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$\lfloor a_n \rfloor$=? where $a_n=\sqrt{2005+\sqrt{2005+\sqrt{2005+...+\sqrt{2005}}}}$ $\lfloor a_n \rfloor$=?, where $$a_n=\sqrt{2005+\sqrt{2005+\sqrt{2005+...+\sqrt{2005}}}}$$
and the number of radicals is equal to n.
$\lfloor a_1]=\lfloor \sqrt{2005} \rfloor=44$, $\lfloor a_2 \rfloor=\lfloor \sqrt{2005+\sqrt{2005}} \rfloor=45$, $\lfloor a_3 \rfloor=\lfloor \sqrt{2005+\sqrt{2005+\sqrt{2005}}} \rfloor=45$ and so on,
but at some point becomes $\lfloor a_k \rfloor=46$ and then $47$ and so on.
So we can say that $a_n$ depends on the value of $n$, which in my opinion has no rule.
In order, how it should be defined $\lfloor a_n \rfloor$=?
Thank you!
|
$\def\e{\varepsilon}$Here we bound the sequence by examining the sequence of successive differences.
Let $x=2005$.
We have the recursion
\begin{align*}
a_0 &= 0 \\
a_n &= \sqrt{x+a_{n-1}},\quad n\ge1.
\end{align*}
Note that $a_n\ge0$ for $n\ge 0$.
Let $d_n = a_n-a_{n-1}$ for $n\ge 1$.
One can verify that
\begin{align*}
d_n &= \frac{d_{n-1}}{\sqrt{x+a_{n-1}}+\sqrt{x+a_{n-2}}},
\quad n\ge 2.
\end{align*}
Since $d_2 = \sqrt{x+\sqrt x}-\sqrt{x} > 0$, this implies that $d_n > 0$ for $n\ge 2$.
Note that $d_1=a_1-a_0 = \sqrt x>0$.
Thus, $a_n$ is increasing and so
$$a_n > a_{n-1} > \cdots > a_0 = 0.$$
Thus, $d_n < \e d_{n-1}$ for $n\ge 2$, where $\e = 1/(2\sqrt{x}).$
More generally,
$$d_n < \e d_{n-1} < \e^2 d_{n-2} < \cdots < \e^{n-2}d_2 < \e^{n-1}d_1.$$
The last inequality follows since $d_2<\e d_1$.
Thus,
\begin{align*}
a_n &= a_0 + \sum_{k=1}^n d_k
< d_1\sum_{k=1}^n \e^{k-1}
< \sqrt x \sum_{k=1}^\infty \e^{k-1}
= \frac{\sqrt{x}}{1-\e}
= \frac{2x}{2\sqrt{x}-1}.
% &= 45.2829\ldots
\end{align*}
Therefore, $a_n < 45.2829\ldots$.
(This technique can be extended to find accurate closed form approximations for $a_n$.)
Since
\begin{align*}
a_0 &= 0 \\
a_1 &= \sqrt{x} = 44.7772 \\
a_2 &= \sqrt{x+\sqrt x} = 45.2745
\end{align*}
we find
$$\lfloor a_n \rfloor = \begin{cases}
0, & n = 0 \\
44, & n = 1 \\
45, & n\ge 2
\end{cases}$$
agreeing with the result of @MohammadRiazi-Kermani.
|
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|
Prove that the limit $\lim_{(x,y)\to(0,0)} \frac{y^4}{\sqrt{3x^2+y^2}}\cos(\frac{1}{x})$ exists using $\epsilon \text{ and } \delta$ definition Prove that the limit $\lim_{(x,y)\to(0,0)} \frac{y^4}{\sqrt{3x^2+y^2}}\cos(\frac{1}{x})$ exists using $\epsilon \text{ and } \delta$ definition.
I have so far reduced it algebraically to the point $\frac{y^4}{\sqrt{x^2+y^2}}$ but I am not sure if I have to use polar coordinates or something to continue, kinda lost from here. I eliminated the cosine because $\cos(\frac{1}{x})\leq 1$ so
$$\frac{y^4}{\sqrt{3x^2+y^2}}\cos(\frac{1}{x}) \leq \frac{y^4}{\sqrt{3x^2+y^2}}$$
And likewise I eliminated the three from $3x^2$. Do I have to use polar coordinates like $r=x^2 + y^2$???
I did some more working and got $\delta=\epsilon^{\frac{1}{3}}$...
|
Hint :
Observe that $y^2 \le x^2 +y^2$ so $y^4 \le (x^2 + y^2)^2$.
Also $3x^2 +y^2 \ge x^2 +y^2$ which implies that $\frac{1}{\sqrt{3x^2 + y^2}} \le \frac{1}{\sqrt{x^2+y^2}}$
Combining both we get $$\frac{y^4}{\sqrt{3x^2 + y^2}} \le \frac{(x^2+y^2)^2}{\sqrt{x^2+y^2}} = (x^2+y^2)\sqrt{(x^2+y^2)}$$
So for given $\epsilon > 0$ you can choose $\delta <\sqrt[3]{\epsilon}$ to conclude the proof.
|
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|
Prove the inequality $\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\geq\frac 32$ where $a,b,c$ are positive reals.
Let $a,b,c$ be positive real numbers, prove that $$\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\geq\frac 32.$$
The problem is from an inequality handout. Here is my attempt to solve the problem:
I first rewrote the inequality as $$\sum_{cyc}\frac{a^2}{(b+c)\sqrt{b^2-bc+c^2}}\geq\frac 32.$$
Here I noticed that the inequality has some similarities with Nesbitt's inequality, because that says $$\sum_{cyc}\frac{a}{b+c}\geq\frac 32.$$
But I don't know how to use this to reach the given inequality.
I also used Cauchy-Schwarz and got $$\sum_{cyc}\frac{a^2}{(b+c)\sqrt{b^2-bc+c^2}}\geq\frac{(a+b+c)^2}{\sum_{cyc}(b+c)\sqrt{b^2-bc+c^2}}.$$
But this also makes the problem too complicated that involves too many square roots. I think the square root on the denominator should be removed first to prove the inequality. But I am unable to do that.
So, I need a solution to the problem. Also it would be helpful for me if the answerer provides some motivation for the solution.
|
Another way.
By Holder $$\left(\sum_{cyc}\frac{a^2}{\sqrt{(b+c)(b^3+c^3)}}\right)^2\sum_{cyc}a^2(b+c)(b^3+c^3)\geq(a^2+b^2+c^2)^3$$ and it's enough to prove that:
$$4(a^2+b^2+c^2)^3\geq9\sum_{cyc}(a^2b+a^2c)(b^3+c^3)$$ or
$$\sum_{cyc}(4a^6+3a^4b^2+3a^4c^2-9a^3b^2c-9a^3c^2b+8a^2b^2c^2)\geq0,$$ which is true by Schur, AM-GM, Schur and Muirhead:
$$\sum_{cyc}(4a^6+3a^4b^2+3a^4c^2-9a^3b^2c-9a^3c^2b+8a^2b^2c^2)\geq$$
$$\geq\sum_{cyc}(7a^4b^2+7a^4c^2-9a^3b^2c-9a^3c^2b+4a^2b^2c^2)\geq$$
$$\geq\sum_{cyc}(14a^4bc-9a^3b^2c-9a^3c^2b+4a^2b^2c^2)=$$
$$=9abc\sum_{cyc}(a^3-a^2b-a^2c+abc)+5abc\sum_{cyc}(a^3-abc)\geq0.$$
|
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|
If $a^{2}+b^{2} \leq 4$, prove that $a+b \leq 4$ \begin{equation}
\text { If } a^{2}+b^{2} \leq 4 \text { prove that } a+b \leq 4 \text { }
\end{equation} What I have tried:
\begin{equation}
a^{2} \leq a^{2}+b^{2} \leq 4 \text { and } b^{2} \leq a^{2}+b^{2} \leq 4
\end{equation}
\begin{equation}
|a| \leq 2 \text { and }|b| \leq 2
\end{equation}
\begin{equation}\text { So } a+b
\text { can get the maximum value, then } a \geq 0 \text { and } b \geq 0 \text { }
\end{equation}\begin{equation}
\text { } 0 \leq a \leq 2 \text { and } 0 \leq b \leq 2 \text {}
\end{equation}
\begin{equation}
(a-b)^{2}=a^{2}+b^{2}-2 a b \geq 0
\end{equation}\begin{equation}
a^{2}+b^{2} \geq 2 a b \text { So at this step I am not certain what to do next. }
\end{equation}
|
From Cauchy-Schwartz inequality,
$a^2 + b^2 = (a^2 + b^2) ( \dfrac{1}{2} + \dfrac{1}{2} ) \ge (\dfrac{1}{\sqrt{2}} a + \dfrac{1}{\sqrt{2}} b )^2 = \dfrac{1}{2} (a + b)^2$
Hence $(a + b)^2 \le 2 (a^2 + b^2) = 8 $
Thus $|a + b| \le \sqrt{8} $
The last inequality is equivalent to $ -\sqrt{8} \le a + b \le \sqrt{8} \lt 4 $
|
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|
Evaluate : $S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$ Evaluate:$$S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7} + \frac{1}{9\cdot10\cdot11}+\cdots$$to infinite terms
My Attempt:
The given series$$S=\sum_{i=0}^\infty \frac{1}{(4i+1)(4i+2)(4i+3)} =\sum_{i=0}^\infty \left(\frac{1}{2(4i+1)}-\frac{1}{4i+2}+\frac{1}{2(4i+3)}\right)=\frac{1}{2}\sum_{i=0}^\infty \int_0^1 \left(x^{4i}-2x^{4i+1}+x^{4i+2}\right) \, dx$$
So,$$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4} + \frac{x^2}{1-x^4}\right)dx=\frac{1}{2} \int_0^1 \left(\frac{1+x^2}{1-x^4}-\frac{2x}{1-x^4}\right)\,dx = \frac{1}{2} \int_0^1 \left(\frac{1}{1-x^2}-\frac{2x}{1-x^4}\right)\,dx$$
$$=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\int_{0}^{1}\frac{2x}{1-x^4}dx=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\frac{1}{2}\int_{0}^{1}\frac{1}{1-y^2}dy=0(y=x^2)$$
which is obviously absurd since all terms of $S$ are positive.
But if I do like this then I am able to get the answer, $$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4}+\frac{x^2}{1-x^4}\right)dx=\frac{1}{2}\int_{0}^{1}\frac{(1-x)^2}{1-x^4}dx=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1+x}-\frac{x}{1+x^2}\right)dx=\frac{\ln2}{4}$$
What is wrong with the previous approach
|
As was mentioned in the comments, the integral $$\int_0^1\frac{dx}{1-x^2}$$
diverges, so you cannot split the integral $$\int_0^1\left(\frac{1}{1-x^2}-\frac{2x}{1-x^4}\right)dx$$
up into parts.
|
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|
Solving the equation $2x^2-\left(4x-7\right)\sqrt{x-1}-5x+4=0$
Solve the equation $$2x^2-\left(4x-7\right)\sqrt{x-1}-5x+4=0$$
Here is my work:
$$2x^2-5x+4=(4x-7)\sqrt{x-1}$$
$$2(x-1)^2-(x-1)+1=(4(x-1)-3)\sqrt{x-1}$$
Using the substitution $\sqrt{x-1}=t$,
$$2t^4-t^2+1=t(4t^2-3)$$
$$2t^4-4t^3-t^2+3t+1=0$$
Here I plugged in some natural numbers like $1,2,3$ but neither of them worked. So I don't know how to factor the polynomial to find the values of $t$.
|
What you did is fine. Now, let\begin{align}p(t)&=2t^4-4t^3-t^2+3t+1\\&=2\left(t^4-2t^3-\frac12t^2+\frac32t+\frac12\right)\end{align}and let $r(t)$ be the depressed quartic of $t^4-2t^3-\frac12t^2+\frac32t+\frac12$, that is\begin{align}\left(t+\frac12\right)^4-2\left(t+\frac12\right)^3-\frac12\left(t+\frac12\right)^2+\frac32\left(t+\frac12\right)+\frac12&=t^4-2t^2+\frac{15}{16}\\&=\left(t^2-\frac54\right)\left(t^2-\frac34\right),\end{align}since the roots of the quadratic equation $T^2-2T+\frac{15}{16}=0$ are $\frac34$ and $\frac54$. So,\begin{align}p(t)&=2r\left(t-\frac12\right)\\&=2(t^2-t-1)\left(t^2-t-\frac12\right).\end{align}
|
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|
Infinite product involving triangular numbers $\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}$ The question is about the value for the following infinite product involving triangular numbers $T_n=\frac{n(n+1)}2$:
$$\prod_{n=1}^{\infty} \frac{T_n}{T_{n}+1}=\prod_{n=1}^{\infty} \frac{n(n+1)}{n(n+1)+2}=\prod_{n=1}^{\infty} \frac{1}{1+\frac 2{n(n+1)}}$$
Searching on MSE I've found solutions for other similar problems related to Weierstrass' product for the sine function:
$$\frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right)$$
but nothing specific on that particular case.
I've tried to use Weierstrass' product but without success.
Also expansion in logarithmic sum allow me to find good estimation with a small number of terms but nothing more than that.
According to wolfram the value for the infinite product should be: $\frac{2\pi}{\cosh\left(\frac{\sqrt 7}2\pi\right)} \approx 0.197$.
How can we determine the value for the partial and the infinite product in a closed form?
|
You can also do it using Pochhammer symbols
$$\frac{n(n+1)}{n(n+1)+2}=\frac{n(n+1)}{(n-a)(n-b)}$$
$$P_p=\prod_{n=1}^p\frac{n(n+1)}{(n-a)(n-b)}=\frac{\prod_{n=1}^p n(n+1) }{{\prod_{n=1}^p (n-a)(n-b)} }=\frac{\Gamma (p+1) \Gamma (p+2)}{(1-a)_p (1-b)_p}$$
Take logarithms and use Stirling approximation
$$\log(P_p)=((a+b) \log (p)+\log (p \Gamma (1-a) \Gamma (1-b)))+\frac{-a^2+a-b^2+b+2}{2
p}+O\left(\frac{1}{p^2}\right)$$ Now, using $a=\frac{1}{2} \left(-1-i \sqrt{7}\right)$ and $b=\frac{1}{2} \left(-1+i \sqrt{7}\right)$
$$\log(P_p)=\log \left(2 \pi \text{sech}\left(\frac{\sqrt{7} \pi
}{2}\right)\right)+\frac{2}{p}+O\left(\frac{1}{p^2}\right)$$
$$P_p=e^{\log(P_p)}=2 \pi \text{sech}\left(\frac{\sqrt{7} \pi }{2}\right)+\frac{4 \pi
\text{sech}\left(\frac{\sqrt{7} \pi }{2}\right)}{p}+O\left(\frac{1}{p^2}\right)$$
Edit
Making the problem more general
$$Q_p=\prod_{n=1}^p\frac{n(n+1)}{n(n+1)+k}$$ and using the same procedure
$$Q_p=\pi k \,\text{sech}\left(\frac{\pi }{2} \sqrt{4 k-1}\right)\exp\Bigg[ \frac{k}{p}-\frac{k}{p^2}-\frac{(k-6) k}{6 p^3}+\frac{(k-2) k}{2 p^4}+O\left(\frac{1}{p^5}\right)\Bigg]$$
|
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Let $a,b,c$ be non-negative real numbers .Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $ Let $a,b,c$ be non-negative real numbers
Prove that : $ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
My idea is to use the $(p,q,r)$ method:
$p=a+b+c$
$q=ab+bc+ca$
$r = abc $
$\Rightarrow a^2+b^2+c^2 = p^2-2q $
$ a^2+b^2+c^2 +\sqrt{2} abc + 2\sqrt{2} +3 \geq (2+\sqrt {2} )(a+b+c) $
or
$ p^2-2q +\sqrt{2} r + 2\sqrt{2} +3 \geq (2+\sqrt {2} )p $
or
$ (p - (1+\sqrt{2} ))^2 -2q +\sqrt{2} r \geq 0 $
or
The problem I am facing is exactly what I want to prove : $\sqrt{2} r \geq 2q$ is completely wrong .
I hope to get help from everyone. Thanks very much !
|
Alternative proof:
The desired inequality is easily written as
$$\left(a + \frac{\sqrt2\, bc - 2 - \sqrt2}{2}\right)^2 + \frac{2 - c^2}{2}\left(b - \frac{2 + \sqrt2 - c - c\sqrt2}{2 - c^2}\right)^2 + \frac{c(c - 1)^2}{c + \sqrt2} \ge 0.$$
So, if $c < \sqrt2$, the desired inequality is true.
If $c \ge \sqrt2$, it suffices to prove that
$$a^2 + b^2 + c^2 + \sqrt2\, ab \cdot \sqrt2 + 2\sqrt2 + 3 \ge (2 + \sqrt2)(a + b + c)$$
or
$$\frac14(2c - 2 - \sqrt2)^2 + \frac14(2a + 2b - 2 - \sqrt2)^2 \ge 0$$
which is true.
We are done.
|
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Evaluating $ \sum_{r=0}^\infty { 2r \choose r} x^r $ I recently came across this question during a test
Evaluate the following $$ \sum_{r=0}^ \infty \frac {1\cdot 3\cdot 5\cdots(2r-1)}{r!}x^r $$
I converted it to the follwing
into $\sum_{r=0}^\infty { 2r \choose r} x^r $
Then I tried using Beta function to further expand the binomial coefficient and tried to write it as sum of integrals but a pesky "r" term ends up being multiplied
How can we evaluate this with and without using calculus?
|
Yet another way.
Lemma 1:
$$ \int_{0}^{\pi/2}\left(\cos\theta\right)^{2n}\,d\theta = \frac{\pi}{2\cdot 4^n}\binom{2n}{n} \tag{1}$$
Proof of Lemma 1: by parity and De Moivre's formula
$$ \int_{0}^{\pi/2}(\cos\theta)^{2n}\,d\theta = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\left(\frac{e^{i\theta}+e^{-i\theta}}{2}\right)^{2n}\,d\theta =\frac{1}{2\cdot 4^n}\sum_{k=0}^{2n}\binom{2n}{k}\int_{-\pi/2}^{\pi/2} e^{ki\theta} e^{-(2n-k)i\theta}\,d\theta $$
where the last integrals always equals zero, unless $k=n$.
Corollary:
$$ \sum_{r\geq 0}\binom{2r}{r} x^r = \frac{2}{\pi}\sum_{r\geq 0}\int_{0}^{\pi/2}(4x\cos^2\theta)^{r}\,d\theta=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{1-4x\cos^2\theta}.\tag{2} $$
Letting $\theta=\arctan u$ in the last integral we have
$$\sum_{r\geq 0}\binom{2r}{r}x^r = \frac{2}{\pi}\int_{0}^{+\infty}\frac{du}{1+u^2-4x}\stackrel{u\mapsto v\sqrt{1-4x}}{=}\frac{2}{\pi\sqrt{1-4x}}\int_{0}^{+\infty}\frac{dv}{v^2+1}=\frac{1}{\sqrt{1-4x}}.\tag{3} $$
And another one. Since by Vandermonde's identity
$$\binom{2n}{n}=\sum_{k=0}^{n}\binom{n}{k}^2 $$
the Cauchy-Schwarz inequality gives $\binom{2n}{n}\geq \frac{4^n}{n+1}$. On the other hand $\binom{2n}{n}\leq \sum_{k=0}^{n}\binom{2n}{k}=4^n$, so the radius of convergence of our power series is exactly $\frac{1}{4}$. As a consequence, the radius of convergence of
$$ f(z)=\sum_{n\geq 0}a_n z^n\quad\text{with}\; a_n=\frac{1}{4^n}\binom{2n}{n} $$
is exactly one. The coefficients $a_n$ satisfy a nice recurrence relation: since
$$ \frac{1}{4^{n+1}}\binom{2n+2}{n+1} = \frac{(2n+2)(2n+1)}{4(n+1)^2}\binom{2n}{n}=\frac{(2n+1)}{2(n+1)}\binom{2n}{n} $$
we have $2(n+1)a_{n+1}=(2n+1)a_n$, which can be turned into a differential equation. Indeed
$$ f'(z) = \sum_{n\geq 0} (n+1) a_{n+1} z^{n},\qquad z\cdot f'(z)=\sum_{n\geq 1} n a_n z^n $$
hence
$$ 2(1-z) f'(z) = f(z) $$
with $f(0)=a_0=1$. The last differential equation is a separable differential equation:
$$ \frac{f'(z)}{f(z)}=\frac{1}{2(1-z)}\quad\Longrightarrow\quad \log f(x) = \int_{0}^{x}\frac{dz}{2(1-z)} = -\frac{1}{2}\log(1-x)$$
so by exponentiating both sides of the the last identity we get
$$ \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}x^n = \frac{1}{\sqrt{1-x}} $$
as wanted.
|
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|
What is the value of the $\measuredangle x$ in the figure below? For reference:
If ABCD is a square calculate "x" being "M" and "N" midpoints of CD and AD respectively.
My progress:
$\triangle ABN \cong \triangle BC (2:1- special~ right~ triangle) \implies \measuredangle ABN = \frac{53^o}{2}=\measuredangle MBC\\
\therefore \measuredangle NBM = 90-53 = 47^o$
is there any property that shows that $x$ and $47^o$ are complementary?
|
WLOG, we assume $ABCD$ is a unit square. Then,
$BN = NP = \frac{\sqrt5}{2}$
Using power of point of $B$,
$BH \cdot BP = AB^2 \implies BH \cdot \sqrt5 = 1$
$BH = \frac{1}{\sqrt5}$
Now, $\frac{BQ}{BR} = \frac{HQ}{NR} = \frac{BH}{BN}$
$\frac{BQ}{3 / 2 \sqrt2} = \frac{HQ}{1/2\sqrt2} = \frac{1/\sqrt5}{\sqrt5/2}$
$BQ = \frac{3}{5 \sqrt2}, HQ = \frac{1}{5 \sqrt2}$
$GQ = BG - BQ = \frac{1}{\sqrt2} - \frac{3}{5 \sqrt2}
= \frac{2}{5 \sqrt2}$
In right triangle $\triangle GQH$, perpendicular sides are in ratio $1:2$, so $\angle HGQ \approx 26.5^\circ$
$\therefore x \approx 53^\circ$
|
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|
Elementary solutions of the equation of a quadratic formula We know that when $A=0$, the quadratic equation $Ax^2+Bx+C=0$ has one solution, $-\frac{C}{B}$.
Since for every quadratic equation, a quadratic equation has at most two solutions: $\frac{-B + \sqrt{B^2 - 4AC}}{2A}$ and $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$, I think that in the case $A=0$ one of the results of the two formulas should match $-\frac{C}{B}$. Yet, if $A=0$, $\frac{-B + \sqrt{B^2 - 4AC}}{2\times0}$, we will get $\frac{0}{0}$, and with $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$ we will get $-\frac{B}{0}$, so none of the two possible solutions will match $-\frac{C}{B}$. Shouldn't there be at least one solution that has the same value as $-\frac{C}{B}$?
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You can use the conjugate quantity to get an alternative expression of the roots:
Suppose that $B > 0$. Then:
$\frac{-B + \sqrt{B^2-4AC}}{2A} = \frac{(-B + \sqrt{B^2-4AC})(-B - \sqrt{B^2-4AC})}{2A(-B - \sqrt{B^2-4AC})} = \frac{(-B)^2 - (\sqrt{B^2-4AC})^2}{2A(-B - \sqrt{B^2-4AC})} = \frac{2C}{-B - \sqrt{B^2-4AC}}$
If $A = 0$, the denominator is $-B - \sqrt{B^2} = -B - |B| = -2B$.
Therefore the root is $-\frac{C}{B}$.
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|
Frustrated because used standard R addition formula for mechanics problem, but the answer came out wrong. How to resolve this so doesn't happen again? In the middle of quite a lengthy mechanics problem, I got to:
$$\cos\alpha-\sin\alpha = 0.05\ $$
and I wanted to find the value of $\ \alpha,\ 0<\alpha<90^{\circ}\ $ that satisfies this equation using an R addition formulae, as that is the usual way to go here.
So I tried:
$$\cos\alpha-\sin\alpha = R\sin\left(\alpha+\gamma\right) = R\sin\alpha\cos\gamma\ + R\cos\alpha\sin\gamma $$
$$\implies R\cos\gamma = -1\quad \text{and}\quad R\sin\gamma = 1$$
$$\implies R^2 = 2\quad \text{and}\quad \tan\gamma = -1$$
At this point, I usually assume $R>0$ and $0<\gamma<90^{\circ}\ $ will give the correct evaluation of $\cos\alpha-\sin\alpha\ $, but that is not the case here.
In fact the correct evaluation of $\cos\alpha-\sin\alpha\ $ is:
$\cos\alpha-\sin\alpha\ = -\sqrt{2}\sin(\alpha-45^{\circ}).\ $
So I got the answer wrong because I didn't even think to check if $R$ and $\gamma$ were correct because like I said, I just assumed $R>0$ and $0<\gamma<90^{\circ}\ $ should always work.
But when I get to
$$\implies R^2 = 2\quad \text{and}\quad \tan\gamma = -1$$
in my working, I don't want to check the four combinations: $\ R = \pm \sqrt{2}\ $ and
$\ \left( \gamma = -45^{\circ}\ \text{or}\ \gamma = 135^{\circ}\ \right).$
So instead of having to check these four options every time, what is a quicker way of ensuring I get to the right answer every time? Probably I should use a different R addition formula for this circumstance, but I don't know a good summary of which R addition formula I should use for each of the different circumstances...
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\begin{align}
\cos\alpha - \sin\alpha &= \frac{1}{20}\\
\cos\alpha \cdot \frac{\sqrt 2}{2} -\sin\alpha\cdot \frac{\sqrt 2}{2}
&= \frac{\sqrt 2}{40}\\
\cos\alpha \cdot \cos\left(\frac{\pi}{4}\right)
-\sin\alpha \cdot \sin\left(\frac{\pi}{4}\right)
&= \frac{\sqrt 2}{40}\\
\cos\left(\alpha + \frac{\pi}{4}\right) &= \frac{\sqrt 2}{40}\\
\alpha + \frac{\pi}{4} &=\arccos \frac{\sqrt 2}{40} \\
\text{and so on}
\end{align}
Another example
\begin{align}
5\cos\alpha - 7\sin\alpha &= z\\
\frac{5}{\sqrt{74}}\cos\alpha - \frac{7}{\sqrt{74}}\sin\alpha
&= \frac{z}{\sqrt{74}}\\
\end{align}
$$\left\{\text{Let $\cos \beta = \frac{5}{\sqrt{74}}$ and
$\sin \beta = \frac{7}{\sqrt{74}}$}\right\}$$
\begin{align}
\cos\alpha \cos \beta - \sin\alpha \sin \beta
&= \frac{z}{\sqrt{74}}\\
\cos(\alpha + \beta) &= \frac{z}{\sqrt{74}}\\
&\cdots
\end{align}
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|
Proving a curve has no curve point I am trying to show that the curve $x^2 + y^2 - 3 = 0$ has no rational point, where a rational point is defined as a solution $(x,y) \in \mathbb{Q}^2$ to this equation.
My attempt was to proceed by contradiction. Assume there is such a point $(x,y) \in \mathbb{Q}^2$ and let $x = \frac{a}{b}$, $y = \frac{c}{d}$ for $a,b,c,d \in \mathbb{Z}$, $b,d \neq 0$ where we assume without loss of generality that these fractions are fully reduced. Then we substitute:
\begin{align*}
\frac{a^2}{b^2} + \frac{c^2}{d^2} - 3 = 0.
\end{align*}
Rearranging and finding a common denominator of $b^2 d^2$ on the left:
\begin{align*}
\frac{a^2 d^2 + b^2 c^2}{b^2 d^2} = 3.
\end{align*}
We multiply through by $b^2 d^2$:
\begin{align*}
a^2 d^2 + b^2 c^2 = 3b^2 d^2.
\end{align*}
At this point, I'm stuck. I could try to argue that $3$ divides the left-hand side and apply Euclid's lemma, but I can't figure out how to get a contradiction out of the assumption that $\frac{a}{b}$ and $\frac{c}{d}$ have no common factors.
|
you want a solution to the equation:
$$x^2+y^2=3$$
you could try proving that if this is a point on a circle, it cannot have both coordinates be integer, so say:
$$x=\sqrt{3}\cos\theta,y=\sqrt{3}\sin\theta$$
for $x$ to even be rational $\cos\theta$ must be be a rational multiple of $\sqrt3$ and similar for $y$. Now show there exists no value $0\le\theta< 2\pi$ where both $\cos,\sin$ are multiples of $\sqrt3$
|
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|
Prove that for every positive integer $n$, $1/3 + 1/9 + \cdots + 1/{3^n} < 1/2$ Base case is $n=1$: $\frac {1}{3} < \frac{1}{2}$. So the base case holds.
Inductive hypothesis for $n = k$: $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} < \frac{1}{2}$
Inductive Step for $n = k + 1$:
$$ \left( \frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} \right) + \frac {1}{3^{k+1}}< \frac{1}{2}.$$
Multiplying the $n = k + 1$ step by $3$:
$$ \left( 1 + \frac{1}{3} + \cdots + \frac {1}{3^{k-1}} \right) + \frac {1}{3^{k}}< \frac{3}{2}$$
$$\implies \frac{1}{3} + \cdots + \frac {1}{3^{k-1}} + \frac {1}{3^{k}} < \frac{3}{2} - 1 = \frac {1}{2}.$$
We know this to be true from our inductive hypoethsis. Hence, $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^n} < \frac{1}{2}.$
Is this proof correct?
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Consider a square of area $1$ and divide it into 3 congruent rectangles.
The area of each rectangle is $\frac{1}{3}$.
Select one of the three rectangles.
Divide one of the remaining two rectangles into three congruent rectangles, select only, as done before. Continue like this.
Irrespective of the number of iterations, the sum never reaches half of the original square.
|
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|
Prove $\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\ge 4(\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x})$ with Popoviciu's inequality. I want to prove the inequality $$\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\ge 4(\frac{z}{x+y}+\frac{x}{y+z}+\frac{y}{z+x})$$, for positive reals. I have a hint that it is solved with Popoviciu's inequality applied to $f(x)=x+\frac{1}{x}$. I can show the the function $f$ is convex. Moreover, I see that the LHS of the inequality can be expressed as $f(\frac{x}{y})+f(\frac{x}{z})+f(\frac{z}{y})$. There are other representations as we can choose $2^3=8$ different ways to express the LHS as the sum of $f$'s. However, none of these seem to lead to the desired inequality. I considered rewriting the Popoviciu's inequality as $$f(a)+f(b)+f(c)\ge 2(f(\frac{a+b}{2})+f(\frac{a+c}{2})+f(\frac{b+c}{2}) - \frac{3}{2}f(\frac{a+b+c}{3}))$$, where $a,b,c$ equal $x/y, x/z, z/y$ respectively.
However, it doesn't seem to be the solution. After simplifying the RHS becomes: $$\frac{4}{a+b}+\frac{4}{a+c}+\frac{4}{b+c}-\frac{9}{a+b+c}+a+b+c$$.
Trying slightly different $a, b, c$, namely $x/z, y/z, y/x$ still gives the same result that I can't simplify further. Note, however, that in this case $\frac{1}{a+b}=\frac{z}{x+y}$.
I thought that maybe the RHS of my Popoviciu's inequality can be shown to be greater than or equal than the desired inequality, but I couldn't come up with anything.
