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Solve this integral:$\int_0^\infty\frac{\arctan x}{x(x^2+1)}\mathrm dx$ I occasionally found that $\displaystyle\int_0^{\Large\frac{\pi}{2}}\dfrac{x}{\tan x}=\dfrac{\pi}{2}\ln 2$.
I tried that
$$\int_0^{\Large\frac{\pi}{2}}\dfrac{x}{\tan x}=\int_0^{\Large\frac{\pi}{2}}x \ \mathrm d(\ln \sin x)=-\int_0^{\Large\frac{\pi}{2}}\ln (\sin x)=\dfrac{\pi}{2}\ln 2$$
Then I tried another method
$$\int_0^{\Large\frac{\pi}{2}}\dfrac{x}{\tan x}=\int_0^\infty\dfrac{\arctan x}{x(x^2+1)}\mathrm dx$$
I tried to expand $\arctan x$ and $\dfrac{1}{1+x^2}$, but got nothing, also I was confused that whether $\displaystyle\int_0^\infty$ and $\displaystyle\sum_{i=0}^\infty$ can exchange or not? If yes, on what condition?
Sincerely thanks your help!
|
I just want to seek ways that have nothing to do with $\ln (\sin x)$.
Hint. You may consider
$$
I(a):=\int_0^\infty\frac{\arctan (ax)}{x(x^2+1)}\:\mathrm dx,\quad 0<a<1, \tag1
$$
and obtain
$$
I'(a)=\int_0^\infty\frac1{(x^2+1)(a^2x^2+1)}\:\mathrm dx.
$$ By using partial fraction decomposition, we have
$$
\frac1{(x^2+1)(a^2x^2+1)}=\frac1{\left(1-a^2\right) \left(x^2+1\right)}-\frac{a^2}{\left(1-a^2\right) \left(a^2 x^2+1\right)}
$$ giving
$$
\begin{align}
I'(a)&=\frac1{\left(1-a^2\right)}\int_0^\infty\!\frac1{x^2+1}\:\mathrm dx-\frac{a^2}{\left(1-a^2\right)}\int_0^\infty\frac1{a^2x^2+1}\:\mathrm dx\\\\
&=\frac1{\left(1-a^2\right)}[\arctan x]_0^\infty-\frac{a^2}{\left(1-a^2\right)}\left[\frac{\arctan (ax)}a\right]_0^\infty\\\\
&=\frac1{\left(1-a^2\right)}\frac{\pi}2-\frac{a}{\left(1-a^2\right)}\frac{\pi}2\\\\
&=\frac{\pi}2\frac1{1+a} \tag2
\end{align}
$$ Since $I(0)=0$, by integrating $(2)$, you easily get
$$
\int_0^\infty\frac{\arctan (ax)}{x(x^2+1)}\:\mathrm dx=\frac{\pi}2\: \ln (a+1), \qquad 0\leq a <1,
$$
from which, by letting $a \to 1^-$, you deduce
$$
\int_0^\infty\frac{\arctan x}{x(x^2+1)}\:\mathrm dx=\frac{\pi}2 \ln 2
$$
as announced.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to solve $\sum_{k=1}^n\frac{k}{n^2+k}$? Can someone show me what is wrong with the expression I got for evaluating $\sum_{k=1}^n\frac{k}{n^2+k}$?
Steps:
$\sum_{k=1}^n\frac{k}{n^2+k} = \frac{\sum_{k=1}^nk}{\sum_{k=1}^{n}n^2+k} = \frac{\frac{1}{2}n(n+1)}{\sum_{k=1}^{n}n^2+\sum_{k=1}^{n}k} = \frac{\frac{1}{2}n(n+1)}{n^3+\frac{1}{2}n(n+1)} = \frac{n(n+1)}{2n^3+n(n+1)} = \frac{n+1}{2n^2+n+1}$
However, when I plug in something like $\sum_{k=1}^{10}\frac{k}{10^2+k}$ into this online summation calculator, the answer I get is $\sim0.514$, which is not the same answer I get with $\frac{10+1}{2(10)^2+10+1} = \frac{11}{211} \approx 0.0521$
| It's equal to $n - n^2 \sum_{k=1}^{n} \frac{1}{n^2 + k}$. Now use Harmonic sum to solve the rest.
| {
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Determine polynomial whose roots are a linear combination of roots of another polynomial Let $\alpha_1, \alpha_2, \alpha_3$ be the roots of the polynomial $p(x)=x^3+5x^2+7x+11$. Find a polynomial whose roots are $\frac{\alpha_1+\alpha_2}{2}, \frac{\alpha_2+\alpha_3}{2}, \frac{\alpha_1+\alpha_3}{2}.$ Calculating the exact roots doesn't really help because only one root is real. I've tried expanding
$$
\left(x-\frac{\alpha_1+\alpha_2}{2}\right)\left(x-\frac{\alpha_2+\alpha_3}{2}\right)\left(x-\frac{\alpha_1+\alpha_3}{2}\right)
$$
but it gets kind of messy. Any "nicer" ideas?
| $\bf{Alternatively::}$ If $x=\alpha\;,x=\beta\;,x=\gamma$ are the roots
of the equation $x^3+5x^2+7x+11=0.\;,$ Then $\alpha+\beta+\gamma = -5$
Now have to find an equation whose roots are $\displaystyle \frac{\alpha+\beta}{2}$ and $\displaystyle \frac{\beta+\gamma}{2}$ and $\displaystyle \frac{\gamma+\alpha}{2}$
Now Using above relation $$\alpha+\beta+\gamma = -5$$
We can write $$\alpha+\beta = -5-\gamma$$ and $$\beta+\gamma = -5-\alpha$$ and $\gamma+\alpha = -5-\beta$
So now we have to find an equation whose roots are $$\displaystyle \frac{-5-\alpha}{2}\;\;,\frac{-5-\beta}{2}\;\;,\frac{-5-\gamma}{2}.$$
so Let $$\displaystyle y = \frac{-5-x}{2}\Rightarrow 2y=-5-x\Rightarrow x=-\left(5+2y\right).$$
Now Put value of $y$ into $x^3+5x^2+7x+11=0\;,$
We get $$\displaystyle -(2y+5)^3+5(2y+5)^2-7(2y+5)+11=0$$
So our polynomial equation is $$(2y+5)^3-5(2y+5)^2+7(2y+5)-11=0$$
So we get $$8y^3+40y^2+64y+24 = 0\Rightarrow y^3+5y^2+8y+3=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Existence of integer solution of $a^2 -17b^2 = $ any constant When checking whether if $9-\sqrt{17}$ in the ring $\{a+b\sqrt17: a,b \in \mathbb{Z}\}$ is a prime.
Suppose
$$\alpha\cdot \beta = 9-\sqrt{17},$$
using norm argument
$$N(\alpha)N(\beta) = N(9-\sqrt{17}) = -8.$$
If say $N(\alpha) = N(a+b\sqrt17) = a^2 - 17b^2$, I need to know if there exist integer solutions for
$a^2 - 17b^2= 2$ or $a^2 - 17b^2 = 4$. So in general, how can I check if there exist integer solutions of $a^2 -17b^2 = $ any constant?
Thank you very much!
| Let the "any constant" be $\mu$, $\mu \geq 0$. Then $a^{2} - 17 \, b^{2} = \mu$ becomes
$$x^{2} - 17 \, y^{2} = 1 \tag{1}$$
where $a = \sqrt{\mu} \, x$ and $b = \sqrt{\mu} \, y$. Equation (1) is a Pell equation, see Pell Equations, and has solutions
\begin{align}
x_{n} &= \frac{1}{2} \, \left[ (33+8 \, \sqrt{17})^{n} + (33 - 8 \, \sqrt{17})^{n} \right] \\
y_{n} &= \frac{1}{2 \, \sqrt{17}} \, \left[ (33+8 \, \sqrt{17})^{n} - (33 - 8 \, \sqrt{17})^{n} \right]
\end{align}
and in terms of $(a,b)$ the result becomes
\begin{align}
a_{n} &= \frac{\sqrt{\mu}}{2} \, \left[ (33+8 \, \sqrt{17})^{n} + (33 - 8 \, \sqrt{17})^{n} \right] \\
b_{n} &= \frac{\sqrt{\mu}}{2 \, \sqrt{17}} \, \left[ (33+8 \, \sqrt{17})^{n} - (33 - 8 \, \sqrt{17})^{n} \right]
\end{align}
for $n\geq 0$. For the case of negative "any constant" a similar path may be taken.
| {
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How many ways to write $2010$?
Let $ N$ be the number of ways to write $ 2010$ in the form $ 2010 = a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10 + a_0$, where the $ a_i$'s are integers, and $ 0 \le a_i \le 99$. An example of such a representation is $ 1\cdot10^3 + 3\cdot10^2 + 67\cdot10^1 + 40\cdot10^0$. Find $ N$.
I picked the biggest $a_1$ so: $a_1 = 2$, there are only two ways to form $2010$.
Take $a_1 = 1$ now. This opens up to a lot of possibilities.
Specific Casework should work:
Cases 1-1: $a_2 = 10$, then possibilities are: $a_1 = 1, a_0 = 0$ or $a_1 = 0, a_0 = 10$
Actually, I think a number-theoretic way is easier.
But still.
Case 1: $a_1 = 1$ then we must solve:
$100x + 10y + z = 1010$.
Since $0 \le x \le 10$, we can casework $x$ so that:
Case 1-1:$x = 0$. So that:
$10y + z = 1010 \implies z \equiv 0 \pmod{10}, z = 10k$ and $y = 101 - k$.
Hence, $(0, 101 - k, 10k)$. $\min{k} = 0 $ and we need to find the max of $k$. We must have, $101 - k \le 99$ and $10k \le 99$. This suggests, $k \le 9$.
Cases 1-2: $x=1$. So that:
$10y + z = 910 \implies z \equiv 0 \pmod{10}$ Again, $z = 10k$ and $y = 91 - k$. Giving a set of $(1, 91 - k, 10k)$.Here again, $\min{k} = 0$ and $10k \le 99$ so $k \le 9$.
I am conjecturing that since we are always increasing $x$ values, the value on the RHS will always be divisible by $10$.
$x = 9$ so that:
$10y + z = 110 \implies z = 10k$ and $y = 11 - k$, which again there are $9$ values.
Except if $x=10$ then there is: $10y + z = 10$ then $z = 10k$ and $y = 1 - k$. Then $k$ must be $1$.
So there are: $10(9) + 1 + 2 = 93$ solutions total.
This is just an attempt!
Bump: anybody have anything?
| Here's the bijection approach, which is hinted at by Joriki's solution.
Let $a_i = 10 b_i + c_i$, where $ 0 \leq c_i \leq 9$, $ 0 \leq b_i \leq 9$.
Then, we have
$$2010 = 10 (1000b_3 + 100 b_2 + 10b_1 + b_0) + ( 1000 c_3 + 100 c_2 + 10 c_1 + c_0)$$
Thus, given any representation of $2010 = 10 B + C $, there is a unique corresponding $b_i, c_i$ (clearly the digits of B and C) that we can create, and vice versa. This establishes the bijection. Hence, the answer is 202.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can one go around dividing by zero when simplifying? I have the expression $$\frac{1}{2}i\log\left(1-e^{\frac{2i\pi{x}}{b}}\right)-\frac{1}{2}i\log\left(1-e^{-\frac{2i\pi{x}}{b}}\right)$$ and I want to simplify it without getting any division-by-zero errors. Like when I convert this to $$\frac{1}{2}i\log\left(\frac{1-e^{\frac{2i\pi{x}}{b}}}{1-e^{-\frac{2i\pi{x}}{b}}}\right)$$ I get division by zero whenever $e^{\pm\frac{2i\pi{x}}{b}}=1$. Although, I know both expressions give singularities when $e^{\pm\frac{2i\pi{x}}{b}}=1$, the first one cancels out the singularity and outputs zero. Is there a mathematical way to simplify an expression such as this and go around the division-by-zero problem? Perhaps we can say that the $\log(\frac{n}{0})=0$? However that wouldn't satisfy me because no computation applications accept that.
| For
$$\frac{1}{2}i\log\left(\frac{1-e^{\frac{2i\pi{x}}{b}}}{1-e^{-\frac{2i\pi{x}}{b}}}\right)$$
then let $y = (\pi x)/b$ to obtain
\begin{align}
\ln\left( \frac{1 - e^{2i y}}{1- e^{-2i y}} \right) &= \ln\left( \frac{e^{i y} \, (e^{-i y} - e^{i y})}{e^{- i y} \, (e^{i y} - e^{-i y}) } \right) \\
&= \ln\left( - e^{2 i y} \right) = \ln\left( e^{i (\pi + 2y)}\right) \\
&= i \, (\pi + 2y) = i \, \pi \, \left( 1 + \frac{2 x}{b}\right).
\end{align}
Now,
$$\frac{i}{2} \, \ln\left( \frac{1 - e^{2i y}}{1- e^{-2i y}} \right) = - \frac{\pi}{2} \, \left(1 + \frac{2 x}{b}\right).$$
Since $1 = e^{\pm 2 \, n \, \pi \, i}$, $n \geq 0$, then $e^{\pm \frac{2 \pi \, i \, x}{b}} = 1 = e^{2 \, n \, \pi \, i}$ leads to $\frac{x}{b} = n$ and the expression for the logarithm becomes
$$\frac{i}{2} \, \ln\left( \frac{1 - e^{2i y}}{1- e^{-2i y}} \right) = - \frac{\pi}{2} \, \left(1 + \frac{2 x}{b}\right) = - \frac{\pi \, (1 + 2n)}{2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$n-\frac{n(n^2-1)}{2!}+\frac{n(n^2-1)(n^2-4)}{2!3!}-\frac{n(n^2-1)(n^2-4)(n^2-9)}{3!4!}+.....$ If $n\in \mathbb{N}$, and
$n-\dfrac{n(n^2-1)}{2!}+\dfrac{n(n^2-1)(n^2-4)}{2!3!}-\dfrac{n(n^2-1)(n^2-4)(n^2-9)}{3!4!}+\dots$
$=s_1$ when $n$ is even and $s_2$ when $n$ is odd
then prove that $s_1+s_2=0$
I know that when I put any even value (say $n=2$), I get $s_1$ and when I put any odd value (say $n=5$), I get $s_2$. And these add up to zero. But this is just verification, not a proof. I want to know how should I prove this.
| A neat question.
In fact, this is a particular case of a more general identity. The left-hand side may be rewritten as
\begin{gather}
n-\dfrac{n(n^2-1)}{2!}+\dfrac{n(n^2-1)(n^2-4)}{2!3!}-\dfrac{n(n^2-1)(n^2-4)(n^2-9)}{3!4!}+\cdots\\ = \sum_{k=0}^{n-1} (-1)^k{n\choose k+1} {n+k \choose k} = \sum_{k=0}^{n-1} {n \choose n-1-k} {-n-1 \choose k},
\end{gather}
where ${x\choose k} = x(x-1)\cdots(x-k+1)/k!$, $k\ge 1$; ${x\choose 0} = 1$.
Now this is a standard fact that
$$\sum_{k=0}^m {a \choose k}{b \choose m-k} = {a+b \choose m}.$$
It is well known for non-negative integers $a$, $b$, $m$, but can be easily shown for any $a,b\in \mathbb R$, $m\in \mathbb Z_+$ by noting that $(1+z)^a = \sum_{k\ge 0} {a\choose k} z^k$.
Returning to your problem, we can take $a=n$, $b=-n-1$, $m=n-1$, so
\begin{gather}
\sum_{k=0}^{n-1} {n \choose n-1-k} {-n-1 \choose k} = {-1 \choose n-1} = (-1)^{n-1},
\end{gather}
as required.
| {
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What is the expansion in power series of ${x \over \sin x}$ How can I expand in power series the following function:
$$
{x \over \sin x}
$$
? I know that:
$$
\sin x = x - {x^3 \over 3!} + {x^5 \over 5!} - \ldots,
$$
but a direct substitution does not give me a hint about how to continue.
| A simple way is to manipulate the generating function for Bernoulli numbers, but I will follow another approach. We have:
$$ \frac{1}{\sin x}=\frac{d}{dx}\log\tan\frac{x}{2}\tag{1}$$
as well as:
$$ \frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{\pi^2 n^2}\right),\qquad \cos x=\prod_{n\geq 0}\left(1-\frac{4x^2}{(2n+1)^2 \pi^2}\right)\tag{2}$$
hence by considering the logarithmic derivatives:
$$ -\frac{1}{x}+\cot(x) = \sum_{n\geq 1}\frac{2x}{x^2-n^2\pi^2},\qquad -\tan(x)=\sum_{n\geq 0}\frac{8x}{4x^2-(2n+1)^2 \pi^2}\tag{3}$$
so:
$$ \frac{1}{\sin x}=\frac{1}{2}\cot\left(\frac{x}{2}\right)+\frac{1}{2}\tan\left(\frac{x}{2}\right)=\frac{1}{x}+\sum_{n\geq 1}\frac{2x}{x^2-4n^2\pi^2}-\sum_{n\geq 0}\frac{2x}{x^2-(2n+1)^2 \pi^2}$$
and:
$$ \frac{x}{\sin x}=1+2\sum_{m\geq 1}x^{2m}\left(-\sum_{n\geq 1}\frac{1}{(2\pi)^{2m} n^{2m}}+\sum_{n\geq 0}\frac{1}{\pi^{2m}(2n+1)^{2m}}\right)\tag{4} $$
that gives:
$$\begin{eqnarray*} \frac{x}{\sin x}&=&1+2\sum_{m\geq 1}x^{2m}\left(-\frac{\zeta(2m)}{(2\pi)^{2m}}+\frac{(4^m-1)\zeta(2m)}{(2\pi)^{2m}}\right)\\&=& 1+2\sum_{m\geq 1}\frac{(4^m-2)\zeta(2m)}{(2\pi)^{2m}}x^{2m}.\tag{5}\end{eqnarray*}$$
As expected, we have an even analytic function in a neighbourhood of the origin, and the radius of convergence of the previous power series is $\pi$.
| {
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Solving complex equation ${{\left( z+2\bar{z} \right)}^{3}}=1$ Hey i am stuck tring to solve :
$${{\left( z+2\bar{z} \right)}^{3}}=1$$
I used Binomial theorem to expand the equation to :
$${{z}^{3}}+6{{z}^{2}}\bar{z}+12z{{\bar{z}}^{2}}+8{{\bar{z}}^{3}}=1$$
and i am not sure how should i continue
| Here $$(z+2\bar{z})^3 = 1\Rightarrow (z+2\bar{z}) = (1)^{\frac{1}{3}} = 1,\omega,\omega^2.$$
Where $\omega,\omega^2$ are cube root of unity.
So $$z+2\bar{z} = 1,\omega,\omega^2.$$
Now Put $z=x+iy$ and $\bar{z} = x-iy$
$\bullet\; $ If $z+2\bar{z} = 1\;,$ Then $x+iy+2(x-iy) = 1\Rightarrow 3x-iy=1+0\cdot i$
So we get $\displaystyle x = \frac{1}{3}$ and $y=0$
$\bullet\; $ If $z+2\bar{z} = \omega\;,$ Then $\displaystyle x+iy+2(x-iy) = -\frac{1}{2}+i\frac{\sqrt{3}}{2}\Rightarrow 3x-iy=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$
So we get $\displaystyle x = -\frac{1}{6}$ and $\displaystyle y=-\frac{\sqrt{3}}{2}$
$\bullet\; $ If $z+2\bar{z} = \omega\;,$ Then $\displaystyle x+iy+2(x-iy) = -\frac{1}{2}-i\frac{\sqrt{3}}{2}\Rightarrow 3x-iy=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$
So we get $\displaystyle x = -\frac{1}{6}$ and $\displaystyle y=+\frac{\sqrt{3}}{2}$
So we get $$\displaystyle z = \left\{\frac{1}{3}+0\cdot i\;\;,-\frac{1}{6}-\frac{\sqrt{3}}{2}\cdot i\;\;,-\frac{1}{6}+\frac{\sqrt{3}}{2}\cdot i\right\}$$
| {
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Singular points of algebraic curve, multiplicity, ordinary? Let $C \in \mathbb{P}_2$ be the curve defined by the polynomial$$P(x, y, z) = x^2z^2 + y^2z^2 + y^4.$$Find the singular points of $C$. For each one, calculate the multiplicity and say whether it is ordinary or nonordinary.
| We calculate$$P_x = 2xz^2,\text{ }P_y = 2yz^2 + 4y^3 =2y(z^2 + 2y^2),\text{ }P_z = 2x^2z + 2y^2z = 2z(x^2 + y^2).$$$P_x$ vanishes if $x = 0$ or $z = 0$. If $x = 0$, then $P_z$ vanishes if and only if $z = 0$ or $y = 0$. If $x = z = 0$, then $P = 0$ implies $y = 0$, which is impossible. Thus, if $x = 0$ then $y = 0$, and we find $[0, 0, 1]$ is a common zero of $P$ and its first derivatives. We have$$P(x, y, 1) = x^2 + y^2 + y^4.$$The lowest order part is $x^2 + y^2 = (x + iy)(x-iy)$, so $[0, 0, 1]$ is an ordinary double point.
If $z = 0$, then $P_y$ vanishes only if $y = 0$. We find that $[1, 0, 0]$ is a point on $C$ which is a common zero of all the first derivatives of $P$, so this is also a singular point. We have$$P(1, y, z) = z^2 + y^2z^2 + z^4.$$The lowest order part is $z^2$, which has a repeated factor, so $[1, 0, 0]$ is a nonordinary double point.
| {
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When the $a^2+b+c+d,b^2+a+c+d,c^2+a+b+d,d^2+a+b+c$ are all perfect squares? Find all $a,b,c,d\in \mathbb{Z}^+$, which $a^2+b+c+d,b^2+a+c+d,c^2+a+b+d,d^2+a+b+c$ are all perfect squares.
I found $(1,1,1,1)$, but I can't find more.
Is $a=b=c=d$ true?
| Since the expressions are symmetric, then we can set $a\leq b\leq c\leq d$ without loss of generality. There is an infinite number of positive integer solutions if we assume $a=1$ and $b=c=d=n$ which yields,
$$\begin{aligned}
a^2+b+c+d &= 3n+1\\
a+b^2+c+d &= (n+1)^2\\
a+b+c^2+d &= (n+1)^2\\
a+b+c+d^2 &= (n+1)^2
\end{aligned}\tag1$$
and it is easy to find solve $3n+1 = y^2$.
Excluding this infinite family, it seems the only solutions with all variables $1\leq a\leq b\leq c\leq d < 100$ are,
$$a,b,c,d = 6,6,11,11$$
$$a,b,c,d = 40,57,96,96$$
though I am not sure of this result. (Anyone can verify it?)
| {
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Fibonacci infinite sum resulting in $\pi$ I found the following identity. While trying to prove it, I found some things that I don’t quite understand:
$$\frac{\pi}{4}=\sqrt{5} \sum_{n=0}^{\infty} \frac{(-1)^n F_{2n+1}}{(2n+1) \phi^{4n+2}}$$
(where $\phi=\frac{\sqrt{5}+1}{2}$).
What I tried
I first considered the series:
$$F(x)=\sum_{n=0}^{\infty}F_{2n+1}x^{n}=\frac{1-x}{x^2-3x+1}$$ (when it converges).
Then I replaced $x$ with $x^2$ and tried integrating to get something like:
$$A(x)=\sum_{n=0}^{\infty}F_{2n+1} \frac{x^{2n+1}}{2n+1}=\int \frac{1-x^2}{x^4-3x^2+1}$$
This is where one question arises:
*
*Is this integration a valid thing to do? The sum on the left has a value, but the integral on the right has some constant added to it. In that case, how should I choose the constant?
Now, making $x=\frac{1}{\phi^2}$ would make the $\phi^{4n+2}$ term appear, but we still need to put a $(-1)^n$ in there, so I thought of putting an $i$ there (because $i^{2n+1}=i \cdot (-1)^n$) this way:
$x=\frac{i}{\phi^2}$, and then $$A(x)=i \sum_{n=0}^{\infty} \frac{(-1)^n F_{2n+1}}{(2n+1) \phi^{4n+2}}.$$ So now I need to prove that the integral for this $x$ is exactly $\frac{i \pi}{4\sqrt{5}}$, but the problem I have is that the integral has logarithms and I don’t know how to find logarithms of complex numbers like $\log(5+2i)$. (I found on Wikipedia the Taylor series for logarithms, but I can’t see how this makes the problem simpler.)
More questions
*Does it make sense to plug this complex value into the power series and the integral? If so, then how one would evaluate the integral for this specific $x=\frac{i}{\phi^2}$?
*Is there any other path to prove this intriguing identity?
| Start with
$$
\begin{align}
\sum_{k=0}^\infty F_{2k+1}x^{2k}
&=\frac{1-x^2}{1-3x^2+x^4}\\
&=\frac{1-x^2}{(x-\phi)(x+\phi)(x+\frac1\phi)(x-\frac1\phi)}\\
&=\frac1{2\sqrt5}\frac1{x+\phi}-\frac1{2\sqrt5}\frac1{x-\phi}+\frac1{2\sqrt5}\frac1{x+\frac1\phi}-\frac1{2\sqrt5}\frac1{x-\frac1\phi}\tag{1}
\end{align}
$$
Integration yields
$$
\begin{align}
\sum_{k=0}^\infty\frac{F_{2k+1}}{2k+1}x^{2k+1}
&=\frac1{2\sqrt5}\log\left(\frac{(x+\phi)(x+\frac1\phi)}{(x-\phi)(x-\frac1\phi)}\right)\\
&=\frac1{2\sqrt5}\log\left(\frac{x^2+\sqrt5x+1}{x^2-\sqrt5x+1}\right)\tag{2}
\end{align}
$$
Substituting $x\mapsto ix$ then multiplying by $-i\sqrt5$ gives
$$
\begin{align}
\sqrt5\sum_{k=0}^\infty(-1)^k\frac{F_{2k+1}}{2k+1}x^{2k+1}
&=\frac1{2i}\log\left(\frac{1-x^2+i\sqrt5x}{1-x^2-i\sqrt5x}\right)\\
&=\frac1{2i}\left[\frac12\log\left(1+3x^2+x^4\right)+i\arctan\left(\frac{\sqrt5\,x}{1-x^2}\right)\right]\\
&-\frac1{2i}\left[\frac12\log\left(1+3x^2+x^4\right)-i\arctan\left(\frac{\sqrt5\,x}{1-x^2}\right)\right]\\
&=\arctan\left(\frac{\sqrt5\,x}{1-x^2}\right)\tag{3}
\end{align}
$$
since $\log(a+ib)=\tfrac12\log(a^2+b^2)+i\arctan\left(\tfrac ba\right)$ for $a\gt0$; that is $|x|\lt1$.
