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Prove $\displaystyle\frac{H(x^2)}{H(x)}$ increases. For $x\in[0,1]$, let $f(x):=-x\ln x$ and the two-sample entropy function $H(x)=f(x)+f(1-x)$. Prove $h(x):=\displaystyle\frac{H(x^2)}{H(x)}$ increases.
Here is my proof which is a bit cumbersome. I am seeking a much more elegant approach.
The numerator of the derivative of the sought fraction is
\begin{align}
&H(x)^2\frac{dh(x)}{dx} \\
=&\frac d{dx}H(x^2)H(x)-H(x^2)\frac d{dx}H(x) \\
=& 2x\ln\frac{x^2}{1-x^2}\,\big(x\ln x+(1-x)\ln(1-x)\big)-\big(x^2\ln x^2+(1-x^2)\ln(1-x^2)\big)\ln\frac x{1-x} \\
=& 2x^2\ln^2 x+2x(2-x)\ln x\ln(1-x)-(x^2+1)\ln x\ln(1-x^2)+(1-x)^2\ln(1-x)\ln(1-x^2). \tag1\label1
\end{align}
All four terms above except the third are positive. I combine the second and the third term together and divide it by $-\ln x$ which is positive, and get
\begin{align}
g(x):&=-2x(2-x)\ln(1-x)+(x^2+1)\ln(1-x^2) \\
&=(3x-1)(x-1)\ln(1-x)+(x^2+1)\ln(1+x) \tag2\label2 \\
&= \int_0^x \Big(g''(a)-\int_t^a g'''(s)ds\Big)(x-t)dt
\end{align}
for some $a\in[0,x]$. So we only need to show $g(x)>0, \forall x\in\big(0,\frac13\big]$.
$$\frac{d^3g(x)}{dx^3}= \frac{4x(2x^3 +3x^2-2x-7)}{(1-x)^2(1+x)^3}.$$
Let $p(x):=2x^3+3x^2-2x-7$. It can be shown that $p(x)\le p(1)=-4, \forall x\in[0,1]$. We can take $a=\frac13$ since we can show, with a bit of work, $f''(\frac13)>0$.
(to be continued)
| Partial Hints :
We broke the fraction into two part :
Show that :
$$f(x)=x^{-1}\left(-x^{2}\ln\left(x^{2}\right)-\left(1-x^{2}\right)\ln\left(1-x^{2}\right)\right)$$
Is increasing on $[0,2/5]$
And :
$$g(x)=x^{-1}\left(-x\ln\left(x\right)-\left(1-x\right)\ln\left(1-x\right)\right)$$
Is decreasing on $(0,1)$
Obviously a increasing function divided by a decreasing function both positive is increasing .
We can make the same thing on $x\in[2/5,3/5]$ with :
$$\frac{1}{1+x}\left(-x^{2}\ln\left(x^{2}\right)-\left(1-x^{2}\right)\ln\left(1-x^{2}\right)\right),\frac{1}{1+x}\left(-x\ln\left(x\right)-\left(1-x\right)\ln\left(1-x\right)\right)$$
Next we use symmetry to get :
$$r\left(x\right)=\frac{\left(-x^{2}\ln\left(x^{2}\right)-\left(1-x^{2}\right)\ln\left(1-x^{2}\right)\right)}{x},\frac{r\left(1-x^{2}\right)}{x},\frac{t\left(1-x^{2}\right)}{x},t\left(x\right)=\frac{-x\ln\left(x^{2}\right)-\left(1-x\right)\ln\left(\left(1-x\right)^{2}\right)}{x}$$
Currently the function $\frac{r\left(1-x^{2}\right)}{x}$ is decreasing on $[0.5,1)$ and $\frac{t\left(1-x^{2}\right)}{x}$ is increasing both positive so the fraction is decreasing remains to substitute $y=1-x^2$.
For a end see RiverLi's answer I stop here .
| {
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"url": "https://math.stackexchange.com/questions/4613282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Given the curve $y=\frac{5x}{x-3}$. Find its asymptotes, if any. Given the curve $y=\frac{5x}{x-3}$. To examine its asymptotes if any.
We are only taking rectilinear asymptotes in our consideration
My solution goes like this:
We know that, a straight line $x=a$, parallel to $y$ axis can be a vertical asymptote of a branch of the curve $y=f(x)$, iff $f(x)\to\infty$ when $x\to a+0$, $x\to a-0$ or $x\to a$. Similarly, a straight line $y=a$, parallel to $x$ axis can be a horizontal asymptote of a branch of the curve $x=\phi(y)$, iff $x\to\infty$ when $y\to a+0$, $y\to a-0$ or $y\to a$. Using these lemmas, we obtain $x=3$ and $y=5$ as the rectilinear asymptotes. This is because, $y=\frac{5x}{x-3}=f(x)$, then as $x\to 3$, $f(x)\to \infty$. Also, since $y=\frac{5x}{x-3}$, thus,$xy-3y=5x$ or $xy-5x=3y$, hence $x=\frac{3y}{y-5}=\phi(y)$. Due to which $\phi(y)\to \infty$ as $y\to 5$.
Now, I tried checking, if there is any oblique asymptote. We know that, $y=mx+c$, is an oblique asymptote of $y=f(x)$, iff $\exists $ a finite $m=\lim_{|x|\to\infty}\frac{y}{x}$ and $c=\lim_{|x|\to\infty} y-mx$. Now, we have the function $y=\frac{5x}{x-3}$ and hence, $\frac{y}{x}=\frac{5}{x-3}$ and hence, $m=\lim_{|x|\to\infty}\frac{y}{x}=\lim_{|x|\to\infty}\frac{5}{x-3}=0$. Hence, $c=\lim_{|x|\to\infty} y-mx=\frac{5x}{x-3}=\frac{5}{1-\frac 3x}=5$. So, the oblique asymptote of the given curve is $y=mx+c=5$, which is same as the horizontal asymptote, parallel to $x-axis$ as shown above. Is the solution correct? If not, where is it going wrong?
| You have correctly found the asymptotes of the function. Observe that we can express the function in the form
$$f(x) = \frac{5x}{x - 3} = \frac{5x - 15 + 12}{x - 3} = \frac{5x - 15}{x - 3} + \frac{12}{x - 3} = \frac{12}{x - 3} + 5$$
which tells us that the graph of $f$ can be obtained by dilating the graph of $y = 1/x$ by a factor of $12$, shifting it to the right by three units, and up by $5$ units. Since the curve $y = 1/x$ has horizontal asymptote $y = 0$ and vertical asymptote $x = 0$, it follows that the function $f$ has horizontal asymptote $y = 5$ and horizontal asymptote $x = 3$.
A function $f$ has horizontal asymptote $y = k$ if
$$\lim_{x \to \infty} f(x) = k$$
or
$$\lim_{x \to -\infty} f(x) = k$$
Observe that
\begin{align*}
\lim_{x \to \infty} f(x) & = \lim_{x \to \infty} \frac{4x}{x - 3} = \lim_{x \to \infty} \left(\frac{12}{x - 3} + 5\right) = 0 + 5 = 5\\
\lim_{x \to -\infty} f(x) & \lim_{x \to -\infty} \frac{4x}{x - 3} = = \lim_{x \to \infty} \left(\frac{12}{x - 3} + 5\right) = 0 + 5 = 5
\end{align*}
which tells us that the only horizontal asymptote is $y = 5$.
In fact, a rational function can have at most one horizontal asymptote. This is not true for all functions. For instance, the function $g: \mathbb{R} \to \mathbb{R}$ defined by $g(x) = \frac{x}{|x| + 1}$ has two horizontal asymptotes since
\begin{align*}
\lim_{x \to \infty} g(x) & = \lim_{x \to \infty} \frac{x}{|x| + 1} = \lim_{x \to \infty} \frac{x}{x + 1} = \lim_{x \to \infty} \frac{1}{1 + \frac{1}{x}} = 1\\
\lim_{x \to \infty} g(x) & = \lim_{x \to \infty} \frac{x}{|x| + 1} = \lim_{x \to \infty} \frac{x}{-x + 1} = \lim_{x \to \infty} \frac{1}{-1 + \frac{1}{x}} = -1
\end{align*}
A function $f$ has a vertical asymptote at $x = a$ if at least one of the following four statements is true:
\begin{align*}
\lim_{x \to a^+} f(x) & = \infty & \lim_{x \to a^+} f(x) & = -\infty\\
\lim_{x \to a^-} f(x) & = \infty & \lim_{x \to a^-} f(x) & = -\infty
\end{align*}
In this case,
\begin{align*}
\lim_{x \to 3^+} f(x) & = \lim_{x \to 3^+} \frac{5x}{x - 3} = \lim_{x \to 3^+} \left(\frac{12}{x - 3} + 5\right) = \infty\\
\lim_{x \to 3^-} f(x) & = \lim_{x \to 3^-} \frac{5x}{x - 3} = \lim_{x \to 3^-} \left(\frac{12}{x - 3} + 5\right) = -\infty
\end{align*}
so $x = 3$ is a vertical asymptote of the graph.
A rational function can have a horizontal asymptote when the degree of the numerator is at most that of the denominator, which is the case here since both the numerator and denominator have degree one. A rational function can have an oblique asymptote when the degree of the numerator exceeds that of the denominator by one. A rational function can have a curvilinear asymptote when the degree of the numerator exceeds that of the denominator by more than one.
| {
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"url": "https://math.stackexchange.com/questions/4614269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is there any other simpler method to evaluate $\int_0^{\infty} \frac{x^n-2 x+1}{x^{2 n}-1} d x,$ where $n\geq 2?$ Though $n\geq 2$ is a real number, which is not necessarily an integer, we can still resolve the integrand into two fractions,
$$\displaystyle I=\int_0^{\infty} \frac{x^n-2 x+1}{\left(1+x^n\right)\left(1-x^n\right)} d x=\int_0^{\infty}\left(\frac{x}{1+x^n}-\frac{1-x}{1-x^n}\right) d x=J-K\tag*{} $$
For the integral $J$, we are going to transforms it into a Beta function by letting $y=\frac{1}{1+x^n}$.
$$\displaystyle \begin{aligned}J & =\frac{1}{n} \int_0^1 y^{-\frac{2}{n}}(1-y)^{\frac{2}{n}-1} d y \\& =\frac{1}{n} B\left(-\frac{2}{n}+1, \frac{2}{n}\right) \\& =\frac{\pi}{n} \csc \left(\frac{2 \pi}{n}\right) \quad \textrm{ (By Euler Reflection Formula)}\end{aligned}\tag*{} $$
Next, we are going to evaluate the integral
$ \displaystyle K=\displaystyle \int_{0}^{\infty} \frac{1-x}{1-x^{n}} d x \tag*{} $
by the theorem
$ \displaystyle \displaystyle \sum_{k=-\infty}^{\infty} \frac{1}{k+z}=\pi \cot (\pi z), \textrm{ where } z\notin Z.\tag*{} $
We first split the integral into two integrals
$$ \displaystyle \displaystyle \int_{0}^{\infty} \frac{1-x}{1-x^{n}} d x=\int_{0}^{1} \frac{1-x}{1-x^{n}} d x+\int_{1}^{\infty} \frac{1-x}{1-x^{n}} d x \tag*{}$$
Transforming the latter integral by the inverse substitution $ x\mapsto \frac{1}{x}$ m, we have
$$\displaystyle \displaystyle \int_{1}^{\infty} \frac{1-x}{1-x^{n}} d x=\int_{0}^{1} \frac{x^{n-3}-x^{n-2}}{1-x^{n}} d x \tag*{} $$
Putting back yields
$ \begin{aligned}\displaystyle K&=\int_{0}^{1} \frac{1-x+x^{n-3}-x^{n-2}}{1-x^{n}} d x\\\displaystyle &=\int_{0}^{1}\left[\left(1-x+x^{n-3}-x^{n-2}\right) \sum_{k=0}^{\infty} x^{n k}\right] d x\\ \displaystyle & =\sum_{k=0}^{\infty} \int_{0}^{1}\left[x^{n k}-x^{n k+1}+x^{n(k+1)-3}-x^{n(k+1)-2}\right] d x\\ & =\sum_{k=0}^{\infty}\left(\frac{1}{n k+1}-\frac{1}{n k+2}+\frac{1}{n(k+1)-2}-\frac{1}{n(k+1)-1}\right)\\ & =\sum_{k=0}^{\infty}\left[\frac{1}{n k+1}-\frac{1}{n(k+1)-1}\right]+\sum_{k=0}^{\infty}\left[\frac{1}{n(k+1)-2}-\frac{1}{n k+2}\right] \end{aligned}\tag*{} $
Modifying yields
$$\displaystyle \begin{aligned} K&=\frac{1}{n}\left[\sum_{k=0}^{\infty} \frac{1}{k+\frac{1}{n}}+\sum_{k=-1}^{-\infty} \frac{1}{k+\frac{1}{n}}\right]+\frac{1}{n}\left(\sum_{k=1}^{\infty} \frac{1}{k-\frac{2}{n}}+\sum_{k=0}^{-\infty} \frac{1}{k-\frac{2}{n}}\right)\\& =\frac{1}{n}\left(\sum_{k=-\infty}^{\infty} \frac{1}{k+\frac{1}{n}}+\sum_{k=-\infty}^{\infty} \frac{1}{k-\frac{2}{n}}\right)\end{aligned} \tag*{} $$
By the Theorem,
$$ \displaystyle \displaystyle \sum_{k=-\infty}^{\infty} \frac{1}{k+z}=\pi \cot (\pi z), \tag*{} $$
where $ \displaystyle z\notin Z,$
we have
$$ \displaystyle \displaystyle K=\frac{1}{n}\left[\pi \cot \left(\frac{\pi}{n}\right)+\pi \cot \left(\frac{-2 \pi}{n}\right)\right]=\frac{\pi}{n}\left[\cot \left(\frac{\pi}{n}\right)-\cot \left(\frac{2 \pi}{n}\right)\right] =\frac{\pi}{n} \csc \frac{2 \pi}{n} =J\tag*{} $$
We can now conclude that
$\displaystyle \boxed{I=J-K=0 }\tag*{} $
Is there any other simpler method to evaluate $\int_0^{\infty} \frac{x^n-2 x+1}{x^{2 n}-1} d x,$ where $n\geq 2?$
| It is debatable whether the following method is simpler, but it leads to a stronger result:
For any $z \in \mathbb{C}$ with $\operatorname{Re}(z) > 0$ we have
\begin{align}
\int \limits_0^1 \frac{1 - 2 x + x^z}{1 - x^{2z}} \, \mathrm{d} x &\stackrel{x=t^{1/2z}}{=} \frac{1}{2z} \int \limits_0^1 \frac{t^{\frac{1}{2z} - 1} + t^{\frac{1}{2z} - \frac{1}{2}} - 2 t^{\frac{1}{z} - 1}}{1 - t} \, \mathrm{d} t \\
&\hspace{7pt}= \frac{1}{2z} \left[2 \operatorname{\psi}_0 \left(\frac{1}{z}\right) - \operatorname{\psi}_0 \left(\frac{1}{2z}\right) - \operatorname{\psi}_0 \left(\frac{1}{2z} + \frac{1}{2}\right)\right] \\
&\hspace{7pt}= \frac{\log(2)}{z} \, ,
\end{align}
where the last step follows from the Legendre duplication formula for the digamma function.
The same formula (along with the substitution $x = t^{-1/2z}$) yields
$$ \int \limits_1^\infty \frac{1 - 2 x + x^z}{1 - x^{2z}} \, \mathrm{d} x = - \frac{\log(2)}{z} $$
for $z \in \mathbb{C}$ with $\operatorname{Re} (z) > 1$, so
$$ \int \limits_0^\infty \frac{1 - 2 x + x^z}{1 - x^{2z}} \, \mathrm{d} x = 0 $$
holds for any such $z$.
| {
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Is this logical deduction regarding some modular restrictions on odd perfect numbers valid? - Part II Let $p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
Here is the Abstract of Dris and San Diego's article titled "Some modular considerations regarding odd perfect numbers – Part II", published in NNTDM:
In this article, we consider the various possibilities for $p$ and $k$ modulo $16$, and show conditions under which the respective congruence classes for $\sigma(m^2)$ (modulo $8$) are attained, if $p^k m^2$ is an odd perfect number with special prime $p$. We prove that:
(1) $\sigma(m^2) \equiv 1 \pmod 8$ holds only if $p + k \equiv 2 \pmod {16}$.
(2) $\sigma(m^2) \equiv 3 \pmod 8$ holds only if $p - k \equiv 4 \pmod {16}$.
(3) $\sigma(m^2) \equiv 5 \pmod 8$ holds only if $p + k \equiv {10} \pmod {16}$.
(4) $\sigma(m^2) \equiv 7 \pmod 8$ holds only if $p - k \equiv 4 \pmod {16}$.
From this answer by mathlove and this earlier MSE question, we get the biconditional
$$\sigma(m^2) \equiv 3 \pmod 4 \iff p \equiv {k+4} \pmod {16}.$$
Also, we have the biconditional
$$\sigma(m^2) \equiv 1 \pmod 4 \iff p \equiv k \pmod 8,$$
per the characterization theorem of Chen and Luo.
Assume that the Descartes-Frenicle-Sorli Conjecture that $k=1$ holds.
If $p=5$, then this evidently implies that $\sigma(m^2) \equiv 3 \pmod 4$.
Here is our:
QUESTION: If $p=13$, then since $13 = p \equiv {k+4=5} \pmod 8$ holds, but the conjunction
$$(p = 13) \land (k=1)$$
obviously does not satisfy $p \equiv {k+4} \pmod {16}$ nor $p \equiv k \pmod 8$, does it follow that the implication $k = 1 \implies p \neq 13$ holds, since otherwise
$$\lnot\Bigg(p \equiv {k+4} \pmod {16}\Bigg) \iff \lnot\Bigg(\sigma(m^2) \equiv 3 \pmod 4\Bigg)$$
and
$$\lnot\Bigg(p \equiv k \pmod 8\Bigg) \iff \lnot\Bigg(\sigma(m^2) \equiv 1 \pmod 4\Bigg),$$
contradicting the fact that $\sigma(m^2)$ must be odd?
| This is not a comprehensive proof, it is a mere observation that there are other prime components that do not satisfy the condition you stated. $13^1$ is not the only prime component that violates this rule. Let us take $p^kn^2$ to be an odd perfect number and k=1.
We know that $\sigma (p^k)/2\times gcd(n^2,\sigma (n^2))=n^2$.
We also know that $n^2\equiv 1\quad mod\quad 8$. Therefore if $\sigma (p^k)/2\equiv 3\quad mod\quad 4$ then $gcd(n^2,\sigma (n^2))$ must also be congruent to 3 mod 4. Therefore we can conclude that $gcd(n^2,\sigma (n^2))\equiv \sigma (p^k)/2$ if $gcd(n^2,\sigma (n^2))\equiv 3\quad mod\quad 4$
Now, $\sigma (p^1)/2\equiv 3\quad mod\quad 4$ if p is a prime of the form 5+8t where t is any positive integer. I will list a few examples below.
*
*[$\sigma (5^1)/2=3]\equiv 3\quad mod\quad 4,\quad$ $5=5+8(0)$
*[$\sigma (13^1)/2=7]\equiv 3\quad mod\quad 4,\quad$ $13=5+8(1)$
*[$\sigma (29^1)/2=15]\equiv 3\quad mod\quad 4,\quad$ $29=5+8(3)$
*[$\sigma (37^1)/2=19]\equiv 3\quad mod\quad 4,\quad$ $37=5+8(4)$
Now, a pattern emerges where we notice that $\sigma (n^2)\equiv gcd(n^2,\sigma (n^2))\equiv \sigma (p^k)/2\equiv 3\quad mod\quad 4\iff p\equiv k+4\quad (mod\quad 16)$ does not hold whenever prime p in the list above is of the form $5+8t$, where t is an odd integer. A more comprehensive list of values of p that do not satisfy the condition above is listed below
*
*$p^k$=(5+8(1))^1=13^1
*$p^k$=(5+8(3))^1=29^1
*$p^k$=(5+8(7))^1=61^1
*$p^k$=(5+8(13))^1=109^1
*$p^k$=(5+8(19))^1=157^1
*$p^k$=(5+8(21))^1=173^1
*$p^k$=(5+8(33))^1=269^1
The reader can test and satisfy himself or herself that the prime components in the above list violate the following condition
$\sigma (n^2)\equiv gcd(n^2,\sigma (n^2))\equiv \sigma (p^k)/2\equiv 3\quad mod\quad 4\iff p\equiv k+4\quad (mod\quad 16)$
| {
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"url": "https://math.stackexchange.com/questions/4614690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of $\frac 12$, $\frac{1\cdot 4}{2 \cdot 3}$, $\frac{1\cdot 4\cdot 5}{2\cdot 3\cdot 6}$, ... The following is not a homework, just curiosity.
Consider the integers grouped by consecutive pairs : $(1,2)$, $(3,4)$, ...
What is the limit of the "switching fractions" where we alternatively use the largest number in a pair either upward or downward :
$$\frac 12, \frac{1\cdot 4}{2 \cdot 3}, \frac{1\cdot 4\cdot 5}{2\cdot 3\cdot 6}, \frac{1\cdot 4\cdot 5\cdot 8}{2\cdot 3\cdot 6\cdot 7},\ldots?$$
A proof as elementary as possible would be nice, if not it could use standard results on prime distribution.
Also, was it considered before? Any reference welcomed.
Edit
Numerically we have:
$0.5, 0.6666... , 0.5555... , 0.6349206..,0.5714286..,0.6233766...$ More terms would certainly help.
| A more simpler, yet heavier way to look at this is,
A more common series is given by $\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\ldots=\prod\limits^{n}_{k=1}{\dfrac{2k-1}{2k}}=\dfrac{\left(n-\frac{1}{2}\right)!}{n!\cdot\left(\frac{-1}{2}\right)!}$.
Your problem is slight altered with every alternate term inversed, $\dfrac{1}{2}\cdot\left(\dfrac{3}{4}\right)^{-1}\cdot\dfrac{5}{6}\cdot\left(\dfrac{7}{8}\right)^{-1}\cdot\ldots$
In fact we factor with respect to the inverse power, $$\left(\frac{1}{2}\cdot \frac{5}{6}\cdot\ldots\right)\cdot\left(\frac{3}{4}\cdot\frac{7}{8}\cdot\ldots\right)^{-1}$$
$$=\left(\prod^{n}_{k=1}{\frac{4k-3}{4k-2}}\right)\cdot\left(\prod^{n}_{k=1}{\frac{4k-1}{4k}}\right)^{-1}$$
$$=\left(\frac{\left(n-\frac{1}{4}\right)!}{n!\cdot \left(\frac{-1}{4}\right)!}\right)^{-1}\cdot\frac{\left(n-\frac{3}{4}\right)!\cdot\left(\frac{-1}{2}\right)!}{\left(n-\frac{1}{2}\right)!\cdot\left(\frac{-3}{4}\right)!}$$
Funnily enough, the limit does converge for the above expression ($n\to\infty$); $$\frac{\left(\frac{-1}{4}\right)!\cdot \left(\frac{-1}{2}\right)!}{\left(\frac{-3}{4}\right)!}\approx 0.599195 \text{ as mentioned in answers above}$$
| {
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Maximize $x^2 + y^2$ subject to $x^2 + y^2 = 1 - xy$ The equation $x^2 + y^2 = 1 - xy$ represents an ellipse. I am trying to show that its major axis is along $y=-x$ and find the vertex. To find the vertex I tried to find the vector in the ellipse with the greatest norm, which is equivalent to maximizing $x^2 + y^2$ subject to $x^2 + y^2 = 1 - xy$. How can I approach this?
(The maximum value of the objective function $x^2 + y^2$ seems to be 2. If I start out by assuming this I am able to show that the function cannot exceed this value. But this is kind of doing it in reverse as I start by evaluating the function at $y=-x=1.$ This also doesn't prove that $y=-x$ is the only optimal solution. How would I do it without "guessing" the solution first?)
| Using polar coordinates, then
$ x = r \cos \theta $
$ y = r \sin \theta $
So we want to maximize $ r^2 = x^2 + y^2 $ subject to $ r^2 = 1 - \dfrac{1}{2} r^2 \sin(2 \theta) $
Hence,
$ r^2 = \dfrac{1}{1 + \frac{1}{2} \sin(2 \theta) } $
Clearly the maximum is when $ \sin(2 \theta) = -1 $, which corresponds to $ \theta = \dfrac{3 \pi}{4} $ and $ \theta = \dfrac{7 \pi}{4} $
At which points , $ r^2 = 2 $ (and this is the maximum value) and for the first value of $\theta$, we get
$ x = -1 , y = 1 $
And for the second value of $\theta$ we get
$ x = 1, y = -1 $
| {
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"timestamp": "2023-03-29T00:00:00",
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What does $\sin(4\theta)$ equal? I want to write $\sin(4\theta)$ in terms of $\cos(\theta)$ and $\sin(\theta)$. My work:
\begin{align*}\sin(4\theta) & = 2\sin(2\theta) \cos(2\theta)
\\ &= 2[(2\sin(\theta) \cos(\theta)(2\cos^2(\theta)-1)]
\\ &= 2(4\sin(\theta) \cos^3(\theta) - 2\sin(\theta) \cos(\theta))
\\ &= 8\sin(\theta) \cos^3(\theta) - 4\sin(\theta) \cos(\theta)
\end{align*}
This is wrong. Why? The solution manual says it's $\cos(\theta)(4\sin(\theta) - 8\sin^3(\theta))$.
Thanks.
| Those two equations are the same. Using $\cos^2(\theta) = 1 - \sin^2(\theta)$ you get your equation:
$= 4\cos(\theta)\sin(\theta) - 8\sin(\theta)^3\cos(\theta) \\
= 4\cos(\theta)\sin(\theta) - 8\sin(\theta)\cos(\theta) \cdot (1-\cos^2(\theta)) \\
= -4\cos(\theta)\sin(\theta) + 8\sin(\theta)\cos^3(\theta) \\
= 8\sin(\theta)\cos^3(\theta) - 4\cos(\theta)\sin(\theta)$
| {
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"url": "https://math.stackexchange.com/questions/4617244",
"timestamp": "2023-03-29T00:00:00",
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Arithmetic Progression with a Sum Property Question I had a question in a exam, and I had no idea how to do it.
The question is:
Let $(a_k)_{k\geq1}$ a sequence of not null real numbers such that:
$$\displaystyle \sum_{k=1}^{n-1}\frac{1}{a_k\cdot a_{k+1}}=\frac{n-1}{a_1\cdot a_n} \, \forall \,n\geq 2$$
Show that this sequence is an arithmetic progression.
I know that to show that this is an arithmetic progression, I have to show that $a_n-a_{n-1}$ is constant. But I don't know how to do it with the sum hypothesis.
Appreciate your help.
| Let $r = a_2 - a_1$. To prove that $(a_n)$ is arithmetic is equivalent to prove that:
$$\forall n \geq 1, a_n = (n - 1) r + a_1$$
Let's proceed by strong induction.
*
* It's clear, it's true for $n = 1$.
* Assume it's true for $k \in \{1, \ldots, n - 1\}$ then :
$$\begin{array}{lcl}
\displaystyle \dfrac{n - 1}{a_1 a_n} & = & \displaystyle \sum_{k = 1}^{n - 1} \dfrac{1}{a_k a_{k + 1}} \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \sum_{k = 1}^{n - 2} \dfrac{1}{a_k a_{k + 1}} \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \sum_{k = 1}^{n - 2} \dfrac{a_{k + 1} - a_k}{a_k a_{k + 1}} \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \sum_{k = 1}^{n - 2} \left(\dfrac{1}{a_k} - \dfrac{1}{a_{k + 1}}\right) \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \left(\dfrac{1}{a_1} - \dfrac{1}{a_{n - 1}}\right) \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \dfrac{a_{n - 1} - a_1}{a_1 a_{n - 1}} \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \dfrac{(n - 2) r + a_1 - a_1}{a_1 a_{n - 1}} \\[3mm]
& = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{n - 2}{a_1 a_{n - 1}}
\end{array}$$
We deduce that :
$$\begin{array}{lcl}
\dfrac{n - 2}{a_1 a_{n - 1}} & = & \dfrac{1}{a_n} \left(\dfrac{n - 1}{a_1} - \dfrac{1}{a_{n - 1}}\right) \\[3mm]
& = & \dfrac{1}{a_n} \left(\dfrac{(n - 1) a_{n - 1} - a_1}{a_1 a_{n - 1}}\right) \\[3mm]
& = & \dfrac{1}{a_n} \left(\dfrac{(n - 1) ((n - 2) r + a_1) - a_1}{a_1 a_{n - 1}}\right) \\[3mm]
& = & \dfrac{1}{a_n} \left(\dfrac{(n - 1) (n - 2) r + (n - 2) a_1}{a_1 a_{n - 1}}\right) \\[3mm]
& = & \dfrac{n - 2}{a_n} \left(\dfrac{(n - 1) r + a_1}{a_1 a_{n - 1}}\right) \\[3mm]
\end{array}$$
then :
$$\dfrac{(n - 1) r + a_1}{a_n} = 1$$
so $a_n = (n - 1) r + a_1$.
