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Consecutive Prime Gap Sum (Amateur) List of the first fifty prime gaps:
1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 4, 14, 4, 6, 2, 10, 2, 6, 6, 4, 6, 6, 2, 10, 2, 4, 2, 12, 12, 4, 2, 4.
My conjecture is that the sum of consecutive prime gaps is always prime whenever a prime gap of 2 is added.
$$ 1 + 2 = 3 $$
$$ 1 + 2 + 2 = 5 $$
$$ 1 + 2 + 2 + 4 + 2 = 11 $$
$$ 1 + 2 + 2 + 4 + 2 + 4 + 2 = 17 $$
$$ 1 + 2 + 2 + 4 + 2 + 4 + 2 + 4 + 6 + 2 = 29 $$
I don't know if this is meaningful or how to go about testing it completely (I've tested it up to 461) so I'll just leave this here and see what comes of it.
| Let $p_n$ denote the $n^{\text{th}}$ prime. Then $p_{n+1}-p_n$ is the $n^{\text{th}}$ prime gap. The sum of the first $k$ prime gaps is $$\sum_{n=1}^k(p_{n+1}-p_n) = p_{k+1} - p_1 = p_{k+1} - 2.$$ Now, if the $k^{\text{th}}$ prime gap is $2$, that is $p_{k+1} - p_k = 2$, then $p_{k+1} - 2 = (p_k + 2) - 2 = p_k$ which is prime as you've noticed.
A pair of consecutive primes $p_n$, $p_{n+1}$ which differ by two (i.e. the $n^{\text{th}}$ prime gap is two) are said to be twin primes. It is still unknown as to whether there are infinitely map twin primes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Finding the asymptotes of a general hyperbola I'm looking to find the asymptotes of a general hyperbola in $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ form, assuming I know the center of the hyperbola $(h, k)$. I came up with a solution, but it's too long for me to be confident that I didn't make a mistake somewhere, so I was wondering if I could run it by someone and see if it works. It's mostly algebraic, and I'm prone to making tiny errors in algebra that throw off the entire problem.
So to start, since we know the center $(h, k)$, we can first translate the hyperbola by $(-h, -k)$ using the transform $x_0 = x - \Delta{x}, y_0 = y - \Delta{y}$ with $\Delta{x} = -h$ and $\Delta{y} = -k$. Assuming $F'$ is the translated $F$, we can divide the entire equation by $-F'$ to put it in the following form:
$$
ax^2 + bxy + cy^2 + dx + ey = 1
$$
With $a = -A'/F'$ and $A'$ the translated $A$, $b = -B'/F'$ and $B'$ the translated $B$, and so on.
Next we convert to polar coordinates to get the following:
$$
r^2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta}) + r(d\cos{\theta} + e\sin{\theta}) - 1 = 0
$$
Solving for $r$ will give us
$$
r = \frac{-d\cos{\theta} - e\sin{\theta} \pm \sqrt{(d\cos{\theta} + e\sin{\theta})^2 + 4(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta})}}{2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta})}
$$
Now assume $\theta_0 = 2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta})$, this means that as $\theta_0 \rightarrow 0^{\pm}$, $r \rightarrow \pm\infty$. The angles at which $r \rightarrow \pm\infty$ are the asymptotes of the hyperbola, so now it's just a matter of solving for where $\theta_0 = 0$. This is where the majority of the algebra takes place and this is where I'm worried I made some miniscule mistake.
$$
\begin{align}
& 2(a\cos^2{\theta} + b\cos{\theta}\sin{\theta} + c\sin^2{\theta}) = 0\\
& \Longleftrightarrow a(1 - \sin^2{\theta}) + b\cos{\theta}\sin{\theta} + c\sin^2{\theta} = 0 \\
& \Longleftrightarrow \sin^2{\theta}(c - a) + b\cos{\theta}\sin{\theta} = -a \\
& \Longleftrightarrow \frac{1 - \cos{2\theta}}{2}(c - a) + \frac{b}{2}\sin{2\theta} = -a \\
& \Longleftrightarrow (c - a)(1 - \cos{2\theta}) + b\sin{2\theta} = -2a \\
& \Longleftrightarrow (c - a)(1 - \cos{2\theta}) + 2a = -b\sqrt{1 - \cos^2{2\theta}} \\
& \Longleftrightarrow \frac{a - c}{b}(1 - \cos{2\theta}) - \frac{2a}{b} = \sqrt{1 - \cos^2{2\theta}} \\
& \Longleftrightarrow (\frac{a - c}{b}(1 - \cos{2\theta}))^2 - 4a\frac{a - c}{b^2}(1 - \cos{2\theta}) + (\frac{2a}{b})^2 = 1 - \cos^2{2\theta} \\
& \Longleftrightarrow (\frac{a - c}{b})^2(1 - 2\cos{2\theta} + \cos^2{2\theta}) - 4a\frac{a - c}{b^2}(1 - \cos{2\theta}) + (\frac{2a}{b})^2 = 1 - \cos^2{2\theta} \\
& \Longleftrightarrow [(\frac{a - c}{b})^2 + 1]\cos^2{2\theta} + [2(\frac{a - c}{b})(\frac{a}{b} - \frac{a - c}{b})]\cos{2\theta} + [(\frac{2a}{b})^2 + (\frac{a - c}{b})^2 - 4a\frac{a - b}{b^2}) - 1] = 0 \\
& \Longleftrightarrow [(\frac{a - c}{b})^2 + 1]\cos^2{2\theta} + 2(\frac{a - c}{b})(\frac{c}{b})\cos{2\theta} + [(\frac{2a}{b})^2 + (\frac{a - c}{b})(\frac{-3a - c}{b}) - 1] = 0 \\
& \Longleftrightarrow [(\frac{a - c}{b})^2 + 1]\cos^2{2\theta} + 2(\frac{a - c}{b})(\frac{c}{b})\cos{2\theta} + [\frac{a^2 + 2ac + c^2}{b^2} - 1] = 0 \\
\end{align}
$$
Now let
$$
U = [(\frac{a - c}{b})^2 + 1] \\
V = 2(\frac{a - c}{b})(\frac{c}{b}) \\
W = \frac{a^2 + 2ac + c^2}{b^2} - 1
$$
So that the above equation becomes
$$
U\cos^2{2\theta} + V\cos{2\theta} + W = 0.
$$
Solving for $\cos{2\theta}$ gives us
$$
\cos{2\theta} = \frac{-V \pm \sqrt{V^2 - 4UW}}{2U}.
$$
Finally we may solve for theta like so,
$$
\theta = \frac{1}{2}\arccos{\frac{-V \pm \sqrt{V^2 - 4UW}}{2U}}.
$$
This gives us two numbers, $\theta_1$ and $\theta_2$, each corresponding to the slopes $m_1$ and $m_2$ of the asymptotes.
The relationship between the slope of a line $m$ and the angle $\theta$ between the line and the positive x-axis is $m = \tan{\theta}$. You can use the identity $\tan{(\frac{1}{2}\arccos{x})} = \sqrt{\frac{1 - x}{x + 1}}$ to solve for $m_1$ and $m_2$ in terms of non-trig functions, but I think this answer is sufficient enough.
Now that we have the slopes of the asymptotes, we can find the y-intercepts $b_1$ and $b_2$ of each line by simply plugging in the original center $(h, k)$ into each equation for the line and solving.
$$
b_1 = k - m_1h \\
b_2 = k - m_2h
$$
Thus, the asymptotes of the hyperbola with general equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ have equations $y = m_1x + b_1$ and $y = m_2x + b2$.
Does this look correct? Also, did I overcomplicate things? Was there an easy solution all along that I was missing?
| So a fairly simple calculus based solution arises from knowing that the Asymptotes are the points where the slope tends toward being constant
That is given
$$Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$$
We begin by deriving with respect to $x$
$$2Ax + 2By \frac{dy}{dx} + Cy+Cx\frac{dy}{dx} + D + E\frac{dy}{dx} = 0 $$
And now solve for $\frac{dy}{dx}$
$$\frac{dy}{dx} = -\frac{2Ax + Cy + D}{2By + Cx + E} $$
Now we can take the limit as $x \rightarrow \pm \infty$ to determine the slopes of the Asymptotes and call this values $m$
Now take the original curve, and solve it for $y$ (it won't be too pretty)
$$y = \frac{-E- Cx \pm \sqrt{(E + Cx)^2 - 4B(Dx + F)}}{2B} $$
And evaluate $$ \lim_{x \rightarrow \pm \infty} y(x) - mx $$
to recover the y intercepts of the asymptotes.
This ENTIRE process can now be repeated with $\frac{dx}{dy}$ in place of $\frac{dy}{dx}$ and y exchanged with x to interpret the asymptotes in a reflected coordinate systems
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Determining if a recursively defined sequence converges and finding its limit Define $\lbrace x_n \rbrace$ by $$x_1=1, x_2=3; x_{n+2}=\frac{x_{n+1}+2x_{n}}{3} \text{if} \, n\ge 1$$ The instructions are to determine if this sequence exists and to find the limit if it exists.
I prove that the sequence is eventually decreasing and that it is bounded below (see work below). Hence, if my work is correct, the sequence will have a limit. The problem is that if the limit is, say, $L$ then it should satisfy $L=\frac{L+2L}{3}$; however, every real number satisfies the latter equality. So, if this limit exists, how do I actually find it?
Proving the limit exists:
1) The sequence is bounded below by $1$.
Base Case: $x_1=1\ge 1$.
Induction: Suppose $x_n\ge 1$ for all $k=1,\ldots ,n+1$. Then $x_{n+2}=\frac{x_{n+1}+2x_{n}}{3}\ge \frac{3}{3}=1$.
2) The sequence is eventually decreasing: It actually starts decreasing after the 4th term. So I take as my base case $x_5<x_4$.
Base Case: Note that $$x_3=\frac{x_2+2x_1}{3}=\frac{3+2}{3}=5/3$$ and hence $$x_4=\frac{x_{3}+2x_{2}}{3}=\frac{5/3+2\times3}{3}=23/9$$ so $$x_5=\frac{x_{4}+2x_{3}}{3}=\frac{23/9+2\times5/3}{3}=\frac{53}{27}<x_4$$
Induction: Suppose $x_{n}<x_{n-1}.$ Then $$x_{n+1}-x_n=\frac{x_{n}+2x_{n-1}}{3}-x_n \\=2\frac{x_{n-1}-x_n}{3}<0$$
Therefore, the sequence is bounded below and is eventually decreasing.
| This is a linear recurrent sequence of order $2$. The characteristic
polynomial $X^2-\frac{X+2}{3}$ has roots $1$ and $-\frac{2}{3}$. So there
are two constants $a$ and $b$ such that $x_n=a(1^n)+b(\frac{-2}{3})^n$.
Using the initial conditions, we obtain $a=\frac{11}{5}$ and $b=\frac{9}{5}$.
The limit is therefore $\frac{11}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Integrate $\int \left(A x^2+B x+c\right) \, dx$ I am asked to find the solution to the initial value problem:
$$y'=\text{Ax}^2+\text{Bx}+c,$$
where $y(1)=1$,
I get:
$$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$
But the answer to this is:
$$y=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1.$$
Could someone show me what has been done and explain why?
| Use the condition $y(1)=1$, i.e. substitute $x=1$ and then you have
\begin{equation}
\frac{A}{3} + \frac{B}{2} + c +d = 1
\end{equation}
So, you get
\begin{equation}
d = 1 -\frac{A}{3} - \frac{B}{2} - c
\end{equation}
By substituting in $\frac{A}{3}x^3+\frac{B}{2}x^2+cx+d$, you get that equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Partial fraction decomposition of type $1/(x^2+k)$ I know that partial fraction of this can be written as:
$$\frac{3x}{(1+x)(2+x)}=\frac{-3}{1+x}+\frac{6}{2+x}$$
Which can be done in these ways:
$$\frac{3x}{(1+x)(2+x)}=\frac{A}{1+x}+\frac{B}{2+x}\implies3x=A(2+x)+B(1+x),\forall\;x$$
And now solving it to get A and B.Also someone told me that write the fraction without that denominator whose coeffecient you want to know and put root of denominator, for e.g.:
$$\text{coefficient of $1/(1+x)$ i.e. A}=\frac{3x}{2+x}|_{x+1=0\implies x=-1}=\frac{-3}{1}$$
I know that this is just the first way disguised in an easy form, now what when I need to decompose:
$$\frac1{(x-3)(x^2-1)}\text{ or }\frac1{(x-3)(x^2+1)}$$
There exist multiple root of first and root to second doesn't exist.
Now what to do, adopt the first method? Is there any easy method for these?
| Decomposing $1/(x+n)(x^2 + k)$ is only difficult (for me) when k is a positive quantity. Otherwise just use $a^2 - b^2$. For example:
$$\frac1{(x-3)(x^2-1)} = \frac1{(x-3)(x-1)(x+1)} = \frac{A}{x-3} + \frac{B}{x-1} + \frac{C}{x+1}$$
Here,
$$A = \frac1{(3-1)(3+1)} = \frac1{8}\\
B = \frac1{(1-3)(1+1)} =\frac{-1}{4} \\
C = \frac1{(-1-3)(-1-1)} = \frac{1}{8}$$
Now, for the guy I don't like:
$$\frac{1}{(x^2 +1)(x-3)} = \frac{Ax+B}{x^2 +1} + \frac{C}{x-3}$$
Try some easy values of $x$, I'll take $x=0$,
$$ B + \frac{C}{3} = \frac{-1}{3} \quad\dots(1)$$
Now, I'll take $x= 1$,
$$\frac{A + B}{2} - \frac{C}{2} = -\frac{1}{4} \quad\dots(2)$$
Let $x=2$,
$$\frac{2A + B}{5} - C= -\frac1{5}\quad\dots(3)$$
Solving $(1)$ and $(2)$ and $(3)$, we get
$$A = -\frac{1}{10},\, B = -\frac{3}{10},\, C = \frac{1}{10} $$
This is the easiest (and only) method I know to find partial fractions and it helps me immensely in indefinite integration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/902874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simple use of log I am struggling to see how we can go from the first expression to the second:
$$\begin{align}
2\log_3 12 - 4\log_3 6 &= \log_3 \left ( \frac{4^2 \cdot 3^2}{2^4 \cdot 3^4} \right )\\
&= \log_3 (3^{-2}) = -2
\end{align}$$
| Remember, $$\log ab = \log a + \log b$$
and
$$m\log a=\log a^m.$$
HINT:
$$12^2 = (4\cdot 3)^2=4^2\cdot 3^2.$$
Similarly,
$$6^4 = (2\cdot 3)^4 = 2^4\cdot 3^4$$
I think you can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/903168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find $\int \frac{x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$ $$I=\int x.\frac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\mathrm dx$$
Try 1:
Put $z= \ln(x+\sqrt{1+x^2})$, $\mathrm dz=1/\sqrt{1+x^2}\mathrm dx$
$$I=\int \underbrace{x}_{\mathbb u}\underbrace{z}_{\mathbb v}\mathrm dz=x\int zdz-\int (z^2/2)\mathrm dz\tag{Wrong}$$
Try 2:
Put
$z= x+\sqrt{1+x^2}$
$$\implies x-z
=\sqrt{1+x^2}\implies x^2+z^2-2xz
=1+x^2\implies x
=\frac{z^2-1}{2z}$$
$$\mathrm dz
=\left(1+\frac{x}{\sqrt{1+x^2}}\right)\mathrm dx
=\frac{z\mathrm dx}{x-z}=\frac{-2z^2\mathrm dx}{1+z^2}$$
$$I
=\int\frac{(z^2-1)\ln z}{2z}.\frac{(1+z^2)\mathrm dz}{-2z^2}$$
$$=\int\frac{(z^4-1)\mathrm dz}{4z^3}
=\frac14\int\left(z-\frac1{z^3}\right)\mathrm dz
=z^2/2+2/z^2+C\tag{Wrong}$$
Try 3:
Put
$z
=\sqrt{1+x^2},\mathrm dx
=x/\sqrt{1+x^2}\mathrm dx$
$$I
=\int \ln(x+z)\mathrm dz
=\int \ln(z+\sqrt{z^2-1})\mathrm dz$$
Don't know how to solve this integral.
[Note that if I take $u=z+\sqrt{z^2-1}$, it is $=\sqrt{1+x^2}+\sqrt{1+x^2-1}=x+\sqrt{1+x^2}$; same as first try.]
What's wrong in try 1 & 2, how to further solve try 3 and the best method to solve this question?
Update: Sorry, I don't know hyperbolic/inverse hyperbolic trigonometry.
| Hint :
Use IBP by setting $u=x$ and $dv=\dfrac{\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\ dx$. Note that: $\text{arsinh}\ x=\ln(x+\sqrt{1+x^2})$ and $\dfrac{d}{dx}\text{arsinh}\ x=\dfrac{1}{\sqrt{1+x^2}}$, then
\begin{align}
\int \frac{x\ln(x+\sqrt{1+x^2})}{\sqrt{1+x^2}}\ dx&=x\left[\int\text{arsinh}\ x\ d(\text{arsinh}\ x)\right]-\int\left[\int\text{arsinh}\ x\ d(\text{arsinh}\ x)\right]\ dx.
\end{align}
The rest part, you may refer to Wikipedia: hyperbolic function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
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Inequality involving Pochhammer symbols Let $m,S$ be integers satisfying $2\leq m\leq S$. I would like to show that
$$h_1\left(x\right) h_3\left(x\right) \leq h_2^2\left(x\right)$$
for all $x\geq 0$ where
$$h_k\left(x\right) \equiv {}_2F_1\left(k,k-1-m;S-m+k,-x\right).$$
${}_2F_1$ denotes the hypergeometric function.
What I have so far...
Note that \begin{multline*}
h_{1}\left(y\right)h_{3}\left(y\right)=\sum_{d=0}^{\infty}\left(-y\right)^{d}\sum_{p=0}^{d}\frac{\left(1\right)^{\left(p\right)}\left(-m\right)^{\left(p\right)}}{\left(S-m+1\right)^{\left(p\right)}p!}\frac{\left(3\right)^{\left(q\right)}\left(2-m\right)^{\left(q\right)}}{\left(S-m+3\right)^{\left(q\right)}q!}\\
=\sum_{d=0}^{\infty}y^{d}\sum_{p=0}^{d}\frac{m_{p}}{\left(S-m+1\right)^{\left(p\right)}}\frac{\left(q+1\right)\left(q+2\right)\left(m-2\right)_{q}}{2\left(S-m+3\right)^{\left(q\right)}}
\end{multline*} where $q\equiv d-p$ and the Pochhammer symbols are
defined here: https://en.wikipedia.org/wiki/Pochhammer_symbol.
Similarly, \begin{multline*}
h_{2}^{2}\left(y\right)=\sum_{d=0}^{\infty}\left(-y\right)^{d}\sum_{p=0}^{d}\frac{\left(2\right)^{\left(p\right)}\left(1-m\right)^{\left(p\right)}}{\left(S-m+2\right)^{\left(p\right)}p!}\frac{\left(2\right)^{\left(q\right)}\left(1-m\right)^{\left(q\right)}}{\left(S-m+2\right)^{\left(q\right)}q!}\\
=\sum_{d=0}^{\infty}y^{d}\sum_{p=0}^{d}\frac{\left(p+1\right)\left(m-1\right)_{p}}{\left(S-m+2\right)^{\left(p\right)}}\frac{\left(q+1\right)\left(m-1\right)_{q}}{\left(S-m+2\right)^{\left(q\right)}}.
\end{multline*}
So it is sufficient to show $$
\sum_{p=0}^{d}\frac{\left(q+1\right)\left(q+2\right)m_{p}\left(m-2\right)_{q}}{2\left(S-m+1\right)^{\left(p\right)}\left(S-m+3\right)^{\left(q\right)}}-\frac{\left(q+1\right)\left(p+1\right)\left(m-1\right)_{p}\left(m-1\right)_{q}}{\left(S-m+2\right)^{\left(p\right)}\left(S-m+2\right)^{\left(q\right)}}\leq0, $$ for all $d\geq0$.
Note that for $d \geq 2m - 1$, the above is zero (see also
@Semiclassical's comment below), so that we only have to verify the
claim for $0 \leq d \leq 2\left(m -1\right)$.
| I'll prove here some other easy cases, the first being in the comments.
My numerical experiments indicate that one should require $2 \le k \le m \le S$ to show
\begin{align}
q_k(x) := h_{k-1}(x)h_{k+1}(x) - h_{k}^2(x) \le 0
\end{align}
Step 0, one sees immediately that $h_k(x) = h_{k,S,m}(x)$ is a polynomial of degree $(m-k+1)$.
Step 1, $m=k$, one has
\begin{align}
h_{k-1,S,k}(x) &= 1+\frac{(2k-2)}{(S-1)}x+\frac{k(k-1)}{S(S-1)}x^2\\
h_{k,S,k}(x) &= 1+\frac{k}{S}x\\
h_{k+1,S,k}(x) &= 1
\end{align}
which gives
\begin{align}
q_k(x)
&= h_{k-1,S,k}(x)\:h_{k+1,S,k}(x)-h_{k,S,k}(x)^2\\
&= \left(1+\frac{(2k-2)}{(S-1)}x+\frac{k(k-1)}{S(S-1)}x^2\right)\cdot 1 -\left(1+\frac{k}{S}x\right)^2\\
&= -x(2S+kx)\frac{(S-k)}{(S-1)S^2}.
\end{align}
This is clearly a degree 2 polynomial in x with zeros $\{-2S/k,0\}$.
So $x\ge0$ gives $q_k(x) \le 0$. $\square$
Step 2, $m=k+1$, $2 \le k \le k+1 \le S$, we have
\begin{equation}
(5kS+3S-k-3)\ge (5k(k+1)+3(k+1)-k-3) = 5k^2+7k+6 > 0,
\end{equation}
which gives
\begin{align}
q_k(x) = -x(2S(S^2-1)+S(5kS+3S-k-3)x
+4k(k+1)Sx^2+k(k+1)^2x^3)\frac{(S-k-1)}{(S-2) (S-1)^2 S^2} \le 0. \square
\end{align}
Now the easy coefficients of the general case. Step 0, gives that $\text{deg}(q_k(x)) = 2m-2k+2$.
Step 3, $r\ge 2$, $m=k+r$, $2\le k\le k+r \le S$, one has, without proving the equality,
\begin{align}
[x^0](q_k(x)) &= 0,\\
[x^1](q_k(x)) &= -2\cdot\frac{(S+1)(S-k-r)}{(S-r-1)(S-r+0)(S-r+1)} \le 0,\\
[x^{2r+1}](q_k(x)) &= -(2r+2)\cdot\frac{
(S-k-r)
\prod\limits_{\nu=0}^{r-1}(k+\nu)
\prod\limits_{\nu=1}^{r}(k+\nu)
}{
\prod\limits_{\nu=0}^{r}(S-\nu)
\prod\limits_{\nu=1}^{r+1}(S-\nu)
} \le 0,\\
[x^{2r+2}](q_k(x)) &= -\frac{
(S-k-r)
\prod\limits_{\nu=0}^{r}(k+\nu)
\prod\limits_{\nu=1}^{r}(k+\nu)
}{
\prod\limits_{\nu=0}^{r}(S-\nu)
\prod\limits_{\nu=0}^{r+1}(S-\nu)
} \le 0.\square
\end{align}
Next the remaining three cases for $r=2$.
Step 4, $r=2$, $m=k+r$, $2\le k\le k+r \le S$, we have
\begin{align}
(3kS+2S-4k-4) & \ge (3k(k+2)+2(k+2)-4k-4) \ge 3k^2+4k > 0,\\
(4kS+2S+k-4) & \ge (4k(k+2)+2(k+2)+k-4) \ge 4k^2+11k > 0, \\
(7kS+11S-2k-7) & \ge (7k(k+2)+11(k+2)-2k-7) \ge 7k^2+23k+15 > 0,
\end{align}
which gives, without proving the equality,
\begin{align}
[x^2](q_k(x)) &= -3(3kS+2S-4k-4)\cdot\frac{(S-k-2)(S+1)}{(S-3)(S-2)^2(S-1)S} \le 0,\\
[x^3](q_k(x)) &= -4(4kS+2S+k-4)\cdot\frac{(S-k-2)(k+1)}{(S-3)(S-2)^2(S-1)S}
\le 0,\\
[x^4](q_k(x)) &= -2(7kS+11S-2k-7)\cdot\frac{(S-k-2)k(k+1)}{(S-3)(S-2)^2(S-1)^2S} \le 0. \square
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
How do you solve for x in this equation? I tried expanding, but I still can't get rid of the exponents to isolate x.
$$\frac{(1+x)^4-1}{x}=4.374616$$
Thank you in advance for your help.
| \begin{align}
(1+x)^{4} - 1 &= [(1+x)^{2} - 1] [ (1+x)^{2} + 1] = (1+x-1)(1+x+1)(x^{2} + 2x +1)
\nonumber\\
&= x(x+2)(x^{2}+2x+2)
\end{align}
or
\begin{align}
\frac{(1+x)^{4} -1}{x} = (x+2) (x^{2} + 2x +2).