Evaluating $(3)$ at $x=\frac1{\phi^2}$ yields
$$
\begin{align}
\sqrt5\sum_{k=0}^\infty(-1)^k\frac{F_{2k+1}}{2k+1}\frac1{\phi^{4k+2}}
&=\arctan\left(\frac{\sqrt5}{\phi^2-\frac1{\phi^2}}\right)\\
&=\arctan(1)\\[9pt]
&=\frac\pi4\tag{4}
\end{align}
$$
| {
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"url": "https://math.stackexchange.com/questions/1424554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "55",
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Pairs of cards from 1,2,...,n arranged according to rules on pairwise separation - is it possible for various n?
A deck of $2n+1$ cards consists of a joker and, for each number between $1$ and $n$ inclusive, two cards marked with that number. The $2n + 1$ cards are placed in a row, with the joker in the middle. For each $k$ with $1\le k\le n$, the two cards numbered $k$ have exactly $k−1$ cards between them. Determine all the values of $n$ not exceeding 10 for which this arrangement is possible. For which values of $n$ is it impossible?
The problem is from the 1992 Canada National Olympiad. I can show by parity arguments that the arrangement is impossible for some $n$. This approach is inconclusive for other values, though.
Numbering positions from $0$ at the left, the joker occupies position $n$ and the rightmost card occupies position $2n$.
Then the sum of positions is given by:
$$S_n = -n + \sum_{k=0}^{2n}{k} = -n + \frac{2n(2n+1)}{2} = 2n^2 \tag{1}$$
Now let the leftmost card of each pair of $k$ be at position $\alpha_k$. Then the rightmost card in the pair is at $\alpha_k + k$, so we have:
$$S_n = \sum_{k=1}^{n}{(2\alpha_k + k)} = \frac{n(n+1)}{2}+2\sum_{k=1}^{n}{\alpha_k}$$
Letting the sum of $\alpha$ values be $A$, then we have:
$$S_n = 2A + \frac{n(n+1)}{2} \tag{2}$$
By (1), $S_n$ is even for all $n$, and by (2) $S_n$ is even only if and only if the second term is even, so we deduce that the arrangement of cards is only possible if $\frac{n(n+1)}{2}$ is even.
This immediately rules out $n=1,2,5,6,9,10$.
For $n=3$, a possible arrangement is:
$$\begin{array}{ccccccc} 1&1&3&\bigstar&2&3&2 \end{array}$$
while for $n=4$, a possible arrangement is:
$$\begin{array}{ccc} 1&1&3&4&\bigstar&3&2&4&2 \end{array}$$
I am not sure how to resolve the question of whether there is a viable arrangement for $n=7$ or for $n=8$.
| Possible arrangement for $n=7$:$$\begin{array}{ccc} 1&1&2&5&2&6&7&\bigstar&5&3&4&6&3&7&4 \end{array}$$
Possible arrangement for $n=8$:$$\begin{array}{ccc} 1&1&2&4&2&7&8&4&\bigstar&6&3&5&7&3&8&6&5 \end{array}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I got a negative delta when computing deltas algebraically I am trying to find $\delta$ of $f(x)=4-x^2$ for $c=-1, L=3, \epsilon=\frac{1}{4}$. Here is what I did:
$$|f(x)-L|<\epsilon$$
$$|4-x^2-3|<\frac{1}{4}$$
$$-\frac{1}{4}<1-x^2<\frac{1}{4}$$
$$-\frac{5}{4}<-x^2<-\frac{3}{4}$$
$$\frac{5}{4}>x^2>\frac{3}{4}$$
$$\frac{\sqrt{5}}{2}>x>\frac{\sqrt{3}}{2}$$
$$\frac{\sqrt{3}}{2}<x<\frac{\sqrt{5}}{2}$$
Then the second step:
$$|x-c|<\delta$$
$$|x+1|<\delta$$
$$-\delta<x+1<\delta$$
$$-\delta-1<x<\delta-1$$
Left substitution:
$$-\delta-1=\frac{\sqrt{3}}{2}$$
$$\delta+1=-\frac{\sqrt{3}}{2}$$
$$\delta=-1-\frac{\sqrt{3}}{2}$$
Right substitution:
$$\delta-1=\frac{\sqrt{5}}{2}$$
$$\delta=\frac{\sqrt{5}}{2}+1$$
As you can see, I got a negative delta when doing the left substitution of the inequality ($\delta=-1-\frac{\sqrt{3}}{2}$). I heard it said somewhere that if you get a negative delta, you have an error in your algebra. Is this true? If so, what did I do wrong?
| The problem in your proof is the step where you square root the inequality. While the steps in deducing an interval for $x$ are correct, it's not the only interval. The interval you seek is a punctured interval around $x = -1$, so that should hint that you take the negative roots. Picking from there,
$$\frac{3}{4} < x^2 < \frac{5}{4}$$
$$- \frac{\sqrt 5}{2} < x < -\frac{\sqrt 3}{2}.$$
This means that so long as $x$ is between these numbers, $|f(x) - L| < \varepsilon$. Now we need to find a $\delta$ so that the interval $(-1 - \delta, -1 + \delta)$ is contained entirely within the inequality required for $x$. Unfortunately we have to cheat a bit -- numerically, $-\frac{\sqrt 5}{2} \approx -1.118$, and $-\frac{\sqrt 3}{2} \approx -0.866$. This hints that as $\delta$ grows, we will "hit" $-\frac{\sqrt 5}{2}$ first, so we only have to worry about the left side. That is, we should choose $\delta = -1 + \frac{\sqrt 5}{2}$.
Tackling the more general case, let $\varepsilon > 0$ be given. We seek $\delta$ so that whenever $0 < |x - c| < \delta$, we have
$$|f(x) - L| = |f(x) - f(-1)| = |4 - x^2 - (4 - c^2)| = |x^2 - c^2| = |x+c||x - c| < \varepsilon.$$
The $|x - c|$ in the last expression can be made as small as we want (since $|x - c| < \delta$), so we have to control $|x + c|$. Arbitrarily, choose $\delta < 1$. Then by the triangle inequality,
$$|x + c| = |x - c + 2c| \leq |x - c| + |2c| < 1 + |2c|.$$
In words, as long as $\delta < 1$, the $|x + c|$ term can only get so large. With this,
$$|x + c||x - c| < (1 + |2c|)\delta < \varepsilon,$$
suggesting we also choose $\delta < \frac{\varepsilon}{1 + |2c|}$. But how can we have two expressions for $\delta$? Simple, just pick the smaller of the two. Now letting $\delta = \min\left(1,\frac{\varepsilon}{1 + |2c|}\right)$, we are guaranteed that
$$|f(x) - L| = |x + c||x - c| < (1 + |2c|)\frac{\varepsilon}{1 + |2c|} = \varepsilon,$$
as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
modular exponentiation $2^{103}\equiv 63\pmod {143}$ I am trying to prove that $2^{103}\equiv 63\pmod {143}$.One way is to ''break'' the power of $2$ but $103$ is prime.Is there any way or theorem?
| One very common method for calculating large powers is repeated squaring.
To use it for your example here, we have:
$$
2 \equiv 2 \pmod{143} \\
2^2 \equiv 4 \pmod{143} \\
2^4 \equiv 16 \pmod{143} \\
2^8 \equiv 256 \equiv 113 \pmod{143} \\
2^{16} \equiv (-30)^2 \equiv 900 \equiv 42 \pmod{143} \\
2^{32} \equiv 42^2 \equiv 1764 \equiv 48 \pmod{143} \\
2^{64} \equiv 48^2 \equiv 2304 \equiv 16 \pmod{143} \\
$$
We therefore have:
$$
\begin{align}
2^{103} &\equiv 2^{64} \cdot 2^{32} \cdot 2^4 \cdot 2^2 \cdot 2 \pmod{143} \\
&\equiv (16 \cdot 48) \cdot 16 \cdot 4 \cdot 2 \pmod{143} \\
&\equiv (53 \cdot 16) \cdot (4 \cdot 2) \pmod{143} \\
&\equiv 133 \cdot 8 \pmod{143} \\
&\equiv 63 \pmod{143}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Minimum of $\frac{1}{x+y}+\frac{1}{x+z}-\frac{1}{x+y+z}$ Let $0\leq x,y,z\leq 1$. What is the minimum of $$F(x,y,z)=\frac{1}{x+y}+\frac{1}{x+z}-\frac{1}{x+y+z}?$$
We have $F(1,1,1)=2/3$, $F(1,1,0)=F(1,0,1)=F(1,0,0)=1$, and $F(0,1,1)=3/2$.
Is the minimum $2/3$?
| This can be done in an elementary way.
Consider the last two terms, which contain $z$ - they sum to $$\frac y{(x+z)(x+y+z)}$$
Increasing $z$ clearly decreases this term, so for a minimum you want $z$ as large as possible - changing $z$ doesn't change the first term at all. You can do a similar analysis for $y$. So with the constraints you have $y=z=1$
$$F(x,1,1)=\frac 2{x+1}-\frac 1{x+2}=\frac 1{x+1}+\frac 1{(x+1)(x+2)}$$
This is the sum of two positive terms which decrease with increasing $x$ (in the range you have) so the minimum is when $x=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $4\cos\frac\pi5\cdot\sin\frac\pi{10}=1$ We have to prove this identity : $$4\cos\frac\pi5\cdot\sin\frac\pi{10}=1$$
The book's hint is we somehow find out that $\displaystyle1+2\left(\cos\frac\pi5-\cos\frac{2\pi}5\right)$ equals something (from the geometry of a regular pentagon) and the result follows. But I can neither see why the book's hint is true nor how it could yield a result for the original problem. So could you help me proving it either by the book's hint or another way?
| Given $\displaystyle 2\cos \frac{\pi}{5}\cdot \sin \frac{\pi}{10}\;,$ Now Let $\displaystyle \frac{\pi}{10} = \theta \;,$ and using $$ 2\sin A\cdot \cos A = \sin 2A$$
Then $$\displaystyle 2\cos 2\theta \cdot \sin \theta=\frac{2\cos 2\theta\cdot \sin \theta \cdot \cos \theta}{\cos \theta} = \frac{2\sin 2\theta \cdot \cos 2\theta}{2\cos \theta} =\frac{\sin 4\theta}{2\cos \theta}$$
So we get $$\displaystyle \frac{\sin 4 \theta}{2\cos \theta} = \frac{\sin \left(\frac{4\pi}{10}\right)}{2\cos \frac{\pi}{10}} = \frac{\cos \frac{\pi}{10}}{2\cos \frac{\pi}{10}} = \frac{1}{2}$$
Bcz $$\displaystyle \sin \frac{4\pi}{10} = \sin \left(\frac{\pi}{2}-\frac{\pi}{10}\right) = \cos \frac{\pi}{10}$$
| {
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"url": "https://math.stackexchange.com/questions/1436565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$y=\sqrt{\cot 2x} \;$ revolved around x-axis between $x = \frac{\pi }{12} \;$ and $x=\frac{\pi }{4} \;$ find volume The region between the curve $y=\sqrt{\cot 2x} \;\;$ and the x-axis, from $x=\frac{\pi }{12} \;\;$ to $x=\frac{\pi }{4} \;\;$ is revolved around the x-axis to generate a solid. Find the volume of the solid.
I tried finding critical points, by taking derivative, but it seems y' ≠ 0
so I just plugged in $\frac{\pi }{12}$ and $\frac{\pi }{4}$ into $\sqrt{\cot 2x}$ and it seems that at $\frac{\pi }{12}$ function has the maximum y-value of $3^\frac{1}{4}$.
My radius will vary, r = $(3^\frac{1}{4} - \sqrt{\cot2x})$
Area = $\pi \cdot r^2$
Volume = $\pi \cdot r^2 \cdot dx$
Integration:
$$ \int^{3^\frac{1}{4}}_{0} \pi (\sqrt{3} - 2\cdot3^\frac{1}{4} \cdot \sqrt{\cot2x} + \cot2x) $$
but wolfram can't integrate it...
is it even correct?
| Both your bounds and integrand are incorrect. $x$ goes from $\dfrac{\pi}{12}$ to $\dfrac{\pi}{4}$. These are our bounds of integration; there is no need to calculate $y$ at these values because we are integrating with respect to $x$. The radius is given by the function itself. $y=r.$ So $$V=\int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \pi \cdot \left(\sqrt{\cot 2x}\right)^2\ \mathrm{d}x = \pi \cdot\int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \cot 2x\ \mathrm{d}x$$
This is easy to integrate.
| {
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"timestamp": "2023-03-29T00:00:00",
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Image of the Riemann-sphere Let $S$ be the Riemann-sphere (the unit sphere in $\mathbb{R}^3$) and $\psi: S \rightarrow \mathbb{C}$ be defined by $$\psi(x_1, x_2, x_3)=\frac{x_1 + ix_2}{1-x_3}.$$ Let $\pi$ be a plane in $\mathbb{R}^3$ such that the intersection with $S$ is not empty. Show that $\psi(\pi \cap S)$ is a line or a circumference in $\mathbb{C}$.
It's easy to see that if the plane has normal vector $(0,0,k)$, the image is a circumference. But i can't generalize the result.
| We first note that $\psi$ has an inverse function, given by
$$
\psi^{-1}(x + iy) = \left(\frac{2x}{x^2 + y^2 + 1}, \frac{2y}{x^2 + y^2 + 1}, \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1}\right).
$$
Let now $P$ be a plane in $\mathbb R^3$ given by
$$
P := \left\{ (x_1, x_2, x_3) \in \mathbb R^3 \middle| \sum_{j=1}^3 a_j x_j = b \right\}.
$$
First, we cover the case where the north pole (i. e. the vector $(0, 0, 1)$) is not contained within the plane. This implies that $a_3 \neq b$, since otherwise the north pole would be contained. Let $x \in P \cap S_2$. If $a := (a_1, a_2, a_3)$, we have due to the Cauchy-Schwarz inequality
$$
a \cdot x \le \|a\| \|x\| = \|a\|
$$
and therefore, since $x \in P$,
$$
\sqrt{\sum_{j=1}^3 a_j^2} \ge b \Rightarrow \sum_{j=1}^3 a_j^2 \ge b^2 \\
\Rightarrow a_1^2 + a_2^2 \ge b^2 - a_3^2 \ge (b - a_3)(b + a_3) \\
\Rightarrow \frac{a_1^2 + a_2^2}{(a_3 - b)^2} + \frac{b + a_3}{a_3 - b} \ge 0.
$$
Now if $x + iy$ is contained in $\psi(P)$, then we have
$$
a_1 \frac{2x}{x^2 + y^2 + 1} + a_2 \frac{2y}{x^2 + y^2 + 1} + a_3 \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1} = b.
$$
We want this to be a circle equation, i. e. an equation of the form
$$
(x - \lambda)^2 + (y - \mu)^2 = r
$$
for some $\lambda, \mu \in \mathbb R$ and $r \in \mathbb R_{\ge 0}$. Hence, we group all the terms of $x$ and $y$ together and complete the squares:
$$
a_1 \frac{2x}{x^2 + y^2 + 1} + a_2 \frac{2y}{x^2 + y^2 + 1} + a_3 \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1} = b \\
\Leftrightarrow 2 a_1 x + 2 a_2 y + a_3 x^2 + a_3 y^2 - a_3 = b (x^2 + y^2 + 1) \\
\Leftrightarrow (a_3 - b) x^2 + 2 a_1 x + (a_3 - b) y^2 + 2 a_2 y = b + a_3 \\
\Leftrightarrow x^2 + 2 x \frac{a_1}{a_3 - b} + y^2 + 2 y \frac{a_2}{a_3 - b} = \frac{b + a_3}{a_3 - b} \\
\Leftrightarrow
\left( x - \frac{a_1}{a_3 - b} \right)^2 + \left( y - \frac{a_2}{a_3 - b} \right)^2 = \frac{b + a_3}{a_3 - b} + \frac{a_1^2 + a_2^2}{(a_3 - b)^2}
$$
Now we consider the case $b = a_3$. Then we have
$$
a_1 \frac{2x}{x^2 + y^2 + 1} + a_2 \frac{2y}{x^2 + y^2 + 1} + a_3 \frac{x^2 + y^2 - 1}{x^2 + y^2 + 1} = b \\
\Leftrightarrow 2 a_1 x + 2 a_2 y + a_3 x^2 + a_3 y^2 - a_3 = b (x^2 + y^2 + 1) \\
\Leftrightarrow 2 a_1 x + 2 a_2 y = b + a_3,
$$
which defines a line.
| {
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If $p\equiv 3\pmod{4}$ and $p\mid x^2+y^2$, prove $p\mid x,y$. I have to prove that if $p$ is a prime number of the form $p = 4n - 1$, $n\in N$ and $x^2+y^2\equiv 0\pmod{p}$, then $x\equiv 0\pmod{p}$ and $y\equiv 0\pmod{p}$.
I have gone about this as follows and I hope you will correct me if I made any errors:
First of $x^2 + y^2$ is divisible by p so we can write it as being equal to some multiple $k$ of $p$ $+$ a remainder of $0$.
Then let's assume the opposite that neither $x$ or $y$ are congruent to $p$ ergo they aren't divisible by $p$ so we can represent them as being equal to some multiples $l$ and $m$ of $p$ (respectively) and with remainders (nonzero of course) $a$ and $b$ (respectively).
So we get:
$$ x^2 + y^2 = kp + 0$$
$$ x = lp + a $$
$$ y = mp + b$$
If we take the second and third expression and square them we get:
$$ x^2 = l^2p^2 + 2lpa + a^2 $$
$$ y^2 = m^2p^2 + 2mpb + b^2$$
Then we add them up:
$$ x^2 + y^2 = l^2p^2 + 2lpa + a^2 + m^2p^2 + 2mpb + b^2$$
Factorize:
$$ x^2 + y^2 = \left(l^2p + 2la + m^2p + 2mb\right)p + a^2 + b^2$$
We see that $k = l^2p + 2la + m^2p + 2mb$, which gives us:
$$ x^2 + y^2 = kp + a^2 + b^2$$
At the beginning it is stated that the remainder $a$ is equal to zero since $x^2 + y^2$ is divisible by $p$ so get to the conclusion that the sum of two nonzero, positive numbers, more precisely the sum of two squares $a^2 + b^2$ has to be $0$. This is a contradiction.
The contradiction was brought on by the assumption that $x$ and $y$ aren't divisible by $p$ therefore we conclude that $x$ and $y$ must be divisible by $p$.
| I.e. you want to prove:
$p\equiv 3\pmod{4},\ p\mid x^2+y^2\implies p\mid x,y$.
If $p\mid x$, then $p\mid y$, and vice versa.
So for contradiction assume $\gcd(p,x)=\gcd(p,y)=1$. See Modular Inverse.
$x^2\equiv -y^2\pmod{p}\iff (xy^{-1})^2\equiv -1\pmod{p}$
This can give a contradiction in several ways:
$1)\ $ Just remember Quadratic Reciprocity.
$2)\ $ Raise both sides by $\frac{p-1}{2}$ (which is odd):
$\implies \left(xy^{-1}\right)^{p-1}\equiv (-1)^{\frac{p-1}{2}}\equiv -1\pmod{p}$,
which contradicts Fermat's Little Theorem.
$3)\ $ Square both sides: $\left(xy^{-1}\right)^4\equiv 1\pmod{p}$.
Then $\text{ord}_p\left(xy^{-1}\right)\mid 4$. But it cannot be $1$ or $2$, because $\left(xy^{-1}\right)^2\equiv -1\pmod{p}$, therefore $\text{ord}_p\left(xy^{-1}\right)=4$, so $4\mid p-1$ by Fermat's Little Theorem (see below).
Theorem: If $\text{ord}_m(a)=d$ and $a^k\equiv 1\pmod{m}$, then $d\mid k$.
Proof: For contradiction, assume $k=dl+r$ for some $l,r\in\Bbb Z^+,\, 0<r<d$.
$a^{k}\equiv \left(a^d\right)^la^r\equiv 1^la^r\equiv a^r\equiv 1\pmod{m}$, contradiction.
| {
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How do I evaluate $\int \frac {x+4}{ 2x+6 } dx $? $$\int \frac {x+4}{ 2x+6 } dx$$
This is a problem from Khan Academy that I was reading about how to solve when I accidentally clicked next and lost the explanation. I was reading something about how there is a clever way to divide the function to make it easier to integrate. Can someone please explain this to me?
No actual solution, please. I want to get it by myself.
| By chain rule and fundamental theorem of calculus we have
$$
\int_{x} \frac{x+4}{2x+6} = \frac{1}{2} \int_{x} \frac{2x+8}{2x+6} = \frac{1}{2} \int_{x} \left( 1 + \frac{1}{x+3} \right) = \frac{x}{2} + \frac{1}{4}\int_{x}\frac{D(x+3)}{x+3} + \text{some constant} \\= \frac{x}{2} + \frac{1}{2}\ln |x+3| + \text{some constant}.
$$
| {
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} |
Limit of $(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}}$ I assume it is $1$.
Though $1$ is a limit of the function $(\frac{2^x+3^x+4^x}{5^x+6^x})^{\frac{1}{x}}$. Besides, even if it is $1$, I still need to solve the following inequality
$|(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}}-1|< \epsilon$
for arbitrary $\epsilon > 0$. That I don't know how to do as well.
| Notice that
$$(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}} \geq (\frac{4^n}{2 \cdot 6^n})^{\frac{1}{n}} = \frac{2}{3}\cdot (\frac{1}{2})^{\frac{1}{n}}$$
While
$$(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}} \leq (\frac{3 \cdot 4^n}{6^n})^{\frac{1}{n}}=\frac{2}{3}\cdot 3^{\frac{1}{n}}$$
And
$$\lim \limits_{n \to \infty} (\frac{1}{2})^{\frac{1}{n}}=\lim \limits_{n \to \infty} 3^{\frac{1}{n}}=1.$$
So the answer is $\frac{2}{3}$.
However I assume you are a beginner like me, thus maybe prefer to prove it by definition. It is not difficult, you only need to repeat trick I have shown.
$$|(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}}-\frac{2}{3}| \leq |(\frac{3 \cdot 4^n}{6^n})^{\frac{1}{n}}-\frac{2}{3}| \\
=\frac{2}{3} \cdot (3^{\frac{1}{n}}-1)$$
Thus for any positive real number $\epsilon$, there is a positive integer $N=\max \{1, \lfloor \frac{\ln 3}{\ln (\frac{3\epsilon}{2}+1)} \rfloor \}$ such that when $n>N$, $$|(\frac{2^n+3^n+4^n}{5^n+6^n})^{\frac{1}{n}}-\frac{2}{3}|<\epsilon.$$
That's what we want to prove.
Edited: So far I think the most nature way in this kind of approach is to notice the result here: Prove the following limit $\lim \limits_{n \to \infty} (3^n+4^n)^{\frac{1}{n}} =4$.
I think one can prove it without squeeze lemma.
So
$$\lim \limits_{n \to \infty} (2^n+3^n+4^n)^{\frac{1}{n}}=4$$
And
$$\lim \limits_{n \to \infty} (5^n+6^n)^{\frac{1}{n}}=6$$
Thus the original limit is $\frac{2}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate this particular improper intergral I am studying for upcoming exams and I am wondering how I would evaluate this question? (I have used a calculator and gotten an answer of 0.43, this answer is acceptable for me but I do not know how to get to there manually)
$$\int_1^\infty\dfrac{18}{8x(x+1)^2}\,dx$$
Thanks!
| The canonical way is to perform partial fraction decomposition. For this particular integrand, one trick that is useful is
$$\frac{1}{x(x+a)} = \frac{1}{a}\left(\frac{1}{x} - \frac{1}{x+a}\right)$$
With this, you can basically read off the partial fraction decomposition of the integrand as:
$$\frac{1}{x(x+1)^2} = \left(\frac{1}{x} - \frac{1}{x+1}\right)\frac{1}{x+1} =
\frac{1}{x(x+1)} - \frac{1}{(x+1)^2} = \frac{1}{x} - \frac{1}{x+1} - \frac{1}{(x+1)^2}$$
So the integral becomes
$$\frac94 \int_1^\infty \frac{dx}{x(x+1)^2}
= \frac94 \lim_{\Lambda\to\infty}\int_1^\Lambda \frac{dx}{x(x+1)^2}
= \frac94 \lim_{\Lambda\to\infty} \left[ \log\frac{x}{x+1} + \frac{1}{1+x}\right]_1^\Lambda
= \frac94 \left( \left(\log 1 + \frac{1}{\infty} \right) - \left(\log\frac12 + \frac12\right)\right)
= \frac98\left(2\log 2 - 1\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluation of $\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx$
Evaluation of $\displaystyle \int_{0}^{1}\left(1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots\right)dx$
$\bf{My\; try::}$ We can write $\displaystyle 1-x^3+x^5-x^8+x^{10}-x^{13}+\ldots$ as
$$\displaystyle (1-x^3)\cdot (1+x^5+x^{10}+\ldots ) = \frac{(1-x^3)(1)}{1-x^5}$$
So we can write it as $\displaystyle \frac{(1-x)(x^2+x+1)}{(1-x)(x^4+x^3+x^2+x+1)}$
So our Integral Convert into $\displaystyle \int_{0}^{1}\frac{x^2+x+1}{x^4+x^3+x^2+x+1}dx$
Now How can I solve it, Help me
Thanks
| $$\begin{eqnarray*}\sum_{k=0}^{+\infty}\left(\frac{1}{5k+1}-\frac{1}{5k+4}\right)&=&\int_{0}^{1}\frac{1-x^3}{1-x^5}\,dx\\&=&\int_{0}^{1}\frac{1+\frac{1}{\sqrt{5}}}{2+(1-\sqrt{5})x+2x^2}\,dx+\int_{0}^{1}\frac{1-\frac{1}{\sqrt{5}}}{2+(1+\sqrt{5})x+2x^2}\,dx\end{eqnarray*} $$
where the last identity comes from the residue theorem.
