Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ I want to solve $\frac{5}{x+3}+\frac{2}{2x+6}=4$ for $x$
I try this;
$5(2x+6)+2(x+3)=4(x+3)(2x+6)$
$12x+36 = 4(2x^2+12x+18)$
$8x^2+36x+36=0$
Where I would then go to the formula, however the answer says -1.5, what am I doing wrong.
| The solutions of $8x^2+36x+36=0$ are $-1.5$ and $-3$, but the latter is not a solution of the original equation. This new, false solution comes when you multiply the equation by $(x+3)(2x+6)$, which is not the lcm of the denominators.
Note that $2x+6=2(x+3)$.
An example to show what's happening:
Take the equation
$$3x+5=-1$$
The solution is $x=-2$. If you multiply both sides of any equation by a non zero number, you get another equation, whose solutions are the same. But if you multiply both sides by $0$ you get $0=0$.
Now, if you multiply both sides by, say, $x+4$, we don't know if $x+4$ is zero or not. So we obtain another equation with two solutions:
$$3x^2+17x+20=-x-4$$
$$3x^3+18x+24=0$$
$$x=\left\lbrace\begin{align}&-2\\&-4\end{align}\right.$$
One of the solutions is the original one. The other is the value for which the factor $x+4$, that we multiplied by, vanishes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 2
} |
the sum of consecutive odd numbers If the sum of consecutive odd numbers starting with $-3$ until $2k+1$ equals $21$
What is the value of $k$ ?
I can solve this by trying the numbers $-3-1+1+3+5+7+9=21$ , so the last term is $7th$ so the $k$ value is $3$
But I could not solve this with formula, I know the odd numbers come in the form of $2k+1$ but could not get much further.
| Notice that
$$\begin{align}1^2&=1\\2^2&=1+3\\3^2&=1+3+5\\\vdots&\qquad\qquad\qquad\quad\ddots\\k^2&=1+3+5+\dots+(2k-1)\\(k+1)^2&=1+3+5+\dots+(2k-1)+(2k+1)\end{align}$$
Thus,
$$-3-1+1+3+\dots+(2k+1)=(k+1)^2-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Limit of a sequence Calculate the limit of the sequence $\{x_n\},$ defined as follows: $$x_n=\dfrac{a+aa+aaa+aaaa\cdots+aaaaaaa..aaa }{10^n},$$ where $a\in\{1,2\ldots,9\}.$
$aaaaaaa..aaa = a , n$ time
Can anyone help ?
| It looks like you mean the following:
$$x_n = \frac{a}{10^n}\sum_{k=1}^{n}\sum_{j=0}^{k-1}10^j$$
Assuming this is the case, first note that the inner sum can be computed in closed form. For any constant $c \neq 1$, we have
$$\sum_{j=0}^{k-1}c^j = \frac{c^k - 1}{c - 1}$$
so for $c = 10$ this is
$$\sum_{j=0}^{k-1}10^j = \frac{1}{9}(10^k - 1)$$
Substituting this back into the original expression gives us
$$x_n = \frac{a}{9\cdot 10^n}\sum_{k=1}^{n}(10^k - 1)$$
We can evaluate the sum as follows:
$$\begin{aligned}
\sum_{k=1}^{n}(10^k - 1)
&= \sum_{k=1}^{n}10^k - \sum_{k=1}^{n}1 \\
&= 10\sum_{k=0}^{n-1}10^k - n \\
&= \frac{10}{9}(10^n - 1) - n \\
\end{aligned}$$
Substituting this into the previous expression, we end up with
$$\begin{aligned}
x_n &= \frac{a}{9\cdot 10^n}\left( \frac{10}{9}(10^n - 1) - n \right) \\
&= \frac{10a}{81}\left( 1 - \frac{1}{10^n} \right) - \frac{na}{9 \cdot 10^n}\\
\end{aligned}$$
In the limit as $n \to \infty$, the second term converges to zero, and the first term converges to $10a/81$. Therefore we conclude that
$$\lim_{n \to \infty}x_n = \frac{10a}{81}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Which would you rather have? Q. Which would you rather have, a piece of an 8-inch pie that's been cut into sixths or a piece of a 10-inch pie that's been cut into eights?
A. This is a problem involving sectors. One-sixth of a pie is $\frac {1}{6}$ of $ 2\pi$ radians. The measure of the central angle is $\frac{1}{6} \cdot 2\pi = \frac{\pi}{3}$. An 8-inch pie has a 4-inch radius. Putting the angle and radius into the formula, for the area of a sector, $A= \frac{1}{2} \cdot \frac{\pi}{3} \cdot 4^2 = \frac{16\pi}{6} = \frac{8\pi}{3} \approx 8.38$ square inches. One-eighth of a pie is $\frac{1}{8}$ of $2\pi$ radians. The measure of the angle is $\frac{1}{8} \cdot 2\pi = \frac{2\pi}{8} = \frac{\pi}{4}$. A 10-inch pie has a 5-inch radius. Putting the angle and radius into the formula, for the area of a sector, $A= \frac{1}{2} \cdot \frac{\pi}{4} \cdot 5^2 = \frac{25\pi}{8} \approx 9.82$ square inches. There isn't too much difference, but it looks like the smaller part of the bigger pie has the larger piece, in terms of area.
My question:
Is how do one, get, $2\pi$ radians?
| You don't need to even use $\pi$ here. Circles behave like squares when it comes to area, so you can just calculate that
$$\frac{8^2}6 < \frac{10^2}8.$$
| {
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"url": "https://math.stackexchange.com/questions/2048416",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^{\infty}\frac{e^x-1}{xe^x(e^x+1)}dx.$
Evaluate the integral
$$
\int_{0}^{\infty}{\mathrm{e}^{x} - 1 \over x\,\mathrm{e}^{x}\left(\mathrm{e}^x+1\right)}\,\mathrm{d}x\,.
$$
I have no idea how to approach integrals like this and I can't get any valuable result. Any hints will be appreciated.
| Notice that
\begin{align*}
\int_{0}^{\infty} \frac{e^x - 1}{x e^x(e^x + 1)} \, dx
&= \int_{0}^{\infty} \left( \int_{0}^{1} e^{xt} \, dt \right) \frac{dx}{e^x(e^x+1)} \\
&= \int_{0}^{1} \int_{0}^{\infty} \frac{e^{tx}}{e^x(e^x+1)} \, dxdt.
\end{align*}
In order to compute the inner integral, we utilize the geometric series as follows:
\begin{align*}
\int_{0}^{\infty} \frac{e^{tx}}{e^x(e^x+1)} \, dx
&= \int_{0}^{\infty} \frac{e^{-(2-t)x}}{1+e^{-x}} \, dx \\
&= \sum_{n=0}^{\infty} (-1)^n \int_{0}^{\infty} e^{-(n+2-t)x} \, dx \\
&= \sum_{n=0}^{\infty}\frac{(-1)^n}{n+2-t}. \tag{1}
\end{align*}
Plugging this back, we have
\begin{align*}
\int_{0}^{\infty} \frac{e^x - 1}{x e^x(e^x + 1)} \, dx
&= \sum_{n=0}^{\infty} (-1)^n\int_{0}^{1} \frac{dt}{n+2-t} \\
&= \sum_{n=0}^{\infty} (-1)^n ( \log(n+2) - \log(n+1) ) \\
&= \log \left( \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots \right) \\
&= \log \left(\frac{\pi}{2}\right),
\end{align*}
where in the last line we utilized the Wallis product. Alternatively, we can identify $\text{(1)}$ as
$$ \int_{0}^{\infty} \frac{e^{tx}}{e^x(e^x+1)} \, dx = \frac{1}{2}\left( \psi_0\left(\frac{3-t}{2}\right) - \psi_0\left(\frac{2-t}{2}\right) \right), $$
where $\psi_0$ is the digamma function. Plugging this back gives
$$ \int_{0}^{\infty} \frac{e^x - 1}{x e^x(e^x + 1)} \, dx = \log\left( \frac{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1)^2} \right) = \log\left(\frac{\pi}{2}\right). $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Let $a^2b^2+b^2c^2+a^2c^2=abc(a+b+c)$. Why does $a=b=c$? Let $a,b,c\in \mathbb{R}$ and $a,b,c \ne 0$ and $a^2b^2+b^2c^2+a^2c^2=abc(a+b+c)$.
Why does $a=b=c$?
| $$2(a^2b^2+b^2c^2+c^2a^2)=2a^2bc+2b^2ca+2c^2ab\\\implies (a^2b^2+b^2c^2-2ab^2c)+(a^2b^2+c^2a^2-2a^2c)+(b^2c^2+c^2a^2-2abc^2)=0\\\implies (ab-bc)^2+(bc-ca)^2+(ca-ab)^2=0$$
From here you can only say $$ab=bc=ac$$
However since $a,b,c\neq0$ you get $a=b=c$
Bingo!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051643",
"timestamp": "2023-03-29T00:00:00",
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Given that $abc=1$, prove that $\frac{b}{c} + \frac{c}{a}+ \frac{a}{b} \ge \frac{1}{a} + \frac{1}{b} +\frac{1}{c}$. I have tried to do a bunch of different versions of AM-GM on this and the equivalent versions that come from substituting in $abc=1$ and I always flip the inequality by doing it.
