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Evaluate surface integral over unit sphere
Evaluate $\iint (ax^2+by^2+cz^2)dS$, where $S$ is surface of unit sphere $x^2+y^2+z^2 = 1$.
My attempt. Using Gauss divergence theorem, $F.n = (ax^2+by^2+cz^2) \implies F=\langle ax,by,cz\rangle$
$\text{div} F = (a+b+c)$. by gauss divergence theorem, $$\iint_S F.n ds = \iiint_V \text{div} F dV = \iiint (a+b+c)dV = \frac{4}{3}\pi(a+b+c)$$
But if I calculate by surface integral I am getting wrong answer.
My try:
Changing to spherical coordinates $x=\sin \phi \cos \theta, y = \sin \phi \sin \theta, z= \cos \phi$ (I am taking radius $r = 1$ over sphere)
$$\iint_{S: \phi: 0 - \pi,\; \theta:0-2\pi} (a\sin^3 \phi \cos^2 \theta+b\sin^3 \phi \sin^2 \theta + c\cos^2\phi \sin \phi)\ d\phi d\theta = \Big(2\pi a+2\pi b +\frac{2}{3}c\Big).$$ Both the answers are different. I know that I made some mistake in latter part. Please help.
| Check your last evaluation. Note that
$$\int_{\phi=0}^{\pi}\int_{\theta=0}^{2\pi} \sin^3 \phi \cos^2 \theta d\theta d\phi=\int_{\phi=0}^{\pi} \sin^3 \phi d\phi\int_{\theta=0}^{2\pi}\cos^2 \theta d\theta = \frac{4}{3}\cdot \pi$$
(the same result if you replace $\cos^2 \theta$ with $\sin^2 \theta$) and
$$\int_{\phi=0}^{\pi}\int_{\theta=0}^{2\pi} \cos^2 \phi \sin \phi d\theta d\phi=2\pi\int_{\phi=0}^{\pi} \cos^2 \phi \sin \phi d\phi = 2\pi\cdot\frac{2}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2509291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What are the possible prime factors of $3^n+2$ , where $n$ is a positive integer?
What are the possible prime factors of $3^n+2$, where $n$ is a positive integer ?
It is clear that a prime $p$ for which neither $-2$ nor $-6$ is a quadratic residue modulo $p$, cannot be a prime factor of $3^n+2$, so the primes congruent to $13$ or $23$ modulo $24$ can be ruled out. But is there a simple necessary and sufficient condition whether $p$ can be a prime factor of $3^n+2$ ?
| Just an observation not a complete answer.
For a prime $p>3$, applying Fermat's little theorem
$$2^p \equiv 2 \pmod{p}$$
and if there is a minimal $n_0 \in \mathbb{N}$ such that ($p$ - odd!):
$$3^{n_0} \equiv -2 \pmod{p} \Rightarrow 3^{n_0\cdot p} \equiv (-2)^p \equiv -2 \pmod{p}$$
Also $n_0<p-1$, otherwise $n_0=(p-1)\cdot q + r, 0\leq r < p-1$ and
$$3^{n_0}=3^{(p-1)\cdot q + r}=3^{(p-1)\cdot q}\cdot 3^{r}\equiv 3^r \equiv -2 \pmod{p}$$
since $3^{p-1}\equiv 1 \pmod{p}$. So, we can look for such mininal $n_0$ within $\{0,1,2,..,p-1\}$ range.
For example for
*
*$p=5$, $n_0=1$ because $3 \equiv -2 \pmod{5} \Rightarrow 3^{5^k} \equiv -2 \pmod{5}$.
*$p=7$, $n_0=5$ because $3^5 \equiv -2 \pmod{7} \Rightarrow 3^{5\cdot7^k} \equiv -2 \pmod{7}$.
*$p=11$, $n_0=2$ because $3^2 \equiv -2 \pmod{11} \Rightarrow 3^{2\cdot11^k} \equiv -2 \pmod{11}$.
*$p=13$ there is no such $n_0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Find the Fourier Series for $f(x)=$ { $1$ from $-\frac{\pi}{2} < x < \frac{\pi}{2}$ , $-1$ from $\frac{\pi}{2} < x < \frac{3\pi}{2}$ } I'm trying to find the Fourier Series for the following function, but I'm having trouble at some point. I'm hoping someone can give me a hand...
$
f(x) =
\begin{cases}
1 \text{,} & \text{if }\quad -\frac{\pi}{2} < x < \frac{\pi}{2}\\
-1 \text{,} & \text{if }\quad \frac{\pi}{2} < x < \frac{3\pi}{2}\\
\end{cases}
$
Here is what I tried so far:
$f(x) = a_0 + \sum_{n=1}^{\inf}[a_n \cdot cos(n \cdot x) + b_n \cdot sin(n \cdot x)]$
Part (1): Finding $a_0$
$a_0 = \frac{1}{2\pi} \cdot \int_{-\pi}^{\pi}f(x) dx$
$a_0 = \frac{1}{2\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(x)dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}f(x)dx]$
$a_0 = \frac{1}{2\pi} \cdot [(1) \cdot(\frac{\pi}{2}) - (1) \cdot(-\frac{\pi}{2}) + (-1) \cdot (\frac{3\pi}{2}) - (-1) \cdot(\frac{\pi}{2})]$
$a_0 = \frac{1}{2\pi} \cdot [\frac{\pi}{2} + \frac{\pi}{2} - \frac{3\pi}{2} + \frac{\pi}{2}]$
$a_0 = 0$
Part (2): Finding $a_n$
$a_n = \frac{1}{\pi} \cdot \int_{-\pi}^{\pi}f(x) \cdot cos(n \cdot x) dx$
$a_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(x) \cdot cos(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}f(x) \cdot cos(n \cdot x) dx]$
$a_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(1) \cdot cos(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}(-1) \cdot cos(n \cdot x) dx]$
$a_n = \frac{1}{\pi} \cdot [\frac{1}{n} \cdot sin(n \cdot \frac{\pi}{2}) - \frac{1}{n} \cdot sin(n \cdot \frac{-\pi}{2}) + (-1) \cdot \frac{1}{n} \cdot sin(n \cdot \frac{3\pi}{2}) - (-1) \cdot \frac{1}{n} \cdot sin(n \cdot \frac{\pi}{2})]$
$a_n = \frac{1}{n\pi} \cdot [sin(n \cdot \frac{\pi}{2}) - sin(n \cdot \frac{-\pi}{2}) - sin(n \cdot \frac{3\pi}{2}) + sin(n \cdot \frac{\pi}{2})]$
since
$sin(n \cdot \frac{-\pi}{2}) = sin(n \cdot \frac{\pi}{2})$
and
$sin(n \cdot \frac{3\pi}{2}) = sin(n \cdot \frac{\pi}{2})$
then
$a_n = 0$
Part (3): Finding $b_n$
$b_n = \frac{1}{\pi} \cdot \int_{-\pi}^{\pi}f(x) \cdot sin(n \cdot x) dx$
$b_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(x) \cdot sin(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}f(x) \cdot sin(n \cdot x) dx]$
$b_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(1) \cdot sin(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}(-1) \cdot sin(n \cdot x) dx]$
$b_n = \frac{1}{\pi} \cdot [\frac{1}{n} \cdot -cos(n \cdot \frac{\pi}{2}) - \frac{1}{n} \cdot -cos(n \cdot \frac{-\pi}{2}) + (-1) \cdot \frac{1}{n} \cdot -cos(n \cdot \frac{\pi}{2}) - (-1) \cdot \frac{1}{n} \cdot -cos(n \cdot \frac{3\pi}{2})]$
$b_n = \frac{1}{n\pi} \cdot [-cos(n \cdot \frac{\pi}{2}) + cos(n \cdot \frac{-\pi}{2}) + cos(n \cdot \frac{\pi}{2}) - cos(n \cdot \frac{\pi}{2})]$
since
$cos(n \cdot \frac{-\pi}{2}) = cos(n \cdot \frac{3\pi}{2})$
then
$b_n = 0$
I can't figure out where did I missed something. I was able to do a similar problem, but in the case where $f(x)$ is defined in two sets from $-\pi < x < 0$ and $0 < x < \pi$, but in such case, some terms got simplified because of the presence of the $0$. But here, this doesn't happen...
| Thanks to MyGlasses I was able to finish the exercise. I'll leave the full answer here for future reference.
$
f(x) =
\begin{cases}
1 \text{,} & \text{if }\quad -\frac{\pi}{2} < x < \frac{\pi}{2}\\
-1 \text{,} & \text{if }\quad \frac{\pi}{2} < x < \frac{3\pi}{2}\\
\end{cases}
$
$f(x) = a_0 + \sum_{n=1}^{\inf}[a_n \cdot cos(n \cdot x) + b_n \cdot sin(n \cdot x)]$
Part (1): Finding $a_0$
$a_0 = \frac{1}{2\pi} \cdot \int_{-\pi}^{\pi}f(x) dx$
$a_0 = \frac{1}{2\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(x)dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}f(x)dx]$
$a_0 = \frac{1}{2\pi} \cdot [(1) \cdot(\frac{\pi}{2}) - (1) \cdot(-\frac{\pi}{2}) + (-1) \cdot (\frac{3\pi}{2}) - (-1) \cdot(\frac{\pi}{2})]$
$a_0 = \frac{1}{2\pi} \cdot [\frac{\pi}{2} + \frac{\pi}{2} - \frac{3\pi}{2} + \frac{\pi}{2}]$
$a_0 = 0$
Part (2): Finding $a_n$
$a_n = \frac{1}{\pi} \cdot \int_{-\pi}^{\pi}f(x) \cdot cos(n \cdot x) dx$
$a_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(x) \cdot cos(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}f(x) \cdot cos(n \cdot x) dx]$
$a_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(1) \cdot cos(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}(-1) \cdot cos(n \cdot x) dx]$
$a_n = \frac{1}{\pi} \cdot [\frac{1}{n} \cdot sin(n \cdot \frac{\pi}{2}) - \frac{1}{n} \cdot sin(n \cdot \frac{-\pi}{2}) + (-1) \cdot \frac{1}{n} \cdot sin(n \cdot \frac{3\pi}{2}) - (-1) \cdot \frac{1}{n} \cdot sin(n \cdot \frac{\pi}{2})]$
$a_n = \frac{1}{n\pi} \cdot [sin(n \cdot \frac{\pi}{2}) - sin(n \cdot \frac{-\pi}{2}) - sin(n \cdot \frac{3\pi}{2}) + sin(n \cdot \frac{\pi}{2})]$
since
$sin(-x) = -sin(x) \implies sin(n \cdot \frac{-\pi}{2}) = -sin(n \cdot \frac{\pi}{2})$
and
$sin(n \cdot \frac{3\pi}{2}) = sin(n \cdot 2 \cdot \pi - n \cdot \frac{\pi}{2}) = sin(n \cdot 2 \cdot \pi) \cdot cos(n \cdot \frac{\pi}{2}) - cos(n \cdot 2 \cdot \pi) \cdot sin(n \cdot \frac{\pi}{2}) = -sin(n \cdot \frac{\pi}{2}) $
then
$a_n = \frac{1}{n\pi} \cdot [sin(n \cdot \frac{\pi}{2}) - (-sin(n \cdot \frac{\pi}{2})) - (-sin(n \cdot \frac{\pi}{2})) + sin(n \cdot \frac{\pi}{2})]$
$a_n = \frac{4}{n \cdot \pi}*sin(n \cdot \frac{\pi}{2})$
Part (3): Finding $b_n$
$b_n = \frac{1}{\pi} \cdot \int_{-\pi}^{\pi}f(x) \cdot sin(n \cdot x) dx$
$b_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}f(x) \cdot sin(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}f(x) \cdot sin(n \cdot x) dx]$
$b_n = \frac{1}{\pi} \cdot [\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(1) \cdot sin(n \cdot x) dx + \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}(-1) \cdot sin(n \cdot x) dx]$
$b_n = \frac{1}{\pi} \cdot [\frac{1}{n} \cdot -cos(n \cdot \frac{\pi}{2}) - \frac{1}{n} \cdot -cos(n \cdot \frac{-\pi}{2}) + (-1) \cdot \frac{1}{n} \cdot -cos(n \cdot \frac{\pi}{2}) - (-1) \cdot \frac{1}{n} \cdot -cos(n \cdot \frac{3\pi}{2})]$
$b_n = \frac{1}{n\pi} \cdot [-cos(n \cdot \frac{\pi}{2}) + cos(n \cdot \frac{-\pi}{2}) + cos(n \cdot \frac{\pi}{2}) - cos(n \cdot \frac{3\pi}{2})]$
since
$cos(-x) = cos(x) \implies cos(n \cdot \frac{-\pi}{2}) = cos(n \cdot \frac{\pi}{2})$
and
$cos(n \cdot \frac{3 \pi}{2}) = cos(n \cdot 2\pi - n \cdot \frac{\pi}{2}) = cos(n \cdot 2\pi) \cdot cos(n \cdot \frac{\pi}{2}) - sin(n \cdot 2\pi) \cdot sin(n \cdot \frac{\pi}{2}) = cos(n \cdot \frac{\pi}{2})$
then
$b_n = \frac{1}{n\pi} \cdot [-cos(n \cdot \frac{\pi}{2}) + cos(n \cdot \frac{\pi}{2}) + cos(n \cdot \frac{\pi}{2}) - cos(n \cdot \frac{\pi}{2})]$
$b_n = 0$
Solution plot
from matplotlib import pyplot as plt
import numpy as np
def f(x):
return np.where(
np.logical_and(-np.pi / 2.0 < x, x < np.pi / 2.0),
1.0,
-1.0
)
def f_(x, N):
a0 = 0.0
r = a0
for n in range(1,N):
an = 4. / (n * np.pi) * np.sin(n * np.pi / 2.)
bn = 0.0
r += an * np.cos(n * x) + bn * np.sin(n * x)
return r
x = np.linspace(-np.pi / 2.0, 3 * np.pi / 2.0, 100)
y = f(x)
plt.plot(x, y)
y = f_(x, N=3)
plt.plot(x, y, '--')
y = f_(x, N=10)
plt.plot(x, y, '--')
y = f_(x, N=100)
plt.plot(x, y, '--')
plt.show()
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For Which $a$ And $b$ Above $\mathbb{Z}_5$ There Are Solutions?
For which values of $a$ and $b$ above $\mathbb{Z_5}$ the following equations have no solution/one solution, infinite solutions
\begin{cases}
ax+4y+3z=0 \\
2y+3z=1\\
3x-bz=3
\end{cases}
So the matrix is
$$
\left[\begin{array}{rrr|r}
a & 4 & 3 & 0 \\
0 & 2 & 3 & 1 \\
3 & 0 & -b & 3
\end{array}\right]
$$
After $-\frac{3}{a}R_1+R_3\rightarrow R_3$ and $\frac{6}{a}R_2+R_3\rightarrow R_3$ and assuming $a\neq 0$
$$
\left[\begin{array}{rrr|r}
a & 4 & 3 & 0 \\
0 & 2 & 3 & 1 \\
0 & 0 & -b+\frac{9}{a} & 3+\frac{6}{a}
\end{array}\right]
$$
So I have no solution if $-b+\frac{9}{a}=0$ and $3+\frac{6}{a}\neq 0$
Infinite solution if $-b+\frac{9}{a}=0$ and $3+\frac{6}{a}=0$
And one solution in all the other cases?
| from the equation (III) we get
$$x=1+\frac{b}{3}z$$
plugg in (I)
$$a+z\left(\frac{ab}{3}+3\right)+4y=0$$
from (II) we have
$$4y=2-6z$$ and we obtain
$$z\left(\frac{ab}{3}-3\right)=-(2+a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A box contains a penny, two nickels, and a dime. If two coins are selected randomly from the box, without replacement, and if X is the sum... A box contains a penny (1¢), two nickels (5¢), and a dime (10¢). If two coins are selected randomly from the box, without replacement, and if $X$ is the sum of the values of the two coins,
*
*What is the probability distribution table of $X$?
$$\begin{array}{|c|c|c|c|c|}\hline X & 6¢ & 10¢ & 11¢ & 15¢ \\ \hline f(x) & 2/6 & 1/6 & 1/6 & 2/6\\\hline\end{array}$$
*What is the cumulative distribution function $F(x)$ of $X$?
The cumulative distribution function, $F(x)$ of $X$ is defined as: $F(x) = P(X ≤ x)$
So would that mean I just write:
$P(X ≤ 6) = 2/6$
$P(X ≤ 10) = 1/6$
$P(X ≤ 11) = 1/6$
$P(X ≤ 15) = 2/6$
\begin{align*}
P(X \leq 6) & = P(X = 6)=2/6\\
P(X \leq 10) & = P(X = 6) + P(X = 10)=2/6+1/6=1/2\\
P(X \leq 11) & = P(X = 6) + P(X = 10) + P(X = 11)=1/2+1/6=2/3\\
P(X \leq 15) & = P(X = 6) + P(X = 10) + P(X = 11) + P(X = 15)=2/3+2/6=1
\end{align*}
| A good notation for the $CDF$ would be
$$ F_{X}(x)=
\begin{cases}
1 & x \geq 15 \\
\frac{4}{6} & 11 \leq x \lt 15 \\
\frac{3}{6} & 10 \leq x \lt 11 \\
\frac{2}{6} & 6 \leq x \lt 10 \\
0 & x \lt 6
\end{cases} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2517175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Given $a^3+b^3=1$ and $(a+b)(a+1)(b+1)=2$, find the value of $(a+b)$
Given: $(a,b)\subset \mathbb R^2$, $a^3+b^3=1$ and $(a+b)(a+1)(b+1)=2$.
Find: The value of $(a+b)$
Question on the Brazilian Math Olympic (OBM), level 2, phase 3, 2012. No answer provided.
By inspection I can easily see that $(a,b)=(1,0)$, $(a,b)=(0,1)$, are possible solutions, leading to $(a+b)=1$. But are there other solutions? I developed both terms using usual identities, but I'm missing something.
Hints and solutions are appreciated. Sorry if this is a duplicate.
| $(a+b)^3 = a^3+b^3+3a^2b+3ab^2$
$
2=(a+b)(a+1)(b+1)
=(a^2 b + a b^2) + (a+b)^2+ (a+b)
$
Let $u=a+b$ and $v=a^2 b + a b^2$.
Then $u^3=1+3v, 2=v+u^2+u$ and so
$u^3 + 3 u^2 + 3 u - 7=0$.
Now, $u^3 + 3 u^2 + 3 u - 7 = (u+1)^3-8$. Therefore, $u+1=2$ and $u=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2518270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If polynomial is divisible by quadratic, find values of $a$ and $b$ Equation is $z^4+(a+b)z^3+4az^2+(a+b+32)z+45$ which is divisible by $z^2+6z+9$.
We're meant to find values of $a$ and $b$ and then solve the equation itself!
I figured $z^2+6z+9=(z+3)^2$, but was wondering if this automatically implies that it is a double root(?) If so, would differentiating the equation and do the normal simultaneous equations thing suitable for this question?
| use that $${z}^{2}+ \left( a+b-6 \right) z-2\,a-6\,b+27+{\frac {4\,a+28\,b-76}{z+
3}}+{\frac {6\,a-30\,b+30}{ \left( z+3 \right) ^{2}}}=\frac{z^4+(a+b)z^3+4az^2+(a+b+32)z+45}{z^2+6z+9}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to solve this limit using Taylor formula? I have difficulties in solving this limit using Taylor formula.
$$ \lim_{x \rightarrow+ \infty} \big((x+a)^{1+\frac {1}{x}} - x^{1+\frac{1}{x+a}}\big) $$
Can you give me some advice? I will be very thankful for your help.
| Let $a = bx$,
$(x+a)^{1+\frac {1}{x}} - x^{1+\frac{1}{x+a}}=((b+1)x)^{1+\frac{1}{x}}-x^{1+\frac{1}{(1+b)x}}$
fixing $x$, expand at $b=0$
$x^{1+\frac{1}{x}}(1+b+o(b^2))-(x^{1+\frac{1}{x}}-\log(x)x^{1+\frac{1}{x}}\frac{1}{x}b+o(b^2))=x^{1+\frac{1}{x}}(1+\frac{\log(x)}{x})b+o(b^2)=ax^{\frac{1}{x}}(1+\frac{\log(x)}{x})+o(b^2)$
This converges to $a$ as $x\rightarrow\infty, b\rightarrow0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sum of infinite sequence Let $$T_r=\frac{rx}{(1-x)(1-2x)(1-3x)\cdots(1-rx)}$$
Can someone please tell me how to break this expression into partial fractions (because I am a bit weak at it) to find the following $$\sum_{r=2}^\infty T_r$$
| This sum is telescopic with a different decomposition as follows:
$$\begin{align}
T_r&=\frac{rx}{\prod\limits_{k=1}^r1-kx}=\frac{1}{\prod\limits_{k=1}^{r}1-kx}-\frac{1-rx}{\prod\limits_{k=1}^{r}1-kx}=\frac{1}{\prod\limits_{k=1}^{r}1-kx}-\frac{1}{\prod\limits_{k=1}^{r-1}1-kx}
\end{align}$$
It should be fairly clear that $-\frac 1{1-x}$ stays and all other terms telescope to $0$. It is also interesting to note that if we begin the sum at $r=1$, we get the result $-1={x-1\over1-x}$. For an approach on how to decompose all the terms directly, read on:
$$\begin{align}T_2&=\frac {2x}{(1-x)(1-2x)}=\frac A{1-2x}+\frac B{1-x}\\
2x &= A(1-x)+B(1-2x)\to A=-B=2\\
T_2&=\frac 2{1-2x}-\frac 2{1-x}\\
T_3=\frac 32T_2\frac 1{1-3x}&=\frac 3{(1-2x)(1-3x)}-\frac 3{(1-x)(1-3x)}\\
&=\frac 9{1-3x}-\frac 6{1-2x}-\frac 9{2(1-3x)}+\frac 3{2(1-x)}\\
T_4=\frac 43T_3\frac 1{1-4x}&=\frac 6{(1-3x)(1-4x)}-\frac 8{(1-2x)(1-4x)}+\frac 2{(1-x)(1-4x)}\\
&=\dots
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
An atypical inequality If $a, b, c \in (0, \infty)$ with $abc=1$ prove: $$a^2+b^2+c^2+3(ab+bc+ca)\leq\frac{1}{2}(\sqrt{a}+\sqrt{b}+\sqrt{c})(a+b)(b+c)(c+a).$$
On atypical inequality (a smaller amount than a product) that resisted my attempts to prove it.
| We need to prove that
$$\prod_{cyc}(a^2+b^2)\sum_{cyc}a\geq2abc\sum_{cyc}(a^4+3a^2b^2)$$ for positives $a$, $b$ and $c$ or
$$\sum_{cyc}\left(a^4b^2+a^4c^2+\frac{2}{3}a^2b^2c^2\right)\sum_{cyc}a\geq2\sum_{cyc}(a^5bc+3a^3b^3c)$$ or
$$\sum_{cyc}(a^5b^2+a^5c^2+a^4b^3+a^4c^3+a^4b^2c+a^4c^2b+2a^3b^2c^2-2a^5bc-6a^3b^3c)\geq0,$$
which is true because $$\sum_{cyc}(a^5b^2+a^5c^2-2a^5bc)=\sum_{cyc}a^5(b-c)^2\geq0$$ and by AM-GM
$$\sum_{cyc}(a^4b^3+a^4c^3+a^4b^2c+a^4c^2b+2a^3b^2c^2-6a^3b^3c)=$$
$$=\sum_{cyc}(a^4b^3+a^3b^4+a^4b^2c+b^4a^2c+a^3b^2c^2+b^3a^2c^2-6a^3b^3c)\geq$$
$$\geq\sum_{cyc}\left(6\sqrt[6]{a^4b^3\cdot a^3b^4\cdot a^4b^2c\cdot b^4a^2c\cdot a^3b^2c^2\cdot b^3a^2c^2}-6a^3b^3c\right)=0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2523020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Locating the third vertex of an equilateral triangle Two vertices of an equilateral triangle are at $A=(10,-4)$ and $B=(0,6)$.
How can one locate the third vertex?
Maybe someone could give me the easy way please.
My attempt:
*
*Find the average of $M=(x , y)$ of $A$ and $B$, which is $(10+0)/2 = 5$, $(-4+6)/2 = 1$. So M(5,1)$.
*Find the equation of the perpendicular bisector. Slope of $AB = 10/-10 = -1$, Slope $m$ of the perpendicular bisector $= +1$
$$ \Rightarrow y-1 = 1*(x-5)\Rightarrow y = x-4 .$$
*There are $2$ vertices, $C$ and $D$, both on the line $y = x-4$. The altitude of the triangle is $10$. Find the $2$ points on $y = x-4$ which is $10$ units distance from $M(5,1)$:
$$d^2 = (\text{difference of } x)^2 + (\text{difference of } y)^2 \Rightarrow 100 = (x-5)^2 + (y-1)^2$$
Sub for $y = x-4$:
$$ 100 = (x-5)^2 + (x-5)^2\Rightarrow (x-5)^2 = 50 \Rightarrow x-5 = \pm\sqrt{50}$$
Thus $x = 5 + \sqrt{50}$, $y = 1 + \sqrt{50}$ and so the Vertices are $C$: $x = 5 - \sqrt{50}$, $y = 1 - \sqrt{50}$ --> Vertex $D$
| First, find the length of a side of the triangle. $$l=\sqrt{(10-0)^2 +(-4-6)^2}=10\sqrt{2}$$
Next, we draw two circles of radius $l$ and centers at the given points. The intersection of these circles will give us the third point (well, there will be two points so two solutions to this problem). We have two equations:
$$(x-10)^2+(y+4)^2=200$$
and
$$x^2+(y-6)^2=(x-10)^2+(y+4)^2$$
From the second equation we have $x^2+y^2-12y+36=x^2-20x+100+y^2+8y+16$ which will give us $x=y+4$. Now substitute $x$ in the first equation to get a quadratic equation for $y$. $$(y-6)^2+(y+4)^2=2y^2+4y+52=200$$
This has two solutions, $y_1=-1-5\sqrt{3}$ and $y_2=-1+5\sqrt{3}$. Thus, we have two possible locations $(3-5\sqrt{3}, -1-5\sqrt{3})$ and $(3+5\sqrt{3}, -1+5\sqrt{3})$ for the third vertex.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solving $3^x = 2^y + 1$ with $x,y \in \mathbb{N}^2$ I consider the following equation
$$ 3^x = 1 + 2^y \tag{$\star$} $$
with $(x,y) \in \mathbb{N}^2$ and $y \geq 3$.
