^2 x\end{aligned}
$$
We then deal the last integral using an infinite series on the integral
$$
\begin{aligned}
\int_0^1 \frac{y^a}{1-y} d y=\sum_{k=0}^{\infty} \int_0^1 y ^a\cdot y^k d y =\sum_{k=1}^{\infty} \frac{1}{a+k}
\end{aligned}
$$
Differentiating both sides w.r.t. $a$ by $n$ times yields $$
\int_0^1 \frac{\ln^{n+1} y}{1-y} d y=\left.\frac{\partial^{n+1}}{\partial a^{n+1}} \int_0^1 \frac{y^a}{1-y} d t\right|_{a=0} =(-1)^{n+1}(n+1) ! \sum_{k=1}^{\infty} \frac{1}{k^{n+2}}= (-1)^{n+1}(n+1) !\zeta(n+2)
$$
Then we can conclude that
$$\boxed{\int_0^{\frac{\pi}{2}} \frac{\ln^n (\sin x) \ln (\cos x)}{\tan x} d x = \frac{(-1)^{n+1} n !}{2^{n+2}} \zeta(n+2)} $$
For examples,
$$
\begin{aligned}
&I_1=\frac{1}{8} \zeta(3);\quad I_2=-\frac{1}{16} \cdot 2 \cdot \zeta(4)=-\frac{\pi^4}{720} ;\quad I_{12}=-\frac{12 !}{2^{14} }\zeta(12)=-\frac{691 \pi^{12}}{21840}
\end{aligned}
$$
Similarly, I want to generalise the integral further as
$$I(n,m)=\int_0^{\frac{\pi}{2}} \frac{\ln ^n(\sin x) \ln ^m(\cos x)}{\tan x} d x =\frac{m}{2^{n+m}(n+1)} \int_0^1 \frac{\ln ^{n+1} y \ln ^{m-1}(1-y)}{1-y} d y $$
Noticing that
$$
\begin{aligned}
\int_0^1 y^a(1-y)^bd y =& \int_0^1 y^a(1-y)^{b} d y =B(a+1, b+1)
\end{aligned}
$$
Therefore
$$
\boxed{I(n, m)= \frac{1}{2^{n+2}(n+1)} \left.\frac{\partial^{n+1}}{\partial a^{n+1}} \frac{\partial^{m-1}}{\partial b^{m-1}}\left(\frac{\Gamma(a+1) \Gamma(b+1)}{\Gamma(a+b+2)}\right)\right|_{a=0,b=-1}}
$$
My question: Is there a closed form for the last high derivative?
Your comments and suggestions are highly appreciated.
|
After submitting the post, I suddenly recognise that $I(n,m) $ is actually a high derivative of a Beta function as
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{\sin ^a x \cos ^b x}{\tan x} d x
&=\int_0^{\frac{\pi}{2}} \sin ^{a-1} x \cos ^{b+1} x d x
=\frac{1}{2} B\left(\frac{a}{2} , \frac{b}{2} +1\right)
\end{aligned}
$$
Then we can write down
$$
\begin{aligned}\int_0^{\frac{\pi}{2}} \frac{\ln ^n(\sin x) \ln ^m(\cos x)}{\tan x} d x & =\left.\frac{1}{2} \frac{\partial^{n+m}}{\partial a^n \partial b ^m}\left(B\left(\frac{a}{2} , \frac{b}{2} +1\right)\right)\right|_{a,b \rightarrow 0}\\&=\frac{1}{2^{n+m+1}} \lim _{a \rightarrow 0} \lim _{b \rightarrow 1}\left[\frac{\partial^{n+m}}{\partial a^n\partial b^m} B(a, b)\right]\end{aligned}
$$
|
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Is there any natural number between $(N^2 + N)$ and $(N^2 + 1)$ that can divide $(N^2 + N)\times(N^2 + 1)$ My question is fairly simple and explained in title. I'm trying to prove that there are no natural number(s) between $(n^2+n)$ and $(n^2+1)$ that can divide $(n^2+1) \times (n^2 + n)$
[EDIT]
I was trying to solve the problem $n^4 + n^3 + n^2+n+1 = a^2$.
I said that $n^4 + n^3 + n^2+n = (a-1)\times(a+1)$. Refactoring left side, i got:
$(n^2+n)\times(n^2+1) = (a-1)+(a+1)$. It is obvious that $a-1$ and $a+1$ are the closest dividers of $a^2-1$ so i need to prove that, in natural numbers, $n^2+n$ and $n^2+1$ are the closest dividers of $a^2-1$ so i can assume that $(n^2+n) = a+1$ and $(n^2+1) = a-1$ and continue to solve the problem in proper way. Thanks for any help or hint in front.
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Let $N=(n^2+1)(n^2+n)$. Assume that there exists an integer $A$ such that $A\mid N$ and $(n^2+1)<A<(n^2+n)$. Then the complementary factor $B=N/A$ is also in that interval.
Because $A$ and $B$ are closer to each other than $n^2+1$ and $n^2+n$, and yet $AB=(n^2+1)(n^2+n)$, we must have
$A+B<(n^2+1)+(n^2+n),$ or $A+B\le 2n^2+n$.
But then
$$(A-B)^2=(A+B)^2-4AB\le (2n^2+n)^2-4(n^2+1)(n^2+n)=-4n-3n^2,$$
where the right hand side is negative. This is a contradiction.
|
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Big-O analysis of a combination function Prove that $$\binom{2n}{n} \in O\left(4^n/\sqrt{n}\right) $$
I am not familiar with markdown syntax so sorry for the formula display
|
In the comments you can find links to an argument using Stirling's approximation and an argument using an integral formula; here's a much more elementary argument (which gives a worse bound). Write $a_n = {2n \choose n}$. We have
$$\frac{a_{n+1}}{a_n} = \frac{(2n+2)! n!^2}{(2n)! (n+1)!^2} = \frac{4n+2}{n+1} = 4 - \frac{2}{n+1}$$
which gives
$$a_n = \prod_{i=1}^n \left( 4 - \frac{2}{i} \right) = 4^n \prod_{i=1}^n \left( 1 - \frac{1}{2i} \right) = 4^n \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{4} \right) \dots \left( 1 - \frac{1}{2n} \right).$$
Squaring $\frac{a_n}{4^n}$ gives
$$\begin{align*} \prod_{i=1}^n \left( 1 - \frac{1}{2i} \right)^2 & \le \prod_{i=1}^n \left( 1 - \frac{1}{2i-1} \right) \left( 1 - \frac{1}{2i} \right) \\
&= \left( 1 - \frac{1}{2} \right) \left( 1 - \frac{1}{3} \right) \dots \left( 1 - \frac{1}{2n-1} \right) \left( 1 - \frac{1}{2n} \right) \\
&= \frac{1}{2n} \end{align*}.$$
This gives
$$\boxed{ {2n \choose n} \le \frac{4^n}{\sqrt{2n}} }$$
although that constant $2$ in the denominator is not best possible. Doing this argument more carefully can get the best constant, which is $\pi$, using the Wallis product, as explained on Wikipedia.
Alternatively, the inequality $1 - x \le \exp(-x)$ gives
$$a_n \le 4^n \exp \left( \sum_{i=1}^n - \frac{1}{2i} \right) = 4^n \exp \left( - \frac{H_n}{2} \right)$$
where $H_n$ is the $n^{th}$ harmonic number. The elementary inequality $H_n \ge \log n$ gives ${2n \choose n} \le \frac{4^n}{\sqrt{n}}$ which is a bit worse, but this argument is very general. We could do a more careful analysis here by taking logarithms.
There's also a conceptual interpretation of the true asymptotic ${2n \choose n} \sim \frac{4^n}{\sqrt{\pi n}}$ that comes from applying the central limit theorem to the binomial distribution $\text{Bin} \left( 2n, \frac{1}{2} \right)$. I think this is in some sense the best explanation of that funny $\sqrt{\pi n}$.
|
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prove that $\{\det(A) : A\in S\} = \{-4,-3,-2,-1,0,1,2,3,4\}$
Let $S$ be the set of all $3\times 3$ matrices with entries in $\{0,1,-1\}$. Prove that $\{\det(A) : A\in S\} = \{-4,-3,-2,-1,0,1,2,3,4\}$.
Let $R = [-4,4]\cap \mathbb{Z}$, $T =\{\det(A) : A\in S\}$. To show $R\subseteq T,$ it suffices to show that $R\cap \mathbb{Z}^+\subseteq T,$ since clearly any matrix with an all-zero row or column (or a repeated row or column) has determinant zero and swapping any two distinct rows or columns of a matrix multiplies the determinant by $-1$ (the determinant is an alternating linear map). I know the determinant of an n by n matrix is also $n$-linear, meaning that it's linear with respect to any single row or column. Obviously the identity matrix has determinant 1. Let $A_2 = \begin{pmatrix}1 & 1 & 0\\
-1 & 1 & 0\\
0 & 0 & 1\end{pmatrix}, A_3 = \begin{pmatrix}1 & 1 & 0\\
-1 & 1 & 1\\
1 & 0 & 1\end{pmatrix},A_4 = \begin{pmatrix}1 & 1 & 0\\
-1 & 1 & 1\\
1 & -1 & 1\end{pmatrix}.$ Then using Laplace expansion along the first column, we can easily see that $\det A_i = i$ for $2\leq i\leq 4.$ Thus we just need to show $T\subseteq R$ to solve the question. Let $A = \begin{pmatrix}a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}\end{pmatrix}$. If $a_{i1}=0$ for $1\leq i\leq 3,$ then $\det(A) = 0\in R,$ so suppose otherwise. We just need to show that $|\det(A)|\leq 4$ since it is an integer (being a (multivariate) polynomial in the entries). WLOG, assume $a_{11} = 1$ (we may swap row $i\neq 1$ with row 1 and multiply the first row by $-1$ if necessary, which doesn't change the absolute value of the determinant). Perform the elementary row operations $R_i\mapsto R_i - R_1$ for $i=2,3$. Now note that if $a_{22} = a_{32} = 0,$ rows 2 and 3 are scalar multiples of each other or one of them equals the zero row, so the determinant is zero, which is in S. Hence, by swapping the two and multiplying row 2 by $-1$ if necessary, assume WLOG that $a_{22} > 0.$ Suppose first that $a_{22} = 1.$ Then $a_{12} = 0$ and $|a_{32}|\leq 1.$ Note that $|a_{23}|\leq 2,$ since $a_{13}$ was subtracted from it and has absolute value at most 1 and $a_{23}$ initially had absolute value at most 1. Also, we similarly have $|a_{33}|\leq 2.$ Subtracting row 2 from row 3 if necessary (i.e. if $a_{32}$ is nonzero), we get that $|a_{33}|\leq 4.$ But now the determinant equals $|a_{33}|\leq 4.$ Finally suppose $a_{22}= 2.$ Then $|a_{12}|=1$. Note that $|a_{23}| = 2$ only if $a_{23}' a_{13} = -1,$ where $a_{23}'$ is the original value of $a_{23}$ before subtracting from row 1. Similarly, $a_{22} = 2\Rightarrow a_{12}a_{22}' = -1.$ So $|a_{33}|=2$ and $|a_{23}|=2$ implies that $a_{33}' = a_{23}'$ and so $a_33 = a_{23}$. However, this approach seems very tedious and error-prone, so there's probably a better method.
|
The main task is to prove that $T\subseteq R$.
$\det A$ is the sum of six elements of $\{0,1,-1\}$ but if some entry of $A$ is $0$, at least two of these six elements are $0$ and you have won.
If all entries of $A$ are $\pm1$ then $\det A$ is even and wlog (up to changing some rows to their opposite), $A = \begin{pmatrix}1 &a&x\\
1&b&y\\
1 &c&z\end{pmatrix}$.
Then for $\det A$ to be $6$, we should have $b=z$, $c=x$, $a=y$, $b=-x$, $c=-y$, $a=-z$, but theses six equations are incompatible. Similarly, the six equations necessary for $\det A$ to be $-6$ are incompatible, which ends the proof.
|
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Characterize all primes p such that 15 is a square modulo p I was having some difficulty understanding the following parts of this proof:
Proof:
Obviously, 15 is a square mod 2, 3, 5. So suppose p > 5. We compute the Jacobi
symbol:
($\frac{15}{p}$) = ($\frac{3}{p})(\frac{5}{p}$) = $(-1)^\frac{p-1}{2}(\frac{p}{3})(\frac{p}{5}) $
Up to here, all is fine, but now is where I got confused:
"So the answer will depend on p modulo 4 · 15 = 60. Looking at the φ(60) = 2 · 2 · 4 = 16
residue classes mod 60, we see that the RHS is +1 exactly when
$p \equiv \pm1 , \pm7 , \pm 11, \pm17 \pmod{60}$ "
I have absolutely no clue why we are working mod 60 nor how they obtained the above numbers.
Any help would be appreciated
|
This is a fairly standard exercise. As you note, using the Legendre (or Jacobi) symbol and Quadratic Reciprocity, we have
$$\begin{align*}
\left(\frac{15}{p}\right) &= \left(\frac{3}{p}\right)\left(\frac{5}{p}\right)\\
&=(-1)^{(\frac{p-1}{2})(\frac{3-1}{2})}\left(\frac{p}{3}\right)(-1)^{(\frac{p-1}{2})(\frac{5-1}{2})}\left(\frac{p}{5}\right)\\
&= (-1)^{\frac{p-1}{2}}\left(\frac{p}{3}\right)\left(\frac{p}{5}\right).