By induction :
$$\forall n \geq 1,a_n = (n - 1) r + a_1$$
so $(a_n)$ is arithmetic.
| {
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"timestamp": "2023-03-29T00:00:00",
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Studying $\lim_{x \rightarrow 0^+} \frac{xe^{-2x^2}-\sin(x)+\beta x^3}{x^{\alpha}\cos(x^2)}$ Study the following limit with $\alpha$ and $\beta$ parameters:
$$\lim_{x \rightarrow 0^+} \frac{xe^{-2x^2}-\sin(x)+\beta x^3}{x^{\alpha}\cos(x^2)}$$
My attempt:
$$\frac{xe^{-2x^2}-\sin(x)+\beta x^3}{x^{\alpha}\cos(x^2)} \;\sim_{0}\; \frac{xe^{-2x^2}-\sin(x)+\beta x^3}{x^{\alpha}}$$
Now I use Hospital three times and I have:
$$\frac{(-12+96x^2-64x^4)e^{-2x^2}+\cos(x)+6 \beta}{\alpha (\alpha-1)(\alpha-2)x^{\alpha-3}}\;\sim_0\; \frac{-11 +6 \beta}{\alpha (\alpha-1)(\alpha-2)x^{\alpha-3}}$$
Now if $\beta = \frac{11}{6}$ and $\alpha \gt 3$, I have $0 \cdot \infty$ indeterminate form. What can we say about the case $\beta = \frac{11}{6}$ and $\alpha \gt 3$?
| $$xe^{-2x^2}-\sin x+\beta x^3$$$$=x\left(1-2x^2+\frac{(-2x^2)^2}2\right)-\left(x-\frac{x^3}6+\frac{x^5}{120}\right)+\beta x^3+o(x^5)$$$$\sim_{(x\to0)}\begin{cases}\left(\beta-\frac{11}6\right)x^3&\text{if }\beta\ne\frac{11}6\\\frac{239}{120}x^5&\text{if }\beta=\frac{11}6.\end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding The Square mean of A Gaussian Function. I've been trying to find the square mean of a gaussian function using the limits of $+/-$ infinity.
$$\int_{-\infty}^{\infty} x^2e^{-2x^2}\mathrm{d}x$$
Why does splitting the function into a $$ u = x$$ and $$v'=xe^{-2x^2}$$ and integrating by parts give a different answer to $$u = x^2$$ and $$v' = e^{-2x^2}$$
| If you integrate by parts it doesn't matter in which functions you split your original integral. Both ways will resolve to the same solution but one way might be easier to evaluate. Lets look at both of your proposals and compare the result:
$$
\begin{align}
\int \underbrace{x}_u\cdot \underbrace{xe^{-2x^2}}_{v'}\text{d}x &= \underbrace{x}_u\cdot\underbrace{\left(-\frac{1}{4}e^{-2x^2}\right)}_v-\int \underbrace{1}_{u'}\cdot\underbrace{\left(-\frac{1}{4}e^{-2x^2}\right)}_v\text{d}x \\
&= -x\frac{1}{4}e^{-2x^2} + \frac{1}{8}\sqrt{\frac{\pi}{2}}\operatorname{erf}\left(\sqrt{2}x\right) \\
&= \frac{1}{8}\sqrt{\frac{\pi}{2}}\operatorname{erf}\left(\sqrt{2}x\right)-\frac{1}{4}e^{-2x^2}x
\\
\int \underbrace{x^2}_{u_1}\cdot\underbrace{e^{-2x^2}}_{v_1'}\text{d}x &= \underbrace{x^2}_{u_1}\cdot\underbrace{\frac{1}{2}\sqrt{\frac{\pi}{2}}\operatorname{erf}\left(\sqrt{2}x\right)}_{v_1} - \int \underbrace{2x}_{u_1',u_2}\cdot\underbrace{\frac{1}{2}\sqrt{\frac{\pi}{2}}\operatorname{erf}\left(\sqrt{2}x\right)}_{v_1,v_2'}\text{d}x \\
&= x^2\frac{1}{2}\sqrt{\frac{\pi}{2}}\operatorname{erf}\left(\sqrt{2}x\right) - \left(\underbrace{2x}_{u_2} \cdot \underbrace{\frac{1}{4}\left(\sqrt{2\pi}x\operatorname{erf}\left(\sqrt{2}x\right)+e^{-2x^2}\right)}_{v_2} - \int \underbrace{2}_{u_2'}\cdot\underbrace{\frac{1}{4}\left(\sqrt{2\pi}x\operatorname{erf}\left(\sqrt{2}x\right)+e^{-2x^2}\right)}_{v_2}\text{d}x\right)\\
&\;\; \vdots \\
&= x^2\frac{1}{2}\sqrt{\frac{\pi}{2}}\operatorname{erf}\left(\sqrt{2}x\right)-\frac{1}{16}\left(\sqrt{2\pi}\left(4x^2-1\right)\operatorname{erf}\left(\sqrt{2}x\right)+4e^{-2x^2}x\right) \\
&= \frac{1}{8}\sqrt{\frac{\pi}{2}}\operatorname{erf}\left(\sqrt{2}x\right)-\frac{1}{4}e^{-2x^2}x
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Change of variables example in an ODE I do not understand how this was obtained
Why the $(y')^2$ becomes $y'$ and the $y'$ becomes $(y')^2$ and why $y^2$ becomes $x^2 (y')^3$ in the new ode?
How does $\frac{d y(x)}{d x}$ change under this change of variables?
| IMO Leibniz's notation is a bit easier to follow.
$$\frac{d^2y}{dx^2} - \frac1y \left(\frac{dy}{dx}\right)^2 + \sin(x) y \frac{dy}{dx} + \cos(x) y^2 = 0$$
By the chain rule,
$$\frac{dy}{dx} = \frac1{\frac{dx}{dy}} \implies \frac{d^2y}{dx^2} = -\frac1{\left(\frac{dx}{dy}\right)^2} \frac{d\frac{dx}{dy}}{dx} = -\frac1{\left(\frac{dx}{dy}\right)^2} \frac{d^2x}{dy^2} \frac{dy}{dx} = -\frac1{\left(\frac{dx}{dy}\right)^3} \frac{d^2x}{dy^2}$$
The ODE transforms to
$$-\frac1{\left(\frac{dx}{dy}\right)^3} \frac{d^2x}{dy^2} - \frac1y \left(\frac1{\frac{dx}{dy}}\right)^2 + \sin(x) y \left(\frac1{\frac{dx}{dy}}\right) + \cos(x) y^2 = 0,$$
simplifies to
$$- \frac{d^2x}{dy^2} - \frac1y \frac{dx}{dy} + \sin(x) y \left(\frac{dx}{dy}\right)^2 + \cos(x) y^2 \left(\frac{dx}{dy}\right)^3 = 0,$$
and with the last variable change, we end up with
$$-\frac{d^2y}{dx^2} - \frac1x \frac{dy}{dx} + x \sin(y) \left(\frac{dy}{dx}\right)^2 + x^2 \cos(y) \left(\frac{dy}{dx}\right)^3 = 0$$
Multiply by $-1$ and swap out $\frac{dy}{dx}$ for $y'$ to get the exact same ODE.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sum_{k=0}^n\frac{3^{k+4}\binom{n}{k}}{\binom{k+4}4}+\sum_{m=0}^3\frac{\binom{n+4}{m}3^m}{\binom{n+4}4}=\frac{4^{n+4}}{\binom{n+4}4}$
Prove that
$$\sum_{k=0}^n\frac{3^{k+4}\binom{n}{k}}{\binom{k+4}{4}}
+\sum_{m=0}^3\frac{\binom{n+4}{m}3^m}{\binom{n+4}{4}}
=\frac{4^{n+4}}{\binom{n+4}{4}}$$
Wolfram Alpha shows that both expressions are equal but I can't prove it mathematically.
I am not able to calculate each series individually let alone the whole expression.
Any help is greatly appreciated.
| As Bruno B suggested in a comment, multiply by $\binom{n+4}4$ and simplify the terms in the first sum:
$$
\binom{n+4}4\frac{\binom nk}{\binom{k+4}4}=\frac{(n+4)!}{n!4!}\frac{\frac{n!}{k!(n-k)!}}{\frac{(k+4)!}{k!4!}}=\frac{(n+4)!}{(k+4)!(n-k)!}=\binom{n+4}{k+4}\;.
$$
Then merge the two sums, with $m=k+4$, to get the sum for the binomial expansion of $(3+1)^{n+4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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On factoring $2+128(a+b)^3$ Factor $2+128(a+b)^3$
$=2[1+64(a+b)^3]= 2[1+64(a^3+3a^2b+3ab^2+b^3)]$
Then I'm at a loss for what to do.
$=2[1+64(a^3+b^3)+192ab(a+b)]= 2[1+64(a+b)(a^2-ab+b^2)+192ab(a+b)]=2[1+64(a+b)(a^2-ab+b^2+3ab)]=2[1+64(a+b)^3]$
Back to where I began...
But the answer is given as $2[1+4a+4b][1-4(a+b)+16(a+b)^2]$
I cannot comprehend how it's possible to factor into this answer. Please help me. Thank you.
| You're almost there, in fact consider
$$2[1+64(a+b)^3] $$
and recall that $x^3+y^3=(x+y)(x^2-xy+y^2)$. In your case, take $x=1$ and $y=4(a+b)$.
Hence you have:
$$2[1+64(a+b)^3]=2[1+4(a+b)][1-4(a+b)+16(a+b)^2] $$
| {
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Showing that $\sum_{n \geq 1} \frac{a}{n^2 + a^2} $ is bounded by integrals. For $ a > 0$, we have: $ S(a) = \sum_{n \geq 0} \frac{a}{n^2 + a^2} $
Prove that $ S(a) $ is bounded between 2 integrals.
If we take: $ k \leq x \leq k + 1 $, we will have:
$ S(k+1) \leq \frac{a}{n^2 + a^2} \leq S(k) $
$ \implies \int_{k}^{k+1} S(k+1) da \leq \int_{k}^{k+1}\frac{a}{n^2 + a^2} da \leq \int_{k}^{k+1} S(k) da $
$ \implies \int_{0}^{n+1} S(k+1) da \leq \int_{0}^{n+1} S(a) da \leq \int_{0}^{n+1} S(k) da $
I don't see how to proceed to get the sum between integrals.
| What is $x$ supposed to be in the first line ? What you want to have is
$$ \int_k^{k+1}\frac{a}{x^2+a^2}dx\leqslant\frac{a}{k^2+a^2}\leqslant\int_{k-1}^k\frac{a}{x^2+a^2}dx $$
for all $k\geqslant 1$, which follows from the fact that $x\mapsto\frac{a}{x^2+a^2}$ is decreasing on $[0,+\infty)$. Summing this gives
$$ \int_0^{n+1}\frac{a}{x^2+a^2}dx\leqslant\sum_{k=0}^n\frac{a}{k^2+a^2}\leqslant\frac{1}{a}+\int_0^n\frac{a}{x^2+a^2}dx $$
and taking the limit $n\rightarrow +\infty$,
$$ \int_0^{+\infty}\frac{a}{x^2+a^2}dx\leqslant S(a)\leqslant\frac{1}{a}+\int_0^{+\infty}\frac{a}{x^2+a^2}dx. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ y^2(y^2-6)+x^2-8x+24=0$ and the minimum value of $x^2+y^4$ is $m$ and maximum value is $M$, then find the value of $m$ and $M$ If $ y^2(y^2-6)+x^2-8x+24=0$ and the minimum value of $x^2+y^4$ is $m$ and maximum value is $M$, then find the value of $m$ and $M$
My Attempt:
I converted them to perfect square i.e., $(x-4)^2+(y^2-3)^2=1$.
Any point on above curve can be written as $x=4+\cos\theta$ and $y^2=3+\sin\theta$
So $x^2+y^4=26+8\cos\theta+6\sin\theta$
As we know $8\cos\theta+6\sin\theta$ varies from $-10$ to $10$
So $x^2+y^4$ must vary from $16$ to $36$.
But given maximum value is $36$ and minimum value is $1$.
can you help me finding minimum value correctly?
| Your work seems fine to me.
A geometric approach for the second half (since the bounds on $8 \cos \theta + 6 \sin \theta$ were not obvious to me): Note that if you let $u=y^2$, then you are basically finding the points of the circle defined by $(x-4)^2 + (u-3)^2 = 1$ that are closest/farthest from the origin, which are $\frac{4}{5} (4, 3)$ and $\frac{6}{5}(4, 3)$ respectively, and have squared distances $4^2=16$ and $6^2=36$ from the origin respectively.
| {
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Line integral of $\int_\gamma\left(\frac{1}{x+1}-\frac{y}{(x+y)^2}\right)dx+\left(\frac{1}{y+1}-\frac{x}{(x+y)^2}\right)dy$ Determine the line integral $\int_\gamma\left(\frac{1}{x+1}-\frac{y}{(x+y)^2}\right)dx+\left(\frac{1}{y+1}-\frac{x}{(x+y)^2}\right)dy$ where $\gamma$ follows the circle from $(1,0)$ to $(0,1)$.
Let $x=\cos\theta$ and $y=\sin\theta$ where $0\leq\pi\leq\pi/2$. Then we have $\int_0^{\pi/2}\left(\left(\frac{1}{\cos\theta+1}-\frac{\sin\theta}{(\cos\theta+\sin\theta)^2}\right)(-\sin\theta)+\left(\frac{1}{\sin\theta+1}+\frac{\cos\theta}{(\cos\theta+\sin\theta)^2}\right)\cos\theta \right)d\theta$. Now putting this into symbolab I get $-0$ while the answer should be $1$.
But continuing $\int_0^{\pi/2}\left(\frac{-\sin\theta}{\cos\theta+1}+\frac{\sin^2\theta}{(\cos\theta+\sin\theta)^2}\right)+\left(\frac{\cos\theta}{\sin\theta+1}+\frac{\cos^2\theta}{(\cos\theta+\sin\theta)^2}\right)\ d\theta=\int_0^{\pi/2}\frac{-(cos\theta-1)}{sin\theta}+\frac{1}{sin2\theta}+\frac{sin\theta-1}{cos\theta}=\int^{\pi/2}_{0}-\cot\theta+\csc\theta+\csc 2\theta+\tan\theta-\sec\theta d\theta$. Putting this into symbolab I get that it diverges. I don't know what I'm doing wrong.
| As it is stated the result is 0. To obtain $1$ it should be
$$\int_\gamma\left(\frac{1}{x+1}-\frac{y}{(x+y)^2}\right)dx+\left(\frac{1}{y+1}+\frac{x}{(x+y)^2}\right)dy.$$
Then
$$
\begin{align}\int_0^{\pi/2}&\left(\left(\frac{1}{\cos\theta+1}-\frac{\sin\theta}{(\cos\theta+\sin\theta)^2}\right)(-\sin\theta)+\left(\frac{1}{\sin\theta+1}+\frac{\cos\theta}{(\cos\theta+\sin\theta)^2}\right)\cos\theta \right)d\theta\\
&=\int_0^{\pi/2}\left(\frac{-\sin\theta}{\cos\theta+1}+\frac{\cos(\theta)}{\sin\theta+1}+\frac{1}{(\cos\theta+\sin\theta)^2}\right)d\theta\\
&=\left[\ln(\cos\theta+1)+\ln(\sin\theta+1)+\frac{\sin\theta}{\cos\theta+\sin\theta}\right]_0^{\pi/2}\\
&=\ln(2)+1-\ln(2)=1\end{align}$$
Note that in your work $(\cos\theta+\sin\theta)^2$ should be equal to $1+\sin(2\theta)$ and not $\sin(2\theta)$.
Actually, there is no need to compute the line integral because the given differential form is exact with $f(x,y)=\ln(x+1)+\ln(y+1)+\frac{y}{x+y}$ and
$$\int_\gamma\left(\frac{1}{x+1}-\frac{y}{(x+y)^2}\right)dx+\left(\frac{1}{y+1}+\frac{x}{(x+y)^2}\right)dy=f(0,1)-f(1,0)=1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for all $x$ such that $x^3 = 2x + 1, x^4 = 3x + 2, x^5 = 5x + 3, x^6 = 8x +5 \cdots$ Question:
Solve for all $x$ such that $\begin{cases}&{x}^{3}=2x+1\\&{x}^{4}=3x+2\\&{x}^{5}=5x+3\\&{x}^{6}=8x+5\\&\vdots\end{cases}$.
My attempt:
I sum up everything.
$$\begin{aligned}\sum_{i=1}^n x^{i+2} &= (2 + 3 + 5 + 8 + \cdots)x + (1 + 2 + 3 + 5 +...)\\&= (S_n - 2)x + (S_n - 1)\\& = (a_{n+2} - 3)x + (a_{n+2} - 2)\end{aligned}$$
where $S_n$ is sum of first $n$ terms of Fibonacci series and $a_n$ is its $n$th term.
I'm not getting any idea how to solve it further.
| $$x^4=x^3x=2x^2+x\implies3x+2=2x^2+x\implies-2x^2+2x+2=0\implies x^2-x-1=0$$
So $x=\frac{1\pm\sqrt{5}}{2}=\varphi,\varphi^{\dagger}$. You could use induction to prove that they are solutions to all other equations.
| {
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Solve for $x$: $\frac{x-a+b}{x-a}+\frac{x-b}{x-2b}=\frac{x}{x-b}+\frac{x-a}{x-a-b}$ Solve for $x$: $\dfrac{x-a+b}{x-a}+\dfrac{x-b}{x-2b}=\dfrac{x}{x-b}+\dfrac{x-a}{x-a-b}$
$\Rightarrow \dfrac{(x-a)+b}{x-a}+\dfrac{(x-2b)+b}{x-2b}=\dfrac{(x-b)+b}{x-b}+\dfrac{(x-a-b)+b}{x-a-b} \ \ \ ...(1)$
$\Rightarrow 1+\dfrac{b}{x-a}+1+\dfrac{b}{x-2b}=1+\dfrac{b}{x-b}+1+\dfrac{b}{x-a-b} \ \ \ ...(2)$
$\Rightarrow \dfrac{b}{x-a}+\dfrac{b}{x-2b}=\dfrac{b}{x-b}+\dfrac{b}{x-a-b} \ \ \ ...(3)$
$\Rightarrow \dfrac{b(x-2b)+b(x-a)}{(x-a)(x-2b)}=\dfrac{b(x-a-b)+b(x-b)}{(x-b)(x-a-b)} \ \ \ ...(4)$
$\Rightarrow \dfrac{b(2x-2b-a)}{(x-a)(x-2b)}=\dfrac{b(2x-2b-a)}{(x-b)(x-a-b)} \ \ \ ...(5)$
$\Rightarrow b(2x-2b-a)(x-b)(x-a-b)=b(2x-2b-a)(x-a)(x-2b)\ \ \ ...(6)$
$\Rightarrow (x-a)(x-2b)=(x-b)(x-a-b) \ \ \ ...(7)$
$\Rightarrow x^2-2bx-ax+2ab=x^2-ax-bx-bx+ab+b^2 \ \ \ ...(8)$
$\Rightarrow 2ab=ab+b^2 \ \ \ ...(9)$
$\Rightarrow a=b \ \ \ $
So the result I got does not solve for $x$. Clearly something went wrong. I can't see it however because to me everything feels correct computationally. Thanks for the assistance.
| From (6) to (7) you lost the solution $x=(2b+a)/2.$ And from (8) to (9) you lost $b=0.$ You must present the set of solutions $x$ in the case $b=0$ or $a$, and in the case $b\ne0,a.$ And in each case, exclude the values which would make some of the 4 denominators equal to $0.$
So you need to split in three cases:
*
*If $b\in\{0,a\},$ every $x$ distinct from $a$ and $2b$ is a solution.
*If $b\notin\{0,a\}$ and $a\notin\{0,2b\},$ the unique solution is $x=b+\frac a2.$
*If $b\ne0$ and $a\in\{0,2b\},$ there is no solution.
| {
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Evaluate $\int \sqrt{ a^2 - x^2} dx$ Problem:
Evaluate the integral
$$
\int \sqrt{ a^2 - x^2 } dx.
$$
Solution:
Let us put
$$
x = a \sin \theta . \tag{0}
$$
Then we have
$$
dx = a \cos \theta d \theta, \tag{1}
$$
and
$$
a^2 - x^2 = a^2 \left( 1 - \sin^2 \theta \right) = a^2 \cos^2 \theta. \tag{2}
$$
So we have
$$
\begin{align}
& \ \ \ \int \sqrt{ a^2 - x^2 } dx \\
&= \int \sqrt{ a^2 - a^2 \sin^2 \theta } \, a \cos \theta d \theta \\
&= \int \sqrt{ a^2 \left( 1 - \sin^2 \theta \right) } \, a \cos \theta d \theta \\
&= \int \sqrt{ a^2 \cos^2 \theta }\, a \cos \theta d \theta \\
&= \int \big\lvert a \cos \theta \big\rvert \, a \cos \theta d \theta \\
& \mbox{[ note that for any real number $r$, we have $\lvert r \rvert = \sqrt{r^2}$ ]} \\
&= \int (a \cos \theta ) a \cos \theta d \theta \\
& \mbox{[ we assume that $0 < \theta < \frac{\pi}{2}$ so that $\cos \theta > 0$, } \\
& \qquad \mbox{ and we also assume that $a > 0$ ]} \\
&= \int a^2 \cos^2 \theta d \theta \\
&= a^2 \int \cos^2 \theta d \theta \\
&= a^2 \int \frac{2 \cos^2 \theta }{2} d \theta \\
&= a^2 \int \frac{2 \cos^2 \theta -1 + 1 }{2} d \theta \\
&= a^2 \int \frac{ \cos 2 \theta + 1 }{2} d \theta \\
&= \frac{a^2}{2} \int (\cos 2 \theta + 1) d \theta \\
&= \frac{a^2}{2} \big( \int \cos 2 \theta d \theta + \int d \theta \big).
\end{align}
$$
Thus we have
$$
\int \sqrt{ a^2 - x^2 } dx = \frac{a^2}{2} \big( \int \cos 2 \theta d \theta + \int d \theta \big). \tag{3}
$$
Now we evaluate $\int \cos 2 \theta d \theta$ as follows:
Let us put $2 \theta = \psi$. Then we have
$$
2 d \theta = d \psi,
$$
and hence
$$
d \theta = \frac{1}{2} d \psi.
$$
Then
$$
\begin{align}
\int \cos 2 \theta d \theta &= \int \cos \psi \frac{1}{2} d \psi \\
&= \frac12 \int \cos \psi d \psi \\
&= \frac12 \sin \psi \\
&= \frac12 \sin 2 \theta. \tag{4}
\end{align}
$$
Now putting the value of $\int \cos 2 \theta d \theta$ from (4) into (3) we get
$$
\begin{align}
& \ \ \ \int \sqrt{ a^2 - x^2 } dx \\
&= \frac{a^2}{2} \big( \int \cos 2 \theta d \theta + \int d \theta \big) \qquad \mbox{[ using (3) above ]} \\
&= \frac{a^2}{2} \left( \frac12 \sin 2 \theta \ + \theta \right) + C \\
& \qquad \mbox{[ using (4) above; } \\
& \qquad \qquad \mbox{ note that $C$ is an arbitrary constant of integration ]}\\
&= \frac{a^2}{2} \left( \frac12 (2 \sin \theta \cos \theta ) + \theta \right) + C \\
&= \frac{a^2}{2} \big( \sin \theta \cos \theta + \theta \big) + C \\
&= \frac{1}{2} \big( (a \sin \theta) (a \cos \theta) + a^2 \theta \big) +C \\
&= \frac12 \left( x \sqrt{a^2 - x^2} + a^2 \sin^{-1} \frac{x}{a} \right) + C \\
& \qquad \mbox{[ refer to (0) and (2) above; } \\
& \qquad \qquad \mbox{ note that since $x = a \sin \theta$, therefore } \\
& \qquad \qquad \mbox{ we have $\sin \theta = \frac{x}{a}$ and so $\theta = \sin^{-1} \frac{x}{a}$. ] } \\
&= \frac{x}{2} \sqrt{ a^2 - x^2} + \frac{a^2}{2} \sin^{-1} \frac{x}{a} + C.
\end{align}
$$
Is this procedure correct, clear and rigorous enough?
| You obtained the correct answer. However, your assumption that $0 < \theta < \pi/2$ is not necessary.
Since you let $x = a\sin\theta$, where $a > 0$, then
$$\theta = \arcsin\left(\frac{x}{a}\right)$$
is defined to be the unique angle in the interval $[-\pi/2, \pi/2]$ such that $\sin\theta = x/a$. Since $\theta \in [-\pi/2, \pi/2]$, $\cos\theta \geq 0$. Hence, $|\cos\theta| = \cos\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the differential equation: $(x^2-y^2)dx+2xydy=0$. Given $(x^2-y^2)dx+2xydy=0$
My solution-
Divide the differential equation by $dx$
$\Rightarrow x^2-y^2+2xy\frac{dy}{dx}=0$
$\Rightarrow 2xy\frac{dy}{dx}=y^2-x^2$
Divide both sides by $2xy$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2}[\frac{y}{x}-\frac{x}{y}]$
This is a homogenous differential equation. Substitute $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}=\frac{1}{2}[v-\frac{1}{v}]$
$\Rightarrow x\frac{dv}{dx}=-\frac{v^2+1}{2v}$
$\Rightarrow -\frac{2v}{v^2+1}dv=\frac{dx}{x}$
Integrating both sides
$\Rightarrow -\log|v^2+1|=\log x+\log c$
$\Rightarrow -\log|\frac{y^2}{x^2}+1|=\log xc$
$\Rightarrow -\log|\frac{x^2+y^2}{x^2}|=\log xc$
$\Rightarrow \frac{x^2}{x^2+y^2}= xc$
$\Rightarrow x= c(x^2+y^2)$
$\Rightarrow y=\pm \sqrt{xc-x^2}$
Kindly review my solution and let me know if there are other methods of solving such problems.
| Your solution it is correct. Answering your second question, here is another slightly different solution.
If we can write the ODE as a Bernoulli equation: $y'+a(x)y=b(x)y^{\alpha}$ for some real $\alpha$ so the substitution $v=y^{1-\alpha}$ transforms the Bernoulli equation into a linear equation $\frac{1}{1-\alpha}v'+a(x)v=b(x)$ which can be solved using integrating factor.