I would like to show that :
$$ 3^x \equiv 1 \; [2^y] \; \Leftrightarrow \; 2^{y-2} \mid x. $$
I assume that $3^x \equiv 1 \; [2^y]$. Writing
$$ 3^x = \sum_{k=0}^{x} C_{x}^{k} 2^k = 2^x + x 2^{x-1} + \ldots + 2x + 1 $$
I have that $3^x \equiv 1 \; [2^y]$ is equivalent to :
$$ 2^x + x 2^{x-1} + \ldots + 2x \equiv 0 \; [2^y]. $$
This is also equivalent to :
$$ 2^{x-1} + x 2^{x-2} + \ldots + x \equiv 0 \; [2^{y-1}]. $$
Then, I am unsure about how to proceed. I would like to assume that $x > y$ and then, it would follow that $2^{x-1} \equiv 0 \; [2^{y-1}]$. Therefore, I would end up with :
$$ x \big( 2^{x-2} + \ldots + 1 ) \equiv 0 \; [2^{y-1}]. $$
Since $\operatorname{gcd}\big( 2^{y-1}, 2^{x-2} + \ldots + 1 \big) = 1$, it follows that $2^{y-1} \mid x$. But this is not the expected result. Where did I make a mistake ?
| First of all $x$ can't bigger than $y$, as then obviously $3^x > 1 + 2^y$.
Another way to solve the question is to note that if $y \ge 2$ then $x$ is even, because $4 \mid 2^y = 3^x - 1$. Therefeore if $x=2k$ we have:
$$2^y = 3^{2k} - 1 = (3^k - 1)(3^k + 1)$$
But now $\gcd(3^k - 1, 3^k + 1) = 2$ and from the above equation we must have $3^k - 1 = 2$ and $3^k + 1 = 2^{y-1}$. From the first equation we can conclude $k=1$ and so $x=2$ and plugging in the second one we get $y=3$. These are the only possibilities for $y \ge 2$
Now look at the case $y=1$ separately and you will obtain another solution, namely $x=1; y=1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Computing limit of $\sqrt{n^2+n}-\sqrt[4]{n^4+1}$ I have tried to solve this using conjugate multiplication, but I got stuck after factoring out $n^2$.
$\begin{align}
\lim_{n\rightarrow\infty}\dfrac{n^2+n-\sqrt{n^4+1}}{\sqrt{n^2+n}+\sqrt[4]{n^4+1}}
&=\lim_{n\rightarrow\infty}\dfrac{n(1+\dfrac{1}{n}-\sqrt{1+\dfrac{1}{n^4}})}{\sqrt{1+\dfrac{1}{n}}+\sqrt[4]{1+\dfrac{1}{n^4}}}\\&
=\lim_{n\rightarrow\infty}\dfrac{n+1-n\sqrt{1+\dfrac{1}{n^4}}}{\sqrt{1+\dfrac{1}{n}}+\sqrt[4]{1+\dfrac{1}{n^4}}}
\end{align}$
Given that $\dfrac{1}{n}$ tends to $0$ (so denominator is 2), can I reduce $n$ and $-n\sqrt{1+\dfrac{1}{n^4}}$ and say that the limit is $\dfrac{1}{2}$?
I mean $\dfrac{1}{n^4}$ tends to $0$, so $\sqrt{1+\dfrac{1}{n^4}}$ tends to $1$ and in this case $n-n\sqrt{1+\dfrac{1}{n^4}}$ can be simplified to $n-n$.
Solution given in my book uses conjugate multiplication twice to get rid of all the roots in nominator, but I am curious if my answer is correct or my teacher will tell me that simplifying the way I did it is incorrect.
| You need to show that $\lim\limits_{n\to\infty}\left(n-n\sqrt{1+\frac1{n^4}}\right)=0$. You cannot simply use that $\lim\limits_{n\to\infty}\sqrt{1+\frac1{n^4}}=1$ since $\lim\limits_{n\to\infty}\sqrt{1+\frac1n}=1$, yet $\lim\limits_{n\to\infty}\left(n-n\sqrt{1+\frac1n}\right)=-\frac12$.
If you are looking for an alternate approach, note that $a^4-b^4=(a-b)\left(a^3+a^2b+ab^2+b^3\right)$. Therefore,
$$
\begin{align}
\sqrt{n^2+n}-\sqrt[4]{n^4+1}
&=\frac{\left(n^2+n\right)^2-\left(n^4+1\right)}{\scriptsize\left(n^2+n\right)^{3/2}+\left(n^2+n\right)\left(n^4+1\right)^{1/4}+\left(n^2+n\right)^{1/2}\left(n^4+1\right)^{1/2}+\left(n^4+1\right)^{3/4}}\\
&=\frac{2+\frac1n-\frac1{n^3}}{\scriptsize\left(1+\frac1n\right)^{3/2}+\left(1+\frac1n\right)\left(1+\frac1{n^4}\right)^{1/4}+\left(1+\frac1n\right)^{1/2}\left(1+\frac1{n^4}\right)^{1/2}+\left(1+\frac1{n^4}\right)^{3/4}}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Find limit $\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}$
Find the following limit without using L'Hopital's rule:
$$\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}$$
My attempt:
$$x-1=u\implies x=u+1$$
So we have:
$$\lim_{u\to 0} \dfrac{4(u+1)^2\sqrt{u+4}-17(u+1)+9}{(u)^2}$$
What now?
| Let $\sqrt{x+3}-2=y\implies x=y^2+4y+1$
$$\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}=\lim_{y\to0}\dfrac{4(1+4y+y^2)^2(y+2)-17(1+4y+y^2)+9}{y^2(y+4)^2}$$
Now,
$$4(1+4y+y^2)^2(y+2)-17(1+4y+y^2)+9$$
$$=(8+4y)(1+8y+16y^2+2y^2+8y^3+y^4)-17(1+4y+y^2)+9$$
$$=(8-17+9)+y(64+4-68)+y^2(8\cdot18+32-17)+O(y^3)=159y^2+O(y^3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2532738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How do I rationalize the following fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$? As the title says I need to rationalize the fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$. I wrote the denominator as: $\sqrt[3]{9^4}-\sqrt[3]{9^2}-3^3$ but I do not know what to do after. Can you help me?
| Let $x=3^\frac13$ then the fraction becomes
$$-\frac{1}{x^4}\cdot\frac{1}{x^5-x^4+1}$$
What we need is to multiply nominator and denominator with some polynomial of $x$ in order to leave only terms with exponents divisible by $3$ in the denominator. How can we find such polynomial?
The answer is simple: find remained when dividing a polynomial by $x^5-x^4+1$ and check if we can obtain such polynomial. Maybe it is confusing, here is an example.
Let's start with $x^6+ax^3+b$. Suppose we want to multiply $x^5-x^4+1$ with something to obtain $x^6+ax^3+1$ (because it will cancel out all cube roots). Here $a,b$ are constants. So, the remainder of the division is
$$x^4+ax^3-x+b-1$$
Then, check if we can chose $a,b$ in such a way to make this remainder $0$. If we can, then we are done. Otherwise check the next polynomial $x^9+ax^6+bx^3+c$. The ramainder is now
$$a x^4+(-a-1) x-a+(b-1) x^3+c-x^2-1$$
Can we chose some $a,b,c$ to make this $0$? No, because $a$ must be $0$ to cancel out $x^4$, but at the same time must be $-1$ to cancel out $x$. So, continue with checking polynomials. x^12+ax^9+bx^6+cx^3+d is the next one we want to check. Here is the remainder
$$x (-a-b+1)+x^3 (-a+c-1)-a x^2-a+(b-3) x^4-b+d+2$$
What we have is basically a system of equations
$$a+b=1\\a-c=-1\\-a=0\\b-3=0\\a-b-d=2$$
This system of equations has no solutions. Let's continue. Remainder for degree $15$
$$x (a-b-c+4)+x^3 (-a-b+d+2)+x^4 (-3 a+c-3)+2 a+(3-b) x^2-b-c+e+4$$
Good news! This system has a solution! We have
$$a=-1\\b=3\\c=0\\d=0\\e=1$$
which corresponds to the polynomial
$$x^{15} - x^{12} + 3 x^9 + 1$$
and we want to multiply denominator with
$$1 + x^4 - x^5 + x^8 + x^9 + x^{10}$$
Solution
As explained above, our polynomial is
$$1 + x^4 - x^5 + x^8 + x^9 + x^{10}$$
But, notice the $\frac1{x^4}$, so we need to multiply our polynomial with $x^2$.
Verification
Mathematica can verify that this actually works
| {
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"url": "https://math.stackexchange.com/questions/2535817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Finding function $f(x)$ which satisfy given functional equation
Find all function $f:\mathbb{R}-\{0,1\}$ in $$f(x)+2f\left(\frac{1}{x}\right)+3f\left(\frac{x}{x-1}\right)=x$$
Attempt: put $\displaystyle x = \frac{1}{x}$, then $$f\left(\frac{1}{x}\right)+2f(x)+3f\left(\frac{1}{1-x}\right) = \frac{1}{x}$$
could some help me how to solve it , thanks
| Subtract twice from what you get from the given equation to get:
$$-f(x) + f\left(\frac{x}{x-1}\right) - 2f\left(\frac{1}{1-x} \right) = \frac{x^2 - 2}{3x}$$
Then in this equation plug $x= \frac {x}{x-1}$ to get:
$$-f\left(\frac {x}{x-1}\right) + f\left(x \right) - 2f\left(1-x\right) = \frac{-x^2+4x-2}{3x(x-1)}$$
Adding the two equations you will get:
$$- 2f\left(\frac{1}{1-x} \right) - 2f(1-x) = \frac{x^3 - 2x^2 + 2x}{3x(x-1)}$$
Substituting $x = 1-x$ you should be able to get:
$$f(x) + f\left(\frac{1}{x} \right) = \frac{-x^3 + x^2 - x +1}{6x(1-x)}$$
Substitute in the main equation to get:
$$f\left(\frac{1}{x} \right) + 3f\left(\frac{x}{x-1}\right) = x + \frac{x^3 - x^2 + x -1}{6x(1-x)}$$
Substitute $x = \frac 1x$ to get:
$$f\left(x\right) + 3f\left(\frac{1}{1-x}\right) = \frac{-x^3 + x^2 + 5x -5}{6x(x-1)}$$
From now on I will stop writing the RHS, as the caclualtions get messy. We substitute $x = \frac{1}{1-x}$ and subtract three times the new equations from the one above the get:
$$f(x) - 9f\left(\frac{x-1}{x}\right) = \cdots$$
Substitute $x = \frac{x-1}{x}$ and add nine times the new equation to the one above to get:
$$f(x)- 81f\left(\frac{1}{1-x}\right) = \cdots$$
Now combining the last equation with the one just above the line I draw you can get an expression for $f(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2536550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Convergence of $ \sum_{k=1}^{\infty} \frac {3^k}{5^k + 1}$
Show the convergence of the following series:
$$\sum_{k=1}^{\infty}\frac {3^k}{5^k + 1}$$
*
*a) Show the monotony of the partial sums
*b) estimate upwards
*c) remember the geometric series (I do not know how to use that here.)
The following is what I have done so far:
To show by induction: $a_{k+1} < a_k \forall k \in \mathbb N_0$
Induction start: $n=1$
$a_2= \frac{3^2}{5^2+1}=\frac{9}{25+1}=\frac{9}{26}=\frac{18}{52} < \frac{26}{52}=\frac{1}{2}=\frac{3}{6}=\frac{3^1}{5^1+1}=a_1$
Induction step:
$$\begin{align}
a_{k+1}&<a_k \\
\equiv \frac{3^{k+1}}{5^{k+1}+1} &< \frac{3^k}{5^k+1} \\
\equiv \frac{3^{k+1}}{5^{k+1}} &< \frac{3^k}{5^k} \\
\equiv \frac{3^{k+2}}{5^{k+1}} &< \frac{3^{k+1}}{5^k} \\
\equiv \frac{3^{k+2}}{5^{k+2}} &< \frac{3^{k+1}}{5^{k+1}} \\
\equiv \frac{3^{k+2}}{5^{k+2}+1} &< \frac{3^k+1}{5^{k+1}+1} \\
\equiv a_{k+2} &< a_{k+1} \\
\end{align}$$
To Show: $|a_k|= a_k$
$$
\begin{align}
|a_k| &= |\frac{3^k}{5^k+1}| \\
&= \frac{|3^k|}{|5^k+1|} \\
&= \frac{3^{|k|}}{5^{|k|}+|1|} \\
&= \frac{3^k}{5^k+1} \\
&= a_k
\end{align}$$
Because of the induction I can conclude that the sequence $\lim_{k \to \infty}a_k$ becomes smaller and smaller.
And because of $|a_k|=a_k$ are all values $\forall k \in \mathbb N$ positive.
$\Rightarrow $ The sequence $a_k$ is monotically decreasing.
$\Rightarrow $ The series $\sum_{k=1}^{\infty}a_k$ is monotically increasing.
$$
\begin{align}
a_k = \frac{3^k}{5^k+1} &< \frac{3^k}{5^k} \\
&< \frac{3^k}{3^k} \\
&= 1 \\
\end{align}$$
$$\lim_{k \to \infty} 1 = 1$$
And thus: $\exists N \in \mathbb N$, such that
$$|a_k| \le 1, \forall \quad k \ge N$$
Using the direct comparison test there can be concluded that $\sum_{k=1}^{\infty}{a_k}$ converges.
Question: Is my proof correct?
| We know that a geometric series $\displaystyle \sum_{k=0}^{\infty} r^k$ converges $\iff$ $|r|<1$.
Let $b_k=\left(\dfrac 35 \right)^k$.
For every $k \ge 0$, we have $0<a_k \le b_k$. Thus if $\displaystyle \sum_{k=0}^{\infty} b_k$ converges, by the comparison test $\displaystyle \sum_{k=0}^{\infty}a_k$ converges too. And we know that $b_k$ converges because it is a geometric series with $|r|<1$.
A few comments about your work. You wrote
$\dfrac{|3^k|}{|5^k+1|} =\dfrac{3^{|k|}}{5^{|k|}+|1|}$
But $|3^k|=3^{|k|}$ is in general not true. Consider that $|3^{-2}|=\dfrac 19$ , while $3^{|-2|}=9$. What should have been written is $|3^k|=3^{k}$, because for any $k$ we have $3^k>0$.
There is also a mistake in the denominator; you seem to think that $|a+b|=|a|+|b|$. But this is again not true in general. Consider $a=1, b=-1$. It is actually true that $|a+b| \le |a|+|b|$, with equality holding $\iff$ $a, b$ are both the same sign, or if both or one of them is zero.
You also wrote at the bottom that there is some $N$ such that when $k\ge N$, $a_k<1$, so therefore the series converges by the comparison test. But that is not at all what the comparison test says. Consider $c_k = \dfrac 1k$. There does exist an $N$ such that $k \ge N \implies c_k<1$ (namely $N=2$), but $\displaystyle \sum_{k=1}^{\infty} \dfrac 1k = 1 + \dfrac 12 + \dfrac 13 + \dfrac 14 + \cdots$ is the harmonic series, which famously diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2538022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Solve over integers $\frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} = 3$ I am trying to solve $\frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} = 3$ over integers, but I have no idea what's the best way to do it. I have tried multiplying both sides by $xyz$ and then figuring out that $z$ divides $xy$, $x$ divides $yz$, etc, but with no effect.
| It appears there are no non-trivial (i.e. not $x,y,z = \pm 1$) solutions: sorry, no number theory to do here. (Just some analysis.) Dividing everything by $xyz$ puts the equation in a nice form:
$\frac{1}{x^2} + \frac{1}{y^2}+\frac{1}{z^2} = \frac{3}{xyz}$.
Staring at that for a moment, it seems like the LHS is always larger than the RHS, and this is easy to show. Suppose WLOG $x \leq y \leq z$.
Case 1: $x < z^{2/3}$. Then
$\frac{1}{x^2} + \frac{1}{y^2}+\frac{1}{z^2} \geq \frac{3}{z^2} >\frac{3}{x^3} \geq \frac{3}{xyz}.$
So there are no solutions with $x < z^{2/3}$.
Case 2: $x \geq z^{2/3}$. Then
$\frac{1}{x^2} + \frac{1}{y^2}+\frac{1}{z^2} \geq \frac{3}{z^2} > \frac{3}{z^{7/3}} \geq \frac{3}{xyz}$,
since the strict inequality $z^{-2} > z^{-7/3}$ holds for $z > 1.$ So there are no solutions in this case either.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2539902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
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$\sin3x-\sin2x-\sin x=0$ I have issue solving this equation.
So I wrote $$\sin3x= 3\sin x-4\sin^3x$$ and $$\sin2x = 2\sin x\cos x$$
So we have $$3\sin x-4\sin^3x-(2\sin x\cos x)-\sin x = 0$$
But now I have $$\sin x$$ and $$\cos x$$ as unknown, and I don't know how to finish this.
| $$\sin { 3x-\sin { 2x } -\sin { x } =0 } \\ \left( \sin { 3x-\sin { x } } \right) -\sin { 2x } =0\\ 2\sin { \frac { 3x-x }{ 2 } \cos { \frac { 3x+x }{ 2 } } -\sin { 2x } } =0\\ 2\sin { x\cos { 2x-2\sin { x\cos { x } } =0 } } \\ \sin { x } \left( \cos { 2x-\cos { x } } \right) =0\\ \sin { x } =0,\cos { 2x-\cos { x } =0 } \Rightarrow 2\cos ^{ 2 }{ x } -\cos { x } -1=0\\ $$
Can you finish?
| {
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"timestamp": "2023-03-29T00:00:00",
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Generating fuctions: $\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k=2^{-2n}\binom{2n}{n}$ I'm trying to solve one of my combinatorics exercise but I struggle a bit.
Is the equality correct for all the $n\ge 0$?
$$\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k=2^{-2n}\binom{2n}{n}$$
First of all:$$\sum_{n=0}^\infty \left(2^{-2n}\binom{2n}{n}\right)x^n=\sum_{n=0}^\infty \left( \frac{x}{4}\right)^n\binom{2n}{n}=\frac{1}{\sqrt{1-x}}$$
Now
$$\sum_{n=0}^\infty \left(\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k \right)x^n=\sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k\sum_{n=k}^\infty \binom{2k}{k}\binom{n}{k} x^n=[n-k=m]=\sum_{k=0}^\infty \left(-\frac{1}{4}\right)^k\sum_{m=0}^\infty \binom{2k}{k}\binom{m+k}{m} x^{m+k}=\sum_{k=0}^\infty \left(-\frac{x}{4}\right)^k\binom{2k}{k}\sum_{m=0}^\infty \binom{m+k}{m} x^{m}$$
and here I don't know what to do next. Can anyone help me? Thanks in advice!
| You may consider that
$$ \frac{1}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{\pi/2}\sin^{2k}(\theta)\,d\theta \tag{A}$$
from which it follows that:
$$ \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{\pi/2}\sum_{k=0}^{n}\binom{n}{k}(-\sin^2\theta)^k =\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta\tag{B}$$
and the conclusion is straightforward through the substitution $\theta\mapsto\frac{\pi}{2}-\varphi$.
With generating functions, from
$$ \sum_{k\geq 0}\binom{2k}{k}\frac{z^k}{4^k}=\frac{1}{\sqrt{1-z}} \tag{C}$$
by replacing $z$ with $\frac{x}{1+x}$, then by multiplying both sides by $\frac{1}{1+x}$, we get
$$ \sum_{k\geq 0}\binom{2k}{k}\frac{x^k}{4^k(1+x)^{k+1}} = \frac{1}{\sqrt{1-x}}\tag{D}$$
then by applying $[x^n]$ to both sides:
$$ \sum_{k\geq 0}\binom{2k}{k}\frac{(-1)^k}{4^k}\binom{n}{k} = \frac{1}{4^n}\binom{2n}{n}\tag{E}$$
where the RHS has been managed through $(C)$ and the LHS through stars and bars.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integral with a fixed parameter For a parameter $a\ge 1$
How to calculate
$$\int_0^1 \frac{2\sqrt2}{(\sqrt2+1)(1-u^2)+2au}\ du$$?
I note $p(u) = (\sqrt{2}+1)(1-u^2)+2au$
The roots of $p$ are
$$u=\frac{-2a-2\sqrt{a^2+(\sqrt2+1)^2}}{-2(\sqrt2+1)}=\frac{-a-\sqrt{a^2+(\sqrt2+1)^2}}{-(\sqrt2+1)}=\frac{-a-\sqrt{a^2+2\sqrt2+3}}{-(\sqrt2+1)}$$
and
$$u=\frac{-2a+2\sqrt{a^2+(\sqrt2+1)^2}}{-2(\sqrt2+1)}=\frac{-a+\sqrt{a^2+(\sqrt2+1)^2}}{-(\sqrt2+1)}=\frac{-a+\sqrt{a^2+2\sqrt2+3}}{-(\sqrt2+1)}$$
So
$$\begin{align*}\frac{1}{(\sqrt2+1)(1-u^2)+2au}
&=\frac{1}{-(\sqrt2+1)\left(u-\frac{-a-\sqrt{a^2+2\sqrt2+3}}{-(\sqrt2+1)}\right)\left(u-\frac{-a+\sqrt{a^2+2\sqrt2+3}}{-(\sqrt2+1)}\right)}\\
&=\frac{1-\sqrt{2}}{\left(u-\frac{-a-\sqrt{a^2+2\sqrt2+3}}{-(\sqrt2+1)}\right)\left(u-\frac{-a+\sqrt{a^2+2\sqrt2+3}}{-(\sqrt2+1)}\right)}\\
&=\frac{1-\sqrt{2}}{\left(u+\frac{-a-\sqrt{a^2+2\sqrt2+3}}{(\sqrt2+1)}\right)\left(u+\frac{-a+\sqrt{a^2+2\sqrt2+3}}{(\sqrt2+1)}\right)}\\
&=\frac{2a}{u+\frac{-a-\sqrt{a^2+2\sqrt2+3}}{(\sqrt2+1)}}+\frac{2a}{u+\frac{-a+\sqrt{a^2+2\sqrt2+3}}{(\sqrt2+1)}}\end{align*}$$
What I wrote is it right? and how can I continue?
Thanks
| Some Preliminary Steps:
$\require{amsmath}$
$\DeclareMathOperator\arctanh{arctanh}$
$$\begin{aligned}
&(\sqrt2+1)(1-u^2)+2au=-(\sqrt2+1)\left(u^2-\frac{2au}{\sqrt2+1}-1\right)=\\&-(\sqrt2+1)\left(u^2-\frac{2au}{\sqrt2+1}+\frac{a^2}{(\sqrt2+1)^2}-\left(1+\frac{a^2}{(\sqrt2+1)^2}\right)\right)=\\&-(\sqrt2+1)\left(\left(u-\frac{a}{\sqrt2+1}\right)^2-\left(1+\frac{a^2}{(\sqrt2+1)^2}\right)\right)=\\&-(\sqrt2+1)\left(1+\frac{a^2}{(\sqrt2+1)^2}\right)\left(\left(\frac{(\sqrt{2}+1)u-a}{\sqrt{a^2+2\sqrt2+3}}\right)^2-1\right).
\end{aligned}$$
Your integral can now be expressed as:
$$-\frac{2\sqrt2(\sqrt2+1)}{a^2+2\sqrt2+3}\operatorname{\LARGE\int}\frac{1}{\left(\frac{(\sqrt{2}+1)u-a}{\sqrt{a^2+2\sqrt2+3}}\right)^2-1}\,du.$$
Let
$$\begin{aligned}
& s =\frac{(\sqrt{2}+1)u-a}{\sqrt{a^2+2\sqrt2+3}}\\
& ds=\frac{\sqrt{2}+1}{\sqrt{a^2+2 \sqrt{2}+3}}\,du
\end{aligned}$$
and by substituting for $s$, the integral becomes
$$\frac{2\sqrt2}{\sqrt{a^2+2\sqrt2+3}}\operatorname{\LARGE\int}\frac{1}{1-s^2}\,ds=\frac{2\sqrt2\,\arctanh(s)}{\sqrt{a^2+2\sqrt2+3}}+C=\frac{2\sqrt2\,\arctanh\left(\frac{(\sqrt{2}+1)u-a}{\sqrt{a^2+2\sqrt2+3}}\right)}{\sqrt{a^2+2\sqrt2+3}}+C$$
Now, to find the indefinite just plug in the endpoints, to obtain:
$$\boxed{\frac{2 \sqrt{2} \left(\arctanh\left(\frac{\sqrt{2}+1-a}{\sqrt{a^2+2
\sqrt{2}+3}}\right)+\arctanh\left(\frac{a}{\sqrt{a^2+2
\sqrt{2}+3}}\right)\right)}{\sqrt{a^2+2 \sqrt{2}+3}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve $|x^2+6x+8|=|x^2+4x+5|+|2x+3|$ Question : Find solution for $x \in \Bbb R$, $|x^2+6x+8|=|x^2+4x+5|+|2x+3|$
I considered 8 different cases and arrived at the answer $\big [ \frac{-3}{2}, \infty)$
I dont know if its correct. Also, considering 8 different cases is tedious. Is there any other method to solve problems like these?
| The equation is piecewise quadratic and the pieces are delimited by the zeros of the arguments of the absolute values.