Help is greatly appreciated.
| Put $a = \dfrac{x}{y}, b = \dfrac{y}{z}, c = \dfrac{z}{x}$, then $x,y,z > 0$, and $abc = 1$, the inequality to prove is: $\dfrac{xy}{z^2}+\dfrac{yz}{x^2}+\dfrac{xz}{y^2} \ge \dfrac{x}{z}+\dfrac{z}{y}+\dfrac{y}{x}$, or equivalently:$(xy)^3+(yz)^3+(xz)^3 \ge (xy)^2(xz)+(yz)^2(xy)+(xz)^2(yz)$. Put $m = xy, n = yz, p = xz$, then we finally prove the inequality: $m^3+n^3+p^3 \ge m^2p+p^2n+n^2m$. By AM-GM inequality: $m^2p = m\cdot m\cdot p \le \dfrac{m^3+m^3+p^3}{3}, n^2m = n\cdot n\cdot m \le \dfrac{n^3+n^3+m^3}{3}, p^2n = p\cdot p\cdot n \le \dfrac{p^3+p^3+n^3}{3}$. Adding these $3$ inequalities we have the desire result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2054232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving one of the Binomial Identities The binomial identity is
$$\sum^{\left \lfloor{n/3}\right \rfloor }_{k=-\left \lfloor{n/3}\right \rfloor} (-1)^k \binom{2n}{n+3k} = 2 \cdot 3^{n-1} $$
and this is valid for all positive integers $n$. What would be some proofs to show that this identity is true?
| Suppose we seek to prove that
$$\sum_{k=-\lfloor n/3\rfloor}^{\lfloor n/3\rfloor} (-1)^k {2n\choose n+3k}
= 2\times 3^{n-1}.$$
We start by introducing the integral
$${2n\choose n+3k} = {2n\choose n-3k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-3k+1}} (1+z)^{2n}
\; dz.$$
Observe that this vanishes for $3k\gt n$ (pole canceled) and for
$3k\lt -n$ (upper range of polynomial term exceeded) so we may extend
the summation to $[-n, n]$ getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{2n}
\sum_{k=-n}^{n} (-1)^k z^{3k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^{2n} (-1)^n z^{-3n}
\sum_{k=0}^{2n} (-1)^k z^{3k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{4n+1}} (1+z)^{2n} (-1)^n
\frac{1-(-1)^{2n+1} z^{3(2n+1)}}{1+z^3}
\; dz
.$$
Only the first piece from the difference due to the geometric series
contributes and we get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{4n+1}} (1+z)^{2n} (-1)^n
\frac{1}{1+z^3}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{4n+1}} (1+z)^{2n-1} (-1)^n
\frac{1}{1-z+z^2}
\; dz
.$$
We have two poles other than zero and infinity at $\rho$ and $1/\rho$ where
$$\rho = \frac{1+\sqrt{3}i}{2}$$
and using the fact that residues sum to zero we obtain
$$S + \frac{(-1)^n}{\rho(1+\rho)} \frac{1}{\rho-1/\rho}
\left(\frac{(1+\rho)^2}{\rho^4}\right)^n
+ \frac{(-1)^n}{1/\rho(1+1/\rho)} \frac{1}{1/\rho-\rho}
\left(\frac{(1+1/\rho)^2}{1/\rho^4}\right)^n
\\ + \mathrm{Res}_{z=\infty} \frac{1}{z^{4n+1}} (1+z)^{2n-1} (-1)^n
\frac{1}{1-z+z^2} = 0.$$
We get for the residue at infinity
$$-\mathrm{Res}_{z=0} \frac{1}{z^2}
z^{4n+1} (1+1/z)^{2n-1} (-1)^n \frac{1}{1-1/z+1/z^2}
\\= -\mathrm{Res}_{z=0}
z^{2n+2} (1+z)^{2n-1} (-1)^n \frac{1}{z^2-z+1} = 0.$$
Now if $z^2 = z-1$ then $z^4 = z^2-2z+1 = -z$ and thus
$$\frac{(1+1/\rho)^2}{1/\rho^4} = \frac{(1+\rho)^2}{\rho^4}
= \frac{\rho-1+2\rho+1}{-\rho} = -3$$
and furthermore with $z(1+z)(z-1/z) = (1+z)(z^2-1)$ and
$(1+z)(z-2) = z^2-z-2 = -3$ we finally get
$$S + (-1)^n \times \left(-\frac{1}{3}\right) (-3)^n
+ (-1)^n \times \left(-\frac{1}{3}\right) (-3)^n = 0$$
or
$$\bbox[5px,border:2px solid #00A000]{S = 2\times 3^{n-1}.}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the divergence of the field F: $\vec{F} = \frac{yj - zk}{\sqrt{y^2+ z^2}}$
Find the divergence of the field F:
$$\vec{F} = \frac{yj - zk}{\sqrt{y^2+ z^2}}$$
Normally I find dot the gradient with F like this:
$$\langle 0,y,-z\rangle\cdot\langle F_x, F_y, F_z\rangle$$
And this simplifies to:
$$\langle 0, 1, -1\rangle$$
And then I tried dividing each component of the answer by:
$$\sqrt{y^2 +z^2},$$
but that isn't correct according to the answer in the back of my book.
How do I go about properly solving this? Thanks
| $\nabla \cdot (0\mathbf i + \frac {y}{\sqrt{y^2+z^2}}\mathbf j - \frac {z}{\sqrt{y^2+z^2}}\mathbf k) = \frac {\partial}{\partial x}0+\frac {\partial}{\partial y}\frac {y}{\sqrt{y^2+z^2}} + \frac {\partial}{\partial z} \frac {-z}{\sqrt{y^2+z^2}}\\
0+\frac {2y^2 + z^2}{(y^2+z^2)^\frac 32}-\frac {y^2 + 2z^2}{(y^2+z^2)^\frac 32} = \frac {y^2-z^2}{(y^2+z^2)^\frac 32}$
alternate
$\nabla \cdot \left((0\mathbf i + y\mathbf j - z\mathbf k) \left(\frac 1{(y^2+z^2)^\frac 12}\right)\right)\\
\left(\nabla\cdot(0,y,-z)\right)\left(\frac 1{(y^2+z^2)^\frac 12}\right) +(0,y,-z)\cdot \nabla \left(\frac 1{(y^2+z^2)^\frac 12}\right) \\
0 + \left(\frac {(0,y,-z)\cdot(0,y,z)}{(y^2+z^2)^\frac 32}\right)\\
\frac {y^2-z^2}{(y^2+z^2)^\frac 32}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the Sum of an Infinite Series Compute
$${1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.}$$
I tried creating a partition for this but no such luck.
What would the the equation to generate the sum for the nth term
| 1/2 + 1/4 + 1/8 + 1/16 + ..... + 1/(2^n) + ..... = (1/2)*2 = 1
1/4 + 1/8 + 1/16 + ..... + 1/(2^n) + ..... = [1/(2^2)]*2 = 1/2
1/8 + 1/16 + ..... + 1/(2^n) + ..... = [1/(2^3)]*2 = 1/(2^2)
So on and so forth.
Hence, S = 1 + 1/2 + 1/(2^2) + ..... + 1/(2^n) + ..... = 1 + 1 = 2
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I find the quadratic equation from highest point on parabola? If I have a parabola, where the vertex is in $P=(2,0)$, and a point on the parabola is $Q=(-1,-6)$, how can I find the quadratic equation of the form:
$$f(x) = ax^2 + bx + c$$
| The vertex is the
extreme point,
at which $f'(x) = 0$.
Since
$f'(x) = 2ax+b$,
this is zero when
$x = -\frac{b}{2a}
$.
At this point
$\begin{array}\\
f(x)
&=a(-\frac{b}{2a})^2+b(-\frac{b}{2a})+c\\
&=\frac{ab^2}{4a^2}-\frac{b^2}{2a}+c\\
&=\frac{b^2}{4a}-\frac{b^2}{2a}+c\\
&=-\frac{b^2}{4a}+c\\
\end{array}
$
Since this point is
$(2, 0)$,
$-\frac{b}{2a}=2$
and
$-\frac{b^2}{4a}+c = 0$,
or
$b = -4a$
and
$c = \frac{b^2}{4a}
=\frac{16a^2}{4a}
=4a
$.
Since the parabola
also passes through
$(-1, -6)$,
we get
$-6
= a(-1)^2+b(-1)+c
= a-b+c
=a-(-4a)+4a
=9a
$
or
$a=-\frac{6}{9}=-\frac{2}{3}$.
$b$ and $c$ follow.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $5x^2+5y^2-8xy-2x-4y+5=0$ While I was solving the question given by Harsh Kumar I could not find the real roots just because I changed the sign of $5y^2$. I also proved that there don't exist any real solution of that equation.
The equation I made by mistake was: $$5x^2+5y^2-8xy-2x-4y+5=0.$$
Please help me to find the solutions if exist.
| We may rewrite this as
$$
4x^2 - 8xy + 4y^2 + x^2 -2x + 1 + y^2 - 4y + 4 = 0\\
(2x - 2y)^2 + (x-1)^2 + (y-2)^2 = 0
$$
and we see that this is a sum of three (real) squares that equals $0$. If the sum of three squares equals $0$, then that means that each of the squares are zero. But $(x-1)^2 = 0$ means $x = 1$ and $(y-2)^2 = 0$ means $y = 2$, and then $(2x-2y)^2 \neq 0$. Therefore there can be no real solutions to this polynomial equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{a}{2}x^2+bx+c=0$ has a root between $x_1$ and $x_2$. If $x_1$ and $x_2$ are non-zero roots of the equations $ax^2+bx+c=0$ and $-ax^2+bx+c=0$
respectively. Prove that $\frac{a}{2}x^2+bx+c=0$ has a root between $x_1$ and $x_2$.
Please help me ..
| Let $f(x)=\frac{a}{2}x^2+bx+c$, then $f(x_1)=-\frac{a}{2}x_1^2$ and $f(x_2)=\frac{3a}{2}x_2^2$. Consequently $f(x_1)f(x_2)<0$. By continuity it should have a root in $(x_1,x_2)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Calculate the limit $\lim_{(x,y,z)->(0,0,0)}{\frac{xyz}{\sqrt{x} + \sqrt{y} + \sqrt{z}}}$ $$\lim_{(x,y,z)->(0,0,0)}{\frac{xyz}{\sqrt{x} + \sqrt{y} + \sqrt{z}}}$$
I tried thinking about a solution using the epsilon-delta definition: I assumed that the limit is 0 but I couldn't figure out a inequality that would lead me to the solution.
| Using the AM-GM inequality
$$0 \le \frac{xyz}{\sqrt{x}+\sqrt{y}+\sqrt{z}} = \frac{\left( \sqrt[3]{\sqrt{x}\sqrt{y}\sqrt{z}} \right)^{3/2}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} \le \frac{\left( \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3} \right)^{3/2}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} = \frac{1}{3\sqrt{3}}\sqrt{\sqrt{x}+\sqrt{y}+\sqrt{z}}$$
so by the squeeze theorem the limit is $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of csc$\frac{16\pi}{3}$. Q. Find the value of csc$\frac{16\pi}{3}$.
A. This angle is equal to two full revolutions plus $\frac{4\pi}{3}$. I got this by subtracting, $\frac{16\pi}{3} - 4\pi = \frac{16\pi}{3} - \frac{12\pi}{3} = \frac{4\pi}{3}$. The terminal side is in Quadrant III. The reference angle is $\frac{4\pi}{3} - \pi = \frac{\pi}{3}$. The cosecant of $\frac{\pi}{3}$ is $\frac{2\sqrt{3}}{3}$. Using the negative, the answer is $-\frac{2\sqrt{3}}{3}$.
My questions:
is how do one tell how many revolutions it takes, based off the fraction they give?
And how does one find the cosecant of $\frac{\pi}{3}$? On the calculator I use sine and then take the reciprocal of sine, but sine come out to $.018276028$. And I'm not sure on how to get the reciprocal from that or even find the cosecant of that number.
| Consider the angle $960^\circ$. I can easily break this down into
$$360^\circ + 360^\circ + 180^\circ + 60^\circ$$
knowing that $360^\circ$, $180^\circ$, and $60^\circ$ are a full revolution, half revolution, and an acute angle, respectively. Instead of making a "note to self" regarding the number of revolutions, let's just make it a bit more explicit by expressing the angle as a multiple of $180^\circ$.
$$960^\circ = 720^\circ + 180^\circ + 60^\circ = \frac{720^\circ + 180^\circ + 60^\circ}{180^\circ}\cdot 180^\circ = (4+1+\frac{1}{3})180^\circ$$
And instead of saying to myself $(4+1+\frac{1}{3})180^\circ$, I'll say instead, $(4+1+\frac{1}{3})\text{"Half-Revolutions"}$.
