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Conjecture: $\lim_{x\to 0^+}f(2x)\log f(x)=0$. I encountered the following problem:
Suppose $f$ is a strictly increasing $C^1$ function defined on $[0,a]$ for some $a>0$, such that $f(0)=f'(0)=0$. Is it true that $\lim_{x\to 0^+}f(2x)\log f(x)=0$?
I tried to use the fact that $z^{\alpha}\log z\to 0$ as $z\to 0^+$ for any $\alpha>0$, and write
$f(2x)\log f(x) = \frac{f(2x)}{f(x)^{\alpha}}f(x)^{\alpha}\log f(x)$. It would then be sufficient to show that there always exists an $\alpha>0$ such that $\frac{f(2x)}{f(x)^{\alpha}}$ is bounded at zero. However, it turns out this is not the case. Here's a counterexample: $f(x) = e^{-e^{1/x}}$ (augmented with $f(0)=0$). But even for this counterexample the initial conjecture works.
Any help would be appreciated.
| We want $f$ to decrease really fast, so that decreasing it's argument two times increases even $\log$ by some large quantity. Also, note that it's enough to have such behavior only on points $x = 2^{-n}$, as long as function will remain in $C^1$.
Let $f\left(\frac{1}{2^n}\right) = \frac{1}{^n2}$, where $^n2$ is tetration, $^02 = 1$, $^{n+1}2 = 2^{(^{n}2)}$, and interpolate $f$ smoothly between with something like
$$f(x) = \frac{1}{^n2} + g\left(2^n \cdot \left(x - \frac{1}{2^n}\right)\right) \cdot \left(\frac{1}{^{n-1}2} - \frac{1}{^n2}\right)$$
when $x \in \left[\frac{1}{2^n}, \frac{2}{2^n}\right)$ and $g$ is some increasing smooth function on $[0, 1]$ with $g(0) = 0$, $g(1) = 1$ and having all derivatives in $0$ and $1$ equal to $0$, and having first derivative bounded by some constant $M$.
We have $f(x) < \frac{1}{^n2} < \frac{1}{2^{2n}}$ (for large enough $n$) if $x < \frac{1}{2^n}$, so $f(x) \to 0$ and $f'(0) = 0$.
For derivatives, if $x = \frac{1}{2^n}$, then $f'(x) = 0$ both from left and right by property of $g$.
Otherwise, if $x \in \left(\frac{1}{2^n}, \frac{2}{2^n}\right)$, we have
$$f'(x) =
g'\left(2^n \cdot \left(x - \frac{1}{2^n}\right)\right) \cdot 2^n \cdot \left(\frac{1}{^{n-1}2} - \frac{1}{^n2}\right) < \frac{M\cdot 2^n}{^{n-1}2}
$$
which goes to $0$ as tetration growths much faster than exponent, so $f'(x) \to 0$.
Important property of tetration: $\log \left({^{n+1}2}\right) = {^{n}2}$.
For $x = \frac{1}{2^{n+1}}$ we have
$$f(2x)\log_2 f(x) = \frac{1}{^n2} \cdot \log \frac{1}{^{n+1}2} = \frac{1}{^n2} \cdot \left(-{^n2}\right) = -1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441757",
"timestamp": "2023-03-29T00:00:00",
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I got different answer of $~\int \frac{1}{(x^2+1)^2}\mathrm{dx}$ $$\begin{align}
&\int {1 \over (x^2+1)^2 } \mathrm{dx} ~~ \leftarrow~~ \text{I assume}~~x\ne0 ~~ \text{since it is trivial as it is held} \\
&=\int {x^2+1-x^2 \over (x^2+1)^2 } \mathrm{dx}\\
&=\int \left\{ {(x^2+1) \over (x^2+1)^2 }- {x^2 \over (x^2+1)^2 } \right\} \mathrm{dx} \\&=
\int {1 \over (x^2+1) } \mathrm{dx}- \int {x^2 \over (x^2+1)^2 } \mathrm{dx}\\&=
\arctan(x)+\mathrm{const_1}-\int x^2 (x^2+1)^{-2} \mathrm{dx} \\&=
\arctan(x)+ \mathrm{const_1}-\left\{ x^2\cdot {(-1)(x^2+1)^{-1} \over 2x } - \int (2x) \cdot {(-1)(x^2+1)^{-1} \over 2x } \mathrm{dx} \right\}\\&=\arctan(x)+ \mathrm{const_1} - \left\{ -{1 \over 2 }x {1 \over (x^2+1) }+\int {1 \over (x^2+1) } \mathrm{dx} \right\} \\&=\arctan(x)+ \mathrm{const_1}+ {x \over 2 (x^2+1) }-\int {1 \over x^2+1 } \mathrm{dx}\\&=
\arctan(x)+ \mathrm{const_1}+ {x \over 2(x^2+1) }- \left(\arctan(x)+ \mathrm{const_2} \right) \\&={x \over 2(x^2+1) }+ \underbrace{\mathrm{const_3} }_{ \mathrm{const_1}-\mathrm{const2} } \end{align}$$
But the answer in the book(A First Course in Calculus by Serge Lang) says the correct form is
$$ \int {1 \over (x^2+1)^2 } \mathrm{dx}= \underbrace{\color{fuchsia}{{x \over 2(x^2+1) } + {1 \over 2 }\arctan(x)}}_{\text{I assume arbitrary const ommited} } $$
Where I've made mistake(s)?
ADD
I am currently in outside so I will to be late to respond.
| When you applied by-parts on $x^2(x^2+1)^{-2}$, you (wrongly) wrote the antiderivative of $(x^2+1)^{-2}$ as $\dfrac{(-1)(x^2+1)^{-1}}{2x}$ which is wrong. The correct one is, well, what you want to find in the first place.
You may want to (and should) try this with a different way. I'm putting a hint here if you're stuck:
which substitution does $1+x^2$ remind you of
Hope this helps. :)
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrating $\int \frac{dx}{\sin^2 x \cos^2x-6\sin x\cos x}$ Our goal is to find
$$\int \frac{dx}{\sin^2 x \cos^2x-6\sin x\cos x}$$
Here is my approach:
We divide both numerator and denominator by $\cos^2x$ and the letting $\tan x=u$,our integrand becomes $-\int \frac{u^2+1}{u(6u^2-u+6)} du$. It is do-able but it seems to need partial fractions,but that would be really tiresome and tedious. Is there any clever way by using further substitution to finish from here? Or do we need to change our approach entirely for a nicer solution?
| Using you substitution :
Performing a partial fraction decomposition we'll have :
$$F(u) = \frac{a_1}{u} +\frac{a_2u+a_3}{6u^2-u+6}$$
Where :
$$a_1 = uF(u)\Big\vert_{u=0} = \frac{1}{6}$$
Let's calculate $F(1)$ and $F(-1)$ :
$$\begin{cases} F(1)&=\frac{1}{6} +\frac{a_2+a_3}{11} &= \frac{2}{11} \\ F(-1)&=-\frac{1}{6}+\frac{-a_2+a_3}{13} &=-\frac{2}{13} \end{cases} \Longleftrightarrow \begin{cases} a_2+a_3 &=\frac{1}{6} \\ -a_2+a_3&=\frac{1}{6} \end{cases} \Longleftrightarrow \begin{cases}a_2&=0\\ a_3&=\frac{1}{6} \end{cases} $$
Hence you'll get :
$$\frac{u^2+1}{u(6u^2-u+6)} = \frac{1}{6u} +\frac{1}{6(6u^2-u+6)}$$
Now I think the only "messy" part is the second term, so let's antidifferentiate it (dunno if this expression exists lol) :
$$\frac{1}{6} \int \frac{\mathrm{d}u}{6u^2-u+6} =\frac{1}{6}\int \frac{\mathrm{d}u}{\left(\sqrt{6} u -\frac{1}{2\sqrt{6}}\right)^2+\frac{143}{24}} \qquad \text{Complete the square} $$
Now let $$t=\frac{12u-1}{\sqrt{143}} \Longleftrightarrow \mathrm{d}u = \frac{\sqrt{143}}{12}\mathrm{d}t $$ Therefore :
\begin{align} \frac{1}{6} \int \frac{\sqrt{143}}{12\left(\frac{143}{24}t^2+\frac{143}{24}\right)}\mathrm{d}t&= \frac{1}{3\sqrt{143}}\int \frac{\mathrm{d}t}{t^2+1}\\ &=\frac{1}{3\sqrt{143}} \arctan(t) \\ &=\frac{1}{3\sqrt{143}} \arctan\left( \frac{12u-1}{\sqrt{143}} \right)\end{align}
Plugging this result to what we got initially :
\begin{align}
-\int \frac{u^2+1}{u(6u^2-u+6)} \mathrm{d}u &= -\frac{1}{3\sqrt{143}} \arctan\left( \frac{12u-1}{\sqrt{143}} \right)+\frac{1}{6}\int \frac{\mathrm{d}u}{u} \\ &=-\frac{1}{3\sqrt{143}} \arctan\left( \frac{12u-1}{\sqrt{143}} \right)-\frac{1}{6}\ln(u) \\
&=-\frac{1}{3\sqrt{143}} \arctan\left( \frac{12\tan(x)-1}{\sqrt{143}} \right)-\frac{1}{6}\ln(\tan(x))\end{align}
Whoops I almost forget the costant :
$$\int \frac{\mathrm{d}x}{\sin^2 x \cos^2x-6\sin x\cos x}=-\frac{1}{6}\ln(\tan(x))-\frac{1}{3\sqrt{143}} \arctan\left( \frac{12\tan(x)-1}{\sqrt{143}} \right)+\text{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let m be the smallest number among: $(x-y)^2, (y-z)^2, (z-x)^2$ Prove $m \le \frac{1}{2}(x^2+y^2+z^2)$ Let m be the smallest number among: $(x-y)^2, (y-z)^2, (z-x)^2$
Prove $m \le \frac{1}{2}(x^2+y^2+z^2)$
My attempts: $3m \le (x-y)^2+(y-z)^2+(z-x)^2$ so I tried to prove $(x-y)^2+(y-z)^2+(z-x)^2 \le \frac{3}{2}(x^2+y^2+z^2) \rightarrow 4(xy+yz+zx)\le x^2+y^2+z^2$ but it wrong
Can anyone help me with this problem? Thank you
| Wlog assume that $x\ge y \ge z $ , then $m=min\{(x-y)^2, (y-z)^2\} \le (x-y)(y-z)$
Note that
$$x^2+y^2+z^2-2(x-y)(y-z)=(x-y+z)^2+2y^2 \ge 0$$
Hence
$$x^2+y^2+z^2 \ge 2(x-y)(y-z) \ge 2m$$
Equality holds when $y=0, x=-z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454121",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove $\sqrt{x-1} + \sqrt{y-1} \le xy, x \ge 1, y \ge 1$ Let $x$ and $y$ be real numbers, such that $x \ge 1$ and $y \ge 1$.
Prove that this inequality is true: $\sqrt{x-1} + \sqrt{y-1} \le xy$
Can someone show me steps to solve it.
PS:I need to give steps on how to solve it.
| you can use schwartz inequality directly
we know ;$a_1 b_1+a_2b_2 \leq \sqrt{a_{1}^2+a_{2}^2} \sqrt{b_{1}^2+b_{2}^2}$
just put $a_1 =\sqrt{x-1}$
$a_2 =1$
$b_1 =1$
and
$b_2 =\sqrt{y-1}$
so you will get ;$\sqrt{x-1} .1+1.\sqrt{y-1} \leq \sqrt{x-1+1}\sqrt{y-1+1}=\sqrt{xy}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Collatz conjecture but with $n^2-1$ instead of $3n+1.$ Does the sequence starting with $13$ go to infinity? Let's consider the following variant of Collatz $(3n+1) : $
If $n$ is odd then $n \to n^2-1.$
$1\to 0.$
$3\to 8\to 1\to 0.$
$5\to 24\to 3\to 0.$
$7\to 48\to 3\to 0.$
$9\to 80\to 5\to 0.$
$11\to 120\to 15\to 224\to 7\to 0.$
$\color{red}{13\to 168\to 21\to 440\to 55\to 3024\to 189\to 35720\to 4465\to 19936224\to 623007\to\ldots\ ?}\ $
Does the sequence starting with $13$ go off to infinity? If yes, what
is a proof? If no, is there a starting number whose sequence does go off to infinity, and
how do we prove either that such a number must exist, or even better that a specific starting number goes off to infinity?
Here is some Python code I ran, which suggests that the numbers in the sequence starting at $13$ quickly become large:
n=13
num_loops=0
print(n)
while n!=0:
if n%2==0:
n//=2
else: n=n**2-1
print('\n', n)
num_loops+=1
if num_loops==70:
print("too many loops")
break
| About Eric Syder's comment: suppose $ x^2 - 1 = 2^k y$ with $x,y$ odd. We wish to investigate what happens when $y \leq x \; . \; \;$ Note that $\gcd(x+1, x-1) = 2$ because $x \equiv 1,3 \pmod 4.$ One of the $x \pm 1 $ is $\equiv 2 \pmod 4. $
Let us make the name $ \delta = \pm 1.$ Then we may demand
$$ x + \delta \equiv 2 \pmod 4$$
This tells us that the integer $ \frac{x + \delta}{2} $ is odd. We also have
$$ \frac{x + \delta}{2} \frac{x - \delta}{2} = 2^{k-2}y$$
By repeated division by $2$ it follows that
$$ \frac{x - \delta}{2^{k-1}} = w $$ is an odd positive integer, with
$$\frac{x + \delta}{2} \; \frac{x - \delta}{2^{k-1}} = y $$
IF WE ASSUME $$ w = \frac{x - \delta}{2^{k-1}} \geq 3, $$
we find
$$ y = w \frac{x + \delta}{2} \geq 3 \frac{x + \delta}{2} \geq \frac{3x - 3}{2}$$
The assumption $ w \neq 1$ has led us to $y \geq \frac{3x - 3}{2}$ The hypothesis that $x \geq y$ now says $x \geq \frac{3x - 3}{2},$ or
$ 2x \geq 3x - 3,$ or $3 \geq x$
If $ x \geq y$ and $ x \geq 5$ in $ x^2 - 1 = 2^k y,$ we find that $w=1$ in $ \frac{x - \delta}{2^{k-1}} = w .$ So that $ x - \delta = 2^{k-1} .$ or
$$ x = \pm 1 + 2^{k=1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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Find all $4$ digit numbers such that sum of digits is $11$. Find all $4$ digit numbers such that sum of digits is $11$.
\begin{cases} x_1+x_2+x_3+x_4=11\\ 1\leq x_1{\leq 9}\\ 0\leq x_2{\leq 9}\\0\leq x_3{\leq 9}\\ 0\leq x_4{\leq 9}\end{cases}.
Using stars and bars there is 13 choose 3 ways to do. So we get $286$ but answer is $279$. I saw this same question in the forum but didn't fully understand solution. Where we are overcounting when solving in this way? And mainly what we are overcounting?
| In this answer it is described how to solve this by generating functions. The solution is the coefficient of the degree $11$ monomial of the polynomial
$$(x+x^2+\cdots+x^9)(1+x+x^2+x^3+\cdots+x^9)^3$$
This coefficient can be calculated by a CAS like Mathematica as shown in the link above. It can even calculated by a numerical calculator that can handle arbitrary length integers as describe in the second half of my answer here. But in this answer I learned in one of the comments how to calculate this by hand.
Note that
$$\frac 1 {1-x}=\sum_{n=0}^\infty x^n$$
and by differentiation using induction you get
$$\frac 1 {(1-x)^k}=\sum_{n=0}^\infty{ n+k-1 \choose k-1} x^n$$
So we have
$$(x+x^2+\cdots+x^9)(1+x+x^2+x^3+\cdots+x^9)^3\\
=x\frac{1-x^9}{1-x}\frac{(1-x^{10})^3}{(1-x)^3}\\
=x(1-x^9)(1-x^{10})^3 \sum_{n=0}^\infty{ n+3 \choose 3} x^n\\
=(x-x^{10}-3x^{11}+3x^{20}+\cdots)\frac 1 6 (1\cdot2\cdot3+2\cdot 3 \cdot4\cdot x+\cdots+11\cdot 12\cdot 13\cdot x^{10} \cdots )\\
=\cdots +\cdot \frac 1 6 (-3\cdot 1\cdot 2 \cdot 3-2\cdot 3\cdot 4+11\cdot 12 \cdot 13)x^{11}+\cdots\\
=\cdots+279x^{11}+\cdots$$
The coefficients of the monomials of degree greater than $36$ cancel out.
We have
$$
\frac{\left( {{x}^{40}}-{{x}^{31}}-3 {{x}^{30}}+3 {{x}^{21}}+3 {{x}^{20}}-3 {{x}^{11}}-{{x}^{10}}+x\right) \, \sum_{n=0}^{\infty }{\left. \left( {{n}^{3}}+6 {{n}^{2}}+11 n+6\right) {{x}^{n}}\right.}}{6}
$$
We abbreviate
$$f(n):={{n}^{3}}+6 {{n}^{2}}+11 n+6$$
and verify that
$$f(n+39)-f(n+30)-3f(n+29)+3f(n+20)+3f(n+19)-3f(n+10)-f(n+9)+f(n)=0$$
and
$$f(n+39)-f(n+30)-3 f(n+29)+3 f(n+20)+3 f(n+19)-3 f(n+10)-f(n+9)$$
for $n=-1,-2,-3$ because $f(n)=0$ for this values
| {
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"timestamp": "2023-03-29T00:00:00",
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Solution (recurrence relation) of non-linear DE using the method of power series I have to solve this non-linear DE
$y' -e^y -x^2 = 0 , y(0)=c$
using powerseries.
$y(x) = \sum_{n=0}^\infty a_{n}x^n $
$y'(x) = \sum_{n=1}^\infty na_{n}x^{n-1} $
so we get
$\sum_{n=1}^\infty na_{n}x^{n-1} -e^{\sum_{n=0}^\infty a_{n}x^n} -x^2 = 0 $
changing the sum indexes
$\sum_{n=0}^\infty (n+1)a_{n+1}x^{n} -e^{\sum_{n=0}^\infty a_{n}x^n} -x^2 = 0 $
I want to find the recurrence relation of the coefficients.
But i cannot merge/group them into the same sum, meaning, the sum of $y'(x) = \sum_{n=1}^\infty na_{n}x^{n-1} $ and the sum of $y(x) = \sum_{n=0}^\infty a_{n}x^n $ cause it is power of $e$ in order to get the equality of them with $0$.
| The differential equation to be solved is
$$ y' -e^y -x^2 = 0 ,\qquad y(0)=c $$
and the suggested approach is with power series.
It may not be immediately obvious, but after working
with the power series for $\,y\,$ a good step is to
define $\, u := e^{-y}.\,$ Rewrite$\,y\,$ in terms
of $\,u\,$ in the D.E. to get
$$ -\log(u)' - 1/u - x^2 = 0, \qquad
u(0) = c_0 := e^{-c}. $$
Simplify to get a new differential equation
$$ u' = -1 - u\,x^2. $$
Expand $\,u\,$ as a power series to get
$$ u \!=\! \sum_{n=0}^\infty c_n \frac{x^n}{n!},
u' \!=\! \sum_{n=0}^\infty c_{n+1} \frac{x^n}{n!},
u\,x^2 \!=\! \sum_{n=0}^\infty c_{n-2}n(n\!-\!1) \frac{x^n}{n!}. $$
Substitute these series into the new D.E. to get
$$ c_1 = -1,\quad c_{n+1} = -c_{n-2}n(n-1) \quad\text{ unless } \quad n=0. $$
Use this recursion relation to get
$$ u = c_0-x-2c_0\frac{x^3}{3!} +6\frac{x^4}{4!}
+40c_0\frac{x^6}{6!}-180\frac{x^7}{7!}+\cdots.$$
Use $\,y = -\log(u/c_0)+c\,$ to get the solution
$$ c + \frac1{c_0}\frac{x^1}{1!} +
\frac1{c_0^2} \frac{x^2}{2!} +
2\frac{1+c_0^3}{c_0^3}\frac{x^3}{3!} +
2\frac{3+c_0^3}{c_0^4}\frac{x^4}{4!} +
\cdots. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the focus and directrix of the general parabola? Q: Let us be given the parabola $(Ax + By)^2 + 2gx + 2fy + c = 0$ s.t. $A^2 + B^2 \neq 0$. Then what is the directrix and focus for the parabola when $Af - Bg = 0$?
Background. Consider that equation of the parabola, we want to write it as the distance of $(x,y)$ from some line equals distance from some point (the line and point being the directrix and focus respectively). [Let K be a parameter which can take any value.]
\begin{align*}
(Ax + By)^2 + 2gx + 2fy + c &= 0 \\
\Leftrightarrow B^2 x^2 + A^2 y^2 - 2AB xy + 2BKx - 2AKy + K^2 + \Big(\frac{(g+BK)^2}{A^2 + B^2} + \frac{(f-AK)^2}{A^2 + B^2} - c - K^2 \Big) = \\
(A^2+B^2)\Big( (x + \frac{g + BK}{A^2 + B^2})^2 + (y + \frac{f - AK}{A^2 + B^2})^2 \Big) \\
\Leftrightarrow (Bx - Ay + K)^2 + \epsilon(K) = (A^2+B^2)\Big( (x + \frac{g + BK}{A^2 + B^2})^2 + (y + \frac{f - AK}{A^2 + B^2})^2 \Big) \\
\end{align*}
where $\epsilon(K) := \Big(\frac{(g+BK)^2}{A^2 + B^2} + \frac{(f-AK)^2}{A^2 + B^2} - c - K^2 \Big)$. Now above equation holds for any value of $K$, and we chose that value of K s.t. $\epsilon(K) = 0$ (as notice once we have $\epsilon(K) = 0$, we have found the directrix and focus), and we get the following result:
If $Af - Bg \neq 0$, then $K = \frac{(A^2 +B^2)c - (g^2 + f^2)}{2(Af - Bg)} \Rightarrow \epsilon(K) = 0$.
But this now leaves the important question, what if $Af - Bg = 0$, how do we find the directrix and focus?
| Let $r = [ x, y]^T $ then your parabola is given by
$ r^T Q r + r^T L + c = 0 $
where
$ Q = \begin{bmatrix} A^2 && A B \\ A B && B^2 \end{bmatrix} $
$L = \begin{bmatrix} 2 g \\ 2 f \end{bmatrix} $
Diagonalize $Q$. To do that, calculate the angle $\theta$ as follows
$ \tan (2 \theta) = \dfrac{ 2 AB }{A^2 - B^2} $
So that
$R = \begin{bmatrix} \cos \theta && - \sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $
Now,
$ \cos^2(2 \theta) = \dfrac{1}{\tan^2(2 \theta) + 1} = \dfrac{ (A^2 - B^2)^2 }{(A^2 + B^2)^2 } $
So, $\cos(2 \theta) = \dfrac{ |A^2 - B^2| }{A^2 + B^2} $
We can simply take $ \cos(2 \theta) = \dfrac{ A^2 - B^2 }{A^2 + B^2} $ and maintain $\tan(2 \theta) $ at the above value,then
$\sin(2 \theta) = \tan(2 \theta) \cos(2 \theta) = \dfrac{2 A B}{A^2 + B^2} $
If $Q = R D R^T$. The diagonal entries of $D$ are
$D_{11} = \dfrac{1}{2} (A^2 + B^2) + \dfrac{1}{2} (A^2 - B^2) \cos(2 \theta) + AB \sin(2 \theta) $
And this is equal to
$ D_{11} = \dfrac{1}{2}(A^2 + B^2) + \dfrac{1}{2} \dfrac{(A^2 - B^2)^2}{A^2 + B^2} + \dfrac{ 2 A^2 B^2}{A^2 + B^2}$
And this simplifies to
$D_{11} = \dfrac{1}{2} \dfrac{A^4 + 4 A^2 B^2 + B^4}{A^2 + B^2} $
while
$D_{22} = 0 $
Back to the expression for $\cos(2\theta) = \dfrac{A^2 - B^2}{A^2 + B^2} $
It follows that
$ \cos(\theta) = \sqrt{ \dfrac{1 + \cos(2 \theta)}{2} } = \dfrac{A}{\sqrt{A^2 + B^2}} $
$ \sin(\theta) = \sqrt{ \dfrac{ 1 - \cos(2 \theta)}{2} } = \dfrac{B}{\sqrt{A^2+B^2} } $
From this,
$ \tan(2 \theta) = \dfrac{2 \tan(\theta)}{1 - \tan^2(\theta)} = \dfrac{2 \sin(\theta)\cos(\theta) }{\cos^2(\theta) - \sin^2(\theta)} = \dfrac{2 AB}{A^2 - B^2} $
Back to the original equation
$ r^T Q r + r^T L + c = 0 $
with $Q = R D R^T $ this becomes
$ r^T R D R^T r + r^T L + c = 0 $
Let $w = R^T r$ , then
$ w^T D w + w^T R^T L + c = 0 $
Now
$ R^T L = \dfrac{1}{\sqrt{A^2 + B^2}} \begin{bmatrix} A && B \\ -B && A \end{bmatrix} \begin{bmatrix} 2 g\\ 2 f \end{bmatrix} = \dfrac{ 2}{\sqrt{A^2 + B^2}} \begin{bmatrix} A g + B f \\ 0 \end{bmatrix}$
So in this case, we get a degenerate equation
$ D_{11} w_1^2 + \dfrac{2}{\sqrt{A^2 + B^2}} (A g + B f) w_1 + c = 0 $
which can have 0, 1, or 2 solutions, while $w_2$ is arbitrary, and this corresponds to no solution, one line, or two parallel lines.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $n^{12} + 64$ has at least 4 non-trivial factors.
Show that for every integer $n>1$, $n^{12} + 64$ has at least four factors different from $1$ and the number.(It has at least four non-trivial factors.)
My attempt:
We can expand $n^{12} + 64$ in the following way :
$=>n^{12} + 64 = (n^4)^3 + (2^2)^3$
$=>n^{12} + 64 = (n^4 + 2^2)((n^4)^2 + (2^2)^2 - (2^2)(n^4))$
$=>n^{12} + 64 = ((n^2 + 2)^2 - 2(2)(n^2))((n^4)^2 + (2^2)^2 - (2^2)(n^4)) $
$=>n^{12} + 64 = (n^2 + 2 + 2n)(n^2 + 2 - 2n)(n^8 + 2^4 - 4n^4)$
i.e there are 3 different factors.
How can I prove that there is one more factor ?.
Update: I think we are already done, because we got the following factors:
1.$(n^2 + 2 + 2n)$
2.$(n^2 + 2 - 2n)$
3.$(n^8 + 2^4 - 4n^4)$
4.$(n^2 + 2 + 2n)(n^2 + 2 - 2n)$
5.$(n^2 + 2 + 2n)(n^8 + 2^4 - 4n^4)$
6.$(n^2 + 2 - 2n)(n^8 + 2^4 - 4n^4)$
Clearly, we have got 6 non-trivial factors.(None of them can be one.)
| Well we can write-
$n^{12}+64=(n^4)^3+4^3$
$=(n^4+4)(n^8-4n^4+16)$
$=[(n^2)^2+2(n^2)(2)+2^2-(2n)^2](n^8-4^2+4n^6-4n^6+12n^4-16n^4+16n^2-16n^2)$
$=[(n^2+2)^2-(2n)^2][(n^8+4^2+4n^4+4n^6+8n^4+16n^2)-(4n^6+16n^2+16n^4)]$
$=(n^2+2n+2)(n^2-2n+2)[(n^4+2n^2+4)^2-(2n^3+4n)^2]$
$=(n^2+2n+2)(n^2-2n+2)(n^4+2n^3+2n^2+4n+4)(n^4-2n^3+2n^2-4n+4)$
| {
"language": "en",
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"answer_id": 0
} |
Evaluating $\int_{0}^{\frac{\pi}{2}} x^{2} \ln (\cos x) d x$ Latest Edit
As suggested by Mr Claude Leibovici, I go further to investigate the integral in general,
$$I_{m}:=\int_{0}^{\frac{\pi}{2}} x^{m} \ln (\cos x) d x$$
By the Fourier Series of ln(cos x)
,
$ \displaystyle \ln (\cos x)=-\ln 2+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cos (2 n x) \tag*{(*)} $
Multiplying (*) by $ x^m$ and then Integrating both sides from $0 $ to $\frac{\pi}{2}$ yields
$$\begin{aligned}
I_m&=-\int_{0}^{\frac{\pi}{2}} x^{m} \ln 2 d x+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n !} \int_{0}^{\frac{\pi}{2}} x^{m} \cos (2 n x) d x\\&= -\frac{\pi^{m+1}\ln 2}{(m+1) 2^{m+1}}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} J(m, n)\end{aligned}
$$
Next we need a reduction formula on $J(m,n)$.
Applying integration by parts twice, we obtain
$$
\begin{aligned}J(m, n)&=\frac{1}{2 n} \int_{0}^{\frac{\pi}{2}} x^{m} d(\sin 2 n x)\\ &= -\frac{m}{2 n} \int_{0}^{\frac{\pi}{2}} x^{m-1} \sin 2 n x d x\\&=\frac{m}{4 n^{2}} \int_{0}^{\frac{\pi}{2}} x^{m-1} d(\cos 2 n x)\\&=\frac{(-1)^{n} \pi^{m-1} m}{2^{m+1} n^{2}} -\frac{m(m-1)}{4 n^{2}} J (m-2, n)\end{aligned}
$$
By the reduction formula of $J(m,n)$, we can find $I_m$ by plugging $J(m,n)$.
By the Fourier Series of ln(cos x)
,
$ \displaystyle \ln (\cos x)=-\ln 2+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cos (2 n x) \tag*{(*)} $
Multiplying (*) by $ x^2$ and then Integrating both sides from $0 $ to $\frac{\pi}{2}$ yields
$\displaystyle I=-\underbrace{\int_{0}^{\frac{\pi}{2}} x^{2} \ln 2 d x}_{\frac{\pi^{3} \ln 2}{24}}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \underbrace{ \int_{0}^{\frac{\pi}{2}} x^{2} \cos (2 n x) d x}_{J_n}\tag*{} $
Integrating by parts twice yields
$\displaystyle \begin{aligned}J_{n} &=\frac{1}{2 n} \int_{0}^{\frac{\pi}{2}} x^{2} d(\sin 2 n x) \\&=\frac{1}{2 n}\left[x^{2} \sin 2 n x\right]_{0}^{\frac{\pi}{2}}-\frac{1}{n} \int_{0}^{\frac{\pi}{2}} x \sin 2 n x d x \\&=\frac{1}{2 n^{2}} \int_{0}^{\frac{\pi}{2}} x d(\cos 2 n x) \\&=\left[\frac{1}{2 n^{2}} x \cos 2 n x\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2 n^{2}} \int_{0}^{\frac{\pi}{2}} \cos 2 n x d x \\&=\frac{\pi}{4 n^{2}} \cos n \pi-\frac{1}{2 n^{2}}\left[\frac{\sin 2 n x}{2 n}\right]_{0}^{\frac{\pi}{2}} \\&=\frac{\pi}{4 n^{2}} \cos n \pi\end{aligned}\tag*{} $
We can now conclude that
$\displaystyle \begin{aligned}I &=-\frac{\pi^{3} \ln 2}{24}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cdot \frac{\pi}{4 n^{2}} \cos n \pi \\&=-\frac{\pi^{3} \ln 2}{24}+\frac{\pi}{4} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(-1)^{n}}{n^{3}} \\&=-\frac{\pi^{3} \ln 2}{24}-\frac{\pi}{4} \sum_{n=1}^{\infty} \frac{1}{n^{3}} \\&=\boxed{-\frac{\pi^{3} \ln 2}{24}-\frac{\pi}{4} \zeta(3)}\end{aligned}\tag*{} $
Is there any other method to evaluate $I$?
| As Mr Claude Leibovici suggested, I go further to investigate the integral in general,
$$I_{m}:=\int_{0}^{\frac{\pi}{2}} x^{m} \ln (\cos x) d x$$
By the Fourier Series of ln(cos x)
,
$ \displaystyle \ln (\cos x)=-\ln 2+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \cos (2 n x) \tag*{(*)} $
Multiplying (*) by $ x^m$ and then Integrating both sides from $0 $ to $\frac{\pi}{2}$ yields
$$\begin{aligned}
I_m&=-\int_{0}^{\frac{\pi}{2}} x^{m} \ln 2 d x+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n !} \int_{0}^{\frac{\pi}{2}} x^{m} \cos (2 n x) d x\\&= -\frac{\pi^{m+1}\ln 2}{(m+1) 2^{m+1}}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} J(m, n)\end{aligned}
$$
Next we need a reduction formula on $J(m,n)$.
Applying integration by parts twice, we obtain
$$
\begin{aligned}J(m, n)&=\frac{1}{2 n} \int_{0}^{\frac{\pi}{2}} x^{m} d(\sin 2 n x)\\ &= -\frac{m}{2 n} \int_{0}^{\frac{\pi}{2}} x^{m-1} \sin 2 n x d x\\&=\frac{m}{4 n^{2}} \int_{0}^{\frac{\pi}{2}} x^{m-1} d(\cos 2 n x)\\&=\frac{(-1)^{n} \pi^{m-1} m}{2^{m+1} n^{2}} -\frac{m(m-1)}{4 n^{2}} J (m-2, n)\end{aligned}
$$
By the reduction formula of $J(m,n)$, we can find $I_m$ by plugging $J(m,n)$.
For example,
$$
\begin{aligned}
I_{3} &=-\frac{\pi^{4} \ln 2}{64}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} J(3, n) \\
&=-\frac{\pi^{4} \ln 2}{64}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\left[\frac{3(-1)^{n} \pi^{2}}{16 n^{2}}-\frac{3}{2 n^{2}} J(1, n)\right] \\
&=-\frac{\pi^{4} \ln 2}{64}-\frac{3 \pi^{2}}{16} \zeta(3)-\frac{3}{2} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{3}}\left[\frac{1}{4 n^{2}}(1-(-1)^n)\right] \\
&=-\frac{\pi^{4} \ln 2}{64}-\frac{3 \pi^{2}}{16} \zeta(3)+\frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{(2 n+1)^{5}} \\
&=-\frac{\pi^{4} \ln 2}{64}-\frac{3 \pi^{2}}{16} \zeta(3)+\frac{93}{128}\zeta(5)
\end{aligned}
$$
$$
\begin{aligned}
I_{4}=&-\frac{\pi^{5} \ln 2}{160}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\left(\frac{(-1)^{n} \pi^{3}}{8 n^{2}}-\frac{3}{n^{2}} J(2, n)\right) \\
=&-\frac{\pi^{5} \ln 2}{160}-\frac{\pi^{3}}{8} \sum_{n=1}^{\infty} \frac{1}{n^{3}}+3 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{3}} \left(\frac{(-1)^{n} \pi}{4 n^{2}}-\frac{1}{2 n^{2}} J(0, n)\right) \\
=&-\frac{\pi^{5} \ln 2}{160}-\frac{\pi^{3}}{8} \zeta(3)+\frac{3 \pi}{4} \zeta(5) \end{aligned}
$$
Is there a closed form for the integral $I_m$ ?
| {
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"question_score": "4",
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Let $z$ be a complex number, s.t. $z^2-z+1=0$. Find $z^5 + z$.
Let $z$ be a complex number, s.t. $z^2-z+1=0$.
Find $z^5 + z$.
As $z= z^2+1$, so $z\neq 0$, also $z^2\neq 0$, as $z\neq 1$.
Anyway, we can divide $z^2-z+1=0$ by $z$ to get:
$z+\frac1z-1=0$
We need to derive $z^5 + z$ in terms of $z+\frac1z$, so on division get the quotient.
$$
\require{enclose}
\begin{array}{rll}
z^4 -z^2 +2 && \hbox{} \\[-3pt]
z+\frac1z \enclose{longdiv}{z^5 + z}\kern-.2ex \\[-3pt]
\underline{z^5+ z^3} && \hbox{} \\[-3pt]
-z^3 +z && \hbox{} \\[-3pt]
\underline{-z^3-z} && \hbox{} \\[-3pt]
2z && \hbox{} \\[-3pt]
\underline{2z+\frac2z} && \hbox{} \\[-3pt]
-\frac2z
\end{array}
$$
So, $z^5 + z= (z+\frac1z)(z^4 -z^2 +2)-\frac2z$
Seems it needs to further find division of $(z^4 -z^2 +2)$ too by $z+\frac1z$. Still remainders at each step (like, $\frac2z$) will be a further issue.
Alternatively (second approach), multiply by $(z+1)$ to get: $(z+1)(z^2-z+1)= z^3 +1=0\implies z^3=1$
$z(z^4+1)= z((z^2+1)(z^2 -1)+2)$
This approach fails too as remainder $2$ is obtained.
Next (third approach), $z+\frac1z=1\implies z^2 +\frac1{z^2}= -1$
This too fails to be useful in solving $z^5+z$.
Next (fourth approach)
, $z^2=z-1\implies z^4 = z^2-2z +1\implies z^4+1= z^2-2z +2\implies z(z^4+1)= z^3-2z^2 +2z$.
Can use value of $z^3=1$, to get:
$z^3-2z^2 +2z\implies 1-2z^2+2z\implies 1-2(z^2 -z) = 3$.