We have
$$|x^2+6x+8|=\begin{cases}x\le-4\lor x\ge-2&\to x^2+6x+8\\-4\le x\le-2&\to-(x^2+6x+8),\end{cases}$$
$$|x^2+4x+5|=x^2+4x+5,$$
$$|2x+3|=\begin{cases}x\le-\dfrac32&\to-(2x+3)\\x\ge-\dfrac32&\to2x+3.\end{cases}$$
There are actually four pieces to consider:
$$\begin{cases}
x\le-4&\to x^2+6x+8=x^2+4x+5-(2x+3)\to\text{incompatible} \left(x=-\dfrac32\right)\\
-4\le x\le-2&\to-(x^2+6x+8)=x^2+4x+5-(2x+3)\to\text{no roots}\\
-2\le x\le-\dfrac32&\to x^2+6x+8=x^2+4x+5-(2x+3)\to x=-\dfrac32\\
-\dfrac32\le x&\to x^2+6x+8=x^2+4x+5+2x+3\to\text{any }x.
\end{cases}$$
Note that this is a general approach. As noticed by Jaideep Khare, there is a shortcut for this particular problem.
In the case of $q$ quadratic terms with real roots and $l$ linear terms, the combinations lead to $2q+l+1$ intervals (if you handled all bounds independently, you would have to consider $3^q2^l$ cases !)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Find the real solutions for the system: $ x^3+y^3=1$,$x^2y+2xy^2+y^3=2.$
Find the real solutions for the system:
$$\left\{
\begin{array}{l}
x^3+y^3=1\\
x^2y+2xy^2+y^3=2\\
\end{array}
\right.
$$
From a book with exercises for math contests. The solutions provided are: $(x,y)=(\dfrac{1}{\sqrt[3]{2}},\dfrac{1}{\sqrt[3]{2}})$ and $(\dfrac{1}{3^{\frac{2}{3}}},\dfrac{2}{3^{\frac{2}{3}}})$.
Working with the expressions I could find that an equivalent system is
$$\left\{
\begin{array}{l}
(x+y)(x^2-xy+y^2)=1\\
y(x+y)^2=2\\
\end{array}
\right.
$$
Developing these expressions I got stuck.
Hints and answers are appreciated. Sorry if this is a duplicate.
| Hint:
Observe that $xy\ne0$
Set $y=mx$
divide the resultant one equation by the other to form a cubic equation in $m$ with $m+1\ne0$ being one factor
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to show $x^{10}-2x^9+3x^8-...-10x+11=0$ has no real root? I tried to solve below equation $$x^{10}-2x^9+3x^8-...-10x+11=0$$ I plot the graph and see there is no real root , But I get stuck how to show analytical .Can some one help me or give an idea ?
Thanks in advance.
This is graph of the function $$f(x)=x^{10}-2x^9+3x^8-...-10x+11$$
https://www.desmos.com/calculator/6k1lj498ra
| There is some confusion over the sign of the coefficients. But it might prove useful to consider $$x^8(x-1)^2+2x^6(x-1)^2+3x^4(x-1)^2+4x^2(x-1)^2 =$$$$=x^{10}-2x^9+x^8+2x^8-4x^7+2x^6+3x^6-6x^5+3x^4+4x^4-8x^3+4x^2 =$$$$=x^{10}-2x^9+3x^8-4x^7+5x^6-6x^5+7x^4-8x^3+4x^2$$which is clearly non-negative, or some variant of this. You should be able to conclude from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2547766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Exercise on elementary number theory Let $a,b,c,d$ integers, $c\not=0$ such that $ad-bc=1$ and $c\equiv 0 \pmod p$ for some prime $p>3$.
Show that if $a+d=\pm1$ then $p\equiv 1\pmod 3$
I don't know how to approach this problem because when I take the expression $ad-bc$ modulo $p$ we have that either $d-d^2\equiv 1\pmod p$ or $d+d^2\equiv 1\pmod p$.
| Given $ad-bc=1$ with $c\equiv 0 \pmod p$ for $p$ a prime exceeding $3$. If in addition we have $a+d = \pm 1$ then using $a = \pm 1 - d$ in $ad-bc=1$
$$
1 \equiv ad - bc \equiv (\pm 1 - d)d - 0 \equiv \pm d -d^2
\quad \text{or}\quad d^2 \equiv \pm d -1 \pmod p
$$
$$
d^3 \equiv \pm d^2-d \equiv \pm (\pm d - 1)-d \equiv \mp 1 \implies d^6 \equiv 1 \pmod p.
$$
The order of $d \pmod p$ is $1$, $2$, $3$, or $6$. The above also applies to $a$ so the order of $a$ is also restricted to those four possibilities. If we knew that there was an element of order $3$ (or $6$) we would be done because then $3 | (p-1)$.
We need to exclude the cases where the order of $d$ is $1$ or $2$.
If the order of $d$ is $1$ then $d\equiv 1\pmod p$.
Using this together with $a+d = \pm 1$ and the observation that $a\not\equiv 0$ gives that $a\equiv -2$. So if the order of $d \pmod p$ is $1$ then the order of $a$ is not $1$ or $2$ so $3 | (p-1)$.
Similarly if the order of $d$ is $2$ then $d\equiv -1$ and $a\equiv 2$ and again $3 | (p-1)$.
An all cases $3|(p-1)$ and $p\equiv 1\pmod 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Coefficient of $x^3$ in $(1-2x+3x^2-4x^3)^{1/2}$ The question is to find out the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$
I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?
| Hint:
The coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$
$=$
the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3+\cdots)^{1/2}$
Using Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
$$1-2x+3x^2-4x^3+\cdots=(1+x)^{-2}$$
$$(1-2x+3x^2-4x^3+\cdots)^{1/2}=(1+x)^{-1}=1-x+x^2-x^3+\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2555399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the sum of the series of $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+...$ Find the sum of the series
$$\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots$$
My attempt solution:
$$\frac13\cdot\left(1+\frac15\right)+\frac17\cdot\left(\frac15+\frac19\right)+\frac1{11}\cdot\left(\frac19+\frac1{13}\right)+\cdots$$
$$=\frac13\cdot\left(\frac65\right)+\frac17\cdot \left(\frac{14}{45}\right)+\frac1{11}\cdot\left(\frac{22}{117}\right)+\cdots$$
$$=2\cdot\left(\left(\frac15\right)+\left(\frac1{45}\right)+\left(\frac1{117}\right)+\cdots\right)$$
$$=2\cdot\left(\left(\frac15\right)+\left(\frac1{5\cdot9}\right)+\left(\frac1{9\cdot13}\right)+\cdots\right)$$
It is here that I am stuck. The answer should be $\frac12$ but I don't see how to get it. Any suggestions?
Also, a bit more generally, are there good books (preferably with solutions) to sharpen my series skills?
| This is a general approach to evaluate the sum of series, like these.
First find $n^{th}$ term of series.
Let $T_n$ denote the $n^{th}$ term.
We see that,
$T_1 = \frac{1}{\color{green}{1} \cdot \color{teal}{3}} $
$T_2 = \frac{1}{\color{green}{3} \cdot \color{teal}{5}} $
And so on.
Let the numbers in $\color{green}{green} $ be
$$\color{green}{X_1,X_2,X_3,X_4,..=1,3,5,7...}$$
Clearly they form an A.P. with common difference $=2$
So, $n^{th} $ term of this AP is $ 1 + (n-1) × 2 = \color{green}{2n-1} $
Similarly,
Let the numbers in $\color{teal}{teal} $ be $$\color{teal}{Y_1,Y_2,Y_3,Y_4,..=3,5,7,9...}$$
Clearly they form an A.P. with common difference $=2$
So, $n^{th} $ term of this AP is $ 3 + (n-1) × 2 =\color{teal}{ 2n+1 }$
So, the $n^{th}$ term of the main question is just
$$ T_n = \frac{1}{\color{green}{(2n-1)} \cdot \color{teal}{(2n+1)}} $$
Now, taking summation from $ 1 $ to $ n $ , we have,
$$ \sum_{n=1}^n \frac{1}{(2n-1) \cdot (2n+1)} $$
$$= \sum_{n=1}^n = \frac{1}{2} \cdot \frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)} $$
$$= \sum_{n=1}^n = \frac{1}{2} \cdot \frac{1}{(2n-1)} - \frac{1}{(2n+1)} $$
$$= \sum_{n=1}^n = \frac{1}{2} \cdot ( 1 - \frac{1}{(2n+1)} ) $$
While $ n = ∞ $,
$$ \sum_{n=1}^∞ = \frac{1}{2} \cdot ( 1 - \frac{1}{(2(∞)+1)} ) $$
$$= \sum_{n=1}^∞ = \frac{1}{2} \cdot ( 1 - \frac{1}{∞} ) $$
Since $ \frac{1}{∞} = 0 $,
$$ \sum_{n=1}^∞ = \frac{1}{2} \cdot ( 1 - 0 ) $$
Which is
$$ \sum_{n=1}^∞ = \frac{1}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
How to solve a matrix equation of type $XAX=B$ This is the question:
Find out the matrix $X\in M_2(\mathbb R)$, so that
$X*A*X=B$, where $A,B \in M_2(\mathbb R)$ and $A=(a_{ij}), a_{11}=1, a_{12}=2, a_{21}=1, a_{22}=5$ and $B=(b_{ij}), b_{11}=2, b_{12}=-2, b_{21}=-2, b_{22}=2$.
How can I solve it and what is\are the solution\s?
Consider the following two matrices
$$
\begin{array}{cc}
A:=
\left(
\begin{array}{cc}
1&2 \\
1& 5
\end{array}
\right), &
B:=
\left(
\begin{array}{cc}
2&-2 \\
-2& 2
\end{array}
\right)
\end{array}
$$
My question: How to obtain the matrix $X\in M_2(\mathbb R)$ from the next matrix equation
$$
X\,A\,X:=B \tag{1}
$$
My try: I assumed that $X:=\left(
\begin{array}{cc}
x_1&x_2 \\
x_3& x_4
\end{array}
\right)$ where $x_i$'s are variables,
but could not obtain the values of $x_i$, $1\leq i \leq 4$ from $(1)$.
Thanks for any suggestions.
| Taking the determinant of both sides of relationship $XAX=B$, we find
$\det(X)^2 \det(A)=\det(B)=0$. As $\det(A) \neq 0$. We must have $\det(X)=0$ which is equivalent to the fact that the 2 columns of $X$ are proportional.
We can assume thus, WLOG, that $X$ can be written :
$$X=\left(
\begin{array}{rr}
a&ca \\
b&cb
\end{array}
\right)$$
Expanding the following condition:
$$
\left(
\begin{array}{rr}
a&ca \\
b&cb
\end{array}
\right)
\left(
\begin{array}{rr}
1&2 \\
1& 5
\end{array}
\right)
\left(
\begin{array}{rr}
a&ca \\
b&cb
\end{array}
\right)
=
\left(
\begin{array}{rr}
2&-2 \\
-2&2
\end{array}
\right)$$
gives rise to the following system of equations:
$$\begin{cases}ab(2 + 5c) + a^2(1 + c)=2\\
a^2c(1 + c) + abc(2+ 5c)=-2\\
b^2(2 + 5c) + ab(1 + c)=-2\\
abc(1 + c) + b^2c(2 + 5c)=2\end{cases} \ \iff \ \begin{cases}a S =2 \\
acS=-2\\
bS=-2\\
bcS=2\end{cases}$$
by setting
$$\tag{1}S:=b(2 + 5c) + a^2(1 + c).$$
Out of these 4 compatible equations, taking into account the fact that $S \neq 0$, it is easy to see that necessarily $c=-1$, $b=-a$ , and thus, plugging these values in (1), we get $\frac{2}{a}=-a(2-5)$, thus
$$a=\sqrt{\tfrac{2}{3}} \ \ \text{ or} \ \ a=-\sqrt{\tfrac{2}{3}}.$$
It suffices now to check that, with one of these values of $a$:
$$
\left(
\begin{array}{rr}
a&-a \\
-a&a
\end{array}
\right)
\left(
\begin{array}{rr}
1&2 \\
1& 5
\end{array}
\right)
\left(
\begin{array}{rr}
a&-a \\
-a&a
\end{array}
\right)
=
\left(
\begin{array}{rr}
3a^2&-3a^2 \\
-3a^2& 3a^2
\end{array}
\right)$$
indeed coincides with matrix $B$, giving two solutions for $X$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluating a tricky limit I'm trying to calculate the radius of convergence for a power series and I'm stuck at the following step:
$$\lim_{n\to \infty}\Bigg\lvert\left(\frac{3^n+\left(-4\right)^n}{5}\right)^{\frac{1}{n}}\Bigg\rvert$$
How would you evaluate this limit?
| For $n$ odd, we see that
\begin{align*}
\left(\dfrac{4^{2k+1}-3^{2k+1}}{5}\right)^{1/(2k+1)}&=\dfrac{3}{5^{1/(2k+1)}}\left(\left(\dfrac{4}{3}\right)^{2k+1}-1\right)^{1/(2k+1)},
\end{align*}
where
\begin{align*}
\lim_{k\rightarrow\infty}\log\left(\left(\dfrac{4}{3}\right)^{2k+1}-1\right)^{1/(2k+1)}&=\lim_{k\rightarrow\infty}\dfrac{1}{2k+1}\log\left(\left(\dfrac{4}{3}\right)^{2k+1}-1\right)\\
&=\lim_{k\rightarrow\infty}\dfrac{1}{2}\dfrac{2\cdot(4/3)^{2k+1}\log(4/3)}{(4/3)^{2k+1}-1}\\
&=\log(4/3),
\end{align*}
so
\begin{align*}
\left(\dfrac{4^{2k+1}-3^{2k+1}}{5}\right)^{1/(2k+1)}\rightarrow\dfrac{3}{1}\cdot e^{\log(4/3)}=4.
\end{align*}
For $n$ even, the same procedure leads to the limit $4$.
| {
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"url": "https://math.stackexchange.com/questions/2561225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prime and primary ideals in $\mathbb{Z}[\sqrt{5}]$
Let $R=\mathbb{Z}[\sqrt{5}]$. The ideal $(2, 1-\sqrt{5})$ is prime in $R$, right (as $R/(2, 1-\sqrt{5})=\mathbb{F}_2$)?
1. Is then $(2^n, 1-\sqrt{5})$ primary for some $n\geq2$?
2. Are $(2)$ and $(3)$ prime ideals and $(2^n)$ and $(3^k)$ primary ideals in $R$?
I was thinking that at least $(2^n)$ should be primary as the zero divisors of $R/(2^n)$ are $2, 1+\sqrt{5}, 1-\sqrt{5}$, i.e. nilpotent...
Or are there some problems as $-4=(1+\sqrt{5})(1-\sqrt{5})$...
| $P=(2,1-\sqrt{5})$ is a prime:
First show it is not $R$. Suppose $1=2(a+b\sqrt{5})+(1-\sqrt{5})(c+d\sqrt{5})$ for some $a,b,c,d\in \mathbb{Z}$. Simplify we have that
$$
2a+c-5d=1, \quad 2b-c+d=0,
$$
add these two we have $2a+2b-4d=1$, contradiction. (By this argument, we can show that $2\notin (4,1-\sqrt{5})$, so $(2,1-\sqrt{5})\neq (4,1-\sqrt{5})$.)
Then as you said, we prove $R/P=\{\bar{0},\bar{1}\}$. Let $a+b\sqrt{5}+P\in R/P$. Then $a,b$ can only be $0$ or $1$. Since $1+\sqrt{5}\in P$, $1+P=\sqrt{5}+P$, we have $R/P=\{P,1+P\}$. These two elements are different since $1\notin P$.
$(2)$ is $P$-primary:
$(2)$ is not prime as mentioned in the comment above. Let $(a+b\sqrt{5})(c+d\sqrt{5})=2(e+f\sqrt{5})$, and $a+b\sqrt{5}\notin (2)$, which means $a,b$ not both even. Then we have
$$
ac+5bd+\sqrt{5}(ad+bc)=2e+2f\sqrt{5}.
$$
Case 1: $a$ even, $b$ odd.
Then $d$ even and $c$ even. So $c+d\sqrt{5}\in (2)$, done.
Case 2: $a$ odd, $b$ even. (similar as case 1)
Case 3: $a,b$ both odd.
If $c$ is even, then $d$ is even, done. If $c$ is odd, then $d$ is odd. Now, $(c+d\sqrt{5})^{2}=c^{2}+5d^{2}+2cd\sqrt{5}\in (2)$, done. So, $(2)$ is primary.
$\sqrt{(2)}=(2,1-\sqrt{5})$: $2\in\sqrt{(2)}$. Since $(1-\sqrt{5})^{2}\in (2)$, $1-\sqrt{5}\in \sqrt{(2)}$. So, $P\subseteq\sqrt{(2)}$. Since $P$ is maximal and $\sqrt{(2)}$ is prime (so not $R$), $\sqrt{(2)}=P$.
Edit:
$(2^{n})$ is primary:
I just saw a theorem: if $\sqrt Q$ is maximal, then $Q$ is primary. (Dummit Foote page 682 prop 19).
$\sqrt{(2^{n})}=\sqrt{(2)}=(2,1-\sqrt{5})$ is maximal, so $(2^{n})$ is primary. Its associated prime is also $(2,1-\sqrt{5})$.
Edit 2:
$(2^{n}, 1-\sqrt{5})$ is primary:
first show $\sqrt{(2^{n},1-\sqrt{5})}=(2,1-\sqrt{5})$. We have $(2,1-\sqrt{5})\subseteq \sqrt{(2^{n},1-\sqrt{5})}$. Conversely, since $(2^{n},1-\sqrt{5})\subseteq (2,1-\sqrt{5})$, $\sqrt{(2^{n},1-\sqrt{5})}\subseteq \sqrt{(2,1-\sqrt{5})}=(2,1-\sqrt{5})$. Now, $\sqrt{(2^{n},1-\sqrt{5})}=(2,1-\sqrt{5})$ is maximal, so by the theorem above, $(2^{n}, 1-\sqrt{5})$ is primary with associated prime $(2,1-\sqrt{5})$.
$(3)$ is prime:
First, $(3)\neq R$ since otherwise $1=3a+3b\sqrt{5}$, contradiction. Next, suppose $(a+b\sqrt{5})(c+d\sqrt{5})=3(e+f\sqrt{5})$, and $a,b$ not both multiple of $3$. Then
$$
ac+5bd=3e, \quad ad+bc=3f.
$$
If $3\mid a$ and $3\nmid b$, then $3\mid c,d$, done. (Similarly, we are done if $3\mid b,3\nmid a$.)
If $3\nmid a$ and $3\nmid b$, then if $3\mid c$ then $3\mid d$, and if $3\mid d$ then $3\mid c$, so we can assume $3\nmid c$ and $3\nmid d$.
case 1: $a\equiv 1 \bmod 3$, $b\equiv 1 \bmod 3$.
If $c\equiv 1$, then $d\equiv 1$ by the first equation above. This contradicts the second equation. If $c\equiv 2$, then $d\equiv 2$, also contradiction. The other $3$ cases are similar. So, $(3)$ is prime.
$(3^{k})$ is primary:
$R/(3)=\{a+b\sqrt{5}+(3):a,b\in\{0,1,2\}\}$ finite integral domain, so a field. Hence $(3)$ is maximal. Also, $\sqrt{(3^{k})}=(3)$ maximal, so $(3^{k})$ is primary with associated prime $(3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Image and kernel of a linear transformation I have a linear transformation:
$$f:\mathbb{R}^3 \rightarrow \mathbb{R}[x]_2$$
$$f((1,1,1))=2x^2-3x,\ f((1,2,3))=-3x,\ f((1,2,4))=2x^2-4x$$
I need to find kernel, image, dimension of image and dimension of kernel.
I tried to find formula describing this transformation:
$f((a,b,c))=(x_1a+y_2b+z_1c)x^2+(x_2a+y_2b+z_2c)x+x_3a+y_3b+z_3c$
$$\begin{cases} x_1+y_1+z_1=2 \\ x_1+2y_1+3z_1=0 \\x_1+2y_1+4z_1=2 \end{cases}$$
$$\begin{cases}x_2+y_2+z_2=-3 \\x_2+2y_2+3z_2=-3 \\x_2+2y_2+4z_2=-4 \end{cases}$$
$$\begin{cases}x_3+y_3+z_3=0 \\x_3+2y_3+3z_3=0 \\x_3+2y_3+4z_3=0\end{cases}$$
$$\begin{cases}x_1=6 \\ y_1=-6 \\ z_1=2 \\ x_2=-4 \\ y_2=2 \\ z_2=-1 \\ x_3=0 \\ y_3=0 \\ z_3=0 \end{cases}$$
$$f((a,b,c))=(6a-6b+2c)x^2+(-4a+2b-c)x$$
Kernel:
$$\begin{cases}6a-6b+2c=0 \\ -4a+2b-c=0\end{cases}$$
$$\begin{cases}b=-a \\ c=-6a\end{cases}$$
$$\ker(f)=\operatorname{Lin}((1,-1,6)) \\\dim\ker(f)=1$$
Unfortunately, I don't know how it is possible to find image (and its dimension) of $f$.
I think that $\dim\operatorname{Im}(f)=3-1=2$, but I am not sure about it.
| Your expression for $f$ is correct. Here is a slightly easier way to get it:
$$f(0,0,1) = f(1,2,4) - f(1,2,3) = 2x^2 - x$$
$$f(0,1,0) = f(1,2,3) - f(1,1,1)-2\cdot f(0,0,1) = -6x^2 + 2x$$
$$f(0,0,1) = f(1,1,1) - f(0,1,0) - f(0,0,1) = 6x^2 - 4x$$
So
$$f(a,b,c) = a\cdot f(1,0,0) + b\cdot f(0,1,0) + c\cdot f(0,0,1) = (6a-6b+2c)x^2+(-4a+2b-c)x$$
Your calculations for the kernel are correct. $\{(1,-1,6)\}$ is the basis for $\operatorname{Ker} f$, and hence $\dim\operatorname{Ker} f = 1$.
Now, for the image, the set
$$\{f(1,0,0), f(0,1,0), f(0,0,1)\} = \{2x^2 - x, -6x^2 + 2x, 6x^2 - 4x\}$$
certainly spans $\operatorname{Im} f$. It only remains to be seen whether it is linearly independent. We obtain:
$$6x^2 - 4x = 6\cdot(2x^2-2) + (-6x^2 + 2x)$$
So
$$\operatorname{Im} f = \operatorname{span} \{2x^2 - x, -6x^2 + 2x, 6x^2 - 4x\} = \operatorname{span} \{2x^2 - x, -6x^2 + 2x\}$$
The set $\{2x^2 - x, -6x^2 + 2x\}$ is linearly independent so it is a basis for $\operatorname{Im}f$. We conclude that $\dim\operatorname{Im}f = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solution of $\int\frac{\sqrt {x^3-4}}{x}dx$ : Need Hints $$\int\frac{\sqrt {x^3-4}} x \, dx$$
My attempt: $ \displaystyle \int\frac{3x^2\sqrt {x^3-4}}{3x^3}\,dx$
Then, substituting $u=x^3$; $\displaystyle \int\frac{\sqrt {u-4}}{3u} \, du$
$$\int\frac{u-4}{3u\sqrt{u-4}} \, du$$
$$\int\frac{1}{3\sqrt{u-4}}\,du-4\int\frac{1}{3u\sqrt{u-4}}\,du $$
I am having trouble with the 2nd part. And Wolfram Alpha says
Can you give me some hints on how to get arctanh function here?
| \begin{align}
w & = \sqrt{u-4} \\
w^2 & = u-4 \\
2w\,dw & = du \\
w^2+4 & = u\\[15pt]
\int \frac{\sqrt {u-4}}{3u} \, du & = \int\frac{w}{3(w^2+4)} (2w\,dw) = \frac 2 3 \int \left( 1 - \frac 4 {w^2+4} \right) dw
\end{align}
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find : $\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)$ with l'Hôpital's rules I want to calculate :
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)$$
with l'Hôpital's rules.
I get
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{x-\sin(x)}{x\sin(x)}\right) = \lim\limits_{x \rightarrow 0}\left(\frac{1-\cos(x)}{\sin(x)+x\cos(x)}\right)$$
but I don't see useful next steps.
| As it turns out, you just have to use L'Hospital's rule twice and observe that on the second try the denominator is no longer zero when you use directly substitution:
\begin{align}
\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin x} - \frac{1}{x}\right)
&=\lim\limits_{x \rightarrow 0}\frac{x -\sin x}{x\sin x}\\
&=\lim\limits_{x \rightarrow 0}\frac{(x -\sin x)'}{(x\sin x)'}\\
&=\lim\limits_{x \rightarrow 0}\frac{x' -(\sin x)'}{x'\sin x+x(\sin x)'}\\
&=\lim\limits_{x \rightarrow 0}\frac{1 - \cos x}{\sin x+x \cos x}\\
&=\lim\limits_{x \rightarrow 0}\frac{(1 - \cos x)'}{(\sin x+x \cos x)'}\\
&=\lim\limits_{x \rightarrow 0}\frac{1' - (\cos x)'}{(\sin x)'+x' \cos x +x(\cos x)'}\\
&=\lim\limits_{x \rightarrow 0}\frac{\sin x}{\cos x + \cos x - x \sin x}\\
&=\lim\limits_{x \rightarrow 0}\frac{\sin x}{2\cos x - x \sin x}\\
&=\frac{0}{2\cdot 1 - 0 \cdot 0}\\
&=\frac{0}{2}\\
&=0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
Use Viete's relations to prove the roots of the equation $x^3+ax+b=0$ satisfy $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3-27b^2$ Use Viete's relations to prove that the roots $x_1$, $x_2$, and $x_3$ of the equation $x^3+ax+b=0$ satisfy the identity $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3-27b^2$.