Were the angle expressed in Radians, instead, I would do much the same thing.
$$\frac{16\pi}{3} = \frac{16}{3}\pi = (\frac{12 + 3 + 1}{3})\pi= (4 + 1 + \frac{1}{3})\pi = (4+1+\frac{1}{3})\text{"Half-Revolutions"}$$
I prefer the "half-revolution" approach because of it's one-to-one correspondence with the angle $\pi$ (when expressed in Radians).
Now that I have an idea of what the angle is (in revolutions), I may try to plot it.
The green spiral shows my 4 half-revolutions, the red curve my single half revolution, and the blue my reference angle. Now, to find the $\csc \frac{16}{3}\pi$, just find the $\sin \frac{16}{3}\pi$ and take its reciprocal.
$$\csc \frac{16}{3}\pi = \frac{1}{\sin \frac{16}{3}\pi}=\frac{1}{-\frac{\sqrt{3}}{2}} = \frac{1}{-\frac{\sqrt{3}}{2}}\cdot \frac{-\frac{2}{\sqrt{3}}}{-\frac{2}{\sqrt{3}}} = \frac{-\frac{2}{\sqrt{3}}}{1} = -\frac{2}{\sqrt{3}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do you find the derivative of $\sqrt{x^2+9}$ using the definition of a derivative? I had this question on a not-so-recent math test. It asked to find the derivative of $\sqrt{x^2+9}$ with the definition of a derivative. Using the chain rule, I can figure out that the derivative is $\frac x{\sqrt{x^2+9}}$, but how can it be done with only the definition of a derivative? I tried multiplying both sides of the fraction by the square roots, but that just makes a mess of everything.
| By the definition of the derivative, what we want to evaluate is
$$\lim_{h \to 0} \frac{\sqrt{(x+h)^2+9}-\sqrt{x^2+9}}{h}$$
We now multiply the top and bottom by the conjugate of the numerator (to simplify the square roots) and simplify to get
$$=\lim_{h \to 0} \frac{\sqrt{(x+h)^2+9}-\sqrt{x^2+9}}{h}\frac{\sqrt{(x+h)^2+9}+\sqrt{x^2+9}}{\sqrt{(x+h)^2+9}+\sqrt{x^2+9}}$$
$$=\lim_{h \to 0} \frac{h (h + 2 x)}{h(\sqrt{(x+h)^2+9}+\sqrt{x^2+9})}$$
$$=\lim_{h \to 0} \frac{h + 2 x}{\sqrt{(x+h)^2+9}+\sqrt{x^2+9}}$$
$$=\frac{2 x}{\sqrt{x^2+9}+\sqrt{x^2+9}}$$
$$=\frac{2 x}{2\sqrt{x^2+9}}$$
$$={x\over {\sqrt{x^2+9}}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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solve for $x$ in inverse trignometry $$
\operatorname{arccot} x + \operatorname{arccot} (n^2-x + 1) = \operatorname{arccot }(n - 1)
$$
In this we have to solve for value of $x$ .
I thought to convert arccot into arctan . Then add using the identity .
But its getting too long .
Is there any short method to solve it.
|
1) Take the cotangent from both side:
$cot(arccot(x)+arccot(n^2-x+1))=cot(arccot(n-1))$
2) Use $cot(A+B)={cot(A)cot(B)-1\over cotA+cotB}$
${cot(arccot(x))cot(arccot(n^2-x+1))-1\over cot(arccot(x))+cot(arccot(n^2-x+1))}=n-1 $
${x(n^2-x+1)-1\over x+n^2-x+1 }=n-1$
$ -x^2+(n^2+1)x=(n^2+1)(n-1)+1$
$A:=n^2+1$
$x^2-Ax+(A(n-1)+1)$
$x= {A\pm\sqrt{A^2-4(A(n+1)+1)}\over2}$=$ {(n^2+1)\pm\sqrt{(n^2+1)^2-4((n^2+1)(n+1)+1)}\over2}$
$x_{1}=n$
$x_{2}=n^2-n+1$
| {
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Finding out the number of solutions If $a,b,c$ are three distinct real numbers for which there exists $x,y,z$ such that $$x+y+z=1$$$$ax+by+cz=t$$$$a^2x+b^2y+c^2z=t^2$$then $a^3x+b^3y+c^3z-t^3=P(t)$,be a polynomial in $t$,then the question is to find out the number of solutions to P(t)=0 .
My attempt at the solution
We are given three equations.From there we can solve for $x,y,z$ using Cramer's rule.by solving I got $$x=\frac{(t-b)(b-c)(c-t)}{(a-b)(b-c)(c-a)}$$ $$y=\frac{(a-t)(c-a)(t-c)}{(a-b)(b-c)(c-a)}$$ $$z=\frac{(a-b)(b-t)(t-a)}{(a-b)(b-c)(c-a)}$$ however plugging it in the equation I got $$P(t)=a^3( t-b)(b-c)(c-t)+b^3 (a-t)(c-a)(t-c)+c^3 (a-b)(b-t)(t-a)-t^3(a-b)(b-c)(c-a)=0$$ from here I couldn't proceed.please help me in this regard.thanks.
| you are given $a^{3}x+b^{3}y+c^{3}z−t^{3}=P(t)$.Equate $P(t)=0$.Then multiply $(a+b+c)$ on both sides and then simplify to get
$t^{3}-(a+b+c)t^{2}+(ab+bc+ca)t-abc=0 $ which shows that the roots are $a,b,c$.
| {
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Number of positive integers $n$ The question is to find out the number of positive integers $n$ such that $nx^4+4x+3 \leq 0 $ for some real $x$ (without using a graphic calculator).
My attempt at the solution:
We have $nx^4+4x+3 \leq 0 $ which is $nx^4 \leq -3-4x$ or $n \leq -\frac{(3+4x)}{x^4}$. It can easily be observed that for $x=-1$ the condition of $n$ being a positive integer is satisfied. But I could not get others. Moreover to satisfy that $n$ is positive we must have $x<-3/4$.
Please help me in this regard. Thanks.
| Differentiate to get $4 n x^{3} + 4$. This has the single root $x_{0} = - \dfrac{1}{\sqrt[3]{n}}$ which is a minimum for $f(x) = nx^4+4x+3$. We want $f(x_{0}) \le 0$, or
$$
0 \ge n x_{0}^{4} + 4 x_{0} + 3 = n (-\dfrac{1}{n}) x_{0} + 4 x_{0} + 3 = 3 (x_{0} + 1).
$$
Thus we must have
$$
- \dfrac{1}{\sqrt[3]{n}}= x_{0} \le -1,
$$
or $n \le 1$.
| {
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prove $\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\le\frac{1}{64abcd}$ Let $a,b,c,d$ be positive real no such that $a+b+c+d\le1$
prove that$$\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\le \frac{1}{64abcd}$$
I did this way$$\frac{a^2cd+b^2cd+c^2ab+d^2ab}{abcd}\le\frac{1}{64abcd}$$
$\implies$ $a^2cd+b^2cd+c^2ab+d^2ab\le\frac{1}{64}$
$\implies$ $(ac+bd)(ad+bc)\le\frac{1}{64}$
now I don't know how to proceed.
| $\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}=\frac{a^2cd+b^2cd+c^2ab+d^2ab}{abcd}$
$\implies(ac+bd)(ad+bc)\le\frac{1}{64}$
Now $(ac+bd)(ad+bc)\le\frac{ac+bd+ad+bc}{4}$
$=\frac{(a+b)(c+d)}{4}\le\frac{(a+b+c+d)}{64}$
As $a+b+c+d\le1$
$\implies\frac{(a+b+c+d)^4}{64}\le\frac{1}{64}$
Hence proved
| {
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"timestamp": "2023-03-29T00:00:00",
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Natural number which can be expressed as sum of two perfect squares in two different ways? Ramanujan's number is $1729$ which is the least natural number which can be expressed as the sum of two perfect cubes in two different ways. But can we find a number which can be expressed as the sum of two perfect squares in two different ways. One example I got is $50$ which is $49+1$ and $25+25$. But here second pair contains same numbers. Does any one have other examples ?
| The Brahmagupta–Fibonacci identity says that every product of two sums of two squares is a sum of two squares in two different ways:
\begin{align}
(a^2+b^2)(c^2+d^2) & = (ac+bd)^2 + (ad-bc)^2 \\[6pt]
& = (ac-bd)^2 + (ad+bc)^2
\end{align}
For example:
\begin{align}
(2^2+3^2)(1^2+7^2) & = 23^2 + 11^2 \\[6pt]
& = 19^2 + 17^2
\end{align}
| {
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Prove by induction that $n^3-n$ is divisible by $24$ for all odd positive integers Q: Prove by induction that $n^3-n$ is divisible by 24 for all odd positive integers
After proving the first part for n=1
Assume true for some positive integer $n=k$
ie $k^3-k=24x$ where x is an integer
Prove true for $n=k+2$
ie $(k+2)^3-(k+2)=24y$ where y is an integer
$=k^3+6k^2+11k+6$
$=24x+12k+6k^2+6$
But how do I get this in the form $24y$? Am i supposed to use $2k+1$ instead?
| We can see that when $n = 1 -----> n (n^2 − 1) = 1 (1 − 1) = 1 (0) = 0$
$n = 3 -----> n (n^2 − 1) = 3 (9 − 1) = 3 (8) = 24$
We can see that this is true for $n = 1$ and $3$.
Assume statement is true for $n = 2k−1$. Then we must show it is true for $n = 2k+1$.
Then for $n = 2k-1 -----> (2k−1) ((2k−1)^2 − 1) = 24m$, for some integer m, giving us,
$(2k−1) ((2k−1)^2- 1) = 24m$
$\implies (2k−1) ((2k−1−1)(2k−1+1)) = 24m$
$\implies (2k−1) (2k−2) (2k) = 24m$. Thus,
$8k^3 − 12k^2 + 4k = 24m$.
When $n = 2k+1:$
\begin{align}
n (n^2 + 1) = (2k+1) ((2k+1)^2 − 1)
\\& = (2k+1) ((2k+1−1)(2k+1+1))
\\& = (2k+1) (2k) (2k+2)
\\& = 8k^3 + 12k^2 + 4k
\\& = (8k^3 − 12k^2 + 4k) + 24k^2
\\& = 24m + 24k^2
\\& = 24 (m + k^2) \text{divisible by 24}
\end{align}
Statement true for $n = 2k−1 \Rightarrow$ statement true for $n = 2k+1$.