\end{align}
Now, $4.374616$ can be factored into
\begin{align}
4.374616 = (2.06)(2.1236) = (2 + .06)((.06)^{2} + 2(.06) + 2).
\end{align}
This yields
\begin{align}
(x+2) (x^{2} + 2x +2) = (2 + .06)((.06)^{2} + 2(.06) + 2)
\end{align}
which gives $x = .06$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
To test the convergence of series 2 To test the convergence of series 2
$\displaystyle \frac{a+x}{1!}+\frac{(a+2x)^2}{2!}\frac{(a+3x)^3}{3!}+...\infty$
My Attempt:
$$\begin{align}
\frac{u_n}{u_{n+1}} & = \frac{(a+nx)^n}{n!}\frac{(n+1)!}{(a+(n+1)x)^{n+1}} \\
& = \frac{(a+nx)^n}{1}\frac{n+1}{(a+nx+x)^{n+1}} \\
& = \frac{(x+a/n)^n}{1}\frac{1+1/n}{(x+a/n+x/n)^{n+1}} \\
& = \frac{x^n(1+a/nx)^n}{1}\frac{1+1/n}{x^{n+1}(1+a/nx+1/n)^{n+1}} \\
& = \frac{(1+a/nx)^n}{1}\frac{1+1/n}{(1+a/nx+1/n)^{n+1}}\frac{1}{x} \\
& = ??
\end{align}$$
I see an "e" in the limit $n\to \infty$ here, but can't quite get "there". Any hints?
| $$\begin{align}
lim_{n \to \infty}\frac{u_n}{u_{n+1}}
& = lim_{n \to \infty} (1+a/nx)^n\frac{1+1/n}{(1+a/nx+1/n)^{n+1}}\frac{1}{x} \\
& =lim_{n \to \infty} (1+a/nx)^n\frac{1}{ lim_{n \to \infty}(1+(a+x)/nx)^n}\frac{1}{x}\\
&= e^{a/x} \dfrac{1}{xe^{a+x/x}}\\ &=\dfrac{e^{-1}}{x}\end{align}$$
Therefore your series is convergent for all $e^{-1}>|x|.$ But I don't have any idea about how to find the sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/904933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integral of $\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}$ How do I find $$\int\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}\mathrm dx$$
I used partial fractions by breaking up $x^2 + 2x - 2$ into $(x+1)^2 - 3$ and split it into $(a+b)(a-b)$ but as u can see it's extreme tedious. I was wondering if there is a faster technique to resolve this.
| $$\int\frac{2x^3 + 5x^2 +2x +2}{(x^2 +2x + 2)(x^2 + 2x - 2)}\mathrm dx\\
=\int\left(+\frac{2x+2}{2(x^2+2x+2)}+\frac{2x+2}{2(x^2+2x-2)}-\frac1{\frac{(2x+2)^2}4+1}\right)\mathrm dx\\
=\frac12\ln(x^2+2x+2)+\frac12\ln(x^2+2x-2)-\arctan\left(\frac{2x+2}2\right)+\mathbb C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/905084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Show that $ax^2+2hxy+by^2$ is positive definite when $h^2The question asks to "show that the condition for $P(x,y)=ax^2+2hxy+by^2$ ($a$,$b$ and $h$ not all zero) to be positive definite is that $h^2<ab$, and that $P(x,y)$ has the same sign as $a$."
Now I've seen questions similar to this before where it's a two variable quadratic and I'm not too sure how to go about it. Normally with one variable you could just show the discriminant is less than $0$ but since there's two variables I can't use the same process (since I'm not sure how the discriminant is defined for a two variable quadratic). I also tried completing the square but that doesn't seem to make the algebra anything near as simple as the answer (I got a rather complicated fraction with cubes). Is there a way to go about this type of problem?
Thanks.
| In higher dimensions (than 1), there is a elegant matrix based approach. In this approach, the sign is determined by the diagonal matrix. Specifically, note that for your question $$a{x^2} + 2hxy + b{y^2} = \left[ {\begin{array}{*{20}{c}}x&y\end{array}} \right]\left[ {\begin{array}{*{20}{c}}a&h\\h&b\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right]$$But the matrix could be diagonalized as $$\left[ {\begin{array}{*{20}{c}}a&h\\h&b\end{array}} \right] = M\left[ {\begin{array}{*{20}{c}}{\frac{{a + b - \sqrt {{{(a - b)}^2} + 4{h^2}} }}{2}}&0\\0&{\frac{{a + b + \sqrt {{{(a - b)}^2} + 4{h^2}} }}{2}}\end{array}} \right]{M^T}$$ (however irrelevant to your question, note that $$M = \left[ {\begin{array}{*{20}{c}}{\frac{{a - b - \sqrt {{{(a - b)}^2} + 4{h^2}} }}{{2h\sqrt {1 + {{\left( {\frac{{a - b - \sqrt {{{(a - b)}^2} + 4{h^2}} }}{{2h}}} \right)}^2}} }}}&{\frac{{a - b + \sqrt {{{(a - b)}^2} + 4{h^2}} }}{{2h\sqrt {1 + {{\left( {\frac{{a - b + \sqrt {{{(a - b)}^2} + 4{h^2}} }}{{2h}}} \right)}^2}} }}}\\{\frac{1}{{\sqrt {1 + {{\left( {\frac{{a - b - \sqrt {{{(a - b)}^2} + 4{h^2}} }}{{2h}}} \right)}^2}} }}}&{\frac{1}{{\sqrt {1 + {{\left( {\frac{{a - b + \sqrt {{{(a - b)}^2} + 4{h^2}} }}{{2h}}} \right)}^2}} }}}\end{array}} \right]$$)Now, by introducing the new variables $${M^T}\left[ {\begin{array}{*{20}{c}}x\\y\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{x'}\\{y'}\end{array}} \right]$$we have $$P(x',y') = \left[ {\begin{array}{*{20}{c}}{x'}&y\end{array}'} \right]\left[ {\begin{array}{*{20}{c}}{\frac{{a + b - \sqrt {{{(a - b)}^2} + 4{h^2}} }}{2}}&0\\0&{\frac{{a + b + \sqrt {{{(a - b)}^2} + 4{h^2}} }}{2}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{x'}\\{y'}\end{array}} \right]$$ Clearly for the polynomial to be of determined sign, both values on the diagonal should have the same sign. Now you should be able to figure out the conditions. Good luck ;)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/905293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Evaluation of a dilogarithmic integral
Problem. Prove that the following dilogarithmic integral has the indicated value:
$$\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(x)}\operatorname{Li}_2{(x)}}{1-x}\stackrel{?}{=}-11\zeta{(5)}+6\zeta{(3)}\zeta{(2)}.$$
My attempt:
I began by using the polylogarithmic expansion in terms of generalized harmonic numbers,
$$\frac{\operatorname{Li}_r{(x)}}{1-x}=\sum_{n=1}^{\infty}H_{n,r}\,x^n;~~r=2.$$
Then I switched the order of summation and integration and used the substitution $u=-\ln{x}$ to evaluate the integral:
$$\begin{align}
\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(x)}\operatorname{Li}_2{(x)}}{1-x}
&=\int_{0}^{1}\mathrm{d}x\ln^2{(x)}\sum_{n=1}^{\infty}H_{n,2}x^n\\
&=\sum_{n=1}^{\infty}H_{n,2}\int_{0}^{1}\mathrm{d}x\,x^n\ln^2{(x)}\\
&=\sum_{n=1}^{\infty}H_{n,2}\int_{0}^{\infty}\mathrm{d}u\,u^2e^{-(n+1)u}\\
&=\sum_{n=1}^{\infty}H_{n,2}\frac{2}{(n+1)^3}\\
&=2\sum_{n=1}^{\infty}\frac{H_{n,2}}{(n+1)^3}.
\end{align}$$
So I've reduced the integral to an Euler sum, but unfortunately I've never quite got the knack for evaluating Euler sums. How to proceed from here?
| It is easy to see that
$$2\sum^\infty_{n=1}\frac{H_n^{(2)}}{(n+1)^3}=2\sum^\infty_{n=1}\frac{H_{n+1}^{(2)}}{(n+1)^3}-2\sum^\infty_{n=1}\frac{1}{(n+1)^5}=2\sum^\infty_{n=1}\frac{H_n^{(2)}}{n^3}-2\zeta(5)$$
Consider $\displaystyle f(z)=\frac{\pi\cot{\pi z} \ \Psi^{(1)}(-z)}{z^3}$. We know that
$$\pi\cot{\pi z}=\frac{1}{z-n}-2\zeta(2)(z-n)+O((z-n)^3)$$
and
$$\Psi^{(1)}(-z)=\frac{1}{(z-n)^2}+\left(H_n^{(2)}+\zeta(2)\right)+O(z-n)$$
At the positive integers,
\begin{align}
{\rm Res}(f,n)
&=\operatorname*{Res}_{z=n}\left[\frac{1}{z^3(z-n)^3}+\frac{H_n^{(2)}-\zeta(2)}{z^3(z-n)}\right]\\
&=\frac{H_n^{(2)}}{n^3}-\frac{\zeta(2)}{n^3}+\frac{6}{n^5}\\
\end{align}
At the negative integers,
\begin{align}
{\rm Res}(f,-n)&=-\frac{\Psi^{(1)}(n)}{n^3}\\&=\frac{H_{n-1}^{(2)}-\zeta(2)}{n^3}\\&=\frac{H_{n}^{(2)}}{n^3}-\frac{\zeta(2)}{n^3}-\frac{1}{n^5}\tag1
\end{align}
At $z=0$,
\begin{align}
{\rm Res}(f,0)&=[z^2]\left(\frac{1}{z}-2\zeta(2)z\right)\left(\frac{1}{z^2}+\zeta(2)+2\zeta(3)z+3\zeta(4)z^2+4\zeta(5)z^3\right)\\
&=4\zeta(5)-4\zeta(2)\zeta(3)
\end{align}
Since the sum of the residues $=0$, we conclude that
\begin{align}
\color\red{\int^1_0\frac{\log^2{x} \ {\rm Li}_2(x)}{1-x}{\rm d}x}
&=2\sum^\infty_{n=1}\frac{H_n^{(2)}}{n^3}-2\zeta(5)\\
&=\zeta(2)\zeta(3)-6\zeta(5)+\zeta(2)\zeta(3)+\zeta(5)-4\zeta(5)+4\zeta(2)\zeta(3)-2\zeta(5)\\
&\large{\color\red{=6\zeta(2)\zeta(3)-11\zeta(5)}}
\end{align}
Explanation
$(1):$ Use the functional equation $\displaystyle \Psi^{(1)}(z+1)=-\frac{1}{z^2}+\Psi^{(1)}(z)$ which is derived by differentiating the functional equation of the digamma function, as well as the fact that $\displaystyle H_n^{(2)}=\frac{1}{n^2}+H_{n-1}^{(2)}$.
As for how to obtain the laurent series, the series for $\Psi(z)$ was cleverly derived here by Random Variable. In essence,
$$\color{blue}{\gamma+\Psi(-z)=\frac{1}{z-n}+H_n+\sum^\infty_{k=1}(-1)^k\left(H_n^{(k+1)}+(-1)^{k+1}\zeta(k+1)\right)(z-n)^k}$$
Differentiating yields
$$\color{blue}{\Psi^{(1)}(-z)=\frac{1}{(z-n)^2}+\sum^\infty_{k=1}(-1)^{k+1}k\left(H_n^{(k+1)}+(-1)^{k+1}\zeta(k+1)\right)(z-n)^{k-1}}$$
For $\pi\cot{\pi z}$,
\begin{align}
\color{blue}{\pi\cot{\pi z}}
&=\Psi(1-z)-\Psi(z) \ \ \ \ \ \text{(reflection formula for digamma function)}\\
&=\int^1_0\frac{t^{z-1}-t^{-z}}{1-t}{\rm d}t \ \ \ \ \ \text{(recall that $\Psi(z)=-\gamma+H_{z-1}$)}\\
&=\sum_{k=0}^\infty\int^1_0\left(t^{z+k-1}-t^{-z+k}\right){\rm d}t\\
&=\sum_{k=0}^\infty\left(\frac{1}{z+k}+\frac{1}{z-k-1}\right)\\
&=\frac{1}{z}+\frac{1}{z-1}+\frac{1}{z+1}+\frac{1}{z-2}+\frac{1}{z+2}+\cdots\\
&=\frac{1}{z}+\sum^\infty_{k=1}\left(\frac{1}{z-k}+\frac{1}{z+k}\right)\\
&=\frac{1}{z}+\sum^\infty_{k=1}\frac{2z}{z^2-k^2}\\
&=\frac{1}{z}-2\sum^\infty_{k=1}\sum^\infty_{m=1}\frac{z^{2m-1}}{k^{2m}}\\
&=\color{blue}{\frac{1}{z}-2\sum^\infty_{m=1}\zeta(2m)z^{2m-1}}\\
&=\pi\cot(\pi (z-n)) \ \ \ \ \ \text{(since cotangent has a period of $\pi$)}\\
&=\color{blue}{\frac{1}{z-n}-2\sum^\infty_{m=1}\zeta(2m)(z-n)^{2m-1}}\\
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/906579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to express $\log_2 (\sqrt{9} - \sqrt{5})$ in terms of $k=\log_2 (\sqrt{9} + \sqrt{5})$? If $$k=\log_2 (\sqrt{9} + \sqrt{5})$$
express $\log_2 (\sqrt{9} - \sqrt{5})$ in terms of $k$.
| Add to both sides the term $\log_2 (\sqrt{9} - \sqrt{5})$, then you have
$$
\log_2 (\sqrt{9} - \sqrt{5}) + k = \log_2 (4),
$$
so that
$$
\log_2 (\sqrt{9} - \sqrt{5}) = 2 - k.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/907687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Limit of a quotient with a radical in the numerator I have a limit but I'm so confused in how to rationalize the numerator because it has two numbers separated. How should I change the signs, please help me out.
$$\lim \limits_{x \to 6}{\sqrt{x+10}-x+2\over 3x-18}$$
| The numerator is $\sqrt{x+10} - (x-2)$, so let's multiply the numerator and denominator by $\sqrt{x+10} + (x-2)$. This gives us:
$\dfrac{\sqrt{x+10} - (x-2)}{3x-18}$
$= \dfrac{\sqrt{x+10} - (x-2)}{3x-18} \cdot \dfrac{\sqrt{x+10} + (x-2)}{\sqrt{x+10} + (x-2)}$
$= \dfrac{(x+10)-(x-2)^2}{(3x-18)(\sqrt{x+10} + (x-2))}$
$= \dfrac{-x^2+5x+6}{(3x-18)(\sqrt{x+10} + (x-2))}$
$= \dfrac{-(x+1)(x-6)}{3(x-6)(\sqrt{x+10} + (x-2))}$
Can you finish from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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help in Laplace and partial fractions Can any one teach me how to solve C2.(a) and (b) step by step?
C2. (a) Resolve $\frac{1}{s^2(s^2+s+1)}$ into partial fractions of the form $\frac{A}{s}+\frac{B}{s^2}+\frac{Cs+D}{s^2+s+1}$.
Hence, resolve $\frac{1}{s(s^2+s+1)}$ into its partial fractions.
C2. (b) Find
$$\mathcal{L}^{-1}\left\{\frac{s}{s^2+s+1}\right\}.$$
| (a) The function
$$ f(z)=\frac{1}{z^2(z^2+z+1)}$$
has a double pole in $z=0$ and a simple pole in $z=e^{\pm 2\pi i/3}$, since $z^2+z+1=\frac{z^3-1}{z-1}$. This gives:
$$ f(z)=\frac{A}{z^2}+\frac{B}{z}+\frac{C}{z-\omega}+\frac{D}{z-\omega^2}+g(z)$$
where $g(z)$ is a holomorphic function. By computing:
$$\lim_{z\to 0}z^2 f(z)=1,\quad \lim_{z\to\omega}f(z)(z-\omega)=\frac{1}{\sqrt{3}}e^{i\pi/6},\quad \lim_{z\to\omega}f(z)(z-\omega^2)=\frac{1}{\sqrt{3}}e^{-i\pi/6}$$
we have:
$$ f(z)=\frac{1}{z^2}+\frac{z}{z^2+z+1}+\frac{B}{z}+g(z) $$
where $B=-1$ since the sum of the residues must be zero.
Rearranging and checking that $g(z)=0$, we get:
$$ f(z) = \frac{1}{z^2}-\frac{1}{z}+\frac{e^{i\pi/6}}{\sqrt{3}(z-\omega)}+\frac{e^{-i\pi/6}}{\sqrt{3}(z-\omega^2)}=\frac{1}{z^2}-\frac{1}{z}+\frac{z}{z^2+z+1}.$$
Multypling by $z$ and rearranging we have:
$$ \frac{1}{z(z^2+z+1)}=\frac{1}{z}+\frac{-3+i\sqrt{3}}{6(z-\omega)}+\frac{-3-i\sqrt{3}}{6(z-\omega^2)}=\frac{1}{z}-\frac{z+1}{z^2+z+1}.$$
(b) Since
$$\frac{z}{z^2+z+1}=\frac{e^{i\pi/6}}{\sqrt{3}(z-\omega)}+\frac{e^{-i\pi/6}}{\sqrt{3}(z+\omega)}$$
and
$$\mathcal{L}^{-1}\left(\frac{1}{z-\xi}\right)=e^{s\xi},$$
by linearity it follows that
$$\color{red}{\mathcal{L}^{-1}\left(\frac{z}{z^2+z+1}\right)} = \frac{1}{\sqrt{3}}\left(e^{i\pi/6}e^{\omega s}+e^{-i\pi/6}e^{\omega^2 s}\right)=\color{red}{e^{-s/2}\left(\cos\frac{\sqrt{3}\,s}{2}-\frac{1}{\sqrt{3}}\sin\frac{\sqrt{3}\,s}{2}\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Check if two vector equations of parametric surfaces are equivalent Give the vector equation of the plane through these lines:
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}\,\,\,$ and $\,\,\,\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\0\\3\end{pmatrix}+\mu\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}$.
My answer is: $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\-1\\2\end{pmatrix}+\mu\cdot\begin{pmatrix}0\\-3\\1\end{pmatrix}$. The solutions manual suggests the following equation, which is the equation of a straight line (probably a typo): $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}+\mu\cdot\begin{pmatrix}0\\2\\1\end{pmatrix}$
Is my solution the right solution? Could someone provide a general way to check?
| Since the first equation describes a plane whereas the second one describes a straight line the equations aren't equivalent.
| {
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"timestamp": "2023-03-29T00:00:00",
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Equation $3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$ Solve the equation
$$3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$$
Having a complex root of modulus $1$.
To get the solution, I tried to take a complex root $\sqrt{\frac{1}{2}} + i \sqrt{\frac{1}{2}}$ but couldn't get the solution right. Please help me.
| Note that you have an equation with integer coefficients (real coefficients would do), so that any complex root $a$ of modulus $1$ implies a conjugate root $\bar {a }$ with $a\bar a=1$. You will therefore have a real factor of the form $x^2-bx+1$ with $b=a+\bar a$. Making obvious deductions from the leading and constant terms, there is a real factorisation of the original quartic of the form $$3x^4 + 2x^3 + 9x^2 + 4x + 6 = (x^2-bx+1)(3x^2+cx+6)$$
Equating coefficients of $x^3$ and $x$ gives $c-3b=2$ and $c-6b=4$, so that $c=0, b=-\frac 23$.
That gives two quadratic equations to be solved. [Note the factor $3$ can be redistributed to give a factorisation with integer coefficients]
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Closed-forms of infinite series with factorial in the denominator How to evaluate the closed-forms of series
\begin{equation}
1)\,\, \sum_{n=0}^\infty\frac{1}{(3n)!}\qquad\left|\qquad2)\,\, \sum_{n=0}^\infty\frac{1}{(3n+1)!}\qquad\right|\qquad3)\,\, \sum_{n=0}^\infty\frac{1}{(3n+2)!}\\
\end{equation}
Of course Wolfram Alpha can give us the closed-forms
\begin{align}
\sum_{n=0}^\infty\frac{1}{(3n)!}&=\frac{e}{3}+\frac{2\cos\left(\frac{\sqrt{3}}{2}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+1)!}&=\frac{e}{3}+\frac{2\sin\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+2)!}&=\frac{e}{3}-\frac{2\sin\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right)}{3\sqrt{e}}
\end{align}
but how to get those closed-forms by hand? I can only notice that
\begin{equation}
\sum_{n=0}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{(3n)!}+\sum_{n=0}^\infty\frac{1}{(3n+1)!}+\sum_{n=0}^\infty\frac{1}{(3n+2)!}=e
\end{equation}
Could anyone here please help me? Any help would be greatly appreciated. Thank you.
PS: Please don't work backward.
| This problem can be solved without advanced techniques. We have the Taylor series
$$
e^x = \sum_{n\geq 0} \frac{x^n}{n!}.
$$
As the questioner noted, plugging $x=1$ yields the equation $A+B+C=e$ connecting the three unknown sums. However, plugging in any cube root of unity also sheds light on the question because the numerators $x^n$ will repeat with period $3$. Let $\omega = e^{2\pi i/3}$; plugging $x = \omega$ yields $A + \omega B + \omega^2 C = e^{\omega}$, and plugging $x = \omega^2$ yields $A + \omega^2 B + \omega C = e^{\omega^2}$. We now have three equations that can be solved easily for the three unknowns $A$, $B$, and $C$. For instance, adding all three equations together yields
$$
3A = e + e^\omega + e^{\omega^2} = e + e^{-1/2}\left(\cos \frac{\sqrt{3}}{2} + i \sin \frac{\sqrt{3}}{2}\right) + e^{-1/2}\left(\cos \frac{-\sqrt{3}}{2} + i \sin \frac{-\sqrt{3}}{2}\right)
$$ $$
= e + \frac{2}{\sqrt{e}} \cos \frac{\sqrt{3}}{2},
$$
in accordance with Wolfram Alpha.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/914176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
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The line is tangent to a parabola The line $y = 4x-7$ is tangent to a parabola that has a $y$-intercept of $-3$ and the line $x=\frac{1}{2}$ as its axis of symmetry. Find the equation of the parabola. I really need help solving this question. THx
| Let the formula for the parabola be $f(x)=ax^2+bx+c$
1. y-intercept: $y$-intercept is @ $(0,-3)$ so we have: $f(0)=-3 \Rightarrow c=-3$.
2. Vertex: The vertex is @ $(x,\frac{1}{2})$ and we now that at the vertex, the derivative is equal to zero, so we have:
$f'(\frac{1}{2})=0 \Leftrightarrow 2a\cdot(\frac{1}{2})+b=0 \Rightarrow a+b=0$.
3. Line tangent to parabola: Furthermore, the line $y=4x-7$ is tangent to the parabola (and therefore $4x-7=ax^2+bx+c$ has only one solution).
Intersection point: $ax^2+bx+c=4x-7 \Rightarrow ax^2+x(b-4)+(c+7)$. Discriminant must be zero in order to have only one intersection point, so we have $D=b^2-4ac=(b-4)^2-4a(c+7)=0$.
Now, we substitute $c=-3$ and $a=-b$ and solve for $a$:
$(b-4)^2+4b(-3+7)=0 \Rightarrow (b-4)^2+16b=0 \Rightarrow b=-4$. $a=-b$ so $a=4$.
The formula is $\boxed{y=4x^2-4x-3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/915882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\int\frac{1-5\sin^2 x}{\cos^5 x\cdot \sin^2 x}dx = \frac{f(x)}{\cos^5 x}+\mathcal {C}$. Then value of $f(x)$. If $\displaystyle \int\frac{1-5\sin^2 x}{\cos^5 x\cdot \sin^2 x}dx = \frac{f(x)}{\cos^5 x}+\mathcal {C}$. Then value of $f(x)$.
$\bf{My\; Try::}$ Given $$\displaystyle \int\frac{1-5\sin^2 x}{\cos^5 x\cdot \sin ^2 x}dx = \underbrace{\int \frac{\csc^2 x}{\cos^5 x}dx}_{=I}-5\int\frac{1}{\cos^5 x}dx$$
So $$\displaystyle I = \int\frac{1}{\cos ^5 x}\cdot \csc^2 xdx = \frac{1}{\cos^5 x}\cdot -\cot x+5\int\frac{-\sin x}{\cos^6 x}\cdot -\cot xdx$$
$$\displaystyle = -\frac{\cot x}{\cos^5 x}+5\int\frac{1}{\cos^5 x}dx+\mathcal{C}$$
So $$\displaystyle \int\frac{1-5\sin^2 x}{\cos^5 x\cdot \sin^2 x}dx = -\frac{\cot x}{\cos^5 x}+\mathcal{C}=\frac{f(x)}{\cos^5 x}+\mathcal{C}$$
So $f(x) = -\cot x$.