\end{align*}$$
So we have a number of cases. Excluding $2$, $3$, and $5$ (for which $15$ is a square), we have:
*
*If $p\equiv 1\pmod{4}$, then for $15$ to be a square modulo $p$ we need $\left(\frac{p}{3}\right)=\left(\frac{5}{p}\right)$. That means either $p\equiv 1\pmod{3}$ and $p\equiv \pm 1\pmod{5}$; or $p\equiv 2\pmod{3}$ and $p\equiv \pm2\pmod{5}$. Using the Chinese Remainder Theorem for the three congruences, we get: if $p\equiv 1\pmod{4}$, $p\equiv1 \pmod{3}$, and $p\equiv 1\pmod{5}$, then $p\equiv 1\pmod{60}$. If $p\equiv 1\pmod{4}$, $p\equiv 1\pmod{3}$, and $p\equiv 4\pmod{5}$, then $p\equiv 49\equiv -11\pmod{60}$. Etc. We end up with
$$p\equiv 1,\ -11,\ 17,\ -7\pmod{60}.$$
*If $p\equiv 3\pmod{4}$, then we need $\left(\frac{p}{3}\right)=-\left(\frac{p}{5}\right)$, so either $p\equiv 1\pmod{3}$ and $p\equiv \pm 2\pmod{5}$, or else $p\equiv 2\pmod{3}$ and $p\equiv \pm1\pmod{5}$. These give the other four values modulo $60$:
$$p\equiv 7,\ -17, 11,\ -1\pmod{60}.$$
So this gives that the primes for which $15$ is a square are precisely $p=2$, $3$, $5$, and all primes $p$ such that $p\equiv \pm 1,\ \pm7,\ \pm 11,\ \pm 17\pmod{60}$, as claimed.
|
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|
A smarter (not bashy) way to solve this roots of unity problem? (Mandelbrot) Let $\xi = \cos \frac{2\pi}{5} + i \sin \frac{2pi}{5}$ be a complex fifth root of unity. Set $a = 20\xi^2 + 13 \xi, b = 20\xi^4 + 13\xi^2, c = 20\xi^3 + 13\xi^4, \text{and } d = 20\xi + 13\xi^3$. Find $a^3 + b^3 + c^3 + d^3$
Immediately what comes to mind is finding $(a + b + c + d)^3$ and subtracting whatever we don't need to get $a^3 + b^3 + c^3 + d^3$. However,
\begin{equation*}
(a+b+c+d)^3 = a^3+3a^2b+3a^2c+3a^2d+3ab^2+6abc+6abd+3ac^2+6acd+3ad^2+b^3+3b^2c+3b^2d+3bc^2+6bcd+3bd^2 + c^3 + 3c^2d + 3cd^2 + d^3
\end{equation*}
There is simply no good way to calculate $6abc + 6abd + 6acd + 6bcd $ without expanding everything.
|
Let $\varphi:\mathbb{Q}[\xi]\to\mathbb{Q}[\xi]$ be the field homomorphism which is the identity on $\mathbb{Q}$ and $\varphi(\xi)=\xi^2$.
Since $\xi+\xi^2+\xi^3+\xi^4=-1$, we have
$$
\begin{align}
a^3&=\left(20\xi^2+13\xi\right)^3&&=8000\xi+15600+10140\xi^4+2197\xi^3\\
b^3&=\varphi\!\left(a^3\right)&&=8000\xi^2+15600+10140\xi^3+2197\xi\\
c^3&=\varphi\circ\varphi\!\left(a^3\right)&&=8000\xi^4+15600+10140\xi+2197\xi^2\\
d^3&=\varphi\circ\varphi\circ\varphi\!\left(a^3\right)&&=8000\xi^3+15600+10140\xi^2+2197\xi^4\\\hline
&\!\!\!\!\!a^3+b^3+c^3+d^3&&=-8000+62400-10140-2197\\
&&&=42063
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4549205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Prove that $\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln\big(\frac mn \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}\dots\big)$ page 371 in ‘Synopsis Of Elementary Results In Pure Mathematics’ contains the following result,
$$\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln \left(\frac{m}{n} \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}…\right)$$
How does one prove this step-by-step?
|
I am going to evaluate the integral by differentiating its partner
$$
\begin{aligned}
I(a) &=\int_0^1 \frac{x^a-x^{n-1}}{\left(1+x^p\right) \ln x} dx\\
I^{\prime}(a) &=\int_0^1 \frac{x^a}{1+x^p} d x \\
&=\sum_{k=0}^{\infty} \frac{(-1)^k}{a+pk+1}
\end{aligned}
$$
Integrating back gives our integral
$$
\begin{aligned}
I &=I(m-1)-I(n-1) \\
&=\int_{n-1}^{m-1} \sum_{k=0}^{\infty} \frac{(-1)^k}{a+p k+1} d a \\
&=\sum_{k=0}^{\infty}(-1)^k \ln \left(\frac{m+p k}{n+p k}\right)\\&= \ln \left(\frac{m}{n} \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}…\right)
\end{aligned}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4553487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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|
Compute $I=\int_0^2(3x^2-3x+1)\cos(x^3-3x^2+4x-2)\,dx$ If the value of $I = \int\limits_0^2 {\left( {3{x^2} - 3x + 1} \right)\cos \left( {{x^3} - 3{x^2} + 4x - 2} \right)dx} $ can be expressed as $p(\sin q)$ , where $p,q\in \mathbb N$, then $p+q-1=$?
My approach is as follow ${x^3} - 3{x^2} + 4x - 2 = {x^3} - {x^2} - 2\left( {{x^2} - 2x + 2} \right) \Rightarrow {x^3} - 3{x^2} + 4x - 2 = {x^2}\left( {x - 1} \right) - 2{\left( {x - 1} \right)^2}$
${x^3} - 3{x^2} + 4x - 2 = \left( {x - 1} \right)\left( {{x^2} - 2\left( {x - 1} \right)} \right) \Rightarrow {x^3} - 3{x^2} + 4x - 2 = \left( {x - 1} \right)\underbrace {\left( {{x^2} - 2x + 2} \right)}_{ > 0}$
Unable to proceed as cannot find the substitution of ${x^3} - 3{x^2} + 4x - 2$
|
Let $y=x-1$, then $3x^2 - 3x + 1=3y^2+3y+1$, and $x^3-3x^2+4x-2=y^3+y$. Hence
$$I=\int_0^2(3x^2-3x+1)\cos(x^3-3x^2+4x-2)\,dx=\int_{-1}^1(3y^2+3y+1)\cos(y^3+y)\,dy.$$
Since $\frac{d}{dy}(y^3+y)=3y^2+1$, we have
$$\int_{-1}^1(3y^2+1)\cos(y^3+y)\,dy=\sin(y^3+y)\Big|_{-1}^1=2\sin 2.$$
Since $3y\cos(y^3+y)$ is an odd function in $y$, we have
$$\int_{-1}^13y\cos(y^3+y)\,dy=0.$$
Therefore, $I=2\sin 2.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4554753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Solve the equation $(x-1)^5+(x+3)^5=242(x+1)$ Solve the equation $$(x-1)^5+(x+3)^5=242(x+1)$$ My idea was to let $x+1=t$ and use the formula $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ but I have troubles to implement it. The equation becomes $$(t-2)^5+(t+2)^5=242t\\(t-2+t+2)\left[(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-\\-(t-2)(t+2)^3+(t+2)^4\right]=242t$$ Let $A=(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-(t-2)(t+2)^3+(t+2)^4.$
Then $$A=(t-2)^4-(t-2)^2(t^2-4+t^2+4t+4)-(t+2)^3(2-t+t+2)\\=(t-2)^4-2t(t+2)(t-2)^2-4(t+2)^3.$$
|
We have
$$0=(x-1)^5+(x+3)^5-242(x+1)=2(x^2 + 2x + 42)(x + 2)(x + 1)x
$$
by applying the rational root theorem to the polynomial equation
$$
2x^5 + 10x^4 + 100x^3 + 260x^2 + 168x=0,
$$
which yields the linear factors $x$, $x+1$ and $x+2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4555212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Number of strings of length $n$ with no consecutive $y$'s Suppose we have a set $S$ such that $\lvert S\rvert=k+1$. Fix an element $y$ in $S$. We want to find the recurrence relation on the number of $S$-strings of length $n$ that don't have two consecutive $y$'s, namely $yy$. Use $f(n)$ to denote the answer.
I already find some initial conditions: for $f(1)$, that is the number of $S$-strings of length $1$ without two consecutive $y$'s, which is obviously $f(1)=k+1$; For $f(2)$, there is only one string that has two consecutive $y$'s, which is $yy$, so $f(2)=(k+1)^2-1$; For $f(3)$, If we place the two consecutive $y$'s at the first and the second positions, there are $k+1$ strings and $yyy$ are among the $k+1$ strings; If we place $yy$ at the second and the third positions, there are another $k+1$ strings and $yyy$ are also among these $k+1$ strings. So we conclude $f(3)=(k+1)^3-(2k+1)$.
Yet I have no idea what the general case is like. Any help please?
|
The empty string does not contain two consecutive $y$s, so $f(0) = 1$. None of the $k + 1$ strings of length $1$ contain two consecutive $y$s, so $f(1) = k + 1$. Every string of length $2$ is admissible except $yy$, so $f(2) = (k + 1)^2 - 1 = k^2 + 2k + 1 - 1 = k^2 + 2k$.
An admissible string of length $n \ge 2$ must either begin with an element $x \in S$ other than $y$ or must begin $yx$, where $x \in S$ and $x \neq y$. If a string begins with $x \in S$, where $x \neq y$, it can be extended to an admissible string of length $n$ by appending an admissible string of length $n - 1$ to the end of the string $x$. There are $k$ ways to choose $x \in S$ such that $x \neq y$ and $f(n - 1)$ admissible strings of length $n - 1$, so there are $kf(n - 1)$ such words. A string that begins with $yx$, where $x \in S$ and $x \neq y$, can be extended to an admissible string of length $n$ by appending an admissible string of length $n - 2$ to the end of the string $yx$. There are $k$ ways to choose $x$ and $f(n - 2)$ admissible strings of length $n - 2$. Hence, there are $kf(n - 2)$ such strings. Thus, we have
\begin{align*}
f(0) & = 1\\
f(1) & = k + 1\\
f(n) & = kf(n - 1) + kf(n - 2), n \ge 2
\end{align*}
Notice that we obtain
\begin{align*}
f(2) & = kf(1) + kf(0)\\
& = k(k + 1) + k \cdot 1\\
& = k^2 + k + k\\
& = k^2 + 2k
\end{align*}
which agrees with your calculation $f(2) = (k + 1)^2 - 1$, and
\begin{align*}
f(3) & = kf(2) + kf(1)\\
& = k(k^2 + 2k) + k(k + 1)\\
& = k^3 + 2k^2 + k^2 + k\\
& = k^3 + 3k^2 + k
\end{align*}
which agrees with your calculation $f(3) = (k + 1)^3 - (2k + 1)$.
|
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"url": "https://math.stackexchange.com/questions/4557916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
In how many ways can points be assigned to five questions with one to four points per question so that the total is $14$? A professor must write an exam with $5$ questions. Question number i should give $p_i \in \mathbb{Z}$ points. The sum of the points must be $14$ and each question must give at least $1$ point and a maximum of $4$ points. In how many ways can he distribute the points on the questions?
I first turned $p_i= x_i - 1$ so I can get $0 \le x_i \le 3$ instead. Then I put the variable $x_i$ in the equation and got $x_1 + x_2 + x_3 + x_4 + x_5 = 9$. The total amount of combination is $C(tot) = C(5+9-1, 9)= C(13, 4)$. What I then did was find how many combinations there are when one $x$ can be greater than $3$, and then when $2$ can be greater than $3$. $C_1 = C(9,4)C(5,1)$ and $C_2 = C(5,4)C(5,2)$. So the answer is $C(13,4) - C(9,4)C(5,1) - C(5,4) C(5,2)$
What my answer sheet says is, $C = C(13,4) - C(9,4)C(5,1) + C(5,4) C(5,2) = 715-630+50= 135$ What I answered was $715 - 105 - 50 = 560$ How is $C(9,4)C(5,1)= 630$? Besides why did they do $+50$ instead of $-50$??
Can anyone help me ?
|
If $p_i$ points are assigned to the $i$th question, with each $p_i$ a positive integer satisfying $1 \leq p_i \leq 4$, and the examination has a total of $14$ points, $1 \le i \le 5$, then
$$p_1 + p_2 + p_3 + p_4 + p_5 = 14 \tag{1}$$
is an equation in the positive integers subject to the restrictions $p_i \leq 4$, $1 \le i \le 5$.
Following your method of converting the equation into the nonnegative integers, we let $x_i = p_i - 1$, $1 \le i \le 5$. Then each $x_i$ is a nonnegative integer. Substituting $x_i + 1$ for $p_i$, $1 \le i \le 5$, in equation $1$ and simplifying yields
$$x_1 + x_2 + x_3 + x_4 + x_5 = 9 \tag{2}$$
which is an equation in the nonnegative integers subject to the restrictions that $x_i = p_i - 1 \leq 4 - 1 = 3$, $1 \leq i \leq 4$.