Therefore, by principles of mathematical induction, statement is true for all odd positive integers $n$. Hope it helps.
| {
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Proving $\sqrt{x+y-x^2}+\sqrt{y+z-y^2}+\sqrt{z+w-z^2}+\sqrt{w+x-w^2}\le4\sqrt2-3$ I found this inequality using unusual calculations in maths Olympics
and I wonder if some clever teenager could prove it using their elementary knowledge of mathematics.
Let $x,y,z,w$ be non-negative numbers such that $$x+y+z+w=1$$ Prove that
$$\sqrt{x+y-x^2}+\sqrt{y+z-y^2}+\sqrt{z+w-z^2}+\sqrt{w+x-w^2}\le4\sqrt2-3$$
| By applying the generalized mean inequality twice,
$$\left(\frac{\sqrt{x + y - x^2} + \sqrt{y + z - y^2} + \sqrt{z + w - z^2} + \sqrt{w + x - w^2}}{4}\right)^2\\
\le \frac{1}{4}(2x + 2y + 2z + 2w - x^2 - y^2 - z^2 - w^2)\\
= \frac{1}{2} - \frac{x^2 + y^2 + z^2 + w^2}{4}\\
\le \frac{1}{2} - \left(\frac{x + y + z + w}{4}\right)^2\\
=\frac{1}{2} - \frac{1}{16}\\
= \frac{7}{16}$$
so that
$$\sqrt{x + y - x^2} + \sqrt{y + z - y^2} + \sqrt{z + w - z^2} + \sqrt{w + x - w^2}\le\sqrt7 $$
with equality at $$x = y = z = w = \frac{1}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Factoring two complicated polynomials I have two expressions which I've derived by taking the derivative. Both of these expressions are from separate problems, and in both of these instances I'm having difficulty understanding the algebra that brings the expression to a factorized form. They are below:
Problem #1
Given:
$$f(x)=5(2x+1)(2x-3)^4(x^2+x+1)^4+8(2x-3)^3(x^2+x+1)^5$$
Factor/Simplify to:
$$f(x)=(2x-3)^3(x^2+x+1)^4(28x^2-12x-7)$$
Problem #2
Given:
$$h(t) = (t+1)^{2/3}(3)(2t-1)^2(4t)+\frac{2}{3}(t+1)^{-1/3}(2t-1)^3$$
Factor/Simplify to:
$$\frac{2}{3}(t+1)^{-1/3}(2t-1)^2(18t^2+20t-1)$$
If you could please help explain how to get from what I've given to the factorized form of the answer, I would really appreciate it. I know part of the process is to identify the greatest common factor (if it exists), group the like terms, and multiply by their factor. But I don't have anything close enough to the answer worthy of putting into mathjax for either of these problems. Thanks for your help.
| Problem 1
$f(x)=5(2x+1)(2x-3)^4(x^2+x+1)^4+8(2x-3)^3(x^2+x+1)^5$
As in above equation we have $(2x-3)^3$ and $(x^2+x+1)^4$ common in both terms so take them out. And solve the remaining.
$f(x)=(2x-3)^3(x^2+x+1)^4\left[(5(2x+1)(2x-3)+8(x^2+x+1)\right]$
Problem 2
$h(t) = (t+1)^{2/3}(3)(2t-1)^2(4t)+\frac{2}{3}(t+1)^{-1/3}(2t-1)^3$
As in above equation we have $(t+1)^{-1}{3}$, 2 and $(2t-1)^3$ common in both terms so take them out. And solve the remaining.
$2(t+1)^{-1/3}(2t-1)^2\left[6t(t+1)+\frac{1}{3}(2t-1)\right]$
| {
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How to express a power series in closed form I am trying to express the power series $x + x^4/4! + x^7/7! + \cdots$ in closed form;
I have already worked out the power series $1 + x^3/3! + x^6/6! + \cdots$ to be $(e^x + e^{x(2\pi i/3)} + e^{x(2\pi i/3)^2})/3$ using the roots of unity and the series for $e^x$ to get there, and I think I need to somehow "cycle" through to get to my desired closed form but I am not sure how to go about doing this.
Thanks!
| Apologies to the two earlier answers, I made a typo - as they noticed - in my original question; the question now reflects what the closed form for $1 + x^3/3! + x^6/6! + \cdots$ should be. From that we can either integrate or differentiate twice to get my desired closed form for $x + x^4/4! + x^7/7! + \cdots$ as follows:
By integration:
Notice that the infinite sequence $x + x^4/4! + x^7/7! + \cdots = \int^x_0 1 + x^3/3! + x^6/6! + \cdots dx$
For efficiency, let $\omega=2\pi i/3$, then,
$$\int^x_0 1 + x^3/3! + x^6/6! + \cdots dx = \int^x_0 \frac 1 3 (e^x+e^{\omega x} + e^{\omega^2 x}) \, dx$$
$$=[e^x/3 +e^{\omega x}/3\omega +e^{\omega^2 x}/3\omega^2]^x_0$$
$$=e^x/3 +e^{\omega x}/3\omega +e^{\omega^2 x}/3\omega^2-1/3-1/3\omega-1/3\omega^2$$
$$=\omega^2e^x/3\omega^2+\omega e^{\omega x}/3\omega^2+e^{\omega^2 x}/3\omega^2-(\omega^2+\omega+1(1/3))$$
$$=\omega^2e^x/3\omega^2+\omega e^{\omega x}/3\omega^2+e^{\omega^2 x}/3 \omega^2$$
By differentiating twice:
Notice that the infinite sequence $$x + x^4/4! + x^7/7! + \cdots = d/dx(d/dx( 1 + x^3/3! + x^6/6! + \cdots ))$$
For efficiency, let $\omega=2\pi i/3$, then,
$$\frac d {dx} \left(\frac d {dx}( 1 + x^3/3! + x^6/6! + \cdots )\right) = \frac d {dx}\left(\frac d {dx} \left(\frac 1 3(e^x+e^{\omega x} + e^{\omega^2 x}) \right) \right)$$
$$= \frac d {dx}(e^x/3+\omega e^{\omega x}/3+\omega^2 e^{\omega^2 x}/3) = e^x/3+\omega^2e^{\omega x}/3 +\omega e^{\omega^2 x}/3$$
The answers from both are of course equivalent.
| {
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Limits - a different approach $\lim_{x \to\infty }(\frac{x^3+4x^2+3x+5}{x^2+2x+3}+lx+m)=10$.
How do I calculate the value of l and m?
My try: I know questions having limit tending to infinity can be solved by dividing the numerator and denominator by greatest power of $x$.But it got me nowhere in this question. Any help appreciated.
| Indeed, we have
$$\lim_{x\to\infty }\left(\frac{x^3+4x^2+3x+5}{x^2+2x+3}-[-lx+10-m]\right)=0$$.
$$-l=\lim_{x\to\infty }\frac{\frac{x^3+4x^2+3x+5}{x^2+2x+3}}{x}=1$$
thus
$l=-1$
and
$$10-m=\lim_{x\to\infty }\left(\frac{x^3+4x^2+3x+5}{x^2+2x+3}-x\right)=2$$
therefore $m=8$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that:$\sum_{n=0}^{\infty}{2^{n+3}(n^2+n+\phi)\over (n+1)(2n+1)(2n+3){2n\choose n}}=\phi\pi^2+8\phi^2\pi-8\phi^3\sqrt{5}$ $\phi$ is the golden ratio
$$\sum_{n=0}^{\infty}{2^{n+3}(n^2+n+\phi)\over (n+1)(2n+1)(2n+3){2n\choose n}}=\phi\pi^2+8\phi^2\pi-8\phi^3\sqrt{5}$$
I try:
$$S=\sum_{n=0}^{\infty}{2^{n+3}\over {2n\choose n}}\left[{A\over n+1}+{B\over 2n+1}+{C\over 2n+3}\right]$$
$n^2+n+\phi=A(2n+1)(2n+3)+B(n+1)(2n+3)+C(n+1)(2n+1)$
$A=-\phi$, $B=\phi-0.25$ and $C=\phi+0.75$
$$S=\sum_{n=0}^{\infty}{2^{n+3}\over {2n\choose n}}\left[{-\phi\over n+1}+{\phi-0.25\over 2n+1}+{\phi+0.75\over 2n+3}\right]$$
If we know the closed form
$$\sum_{n=0}^{\infty}{2^n\over {2n\choose n}(an+b)}=F(a,b)$$
Then it is easy to prove the above series.
I manage to find
$$\sum_{n=0}^{\infty}{2^n\over {2n\choose n}(2n+1)}={\pi\over 2}$$
Any help? Thank you.
| Too long for a comment.
$$F(a,b)=\sum_{n=0}^{\infty}{2^n\over {2n\choose n}(an+b)}=\frac 1b \,\, _3F_2\left(1,1,\frac{b}{a};\frac{1}{2},\frac{b}{a}+1;\frac{1}{2}\right)$$ where appears the generalized hypergeometric function.
$$F(2,1)=\frac \pi 2$$ $$F(2,3)=\frac{3 \pi }{2}-4$$ $$F(1,1)=\pi -\frac{\pi ^2}{8}$$
| {
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How does this separation of the variable from the fraction result in $3\cdot \frac{1}{x}$ & not $\frac{3}{x}$ I wish to write the following expression as the product of a whole number or a fraction and variable expression. The answer given in my textbook is as follows:
$\frac{3}{x} = \frac{3\cdot 1}{1\cdot x} = \frac{3}{1} \cdot \frac{1}{x} = 3\cdot \frac{1}{x} $
However, I am not sure how to get to $3\cdot \frac{1}{x} $. When I complete the multiplication I get to $\frac{3}{1} \cdot \frac{1}{x} = \frac{3}{x}$, which of course loops me back round!
| The key is simply to notice that
$$\frac31=3$$
Thus,
$$\frac31\cdot\frac1x=3\cdot\frac1x$$
| {
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Prove that if a prime $p$ divides $5^n-2$ and $2^n-5$, then $p = 3$ The result below has been disproven.
Let $n \in \mathbb{N}$. Prove that if a prime $p$ divides $5^n-2$ and $2^n-5$, then $p = 3$.