Seems the least intuitive approach works only.
Alternatively (fifth approach) to have a working intuitive approach, use the trigonometric /polar form approach:
the roots of $z^2-z+1$ are: $z=\frac12 \pm \frac{\sqrt3}{2}i$, with two roots $z_1, z_2$.
$e^{i x} = \cos(x) + i \sin(x)$
$z_1 = \frac12 + \frac{\sqrt3}{2}i = e^{ i\pi/3}$
$z_2 = \frac12 - \frac{\sqrt3}{2}i = e^{- i\pi/3}$
\begin{align}
z_1^5 + z_1 &= e^{5 i\pi/3} + e^{i\pi/3} \\
&= (sin(-\pi/3)+\cos (\pi/3))+(sin(\pi/3)+\cos (\pi/3))
= 2\cos (\pi/3)
= 1
\end{align}
\begin{align}
z_2^5 + z_2 &= e^{-5 i\pi/3} + e^{-i\pi/3} \\
&= (sin(\pi/3)+\cos (\pi/3))+(sin(-\pi/3)+\cos (\pi/3))
= 2\cos (\pi/3)
= 1
\end{align}
Now, the answer differs from earlier of $3$.
| $z^2 - z + 1 = 0$ implies $$z^2 = z-1,$$ hence
$$\begin{align}
z^5 + z &= z((z^2)^2 + 1) \\
&= z((z-1)^2 + 1) \\
&= z(z^2 - 2z + 2) \\
&= z(z-1 - 2z + 2) \\
&= z(-z+1) \\
&= -z^2 + z \\
&= 1-z + z \\
&= 1.
\end{align}$$
All your other calculations either have errors or are essentially failing to simplify the expression. The only one that is correct is your fifth approach. The aim is to reduce the power of $z^5$, so the above method is how you would do it.
Multiplying $z^2 - z + 1$ by $z+1$ also works, which gives you $z^3 = -1$, thus
$$z + z^{-1} = 1$$ implies $$z^5 + z = z^3 (z^2 + z^{-2}) = -(z^2 + z^{-2}) = -(z + z^{-1})^2 + 2 = -1 + 2 = 1.$$
| {
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} |
Denesting radicals $\sqrt[3]{-22+15\sqrt[3]{3}+9\sqrt[3]{9}}$ and $\sqrt[3]{8-9\sqrt[3]{3}+3\sqrt[3]{9}}$ I am trying to do denesting radicals:$$\sqrt[3]{-22+15\sqrt[3]{3}+9\sqrt[3]{9}}$$ and $$\sqrt[3]{8-9\sqrt[3]{3}+3\sqrt[3]{9}}$$
I tried to find Ramanujan polynomial like this link denesting radicals
But it doesn't work. I also tried to solve the system of equation $\sqrt[3]{-22+15\sqrt[3]{3}+9\sqrt[3]{9}}=a+b\sqrt[3]{3}+c\sqrt[3]{9}$ but it led to a scary-looking one.
| Let $\mathbb{Q}(\sqrt[3]{3})$ be the set of all numbers of the form $a + b\sqrt[3]{3} + c\sqrt[3]{9}$ for $a, b, c \in \mathbb{Q}$, and define the function $N(\cdot)$ on $\mathbb{Q}(\sqrt[3]{3})$ by
\begin{align*}
N(a + b\sqrt[3]{3} + c\sqrt[3]{9})
&= \prod_{k=0}^{2}(a + b \sqrt[3]{3}\,e^{2\pi i k/3} + c \sqrt[3]{9} \,e^{4\pi i k/3}) \\
&= a^3 + 3b^3 + 9c^3 - 9abc,
\end{align*}
Then it is not hard to check that, for $x, y \in \mathbb{Q}(\sqrt[3]{3})$,
$$ N(xy) = N(x)N(y). $$
Now, suppose that $x = \sqrt[3]{-22 + 15\sqrt[3]{3} + 9\sqrt[3]{9}}$ is actually of the form $x = a + b\sqrt[3]{3} + c\sqrt[3]{9}$ for some integers $a, b, c$. Then we have
\begin{align*}
N(x)^3 = N(x^3) &= N(-22 + 15\sqrt[3]{3} + 9\sqrt[3]{9}) = 32768 = 2^{15}
\end{align*}
So, this and the equation $x^3 = -22 + 15\sqrt[3]{3} + 9\sqrt[3]{9}$ together yield
\begin{align*}
a^3 + 3b^3 + 9c^3 - 9abc &= N(x) = 32, \\
a^3 + 3b^3 + 9c^3 + 18abc &= -22, \\
3 a^2b + 9b^2 c + 9ac^2 &= -15, \\
3 ab^2 + 3a^2 c + 9bc^2 &= 9.
\end{align*}
From the first two equations, we get $27abc = -54$, or equivalently, $abc = -2$. This leaves only a handful of cases to be checked, and then it is not hard to find that $(a, b, c) = (2, -1, 1)$.
Generalizing this observation, we get:
Observation. Suppose that rational numbers $a, b, c, A, B, C$ satisfy the equation
$$ a + b\sqrt[3]{3} + c\sqrt[3]{9} = \sqrt[3]{A + B\sqrt[3]{3} + C\sqrt[3]{9}}. $$
Then we have
$$ 27abc = A - \sqrt[3]{A^3 + 3B^3 + 9C^3 - 9ABC}. $$
Likewise, assume that $\sqrt[3]{8-9\sqrt[3]{3}+3\sqrt[3]{9}} = a + b\sqrt[3]{3} + c\sqrt[3]{9}$ for some integers $a, b, c$. Then by the observation above, we get
$$ 27abc = 8 - \sqrt[3]{512} = 0 $$
and hence at least one of $a, b, c$ is zero. Indeed, when $b = 0$, we are led to solve the system of equations
\begin{align*}
a^3 + 9c^3 &= 8, &
9ac^2 &= -9, &
3a^2c &= 3.
\end{align*}
The last two equations shows that $c = -a$, and then the first equation gives $-8a^3 = 8$, hence $ a = -1$ and $c = 1$. Therefore $(a, b, c) = (-1, 0, 1)$.
| {
"language": "en",
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In the field $F_{3^2} = F_3[x]/(2x^2 + x + 1)$, let $\alpha$ be the root and primitive element. The elements of $F_{3^2}$ can be represented in 2 ways:
*
*As polynomials (degree $\leq 1$) in $\alpha$ with coefficients in $F_3$
*As $0, 1, \alpha, \alpha^2, \dots, \alpha^{3^2 - 2}$.
And the $2$ ways can be related by using the fact that
$2\alpha^2 + \alpha + 1 = 0$ in $F_3[\alpha]$.
But that is true only in $F_3[\alpha]$?
For e.g. I know $\alpha^5 = 2\alpha$ in $F_3[\alpha]$. But if I want to find $\alpha + \alpha$ in $F_{3^2}$, that will be $2\alpha$, where $2$ is just an integer, that will not be $\alpha^5$ right?
| First of all, your skepticism about $\alpha^5 = 2\alpha$ is misplaced. You are absolutely correct about that equation.
We can simplify the minimal polynomial for $\alpha$ over $F_3$. Since $2 \equiv -1 \pmod{3}$, the equation $2\alpha^2 + \alpha + 1 = 0$ in $F_3[\alpha]$ is equivalent to the equation
$$
\alpha^2 = \alpha + 1.
$$
We can treat this as a power-reducing identity to inductively express each power of $\alpha$ (your second representation) as an $F_3$-linear combination of $1$ and $\alpha$ (your first representation).
Here's what this looks like:
$$
\alpha^3 = \alpha \, \alpha^2 = \alpha (\alpha + 1)
= \alpha^2 + \alpha = (\alpha + 1) + \alpha = 2\alpha + 1
$$
and
$$
\alpha^4 = \alpha \, \alpha^3 = \alpha (2\alpha + 1)
= 2\alpha^2 + \alpha = 2(\alpha + 1) + \alpha = 2
$$
etc. It's interesting to recalculate $\alpha^4$ as $\alpha^2 \, \alpha^2$ instead:
$$
\alpha^2 \, \alpha^2 = (\alpha + 1) (\alpha + 1) = \alpha^2 + 2\alpha + 1
= (\alpha + 1) + 2\alpha + 1 = 2
$$
Repeating this process produces the dictionary:
\begin{array}{l|r}
\alpha^n & b\alpha + c \\
\hline
0 & 0 \\
1 & 1 \\
\alpha & \alpha \phantom{{}+1} \\
\alpha^2 & \alpha + 1 \\
\alpha^3 & 2\alpha + 1 \\
\alpha^4 & 2 \\
\alpha^5 & 2\alpha \phantom{{}+1} \\
\alpha^6 & 2\alpha + 2 \\
\alpha^7 & \alpha + 2
\end{array}
Of course, $\alpha^8 = 1$, showing that all of the nonzero elements of the field form the cyclic group of order $8 = 3^2 - 1$. This is true in general.
| {
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Consider sequence of numbers $a_r,\;r\geq0\;$ with $a_0=1\;$ and $a_{r+1}^2=1+a_r\cdot a_{r+2}.\;$ Then which of the following is/are true? Let $\alpha, \beta$ are roots of equation $x^2-a_1 x+1=0$ and consider sequence of numbers $a_r,\;r\geq0\;$ with $a_0=1\;$ and $a_{r+1}^2=1+a_r\cdot a_{r+2}.\;$ Then which of the following is/are true?
(A) $a_r+a_{r+2}=a_1 \cdot a_{r+1}$
(B) $a_r+a_{r+2}=2\cdot a_{r+1}$
(C) $a_n=\dfrac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta}$
(D) $a_n=\dfrac{\alpha^{n+1}+\beta^{n+1}}{\alpha+\beta}$
My Approach:
Since it is given that $a_{r+1}^2=1+a_r\cdot a_{r+2}.\; \implies a_{r+1}^2-1=a_r\cdot a_{r+2}.\;$
Now using option (A) $\;a_r+a_{r+2}=a_{r}+\dfrac{a_{r+1}^2-1}{a_r}\implies a_r+a_{r+2}=\dfrac{a_{r}^2-1+a_{r+1}^2}{a_r}\implies a_r+a_{r+2}=\dfrac{(a_{r-1}+a_{r+1})\cdot a_{r+1}}{a_r}.$
Now $a_{r-1}+a_{r+1}=a_{r-1}+\dfrac{a_{r}^2-1}{a_{r-1}}\implies a_{r-1}+a_{r+1}=\dfrac{a_{r-1}^2-1+a_{r}^2}{a_{r-1}}\implies a_{r-1}+a_{r+1}=\dfrac{(a_{r-2}+a_{r})\cdot a_{r}}{a_{r-1}}$
When i continue this process i obtain $a_{r}+a_{r+2}=\dfrac{(a_0+a_2)a_2}{a_1}$
But answer given is (A), How do i obtain answer as option (A) which depend upon $\;a_{r+1}?\;$
Also how to obtain option (C) and (D)?
| Proof for (A) follows. I've pretty sure induction is the best method; you're already dealing with recurrences, why not add in one more?
I'm going to designate $a_1$ as just $a$ for the sake of simplicity, because it'll show up a lot more than the other $a_r$.
As you already showed, $a_{r+2} = (a_{r+1}^2 - 1)/a_r$. For $r=0$, this is just $a_2 = (a^2-1)/1 = a^2 - 1$.
Since we have an expression in just $a$ for $a_2$, we can see $a_3 = ((a^2-1)^2-1)/a = a^3 - 2a$. We could continue this*, but it's not necessary.
First, we can see quickly that (A) is true for $r=0,1$:
$$a_2 + a_0 = a^2-1+1 = a^2 = a_1 \cdot a_1 \\
a_3 + a_1 = a^3-2a+a = a(a^2-1) = a_1 \cdot a_2$$
Let's do some induction. We want to show that:
$$a_r + a_{r+2} = a \cdot a_{r+1} \implies a_{r+1} + a_{r+3} = a \cdot a_{r+2}$$
Or, having shown $P(0), P(1)$ to be true, we must show that $P(r) \implies P(r+1)$.
I hate big subscripts**, so I'm setting $x = a_r, \ y= a_{r+1}, \ z= a_{r+2}, \ w = a_{r+3}$. This means we want to show:
$$x+z=ay \implies y+w=az$$
Let's begin by ridding ourselves of the $w$. We know $w = (z^2 - 1)/y$, and we can rewrite the RHS of our implication as $y^2+z^2-1 = ayz$. But we also know that $y^2-1=xz$. Making that substitution gives us a RHS of $xz+z^2=ayz$
Now, none of these values can be $0$, as $a_r = 0 \implies a_{r+2} = (a_r^2-1)/0$, and division by zero is a nicely instant contradiction. Since $z \ne 0$, we can divide by it to get $x+z=ay$ on the RHS, which is also the LHS of our implication. Since the LHS and RHS are the same, the implication must be true, and thus $P(r) \implies P(r+1)$.
Because $P(0)$ and $P(1)$ are true, and $P(r) \implies P(r+1)$ is true, $P(r)$ is true for all $r \ge 0$, and statement (A) is proven. $\square$
(Note the importance of not using the LHS as a substitution in the RHS, which would cause logical loops.)
*If we had continued with the resubstitutions, we actually get a very nice pattern:
$$
\begin{array}{l l l l l}
a_0 = & 1 \\
a_1 = & a \\
a_2 = & a^2 & -1 \\
a_3 = & a^3 & -2a \\
a_4 = & a^4 & -3a^2 & +1 \\
a_5 = & a^5 & -4a^3 & +3a \\
a_6 = & a^6 & -5a^4 & +6a^2 & -1 \\
a_7 = & a^7 & -6a^5 & +10a^3 & -4a
\end{array}
$$
The coefficients, if you read going southeast, are the rows of Pascal's Triangle. So column $2$ is the integers, column $3$ is the triangular numbers, etc. It sure looks like each line is:
$$a_r = \binom{r}{0}a^r - \binom{r-1}{1}a^{r-2} + \binom{r-2}{2} a^{r-4} - \binom{r-3}{3}a^{r-6} + \cdots = \sum_{i=0}^{\lfloor r/2 \rfloor} (-1)^i \binom{r-i}{i}a^{r-2i}$$
...but I'm not going to try proving it.
**$\tiny\text{(And I cannot lie)}$
| {
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"url": "https://math.stackexchange.com/questions/4468771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Solving $a^3 - 33 ab^2 = -217$ and $3a^2 b - 11b^3 = 18$? I am looking for ways of solving systems like that:
$$\left\{\begin{array}{lcl} a^3 - 33 ab^2 = -217 \\ 3a^2 b - 11b^3 = 18 \end{array} \right.$$
I've tried turning it into a system of 2 equations with 4 variables but then, having the relations and 4 degrees of freedom makes it equally difficult. Are there any catches one could apply here?
| Taking the resultant of $p=a^3-33ab^2+217$ and $q=3a^2b-11b^3-18$ with respect to $b$ immediately gives
$$
(4a^4 - 14a^3 + 111a^2 + 217a + 961)(4a^2 + 14a + 49)(2a^2 + 7a - 31)(2a - 7)=0.
$$
Hence the only rational solution is
$$
(a,b)=\left( \frac{7}{2},\frac{3}{2}\right).
$$
The quadratic equation $2a^2+7a-31=0$ yields
$$
(a,b)=\left( \frac{-7\pm 3\sqrt{33}}{2},\frac{-33\pm 7\sqrt{33}}{2}\right)
$$
Similarly we can solve the quadratic equation $4a^2 + 14a + 49=0$, which has non-real solutions. The quartic equation has four non-real roots.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$
Solve the following equation: $\sqrt{x+3}+\sqrt{5-x}-2\sqrt{15+2x-x^2}=-4$
Since real solutions are to be found, the domain of $x$ is $[-3; 5]$.
I immediately found that $15+2x-x^2$ can be factored to $(5-x)(x+3)$, this gave:
$\sqrt{x+3}+\sqrt{5-x}-2\sqrt{(5-x)(x+3)}=-4$
Squaring both sides was too complicated so I tried to substitute $a=\sqrt{5-x}$ for $a \in [0; \sqrt{8}]$ and $b=\sqrt{x+3}$ for $b \in [0; \sqrt{8}]$, this gave the new multivariable equation: $a+b-2ab=-4$
Another thing I noticed was $a^2+b^2=8$, together with the above equation, I got this system of equations: $\left\{ \begin{array}{l}
a + b - 2ab = - 4\\
{a^2} + {b^2} = 8
\end{array} \right.$
. Solving this by elimination is quite difficult for me.
This problem needs to be solved using algebra so I wonder how do I continue with this or are there any better way to solve this algebraically?
| Using dxiv's hint.
$a^2+b^2 = (a+b)^2 - 2ab\\
8 = (a+b)^2 - 2ab\\
-2ab = 8 - (a+b)^2$
Substitute into
$a+b - 2ab = -4\\
(a+b) + 8 - (a+b)^2 = -4\\
(a+b)^2 -(a+b) - 12 = 0\\
(a+b - 4)(a+b + 3) = 0$
$a+b>0$ so $a+b = 4$
using this value in the equation above
$-2ab = 8 - 16\\
ab = 4$
$(y-a)(y-b) = y - (a+b)y + (ab) = y^2 -4y +4 = (y-2)(y-2)$
$a = 2, b = 2$ and $x = 1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Matrix of a linear transformation in a different basis Let $A : \mathbb{R_2}[x] \rightarrow \mathbb{R_2}[x]$be a linear transormation that has in a basis $\{1,x,x^2\}$ a given matrix:
$$
\begin{bmatrix}
1 & 1 & 0\\
0 & 1 & 2\\
0 & 0 & 1
\end{bmatrix}
$$
Find the matrix of the transformation in a basis $\{1+x,x+x^2,1+x^2\}$.
Attempted solution:
I've calculated from the given matrix: $A(1) = x^2, A(x) = x^2 + x, A(x^2) = 2x+1$
Then I've calculated from these: $A(1+x) = 2x^2+x, A(x+x^2) = x^2+3x+1, A(1+x^2) = x^2+x+x+1$
I'm stuck here since i can't seem to factor polynomials from the second basis out of all of these expresions. Any help would be appreciated
| I think that you may have mixed up how your basis is acting a little bit. The standard convention is to have the transformation act on a column vector by matrix multiplication from the left.
$$\begin{array}{||c||ccc||}
\hline\hline
&1 &x &x^2 \\
\hline\hline
1 &1&1&0\\
x &0 &1&2\\
x^2 &0&0&1 \\
\hline\hline
\end{array}$$
and we can compute
$$A(1) = \begin{pmatrix}1&1&0\\ 0&1&2\\ 0&0&1\end{pmatrix}\begin{pmatrix}1\\ 0\\ 0\end{pmatrix} = 1.$$
And similarly, $A(x) = 1 + x$ and $A(x^2) = 2x + x^2$. Now we need to express these basis vectors with respect to our new desired basis. If we let $e_1 = 1 + x$, $e_2 = x + x^2$, and $e_3 = 1 + x^2$, then we can see that $1 = \frac{1}{2}(e_1 - e_2 + e_3),\, x = \frac{1}{2}(e_1 + e_2 - e_3),$ and $x^2 = \frac{1}{2}(-e_1 + e_2 + e_3)$. We also have that $A(1) = \frac{1}{2}(e_1 - e_2 + e_3),\, A(x) = e_1,$ and $A(x^2) = \frac{1}{2}(-e_1 + 3e_2 + e_3)$. This means that if we wanted to write the change of basis matrix from our basis $\{1,x,x^2\}$ to the basis $\{1 + x, x + x^2, 1 + x^2\}$, we could represent it by
$$B = \begin{pmatrix}\frac{1}{2}&\frac{1}{2}&-\frac{1}{2}\\ -\frac{1}{2}&\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&-\frac{1}{2}&\frac{1}{2}
\end{pmatrix}$$
(to get this, just note that the columns come from rewriting the old basis vectors in terms of the new basis vectors as shown above). Now, the fastest way to compute the transformation would be to just take any vector that is written with respect to our new basis, transform it back to the original basis, apply $A$, and then transform it back to the desired basis once more. So, we can just conjugate by $B$ to get that the transformation matrix with respect to our new basis is given by
$\tilde{A} = BAB^{-1}$. A quick computation shows that
$$B^{-1} = \begin{pmatrix} 1 & 0 & 1\\ 1 & 1 & 0\\ 0 & 1 & 1\end{pmatrix},$$
and once we multiply everything out, we get the final answer
$$\tilde{A} = \begin{pmatrix}\frac{3}{2}&\frac{3}{2}&1\\-\frac{1}{2}&\frac{3}{2}&1\\\frac{1}{2}&-\frac{1}{2}&0\end{pmatrix}.$$
Fun math note: Something that you might realize after this presentation is that to compute a transformation with respect to any coordinate system you will need to use conjugation. This is not an accident! In fact, with this, we have made a profound observation about the structure of the space itself. In group theory, we can see this as saying something about the set of automorphisms on a group, and in differential geometry, you can use this structure to derive coordinate-independent formulas for geometric structures. Long story short, conjugation is a super common operation that has a lot of nice properties, and you will want to keep it in mind as you move forward in mathematics.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving a closed form of an integral Is there any proof for this integral?$$\int \limits _0^1\frac{1}{a^2x^2+1}\left [\left (1-\frac{x}{2}\ln \frac{1+x}{1-x}\right )^2+\frac{\pi^2x^2}{4}\right ]^{-1}\,dx=\frac{\arctan a}{a-\arctan a}-\frac{3}{a^2},\quad \operatorname{Re}(a)>0.$$I tried substituting $x=\frac{1-x}{1+x}$ but the integral seems to be harder.
| Let $f$ be an analytic function defined on $|z|>1$ via
$$f(z) = 1-\frac{z}{2}\log \frac{1+z^{-1}}{1-z^{-1}}$$
then it's easy to show:
*
*$f$ can be analytic continued to $\mathbb{C}-[-1,1]$.
*$f \sim -1/(3z^2)$ for $|z|$ large.
*For $-1<x<1$,
$$f_\pm (x) := \lim_{y\to 0^\pm} f(x+yi) = 1-\frac{x}{2}\left(\log\frac{1-x}{1+x} \mp \pi i\right) $$
*$f$ has no zero on $\mathbb{C}-[-1,1]$.
The fourth one is actually an important point: it makes the following discussion invalid if we replace $2$ by $3$ in definition of $f$, for example.
For $\varepsilon > 0$ small, integrate $I = \int_\gamma \frac{1}{(z+\varepsilon i)(1+a^2 z^2)f(z)} dz$ with $\gamma$ dog-bone contour.
Integral along big circle is $-3a^{-2}(2\pi i)$, so
$$\begin{aligned} I &= 2\pi i (-\frac{1}{\left(a^2 \epsilon ^2-1\right) f(-i \epsilon )}+\frac{1}{2 f\left(-\frac{i}{a}\right) (a \epsilon -1)}-\frac{1}{2 f\left(\frac{i}{a}\right) (a \epsilon +1)}) \\ &= -\frac{3}{a^2}(2\pi i) + \int_{-1}^1 \frac{1}{(x+\varepsilon i)(1+a^2 x^2)} (\frac{1}{f_+(x)} - \frac{1}{f_-(x)}) dx \\
&= -\frac{3}{a^2}(2\pi i) + \int_{-1}^1 \frac{-\pi i x}{(x+\varepsilon i)(1+a^2 x^2)} \left [\left (1-\frac{x}{2}\ln \frac{1+x}{1-x}\right )^2+\frac{\pi^2x^2}{4}\right ]^{-1} dx\end{aligned}$$
Let $\varepsilon \to 0$ gives
$$\int_{-1}^1 \frac{1}{1+a^2 x^2} \left [\left (1-\frac{x}{2}\ln \frac{1+x}{1-x}\right )^2+\frac{\pi^2x^2}{4}\right ]^{-1} dx = \frac{1}{f\left(\frac{i}{a}\right)}+\frac{1}{f\left(-\frac{i}{a}\right)}-2 - \frac{6}{a^2}$$
which you can verify equals $\frac{2\arctan a}{a-\arctan a}-\frac{6}{a^2}$ for $a>0$.
| {
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Let $x^2 + 3x +1 = 0$. Is $x^{2048} + \dfrac{1}{x^{2048}}$ divisible by 3? Let $x^2 + 3x +1 = 0$. Solve for $x^{2048} + \dfrac{1}{x^{2048}}$. Is it divisible by 3?
$x^2 + 1 = -3x \Rightarrow x+ \dfrac{1}{x} = -3$
$x^2 + \dfrac{1}{x^2} = (x + \dfrac{1}{x})^2 - 2 = 9 -2 = 7$
$x^4 + \dfrac{1}{x^4} = (x^2 + \dfrac{1}{x^2})^2 - 2 = 49 - 2 = 47$
seeing the pattern, let $s_n = x^{2^n} + \dfrac{1}{x^{2^n}}$
I need $s_{11}$
$s_1 = -3$
$s_2 = 7$
$s_3 = 47$
$s_4 = 47^2 - 2 = 2207$
$\vdots$
$((2207^2 -2)^2 - 2)^2 -2)^2 ... - 2)$
quite big
But I realized I didn't have to simplify it, I just have to check if
$((((((2207^2 - 2)^2 -2)^2 -2)^2 -2)^2 -2)^2 - 2)^2 -2$ is divisible by 3
| To get a "closed form" solution, you can do the following:
*
*Simply evaluate roots, and write the answer as $\big(\frac{-3+\sqrt{5}}{2}\big)^{2048}+\big(\frac{-3-\sqrt{5}}{2}\big)^{2048}$
*Or, you could solve the recurrence relation as follows:
$s_n$ is defined in the question. $s_{n+1}=(s_n)^2-2$, where $s_0=-3$.
A solution is: $$s_n=2\cos(\cos^{-1}(-1.5)\cdot2^n)$$
Note the use of the identity $\cos(2\theta)=2\cos^2(\theta)-1$.
To compute $\cos^{-1}(-1.5)$, let $\cos(x+iy)=-1.5$, and proceed ($i$ is the imaginary unit). This gives a closed form.
| {
"language": "en",
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"source": "stackexchange",
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Finding the set of parameters for which the inequality holds for all $x,y$ I encountered the following question about inequalities which I am curious how to solve.
The simplest case is to consider the inequality
$$|x|+|y|+|x+y|+ax+by\geq 0$$
where $x,y,a,b\in\mathbb{R}$.
The question is which values of parameters $a,b\in\mathbb{R}$ would guarantee that the inequality holds for all $x,y\in\mathbb{R}$.
I tried playing with such inequalities and was surprised that this question is harder than I anticipated.
I first found that necessarily $a\in [-2,2]$:
for $x=1$ and $y=0$ we get $2+a\geq 0$ so $a\geq -2$ and for $x=-1$ and $y=0$ we get $2-a\geq 0$ so $a\leq 2$.
Similarly, $b\in [-2,2]$ must hold.
Furthermore, I did find out that if $a=b\in [-2,2]$ then the inequality certainly holds for all $x,y\in\mathbb{R}$, because
$$|x|+|y|+|x+y|+ax+by=(|x|+ax/2)+(|y|+ay/2)+(|x+y|+a(x+y)/2)$$
and each element in the parentheses is non-negative because $a/2\in [-1,1]$ and so $$|z|+az/2=\begin{cases} (1+a/2)z & z\geq 0\\ (-1+a/2)z & z<0\end{cases}\geq 0$$
However, I am sure that there are also other parameters $a,b\in [-2,2]$ for which the inequality holds. For instance $a=2$ and $b=1$ turn out to be a valid solution for all $x,y\in\mathbb{R}$ if you consider the various different cases.
My question is how would I go about finding all such $a,b$?
More generally, is there some theory about such inequalities? Any good reference (paper/book) that is recommend to read in order to better grasp these inequalities?
I am also interested in more general inequalities, with more parameters and variables, for instance
$|x|+|y|+|z|+|x+y|+|x+z|+ax+by+cz\geq 0$
(with the task of finding $a,b,c$ so that it holds for all $x,y,z$).
I am interested to which extent there is a theory of how to solve these questions generally.
| Letting $x = 1, y = 0$, we have
$2 + a \ge 0$.
Letting $x = -1, y = 0$, we have
$2 - a \ge 0$.
Letting $x = 0, y = 1$, we have
$2 + b \ge 0$.
Letting $x = -1, y = 1$, we have
$2 - a + b \ge 0$.
Letting $x = 0, y = -1$, we have
$2 - b \ge 0$
Letting $x = 1, y = -1$, we have
$2 + a - b \ge 0$.
Thus, we have $a, b \in [-2, 2]$ and $|a - b| \le 2$.
$\phantom{2}$
On the other hand, suppose that
$a, b \in [-2, 2]$ and $|a - b| \le 2$.
Let us prove that
$$|x| + |y| + |x + y| + ax + by \ge 0,
\, \forall x, y \in \mathbb{R}.$$
We split into three cases:
Case 1: If $y = 0$, we have
$$\mathrm{LHS} = 2|x| + ax \ge 2|x| - |a|\cdot |x| = (2 - |a|)|x| \ge 0.$$
Case 2: If $y > 0$, since the inequality is homogeneous, WLOG, assume that $y = 1$.
It suffices to prove that
$$|x| + 1 + |x + 1| + ax + b \ge 0, \, \forall x\in \mathbb{R}.$$
(1) If $x \le -1$, we have
$$\mathrm{LHS} = - (2 - a)x + b \ge (2 - a) + b \ge 0.$$
(2) If $-1 < x \le 0$, we have
$$\mathrm{LHS} = 2 + ax + b.$$
If $a \le 0$, we have $2 + ax + b \ge 0$.
If $a > 0$, we have $2 + ax + b \ge 2 - a + b \ge 0$.
(3) If $x > 0$, we have
$$\mathrm{LHS} = (2 + a)x + 2 + b \ge 0.$$
Case 3: If $y < 0$, since the inequality is homogeneous, WLOG, assume that $y = -1$.
It suffices to prove that
$$|x| + 1 + |x - 1| + ax - b \ge 0, \, \forall x\in \mathbb{R}.$$
(1) If $x \le 0$, we have
$$\mathrm{LHS}
= - (2 - a)x + 2 - b \ge 0.$$
(2) If $0 < x \le 1$, we have
$$\mathrm{LHS} = 2 + ax - b.$$
If $a \ge 0$, we have $2 + ax - b \ge 0$.
If $a < 0$, we have $2 + ax - b \ge 2 + a - b \ge 0$.
(3) If $x > 1$, we have
$$\mathrm{LHS}
= (2 + a)x - b \ge 2 + a - b \ge 0.$$
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Finding $\lim _{n\to \infty}n\sin (2\pi \sqrt{1+n^2})$ Question:
Find $\lim \limits _{n\to \infty}n\sin \left (2\pi \sqrt{1+n^2}\right )$, where $n\in \mathbb{N}$.
Context:
This was a question given in my workbook, where it mentioned a hint that $n$ belonging to natural numbers is an important piece of information.
Approach:
I tried expanding the inner part using binomial expansion and then using Taylor series to expand the $\sin (\cdot )$ out, but I seem to keep hitting an indeterminate form. Help in understanding would be much appreciated!
| As said in comment and answers $$\sin \left(2 \pi \sqrt{n^2+1}\right)=\sin \left(2 \pi \sqrt{n^2+1}-2 \pi n\right)$$
When $n$ is large
$$\sqrt{n^2+1}-n=\frac{1}{2 n}-\frac{1}{8 n^3}+\frac{1}{16
n^5}+O\left(\frac{1}{n^7}\right)$$
$$\sin \left(2 \pi \sqrt{n^2+1}\right)=\frac{\pi }{n}-\frac{\pi \left(3+2 \pi ^2\right)}{12 n^3}+\frac{\pi
\left(15+15 \pi ^2+\pi ^4\right)}{120
n^5}+O\left(\frac{1}{n^7}\right)$$
$$n\sin \left(2 \pi \sqrt{n^2+1}\right)=\pi -\frac{\pi \left(3+2 \pi ^2\right)}{12 n^2}+\frac{\pi \left(15+15 \pi
^2+\pi ^4\right)}{120 n^4}+O\left(\frac{1}{n^6}\right)$$ which, for sure, gives the desired limit.
But, it also gives a simple way to approximate the value of the sine.
Just use $n=7$; the truncated series gives, as an approximation,
$$\sin \left(10 \sqrt{2} \pi \right) \sim \frac{\pi \left(286665-965 \pi ^2+\pi ^4\right)}{2016840}=0.4318486$$ while the exact value is $0.4318404$
| {
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Optimize $xyz$ where $x+y+z=1$ and $x^2+y^2+z^2=1$? Im trying to optimize
$f(x,y,z)=xyz$ restricted to $g(x,y,z)=x+y+z=1$ and $h(x,y,z)=x^2+y^2+z^2=1$.
$∇f=(yz,xz,xy)$, $∇g=(1,1,1)$ and $∇h=(2x,2y,2z)$.
I tried using the determinant $det(∇f,∇g,∇h)=yz(2z-2y)-xz(2z-2x)+xy(2y-2x)=0$ which I dont know what to do with and I cant simplify the determinant in a good way with row operations.
I also tried solving $z$ from $g=1$. $z=1-x-y$.
$f(x,y,1-x-y)=xy-x^2y-xy^2$ restricted to $h(x,y)=2x^2+2y^2-2x-2y+2xy+1$ with Lagrange multiplier but I made no progress there either as the partial derivatives got too messy.
| Notice that $$1=1^2=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xy)=1+2(xy+yz+xy)$$
Solving for the products, we learn all elementary symmetric polynomials in $(x,y,z)$: \begin{align*}
x+y+z&=1 \\
xy+xz+yz&=0 \\
xyz&=f
\end{align*} where $f$ is what we want to maximize. By Vietá's formulae, $x$, $y$, and $z$ are the roots of $$g(t)=t^3-t^2-f=0$$
We must maximize $f$ so that $g$ has $3$ real roots. To do this, analyze $g$ geometrically.
At local extrema, $$0=g'(t)=3t^2-2t=t(3t-2)$$ There are two extrema: one at $0$ and one at $\frac{2}{3}$. Since $g(t)$ is increasing at large $|t|$, the former extremum is a maximum; the latter a minimum.
Thus $g$ has three real roots iff $$g(0)\geq0\geq g\left(\frac{2}{3}\right)$$ From the left-hand inequality, $f\leq0$, which is clearly sharp (take $x=y=0$, $z=1$).
| {
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Show that $(1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}}$ is $1$ as $(x,y) \to(0,0)$ I want to show that:
$$\lim_{(x,y)\to(0,0)} (1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}} = 1$$
Direct substitution doesn't work, because of the denominator of $\frac{1}{x^2 + y^2}$.
Usually these problems are solved by bounding the following expression:
$$\left|(1 + x^2 y^2)^{-\frac{1}{x^2 + y^2}} -1 \right| = \left| \frac{1 - (1 + x^2 y^2)^{\frac{1}{x^2 + y^2}} }{ (1 + x^2 y^2)^{\frac{1}{x^2 + y^2}}} \right|$$
I don't know how to continue.
| An alternative using polar coordinates:
Taking $x=r\cos \theta\;,y=r\sin\theta$
$$\lim_{r\to 0}(1+r^4\sin^2\theta\cos^2\theta)^{-1/r^2}=e^{\lim_{r\to 0}\frac{-1}{r^2}(r^4\sin^2\theta\cos^2\theta)}=e^{\lim_{r\to 0}(-r^2\sin^2\theta\cos^2\theta)}=e^0=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Follow up question to question involving $\lim_{n\to \infty}\frac{\int_0^1 \left(x^2-x-2\right)^n \, dx}{\int_0^1 \left(4 x^2-2 x-2\right)^n \, dx}$ I tried to answer Q 4482921 by elementary calculus but got stuck :
The first step in my solution is to replace $x$ by $2u$ so that $dx=2du$. The integral becomes $$\lim_{n\to \infty}\frac{\int_0^1 \left(x^2-x-2\right)^n \, dx}{\int_0^1 \left(4 x^2-2 x-2\right)^n \, dx}= \lim_{n\to \infty}\frac{\int_0^\frac{1}{2}\left(4u^2-2u-2\right)^n \ 2du}{\int_0^1 \left(4 x^2-2 x-2\right)^n \, dx}$$$$=2\left(1-\lim_{n\to \infty}\frac{\int_
\frac{1}{2}^1 \left(4x^2-2x-2\right)^n \, dx}{\int_0^1 \left(4 x^2-2 x-2\right)^n \,
dx}\right)$$
Now I need to prove that $\displaystyle\lim_{n\to \infty}\frac{\int_
\frac{1}{2}^1 \left(4x^2-2x-2\right)^n \, dx}{\int_0^1 \left(4 x^2-2 x-2\right)^n\, dx} =0.$ It certainly looks like it in the graph, and I also found that $$\int_{\frac 12}^1 \left(4 x^2-2 x-2\right)^n\, dx= \int_{-1}^{0} \left(4 x^2-2 x-2\right)^n\, dx.$$
| $f(x)=x^2-x-2$ has a minimum at $x=1/2,$ decreases for $x<1/2$ and increases for $x>1/2.$
In particular, $f(x)<0$ on $(0,2).$
So for $1<2x<2,$ $0=f(2)>f(2x)>f(1)=-2.$ So:
$$\left|\int_{1/2}^1 f(2x)^n\,dx \right|\leq 2^{n-1}.$$
On the other hand, when $\frac16<x<\frac13,$ $f(2x)\leq f(1/3)=f(2/3)=-\frac{20}{9}.$ So:
$$\left|\int_{0}^1 f(2x)^n\,dx \right|\geq \frac16\left(\frac{20}9\right)^n.$$
So the absolute value of your quotient is less than: $$3\left(\frac{9}{10}\right)^n,$$ which converges to $0.$
We could have picked any symmetric interval around $\frac{1}{4}$ of length less that $\frac{1}{4},$ $(1/6,1/3)$ was just the simplest.