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For positives $x$, $y$ and $z$ we need to prove that
$$\sum_{cyc}\frac{x+y+z}{x}-3\geq4\sum_{cyc}\frac{x+y+z}{x+y}-12$$ or
$$\sum_{cyc}\frac{1}{x}+\frac{9}{x+y+z}\geq4\sum_{cyc}\frac {1}{x+y}$$
or
$$\sum_{cyc}\frac{1}{x}+\frac{3}{\frac{x+y+z}{3}}\geq2\sum_{cyc}\frac {1}{\frac{x+y}{2}},$$which is Popovicui for the convex function $f(x)=\frac{1}{x}.$
There are another ways for the proof of your inequality (I see at least 10 of them).
If you want, I can show something.
|
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|
Factoring cubic equation Trying to understand this passage where this:
$$\left(\frac{11}{12}\beta-I\right)\left[\left(\frac{2}{3}\beta-I\right)^2 -\frac{1}{8}\beta^2-\frac{1}{4}\beta\left(\frac{2}{3}\beta-I\right)\right]=0$$
gets factored into this:
$$\left(\frac{1}{6}\beta-I\right) \left(\frac{11}{12}\beta-I\right) \left(\frac{11}{12}\beta-I\right)=0$$
I is a real variable and $\beta$ is a constant. What method is used to achieve this factorization?
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We have by expanding and simplifying
$$\left(\frac{11}{12}\beta-I\right)\left[\left(\frac{2}{3}\beta-I\right)^2 -\frac{1}{8}\beta^2-\frac{1}{4}\beta\left(\frac{2}{3}\beta-I\right)\right]$$
$$=\left(\frac{11}{12}\beta-I\right)\left[\frac{4}{9}\beta^2-\frac{4}{3}\beta I+I^2-\frac{1}{8}\beta^2-\frac{1}{6}\beta^2+\frac{1}{4}\beta I\right]$$
$$=\left(\frac{11}{12}\beta-I\right)\left[\frac{11}{72}\beta^2-\frac{13}{12}\beta I+I^2\right]$$
$$=\left(\frac{11}{12}\beta-I\right)\left[\frac{11}{72}\beta^2-\frac{1}{6}\beta I-\frac{11}{12}\beta I +I^2\right]$$
$$\left(\frac{11}{12}\beta-I\right)\left[\frac{1}{6}\beta\left(\frac{11}{12}\beta-I\right)-I\left(\frac{11}{12}\beta-I\right)\right]$$
$$=\left(\frac{1}{6}\beta-I\right)\left(\frac{11}{12}\beta-I\right)\left(\frac{11}{12}\beta-I\right)$$
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|
Is there any other method to find the remainder when $x+x^3+x^5+\cdots+x^{2n-1}$ is divided by $x^3-x$? A method I got from somewhere.
\begin{align}
f(x)
&=x+x^3+x^5+\cdots+x^{2n-1}\\
&= x\Bigg[\Big(x^{2(n-1)}-1\Big)+\Big(x^{2(n-2)}-1\Big)+\cdots+\Big(x^{2\times 1}-1\Big)+\Big(x^{2\times 0}-1\Big)+n\Bigg]\\
&= x\Big(x^{2(n-1)}-1\Big)+x\Big(x^{2(n-2)}-1\Big)+\cdots+x\Big(x^{2\times 1}-1\Big)+x\Big(x^{2\times 0}-1\Big)+nx
\end{align}
As the last term $nx$ cannot be divided by $x^3-x$, the remainder is $nx$.
My Attempt
I attempted to find another method as follows.
\begin{align}
f(x)
&=x+x^3+x^5+\cdots+x^{2n-1}\\
&= x\Big[x^{2\times 0}+x^{2\times 1}+x^{2\times 2}+\cdots+x^{2(n-1)}\Big]\\
&= x\frac{1\times(x^{2n}-1)}{x^2-1}\\
&=\frac{x(x^n-1)(x^n+1)}{(x-1)(x+1)}
\end{align}
From this, how can we find the remainder $nx$ when $\displaystyle\frac{x(x^n-1)(x^n+1)}{(x-1)(x+1)}$ is divided by $x^3-x$?
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First, we can eliminate a factor of $x$ , and find the remainder of
$P(x) = 1+x^2+x^4+\cdots+x^{2n-2}$ divided by $x^2-1$
Write $P(x)$ as
$P(x) = (x^2-1)Q(x) + ax + b$
Now substitute the values $x=1$, $x=-1$ to obtain
$P(1) = a+b$
$P(-1) = b-a$
and note that $P(1)=P(-1) = n $, and solve the simultaneous equations...
|
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|
Why do we consider the lower bound in this Inequality? Given that $a,b,c,d,e$ are real numbers such that
$a+b+c+d+e=8$,
$a^2+b^2+c^2+d^2+e^2=16$.
Determine the maximum value of $e$.
Let $t$ be a real number. Now consider the expression $$ \sum_{cyc} (t-a)^2 = \sum_{cyc} a^2-2t \sum_{cyc} a
+ 5t^2$$
Now obviously $$(t-a)^2+(t-b)^2+(t-c)^2+(t-d)^2 \ge 0 $$ with equality at $a = b= c = d = t$ so $(t-e)^2 \le \sum_{cyc} a^2-2t \sum_{cyc} a + 5t^2$ or equivalently $e \le \sqrt{16-16t+5t^2} + t = f(t)$.
Now $f'(t) = 1+ \frac{5t-8}{16-16t+5t^2}$.
The resultant quadratic is easily solved for $t =6/5$ and $t = 2$, with the latter being an extraneous root introduced by squaring. The largest
possible $e $ and greatest lower bound of $f(t)$ is then $f(6/5) = 16/5$ , which occurs when
$a = b = c = d = 6/5 $ and $e = 16/5$ .
I had a small confusion here, why does the maximum possible value of $e$ where $f(t)$ is minimum, if $e≤f(t)$? Shouldn't it have been the maximum possible value of $f(t)$ in the possible interval of $t$ that would be allowed by the constraints, which here would just be at the endpoints since $f(t)$ has no maxima where it's derivative is $=0$.
In particular, from the second condition I am getting $-4≤e≤4$. But from first constraint, when $a=b=c=d=r$, $e=8-4t$ and thus equality can only hold when $t≤3$.
So why not just look at the values of $e$ at the endpoints of $t$ in this interval (ie at $3$ itself)?
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Because if $e>\min\limits_{t}f$ so for $t=\frac{6}{5}$ and $a=b=c=d=\frac{6}{5}$ we'll obtain a contradiction.
Another way.
By C-S
$$16=a^2+b^2+c^2+d^2+e^2\geq\frac{1}{4}(a+b+c+d)^2+e^2=\frac{1}{4}(8-e)^2+e^2,$$ which gives $0\leq e\leq\frac{16}{5}.$
The equality occurs for $a=b=c=d$, id est occurs, which says that $\frac{16}{5}$ is a maximal value.
We got also that $0$ is a minimal value of $e$.
|
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How to prove that $\varphi$ is an injective homomorphism and why it is surjective? Here is the question I want to solve:
Prove that the subgroup of $SL_2(\mathbb F_3)$ generated by $\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}$ and $\begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix}$ is isomorphic to the quaternion group of order $8.$[use a presentation for $\mathcal{Q}_8$]
Here is a solution I found online:
A presentation for $\mathcal{Q}_8,$ is $$\langle i,j | i^2 = j^2, i^4 = 1, ij = -ji \rangle$$
Which means that $i,j$ generates $\mathcal{Q}_8$ and they satisfy the relations $i^2 = j^2, i^4 = 1, ji = -ij = i^3j.$
Let $A = \begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}$ and $B = \begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix}.$ Define $\varphi: \{i,j\} \rightarrow \langle A, B \rangle $
as $\varphi(i) = A$ and $\varphi(j) = B.$
Now, note that by direct calculation , the subgroup generated by $A$ is $$\langle A \rangle = \{ \begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}, \begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix}, \begin{pmatrix}
0 & 1 \\
-1 & 0
\end{pmatrix}, \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix} \}$$
And also, by direct calculation , the subgroup generated by $B$ is $$\langle B \rangle = \{ \begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix}, \begin{pmatrix}
-1 & 0 \\
0 & -1
\end{pmatrix}, \begin{pmatrix}
-1 & -1 \\
-1 & 1
\end{pmatrix}, \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix} \}$$
And it is clear that $|A|=|B| = 4$ and $A^2 = B^2.$\
Also, by direct calculation, we have that $BA = \begin{pmatrix}
1 & -1 \\
-1 & -1
\end{pmatrix} = A^3B.$
Here by the lemma $\varphi$ extends to an injective homomorphism $$\bar{\varphi} : \mathcal{Q}_8 \rightarrow \langle A, B \rangle $$
Also, by construction $\bar{\varphi}$ is surjective. Hence the required is proved.
My questions are:
1-I do not know what lemma says that $\varphi$ is an injective homomorphism, tthis problem is #10 in section 2.4 of Dummit and Foote (3rd edition) and I could not find any lemma that said this. Could anyone help me in proving that $\varphi$ is an injective homomorphism please?
2- Why by construction is $\bar{\varphi}$ is surjective? what confuses me is that $\langle A, B \rangle $ is the subgroup generated by $A$ and $B$ and not the set that contains $A$ and $B,$ Could anyone explain to me this please?
|
You know that $A^2=B^2$, $A^4=\operatorname{Id}_2$, and that $AB=-BA$. So, since $\mathcal Q_8$ since the group generated by two generators $i$ and $j$ and by the relations $i^2=j^2$, $i^4=e$, and $ij=-ji$, there is a surjective group homomorphism $\varphi$ from $\mathcal Q_8$ onto the subgroup $G$ of $SL_2(\Bbb F_3)$ spanned by $A$ and $B$. But $\mathcal Q_8$ and $G$ have $8$ elements each. Therefore, $\varphi$ is a bijection.
|
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|
Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\frac{7}{x+4}\\
-1(x+4)<7\\-x < 11\\
x>-11.$$
This results in the answer $x>-11, x>3,$ which doesn't make sense.
|
As noticed you went wrong multiplying for a quantity which can also be negative.
Using that for $A,B>0$ then
$$|A|<B \iff \left|\frac1{A}\right|>\frac1B $$
a very effective way to proceed in this case is the following
$$\left|\frac{7}{x+4}\right|<1 \iff \left|\frac{x+4}{7}\right|>1 \iff |x+4|>7$$
$$\iff x+4<-7 \quad \lor \quad x+4 >7 $$
$$\iff x<-11 \quad \lor \quad x>3 $$
We can also proceed by your way but we need to consider two cases, that is
*
*For $x+4>0 \iff x>-4$
$$\left|\frac{7}{x+4}\right|<1\iff \frac{7}{x+4}<1\iff x+4>7 \iff x>3$$
*
*For $x+4<0 \iff x<-4$
$$\left|\frac{7}{x+4}\right|<1\iff \frac{7}{-(x+4)}<1\iff -(x+4)>7\iff x+4<-7 \iff x<-11$$
|
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|
How does $\left(2 - \frac{1}{2^{k}}\right) + \frac{1}{2^{k + 1}}$ become $2 - \frac{1}{2^{k+1}}$? I'm studying induction here and throughout the proof we got from here:
$$ \begin{align}
&\left(2 - \frac{1}{2^{k}}\right) + \frac{1}{2^{k + 1}} \tag1\\[0.75em]
=\; &2 - \frac{1}{2^{k+1}} ( 2-1) \tag2\\[0.75em]
=\; &2 - \frac{1}{2^{k+1}} \tag3
\end{align}$$
How was that done?
|
They did this by adding up the two fractions:
$\frac{1}{2^k}$ - $\frac{1}{2^{k+1}}$
When dealing with the denominators, you can write $2^{k+1}$ as ${2^k}$$\cdot$${2^1}$
Then the $2^k$ cancels out with the first fraction's denominator and you are left with $2$$\cdot$$1$ which totals to 2. In the second fraction you are just left with 1.
Then this all adds upto: $\frac{2}{2^{k+1}}$ - $\frac{1}{2^{k+1}}$. You can pull out $2^{k+1}$ which equals $\frac{1}{2^{k+1}}$$(2-1)$ = $\frac{1}{2^{k+1}}$.
With the complete expression this equals to what we needed, $2$ - $\frac{1}{2^{k+1}}$
|
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|
Simultaneous equations to solve unknown exponents Solved the following simultaneous equations for $a$ and $b$:
(1)
$$a + 2b = 2\\3a + b = 10$$
$$3a + 6b = 6\\3a + b = 10$$
$$5b = -4$$
$$b = \frac{-4}{5}\\a = \frac{18}{5}$$
(2)
How do we use the answers/information in (1) to solve the following equations for $x$ and $y$?
$$2^{x+2y} = 2^{20}\\5^{5x+y} = 25^{x+50}$$
In general how do these types of equations relate?
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Note that $25^{x+50}=(5^2)^{x+50}=5^{2x+100}$ so the system can be written as
$$2^{x+2y} = 2^{20}\\5^{5x+y} = 5^{2x+100}$$
or
$$2^{x+2y} = 2^{20}\\5^{3x+y} = 5^{100}$$
and comparing exponents we obtain
$$x+2y=20$$
$$3x+y=100$$
This is very similar to the system
$$a + 2b = 2\\3a + b = 10$$
but the RHS is multiplied by $10$. So multiplying by $10$ we have
$$10a + 2(10b) =x+2y= 20\\3(10a) + 10b =3x+y= 100$$
where $x=10a$ and $y=10b$ and you have already found $a$ and $b$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
For all diagonals $a$ of Pascal's triangle (figurate numbers), $\sum_{k=a}^\infty {k\choose a}\frac{1}{2^k}=2$? I am looking for the derivation of the closed form along any given diagonal $a$ of Pascal's triangle,
$$\sum_{k=a}^n {k\choose a}\frac{1}{2^k}=?$$
Numbered observations follow. As for the limit proposed in the title given by:
Observation 1
$$\sum_{k=a}^\infty {k\choose a}\frac{1}{2^k}=2,$$
when I calculate the sums numerically using MS Excel for any $a$ within the domain ($0\le a \le100$) the sum approaches 2.000000 in all cases within total steps $n\le285$. The first series with $a=0$ is a familiar geometric series, and perhaps others look familiar to you as well:
$$\sum_{k=0}^\infty {k\choose 0}\frac{1}{2^k}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+... =2,$$
$$\sum_{k=1}^\infty {k\choose 1}\frac{1}{2^k}=\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{1}{4}+\frac{5}{32}... =2,$$
$$\sum_{k=2}^\infty {k\choose 2}\frac{1}{2^k}=\frac{1}{4}+\frac{3}{8}+\frac{3}{8}+\frac{5}{16}+\frac{15}{64}... =2,$$
but it is both surprising and elegantly beautiful that these sums across all diagonals appear to approach the same value. Some additional observations from the numerically determined sums:
Observation 2
The maximum value of any term ${k\choose a}\frac{1}{2^k}$ within a diagonal $a$ for the domain $(a>0)$ is attained at $k=2a-1$ and repeated for the term immediately following ($k=2a$).
Observation 3
$$\sum_{k=a}^{2a} {k\choose a}\frac{1}{2^k}=1$$
Observation 4
$$\sum_{k=a}^{n} {k\choose a}\frac{1}{2^k} + \sum_{k=n-a}^{n} {k\choose n-a}\frac{1}{2^k}=2$$
It's very likely that the general closed form has been derived before, but searching for the past several days has produced no results. It appears that setting up the appropriate generating function may play a role, but I am at a loss as to how to proceed. Looking forward to the responses.
|
The generating function proof of Observation 1.
Write $$f(x,z)=\sum_{n=0}^{\infty}(1+x)^nz^n=\frac{1}{1-z(x+1)}$$ Then the coefficient of $x^a$ is $$h_a(z)=\sum_{k=a}^\infty \binom ka z^k$$
When $z=1/2,$ then $h_a(1/2)$ is your value.
But $$f(x,1/2)=\frac1{1/2-x/2}=\frac{2}{1-x}$$
So $h_a(1/2)=2,$ the result you want.
For any particular value $|z_0|<1,$ you get:
$$
\begin{align}
f(x,z_0)&=\frac{1}{1-z_0}\frac1{1-x\frac{z_0}{1-z_0}}\\&=\sum_{n=0}^{\infty}\frac{z_0^n}{(1-z_0)^{n+1}}x^n
\end{align}
$$
So $$h_a(z_0)=\frac{z_0^a}{(1-z_0)^{a+1}}\tag1$$
The probability proof of Observation 1.
Let $p_n$ be the probability that in an infinite sequence of tosses of a fair coin, that after $n$ toss you have $a$ heads and the next toss is a another heads. So $$p_n=\binom na \frac1{2^{n+1}}.$$ Since it is probability zero that you will never get $a+1$ heads, this means:
$$\sum_{n} p_n=1.$$
If the probality of heads is $p,$ then:
$$p_n=\binom na p^{a+1}(1-p)^{n-a}$$
This lets you get, with a little manipulation, the same result as $(1)$ with $z_0=1-p.$
Observation 2.
If $x_n=\binom na\frac1{2^n},$ then when $n\geq a,$ we get a recurrence:
$$x_{n+1}=\frac{n+1}{2(n+1-a)}x_n.$$
Show that if $a\leq n<2a-1,$ then $2(n+1-a)<n+1,$ so $x_{n+1}>x_n.$
When $n=2a-1,$ you get $2(n+1-a)=n+1.$ So $x_{2a-1}=x_{2a}.$
When $n>2a-1$ you likewise get $x_{n+1}<x_n.$
Observation 3.
$$\sum_{k=a}^{2a} \binom ka \frac1{2^{k+1}}\tag2$$ is the probability after $2a+1$ tosses of a fair coin, that you’ve gotten $a+1$ or more heads, with $k$ being the point where the $k+1$st toss was when you got the $a+1$st heads.
But $(2)$ is also equal to the probability that you got $a+1$ or more tails.
And you always get $a+1$ or more heads or $a+1$ or more tails in $2a+1$ tosses, but not both. So (2) is equal to $\frac12.$ Multiply by $2$ and you get your identity.
Observation 4.
This is roughly the same as for observation $(3).$
Let $b=n-a.$
Then $$\sum_{k=a}^{a+b} \binom ka \frac1{2^{k+1}}\tag3$$
is the probability in $a+b+1$ tosses of a fair coin that you get $a+1$ or more heads.
And:
$$\sum_{k=b}^{a+b}\binom kb\frac1{2^{k+1}}\tag4$$
is the probability of $b+1$ or more tails in $a+b+1$ tosses of a fair coin.
As with Observation 3, exactly one of the events counted in $(3)$ and $(4)$ is true, so the sum of these two values is $1.$
Multiplying that sum by $2$ gives you the Observation 4.
If $p$ is the probability of heads in a coin, you get a generalization of Observation 4:
$$p^{a+1}\sum_{i=0}^{b}\binom {a+i}a (1-p)^{k-a}+(1-p)^{b+1}\sum_{j=0}^{a}\binom {b+j}b p^{j}=1.$$
Another way to write the probability that you get more than $a$ heads win $a+b+1$ tosses is:
$$\sum_{k=a+1}^{a+b+1}\binom{a+b+1}{k}p^k(1-p)^{a+b+1-k}\\=p^{a+1}\sum_{j=0}^{b}\binom {a+b+1}jp^{b-j}(1-p)^j$$ where $k$ goes over all the possible numbers of heads $>a,$ and $j=a+b+1-k$ loops over the possible numbers of tails.
That gives a new identity, swapping $p$ and $1-p,$ you get:
$$\sum_{j=0}^{b}\binom {a+b+1}jp^j(1-p)^{b-j}= \sum_{i=0}^{b}\binom {a+i}a p^{i}$$
When $p=\frac12,$ this gives you: $$\frac{1}{2^{a+b}}\sum_{j=0}^b \binom{a+b+1}j=\sum_{k=a}^{a+b}\binom ka\frac{1}{2^k}.\tag 5$$
Indeed, the identity $(5)$ can be seen as the heart of the all the observations other than $2.$
For example, looking at the left side, we see it is $\frac{1}{2^{a+b}}\left(2^{a+b+1}-g(b)\right),$ where $g(b)=\sum_{i=0}^a\binom{a+b+1}i$ is a polynomial of degree $a.$ So we get as $b\to \infty,$ that the right side converges to $2.$
You can prove $(5)$ directly by induction.
When $b=0,$ it is easy.
$$
\begin{align}
\sum_{j=0}^{b+1}\binom{a+b+2}j&=\sum_{j=0}^{b+1}\left(\binom{a+b+1}j+\binom{a+b+1}{j-1}\right)\\&=\binom{a+b+1}{b+1}+2\sum_{j=0}^b\binom {a+b+1}j\\&= \binom{a+b+1}{a}+2\sum_{j=0}^b\binom {a+b+1}j
\end{align}
$$ Dividing by $2^{a+b+1}$ on both sides, you get the left side for $(5)$ increases from $b$ to $b+1$ by $\frac1{2^{a+b+1}}\binom{a+b+1}{a}.$
I’m not sure the induction proof of $(5)$ is the most edifying - it is clearer if you understand the underlying probabilities. But it might be the most direct.
|
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|
Given $a + b = c + d$ and $a^2 + b^2 = c^2 + d^2$ prove $a=c, b=d$ or $a=d, b=c$
Given $$a + b = c + d\space \text{and}\space a^2 + b^2 = c^2 + d^2\quad\forall\space a,b,c,d \in \mathbb{R}$$ Prove: $$a=c, b=d\space \text{or} \space a=d, b=c $$
*
*I managed to get $ab=cd$. Don't know how to proceed further.
|
Here's another method.
Note that by re-arranging, we may write $a+b=c+d$ either as $a-c=d-b$ or as $b-c = d-a.$ Just keep that in mind as we manipulate the second equation:
$$\begin{align*}0 &= (c^2 + d^2) - (a^2 + b^2) \\ &= (d^2 - b^2) - (a^2 - c^2) \\ &= (d-b)(d+b) - (a-c)(a+c) \\ &= (a-c)(d+b) - (a-c)(a+c) \\ &= (a-c)((d+b)-(a+c)) \\ &= (a-c)((d-a)+(b-c)) \\ &= (a-c)((d-a)+(d-a)) \\ &= 2(a-c)(d-a)\end{align*}$$
[Do you see the steps I've used each of $a-c=d-b$ and $b-c=d-a$ to simplify the expression?]
At this point, we're effectively done, as the factorization gives us only two solutions:
*
*$a-c=0$: Then $d-b=a-c = 0$ as well, which gives $a=c$ and $b=d$.
*$d-a=0$: Then $b-c=d-a = 0$ as well, which gives $a=d$ and $b=c$.
|
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|
What is the correct general solution of $\sin x=-\frac{1}{2}$? My attempt:
$$\sin x=-\frac{1}{2}$$
$$x=n\pi+(-1)^{n}\sin^{-1}(-\frac{1}{2})$$
$$x=n\pi+(-1)^{n}(-\frac{\pi}{6})$$
$$x=n\pi-(-1)^{n}(\frac{\pi}{6})$$
My book's solution:
$$x=n\pi+(-1)^{n}(\frac{7\pi}{6})$$
Who is correct?
|
We have that
$$\sin x=-\frac{1}{2} \iff x=-\frac \pi 6+2k\pi \quad \lor \quad \frac {7\pi} 6+2k\pi$$
which is equivalent to both solutions indeed
*
*for $n=2k$
$$x=2k\pi-(-1)^{2k}\left(\frac{\pi}{6}\right)=2k\pi-\frac{\pi}{6}$$
$$x=2k\pi+(-1)^{2k}\left(\frac{7\pi}{6}\right)=2k\pi+\frac{7\pi}{6}$$
*
*for $n=2k+1$
$$x=(2k+1)\pi-(-1)^{2k+1}\left(\frac{\pi}{6}\right)=2k\pi+\frac{7\pi}{6}$$
$$x=(2k+1)\pi+(-1)^{2k+1}\left(\frac{7\pi}{6}\right)=2k\pi-\frac{\pi}{6}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the expected number of $\min\{i \mid a_i\geq a_{i+1}\}$? I am solving a computer programming problem.
I wonder if an easy mathematical solution to the following problem exists or not.
Suppose we throw a fair die sufficiently number of times.
Let $a_i$ be the outcome of $i$-th roll.
What is the expected number of $\min\{i \mid a_i\geq a_{i+1}\}$?
My attempt is here:
The probability of $\min\{i \mid a_i\geq a_{i+1}\}=1$ is $\frac{1}{6}\frac{6}{6}+\frac{1}{6}\frac{5}{6}+\frac{1}{6}\frac{4}{6}+\frac{1}{6}\frac{3}{6}+\frac{1}{6}\frac{2}{6}+\frac{1}{6}\frac{1}{6}$.
|
Let $X=\min\{i \mid a_i\geq a_{i+1}\}$
Then $P(X\ > i)=P(a_1 < a_2, a_2 < a_3 , ..., a_i < a_{i+1})$
Now, if we throw a $6-$faces die $k$ times, there are $\binom{6}{k}$ possible strictly increasing sequences.
Hence the above probability is
$$P(X\ > i) = \frac{\binom{6}{i+1}}{6^{i+1}}$$
for $i =1,2 \cdots 5$. And the expected value is
$$\begin{align}
E[X] &= \sum_{i=0}^\infty P(X\ > i)
\\ &= 1 + \sum_{i=1}^5 \frac{\binom{6}{i+1}}{6^{i+1}} \\ &=\sum_{k=0}^6 \binom{6}{k}{(1/6)^{k}} -1\\
&= \left(1 + \frac{1}{6}\right)^6-1\\
&=1.5216\cdots\end{align}$$
where we've used the Binomial theorem.
|
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|
Solve the following equation: $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$ I have tried many ways but not success
$\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$
$\Leftrightarrow \sqrt{x}(\sqrt{x+1}-\sqrt{x-1})=x+1$
or $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1 \Leftrightarrow \sqrt{\dfrac{x}{x+1}}-\dfrac{\sqrt{x(x-1)}}{x+1}=1 \Leftrightarrow \dfrac{x}{x+1}=1+\dfrac{2\sqrt{x(x-1)}}{x+1}+\dfrac{x(x-1)}{(x+1)^2} \Leftrightarrow \dfrac{(x+1)^2+2\sqrt{x(x-1)}(x+1)+x(x-1)-x(x+1)}{(x+1)^2}=0 \Leftrightarrow x^2+1+2\sqrt{x(x-1)}(x+1)=0 $
I need your help!
|
After squaring, you face the quartic equation
$$3 x^4+4 x^3-6 x^2-4 x-1=0$$
Since $\Delta=-94208$, there are two real roots. Now, forget the nasty formulae and use inspection or graphing; there are two roots "close" to $-2$ and $1.5$. So, use Newton method and obtain the following iterates
$$\left(
\begin{array}{cc}
n & x_n \\
0 & -2.00000 \\
1 & -2.03571 \\
2 & -2.03396
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
0 & 1.50000 \\
1 & 1.32005 \\
2 & 1.26513 \\
3 & 1.26015 \\
4 & 1.26011
\end{array}
\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Spivak Proof: $\left| \frac{1}{x} \right| = \frac{1}{|x|}$ ..... if $x \neq 0$
Spivak's Proof:
$\left| \frac{1}{x} \right| \cdot \left| x \right| = \left| \left(\frac{1}{x}\right) \cdot x \right | = 1$ so $\left| \frac{1}{x} \right| = \frac{1}{\left|x\right|} $
I am fine until the 'so'. Is the 'so'? $$ 1 \cdot \left| {x}^{-1} \right| = \frac{1}{\left| x \right|}$$
This strategy seems to be used a lot by Spivak.
|
By this
$$\left| \frac{1}{x} \right| \cdot \left| x \right| = \left| \left(\frac{1}{x}\right) \cdot x \right | = 1$$
we have shown that $\left| \frac{1}{x} \right|$ and $\left| x \right|$ are one the inverse of the other, that is $\left| \frac{1}{x} \right| = \frac{1}{\left|x\right|}$.
Edit
As an alternative
*
*for $x>0 \implies \left| \frac{1}{x} \right| = \frac1 x =\frac{1}{\left|x\right|}$
*for $x<0 \implies \left| \frac{1}{x} \right| =- \frac1 x =\frac{1}{\left|x\right|}$
and for $x=0$ they are both not defined, therefore they are equivalent.
|
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|
Help with convergence sequence by cases I have to study the converge or diverge or this sequence:
\begin{eqnarray}
c_{n}=\left\{\left(\cos (\frac{a}{n})\right)^{\frac{1}{\ln(|\cos(\frac{b}{n})|)}}\right\} \hspace{0.3cm} \text{for a,b > 0}
\end{eqnarray}
So I noticed that:
\begin{eqnarray}
(\cos (\frac{a}{n}))^{\frac{1}{\ln(|\cos(\frac{b}{n})|)}} &=& e^{\frac{1}{\ln(|\cos(\frac{b}{n})|)}\ln(\cos(\frac{a}{n}))}\\
&=& e^{\frac{\ln(\cos(\frac{a}{n}))}{\ln|(\cos(\frac{b}{n})|}}\\
&=& e^{\log_{|\cos(\frac{b}{n})|}{\cos}(\frac{a}{n})}\\
&=& e^{\log_{\cos(\frac{b}{n})}{\cos(\frac{a}{n})}}
\end{eqnarray}
So, if $a=b$ :
\begin{eqnarray}
\lim_{n\rightarrow \infty} c_{n}= e
\end{eqnarray}
If $b>a$
\begin{eqnarray}
e>e^{\log_{\cos(\frac{b}{n})}{\cos(\frac{a}{n})}}>1
\end{eqnarray}
Can I find a specific value of this limit in this case? and If $a > b$ diverges.
|
HINT
Since $\cos \frac{b}{n}\to 1$ we can eliminate absolute value and we can use that
$$\frac{\ln\left(\cos \frac{a}{n}\right)}{\ln\left(\cos \frac{b}{n}\right)}=$$
$$= \frac{\ln(1+(\cos \frac{a}{n}-1))}{\cos \frac{a}{n}-1} \frac{\cos \frac{b}{n}-1}{\ln(1+(\cos \frac{b}{n}-1))}\frac{\cos \frac{a}{n}-1} {\left(\frac{a}{n}\right)^2}\frac{\left(\frac{b}{n}\right)^2}{\cos \frac{b}{n}-1}\frac{a^2}{b^2}$$
then apply standard limits.
|
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|
Issue solving an integral I'm having issues solving the following integral:
$$ \int{\frac{e^x}{e^{2x}-e^x-2}}dx $$
From what I can tell, I should substitute $u = e^x$ and $du = e^xdx$ and end up with
$$ \int{\frac{1}{u^2 - u - 2}}du $$
where I can then use partial fraction decomposition to simplify the integrand as follows
$$ \frac{1}{u^2 - u - 2} = \frac{A}{u-2} + \frac{B}{u+1} $$
$$ A(u+1) + B(u-2) = 1 $$
$$ Au + A + Bu -2B = 1 $$
$$\begin{pmatrix}1 & 1 & 0\\1 & -2 & 1\end{pmatrix} \to \begin{pmatrix} 1 & 0 & \frac{1}{3} \\ 0 & 1 & -\frac{1}{3} \end{pmatrix} \to A=\frac{1}{3}, B=-\frac{1}{3} $$
So I have
$$ \int{ \left( \frac{ \frac{1}{3} }{u-2} - \frac{ \frac{1}{3} }{u + 1} \right) } du = \frac{1}{3} \int{\frac{1}{u - 2}du} - \frac{1}{3}\int{\frac{1}{u+1}}du $$
which gives me
$$ \frac{1}{3}\ln(u-2) - \frac{1}{3}\ln(u + 1) + C, \textrm{ where } u =e^x $$
so my final answer is
$$ \frac{1}{3}( \ln(e^x-2) - \ln(e^x + 1) ) + C $$
However, wolframalpha is telling me that the answer is
$$ \frac{2}{3}\tanh^{-1}\left(\frac{1}{3} - \frac{2e^x}{3}\right) + C $$
I really have no idea how the inverse tangent hyperbolic function would even get there. We haven't even covered any integrals that give $\tanh^{-1}$ in class.
|
Answer in the Question
Your answer
$$
\frac13\log\left(\frac{e^x-2}{e^x+1}\right)+C\tag1
$$
is partially correct; another possibility is
$$
\frac13\log\left(\frac{2-e^x}{1+e^x}\right)+C\tag2
$$
Which one is applicable depends on the initial value of $x$. If $x\gt\log(2)$, use $(1)$; if $x\lt\log(2)$, use $(2)$.