I know that viete's relations state that the roots $x_1$, $x_2$, and $x_3$ of the equation $x^3-px^2+qx-r=0$ have the property $p=x_1+x_2+x_3$, $q=x_1x_2+x_1x_3+x_2x_3$ and $r=x_1x_2x_3$.
My question is whether or not there is a way to do this without multiplying out $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2$ and showing that it factors into $4(x_1+x_2+x_3)^3+27(x_1x_2x_3)$ because that algebra involved in that looks like it will be nasty.
| Let's first prove Vieta's cubic relations.
$\underline {\text{Proof}}$: Let the roots of a cubic polynomial, $f(x)$, be $\alpha$, $\beta$, $\gamma$. Then $f(x) = (x-\alpha)(x-\beta)(x-\gamma)$. Let $f(x) = x^3 - px^2 + qx - r$.
\begin{align}
f(x) & = (x-\alpha)(x-\beta)(x-\gamma) \\
& = (x^2 - \beta x - \alpha x + \alpha \beta) (x - \gamma) \\
& = (x^3 - \gamma x^2 - \beta x^2 + \beta \gamma x - \alpha x^2 + \alpha \gamma x + \alpha \beta x - \alpha \beta \gamma) \\
& = x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha \beta + \alpha \gamma + \beta \gamma)x - (\alpha \beta \gamma) \\
\implies p & = \alpha + \beta + \gamma \\
q & = \alpha \beta + \alpha \gamma + \beta \gamma \\
r & = \alpha \beta \gamma
\end{align}
The proof of a cubic discriminant is quite long and there is no easy way to do it, so I'll link a proof here.
Now with the question.
$\underline {\text{Proof}}$: A cubic polynomial $f(x) = (x-\alpha)(x-\beta)(x-\gamma) = x^3 - px^2 + qx - r$ has a discriminant
\begin{align}
\Delta_3 & = (\alpha-\beta)^2 (\alpha-\gamma)^2 (\beta-\gamma)^2 \\
\end{align}
Assume $p=0$. We have that $q$ and $r$ are polynomials of degrees $2$ and $3$. The discriminant is a polynomial of degree $6$ (look at the definition just above), and hence the discriminant must be a linear combination of the one-term polynomials, $q^3$ and $r^2$.
$$\Delta_3 = mq^3 + nr^2$$
where $m$ and $n$ are two constants. Now, we can do something clever to get our final result. Let $q = -1$ and $r=0$. Now we have a new polynomial
\begin{align}
0 & = x^3 - x \\
& = x(x^2 - 1)\\
& = x(x-1)(x+1)\\
\end{align}
with roots $-1, 0, 1$.
\begin{align}
\Delta_3 & = (\alpha-\beta)^2 (\alpha-\gamma)^2 (\beta-\gamma)^2 \\
& = (-1 - 0)^2 (-1 - 1)^2 (0 - 1)^2\\
& = 1 \cdot 4 \cdot 1\\
& = 4 \\
\end{align}
So far, if you've been keeping up, we have $\Delta = 4q^3 + br^2$.
In a similar method, if we set $q=0$, $r=−1$ we get the polynomial $x^3−1=0$. Now we are in the territory of complex numbers! Solving for roots of $x^3-1=0$ we get $1, \omega, \omega^2$. $\omega$ is a third root of unity.
\begin{align}
\Delta_3 & = (\alpha-\beta)^2 (\alpha-\gamma)^2 (\beta-\gamma)^2 \\
& = (1 - \omega)^2 (1 - \omega^2)^2 (\omega - \omega^2)^2 \\
\end{align}
Finishing this we get the discriminant is equal to $27$.
Hence, we have
\begin{align}
\Delta_3 &= 4q^3 + 27r^2 \\
& = 4(\alpha \beta + \alpha \gamma + \beta \gamma)^3 + 27(\alpha \beta \gamma)^2 \\
\end{align}
And we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Convergence and divergence of an infinite series The series is
$$1 + \frac{1}{2}.\frac{x^2}{4} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6}.\frac{x^4}{8} + \frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10}.\frac{x^6}{12}+... , x\gt0$$
I just stuck over the nth term finding and once I get nth term than I can do different series test but here I am unable to find the nth term of the given series please help me out of this.
The question is different as it contains x terms and its nth term will be totally different from marked as duplicate question
| Notice
$$
\begin{align}
\frac{1}{2\cdot 4}x^2 &= \frac{2!}{(2^1 1!)^2(2 \cdot 2)} x^2
= \frac12 \binom{2}{1} \int_0^x \left(\frac{t}{4}\right)^1 dt
\\
\frac{1\cdot 3\cdot 5}{2\cdot 4 \cdot 6 \cdot 8}x^4
&= \frac{6!}{(2^3 3!)^2 (2\cdot 4)} x^4
= \frac12 \binom{6}{3} \int_0^x \left(\frac{t}{4}\right)^3 dt\\
\frac{1\cdot 3\cdot 5\cdot 7\cdot 9}{2\cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12}x^6
&= \frac{10!}{(2^5 5!)^2 (2\cdot 6)} x^6
= \frac12 \binom{10}{5} \int_0^x \left(\frac{t}{4}\right)^5 dt
\end{align}
$$
So aside from the first constant term $1$, the $k^{th}$ non-constant term has the form
$$a_k = \frac12\binom{2\ell}{\ell}\frac{x^{\ell+1}}{2^{2\ell+1}(\ell+1)} = \frac12\binom{2\ell}{\ell}\int_0^x \left(\frac{t}{4}\right)^\ell dt
$$ where $\ell = 2k-1$, the $k^{th}$ odd number.
If one compute the ratio of successive terms of $a_k$, one find
$$\frac{a_{k+1}}{a_k} = \frac{\binom{4k+2}{2k+1}(2k)}{\binom{4k-2}{2k-1}(2k+2)} \frac{x^4}{2^4}
= \frac{(4k+2)(4k+1)(4k)(4k-1)}{(2k+1)^2(2k)(2k+2)}\frac{x^4}{2^4}
= \frac{\left(1-\frac{1}{4k}\right)^2}{\left(1 + \frac{1}{2k}\right)\left(1+\frac1k\right)} x^4$$
Since this ratio converges to $x^4$ as $k \to \infty$, the radius of convergence of the series is $1$.
In fact, for $|x| < 1$, the series has following representation
$$1 + \frac12\sum_{\ell\text{ odd}}\binom{2\ell}{\ell}\int_0^x \left(\frac{t}{4}\right)^\ell dt
= 1 + \frac14\sum_{\ell=0}^\infty \binom{2\ell}{\ell} (1 - (-1)^{\ell})\int_0^x \left(\frac{t}{4}\right)^\ell dt
$$
For $|z| < \frac14$, we have following identity:
$$\sum_{\ell=0}^\infty \binom{2\ell}{\ell} z^\ell = \frac{1}{\sqrt{1-4z}}$$
From this, we find for $|x| < 1$, the series evaluates to
$$1 + \frac14 \int_0^x \left(\frac{1}{\sqrt{1-t}} - \frac{1}{\sqrt{1+t}}\right) dt
= 2 - \frac12\left( \sqrt{1+x} + \sqrt{1-x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Find all real solutions for the system: $x^3=y+y^5$, $y^5=z+z^7$, $z^7=x+x^3.$
Given: $$\left\{
\begin{array}{l}
x^3=y+y^5\\
y^5=z+z^7\\
z^7=x+x^3
\end{array}
\right.
$$
Find: all real solutions for the system.
From a book on preparation for math contests. The answer states there is just one solution. My problem is showing that this is indeed the case.
My attempt: it is easy to see that one solution is $(x,y,z)=(0,0,0)$. And, adding the equations, we get to $x+y+z=0$. But my problem is showing that this solution is indeed unique, if the answer provided in the book is right.
Hints and answers are appreciated. Sorry if this is a duplicate.
| Multiplying the three equations gives $x^3 y^5 z^7 = x y z (1+x^2)(1+y^4)(1+z^6)\,$.
If $xyz \ne 0$ then it follows that $x^2 y^4 z^6 = (1+x^2)(1+y^4)(1+z^6)\,$, but the latter is not possible since $0 \le x^2 \lt 1+ x^2, 0 \le y^4 \lt 1+y^4, 0 \le z^6 \lt 1+ z^6\,$.
Therefore $xyz=0\,$, and it is easy to show that any one of $x,y,z$ being $0$ implies all of them being $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2566058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Polynomial long division - intermediate steps missing I'm trying to find the intermediate step:
$$\frac{1}{(x-\frac{x^2}{2}+\frac{x^3}{12}-\frac{x^4}{144}+...)^2} = \frac{1}{x^2}+\frac{1}{x}+\frac{7}{12}+\frac{19x}{72} ...$$
Is there a quick way to find these first few terms that I'm missing here? Thanks
| The first terms of the doninator of the LHS are$$x^2-x^3+\frac{5 x^4}{12}-\frac{7 x^5}{72}+\cdots=x^2\left(1-x+\frac{5 x^2}{12}-\frac{7 x^3}{72}+\cdots\right).$$Therefore, the RHS can be written as$$\frac1{x^2}\left(b_0+b_1x+b_2x^2+b_3x^3+\cdots\right)$$and we must have$$\left(1-x+\frac{5 x^2}{12}-\frac{7 x^3}{72}+\cdots\right)\left(b_0+b_1x+b_2x^2+b_3x^3+\cdots\right)=1$$and therefore$$\left\{\begin{array}{l}b_0=1\\b_1-b_0=0\\b_2-b_1+\frac5{12}b_0=0\\b_3-b_2+\frac5{12}b_1-\frac7{72}b_0=0,\end{array}\right.$$from which we get that$$\left\{\begin{array}{l}b_0=1\\b_1=1\\b_2=\frac7{72}\\b_3=\frac{19}{72}.\end{array}\right.$$But that's all you can get from the information provided.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How do you go about factoring $x^3-2x-4$? How do you factor $x^3-2x-4$?
To me, this polynomial seems unfactorable.
But I check my textbook answer key, and the answer is $(x-2)(x^2+2x+2)$.
So I got to solve backward:
$$x(x^2-2)-4$$
And add some terms and subtract them later,
$$x(x^2+2x+2-2x-4)-4=0$$
$$x(x^2+2x+2)-4-2x^2-4x=0$$
$$x(x^2+2x+2)-2(x^2+2x+2)=0$$
But I would have thought of this way to factor if I didn't look at the answer key.
| The way I factor
$x^3 - 2x - 4 \tag 1$
is to notice that $2$ is a root; this I do by intuition and a good amount of experience. $2$ looks like a good guess to me since the non-leading coefficients are divisible by $2$; once I find that $2$ is a zero of (1), I use synthetic division to find $q(x)$ such that
$x^3 - 2x - 4 = (x - 2)q(x), \tag 2$
thus: $x - 2$ into $x^3$ yields $x^2$;
$x^3 - 2x - 4 - x^2(x - 2) = 2x^2 - 2x - 4; \tag 3$
$x -2$ in to $2x^2$ yields $2x$:
$2x^2 - 2x - 4 - 2x(x - 2) = 2x - 4; \tag 4$
$x - 2$ into $2x - 4$ yields $2$, and since
$2x - 4 - 2(x - 2) = 0, \tag 5$
we are done. Gathering the partial quotients yields $x^2 + 2x + 2$, and it is easily checked that
$x^3 -2x -4 = (x - 2)(x^2 + 2x + 2); \tag 6$
now if we want to go further we can use the quadratic formula to find the roots $\mu_\pm$ of $x^2 + 2x + 2$:
$\mu_\pm = \dfrac{-2 \pm \sqrt{-4}}{2} = \dfrac{-2 \pm 2i}{2} = -1 \pm i; \tag 7$
we easily check
$x^2 + 2x + 2 = (x + (1 + i))(x + (1 - i)); \tag 8$
therefore
$x^3 - 2x - 4 = (x - 2)(x + (1 + i))(x + (1 - i)). \tag 9$
And that's the way I factor it!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2567246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$? Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase i.e.,
Is
$$\sum_{i=1}^n i^5+i^7=2\left( \sum_{i=1}^ni\right)^4$$ true $\forall $ $n \in \mathbb{N}$
| Both sides are polynomials in $n$ of degree $8$. Since they coincide for $n=0,\dots,8$, they are equal.
Any $9$ points will do. Taking $n=-4,\dots,4$ is probably easier to do by hand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
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If $\log_{10}15=a$ and $\log_{20}50=b$, express $\log_940$ in terms of $a$ and $b$. I've done this so far: $$\log_940=\frac{\log_{10}40}{\log_{10}9}=\frac{1+\log_{10}4}{a+\log_{10}6-1}.$$ How do I proceed?
| Well, this will be little messy but here is an attempt:
$$b = \frac{\log_{10}50}{\log_{10}20} = \frac{1+\log_{10}5}{1+\log_{10}2} = \frac{2-\log_{10}2}{1+\log_{10}2} \implies \log_{10}2 = \frac{2-b}{1+b}$$
and
$$a = \log_{10}15 = \log_{10}3+\log_{10}5 = \log_{10}3 + 1 - \log_{10}2 \Rightarrow \log_{10}3 = a+\log_{10}2-1$$
and notice that
$$\log_{9}40 = \frac{\log_{10}40}{\log_{10}9} = \frac{1+2\log_{10}2}{2\log_{10}3}$$
By the above equations, you have $\log_{10}2$ and $\log_{10}3$ in terms of $a$ and $b$ so you can find the answer from that.
Also, note that in above equations, we used the fact that $\log_{10}5 = 1-\log_{10}2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let$A$ be a $3\times3$ real symmetric matrix such that $A^6=I$ . Then $A^2=I$ Let$A$ be a $3\times3$ real symmetric matrix such $ A^6=I$ . Then $A^2=I$.
How can I prove this statement is true or false?
As it is given $A$ is symmetric so $A=A^T$. Also $ A^6=I$.
But the main problem is that I can't operate $A^{-1}$ on both sides whether it is invertible or not. Can any help me what should I do?
| $A^6=I$ gives $x^6-1$ $=(x^2-1)(x^2-x+1)(x^2+x+1)$ is an annihilating polynomial of $A$. Let $m(x)$ be the minimal polynomial of $A$ then $m(x)$ divides $x^6-1$.
Since all eigenvalues of a real symmetric matrix are real and $x^2\pm x+1$ does not contain any real root, $m(x)$ cannot contain any factor of $x^2-x+1$ or $x^2+x+1$. Thus $m(x)$ divides $x^2-1$, which gives $x^2-1$ is an annihilating polynomial of $A$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Compare sum of radicals I am stuck in a difficult question:
Compare $18$ and
$$
A=\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}
$$
without using calculator.
Thank you for all solution.
| $\sqrt{7}$ and $\sqrt{11}$ can be written as $\sqrt{9\pm 2}$ and similarly $\sqrt{32}$ and $\sqrt{40}$ can be written as $\sqrt{36\pm 4}$.
Since $\sqrt{x}$ is a concave function on $\mathbb{R}^+$,
$$ \sqrt{9-2}+\sqrt{9+2}+\sqrt{36-4}+\sqrt{36+4} \color{red}{\leq} 2\sqrt{9}+2\sqrt{36} = 18.$$
We may also estimate the difference between the RHS and the LHS:
$$\begin{eqnarray*} 2n-\sqrt{n^2+x}-\sqrt{n^2-x}&=&\frac{x}{n+\sqrt{n^2-x}}-\frac{x}{n+\sqrt{n^2+x}}\\&=&x\cdot \frac{\sqrt{n^2+x}-\sqrt{n^2-x}}{(n+\sqrt{n^2-x})(n+\sqrt{n^2+x})}\\&=&\frac{2x^2}{(n+\sqrt{n^2-x})(n+\sqrt{n^2+x})(\sqrt{n^2+x}+\sqrt{n^2-x})}\\&\geq&\frac{x^2}{n(3n^2+\sqrt{n^4-x^2})}\geq\frac{x^2}{n\left(4n^2-\frac{x^2}{2n^2}\right)}.\end{eqnarray*} $$
This leads to $18-\left(\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}\right)\geq \frac{1}{18}$.
| {
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Find $x$ given $\sqrt{x+14-8\sqrt{x-2}}$ + $\sqrt{x+23-10\sqrt{x-2}} = 3$ If $\sqrt{x+14-8\sqrt{x-2}}$ + $\sqrt{x+23-10\sqrt{x-2}} = 3$, then what is the value of $x$?
Is there an easy way to solve such equation, instead of squaring on both sides and replacing $\sqrt{x-2}$ with a different variable?
| Condition: $x\geq 2$.
You have $$LHS = \sqrt{(\sqrt{x-2}-4)^2} + \sqrt{(\sqrt{x-2}-5)^2}.$$
So, if $\sqrt{x-2} < 4$, or $x<18$, then $$LHS = 4-\sqrt{x-2} + 5-\sqrt{x-2} = 9-2\sqrt{x-2} = 3 = RHS.$$
If $4 \leq \sqrt{x-2} < 5$, or $18\leq x < 27$, then $$LHS = \sqrt{x-2} -4 + 5-\sqrt{x-2} = 1 \neq 3 = RHS.$$
If $\sqrt{x-2} \geq 5$, or $x \geq 27$, then $$LHS = \sqrt{x-2} - 4+ \sqrt{x-2} -5= 2\sqrt{x-2} - 9 = 3 = RHS.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving whether the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. I've updated my proof to be complete now, edited for proof-verification!
We know that for the partial sums with even an uneven terms, the following holds:
$S_{2N}=\sum_{n=1}^{2N} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\frac{1}{4} +\frac{1}{3} -\frac{1}{6} +\frac{1}{5}-\dots+\frac{1}{2N-1}$
$= \frac{1}{2\times1} +\frac{1}{4\times3} + \frac{1}{6\times5}+\dots+\frac{1}{2N(2N-1)} = \sum_{n=1}^{2N} \frac{1}{2n(2n-1)}$
We may rewrite the series in pairs as we know it will have an even amount of terms.
$S_{2N+1} = \sum_{n=1}^{2N+1} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\frac{1}{4} +\frac{1}{3}-\dots+\frac{1}{2N-1} -\frac{1}{2N+2}$
$=\frac{1}{2\times1} +\frac{1}{4\times3} + \frac{1}{6\times5}+\dots+\frac{1}{2N(2N-1)} - \frac{1}{2N+2} = \sum_{n=1}^{2N} \frac{1}{2n(2n-1)}-\frac{1}{2N+2}$
As $n\in\mathbb{N}$, we know that $n\geq1$ so:
$n\geq1 \iff 3n\geq3 \iff 3n^2\geq3n \iff 3n^2-3n\geq0 \iff 4n^2-2n \geq n^2+n$
So: $2n(2n-1)\geq n(n+1) \iff \frac{1}{2n(2n-1)}\leq \frac{1}{n(n+1)}$ for all $n\geq1$.
As the series of the latter sequence converges, we can conclude, by the comparison, test that the series $\sum_{n=1}^{2N} \frac{1}{2n(2n-1)}$ converges.
Suppose it converges to $s$, then we know $^{\lim S_{2N}}_{N\to\infty} = s$ and thus $\lim_{N\to\infty}[S_{2N+1}] = s - (\lim_{N\to\infty}[\frac{1}{2N+2}]) = s-0 = s.$ As the partial sums ending with even and uneven terms both converge to the same limit, the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. $\tag*{$\Box$}$
| Let:
$$a_n=\frac{1}{n-(-1)^n}$$
Note that:
*
*for $n=2k-1 \implies a_{2k-1}=\frac{1}{2k}$
*for $n=2k \implies a_{2k}=\frac{1}{2k-1}$
*for $n=2k+1 \implies a_{2k+1}=\frac{1}{2k+2}$
*for $n=2k+2 \implies a_{2k+2}=\frac{1}{2k+1}$
Therefore we can reorder the series $a_n\to b_n$ in such way that $b_n$ is monolitically decreasing, since:
$$\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}=\sum_{n=1}^\infty (-1)^na_n=\sum_{n=1}^\infty (-1)^nb_n<+\infty$$
the series converges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Hypergeometric Function Value Wolfram gives the following:
$$_2F_1 \left(a,b;\frac{a+b+1}{2};\frac{1}{2}\right) = \frac{\sqrt{\pi}
\Gamma \left(\frac{a+b+1}{2} \right)} {\Gamma \left(\frac{a+1}{2} \right) \Gamma \left(\frac{b+1}{2} \right) }$$ and I am struggling to rederive it.
My ultimate aim is to allow the final parameter (the 1/2) to vary. I am fairly certain that the function is increasing up to 1/2, decreasing afterwards, but I am struggling to prove it. Any help in that direction would also be great.
Thanks!
| Here is a derivation of your formula using a transformation and the
Gauss's Hypergeometric Theorem. The quadratic transformation
http://dlmf.nist.gov/15.8.E18
$${_2}F_1 \left(a,b;\frac{a+b+1}{2};z\right)={_2}F_1 \left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};4z(1-z) \right)$$
gives
$${_2}F_1 \left(a,b;\frac{a+b+1}{2};\frac{1}{2}\right)={_2}F_1 \left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};1 \right)$$
With the well known Gauss's Hypergeometric Theorem
$${_2}F_1\left(a,b;c;1\right)=\frac{\Gamma\left(c\right)\Gamma\left(c-a-b\right)}{%
\Gamma\left(c-a\right)\Gamma\left(c-b\right)}.$$
you get
$$
{_2}F_1 \left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};1 \right) =
\frac{\Gamma\left(\frac{a+b+1}{2}\right)\Gamma\left(\frac{a+b+1}{2}-\frac{a}{2}-\frac{b}{2}\right)}
{\Gamma\left(\frac{a+b+1}{2}-\frac{a}{2}\right)\Gamma\left(\frac{a+b+1}{2}-\frac{b}{2}\right)}$$
$$
{_2}F_1 \left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};1 \right) =
\frac{\Gamma\left(\frac{a+b+1}{2}\right)\Gamma\left(\frac{1}{2}\right)}
{\Gamma\left(\frac{b+1}{2}\right)\Gamma\left(\frac{a+1}{2}\right)}$$
$$
{_2}F_1 \left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};1 \right) =
\frac{\sqrt{\pi}\;\Gamma\left(\frac{a+b+1}{2}\right)}
{\Gamma\left(\frac{b+1}{2}\right)\Gamma\left(\frac{a+1}{2}\right)}$$
Since $4z(1-z)$ has a maximum for $z=\frac{1}{2}$ the first formula should be
a good starting point for your conjecture, "that the function is increasing up to 1/2, decreasing afterwards".
| {
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Convergence/absolute convergence of $\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin \frac{1}{2n+1}\right)$
Does the following sum converge? Does it converge absolutely?
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}\right)$$
I promise this is the last one for today:
Using Simpson's rules:
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}\right) = \sum_{n=1}^\infty 2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}$$
Now, $$\left|2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}\right| \leq \frac{2}{8n² + 4n}$$
hence by the comparison test, the series converges absolutely, and hence it also converges. Is this correct?
| As an alternative, since:
$$\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}=\frac{1}{2n}-\frac{1}{2n+1}+o\left(\frac{1}{n^2}\right)=\frac{1}{2n(2n+1)}+o\left(\frac{1}{n^2}\right)$$
the series converges by limit comparison with:
$$\sum_{n=1}^\infty \frac{1}{2n(2n+1)}$$
| {
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Is it possible to compute $\int_{-\infty}^\infty {x \sin x\over x^4+1}dx$ without using complex analysis?
$$\int_{-\infty}^\infty {x\sin x\over x^4+1}dx$$
By using things like Feynman's trick (Differentiating under the integral sign) or other high school level calculus concepts? I do not know things like residue theorem.
| The same answer as @JackD'Aurizio with slightly more details. I was doing this after he sent his answer so I finish anyway.
Let $\alpha=\frac{1}{2\sqrt[]{2}}$. After partial fraction decomposition we get:
\begin{align}
\int^\infty_{-\infty}\frac{x\sin(x)}{x^4+1}\,dx=-\alpha\int^\infty_{-\infty}\frac{\sin(x)}{x^2+\sqrt[]{2}x+1}\,dx+\alpha\int^\infty_{-\infty}\frac{\sin(x)}{x^2-\sqrt[]{2}x+1}\,dx
\end{align}
Now define:
\begin{align}
I:=\int^\infty_{-\infty}\frac{\sin(x)}{x^2+\sqrt[]{2}x+1}\,dx,\ \ \ \ J:=\int^\infty_{-\infty}\frac{\sin(x)}{x^2-\sqrt[]{2}x+1}\,dx
\end{align}
First we do $I$. Completing the squares in the denominator yields:
\begin{align}
I=\int^\infty_{-\infty}\frac{\sin(x)}{(x+\frac{\sqrt[]{2}}{2})^2+\frac{1}{2}}\,dx
\end{align}
Substitute $u=x+\frac{\sqrt[]{2}}{2}$ and use the formula for $\sin(x+y)$:
\begin{align}
I&=\int^\infty_{-\infty}\frac{\sin\left(u-\frac{\sqrt[]{2}}{2}\right)}{u^2+\frac{1}{2}}\,du\\& = \int^\infty_{-\infty}\frac{\sin\left(u\right)\cos\left(-\frac{\sqrt[]{2}}{2}\right)}{u^2+1}\,du+\int^\infty_{-\infty}\frac{\cos(u)\sin\left(-\frac{\sqrt[]{2}}{2}\right)}{u^2+\frac{1}{2}}\,du\\
\end{align}
The integral with $\sin(u)$ is convergent and the ingerand is an odd function, so it vanishes. Now we get:
\begin{align}
I=\sin\left(-\frac{\sqrt[]{2}}{2}\right)\int^\infty_{-\infty}\frac{\cos(u)}{u^2+\frac{1}{2}}\,du
\end{align}
This integral can be done with the explicitly mentioned trick by you, namely Feynman's trick. It is done nicely with full details by user Mark Viola in this answer.