More generally, if $h(x)$ is a continuous non-negative function on $[a,b]$ which is not zero everywhere, and with all its maximum values taken in $[a,c)$ for some $c\in (a,b),$ then $$\frac{\int_c^b h(x)^n\,dx}{\int_a^b h(x)^n\,dx}\to 0.$$
Essentially, if $M=\sup_{x\in [a,b]} f(x)$ is the maximum, and $M_1=\sup_{x\in [c,b]} f(x)$ satisfies $M_1<M,$ then for some interval $(u,v)$ in $(a,c)$ we have the $h(x)>\frac{M_1+M}{2}$ for $x\in(u,v)$ and this:
$$\int_{c}^b f(x)^n\,dx \leq (b-c)M_1^n.$$
and:
$$\int_{a}^b f(x)^n\,dx \geq (v-u)\left(\frac{M_1+M}2\right)^n.$$
So you quotient is bounded above by $$\frac{b-c}{v-u}\left(\frac{2M_1}{M+M_1}\right)^n$$
We have $0<\frac{2M_1}{M+M_1}<1,$ so this goes to zero.
In your case, $h(x)=2+2x-4x^2,$ $a=0,b=1,c=1/2,$ $M=2\frac12$ at $x=1/4,$ and $M_1=h(1/2)=2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$ \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx $ $$ \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx $$
is there more efficient and elegant way? Other than the solution below.
a solution:
Let $ I = \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx $,
$$ \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx = \int \frac{\sin^{8}(x) + \cos^{8}(x) - \cos^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx = x - \int \frac{\cos^{8}(x)}{ \sin^{8}(x) + \cos^{8}(x) }dx $$
so we have $2I$ is:
$$ 2I = x + \int \frac{\sin^{8}(x)-\cos^{8}(x)}{ \sin^{8}(x) + \cos^{8}(x) }dx $$
Now,
$$ \sin^{8}(x) - \cos^{8}(x) = \left[ \sin^{4}(x) - \cos^{4}(x) \right] \left[ \sin^{4}(x) + \cos^{4}(x) \right] $$
$$ = \left[ \sin^{2}(x) - \cos^{2}(x) \right]\left[ \sin^{2}(x) + \cos^{2}(x) \right] \left[ (\sin^{2}(x) + \cos^{2}(x))^{2} - 2 \sin^{2}(x)\cos^{2}(x) \right] $$
$$ = \left[ \sin^{2}(x) - \cos^{2}(x) \right] \left[1 - 2 \sin^{2}(x)\cos^{2}(x) \right] $$
For the denominator:
$$ \sin^{8}(x) + \cos^{8}(x) = \left[ \sin^{4}(x) + \cos^{4}(x) \right]^{2} - 2 \sin^{4}(x)\cos^{4}(x) $$
$$ = \left[ (\sin^{2}(x) + \cos^{2}(x))^{2} - 2 (\sin(x)\cos(x))^{2} \right]^{2} - 2 \sin^{4}(x)\cos^{4}(x) $$
$$ = \left[1 - 2 (\sin(x)\cos(x))^{2} \right]^{2} - 2 (\sin(x)\cos(x))^{4} $$
Now
$$ 2I =x+ \int \frac{ \left[ \sin^{2}(x) - \cos^{2}(x) \right] \left[1 - 2 \sin^{2}(x)\cos^{2}(x) \right] }{\left[1 - 2 (\sin(x)\cos(x))^{2} \right]^{2} - 2 (\sin(x)\cos(x))^{4}} dx $$
Let $U = \sin(x) \cos(x)$, then $U'(x) = \cos^{2}(x) - \sin^{2}(x)$, so
$$ 2I = x- \int \frac{\left[1 - 2 U^{2} \right] }{\left[1 - 2 U^{2} \right]^{2} - 2 U^{4}} dU $$
$$ = x- \int \frac{\left[ 2 U^{2} - 1 \right] }{ (1 - (2 + \sqrt{2})U^{2})(1 + (\sqrt{2} - 2)U^{2})} dU $$
And from here I can continue using Partial Fraction:
$$- \int \frac{\left[ 2 U^{2} - 1 \right] }{ (1 - (2 + \sqrt{2})U^{2})(1 + (\sqrt{2} - 2)U^{2})} dU = -\frac{1}{2}\int \frac{1}{ (1 - (2 + \sqrt{2})U^{2})} + \frac{1}{(1 - (2 - \sqrt{2})U^{2})} dU $$
and then for each of the 2 terms we use Partial Fraction again. First, let $\alpha = \sqrt{2+\sqrt{2}}U$, $\beta = \sqrt{2-\sqrt{2}}U$, so we have
$$ = -\frac{1}{2\sqrt{2+\sqrt{2}}}\int \frac{1}{ 1 - \alpha^{2}} d \alpha - \frac{1}{2\sqrt{2-\sqrt{2}}} \int \frac{1}{(1 - \beta^{2})} d \beta $$
then we have
$$ = -\frac{1}{2\sqrt{2+\sqrt{2}}} \left( \int \frac{1/2}{1-\alpha} d \alpha + \int \frac{1/2}{1+\alpha} d\alpha \right) - \frac{1}{2\sqrt{2-\sqrt{2}}} \left( \int \frac{1/2}{1-\beta } d \beta + \int \frac{1/2}{1 + \beta } d \beta \right) $$
$$ = -\frac{1}{4\sqrt{2+\sqrt{2}}} \left( -\ln(1-\alpha) + \ln(1+\alpha) \right) - \frac{1}{4\sqrt{2-\sqrt{2}}} \left( -\ln(1-\beta) + \ln(1+\beta) \right) $$
| Another option would be to enforce the substitution $x=\frac {\pi}2-y$ and rewrite each sine and cosine term using their respective double angle identity.
The numerator is a direct application of
$$\cos 2y=2\cos^2y-1$$
Whereas the denominator can be expressed using a bit of algebraic manipulation
$$\begin{align*}\sin^8 y+\cos^8 y & =\left(\sin^4 y+\cos^4 y\right)^2-2\sin^4 y\cos^4 y\\ & =\left(1-\frac 12\sin^2 2y\right)^2-\frac 18\sin^4 2y\\ & =\color{red}{1-\sin^2 2y+\frac 18\sin^4 2y}\\ & =\color{blue}{\frac 18+\frac 34\cos^2 2y+\frac 18\cos^4 2y}\end{align*}$$
The colored expressions will be the relations used in the computation below
$$\begin{align*}I & =-\int\frac {\cos^8 y}{\sin^8y+\cos^8y}\,\mathrm dy\\ & =-\frac 12\int\frac {(1+\cos 2y)^4}{1+6\cos^22y+\cos^42y}\,\mathrm dy\\ & =-\frac y2-2\int\frac {\cos 2y(2-\sin^22y)}{8-8\sin^22y+\sin^42y}\,\mathrm dy\\ & =-\frac y2-\int\frac {2-v^2}{8-8v^2+v^4}\,\mathrm dv\end{align*}$$
Where the substitution $v=\sin2y$ was made between the penultimate and the ultimate line. The remaining integral can be tackled in whatever way the reader sees fit.
Note that for the general case of the integral
$$J=\int\frac {\mathrm dv}{v^4+av^2+b^2}$$
The result can be obtained as such
$$\begin{align*}J & =\frac 12\int\frac {2+\frac b{v^2}-\frac b{v^2}}{v^2+\frac {b^2}{v^2}+a}\,\mathrm dv\\ & =\frac 12\int\frac {1+\frac b{v^2}}{\left(v-\frac bv\right)^2+a+2b}\,\mathrm dv+\frac 12\int\frac {1-\frac b{v^2}}{\left(v+\frac bv\right)^2+a-2b}\,\mathrm dv\\ & =\frac 12\int\frac {\mathrm dw_-}{w_-^2+a+2b}+\frac 12\int\frac {\mathrm dw_+}{w_+^2+a-2b}\end{align*}$$
Where the sign of the $a\pm 2b$ term determines whether the two resultant integrals can be expressed in terms of the arctangent function or natural logarithms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find $n \in \mathbb{N}$ such that $(-2\sqrt{3}+i)^n=2^{n-1}(1-i\sqrt{3})$ I was requested to find all $n \in \mathbb{N}$ such that $(-2\sqrt{3}+i)^n=2^{n-1}(1-i\sqrt{3})$. My approach was
$$(-2\sqrt{3}+i)^n=2^{n-1}(1-i\sqrt{3})$$
$$\implies 2^n(-\sqrt{3}+i)^n= \frac{2^n(1-i\sqrt{3})}{2}$$
$$\implies (-\sqrt{3}+i)^n = \frac{1-i\sqrt{3}}{2}$$
However, I was unable to proceed from here. Am I missing something? Thanks in advance.
| Suppose $(-2\sqrt{3}+i)^n=2^{n-1}(1-i\sqrt{3})$, then_
$$(-2\sqrt{3}+i)^n=2^{n-1}(1-i\sqrt{3})$$
$$\implies 2(-\sqrt{3}+i)^n= 2^n(1-i\sqrt{3})$$
$$\implies (\frac{-\sqrt{3}+i}{2})^n = \frac{1-i\sqrt{3}}{2}$$
Take $z=\frac{-\sqrt{3}+i}{2}$. Then $\frac{1-i\sqrt{3}}{2}=i\bar{z}$
So we have the complex equation
$$z^n=i\bar{z}$$
Since $\vert z\vert =1$, we have $z^{n+1}=i$. Can you conclude from here?
Edit: Suppose there exist some $n\in\mathbb{N}$ such the inequality holds. Applying the modulus both sides we Will get that $\sqrt{13}^n=2^n$, which is clearly impossible
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$a+b+c=1$, prove $(1+a)(1+b)(1+c) \ge 8(1-a)(1-b)(1-c) $, without AM-GM $a,b,c \in \mathbb{R}^{+}$, if $a+b+c=1$, prove that
$$ (1+a)(1+b)(1+c) \ge 8(1-a)(1-b)(1-c) $$
but without AM-GM as the only tool.
Collecting data:
Since $a+b+c=1$, we cannot have one of the variable to be greater than or equal to 1. So
$$ \boxed{0 < a,b,c < 1} $$
Also by AM-GM: $1=a+b+c \ge 3(abc)^{1/3}$, or
$$ \boxed{abc \le \frac{1}{27}} $$
Next,
$$ (1-a)(1-b)(1-c) = [1 - (a+b) + ab](1-c) = \boxed{ 1 - (a+b+c) + (ab + ac + bc) - abc }$$
$$ =\boxed{ (ab+ac+bc)-abc} $$
Next,
$$ (1+a)(1+b)(1+c) = [1 + (a+b) + ab](1+c) = 1 + (a+b+c) + (ac+ab+bc) + abc$$
$$ = \boxed{2 + (ac+bc+bc) + abc} $$
so the inequality is equivalent with:
$$ 2 + ac+bc+ab + abc \ge 8(ab+ac+bc) - 8abc $$
or
$$ \boxed{ 2 + 9abc \ge 7 (ab+ac+bc) } \:\: \leftarrow \:\: \text{to be proven} $$
Next, $(a+b+c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab+ac+bc) = 1$,
$$ \boxed{ab+ac+bc = \frac{1- (a^{2} + b^{2}+ c^{2})}{2}} $$
which means
$$ 7(ab+ac+bc) = \frac{7}{2} -\frac{7}{2}(a^{2}+b^{2}+c^{2}) = 2 + \frac{3 - 7(a^{2}+b^{2}+c^{2})}{2}$$
and it is left to prove
$$ \frac{3-7(a^{2}+b^{2}+c^{2})}{2} \le 9abc $$
$$ \boxed{3 \le 18 abc + 7(a^{2}+b^{2}+c^{2})} \:\: \leftarrow \:\: \text{to be proven}$$
|
$a+b+c=1.$ Prove that $(1+a)(1+b)(1+c)\geq 8(1-a)(1-b)(1-c).$
ETS) $(2a+b+c)(a+2b+c)(a+b+2c) \geq 8(a+b)(b+c)(c+a).$
Let $b+c=x, c+a=y, a+b=z.$
$\Leftrightarrow (x+y)(y+z)(z+x) \geq 8xyz.$
You may try from here.
Hint: $(x+y)^2\geq4xy.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Finite Power Sum Derivation TL;DR: "I've made something work but I do not know why it works - please explain it to me (or disprove my finding)"
DISCLAIMER: A lot of things here are done in a non-rigorous way - which is all the more surprising that in the end it "all works".
Let's define $f_m(n) = \sum_{k=0}^nk^m$ . For example, $f_1(n) = \sum_{k=0}^nk$ which is just a triangular numbers summation. What I wanted to find is the "closed" form for $f_m(n)$ meaning a polynomial form. There are two things I used here as a "mathematical intuition":
*
*The resulting polynomial must be of m+1 degree. That's because we're adding the m-th powers of n exactly n times
*The differentiation "works" for discrete sums (with some "adjustments" - probably the crux of this question)
Assuming these two points I came out with an idea that - well, $f'_{m+1}(n) = (m+1)\sum_{k=0}^nk^m = (m+1)f_m(n)$ so we can "integrate" $f_m(n)$ to get the expression for $f_{m+1}(n)$ . However if used directly, it will not yield correct results. For instance, if we use it on $f_0(n)$ and $f_1(n)$ we run into a problem: $\int{f_0(n)} = \int{n} = \frac{n^2}{2}$ (I intentionally don't write "dn" here as it makes no sense). As we know the triangular sum is $\frac{n(n+1)}{2} = \frac{n^2}{2} + \frac{n}{2}$, i.e. we are missing a term here.
But wait - when we're integrating, we have to also add a "constant term" - which in this case I will denote $L(n)$. The $L(n)$ is a linear function of $n$, i.e. $L(n) = An+B$ - furthermore, since for all m we have $f_m(0) = 0$ we get that $B = 0$ so $L(n) = An$ for some number $A$. To get the number $A$ we just substitute the $n=1$ to a given $f_m(n)$ and work out what it should be. For example, $f_1(n) = \frac{n^2}{2} + L(n)$, so $L(1) = A\cdot 1 = f_1(1) - \frac{1^2}{2}$ and since $f_1(1) = 1$ we get that $A = \frac{1}{2}$ thus a "correct" expression is: $f_1(n) = \frac{n^2}{2} + L(n) = \frac{n^2}{2} + \frac{n}{2}$ which is exactly what we would expect. How convenient!
Now if all of this sounds like borderline nonsense to you - that's okay - because I also think it is. But the interesting thing is - it works for any $f_m(n)$ (at least I think it does, I've checked it for several higher values). For example:
$$f_1(n) = \frac{n^2}{2} + \mathbf{\frac{n}{2}}
\\f_2(n) = \frac{n^3}{3} + \frac{n^2}{2} + \mathbf{\frac{n}{6}}
\\f_3(n) = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4} + \mathbf{0}
\\f_4(n) = \frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} \mathbf{- \frac{n}{30}}
\\f_5(n) = \frac{n^6}{6} + \frac{n^5}{2} + \frac{5n^4}{12} - \frac{n^2}{12} + \mathbf{0}$$
So, in general, $f_{m+1}(n) = (m+1)\int{f_m(n)} + L(n)$ and of course, in cases when we find that $A=0$ we obviously get $L(n)=0$
So - why does it work? The method that I used is .. dubious to say the least. There was also no indication that $L(n)$ is a linear function - because for instance, for cases like $f_4(n)$ we have our sum of known members terminating at a 3rd power (i.e. why can't $L(n)$ be a quadratic?)
Or maybe it doesn't work at all and the expressions I found are just a lucky coincidence?
| Here we use some results from the calculus of finite differences. At first we derive discrete versions of differentiation and integration. Then we use these results to calculate
\begin{align*}
\color{blue}{f_m(n)=\sum_{k=0}^nk^m}
\end{align*}
Difference operator $\Delta$:
Since $f_m(n)=\sum_{k=0}^nk^m$ are polynomials in $n$, we take the vector space of polynomials over the field $\mathbb{R}$ and consider the function $f(x)=x^n$ and the differential operator $D$ which we apply on $f$. We obtain
\begin{align*}
f(x)=x^n\qquad\qquad\qquad Df(x)=Dx^n=nx^{n-1}\tag{1.1}
\end{align*}
A discrete analogon of the differential operator $D$ is the forward difference operator $\Delta$ defined as
\begin{align*}
\color{blue}{\Delta f(x)=f(x+1)-f(x)}
\end{align*}
The role of $x^n$ in the continuous case is given in the discrete case by the falling factorial $x^{\underline{n}}=x(x-1)\cdots (x-n+1)$, since
\begin{align*}
\color{blue}{\Delta x^{\underline{n}}}&=(x+1)^{\underline{n}}-x^{\underline{n}}\\
&=(x+1)x\cdots(x+1-n+1)\\
&\qquad-x(x-1)\cdots(x-n+1)\\
&=x(x-1)\cdots(x-n+2)\left(x+1-\left(x-n+1\right)\right)\\
&=nx(x-1)\cdots(x-n+2)\\
&\,\,\color{blue}{=nx^{\underline{n-1}}}\tag{1.2}
\end{align*}
we have with (1.2) a convenient discrete pendant to (1.1).
\begin{align*}
g(x)=x^{\underline{n}}\qquad\qquad\qquad \Delta g(x)=\Delta x^{\underline{n}}=nx^{\underline{n-1}}\tag{1.3}
\end{align*}
Summation operator $\sum$:
Next we derive a discrete version of the Fundamental theorem of calculus. We consider functions $F$ and $f$ given in two flavors: $\Delta F=f$ resp. $DF=f$
\begin{align*}
\Delta F&=f\qquad\mathrm{domain }\ \mathbb{N}& \Longleftrightarrow &\qquad F(n)-F(0)=\sum_{0\leq k<n}f(k)\tag{2.1}\\
D F &=f\qquad\mathrm{domain }\ \mathbb{R}& \Longleftrightarrow &\qquad F(x)-F(0)=\int_{0}^xf(t)\,dt\tag{2.2}\\
\end{align*}
Just to make the right-hand side of (2.1) plausible we consider from the left-hand side of (2.1)
\begin{align*}
\Delta F(n-1)&=F(n)-F(n-1)=f(n-1)\\
\Delta F(n-2)&=F(n-1)-F(n-2)=f(n-2)\\
&\ \ \vdots\\
\Delta F(1)&=F(2)-F(1)=f(1)\\
\Delta F(0)&=F(1)-F(0)=f(0)\\
\end{align*}
and obtain by adding up these rows
\begin{align*}
&\color{blue}{F(n)-F(0)=\sum_{0\leq k<n}f(k)}\tag{3.1}\\
\end{align*}
which can be seen as discrete analogon of the fundamental theorem of calculus. Donald Knuth introduced the following notation $\sum_{a}^{b}f(x)\,\delta x$ which makes the analogy even more evident
\begin{align*}
\int_{a}^b f(t)\,dt&=F(x)\big|_{a}^{b}=F(b)-F(a)\\
\color{blue}{\sum_{a}^{b}f(x)\,\delta x}&\color{blue}{:= F(x)\big|_{a}^{b}}=F(b)-F(a)\tag{3.2}\\
\end{align*}
Using (3.1) and (3.2) we can write
\begin{align*}
\color{blue}{\sum_{0}^n f(x)\delta x}&=F(x)\big|_{0}^{n}=F(n)-F(0)=\sum_{0\leq k<n}f(k)\\
&\,\,\color{blue}{=\sum_{k=0}^{n-1}f(k)}\tag{3.3}
\end{align*}
We are now well prepared to calculate $f_m(n)=\sum_{k=0}^nk^m$.
Calculation of $f_m(n)$:
We obtain from (3.3) and (1.2)
\begin{align*}
\color{blue}{S_m(n)=\sum_{0\leq k<n}k^{\underline m}}&=\sum_{0}^{n}x^{\underline{m}}\delta x
=\sum_{0}^n\frac{\Delta x^{\underline{m+1}}}{m+1}\delta x\color{blue}{=\frac{n^{\underline{m+1}}}{m+1}}\tag{4.1}
\end{align*}
which is a discrete analogon to
\begin{align*}
\int_{0}^xt^m\,dt=\frac{x^{m+1}}{m+1}
\end{align*}
and can now iteratively calculate $f_m(n)$ as follows:
*
*$f_1(n)=\sum_{k=0}^n k$
Since $k=k^{\underline{1}}$ we obtain according to (4.1)
\begin{align*}
\color{blue}{f_1(n)}&=\sum_{k=0}^n k^1=\sum_{k=0}^n k^{\underline{1}}=S_1(n+1)\\
&=\frac{1}{2}(n+1)^{\underline{2}}=\frac{1}{2}(n+1)n\\
&\,\,\color{blue}{=\frac{n^2}{2}+\frac{n}{2}}
\end{align*}
*$f_2(n)=\sum_{k=0}^n k^2$
Since $k^2=k(k-1)+k=k^{\underline{2}}+k^{\underline{1}}$ we obtain according to (4.1) and (5.1)
\begin{align*}
\color{blue}{f_2(n)}&=\sum_{k=0}^n k^2
=\sum_{k=0}^n k^{\underline{2}}+\sum_{k=0}^n k^{\underline{1}}
=S_2(n+1)+S_1(n+1)\\
&=\frac{1}{3}(n+1)^{\underline{3}}+\frac{1}{2}\left(n^2+n\right)\\
&=\frac{1}{3}(n+1)n(n-1)+\frac{1}{2}\left(n^2+n\right)\\
&\,\,\color{blue}{=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}}
\end{align*}
This way we can iteratively calculate $f_m(n), m=3,4,\ldots$. Here it is useful to know that
\begin{align*}
k^m=\sum_{j=0}^m\begin{Bmatrix} m \\ j \end{Bmatrix} k^{\underline{j}}
\end{align*}
with $\begin{Bmatrix} m \\ j \end{Bmatrix}$ the Stirling numbers of the second kind.
Note: This answer follows closely chapter 1 Differenzen und Summen in Konkrete Analysis by J. Cigler.
| {
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"url": "https://math.stackexchange.com/questions/4493382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
(Absolute) convergence of $\sum_{k=1}^\infty \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1}}$ We want to check if the following series converges (absolutely).
$$\sum_{k=1}^\infty \frac{\sqrt{k+1} - \sqrt{k}}{\sqrt{k+1}}$$
This is what I have so far:
\begin{align}
\sum_{k=1}^\infty\frac{\sqrt{k+1}-\sqrt k}{\sqrt{k+1}}
&=\sum_{k=1}^\infty\frac{(\sqrt{k+1}-\sqrt k)(\sqrt{k+1}+\sqrt k)}{\sqrt{k+1}(\sqrt{k+1}+\sqrt k)}\\
&=\sum_{k=1}^\infty\frac1{\sqrt{k+1}(\sqrt{k+1}+\sqrt k)}\\
&= \sum_{k=1}^\infty \frac{1}{k+1+\sqrt{k^2+k}}
\end{align}
But how do we continue from here?
| We have
$$\frac{1}{k+1+\sqrt{k^2+k}} > \frac{1}{k+1+\sqrt{k^2+2k+1}} = \frac{1}{2(k+1)} .$$
Thus the comparison test shows that $\sum_{k=1}^\infty\frac{\sqrt{k+1}-\sqrt k}{\sqrt{k+1}}$ diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Showing $\sum_{cyc} \frac{\cos(\frac{\alpha+\beta}{2})}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}=2$ when $\alpha+\beta+\gamma=\pi$ I saw this problem in a math magazine:
Let $\alpha,\beta$ and $\gamma$ be the angles of a triangle, so that
$$
\alpha+\beta+\gamma=\pi
$$
Show that
$$
\frac{\cos\left(\frac{\alpha+\beta}{2}\right)}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}+\frac{\cos\left(\frac{\alpha+\gamma}{2}\right)}{\cos \frac{\alpha}{2}\cos \frac{\gamma}{2}}+
\frac{\cos\left(\frac{\gamma+\beta}{2}\right)}{\cos \frac{\gamma}{2}\cos \frac{\beta}{2}}=2
$$
I tried to rewrite $\alpha+\beta$ with $\pi-\gamma$ so that
$$
\cos\left(\frac{\alpha+\beta}{2}\right)=\cos\left(\frac{\pi}{2}-\frac{\gamma}{2}\right)= \cos \frac{\pi}{2}\cos \frac{\gamma}{2}+ \sin\frac{\pi}{2}\sin \frac{\gamma}{2}= \sin\frac{\gamma}{2}
$$
Now the above equation equals
$$
\frac{\sin\frac{\gamma}{2}}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}+\frac{\sin\frac{\beta}{2}}{\cos \frac{\alpha}{2}\cos \frac{\gamma}{2}}+
\frac{\sin\frac{\alpha}{2}}{\cos \frac{\gamma}{2}\cos \frac{\beta}{2}}=2
$$
But now I don't have ideas for continuation. Any hints to tackle this?
|
$$
\frac{\sin\frac{\gamma}{2}}{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}}+\frac{\sin\frac{\beta}{2}}{\cos \frac{\alpha}{2}\cos \frac{\gamma}{2}}+
\frac{\sin\frac{\alpha}{2}}{\cos \frac{\gamma}{2}\cos \frac{\beta}{2}}=2
$$
I start from your work, (Note the LHS of your equation in the last line should be all positive sum). First, do common denominator and use $~~\sin(\frac{x}2)\cos(\frac{x}2)=\frac{1}2\sin(x)$
$$\begin{align}
LHS&=\frac{\frac{1}{2}\cos \frac{\alpha}{2}\cos \frac{\beta}{2}\cos \frac{\gamma}{2}\sin\gamma
+\frac{1}{2}\cos \frac{\alpha}{2}\cos \frac{\beta}{2}\cos \frac{\gamma}{2}\sin\beta
+\frac{1}{2}\cos \frac{\alpha}{2}\cos \frac{\beta}{2}\cos \frac{\gamma}{2}\sin\alpha }{\cos^2 \frac{\alpha}{2}\cos^2 \frac{\beta}{2}\cos^2 \frac{\gamma}{2}}\\
\\
&=\frac{\frac{1}{2}\sin\gamma
+\frac{1}{2}\sin\beta
+\frac{1}{2}\sin\alpha }{\cos \frac{\alpha}{2}\cos \frac{\beta}{2}\cos \frac{\gamma}{2}}=\frac{\sin\alpha+\sin\beta+\sin(\alpha+\beta)}{2\cos\frac{\alpha}2\cos\frac{\beta}2\sin\frac{\alpha+\beta}2}\\
\end{align}$$
Now we work on the denominator
$$\begin{align}
2\cos\frac{\alpha}2\cos\frac{\beta}2\sin\frac{\alpha+\beta}2&=2\cos\frac{\alpha}2\cos\frac{\beta}2\left(\sin\frac{\alpha}2\cos\frac{\beta}2+\cos\frac{\alpha}2\sin\frac{\beta}2\right)\\
\\
&=\sin\alpha\cos^2\frac{\beta}2+\sin\beta\cos^2\frac{\alpha}2\\
\\
&=\sin\alpha~\frac{1+\cos\beta}2+\sin\beta~\frac{1+\cos\alpha}2\\
\\
&=\frac{1}2\left( \sin\alpha+\sin\beta+\sin(\alpha+\beta)\right)
\end{align}$$
$$$$
$$\Rightarrow LHS=2=RHS$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Without a calculator, determine whether chords $AB, AC \text{ and }BD$ can divide a circle into five equal-area regions. Without a calculator, determine whether chords $AB, AC \text{ and }BD$ can divide a circle into five equal-area regions.
Context: I just made up this question.
I have only been able to answer the question with a calculator. Surely there must be some clever way to answer the question without a calculator.
Answer that uses a calculator:
Proof by contradiction: assume the answer is yes.
Let radius $=1$.
Call the centre $O$.
$\alpha=\angle{AOB}$
$\beta=\angle{AOC}$
$\text{Area}_{\text{minor segment }AB}=\dfrac12(\alpha-\sin{\alpha})=\dfrac{\pi}{5}\implies \alpha\approx2.1131$
$\text{Area}_{\text{segment }ADC}=\dfrac12(\beta-\sin{\beta})=\dfrac{2\pi}{5}\implies \beta\approx2.8248$
$\text{Area}_{\text{triangle}}=\dfrac{\pi}{5}\text{ and }AB=2\sin{\dfrac{\alpha}{2}}\implies ...\implies \angle{BAC}=\arctan{\left(\dfrac{\pi}{5\sin^2{\dfrac{\alpha}{2}}}\right)}\approx{0.692}$
But $\angle{BAC}=\dfrac12{\angle{BOC}}=\dfrac12{\left(2\pi-\alpha-\beta\right)}\approx{0.673}$, contradiction.
Therefore the answer is no.
UPDATE: I found essentially the same question, except it does not request a calculator-free solution. So it doesn't answer my question.
|
Let us assume that the radius of the circle is $1$ and let $\widehat{BOA}=2\theta$.
Then $\theta$ is the unique solution of $\theta-\frac{1}{2}\sin(2\theta)=\frac{\pi}{5}$ and $AB=\sin(2\theta)$.
$[AEB]=\frac{\pi}{5}$ then implies $ME=\frac{\pi}{5\sin\theta}$ and $AE=EB=\sqrt{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}$.
The height of the circle segment with base $AB$ is $1-\cos\theta$, so by naming $P$ the antipode of $N$ we have $EP=1-\cos\theta+\frac{\pi}{5\sin\theta}$ and $EN=1+\cos\theta-\frac{\pi}{5\sin\theta}$. By the chords theorem $EN\cdot EP = EC\cdot EA$, hence
$$ EC = \frac{1-\left(\cos\theta-\frac{\pi}{5\sin\theta}\right)^2}{\sqrt{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}}$$
and by the similarity between $EAB$ and $ECD$ we have
$$ CD = \frac{1-\left(\cos\theta-\frac{\pi}{5\sin\theta}\right)^2}{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}\cdot 2\sin\theta$$
and
$$ \widehat{DOC} = 2\arcsin\left(\frac{1-\left(\cos\theta-\frac{\pi}{5\sin\theta}\right)^2}{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}\cdot \sin\theta\right).$$
The length of $OE$ is $\frac{\pi}{5\sin\theta}-\cos\theta$, hence the area of the quadrilateral $DOCE$ equals
$$ [DOCE] = \left(\frac{\pi}{5\sin\theta}-\cos\theta\right)\left(\frac{1-\left(\cos\theta-\frac{\pi}{5\sin\theta}\right)^2}{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}\cdot \sin\theta\right) $$
and the area of the portion of the circle bounded by $E,D,C$ equals
$$\arcsin\left(\frac{1-\left(\cos\theta-\frac{\pi}{5\sin\theta}\right)^2}{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}\cdot \sin\theta\right)-\left(\frac{\pi}{5\sin\theta}-\cos\theta\right)\left(\frac{1-\left(\cos\theta-\frac{\pi}{5\sin\theta}\right)^2}{\sin^2\theta+\frac{\pi^2}{25\sin^2\theta}}\cdot \sin\theta\right).$$
Few steps of Newton's method give $\theta\approx 1.05657$, hence the last expression is $\approx 0.552681$ while $\pi/5\approx 0.628319$, proving that the wanted pentasection is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
How does $\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$ become $\sqrt{2(2+\sqrt{2})}$? I'd like to know how can one simplify the following expression
$$\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$$
into
$$\sqrt{2(2+\sqrt{2})}.$$
Wolfram alpha suggests it as an alternative form, and numerically it's easy to verify, but I can't find the right algebra to show they are indeed equivalent.
Note I ran into this problem, trying to do: $2\cos(\pi/8)+2\sin(\pi/8)$, where
$$2\cos(\pi/8)=\sqrt{2+\sqrt{2}},$$
$$2\sin(\pi/8)=\sqrt{2-\sqrt{2}}.$$
| Following @Yousef's answer:
\begin{align} \sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}} &= x \\ \left(\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}} \right)^2 &= x^2 \\ 2+\sqrt{2}+2-\sqrt{2}+2\sqrt{4-2} &= x^2 \\ 4+2\sqrt{2} &= x^2 \implies x=\sqrt{4+2\sqrt{2}} \end{align} as $x>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 4
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Generalization of $1988$ IMO #6 $1988$ IMO Problem 6 states:
Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Show that
$$\frac{a^2 + b^2}{ab + 1}$$
is the square of an integer.
My question is: what if $a$ and $b$ don't need to be positive? What other values of $\frac{a^2 + b^2}{ab + 1}$ are possible? I found a way to get $-5$ by setting $a = -1$ and $b = 2$. Is there a simple classification of such reachable integers?
| I prefer to graph everything. The thing that Vieta Jumping does is to guarantee a point (if there are any integer points) on a short arc of the hyperbola. Then inequalities based on the size of $n$ tell us whether there can be any such integer points.
Here $$ \frac{x^2 + y^2}{xy-1} = n $$
with $x,y > 0.$
If there are any integer points at all, there are jumped integer points that lie on the arc between the endpoints
$$ \left( \; \sqrt{\frac{4n}{n^2-4}} \; , \; \; \sqrt{\frac{n^3}{n^2-4}} \; \; \right) $$
and
$$ \left( \; \sqrt{\frac{n^3}{n^2-4}} \; , \; \; \sqrt{\frac{4n}{n^2-4}} \; \; \right) $$
One must consider $n=3,4$ separately. For $n= 5$ there are indeed integer points at $(1,2) , \; \; (2,1).$
For $n \geq 6$ we see the picture below (done with $n=16.$ The axis of symmetry $y=x$ meets the hyperbola at $(t,t)$ where $t = \sqrt{ \frac{n}{n-2}},$ between $1$ and $2$
For $x \geq \sqrt{ \frac{n}{n-2}},$ but $x \leq \sqrt{\frac{n^3}{n^2-4}} , $ we see that $y$ is decreasing. Furthermore, for $x = 2$ we find
$$ y_2 = \frac{n+4}{n+\sqrt{n^2 - n - 4}} .$$
Thus, for
$$ 2 \leq x \leq \sqrt{\frac{n^3}{n^2-4}}$$
we see $0 < y \leq \frac{n+4}{n+\sqrt{n^2 - n - 4}} .$ For $n \geq 6$ this is smaller than $1$
So that's it: we can jump an integer point to an integer point along a certain bounded arc of the hyperbola. Then, for $n \geq 6$ we see there are no integer points within that arc. So, no integer points at all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there any other method to show that $\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x =-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3)?$ Noting that the evaluation of the integral can be simplified by the Fourier series of $\ln(\sin x)$,
$$\ln (\sin x)+\ln 2=-\sum_{k=1}^{\infty} \frac{\cos (2 k x)}{k}$$
Multiplying the equation by $x$ followed by integration from $0$ to $\infty$, we have
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\int_{0}^{\frac{\pi}{2}} x\ln 2 d x&=-\sum_{k=1}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{x \cos (2 k x)}{k} d x\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\left[\frac{x^{2}}{2} \ln 2\right]_{0}^{\frac{\pi}{2}}&=-\sum_{k=1}^{\infty} \frac{1}{2 k^{2}} \int_{0}^{\frac{\pi}{2}} x d(\sin 2 k x)\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k^{2}}\left[\frac{\cos 2(x)}{2 k}\right]_{0}^{\frac{\pi}{2}} \\
&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{4} \sum_{k=1}^{\infty} \frac{(-1)^{k}-1}{k^{3}}\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{4}\left(\sum_{k=1}^{\infty} \frac{2}{(2 k+1)^{3}}\right)\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{2}\left[\sum_{k=1}^{\infty} \frac{1}{k^{3}}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{3}}\right]\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3) \blacksquare
\end{aligned}
$$
Furthermore, $$
\begin{aligned}
\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x&=\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \ln (\sin x)-\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x \\
&=-\frac{\pi^2}{4} \ln 2-\left(-\frac{\pi^2}{8} \ln 2+\frac{7}{16}\zeta(3)\right) \\
&=-\frac{\pi^2}{8} \ln 2-\frac{7}{16} \zeta(3)
\end{aligned}
$$
and $$
\int_0^{\frac{\pi}{2}} x \ln (\tan x) d x=\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x-\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x=\frac{7}{8}\zeta(3)
$$
| Enforcing the substitution $x\mapsto\frac {\pi}2-x$, then we get
\begin{align*}
\int\limits_0^{\pi/2}x\log\sin x\,\mathrm dx & =\int\limits_0^{\pi/2}\left(\frac {\pi}2-x\right)\log\cos x\,\mathrm dx\\ & =\frac {\pi}2\int\limits_0^{\pi/2}\log\cos x\,\mathrm dx-\int\limits_0^{\pi/2}x\log\cos x\,\mathrm dx\\ & =-\frac {\pi^2}4\log 2-\int\limits_0^{\pi/2}x\log\cos x\,\mathrm dx
\end{align*}
Now, express $\cos x$ as
$$2\cos x=e^{ix}+e^{-ix}$$
And expand the logarithm into three separate terms. Since the integral is entirely real, the complex components will cancel out in the end.
\begin{align*}
\int\limits_0^{\pi/2}x\log\cos x\,\mathrm dx & =\int\limits_0^{\pi/2}x\log\left(e^{ix}+e^{-ix}\right)\,\mathrm dx-\log 2\int\limits_0^{\pi/2}x\,\mathrm dx\\ & =i\int\limits_0^{\pi/2}x^2\,\mathrm dx+\int\limits_0^{\pi/2}x\log\left(1+e^{-2ix}\right)\,\mathrm dx-\frac {\pi^2}8\log 2\\ & =\frac {\pi^3i}{24}-\frac {\pi^2}8\log 2+\sum\limits_{n=1}^{+\infty}\frac {(-1)^{n-1}}n\int\limits_0^{\pi/2}xe^{-2nix}\,\mathrm dx
\end{align*}
Focusing on the remaining sum, we use integration by parts
\begin{align*}
\sum\limits_{n=1}^{+\infty}\frac {(-1)^{n-1}}n\int\limits_0^{\pi/2}xe^{-2nix}\,\mathrm dx & =-\frac 14\sum\limits_{n=1}^{+\infty}\frac 1{n^3}-\frac {\pi i}4\sum\limits_{n=1}^{+\infty}\frac 1{n^2}-\frac 14\sum\limits_{n=1}^{+\infty}\frac {(-1)^{n-1}}{n^3}\\ & =-\frac 7{16}\zeta(3)-\frac {\pi^3 i}{24}
\end{align*}
So the integral is
\begin{align*}
\int\limits_0^{\pi/2}x\log\sin x\,\mathrm dx=-\frac {\pi^2}4\log 2+\frac {\pi^2}8\log 2+\frac 7{16}\zeta(3)\color{blue}{=-\frac {\pi^2}8\log 2+\frac 7{16}\zeta(3)}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Should we consider the line passing through the origin is having equal intercepts? I am a grade 11 student and I had the following question on my math test.