We can re-write the ODE as
$$(x^2-y^2)dx+2xydy=0$$
$$(x^2-y^2)+2xy\frac{dy}{dx}=0$$
$$2xy\frac{dy}{dx}-y^2=-x^2$$
$$\frac{dy}{dx}-\frac{1}{2x}y=-\frac{x}{2}y^{-1}$$
Consider $a(x)=\frac{-1}{2x}$ and $b(x)=\frac{-x}{2}$ so the equation is Bernoulli and we can make the substitution $v=y^{1-(-1)}=y^{2}$ and then we can write the ODE as
$$\frac{1}{2}\frac{dv}{dx}-\frac{1}{2x}v=-\frac{x}{2}$$
$$\frac{dv}{dx}-\frac{1}{x}v=-x.$$
The equation it is linear and using your integrating factor $\mu(x)=e^{\int -\frac{1}{x}dx}=\frac{1}{x}$ we can transforms the linear equation into a separable equation.
$$\frac{d}{dx}\left(\mu(x)v\right)=-x\mu(x)$$
Integrating,
$$\frac{v}{x}=\int-1dx=-x+c$$
Thus,
$$v=-x^2+cx$$
Substitution back
$$y^2=-x^2+cx$$
Therefore, the general solution is given by
$$ y=\pm\sqrt{-x^2+cx}$$
as you said.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4642892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Evaluate: $\int \frac{2}{(1-x)(1+x^2)}dx$ Given $\int \frac{2}{(1-x)(1+x^2)}dx$
The most obvious approach is to use Partial fractions
Let $\frac{2}{(1-x)(1+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{1+x^2}$
$\Rightarrow \frac{2}{(1-x)(1+x^2)}=\frac{A+Ax^2+Bx-Bx^2+C-Cx}{(1-x)(1+x^2)}$
We get $A=1, B=1, C=1$
The integral now becomes
$\int[\frac{1}{1-x}+\frac{x+1}{1+x^2}]dx$
$\Rightarrow \int[\frac{1}{1-x}+\frac{x}{1+x^2}+\frac{1}{1+x^2}]dx$
Now, we can simply integrate term by term
$\Rightarrow -\log|1-x|+\frac{\log(1+x^2)}{2}+\tan^{-1}x+c $
$\Rightarrow \log |\frac{\sqrt{1+x^2}}{1-x}|+\tan^{-1}x+c $
Please review my solution and if you have any other way of integrating then please share your solution.
| $$\Im \left(\frac{1}{x-i}\right)=\frac{1}{1+x^2}$$
$$\int \frac{2}{(1-x)(1+x^2)}dx=\Im\int\frac{2}{(1-x)(x-i)}dx$$$$=\Im\int\frac{(1+i)((1-x)+(x-i))}{(1-x)(x-i)}dx=\Im\left[\int\frac{1+i}{x-i}dx+\int\frac{1+i}{1-x}dx\right]$$
$$=\Im\left[(1+i)\ln(x-i)-(1+i)\ln(1-x)\right]+c$$$$=\frac{1}{2}\ln(x^2+1)+\arctan x-\ln(1-x)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Compute the integral $\int_{-\infty}^\infty \frac{\sin(x) (1+\cos(x))}{x(2+\cos(x))} dx$ Background
I recently found out about Lobachevsky's integral formula, so I tried to create a problem on my own for which I'd be able to apply this formula. The problem is presented below.
Problem
Compute the integral $\int_{-\infty}^\infty \frac{\sin(x) (1+\cos(x))}{x(2+\cos(x))} dx$
Attempt
If we define $f(x) := \frac{1+\cos(x)}{2+\cos(x)}$, which most definitely is $\pi$ - periodic.
The integral is, using our notation above, on the form
$$I = \int_{-\infty}^\infty \frac{\sin(x)}{x}f(x) dx.$$
The integrand is even, so we might as well compute
$$ I = 2 \int_{0}^\infty \frac{\sin(x)}{x}f(x) dx.$$
We will now have to make use of a theorem.
Lobachevsky's integral formula states that if $f(x)$ is a continous $\pi$ - periodic function then we have that $$ \int_0^\infty \frac{\sin(x)}{x}f(x) dx= \int_0^{\pi/2} f(x) dx.$$
Substituing our $f(x)$ yields us
$$ \int_0^{\pi/2} \frac{1+\cos(x)}{2+\cos(x)} dx = \pi/2 - \int_0^{\pi/2}\frac{1}{2+\cos(x)}dx $$
where
$$I_2 = \int_0^{\pi/2}\frac{1}{2+\cos(x)}dx = \int_0^{\pi/2}\frac{\sec^2(x/2)}{3+\tan^2(x/2)}dx.$$
Letting $ u = \tan(x/2)/\sqrt{3}$, for which $du = \sec^2(x/2)/(2\sqrt{3})dx$, therefore gives us:
$$ I_2 = \int_0^{1/\sqrt{3}}\frac{2\sqrt{3}}{3u^2+3} = \frac{\pi}{3\sqrt{3}}.$$
Finally we can compute $I$ to
$$I = 2\left(\frac{\pi}{2} - \frac{\pi}{3\sqrt{3}}\right) = \frac{\pi(3\sqrt{3}-2)}{3\sqrt{3}}.$$
I've tried calculating this integral in Desmos where it gives me $0$ when I calculate the integrand on the interval $(-\infty, \infty)$, and something negative for $(0,\infty)$. This contradicts my answer.
I also tried typing it into Wolfram, without success. Can anyone confirm the validity of my result?
| There is a generalization of Lobachevsky's integral formula that is applicable.
The formula states that if $f(x)$ is an odd periodic function of period $a$, then $$\int_{0}^{\infty} \frac{f(x)}{x} \, \mathrm dx = \frac{\pi}{a} \int_{0}^{a/2} f(x) \cot \left(\frac{\pi x}{a} \right) \, \mathrm dx \tag{1}$$ if the integral on the left converges.
A proof can be found in the paper The integrals in Gradshteyn and Ryzhik.
Part 16: Complete elliptic integrals by Boettner and Moll. It's Lemma 3.1.
The proof says to use the partial fraction expansion of $\tan (z)$. That's clearly a typo. It should be the partial fraction expansion of $\cot (z)$ since $$\sum_{k=0}^{\infty} \left( \frac{1}{x+ka}- \frac{1}{(k+1)a-x} \right) = \frac{1}{x} + \sum_{k=1}^{\infty} \frac{2x}{x^{2}-k^{2}a^{2}} = \frac{\pi}{a} \cot \left(\frac{\pi x}{a} \right).$$
Using $(1)$ and the tangent half-angle substitution, we get $$\begin{align} \int_{-\infty}^{\infty} \frac{\sin(x)}{x} \frac{1+ \cos x}{2+ \cos x} \, \mathrm dx &= 2 \int_{0}^{\infty} \frac{\sin(x)}{x} \frac{1+ \cos x}{2+ \cos x} \, \mathrm dx \\ &= \int_{0}^{\pi} \sin(x) \, \frac{1+ \cos x}{2+ \cos x} \, \cot\left(\frac{x}{2} \right) \, \mathrm dx \\ &= 8 \int_{0}^{\infty} \frac{\mathrm dt}{(t^{2}+1)^{2}(t^{2}+3)} \\&= 4 \int_{0}^{\infty} \frac{\mathrm dt}{(t^{2}+1)^{2}} -2 \int_{0}^{\infty} \frac{\mathrm dt}{t^{2}+1} + 2 \int_{0}^{\infty}\frac{\mathrm dt}{t^{2}+3} \\ &= 4 \left(\frac{\pi}{4} \right) - 2 \left( \frac{\pi}{2} \right) + 2 \left( \frac{\pi}{2\sqrt{3}} \right) \\ &= \frac{\pi}{\sqrt{3}}. \end{align} $$
To emphasize what is stated in the paper, if $g(x)$ is an even $\pi$-periodic function, then $g(x) \sin(x)$ is an odd $2 \pi$-periodic function, and $$ \begin{align} \int_{0}^{\infty} \frac{g(x) \sin (x)}{x} \, \mathrm dx &= \frac{1}{2} \int_{0}^{\pi} g(x) \sin(x) \cot \left(\frac{x}{2} \right) \, \mathrm dx \\ &= \frac{1}{2} \int_{0}^{\pi} g(x) \sin(x) \, \frac{1+ \cos(x)}{\sin(x)} \, \mathrm dx \\ &= \frac{1}{2} \int_{0}^{\pi} g(x) \, \mathrm dx + \frac{1}{2} \int_{0}^{\pi} g(x) \cos (x) \, \mathrm dx \\ &= \int_{0}^{\pi/2} g(x) \, \mathrm dx + \frac{1}{2} (0) \\ &= \int_{0}^{\pi/2} g(x) \, \mathrm dx \end{align}$$ which is Lobachevsky's integral formula.
It wasn't immediately clear to me why $$\int_{0}^{\pi} g(x) \cos(x) \, \mathrm dx =0. $$
But since $g(x) \cos(x)$ is an even function, we have $$\int_{0}^{\pi} g(x) \cos(x) \, \mathrm dx = \int_{-\pi}^{0} g(x) \cos (x) \, \mathrm dx. $$
And making the substitution $u = x- \pi$, we have $$\int_{0}^{\pi} g(x) \cos(x) \, \mathrm dx = -\int_{-\pi}^{0} g(u) \cos (u) \, \mathrm du. $$
The only way both equations can be simultaneoulsy true is if $$\int_{0}^{\pi} g(x) \cos(x) \, \mathrm dx=0. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4645216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
How to solve this series $f(n) = f(n/2) + n$? Can I solve this as:
$f(n) = f(n/2) + n$
let find, $$f(n/2) = f(n/2/2) + n/2\\
f(n/2) = f(n/4) + n/2$$
Now,
$$\begin{split}
f(n) &= f(n/4) + n/2 + n\\
f(n) &= f(n/8) + n/4 + n/2 + n
\end{split}$$
hence so on.
$$\vdots$$
Now, $n = 2^i$.
$$\begin{split}
f(2^i) &= f(2^i/2^i) + 2^i/2^{i-1} + 2^i/2^{i-2} + \cdots + 2^i\\
f(2^i) &= f(1) + 2^1 + 2^2 + \cdots + 2^i\\
f(2^i) &= 2^0 + 2^1 + 2^2 + \cdots + 2^i\\
f(2^i) &= 2^{i+1} - 1\\
f(2^i) &= 2^i\cdot 2^1 - 1\\
f(n) &= 2n - 1\\
f(2^k) &= f(2^{k-k}) + k\\
f(2^k) &= f(2^0) + k\\
\cdots &= f(1) + k\\
\cdots &= 1 + k
\end{split}$$
As we know
$$\begin{split}
n &= 2^k\\
\log (n) &= k \log(2)\\
k &= \log (n) / \log(2)\\
k &= \log_2 (n)\\
f(n) &= \log_2(n) + 1
\end{split}$$
| At some point you need to have the value of say $f(1)$ or something. The classical way of solving this kind of questions would be to do as you did until:
$$ f(n) = f(1) + \sum_{i=0}^{k} \frac{n}{2^i}$$
where $2^k = n$. Then you should be able to solve the problem by computing $\sum_{i=0}^{k} \frac{n}{2^i}$ using geometric series.
In the case you $n$ is not a power of $2$, you must apply the reccurence formula until step $k$ with $\frac{n}{2^k} \le 1$, and the you must know what is the value of $f$ of say, interval $[0,1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Variation of the Polya Urn Model An urn contains ${b \ge 1}$ blue balls and one red ball. At each stage a ball is randomly chosen from the urn and then replaced along with a new ball of the same color. Let $T$ denote the first step when a red ball is chosen. What is ${P[T \ge i]}$ and what is $E[T]$?
I calculated $1- P[T < i]$ to get $\frac{b}{b+i-2}$. However, this doesn't allow for me to calculate the infinite sum that is required for $E[T]$ which by my current answer would be $b(\frac{1}{b-1}+\frac{1}{b}+\frac{1}{b+1}+..)$
Would appreciate some help. Thanks!
| For $T \ge i$ you must draw a blue ball for each of the first $i-1$ steps.
The probability of this occuring is
$$\frac{b}{b+1}\times\frac{b+1}{b+2}\times\frac{b+2}{b+3}\ \times\ ... \ \times\frac{b+i-2}{b+i-1}$$
Hence, $P[T \ge i]$ is equal to $\frac{b}{b+i-1}$
The expected value $E[T]$ is given by
$$(1\times\frac{1}{b+1})+(2\times\frac{b}{b+1}\frac{1}{b+2})+(3\times\frac{b}{b+1}\frac{b+1}{b+2}\frac{1}{b+3})+\ ... \ =$$
$$b\sum_{i=1}^{\infty} \frac{i}{(b+i-1)(b+i)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $|z-1|=1$, where $z$ is a point on the argand plane, show that $\dfrac{z-2}{z}=i\cdot\tan(\arg z)$, where $i=\sqrt{-1}$ If $|z-1|=1$, where $z$ is a point on the argand plane, show that $\dfrac{z-2}{z}=i\cdot\tan(\arg z)$, where $i=\sqrt{-1}$
M y Approach: Let $z=x+iy$
$|z-1|^2=1\implies (z-1)(\bar z-1)=1\implies x^2+y^2-2x=0$
Now, $\dfrac{z-2}{z}=\dfrac{x+iy-2}{x+iy}=\dfrac{(x^2+y^2-2x+2iy)}{x^2+y^2}$
$\implies$ $\dfrac{z-2}{z}=i\cdot \dfrac{2y}{x^2+y^2}$
Now do I prove that $\dfrac{2y}{x^2+y^2}=\tan(\arg z)$
I know other methods to solve this problem. I Just want to know flaw in my approach.
| You've already got $\dfrac{z-2}{z}=i\cdot \dfrac{2y}{x^2+y^2}$ (for $z\not=0$).
Now you can use $x^2+y^2=2x$ again to have $\dfrac{z-2}{z}=i\cdot \dfrac{2y}{2x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4650758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $\int\frac{dx}{1+\tan x}$ How do we solve $$\int\frac{dx}{1+\tan x}$$? I found two ways to solve it, one with the Weierstrass substitution ($t=\tan(\dfrac x2)$) and one with simply substituting $u=\tan x$ (which is better in my opinion). To solve it with the Weierstrass substitution, we get $$\int\dfrac{\frac{2}{1+t^2}}{1+\dfrac{2t}{1-t^2}}dt$$ Which is very complicated. But if we do $u=\tan x$ we would need to solve $$\int\frac{1}{(u+1)\left(u^2+1\right)}du$$Which is easier. Are there any other ways to solve this problem?
| Solution using Euler substitution, provided $\cos(x)>0$:
$$\begin{align*}
& \int \frac{dx}{1+\tan(x)} \\
&= \int \frac{\cos(x)}{\sin(x)+\cos(x)} \, dx \\
&= \int \frac{du}{u+\sqrt{1-u^2}} & u=\sin(x) \\
&= -2 \int \frac{1-t^2}{(1+t^2) \left(1-2t -t^2\right)}\,dt & u=-\frac{2t}{1+t^2}\\
&= \int \left(\frac{-1-t}{1-2t-t^2} - \frac{1}{1+t^2} - \frac t{1+t^2}\right) \, dt \\
&= \frac12 \ln\left|1-2t-t^2\right| - \arctan(t) - \frac12 \ln\left(1+t^2\right) + C \\
&= \frac12 \ln\left|\frac{(1-u)\left(1-\sqrt{1-u^2}\right)-u^2}{1-\sqrt{1-u^2}}\right| - \arctan\left(\frac{\sqrt{1-u^2}-1}u\right) + C \\
&= \frac12 \ln\left|\cos(x)+\sin(x)\right| + \arctan\left(\frac{1-\cos(x)}{\sin(x)}\right) + C \\
\end{align*}$$
and we recall
$$\arctan\left(\frac{1-\sqrt{1-z^2}}z\right) = \arctan\left(\frac{z}{1+\sqrt{1-z^2}}\right) = \frac12 \arcsin(z)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4650904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Recurrence relation for $a_{n}$ where $6a_{n}$ and $10a_{n}$ are both triangular According A180926, the elements of the set
{$a:\exists m,n|60a=5n^2+5n=3m^2+3m$} satisfy the following recurrence relation:
$$a_{n}=\frac{62a_{n-1}+1+\sqrt{(48a_{n-1}+1)(80a_{n-1}+1)}}{2}$$
$$a_1=0$$
How this can be derived?
| Here's a complete solution, including a derivation of the recurrence relation in A180926. First, notice
$\rm\ 60\:a \:=\: 5\:(n^2+n)\ =\ 3\:(m^2+m)\ \ $ is easily transformed into the following Pell equation
$\rm\quad\quad\quad\quad\ \: 5\:x^2-3\:y^2 \:= \ 2\ \ \ $ for $\rm\ \ \ (x,\:y)\: =\: (2\:n+1,\:2\:m+1)$
Hence $\rm\ \ \ \: x^2 - 1\ =\ 4\: (n^2+n)\ =\ 48\:a,\quad\ y^2 - 1\ =\ 4\: (m^2+m)\ =\ 80\:a\quad\quad\quad\ \ (*)$
By standard Pell theory $\rm\ \: (x,y) \:\to\: (X,Y)\:=\: (4\:x+3\:y\ \ ,\ \ 5\:x+4\:y)\ $ is a solution too, since
$\rm\phantom{\quad\Rightarrow\quad}\ X+Y\omega \: =\: (4+5\:\omega)\:(x+y\:\omega) \: = \: (4\:x+3\:y + (5\:x+4\:y)\:\omega) \: ,\ \ \ \omega \:=\: \sqrt{3/5}$
$\rm\quad\Rightarrow\quad X- Y\omega \: =\: (4-5\:\omega)\:(x-y\:\omega)\quad\quad\:$ via conjugate prior equation
$\rm\quad\Rightarrow\quad 5\: X^2 - 3\: Y^2 \: =\ 5\: x^2 - 3\: y^2\: =\: 2\ \ $ via multiply the prior two equations, then scale by 5
This yields a recurrence to generate a new solution $\rm (X,Y) = (x_{n+1},\:y_{n+1})$ from a known solution $\rm\: (x,\:y) = (x_n,\:y_n)\:.\:$ We square this recurrence to obtain a recurrence for $\rm (A,\:a) = (a_{n+1},\:a_n)$
Namely $\rm\ \: (4\:x+3\:y)^2 =\ X^2$
$\rm\quad\quad\quad\quad \Rightarrow\quad\ 24\ x\:y\ =\ X^2-1 - 16\:(x^2-1) - 9\:(y^2-1) - 24$
$\rm\quad\quad\quad\quad\quad \phantom{\Rightarrow\ 24\ x\:y}\ =\ 48\:A - 16\:(48\:a) - 9\:(80\:a) - 24\quad$ by $\ (*)$
$\rm\quad\quad\quad\quad\quad \phantom{\Rightarrow\ 24\ x\:y}\ =\ 24\ (2\:A - 62\: a - 1)$
$\rm\quad\quad\quad\quad \Rightarrow\quad\ x^2\ y^2\ =\ (2\:A - 62\: a - 1)^2 $
But also $\rm\quad\quad\quad\: x^2\ y^2\ =\ (48\: a + 1)\ (80\: a + 1)\quad$ by $\ (*)$
Hence $\rm\ \ (2\ a_{n+1} - 62\ a_n - 1)^2 \ = \ (48\ a_n+1)\ (80\ a_n+1)\ \ $ as was to be proved.
| {
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How to prove $\cos \frac{2\pi }{5}=\frac{-1+\sqrt{5}}{4}$? I would like to find the apothem of a regular pentagon. It follows from
$$\cos \dfrac{2\pi }{5}=\dfrac{-1+\sqrt{5}}{4}.$$
But how can this be proved (geometrically or trigonometrically)?
|
Consider a $\triangle ABC$ with $AB=1$, $\mathrm{m}\angle A=\frac{\pi}{5}$ and $\mathrm{m}\angle B=\mathrm{m}\angle C=\frac{2\pi}{5}$, and point $D$ on $\overline{AC}$ such that $\overline{BD}$ bisects $\angle ABC$. Now, $\mathrm{m}\angle CBD=\frac{\pi}{5}$ and $\mathrm{m}\angle BDC=\frac{2\pi}{5}$, so $\triangle ABC\sim\triangle BCD$. Also note that $\triangle ABD$ is isosceles so that $BC=BD=AD$.
Let $x=BC=BD=AD$. From the similar triangles, $\frac{AB}{BC}=\frac{BC}{CD}$ or $\frac{1}{x}=\frac{x}{1-x}$, so $1-x=x^2$ and $x=\frac{\sqrt{5}-1}{2}$ (the other solution is negative and lengths cannot be negative).
Now, apply the Law of Cosines to $\triangle ABC$:
$$\begin{align}
\cos\frac{2\pi}{5}=\cos C&=\frac{a^2+b^2-c^2}{2ab}
\\\\
&=\frac{\left(\frac{\sqrt{5}-1}{2}\right)^2+1^2-1^2}{2\cdot\frac{\sqrt{5}-1}{2}\cdot 1}
\\\\
&=\frac{\frac{\sqrt{5}-1}{2} \cdot \frac{\sqrt{5}-1}{2}}{2\cdot\frac{\sqrt{5}-1}{2}}
\\\\
&=\frac{\sqrt{5}-1}{4}.
\end{align}$$
| {
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If $2 x^4 + x^3 - 11 x^2 + x + 2 = 0$, then find the value of $x + \frac{1}{x}$?
If $2 x^4 + x^3 - 11 x^2 + x + 2 = 0$, then find the value of $x + \frac{1}{x}$ ?
I would be very grateful if somebody show me how to factor this polynomial by hand, as of now I have used to Mathematica to get $(x-2) (2x - 1) (1 + 3x + x^2)$, but I am having trouble to factor this manually.
Best regards,
| You don't need to factor the polynomial to find the value of $x + \frac{1}{x}$. Note that the given condition implies $\left( 2x^2 + \frac{2}{x^2} \right) + \left( x + \frac{1}{x} \right) - 11 = 0$ by dividing by $x^2$ and collecting symmetric terms. Now observe that $\left( x + \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} + 2$...
But if you want to factor it by hand anyway, your best friend is the rational root test.
| {
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Solving the equation $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$ I wanna know how to solve this equation: $-2x^3 +10x^2 -17x +8=(2x^2)(5x -x^3)^{1/3}$
I have some trouble to do that and I'd glad with any help I may get.
| Cube both sides and collect terms. You should get
\begin{eqnarray}
512 - 3264 x + 8856 x^2 - 13457 x^3 + 12702 x^4 - 7794 x^5 +
3136 x^6 - 844 x^7 + 120 x^8 = 0
\end{eqnarray}
which factorizes into
\begin{eqnarray}
(8 - 17 x + 6 x^2) (64 - 272 x + 481 x^2 - 456 x^3 + 258 x^4 -
84 x^5 + 20 x^6) = 0.
\end{eqnarray}
Since this is an $8^{\text{th}}$-degree polynomial equation, you can use Mathematica or a calculator to determine six of the eight solutions (by solving the $6^{\text{th}}$-degree polynomial equation numerically) and the other two by solving the quadratic equation: $\frac{1}{12}(17 \pm \sqrt{97})$. The 6 solutions are 3 pairs of complex conjugates:
\begin{eqnarray}
& & 0.647522 \dots \pm i 2.21209 \dots \
& & 0.657656 \dots \pm i 0.218497 \dots \
& & 0.794822 \dots \pm i 0.788971 \dots
\end{eqnarray}
Of course, you'll have to check that these solutions work. At most one real solution and two complex solutions are spurious.
| {
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Find the value of $\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ}) $ How to find the value of
$$\displaystyle\sqrt{3} \cdot \cot (20^{\circ}) - 4 \cdot \cos (20^{\circ})$$
manually ?
| For $\sin x\ne0,$
$$2\sin3x\cot x-4\cos x$$
$$=\dfrac{2\sin3x\cos x-2(2\cos x\sin x)}{\sin x}$$
$$=\dfrac{\sin4x+\sin2x-2\sin2x}{\sin x}$$
$$=\dfrac{\sin4x-\sin2x}{\sin x}$$
$$=2\cos3x$$
Now set $\ 2\sin3x=\dfrac{\sqrt3}2=\sin60^\circ\implies 3x=n180^\circ+(-1)^n60^\circ\text{ where } n \text{ is any integer}$
So, $x=60^\circ n+(-1)^n20^\circ$ where $n\equiv-1,0,1\pmod3$
Here $n=0$
| {
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Partial Derivative proof For
$$f(x,y) = \begin{cases}
\frac{xy}{x^2+y^2} &\text{if }(x,y)\neq (0,0);\\
0 &\text{if }(x,y)=(0,0).\end{cases}$$
I'm trying to prove that $\frac{\partial f}{\partial x}$, the partial derivative with respect to $x$, exists.
Taking the partial derivative freehand I get
$$\frac{y(x^2+y^2)-xy(2x)}{(x^2+y^2)^2}=
\frac{y(y^2-x^2)}{(x^2+y^2)^2}.$$
But using the limits I get
$$\lim_{x\to c}\frac{f(x,y)-f(c,y)}{x-c} = y(y^2-c^2).$$
I'm not sure what I'm doing wrong...
| I have no idea what you did with the limit. For $y\neq 0$ or $c\neq 0$, we have:
\begin{align*}
\lim{x\to c}\frac{f(x,y)-f(c,y)}{x-c} &= \lim_{x\to c}\frac{\frac{xy}{x^2+y^2} - \frac{cy}{c^2+y^2}}{x-c}\\
&= \lim_{x\to c}\frac{xy(c^2+y^2)-cy(x^2+y^2)}{(c-y)(x^2+y^2)(c^2+y^2)}\\
&= \lim_{x\to c}\frac{(x-c)(y^3-xyc)}{(x-c)(x^2+y^2)(c^2+y^2)}\\
&= \lim{x\to c}\frac{y^3-xyc}{(x^2+y^2)(c^2+y^2)}\\
&= \frac{y(y^2-c^2)}{(c^2+y^2)^2}.
\end{align*}
So, how did you get what you claim you got?
Added. For the partial at $(x,y)=(0,0)$, you have:
$$\lim_{x\to 0}\frac{f(x,0)-f(0,0)}{x-0} = \lim_{x\to 0}\frac{0 - 0}{x} = 0.$$
Since the limits exist, the partial derivative exists at each point.
| {
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Quadratic Congruence and Sum of Two Squares How would one go about showing how many solutions the following congruence has?
$$x^2 + y^2 \equiv 23 \pmod{93}.$$
| If you are working modulo a prime $p$, then you can obtain the result using the Legendre symbol or by simple trial and error if it is small. For instance, working modulo $3$, as CJost suggests, you are trying to solve $x^2 + y^2 \equiv 23 \equiv 2 \pmod{3}$. then you want to solve $x^2 \equiv 2-y^2\pmod{3}$. Since $y^2$ can only be congruent to $0$ or to $1$ modulo $3$, you only have solutions when $x^2\equiv y^2\equiv 1\pmod{3}$. That gives you four solutions modulo $3$: $(1,1)$, $(1,-1)$, $(-1,1)$, and $(-1,-1)$.
You need to do something similar for $31$: for each possibly value of $y^2$, decide if $23-y^2$ is a square modulo $31$; if it is, that gives you two possible values for $x$ and two possible values for $y$ (unless one of the two needs to be $0$ modulo $23$). But remember that you are working modulo $31$; for example, with $y=5$, there is no problem with the fact that $23-y^2 = -2$, because you are trying to solve $x^2\equiv -2\pmod{31}$, and this is the same as $x^2\equiv 29 \pmod{31}$; the negative number is not a problem. And of course, to decide if $x^2\equiv 29\pmod{31}$ has a solution, you can use the Legendre symbol and quadratic reciprocity to compute it.
Added: As it happens, $23$ is not a square modulo $31$, so any solutions to $x^2+y^2\equiv 23\pmod{31}$ will necessarily have both $x$ and $y$ not congruent to zero modulo $31$; so you don't need to worry about that when you are counting solutions.
Now, suppose you find a solution $x=a$ and $y=b$ modulo $31$. How do you get solutions modulo $93$? You take one solution modulo $3$, say $(1,1)$, and the solution modulo $31$, say $(a,b)$, and you use the Chinese Remainder Theorem: you can find an integer $x$ such that $x\equiv 1 \pmod{3}$ and $x\equiv a\pmod{31}$; then find an integer $y$ such that $y\equiv 1\pmod{3}$ and $y\equiv b\pmod{31}$. Then the pair $(x,y)$ will be a solution modulo $93$ that corresponds to the solution $(1,1)$ modulo $3$ and the solution $(a,b)$ modulo $31$. So each choice of a solution modulo $3$ and a solution modulo $31$ will give you a solution modulo $93$.
| {
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Geometry problem on the incentre and circumcenter of a triangle I have the following problem:
In a triangle $ABC$ the line joining incentre and circumcentre is parallel to side $BC$. Prove that $\cos B + \cos C=1$.
Could someone help me solve it?
| This one has been untouched for a few days now, so here's a solution. It does the job but it feels like I beat it into submission with a stick.
I would be interested to know if anyone has a more elegant approach, as it was tougher than I was expecting.
In the above diagram $I$ is the incentre and $K$ is the circumcentre of $\Delta$ ABC.
$PS \parallel BC $ therefore
$$ \frac{CP}{BS}=\frac{AC}{AB}= \frac{ \sin B}{\sin C}, \quad (1)$$
by the sine rule.