So we get:
\begin{align}
I=\sin\left(-\frac{\sqrt[]{2}}{2}\right) \frac{\pi\sqrt[]{2}}{e^{1/\sqrt[]{2}}}=-\sin\left(\frac{\sqrt[]{2}}{2}\right) \frac{\pi\sqrt[]{2}}{e^{1/\sqrt[]{2}}}
\end{align}
The same can be done for $J$ (do it yourself!). We get:
\begin{align}
J=\sin\left(\frac{\sqrt[]{2}}{2}\right)\frac{\pi\sqrt[]{2}}{e^{1/\sqrt[]{2}}}
\end{align}
Finally we get:
\begin{align}
\int^\infty_{-\infty}\frac{x\sin(x)}{x^4+1}\,dx = \frac{\pi}{e^{1/\sqrt[]{2}}}\sin\left(\frac{\sqrt[]{2}}{2}\right)
\end{align}
| {
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$\max a^3+b^3+c^3+4abc$ sub $0\leq a,b,c \le 3/2$ and $a+b+c=3$ Let $S$ be the set of $(a,b,c) \in \mathbb{R}^3$ such that $0\leq a,b,c \leq \frac{3}{2}$ and $a+b+c=3$. Find
$$
\max_{(a,b,c) \in S} a^3+b^3+c^3+4abc.
$$
| I don't know a nice solution. However, the gradient of $f(a,b,c)=a^3+b^3+c^3+4abc-\lambda(a+b+c-3)$ is
$$
\nabla f(a,b,c,\lambda)=\left(3a^2+4bc-\lambda , 3b^2+4ac-\lambda , 3c^2+4ab-\lambda , 3-a-b-c\right).
$$
By Weierstrass' theorem, the maximum exists on, let's say, $(x,y,z,w)$. Hence, a necessary condition is $\nabla f(x,y,z,w)=(0,0,0,0)$. From the first two conditions here we get
$$
3x^2+4yz-w =3y^2+4 xz-w \implies (x-y)(3x+3y-4z)=0\,\,\,\,(\star)
$$
and the other two cyclic conditions.
We claim that at least two variables need to be equal. This is clear in the case that $x-y=0$ (resp. $y=z$ ro $z=x$) in $(\star)$. Otherwise we have wlog $3x+3y-4z=3y+3z-4x$, which implies $x=z$ (and similarly the others). Hence, we reduce to the two variable problem: maximize
$$
g(x,y)=2x^3+y^3+4x^2y\,\,\, \text{ sub } 2x+y=3 \text{ and }0\le x,y \le \frac{3}{2}.
$$
The condition $0\le y=3-2x \le \frac{3}{2}$ implies that, substituing $y=3-2x$, we rewrite the problem as to maximize, under the condition $3/4 \le x \le 3/2$,
$$
h(x):=g(x,3-2x)=2x^3+(3-2x)^3+4x^2(3-2x) \implies h^\prime(x)=6(-7x^2+16x-9).
$$
Considering that $-7x^2+16x-9$ has the maximum in $8/7$ and has a $0$ in $1$, the other zero is in $9/7$, which is smaller than $3/2$. On the other hand, $h(3/4)$ is greater than $h(9/7)$, hence the maximum of the function is at the point $(3/4, 3/4, 3/2)$, that is,
$$
2\left(\frac{3}{4}\right)^3+\left(\frac{3}{2}\right)^3+4\cdot \frac{3^3}{32}=\frac{243}{32}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral of rational function - which contour to use? Evaluate : $$\int_{-\infty}^{+\infty} \frac {x}{(x^2+2x+2)(x^2+4)}$$
I found that the integrand can be extended to a function on a complex plane has simple poles at $\pm 2i$ and $-1\pm i$. Now I want to compute the integral by contour integration but I am unable to assume any contour here.
Do excuse me , if my approach is wrong.
| The computation of residues leads to the partial fraction decomposition
$$ \frac{x}{(x^2+2x+2)(x^2+4)}=\frac{1}{10}\left[\frac{4-x}{x^2+4}+\frac{x-2}{x^2+2x+2}\right] $$
and now you don't have to pick any contour, since
$$\int_{\mathbb{R}}\frac{4\,dx}{x^2+4}=2\pi,\qquad \int_{\mathbb{R}}\frac{3\,dx}{(x+1)^2+1}=\int_{\mathbb{R}}\frac{3\,dx}{x^2+1}=3\pi$$
and
$$ \lim_{M\to +\infty}\int_{-M}^{M}\left[\frac{x+1}{(x+1)^2+1}-\frac{x}{x^2+4}\right]=0$$
ensure the following identity:
$$ \int_{-\infty}^{+\infty}\frac{x\,dx}{(x^2+2x+2)(x^2+4)} = \frac{2\pi-3\pi}{10} = \color{red}{-\frac{\pi}{10}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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product of terms taken $3$ at a time in polynomial expression
Finding product of terms taken $3$ at a time in $\displaystyle \prod^{100}_{r=1}(x+r)$
Try:
$$\displaystyle \prod^{100}_{r=1}(x+r)=x^{100}+(1+2+3+\cdots +100)x^{99}+(1\cdot 2+1\cdot 3+\cdots+100\cdot 99)x^{98}+(1\cdot 2\cdot 3+2\cdot 3 \cdot 4+\cdot\cdots+98\cdot 99\cdot 100)x^{98}+\cdots$$
for $1$ at a time (Coefficient of $x^{99}$) is $\displaystyle \sum^{100}_{i=1}i = 50\cdot 101$
for $2$ at a time (Coefficient of $x^{98}$) is $\displaystyle \sum^{100}_{i=1}\sum^{100}_{j=1\;, (1\leq i<j \leq 100)}i \cdot j= \frac{1}{2}\bigg[\bigg(\sum^{100}_{i=1}i\bigg)^2-\sum^{100}_{i=1}i^2\bigg]$
But could some help me how to find coefficient of $x^{97},$ thanks
| If you try to expand the multiplications, you'll find that to produce terms with $x^{98}$, you'll need $2$ distinct constant factors (from the set $\{1,2,\dots,100\})$ and $98$ number of $x$ factors. So the sum of these terms, $a_{98}$, would be $$a_{98}=\sum_{n=1}^{100}\sum_{m=n+1}^{100}mn$$
How can we find this sum? Note that $$(1+2+…100)^2= 1^2+2^2+… 100^2 + 2a_{98}$$ $$\implies a_{98} = \frac12 \left[(5050)^2- \sum_{i=1}^{100} i^2 \right]$$ $$\implies \boxed{ a_{98} = 12582075}$$
| {
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Prove by induction that for every $n \ge 0$ the term $2^{2n+1} - 9n^2 + 3n -2$ is divisible by $54$ What I get is $4 \cdot 54k + 27(n(n-1)) - 2$.
I understand first two terms but $-2$ is giving me a headache. I suppose to be true every term should be divisible by $54$. Or did I make wrong conclusion?
P. S: Is there mathjax tutorial?
| $2^{2n+1}-9n^2 +3n -2 = 54k$
$2^{2n+1} =54k +9n^2 -3n +2$
$2^{2(n+1)+1} -9(n+1)^2 +3(n+1) -2=$
$2^2(54k +9n^2 -3n +2) -9(n+1)^2 +3(n+1) -2$
After I compute everything, I get this: $4\cdot54k +27(n(n+1)) -2.$
| {
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"timestamp": "2023-03-29T00:00:00",
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For $a,b>0$, $c>1$, and $n=3,4,\dots$ is $|ca^n+(-b)^n|\leq(ca^2+b^2)^{n/2}$ always true? Let $a,b>0$, $c>1$, and $n=3,4,\dots$, then is
\begin{equation}
|ca^n+(-b)^n|\leq(ca^2+b^2)^{n/2},
\end{equation}
true for all possible combinations of $a$, $b$, $c$, and $n$?
| Hint: prove first that $\,\big(c+x^n\big)^2 \le \big(c+x^2\big)^n\,$ for $\,c \ge 1, x \ge 0, n \ge 2\,$, using for example the binomial expansion, noting that $\,\color{red}{c^2} \le \color{red}{c^n}\,$ and $\,\color{blue}{2cx^n}\,$ is no larger than either of the two blue terms on the RHS depending on whether $\,x \ge 1\,$ (when $\,x^n \le x^{2n-2}\,$) or $\,x \le 1\,$ (when $\,x^n \le x^{2}\,$):
$$\require{cancel}
\color{red}{c^2} + \color{blue}{2cx^n} + \cancel{x^{2n}} \le \color{red}{c^n} + \color{blue}{n \cdot c^{n-1}x^2} + \ldots + \color{blue}{n \cdot c x^{2n-2}} + \cancel{x^{2n}}
$$
Then, use the above for $\,x=b/a\,$, multiply the inequality by $\,a^{2n}\,$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Find $f(f(f(f(f(f(\cdots f(x)))))))$ $2018$ times
I was given a problem to calculate $f(f(f(\dots f(x))))$ $2018$ times if $f(x)=\frac{x}{3} + 3$ and $x=4$.
Is it even possible to be done without a computer?
| if $f(x)=\frac{x}{3}+3$
Then $f^2(x)=\frac{f(x)}{3}+3=\frac{x}{9}+\frac{3}{3}+3$
and $f^3(x)=\frac{f^2(x)}{3}+4=\frac{x}{27}+\frac{3}{9}+\frac{3}{3}+3$
By induction the general form is emerging as:
$$f^n(x)=3^{-n}x+3\left(\sum_{k=1}^n3^{1-k}\right)$$
Use the formula for a geometric sum and plug in $x=4, n=2018$:
$f^{2018}(4)=4+\frac{1}{2}\left(1-3^{-2018}\right)$
| {
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"url": "https://math.stackexchange.com/questions/2590631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
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} |
$\lim_{n\rightarrow \infty}\left[\frac{\left(1+\frac{1}{n^2}\right)\cdot \cdots\cdots \left(1+\frac{n}{n^2}\right)}{\sqrt{e}}\right]^n$
$$\lim_{n\rightarrow \infty}\Bigg[\frac{\bigg(1+\frac{1}{n^2}\bigg)\bigg(1+\frac{2}{n^2}\bigg)\cdots\cdots \bigg(1+\frac{n}{n^2}\bigg)}{\sqrt{e}}\Bigg]^n$$
Try: $$y=\lim_{n\rightarrow \infty}\Bigg[\frac{\bigg(1+\frac{1}{n^2}\bigg)\bigg(1+\frac{2}{n^2}\bigg)\cdots\cdots \bigg(1+\frac{n}{n^2}\bigg)}{\sqrt{e}}\Bigg]^n$$
$$\log_{e}(y) =n\sum^{n}_{r=1}\log_{e}\bigg(1+\frac{r}{n^2}\bigg)-\frac{n}{2}$$
could some help me to solve it , thanks
| We have that
$$\begin{align}
\left[\frac{\left(1+\frac{1}{n^2}\right)\cdots \left(1+\frac{n}{n^2}\right)}{\sqrt{e}}\right]^n&=\frac{\left(\frac{(n^2+n)!}{(n^2)!}\right)^n}{n^{2n^2} e^{n/2}}
\sim\frac{\left(1+\frac{1}{n}\right)^{n^3+n^2+\frac{n}{2}}}{e^{n^2+n/2}}
\\
&\sim\frac{\exp\left((n^3+n^2+\frac{n}{2})\ln\left(1+\frac{1}{n}\right)\right)}{e^{n^2+n/2}}\\
&\sim\frac{\exp\left((n^2+n+\frac{1}{2})-\frac{n+1}{2}+\frac{1}{3}\right)}{e^{n^2+n/2}}\to e^{1/3}
\end{align}$$
where we used the Stirling approximation for factorials and the expansion $$\ln(1+t)= t-\frac{t^2}{2}+\frac{t^3}{3}+o(t^3).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A problem of determinants Suppose that $a,b,c,d \in \mathbb R$. Then
$$\begin{vmatrix} a& -b & -c& -d\\b& a& d& -c\\c& -d& a& b\\d& c& -b& a \end{vmatrix} = 0 \iff a＝b＝c＝d＝0$$
I can prove this by computing this determinant exactly. But I wonder if there is any brief approach since the matrix has the special structure
$$\begin{bmatrix}A& -B^t\\B& A^t \end{bmatrix}$$
| In this case your matrix $M$ satisfies
$$MM^t=\pmatrix{a^2+b^2+c^2+d^2&0&0&0\\
0&a^2+b^2+c^2+d^2&0&0\\
0&0&a^2+b^2+c^2+d^2&0\\
0&0&0&a^2+b^2+c^2+d^2\\
}$$
so that $\det(M)^2=\det(MM^t)=(a^2+b^2+c^2+d^2)^2$. If
$M=\pmatrix{A&-B^t\\B&A^t}$ then
$$MM^t=\pmatrix{AA^t+B^tB&AB^t-B^tA\\BA^t-A^tB&BB^t+A^tA}$$
which will simplify if (as in your example) $A$ and $B^t$ commute.
| {
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"url": "https://math.stackexchange.com/questions/2595100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Absolute value definition Is it true that $\dots$
$$
\left| y \right| =
\begin{cases}
y \hspace{1cm} y \geq 0 \\ -y \hspace{0.7cm} y < 0
\end{cases}
$$
I'm a little bit confused with the second case, where $|y| = -y$ then $y<0$,
for example :
$$
\left| 2x-4 \right|=-(2x-4)
$$
if we assume that $ y=2x-4 $ then
$$
\begin{align*}
y&<0 \\
2x-4&<0 \\
2x&<4 \\
x&<2
\end{align*}
$$
in the other way, we can solve it like this
$$
\begin{align*}
|y| \geq 0 \\
|2x-4| \geq 0 \\
-(2x-4) \geq 0 \\
2x-4 \leq 0 \\
x \leq 2
\end{align*}
$$
why is it giving the different answers?
| It is a bit confusing to use the same letter $x$ for your substitution.
Let set $y=2x-4$.
Then $|y|$ is $\begin{cases}+y&=2x-4&\quad\text{when}\quad y\ge0\iff 2x-4\ge 0\iff x\ge 2\\-y&=4-2x&\quad\text{when}\quad y<0\iff 2x-4<0\iff x<2\end{cases}$
Indeed for $x=1$ then $x<2$ then $|y|=|2\times 1-4|=|-2|=2=4-2\times 1$
And for $x=3$ then $x\ge 2$ then $|y|=|2\times 3-4|=|2|=2=2\times 3-4$
| {
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"source": "stackexchange",
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How to prove that $\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ for non-negative $a,b$? If $a, b$ are non-negative real numbers, prove that
$$
\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}
$$
I am trying to prove this result. To that end I added $ab$ to both denominator and numerator as we know
$$
\frac{a+b}{1+a+b} \leq \frac{a+b+ab}{1+a+b+ab}
$$
which gives me
$$
\frac{a}{1+a} + \frac{b}{(1+a)(1+b)}
$$
How can I reduce the second term further, and get the required result?
| Since both $a$ and $b$ are nonnegative so are ${{1}\over{1+a}}$ and ${{b}\over{1+b}}+a$ and their multiplication.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Recurrent equation system How to solve this recurrent equation system?
$$\begin{cases}a_{n+1}=a_n+2b_n\\
b_{n+1}=2a_n+b_n\end{cases}$$
The possible solutions should be
\begin{cases}a_n=\frac{1}{2}3^n+\frac{1}{2}(-1)^n\\
b_n=\frac{1}{2}3^n-\frac{1}{2}(-1)^n\end{cases}
| $a_{n+1}=a_n+2(2a_{n-1}+b_{n-1})=a_n+4a_{n-1}+(a_n-a_{n-1})$
So, $a_{n+1}-2a_n-3a_{n-1}=0$.
The roots of $x^2-2x-3=0$ are $-1$ and $3$.
$a_n=\alpha (-1)^n+\beta(3)^n$ for some $\alpha$ and $\beta$.
$2b_n=a_{n+1}-a_n=\alpha(-1)^{n+1}+\beta(3)^{n+1}-\alpha (-1)^n-\beta(3)^n=-2\alpha(-1)^n+2\beta (3)^n$
$b_n=-\alpha(-1)^n+\beta (3)^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Determine the Taylor expansion of $f$ defined by $f(x)=e^{-1/x^2}$ if $x \ne 0$ and $f(0)=0$ I need to find the Taylor series, at $x=0$ for the piece-wise function:
$$f(x)=\begin{cases} \displaystyle e^{\frac{-1}{x^2}} &\text{if}\, x \ne 0\\ 0 &\text{if}\, x = 0\end{cases}$$
I know that the Taylor series formula is $\sum\limits_{k=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$. So I found the first 3 derivative of $f(x)$
$$f'(x)=\begin{cases} \displaystyle \frac{2e^{\frac{-1}{x^2}}}{x^2} &\text{if}\, x \ne 0\\ 0 &\text{if}\, x = 0\end{cases}$$
$$f''(x)=\begin{cases} \displaystyle \frac{2(2e^{\frac{-1}{x^2}}-3e^{\frac{-1}{x^2}}x^2)}{x^6} &\text{if}\, x \ne 0\\ 0 &\text{if}\, x = 0\end{cases}$$
$$f'''(x)=\begin{cases} \displaystyle \frac{2(12x^4e^{\frac{-1}{x^2}}-18e^{\frac{-1}{x^2}}+4e^{\frac{-1}{x^2}})}{x^9} &\text{if}\, x \ne 0\\ 0 &\text{if}\, x = 0\end{cases}$$
After that I bascially reformed the Taylor series equation as
$$f(0)+\frac{f'(0)(x-0)^1}{1!}+\frac{f''(0)(x-0)^2}{2!}+\frac{f'''(0)(x-0)^3}{3!}+...$$
Since $f(0)=0=f'(0)=f''(0)=f'''(0)$ is the Taylor series just $0+0+0+...$?
| Yes, the Taylor series is just $0$ as you had said.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Prove $\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}=(m^2+2) \sqrt{m^2-1}$ Im trying to get from this expression into:
$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}$$
this expression:
$$(m^2+2) \sqrt{m^2-1}$$
someone know how to do it?
i tried it for hours and can't get from the first expression into the second expression.
please explain to me step by step. I'm newbie.
| Write $(m-1)^2(m+1)^2$ as $(m^2-1)^2$.
Then you have $$\frac{(m^2-1)^2}{(m^2-1)^\frac{3}{2}}=\frac{(m^2-1)^\frac{4}{2}}{(m^2-1)^\frac{3}{2}}=(m^2-1)^{\frac{4}{2}-\frac{3}{2}}=(m^2-1)^\frac{1}{2}=\sqrt{m^2-1}$$
The $(m^2+2)$ part stays intact.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can we solve this without cubic formula? I'm looking for an algebraic solution for $x$.
$$ \frac{x}{x+2} -3 = \frac{5x}{x^2-4}+x$$
My first go at this involved converting this into an expression with a cubic numerator and $(x+2)(x-2)$ as the denominator. To find the roots, I then tried to divide each factor in the denominator into to cubic. No success.
I've converted this expression to:
$$ x(x-7)= (x+3)(x+2)(x-2)$$
which illustrates the futility of my first approach. There are no common factors. Can I solve this without invoking the cubic formula?
Edit: To clarify, the the Precalculus textbook calls for an algebraic and graphic solution. If you have an algebraic solution that would be accessible to a precalculus student, please provide it.
| Assuming $x^2\neq 4$, you end with the cubic equation
$$x^3+2 x^2+3 x-12=0$$ So, consider the function $$f(x)=x^3+2 x^2+3 x-12 \implies f'(x)=3 x^2+4 x+3$$ The first derivative does not cancel which means that there is only one real root.
Now, use inspection : $f(0)=-12$, $f(1)=-6$, $f(2)=10$. So, the root is between $1$ and $2$. Looking deeper $f(\frac 32)=\frac 38$ telling that the root is slightly below $1.5$.
Make $x=y+\frac 32$ which makes the equation to be
$$g(y)=y^3+\frac{13 y^2}{2}+\frac{63 y}{4}+\frac{3}{8}$$ if we admit that $y$ is small, then $$g(y) \approx \frac{63 y}{4}+\frac{3}{8}=0 \implies y=-\frac{1}{42}$$ So, an approximate solution is
$$x \approx \frac 32-\frac{1}{42}=\frac{31}{21}\implies f(\frac{31}{21})=\frac{34}{9261}$$ which is now quite small.
Repeat the process making now $x=y+\frac{31}{21}$ giving
$$g(y)=y^3+\frac{45 y^2}{7}+\frac{2270 y}{147}+\frac{34}{9261}$$ then $$g(y) \approx \frac{2270 y}{147}+\frac{34}{9261}=0 \implies y=-\frac{17}{71505}$$
So, an approximate solution is
$$x \approx \frac{31}{21}-\frac{17}{71505}=\frac{105538}{71505}\approx 1.47595$$ while the exact solution (solving the cubic) would be $\approx 1.47595$ !
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the angle between the two tangents on a parabola
Two tangents are drawn from the point $(-2,-1)$ to the parabola $y^2=4x$.
If $\alpha$ is the angle between them then find the value of $\tan \alpha$ .
My try:
Eqn of tangent $1,T_1$ say at the point $(x_1,y_1)$ is $yy_1=2a(x+x_1)$,
Eqn of tangent $2,T_2$ say at the point $(x_2,y_2)$ is $yy_2=2a(x+x_2)$
Since they both pass through $(-2,-1)$ hence we have $y_1=2(2-x_1),y_2=2(2-x_2)$.
Angle between two tangents is $\tan \alpha=\vert\dfrac{m_1-m_2}{1+m_1m_2}\lvert$
Now $m_1=\dfrac{y_1+1}{x_1+2},m_2=\dfrac{y_2+1}{x_2+1}$
Replacing $y_1,y_2$ is not giving the required angle.
How to do it?Please help.
| Taking the derivative wrt $x$, $y^2=4x$ becomes $2yy'= 4$, or $y' = \dfrac{2}{y}$.
So, at the point $(\frac 14t^2,t)$ on the parabola, the slope of the tangent is
$m = \dfrac 2t$.
So the equation of the tangent line through the point $(\frac 14t^2,t)$
is $y-t = \frac 2t(x-\frac 14t^2)$.
For what values of $t$ does this line pass through the point $(-2,-1)$?
\begin{align}
-1-t &= \frac 2t(-2-\frac 14t^2) \\
1+t &= \frac 4t + \frac t2\\
2t + 2t^2 &= 8 + t^2\\
t^2+2t-8 &= 0 \\
(t+4)(t-2) &= 0 \\
t &\in \{2, -4\} \\
(x,y) &\in \{(1,2), (4,-4)\} \\
m &\in \left\{1, -\frac 12 \right\}
\end{align}
\begin{align}
\tan\alpha
&=\left|\dfrac{m_1-m_2}{1+m_1 m_2}\right| \\
&=\left|\dfrac{1+\frac 12}{1-\frac 12}\right| \\
&= 3
\end{align}
$$\alpha \approx 71.57^\circ$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$ABCD$ is parallelogram. $\angle ABC=105^{\circ}$. $\angle CMD=135^\circ$. Find $\angle BKC$
$ABCD$ is parallelogram. $\angle ABC=105^{\circ}$. $BMC$ is equilateral triangle and $\angle CMD=135^\circ$. $K$ is midpoint of $AB$. Find $\angle BKC$
My attempts
1) $\angle MBC=\angle MCB=\angle BMC=60^{\circ}$
2) $\angle MCD=15^{\circ}$, $\angle MDB=30^{\circ}$
I want to prove that $MB \perp CK$ but I need help here
| Let $N$ be the point on $CD$ such that $\angle CMN=15^\circ$.
Then $\angle CMN=15^\circ=\angle MCN$ and $\angle MDN=30^\circ=\angle MND$.
So $DM=MN=NC$.
As $BC=BM$, $MN=CN$ and $BN=BN$, $\triangle BCN\cong \triangle BMN$ and therefore, $\angle CBN=\angle MBN=30^\circ$.
So, $\angle BNC=75^\circ=\angle BCN$ and hence $BC=BN=BM$.
As $\angle ADM=75^\circ=\angle BCN$, $BC=AD$ and $DM=NC$, $\triangle BCN\cong \triangle ADM$.
So, $AM=BN=BM$.
Therefore, $\angle BAM=\angle ABM=45^\circ$.
Since $K$ is the midpoint of $AB$, $MK\perp AB$.
$\angle KMB=180^\circ-90^\circ-45^\circ=45^\circ=\angle KBM$.
Therefore, $BK=KM$.
$\triangle KBC\cong\triangle KMC$.
$\angle BKC=\angle MKC=90^\circ\div 2=45^\circ$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the angle between the two tangents drawn from the point $(1,2)$ to the ellipse $x^2+2y^2=3$.
Find the angle between the two tangents drawn from the point $(1,2)$ to the ellipse $x^2+2y^2=3$.
The given ellipse is $\dfrac{x^2}{3}+\dfrac{y^2}{\frac{3}{2}}=1$
Any point on the ellipse is given by $(a\cos \theta,b\sin \theta)$ where $a=\sqrt 3,b=\frac{\sqrt 3}{\sqrt 2}$.