$x^2 +y^2+4x-8y+2=0$
is a circle and we have to find out the equations of the tangents of the circle that cuts off equal intercepts of the same sign from the axes of coordinates.
Naturally, I started with the equation being $\frac{x}{a}+\frac{y}{a}=1$ and found 2 equations. I then thought that I was assuming that the intercepts can never be 0 if I only used the formula.
So I then used the formula $y=mx$ to find two more equations and then kept the 4 equations as the answer. I didn't get any marks; so I gave back the copy for a recheck as I still , to some extent, believe that I was right.
Was I right about this? Or I should have kept the 1st two equations only??
Some people wanted to know the process of how I found the equation more descriptively. Here it is:
Given, $x^2 +y^2+4x-8y+2=0$ is a circle.
$\therefore$ centre, $C \equiv (-2,4)$, Radius, $R =\sqrt{18}$
Now, let the equation be $\frac{x}{a}+\frac{y}{a}=1$ where $a \neq 0$
$\implies x+y-a=0 \cdots (1)$.
$\therefore $ perpendicular distance to (1) from the centre
$ = \frac{|-2+4-a|}{\sqrt{1^2+1^2}} $ which should be equal to the radius if the line I tangent.
$\therefore \frac{|-2+4-a|}{\sqrt{1^2+1^2}} = \sqrt{18}\cdots (2)$
Solving 2 we get, $a=8,-4$
Again let,
Equation be $y-mx=0$ if the intercept is 0
Perpendicular distance from centre = $\frac{|2m+4|}{\sqrt{m^2+1}}=\sqrt{18}$[Radius] $\cdots (3)$
solving 3 we get, $m=1,-\frac{1}{7}$
Thus the required equations of tangent are:
$x+y-8=0$
$x+y+4=0$
$y=x$
$7y=x$
| We have the circle $x^2+y^2+4x-8y+2 = 0$. We can rewrite that as $x^2+y^2+2gx+2fy+c = 0$, where $g=2$, $f=-4$, and $c=2$. This means the center of the circle is $\left(-g,-f\right) = \left(-2,4\right)$ and the radius $r$ is $\sqrt{g^2+f^2-c} = \sqrt{(-2)^2+(4)^2-2} = 3\sqrt2$.
Consider the equation of a line $y=mx+c$, which we can rewrite as $mx-y+c = 0$. Notice how the intercepts are $\left(0,c\right)$ and $\left(-\frac{c}{m},0\right)$. If we want to find the "equal intercepts of the same sign from the axes of coordinates," then it makes sense to let $m=1$ or $m=-1$.
(The wording in quotes is confusing, so from here I just decided to find all the tangent lines in general.)
The distance we are looking for, which is also the radius, is given by
$$\frac{\left|m\alpha-\beta+c\right|}{\sqrt{1+m^2}} = r \iff r^2 = \frac{(m\alpha-\beta+c)^2}{1+m^2}.$$
Since the center is $\left(-2,4\right)$, we can plug in $-2$ and $4$ for $\alpha$ and $\beta$, respectively, to get
$$\frac{(m(-2)-(4)+c)^2}{1+m^2} = 18.$$
Now we have to choose our $m$. If $m=-1$, then we can solve for $c$ to get $c=-4$ or $c=8$. If $m=1$, then we can solve for $c$ to get $c=0$ or $c=12$. This means the tangent lines are
$$\eqalign{
y&=-x+8 \cr
y &= -x-4 \cr
y &= x+12 \cr
y &= x+0
}$$
and notice how if you graph them along with the circle, they are lines tangent to the circle.
Does that answer your question?
| {
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"timestamp": "2023-03-29T00:00:00",
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Laurent Series of $f(z) = \frac{1}{(1-z)(1-cz)}$ in the point $z_0 = \frac{1}{c}$, where $c = \frac{3}{4}i, z \in \mathrm{C}$. I am currently trying to find the Laurent Series to $f(z) = \frac{1}{(1-z)(1-cz)}$ in the point $z_0 = \frac{1}{c}$. Therefore I calculated the partial fractions and found that $f(z) = \frac{A}{1-z} + \frac{B}{1-cz}$, with $A = \frac{16}{25} + \frac{12}{25}i$ and $B=\frac{9}{25} - \frac{12}{25}i$. Now when I use the geometric sum to find the Laurent Series I get $f(z) = \sum_{k=0}^{\infty} z^k A + \sum_{k=0}^{\infty} (cz)^k B = \sum_{k=0}^{\infty} z^k(A+c^kB)$. But that means that $|cz|$ must be less then 1 and than $|z| < |1/c| = |z_0|$. Doesn't than mean, I can't use that approach or is it not possible to develop the series in that point?
| You computed the Laurent series of $f$ in $z_0 = 0$, which then obviously only converges for $|z| < |1/c|$. If you want to compute the Laurent series in $z_0$ = 1/c, you can factorize $f$ as
$$f(z) = \frac{1}{z - \frac{1}{c}} \frac{1}{c(z-1)}.$$
The first factor is in the correct form. Now expand the second factor in $z_0 = \frac{1}{c}$ and you get the desired Laurent series.
EDIT: Because there seems to be some confusion, here is the Laurent series around $z_0 = 1/c$. We have for $0<|z-1/c|<|1-1/c|=5/3$
$$\frac{1}{1-z} = \frac{1}{\frac{c-1}{c} - (z - \frac{1}{c})} = \frac{c}{c-1}\frac{1}{1 - \frac{c}{c-1}(z - \frac{1}{c})} = \frac{c}{c-1}\sum\limits_{n = 0}^{\infty} \frac{c^n}{(c-1)^n}(z-\frac{1}{c})^n.$$
Thus
$$f(z) = (z-\frac{1}{c})^{-1}\frac{1}{1-c}\sum\limits_{n = 0}^{\infty} \frac{c^n}{(c-1)^n}(z-\frac{1}{c})^n = \frac{1}{1-c}\sum\limits_{n = 0}^{\infty} \frac{c^n}{(c-1)^n}(z-\frac{1}{c})^{n-1}.$$
Shifting the index by one, you get
$$f(z) = -\sum\limits_{n = -1}^{\infty} \frac{c^{n+1}}{(c-1)^{n+2}}(z-\frac{1}{c})^n.$$
| {
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"url": "https://math.stackexchange.com/questions/4506589",
"timestamp": "2023-03-29T00:00:00",
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Solving a Diophantine equation with power of $24$. We are going to prove that there is no integer solution for the Diophantine equation
$$\displaystyle 24^{a}=3 b^{2}+3 b \tag*{(*)} \\$$
$\textrm{Splitting }24=3\times 8 \text{ and factorising }3b^2+3b \textrm{ yields }$$ 3^{a-1} \cdot 8^{a}=b(b+1).$
$ (b, b+1)=(b,1)=1 \textrm{ and } 3^{a-1}<8^a \Rightarrow b = 3 ^ { a - 1 } \textrm{ and } b + 1 = 8 ^ { a } \Rightarrow 3^{a-1} +1=8^{a} \\ 3^{a-1}=8^{a}-1 =7\left(8^{a-1}+8^{a-2}+\cdots+1\right) \Rightarrow \quad 7 \mid 3^{a-1}, \text { which is a contradiction. } \tag*{} $
Hence we can conclude that the Diophantine equation (*) has no integer solution.
Your comments and alternate solutions are highly appreciated.
| Treat the equation as a quadratic equation in variable $b$. Thus: $b^2+b - 2^{3a}\cdot 3^{a-1} = 0$. Taking usual $\triangle = 1-4(1)(-2^{3a}\cdot 3^{a-1})=1+2^{3a+2}\cdot 3^{a-1}=k^2$, for some integer $k$. Hence: $(k-1)(k+1) = 2^{3a+2}\cdot 3^{a-1}$. Observe that $\text{gcd}(k-1,k+1) = 1$ or $2$. If $\text{gcd}(k-1,k+1) = 1$, then these are coprime integers. Thus: $k-1 = 2^{3a+2}, k+1 = 3^{a-1}$, or $k-1 = 1, k+1 = 2^{3a+2}\cdot 3^{a-1}$ or $k-1 = 3^{a-1}, k+1 = 2^{3a+2}$. From this, you can show either case yields no integer solutions in $k, a$. If $\text{gcd}(k-1,k+1) = 2$, then it must be the case that: $k-1 = 2^{3a+1}, k+1 = 2\cdot 3^{a-1}$ or $k-1 = 2\cdot 3^{a-1}, k+1 = 2^{3a+1}$. And again, either case yields no integer solutions in $a,k$. So overall, there is no integer solutions in $a,b$.
Note: We can demonstrate one example above that for the case $k-1 = 2^{3a+2}, k+1 = 3^{a-1}\implies 2k = 2^{3a+2}+3^{a-1}$. This has no solution since the left is even while the right is odd. The remaining cases are done similarly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solving $x+y+z=4$, $x^2+y^2+z^2=14$, $x^3+y^3+z^3=34$
Solve the system $$\begin{equation} \label{equation1}
\begin{split}
x+y+z=4 \\
x^2+y^2+z^2=14 \\
x^3+y^3+z^3=34
\end{split}
\end{equation}$$
My work:
I found out that $$xy+yz+xz=1$$ and $$x^2y+x^2z+y^2x+y^2z+z^2x+z^2y=22$$
After this I'm stuck. Any help is greatly appreciated.
EDIT This not a duplicate. I'm looking for a detailed solution and not a solution just by inspection. Also I thought of a new idea. Maybe e should consider a cubic polynomial whose roots are $x,y,z$
| $x+y+z=4$
$x^2+y^2+z^2=14$
$xy+yz+zx=1$(you already established)....(1)
$x^3+y^3+z^3-3xyz+3xyz=34$
$\Rightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-zx)+3xyz=34$
$\Rightarrow 4(14-1)+3xyz=34$
So, $xyz=-6$....(2)
From (1) and (2)
$(1/x)+(1/y)+(1/z)=-(1/6)$
$\Rightarrow (x+y)/xy+(1/z)=-(1/6)$
$\Rightarrow (4-z)/(-6/z)+(1/z)=-(1/6)$
Simplifying we get
$z^3-4z^2+z+6=0$
$\Rightarrow (z-3)(z-2)(z+1)=0$
$\therefore, z=3,2,-1$
Put these values in $x+y+z=4$ and $xyz=-6$, you can easily get the values of $x$ and $y$.
There will be 6 solutions, permutations of
$(3,2,-1)$.
| {
"language": "en",
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find the Laurent series expansion of $f(z) = \frac{1}{(z + 1)(z +3)}$ for the region $0 \lt |{z + 1}| \lt 2$ I have to find the Laurent series expansion of $$f(z) = \frac{1}{(z + 1)(z +3)}$$ for the region $$0 \lt |{z + 1}| \lt 2$$
Using partial fration f(z) can written as:
$$f(z) = \frac{1}{2} \frac{1}{z + 1} - \frac{1}{2} \frac{1}{z + 3}$$
since we have from Maclurin's series,
$$\sum_0^\infty \frac{1}{1+z} = (-z)^n for |z| <1$$
so ,
$$f(z) = \frac{1}{2} [\sum (-1)^n z^n - \frac{1}{1 + (z + 2)}]$$
How to proceed from here?
| we have,
$$0 < | z + 1| < 2 \implies |z| \lt 1 , \frac{|z + 1|}{2} \lt 1$$
and Using partial fraction we have,
$$f(z) = \frac{1}{2}\frac{1}{z+1} - \frac{1}{2}\frac{1}{z+3}$$
$$\implies \frac{1}{2} \sum(-1)^n z ^n - \frac{1}{4}\frac{1}{1 + \frac{z + 1}{2}}$$
$$\implies \frac{1}{2} \sum(-1)^n z ^n - \frac{1}{4} \sum(-1)^n \frac{(z + 1)^n}{2^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508431",
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} |
Finding the sum of the real roots of $(\tan(x)-\sqrt{3})\cdot \arcsin \dfrac{2(x-\pi)}{\pi}=0$
Find the sum of the real roots of
$$(\tan(x)-\sqrt{3})\cdot \arcsin \dfrac{2(x-\pi)}{\pi}=0$$
can anybody assist with my solution.
*
*$$\tan(x)=\sqrt{3}$$
$$x=\frac{\pi}{3}$$ and $$x=\frac{4\pi}{3}$$. On the other hand $$x=\frac{\pi}{3}+ \frac{k}{\pi}$$, where $k∈\mathbb Z$.
*$$\arcsin(\frac{2(x-\pi)}{\pi})=0\implies
x=\pi$$
Answer: $$\frac{\pi}{3}+\pi=\frac{4\pi}{3} $$(got negative result with this. No instructions about $k∈\mathbb Z$)
$$\pi+\frac{4\pi}{3}+\pi=\frac{8\pi}{3} \tag{?}$$
Thank you for help!
| We need to ensure that the roots lie within the domain of definition of arcsin.
That is, we need
$$-1 \leq \frac{2(x-\pi)}{\pi} \leq 1.$$
We know that $x = \pi$ is a solution that makes this equal to zero. But, it can also be the case that the tangent evaluates to zero, but not the arcsin. We substitute $x = \pi/3+k\pi$ into the above equation. This gives (after simplification),
$$\frac{1}{6} \leq k \leq \frac{7}{6},$$
from which we conclude that $k$ can only be $1$.
Thus, the roots are $\pi$ and $\pi/3+\pi$.
The sum is then $2\pi+\pi/3 = 7\pi/3$.
| {
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Solution to $\frac{n(n+1)}{2} \leq x<\frac{(n+1)(n+2)}{2}$ when given $x$ I tried to find a solution to
$$\frac{n(n+1)}{2} \leq x<\frac{(n+1)(n+2)}{2} \tag0$$
when $x$ is given $(n \in \mathbb N_0, x \geq 0)$.
This inequality means $n$-th triangular number is the greastest triangular number less than or equal to $x$.
There is one and only one $n \in \mathbb N_0$ given $x \geq 0$.
$$\text{if}\; x \in [0, 1),\; n = 0$$
$$\text{if}\; x \in [1, 3),\; n = 1$$
$$\text{if}\; x \in [3, 6),\; n = 2$$
$$...$$
Split the inequality in two, we get
$$\frac{n(n+1)}{2} \leq x \tag1$$
$$x<\frac{(n+1)(n+2)}{2} \tag2$$
Focus on $(1)$. Using quadratic formula for $(1)$,
$$\frac{-1 - \sqrt{8x+1}}{2} \leq n \leq \frac{-1 + \sqrt{8x+1}}{2} \tag{$1.1$}$$
Note that $n \in \mathbb N_0$, maximum value of $n$ is
$$n = \lfloor \frac{-1 + \sqrt{8x+1}}{2} \rfloor \tag{1.2}$$
and this is the solution to the original inequality $(0)$. Only maximum value is allowed, because if $n=$ (value less than the maximum) is the solution, then $n=$ (the maximum value) is another solution, which breaks the uniqueness of $n$.
Similarly using quadratic formula for $(2)$,
$$n < \frac{-3 - \sqrt{8x+1}}{2}\; \lor\; n > \frac{-3 + \sqrt{8x+1}}{2} \tag{$2.1$}$$
And the solution (the minimum value) is
$$ n =
\begin{cases}
\lceil \frac{-3 + \sqrt{8x+1}}{2} \rceil &\quad\text{if}\; \frac{-3 + \sqrt{8x+1}}{2} \not \in \mathbb N_0\\
\frac{-3 + \sqrt{8x+1}}{2} + 1 &\quad\text{if}\; \frac{-3 + \sqrt{8x+1}}{2} \in \mathbb N_0\\
\end{cases}\tag{$2.2$}$$
Actually $(1.2)$ and $(2.2)$ are the same thing with different notation.
Interestingly, I found
$$n = \lfloor \sqrt{\frac{\lfloor \sqrt{2x} \rfloor}{\lfloor \sqrt{2x} \rfloor +1}} \cdot \sqrt{2x} \rfloor \tag3$$
I cannot find any relationship between $(3)$ and $(1.2)$ or $(3)$ and $(0)$, but it seems like $(3)$ is also a solution.
(I brute-forced for $x \in \mathbb{N}_0$, $x \leq 50000000$)
So this is my question: Is $(3)$ the solution to $(0)$?
P.S.
Modify $(0)$ slightly,
$$n(n+1) \leq 2x < (n+1)(n+2) \tag{0.1}$$
and substitude $X = 2x$
$$n(n+1) \leq X < (n+1)(n+2) \tag{0.2}$$
$$n = \lfloor \sqrt{\frac{\lfloor \sqrt{X} \rfloor}{\lfloor \sqrt{X} \rfloor +1}} \cdot \sqrt{X} \rfloor \tag{3.1}$$
Now the problem is more simple: Is $(3.1)$ the solution to $(0.2)$?
| Well, by solving:
*
*$$x=\frac{\text{n}\left(\text{n}+1\right)}{2}\space\Longleftrightarrow\space\text{n}=\frac{\pm\sqrt{1+8x}-1}{2}\tag1$$
*$$x=\frac{\left(\text{n}+1\right)\left(\text{n}+2\right)}{2}\space\Longleftrightarrow\space\text{n}=\frac{\pm\sqrt{1+8x}-3}{2}\tag2$$
We can see that we have the following solutions:
*
*$$\text{n}=-\frac{1}{2}\space\wedge\space x=-\frac{1}{8}\tag3$$
*For $-\frac{1}{8}<x<0$:
$$-\frac{\sqrt{1+8x}+1}{2}\le\text{n}\le\frac{\sqrt{1+8x}-1}{2}\tag4$$
*For $x\ge0$:
$$\frac{\sqrt{1+8x}-3}{2}<\text{n}\le\frac{\sqrt{1+8x}-1}{2}\tag5$$
And for solving $\text{k}^2=1+8x$, such that $\sqrt{1+8x}\in\mathbb{N}$, you can see that:
*
*$$x=\text{c}_1\left(1+2\text{c}_1\right)\space\wedge\space\text{k}=1+4\text{c}_1\tag6$$
Where $\text{c}_1\in\mathbb{Z}$
*$$x=2\text{c}_2^2+3\text{c}_2+1\space\wedge\space\text{k}=3+4\text{c}_2\tag7$$
Where $\text{c}_2\in\mathbb{Z}$
So, we can see two cases for $x\ge0$:
*
*$$\frac{1+4\text{c}_1-3}{2}=2\text{c}_1-1<\text{n}\le2\text{c}_1=\frac{1+4\text{c}_1-1}{2}\tag8$$
*$$\frac{3+4\text{c}_2-3}{2}=2\text{c}_2<\text{n}\le1+2\text{c}_2=\frac{3+4\text{c}_2-1}{2}\tag9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Binomial related problem $\left( {\begin{array}{*{20}{c}}
5\\
0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{50}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
5\\
1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{40}\\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{30}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
5\\
3
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
5\\
4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{10}\\
5
\end{array}} \right) = $
Where ${}^n{C_r} = \left( {\begin{array}{*{20}{c}}
n\\
r
\end{array}} \right)$
My approach is as follow $\left( {\begin{array}{*{20}{c}}
5\\
0
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{50}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
5\\
1
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{40}\\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
5\\
2
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{30}\\
5
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
5\\
3
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{20}\\
5
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
5\\
4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
{10}\\
5
\end{array}} \right) = \sum\limits_{r = 0}^4 {{{\left( { - 1} \right)}^r}.{}^5{C_r}.{}^{50 - 10r}{C_5}} $
Using the technique of some previous problem I tried to solve by absorption
$\sum\limits_{r = 0}^4 {{{\left( { - 1} \right)}^r}.\frac{{5!}}{{r!\left( {5 - r} \right)!}}.\frac{{\left( {50 - 10r} \right)!}}{{5!\left( {45 - 10r} \right)!}}} = \sum\limits_{r = 0}^4 {{{\left( { - 1} \right)}^r}.\frac{{\left( {45 - 9r} \right)!}}{{r!\left( {45 - 10r} \right)!}}.\frac{{\left( {50 - 10r} \right)!}}{{\left( {5 - r} \right)!\left( {45 - 9r} \right)!}}} \Rightarrow \sum\limits_{r = 0}^4 {{{\left( { - 1} \right)}^r}.{}^{45 - 9r}{C_r}.{}^{50 - 10r}{C_{5 - r}}} $
Not able to proceed further
| HINT:
In your expression , what we are looking for is the sum of the coefficients of the term $x^5$ in the expansion of $$-\bigg[\bigg(1-(1+x)^{10}\bigg)^5 -1\bigg]$$
Because :
*
*$$[x^5][\bigg(1-(1+x)^{10}\bigg)^5 =\sum_{k=0}^{5}-\binom{5}{0}[(1+x)^{10}]^5+\binom{5}{1}[(1+x)^{10}]^4-\binom{5}{2}[(1+x)^{10}]^3...+\binom{5}{5}[(1+x)^{10}]^0$$
To render this expression into yours , subtract the last term , and multiply by $(-1)$
| {
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"source": "stackexchange",
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Proving polynomial division examples through Mathematical Induction While proving the examples of polynomial division through mathematical induction, I am led to a curious conclusion, here is an overview:
The following statement holds true for all whole number values of 'n'.
$ \frac{x^n - y^n}{1} $ = p(x-y) -------------(1)
1. For now, let us assume 'x', 'y' to be integers
So, through the closure property of integers under multiplication and subtraction, we can state:
'p' is an integer
After substituting several whole number values of 'n' and integer values of 'x' and 'y', we find that the statement still holds true.
Thus:
Proving (1) through MATHEMATICAL INDUCTION:
To prove : $ \frac{x^(n+1) - y^(n+1)}{1} $ = a(x-y) ; or to put in other words : $ \frac{x^(n+1) - y^(n+1)}{1} $ must be represented in the form of 'a(x-y)' where 'a' must be a integer. (prove: 'a' is an integer.
Equation : $ \frac{x^n - y^n}{1} $ = p(x-y)
Assuming (1) to be true,
for n = n+1, we have:
$ \frac{x^(n+1) - y^(n+1)}{1} $ = a(x-y) -----------------(2)
also we have: $ \frac{x^n - y^n}{1} $ = p(x-y) -----------------------------------(1)
Multiplying LHS and RHS of (1) with : (x+y) -
(x+y) $ \frac{x^n - y^n}{1} $ = p(x-y)(x+y)
Simplifying through index law of multiplication:
= $ \frac{x^(n+1) - y^nx + x^ny - y^(n+1) }{1} $ = $ \frac{p(x+y)(x-y)}{1} $
Rearranging the terms in the numerator of the fraction:
= $ \frac{x^(n+1)- y^(n+1) - y^nx + x^ny}{1} $ = $ \frac{p(x+y)(x-y)}{1} $
Adding LHS and RHS of (1) to : $ \frac{(y^nx)}{1}$ -
= $ \frac{x^(n+1)- y^(n+1) + x^ny}{1} $ = $ \frac{p(x+y)(x-y)+(yn^x)}{1} $
Subtracting ($ \frac{x^ny}{1}$) from LHS and RHS of (1) -
$ \frac{x^(n+1)- y^(n+1)}{1} $ = $ \frac{p(x+y)(x-y)+(y^nx)-(x^ny)}{1} $ ---------(3)
we also have another equation : $ \frac{x^(n+1) - y^(n+1)}{1} $ = a(x-y) -----------(2)
After observing (3) and (2), we can clearly state the following:
a(x-y) = $ \frac{p(x+y)(x-y)+(y^nx)-(x^ny)}{1} $ ------------------------(3)
a(x-y) = $ \frac{p(x+y)(x-y)+(y^nx)-(x^ny)}{1} $
Taking '-xy' as a common factor:
a(x-y) = $ \frac{p(x+y)(x-y)-xy(x^(n-1)-y^(n-1))}{1} $
We have tested that for all integers less than 'n' that the equation : $ \frac{x^n - y^n}{x-y} $ = (x+y), takes the form of k(x-y), where 'k' is an integer. So,
a(x-y) = $ \frac{p(x+y)(x-y)-xy(k[x-y])}{1} $
On removing the brackets and taking '(x-y)' as a common factor, we have the following:
a(x-y) = $ \frac{(x-y)[p(x+y)-xyk]}{1} $ --------------------(4)
Before going further, thoroughly read the following -
Through the closure property of integers under addition, subtraction and multiplication, we can state the following:
i). (x+y) = an integer
.....also (x-y) = an integer
ii). $ \frac{(xyk)}{1} $ = an integer
iii). p = an integer [assumed at the beginning]
Thus we can say that the expression on the RHS is an integer.
= a(integer) = integer
Now, we can state that 'a' is an integer, as an integer multiplied by an integer (x+y) must be equivalent to an integer (on the LHS).
Coming back to (4):
a(x-y) = $ \frac{(x-y)[p(x+y)-xyk]}{1} $ --------------------(4)
On Dividing both sides by the expression '(x-y)', we have the followwing:
a = p(x+y)-xyk
Substituting the value of 'a' in the (2), we have:
$ \frac{x^(n+1) - y^(n+1)}{1} $ = [p(x+y)-xyk] (x-y)
Now, we have successfully represented $ \frac{x^(n+1) - y^(n+1)}{1} $ in the form of 'a(x-y)' where 'a' is an integer.
On this wise, we can conclude that $ \frac{x^n - y^n}{1} $ is divisible by (x-y) for 'x', being integers and 'n' being a whole number; because 'a' is an integer
2. Now lets assume 'x' and 'y' to be rational numbers:
Our equation: $ \frac{x^n - y^n}{1} $ = p(x-y)
x,y = rational numbers
n = (remains)a whole number
On proving the equation for 'n+1' through mathematical induction, we get the following result:
$ \frac{x^n - y^n}{1} $ is divisible by (x-y) for 'x' and 'y' being rational numbers and 'n' being a whole number.
However, this statement is too argumentative as every non-zero rational number will divide every rational number.
Instance-1:
Let us assume we have to find whether $ \frac{21}{4} $ is divisible by $ \frac{2}{3} $
So, let the number to be multiplied with $ \frac{2}{3} $ be 'b':
($ \frac{2}{3} $)(b) = $ \frac{21}{4} $
*If 'b' is and integer, then, we can safely conclude that it $ \frac{21}{4} $ is divisible by $ \frac{2}{3}.*
<br />
*But if 'b' is a rational, then it means that $ \frac{21}{4} $ isn't divisible by $ \frac{2}{3} - Hopefully, I am correct.
Solving the equation in order to obtain the value of 'b':
($ \frac{2}{3} $)(b) = $ \frac{21}{4} $
Multiplying both sides of the equation by ($ \frac{3}{2} $):
b = $ \frac{84}{8} $
Simplifying the value of 'b':
b = $ \frac{84/4}{8/4} $
b = $ \frac{21}{2} $
We see that the result obtained is a rational number, thus $ \frac{21}{4} $ isn't divisible by $ \frac{2}{3}$
Instance-2:
Now, let us find whether $ \frac{21}{12} $ is divisible by $ \frac{7}{4} $
So, let the number to be multiplied with $ \frac{7}{4} $ be 'c':
($ \frac{7}{4} $)(c) = $ \frac{21}{12} $
Solving the equation in order to obtain the value of 'c':
($ \frac{7}{4} $)(c) = $ \frac{21}{12} $
Through cross-multiplication:
c = 3
We see that the result obtained is an integer(although it is also a rational number), thus $ \frac{21}{12} $ is divisible by $ \frac{7}{4}$.
So, I think, every non-zero rational number is divisible by every other rational number. Such a division for polynomial expression where 'x', 'y' are rational numbers isn't logical.
Now, the question arises, should the equation:'$ \frac{x^n - y^n}{1} $ = p(x-y)' be considered for 'x' and 'y' being rational numbers?
Also can I safely assume that 'x' and 'y' are integers in the equation : '$ \frac{x^n - y^n}{1} $ = p(x-y)' ?
| To show that $x-y\mid x^n-y^n$, you can factorise $x^n-y^n$ or use the Fundamental theorem of algebra. Observe that, $y=x$ is the common root of the polynomials $x-y$ and $x^n-y^n$. But you want to prove this using mathematical induction. For this, we will use the method known as "strong induction".
Let, $$p(n)=\frac{x^n-y^n}{x-y}$$
We want to prove that, $p(n)$ is a polynomial. Note that, $n=1$ is trivial. For $n=2$ you get $$\frac{x^2-y^2}{x-y}=x+y$$
Suppose that for $n=k-1$ and $n=k$, $p(k-1)=\frac{x^{k-1}-y^{k-1}}{x-y}$ and $p(k)=\frac{x^k-y^k}{x-y}$ are polynomials. Then for $n=k+1$ you have,
\begin{align}p(k+1)&=\frac{x^{k+1}-y^{k+1}}{x-y}\\
&=\frac{p(k)(x-y)(x+y)-xy(x^{k-1}-y^{k-1})}{x-y}\\
&=p(k)(x+y)-xyp(k-1).\end{align}
This completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of this trigonometric expression
If $$\operatorname{sin}(x)+\operatorname{sin}(y)\ge\operatorname{cos}(x)\cdot\operatorname{cos}(\alpha)$$ $\forall x\in\mathbb{R}$ then find the value of $$\operatorname{sin}(y)+\operatorname{cos}(\alpha)$$
I have only been able to prove the fact that $$\operatorname{sin}(y)\ge\operatorname{cos}(\alpha)$$ by plugging $x=\pm\frac{\pi}{2}$
In other words we can say that $$2\ge\operatorname{sin}(y)+\operatorname{cos}(\alpha)\ge2\operatorname{cos}(\alpha)\ge-2$$
How to proceed further$?$
| When $x$ is a multiple of $\frac{\pi}{2}$, the inequality simplifies to:
$$x = 0 \implies \sin(y) \ge \cos(\alpha)$$
$$x = \frac{\pi}{2} \implies 1 + \sin(y) \ge 0$$
$$x = \pi \implies \sin(y) \ge -\cos(\alpha)$$
$$x = \frac{\pi}{2} \implies -1 + \sin(y) \ge 0$$
The last of these is equivalent to $\sin(y) \ge 1$. But since the sine of a real number is never greater than 1, we must have $\boxed{\sin(y) = 1}$. Plugging this back into the original inequality gives:
$$\sin(x) + 1 \ge \cos(x)\cos(\alpha)$$
If $\cos(x) = 0$, then we just get the tautology $\pm 1 + 1 \ge 0$. Otherwise, let's divide by $\cos(x)$, but remember that dividing by a negative number flips the order.
$$\cos(x) > 0 \implies \frac{\sin(x) + 1}{\cos(x)} \ge \cos(\alpha)$$
$$\cos(x) < 0 \implies \frac{\sin(x) + 1}{\cos(x)} \le \cos(\alpha)$$
Now, $\frac{d}{dx} \frac{\sin(x) + 1}{\cos(x)} = \frac{1 + \sin(x)}{\cos^2(x)}$, so the left-hand expression has a critical point when $\sin(x) = -1$, or $x = \frac{3\pi}{2}$. But this
means $\cos(x) = 0$, making the expression an undefined $\frac{0}{0}$. But we can use L'Hôpital's Rule.
$$\lim_{x \rightarrow \frac{\pi}{2}} \frac{\sin(x) + 1}{\cos(x)} = \lim_{x \rightarrow \frac{\pi}{2}} \frac{\cos(x)}{-\sin(x)} = \frac{\cos(\frac{3\pi}{2})}{-\sin(\frac{3\pi}{2})} = \frac{0}{-1} = 0$$
Plugging in this limit gives:
$$\cos(x) \rightarrow 0^+ \implies 0 \ge \cos(\alpha)$$
$$\cos(x) \rightarrow 0^- \implies 0 \le \cos(\alpha)$$
Which can only be simultaneously true for both positive and negative values of $\cos(x)$ if $\boxed{\cos(\alpha) = 0}$.
Therefore, $\boxed{\sin(y) + \cos(\alpha) = 1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sum_{a+b+c=n}ab \binom{n}{a,b,c}=n(n-1)3^{n-2}$ I want to prove $\sum_{a+b+c=n}ab \binom{n}{a,b,c}=n(n-1)3^{n-2}$.
Considering the multinomial theorem, setting two of the variables equal to $1$ and finally differentiating both sides twice, we get another identity, which is $\sum_{a+b+c=n}a(a-1) \binom{n}{a,b,c}=n(n-1)3^{n-2}$, which looks different from the former.
Actually I also proved the former identity combinatorially but I don’t know how these two different looking results are related to each other.
| Another approach reducing multinomial coefficients to binomial coefficients.
$$
\begin{align}
\sum_{a+b+c=n}ab\binom{n}{a,b,c}
&=\sum_{a=1}^n\sum_{b=1}^{n-a}ab\binom{n}{a}\binom{n-a}{b}\tag{1a}\\
&=\sum_{a=1}^n\sum_{b=1}^{n-a}a(n-a)\binom{n}{a}\binom{n-a-1}{b-1}\tag{1b}\\
&=\sum_{a=1}^na(n-a)\binom{n}{a}2^{n-a-1}\tag{1c}\\
&=\sum_{a=1}^na(n-1)\binom{n}{a}2^{n-a-1}-\sum_{a=1}^na(a-1)\binom{n}{a}2^{n-a-1}\tag{1d}\\
&=\sum_{a=1}^nn(n-1)\binom{n-1}{a-1}2^{n-a-1}-\sum_{a=1}^nn(n-1)\binom{n-2}{a-2}2^{n-a-1}\tag{1e}\\
&=\frac{n(n-1)}2\,3^{n-1}-\frac{n(n-1)}2\,3^{n-2}\tag{1f}\\[9pt]
&=n(n-1)\,3^{n-2}\tag{1g}
\end{align}
$$
Explanation:
$\text{(1a):}$ when $n=a+b+c$, $\binom{n}{a,b,c}=\binom{n}{a}\binom{n-a}{b}$
$\text{(1b):}$ $b\binom{n-a}{b}=\binom{n-a-1}{b-1}$
$\text{(1c):}$ sum in $b$ using the Binomial Theorem
$\text{(1d):}$ $n-a=(n-1)-(a-1)$
$\text{(1e):}$ $a\binom{n}{a}=n\binom{n-1}{a-1}$
$\phantom{\text{(1e):}}$ apply once to the left sum and twice to the right sum
$\text{(1f):}$ sum in $a$ using the Binomial Theorem
$\text{(1g):}$ simplify
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show inequality for positive real numbers If $x,y$ are positive real numbers then we have that $$ \frac{1}{\sqrt{x+y}}<\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}$$ right? But how can we show that?
I have tried the following but I don't think that this is the way we should go.
\begin{align*}\frac{1}{\sqrt{x + y}} < \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}}=\frac{\sqrt{y}+\sqrt{x}}{\sqrt{xy}} & \iff \left(\frac{1}{\sqrt{x + y}}\right)^2 < \left(\frac{\sqrt{y}+\sqrt{x}}{\sqrt{xy}}\right)^2 \\ & \iff \frac{1}{x + y} <\frac{x+y+2\sqrt{xy}}{xy} \\ & \iff xy<(x+y)(x+y+2\sqrt{xy}) \\ & \iff xy<x^2+xy+2x\sqrt{xy}+xy+y^2+2y\sqrt{xy}\\ & \iff 0<x^2+2(x+y)\sqrt{xy}+xy+y^2\end{align*}
| In fact, for $x,y>0$, one has
$$ x+y>x \Longrightarrow \sqrt{x+y}>\sqrt x \Longrightarrow\frac{1}{\sqrt{x+y}}<\frac1{\sqrt x}<\frac1{\sqrt x}+\frac1{\sqrt y.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Multiplication in $S_4$ with cycle notation and permutation matrices
Solve the equation $(23)x^{-1}(124)=(34)^4$ in $S_4$.