By using the reflection formula for the $\psi$ function, we have that the original series equals:
$$ \frac{1}{5}\left(\psi\left(\frac{4}{5}\right)-\psi\left(\frac{1}{5}\right)\right) =\frac{\pi}{5}\cot\frac{\pi}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
prove $2\sin^4(x)+2\cos^4(x)+\sin^2(2x)=2$ \begin{align}2\sin^4(x) + 2\sin^2(x)\cos^2(x) + 2\cos^4(x)&=2(\sin^4(x) + \sin^2(x)\cos^2(x) + \cos^4(x))\\
&=2(\sin^2(x) + \cos^2(x))^2 -\sin^2(x)\cos^2(x)\\
&=2(1) -\sin^2(x)\cos^2(x) \\
&=2-\sin^2(x)\cos^2(x)
\end{align}
And I'm stuck
| Using the formula $\sin (2x)=2\sin x\cos x$ we can expand/reduce
\begin{align}
2\sin^4 x+2\cos^4 x+\sin^2(2x)&=2\sin^4 x + 2\cos^2 x + 4\sin^2 x\cos^2 x\\
&=2(\sin^2 x+\cos^2x)^2\\
&=2(1)^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1445653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Middle School Probability Problem
Sven and Ole take turns rolling a standard six-sided die. The first person to roll
a six wins. If Sven goes first, what is the probability that he will win the game?
This is a problem from a worksheet for middle schoolers. I see that one could solve it by summing the infinite series
$$P(\text{Sven wins}) = P(\text{Sven wins in 1st round}) + P(\text{Sven wins in 2nd round}) + \cdots$$ but that's surely not how it is supposed to be solved. How else could one solve this?
| Let $A$ be the event in which Sven roll $6$ and $B$ be the event in which Ole rool $6$
So $\displaystyle P(A)=\frac{1}{6}$ and $\displaystyle P(\bar{A}) = \frac{5}{6}$ and $\displaystyle P(B)=\frac{1}{6}$ and $\displaystyle P(\bar{B}) = \frac{5}{6}$
So Required Probability
$$\displaystyle = P(A)+P(\bar{A})\cdot P(\bar{B})\cdot P(A)+P(\bar{A})\cdot P(\bar{B})\cdot P(\bar{A)}\cdot p(\bar{B})\cdot P(A)+.......\infty$$
So required probability $$\displaystyle = \frac{1}{6}+\frac{5}{6}\cdot \frac{5}{6}\cdot \frac{1}{6}+\frac{5}{6}\cdot \frac{5}{6}\cdot \frac{5}{6}\cdot \frac{5}{6}\cdot \frac{1}{6}+.....\infty$$
So Required Probability $$\displaystyle = \frac{1}{6}\left[\frac{1}{1-\left(\frac{5}{6}\right)^2}\right] = \frac{6}{11}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to Compute $\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}$
How to compute
$$\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}=-\dfrac{3}{2}$$
I'm interested in more ways of computing limit for this expression.
My Thoughts
at least i tried to use L'Hospital's rule but with no luck
\begin{align}
&\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2}=\lim _{x\to \:0}\dfrac{2x\left(\dfrac{\cos \left(x^2\right)}{\sin \left(x^2\right)+1}-1\right)}{\dfrac{2\arcsin \left(x\right)}{\sqrt{1-x^2}}-2x}\\
&=\lim _{x\to \:0}\dfrac{2\left(\dfrac{\cos \left(x^2\right)\left(\sin \left(x^2\right)+1\right)-2x^2\left(\cos ^2\left(x^2\right)+\sin ^2\left(x^2\right)+\sin \left(x^2\right)\right)}{\left(\sin \left(x^2\right)+1\right)^2}-1\right)}{2\left(\dfrac{x\arcsin \left(x\right)}{\left(1-x^2\right)^{\frac{3}{2}}}+\frac{1}{1-x^2}-1\right)}
\end{align}
Note this limit was taken from competition mathematics so i can't go with L'Hopital because each time i use it i got big terms that i have to derivative for the next time
| \begin{align}
\lim _{x\to \:0}\frac{\ln \left(1+\sin \left(x^2\right)\right)-x^2}{\left(\arcsin \:x\right)^2-x^2} &= \lim _{x\to \:0}\frac{\sin \left(x^2\right)-\frac{\sin^2(x^2)}{2} + O(x^6)-x^2}{\left(\arcsin \:x\right)^2-x^2}
\end{align}
(The above is true as $\ln(1+x) \approx x-x^2/2 +O(x^3)$ when $x$ is small)
\begin{align}
\lim _{x\to \:0}\frac{\sin \left(x^2\right)-\frac{\sin^2(x^2)}{2} + O(x^6)-x^2}{\left(\arcsin \:x\right)^2-x^2} &= \lim _{x\to \:0}\frac{x^2 -x^4/2 + O(x^{6})-x^2}{\left(x + \frac{x^3}{6}+O(x^5)\right)^2-x^2} \\
&=\lim _{x\to \:0}\frac{\frac{-x^4}{2}}{\frac{2x^4}{6}}\\
&= -\frac{3}{2}
\end{align}
(I used Taylor expansions)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find the maximum and minimum value of $2^{\sin x}+2^{\cos x}$ My try: Let $y$ = $2^{\sin x}+2^{\cos x}$
*
*Applying AM GM inequality I get
$y$ $> 2.2^{(\sin x+\cos x)/2}$.
Now, the highest value of R.H.S is $2^{\frac{\left(2+\sqrt{2}\right)}{2}}$.
Should this mean that $y$ is always greater than $2^{\frac{ \left ( 2+\sqrt{2}\right ) }{2}}$?
But this is not true (we can see in the graph).
*Calculus method:
$dy/dx$ = $\ln\left(2\right){\cdot}{2}^{\sin\left(x\right)}\cos\left(x\right){-\ln\left(2\right){\cdot}{2}^{\cos\left(x\right)}\sin\left(x\right)}$
When $dy/dx$ =0,
$\tan x = 2^{\sin x- \cos x}$ and I am stuck here.
https://www.desmos.com/calculator/p3zfvkq2mn
| Let $f(x)=2^{\sin x}+2^{\cos x},\;$ so $f^{\prime}(x)=(2^{\sin x}\cos x-2^{\cos x}\sin x)(\ln 2)$.
$\textbf{A)}$ To find the maximum of $f(x)$, it is enough to consider $f$ on $[0,\frac{\pi}{2}]$.
$\;\;\;$Then $f(0)=3$ and $f(\frac{\pi}{2})=3$, and
$\;\;\;f^{\prime}(x)=0\iff 2^{\sin x}\cos x=2^{\cos x}\sin x\iff\frac{2^{\sin x}}{\sin x}=\frac{2^{\cos x}}{\cos x}\iff x=\frac{\pi}{4}$,
$\;\;\;$since if $\displaystyle g(t)=\frac{2^t}{t},\;$ $\displaystyle g^{\prime}(t)=\frac{2^t(t\ln 2-1)}{t^2}<0$ for $0< t<1$ so $g$ is 1-1 on $(0,1]$.
$\;\;\;$Since $f(\frac{\pi}{4})=2\cdot2^{\sqrt{2}/2}=2^{1+\frac{\sqrt{2}}{2}}>3,\;$ $f(\frac{\pi}{4})=2^{1+\frac{\sqrt{2}}{2}}$ is the maximum value.
$\textbf{B)}$ Similarly, to find the minimum of $f(x)$ it is enough to consider $f$ on $[\pi, \frac{3\pi}{2}]$.
$\;\;\;$Then $f(\pi)=\frac{3}{2}$ and $f(\frac{3\pi}{2})=\frac{3}{2}$, and $f^{\prime}(x)=0\iff\frac{2^{\sin x}}{\sin x}=\frac{2^{\cos x}}{\cos x}\iff x=\frac{5\pi}{4}$
$\;\;\;$ since $\displaystyle g(t)=\frac{2^t}{t}$ is 1-1 on $[-1,0)$ as above.
$\;\;\;$Since $f(\frac{5\pi}{4})=2\cdot2^{-\frac{\sqrt{2}}{2}}=2^{1-\frac{\sqrt{2}}{2}}<\frac{3}{2},\;\;$ $f(\frac{5\pi}{4})=2^{1-\frac{\sqrt{2}}{2}}$ is the minimum value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Limit of $x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)$ when $x\to+\infty$
Find $$\lim_{x\to\infty}x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)$$
I calculated $\lim\limits_{x\to\infty}\left(\frac{x+2}{x+1}\right)^x=e$ but then the limit in question becomes $0\times \infty $ form, and further solution becomes messy.
Please tell a solution without the use of series expansions because I have no knowledge of these.
| Rewriting $\left(\frac{x+2}{x+1}\right)^x=\left(1+\frac{1}{x+1}\right)^x$ with Laurent series expansion rather than calculating out the limit you get:
$$\left(\frac{x+2}{x+1}\right)^x=\left(1+\frac{1}{x+1}\right)^x=e - \frac{3e}{2x} +\frac{83e}{24x^2} +\text{some higher order terms}$$
Thus,
$$x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)=xe-x\left(e - \frac{3e}{2x} +\frac{83e}{24x^2} +\text{some h.o.t.}\right)=\frac{3e}{2} +\frac{83e}{24x} +\text{some h.o.t.}$$
hence
$$
\lim_{x\to\infty}x\left(e-\left(\frac{x+2}{x+1}\right)^x\right)=\frac{3e}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Shortest method for $\int_{0}^{1}\frac{x^{4}\left(1-x\right)^{4}}{1+x^{2}}$ I don't want to solve by expanding it and all, I tried corollary but denominator becomes messy, also in the options there is $\pi$ so tried several trigonometric substitutions too
| Note that $(1 \pm i)^4 = -4$, hence $(1-x)^4 + 4 = p(x)(1+x^2)$ for some normalized second degree polynomial $p$. To find $p$, we plug in $0$, $1$, giving
$$ 5 = p(0), 4 = 2p(1) $$
With the ansatz $p(x) = x^2 + ax + b$, we have
$$ 5 = b,\quad 2+2a + 2b = 4 \iff a = -4 $$
Hence
$$ \frac{(1-x)^4}{1+x^2} = x^2 - 4x + 5 - \frac{4}{1+x^2} $$
Multiplying by $x^4$ gives
$$ \frac{x^4(1-x)^4}{1+x^2} = (x^6 - 4x^5 + 5x^4) - \frac{4x^4}{1+x^2} $$
Now we use
$$ \frac{x^4}{1+x^2} = \frac{x^4 - 1}{x^2 + 1} + \frac{1}{x^2+ 1} = x^2 - 1 + \frac{1}{x^2 + 1}$$
giving
$$ \frac{x^4(1-x)^4}{1+x^2} = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac 4{x^2 + 1} $$
So
$$ \int \frac{x^4(1-x)^4}{1+x^2}\, dx = \frac{x^7}7 - \frac 23x^6 + x^5 - \frac 43x^3 + 4x - 4\arctan x. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluating an integral in mathematica What does this interal evaluate to $$\int_m^n\int_m^n\frac{1}{4\sqrt{u^2+v^2}}\ du\ dv?$$
Mathematica seems to crap out on this.
Assume $m,n>0$.
| This is another approach which convert the two dimensional integral to a line integral first.
Let $r = \sqrt{u^2+v^2}$, $\displaystyle\;P = -\frac{v}{r}$ and $\displaystyle\;Q = \frac{u}{r}$, it is easy to check
$$\frac{\partial Q}{\partial v} - \frac{\partial P}{\partial u} = \frac{1}{r}$$
By Green's theorem, we have
$$\frac14 \int_{[m,n]^2} \frac{du dv}{r}
= \frac14 \int_{\partial [m,n]^2} P du + Qdv
= \frac14 \int_{\partial [m,n]^2} \frac{u dv - vdu}{\sqrt{u^2+v^2}}
$$
The boundary $\partial [m,n]^2$ consists of 4 line segments
*
*$v = m$, $u : m \to n$.
$$\verb/RHS/ = -\frac{m}{4} \int_m^n \frac{du}{\sqrt{u^2+m^2}} = -\frac{m}{4}\left[\sinh^{-1}\left(\frac{u}{m}\right)\right]_m^n
= -\frac{m}{4}\left(\sinh^{-1}\left(\frac{n}{m}\right) - \sinh^{-1}(1)\right)$$
*
*$u = n$, $v : m \to n$.
$$\verb/RHS/ = \frac{n}{4}\int_m^n \frac{dv}{\sqrt{n^2 + v^2}} =\frac{n}{4}\left[\sinh^{-1}\left(\frac{v}{n}\right)\right]_m^n = \frac{n}{4}\left(\sinh^{-1}(1) - \sinh^{-1}\left(\frac{m}{n}\right)\right)$$
*
*$v = n$, $u : n \to m$.
$$\verb/RHS/ = -\frac{n}{4}\int_n^m \frac{du}{\sqrt{u^2+n^2}}
= -\frac{n}{4}\left[\sinh^{-1}\left(\frac{u}{n}\right)\right]_n^m = \frac{n}{4}\left(\sinh^{-1}(1) - \sinh^{-1}\left(\frac{m}{n}\right)\right)$$
*$u = m$, $v : n \to m$.
$$\verb/RHS/ = \frac{m}{4}\int_n^m \frac{dv}{m^2 + v^2} =
\frac{m}{4}\left[\sinh^{-1}\left(\frac{v}{m}\right)\right]_n^m = -\frac{m}{4}\left(\sinh^{-1}\left(\frac{n}{m}\right)-\sinh^{-1}(1)\right)$$
Combine these four terms, we get
$$\frac14 \int_{[m,n]^2} \frac{du dv}{r}
= \frac{m+n}{2}\sinh^{-1}(1) - \frac{m}{2}\sinh^{-1}\left(\frac{n}{m}\right) - \frac{n}{2}\sinh^{-1}\left(\frac{m}{n}\right)\\
= \frac{m+n}{2}\log(\sqrt{2}+1)
- \frac{m}{2}\log\left(\frac{\sqrt{n^2+m^2}+ n}{m}\right)
- \frac{n}{2}\log\left(\frac{\sqrt{m^2+n^2}+ m}{n}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1452107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Precise definition of limits and $\lim_{x\to-1}\frac1{\sqrt{x^2+3}}=\frac12$
Prove using the epsilon-delta definition,
$$
\lim_{x\to-1}\frac1{\sqrt{x^2+3}}=\frac12
$$
what I came up so far was:
$0<|x+1|<δ$ implies $\left|\frac1{\sqrt{x^2+3}} - \frac12\right|<ϵ$
\begin{align}
&\frac1{\sqrt{x^2+3}} - \frac12 \\
&=\frac{2\sqrt{x^2+3}-(x^2+3)}{2(x^2+3)} \\
&= \frac{2\sqrt{(x+1)^2-2(x+1)+4} - ((x+1)^2-2(x+1)+4)}{2[(x+1)^2-2(x+1)+4]}
\end{align}
I'm unable to proceed because there are no real roots for $y^2-2y+4$
Can some please help me with this question? Thanks!
| Perhaps it is not the best idea to make the denominator square free
\begin{align}
\frac{1}{\sqrt{x^2+3}}-\frac12
&= \frac{2-\sqrt{x^2+3}}{2\sqrt{x^2+3}}=\frac{4-(x^2+3)}{2\sqrt{x^2+3}(2+\sqrt{x^2+3})}
\\
&=\frac{(1-x)(1+x)}{2\sqrt{x^2+3}(2+\sqrt{x^2+3})}
\end{align}
using binomial formulas twice.
| {
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"url": "https://math.stackexchange.com/questions/1452318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculating $\lim_{x\to 0}\frac{\sec (x)-1}{x^2\sec(x)}$ $$\lim_{x\to 0}\frac{\sec (x)-1}{x^2\sec(x)}$$
The first thing to do, as I was taught, was to rewrite this in terms of sine and cosine. Since $\sec(x) = \frac{1}{\cos(x)}$ we have
$$\frac{\frac{1}{\cos(x)}-1}{x^2\frac{1}{\cos(x)}}$$
And that is
$$\frac{\frac{1-\cos(x)}{\cos(x)}}{\frac{x^2}{\cos(x)}}$$
Then
$$\frac{(1-\cos(x))\cdot\cos(x)}{\cos(x) \cdot x^2}$$
And
$$\frac{(1-\cos(x))}{x^2}$$
Alright. It is my understanding that there is a formula or something that states $\frac{1-\cos(x)}{x} = 0$, so I can apply it here because we actually have
$$\frac{\color{green}{(1-\cos(x))}}{\color{green}{x}\cdot x}$$
Leaving us with
$$\color{green}0\cdot\frac{1}{x}$$
But if we evaluate, we get
$$0\cdot\frac{1}{0}$$
Which I guess is indeterminate because of $\frac{1}{0}$.
Also, the correct answer should be $\frac{1}{2}$.
What was wrong with my procedure?
| $1-\cos{x} = 2\sin^2{\frac{1}{2}x}$, so
$$ \frac{1-\cos{x}}{x^2} = \frac{1}{2}\left(\frac{\sin{\frac{1}{2}x}}{x/2}\right)^2, $$
and you probably know that
$$ \lim_{y \to 0} \frac{\sin{y}}{y} = 1, $$
so the limit is $1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1452853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Finding center of mass of a thin plate of constant density in specific region The region is between $ y= \frac{1}{1+x^2}$ and $ y= -\frac{1}{1+x^2} , x=0 , x=1$
What I currently have is this:
The answer is $ \bar {x} = \frac{\ln 4}{\pi} $ but I cannot seem to get that. Was I supposed to have used $\tan^{-1}x$ ? I'm lost.
| Notice, the area bounded by the given curves: $y=\frac{1}{1+x^2}$ & $y=-\frac{1}{1+x^2}$ between $x=0$ & $x=1$, is symmetrical about the x-axis hence the center of mass will lie on the x-axis.
Now, the x-coordinate of the center of mass of the bounded region is given as
$$\bar{x}=\frac{\int x\ dA}{\int dA}=\frac{\int x(ydx)}{\int (ydx)}$$
Setting the value of $y$ & applying proper limits,
$$=\frac{2\int_{0}^{1}x(ydx)}{2\int_{0}^{1}(ydx)}$$$$=\frac{\int_{0}^{1}x\frac{1}{1+x^2}\ dx}{\int_{0}^{1}\frac{1}{1+x^2}\ dx}$$
$$=\frac{\int_{0}^{1}\frac{1}{2}\frac{2x}{1+x^2}\ dx}{\int_{0}^{1}\frac{1}{1+x^2}\ dx}$$$$=\frac{\frac{1}{2}\int_{0}^{1}\frac{d(x^2)}{1+x^2}}{\int_{0}^{1}\frac{1}{1+x^2}\ dx}$$
$$=\frac{1}{2}\frac{[\ln(1+x^2)]_{0}^{1}}{[\tan^{-1}(x)]_{0}^{1}}$$
$$=\frac{1}{2}\frac{[\ln(2)-0]}{[\tan^{-1}(1)-0]}=\frac{1}{2}\frac{\ln 2}{\frac{\pi}{4}}=\frac{2\ln 2}{\pi}$$$$\color{red}{\bar x=\frac{\ln 4}{\pi}}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\lim_{x \to 0} x \lfloor x - \frac{1}{x} \rfloor$ Let $f: \mathbb{R} \to \mathbb{R}$ be the following function:
$f(x) = x \lfloor x - \frac{1}{x} \rfloor$
Show that $f(x)$ admits a limit at zero, and compute its value.
Using epsilon delta, I can prove that $\lim_{x \to 0} x \lfloor x - \frac{1}{x} \rfloor = - 1$:
$|x \lfloor x - \frac{1}{x} \rfloor + 1|$
$\leq |x \lfloor x - \frac{1}{x} \rfloor| + |1|$
$\leq |x| |\lfloor x - \frac{1}{x} \rfloor| + |1|$
$\leq |x| |x| + |1|$
$\leq |x|^2 + |1| < \epsilon$ if $|x| < \delta = \sqrt{\epsilon - 1}$
Is that right? If not, how can I do prove it?
But how do I show that $\lim_{x \to 0} x \lfloor x - \frac{1}{x} \rfloor$ exists?
| By definition of the floor function
$$x-\frac{1}{x}-1<\left\lfloor x-\frac{1}{x}\right\rfloor \le x-\frac{1}{x}$$
So for $x>0$
$$x^2-1-x < x \left\lfloor x-\frac{1}{x}\right\rfloor \le x^2-1$$
so by the Squeeze Theorem,
$$\lim_{x\to 0^+}{x \left\lfloor x-\frac{1}{x}\right\rfloor}=-1$$
Now for $x<0$ the direction of the inequality reverses so
$$x^2-1-x > x \left\lfloor x-\frac{1}{x}\right\rfloor \ge x^2-1$$
and again by the Squeeze Theorem,
$$\lim_{x\to 0^-}{x \left\lfloor x-\frac{1}{x}\right\rfloor}=-1$$
Since the two one-sided limits are the same, the limit as $x\to 0$ is $-1$.
| {
"language": "en",
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"answer_id": 0
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Three sides of a trapezium are each equal to $k$ cm.Prove that the greatest possible area of the trapezium is $\frac{3\sqrt3 k^2}{4}$ sq cm. Three sides of a trapezium are each equal to $k$ cm.Prove that the greatest possible area of the trapezium is $\frac{3\sqrt3 k^2}{4}$ sq cm.
I let two non-parallel sides and one of the parallel sides as $k$(shorter one).I know that area of the trapezium is $\frac{1}{2}\times $sum of parallel sides $\times$ height.But in this question,neither height is given nor longest side is given.How should i formulate the equation of the area?
|
Since we have an isosceles trapezium (non-parallel sides of equal length), we can "clip off" a triangular section and arrange the portions into a rectangle of dimensions $ \ k + x \ $ by $ \ h \ \ , $ with $ \ k \ $ being the hypotenuse of a right triangle with legs $ \ x \ $ and $ \ h \ = \ \sqrt{k^2 - x^2} \ \ . $ The area of the trapezium is then given by
$$ A \ \ = \ \ (k + x)·\sqrt{k^2 - x^2} \ \ = \ \ \sqrt{(k + x)^2·(k^2 - x^2)} \ \ = \ \ \sqrt{(k + x)^3·(k - x )} \ \ . $$
Without the use of calculus, we can take $ \ x \ = \ r·k \ $ to express the area as $ \ \sqrt{(k + rk)^3·(k - rk )} $ $ = \ k^2·\sqrt{(1 + r )^3·(1 - r )} \ = \ k^2·\sqrt{2·(1 + r )^3 \ - \ (1 + r )^4} \ \ . $ Making the further substitution $ \ s \ = \ 1 + r \ \ , $ the area is then $ \ k^2·\sqrt{2 s^3 - s^4} \ = \ k^2·\sqrt{ s^3 · (2 - s)} \ \ . $ Applying the AM-GM inequality, we obtain
$$ \sqrt[4]{\left(\frac{s}{3} \right)^3·(2 - s )}
\ \ \le \ \ \frac{\frac{s}{3} \ + \ \frac{s}{3} \ + \ \frac{s}{3} \ + \ (2 \ - \ s)}{4} \ \ = \ \ \frac{1}{2} $$ $$ \Rightarrow \ \ A \ = \ k^2 \ · \ \sqrt{3^3} \ · \ \left[ \ \sqrt[4]{\left(\frac{s}{3} \right)^3·(2 - s )} \ \right]^2 \ \ \le \ \ k^2 \ · \ 3\sqrt{3} \ · \left( \frac{1}{2} \right)^2 \ \ , $$
with equality occurring for $ \ s \ = \ \frac32 \ \Rightarrow \ r \ = \ \frac12 \ \Rightarrow \ x \ = \ \frac{1}{2}·k \ \ . $
With the use of differentiation, we have
$$ \frac{dA}{dx} \ \ = \ \ \frac{3}{2}·(k + x)^{1/2}·(k - x)^{1/2} \ - \ \frac{1}{2}·(k + x)^{3/2}·(k - x)^{-1/2} \ \ = \ \ 0 $$
$$ \Rightarrow \ \ 3·(k - x) \ \ = \ \ k + x \ \ \Rightarrow \ \ x \ = \ \frac{k}{2} \ \ . $$
[We then have $ \frac{d^2A}{dx^2} \ = \ \frac{2x^2-2kx-k^2}{(k - x)^{3/2}·(k + x)^{ 1/2}} \ \ < \ \ 0 \ \ $ for $ \ x \ = \ \frac{k}{2} \ \ . \ ] $
So the maximum area of the trapezium is $ \ \sqrt{\left(\frac{3k}{2} \right)^3·\left(\frac{ k}{2} \right) } \ = \ \frac{3\sqrt3}{4}·k^2 \ \ . $
(The factor of $ \ \frac{ \sqrt3}{4}·k^2 \ $ is perhaps a bit of a tip-off that a $ \ 60º \ $ angle is involved (as seen in Lai's solution), as this is the area of an equilateral triangle of side $ \ k \ \ . $ This actually suggests another approach: as the equilateral triangle maximizes the area of an isosceles triangle, Lai's parallelogram can be subdivided into three equilateral triangles with the indicated maximal total area.