Furthermore, $\Delta$ CIP and $\Delta$ BIS are isosceles and so
$$ \frac{PS}{BC}=\frac{CP+BS}{BC}= 1 - \frac{BS}{AB},$$
the last equality is implied by $PS \parallel BC .$ And so
$$ CP = \left( 1 - \frac{BS}{AB} \right) BC - BS = \frac{\sin B}{\sin C} \cdot BS,$$
using $(1).$
Hence
$$ BC = BS \left( 1+ \frac{\sin B}{\sin C} + \frac{BC}{AB} \right)
= BS \left( 1+ \frac{\sin B}{\sin C} + \frac{\sin (B+C)}{\sin C} \right), \quad (2)$$
by the sine rule.
Let $ \angle CBK=\theta$ then, again by the sine rule and noting that $BK=CK,$
we have
$$\frac{\sin B}{\sin (B+C)} = \frac{AC}{BC} = \frac{\cos (C-\theta)}{\cos \theta} =
\cos C + \sin C \tan \theta.$$
Hence
$$ \tan \theta = \frac{ \frac{\sin B}{\sin (B+C)} - \cos C}{\sin C}. $$
Also
$$BC = 2 RK \cot \theta = 2 BS \sin B \cot \theta
= \frac{2BS \sin B \sin C}{ \frac{\sin B}{\sin(B+C)} - \cos C}, \quad (3)$$
using $RK = BS \sin B$ and the expression for $\tan \theta.$
Combining $(2)$ and $(3)$ we obtain
$$ 1+ \frac{\sin B}{\sin C} + \frac{\sin (B+C)}{\sin C}
=\frac{2\sin B \sin C}{ \frac{\sin B}{\sin(B+C)} - \cos C}. \quad (4)$$
Now we need to show that $(4) \Rightarrow \cos B + \cos C = 1.$
ABC is a non-degenerate triangle and so $C \ne 0$ and $\cos B + \cos C \ne -1$ and so
$\cos B + \cos C = 1 \Longleftrightarrow$
$$2\sin B \sin^2 C = 2 \sin B (1- \cos^2 C) + \sin C
\lbrace 1 - (\cos B + \cos C)^2 \rbrace $$
$$= 2\sin B - 2\sin B \cos^2 C + \sin C (-2\cos B \cos C + \sin^2 B - \cos^2 C)$$
$$=\lbrace 2 + \cos(B-C) \rbrace \lbrace \sin B - \cos C \sin(B+C) \rbrace, \quad (5)$$
noting that $\sin(B+C)\cos(B-C) = \sin B \cos B + \sin C \cos C.$ When we expand the RHS of $(5)$ the two terms $\sin B \cos B \cos C$ and $-\sin B \cos B \cos C$ cancel, leaving precisely the RHS of the second to last line of $(5).$
Hence we have $\cos B + \cos C = 1 \Longleftrightarrow$
$$ \frac{2+ \cos(B-C)}{\sin C} = \frac{2 \sin B \sin C}{\sin B - \cos C \sin (B+C)}. \quad (6)$$
Now
$$\sin (B+C) \lbrace 1+ \cos(B-C) \rbrace = \sin(B+C) + \sin(B+C) \cos(B-C)$$
$$=\sin B \cos C + \cos B\sin C + \sin B \cos B + \sin C \cos C $$
$$=(\sin B + \sin C)(\cos B + \cos C) = \sin B + \sin C \quad (7)$$
$\Longleftrightarrow \cos B + \cos C = 1,$ since $\sin B + \sin C \ne 0.$
And so the LHS of $(4)$
$$ 1+ \frac{\sin B}{\sin C} + \frac{\sin (B+C)}{\sin C} =
\frac{\sin B + \sin C + \sin(B+C)}{\sin C}$$
and using $(7)$ to substitute for $\sin B + \sin C$
$$=\frac{2\sin(B+C)+ \sin(B+C)\cos(B-C)}{\sin C}
=\sin(B+C)\frac{2+\cos(B-C)}{\sin C}$$
and using $(6)$ to substitute for the fraction
$$=\frac{2\sin B \sin C \sin(B+C)}{\sin B - \cos C \sin(B+C)}
=\frac{2\sin B \sin C}{ \frac{\sin B}{\sin(B+C)} - \cos C}.$$
And so we have shown that when $B$ and $C$ are angles of a triangle then
$(4) \Rightarrow \cos B + \cos C = 1$ and the result is proved.
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Probability of 3 of a kind with 7 dice Similar questions:
Chance of 7 of a kind with 10 dice
Probability of getting exactly $k$ of a kind in $n$ rolls of $m$-sided dice, where $k\leq n/2$
Probability was never my thing, so please bear with me.
I've reviewed the threads above to the best of my ability, but I still wonder how to go about finding a match of 3 from 7 dice.
At least three match, but no more (two sets of three is okay, a set of three and a set of four is not):
(a) : $ \frac{6 \binom{7}{3} 5^4}{6^7} $
In the other discussions, this wasn't desired since it would allow for a second triple to occur, or even a quadruple.
Odds of a quadruple with the remaining 4 dice:
(b) : $(1/5)^4 $
Then, the probability that from rolling 7 dice that there is at least three that match, and no more than three, would be:
(c) : $ \frac{6 \binom{7}{3} * 5^4}{6^7}- (1/5)^4 $
Exactly two sets of three:
(d) : $ \frac{6 \binom{7}{3} \binom{4}{3} \binom{1}{1}}{ 6^7} $
Maybe? My thought process was that if $\binom{7}{3}$ will give me a set of three, then with the remaining 4, I could pick 3 yielding $\binom{4}{3}$ with 1 leftover. I realize this is probably wrong. Why? What would be the proper way to go about this?
Exactly one set of three:
Then to find the probability that there is one and only one set of three from 7 dice, we could take the probability of one or more sets of three (c) and subtract the probability of exactly two sets (d), for:
$ \frac{6 \binom{7}{3} 5^4}{ 6^7} - (1/5)^4 - \frac{6 \binom{7}{3} \binom{4}{3} \binom{1}{1} }{ 6^7} $
(e) : $ \left(\frac{6 \binom{7}{3}}{ 6^7}\right) \left( 5^4 - \binom{4}{3} \binom{1}{1} \right) - (1/5)^4 $
Is this at all on the right path?
Thank you!
PS.
Sorry about the syntax, but I couldn't figure out how to make the standard nCr() symbol with MathJaX.
| Carrying around the $6^7$ is just complicating your life. Instead, just count how many distinct rolls have exactly one triple; the end probability will be that count, divided by the total number of possible rolls (namely, $6^7$).
There is also the issue of distinguishable and non-distinguishable dice. But let's assume you roll a single dice in sequence and write down the results.
Proceeding along the lines you have, and hoping I'm not making a mistake:
*
*At least 3-of-a-kind of x, with the other four dice being anything except x. This one you compute correctly: pick what is x (six possibilities), then pick which three dice fall come up x ($\binom{7}{3}$ ways), then choose any of five possible outcomes for each of the remaining four dice, $5^4$. So you have $\binom{6}{1}\binom{7}{3}5^4$.
Edit. However, this overcounts! (Thanks to Henry for spotting it) If a roll has two different three-of-a-kind, then I count that roll twice; once when I'm counting the lower three-of-a-kind, and again when I'm counting the higher one. So we need to subtract those rolls that have two three-of-a-kinds. We select the two ranks by using $\binom{6}{2}$, then select the dice that show the lower rank with $\binom{7}{3}$, and then select the dice, out of the remaining $4$, that have the higher rank three-of-a-kind with $\binom{4}{3}$. Then the remaining die can be any of the remaining $4$ ranks; so we overcounted by $\binom{6}{2}\binom{7}{3}\binom{4}{3}4$. So the real number of rolls that have at least one three-of-a-kind is
$$\binom{6}{1}\binom{7}{3}5^4 - \binom{6}{2}\binom{7}{3}\binom{4}{3}4.$$
*Discard the case in which we get 3-of-a-kind of x, but the other four dice are equal to y, with x$\neq$ y. There are $\binom{6}{1}$ ways of picking x, and then you need to pick what the other $4$ are; there are $\binom{5}{1}$ ways of doing it; then pick the three dice that will come up x, $\binom{7}{3}$ ways. So you have $\binom{6}{1}\binom{5}{1}\binom{7}{3}$ outcomes that you don't want.
*Discard the case in which you get two sets of 3-of-a-kind. We already figured out, when we computed the overcount, that there are
$$4\binom{6}{2}\binom{7}{3}\binom{4}{3}$$
rolls with two 3-of-a-kind. So we must subtract this total again.
So: the total number of rolls in which the largest number of dice that come up the same is $3$, and there is one and only one set that does this, is:
$$\binom{6}{1}\binom{7}{3}5^4 - \binom{6}{1}\binom{7}{3}\binom{5}{1} - 2\times\left(\binom{6}{2}\binom{7}{3}\binom{4}{3}4\right) = 113400.$$
Given that there are $6^7$ sequential rolls possible, you get a probability of
$$\frac{\mbox{number of rolls with exactly one three-of-a-kind, no 4-of-a-kind}}{6^7} = \frac{113400}{6^7} \approx 0.4050925926.$$
| {
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Dot product of two vectors How does one show that the dot product of two vectors is $ A \cdot B = |A| |B| \cos (\Theta) $?
| Think about a triangle with sidelengths $|\textbf{a}|,|\textbf{b}|,|\textbf{c}|$. Then we can use the law of cosines.
$$
\begin{align}
|\textbf{c}|^2&=|\textbf{a}|^2+|\textbf{b}|^2-2|\textbf{a}||\textbf{b}| \cos \theta \\
\implies 2|\textbf{a}||\textbf{b}| \cos \theta &= |\textbf{a}|^2+|\textbf{b}|^2-|\textbf{c}| = \textbf{a}\cdot \textbf{a} + \textbf{b} \cdot \textbf{b} - \textbf{c}\cdot \textbf{c}
\end{align}
$$
By the properties of the dot product and from the fact that $\textbf{c}=\textbf{b}-\textbf{a}$ we find that
$$
\begin{align}
\textbf{c}\cdot \textbf{c} &=(\textbf{b}-\textbf{a}) \cdot (\textbf{b}-\textbf{a}) \\
&=(\textbf{b}-\textbf{a})\cdot \textbf{b} - (\textbf{b}-\textbf{a}) \cdot \textbf{a} \\
&= \textbf{b}\cdot \textbf{b} - \textbf{a}\cdot \textbf{b} - \textbf{b}\cdot \textbf{a} + \textbf{a}\cdot \textbf{a}.
\end{align}
$$
By substituting $\textbf{c}\cdot \textbf{c}$, we get
$$
\begin{align}
2|\textbf{a}||\textbf{b}| \cos \theta &= \textbf{a}\cdot \textbf{a} + \textbf{b} \cdot \textbf{b} - (\textbf{b}\cdot \textbf{b} - \textbf{a}\cdot \textbf{b} - \textbf{b}\cdot \textbf{a} + \textbf{a}\cdot \textbf{a}) \\
&=2 \textbf{a}\cdot \textbf{b}.
\end{align}
$$
| {
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Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$ This is being asked in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions, and here: List of abstract duplicates.
What methods can be used to evaluate the limit $$\lim_{x\rightarrow\infty} \sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x.$$
In other words, if I am given a polynomial $P(x)=x^n + a_{n-1}x^{n-1} +\cdots +a_1 x+ a_0$, how would I find $$\lim_{x\rightarrow\infty} P(x)^{1/n}-x.$$
For example, how would I evaluate limits such as $$\lim_{x\rightarrow\infty} \sqrt{x^2 +x+1}-x$$ or $$\lim_{x\rightarrow\infty} \sqrt[5]{x^5 +x^3 +99x+101}-x.$$
| Possibly more elementary proof based on $\frac{c^{n}-d^{n}}{c-d} = \sum_{k=0}^{n-1} c^{n-1-k} d^k$. Using this for $c = \sqrt[n]{ x^n+ \sum_{m=0}^{n-1} a_m x^m }$ and $d=x$.
$$
c - d = \frac{c^{n}-d^{n}}{ \sum_{k=0}^{n-1} c^{n-1-k} d^k } = \frac{ a_{n-1} x^{n-1} + \ldots + a_1 x + a_0}{ x^{n-1} \sum_{k=0}^{n-1} (\frac{c}{d})^{n-1-k} } = \frac{ a_{n-1} + a_{n-2} x^{-1} + \ldots + a_0 x^{1-n}}{\sum_{k=0}^{n-1} (\frac{c}{d})^{n-1-k} }
$$
Now $\lim_{x\to \infty} \frac{c}{d} = \lim_{x \to \infty} \sqrt[n]{ 1 + \frac{a_{n-1}}{x} + \ldots + \frac{a_0}{x^n} } = 1$. This gives $\frac{a_{n-1}}{n}$.
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Sum of series $\sum_{k=1}^\infty{-\frac{1}{2^k-1}}$ What is the sum of the series $\displaystyle\sum_{k=1}^\infty{\frac{-1}{2^k-1}}$?
Also, more generally, can we find $\displaystyle\sum_{k=1}^\infty{\frac{-1}{c^k-1}}$ for some $c$?
| Ignoring the negative sign, you're looking for
$$ {1 \over 2^1 - 1} + {1 \over 2^2 - 1} + {1 \over 2^3 - 1} + \cdots $$
but we have
$$ {1 \over 2^{jk} - 1} = {1 \over 2^{jk}} + {1 \over 2^{2jk}} + {1 \over 2^{3jk}} + \cdots. $$
If we rewrite the first expression using the second one, then your sum is
$$ \left( {1 \over 2^1} + {1 \over 2^2} + {1 \over 2^3} + \cdots \right) + \left( {1 \over 2^2} + {1 \over 2^4} + {1 \over 2^6} + \cdots \right) + \left( {1 \over 2^3} + {1 \over 2^6} + {1 \over 2^9} + \cdots \right) + \cdots $$
and $1/2^r$ appears $\tau(r)$ times, where $\tau(r)$ is the number of divisors of $r$. Therefore your sum is
$$ \sum_{r \ge 1} \tau(r) 2^{-r} $$
and this should allow you to compute it to any desired level of numerical accuracy fairly quickly. For example,
$$ \sum_{r=1}^{40} \tau(r) 2^{-r} = {27602812537 \over 17179869184} = 1.606695152 $$
and the error here is less than $\sum_{r \ge 41} r 2^{-r} = 84/2^{41} < 4 \times 10^{-11}$ since $\tau(r) < r$ for all $r \ge 3$.
Of course this method works when $2$ is replaced by any constant $c > 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I sum the infinite series $\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad$ How can I find the sum of the infinite series
$$\frac{1}{5} - \frac{1\cdot4}{5\cdot10} + \frac{1\cdot4\cdot7}{5\cdot10\cdot15} - \cdots\qquad ?$$
My attempt at a solution - I saw that I could rewrite it as
$$\frac{1}{5}\left(1 - \frac{4}{10} \left( 1 - \frac{7}{15} \left(\cdots \left(1 - \frac{3k - 2}{5k}\left( \cdots \right)\right)\right.\right.\right.$$
and that $\frac{3k - 2}{5k} \to \frac{3}{5}$ as $k$ grows larger. Using this I thought it might converge to $\frac{1}{8}$, but I was wrong, the initial terms deviate significantly from $\frac{3}{5}$.
According to Wolfram Alpha it converges to $1-\frac{\sqrt[3]{5}}{2}$. How can I get that ?
| \begin{align*}(-1)^{n-1}\frac{1\cdot 4 \dots (3n-2)}{5\cdot 10 \dots 5n}&=
(-1)^{n-1}\frac{3^n}{5^n}(-1)^n\frac{(-\frac{1}{3})\cdot (-\frac{4}{3}) \dots (-\frac{3n-2}{3})}{1\cdot 2 \dots n}\\ &= -(3/5)^n\binom{-1/3}{n}\end{align*}
Therefore, you can obtain
$$\sum_{n=1}^{\infty} -(3/5)^n\binom{-1/3}{n} = 1 - \sum_{n=0}^{\infty} (3/5)^n\binom{-1/3}{n} = 1- (1+\frac35)^{-1/3} = 1- \sqrt[3]{5/8} =1-\frac{\sqrt[3]5}{2}$$
| {
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Writing a percent as a decimal and a fraction I am having a problem understanding some manipulations with recurring decimals. The exercise is
Write each of the following as a
decimal and a fraction:
(iii) $66\frac{2}{3}$%
(iv) $16\frac{2}{3}$%
For (iii), I write a decimal $66\frac{2}{3}\% = 66.\bar{6}\% = 0.66\bar{6} = 0.\bar{6}$ and a fraction $66\frac{2}{3}\% = \frac{66*3 + 2}{3*100}\% = \frac{200}{3}\times\frac{1}{100} =\frac{2}{3}$.
For (iv), I follow a similar path to establish that the fraction is $\frac{1}{6}$ and the decimal is $0.1\bar{6}$.
What I don't understand is a part of the model answer for this exercise. They say
$66\frac{2}{3}\% = 66.\bar{6}\% = 0.\bar{6} = \frac{6}{9} = \frac{2}{3}$
and
$16\frac{2}{3}\% = 16.\bar{6}\% = 0.1\bar{6} = \frac{16-1}{90} = \frac{15}{90} = \frac{1}{6}$
I did not do much work with recurring decimals before and I don't know how to justify the 9 and 90 in the denominator.
Could you please explain?
| The model answer for this exercise is too complicated: $66 {2\over 3} = 66 + {2\over 3} = {200 \over 3}$. So $66 {2\over 3} \% = {200 \over 300} = {2 \over 3}$. Similarly, $16 {2\over 3} = {50 \over 3}$ and $16 {2\over 3}\% = {50 \over 300} = {1 \over 6}$.
| {
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Real and imaginary parts of the Möbius transformation Given that the Möbius transformation is:
$f(z) = \dfrac{az+b}{cz+d} ,\, (a d-b c) \neq 0$ and with $a,b,c$ and $d$ complex numbers written $a= a_1 + a_2i$ etc.
I think I must be missing something because when separating the Möbius transformation in to its real and imaginary parts I got this:
$\dfrac{(a_1c_1+a_2c_2)z^2 + (b_1c_1+b_2c_2)z +(a_1d_1+a_2d_2)z +(b_1d_1+b_2d_2)}{|c|^2z^2 + |d|^2}$ for the real part... and
$\dfrac{(a_1c_1-a_2c_2)z^2 + (b_1c_1-b_2c_2)z +(a_1d_1-a_2d_2)z +(b_1d_1-b_2d_2)}{|c|^2z^2 + |d|^2}i$ as the imaginary.
It really looks awful, is there a better way to write this?
Update. Duh. Dumb mistake, reworking them I get this, the point is it's still ugly.
Real:
$\dfrac{(a_1x-a_2y+b_1)(c_1x-c_2y+d_1) -(a_2x+a_1y+b_2)(c_2x+c_1y+d_2)}{(c_1x-c_2y+d_1)^2 - (c_2x+c_1y+d_2)^2}$
$\dfrac{R(ac)x^2 - 2(a_1c_2 + a_2c_1)xy + (R(ad) + R(bc))x - (R(ab) + R(bc))y - R(ac)y^2 + R(bd)}{(c_1x-c_2y+d_1)^2 - (c_2x+c_1y+d_2)^2}$
Imag:
$\dfrac{(a_2x-a_1y+b_2)(c_1x-c_2y+d_1) -(a_1x+a_2y+b_1)(c_2x+c_1y+d_2)}{(c_1x-c_2y+d_1)^2 - (c_2x+c_1y+d_2)^2}$
| Edit: Sorry - this only holds for $a,b,c,d \in \mathbb{R}$!
Thanks to some help in a question I asked:
\begin{align*}
\textrm{Re} (f(z))
&=\textrm{Re} \left(\frac{az+b}{cz+d}\right) \\
&=\textrm{Re} \left( \frac{(az+b)(c \overline{z}+d)}{(cz+d)(c \overline{z}+d)}\right) \\
&=\frac{1}{|cz+d|^2} \textrm{Re} \left( acz\overline{z}+bc\overline{z}+adz+bd\right) \\
&= \frac{\left(ac|z|^2+bd\right)+(ad+bc)\textrm{Re}(z)}{|cz+d|^2}
\end{align*}
and
\begin{align*}
\textrm{Im} (f(z))
&=\textrm{Im} \left(\frac{az+b}{cz+d}\right) \\
&=\textrm{Im} \left( \frac{(az+b)(c \overline{z}+d)}{(cz+d)(c \overline{z}+d)}\right) \\
&=\frac{1}{|cz+d|^2} \textrm{Im} \left( acz\overline{z}+bc\overline{z}+adz+bd\right) \\
&= \frac{(ad-bc)}{|cz+d|^2} \textrm{Im}(z)
\end{align*}
| {
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Finding double root. An easier way? Given the polynomial $f = X^4 - 6X^3 + 13X^2 + aX + b$ you have to find the values of $a$ and $b$ such that $f$ has two double roots.
I went about this by writing the polynomial as:
$$f = X^4 - 6X^3 + 13X^2 + aX + b = (X - r)^2(X - s)^2$$
... and from there I arrived at a system of four equations from where I could find the desired values. However, there's quite a bit of work to do it this way; is there a shortcut?
| HINT $\ $ Simpler, first solve $\rm\ f\: =\: (x^2 + c\ x + d)^2\:.\:$ This yields
$$\rm x^4 - 6\ x^3 +13\ x^2 +\cdots\ =\ x^4 + 2\:c\ x^3 + (2\:d+c^2)\ x^2\:+\cdots$$
hence $\rm\ 2\:c = -6\ $ so $\rm\ c = -3\:.\ $ And $\rm\ 2\:d+c^2 = 2\:d+9 = 13\ $ so $\rm\ d = 2\:.$
| {
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How to factor quadratic $ax^2+bx+c$? How do I shorten this? How do I have to think?
$$ x^2 + x - 2$$
The answer is
$$(x+2)(x-1)$$
I don't know how to get to the answer systematically. Could someone explain?
Does anyone have a link to a site that teaches basic stuff like this? My book does not explain anything and I have no teacher; this is self-studies.
Please help me out; thanks!
| Still one more way: if you don't want to remember the quadratic formula, then just remember how it is proved. Let's do an example: Factor $3x^2 + 4x - 5$.
1) What gives $3x^2$ when squared? Answer: $\sqrt{3}x$.
2) What number $a$ will I have to add to this to get $4x$ as one of the terms in $(\sqrt{3}x + a)^2$? Answer: The coefficient of $x$ in $(\sqrt{3}x + a)^2$ is $2\sqrt{3}a$, so the answer is $\frac{2}{\sqrt{3}}$.
3) What is $\left(\frac{2}{\sqrt{3}}\right)^2$? It is $\frac{4}{3}$.
From these considerations we get
$$3x^2 + 4x - 5 = (\sqrt{3}x + \frac{2}{\sqrt{3}})^2 - \frac{4}{3} - 5 = (\sqrt{3}x + \frac{2}{\sqrt{3}})^2 - \frac{19}{3}.$$
Finally you can use the factorization $a^2 - b^2 = (a+b)(a-b)$ to conclude
$$3x^2 + 4x - 5 = (\sqrt{3}x + \frac{2+\sqrt{19}}{\sqrt{3}})(\sqrt{3}x + \frac{2-\sqrt{19}}{\sqrt{3}}).$$
| {
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Evaluate $\sum\limits_{k=1}^n k^2$ and $\sum\limits_{k=1}^n k(k+1)$ combinatorially
$$\text{Evaluate } \sum_{k=1}^n k^2 \text{ and } \sum_{k=1}^{n}k(k+1) \text{ combinatorially.}$$
For the first one, I was able to express $k^2$ in terms of the binomial coefficients by considering a set $X$ of cardinality $2k$ and partitioning it into two subsets $A$ and $B$, each with cardinality $k$. Then, the number of ways of choosing 2-element subsets of $X$ is $$\binom{2k}{2} = 2\binom{k}{2}+k^2$$ So sum $$\sum_{k=1}^n k^2 =\sum_{k=1}^n \binom{2k}{2} -2\sum_{k=2}^n \binom{k}{2} $$ $$ \qquad\qquad = \color{red}{\sum_{k=1}^n \binom{2k}{2}} - 2 \binom{n+1}{3} $$ I am stuck at this point to evaluate the first of the sums. How to evaluate it?
I need to find a similar expression for $k(k+1)$ for the second sum highlighted above. I have been unsuccessful this far. (If the previous problem is done then so is this, but it would be nice to know if there are better approaches or identities that can be used.)
Update: I got the second one. Consider $$\displaystyle \binom{n+1}{r+1} = \binom{n}{r}+\binom{n-1}{r}+\cdots + \binom{r}{r}$$ Can be shown using recursive definition. Now multiply by $r!$ and set $r=2$
| We show bijectively that
$$2^2+4^2+6^2+\cdots +(2n)^2=\binom{2n+2}{3}.$$
That does not quite give a purely combinatorial expression for $1^2+2^2+ \cdots +n^2$, since we still need to divide by $4$, which is "algebra." But the sin of multiplying or dividing by a constant seems to be a small one in this game. And the result is interesting.
Imagine choosing $3$ numbers from the numbers $1, 2, \dots, 2n+2$. This can be done in $\binom{2n+2}{3}$ ways.
We will show that the number of choices is also
$$2^2+4^2+6^2+\cdots +(2n)^2.$$
Maybe the largest number chosen is $2k+1$ or $2k+2$. If it is $2k+1$, the other two can be chosen in $\binom{2k}{2}$ ways, and if it is $2k+2$, then the other two can be chosen in $\binom{2k+1}{2}$ ways. Using "algebra", we find that
$$\binom{2k}{2}+\binom{2k+1}{2}= (2k)^2.$$
Take $k=1, 2, 3, \dots, n$. We conclude that the number of ways of choosing $3$ numbers from $1, 2, \dots, 2n+2$ is
$$2^2+4^2+6^2+\cdots +(2n)^2.$$
Since the number of choices is clearly $\binom{2n+1}{3}$, this almost completes the argument. (In a note at the end of this post, we deal with the minor task of eliminating the use of algebra.)
Comment: If we allow division by $4$, we can conclude that
$$1^2+2^2+\cdots +n^2=\frac{1}{4}\binom{2n+2}{3}.$$
I used to think that in the usual expression $\frac{(n)(n+1)(2n+1)}{6}$ for the sum of the first $n$ squares, the $2n+1$ somehow is the weird term, it does not fit in. But it does! Actually, it is $n$ and $n+1$ that are strange. We can think of $(n)(n+1)(2n+1)$ as ``really'' being $(1/4)[(2n)(2n+1)(2n+2)]$.
Cleaning up: There is a standard way to avoid the ``calculation''
$$\binom{2k}{2}+\binom{2k+1}{2}= (2k)^2.$$
For completeness we show bijectively that for any $m$, in particular $m=2k$, we have
$$\binom{m}{2} +\binom{m+1}{2}=m^2.$$
For let $M$ be the collection $\left\{1,2,3,\dots,m\right\}$. Then the number of ordered pairs $(a,b)$ with $a$ and $b$ in $M$ is $m^2$. These ordered pairs are of two types: (i) the ones with $a<b$ and (ii) the ones with $a \ge b$. The number of ordered pairs $(a,b)$ with $a<b$ is just $\binom{m}{2}$. The ordered pairs $(a,b)$ with $a \ge b$ are in a natural bijection with the ordered pairs $(b,a+1)$ with $b<a+1$, and there are $\binom{m+1}{2}$ of these.
| {
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How do I get the square root of a complex number? If I'm given a complex number (say $9 + 4i$), how do I calculate its square root?
| Claim 1. Suppose $b\neq 0$. Then the two roots to the equation $x^2 = a +bi$ are: $$\pm\frac{\sqrt{2}}{2}\left(\sqrt{\sqrt{a^{2}+b^{2}}+a}+\mathrm{i}\frac{b}{\left|b\right|}\sqrt{\sqrt{a^{2}+b^{2}}-a}\right).$$
Claim 2. Suppose $b>0$. Then:
(a) The two roots to the equation $x^2 = a +bi$ are $$x = \pm\frac{\sqrt{2}}{2}\left(\sqrt{\sqrt{a^{2}+b^{2}}+a}+\textrm{i}\sqrt{\sqrt{a^{2}+b^{2}}-a}\right). $$
(b) And the two roots to the equation $x^2 = a -bi$ are $$x=\pm\frac{\sqrt{2}}{2}\left(\sqrt{a+\sqrt{a^{2}-b^{2}}}-\textrm{i}\sqrt{a-\sqrt{a^{2}-b^{2}}}\right).$$
Proof of Claim 1. \begin{alignat*}{1}
& \left[\pm\frac{\sqrt{2}}{2}\left(\sqrt{\sqrt{a^{2}+b^{2}}+a}+\mathrm{i}\frac{b}{\left|b\right|}\sqrt{\sqrt{a^{2}+b^{2}}-a}\right)\right]^{2}\\
=\ & \frac{1}{2}\left[\sqrt{a^{2}+b^{2}}+a-\left(\sqrt{a^{2}+b^{2}}-a\right)+2\mathrm{i}\frac{b}{\left|b\right|}\sqrt{\left(\sqrt{a^{2}+b^{2}}+a\right)\left(\sqrt{a^{2}+b^{2}}-a\right)}\right]\\
=\ & \frac{1}{2}\left(2a+2\mathrm{i}\frac{b}{\left|b\right|}\sqrt{a^{2}+b^{2}-a^{2}}\right)=a+\mathrm{i}\frac{b}{\left|b\right|}\left|b\right|=a+\mathrm{i}b.