If we temporarily ignore the restrictions, equation $2$ has
$$\binom{9 + 5 - 1}{5 - 1} = \binom{13}{4}$$
solutions in the nonnegative integers.
From these, we must subtract those cases in which one or more of the $x_i$s exceeds $3$. Since the $x_i$s are nonnegative integers, at most two variables can exceed $3$ since $3 \cdot 4 = 12 > 9$.
There are five ways to choose a variable that exceeds $3$. Suppose it is $x_1$. Then $x_1 \geq 4$. Let $x_1' = x_1 - 4$. Then $x_1'$ is a nonnegative integer. Substituting $x_1' + 4$ for $x_1$ in equation $2$ and simplifying yields
$$x_1' + x_2 + x_3 + x_4 + x_5 = 5 \tag{3}$$
which is an equation in the nonnegative integers with
$$\binom{5 + 5 - 1}{5 - 1} = \binom{9}{4}$$
solutions. Therefore, there are
$$\binom{5}{1}\binom{9}{4}$$
solutions in which a variable exceeds $3$.
However, if we subtract this amount from the total, we will have subtracted too much. This is because we will have subtracted each case in which two of the variables exceed $3$ twice, once for each way we could have designated one of the variables as the variable that exceeds $3$. We only want to subtract such cases once, so we must add them to the total.
There are $\binom{5}{2}$ ways to select two of the five variables to exceed $3$. Suppose they are $x_1$ and $x_2$. Then $x_1 \geq 4$ and $x_2 \geq 4$. Let $x_1' = x_1 - 4$ and $x_2' = x_2 - 4$. Then $x_1'$ and $x_2'$ are nonnegative integers. Substituting $x_1' + 4$ for $x_1$ and $x_2' + 4$ for $x_2$ in equation $2$ and simplifying yields
$$x_1' + x_2' + x_3 + x_4 + x_5 = 1 \tag{4}$$
which is an equation in the nonnegative integers with five solutions. Hence, there are
$$\binom{5}{2}\binom{1 + 5 - 1}{5 - 1} = \binom{5}{2}\binom{5}{4}$$
solutions in which two variables exceed $3$.
By the Inclusion-Exclusion Principle, the number of ways the professor can distribute the points to the five questions so that each question is worth a positive integer number of points that is at most $4$ is
$$\binom{13}{4} - \binom{5}{1}\binom{9}{4} + \binom{5}{2}\binom{5}{4}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to show $\sum_{n=0}^\infty (-1)^n\, \frac{\Gamma \left(\frac{n}{2}+1\right)}{n! \,\Gamma \left(2-\frac{n}{2}\right)}=\frac{3-\sqrt{5}}{2}$ The following sum was used in an unrelated answer on math.stackexchange.com.
$$\sum_{n=0}^\infty (-1)^n \frac{\Gamma \left(\frac{n}{2}+1\right)}{n! \Gamma \left(2-\frac{n}{2}\right)}=\frac{3-\sqrt{5}}{2}$$
How can you show this to be true?
|
We will use the facts that
$$
\frac{1}{{\Gamma\! \left( {2 - \frac{n}{2}} \right)}} = \frac{{\Gamma \!\left( {\frac{n}{2} - 1} \right)}}{{\Gamma \!\left( {2 - \frac{n}{2}} \right)\Gamma\! \left( {\frac{n}{2} - 1} \right)}} = - \frac{1}{\pi }\sin \left( {\frac{{\pi n}}{2}} \right)\Gamma \!\left( {\frac{n}{2} - 1} \right)
$$
for $n\geq 3$,
$$
\Gamma\! \left( {m + \frac{1}{2}} \right) = \frac{{(2m)!}}{{4^m m!}}\sqrt \pi
$$
for $m\geq 0$, and
$$
\sqrt {1 + x} = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^{k + 1} }}{{4^k (2k - 1)}}} \binom{2k}{k}x^k
$$
for $|x|<1$. Thus
\begin{align*}
\sum\limits_{n = 0}^\infty {( - 1)^n \frac{{\Gamma\! \left( {\frac{n}{2} + 1} \right)}}{{n!\Gamma\! \left( {2 - \frac{n}{2}} \right)}}} & = \frac{1}{2} + \sum\limits_{n = 3}^\infty {( - 1)^n \frac{{\Gamma\! \left( {\frac{n}{2} + 1} \right)}}{{n!\Gamma\! \left( {2 - \frac{n}{2}} \right)}}} \\ & = \frac{1}{2} + \frac{1}{\pi }\sum\limits_{n = 3}^\infty {( - 1)^{n + 1} \sin \left( {\frac{{\pi n}}{2}} \right)\frac{{\Gamma\! \left( {\frac{n}{2} - 1} \right)\Gamma\! \left( {\frac{n}{2} + 1} \right)}}{{n!}}} \\ & = \frac{1}{2} + \frac{1}{\pi }\sum\limits_{k = 1}^\infty {( - 1)^k \frac{{\Gamma\! \left( {k - \frac{1}{2}} \right)\Gamma\! \left( {k + \frac{3}{2}} \right)}}{{(2k + 1)!}}}
\\ &
= \frac{1}{2} + \sum\limits_{k = 1}^\infty { \frac{( - 1)^k}{{(2k + 1)!}}\frac{1}{{16^k }}\frac{{(2k - 2)!}}{{(k - 1)!}}\frac{{(2k + 2)!}}{{(k + 1)!}}} \\ & = \frac{1}{2} + \sum\limits_{k = 1}^\infty {\frac{{( - 1)^k }}{{16^k (2k - 1)}}} \binom{2k}{k}
\\ &
= \frac{3}{2} + \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }}{{16^k (2k - 1)}}} \binom{2k}{k} \\ & = \frac{3}{2} - \sqrt {1 + \frac{1}{4}} = \frac{3}{2} - \frac{{\sqrt 5 }}{2}.
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Calculate $\sum_{n=2}^{\infty}\left (n^2 \ln (1-\frac{1}{n^2})+1\right)$ I am interested in evaluating
$$\sum_{n=2}^{\infty}\left (n^2 \ln\left(1-\frac{1}{n^2}\right)+1\right)$$
I am given the solution for the question is $\,\ln (\pi)-\frac{3}{2}\,.$
$$\sum_{n=2}^{\infty}\left(n^2\ln\left(\!1\!-\!\frac{1}{n^2}\!\right)+1\right)=4\ln\left(\!\frac{3}{4}\!\right)+1+9\ln\left(\!\frac{8}{9}\!\right)+1+\ldots$$
Any tricks to solve it?
|
(I have partially copied @Hamdiken's answer to make it more complete)
Define the sum
$$S(x):=\sum_{n=2}^{\infty}\left(n^2\ln\left(1-\frac{x^2}{n^2}\right)+x^2\right)$$
and differentiate with respect to $x$ to obtain
\begin{align*}
S'(x)=\sum_{n=2}^{\infty}\left(-2x{\frac{n^2}{n^2-x^2}}+2x\right)
&=2x\sum_{n=2}^{\infty}\left({\frac{n^2}{x^2-n^2}}+1\right)\\
&=2x\sum_{n=2}^{\infty}\left({\frac{n^2+x^2-n^2}{x^2-n^2}}\right)\\
&=x^2\left[2x \sum_{n=2}^{\infty}{\frac{1}{x^2-n^2}}\right]\\
&=\pi x^2\left[2\pi x \sum_{n=2}^{\infty}{\frac{1}{(\pi x)^2-(\pi n)^2}}\right]\\
&=\pi x^2\left[-\frac{2x}{\pi(x^2-1)}-\frac{1}{\pi x}+\underbrace{\frac{1}{\pi x}+2\pi x \sum_{n=1}^{\infty}{\frac{1}{(\pi x)^2-(\pi n)^2}}}_{\cot(\pi x)}{}\right]\\
\end{align*}
by the Mittag-Leffler pole expansion of $\cot(\pi x)$.
We finally get that
$$S'(x)=-x+\left(\pi x^2\cot(\pi x)-\frac{2x^3}{x^2-1}\right)$$
Upon integrating, we determine
$$S(x)=-\frac{x^2}2+\int_0^x\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)\,dt$$
For $x=1$, we get
$$S(1)=-\frac{1}{2}+\int_0^1\left(\pi t^2\cot(\pi t)-\frac{2t^3}{t^2-1}\right)\,dt\stackrel{t\to 1-t}{=}-\frac{3}{2}-\ln(2)+\int_{0}^{1} \left(\frac{1}{t} -\pi (1-t)^2\cot(\pi t)\right)\, dt
$$
Now by the Laurent series for $\cot(\pi x)$, namely, $$\cot(\pi x) = \frac{1}{\pi x} - \frac{2}{\pi x} \sum_{k=1}^{\infty}\zeta(2k) x^{2k}$$
we have $$S(1)=-\ln(2)+\sum_{k=1}^{\infty}\frac{\zeta(2k)}{k(k+1)(2k+1)}$$
Now since for $\Re(s)>1$ we have the following Mellin transform $$\zeta(s)\Gamma(s)=\int_{0}^{\infty}\frac{x^{s-1}}{e^x-1}\,dx$$ we get $$S(1)=-\ln(2)+\int_{0}^{\infty}\frac{1}{e^x-1}\sum_{k=1}^{\infty}\frac{x^{2k-1}}{\Gamma(2k)k(k+1)(2k+1)}=-\ln(2)-2\int_{0}^{\infty}\frac{2+x^2-2\cosh(x)}{x^3(e^x-1)}\,dx$$
We shall now evaluate the integral $$I:=-2\int_{0}^{\infty}\frac{2+x^2-2\cosh(x)}{x^3(e^x-1)}\,dx$$
Write $$\frac{1}{e^x-1}=\sum_{n=1}^{\infty}e^{-nx}$$
so $$I=-2\sum_{n=1}^{\infty}\int_{0}^{\infty} \frac{e^{-n x}}{x^3}\left(2+x^2-2\cosh(x)\right)\,dx$$
Now using the convolution property of the Laplace Transform:
$$\int_{0}^{\infty} f(x)\cdot g(x)\, dx=\int_{0}^{\infty} (\mathcal{L}f)(s)\cdot (\mathcal{L}^{-1}g)(s)\, ds$$ with $f(x) = e^{-nx}\left(2+x^2-2\cosh(x)\right)$ and $g(x)=\frac{-2}{x^3}$ we have $$I=\sum_{n=1}^{\infty}\int_{0}^{\infty}\frac{2 s^2}{(n+s-1) (n+s)^3 (n+s+1)}\, ds=\int_{0}^{\infty}\frac{s^2+2 s+2}{s(s+1)}+s^2 \psi ^{(2)}(s)\, ds\stackrel{\text{IBP}}{=}\ln(2\pi)-\frac{3}{2}$$
where $\psi$ is the polygamma function. So $S(1)=\ln(\pi)-\frac{3}{2}$ as required. $\square$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Better approach to evaluate the limit $\lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})$ I solved it by rewriting the limit as indeterminate form $0/0$, then apply L'Hopital's rule 4 times, It was really lengthy and easy to make mistakes, If anyone got a better approach, please tell me!
$$
\begin{align}
\lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})&=\lim_{x\to0^+}\frac{x^2\cos^2x-\sin^2x}{x^2\sin^2 x}\\
\vdots\\
&=\lim_{x\to0^+}\frac{-16\cos 2x+\ldots}{24\cos 2x+\ldots}\\
&=-\frac{2}{3}
\end{align}
$$
|
Using $$\tan(x)=x+\frac{x^3}{3}+o(x^3)\quad \text{and}\quad \frac{1}{1+x}=1-x+o(x),$$ yields
\begin{align*}
\left(\cot(x)-\frac{1}{x}\right)\left(\cot(x)+\frac{1}{x}\right)&=\cot^2(x)-\frac{1}{x^2}\\
&=\frac{1}{\tan^2(x)}-\frac{1}{x^2}\\
&=\frac{1}{x^2+\frac{2}{3}x^4+o(x^4)}-\frac{1}{x^2}\\
&=\frac{1}{x^2}\left(\frac{1}{1+\frac{2}{3}x^2+o(x^2)}-1\right)\\
&=\frac{1}{x^2}\left(1-\frac{2}{3}x^2+1+o(x^2)\right)\\
&=-\frac{2}{3}+o(1).
\end{align*}
|
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|
On an infinite harmonic series. Let $n \in \mathbb{N}$ and let $\mathcal{H}_m$ denote the $m$-th harmonic number. Evaluate in a closed form the sum
$$\mathcal{S}_n = \sum_{m=1}^{\infty} \mathcal{H}_m \left ( \frac{1}{m+1} - \frac{1}{m+n+2} \right )$$
I do not know how to tackle this. My instict tells me that there should be a telescopic sum somewhere inside but I cannot figure it out. The sum should calculate to
$$\mathcal{S}_n = \frac{\mathcal{H}_{n+1}^2 + \mathcal{H}_{n+1}^{(2)}}{2}$$
(where $\mathcal{H}^{(2)}_m$ is the $m$-th generalized harmonic number of order $2$) since this particular series comes from evaluating the integral
$$\mathcal{J}_n = \int_0^1 x^n \log^2(1-x) \, \mathrm{d}x$$
Any help finishing $\mathcal{S}_n$?
|
It seems I got preempted by a comment, but since nobody has posted an answer, I see no reason to not post my own answer, though it does use a very similar method.