WolframAlpha's Answer
Since
$$\newcommand{\arctanh}{\operatorname{arctanh}}
\arctanh(u)=\frac12\log\left(\frac{1+u}{1-u}\right)\tag3
$$
we have that
$$
\begin{align}
\frac23\arctanh\left(\frac3{1-2e^x}\right)
&=\frac23\cdot\frac12\log\left(\frac{1+\frac3{1-2e^x}}{1-\frac3{1-2e^x}}\right)\\
&=\frac13\log\left(\frac{e^x-2}{e^x+1}\right)\tag4
\end{align}
$$
and
$$
\begin{align}
\frac23\arctanh\left(\frac{1-2e^x}3\right)
&=\frac23\cdot\frac12\log\left(\frac{1+\frac{1-2e^x}3}{1-\frac{1-2e^x}3}\right)\\
&=\frac13\log\left(\frac{2-e^x}{1+e^x}\right)\tag5
\end{align}
$$
However, the answer WA gave you was for $x\lt\log(2)$, whereas the answer in your question is for $x\gt\log(2)$.
|
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|
Finding all pairs of functions $f, g: \mathbb{R} \to \mathbb{R} $ which satisfy $f(x+y) = g\left(\frac{1}{x}+\frac{1}{y}\right)(xy)^{2008}$ Find two functions $f, g: \mathbb{R} \to \mathbb{R} $ which satisfy the condition:
$$
f(x+y) = g\left(\frac{1}{x}+\frac{1}{y}\right)(xy)^{2008}
$$
The first one I thought of was this:
$$
f(x)=x^{2008}, g(x)=x^{2008}
$$
or
$$
f \equiv 0, g \equiv 0
$$
Then, I tried:
$$
P(x, y): f(x+y) = g\left(\frac{1}{x}+\frac{1}{y}\right)(xy)^{2008} \\
P(1, 1): f(2)=g(2). \\
P(x, -x): f(0)=x^{4016}g(0) \\
\implies f(0)=0, g(0)=0 \ (\because P(x, -x) \text{ should be true for all } x \ \in \mathbb{R} \text{.}) \\
P\left(x, \frac{1}{x}\right): f\left( x + \frac {1}{x} \right) = g\left(x + \frac {1}{x}\right)
$$
Help me solving this.
p.s. I'm looking for the process of finding the function, and the functions which I am finding are "all functions that can exist".
|
Let find $y$ such that $x+y=xy\Rightarrow y=\frac{x}{x-1}$.
One can take $y=\frac{x}{x-1}$ for $x\neq 0$, $x\neq 1$, then
$$f\left(x+\frac{x}{x-1}\right)=g\left(\frac{1}{x}+\frac{x-1}{x}\right)\cdot\left(x\cdot \frac{x}{x-1}\right)^{2008}\Rightarrow$$
$$f\left(\frac{x^2}{x-1}\right)=g(1)\cdot \left(\frac{x^2}{x-1}\right)^{2008}$$
$t=\frac{x^2}{x-1}$ could be any in $(-\infty;0)\cup[4;+\infty)$, so at least for such $t$: $f(t)=g(1)\cdot t^{2008}$.
Let find $y$ such that $x+y=-xy\Rightarrow y=-\frac{x}{x+1}$.
One can take $y=-\frac{x}{x+1}$ for $x\neq 0$, $x\neq -1$, then
$$f\left(x-\frac{x}{x+1}\right)=g\left(\frac{1}{x}-\frac{x+1}{x}\right)\cdot\left(-x\cdot \frac{x}{x+1}\right)^{2008}\Rightarrow$$
$$f\left(\frac{x^2}{x+1}\right)=g(-1)\cdot \left(\frac{x^2}{x+1}\right)^{2008}$$
$t=\frac{x^2}{x+1}$ could be any in $(-\infty;-4]\cup(0;+\infty)$, so at least for such $t$: $f(t)=g(-1)\cdot t^{2008}$.
For any $t>4$: $f(t)=g(-1)\cdot t^{2008}=g(1)\cdot t^{2008}$, therefore $g(-1)=g(1)=a$.
For any $t\neq 0$: $f(t)=a t^{2008}$.
Taking $y=-x$ one can obtain $f(0)=g(0)=0$, so $f(t)=a t^{2008}$ for any $t$.
Taking $y=1/x$: $f(x^2+1/x)=g(x^2+1/x)$
Equation $x^2+1/x=t$ has solution for $x$ for any $t\in(-\infty;-2]\cup[2;\infty)$, so $f(t)=g(t)$ for any $t$ such that $|t|\geq 2$.
Taking $y=x$: $f(2x)=g(2/x)\cdot x^{4016}$.
Using $x=t$ and $x=1/t$ in last equation, one can get $f(2t)=g(2/t)\cdot t^{4016}$ and $f(2/t)=g(2t)\cdot (\frac{1}{t})^{4016}$. Multiplying these equations gives $f(2t)\cdot f(2/t)=g(2/t)\cdot g(2t)$.
Using $t$ such that $|2t|\geq 2$ last equation transforms to $f(2t)\cdot f(2/t)=g(2/t)\cdot f(2t)$. $a\neq 0$, then $f(2/t)=g(2/t)$. If $|2t|\geq 2$ then $|t| \geq 1$, $|1/t| \leq 1$, $|2/t| \leq 2$. So $f(x)=g(x)$ for any $x\neq 0$ such that $|x|\leq 2$.
So $g(x)=f(x)$ for any $x$ except case $a=0$.
If $a=0$ then $f(x)=0$ and $g(x)=0$ because $t=\dfrac{1}{x}+\dfrac{1}{y}$ can take any value $t\neq 0$ and $g(0)=0$.
The final answer: $f(x)=g(x)=a x^{2008}$, $a$ is arbitrary real constant.
|
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|
Solve the elliptic integral $\int \frac1{\sqrt{C-x^2+\frac{x^4}2}}$ as the solution to $\ddot x + x - x^3 = 0.$ It is said that the elliptic integral $\int \frac1{\sqrt{C-x^2+\frac{x^4}2}}$ (whose solution is in terms of cnoidal and snoidal functions) is the solution to $\ddot x + x - x^3 = 0.$ (Multiply the left side by $\dot x$ and get the first integral.)
How to calc the integral? Why such a function is called an elliptic integral, and is important for solving non-linear ode's?
|
First we have to express the radicand as a product of its roots:
The roots of the polynomial are
$$x_{1}= \pm \sqrt{1-\sqrt{2C+1}} \quad C>-\frac{1}{2}$$
$$x_{2}= \pm \sqrt{1+\sqrt{2C+1}} \quad C>-\frac{1}{2}$$
For simplicity denote $a=x_{1}^2, b=x_{2}^2$. Then
$$I= \int \frac1{\sqrt{C-x^2+\frac{x^4}2}}dx = \int \frac1{\sqrt{(x^2-a)(x^2-b)}}dx = \frac{1}{\sqrt{ab}}\int \frac1{\sqrt{\left(\frac{x^2}{a}-1\right)\left(\frac{x^2}{b}-1\right)}}dx$$
If we make the substitution $\displaystyle w^2=\frac{x^2}{a}$ then
$$I= \frac{1}{\sqrt{b}}\int \frac1{\sqrt{\left(w^2-1\right)\left(\frac{a}{b}w^2-1\right)}}dx$$
Denote $$ k^2 = \frac{a}{b}$$
and make the change of variable $$ w = \frac{1}{k\sin^2\theta} \Longrightarrow dw = -\frac{\cos\theta}{k\sin^2 \theta} = -\frac{\sqrt{1-\sin^2\theta}}{k\sin^2 \theta} $$
$$I = \frac{1}{\sqrt{b}}\int \frac1{\sqrt{\left(w^2-1\right)\left(k^2w^2-1\right)}}dx = -\frac{1}{\sqrt{b}} \underbrace{\int \frac{1}{\sqrt{1-k^2\sin^2\phi}}d\phi}_{J}$$
this integral $J$ of the right hand side is the Elliptic integral of the first kind expressed in Legendre notation
The integral is taken from $\theta =0$ up to an arbitraty value $\theta = \phi$
$$F(\phi,k) = \int_{0}^{\phi} \frac{1}{\sqrt{1-k^2\sin^2\phi}}d\phi$$
The parameter $k$ is assumed $|k|<1$ and is called the modulus while the parameter $k'=\sqrt{1-k^2}$ is called the complementary modulus.
The theory of elliptic integrals is important because there is no closed form for this kind integrals in terms of elementary functions (there are numerical solutions to them), so a new theory had to be developed to study them. If $k=0$ or $k=1$ the Elliptic integrals are just normal functions, but the power of this functions appears when $0<k<1$.You can find applications in the theory of numbers, differential equations, geometry, etc.
|
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|
$(a^2+b^2)\sqrt{a^2b^2+c^2}+(b^2+c^2)\sqrt{b^2c^2+a^2}+(c^2+a^2)\sqrt{c^2a^2+b^2}+(a-b)(b-c)(c-a)\ge2(a^2b^2+b^2c^2+c^2a^2)$ Problem: Let $a,b,c\ge0.$ Prove that: $$(a^2+b^2)\sqrt{a^2b^2+c^2}+(b^2+c^2)\sqrt{b^2c^2+a^2}+(c^2+a^2)\sqrt{c^2a^2+b^2}+(a-b)(b-c)(c-a)\ge2(a^2b^2+b^2c^2+c^2a^2)$$
This is strange form for me.
My approach: By AM-GM inequality:$(ab)^2+c^2\ge\frac{1}{2}(ab+c)^2$ and $a^2+b^2\ge2ab$
It is enough to show: $\sqrt{2}\sum_{cyc}{ab(ab+c)}+(a-b)(b-c)(c-a)\ge2(a^2b^2+b^2c^2+c^2a^2)$
But it seems not true by calculation example. I think Equality holds at (0,0,0). Any one help me? Thanks!
|
We will prove the stronger inequality
$$
\sum (a^2 + b^2)\sqrt{a^2b^2 + c^2} \ge 2\sum(ab)^2 + |(a-b)(b-c)(c-a)|
$$
The inequality is equivalent to
$$
\sum \left[(a-b)^2\sqrt{a^2b^2+c^2} + 2(\sqrt{a^2b^2 + c^2} - ab)ab\right] \ge |(a-b)(b-c)(c-a)|
$$
However, this is true as
$$
\sum \left[(a-b)^2\sqrt{a^2b^2+c^2}\right] \ge \sum \left[(a-b)^2c\right] \ge |(a-b)(b-c)(c-a)|
$$
and
$$
\sum 2(\sqrt{a^2b^2 + c^2} - ab)ab \ge 0
$$
This completes our proof. Hopefully it is okay this time round :)
|
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|
Simplifying a derivative of a function involving roots. I am working on a question where I need to differentiate $$y=\{x+\sqrt{1+x²}\}^\frac{3}{2}\tag1$$
Using the chain rule I have found the first derivative $$\frac{dy}{dx} =\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{1}{2}\{1+\frac{x}{\sqrt{1+x²}}\}\tag2$$
Which can be written as
$$\frac{dy}{dx} = \frac{3y}{2\{x+\sqrt{1+x²}\}}\{1+\frac{x}{\sqrt{1+x²}}\}\tag3$$
This is as far as I have got in terms of simplifying. The solution says that $(2)$ is equal to $(4)$ $$\frac{\frac{3}{2}\{x+\sqrt{1+x²}\}^\frac{3}{2}}{\sqrt{1+x²}}=\frac{3y}{2\sqrt{1+x^2}}\tag4$$
How do you get from $(2)$ to $(4)$? I have included $(3)$ in order to show my workings.
|
A quick way of differentiation is to use logarithms on both sides of the equation:
$$
y = \left[ x + \sqrt{1 + x^2} \right]^{3 \over 2} \tag{1}
$$
We get
$$
\ln|y| = {3 \over 2} \ln\left| x + \sqrt{1 + x^2} \right| \tag{2}
$$
Differentiating both sides of (2) with respect to $x$, we get
$$
{1 \over y} \, {dy \over dx} = {3 \over 2} \ {1 + {2 x \over 2 \sqrt{1 + x^2}}
\over x + \sqrt{1 + x^2}}
$$
which can be simplified as
$$
{1 \over y} {dy \over dx} = {3 \over 2} \ {x + \sqrt{1 + x^2} \over
\sqrt{1 + x^2} \ [ x + \sqrt{1 + x^2} ]} \tag{3}
$$
Simplifying (3), we get
$$
{1 \over y}{dy \over dx} = {3 \over 2 \sqrt{1 + x^2}}
$$
Equivalently, we get
$$
{dy \over dx} = {3 y \over 2 \sqrt{1 + x^2}}
$$
|
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|
Find closed formula for the exponential generating function of sequence counting permutations with 2 cycles. Find a closed formula for the exponential generating function of $\{c(n,2)\}_{n\geq2}$, that is the sequence counting permutations of length $n$ that consist of $2$ cycles.
My approach: The exponential generating function for Stirling Numbers of the First Kind,
$$
C(n,k) = \sum_{n=k}^{+\infty}{c(n,k)} \, \frac{x^n}{n!}
= \frac{\log^k(1/(1-x))}{k!}
$$
Replacing $k$ with $2$,
$$
\sum_{n=2}^{+\infty} c(n,2) \, \frac{x^n}{n!}
= \frac{1}{2!} \biggl( \log \frac{1}{1-x} \biggr)^2
$$
Using
$$
\log \frac{1}{1-x} = x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots{}
$$
we have
$$
\biggl( \log \frac{1}{1-x} \biggr)^2
= \biggl( x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots{} \biggr)
\biggl( x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots{} \biggr)
$$
Now $c(n,2)$ is given by the coefficient of $\dfrac{x^n}{n!}$ in $C(x)$
Hence,
$$
c(n,2) = \frac{1}{2!} \sum_{k\geq 1} \frac{n!}{k(n-k)}
$$
I may be dead wrong about this. Any help is appreciated.
|
Your approach is correct, to finish notice that $\frac{1}{k(n-k)}=\frac{1}{n}\left (\frac{1}{k}+\frac{1}{n-k}\right )$ and adding over $k$ you get twice $H_{n-1}.$
For a combinatorial way, choose the elements that you want in the cycle containing one, say of size $k$ in $\binom{n-1}{k}$ ways. Notice that when you pick the elements there are $k!$ possible cycles (by fixing $1$ as the minimal element and permuting the other elements in all ways), the rest $n-k$ elements are in the other cycle and there are $(n-k-1)!$ to arrange it. Adding over $k$ you get
$$\sum _{k=1}^{n-1}\binom{n-1}{k}k!(n-k-1)!=\sum _{k=1}^{n-1}\frac{(n-1)!}{(n-k)}=(n-1)!H_{n-1}.$$
|
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|
Write with one radical $\sqrt[4]{2^6}\cdot\sqrt{3^3}$
Write with one root (radical) $$\sqrt[4]{2^6}\cdot\sqrt{3^3}$$
In this lesson we have learnt that when the roots exist then $$\sqrt[n]{a}=\sqrt[nk]{a^k}$$
Using that here, we have $\gcd(4,2)=2,$ so
$$\sqrt[4]{2^6}\cdot\sqrt[2\cdot2]{\left(3^{3}\right)^2}=\sqrt[4]{2^6}\cdot\sqrt[4]{3^6}=\sqrt[4]{2^6\cdot3^6}=\sqrt[4]{\left(2\cdot3\right)^6}=\sqrt[4]{6^6}$$
Their solution, though, goes as
$$\sqrt[4]{2^6}\cdot\sqrt{3^3}=\sqrt{2^3}\sqrt{3^3}=\sqrt{6^3}=6\sqrt6$$
They haven't calculated the $\gcd(4,2)$ of the indices $4$ and $2$ as in the previous examples (e.g. $\sqrt[4]{3}\cdot\sqrt[3]{2}=...=\sqrt[12]{432}$). What am I missing? Thank you!
|
The author of the solution notices that $6=2\cdot 3$, so $$\sqrt[4]{2^6}=\sqrt[4]{2^{2\cdot 3}}=\sqrt[4]{\left(2^3\right)^2}=\sqrt[4]{8^2}$$
By your provided equality, if we let $a=8, n=k=2$, then we have
$$\sqrt{8}=\sqrt[4]{8^2}$$
Now, $8=2^3$, so we can conclude
$$\sqrt[4]{2^6}\sqrt{3^3}=\sqrt{2^3}\sqrt{3^3}$$
Also, we don't really need to write $2^{2\cdot 3}$ as $8^2$ to prove that $\sqrt[4]{2^6}=\sqrt{2^3}$. In fact, we have
$$\sqrt[nk]{a^{ck}}=\sqrt[n]{a^c}$$
This means we only need to compute the GCD of the radical and exponent, or in our case, $\gcd(4,6)$, to simplify radical expressions. I'll leave it to you to prove this yourself. (Hint: write $a^{ck}=\left(a^c\right)^k$ and let $b=a^c$)
|
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|
Series $\sum_{n=1}^\infty \frac{\cos n}{n^2}$ This is the series I want to find:
$$\sum_{n=1}^\infty \frac{\cos n}{n^2}$$
WolframAlpha gives the answer of $\frac14 + \frac16 (\pi - 3)\pi$.
I do not know anything about this problem other than to prove that it is convergent.
Thanks.
|
Here's a solution using Fourier series.
For $a\in\mathbb C-\mathbb Z$, consider the $2\pi$-periodic function $f$ defined as $f_a(x)=\cos (ax)$ for $x\in [-\pi,\pi)$.
Let's expand it as a Fourier series (since it's periodic). The Fourier coefficients are
$$a_n = \int_{-\pi}^{\pi} \cos(ax)\cos(nx)dx = \frac{(-1)^{n+1}}{\pi}\frac{a \sin{\pi a}}{n^2-a^2}$$
So the Fourier expansion is $$f_a(x)=\cos(ax)=\frac{\sin{\pi a}}{\pi a} \left [1+2 a^2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \cos{n x}}{n^2-a^2} \right ]$$
In other words
$$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} \cos{n x}}{n^2-a^2} = \frac 1 {2a^2}\left ( \frac{\pi a \cos(ax)}{\sin(\pi a)} - 1\right)$$
Evaluating at $x=1-\pi\in [-\pi, \pi)$ gives
$$\sum_{n=1}^{\infty} \frac{ \cos n}{n^2-a^2} = \frac 1 {2a^2}\left (1 - \frac{\pi a \cos(a(1 - \pi))}{\sin(\pi a)} \right)$$
Now letting $a\rightarrow 0$ gives
$$\begin{split}
\sum_{n=1}^{\infty} \frac{ \cos n}{n^2} &= \lim_{a\rightarrow 0}\frac 1 {2a^2}\left ( 1 - \frac{\pi a \cos(a (1-\pi))}{\sin(\pi a)} \right)\\
&=\lim_{a\rightarrow 0}\frac 1 {2a^2}\left (1 - \frac{\pi a \left(1-\frac{a^2(1-\pi)^2}{2} + o(a^2)\right)}{\pi a -\frac{\pi^3 a^3}6 + o(a^2)} \right)\\
&= \lim_{a\rightarrow 0}\frac 1 {2a^2}\left (\frac{a^2}2 - a^2\pi + a^2\frac{\pi^2}3 +o(a^2)\right)\\
&=\frac 1 4 -\frac {\pi} 2 + \frac {\pi^2} 6
\end{split}$$
|
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|
Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now I need to factor the quadratic $y^2-12y+12$.
So, I calculated discriminant
$$D=12^2-4\times 12=96\implies \sqrt D=4\sqrt 6.$$
This means that the multipliers of quadratic are not rational. So I don't know how to proceed anymore.
|
$4x^2-2xy-4x+3y-3 = 4x^2+(-2y-4)x+3y-3=4x^2+[-2(y-1)-2(3)]x+3(y-1)=(2x-y+1)(2x-3).$
|
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|
How to find the exact value of the integral $ \int_{0}^{\infty} \frac{d x}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}}$? $\textrm{I first reduce the power two to one by Integration by Parts.}$
$\begin{aligned}\displaystyle \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x &=\int_{0}^{\infty} \frac{x^{6}}{\left(x^{6}+1\right)^{2}} d x\\&=\displaystyle -\frac{1}{6} \int_{0}^{\infty} x d\left(\frac{1}{x^{6}+1}\right)\\&
=\displaystyle -\left[\frac{x}{6\left(x^{6}+1\right)}\right]_{0}^{\infty}+\frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x \quad \textrm{ (Via Integration by Parts})\\&=\displaystyle \frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x\end{aligned}$
$\textrm{Then I am planning to evaluate }\displaystyle I= \int_{0}^{\infty} \frac{1}{x^{6}+1}\text{ by resolving }\frac{1}{x^{6}+1} \text{ into partial fractions.}$
But after noticing that $$I=\int_{0}^{\infty} \frac{d x}{x^{6}+1}\stackrel{x\mapsto\frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{4}}{x^{6}+1} d x,$$
I changed my mind and started with $3I$ instead of $I$ as below:
$$
\begin{aligned}
3 I &=\int_{0}^{\infty} \frac{x^{4}+2}{\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)} d x \\
&=\int_{0}^{\infty}\left(\frac{1}{x^{2}+1}+\frac{1}{x^{4}-x^{2}+1}\right) d x \\
&=\left[\tan ^{-1} x\right]_{0}^{\infty}+\int_{0}^{\infty} \frac{\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x \\
&=\frac{\pi}{2}+\frac{1}{2} \int_{0}^{\infty} \frac{\left(1+\frac{1}{x^{2}}\right)-\left(1-\frac{1}{x^{2}}\right)}{x^{2}+\frac{1}{x^{2}}-1} d x\\
&=\frac{\pi}{2}+\frac{1}{2}\left[\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1}-\int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^{2}-3}\right] \\
&=\frac{\pi}{2}+\frac{1}{2}\left[\tan ^{-1}\left(x-\frac{1}{x}\right)\right]_{0}^{\infty}-0 \\
&=\frac{\pi}{2}+\frac{1}{2}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right] \\
&=\pi \\ \therefore I &=\frac{\pi}{3}
\end{aligned}
$$
Now I can conclude that
$$\boxed{\displaystyle \quad \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x=\frac{\pi}{18} }.$$
:|D Wish you enjoy the solution! Opinions and alternative methods are welcome.
|
In general
$$ \int_{0}^{\infty} \frac{1}{1+x^s} dx = \frac{\pi}{s}\csc\left(\frac{\pi}{s}\right) \quad \Re(s)>1$$
$$\int_{0}^{\infty} \frac{1}{1+x^s}dx = \frac{2}{s}\int_{0}^{\infty} \frac{w^{\frac{2}{s}-1}}{1+w^2}dw \quad (w^2 \mapsto x^s)$$
Recall the integral representation of $\displaystyle \sec z$:
$$ \sec z = \frac{2}{\pi}\int_{0}^{\infty} \frac{t^{\frac{2z}{\pi}}}{1+t^2} dt \quad |\Re(z)|<\frac{\pi}{2}$$
Putting $ \displaystyle z = \frac{\pi}{s} -\frac{\pi}{2}$
$$ \frac{\pi}{2} \sec\left(\frac{\pi}{s} -\frac{\pi}{2} \right) = \frac{\pi}{2} \csc\left(\frac{\pi}{s}\right)= \int_{0}^{\infty} \frac{t^{\frac{2}{s}-1}}{1+t^2} dt $$
$$ \Longrightarrow \frac{\pi}{s}\csc\left(\frac{\pi}{s}\right) = \frac{2}{s} \int_{0}^{\infty} \frac{t^{\frac{2}{s}-1}}{1+t^2} dt \quad \Re(s)> 1 $$
Hence
$$\boxed{ \int_{0}^{\infty} \frac{1}{1+x^s}dx =\frac{\pi}{s}\csc\left(\frac{\pi}{s}\right) \quad \Re(s)> 1 }$$
Therefore
$$\boxed{ \int_{0}^{\infty} \frac{1}{1+x^6}dx =\frac{\pi}{6}\csc\left(\frac{\pi}{6}\right) = \frac{\pi}{3} }$$
|
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|
Proving $\arctan x = 2 \arctan \frac{x}{1 + \sqrt{1 + x^2}}$, starting from the integral representation of $\arctan$ Starting from the integral representation of the arctan function, $$\arctan x = \int_0^x \frac{dt}{1+t^2},$$how does one prove the (half-angle) identity? $$\arctan x = 2 \arctan \frac{x}{1 + \sqrt{1 + x^2}}$$
I am interested in a clever change of variables that should do the trick. However, I am unable to find it.
|
If you differentiate$$2\arctan\left(\frac t{1+\sqrt{1+t^2}}\right),$$then you get $\frac1{1+t^2}$. So\begin{align}2\arctan\left(\frac x{1+\sqrt{1+x^2}}\right)&=2\arctan\left(\frac x{1+\sqrt{1+x^2}}\right)-2\arctan\left(\frac0{1+\sqrt{1+0^2}}\right)\\&=\int_0^x\frac{\mathrm dt}{1+t^2}\\&=\arctan(x).\end{align}
|
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|
Non-linear 1st order differential equation of $~x\frac{dy}{dx}+y=x\sqrt{y}~~~$where$~\sqrt{y}~$exists I think this is the first time when I handle of non-linear differential equation.
$$ \underbrace{x \frac{dy}{dx} + y = x\sqrt{ y } }_{x > 0} \tag{1} $$
$$ \frac{dy}{dx} + \frac{1}{ x } y = \sqrt{ y } $$
$$ \underbrace{\frac{d}{dx}\left(yx\right)}_{\text{LHS of eqn1} } = x \sqrt{ y } ~~ \leftarrow~~ \text{What can I do for next?} $$
Which website is suitable for this problem?
ADD
I got the following after I have read the post of @Gerd
$$ \frac{d}{dx}\left(yx\right) = x \sqrt{ y } \tag{2} $$
$$ \frac{d}{dx}\left(xy\right) = x^{\frac{1}{2} } \cdot x^{\frac{1}{2} } \sqrt{ y } $$
$$ = \sqrt{ x } \sqrt{ xy } $$
$$ z:=xy \tag{3} $$
$$ \frac{ dz }{ dx } = \sqrt{ x } \sqrt{ z } \tag{4} $$
$$ \frac{ 1 }{ \sqrt{ z } } \frac{dz}{dx} = \sqrt{ x } $$
$$ \int_{ }^{ } \frac{1}{ \sqrt{ z } } \frac{dz}{dx} \,dx = \int_{ }^{ } \sqrt{ x } \,dx $$
$$ \int_{ }^{ } \frac{1}{ \sqrt{ z } } \,dz = \int_{ }^{ } x^{\frac{1}{ 2 } } \,dx $$
$$ \int_{ }^{ } \frac{ 1 }{ z^{\frac{1}{2} } } \,dz = \frac{ x^{\frac{ 3 }{ 2 } } }{ \frac{ 3 }{ 2 } } + \text{const}_{1} $$
$$ \int_{ }^{ } z^{-1/2} \,dz = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ \frac{ z^{\frac{1}{2}} }{ \frac{1}{2} } = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ 2z^{\frac{1}{2}} = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ 2 \sqrt{ z } = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ 2 \sqrt{ yx } = \frac{ 2 }{ 3 } x\sqrt{ x } + \text{const}_{1} $$
$$ 2 \sqrt{ yx } - \frac{ 2 }{ 3 } x\sqrt{ x } = \text{const}_{1} $$
$$ \underbrace{\sqrt{ x } \left( 2 \sqrt{ y } - \frac{ 2 }{ 3 } x \right) = \text{const}_{1} }_{\text{general solution} } ~~ \leftarrow~~ \text{Is this also correct?} $$
|
For example, if you are interested in solutions on intervals $I\subseteq (0, \infty)$ write
$$
\frac{d}{dx}(xy(x)) =x \sqrt{y(x)} = \sqrt{x}\sqrt{xy(x)}.
$$
Then substitute $z(x):=xy(x)$ and solve $z'(x)=\sqrt{x}\sqrt{z(x)}$ by separation of variables (initial values $>0$ are adequate here). Then $y(x)=z(x)/x$ is a solution of the original problem.
|
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|
Show that $|f(x)| \leq 2 \pi^2/9$ The problem is stated as:
Show that $|f(x)| \leq 2 \pi^2/9$ for $ |x| \leq \pi/3$ where $f(x) = \int_{0}^{x}\frac{t}{cos(t)}dt$
Solution:
So, first of all, I try to form a Maclaurin expression around $x = 0$. Without digging into the calculations too deep, I get the following:
$f(0) = 0 \\ f'(0) = 0 \\ f''(0) = 1$
For $f^{(3)}(\theta x) = \dfrac{2\sin\left(\theta x\right)\left(\theta x\sin\left(x\right)+\cos\left(\theta x\right)\right)}{\cos^3\left(\theta x\right)}+\dfrac{\theta x}{\cos\left(\theta x\right)}$
Where $\theta \in (0,1)$.
Therefore, we have that our Maclaurin expansion can be written in the following form: $$f(x) = \frac{x^2}{2} +f^{(3)}(\theta x) = \dfrac{2\sin\left(\theta x\right)\left(\theta x\sin\left(x\right)+\cos\left(\theta x\right)\right)}{\cos^3\left(\theta x\right)}+\dfrac{\theta x}{\cos\left(\theta x\right)}$$
Taking the absolute value of both sides, and applying the triangle ineqality, we get:
$|f(x)| \leq \frac{x^2}{2!} + \frac{1}{3!}|\dfrac{2\sin\left(\theta x\right)\left(\theta x\sin\left(x\right)+\cos\left(\theta x\right)\right)}{\cos^3\left(\theta x\right)}| + |\dfrac{\theta x}{\cos\left(\theta x\right)}|$
Now, we notice that we have:
$ 1/2 \leq |\cos(\theta x)| \leq 1 \\ 0 \leq |\sin(\theta x)| \leq \sqrt{3}/2 \\ 0 \leq |\theta x |\leq \pi/3$
for $|x| \leq \pi/3$
Applying these to our inequality above, we have:
$|f(x)| \leq \pi^2/18 + \frac{1}{3}\frac{\sqrt{3}/2(\pi/3 *\sqrt{3}/2 + 1)}{1/8}+\frac{\pi/3}{1/2}$
However, when I calculate the RHS, I get a much bigger number than $2\pi^2 / 9$. How can I proceed from this?
Thanks!
|
Here is a simpler way:
For $|x| \leq \frac{\pi}{3}, \cos x \geq \frac{1}{2}$ or $\frac{1}{\cos x} \leq 2$
So, $ \displaystyle |f(x)| = \left|\int_{0}^{x}\frac{t}{cos(t)} ~ dt \right| \leq \int_{0}^{|x|} 2t ~ dt = x^2$.