We know that $p \neq 2,5$. We need to have \begin{align*}5^n &\equiv 2 \pmod{p}\\2^n &\equiv 5 \pmod{p}.\end{align*} This gives us $10^{n-1} \equiv 1 \pmod{p}$. Thus $\text{ord}_{p}(10) \mid (n-1)$. How do we continue?
| Proof that $1409 \mid \gcd(5^{65}-2,2^{65}-5$):
We have \begin{align*}5^5 &\equiv 307 \pmod{1409}\\5^{10} &\equiv 1255 \pmod{1409}\\5^{20} &\equiv 1172 \pmod{1409}\\5^{40} &\equiv 1218 \pmod{1409}\\5^{65} &\equiv 2 \pmod{1409}\end{align*} and \begin{align*}2^{12} &\equiv 1278 \pmod{1409}\\2^{24} &\equiv 253 \pmod{1409}\\2^{48} &\equiv 604 \pmod{1409}\\2^{60} &\equiv 1189 \pmod{1409}\\2^{65} &\equiv 5 \pmod{1409}.\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
If $x = y$, $p$ = what? It is given that,
$$
\frac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p
$$
If $x=y$, $p$ =?
| $
\frac{\sqrt{2x+3y}+\sqrt{2x-3y}}{\sqrt{2x+3y}-\sqrt{2x-3y}}=p
$
Apply componendo and dividendo,
$
\frac{\sqrt{2x+3y}+\sqrt{2x-3y}+\sqrt{2x+3y}-\sqrt{2x-3y}}{\sqrt{2x+3y}+\sqrt{2x-3y}- \sqrt{2x+3y}+\sqrt{2x-3y}}=\frac{p+1}{p-1}
$
$
\frac{\sqrt{2x+3y}}{\sqrt{2x-3y}} = \frac{p+1}{p-1}
$
Put x=y
$
\frac{\sqrt{5y}}{\sqrt{-y}} = \frac{p+1}{p-1}
$
Squaring both sides,
$5.(p-1)^2=-1(p+1)^2$
Sorry I have little mistake here,
$5p^2-10p+5 = -p^2-2p-2$
$5p^2-10p+5 = -p^2-2p-1$
$6p^2-8p+6=0$
$3p^2-4p+3=0$
$D = b^2 - 4ac$
= $(-4)^2 - 4.3.3$ = 16 - 36 = - 20
$ p = \frac{-b \pm \sqrt D}{2a}$
= $\frac{-(-4) \pm \sqrt{-20}}{2.3}$
= $\frac{4 \pm \sqrt{20i^2}}{6}$
= $\frac{4 \pm 2i\sqrt5}{6}$
p = $\frac{2 + i\sqrt5}{3}, \frac{2 - i\sqrt5}{3}$
Edit-
In above formula D is discriminat. It is used when we can't factorise equation using factorisation method.
Here's the link when you can't find root using factorisation use discrimat method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $x = 2\log_39 + \log_{27}5,$ then $3^x = ??$ If
$$x = 2\log_39 + \log_{27}5,$$
then $3^x = ??$
I know $2\log_39$ is $4$, however how do I find an exact answer to the whole equation??
I tried to change the base of the second one to $ 3 $, however it did not work out.
| Use the fact that $\log_{27}(5) = \frac{\log_3(5)}{\log_3(27)} = \frac{\log_3(5)}{3} = \log_3(5^{\frac{1}{3}})$.
Then $x = 2\log_3(9) + \log_{27}(5) = \log_3(81) + \log_3(5^\frac{1}{3}) = \log_3(81 \cdot 5^{\frac{1}{3}})$.
You obtain that $3^x = 81 \cdot 5^\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Solve equation in determinant Let $ a,b,c,m,n,p\in \mathbb{R}^{*} $, $ a+m+n=p+b+c $. Solve the equation:
$$
\begin{vmatrix}
x & a & b &c \\
a & x & b &c \\
m &n & x &p \\
m& n& p& x
\end{vmatrix}
=0
$$
I had used the Schur complement ($\det(M)=\det(A)\cdot (D-C\cdot A^{-1}\cdot B)$, for $ M= \begin{bmatrix}
A &B \\
C & D
\end{bmatrix}) $ but it didn't help me.
| $$ \begin{vmatrix}
x & a & b &c \\
a & x & b &c \\
m &n & x &p \\
m& n& p& x
\end{vmatrix}=0 $$
Subtract the first row to the second row, subtract the third row from the fourth row:
$$ \begin{vmatrix}
x & a & b &c \\
a-x & x-a & 0 &0 \\
m &n & x &p \\
0& 0& p-x& x-p
\end{vmatrix}=0 $$
Factorize $(a-x)$ and $(x-p)$ out:
$$ (a-x)(x-p)\begin{vmatrix}
x & a & b &c \\
1 & -1 & 0 &0 \\
m &n & x &p \\
0& 0& -1& 1
\end{vmatrix}=0 $$
Add the first column to the second column:
$$ (a-x)(x-p)\begin{vmatrix}
x & a+x & b &c \\
1 & 0 & 0 &0 \\
m &n+m & x &p \\
0& 0& -1& 1
\end{vmatrix}=0 $$
Compute the determinant by using the second row:
$$ (a-x)(x-p)\begin{vmatrix}
a+x & b &c \\
n+m & x &p \\
0& -1& 1
\end{vmatrix}=0 $$
Add the third column to the second column:
$$ (a-x)(x-p)\begin{vmatrix}
a+x & b+c &c \\
n+m & p+x &p \\
0& 0& 1
\end{vmatrix}=0 $$
Expand the determinant by the last row:
$$ (a-x)(x-p)\begin{vmatrix}
a+x & b+c \\
n+m & p+x \\
\end{vmatrix}=0 $$
Adding the first row to second row:
$$ (a-x)(x-p)\begin{vmatrix}
a+x & b+c \\
x+a+n+m & b+c+p+x \\
\end{vmatrix}=0 $$
Factorize $(x+a+n+m)$ out since $a+m+n=b+c+p$:
$$ (a-x)(x-p)(x+a+n+m)\begin{vmatrix}
a+x & b+c \\
1 & 1 \\
\end{vmatrix}=0 $$
$$(a-x)(x-p)(x+a+n+m)(x+a-b-c)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
If $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle $ABC$ is equilateral In a triangle $ABC,$ if $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle
$ABC$ is equilateral
$\displaystyle \frac{2(bc^2+ca^2+ab^2)}{abc} = \frac{b^2c+c^2a+a^2b+3abc}{abc}$
$\displaystyle 2\left(\frac{c}{a}+\frac{a}{b}+\frac{b}{c}\right) = \left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)+3$
let $\displaystyle \frac{a}{b} = x,\frac{b}{c}=y,\frac{c}{a}=z$
so $\displaystyle 2(x+y+z) = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3$
wan,t be able to proceed after that, could some help me
| $x+y+z\geq 3(xyz)^{\frac{1}{3}}$,when $x=y=z$ ,equal sign is established.So:$$2\sum\frac{c}{a}\geq\sum\frac{c}{a}+3$$ i.e.$$\frac{c}{a}+\frac{b}{c}+\frac{a}{b}\leq\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$$Without loss of generality, we assume:$c\geq b\geq a$,so $\frac1{a}\geq\frac1{b}\geq\frac1{c},$then we have:$$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\leq\frac{c}{a}+\frac{b}{c}+\frac{a}{b},$$so $a=b=c$.I'm using an incomplete sort to derive a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Maximum value of $f(x)=2(a-x)(x+\sqrt{x^2+b^2})$ If $a,b,x$ are real and $$f(x)=2(a-x)(x+\sqrt{x^2+b^2}),$$ then find the maximum value of $f(x)$.
Is there any method to solve this question without differentiation because using differentiation I am getting an ugly expression.
| This is still differentiation, but less ugly:
$$f(x)=2(a-x)(x+\sqrt{x^2+b^2})\implies\log f = \log 2+\log(a-x)+\log(x+\sqrt{x^2+b^2})$$
$$\implies\frac{f'(x)}{f(x)}=\frac{-1}{a-x}+\frac{1}{\sqrt{x^2+b^2}}$$
$$f'(x_*)=0\implies a-x_*=\sqrt{x_*^2+b^2}\implies(a-x_*)^2=x_*^2+b^2$$
$$\implies x_*=\frac{a^2-b^2}{2a}$$
So, $f(x)\le f(x_*)$
$$=2(a-x_*)(x_*+\sqrt{x_*^2+b^2})=2(a-x_*)(x_*+a-x_*)$$
$$=2a(a-x_*)=2a^2-2ax_*=a^2+b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Deriving $\cosh^{-1}{x}=\ln\left(x+\sqrt{x^2-1}\right)$ Let $y=\cosh^{-1}{x}$. Then, $x=\cosh{y}=\frac{1}{2}\left(e^y+e^{-y}\right)$. Multiplying by $2e^y$, we get $2xe^y=e^{2y}+1$. Solving $e^{2y}-2xe^y+1=0$ by the quadratic formula, we have $e^y=\frac{2x\pm\sqrt{4x^2-4}}{2}=x\pm\sqrt{x^2-1}$. We find that both roots are possible. Thus, $y=\ln\left(x\pm\sqrt{x^2-1}\right)$. Why did we fail to eliminate the minus?