Yet another way -- which would need to be made rigorous -- is to surmount the isosceles trapezium with an isosceles triangle. The area is maximized when the large "total" triangle is equilateral, which occurs for $ \ y \ = \ z \ = \ k \ \ $ in the diagram below. This maximal area is $ \ \frac{\sqrt{3}}{4}·(2k)^2 \ \ , $ of which the maximal-area isosceles trapezium has $ \ \frac34 \ $ of that area.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1457865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
How to prove this inequality $\frac{a^2}{a+b}+\frac{b^2}{b+c}\ge\frac{3a+2b-c}{4}$ Let $a,b,c\ge 0$,show that
$$\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}\ge\dfrac{3a+2b-c}{4}$$
| $$\begin{align}
(a+b+b+c)\left(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}\right)&\ge (a+b)^2 \\
&\ge (a+b)^2 - \left(\frac{a -c}{2}\right)^2 \\
&\ge (a+b - \frac{a -c}{2})(a+b + \frac{a -c}{2}) \\
&= \frac{2a+2b-a+c}{2}\cdot \frac{2a+2b+a-c}{2} \\
&= (a+2b+c)(\frac{3a+2b-c}{4})
\end{align}$$
So
$$ \left(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}\right)\ge \frac{3a+2b-c}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do we know that $x^2 + \frac{1}{x^2}$ is greater or equal to $2$? For one problem, we were supposed to know that:
$$x^2 + \frac{1}{x^2}\geq 2.$$
How do you deduce this instantly when looking at the expression above?
| Since you tagged this with precalculus, we'll try the following. Start with the inequality $(x^2 - 1)^2 \geq 0$. Then,
\begin{align*}
(x^2 - 1)^2 \geq 0 && \implies && x^4 - 2x^2 + 1 &\geq 0 \\
&& \implies && x^4 + 1 &\geq 2x^2 \\
&& \implies && \frac{x^4 + 1}{x^2} & \geq 2 & \text{for } x \neq 0 \\
&& \implies && x^2 + \frac{1}{x^2} & \geq 2.
\end{align*}
This was the inequality we wanted, except we have to be sure to exclude the case $x = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1460256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 7,
"answer_id": 2
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How to solve $\int_{0}^{\frac{π}{2}} \frac{dx}{4\sin^2(x) +5\cos^2(x)} $ $?$ I apply the substitutions:
$$t=\tan(x), \sin(x)=\frac{t}{\sqrt{1+t^2}}, \cos(x)=\frac{1}{\sqrt{1+t^2}}\ \&\ dx=\frac{dt}{1+t^2}$$
(using $t=\tan(x)$ you can draw a right angled triangle to find the other substitutions)
So we get:
$$\int_{0}^{\frac{π}{2}} \frac{1}{\frac{4t^2}{1+t^2}+\frac{5}{1+t^2}}\frac{dt}{1+t^2}=\frac{1}{4}\int_{0}^{\frac{π}{2}} \frac{1}{t^2+\Big(\frac{\sqrt{5}}{2}\Big)^2} dt$$
This is in a standard integral form, thus:
$$\frac{1}{4}\cdot\frac{2}{\sqrt{5}}\tan^{-1}\Bigg(\frac{2t}{\sqrt{5}}\Bigg)=\frac{1}{2\sqrt{5}}\tan^{-1}\Bigg(\frac{2\tan(x)}{\sqrt{5}}\Bigg)$$
this is from $0$ to $π/2$.
But I can't substitute for $π/2$, because $\tan(x)$ is undefined for $π/2$. If I input this integral into Mathcad I get $\frac{π\sqrt{5}}{20}$. How can I get the right answer out of this? Thanks in advance!
| Notice, $$\int_{0}^{\pi/2}\frac{dx}{4\sin^2x+5\cos^2 x}$$
$$=\int_{0}^{\pi/2}\frac{dx}{\cos^2 x\left(4\frac{\sin^2x}{\cos^2 x}+5\right)}$$
$$=\int_{0}^{\pi/2}\frac{\sec^2 x\ dx}{5+4\tan^2 x}$$
$$=\frac{1}{4}\int_{0}^{\pi/2}\frac{d(\tan x)}{\left(\frac{\sqrt 5}{2}\right)^2+(\tan x)^2}$$
$$=\frac{1}{4}\frac{2}{\sqrt 5}\left[\tan^{-1}\left(\frac{2\tan x}{\sqrt 5}\right)\right]_{0}^{\pi/2}$$
$$=\frac{1}{2\sqrt 5}\left[\frac{\pi}{2}-0\right]=\color{blue}{\frac{\pi}{4\sqrt 5}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1462259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
Summing up trigonometric series
By considering $\sum_{r=1}^n z^{2r-1}$ where z= $\cos\theta + i\sin\theta$, show that if $\sin\theta$ $\neq$ 0, $$\sum_{r=1}^n \sin(2r-1)\theta=\frac{\sin^2n\theta}{\sin\theta}$$
I couldn't solve this at first but with some hints some of you gave, I was able to come up with my own solution. Here it is:
First, we want to consider what is given in the question,
$$\begin{align}
\sum_{r=1}^n z^{2r-1} & =z+z^3+z^5...\\
& = z + z(z^2)+z^3(z^2)+...\\
\end{align}$$
In Geometric Progression, sum is given by
$$\sum_{r=1}^n z^{2r-1}=S_n = \frac{a (1-r^n)}{1-r}=\frac{z(1-(z^2)^n)}{1-z^2} = \frac {z-z^{2n+1}}{1-z^2}=\frac {1-z^{2n}}{z^{-1}-z} $$
Now, substitute $z = \cos \theta +i \sin \theta$
$$\begin{align}
\sum_{r=1}^n z^{2r-1} & =\frac{1-(\cos (2n\theta) + i \sin(2n\theta))}{\cos\theta - i\sin\theta-(\cos\theta+i\sin\theta)} \\
& = \frac{1-\cos (2n\theta) - i \sin(2n\theta)}{-2i\sin\theta} \cdot \frac{(i\sin\theta)}{(i\sin\theta)}\\
& = \frac{i\sin\theta-i\sin\theta \cos (2n\theta) + \sin\theta \sin(2n\theta)}{2\sin^2\theta} \\
\end{align}$$
Equating imaginary parts,
$$\sum_{r=1}^n \sin(2r-1)\theta = \frac{\sin\theta (1 - \cos(2n\theta))}{2\sin^2\theta} = \frac{\sin^2n\theta}{2\sin\theta} $$
| An alternative (maybe easier) approach. Since:
$$ 2\sin(\theta)\sin((2r-1)\theta) = \cos((2r-2)\theta)-\cos(2r\theta), $$
the original sum, multiplied by $2\sin\theta$, becomes a telescopic sum, whose value equals $1-\cos(2n\theta) = 2\sin^2(n\theta).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1463881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sum_{r=1}^n \frac 1{r}\binom{n}{r} = \sum_{r=1}^n \frac 1{r}(2^r - 1)$
Let $n$ be a nonnegative integer. Prove that $\sum\limits_{r=1}^n \dfrac 1{r}\dbinom{n}{r} = \sum\limits_{r=1}^n \dfrac 1{r}(2^r - 1)$.
One thing I have tried is to represent both $\binom{n}{r}$ and $2^r$ as sums of binomial coefficients, i.e. $\sum \binom{i}{r-1}$ and $\sum \binom{r}{i}$ respectively, but it does not seem to be helpful. I have also tried to use binomial identities but I do not see how they can be applied to the problem.
| Another solution written for one of my classes.
We must prove that
\begin{equation}
\sum_{r=1}^{n}\dfrac{1}{r}\dbinom{n}{r}=\sum_{r=1}^{n}\dfrac{2^{r}-1}
{r}
\label{darij1.eq.1}
\tag{1}
\end{equation}
holds for each nonnegative integer $n$.
Here is a proof of \eqref{darij1.eq.1} by induction on $n$:
The base case (the case $n=0$) is trivial, since the equality \eqref{darij1.eq.1} boils down to $0=0$ in this case (recall that empty sums are $0$ by definition).
For the induction step, we fix some positive integer $m$, and we assume that
\eqref{darij1.eq.1} holds for $n=m-1$. We must now prove that
\eqref{darij1.eq.1} holds for $n=m$.
We recall the following basic facts:
*
*Sum of a row in Pascal's triangle: We have
\begin{equation}
\sum_{r=0}^{n}\dbinom{n}{r}=2^{n}\qquad\text{for each integer }n\geq
0.
\label{darij1.eq.2n}
\tag{2}
\end{equation}
(This follows from substituting $x=1$ and $y=1$ into the binomial formula
$\left( x+y\right) ^{n}=\sum\limits_{r=0}^{n}\dbinom{n}{r}x^{r}y^{n-r}$.)
*Absorption identity: We have
\begin{align}
\dfrac{n}{k}\dbinom{n-1}{k-1}=\dbinom{n}{k}\qquad\text{for any integers
}n\text{ and }k>0.
\end{align}
(This follows by recalling the definitions of the two binomial coefficients:
\begin{align*}
\dbinom{n}{k} & =\dfrac{n\left( n-1\right) \left( n-2\right)
\cdots\left( n-k+1\right) }{k!}\qquad\text{and}\\
\dbinom{n-1}{k-1} & =\dfrac{\left( n-1\right) \left( n-2\right)
\cdots\left( n-k+1\right) }{\left( k-1\right) !},
\end{align*}
and comparing the left and right hand sides.)
*Pascal recursion: We have
\begin{align}
\dbinom{n}{k}=\dbinom{n-1}{k-1}+\dbinom{n-1}{k}\qquad\text{for any integers
}n\text{ and }k>0.
\end{align}
Furthermore, recall that $\dbinom{n}{k}=0$ whenever $n$ is a nonnegative
integer and $k$ is an integer satisfying $k>n$. Applying this to $n=m-1$ and
$k=m$, we obtain $\dbinom{m-1}{m}=0$ (since $m-1$ is a nonnegative integer and
$m>m-1$).
However, the Pascal recursion yields that
\begin{align}
\dbinom{m}{r}=\dbinom{m-1}{r-1}+\dbinom{m-1}{r}
\end{align}
for each integer $r>0$. Thus,
\begin{align*}
& \sum_{r=1}^{m}\dfrac{1}{r}\underbrace{\dbinom{m}{r}}_{=\dbinom{m-1}
{r-1}+\dbinom{m-1}{r}}\\
& =\sum_{r=1}^{m}\dfrac{1}{r}\left( \dbinom{m-1}{r-1}+\dbinom{m-1}{r}\right)
\\
& =\sum_{r=1}^{m}\dfrac{1}{m}\cdot\underbrace{\dfrac{m}{r}\dbinom{m-1}{r-1}
}_{\substack{=\dbinom{m}{r}\\\text{(by the absorption identity)}
}}+\underbrace{\sum_{r=1}^{m}\dfrac{1}{r}\dbinom{m-1}{r}}_{=\sum
\limits_{r=1}^{m-1}\dfrac{1}{r}\dbinom{m-1}{r}+\dfrac{1}{m}\dbinom{m-1}{m}}\\
& =\underbrace{\sum_{r=1}^{m}\dfrac{1}{m}\cdot\dbinom{m}{r}}_{=\dfrac{1}
{m}\sum\limits_{r=1}^{m}\dbinom{m}{r}}+\sum\limits_{r=1}^{m-1}\dfrac{1}
{r}\dbinom{m-1}{r}+\dfrac{1}{m}\underbrace{\dbinom{m-1}{m}}_{\substack{=0}}\\
& =\dfrac{1}{m}\underbrace{\sum\limits_{r=1}^{m}\dbinom{m}{r}}_{=\sum
\limits_{r=0}^{m}\dbinom{m}{r}-\dbinom{m}{0}}+\underbrace{\sum\limits_{r=1}
^{m-1}\dfrac{1}{r}\dbinom{m-1}{r}}_{\substack{=\sum\limits_{r=1}^{m-1}
\dfrac{2^{r}-1}{r}\\\text{(because we assumed }\\\text{that
\eqref{darij1.eq.1} holds for }n=m-1\text{)}}}+\underbrace{\dfrac{1}{m}0}
_{=0}\\
& =\dfrac{1}{m}\left( \underbrace{\sum\limits_{r=0}^{m}\dbinom{m}{r}
}_{\substack{=2^{m}\\\text{(by \eqref{darij1.eq.2n})}}}-\underbrace{\dbinom
{m}{0}}_{=1}\right) +\sum\limits_{r=1}^{m-1}\dfrac{2^{r}-1}{r}\\
& =\underbrace{\dfrac{1}{m}\left( 2^{m}-1\right) }_{=\dfrac{2^{m}-1}{m}
}+\sum\limits_{r=1}^{m-1}\dfrac{2^{r}-1}{r}=\dfrac{2^{m}-1}{m}+\sum
\limits_{r=1}^{m-1}\dfrac{2^{r}-1}{r}\\
& =\sum\limits_{r=1}^{m}\dfrac{2^{r}-1}{r}.
\end{align*}
In other words, \eqref{darij1.eq.1} holds for $n=m$. This completes the
induction step. Thus, \eqref{darij1.eq.1} is proved by induction.
Remark: Another known identity analogous to \eqref{darij1.eq.1} is
\begin{equation}
\sum_{r=1}^{n}\dfrac{\left(-1\right)^{r-1}}{r}\dbinom{n}{r} = \sum_{r=1}^{n}\dfrac{1}{r} .
\label{darij1.eq.3}
\tag{3}
\end{equation}
It is a nice exercise to unite \eqref{darij1.eq.1} and \eqref{darij1.eq.3} under a common generalization.
| {
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"url": "https://math.stackexchange.com/questions/1466769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ A question on submultiple angles states...
Prove that:$$\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$$
My efforts
I tried using the formula
$$a^3+b^3=(a+b)^3-3ab(a+b)$$
and
$$\cos^2{\frac{\theta}{2}=\frac{\cos\theta + 1}{2}}$$
Then I tried simplifying it:
$$\require{cancel} \begin{align} \sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}} & = (\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})^3 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}(\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}})\\
& = 1 - 3\sin^2{\frac{\theta}{2}}\cos^2{\frac{\theta}{2}}\\
& = 1 - 3(1 - \cos^2{\frac{\theta}{2}})\cos^2{\frac{\theta}{2}}\\
& = 1 - \frac{3}{2}(\cos\theta+1) + \frac{3}{4}(\cos\theta+1)^2\\
& = \frac{4\cancel{-6\cos\theta}-2+3\cos^2+3+\cancel{6\cos\theta}}{4}\\
& = \frac{1}{4}(5+3\cos^2\theta)\end{align} $$
I suspect I must have messed up with some sign somewhere. The trouble is, I can't seem to find where. Please help me in this regard.
Update: I am not accepting an answer because all the answers are equally good. It would be an injustice to the other answers.
Note to the editors: I also suspect that my post is a little short on appropriate tags. If required, please do the needful.
| You may begin factoring:
\begin{align*}
\sin^6\frac\theta2+\cos^6\frac\theta2 &=\Bigl(\sin^2\frac\theta2+\cos^2\frac\theta2\Bigr)\Bigl(\sin^4\frac\theta2-\sin^2\frac\theta2\cos^2\frac\theta2+\cos^4\frac\theta2\Bigr)\\
&=\Bigl(\cos^2\frac\theta2-\sin^2\frac\theta2\Bigr)^2+\sin^2\frac\theta2\cos^2\frac\theta2=\cos^2\theta+\frac14\sin^2\theta\\
&=\frac14(3\cos^2\theta+1).
\end{align*}
You also may linearise:
$$\cos^2\theta+\frac14\sin^2\theta=\frac12(1+\cos2\theta)+\frac18(1-\cos2\theta)=\frac18(3\cos2\theta+5).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Fourth complex roots of $i$ Solve $z^4 = i$. I cannot figure out why the angle of $i$ is $\frac{\pi}{2}$ and how to determine the values of $k$. If someone could show step-by-step that would be great!
Thanks.
| for $z=x+iy$ you can write its Polar Representation $z=re^{i\theta}=r(\cos\theta+i\sin\theta)=\color{red}{r\cos\theta}+i\color{blue}{r\sin\theta}$(respectively real part and imaginary part red and blue) where $r=\sqrt{x^2+y^2}$ and $\theta=tan^{-1}\frac{y}{x} $ for $x>0$ and $\theta=tan^{-1}\frac{y}{x}+\pi $ for $x<0$.
For $w = \rho e^{i \phi}$ that $w^n = z$, $\rho = r^{1/n}$, $\phi = \frac{\theta}{n} + \frac{2\pi k}{n}$ for $k=0, 1, ..,n-1$.
here for $z=i$, we have $r=\sqrt{0^2+1^2}=1$ and $\theta=tan^{-1}\frac{1}{0}=\frac{\pi}{2} $
thus answers will be $w=(1)^\frac{1}{4}e^{i(\frac{\frac{\pi}{2}}{4} + \frac{2\pi k}{4})}, k=0,1,2,3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1467905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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second order differential equation to hypergeometric equation I know how to transform a general second order differential equation of the form
$$\frac{d^2w}{dz^2} + \left(\frac{A}{z-\xi}+\frac{B}{z-\eta}\right)\frac{dw}{dz} + \frac{1}{z-\xi}\frac{1}{z-\eta}\left(\frac{D}{z-\xi}+\frac{E}{z-\eta}\right)w=0$$
into a hypergeometric equation. But now I have to solve
$$\frac{d^2w}{dz^2} + \left(\frac{A}{z-\xi}+\frac{B}{z-\eta}\right)\frac{dw}{dz} + \frac{1}{z-\xi}\frac{1}{z-\eta}\left(\frac{D}{z-\xi}+\frac{E}{z-\eta}+C\right)w=0$$
in terms of hypergeometric functions and I can't seem to find the change of variables adequate to obtain an equation like the first one. I'm sure I'm missing something very simple here.
| Let $w=(z-\xi)^a(z-\eta)^by$ ,
Then $\dfrac{dw}{dz}=(z-\xi)^a(z-\eta)^b\dfrac{dy}{dz}+(z-\xi)^a(z-\eta)^b\left(\dfrac{a}{z-\xi}+\dfrac{b}{z-\eta}\right)y$
$\dfrac{d^2w}{dz^2}=(z-\xi)^a(z-\eta)^b\dfrac{d^2y}{dz^2}+(z-\xi)^a(z-\eta)^b\left(\dfrac{a}{z-\xi}+\dfrac{b}{z-\eta}\right)\dfrac{dy}{dz}+(z-\xi)^a(z-\eta)^b\left(\dfrac{a}{z-\xi}+\dfrac{b}{z-\eta}\right)\dfrac{dy}{dz}+(z-\xi)^a(z-\eta)^b\left(\dfrac{a(a-1)}{(z-\xi)^2}+\dfrac{2ab}{(z-\xi)(z-\eta)}+\dfrac{b(b-1)}{(z-\eta)^2}\right)y=(z-\xi)^a(z-\eta)^b\dfrac{d^2y}{dz^2}+2(z-\xi)^a(z-\eta)^b\left(\dfrac{a}{z-\xi}+\dfrac{b}{z-\eta}\right)\dfrac{dy}{dz}+(z-\xi)^a(z-\eta)^b\left(\dfrac{a(a-1)}{(z-\xi)^2}+\dfrac{2ab}{(z-\xi)(z-\eta)}+\dfrac{b(b-1)}{(z-\eta)^2}\right)y$
$\therefore(z-\xi)^a(z-\eta)^b\dfrac{d^2y}{dz^2}+2(z-\xi)^a(z-\eta)^b\left(\dfrac{a}{z-\xi}+\dfrac{b}{z-\eta}\right)\dfrac{dy}{dz}+(z-\xi)^a(z-\eta)^b\left(\dfrac{a(a-1)}{(z-\xi)^2}+\dfrac{2ab}{(z-\xi)(z-\eta)}+\dfrac{b(b-1)}{(z-\eta)^2}\right)y+(z-\xi)^a(z-\eta)^b\left(\dfrac{A}{z-\xi}+\dfrac{B}{z-\eta}\right)\dfrac{dy}{dz}+(z-\xi)^a(z-\eta)^b\left(\dfrac{A}{z-\xi}+\dfrac{B}{z-\eta}\right)\left(\dfrac{a}{z-\xi}+\dfrac{b}{z-\eta}\right)y+\dfrac{1}{(z-\xi)(z-\eta)}\left(\dfrac{D}{z-\xi}+\dfrac{E}{z-\eta}+C\right)(z-\xi)^a(z-\eta)^by=0$
$\dfrac{d^2y}{dz^2}+\left(\dfrac{A+2a}{z-\xi}+\dfrac{B+2b}{z-\eta}\right)\dfrac{dy}{dz}+\left(\dfrac{a(a-1)}{(z-\xi)^2}+\dfrac{2ab}{(z-\xi)(z-\eta)}+\dfrac{b(b-1)}{(z-\eta)^2}\right)y+\left(\dfrac{Aa}{(z-\xi)^2}+\dfrac{Ba+Ab}{(z-\xi)(z-\eta)}+\dfrac{Bb}{(z-\eta)^2}\right)y+\left(\left(\dfrac{1}{(\xi-\eta)(z-\xi)}-\dfrac{1}{(\xi-\eta)(z-\eta)}\right)\left(\dfrac{D}{z-\xi}+\dfrac{E}{z-\eta}\right)+\dfrac{C}{(z-\xi)(z-\eta)}\right)y=0$
$\dfrac{d^2y}{dz^2}+\left(\dfrac{A+2a}{z-\xi}+\dfrac{B+2b}{z-\eta}\right)\dfrac{dy}{dz}+\left(\dfrac{a(a-1)+Aa}{(z-\xi)^2}+\dfrac{2ab+Ba+Ab}{(z-\xi)(z-\eta)}+\dfrac{b(b-1)+Bb}{(z-\eta)^2}\right)y+\left(\dfrac{D}{(\xi-\eta)(z-\xi)^2}+\dfrac{(\xi-\eta)C-D+E}{(\xi-\eta)(z-\xi)(z-\eta)}-\dfrac{E}{(\xi-\eta)(z-\eta)^2}\right)y=0$
$\dfrac{d^2y}{dz^2}+\left(\dfrac{A+2a}{z-\xi}+\dfrac{B+2b}{z-\eta}\right)\dfrac{dy}{dz}+\left(\dfrac{(\xi-\eta)a(a-1)+(\xi-\eta)Aa+D}{(\xi-\eta)(z-\xi)^2}+\dfrac{2(\xi-\eta)ab+(\xi-\eta)Ba+(\xi-\eta)Ab+(\xi-\eta)C-D+E}{(\xi-\eta)(z-\xi)(z-\eta)}+\dfrac{(\xi-\eta)b(b-1)+(\xi-\eta)Bb-E}{(\xi-\eta)(z-\eta)^2}\right)y=0$
Choose $(\xi-\eta)a(a-1)+(\xi-\eta)Aa+D=0$ and $(\xi-\eta)b(b-1)+(\xi-\eta)Bb-E=0$ , the terms of coefficient $\dfrac{1}{(z-\xi)^2}$ and $\dfrac{1}{(z-\eta)^2}$ can eliminate, and the ODE can reduces to Gaussian hypergeometric equation.
| {
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"url": "https://math.stackexchange.com/questions/1467981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Show $\sum \frac{n}{2^n + n} <\frac 32$ Let $n\in \mathbb N^{+}$,show that
$$\dfrac{1}{2+1}+\dfrac{2}{2^2+2}+\dfrac{3}{2^3+3}+\cdots+\dfrac{n}{2^n+n}<\dfrac{3}{2}$$
| Using that $\sum_{n=k}^\infty n 2^{-n} = (k+1)2^{1-k}$,
$$
\sum_{n=1}^\infty \frac{n}{2^n + n} < \frac{1}{3} + \frac{2}{6} + \frac{3}{11} + \frac{4}{20} + \frac{5}{37} + \sum_{n=6}^\infty \frac{n}{2^n}\\ = \frac{1}{3} + \frac{2}{6} + \frac{3}{11} + \frac{4}{20} + \frac{5}{37} + \frac{7}{32} = \frac{291727}{195360} < \frac{3}{2}.
$$
Also ugly, but also works.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $2x^3+4x^2+6x+8=2468$ and $x$ is a positive number, what is the value of $x^3+9x^2+8x+8$ If $2x^3+4x^2+6x+8=2468$ and $x$ is a positive number, what is the value of $x^3+9x^2+8x+8$
I'm trying to find $x$:
this is what i did so far:
$2x^3+4x^2+6x+8=2468$
Divide by $2$
$x^3+2x^2+3x+4=1234$
Minus $4$, $x^3+2x^2+3x=1230$
factor out an $x$
$x(x^2+2x+3)=1230$
and now, i'm stuck.
| If you notice that $2,4,6,8$ correspond to the coefficients of the polynomial $2x^{3} + 4x^{2} + 6x + 8$, then $2x^{3} + 4x^{2} + 6x + 8 = 2468$ is implied by $x =$ ?
Using this root and the factor theorem, we can then arrive at $(x-10)(x^{2} + 12x + 123) = 0$, which shows that $x=10$ is the only (real; for complex numbers cannot be ordered like the reals) solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Generating functions for finding the coefficients I am new to the field of combinatorics and I recently came across a problem where it was asked to find the number of integer solutions to ${c_1 + c_2 + c_3 + c_4=20 }$ where ${c_i\ge 0}$ for all ${1\le i\le4}$ with ${c_2}$ and ${c_3}$ being even integers.
Using the generating function, we know that the solution would be the coefficient of ${x^{20}}$ in the expansion of ${(1 +x+x^2+x^3+...)^2*(1+x^2+x^4+x^6+...)^2}$ which is equivalent to: ${(1-x)^{-2}*(1-x^2)^{-2}}$
Now, we know that the first part could be solved by computing:${(-1)^{20}\dbinom{-2}{20} = \dbinom{2+20-1}{20}=21}$
Since we know that: ${(1+x^m)^n = \dbinom{n}{0} + \dbinom{n}{1}x^m + \dbinom{n}{2}x^{2m}+...+\dbinom{n}{n}x^{nm}}$
Would it be correct to assume that the coefficient of ${x^{20}}$ in ${{(1-x^2)^{-2}}}$ would be: ${\dbinom{-2}{10}}$
| This is an answer with a slightly different approach. We use the the coefficient of operator $[x^t]$ to denote the coefficient $a_t$ of $x^t$ in a series $A(x)=\sum_{j=0}^{\infty}a_jx^j$.