As $ \displaystyle |x| \leq \frac{\pi}{3}, |f(x)| \leq \frac{\pi^2}{9} \leq \frac{2\pi^2}{9}$
|
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|
Solving a problem in Coordinate Geometry Suppose, $2x \cos \alpha - 3y \sin \alpha = 6$ is equation of a variable straight line. From two point $A(\sqrt{5},0$) and $B(-\sqrt{5},0)$ foot of the altitude on that straight line is $P$ and $Q$ respectively. Show that the product of the length of two line segment $AP$ and $BQ$ is free of $\alpha$.
This question appeared in my exam today. The way I did it is first constructed the equation of two perpendicular line to $2x \cos \alpha - 3y \sin \alpha = 6$ which goes through the points $A(\sqrt{5},0$) and $B(-\sqrt{5},0)$. In this way, for $A$ and $B$, I got the following equations respectively-
$$3x \sin \alpha + 2y \cos \alpha -3 \sqrt{5}\sin \alpha=0 \qquad(1)$$$$3x \sin \alpha + 2y \cos \alpha +3 \sqrt{5}\sin \alpha=0 \qquad(2)$$
Then I found that the line $(1)$ intersects the line $2x \cos \alpha - 3y \sin \alpha = 6$ at point $$P\left ( \frac{9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}, \frac{-18 \sin \alpha+6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )$$
and the line $(2)$ intersects the line $2x \cos \alpha - 3y \sin \alpha = 6$ at point $$Q\left ( \frac{-9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}, \frac{-18 \sin \alpha-6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )$$
Now using distance formula $$AP = \sqrt{ \left ( \frac{9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha}- \sqrt5 \right )^2 + \left ( \frac{-18 \sin \alpha+6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )^2}$$ and $$ BQ= \sqrt{ \left ( \frac{-9\sqrt5 \sin^2 \alpha+12 \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} +\sqrt5 \right )^2 + \left ( \frac{-18 \sin \alpha-6\sqrt{5} \sin \alpha \cos \alpha}{4 \cos^2 \alpha + 9 \sin^2 \alpha} \right )^2}$$
Now the multiplication product $AP \cdot BQ$ indeed gives a constant value of $4$ which is free of the arbitrary variable $\alpha$ as you can see here is the simplified version of product of those two quantity. But this is tedious and I do not think this is the only way to do it and an appropriate way to follow in exam with limited time. So, I am looking for an alternative, time saving proof of it. I was wondering if using parametric form would help me, but I think it would get as difficult equally.
I created a visualisation for you on desmos to help my problem understand better.
|
Well I have another solution which is very helpful for exams based on mcq pattern, like the one we give it in india.
I have learnt and derived a property in conic sections which says that in a ellipse, the product of perpendiculars from both foci of ellipse to tangent is equal to b$^2$ ,b is semi minor axis.
equation of a general ellipse is
$\frac{x^2}{a^2}$ + $\frac{y^2}{b^2}$=1
it's tangent equation is-
$\frac{xx1}{a^2}$ + $\frac{yy1}{b^2}$=1 (x1 and y1 are points on ellipse, which is taken x1=acos$\phi$ and y1=bsin$\phi$)
now tangents is revolving will all possible slopes around ellipse...now to ease our work assume slope =o for a tangent... $\alpha$ in your equation will be $\pi$/2 for zero slope with y intercept +2 or-2
.put that in your equation and find the product of perpendiculars, as line is horizontal ,perpendicular will be of of equal length. you will find it to be 4.
now as I told product of perpendiculars is $b^2$, so b=2 ,now use that general tangent equation and given equation to find value of a=3.you will see that it satisfies the equation and a ellipse. It means we were assuming it right. The parameter in your equation was making the arbitrary line a tangent to ellipse.
Here is illustration for better understanding
image
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|
Solve $|2x−3| < |x+5|$ I've tried solving $$|2x-3|=|x+5|$$ but the same method does not look applicable here.
$$2x-3 = x+5, \text{if}\; x>3/2$$
$$2x-3= -(x+5), \text{if} \; x<3/2$$
This method was not working here.
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There are two nodes $x=3/2$ and $x=-5$, so we have to consider the in-equation $|2x-3|< |x+5|$ in three regions three regions
I: $x\le -5 \implies 3-2x<-(x+5) \implies x>8$ (A contradiction).
II:$-5 <x\le 3/2 \implies 3-2x<x+5 \implies x>-2/3 \implies -2/3<x\le 3/2$
III: $x>3/2 \implies 2x-3 <x+5 \implies x<8 \implies 3/2<x \le 8$
Finally, by combining I and II we get the total solution as $x\in (-2/3,8].$
|
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|
Alternative proof of an intereting identity of Catalan's Numbers and central binomial coefficients Some time ago i got from Polya's Urn Scheme that for the n-th Catalan number $C_n = \frac{1}{n+1}\binom{2n}{n}$ and the central binomial coefficient takes place the identity
$$\sum_{n = 0}^\infty\frac{C_{n+k}}{4^n} = 2\binom{2k}{k}$$
I'm looking for a non-probabilistic proof of that result.
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The Catalan numbers generating function
power series multiplied by $x$ is
$$ \frac{1-\sqrt{1-4x}}2 = \sum_{n=0}^\infty C_n x^{n+1}. \tag{1} $$
Substitute $\,y\,$ for $\,x\,$ and subtract both equations to get
$$ \frac{1-\sqrt{1-4y}}2- \frac{1-\sqrt{1-4x}}2 =
\sum_{n=0}^\infty C_n (y^{n+1}-x^{n+1}). \tag{2} $$
Divide both sides by $\,y-x\,$ and express
$\,\frac{y^{n+1}-x^{n+1}}{y-x}\,$ as a sum to get
$$ \frac{\sqrt{1-4x}-\sqrt{1-4y}}{2(y-x)} =
\sum_{n=0}^\infty C_n \frac{y^{n+1}-x^{n+1}}{y-x} =
\sum_{n=0}^\infty C_n\sum_{m=0}^n y^m x^{n-m}. \tag{3} $$
Simplify the left side and change summation indices on the right
to get
$$ \frac2{\sqrt{1-4y}+\sqrt{1-4x}} = \sum_{k=0}^\infty\sum_{n=0}^\infty
C_{n+k}\,y^n x^k. \tag{4}$$
Substitute $\,y=\frac14\,$ and expand $\,\frac2{\sqrt{1-4x}}\,$
with binomial coefficients to get
$$ \frac2{\sqrt{1-4x}} = \sum_{k=0}^\infty 2\binom{2k}{k}x^k =
\sum_{k=0}^\infty\sum_{n=0}^\infty\frac{C_{n+k}}{4^n} x^k. \tag{5}$$
Finally, equate corresponding coefficients of $\,x^k\,$ to get
$$ 2\binom{2k}{k} = \sum_{n=0}^\infty\frac{C_{n+k}}{4^n}. \tag{6} $$
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|
Sign of $P(Δ)$ with $P(x)=ax²+bx+c$ and $Δ=b²-4ac$ Let $P(x)=ax^2+bx+c$ be a polynomial with $a≥0$ and $b≥\frac{1}{8a}$, and $Δ=b^2-4ac$. Show that $P(Δ)≥0$.
My answer: We have $Δ=b^2-4ac$. If$Δ≤0$, then $P(Δ)≥0$. If $Δ>0$, there are two cases:
If $c≥0$, $P(Δ)=aΔ^2+bΔ+c≥0$. If $c<0,Δ=b^2-4ac$ then $Δ>b^2$ and $Δ>-4ac$, then
$$P(Δ)=aΔ^2+bΔ+c>a×(-4ac)^2+b^3+c=16a^3c^2+c+b^3$$
is of degree two of variable $c$. We have $Δ_1=1-4×16a^3b^3=1-(4ab)^3$, which is negative if $b≥\frac{1}{4a}$ or we have $b≥\frac{1}{8a}$.
I want the correct solution of the exercise.
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Completing the square differently, this is non-negative when $\,a \gt 0\,$, $\,\Delta \ge 0\,$, $\,2b - \dfrac{1}{4a} \ge 0\,$:
$$
\begin{align}
\frac{1}{a} P(\Delta) &= \Delta^2+\frac{b}{a}\Delta \color{red}{-2 \frac{b}{a}\Delta + \frac{b^2}{4a^2} + 2 \frac{b}{a}\Delta - \frac{b^2}{4a^2}}+\frac{c}{a}
\\ &= \Delta^2 - 2 \frac{b}{2a}\Delta + \frac{b^2}{4a^2}+\frac{2b}{a}\Delta-\frac{b^2-4ac}{4a^2}
\\ &= \left(\Delta - \frac{b}{2a}\right)^2 + \left(2 b - \frac{1}{4a}\right)\frac{\Delta}{a}
\end{align}
$$
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|
Proving that $x^2-2xy+6y^2-12x+2y+41\ge 0$ where $x,y \in\Bbb{R}$ We use the result that $AX^2+BX+C \ge 0, \forall X ~ \in \Bbb{R} $ if $A>0$ and $B^2-4AC \le 0$
for proving that $f(x,y)=x^2-2xy+6y^2-12x+2y+41\ge 0$, where $x,y \in\Bbb{ R}.$
Let us re-write the quadratic of $x$ and $y$ as a quadratic of $x$ as
$$f(x,y)=x^2+x(-2y-12)+6y^2+2y+41\ge 0, \forall x \in \Bbb{R}.$$
$$\implies B^2-4AC=(2y+12)^2-4(6y^2+2y+41)=-20(y-1)^2 \le 0,$$
which is true and the equality holds if $y=1$ this further means that $f(x)=(x-7)^2.$
So $f(x,y) \ge 0$, the equality holds if $x=7$ and $y=1$.
The question is: What could be other ways of proving this.
Edit: it will be interesting to note that this quadratic of $x$ and $y$ would represent just a point that is $(7,1)$. It is more clear by the solution of @Aqua given below.
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We can write given Polynomial in Quadratic equation in (y-1) as-
f(x,y)=$x^2−2xy+6y^2−12x+2y+41$
=$x^2−2xy+y^2+36−12x+12y+5y^2-10y+5$
=${(x-y)}^2$ + ${6}^2$ -2.6.(x-y) + 5${(y-1)}^2$
=${(x-y-6)}^2$ + 5${(y-1)}^2$
=${[(x-7)-(y-1)]}^2$ + 5${(y-1)}^2$
=6${(y-1)}^2$ - 2(x-7)(y-1) + ${(x-7)}^2$
which is Quadratic equation in (y-1).
Now $B^2$-4AC=${[-2(x-7)]}^2$ - 4.6.${(x-7)}^2$
=-20.${(x-7)}^2$ <0 or 0 (at x=7)
= -ve definite or 0,
which implies that f(x,y)=$x^2−2xy+6y^2−12x+2y+41$>0 or 0 i.e. f(x,y)is positive definite & the equality holds if x=7 & y=1 [as it is quadratic also in (y-1)].
Hence proved.
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|
System of non-linear equations (ISNMO, 1991) Let $x,y,z$ be real numbers. Solve the following system of equations: $$\begin{cases}
\frac{3(x^2+1)}x=\frac{4(y^2+1)}y=\frac{5(z^2+1)}z; \\
xy+xz+yz=1.
\end{cases}$$
I tried to solve this system using the following method:
Since $$\frac{3(x^2+1)}x=\frac{4(y^2+1)}y; \Rightarrow 3y(x^2+1)=4x(y^2+1).$$ Now, from the second row, we have $$3y(x^2+xy+xz+yz)=4x(y^2+xy+xz+yz); \Rightarrow (x+y)(xy+4xz-3yz)=0.$$
Thus, $$x=-y; z=\frac {xy}{3y-4x}.$$
But if we replace $y$ with $-x$, we obtain a complex solution. The same happens when we replace $z$ with $\frac {xy}{3y-4x}$ in the second row because it takes us back to the first expression $x=-y$.
Finally, in case we try to deal with $$\frac{3(x^2+1)}x=\frac{5(z^2+1)}z, z=\frac {xy}{3y-4x},$$
then it complicates the problem with huge expressions.
Wolfram Alpha shows that the real solutions are $(x;y;z)=(\frac 13;\frac 12;1)\cup(-\frac 13;-\frac 12;-1)$, but I do not know how to get to them. I would appreciate any help.
Thank you in advance.
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HINT.-Equalizing to $t$ each of the expressions in first line we have
$$3(x^2+1)=tx\\4(y^2+1)=ty\\5(z^2+1)=tz$$ we deduce
$$x=\frac{t\pm\sqrt{t^2-36}}{6}\\y=\frac{t\pm\sqrt{t^2-64}}{8}\\z=\frac{t\pm\sqrt{t^2-100}}{10}$$Now taking into account the second line of the proposed system we have the solution given by Wolfram:
$$\left(\frac{t\pm\sqrt{t^2-36}}{6}\right)\left(\frac{t\pm\sqrt{t^2-64}}{8}\right)+\left(\frac{t\pm\sqrt{t^2-36}}{6}+\frac{t\pm\sqrt{t^2-64}}{8}\right)\left(\frac{t\pm\sqrt{t^2-100}}{10}\right)=1$$ but it is not necessary if we before that $t=10$ is a good value.
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|
How to determine the positive and negative nature of $a\sec\theta-b\tan\theta$?
Given $a,b\in\mathbb R^+$ and $a\gt b$, find the minimum value of $a\sec\theta-b\tan\theta$.
My approach:
Let $$x=a\sec\theta-b\tan\theta\\\implies x+b\tan\theta=a\sec\theta\\\implies x^2+b^2\tan^2\theta+2bx\tan\theta=a^2(1+\tan^2\theta)\\\implies(b^2-a^2)\tan^2\theta+2bx\tan\theta+x^2-a^2=0$$
Since $\tan\theta\in\mathbb R$
$$\implies D\ge0 \\ \implies4b^2x^2-4(b^2-a^2)(x^2-a^2)\ge0 \\ \implies b^2x^2-(b^2x^2-a^2b^2-a^2x^2+a^4)\ge0\\ \implies a^2b^2+a^2x^2-a^4\ge0\\ \implies b^2+x^2-a^2\ge0\\\implies x^2\ge a^2-b^2\\\implies x\le-\sqrt{a^2-b^2} \text{ or }x\ge\sqrt{a^2-b^2}$$
This is where my doubt is. If we consider $x\le-\sqrt{a^2-b^2}$ to be true then the minimum value of $x$ is not defined. If we consider $x\ge\sqrt{a^2-b^2}$ to be true then the minimum value is $\sqrt{a^2-b^2}$.
In the given statement, is it implied anywhere that the given expression is positive?
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Your working shows you that the range of the function is $\mathbb{R}$ excluding the set $\{x:-\sqrt{a^2-b^2}<x<\sqrt{a^2-b^2}\}$
The local minimum is $\sqrt{a^2-b^2}$ and the local maximum is $-\sqrt{a^2-b^2}$.
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How can i evaluate this integral with or without using CAS $\int_0^{\infty}\frac{\mathrm dx}{(x^2+1)\cosh(\pi x)}$ $$\int_0^{\infty}\frac{\mathrm dx}{(x^2+1)\cosh(\pi x)}$$
Firstly i used substitution $\pi x=t; \mathrm dx=\frac{\mathrm dt}{\pi}$
$$\pi \int_0^{\infty}\frac{\mathrm dt}{(t^2+\pi^2)\cosh t}$$
writing $\cosh t =\frac{e^{t}+e^{-t}}{2}=\frac{1+e^{-2t}}{2e^{-t}}$
$$2\pi\int_0^{\infty}\frac{e^{-t}}{(t^2+\pi^2)(1+e^{-2t})}$$
substituting $e^{-t}=v;-e^{-t}\mathrm dt=\mathrm dv$
$$2\pi\int_0^{1}\frac{\mathrm dv}{(\ln^2 v+\pi^2)(1+v^2)}$$
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Recall the following two results:
$$\int_0^\infty e^{-bt}\cos\left(xt \right)\,dt=\frac{b}{b^2+x^2}\tag{1}$$
$$\int_0^\infty \frac{\cos\left(xt \right)}{\cosh\left(ax \right)}\,dx=\frac{\pi} {2a \cosh\left(\frac{\pi t}{2a} \right)} \tag{2}$$
$$
\begin{aligned}
\int_0^\infty \frac{1}{(b^2+x^2)\cosh\left(ax \right)}\,dx&= \frac 1b\int_0^\infty \frac{1}{\cosh\left(ax \right)}\,\int_0^\infty e^{-bt}\cos\left(xt \right)\,dt\,dx\\
&= \frac 1b\int_0^\infty e^{-bt}\int_0^\infty \frac{\cos\left(xt \right)}{\cosh\left(ax \right)}\,dx\,dt\\
&= \frac 1b\int_0^\infty e^{-bt}\left(\frac{\pi} {2a \cosh\left(\frac{\pi t}{2a} \right)} \right)\,dt\\
&= \frac{\pi}{2 a b}\int_0^\infty \frac{ e^{-bt}} {\cosh\left(\frac{\pi t}{2a} \right)} \,dt\\
&= \frac{1}{b}\int_0^\infty \frac{ e^{-\frac{2ab}{\pi}t}} {\cosh\left(t \right)} \,dt \qquad \left(\frac{2ab}{\pi}w=t\right)\\
&= \frac{2}{b}\int_0^\infty \frac{ e^{-ct}} {e^{t}+e^{-t}} \,dt \qquad \left(c=\frac{2ab}{\pi}\right)\\
&= \frac{2}{b}\int_0^\infty \frac{ e^{-ct}e^{-t}} {1+e^{-2t}} \,dt\\
&= \frac{2}{b}\int_0^\infty e^{-ct}e^{-t}\left(\sum_{k=0}^\infty (-1)^{k}e^{-2kt} \right) \,dt\\
&= \frac{2}{b}\int_0^\infty e^{-ct}\left(\sum_{k=0}^\infty (-1)^{k}e^{-(2k+1)t} \right) \,dt\\
&= \frac{2}{b}\int_0^\infty e^{-ct}\left(\sum_{k=1}^\infty (-1)^{k-1}e^{-(2k-1)t} \right) \,dt\\
&= \frac{2}{b}\sum_{k=1}^\infty (-1)^{k-1}\,\int_0^\infty e^{-\left(c+(2k-1)\right)t} \,dt\\
&= \frac{2}{b}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{\frac{2ab}{\pi}+(2k-1)}\\
&=\frac{2 \pi}{b}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2ab+(2k-1)\pi} \qquad \blacksquare
\end{aligned}
$$
$$
\begin{aligned}
\int_0^\infty \frac{1}{(1+x^2)\cosh\left( \pi x \right)}\,dx&=2 \pi\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2 \pi+(2k-1)\pi}\\
&=2 \sum_{k=1}^\infty \frac{(-1)^{k-1}}{2k+1}\\
&=-2 \sum_{k=1}^\infty \frac{(-1)^{k}}{2k+1}\\
&=2-2 \sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1}\\
&=2-2 \frac \pi 4 \\
&=2-\frac \pi 2 \qquad \blacksquare
\end{aligned}
$$
$$
\begin{aligned}
\int_0^\infty \frac{1}{(\pi^2+x^2)\cosh\left( x \right)}\,dx&=\frac{2 \pi}{\pi}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{2 \pi+(2k-1)\pi}\\
&=\frac 2 \pi \sum_{k=1}^\infty \frac{(-1)^{k-1}}{2k+1}\\
&=-\frac 2 \pi \sum_{k=1}^\infty \frac{(-1)^{k}}{2k+1}\\
&=\frac 2 \pi-\frac 2 \pi \sum_{k=0}^\infty \frac{(-1)^{k}}{2k+1}\\
&=\frac 2 \pi-\frac 2 \pi \cdot \frac \pi 4 \\
&=\frac 2 \pi-\frac 1 2 \qquad \blacksquare
\end{aligned}
$$
$$
\begin{aligned}
&\sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}\\
&=\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{k+\frac{1}{2}}\\
&=\frac{1}{2}\sum_{k=0}^{\infty}(-1)^k\int_0^1t^{k+1/2-1}dt\\
&=\frac{1}{2}\int_0^1t^{-1/2}\left(\sum_{k=0}^{\infty}(-1)^kt^k\right)dt\\
&=\frac{1}{2}\int_0^1\frac{t^{-1/2}}{1+t}dt\\
&=\frac{1}{4}\left(\psi\left( \frac{3}{4}\right)-\psi\left( \frac{1}{4}\right) \right)\\
&=\frac{1}{4}\left(\pi \cot\left( \frac{\pi}{4}\right) \right)\\
&=\frac{\pi}{4}\\
\end{aligned}
$$
Where We used the results
$$\int_0^1\frac{t^{x-1}}{1+t}dt=\frac{1}{2}\left(\psi\left( \frac{x+1}{2}\right)-\psi\left( \frac{x}{2}\right) \right)$$
and
$$\psi(1-x)-\psi(x)=\pi \cot(\pi x)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4321734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Showing there exists a unique $\theta$ for 2D rotation matrix I'm trying to solve some exercises in the book I'm reading. Take a look at this question:
Suppose A is a $2\times2$ rotation matrix. In other words $A^T A = I$
and $\det A = 1$. Show that there exists a unique $\theta$ such that
$A$ is of the form $$ A = \begin{bmatrix} \cos\theta & -\sin\theta \\
\sin\theta &\cos\theta\end{bmatrix} $$
My attempt is first to compute the determinant
$$
\det(A) = \cos^2 \theta + \sin^2\theta = 1
$$
From the determinant, it clearly holds if the matrix $A$ has the form expressed in the question but I couldn't understand what does exactly the author mean by "a unique $\theta$". Any suggestions
|
The question says, or rather, should say:
Suppose we are given a matrix $A$ with the properties: $A^T A = I$ and $\det A = 1$.
Show that there exists a unique $\theta\in [0,2\pi)$ such that $A$ is
of the form $$ A = \begin{bmatrix} \cos\theta & -\sin\theta \\
\sin\theta &\cos\theta\end{bmatrix}. $$
$$$$
Let $$A = \begin{bmatrix} a & b \\
c &d\end{bmatrix}.$$
$A^T A = I\ \implies$
$a^2 + c^2 = b^2 + d^2 = 1\quad (1),\ $ and
$ab+cd = 0\quad (2).$
$\det A = 1\implies$
$ad-bc = 1\quad (3).$
If $\ b=0,\ (1)\implies d=\pm 1 \overset{(2)}{\implies} c=0\overset{(3)}{\implies} a=\pm 1.$
If $\ b\neq 0,\ $ then $\ (2)\implies a= \frac{-cd}{b} \overset{(1)}{\implies} \frac{c^2d^2}{b^2} + c^2 = 1 \implies c^2 \left( \frac{d^2+b^2}{b^2}\right) = 1 \implies c^2 = b^2$
$\implies c=\pm b \implies a = \mp d \overset{(3)}{\implies} c=-b\ $ and $\ a=d.$
In either case,
$$A = \begin{bmatrix} a & b \\
-b &a\end{bmatrix},$$
where $\ a^2+b^2=1\ $ is the only restriction on $\ a\ $ and $\ b.$
They want us to write this as
$$ A = \begin{bmatrix} \cos\theta & -\sin\theta \\
\sin\theta &\cos\theta\end{bmatrix} $$
and show that for each pair $\ (a,b)\ $ with $\ a^2+b^2=1,\ $ this can only be done with one value of $\ \theta\in [0,2\pi).$
To this end:
If $\ -b\ (=\sin\theta) >0,\ $ then $\ 0<\theta<\pi,\ $ and since $\ \cos\ $ is injective on this interval, there is only one $\ \theta\ $ such that $\ \cos\theta = a\ $ also.
Else if $\ b\ (=\sin\theta) = 0\ $ then $\ \theta\ $ can only equal $\ 0\ $ or $\ \pi.\ a=\cos\theta\ $ has a different value in both cases.
Else $\ -b\ (=\sin\theta) <0,\ $ in which case $\ \pi<\theta<2\pi,\ $ and since $\ \cos\ $ is injective on this interval, there is only one $\ \theta\ $ such that $\ \cos\theta = a\ $ also.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4322783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Calculating the volume of the solid defined by $x^2+y^2+z^2 \leq 2a^2$ and $z \leq \frac{x^2+y^2}{a}$, with $a>0$
I have to calculate the volume of solid geometry for $a>0$
$$T=\{(x,y,z)\in \mathbb{R}; x^2+y^2+z^2 \leq 2a^2; z \leq \frac{x^2+y^2}{a}\}$$
I know that first formula is inside of sphere with radium $\sqrt2a$. So I have to use triple integral to calculate this. I use cylindric coordinates $$x = rcos\phi, y=rsin\phi, z = z$$
I tried to calculate like this $$\int_{0}^{2\pi}d\phi \int_{0}^{a} dr \int_{\sqrt{2a^2-r^2}}^{\frac{r^2}{a}}r\,dz $$
but I do not know if is right.
|
Using cylindrical coordiantes, you have two surfaces:
Surface 1: $r^2 + z^2 = 2 a^2$
Surface 2: $ z = r^2 / a $
We want to find where these two surfaces intersect. Plugging the second equation into the first,
$ r^2 + r^4 / a^2 = 2 a^2 $
whose solution is by inspection, $r = a$
So, now if $ r \le a $ then $ r^2 / a \le \sqrt{ 2 a^2 - r^2 } $ and vice versa.
The volume is computed by taking all the range of $r \in [0, \sqrt{2} a ] $ and segmenting the integral to take the two cases
$V = \displaystyle \int_{\phi = 0}^{2 \pi} \left( \int_{r = 0 }^{a} \int_{z= -\sqrt{2 a^2 - r^2}}^{r^2 / a} dz r dr + \int_{r = a }^{\sqrt{2} a} \int_{z= -\sqrt{2 a^2 - r^2}}^{\sqrt{2 - r^2}} dz r dr \right) d\phi $
Integrating with respect to $\phi$ and $z$, the above simplifies to
$V = \displaystyle 2 \pi \left( \int_{0}^{a} \left( r^2/a + \sqrt{2a^2 - r^2} \right) r dr + \int_{a}^{\sqrt{2}a} 2 \sqrt{2 a^2 - r^2} r dr \right) $
And this integrates to
$V = \displaystyle 2 \pi \left( a^3 / 4 + \frac{1}{3} ( 2^\frac{3}{2} - 1) a^3 + \frac{1}{3} ( 2 a^3 ) \right) $
Which reduces to
$V = \displaystyle 2 \pi a^3 \left( \dfrac{7}{12} + \dfrac{ 2^\frac{3}{2} }{3} \right)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4324316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
What is $ \lim_{x \rightarrow 0}\left(\frac{1}{\ln\cos (x)}+\frac{2}{\sin ^{2}(x)}\right) $? The answer of the following limit:
$$
\lim_{x \rightarrow 0}\left(\frac{1}{\ln \cos (x)}+\frac{2}{\sin ^{2}(x)}\right)
$$
is 1 by Wolfram Alpha.
But I tried to find it and I got $2/3$ :
My approach :
$1)$
$
\ln(\cos x)=\ln\left(1-\frac{x^{2}}{2}+o\left(x^{3}\right)\right)=-\frac{x^{2}}{2}+o\left(x^{3}\right)
$
$2)$
$
\sin ^{2}(x)=\left(x-\frac{x^{3}}{3!}+o\left(x^{3}\right)\right)^{2}=x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)
$
$3)$
$\begin{aligned} \frac{1}{-\frac{x^{2}}{2}+o\left(x^{3}\right)}+\frac{2}{x^{2}-\frac{x^{4}}{3}+o\left(x^{4}\right)}=\frac{-x^{2}+x^{2}-\frac{x^{4}}{3}+o\left(x^{3}\right)}{-\frac{x^{4}}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)} \end{aligned}$
$4)$
$\lim _{x \rightarrow 0} \frac{-\frac{1}{3}+o\left(x^{3}\right)}{-\frac{1}{2}+o\left(x^{5}\right)}=\frac{-\frac{1}{3}}{-\frac{1}{2}}=\frac{2}{3}$
So where is the mistake in my approach?
Note: $o$ denotes the little-o notation
Edit : I've understood where's my mistake is, but another question popped up reading the answers which is : does $o(1/x)$ tends to zero as x tends to zero?
|
It is because when you expand $\cos(x)$ and $\ln(\cos(x))$, you need to consider more fourth-order term. Specifically,
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+o(x^5).$$
Then
$$\ln\left(\cos(x)\right)=-\frac{x^2}{2}+\frac{x^4}{24}-\frac{\left(-\frac{x^2}{2}+\frac{x^4}{24}\right)^2}{2}+o(x^5)=-\frac{x^2}{2}-\frac{x^4}{12}+o(x^5).$$
With the same expression for $\sin(x)$ as you have written, we obtain
\begin{align*}
\frac{1}{\ln\left(\cos(x)\right)}+\frac{2}{\sin^2(x)}&=\frac{1}{-\frac{x^2}{2}-\frac{x^4}{12}+o(x^5)}+\frac{2}{x^2-\frac{x^4}{3}+o(x^4)}\\
&=\frac{x^2-\frac{x^4}{3}-x^2-\frac{x^4}{6}}{-\frac{x^4}{2}+o(x^4)}\\
&=\frac{x^4}{x^4+o(x^4)}.
\end{align*}
|
{
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"url": "https://math.stackexchange.com/questions/4324786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Minimizing $(x+y)(z+t)(\frac{a}{x} + \frac{c}{z})(\frac{b}{y} + \frac{d}{t}).$ Let $a,b,c,d$ be positive constants and $x,y,z,t$ be positive variables. Find the minimum value of
\begin{equation}
(x+y)(z+t)\left(\frac{a}{x} + \frac{c}{z}\right)\left(\frac{b}{y} + \frac{d}{t}\right).
\end{equation}
I came up with this nice problem recently but haven't found a solution yet, so I hope somebody here could help.
Using AM-GM one can easily obtain a lower bound of $16\sqrt{abcd}$, which is not necessarily the minimum value unless $ad = bc$.
|
Using algebra
$$F=(x+y)(z+t)\left(\frac{a}{x} + \frac{c}{z}\right)\left(\frac{b}{y} + \frac{d}{t}\right)$$ Compute the partial derivatives
$$\frac{\partial F}{\partial x}=\frac{(t+z) (b t+d y) }{t x^2 y z}\color{red}{\left(c x^2-a y z\right)}=0$$
$$\frac{\partial F}{\partial y}=\frac{(t+z) (a z+c x)}{t x y^2 z} \color{red}{\left(d y^2-b t x\right)}=0$$
$$\frac{\partial F}{\partial z}=\frac{(x+y) (b t+d y) }{t x y z^2}\color{red}{\left(a z^2-c t x\right)}=0$$
$$\frac{\partial F}{\partial t}=\frac{(x+y) (a z+c x)}{t^2 x y z} \color{red}{\left(b t^2-d y z\right)}=0$$
Since all variables and parameters are positive, you need to solve for $(x,y,z,t)$ the "red" equations. Using the first, second and third, we have
$$y=\sqrt[4]{\frac{b c}{a d}}x \qquad z=\sqrt[4]{\frac{c^3 d}{a^3 b}}x\qquad t=\sqrt[2]{\frac{c d}{a b}}x$$ Plugged in the fourth equation, we just find the interesting $0=0$ result.
All of the above plugged in $F$ gives as a result
$$\Big[\sqrt[4]{ad}+\sqrt[4]{bc}\Big]^4$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4327621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Proving continuity of the function $ f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \frac{x^{3}}{1+x^{2}} $ Using only the epsilon delta criterion, show that the function $ f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \frac{x^{3}}{1+x^{2}} $ is continuous over its entire domain of definition.