| The reason why we can dispose of the minus sign is that, firstly, $$(x+\sqrt{x^2-1})^{-1}=(x-\sqrt{x^2-1})$$
Therefore $$\ln(x\pm \sqrt{x^2-1})=\ln(x+\sqrt{x^2-1})^{\pm1}=\pm\ln(x+\sqrt{x^2-1})$$
But $\operatorname{arcosh}(x)$ is defined as non-negative, so the minus sign is not required
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Prove that:$\int_{0}^{1}{x+x^2+\cdots+x^{2n}-2nx\over (1+x)\ln{x}}dx=\ln{\left[\left({2\over \pi}\right)^n(2n)!!\right]}$ Prove that
$$\int_{0}^{1}{x+x^2+x^3+\cdots+x^{2n}-2nx\over (1+x)\ln{x}}dx=\ln{\left[\left({2\over \pi}\right)^n(2n)!!\right]}$$
$n\ge1$
My try:
$${x+x^2+\cdots+x^{2n}\over 1+x}={1\over 1+x}\cdot{x(1-x^{2n})\over 1-x}=x\cdot{1-x^{2n}\over 1-x^2}$$
$${x+x^2+\cdots+x^{2n}-2nx\over 1+x}=x\cdot{1-x^{2n}\over 1-x^2}-{2nx\over 1+x}$$
$$\int_{0}^{1}\left(x\cdot{1-x^{2n}\over 1-x^2}-{2nx\over 1+x}\right)\cdot{1\over \ln{x}}dx$$
I am stuck not quite sure what to do next, please help
Edit
$$\int_{0}^{1}\left(x\cdot{1-x^{2n}\over 1-x^2}-{2nx\over 1+x}\right)\cdot{1\over \ln{x}}dx$$
| In THIS ANSWER, I showed that
$$\bbox[5px,border:2px solid #C0A000]{\int_0^1 \left(\frac{x^{2n}-x}{1+x}\right)\,\frac{1}{\log(x)}\,dx=\log\left(\frac{2}{\pi}\frac{(2n)!!}{(2n-1)!!}\right)}$$
We can use this result to evaluate the integral
$$\begin{align}
\int_0^1 \left(\frac{x^{2n+1}-x}{1+x}\right)\,\frac{1}{\log(x)}\,dx&=\int_0^1 \left(\frac{x^{2n+1}-x^2+(x^2-x)}{1+x}\right)\,\frac{1}{\log(x)}\,dx\\\\
&=\int_0^1 \left(\frac{x^{2n+1}-x^2}{1+x}\right)\,\frac{1}{\log(x)}\,dx+\log(4/\pi)\\\\
&=\int_0^1 \frac{x^{2n}-x}{\log(x)}\,dx+\log(4/\pi)-\log\left(\frac{2}{\pi}\frac{(2n)!!}{(2n-1)!!}\right)\tag1
\end{align}$$
The integral on the right-hand side of $(1)$ can be evaluated by enforcing the substitution $x\to e^{-x}$ and evaluating the resulting FRULLANI INTEGRAL. Proceeding, we have
$$\int_0^1 \frac{x^{2n}-x}{\log(x)}\,dx=\log\left(\frac{2n+1}{2}\right) \tag 2$$
Substituting $(2)$ into $(1)$ reveals
$$\begin{align}
\int_0^1 \left(\frac{x^{2n+1}-x}{1+x}\right)\,\frac{1}{\log(x)}\,dx&=\log\left(\frac{2n+1}{2}\right)+\log(4/\pi)-\log\left(\frac{2}{\pi}\frac{(2n)!!}{(2n-1)!!}\right)\\\\
&= \log\left(\frac{2}{\pi}(2n+1)\right)-\log\left(\frac{2}{\pi}\frac{(2n)!!}{(2n-1)!!}\right) \tag 3\\\\
&=\log\left(\frac{(2n+1)!!}{(2n)!!}\right)
\end{align}$$
Finally, we see that
$$\begin{align}
\int_0^1 \left(\frac{x+x^2+\cdots +x^{2n}-2nx}{1+x}\right)\,\frac{1}{\log(x)}\,dx&=\sum_{k=1}^n \int_0^1 \frac{x^{2k}-x}{1+x}\frac{1}{\log(x)}\,dx\\\\
&+\sum_{k=1}^{n-1}\int_0^1 \frac{x^{2k+1}-x}{1+x}\frac{1}{\log(x)}\,dx\\\\
&=\sum_{k=1}^n\log\left(\frac{2}{\pi}\frac{(2k)!!}{(2k-1)!!}\right)\\\\
&+\sum_{k=1}^{n-1}\log\left(\frac{2}{\pi}(2n+1)\right)\\\\
&-\sum_{k=1}^n\log\left(\frac{2}{\pi}\frac{(2k)!!}{(2k-1)!!}\right)\\\\
&=\log\left(\left(\frac{2}{\pi}\right)^n (2n)!!\right)
\end{align}$$
as was to be shown!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\ln(x+\sqrt{x^2 + 4}) - \ln2$ is odd I know that a function is odd when
$$f(-x) = -f(x)$$
Therefore I can say that if for a function $$-f(x) + f(x) = f(-x) + f(x) = 0$$
Then the function is odd!
I tried to use this trick to prove that $f(x) = \ln\left(x+\sqrt{x^2 + 4}\right) - \ln2$ is odd.
However, I would want to prove directly that $$f(-x) = -f(x)$$
In other words, I want to solve $$\ln\left(-x+\sqrt{(-x)^2 + 4}\right) - \ln2$$
and to come at the end to this:
$$-\ln\left(x+\sqrt{x^2 + 4}\right) + \ln2$$
This was my approach:
$$\ln\left(-x+\sqrt{(-x)^2 + 4}\right) - \ln2$$
$$\ln\left(-x+\sqrt{x^2 + 4}\right) - \ln2$$
$$\ln\left(\frac{-x+\sqrt{x^2 + 4}}{2}\right)$$
$$\ln\left(\left(\frac{2}{-x+\sqrt{x^2 + 4}}\right)^{-1}\right)$$
$$-\ln\left(\frac{2}{-x+\sqrt{x^2 + 4}}\right)$$
Here I got stuck. I want to get to $-\ln\left(x+\sqrt{x^2 + 4}\right) + \ln2$ but if I use $\ln\left(\frac ab\right) = \ln a - \ln b$ then I will get back to $f(-x)$ and not to $-f(x)$.
Any help?
| It simply means
\begin{align}
&\ln(-x+\sqrt{x^2+4})-\ln2=-\ln(x+\sqrt{x^2+4})+\ln2\\
\iff &\ln(x+\sqrt{x^2+4})+\ln(-x+\sqrt{x^2+4})=2\ln 2\\
\iff&\ln[(\sqrt{x^2+4}+x)(\sqrt{x^2+4}-x)]=\ln4\\
\iff &(\sqrt{x^2+4}+x)(\sqrt{x^2+4}-x)=4.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
$x-\frac1x+y-\frac1y=4\,$ has no rational solutions? Do there exist rational numbers $x,y$ such that $x-\frac1x+y-\frac1y=4?$
I think no, because the equation reduces to $\frac{x^2-1}{x}+\frac{y^2-1}{y}=4\implies x^2y+x(y^2-4y-1)-y=0$ in two variables is irreducible in the field $\mathbb{Q}$(Eisenstein) Is my reasoning right? Thanks beforehand.
| Here is a sketch of a naive proof that it has no solutions (no elliptic curves).
We can assume x is positive, otherwise we would change x to -1/x. The same with y.
$x^2y+x(y^2-4y-1) -y =0$
$D=(y^2-4y-1)^2+4y^2$ should be a square of a rational number.
If $y=\frac{m}{n}$ (assuming m and n are co-prime), then multiplying by $n^4$ we get
$(m^2-4mn-n^2)^2 + 4m^2n^2 = v^2$, where $v$ is natural.
*
*If either $m$ or $n$ is even, than $2mn$ and $m^2-4mn-n^2$ are co-prime.
Hence
$2mn=2xy$
and $m^2-4mn-n^2 = x^2-y^2$ for some natural $x$ and $y$.
Or the last one is $m^2-2mn +n^2-2n^2 = x^2+2xy + y^2-2y^2$ since $mn=xy$.
$(m-n)^2-2n^2=(x+y)^2-2y^2$, or $(m-n)^2 - (x+y)^2= 2 (n^2-y^2)$
$(m-n-x-y)(m-n+x+y)=2(n-y)(n+y)$.
Using the fact that numbers $m-n-x-y$ and $m-n+x+y$ are both odd or even and the same with numbers $n-y$ and $n+y$ we obtain that all of those numbers are even.
Making similar step infinitely many times we obtain that both sides are divisible by infinite number of powers of 2, hence both sides are 0 which is impossible.
UPD: this part is wrong. I will try to change it later. Sorry.:(
*If both $m$ and $n$ are odd, then $mn=x^2-y^2$ and $m^2-4mn-n^2 = 8xy$ for some natural $x$ and $y$.
Then $5m^2-(2m+n)^2$ is divisible by 8 for some odd $m$ and $n$, which is impossible since odd square always has remainder 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Integral $\int \frac{dx}{\sin^2 (x) + \sin(2x)}$ We have to do the following integration
$$\int \frac{dx}{\sin^2 (x) + \sin(2x)}$$
I tried as
In the numerator, $1 = \sin^2(x)+\sin(2x)+\cos^2(x)-\sin(2x)$
But now how to proceed?
| First use the fact that $\sin(2x) = 2\sin(x)\cos(x)$ to obtain
$$\int \frac{1}{\sin^2(x) + 2\cos(x)\sin(x)}dx$$
Divide numerator and denominator by $\sin^2(x)$:
$$\int \frac{\frac{1}{\sin^2(x)}}{1 + 2\frac{\cos(x)}{\sin(x)}} = \int \frac{\csc^2(x)}{1 + 2\cot(x)}dx$$
Then you can use the substitution $u = 2\cot(x)+1$; $du = -2\csc(x)dx$:
$$-\frac{1}{2}\int\frac{1}{u}du = -\frac{\ln(u)}{2}+c$$
And substitute $u$ for $2\cot(x)+1$ again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Find the sum of the reciprocals
Let $A$ be the sum of the reciprocals of the positive integers that can be formed by only using the digits $0,1,2,3$. That is, $$A = \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+\dfrac{1}{30}+\cdots.$$ What is $\lfloor A\rfloor$?
The sum is greater than $2$ since $\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{11} > 2$. Can we show it is less than $3$?
| There are $3$ single digit inverses, for a total of $\frac{11}{6}$.
There are $3\cdot4=12$ two digits inverses, which don't exceed a total of $\frac{11}{6}\cdot\frac4{10}$, (as $\frac1{10}+\frac1{11}+\frac1{12}+\frac1{13}\le\frac4{10}, \frac1{20}+\frac1{21}+\frac1{22}+\frac1{23}\le\frac4{20}, \cdots$).
There are $3\cdot4^2=48$ three digits inverses, which don't exceed a total of $\frac{11}{6}\cdot\frac{4^2}{10^2}$.
We have a geometric progression, hence the total sum does not exceed
$$\frac{11}6\frac1{1-\dfrac4{10}}=\frac{55}{18}=3+0.055555\cdots.$$
To tighten the bound, it suffices to correct sufficiently many terms that were over-estimated.
Deducing
$$\frac1{10}-\frac1{11}+\frac1{10}-\frac1{12}+\frac1{10}-\frac1{13}=\frac{419}{8580}=0.048834\cdots$$ we are almost there.
Then with a few more terms
$$\cdots+\frac1{20}-\frac1{21}+\frac1{20}-\frac1{22}=\frac{3349}{60060}=0.055760\cdots$$
we are done.
This establishes the upper bound
$$A<\frac{540503}{180180}<3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How to prove this sum $\sum_{k=1}^{n-1}\frac{1}{k(n-k)}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\binom{2(n-1)}{n-1}$ Let $n\ge 2$ postive integer, show that
$$I=\sum_{k=1}^{n-1}\dfrac{1}{k(n-k)}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\binom{2(n-1)}{n-1}$$
I have done this works
$$I=\sum_{k=1}^{n-1}\dfrac{1}{k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}+\sum_{k=1}^{n-1}\dfrac{1}{n-k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\sum_{k=1}^{n-1}\dfrac{2}{k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}$$
| Note the Catalan numbers are defined as
\begin{align*}
C_{n}&=\frac{1}{n+1}\binom{2n}{n}\qquad\quad n\geq 0\\
\end{align*}
with generating function
\begin{align*}
C(z)=\sum_{n=0}^\infty C_nz^n=\frac{1}{2z}\left(1-\sqrt{1-4z}\right)
\end{align*}
The LHS of OPs binomial identity is a Cauchy-product of Catalan numbers
\begin{align*}
\sum_{k=1}^{n - 1} C_{k-1}C_{n -k-1}\tag{1}
\end{align*}
Since Cauchy products occur when multiplying series we could work with generating functions:
\begin{align*}
C^2(z)=\sum_{n=0}^\infty\left(\sum_{k=0}^{n}C_kC_{n-k}\right)z^n\tag{2}
\end{align*}
Let $[z^n]$ denote the coefficient operator.