We obtain
\begin{align*}
[x^t]&(1-x^2)^{-2}(1-x)^{-2}\\
&=[x^t]\sum_{k=0}^{\infty}\binom{-2}{k}(-x^2)^k\sum_{j=0}^{\infty}\binom{-2}{j}(-x)^j\tag{1}\\
&=[x^t]\sum_{k=0}^{\infty}\binom{k+1}{k}x^{2k}\sum_{j=0}^{\infty}\binom{j+1}{j}x^j\tag{2}\\
&=[x^t]\sum_{k=0}^{\infty}(k+1)x^{2k}\sum_{j=0}^{\infty}(j+1)x^j\\
&=\sum_{k=0}^{\left\lfloor t/2\right\rfloor}(k+1)[x^{t-2k}]\sum_{j=0}^{\infty}(j+1)x^j\tag{3}\\
&=\sum_{k=0}^{\left\lfloor t/2\right\rfloor}(k+1)(t-2k+1)\tag{4}\\
&=\sum_{k=1}^{\left\lfloor t/2\right\rfloor+1}\left((t+3)k-2k^2\right)\tag{5}\\
&=\frac{1}{6}\left(\left\lfloor t/2\right\rfloor+1\right)\left(\left\lfloor t/2\right\rfloor+2\right)
\left(3t-4\left\lfloor t/2\right\rfloor+3\right)\tag{6}
\end{align*}
Comment:
*
*In (1) we use the series expansion of the binomial series
*In (2) we use the identity $\binom{r}{s}=\binom{-r+s-1}{s}(-1)^s$
*In (3) we use the linearity of the coefficient of operator and the identity $[x^t]x^uA(x)=[x^{t-u}]A(x)$. We also set the upper index of the first sum to $\left\lfloor t/2\right\rfloor$, since the exponent of $x^{t-2k}$ has to be non-negative.
*In (4) we set $j=t-2k$ in order to select the proper coefficient of $x^{t-2k}$.
*In (5) we shift the index by one for simplification only and rearrange the summands according to increasing powers of $k$.
*In (6) we use the summation formula $\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ and $\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$.
Since we need the coefficient of $x^{20}$ we obtain from (6)
\begin{align*}
[x^{20}](1-x^2)^{-2}(1-x)^{-2}=\frac{1}{6}\cdot11\cdot12\cdot23=405
\end{align*}
Note, that we do not multiply the series representation of $(1-x^2)^{-2}$ and $(1-x)^{-2}$ using Cauchy multiplication. We walk instead with the coefficient of operator through the series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Conical Equation A quadric surface has the following equation: $2 x^2+3 y^2+3 z^2+J+16 x−18 y−6 z=0$
Enter a value of $J$ for which the quadric is :
a) A single point
b) The empty set
Working -
I honestly have no clue on how to answer either of the two questions. All that I do know is that I can complete the square to get $2(x+4)^2+3(y-3)^2+3(z-1)^2+J=0$ but do not know how to find a value of J such that the equation equals both a single point or the empty set. I do however also know that the value for J to make the equation a single point is a number however. Any suggestions are appreciated. Please edit my question for any clarification. Thank you
| $$2(x+4)^2+3(y-3)^2+3(z-1)^2+J=0$$
is not correct.
We have
$$\begin{align}&2x^2+3y^2+3z^2+J+16x-18y-6z=0\\&\iff 2x^2+16x+3y^2-18y+3z^2-6z+J=0\\&\iff 2x^2+16x+32+3y^2-18y+27+3z^2-6z+3=32+27+3-J\\&\iff 2(x+4)^2+3(y-3)^2+3(z-1)^2=62-J\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Numerical Series nature I need someone to approve if my way to prove that the following numerical series is divergent.
*
*Consider the Series $\sum_{n=1} \frac{1}{ln(n+n^2+n^3)}$. following the comparison test i found that the series diverge.
*My approach:
since $e^n +\frac{n^2}{2!} +\frac{5n^3}{3!} \ge n+n^2+n^3$ then $ \ln(e^n +\frac{n^2}{2!} +\frac{5n^3}{3!}) \ge ln(n+n^2+n^3) $
i.e $\frac{1}{ln(e^n +\frac{n^2}{2!} +\frac{5n^3}{3!})} \le \frac{1}{ln(n+n^2+n^3)}$ but since $e^n +\frac{n^2}{2!} +\frac{5n^3}{3!} \sim e^n$ as $n\rightarrow \infty$
hence$\frac{1}{ln(e^n +\frac{n^2}{2!} +\frac{5n^3}{3!})} \sim \frac{1}{n}$ where $\sum\frac{1}{n}$ is divergent according to Reimann and so we get that the sum $\sum_{n=1} \frac{1}{ln(n+n^2+n^3)}$ is also divergent.
Am i right doing so ?
| $\log(n+n^2+n^3)\leq\log(n^3+n^3+n^3)=\log(3n^3)\leq\log(n^4)=4\log(n)\leq 4n$ if $n\geq 3$. Hence
$$
\sum_{n=1}^{\infty}\frac{1}{\log(n+n^2+n^3)}\geq\frac{1}{4}\sum_{n=3}^{\infty}\frac{1}{n}
$$
which is divergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1473247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $z^5-32$ can be factorised into linear and quadratic factors over real coefficients as $(z^5-32)=(z-2)(z^2-pz+4)(z^2-qz+4)$,then find $p^2+2p.$ If the expression $z^5-32$ can be factorised into linear and quadratic factors over real coefficients as $(z^5-32)=(z-2)(z^2-pz+4)(z^2-qz+4)$,then find the value of $p^2+2p.$
I used $a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$ to get $z^5-2^5=(z-2)(z^4+2z^3+4z^2+8z+16)$
Then i compared $z^4+2z^3+4z^2+8z+16$ with $(z^2-pz+4)(z^2-qz+4)$
to get $p+q=-2$ and $p^2q^2=2$
$pq=\pm\sqrt2$
But when i solve the two equations $pq=\pm\sqrt2$ and $p+q=-2$,i get the value of $p$ and $q$ whch are non-manageable difficult to simplify.And i am not able to calculate the final desired result $p^2+2p$.
Is my approach not correct?What is the simple and elegant method to do this?Please help me.
| HINT:
$z^5=32=2^5,$
$z_r=2e^{2r\pi/5}$ where $r=0,\pm1,\pm2$
Now $(z-z_s)(z-z_{-s})=z^2-2z\cos\dfrac{2s\pi}5+4$
So, the values of $p,q$ will be reached by setting $s=1,2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How Evaluate This Integral $\int x^{3} \sin 2x dx$?
How can find the following indefinite integral?
$$\int x^{3} \sin 2x dx$$
Thanks For the Help!
| Hint: Integration by parts:
$\int_{x} x^{3}\sin 2x = \frac{-1}{2} \int_{x} x^{3} D\cos 2x = \frac{-1}{2}\cos (2x)x^{3} + \frac{3}{2} \int_{x} \cos (2x) x^{2} + \text{some constant}$; $\int_{x}\cos (2x)x^{2} = \frac{1}{2}\int_{x}x^{2}D\sin 2x = \frac{1}{2}\sin (2x)x^{2} - \frac{1}{2}\int_{x} \sin (2x)2x + \text{some constant}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1474628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I've formulated a proof, but also a counterexample? show that f(x) is injective.
$f(x)= \frac{x^2}{1+x^2}$
if $f(x)=f(y)$ then $\frac{x^2}{1+x^2}=\frac{y^2}{1+y^2}$
$(x^2)(1+y^2)=(y^2)(1+x^2)$
$x^2+x^2y^2=y^2+x^2y^2$
$x^2 = y^2$
$x = y$
but $f(1)=f(-1)$ due to the square roots. Where did I go wrong in the original proof?
| $x^2 = y^2$ does not imply $x = y$.
$x^2 = y^2 \Leftrightarrow (x = y \mbox{ or } x = -y)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1476061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $x^2 + x = 0 (\mod 10^5)$? I've tried to solve it in the following way:
First let's solve $x^2+x\equiv 0\ (\mod10)$. Solutions are $-1, 0, 4, 5$.
So $-1$ and $0$ are obviously solutions of $x^2 + x \equiv 0\ (\mod 10^5)$.
Now let's solve this for $4$ and $5$:
Let $x = 10k + 4$, then put this $x$ into $x^2 + x \equiv 0\ (\mod 100)$. So after a few operation we get $k=-2\ (\mod 10)$. Now let $l = 10k + 2$ and $x = 10\cdot (10k + 2) + 4$. Now we can solve our equation by modulo $1000$. Then we keep doing similar steps until modulo $10^5$ solution is found.
The problem is we need to operate really big numbers. And we need to do it twice (for both roots $4$ and $5$). Is there any less complicated solution?
| $10 = 2 \times 5$, so you should really do this mod $2^5$ and mod $5^5$ and use CRT.
Since $x^2 + x = x (x + 1)$ and $\gcd(x,x+1) = 1$, $x^2+x \equiv 0 \mod p^n$ (where $p$ is prime) iff either $x \equiv 0 \mod p^n$ or $x \equiv -1 \mod p^n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the derivative for a product of two polynomial functions? In my problem, I am attempting to find $f'(x)$ when $f(x)=(5x^2-2x+8)(4x^2+7x-3)$. For my work I have:
\begin{align}
& \frac{d}{dx} (uv) = u\frac {dv}{dx} + v\frac {du}{dx} \\[8pt]
= {} & (5x^2-2x+8)(8x+7)+(4x^2+7x-3)(10x-2) \\[8pt]
= {} & 40x^3+35x^2-16x^2-14x+64x+56+40x^3-16x^2+70x^2-14x-30x+6 \\[8pt]
= {} & 80x^3 + 73x^2+6x+62
\end{align}
But when I plugged my original equation [$f(x)=(5x^2-2x+8)(4x^2+7x-3)$] into an online derivative calculator to check my answer, it comes out as:
$$=80x^3+81x^2+6x+62\ldots\text{ ?}$$
Can anyone spot where I am going wrong (if I am)?
| HINT: after the product rule we get
$$f'(x)=(10x-2)(4x^2+7x-3)+(5x^2-2x+8)(8x+7)$$
after expanding we get
$$f'(x)=80 x^3+81 x^2+66 x+50$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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"answer_id": 3
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Probability of drawing numbers based on past drawing counts? Just a mathematical thought I had, and I couldnt find a relevant answer.
Lets assume that 3 numbers are drawn from the following list: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. The drawing is without replacement (if we draw 1, we have 2, 3, 4, 5, 6, 7, 8, 9, 10 left, and so on). Now, if we know that there were 10 previous drawings with the following statistics:
*
*2, 4, 7
*1, 4, 8
*3, 8, 10
*1, 5, 6
*2, 7, 10
*5, 7, 9
*6, 7, 10
*1, 4, 7
*4, 8, 9
*8, 9, 10
So now we know that the following numbers were drown x times as follows:
*
*1 = 3 times total
*2 = 2 times total
*3 = 1 times total
*4 = 4 times total
*5 = 2 times total
*6 = 2 times total
*7 = 5 times total
*8 = 4 times total
*9 = 3 times total
*10 = 4 times total
How can one calculate the probability of drawing, X, Y, Z on the 11th drawing?
Is this probability purely based on the count of previous number drawings? Or also on the number combinations of each individual drawings?
| You don't have nearly enough data to estimate dependencies.
The best you can do is estimate weighted probabilities. Such as:
$$\Pr(\{1, 7,10\}) \simeq \dfrac{3}{30}\dfrac{5}{27}\dfrac{4}{22}+\dfrac{3}{30}\dfrac{4}{27}\dfrac{5}{23}+\dfrac{4}{30}\dfrac{3}{26}\dfrac{5}{23}+\dfrac{4}{30}\dfrac{5}{26}\dfrac{3}{21}+\dfrac{5}{30}\dfrac{3}{25}\dfrac{4}{22}+\dfrac{5}{30}\dfrac{4}{25}\dfrac{3}{21}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Calculate expression: $\cos\alpha-\cos2\alpha$ I attempt calculate this expression:
$$\cos\dfrac{\pi}{5}-\cos\dfrac{2\pi}{5}$$
Please, help me, somebody. Thanks.
| Let $\alpha=\pi/5$ for simplicity. By De Moivre formulas,
$$
(\cos\alpha+i\sin\alpha)^5=\cos\pi+i\sin\pi=-1
$$
and, if we set $z=\cos\alpha+i\sin\alpha$, $z^5+1=0$ that means
$$
(z+1)(z^4-z^3+z^2-z+1)=0
$$
Since $z\ne-1$, the second factor vanishes. Dividing by $z^2$ we get
$$
z^2+\frac{1}{z^2}-z-\frac{1}{z}+1=0
$$
And, recalling that $(z+\frac{1}{z})^2=z^2+2+\frac{1}{z^2}$,
$$
\left(z+\frac{1}{z}\right)^{\!2}-\left(z+\frac{1}{z}\right)-1=0
$$
Now
$$
z+\frac{1}{z}=z+\bar{z}=2\cos\alpha
$$
so $2\cos\alpha$ is the positive root of the equation $t^2-t-1=0$ and thus
$$
2\cos\alpha=\frac{1+\sqrt{5}}{2}
$$
Now compute
\begin{align}
\cos\alpha-\cos2\alpha
&=\cos\alpha-2\cos^2\alpha+1\\[6px]
&=\frac{1+\sqrt{5}}{4}-2\frac{1+2\sqrt{5}+5}{16}+1\\[6px]
&=\frac{1+\sqrt{5}-3-\sqrt{5}+4}{4}\\[6px]
&=\frac{1}{2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Prove $∑^{(p−1)/2}_{k=1} \left \lfloor{\frac{2ak}{p}}\right \rfloor \equiv ∑^{(p−1)/2}_{k=1} \left \lfloor{\frac{ak}{p}}\right \rfloor ($mod $2)$ If $p$ is an odd prime number and $a$ is an odd integer not divisible by $p$, then why does
$∑^{(p−1)/2}_{k=1} \left \lfloor{\frac{2ak}{p}}\right \rfloor \equiv ∑^{(p−1)/2}_{k=1} \left \lfloor{\frac{ak}{p}}\right \rfloor ($mod $2)$ ?
| If for an odd prime $p$ and an arbitrary natural number $b$ not divisible by $p$ we write out:
$$\mu_p(b)=\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{2bk}{p}}\right\rfloor$$
Then using Eisenstein's lemma in his geometric proof of quadratic reciprocity we have:
$$b^{\frac{p-1}{2}}\equiv (-1)^{\mu_p(b)} \text{ mod } p$$
Which as a result gives us:
$$(-1)^{\mu_p(uv)}\equiv (uv)^{\frac{p-1}{2}}\equiv u^{\frac{p-1}{2}}v^{\frac{p-1}{2}}\equiv (-1)^{\mu_p(u)}(-1)^{\mu_p(v)}\equiv (-1)^{\mu_p(u)+\mu_p(v)} \text{ mod } p$$
So that we get:
$$\mu_p(uv)\equiv \mu_p(u)+\mu_p(v)\text{ mod } 2$$
Thus sending $u\to b$ and $v\to \frac{p+1}{2}$ gives us:
$$\mu_p\left(\frac{b(p+1)}{2}\right)\equiv \mu_p(b)+\mu_p\left(\frac{p+1}{2}\right)\text{ mod } 2\\\implies \mu_p(b)\equiv \mu_p\left(\frac{b(p+1)}{2}\right)-\mu_p\left(\frac{p+1}{2}\right)\text{ mod } 2$$
However by definition we have:
$$\mu_p\left(\frac{b(p+1)}{2}\right)-\mu_p\left(\frac{p+1}{2}\right)=\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{2b(\frac{p+1}{2})k}{p}}\right\rfloor-\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{2(\frac{p+1}{2})k}{p}}\right\rfloor\\=\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{b(p+1)k}{p}}\right\rfloor-\left\lfloor{\frac{(p+1)k}{p}}\right\rfloor=\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\left(1+\frac{1}{p}\right)bk}\right\rfloor-\left\lfloor{\left(1+\frac{1}{p}\right)k}\right\rfloor\\=\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{bk+\frac{bk}{p}}\right\rfloor-\left\lfloor{k+\frac{k}{p}}\right\rfloor=\sum_{k=1}^{\frac{p-1}{2}}bk+\left\lfloor{\frac{bk}{p}}\right\rfloor-k=\sum_{k=1}^{\frac{p-1}{2}}k(b-1)+\left\lfloor{\frac{bk}{p}}\right\rfloor\\=\sum_{k=1}^{\frac{p-1}{2}}k(b-1)+\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{bk}{p}}\right\rfloor=(b-1)\frac{(\frac{p-1}{2})(\frac{p-1}{2}+1)}{2}+\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{bk}{p}}\right\rfloor\\=(b-1)\frac{p^2-1}{8}+\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{bk}{p}}\right\rfloor$$
Which by our previous congruence modulo $2$ means:
$$\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{2bk}{p}}\right\rfloor=\mu_p(b)\equiv \mu_p\left(\frac{b(p+1)}{2}\right)-\mu_p\left(\frac{p+1}{2}\right)=(b-1)\frac{p^2-1}{8}+\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{bk}{p}}\right\rfloor\text{ mod } 2$$
So that we have:
$$\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{2bk}{p}}\right\rfloor\equiv (b-1)\frac{p^2-1}{8}+\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{bk}{p}}\right\rfloor\text{ mod } 2$$
Now if $b=a$ is an odd integer then $(b-1)\equiv (a-1)\equiv 0 \text{ mod } 2$ thus by the above equality:
$$\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{2ak}{p}}\right\rfloor\equiv\sum_{k=1}^{\frac{p-1}{2}}\left\lfloor{\frac{ak}{p}}\right\rfloor\text{ mod } 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1483358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I prove that the inequality $\frac { 2xy }{ x+y } \le \sqrt { xy } \le \frac { x+y }{ 2 } $ holds for every $x,y>0$ Prove that for every $x,y>0$, the following inequality holds. Begin by proving the right side first and assume that $(\sqrt { x } -\sqrt { y } )^2\ge 0$
$$\frac { 2xy }{ x+y } \le \sqrt { xy } \le \frac { x+y }{ 2 } $$
What I did thus far:
Assuming: $x,y>0$ and $(\sqrt { x } -\sqrt { y } )^2\ge 0$, I will prove that $\sqrt { xy } \le \frac { x+y }{ 2 } $
1) By expanding $(\sqrt { x } -\sqrt { y } )^2\ge 0$, we get $x-2\sqrt { xy } +y\ge 0$
2) Then, $x+y\ge 2\sqrt { xy }$
3) Then, $\frac { x+y }{ 2 } \ge \sqrt { xy } $
4) Therefore, $\sqrt { xy } \le \frac { x+y }{ 2 } $
Assuming: $x,y>0$ and $(\sqrt { x } -\sqrt { y } )^2\ge 0$, I will prove that $\frac { 2xy }{ x+y } \le \sqrt { xy } $
At this point, I get stuck. I don't know what do to prove the other side of the inequality. I also don't know if I am going about this proof the right way.
Hints are much more appreciated than the actual solution.
Edit: (added this after getting advice from answers below)
1) By expanding $(\sqrt { x } -\sqrt { y } )^2\ge 0$, we get $x-2\sqrt { xy } +y\ge 0$
2) Then, $x+y\ge 2\sqrt { xy } $
3) By taking the reciprocal of both sides, we get: $\frac { 1 }{ x+y } \le \frac { 1 }{ 2\sqrt { xy } } $
4) Then, $(\sqrt { xy } )\frac { 1 }{ x+y } \le (\sqrt { xy } )\frac { 1 }{ 2\sqrt { xy } } \Rightarrow \frac { \sqrt { xy } }{ x+y } \le \frac { \sqrt { xy } }{ 2\sqrt { xy } } $
5) Then, $(2\sqrt { xy } )\frac { \sqrt { xy } }{ x+y } \le (2\sqrt { xy } )\frac { \sqrt { xy } }{ 2\sqrt { xy } } $
6) Therefore, $\frac { 2xy }{ x+y } \le \sqrt { xy } $
QED
| \begin{align*}
\frac { 2xy }{ x+y } \le \sqrt { xy } \le \frac { x+y }{ 2 }\\
\text{assuming $x,y>0, \space$ we square all terms }\\
\frac{4x^2y^2}{(x+y)^2 } \le xy \le \frac{(x+y)^2}{4}
\\
\end{align*}
\begin{align*}
\text{we eliminate denominators }
\\
(4x^2y^2(4) &=16x^2y^2 \\
\le xy(4)(x+y)^2 &=4 x^3 y + 8 x^2 y^2 + 4 x y^3\\
\le
(x+y)^2(x+y)^2&=x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4
\\
\\
\text{subtract }\space 4x^2y^2+4x^3y+4xy^3\space& \text{ from all sitermses}
\\
\space -4 x y (x^2 - 3 x y + y^2)
&\le (2xy)^2 \le (x^2+y^2)^2
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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General form of $\sqrt{a - b} - \sqrt{a + b}$?
General form of $\sqrt{a - b} - \sqrt{a + b}$?
What I would do is: let
$x = \sqrt{a - b} - \sqrt{a + b}$
$x^2 = 2a - 2\sqrt{a^2 - b^2}$
Then since $a + b > a - b$
$x = -\sqrt{2a - 2\sqrt{a^2 - b^2}}$
This is particularly useful in solving problem with nested square roots (like, let $a = 11, b = 2\sqrt{10}$
I am wondering, is there any other derivation and MORE WANTED: any simpler formula?
| "Simpler" is dependent on the use made of the formula. For approximate calculations, the form
$$\frac{-2b}{\sqrt{a - b} \; + \; \sqrt{a + b}}$$
(obtained by multiplying numerator and denominator by $\sqrt{a - b} \; + \; \sqrt{a + b}\,)$ is better when $|b|$ is very small in comparison to $|a|.$
| {
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"url": "https://math.stackexchange.com/questions/1494304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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common denominator What is the common denominator of the following:
$x^2-x$ and $2x$, $x^2$
(the first is the one side, and the other 2 are another side of the formula)
The full formula is:
$1/(x^2-x) = 1/2x + 1/x^2$
P.G
That's not homework. I am a mother of 3 trying to enjoy math again
| Your equation is
$$\frac{1}{x^2-x} = \frac{1}{2x} + \frac{1}{x^2}.$$
To find a common denominator, the first thing to look for is if any of the current denominators can be factored. The first one can, as $x^2-x = x(x-1)$, so your equation becomes
$$\frac{1}{x(x-1)} = \frac{1}{2x} + \frac{1}{x^2}.$$
You then look for what each denominator is missing compared to the others. When I compare the first denominator $x(x-1)$ to the second $2x$, then the first one already has the $x$, but it is missing the factor of $2$, and when I compare the first denominator to the third $x^2$, I see that I am missing one factor of $x$ in the first to match the $x^2$ in the third. In total, I multiply the numerator and denominator of the first fraction by $2x$ to get
$$\frac{2x}{2x^2(x-1)} = \frac{1}{2x} + \frac{1}{x^2}.$$
The second denominator $2x$ needs to change to become equal to the new first denominator, because this is our guess of a common denominator - but to change $2x$ to $2x^2(x-1)$, I need only multiply by $x(x-1)$, because this is what is missing. Similarly, to change the third denominator $x^2$ to $2x^2(x-1)$, I multiply by $2(x-1)$. In total, you equation becomes
$$\frac{2x}{2x^2(x-1)} = \frac{x(x-1)}{2x^2(x-1)} + \frac{2(x-1)}{2x^2(x-1)},$$
and your common denominator is $2x^2(x-1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove: ${n\choose 1}-3{n\choose 3}+9{n\choose 5}-...=\frac{-1}{\sqrt{3}}(-2)^n\sin\frac{2n\pi}{3}$ Prove: ${n\choose 1}-3{n\choose 3}+9{n\choose 5}-...=\frac{-1}{\sqrt{3}}(-2)^n\sin\frac{2n\pi}{3}$
How to use binomial theorem on a left sum?
| We could also start with $\sin\left(\frac{2\pi n}{3}\right)$ and use de Moivre's formula
in order to derive the binomial expression on the LHS.
The following is valid
\begin{align*}
\sum_{k}\binom{n}{2k+1}(-1)^k3^k=-\frac{1}{\sqrt{3}}(-2)^n\sin\left(\frac{2\pi n}{3}\right)\qquad n>0\\
\end{align*}
$$ $$
We obtain for $n>0$
\begin{align*}
\sin\left(\frac{2\pi n}{3}\right)&=\Im\left(\cos\left(\frac{2\pi n}{3}\right)+i\sin\left(\frac{2\pi n}{3}\right)\right)\tag{1}\\
&=\Im\left(\cos\left(\frac{2\pi}{3}\right)+i\sin\left(\frac{2\pi}{3}\right)\right)^n\\
&=\Im\left(\sum_{k=0}^{n}\binom{n}{k}i^k\sin^k\left(\frac{2\pi}{3}\right)\cos^{n-k}\left(\frac{2\pi}{3}\right)\right)\\
&=\sum_{k}\binom{n}{2k+1}(-1)^k\sin^{2k+1}\left(\frac{2\pi}{3}\right)\cos^{n-(2k+1)}\left(\frac{2\pi}{3}\right)\tag{2}\\
&=\sum_{k}\binom{n}{2k+1}(-1)^k\left(\frac{\sqrt{3}}{2}\right)^{2k+1}\left(-\frac{1}{2}\right)^{n-(2k+1)}\\
&=-\sqrt{3}\left(-\frac{1}{2}\right)^n\sum_{k}\binom{n}{2k+1}(-1)^k3^k\\
\end{align*}
and the claim follows.
Comment:
*
*In (1) we apply de Moivre's formula.
*In (2) we take odd index values $2k+1$ since we need the imaginary part only.
| {
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"url": "https://math.stackexchange.com/questions/1496756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find last digit of $7^{7^{7^7}}$ Find last digit of $7^{7^{7^7}}$
I know that the last digit of $7^x$ depends on the remainder $x$ leaves when divided by $4$: ($x = 7^{7^7}$)
$$7^{4k} \equiv 1 \bmod 10$$
$$7^{4k+1} \equiv 7 \bmod 10$$
$$7^{4k+2} \equiv 9 \bmod 10$$
$$7^{4k+3} \equiv 3 \bmod 10$$
And also that
$$7 \equiv -1 \bmod 4$$
So for all odd positive numbers, $7^{n} \equiv -1 \bmod 4$ and $7^{7}$ is odd. But which congruence relation should I use?