I'm a bit confused about which way I should go when multiplying. Using cycle notation I've been told to go from right to left, but I find it much easier to do these using matrix notation. So I write the equation as $$\begin{pmatrix}1&2&3&4\\ 1&3&2&4\end{pmatrix}\begin{pmatrix}1&2&3&4\\ &&&\end{pmatrix}\begin{pmatrix}1&2&3&4\\ 2&4&3&1\end{pmatrix}=\begin{pmatrix}1&2&3&4\\ 1&2&3&4\end{pmatrix}$$ since I figured out that $(34)^4=(1)$.
Now I need to figure out what to put in the blank spots. Going from left to right I figured that I need to have $$\begin{pmatrix}1&2&3&4\\ 1&3&2&4\end{pmatrix}\begin{pmatrix}1&2&3&4\\ 4&3&1&2\end{pmatrix}\begin{pmatrix}1&2&3&4\\ 2&4&3&1\end{pmatrix}=\begin{pmatrix}1&2&3&4\\ 1&2&3&4\end{pmatrix}$$
but when going from right to left I ended up with
$$\begin{pmatrix}1&2&3&4\\ 1&3&2&4\end{pmatrix}\begin{pmatrix}1&2&3&4\\ 4&1&2&3\end{pmatrix}\begin{pmatrix}1&2&3&4\\ 2&4&3&1\end{pmatrix}=\begin{pmatrix}1&2&3&4\\ 1&2&3&4\end{pmatrix}$$ which is wrong. Why is the multiplication reversed in this case?
| You are right that $(34)^4=e.$
What I would do next is take the inverse of the whole of the LHS, since $(ab^{-1}c)^{-1}=c^{-1}ba^{-1}$; thus
$$(124)^{-1}x(23)^{-1}=e^{-1}=e,$$
which gives
$$\begin{align}
x&=(124)e(23)\\
&=(1234)\\
&=\begin{pmatrix}
1 & 2 & 3 & 4\\
2 & 3 & 4 & 1
\end{pmatrix},
\end{align}$$
whose inverse is
$$x^{-1}=\begin{pmatrix}
1 & 2 & 3 & 4\\
4 & 1 & 2 & 3
\end{pmatrix},$$
so you were right; you just forgot to take inverses.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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When do I add a negative sign in front of a square root with the limit approaching negative infinity I am confused on when to add a negative and when not to add a negative when doing limit at negative infinity questions.
Consider this
$$\lim_{x \to -\infty} \frac{\sqrt{4x^4-x}}{2x^2+3}$$
I would solve like this:
$$\lim_{x \to -\infty} \frac{(1/x^2)\sqrt{4x^4-x}}{(1/x^2)({2x^2+3})}$$
$$\lim_{x \to -\infty} \frac{\sqrt{{4x^4/x^4}-{x/x^4}}}{(1/x^2)({2x^2+3})}$$
Since
$$\sqrt{x^2} = −x \text{ if } x<0 \\
\sqrt{x^2} = x \text{ if } x>0$$
Add a negative to the numerator ?? (WRONG)
$$\lim_{x \to -\infty} \frac{- \sqrt{{4}-{1/x^3}}}{({2+3/x^2})}$$
$$\lim_{x \to -\infty} \frac{- \sqrt{{4}-0}}{({2+0})}$$
$$\frac{- 2}{({2})}$$
$$=-1 \text{ Wrong answer. Right answer is 1 }$$
The right answer is 1 but I got it wrong because I added negative to the numerator. So can you please tell me when to add the negative sign in front of a square root of x to the power of n?
On the other hand, in this example, the negative seems to be the right thing to do:
$$\lim_{x \to -\infty} \frac{\sqrt{9x^6-x}}{x^3+6}$$
$$\lim_{x \to -\infty} \frac{(1/x^3)\sqrt{9x^6-x}}{(1/x^3)(x^3+6)}$$
$$\lim_{x \to -\infty} \frac{\sqrt{9x^6/x^6-x/x^6}}{(x^3/x^3+6/x^3)}$$
Adding negative to numerator
$$\lim_{x \to -\infty} \frac{-\sqrt{9-1/x^5}}{(1+6/x^3)}$$
$$\frac{-\sqrt{9-0}}{(1+0)}$$
$$-\sqrt{9}$$
$$=-3$$
| The key difference is that $x^2$ is even but $x^3$ is odd. Specifically you, in the first example, multiplied by $1/x^2$ and converted this to $\sqrt{1/x^4}$. That’s correct because $x^2$ is always positive. Adding a minus sign is not necessary or correct. For example $(-2)^2=\sqrt{(-2)^4}=\sqrt{16}$ is completely ok since $(-2)^2=4\ge0$. But, for the $x^3$ example, $x^3<0$ when $x<0$ so in fact: $$x^3=-(-x^3)=-\sqrt{(-x^3)^2}=-\sqrt{x^6}$$Is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why $(2m)^2 + (m^2 - 1)^2 = (m^2 + 1)^2$ results in pythagorean triples? As you increase the value of n, you will generate all pythagorean triples whose first square is even. Is there any visual proof of the following explicit formula and where does it come from or how to derive it?
$(2n)^2 + (n^2 - 1)^2 = (n^2 + 1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2n)^2+(n^2-1)^2=(n^2+1)^2$
$(2*0)^2+(0^2-1)^2=(0^2+1)^2$
$(2*1)^2+(1^2-1)^2=(1^2+1)^2$
$(2*2)^2+(2^2-1)^2=(2^2+1)^2$
$(2*0)^2+(0-1)^2=(0+1)^2$
$(2*1)^2+(1-1)^2=(1+1)^2$
$(2*2)^2+(4-1)^2=(4+1)^2$
$0^2+1^2=1^2$
$2^2+0^2=2^2$
$4^2+3^2=5^2$
$0+1=1$
$4+0=4$
$16+9=25$
$1=1$
$4=4$
$25=25$
| The formula that you are referring is a sub-case of the Euclid's formula
Accroding to the Euclid's formula it is true that:
$$\text{Given an arbitrary pair of integers m and n with m$\gt$n and m,n$\gt$0}$$
$$a=m^2-n^2 \text{ , } b=2mn \text{ , } c=m^2+n^2 \text{ form a pythagorean triple}$$
$\text{In your case for:} n=1 \text{ and } m\epsilon\mathbb{N} $
You can found more about this forumla in this links:
proof of euclid's formula MathExchange
Pythagorean triple-wikipedia
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral: $\int_{0}^{2\pi}\arctan\left(\frac{1+2\cos x}{\sqrt{3}}\right)dx$ (Context) While working on an integral for fun, I stumbled upon the perplexing conjecture:
$$\int_{0}^{2\pi}\arctan\left(\frac{1+2\cos x}{\sqrt{3}}\right)dx = 2\pi\operatorname{arccot}\left(\sqrt{3+2\sqrt{3}}\right).$$
(Attempt) I tried multiple methods. One method that stuck out to me was using the formula $$\arctan(\theta) = \frac{1}{2i}\ln{\left(\frac{1+i\theta}{1-i\theta}\right)}$$ so that my integral becomes
$$\frac{1}{2i}\int_{0}^{2\pi}\ln\left(1+i\left(\frac{1+2\cos x}{\sqrt{3}}\right)\right)dx-\frac{1}{2i}\int_{0}^{2\pi}\ln\left(1-i\left(\frac{1+2\cos x}{\sqrt{3}}\right)\right).$$
Both of these look similar to the integral
$$\int_{0}^{2\pi}\ln\left(1+r^2-2r\cos(x)\right)dx=\begin{cases}
0, &\text{for}\; |r|<1,\\
2\pi\ln \left(r^2\right), &\text{for}\; |r|>1,
\end{cases}\tag{2}$$
and its solution can be found here.
I tried to get my integrals to "look" like the above result but to no avail. Not wanting to give up, I searched on this site for any ideas, and it seems like a few people have stumbled upon the same kind of integral, such as here and here.
In the first link, the user @Startwearingpurple says,
"Now we have
\begin{align}
4\sqrt{21}\pm i(11-6\cos\varphi)=A_{\pm}\left(1+r_{\pm}^2-2r_{\pm}\cos\varphi\right)
\end{align}
with
$$r_{\pm}=\frac{11-4\sqrt7}{3}e^{\pm i\pi/3},\qquad A_{\pm}=(11+4\sqrt7)e^{\pm i\pi /6}."$$
I tried to replicate his method but even after doing messy algebra, I couldn't figure out how to manipulate the inside of my logarithm such that it looked like what he did. I also tried letting $\operatorname{arg}\left(1+i\left(\frac{1+2\cos x}{\sqrt{3}}\right)\right) \in \left(-\pi/2, \pi/2\right)$, if that helps.
(Another method I tried was noticing that the original integral's function is periodic, so I tried using residue theory by letting $z=e^{ix}$, but I wasn't able to calculate the residues.)
(Question) Can someone help me approach this integral (preferably finding a closed form)? Any methods are absolutely welcome. And if someone could figure out how to get my logarithms to look like $\ln{\left(1+r^2-2r\cos{(x)}\right)}$, that would be nice.
(Edit) After using @SangchulLee's integral,
$$ \int_{0}^{\pi} \arctan(a + b\cos\theta) \, d\theta = \pi \arg \left(1 + ia + \sqrt{b^2 + (1+ia)^2}\right), $$
found here, I was able to deduce that
$$\int_{0}^{2\pi}\arctan\left(\frac{1+2\cos x}{\sqrt{3}}\right)dx\ =\ 2\pi\operatorname{arccot}\left(\sqrt{3+2\sqrt{3}}\right).$$
I still have no idea how they proved it though.
| $$I=\int_0^{2\pi}\arctan\left(\frac{1+2\cos x}{\sqrt 3}\right)dx\overset{\tan \frac{x}{2}\to x}=4\int_0^\infty \frac{\arctan(\sqrt 3)-\arctan\left(\frac{x^2}{\sqrt 3}\right)}{1+x^2}dx$$
$$I(t)=\int_0^\infty \frac{\arctan\left(tx^2\right)}{1+x^2}dx\Rightarrow I'(t)=\int_0^\infty \frac{x^2}{1+t^2x^4}\frac{1}{1+x^2}dx$$
$$=\frac{\pi}{2\sqrt 2}\frac{1}{1+t^2}\left(\sqrt t+\frac{1}{\sqrt t}\right)-\frac{\pi}{2}\frac{1}{1+t^2}$$
$$I\left(\frac{1}{\sqrt 3}\right)=\frac{\pi}{2\sqrt 2}\int_0^\frac{1}{\sqrt 3}\frac{1}{1+t^2}\left(\sqrt t+\frac{1}{\sqrt t}\right)dt-\frac{\pi}{2}\int_0^\frac{1}{\sqrt 3}\frac{1}{1+t^2}dt$$
$$=\frac{\pi}{2}\arctan\left(\frac{\sqrt {2t}}{1-t}\right)\bigg|_0^\frac{1}{\sqrt 3}-\frac{\pi^2}{12}=\boxed{\frac{\pi}{2}\arctan\left(\sqrt{3+2\sqrt 3}\right)-\frac{\pi^2}{12}}$$
$$\Rightarrow I=4\left(\frac{\pi^2}{6}-\mathcal J\left(\frac{1}{\sqrt 3}\right)\right)=\boxed{2\pi\operatorname{arccot}\left(\sqrt{3+2\sqrt 3}\right)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Exercise 13, Section 5.2 of Hoffman’s Linear Algebra
Let $\Bbb{R}$ be the field of real numbers, and let $D$ be a function on $2 \times 2$ matrices over $\Bbb{R}$, with values in $\Bbb{R}$, such that $D(AB) = D(A)D(B)$ for all $A, B$. Suppose
also that $$D \left(\begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix}\right)\neq D\left(\begin{bmatrix}1&0\\ 0&1\\ \end{bmatrix}\right)$$ Prove the following.
(a) $D(0)=0$;
(b) $D(A)=0$ if $A^2 = 0$;
(c) $D(B) = -D(A)$ if $B$ is obtained by interchanging the rows (or columns) of $A$;
(d) $D(A) = 0$ if one row (or one column) of $A$ is $0$;
(e) $D(A) = 0$ whenever $A$ is singular.
My attempt: (a) Since $D(AB)=D(A)D(B)$, $\forall A,B\in M_2(\Bbb{R})$ and $D \left(\begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix}\right)\neq D\left(\begin{bmatrix}1&0\\ 0&1\\ \end{bmatrix}\right)$, we have $D(I_2)=1$, by exercise 12 section 5.2. By hypothesis, $D(0)=D(0)\cdot D(0)$. By elementary properties of field, $D(0)\cdot (1-D(0))=0$. Which implies $D(0)=0$ or $D(0)=1$. If $D(0)=1$, then $$D(0) =D\left(\begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix}\right)\cdot D(0)=1= D\left(\begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix}\right)= D\left(\begin{bmatrix}1&0\\ 0&1\\ \end{bmatrix}\right).$$ Thus we reach contradiction. Hence $D(0)=0$.
(b) Suppose $A^2=0$. Then $D(A^2)$ $=D(A)\cdot D(A)$ $=D(0)$ $=0$. It’s easy to check, $(D(A))^2=0$$\iff$$D(A)=0$.
(c) Let $A= \begin{bmatrix}a&b\\ c&d\\ \end{bmatrix}$ and $B= \begin{bmatrix}c&d\\ a&b\\ \end{bmatrix}$. Since $\begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix}\cdot \begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix}$ $=\begin{bmatrix}1&0\\ 0&1\\ \end{bmatrix}$ $=I_2$, we have $D(I_2)$ $=\left[D\left(\begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix} \right) \right]^2=1$. Which implies $D\left(\begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix} \right)$ $=\pm1$. Since $D(I_2)=1$ and $D \left(\begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix}\right)\neq D(I_2)$, we have $D \left(\begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix}\right)=-1$. Since $\begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix}\cdot \begin{bmatrix}a&b\\ c&d\\ \end{bmatrix}= \begin{bmatrix}c&d\\ a&b\\ \end{bmatrix} $, we have $$D\left(\begin{bmatrix}c&d\\ a&b\\ \end{bmatrix}\right)= D\left(\begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix}\right)\cdot D\left(\begin{bmatrix}a&b\\ c&d\\ \end{bmatrix}\right)=-D\left(\begin{bmatrix}a&b\\ c&d\\ \end{bmatrix}\right).$$ Hence $D(B)=-D(A)$. Let $A= \begin{bmatrix}a&b\\ c&d\\ \end{bmatrix}$ and $B= \begin{bmatrix}b&a\\ d&a\\ \end{bmatrix}$. Since $\begin{bmatrix}a&b\\ c&d\\ \end{bmatrix}\cdot \begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix}= \begin{bmatrix}b&a\\ d&c\\ \end{bmatrix} $, we have $$D\left(\begin{bmatrix}b&a\\ d&c\\ \end{bmatrix}\right)= D\left(\begin{bmatrix}a&b\\ c&d\\ \end{bmatrix}\right)\cdot D\left(\begin{bmatrix}0&1\\ 1&0\\ \end{bmatrix}\right)=-D\left(\begin{bmatrix}a&b\\ c&d\\ \end{bmatrix}\right).$$ hence $D(B)=-D(A)$. Is my proof correct?
I tried very hard to do exercise (d) and (e), but can’t prove it. Observating pattern of proofs of (a), (b) and (c), I think we have to multiply some “clever” matrix with given matrix in such a way that $D$ of result matrix is familiar. In (d), we have to find matrix $\begin{bmatrix}p&q\\ r&s\\ \end{bmatrix}$ such that $\begin{bmatrix}a&b\\ 0&0\\ \end{bmatrix} \cdot \begin{bmatrix}p&q\\ r&s\\ \end{bmatrix}$ $= \begin{bmatrix}0&0\\ 0&0\\ \end{bmatrix}$, I guess. If we take $\begin{bmatrix}b&b\\ -a&-a\\ \end{bmatrix}$, then that works but what is value of $D\left(\begin{bmatrix}b&b\\ -a&-a\\ \end{bmatrix}\right)$.
| Too long for a comment:
For (d): Let $E_{ij}$ be the matrix of zeros except with one in the $ij$ position. Let $S=E_{12}+E_{21}$ be the 'swap' matrix.
Show that $D(E_{11})$ is either zero or one.
Show that $D(SE_{11}) = D(E_{22}S)$ and since $E_{11} E_{22} = 0$ conclude that $D(E_{11})=0$.
If the top row of $A$ is zero then $E_{11}A = A$. Similar considerations work for the other situations.
For (e): If $A$ is singular then it can be reduced to a row echelon form (may need to swap rows) in which the lower row is zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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Assistance with idempotent matrices I am taking linear algebra for the first time and am struggling with the concept of idempotent matrices. I know that $A = A^2$ is the concept behind it, but I can't seem to understand HOW one would find the entries, and the explanations given confused me quite a lot. I was hoping someone could give me the gist of this concept and point me in the right direction.
For example, I am faced with the question of "Find all $2\times 2$ matrices such that $A^2=A$"
Currently, I know that with the matrix $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ that:
$a = a^2 + bc\\
b = ab + bd\\
c = ca + cd\\
d = bc + d^2$
but I'm now unsure of how to determine entries off of that.
| Consider the system
$$a = a^2 + bc \tag{1}$$
$$b = ab + bd \tag{2}$$
$$c = ca + cd \tag{3}$$
$$d = bc + d^2 \tag{4}$$
Then, $(1) - (4)$ produces
$$a - d = a^2 - d^2 = (a - d)(a + d),$$
so $a = d$ or $a + d = 1$.
Assume, as one possible case, $a + d = 1$. Note that this implies $(2)$ and $(3)$ are automatically satisfied. As expected, if you substitute $d = 1 - a$ into $(4)$, you just get back $(1)$, so we now have only two equations (in this case):
\begin{align*}
bc &= a - a^2 \\
d &= 1 - a.
\end{align*}
We should let $a$ be a free variable; no matter what value $a$ takes, there will always be a solution, from which we can uniquely determine $d$. If $a - a^2 \neq 0$, then we can also let $b$ be a free variable, with the caveat that $b \neq 0$, and $c = \frac{a - a^2}{b}$. This produces a family of solutions:
$$\fbox{$\begin{pmatrix}
a & b \\
\frac{a - a^2}{b} & 1 - a
\end{pmatrix}, \qquad a, b \in \Bbb{R}, \quad b \neq 0$}.$$
If $a - a^2 = 0$, i.e. $a = 0$ or $a = 1$ (and $d = 1$ or $d = 0$ respectively), then we have to be more careful. We would have two cases: $b = 0$ or $c = 0$. This gives us four families of solutions (which, as you should verify, are all solutions:
$$\fbox{$\begin{pmatrix}
1 & b \\
0 & 0
\end{pmatrix},
\begin{pmatrix}
1 & 0 \\
c & 0
\end{pmatrix},
\begin{pmatrix}
0 & b \\
0 & 1
\end{pmatrix},
\begin{pmatrix}
0 & 0 \\
c & 1
\end{pmatrix}, \qquad b, c \in \Bbb{R}$}.$$
This exhausts all the cases where $a + d = 1$. Otherwise, $a = d$. Then $(2)$ becomes
$$b = 2ab$$
i.e. $b = 0$ or $a = d = \frac{1}{2}$. Since $a = d = \frac{1}{2}$ is covered by the $a + d = 1$ case, we dismiss it. So, we assume $b = 0$. Similar analysis of $(3)$ shows us that $c = 0$. Hence, $(1)$ and $(4)$, along with our assumption that $a = d$, tells us that $a = d = 0$ or $a = d = 1$. That is, we have just two remaining solutions:
$$\fbox{$\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix},
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}$},$$
i.e. the zero and identity matrices.
| {
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"url": "https://math.stackexchange.com/questions/4539018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Trouble with tedious algebra (Oxford 1992 Admissions Test 2 1992)
(i) Show that the condition that the points $P$ $(a\cos A,b\sin A)$ and $Q$ $(a\cos B,b\sin B )$
should subtend a right angle at O is
$$a^2\cos A\cos B+b^2\sin A\sin B=0$$
(ii) Let S be a circle centre $O$ and radius $C$. Find the equation of the tangent to S at
the point $(C\cos t, C\sin t)$.
(iii) If $C = \dfrac{ab}{\sqrt{a^2+b^2}}$, show that the points where a tangent to S cuts the ellipse $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
subtend a right angle at $O$.
Part (i) can be done by considering gradients of lines from $O$ to each of the points, multiplying them and setting equal to $-1$
Part (ii) gives the result $x\cos t+y\sin t=C$
With part (iii), I have attempted to simply rearrange for $y$ in the tangent equation, subbing into the ellipse equation and trying to solve for $x$ but this results in a huge amount of tedious algebra and I am unable to simplify it properly.
I then attempted to parameterize the ellipse in the form $x=a\cos T, y=b\sin T$ but I am unsure how to go from there to solve for the points. I did attempt to use harmonic addition and this does give some exact solutions in terms of $a,b,t$ but to simplify requires identities for $\sin(\arccos(x))$ and other similar identities like $\arctan\left(\frac{b}{a}\tan t\right)$.
Some useful information may be that we can let the points of intersections be $R(a \cos P,b\sin P)$ and $L(a\cos Q,b\sin Q)$ and substitute these points into the equation for the tangent. Now my problem is taking those equations and getting the equation we want from (i)
| Another way: I might say this way is "more obvious" in the sense that it is easy to think of. You "just" sit down and crunch through the algebra as if it were an ordinary simultaneous equation job. Though, one must be very careful with the algebra.
The cases where $\cos(t)=0$ or $\sin(t)=0$ can be very easily handled. Suppose now that neither is zero. Then we arrange the tangent line as: $$y=C\csc(t)-x\cot(t)$$And the ellipse equation can be arranged as: $$b^2x^2+a^2y^2-a^2b^2=0$$Let $y=C\csc(t)-x\cot(t)$ in the ellipse equation: $$\begin{align}b^2x^2+a^2[C^2\csc^2(t)+x^2\cos^2(t)\csc^2(t)-2C\cos(t)\csc^2(t)\cdot x]-a^2b^2&=0\\x^2[b^2\sin^2(t)+a^2\cos^2(t)]+a^2C^2-a^2b^2\sin^2(t)-2a^2C\cos(t)\cdot x&=0\\\frac{x^2}{a^2}[b^2\sin^2(t)+a^2\cos^2(t)]+\frac{a^2b^2}{a^2+b^2}-b^2\sin^2(t)-\frac{2ab\cos(t)}{\sqrt{a^2+b^2}}\cdot x&=0\end{align}$$Now let $x=a\cos\vartheta$ for some $\vartheta$ to be determined: $$\cos^2\vartheta[b^2\sin^2(t)+a^2\cos^2(t)]+b^2\cdot\frac{a^2\cos^2(t)-b^2\sin^2(t)}{a^2+b^2}-\frac{2a^2b\cos(t)}{\sqrt{a^2+b^2}}\cos\vartheta=0$$We use the quadratic formula. The "$b^2-4ac$" term involves a very nice difference-of-two-squares: $$\begin{align}\frac{4b^2}{a^2+b^2}a^4\cos^2(t)-&\frac{4b^2}{a^2+b^2}[b^2\sin^2(t)+a^2\cos^2(t)][a^2\cos^2(t)-b^2\sin^2(t)]\\&=\frac{4b^2}{a^2+b^2}[a^4\cos^2(t)(1-\cos^2(t))+b^4\sin^4(t)]\\&=\frac{4b^2}{a^2+b^2}\sin^2(t)\cdot[a^4\cos^2(t)+b^4\sin^2(t)]\end{align}$$The quadratic formula then yields: $$\begin{align}\cos\vartheta&=\frac{\frac{2a^2b\cos(t)}{\sqrt{a^2+b^2}}\pm\frac{2b\sin(t)}{\sqrt{a^2+b^2}}\sqrt{a^4\cos^2(t)+b^4\sin^2(t)}}{2[b^2\sin^2(t)+a^2\cos^2(t)]}\\&=\frac{b}{\sqrt{a^2+b^2}}\cdot\frac{a^2\cos(t)\pm\sin(t)\sqrt{a^4\cos^2(t)+b^4\sin^2(t)}}{b^2\sin^2(t)+a^2\cos^2(t)}\end{align}$$A near-identical computation, arising from letting $x=C\sec(t)-y\tan(t)$, cancelling the $\sec^2$ terms, and dividing by $b^2$ to get a quadratic formula in $\sin\vartheta$, then solving, gives: $$\sin\vartheta=\frac{a}{\sqrt{a^2+b^2}}\cdot\frac{b^2\sin(t)\pm\cos(t)\sqrt{a^4\cos^2(t)+b^4\sin^2(t)}}{b^4\sin^2(t)+a^4\cos^2(t)}$$Now we use the hint of part $(i)$ and the difference of two squares: $$\begin{align}a^2\cos(\vartheta_+)\cos(\vartheta_-)+b^2\sin(\vartheta_+)\sin(\vartheta_-)&=\underset{K}{\underbrace{\left(\frac{ab}{[b^4\sin^2(t)+a^4\cos^2(t)]\sqrt{a^2+b^2}}\right)^2}}\\&\times\Big[a^4\cos^2(t)-\sin^2(t)[b^4\sin^2(t)+a^4\cos^2(t)]\\&+b^4\sin^2(t)-\cos^2(t)[b^4\sin^2(t)+a^4\cos^2(t)]\Big]\\&=K\cdot\Big[a^4\cos^2(t)(1-\sin^2(t))+b^4\sin^2(t)(1-\cos^2(t))\\&-a^4\cos^4(t)-b^4\sin^4(t)\Big]\\&=0\end{align}$$
So the intersection points involving $a\cos\vartheta_{\pm},b\sin\vartheta_{\pm}$ (the pairing of which will vary in general) subtend a right angle, as desired.
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 4,
"answer_id": 1
} |
Maximum of $\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}$ for $a, b, c \ge 0$; $a+b+c=1$
If $a+b+c=1$ and $a, b, c\geq0$,
what is the maximum value of $\frac{1}{a^2-4a+9}+\frac{1}{b^2-4b+9}+\frac{1}{c^2-4c+9}$ ?
I found its answer by using CAS Calculator. The answer is $\frac{7}{18} (a=b=0, c=1)$.
But i wonder not only the answer but also the solution of it.
Please help me.
| This is something of a brute force method.
Let $f(a,b,c)=\sum_{a,b,c}1/ (a^2-4a+9)$ and $g(a,b,c)=a+b+c-1$ then calculate $\nabla f-\lambda \nabla g=0$ to find
$$
{2a-4\over (a^2-4a+9)^2}-\lambda = 0
$$
and similar for $b$, $c$. On the range $[0,1]$, this is a strictly increasing function of $a$, so equating the similar equations for $b$ and $c$ gives us a critical point where $a=b=c=\frac13$. Because we can set $a=b=0, c=1$ and get a larger value $f(0,0,1)={7\over18}$ than the value of $f(\frac13,\frac13,\frac13)={27\over70}$, it means this critical point is a minimum, so the maximum is achieved on the boundary where one or two of $a, b$ and $c$ are zero.
Let $a=0$ then $b+c=1$ and we have
$$
f(0,1-c,c)={1\over9}+{1\over(c+1)^2+5}+{1\over(c-2)^2+5}
$$
By graphing or inspecting values we have that this is a maximum on $[0,1]$ when $c=0$ or $c=1$, hence the maximum is at $(0,0,1)$.
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$|\sin\phi_{n}(x)| = \sqrt{\frac{1}{2} - \dfrac{1}{2[1+ (\omega_{n}a(x))^{2}]^{1/2}}}$ Let a sequence of real numbers $\omega_{n}$ with $\omega_{n} \to \infty$ with $a(\cdot) \in L^{\infty}(0,1)$, $a(x) > 0$ for all $x \in (0,1]$ and $a(0) = 0$.
Let $\phi_{n}(x) = -\frac{1}{2}\text{arg}[1+i\omega_{n}a(x)]$, then
$$
|\sin\phi_{n}(x)| = \sqrt{\frac{1}{2} - \dfrac{1}{2[1+ (\omega_{n}a(x))^{2}]^{1/2}}}
$$
I am thinking that the author is using $\sin(\text{arg}(z)) = \dfrac{\text{Im}(z)}{|z|}$, but I do not understand.
| We need the formulas $\arg(1+iy)=\arctan y$, $\sin\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{2}}$ and $\cos\arctan x=\frac{1}{\sqrt{1+x^2}}$. Then,
\begin{align}
|\sin\phi_n (x)|&=|\sin(-\frac{1}{2}\arctan(w_na(x))|\\
&=\left|-\sin\left(\frac{\arctan(w_na(x))}{2}\right)\right|\\
&=\left|\sin\left(\frac{\arctan(w_na(x))}{2}\right)\right|\\
&=\sqrt{\frac{1-\cos\arctan(w_na(x))}{2}}\\
&=\sqrt{\frac{1}{2}-\frac{1}{2\sqrt{1+(w_n a(x))^2}}}\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4540368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\sqrt[3]{3} + \sqrt{3} + \sqrt{2}$ is algebraic? How to show that $\sqrt[3]{3} + \sqrt{3} + \sqrt{2}$ is algebraic?
I know how to prove x = $\sqrt[3]{3} + \sqrt{2}$ and x = $\sqrt{3} + \sqrt{2}$ , but the root doesn't disappear from $\sqrt[3]{3} + \sqrt{3} + \sqrt{2}$ .
I want to make an algebraic equation that has this as an solution.
| You can show that the polynomial of degree $2\cdot 2 \cdot 3 = 12$
$$\prod_{\epsilon^2 = \epsilon'^2 = \omega^3=1} ( x -\epsilon \sqrt{2} -\epsilon' \sqrt{3} - \omega \sqrt[3]{3})$$
has integral coefficients and your number as a root.
In general you cook up a polynomial formed from all possible sums of roots in a similar way ( here use roots of $x^2-2$, $x^2-3$, $x^3-3$ ).
More hints: the above polynomial equals
$$\prod ((x\pm \sqrt{2}\pm \sqrt{3})^3 - 3)$$
| {
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An interesting sum involving binomial coefficients $\sum_{k=1}^n\frac{(-1)^{k+1} 2^{2k}}{k}\binom{n}{k}\binom{2k}{k}^{-1}$ The following sum appears in Brychkov, Marichev, and Prudnikov, Integrals and Series, Vol 1, 4.2.8, #25
$$\sum_{k=1}^n\frac{(-1)^{k+1} 2^{2k}}{k}\binom{n}{k}\binom{2k}{k}^{-1}= 2 H_{2n} - H_n$$
where $H_n$ are the harmonic numbers.
Here is what I've tried:
We have the series expansion:
$$\sum_{n\ge 1} \frac{2^{2n-1} }{n \binom{2n}{n}} x^{2n-1} = \frac{\arcsin x}{\sqrt{1-x^2}}$$
and so
$$\sum_{n\ge 1} \frac{2^{2n-1} }{n \binom{2n}{n}} x^{n} = \frac{\sqrt{x} \arcsin \sqrt{x}}{\sqrt{1-x}}$$
Now we can use the relation between the generating functions for the binomial transform
$$G(x) = \frac{1}{1-x}\cdot F(-\frac{x}{1-x})$$
Now things get a bit fuzzy : it's not clear what will be the expansion after the transformation, and how that involves harmonic numbers.
Also, here is the Taylor expansions involving harmonic numbers
$$\sum_{n\ge 1} H_n x^n= \frac{1}{1-x} \log \frac{1}{1-x}$$
and from here using $x \mapsto \pm \sqrt{x}$ and averaging, we can get the series
$$\sum_{n \ge 1} H_{2n} x^n $$
Any feedback is appreciated!
$\bf{Added:}$ Thank you all for all the great answers! I've learned a lot.
I will try to share a part of what I've learned from your answers:
*
*The binomial transform at the level of (exponential) generating series is very powerful. I have to get more comfortable with formulas.
*There are other interesting formulas that use the Beta integrals to express the inverse of binomial coefficients as an integral--- very useful.
*I've learned a new formula from this question, indicated by @Marko Riedel:
We have the identity in $\alpha$
$$\sum_{k=1}^n \frac{\binom{\alpha + n-k}{n-k}}{\binom{\alpha + n}{n}} \cdot \frac{1}{k} = \sum_{k=1}^n \frac{1}{\alpha + k}$$
This can also be rewritten as
$$\sum_{k=1}^n \frac{\binom{n}{k}}{\binom{\alpha + n}{k}} \cdot \frac{1}{k} = \sum_{k=1}^{n} \frac{1}{\alpha + k}$$
or with $\alpha = \beta- n$,
$$\sum_{k=1}^n \binom{n}{k} \frac{1}{k \binom{\beta}{k} } = \sum_{k=0}^{n-1} \frac{1}{\beta- k}$$
If we take $\alpha = -\frac{1}{2}-n$ in the formula we get our formula ( we have
$\binom{-\frac{1}{2}}{k} = \frac{(-1)^k \binom{2k}{k}}{2^{2k}}$)
$\bf{Added:}$ This formula (slightly modified) at 4.2.8. #27 appears in the Volume but only for natural values of $m$.
| We seek to verify that
$$\sum_{k=1}^n \frac{(-1)^{k+1} 2^{2k}}{k}
{n\choose k} {2k\choose k}^{-1}
= 2 H_{2n} - H_n.$$
Recall from MSE
4316307
the following identity which was proved there: with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1}
= [z^n] \log\frac{1}{1-z} (z-1)^{n-k}.$$
We get for our sum
$$\sum_{k=1}^n (-1)^{k+1} 2^{2k} {n\choose k}
[z^{2k}] \log\frac{1}{1-z} (z-1)^k
\\ = (-1)^n 2^{2n}
\sum_{k=0}^{n-1} {n\choose k} (-1)^{k+1} 2^{-2k}
[z^{2n-2k}] \log\frac{1}{1-z} (z-1)^{n-k}
\\ = (-1)^{n+1} 2^{2n} [z^{2n}] \log\frac{1}{1-z} (z-1)^n
\sum_{k=0}^{n-1} {n\choose k} (-1)^{k} 2^{-2k}
z^{2k} (z-1)^{-k}.$$
We see that we may raise $k$ to $n$ because this is a zero
contribution owing to the fact that $\log\frac{1}{1-z}$ does not have a constant term, getting
$$(-1)^{n+1} 2^{2n} [z^{2n}] \log\frac{1}{1-z} (z-1)^n
\left[1-\frac{z^2}{4(z-1)}\right]^n
\\ = (-1)^{n+1} [z^{2n}] \log\frac{1}{1-z}
\left[4z-4-z^2\right]^n
\\ = - [z^{2n}] \log\frac{1}{1-z} (z-2)^{2n}.$$
This is
$$- \; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{2n+1}} \log\frac{1}{1-z} (z-2)^{2n}.$$
Now put $z/(z-2) = v$ so that $z=2v/(v-1)$ and $dz = - 2/(v-1)^2 \; dv$
to obtain
$$\; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{2n+1}} \log\frac{1}{1-2v/(v-1)} \frac{v-1}{2}
2 \frac{1}{(v-1)^2}
\\ = - \; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{2n+1}} \frac{1}{1-v}
\log\frac{v-1}{-v-1}
\\ = - \; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{2n+1}} \frac{1}{1-v}
\log\frac{1-v}{1+v}.$$
We get two pieces, the first is
$$\; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{2n+1}} \frac{1}{1-v}
\log\frac{1}{1-v} = H_{2n}.$$
The second is
$$- \; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{2n+1}} \frac{1}{1-v}
\log\frac{1}{1+v}
= - \sum_{q=1}^{2n} \frac{(-1)^q}{q}
= - \left[\sum_{p=1}^n \frac{1}{2p}
- \sum_{p=0}^{n-1} \frac{1}{2p+1} \right]
\\ = - \frac{1}{2} H_n
+ H_{2n} - \sum_{q=1}^n \frac{1}{2q}
= H_{2n} - H_n.$$
Collecting the two pieces we obtain
$$\bbox[5px,border:2px solid #00A000]{
2 H_{2n} - H_n}$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find the first derivative of $y=\sqrt[3]{\frac{1-x^3}{1+x^3}}$ Find the first derivative of $$y=\sqrt[3]{\dfrac{1-x^3}{1+x^3}}$$
The given answer is $$\dfrac{2x^2}{x^6-1}\sqrt[3]{\dfrac{1-x^3}{1+x^3}}$$ It is nice and neat, but I am really struggling to write the result exactly in this form. We have $$y'=\dfrac13\left(\dfrac{1-x^3}{1+x^3}\right)^{-\frac23}\left(\dfrac{1-x^3}{1+x^3}\right)'$$ The derivative of the "inner" function (the last term in $y'$) is $$\dfrac{-3x^2(1+x^3)-3x^2(1-x^3)}{\left(1+x^3\right)^2}=\dfrac{-6x^2}{(1+x^3)^2},$$ so for $y'$ $$y'=-\dfrac13\dfrac{6x^2}{(1+x^3)^2}\left(\dfrac{1+x^3}{1-x^3}\right)^\frac23=-\dfrac{2x^2}{(1+x^3)^2}\left(\dfrac{1+x^3}{1-x^3}\right)^\frac23$$ Can we actually leave the answer this way?
| HINT
I would start with noticing that
\begin{align*}
y^{3} = \frac{1 - x^{3}}{1 + x^{3}} = -1 + \frac{2}{1 + x^{3}} \Rightarrow 3y^{2}y' = -\frac{6x^{2}}{(1 + x^{3})^{2}}
\end{align*}
Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
limits of 2 variables with trigonometric terms I'm trying to determine the following limit
$$ \lim_{(x,y) \to (0,0)} \frac{x^2-\sin(x^2y^2)+y^2}{x^2+\sin(x^2y^2)+y^2}$$
I tried to use polar coordinates and then Taylor and got
$$ \lim_{r \to 0} \frac{2-2 \cdot \cos^2(\theta) \cdot \sin^2(\theta) \cdot \cos(r^2\cos^2(\theta)\sin^2(\theta))}{2+2 \cdot \cos^2(\theta) \cdot \sin^2(\theta) \cdot \cos(r^2\cos^2(\theta)\sin^2(\theta))}$$
Which does not say anything I guess? How can I solve this type of questions?
| We have
$ 0\le \sin t\le t,\ 0\le t\le {\pi \over 2}.$ Thus
$${x^2+y^2-x^2y^2\over x^2+y^2+x^2y^2}\le h(x,y)\le 1\quad (*)$$
As $x^2y^2\le (x^2+y^2)^2$ we get
$${1-(x^2+y^2)\over 1+(x^2+y^2) }\le h(x,y)\le 1$$
Thus $\displaystyle\lim_{(x,y)\to (0,0)}h(x,y)=1.$
Remark We can as well apply the polar coordinates to LHS of $(*).$ Then
$${x^2+y^2-x^2y^2\over x^2+y^2+x^2y^2}={1-r^2\sin^2\theta\cos^2\theta\over 1+r^2\sin^2\theta\cos^2\theta}\to 1, \ r\to 0^+$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
prove $\exists N,\forall n\ge N$, there are more powers of $2$ within $2^1,2^2,\cdots, 2^n$ that start with $7$ than $8$.