\end{alignat*}
Proof of Claim 2. \begin{alignat*}{1}
\textbf{(a)} \ & \left[\pm\frac{\sqrt{2}}{2}\left(\sqrt{\sqrt{a^{2}+b^{2}}+a}+\mathrm{i}\sqrt{\sqrt{a^{2}+b^{2}}-a}\right)\right]^{2}\\
= \ & \frac{1}{2}\left[\sqrt{a^{2}+b^{2}}+a-\left(\sqrt{a^{2}+b^{2}}-a\right)+2\mathrm{i}\sqrt{\left(\sqrt{a^{2}+b^{2}}+a\right)\left(\sqrt{a^{2}+b^{2}}-a\right)}\right]\\
= \ & \frac{1}{2}\left(2a+2\mathrm{i}\sqrt{a^{2}+b^{2}-a^{2}}\right)=a+\mathrm{i}\sqrt{b^{2}}=a+\mathrm{i}b.
\end{alignat*}
\begin{alignat*}{1}
\textbf{(b)} \ & \left[\pm\frac{\sqrt{2}}{2}\left(\sqrt{a+\sqrt{a^{2}-b^{2}}}-\mathrm{i}\sqrt{a-\sqrt{a^{2}-b^{2}}}\right)\right]^{2}\\
= \ & \frac{1}{2}\left[a+\sqrt{a^{2}-b^{2}}+a-\sqrt{a^{2}+b^{2}}-2\mathrm{i}\sqrt{\left(a+\sqrt{a^{2}-b^{2}}\right)\left(a-\sqrt{a^{2}-b^{2}}\right)}\right]\\
= \ & \frac{1}{2}\left[2a-2\mathrm{i}\sqrt{a^{2}-\left(a^{2}-b^{2}\right)}\right]=a-\mathrm{i}\sqrt{b^{2}}=a-\mathrm{i}b.
\end{alignat*}
| {
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Prove that $6|2n^3+3n^2+n$ My attempt at it: $\displaystyle 2n^3+3n^2+n= n(n+1)(2n+1) = 6\sum_nn^2$
This however reduces to proving the summation result by induction, which I am trying to avoid as it provides little insight.
| to check for divisibility by 6 a number must be divisible by both 2 and 3 so we will prove that
$2n^3 + 3n^2 + n
= n (2n^2 + 3n +1)
= n (n+1) (2n+1)$
If $n$ is even then 2 divides $n$ and $n+1$ will be odd so $n+1$ can be $3k+2$ or $3k$ where $k$ is some integer.
If $3k+1 = n+1$ as it would make $n$ itself a multiple of 6 as our assumption that $n$ is even
So if $n+1$ is $3k$ it can be divided by 3 so no problem.
But if $n+1 = 3k+2$ then $2n+1$ will be $2(3k+1) +1 = 6k+3$ which is divisible by 3
and hence by 6.
Case 2: If $n$ is odd then $n+1$ is even and thus divisible by 2.
*
*if $n$ is $3k$ then it is divisible by 3 so no issues
*if $n$ is $3k+2$ as it makes $n+1=3k+3$ which is itself a multiple of 6 as our earlier assumption is that $n+1$ is even
*if $n$ is $3k+1$ then $2n+1$ becomes $2(3k+1)+1=6k+3$ which is divisible by 3
and hence proved
| {
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Seriously: What is the inverse of this 2-by-2 matrix? Okay, I have a hangover and it must be a stupid error, but I just don't get it:
The inverse of a 2-by-2 matrix $A=\left( \matrix{a & b \\ c& d} \right )$ is $\frac{1}{det A}\left( \matrix{d & -b \\ -c& a} \right ) = \frac{1}{ad-bc}\left( \matrix{d & -b \\ -c& a} \right )$. Now when I apply this recipe to the matrix
$$G:=\frac{1}{1+x^2+y^2}\left( \matrix{1+y^2 & -xy \\ -xy & 1+x^2} \right ) = \left( \matrix{\frac{1+y^2}{1+x^2+y^2} & \frac{-xy}{1+x^2+y^2} \\ \frac{-xy}{1+x^2+y^2} & \frac{1+x^2}{1+x^2+y^2}} \right )$$ I get
$$det G = \frac{1+x^2+y^2+x^2y^2 -x^2y^2}{(1+x^2+y^2)^2} = \frac{1}{1+x^2+y^2}$$
and hence
$$G^{-1} = (1+x^2+y^2)\left( \matrix{1+x^2 & xy \\ xy & 1+y^2} \right )$$
But direct calculation shows that the inverse of $G$ is $\left( \matrix{1+x^2 & xy \\ xy & 1+y^2} \right )$, without the factor in front. Am I mad or still drunk? Can someone help my addled brain? Thanks - seriously!
| Here is another way to solve the problem altogether. Notice that $$\det\left(\begin{array}{cc}
1+y^{2} & -xy\\
-xy & 1+x^{2}\end{array}\right)=1+x^{2}+y^{2},$$ and that $$\det\left(\begin{array}{cc}
d & -b\\
-c & a\end{array}\right)=\det\left(\begin{array}{cc}
a & b\\
c & d\end{array}\right).$$
This means $G$ is of the form $$G=\frac{1}{ad-bc}\left(\begin{array}{cc}
d & -b\\
-c & a\end{array}\right).$$
So the formula for $G$ really looks like the formula for the inverse of a matrix. It has $\frac{1}{\det}$ and the minus signs in the correct spots.
Then we see that
$$G^{-1}=\left(\begin{array}{cc}
1+x^{2} & xy\\
xy & 1+y^{2}\end{array}\right)$$
because applying the inversion formula to $G^{-1}$ will give $G$ directly.
| {
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How to add compound fractions? How to add two compound fractions with fractions in numerator like this one:
$$\frac{\ \frac{1}{x}\ }{2} + \frac{\ \frac{2}{3x}\ }{x}$$
or fractions with fractions in denominator like this one:
$$\frac{x}{\ \frac{2}{x}\ } + \frac{\ \frac{1}{x}\ }{x}$$
| One easy way to figure this out is that dividing by $x$ is the same as multiplying by $1/x$ (but all bets are off when $x=0$, as division by $0$ is undefined). So
$$
\begin{align*}
\frac{ \frac{a}{b} }{c} &= \frac{1}{c} \frac{a}{b} = \frac{a}{bc} \\ \\
\frac{a}{\frac{b}{c}} &= a \frac{1}{\frac{b}{c}} = a \frac{c}{b} = \frac{ac}{b}
\end{align*}
$$
| {
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number of combinations without repetition with limited supply Say, I got $1$ red ball, $1$ blue, $2$ yellow, $3$ green, totally $7$ balls.
I wanna select $3$ balls from them. How many ways I can do this?I counted manually
$$123, 124, 133, 134, 144, 233, 234, 244, 334, 344, 444,$$ so $11$ combinations.Is there a formula for it?
| Your problem is equivalent with this one.
$x_1+x_2+x_3+x_4=3$ where
$0\le x_1\le 1,~0\le x_2\le 1, 0\le x_3\le 2,~0\le x_4\le 3$
$0+0+0+3=3$
$0+0+1+2=3$
$0+1+0+2=3$
$1+0+0+2=3$
$0+0+2+1=3$
$0+1+2+0=3$
$1+0+2+0=3$
$0+1+1+1=3$
$1+0+1+1=3$
$1+1+0+1=3$
$1+1+1+0=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/52135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Limit of This Complicated Formula My brain had twisted because of this nasty problem.
let
$$r_{n}=\sqrt{n^2+n+\frac{3}{4}}$$
$$x_{n}=\left \lfloor \frac{r_{n}}{\sqrt{2}}-\frac{1}{2} \right \rfloor$$
$$a_{n}=\sum_{k=1}^{\left \lfloor r_{n}-x_{n} \right \rfloor} \left \lfloor \sqrt{n^2+n-k^2-k+\frac{1}{2}-(x_{n})^2-(2k+1)x_{n}}-\frac{1}{2} \right \rfloor $$
$$A_{n}=4 ( (x_{n})^2+2a_n+n )+1$$
Question. How can I find the limit of below?
$$\lim_{n\to\infty}\frac{A_{n}}{n^2}$$
| Let's use some little-O notation and hand-waving:
$$r_n = n + o(n)$$
$$x_n = \frac{n}{\sqrt{2}} + o(n)$$
$$\lfloor r_n - x_n\rfloor = \frac{2 - \sqrt{2}}{2}n + o(n)$$
$$a_n = \sum_{k = 1}^{\frac{2 - \sqrt{2}}{2}n + o(n)} \Big(\sqrt{1 - \frac{1}{\sqrt{2}}}\Big) n + o(n) - k + o(k) - \sqrt{2kn} $$
Thus,
$$ a_n = \Big(\frac{2 - \sqrt{2}}{2}\Big)^{3/2} n^2 + o(n^2) - \Big(\frac{2 - \sqrt{2}}{4} \Big) n^2 + o(n^2) - \sqrt{2n} H_{\frac{2 - \sqrt{2}}{2} n + o(n),-1/2} $$
Unfortunately I can only bound the last term:
$$ -\Big(\frac{2 - \sqrt{2}}{4}\Big) n^2 \leq a_n - o(n^2) \leq \Bigg(\Big(\frac{2 - \sqrt{2}}{2}\Big)^{5/2} - \frac{2 - \sqrt{2}}{4}\Bigg) n^2 $$
$$ - 0.14645 n^2 \leq a_n - o(n^2) \leq -0.1 n^2$$
Thus, we get:
$$ 0.8284 \leq \lim_{n \rightarrow \infty} \frac{A_n}{n^2} \leq 1.2$$
Hope that helps if Christian Blatter's comment doesn't get you the answer already.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/53257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is this Batman equation for real? HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?
Batman Equation in text form:
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\
&\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\
&\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\
&\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\
&\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0
\end{align}
| Sorry but this is not the answer but too long for a comment:
Probably the easiest verification is to type the equation on Google you'l be surprised :
The easiest way is to Google :2 sqrt(-abs(abs(x)-1)abs(3-abs(x))/((abs(x)-1)(3-abs(x))))(1+abs(abs(x)-3)/(abs(x)-3))sqrt(1-(x/7)^2)+(5+0.97(abs(x-.5)+abs(x+.5))-3(abs(x-.75)+abs(x+.75)))(1+abs(1-abs(x))/(1-abs(x))),-3sqrt(1-(x/7)^2)sqrt(abs(abs(x)-4)/(abs(x)-4)),abs(x/2)-0.0913722(x^2)-3+sqrt(1-(abs(abs(x)-2)-1)^2),(2.71052+(1.5-.5abs(x))-1.35526sqrt(4-(abs(x)-1)^2))sqrt(abs(abs(x)-1)/(abs(x)-1))+0.9
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "466",
"answer_count": 10,
"answer_id": 3
} |
Inequality involving sides of a triangle How can one show that for triangles of sides $a,b,c$ that
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} < 2$$
My proof is long winded, which is why I am posting the problem here.
Step 1: let $a=x+y$, $b=y+z$, $c=x+z$, and let $x+y+z=1$ to get
$\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}<2$
Step 2: consider the function $f(x)=\frac{1-x}{1+x}$, and note that it is convex on the interval (0,1), so the minimum of
$\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}$
is reached when the function takes the extreme points. i.e. $x=y=0, z=1$.
But going back to the fact that this is a triangle, we note that $x=y=0 \implies a=0$ which is not possible, so the inequality is strict.
| Let $c$ be the largest of $a$, $b$, and $c$. Then
\begin{align}
\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&\le\frac{a}{a+b}+\frac{b}{a+b}+\frac{c}{a+b}\\
&=1+\frac{c}{a+b}\\
&< 2
\end{align}
since $c<a+b$.
For a degenerate triangle with $a=0$, we have $c=a+b$ and the sum equals $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/54627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Surface area of revolution Given the function $y=\frac{1}{3}x^3$ on the interval [0, 3], find the surface area of the revolution on the x-axis.
The derivative is $y'=x^2$, so plugging everything in the integral gives
$$2\pi\int_0^3\frac{x^3}{3}\sqrt{1+(x^2)^2}dx$$
$$2\pi\int_0^3\frac{x^3}{3}\sqrt{1+x^4}dx$$
I got a little stuck here, so looking at the book work, the next step is shown as
$$\frac{\pi}{6}\int_0^3(1+x^4)^{\frac{1}{2}}\cdot(4x^3)dx$$
After this step, I can figure out the integral, but this setup has me confused on two fronts.
First, how did $2\pi$ transform into $\frac{\pi}{6}$?
Second, where did $4x^3$ come from?
| $$
2\pi \int_0^3 {\frac{{x^3 }}{3}\sqrt {1 + x^4 } dx} = \frac{{2\pi }}{3}\int_0^3 {x^3 (1 + x^4 )^{1/2} dx} = \frac{{2\pi }}{{3 \cdot 4}}\int_0^3 {4x^3 (1 + x^4 )^{1/2} dx}$$
$$
= \frac{\pi }{6}\int_0^3 {(1 + x^4 )^{1/2} 4x^3 dx} .
$$
Hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/55335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the perimeter of any triangle given the three altitude lengths The altitude lengths are 12, 15 and 20. I would like a process rather than just a single solution.
| Consider the triangle
Triangle with angles $A,B,C$ and opposite sides $a,b,c$
Its area is given e.g. by $$S=\dfrac{bc\sin A}{2}.\tag{1}$$ From the following trigonometric relations valid for a triangle
$$\sin A=2\sin \frac{A}{2}\cos \frac{A}{2}\tag{2}$$
and
$$\sin \frac{A}{2}=\sqrt{\dfrac{(p-b)(p-c)}{bc}}\tag{3}$$
$$\cos \frac{A}{2}=\sqrt{\dfrac{p(p-a)}{bc}},\tag{4}$$
we obtain the Heron's formula
$$S=\sqrt{p\left( p-a\right) \left( p-b\right) \left( p-c\right) },\tag{5}$$
where $$2p=a+b+c\tag{6}$$ is the perimeter. A geometric proof of $(5)$ can be found in this post, in Portuguese. In this case $S=\dfrac{12a}{2}=\dfrac{15b}{2}=\dfrac{20c}{2}$. Then $a=\dfrac{1}{6}S,b=\dfrac{2}{15}S$ and $c=\dfrac{1}{10}S$. Hence
$$2p=a+b+c=\frac{1}{6}S+\frac{2}{15}S+\frac{1}{10}S=\frac{2}{5}S,\qquad p=\frac{1}{5}\tag{7}S$$
and
$$S=\sqrt{\frac{1}{5}S\left(\frac{1}{5}S-\frac{1}{6}S\right)\left(\frac{1}{5}S-\frac{2}{15}S\right)\left(%
\frac{1}{5}S-\frac{1}{10}S\right)}=\frac{1}{150}S^{2}.\tag{8}$$
Solving for $S$, we get $S=150$. So the perimeter is $$\boxed{2p=\frac{2}{5}150=60.}\tag{9}$$
Note: the sides
are $a=\dfrac{1}{6}150=25,b=\dfrac{2}{15}150=20,c=\dfrac{1}{10}150=15$.
Added:
Triangle with angles $A,B,C$, opposite sides $a,b,c$ and altitudes $h_1,h_2,h_3$
For the general case of a triangle with altitudes $h_{1},h_{2},h_{3}$
perpendicular respectively to sides $a,b,c$ its area is $S=\dfrac{ah_{1}}{2}=%
\dfrac{bh_{2}}{2}=\dfrac{ch_{3}}{2}$. Consequently $a=\dfrac{2S}{h_{1}}$, $b=%
\dfrac{2S}{h_{2}}$, $c=\dfrac{2S}{h_{3}}$. So the perimeter $2p$ and
semi-perimeter $p$ of the triangle are given by
$$\begin{eqnarray}
2p &=&a+b+c=\frac{2S}{h}, \\
p &=&\frac{S}{h},
\end{eqnarray}\tag{A}$$
where $h$ is such that
$$\begin{equation}
\frac{1}{h}=\frac{1}{h_{1}}+\frac{1}{h_{2}}+\frac{1}{h_{3}}
\end{equation}\Leftrightarrow h=\frac{h_{1}h_{2}h_{3}}{h_{2}h_{3}+h_{1}h_{3}+h_{1}h_{2}}.\tag{B}$$
Thus
$$\begin{eqnarray*}
S &=&\sqrt{\frac{S}{h}\left( \frac{S}{h}-\frac{2S}{h_{1}}\right) \left(
\frac{S}{h}-\frac{2S}{h_{2}}\right) \left( \frac{S}{h}-\frac{2S}{h_{3}}%
\right) } \\
&=&\frac{S^{2}}{h^{2}}\sqrt{\frac{\left( h_{1}-2h\right) \left(
h_{2}-2h\right) \left( h_{3}-2h\right) }{h_{1}h_{2}h_{3}}}
\end{eqnarray*}\tag{C}$$
Solving for $S$, we now get
$$\begin{equation}
S=\dfrac{h^{2}}{\sqrt{\dfrac{\left( h_{1}-2h\right) \left( h_{2}-2h\right)
\left( h_{3}-2h\right) }{h_{1}h_{2}h_{3}}}}
\end{equation}\tag{D}$$
and finally the perimeter as a function of $h,h_1,h_2$ and $h_3$, with $h$ being a function of $h_1,h_2,h_3$ given by $(\text{B})$:
$$\boxed{\begin{equation}
2p=\dfrac{2h}{\sqrt{\dfrac{\left( h_{1}-2h\right) \left( h_{2}-2h\right)
\left( h_{3}-2h\right) }{h_{1}h_{2}h_{3}}}}.
\end{equation}}\tag{E}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/55440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
How to prove the following identities?
Prove:
\begin{align}
\tan(A) + \cot(A) & = 2 \text{cosec}(2A)\\
\tan(45^{\circ}+A^{\circ}) - \tan(45^{\circ}-A^{\circ}) & = 2 \tan(2A^{\circ})\\
\text{cosec}(2A) + \cot(2A) & = \cot(A)
\end{align}
I have got all the formulas that I need but I just couldn't solve these. Some help, please?
| For the second one:
*
*$2)$ Call $\alpha = \frac{\pi}{4} +A$ and $\beta= \frac{\pi}{4}-A$. So you have $\alpha + \beta = \frac{\pi}{2} \Rightarrow \cot(\alpha+\beta) = 0$. From this you have $$\frac{1}{\tan(\alpha+\beta)} = \frac{1 - \tan{\alpha}\tan{\beta}}{\tan{\alpha} + \tan{\beta}}=0 \Rightarrow \tan{\alpha} \cdot \tan{\beta}=1$$ Now note that $\alpha - \beta= 2A$ and so we again have
\begin{align*}
\tan(\alpha-\beta) &= \frac{\tan(\alpha)-\tan(\beta)}{1+\tan(\alpha)\cdot \tan(\beta)} \\ &= \frac{\tan(\alpha)-\tan(\beta)}{2} \\ \Rightarrow \ 2\tan(2A)&= \tan(\alpha)- \tan(\beta)
\end{align*}
*$3)$ You have : \begin{align*} \csc{2A} + \cot{2A} &= \frac{1}{\sin{2A}} + \frac{\cos{2A}}{\sin{2A}} \\ & = \frac{1+\cos{2A}}{\sin{2A}} \\ &= \frac{1+\cos^{2}{A} - \sin^{2}{A}}{2 \cdot \sin{A} \cdot \cos{A}} \\ &= \frac{2 \cdot\cos^{2}{A}}{2 \cdot \sin{A} \cdot \cos{A}} = \cot(A)\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/56122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the minimal polynomial of a $A$, given a diagonal matrix equivalent to $XI_{10}-A$ Let $A\in M_{10}(\mathbb{C})$. We know that $XI_{10}-A $ is equivalent(By finite number of operations on the rows) to the diagonal matrix:
$$C=\begin{pmatrix}
1& & & & & & & & & \\
& 1 & & & & & & & & \\
& &x & & & & & & & \\
& & &x+2 & & & & & & \\
& & & &x^2 & & & & & \\
& & & & &x+2 & & & & \\
& & & & & & (x^2+4)x & & & \\
& & & & & & & (x+2)^2 & & \\
& & & & & & & &1 & \\
& & & & & & & & & 1
\end{pmatrix}.$$
(the blank entries are $0$).
I need to find the minimal polynomial and Jordan form of $A$.
So I know know that the characteristic polynomial of A is: $F_A(x)= x^4(x+2)^4(x^2+4)$, and I'm having a hard realize what I can know in addition, to solve the problem. any hints? thanks!
| Let $K$ be a field, $X$ an indeterminate, and $V$ a finite dimensional $K[X]$-module.
Recall that the multiset of elementary divisors of $V$ is the multiset
$$f_1^{n(1,1)},\dots,f_1^{n(1,k(1))},$$
$$\vdots$$
$$f_r^{n(r,1)},\dots,f_r^{n(r,k(r))},$$
defined by the following conditions: the $f_i$ are distinct monic irreducible polynomials, the $n(i,j)$ satisfy $$n(i,1)\ge\cdots\ge n(i,k(i)),$$
and we have
$$V\simeq\bigoplus_{i,j}\frac{K[X]}{\big(f_i^{n(i,j)}\big)}\quad.$$
The characteristic polynomial is the product of the elementary divisors, whereas the minimal polynomial is the product of the $f_i^{n(i,1)}$.
If $W$ is also a finite dimensional $K[X]$-module, then the multiset of elementary divisors of $V\oplus W$ is the "disjoint union" (in the obvious sense) of the multisets of elementary divisors of $V$ and $W$.
Some diagonal entries of Nir's matrix are equal to 1, and the others are
$$X,\quad X+2,\quad X^2,\quad X+2,\quad (X+2i)(X-2i)X,\quad (X+2)^2.$$
The multiset of elementary divisors is thus
$$\begin{matrix}
(X+2)^2,&X+2,&X+2,\\ \\
X^2,&X,&X,\\ \\
X+2i,&&\\ \\
X-2i.&&
\end{matrix}$$
The characteristic polynomial is $$(X+2)^4X^4(X+2i)(X-2i).$$ The minimal polynomial is $$(X+2)^2X^2(X+2i)(X-2i).$$
The Jordan blocks $J(\lambda,n)$ are
$$\begin{matrix}
J(-2,2),&J(-2,1),&J(-2,1),\\ \\
J(0,2),&J(0,1),&J(0,1),\\ \\
J(-2i,1),&&\\ \\
J(2i,1).&&
\end{matrix}$$
Let's compute the invariant factors (even if it wasn't required). In the array of elementary divisors, there are four blank entries. We put a $1$ in each of them:
$$\begin{matrix}
(X+2)^2,&X+2,&X+2,\\ \\
X^2,&X,&X,\\ \\
X+2i,&1,&1,\\ \\
X-2i,&1,&1.
\end{matrix}$$
We get the first invariant factor by multiplying together the polynomials in the first row, and so on. Thus the invariant factors are
$$(X+2)^2X^2(X+2i)(X-2i),\quad (X+2)X,\quad (X+2)X.$$
[Reference: Bourbaki, Algèbre, VII.4.8.]
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The rank of skew-symmetric matrix is even I know that the rank of a skew-symmetric matrix is even. I just need to find a published proof for it. Could anyone direct me to a source that could help me?
| $\newcommand{\rank}{\mathrm{rank}}$
$\newcommand{\diag}{\mathrm{diag}}$
If $K$ is a real skew symmetric matrix, typically it is proven by showing $K$ has even non-zero eigenvalues. Below is a proof using the Hermitian form of $K$.
Since $\rank(K) = r$, there exist order $n$ invertible matrices $P$ and $Q$, such that $K = P\diag(I_{(r)}, 0)Q$. By $K = -K'$, $Q'\diag(-I_{(r)}, 0)P' = P\diag(I_{(r)}, 0)Q$, i.e.,
\begin{align*}
P^{-1}Q\diag(-I_{(r)}, 0) = \diag(I_{(r)}, 0)(P^{-1}Q)'.
\end{align*}
Let $P^{-1}Q' = R = \begin{pmatrix} R_{11} & R_{12} \\ R_{21} & R_{22} \end{pmatrix}$, where $R_{11}$ is an order $r$ matrix. Substituting it to the above equation yields
$R_{11}' = -R_{11}, R_{21} = 0$. Therefore,
\begin{align*}
P^{-1}Q' = \begin{pmatrix} R_{11} & R_{12} \\ 0 & R_{22} \end{pmatrix}, \text{ i.e., } Q' = P\begin{pmatrix} R_{11} & R_{12} \\ 0 & R_{22} \end{pmatrix},
\end{align*}
whence $Q = \begin{pmatrix} R_{11}' & 0 \\ R_{12}' & R_{22}' \end{pmatrix}$. Consequently,
\begin{align*}
K = P\begin{pmatrix} I_{(r)} & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} R_{11}' & 0 \\ R_{12}' & R_{22}' \end{pmatrix}P'
= P\begin{pmatrix} R_{11}' & 0 \\ 0 & 0 \end{pmatrix}P'.
\end{align*}
Since $P$ is invertible, $\rank(K) = \rank(\diag(R_{11}, 0)) = \rank(R_{11}) = r$, whence $R_{11}$ is an order $r$ invertible matrix. Taking
determinants on both sides of $R_{11}' = -R_{11}$ yields $\det(R_{11}') = (-1)^r\det(R_{11})$, hence $r$ must be an even number, otherwise $\det(R_{11}) = 0$, which contradicts with $R_{11}$ is invertible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/57696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Tangent of an ellipse to an outside point Let $C$ be a curve that is given by the equation:
$$
2x^2 + y^2 = 1
$$
and let P be a point $(1,1)$, which lies outside of the curve.
We want to find all lines that are tangent to $C$ and intersect $P$, and have found $y=1$, but are not sure how to find the other line.
| Let the equation of the line be $y = mx+c$. Since the line passes through $(1,1)$, we have $1 = m + c$ i.e. $c = 1 - m$. Hence, the equation of the line is $y = mx + 1 - m$.
Now we want the above line to be a tangent to $2x^2 + y^2 = 1$. This means the line should intersect the ellipse at exactly one point. Plug in $y = mx + 1 - m$ and the condition that the line intersects the ellipse at only one point means that the quadratic in $x$ must have only one root which means that the discriminant of the quadratic must be zero.
The quadratic we get is $2x^2 + (mx + 1 - m)^2 = 1$. Rearranging we get $$(m^2+2)x^2 + 2m(1-m)x + m^2 - 2m = 0$$
The discriminant is $4m^2(1-m)^2 - 4(m^2-2m)(m^2+2) = -4m(m-4)$
Setting this to zero, we get the two tangent from $(1,1)$ to the ellipse $2x^2 + y^2 = 1$ are
$$y= 1$$ $$y = 4x - 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/58055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Elementary central binomial coefficient estimates
*
*How to prove that $\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}<\binom{2n}{n}<\frac{4^{n}}{\sqrt{3n+1}}\quad$ for all $n$ > 1 ?
*Does anyone know any better elementary estimates?