$4k+3$ and $4k+1$ are both odd
Please explain, thanks.
| You're getting yourself confused. $7^7$ is odd, correct; but $7^7$ isn't the exponent you care about. To determine what form to use, the exponent you're looking at is $7^{7^7}$. By your observations, $7^7$ is odd, so $7^{7^7} \equiv -1 \mod 4$. $4n+1$ and $4n+3$ are both odd, yes, but only one of them is $-1\mod 4$: $4n+3$. So $7^{7^7} \equiv 3\mod 4$, and so $7^{7^{7^7}} \equiv 3\mod 10$ according to your table.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Questions concerning $ \det ({}^A_{C\,}{}^B_D) = \det ({}^D_{B\,}{}^C_A)$ Let $A, B, C, D $ be $n \times n $ matrices. Using Schur complements I have found that
$$ \begin{pmatrix} A & B \\ C & D \end{pmatrix} =
\begin{pmatrix} A & 0 \\ 0 & I \end{pmatrix}
\begin{pmatrix} I & 0 \\ C & I \end{pmatrix}
\begin{pmatrix} I & A^{-1}B \\ 0 & D-CA^{-1}B \end{pmatrix} $$
and
$$ \begin{pmatrix} D & C \\ B & A \end{pmatrix} =
\begin{pmatrix} I & CA^{-1} \\ 0 & I \end{pmatrix}
\begin{pmatrix} D-CA^{-1}B & 0 \\ 0 & A \end{pmatrix}
\begin{pmatrix} I & 0 \\ A^{-1}B & I \end{pmatrix}, $$
from which the determinant equality follows, as long as $ A^{-1} $ exists. However, how do I tackle this when $A$ is singular? Here, I could just switch decompositions, but then I will get the same problem when $D$ is singular. Maybe one could derive two more decompositions using Schur complements, involving $B^{-1}$ and $C^{-1}$, respectively, and then one could say that the equality holds if at least one of the submatrices is nonsingular? Then, if all four submatrices are singular, the determinant must be zero -- from which equality follows trivially.
Also, does the equality hold when $A, B, C, D$ are not necessarily square but of matching sizes? Here, it doesn't seem like the decompositions will be valid, as $A$ or $D$ aren't necessarily square matrices (although then $B$ and $C$ must be).
| Wouldn't it be much easier to consider row/column swaps? By performing $n$ row swaps, you transform
$$\begin{pmatrix} A & B\\C & D \end{pmatrix} \to \begin{pmatrix} C&D\\A&B\end{pmatrix},$$
then $n$ column swaps transforms
$$\begin{pmatrix} C&D\\A&B\end{pmatrix} \to \begin{pmatrix} D&C\\B&A \end{pmatrix}.$$
Each row/column swap multiplies the determinant by $-1$, so this whole process multiplies the determinant by $(-1)^{2n} = 1$.
| {
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"url": "https://math.stackexchange.com/questions/1500366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$ Prove by induction: $2!\cdot 4!\cdot 6!\cdot\cdot\cdot (2n)!\ge ((n+1)!)^n$
For $n=1$ inequality holds.
$(*)$For $n=k$
$2!\cdot\cdot\cdot (2k)!\ge ((k+1)!)^k$
Multiplying LHS and RHS with $(2k+2)!$ gives
$$2!\cdot\cdot\cdot (2k)!(2k+2)!\ge ((k+1)!)^k(2k+2)!$$
Assume (by contradiction)$$2!\cdot\cdot\cdot (2k)!(2k+2)!< ((k+1)!)^k(2k+2)!$$
$$2!\cdot\cdot\cdot (2k)!(2k+2)!-((k+1)!)^k(2k+2)!<0$$
$$(2k+2)!(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)<0$$
$(2!\cdot\cdot\cdot (2k)!-((k+1)!)^k)\ge 0$ by $(*)$, thus inequality holds $\forall n\in\mathbb{N}$
Is this proof correct?
| First, show that this is true for $n=1$:
$\prod\limits_{k=1}^{1}(2k)!\geq(1+1)!^{1}$
Second, assume that this is true for $n$:
$\prod\limits_{k=1}^{n}(2k)!\geq(n+1)!^{n}$
Third, prove that this is true for $n+1$:
$\prod\limits_{k=1}^{n+1}(2k)!=$
$\color\red{\prod\limits_{k=1}^{n}(2k)!}\cdot(2(n+1))!\geq$
$\color\red{(n+1)!^{n}}\cdot(2(n+1))!=$
$(n+1)!^{n}\cdot(2n+2)!=$
$(n+1)!^{n}\cdot(n+1)!\cdot\dfrac{(2n+2)!}{(n+1)!}=$
$(n+1)!^{n+1}\cdot\dfrac{(2n+2)!}{(n+1)!}=$
$(n+1)!^{n+1}\cdot\underbrace{(n+2)\cdot(n+3)\cdot\ldots\cdot(2n+2)}_{n+1\text{ times}}\geq$
$(n+1)!^{n+1}\cdot\underbrace{(n+2)\cdot(n+2)\cdot\ldots\cdot(n+2)}_{n+1\text{ times}}=$
$(n+1)!^{n+1}\cdot(n+2)^{n+1}=$
$(n+2)!^{n+1}$
Please note that the assumption is used only in the part marked red.
| {
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"timestamp": "2023-03-29T00:00:00",
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Question referring to the exponential function defined as a limit of a sequence Suppose that $(x_n)_{n \in \mathbb{N}}$ is a sequence with $\lim_{n \to \infty} x_n = x \in \mathbb{R}$.
I want to show that
$$
\lim_{n \to \infty} \left( 1 + \frac{x_n}{n} \right)^n = \text{e}^x.
$$
My attempt: We have that
$$
\text{e}^x = \lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^m = \lim_{m \to \infty} \left( 1 + \frac{\lim_{n \to \infty }x_n}{m} \right)^m = \lim_{m \to \infty}\lim_{n \to \infty } \left( 1 + \frac{x_n}{m} \right)^m.
$$
Now I am not sure if (or why) it is allowed to replace the limits $m,n \to \infty$ by one limit $n \to \infty$.
| Since $\lim\limits_{n\to \infty} (1 + \frac{x}{n})^n = e^x$, the limit in question can be proven by showing
$$\lim_{n\to \infty} \left(\dfrac{1 + \dfrac{x_n}{n}}{1 + \dfrac{x}{n}}\right)^n = 1.\tag{1}$$
Let $a_n = x_n - x$. Then
$$\frac{1 + \dfrac{x_n}{n}}{1 + \dfrac{x}{n}} = \dfrac{1 + \dfrac{x}{n} + \dfrac{a_n}{n}}{1 + \dfrac{x}{n}} = 1 + \frac{\dfrac{a_n}{n}}{1 + \dfrac{x}{n}} = 1 + \frac{a_n}{n + x}.$$
So $(1)$ is equivalent to
$$\lim_{n\to \infty} \left(1 + \frac{a_n}{n + x}\right)^n = 1.\tag{2}$$
Let $n > 2|x|$. By the binomial theorem and triangle inequality,
$$\left|\left(1 + \frac{a_n}{n+x}\right)^n - 1\right| \le \sum_{k = 1}^n \binom{n}{k}\frac{|a_n|^k}{(n - |x|)^k}.$$
Since $n > 2|x|$, $n - |x| > \frac{n}{2}$. Furthermore, $\binom{n}{k} < \frac{n^k}{2^{k-1}}$ for $1\le k \le n$ since $n(n-1)\cdots (n-k+1) < n^k$ and $k! \ge 2^{k-1}$ for all such $k$. Therefore
$$\sum_{k = 1}^n \binom{n}{k}\frac{|a_n|^k}{(n - |x|)^k} <\sum_{k = 1}^n \frac{n^k}{2^{k-1}}\cdot \frac{|a_n|^k}{(n/2)^k} = 2\sum_{k=1}^n |a_n|^k.$$
Let $\epsilon > 0$, and choose a positive number $r < \min\{\frac{1}{2}, \frac{\epsilon}{4}\}$. Since $|a_n| \to 0$, there exists a positive integer $N$ such that $|a_n| < r$ for all $n > N$. So for $n > \max\{2|x|, N\}$,
$$\left|\left(1 + \frac{a_n}{n+x}\right)^n - 1\right| \le 2\sum_{k = 1}^n |a_n|^k = 2\sum_{k = 1}^{N} |a_n|^k + 2\sum_{k = N+1}^n r^k \le 2\sum_{k = 1}^N |a_n|^k + \frac{2r^{N+1}}{1 - r}.\tag{3}$$
Since $r < \frac{1}{2}$, $r < 1 - r$ and $r^N < \frac{r}{2^{N-1}}$, so $$\frac{2r^{N+1}}{1-r} < 2r^N < \frac{r}{2^{N-2}} < \frac{\epsilon}{2^N} \le \frac{\epsilon}{2}.$$ Since $2\sum_{k = 1}^N |a_n|^k \to 0$, there exists a positive integer $N'$ for which $2\sum_{k = 1}^N |a_n|^k < \frac{\epsilon}{2}$ for all $n > N'$. Hence, for $n > \max\{2|x|, N, N'\}$, the right-most expression of $(3)$ is less than $\epsilon$. Since $\epsilon$ was arbitrary, $(2)$ holds.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: ${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}$
Prove: $${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...+(-1)^n\frac{1}{2n+1}{n\choose n}=\frac{n!2^n}{(2n+1)!!}.$$
Here, $(2n+1)!!$ is an "odd factorial": $(2n+1)!! = 1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n+1$).
How to prove this equation?
Is it possible to use induction?
$${n\choose 0}-\frac{1}{3}{n\choose 1}+\frac{1}{5}{n\choose 2}-...(-1)^n\frac{1}{2n+1}{n\choose n}=\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1};$$
$$(2n+1)!!=\frac{(2n)!(2n+1)}{2^nn!}\Rightarrow \frac{n!2^n}{(2n+1)!!}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)};$$
$$\sum\limits_{k=0}^{n}{n\choose k}(-1)^k\frac{1}{2k+1}=\frac{2^{2n}(n!)^2}{(2n)!(2n+1)}$$
What now?
| We can rewrite the sum as $$\sum_{k=0}^n \binom{n}{k}(-1)^k\int_0^1x^{2k}dx = \int_0^1(1-x^2)^ndx.$$
Denote the last integral as $I(n)$. We integrate by parts:
$$I(n) = x(1-x^2)\big|_{x=0}^{x=1} - \int_0^1 x\cdot (-2nx)\cdot (1-x^2)^{n-1}dx = 2n \int_0^1 x^2(1-x^2)^{n-1}dx = 2n (I(n-1)-I(n)),$$
therefore
$$I(n) = \frac{2n}{2n+1}I(n-1).$$
Given that $I(1) = \frac{2}{3}$, a reasoning by recurrence allows us to conclude.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to have idea to prove trigonometric identities Hi to explain this better I'll take an example.
I have this identity that's giving me a hard time.
$$\frac{\cos^2(a)-\sin^2(b)}{\sin^2(a)\sin^2(b)} = \cot^2(a)\cot^2(b)-1$$
This is what i would do
$$\cos^2(a)/(\sin^2(a)·\sin^2(b))-\sin^2(b)/(\sin^2(a)·\sin^2(b)) \\
\cot^2(a)·1/\sin^2(b)-1/\sin^2(a)$$
then, we know that
$$1=\cos^2(b)+\sin^2(b) \\
\vdots \\
\cot^2(a)·\cot^2(b)+\cot^2(a)-1/\sin^2(a)$$
Which of course is wrong but just wanted to show you guys how my mind thinks.
Is there any right way of solving this or do I just have to keep trying.
THANKS.
| $$\frac{\cos^2(a)-\sin^2(b)}{\sin^2(a)\sin^2(b)} = \cot^2(a)\cot^2(b)-1\Longleftrightarrow$$
$$\cos^2(a)-\sin^2(b) = \sin^2(a)\sin^2(b)\left(\cot^2(a)\cot^2(b)-1\right)\Longleftrightarrow$$
$$\cos^2(a)-\sin^2(b) = \sin^2(a)\sin^2(b)\left(\left(\frac{\cos(a)}{\sin(a)}\right)^2\cdot\left(\frac{\cos(b)}{\sin(b)}\right)^2-1\right)\Longleftrightarrow$$
$$\cos^2(a)-\sin^2(b) = \sin^2(a)\sin^2(b)\left(\frac{\cos^2(a)\cos^2(b)}{\sin^2(a)\sin^2(b)}-1\right)\Longleftrightarrow$$
$$\cos^2(a)-\sin^2(b) = \cos^2(a)\cos^2(b)-\sin^2(a)\sin^2(b)\Longleftrightarrow$$
$$\cos^2(a)-1-\cos^2(b) = \cos^2(a)\cos^2(b)-\sin^2(a)\sin^2(b)\Longleftrightarrow$$
$$\cos^2(a)+\cos^2(b)-1 = \cos^2(a)\cos^2(b)-\sin^2(a)\sin^2(b)\Longleftrightarrow$$
$$-1+\cos^2(a)+\cos^2(b)=-1+\cos^2(a)+\cos^2(b)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to reconcile the fact that the antiderivative of $\sin(x)\cos(x)$ has two possible answers? So here is the problem.
$$
\int\sin(x)\cos(x)dx = \frac{1}{2}\sin^2(x) + c_1 \\
= -\frac{1}{2}\cos^2(x) + c_2
$$
This fact doesn't make much sense to me. How do you reconcile the fact that there are two possibilities (notice that these answers aren't only different by a constant)?
What are some other functions that have more than one antiderivatives (not only different by a constant)?
| Since $\ \sin^2(x)+\cos^2(x)=1$, if we let $c_1=c_2-\frac 12$ so we'd have
$$\begin{align}
\frac{1}{2}\sin^2(x) + c_1 &= \frac{1}{2}\sin^2(x) + (c_2-\frac 12) \\
&= \frac{1}{2}\sin^2(x) + (c_2-\frac 12) \\
&= \frac{1}{2}(1-\cos^2(x)) + (c_2-\frac 12) \\
&= c_2-\frac{1}{2}\cos^2(x)
\end{align}$$
| {
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Why is this sum what it is? I am trying to prove that
$$\frac{1}{1} + \frac{1}{1+2} + \frac{1}{1+2+3} + \cdots + \frac{1}{1+2+3+\cdots+n} = \frac{2n}{n+1}$$
I know the denominators have a closed form $\sum_{i=1}^{k} i = \frac{k(k+1)}{2}$ implying
$$\frac{1}{1} + \frac{1}{1+2} + \frac{1}{1+2+3} + \cdots + \frac{1}{1+2+3+\cdots+n} = 2\sum_{k=1}^{n}\frac{1}{k(k+1)}$$
Can't figure out where to go from here.
| To continue sheol's answer.
A telescope sum cancel out.
$2\sum_{k=1}^{n}\frac{1}{k(k+1)} = 2( \sum_{k=1}^{n}(\frac{1}{k} - \frac{1}{k+1}))$
$= 2(\sum_{k=1}^{n}\frac{1}{k} - \sum_{k=1}^{n}\frac{1}{k + 1})$
$= 2(\sum_{k=0}^{n-1}\frac{1}{k+1} - \sum_{k=1}^{n}\frac{1}{k + 1})$
$= 2(\frac 1{0 + 1} + \sum_{k=1}^{n-1}\frac{1}{k + 1} - \sum_{k=1}^{n-1}\frac{1}{k + 1} + \frac 1{n+1})$
| {
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How to Find the value of a trigonometric function if other very complicated trigonometric equation is given?
Q)If $\operatorname{sin}\alpha+\operatorname{cos}\alpha=\frac{\sqrt7}2$, $0 \lt \alpha \lt\frac{\pi}{6}, then \operatorname{tan}\frac{\alpha}2 $ is:(1)$\sqrt{7}-2$(2)$(\sqrt{7}-2)/3$(3)$-\sqrt{7}+2$(4)$(-\sqrt{7}+2)/3$
i did this question by two ways but both the was were getting so complex however on solving them completely which took $4.25$ pages the answers were ridiculously different.
$I^{st}$ Method:
squaring both sides $$(\operatorname{sin}\alpha+\operatorname{cos}\alpha)^2=\frac{7}4=1+ \operatorname{sin}2\alpha=\frac 74\implies\operatorname{sin}2\alpha=\frac 34\implies\operatorname{cos}2\alpha=\frac{\sqrt{7}}{4}$$
$$\text{hence, }\operatorname{tan}2\alpha=\frac{3}{\sqrt{7}}\implies 3tan^2\alpha+2\sqrt{7}tan \alpha -3=0\implies tan\alpha=\frac{\surd7 \pm 4}3 $$
no from this if i want to calculate the value of $\tan\alpha/2 $ then you can think how different the answer could be!
$II^{nd} $ Method
$$1+2\sin\alpha\cos\alpha=\frac74\implies \sin^2\alpha\cos^2\alpha=\frac{9}{64}$$
$$64\sin^4\alpha-64\sin^2\alpha+9=0\implies \sin^2\alpha=\frac{64\pm 64\sqrt{7}}{2\times 64}$$
even from this you can have the idea how different the answer will be so different because the value of $$\tan\alpha=\sqrt{\frac{64 \pm16\surd7}{64 \mp 16\surd7}}$$.
| Good news: they are the same!
$$\sqrt{\frac{64 \pm16\surd7}{64 \mp 16\surd7}}$$
$$=\sqrt{\frac{64 \pm16\surd7}{64 \mp 16\surd7}}\sqrt{\frac{64 \pm16\surd7}{64 \pm 16\surd7}}$$
$$=\frac{64 \pm16\surd7}{16\sqrt{9}}$$
$$=\frac{4 \pm\surd7}{3}$$
| {
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Is the argument used by the Numberphile video to show that $1 + 2 + 3 + 4 + \dots = -1/12$ valid? At the end of the Wikipedia article on:
$$1+2+3+4 +\dots$$
an argument is present that the sum adds up to $-\frac{1}{12}$.
Here is the numberphile video.
Here's my attempt to fill in the details from the argument:
(1) Let $S_1 = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + \dots$
(2) Let $S_2 = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + \dots$
(3) $2S_2 = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + \dots$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\, 1 - 2 + 3 - 4 + 5 - 6 + 7 + \dots$
(4) $\,\,\,\,\,\,\,\,\,\, = 1 -1 + 1 -1 + 1 - 1 + 1 - 1 + \dots$
(5) $S_1$ is the Grandi's series whose Cesaro's sum is $1/2$
(6) So, $S_2 = 1/4$
(7) Let $S = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + \dots$
(8) $S - S_2 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + \dots$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-1 + 2 - 3 + 4 - 5 + 6 -7 + 8 + \dots$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= 0 + 4 + 0 + 8 + 0 + 12 + 0 + 16 + \dots$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= 4 + 8 + 12 + 16 + 20 + 24+ 28 + \dots$
(9) $4S = 4 + 8 + 12 + 16 + 20 + 24 + 28 + \dots$
(10) $S - S_2 = S - 1/4 = 4S$
(11) Solving for $S - 1/4 = 4S$ gives $S = -\frac{1}{12}$
Is this argument valid? Is there a flaw in the logic that leads to the right answer in the wrong way?
| The argument is invalid for one simple reason:
The operator "$+$" is not defined for infinite sums.
Let me elaborate:
The symbol $+$ is a symbol that means something to us. It is a symbol that means "take whatever whole/rational/real/complex number is to the left of me and add it to whatever number is to the right of me."
In mathematical terms, $+$ is a mapping from $X\times X$ to $X$, i.e., if $X=\mathbb R$, then $+$ is a mapping that maps the pair $(x,y)$ into $x+y$. This would be more clear if we would write $$+(5,10) = 15,$$ but for historic reasons, we write $$5+10=15$$
Now, $+$ has some properties. For example, we know that $a+(b+c) = (a+b)+c$ for any three numbers $a,b,c$. This means that for any finite series of numbers $a_1,a_2,\dots a_n$, we know that no matter what the order in which we add the numbers, their sum is the same. That's why we are allowed to simply write
$$a_1+a_2+\dots + a_n$$
to denote the sum of the numbers.
However there is no trivial way in which we can define the sum of an infinte number of numbers!!!
The standard definition of a sum of real numbers is:
$$a_1 + a_2 + a_3 + \cdots = \lim_{n\to \infty} (a_1 + a_2 + \cdots + a_n),$$
and using this definition, the sum of all ones is $\infty.$
| {
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If $\int_{0}^{\frac{\pi}{4}}\tan^6(x) \sec(x) dx = I$ then express $\int_{0}^{\frac{\pi}{4}} \tan^8(x) \sec(x) dx$ in terms of $I$ how can I proceed with this exercise?
If
$$\int_{0}^{\frac{\pi}{4}} \tan^6(x) \sec(x) dx = I$$
then express
$$\int_{0}^{\frac{\pi}{4}} \tan^8(x) \sec(x) dx$$
in terms of $I$.
What I've got so far:
$$\int_{0}^{\frac{\pi}{4}} \tan^8(x) \sec(x) dx = \int_{0}^{\frac{\pi}{4}} \tan^2(x) \tan^6(x) \sec(x) dx = \int_{0}^{\frac{\pi}{4}} \left( \sec^2(x) - 1 \right) \tan^6(x) \sec(x) dx = \\ = - \int_{0}^{\frac{\pi}{4}} \tan^6(x) \sec(x) dx + \int_{0}^{\frac{\pi}{4}} \tan^6(x) \sec^3(x) dx = -I + \cdots$$
Any help is highly appreciated.
$\\$
(Exercise 50 from Stewart's Calculus book section chapter 7.2 7th edition)
| Short answer: intergration by parts.
Long answer: denote the integral you want to find as $J$. Then you have
$J = \int_0^{\frac{\pi}{4}}\tan^8(x)\sec(x)dx = \int_0^{\frac{\pi}{4}}\tan^7(x)d(\sec(x)) = \left.\tan^7(x)\sec(x)\right|_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}}\sec(x)d(\tan^7(x)) = \sqrt{2} - 7\int_0^{\frac{\pi}{4}}\tan^6(x)\sec^3(x)dx = \sqrt{2}-7\int_0^{\frac{\pi}{4}}\tan^6(x)\sec(x)(1+\tan^2(x))dx = \sqrt{2} - 7I - 7J = J$
$\sqrt{2} - 7I - 7J = J$
Hope all the steps are clear.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maximum Value satisfying a variation of the Triangle Inequality The Triangle Inequality states that:
$$||\vec{x}+\vec{y}||\le ||\vec{x}||+||\vec{y}||$$
Now, suppose that we have a triangle with side lengths of $a,b,c$ such that $a+b+c=2$. Avoiding any assumptions of what kind of triangle we have, what the angle measures are, etc.,
Now suppose we have the following inequality:
$$k\le\dfrac{1-a}{b}+\dfrac{1-b}{c}+\dfrac{1-c}{a}$$
My question is that what can $k$ be. Meaning what is the maximum possible $k$ which satisfies the inequality I gave. I just don't want an answer, but a proof as to what the maximum value of $k$ will be.
| The best bound is given by:
$$\frac{1 - a}{b} + \frac{1 - b}{c} + \frac{1 - c}{a} > 1$$
Proof:
Since $a$, $b$, and $c$ are the sides of a triangle, there exists positive real numbers such that $a = y + z$, $b = x + z$, and $c = x +y$. Therefore, $2x + 2y + 2z = a + b + c = 2\implies x + y + z = 1$, and
$$ \begin{align*} \frac{1 - a}{b} + \frac{1 - b}{c} + \frac{1 - c}{a} &= \frac{1 - y - z}{x + z} + \frac{1 - x - z}{x + y} + \frac{1 - x - y}{y + z} \\ &= \frac{x}{z + x} + \frac{y}{x + y} + \frac{z}{y + z}. \end{align*} $$
Note that this expression is homogeneous in $x$, $y$, and $z$, so we can discard the condition $x + y + z = 1$.
We see that
$$\frac{x}{z + x} + \frac{y}{x + y} + \frac{z}{y + z} > \frac{x}{x + y + z} + \frac{y}{x + y + z} + \frac{z}{x + y + z} = 1$$
Now, let $x = 1$, $y = \epsilon$, and $z = \epsilon^2$, where $\epsilon$ is some positive real number. As $\epsilon \to 0^+$,
$$ \begin{align*} \frac{x}{z + x} + \frac{y}{x + y} + \frac{z}{y + z} &= \frac{1}{\epsilon^2 + 1} + \frac{\epsilon}{1 + \epsilon} + \frac{\epsilon^2}{\epsilon + \epsilon^2} \\ &= \frac{1}{\epsilon^2 + 1} + \frac{\epsilon}{1 + \epsilon} + \frac{\epsilon}{1 + \epsilon} \\ &\to 1, \\ \end{align*} $$
so the bound of $1$ cannot be replaced by a larger number.
| {
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Asymptotes of a rational function We have a function $$f(x) =\dfrac{2x^4+4x^3+3x^2+4x-4}{x^3-x^2-6x}$$
How would I systematically go about finding the asymptotes of this function? I know how to find the asymptotes of for example log functions or functions with a square root in it, but I don't really know how to find them for this function.
| The problem can be simplified by dividing the numerator and denominator by $(x+2)$:
$$
\dfrac{2x^3+3x-2}{x^2-3x} \\
$$
*
*Vertical asymptotes occur where $y$ tends to infinity, which happens at the roots of the denominator:
$$
x^2 - 3x = x (x - 3) = 0 \\
$$
There are two roots so the vertical asymptotes are $x = 0$ and $x = 3$.
*Horizontal asymptotes occur where $x$ tends to infinity, which becomes clear by dividing the numerator and the denominator by the highest power of $x$:
$$
\lim_{x\rightarrow \infty} \dfrac{2x^3+3x-2}{x^2-3x} = \lim_{x\rightarrow \infty} \dfrac{2+3/x^2-2/x^3}{1/x-3/x^2} = \infty \\
$$
There is no limit so there are no horizontal asymptotes.