Can we solve it any other Method, If Yes then plz explain here
Thanks
| Consider the differentiation of both sides of the integral to find $f(x)$. In this view differentiating
\begin{align}
\int \frac{1-5 \sin^{2}(x)}{\cos^{5}(x) \, \sin^{2}(x)} \, dx = \frac{f(x)}{\cos^{5}(x)} + c
\end{align}
leads to
\begin{align}
f'(x) + 5 \tan(x) \, f(x) = \csc^{2}(x) -5.
\end{align}
Now, the right-hand side of this equation can be written in such a way that
\begin{align}
f'(x) + 5 \tan(x) \, f(x) &= \frac{d}{dx}( - \cot(x)) + 5 \tan(x) \, (-\cot(x)) \\
\left[ \frac{d}{dx} + 5 \tan(x) \right] f(x) &= \left[ \frac{d}{dx} + 5 \tan(x) \right](-\cot(x))
\end{align}
which yields that $f(x) = - \cot(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/916239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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The limit of $(4-\sqrt{16-7\sin(x)})/(8x)$ at zero without using L'Hôpital I stumbled across this silly limit and I am perplexed at how I can arrive to a solution by only relying on the simplest rules of limits.
$$
\lim_{x \to 0}\frac{4-\sqrt{16-7\sin(x)}}{8x}
$$
Any help is appreciated, thanks in advance.
| $$ \lim_{x\rightarrow 0}\frac{4-\sqrt{16-7\sin x}}{8x}=\\
\lim_{x\rightarrow 0}\frac{4-\sqrt{16-7\sin x}}{8x}\frac{4+\sqrt{16-7\sin x}}{4+\sqrt{16-7\sin x}}=\\
\lim_{x\rightarrow 0}\frac{16-{16+7\sin x}}{8x}\frac{1}{4+\sqrt{16-7\sin x}}=\\
\lim_{x\rightarrow 0}\frac{{7\sin x}}{8x}\frac{1}{4+\sqrt{16-7\sin x}}=\\
\frac{7}{8}\lim_{x\rightarrow 0}\frac{{\sin x}}{x}\frac{1}{4+\sqrt{16-7\sin x}}=\\(\lim_{x\rightarrow 0}\frac{{\sin x}}{x}=1)\\
=\frac{7}{8}\lim_{x\rightarrow 0}\frac{1}{4+\sqrt{16-7\sin x}}=\\
\frac{7}{8}\lim_{x\rightarrow 0}\frac{1}{4+\sqrt{16-7(0)}}=\\
\frac{7}{8}\frac{1}{8}=\\
\frac{7}{64}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/916332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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General formula of repeated roots. Prove that $$\underbrace{\sqrt{k\sqrt{k\sqrt{k\sqrt{\cdots\sqrt{k}}}}}}_{n\text { times}}=k^{1-1/2^n}$$
How do I derive this formula?
| $$\underbrace{\sqrt{k\sqrt{k\sqrt{k\sqrt{\cdots\sqrt{k}}}}}}_{n\text { times}} = \sqrt{k}\times\sqrt[4]{k}\times \ldots \times \sqrt[2^n]{k} = k^{1/2}\times k^{1/4}\times \ldots \times k^{1/2^n} = \\ = k^{\Large\frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^n}} = k^{\Large1-\frac{1}{2^n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/919313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proof by Induction - Algebra Problem (Steps included but not understood) I do not quite understand this proof, if anyone could explain the steps for me it would be greatly appreciated. It's probably something glaringly obvious I'm not seeing, thanks in advance.
Prove that for every integer $n \ge 0,$ the number $4^{2n+1}+3^{n+2} $ is a multiple of 13.
Proof. We use induction on n, starting with $n=0$
$P(0):4^{2(0)+1}+3^{0+2}=4+9=13=13\cdot1$
Assume $P(k):4^{2k+1}+3^{k+2}=13t$ for some integer $t$. We must prove
$P(k+1): 4^{2(k+1)+1}+3^{(k+1)+2}$ is a multiple of $13$.
We have
$4^{2(k+1)+1}+3^{(k+1)+2}=4^{(2k+1)+2}+3^{(k+2)+1}$
$=4^2(4^{2k+1})+4^2(3^{k+2}-3^{k+2})+3\cdot3^{k+2}$
$=4^2(4^{2k+1}+3^{k+2})+3^{k+2}(-4^2+3)$
$=16\cdot13t+3^{k+2}\cdot(-13)$ (by $P(k)$)
$=13(16t-3^{k+2})$, proven.
| Here is a variant on the same idea. $4^{2n+1}+3^{n+2}=4(13+3)^n+9\times 3^n$. Expand the bracket by the binomial theorem. Every term has a factor 13 except $3^n$, so we have that $4^{2n+1}+3^{n+2}$ is a multiple of 13 plus $3^n(4+9)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/919380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To find the determinant of a matrix Given $A_{n\times n}$=$(a_{ij}),$ n $\ge$ 3, where $a_{ij}$ = $b_{i}^{2}$-$b_{j}^2$ ,$i,j = 1,2,...,n$ for some distinct real numbers $b_{1},b_{2},...,b_{n}$. I have to find the determinant of A. I can see that A is a skew-symmetric matrix. So determinant of A is $0$ when n is odd. But how to find it when n is even? Is the condition given for A implies anything other than A is skew-symmetric?
| Let $v := \begin{bmatrix}
\dfrac{b_2^2 - b_3^2}{b_2^2 - b_1^2} &
\dfrac{b_1^2 - b_3^2}{b_1^2 - b_2^2} &
-1 &
0 &
\ldots &
0
\end{bmatrix}^T$ if $b_1^2 \ne b_2^2$, or else
$v := \begin{bmatrix}
1 &
-1 &
0 &
\ldots &
0
\end{bmatrix}^T$. Then $Av = 0$, so $\ker A \ne 0 \implies \det A = 0$.
Proof of $Av = 0$: Let $r_j := \begin{bmatrix}b_j^2 - b_1^2 & b_j^2 - b_2^2 & \ldots & b_j^2 - b_n^2\end{bmatrix}$ be the $j^\text{th}$ row of $A$. If $b_1^2 \ne b_2^2$,
$$r_j v = (b_j^2 - b_1^2)\left(\frac{b_2^2 - b_3^2}{b_2^2 - b_1^2}\right) + (b_j^2 - b_2^2)\left(\frac{b_1^2 - b_3^2}{b_1^2 - b_2^2}\right) + (b_j^2 - b_3^2)(-1)$$
is a linear polynomial in the variable $b_j^2$ which vanishes at $b_1^2$ and $b_2^2$, hence is identically zero. If $b_1^2 = b_2^2$, then
$r_jv = (b_j^2 - b_1)^2(1) + (b_j^2 - b_2^2)(-1) = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/921700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\int \:\sqrt{1+e^x}dx$ , why I got different answers? I've got 2 steps to evaluating $\int \:\sqrt{1+e^x}dx$ which lead to different values
first step :
$\int \:\sqrt{1+e^x}dx$
let $u\:=\:\sqrt{1+e^x}$ , $du\:=\:\frac{e^x}{2\sqrt{1+e^x}}dx$ , but $e^x\:=\:u^2-1$ and substitute $u$ into $du$, i get $dx\:=\:\frac{2u}{u^2-1}du$
thus, $\int \:\sqrt{1+e^x}dx$ $=2\int \frac{u^2}{u^2-1}du\:$
apply the following algebraic property $\frac{a}{1+a}=-\frac{1}{1+a}+1$ , then the integrand becomes
$\frac{u^2}{u^2-1}=\frac{1}{\:u^2-1}+1$ ,thus
$=2\int \:\:\left(\frac{1}{u^2-1}+1\right)du\:=\:-2arctanh\left(u\right)\:+\:2u\:+C$
substitute back $u$, thus
$\int \:\sqrt{1+e^x}dx=2\sqrt{1+e^x}-2arctanh\left(\sqrt{1+e^x}\right)+C$
2nd step :
using the same substitution for $u$ , $du$, and $dx$
$\int \:\sqrt{1+e^x}dx$ $=\int \frac{2u^2}{u^2-1}du\:$ $=\int \:\left(2\:+\:\frac{1}{u-1}-\frac{1}{u+1}\right)du$
evaluating the integrand one by one, then i get
$=2u\:+ln\left(u-1\right)-ln\left(u+1\right)\:+C\:=\:2u\:+\:ln\left(\frac{u-1}{u+1}\right)\:+\:C$
substitute back $u$, thus the answer becomes
$\int \:\sqrt{1+e^x}dx=2\sqrt{1+e^x}+ln\left(\frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\right)+C$
when i do derivative to both answers, i got the same answer which is $\sqrt{1+e^x}$ , it makes me confused which step is the right one? thanks for you answer. :)
EDIT :
Thank you for the answers, now i get it that both answers are correct. :)
| Suppose $y = \tanh^{-1} x$. Then $$x = \tanh y = \frac{\sinh y}{\cosh y} = \frac{(e^y - e^{-y})/2}{(e^y + e^{-y})/2} = \frac{e^{2y} - 1}{e^{2y}+1}.$$ Hence $(e^{2y}+1)x = e^{2y}-1$, or $e^{2y}(1-x) = x+1$, or $$e^{2y} = \frac{1+x}{1-x}.$$ It follows that $$y = \frac{1}{2} \log \frac{1+x}{1-x} = \tanh^{-1}x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Polynomial factοrisation Problem
Let $P(x)=x^n+64$ be a polynomial. Find the form of the natural number $n$ such that $P(x)=P_1(x)\times P_2(x)$, $\deg P_1(x),\deg P_2(x)\geq1$.
I thought of taking $n \mod 4$. For $n=0\mod 4$ we can use Sophie Germain's identity to show that $P(x)$ is reducible but after this point I'm stuck.
Any ideas?
| Here are the the ones I found
\begin{align}
x^{12k+0}+64 =\\ (x^{2k}-2x^k+2)(x^{2k}+2x^k+2)(x^{4k}-2x^{3k}+2x^{2k}-4x^k+4)(x^{4k}+2x^{3k}+2x^{2k}+4x^k+4)\\
x^{12k+3}+64 = (x^{4k+1}+4)(x^{8k+2}-4x^{4k+1}+16)\\
x^{12k+4}+64 = (x^{6k+2}-4x^{3k+1}+8)(x^{6k+2}+4x^{3k+1}+8)\\
x^{12k+6}+64 = (x^{4k+2}+4)(x^{8k+4}-4x^{4k+2}+16)\\
x^{12k+8}+64 = (x^{6k+4}-4x^{3k+2}+8)(x^{6k+4}+4x^{3k+2}+8)\\
x^{12k+9}+64 = (x^{4k+3}+4)(x^{8k+6}-4x^{4k+3}+16),
\end{align}
and the ones with the imaginary unit are
\begin{align}
x^{12k+2}+64 = (x^{6k+1}-8i)(x^{6k+1}+8i)\\
x^{12k+10}+64 = (x^{6k+5}-8i)(x^{6k+5}+8i).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/926774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a+b+c=1$ and $abc>0$, then $ab+bc+ac<\frac{\sqrt{abc}}{2}+\frac{1}{4}.$ Question:
For any $a,b,c\in \mathbb{R}$ such that $a+b+c=1$ and $abc>0$, show that
$$ab+bc+ac<\dfrac{\sqrt{abc}}{2}+\dfrac{1}{4}.$$
My idea: let
$$a+b+c=p=1, \quad ab+bc+ac=q,\quad abc=r$$
so that
$$\Longleftrightarrow q<\dfrac{\sqrt{r}}{2}+\dfrac{1}{4}$$
Note this $a,b,c\in \mathbb{R}$, so we can't use schur inequality such
$$p^3-4pq+9r\ge 0, \quad pq\ge 9r$$
and so on
maybe can use AM-GM inequality to solve it.
| OK. Let me try to complete the answer.
We want to prove that for $a,b,c\in \mathbb{R},abc>0$,
$$ab+bc+ac<\dfrac{\sqrt{abc}}{2}+\dfrac{1}{4}. \tag{1}$$
In a previous post, it is proved for $a,b,c\ge 0$
Because $abc>0$, the only other possibility is that two of the numbers are negative and one is positive. We can assume that $a=A>0,-b=B>0,-c=C>0$.
It is then suffice to prove that
$$ab+bc+ac=AB-C(A+B)=AB-(1+A+B)(A+B)<0 \tag{2}$$
Because
$$(1+A+B)(A+B)>(A+B)(A+B)\ge 4AB > AB \tag{3}$$
We know that (2) is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 0
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Calculus limit epsilon delta Prove using only the epsilon , delta - definition
$\displaystyle\lim_{x\to 2}\dfrac{1}{x} = 0.5$
Given $\epsilon > 0 $, there exists a delta such that
$ 0<|x-2|< \delta$ implies $|(1/x) – 0.5| < \epsilon$.
Therefore, $$\begin{align}|(1/x) – 0.5| & = |(2-x)/2x| \\
& =|\frac{-(x-2)}{2(x-2+2)}| \\ & < |\frac{(x-2)}{2(x-2)} |=\frac{|x-2|}{2|x-2|}=\delta/2\delta=0.5
\end{align}$$
But epsilon is $> 0$. My $0.5$ is causing issue.
| Hint: You want the following to hold for all $x$ such that $2-\delta < x < 2+\delta$:
$\left|\dfrac{1}{x} - \dfrac{1}{2}\right| < \epsilon$
$-\epsilon < \dfrac{1}{x} - \dfrac{1}{2} < \epsilon$
$\dfrac{1}{2} -\epsilon < \dfrac{1}{x}< \dfrac{1}{2} + \epsilon$
For what range of values for $x$ is this inequality true? That will help you pick $\delta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The area not covered by six pointed star In a circle with radius $r$, two equi triangles overlapping each other in the form of a six pointed star touching the circumference is inscribed! What is the area that is not covered by the star?
Progress
By subtracting area of the star from area of circle , the area of the surface can be found! But how to calculate the area of the star?
| If the six-pointed star is regular, then the answer is $r^2(\pi-\sqrt{3})$. If it is not, then the answer can be larger, up to a limit of $r^2\big(\pi-\frac{3}{4}\sqrt{3}\big)$.
Proof
The required area is the area of the circle ($\pi r^2$) minus the area of the star.
The area of the star is the area of a large equilateral triangle (A) plus the area of three small ones. Each small one has a side-length $\frac{1}{3}$ of the large triangle's and therefore an area $\frac{1}{9}$ of its area. So the area we want is $\pi r^2 - A(1+\frac{3}{9}) = \pi r^2 - \frac{4A}{3}$.
A is equal to 6 times the area of the right-angled triangle shown. Stand it on its short side. Its area is $B=\frac{\mathrm{base * height}}{2} = \frac{r^2\sin{30^\circ}\cos{30^\circ}}{2}=r^2(\frac{1}{2})\frac{\sqrt{3}}{2}\frac{1}{2} = r^2(\frac{\sqrt{3}}{8})$. So $A=6B=\frac{3r^2\sqrt{3}}{4}$
So the required area is $\pi r^2 - \big(\frac{3r^2\sqrt{3}}{4}\big)\frac{4}{3} = r^2(\pi-\sqrt{3})$
However, all you say is that the triangles overlap to make a star. You do not define a star. If we allow the overlap such that the points on the circle are unequally spaced, then we might get a shape like this:
At the limit, the area of the star is the area of a large triangle, A, giving required area $\pi r^2 - A = r^2\big(\pi-\frac{3}{4}\sqrt{3}\big)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{1}{1^4}+\frac{1}{2^4}+\cdots+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$ I have to show: $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{n^4} \le 2-\frac{1}{\sqrt{n}}$ for natural $n$
I tried to show it by induction (but I think it could be possible to show it using some ineqaulity of means) so for $n=1$ we have $1=1$ so inequality holds then I assume it's true for $n$ and for $n+1$ my thesis is $\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{(n+1)^4} \le 2-\frac{1}{\sqrt{n+1}}$
I know that:
$\displaystyle\frac{1}{1^4}+\frac{1}{2^4}+...+\frac{1}{(n+1)^4} \le 2-\frac{1}{\sqrt{n}}+\frac{1}{(n+1)^4}$ but later I'm not sure if I have to show
$\displaystyle2-\frac{1}{\sqrt{n}}+\frac{1}{(n+1)^4}\le2-\frac{1}{\sqrt{n+1}}$ or should be $\ge$
| You want to show that
$$2-\frac{1}{\sqrt{n}}+\frac1{(n+1)^4}\le 2-\frac1{\sqrt{n+1}}.\tag{1}$$
If (1) is true, then using induction hypothesis one gets
$$1+\frac1{2^4}+\cdots+\frac{1}{n^4}+\frac{1}{(n+1)^4}\le 2-\frac1{\sqrt{n}}+\frac1{(n+1)^4}\le 2-\frac1{\sqrt{n+1}},$$
and we are done by induction.
Showing (1) should be easy since
$$\frac{1}{\sqrt{n}}-\frac1{\sqrt{n+1}}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=\frac1{\sqrt{n(n+1)}(\sqrt{n+1}+\sqrt{n})}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Gosper's Identity $\sum_{k=0}^n{n+k\choose k}[x^{n+1}(1-x)^k+(1-x)^{n+1}x^k]=1 $ The page on Binomial Sums in Wolfram Mathworld http://mathworld.wolfram.com/BinomialSums.html (Equation 69) gives this neat-looking identity due to Gosper (1972):
$$\sum_{k=0}^n{n+k\choose k}[x^{n+1}(1-x)^k+(1-x)^{n+1}x^k]=1 $$
Would anyone know if there is a simple proof of this identity without using induction?
| Consider the expression
\begin{align}
\sum_{k=0}^n{n+k\choose k}[x^{n+1}(1-x)^k+(1-x)^{n+1}x^k]
\end{align}
in the following way. First consider the generating function of the series
\begin{align}
S_{n} = \sum_{k=0}^{n} \binom{k+n}{k} \, x^{k}
\end{align}
for which it is seen that
\begin{align}
\phi(x, t) &= \sum_{n=0}^{\infty} S_{n} \, t^{n} \\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{n} \binom{n+k}{k} x^{k} t^{n} \\
&= \sum_{n,k=0}^{\infty} \binom{2k+n}{k} (xt)^{k} t^{n} \\
&= (1-4xt)^{-1/2} \sum_{n=0}^{\infty} \left( \frac{2t}{1+\sqrt{1-4xt}} \right)^{n} \\
&= \frac{1}{\sqrt{1-4xt}} \cdot \frac{1+\sqrt{1-4xt}}{1-2t + \sqrt{1-4xt}},
\end{align}
where the series
\begin{align}
\sum_{k=0}^{\infty} \binom{2k+n}{k} x^{k} = \frac{2^{n}}{\sqrt{1-4x} \, (1+\sqrt{1-4x})^{n}}
\end{align}
was used. This is equation (66) of the site referenced in the proposed problem.
Now, for $A = 1-4x(1-x)t$,
\begin{align}
\theta(x,t) &= x \phi(1-x, xt) + (1-x) \phi(x, (1-x)t) \\
&= \frac{1+\sqrt{A}}{\sqrt{A}} \left[ \frac{x}{1-2xt+ \sqrt{A}} + \frac{1-x}{1-2(1-x)t + \sqrt{A}} \right].
\end{align}
Making use of the definition of $A$ then
\begin{align}
1-2xt+\sqrt{A} = \frac{1+\sqrt{A}}{2(1-x)} \, (1-2x + \sqrt{A})
\end{align}
and leads to
\begin{align}
\theta(x,t) &= \frac{1+\sqrt{A}}{\sqrt{A}} \left[ \frac{2x(1-x)}{(1+\sqrt{A})(1-2x+\sqrt{A}) } + \frac{2x(1-x)}{(1+\sqrt{A})(1-2(1-x)+\sqrt{A})} \right] \\
&= \frac{2x(1-x)}{\sqrt{A}} \left[ \frac{1}{1-2x+\sqrt{A}} + \frac{1}{1-2(1-x)+\sqrt{A}} \right] \\
&= \frac{4x(1-x)}{-1+4x(1-x) + A} \\
&= \left(\frac{1-A}{t}\right) \, \frac{1}{\left(\frac{1-A}{t}\right) - (1-A)} \\
&= \frac{1}{1-t} = \sum_{n=0}^{\infty} t^{n}.
\end{align}
Since
\begin{align}
\theta(x,t) &= x \phi(1-x, xt) + (1-x) \phi(x, (1-x)t) \\
&= \sum_{n=0}^{\infty} \left[ \sum_{k=0}^n{n+k\choose k}[x^{n+1}(1-x)^k+(1-x)^{n+1}x^k] \, \right] \, t^{n}\\
&= \sum_{n=0}^{\infty} t^{n}
\end{align}
then by comparison of the coefficients of $\theta(x,t)$ the result follows as
\begin{align}
\sum_{k=0}^n{n+k\choose k}[x^{n+1}(1-x)^k+(1-x)^{n+1}x^k] = 1.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Why this gamma function reduces to the factorial? $$\Gamma(m+1) = \frac{1\cdot2^m}{1+m}\frac{2^{1-m}\cdot3^m}{2+m}\frac{3^{1-m}\cdot4^m}{3+m}\frac{4^{1-m}\cdot5^m}{4+m}\cdots$$
My books says that in a letter from Euler to Goldbach, this expression reduces to $m!$ when $m$ is a positive integer, but that Euler verified it only for $m=2$ and $m=3$
How can I verify it?
Also, the book shows this other form:
$$\frac{1\cdot2\cdot3\cdots n\cdot(n+1)^m}{(1+m)(2+m)\cdots(n+m)}$$
| The Product
$$
\begin{align}
\prod_{k=1}^n\frac{\color{#C00000}{k^{1-m}}\color{#00A000}{(k+1)^m}}{\color{#0000FF}{k+m}}
&=\color{#C00000}{n!^{1-m}}\color{#00A000}{(n+1)!^m}\color{#0000FF}{\frac{m!}{(n+m)!}}\\
&=n!\left(\frac{(n+1)!}{n!}\right)^m\frac{m!}{(n+m)!}\\
&=n!(n+1)^m\frac{m!}{(n+m)!}\\
&=m!\color{#00A000}{(n+1)^m}\color{#0000FF}{\frac{n!}{(n+m)!}}\\
&=m!\prod_{k=1}^m\frac{\color{#00A000}{n+1}}{\color{#0000FF}{n+k}}\\
&=m!\prod_{k=1}^m\frac{1+\frac1n}{1+\frac kn}\tag{1}
\end{align}
$$
Taking the limit as $n\to\infty$ yields
$$
\prod_{k=1}^\infty\frac{k^{1-m}(k+1)^m}{k+m}=m!\tag{2}
$$
Comment 1
Bernoulli's Inequality says that
$$
1+\frac mk\le\left(1+\frac1k\right)^m\tag{3}
$$
and each term in the product of $(2)$ is
$$
\begin{align}
\frac{k^{1-m}(k+1)^m}{k+m}
&=\frac{\left(1+\frac1k\right)^m}{1+\frac mk}\\
&\ge1\tag{4}
\end{align}
$$
Thus, $(4)$ says that each term of the product is at least $1$.
Comment 2
The Binomial Theorem says that for $k\ge1$,
$$
\begin{align}
\left(1+\frac1k\right)^m
&=1+\frac mk+\sum_{j=2}^m\binom{m}{j}\frac1{k^j}\\
&\le1+\frac mk+\frac{2^m}{k^2}\tag{5}
\end{align}
$$
Therefore, each term in the product $(2)$ is
$$
\begin{align}
\frac{k^{1-m}(k+1)^m}{k+m}
&=\frac{\left(1+\frac1k\right)^m}{1+\frac mk}\\
&\le1+\frac{2^m}{k^2}\\
&\le e^{2^m/k^2}\tag{6}
\end{align}
$$
Taking the product of $(6)$ gives
$$
\prod_{k=1}^\infty\frac{k^{1-m}(k+1)^m}{k+m}\le e^{2^m\pi^2/6}\tag{7}
$$
Thus, $(7)$ says that the product is finite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/934767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating an indefinite integral $\int\sqrt {x^2 + a^2} dx$ indefinite integral $$\int\sqrt {x^2 + a^2} dx$$
After some transformations and different substitution, I got stuck at this
$$a^2\ln|x+(x^2+a^2)| + \int\sec\theta\tan^2\theta d\theta$$
I am not sure I am getting the first step correct. Tried substituting $ x=a\tan \theta$ but that doesn't help either.
| Let $x=a\tan\theta$, so $dx=a\sec^{2}\theta d\theta$ to get
$\int\sqrt{x^2+a^2}\;dx=a^2\int\sec^{3}\theta\;d\theta$. Using integration by parts with $u=\sec\theta$ and $dv=\sec^{2}\theta\;d\theta$ gives
$\hspace{.6 in}\sec^{3}\theta\;d\theta=\frac{1}{2}\left(\sec\theta\tan\theta+\ln|\sec\theta+\tan\theta|\right)+C$,
so $\int\sqrt{x^2+a^2}=\frac{a^2}{2}\left(\frac{\sqrt{x^2+a^2}}{a}\frac{x}{a}+\ln\left|\frac{\sqrt{x^2+a^2}}{a}+\frac{x}{a}\right|\right)+C$
$\hspace{.9 in}=\frac{1}{2}\left({x\sqrt{x^2+a^2}}+a^2\ln\left(\sqrt{x^2+a^2}+x\right)\right)+C$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
How do I solve $\lim_{x\to -\infty}(\sqrt{x^2 + x + 1} + x)$? I'm having trouble finding this limit:
$$\lim_{x\to -\infty}(\sqrt{x^2 + x + 1} + x)$$
I tried multiplying by the conjugate:
$$\lim_{x\to -\infty}(\frac{\sqrt{x^2 + x + 1} + x}{1} \times \frac{\sqrt{x^2 + x + 1} - x}{\sqrt{x^2 + x + 1} - x}) = \lim_{x\to -\infty}(\frac{x + 1}{\sqrt{x^2 + x + 1} - x})$$
And multiplying by $\frac{\frac{1}{x}}{\frac{1}{x}}$
$$\lim_{x\to -\infty}(\frac{x + 1}{\sqrt{x^2 + x + 1} - x} \times \frac{\frac{1}{x}}{\frac{1}{x}}) = \lim_{x\to -\infty}(\frac{1 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x} + \frac{1}{x^2}} - 1})$$
That gives me $\frac{1}{0}$. WolframAlpha, my textbook, and my estimate suggest that it should be $-\frac{1}{2}$. What am I doing wrong?