We observe with the help of (1) and (2) for $n\geq 2$
\begin{align*}
[z^{n-2}]C^2(z)&=[z^{n-2}]\sum_{n=0}^\infty\left(\sum_{k=0}^{n}C_kC_{n-k}\right)z^n\\
&=\sum_{k=0}^{n-2}C_kC_{n-2-k}\\
&=\sum_{k=1}^{n-1}C_{k-1}C_{n-1-k}\\
\end{align*}
on the other hand we obtain
\begin{align*}
[z^{n-2}]C^2(z)&=[z^{n-2}]\left(\frac{1}{2z}\left(1-\sqrt{1-4z}\right)\right)^2\\
&=[z^{n-2}]\frac{1}{2z^2}\left(1-\sqrt{1-4z}\right)-1\\
&=[z^{n-1}]\frac{1}{2z}\left(1-\sqrt{1-4z}\right)\\
&=[z^{n-1}]C(z)\\
&=C_{n-1}\\
&=\frac{1}{n}\binom{2n-2}{n-1}\\
\end{align*}
and the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Divisibility of $(4^{2^{2n+1}}-3)$ by 13. The question is to simply proof that $13\space|(\space4^{2^{2n+1}}-3)$ for all $n\in\mathbb{N}$. Since the chapter in my book is about the Euler-phi function and Euler's theorem, my guess would be to look at $4^{2^{2n+1}}\equiv3\space \pmod{13}$ and do something with Euler's theorem. Another guess is to solve this by induction, but in both approaches I am very stuck.
| As you correctly note, to compute
$$
4^{2^{2n+1}} \pmod{13}
$$
you can compute
$$
2^{2n+1} \pmod{12}.
$$
It is advisable to use CRT, and compute separately modulo $3$ and modulo $4$.
Modulo $3$ you have
$$
2^{2n+1} \equiv (-1)^{2n+1} \equiv -1 \pmod{3},
$$
whereas modulo $4$ you have
$$
2^{2n+1} \equiv 0 \pmod{4},
$$
for $n \ge 1$.
Therefore
$$
2^{2n+1} \equiv 8 \pmod{12}.
$$
(This usually implies solving the system of congruences
$$
\begin{cases}
x \equiv -1 \pmod{3}\\
x \equiv 0 \pmod{4}\\
\end{cases}
$$
but in this case it is just a matter of checking which of $x = 0, 4, 8$ is a solution.)
Thus
$$
4^{2^{2n+1}} \equiv 4^8 \equiv 3^4 \equiv 3 \pmod{13}
$$
as $4^2 = 16 \equiv 3 \pmod{13}$ and $3^3 = 27 \equiv 1 \pmod{13}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
If, in a triangle, $\cos(A) + \cos(B) + 2\cos(C) = 2$ prove that the sides of the triangle are in AP By using the formula :
$$
\cos(A)+\cos(B)+\cos(C) = 1 + 4 \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)
$$
I've managed to simplify it to :
$$
2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)=\sin\left(\frac{C}{2}\right)$$
But I have no idea how to proceed.
| Inspired by Dr. Sonnhard's answer.
$$cos A + cos B + 2cos C = 2$$
=> $$cos A + cos B = 2 (1 - cos C)$$
=> $$(b²+c²-a²)/2bc + (c²+a²-b²)/2ca = 2[1 -
(a²+b²-c²)/2ab]$$
=> $$a(b²+c²-a²) + b(c²+a²-b²) = 2c[2ab - (a²
+b²-c²)]$$
=>$$ ab² + ac² -a³ + bc² + ba² - b³ = 2c[2ab -
(a²+b²-c²)]$$
=> $$ab² + ba² + ac² + bc² - a³ - b³ = 2c[2ab -
(a²+b²-c²)]$$
=> $$ab(a+b) + c²(a+b) - (a+b)(a²-ab+b²) = 2c
[2ab - (a²+b²-c²)]$$
=>$$ (a+b)(ab+c² - a²+ab-b²) = 2c[2ab - (a²
+b²-c²)]$$
=> $$(a+b)[2ab - (a²+b²-c²)] = 2c[2ab - (a²
+b²-c²)]$$
=>$$ a+b = 2c$$
=> sides of the triangle are in A.P.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Help in understanding a simple proof. Let $Ax + By + C = 0$ be a general equation of a line and $x\cos \alpha + y\sin \alpha - p = 0$ be the normal form of the equation.
Then,
$${-p\over C } = { \cos \alpha\over A} = { \sin\alpha\over B}\tag{1}$$
$${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} \tag{2}$$
$${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} = {\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} = {1\over \sqrt{A^2 + B^2}} \tag{3}$$
$$\therefore \bbox[ #FFFDD0, 10px, Border:2px solid #DC143C]{p = {C\over \sqrt{A^2 + B^2}}, \cos \alpha = {-A\over \sqrt{A^2 + B^2}},\sin\alpha = {-B\over \sqrt{A^2 + B^2}}} $$
I did not get the $(3)$ part. Where does $\displaystyle{\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}}$ come from ?
| Let $\frac{p}{C}=\frac{\cos\alpha}{A}=\frac{\sin\alpha}{B} = k$
Then we have,
$\cos\alpha = kA$
On squaring we have,
$\cos^2\alpha = k^2A^2$
Also $\sin\alpha = kB$
On squaring $\sin^2\alpha = k^2B^2$
Adding these two squared equations,
$\sin^2\alpha + \cos^2\alpha = k^2A^2 + k^2B^2$
$\sin^2\alpha + \cos^2\alpha= k^2(A^2 + B^2)$
$\frac{\sin^2\alpha + \cos^2\alpha}{(A^2 + B^2)} = k^2$
$\frac{\sqrt{\sin^2\alpha + \cos^2\alpha}}{\sqrt{(A^2 + B^2)}} = k$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Alternative approaches to showing that $\gamma=\int_0^\infty \left(\frac{1}{1+x^a}-\frac{1}{e^x}\right)\,\frac1x\,dx$, $a>0$ Starting from the limit definition of the Euler-Mascheroni constant $\gamma$ as given by
$$\gamma=\lim_{n\to \infty}\left(-\log(n)+\sum_{k=1}^n\frac1k\right)\tag 1$$
we can show that $\gamma$ has an integral representation
$$\gamma=\int_0^\infty\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right)\,dx \tag 2$$
Proof of $(2)$: This is provided for completeness only and one can skip this part without losing context.
To show that the integral in $(2)$ is equivalent to $(1)$, we can proceed as follows.
$$\begin{align}
\int_0^\infty\left(\frac{1}{e^x-1}-\frac{1}{xe^x}\right)\,dx&=
\int_0^\infty \frac{e^{-x}}{1-e^{-x}}\left(1-\frac{1-e^{-x}}{x}\right)\,dx\\\\
&=\sum_{k=1}^\infty \int_0^\infty\left(e^{-kx}-\frac{e^{-kx}-e^{-(k+1)x}}{x}\right)\,dx\\\\
&=\sum_{k=1}^\infty \left(\frac{1}{k}-\log\left(\frac{k+1}{k}\right)\right)\\\\
&=\lim_{n\to \infty}\sum_{k=1}^n \left(\frac1k -\log\left(\frac{k+1}{k}\right)\right)\\\\
&=\lim_{n\to \infty}\left(-\log(n+1)+\sum_{k=1}^n\frac1k\right)\\\\
&=\lim_{n\to \infty}\left(-\log(n)+\sum_{k=1}^n\frac1k\right)\\\\
\end{align}$$
Another integral representation for $\gamma$ is given by
$$\gamma=\int_0^\infty \left(\frac{1}{x(1+x^a)}-\frac{1}{xe^x}\right) \,dx \tag 3$$
for $a>0$.
Equipped with $(2)$, we can show the equivalence of $(3)$ with $(1)$ by showing that
$$\int_0^\infty \left(\frac{1}{x(1+x^a)}-\frac{1}{e^x-1}\right)\,dx=0\tag 4$$
To prove $(4)$, I proceeded as follows.
$$\begin{align}
\lim_{\epsilon\to 0^+}\int_{\epsilon}^\infty\left(\frac{1}{x(1+x^a)}-\frac{1}{e^x-1}\right)\,dx &=\lim_{\epsilon\to 0^+}\left.\left(-\frac1a \log(1+x^{-a})-\log(1-e^{-x})\right)\right|_{\epsilon}^{\infty}\\\\
&=\lim_{\epsilon\to 0^+}\left(\frac1a \log(1+\epsilon^{-a})+\log(1-e^{-\epsilon})\right)\\\\
&=0
\end{align}$$
And we are done!
This approach seemed a bit cumbersome and indirect.
QUESTION: So, what are alternative approaches to establishing equivalence of $(3)$ and $(1)$?
| The same idea as in my previous answer applies. Notice that
$$ f(x) = \frac{1}{1+x} \quad \Rightarrow \quad \int_{\epsilon}^{\infty} \frac{f(x)}{x} \, dx = -\log\epsilon + \log(1+\epsilon) = -\log\epsilon + o(1) $$
as $\epsilon \to 0^+$. Now from the linked answer above, we find that
\begin{align*}
f(x) = e^{-x}
&\quad \Rightarrow \quad c(f) = \lim_{R\to\infty} \left( \int_{0}^{R} \frac{ds}{1+s} - \log R \right) - \gamma = -\gamma \\
&\quad \Rightarrow \quad \int_{\epsilon}^{\infty} \frac{e^{-x}}{x} \, dx = -\log\epsilon - \gamma + o(1).
\end{align*}
It is worth to remark that the $\gamma$ term above is computed from the identity $\gamma = -\int_{0}^{\infty} e^{-x}\log x \, dx$, which you are already aware of. From this,
\begin{align*}
\int_{\epsilon}^{\infty} \frac{1}{x^a + 1} \, \frac{dx}{x} - \int_{\epsilon}^{\infty} e^{-x} \, \frac{dx}{x}
&= \frac{1}{a}\int_{\epsilon^a}^{\infty} \frac{1}{x + 1} \, \frac{dx}{x} - \int_{\epsilon}^{\infty} e^{-x} \, \frac{dx}{x}\\
&= \gamma + o(1)
\end{align*}
and taking $\epsilon \to 0^+$ gives the result.