Attempt. We have
$$\frac1{2^n}\binom{2n}{n}=\prod_{k=0}^{n-1}\frac{2n-k}{2(n-k)}=\prod_{k=0}^{n-1}\left(1+\frac{k}{2(n-k)}\right).$$
Then we have
$$\left(1+\frac{k}{2(n-k)}\right)>\sqrt{1+\frac{k}{n-k}}=\frac{\sqrt{n}}{\sqrt{n-k}}.$$
So maybe, for the lower bound, we have
$$\frac{n^{\frac{n}{2}}}{\sqrt{n!}}=\prod_{k=0}^{n-1}\frac{\sqrt{n}}{\sqrt{n-k}}>\frac{2^n}{\sqrt{4n}}.$$
By Stirling, $n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$, so the lhs becomes
$$\frac{e^{\frac{n}{2}}}{(2\pi n)^{\frac14}},$$
but this isn't $>\frac{2^n}{\sqrt{4n}}$.
| You can get an even more precise answer than those already provided by using more terms in the Stirling series. Doing so yields, to a relative error of $O(n^{-5})$,
$$\binom{2n}{n} = \frac{4^n}{\sqrt{\pi n}} \left(1 - \frac{1}{8n} + \frac{1}{128n^2} + \frac{5}{1024n^3} - \frac{21}{32768 n^4} + O(n^{-5})\right).$$
To the same relative error of $O(n^{-5})$, the Stirling series is
$$n!=\sqrt{2\pi n}\left({n\over e}\right)^n
\left(
1
+{1\over12n}
+{1\over288n^2}
-{139\over51840n^3}
-{571\over2488320n^4}
+ O(n^{-5})
\right).$$
Then we have
$$\binom{2n}{n} = \frac{(2n)!}{n! n!} = \frac{4^n}{\sqrt{\pi n}} \frac{1 + \frac{1}{12(2n)} + \frac{1}{288(2n)^2} - \frac{139}{51840(2n)^3} - \frac{571}{2488320(2n)^4} + O(n^{-5})}{\left(1 + \frac{1}{12n} + \frac{1}{288n^2} - \frac{139}{51840n^3} - \frac{571}{2488320n^4} + O(n^{-5})\right)^2}.$$
Crunching through the long division with the polynomial in $\frac{1}{n}$ (which Mathematica can do immediately with the command Series[expression, {n, ∞, 4}]) yields the expression for $\binom{2n}{n}$ at the top of the post.
See also Problem 9.60 in Concrete Mathematics (2nd edition).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/58560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 7,
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tough integral involving $\sin(x^2)$ and $\sinh^2 (x)$ I ran across this integral I get no where with. Can someone suggest a method of attack?.
$$\int_0^{\infty}\frac{\sin(\pi x^2)}{\sinh^2 (\pi x)}\mathrm dx=\frac{2-\sqrt{2}}{4}$$
I tried series, imaginary parts, and so forth, but have made no progress.
Thanks very much.
| Although this question is two years old, the integral was mentioned in chat recently, I evaluated it, and then found this question. Since there is no complete solution, although Hans Lundmark's suggestion is excellent and similar in nature, I am posting what I have done.
Contours
Since the integrand is even,
$$
\begin{align}
\int_0^\infty\frac{\sin(\pi x^2)}{\sinh^2(\pi x)}\,\mathrm{d}x
&=\frac12\int_{-\infty}^\infty\frac{\sin(\pi x^2)}{\sinh^2(\pi x)}\,\mathrm{d}x
\end{align}
$$
Define
$$
f(z)=\frac{\cos\left(\pi z^2\right)}{\sinh(2\pi z)\sinh^2(\pi z)}
$$
Note that because
$$
f(x\pm i)
=\frac{-\cos\left(\pi x^2\right)\cosh(2\pi x)\pm i\sin\left(\pi x^2\right)\sinh(2\pi x)}{\sinh(2\pi x)\sinh^2(\pi x)}\\
$$
we have
$$
\begin{align}
\int_\gamma f(z)\,\mathrm{d}z
&=\int_{-\infty}^\infty\big[f(x-i)-f(x+i)\big]\,\mathrm{d}x\\
&=-2i\int_{-\infty}^\infty\frac{\sin(\pi x^2)}{\sinh^2(\pi x)}\,\mathrm{d}x\\
&=2\pi i\times\begin{array}{}\text{the sum of the residues}\\\text{inside the contour}\end{array}
\end{align}
$$
where $\gamma$ is the contour
$\hspace{3.2cm}$
Therefore,
$$
\int_0^\infty\frac{\sin(\pi x^2)}{\sinh^2(\pi x)}\,\mathrm{d}x
=-\frac\pi2\times\begin{array}{}\text{the sum of the residues}\\\text{inside the contour}\end{array}
$$
Residues
near $0$ :
$$
\begin{align}
f(z)
&=\frac{\cos\left(\pi z^2\right)}{\sinh(2\pi z)\sinh^2(\pi z)}\\
&=\frac{1-\frac12\pi^2z^4+O(z^8)}{2\pi z\left(1+\frac23\pi^2z^2+O(z^4)\right)\pi^2 z^2\left(1+\frac13\pi^2z^2+O(z^4)\right)}\\
&=\frac{1-\pi^2z^2}{2\pi^3z^3}+O(z)\\[10pt]
&\implies\text{residue}=-\frac1{2\pi}
\end{align}
$$
at $\pm i/2$, use L'Hosptal :
$$
\begin{align}
\text{residue}
&=\lim_{z\to\pm i/2}\frac{(z\mp i/2)\cos\left(\pi z^2\right)}{\sinh(2\pi z)\sinh^2(\pi z)}\\
&=\frac1{2\pi\cosh(\pm\pi i)}\frac{\cos(-\pi/4)}{\sinh^2(\pm\pi i/2)}\\
&=\frac1{2\pi\cos(\pm\pi)}\frac{\sqrt2/2}{-\sin^2(\pm\pi/2)}\\[4pt]
&=\frac{\sqrt2}{4\pi}
\end{align}
$$
near $\pm i$ :
$$
\begin{align}
f(z\pm i)
&=\frac{-\cos\left(\pi z^2\right)\cosh(2\pi z)\pm i\sin\left(\pi z^2\right)\sinh(2\pi z)}{\sinh(2\pi z)\sinh^2(\pi z)}\\
&=\frac{-\left(1-\frac12\pi^2z^4+O(z^8)\right)\left(1+2\pi^2z^2+O(z^4)\right)+O(z^3)}{2\pi z\left(1+\frac23\pi^2z^2+O(z^4)\right)\pi^2 z^2\left(1+\frac13\pi^2z^2+O(z^4)\right)}\\
&=-\frac{1+\pi^2z^2}{2\pi^3z^3}+O(1)\\[10pt]
&\implies\text{residue}=-\frac1{2\pi}
\end{align}
$$
Result
Thus,
$$
\begin{align}
\int_0^\infty\frac{\sin(\pi x^2)}{\sinh^2(\pi x)}\,\mathrm{d}x
&=-\frac\pi2\left(-\frac1{2\pi}-\frac1{2\pi}+\frac{\sqrt2}{4\pi}+\frac{\sqrt2}{4\pi}\right)\\[6pt]
&=\frac{2-\sqrt2}{4}
\end{align}
$$
| {
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"question_score": "31",
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} |
Probability of balls in the box In a box there are 12 balls; 4 defective, 8 not defective.
What is the probability that when 3 balls are drawn, at least two of them are defective.
I know the answer is $$\frac{{4 \choose 2}{8 \choose 1} + {4 \choose 3}}{{12 \choose 3}}$$
But why isn't the answer also $$\frac{{4 \choose 2}{10 \choose 1}}{{12 \choose 3}} $$?
Because I choose 2 balls from the 4 defective, and 1 from the remaining 10 (it will either be defective or not).
| Another way is to look at the possible sequences. Let's say $D$ is a defect ball and $G$ is a nondefect (good ball). If you sample 3 balls without replacement the event 'to get at least 2D' is $\{DDD,GDD,DGD,DDG\}$. The probability of the first event is
$P(DDD)= \frac{4 \cdot 3 \cdot 2}{12 \cdot 11 \cdot 10}$
It seems that the probability of the other 3 events has to be different, but it is actually not:
$P(GDD)=P(DGD)=P(DDG)=\frac{4 \cdot 3 \cdot 8}{12 \cdot 11 \cdot 10}$ because the denominator does not change and the numerator is the product of the same 3 numbers (and multiplication is commutative), therefore:
$P(D \geq 2)=P(DDD)+P(GDD)+P(DGD)+P(DDG)=\frac{4 \cdot 3 \cdot 2}{12 \cdot 11 \cdot 10}+3 \cdot \frac{4 \cdot 3 \cdot 8}{12 \cdot 11 \cdot 10}=.2364$
Control:$1-P(D<2)=1-\Big( \frac{8 \cdot 7 \cdot 6}{12 \cdot 11 \cdot 10}+3 \cdot \frac{8 \cdot 7 \cdot 4}{12 \cdot 11 \cdot 10} \Big)=.2364$
| {
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"answer_count": 2,
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} |
Simplifying the expression $\sqrt[4]{\frac{g^3h^4}{4r^{14}}}$? How would I simplify this expression?
$$\sqrt[4]{\frac{g^3h^4}{4r^{14}}}\ ?$$
I did this
$$\begin{align*}
\sqrt[4]{g^3h^3h^4}\\
h\sqrt[4]{g^3h^3}\\
\sqrt[4]{4r^{14}}\\
\sqrt[4]{2r^2r^{12}}\\
r^3\sqrt[4]{2r^2}\\
\end{align*}$$
But I am stuck?
Yes that is correct
| If the initial expression is $$\sqrt[4]{\dfrac{g^3 h^4}{4 r^{14} } }$$ then you have made slight errors as $\sqrt[4]{g^3 h^4} = h\sqrt[4]{g^3}$ not $h\sqrt[4]{g^3 h^3}$, while $\sqrt[4]{4 r^{14}} = r^3 \sqrt[4]{4 r^2}$ not $r^3 \sqrt[4]{2 r^2}$.
But otherwise you seem to have done sensible things.
So you could end up with $$\frac{h}{r^3}\sqrt[4]{\dfrac{g^3 }{4 r^{2} } }$$ or write it some other way, such as $$2^{-0.5} g^{0.75} h^1 r^{-3.5}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Need help solving a particular system of non-linear equations analytically How would one go about analytically solving a system of non-linear equations of the form:
$a + b + c = 4$
$a^2 + b^2 + c^2 = 6$
$a^3 + b^3 + c^3 = 10$
Thanks!
| Hint: Newton's identities.
$$\begin{align*}a+b+c&=4\\ ab+bc+ca=\frac12((a+b+c)^2-(a^2+b^2+c^2))&=\frac12(4^2-6)=5\\abc=\frac13((ab+bc+ca-a^2-b^2-c^2)(a+b+c)+(a^3+b^3+c^3))&=\frac13((5-6)\cdot 4+10)=2\end{align*}$$
Remember that $a,\ b,\ c$ are the three roots of a polynomial $P(t)=(t-a)(t-b)(t-c)=t^3-(a+b+c)t^2+(ab+bc+ca)t-abc$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
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Proving this identity $\sum_k\frac{1}{k}\binom{2k-2}{k-1}\binom{2n-2k+1}{n-k}=\binom{2n}{n-1}$ using lattice paths How can I prove the identity $\sum_k\frac{1}{k}\binom{2k-2}{k-1}\binom{2n-2k+1}{n-k}=\binom{2n}{n-1}$?
I have to prove it using lattice paths, it should be related to Catalan numbers
The $n$th Catalan number $C_n$ counts the number of monotonic paths along the edges of a grid with $n\times n$ square cells, which do not pass above the diagonal.
See for example this link
For example $\frac{1}{k}\binom{2k-2}{k-1}$ is exactly $C_{k-1}$, and the other terms can also be expressed in terms of the Catalan numbers.
The second part of the exercise ask to prove the recurrence formula $C_n=\sum_{k=1}^n C_{k-1}C_{n-k}$ using similar reasoning (i.e. lattice paths). So we can't use this formula to prove the first.
Could you help me please?
| This one can also be done using complex variables.
Suppose we seek to evaluate
$$\sum_{k=1}^n \frac{1}{k} {2k-2\choose k-1}
{2n-2k+1\choose n-k}.$$
Introduce the integral representation
$${2n-2k+1\choose n-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n-2k+1}}{z^{n-k+1}} \; dz.$$
This has the property that it is zero when $k\gt n.$
We obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{n+1}}
\sum_{k\ge 1} \frac{1}{k} {2k-2\choose k-1}
\frac{z^k}{(1+z)^{2k}} \; dz.$$
Recall the generating function for the Catalan numbers
$$\sum_{q\ge 0} \frac{1}{q+1} {2q\choose q} w^q
= \frac{1-\sqrt{1-4w}}{2w}$$
This is equal to
$$\sum_{q\ge 1} \frac{1}{q} {2q-2\choose q-1} w^{q-1}$$
so that
$$\sum_{q\ge 1} \frac{1}{q} {2q-2\choose q-1} w^q
= \frac{1-\sqrt{1-4w}}{2}.$$
Substitute this into the integral to obtain
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{n+1}}
\frac{1-\sqrt{1-4z/(1+z)^2}}{2} \; dz.$$
This has two components, the first is
$$\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{n+1}} \; dz
= \frac{1}{2} {2n+1\choose n}.$$
The second is
$$-\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n+1}}{z^{n+1}}
\sqrt{1-4z/(1+z)^2} \; dz
\\ = -\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}}
\sqrt{(1+z)^2-4z} \; dz
\\ = -\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}}
\sqrt{(1-z)^2} \; dz
\\ = -\frac{1}{2} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{2n}}{z^{n+1}}
(1-z) \; dz.$$
This evaluates to
$$\frac{1}{2} {2n\choose n-1}
- \frac{1}{2} {2n\choose n}.$$
Factoring the sum of the two contributions to reveal the target term
we obtain
$$\frac{1}{2}
\left(\frac{2n+1}{n} + 1 - \frac{n+1}{n}\right)
{2n\choose n-1}
= {2n\choose n-1}.$$
| {
"language": "en",
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"question_score": "11",
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} |
evaluating $ \int_0^{\sqrt3} \arcsin(\frac{2t}{1+t^2}) \,dt$ $$\begin{align*}
\int \arcsin\left(\frac{2t}{1+t^2}\right)\,dt&=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int\frac{2t}{1+t^2}\,dt\\
&=t\arcsin\left(\frac{2t}{1+t^2}\right) + \ln(1+t^2)+C
\end{align*}$$
So
$$ \int\nolimits_0^{\sqrt3} \arcsin\left(\frac{2t}{1+t^2}\right)=\pi/\sqrt3+2\ln2.$$
However the result seems to be $ \pi/\sqrt3 $ only.
Why is there this $ 2\ln2 $?
Detail:
$$ \begin{align*}
t \arcsin\left(\frac{2t}{1+t^2}\right)&- \int t \left(\frac{2(1-t^2)}{(1+t^2)^2}\right)\frac{1}{\sqrt{1-\frac{4t^2}{(1+t^2)^2}}}\,dt\\
&= t\arcsin\left(\frac{2t}{1+t^2}\right)- \int \frac{2(1-t^2)t}{(1+t^2)\sqrt{(t^2-1)^2}}\,dt\\
&=t\arcsin\left(\frac{2t}{1+t^2}\right)+\int \frac{2t}{1+t^2}\,dt
\end{align*} $$
| part of integral solution is $\ln(1+t^2)$. When you insert integral bounds you get $\ln(1+(\sqrt{3})^2)-\ln(1+(0)^2)$$=\ln(4)-ln(1)$$=2\ln(2)$
| {
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Proof of dividing fractions $\frac{a/b}{c/d}=\frac{ad}{bc}$ For dividing two fractional expressions, how does the division sign turns into multiplication? Is there a step by step proof which proves
$$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c}=\frac{ad}{bc}?$$
| There is something assumed about the order of operations in the notation
$a/b \div c/d$. I see this as a problem in pedagogy when fractions are first introduced in schools. Somehow we have to assume that the slash bars are to be done before the division symbol, even though each of those symbols stands for "divide". If I were you, I'd always stay in the habit of writing fractions as $\frac{a}{b}$ rather than $a/b$.
end rant.
Traditionally, $\frac{a}{b} \div \frac{c}{d}$ means: Beginning with the quantity $\frac{a}{b}$, divide this by the quantity $\frac{c}{d}$. So let's think of a small example not dealing with fractions first:
$$ 8 \div 2.$$
Now, we all know the answer is $4$. But I want to draw your attention to the fact that this expression is the same as:
$$ 8 \cdot \frac{1}{2}.$$
What happened? Well division by $2$ is the same as multiplication by $\frac{1}{2}$, because $\frac{1}{2}$ is the multiplicative inverse, or reciprocal of $2$. By defintion, the reciprocal of a number $x$ is a number $y$ such that $x\cdot y = 1$.
Every nonzero number has a reciprocal. If $x \neq 0$, then its reciprocal is $\frac{1}{x}$ (because $x \cdot \frac{1}{x} = 1$). So, for $x \neq 0$:
$$ 8 \div x = 8 \cdot \frac{1}{x}.$$
Now, if $x$ is a fraction, such as $\frac{c}{d}$, then the most natural way to write the reciprocal of $x$ is by "flipping" the fraction. The reciprocal of $\frac{c}{d}$ is $\frac{d}{c}$ (well... because $\frac{c}{d}\cdot \frac{d}{c} = \frac{cd}{cd} = 1$). Thus:
$$ 8 \div \frac{c}{d} = 8 \cdot \frac{d}{c}. $$
Finally, it does not matter so much what the first factor of the expression is. It could be another fraction:
$$ \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \cdot \frac{d}{c} = \frac{ad}{bc}.$$
Hope this helps!
| {
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Partial Fractions of form $\frac{1}{(ax+b)(cx+d)^2}$ When asked to convert something like $\frac{1}{(ax+b)(cx+d)}$ to partial fractions, I can say
$$\frac{1}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$$
Then why can't I split $(cx+d)^2$ into $(cx+d)(cx+d)$ then do
$$\frac{1}{(ax+b)(cx+d)^2} = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{cx+d}$$
The correct way is
$$\frac{1}{(ax+b)(cx+d)^2} = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{(cx+d)^2}$$
| HINT $\rm\displaystyle\quad \frac{f(x)}{g(x)\ (c\:x+d)^2}\: =\: \frac{h(x)}{(a\:x+b)\:(c\:x+d)}\ \ \Rightarrow\ \ f(x)\:(a\:x+b)\: =\: g(x)\:h(x)\:(c\:x+d)\:$
hence, evaluating at $\rm\:x = -d/c\:$ yields that either $\rm\:f(-d/c) = 0\:$ (so the LHS isn't in lowest terms) or $\rm\ a\:x+b\ $ has root $\rm\:x = -d/c\:$ (so the RHS has denominator $\rm\:(c\:x+d)^2\:$ times constant).
NOTE $\ $ This is simply a rational function analog of the fact that rational numbers have unique prime factorization. Hence above, if the LHS is in lowest terms and the prime $\rm\:c\:x+d\:$ occurs to power $\:2\:$ in its denominator, then the same must be true of the RHS when in lowest terms. Therefore $\rm\ a\:x+b\: =\: e\:(c\:x+d)\:$ for some constant $\rm\:e\:.$
| {
"language": "en",
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$1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = n(n+1)(2n+7)/6$ by mathematical induction I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side.
Please guide me how to do it further.
$1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.
Sol:
$P(n):\ 1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.
$P(1):\ \frac{1}{6}(2)(9) = \frac{1}{2}(2)(3)$.
$P(1): 3$.
Hence it is true for $n=n_0 = 1$.
Let it be true for $n=k$.
$P(k):\ 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) = \frac{1}{6}k(k+1)(2k+7)$.
We have to prove
$P(k+1):\ 1\cdot 3 + 2\cdot 4 + \cdots + (k+1)(k+3)= \frac{1}{6}(k+1)(k+2)(2k+9)$.
Taking LHS:
$$\begin{align*}
1\cdot 3 + 2\cdot 4+\cdots + (k+1)(k+2) &= 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) + (k+1)(k+3)\\
&= \frac{1}{6}(k+1)(k+2)(2k+9) + (k+1)(k+3)\\
&= \frac{1}{6}(k+1)\left[(k+2)(2k+9) + 6k+18\right]\\
&= \frac{1}{6}(k+1)\left[2k^2 + 13k + 18 + 6k + 18\right]\\
&= \frac{1}{6}(k+1)\left[2k^2 + 19k + 36\right].
\end{align*}$$
| HINT $\: $ First trivially inductively prove the Fundamental Theorem of Difference Calculus
$$\rm\ F(n)\ =\ \sum_{i\: =\: 1}^n\:\ f(i)\ \ \iff\ \ \ F(n) - F(n-1)\ =\ f(n),\quad\ F(0) = 0$$
Your special case now follows immediately by noting that
$$\rm\ F(n)\ =\ \dfrac{n\ (n+1)\ (2\:n+7)}{6}\ \ \Rightarrow\ \ F(n)-F(n-1)\ =\: n\ (n+2)\:.\ $$
Note that by employing the Fundamental Theorem we have reduced the proof to the trivial mechanical verification of a polynomial equation. Absolutely no ingenuity is required.
Note that the proof of the Fundamental Theorem is much more obvious than that for your special case because the telescopic cancellation is obvious at this level of generality, whereas it is usually obfuscated in most specific instances. Namely, the proof of the Fundamental Theorem is just a rigorous inductive proof of the following telescopic cancellation
$$\rm - F(0)\!+\!F(1) -F(1)\!+\!F(2) - F(2)\!+\!F(3)-\:\cdots - F(n-1)\!+\!F(n)\ =\:\: -F(0) + F(n) $$ where all but the end terms cancel out. For further discussion see my many posts on telescopy.
| {
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Find $x$ if $(7+4\sqrt 3)^{x^2-8}+(7-4\sqrt 3)^{x^2-8}=14$
Known that:
$$(7+4\sqrt 3)^{x^2-8}+(7-4\sqrt 3)^{x^2-8}=14$$
What is the value of $x$
| Note that $$(7+4 \sqrt{3}) \cdot (7-4\sqrt{3}) = 49-48 =1$$
| {
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Pythagorean quadruples Another Project Euler problem has me checking the internet again. Among other conditions, four of my variables satisfy:
$$a^2+b^2+c^2=d^2 .$$
According to Wikipedia, this is known as a Pythagorean Quadruple. It goes on to say all quadruples can be generated from an odd value of $a$ and an even value of $b$ as:
$$c=\frac{a^2+b^2-p^2}{2}, \quad d=\frac{a^2+b^2+p^2}{2} ,$$
where $p$ is any factor of $a^2+b^2$ that satisfies $p^2<a^2+b^2$.
However, I can't see how or why this works. I also can't seem to see how this works for $\lbrace 4,4,7,9 \rbrace$. Am I missing something here?
| Sum3Squares
I thought of extremely simple derivation of the parametrization of three squares which sum to square.
Suppose $a^2 + b^2 + c^2 = d^2$
then $a^2 + b^2 = d^2 - c^2$
As is well known, any number which is the sum of two squares is the product of only primes which are the sum of two squares.
We can thus easily derive from complex number arithmetic the parametrization of composite numbers which are the sum of two squares.
$$(a_1 + i b_1)(a_1-i b_1)(a_2 + i b_2)(a_2 - i b_2) = (a_1 + i b_1)(a_2 + i b_2) (a_1 - i b_1) (a_2 - i b_2)$$
$$(a_1^2 + b_1^2) (a_2^2 + b_2^2)
= ( (a_1 a_2 - b_1 b_2) + i (a_1 b_2 + b_1 a_2) )( (a_1 a_2 - b_1 b_2) - i (a_1 b_2 + b_1 a_2) )$$
$$(a_1^2 + b_1^2) (a_2^2 + b_2^2) = (a_1 a_2 - b_1 b_2)^2 + (a_1 b_2 + b_1 a_2)^2$$
Also, since the product of two numbers is the difference of the squares of half their sum and half their difference,
$$(a_1^2 + b_1^2)(a_2^2 + b_2^2)
= ((a_1^2 + b_1^2 + a_2^2 + b_2^2)/2)^2 - ((a_1^2 + b_1^2 - a_2^2 -b_2^2)/2)^2$$
Thus
$$(a_1 a_2 - b_1 b_2)^2 + (a_1 b_2 + b_1 a_2)^2
= ((a_1^2 + b_1^2 + a_2^2 + b_2^2)/2)^2 - ((a_1^2 + b_1^2 - a_2^2 -b_2^2)/2)^2$$
$$(a_1 a_2 - b_1 b_2)^2 + (a_1 b_2 + b_1 a_2)^2 + ((a_1^2 + b_1^2 - a_2^2 -b_2^2)/2)^2
= ((a_1^2 + b_1^2 + a_2^2 + b_2^2)/2)^2$$
$(a_1 a_2 - b_1 b_2)^2 + (a_1 b_2 + b_1 a_2)^2$ not have remainder 2 when divided by 4.
That is why it is required that not both $(a_1 a_2 - b_1 b_2)$ and $(a_1 b_2 + b_1 a_2)$
be odd.
Kermit
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/76892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Decomposition of a nonsquare affine matrix I have a $2\times 3$ affine matrix
$$
M = \pmatrix{a &b &c\\ d &e &f}
$$
which transforms a point $(x,y)$ into $x' = a x + by + c, y' = d x + e y + f$
Is there a way to decompose such matrix into shear, rotation, translation,and scale ? I know there's something for $4\times 4$ matrixes, but for a $2\times 3$ matrix ?
| If $(x, y, 1)$ is a vector in homogeneous coordinates, we have, by decomposing $M$ into blocks, that
$$M \left[\begin{array}{c}x\\y\\1\end{array}\right] = \left[\begin{array}{cc} a& b\\ d&e\end{array}\right]\left[\begin{array}{c}x\\y\end{array}\right] + \left[\begin{array}{c}c\\f\end{array}\right].$$
Here $(c,f)$ is the translation component. We can decompose the 2x2 matrix into a composition of a rotation, shear, and scale by using the QR decomposition:
$$\begin{align*}\left[\begin{array}{cc}a & b\\d & e\end{array}\right] &= \left[\begin{array}{cc} \cos \theta &-\sin \theta \\ \sin\theta &\cos \theta\end{array}\right]\left[\begin{array}{cc} \sqrt{a^2+d^2} & b\cos \theta + e\sin \theta\\0 & e\cos \theta - b\sin \theta\end{array}\right]\\
&=\left[\begin{array}{cc} \cos \theta &-\sin \theta \\ \sin\theta &\cos \theta\end{array}\right]\left[\begin{array}{cc}1 & \frac{b\cos \theta + e\sin\theta}{e\cos \theta-b\sin\theta}\\0 & 1\end{array}\right]\left[\begin{array}{cc}\sqrt{a^2+d^2} & 0\\0 & e\cos\theta - b\sin\theta\end{array}\right],\end{align*}$$
where $\theta = \arctan\left(\frac{d}{a}\right).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/78137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
How can I prove the inequality $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$? For $x > 0$, $y > 0$, $z > 0$, prove:
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z} .$$
I can see that this is true, I also checked it with a few numbers. But I guess that is not enough to prove it. So how can I prove it?
(You don't need to show the whole proof, I think just a hint will be enough. I simply don't know how to start.)
| Since $w+\frac{1}{w} \geq 2$ for all $w>0$, we have
$$
\begin{split}
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)
&= 3 + \frac{x}{y}+\frac{y}{x} + \frac{x}{z}+ \frac{z}{x} +\frac{y}{z}+\frac{z}{y} \\&\geq 3+2+2+2 = 9
\end{split}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/78406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Asymptotic formula of $\sum_{n \le x} \frac{d(n)}{n^a}$ As the title says, I'm trying to prove
$$\sum_{n \le x} \frac{d(n)}{n^a}= \frac{x^{1-a} \log x}{1-a} + \zeta(a)^2+O(x^{1-a}),$$
for $x \ge 2$ and $a>0,a \ne 1$, where $d(n)$ is the number of divisors of $n$. There is a post here dealing with the case $a=1$. This is what I have done so far:
\begin{align*}
\sum_{n \le x} \frac{d(n)}{n^a} &= \sum_{n \le x} \frac{1}{n^a} \sum_{d \mid n} 1 = \sum_{d \le x} \sum_{\substack{n \le x \\\ d \mid n}} \frac{1}{n^a} = \sum_{d \le x} \sum_{q \le x/d } \frac{1}{(qd)^a} = \sum_{d \le x} \frac{1}{d^a} \sum_{q \le x/d} \frac{1}{q^a} \\\
&= \sum_{d \le x} \frac{1}{d^a} \left( \frac{(x/d)^{1-a}}{1-a} + \zeta(a) + O((x/d)^{-a}) \right) \\\
&= \sum_{d \le x} \left( \frac{x^{1-a}}{d(1-a)} + \frac{\zeta(a)}{d^a} \right) + O(x^{1-a}),
\end{align*}
from here things start to go out of hand... I've tried using the relevant formulas from this page, but I can't get it to "fit". Any help would be appreciated.
| What you're trying to show isn't true. You should have $\zeta(a)^2$ rather than $\zeta(a)$. (See Exercise 3.3 in Apostol's number theory book in the link above.)