(Problem from the 2nd chapter of Early Transcendentals by James Stewart)
| As $\;x<0\;$
$$\frac{x+1}{\sqrt{x^2+x+1}-x}\cdot\frac{-\frac1x}{-\frac1x}=\frac{-1-\frac1x}{\sqrt{1+\frac1x+\frac1{x^2}}+1}\xrightarrow[x\to-\infty]{}-\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/937182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Solving Diophantine equations involving $x, y, x^2, y^2$ My father-in-law, who is 90 years old and emigrated from Russia, likes to
challenge me with logic and math puzzles. He gave me this one:
Find integers $x$ and $y$ that satisfy both $(1)$ and $(2)$
$$x + y^2 = 8 \tag{1} $$
$$x^2 + y = 18 \tag{2}$$
I found one solution by deriving this equation:
$$ x(x+1) + y(y+1) = 26$$
which means, I need to find two numbers that add up to $26$, where each number is the product of two consecutive numbers. Through trial and error I found a solution. From college days I seem to remember that there is a more elegant way using the theory of Diophantine equations.
Can someone remind me or point to me to an easy to follow explanation of the method for solving problems like these?
| Here is an alternative way to solve for the original problem.
$$\left\{\begin{array}{ll}x+y^2&=&8\\x^2+y&=&18\end{array}\right. \tag{1}$$
Instead of adding, subtract the two equations and get
$$x^2-y^2+y-x=10\Rightarrow\\(x-y)(x+y-1)=10$$
For two integers $a$ and $b$ to have product $10$, list all possibilities. $$1\times 10,2\times 5,2\times 5, 10\times 1,-1\times -10,-2\times -5,-2\times -5, -10\times -1, $$
Now for each $a$ and $b$, obtain the system $$\left\{\begin{array}{lll}x-y&=&a\\x+y-1&=&b\end{array}\right. \tag{2}$$
This is a linear equation and has solution $x=\dfrac{a+b+1}{2}$ and $y=\dfrac{b-a+1}{2}$ (if instead of $10$, say, we have $16$; use the constraint that $x$ and $y$ are integers to eliminate certain choices. In particular, we can eliminate cases where $a$ and $b$ have same parity because $a+b+1$ and $a-b+1$ have to be even).
Plug in values of $a$ and $b$ obtained from $(2)$ to get possible values for $x$ and $y$ (that are integers). And recheck the values for $x$ and $y$ so obtain matches with the original system (because we are essentially solving $(2)$ and not $(1)$, and we haven't shown that $(2)$ is necessary and sufficient condition for $(1)$ to hold).
$$\begin{array}{rr|rr}
a&b&x&y\\ \hline
1&10&6&5\\
2&5&4&2\\
5&2&2&4\\
-1&-10&-5&-4\\
-2&-5&-3&-1\\
-5&-2&-1&-3\\
-10&-1&-4&-5
\end{array}$$
See that $(4,2)$ is the only possible answer for $(x,y)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/937891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
How find this integral $\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$ let $$D=\{(x,y)|y\ge x^3,y\le 1,x\ge -1\}$$
Find the integral
$$I=\dfrac{1}{2}\iint_{D}(x^2y+xy^2+2x+2y^2)dxdy$$
My idea:
$$I=\int_{0}^{1}dx\int_{x^3}^{1}(x^2y+2y^2)dy+\int_{-1}^{0}dx\int_{0}^{-x^3}(xy^2+2x+2y^2)dy$$
so
$$I=\int_{0}^{1}[\dfrac{1}{2}x^2y^2+\dfrac{2}{3}y^3]|_{x^3}^{1}dx+\int_{-1}^{0}[\dfrac{1}{3}xy^3+2xy+\dfrac{2}{3}y^3]|_{0}^{-x^3}dx$$
$$I=\int_{0}^{1}[\dfrac{1}{2}x^2+\dfrac{2}{3}-\dfrac{1}{2}x^8-\dfrac{2}{3}x^9]dx+\int_{-1}^{0}[-\dfrac{1}{3}x^{10}-2x^4-\dfrac{2}{3}x^9]dx$$
so
$$I=\dfrac{5}{6}-\dfrac{1}{18}-\dfrac{2}{30}+\dfrac{1}{33}+\dfrac{1}{10}-\dfrac{2}{30}=\dfrac{67}{90}$$
My question: my reslut is true? can you someone can use computer find it value?
because I use Tom methods to find this reslut is
$$\dfrac{79}{270}$$
which is true? so someone can use maple help me?Thank you
| The domain $D$ looks roughly like a right triangle $ABC$ with the right angle $B$ at $(-1,1)$, $A$ at $(-1,-1)$, $C$ at $(1,1)$ and a curve $y=x^3$ instead of a straight line from $A$ to $C$. Since the curve does not do anything tricky (one value of $x$ maps to one value of $y$ and vice-versa) you can do this as a single integral, integrating either $x$ or $y$ first.
Your tactical choice of breaking the integral into two regions separated by the Y axis led you to make a simple error in the limits in the second region, which should still have been $\int_{x^3}^{1}$. You also left out some of the terms in the integrand.
The correct answer is
$$
I = \int_{x=-1}^{1} dx \int_{y=x^3}^{1}\left( x^2y + xy^2 + 2x + 2y^2 \right) dy
$$
$$
I = \int_{x=-1}^{1} dx \left[ \frac{1}{2}y^2 x^2 + \frac{1}{3} y^3 x + 2yx + \frac{2}{3} y^3 \right]_{x^3}^1
$$
$$
I = \int_{x=-1}^{1} dx \left[ \frac{1}{2}x^2 + \frac{1}{3} x + 2x + \frac{2}{3}
- \frac{1}{2}x^8 - \frac{1}{3} x^{10} - 2x^4 - \frac{2}{3}x^9
\right]
$$
$$
I = 2 \left[ \frac{1}{6} + 0 + 0 + \frac{2}{3}
- \frac{1}{18} - \frac{1}{33} - \frac{2}{5} - 0
\right]
$$
(here, we use the fact that the integral from $-1$ to $1$ of an odd power of $x$ is zero to drop terms with $x$ and $x^9$)
$$
I = \frac{344}{495}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/937955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Closed Form of Recursion $a_n = \frac{6}{a_{n-1}-1}$ Given that $a_0=2$ and $a_n = \frac{6}{a_{n-1}-1}$, find a closed form for $a_n$.
I tried listing out the first few values of $a_n: 2, 6, 6/5, 30, 6/29$, but no pattern came out.
| I think that Semiclassical proposed a very nice solution rewriting $$a_n = \frac{6}{a_{n-1}-1}$$ $$\dfrac{1}{a_n+2}=\dfrac{1}{2}-\dfrac{3/2}{a_{n-1}+2}$$ So, let us define $$b_n=\dfrac{1}{a_n+2}$$ (with $b_0=\dfrac{1}{4}$); so the recurrence equation is simply $$b_n=\dfrac{1}{4}-\dfrac{3}{2}b_{n-1}$$ from which $$b_n=\frac{1}{20} \left(4+\left(-\frac{3}{2}\right)^n\right)$$ and then $$a_n=\frac{20}{4+\left(-\frac{3}{2}\right)^n}-2$$ as shown by Git Gud.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/939725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Probably of 2 six in 5 dice rolls What is the probability of obtaining exatcly 2 six when rolling a dice 5 times?
In order to obtain this probability, I will need to devide the number of favorable events by the number of possible events. I believe the denominator is $6^5$. But I am having troubles figuring out the numerator.
| The generating function for the five throws is
$$ \left(\frac{s}{6}+\frac{5 o}{6}\right)^5 $$ where o are the other faces (1,2,3,4,5) and s represents the six. You need the coefficient of $$ s^2o^3 $$. This you get from Pascals triangle which tells us that the expansion of $$(a+b)^5$$ is
$$(a+b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5 $$
So $$ a = \frac{5}{6} $$ and $$ b = \frac{1}{6} $$ gives us using the third term
$$ 10 \left(\frac{5}{6}\right)^3 \left(\frac{1}{6}\right)^2 = \frac{625}{3888} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/945277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove $\frac{n^2+2}{(2 \cdot n^2)-1} \to \frac{1}{2}$
Prove $\frac{n^2+2}{(2 \cdot n^2)-1} \to \frac{1}{2}$ for $n \to \infty$.
I've been looking at this for hours! Also, sorry I don't have the proper notation.
This is where I'm at:
$$
\left| \frac{n^2 + 2}{2 \cdot n^2 - 1} - \frac{1}{2}\right| = \left| \frac{5}{4 \cdot n^2 - 2} \right|
$$
I thought I was supposed to get to the point where I can say $1/n <n$, but I can only get to $1/n-1$ so that can't be the right approach or I'm missing something.
A friend says to make $n > \frac{5}{\varepsilon^2}$ but I''m not sure what to do with that tip.
Any help would be greatly appreciated!
| You might find it easier to write
$$\dfrac{n^2+2}{2n^2-1} = \dfrac{1}{2} \times \dfrac{2n^2+4}{2n^2-1} = \dfrac{1}{2} + \dfrac{5}{4n^2-2}$$ and note you can make $\dfrac{5}{4n^2-2}$ as small as you want by making $n$ large enough.
If you must use epsilon-delta or epsilson-$N_0$, then note that $\dfrac{5}{4n^2-2} \lt \varepsilon $ if $\dfrac5\varepsilon \lt 4n^2 -2$ which is true if $n \gt \sqrt{\dfrac12+\dfrac{5}{4\varepsilon}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/947452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I solve this geometric series I have this geometric series $2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{128}$to solve. So I extract the number two and get $2(\frac{1}{2}^0+ \frac{1}{2}^1+...+ \frac{1}{2}^7)$
I use the following formula $S_n= \frac{x^{n+1}-1}{x-1}$ so I plug in the values in this formula and get $S_n= 2\frac{\frac{1}{2}^{7+1}-1}{\frac{1}{2}-1}$ but the result is not correct.
What did I do wrong?
Thanks!!
| $$S=2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{128}$$
Now if we subtract $2$ from both sides of the equation we get
$$S-2=1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{128}$$
$$S-2=\frac{1}{2}\cdot \left(2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{64}\right)$$
Now notice that $2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{64}$ is $S-\frac1{128}$ so:
$$2S-4=S-\frac1{128} \rightarrow S=4-\frac1{128}=\frac{511}{128}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/948146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral of voltage, $\int_{-a}^a \frac{dy}{\sqrt{x^2 + y^2}}$ This is (probably) a very easy integral to solve, but for some reason the answer just isn't coming to me (or at least the one my professor got isn't). He gave us a formula for voltage along the x-axis from a charged wire of length 2a, but none of the details are particularly important to my question. The integral I'm looking to solve is: $$ \int_{-a}^a \frac{dy}{\sqrt{x^2 + y^2}}$$
The answer my professor got is $$\ln\left(\frac{\sqrt{a^2+x^2}+a}{\sqrt{a^2+x^2}-a}\right) $$
and I'm not entirely sure how. Can someone go over this integral step-by-step so I know what I'm missing?
| The ideal substitution in this case would be $\sqrt{x^2+y^2}+y=t$. The substitution relation gives $t$ as a function of $y$, but we also need to know $y$ as a function of $t$. We can isolate $y$ as follows:
$$\begin{align}
\sqrt{x^2+y^2}+y&=t\\
\sqrt{x^2+y^2}&=t-y\\
x^2+y^2&=(t-y)^2\\
x^2+y^2&=t^2-2ty+y^2\\
x^2&=t^2-2ty\\
2ty&=t^2-x^2\\
y&=\frac{t^2-x^2}{2t}.
\end{align}$$
Differentiating, we also find:
$$\mathrm{d}y=\frac{x^2+t^2}{2t^2}\,\mathrm{d}t.$$
Thus,
$$\begin{align}
\int_{-a}^{a}\frac{\mathrm{d}y}{\sqrt{x^2+y^2}}
&=\int_{\sqrt{x^2+a^2}-a}^{\sqrt{x^2+a^2}+a}\frac{2t}{x^2+t^2}\cdot\frac{x^2+t^2}{2t^2}\,\mathrm{d}t\\
&=\int_{\sqrt{x^2+a^2}-a}^{\sqrt{x^2+a^2}+a}\frac{\mathrm{d}t}{t}\\
&=\ln{\left(\sqrt{x^2+a^2}+a\right)}-\ln{\left(\sqrt{x^2+a^2}-a\right)}\\
&=\ln{\left(\frac{\sqrt{x^2+a^2}+a}{\sqrt{x^2+a^2}-a}\right)}.~~\blacksquare
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int\frac{2x+1}{x^2+2x+5}dx$ by partial fractions $$\int\frac{2x+1}{x^2+2x+5}dx$$
I know I'm supposed to make the bottom a perfect square by making it $(x+1)^2 +4$ but I don't know what to do after that. I've tried to make $x+1= \tan x$ because that's what we did in a class example but I keep getting stuck.
| $$\frac{d}{dx} \ln(x^2 + 2x + 5) = \frac{2x + 2}{x^2 + 2x + 5}dx$$
$$\int\,\left(\dfrac{2x+2}{x^2 + 2x + 5}- \dfrac{1}{x^2 + 2x + 5}\right)\,dx = \ln(x^2 + 2x + 5) + C - \int\,\dfrac{1}{x^2+2x+5}dx$$
=
Let $u=x+1$, so $du = dx$, $x = u - 1$, and $$x^2 + 2x + 5= u^2 - 2u + 1 + 2u - 2 + 5 = u^2 + 4$$
$$\begin{align}
\int\,\dfrac{1}{x^2+2x+5}dx & = \int\,\dfrac{1}{u^2 + 4}du\\
& = \frac{\arctan\left(\frac{u}2\right)}{2} + C\\
& = \frac{\arctan\left(\frac{x+1}{2}\right)}{2} + C\\
\end{align}$$
$$\int\,\dfrac{2x+1}{x^2+2x+5}\,dx = \ln(x^2 + 2x + 5) - \frac{\arctan\left(\frac{x+1}{2}\right)}{2} + C$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $a+b+c=4$ find $\max(ab+ac+bc)$ $a+b+c = 4$.
What is the maximum value of $ab+ac+bc$?
Could this be solved by a simple application of Jensen's inequality? If so, I am unsure what to choose for $f(x)$. If $ab+ac+bc$ is treated as a function of $a$ there seems no easy way to express $bc$ in terms of $a$.
EDIT: The context of the question is maximising the surface area of a rectangular prism. Also I might have misinterpreted the question, because it says "the sum of the length of the edges (side lengths are a,b,c) is 4", and gives the options $\frac1{3}, \frac{2}{3}, 1, \frac{4}{3}$.
Otherwise, how would this be done?
| @James, first we square both sides:
$$(a+b+c)^2=1 \implies 1=a^2+b^2+c^2+2(ab+bc+ac)\tag{1}$$
$$\because a^2+b^2+c^2 \geq ab +ac + bc\tag{2}$$
For a,b,c>0
$$ \\ 1 \geq 3(ab+bc+ac) \tag{3}$$ $$ \frac{1}{3} \geq ab +bc+ca\tag{4} \\ \square$$
$(2)$ can be easily proven by considering: $(a-b)^2 \geq 0 \\ (b-c)^2 \geq 0 \\ (a-c)^2 \geq 0 $, and adding the 3 inequalities and rearranging.
$(3)$ follows by using $(2)$ to replace $a^2+b^2+c^2$ in $(1)$ with the smaller $ab+ac+bc$.
Surface area is $2(ab+bc+ca)$, so maximum surface area is simply $\frac{2}{3}$
| {
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Prove that $ \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x $ In a related question the following integral was evaluated
$$
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
=\int_0^{\pi} \frac{\mathrm{d}x/2}{1 + \cos x \sin x}
=\int_0^{2\pi} \frac{\mathrm{d}x/2}{2 + \sin x} \,\mathrm{d}x
=\int_{-\infty}^\infty \frac{\mathrm{d}x/2}{1+x+x^2}
$$
I noticed something interesting, namely that
$$
\begin{align*}
\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x
& = \int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \\
& = \int_0^{\pi} \frac{(\cos x)^2}{1 - \cos x \sin x} \,\mathrm{d}x
= \int_0^{\pi} \frac{(\sin x)^2}{1 - \cos x \sin x} \,\mathrm{d}x
\end{align*}
$$
The same trivially holds if the upper limits are changed to $\pi/2$ as well ($x \mapsto \pi/2 -u$).
But I had problems proving the first equality. Does anyone have some quick hints?
| The integrands are both periodic with period $\pi$, so it suffices to verify the identity over any interval of length $\pi$, including $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Then, since both integrands are even functions, it suffices to check that the identity holds when integrating over $\left[0, \frac{\pi}{2}\right]$, that is, that
$$ \int_0^{\frac{\pi}{2}} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\frac{\pi}{2}} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x.$$
But one can show this simply by substituting $x = \frac{\pi}{2} - u$ on either side.
| {
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Any shorter way to solve trigonometric problem? If $10 \sin^4\theta + 15 \cos^4 \theta=6$, then find value of $27 \csc^2 \theta + 8\sec^2 \theta$
I know the normal method o solve this problem in which we need to multiply L.H.S. of $10 \sin^4\theta + 15 \cos^4 \theta=6$ by $(\sin^2\theta + \cos \theta^2)^2$ and then simplifying it. It is kind of simplification in which we can commit silly mistakes.
So, is there any shorter way to solve this trigonometric problem?
| Following your way:
$$10 \sin^4\theta + 15 \cos^4 \theta=6(\sin^2\theta + \cos \theta^2)^2$$
$$\iff4(\sin^2\theta)^2-12\sin^2\theta\cos^2\theta+9(\cos^2\theta)^2=0$$
Dividing either sides by $\cos^4 \theta,$
$$4(\tan^2\theta)^2-12\tan^2\theta+9=0\iff(2\tan^2\theta-3)^2=0$$
We can find $\sec^2\theta=1+\tan^2\theta, \cot^2\theta=\dfrac1{\tan^2\theta},\csc^2\theta=1+\cot^2\theta$
Hope you can take it home from here
| {
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Does an elementary solution exist to $x^2+1=y^3$? Prove that there are no positive integer solutions to $$x^2+1=y^3$$
This problem is easy if you apply Catalans conjecture and still doable talking about Gaussian integers and UFD's. However, can this problem be solved using pre-university mathematics?
I am talking about elementary number theoretical solutions. Do they exist?
| An alternative proof and a generalisation
As stated in the accepted answer, it is sufficient to prove that $3a^4+3a^2+1=b^2$ has no positive integer solution. With a change of notation we shall prove the following generalisation by using the same method used for Does the equation $y^2=3x^4-3x^2+1$ have an elementary solution?
Theorem
No equation of either of the forms
$$ Ax^4-6x^2y^2+Cy^4=z^2,AC=-3\tag{1}$$ $$ ax^4+3x^2y^2+cy^4=z^2,ac=3\tag{2}$$
has a positive integer solution.
Proof
First note that, for either equation, we can suppose that $x,y,z$ are pairwise coprime since a common factor of any pair of $x,y,z$ would be a factor of all and cancellation can occur.
Also note that precisely one of $A$ and $C$ is divisible by $3$. Without loss of generality we can suppose that $3$ is a factor of $C$ and that $A\in \{\pm 1\}.$ Then $z^2\equiv A\pmod 3$ and so $A=1$. Similarly, we can suppose $a=1$.
An equation of form (1)
$x^4-6x^2y^2-3y^4=z^2$ can be rewritten, using completing the square, as $$\left (\frac{x^2-3y^2-z}{2}\right )\left (\frac{x^2-3y^2+z}{2}\right)=3y^4.$$
Since the two bracketed factors, $L$ and $M$ say, differ by the integer $z$ and have integer product, they are both integers. Furthermore, if $q$ is a prime common factor of $L$ and $M$, then $q$ would be a factor of both $z$ and $y$, a contradiction.
Therefore $\{L,M\}=\{au^4,cv^4\}$, where $ac=3$ and $y=uv$, with $u$ and $v$ coprime. Then $$au^4+cv^4=x^2-3y^2=x^2-3u^2v^2.$$ Therefore $au^4+3u^2v^2+cv^4=x^2$, $ac=3$, an equation of form (2).
It is important to note that the mapping $(x,y,z)\rightarrow (u,v,w)$ is invertible. Only one solution set can map to $(u,v,w)$ by this process.
An equation of form (2)
Let $u,v,x$ be a pairwise coprime solution of $ u^4+3u^2v^2+3v^4=x^2$ and let $t$ be the greatest common divisor of $v$ and $2$. Then $(U,V,W)=(\frac{2u}{t},\frac{v}{t},\frac{4x}{t^2})$ is a pairwise coprime solution of $U^4+12U^2V^2+48V^4=W^2$.
This can be rewritten, using completing the square, as $$\left (\frac{U^2+6V^2-W}{2}\right )\left (\frac{U^2+6V^2+W}{2}\right)=-3V^4.$$
The bracketed factors, $L$ and $M$, are again coprime integers.Therefore $\{L,M\}=\{aX^4,cY^4\}$, where $ac=-3$ and $V=XY$, with $X$ and $Y$ coprime. Then $$aX^4-6X^2Y^2+cY^4=U^2,ac=-3,$$ an equation of form (1). Again, this mapping of solutions is invertible.
Conclusion
We have seen that any positive integer solution $(x,y,z)$ of an equation of form (1) leads to another positive integer solution $(X,Y,Z)$, where $$Y=\frac{y}{tuX}.$$
Any monotonically strictly decreasing series of positive integers must terminate and so the above process must lead to solutions with $tuX=1$.
It is now straightforward to plug $u=X=1$ and thus $U=2$ back into the equations to obtain the contradiction
$$1-6Y^2-3Y^4=4.$$
| {
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Show that $x(\pi - x)= \frac{\pi^2}{6}-\sum_{k=1}^{\infty} \frac{\cos(2kx)}{k^2}$ Show that
$$x(\pi - x)= \frac{\pi^2}{6}-\sum_{k=1}^{\infty} \frac{\cos(2kx)}{k^2}$$
for $ 0<x<\pi$
My idea:
I've defined the periodic function
$$f(x) = 0 \text{ if } x \in [- \pi, 0) \text{ and }$$
$$f(x)=x(\pi-x) \text{ if } x \in [0, \pi)$$
and make the Fourier Series of f.But, imediatly we have
$$a_o = \frac{1}{\pi} \int_0^{\pi} x(\pi -x) dx = \frac{\pi^2}{6}$$
So I am seeing a problem since we should use $ a_o/2. $Also,
$$b_n = \frac{1}{\pi} \int_0^{\pi} x(\pi-x) sin(nx) dx = \frac{-n \pi \sin(n \pi) - 2 cos(n \pi) +2}{n^3 \pi} = \frac{4}{n^3 \pi} $$
If $ n = 2k+1$. So the series term with $\sin(n x) $ cannot be zero.
Since the f is continuous over $ (0, \pi)$, I thought the series should converge tl the given function.What am I doing wrong?
P.s: for the cosine part, I've obtained exactly the series part given an answer.
P.S.S: we usethat the fourier series is
$$SF_f = \frac{a_o}{2} + \sum_{n=1}^{\infty} a_n \cos(nx) + b_n \sin(nx)$$
Edit:
I've found this searching about Fourier series. Since the function f is even on $(0,\pi$), does the theorem holds or it must be even in $R$?