Now slope of the tangent to the curve at $(a\cos \theta,b\sin \theta)$ is $\dfrac{-a\cos \theta}{2b\sin \theta}$.
Hence we have $\dfrac{b\sin \theta- 2}{a\cos \theta -1}=\dfrac{-a\cos \theta}{2b\sin \theta}$.
On simplifying we get $4b\sin \theta +a\cos\theta =3$
If we can find the value of $\theta $ from above then we can find the two points on the ellipse where the tangents touch them but I am unable to solve them.
Please help to solve it.
Any hints will be helpful
| Let $m$ be the slope of a tangent to the ellipse from $(1,2)$.
The equation of the tangent is $y=mx-m+2$.
Put it into the equation of the ellipse,
\begin{align*}
x^2+2(mx-m+2)^2&=3\\
(1+2m^2)x^2-4m(m-2)x+2m^2-8m+5&=0
\end{align*}
As the tangent meets the ellipse at only one point.
\begin{align*}
[4m(m-2)]^2-4(1+2m^2)(2m^2-8m+5)&=0\\
(4m^4-16m^3+16m^2)-(4m^4-16m^3+12m^2-8m+5)&=0\\
4m^2+8m-5&=0
\end{align*}
If $m_1$ and $m_2$ are the slopes of two tangents, then $\displaystyle m_1+m_2=-2$ and $\displaystyle m_1m_2=-\frac{5}{4}$.
Let $\theta$ be the acute angle between the two tangents.
\begin{align*}
\tan\theta&=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\\
\tan^2\theta&=\frac{(m_1+m_2)^2-4m_1m_2}{(1+m_1m_2)^2}\\
&=\frac{(-2)^2-4(\frac{-5}{4})}{(1+\frac{-5}{4})^2}\\
&=144\\
\theta&=\tan^{-1}(12)
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Which sets of base 10 digits have the property that, for every $n$, there is a $n$-digit number made up of these digits that is divisible by $5^n$? Which sets of non-zero base 10 digits have the property that, for every $n$, there is a $n$-digit number made up of these digits that is divisible by $5^n$?
This is an extension of
Prove that for any integer $n>0$, there exists a number consisting of 1's and 2's only, which is divisible by $2^n$.
Here is my answer:
Any set of digits
which form
a complete residue set
modulo 5.
In particular,
any 5 consecutive digits.
A proof by induction
is not too difficult.
| Partial result . . .
Claim:
If $S \subseteq \{1,2,3,4,5,6,7,8,9\}$ contains a complete residue system, mod $5$, then for all positive integers $n$, there is an $n$-digit number $x$ such that
*
*All digits of $x$ are elements of $S$.$\\[4pt]$
*$5^n{\mid}x$.
Proof:
Assume $S \subseteq \{1,2,3,4,5,6,7,8,9\}$ contains a complete residue system, mod $5$.
Necessarily $5 \in S$, hence the claim holds for $n=1$.
Proceed by induction on $n$.
Fix $n \ge 1$, and let $x$ be an $n$-digit number such that all digits of $x$ are elements of $S$, and $5^n{\mid}x$.
Let $y ={\large{\frac{x}{5^n}}}$.
Choose $d \in S$ such that $d(2^n) + y \equiv 0\;(\text{mod}\;5)$.
Let $x'=d(10^n)+x$.
Then $x'$ is an $(n+1)$-digit number, all of whose digits are elements of $S$.
\begin{align*}
\text{Also,}\;\;x'&=d(10^n)+x\\[4pt]
&=(5^n)(d(2^n)+y)\\[4pt]
\end{align*}
which is a multiple of $5^{n+1}$.
Thus, the induction is complete.
Update:
For $S \subseteq \{1,2,3,4,5,6,7,8,9\}$, call $S$ qualifying if for every positive integer $n$, there is an $n$-digit number $x$ such that
*
*All digits of $x$ are in $S$.$\\[4pt]$
*$5^n{\mid}x$.
As was shown, if $S \subseteq \{1,2,3,4,5,6,7,8,9\}$, and $S$ contains a complete residue system, mod $5$, then $S$ is qualifying.
For $S \subseteq \{1,2,3,4,5,6,7,8,9\}$, call $S$ a minimal exception if
*
*$S$ is a qualifying set.$\\[4pt]$
*$S$ does not contain a complete residue system, mod $5$.$\\[4pt]$
*No proper subset of $S$ is qualifying.
Conjecture:
There are exactly $11$ minimal exceptions, as listed below:
\begin{align*}
&\{1, 2, 3, 5, 6, 7\}\\[4pt]
&\{1, 2, 3, 5, 6, 8\}\\[4pt]
&\{1, 2, 3, 5, 7, 8\}\\[4pt]
&\{1, 2, 5, 6, 7, 8\}\\[4pt]
&\{1, 2, 5, 6, 7, 9\}\\[4pt]
&\{1, 3, 5, 6, 7, 8\}\\[4pt]
&\{2, 3, 4, 5, 7, 9\}\\[4pt]
&\{2, 3, 5, 6, 7, 8\}\\[4pt]
&\{2, 3, 5, 7, 8, 9\}\\[4pt]
&\{2, 4, 5, 6, 7, 9\}\\[4pt]
&\{3, 4, 5, 7, 8, 9\}\\[4pt]
\end{align*}
Remarks: All of the above sets survived testing for $1 \le n \le 10000$.
| {
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linear system of two ODEs Consider the following system of ODE:
$$\begin{cases} \dfrac{df}{dt} = \dfrac{\alpha \beta}{1+t^2}g(t), \\ \\
\dfrac{dg}{dt} = \dfrac{\beta}{\alpha}f(t), \\ \\
t \in [-\alpha, \alpha], \\ \\
f(-\alpha) = g(-\alpha) = 1. \end{cases}$$
With $\alpha \ge 1$ and $\beta > 0$ as small as you like. Can this system be solved explicitly?
| Another way:
$$\frac{d^2g}{dt^2}=\frac{\beta}{\alpha} \frac{df}{dt}=\frac{\beta^2}{1+t^2}g(t)$$
We obtain a 2nd order ODE:
$$(1+t^2)\frac{d^2g}{dt^2}-\beta^2g=0$$
$$g(-\alpha)=1$$
$$\frac{dg}{dt}(-\alpha)=\frac{\beta}{\alpha}$$
The general solution is (we can check with Wolfram Alpha ):
$$g(t)=C_1 \cdot {_2F_1} \left(-\frac{1+\sqrt{1+4\beta^2}}{4},-\frac{1-\sqrt{1+4\beta^2}}{4};\frac{1}{2};-t^2 \right)+ \\ + C_2 \cdot t \cdot {_2F_1} \left(\frac{1+\sqrt{1+4\beta^2}}{4},\frac{1-\sqrt{1+4\beta^2}}{4};\frac{3}{2};-t^2 \right)$$
Now we need to substitute the initial conditions to obtain the explicit formula in terms of Hypergeometric function.
$$C_1 \cdot {_2F_1} \left(-\frac{1+\sqrt{1+4\beta^2}}{4},-\frac{1-\sqrt{1+4\beta^2}}{4};\frac{1}{2};-\alpha^2 \right)- \\ - C_2 \cdot \alpha \cdot {_2F_1} \left(\frac{1+\sqrt{1+4\beta^2}}{4},\frac{1-\sqrt{1+4\beta^2}}{4};\frac{3}{2};-\alpha^2 \right)=1$$
$$C_1=\frac{1+C_2 \alpha {_2F_1} \left(\frac{1+\sqrt{1+4\beta^2}}{4},\frac{1-\sqrt{1+4\beta^2}}{4};\frac{3}{2};-\alpha^2 \right)}{{_2F_1} \left(-\frac{1+\sqrt{1+4\beta^2}}{4},-\frac{1-\sqrt{1+4\beta^2}}{4};\frac{1}{2};-\alpha^2 \right)}$$
I'll leave the second condition to the OP, I'll just provide the formula for the derivative of Hypergeometric function:
$$\frac{\partial}{\partial z} {_2F_1} (a,b;c;z)=\frac{a b}{c} {_2F_1} (a+1,b+1;c+1;z)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2606993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does the determinant always equal zero for a square matrix of consecutive numbers? This works if the integers are listed in consecutive order either along the rows or columns. Why does the determinant of the square matrix always equal $0$ for $ n > 2 $?
| Alternatively, using arithmetic mean, for $3\times 3$ matrix determinant:
$$\begin{vmatrix}
x+1 & x+2 & x+3 \\
x+4 & x+5 & x+6 \\
x+7 & x+8 & x+9 \end{vmatrix}=0,$$
because: $C_1+C_3=2C_2$, which shows linear dependence of the column vectors.
For $4\times 4$ matrix determinant:
$$\begin{vmatrix}
x+1 & x+2 & x+3 & x+4 \\
x+5 & x+6 & x+7 & x+8 \\
x+9 & x+10 & x+11 & x+12 \end{vmatrix}=0, \ \text{because: }
C_1+C_4=C_2+C_3.$$
In geneal, for $(2n+1)\times (2n+1), n>0,$ matrix determinant:
$$\begin{vmatrix}
x+1&\cdots &x+(n+1) &\cdots &x+(2n+1)\\
x+(2n+1)+1&\cdots&x+(2n+1)+n+1&\cdots&x+2(2n+1)\\
\vdots&\vdots&\vdots &\vdots&\vdots\\
x+2n(2n+1)+1&\cdots &x+2n(2n+1)+n+1&\cdots&x+(2n+1)^2
\end{vmatrix}=0,\\
\text{because: } C_1+C_{2n+1}=C_{n+1}.$$
Can you write the generalization for $(2n)\times (2n),n>0,$ matrix determinant?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Computing the series$\sum\limits_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} $ Show convergence of
$\begin{align}
\sum_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} &= \sum_{k=1}^{\infty}{\frac{3^{k-1}}{5^k}}+ \sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}} \\
&= \sum_{k=1}^{\infty}{\frac{1}{3} \cdot \left( \frac{3}{5} \right) ^k} + \sum_{k=1}^{\infty}{ \left( \frac{2i}{5} \right )^k} \\
&= \frac{1}{3} \cdot \sum_{k=1}^{\infty}{ \left( \frac{3}{5} \right) ^k} + \sum_{k=1}^{\infty}{ \left( \frac{2i}{5} \right )^k} \\
\end{align}$
The first part of the sum converges because it is the geometric series with $q= \frac{3}{5}, 1> \left| \frac{3}{5} \right|$.
$$\sum_{n=1}^{\infty}{\left(\frac{2i}{5} \right)^k}$$
Question: Why does that series diverge (WolframAlpha)?
I mean, if $q=\frac{2i}{5}, |q|<1$ then it should be the geometric series and thus converge?
| If $i$ satisfies $i^2=-1$ then
$$\sum_{k=1}^\infty\left(\frac{2i}5\right)^k$$
converges in $\Bbb C$ to
$$\frac{2i/5}{1-2i/5}=\frac{2i}{5-2i}=\frac{-4+10i}{29}$$
(geometric series) as $|2i/5|<1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the indefinite integral $\int \left(\frac{\arctan x}{\arctan x - x}\right)^2 dx$. I am wondering if there is a systematic way to find the indefinite integral
$$\int \left(\frac{\arctan x}{\arctan x - x}\right)^2 dx.$$
It indeed has a clean closed form
$$\frac{x^2 + 1}{\arctan x - x} + x.$$
But I am not able to reach it in a logical way. The transformation $t = \arctan x$ doesn't seem to help much.
| The integrand is
$$ \left(1+\frac{x}{\arctan x - x}\right)^2 = 1 + \frac{2x}{\arctan x - x} + \frac{x^2}{(\arctan x - x)^2} $$
Observe that
$$ (\arctan x - x)' = \frac{1}{x^2+1} - 1 = -\frac{x^2}{x^2+1} $$
So we can perform IBP on the last term
$$ \int (x^2+1)\left(\frac{1}{(\arctan x - x)^2}\frac{x^2}{x^2+1}\right)dx = \frac{x^2+1}{\arctan x - x} - \int\frac{2x}{\arctan x - x} dx $$
Thus
$$ \int \frac{x^2}{(\arctan x - x)^2}dx + \int\frac{2x}{(\arctan x - x)^2} dx = \frac{x^2 + 1}{\arctan x - x} $$
The rest is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Using complete the square to determine positive definite matrices I realize this may be a basic question but I am having trouble following my notes.
I have the matrix $$\begin{bmatrix}16&12\\12&9\end{bmatrix} .$$ So I've got my equation from the matrix to be $(16x)^2 - 12xy - 12yx + 9y^2.$ Not sure where to go from here
| The direction that would usually be called completing the square is
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rr}
1 & 0 \\
\frac{ 3 }{ 4 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
16 & 0 \\
0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rr}
1 & \frac{ 3 }{ 4 } \\
0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rr}
16 & 12 \\
12 & 9 \\
\end{array}
\right)
$$
which means semidefinite, not definite. In particular
$$ 16 (x + \frac{3}{4}y)^2 = 16x^2 + 24 xy + 9 y^2 $$
=================================
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = \left(
\begin{array}{rr}
16 & 12 \\
12 & 9 \\
\end{array}
\right)
$$
==============================================
$$ E_{1} = \left(
\begin{array}{rr}
1 & - \frac{ 3 }{ 4 } \\
0 & 1 \\
\end{array}
\right)
$$
$$ P_{1} = \left(
\begin{array}{rr}
1 & - \frac{ 3 }{ 4 } \\
0 & 1 \\
\end{array}
\right)
, \; \; \; Q_{1} = \left(
\begin{array}{rr}
1 & \frac{ 3 }{ 4 } \\
0 & 1 \\
\end{array}
\right)
, \; \; \; D_{1} = \left(
\begin{array}{rr}
16 & 0 \\
0 & 0 \\
\end{array}
\right)
$$
==============================================
$$ P^T H P = D $$
$$\left(
\begin{array}{rr}
1 & 0 \\
- \frac{ 3 }{ 4 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
16 & 12 \\
12 & 9 \\
\end{array}
\right)
\left(
\begin{array}{rr}
1 & - \frac{ 3 }{ 4 } \\
0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rr}
16 & 0 \\
0 & 0 \\
\end{array}
\right)
$$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rr}
1 & 0 \\
\frac{ 3 }{ 4 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
16 & 0 \\
0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rr}
1 & \frac{ 3 }{ 4 } \\
0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rr}
16 & 12 \\
12 & 9 \\
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Evaluate limit containing $\sum{n^6}$ Evaluate:
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6+\ldots+n^6}{(1^2+2^2+3^2+\ldots+n^2)(1^3+2^3+3^3+\ldots+n^3)}}$$
I can solve the denominator as:
$$\frac{n(n+1)(2n+1)}{6}\cdot\frac{n^2(n+1)^2}{4}$$
$$n^7\cdot\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}\cdot\frac{(1+\frac{1}{n})}{4}$$
$$=\frac{n^7}{12}$$
How can I reduce the numerator?
| The sum of the $k^{th}$ power of the integers is a polynomial of degree $k+1$ in $n$, with leading term $\dfrac{n^{k+1}}{k+1}$.
Indeed, by the binomial theorem,
$$n^k=\sum_{i=1}^ni^k-\sum_{i=1}^{n-1}i^k\sim\frac{n^{k+1}}{k+1}-\frac{(n-1)^{k+1}}{k+1}=n^k+\text{lower degree terms}.$$
The given limit equals that of
$$\frac{n^7\cdot3\cdot4}{7\cdot n^3\cdot n^4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Show that $Q=X^2+5X+7$ divides $P=(X+2)^m+(X+3)^{2m+3}$ for any $m\in\Bbb N$ Let $$P=(X+2)^m+(X+3)^{2m+3}$$ and $$Q=X^2+5X+7.$$ I need to show that $Q$ divides $P$ for any $m$ natural.
I said like this: let $a$ be a root of $X^2+5X+7=0$. Then $a^2+5a+7=0$.
Now, I know I need to show that $P(a)=0$, but I do not know if it is the right path since I have not found any way to do it.
| We can also prove it by induction.
$P_0(x)=1+(x+3)^3=x^3+9x^2+27x+28=(x^2+5x+7)(x+4)\quad\checkmark$
$\begin{align}
P_{m+1}(x) &=(x+2)^{m+1}+(x+3)^{2m+5}\\
&=(x+2)^m(x+2)+(x+3)^2\overbrace{\big((x^2+5x+7)Q_m(x)-(x+2)^m\big)}^{\text{induction hypothesis}}\\\\
&=(x+2)^m\underbrace{(x+2-x^2-6x-9)}_{-x^2-5x-7}+(x+3)^2(x^2+5x+7)Q_m(x)\\\\
&= (x^2+5x+7)Q_{m+1}(x)\quad\checkmark
\end{align}$
| {
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"url": "https://math.stackexchange.com/questions/2619185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Real ordered pair $(a,b)$ in equation
If $$a^2+5b^2+2b=6a+2ab-10$$
then all real ordered pair of $(a,b)$ is?
Try: I am trying to convert it into $$(..)^2+(..)^2=0$$ or $$(..)^2+(..)^2+(..)^2=0.$$
So $$(a-b)^2+(2a-0.75)^2+10-(0.75)^2-6a=0.$$
Could some help me to solve it? Thanks.
| Consider the equation as a quadratic in $\,a\,$:
$$
a^2 - 2a(b+3) +5b^2+2b + 10 = 0
$$
Then its reduced discriminant is:
$$
\frac{1}{4} \Delta = (b+3)^2-(5b^2+2b +10) = -4b^2 + 4b -1 = -(2b-1)^2 \;\;\le\;\; 0
$$
It follows that $\,\Delta = 0\,$ for real roots to exist, so $\,b=\dfrac{1}{2}\,$, then substituting back $\,a=b+3=\dfrac{7}{2}\,$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the following limit algebraically? I've been trying to answer this for a while and I know it's a simple question relative to most questions that are posted here.
$$
\lim_{x\rightarrow -2}\: \frac{x^4+5x^3+6x^2}{x^2(x+1)-4(x+1)}
$$
If we substitute -2 for $ x $ we get $ 0/0 $, an indeterminate form. I figured that the denominator can be rewritten as $ (x^2-4)(x+1) $. And then I tried to factor something in the numerator but couldn't see anything interesting. How do I find the limit algebraically?
I know that the answer is supposed to be 1, but I don't know how they got there.
Thanks in advance!
| Notice that:
$$x^4+5x^3+6x^2 = x^2(x^2+5x+6) = x^2(x+2)(x+3)$$
and
$$x^2(x+1)-4(x+1) = (x^2-4)(x+1) = (x+2)(x-2)(x+1).$$
Then, you have:
$$\lim_{x\rightarrow -2}\: \frac{x^2(x+2)(x+3)}{(x+2)(x-2)(x+1)} = \\
= \lim_{x\rightarrow -2}\: \frac{x^2(x+3)}{(x-2)(x+1)} = \\
= \frac{(-2)^2(-2+3)}{(-2-2)(-2+1)} = \frac{4}{4} = 1.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to solve the integral $\int \frac{4+12x}{({1-2x-3x^2})^\frac{2}{3}}\,dx$ $$\int \frac{4+12x}{({1-2x-3x^2})^\frac{1}{3}}\,dx$$
So first I tried setting $u = 1-2x-3x^2, du = -2-6x\,dx$
Which gives
$$-2\int u^\frac{-1}{3}du = \frac{-6 u^\frac{2}{3}}{2} =-3u^\frac{2}{3}$$
And ultimately:
$$-3(1-2x-3x^2)^\frac{2}{3}$$
For some odd reason when I take the derivative I'm getting something completely different, what's wrong with my substitution?
| \begin{align}
& \frac d {dx} \left( -3(1-2x-3x^2)^{2/3} \right) \\[10pt]
= {} & -3\cdot \frac 2 3(1-2x-3x^2)^{(2/3)-1}(-2-6x) & & \text{by the chain rule} \\[10pt]
= {} & -2(1-2x-3x^2)^{-1/3}(-2-6x) \\[10pt]
= {} & (1-2x-3x^2)^{-1/3} (4+12x)
\end{align}
| {
"language": "en",
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How do I interpret this sum? So if the sum of $n$ integers $\ge 1$ equal $\frac{n(n+1)}2$. Then my book goes on and says $1 + 2 + 3 +\ldots + 2n = \frac{2n(2n + 1) }2$.
I'm confused about what $1 + 2 + 3 + \ldots +2n$ means. If the sequence is $1, 2, 3, 4$ then where does $2n$ have to do with the $n$th number?
| The formula $1+2+3+...+n$ mean the "sum of all integers starting with $1$ and going up to $n$". You can substitute any positive integer you want in place of $n$, for instance by substituting $n=6$ you get $1 + 2 + 3 + ... + 6$, which ordinarily we would write out more explicitly as $1+2+3+4+5+6$, and them sum to get $21$.
Similarly, the formula $1+2+3+....+BLAH$ to represent the "sum of all integers starting with $1$ and going up to BLAH. You can substitute any positive integer you want in place of BLAH.
Finally the formula $1+2+3+...+2n$ mean "the sum of all integers starting with $1$ and going up to $2n$". You can substitute any positive integer you want in place of $n$, and then do the arithmetic to get $2n$, and then substitute that amount. For instance, if you take $n=3$ and substitute that into the expression $2n$ then you get $2n = 2 \cdot 3 = 6$, and then when you substitute $2n=6$ into the formula $1+2+3+...+2n$ you get $1+2+3+...+6 = 21$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Coefficients of the stirling's series expansion for the factorial. Knowing the Stirling's approximation for the Gamma function (factorial) for integers:
$$\Gamma(n+1)=n!\approx \sqrt{2\pi n}n^ne^{-n}\bigg(1+\frac{a_1}{n}+\frac{a_2}{n^2}+\cdots\bigg)$$
Using the above approximation one can write:
$$(n+1)!=\sqrt{2\pi(n+1)}(n+1)^{n+1}e^{-(n+1)}\bigg(1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\cdots\bigg)$$
We know that following recursion holds:
$$(n+1)!=(n+1)n!$$
One can rewrite this:
$$(n+1)!=(n+1)\sqrt{2\pi n}n^ne^{-n}\bigg(1+\frac{a_1}{n}+\frac{a_2}{n^2}+\cdots\bigg)$$
All this comes from: https://www.csie.ntu.edu.tw/~b89089/link/gammaFunction.pdf (Page 8-9)
Then the author gives this expansion to calculate the $a_k$ coefficients when $n$ becomes large.
Comparing these two expressions for $(n+1)!$ gives
$$1+\frac{a_1}{n}+\frac{a_2}{n^2}+\cdots=\bigg(1+\frac{1}{n}\bigg)^{n+1/2}e^{-1}\bigg(1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\cdots\bigg)$$
Then he says, that after "classical series expansion" this equals:
$$1+\frac{a_1}{n}+\frac{a_2-a_1+\frac{1}{12}}{n^2}+\frac{\frac{13}{12}a_1-2a_2+a_3+\frac{1}{12}}{n^3}+\cdots$$
I don't understand how he got there. Only thing that came to my mind was, that as $n\to\infty$ $\big(1+\frac{1}{n}\big)^n$ goes to $e$ but then we are left with $\big(1+\frac{1}{n}\big)^{1/2}$. When i expand this into binomial series, i get:
$$\bigg(1+\frac{1}{2n}-\frac{1}{8n^2}+\frac{1}{16n^3}\cdots\bigg)\cdot\bigg(1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\cdots\bigg)$$
And I'm stuck here.
Or is there any other elementary way how to compute the coefficients $a_k$ of the stirling's series expansion for factorial/Gamma function?
| You were close. Note that1
$$
e^{-1}\left(1+\frac{1}{n}\right)^{n+1/2}=1+\frac{1}{12n^2}-\frac{1}{12n^3}+\frac{113}{1440n^4}+\cdots\tag1
$$
Therefore, you have
$$
\bigg(1+\frac{1}{12n^2}-\frac{1}{12n^3}+\cdots\bigg)\cdot\bigg(1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\frac{a_3}{(n+1)^3}+\cdots\bigg)=\\
1+\frac{a_1}{n}+\frac{a_2-a_1+\frac{1}{12}}{n^2}+\frac{a_3-2a_2+\frac{13}{12}a_1+\frac{1}{12}}{n^3}+\cdots\tag2
$$
as per
$$
\frac{1}{n+1}=\frac{1}{n}-\frac{1}{n^2}+\cdots\\
\frac{1}{(n+1)^2}=\frac{1}{n^2}-\frac{2}{n^3}+\cdots\\
\frac{1}{(n+1)^3}=\frac{1}{n^3}-\frac{3}{n^4}+\cdots\\
\text{etc.}\tag3
$$
Easy (but slightly cumbersome) exercise:
\begin{equation}
\begin{aligned}
{}&\lim_{n\to\infty}e^{-1}\left(1+\frac{1}{n}\right)^{n+1/2}=1\\
{}&\lim_{n\to\infty}n^2\left[e^{-1}\left(1+\frac{1}{n}\right)^{n+1/2}-1\right]=\frac{1}{12}\\
{}&\lim_{n\to\infty}n^3\left[e^{-1}\left(1+\frac{1}{n}\right)^{n+1/2}-1-\frac{1}{12n^2}\right]=-\frac{1}{12}\\
{}&\text{etc.}
\end{aligned}\tag4
\end{equation}
These are basic limits that are left to the reader.
| {
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Trigonometry problem, Evaluate: $\frac {1}{\sin 18°}$ My problem is,
Evaluate: $$\frac {1}{\sin 18°}$$
I tried to do something myself.