Definition of the epsilon delta criterion:
A function $f$ from $\Bbb R$ to $\Bbb R$ is continuous at a point $x_0 \in \mathbb{R} $ if given $ε > 0$ there exists $δ > 0$ such that if $|x-x_0| < \delta $ then $|f(x)-f(x_0)|< \epsilon$.
I have already started looking for delta, but get stuck here, and don't know how to further transform or estimate to find the appropriate delta:
$ \begin{aligned}\left|f(x)-f\left(x_{0}\right)\right| &=\left|\frac{x^{3}}{1+x^{2}}-\frac{x_{0}^{3}}{1+x_{0}^{2}}\right|=\\ &=\left|\frac{x^{3}\left(1+x_{0}^{2}\right)}{\left(1+x^{2}\right)\left(1+x_{0}^{2}\right)}-\frac{x_{0}^{3}\left(1+x^{2}\right)}{\left(1+x_{0}^{2}\right)\left(1+x^{2}\right)}\right|=\\ &=\left|\frac{x^{3}\left(1+x_{0}^{2}\right)-x_{0}^{3}\left(1+x^{2}\right)}{\left(1+x^{2}\right)\left(1+x_{0}^{2}\right)}\right|=?\end{aligned} $
|
We have
\begin{align}
|x^3(1+y^2)- y^3(1+x^2)| & = |(x-y)(x^2+x y + y^2 + x^2 y^2)|\\
&\le |x-y|(x^2 + \frac{1}{2}(x^2+y^2) + y^2 + x^2 y^2)\\
&\le \frac{3}{2}|x-y|(x^2 + y^2 + x^2 y^2)\\
&\le \frac{3}{2}|x-y|(1 + x^2)(1+y^2)\\
\end{align}
Hence
\begin{equation}
\frac{|x^3(1+y^2)- y^3(1+x^2)|}{(1 + x^2)(1+y^2)}
\le \frac{\frac{3}{2}|x-y|(1 + x^2)(1+y^2)}{(1 + x^2)(1+y^2)}
\le\frac{3}{2}|x-y|
\end{equation}
Hence
\begin{equation}
|f(x)-f(x_0)|\le \frac{3}{2} |x-x_0|
\end{equation}
|
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"url": "https://math.stackexchange.com/questions/4328172",
"timestamp": "2023-03-29T00:00:00",
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|
When is the projection of an ellipsoid a circle? Consider an ellipsoid in the three dimensional Euclidean space, say
$$\frac{x^2}{a^2}+\frac{y^2}{b^2} + \frac{z^2}{c^2} =1 $$
where $a$, $b$, $c$ are positive reals.
I'm counting the number of planes through the origin so that the image is a perfect circle. There may be divergent cases if we consider the case that some of $a$, $b$, $c$ are coincide. But at first, let us focus on the case that $a$, $b$, $c$ are all different, say $a>b>c$.
I guess the answer would be $4$. I have made many efforts but failed.
What I have observed is the that at least two such planes exists and the radius of the circle is $b$. Just consider rotating plane possesses $y$ axis and apply intermediate value theorem.
Causion! We are concerning projection, not intersection.
PS. Now I guess there are infinitely many...
PS2. According to one suggested answer, there are just two such planes for the non-degenerate case. I'm checking if it is correct.
PS3. Another opinion appeared that the selected answer may have fault. And it seems making sense. I think somewhat stronger analysis is required.
PS4. The above PS3 is about another answer which now have disappeared.
|
For convenience, we use $$ax^2+by^2+cz^2=1.$$
Consider the parametric lines of constant direction $(u,v,w)$ through some point $(x,y,z)$ on the ellipsoid.
$$a(x+tu)^2+b(y+tv)^2+c(z+tw)^2=1$$
or, after simplification,
$$(au^2+bv^2+cw^2)t^2+2(aux+bvy+cwz)t=0.$$
We will obtain the silhouette line on the ellipsoid by expressing that the rays are tangent, i.e. that the $t$ equation has a double root at $t=0$, which simply gives us the plane
$$aux+bvy+cwz=0.$$
That plane intersects the ellipsoid along an ellipse and one can always find an oblique projection that makes it a circle !
If you mean an orthogonal projection, then we must find the plane(s) that intersect along a circle. We can do this by rotating the ellipsoid so that the plane goes to $Z=0$.
We use the rotation relations
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=
\begin{pmatrix}0&\frac{B^2+C^2}{\sqrt{(B^2+C^2)^2+(AB)^2+(AC)^2}}&\frac{A}{\sqrt{A^2+B^2+C^2}}\\\frac{C}{\sqrt{B^2+C^2}}&\frac{-AB}{\sqrt{(B^2+C^2)^2+(AB)^2+(AC)^2}}&\frac{B}{\sqrt{A^2+B^2+C^2}}\\\frac{-B}{\sqrt{B^2+C^2}}&\frac{-AC}{\sqrt{(B^2+C^2)^2+(AB)^2+(AC)^2}}&\frac{C}{\sqrt{A^2+B^2+C^2}}\end{pmatrix}
\begin{pmatrix}X\\Y\\Z\end{pmatrix}.$$
When plugged in the ellipsoid equation, this gives the equation of a conic in the $XY$ plane. This conic is a circle when the quadratic coefficients are equal and the is no mixed term. This is more conveniently done by rewriting the matrix in spherical coordinates.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $(y+1)dx+(x+1)dy=0$ Solve $(y+1)dx+(x+1)dy=0$
$$\frac{dy}{y+1}=-\frac{dx}{x+1}$$
then we get $\ln|y+1|=-\ln|x+1|+c$
$$\ln(|(y+1)(x+1)|)=c$$
$$|(y+1)(x+1)|=e^c=c_1$$
but answer is $y+1=\frac{c}{x+1}$ can you help to find where is my mistake?
|
Simpler:
$(y+1)dx + (x+1)dy = 0$
$ydx + xdy = -(dx+dy)$
$d(xy) = d(-(x+y))$
$xy = -(x+y) + c$
$xy + x + y = c$, which is equivalent to your given answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Applying " divide by highest denominator power" to $ f(x)= \frac {4x+1} {\sqrt{x^2+9}}$ ( Context : limits at infinity and asymptotes). Source : Problem $4.2.$ , $n° 4 $
http://www.gycham.vd.ch/favre/2M/asymptotes/Asymptotes.pdf
The question I'm trying to solve is " determine, in case it exists, the equation of the horizontal asymptote of the following function"
$ \LARGE f(x)= \frac {4x+1} {\sqrt{x^2+9}}$
I do not understand the way Symbolab applies the rule " divide by the highest power of the denominator" to this question.
According to Symbolab, the result of applying this rule yields :
$ \LARGE f(x)= \frac {4+ \frac {1}{x}} {\sqrt {1+\frac {9} {x^2}}}$
This raises some questions :
(1) what is supposed to be the power of the denominator, in case the denominator is a polynomial placed under a radical sign? is there an official rule regarding this point?
(2) how does the transformed expression result from dividing by the same quantity both the numerator and the denominator? the numerator has been dvided by $x$, is it also the case of the denominator?
|
what is supposed to be the power of the denominator, in case the denominator is a polynomial placed under a radical sign?
Informally, we can think of this by reasoning "if $x$ is very large, then $9$ is insignificant compared to $x^2$, so $x^2+9$ acts like $x^2$ and $\sqrt{x^2+9}$ acts like $\sqrt{x^2}=|x|$. This doesn't prove anything in particular, especially since I haven't clarified exactly what "acts like" means, but it justifies the idea of dividing numerator and denominator by $|x|$.
how does the transformed expression result from dividing by the same quantity both the numerator and the denominator? the numerator has been dvided by $x$, is it also the case of the denominator?
Yes, dividing any fraction's numerator and denominator by any non-zero number (or formula resulting in a number) does not change the value of the fraction:
$$ \frac{A / R}{B / R} = \frac{A}{B} \cdot \frac{1/R}{1/R} = \frac{A}{B} \cdot 1 = \frac{A}{B} $$
But there is something strange in the result
$$ f(x) = \frac{4 + \frac{1}{x}}{\sqrt{1+\frac{9}{x^2}}} $$
-- it's correct when $x>0$ but incorrect when $x<0$. Since $\sqrt{x^2}=|x|$, we need to divide numerator and denominator by $|x|$, not by $x$:
$$ \begin{align*}
f(x) &= \frac{\frac{4x+1}{|x|}}{\frac{1}{|x|}\sqrt{x^2+9}} \\
f(x) &= \frac{4 \frac{x}{|x|} + \frac{1}{|x|}}{\sqrt{\frac{1}{x^2}}\sqrt{x^2+9}} \\
f(x) &= \frac{4 \frac{x}{|x|} + \frac{1}{|x|}}{\sqrt{1+\frac{9}{x^2}}}
\end{align*} $$
When $x>0$, $|x|=x$ and this gives the formula you quoted. When $x<0$, $|x|=-x$ and it instead gives
$$ f(x) = \frac{-4 - \frac{1}{x}}{\sqrt{1+\frac{9}{x^2}}} $$
So the graph approaches two different horizontal asymptotes, at $y=+4$ and $y=-4$:
$$ \lim_{x \to +\infty} f(x) = 4 $$
$$ \lim_{x \to -\infty} f(x) = -4 $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Ways to find $\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\cdots$
$$\frac{1}{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10}+\cdots$$
is equal to?
My approach:
We can see that the $n^{th}$ term is \begin{align}a_n&=\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)\cdot(2n+2)}\\&=\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)\cdot(2n+2)}\color{red}{[(2n+2)-(2n+1)}]\\&=\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)}-\frac{1\cdot3\cdot5\cdot\space\dots\space\cdot(2n-3)\cdot(2n-1)\cdot(2n+1)}{2\cdot4\cdot6\cdot\space\dots\space\cdot(2n)\cdot(2n+2)}\\
\end{align}
From here I just have a telescopic series to solve, which gave me $$\sum_{n=1}^{\infty}a_n=0.5$$
Another approach : note : $$\frac{(2n)!}{2^nn!}=(2n-1)!!$$
Which gives $$a_n=\frac{1}{2}\left(\frac{(2n)!\left(\frac{1}{4}\right)^n}{n!(n+1)!}\right)$$
So basically I need to compute
$$\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{(2n)!\left(\frac{1}{4}\right)^n}{n!(n+1)!}\right) \tag{*}$$
I'm not able to determine the binomial expression of $(*)$ (if it exists) or else you can just provide me the value of the sum
Any hints will be appreciated, and you can provide different approaches to the problem too
|
We can write the series as
\begin{align*}
\color{blue}{\frac{1}{2}\sum_{n=1}^{\infty}\left(\frac{(2n)!\left(\frac{1}{4}\right)^n}{n!(n+1)!}\right)}
&=\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n+1}\binom{2n}{n}\left(\frac{1}{4}\right)^n\tag{1}\\
&=\frac{1}{2}\sum_{n=1}^{\infty}C_n\left(\frac{1}{4}\right)^n\tag{2}\\
&=\frac{1}{2}\left(\left.\frac{1-\sqrt{1-4x}}{2x}\right|_{x=\frac{1}{4}}\right)-\frac{1}{2}\tag{3}\\
&=\frac{1}{2}\cdot 2-\frac{1}{2}\tag{4}\\
&\,\,\color{blue}{=\frac{1}{2}}
\end{align*}
Comment:
*
*In (1) we write the coefficient using binomial coefficients.
*In (2) we note that $C_n=\frac{1}{n+1}\binom{2n}{n}$ are the ubiquituous Catalan numbers.
*In (3) we use the generating function of the Catalan numbers evaluated at $x=\frac{1}{4}$. Since the series expansion of the generating function starts with $n=0$ we compensate it by subtracting $\frac{1}{2}$.
*in (4) we evaluate the series at $x=\frac{1}{4}$ and simplify in the last step.
|
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|
Why am I getting a different value for $\sin\left(2\tan^{-1}\frac{4}{3}\right)$ than my calculator? The expression:
$$\sin\left(2\tan^{-1}\left(\frac{4}{3}\right)\right)$$
Way 1:
If I punch the above expression in my calculator, I get $\frac{24}{25}$.
Way 2:
$$\sin\left(2\tan^{-1}\left(\frac{4}{3}\right)\right)$$
$$\sin\left(\tan^{-1}\left(\frac{2\times\frac{4}{3}}{1-(\frac{4}{3})^{2}}\right)\right)$$
$$[\text{Using the formula $2\arctan(x)=\arctan\left(\frac{2x}{1-x^2}\right)$}]$$
$$\sin\left(\tan^{-1}\left(\frac{-24}{7}\right)\right)$$
$$-\sin\left(\tan^{-1}\left(\frac{24}{7}\right)\right)$$
$$-\sin\left(\sin^{-1}\left(\frac{24}{25}\right)\right)$$
$$-\frac{24}{25}$$
Why am I getting a different answer than that of my calculator?
|
Notice that your formula to simplify the arctan gives a positive answer on the left hand side
$$2\arctan \frac{4}{3} > 0$$
but a negative on the right
$$\arctan \frac{\frac{8}{3}}{1-\left(\frac{4}{3}\right)^2} < 0$$
The formula itself is the mistake.
|
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|
Find the point in a tetrahedron that minimizes the sum of distances between vertices of a tetrahedron
Tetrahedron $ABCD$ has $AD=BC, AC=BD,$ and $AB=CD$. Find the point $X$ that minimizes the sum of the distances to the four vertices (i.e. it minimizes $f(X) = AX+BX+CX+DX$).
I found the following proof online, but I can't understand how to justify some of its claims:
Claim: $X$ is the centroid, or $X = \frac{A+B+C+D}4$, where $X,A,B,C,D$ are considered as vectors from some arbitrary point $O$.
Proof: Let $M$ and $N$ be the midpoints of $AB$ and $CD$. Observe that $\Delta ABD\cong \Delta BAC$ and $\Delta CDA\cong \Delta DCB$. Thus $MC=MD$ and $NA=NB$.
Now treating all variables as vectors (e.g. $MN$ represents $\vec{MN}$ and $A^2 = A\cdot A$ for a vector $A$), we have $MN = \frac{C+D}2 - \frac{A+B}2$. Also $MC^2 = MD^2 \Rightarrow (C - \frac{A+B}2)^2 = (D-\frac{A+B}2)^2$ and $NA^2 = NB^2\Rightarrow (A - \frac{C+D}2)^2 = (B- \frac{C+D}2)^2$). Thus, it suffices to show that $MN\cdot AB = 0$. We have $C^2 - C\cdot (A+B) + \frac{(A+B)^2}4 = D^2 - D\cdot (A+B) + \frac{(A+B)^2}4\Rightarrow C(C - A-B) = D\cdot (D-A-B)$ and thus $(D-C)\cdot (A+B) = D^2 - C^2\tag{1}$
Similarly, we have $(B-A)\cdot (C+D) = B^2 - A^2$. Thus, $MN\cdot AB = \frac{1}2 ((C+D)\cdot (B-A) - (B^2 - A^2)) = 0$, as required. Similarly, using $(1)$ we get $MN\cdot CD=0$.
But why is it the case that if $X$ is not on the line $MN$, then its orthogonal projection onto line $MN$ results in a smaller value of $f$ (claim (*))?
Assuming claim (*) holds, by similar reasoning, $X$ must lie on the line joining the midpoints of $AC$ and $BD$.
Now let $J$ and $K$ be the midpoints of $AC$ and $BD$ respectively.
Assuming the lines intersect, let $X$ be their intersection point, so that $MX = c_1MN$ for some scalar $c_1$ and $JX = c_2 JK$ for some scalar $c_2$. Then using the definitions of $M,N,J,K$ we have $X - \frac{A+B}2 = c_1(\frac{C+D}2 - \frac{A+B}2)$ and $X - \frac{B+C}2 = c_2(\frac{A+D}2 - \frac{B+C}2)$. Then $X= (1-c_1)\frac{A+B}2 + c_1\frac{C+D}2 = (1-c_2)\frac{B+C}2 + c_2\frac{A+D}2.$
Why is it that since $A,B,C,D$ can have an arbitrary origin (but are fixed as points), we must "equate coefficients"?
Equating coefficients gives $1-c_1 = c_2$ for $A$ and $1-c_1 = 1-c_2$ for $B$, so $c_1 = c_2 = \frac{1}2$, from which we obtain $X = \frac{A+B+C+D}4$, as desired.
For this question, to clarify, it suffices to justify/prove the questions highlighted in gray using the ">" symbol.
|
Hint: the desired point $X$ is required to have the following property - from the point $X$ each of the faces $ABC, ABD, ACD, BCD$ of the tetrahedron should be viewed by the same solid angle (which is equal to $\pi$). This is equivalent to the following property of the required $X$: if one draws unit vectors from $X$ toward each of the vertices $A,B,C,D$, then the ends of those unit vectors form a regular tetrahedron.
The proof is analogues to the construction of Fermat-Toriccelli point for triangles.
Solid angle:
When there are 3 points $A,B,C$ on a plane and a point $X$ putside that plane, a solid angle is formed between $X$ and the other points. If we draw a sphere with $X$ as its center, and each of the lines $XA,XB,XC$ intersect that sphere at points $A',B',C'$ respectively, the solid angle is defined to be the fraction of the area of the spherical triangle $A'B'C'$ relative to the area of the whole sphere, multiplied by $4\pi$.
Fermat point of a triangle
A famous result of Fermat states that for any triangle $ABC$, the point for which $XA+XB+XC$ is minimal is the point from which one views each the sides with equal angle 120 degrees. There is a nice physical construction that proves it: suppose we hang three weights of equal mass by wires, and those three wires connect in a point inside the triangle. The system stabilizes itself in a state of minimal total potential energy of the weights; and this is also the state where the sum of total wire lengths from $X$ to $A,B,C$ is minimal. Since the tension force is the same
at each wire, the only possible configuration for this to occur is when all the view angles are 120.
|
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$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx.$ Definite Integral: $$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx,$$
where $0 < a <1$.
I tried with integration by parts taking $\log\left(\frac{1+x}{1-x} \right) $ as the first function and $\frac{1}{1-ax} $ as the second function, but it did not work.
Need some help to compute it.
|
Note that
$$\frac{d}{dx}\text{Li}_2\left(\frac{1-ax}{1+a}\right)=\frac{a\ln\left(\frac{a}{1+a}\right)+a\ln(1+x)}{1-ax},$$
$$\frac{d}{dx}\text{Li}_2\left(\frac{1-ax}{1-a}\right)=\frac{a\ln\left(\frac{a}{a-1}\right)+a\ln(1-x)}{1-ax}.$$
On subtracting, we have
$$\frac{d}{dx}\left(\text{Li}_2\left(\frac{1-ax}{1+a}\right)-\text{Li}_2\left(\frac{1-ax}{1-a}\right)\right)=\frac{a}{1-ax}\left(\ln\left(\frac{a-1}{1+a}\right)+\ln\left(\frac{1+x}{1-x}\right)\right).$$
Integrating both sides,
$$\text{Li}_2\left(\frac{1-ax}{1+a}\right)-\text{Li}_2\left(\frac{1-ax}{1-a}\right)=-\ln\left(\frac{a-1}{1+a}\right)\ln(1-ax)+a\int\frac{\ln\left(\frac{1+x}{1-x}\right)}{1-ax}dx.$$
Then
$$\int_{-1}^1\frac{\ln\left(\frac{1+x}{1-x}\right)}{1-ax}dx=\left.\frac{\text{Li}_2\left(\frac{1-ax}{1+a}\right)-\text{Li}_2\left(\frac{1-ax}{1-a}\right)+\ln\left(\frac{a-1}{1+a}\right)\ln(1-ax)}{a}\right]_{-1}^1$$
$$=\frac{\text{Li}_2\left(\frac{1-a}{1+a}\right)+\text{Li}_2\left(\frac{1+a}{1-a}\right)+\ln\left(\frac{a-1}{1+a}\right)\ln\left(\frac{1-a}{1+a}\right)-2\zeta(2)}{a}.$$
By setting $z=\frac{a-1}{1+a}$ in the dilogarithm inversion formula:
$$\text{Li}_2(-z)+\text{Li}_2(-1/z)=-\frac12 \ln^2(z)-\zeta(2),$$
we get
$$\text{Li}_2\left(\frac{1-a}{1+a}\right)+\text{Li}_2\left(\frac{1+a}{1-a}\right)=-\frac12\ln^2\left(\frac{a-1}{1+a}\right)-\zeta(2).$$
Thus,
$$\int_{-1}^1\frac{\ln\left(\frac{1+x}{1-x}\right)}{1-ax}dx=\frac{-\frac12\ln^2\left(\frac{a-1}{1+a}\right)+\ln\left(\frac{a-1}{1+a}\right)\ln\left(\frac{1-a}{1+a}\right)-3\zeta(2)}{a}=\frac{\ln^2\left(\frac{1-a}{1+a}\right)}{2a}.$$
The last equality follows from writing $\ln\left(\frac{a-1}{1+a}\right)=i\pi+\ln\left(\frac{1-a}{1+a}\right)$, which follows from $\ln(-z)=i\pi+\ln(z)$.
|
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|
Solve the equation over the integers $2^a b + 2^b a = a^2+b^2$
Clearly all possible solutions are
$(0,0), (0,1), (1,0)$
What is a proper method of solving it?
Thank you.
EDIT:
Following Dietrich Burde advice, I will make an attempt:
For every $a, b \geq 0, 2^a>a$
Without loss of generality and since the equation is totally symmetric, we can assume $a \geq b$.
So we have:
$2^a>a \geq b \Rightarrow 2^a.b \geq b^2$.
Similarly,
$2^b.a \geq b.a$.
but this is as far as I can go.
|
Following an observation by Dietrich Burde in the comments, we will prove the following:
Lemma: For all integers $a, b \geq 2,$ we have
$$2^{a} \cdot b + 2^{b} \cdot a > a^{2} + b^{2}.$$
Proof: We say a pair of integers $(a, b)$ is good if $a, b \geq 2$ and
$$2^{a} \cdot b + 2^{b} \cdot a > a^{2} + b^{2}.$$
Claim: if $(a, b)$ is good, $(a+1, b)$ is good. Note that
$$2^{a+1} \cdot b + 2^{b} \cdot (a+1) = 2^{a} \cdot b + 2^{a} \cdot b + 2^{b} \cdot a + 2^{b} = (2^{a} \cdot b + 2^{b} \cdot a) + 2^{a} \cdot b + 2^{b}.$$ Since $(a, b)$ is good, we have $$2^{a} \cdot b + 2^{b} \cdot a > a^{2} + b^{2}.$$ Furthermore, since $a, b \geq 2,$ we have $$2^{a} \cdot b > ab \geq 2a,$$ and $$2^{b} > 1.$$ Therefore, it follows that
$$2^{a+1} \cdot b + 2^{b} \cdot (a+1) = (2^{a} \cdot b + 2^{b} \cdot a) + 2^{a} \cdot b + 2^{b} > a^{2} + b^{2} + 2a + 1 = (a+1)^{2} + b^{2}.$$ So, $(a+1, b)$ is good, as claimed.
Since the inequality is symmetric with respect to $a$ and $b$, the same argument shows that if $(a, b)$ is good, $(a, b+1)$ is also good. Finally, note that $(2, 2)$ is good: $$2^{2} \cdot 2 + 2^{2} \cdot 2 = 16 > 8 = 2^{2} + 2^{2}.$$ So, it follows that $(a, b)$ is good for all integers $a, b \geq 2,$ which proves the desired lemma. $\blacksquare$
So, if $a, b \geq 2,$ there are no solutions to $2^{a} \cdot b + 2^{b} \cdot a = a^{2} + b^{2}.$ For completeness, we will also rule out the cases where one or both of $a, b$ is negative. If $a$ is negative and $b$ is nonnegative, then we have $$2^{a} \cdot b = \frac{b}{2^{-a}} < b \leq b^{2},$$ and $$2^{b} \cdot a < 0 < a^{2}.$$ Clearly, $$2^{a} \cdot b + 2^{b} \cdot a < a^{2} + b^{2},$$ and there are no solutions to the equality. Similarly, if $a$ is nonnegative and $b$ is negative, by symmetry there are also no solutions. Finally, if $a, b$ are both negative, we have
$$2^{a} \cdot b + 2^{b} \cdot a < 0 < a^{2} + b^{2},$$ and there are no solutions.
Therefore, if there are solutions $(a, b)$ to
$$2^{a} \cdot b + 2^{b} \cdot a = a^{2} + b^{2}$$ in the integers, then neither of $a, b$ is negative, and at least one of $a, b$ is less than 2. Suppose without loss of generality (by symmetry) that $a < 2.$ If $a = 0,$ we have
$$2^{0} \cdot b + 2^{b} \cdot 0 = 0^{2} + b^{2},$$ which simplifies to
$$b = b^{2}.$$ Clearly, $b = 0$ and $b = 1$ are the only solutions here, giving us the pairs $(0, 0)$, $(0, 1).$
Now, suppose that $a =1$. We have $$2^{1} \cdot b + 2^{b} \cdot 1 = 1^{2} + b^{2}.$$
This simplifies to $$2^{b} = (b-1)^{2}.$$ It is not difficult to show that the only solution to this in the integers is $b = 0$ (check this for yourself), which gives us the pair $(1, 0)$.
Therefore, in the case that $a < 2,$ we have the pairs $(0, 0)$, $(0, 1)$, $(1, 0)$. We should expect the case where $b < 2$ to give us these pairs but with $a$ and $b$ flipped. However, note that this just gives us the same three pairs.
So, we conclude that the solution pairs in the integers are $(0, 0)$, $(0, 1)$, and $(1, 0)$, which matches what you found.
|
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Let $a,b\in \mathbb{R}$ such that $ab=2$. Find the max value of $\frac{3}{2\left(a+b\right)^2}$ and $a, b$ where max is attained, without calculus. My thinking:
By Arithmetic and Geometric mean inequality (AGM):
$ab\le \left(\frac{a+b}{2}\right)^2$
We know $ab=2$
$\rightarrow$ $2\le \left(\frac{a+b}{2}\right)^2$
$\rightarrow$ $2\le \frac{a^2+b^2}{4}$
$\rightarrow$ $\frac{1}{2}\le \:\frac{1}{\frac{a^2+b^2}{4}}$
$\rightarrow$ $\frac{1}{2}\le \:\frac{4}{a^2+b^2}$
$\rightarrow$ $\frac{1}{2}\left(\frac{3}{a^2+b^2}\right)\le \:\frac{4}{a^2+b^2}\left(\frac{3}{a^2+b^2}\right)$
$\rightarrow$ $\frac{3}{2\left(a+b\right)^2}\le \frac{12}{\left(a+b\right)^4}$
Therefore the max value is $\frac{12}{\left(a+b\right)^4}$
Maximum value is attained when $a=b$:
So we can write:
$\rightarrow$ $\frac{3}{2\left(a\right)^2}=\frac{12}{\left(a\right)^4}$
$\rightarrow$ $3a^4=24a^2$
$\rightarrow$ $a=0, -2\sqrt{2},+2\sqrt{2}$
Therefore the max value of $a,b$ are $0,-2\sqrt{2},+2\sqrt{2}$
I'm not completely sure about my answer, if anyone could provide some feedback, that would be amazing! Thanks in advance.
EDIT:
$ab\le \frac{\left(a+b\right)^2}{2}\:\rightarrow \:2\le \frac{\left(a+b\right)^2}{4}\:\rightarrow \:\frac{1}{2}\le \frac{\left(a+b\right)^2}{16}\:\rightarrow \:\frac{3}{2\left(a+b\right)^2}\le \:\frac{3}{16}$
|
Substitute the given restriction into the proposed expression.
Then your problem reduces to study a single-variable real-valued function:
\begin{align*}
\frac{3}{2(a + b)^{2}} & = \frac{3}{2\left(a + \dfrac{2}{a}\right)^{2}}\\\\
& = \frac{3a^{2}}{2(a^{2} + 2)^{2}}\\\\
& = \frac{3}{2}\left(\frac{a}{a^{2} + 2}\right)^{2}
\end{align*}
Since the quadratic function is increasing, it suffices to study when the argument attains its max.
More precisely, according to the AM-GM inequality, we can conclude that
\begin{align*}
a^{2} + 2 \geq 2\sqrt{2a^{2}} = 2\sqrt{2}|a| \Longleftrightarrow \left|\frac{a}{a^{2} + 2}\right| \leq \frac{1}{2\sqrt{2}}
\end{align*}
Hence the maximum value attained is given by:
\begin{align*}
\frac{3}{2(a + b)^{2}} \leq \frac{3}{2}\left(\frac{1}{2\sqrt{2}}\right)^{2} = \frac{3}{16}
\end{align*}
In order to determine $a$ which corresponds to the maximum, you can solve the equation:
\begin{align*}
a^{2} - 2\sqrt{2}|a| + 2 = 0
\end{align*}
Once you know $a$, you also know $b$ and you are done.
Hopefully this helps !
|
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Consider the non-homogeneous linear recurrence relations $a_n = 2a_{n−1} + 2^n$ find all solutions. I can show that $a_n^{(h)}$characteristic equation $r-2=0 \to a_n^{(h)}=\alpha2^n$
But I'm stuck on $a_n^{(p)}$ characteristic equation $A2^n = 2A2^{n-1} + 2^n$
Simplifies to $-A = 2$
$A = 2A + 2$
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Let $A(z)=\sum_{n=0}^\infty a_n z^n$ be the ordinary generating function. Then
\begin{align}
A(z)
&= a_0 + \sum_{n=1}^\infty (a_{n-1} + 2^n)z^n \\
&= a_0 + z \sum_{n=1}^\infty a_{n-1} z^{n-1} + \sum_{n=1}^\infty (2z)^n \\
&= a_0 + z A(z) + \frac{2z}{1-2z},
\end{align}
so
\begin{align}
A(z)
&= \frac{a_0}{1-z} + \frac{2z}{(1-z)(1-2z)} \\
&= \frac{a_0}{1-z} - \frac{2}{1-z} + \frac{2}{1-2z} \\
&= \frac{a_0-2}{1-z} + \frac{2}{1-2z} \\
&= (a_0-2) \sum_{n=0}^\infty z^n + 2\sum_{n=0}^\infty (2z)^n \\
&= \sum_{n=0}^\infty (a_0-2 + 2^{n+1}) z^n,
\end{align}
which immediately implies that
$$a_n = a_0-2 + 2^{n+1}.$$
|
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|
How does the theorem 2 of algebra of limits contradict in this Q? Q: $\lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1}$
Solution in my textbook using theorem 2 of algebra of limits I.e $\lim _{x \rightarrow a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}$ .
$\begin{aligned} \lim _{x \rightarrow 1} \frac{x^{15}-1}{x^{10}-1} &=\lim _{x \rightarrow 1}\left[\frac{x^{15}-1}{x-1} \div \frac{x^{10}-1}{x-1}\right] \\ &=\lim _{x \rightarrow 1}\left[\frac{x^{15}-1}{x-1}\right] \div \lim _{x \rightarrow 1}\left[\frac{x^{10}-1}{x-1}\right] \\ &=15(1)^{14} \div 10(1)^{9} \text {} \\ &=15 \div 10=\frac{3}{2} \end{aligned}$
I agree with it using the formula but I also thought of thinking how would it be solved from the long way [Inserting x = 1]
We get :
$\frac{(1)^{15}-1}{1-1} \div\left[\frac{(1)^{10}-1}{1-1}\right]$
$\rightarrow \frac{0}{0} \div \frac{0}{0}=0$
My answer is completely different from the solution of my textbook. How is that possible ?
If anyone wants to argue that we cannot put value of x = 1 just like that. There is another example in my textbook according to which , we can.
In all of them , the Q simply says: Find the limit.