Edit 2:
$$f(x) = -x(\pi+x) \text{ if } x \in [- \pi, 0) \text{ and }$$
$$f(x)=x(\pi-x) \text{ if } x \in [0, \pi)$$
Is even and the equality, now, holds.
| Target function
$$
f(x) = x \left(\pi - x \right)
$$
To reproduce the answer, use an even reflection of the function:
$$
\begin{align}
r(x) &= x \left(\pi - x \right), \quad x\ge 0, \\
l(x) &= -x \left(\pi + x \right), \quad x < 0.
\end{align}
$$
Fourier amplitudes
$$
\begin{align}
%
a_{0} &= \frac{1}{\pi} \left(
\int_{-\pi}^{0} l (x) \, dx +
\int_{0}^{\pi} r (x) \, dx
\right)
= \frac{\pi ^3}{3} \\[5pt]
%
a_{k} &= \frac{1}{\pi} \left(
\int_{-\pi}^{0} l (x) \, dx +
\int_{0}^{\pi} r (x) \, dx
\right) = -\frac{2 \left((-1)^k+1\right)}{k^2}
\\[5pt]
%
b_{k} &= \frac{1}{\pi} \left(
\int_{-\pi}^{0} l (x) \, dx +
\int_{0}^{\pi} r (x) \, dx
\right) = 0
%
%
\end{align}
$$
The first few terms in the series look like
$$
\begin{array}{rr}
k & a_{k} \\\hline
1 & 0 \\
2 & -1 \\
3 & 0 \\
4 & -\frac{1}{4} \\
5 & 0 \\
6 & -\frac{1}{9} \\
\end{array}
$$
Fourier Series
$$
\begin{align}
f(x)
&= \frac{a_{0}}{2} + \sum_{k=1}^{\infty} \frac{-2 \left((-1)^k+1\right)}{k^2}\cos \left( kx \right) \\[3pt]
&= \frac{\pi^{3}}{6} - \sum_{k=1}^{\infty} k^{-2} \cos \left( 2kx \right)
\end{align}
$$
Building the approximation
| {
"language": "en",
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Sum the series (real analysis) $$\sum_{n=1}^\infty {1 \over n(n+1)(n+2)(n+3)(n+4)}$$
I tried to sum the above term as they way I can solve the term $\sum_{n=1}^\infty {1 \over (n+3)}$ by transforming into ${3\over n(n+3)} ={1\over n}-{1\over(n+3)}$ but I got stuck while trying to transform $12\over n(n+1)(n+2)(n+3)(n+4)$ into something solvable.
| Hint
Notice that
$${1 \over n(n+1)(n+2)(n+3)(n+4)}=\frac14{(n+4 )-n\over n(n+1)(n+2)(n+3)(n+4)}\\=\frac14\left({1\over n(n+1)(n+2)(n+3)}-{1\over (n+1)(n+2)(n+3)(n+4)}\right)$$
and then telescope.
| {
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How did we find the solution? In my lecture notes, I read that "We know that $$x^2 \equiv 2 \pmod {7^3}$$ has as solution $$x \equiv 108 \pmod {7^3}$$"
How did we find this solution?
Any help would be appreciated!
| For example, $3^2=2\bmod 7$. So $(3+7k)^2=2+7+42k\bmod 7^2$. So $k=1$ and $10^2=2\bmod 7^2$. So $(10+7^2h)^2=2+2\cdot7^2(1+10h)\bmod 7^3$. So $h=2$. Bother! Is the crummy formatting my fault or the iPhone 6+ ?
| {
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Sums with squares of binomial coefficients multiplied by a polynomial It has long been known that
\begin{align}
\sum_{n=0}^{m} \binom{m}{n}^{2} = \binom{2m}{m}.
\end{align}
What is being asked here are the closed forms for the binomial series
\begin{align}
S_{1} &= \sum_{n=0}^{m} \left( n^{2} - \frac{m \, n}{2} - \frac{m}{8} \right) \binom{m}{n}^{2} \\
S_{2} &= \sum_{n=0}^{m} n(n+1) \binom{m}{n}^{2} \\
S_{3} &= \sum_{n=0}^{m} (n+2)^{2} \binom{m}{n}^{2}.
\end{align}
| Lemma:
$$
\begin{align}
\sum_{n=0}^m\binom{n}{k}\binom{m}{n}^2
&=\sum_{n=0}^m\binom{n}{k}\binom{m}{n}\binom{m}{m-n}\tag{1}\\
&=\sum_{n=0}^m\binom{m}{k}\binom{m-k}{n-k}\binom{m}{m-n}\tag{2}\\
&=\binom{m}{k}\binom{2m-k}{m-k}\tag{3}\\
&=\binom{m}{k}\binom{2m-k}{m}\tag{4}\\
&=\binom{2m-k}{k}\binom{2m-2k}{m-k}\tag{5}
\end{align}
$$
Explanation:
$(1)$: $\binom{m\vphantom{k}}{n}=\binom{m\vphantom{k}}{m-n}$
$(2)$: $\binom{n\vphantom{k}}{k}\binom{m\vphantom{k}}{n}=\binom{m\vphantom{k}}{k}\binom{m-k}{n-k}$
$(3)$: Vandermonde's Identity$\vphantom{\binom{k}{n}}$
$(4)$: $\binom{m\vphantom{k}}{n}=\binom{m\vphantom{k}}{m-n}$
$(5)$: $\binom{n\vphantom{k}}{k}\binom{m\vphantom{k}}{n}=\binom{m\vphantom{k}}{k}\binom{m-k}{n-k}$
Apply the Lemma to
$$
\color{#C00000}{n^2}-\frac{m}2\color{#00A000}{n}-\frac{m}8\color{#0000FF}{1}=\color{#C00000}{2\binom{n}{2}+\binom{n}{1}}-\frac{m}2\color{#00A000}{\binom{n}{1}}-\frac{m}8\color{#0000FF}{\binom{n}{0}}
$$
and
$$
n(n+1)=2\binom{n}{2}+2\binom{n}{1}
$$
and
$$
(n+2)^2=2\binom{n}{2}+5\binom{n}{1}+4\binom{n}{0}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/957605",
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Calculation of $\int\frac1{\tan \frac{x}{2}+1}dx$ Calculation of $\displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$
$\bf{My\; Try}::$ Let $\displaystyle I = \displaystyle \int\frac{1}{\tan \frac{x}{2}+1}dx$, Now let $\displaystyle \tan \frac{x}{2}=t\;,$ Then $\displaystyle dx=\frac{2}{1+t^2}dt$
So $\displaystyle I = 2\int\frac{1}{(1+t)\cdot (1+t^2)}dt$
Now Using Partial fraction,
$\displaystyle \frac{1}{(1+t)\cdot (1+t^2)} = \frac{A}{1+t}+\frac{Bt+C}{1+t^2}\Rightarrow 1=A(1+t^2)+(Bt+C)(1+t)$
Now put $(1+t)=0\Rightarrow t=-1\;,$ We get $\displaystyle 1=2A\Rightarrow A = \frac{1}{2}.$
Now Put $(1+t^2)=0\Rightarrow t^2=-1\;,$ We Get $1=Bt^2+(B+C)t+C$
So $\displaystyle 1=\left(-B+C\right)+(B+C)$. So Solving equation...$B+C=0$ and $-B+C=1$
So $\displaystyle B=-\frac{1}{2}$ and $\displaystyle C=\frac{1}{2}$
So $\displaystyle I = 2\int\frac{1}{(1+t)\cdot (1+t^2)}dt = \int\frac{1}{1+t}dt+\int\frac{-t+1}{1+t^2}dt$
So $\displaystyle I = \frac{1}{1+t}dt-\frac{1}{2}\int\frac{2t}{1+t^2}dt+\int \frac{1}{1+t^2}dt$
So $\displaystyle I = \ln \left|1+t\right|-\frac{1}{2}\ln \left|1+t^2\right|+\tan^{-1}(t)+\mathcal{C}$
So $\displaystyle I = \ln \left|1+\tan \frac{x}{2}\right|-\frac{1}{2}\ln \left|1+\tan^2 \frac{x}{2}\right|+\frac{x}{2}+\mathcal{C}$
Can we solve it without using partial fraction? If yes then please explain to me.
Thanks
| Partial fractions can be avoided, but we need to integrate both $\sec x$ and $\tan x$:
\begin{align}
\frac{1}{\tan(x/2)+1} & = \frac{1}{\tan(x/2)+1} \cdot \frac{\cos(x/2)}{\cos(x/2)} \cdot \frac{\cos(x/2)-\sin(x/2)}{\cos(x/2)-\sin(x/2)} \\[0.1in]
& = \frac{\cos(x/2)}{\sin(x/2)+\cos(x/2)} \cdot \frac{\cos(x/2)-\sin(x/2)}{\cos(x/2)-\sin(x/2)} \\[0.1in]
& = \frac{\cos(x/2)\big(\cos(x/2)-\sin(x/2)\big)}{\cos^2(x/2)-\sin^2(x/2)} \\[0.1in]
& = \frac{\cos^2(x/2)-\cos(x/2)\sin(x/2)}{\cos x} \\[0.1in]
& = \frac{\frac{1}{2}(\cos x+1)-\frac{1}{2}\sin x}{\cos x} \\[0.1in]
& = \frac{1}{2} \bigg(1+\sec x - \tan x\bigg).
\end{align}
| {
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How to derive the closed form of this recurrence? For the recurrence, $T(n) = 3T(n-1)-2$, where $T(0)= 5$, I found the closed form to be $4\cdot 3^n +1$(with help of Wolfram Alpha). Now I am trying to figure it out for myself. So far, I have worked out: $T(n-1) = 3T(n-2)-2, T(n-2) = 3T(n-3)-2, T(n-3) = 3T(n-4)-2$ leading me to: $T(n)=81\cdot T(n-4)-54-18-6-2 $ etc. I have noticed the constants follow a Geometric Series whose sum is given by $1-3^n$. I am having trouble putting this together to end up with the final closed form.
| There are many ways to solve such linear recurrences. We describe one of them.
Let the $n$-th term be $T_n$. We are told that
$$T_n=3T_{n-1}-2.$$
It would be nice to get rid of the constant term $-2$. Let $T_n=U_n+d$, where we will choose $d$ soon. Then
$$U_n+d=3(U_{n-1}+d)-2.$$
The constant term disappears if we choose $d=1$, and we obtain the recurrence
$$U_n=3U_{n-1}.$$
Since $T_0=5$, it follows that $U_0=4$, and therefore $U_n=4\cdot 3^n$. It follows that $T_n=4\cdot 3^n+1$.
Remark: We analyze along your lines. We have $T_n=3T_{n-1}-2$. Thus $$T_n=3(3T_{n-2}-2)-2=3^2T_{n-2}-3.\cdot 2-2.$$ Substituting again, we get $$T_n=3^3T_{n-2}-3^2\cdot 2-3\cdot 2-2,$$ and then
$$T_n=3^4T_{n-4}-3^3\cdot 2-3^2\cdot 2-3\cdot 2-2,$$ and so on. Ultimately, we end up with
$$T_n=3^nT_0 -2(1+3+3^2+\cdots +3^{n-1}).$$
The sum of the geometric series $1+3+\cdot +3^{n-1}$ is $\frac{3^n-1}{3-1}$.
Using the fact that $T_0=5$, we find that
$$T_n=3^n \cdot 5-(3^n-1)=4\cdot3^n+1.$$
| {
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How find this sum $\frac{1}{1^2}+\frac{2}{2^2}+\frac{2}{3^2}+\frac{3}{4^2}+\frac{2}{5^2}+\frac{4}{6^2}+\cdots+\frac{d(n)}{n^2}+\cdots$ Question:
Find the value
$$\dfrac{1}{1^2}+\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{3}{4^2}+\dfrac{2}{5^2}+\dfrac{4}{6^2}+\cdots+\dfrac{d(n)}{n^2}+\cdots$$
where $d(n)$ is The total number of positive divisors of $n$
I think we can use
$$\zeta{(2)}=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\cdots+\dfrac{1}{n^2}+\cdots=\dfrac{\pi^2}{6}$$
I know If the prime factorization of is given by
$$n=p^{a_{1}}_{1}\cdot p^{a_{2}}_{2}\cdots p^{a_{n}}_{n}$$
then the number of positive divisors of is
$$d(n)=(a_{1}+1)(a_{2}+1)\cdots(a_{n}+1)$$
But follow is very ugly,I don't understand @Nate idea(my English is poor),can you post detail? Thank you
| Let $p_1, p_2, \ldots$ be the list of primes and
$\mathcal{E}$ be the collection of non-negative integer sequences
$( e_k )_{k\in \mathbb{Z}_{+}}$ with finitely many non-zero terms.
Evey $n \in \mathbb{Z}_{+}$ can be represented as a product of the form
$\prod\limits_{k=1}^\infty p_k^{e_k}$ for an unique $e \in \mathcal{E}$.
We have
$$\sum_{n\in\mathbb{Z}_{+}} \frac{d(n)}{n^2}
= \sum_{e\in \mathcal{E}}
\frac{d\left(\prod_{k=1}^\infty p_k^{e_k}\right)}{\left(\prod_{k=1}^\infty p_k^{e_k}\right)^2}
= \sum_{e\in \mathcal{E}} \prod_{k=1}^\infty \frac{e_k+1}{p_k^{2e^k}}
= \prod_{k=1}^{\infty} \left[ \sum_{e_k=0}^\infty \frac{e_k+1}{p_k^{2e^k}} \right]\\
= \prod_{k=1}^{\infty} \left[ 1 - \frac{1}{p_k^2}\right]^{-2}
= \left\{\prod_{k=1}^{\infty} \left[ 1 - \frac{1}{p_k^2}\right]^{-1}\right\}^2
= \left\{\prod_{k=1}^{\infty} \sum_{e_k=0}^{\infty} \frac{1}{p_k^{2^{e_k}}}\right\}^2\\
= \left\{\sum_{e\in \mathcal{E}}
\frac{1}{\left(\prod_{k=1}^\infty p_k^{e_k}\right)^2}
\right\}^2
= \left\{\sum_{n\in\mathbb{Z}_{+}}\frac{1}{n^2}\right\}^2
= \zeta(2)^2
= \frac{\pi^4}{36}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sketch the set of points determined by $\lvert z-i \rvert^2 + \lvert z+i \rvert^2 \le 4$ Sketch the set of points determined by the following conditions $\lvert z-i \rvert^2 + \lvert z+i \rvert^2 \le 4$ So I started by $(z-i)(\bar{z} -\bar{i}) + (z+i)(\bar{z} + \bar{i}) \le 4$
$z\bar{z} - z\bar{i} - i\bar{z} + i\bar{i} + z\bar{z} + z\bar{i} + i\bar{z} + i\bar{i}\le 4$
$=2z\bar{z} + i\bar{i}\le 4$
$=(x+iy)(x-iy) \le 2$
$=(x^2 + y^2)\le2$
$=x^2 + y^2 \le 2$ which gives me a circle with center at the origin with $r= \sqrt2$
But then I thought to do it a different way and then got confused on which way was correct. The second way is as follows.
$\lvert z-i\rvert^2 + \lvert z+i\rvert^2 \le4$
$=\lvert a+(b-1)i\rvert^2 + \lvert a+(b+1)\rvert^2 \le 4$
$=a^2 + (b-1)^2 + a^2 + (b+1)^2 \le 4$
$=a^2 + b^2 -2b+1 + a^2 + b^2 + 2b+1 \le 4$
$=2a^2 +2b^2 \le 4$
$=a^2 + b^2 \le 1$ which gives me a circle with $r=1$
I guess my question is, which one is the right process?
| In your first derivation, you have a sign error:
$$z\bar{z} - z\bar{i} - i\bar{z} \oplus i\bar{i} + z\bar{z} + z\bar{i} + i\bar{z} + i\bar{i}\le 4$$
The plus sign is circled for clarity. From there on, it works.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\gcd(a, 42) = 1$, then $a^6 \equiv 1 \pmod{168}$. I have just learned Fermat's little theorem.
That is,
If $p$ is a prime and $\gcd(a,p)=1$, then $a^{p-1} \equiv 1 \mod p$
Well, there's nothing more explanation on this theorem in my book.
And there are exercises of this kind
If $\gcd(a,42)=1$, then show that $a^6\equiv 1 \mod 168$
I don't have any idea how to approach this problem.
The only thing I know is that $168=3\cdot 7\cdot 8$.
Hence, I get $[a^2\equiv 1 \mod 3]$ and $[a^6\equiv \mod 7]$ and $[a\equiv 1\mod2]$.
That's it.
I don't know what to do next.
Please help.
| If $x \equiv 1 \bmod y$ and $x \equiv 1 \bmod z$, then $x \equiv 1 \bmod yz$ if $y$ and $z$ are coprime.
It also turns out that $a \equiv 1 \bmod 2 \Rightarrow a^2 \equiv 1 \bmod 2^3$. This is because if $a = 2k + 1$, $a^2 = 4(k^2 + k) + 1$. Since $k^2 \equiv k \bmod 2$, $k^2 + k$ is even. Thus $8 \mid 4(k^2+k)$.
Then $a^6 \equiv 1 \bmod 3$, $a^6 \equiv 1 \bmod 7$, and $a^6 \equiv 1 \bmod 8$. Hence $a^6 \equiv 1 \bmod (3 \cdot 7 \cdot 8)$, or $a^6 \equiv 1 \bmod 168$.
| {
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If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$. and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $ If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $.
Progress
We have $\gcd(a,b)=1\implies \exists u,v\in \mathbb Z$ such that $au+bv=1\implies a^2u'+bv'=1, u',v'\in \mathbb Z$.
Let $\gcd(a+b, a-b)=d$. Then $\mid (a+b)x+(a-b)y, \forall x,y\in \mathbb Z$.
How to show $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $.
| Let $d=\text{gcd}(a+b,a-b)$ then $d$ divides $2a$ and $2b$ so that $d|\text{gcd}(2a,2b)=2$. The latter equality is because $\text{gcd}(a,b)=1$ so, in particular, $a$ and $b$ cannot be both even. In any case, $d|2$ means $d=1$ or $2$.
Edit: this is an edit to respond to your own edit:
Let $e=\text{gcd}(a^2+b^2,a^2-b^2)$ then $e$ divides $2a^2$ and $2b^2$ so that $e|\text{gcd}(2a^2,2b^2)$. Again, the latter equals to $2$: $a$ and $b$ cannot be both even and by the Prime Factorization Theorem, $\text{gcd}(a,b)=1$ means $a^2$ and $b^2$ share no common prime factor. Again, the conclusion follows.
| {
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Prove the gcd $(4a + b, a + 2b) $ is equal to $1$ or $7$. So in the question it says to let $a$ and $b$ be nonzero integers such that $\gcd(a,b) = 1$.
So based on that I know that $a$ and $b$ are relatively prime and that question is basically asking if the GCD divides $7$.
So I tried to prove through the Euclidean algorithm but I think I messed up.
$a+2b = 2\times(4a+b) - 7a$
$4a+b = \frac{-4}{7}\times(-7b) +a$
$7b = 0\times a +7b$
$a = 0\times 7b +a$
$7b = 0\times a + 7b$
and so on and so on.
The fact that it is repeating tells me that I am doing something wrong.
Thanks you in advance for all of your help :)
| As you saw, we have $2(4a+b)-(a+2b)=7a$. Also, $4(a+2b)-(4a+b)=7b$.
So any common divisor of $4a+b$ and $a+2b$ is a common divisor of $7a$ and $7b$.
Since $\gcd(a,b)=1$, there are integers $x$ and $y$ such that $ax+by=1$. It follows that $(7a)x+(7b)y=7$, so any common divisor of $7a$ and $7b$ divides $7$.
Remark: We have shown that if $a$ and $b$ are relatively prime, then no positive integer other than $1$ or $7$ can be the gcd of $4a+b$ and $a+2b$.
Each of these can arise. For gcd equal to $1$, let $a=b=1$. For gcd equal to $7$, let $a=1$ and $b=3$.
| {
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How to find all solutions of $\tan(x) = 2 + \tan(3x)$ without a calculator? Find all solutions of the equation $\tan(x) = 2 + \tan(3x)$ where $0<x<2\pi$.
By replacing $\tan(3x)$ with $\dfrac{\tan(2x) + \tan(x)}{1-\tan(2x)\tan(x)}$
I've gotten to $\tan^3 (x) - 3 \tan^2 (x) + \tan(x) + 1 =0$.
I am not sure how to proceed from there without the use of a calculator.
| Put $t=\tan(x)$, so that your equation become:
$t^3-3t^2+t+1=0$.
Since $t^3-3t^2+t+1=(t-1)(t^2-2t-1)$, it follows that the solutions are $t=1$, $t=1\pm\sqrt 2$, from which
$x\equiv\arctan(1)\equiv\frac\pi 4\pmod \pi$,
$x\equiv\arctan(1+\sqrt 2)\equiv\frac{3\pi}8\pmod\pi$,
$x\equiv\arctan(1-\sqrt 2)\equiv-\frac\pi8\pmod\pi$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Possible eigenvalues of a matrix $AB$ Let matrices $A$, $B\in{M_2}(\mathbb{R})$, such that $A^2=B^2=I$, where $I$ is identity matrix.
Why can be numbers $3+2\sqrt2$ and $3-2\sqrt2$ eigenvalues for the Matrix $AB$?
Can be numbers $2,1/2$ the eigenvalues of matrix $AB$?
| Set
$$
A=\left(\begin{matrix}0 & 3-2\sqrt{2} \\ 3+2\sqrt{2} & 0\end{matrix}\right),\quad
B=\left(\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right).
$$
Then
$$
AB=\left(\begin{matrix} 3-2\sqrt{2} & 0 \\ 0 & 3+2\sqrt{2}\end{matrix}\right).
$$
The eigenvalues of $A,B$ are $\pm 1$, and hence $A^2=B^2=I$, while the eigenvalues of $AB$
are $3\pm2\sqrt{2}$.
Next, set
$$
A=\left(\begin{matrix}0 & 2 \\ 1/2 & 0\end{matrix}\right),\quad
B=\left(\begin{matrix}0 & 1 \\ 1 & 0\end{matrix}\right).
$$
Then
$$
AB=\left(\begin{matrix} 2 & 0 \\ 0 & 1/2\end{matrix}\right).
$$
The eigenvalues of $A,B$ are $\pm 1$, and hence $A^2=B^2=I$, while the eigenvalues of $AB$
are $2,1/2$.
In particular, every pair $a,1/a$ can be eigenvalues of $AB$!
| {
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If $\sum_{n=1}^\infty \frac{1}{a_n}$ converges, must $\sum_{n=1}^\infty \frac{n}{a_1 + \dots + a_n}$ converge? Suppose $\sum_{n=1}^\infty \frac{1}{a_n} = A$ is summable, with $a_n > 0,$ $n = 1,2,3,\cdots.$ How can we prove that $\sum_{n=1}^\infty \frac{n}{a_1 + \dots + a_n}$ is also summable?
This question came from a problem-solving seminar, but I'm quite stuck without a push in the right direction. I tried a few things, including Cauchy-Schwarz (which says $\sum_{n=1}^\infty \frac{n}{a_1 + \dots + a_n} < \sum_{n=1}^\infty \frac{A}{n}$) and also the idea of assuming the latter series diverges and attempting to deduce the divergence of the former series from that, using facts such as $\sum a_n = \infty \implies \sum \frac{a_n}{a_1 + \cdots + a_n} = \infty$. Nothing has worked so far.
| Allow me to present a short, "magical" solution by thoughtful application of Cauchy-Schwarz.
Define $A_n := a_1 + \cdots + a_n.$ We will first prove the convergence of $\sum_{n=1}^\infty \frac{n^2}{A_n^2}a_n$. Now,
\begin{eqnarray} \sum_{n=1}^N \frac{n^2}{A_n^2}a_n &<& \sum_{n=2}^N \frac{n^2}{A_nA_{n-1}}a_n + \frac{1}{{a_1}} \\ &=& \sum_{n=2}^N \frac{a_n}{A_nA_{n-1}}n^2 + \frac{1}{{a_1}} \\ &=& \sum_{n=2}^N \left( \frac{1}{A_{n-1}} - \frac{1}{A_n}\right)n^2 + \frac{1}{{a_1}} \\ &<& \sum_{n=1}^N \frac{(n+1)^2 - n^2}{A_n} + \frac{2}{a_1} \\ &=& \sum_{n=1}^N \frac{2n + 1}{A_n} + \frac{2}{a_1} = 2\sum_{n=1}^N \frac{n}{A_n} + C \end{eqnarray}
where in the last step we added the sum $\sum \frac{1}{A_n}$ to the constant term. Now,
$$\left(\sum_{n=1}^N \frac{n}{A_n}\right) \le {\left(\sum_{n=1}^N \frac{n^2}{A_n^2}a_n\right)}^{1/2}{\left(\sum_{n=1}^N \frac{1}{a_n} \right)}^{1/2} $$
by Cauchy-Schwarz, so, writing $S_N = \sum_{n=1}^\infty \frac{n^2}{A_n^2}a_n$, we will have
\begin{equation} S_N \le B \sqrt{S_N} + C \end{equation}
for all $N$. This implies convergence of $S_N$, since $\sqrt{S_N} \le B + \frac{C}{\sqrt{S_N}}$ could not hold for large $N$ if $S_N$ diverged.