*Oblique asymptotes occur where the graph approaches the line $(mx+b)$, which becomes clear by dividing the numerator and denominator:
$$
\dfrac{2x^3+3x-2}{x^2-3x} = 2x + 6 \text{ remains } 21x - 2 \\
$$
There is a whole part so the oblique asymptote is $y = 2x + 6$.
| {
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Find the sum of n terms of sequence Given is the sequence $S[i]=(i-1)%9+1$, that is
$$1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9\dots$$
Also given are three numbers $a$, $d$ and $n$.
The task is to find the sum of first $n$ terms starting from $S[a]$, separated by $d$ terms in between.
For example, for $a=8$, $d=1$, $n=5$, the sum will be $8+1+3+5+7$.
| This answer is a sketch of one approach.
For each $i^\text{th}$ number in your sequence ($i=0$ for the first term), define $g_i = \lfloor \frac{i}{9} \rfloor$.
Given a term (say $t_i$) in your original sequence, let $T_i = t_i + 9 g_i$. This produces the familiar sequence of positive integers, for which you can work out the modified sum (say $S$) in the usual manner.
Given the set of selected $i$, you can then calculate $G = \sum 9 g_i = 9 \sum g_i$, which I'll leave as an exercise for you.
Your original sum would then be $S - G$.
Here's an example with $a=3, d=1, n=5$ (first index is 2, last index is 10) :
$$
\begin{matrix}
i & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
g_i & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\
t_i & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 1 & 2 \\
T_i & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11
\end{matrix}
$$
We now calculate $S$:
$$S = \sum_{j=0}^{4} (3 + 2j) = 3 \cdot 5 + 2 \sum_{j=0}^{4} j = 15 + 2 \cdot \frac{4 \cdot 5}{2} = 35$$
We also have $i \in \{2,4,6,8,10\}$, and calculate $G = 9\sum g_i = 9 \cdot 1 = 9$.
So the required sum is $S-G = 35 - 9 = 26$.
(Check using the original $t_i$: $3 + 5 + 7 + 9 + 2 = 26$.)
| {
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What is the circumference (arc length) of $x^4 + x^2 + y^4 + y^2 = 2$? Consider
$$x^4 + x^2 + y^4 + y^2 = 2$$
It is a smooth non-intersecting circle like curve in the plane.
A bit like a Hyperellipse.
See https://en.m.wikipedia.org/wiki/Superellipse
What is the circumference (arc length) of $x^4 + x^2 + y^4 + y^2 = 2$ ?
Wolfram|Alpha plot
| Let us first rewrite the curve in the hope to perhaps parametrize it.
\begin{align*}
x^4+x^2+y^4+y^2=2\iff{}&\left(x^2+\frac12\right)^2+\left(y^2+\frac12\right)^2-\frac12=2\iff{} \\
{}\iff{}&\frac25\left(x^2+\frac12\right)^2+\frac25\left(y^2+\frac12\right)^2=1.
\end{align*}
So we must have $t$ such that:
$$\left\{\begin{array}{c}
\sqrt{\frac25}\left(x^2+\frac12\right)=\cos t \\
\sqrt{\frac25}\left(y^2+\frac12\right)=\sin t \\
\end{array}\right.
\iff
\left\{\begin{array}{c}
x^2=-\frac12+\sqrt{\frac52}\cos t \\
y^2=-\frac12+\sqrt{\frac52}\sin t \\
\end{array}\right.$$
Split a couple of cases depending on sign of $x,y$ (4 cases, I should think), then calculate the integrals of:
$$\gamma(t)=\left(\pm\sqrt{-\frac12+\sqrt{\frac52}\cos t},\pm\sqrt{-\frac12+\sqrt{\frac52}\sin t}\right)$$
with signs depending on the cases, sum them, and you're done.
| {
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Find specific 4 curves touching $y=\cos10x+\cos21x$. The following is the graph of $y=\cos10x+\cos21x$.
You can see that there seems to be four curves that can touch this graph. I tried $y=\cos(x/2+\pi/2\pm\pi)+1$ and $y=-\cos(x/2\pm\pi/2)-1$:
But unfortunately, they cut the graph. What actually are that four curves touching the graph? Thanks.
| Take the one that goes through $(0,2)$. The peaks are roughly at $x=4n\pi/21$, call them $x_n=4n\pi/21+\beta_n$. I will assume $\beta_n$ are small.
At these peaks,
$$\frac{dy}{dx}=-10\sin10x-21\sin21x=0\\
10\sin\left(\frac{40n\pi}{21}+10\beta_n\right)+21\sin(4n\pi+21\beta_n)=0\\
10\left(\sin\frac{40n\pi}{21}+10\beta_n\cos\frac{40n\pi}{21}\right)+21^2\beta_n\approx0\\
\beta_n\approx\frac{10\sin(2n\pi/21)}{10^2\cos(2n\pi/21)+21^2}<\frac1{40}$$
The height of the peak is $\cos10x+\cos21x$, which is roughly
$$y=\cos\left(\frac{40n\pi}{21}+10\beta_n\right)+\cos(21\beta_n)\\
\approx\cos\left(\frac{2n\pi}{21}-10\beta_n\right)+1-(21\beta_n)^2/2\\
\approx\cos\left(\frac x2-\frac{21}2\beta_n\right)+1-5\beta_n\sin(x/2)\\
\approx\cos\left(\frac x2-\frac{11}2\beta_n\right)+1\\
\approx1+\cos\left(\frac x2-\frac18\sin\frac x2\right)\\
\approx1+\cos(\frac x2)+\frac18\sin^2\frac x2$$
| {
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Evaluating $\int\frac{x^4}{x-1} \, dx$ This question involves long division.
I calculated the value. However, I want to ask two concept questions:
1) Why am I doing long division rather than writing out the form of the partial fraction decomposition of the function.
2) When utilizing long division do I place a constant at the end and if not why?
Here is my set up $\dfrac{x^4+0x^3+0x^2+0x+1}{x-1}$
| Partial fractions will only work if the degree of the numerator is strictly less than the degree of the denominator. Hence why you need to do long division. You should not add the +1 at the end on the numerator. This changes the value of the fraction.
If you remember differences of squares you can do the division without long division as:
$$\frac{x^4}{x-1}=\frac{x^4-1}{x-1}+\frac{1}{x-1}$$
$$=\frac{(x^2-1)(x^2+1)}{x-1}+\frac{1}{x-1}$$
$$=\frac{(x-1)(x+1)(x^2+1)}{x-1}+\frac{1}{x-1}$$
$$=(x+1)(x^2+1)+\frac{1}{x-1}$$
Alternatively you could substitution to make the division easier. Let $y=x-1$, hence $dy=dx$ and the integral becomes:
$$\int{\frac{(y+1)^2}{y}dy}=\int\frac{y^4+4y^3+6y^2+4y+1}{y}dy$$
$$=\int y^3+4y^2+6y+4+\frac{1}{y}dy$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1524315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to prove that $\frac{1}{3}e^{2t} + \frac{2}{3}e^{-t}\leq e^{2t^2}$ How to prove that, for every real $t$, one has
$$\frac{1}{3}e^{2t} + \frac{2}{3}e^{-t}\leq e^{2t^2}?$$
| Elementary but not entirely trivial: Note that for real $t$ both $t-t^2$ and $-t-2t^2$ are at most $\tfrac{1}{4}$ and therefore we can use the inequality $$e^x\leq\frac{1}{1-x}$$ that holds for all $x<1$ to obtain
$$\tfrac{1}{3}\left(e^{t-t^2}\right)^2+\tfrac{2}{3}e^{-t-2t^2}\leq \frac{1}{3(1-t+t^2)^2}+\frac{2}{3(1+t+2t^2)}.$$ The right hand side is in fact $\leq 1$, proving your original inequality. One (tedious) way to see this is to compute
$$1-\frac{1}{3(1-t+t^2)^2}-\frac{2}{3(1+t+2t^2)}=\frac{t^2(1+(2t^2-t)(5-3t+3t^2))}{3(1-t+t^2)^2(1+t+2t^2)}$$
and show that numerator and denominator are both positive for $t\neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
An inequality using convex functions: If $\frac{a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2} = 1$ then $abc \le \frac1{2\sqrt 2}$ I see this in a Chinese convex analysis book.
Suppose that $a$,$b$,$c$ are positive real numbers satisfying
\begin{equation}
\frac{a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2} = 1.
\end{equation}
Show that $abc \le \dfrac1{2\sqrt 2}$.
Since it's from a convex analysis book, I tried proving this using Jensen's inequality. However, I can't think of a suitable convex function. Therefore, I tried AM–GM, but I can't get a product $abc$.
$$(abc)^2\le\frac18\iff8(abc)^2\le1$$
$$\iff\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}\right)^3 (abc)^2 \le 1$$
Finally, I used Lagrange multiplier to solve the problem, but I think there is some more elementary solution.
$$f(a,b,c)=abc$$
$$g(a,b,c)=\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}-2=0$$
$$\nabla f(a,b,c)=(bc,ca,ab)$$
$$\nabla g(a,b,c)=\left(-\frac{2a}{(1+a^2)^2},-\frac{2b}{(1+b^2)^2},-\frac{2c}{(1+c^2)^2}\right)$$
$$\because \nabla f = \lambda \nabla g$$
$$\therefore bc = -\frac{2a\lambda}{(1+a^2)^2} \iff abc = -\frac{2a^2\lambda}{(1+a^2)^2}$$
$$abc = -\frac{2a^2\lambda}{(1+a^2)^2} = -\frac{2b^2\lambda}{(1+b^2)^2} = -\frac{2c^2\lambda}{(1+c^2)^2}$$
$$\frac{a}{1+a^2}=\frac{b}{1+b^2}=\frac{c}{1+c^2}=\mu$$
$$a+\frac1a=b+\frac1b=c+\frac1c$$
$$\because \frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}=2$$
$$\therefore \frac{\mu}{a}+\frac{\mu}{b}+\frac{\mu}{c}=2$$
$$\because \frac{a^2}{1+a^2}+\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2} = 1$$
$$\therefore a\mu + b\mu + c\mu = 1$$
$$\frac1a+\frac1b+\frac1c=2(a+b+c)$$
$$3(a+b+c)=a+\frac1a+b+\frac1b+c+\frac1c=3\left(a+\frac1a\right)$$
$$b+c=\frac1a$$
Similarly, $c+a=\dfrac1b$ and $a+b=\dfrac1c$. Substitute $a=\dfrac1{b+c}$ into the other two equations.
$$c+\frac1{b+c}=\frac1b$$
$$\frac1{b+c}+b=\frac1c$$
$$b(b+c)c+b=b+c$$
$$c+b(b+c)c=b+c$$
Subtracting one equation from another, we get $b=c$. Similarly, we have $a=b=c$. It remains to substitute it back to the original constraint and calculate the product.
Any alternative solution is appreciated.
| I believe I can get you started at least. Break down your equation a little.
$a,b,c \in \mathbb{R}^+ \rightarrow \frac{1}{1+n^2} \in [0,1],\;\;\; n = a,b,c$
This means that your original equation could be considered a convex combination as follows: $a^2\lambda_1 + b^2\lambda_2 +c^2\lambda_3 = 1$, where the $\lambda$ values are the fractions defined above.
That, however, is just part of the setup. Basically you've determined that $\frac{1}{1+a^2} + \frac{1}{1+b^2} + \frac{1}{1+c^2} = 1$, by definition of convex combination. My interpretation is that what you're really being asked to prove is what is the minimum value that each $\lambda_i$ can take in a convex combination. This is because, given the definition of $\lambda_i$ for this problem, when a $\lambda$ reaches a minimum, it is because the variable ($a,b,c$ for a given $\lambda$) will be at a maximum. When you maximize $a,b,$ and $c$ then you also maximize their product $abc$.
Consider the case with just two variables between 0 and 1 that must also sum to 1. You want to minimize them both, so the only possible solution is that they equal each other. Neither can be improved without departing from the constraint of the problem (namely that they sum to 1). This extends quite easily to 3 variables, and there are plenty of proofs to look up online.
Back to the specifics of your problem: Since all of the coefficient expressions are equivalent, you can assume $a = b = c$ at the extreme value. Knowing this:
$\frac{3a^2}{1+a^2} = 1 \rightarrow a = \frac{1}{\sqrt{2}}$ and $abc = (\frac{1}{\sqrt{2}})^3 = \frac{1}{2\sqrt{2}} $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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To find the nth Derivative of $9\sqrt{x}$ Can you help me to proof that the nth derivative of $9\sqrt{x}$ is $$ (-1)^{(n-1)} \cdot \frac{9(2n-2)!}{(n-1)!} \cdot (4x)^{\frac{1-2n}{2}}$$
I've tried induction but didn't go very far.
Many thanks
| Induction is a very good way. For the induction step,
Say $P(n)$ is true for $n=k$
For $P(k+1)$, $$\frac{d}{dx^{k+1}}(9\sqrt x)$$
$$=\frac{d}{dx}\left(\frac{d}{dx^{k}}(9\sqrt x)\right)$$
$$=\frac{d}{dx} \left[(-1)^{(k-1)} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot (4x)^{\frac{1-2k}{2}}\right]$$
$$=(-1)^{(k-1)} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot \frac{d}{dx} \left[(4x)^{\frac{1-2k}{2}}\right]$$
$$=(-1)^{(k-1)} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot \frac{1-2k}{2} (4x)^{\frac{-1-2k}{2}}\cdot 4$$
$$=(-1)^{[(k+1)-1]} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot (2k-1) (4x)^{\frac{-1-2k}{2}}\cdot 2$$
$$=(-1)^{[(k+1)-1]} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot (2k-1) \cdot \frac{2k}{k} \cdot (4x)^{\frac{-1-2k}{2}}$$
$$=(-1)^{[(k+1)-1]} \cdot \frac{9[2(k+1)-2]!}{[(k+1)-1]!} \cdot (4x)^{\frac{1-2(k+1)}{2}}$$
$P(k+1)$ is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2}$ Find the following limit
$$
\lim_{x\to0}\left(\frac{1+x2^x}{1+x3^x}\right)^\frac1{x^2}
$$
I have used natural logarithm to get
$$
\exp\lim_{x\to0}\frac1{x^2}\ln\left(\frac{1+x2^x}{1+x3^x}\right)
$$
After this, I have tried l'opital's rule but I was unable to get it to a simplified form.
How should I proceed from here? Any here is much appreciated!
| $\lim_{x\to0} f(x)\\
\text{where } f(x)
=\left(\frac{1+xa^x}{1+xb^x}\right)^\frac1{x^2}
$
Let
$g(x, a)
=x a^x
$.
For small $x$,
$g(x, a)
=xe^{x \ln a}
\approx x(1+x \ln a + x^2 \ln^2a/2 + O(x^3))
= x(1+x \ln a + O(x^2))
$
so
$\begin{array}\\
\frac{1+xa^x}{1+xb^x}
&\approx \frac{1+x(1+x \ln a + O(x^2))}{1+x(1+x \ln b + O(x^2))}\\
&\approx (1+x(1+x \ln a + O(x^2)))(1-(x(1+x \ln b + O(x^2)))+x^2+O(x^2))\\
&= (1+x+x^2 \ln a + O(x^3))(1-x-x^2 (\ln b-1) + O(x^3))\\
&=1+x^2(\ln a-1 - (\ln b -1))+O(x^3)\\
&=1+x^2(\ln(a/b))+O(x^3)\\
\end{array}
$
so,
since
$(1+ax)^{1/x}
\approx e^{a}
$
as
$x \to 0$,
$f(x)
\approx (1+x^2(\ln(a/b))+O(x^3))^{1/x^2}
\approx e^{\ln(a/b)}
=\frac{a}{b}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Prove a binary quadratic equation has specific number of solutions How do I show that the binary quadratic equation $f(x, y) = x^2 + xy + y^2 = 1$ has exactly $6$ solutions?
The discriminant is $-3$, so I cannot use Pell's Equation ($x^2 - dy^2 = p$, where $d>0$ is an integer and $p$ prime).
I know that $4 f(x,y) = (2x+y)^2+3y^2$ is equivalent to $f(x, y) = x^2 + xy + y^2$. But I am unsure how this helps show there are exactly $6$ solutions.
Can anyone please help?
| The integer solutions of $(2x+y)^2+3y^2=4$ are not hard to find. We must have $y^2=0$ or $y^2=1$.
If $y^2=0$, we need $(2x)^2=4$, giving $x=\pm 1, y=0$.
If $y^2=1$, there are two cases. If $y=1$, we want $(2x+1)^2=1$, that is, $x=0$ or $x=-1$. If $y=-1$, we want $(2x-1)^2=1$, giving $x=0$ or $x=1$.
Another way: Note that
$$2(x^2+xy+y^2)=x^2+y^2+ (x+y)^2.$$ So we are solving the Diophantine equation $x^2+y^2+(x+y)^2=2$. This holds if precisely two of $x^2$, $y^2$, and $(x+y)^2$ is equal to $1$. Each of the three possibilities gives rise to two solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1532493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the joint PMF of X and Y, are they independent? A fair die is rolled, and then a coin with probability $p$ of Heads is flipped as many times as the die roll says, e.g., if the result of the die roll is a 3, then the coin is flipped 3 times. Let $X$ be the result of the die roll and $Y$ be the number of times the coin lands Heads.
Find the joint PMF of $X$ and $Y$. Are they independent?
My issue is with finding the joint PMF. I started by finding the supports of both variables. The support of X is {1,2,3,4,5,6} and the support of Y is {0,1,2,3,4,5,6}. Since the value of the roll dictates how many heads can be flipped, the joint support is of size 27. I am trying to construct a table of all the different probabilities but I can't seem to figure out what the specific probabilities are. Once I find the joint PMF and marginals I can figure out if the variables are independent or not.
Here is my work so far...
$$\begin{array}{c|c|c|c|c|c|c|}
& \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6} & f_Y(y) \\ \hline
\text{0} & p & p & p & p & p & p & 6p \\ \hline
\text{1} & p & p & p & p & p & p & 6p \\ \hline
\text{2} & 0 & p & p & p & p & p & 5p \\ \hline
\text{3} & 0 & 0 & p & p & p & p & 4p \\ \hline
\text{4} & 0 & 0 & 0 & p & p & p & 3p \\ \hline
\text{5} & 0 & 0 & 0 & 0 & p & p & 2p \\ \hline
\text{6} & 0 & 0 & 0 & 0 & 0 & p & 1p \\ \hline
f_X(x) & 0 & 0 & 0 & 0 & 0 & p & 1p \\ \hline
\end{array}$$
| $\newcommand{\rchi}{\raise{0.5ex}\chi}$
You have been given (effectively) that: $X\sim \mathcal U\{1,2,3,4,5,6\}$ and $Y\mid X\sim \mathcal{Bin}(X, p)$.
That is that $X$ is discrete uniformly distributed (the roll of an unbiased die), and $Y$ when conditioned on $X$ is binomially distributed (the count of successes in $X$ iid Bernoulli events).
So you know the marginal pmf of $X$ is $\mathsf P_X(k) = \frac 1 6 \;\rchi_{k\in\{1;6\}}$ and the conditional pmf of $Y$ is $\mathsf P_{Y\mid X}(h\mid k) = \binom{k}{h} p^h(1-p)^{k-h} \;\rchi_{h\in\{0;k\}} $
From this you can determine the joint pmf of $X,Y$, and from that the marginal pmf of $Y$.
$$\begin{array}{|c|c:c:c:c:c|c|} \hline
& \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6} & f_Y(y) \\ \hline
\text{0} & (1-p)/6 & (1-p)^2/6 & (1-p)^3/6 & (1-p)^4/6 & (1-p)^5/6 & (1-p)^6/6 & \tfrac 1 6 \sum\limits_{k=1}^6 (1-p)^k \\ \hdashline
\text{1} & p/6 & p(1-p)/3 & & & & & \\ \hdashline
\text{2} & 0 & p^2/6 & & & & & \\ \hdashline
\text{3} & 0 & 0 & & & & & \\ \hdashline
\text{4} & 0 & 0 & 0 & & & & \\ \hdashline
\text{5} & 0 & 0 & 0 & 0 & & & \\ \hdashline
\text{6} & 0 & 0 & 0 & 0 & 0 & & \\ \hline
f_X(x) & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1/6 & 1 \\ \hline
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving Cubic when There are Known to be 3 Real Roots When solving for roots to a cubic equation, the sign of the $\Delta$ tells us when there will be 3 distinct real roots (as long as the first terms coefficient, $a$, is non-zero.) Namely when $\Delta$ is positive.
The equations to find the 3 roots are:
*
*$x_1 = -\frac{1}{3a}(b + C + \frac{\Delta_0}{C})$
*$x_2 = -\frac{1}{3a}(b + \frac{C(-1 + i\sqrt 3)}{2} + \frac{2\Delta_0}{C(-1 + i\sqrt 3)})$
*$x_3 = -\frac{1}{3a}(b + \frac{C(-1 - i\sqrt 3)}{2} + \frac{2\Delta_0}{C(-1 - i\sqrt 3)})$
Where: $$C = \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4\Delta_0^3}}{2}}$$
By the given equation, $\Delta_1^2 - 4\Delta_0^3 = -27a^2\Delta$, I know that when $\Delta$ is positive the square root in $C$ will produce an $i$. So when $\Delta$ is positive $C$ is effectively: $$C = \sqrt[3]{\frac{\Delta_1 + i\sqrt{4\Delta_0^3 - \Delta_1^2}}{2}}$$
So obviously the $i$ in $C$ cancels with the $i$s in the $x_2$ and $x_3$ roots, and we get 3 real roots. But for the life of me I cannot work out how. Can someone help me break those steps down?
| Given $x^3+a x^2+b x+c x+d=0$, substitute $x=y-a/3. $ This gives $y^3+p y+q=0$ where $p,q$ are determined by $a,b,c,d$ and the substitution. Now let $y=u+v$. We have $$0=(u+v)^3+p(u+v)+q=(u^3+v^3+q)+(u+v)(3u v+p).$$ Suppose there are non-zero $u,v$ such that $$u^3+v^3+q=0=3u v+p.$$ Substituting $v=-p/3u$ into $0=u^3+v^3+q$ and letting $w=u^3$, we have $$w^2+q w+(-p^3/27)=u^3(u^3+q+v^3)=0.$$ Solve the quadratic equation in $w . $ Let $u$ be any complex cube root of (either value of) $w$. Then $x=y-a/3=u+v-a/3=u-p/3u-a/3$ solves the original cubic....When there are 3 real solutions for $x$ and they are not an arithmetic sequence, the values of $w$ will not be real numbers and you need the cube root of one of them. One way is to put $w=r. cis (t)=r(\cos t+i\sin t)$ with $r>0$ and $t\in (0,2\pi). $ The cube roots of $w$ are $r^{1/3}cis ((t+2\pi n)/3)$ for $n\in \{0,1,2\}.$.....The main idea in this method of solving the cubic equationis to change the variable to a sum $u+v$ and see if the equation can be split into 2 simpler equations.
| {
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"url": "https://math.stackexchange.com/questions/1535552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Series expansion for $x$, when $x$ is small Suppose that we are given the series expansion of $y$ in terms of $x$, where $|x|\ll 1$. For example, consider $$y=x+x^2+x^3+\cdots\qquad\qquad\qquad (1).$$ From this I would like to derive the series expansion of $x$ in terms $y$. Given that $|x|\ll 1$ we can take $y=x+O(x^2)$ such that $x\sim y$.
It then follows from $(1)$ that $$x=y-(x^{2}+x^{3}+\cdots)\qquad\qquad\qquad (2).$$ Given that $x\sim y$, $(2)$ becomes $$x=y-y^2-y^3+\cdots\qquad\qquad\qquad (3).$$
While the first two terms of $(3)$ are correct I suspect that the remaining terms are incorrect.
For instance given that $y=x+O(x^2)$ implies that $y^2=x^2+2xO(x^{2})+(O(x^{2}))^{2}$. Therefore, $2xO(x^{2})$ will yield a $y^{3}$ term in $(3)$ when computing $x^{2}$ in $(2)$.
How do I find the correct term containing $y^{3}$ in $(3)$, and so on.
| This is not an answer but it is too long for a comment.
Just as Will Jagy commented, you can inverse the series building one coefficient at the time. This is a quite tedious task but it is doable.
Suppose that you have
$$y=a_0+a_1+a_2x^2+a_3x^3+O\left(x^4\right)$$, replacing $x$ by $b_0+b_1+b_2y^2+b_3y^3$, replacing, expanding and identifying one coefficient at the time, you will then arrive to $$x=\frac{1}{a_1}(y-a_0)-\frac{a_2 }{a_1^3}(y-a_0)^2+\frac{2 a_2^2-a_1
a_3 }{a_1^5}(y-a_0)^3+O\left((y-a_0)^4\right)$$
If we set all the $a_i$'s equal to $1$ as in the example in the post, you will then have $$x=y-y^2+y^3+O\left(y^4\right)$$ If we replace $x$ by $y-y^2+y^3$ in $y=x+x^2+x^3$, we should get $$y=y+4y^5-6y^6+\cdots$$ which shows the error.
| {
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"url": "https://math.stackexchange.com/questions/1536118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove from first principles: $\lim_{n\rightarrow\infty}\frac{n^2+2n-2}{n^2-6}=1$ Prove using $\epsilon - n_0$:
$$\lim_{n\rightarrow\infty}\frac{n^2+2n-2}{n^2-6}=1$$
I made an attempt on this question but I'm not sure if this is the correct path.
Firstly I did
$\frac{n^2+2n-2}{n^2-6}-1$ < $\epsilon$ ?
after that |$\frac{n^2+2n-2}{n^2-6}-1$| = |$\frac{2(n+2)}{n^2-6}$| which =|$\frac{2n+4}{n^2-6}$|.
What do I do after this?