Prove that there exists an integer $N$ so that for any $n\ge N,$ there are more powers of $2$ within $2^1,2^2,\cdots, 2^n$ that start with $7$ than $8$.
Note that there are infinitely many powers of 2 starting with any given positive integer. Also that $0.8 \cdot 0.875 = 0.7, 0.9\times 0.876 = 0.7884.$ The statement that a power of $2$, say $2^m$, starts with $7$ is equivalent to saying that there exists a power of $10$, say $10^k$ so that $7\cdot 10^k \leq 2^m < 8\cdot 10^k,$ or that $\dfrac{2^m}{10^k}\in [7,8)$ or equivalently that $m' - k\log_2 5 \in [\log_2 7, 3)$ for $m'=m-k$. It might be useful to try a few specific cases of powers of two to get an idea of how to select $N$. Note that by this link the first power of $2$ starting with 7 is $2^{46}$ and the first one starting with $9$ is $2^{64}$. Or it might be easier to use an existence proof similar to the group-theoretic proof that there are infinitely many powers of 2 starting with any given positive integer.
| A digit $d$ is the starting digit of $2^n$ if and only if there exists some $m$ for which
$$d10^m\leq 2^n<(d+1)10^m.$$
This $m$ is exactly $\lfloor n\log_{10}2\rfloor$, and so one has
$$d=\left\lfloor\frac{2^n}{10^m}\right\rfloor=\left\lfloor\frac{2^n}{10^{\lfloor n\log_{10}2\rfloor}}\right\rfloor=\left\lfloor 10^{\{n\log_{10}2\}}\right\rfloor,$$
where $\{x\}=x-\lfloor x\rfloor$ is the fractional part of $x$. Put another way, $d$ is the starting digit of $2^n$ if and only if
$$\log_{10}d\leq \{n\log_{10}2\}<\log_{10}(d+1).$$
Now, note that $\log_{10}2$ is irrational, so the sequence $a_n=\{n\log_{10}2\}$ is uniformly distributed within $[0,1)$. That is, as $N\to\infty$, for any $0\leq a<b<1$,
$$\frac{\text{# of }1\leq n\leq N\text{ with }a\leq \{n\log_{10}2\}<b}{N}$$
tends to $b-a$. In particular, the proportion of $n$ for which $2^n$ has first digit $7$ tends to $\log_{10}(8/7)$, while the proportion of $n$ for which $2^n$ has first digit $8$ tends to $\log_{10}(9/8)$. Since $\log_{10}(9/8)<\log_{10}(8/7)$, eventually the first quantity is larger than the second, as desired.
| {
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"source": "stackexchange",
"question_score": "2",
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Conditions for generalized projection matrix of size (2x2)? My results seem incorrect... I am trying to derive the general conditions that must be imposed upon the real-valued entries of a $2 \times 2$ orthogonal projection matrix. However, I am coming to a conclusion that seems wrong and hope someone can point me in the right direction.
The prompt reads as follows:
Find conditions on $a,b,c,d \in \mathbb{R}$ that guarantee the matrix $\begin{pmatrix} a & b \\ c & d\end{pmatrix}$ defines a rank-1 orthogonal projection.
I begin by imposing the conditions that all projection matrices $P$ must fulfill, namely
(i) $P = P^T$
(ii) $P = P^2$
Applying condition (i),
$\begin{pmatrix} a & b \\ c & d\end{pmatrix} \overset{!}{=} \begin{pmatrix} a & c \\ b & d\end{pmatrix}$
I get $b=c$ and so I continue working with $\begin{pmatrix} a & c \\ c & d\end{pmatrix}$ and move on to condition (ii):
$\begin{pmatrix} a & c \\ c & d\end{pmatrix} \overset{!}{=} \begin{pmatrix} a & c \\ c & d\end{pmatrix}\begin{pmatrix} a & c \\ c & d\end{pmatrix}=\begin{pmatrix} a^2 +c^2 & ac + cd \\ ac +cd & d^2 + c^2\end{pmatrix}$
From that, I get three equations:
(1) $a = a^2 +c^2$
(2) $c=ac+cd=c(a+d)$
(3) $d=d^2 + c^2$
Further working out (2) by striking $c$ from each side, the system is then
(1) $a = a^2 +c^2$
(2) $1=a+d$
(3) $d=d^2+c^2$
I re-arrange (1) and (3) to get
(1) $c^2 = a - a^2$
(3) $c^2 = d - d^2$
and thus $a-a^2 = d-d^2$ from which I conclude that $a = d$. Returning to equation (2), I then get
(2) $1 = a+d = 2a$
and so $a = \frac{1}{2}$ and $d = \frac{1}{2}$.
Plugging either one of these into equation (1) or (3) then allows me to solve for $c$,
$\frac{1}{2} = \left ( \frac{1}{2} \right )^2+c^2$
or
$c^2 = \frac{1}{4}$
And thus $c=\pm \frac{1}{2}$.
So, according to these results, to be guaranteed an orthogonal projection matrix of rank-1, my matrix $P$ must be one of two possibilities (as indicated by $\pm$):
$P = \frac{1}{2} \begin{pmatrix} 1 & \pm 1 \\ \pm 1 & 1\end{pmatrix}$
This result is unsettling. First of all, I was expecting a broader range of possibilities for $a,b,c,d$ or at least for one or two variables. It seems odd that $a$ and $d$ are wholly constrained to one value and that $c$ only has two options.
Is this correct? It seems like there should be more here that is possible. Aren't matrices such as $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ or $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ also to be included here?
For example, when I apply $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ to a general vector $\begin{pmatrix} x \\ y \end{pmatrix}$, it's easy for me to see the orthogonal projection:
$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} x \\ 0\end{pmatrix}$
But when I apply the resulting projection matrix from above to the same generalized vector, I get a result that does reveal orthogonal projection to me (using $c = + \frac{1}{2}$):
$\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{pmatrix} \begin{pmatrix} x \\ y\end{pmatrix} = \frac{1}{2} \begin{pmatrix} x+y \\ x+y\end{pmatrix}$
How can that be an orthogonal projection? Isn't that simply a change in length along the same direction? Am I not understanding the fundamental concept?
Thanks for any help you might be able to provide.
| According to this source, the projection matrix is given by
$ P = A (A^T A)^{-1} A^T $
So you can select $A$ to be any vector, and in particular a unit vector in $\mathbb{R}^2$, i.e.
$ A = [\cos \theta, \sin \theta ]^T $
The projection matrix parametrized by $ \theta $ is
$ P = \begin{bmatrix} \cos \theta \\ \sin \theta \end{bmatrix} \begin{bmatrix} \cos \theta && \sin \theta \end{bmatrix} = \begin{bmatrix} \cos^2 \theta && \cos \theta \sin \theta \\ \cos \theta \sin \theta && \sin^2 \theta \end{bmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4550807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Sum of reciprocal of primes failed computation Set $$X:=\sum_p \dfrac{1}{p^2}=\dfrac{1}{2^2}+\dfrac{1}{3^2} +\dfrac{1}{5^2}+\cdots$$
As $X$ is absolute convergent and less than $1$, we have (not sure for infinite rearrangement) naive calculation implies
$$\dfrac{\pi^2}{6}-1=\sum_{\substack{n\\ \Omega(n)=1}}\dfrac{1}{n^2}+\sum_{\substack{n\\ \Omega(n)=2}}\dfrac{1}{n^2}+\cdots,$$
where $\Omega(n)$ denotes the number of prime factors of $n$ counting multiple.
The right term above is
$$X+X^2+\cdots.$$
Thus, $X=1-6/\pi^2<0.4$, but with the answer in the following link this is false.
https://mathoverflow.net/questions/53443/sum-of-the-reciprocal-of-the-primes-squared
Is it possible to fill the gap of the calculation?
| $X^2$ and $\sum_{n, \Omega(n) = 2} \frac{1}{n^2}$ are distinct: the second one only counts each $\frac{1}{p^2q^2}$ once, while $X^2$ will count both the pair $\frac{1}{p^2q^2}$ and $\frac{1}{q^2p^2}$, thanks to distributivity, which adds up to $\frac{2}{p^2q^2}$.
The same logic works for all the higher $k$, the number of times each $\frac{1}{p_1^2 \dots p_k^2}$ will appear in $X^k$ is $k!$, meaning:
$$X^k \geq k! \sum_{n, \Omega(n) = k} \frac{1}{n^2}$$
Why the $\geq$? Because $X^k$ will also contain numbers of the form $\frac{1}{p^{2k}}$, $\frac{1} {p^{2(k-1)}q^2} $ and so on! Meaning it's going to be a more complicated expression sadly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4551701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving the inequality with finite sums I would like to show the following inequality involving finite sums ($x>0$):
$$2 \left(\sum_{n=1}^k nx^n\right)^2 + \sum_{n=0}^k x^n \sum_{n=1}^k nx^n - \sum_{n=0}^k x^n \sum_{n=1}^k n^2x^n \geq 0 $$
I thought an induction would work, but then realized what really matters is the first few terms in series. In my view, induction will be of no use. After plugging in some value for $k$, I noticed a nice pattern. For ex: when $k=3$, first $3$ terms get cancelled. When $k=4$, first 4 terms get cancelled. However, I am unable to write things precisely. For a given $k$, the coefficients of first $k$ terms are $1,5,14,30,55...$. The hard part is actually to figure out the stuff after $k$ terms.
I was wondering if there would be another way to go about showing this. Any help/hint would be great.
A direct approach will definitely work. Is there a way to write these products of the form $\sum_{n=1}^{2k} c_n x^n$?.
| We obtain
\begin{align*}
\color{blue}{2}&\color{blue}{\left(\sum_{n=1}^knx^n\right)^2+\sum_{n=0}^kx^n\sum_{n=1}^knx^n
-\sum_{n=0}^kx^n\sum_{n=1}^kn^2x^n}\\
&=2\left(\sum_{n=1}^knx^n\right)^2
+\sum_{n=0}^kx^n\left(\sum_{n=0}^knx^n-\sum_{n=0}^kn^2x^n\right)\tag{1}\\
&=2\left(\sum_{n=1}^knx^n\right)^2
+\left(\sum_{n=0}^kx^n\right)\left(\sum_{n=0}^kn(1-n)x^n\right)\tag{2}\\
&=2\sum_{q=0}^{2k}\left(\sum_{{m+n=q}\atop{0\leq m,n\leq k}}mn\right)x^q
+\sum_{q=0}^{2k}\left(\sum_{{m+n=q}\atop{0\leq m,n\leq k}}n(1-n)\right)x^q\tag{3}\\
&=\sum_{q=0}^{2k}\left(\sum_{{m+n=q}\atop{0\leq m,n\leq k}}2mn+n(1-n)\right)x^q\\
&\,\,\color{blue}{=\sum_{q=0}^{k}\left(\sum_{n=0}^q\left(2n(q-n)+n(1-n)\right)\right)x^q}\\
&\,\,\color{blue}{\qquad+\sum_{q=k+1}^{2k}\left(\sum_{n=q-k}^k\left(2n(q-n)+n(1-n)\right)\right)x^q}\tag{4}\\
\end{align*}
Comment:
*
*In (1) we factor out $\sum_{n=0}^kx^k$ and start indices with zero which is simply an addition of zero terms.
*In (2) we merge the inner sums.
*In (3) we apply the Cauchy product of series.
*In (4) we eliminate the index variable $m$ by substituting $m=q-n$. We also split the sum due to the restriction $0\leq m,n\leq k$.
The left-hand inner sum of (4) simplifies to
\begin{align*}
\color{blue}{\sum_{n=0}^q}&\color{blue}{\left(2n(q-n)+n(1-n)\right)}\\
&=\sum_{n=0}^q\left((2q+1)n-3n^2\right)\\
&=(2q+1)\sum_{n=0}^qn-3\sum_{n=0}^qn^2\\
&=(2q+1)\frac{q(q+1)}{2}-3\,\frac{q(q+1)(2q+1)}{6}\\
&\,\,\color{blue}{=0}\tag{5}
\end{align*}
The right-hand inner sum of (4) simplifies to
\begin{align*}
\color{blue}{\sum_{n=q-k}^k}&\color{blue}{\left(2n(q-n)+n(1-n)\right)}\\
&=\sum_{n=q-k}^k\left((2q+1)n-3n^2\right)\\
&=(2q+1)\frac{1}{2}\,k(k+1)-3\frac{1}{6}\,k(k+1)(2k+1)\\
&\qquad-(2q+1)\frac{1}{2}\,(q-k-1)(q-k)+3\frac{1}{6}\,(q-k-1)(q-k)(2q-2k-1)\\
&\,\,\qquad\color{blue}{=(k+1)\left(3kq-2k^2-q^2+q-k\right)}\tag{6}
\end{align*}
From (1) and (4) to (6) we finally derive
\begin{align*}
\color{blue}{2}&\color{blue}{\left(\sum_{n=1}^knx^n\right)^2+\sum_{n=0}^kx^n\sum_{n=1}^knx^n
-\sum_{n=0}^kx^n\sum_{n=1}^kn^2x^n}\\
&\qquad\,\,\color{blue}{=(k+1)\sum_{q=k+1}^{2k}\left(3kq-2k^2-q^2+q-k\right)x^q}\tag{7}
\end{align*}
A plausibility check for small values $k$ verifies (7) with somewhat help from Wolfram Alpha.
\begin{align*}
\begin{array}{c|c}
k&\mathrm{polynomial}\\
\hline
1&2x^2\\
2&3\left(2x^3+2x^4\right)\\
3&4\left(3x^4+4x^5+3x^5\right)\\
4&5\left(4x^5+6x^6+6x^7+5x^8\right)\\
5&6\left(5x^6+8x^7+9x^8+8x^9+5x^{10}\right)\\
6&7\left(6x^7+10x^8+12x^9+12x^{10}+10x^{11}+6x^{12}\right)\\
\end{array}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4557407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
For c = 3 find the complete solution $x$ to $Ax = \begin{bmatrix}1\\c\\0\end{bmatrix} $ the matrix is: $$\color{green}{ \begin{bmatrix}1&1&2&4\\3&c&2&8\\0&0&2&2\end{bmatrix} }\cdot$$
reducing the matrix gives me:
$$\color{green}{ \begin{bmatrix}1&1&2&4\\0&c-3&-4&-4\\0&0&2&2\end{bmatrix} }\cdot$$
I already calculated the solution for the nullspace of A
$$\color{green}{ \begin{bmatrix}-2\\0\\-1\\1\end{bmatrix} }\cdot$$
the solution is one vector + the solution for the nullspace help me find that first vector by inspection
| $\begin {pmatrix}0\\1\\0\\0\end {pmatrix} $ works nicely as a particular solution, and the solution set is thus $$\{\begin {pmatrix}0\\1\\0\\0\end {pmatrix}+s\cdot \begin {pmatrix}-2\\0\\-1\\1\end {pmatrix}+t\cdot \begin {pmatrix}1\\-1\\0\\0\end {pmatrix}\mid s,t\in\Bbb F\}$$.
That's the null space is actually $\color{red}2$-dimensional.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4557725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\sqrt{1+x^2+y^2}$ Calculate $$I=\int_{-1}^1\int_{-1}^1\sqrt{1+x^2+y^2}\,\mathrm{d}y\,\mathrm{d}x.$$
It's a problem from a book about calculus. My attempt:
$$\begin{align}
I &= \int_{-1}^1\int_{-1}^1\sqrt{1+x^2+y^2}\,\mathrm{d}y\,\mathrm{d}x \\
&= \int_{-1}^1\left.\frac{x}{2}\sqrt{1+x^2+y^2}+\frac{1+y^2}{2}\log\left(x+\sqrt{1+x^2+y^2}\right)\right|_{-1}^{1}\,\mathrm{d}y \\
&= \int_{-1}^1\sqrt{2+y^2}+\frac{1+y^2}{2}\left(\log \left(\sqrt{2+y^2}+1\right)-\log\left(\sqrt{2+y^2}-1\right)\right)\,\mathrm{d}y \\
&= \sqrt{3}+2\operatorname{arsinh}\frac{1}{\sqrt{2}}+\int_{-1}^1(y^2+1) \operatorname{arsinh} \frac{1}{\sqrt{y^2+1}}\,\mathrm{d}y \\
&= \color{red}\ldots \\
&= -\frac{2}{9} (\pi + 12 \log 2 - 6 \sqrt{3} - 24 \log (1+ \sqrt{3}))
\end{align}$$
(answer taken from the solutions, no idea how to reach it).
[edit] Here is an attempt with polar coordinates. Due to symmetry, it's enough to integrate over $0 \le x \le 1$ and $0 \le y \le x$, 1/8th of the initial square.
$$\begin{align}
I &= 8\int_0^{\pi/4} \int_0^{1/\cos \theta}r \sqrt{1+r^2}\,\mathrm{d}\theta\\
&= 8\int_0^{1/\cos \theta} \frac{(1+1/\cos^2\theta)^{3/2}-1}{3}\,\mathrm{d}\theta\\
&= {?}
\end{align}$$
| Not an answer, but a start. Might not work.
In polar coordinates, it is $$4\int_{0}^{\sqrt2} f(r)\sqrt{1+r^2}\,dr$$ where $f(r)=\pi r/2$ for $0\leq r\leq 1.$
When $1<r\leq \sqrt2,$ we need $f(r)$ to be the length of the arc of radius $r$ around $(0,0)$ contained in the square $[0,1]^2.$
This is $$f(r)=r\left(\frac\pi2-2\arctan\left(\sqrt{r^2-1}\right)\right).$$
So the integral is:
$$4\frac{\pi}2\int_0^1 r\sqrt{1+r^2}\,dr+4\int_1^{\sqrt2}r\left(\frac\pi2-2\arctan\left(\sqrt{r^2-1}\right)\right)\sqrt{1+r^2}\,dr$$
The first integral is easy. Don't know what to do about the second, however.
When $x\leq 1,$ $2\arctan x =\arctan\frac{2x}{1-x^2}$ and $\frac{\pi}2-\arctan(y)=\arctan(1/y).$
So $$\frac{\pi}2-2\arctan\left(\sqrt{r^2-1}\right)=\arctan\left(\frac{2-r^2}{2\sqrt{ r^2-1}}\right)$$
If we set $r=\sec\theta,$ then $$\arctan\left(\frac{2-r^2}{2\sqrt{ r^2-1}}\right)=\frac\pi2-2\theta.$$
So the integral becomes:
$$\int_{0}^{\pi/4}\left(\frac\pi2-2\theta\right)\sec^2\theta\tan(\theta)\sqrt{1+\sec^2\theta}\,d\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Find the real root of the almost symmetric polynomial $x^7+7x^5+14x^3+7x-1$
Find the real root of following almost symmetric polynomial by radicals $$p(x)=x^7+7x^5+14x^3+7x-1$$
Here are my attempts.
The coefficients of $p(x)$ are : $1,7,14,7,-1$.
I wanted to try possible factorizations. But Wolfram Alpha can not factorise this polynomial. This can be a reason of our case, so factorisation over $\Bbb R$ seems impossible.
The Rational root theorem also failed.
Again I tried
$$\begin{align}
x^7+7x^5+14x^3+7x-1
&=x^7+7x^5+7x^3+7x^3+7x-1 \\
&=x^7+7x^3(x^2+1)+7x(x^2+1)-1 \\
&=x^7+7x(x^2+1)^2-1
\end{align}$$
But, this manipulation also didn't work.
| Remarks: For a cubic equation $x^3 + px + q = 0$, we use the identity $(u + v)^3 \equiv 3uv(u + v) + u^3 + v^3$ and let $x = u + v$.
Similarly, we have the identities
$$(a+b)^4 - 4ab(a + b)^2 - (a^4 + b^4 - 2a^2b^2) \equiv 0,$$
$$(a+b)^5 - 5ab(a+b)^3 + 5a^2b^2(a+b) - (a^5+b^5) \equiv 0.$$
In general, $a^n + b^n$ ($n\in \mathbb{Z}_{>0}$) can be expressed in terms of $ab$ and $a + b$. See: 1, and 2.
We use the identity
$$(u + v)^7 - 7uv(u+v)^5 + 14u^2v^2(u + v)^3 - 7u^3v^3(u+v) - (u^7+v^7) \equiv 0. \tag{1}$$
Let $x = u + v$. From (1), we have
$$x^7 - 7uvx^5 + 14u^2v^2x^3 - 7u^3v^3x - (u^7+v^7) = 0. \tag{2}$$
If $uv = -1$ and $u^7 + v^7 = 1$, then
(2) gives the equation $x^7+7x^5+14x^3+7x-1 = 0$.
Since $u^7, v^7$ are roots of $y^2 - y - 1 = 0$, we have
$$u = \sqrt[7]{\frac{\sqrt 5 + 1}{2}}, \quad
v = - \sqrt[7]{\frac{\sqrt 5 - 1}{2}}.$$
Thus, one root of the equation is given by
$$x = u + v = \sqrt[7]{\frac{\sqrt 5 + 1}{2}} - \sqrt[7]{\frac{\sqrt 5 - 1}{2}} \approx 0.1375974100.$$
One can prove that
$$x_k = \mathrm{e}^{\mathrm{i}2\pi k/7} u + \mathrm{e}^{- \mathrm{i}2\pi k/7} v$$
are roots of the equation for $k = 0, 1, \cdots, 6$, which are all roots of the equation.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4559276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
A certain subgroup is isomorphic to Klein group I'm given this presentation
$$G=\langle x,y \mid x^2=1, y^3=1, (xy)^4=1\rangle,$$
and I have to prove that $N=\langle (xy)^2, (yx)^2\rangle$ is isomorphic to the Klein group $V=\{1, a, b, ab\}$.
I've tried identifying $(xy)^2$ with $a$ and $(yx)^2$ with $b$, which until now is ok because $((xy)^2)^2=1, ((yx)^2)^2=1$, but I have no idea how to prove that $((xy)^2(yx)^2)^2=1$
Can anyone help?
| We have $(xy)^2 = (xy)^{-2} = (y^{-1}x^{-1})^2 = (y^2x)^2$. And $(yx)^2 = (yx)^{-2}=(x^{-1}y^{-1})^2 = (xy^2)^2$.
So
$$\begin{align*}
\Bigl( (xy)^2(yx)^2\Bigr)^2 &= \Bigl( (y^2x)^2(xy^2)^2\Bigr)^2\\
&= \Bigl( y^2xy^2xxy^2xy^2\Bigr)^2\\
&= \Bigl( y^2xy^4xy^2\Bigr)^2\\
&= \Bigl( y^2xyxy^2\Bigr)^2\\
&= y^2xyxy^2y^2xyxy^2\\
&= y^2xyxyxyxy^2\\
&= y(yx)(yx)(yx)(yx)y^2\\
&= y(yx)^4y^2\\
&= yy^2\\
&=e
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4562288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Do we have a closed form for $\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^n} d t $? Latest Edit
We are glad to see there are 4 alternative solutions which give the same closed form to the integral:
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a^2+t^2\right)^n} d t = \frac{a^{1-2n}\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left[2\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right] },$$
where $\psi$ denotes the Digamma Function.
In the post, we found that
$$\int_0^{\infty} \frac{\ln x}{a^2+t^2} d x =\frac{\pi \ln a}{2 a }$$
Now I want to generalise the integral as
$$
I_n=\int_0^{+\infty} \frac{\ln t}{\left(a+t^2\right)^n} d t
$$
where $n\in N$.
Replacing a by $\sqrt{a}$ and differentiating both sides w.r.t. $a$ by $n$ times yields
$$
\begin{aligned}
&J(a)=\int_0^{\infty} \frac{\ln t}{a+t^2} d t=\frac{\pi}{4 \sqrt{a}} \ln a\\
&\frac{d^n}{d a^n}(J(a))=\frac{\pi}{4} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right) \\ & (-1)^n n ! \int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)} d t= \frac{\pi}{4} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)\\& \boxed{\int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)^{n+1}} d t=\frac{(-1)^n \pi}{4 n !} \frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)}
\end{aligned}
$$
By Wolfram-alpha, we have
$$
\frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right)=(-1)^n a^{-\frac{1}{2}-n}\left(\frac{1}{2}\right)_n\left(\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)\cdots (*)
$$
Hence
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a+t^2\right)^{n+1}} d t=\frac{\pi}{4 n !} a^{-\frac{1}{2}-n}\left(\frac{1}{2}\right)_n\left(\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)}$$
In particular, when $a=1$, we have
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^{n+1}} d t=\frac{\pi}{4 n !} \left(\frac{1}{2}\right)_n\left(\psi\left(\frac{1}{2}\right)-\psi\left(\frac{1}{2}-n\right)\right)}$$
For example,
$$$$
\begin{aligned}
\int_0^\infty \frac{\ln t}{\left(1+t^2\right)^4} d t=& \frac{\pi}{24} \cdot \frac{15}{8}\left(-\gamma -\ln 4-\frac{46}{15}+\gamma +\ln 4\right)
=-\frac{23 \pi}{96}
\end{aligned}
$$
$$
My Question: How to find a closed form for $\frac{d^n}{d a^n}\left(\frac{\ln a}{\sqrt{a}}\right) $?
| As @metamorphy suggested, let $I\left(\lambda\right)=\int_0^{\infty} \frac{t^{2 \lambda-1}}{\left(a^2+t^2\right)^n} d t,$ then
$$I_n=\int_0^{\infty} \frac{\ln t}{\left(a^2 +t^2\right)^n} d t= \frac{1}{2} I^{\prime}\left(\frac{1}{2}\right) .$$
Now we are going to express $I\left(\lambda\right)$ as a beta function by letting $t=a\tan \theta$, then
$$
\begin{aligned}
I(\lambda) &= a^{2(\lambda-n)}\int_0^{\frac{\pi}{2}} \sin ^{2 \lambda-1} \theta \cos ^{2(n-\lambda)-1} \theta d \theta \\
&=\frac{a^{2(\lambda-n)}}{2} B(\lambda, n-\lambda) \\
&=\frac{a^{2(\lambda-n)}}{2\Gamma(n) } \Gamma(\lambda) \Gamma(n-\lambda)
\end{aligned}
$$
By logarithmic differentiation, we get
$$
\begin{aligned}
&\frac{I^{\prime}(\lambda)}{I(\lambda)}=2\ln a+\psi(\lambda)-\psi(n-\lambda) \\
&I^{\prime}(\lambda)=\frac{a^{2(\lambda-n)}}{2 \Gamma(n)} \Gamma(\lambda) \Gamma(n-\lambda)[ 2\ln a+\psi(\lambda)-\psi(n-\lambda)]
\end{aligned}
$$
Putting $\lambda=\frac{1}{2} $ yields
$$
I^{\prime}\left(\frac{1}{2}\right)=\frac{a^{1-2n}\sqrt{\pi}}{2} \cdot \frac{\Gamma\left(n-\frac{1}{2}\right)}{\Gamma(n)}\left[2\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right]
$$
Hence $$\boxed{\int_0^{\infty} \frac{\ln t}{\left(a^2+t^2\right)^n} d t= \frac{1}{2} I^{\prime}\left(\frac{1}{2}\right) = \frac{a^{1-2n}\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left[2\ln a+\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right] },$$
which is same as the answer provided by @Felix Marin.
Back to our integral,
$$\boxed{\int_0^{\infty} \frac{\ln t}{\left(1+t^2\right)^n} d t= \frac{1}{2} I^{\prime}\left(\frac{1}{2}\right) = \frac{\sqrt{\pi}\Gamma \left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left[\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right)\right] }$$
By Wolfram-Alpha, $\psi\left(\frac{1}{2}\right)-\psi\left(n-\frac{1}{2}\right) = -H_{n-\frac{3}{2}}-2 \log (2) $ yields
$$I_n= -\frac{\sqrt{\pi}\Gamma\left(n-\frac{1}{2}\right)}{4\Gamma(n)}\left(H_{n-\frac{3}{2}}+2\ln 2\right),$$
which is the same as the answer provided by @Claude Leibovici.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4562577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How to take the square root of a dual number ($\sqrt{a + b \cdot \varepsilon}$ with $a, ~b \in \mathbb{R}$ and $\varepsilon^{2} = 0$)? How to take the square root of a dual number:
$\sqrt{\Xi}$ with $\begin{align*}
a, ~b &\in \mathbb{R}\\
\varepsilon^{2} &= 0\\
\varepsilon &\ne 0\\
\\
\Xi &:= a + b \cdot \varepsilon\\
\end{align*}$
My first attempt worked (at least I think so), but the second one came to nothing, but it seems to me that taking the square root of binary numbers can also be derived in a different way, which is why I am here under the question in the question and the answers to show all possibilities to the question. (I wrote my first attempt as an answer under the question.)
Surely the question sounds strange, because why should one calculate that at all, but let's just ask ourselves how that would work?
My attempt $2$
My attempt $2$ was "try and hope that it works":
$$
\begin{align*}
\Xi^{2} &= \left( a + b \cdot \varepsilon \right)^{2}\\
\Xi^{2} &= a^{2} + 2 \cdot a \cdot + b \cdot \varepsilon + \left( b \cdot \varepsilon \right)^{2}\\
\Xi^{2} &= a^{2} + 2 \cdot a \cdot b \cdot \varepsilon + b^{2} \cdot \varepsilon^{2}\\
\Xi^{2} &= a^{2} + 2 \cdot a \cdot b \cdot \varepsilon + b^{2} \cdot 0\\
\Xi^{2} &= a^{2} + 2 \cdot a \cdot b \cdot \varepsilon + 0\\
\Xi^{2} &= a^{2} + 2 \cdot a \cdot b \cdot \varepsilon \\
\Xi &= \sqrt{a^{2} + 2 \cdot a \cdot b \cdot \varepsilon}\\
a + b \cdot \varepsilon &= \sqrt{a^{2} + 2 \cdot a \cdot b \cdot \varepsilon}\\
c := a^{2} &\wedge d := 2 \cdot a \cdot b \cdot \varepsilon\\
a = \sqrt{c} &\wedge b = \frac{d}{2 \cdot a} \cdot \varepsilon^{-1}\\
a = \sqrt{c} &\wedge b = \frac{d}{2 \cdot \sqrt{c}} \cdot \varepsilon^{-1}\\
\\
\Xi &= \sqrt{a^{2} + 2 \cdot a \cdot b \cdot \varepsilon}\\
a + b \cdot \varepsilon &= \sqrt{c + d \cdot \varepsilon} \quad\mid\quad a = \sqrt{c} \wedge b = \frac{d}{2 \cdot \sqrt{c}} \cdot \varepsilon^{-1}\\
\sqrt{c} + \frac{d}{2 \cdot \sqrt{c}} \cdot \varepsilon^{-1} \cdot \varepsilon &= \sqrt{c + d \cdot \varepsilon}\\
\sqrt{c} + \frac{d}{2 \cdot \sqrt{c}} \cdot 1 &= \sqrt{c + d \cdot \varepsilon}\\
\sqrt{c} + \frac{d}{2 \cdot \sqrt{c}} &= \sqrt{c + d \cdot \varepsilon}\\
\sqrt{c + d \cdot \varepsilon} &= \sqrt{c} + \frac{d}{2 \cdot \sqrt{c}}\\
\sqrt{a + b \cdot \varepsilon} &= \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \quad\mid\quad \left( ~~ \right)^{2}\\
a + b \cdot \varepsilon &= \left( \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \right)^{2}\\
a + b \cdot \varepsilon &= a + 2 \cdot \sqrt{a} \cdot \frac{b}{2 \cdot \sqrt{a}} + \frac{b^{2}}{4 \cdot a}\\
a + b \cdot \varepsilon &= a + b + \frac{b^{2}}{4 \cdot a}\\
b \cdot \varepsilon &= b + \frac{b^{2}}{4 \cdot a}\\
\end{align*}
$$
But that makes no sense...
| My attempt $1$:
I wanted to try to derive the method of taking the square root of a dual number analogously to taking the square root of complex numbers.