From the Cauchy-Schwarz inequality derived above, we can thus immediately infer the convergence of our original series $\sum_{n=1}^\infty \frac{n}{A_n}$.
| {
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Sum of roots of an equation $\sqrt{x-1}+\sqrt{2x-1}=x$ Find the sum of the roots of the equation $\sqrt{x-1}+\sqrt{2x-1}=x$
My attempt: Squaring the equation: $(x-1)+(2x-1) +2\sqrt{(x-1)(2x-1)}=x^2$
$\implies x^2-3x+2=2\sqrt{(x-1)(2x-1)} $
$\implies (x-1)(x-2)=2\sqrt{(x-1)(2x-1)} $
$\implies (x-2)=2\sqrt{\displaystyle \frac{(2x-1)}{(x-1)}} $
Squaring, $(x^2-4x+4)(x-1)=8x-4$
$\implies x^2(x-5)=0$. So, the sum of roots should be five.
The given answer is 6.
Could anyone look at my attempt to find where I went wrong. Thanks.
| HINT:
$x=1$ is also a valid root where $x-1$ has been cancelled out assuming $x-1\ne0$
Observe that, though $0$ being additive identity, $x=0$ does not effect the sum is not a valid root
| {
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Integrate $\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$ I would like some guidance regarding the following integral:
$$\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$$
EDIT: The upper problem was derived from the following integral $$\int\frac{\sqrt{x^2+2}}{x^2+1}dx$$
Where I rationalized the numerator which followed into: $$\int\frac{dx}{\sqrt{x^2+2}}+\int\frac{dx}{(x^2+1)\sqrt{x^2+2}}$$
| If you aren't aware of the posted formula for calculating an integral of this form (which would completely understandable!), we can start with more familiar approaches, such as using trigonometric substitution.
Try putting $x = \sqrt 2\tan \theta$. Then $$dx = \sqrt 2 \sec^2 \theta\,d\theta$$
$$\text{ and }\,x^2 + 1 = 2\tan^2 \theta + 1 = \tan^2 \theta + \tan^2 \theta + 1 = \tan^2 \theta + \sec^2 \theta$$
$$\begin{align}\int \frac{dx}{(x^2 + 1)\sqrt {x^2 + 2}} &=\int \frac{\sqrt 2\sec^2 \theta\,d\theta}{(\tan^2 \theta+\sec^2 \theta)\sqrt{2\tan^2 \theta + 2}}\\ \\
&= \int \frac{\sec^2 \theta\,d\theta}{(\tan^2 \theta+\sec^2 \theta)\underbrace{\sqrt{\tan^2 \theta + 1}}_{\large = \,\sqrt{\sec^2 x}}}\\ \\ &=\int \frac{ \sec^2 \theta\,d\theta}{(\tan^2 \theta + \sec^2 \theta)\sec\theta} \\ \\
&= \int \dfrac{\sec \theta\,d\theta}{\tan^2 \theta + \sec^2 \theta}\cdot\left(\frac{\cos^2 \theta}{\cos^2 \theta}\right) \\ \\ &= \int\frac{\cos \theta\,d\theta}{\sin^2\theta + 1} \\ \\
&= \int\frac{d(\sin \theta)}{(\sin \theta)^2 + 1}\\ \\ &= \tan^{-1}(\sin \theta) + C \end{align}$$
Now, in terms of $x$, we have $$x = \sqrt 2\tan\theta \implies \tan\theta = \dfrac x{\sqrt 2} \implies \sin \theta = \frac{x}{\sqrt{x^2 + 2}}$$
That gives us the final answer in terms of $x$: $$\tan^{-1}(\sin \theta) + C = \tan^{-1}\left(\frac{x}{\sqrt{x^2 + 2}}\right) + C$$
| {
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If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
If $(a + ib)^3 = 8$, prove that $a^2 + b^2 = 4$
Hint: solve for $b^2$ in terms of $a^2$ and then solve for $a$
I've attempted the question but I don't think I've done it correctly:
$$
\begin{align*}
b^2 &= 4 - a^2\\
b &= \sqrt{4-a^2}
\end{align*}
$$
Therefore,
$$
\begin{align*}
(a + ib)^3 &= 8\\
a + \sqrt{4-a^2} &= 2\\
\sqrt{4-a^2} &= 2 - a\\
2 - a &= 2 - a
\end{align*}
$$
Therefore if $(a + ib)^3 = 8$, then $a^2 + b^2 = 4$.
| Let $z=a+bi$. Then $z^3=8$ and $8=|z^3|=|z|^3$, and since $|z|$ is a positive real number, $|z|=2$, that is, $a^2+b^2=|z|^2=4$.
| {
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Find the continued fraction of the square root of a given integer How to find the continued fraction of $\sqrt{n}$, for an integer $n$?
I saw a site where they explained it, but it required a calculator. Is it possible to do it without a calculator?
| The easiest way to find a continued fraction for $\sqrt{n}$ is to use a generalized continued fraction (continued fractions where the numerator is something besides 1).
Starting from $\sqrt{n}$, partition ${n}$ as the sum of the perfect square which is less than or equal to ${n}$, ${s^{2}}$, and a remainder, ${r}$, like this: $\sqrt{s^{2} + r}$.
The generalized continued fraction of this answer will always be:
$$
\sqrt{s^{2} + r} = s + \frac{r}{2s+\frac{r}{2s+\frac{r}{2s+\frac{r}{2s+\frac{r}{2s+\ddots}}}}}
$$
...or, in Gauss' Kettenbruch notation:
$$
\sqrt{s^{2} + r} = s + \underset{i=1}{\overset{\infty}{\LARGE\mathrm K}}\frac{r}{2s}
$$
For example, what's the generalized continued fraction for $\sqrt{21}$? $\sqrt{21} \ = \ \sqrt{16 + 5} \ = \ \sqrt{4^{2} + 5}$, so the generalized continued fraction is:
$$
\sqrt{21} = 4 + \frac{5}{8+\frac{5}{8+\frac{5}{8+\frac{5}{8+\frac{5}{8+\ddots}}}}}
$$
..or..
$$
\sqrt{21} = 4 + \underset{i=1}{\overset{\infty}{\LARGE\mathrm K}}\frac{5}{8}
$$
Here's how to to verify this in Wolfram|Alpha: http://www.wolframalpha.com/input/?i=%284%2B%28ContinuedFractionK%5B5%2C+8%2C+%7Bi%2C+1%2C+infinity%7D%5D%29%29%5E2
You don't have to choose a perfect square less than or equal to $n$, however. You can choose a perfect square larger than $n$. $\sqrt{21} \ = \ \sqrt{25 - 4} \ = \ \sqrt{5^{2} - 4}$, which becomes:
$$
\sqrt{21} = 5 - \frac{4}{10-\frac{4}{10-\frac{4}{10-\frac{4}{10-\frac{4}{10-\ddots}}}}}
$$
You could even do it this way: $\sqrt{21} \ = \ \sqrt{\pi^{2} + (21 - \pi^{2})}$, which gives the result:
$$
\sqrt{21} = \pi + \frac{21 - \pi^{2}}{2\pi+\frac{21 - \pi^{2}}{2\pi+\frac{21 - \pi^{2}}{2\pi+\frac{21 - \pi^{2}}{2\pi+\frac{21 - \pi^{2}}{2\pi+\ddots}}}}}
$$
For more information on continued fractions and roots, read Manny Sardina's General Method for Extracting Roots using (Folded) Continued Fractions: http://myreckonings.com/Dead_Reckoning/Online/Materials/General%20Method%20for%20Extracting%20Roots.pdf
| {
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How to prove this inequality relating to trigonometric function? In a triangle, A, B, C are three corners of the triangle, try to prove that :
$$\root 3 \of {1 - \sin A\sin B} + \root 3 \of {1 - \sin B\sin C} + \root 3 \of {1 - \sin C\sin A} \geqslant {3 \over 2}\root 3 \of 2 $$
So complicated that I have no idea...
| \begin{align}
\sin^2C &= \sin^2(A+B) = (\sin A \cos B + \cos A \sin B)^2 \\
&= \sin^2A \cos^2B + \sin^2B \cos^2A + 2\sin A \sin B \cos A \cos B \\
&\leqslant \sin^2A \cos^2B + \sin^2B \cos^2A + \sin A \sin B (cos^2A + cos^2B) \\
&= \sin^2A(1-\sin^2B)+\sin^2B(1-\sin^2A)+\sin A \sin B(2-\sin^2A - \sin^2B) \\
&= (\sin A + \sin B)^2 - 2\sin A \sin B(\sin A + \sin B)^2 \\
&= (\sin A + \sin B)^2(1-\sin A \sin B) \\
\Longrightarrow 1 - \sin A \sin B &\geqslant {\left( {{{\sin C} \over {\sin A + \sin B}}} \right)^2} = 2\left( {{{\sin C} \over {\sin A + \sin B}} \cdot {{\sin C} \over {\sin A + \sin B}} \cdot {1 \over 2}} \right) \\
&= {2 \over {{{\sin A + \sin B} \over {\sin C}} \cdot {{\sin A + \sin B} \over {\sin C}} \cdot 2}} \\
&\geqslant {2 \over {{{\left[ {{{{{\sin A + \sin B} \over {\sin C}} + {{\sin A + \sin B} \over {\sin C}} + 2} \over 3}} \right]}^3}}} \\
&= {{27} \over 4}{\left( {{{\sin C} \over {\sin A + \sin B + \sin C}}} \right)^3} \\
\Longrightarrow \root 3 \of {1 - \sin A\sin B} &\geqslant {3 \over 2}\root 3 \of 2 \cdot {{\sin C} \over {\sin A + \sin B + \sin C}}
\end{align}
$$ \Longrightarrow \root 3 \of {1 - \sin A\sin B} + \root 3 \of {1 - \sin B\sin C} + \root 3 \of {1 - \sin C\sin A} \ge {3 \over 2}\root 3 \of 2 $$
| {
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Regular Octagon Area Doing some maths homework I came across the area of a regular octagon on Google. This was given by:
$$ A=2(1+\sqrt{2})a^2 $$
I thought this looked rather ugly and slightly complicated and so began to look at regular octagons myself (Yes, I'm a nerd :)!). I managed to re-write the equation to
$$ A=\frac{x^2} {\sqrt{2}\cdot\sin^2(22.5)} $$
I could not find this equation anywhere on the internet so I don't know if it's correct. Has it been discovered before? Is it correct?
Thank you,
Sam.
P.S. I could post the proof if you need it?
| This result is known. More generally, a regular $n$-gon with side-length $x$ has area $\frac{n}{4}x^2\cot(\frac{\pi}{n})$.
To see this: by translating and rotating we can assume that the vertices of the regular $n$-gon are at $(r\cos k\theta, r\sin k\theta)$ for $0 \le k < n$, where $\theta = \frac{2\pi}{n}$ and $r$ is the distance from the centre of the $n$-gon to its vertices.
Thus the area is $n$ times the area of the triangle with vertices $(0,0)$, $(r,0)$ and $(r\cos \theta, r\sin \theta)$. Thus
$$A = \frac{n}{2}r^2\sin \frac{2\pi}{n}$$
Now the side length is given by
$$x = \lVert (r\cos \theta, r\sin\theta) - (r,0) \rVert = \sqrt{(r\cos \theta-1)^2+\sin^2\theta} = r\sqrt{2-2\cos \theta}$$
So substituting for $r$ in the formula above gives
$$A = \frac{n}{4(1-\cos \frac{2\pi}{n})}x^2\sin \frac{2\pi}{n}$$
Finally we use $\cos 2\theta \equiv 1-2\sin^2 \theta$ and $\sin 2\theta = 2\sin \theta \cos \theta$ to get
$$A = \frac{n}{8\sin^2\frac{\pi}{n}}x^2 \cdot 2 \sin \frac{\pi}{n} \cos \frac{\pi}{n} = \frac{n}{4}x^2\cot\frac{\pi}{n}$$
Your expression is equivalent in the case when $n=8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/981754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Scaling variables in homogeneous equation of degree two in a,b,c The problem I'm having trouble with is:
Let $a,b,c$ be nonzero real numbers and let $a^2 - b^2 = bc$ and $b^2 - c^2 = ca$. Prove that $a^2 - c^2 = ab$.
The solution strategy given in the course was to scale the two given equations by $s=\frac{1}{c}$, resulting in $a^2 - b^2 = bc$ becoming $a^2 - b^2 = b$ and $b^2 - c^2 = ca$ becoming $b^2 - 1 = a$. I see that $c$ is basically being set to 1, but I don't understand the justification. Doesn't scaling by $\frac{1}{c}$ by definition not change the equations, since
$(a/c)^2 - (b/c)^2 = (b/c)(c/c) \Longleftrightarrow \dfrac{a^2 - b^2}{c^2} = \dfrac{bc}{c^2} \Longleftrightarrow a^2 - b^2 = bc$
What am I missing?
| You can think of this scaling as dividing all equalities by $c^2$, which is valid because $c\neq 0$. This division basically reduces a 3-variables problem into a 2-variables problem, which should be more tractable.
Let $A=a/c$, $B=b/c$ and $C=c/c=1$. Then, we have
$$
a^2-b^2=bc\implies A^2-B^2=B;\\
b^2-c^2=ca\implies B^2-1=A.
$$
It follows that
$$
A^2-1=(A^2-B^2)+(B^2-1)=B+A.
$$
Thus, it remains that we show $A+B=AB$. Note that
$$
B=A^2-B^2=A^2-(A+1)=A^2-A-1
$$
so
$$
AB=A+B\iff A^3-A^2-A=A^2-1\iff A^3-2A^2-A+1=0.\tag{i}
$$
To show that (i) holds, we use
$$
1+A=B^2=(A^2-A-1)^2=A^4-2A^3-A^2+2A+1
$$
which implies
$$
0=A^4-2A^3-A^2+A=A(A^3-2A^2-A+1).\tag{ii}
$$
Now we are done: (ii) implies that (i) is true, therefore we indeed have $AB=A+B$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Cardinality of a set of polynomials where the sum of the cubes of the roots is zero Let $C\subseteq \mathbb Z\times \mathbb Z$ be the set of integer pairs $(a,b)$ for which the 3 complex roots $r_1,r_2,r_3$ of the polynomial $p(x)=x^3-2x^2+ax+b$ satisfy $r_1^3+r_2^3+r_3^3=0$ .Then what will be the cardinality of $ C $ ?
| Applying Vieta's formulas, we know that $$\begin{align}
r_1+r_2+r_3 & = 2 \tag{A}\\
r_1r_2 + r_1r_3 + r_2r_3 & = a\tag{B} \\
r_1r_2r_3 & = -b \tag{C}
\end{align}$$
Multiplying $(A)$ by $(B)$ yields:
$$r_1^2r_2 + r_1^2r_3 + r_1r_2^2 + r_1r_3^2 + r_2^2r_3 + r_2r_3^2 +
3r_1r_2r_3 = 2a \tag{D}$$
and then subtracting three times $(C)$ we get:
$$r_1^2r_2 + r_1^2r^3 + r_1r_2^2 + r_1r_3^2 + r_2^2r_3 + r_2r_3^2 = 2a+3b.\tag{E}$$
Now if we cube $(A)$, we get $$\begin{align}
r_1^3 +r_2^3 + r_3^3 + \\
3(r_1^2r_2 + r_1^2r_3 + r_1r_2^2 + r_1r_3^2 + r_2^2r_3 + r_2r_3^2) + \\
6r_1r_2r_3 &= 8\tag{F}\end{align}$$
and by combining this with $(E)$ and $(C)$, we obtain finally
$$\begin{align}
r_1^3 +r_2^3 + r_3^3& + 3(2a+3b) + 6(-b) &&=8\\
r_1^3 +r_2^3 + r_3^3 & && = 8-6a-3b\tag{$\ast$}
\end{align}$$
(Taking $a=b=0$ and $a=1, b=0$ provides a check that this result is correct.)
In order to have $r_1^3 +r_2^3 + r_3^3=0$, therefore, we need $8-6a-3b = 0$, or equivalently $6a+3b = 8$. But when $a$ and $b$ are integers, the left side is a multiple of 3, while $8$ is not a multiple of 3, so there is no solution; the cardinality of the set in question therefore is $0$.
| {
"language": "en",
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"source": "stackexchange",
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Find series expansion of 1/cosx Find the series expansion of 1/cosx from basic series expansions.
I tried to find 1/cosx from the expansion of cosx but was unsure how to continue. When I found 1/cosx from the basic formula for finding series expansions I didn't get the same answer.
| $\sec\left(x\right)=\dfrac{1}{\cos \left(x\right)}$ is well-behaved about $x=0$, so assume
$$
\frac{1}{\cos \left(x\right)} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots
$$
Also, $\cos \left(x\right)$ is
$$
\cos \left(x\right) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots
$$
So, on performing multiplication of the two series,
$$
1 \equiv a_0 + a_1 x + \left(a_2 - \frac{1}{2!}\right)x^2 + \left(a_3 - \frac{a_1}{2}\right)x^3 + \left(a_4 - \frac{a_2}{2} + \frac{1}{24}\right)x^4 \ldots
$$
which, \left(on equating the coefficients on powers of x on the left and right hand sides\right) gives the first few terms as
$a_0=1, a_1=0, a_2=\dfrac{1}{2}, a_3=0, a_4=\dfrac{1}{4}-\dfrac{1}{24} = \dfrac{5}{24}$. You can proceed to get more terms or a general formula for the nth term.
| {
"language": "en",
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Using a recursion tree to obtain an algorithm classification with n^2 time I'm having trouble getting the classification of this recurrence relation using a recursion tree.
$$T(n) = 3T(n/2) + n^2$$
I have the tree written out correctly (I hope):
n^2
/ | \
/ | \
(n/2)^2 (n/2)^2 (n/2)^2
/ | \ / | \ / | \
/ | \ / | \ / | \
(n/4)^2 (n/4)^2 ... ... ... (n/4)^2
...
...
T(1) T(1) ... ... ... ... ... ... ... ... ... ... T(1) T(1)
And for each "level" I have the running times:
Level Time
0 n^2
1 3(n/2)^2
2 9(n/4)^2
log_2{n} ???
So I'm having trouble getting the time for the base case at level $log_2{n}$, and after that I'm unsure of what I'm supposed to do with all the different times to get the overall running time. Do I add them all up? Do I just take the running time of the $log_2{n}$ level?
Thank you for any help you can give.
| At each level, the number of nodes triples and $n$ is cut in half. Extending the pattern in your table, we see that at level $k$, the time is $3^k(\tfrac{n}{2^k})^2 = (\tfrac{3}{4})^kn^2$. At the second-to-last level, we have $k = \log_2 n - 1$, which corresponds to the term $(\tfrac{3}{4})^{\log_2 n - 1}n^2$. At the last level when $k = \log_2 n \iff n = 2^k$, the base case kicks in. Assuming that $T(1) = c$ for some constant $c$, this level has running time $3^kT(1) = 3^{\log_2 n}c = cn^{\log_2 3}$.
Adding up all the levels, we observe that (other than the last level) we have a finite geometric series with initial term $a = n^2$, common ratio $r = \frac{3}{4}$, and a total of $N = \log_2 n$ terms. By applying the usual formula, we obtain:
\begin{align*}
T(n)
&= [n^2 + (\tfrac{3}{4})n^2 + (\tfrac{3}{4})^2n^2 + \cdots + (\tfrac{3}{4})^{\log_2 n - 1}n^2] + cn^{\log_2 3} \\
&= \frac{n^2(1 - (\frac{3}{4})^{\log_2 n})}{1 - \frac{3}{4}} + cn^{\log_2 3} \\
&= 4n^2\left(1 - \frac{3^{\log_2 n}}{4^{\log_2 n}}\right) + cn^{\log_2 3} \\
&= 4n^2\left(1 - \frac{n^{\log_2 3}}{n^{\log_2 4}}\right) + cn^{\log_2 3} \\
&= \left(4n^2 - 4n^2 \cdot \frac{n^{\log_2 3}}{n^2}\right) + cn^{\log_2 3} \\
&= 4n^2 + (c - 4)n^{\log_2 3} \\
\end{align*}
Notice that since $2 = \log_2 4 > \log_2 3$, it follows that $T(n) = \Theta(n^2)$.
| {
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"source": "stackexchange",
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} |
Exact value of polynomial at trigonometric argument
Given that $$\cos 8\theta= 128\cos^8 \theta −256\cos^6 \theta +160 \cos^4 \theta −32\cos^2 \theta +1$$
Find the exact value of:
$$4x^4 −8x^3 +5x^2 −x$$
where $x=\cos^2 (\frac{\pi}{8})$
My try:
$$\cos 8\theta= 128\cos^8 \theta −256\cos^6 \theta +160 \cos^4 \theta −32\cos^2 \theta +1$$
$$\cos 8\theta= 32(4\cos^8 \theta -8\cos^6 \theta +5 \cos^4 \theta -\cos^2 \theta) +1$$
$$\frac{\cos 8\theta-1}{32}= 4\cos^8 \theta -8\cos^6 \theta +5 \cos^4 \theta -\cos^2 \theta$$
Let $x=\cos^2 \theta$ ? (*)
Therefore:
$$\frac{\cos 8\theta-1}{32}= 4x^4 -8x^3 +5x^2 -x$$
$\theta=\frac{\pi}{8}$
$$\frac{\cos \pi -1}{32}= 4x^4 -8x^3 +5x^2 -x$$
Which gives me : $\frac{-1}{16}$
Im not happy with my presentation, is it missing something? Specially (*)? I wrote $x=\cos^2 \theta$ when they have told $x=\cos^2 (\frac{\pi}{8})$
Can someone helps me prefect out my working. I want to improve my working. I know my result is correct. But always in my working im not happy with my proofs. I want to try and perfect out the method at which I get the result.
| Your approach is correct, but you should probably say that you let $\theta=\frac{\pi}{8}$ before you introduce $x$. Continuing after your third line, you could write something like this:
Let $\theta = \frac{\pi}{8}$, and denote $x=\cos^2\left(\frac{\pi}{8}\right)$. This gives
$$ \frac{\cos\pi-1}{32}=4x^4-8x^3+5x^2-x$$
so $4x^4-8x^3+5x^2-x=-\frac{1}{16}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Optimization problem: rowing across a lake
A woman at a point A on the shore of a circular lake with radius $r=3$ wants to arrive at the point $C$ diametrically opposite $A$ on the other side of the lake in the shortest possible time. She can walk at the rate of 10 mph and row a boat at 5 mph. What is the shortest amount of time it would take her to reach point $C$?
My Working
Let $C\hat{A}B=\theta$
$\begin{align*}\therefore AB+arcBC&=\sqrt{3^2+3^2-2\cdot3\cdot3\cdot\cos(\pi - 2\theta)}+3\cdot2\theta\\
&=3\sqrt{2-2\cos(\pi-2\theta)}+6\theta\end{align*}$
$\begin{align*}\therefore \text{time taken} =t=\frac{3\sqrt{2-2\cos(\pi-2\theta)}}{5}+\frac{6\theta}{10}\end{align*}$
To find the point $B$ at which $t$ is minimized, we make $\frac{\operatorname d t}{\operatorname d\theta}=0$:
$\begin{align*}\frac{\operatorname d t}{\operatorname d\theta}=\frac{3-4\sin(\pi-2\theta)}{10\sqrt{2-2\cos(\pi-2\theta)}}+\frac{3}{5}=0\end{align*}$
And the rest of my working gets messy from here. Is there a better method than this?
| If you call the center of the circle $O$ and let $\alpha=\angle AOB$, then the woman's rowing distance is $2r\sin \frac{\alpha}{2}=6\sin \frac{\alpha}{2}$; and her walking distance is $r\cdot (\pi-\alpha)=3(\pi-\alpha)$.
So her travelling time is $T(\alpha)=\frac{6}{5}\sin\frac{\alpha}{2}+\frac{3}{10}(\pi-\alpha)$.
I think you will find that this minimizes nicely with calculus methods.
| {
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Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}$ or $\int_0^{2\pi}\frac{\cos\theta}{(1-a\cos\theta)^3}=\frac{3a\pi}{(1-a^2)^{5/2}}$ While doing some mathematical modelling of planetary orbits I have come up with two definite integrals $D_1$ and $D_2$ which appear to produce the same result R when tested with various values of $a$ ( where $0<a<1$).
$$
D_1
\, =\,
\int_0^{2\pi}f_1\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R
$$
and
$$D_2\, =\,\int_0^{2\pi}f_2\,\mathrm{d}\theta
\, =\,
\int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
\frac{3a\pi}{(1-a^2)^{5/2}}
\, =\,R$$
The hypothesis: $D_1$ = $D_2$ has been proved in a separate question Prove $\int_0^{2\pi}\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta = \int_0^{2\pi}\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta$ .