First, we let
$$T_n=\frac{1}{2}[H_{n+1}^2+H^{(2)}_{n+1}]$$
We see that for $n\geq 0$,
\begin{equation}
\begin{split}
T_n-T_{n-1}&=\frac{1}{2}[H_{n+1}^2-H_n^2+H_{n+1}^{(2)}-H_{n}^{(2)}]\\
&=\frac{1}{2}[(H_{n+1}-H_n)(H_{n+1}+H_n)+H_{n+1}^{(2)}-H_{n}^{(2)}]\\
&=\frac{1}{2}\left[\frac{1}{n+1}\left(2H_{n+1}-\frac{1}{n+1}\right)+\frac{1}{(n+1)^2}\right]\\
&=\frac{H_{n+1}}{n+1}
\end{split}
\end{equation}
and furthermore,
\begin{equation}
\begin{split}
S_n-S_{n-1}&=\sum_{m=1}^\infty\left[\frac{H_m}{m+n+1}-\frac{H_m}{m+n+2}\right]\\
&=\sum_{m=1}^\infty\left[\frac{H_m}{m+n+1}-\frac{H_{m+1}}{m+n+2}+\frac{1}{(m+n+2)(m+1)}\right]\\
&=\sum_{m=1}^\infty\left[\frac{H_m}{m+n+1}-\frac{H_{m+1}}{(m+1)+n+1}\right]+\frac{1}{n+1}\sum_{m=1}^\infty\left[\frac{1}{m+1}-\frac{1}{(m+n+1)+1}\right]\\
\end{split}
\end{equation}
Notice that both series exhibit telescoping-like behaviour, so
$$\sum_{m=1}^\infty\left[\frac{H_m}{m+n+1}-\frac{H_{m+1}}{(m+1)+n+1}\right]=\frac{H_1}{n+2}=\frac{1}{n+2}$$
and furthermore,
$$\sum_{m=1}^\infty\left[\frac{1}{m+1}-\frac{1}{(m+n+1)+1}\right]=\sum_{m=1}^{n+1}\frac{1}{m+1}=H_{n+2}-1$$
Therefore,
$$S_n-S_{n-1}=\frac{1}{n+2}+\frac{H_{n+2}-1}{n+1}=\frac{H_{n+1}}{n+1}$$
We have proven that $T_n$ and $S_n$ satisfy the same recurrence relation, and since $T_{-1}=0=S_{-1}$, then $T_n=S_n$ for all $n\geq -1$, as desired.
Please do ask if anything needs clarification.
|
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|
Prove that $\frac{\ln x}{x}+\frac{1}{e^x}<\frac{1}{2}$ for $x > 0$
Given $x>0$ , prove that $$\frac{\ln x}{x}+\frac{1}{e^x}<\frac{1}{2}$$
I have tried to construct $F(x)=\frac{\ln x}{x}+\frac{1}{e^x}$ and find the derivative function of $F(x)$ to find the maximun value, but I can't solve the transcendental equation.
So I tried another way.I tried to use the inequality $e^{-x}\le \frac{1}{x+1}$ (when $x>0$ ) to prove the inequality $\frac{\ln x}{x}<\frac{1}{2}-\frac{1}{e^x}$ , but I can't connect these two inequalities well, and I did't solve the problem in the end.
How can I do?
|
Let
\begin{align*}
f(x) &= e^{-x} + \frac{\log (x)}{x}, \\
f'(x) &= -e^{-x} + \frac{1 - \log (x)}{x^2}, \\
f''(x) &= e^{-x} + \frac{2 \log (x) - 3}{x^3}.
\end{align*}
We would like to show that $f(x) < 1/2$ for all $x > 0$.
Case 1. $0 < x \le 1$
Using the inequalities $e^x > x + 1$ and $\log(x) \le x - 1$,
$$ \begin{split}
\frac{x e^x}{2} (1 - 2 f(x))
&= \frac{1}{2} e^x (x-2 \log (x))-x \\
&> \frac{1}{2} (x + 1) (x - 2(x - 1)) - x \\
&= - \frac{1}{2} (x-1) (x+2) \\
&< 0.
\end{split} $$
Therefore, $f(x) < 1/2$.
Case 2. $1 < x \le 1.7$
We have $2 \log (x) - 3 < 0$. Hence,
$$ \begin{split}
f''(x) &= e^{-x} + \frac{2 \log (x) - 3}{x^3} \\
&< e^{-1} + \frac{2 \log(1.7) - 3}{(1.7)^3}
\approx -0.0267 < 0.
\end{split} $$
(Such approximations can obviously be made rigorous by using interval arithmetic.)
Thus, $f'$ is strictly decreasing on $[1, 1.7]$.
Since
$$
f'(1) = 1 - \frac{1}{e} \approx 0.632
$$
and
$$
f'(1.7) = -e^{-1.7} + \frac{1 - \log(1.7)}{1.7} \approx -0.020,
$$
there is a unique $\alpha \in (1, 1.7)$ such that
$$ f'(\alpha) = -e^{-\alpha} + \frac{1 - \log (\alpha)}{\alpha^2} = 0 $$
and $f$ reaches its maximum on $[1, 1.7]$ at $\alpha$. It can then be shown via interval arithmetic that $\alpha \approx 1.618$ and $f(\alpha) \approx 0.496 < 1/2$.
Case 3. $x > 1.7$
Let
$$
g(x) = x^2 f'(x) = -x^2 e^{-x} + 1 - \log(x).
$$
Then,
$$
g'(x) = \frac{e^{-x} \left(x^3 - 2x^2 - e^x\right)}{x}.
$$
It is left as an exercise for the reader to show that $x^3 - 2x^2 - e^x < 0$, so $g'(x) < 0$ for all $x > 0$. But $g(1.7) < 0$, so for all $x > 1.7$, $g(x) < 0$ and thus $f'(x) < 0$. Therefore, $f$ is decreasing on $[1.7, +\infty)$.
|
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|
short weierstrass form for cubic I have some question about the derivation of the short Weierstrass form. In https://www.staff.uni-mainz.de/dfesti/EllipticCurvesNotes.pdf this note, I follow the derivation till the point $y^2 = β_0x^3 + β_1x^2 + β_2x + β_3$, but then it says that using the transformation $x' = x + β_1/3, y'=y$, we then arrive at $y^2 = x^3 + ax + b$. But my question is how it turns the coefficient $\beta_0$ into $1$? One method I come up with is to let $y^2 = β_0(x^3 + \gamma_1x^2 + \gamma_2x + \gamma_3)$, then applying $x' = x + \gamma_1/3, y'=y$, we get $y^2 = \beta_0(x^3 + ax + b)$ then change by a scalar $x' = \beta_0^{1/3}x, y'=y$, we get the form $y^2 = x^3 + ax + b$. but the problem is the final step would need to assume that $\beta_0$ has a cube root, which is not true from general rational numbers. Does anyone knows how to solve this? for a specific example, how could one turn the equation $y^2 = 5x^3+x+1$ into the short Weierstrass form?
|
You can get rid of the $\beta_0$ by multiplying both sides by $\beta_0^2$ and then letting $u = \beta_0 x$ and $v = \beta_0 y$. So in your example $y^2 = 5x^3+x+1$, we have
\begin{align*}
y^2 &= 5x^3+x+1\\
5^2 y^2 &= 5^3 x^3 + 5^2 x + 5^2\\
(5y)^2 &= (5x)^3 + 5(5x) + 25\\
v^2 &= u^3 + 5 u + 25 \, .
\end{align*}
|
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|
What is the value of $\int_{-1}^{2} 4 \, x^2 \, (2+x-x^2) \, dx$? $$\int_{-1}^{2} 4 \, x^2 \, (2+x-x^2) \, dx$$
*
*When I solved this took the constant out of the integral and then multiplied $x^2$
by the bracket and evaluated the integral to get:
$$4 \, \left[ \frac{x^4}{4} +\frac{2x^3}{3}-\frac{x^5}{5} \right]^2_{-1}=\frac{63}{20}$$
*But when I solve this integral on my calculator on (Casio 991ERX) or Wolfram I get $ \frac{63}{5}$. Why is that?
|
The first step should be simplifying the integral, like this:
$$\displaystyle \int -4x^{4} + 4x^{3} + 8x^2 \,dx$$
$$-4\int x^4dx + 4\int x^3 dx + 8\int x^2 dx$$
use the power rule, $$\int x^n \,dx = \frac{x^{n+1}}{n+1}$$
and apply to the problem
$$-4 \times\frac{x^5}{5} + 4\times \frac{x^4}{4} + 8 \times\frac{x^3}{3}$$
Now just plug in the values and you should get 12.6 :)
|
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|
How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$
How is $\ln (x-2) - \frac{1}{2} \ln (x-1) = \frac{1}{2} \ln \frac{(x-2)^2}{x-1}$
Can someone enlighten me on how is these 2 actually equals and the steps taken? the left hand side is actually the answer for $\int \frac{x}{2 (x-2)(x-1)} dx$ but I need to combine the expression to the right hand side to continue with the steps in the question I am attempting
|
HINT:
Multiply both side by $2.$ You need to show that:
$$2 \ln (x-2) -2 \cdot \frac{1}{2} \ln (x-1) = 2\cdot\frac{1}{2} \ln \frac{(x-2)^2}{x-1}$$
Use the rule
$$ \ln x - \ln y= \ln~(x/ y)$$
Next get rid of $\ln ..$ ( by exponentiation of both sides). What went before integration is not asked, but only about last step redoing..
|
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|
How do you solve these kinds of system of equations? Recently, I came across this system of equations,
\begin{align}
\begin{cases} \dfrac{2x}{p(p+4)}=\dfrac{2x}{q(q-4)}=3 & (1) \\ \dfrac{x}{p+q} = 4 & (2) \end{cases}
\end{align}
or
\begin{align}
\begin{cases} \dfrac{2x}{p(p+4)}=\dfrac{2x}{q(q-4)}=3 & (1) \\ \dfrac{y}{p+4} = \dfrac{x-y}{q-4} = 4 & (2) \end{cases}
\end{align}
It would be a great help if someone can help me solve these.
Thanks!
|
Consider the first set of fractions:
\begin{align}
\frac{2 \, x}{p (p+4)} &= \frac{2 \, x}{q (q-4)} = 3 \\
\frac{x}{p+q} &= 4
\end{align}
it would seem that the relations lead to an equation for $p$ and $q$. This follows from solving for $x$ in the second equation, $x = 4 \, (p+q)$, and using it in the first which gives
$$ \frac{8 \, (p+q)}{p (p+4)} = 3 \hspace{10mm} \frac{8 \, (p+q)}{q (q-4)} = 3 \to \frac{8}{3} \, (p+q) = p (p+4) = q (q-4). $$
This gives $p^2 + 4 \, p = q^2 - 4 \, q$ and factors to $(p+q) (p - q + 4) = 0$ which gives $q = -p$ or $q = p + 4$.
The second set of equations follows a similar pattern. First solve for $y$ to obtain $y = 4 \, (p+4)$ and then solve for $x$ which is $x = 4 \, (p+q)$. Then find the relation between $p$ and $q$ from the first line (which is $p (p+4) = q (q-4)$).
|
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|
Find limit of trigonometric function with indeterminacy Find limit of the given function:
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
$$
I tried putting 0 instead of x
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
=
\frac{(4^{\arcsin(0)} - 1)(\sqrt[10]{1 - \arctan(0)} - 1)}{(1-\cos\tan0)\ln(1-\sqrt{\sin 0})}
=
\frac{(1 - 1)(\sqrt[10]{1 - 0} - 1)}{(1-1)\ln(1-\sqrt{0})}
=
\frac{0*0}{0*0}
=
\frac{0}{0}
$$
But as you can see at $0$ there is limit Indeterminacy ($0/0$). How to play around and solve it?
|
$$\lim_{x\to 0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
$$
$4^{\arcsin(x^2)} - 1=e^{\arcsin(x^2) \log 4}-1\sim x^2\log 4 $
$\sqrt[10]{1 - \arctan(3x^2)} - 1\sim 1-\dfrac{1}{10}\, \arctan(3x^2)-1\sim -\dfrac{3}{10}x^2 $
$1-\cos\tan6x \sim 1-1+\dfrac{\tan^2 6x}{2}\sim 18 x^2$
$\ln(1-\sqrt{\sin x^2})\sim \sqrt{\sin x^2} \sim |x|$
Therefore the limit is $0$.
|
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|
Prove $^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,... How can we prove that
$^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,...
I attempted to solve this with Mathematical Induction as follows:
Let s(n) = $x^2 - x + 1$ | $^{6n+2} − ^{6n+1} + 1$; = 1, 2, 3,..
Basic Step
Let n = 1
⇒ $x^2 - x + 1$ | $^8 − ^7 + 1$
I then proved that the remainder is 0 using polynomial long division.