It is obvious,
$$\cos 18°= \sin 72°$$
I accept $\left\{18°=x \right\}$ for convenience and here, $\sin (x)>0$
$$\cos (x)=\sin (4x)$$
$$\cos (x)=2× \sin(2x) \cos (2x)$$
$$\cos (x)=2× 2\sin(x) \cos (x)×(1-2\sin^2(x)), \cos(x)>0$$
$$8\sin^3(x)-4\sin(x)+1=0$$
$$(2\sin(x)-1)(4\sin^2(x)+2\sin (x)-1)=0$$
$$4\sin^2(x)+2\sin (x)-1=0, \sin(x)≠\frac 12$$
$$4t^2+2t-1=0$$
$$t_{1,2}=\frac {-1±\sqrt 5}{4}$$
$$t=\frac {\sqrt5-1}{4} ,t>0$$
$$\sin 18°=\frac {\sqrt5-1}{4} .$$
Finally, $$\frac {1}{\sin 18°}=\frac {4}{\sqrt5-1}=\sqrt5+1$$
Is this way correct and is there a better/elegant way to do it? As always, it was an ugly solution.
Thank you!
| Take $\Delta ABC$ with $AB=AC=1$ and $BC=x,$ and with $\angle ABC=\angle ACB=2\pi/5.$ Then $\angle BAC=\pi/5$ and $x=2AB\sin \frac {1}{2}\angle BAC=2\sin \pi/10.$
Take $D$ on side $AC$ with $\angle DBC=\pi/5.$ Then $ABC$ and $BDC$ are similar triangles so $CD/x= CD / CB= CB/CA=x/1$ . So $CD=x^2.$
And $\Delta BDA$ is isosceles because $\angle DCA=\angle DAC=\pi/5 .$ So $DA=DB=CB=x.$
$$\boxed{ \therefore 1=AC=AD+DC=x+x^2=2\sin \frac{\pi}{10} +4\sin^2\frac{\pi}{10} \ }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Solving $\int(x^2 + 1)^7x^3{\mathrm d}x$ without expansion What is $\int(x^2 + 1)^7x^3{\mathrm d}x$?
I wasn't able to come up with a substitution so I attempted integration by parts:
$$u = x^2 + 1, u' = 2x, v' = x^3, v = \frac{x^4}{4}$$
$$(x^2+1)\frac{x^4}{4} - \frac{x^6}{12}$$
The derivative clearly shows that this is wrong. How can I solve this without using the binomial theorem?
| You have chosen the wrong functions for the integration by parts. Take: $$f'(x)=x(x^2+1)^7\\ f(x)=\frac{1}{16}(x^2+1)^8$$
and $$g(x)=x^2\\ g'(x)=2x$$ Hence:
\begin{align}
\int x^3(x^2+1)^7\,dx = \frac{1}{16}(x^2+1)^8 x^2-\int \frac{x}{8}(x^2+1)^8\,dx
\end{align}
And the last one is easy to do.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Showing that $\sum_{j=0}^{2n-1}{\cos^n(\frac{j\pi}{2n})(2\cos(\frac{2j\pi}n)+1)\cos(\frac{j\pi}2-\frac{2j\pi}n)}$ is never an integer for $n>10$ I want to show that
$$f(n) = \sum_{j=0}^{2n-1}{\cos^n\left( \frac{j \pi}{2n}\right) \left( 2\cos \left( \frac{2 j \pi}{n} \right) + 1\right) \cos \left( \frac{j \pi}{2} - \frac{2 j \pi}{n} \right)}$$
is an integer only for $n \in \{0,1,2,3,4, 5, 10\}$.
What I have tried so far is:
*
*Calculate $f(n)$ for each $n$ up to $16$. The values above are the only values I have found that are integers.
*At $n = 16$ it's value is $\frac{1941}{2048}$, just under $1$. By computing $f(n)$ for $16$ and up, I can see that it is always decreasing and therefore is $\lt 1$.
If I can show $\forall n \in \mathbb{N}, n\ge 16, f(n) \lt 1$, then it follows that it can't be an integer, and I am done.
How do I show $f(n)$ is always decreasing (once $n$ gets big enough)?
| Claim: for $n\geq 3$,
$$
f(n)= 2^{-n}\cdot 2n\left(\binom{n}{4}+\binom{n}{2}+\binom{n}{0}\right)
$$This immediately gives that for $n\geq 16$, $0<f(n)<1$; then the only integer solutions are the ones shown in the problem statement.
Start with
$$
f(n) = \sum _{j=0}^{2 n-1} \left(2 \cos \left(\frac{2 \pi j}{n}\right)+1\right)
\cos \left(\frac{\pi j}{2}-\frac{2 \pi j}{n}\right)\cos ^n\left(\frac{\pi j}{2 n}\right)
$$A key observation: each term is real. Use $\cos(\theta)= 1/2 (e^{i\theta}+e^{-i \theta})$:
$$
f(n)=2^{-n} \sum _{j=0}^{2 n-1} \left(2 \cos \left(\frac{2 \pi j}{n}\right)+1\right) \cos \left(\frac{\pi j}{2}-\frac{2 \pi
j}{n}\right)\left(\exp \left(\frac{\pi i j}{2 n}\right)+\exp \left(\frac{-\pi i j}{2
n}\right)\right)^n
$$
$$
=2^{-n} \sum _{j=0}^{2 n-1} \left(2 \cos
\left(\frac{2 \pi j}{n}\right)+1\right) \cos \left(\frac{\pi j}{2}-\frac{2 \pi j}{n}\right)(-i)^j \left(1+\exp \left(\frac{\pi i j}{n}\right)\right)^n
$$Now use the Binomial Theorem and switch the order of summation:
$$
f(n) = 2^{-n} \sum _{j=0}^{2 n-1} \left(2 \cos \left(\frac{2 \pi j}{n}\right)+1\right) \cos \left(\frac{\pi
j}{2}-\frac{2 \pi j}{n}\right) (-i)^j \sum _{k=0}^n \binom{n}{k} \exp \left(\frac{\pi i j k}{n}\right)
$$
$$
=2^{-n} \sum _{k=0}^n \binom{n}{k} \sum _{j=0}^{2 n-1} \left(2 \cos \left(\frac{2 \pi
j}{n}\right)+1\right) \cos \left(\frac{\pi j}{2}-\frac{2 \pi j}{n}\right) (-i)^j \exp \left(\frac{\pi i j k}{n}\right)
$$Playing around with the summand gives
$$
\cos \left(\frac{\pi j}{2}-\frac{2 \pi j}{n}\right) (-i)^j \exp \left(\frac{\pi i j k}{n}\right)
$$
$$
= \cos \left(\frac{\pi j (n-4)}{2 n}\right) \left(\cos \left(\frac{\pi j
(n-2 k)}{2 n}\right)-i \sin \left(\frac{\pi j (n-2 k)}{2 n}\right)\right)
$$But now, by the fact that everything in the problem is real, we can discard the imaginary term. Thus
$$
f(n) = \sum _{k=0}^n 2^{-n} \binom{n}{k} \sum _{j=0}^{2 n-1}\left(2 \cos \left(\frac{2 \pi j}{n}\right)+1\right) \cos
\left(\frac{\pi j (n-4)}{2 n}\right) \cos \left(\frac{\pi j (n-2 k)}{2 n}\right)
$$Use the product-to-sum formula for cosine:
$$
f(n) = \sum _{k=0}^n 2^{-n} \binom{n}{k} \sum _{j=0}^{2 n-1}\frac{1}{2} \left(\cos \left(\frac{\pi j (k-4)}{n}\right)+\cos \left(\frac{\pi j (k-2)}{n}\right)+\cos \left(\frac{\pi j
k}{n}\right)+\cos \left(\frac{\pi j (k-n+4)}{n}\right)+\cos \left(\frac{\pi j (-k+n-2)}{n}\right)+\cos \left(\frac{\pi j
(n-k)}{n}\right)\right)
$$
$$
=\sum _{k=0}^n 2^{-n-1} \binom{n}{k} \underbrace{\sum _{j=0}^{2 n-1} \cos \left(\frac{\pi j (k-4)}{n}\right)+\cos \left(\frac{\pi j (k-2)}{n}\right)+\cos \left(\frac{\pi j
k}{n}\right)}_{S_1}
$$
$$
+\sum _{k=0}^n 2^{-n-1} \binom{n}{k} \underbrace{\sum _{j=0}^{2 n-1}\cos \left(\frac{\pi j (k-n+4)}{n}\right)+\cos \left(\frac{\pi j (-k+n-2)}{n}\right)+\cos \left(\frac{\pi j
(n-k)}{n}\right)}_{S_2}
$$Now I claim $S_1=S_2$. Indeed, if you sum $S_2$ in the opposite direction, you get exactly $S_1$. Thus we can do away with it, leaving:
$$
f(n)=\sum _{k=0}^n 2^{-n} \binom{n}{k}\sum _{j=0}^{2 n-1} \cos \left(\frac{\pi j (k-4)}{n}\right)+\cos \left(\frac{\pi j (k-2)}{n}\right)+\cos \left(\frac{\pi j
k}{n}\right)
$$Here's the rub: the inner sums cancel by symmetry, except when $k=4,2,0$, when they are $2n$ (because $\cos(0)=1$). Then we have
$$
f(n) = 2^{-n}\cdot 2n\left(\binom{n}{4}+\binom{n}{2}+\binom{n}{0}\right)
$$I double-checked the closed-form against the original series for $3\leq n\leq 16$ and everything worked!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Factor of a Mersenne number Why is it true that if 7 divides 91 then $(2^7-1) $ divides $(2^{91}-1)$?
1) $2^{91}-1$
$7|91 \implies (2^7-1)|(2^{91}-1)$
$\implies 2^7-1$ is factor
2) $2^{1001}-1$
$7|1001 \implies (2^7-1)|(2^{1001}-1)$
$\implies 2^7-1$ is factor
| We have $$2^{91}-1 = (2^7)^{13}-1 = (2^7-1)((2^7)^{12}+...+2^7+1)$$
More generally:
$$2^{ab}-1 = (2^a)^{b}-1 = (2^a-1)((2^a)^{b-1}+...+2^a+1)$$
so $2^a-1\mid 2^{ab}-1$
And we can replace $2$ with any other number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
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Are reduced SVD and truncated SVD the same thing? Truncated SVD: http://scikit-learn.org/stable/modules/generated/sklearn.decomposition.TruncatedSVD.html
Reduced SVD, I thought this is essentially the same thing, and it appears to be actually more commonly called this way.
If you could provide reference, that'll be great.
| Suppose the shape of $A$ is $m \times n$, written as $\underset{m \times n}{A}$, with rank $r$.
In full SVD:
*
*$U$ is composed of $r$ orthonormal columns that span the column space of $A$ and $m - r$ orthonormal columns that span the left null space of $A$.
*$\Sigma$ is diagonal and composed of the square root of eigenvalues of $A^T A$ (or $A A^T$) padded with zero rows and columns to be of shape $m \times n$. The diagonal elements are also called the singular values of $A$.
*$V$ is composed of $r$ orthonormal columns that span the row space of $A$ and $n - r$ orthonormal columns that span the null space of $A$.
\begin{align}
\underset{m\times n}{A} &= \underset{m \times m,}{U} \underset{m \times n,}{\Sigma} \underset{n \times n}{V^{T}}
\end{align}
In reduced SVD:
*
*the $m-r$ columns that span the left null space are removed from $U$.
*the padded rows and columns of zeros are removed from $\Sigma$.
*the $n-r$ columns that span the null space are removed from $V$.
so it becomes
\begin{align}
\underset{m\times n}{A} &= \underset{m \times r,}{U_r} \underset{r \times r,}{\Sigma_r} \underset{r \times n}{V_r^{T}}
\end{align}
Note, both reduced SVD and full SVD results in the original $A$ with no information loss.
In truncated SVD, we take $k$ largest singular values ($0 \lt k \lt r$, thus truncated) and their corresponding left and right singular vectors,
\begin{align}
\underset{m\times n}{A} &\approx \underset{m \times k,}{U_t} \underset{k \times k,}{\Sigma_t} \underset{k \times n}{V_t^{T}}
\end{align}
$A$ constructed via truncated SVD is an approximation to the original A.
Example 1
For $A = \begin{bmatrix}
1 & 1 & 0 \\
2 & 2 & 0 \\
\end{bmatrix}$, where $m = 2$, $n = 3$, and $r = 1$.
Full SVD:
\begin{align*}
\begin{bmatrix}
1 & 1 & 0 \\
2 & 2 & 0 \\
\end{bmatrix} =
\begin{bmatrix}
\frac{1}{\sqrt{5}} & \color{Blue}{- \frac{2}{\sqrt{5}}} \\
\frac{2}{\sqrt{5}} & \color{Blue}{\frac{1}{\sqrt{5}}}
\end{bmatrix}
\begin{bmatrix}
\sqrt{10} & \color{Gray}{0} & \color{Gray}{0} \\
\color{Gray}{0} & \color{Gray}{0} & \color{Gray}{0}
\end{bmatrix}
\begin{bmatrix}
\frac{1}{\sqrt{2}} & \color{Red}{- \frac{1}{\sqrt{2}}} & \color{Red}{0} \\
\frac{1}{\sqrt{2}} & \color{Red}{ \frac{1}{\sqrt{2}}} & \color{Red}{0}
\\0 & \color{Red}{0} & \color{Red}{1}
\end{bmatrix}^T
\end{align*}
Note,
*
*the left null space columns in $U$ are colored blue.
*the padded zero rows and columns in $\Sigma$ are colored gray.
*the null space columns in $V$ are colored red.
Reduced SVD
just remove the colored rows and columns, and it ends with reduced SVD.
\begin{align*}
\begin{bmatrix}
1 & 1 & 0 \\
2 & 2 & 0 \\
\end{bmatrix} =
\begin{bmatrix}
\frac{1}{\sqrt{5}} \\
\frac{2}{\sqrt{5}}
\end{bmatrix}
\begin{bmatrix}
\sqrt{10} \\
\end{bmatrix}
\begin{bmatrix}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}}
\\0
\end{bmatrix}^T
\end{align*}
Since A has only one positive singular value, we can't demonstrate truncated SVD with it.
Example 2
We use another example $B = \begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 1 \\
\end{bmatrix}$ with $m=2$, $n=3$, and $r=2$ to show truncated SVD.
Full SVD:
\begin{align*}
\begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 1 \\
\end{bmatrix} = \begin{bmatrix}
\frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{bmatrix} \begin{bmatrix}
\sqrt{3} & 0 & \color{Gray}{0} \\
0 & 1 & \color{Gray}{0} \\
\end{bmatrix} \begin{bmatrix}
\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} & \color{Red}{\frac{1}{\sqrt{3}}} \\
\frac{2}{\sqrt{6}} & 0 & \color{Red}{- \frac{1}{\sqrt{3}}} \\
\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & \color{Red}{\frac{1}{\sqrt{3}}}
\end{bmatrix}^T
\end{align*}
Reduced SVD:
\begin{align*}
\begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 1 \\
\end{bmatrix} = \begin{bmatrix}
\frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{bmatrix} \begin{bmatrix}
\sqrt{3} & 0 \\
0 & 1 \\
\end{bmatrix} \begin{bmatrix}
\frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\
\frac{2}{\sqrt{6}} & 0 \\
\frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}}
\end{bmatrix}^T
\end{align*}
Truncated SVD:
\begin{align*}
\begin{bmatrix}
\frac{1}{2} & 1 & \frac{1}{2}\\
\frac{1}{2} & 1 & \frac{1}{2}
\end{bmatrix} = \begin{bmatrix}
\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}}
\end{bmatrix} \begin{bmatrix}
\sqrt{3}
\end{bmatrix} \begin{bmatrix}
\frac{1}{\sqrt{6}} \\
\frac{2}{\sqrt{6}} \\
\frac{1}{\sqrt{6}} \\
\end{bmatrix}^T
\end{align*}
Only the largest singular value $\sqrt{3}$ is taken. $\begin{bmatrix}
\frac{1}{2} & 1 & \frac{1}{2}\\
\frac{1}{2} & 1 & \frac{1}{2}
\end{bmatrix}$ is an approximation of the original $\begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 1 \\
\end{bmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Consider polynomial $X^3-3X+1$ If $\alpha$ is a root $\alpha^3-3 \alpha+1=0 $ Consider polynomial $$ X^3-3X+1$$
If $\alpha$ is a root
$$\alpha^3-3 \alpha+1=0 $$
showing $\alpha^2-2$ is also a root
set $X=\alpha^2-2$
$$ (\alpha^2-2)^3-(\alpha^2-2)+1=\alpha^6-9\alpha^4+26 \alpha^2-24$$
Let us look at $\alpha^6$
$$\begin{aligned}
\alpha^6&= \alpha^3 \alpha^3
\\&=(3\alpha-1) (3\alpha-1)
\\&= 3\alpha(3\alpha-1)-1(3\alpha-1)
\\&= 9\alpha^2-3\alpha-3\alpha+1
\\&=9 \alpha^2 -6 \alpha+1
\end{aligned} $$
Now looking at $\alpha^4$
$$
\begin{aligned}
\alpha^4= \alpha^3 \alpha^1
&=(3\alpha-1) \alpha
&=3 \alpha^2-\alpha
\end{aligned}
$$
Let us go back
$$ \begin{aligned}
&\alpha^6-9\alpha^4+26 \alpha^2 -24
\\
&=( 9 \alpha^2-6 \alpha+1 )
-9(3\alpha^2-\alpha)
+26 \alpha^2 -24
\\ &=9\alpha^2-6\alpha+1-27\alpha^2+9 \alpha+26 \alpha^2 -24
\\&=18 \alpha^2 + 3\alpha -23
\\& = \vdots?
\\&=0
\end{aligned}$$
| Expand and do long division to get:
$$
(x^2-2)^3-3(x^2-2)+1 = x^6 - 6 x^4 + 9 x^2 - 1 = (x^3 - 3 x - 1) (x^3 - 3 x + 1)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Using Fourier series to calculate infinite series. I am asked to devolop the function $f(x)=x^2$ in a series of sine and cosine (Fourier series) in the interval $[\frac{-1}{2},\frac{1}{2}]$. And use one of these series to calculate $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} \space\space\space \sum_{n=1}^{\infty} \frac{1}{n^2}$$
Now, the Fourier series is
$S(f)(x)=a_0+2\sum_{n=1}^{\infty}a_ncos(2\pi nx)+2\sum_{n=1}^{\infty}b_nsin(2\pi nx)$
Since $f(x)=x^2$ is even, $b_n=0$. I understand this means that I have to use the cosine series the infinite series.
Using $a_0=\int_{0}^{1} f(x) dx$ and $a_n=\int_{0}^{1} f(x)cos(2\pi nx) dx$
I got that $S(f)(x)=\frac{1}{12}+2\sum_{n=1}^{\infty}\frac{(-1)^n}{\pi^2n^2}cos(2\pi nx)$
Now, how can I calculate the initial series?
| The function $x^2$ is an even function in the given interval, $[\frac{-1}{2},\frac{1}{2}]$, thus the sine terms will vanish.
I calculated the Fourier series to be:
$$f(x)=x^2=\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}cos({2\pi}nx)}$$
NOTE: There is a mistake in your equation above where you multiplied the summation by 2.
*
*In solving for $\sum_{n=1}^\infty {\frac{(-1)^{n+1}}{n^2}}$:
Let $x=0$. The Fourier series becomes:
$$f(0)=(0)^2=\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}cos({2\pi}n{0})}$$
Simplifying:
$$0=\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}cos(0)} = \frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}}$$
Multiplying by $-1$ gives:
$$0=-\frac{1}{12}-\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}} = -\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^{n+1}}{\pi^2n^2}}$$
Simplifying further:
$$\sum_{n=1}^\infty {\frac{(-1)^{n+1}}{\pi^2n^2}}=\frac{1}{12}$$
Finally, multiplying throughout by $\pi^2$ gives:
$$\sum_{n=1}^\infty {\frac{(-1)^{n+1}}{n^2}}=\frac{\pi^2}{12}$$
*In solving for $\sum_{n=1}^\infty {\frac{1}{n^2}}$:
If we substitute for $x=\frac{1}{2}$, the Fourier series gives:
$$f(\frac{1}{2})=(\frac{1}{2})^2=\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}cos({2\pi}n\frac{1}{2})}$$
Simplifying:
$$\frac{1}{4}=\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}cos(n\pi)} = \frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}(-1)^n}$$
But
$$\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}(-1)^n}=\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^{2n}}{\pi^2n^2}}=\frac{1}{12}+\sum_{n=1}^\infty {\frac{1}{\pi^2n^2}}$$
Therefore
$$\frac{1}{4}=\frac{1}{12}+\sum_{n=1}^\infty {\frac{1}{\pi^2n^2}}$$
Again simplifying gives:
$$\frac{1}{6}=\sum_{n=1}^\infty {\frac{1}{\pi^2n^2}}$$
And finally, multiplying both sides by $\pi^2$ gives:
$$\frac{\pi^2}{6}=\sum_{n=1}^\infty {\frac{1}{n^2}}$$
| {
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"source": "stackexchange",
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"answer_count": 1,
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Proving that $\binom{2n}{n} > n2^n , \forall n \ge 4 $ I'm trying to prove the following statement: $\binom{2n}{n} > n2^n , \forall n \ge 4 $
This is my attempt at an inductive proof:
Let $P(n)$ be the following proposition: "$\binom{2n}{n} > n2^n , \forall n \ge 4 $"
Base case:
$\binom{2*4}{4} = 70 > 4*2^4 = 64$ so $P(4)$ is true.
Inductive step:
(I assume as an inductive hypothesis that P(n) is true and try to show that P(n+1) is true)
$\binom{2(n+1)}{n+1} = \binom{2n+2}{n+1} = \binom{2n+1}{n+1} + \binom{2n+1}{n}$
$= \binom{2n}{n} + \binom{2n}{n+1} + \binom{2n}{n} + \binom{2n}{n-1}$
$= 2 \binom{2n}{n} + \binom{2n}{n+1} + \binom{2n}{n-1}$
$= 2 \binom{2n}{n} + \frac{(2n)!}{(n+1)!(2n-(n+1))!} + \frac{(2n)!}{(n-1)!(2n-(n-1))!}$
$= 2 \binom{2n}{n} +2 \frac{(2n)!}{(n+1)!(n-1)!} $
By the inductive hypothesis we have:
$ 2 \binom{2n}{n} +2 \frac{(2n)!}{(n+1)!(n-1)!} > 2n2^n +2 \frac{(2n)!}{(n+1)!(n-1)!} = (2^{n+1}) n + 2 \frac{(2n)!}{(n+1)!(n-1)!} $
And here I'm stuck.
If I could prove that $\frac{(2n)!}{(n+1)!(n-1)!} \ge 2^n , \forall n \ge 4 $ the last step would show that P(n+1) is true. But I haven't been able to prove that. Can anyone help me ?
| You can actually do this without explicit induction.
$$\binom{2n}{n} = \frac{(2n)(2n-1)\dots(2n-(n-1))}{n(n-1)\dots2 \cdot 1} = \frac{2n}{n} \cdot \frac{2n-1}{n-1} \cdot \dots \cdot \frac{n+2}{2} \cdot \frac{n+1}{1}$$
Each of those $n$ fractions is bigger than or equal to $2$, thereby providing our $2^n$; but we can do better by noting that in fact the final two fractions come to $\frac{1}{2} (n+2)(n+1)$ which is bigger than $4n$, not just bigger than $4$, when $n > 4.56$. We could just leave it there after verifying the $n=4$ case manually, or we can do even better by accounting for the final three fractions, whereupon we have that $\frac{1}{6} (n+1)(n+2)(n+3) > 8n$ when $n > 3.66$.
So as long as there are at least a final three fractions (i.e. $n \geq 3$), and as long as $n > 3.66$, the result holds, and in fact it holds pretty massively.
| {
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"url": "https://math.stackexchange.com/questions/2630356",
"timestamp": "2023-03-29T00:00:00",
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Given a root, determine $a$ and $b$.
Given $p(x)= x^3 + ax^2 + bx -6$ with a zero at $x=1+i$. Determine the values of $a$ and $b$.
With the question given in its current form, would it be reasonable to assume that both $a$ and $b$ are real? Since, if so, one may use the fact that $x= 1-i$ is also a zero, and therefore find a quadratic factor and use that to proceed.
Or is there another way to proceed to answer this question without having to assume real coefficients?
| Which says that $x^3+ax^2+bx-6$ is divisible by $$(x-1+i)(x-1-i)=(x-1)^2-i^2=x^2-2x+2$$ and since
$$x^3+ax^2+bx-6=x^3-2x^2+2x+(a+2)x^2+(b-2)x-6,$$ we obtain $a+2=-3$ and $b-2=6$,
which gives $$(a,b)=(-5,8).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral of $\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ How do we integrate the following?