The remaining hypotheses $D_1$ = $R$ and $D_2$ = $R$ have not been proved. So the question is:-
Prove $D_1$ = $R$ or $D_2$ = $R$.
Only one proof is required because the other can then be obtained from $D_1$ = $D_2$.
For information
WolframAlpha computes expressions for the indefinite integrals $I_1,I_2$ as follows:-
$$I_1
\, =\,
\int\frac{3a\sin^2\theta}{(1-a\cos \theta)^4}\mathrm{d}\theta
\,=\,
$$
$$constant1 + \frac
{a\,\sqrt{a^2-1}\sin\theta\,[-(2a^3+a)\cos^2\theta+3(a^2+1)cos\theta+a(2a^2-5)]}
{2(a^2-1)^{5/2}(a\cos\theta-1)^3}
$$
$$-\frac
{6a\,(a\cos\theta-1)^3\,\tanh^-1
\left(
\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}
\right)
}
{(2(a^2-1)^{5/2}\,(a\cos\theta-1)^3}
$$
and
$$I_2
\, =\,
\int\frac{\cos \theta}{(1-a\cos \theta)^3}\,\mathrm{d}\theta
\, =\,
$$
$$constant2 -
\frac
{2a^2\sin\theta-sin\theta}
{2(a^2-1)^2(a\cos\theta-1)}
-\frac
{\sin\theta}
{2(a^2-1)(a\cos\theta-1)^2}
$$
$$
-\frac
{3a\tanh^-1\left(\frac{(a+1)\tan(\theta/2)}{\sqrt{a^2-1}}\right)}
{(a^2-1)^{5/2}}
$$
Note that the final terms of each expression ( i.e. the terms involving $\tanh^{-1} $ and $\tan$ ) are equivalent to each other.
Also, note that
$$\int\frac{\cos\theta}{(1-a\cos\theta)^3}\,d\theta=
\frac{-\sin\theta}{(1-a\cos\theta)^3}
+\int \frac{3a\sin^2\theta}{(1-a\cos\theta)^4}\,d\theta.
$$
Written with StackEdit.
UPDATE 20141028
I have accepted TenaliRaman's answer. I don't yet understand all the steps but his helpful exposition gives me confidence that with time I can understand it because the methods cited (binomials, series) are ones I have learned (at high school).
The answer of M.Strochyk also appears to give a good proof. But the residue method is too advanced for me to understand at present.
UPDATE 20220713
I have now accepted Quanto's answer (because it is simple enough for me to understand). I have also added an answer based on Quanto's but with the intermediate steps written out.
| I try to tackle the second integral
$$
\int_{0}^{2\pi} \frac{\cos \theta}{(1-a \cos \theta)^{3}} d \theta
$$
by the result in my post
$$I_n(a)=\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}=\frac{(-1)^{n-1} \pi}{(n-1) !} \frac{d^{n-1}}{d a^{n-1}}\left(\frac{1}{\sqrt{a^{2}-1}}\right) $$
I start with the integral
$$
\int_{0}^{\pi} \frac{1}{(1-a \cos \theta)^{2}} d \theta =\frac{1}{a^{2}} \int_{0}^{\pi} \frac{1}{\left(\frac{1}{a}-\cos \theta\right)^{2}} d \theta=\frac{1}{a^{2}}\left(\frac{\pi\left(\frac{1}{a}\right)}{\left(\frac{1}{a^{2}}-1\right)^{\frac{3}{2}}}\right)
=\frac{\pi}{\left(1-a^{2}\right)^{\frac{3}{2}}}
$$
Differentiating both sides w.r.t. $a$ yields
$$
2 \int_{0}^{\pi} \frac{\cos \theta}{(1-a \cos \theta)^{3}} d \theta =\frac{3 \pi a}{\left(1-a^{2}\right)^{\frac{5}{2}}}
$$
By symmetry, we get
$$
\boxed{\int_{0}^{2 \pi} \frac{\cos \theta}{(1-a \cos \theta)^{3}} d \theta=2 \int_{0}^{\pi} \frac{\cos \theta}{(1-a \cos \theta)^{3}} d \theta= \frac{3 \pi a}{\left(1-a^{2}\right)^{\frac{5}{2}}}}
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
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How to express $z^8 − 1$ as the product of two linear factors and three quadratic factors
Verify
$$(z-e^{i \theta} ) (z - e^{-i \theta} ) ≡ z^2 - 2\cos \theta + 1$$
Hence express $z^8 − 1$ as the product of two linear factors and three quadratic factors, where all
coefficients are real and expressed in a non-trigonometric form.
For the first part I just expanded the LHS and showed its equal to RHS.
Roots of $z^8-1$:
$$z= e^{i \frac{\pi k}{4}}$$
Where $k=0,1,2,3,4,5,6,7$
Ok since they want non-Trignometric so :
I know two roots, which are obvious:
$$z=1,-1$$
$$z^8-1=(z-1)(z+1)(z^6+z^5+z^4+z^3+z^2+z+1)$$
They want two linear factors which I believe I have found: $(z+1)$ and $(z-1)$
How do I make
$$(z^6+z^5+z^4+z^3+z^2+z+1)$$
to three quadratic factors? And they mentioned Hence , so I have to use the identity I verified. Please help.
| $$z^8-1=(z^2-1)(z^6+z^4+z^2+1)=(z-1)(z+1)(z^2+1)(z^4+1)$$
$$z^4+1=(z^2+1)^2-2z^2=(z^2+\sqrt2 z+1)(z^2-\sqrt2 z+1)$$
| {
"language": "en",
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"source": "stackexchange",
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Solve $3x^2-y^2=2$ for Integers If $x$ and $y$ are integers, then solve (using elementary methods)
$$3x^2-y^2=2$$
I tried the following
If $y$ is even, then $4|y^2$ and hence $2|y^2+2$ (and $4$ doesn't divide it), but $3x^2=y^2+2$ and since for R.H.S. to be even, $2|x^2 \implies 4|x^2$, and we get a contradiction. So $x$ and $y$ both are odd. If I substitute $x=2a+1$ and $y=2b+1$ for some integers $a$ and $b$, I get a more tedious equation and can't solve further. Please Help!
Thanks!
| For the equation: $$3y^2-x^2=2$$
Knowing the first solution: $(y,x) - (3;5)$
The equation to find the rest:
$$y_2=x+2y$$
$$x_2=2x+3y$$
| {
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"source": "stackexchange",
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An equation for the third powers of the roots of a given quadradic polynomial The roots of the equation $3x^2-4x+1=0$ are $\alpha$ and $\beta$. Find the equation with integer coefficients that has roots $\alpha^3$ and $\beta^3$.
GIVEN
SR: $\alpha + \beta = \frac43$
PR: $\alpha\beta = \frac13$
REQUIRED
SR: $\alpha^3 + \beta^3 = (\alpha + \beta) (\alpha^2 + 2\alpha \beta + \beta^2)$
$\alpha^3 + 3\alpha^2 \beta + \alpha \beta^2 + \alpha^2 \beta + 3\alpha \beta^2 + \beta^3$
$\alpha^3 + 3\alpha^2 \beta + \alpha \beta^2 + \alpha^2 \beta + 3\alpha \beta^2 + \beta^3 = (\alpha + \beta)^3$
$\alpha^3+ \beta^3 = (\alpha + \beta)^3 - 3(\alpha \beta)^2 - (\alpha \beta)^2$
$\alpha^3 + \beta^3 = (\frac43)^2 - 3(\frac13)^2 - (\frac13)^2$
$\alpha^3 + \beta^3 = \frac{16}{9} - \frac13 - \frac19$
$\alpha^3 + \beta^3 = \frac43$
PR: $\alpha^3 \beta^3$
$(\alpha \beta)^3 = (\frac13)^3 = \frac{1}{27}$
EQUATION
$x^2 - \frac43 x + \frac{1}{27} = 0$
$27x^2 - 36x + 1 = 0$
I am not sure if it's suppose to be:
$(\alpha + \beta) (\alpha^2 + 2\alpha \beta + \beta^2)$
Or
$(\alpha + \beta) (\alpha^2 + 3\alpha \beta + \beta^2)$
Also sorry about the formatting, if you don't mind fixing it for me. Not sure how to do it.
Thank You.
| solving the equation we get
$x_1=1$ and $x_2=\frac{1}{3}$ thus we have $\alpha^3=1$ and $\beta^3=\frac{1}{27}$
and our equation is given by
$(x-1)(x-1/27)=0$ therefore we get
$27x^2-28x+1=0$
the equation $3x^2-4x+1=0$ is equivalent to
$x^2-\frac{4}{3}x+\frac{1}{3}=0$
after the formula for a quadratic equation we obtain
$x_{1,2}=\frac{2}{3}\pm\sqrt{\frac{4}{9}-\frac{3}{9}}$ thus we get $x_1=1$ or $x_2=\frac{1}{3}$
| {
"language": "en",
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"source": "stackexchange",
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Show $1/(1+ x^2)$ is uniformly continuous on $\Bbb R$.
Prove that the function $x \mapsto \dfrac 1{1+ x^2}$ is uniformly continuous on $\mathbb{R}$.
Attempt: By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ implies $|f(x) - f(a)| < ε.$
Then suppose $x, a$ are elements of $\Bbb R. $
Now
\begin{align}
|f(x) - f(a)|
&= \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right|
\\&= \left| \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right|
\\&= |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)}
\\&≤ |x - a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)}
\\&= |x - a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right]
\end{align}
I don't know how to simplify more. Can someone please help me finish? Thank very much.
| You are nearly finishing the proof.
$$|x - a| (\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)})\le |x - a| (\frac{1}{2(1 + a^2)} + \frac{1}{2(1 + x^2)})\le |x-a|$$
Take $\delta=\epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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} |
Quadratic equation, math olympiad question So this is a 9-10th grade, math olympiad problem I found. Define the parabola $y=ax^2+bx+c$ such that $a,b,c$ are positive integers. Suppose that the roots of the quadratic equation $ax^2+bx+c=0$ are $x_1,x_2$ such that $|x_1|,|x_2|>1$, then compute the minimum value of $abc$ and when the minimum occur what is the value of $a+b+c$. Maybe there is a way to solve this using advanced math, calculus, but I am sure there is an elementary way of solving this without using any calculus.
| $f(x) = ax^2+bx+c$ cannot have any positive root when $a$, $b$, $c$ are all positive. So we have $x_1, x_2<-1$.
The midpoint between the roots is $-b/2a$. That must be less than $-1$, so we have $$ \tag{1} \frac{-b}{2a}<-1 \implies \frac{b}{2a}>1 \implies b>2a $$
Furthermore the larger root is less than $-1$ too, so
$$ \tag{2} f(-1) >0 \implies a-b+c>0 \implies c > b-a $$
Given these conditions, the smallest values $a$, $b$ and $c$ can have are $a=1$, $b=c=3$. Unfortunately $x^2+3x+3$ doesn't have roots at all. Increasing $c$ can't help, so $b$ must be at least $4$, and so $c=4$ too.
$x^2+4x+4$ has a double root at $x_1=x_2=-2$ and if that is allowed by the conditions we have $abc=16$ and $a+b+c=9$.
If $a$ was larger, say $a=2$ conditions $(1)$ and $(2)$ would give $b\ge 5$ and $c\ge 4$, which would give $abc\ge40$, which is much larger so we can discount that.
If the double root is not allowed, we would need to go to higher $b$ still, so consider $b=5$ -- then we get $c=5$ too. Now at last there are two different roots (we don't need to compute them, only observe that the discriminant is positive). The product $abc$ is now $25$ which is still less than $40$, and the answer for $a+b+c$ is then $11$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/999221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
derivative of $y=\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$ $y=\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}$
The answer is $\dfrac{x^2\sqrt{x+1}}{(x+2)(x-3)^5} \left(\dfrac{2}{x}+\dfrac{1}{2(x+1)}-\dfrac{1}{x+2}-\dfrac{5}{x-3}\right)$
I know that the quotient rule is used but I don't know how to do this problem. Would you multiply together all the terms and then differentiate?
| $$y=\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}\\ln (y)=ln(\frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5})=\\ln(y)=2lnx +\frac{1}{2}ln(x+1)-ln(x+2)-5ln(x-3)\\\frac{y'}{y}=(2\frac{1}{x}+\frac{1}{2(x+1)}-\frac{1}{x+2}-5\frac{1}{x-2})\\so\\y'=y*(2\frac{1}{x}+\frac{1}{2(x+1)}-\frac{1}{x+2}-5\frac{1}{x-2})\\y'= \frac{x^2\sqrt{x+1}}{(x+2)(x-3)^5}(2\frac{1}{x}+\frac{1}{2(x+1)}-\frac{1}{x+2}-5\frac{1}{x-2})$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}{4\sqrt{2}}$
Indeed,$\frac{{{a}^{2}}}{1+a}\ge \frac{9{{a}^{2}}+6a+1}{32}$
$\Leftrightarrow 32{{a}^{2}}\ge 9{{a}^{2}}+6a+1+9{{a}^{3}}+6{{a}^{2}}+a$
$\Leftrightarrow 9{{a}^{3}}-17{{a}^{2}}+7a+1\le 0$
$\Leftrightarrow 9{{\left( a-1 \right)}^{2}}\left( a+\frac{1}{9} \right)\le 0$ (!)
It is wrong. Advice on solving this problem.
| Some observations:
$$A=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \tag{1}$$
$$A=\sqrt{1+a}+\sqrt{1+b}+\sqrt{1+c} +\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} -\left(\frac{2}{\sqrt{1+a}}+\frac{2}{\sqrt{1+b}}+\frac{2}{\sqrt{1+c}} \right)$$
$$A\ge(2+2+2)-\left(\frac{2}{\sqrt{1+a}}+\frac{2}{\sqrt{1+b}}+\frac{2}{\sqrt{1+c}} \right)\tag{2}$$
Comparing with the original definition:
$$A=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2},\tag{3}$$
We conclude that we need to prove:
$$\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} \le 3-\frac{3\sqrt{2}}{4}\tag{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 6
} |
solve quadratic equation I'm trying to solve the following equation $2t^2 + t - 3 = 0$
I start by dividing by 2, $t^2 + \frac {t}{2} - \frac {3}{2} = 0$
Then I solve for t $t = - \frac{ \frac {1}{2} }{2} \binom{+}{-} \sqrt{(\frac {1}{2})^2 + \frac {3}{2}}$
$t = - \frac{1}{4} \binom{+}{-} \sqrt{(\frac {1}{4}) + \frac {6}{4}}$
I calculate $t = - \frac{1}{4} \binom{+}{-} \frac {\sqrt7}{4}$
$t_1 = - \frac{1}{4} + \frac {\sqrt7}{4}$
$t_2 = - \frac{1}{4} - \frac {\sqrt7}{4}$
But according to wolframalpha it's suppose to be
$t_1 = 1$
$t_2 = - \frac {3}{2}$
Can't figure out where did I go wrong in my calculation?
| we get $t_{1,2}=-\frac{1}{4}\pm\sqrt{\frac{1}{16}+\frac{24}{16}}$ and we get
$t_1=-\frac{1}{4}+\frac{5}{4}=\frac{4}{4}=1$ and
$t_2=-\frac{1}{4}-\frac{5}{4}=-3/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $\int_0^1\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\,dx=\frac{5\pi^2}{48}-\frac{\ln^22}{4}$ How does one prove the following integral
\begin{equation}
\int_0^1\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\,dx=\frac{5\pi^2}{48}-\frac{\ln^22}{4}
\end{equation}
Wolfram Alpha and Mathematica can easily evaluate this integral. This integral came up in the process of finding the solution this question: Evaluating $\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}$. There are some good answers there but avoiding this approach. I have been waiting for a day hoping an answers would be posted using this approach, but nothing shows up. The integral cannot be evaluated separately since each terms doesn't converge. I tried integration by parts but the problem arises when substituting the bounds of integration.
I would appreciate if anyone here could provide an answer where its approach using integral only preferably with elementary ways.
| Start with subbing $x=\frac{1-y}{1+y}$
$$\mathcal{I}=\int_0^1\frac{\ln(2)-\ln(1+x^2)}{1-x}dx=\int_0^1\frac{2\ln(1+y)-\ln(1+y^2)}{y(1+y)}dy$$
$$=2\int_0^1\frac{\ln(1+y)}{y}dy-2\int_0^1\frac{\ln(1+y)}{1+y}dy-\int_0^1\frac{\ln(1+y^2)}{y(1+y)}dy$$
$$=-2\operatorname{Li}_2(-y)|_0^1-\ln^2(1+y)|_0^1-\mathcal{J}$$
$$=\frac{\pi^2}{6}-\ln^2(2)-\left(\frac{\pi^2}{16}-\frac34\ln^2(2)\right)=\boxed{\frac{5\pi^2}{48}-\frac14\ln^2(2)}$$
Where the last integral $\mathcal{J}=\int_0^1\frac{\ln(1+y^2)}{y(1+y)}dy$ can be proved using the common Feynman's method:
Let $$I(a)=\int_0^1\frac{\ln(1+a^2y^2)}{y(1+y)}dy, \quad I(0)=0, \quad I(1)=\mathcal{J}$$
$$I'(a)=\int_0^1\frac{2ay}{(1+y)(1+a^2y^2)}dy=2\frac{\tan^{-1}(a)}{1+a^2}+\frac{a\ln(1+a^2)}{1+a^2}-\ln(2)\frac{2a}{1+a^2}$$
$$\Longrightarrow \mathcal{J}=2\int_0^1\frac{\tan^{-1}(a)}{1+a^2}da+\int_0^1\frac{a\ln(1+a^2)}{1+a^2}da-\ln(2)\int_0^1\frac{2a}{1+a^2}da\\=\left(\frac{\pi}{4}\right)^2+\frac14\ln^2(2)-\ln^2(2)=\boxed{\frac{\pi^2}{16}-\frac34\ln^2(2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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If $x,y,z.\geq 0$ and $x+y+z = 10$ , Then Max. value of $xyz+xy+yz +zx$, If $x,y,z.\geq 0$ and $x+y+z = 10$ , Then Maximum value of $xyz+xy+yz +zx$, is
$\bf{My\; Try::}$ First we can write the given expression
$xyz+xy+yz+zx = (x+1)(y+1)(z+1)-(x+y+z)-1 = $
$\displaystyle = (x+1)(y+1)(z+1)-\left\{(x+1)+(y+1)+(z+1)\right\}+2$.
So we can write $x+y+z = 10$ as $(x+1)+(y+1)+(z+1) = 13$
Now Let $\displaystyle (x+1)=a\geq 1\;\;,(y+1)=b\geq 1$ and $(z+1)=c\geq 1$
So expression convert into $a+b+c=13\;\;,a,b,c\geq 1$ and Maximize $abc-(a+b+c)+2$
Now How can i solve after that, Help me
Thanks
| From there: you know that $a+b+c = 13$, so it simplifies into $abc-14$. Then, since the max of the product of three numbers is if all the numbers are equal, so the max of $abc$ is $\left(\frac{13}{3}\right)^3 = \frac{2107}{27}$. So, the answer would be $\frac{2107}{27}-14 = \frac{1729}{27}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1001330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Number of positive integral solutions to $x+y+z+w=20$ with $xWhat is the number of positive unequal integral solution of the equation $x+y+z+w=20$, if $\,x<y<z<w\,$ and $\,x,y,z,w\ge1\;?$
How to solve this question?
| There are following posible solution:
1,2,3,14 $\,\,\,\,\,\, $ 1,3,4,12, $\,\,\,\,\,\, $ 1,4,5,10 $\,\,\,\,\,\, $ 1,5,6,8
1, 2,4,13 $\,\,\,\,\,\, $ 1,3,5,11 $\,\,\,\,\,\, $ 1,4,6,9
1,2,5,12 $\,\,\,\,\,\, $ 1,3,6,10 $\,\,\,\,\,\, $ 1,4,7,8
1,2,6,11 $\,\,\,\,\,\, $ 1,3,7,9
1,2,7,10
1,2,8,9
So there is 14 $\times $ 3 = 42 if $x=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How prove this $x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$ Question:
let $x,y,z>0$ and such $xyz=1$, show that
$$x^3+y^3+z^3+3\ge 2(x^2+y^2+z^2)$$
My idea: use AM-GM inequality
$$x^3+x^3+1\ge 3x^2$$
$$y^3+y^3+1\ge 3y^2$$
$$z^3+z^3+1\ge 3z^2$$
so
$$2(x^3+y^3+z^3)+3\ge 3(x^2+y^2+z^2)$$
But this is not my inequality,so How prove it? I know this condition is very important.but how use this condition? and this inequality is stronger
| This is hardly a pretty answer, but then again this inequality seems rather tight.
Consider the function $f(t)=e^{3t}-2e^{2t}+1+t$. It has the following properties:
*
*$f'(t)=e^{2t}(3e^t-4)+1$, so $f(t)$ is tangent to $y=0$ at $t=0$;
*$f''(t)=e^{2t}(9e^t-8)$, so $f(t)$ is concave down on $(-\infty,\ln\frac89)$ and concave up on $(\ln\frac89,\infty)$. This implies that $f(t)$ has at most one zero on $(-\infty,\ln\frac89)$ and no zeroes on $(\ln\frac89,0)$ and $(0,\infty)$. In other words, $f(t)$ changes sign at most once.
*We have $$\begin{align*}f\left(\ln\frac23\right)&=\frac{11}{27}-\ln\left(1+\frac12\right)\\
&>\frac{11}{27}-\left(\frac12-\frac12(\frac12)^2+\frac13(\frac12)^3-\frac14(\frac12)^4+\frac15(\frac12)^5\right)\\
&=\frac1{8640}>0,\end{align*}$$
so $f(t)\geq0$ on $[\ln\frac23,\infty)$.
Returning to the original question, let $xyz=1$, and assume $x\leq y\leq z$. We want to show that $\sum(x^3-2x^2+1)\geq0$.
Case 1: $x\leq\frac23$. Then we can easily check that $g(t)=t^3-2t^2+1$ is decreasing on $(0,\frac43)$ and increasing on $(\frac43,\infty)$. Hence
$$\begin{align*}\sum(x^3-2x^2+1)
&=g(x)+g(y)+g(z)\\
&\geq g\left(\frac23\right)+2g\left(\frac43\right)\\
&=\frac{11}{27}+2\left(\frac{-5}{27}\right)\\
&=\frac1{27}>0.\end{align*}$$
Case 2: $x>\frac23$. Set $x=e^a,y=e^b,z=e^c$. Then $a+b+c=0$, $a,b,c>\ln\frac23$, so
$$\sum(x^3-2x^2+1)=f(a)+f(b)+f(c)-(a+b+c)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Solve $\sqrt{3}\cos2\theta+\sin2\theta-1=0$ I tried using the identities $\cos2\theta=1-2\sin^2\theta$ and $\sin2\theta=2\sin\theta\cos\theta$. These give
$\sqrt{3}(1-2\sin^2\theta)+2\sin\theta\cos\theta-1=0$
which doesn't seem to lead anywhere. Perhaps I must equate the function to something like $R\sin(2\theta+\alpha)$?
| In general you can write:
\begin{align}
a\cos\theta+b\cos\theta&=
\sqrt{a^2+b^2}\left(
\frac{a}{\sqrt{a^2+b^2}}\cos\theta+\frac{b}{\sqrt{a^2+b^2}}\sin\theta
\right)\\[2ex]
&=\sqrt{a^2+b^2}(\sin\alpha\cos\theta+\cos\alpha\sin\theta)
\end{align}
Hope this is helpful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that: $\lim_{x\to 0}\frac{x}{\sin^2(x) + 1} = 0$ Prove
$$\displaystyle \lim_{x\to 0} \frac{x}{\sin^2(x) + 1} = 0$$
The proof:
Let $$|x| \le 1 \implies -1 \le x \le 1$$
$$\displaystyle \frac{|x|}{|\sin^2(x) + 1|} < \epsilon\text{ for }\displaystyle |x| < \delta$$
$$-1 \le x \le 1
\\\implies \sin(-1) \le \sin(x) \le \sin(1) \implies -\sin(1) \le \sin(x) \le \sin(1)
\\\implies \sin^2(1) \le \sin^2(x) \le \sin^2(1) \implies |\sin^2(x) + 1| = |\sin^2(1) + 1| \implies \displaystyle |\frac{1}{\sin^2(x) + 1}| = |\frac{1}{\sin^2(1) + 1}|$$
(1) $$|x| < \delta_1$$
(2) $$\displaystyle |\frac{1}{\sin^2(x) + 1}| = |\frac{1}{\sin^2(1) + 1}|$$
(3) $$\displaystyle \frac{|x|}{|\sin^2(x) + 1|} < \frac{\delta_1}{|\sin^2(1) + 1|}$$
(4) $$\displaystyle \frac{|\delta_1|}{|\sin^2(1) + 1|} = \epsilon \implies \delta_1 = (|\sin^2(1) + 1|)(\epsilon) $$
Finally, $\delta = \min(1, (|\sin^2(1) + 1|)(\epsilon)) \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \blacksquare$
Thoughts?