With that correction, you're almost there! Picking up where you left off we have, and using two of the formulas in Apostol's text in the link above,
$$
\begin{align}
&\sum_{d \le x} \left( \frac{x^{1-a}}{d(1-a)} + \frac{\zeta(a)}{d^a} \right) + O\left(x^{1-a}\right) \\
&= \frac{x^{1-a}}{1-a} \sum_{d \le x} \frac{1}{d} + \zeta(a) \sum_{d \le x}\frac{1}{d^a} + O\left(x^{1-a}\right) \\
&= \frac{x^{1-a}}{1-a} \Big(\log x + O(1)\Big) + \zeta(a) \left(\frac{x^{1-a}}{1-a} + \zeta(a) + O(x^{-a})\right) + O\left(x^{1-a}\right) \\
&= \frac{x^{1-a} \log x}{1-a} + \zeta(a)^2 + O\left(x^{1-a}\right). \\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/79233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Probability that a $(n, \frac12)$-binomial random variable is even Let $X$ be a $(n, \frac12)$-Binomial RV. Show that $X$ is even with probability $\frac12$.
| A more general question might be "What is the probability that a $(n,p)$-Binomial random variable $X$ is even?" Generalizing on my hint on the main question,
$$
\begin{align*}
\left((1-p) + p\right)^n &= \sum_{k=0}^n\binom{n}{k}p^k(1-p)^{n-k}\\
&= \sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}p^{2k}(1-p)^{n-2k} +
\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k+1}p^{2k+1}(1-p)^{n-(2k+1)}\\
&= P\{X~\text{even}\} + P\{X~\text{odd}\}
\end{align*}
$$
and
$$
\begin{align*}
\left((1-p)-p\right)^n &= \sum_{k=0}^n \binom{n}{k}(-p)^k(1-p)^{n-k}\\
&= \sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k}p^{2k}(1-p)^{n-2k} -
\sum_{k=0}^{\lfloor n/2 \rfloor}\binom{n}{2k+1}p^{2k+1}(1-p)^{n-(2k+1)}\\
&= P\{X~\text{even}\} - P\{X~\text{odd}\}
\end{align*}
$$
giving that
$$P\{X~\text{even}\} = \frac{1}{2}(1 + (1-2p)^n)
= \frac{1}{2} + \frac{1}{2}(1-2p)^n$$
which is $\frac{1}{2}$ if $p = \frac{1}{2}$ and converges to $\frac{1}{2}$
as $n \to\infty$ except when $p = 1$, in which case the probability alternates
between $0$ and $1$ depending on the parity of $n$ (as is obvious without
all the above calculation).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/82841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to prove whether a polynomial function is even or odd We know that a function is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$. With this reasoning is it possible to prove that a polynomial function such as $f(x) = a_{2n}x^{2n} + a_{2n-2}x^{2n-2} + ...+a_{2}x^2 + a_{0}$ is even or odd?
What do you suggest? How do we get started?
| Two polynomials have the same value at every real number if and only if they are identical (exact same coefficients in each and every power of $x$); this follows because a polynomial of degree $n\gt 0$ can have at most $n$ roots. If $f(x)$ and $g(x)$ are the same at every value of $x$, then $f-g$ has infinitely many roots, and so must be the zero polynomial.
So write
$$f(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x + a_0.$$
Then
$$\begin{align*}
f(-x) &= a_n(-x)^n + x_{n-1}(-x)^{n-1} + \cdots + a_1(-x) + a_0\\
&= a_n (-1)^nx^n + a_{n-1}(-1)^{n-1}x^{n-1} + \cdots + a_1(-1)x + a_0\\
&= \left((-1)^na_n\right)x^n + \left((-1)^{n-1}a_{n-1}\right)x^{n-1} + \cdots + \left(-a_1\right)x + a_0.
\end{align*}$$
Now, $f(x)$ is even if and only if $f(x) = f(-x)$. That means that we need
$$a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 = (-1)^nx^n + (-1)^{n-1}a_{n-1}x^{n-1}+\cdots +(-a_1)x + a_0.$$
That, in turn, means we need
$$a_n = (-1)^{n}a_n,\quad a_{n-1}=(-1)^{n-1}a_{n-1},\ldots,\quad a_1 = -a_1, \quad a_0=a_0.$$
If $k$ is even, then $(-1)^k = 1$, so we automatically get $a_k=(-1)^ka_k$; that just says we need $a_k=a_k$, which is always true. If $k$ is odd, then $(-1)^k = -1$, so we need $a_k=-a_k$. This can only happen if $a_k=0$.
So $f(x)$ is an even function exactly when all odd terms to have coefficient $0$. So it must be a polynomial in which the only powers of $x$ that "show up" are even powers of $x$ (including $x^0$ which gives the constant term).
For $f(x)$ to be an odd function, we need $-f(x)=f(-x)$. That means that we need:
$$-\Bigl(a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1 x + a_0\Bigr) = (-1)^nx^n + (-1)^{n-1}x^{n-1}+\cdots +(-a_1)x + a_0.$$
That, in turn, means we need
$$-a_n = (-1)^{n}a_n,\quad -a_{n-1}=(-1)^{n-1}a_{n-1},\ldots\quad, -a_1 = -a_1, \quad -a_0=a_0.$$
This time, if $k$ is odd, then we are asking for $-a_k = -a_k$ to be true, which it always is; and if $k$ is even we are asking for $-a_k = a_k$, which is true if and only if $a_k=0$.
So $f(x)$ is an odd function exactly when all the even coefficients are zero; that is, the only powers that "show up" are odd powers of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/85828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Characteristic function of a standard normal random variable The characteristic function of a random variable $X$ is given by
$$\Phi_X(\omega) = \mathbb{E}e^{j\omega X}=\int_{-\infty}^\infty e^{j\omega x}f_X(x) dx.$$
One can easily capture the similarity between this integral and the Fourier transform.
For a standard normal random variable, the characteristic function can be found as follows:
$$\Phi_X(\omega)=
\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}e^{j\omega x} dx =
\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{(x^2-2j\omega x)}{2}\right)dx
$$
I know that the answer must be $\Phi_X(\omega) = \exp(-\omega^2/2)$, but can you explain how to evaluate the integral with a complex number in the exponent?
| I will give two answers:
Do it without complex numbers, notice that
$$ \begin{eqnarray}
\mathcal{F}(\omega) = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x &=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{-\infty}^0 \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x \\
&=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{0}^{\infty} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{-j \omega x} \mathrm{d} x \\
&=& 2 \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \cos(\omega x) \mathrm{d} x
\end{eqnarray}
$$
Now, compute $\mathcal{F}^\prime(\omega)$, and integrate by parts:
$$\begin{eqnarray}
\mathcal{F}^\prime(\omega) &=& -\frac{2}{\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{x^2}{2}} x \sin(\omega x) \mathrm{d} x = \frac{2}{\sqrt{2\pi}} \int_0^\infty \sin(\omega x) \mathrm{d} \left( \mathrm{e}^{-\frac{x^2}{2}} \right) \\
&=& \frac{2}{\sqrt{2\pi}} \left. \mathrm{e}^{-\frac{x^2}{2}} \sin(\omega x) \right|_0^\infty - \frac{2}{\sqrt{2\pi}} \int_0^\infty \mathrm{e}^{-\frac{x^2}{2}} \omega \cos(\omega x) \mathrm{d} x \\
&=& - \omega \mathcal{F}(\omega)
\end{eqnarray}
$$
The solution to so obtained ODE, $\mathcal{F}^\prime(\omega) = - \omega \mathcal{F}(\omega)$ is $\mathcal{F}(\omega) = c \exp\left( - \frac{\omega^2}{2} \right)$, and the integration constant is seen to be one from normalization requirement $\mathcal{F}(0)=1$ of the Gaussian probability density.
Complex integration: As you have started, complete the square:
$$ \left( -\frac{x^2}{2} + j \omega x \right) =
\left( -\frac{x^2}{2} + j \omega x + \frac{\omega^2}{2} \right) - \frac{\omega^2}{2} =
-\frac{1}{2} \left( x - j \omega \right)^2 - \frac{\omega^2}{2}
$$
We then have:
$$
\mathcal{F}(\omega) = \mathrm{e}^{-\frac{\omega^2}{2}} \cdot \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x
$$
The integral $I = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x$ is indeed $1$. To see this, consider
$$
\begin{eqnarray}
I_L &=& \int_{-L}^L \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{1}{2} \left( x - j \omega \right)^2 } \mathrm{d} x = \int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \\
&=& \left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) + \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \\
&=& \left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) + \mathcal{I}_L
\end{eqnarray}
$$
Here we denoted $\mathcal{I}_L = \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z$. Notice that $\lim\limits_{L \to \infty} \mathcal{I}_L = 1$.
Consider a complex contour $\mathcal{C}$, $ -L \to L \to L - j \omega \to -L - j \omega \to -L$:
$$
\begin{eqnarray}
I_L - \mathcal{I}_L &=&\left(\int_{-L-j \omega}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L}^{L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z\right) \\ &=& -\int_\mathcal{C} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{L}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L-j \omega}^{-L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z
\end{eqnarray}
$$
The integral over $\mathcal{C}$ is zero, since the integrand is holomorphic. Therefore:
$$
I-1 = \lim_{L \to \infty} (I_L-\mathcal{I}_L) = \lim_{L \to \infty} \left( - \int_{L}^{L-j \omega} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z - \int_{-L-j \omega}^{-L} \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{z^2}{2} } \mathrm{d} z \right)
$$
And the limit above is easily seen to vanish. Indeed:
$$
\lim_{L\to\infty} \left| \mathrm{e}^{-\frac{-(-L - j \omega t)^2}{2}} \right| =
\lim_{L\to\infty} \left| \mathrm{e}^{-\frac{-(L^2 + \omega^2 t^2)}{2}} \right| =0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/86041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 2,
"answer_id": 0
} |
Determine if the equation is valid/true The equation is:
$$\log_b \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = 2\log_b(\sqrt{3}+\sqrt{2}).$$
I can get as far as:
$$\log_b(\sqrt{3}+\sqrt{2}) - \log_b(\sqrt{3}-\sqrt{2}) = 2\log_b(\sqrt{3}+\sqrt{2})$$
Which looks almost too simple, but I can't get the signs to match up right to solve the problem. Do I need to further break out the logarithmic functions that are there?
| HINT 1. $(a-b)(a+b) = a^2-b^2$ for any numbers $a$ and $b$.
HINT 2. Note that
$$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}\right) = \frac{(\sqrt{3}+\sqrt{2})^2}{3-2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/86860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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$\int \cos^{-1} x \; dx$; trying to salvage an unsuccessful attempt $$
\begin{align}
\int \cos^{-1} x \; dx &= \int \cos^{-1} x \times 1 \; dx
\end{align}
$$
Then, setting
$$\begin{array}{l l}
u=\cos^{-1} x & v=x \\
u' = -\frac{1}{\sqrt{1-x^2}} & v'=1\\
\end{array}$$
Then by the IBP technique, we have:
$$\begin{array}{l l}
\int \cos^{-1} x \times 1 \; dx &=\cos^{-1} (x) \cdot x - \int x \cdot -\frac{1}{\sqrt{1-x^2}} \; dx\\
&= x \cos^{-1} (x) - \int -\frac{x}{\sqrt{1-x^2}} \; dx\\
\end{array}$$
Now at this point suppose I have overlooked the possibility of using integration by substitution (setting $u=1-x^2$) to simplify the second integral. Instead, I attempt to reapply IBP to the second integral $\int -\frac{x}{\sqrt{1-x^2}} \; dx$.
I let
$$\begin{array}{l l}
u= -\frac{1}{\sqrt{1-x^2}} = -(1-x^2)^{-\frac{1}{2}} \qquad & v= \frac{x^2}{2} \\
u' = - \left( -\frac{1}{2} \right) (1-x^2)^{-\frac{3}{2}} \times -2x = -x(1-x^2)^{-\frac{3}{2}} \qquad & v'=x\\
\end{array}$$
Then by IBP again,
$$\begin{array}{l l}
\int \cos^{-1} x \times 1 \; dx &= x \cos^{-1} (x) - \left( -\frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot -x(1-x^2)^{-\frac{3}{2}} \; dx \right) \\
\end{array}$$
At this stage, I can see no way to proceed. Can anyone see a reasonable way to salvage this solution, continuing along this line of reasoning? Or was approaching the second integral by IBP doomed to fail?
| You might have created more work for yourself with the second integration by parts, but have you tried using the substitution you mentioned in the previous step? Assuming your work is correct to that point, the substitution $u=1-x^2$ does appear to lead to something more manageable, even after the second integration by parts.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is this a proper use of induction? ($(n^2+5)n$ is divisible by 6) Just want to get input on my use of induction in this problem:
Question. Use mathematical induction to prove that $(n^2+5)n$ is divisible by $6$ for all integers $n \geqslant 1$.
Proof by mathematical induction.
(1) show base case ($n=1$) is true:
$$
((1)^2 + 5) (1) = 6
$$
$6$ is indeed divisible by $6$, so base case ($n=1$) is true
(2a) Assume case $n$ is true: $(n^2+5)n$ is divisible by $6$.
(2b) Show that case $n$ $\implies$ case $(n+1)$
$$
\begin{align*}
((n+1)^2+5)(n+1)
&\rightarrow
((n^2+2n+1)+5)(n+1)
\\ &\rightarrow
[(n^2+5)+(2n+1)](n+1)
\\ &\rightarrow
(n^2+5)n + (n^2+5)+(2n+1)n+ (2n+1)
\\ &\rightarrow
(n^2+5)n + [(n^2+5)+(2n^2+n)+ (2n+1)]
\\ &\rightarrow
(n^2+5)n + [(3n^2+3n)+6]
\end{align*}
$$
Now we can see case $(n+1)$ $= (n^2+5)n + (3n^2+3n)+6$.
We know $6$ is divisible by $6$ and we are assuming $(n^2+5)n$ is divisible by $6$ already, so all we need to do is show $(3n^2+3n)$ is divisible by $6$:
Letting $n=1$ for $(3n^2+3n)$ gives: $(3(1)^2+3(1)) = 6$
Thus, it has been demonstrated that $(n^2+5)n$ is divisible by $6$ for all integers $n \geqslant 1$.
I'm not sure if letting $n=1$ for that last expression is enough to prove it is divisible by $6$
| You can just say that it is always divisible by 6 since it is divisible by 3 (from the expression) and it is divisible by 2 since it is a multiplication of two consecutive integers- one of which is even.
Also the whole problem could be solved very easily using divisibility rather than induction, but I guess your problem asked you to use it.
| {
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"timestamp": "2023-03-29T00:00:00",
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complex stereographic projection inverse
Wikipedia gives this form of the stereographic Projection from $S^{2} \rightarrow \hat{\mathbb C}$:$$ (1) : z=\frac{x_{1}+ix_{2}}{1-x_{3}}$$ and for the inverse projections the points are supposedly:$$(2): x_{1}=\frac{\overline{z}+z}{z\overline{z}+1}, x_{2}=\frac{z-\overline{z}}{i(z\overline{z}+1)}, x_{3}=\frac{z\overline{z}-1}{z\overline{z}+1}$$
How does he go from $(1)$ to $(2)$ ? I tried calculating the inverse matrix and reading the coefficients from it, looking at $\overline{z}, z$ in $(1)$ and solving it for $x_{1},x_{2},x_{3}$ but I don't get anything of this form
(f.e. I get: $\displaystyle x_{1}=\frac{z(1-x_{3})-2ix_{2}}{\overline{z}(1-x_{3})})$
| If $z=\dfrac{x_{1}+ix_{2}}{1-x_{3}}$, then $$|z|^2=z\overline{z}=\frac{x_{1}+ix_{2}}{1-x_{3}}\cdot\frac{x_{1}-ix_{2}}{1-x_{3}}=\frac{x_{1}^2+x_{2}^2}{(1-x_{3})^2}=\frac{1-x_{3}^2}{(1-x_{3})^2}=\frac{1+x_{3}}{1-x_{3}},$$
since $x_{1}^2+x_{2}^2+x_{3})^2=1$ for $(x_1,x_2,x_3)\in S^2$. From the above equality, we have $|z|^2(1-x_{3})=1+x_{3}$, or equivalently $(|z|^2+1)x_{3}=|z|^2-1$, which implies
$$x_{3}=\frac{|z|^2-1}{|z|^2+1}=\frac{z\overline{z}-1}{z\overline{z}+1}.$$
From $z=\dfrac{x_{1}+ix_{2}}{1-x_{3}}$ again, we have the real part of $z$
$$\Re(z)=\frac{x_1}{1-x_{3}}.$$
Since $\Re(z)=(z+\overline{z})/2$, we get
$$x_1=\frac{z+\overline{z}}{2}\cdot(1-x_{3})=\frac{z+\overline{z}}{2}\cdot(1-\frac{z\overline{z}-1}{z\overline{z}+1})=\frac{\overline{z}+z}{z\overline{z}+1}.$$
I will let you figure out the expression for $x_2$. Here is the hint for $x_2$:
$\Im(z)=\dfrac{x_{2}}{1-x_{3}}$.
| {
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Solving $z^4 + 2z^3 + 6z - 9 = 0$ I'm trying to solve $z^4 + 2z^3 + 6z - 9 = 0$.
$z$ is a complex number.
I usually can solve those equations when they are of second degree.
I don't know what to do, breaking out $z$ doesn't help...
EDIT: Sorry I forgot to mention that $z$ has a solution where $\Re(z) = 0$.
| The Quartic Formula can be very useful to solve this.
But there is an alternate way, which my lecturer suggested me : Summing and Subtracting some powers of $z$ to make expression factorable.
For example $z^4+2z^3+6z-9=0$ can be factorized as:
$z^4+(3z^3-z^3)+(3z^2-3z^2)+(9z-3z)-9=0$
$(z^4+3z^3+3z^2+9z)-(z^3+3z^2+3z+9)=0$
$z(z^3+3z^2+3z+9)-(z^3+3z^2+3z+9)=0$
$z(z^2(z+3)+3(z+3))-(z^2(z+3)+3(z+3))=0$
$(z-1)(z+3)(z^2+3)=0$.
Thus, $z^2+3=0$ or $z=\pm i \sqrt{3}$ are the required solutions for your given conditions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Need help deriving recurrence relation for even-valued Fibonacci numbers. That would be every third Fibonacci number, e.g. $0, 2, 8, 34, 144, 610, 2584, 10946,...$
Empirically one can check that:
$a(n) = 4a(n-1) + a(n-2)$ where $a(-1) = 2, a(0) = 0$.
If $f(n)$ is $\operatorname{Fibonacci}(n)$ (to make it short), then it must be true that $f(3n) = 4f(3n - 3) + f(3n - 6)$.
I have tried the obvious expansion:
$f(3n) = f(3n - 1) + f(3n - 2) = f(3n - 3) + 2f(3n - 2) = 3f(3n - 3) + 2f(3n - 4)$
$ = 3f(3n - 3) + 2f(3n - 5) + 2f(3n - 6) = 3f(3n - 3) + 4f(3n - 6) + 2f(3n - 7)$
... and now I am stuck with the term I did not want. If I do add and subtract another $f(n - 3)$, and expand the $-f(n-3)$ part, then everything would magically work out ... but how should I know to do that? I can prove the formula by induction, but how would one systematically derive it in the first place?
I suppose one could write a program that tries to find the coefficients x and y such that $a(n) = xa(n-1) + ya(n-2)$ is true for a bunch of consecutive values of the sequence (then prove the formula by induction), and this is not hard to do, but is there a way that does not involve some sort of "Reverse Engineering" or "Magic Trick"?
| Let $\alpha$ and $\beta$ be the two roots of the equation $x^2-x-1=0$. Then the $n$-th Fibonacci number is equal to
$$\frac{\alpha^n-\beta^n}{\sqrt{5}}.$$
We are interested in the recurrence satisfied by the numbers
$$\frac{\alpha^{3n}-\beta^{3n}}{\sqrt{5}}.$$
If $x$ is either of $\alpha$ or $\beta$, then $x^2=x+1$. Multiply by $x$. We get $x^3=x^2+x=2x+1$. It follows that $x^4=2x^2+x=3x+2$. But then $x^5=3x^2+2x=5x+3$, and then $x^6=5x^2+3x=8x+5$.
We want $x^6=Ax^3+B$, where $A$ and $B$ are rational, indeed integers. So we want $8x+5=A(2x+1)+B$. Reading off $A$ and then $B$ is obvious: we need $A=4$ and $B=1$.
So the numbers $\alpha^{3n}$ and $\beta^{3n}$ satisfy the recurrence $y_n=4y_{n-1}+y_{n-2}$. By linearity, so do the numbers $\frac{\alpha^{3n}-\beta^{3n}}{\sqrt{5}}$.
Comment: Note that using the same basic strategy, we can write down the recurrence satisfied by $\frac{\alpha^{kn}-\beta^{kn}}{\sqrt{5}}$. The coefficients that we painfully computed by hand, step by step, can be expressed simply in terms of Fibonacci numbers, and therefore so can the recurrence for the numbers $\frac{\alpha^{kn}-\beta^{kn}}{\sqrt{5}}$.
| {
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Difference Equation $y_{n+3} − 3 y_{n+1} + 2 y_n = (−2)^n$ I get the solution to be $y_n = A(-2)^n + Bn + C + \frac{1}{9}n(-2)^{n-1}$ but wolfram alpha gets $y(n) = c_1 (-2)^n+c_2+c_3 n+\frac{1}{27} 2^{n-1} e^{i \pi n} (4-3 n)$
Explanation required.
Help is much appreciated.
| Your answers are the same.
Observe that $e^{i\pi n}=\cos(\pi n)+i\sin(\pi n)=\cos(\pi n)=(-1)^n$. Hence, you have:
$$\begin{align*}
y(n)& = c_1 (-2)^n+c_2+c_3 n+\frac{1}{27} 2^{n-1} e^{i \pi n} (4-3 n) \\
& = c_1 (-2)^n+c_2+c_3 n+\frac{1}{27} (-2)^{n-1} (3n-4) \\
& = c_1 (-2)^n+c_2+c_3 n+\frac{1}{9} n (-2)^{n-1} - \frac{4}{27} (-2)^{n-1} \\
& = \left( c_1+\frac{2}{27}\right)(-2)^n+c_3 n+c_2+\frac{1}{9} n (-2)^{n-1}
\end{align*}$$
The last line is your solution.
| {
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If $n\ge 3$, $4^n \nmid 9^n-1$ Could anyone give me a hint to prove the following?
If $n\ge 3$, $4^n \nmid 9^n-1$
| Possibly I’m missing something obvious, but I don’t see any straightforward way to make pedja’s induction work, and ehsanmo’s answer amounts to waving a magic wand unless you sit down and go through the proof of the lifting-the-exponent lemma in the PDF to which he linked, so I’m going to offer an argument using only tools that you already have; it’s basically a special case of the lifting-the-exponent lemma adapted to your specific problem.
Let’s look at the highest power of $2$ that divides $9^n-1$; call it $t(n)$. $$9^n-1=(9-1)(9^{n-1}+9^{n-2}+\cdots+9+1)=8(9^{n-1}+9^{n-2}+\cdots+9+1)\;,$$ and each term inside the parentheses is odd, so $t(n)=3$ when $n$ is odd. In order for $9^n-1$ to be divisible by $4^n$, we must have $t(n)\ge 2n$, so we’ve at least shown that $4^n\nmid 9^n-1$ when $n\ge 3$ is odd. What about even $n$?
Suppose that $n=2^km$, where $k\ge 1$ and $m$ is odd. Then we can factor $9^n-1$ as a difference of squares to get $$9^n-1=9^{2^km}-1=\left(9^{2^{k-1}m}+1\right)\left(9^{2^{k-1}m}-1\right)\;.$$ If $k>1$ we can keep repeating the process on the last factor until the exponent reaches $m$:
$$\begin{align*}
9^n-1&=9^{2^km}-1\\
&=\left(9^{2^{k-1}m}+1\right)\left(9^{2^{k-1}m}-1\right)\\
&=\left(9^{2^{k-1}m}+1\right)\left(9^{2^{k-2}m}+1\right)\left(9^{2^{k-2}m}-1\right)\\
&\;\vdots\\
&=\left(9^{2^{k-1}m}+1\right)\left(9^{2^{k-2}m}+1\right)\cdots\left(9^{2^0m}+1\right)\left(9^{2^0m}-1\right)\;.\tag{1}\\
\end{align*}$$
Now $9\equiv 1\pmod 4$, so $9^a\equiv 1\pmod 4$ for all $a\ge 0$, and $9^a+1\equiv 2\pmod 4$ for all $a\ge 0$. This means that $9^a+1$ is divisible by $2$ but not by $4$ for $a\ge 0$. The total number of factors of $2$ in the product $(1)$ is therefore $k+t(m)$, where $m$ is odd. We saw in the first paragraph that $t(m)=3$ when $m$ is odd, so $$t(2^km)=k+3\;,\tag{2}$$ and it’s easy to check that $t(n)=k+3<2^{k+1}\le 2\left(2^km\right)=2n$ for $k>1$. This shows that $4^n\nmid 9^n-1$ whenever $4\mid n$, leaving only the case $n=2m$, $m$ odd. In this case $(2)$ implies that $t(n)=4$, and $4<2n$ whenever $n>2$, so we’ve now finally shown that $4^n\nmid 9^n-1$ whenever $n>2$.
The trick of writing a positive integer in the form $2^km$, where $k\ge 0$ and $m$ is odd, is worth remembering: it’s surprisingly useful.
| {
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"timestamp": "2023-03-29T00:00:00",
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The inequality $b^n - a^n < (b - a)nb^{n-1}$ I'm trying to figure out why $b^n - a^n < (b - a)nb^{n-1}$.
Using just algebra, we can calculate
$ (b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) $
$ = (b^n + b^{n-1}a + \ldots + b^{2}a^{n-2} + ba^{n-1}) - (b^{n-1}a + b^{n-2}a^2 + \ldots + ba^{n-1} + a^{n-1}) $
$ = b^n - a^n, $
but why is it necessarily true that $(b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) < (b - a)nb^{n-1}$?
Note: I am interested in an answer to that last question, rather than in another way to prove the general inequality in the title...
| You ask why it is necessarily true that
$$(b - a)(b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1}) < (b - a)nb^{n-1}.$$ A quick answer is that it is not necessarily true. We cannot have strict inequality if $b=a$. And there are other issues. For example, if $n=1$, we always have equality. And the inequality sometimes reverses when one of $a$ or $b$ is negative.
We deal in detail with the cases where $a$ and $b$ are both $\ge 0$. Suppose first that $b>a$, and let $n>1$. Then we want to show that
$$b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1} < nb^{n-1}.$$
The term $b^{n-2}a$ is strictly less than $b^{n-1}$, and there is such a term, since $n>1$. The remaining terms (if any) are also strictly less than $b^{n-1}$. There is a total of $n$ terms, so their sum is strictly less than $nb^{n-1}$.
A similar argument deals with $b<a$. Recall that inequality is reversed when we divide both sides by a negative number. So we want to show that
$$b^{n-1} + b^{n-2}a + \ldots + ba^{n-2} + a^{n-1} > nb^{n-1}.$$
Since $b<a$, the term $a^{n-1}$ is strictly bigger than $b^{n-1}$. We don't even need $n>1$. All the other terms are $\ge b^{n-1}$. There is a total of $n$ terms, and the inequality follows.
We stop at this point. The work is incomplete, since we have not dealt with negative numbers. Let $n=3$, $b=0$, and $a=-1$. Then $(b-a)(b^2+ba+a^2)$ is positive, but $(b-a)(3)b^2$ is $0$, so the inequality does not hold. With some work, one can track down all the situations where the inequality does hold.
| {
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"timestamp": "2023-03-29T00:00:00",
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Remember trig identities like $\cos(\pi/3) = 1/2$ I have started doing complex analysis and I keep having to switch between rectangular co-ordinates and polar form and I keep running into stuff like - $\cos(\pi/3) = 1/2$.