1)$\begin{aligned} \lim _{x \rightarrow 2} \frac{x^{2}-4}{x^{3}-4 x^{2}+4 x} &=\lim _{x \rightarrow 2} \frac{(x+2)(x-2)}{x(x-2)^{2}} \\ &=\lim _{x \rightarrow 2} \frac{(x+2)}{x(x-2)}=\frac{2+2}{2(2-2)}=\frac{4}{0} \end{aligned}$
Here , x is tending to 2. This is a rational function.
EDIT: I will share a few more examples.
*We have $\lim _{x \rightarrow 1} \frac{x^{2}+1}{x+100}=\frac{1^{2}+1}{1+100}=\frac{2}{101}$
*(iii) $\lim _{x \rightarrow-1}\left[1+x+x^{2}+\ldots+x^{10}\right]=1+(-1)+(-1)^{2}+\ldots+(-1)^{10}$
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Let,
$$f(x) = \frac {x^{15}-1}{x^{10}-1}$$
Well, I agree with you! that the value of function $f(x = 1)$ is $ \frac {1^{15}-1}{1^{10}-1}$
But,
The value of limit of function $f(x)$ as $x \to 1$ is $\frac 32$
*
*Limit of function: This means you take the value of the function as $x\to a$ doesn't mean $x = a $ it means it's very very very close to $a$ it's never $a$
*Try this: $f(x = 0.99999) = \frac {(0.99999)^{15} - 1}{(0.99999)^{10} - 1} \approx \frac 32$
|
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|
Laurent series of $f(z) = \frac{z - 4}{(z+1)^2(z - 2)}$ I want to expand function $$f(z) = \frac{z - 4}{(z+1)^2(z - 2)}$$ for $1 < |z| < 2$.
My work so far
By partial fraction decomposition we can get that:
$$\frac{z - 4}{(z + 1)^2(z - 2)} = \frac{-3(z - 4)}{(z + 1)^2} + \frac{\frac 1 9 (z - 4)}{z - 2}$$
Now deriving Laurent expansion for second term is not so hard:
$$\frac{1}{z - 2} = -\frac{1}{2} \frac{1}{1 - \frac z 2} = - \frac 1 2 \sum_{n = 0}^\infty (\frac z 2)^n = - \sum_{n = 0}^\infty \frac{z^2}{2^{n + 1}}$$
So we have that:
$$ \frac{\frac 1 9 (z - 4)}{z - 2} = \sum_{n = 0}^\infty - \frac 1 9 \cdot \frac{z^n}{2^{n + 1}}(z - 4)$$
But I'm quite struggling with expanding $\frac{1}{(z + 1)^2}$. I tried several possibilities but those all led me nowhere. Can I ask you for a hand in expanding form $\frac{1}{(z + 1)^2}$?
EDIT
We know that $\frac{1}{1 + z} = \sum_{n = 0}^\infty \frac{(-1)^n}{z^{n + 1}}$
As I understood the idea with derivative is to:
$$\frac{d(\frac{1}{1 + z})}{dz} = \sum_{n = 0}^\infty \frac{d(\frac{(-1)^n}{z^{n + 1}})}{dz}$$
$$- \frac{1}{(1 + z)^2} = \sum_{n = 0}^\infty - (n + 1)z^{-n-2}(-1)^n$$
$$\frac{1}{(1 + z)^2} = \sum_{n = 0}^\infty (n + 1)z^{-(n + 2)}(-1)^n$$
Does it make any sense to you?
|
Hint:
*
*Compute the Laurent expansion of $\frac1{1+z}$
*Notice that $\frac{d}{dz}\frac{1}{1+z}=-\frac{1}{(1+z)^2}$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Trouble integrating square of a sine I have the following expression which I am trying to evaluate:
$$ \frac{2}{L} \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2 kx$$ where $k = \frac{\pi}{L}$.
According to my calculator, the answer should be $\frac{1}{6}$ but I can't seem to get this result by manual integration.
Here's what I've tried so far:
$$\begin{align} &\frac{2}{L} \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2 kx dx \\
&= \frac{1}{L} \int^{\frac{L}{3}}_{\frac{L}{6}} (1 - \cos 2kx) dx \\
&= \frac{1}{L} \left[ x - \frac{1}{2k} \sin2kx \right]^{\frac{L}{3}}_{\frac{L}{6}} \\
&= \frac{1}{3} + \frac{1}{L} \left[- \frac{L}{2 \pi} \sin \frac{2 \pi x}{L} \right]^{\frac{L}{3}}_{\frac{L}{6}} \\
&= \frac{1}{3} - \left[\frac{1}{2 \pi} \sin \frac{2 \pi}{x} \right]^{3}_{6} \\
&= \frac{1}{3} + \frac{\sqrt{3}}{2 \pi} \neq \frac{1}{6} \end{align}$$
|
There is a trick. Integrate by parts
$$
\int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2(kx)~\mathrm{d}x = -\frac{1}{k}\cos(kx)\sin(kx) \bigg \vert^{\frac{L}{3}}_{\frac{L}{6}} + \int^{\frac{L}{3}}_{\frac{L}{6}} \cos^2(kx)~\mathrm{d}x = \int^{\frac{L}{3}}_{\frac{L}{6}} \cos^2(kx)~\mathrm{d}x
$$
So:
$$
\left(\frac{L}{3}-\frac{L}{6}\right) = \int^{\frac{L}{3}}_{\frac{L}{6}} 1~\mathrm{d}x = \int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2(kx) + \cos^2(kx)~\mathrm{d}x =
$$
$$
\int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2(kx) ~\mathrm{d}x+ \int^{\frac{L}{3}}_{\frac{L}{6}} \cos^2(kx)~\mathrm{d}x =2\int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2(kx) ~\mathrm{d}x
$$
Rearrange to get
$$
\int^{\frac{L}{3}}_{\frac{L}{6}} \sin^2(kx) ~\mathrm{d}x = \frac{1}{2} \left(\frac{L}{3}-\frac{L}{6}\right).
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4359869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Singapore Math 6th grade probability problem Karen has 12 cards, each written with a letter from the word PENNSYLVANIA. She picks a card at random. Without replacing she picks another card at random. Find the probability she picks two vowels.
I was sure the answer is $\frac{5}{12}\times\frac{4}{11}$ :-)
|
With Y as vowel:
There are 5 vowels, picking a vowel first try gives you $\frac{5}{12}$ and then we have 11 letters and 4 vowels left, giving you $\frac{4}{11}$. Therefore, the answer would be $\frac{5}{12}\times\frac{4}{11}$, or $\frac{5}{33}$.
Without Y as vowel:
There are 4 vowels, picking a vowel first try gives you $\frac{4}{12}$, and then we will have 11 letters and 3 vowels left, giving you $\frac{3}{11}$. Therefore, the answer would be $\frac{4}{12}\times\frac{3}{11}$. or $\frac{1}{11}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4361192",
"timestamp": "2023-03-29T00:00:00",
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|
Finding $\int_{1}^{\infty} \frac{1}{1+x^2} \frac{\operatorname{Li}_2\left ( \frac{1-x}{2} \right ) }{\pi^2+\ln^2\left(\frac{x-1}{2}\right)}\text{d}x$ Prove the integral
$$\int_{1}^{\infty} \frac{1}{1+x^2}
\frac{\operatorname{Li}_2\left ( \frac{1-x}{2} \right ) }{
\pi^2+\ln^2\left ( \frac{x-1}{2} \right ) }\text{d}x
=\frac{96C\ln2+7\pi^3}{12(\pi^2+4\ln^2(2))}
-\frac{\pi}{24}(3+4\pi)$$
where $\operatorname{Li}_2$ is dilogarithm and $C$ is Catalan's constant.
(I checked it high precision. So I believe that it's absolutely true.)
How we prove that? I tried to use functional equations of $\operatorname{Li}_2$ but I get nothing useful. Any suggestion will be appreciated.
|
$$I=2\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\pi ^2+\ln ^2x}}\mathrm{d}x$$
Consider
$$f\left( z \right) =\frac{\mathrm{Li}_2\left( z \right)}{1+\left( 2z-1 \right) ^2}\frac{1}{\ln z}$$
Use a key-shaped contour with $0$ and $1$ as the keyholes:
contour
When $x<0$, we have
$$\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\ln x+\pi \mathrm{i}}}\mathrm{d}x-\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\ln x-\pi \mathrm{i}}}\mathrm{d}x\\=-2\pi \mathrm{i}\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\ln ^2x+\pi ^2}}\mathrm{d}x$$
When $x>1$, we have
$$\int_1^{\infty}{\frac{1}{1+\left( 2x-1 \right) ^2}\frac{1}{\ln x}\left( \frac{\pi ^2}{6}-\int_1^x{\frac{\ln \left( t-1 \right)}{t}}\mathrm{d}t+\pi \mathrm{i}\ln x \right)}\mathrm{d}x
\\
-\int_1^{\infty}{\frac{1}{1+\left( 2x-1 \right) ^2}\frac{1}{\ln x}\left( \frac{\pi ^2}{6}-\int_1^x{\frac{\ln \left( t-1 \right)}{t}}\mathrm{d}t-\pi \mathrm{i}\ln x \right)}\mathrm{d}x
\\
=2\pi \mathrm{i}\int_1^{\infty}{\frac{1}{1+\left( 2x-1 \right) ^2}}\mathrm{d}x=\frac{\pi ^2\mathrm{i}}{4}$$
This is because
$$\mathrm{Li}_2\left( x \right) =-\int_0^x{\frac{\ln \left( 1-t \right)}{t}}\mathrm{d}t=\frac{\pi ^2}{6}-\int_1^x{\frac{\ln \left( 1-t \right)}{t}}\mathrm{d}t
\\
=\frac{\pi ^2}{6}-\int_1^x{\frac{\ln \left( t-1 \right)}{t}}\mathrm{d}t\pm \pi \mathrm{i}\ln x\,\,\left( x>1 \right) $$
Then, use the residue theorem, we get
$$-2\pi \mathrm{i}\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\ln ^2x+\pi ^2}}\mathrm{d}x+\frac{\pi ^2\mathrm{i}}{4}=2\pi \mathrm{i}\left[ \mathrm{Res}\left[ f\left( z \right) ,1 \right] +\mathrm{Res}\left[ f\left( z \right) ,\frac{1\pm \mathrm{i}}{2} \right] \right] $$
And
$$\begin{align*}&\mathrm{Res}\left[ f\left( z \right) ,1 \right] =\frac{\pi ^2}{12}
\\
&\mathrm{Res}\left[ f\left( z \right) ,\frac{1-\mathrm{i}}{2} \right] =\frac{\mathrm{i}}{4}\frac{\mathrm{Li}_2\left( \frac{1-\mathrm{i}}{2} \right)}{\ln \left( \frac{1-\mathrm{i}}{2} \right)}=\frac{\mathrm{i}}{4}\frac{\frac{5\pi ^2}{96}-\frac{\ln ^22}{8}+\mathrm{i}\left( \frac{\pi \ln 2}{8}-C \right)}{-\frac{\ln 2}{2}-\frac{\pi \mathrm{i}}{4}}
\\
&\mathrm{Res}\left[ f\left( z \right) ,\frac{1+\mathrm{i}}{2} \right] =-\frac{\mathrm{i}}{4}\frac{\mathrm{Li}_2\left( \frac{1+\mathrm{i}}{2} \right)}{\ln \left( \frac{1+\mathrm{i}}{2} \right)}=-\frac{\mathrm{i}}{4}\frac{\frac{5\pi ^2}{96}-\frac{\ln ^22}{8}-\mathrm{i}\left( \frac{\pi \ln 2}{8}-C \right)}{-\frac{\ln 2}{2}+\frac{\pi \mathrm{i}}{4}}\end{align*}$$
So
$$\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\pi ^2+\ln ^2x}}\mathrm{d}x=\frac{7\pi ^3}{24\left( \pi ^2+4\ln ^22 \right)}+\frac{4C\ln 2}{\pi ^2+4\ln ^22}-\frac{\pi ^2}{12}-\frac{\pi}{16}$$
Finally
$$\boxed{I=\frac{7\pi ^3}{12\left( \pi ^2+4\ln ^22 \right)}+\frac{8C\ln 2}{\pi ^2+4\ln ^22}-\frac{\pi ^2}{6}-\frac{\pi}{8}}$$
|
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|
If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ equals $(a+b)/2$ If $a>0$ and $b>0$, prove that $\lim(\sqrt{(n+a)(n+b)}-n)$ is equal $(a+b)/2$
Multiplying by the conjugate, simplifying and clearing I arrive at the following expression
$$\frac{n}{\sqrt{(n+a)(n+b)}+n} < \frac{\epsilon+(a+b)-ab}{(a+b)}$$
|
HINT
Start with rearranging the proposed expression as follows:
\begin{align*}
\sqrt{(n + a)(n + b)} - n & = \frac{(n + a)(n + b) - n^{2}}{\sqrt{(n + a)(n + b)} + n}\\\\
& = \frac{(a + b)n + ab}{\sqrt{(n + a)(n + b)} + n}\\\\
& = \frac{(a + b) + ab/n}{\sqrt{(1 + a/n)(1 + b/n)} + 1}
\end{align*}
Now it remains to take the limit as $n$ approaches infinity.
Can you take it from here?
|
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|
How many four-digit natural numbers not exceeding the number $4321$ can be formed using the digits $1, 2, 3, 4$ if repetition is allowed?
How many four-digit natural numbers not exceeding the number $4321$ can be formed using the digits $1, 2, 3, 4$ if repetition is allowed?
This is the question and I am solving it like this:
Total no's can be formed - no's not allowed
$4 \cdot 4 \cdot 4 \cdot 4-11=256-11 =245$, but the answer is $229$.
What is wrong in this method?
And my second method is this:
Total no. = no's digit do not repeat + no's in which $2$ digits are same + no's in which $3$ digit are same + no's in which all digit are same - no's not allowed
No repeat$=4!$
2 no's same$= 144$
3 same $=48$
4 same $=4$
Sum $=220-11=209$
What is wrong in both methods?
|
There are actually $27$ numbers larger than $4321$. They are:
Numbers of the form $44\square\square$: There are $1 \cdot 1 \cdot 4 \cdot 4 = 16$ such numbers.
Numbers of the form $433\square$ or $434\square$: There are $1 \cdot 1 \cdot 2 \cdot 4 = 8$ such numbers.
Numbers of the form $432\square$: There are $3$ such numbers, namely $4322, 4323, 4324$.
Hence, you should have $4^4 - (16 + 8 + 3) = 256 - 27 = 229$ in your first method.
In your second method, your forgot to count numbers in which two digits each appear twice, which is why your four cases do not add up to $4^4 = 256$. There are $\binom{4}{2}$ ways to select the two digits which each appear twice and $\binom{4}{2}$ ways to select two positions for the smaller of those digits. Hence, there are
$$\binom{4}{2}\binom{4}{2} = 36$$
numbers which each appear twice.
Thus, you should have a total of $24 + 144 + 36 + 48 + 4 = 256$ numbers, of which $27$ are too large, again giving $256 - 27 = 229$ admissible numbers.
A direct count: We wish to count four-digit numbers formed using the digits $1, 2, 3, 4$ which are at most $4321$ when repetition of digits is permitted.
Numbers of the form $1\square\square\square$, $2\square\square\square$, or $3\square\square\square$: $3 \cdot 4 \cdot 4 \cdot 4 = 192$
Numbers of the form $41\square\square$ or $42\square\square$: $1 \cdot 2 \cdot 4 \cdot 4 = 32$
Numbers of the form $431\square$: $4$
The number $4321$: $1$
Hence, the numbers of admissible numbers is $192 + 32 + 4 + 1 = 229$.
|
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Are there any formula for the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}$, where $m$ and $n$ are natural numbers and $a>0$? As mentioned in my post, I started to investigate the integral $$
I(m,n,a)=\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}},
$$
where $m$ and $n$ are natural number and $a$ is positive.
First of all, let’s start with the ‘simple’ case, $$
I(1, n, 1)=\int_{0}^{\infty} \frac{d x}{x^{n}+1}.
$$
However, $\displaystyle \int_{0}^{\infty} \frac{d x}{x^{n}+1}$ is itself not simple. I was forced to use a ready made formula
$$
\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).
$$
which I can’t prove by elementary method yet. Please tell me if you have any.
Then $$I(m,n,a) =\int_{0}^{\infty} \frac{\sqrt[n]{a} d\left(\frac{x}{\sqrt[n]{a}}\right)}{a\left[\left(\frac{x}{\sqrt[n]{a}}\right)^{n}+1\right]}
=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right) a^{-\frac{n-1}{n}}$$
Now differentiating $I(1, n, a)$ w.r.t. $a$ by $(n-1)$ times yields
$$
\begin{aligned}
\int_{0}^{\infty} \frac{(-1)^{m-1}(m-1) ! d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left(-\frac{n-1}{n}\right)\left(-\frac{2 n-1}{n}\right)\left(-\frac{3 n-1}{n}\right)
\cdots\left(-\frac{m n-n-1}{n}\right) a^{-\frac{m n-1}{n}}
\end{aligned}
$$
Rearranging and simplifying yields
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} a^{\frac{m n-1}{n}} }\prod_{k=1}^{m-1}(k n-1)}
$$
Putting $a=1$ gives our formula
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}}=\frac{\pi \csc \left(\frac{\pi}{n}\right)}{(m-1) ! n^{m} }\prod_{k=1}^{m-1}(k n-1)}
$$
For verification, $$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{3}+1\right)^{10}} =\frac{\pi \csc \left(\frac{\pi}{3}\right)}{9 ! 3^{10}} \cdot 2 \cdot 5 \cdot 8 \cdot 11 \cdot 14 \cdot 17 \cdot 20 \cdot 23 \cdot 26
=\frac{1118260 \pi}{4782969 \sqrt{3}},
\end{aligned}
$$
which is checked by Wolframalpha
$$\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{12}+1\right)^{5}}&=\frac{\pi \csc \left(\frac{\pi}{12}\right)}{4 ! 12^{5}} \cdot 11 \cdot 23 \cdot 35 \cdot 47\\
&=\frac{416185 \pi(\sqrt{6}+\sqrt{2})}{5971968} \\
&\doteq 0.845906950943631,
\end{aligned}
$$
which is checked by Wolframalpha
My question is whether we can prove the formula without using $\int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).$
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Without using the formula $\displaystyle \int_{0}^{\infty} \frac{d x}{x^{n}+1}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right),$ I thought of the Gamma Function.
$$ \begin{aligned}
\text { Let } t &=\frac{1}{x^{n}+1}, \text { then } x=\left(\frac{1}{t}-1\right)^{\frac{1}{n}} \\
d x &=\frac{1}{n}\left(\frac{1}{t}-1\right)^{\frac{1-n}{n}}\left(-\frac{1}{t^{2}} d t\right)=-\frac{(1-t)^{\frac{1}{n}-1}}{n t ^{\frac{1+n}{n }} }d t
\end{aligned}
$$
The integral is transformed into
$$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}}&=\frac{1}{n} \int_{0}^{1} t^{m} \cdot \frac{(1-t)^{\frac{1-n}{n}}}{t^{\frac{1+n}{n}}} d t \\
&=\frac{1}{n} \int_{0}^{1} t^{\frac{m n-1}{n}-1}(1-t)^{\frac{1}{n} -1}dx \\
&=\frac{1}{n} B\left(m -\frac{1}{n}, \frac{1}{n}\right)
\end{aligned}
$$
Now applying the formula $$
B(x, y)=\frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)},
$$
we can now conclude that
$$\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+1\right)^{m}}= \frac{\Gamma\left(m -\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right)}{n(m-1)!}}$$
Letting $x\mapsto\frac{x}{\sqrt[n]a}$ yields the general integral,
$$\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}= \frac{\Gamma\left(m -\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right)}{n a^{m-\frac{1}{n}}(m-1)!}}$$
Taking the same integral for example, we have
$$
\int_{0}^{\infty} \frac{d x}{\left(x^{12}+1\right)^{5}}= \frac{\Gamma\left(\frac{1}{12}\right) \Gamma\left(\frac{59}{12}\right)}{12 \Gamma(5)}= \frac{1}{288} \Gamma\left(\frac{1}{12}\right) \Gamma\left(\frac{59}{12}\right)\approx 0.845907, $$
which is checked by Wolframalpha.
However, I don’t know how to further simplify the formula
$$\Gamma\left(m -\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right),$$
can you help?
$$\textrm{ ********} \tag*{} $$
Thanks to Mr Clathratus who gave me a nice simplification to $$\Gamma\left(m -\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right).$$
Using the identity $$
\Gamma(x+1)=x \Gamma(x),
$$
we have $$
\begin{aligned}
\Gamma\left(m-\frac{1}{n}\right) &=\left(m-1-\frac{1}{n}\right) \Gamma\left(m-\left(-\frac{1}{n}\right)\right.\\
&=\left(m-1-\frac{1}{n}\right)\left(m-1-\frac{1}{n}\right) \Gamma\left(m-2-\frac{1}{n}\right) \\
& \qquad\qquad \vdots \\
&=\left(m-1-\frac{1}{n}\right)\left(m-2-\frac{1}{n}\right) \cdots\left(1-\frac{1}{n} \right)\Gamma\left(1-\frac{1}{n}\right)\\
&= \Gamma\left(1-\frac{1}{n}\right) \prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) \\
\Gamma\left(m-\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right) &=\prod_{k=1}^{m-1}\left(k-\frac{1}{n}\right) \Gamma\left(1-\frac{1}{n} \right)\Gamma\left(\frac{1}{n}\right)
\end{aligned}
$$
By the Euler's Reflection Theorem,
$$
\Gamma\left(1-\frac{1}{n}\right) \Gamma\left(\frac{1}{n}\right)=\pi \csc \left(\frac{\pi}{n}\right)
$$
Putting back, we now obtain the same formula as before
$$
\boxed{\int_{0}^{\infty} \frac{d x}{\left(x^{n}+a\right)^{m}}=\frac{\pi \csc \left(\frac{\pi}{n}\right)}{(m-1) ! na^{\frac{m n-1}{n}} }\prod_{k=1}^{m-1}(k-\frac{1}{n} )}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Basic rotations in $4-D$ Basic rotations in $3-D$ are represented by the matrices:
$$
R^3_{xy}(\theta)=
\begin{bmatrix}
\cos \theta &-\sin \theta & 0\\
\sin \theta &\cos \theta & 0\\
0& 0 & 1\\
\end{bmatrix}\quad \text{rotation in the plane}\; (xy)
$$
$$
R^3_{xz}(\theta)=
\begin{bmatrix}
\cos \theta&0 &\sin \theta\\
0& 1 & 0\\
-\sin \theta & 0&\cos \theta \\
\end{bmatrix}\quad \text{rotation in the plane}\; (xz)
$$
$$
R^3_{yz}(\theta)=
\begin{bmatrix}
1& 0 & 0\\
0 &\cos \theta&-\sin \theta\\
0&\sin \theta & \cos \theta \\
\end{bmatrix}\quad \text{rotation in the plane}\; (yz)
$$
where te sign of the $\sin \theta$ components are determined by the right-hand rule.
In $4-D$ space the basic rotations becomes something as
$$
R^4_{xy}(\theta)=
\begin{bmatrix}
\cos \theta &-\sin \theta & 0&0\\
\sin \theta &\cos \theta & 0&0\\
0& 0 & 1&0\\
0& 0 & 0&1\\
\end{bmatrix}
\quad
R^4_{xz}(\theta)=
\begin{bmatrix}
\cos \theta&0 &\sin \theta&0\\
0& 1 & 0 &0\\
-\sin \theta & 0&\cos \theta&0 \\
0&0&0&1
\end{bmatrix}
$$
$$
R^4_{xt}(\theta)=
\begin{bmatrix}
\cos \theta&0&0 &-\sin \theta\\
0& 1 & 0&0\\
0& 0 & 1&0\\
\sin \theta&0&0 &\cos \theta \\
\end{bmatrix}
\quad
R^4_{yz}(\theta)=
\begin{bmatrix}
1 & 0 &0 &0\\
0&\cos \theta&\sin \theta&0\\
0&-\sin \theta & \cos \theta &0\\
0&0&0&1
\end{bmatrix}
$$
$$
R^4_{yt}(\theta)=
\begin{bmatrix}
1& 0& 0&0\\
0&\cos \theta&0 &\sin \theta\\
0& 0 & 1&0\\
0&-\sin \theta&0 &\cos \theta \\
\end{bmatrix}
\quad
R^4_{zt}(\theta)=
\begin{bmatrix}
1 & 0 &0 &0\\
0&1&0&0\\
0&0&\cos \theta &-\sin \theta\\
0&0&\sin \theta &\cos \theta \\
\end{bmatrix}
$$
But here (as far as i know) we don't have a simple extension of the right-hand rule, so what is the rule for the signs of the $\sin \theta$ components?
|
In $4D$ there are ${4\choose 2}=6$ basic rotation matrices, each one characterized by the two out of four axes that they keep fixed while rotating vectors in the plane perpendicular to those axes. Alternatively, we can characterize them by the two out of four basis vectors they rotate. We know from $3D$ that the right-hand rule (counter clockwise rotation) requires a minus sign at exactly one of the $\sin\theta$ elements in the upper right corner of the $2\times 2$ submatrix that does the rotation.
A plus sign at exactly one of those $\sin\theta$ elements will clearly lead to the left-hand rule (clockwise rotation).
(We don't worry about the $\sin\theta$ in the lower left corner as this always must have the opposite sign to make the matrix orthogonal.)
The $4D$ matrices in the question satisfy the right or left hand rule for every $3D$ subspace:
\begin{array}{|c|c|c|c|}
\hline
\text{ subspace}&\text{matrices}&\text{rule}&\text{rotation}\\
\hline
xyz&R^4_{xy},R^4_{xz},R^4_{yz}&\text{right hand}&\text{counter clockwise }\\
yzt&R^4_{yz},R^4_{yt},R^4_{zt}&\text{right hand}&\text{counter clockwise }\\
xyt&R^4_{xy},R^4_{xt},R^4_{yt}&\text{left hand}&\text{clockwise}\\
xzt&R^4_{xz},R^4_{xt},R^4_{zt}&\text{left hand}&\text{clockwise}\\
\hline
\end{array}
Trying to enforce the right hand rule for every $3D$ subspace seems impossible. For example: change $\sin\theta$ to $-\sin\theta$ in $R^4_{yt}$ would lead to a right hand rule in subspace $xyz$ but break it in the subspace $yzt\,.$
|
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}
|
Find all positive integers s.t. $\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \frac{1}{d^2}$ Find all positive integers $a, b, c, d$ such that :
$$\frac{1}{a^2} - \frac{1}{b^2} = \frac{1}{c^2} - \frac{1}{d^2}$$
The original problem came from atomic electron transitions :
I would like to find out non-trivial positive integer solutions; since it is trivial if $a = b$ and $c = d$, or $a = c$ and $b = d$.
I found some of the solutions, and they look like :
There is some pattern in these integers, but it seems difficult to obtain a gerneralized form of the solution.
|
There are various variants of such equations. For example like this.
https://artofproblemsolving.com/community/c3046h2465588_a_class_of_diophantine_equations_in_three_variables
When solving such equations.
$$\frac{ 1 }{ a^2 } -\frac{ 1 }{ b^2 } = \frac{ 1 }{ c^2 } - \frac{ 1 }{ d^2 } $$
Pythagorean triples can help us. Take any two Pythagorean triples.
$$\left\{\!\begin{aligned}
& p^2=s^2+n^2 \\
& k^2=t^2+j^2
\end{aligned}\right. $$
Then the solutions can be written in this form. Then the truth needs to be reduced by a common divisor.
$a=ktsn$
$b=pstj$
$c=ktpn$
$d=pskj$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4370392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
}
|
Find $T^*\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ in an inner product space $V=\mathbb{R^3}$
$f$ is an inner product ,$f\Bigg (\begin{pmatrix}
x_{1} \\
x_{2} \\
x_{3}
\end{pmatrix} ,\begin{pmatrix}
y_{1} \\
y_{2} \\
y_{3}
\end{pmatrix} \Bigg ) = 4x_1y_1+x_1y_2+x_2y_1+4x_2y_2+x_2y_3+x_3y_2+x_3y_3$
$T\begin{pmatrix}
a \\
b \\
c
\end{pmatrix} =\begin{pmatrix}
b+c \\
a+b \\
a+2c
\end{pmatrix}$
Find $T^*\begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix}$
My solution:
I want to find $T^*\begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix}$ without finiding orthonormal basis. ( Is it possible ? )
Suppose $T^*\begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix}= \begin{pmatrix}
x \\
y \\
z
\end{pmatrix}$
$\langle T\begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix} , \begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix}\rangle$
$=\langle \begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix}, T^*\begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix} \rangle $
$\langle T\begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix} , \begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix}\rangle$ =$\langle \begin{pmatrix}
5 \\
3 \\
7
\end{pmatrix} , \begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix}\rangle = 141$
Then, $\langle \begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix}, T^*\begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix} \rangle = \langle \begin{pmatrix}
1 \\
2 \\
3
\end{pmatrix}, \begin{pmatrix}
x \\
y \\
z
\end{pmatrix} \rangle = 4x+y+2x+8y+2z+3y+3z=141$
Here I get stuck, how can I find $x,y,z$?
Am I supposed to take $x=y=0$ and then get Z and that's it?
Please any help or suggestions. Thanks.
|
If it is the word "orthonormal" that is bugging you then you can do it by taking a basis, say $v_{1},v_{2},v_{3}$. Where you can pick them to be any basis you like. Say $(0,1,1),(1,0,1),(1,1,0)$. Or heck, you might even take the standard basis for $\mathbb{R}^{3}$ , $e_{1},e_{2},e_{3}$. They are not orthonormal wrt the inner product given.
Then if $T^*(1,2,3)=(x,y,z)$. You have to solve the following system of equations for $(x,y,z)$.
$$\langle (1,2,3),T(v_{i})\rangle = \langle (x,y,z),v_{i}\rangle\,\, ,1\leq i\leq 3.$$ (As I said, if you want to make your work easier it suffices to work with $e_{1},e_{2},e_{3}$.)
They will give you three equations. You solve them to find $(x,y,z)$. And hence you will have your image.
|
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"url": "https://math.stackexchange.com/questions/4372173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Given $\varphi$ is golden ratio, how do I prove this $\sum \limits_{j=1}^{\infty}\frac{(1-\varphi)^j}{j^2}\cos{\frac{3j\pi}{5}}=\frac{\pi^2}{100}$? Given $ \varphi$ is golden ratio, how do I prove this:
$ \displaystyle \tag*{}\sum \limits_{j=1}^{\infty}\frac{(1-\varphi)^j}{j^2}\cos{\frac{3j\pi}{5}}=\frac{\pi^2}{100}$
My approach:
We can reduce the sum into simpler parts by using the property of golden ratio, we have:
$\displaystyle \tag*{} \phi^2 - \phi - 1 = 0 \Leftrightarrow 1- \phi = - \dfrac{1}{\phi}$
And I also found the values of $\cos \dfrac{3j\pi}{5}$ in terms of $\varphi$:
$\displaystyle \tag*{} \begin{align} \cos \dfrac{3\pi}{5} &= \dfrac{-1}{2 \varphi} \\\\ \cos \dfrac{6\pi}{5} &= \dfrac{-\varphi}{2} \\\\ \cos \dfrac{9\pi}{5} &= \dfrac{\varphi}{2} \\\\ \cos \dfrac{12\pi}{5} &= \dfrac{1}{2 \varphi} \\\\ \cos \dfrac{15\pi}{5} &= {-1} \end{align}$
And this repeats, periodically with alternate opposite signs. I don't know how to connect these information I found to prove the question. Maybe my approach is wrong. Any help would be appreciated. Thanks.
|
The point is the functional equation $$Li_2(1-z)+Li_2(1-1/z)=-\frac12 \log^2 z$$
of the dilogarithm $$Li_2(z)=\sum_{k\ge 1} \frac{z^k}{k^2}$$ which follows from $z \, Li_2'(z)=-\log(1-z)$,
and that $$1-z= (1-\varphi)e^{3i\pi/5} \implies z=e^{i\pi/5}$$
So $$\sum_{k\ge 1} \frac{(1-\varphi)^k}{k^2}\cos{\frac{3k\pi}{5}}=\frac{-1}4 \log^2 e^{i\pi/5} = \frac{\pi^2}{100}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4379575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
For a natural number $b$, let $N(b)$ denote the number of natural numbers $a$ for which the equation $x^2+ax+b=0$ has integer roots.