$\frac{^{6n+2} − ^{6n+1} + 1}{x^2 - x + 1}$ = $x^6 - x^4 - x^3 + x + 1$ R 0
∴ s(1) is true
Assumption Step
Assume that s(m) is true
⇒ $\frac{^{6m+2} − ^{6m+1} + 1}{x^2 - x + 1}$ = Q(x) where Q(x) is a polynomial
Inductive Step
To prove that s(m+1) is true
⇒ $\frac{^{6(m+1)+2} − ^{6(m+1)+1} + 1}{x^2 - x + 1}$ = T(x) where T(x) is a polynomial(x) is a polynomial
⇒ $\frac{^{6m+8} − ^{6m+7} + 1}{x^2 - x + 1}$ = T(x)
⇒ $\frac{x^6(^{6m+2} − ^{6m+1}) + 1}{x^2 - x + 1}$ = T(x)
However, I'm unsure of how to proceed from here. I would appreciate it if anyone could help me with this. Thanks!
|
Define the polynomial
$$P_n(x):=x^{6n+2}−x^{6n+1} + 1$$
Observe that, $x=-1$ is not a root of $x^2-x+1=0$, then multiplying both sides of the equation by $(x+1)$, yields:
$$(x+1)(x^2-x+1)=0$$
This implies that, $x^3= -1,\;x≠-1$.
Therefore, making $x^3\equiv -1$, by $\mod x^2-x+1$, we have:
$$
\begin{align}P_n(x)&\equiv x^2\cdot \left(x^3\right)^{2n}-x\cdot \left(x^3\right)^{2n}+1\\
&\equiv x^2-x+1.\end{align}
$$
This completes the proof.
More explicit explanation:
We can rewrite the polynomial $P_n(x)$ as follows:
$$P_n(x):=(x^2-x+1)Q(x)+R(x)$$
where, $Q(x)$ and $R(x)=ax+b$ are polynomials with some real coefficients.
Let $z\in\mathbb C\setminus \mathbb R$ be a root of $x^2-x+1=0$. Since $z≠-1$, multiplication both sides by $(z+1)$ yields $z^3+1=0$. This implies that, $z^3=-1$ and $z\not\in \mathbb R.$
Putting $x=z$ in the original polynomial indentity, we get:
$$
\begin{align}&P_n(z):=\frac {z^3+1}{z+1}Q(z)+R(z)\\
\implies &P_n(z)=0+R(z)\\
\implies &P_n(z)=R(z)\end{align}
$$
Then, using $z^3=-1$, we obtain:
$$
\begin{align}P_n(z):&=z^2\cdot \left(z^3\right)^{2n}-z\cdot \left(z^3\right)^{2n}+1\\
&=z^2-z+1\\
&=0\end{align}
$$
This leads to:
$$R(z)=az+b=0.$$
This means, $a=b=0$. Therefore, $R(x)\equiv 0$.
Because, if $a=0$, then $b=0$. Thus $R(x)\equiv 0$.
Otherwise, if $a≠0$, then $z=-\frac ba \in \mathbb R$ which gives a contradiction.
|
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|
Equation has exactly one real solution I am trying to show that the equation
$$
4x^2y^4+12x^2y^2+4x^2+4xy^2+4x+1=0
$$
has exactly one real solution $x,y$ and to determine it.
The first observation is: If $y=0$, we are left with
$$
4x^2+4x+1=0
$$
which is solved by $x=-1/2$.
Thus a real solution is given by
$$
x=-\frac{1}{2}, y=0
$$
and it remains to show that this is the only real solution.
This is where I am stuck...
|
I do not know if it helps, but you can rearrange the proposed equation as follows:
\begin{align*}
4x^{2}y^{4} + 12x^{2}y^{2} + 4x^{2} + 4xy^{2} + 4x + 1 = 0 & \Longleftrightarrow (4x^{2}y^{4} + 4xy^{2} + 1) + (12x^{2}y^{2} + 4x^{2} + 4x) = 0\\\\
& \Longleftrightarrow (2xy^{2} + 1)^{2} + 12x^{2}y^{2} + (4x^{2} + 4x + 1) = 1\\\\
& \Longleftrightarrow (2xy^{2} + 1)^{2} + 12x^{2}y^{2} + (2x + 1)^{2} = 1
\end{align*}
|
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|
Cyclic inequality. $\sum_{cyc} \frac{x+2y}{\sqrt{z(x+2y+3z)}}\geq \frac{3\sqrt{6}}{2}$ Let $x,y,z>0$. Show that
$$\sum_{cyc} \frac{x+2y}{\sqrt{z(x+2y+3z)}}\geq \frac{3\sqrt{6}}{2}$$
Equality case is for $x=y=z$.
A hint I have is that I have to amplify with something convenient, and then apply some sort of mean inequality inequality. Also, how would one go about finding with what to amplify?
I've tried amplifying the given fraction with a lot of random stuff, but no succes in finding something where I can apply the AM $>$ GM inequality.
|
By AM-GM and Cauchy-Schwarz we obtain:
$$\sum_{cyc}\frac{x+2y}{\sqrt{z(x+2y+3z}}=\sum_{cyc}\frac{2\sqrt{6}(x+2y)}{2\sqrt{6z(x+2y+3z})}\geq2\sqrt6\sum_{cyc}\frac{x+2y}{x+2y+9z}=$$
$$=2\sqrt6\sum_{cyc}\frac{(x+2y)^2}{(x+2y)^2+9(x+2y)z}\geq2\sqrt6\cdot\frac{\left(\sum\limits_{cyc}(x+2y)\right)^2}{\sum\limits_{cyc}((x+2y)^2+9z(x+2y))}=$$
$$=\frac{18\sqrt6(x+y+z)^2}{\sum\limits_{cyc}(5x^2+31xy)}\geq\frac{3\sqrt6}{2},$$
where the last inequality it's just $$\sum_{cyc}(x-y)^2\geq0.$$
|
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|
The parabola with the equation $y=-x^2+4x+8$ is shifted so that it passes through the points (1,1) and (3,5). Find the equation of the new parabola. Given the points $(1,1)$ and $(3,5)$, the vertex form would be:
$1=(1-h)^2+k$ for $(1,1)$ and
$5=(3-h)^2+k$ for $(3,5)$.
With a system of equations, I obtain that $h = 1$ and $k = 1$, so the new formula of the shifted parabola would be: $y=(x-1)^2+1$
What I don't understand is the answer of my exercise sheet, which says that the new equation would be $y= -x^2+6x-4$
Can someone help me understand this?
Thanks
|
Here's a slightly different approach.
Assuming that a "shift" strictly means a translation, then the vector between points $A = (1, 1)$ and $B = (3, 5)$, which is $\begin{pmatrix} 2 \\ 4 \end{pmatrix}$ is preserved. Thus the following system of equations holds:
$$\begin{cases} b = -a^2 + 4a + 8 \\ b + 4 = -(a + 2)^2 + 4(a + 2) + 8 \end{cases} \tag{2}$$
and this gives $4 = -(a + 2)^2 - (-a^2) + 4(a + 2) - 4a$ or $4 = -4a - 4 + 8 \implies a = 0$, and $b = 8$.
Hence the point $A$ corresponds to $(0, 8)$ on the original quadratic, and so the new quadratic is a translation of $1$ unit to the right and $7$ units down.
Therefore the equation of the new parabola is:
$$y = -(x - 1)^2 + 4(x - 1) + 8 - 7 = -(x^2 - 2x + 1) + 4x - 4 + 1$$
$$ = \boxed{-x^2+ 6x - 4}.$$
|
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|
Evaluate $\int\frac{1}{x-\sqrt{1-x^2}}dx$ The integral $I$ in question is defined as follows
$$
I \equiv \int\frac{1}{x-\sqrt{1-x^2}}dx
$$
To solve this, I tried the trig substitution $x = \sin\theta$, with $dx = \cos\theta d\theta$, and rewrote the integral as follows
$$
\int\frac{\cos\theta}{\sin\theta-\sqrt{1-\sin^2\theta}}d\theta
$$
I used to identity $1 - \sin^2\theta = \cos^2\theta$ and simplified the denominator as follows
$$
\int\frac{\cos\theta}{\sin\theta-\cos\theta}d\theta
$$
I then rewrote $\cos\theta$ as $\frac{\sin\theta + \cos\theta}{2} - \frac{\sin\theta - \cos\theta}{2}$ and rewrote the integrand as follows
$$
\int\frac{1}{2}\frac{\sin\theta+\cos\theta}{\sin\theta - \cos\theta} - \frac{1}{2}d\theta
$$
I then split the integral as follows
$$
\frac{1}{2}\int\frac{\sin\theta+\cos\theta}{\sin\theta-\cos\theta}d\theta - \frac{1}{2}\int1d\theta
$$
For the first integral, I substituted $\phi = \sin\theta-\cos\theta$, with $d\theta = \frac{1}{\sin\theta+\cos\theta}d\phi$
We can then rewrite our integral as
$$
\frac{1}{2}\int\frac{1}{\phi}d\phi
$$
This is trivial and after undoing the substitutions we have a result of
$$
\frac{\ln({x - \cos(\arcsin(x))})}{2}
$$
The second integral is also trivial and just evaluates to $\frac{x}{2}$
Combining everything together gives us a final simplified answer of
$$
I = \frac{\ln({x - \sqrt{1-x^2}})-x}{2} + C
$$
However, both IntegralCalculator and WolframAlpha give very different answers, so if someone could tell me where I made a mistake or another approach entirely that would be greatly appreciated.
|
You could also proceed in the following way.
By letting $\;t=x-\sqrt{1-x^2}\;,\;$ we get that
$x=\dfrac12\left(t\pm\sqrt{2-t^2}\right)\;,\quad\mathrm dx=\dfrac12\left(\!\!1\mp\dfrac t{\sqrt{2-t^2}}\!\!\right)\mathrm dt\;\;.$
$\displaystyle\int\frac1{x-\sqrt{1-x^2}}\,\mathrm dx=\frac12\!\int\frac1t\left(\!\!1\mp\frac t{\sqrt{2-t^2}}\!\!\right)\mathrm dt=$
$\displaystyle\quad=\frac12\!\int\frac1t\,\mathrm dt\;\mp\;\frac12\!\int\frac{\mathrm dt}{\sqrt{2-t^2}}\;\;.$
Moreover ,
$\mp\dfrac{\mathrm dt}{\sqrt{2-t^2}}=-\dfrac{\mathrm dt}{2x-t}=-\dfrac{t’(x)\,\mathrm dx}{x+\sqrt{1-x^2}}=$
$\quad=-\dfrac{1+\frac x{\sqrt{1-x^2}}}{\sqrt{1-x^2}\left(\!1+\frac x{\sqrt{1-x^2}}\!\right)}\,\mathrm dx=-\dfrac{\mathrm dx}{\sqrt{1-x^2}}\;\;.$
Hence ,
$\displaystyle\int\frac1{x-\sqrt{1-x^2}}\,\mathrm dx=\frac12\!\int\frac1t\,\mathrm dt\;\mp\;\frac12\!\int\frac{\mathrm dt}{\sqrt{2-t^2}}=$
$\displaystyle\quad=\frac12\ln|t|-\frac12\!\int\!\dfrac{\mathrm dx}{\sqrt{1-x^2}}=$
$\quad=\dfrac12\ln\bigg|x-\sqrt{1-x^2}\bigg|-\dfrac12\arcsin x+C\;\;.$
|
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|
A closed form expression of $\sum_{n \ge 0} \biggl( \sum_{k=1}^n \frac{1}{k} \biggr)z^n$
I am working on the following exercise: Use the identity $\frac{1}{1-z} = \sum_{n \ge 0} z^n$ and elementary operations on power series (addition, multiplication, integration, differentiation) to find closed form expressions of the following power series:
*
*$\sum_{n \ge 0} n^2z^n$
*$\sum_{n \ge 0} \frac{n}{n+1}z^n$
*$\sum_{n \ge 0} \biggl( \sum_{k=1}^n \frac{1}{k} \biggr)z^n.$
I have solved the first two points in the following way:
*
*Using differentiation on the identity $\frac{1}{1-z} = \sum_{n \ge 0} z^n$ we obtain
$$\frac{1}{(1-z)^2} = \sum_{n \ge 0} nz^{n-1}$$
and multiplying the above equality with $z$ yields $\frac{z}{(1-z)^2} = \sum_{n \ge 0} nz^{n}$ and again using differentiation and multplication with $z$ on this term yields $\frac{z^2+z}{(1-z)^3} = \sum_{n \ge 0} n^2z^n$ as desired.
*Integrating the term $\frac{z}{(1-z)^2} = \sum_{n \ge 0} nz^{n}$ from before yields
$$\ln(\vert 1-z \rvert)+\frac{1}{1-z} = \sum_{n \ge 0} \frac{n}{n+1}z^{n}$$
I realise that the third part is probably somehow related to the exercises before, but I do not see how. Could you please help me?
|
Expanding the sum: $$\sum_{n\ge 0}H(n)z^n=z+\left(1+\dfrac{1}{2}\right)z^2+\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)z^3+...=\frac{z}{1-z}+\frac{1}{2}z^2+\left(\dfrac{1}{2}+\dfrac{1}{3}\right)z^3+...=\frac{z}{1-z}+\frac{z^2}{2(1-z)}+...$$Where $H(n)$ is the $n$th harmonic number. Using induction, you could prove that: $$\sum_{n\ge 0}H(n)z^n=\sum_{n\ge 0}\frac{z^n}{n(1-z)}$$ Can you take it from here? (The series for $\ln (1-z)$ is the sum of $z^n/n$ from $n$ to $\infty$).