$\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ given that $\cos 2x \gt 0$
I tried to simplify this, but I cannot seem to proceed further than the below form:
$\int{\frac{\sec2x}{\sqrt{2}}dx + \sqrt{2}\int{\frac{\sin^2x\cos^2x}{\cos2x}dx}}$
$\implies \frac{1}{2\sqrt2}\log |\sec 2x + \tan 2x| + \sqrt{2}\int{\frac{\sin^2x\cos^2x}{\cos^2x-\sin^2x}dx} + C$
The answer that I'm supposed to get is:
$\frac{x}{\sqrt2}+C$
Please help, thanks!
| Use $$\frac{\sin^4x+\cos^4x}{\sqrt{1+\cos4x}}=\frac{1-\frac{1}{2}\sin^22x}{\sqrt2\cos2x}=\frac{\frac{1}{2}+\frac{1}{2}\cos^22x}{\sqrt2\cos2x}=\frac{\cos2x}{2\sqrt2(1-\sin^22x)}+\frac{1}{2\sqrt2}\cos2x=$$
$$=\frac{1}{4\sqrt2}\left(\frac{\cos2x}{1+\sin2x}+\frac{\cos2x}{1-\sin2x}\right)+\frac{1}{2\sqrt2}\cos2x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Converting polar equations to cartesian equations. Where $$r=\sin(3\theta)$$ and $$y=r\sin(\theta),~x=r\cos(\theta),~r^2=x^2+y^2$$ I have started by saying that $$ r=\sin (2\theta) \cos (\theta) +\sin (\theta) \cos (2\theta) \\ r=2\sin (\theta) \cos ^2 (\theta) +\sin (\theta) (1-2\sin ^2 (\theta)) \\r=2\sin (\theta) \cos ^2 (\theta) +\sin (\theta)-2\sin^3(\theta) $$ simply making the substitutions $$\sin(\theta)=\frac{y}{r},~\cos(\theta)=\frac{x}{r}$$ noting also that I can square both sides of the above to be substitutions we then can write down $$r=\dfrac{2yx}{r}\cdot \dfrac{x^2}{r^2}+\dfrac{y}{r}-\dfrac{2y^3}{r^3}$$ then multiplying through by $r^3$ we obtain $$r^4=2yx^3+yr^2-2y^3$$ then replacing $r^2$ with $x^2+y^2$ we get $$(x^2+y^2)^2=2xy^2+yx^2+y^3-2y^3 \\(x^2+y^2)^2=y(3x^2-y^2)$$
However I am unsure of where I have made a mistake as the true answer is$$(x^2+y^2)^2=4x^2y-(x^2+y^2)y$$ working backwards I've so far gotten to the point of asking how I would rearrange$$r=\sin (3\theta) \Rightarrow r=4\cos ^2 (\theta) \sin (\theta) -\sin (\theta)$$ so that $$\sin(3\theta)=4\cos ^2 (\theta) \sin (\theta) -\sin (\theta)$$ which I am afraid I'll have to ask help for the next steps. Thanks.
| $$r=\sin (3t)=Im ((\cos (t)+i\sin (t))^3) $$
$$=3\cos^2 (t)\sin (t)-\sin^3 (t) $$
$$=3\sin (t)-4\sin^3 (t) $$
$$=3y/r-4y^3/r^3$$
thus
$$r^4=3yr^2-4y^3$$
now replace $r^2$ by $x^2+y^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Polynomials division Let $p$ be a real polynomial ($p\in\mathbb{R}[x])$, if
*
*5 is the remainder of the division $\frac{p(x)}{x-2}$, and
*2 is the remainder of the division $\frac{p(x)}{x-5}$
What is the reaminder of the division $\frac{p(x)}{(x-2)(x-5)}$?
I have tried it, but I have no idea...
Thank you
| $p(x)=(x−2)(x−5)q(x)+ax+b$, for some $a,b \in \mathbb{R}$
Since we know the remainder of $p(x)$ when divided by $x-2$ and $x-5$
we obtain the following
\begin{align}
p(2) &= (2-2)(2-5)q(2) + a(2)+b\\ &= 2a+b = 5 \label{a}
\end{align}
\begin{align}
p(5) &= (5-2)(5-5)q(5) + a(5)+b\\ &= 5a+b = 2 \label{b}
\end{align}
Solving the simultaneous equations \ref{a} and \ref{b}, we obtain $a = -1 \text{ and } b = 7$.
Hence, the remainder of the division $\frac{p(x)}{(x−2)(x−5)}$ is $-x + 7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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continued fraction of $\sqrt{41}$
Show that $\sqrt{41} = [6;\overline {2,2,12}]$
here's my try:
$$\sqrt{36}<\sqrt{41}<\sqrt{49}\implies6<\sqrt{41}<7\implies\lfloor\sqrt{41}\rfloor=6$$
$$\sqrt{41}=6+\sqrt{41}-6=6+\frac{1}{\frac{1}{\sqrt{41}-6}}$$
$$\frac{1}{\sqrt{41}-6}=\frac{\sqrt{41}+6}{41-36}=\frac{\sqrt{41}+6}{5}=\frac{12+\sqrt{41}-6}{5}=2+\frac{\sqrt{41}-4}{5}$$
So far,
$$\sqrt{41}=6+\frac{1}{2+\frac{\sqrt{41}-4}{5}}=6+\frac{1}{2+\frac{1}{\frac{5}{\sqrt{41}-4}}}$$
But, $$\frac{5}{\sqrt{41}-4}=\frac{5(\sqrt{41}+4)}{41-16}=\frac{\sqrt{41}+4}{5}=\frac{6+\sqrt{41}-2}{5}=\color{red}{1}+\frac{\sqrt{41}-1}{5}$$
It suppose to be $2$ and not $1$.
Where is the mistake? (I triple-checked and it seems fine to me)
| Given $(\sqrt{41}+4)/5$, render $6<\sqrt{41}<7$ which you used at the start. Then your fraction lies between $(6+4)/5$ and $(7+4)/5$ showing the integral part is $2$.
You should be able to get the integer part at every stage just from $6<\sqrt{41}<7$.
| {
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Evaluate $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$ I am trying to evaluate the following integral
$$I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$$
with $0< k < 1$.
My attempt
By performing the substitution
$$y=\frac{1-x}{1+x} \Longleftrightarrow x=\frac{1-y}{1+y}$$
we have
$$ I(k) = \int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{y-1}{(k^2-1)y^2-2y(k^2+1)+(k^2-1)}$$
Now we can decompose
$$\frac{y-1}{(k^2-1)y^2-2y(k^2+1)+(k^2-1)}= \frac{1}{2k(1+k)}\frac{1}{y+a}- \frac{1}{2k(1-k)}\frac{1}{y+a^{-1}}$$
with $a=\frac{1-k}{1+k}$ runs from $0$ to $1$.
Therefore we can write
$$I(k) = \frac{1}{2k(1+k)}\color{blue}{\int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{1}{y+a}} - \frac{1}{2k(1-k)}\color{red}{\int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{1}{y+a^{-1}}}$$
So we have only to evaluate $\color{blue}{I_1}$ and $\color{red}{I_2}$.
Now we consider the following integral
$$ J(\sigma) = \int_0^1 \mathrm{d}x \frac{\log x}{\sqrt{x}}\frac{1}{x-\sigma^2}$$
so that $\color{blue}{I_1}=J(i\sqrt{a})$ and $\color{red}{I_2}=J(i/\sqrt{a})$.
By considering the map $x\mapsto x^2$ we can write
$$ J(\sigma)=4\int_0^1\mathrm{d}x \frac{\log x}{x^2-\sigma^2}= \frac{2}{\sigma}\left[\color{green}{\int_0^1 \frac{\log x}{x-\sigma}}-\color{green}{\int_0^1 \frac{\log x}{x+\sigma}}\right]$$
The problem then reduces to evaluate the $\color{green}{\text{green}}$ integrals. At this point I'm stuck. I think that it needs to be solved by using polylogarithms, but I don't really know how to use these functions.
Mathematica 11.0 says
$$J(\sigma)=4 \left(\frac{\Phi \left(\frac{1}{\sigma ^2},2,\frac{3}{2}\right)}{4 \sigma ^4}+\frac{1}{\sigma ^2}\right)$$
where $\Phi$ is the Lerch transcendent. I don't know if this result is true (numerical integration is somewhat problematic). However, if it is true, I don't know what to do next.
Any hint on how to proceed with the evaluation?
Thanks in advance!
| Another method which gives a closed form result. Using the identity
\begin{equation}
\ln\frac{1-x}{1+x}=-2\operatorname{arctanh} x
\end{equation}
and parity of the integrand, the integral can be written as
\begin{align}
I(k) &= \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)\\
&=-\int_{-1}^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\operatorname{arctanh} x\\
&=-\int_{-\infty}^\infty \frac{u\sinh u}{\cosh^2u-k^2\sinh^2u}du
\end{align}
The last integral is obtained by the substitution $x=\tanh u$.
Then
\begin{align}
I(k)&=-\frac{1}{2k}\left[J(k)-J(-k)\right]\\
J(k)&=\int_{-\infty}^\infty \frac{udu}{\cosh u-k\sinh u}
\end{align}
To evaluate $J(k)$, we enforce the substitution $t=e^u$ and, denoting $m^2=\frac{1+k}{1-k}$, we obtain
\begin{align}
J(k)&=\frac{2}{1-k}\int_0^\infty \frac{\ln v}{v^2+m^2}dv\\
&=\frac{2}{m(1-k)}\int_0^\infty \frac{\ln mt}{t^2+1}dt\\
&=\frac{2}{m(1-k)}\left[\int_0^\infty \frac{\ln t}{t^2+1}dt+\int_0^\infty \frac{\ln m}{t^2+1}dt\right]
\end{align}
We know that $\int_0^\infty \frac{\ln t}{t^2+1}dt=0$, thus
\begin{equation}
J(k)=\frac{\pi \ln m}{m(1-k)}=\frac{\pi}{2}\frac{\ln \frac{1+k}{1-k}}{\sqrt{1-k^2}}
\end{equation}
Finally,
\begin{equation}
I(k)=\frac{\pi}{2k\sqrt{1-k^2}}\ln\frac{1- k}{1+ k}
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
How can I prove that: $\frac{(n+1)(n+2)(n+3)...(3n)}{(1*2)(4*5)(7*8)...(3n-2)*(3n-1)} = 3^n$ How can I prove that:
$$\frac{(n+1)(n+2)(n+3)...(3n)}{(1 * 2)(4*5)(7*8)...(3n-2)*(3n-1)} = 3^n$$
Can you help me and explain me how can I prove it? I thought to prove it by induction, but I don't have idea how to do it in fact.
| Note that
*
*$(n+1)(n+2)(n+3)…(3n)=\frac{(3n)!}{n!}$
*$(1\cdot2)(4\cdot5)(7\cdot8)…(3n-2)\cdot(3n-1)=\frac{(3n-1)!}{3\cdot6\cdot9\cdot...\cdot(3n-3)}=\frac{(3n-1)!}{3^{n-1}(n-1)!}$
thus
$$\frac{(n+1)(n+2)(n+3)…(3n)}{(1\cdot2)(4\cdot5)(7\cdot8)…(3n-2)\cdot(3n-1)} = \frac{(3n)!}{n!}\frac{3^{n-1}(n-1)!}{(3n-1)!}=\\=3^{n-1}\frac{(3n)!}{(3n-1)!}\frac{(n-1)!}{n!}=3^{n-1}\cdot\frac{3n}n=3^n$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ as $x$ goes to $0$ As plugging $0$ in $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ makes the function becomes undetermined form of $\frac{0}{0}$. I tried applying L'Hospital's rule but it became messy and did not look helpful if I do further differentiation. So I tried finding the taylor polynomial of $\log(1+x^2)$, which is $x^2-\frac{1}{2}x^4+$... to see if I could bound the absolute value of the fraction above with something like $|\frac{x^2-x^2}{x^2\sin^2x}|$ and apply the comparison lemma, but it looks like that does not work too...
| As an alternative
$$\frac{x^2-\log(1+x^2)}{x^2\sin^2x}=\frac{x^2-\log(1+x^2)}{x^4}\cdot\frac{x^2}{\sin^2x}\to \frac12$$
indeed
$\frac{x^2}{\sin^2x}\to 1$ by standard limit
and let $y=x^2\to 0$
$$\frac{x^2-\log(1+x^2)}{x^4}=\frac{y-\log(1+y)}{y^2}\stackrel{HR}\implies\frac{1-\frac1{1+y}}{2y}=\frac{1+y-1}{2y(1+y)}=\frac{1}{2(1+y)}\to\frac12$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the area of a region bounded by two parabolas and a line
Given that the parabola $y=x^2+1$ has a tangent line at point $P$, find the area, $S$, of the region enclosed by the parabola $y=x^2+1$, the tangent line, and the parabola $y=x^2$.
The part which I do not understand is how to express the part where it asks to find the region enclosed, as I find no way to express that given region in an integral.
Thanks in advance.
| NOTE: In this method for finding the area enclosed, you will not need to find the actual points of intersection, see my method here.
Point $P$ can be expressed with coordinates $(a, a^2+1)$
Let $f(x)=x^2+a$
$f'(x)=2x$
$f'(a)=2a$
The equation of the tangent line is $y=2a(x-a)+a^2+1=2ax-a^2+1$
Setting this equal to $y=x^2$ to find the coordinates of intersection,
$x^2=2ax-a^2+1$
Here is the part you have gone up to, use this to confirm.
$x^2-2ax+(a^2-1)=0$
By the root coefficient relationship, where $a$ and $b$ are the roots ($a<b$),
$a+b=2a$
$ab=a^2-1$
$S=\int_a^b\left(2ax-a^2+1-x^2\right)dx=-\int_a^b\left(x^2-2ax+a^2-1\right)dx=\frac 16(b-a)^3=\frac 16 \left(\sqrt{(a+b)^2-4ab}\right)^3$
You know the values of $a+b$ and $ab$, can you solve the area enclosed from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Couple of implicit differentiation equations. Check if correct. Wanted to check if I did the implied differentiation correctly:
*
*Find the tangent line at ($\pi, \pi$)
$$ \sin{(x+y)} = 2x -2y$$
$$\cos{(x+y)} \cdot (1+\frac{dy}{dx}) = 2 - (2 \cdot \frac{dy}{dx})$$
$$\cos{(x+y)} + \cos({x+y}) \frac{dy}{dx} = 2 - (2\cdot \frac{dy}{dx})$$
$$\cos{(x+y}) - 2 = -\cos{(x+y)} \frac{dy}{dx} - 2 \cdot \frac{dy}{dx}$$
$$\frac{dy}{dx} (-\cos{(x+y)} + 2) = \cos{(x+y)} - 2$$
$$ \frac{dy}{dx} = -1 \cdot \frac{\cos{(x+y)} - 2}{\cos{x+y} +2}$$
slope at ($\pi, \pi$}: $\frac{- \cos{2\pi} - 2}{\cos{2\pi} + 2} = \frac{1}{3}$
So tangent line: $y - \pi = \frac{1}{3} (x - \pi)$
*Find the tangent line at 1,2
$$y^2 = 5x^4 - x^2$$
$$2y\frac{dy}{dx} = 20x^3 - 2x$$
$$\frac{dy}{dx} = \frac{20x^3 - 2x}{2y}$$
slope at 1,2 = $\frac{20-2}{4}$
tangent line: $y - 2 = \frac{9}{2} \cdot (x-1)$
$$y = \frac{9}{2}x - \frac{5}{2}$$
*I'm confused about how to find the double deriavtive of:
$$x^2 + 4y^2 = 4$$
so
$$y' = 2x +4 \cdot 2y \cdot \frac{dy}{dx} = 0$$
$$2x + 8y \frac{dy}{dx} = 0$$
$$8y \frac{dy}{dx} = -2x$$
$$\frac{dy}{dx} = \frac{-2x}{8y} = \frac{-x}{4y}$$
$$ y'' = \frac{-4y - (-x \cdot 4 \frac{dy}{dx}}{16y^2}$$
$$ \frac{-4y + (4x\frac{dy}{dx})}{16y^2}$$
$$\frac{-4y + \frac{-4x^2}{4y}}{16y^2}$$
| 1) Here you made a sign mistake
$$\frac{dy}{dx} (-\cos{(x+y)} \color{red}{+ 2}) = \cos{(x+y)} - 2$$
It should be negative not positive...
Answer is
$$\frac {dy}{dx}=\frac {2-\cos(x+y)}{2+\cos(x+y)}$$
2) $y - 2 = \frac{9}{2} \cdot x(x-1) \to y =\frac{9}{2}x^2 -\frac92x+2$
Maybe you meant $y - 2 = \frac{9}{2} \cdot (x-1) \to y=\frac 92x-\frac 52$
3)$$x^2 + 4y^2 = 4$$
Differentiate and simplify by 2 $$x + 4yy' = 0$$
Differentiate again $$1 + 4((y')^2+yy'') = 0$$
If you want $y''$ as a fucntion of y and x then substitute the value of $y'=-\frac x {4y}$
Substitution $$1 + 4((\frac {x^2} {16y^2})+yy'') = 0$$
Substitute from your first equation the value of $x^2$
$$1 + (\frac {x^2} {4y^2})= -4yy''$$
$$y''=-\frac 1 {4y} - \frac {1-y^2} {4y^3}$$
$$y''=-\frac {1} {4y^3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove the following algebraically $\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2$ I came across the following proof in my textbook that was used as a end of chapter review. How can I prove the following algebraically?
$$\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2$$
using $\left( \begin{array}{c} n \\ 2\ \end{array} \right) = \dfrac{n(n-1)}{2}$?
$$\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = \frac{2n(2n-1)}{2} = 2n(n-1) = 2 \frac{n(n-1)}{2} + n^2 = 2\left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2 $$
Is this proof correct? if not how do I fix it?
| Since OP asks for an algebraic proof, I type this proof making use of number of roots of a degree two polynomials for fun.
Set
$f(n) = \left( \begin{array}{c} 2n \\ 2\ \end{array} \right) - 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) - n^2
$.
Since $\deg f \le 2$, $f$ has at most two distinct roots.
\begin{align}
f(2) &= \left( \begin{array}{c} 4 \\ 2\ \end{array} \right) - 2 \left( \begin{array}{c} 2 \\ 2\ \end{array} \right) - 4 = 6-2-4=0 \\
f(3) &= \left( \begin{array}{c} 6 \\ 2\ \end{array} \right) - 2 \left( \begin{array}{c} 3 \\ 2\ \end{array} \right) - 9 = 15-6-9=0 \\
f(4) &= \left( \begin{array}{c} 8 \\ 2\ \end{array} \right) - 2 \left( \begin{array}{c} 4 \\ 2\ \end{array} \right) - 16 = 28-12-16=0
\end{align}
Therefore, $f$ is identically zero. Hence $\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer My question is related to using mathematical induction to prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer.
Work so far:
(1) For n = 1:
$2^{5(1) + 1} + 5^{(1) + 2} = 26 + 53 = 64 + 125 = 189$
Check if divisible by $27$: $189$ mod $27$ = $0$
As no remainder is left, the base case is divisible by $27$.
(2) Assume $n = k$, then $2^{5k + 1} + 5^{k + 2} = 27k$
(3) Prove that this is true for n = k + 1:
$$2^{5(k + 1) + 1} + 5^{(k + 1) + 2} $$
$$= 2^{5k + 5 + 1} + 5^{k + 1 + 2} $$
$$ = 32 * 2^{5k + 1} + 5 * 5^{k + 2}$$
$$= ? $$
I know I am supposed to factor out 27 somehow, I just cant seem to figure it out. Any help would be appreciated.
| Use the fact that $32=5+27$ to get
$$\begin{align}2^{5(k+1)+1}+5^{k+1+2}&=2^{5k+6}+5^{k+3}\\&=32\cdot2^{5k+1}+5\cdot5^{k+2}\\&=5\cdot2^{5k+1}+27\cdot2^{5k+1}+5\cdot5^{k+2}\\&=5(2^{5k+1}+5^{k+2})+27\cdot2^{5k+1}\\&=27(5t+2^{5k+1})\end{align}$$ for some positive integer $t$. The result follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine the value of summation form. Determine the value of
$\displaystyle \sum_{k=n}^{2n} \binom{k} {n} 2^{-k}$ for $n \geq 1$
My attempt:
I have asked this question before in another math forum but it leads no solution. My friend suggested me to search for something like series-k. I also looked at my textbook in combinatoric section (which apparently leads to Binomial Theorem), but really I don't have any idea for this one.
Could you help me?
| This answer is based upon the Lagrange inversion formula. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance
\begin{align*}
[z^n]\frac{1}{1-2z}=[z^n]\sum_{j=0}^\infty 2^jz^j=2^n
\end{align*}
We obtain
\begin{align*}
\color{blue}{\sum_{k=n}^{2n}\binom{k}{n}2^{-k}}
&=\sum_{k=0}^n\binom{n+k}{k}2^{-n-k}\tag{1}\\
&=\frac{1}{2^n}\sum_{k=0}^n\binom{-n-1}{k}\left(-\frac{1}{2}\right)^k\tag{2}\\
&=\frac{1}{2^n}[z^n]\frac{1}{\left(1-\frac{z}{2}\right)^{n+1}(1-z)}\tag{3}\\
&=\frac{1}{2^n}[z^n]\left.\left(\frac{1}{\left(1-\frac{w}{2}\right)(1-w)}\cdot\frac{1}{1-\frac{w}{2}\cdot\frac{1}{1-\frac{w}{2}}}\right)\right|_{w=\frac{z}{1-\frac{w}{2}}}\tag{4}\\
&=\frac{1}{2^n}[z^n]\left.\frac{1}{(1-w)^2}\right|_{w=1-\sqrt{1-2z}}\tag{5}\\
&=\frac{1}{2^n}[z^n]\frac{1}{1-2z}\\
&\color{blue}{=1}
\end{align*}
Comment:
*
*In (1) we shift the index $k$ to start with $k=0$.
*In (2) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$ and $\binom{p}{q}=\binom{p}{p-q}$.
*In (3) we observe the sum is the coefficient of the convolution of two series
\begin{align*}
[z^n]\left(\sum_{k}a_kz^k\right)\left(\sum_{l}b_lz^l\right)=[z^n]\sum_{N}\left(\sum_{k=0}^Na_k b_{N-k}\right)z^N
=\sum_{k=0}^na_k b_{n-k}
\end{align*}
Here with $a_k=\binom{-n-1}{k}\left(-\frac{1}{2}\right)^k$ and $b_k=1$.
*In (4) we use the Lagrange Inversion Formula in the form G6 stated in R. Sprugnolis (etal) paper Lagrange Inversion: when and how.
\begin{align*}
[z^n]F(z)\Phi(z)^n=[z^n]\left.\frac{F(w)}{1-z\Phi'(w)}\right|_{w=z\Phi(w)}
\end{align*}
Here with $\Phi(z)=\frac{1}{1-\frac{z}{2}}$ and $F(z)=\frac{1}{\left(1-\frac{z}{2}\right)(1-z)}$. It follows since $w=z\Phi(w)$:
\begin{align*}
z\Phi^{\prime}(w)=\frac{z}{2}\cdot\frac{1}{\left(1-\frac{w}{2}\right)^2}
=\frac{z}{2}\Phi^2(w)=\frac{w}{2}\Phi(w)=\frac{w}{2}\cdot\frac{1}{1-\frac{w}{2}}
\end{align*}
*
*In (5) we simplify the expression and we select the solution $w=w(z)$ from $w=\frac{z}{1-\frac{w}{2}}$ which can be expanded in a power series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can verify this through expansion, but if I were a mathematician without access to CAS, how might I notice that this is the case?
I suppose what I'm asking is how one should "see" that $P$ can be simplified to $(1 + x + x^2 + x^3 + x^4 + x^5)^2$? Is it a multinomial thing (which seems a bit too complicated for someone to "notice"), or is there something simpler about the polynomial that one could use to factor it?
| The coefficient of $x^n$ ($0\leq n\leq 10$) in the expansion is the number of solutions to
$$
x_{1}+x_{2}=n;\quad 0\leq x_i\leq5.
$$
We can compute this via inclusion exclusion. Let $U$ be the set of all solutions to the previous question in non-negative integers and $A_{i}$ be the set of solutions in $U$ which $x_i>5$. Then
$$
\begin{align}
|A_{1}^c\cap A_{2}^c|&=|U|-|A_{1}|-|A_{2}|+|A_{1}\cap A_{2}|\\
&=(n+1)-2(n-5)[n\geq6]+0
\end{align}
$$
where $[\cdot]$ is the iverson bracket since $A_{1}\cap A_{2}=\varnothing$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 5
} |
Evaluate $\sum_{k=1}^n\frac{1}{k}\binom{n}{k}$ I'm interested in finding a nice closed form expression for the sum $\sum_{k=1}^n\frac{1}{k}\binom{n}{k}$. I've tried using the Binomial Theorem to get
\begin{align*}
\sum_{k=1}^n\frac{1}{k}\binom{n}{k}x^k & =\int_0^1\frac{(1+x)^n-1}{x} \, dx\\
&=\int_1^2 (1+u+\cdots+u^{n-1}) \, du
\end{align*}
using the substitution $u=1+x$ but I can't quite to simplify this integral either. I have also not been able to come up with a combinatorial approach, which may not exist since the summation and its terms are in general not integers. Any help in evaluating this sum would be appreciated, thanks!
| $f(n)=∑ ^n_{k=1}\frac{1}{k} C^n_k $
This is equal to term by term multiplication of following series:
$∑^n_{k=1} C^n_k =∑^n_{k=0} C^n_k-1=(1+1)^n-1=2^n-1$
$∑^n_{k=1}\frac{1}{k}= 1+1/2+1/3 + . . . 1/n =H_n $
⇒ $f(n)=∑ ^n_{k=1}\frac{1}{k} C^n_k =(2^n-1)H_n=∑^n_{k=1}\frac{2^k-1}{k}$
Note that $(2^n-1)$represent$(∑ ^n_{k=1} C^n_k)$ and $(H_n)$ represent$(∑ ^n_{k=1}\frac{1}{k})$ that is $(∑ ^n_{k=1}\frac{1}{k} C^n_k)=∑ ^n_{k=1}\frac{1}{k}\times∑ ^n_{k=1} C^n_k$(term by term multiplication).For example for n=5:
$f(5)=∑ ^5_{k=1}\frac{1}{k} C^5_k$ is not$(2^5-1)(1+1/2+1/3+1/4+1/5)=\frac{4247}{60}$, it is calculated as:
$f(5)=∑ ^5_{k=1}\frac{1}{k} C^5_k=\frac{2^1-1}{1}+\frac{2^2-1}{2}+\frac{2^3-1}{3}+\frac{2^4-1}{4}+\frac{2^5-1}{5}=\frac{887}{60}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
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