Would $n\geq3$ due to making the denominator positive?
$$n\geq 3 \implies 2n\geq6 \implies 2n+4 \geq 10 > 0$$
Are these steps correct?
| Notice that $(2n+4)(n-\frac32)=2n^2+n-6>n^2-6$
Then for $n>3/2$ we have that
$$\left|\frac{2(n+2)}{n^2-6}\right|< \frac{2n+4}{(2n+4)(n-3/2)}=\frac1{n-3/2}<\epsilon$$
for an $n>3/2$ large enough.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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System of equations based question in matrix form If $\begin{bmatrix}1^2&2^2&3^2\\2^2&3^2&4^2\\3^2&4^2&5^2\\4^2&5^2&6^2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\4\\10\\\lambda\end{bmatrix}$,then $\lambda=$
$(A)17\hspace{1cm}(B)18\hspace{1cm}(C)19\hspace{1cm}(D)20$
I tried solving this question.
$1^2x+2^2y+3^2z=1$
$2^2x+3^2y+4^2z=4$
$3^2x+4^2y+5^2z=10$
I solved this system to get $x=\frac{25}{8},y=\frac{-5}{2},z=\frac{7}{8}$
$4^2x+5^2y+6^2z=\lambda$
I substituted $x,y,z$ in the above equation to get $\lambda=19$.
But this method is long and tedious.My book gives one method but that i cannot understand.Please help me understand this.Book's method is not self explanatory.
Book's solution:
$(n+3)^2=n^2-3(n+1)^2+3(n+2)^2$
$\lambda=1-3\times4+3\times 10=19$
OR if someone knows any better and elegant way to solve this question,please tell me,like using eigen values etc.Thanks.
| I don't have a complete answer but I do notice that the equation $n^2-3(n+1)^2+3(n+2)^2=(n+3)^2$ shows that when $[x,y,z]=[1,-3,3]$ then the top three rows of the matrix times $[x,y,z]^T$ equals the (transpose of the) fourth row. And it also equals $[1,4,10]^T$. And the second equation is $[1,4,10]\times[1,-3,3]^T=\lambda$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Compute sum with generating functions I need to calculate the following sum with generating functions:
$$\sum_{k=1}^{n-1}\frac{1}{k(n-k)}$$
I tried:
$$\sum_{n \geq 0}(\sum_{k=1}^{n-1}\frac{1}{k}\cdot\frac{1}{n-k})z^k=\sum_{n \geq 0}(\sum_{k=0}^{n}\frac{1}{k+1}\cdot\frac{1}{n-k+1})z^k$$
The inner sum is a Cauchy product, therefore:
$$(\sum_{n \geq 0}\frac{z^n}{n+1})^2$$
Now I'm stuck. How can I calculate a closed form from this sum?
| Method 1: Let $$S(x) = \sum_{n\geq 0} \frac{x^{n}}{n+1}$$ then
\begin{align}
D_{x}\left( x \, S(x) \right) = \sum_{n\geq 0} x^{n} = \frac{1}{1-x}
\end{align}
which leads to
$$S(x) = - \frac{\ln(1-x)}{x}.$$
Now
\begin{align}
\sum_{n=1}^{\infty} a_{n} \, t^{n} &= \sum_{n=1}^{\infty} \sum_{k=1}^{n-1} \frac{1}{k(n-k)} \, t^{n} \\
&= \sum_{n=1}^{\infty} \sum_{n=1}^{\infty} \frac{t^{n+k}}{n \, k} \\
&= \left( \sum_{n=1}^{\infty} \frac{t^{n}}{n} \right)^{2} = \left( - \ln(1-t) \right)^{2} = \ln^{2}(1-t) \\
&= 2 \, \sum_{n=2}^{\infty} \frac{H_{n-1}}{n} \, t^{n}
\end{align}
where
$$\sum_{n=1}^{\infty} H_{n} \, t^{n} = - \frac{\ln(1-t)}{1-t}$$
was used. This yields
$$a_{n} = \sum_{k=1}^{n-1} \frac{1}{k (n-k)} = \frac{2 \, H_{n-1}}{n}.$$
Method 2: Consider $$\frac{1}{k\, (n-k)} = \frac{1}{n} \, \left( \frac{1}{k} + \frac{1}{n-k} \right)$$ for which
\begin{align}
a_{n} &= \sum_{k=1}^{n-1}\frac{1}{k(n-k)} \\
&= \frac{1}{n} \, \left[ \sum_{k=1}^{n-1} \frac{1}{k} + \sum_{k=1}^{n-1} \frac{1}{n-k} \right] \\
&= \frac{1}{n} \, \left[ H_{n-1} + \sum_{k=1}^{n-1} \frac{1}{k} \right]\\
&= \frac{2 \, H_{n-1}}{n},
\end{align}
where $H_{n}$ is the Harmonic number defined by $H_{n} = \sum_{k=1}^{n} \frac{1}{k}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving $\sqrt{2}x-\sqrt{x^{2}+1} \geq \frac{\sqrt{2}}{2}\ln{(x)}$ How can I prove that
$$
x\sqrt{2}-\sqrt{x^{2}+1} \geq \frac{\sqrt{2}}{2}\ln{(x)}
$$
It's a derivation-based process if I remember correctly, however I was unable to prove it correctly.
| Let's get rid of $\sqrt{2}$ by rewriting the inequality as
$$
f(x)=2x-\sqrt{2(x^2+1)}-\ln x\ge0
$$
We have $\lim_{x\to0}f(x)=\infty$. For computing the limit at $\infty$, we do the substitution $t=1/x$, so the limit becomes
$$
\lim_{t\to0^+}\frac{2}{t}-\frac{\sqrt{2(t^2+1)}}{t}+\ln t=
\lim_{t\to0^+}\frac{2-\sqrt{2(t^2+1)}+t\ln t}{t}=\infty
$$
Thus we know that $f$ has at least a point of minimum.
Compute the derivative
$$
f'(x)=2-\frac{2x}{\sqrt{2(x^2+1)}}-\frac{1}{x}
=2-\frac{x\sqrt{2}}{\sqrt{x^2+1}}-\frac{1}{x}
$$
Let's go on:
$$
f''(x)=-\sqrt{2}\,\frac{\sqrt{x^2+1}-\dfrac{x^2}{\sqrt{x^2+1}}}{x^2+1}
+\frac{1}{x^2}
=-\sqrt{2}\,\frac{1}{(x^2+1)\sqrt{x^2+1}}+\frac{1}{x^2}
$$
and we want to evaluate the sign of
$$
(x^2+1)\sqrt{x^2+1}-x^2\sqrt{2}
$$
that is, where $(x^2+1)\sqrt{x^2+1}>x^2\sqrt{2}$. We can square getting
$$
x^6+3x^4+3x^2+1>2x^4
$$
that is
$$
x^6+x^4+3x^2+1>0
$$
which is of course true.
Therefore $f''(x)>0$ for all $x>0$ and so $f'(x)$ is increasing. Since $f'(1)=0$, $f'$ vanishes only at $1$, which is thus the unique minimum point for $f$.
Since $f(1)=0$, we see that $f(x)\ge0$ for all $x>0$ (equality only at $1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1540298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
simplifying the trigonometric expression $\frac{\cot\theta - \tan\theta}{\sin\theta + \cos\theta} + \frac{1}{\cos\theta}$ I'm stuck on this expression for which I don't have any answer.
$$\frac{\cot\theta - \tan\theta}{\sin\theta + \cos\theta}+\frac{1}{\cos\theta}$$
Need some help on the simplifying steps.
Here is what I get from a calculator
| Here is a detailed answer:
$$
\require{cancel}
\begin{align}
\frac{\cot\theta-\tan\theta}{\sin\theta+\cos\theta}+\frac{1}{\cos\theta}&=\frac{\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}}{\sin \theta+\cos \theta}+\frac{1}{\cos \theta}\tag{1}\label{ko-eq1}\\
&=\frac{\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos \theta}}{\sin \theta+\cos \theta}+\frac{1}{\cos \theta}\tag{2}\label{ko-eq2}\\
&=\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos \theta}\cdot\frac{1}{\sin \theta+\cos \theta}+\frac{1}{\cos \theta}\tag{3}\label{ko-eq3}\\
&=\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}+\frac{1}{\cos \theta}\tag{4}\label{ko-eq4}\\
&=\frac{\cos^{2}\theta-\sin^{2}\theta+\sin\theta\left(\sin \theta+\cos \theta\right)}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}\tag{5}\label{ko-eq5}\\
&=\frac{\cos^{2}\theta\cancel{-\sin^{2}\theta}\cancel{+\sin^{2}\theta}+\sin\theta\cos \theta}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}\tag{6}\label{ko-eq6}\\
&=\frac{\cos^{2}\theta+\sin\theta\cos \theta}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}\tag{7}\label{ko-eq7}\\
&=\frac{\cancel{\cos\theta}\bcancel{\left(\sin \theta+\cos \theta\right)}}{\sin\theta\cancel{\cos\theta}\bcancel{\left(\sin \theta+\cos \theta\right)}}\tag{8}\label{ko-eq8}\\
&=\frac{1}{\sin\theta}\\
&=\csc\theta
\end{align}\\
$$
\eqref{ko-eq1}: Expand $\cot\theta$ and $\tan\theta$ as $\frac{\cos\theta}{\sin\theta}$ and $\frac{\sin\theta}{\cos\theta}$, respectively.
\eqref{ko-eq2}: Find common denominator of $\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}$.
\eqref{ko-eq3}: Rewrite $\frac{\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos \theta}}{\sin \theta+\cos \theta}$ as $\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos \theta}\cdot\frac{1}{\sin \theta+\cos \theta}$ and multiply.
\eqref{ko-eq4}: Multiplication resut.
\eqref{ko-eq5}: Find common denominator of $\frac{\cos^{2}\theta-\sin^{2}\theta}{\sin\theta\cos\theta\left(\sin \theta+\cos \theta\right)}+\frac{1}{\cos \theta}$ and continue.
\eqref{ko-eq6}: Cancel the terms.
\eqref{ko-eq7}: After cancellation.
\eqref{ko-eq8}: Final cancellation and finish.
I hope this helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Double radical proof I'm trying to prove that
$$
\sqrt{A+\sqrt{B}}=\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}
$$
With
$$
C=\sqrt{A^2 - B}
$$
How can I handle this?
Edit: obviously is easy that this holds when you know the r.h.s., but my question is: how to get the r.h.s. when you only know the l.h.s.
| $$A+\sqrt{B}=\frac{A}{2}+\frac{C}{2}+\frac{A}{2}-\frac{C}{2}+\sqrt{B},$$
where $C=\sqrt{A^2-B}$. But
$$\sqrt{B}=\sqrt{A^2-(A^2-B)}=\frac{2}{2}\sqrt{(A+\sqrt{A^2-B})(A-\sqrt{A^2-B})}=2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}}.$$
Then
$$A+\sqrt{B}=\frac{A+C}{2}+\frac{A-C}{2}+2\sqrt{\frac{A+C}{2}}\sqrt{\frac{A-C}{2}}=\left(\sqrt{\frac{A+C}{2}}+\sqrt{\frac{A-C}{2}}\right)^2.$$
Taking square roots gives the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
Taylor's Remainder $x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) Prove that
$\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) <x-\frac{x^2}{2}+\frac{x^3}{3}$
My attempt: I can prove this by taking one side at a time and assuming $\displaystyle f(x) = \log (1+x)-x+\frac{x^2}{2}-\frac{x^3}{3}$
and then proving that it is a decreasing function and then the same for the other side.
But I am looking for a better solution using Lagrange's Mean Value Theorem or using Taylor's remainder.
| It is much easier to solve this problem via integration.
If $t > 0$ then we can see that $$1 < 1 + t^{3}$$ and on dividing by $(1 + t) > 0$ we can see that $$\frac{1}{1 + t} < 1 - t + t^{2}$$ and integrating this equation in the interval $[0, x]$ (and noting that $x > 0$) we get $$\log(1 + x) < x - \frac{x^{2}}{2} + \frac{x^{3}}{3}\tag{1}$$ Further note that $$\frac{1 - t^{2}}{1 + t} = 1 - t$$ so that $$\frac{1}{1 + t} = 1 - t + \frac{t^{2}}{1 + t}$$ and integrating this equation on interval $[0, x]$ we get $$\log(1 + x) = x - \frac{x^{2}}{2} + \int_{0}^{x}\frac{t^{2}}{1 + t}\,dt\tag{2}$$ and clearly we can see that for $0 < t < x$ we have $$\frac{t^{2}}{1 + t} > \frac{t^{2}}{1 + x}$$ and hence $$\int_{0}^{x}\frac{t^{2}}{1 + t}\,dt > \int_{0}^{x}\frac{t^{2}}{1 + x}\,dt = \frac{x^{3}}{3(1 + x)}$$ and then from equation $(2)$ we get $$\log(1 + x) > x - \frac{x^{2}}{2} + \frac{x^{3}}{3(1 + x)}\tag{3}$$ Combining equations $(1)$ and $(3)$ we get the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
To prove the limit of $\frac{x^2+2\cos x-2}{x\sin^3 x}$ at zero is $1/12$ To Prove $$\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{1}{12}$$
I tried with L'Hospital rule but in vain.
| from
$$
\cos x = 1 - \frac{x^2}2+\frac{x^4}{24}-\dots
$$
we have
$$
x^2+2\cos x - 2 = \frac{2x^4}{24}-\dots
$$
and
$$
\sin x = x-... \\
x\sin^3 x = x^4-...
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
On the decreasing sequences I do not know how to prove that these sequences are decreasing:
$$
a_n=\frac{1}{\sqrt{n}(n+1)}-\frac{1}{\sqrt{n+1}(n+2)},
$$
$$
b_n=\frac{1}{n(\sqrt{n}+1)}-\frac{1}{(n+1)(\sqrt{n+1}+1)}.
$$
Thank you for all kind help and comments.
My attemption. I considered the following function
$$
f(x)=\frac{1}{\sqrt{x}(x+1)}-\frac{1}{\sqrt{x+1}(x+2)},
$$
and calculated its derivative. But its derivative is rather complicated
$$
\nabla f(x)=\frac{3x+1}{2\sqrt{x}x(x+1)^2}-\frac{3x+4}{2\sqrt{x+1}(x+1)(x+2)^2}
$$
| Use the Mean Value Theorem to show there is an $\eta$ between $n$ and $n+1$ so that
$$
\begin{align}
a_n
&=\frac1{\sqrt{n}(n+1)}-\frac1{\sqrt{n+1}(n+2)}\\
&=\frac{\frac32\sqrt{\eta}+\frac12\frac1{\sqrt{\eta}}}{\eta(\eta+1)^2}\\
&=\frac32\frac1{\sqrt{\eta}(\eta+1)^2}+\frac12\frac1{\sqrt{\eta}^3(\eta+1)^2}
\end{align}
$$
which is obviously a decreasing function.
Likewise, show there is an $\eta$ between $n$ and $n+1$ so that
$$
\begin{align}
b_n
&=\frac1{n(\sqrt{n}+1)}-\frac1{(n+1)(\sqrt{n+1}+1)}\\
&=\frac{\frac32\sqrt{\eta}+1}{\eta^2(\sqrt{\eta}+1)^2}\\
&=\frac32\frac1{\eta^{3/2}(\sqrt{\eta}+1)^2}+\frac1{\eta^2(\sqrt{\eta}+1)^2}
\end{align}
$$
which is obviously a decreasing function.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to prove $\sinh^{-1} (\tan x)=\log \tan (\frac{\pi}{4}+\frac{x}{2})$ Like the question says
How to prove $$\sinh^{-1} (\tan x)=\log \tan (\frac{\pi}{4}+\frac{x}{2})$$
I have tried using many identity but in vain
For reference
$$\tanh ^{-1} x=\frac{1}{2} \log \frac{1+x}{1-x}$$
and $$\sinh^{-1} x=\log (x+\sqrt{x^2+1})$$
| Putting $\tan x$ in place of $x$ in this formula $$\sinh^{-1} x=\log (x+\sqrt{x^2+1})$$ we have,
$$\sinh^{-1} \tan x=\log (\tan x+\sqrt{(\tan x)^2+1})$$
$$=\log (\tan x+\sqrt{(\sec x)^2})$$
$$=\log (\tan x+ \sec x)$$
$$=\log (\frac{1+\sin x}{\cos x})$$
$$=\log \left[\frac{(\sin \frac{x}{2}+\cos \frac{x}{2})^2}{\cos^2 \frac{x}{2}-\sin^2 \frac{x}{2}}\right]$$
$$=\log \left[\frac{\sin \frac{x}{2}+\cos \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right]$$
$$=\log \left[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right]$$
$$=\log \left[\frac{\tan \frac{\pi}{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi}{4}\tan \frac{x}{2}}\right]$$
$$=\log \tan (\frac{\pi}{4}+\frac{x}{2})$$
Hence proved.
| {
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"url": "https://math.stackexchange.com/questions/1552760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Proving inequality with constraint $abc=1$ For positive reals $a,b,c$, with $abc=1$, prove that $$\frac{a^3+1}{b^2+1}+ \frac{b^3+1}{c^2+1}+\frac{c^3+1}{a^2+1} \geq 3$$ I tried the substitution $x/y,y/z,z/x$, but it didn't give me anything. What else to do? Thanks.
| There is already a full and great answer. This is only an alternative
using AM-GM instead of the rearrangement inequality:
From the AM-GM inequality we have
$$
\frac 13 \left(\frac{a^3+1}{b^2+1}+ \frac{b^3+1}{c^2+1}+\frac{c^3+1}{a^2+1} \right) \ge
\left(\frac{a^3+1}{b^2+1} \cdot \frac{b^3+1}{c^2+1} \cdot \frac{c^3+1}{a^2+1} \right)^{1/3}
$$
therefore it suffices to show that
$$
\frac{a^3+1}{a^2+1} \cdot \frac{b^3+1}{b^2+1} \cdot \frac{c^3+1}{c^2+1}
\ge 1 \, .
$$
From
$$
0 \le (a^2 - 1)(a-1) = 2(a^3 +1) - (a^2+1)(a+1)
$$
it follows that
$$
\frac {a^3+1}{a^2+1} \ge \frac{a+1}{2} \, ,
$$
this is the crucial estimate given by Macavity in the comment Proving inequality with constraint $abc=1$ above.
We continue with
$$
\frac{a+1}{2} \ge \sqrt{1 \cdot a} = \sqrt a \,
$$
using AM-GM again.
The same holds for $b$ and $c$, this gives
$$
\frac{a^3+1}{a^2+1} \cdot \frac{b^3+1}{b^2+1} \cdot \frac{c^3+1}{c^2+1}
\ge \sqrt {abc} = 1 \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Volume of solid $W$ delimited by $z=x^2+3y^2$ and $z=8-x^2-y^2$. Volume of solid $W$ delimited by $z=x^2+3y^2$ and $z=8-x^2-y^2$.
I need to solve this with triple integrals, however I'm having a lot of problems. Anyone knows how to solve?
| Let's first find the boundary of the integration region, that is, where both surfaces intersect:
$z=x^2+3y^2=8-x^2-y^2$ thus $2x^2+4y^2=8$ thus it's the ellipse $\frac{x^2}{2^2}+\frac{y^2}{\sqrt 2^2}=1$ in the $xy$ plane.
Now let $E$ be the plane region bounded by that ellipse. The volume of $W$ is then $\displaystyle\iint_E\left(8-x^2-y^2-(x^2+3y^2)\right)\,dx\,dy=\iint_E\left(8-2 x^2-4 y^2\right)\,dx\,dy$.
In order to define the integration limits one should change coordinates as follows: $x=2\,r\cos\theta$, $y=\sqrt2\,r\sin\theta$, so that the bounding ellipse is simply $r=1$, $0\leq\theta\leq2\pi$ in the new coordinates. Now let's calculate the Jacobian:
$\displaystyle J=\begin{vmatrix}\frac{\partial x}{\partial r}&\frac{\partial x}{\partial \theta}\\\frac{\partial y}{\partial r}&\frac{\partial y}{\partial \theta}\end{vmatrix}=\begin{vmatrix}2\cos\theta&-2\,r\sin\theta\\\sqrt2\sin\theta&\sqrt2\,r\cos\theta\end{vmatrix}=2\sqrt2\,r$.
Therefore, the volume is
$\displaystyle\iint_E\left(8-2 x^2-4 y^2\right)\,dx\,dy=\int_0^1\int_0^{2\pi}\left(8-2\cdot4r^2\cos^2\theta-4\cdot2r^2\sin^2\theta\right)2\sqrt2\,r\,d\theta\,dr=8\cdot2\sqrt2\int_0^1(1-r^2)r\,dr\int_0^{2\pi}d\theta=8\cdot2\sqrt2\frac14\cdot2\pi=8\sqrt2\pi$.
| {
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"source": "stackexchange",
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Proof of $\sum\limits_{n=1}^{\infty} \frac{x^n \log(n!)}{n!} \sim x \log(x) e^x$ as $x \to \infty$ Prove that
$$\sum_{n=1}^{\infty} \frac{x^n \log(n!)}{n!} \sim x \log(x) e^x \,\,\,\text{as}\,\,\, x \to \infty$$
and
$$\sum_{n=1}^{\infty} \frac{(-x)^n \log(n!)}{n!} \to 0 \,\,\,\text{as}\,\,\, x \to \infty$$
This question is related to my previous question. My heuristic approach is that the sum's major contribution comes from $n\approx x$ term, so
$$\sum_{n=1}^{\infty} \frac{x^n \log(n!)}{n!} \sim \frac{x^x\log(x!)}{x!}$$
but using the Stirling formula twice leads to
$$\sum_{n=1}^{\infty} \frac{x^n \log(n!)}{n!} \sim \frac{\sqrt{x}\log(x)}{\sqrt{2\pi}}e^x$$
But this reasoning is flawed and I believe I should not ignore all the other terms. (Which I believe to account for $\sqrt{2\pi x}$factor)
What kind of approach may give the wanted asymptotic behavior? Any kind of hint is welcome.
| Your idea is right that the entries near $n=x$ matter the most. In particular, the entries with $n=x+O(\sqrt{x})$ dominate this sum.
The following is not a complete answer, but I believe the holes can be filled in.
Suppose that $N=x + \alpha \sqrt{x},$ where $\alpha$ is fixed. Then, by Stirling's formula
\begin{align*}
\frac{\ln(N!)}{N!} x^N &= \left(1+O\left(\frac{1}{\ln x}\right)\right) \frac{(x+\alpha \sqrt{x}) \ln x}{\sqrt{2 \pi x} (x+\alpha \sqrt{x})^{x+\alpha \sqrt{x}}}e^{x+\alpha \sqrt{x}} x^{x+\alpha \sqrt{x}} \\ &=\left(1+O\left(\frac{1}{\ln x}\right)\right) x (\ln x) e^x \left[e^{\alpha \sqrt{x}} \left( \frac{x}{x+\alpha \sqrt{x}}\right)^{x+\alpha \sqrt{x}} \frac{1}{\sqrt{2\pi x}} \right].
\end{align*}
Now, note that
\begin{align*}
\left( \frac{x}{x+\alpha \sqrt{x}}\right)^{x+\alpha \sqrt{x}}&=\exp \left( -(x+\alpha \sqrt{x}) \cdot \ln\left(1+\frac{\alpha}{\sqrt{x}} \right)\right) \\&=\exp\left( -(x+\alpha \sqrt{x}) \left( \frac{\alpha}{\sqrt{x}} - \frac{\alpha^2}{2x} + O(n^{-3/2}) \right) \right) \\&= \exp \left( -\alpha \sqrt{x} - \alpha^2/2 + O(x^{-1/2}) \right).
\end{align*}
Therefore
$$
\frac{\ln N!}{N!}x^N = \left(1+O\left(\frac{1}{\ln x}\right)\right) [x \ln(x) e^x] \frac{e^{-\alpha^2/2}}{\sqrt{2\pi}} \frac{1}{\sqrt{x}}
$$
Now the sum of these $e^{-\alpha^2/2}{\sqrt{2\pi}} \frac{1}{\sqrt{x}}$ over $N \in [x-C \sqrt{x}, x+C \sqrt{x}]$, $(C$ fixed w.r.t. $x$), is a Riemann sum for the integral $\int_{-C}^C \frac{e^{-t^2/2}}{\sqrt{2\pi}}dt$. This integral can be arbitrarily close to 1 for $C$ large enough.
It remains to be shown that for any $\epsilon >0$ (fixed w.r.t. $x$),
$$
\sum_{n \notin [x-C \sqrt{x}, x+C \sqrt{x}]} \frac{\ln(N!)}{N!} x^N < \epsilon \, x \ (\ln x) \ e^x,
$$
for $C$ large enough.
| {
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"question_score": "11",
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Chain rule with fraction In the case of $$f(x)=\ln\big(x+\sqrt{1+x^2}\big)$$
in the derivative we multiply $$f'(x)=\frac{1}{x+\sqrt{1+x^2}}\bigg(1+\frac{2x}{2\sqrt{1+x^2}}\bigg)$$ when the expression multiply the numerator?
| $$\frac{\text{d}}{\text{d}x}\left(\ln\left(x+\sqrt{1+x^2}\right)\right)=$$
Using the chain rule:
$$\frac{\frac{\text{d}}{\text{d}x}\left(x+\sqrt{1+x^2}\right)}{x+\sqrt{1+x^2}}=$$
$$\frac{\frac{\text{d}}{\text{d}x}\left(x\right)+\frac{\text{d}}{\text{d}x}\left(\sqrt{1+x^2}\right)}{x+\sqrt{1+x^2}}=$$
$$\frac{1+\frac{\text{d}}{\text{d}x}\left(\sqrt{1+x^2}\right)}{x+\sqrt{1+x^2}}=$$
Using the chain rule:
$$\frac{1+\frac{\frac{\text{d}}{\text{d}x}\left(1+x^2\right)}{2\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}=$$
$$\frac{1+\frac{\frac{\text{d}}{\text{d}x}\left(1\right)+\frac{\text{d}}{\text{d}x}\left(x^2\right)}{2\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}=$$
$$\frac{1+\frac{0+\frac{\text{d}}{\text{d}x}\left(x^2\right)}{2\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}=$$
$$\frac{1+\frac{\frac{\text{d}}{\text{d}x}\left(x^2\right)}{2\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}=$$
$$\frac{1+\frac{2x}{2\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}=\frac{1}{\sqrt{1+x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.