Note: I am assuming that the sums converge.
|
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|
Plot of function $(x^\frac{1}{2} + y^\frac{1}{2})^2 = 5$ Can someone explain me how can we make a plot of function:
$$ (x^\frac{1}{2} + y^\frac{1}{2})^2 = 5 $$
I tried to find a first derivative and a second derivative to understand something about this plot of function but it was useless as I understand. What should I do to plot this function by myself not by WolframAlpha. Thank you in advance.
|
Since the equation implies $\sqrt{x}+\sqrt{y}\geq 0$, therefore it is equivalent to $\sqrt{x}+\sqrt{y}=\sqrt{5}$. Therefore because $y\geq 0$, $y=(\sqrt{5}-\sqrt{x})^2=x+5-2 \sqrt{5}\cdot \sqrt{x}$. And we know from the condition of $\sqrt{x}, \sqrt{y}\geq 0$ that $0\leq x, y\leq 5$. Therefore, finally, the original statement is the same as $y=x+5-2\sqrt{5} \sqrt{x} (x\in[0, 5])$.
To do a simple graphical analysis (though may not be necessary), We consider the $1/4$ arc of $(x-5)^2+(y-5)^2=25 (0\leq x, y\leq 5)$, which we can rewrite as
\begin{align*}
y&=5-\sqrt{10x-x^2}\\
&=x+5-\sqrt{10x+2x\sqrt{10x-x^2}}\\
&\geq x+5-\sqrt{10x+2x\cdot \sqrt{25}}\\
&=x+5-2\sqrt{5}\sqrt{x}
\end{align*}
, thus indicating that the whole graph lies 'below' the circle.
|
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|
Show that $b^2+c^2-a^2\leq bc$. Let $a,b,c>0$ such that $b<\sqrt{ac}$, $c<\frac{2ab}{a+b}$. Show that $b^2+c^2-a^2\leq bc$.
I tried to construct a triangle with $a,b,c$ and to apply The cosine rule, but I am not sure that it's possible to construct it and also I have no idea how to prove that an angle it's greater than $60^{\circ}$.
|
Another solution.
First, we find $c<\frac{2ab}{a+b}\leq \frac{2\left(\frac{a+b}{2}\right)^2}{a+b}=\frac{a+b}{2}$ by applying AM-GM to $ab$. We also find $b<\sqrt{ac}\leq \frac{a+c}{2}$.
Now if $(b+c)\leq a$, this implies $(b+c)^2\leq a^2 \Rightarrow b^2+c^2-a^2\leq -2bc<bc$, and we are done.
If not, then $b+c>a$ we'll have a triangle $\triangle ABC$. By the law of cosine,
$$a^2 = b^2+c^2-2bc\cos{A} \Rightarrow b^2+c^2-a^2=2bc\cos{A}$$ we only need to consider cases where $\angle A<60^{\circ}$ which would make our statement false.
Now $\angle A<60^{\circ}$ implies $a$ can't be the longest side, and the condition $b<\sqrt{ac}$ implies $b$ can't be the longest side. Therefore $c$ must be the longest side. However $c<\frac{a+b}{2}$, so that contradicts the fact we have a triangle.
|
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|
Inequality, $\frac{u+v}{2}-\frac{1}{4}$ versus $uv$
On the unit square, where is $f(u,v) = uv$ greater than $g(u,v) = \frac{u+v}{2}-\frac{1}{4}$?
I know that
*
*$f \left( \frac{1}{2} , \frac{1}{2}\right ) = \frac{1}{4} = g \left( \frac{1}{2} , \frac{1}{2}\right ) $
From playing with numbers,
*
*on $\left(0, \frac{1}{2}\right) \times \left(0, \frac{1}{2}\right)$ and $\left(\frac{1}{2}, 1\right) \times \left(\frac{1}{2}, 1\right)$, $f \geq g$
*and $g\leq f$ elsewhere.
How do I show this mathematically?
On $(0, \frac{1}{2}) \times (0, \frac{1}{2})$, is it a valid argument to notice that the partial derivative of $f$ are less than $\frac{1}{2}$ while the partial derivatives of $g$ are both equal to $\frac{1}{2}$, and conclude that $f$ decreases "slower" and is thus larger valued that $g$ (since they are equal at $\left( \frac{1}{2} , \frac{1}{2} \right)$)? (and similarly, on $\left(\frac{1}{2}, 1\right) \times \left(\frac{1}{2}, 1\right)$, $f$ increases "faster" and is thus greater than $g$)?
I have no idea what to do on $\left(\frac{1}{2}, 1\right) \times \left(0, \frac{1}{2}\right)$ and $\left(0, \frac{1}{2}\right) \times \left(\frac{1}{2}, 1\right) $.
|
$$\begin{align}
f(u, v) - g(u, v) &= uv - \left(\frac{u+v}{2} - \frac14\right) \\
&= uv - \frac u2 - \frac v2 + \frac 14 \\
&= \left(u - \frac12\right)\left(v - \frac12\right)
\end{align}$$
|
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|
How do we prove without calculus that $\forall \ x,y \ge 0$, we have $\ 1+x^3+y^3\ge x+x^2y+y^2$ I've been trying to prove an inequality I was given by a friend, but so far my only progress has been calculus bashing:
$$LHS \ge RHS\iff1+x^3+y^3 - x-x^2y-y^2 \ge0$$
Letting $f(x,y) = 1+x^3+y^3 - x-x^2y-y^2$, we want $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0$ for the minimum.
Hence $$3x^2-1-2xy=0 \ \ \ \ \ \ \ \ ... (1)$$ and $$3y^2-x^2-2y=0 \ \ \ \ \ \ \ \ ... (2)$$
which gives one solution$,(x,y)=(1,1),$ in the first quadrant:
The trouble is that proving that $(x,y)=(1,1)$ is the only positive solution to the system of equations is quite cumbersome.
$$\ \ $$
Is there any way we can prove this without calculus, maybe using AM-GM or Cauchy-Schwarz inequalities?
I used AM-GM to obtain $$1+x^3+y^3 \ge 3\sqrt[3]{1^3x^3y^3}=3xy,$$
but I'm not sure how this helps me proceed. Many thanks in advance.
|
Note that
$$
\begin{align}
\frac{1}{3}(1 + 1+x^3) &\ge x \\
\frac{1}{3}(x^3 + x^3 + y^3) &\ge x^2y \\
\frac{1}{3}(1 + y^3 + y^3) &\ge y^2
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Evaluation of $~\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta$ $$
I:=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
$$
My tries
$$\begin{align}
s&:=\sin\theta\\
c&:=\cos\theta\\
I&=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{c^2-s^2\over s^4+c^4}\mathrm d\theta\\
&=\int_{0}^{2\pi}{(1-s^2)-s^2\over s^4+(c^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(1-s^2)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+(s^2-1)^2}\mathrm d\theta\\
&=\int_{0}^{2\pi}{1-2s^2\over s^4+s^4-2s^2+1}\mathrm d\theta\\
&=\int_{0}^{2\pi}\underbrace{\color{red}{\left({1-2s^2\over 2s^4-2s^2+1}\right)}}_{\text{I got stuck here}}\mathrm d\theta\\
\end{align}$$
I need your help.
|
$$
\begin{aligned}
\int_0^{2 \pi} \frac{\cos ^2 \theta-\sin ^2 \theta}{\sin ^4 \theta+\cos ^4 \theta} d \theta
= & \int_0^{2 \pi} \frac{\cos 2 \theta}{\left(\sin ^2 \theta+\cos ^2 \theta\right)^2-4 \sin ^2 \theta \cos ^2 \theta} d \theta \\
= & \int_0^{2 \pi} \frac{\cos 2 \theta}{1-\sin ^2 2 \theta} d \theta \\
= & \frac{1}{2} \int_0^{2 \pi} \frac{d(\sin 2 \theta)}{1-\sin ^2 2 \theta} \\
= & \frac{1}{2}\left[\ln \left|\frac{1+\sin 2 \theta}{1-\sin 2 \theta} \right|\right]_0^{2 \pi}\\
= & 0
\end{aligned}
$$
|
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|
Finding the angles of a right triangle if $\frac{\text{area of triangle}}{\text{area of incircle}}=\frac{2\sqrt{3}+3}{\pi}$
Let $ABC$ be a right triangle with $\measuredangle ACB=90^\circ$. If $k(O;r)$ is the incircle of the triangle and
$$\dfrac{S_{ABC}}{S_k}=\dfrac{2\sqrt3+3}{\pi}$$ find the angles of the triangle.
($S$ denotes the areas of the triangle and the circle.)
The area of the triangle is $S=pr$, where $p$ is the semiperemeter, the area of the circle is $S_k=\pi r^2,$ so $$\dfrac{pr}{\pi r^2}=\dfrac{2\sqrt3+3}{\pi}\\\dfrac{p}{r}=2\sqrt3+3\\\dfrac{\frac{a+b+c}{2}}{\frac{a+b-c}{2}}=2\sqrt3+3\\\dfrac{\frac{a}{c}+\frac{b}{c}+1}{\frac{a}{c}+\frac{b}{c}-1}=2\sqrt3+3\\\ \dfrac{\sin\alpha+\cos\alpha+1}{\sin\alpha+\cos\alpha-1}=2\sqrt3+3$$ I don't know if my approach is reasonable and don't see what can be done from here.
|
Rewrite $$\dfrac{\sin\alpha+\cos\alpha+1}{\sin\alpha+\cos\alpha-1}$$
as
$$\dfrac{2}{\sin\alpha+\cos\alpha-1}+1.$$
Then
$$\dfrac{2}{\sin\alpha+\cos\alpha-1}+1=2\sqrt3+3\implies \dfrac{1}{\sin\alpha+\cos\alpha-1}=\sqrt3+1.$$
Rearranging, we get
$$\sin\alpha + \cos\alpha =\frac{1}{\sqrt{3}+1}+1=\frac{\sqrt{3}-1}{2}+1=\frac{\sqrt{3}}{2}+\frac{1}{2}.$$
We can see that $\alpha = 30^{\circ}, 60^{\circ}$ are solutions.
Using the identity $\sin\alpha + \cos\alpha = \sqrt{2}\sin(\alpha + 45^{\circ})$, we can see that these are also the only solutions.
|
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|
Contest Math Question: simplifying logarithm expression further I am working on AoPS Vol. 2 exercises in Chapter 1 and attempting to solve the below problem:
Given that $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$ (MA$\Theta$ 1991).
My approach is to isolate $n$ and then cube it. Observe:
\begin{align*}
\frac{\log 40\sqrt{3}}{\log 4n} = \frac{\log 45}{\log 3n} \\
\log 40\sqrt{3}\log 3n = \log 45\log 4n\\
\log 40\sqrt{3} \cdot (\log 3 + \log n) = \log 45 \cdot (\log 4 + \log n)\\
\log n \cdot (\log 40\sqrt{3} - \log 45) = \log 45\log 4 - \log 40\sqrt{3}\log 3
\end{align*}
Dividing through and putting the coefficients as powers, we have:
\begin{align*}
\log n &= \frac{\log 45^{\log 4} - \log \left[(40\sqrt{3})^{\log 3}\right]}{\log\left(\frac{40\sqrt{3}}{45}\right)}
=\frac{\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right) }{\log\left(\frac{40\sqrt{3}}{45}\right)} \\
&=\log \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}
which shows that
\begin{align*}
n^3 = \left(\frac{45^{\log 4}}{(40\sqrt{3})^{\log 3}}\right)^{3\cdot\log\left(\frac{40\sqrt{3}}{45}\right)^{-1}}
\end{align*}
Somehow it feels like this answer may be simplified further. Are the steps shown so far correct and can the answer be expressed in a better way?
|
In this answer, we provide another method that uses the definition and basic properties of logarithm.
The original equation states :
$$
\begin{align}&\log_{4n}{40}\sqrt{3}=\log_{3n}{45}\\
\implies &{40}\sqrt 3=\left(4n\right)^{\log_{3n}45}\\
&\thinspace\thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace=\left(3n\cdot \frac 43\right)^{\log_{3n}{45}}\\
&\thinspace\thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace\thinspace\thinspace \thinspace=45\cdot \left(\frac 43\right)^{\log_{3n}{45}}\end{align}
$$
This leads to the following :
$$
\begin{align}\left(\frac {2}{\sqrt 3}\right)^{2\log_{3n}{45}}&=\frac {40\sqrt 3}{45}=\left(\frac {2}{\sqrt 3}\right)^3\end{align}
$$
Finally, again using the definition of logarithm, we get the desired result:
$$
\begin{align}&\log_{3n}{45^2}=3\\
\implies &n^3=\frac {45^2}{3^3}=75\thinspace .\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$.
My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$
Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.
|
It boils down to the comparison of some powers of $2$ and $3$, if I didn't do any mistake, in the following way: $9^{\sqrt{2}}<\sqrt{2}^9$ if $9<2^{\frac{9}{2\sqrt{2}}}$, by using the fact that $\sqrt{2}<1.415$, if $9<2^{\frac{9}{2\times 1.415}}$ if $9<2^{3.18}$ if $3^{100}<2^{159}$ which is true since:
*
*$2^{159}>\frac{1}{2}(1,02\times 10^{3})^{16}>\frac{1.32\times10^{48}}{2}=6,6\times10^{47}$.