For a natural number $b$, let $N(b)$ denote the number of natural numbers $a$ for which the equation $x^2+ax+b=0$ has integer roots. What is the smallest value of $b$ for which $N(b)=20$.
I came across a similar problem at: For a natural number $b$, $N(b)=$ number of natural numbers a such that the equation $x^2+ax+b=0$ has integral roots.
But, there it is asked to find $N(b)$ given value of $b=6$. Here, the opposite is asked, to find $b$ given $N(b)$.
How can I proceed here?
|
First, observe that, for all $a, b \in \mathbb{Z}$, the polynomial $X^{2} +a X +b \in \mathbb{Z}[X]$ has integer roots if and only its discriminant $\Delta = a^{2} -4 b$ is the square of an integer:
*
*if there exists $\delta \in \mathbb{Z}$ such that $\Delta = \delta^{2}$, then $\delta$ and $a$ have the same parity since they both have the same parity as $\Delta$, and hence the roots $x_{1} = \frac{a -\delta}{2}$ and $x_{2} = \frac{a +\delta}{2}$ of this polynomial are integers;
*Conversely, if this polynomial has integer roots $x_{1}, x_{2}$ (repeated according to their multiplicities), then $\Delta = \left( x_{1} -x_{2} \right)^{2}$ is the square of an integer.
Let $b \geq 1$ be an integer. Define $$S(b) = \left\lbrace a \in \mathbb{Z}_{\geq 0} : \exists \delta \in \mathbb{Z}, \, a^{2} -4 b = \delta^{2} \right\rbrace \, \text{,}$$ so that $N(b) = \left\lvert S(b) \right\rvert$ by the discussion above. Let us describe this set $S(b)$.
*
*Suppose that $a \in S(b)$. There exists $\delta \in \mathbb{Z}$ such that $a^{2} -4 b = \delta^{2}$. Then $a$ and $\delta$ have the same parity, and hence $a -\delta$ and $a +\delta$ are both even. Moreover, we have $\left( \frac{a -\delta}{2} \right) \left( \frac{a +\delta}{2} \right) = b$. Therefore, there exists a divisor $d$ of $b$ such that $\frac{a -\delta}{2} = d$ and $\frac{a +\delta}{2} = \frac{b}{d}$, and we have $a = d +\frac{b}{d}$, which also shows that $d$ is positive.
*Conversely, suppose that there exists a positive divisor $d$ of $b$ such that $a = d +\frac{b}{d}$. Then we have $a^{2} -4 b = \left( d -\frac{b}{d} \right)^{2}$, and hence $a \in S(b)$.
Thus, we have proved that $$S(b) = \left\lbrace d +\frac{b}{d} : d \text{ is a positive divisor of } b \right\rbrace \, \text{.}$$
Now, note that, for all positive divisors $d_{1}, d_{2}$ of $b$, we have
\begin{align*}
d_{1} +\frac{b}{d_{1}} = d_{2} +\frac{b}{d_{2}} & \Longleftrightarrow d_{1}^{2} d_{2} +b d_{2} = d_{1} d_{2}^{2} +b d_{1}\\
& \Longleftrightarrow \left( d_{1} -d_{2} \right) \left( d_{1} d_{2} -b \right) = 0\\
& \Longleftrightarrow d_{2} = d_{1} \text{ or } d_{2} = \frac{b}{d_{1}} \, \text{.}
\end{align*}
Moreover, for every positive divisor $d$ of $b$, we have $d = \frac{b}{d}$ if and only if $b = d^{2}$. Therefore, we have $$N(b) = \begin{cases} \frac{\sigma(b) +1}{2} & \text{if } b \text{ is a square}\\ \frac{\sigma(b)}{2} & \text{otherwise} \end{cases} \, \text{,}$$ where $\sigma(b)$ denotes the number of positive divisors of $b$.
Finally, let us find the least integer $b \geq 1$ such that $N(b) = 20$. By the discussion above, for every $b \geq 1$, we have $N(b) = 20$ if and only if $\sigma(b) \in \lbrace 39, 40 \rbrace$ (note that $b$ is a square if and only if $\sigma(b)$ is odd). Recall that, for all pairwise distinct primes $p_{1}, \dotsc, p_{r}$ and all $m_{1}, \dotsc, m_{r} \geq 1$, we have $$\sigma\left( p_{1}^{m_{1}} \dotsm p_{r}^{m_{r}} \right) = \left( m_{1} +1 \right) \dotsm \left( m_{r} +1 \right) \, \text{.}$$ Morevover, $39 = 3 \cdot 13$ and $40 = 2^{3} \cdot 5$. Therefore, the integers $b \geq 1$ such that $N(b) = 20$ are precisely the numbers of the form $$p_{1} p_{2} p_{3} p_{4}^{4} \, \text{,} \quad p_{1} p_{2} p_{3}^{9} \, \text{,} \quad p_{1} p_{2}^{3} p_{3}^{4} \, \text{,} \quad p_{1} p_{2}^{19} \, \text{,} \quad p_{1}^{2} p_{2}^{12} \, \text{,} \quad p_{1}^{3} p_{2}^{9} \, \text{,} \quad p_{1}^{4} p_{2}^{7} \, \text{,} \quad p_{1}^{38} \quad \text{or} \quad p_{1}^{39} \, \text{,}$$ where $p_{1}, p_{2}, p_{3}, p_{4}$ are pairwise distinct prime numbers. Thus, the least integer $b \geq 1$ such that $N(b) = 20$ is $2^{4} \cdot 3 \cdot 5 \cdot 7 = 1680$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4380178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Convert $f(x) = (\cos x+\sin x)(1-\sqrt{3}\tan x)$ into the form of $k(1+\tan x)\cos(x+a).$ I need your help to convert the following into the form of $k(1+\tan(x))\cos(x+a).$
$$f(x) = (\cos(x)+\sin(x))(1-\sqrt{3}\tan x)$$
Here is what I've done so far.
\begin{aligned}
f(x) &= (\cos(x)+\sin(x))(1-\sqrt{3}\tan x)\\
f(x) &= \sqrt{2}(\frac{1}{\sqrt{2}}\cos(x)+\frac{1}{\sqrt{2}}\sin(x))(1-\sqrt{3}\tan x)\\
f(x) &= \sqrt{2}\cos(x-\frac{\pi}{4})(1-\sqrt{3}\tan x)
\end{aligned}
Please help me.
|
We have to find $k$ and $a$ such that
$k(1+\tan x)\cos(x+a)=(\cos x+\sin x)(1-\sqrt3\tan x)$
or
$k(\cos x+\sin x)\dfrac{\cos(x+a)}{\cos x}=(\cos x+\sin x)(1-\sqrt3\tan x)$
or
$k(\cos a-\sin a\tan x)=(1-\sqrt3\tan x).$
Now the similarity is obvious. Can you take it from here?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4380904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Decomposing $\sum_{i = 0}^{2n} x^i$ as a simple sum of squares As we have $\sum_{i = 0}^{2n} x^i = (x^{2n + 1} - 1) / (x - 1)$, the polynomial is positive.
So we know that there is a decomposition as a sum of squares. Is there a closed simple form for such a decomposition? For small value of $n$ we have
$$ x^2 + x + 1 = 1/4 ((2x+1)^2+ 3)$$
$$ x^4 + x^3 + x^2 + x + 1 = 1/16((4 x^2 + 2 x + 1)^2 + (2 x + 3)^2 + 6)$$
$$ x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 = 1/144 ((12x^3+6x^2+4x+3)^2+ 3(2x^2+2x+5)^2 + 5(2x+3)^2 +15)$$
|
Let $\,\omega\,$ be a primitive $\,(2n+1)^{th}\,$ root of unity, so $\,\omega^{2n+1}=1\,$ and $\,\omega^{2n+1-k} = \dfrac{1}{\omega^k}=\overline{\omega^k}\,$. Then:
$$
P(x)=\sum_{k=0}^{2n} x^k = \prod_{k=1}^{2n}(x-\omega^k) = \prod_{k=1}^n\big((x-\omega^k)(x-\overline{\omega^k})\big) = \underbrace{\prod_{k=1}^n(x-\omega^k)}_{Q(x)} \;\underbrace{\prod_{k=1}^n(x-\overline{\omega^k})}_{\overline{\,Q\,}(x)}
$$
$Q\,$ is a polynomial with complex coefficients which can be written as $\,Q(x)=U(x) + i V(x)\,$ where $\,U,V\,$ are polynomials with real coefficients. It follows that:
$$
P(x) = Q(x) \, \overline{Q}(x)=\big(U(x)+i V(x)\big)\big(U(x)-i V(x)\big)=U^2(x) + V^2(x)
$$
[ EDIT ] $\;$ The above proves that $\,P(x)\,$ can always be written as the sum of just $\,2\,$ squares, and gives an explicit form for the two polynomials. However, $\,U(x)\,$ and $\,V(x)\,$ will generally not have rational coefficients, and can be laborious to calculate. A couple of examples below.
*
*$n=1:\;$ then $\,\omega = e^{i\,2\pi/3}=\frac{-1 + i\sqrt{3}}{2}\,$ and $\,Q(x) = \underbrace{\frac{2x+1}{2}}_{\, U(x)} + i \cdot \underbrace{\frac{-\sqrt{3}}{2}}_{\,V(x)}\,$, so:
$$
P(x)=U^2(x)+V^2(x) \quad\iff\quad x^2+x+1 =\frac{1}{4}\big((2x+1)^2 + 3\big)
$$
*
*$n=2:\;$ then $\,\omega = e^{i\,2\pi/5}=\frac{1}{4} (\sqrt{5} - 1) + i \sqrt{\frac{1}{8}\left(5 + \sqrt{5}\right)}\;$ and:
$$
Q(x)=\frac{1}{4} (4 x^2 + 2 x - \sqrt{5} - 1) - \frac{1}{4} i \sqrt{\frac{5 + \sqrt{5}}{2}} \left(\left(\sqrt{5} + 1\right) x + \sqrt{5} - 1\right)
$$
so:
$$
x^4+x^3+x^2+x+1 = \frac{1}{16}(4 x^2 + 2 x - \sqrt{5} - 1)^2 + \frac{5 + \sqrt{5}}{32} \left((\sqrt{5} + 1)x + \sqrt{5} - 1\right)^2
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/4382334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Finding $\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+2p\cos(x)+p^2}\mathrm dx$ Let $f$ be a analytic function in the closed unit circle with its center
at the point $\alpha\in\mathbb{R}$, then:
\begin{equation*}
\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+2p\cos(x)+p^2}\mathrm dx=\frac{2\pi}{1-p^2}f(\alpha+p) ,
\end{equation*}
for $|p|<1$.
My attempt: By
\begin{align*}
\therefore\quad \sum_{n=1}^{\infty}p^{n}\sin(nx)=\frac{p\sin (x)}{1-2p\cos (x)+p^2},\qquad|p|<1
\end{align*}
adjusting $p\to -p$ and highlighting $\displaystyle \frac1{1+2p\cos(x)+p^2}$:
\begin{align*}
\frac1{1+2p\cos(x)+p^2}=-\frac{1}{\sin(x)}\sum_{n=1}^\infty(-p)^{n}\sin(nx).
\end{align*}
Thus:
\begin{align*}
\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+2p\cos(x)+p^2}\mathrm dx=-\sum_{n=1}^\infty(-p)^{n}\int_0^\pi\left\{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)\right\}\frac{\sin(nx)}{\sin(x)}\mathrm dx\tag{1}
\end{align*}
by the \textit{Dirichlet Kernel}: $\displaystyle \sum_{k=0}^{N-1}e^{2ikx}=e^{(N-1)x}\frac{\sin(Nx)}{\sin(x)}$ setting $N\to n$, $n\in\mathbb{N}$ and then taking $(1)$, follows that:
\begin{align*}
&=-\sum_{n=1}^\infty(-p)^{n}\sum_{k=0}^{n-1}\int_0^\pi\left\{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)\right\}e^{-(n-1)x}e^{2ikx}\mathrm dx,\quad\left(e^{ix}\to z\right)\\
&=...
\end{align*}
At this point I'm out of ideas. I would like some light on my last step, or another approach that is similar to this one.
|
Integrate as follows
\begin{align}
&\int_0^\pi\frac{f\left(\alpha+e^{ix}\right)+f\left(\alpha+e^{-ix}\right)}{1+2p\cos x+p^2} dx\\
=&\int_0^\pi\frac{2\Re f\left(\alpha+e^{ix}\right)}{1+2p\cos x+p^2} dx
= 2\int_0^\pi\frac{\sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!} \cos(kx)}{1+2p\cos x+p^2} dx\\
=& \> 2\sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!} \int_0^\pi \frac{\cos(kx)}{1+2p\cos x+p^2} dx\\
=& \> 2\sum_{k=0}^\infty \frac{f^{(k)}(a)}{k!}\cdot \frac{\pi (-p)^k}{1-p^2}=
\frac{2\pi}{1-p^2}f(\alpha-p)
\end{align}
where
$$\eqalign{
\frac{1-p^2}{1+2p\cos x+p^2}
=1+2\sum_{j=1}^\infty (-p)^j\cos(j x)
}
$$
is used to integrate
$\int_0^\pi \frac{\cos(kx)}{1+2p\cos x+p^2} dx=\frac{\pi (-p)^k}{1-p^2}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove $2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} = 1$
Prove $$2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} = 1$$
I got this question when I was going through some basic trigonometric identities as follows
$2(\cos\frac{\pi}{5}-\cos\frac{2\pi}{5}) =1\tag1$
very much straightforward to recognise
$2(\cos\frac{\pi}{7}-\cos\frac{2\pi}{7} + \cos\frac{3\pi}{7})=1\tag2$
Following steps proves the identity
$8\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7}\cdot\cos\frac{4\pi}{7} = -1$
$4\cos\frac{\pi}{7}[2\cos\frac{4\pi}{7}\cdot\cos\frac{2\pi}{7}] = -1$
$4\cos\frac{\pi}{7}[\cos\frac{6\pi}{7}+ \cos\frac{2\pi}{7}] =-1 $
$4\cos\frac{\pi}{7}[\cos\frac{2\pi}{7}-\cos\frac{\pi}{7}] =-1 $
$4\cos\frac{\pi}{7}\cdot\cos\frac{2\pi}{7} - 4\cos^{2}\frac{\pi}{7} = -1$
Further simplification will lead to equality 2
$2(\cos\frac{\pi}{9}-\cos\frac{\pi}{9} + \cos\frac{3\pi}{9}- \cos\frac{4\pi}{9}) = 1$
by using transformation formula we can prove above one also
But when it comes to following equalities
$2(\cos\frac{\pi}{11}-\cos\frac{2\pi}{11} + \cos\frac{3\pi}{11}- \cos\frac{4\pi}{11}+ \cos\frac{5\pi}{11} )= 1$
$2(\cos\frac{\pi}{13}-\cos\frac{2\pi}{13} + \cos\frac{3\pi}{13}- \cos\frac{4\pi}{13}+ \cos\frac{5\pi}{13} -\cos\frac{6\pi}{13} )= 1$
I was able to check the results with brute force in Wolfram|Alpha for above equalities, but not able to get the steps properly even though I tried manually
My question is how to generalise the summation formula and is there any method other than trigonometric approach?
$$2\sum_{k=1}^n{(-1)^{k-1} \cos\frac{k\pi}{2n+1}} = 1$$
|
Let $z=e^{\frac{\pi i}{2 n+1}}$ , then $$
\begin{aligned}
\sum_{k=1}^{n}(-1)^{k-1} z^{k} &=\frac{z\left[1-(-z)^{n}\right]}{1+z} \\
&=\frac{e^{\frac{\pi i}{2 n+1}}\left(1+(-1)^{n+1} e^{\frac{n \pi i}{2 n+1}}\right)}{1+e^{\frac{\pi i}{2 n+1}}}
\end{aligned}
$$
Multiplying both the numerator and denominator by $e^{-\frac{\pi i}{2(2 n+1)}}$ yields
$$
\begin{aligned}\sum_{k=1}^{n}(-1)^{k-1} z^{k}& = \frac{e^{\frac{\pi i}{2(2 n+1)}}+(-1) ^{n+1} e^{\left(\frac{\pi i}{2(2 n+1)}+\frac{n \pi i}{2 n+1}\right)}}{e^{\frac{\pi i}{2(2 n+1)}}+e^{-\frac{\pi i}{2(2 n+1)}}}\\ &= \frac{e^{\frac{\pi i}{2(2 n+1)}}+(-1)^{n+1} e^{\frac{\pi}{2} i}}{e^{\frac{\pi i}{2(2 n+1)}+} e^{-\frac{\pi i}{2(2 n+1)}}}\\&= \frac{\cos \left(\frac{\pi}{2(2 n+1)}\right)+i \sin \left(\frac{\pi}{2(2 n+1)}\right)+(-1)^{n+1} i}{2 \cos \left(\frac{\pi}{2(2 n+1)}\right)} \qquad\qquad (*)
\end{aligned} $$
Comparing the real parts of (*) yields
$$
\sum_{k=1}^{n}(-1)^{k-1} \cos \left(\frac{k \pi}{2 n+1}\right)=\frac{1}{2} $$
By the way, comparing the imaginary parts of (*) gives
$$
\sum_{k=1}^{n}(-1)^{k-1} \sin \left(\frac{k \pi}{2 n+1}\right)=\frac{1}{2}\left[\tan \left(\frac{\pi}{2(2 n+1)}\right)+(-1)^{n+1} \sec \left(\frac{\pi}{2(2 n+1)}\right)\right]
$$
|
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|
Probability of winning a game by rolling the die first Two persons are playing a game where they take turns rolling a die (so A rolls first, then B, then A again and so on). The first person to roll a $6$ wins the game. What is the probability that the person who started the game (rolled the die first) wins?
This was the question that I was given, but I feel like the probability depends on the number of turns played? My approach was that the probability of rolling a 6 at any particular turn will be $\frac{1}{6}$. So, the probability of winning in $n$ turns will be the probability of not rolling a $6$ on the first $n-1$ turns (since if a $6$ had been rolled, that would have been the last, i.e. the $nth$ turn) and then rolling a $6$ on the $nth$ turn. The required probability would then be $$\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdot...(n-1\text{ times})\cdot\frac{1}{6} = \frac{5^{n-1}}{6^n}$$
Is this train of thought correct? Or have I misunderstood something?
|
The probability of winning can be seen as a sum $P = P_1+P_2+P_3+\dots$. Here $P_1$ is the probability of winning at the first turn. Since you are the first to throw the dice, we simply have
$$P_1=\frac{1}{6}\,.$$
Then $P_2$ is the probability of winning at the second term. For this you need to have rolled something which is not a 6 in the first term (which brings a factor $5/6$, than your opponent must have rolled something different from 6 (brings a factor $5/6$) and finally you must roll a 6. So you have
$$P_2= \frac{5}{6}\times\frac{5}{6}\times\frac{1}{6} = \left(\frac{5}{6}\right)^2\frac{1}{6}\,.$$
Similarly $P_3$ means that: no 6 (you), no 6 (opponent), no 6 (you), no 6 (opponent), 6 (you). This brings
$$P_3 = \left(\frac{5}{6}\right)^4\frac{1}{6}\,.$$
In general
$$P_n= \left(\frac{5}{6}\right)^{2(n-1)}\frac{1}{6}\,.$$
Now you conclude that
$$P = \sum_{n=1}^\infty P_n = \frac{1}{6}\sum_{n=0}^\infty\left(\frac{5}{6}\right)^{2n} = \frac{1}{6}\times\frac{1}{1-\frac{25}{36}}=\frac{6}{11}\,.$$
Note that in the sum one should exclude the term $n=\infty$, which is the event in which the game never ends. However $P_\infty=0$.
|
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|
How do you prove: $\dfrac{1}{\sin ^2 (\frac{2\pi}{9})} - \dfrac{1}{\sin ^2(\frac{4 \pi}{9})} = 8\sin(\frac{\pi}{18})$ I have to prove:
$\displaystyle \tag*{} \alpha={\dfrac{1}{\sin ^2 (\frac{2\pi}{9})} - \dfrac{1}{\sin ^2(\frac{4 \pi}{9})}}= 8\sin(\frac{\pi}{18})$
I tried to make a common denominator of $\alpha$ and use a few identities to arrive at:
$\displaystyle \tag*{} \alpha = \dfrac{\sin(\frac{6\pi}{9})}{\sin(\frac{2\pi}{9})\sin^2(\frac{4\pi}
{9})}$
I am not sure how to proceed from this. I even arrived at other 2 forms of $\alpha$, but they make even more complicated. Any hints would be greatly appreciated. Thanks.
|
Good question. For me, I am more customized to write in degrees. Therefore, we can have
$${\frac{1}{\sin ^2 (40^\circ)} - \frac{1}{\sin ^2(80^\circ)}}= 8\sin(10^\circ)$$
Or we can write it in a different way
$${\frac{\sin ^2(80^\circ)}{\sin ^2 (40^\circ)}} - 1= 8\sin(10^\circ)\sin ^2(80^\circ)$$
Notice that $$\sin 20^\circ=2\sin10^\circ\cos10^\circ=2\sin10^\circ\sin 80^\circ$$
Also we have $$\sin 80^\circ=2\sin 40^\circ\cos40^\circ$$
So we ned to show
$$4\cos^2 40^\circ -1=4\sin 20^\circ\sin 80^\circ$$
Notice that $1=2\sin30^{\circ}$. so we have
\begin{align}&4\sin 20^\circ\sin 80^\circ\\=&4\sin (50+30)^\circ\sin (50-30)^\circ\\=&4(\sin 50^\circ\cos30^\circ+\sin 30^\circ\cos50^\circ)(\sin 50^\circ\cos30^\circ-\sin 30^\circ\cos50^\circ)\\=&4\sin^2 50^\circ\cos^230^\circ-\sin^2 30^\circ\cos^250^\circ)\\=&4(\sin^2 50^\circ-\sin^230^\circ)\\=&4\sin^2 50^\circ-4\sin^230^\circ=4\cos^2 40^\circ -1
\end{align}
|
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|
Evaluate $\lim_{n\to\infty} \sqrt{n}\int_{0}^{\pi/2} \sin^{n} x dx$ Question:
Compute $I_n=\lim\limits_{n\to\infty} \sqrt{n}\int_{0}^{\frac{\pi}{2}} \sin^{n} x dx$.
Attempt:
We know the famous result that $\lim\limits_{n\to\infty}\int_{0}^{\frac{\pi}{2}} \sin^{n} x dx=0$,
and by dividing it into two parts $\int_{0}^{\frac{\pi}{2}-n^{-\alpha}}+\int_{\frac{\pi}{2}-n^{-\alpha}}^{\frac{\pi}{2}}(0<\alpha<\frac{1}{2})$ ,
we can show that $\int_{0}^{\frac{\pi}{2}} \sin^{n} x dx=O(\frac{1}{n^{\alpha}})$.
There is a hint which says by substitution we can rewrite the limits with $\Gamma$, and eventually get $I_n\sim c\frac{1}{\sqrt{n}},c>0$.
Is there any method without using the $\Gamma$ function? (Answers using Gamma function are also appreciated.)
|
Proceeding along your approach, without using the gamma function
Let
\begin{align*}
J_n &:= \sqrt{n}\int_{\pi/2 - n^{-1/3}}^{\pi/2} \sin^n x \,\mathrm{d} x, \\
K_n &:= \sqrt{n}\int_0^{\pi/2 - n^{-1/3}} \sin^n x \,\mathrm{d} x.
\end{align*}
We have
$$J_n
= \sqrt{n} \int_0^{n^{-1/3}} \cos^n y \,\mathrm{d} y
= \int_0^{n^{1/6}} \cos^n \frac{z}{\sqrt n}\, \mathrm{d} z
= \int_0^{n^{1/6}}
\mathrm{e}^{n\ln \cos \frac{z}{\sqrt n}}\mathrm{d} z.$$
Fact 1: For $0 \le u \le 1$, it holds that
$$-\frac{u^2}{2} - u^4 \le \ln \cos u \le - \frac{u^2}{2}.$$
(The proof is not difficult. Omitted.)
Using Fact 1, we have, for all $0 \le z \le n^{1/6}$,
$$-\frac{z^2}{2} - \frac{(n^{1/6})^4}{n}\le -\frac{z^2}{2} - \frac{z^4}{n} \le n\ln \cos \frac{z}{\sqrt n} \le -\frac{z^2}{2}.$$
Thus, we have
$$\mathrm{e}^{-n^{-1/3}}\int_0^{n^{1/6}}
\mathrm{e}^{-z^2/2}\mathrm{d} z \le J_n \le \int_0^{n^{1/6}}
\mathrm{e}^{-z^2/2}\mathrm{d} z.$$
Thus,
$$\lim_{n\to \infty} J_n = \int_0^\infty \mathrm{e}^{-z^2/2}\mathrm{d} z = \sqrt{\pi/2}.$$
$\phantom{2}$
On the other hand, we have
\begin{align*}
K_n
&\le \sqrt{n}\, \cdot \frac{\pi}{2} \sin^n \left(\pi/2 - n^{-1/3}\right)\\
&= \frac{\pi}{2} \sqrt{n}\, \cos^n \frac{1}{\sqrt[3]{n}}\\
&\le \frac{\pi}{2} \sqrt{n}\,
\left(1 - \frac{1}{4\sqrt[3]{n^2}}\right)^n
\end{align*}
where we have used $\cos y \le 1 - \frac{y^2}{4}$
for all $y\in [0, \pi/2]$ (easy). Also,
$$\lim_{n\to \infty} \frac{\pi}{2} \sqrt{n}\,
\left(1 - \frac{1}{4\sqrt[3]{n^2}}\right)^n = 0.$$
Thus,
$$\lim_{n\to \infty} K_n = 0.$$
$\phantom{2}$
Finally, we have
$$\lim_{n\to \infty} I_n = \lim_{n\to \infty} J_n + \lim_{n\to \infty} K_n = \sqrt{\pi/2}.$$
|
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|
Why can't I use trig substitution for this integral? $
\int \frac x {\sqrt {1-x^2}}dx
$
I was attempting to solve this integral, and it would appear the solution to it is $-\sqrt{1 -x^2}+C$. When I attempted to solve it, however, I attempted to let $x = \sin\theta$, making $dx=\cos{\theta}d{\theta}$
$
\int \frac {\sin\theta} {\cos^2\theta}\cos\theta{d\theta} = \int \tan \theta d\theta = \ln|\sec\theta| + C = \ln{\frac 1 {\sqrt {1 - x^2}}}+C = -{\frac 1 2}\ln|1-x^2|+C
$
I don't quite understand why this is incorrect. Now, I do understand that what I had to do to get the correct solution is to let $u=\sqrt {1-x^2}$, and everything else works out. But can someone explain where I went wrong with my attempt?
|
Method 1
Let $x=\sin \theta$, where $-\frac{\pi}{2} \leqslant \theta \leqslant \frac{\pi}{2}, $ then
\begin{aligned}
I &=\int \frac{\cos \theta d \theta}{\sqrt{1-\sin ^{2} \theta}} \\
&=\int \frac{\sin \theta \cos \theta d \theta}{\sqrt{\cos ^{2} \theta}}
\end{aligned}
$\text {Since } \cos \theta \geqslant 0 \text { for }-\frac{\pi}{2} \leqslant \theta \leqslant \frac{\pi}{2}$,$
\textrm{ therefore }\sqrt{\cos ^{2} \theta}=|\cos \theta|=\cos \theta.$
\begin{aligned}
I &=\int \frac{\sin \theta \cos \theta d \theta}{\cos \theta} \\
&=\int \sin \theta d \theta \\
&=-\cos \theta+C_1\\
&=-\sqrt{1-\sin ^{2} \theta}+C_1. \\
&=-\sqrt{1-x^{2}}+C_1.
\end{aligned}
Method 2
Let $y=x^{2}$, fhen $d y=2 x d x.$
$$
\begin{aligned}
I &=\frac{1}{2} \int \frac{1}{\sqrt{1-y}} d y \\
&=-\sqrt{1-y}+C_2\\&=-\sqrt{1-x^{2}}+C_2
\end{aligned}
$$
Method 3
Let $y=\sqrt{1-x^{2}}$, fin $y^{2}=1-x^{2}$ and $y d y=-x d x.$
$$
\begin{aligned}
I &=-\int \frac{y d y}{y} \\
&=-\int 1 d y \\
&=-y+C_{3} \\
&=-\sqrt{1-x^{2}}+C_{3}
\end{aligned}
$$
Wish it helps.
|
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Does there exists another approach to solve for the product of such expression? Problem: if the real roots of $x^3-3x+1$ are $\alpha , \beta $ and $\gamma,$ then what is the value of cyclic $(\alpha^2-\gamma)\;?$
Here is my approach, using trigonometry. Is my work correct?
|
The problem can be brute-forced algebraically, but the calculations involve polynomial resultants and are laborious to do by hand, though easily computed using a CAS.
Let $\,x_1=\alpha, x_2=\beta, x_3=\gamma\,$, then the polynomial in $\,y\,$ with roots $\,x_i^2\,$ is:
$$\text{res}(x^3-3x+1, y-x^2, x)=y^3 - 6 y^2 + 9 y - 1 \tag{1}$$
The polynomial in $\,z,y\,$ with roots $\,x_i^2-x_j = y_i-x_j\,$ is :
$$
\text{res}(x^3-3x+1, z-y+x, x) = -y^3 + 3 y^2 z - 3 y z^2 + 3 y + z^3 - 3 z - 1 \tag{2}
$$
Then, the polynomial in $\,z\,$ alone is found by eliminating $\,y\,$ between $(1)$ and $(2)$, which gives:
$$
\begin{align}
& \text{res}(y^3 - 6 y^2 + 9 y - 1, -y^3 + 3 y^2 z - 3 y z^2 + 3 y + z^3 - 3 z - 1, y)
\\ =\, &(z - 2)^3 (z^3 - 6 z^2 + 3 z + 19) (z^3 - 6 z^2 + 3 z + 1) \tag{3}
\end{align}
$$
Three of the $\,z\,$ roots are $\,x_i^2-x_i\,$, which are the roots of:
$$
\text{res}(x^3-3x+1, z-x^2+x, x) = z^3 - 6 z^2 + 3 z + 1 \tag{4}
$$
This corresponds to the last factor in $\,(3)\,$, which leaves the two possibilities $(\dagger)\,$:
*
*$(z-2)^3 = 0\,$ with the triple root $\,2\,$ so $\,z_1z_2z_3=8\,$, which is OP's solution;
*$z^3 - 6 z^2 + 3 z + 19\,$ with the product of the roots $\,z_1z_2z_3=-19\,$, which is the second solution linked in my comment.