I keep having to look these up. Am I expected to memorize these or what, its getting to be a serious pain having to keep looking them up. Do people generally just memorize these identities for complex analysis?
| All you really need to know are the values of $\sin(x)$, where $x$ is $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}$, and $\frac{\pi}{2}$. All of the other values come directly from these:
$$
\begin{align}
\sin(0) &= \frac{\sqrt{0}}{2} = 0
\\
\\
\sin(\pi/6) &= \frac{\sqrt{1}}{2} = \frac{1}{2}
\\
\\
\sin(\pi/4) &= \frac{\sqrt{2}}{2}
\\
\\
\sin(\pi/3) &= \frac{\sqrt{3}}{2}
\\
\\
\sin(\pi/2) &= \frac{\sqrt{4}}{2} = 1
\end{align}
$$
These value are easy to memorize, or you can just find them using Gerry's or Lopsy's answers.
Now, say you want to find $\cos(\pi/3)$. Since $\sin(\pi/3) = \frac{\sqrt{3}}{2}$, and $\sin^2(x) + \cos^2(x) = 1$, you get that
$$
\cos^2(\pi/3) = 1 - \left(\frac{\sqrt{3}}{2}\right)^2 = 1 - \frac{3}{4} = \frac{1}{4}.
$$
Hence, $\cos(\pi/3) = \pm \frac{1}{2}$. Since $\pi/3$ is in the first quadrant, it must be the positive value.
This might be the long way around (and perhaps looking at triangles or the unit circle is easier), but this saves you from looking up any values.
| {
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${a_n}$ series of Fibonacci numbers. $f(x)=\sum_{0}^{\infty}a_nx^n$, show that in the convergence radius: $f(x)= \frac{1}{1-x-x^2}$ I'd really like your help with this following problem:
Let ${a_n}$ be a Fibonacci series $a_1=a_0=1$ and $a_{n+2}=a_n+a_{n+1}$ for every $n \geq 0$.
Let $f(x)=\sum_{0}^{\infty}a_nx^n$, I need to find the radius of convergence and to prove that in the range of this radius $f(x)= \frac{1}{1-x-x^2}$.
we know that the convergence radius $R=\lim_{n\to \infty} |\frac {a_n}{a_{n+1}}| $, How can I apply it in this case?
Any direction to prove that $f(x)$ is as requested?
Thanks alot!
| You can find the radius of convergence by noting the well-known identity:
$$
\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \varphi,
$$
where $\varphi = \frac{\sqrt{5}+1}{2}$ is the golden ratio.
Once you know the radius of convergence, to find the partial sum $S_N$, just write:
$$\begin{align}
S_N := \sum_{n=0}^N a_n x^n &= 1 +x + \sum_{n =2}^N a_n x^n
\\
&= 1+x + \sum_{n=2}^N(a_{n-2} + a_{n-1})x^n
\\
&= 1+ x + x^2\sum_{n=2}^N a_{n-2}x^{n-2} + x\sum_{n=2}^Na_{n-1} x^{n-1}
\\
&= 1 + x + x^2 \sum_{n=0}^{N-2} a_n x^n + x \cdot \sum_{n=1}^{N-1} a_n x^n
\\
&= 1 + x + x^2 S_{N-2} + x\sum_{n=0}^{N-1} a_n x^n - x
\\
&= 1 + x^2 S_{N-2} + x S_{N-1}.
\end{align}$$
Since $S_N = a_Nx^N + S_{N-1} = a_Nx^N + a_{N-1}x^{N-1} + S_{N-2}$, we arrive at
$$
S_N = 1 + x^2 (S_N - a_N x^N - a_{N-1}x^{N-1}) + x(S_N - a_Nx^N)
$$
Solving for $S_N$ gives
$$
S_N = \frac{1 - a_Nx^{N+2} - a_{N-1}x^{N+1} - a_Nx^{N+1}}{1 - x - x^2}.
$$
Finally, taking the limit $\lim_{N \to \infty} S_N$ gives the answer as the terms with powers of $x$ in the numerator go to zero (since the original sum under question converges).
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $x_1^2+x_2^2+x_3^2=1$ yields $ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $ Prove this inequality, if $x_1^2+x_2^2+x_3^2=1$:
$$ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $$
So far I got to $x_1^4+x_2^4+x_3^4\ge\frac{1}3$ by using QM-AM for $(2x_1^2+x_2^2, 2x_2^2+x_3^2, 2x_3^2+x_1^2)$, but to be honest I'm not sure if that's helpful at all.
(You might want to "rename" those to $x,y,z$ to make writing easier):
$$ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2} \le \frac{3\sqrt3}4$$
| Another solution. By the AM-GM inequality, we have
$$
\frac{ 2 \cdot \dfrac{1}{\sqrt{3}} \cdot x_i } {1+x_i^2}
\le
\frac{\dfrac{1}{3} + x_i^2 }{1+x_i^2} = 1 - \frac{2/3}{1+x_i^2},
$$
But $-1/(1+z)$ is a concave function of $z = x^2$, so
\begin{align}
\frac{ \dfrac{2}{\sqrt{3}} x_i } {1+x_i^2}
&\le
\sum_{i=1}^3 \left[1 - \frac{2/3}{1+x_i^2}\right] \\
&\le 3 \left[1 - \frac{2/3}{1 + \dfrac{1}{3}\left(\sum_{i=1}^3 x_i^2\right) } \right] = \frac{3}{2}.
\end{align}
Multiplying both sides by $\sqrt3/2$, we get the desired result.
| {
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zeta of three, question about closed form If $\sum\limits_{n=2}^\infty \frac1{(n^2-n)^3}=10-\pi^2$, then what is the limit in closed form of $\sum\limits_{n=1}^\infty \frac1{n^3}$?
| It's called Apéry's constant, $\zeta(3)$, and it doesn't have any known closed form.
Allow me to demonstrate how the first sum doesn't help. Use $(a-b)^3=a^3-3ab(a-b)-b^3$ below:
$$\sum_{n=2}^\infty \frac{1}{(n^2-n)^3}=\sum_{n=1}^\infty \left(\frac{1}{n}-\frac{1}{n+1}\right)^3=\sum_{n=1}^\infty \color{DarkOrange}{\frac{1}{n^3}}-\frac{3}{\color{Purple}{n(n+1)}}\color{Purple}{\left(\frac{1}{n}-\frac{1}{n+1}\right)}-\color{DarkOrange}{\frac{1}{(n+1)^3}}.$$
The left and rightmost parts of the summand form a $\rm\color{DarkOrange}{telescoping}$ series, so we evaluate this to $1$ and continue on with our derivation with further partial fraction decomposition:
$$=\color{DarkOrange}{1}-3\sum_{n=1}^\infty \color{Purple}{\left(\frac{1}{n}-\frac{1}{n+1}\right)^2}=
1-3\sum_{n=1}^\infty\left(\color{Red}{\frac{1}{n^2}}-\color{Green}{\frac{2}{n(n+1)}}+\color{Blue}{\frac{1}{(n+1)^2}}\right)$$
This time the left and rightmost parts are almost identical copies of $\zeta(2)$ and the middle telescoping:
$$=1-3\big(\color{Red}{\zeta(2)}-\color{Green}2+\color{Blue}{\zeta(2)-1}\big)=10-\pi^2.$$
As you can see, the copies of $\color{DarkOrange}{\zeta(3)}$ cancelled each other out while the $\zeta(2)$s didn't.
| {
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The sum of the coefficients of $x^3$ in $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$ I know how to solve such questions when it's like $(x+y)^n$ but I'm not sure about this one:
In $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$, What's the sum of the
coefficients of $x^3$?
| The formula you're looking for is
$$
(x+y+z)^8 = \sum_{i+j+k = 8} \begin{pmatrix} 8 \\\ i,j,k \end{pmatrix} x^i y^j z^k
$$
with
$$
\begin{pmatrix} 8 \\\ i,j,k \end{pmatrix} = \frac{8!}{i!j!k!}.
$$
This is known as the multinomial expansion (it works for more than $3$ variables too, you just have to add more indices and modify the multinomial coefficient accordingly). Using this, then you find when does $x^3$ appear in the expansion
$$
\left( 1 + \left( \frac {-x}2 \right) + \frac 1{\sqrt x} \right)^8 = \sum_{i+j+k = 8} \begin{pmatrix} 8 \\\ i,j,k \end{pmatrix} (-x/2)^j (1/\sqrt{x})^k
$$
In this case we must have $j-(k/2) = 3$, which means $2j-k = 6$, and $0 \le i,j,k \le 8$, or we can rewrite this as $0 \le j \le 8$, $k = 2j-6$ and $i = 8-j-k$. This leaves the cases $(5,3,0)$ and $(2,4,2)$. Computing, the coefficient in front of $x^3$ is
$$
\begin{pmatrix} 8 \\\ 5,3,0 \end{pmatrix} (-1/2)^3 + \begin{pmatrix} 8 \\\ 2,4,2 \end{pmatrix} (-1/2)^4 = 77/4.
$$
(I got the $77/4$ using WolframAlpha and/or a calculator, no magic there.)
Hope that helps,
| {
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"timestamp": "2023-03-29T00:00:00",
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Consider the sequence $\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\ldots$ Consider the sequence
$$\frac{1}{2},\frac{1}{3},\frac{2}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\ldots$$ For which numbers $b$ is there a subsequence converging to $b$?
| Your sequence includes every rational number in $(0,1)$ infinitely many times. For any $b\in[0,1]$, there exists a sequence of rational numbers in $(0,1)$ converging to $b$. That sequence necessarily occurs as a subsequence of your sequence. Conversely, any $b\notin[0,1]$ cannot have a subsequence of your sequence converge to it, as there exists a positive number $\epsilon$ such that every element of your sequence is more than $\epsilon$ away from $b$.
Thus, the numbers $b$ for which there is a subsequence converging to $b$ are precisely those $b\in[0,1]$.
| {
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Solving $ xy = a + b\cdot\operatorname{lcm}(x,y) + c\cdot\gcd(x,y)$ given $a,b,c$ Given $a$, $b$, and $c$, find the number of pairs of positive integers $(x, y)$ satisfying this equation:
$$ xy = a + b\cdot\operatorname{lcm}(x,y) + c\cdot\gcd(x,y).$$
If $a=2, b= 1, c= 1$, then the answer is 2.
If $a=160, b= 0, c= 90$, then the answer is 8.
If $a=300, b= 7, c= 5$, then the answer is 4.
| First, note that $xy=lcm(x,y)\gcd(x,y)$. So you first want to find solutions to $LG-bL-cG-a = (L-c)(G-b) - (a+bc) = 0$.
So we first need to know the factorizations of $a+bc$.
Even then, we need the conclusion that $G|L$, and then we can get many different $x,y$ for that pair $(L,G)$. Specifically, if $k$ is the number of distinct prime factors of $L/G$, there are $2^k$ different pairs $(x,y)$.
I seriously doubt there is much clever that can be done here.
The case where $L-c=a+bc$ and $G-b=1$ will be yield $G|L$ if and only if $a$ is divisible by $b+1$, in which case, the number of solutions from this pair is $2^k$, where $k$ is the number of distinct prime factors of $\frac{a}{b+1}+c$.
If $a,b,c>0$ and $a+bc$ is prime, these $2^k$ are the only solutions.
For the case $(a,b,c)=(2,1,1)$, $a+bc=3$, and the only factorizations are $(L-c,G-b)=(3,1)$ or $(1,3)$, which yields $(L,G)=(4,2)$ or $(2,4)$. But only the former works, so $(L,G)=(4,2)$ and the number of solutions is $2^1$ since $4/2$ has $1$ prime factor.
For $(a,b,c)=(160,0,90)$, $a+bc=160$, and there are lots of factorizations, the only two of which are relevant are $160=160*1$ and $160=80*2$. Then $(L,G)=(250,1)$ or $(L,G)=(170,2)$. In both cases, $L/G$ has two distinct prime factors, so both factorizations contribute $4=2^2$ solutions.
For $(a,b,c)=(300,7,5)$, $a+bc=335=5*67$. The only relevant factorization is $(L-c,G-b)=(67,5)$, so $(L,G)=(72,12)$, and $L/G=6$ has two prime factors, so there are $4=2^2$, therefore four answers total.
| {
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"timestamp": "2023-03-29T00:00:00",
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modulo question I thought you could only mod positive numbers but then I saw this
and became confused...How does this even work ? how can you have negatives?
$$\begin{align*}
-8 &\equiv 7 \pmod{5}\\
2 &\equiv -3 \pmod{5}\\
-3 &\equiv -8\pmod{5}
\end{align*}$$
Actually....What I thought was that mod just means how many times a number goes into another number while leaving a remainder...
But after some reading I find that negative values can be equal to positive values as long as some value-remainder is divisible by the modulus. But it doesnt make sense to me that the remainderi n these equation is greater than the modulus AND that the remainder is positive while answer on left end is negative.
| In simple terms mod gives you remainder. Also if $a\equiv b\pmod{n}$, then $b\equiv a\pmod{n}$, because $n$ divides $(a-b)$, then $n$ also divides $(b-a)$. With the examples you had, for instance
$$
\begin{align*}
-8 &\equiv 7 \pmod{5}\\
2 &\equiv -3 \pmod{5}\\
-3 &\equiv -8\pmod{5}
\end{align*}
$$
$-8 \equiv 7 \pmod{5} \hspace{5pt} \Rightarrow \hspace{5pt} 7 \equiv -8 \pmod{5}$ or in other words $5$ divides $7-(-8) = 15$, i.e. $5 | 15$.
Just like with positive values you could add $5$, on the negative side you could also subtract $5$. For instance if $x \equiv 1 \pmod{5}$, then $x \equiv -4 \pmod{5}$ and also $x \equiv 6 \pmod{5}$.
In simple words that is all about mod. But if you study Number Theory, there are lot of interesting things you would learn.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding a limit of a series How would you calculate this limit it just blew me off on my midterms i seem to have calculated the limit correctly but my process is bougus < what my friend said.
$$
\lim_{n\to\infty}\frac{n \sqrt{n} +n}{\sqrt{n^3}+2}
$$
How i calculated the limit:
$$
\lim_{n\to\infty}\frac{n \sqrt{n^2 \frac{n}{n^2}{}} +n}{\sqrt{n^4 n^{-1}}+2} =
\lim_{n\to\infty}\frac{n^2 \sqrt{\frac{n}{n^2}{}} +n}{n^2\sqrt{n^2 n^{-1}}+2} =
\lim_{n\to\infty}\frac{n^2 \sqrt{\frac{n}{n^2}{}} +n}{n^2\sqrt{n^2 n^{-1}}+2}\\
\frac{n^2 \sqrt{\frac{1}{n}{}} +n}{n^2\sqrt{n^2 n^{-1}}+2}=
\frac{\frac{n^2\sqrt{0}}{n^2\sqrt{0}}+\frac{n}{n^2\sqrt{0}}}{\frac{n^2\sqrt{0}}{n^2\sqrt{0}}+\frac{2}{n^2\sqrt{0}}}= \frac{1}{1}=1
$$
I followed a book where an example was given where it said you can transform
a expression like so:
$$
\frac{1}{n} \sqrt{n^2 + 2} = \sqrt{\frac{1}{n^2} (n^2+2)}
$$
or
$$
\sqrt{n^2+1} = n \sqrt{1+\frac{1}{n^2}}
$$
| A more readable form would be
$$
\large {
\lim_{n\to\infty}\frac{n \sqrt{n} +n}{\sqrt{n^3}+2} \hspace{4pt} = \hspace{4pt} \lim_{n\to\infty}\frac{n^{\frac{3}{2}}+n}{n^{\frac{3}{2}}+2}
\hspace{4pt} = \hspace{4pt} \lim_{n\to\infty}
\frac
{1+
\frac{1}{\sqrt{n}}
}
{1+\frac{2}
{n^{\frac{3}{2}}}
}
} \hspace{4pt} = \hspace{4pt} 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/112463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Resolve this double integral $\iint_D x(y+x^2)e^{y^2-x^4}dxdy$ over $D=\{ x \geq 0, x^2\leq y \leq x^2+1,2-x^2\leq y \leq 3-x^2 \}$ We can rewrite
$$
I=\iint_D x(y+x^2)e^{(y-x^2)(y+x^2)} dx dy
$$
$$
D= \{ (x,y) \in \mathbb{R}^2:x \geq 0 \land 0 \leq y-x^2 \leq 1 \land 2 \leq y+x^2 \leq 3 \}.
$$
With this new notation we can let
$$
\Phi(x,y) =
\begin{pmatrix}
y+x^2 \\
y-x^2
\end{pmatrix}
=
\begin{pmatrix}
u \\
v
\end{pmatrix}
$$
and
$$
J_{\Phi}(x,y) =
\begin{vmatrix}
2x & 1 \\
-2x & 1
\end{vmatrix}
=
2x +2x = 4x.
$$
Hence
$$\begin{align*}
I &= \int_2^3 \left( \int_0^1 x u e^{uv} \frac{1}{ |4x| } dv \right) du \\
&= \frac{1}{4} \int_2^3 \left( u \int_0^1 e^{uv} dv \right) du \\
&= \frac{1}{4} \int_2^3 \left( u \int_0^1 e^{uv} dv \right) du \\
&= \frac{1}{4} \int_2^3 \left( u \frac{1}{u} e^{uv} \Big\vert_{v=0}^{v=1} \right) du \\
&= \frac{1}{4} \int_2^3 \left( e^{uv} \Big\vert_{v=0}^{v=1} \right) du \\
&= \frac{1}{4} \int_2^3 \left( e^{u}-1 \right) du \\
&= \frac{1}{4} \left( e^{u}-u \right) \Big\vert_{u=2}^{u=3} \\
&= \frac{1}{4} \left( e^{3}-3 -e^2+2 \right) \\
&= \frac{1}{4} \left( e^{3}-e^2-1 \right) \\
\end{align*}$$
Is this correct?
I have some problem to affirm this because if i don't consider the condition $x \geq 0 $ in the domain $D$ the integral must be zero because the integrand function is odd respect the $x-$variable and the domain $D$ is symmetric respect the $y-$axis
| You considered condition $x\geq 0$ implicitly when you canceled out $x$ and $\frac{1}{|4x|}$. Also condition $x\geq 0$ is neccesasry because it guarantees that $\Phi$ is bijective. As for computations - everything is correct.
| {
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Integers and fractions How would I write this as an integer or a fraction in lowest terms?
$(1-\frac12)(1+\frac 12)(1-\frac13)(1+\frac13)(1-\frac14)(1+\frac14).....(1-\frac1{99})(1+\frac1{99})$
I really need to understand where to start and the process if anyone can help me.
| Rewrite it instead as
$$\color{Green}{\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\cdots\left(1-\frac{1}{98}\right)\left(1-\frac{1}{99}\right)}\times \color{Blue}{\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{98}\right)\left(1+\frac{1}{99}\right)}$$
Evaluate left and right parts separately. Hint: Look at how numerator and denominator cancel.
For example,
$$\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)=\frac{\color{DarkOrange}1}{\color{Red}2}\cdot\frac{\color{Red}2}{\color{Green}3}\cdot\frac{\color{Green}3}{\color{Blue}4}\cdot\frac{\color{Blue}4}{\color{DarkOrange}5}=\frac{1}{5}.$$
| {
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"url": "https://math.stackexchange.com/questions/114211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $\binom{2n}{ n}$ is divisible by 2?
Possible Duplicate:
prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer
Show that $\binom{2n}{ n}$ is divisible by 2?
Any help would be appreciated..
| An algebraic approach:
$$ { 2n \choose n } = \frac{(2n)!}{n! \cdot n!} $$
$$ = \frac{ 1 \cdot 2 \cdot 3 \cdots n \cdot (n+1) \cdots (2n-1)\cdot (2n) }{n! \cdot n!}$$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots (2n)]}{n! \cdot n!} $$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [(2.1) \cdot (2.2) \cdot (2.3) \cdots (2.n)]}{n! \cdot n!} $$
$$= \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [ 2^n \cdot (n)! ]}{n! \cdot n!} $$
$$ = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Salem numbers and Lehmer's decic Given,
$$x^{12}-x^7-x^6-x^5+1 = 0\tag1$$
This has Lehmer’s decic polynomial as a factor,
$$x^{10} + x^9 - x^7 - x^6 - x^5 - x^4 - x^3 + x + 1=0\tag2$$
hence one of its roots is the smallest known Salem number. All ten roots obey the beautiful cyclotomic relation,
$$x^{630}-1=\frac{(x^{315}-1)(x^{210}-1)(x^{126}-1)^2(x^{90}-1)(x^{3}-1)^3(x^{2}-1)^5(x-1)^3} {(x^{35}-1)(x^{15}-1)^2(x^{14}-1)^2(x^{5}-1)^6\, x^{68}}$$
found by D. Broadhurst. But this was back in 1999 (paper here). Has anything similar for other Salem numbers been found since then?
| Revisiting this old question, now armed with Mathematica's "Integer Relations", I find that it is quite easy to look for similar polynomial relations. For example, let $x$ be a root of Lehmer's decic, then it is also the case that,
$$x^{630}-1 = \frac{(x^{315}-1)(x^{210}-1)(x^{126}-1)^2(x^{90}-1)(x^{10}-1)(x^{9}-1)}{(x^{70}-1)(x^{63}-1)(x^{45}-1)(x^{42}-1)(x^{30}-1)(x^{6}-1)}$$
Given the fifth smallest known Salem number $y$ and a root of the decic,
$$y^{10}-y^6-y^5-y^4+1=0$$
then,
$$y^{210}-1 =\frac{(y^{105}-1)(y^{70}-1)(y^{42}-1)(y^{30}-1)(y^{14}-1)(y^{6}-1)(y^{3}-1)^2}{ (y^{35}-1)(y^{10}-1)(y^{7}-1)^2(y^{2}-1)\,y}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to eliminate the repeated case from polynomial counting? How to eliminate the repeated case from polynomial counting?
assume a die throw $3$ times, do not allow repeated number appear
$$(x+x^2+x^3+x^4+x^5+x^6)^3 - y$$
how to count the repeated case, that should be minused in above polynomial counting?
what is $y$ in terms of polynomial of $x$?
Mark six maple code
subs(z=0, diff(expand((1+z*x)*(1+z*x^2)*(1+z*x^3)*(1+z*x^4)*(1+z*x^5)*(1+z*x^6)*(1+z*x^7)*(1+z*x^8)*(1+z*x^9)*(1+z*x^10)
*(1+z*x^11)*(1+z*x^12)*(1+z*x^13)*(1+z*x^14)*(1+z*x^15)*(1+z*x^16)*(1+z*x^17)*(1+z*x^18)*(1+z*x^19)*(1+z*x^20)
*(1+z*x^21)*(1+z*x^22)*(1+z*x^23)*(1+z*x^24)*(1+z*x^25)*(1+z*x^26)*(1+z*x^27)*(1+z*x^28)*(1+z*x^29)*(1+z*x^30)
*(1+z*x^31)*(1+z*x^32)*(1+z*x^33)*(1+z*x^34)*(1+z*x^35)*(1+z*x^36)*(1+z*x^37)*(1+z*x^38)*(1+z*x^39)*(1+z*x^40)
*(1+z*x^41)*(1+z*x^42)*(1+z*x^43)*(1+z*x^44)*(1+z*x^45)*(1+z*x^46)*(1+z*x^47)*(1+z*x^48)*(1+z*x^49)), z$6));
| Instead of subtracting out the repeated cases, it's easier to only generate the non-repeated cases in the first place:
$$(1+zx)(1+zx^2)(1+zx^3)(1+zx^4)(1+zx^5)(1+zx^6)\;.$$
Then the coefficient of $z^3x^n$ counts the number of partitions of $n$ into $3$ distinct parts from $1$ to $6$, and the number of ways of getting that sum from the dice is $3!$ times that number, which is the coefficient of $x^n$ in
$$\left.\frac{\partial^3}{\partial z^3}\left((1+zx)(1+zx^2)(1+zx^3)(1+zx^4)(1+zx^5)(1+zx^6)\right)\right|_{z=0}\;.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there something like Cardano's method for a SOLVABLE quintic. So there is no quadratic formula equivalent for a GENERAL fifth degree equation, but is there an equivalent formula for a SOLVABLE fifth degree equation.
| For any solvable polynomial equation $a_px^p+a_{p-1}x^{p-1}+\cdots+a_1x+a_0=0$ of odd prime degree $p$, the roots have the form:
$x=-\frac{a_{p-1}}{pa_p}+\omega\sqrt[p]{r_1}+\omega^2\sqrt[p]{r_2}+\cdots+\omega^{p-1}\sqrt[p]{r_{p-1}}$
Where $r_1,\cdots,r_{p-1}$ are solutions to a polynomial of degree $p-1$ with coefficients in the same base field as the original equation.
A to determine which polynomial, one method is to set each of these expressions equal to a different root and deriving elementary symmetric polynomials in the $r_i$, and solving a linear system.
Example: $x^5-5x+12$ with one real root and Galois Group $D_5$.
We know that the roots are
$x_1=a+b+c+d$
$x_2=a\text{ cis }\frac{2\pi}{5}+b\text{ cis }\frac{4\pi}{5}+c\text{ cis }\frac{6\pi}{5}+d\text{ cis }\frac{8\pi}{5}$
$x_3=a\text{ cis }\frac{4\pi}{5}+b\text{ cis }\frac{8\pi}{5}+c\text{ cis }\frac{2\pi}{5}+d\text{ cis }\frac{6\pi}{5}$
$x_4=a\text{ cis }\frac{6\pi}{5}+b\text{ cis }\frac{2\pi}{5}+c\text{ cis }\frac{8\pi}{5}+d\text{ cis }\frac{4\pi}{5}$
$x_5=a\text{ cis }\frac{8\pi}{5}+b\text{ cis }\frac{6\pi}{5}+c\text{ cis }\frac{4\pi}{5}+d\text{ cis }\frac{2\pi}{5}$
Here, $a,b,c,d$ are fifth-roots of solutions to a quartic.
If you ask Mathematica to solve the quintic in closed form, it will give you root objects. The first one is real, the next two are one complex conjugate pair, and the last two are the other complex conjugate pair.
The expressions of the pair $(x_1,x_4)$ are complex conjugates, and so are expressions for $(x_2,x_3)$. This gives you a limited number of possible candidates for which $x$ corresponds to which root object that Mathematica gave.
For each candidate assignment, ask mathematica to simplify fully with "FullSimplify", the elementary symmetric polynomials in $a^5,b^5,c^5,d^5$ and look for the case where only rational numbers are spat out. These are $\pm$ coefficients of the auxiliary quartic with $a^5,b^5,c^5,d^5$ as the roots. Mathematica may take a while in this step, so be patient.
In our case, the correct assignments are $x_1=r_1$, $x_2=r_3$, $x_3=r_4$, $x_4=r_5$, $x_5=r_2$.
Here, $r_i$ are the root objects Mathematica gave.
The elementary symmetric polynomials in $a^5,b^5,c^5,d^5$ come out:
$e_1=-4, e_2=\frac{4}{5}$, $e_3=\frac{8}{125}$, $e_4=-\frac{1}{3125}$.
The quartic: $u^4+4u^3+\frac{4}{5}u^2-\frac{8}{125}u-\frac{1}{3125}=0$
Solving it and matching up the expressions correctly to $a^5,b^5,c^5,d^5$, we are done.
Note that mathematica evaluates $n$th roots of negative real numbers to complex numbers with argument $\frac{2\pi}{n}$, instead of evaluating them to be equal to negative real when $n$ is odd. You must account for that to check your work by flipping sign on the entire expression in the fifth root and the sign in front of the fifth root if the inside expression is negative real.
Final answer (accounting Mathematica's $n$th root preferences):
$x= -\omega\sqrt[5]{1+\frac{2}{5}\sqrt{5}-\frac{3}{25}\sqrt{125+55\sqrt{5}}}+\omega^2\sqrt[5]{-1+\frac{2}{5}\sqrt{5}+\frac{3}{25}\sqrt{125-55\sqrt{5}}}-\omega^3\sqrt[5]{1-\frac{2}{5}\sqrt{5}+\frac{3}{25}\sqrt{125-55\sqrt{5}}}-\omega^4\sqrt[5]{1+\frac{2}{5}\sqrt{5}+\frac{3}{25}\sqrt{125+55\sqrt{5}}}$
Theoretically, these steps can be repeated for any quintic, and replicated on any polynomial of odd prime degree. The only obstacle is in solving the auxiliary polynomial of degree $p-1$, beginning with $p=7$ onwards, which would require finding an intermediate field in which it factors. This can be done by assuming factorization and solving a system in Mathematica.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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