*$3^{100}<(6\times 10^{4})^{10}=3^{10}\times 2^{10}\times 10^{40}<6\times10^{4}\times 1025\times 10^{40}=6,15\times 10^{47}$
|
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|
residue $\frac{z}{(z-1)(z+1)^2}$ at $z = -1$ My attempt: $$f(z) = \frac{z}{(z-1)(z+1)^2} = \frac{1}{(z+1)^2}\cdot -z\frac{1}{1-z}$$ So at $D(-1;1)$ we have $$-z\frac{1}{1-z} = \sum_{k=0}^\infty-(z+1)^{k+1}$$ and this gives $$f(z) = \sum_{k=-1}^\infty-(z+1)^{k}.$$ So the residue at $z = -1$ is negative one but I feel like this is the wrong way of approaching this problem, because of the open disk I use. Also I think I proofed that $c_{-2} \neq 0$ because of a pole of order two here, so I'm not sure if this can be correct.
|
For each $z\in\Bbb C$ such that $|z+1|<2$, you have\begin{align}\frac z{z-1}&=1+\frac1{z-1}\\&=1-\frac1{2-(z+1)}\\&=1-\frac12\cdot\frac1{1-\frac{z+1}2}\\&=1-\sum_{n=0}^\infty\frac{(z+1)^n}{2^{n+1}}\\&=\frac12-\sum_{n=1}^\infty\frac{(z+1)^n}{2^{n+1}}\end{align}and therefore, if $z\ne-1$,\begin{align}\frac z{(z-1)(z+1)^2}&=\frac1{2(z+1)^2}-\sum_{n=1}^\infty\frac{(z+1)^{n-2}}{2^{n+1}}\\&=\frac1{2(z+1)^2}-\sum_{n=-1}^\infty\frac{(z+1)^n}{2^{n+3}}.\end{align}In particular,$$\operatorname{res}_{z=-1}\frac z{(z-1)(z+1)^2}=-\frac14.$$
|
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|
Prove $\displaystyle\frac{H(x^2)}{H(x)}$ increases. For $x\in[0,1]$, let $f(x):=-x\ln x$ and the two-sample entropy function $H(x)=f(x)+f(1-x)$. Prove $h(x):=\displaystyle\frac{H(x^2)}{H(x)}$ increases.
Here is my proof which is a bit cumbersome. I am seeking a much more elegant approach.
The numerator of the derivative of the sought fraction is
\begin{align}
&H(x)^2\frac{dh(x)}{dx} \\
=&\frac d{dx}H(x^2)H(x)-H(x^2)\frac d{dx}H(x) \\
=& 2x\ln\frac{x^2}{1-x^2}\,\big(x\ln x+(1-x)\ln(1-x)\big)-\big(x^2\ln x^2+(1-x^2)\ln(1-x^2)\big)\ln\frac x{1-x} \\
=& 2x^2\ln^2 x+2x(2-x)\ln x\ln(1-x)-(x^2+1)\ln x\ln(1-x^2)+(1-x)^2\ln(1-x)\ln(1-x^2). \tag1\label1
\end{align}
All four terms above except the third are positive. I combine the second and the third term together and divide it by $-\ln x$ which is positive, and get
\begin{align}
g(x):&=-2x(2-x)\ln(1-x)+(x^2+1)\ln(1-x^2) \\
&=(3x-1)(x-1)\ln(1-x)+(x^2+1)\ln(1+x) \tag2\label2 \\
&= \int_0^x \Big(g''(a)-\int_t^a g'''(s)ds\Big)(x-t)dt
\end{align}
for some $a\in[0,x]$. So we only need to show $g(x)>0, \forall x\in\big(0,\frac13\big]$.
$$\frac{d^3g(x)}{dx^3}= \frac{4x(2x^3 +3x^2-2x-7)}{(1-x)^2(1+x)^3}.$$
Let $p(x):=2x^3+3x^2-2x-7$. It can be shown that $p(x)\le p(1)=-4, \forall x\in[0,1]$. We can take $a=\frac13$ since we can show, with a bit of work, $f''(\frac13)>0$.
(to be continued)
|
Let
$$u = \ln x, \quad v = \ln(1-x), \quad w = \ln(1 + x).$$
We have
\begin{align*}
&[H(x)]^2h'(x)\\[6pt]
=\,& [(1-x)^2v - (1+x^2)u]w + 2u^2x^2 - (1-x)(1-3x)uv + (1-x)^2v^2\\[6pt]
\ge\,& [(1-x)^2v - (1+x^2)u]\cdot (x - x^2/2)\\[6pt]
&\qquad + 2u^2x^2 - (1-x)(1-3x)uv + (1-x)^2v^2 \tag{1}\\[6pt]
=\,& 2u^2x^2 - [x(1-x/2)(1+x^2) + (1-3x)(1-x)v]u\\[6pt]
&\qquad + v(1-x)^2(v + x - x^2/2)\\[6pt]
\ge\,& 0. \tag{2}
\end{align*}
Explanations:
(1): $\ln(1+x) \ge x - x^2/2$ for all $x \ge 0$;
Using $(1-x)\ln(1-x) > - 1$, we have
\begin{align*}
(1-x)^2v - (1+x^2)u &= (1-x)^2\ln(1-x) - (1+x^2)\ln x\\
&\ge -(1-x) - \ln x\\
&\ge 0.
\end{align*}
(2):
$\ln(1-x) + x - x^2/2\le 0$ for all $x\in [0, 1)$;
If $1/3 \le x < 1$, clearly
$$x(1-x/2)(1+x^2) + (1-3x)(1-x)\ln(1-x) \ge 0$$
and if $0 \le x < 1/3$, using $(1-x)\ln(1-x) \ge -x + x^2/2$ for all $0 \le x < 1$, we have
\begin{align*}
&x(1-x/2)(1+x^2) + (1-3x)(1-x)\ln(1-x)\\
\ge\,& x(1-x/2)(1+x^2) + (1-3x)(-x + x^2/2)\\
=\,& x^2(3+x)(1-x/2)\\
\ge\,& 0.
\end{align*}
We are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Limit of $\lim_{(x,y)\to (0,0)}\frac{3x^3y^2+xy^4}{(x^2+y^2)^2}$ I've run across a particular limit in my multivariable calculus class, that being:
\begin{equation}
\lim_{(x,y)\to (0,0)}\frac{3x^3y^2+xy^4}{(x^2+y^2)^2}.
\end{equation}
I'm having quite a bit of trouble finding this limit using the squeeze theorem. Using polar coordinates (and assuming i'm not completely butchering everything), I believe we have that this particular limit evaluates to 0; however, using coordinate changes with limits feels a little uncomfortable and I would like to see if it can be done without it.
|
Using polar coordinates is perfectly alright and does not require any special knowledge. The formal argument can be presented as follows:
Let $\varepsilon > 0$. Take $\delta = \varepsilon / 4$ and assume $|(x,y)| = (x^2+y^2)^{1/2} < \delta$. We can always find $r, \theta$ such that $x = r\cos\theta $ and $y = r\sin\theta$. Then $r^2 = x^2 + y^2$ and our choice of $(x,y)$ implies $r < \delta$.
The numerator satisfies,
\begin{align}
| 3 r^5 \cos^3 \theta \sin^2 \theta + r^5 \cos\theta \sin^4\theta| &= r^5\lvert 3cos^3\theta \sin^2\theta + \cos\theta\sin^4\theta\rvert \\
&\leqslant 4r^5 \\
&<\varepsilon r^4
\end{align}
and the denominator is simply $r^4$, so that,
\begin{align}
\left\lvert \frac{3x^2y^3+xy^4}{(x^2+y^2)^2}\right\rvert < \varepsilon.
\end{align}
Since $\varepsilon$ is arbitrary, the required limit follows.
|
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|
How to solve this functional equation and find the integral of $f(x)$? How to solve this equation and find the integral from $0$ to $1$ of $f(x)$ if
$$f(1-x)=6x^2f(x^3)-\dfrac{6}{\sqrt{3x+1}}?$$
I've tried take integration for two sides of the equation after switching $f$ to one side and $x$ to one side. After that I used integration by parts and got the answer is $4$.
$$A=6\int_0^1{x^2 f(x^3) dx} = \int_0^1{2 f(u) du}$$
$$B=\int_0^1{f(1-x) dx} = \int_0^1{-f(u)du}$$
$$C=\int_0^1{\dfrac{6}{\sqrt{3x+1}}} = 4\sqrt{3x+1}|_0^1=4$$
Then because we have A+B=C
We will get the answer is 4
Did I do it right?
Please help me.
|
Substituting $t$ = $x^3$, we get
$\int{6 x^2 f(x) dx} = \int{2 f(t) dt}$ --- (1)
Using $\int_{a}^{b}f(x)dx = \int_{a}^{b}f(b+a-x)dx\quad$, we get
$\int_{0}^{1}f(x)dx = \int_{0}^{1}f(1-x)dx\quad$ --- (2)
Using (1),(2) and integrating the equation in the question from 0 to 1, we get
$\int_{0}^{1}f(x)dx = 2 \int_{0}^{1}f(x)dx - \int_{0}^{1}\frac{6}{\sqrt{3x+1}} dx$
$\implies \int_{0}^{1}f(x)dx = \int_{0}^{1}\frac{6}{\sqrt{3x+1}} dx = 6 \int_{0}^{1}\frac{1}{\sqrt{3x+1}} dx = 2 \int_{1}^{4}\frac{1}{\sqrt{y}} dy = 4$
|
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"language": "en",
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|
$\lim\limits_{x\to \infty}\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)} $ Compute the limit:
$$\lim_{x\to \infty}\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)} $$
I worked out this to be $\frac{-3}{8}$. I believe this is correct. I used the sandwich theorem but my issue is that the denominator is sometimes negative which isn't helping me. I tried to use absolute values to get rid of this problem:
$$\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)} \leq \frac{3x^3+x^2+1}{5x-8x^3-\frac{\pi}{2}},$$
which I then could show is $\frac{-3}{8}$ using algebra of limits, but I am certain this is wrong.
My other way of thinking is:$$\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)} = \frac{3+\frac{1}{x}+\frac{\sin(e^x)}{x^3}}{\frac{5}{x^2}-8+\frac{\arctan(\log x)}{x^3}}
$$
Note that $-1 \leq \sin(e^x) \leq 1$, so for all $x>0$, we have $\frac{-1}{x^3} \leq \frac{\sin(e^x)}{x^3} \leq \frac{1}{x^3}$, so the limit is $0$.
Note that $\frac{-\pi}{2x^3} \leq \frac{\arctan(\log x)}{x^3} \leq \frac{\pi}{2x^3}$, so the limit is $0$.
Now by taking limits:
$$\frac{3+\frac{1}{x}+\frac{\sin(e^x)}{x^3}}{\frac{5}{x^2}-8+\frac{\arctan(\log x)}{x^3}} =\frac{3+0+0}{0-8+0} = -\frac{3}{8}.$$
I believe my second way of working is correct. Can someone please tell me if I am right ? Also, is there a way of doing with this with absolute values ?
|
Your method is correct, but you can find the limit instantly after reading the text reasoning in term of asymptotic behaviour.
$$\lim_{x\to \infty}\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)} $$
Note that both $\sin e^x$ and $\tan^{-1}(\log x)$ are bounded, so when $x\to +\infty$ larger powers dominate:
$$\frac{3x^3+x^2+\sin(e^x)}{5x-8x^3+\arctan(\log x)}\sim\frac{3x^3}{-8x^3}\to -\frac{3}{8} $$
|
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|
Find the value of $\frac{a+b}{10}$
If $\sin x+\cos x+\tan x+\cot x+\sec x +\csc x=7$, then assume that $\sin(2x)=a-b\sqrt7$, where $a$ and $b$ are rational numbers. Then find the value of $\frac{a+b}{10}$.
How to solve these kind of problems. I can make substitutions and convert all of them to $\sin$ and then solve for it but it'll be very lengthy. Is there any other short and nicer method.
|
$\displaystyle \sin x+\cos x+\tan x+\cot x+\sec x+\text{cosec } x=7$
$\iff\displaystyle \sin x+\cos x+\frac{\sin x}{\cos x} +\frac{\cos x}{\sin x} +\frac{1}{\cos x} +\frac{1}{\sin x} =7$
$\displaystyle ( \sin x+\cos x)\left( 1+\frac{1}{\sin x\cos x}\right) +\frac{1}{\sin x \cos x} =7$
$\iff\displaystyle ( \sin x+\cos x)\left( 1+\frac{2}{\sin(2x)}\right) =7-\frac{2}{\sin( 2x)}$
Now square both the sides:
$\displaystyle ( 1+\sin( 2x))\left( 1+\frac{2}{\sin( 2x)}\right)^{2} =\left( 7-\frac{2}{\sin( 2x)}\right)^{2}$
Now this a cubic in $\sin(2x)$
You can simplify this to: $\displaystyle \sin( 2x)\left( \sin^{2}( 2x) -44\sin( 2x) +36\right)$
Now you can find $\sin(2x)$ using Quadratic formula. (assuming $\displaystyle \sin( 2x) \neq \ 0$)
So $\displaystyle \sin( 2x) =22\pm 8\sqrt{7}$ but we ignore the + result as its greater than 1.
So $ a+b = 30$.
|
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|
Composing function The function is $f: \Bbb{R}\rightarrow\Bbb{R}$ defined as $f(x)= 2/ (x -3).$
I need to find $(f o f)(1).$
I would like to ask which of the following answers are the right one for writing this function.