$(\dagger)\;$ The relevant sextic in $\,(3)\,$ can be factored into two cubics in $\,\binom{6}{3}\,$ ways, but we know that $\,(x_1^2-x_2)+(x_2^2-x_3)+(x_3^2-x_1)=6\,$ and the only factorization where the coefficient of $\,x^2\,$ is $\,-6\,$ in a cubic factor is where the factors are chosen to be $\,(z-2)^3\,$ and $\,z^3 - 6 z^2 + 3 z + 19\,$.
|
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|
$f(x)$ and $g(x)$ are monic cubic polynomials, with $f(x)-g(x)=r$. If $f$ has roots $r+1$ and $r+7$, and $g$ has roots $r+3$ and $r+9$, then find $r$.
Let $f(x)$ and $g(x)$ be two monic cubic polynomials, and let $r$ be a real number. Two of the roots of $f(x)$ are $r+1$ and $r+7$. Two of the roots of $g(x)$ are $r + 3$ and $r + 9,$ and$$f(x) - g(x) = r$$for all real numbers $x.$ Find $r.$
So far, I have $$f(x)=(x-r-1)(x-r-7)(x-p)$$ and $$g(x)=(x-r-3)(x-r-9)(x-q).$$ From $f(x)-g(x)=r$, I know that their constant terms differ by $r$. I expanded the two functions but it was too complicated. I also plugged in $x=r+1,r+7,r+3,r+9$ into $f(x)-g(x)=r$, but it didn't do much.
Thanks in advance!!!!!
|
Your approach is fine. Note that\begin{multline}f(x)-g(x)=\\=(4-p+q)x^2+(2 r p+8p-12 q-2 q r-4 r-20)x-p r^2+q r^2-8 p r+12 q r-7 p+27 q.\end{multline}So, $4-p+q=0$; in other words, $p=q+4$. Replacing $p$ with $q+4$ in the coefficient of $x$ in $f(x)-g(x)$, we get that $4(3-q+r)=0$; in other words, $q=r+3$. And if replace $q$ with $r+3$ in the constant term of $f(x)-g(x)$, we get $32$. But we want this to be equal to $r$. Therefore, $r=32$.
|
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Is there an elementary method to evaluate the indefinite integral $\int \frac{1}{1+\sin ^{6} x} d x?$ Inspired by the post, I want to increase the power of $\sin x$ by $2$ to $6$,
$$
I=\int\frac{1}{1+\sin ^{6} x} d x.
$$
As usual, we multiply both the numerator and denominator by $\sec^6 x$ and get
$$
\begin{aligned}
I &=\int \frac{\sec ^{6} x}{\sec ^{6} x+\tan ^{6} x} d x \\
& \stackrel{t=\tan x}{=} \int\frac{\left(1+t^{2}\right)^{2} d t}{\left(1+t^{2}\right)^{3}+t^{6}}
\end{aligned}
$$
Factorizing the denominator yields $$
I=\int \frac{1+2 t^{2}+t^{4}}{\left(1+t^{2}+t^{4}\right)\left(2 t^{2}+1\right)} d t
$$
Resolving the rational function $$
\frac{1+2 x+x^{2}}{\left(1+x+x^{2}\right)(2 x+1)}=\frac{1}{3(2 x+1)}+\frac{x+2}{3\left(x^{2}+x+1\right)},
$$
into partial fractions yields $$
\int\frac{1+2 t^{2}+t^{4}}{(1 +t^{2}+t^{4})(2t^2+1)} d t=\frac{1}{3} \left(\underbrace{\int\frac{d t}{2 t^{2}+1}}_{K} +\underbrace{\int\frac{t^{2}+2}{t^{4}+t^{2}+1} d t}_{J}\right)
$$
$$
K=\int \frac{d t}{2 t^{2}+1}=\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} t) +C_1 =\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+C_1
$$
We now focus on evaluating the last integral.
\begin{aligned}
J &=\int \frac{t^{2}+2}{t^{4}+t^{2}+1} d t \\
&=\int \frac{1+\frac{2}{t^{2}}}{t^{2}+\frac{1}{t^{2}}+1} d t \\
&=\int \frac{\frac{3}{2}\left(1+\frac{1}{t^{2}}\right)-\frac{1}{2}\left(1-\frac{1}{t^{2}}\right)}{t^{2}+\frac{1}{t^{2}}+1} d t \\
&=\frac{3}{2} \int \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^{2}+3}-\frac{1}{2} \int \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^{2}-1} \\
&=\frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt{3}}\right)+\frac{1}{4 \sqrt{2}} \ln \left|\frac{t+\frac{1}{t}+1}{t+\frac{1}{t}-1}\right|+C \\
&=\frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)-\frac{1}{4 \sqrt{2}} \ln \left|\frac{\tan ^{3} x+\tan x+1}{\tan ^{2} x-\tan x+1}\right|+C_2
\end{aligned}
Now we can conclude that
$$
\begin{aligned}
I=& \frac{1}{3}\left[\frac{1}{\sqrt{2}} \tan ^{-1}(\sqrt{2} \tan x)+\frac{\sqrt{3}}{2} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)\right.\left.-\frac{1}{4 \sqrt{2}} \ln \left|\frac{\tan ^{2} x+\tan x+1}{\tan ^{2} x-\tan x+1}\right|\right]+C
\end{aligned}
$$
My Question
Can we go further with $n\geq 8$,
$$
I_n= \int\frac{1}{1+\sin ^{n} x} d x?
$$
|
Answering by mistake the linked post instead of your, my shortest result for $n=4$ is
$$\int \frac {dx}{1+\sin^4(x)}=\frac{\tan ^{-1}\left(\tan (x)\sqrt{1-i} \right)}{2 \sqrt{1-i}}+\frac{\tan
^{-1}\left(\tan (x)\sqrt{1+i} \right)}{2 \sqrt{1+i}}$$
For $n=6$, doing the same
$$\frac {1}{1+\sin^6(x)}=\frac{2}{(a+1) (b+1) (3-\cos (2 x))}-\frac{2}{(a+1) (a-b) (2 a-1+\cos (2 x))}+$$ $$\frac{2}{(b+1) (a-b) (2 b-1+\cos (2 x))}$$ where $(a,b)$ are the complex roots of $x^2-x+1=0$.
Remember that
$$\int \frac {dx}{\cos(2x)+k}=-\frac{1}{\sqrt{1-k^2}}\tanh ^{-1}\left(\frac{(k-1) }{\sqrt{1-k^2}}\tan (x)\right) $$ I suppose that we could continue in this spirit assuming that we know the roots of $x^n+1=0$
|
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|
Factorial Number System with Repetitions Good Day
I know about how Factorial Number System is useful to find permutations and also find the the index of a permutation.
For example, permutation $[1, 3, 4, 0, 2]$ has index ($0$ - based) $1 \cdot 4! + 2 \cdot 3! + 2 \cdot 2! + 0 \cdot 1! + 0 \cdot 1! = 40$ where the constant is the index ($0$ - based) of the number in the unused numbers.
However, what if the arrangement contains repetitions. Consider $[0, 0, 1, 1, 2, 2]$. Now arrangement $[2, 2, 1, 1, 0, 0]$ should have index $\frac{6!}{2! \cdot 2! \cdot 2!} - 1 = 89$, but if I go the usual way and consider the index in only the distinct unused numbers, I get $2 \cdot 5! + 2 \cdot 4! + \cdots$ which is already larger than $89$. Even doing $2 \cdot \frac{120}{2 \cdot 2} + 2 \cdot \frac{24}{2 \cdot 2} + 1 \cdot \frac{6}{2} + 1 \cdot \frac{2}{2}$ gives $76$.
Thus, is there a simple way to extend this system for repetitions, so that given an array of possibly non-distinct numbers, I can find the lexicographic index of the arrangement and from the (lexicographic) index, find the arrangement without casework?
Thanks
|
The index of a binary string can be found using the combinatorial number system.
For example, $N=100=1100100_2$ can be considered as an element of the k-combination set $\binom{7}{3}$.
As such it can be written $\{6,5,2\}$, where each element is a set bit in $N$. We then have $100=2^6+2^5+2^2$.
Its (zero-based) index is then given by the formula:
$$\binom{6}{3}+\binom{5}{2}+\binom{2}{1}=20+10+2=32$$
(There are $\binom{7}{3}=35$ in total.)
It works by examining each bit separately. For example, the formula states that there are exactly $\binom{6}{3}$ 3-combinations before the $6$, because $0111000 _2< 1000000_2$. Similarly there are $\binom{5}{2}$ 2-combinations before $0100000$, and so on. Therefore every $N$ has a unique representation.
For a permutation with repetition, a similar process can be used.
For example, consider $[3,2,2,0,3,1,3]$. We first break this into its components for each digit:
*
*$[3,-,-,-,3,-,3]$
*$[-,2,2,-,-,-,-]$
*$[-,-,-,-,-,1,-]$
*$[-,-,-,0,-,-,-]$
We follow a similar process as with the binary string for each component, and multiply by a positioning factor which is determined by its associated multinomial.
For example, the $3$ row becomes $\binom{6}{3}+\binom{2}{2}+\binom{0}{1}=20+1+0=21$.
The total number of permutations is $\frac{7!}{3!2!1!1!}=840$. The size of the 3-block is $\binom{7}{3}=35$, and so $\frac{840}{35}=24$ gives the multiplying factor.
The $2$ row is $\binom{3}{2}+\binom{2}{1}=5$ because it is nested into the 3-block. The multiplying factor is $\frac{24}{\binom{4}{2}}=4$.
After working out the last two rows, the index is:
$$21\cdot24+5\cdot4+0\cdot2+0\cdot1=524$$.
|
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|
Finding the maximum of $ \sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x}$ for $0\le x \le 13$ using the Cauchy-Schwarz inequality gives two different answers I am trying to find the maximum of the expression $$\sqrt{x+27} + \sqrt{13-x} + \sqrt{x} \qquad \text{for } 0 \le x \le 13.$$
Clearly using Cauchy is the way to go here, so I tried $$((x + 27) + 2(13-x) + x)(1 + 0.5 + 1) = (x + 27 + 26 - 2x + x)\frac 52 = \frac{265}{2} \ge (\sqrt{x+27} + \sqrt{13-x} + \sqrt{x})^2.$$
But then, we also have that $$\left( 1 + \frac{1}{3} + \frac{1}{2} \right) ((x + 27) + 3(13 - x) + 2x) = 121 \ge (\sqrt{x + 27} + \sqrt{13 - x} + \sqrt{x})^2.$$
Clearly, these are two different values. What gives? How could I “guess” the correct combination of scalars that lead to the correct answer? (as the first attempt is incorrect).
|
Your approach is fine. You start by writing $$(\sqrt{x+27}+\sqrt{13-x}+\sqrt{x})^2\le[(x+27)+a(13-x)+bx]\left(1+\frac1a+\frac1b\right)$$
The factors $a$ and $b$ are to be determined. Since you want the right hand side to be an absolute maximum, that does not depend on $x$, you have
$$1-a+b=0$$
You also know that the Cauchy inequality transforms into an equality if the elements of the two sums on the right hand side are proportional:
$$\frac{x+27}1=\frac{a(13-x)}{1/a}=\frac{bx}{1/b}$$
We can simplify this:
$$x+27=a^2(13-x)\\x+27=b^2x$$
Now find $x$ in terms of $b$, plug it into the equation for $a$ in terms of $x$
$$27=(b^2-1)x\\a^2=\frac{x+27}{13-x}=\frac{27/(b^2-1)+27}{13-27/(b^2-1)}=\frac{27(b^2-1)+27}{13(b^2-1)-27}=\frac{27b^2}{13b^2-40}$$
Finally use the first condition (maximum not dependent on $x$):
$$a^2=1+2b+b^2=\frac{27b^2}{13b^2-40}\\13b^4+26b^3-54b^2-80b-40=0$$
From the rational root theorem, you can see that $b=2$ is a good choice. Then $a=3$
|
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|
How does $\frac{k^2(k+1)^2}{4} + (k+1)^3$ become $\frac{(k+1)^2(k+2)^2}{4}$? As part of an induction proof, the authors of a beginner combinatorics text reduce/factor a polynomial as follows, but do not show the minutiae of their algebraic steps:
$$\frac{k^2(k+1)^2}{4} + (k+1)^3 \tag1$$
manipulated to:
$$(k+1)^2\left(\frac{k^2}{4} + k + 1\right) \tag2$$
manipulated to:
$$\frac{(k+1)^2}{4}(k^2 + 4k + 4) \tag3$$
manipulated to:
$$\frac{(k+1)^2(k+2)^2}{4} \tag4$$
...which was the desired result.
I fear I'm missing some sort of basic algebra required to understand what factoring and reduction steps the authors took to produce the results above. Would anyone be kind enough to explain what algebraic rules/properties/factorization allowed the authors to produce the results as shown? I've been struggling to work it out on my own.
|
For (1), we can separate the $\frac{k^2}{4}$ from $(k+1)^2$ and move out a factor of $k+1$ from $(k+1)^3$, giving us
$$\frac{k^2}{4}(k+1)^2+(k+1)(k+1)^2$$
We can then collect like terms (specifically, the $(k+1)^2$ terms), giving us
$$(k+1)^2\left(\frac{k^2}{4}+k+1\right)$$
Which is (2).
From there, we can divide and multiply by $4$, giving us
$$(k+1)^2\cdot\frac{4}{4}\left(\frac{k^2}{4}+k+1\right)$$
Rearranging slightly gives
$$\frac{(k+1)^2}{4}\cdot4\left(\frac{k^2}{4}+k+1\right)$$
Applying the distributive law gives
$$\frac{(k+1)^2}{4}\left(k^2+4k+4\right)$$
Now we note that $k^2+4k+4=(k+2)^2$ (you can verify this yourself by expanding the RHS), thus we have the end result
$$\frac{k^2(k+1)^2}{4}+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4}$$
|
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|
What´s the length of the segment AC in the triangle below? For reference: In the triangle $\angle A$ is right and $D$ is a point on the side $AC$ such that the segments $BD$ and $DC$ have length equal to $1 m$. Let $F$ be the point on the side $BC$ so that $AF$ is perpendicular to $BC$. If the segment $FC$ measures $1m$, determine the length of $AC$.(Answer$:\sqrt[3]{2})$
My progress:
$ AF = h\\
a=m+1\\
h^2 =m.FC = m.1 = m\\
\triangle AFC: AC^2 =b= h^2+FC^2 \implies b = \sqrt{m+1}\\
\triangle ABC: a^2 = b^2+ c^2 \implies (m+1)^2=c^2+(\sqrt{m+1})^2\\
\therefore c = \sqrt{m^2+m}\\
\triangle BAD: BD^2=AD^2+c^2\implies 1 = (\sqrt{m+1}-1)^2+m^2+m\\
\therefore: m^2+2m-2\sqrt{m+1}+1=0$
...????
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Let $E$ middle of $BC$. Then triangles DEC, AFC and BAC are similar. Then $$\frac{CD}{EC}=\frac{AC}{FC}=\frac{BC}{AC}=k$$
$$EC=\frac{CD}{k}=\frac{1}{k}, AC=k FC=k, BC=k AC=k^2$$
$$BC=2 EC \Rightarrow k^2 = \frac{2}{k} \Rightarrow k=\sqrt[3]{2}$$
$$AC=k=\sqrt[3]{2}$$
|
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|
Finding second derivative of $x^{x^x}$ If $f(x) = x^{x^x} $, then find $f''(1)$.
My attempt:Now,
$\begin{align}y =x^{x^x} \\&\implies ln(y) = x^x \cdot ln(x)
\\&\implies ln(ln(y))= x\cdot ln(x)+ln(ln(x))\\&\implies\frac{y'}{y\cdot ln(y)}
=ln(x) +1+\frac{1}{x\cdot ln(x)}\\&\implies y'=(x^{x^x})\cdot (x^xln(x))\cdot [ln(x) +1+\frac{1}{xln(x)}]
\end{align}$
I find it really tedious to continue in this way to reach desired answer.I have already checked finding second derivative of $x^x$. So, I wonder if I have to continue in the similar manner to get to answer or there is any better method to calculate this?
|
Proceeding along your approach
Let $y = x^{x^x}$. We have $y(1) = 1$.
We have
$$y' = y x^x (\ln^2 x + \ln x + x^{-1}).$$
We have $y'(1) = 1$.
We have
\begin{align*}
y'' &= y' x^x (\ln^2 x + \ln x + x^{-1}) + y \cdot x^x(\ln x + 1) \cdot (\ln^2 x + \ln x + x^{-1})\\
&\qquad + y x^x (2x^{-1}\ln x + x^{-1} - x^{-2} ).
\end{align*}
Letting $x = 1$, using $y(1) = 1$ and $y'(1)=1$, we have
$y''(1) = 2$, i.e. $f''(1) = 2$.
|
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|
Proving that $-\ln(x-\sqrt{x^2-1})=\ln(x+\sqrt{x^2-1})$ Show $-\ln(x-\sqrt{x^2-1})=ln(x+\sqrt{x^2-1})$
Can somebody verify that my solution is correct (or incorrect!) and also show me alternative methods to do this? Thanks!
Let $r=-\ln(x-\sqrt{x^2-1})$ and $s=ln(x+\sqrt{x^2-1})$
$\rightarrow e^r=(x-\sqrt{x^2-1})^{-1}$ and $e^s = x+\sqrt{x^2-1}$
$\rightarrow \frac{e^s}{e^r} = (x+\sqrt{x^2-1})(x-\sqrt{x^2-1}) = x^2 - (x^2 - 1) = 1$
Thus $e^r=e^s$ and so $r=s$
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Your proof looks fine to me. You can also do this: $\ln(x-\sqrt{x^2-1})+\ln(x+\sqrt{x^2-1})=\ln\left((x-\sqrt{x^2-1})(x+\sqrt{x^2-1})\right)=\ln 1 =0$
|
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|
Examples where $(a+b+\cdots)^2 = (a^2+b^2+\cdots)$ Consider the two infinite series
$$
\frac{\pi}{\sqrt{8}} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots
$$
and
$$
\frac{\pi^2}{8} = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{7^2} + \frac{1}{9^2} + \frac{1}{11^2} + \cdots
$$
(Notice that the first series has signs that go two-by-two rather than every-other.)
Squaring the first equality also gives $\pi^2/8$ and so these two, when put together, satisfy the 'highschooler's dream' for squaring a sum: just square each term and sum,
$$
(a + b + c + \cdots)^2 = (a^2 + b^2 + c^2 + \cdots)
$$
with nothing like $2ab + 2ac + 2bc + \cdots$ needed.
A trivial example of this would be
$$
(a + 0)^2 = a^2 + 2a0 + 0^2 = a^2 + 0^2
$$
but it only succeeds because one addend is zero.
My questions are
*
*Are there any other simple nontrivial examples? I believe any other nontrivial example must be an infinite sum. edit: John Omielan provides the simple finite example $(1+1-\frac{1}{2})^2 = 1^2 + 1^2 + \frac{1}{2^2}$.
*Is there an "obvious" demonstration that the above sum (other than the direct evaluation) satisfies the highschooler's dream? Put another way, is there a simple demonstration that the infinite sum of "cross terms" vanishes?
|
It may qualify as trivial, but note that there are infinitely many such series.
Consider a divergent series with positive decreasing terms $\sum_{n=1}^\infty u_n$, such that $\sum_{n=1}^\infty u_n^2$ is convergent to $L$, then it's a known theorem that we can change the signs of $u_n$ so as to get a series convergent to any given real number. Just pick $\sqrt{L}$.
That is, there exist $\sigma_n\in\{+1,-1\}$ such that $\sqrt{L}=\sum_{n=1}^\infty \sigma_nu_n$ while $L=\sum_{n=1}^\infty (\sigma_nu_n)^2$.
The idea of the proof of the aforementioned theorem, assuming $x>0$ (otherwise start with negative terms): sum $u_n$ until you get a value greater than $x$, call it $v_1$. Then sum $-u_n$ until you get a value smaller than $x$, call the sum of the new terms $v_2$, and continue this process to define $v_k$ for all $k$. Then $\sum_k v_k$ is alternating and convergent to $x$.
|
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|
Prove that $\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$
Prove that
$$J_n=\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\frac{1}{n-1}2^\frac{n-1}{2}\sin\frac{(n-1)\pi}{4}$$
For $n=2$, it is OK. For general $n$, it seems impossible by integration by parts. Any other method?
When I calculate an integral $I_n=\int_0^\frac{\pi}{4}\frac{\cos nx}{\cos^nx}dx$, we find $I_n/2^{n-1}-I_{n-1}/2^{n-1}=-1/2^{n-1}J_n$. So we need to find the $J_n$, as the problem states.
The proof of $I_n/2^{n-1}-I_{n-1}/2^{n-1}=-1/2^{n-1}J_n$ is as follows.
\begin{align}
I_n/2^{n-1}-I_{n-1}/2^{n-1}
&=\frac{1}{2^{n-1}} \int_0^\frac{\pi}{4}\left(\frac{\cos nx}{\cos^nx}-\frac{2\cos(n-1)x}{\cos^{n-1}x}\right)dx\\
&=\frac{1}{2^{n-1}}\int_0^\frac{\pi}{4}\frac{\cos[(n-1)x+x]-2\cos(n-1)x\cos x}{\cos^nx}dx\\
&=-\frac{1}{2^{n-1}}\int_0^\frac{\pi}{4}\frac{\cos(n-2)x}{\cos^nx}dx
=-1/2^{n-1}J_n
\end{align}
|
$$J_n=\int_0^\frac{\pi}{4}\frac{\cos (n-2)x}{\cos^nx}dx=\Re \,2^n\int_0^\frac{\pi}{4}\frac{e^{i(n-2)x}}{(e^{ix}+e^{-ix})^n}dx=\Re \,2^{n-1}\int_0^\frac{\pi}{2}\frac{e^{-it}}{(1+e^{-it})^n}dt$$
$$=\Re \,2^{n-1}(-i)\int_{-\frac{\pi}{2}}^0\frac{e^{it}}{(1+e^{it})^n}\,i\,dt$$
Consider the contour in the comples plane: from $0$ to $-i$ (along the axis $Y$), then - along a quarter circle - from $-i$ to $1$, and then from $1$ to $0$ along the axis $X$. There are no singularities inside this closed contour, therefore
$$\oint\frac{dz}{(1+z)^n}=0$$
On the other hand,
$$\oint\frac{z}{(1+z)^n}dz=\int_{-\frac{\pi}{2}}^0\frac{e^{i\phi}}{(1+e^{i\phi})^n}\,i\,d\phi+\int_1^0\frac{dz}{(1+z)^n}+\int_0^{-i}\frac{dz}{(1+z)^n}=0$$
$$\int_{-\frac{\pi}{2}}^0\frac{e^{i\phi}}{(1+e^{i\phi})^n}\,i\,d\phi=\int_0^1\frac{dz}{(1+z)^n}-\int_0^{-i}\frac{dz}{(1+z)^n}$$
Therefore,
$$J_n=\Re \,2^{n-1}(-i)\bigg(\int_0^1\frac{dz}{(1+z)^n}-\int_0^{-i}\frac{dz}{(1+z)^n}\bigg)$$
$$=\Re \,2^{n-1}\frac{i}{n-1}\Big(1-\frac{1}{(1-i)^{n-1}}\Big)=-\Re \,2^{n-1}\frac{i}{n-1}\frac{1}{(\sqrt2)^{n-1}}\frac{1}{e^{-\frac{\pi i(n-1)}{4}}}$$
$$J_n=\frac{2^\frac{n-1}{2}}{n-1}\sin\frac{\pi (n-1)}{4}$$
|
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|
$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}}\text{ for }n\ge 2$ I'm completely lost on this question:
$$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}} \text{ for }n\ge 2$$
alternatively,
$$\prod_{a=2}^{n} {\left(1-\frac{1}{\sqrt{a}}\right)} \lt \frac{2}{n^2} \text{ for } n\geq2$$
I tried to do it by induction but ended up with $\frac{2}{k^2}(1-\frac{1}{\sqrt{k+1}})$ in the RHS which I have no idea what to do with. I also tried to do it with AM/GM inequality but also had little luck. The question did not specifically specify what type of proof to use so I'm assuming you can use any type of proof. The first that came to mind is induction obviously, but I'm really lost on this one. Any help or hints would be appreciated!
With induction, the thing that I'll need to prove is:
$$\frac{2}{k^2}(1-\frac{1}{\sqrt{k+1}}) \lt \frac{2}{(1+k)^{2}}$$
But I'm not sure how to approach it.
SOLVED
$$\text{Prove}\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{n}} \right)\lt \frac{2}{n^{2}} \text{ for }n\ge 2$$
Prove by induction:
A. When $n=2$,
$$\begin{aligned}LHS &= 1-\frac{1}{\sqrt{2}}\newline
&\approx 0.293\newline
RHS&=\frac{2}{2^2}\newline
&=0.5\newline
&\gt LHS\newline
\therefore \text{true for }n&=2\end{aligned}$$
B. Assume the statement is true for the positive integer $n=k$, where $k\in\mathbb{N}$.
That is, assume that
$$\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{k}} \right)\lt \frac{2}{k^{2}}$$
Now prove the statement for $n=k+1$. That is, prove that
$$\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{k+1}} \right)\lt \frac{2}{(k+1)^{2}}$$
By the induction hypothesis:
$$\left( 1-\frac{1}{\sqrt{2}} \right)\left( 1-\frac{1}{\sqrt{3}} \right)...\left( 1-\frac{1}{\sqrt{k}} \right)\left( 1-\frac{1}{\sqrt{k+1}} \right)\lt \frac{2}{k^{2}}\left( 1-\frac{1}{\sqrt{k+1}} \right)$$
Hence, we are required to prove for $k\geq2$ that $$\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)\leq\frac{2}{(k+1)^2}.$$
For $k\geq2,$ \begin{aligned}&\text{LHS-RHS}\\
=&\frac{2}{k^2}\Big(1-\frac{1}{\sqrt{k+1}}\Big)-\frac{2}{(k+1)^2}\\
=&\frac{2\sqrt{k+1}}{1+\sqrt{k+1}}\left(\frac{1}{k^2}\left(1-\frac{1}{\sqrt{k+1}}\right)\left(1+\frac{1}{\sqrt{k+1}}\right)-\frac{1}{(k+1)^2}\left(1+\frac{1}{\sqrt{k+1}}\right)\right)\\
=&\cdots\\ =&\frac{2\sqrt{k+1}}{1+\sqrt{k+1}}\left(\frac{1}{k}-\frac{1}{\sqrt{k+1}}\right)\\
=&\frac{2(\sqrt{k+1}-k)}{k(1+\sqrt{k+1})} \\ \lt&0,\end{aligned} as required.
$\therefore$ It follows from parts A and B by mathematical induction that the result is true for all integers $n\gt2$
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Hint: The induction step is just $\frac{2}{k^2}(1-\frac{1}{\sqrt{k+1}})\leq\frac{2}{(k+1)^2}$, which is equivalent to
$$
\frac{1}{k^2}\frac{k}{k+1}\leq\frac{1}{(k+1)^2}\Big(1+\frac{1}{\sqrt{k+1}}\Big)\iff\frac{k+1}{k}=1+\frac{1}{k}\leq 1+\frac{1}{\sqrt{k+1}}
$$
which boils down to $k\geq\sqrt{k+1}$, for integers $k\geq 2$.
|
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|
Sum of series $\sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{6}$ I first derived the Fourier series for $f(x) = abs(x)$ to be:
$\frac{\pi}{2} - \frac{4}{\pi} \sum_{k=1}^{\infty} \frac{cos((2k-1)x)}{(2k-1)^2}$
Now, I want to use this at x=0 to find $\sum_{k=1}^{\infty} \frac{1}{k^2}$. This is my work:
$abs(x) = 0 = \frac{\pi}{2}- \frac{4}{\pi} \sum_{k=1}^{\infty} \frac{1}{(2k-1)^2}$
$-\frac{\pi}{2}=-\frac{4}{\pi}\sum_{k=1}^{\infty} \frac{1}{(2k-1)^2}$
$\sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} = \frac{\pi^2}{8}$
I know the answer should be $\frac{\pi^2}{6}$, not $\frac{\pi^2}{8}$, but I am not sure where I am going wrong. Is my Fourier series incorrect?
Edit: How can I go from $\sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} = \frac{\pi^2}{8}$
to $\sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{6}$?
|
Let $A=\sum_{k=1}^{\infty}{\frac{1}{k^2}}$. Then we can see that $$\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\cdots=\sum_{k=1}^{\infty}{\frac{1}{(2k)^2}}=\frac{1}{4}\sum_{k=1}^{\infty}{\frac{1}{k^2}}=\frac{A}{4}.$$ It then follows that $$\frac{\pi^2}{8}=\sum_{k=1}^{\infty}{\frac{1}{(2k-1)^2}}=\sum_{k=1}^{\infty}{\frac{1}{k^2}}-\sum_{k=1}^{\infty}{\frac{1}{(2k)^2}}=A-\frac{A}{4}=\frac{3}{4}A,$$ so $A=\pi^2/6$.
|
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|
Finding $\int_C \frac{8z^{11}-3z^6+1}{z^7-1}$ Find $\int_C \frac{8z^{11}-3z^6+1}{z^7-1}$ where $C$ is the positively oriented circle $|z| =4$
How should I go about finding this? We know that we have simple poles at the seventh roots of unity, i.e. $z = e^{\frac{n \pi i}{7}}$ and we also note that these poles are interior to $C: |z| =4$.
Attempt.
$$\frac{8z^{11}-3z^6+1}{z^7-1} = (8z^{11}-3z^6+1) \cdot \Big( - \frac{1}{1-z^7} \Big)$$
$$ = (8z^{11}-3z^6+1) \cdot \Big( - \displaystyle\sum_{n=0}^\infty (z^7)^n\Big) \hspace{0.5cm} |z| <1$$
$$ = (8z^{11}-3z^6+1) \cdot \Big( -1 -z^7-z^{14} + \dots\Big) $$
But the two issues I run into here is that the geometric series is only valid for $|z|<1$ whereas we have $|z| =4$ and also I'm not seeing a $\frac{a_{-1}}{z}$ term popping out of this expansion, and I highly doubt the residue is zero. Any advice on what I should do instead?
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It is easier to compute this integral using one resudue at infinity, due all other singularities are in interior of $C$.
$\int\limits_{|z|=4} \frac{8z^{11}-3z^6+1}{z^7-1} dz = -2 \pi i \mathop{\mathrm{Res}}_{z=\infty} \frac{8z^{11}-3z^6+1}{z^7-1} = (1)$
Using $\mathop{\mathrm{Res}}_{z=\infty} f(z) = -\mathop{\mathrm{Res}}_{z=0} \frac{1}{z^2} f\left(\frac{1}{z}\right)$ we obtain $\mathop{\mathrm{Res}}_{z=\infty} \frac{8z^{11}-3z^6+1}{z^7-1} = -\mathop{\mathrm{Res}}_{z=0}\frac{z^{11}-3 z^5+8}{z^6-z^{13}}$.
$\frac{z^{11}-3 z^5+8}{z^6-z^{13}} = z^{-6}(z^{11}-3 z^5+8)(1-z^{7})^{-1} = z^{-6}(z^{11}-3 z^5+8) \left(1 + \sum\limits_{n=1}^{\infty} z^{7 n}\right)$.
So, it is obviously that $\mathop{\mathrm{Res}}_{z=0}\frac{z^{11}-3 z^5+8}{z^6-z^{13}} = -3$. Then we have
$(1) = -2 \pi i \times (-(-3)) = -6 \pi i$.
|
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.