So I started:
** Derivation **
$$
\begin{align*}
0 &\geq k \in \mathbb{Z}\\
\varepsilon &\ne 0\\
\varepsilon^{2} &= 0\\
\varepsilon^{3} &= \varepsilon^{2} \cdot \varepsilon = 0 \cdot \varepsilon = 0\\
\varepsilon^{4} &= \left( \varepsilon^{2} \right)^{2} \cdot \varepsilon = 0^{2}= 0\\
\varepsilon^{k} &= \varepsilon^{2 + k - 2} = \varepsilon^{2} \cdot \varepsilon^{k - 2} = 0 \cdot \varepsilon^{k - 2} = 0\\
\\
x &\in \mathbb{R}\\
\\
e^{x \cdot \varepsilon} &= \exp\left( x \cdot \varepsilon \right)\\
e^{x \cdot \varepsilon} &= \sum_{n = 0}^{\infty} \frac{\left( x \cdot \varepsilon \right)^{n}}{n!}\\
e^{x \cdot \varepsilon} &= 1 + x \cdot \varepsilon + \frac{\left( x \cdot \varepsilon \right)^{2}}{2!} + \frac{\left( x \cdot \varepsilon \right)^{3}}{3!} + \frac{\left( x \cdot \varepsilon \right)^{4}}{4!} + \cdots\\
e^{x \cdot \varepsilon} &= 1 + x \cdot \varepsilon + \frac{x^{2} \cdot \varepsilon^{2}}{2!} + \frac{x^{3} \cdot \varepsilon^{3}}{3!} + \frac{x^{4} \cdot \varepsilon^{4}}{4!} + \cdots\\
e^{x \cdot \varepsilon} &= 1 + x \cdot \varepsilon + \varepsilon^{2} \cdot \left( \frac{x^{2}}{2!} + \frac{x^{3} \cdot \varepsilon}{3!} + \frac{x^{4} \cdot \varepsilon^{2}}{4!} + \cdots \right)\\
e^{x \cdot \varepsilon} &= 1 + x \cdot \varepsilon + 0 \cdot \left( \frac{x^{2}}{2!} + \frac{x^{3} \cdot \varepsilon}{3!} + \frac{x^{4} \cdot \varepsilon^{2}}{4!} + \cdots \right)\\
e^{x \cdot \varepsilon} &= 1 + x \cdot \varepsilon + 0\\
e^{x \cdot \varepsilon} &= 1 + x \cdot \varepsilon\\
\\
\Xi &= y \cdot \left( 1 + x \cdot \varepsilon \right)\\
\Xi &= y + y \cdot x \cdot \varepsilon \quad\mid\quad \Xi = a + b \cdot \varepsilon\\
a &= y \wedge b = y \cdot x\\
a &= y \wedge \frac{b}{y} = x\\
a &= y \wedge \frac{b}{a} = x\\
\\
\Xi &= y \cdot e^{x \cdot \varepsilon} \quad\mid\quad \sqrt{~~}\\
\sqrt{\Xi} &= \sqrt{y \cdot e^{x \cdot \varepsilon}}\\
\sqrt{\Xi} &= \sqrt{y} \cdot \sqrt{e^{x \cdot \varepsilon}}\\
\sqrt{\Xi} &= \sqrt{y} \cdot \left( e^{x \cdot \varepsilon} \right)^{\frac{1}{2}}\\
\sqrt{\Xi} &= \sqrt{y} \cdot e^{\frac{x}{2} \cdot \varepsilon} \quad\mid\quad
e^{x \cdot \varepsilon} = 1 + x \cdot \varepsilon\\
\sqrt{\Xi} &= \sqrt{y} \cdot \left( 1 + \frac{x}{2} \cdot \varepsilon \right) \quad\mid\quad a = y \wedge \frac{b}{a} = x\\
\sqrt{\Xi} &= \sqrt{a} \cdot \left( 1 + \frac{\frac{b}{a}}{2} \cdot \varepsilon \right)\\
\sqrt{\Xi} &= \sqrt{a} \cdot \left( 1 + \frac{b}{2 \cdot a} \cdot \varepsilon \right)\\
\sqrt{\Xi} &= \sqrt{a} \cdot 1 + \sqrt{a} \cdot \frac{b}{2 \cdot a} \cdot \varepsilon\\
\sqrt{\Xi} &= \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon\\
\end{align*}
$$
** Proof **
$$
\begin{align*}
\sqrt{\Xi} &= \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon\\
\sqrt{\Xi} &= \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon\\
\sqrt{a + b \cdot \varepsilon} &= \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon \quad\mid\quad \left( ~~ \right)^{2}\\
a + b \cdot \varepsilon &= \left( \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon \right)^{2}\\
a + b \cdot \varepsilon &= \left( \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon \right)^{2}\\
a + b \cdot \varepsilon &= a + 2 \cdot \sqrt{a} \cdot \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon + \left( \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon \right)^{2}\\
a + b \cdot \varepsilon &= a + b \cdot \varepsilon + \left( \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon \right)^{2}\\
a + b \cdot \varepsilon &= a + b \cdot \varepsilon + \left( \frac{b}{2 \cdot \sqrt{a}} \right)^{2} \cdot \varepsilon^{2}\\
a + b \cdot \varepsilon &= a + b \cdot \varepsilon + \left( \frac{b}{2 \cdot \sqrt{a}} \right)^{2} \cdot 0\\
a + b \cdot \varepsilon &= a + b \cdot \varepsilon \quad \square\\
\end{align*}
$$
Aka the solution that I find and proved is $\sqrt{a + b \cdot \varepsilon} = \sqrt{a} + \frac{b}{2 \cdot \sqrt{a}} \cdot \varepsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the Laurent series expansion of $\frac{1}{z(e^z-1)}$ and show that it has a pole of second order at $z = 0$. I am following an example solution for finding the Laurent expansion for $\frac{1}{z(e^z-1)}$, $0 <|z|< 2\pi$. The solution is given as:
\begin{align*}\frac{1}{z(e^z-1)} &= \frac{1}{z(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+ \ldots-1)} = \frac{1}{z(z+\frac{z^2}{2!}+\frac{z^3}{3!}+ \ldots)} \\ &= \frac{1}{z^2(1+\frac{z}{2!}+\frac{z^2}{3!}+ \ldots)} = \frac{1}{z^2}[1+(\frac{z}{2!}+\frac{z^2}{3!}+ \ldots)]^{-1} \\ &= \frac{1}{z^2}[1-(\frac{z}{2!}+\frac{z^2}{3!}+ \ldots)+(\frac{z}{2!}+\frac{z^2}{3!}+ \ldots)^2 - \ldots] \\ &= \frac{1}{z^2}[1-\frac{z}{2!} + z^2(-\frac{1}{3!}+ \frac{1}{4}) - z^3(-\frac{1}{4!}+ \frac{1}{3!} -\frac{1}{8}) + \ldots] \\ &= \frac{1}{z^2} - \frac{1}{2z} + \frac{1}{12} + \frac{1}{360z^2} + \ldots\end{align*}
I am able to understand this except the expansion of $[1+(\frac{z}{2!}+\frac{z^2}{3!}+ \ldots)]^{-1}$. My question is how did we ensure that $|(\frac{z}{2!}+\frac{z^2}{3!}+ \ldots)| <1$ so that the binomial formula $(1+z)^{-1} $ is applicable?
| There's another way, using the Bernoulli numbers
$$
\frac{z}{e^z-1}=\sum_{n\geq 0}\frac{B_n}{n!}z^n
$$
Then,
$$
\frac{1}{z(e^z-1)}=\frac{1}{z^2}\frac{z}{e^z-1}=\sum_{n\geq 0}\frac{B_n}{n!}z^{n-2}
$$
Hope it was useful!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4566155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find non-negative integers $n$ such that $9^n + 1$ is divisible by 365 For which non-negative integers $n$ is $9^n + 1$ is divisible by 365?
First approach:
$365 = 5 * 73 \implies 9^n + 1 \equiv (-1)^n + 1 \equiv 0 \; \text{(mod 5)} \Leftrightarrow n = 2k + 1$
So $n$ must be odd. Thus $9^{2k+1} + 1 \equiv 0 \; \text{(mod 73)}$
Unfortunately checking all residues (mod 73) isn‘t an option… How would you proceed from this/is there a better approach?
| You can note that $9\cdot 8 = 72 \equiv -1 \pmod{73}.$ So you know that $-8$ is the inverse of $9 \pmod{73}$. Then notice that $9^2 = 81 \equiv 8 \pmod{73}.$ So $-9^2$ is the inverse of $9 \pmod{73}$. Therefore $9^3\equiv -1 \pmod{73}$ and $9^6 \equiv 1 \pmod{73}.$ Then $9^{6k+3} \equiv -1 \pmod{73}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum value of the expression $\sqrt{\frac{1}{3}x^4+1}+\sqrt{\frac{1}{3}y^4+1}+\sqrt{\frac{1}{3}z^4+1}$ Let $x, y, z$ be positive real numbers such that $x + y + z = xyz$.
Find the minimum value of the expression $$\sqrt{\frac{1}{3}x^4+1}+\sqrt{\frac{1}{3}y^4+1}+\sqrt{\frac{1}{3}z^4+1}$$
My attempt:
By using Cauchy-Schwarz inequality and AM-GM inequality,
$$\sqrt{\frac{1}{3}t^4+1}\cdot\sqrt{3+1}\geq t^2+1$$
$$(x^2+y^2+z^2)(x+y+z)\geq3(xyz)^{\frac{2}{3}}\cdot 3(xyz)^{\frac{1}{3}}=9xyz$$
$$\therefore \sqrt{\frac{1}{3}x^4+1}+\sqrt{\frac{1}{3}y^4+1}+\sqrt{\frac{1}{3}z^4+1}\geq \frac{1}{2}(x^2+y^2+z^2)+\frac{3}{2}$$
$$\geq\frac{1}{2}\left (\frac{9xyz}{x+y+z}+\frac{3}{2} \right)=6$$
with equality if and only if $x=y=z=\sqrt{3}$
What are the alternative methods to solve this question?
| Your method is very straightforward. I am going just to post another approach that may contain a new idea.
Firstly;
$$xyz=x+y+z\ge 3 \sqrt[3] {xyz}\implies (xyz)^3\ge 27xyz\implies x+y+z=xyz\ge 3\sqrt3.$$
Secondly, a simple computation shows that $f(t)=\sqrt{\frac {t^4}{3}+1}$ is a convex function. Therefore, by Jensen's inequality, we get:
$$\sqrt{\frac {x^4}{3}+1}+\sqrt{\frac {y^4}{3}+1}+\sqrt{\frac {z^4}{3}+1}\ge 3\sqrt {\frac {(\frac {x+y+z}{3})^4}{3}+1}\ge 3\sqrt {\frac {(\frac {3\sqrt 3}{3})^4}{3}+1}=6. $$
More information about Jensen's inequality is available here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4572140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$\iiint\limits_{D}\frac{dxdydz}{\sqrt{x^2+y^2+(z-\frac{1}{2})^2} }$ D is given by $x^2+y^2+z^2 \leq 1$ $$\iiint\limits_{D}\frac{dxdydz}{\sqrt{x^2+y^2+(z-\frac{1}{2})^2} }$$
D is given by $x^2+y^2+z^2\leq1$
I try to use $ \left\{\begin{matrix}
x=r\sin \phi \cos \theta \\
y=r\sin \phi \sin \theta \\
z=r\cos\phi
\end{matrix}\right. $ while the Jacobian is $r^2 \sin \phi$ and $ \left\{\begin{matrix} 0\leq \theta \leq 2\pi\\
0\leq \phi \leq \pi \\0\leq r \leq 1\end{matrix}\right. $
$$\iiint\limits_{D}\frac{dxdydz}{\sqrt{x^2+y^2+(z-\frac{1}{2})^2} }\Rightarrow\int _{0}^{2\pi}d\theta \int _{0}^{\pi}d\phi \int _{0}^{1}\dfrac{r^{2}\sin \phi }{\sqrt{r^{2}-r\cos \phi +\dfrac{1}{4}}}dr$$
but it seems tough to do next.
| $$I=\int _{0}^{2\pi}d\theta \int _{0}^{\pi}d\phi \int _{0}^{1}\dfrac{r^{2}\sin \phi }{\sqrt{r^{2}-r\cos \phi +\dfrac{1}{4}}}dr=2\pi\int _{0}^{1}dr \int _{0}^{\pi}\dfrac{r^{2}\sin \phi }{\sqrt{r^{2}-r\cos \phi +\dfrac{1}{4}}}d\phi.$$
Let $\cos\phi=t \implies \sin\phi d\phi=-dt$, then
$$I=2\pi \int_{0}^{1}dr \int_{-1}^{1}\frac{r^2 dt}{\sqrt{r^2-rt+1/4}} dt=2\pi\int_{0}^{1} 2r (r+1/2)-|r-1/2|] dr =2\pi[\int_{0}^{1/2}4r^2 dr +\int_{1/2}^{1}2rdr ]$$ $$=2\pi[1/6+(1-1/4)]=\frac{11}{6}\pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of $\sum_{i=1}^{5}x^5_i+\sum_{i=1}^{5}\frac{1}{x^5_i}$
Suppose $x^5+5x^3+1=0$ and $x_i$ denotes all the complex roots. Find the sum of
$$\sum_{i=1}^{5}x^5_i+\sum_{i=1}^{5}\frac{1}{x^5_i}$$
This polynomial is irreducible over $\Bbb Q[x]$.
I used Vieta's general formulas, however I am not sure if my calculations correct. Here are my calculations.
$$x_1x_2+x_1x_3+x_1x_4+x_1x_5+x_2x_3+x_2x_4+x_2x_5+x_3x_4+x_3x_5+x_4x_5=5$$
$$x_1x_2x_3x_4x_5=-1$$
$$x_1+x_2+x_3+x_4+x_5=0$$
$$x_1x_2x_3+x_1x_2x_4+x_1x_2x_5+x_1x_3x_4+x_1x_3x_5+x_1x_4x_5+x_2x_3x_4+x_2x_3x_5+x_2x_4x_5+x_3x_4x_5=0$$
$$x_1x_2x_3x_4+x_1x_2x_3x_5+x_1x_2x_4x_5+x_1x_3x_4x_5+x_2x_3x_4x_5=0$$
This seems so difficult using these formulas. I couldn't make any solution.
How can I simplify this summation?
$$x_1^5+x_2^5+x_3^5+x_4^5+x_5^5$$
I've tried bring the fractions to common denominator and I got
$$\frac {1}{x_1^5}+\frac {1}{x_2^5}+\frac {1}{x_3^5}+\frac {1}{x_4^5}+\frac {1}{x_5^5}=\frac {(x_1x_2x_3x_4)^5+(x_1x_2x_3x_5)^5+(x_1x_2x_4x_5)^5+(x_1x_3x_4x_5)^5+(x_2x_3x_4x_5)^5}{(x_1x_2x_3x_4x_5)^5}=-(x_1x_2x_3x_4)^5-(x_1x_2x_3x_5)^5-(x_1x_2x_4x_5)^5-(x_1x_3x_4x_5)^5-(x_2x_3x_4x_5)^5$$
But, I couldn't make any further progress from here.
| We have
$$x^5 = -5x^3-1$$
and as t all roots are not $0$, the initial equation is equivalent to
$$1+5\frac{1}{x^2}+\frac{1}{x^5}=0 \Longleftrightarrow \frac{1}{x^5} = -\frac{5}{x^2}-1$$
Then, the sum is equal to
$$L:=-10-5\sum_{i=1}^5(x^3+\frac{1}{x^2})=-10-5\left(\sum_{\text{sym}}x_{\sigma(1)}^3+\sum_{\text{sym}}\frac{1}{x_{\sigma(1)}^2}\right)$$
For the first term $\sum_{\text{sym}}x_{\sigma(1)}^3$, using the Vieta's formula, we have:
$$\begin{align}
&\sum_{\text{sym}}x_{\sigma(1)} &= 0 \\
&\sum_{\text{sym}}x_{\sigma(1)}x_{\sigma(2)} &= 5\\
&\sum_{\text{sym}}x_{\sigma(1)}x_{\sigma(2)}x_{\sigma(3)} &= 0\\
\end{align}$$
We have from this (normally, a real mathematician needs to know this formula by heart ;) )
$$\begin{align}
\sum_{\text{sym}}x_{\sigma(1)}^3&=\left(\sum_{\text{sym}}x_{\sigma(1)} \right)^3 - 3\left(\sum_{\text{sym}}x_{\sigma(1)}\right)\left(\sum_{\text{sym}}x_{\sigma(1)}x_{\sigma(2)} \right)+3\left(\sum_{\text{sym}}x_{\sigma(1)}x_{\sigma(2)}x_{\sigma(3)} \right)\\
& = 0^3 -3\cdot0\cdot5+3\cdot 0 \\
&= 0 \tag{1}
\end{align}$$
For the second term $\sum_{\text{sym}}\frac{1}{x_{\sigma(1)}^2}$, we remark that these $\frac{1}{x}$ are the roots of the equation
$$1 + 5y^2 + y^5 = 0 \tag{2}$$
Let denote $y_i$ the roots of $(2)$, we have $\sum_{\text{sym}}\frac{1}{x_{\sigma(1)}^2} = \sum_{\text{sym}}y_{\sigma(1)}^2$. By the same method, using the Vieta's formula, we have:
$$\begin{align}
&\sum_{\text{sym}}y_{\sigma(1)} &= 0 \\
&\sum_{\text{sym}}y_{\sigma(1)}y_{\sigma(2)} &= 0\\
\end{align}$$
and from that, we have
$$\begin{align}
\sum_{\text{sym}}y_{\sigma(1)}^2&=\left(\sum_{\text{sym}}y_{\sigma(1)} \right)^2 - 2\left(\sum_{\text{sym}}y_{\sigma(1)}y_{\sigma(2)} \right)\\
& = 0^2 -2\cdot 0 \\
&= 0 \tag{3}
\end{align}$$
So, we can compute $L$ from $(1)$ and $(3)$
$$\color{red}{L} = -10 -5 \left( 0 + 0 \right) \color{red}{= -10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Long, complicated implicit differentiation, finding $y'$ Find $y''$ where $y^2 + xy = \ln (x+1)$
Using implicit differentiation,
$\frac{dy}{dx} = \frac{1}{2y+x} (\frac{1}{x+1} -y)$
Now finding the second derivative of this function, using the product rule,
$\frac{d^2y}{dx^2} = (\frac{-2 \frac{dy}{dx} +1}{(2y+x)^2} \times \frac{1}{x+1} - y) + (\frac{1}{2y+x} \times (\frac{-1}{(x+1)^2} - \frac{dy}{dx}))$
This looks very different from what the answer key is telling me:
$\frac{d^2y}{dx^2} = \frac{-1}{2y+x}(\frac{1}{(1+x)^2} + 2\frac{dy}{dx} + 2(\frac{dy}{dx})^2)$
| Just differentiate the entire expression:
$2y y' + xy' + y = (2y+x) y' + y = {1 \over 1+x}$, then again to get
$(2y+x) y'' + (2y' + 1) y' + y' = - {1 \over (1+x)^2}$.
Rearranging gives
$y'' = -{1 \over 2y+x} ({1 \over (1+x)^2} +2(y')^2 +2y')$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Any way to solve $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7}$? I was solving a radical equation $x+ \sqrt{x(x+1)} + \sqrt{(x+1)(x+2)} + \sqrt{x(x+2)} = 2$. I deduced it to $\sqrt{x } + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+7}.$
Answer is $\frac1{24}$.
The first equation has two solutions however the latter one has only one solution. I lossed up one solution but still, I'm interested in solving it.
I wish if someone could help me in solving any of the above equations.
Here's my work:
$$\begin{align}x+ \sqrt{x(x+1)} + \sqrt{(x+1)(x+2)} + \sqrt{x(x+2)} &= 2\\x + \frac{1}{2}\left(2 \sqrt{x}\sqrt{x+1} + 2\sqrt{x+1}\sqrt{x+2} + 2\sqrt{x}\sqrt{x+2}\right) &=2\tag{1}\\x + \frac{1}{2}\Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (x + x + 1 + x + 2)\Big]& = 2\tag{2}\\x + \frac{1}{2}\Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (3x + 3)\Big]& = 2\\2x + \Big[(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 - (3x + 3)\Big]& = 4\\(\sqrt{x} + \sqrt{x+1} + \sqrt{x+2})^2 & = x + 7\\\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} & = \sqrt{x+7}\end{align}$$
Moving from $(1)$ to $(2)$, I used $2(ab + bc + ca) = (a+b+c)^2 - (a^2+b^2+c^2)$.
In general, is it possible to solve $\sqrt{x} + \sqrt{x+1} + \sqrt{x+2} = \sqrt{x+k}$ by hand?
| Let $x+1=y, \;y\geqslant 1$, then you have:
$$
\begin{align}\sqrt y+\sqrt {y-1}+\sqrt {y+1}=\sqrt {y+6}\end{align}
$$
Dividing both side of the equation by $\sqrt y$ and using the substitution $\dfrac 1y=u,\;u>0$, then you obtain:
$$
\begin{align}&\sqrt {1-u}+\sqrt {1+u}=\sqrt {1+6u}-1\\
\implies &\sqrt {1-u^2}=3u-\sqrt {1+6u}\\
\implies &10u^2+6u=6u\sqrt {1+6u}\\
\implies &3\sqrt {6u+1}=5u+3\\
\implies &9(6u+1)=(5u+3)^2\\
\implies &u=\frac {24}{25}\\
\implies &x=\frac {1}{24}.\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4581999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Right angled triangle and its incircle $ABC$ is a right-angled triangle ($\measuredangle ACB=90^\circ$) and $O$ is its incenter. If $BO=m$ and $\measuredangle BAC=\alpha$, find the area of the triangle $ABC$.
the given answer is $$\dfrac{m^2}{2}\cos\alpha\cot\dfrac{\alpha}{2}$$
Let $OP\perp AB, OP=r$. As $\measuredangle ABC=90^\circ-\alpha$, then $\measuredangle OBP=45^\circ-\dfrac{\alpha}{2}$ and $$BP=m\cos(45^\circ-\alpha)\\AP=r\cot\dfrac{\alpha}{2}\\r=m\sin\left(45^\circ-\dfrac{\alpha}{2}\right)$$ (I have used $r$ for simpler calculations.) Because of the equal tangent segments, the semiperimeter is $$p=r\cot\dfrac{\alpha}{2}+m\cos\left(45^\circ-\dfrac{\alpha}{2}\right)+r$$ Now putting $r=m\sin\left(45^\circ-\dfrac{\alpha}{2}\right)$ gives $$p=m\sin\left(45^\circ-\dfrac{\alpha}{2}\right)\cot\dfrac{\alpha}{2}+m\cos\left(45^\circ-\dfrac{\alpha}{2}\right)+m\sin\left(45^\circ-\dfrac{\alpha}{2}\right)\\=m\left(\sqrt2\cos\dfrac{\alpha}{2}+\sin\left(45^\circ-\dfrac{\alpha}{2}\right)\cot\dfrac{\alpha}{2}\right)$$ using $$\sin\left(45^\circ-\dfrac{\alpha}{2}\right)+\cos\left(45^\circ-\dfrac{\alpha}{2}\right)\\=\sin\left(45^\circ-\dfrac{\alpha}{2}\right)+\sin\left(45^\circ+\dfrac{\alpha}{2}\right)=2\sin45^\circ\cos\dfrac{\alpha}{2}$$
I don't really see how to reach the given answer. Any help would be appreciated.
| The area of $\Delta ABC$ is $r(PB+AP)+r^2$ (adding area of the square to the sum of areas of the two quadrilaterals). Pythagorus theorem on $\Delta ABC$ gives that this is just $PB.AP$. Now $PB=r \cot(\pi/4 - \alpha/2)$ and $AP=r\cot(\alpha/2)$. Again, from $r^2+PB^2=m^2$, get $r^2=m^2\sin^2(\pi/4-\alpha/2)$. Now put everything together and expand it out a bit, things cancel and you get what you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $b\ne 0$, orthogonally diagonalize \begin{bmatrix}a&0&b\\0&a&0\\b&0&a\end{bmatrix} I'm trying to calculate the eigenvalues as the first step of this problem, but it's leading me down this rabbit hole of countless computations to find all the eigenvalues. I'm confident I'm supposed to solve this a different way that is less tedious. Anyone able to give me a hand?
| the eigenvalues of
$$
\left(
\begin{array}{cc}
0&1 \\
1&0 \\
\end{array}
\right)
$$
are $1,-1$ with eigenvectors as the columns of (orthogonal)
$$
\left(
\begin{array}{cc}
\frac{1}{\sqrt 2}& \frac{-1}{\sqrt 2}\\
\frac{1}{\sqrt 2}& \frac{1}{\sqrt 2}\\
\end{array}
\right)
$$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
the eigenvalues of
$$
\left(
\begin{array}{ccc}
0&0&1 \\
0&0&0 \\
1&0&0 \\
\end{array}
\right)
$$
are $1,0,-1$ with eigenvectors as the columns of (orthogonal)
$$
\left(
\begin{array}{cc}
\frac{1}{\sqrt 2}&0& \frac{-1}{\sqrt 2}\\
0&1&0 \\
\frac{1}{\sqrt 2}&0& \frac{1}{\sqrt 2}\\
\end{array}
\right)
$$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
the eigenvalues of
$$
\left(
\begin{array}{ccc}
0&0&b \\
0&0&0 \\
b&0&0 \\
\end{array}
\right)
$$
are $b,0,-b$ with eigenvectors as the columns of (orthogonal)
$$
\left(
\begin{array}{cc}
\frac{1}{\sqrt 2}&0& \frac{-1}{\sqrt 2}\\
0&1&0 \\
\frac{1}{\sqrt 2}&0& \frac{1}{\sqrt 2}\\
\end{array}
\right)
$$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
the eigenvalues of
$$
\left(
\begin{array}{ccc}
a&0&b \\
0&a&0 \\
b&0&a \\
\end{array}
\right)
$$
are $a+b,a,a-b$ with eigenvectors as the columns of (orthogonal)
$$
\left(
\begin{array}{cc}
\frac{1}{\sqrt 2}&0& \frac{-1}{\sqrt 2}\\
0&1&0 \\
\frac{1}{\sqrt 2}&0& \frac{1}{\sqrt 2}\\
\end{array}
\right)
$$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4588122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculate $\int\frac{2e^{2x}-e^x}{\sqrt{3e^{2x}-6e^x-1}}dx$ The question is to evaluate I, which is defined as follows:
$$
I \equiv \int\frac{2e^{2x}-e^x}{\sqrt{3e^{2x}-6e^x-1}}dx
$$
The first thing I did was simplify the numerator of the integrand by factoring out an $e^x$ to get
$$
\int\frac{e^x(2e^x-1)}{\sqrt{3e^{2x}-6e^x-1}}dx
$$
We then let $u = e^x$, then $dx = e^{-x}du$, and this enables us to rewrite $I$ as follows after some simplification
$$
\int\frac{2u - 1}{\sqrt{3(u^2-2u)-1}}du
$$
We can then split the integrand as follows by rewriting the numerator as $2u - 2 + 1$
$$
\int\frac{2u-2}{\sqrt{3(u^2-2u)-1}}+\frac{1}{\sqrt{3(u^2-2u)-1}}du
$$
We can then apply the sum rule to split the integral as follows
$$
\int\frac{2u-2}{\sqrt{3(u^2-2u)-1}}du + \int\frac{1}{\sqrt{3(u^2-2u)-1}}du
$$
Then factoring out a 2 from the first integral gives
$$
2\int\frac{u-1}{\sqrt{3(u^2-2u)-1}}du + \int\frac{1}{\sqrt{3(u^2-2u)-1}}du
$$
I'm not sure how to solve these integrals from here though, so any help would be great.
| We need one substitution not two which is $e^x-1=\frac{2}{\sqrt{3}}\sec t$.
Then, with $e^xdx=\frac{2}{\sqrt{3}}\sec t\tan t dt$ and $2e^x-1=\frac{4}{\sqrt{3}}\sec t+1$ we have
$$I=\int\frac{e^x(2e^x-1)}{\sqrt{3(e^x-1)^2-4}}dx=\int\frac{\frac{4}{\sqrt{3}}\sec t+1}{2\tan t}\frac{2}{\sqrt{3}}\sec t\tan t dt=\int (\frac{4}{3}\sec^2 t+\frac{1}{\sqrt{3}}\sec t)dt$$
and
$$I=\frac{4}{3}\tan t +\frac{1}{\sqrt{3}}\ln|\sec t+\tan t|+c.$$
in terms of $x$,
$$I=\frac{2}{3}\sqrt{3e^{2x}-6e^x-1} +\frac{1}{\sqrt{3}}\ln|\sqrt{3}(e^x-1)+\sqrt{3e^{2x}-6e^x-1}|+c.$$
Finally, https://www.wolframalpha.com/input?i=integral+e%5Ex%282e%5Ex-1%29%2Fsqrt%283e%5E%282x%29-6e%5Ex-1%29 I have to check myself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
How do I prove that $\sqrt[5]{80\sqrt 5+176}=1+\sqrt 5$? How do I prove that $\sqrt[5]{80\sqrt 5+176}=1+\sqrt{5}$?
I have no idea how to proceed except just raising both sides to the power of $5$ and expanding $ \left(1+\sqrt{5}\right)^{5}$ using the binomial theorem
| Let's rewrite our radical as $\sqrt[5]{\frac{160\sqrt{5}+352}{2}} $ to enable us to factor out $32$ from the root.
$$\Longrightarrow\sqrt[5]{32}*\sqrt[5]{\frac{5\sqrt{5}+11}{2}}$$
By equating both sides:
$$\Longrightarrow2\sqrt[5]{\frac{5\sqrt{5}+11}{2}}=1+\sqrt{5}$$
$$\Longrightarrow\sqrt[5]{\frac{5\sqrt{5}+11}{2}}=\frac{1+\sqrt{5}}{2}=\phi$$
$$\Longrightarrow\frac{5\sqrt{5}+11}{2}=\phi^5$$
Since $\phi^2=\phi+1$ we can rewrite $\phi^5$ easily as $\phi(\phi+1)^2=\left(\frac{1+\sqrt{5}}{2}\right)\left(\frac{3+\sqrt{5}}{2}\right)^2$
$$\Longrightarrow\left(\frac{1+\sqrt{5}}{2}\right)\left(\frac{7+3\sqrt{5}}{2}\right)$$
$$\Longrightarrow\frac{11+5\sqrt{5}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Find a limit of a sequence with series Find the limit of given sequence, using only the definition and properties of limits of sequences:
$$
x_{n} = \frac{1 + \sqrt{2} + \sqrt[3]{3} + ... +\sqrt[n]{n}}{n}
$$
I've tried to add and substract n to get:
$$
x_{n} = \frac{(1 - 1) + (\sqrt{2} - 1) + (\sqrt[3]{3} - 1) + ... + (\sqrt[n]{n} - 1) + n}{n} \le 1 - \frac{C(n - 1)}{n}, 0<C\le1
$$
| First off, $\sqrt[k]{k}\geq 1$ for all $k=1,2,...$, so
\begin{align*}
x_{n}\geq\frac{n\cdot 1}{n}=1,\quad n=1,2,...
\end{align*}
On the other hand, since $\sqrt[k]{k}\rightarrow 1$ as $k\rightarrow\infty$, given $\varepsilon>0$, there is an $N\in\mathbb{N}$ such that $\sqrt[k]{k}-1<\varepsilon$, $k\geq N$. Subsequently,
\begin{align*}
x_{n}&=\frac{1+\cdots+\sqrt[N-1]{N-1}}{n}+\frac{\sqrt[N]{N}+\cdots+\sqrt[n]{n}}{n}\\
&\leq\frac{1+\cdots+\sqrt[N-1]{N-1}}{n}+\frac{(n-N+1)}{n}(1+\varepsilon).
\end{align*}
For sufficiently large $n\geq N$, $x_{n}-1<\varepsilon$, so $x_{n}\rightarrow 1$ as $n\rightarrow\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4590565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve Volterra Integral Equation I need help solving an Integral Equation that my professor showed without teaching to us. I've tried emulating the textbooks example but I still can't crack it.
$$ f(t) = \cos\, t + 4 \, e^{-2t} - \int \sin\, (t-\tau)f(\tau)\, dt $$
My work so far on the problem:
\begin{align}
f(t) &= \cos\, t + 4 \, e^{-2t} - \int \sin\, (t-\tau)f(\tau)\, dt \\
\mathcal{L}\{f(t)\} &= \mathcal{L}\{\cos\, t\} + \mathcal{L}\{4 \, e^{-2t}\} - \mathcal{L}\left\{\int \sin\, (t-\tau)f(\tau)\, dt \right\} \\
F(s) &= \frac{s}{s^2 + 1} + \frac{4}{s+2} - F(s)\frac{1}{s^2 + 1} \\
F(s) &= \frac{s}{(s^2+2)} + \frac{4 s^2+1}{(s+2)(s^2+2)}
\end{align}
I know that from here I'm supposed to take the inverse Laplace Transform to find $f(t)$ but everything I've tried on the second term (Partial fractions, expanding it out) either fails me or gives me a super ugly equation that I'm sure is false
Any and all help is appreciated.
| First, the equation should read:
$$ F(s) = \frac{s}{s^2+2} + 4 \, \frac{s^2+1}{(s+2)(s^2+2)} $$
which leads to
$$ F(s) = \frac{s}{s^2+2} + \frac{4}{s+2} - \frac{4}{(s+2)(s^2+2)}. $$
Using partial fractions,
$$ \frac{1}{(s+2)(s^2 + 2)} = \frac{A}{s+2} + \frac{B s + C}{s^2+2}$$
and leads to $1 = A \, (s^2 + 2) + (B \, s + C)(s+2) = (A + B) \, s^2 + (C + 2 B) \, s + 2 (A + C)$. From this then
\begin{align}
F(s) &= \frac{s}{s^2+2} + \frac{4}{s+2} - \frac{4}{(s+2)(s^2+2)} \\
&= \frac{4}{s+2} + \frac{s}{s^2 + 2} - \frac{2}{3} \, \left( \frac{1}{s+2} - \frac{s -2}{s^2 + 2} \right) \\
&= \frac{10}{3} \, \frac{1}{s+2} + \frac{5}{3} \, \frac{s}{s^2 + 2} - \frac{\sqrt{2}}{3} \, \frac{\sqrt{2}}{s^2 + 2}.
\end{align}
Inverting this gives
$$ f(t) = \frac{10}{3} \, e^{-2 t} + \frac{5}{3} \, \cos(\sqrt{2} \, t) - \frac{\sqrt{2}}{3} \, \sin(\sqrt{2} \, t). $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4591774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find derivative using implicit function theorem
Let $F(x,y)$ be a function such that
$$F(x,y)=4+4(x -3) +4(y-7)-5(x-3)^2-(x-3)(y-7)-7(y-7)^2 +R_2$$
is the Taylor series.
Using implicit function theorem and find $y'(3),y''(3)$ of the equation $F(x,y)=4$ in $(3,7)$.
My attempt :
Using implicit function theorem , I know that
$$y'(3)=-\frac{f_x}{f_y}=\left(-\frac{4-10\left(x-3\right)-y+7}{4-x+3-14\left(y-7\right)}\right)=-1.$$
Then
$$y''(3)=-\frac{139y-1009}{\left(-x+105-14y\right)^2}=-\frac{139\cdot 7-1009}{\left(-3+105-14\cdot 7\right)^2}=\frac{9}{4}.$$
$y'(3)$ is correct but $y''(3)$ is wrong but I don't get why.
Thanks !
| Let $F=f+R_2$, then, in your work, taking the derivative with respect of $x$ of $-f_x/f_y$, you should consider $y$ as a function of $x$:
$$\frac{d}{dx}\left(-\frac{4-10\left(x-3\right)-(y-7)}{4-(x-3)-14\left(y-7\right)}\right)\\
=-\frac{(-10-y')(4-(x-3)-14\left(y-7\right))-(4-10\left(x-3\right)-(y-7))(-1-14y')}{(4-(x-3)-14\left(y-7\right))^2}.$$
Letting $x=3$, $y=7$ and $y'=-1$, we find
$$y''(3)=-\frac{(-10-(-1))(4)-(4)(-1-14(-1))}{(4)^2}=\frac{11}{2}.
$$
P.S. We should also note that $y'(3)$ and $y''(3)$ DO NOT depend on the remainder $R_2$. Indeed, we have that
$$y'(x)=-\frac{F_x}{F_y}$$
and by the chain rule
$$y''(x) =\frac{\partial y'}{\partial x}\left(-\frac{F_x}{F_y}\right) + \frac{\partial y'}{\partial y}\left(-\frac{F_x}{F_y}\right) y'(x) = \frac{-F_y^2F_{xx}+ 2F_xF_yF_{xy}-F_x^2F_{yy}}{F_y^3}.$$
In our case, from the expansion of $F$ we obtain that at $(3,7)$,
$$F_x=F_y=4,\; F_{xx}=-10,\;F_{xy}=-1,F_{yy}=-14,$$
and therefore
$$y'(3)=-\frac{4}{4}=-1\quad y''(3) = \frac{-4^2(-10)+ 2\cdot 4 \cdot 4 (-1)-4^2 (-14)}{4^3}=\frac{11}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find $\lim_{n\to\infty}\Big(\sum_{k=1}^n\sqrt{\frac{1}{k}-\frac{1}{n}}-\frac{\pi}{2}\sqrt n\Big)$? I'm looking for the asymptotic of the series $\displaystyle S(n)=\frac{1}{2}\sum_{k=1}^n\int_k^n\frac{dx}{x\sqrt{x(n-x)}}\,$ at $\,n\to\infty$
The first term can be found, for example, via switching to Riemann sums, changing the order of integration, and is equal to $\frac{\pi}{2}$. To find the next term I performed integration and got
$$S(n)=\frac{1}{n}\sum_{k=1}^{n-1}\sqrt{\frac{n}{k}-1}=\frac{1}{\sqrt n}\sum_{k=1}^{n-1}\sqrt{\frac{1}{k}-\frac{1}{n}}=\frac{1}{\sqrt n}\sum_{k=1}^{n-1}f(k)\tag{1}$$
To get the flavour of the next asymptotic term I used the Euler-Maclaurin formula
$$\sum_{k=1}^{n-1}f(k)\sim\int_1^{n-1}f(k)dk+\frac{1}{2}\big(f(1)+f(n-1)\big)+\frac{1}{12}\big(f'(n-1)-f'(1)\big)+ ...\tag{2}$$
At $n\to\infty$ the first term (the integral) gives $\frac{\pi}{2}\sqrt n-2$; other terms in (2) give non-zero values at $k=1$. Evaluating a couple of such terms, I got
$$\sum_{k=1}^{n-1}f(k)\sim\frac{\pi}{2}\sqrt n-2+\frac{1}{2}+\frac{1}{24}=\frac{\pi}{2}\sqrt n-1.4583$$
The numeric evaluation at WolframAlpha for $n=1000$ gives
$$ \sum_{k=1}^{n-1}f(k)-\frac{\pi}{2}\sqrt n\,\bigg|_{n=1000}=-1.46046$$
All this strongly resembles $\displaystyle \zeta\Big(\frac{1}{2}\Big)=-1.46035...\,\,$; at $\,n=10000\,$ we get even better agreement.
Questions:
*
*How can we prove that $\,\,\displaystyle \lim_{n\to\infty}\Big(\sum_{k=1}^n\sqrt{\frac{1}{k}-\frac{1}{n}}-\frac{\pi}{2}\sqrt n\Big)=\zeta\Big(\frac{1}{2}\Big)\,\,$?