$( f o f) ( 1 ) = ( f ( f ( 1 ) ) )= ( f ( 2/ 1 - 3) = ( 2 / 1 - 3 - 2 ) = -1/2 $
or
$( f o f ) ( 1 ) = ( f ( f (1) ) )= f( 2/ 1 - 3 ) = 2 / ( 2/ 1 - 3 ) - 3 = 2 / ( - 2 ) - 3 = -4$
|
If $f(x) = \frac{2}{x} -3 $, then
\begin{align*}
(f\circ f)(x) = f(f(x)) = \frac{2}{f(x)}-3 = \frac{2}{\frac{2}{x}-3}-3
\end{align*}
And thus
\begin{align*}
(f\circ f)(1) = \frac{2}{\frac{2}{1}-3}-3 = \frac{2}{-1}-3 = -5
\end{align*}
|
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|
Solve the diophantine equation involving floor function ${\text{Solve for } \forall n \in \mathbb{N}, n \in (1,10^9) \text{ and primes } p \text{ that hold this equation:}}$
$${\sqrt{\lfloor \sqrt{ n^2 }\rfloor+\lfloor \sqrt{ n^2+1 }\rfloor+\lfloor \sqrt{ n^2+2 }\rfloor} = p}$$
Since ${ \forall k \in \mathbb{N}:}$
${0<(\sqrt{k^2+1}-\sqrt{k^2})^2=k^2+1+k^2-2k\sqrt{k^2+1} \lt 1+2k^2-2k\sqrt{k^2}=1}$
$${0<\sqrt{k^2+1}-\sqrt{k^2}<1}$$
$${ \lfloor \sqrt{k^2+1} \rfloor-\lfloor \sqrt{k^2} \rfloor \le 1}$$
$${ \lfloor \sqrt{k^2+1} \rfloor-\lfloor \sqrt{k^2} \rfloor = (0 \lor 1) }$$
${=> (\lfloor \sqrt{ n^2 }\rfloor, \lfloor \sqrt{ n^2+1 }\rfloor,\lfloor \sqrt{ n^2+2 }\rfloor)}$ can either be:
${(n,n,n),(n,n,n+1),(n,n+1,n+1) \text{ or } (n,n+1,n+2)}$
For case ${(n,n,n)}$:
$${p^2=3n => 3|p => p=3, n=3}$$
For case ${(n,n+1,n+2)}$:
$${p^2=3n+3 => 3|p => p=3, n=2}$$
Other two cases are harder to analyse because there numerous solutions since:
If ${p>=5}$, then: ${p=6k+1}$ or ${p=6k-1}$ for some natural number ${k}$
So, in case ${(n,n,n+1)}$: ${p^2=3n+1=3(2k)-1=6k-1}$ and rewriting as:
${(p+1)(p-1)=3n}$ does not make any restrictions except for ${n<=10^9}$.
Similarly for case ${(n,n+1,n+1)}$:
${p^2=3n+2=3(2k-1)+2=6k-1}$ and again, no restrictions whatsoever.
What is the strategy here, in these two harder cases: brute-force or something more clever?
|
You've made a good start, but there's a simpler way to show there are no solutions for most of your cases. First, your equation is
$$\sqrt{\lfloor\sqrt{n^2}\rfloor+\lfloor\sqrt{n^2+1}\rfloor+\lfloor\sqrt{n^2+2}\rfloor} = p \tag{1}\label{eq1A}$$
With the case of $(n,n,n)$, you've found only $p = 3$ is possible, with $n = 3$ in \eqref{eq1A} confirming this is a solution. Since $\lfloor x \rfloor \le x$, note all of your other cases require at least that
$$\begin{equation}\begin{aligned}
n + 1 & \le \sqrt{n^2 + 2} \\
n^2 + 2n + 1 & \le n^2 + 2 \\
2n & \le 1 \\
n & \le \frac{1}{2}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
However, since $n \in \mathbb{N}$ and $n \in (1,10^9)$, this is not possible. Thus, there are no additional solutions to \eqref{eq1A}.
|
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|
I came across this fun pattern, can anyone give me a proof for this? The Pattern
$\begin{align*}1(8)&=(3^2)-1\\
2(8)&= (3+1)^2 \\
3(8)&= (3+2)^2-1\\
4(8)&= (3+3)^2 - 4\\
5(8)&= (3+4)^2- 9\\
6(8)&= (3+5)^2-16\end{align*}$
Conjecture
I think the pattern is that the numbers appear in the following form:
$$8(n+1) = (3+n)^2 - (n-1)^2$$
Please correct me if I'm wrong, I'm not really that good at math.
|
If you are just learning algebra, this might not be obvious. But, the rule to keep in mind is that multiplication distributes over addition. On the left-hand side, you have $8(n+1)$ which after distribution equals $8n+8.$ Now we have to show that the right-hand side gives the same thing.
Let's work out the two terms one at a time.
$(n+3)^2 = (n+3)(n+3) = n(n+3) + 3(n+3) = n^2 + 3n + 3n+9 =
n^2+6n+9$
$(n-1)^2 = (n-1)(n-1) = n(n-1) + (-1)(n-3) = n^2 - n - n +1 = n^2-2n+1$
$(n+3)^2 - (n-1)^2 = (n^2+3n+9)-(n^2-2n+1) = n^2+6n+9 - n^2+2n-1 = 8n+8$
As required.
Hope this helps
|
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|
Can a "sum of two squares" solution be the average of two other solutions? If $n = a^2 + b^2$ and $n$ has more than two unique solutions for $a$ and $b$ (where unique means that $a$ and $b$ are unsigned and unordered), then is it possible, impossible, or unknown whether one solution can be the average of two other solutions? For example, given three solutions:
$n = j^2 + k^2$
$n = s^2 + t^2$
$n = x^2 + y^2$
Can $j^2$ and $k^2$ can be the average of the other two solutions? Is this possible:
$j^2 = \frac{s^2 + x^2}{2}$
$k^2 = \frac{t^2 + y^2}{2}$
Where might I learn more? I ask because I've been playing with the "3x3 magic square of squares" problem and this seems to be a core question. Thanks!
|
Observations towards a solution
*
*$ s^2 + x^2 = 2j^2$ has parametrized solutions for integers $a, b, k$ of
$$ s = k(a^2 + 2ab -b^2), x =k( -a^2 + 2ab + b^2), j = k(a^2 + b^2).$$
*
*This can be algebraically verified.
*In the event that the values are negative, just take the absolute value.
*Note: This is a complete classification, but we don't require this fact. See here for more details.
*Likewise, $ t = c^2 + 2cd - d^2, y = -c^2 + 2cd + d^2$.
*Now, since $s^2 + t^2 = x^2 + y^2$, expanding the expression and simplifying yields
$$ ab(a^2 - b^2) = cd(d^2 - c^2).$$
*This has solutions like $ a = 6, b = 1, c = 2, d = 5$.
*
*There are also other solution sets, like $ a = 7, b = 3, c = 7, d = 8$. (Obtained by setting $ a = c$ as a simplification.)
*I don't know how to get a complete classification of the solutions. These were mainly found by trial and error.
*This yields $n = 47^2 + 1^2 = 37^2 + 29^2 = 23^2 + 41^2, 47^2 + 23^2 = 2\times 37^2, 1^2 + 41^2 = 2 \times 29^2 $.
|
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|
Let $A $ and $B$ be real symmetric matrices of order $n$, satisfying $A^2B=ABA.$ Proof that $AB=BA.$ Let $A $ and $B$ be real symmetric matrices of order $n$, satisfying $A^2B=ABA.$ Proof that $AB=BA.$
I only find that $A^2B=BA^2=ABA.$ Then I don't know what to do next.Can you help me solve this puzzle?
|
If $A$ is invertible or $A=0$ then it is obvious. Assume now that $A\ne 0$ and $\det A=0$.
$A$ is symmetric and hence diagonalizabe by an orthnormal matrix $U$. Set $A=U^{-1}LU$, where$L$ is diagonal. Expressing $L$ and $\hat B=UBU^{-1}$ (which is also symmetric) in block form,
$$
L=\left(
\begin{array}{cc}D & 0 \\ 0 & 0\end{array}
\right),\qquad
UBU^{-1}=\left(
\begin{array}{cc}B_1 & B_2 \\ B_3 & B_4\end{array}
\right)=\hat B,
$$
where $D$ is diagonal and invertible, we have
$$
U^{-1}L^2\hat B U=U^{-1}L^2UB=A^2B=ABA=U^{-1}LUBU^{-1}LU=U^{-1}L\hat BLU
$$
or
$$
L^2\hat B=L\hat BL
$$
or
$$
\left(
\begin{array}{cc}D^2 & 0 \\ 0 & 0\end{array}
\right)\left(
\begin{array}{cc}B_1 & B_2 \\ B_3 & B_4\end{array}
\right)=\left(
\begin{array}{cc}D & 0 \\ 0 & 0\end{array}
\right)\left(
\begin{array}{cc}B_1 & B_2 \\ B_3 & B_4\end{array}
\right)\left(
\begin{array}{cc}D & 0 \\ 0 & 0\end{array}
\right)
$$
or
$$
\left(
\begin{array}{cc}D^2B_1 & D^2B_2 \\ 0 & 0\end{array}
\right)=\left(
\begin{array}{cc}DB_1D & 0 \\ 0 & 0\end{array}
\right).
$$
Hence $B_2=0$ and $DB_1=B_1D$, and also $B_3=B_1^T=0$, in which case
$$
BA=U^{-1}\hat BUU^{-1}\left(
\begin{array}{cc}D & 0 \\ 0 & 0\end{array}
\right)U=
U^{-1}\hat B\left(
\begin{array}{cc}D & 0 \\ 0 & 0\end{array}
\right)U=U^{-1}\left(
\begin{array}{cc}B_1 & 0 \\ 0 & B_4\end{array}
\right)\left(
\begin{array}{cc}D & 0 \\ 0 & 0\end{array}
\right)U=U^{-1}\left(
\begin{array}{cc}B_1D & 0 \\ 0 & 0\end{array}
\right)U=U^{-1}\left(
\begin{array}{cc}DB_1 & 0 \\ 0 & 0\end{array}
\right)U=U^{-1}\left(
\begin{array}{cc}D & 0 \\ 0 & 0\end{array}
\right)\left(
\begin{array}{cc}B_1 & 0 \\ 0 & B_4\end{array}
\right)U=U^{-1}\left(
\begin{array}{cc}D & 0 \\ 0 & 0\end{array}
\right)\hat BU=U^{-1}\left(
\begin{array}{cc}D & 0 \\ 0 & 0\end{array}
\right)UBU^{-1}U=AB
$$
|
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|
Evaluating $\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\mathrm{d}x$ I was able to find
$$\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\mathrm{d}x=-\frac14\sum_{n=1}^\infty\frac{4^n}{{2n\choose n}}\frac{H_{2n}}{n^3}$$
$$=5\operatorname{Li}_4\left(\frac12\right)-\frac{65}{32}\zeta(4)-2\ln^2(2)\zeta(2)+\frac5{24}\ln^4(2)$$
by converting it to the sum above then evaluating this sum but many integrals and sums were involved in the calculations.
Do you have a different idea to find this integral or its sum?
|
Here is one way to break up the integral
\begin{align}
I=\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\ dx
=I_1-2(I_2 -I_3)
\end{align}
where, with $\ln(1-\sin x)=\ln (\cos x) -2\tanh^{-1}(\tan\frac x2 )$ and $ \cot x = \csc x-\tan\frac x2$
\begin{align}
I_1=& \int_0^{\frac{\pi}{2}}x^2 \cot x \ln(\cos x)dx\\
=&\ \text{Li}_4(\frac12)-\frac{\pi^4}{720}-\frac{\pi^2}6\ln^22+\frac1{24}\ln^42 \\
\\
I_2=&\int_0^{\frac{\pi}{2}}x^2 \csc x \tanh^{-1}(\tan\frac x2 ) \ dx \>\>\>\>\> t=\tan\frac x2\\
=& \ 4\int_0^1 \frac{(\tan^{-1}t)^2\tanh^{-1}t}{t}\ dt\\
=& \ 4\left(\pi \Im \text{Li}_3(\frac{1+i}2)+\frac{\pi}2G\ln2 -\frac{3\pi^4}{128} -\frac{\pi^2}{32}\ln^22\right)\\
\\
I_3=&\int_0^{\frac{\pi}{2}}x^2 \tan \frac x2 \ \tanh^{-1}(\tan\frac x2 )\ dx\\
=& \ 8 \int_0^1 \frac{t(\tan^{-1}t)^2\tanh^{-1}t}{1+t^2} dt\\
=& \ 8\bigg( \frac12\text{Li}_4(\frac12)+ \pi \Im \text{Li}_3(\frac{1+i}2)+\frac\pi2\ln2 G \\
&\hspace{20mm}-\frac{601\pi^4}{23040}-\frac{5\pi^2}{96}\ln^22+\frac1{48}\ln^42 \bigg)\\
\end{align}
The evaluation of the three integrals above are still involved, though familiar. Yet, as a by-product
$$\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1+\sin x)\ dx
=-3 \text{Li}_4(\frac12)+\frac{19\pi^4}{960}-\frac1{8}\ln^42
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/4625939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
}
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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.