EDIT:
The original proof was indeed terrible, here's a new approach.
Let $|x| < 1 \implies -1 < x < 1$
$\sin^2(-1) + 1 < \sin^2(x) + 1 <\sin^2(1) + 2$
$\implies \displaystyle \frac{1}{\sin^2(-1) + 1} > \frac{1}{\sin^2(x) + 1} > \frac{1}{\sin^2(1) + 1}$
$\implies \displaystyle \frac{1}{\sin^2(-1) + 1} > \frac{1}{\sin^2(x) + 1} \implies \frac{1}{|\sin^2(-1) + 1|} > \frac{1}{|\sin^2(x) + 1|} \implies \frac{1}{|\sin^2(x) + 1|} < \frac{1} {|\sin^2(-1) + 1|} $
$(1) |x| < \delta_1$
$(2) \displaystyle \frac{1}{|\sin^2(x) + 1|} < \frac{1} {|\sin^2(-1) + 1|}$
$(3) \displaystyle \frac{|x|}{|\sin^2(x) + 1|} < \frac{\delta_1} {|\sin^2(-1) + 1|}$
Finally,
$\epsilon(\sin^2(-1) + 1) = \delta_1$
Therefore,
$\delta = \min(1,\epsilon \cdot (\sin^2(-1) + 1)) \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \blacksquare$
| Remarks on your new approach
Since in your attempt, $x\in[-1, 1]$, lets try $x=0$
$$ \sin^2(-1)+1\lt \sin^2(0)+1 $$
$$ \sin^2(-1)\lt \sin^2(0) $$
$$ \sin^2(1)\lt 0 $$
Is this correct to you? Also if you wish to start from $|x|$, then why not start with the fact that
$$
|x|\le |x|\left|\sin^2(x)+1\right|
$$
$$
\frac{|x|}{\left|\sin^2(x)+1\right|}\le |x|
$$
Method 1
Unless your forced to use the definition of the limit, we could simply show that
$$ \lim_{x\to 0} \frac{x}{\sin^2(x)+1} =\frac{0}{\sin^2(0)+1} = \frac{0}{0+1} = \frac{0}{1} = 0$$
Method 2
If the other answer isn't obvious to you, here's another way to see it. First note that for $x\in\mathbb{R}$
$$ 0\le\sin^2(x) $$
$$ 1\le\sin^2(x)+1 $$
$$ 1\le\left|\sin^2(x)+1\right| $$
$$ \frac{1}{\left|\sin^2(x)+1\right|}\le 1 $$
$$ \frac{|x|}{\left|\sin^2(x)+1\right|}\le |x| $$
$$ \left|\frac{x}{\sin^2(x)+1}-0\right|\le |x-0| $$
Let $\epsilon \gt 0$ and $\delta=\epsilon$, then
$$ \left|\frac{x}{\sin^2(x)+1}-0\right|\lt \epsilon \quad \mbox{whenever} \quad 0\lt\left|x-0\right|\lt\delta$$
Therefore
$$ \left|\frac{x}{\sin^2(x)+1}-0\right|\le |x-0|\lt \delta=\epsilon $$
And
by definition
$$ \lim_{x\to 0} \frac{x}{\sin^2(x)+1} =0 $$
| {
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"url": "https://math.stackexchange.com/questions/1004525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a simple first order differential equation I've got the following first order differential equation that is doing something I can't quite figure out. The question is to solve:
$$y' = 2x^2y^2$$
Here is my solution:
$$y' \frac{1}{y^2} = 2x^2$$
$$\int \frac{1}{y^2} y' dx = \int 2x^2 dx$$
$$\int \frac{1}{y^2}dy = \int 2x^2 dx$$
$$\ln|y^2| = \frac{2}{3}x^3 + c$$
$$y^2 = e^{\frac{2}{3}x^3 + c} = e^{\frac{2}{3}x^3}.e^c = ce^{\frac{2}{3}x^3}$$
$$y = \sqrt{ce^{\frac{2}{3}x^3}}$$
However, I know from the textbook that this answer is incorrect. Please help.
| Instead of keeping $y'$ as $y'$, you should change it into $\frac{dy}{dx}$. Then,$$\frac{dy}{dx} = 2x^2y^2$$ $$y^2 dy = 2x^2 dx$$ $$\int y^2\, dy = \int2x^2\, dx$$ $$\frac{y^3}{3} = \frac{2}{3}x^3 + C_1$$ $$y = \sqrt[3]{2x^3+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Taylor series expansion for $e^{-x}$ could anyone show me the Taylor series expansion for $e^{-x}$.I was trying to find out how
$e^{-i\theta}$=$\cos\theta-i\sin\theta$.
More specifically could you show me how $e^{-i\theta}$=$\cos\theta-i\sin\theta$ is obtained from Taylor series.
| $$e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\frac{x^6}{6!}+\frac{x^7}{7!}+\cdots$$
So, taking into account the fact that $i^0=1$, $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$, etc., we get
\begin{align}
e^{i\theta} &=
1+(i\theta)+\frac{(i\theta)^2}{2!}+\frac{(i\theta)^3}{3!}+\frac{(i\theta)^4}{4!}+\frac{(i\theta)^5}{5!}+\frac{(i\theta)^6}{6!}+\frac{(i\theta)^7}{7!}+\cdots \\
&= 1+i\theta+i^2\frac{\theta^2}{2!}+i^3\frac{\theta^3}{3!}+i^4\frac{\theta^4}{4!}+i^5\frac{\theta^5}{5!}+i^6\frac{\theta^6}{6!}+i^7\frac{\theta^7}{7!}+\cdots \\
&= 1+i\theta-\frac{\theta^2}{2!}-i\frac{\theta^3}{3!}+\frac{\theta^4}{4!}+i\frac{\theta^5}{5!}-\frac{\theta^6}{6!}-i\frac{\theta^7}{7!}+\cdots \\
&= \left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\frac{\theta^6}{6!} + \cdots\right)
+i\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!} \cdots\right) \\
&= \cos(\theta) + i\sin(\theta).
\end{align}
At this point, you can plug $-\theta$ in for $\theta$ to get to
$$e^{-i\theta} = \cos(\theta) - i\sin(\theta).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a$ and $b$ are positive real numbers, then $a + b \geq 2 \sqrt{ab}$. If $a$ and $b$ are positive real numbers, then $a + b \geq 2 \sqrt{ab}$.
I know how to do the direct proof, but in this case, I want to try proving it by contradiction. I have tried manipulating the inequality $a + b < 2 \sqrt{ab}$ after making the assumption that $a,b >0$ to get a contadiction $a,b \leq 0$.
$\begin{align} a + b &< 2 \sqrt{ab} \\a^2-2ab+b^2 &< 0 \\(a-b)^2 &< 0\end{align}$
How do I show that $a,b \leq 0$?
| \begin{align}
a > 0, b > 0, a + b < 2\sqrt{ab}
&\implies a - 2\sqrt{ab} + b < 0 \\
&\implies (\sqrt a - \sqrt b)^2 < 0 \\
\end{align}
Which is a contradiction.
by the way
$a + b < 2 \sqrt{ab}\;$ implies $a^2-2ab+b^2 > 0\;$, not what you said.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Inequality proof by induction, what to do next in the step I have to prove that for $n = 1, 2...$ it holds: $2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}}$
Base: For $n = 1$ holds, because $2\sqrt{2}-2 < 1$
Step: assume holds for $n_0$.
$2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n + 1}}$. But I do not know what to do next? How this can be proved?
| You assume it holds for $n$, i.e.
$$ 2\sqrt{n+1} - 2 < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}}. $$
You now want to prove that it holds for $n+1$, i.e.
$$ 2\sqrt{n+2} - 2 < 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}}. $$
Subtracting the first inequality from the second, you only have to show that
$$ 2(\sqrt{n+2}-\sqrt{n+1}) < \frac{1}{\sqrt{n+1}}. $$
Rearrangement yields
$$ \frac{\sqrt{n+2} - \sqrt{n+1}}{\sqrt{n+1}} = \frac{\sqrt{n+2}}{\sqrt{n+1}} - 1 < \frac{1}{2}, $$
so
$$ \frac{\sqrt{n+2}}{\sqrt{n+1}} < \frac{3}{2}. $$
Squaring both sides gives us
$$ \frac{n+2}{n+1} < \frac{9}{4}, $$
after which cross multiplication results in
$$ 4n+8 < 9n+9, $$
which is obvious for all $n \geq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Finding the positive integer numbers to get $\frac{\pi ^2}{9}$ As we know, there are many formulas of $\pi$ , one of them $$\frac{\pi ^2}{6}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}......
$$
and this $$\frac{\pi ^2}{8}=\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}......$$
Now,find the positive integer numbers $(a_{0}, a_{1}, a_{2}....)$ to get $$\frac{\pi^2 }{9}=\frac{1}{a_{0}^2}+\frac{1}{a_{1}^2}+\frac{1}{a_{2}^2}....$$
| From $$\frac{\pi^2}8=\sum_{2\nmid n}\frac1{n^2}$$
we find that
$$\frac{\pi^2}{72}=\frac1{3^2}\sum_{2\nmid n}\frac1{n^2}=\sum_{2\nmid n, 3\mid n}\frac1{n^2}$$
and hence
$$\begin{align}\frac{\pi^2}9=\frac{\pi^2}8-\frac{\pi^2}{72}&=\sum_{2\nmid n, 3\nmid n}\frac1{n^2}\\&=\frac1{1^2}+\frac1{5^2}+\frac1{7^2}+\frac1{11^2}+\frac1{13^2}+\frac1{17^2}+\frac1{19^2}+\frac1{23^2}+\frac1{25^2}+\ldots\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 3
} |
An annoying Pell-like equation related to a binary quadratic form problem Let $A,B,C,D$ be integers such that $AD-BC= 1 $ and $ A+D = -1 $.
Show by elementary means that the Diophantine equation
$$\bigl[2Bx + (D-A) y\bigr] ^ 2 + 3y^2 = 4|B|$$
has an integer solution (that is, a solution $(x,y)\in\mathbb Z^2$). If possible, find an explicit solution (involving $A,B,C,D$, of course).
Motivation: I arrived at this equation after trying to find explicitly the matrix $g$ suggested by Will Jagy on his answer to this question of mine. Concretely, if $\gamma=\binom{A\ \ B}{C\ \ D}$, then $\gamma$ has order $3$ in $\operatorname{SL_2}(\mathbb Z)$. By indirect methods it can be shown that $\gamma$ is conjugated in $\operatorname{SL_2}(\mathbb Z)$ to one of the matrices $P$ or $P^{-1}$, being $P=\binom{\ \ \,0\quad1}{-1\ \ -1}$ (see studiosus' answer to the same question.). Unfortunately this argument is rather sophisticated to my knowledge, and besides I think that a direct argument is possible.
Because of this I tried to find a explicit matrix $g=\binom{x\ \ y}{z\ \ w}\in\operatorname{SL_2}(\mathbb Z)$ such that $gP=\gamma g$ or $gP^{-1}=\gamma g$. The matricial equalities lead to a system of $4$ linear equations in the unknowns $x,y,z,w$ , which can be easily solved. Plugging these solutions $(x,y,z,w)$ (recall that we are considering the two possibilities of conjugation, to $P$ or $P^{-1}$) into the equation $xy-zw=1$ yields $Bx^2+(D-A)xy+(-C)y^2=\pm1$. Completing the square and using the equalities $AD-BC=1$ and $A+D=-1$ we obtain the required equation. I tried to solve it explicitly, with no success.
| Reduction can also be accomplished in a conjugacy class in $SL_2 \mathbb Z,$ in a manner very similar to that for quadratic forms. As matrices, there is a little difference: for quadratic forms, we start with symmetric $G$ and work with $P^T G P, $ for conjugacy classes we work with $P^{-1} \gamma P.$
Lemma: given real numbers with $|V| \geq |W| > 0,$ then either $|V+W| < |V|$ or $|V-W|< |V|.$
Proof: $$ |V+W| |V-W| = |V^2 - W^2| = |V|^2 - |W|^2 < |V|^2. $$
Now, we have your matrix $\gamma,$ with $(A-D)^2 -4 |BC| = -3,$ or
$$ 4|BC| = (A-D)^2 + 3. $$ Since $A+D = -1,$ we know that $A-D$ is also odd. If both $2|B|, 2|C| \geq |A-D|,$ then actually $2|B|, 2|C| \geq 1 + |A-D|.$ If so,
$4|BC| \geq (A-D)^2 + 2 |A-D| + 1, $ whence
$$ (A-D)^2 + 3 = 4 |BC| \geq (A-D)^2 + 2 |A-D| + 1, $$ or
$$ 3 \geq 2 |A-D| + 1 $$ and
$$ |A-D| = 1. $$ So, the assumption that both $2|B|, 2|C| \geq |A-D|,$ then $|A-D| = 1.$
The other case is either $2|B| < |A-D|$ or $2|C| < |A-D|.$ In this case, we use either of
$$
\left(
\begin{array}{cc}
1 & -n \\
0 & 1
\end{array}
\right)
\left(
\begin{array}{cc}
A & B \\
C & D
\end{array}
\right)
\left(
\begin{array}{cc}
1 & n \\
0 & 1
\end{array}
\right) =
\left(
\begin{array}{cc}
A-Cn & B+(A-D)n - C n^2 \\
C & D+Cn
\end{array}
\right)
$$
or
$$
\left(
\begin{array}{cc}
1 & 0 \\
-n & 1
\end{array}
\right)
\left(
\begin{array}{cc}
A & B \\
C & D
\end{array}
\right)
\left(
\begin{array}{cc}
1 & 0 \\
n & 1
\end{array}
\right) =
\left(
\begin{array}{cc}
A+Bn & B \\
-(Bn^2 + (A-D)n-C) & D-Bn
\end{array}
\right)
$$
with $n=\pm 1$
to strictly reduce $|A-D|,$ as we can have the new difference of diagonal terms either $|A-D-2Cn|$ or $|A-D+2Bn|.$ Keep doing these steps, eventually the difference of the diagonal terms is exactly $1$ in absolute value; so one of them is $0$ and the other is $-1,$ and the off-diagonal terms are $1,-1.$
So far, each possible integer matrix with determinant $1$ and trace $-1$ is conjugate to at least one of these four:
$$
\left(
\begin{array}{cc}
0 & 1 \\
-1 & -1
\end{array}
\right), \;
\left(
\begin{array}{cc}
-1 & 1 \\
-1 & 0
\end{array}
\right), \;
\left(
\begin{array}{cc}
0 & -1 \\
1 & -1
\end{array}
\right) , \;
\left(
\begin{array}{cc}
-1 & -1 \\
1 & 0
\end{array}
\right)
$$
Now, the first and second are conjugate to each other, and the third and fourth are conjugate to each other, because
$$
\left(
\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}
\right)
\left(
\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}
\right) =
\left(
\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}
\right)
$$
and
$$
\left(
\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}
\right)
\left(
\begin{array}{cc}
A & B \\
C & D
\end{array}
\right)
\left(
\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}
\right) =
\left(
\begin{array}{cc}
D & -C \\
-B & A
\end{array}
\right)
$$
As it happens, the first and the third, which are transposes, are not $SL_2 \mathbb Z$ conjugate, so we get two pairs, one pair just the transposes of the other pair.
Why not. IF
$$
\left(
\begin{array}{cc}
s & -q \\
-r & p
\end{array}
\right)
\left(
\begin{array}{cc}
0 & 1 \\
-1 & -1
\end{array}
\right)
\left(
\begin{array}{cc}
p & q \\
r & s
\end{array}
\right) =
\left(
\begin{array}{cc}
0 & -1 \\
1 & -1
\end{array}
\right)
$$
THEN
$$ s^2 + s q + q^2 = -1, $$ which is impossible in real numbers, since $ s^2 + s q + q^2$ is a positive definite quadratic form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Linear independence of $\sin^2(x)$ and $\cos^2(x)$ The Wronskian for $\sin^2x, \cos^2x$ is
\begin{align}
& \left| \begin{array}{cc} \sin^2 x & \cos^2 x \\ 2\sin x\cos x & -2\cos x\sin x \end{array} \right| \\[8pt]
= {} & -2\sin^2x \cos x \sin x - 2 \cos^2 x \sin x \cos x,
\end{align}
with $x = \frac{π}{6},$ this is $=$
$$
-\sqrt{\frac{3}{2}}dx
$$
Does this mean $\sin^2x, \cos^2x$ are linearly independent on the interval from $(-∞, ∞)$?
| No calculus needed --
If $a\sin^2 x + b\cos^2 x=0$, then $a$ must be $0$ because that's the only way to make the sum $0$ at $x=\pi/2$ where $\cos^2 x=0$.
Similarly $b=0$ is the only way to make the sum zero at $x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to compute the smallest integer which is sum of cubes in 13 ways? I would like to know the smallest integer which is sum of three positive cubes in 13 ways, such that $1^3+2^3+3^3=2^3+1^3+3^3$ are same ways of representation? What kind of theory there is behind that? I saw the question Sum of squares in at least 64 ways? but I don't know if that can be generalized as I'm not sure how to generalize Gaussian integers.
| I guess it is
119095488
= 24^3+204^3+480^3
= 48^3+ 85^3+491^3
= 72^3+384^3+396^3
= 113^3+264^3+463^3
= 114^3+360^3+414^3
= 149^3+336^3+427^3
= 176^3+204^3+472^3
= 190^3+279^3+449^3
= 207^3+297^3+438^3
= 226^3+332^3+414^3
= 243^3+358^3+389^3
= 246^3+328^3+410^3
= 281^3+322^3+399^3
The solution was given by Matti K. Sinisalo in http://keskustelu.suomi24.fi/node/12866739
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the $a$ to make the sum of series equal to zero,$\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=0$ Find all values of $a$ which make the sum of series $\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=0$
| Here is one solution: $a = 9/4$.
Then
$$\sum_{n=0}^{\infty }\frac{1}{3n^2+3n-a}=-4/9 + \frac1{3}\sum_{n=1}^{\infty }\frac{1}{n^2+n-3/4}=0$$
Note that
$$\frac1{3}\sum_{n=1}^{\infty }\frac{1}{n^2+n-3/4}=\frac1{6}\sum_{n=1}^{\infty }\left[\frac{1}{n-1/2}-\frac{1}{n+1/2}\right]+\frac1{6}\sum_{n=1}^{\infty }\left[\frac{1}{n+1/2}-\frac{1}{n+3/2}\right]\\=\frac{1}{6}(2+2/3)=\frac{4}{9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$ Evaluation of Integral $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$
$\bf{My\; Try::}$ First we will convert $x^4+x^3+x^2+x+1$ into closed form, which is $\displaystyle \left(\frac{x^5-1}{x-1}\right)$
So Integral is $\displaystyle \int_{0}^{\infty}\frac{\sqrt{x-1}}{\sqrt{x^5-1}}dx$
now i did not understand how can i solve it
Help me
Thanks
| $$
I=\int^{\infty}_{0}\frac{dx}{\sqrt{x^4+x^3+x^2+x+1}}
$$
Making the change of variable $x\rightarrow y/(1+y)$ we get
$$
I=\int^{1}_{0}\frac{dy}{\sqrt{y^4-2y^3+4y^2-3y+1}}=
$$
$$
=\int^{1}_{0}\frac{dy}{\sqrt{1/16(2y-1)^4+5/8(2y-1)^2+5/16}}.
$$
Also setting $(2y-1)\rightarrow u$, then
$$
I=\frac{1}{2}\int_{-1}^{1}\frac{du}{\sqrt{1/16 (u^2+5)^2-5/4}}=\int_{0}^{1}\frac{du}{\sqrt{1/16 (u^2+5)^2-5/4}}
$$
Then setting $(u^2+5)/4\rightarrow w$, we get
$$
I=\int_{5/4}^{6/4}\frac{dw}{\sqrt{w^2-5/4}}\frac{1}{\sqrt{w-5/4}}=\int_{0}^{1/4}\frac{w^{-1/2}dw}{\sqrt{(w+5/4-\sqrt{5/4})(w+5/4+\sqrt{5/4})}}=
$$
$$
=-2i\frac{5-\sqrt{20}}{\sqrt{5}\sqrt{5-\sqrt{20}}}\int^{-1/(5-\sqrt{20})}_{0}\frac{t^{-1/2}}{\sqrt{(1-t)(1-mt)}}dt,
$$
where $m=\frac{5-\sqrt{20}}{5+\sqrt{20}}$. From the deffinition of the elliptic integrals (in Mathematica's notation):
$$
F(\sin\phi,m)=\int_{0}^{\phi}\frac{1}{\sqrt{(1-t^2)(1-mt^2)}}dt=\frac{1}{2}\int_{0}^{ \phi^2}\frac{1}{\sqrt{t(1-t)(1-mt)}}dt\Leftrightarrow
$$
$$
2F\left(\sin\sqrt{\phi},m\right)=\int^{ \phi}_{0}\frac{1}{\sqrt{t(1-t)(1-mt)}}dt.
$$
Hence from (1) and (2) we have
$$
I=-4i\frac{5-\sqrt{20}}{\sqrt{5}\sqrt{5-\sqrt{20}}}\cdot F\left(\arcsin\left(\frac{i}{\sqrt{5-\sqrt{20}}}\right),\frac{5-\sqrt{20}}{5+\sqrt{20}}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Writing $\frac{x^4(1-x)^4}{1+x^2}$ in terms of partial fractions How does one write
$$\frac{x^4(1-x)^4}{1+x^2}$$
in terms of partial fractions?
My Attempt
$$\frac{x^4(1-x)^4}{1+x^2}=\frac{x^4-4x^5+6x^6-4x^7+x^8}{1+x^2}$$
$$=Ax^6+Bx^5+Cx^4+Dx^3+Ex^2+Fx+\frac{G}{1+x^2}$$
Multiplying out and comparing coefficient gives:
$$=x^6-4x^5+5x^4-4x^2+0+\frac{0}{1+x^2}$$
This is obviously not correct, how to do this properly?
| To answer your comment about long division: you want to divide $x^8-4x^7+6x^6-4x^5+x^4$ by $x^2+1$. Then,
$$x^8-4x^7+6x^6=x^6(x^2+1)-4x^7+5x^6$$
$$-4x^7+5x^6-4x^5=-4x^5(x^2+1)+5x^6$$
$$5x^6+x^4=5x^4(x^2+1)-4x^4$$
$$-4x^4=-4x^2(x^2+1)+4x^2$$
$$4x^2=4(x^2+1)-4$$
All in all,
$$x^8-4x^7+6x^6-4x^5+x^4=(x^6-4x^5+5x^4-4x^2+4)(x^2+1)-4$$
Or
$$\frac{x^8-4x^7+6x^6-4x^5+x^4}{x^2+1}=x^6-4x^5+5x^4-4x^2+4-\frac4{x^2+1}$$
Usually you write it like this
First step:
x^8 -4x^7 +6x^6 -4x^5 +x^4 | x^2 + 0x + 1
-4x^7 +5x^6 -4x^5 +x^4 +-----------------------
| x^6
Second step:
x^8 -4x^7 +6x^6 -4x^5 +x^4 | x^2 + 0x + 1
-4x^7 +5x^6 -4x^5 +x^4 +-----------------------
+5x^6 +0x^5 +x^4 | x^6 -4x^5
Then, the complete division:
x^8 -4x^7 +6x^6 -4x^5 +x^4 | x^2 + 0x + 1
-4x^7 +5x^6 -4x^5 +x^4 +-----------------------
+5x^6 +0x^5 +x^4 | x^6 -4x^5 +5x^4 -4x^2 +4
-4x^4 |
4x^2 |
-4 |
It's not very easy to write or to read on computer, so feel free to ask if something is not clear enough.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
When does $x^3+y^3=kz^2$? For which integers $k$ does
$$
x^3+y^3=kz^2
$$
have a solution with $z\ne0$ and $\gcd(x,y)=1$? Is there a technique for counting the number of solutions for a given $k$?
| Consider the equation: $$(a+b)^3+(a-b)^3=2a(a^2+3b^2)$$ Now let $a^2+3b^2=c^2$. We want to find a general equation for this. Note that $$(m-n)^2+2mn=(m+n)^2$$ Therefore, substitute $m\mapsto m^2$ and $n\mapsto n^2k$ then $$(m^2-k\cdot n^2)^2+k(2mn)^2=(m^2+k\cdot n^2)^2$$ Let $k=3$, and it follows that $a=m^2-3n^2$ and $b=2mn$, given that $a^2+b^2=(m^2+3n^2)^2$. $$\therefore (m^2-3n^2+2mn)^3+(m^2-3n^2-2mn)^3=2(m^2-3n^2)(m^2+3n^2)^2$$ We could also substitute $m\to 3m^2$ and $n\mapsto n^2$ to get $$(3m^2-n^2+2mn)^3+(3m^2-n^2-2mn)^3=2(3m^2-n^2)(3m^2+n^2)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Subsets and Splits