*Can we get next asymptotic terms (at least, several of them) analytically ?
| Starting from the equality $ \frac{\pi}{2}\sqrt{n} = \int_{0}^{n} \sqrt{\frac{1}{x} - \frac{1}{n}} \, \mathrm{d}x$, we get
\begin{align*}
\sqrt{n}\left( S(n) - \frac{\pi}{2} \right)
&= \sum_{k=1}^{n} \sqrt{\frac{1}{k} - \frac{1}{n}} - \frac{\pi}{2}\sqrt{n} \\
&= \sum_{k=1}^{n} \sqrt{\frac{1}{k} - \frac{1}{n}} - \int_{0}^{n} \sqrt{\frac{1}{x} - \frac{1}{n}} \, \mathrm{d}x \\
&= \sum_{k=1}^{n} \int_{k-1}^{k} \biggl( \int_{x}^{k} \frac{\partial}{\partial s} \sqrt{\frac{1}{s} - \frac{1}{n}} \, \mathrm{d}s \biggr) \, \mathrm{d}x \\
&= \sum_{k=1}^{n} \int_{k-1}^{k} \biggl( \int_{k-1}^{s} \frac{\partial}{\partial s} \sqrt{\frac{1}{s} - \frac{1}{n}} \, \mathrm{d}x \biggr) \, \mathrm{d}s \\
&= -\frac{1}{2} \int_{0}^{n} \frac{s - \lfloor s \rfloor}{s^{3/2}\sqrt{1 - s/n}} \, \mathrm{d}s \\
&= -\frac{1}{2} \int_{0}^{0.2022 n} \frac{s - \lfloor s \rfloor}{s^{3/2}\sqrt{1 - s/n}} \, \mathrm{d}s + \mathcal{O}(n^{-1/2}).
\end{align*}
Then by the dominated convergence theorem, this converges to
$$ -\frac{1}{2} \int_{0}^{\infty} \frac{s - \lfloor s \rfloor}{s^{3/2}} \, \mathrm{d}s = \zeta\left(\frac{1}{2}\right) $$
as $n \to \infty$. (Note that $\zeta(s) = -s \int_{0}^{\infty} \frac{x-\lfloor x \rfloor}{x^{1+s}} \, \mathrm{d}x $ for $0 < \operatorname{Re}(s) < 1$, see the entry 25.2.8 of DLMF.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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"answer_id": 4
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Showing $\sum_{cyc}\frac{a^2+bc}{b+c}\geq a+b+c$ for positive $a$, $b$, $c$ The following is an inequality which I have trouble solving:
$$\frac{a^2+bc}{b+c}+\frac{b^2+ac}{c+a}+\frac{c^2+ab}{a+b}\geq a+b+c$$
($a, b, c>0$)
I tried multiplying LHS and RHS by 2 and then use
$${2(a+b+c)}={(a+b)+(b+c)+(c+a)}$$
and move to LHS but did not succeed in solving by this means.
Please provide a solution.
| Another way.
By C-S $$\sum_{cyc}\frac{a^2+bc}{b+c}=\sum_{cyc}\frac{(a^2+bc)^2}{(a^2+bc)(b+c)}\geq\frac{\left(\sum\limits_{cyc}(a^2+bc)\right)^2}{\sum\limits_{cyc}(a^2+bc)(b+c)}=$$
$$=\frac{\sum\limits_{cyc}(a^4+2a^2b^2+a^2b^2+2a^2bc+2a^3b+2a^3c+2a^2bc)}{2\sum\limits_{cyc}(a^2b+a^2c)}=$$
$$=a+b+c+\frac{\sum\limits_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+4a^2bc)}{2\sum\limits_{cyc}(a^2b+a^2c)}-\sum_{cyc}a=$$
$$=a+b+c+\frac{\sum\limits_{cyc}(a^4-a^2b^2)}{2\sum\limits_{cyc}(a^2b+a^2c)}=a+b+c+\frac{\sum\limits_{cyc}(a^2-b^2)^2}{4\sum\limits_{cyc}(a^2b+a^2c)}\geq a+b+c.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Consider two circles S1 and S2 given by $x^4 + y^4 +2x^2y^2-10x^2+6y^2+9=0 $? How can I separate the circles? I recently gave a test and had this question there . To be honest I was confused on seeing this question. I was unable to even deduce the approach.
I am a student studying in class 11 so I don't have access to desmos during tests.
| Rewrite the equation as
$$(x^2+y^2)^2-16x^2+6x^2+6y^2+9=0$$
which, using the difference of two squares, is equivalent to
$$(x^2-4x+y^2)(x^2+4x+y^2)+6(x^2+y^2)+9=0.$$
Then rewrite $6(x^2+y^2)$ as $3(x^2-4x+y^2)+3(x^2+4x+y^2)$ to enable us to factorise.
Now let $a=x^2-4x+y^2,b=x^2+4x+y^2$. Then we have $ab+3a+3b+9=0$, which factorises as $(a+3)(b+3)=0$. Hence the equations of the circles are $a+3=0$ and $b+3=0$, or
$$(x-2)^2+y^2=1, (x+2)^2+y^2=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Confused about answer to Spivak's calculus in prologue I have attached the question and answer to this question from Spivak's Calculus.
The proposition we are supposed to prove:
$$
x^n - y^n = (x - y)\,
(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})
$$
The proof of this proposition:
\begin{align}
(x - y) \,
(x^{n-1} + x^{n-2}y &+ \dots + xy^{n-2} + y^{n-1}) \\[2pt]
&= x \, (x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1}) \\
&\qquad{} - y \, (x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1}) \\[2pt]
&= \bigl[ x^n + x^{n-1}y + \dots + x^2y^{n-2} + xy^{n-1} \bigr] \\
&\qquad{} - \bigl[ x^{n-1}y + x^{n-2}y^2 + \dots + xy^{n-1} + y^n \bigr] \\[2pt]
&= x^n - y^n
\end{align}
I do not understand how $x^2\,y^{n-2}$ in the third to last line cancels with $x^{n-2}\,y^2$ in the second to last line (after the negative sign is distributed to $x^{n-2}\,y^2$).
Also, I do not understand how what is within the ellipses is treated. do we conceive of what is in the ellipses as multiplied by $x$ and $-y$ when $x$ and $-y$ are distributed to the expression with which $(x-y)$ is multiplied originally? If so, how can the ellipses cancel with each other as they are multiplied by different variables which we don't know are equal?
thanks, let me know if my question lacks clarity and I will explain my questions better/further.
| You can avoid the $\cdots$ which is confusing you.
You have to use alternatives.
We want to Prove :
$ x^{n}-y^{n} = (x-y)(x^{n-1}y^{0}+x^{n-2}y^{1}+x^{n-3}y^{2}+\cdots+x^{2}y^{n-3}+x^{1}y^{n-2}+x^{0}y^{n-1}) $
Write it alternately like this :
$$Z = (x-y)(x^{n-1}y^{0}+x^{n-2}y^{1}+x^{n-3}y^{2}+\cdots+x^{2}y^{n-3}+x^{1}y^{n-2}+x^{0}y^{n-1})$$
$$Z = (x-y)[\Sigma_{m=0}^{m=n-1} x^{n-1-m}y^{m}]$$
(You can verify that this $ \Sigma $ matches Exactly what we want)
$$Z = (x[\Sigma_{m=0}^{m=n-1} x^{n-1-m}y^{m}]-y[\Sigma_{m=0}^{m=n-1} x^{n-1-m}y^{m}])$$
$$Z = ([\Sigma_{m=0}^{m=n-1} x^{n-1-m+1}y^{m}]-[\Sigma_{m=0}^{m=n-1} x^{n-1-m}y^{m+1}])$$
$$Z = (x^{n}y^{0}+[\Sigma_{m=1}^{m=n-1} x^{n-1-m+1}y^{m}]-[\Sigma_{m=0}^{m=n-2} x^{n-1-m}y^{m+1}]-x^{0}y^{n})$$
( here I have taken out the first & the last terms from the two $ \Sigma $ terms )
$$Z = (x^{n}y^{0}-x^{0}y^{n})+[\Sigma_{m=1}^{m=n-1} x^{n-1-m+1}y^{m}]-[\Sigma_{m=0}^{m=n-2} x^{n-1-m}y^{m+1}]$$
We can see that that 2 $ \Sigma $ terms will cancel ; To make it Explicit , we can change the variables to $ i = m-1 $ ( or $ m = i+1 $ ) in the first $ \Sigma $ term & to $ i = m $ ( or $ m = i $ ) in the second $ \Sigma $ term.
$$0 = [\Sigma_{i=1}^{i=n-2} x^{n-1-i-1+1}y^{i+1}]-[\Sigma_{i=0}^{i=n-2} x^{n-1-i}y^{i+1}]$$
$$0 = [\Sigma_{i=1}^{i=n-2} x^{n-1-i}y^{i+1}]-[\Sigma_{i=0}^{i=n-2} x^{n-1-i}y^{i+1}]$$
$$0 = [\Sigma_{i=1}^{i=n-2} ( x^{n-1-i}y^{i+1} - x^{n-1-i}y^{i+1} )]$$
$$0 = [\Sigma_{i=1}^{i=n-2} ( 0 )]$$
Hence :
$$Z = (x^{n}y^{0}-x^{0}y^{n})+[0]$$
$$(x^{n}-y^{n}) = (x-y)(x^{n-1}y^{0}+x^{n-2}y^{1}+x^{n-3}y^{2}+\cdots+x^{2}y^{n-3}+x^{1}y^{n-2}+x^{0}y^{n-1})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Most efficient way to find the "solution intervals" / "case conditions"? (problem regarding a function with absolute value notation): I am trying to write the function $f(x) = |x^2-1|+|x|-1$ without the notation for the absolute value. "||"
It makes logical sense to consider four cases, because each term in the absolute value notation can either be positive or negative.
Therefore, we get:
*
*Case 1 (positive, positive): $f(x) = x^2-1+x-1 = (x-1)(x+2)$
*Case 2 (positive, negative): $f(x) = x^2-1-x-1 = (x+1)(x-2)$
*Case 3 (negative, positive): $f(x) = -x^2+1+x-1 = -x(x-1)$
*Case 4 (negative, negative): $f(x) = -x^2+1-x-1 = -x(x+1)$
To use the piecewise function notation, I need the intervals for which each term is correct.
$$
f(x) = \cases{ term & $condition\ $ \cr
term & $condition\ $ \cr
term & $condition\ $ \cr
term & $condition\ $ \cr}
$$
What is the most efficient way to find the case conditions (in this particular example, and generally speaking)?
(I have tried putting the term of each case equal to zero to find the roots. However, doing this, I get $5$ roots. Our expression is of order $3$, therefore something has to be wrong/I overlooked something while trying this.)
I am thankful for any input/ideas! Thank you!
| Since the terms inside the absolute value bars are continuous, they can only change sign where they are equal to zero. Since $x^2 - 1 = 0$ when $x = 1$ or $x = -1$ and $x = 0$ when $x = 0$, you should consider the behavior of $f$ in the intervals $(-\infty, -1)$, $[-1, 0)$, $[0, 1)$, and $[1, \infty)$.
If $x \geq 1$, then $|x^2 - 1| = x^2 - 1$ and $|x| = x$.
If $0 \leq x < 1$, then $|x^2 - 1| = -x^2 + 1$ and $|x| = x$.
If $-1 \leq x < 0$, then $|x^2 - 1| = -x^2 + 1$ and $|x| = -x$.
If $x < -1$, then $|x^2 - 1| = x^2 - 1$ and $|x| = -x$.
Hence,
\begin{align*}
f(x) & = |x^2 - 1| + |x| - 1\\
& = \begin{cases}
x^2 - 1 - x - 1 & \text{if $x < -1$}\\
-x^2 + 1 - x - 1 & \text{if $-1 \leq x < 0$}\\
-x^2 + 1 + x - 1 & \text{if $0 \leq x < 1$}\\
x^2 - 1 + x - 1 & \text{if $x \geq 1$}\\
\end{cases}\\
& = \begin{cases}
x^2 - x - 2 & \text{if $x < -1$}\\
-x^2 - x & \text{if $-1 \leq x < 0$}\\
-x^2 + x & \text{if $0 \leq x < 1$}\\
x^2 + x - 2 & \text{if $x \geq 1$}\\
\end{cases}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that the series $\sum^\infty_{k=1}\frac{1}{k^4}$ is bounded. Show that the series $\sum^\infty_{k=1}\frac{1}{k^4}$ is bounded.
What I've tried: Since
$\frac{1}{2^4}+\frac{1}{3^4}\leq \frac{1}{2^4}+\frac{1}{2^4}=\frac{1}{2^3}$
$\frac{1}{4^4}+\frac{1}{5^4}+\frac{1}{6^4}+\frac{1}{7^4}\leq \frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}+\frac{1}{4^4}=\frac{1}{4^3}$
and so on. So the series in question is less than $1+\frac{1}{2^3}+\frac{1}{4^3}+\frac{1}{8^3}...$. If I can show the latter is bounded I'm done. But I can't so maybe there is and easier solution?
| First, be careful with your index of summation - your series is not defined at $k = 0$. Anyways, note that $\frac{1}{k^4} \leq \frac{1}{k^2}$ for each positive integer $k$. Thus,
$$\sum_{k=1}^\infty \frac{1}{k^4} \leq \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4606187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Simplfying a trigonometric expression I would like to simplify:
$$\frac{\cos^2(80)+5\sin^2(80)-3}{\cos(50)}$$
By using the fact that $\sin^2(\theta) + \cos^2(\theta) = 1$,
$$\frac{\cos^2(80)+5\sin^2(80)-3}{\cos(50)} = \frac{\cos^2(80)+5(1-\cos^2(80))-3}{\cos(50)} = -2\biggr(\frac{2\cos^2(80)-1}{\cos(50)}\biggr)$$
Since $2\cos^2(80)-1 = \cos(160)$,
$$-2\biggr(\frac{2\cos^2(80)-1}{\cos(50)}\biggr) = -2\frac{\cos(160)}{\cos(50)}$$
But I am not sure how we could simplify this further.
| Since $\cos(\pi-\theta)=-\cos\theta$, if the angles are given in degrees $\cos(180^{\circ}-\theta)=-\cos \theta$, it follows
\begin{align*}
-2\dfrac{\cos 160^{\circ}}{\cos 50^{\circ}}&=\dfrac{2\cos 20^{\circ}}{\cos 50^{\circ}}\\
&=\dfrac{2\cos 20^{\circ}}{\sin(90^{\circ}-50^{\circ})}\\
&=\dfrac{2\cos 20^{\circ}}{\sin 40^{\circ}}\\
&=\dfrac{2\cos 20^{\circ}}{2\sin 20^{\circ}\cos 20^{\circ}}\\
&=\dfrac1{\sin 20^{\circ}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4609840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to simplify $\sin(n\frac{\pi}{2})$ as we do with $\cos(n\pi)$ When calculating Fourier series we almost always get some expression that look like:
\begin{equation} ...\left[\cos(nx) \right]_0^\pi.
\end{equation}
Then we use $\cos(n\pi) = (-1)^n$.
However, lately I have come across expressions like this:
\begin{equation} ...\left[\sin(nx) \right]_0^{\pi/2}.
\end{equation}
Which is equal to:
\begin{array}{|c|c|c|c|}
\hline
n& 1 & 2 & 3 & 4 \\ \hline
\sin(nx)&\sin(\frac{\pi}{2}) & \sin(\pi) & \sin(\frac{3\pi}{2})& \sin(2\pi)\\ \hline
val &1 &0 &-1 &0\\ \hline
\end{array}
How do we simplify this so that it looks like the $\cos(n\pi) = (-1)^n$.
| You have a sequence with period $4$ (specifically $0,1,0,-1,0,1,\cdots$), but $(-1)^n$ has period $2$, so it can't be used in this form.
However, $(-1)^{\binom n2}=(-1)^{n(n-1)/2}$ does have period $4$:
$$(-1)^0=1,\quad(-1)^1=-1,\quad(-1)^3=-1,\quad(-1)^6=1,$$
$$(-1)^{10}=1,\quad(-1)^{15}=-1,\quad(-1)^{21}=-1,\quad(-1)^{28}=1,$$
$$\cdots$$
To prove periodicity:
$$(-1)^{(n+4)(n+3)/2}=(-1)^{(n^2+7n+12)/2}=(-1)^{4n+6+(n^2-n)/2}\\=((-1)^2)^{2n+3}(-1)^{n(n-1)/2}=(-1)^{n(n-1)/2}$$
Now let's try shifting and combining these:
$$\begin{matrix}n&=&(0,&1,&2,&3,&\cdots) \\ (-1)^n&=&(1,&-1,&1,&-1,&\cdots) \\ (-1)^{n(n-1)/2}&=&(1,&1,&-1,&-1,&\cdots) \\ (-1)^{n(n+1)/2}&=&(1,&-1,&-1,&1,&\cdots) \\ (-1)^{n(n-1)/2}-(-1)^{n(n+1)/2}&=&(0,&2,&0,&-2,&\cdots)\end{matrix}$$
And there's your answer:
$$\sin(n\pi/2)=\frac{(-1)^{n(n-1)/2}-(-1)^{n(n+1)/2}}{2}$$
In fact any $4$-periodic sequence can be expressed in this way:
$$f_n=\frac{1+(-1)^n+(-1)^{n(n-1)/2}+(-1)^{n(n+1)/2}}{4}=(1,0,0,0,1,0,0,0,1,\cdots)$$
$$(a_0,a_1,a_2,a_3,a_0,a_1,a_2,a_3,a_0,\cdots)=a_0f_n+a_1f_{n-1}+a_2f_{n-2}+a_3f_{n-3}$$
$$=a_0\frac{1+(-1)^n+(-1)^{n(n-1)/2}+(-1)^{n(n+1)/2}}{4} \\ +a_1\frac{1-(-1)^n+(-1)^{n(n-1)/2}-(-1)^{n(n+1)/2}}{4} \\ +a_2\frac{1+(-1)^n-(-1)^{n(n-1)/2}-(-1)^{n(n+1)/2}}{4} \\ +a_3\frac{1-(-1)^n-(-1)^{n(n-1)/2}+(-1)^{n(n+1)/2}}{4}$$
See also What is a mathematical expression for the sequence $\{1,1,-1,-1,1,1,-1,-1,\dots\}$? and the linked posts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Calculating the integral $\int \frac{14x+1}{2-7x-4x^2} \, dx$ We have the following integral
$$
\int \frac{14x+1}{2-7x-4x^2} \, dx
$$
By factoring the denominator we get
$$
-\int \frac{14x+1}{4x^2+7x-2} \, dx = -\int \frac{14x+1}{(4x-1)(x+2)} \, dx
$$
And breaking into partial fractions
$$
-\int \frac{2}{4x-1} + \frac{3}{x+2} \, dx = -\int \frac{2}{4x-1} \, dx - \int \frac{3}{x+2} \, dx = - 2\int \frac{1}{4x-1} \, dx - 3 \int \frac{1}{x+2} \, dx
$$
Then by substituting (and resubstituting) we get
$$
- 2\int \frac{1}{4u} \, du - 3 \int \frac{1}{v} \, dv = -\frac{2}{4} \ln(u)-3 \ln (v)=-\frac{1}{2}\ln(4x-1)-3\ln (x+2)
$$
However, the expected result is
$$
-\frac{1}{2}\ln(1-4x)-3\ln (x+2)
$$
Have I made a mistake somewhere? If so, where?
| Too long for a comment:
@Bernkastel is absolutely (haha) right. Your solution $+c$ is valid on a certain (open) portion of the real line, and the advertised solution on an other. With absolute values thrown in, one gets the largest domain for a solution over the reals.
To be more explicit: on the half-line $x<0$, $ |x| = -x$. So, calculating the derivative on that interval, one gets (by the chain rule)
$$ \left(\ln |x|\right)' = \left(\ln \left(-x\right) \right)' = {1\over -x} \cdot (-1) = {1\over x};$$
similarly on the half-line $x>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4612145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Trouble evaluating definite integral with a Riemann sum $$ \int_{1}^{9}(x^2+7) dx $$
That's the integral. And I'm supposed to evaluate it by using:
$$ \lim_ {x \to \infty} \sum_{i=1}^{n} f(x_i)(\Delta x) $$
Also $$ \Delta x = \frac{b-a}{n} $$
And $$ x_i = a + \frac{i(b-a)}{n} $$
The question asks: "Now find the sum of the areas of
n approximating rectangles. (Your answer will include variable n)"
My solution:
$$ \lim_ {n \to \infty} \sum_{i=1}^{n} \left[ \left(\frac{8i}{n}\right)^2 + 7\left(\frac{8i}{n}\right) \right] \cdot \frac{8}{n} $$
$$\lim_ {n \to \infty} \left(\left(\frac{8}{n}\right)^2 \cdot \frac{8}{n}\right) \sum_{i=1}^{n} [i^2] + \left(\left(\frac{8}{n}\right) \cdot \frac{8}{n}\right)\sum_{i=1}^{n} [ i ] $$
$$ \lim_ {n \to \infty} \left(\frac{8}{n}\right)^3 \left(\frac{n(n+1)}{2}\right)^2 + \left(\frac{8}{n}\right)^2\left(\frac{n(n+1)}{2}\right) $$
I've tried a few variations of this but nothing seems to work... The final answer is supposed to be 896/3 and the question I cited earlier calls for a simplified version of what I have up there, but this just isn't making sense at all.
| The formula of the sum of first $n$ squares is mistakenly replaced by that of the sum of first $n$ cubes.
Using the elementary definition of the definite integral,
$$
\int_a^b f(x) d x=\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{b-a}{n} f\left(a+\frac{k(b-a)}{n}\right)
$$
Putting $a=1$ and $b=9$ and $f(x)=x^2+7$ yields
$$
\begin{aligned}
\int_1^9\left(x^2+7\right) d x & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{8}{n}\left[\left(1+\frac{8 k}{n}\right)^2+7\right] \\
& =64 \lim _{n \rightarrow \infty} \sum_{k=1}^n\left(\frac{1}{n}+\frac{2 k}{n^2}+\frac{8 k^2}{n^3}\right)
\end{aligned}
$$
Using the formula for sum of first $n$ positive integers and their squares,
$$
\sum_{k=1}^n k=\frac{n}{2}(n+1) \text { and } \sum_{k=1}^n k^2=\frac{n}{6}(n+1)(2 n+1),
$$
we get $$
\begin{aligned}
\int_1^9\left(x^2+7\right) d x & =64 \lim _{n \rightarrow \infty} \sum_{k=1}^n\left(\frac{1}{n}+\frac{2 k}{n^2}+\frac{8 k^2}{n^3}\right) \\
& =64 \lim _{n \rightarrow \infty}\left(1+\frac{2}{n^2} \cdot \frac{n(n+1)}{2}+\frac{8}{n^3}\cdot \frac{n}{6}(n+1)(2n+1)\right) \\
& =64\left(2+\frac{8}{3}\right) \\
& =\frac{896}{3}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4616731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to approximate a parameter that gives a tangent line to three circles? Three disks are placed on the ground like this:
From left to right, their radii are $\frac{1}{x-1}, \frac{1}{x}, \frac{1}{x+1}$ metres. They lie in a plane perpendicular to the ground. The middle disk touches the other two disks.
Using only paper and pen, approximate the value of $x$ such that the middle disk is tangent to the line that is tangent to and above the other two disks. You may assume that the earth is a sphere of radius $R$ metres.
(Before you read the last sentence, it seems like there's something wrong with the question, because it seems like the middle disk should never touch the line. But the ground is actually a circular arc of the earth, so the middle disk is "pushed up" and touches the line for some value of $x$.)
The answer turns out to be, elegantly, $x\approx R/2$. But the algebra seems to be horrendous and I needed to use my computer to find the answer.
My attempt
Call the angles at the centre of the middle disk $A, B, C, D, E$ with $A$ at the lower-left and going clockwise.
$A=\arccos{\left(\dfrac{\left(\frac{1}{x}+\frac{1}{x-1}\right)^2+\left(R+\frac{1}{x}\right)^2-\left(R+\frac{1}{x-1}\right)^2}{2\left(\frac{1}{x}+\frac{1}{x-1}\right)\left(R+\frac{1}{x}\right)}\right)}$
$B=\arcsin{\left(\dfrac{\frac{1}{x-1}-\frac{1}{x}}{\frac{1}{x-1}+\frac{1}{x}}\right)}$
$C=\dfrac{\pi}{2}$
$D=\arccos{\left(\dfrac{\frac{1}{x}-\frac{1}{x+1}}{\frac{1}{x}+\frac{1}{x+1}}\right)}$
$E=\arccos{\left(\dfrac{\left(\frac{1}{x}+\frac{1}{x+1}\right)^2+\left(R+\frac{1}{x}\right)^2-\left(R+\frac{1}{x+1}\right)^2}{2\left(\frac{1}{x}+\frac{1}{x+1}\right)\left(R+\frac{1}{x}\right)}\right)}$
We assume that the middle disk is tangent to the line that is tangent to and above the other two disks. This implies:
$$A+B+C+D+E=2\pi$$
I am utterly unable to approximate $x$ without a computer, even after attempting to simplify it. And yet the computer-assisted answer is just $x\approx R/2$. Can $x$ be approximated without a computer?
(This question was inspired by a frame challenge.)
| Using the solution given by @Blue in comments, it is not difficult to solve exactly the cubic equation in $y$ provided that we use the hyperbolic solution.
Using only pen and paper, the exact result is
$$\color{blue}{y=\frac 16 +\frac{\sqrt{r^2+6}}{3 r}\cosh \left(\frac{1}{3} \text{sech}^{-1}\left(\frac{2 r
\left(r^2+6\right)^{3/2}}{2 \left(r^2+9\right)
r^2+27}\right)\right)}$$ which must be very close to an hyperbola.
Using a computer, expanding for large values of $r$
$$y=\frac 12 + \frac 1 {r^2}\left( 1-\frac{3}{2
r^2}+\frac{5}{r^4}+O\left(\frac{1}{r^6}\right)\right)$$ that is to say
$$x=\frac r2 + \frac 1 {r}\left( 1-\frac{3}{2
r^2}+\frac{5}{r^4}+O\left(\frac{1}{r^6}\right)\right)$$
as already given by @Empy2 using simple and legitimate approximations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 2,
"answer_id": 1
} |
Area between two equations I want to calculate the area between $y^2=2x+1$ and $x-y-1=0$ . I did this:
$$x_1 = \frac {y^2-1}{2}, x_2 = y+1$$
And the intersection points are, $y=-1$ and $y=3$ . So the area is:
$$\int_{-1}^3|x_1 - x_2| dy$$
$$\implies\int_{-1}^3\left|\frac{y^2-1}{2} - y-1\right| dy = \frac{16}{3}$$
And this is the correct answer. I tried to calculate this, using $dx$. For the area above the x-axis:
$$\int_{-\frac{1}{2}}^4 \sqrt{2x+1}\text{ }dx-\int_1^4(x-1)\text{ }dx = \frac {9}{2}$$
And for the area below the x-axis:
$$\int_{-\frac{1}{2}}^4\left|-\sqrt{2x+1}\right|\text{ }dx-\int_0^4\left|x-1+\sqrt{2x+1}\right|\text{ }dx + \int_1^4(x-1)\text{ }dx=\frac{5}{6}$$
And if we add these numbers we get to $\frac {16}{3}$ . I want to know is there a better way to calculate this area using $dx$? One integration for area above the x-axis and one for another, maybe?
|
Vertical Strips
$$
\begin{aligned}
\textrm{ Area between the curve }& =\int_{-\frac{1}{2}}^0 [\sqrt{2 x+1}-(-\sqrt{2 x+1})] d x+\int_0^4[\sqrt{2 x+1}-(x-1)] d x \\
& =\left[\frac{2}{3}(2 x+1)^{\frac{3}{2}}\right]_{-\frac{1}{2}}^0+\left[\frac{1}{3}(2 x+1)^{\frac{3}{2}}-\frac{(x-1)^2}{2}\right]_0^4 \\
& =\frac{2}{3}+9-\frac{9}{2}-\frac{1}{3}+\frac{1}{2} \\
& =\frac{16}{3}
\end{aligned}
$$
Horizontal Strips
Since the straight line is always on the right of the parabola for $x\in [-1,4]$, therefore we can find conveniently(without absolute sign) the enclosed area by only one integral as below:
$$
\begin{aligned}
\int_{-1}^3\left[1+y-\frac{1}{2}\left(y^2-1\right)\right] d y
= & {\left[\frac{(1+y)^2}{2}-\frac{y^3}{6}+\frac{y}{2}\right]_{-1}^3 } \\
= & 8-\frac{9}{2}+\frac{3}{2}-\frac{1}{6}+\frac{1}{2} \\
= & \frac{16}{3}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $(\sum x^2)^3\ge9\sum x^4yz$
Prove that $\displaystyle\left(x^2+y^2+z^2\right)^3\ge9\left(x^4yz+y^4xz+z^4xy\right)$, for $x$, $y$, $z\in\Bbb R_+$.
The $pqr$ method doesn't seem possible because the power is too high.
$$\iff\left(p^2-2q\right)^2\ge9r\left(p^3-3pq+3r\right).$$
Then expand the expression to get
$$\sum x^6+3\sum\left(x^4y^2+x^2y^4\right)+6x^2y^2z^2\ge9\sum x^4yz.$$
I wanted to use SOS but cannot find the weight of three squares, my progress:
$$3\sum x^4(y-z)^2=3\sum\left(x^4y^2+x^2y^4\right)-6\sum x^4yz.$$
Whats left is $\displaystyle\sum x^6+6x^2y^2z^2-3\sum x^4yz$. I have trouble dealing with it.
| Update: A simpler pqr
It suffices to prove that
$$\left(p^2-2q\right)^3\ge9r\left(p^3-3pq+3r\right).$$
Using $q^2 \ge 3pr$ and $pq \ge 9r$, it suffices to prove that
$$\left(p^2-2q\right)^3\ge 9 \cdot \frac{q^2}{3p}\cdot\left(p^3-3pq+3 \cdot \frac{pq}{9}\right)$$
or
$$(p^2 - 2q)^3 \ge q^2(3p^2 - 8q)$$
or
$$(p^2 - 2q)^3 + (2q)^3 \ge 3p^2 q^2$$
or (using $x^3 + y^3 = (x + y)[(x + y)^2 - 3xy]$)
$$p^2[p^4 - 3(p^2 - 2q)\cdot 2q] \ge 3p^2 q^2$$
or
$$p^2 (p^2 - 3q)^2 \ge 0$$
which is true.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4620974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Find the generating function of the sequence I need to find generating function of sequence : $$\left\{0, 1, \frac12, \frac23, \frac24, \frac45, \frac46, \frac87, 1, \frac{16}9, \frac{16}{10}, \ldots\right\}$$
I saw that odd positions are $\frac{2^n}{2n+1}$ where $n$ goes from $0$ to $\infty$, but nothing else.
Any help would be appreciated.
| Let $a_0=0$, $a_1=1$, $a_2=\frac{1}{2}$, $a_3=\frac{2}{3}$, $a_4=\frac{2}{4},\dots$ and notice that the denominator of $a_n$ is $n$. For the numerator, notice that these are powers of $2$ and come in pairs. The coming in pairs part is ensured by $\lfloor\frac{n-1}{2}\rfloor=\max\{k\in\mathbb Z:k\le\frac{n-1}{2}\}$, i.e. the function rounding down. Combining this gives $a_n=\frac{2^{\lfloor\frac{n-1}{2}\rfloor}}{n}$ and hence the generating function $\sum_{n=1}^\infty\frac{2^{\lfloor\frac{n-1}{2}\rfloor}}{n}x^n$.
The sum for even $n$ is $\sum_{k=1}^{\infty}\frac{2^{k-1}}{2k}x^{2k}=\frac{1}{4}\sum_{k=1}^{\infty}\frac{1}{k}(2x^2)^{k}=-\frac{1}{4}\ln|1-2x^2|$, using the anti-derivative of the geometric series $\sum_k b^k=\frac{1}{1-b}$ to obtain $\sum_k\frac{1}{k+1}b^{k+1}=-\ln|1-x|$. The sum for odd $n$ is $\sum_{k=0}^{\infty}\frac{2^{k}}{2k+1}x^{2k+1}=\frac{1}{\sqrt{2}}\sum_{k=0}^{\infty}\frac{1}{2k+1}(\sqrt{2}x)^{2k+1}=\frac{1}{2\sqrt{2}}\ln\frac{1+\sqrt{2}x}{1-\sqrt{2}x}$, the series expansion of the inverse hyperbolic tangent, which I found using WolframAlpha. So, in total we get
$$\sum_{n=1}^\infty\frac{2^{\lfloor\frac{n-1}{2}\rfloor}}{n}x^n
=-\frac{1}{4}\ln|1-2x^2|+\frac{1}{2\sqrt{2}}\ln\frac{1+\sqrt{2}x}{1-\sqrt{2}x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4621211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $xy+yz+zx=3$ and $x,y,z\geq0$, prove that: $\frac{1}{1+3x-p}+\frac{1}{1+3y-p}+\frac{1}{1+3z-p}\leq\frac{3}{1+2p}$ If $xy+yz+zx=3$ and $x,y,z\geq0$, prove that:
$$\sum_{cyc}\frac{1}{1+3x-p}\leq\frac{3}{1+2p}$$
where $p=xyz$.
*some people are not familiar with the $\sum_{cyc}$ notation, alternative would be
$$\frac{1}{1+3x-p}+\frac{1}{1+3y-p}+\frac{1}{1+3z-p}\leq\frac{3}{1+2p}$$
What I've tried:
*
*Using AM>GM, resulting $p\leq1$, if you use that directly it turns into a false inequality $\sum_{cyc}\cfrac{1}{3x}\leq1$
*Using the fact that $\frac{1}{1+3x-p}<\frac{1}{3x-p}$, resulting again in very clearly false inequality.
*Multipling by random stuff like $\sum_{cyc}(1+3x-p)$ and then applying the CBS inequality
*Deconditioning, first substituting $u=xy, v=yz, w=zx$ and then $u=\cfrac{3a}{a+b+c}, v=\cfrac{3b}{a+b+c}, w=\cfrac{3c}{a+b+c}$
What could be usefull:
$$x+y+z\geq3$$
$$x^2+y^2+z^2\geq3$$
both prooven from $x^2+y^2+z^2\geq xy+yz+zx=3$, because of $(x-y)^2+(y-z)^2+(z-x)^2\geq0$
| The inequality is equivalent to
$$\sum_{cyc}\frac{3x-p}{1+3x-p}+\frac{3}{1+2p}\geq 3.$$
Notice that
$$(3x-p)(y+z)=x(3-yz)(y+z)=x(xy+xz)(y+z)=(xy+xz)^2.$$
Therefore, by Cauchy-Schwarz,
$$\sum_{cyc}\frac{3x-p}{1+3x-p}\cdot\sum_{cyc}(1+3x-p)(y+z)
\geq 4(xy+yz+zy)^2=36$$
and it follows that
$$\sum_{cyc}\frac{3x-p}{1+3x-p}\geq \frac{36}{\sum_{cyc}(1+3x-p)(y+z)}=\frac{18}{9+s(1-p)}$$
where $s=x+y+z$.
Hence it suffices to show that
$$\frac{18}{9+s(1-p)}+\frac{3}{1+2p}\geq 3$$
that is
$$\frac{(1-p)(3-sp)}{(9+s(1-p))(1+2p)}\geq 0$$
which holds because $xy+yz+xz=3$ implies $p\leq 1$ and $sp\leq 3$.
P.S. $sp\leq 3$ iff $9=(xy+yz+zx)^{2} \geq 3xyz(x+y+z)$
iff
$$2(xy+yz+zx)^{2}-6xyz(x+y+z)=x^2(y-z)^2+y^2(z-x)^2+z^2(x-y)^2\geq 0$$
which trivially holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4623148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find all the roots of the equation $z^2 + |z| = 0$ . Find all the roots of the equation $z^2 + |z| = 0$ .
My solution goes like this:
Let $z=a+ib$, and $z^4=|z|^2\implies a^4+4a^3ib+6a^2(ib)^2+4a(ib)^3+(ib)^4=a^4+4a^3ib-6a^2b^2-4ab^3i=a^2+b^2\implies a^4+b^4+4a^3ib-6a^2b^2-4aib^3+b^4=a^2+b^2\implies a^4+b^4+4i(a^3-b^3)=a^2+b^2+6a^2b^2\implies a^4+b^4-2a^2b^2+4i(a^3-b^3)=a^2+b^2+4a^2b^2\implies (a^2-b^2)^2+4i(a-b)(a^2+b^2+ab)=a^2+b^2+4a^2b^2$. Now, since, $\text{R.H.S} \in\Bbb R$ and hence $Im(\text{L.H.S})=0$. Thus, $4(a-b)(a^2+b^2+ab)=0$ so, $a=b$ and $a^2+b^2+ab$. So all $z=a+ib$ are solutions of the equations iff $a=b$ or $\text{if}\space(x=a,y=b)$ satisfies $f(x,y)=x^2+y^2+xy=0$.
Is the above solution correct?
| If $z^2+|z|=0$, then $z^2=-|z|\leqslant0$. The only complex numbers whose square is a real number smaller than or equal to $0$ are those numbers $z$ of the form $\lambda i$, for some $\lambda\in\Bbb R$. But then$$z^2+|z|=0\iff-\lambda^2+|\lambda|=0,$$whose only solutions are $-1$, $0$, and $1$. Therefore, the solutions of your equation are $-i$, $0$, and $i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4623704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Fractions in Questions and Answers
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