Using the quantity $c(f)$, you can compute various integrals (including all the integrals you have asked) together with some tabulated results for $c(f)$:
\begin{align*}
c\left\{\frac{1}{(1+x)^\alpha}\right\} &= -H_{\alpha-1}, &
c\{e^{-x}\} &= -\gamma, \\
c\left\{\frac{x}{e^x-1}\right\} &= 0, &
c\{\cos x\} &= -\gamma,
\end{align*}
where $H_n$ is the harmonic numbers. For instance, if $a > 0$ then
\begin{align*}
\int_{0}^{\infty} \left( \frac{1}{\sqrt{1+a x}} - e^{-x^2} \right) \frac{dx}{x}
&= c\left\{ \frac{1}{\sqrt{1+a x}} - e^{-x^2} \right\} \\
&= c\left\{ \frac{1}{\sqrt{1+a x}} \right\} - c\{e^{-x^2}\} \\
&= c\left\{ \frac{1}{\sqrt{1+x}} \right\} - \log a - \frac{1}{2}c\{e^{-x}\} \\
&= H_{-1/2} - \log a + \frac{\gamma}{2} \\
&= \frac{\gamma}{2} - \log(4a).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
solving Integration of trignometry In the following integral
$$\int \frac {1}{\sec x+ \mathrm {cosec} x} dx $$
My try: Multiplied and divided by $\cos x$ and Substituting $\sin x =t$. But by this got no result.
| A big partial fraction decomposition!!:
We use Weierstrass substitution and use $u =\tan \frac {x}{2} $ to get $$I =\int \frac {1}{\sec x+\mathrm {cosec}x } dx =\int \frac {\sin x\cos x}{\sin x+\cos x} dx= 4\int \frac {u (u^2-1)}{(u^2+1)^2 (u^2-2u-1)} du =4I_1 $$ Now we have $$I_1 =\int \frac {u (u^2-1)}{(u^2+1)^2 (u-\sqrt {2}-1)(u+\sqrt {2}-1)} = -\frac {1}{4} \int \frac {1}{u^2+1} du +\frac{1}{2} \int \frac{u+1}{(u^2+1)^2} du +\frac {4-3\sqrt {2}}{48-32\sqrt {2}} \int \frac {1}{u+\sqrt {2}-1} du +\frac {4+3\sqrt {2}}{32\sqrt {2}+48} \int \frac {1}{u-\sqrt {2}-1} du =I_{11} +I_{12} +I_{13} +I_{14} $$ Hope you can take it from here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{a_1}{a_2}+ \frac{a_2}{a_3}+\frac{a_3}{a_4}+...+\frac{a_{n-1}}{a_n} \le \frac{n}{2}$ Let $n \ge 2$ be a positive integer and let $a_1, a_2, ... a_n$ be positive numbers such that $$ a_1\le a_2, a_1+a_2\le a_3, a_1+a_2+a_3\le a_4, ... ,a_1+a_2+...+a_{n-1}\le a_n$$
prove that
$$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{n-1}}{a_n} \le \dfrac{n}{2} \hspace{2cm} (1)$$
When does the equality holds?
Solution:
I proceed as follows using Mathematical Induction.
For $n=2, \frac{a_1}{a_2} \le 1$. Let the (1) be true for $n=k$ i.e
$$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k-1}}{a_k} \le \dfrac{k}{2} \hspace{2cm} (2)$$
We need to prove
$$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k}}{a_{k+1}} \le \dfrac{k+1}{2} $$
Consider $$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k-1}}{a_k}+\dfrac{a_{k}}{a_{k+1}} \le \dfrac{k}{2} +\dfrac{a_{k}}{a_{k+1}} \hspace{2cm} (3)$$
Since $$a_1+a_2+...+a_{k-1}+a_{k}\le a_{k+1} \hspace{2cm} (4)$$
also $$a_1+a_2+...+a_{k-1}\le a_{k} \hspace{2cm} (5)$$
Using 5 in 4, we get
$$ a_{k}+a_{k}\le a_{k+1}$$
$$\dfrac{a_{k}}{a_{k+1}} \le \dfrac{1}{2}$$
using in (3), we get
$$\dfrac{a_1}{a_2}+ \dfrac{a_2}{a_3}+\dfrac{a_3}{a_4}+...+\dfrac{a_{k-1}}{a_k}+\dfrac{a_{k}}{a_{k+1}} \le \dfrac{k+1}{2}$$
Is the procedure is correct.
And when the equality holds... Thanks for any assistance
| The idea is "local adjustments". If $k$ is the smallest integer such that $a_1+\cdots+a_{k-1} < a_k$, we adjust up $a_1$, $\cdots$, $a_{k-1}$ with a scale factor of $s=\frac{\sum_{i=1}^{k}a_i}{2\sum_{i=1}^{k-1}a_i}$ for all, and adjust down $a_k$ by a scale factor of $t=\frac{\sum_{i=1}^{k}a_i}{2a_k}$, and achieve a large value for LFS, and make $\sum_{i=1}{k}a_i=a_k$. Note that since $\sum_{i=1}^{k}a_i$ is not changed, and $a_1$, $\cdots$, $a_{k-1}$ are scaled up by the same factor, all conditions are still satisfied. We just need to prove that really the left hand side gets larger or equal.
Really there is just one case, but let's do three cases: 1) $k=2$, and 2) $k=n$; and 3) $2<k<n$.
1) $k=2$, which means $a_1<a_2$. Now The only values changed are $\frac{a_{1}}{a_2}$ and $\frac{a_{2}}{a_{3}}$, and we want to show that $\frac{a_{1}}{a_2} + \frac{a_{2}}{a_{3}} \le \frac{sa_{1}}{ta_2} + \frac{ta_{2}}{a_{3}}$. This is easy as $s=\frac{a_1+a_2}{2a_1}$ and $t=\frac{a_1+a_2}{2a_2}$, and $sa_1=ta_2=\frac{a_1+a_2}{2}$:
$$\frac{a_{1}}{a_2} + \frac{a_{2}}{a_{3}} \le \frac{a_{1}}{a_1} + \frac{a_{1}}{a_{3}} = \frac{sa_{1}}{ta_2} + \frac{a_{1}}{a_{3}} < \frac{sa_{1}}{ta_2} + \frac{ta_{2}}{a_{3}}, $$ where the first inequality requires a simple check.
2) $k=n$. This is easy as The only value changed is $\frac{a_{n-1}}{a_n}$ but we scale up $a_{n-1}$ and down $a_n$.
3) $2<k<n$. Now The only values changed are $\frac{a_{k-1}}{a_k}$ and $\frac{a_{k}}{a_{k+1}}$, and we want to show that $\frac{a_{k-1}}{a_k} + \frac{a_{k}}{a_{k+1}} \le \frac{sa_{k-1}}{ta_k} + \frac{ta_{k}}{a_{k+1}}$. Note that since the equality holds for $k-1$, we have $2a_{k-1}=\sum_{i=1}{k-1}a_i$. So $sa_{k-1}= \frac{2a_{k-1}+a_k}{4}$ and $ta_k=\frac{2a_{k-1}+a_k}{2}=2sa_{k-1}$. So we want to prove that
$$\frac{a_{k-1}}{a_k} + \frac{a_{k}}{a_{k+1}} \le \frac{1}{2} + \frac{2a_{k-1}+a_k}{2a_{k+1}}, $$
which is equivalent to the following after multiplying $2a_ka_{k+1}$:
$$2a_{k-1}a_{k+1}+2a_k^2 \le a_ka_{k+1}+(2a_{k-1}+a_k)a_k, $$
which is equivalent to
$$(a_{k+1}-a_k)(a_k-2a_{k-1}) \ge 0,$$
which is true since $a_{k+1}>a_k$ and $a_k \ge \sum_{i=1}^{k-1}a_i = 2a_{k-1}$.
So if we adjust up/down all $a_i$ and make all equalities in the condition hold, we achieve the largest value for $\frac{a_1}{a_2}+\cdots+\frac{a_{n-1}}{a_n}$, which happens to be $\frac{n}{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Absolute maximum of $f(x) = \sqrt{x^4-3x^2-6x+13}-\sqrt{x^4-x^2+1}$ My maths tutor solved this by using $f(x) = \sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+(x-0)^2}$ and treating $A(2,3)$ and $B(1,0)$ as fixed points and $P(x^2,x)$ as a moving point and using the difference between the sides PA and PB of the triangle PAB and relating that to the side AB to find the answer.
i.e. $PA-PB \leq AB = \sqrt{(2-1)^2-(3-0)^2} = \sqrt{10}$
What are other methods to solve this and is there a specific way to turn any such equation into the form of 2 squares as demonstrated above or is that just a special instance wherein that is possible.
| No, it's not so hard.
We need to prove that
$$\sqrt{x^4-3x^2-6x+13}-\sqrt{x^4-x^2+1}\leq\sqrt{10}$$ or
$$\sqrt{x^4-3x^2-6x+13}\leq\sqrt{10}+\sqrt{x^4-x^2+1}$$ or
$$1-3x-x^2\leq\sqrt{10(x^4-x^2+1)},$$
which is obvious for $x^2+3x-1\geq0$, but for $x^2+3x-1\leq0$ we need to prove that
$$(x^2+3x-1)^2\leq10(x^4-x^2+1)$$ or
$$9x^4-6x^3-17x^2+6x+9\geq0$$ or
$$(3x^2-x-3)^2\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How do we show that $\int_{0}^{1}{\arctan(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}=\left({\pi\over 6}\right)^2$ I was going through the the top votes questions and I saw this quite of interesting post posted by @Sangchul Lee. I was just messing around with it and found a slightest Variation of Ahmed's integral.
$$\int_{0}^{1}{\arctan(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}=\left({\pi\over 6}\right)^2\tag1$$
May be substitution might work
$u=x(x^2+2)^{1/2}$ the $du=x(x^2+2)^{-1/2}+(x^2+2)^{1/2}dx$
$$\int_{\sqrt{2}}^{4\sqrt{3}\over3}{x^2\arctan{u}\over x^3+u^2}\cdot{x^2\over u^2-x^2}du$$
More difficult then before. I can't remove the x variable from the integral.
Any suggestion on how to prove $(1)$?
| Substitute : $u=\arctan\left(x\sqrt{x^2+2}\right)$
You will get : $$\dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{\sqrt{x^2+2}+\frac{x^2}{\sqrt{x^2+2}}}{x^2\left(x^2+2\right)+1}$$
Thus :
$$\int_{0}^{1}{\arctan(x\sqrt{x^2+2})\over \sqrt{x^2+2}}\cdot{dx\over x^2+1}=\int_{0}^{\arctan(\sqrt{3})}{u\over 2}\cdot{du}=\left[\frac{u^2}{4}\right]^{\frac{\pi}{3}}_0=\left({\pi\over 6}\right)^2$$
| {
"language": "en",
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"question_score": "5",
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Solve $\cos (2x+3y)=\frac{1}{2}$ and $\cos (3x+2y)=\frac{\sqrt{3}}{2}$ We have to solve for $x$ and $y$:
$$\left \{
\begin{align*}
\cos (2x+3y) &= \frac{1}{2} \\
\cos (3x+2y) &= \frac{\sqrt{3}}{2}
\end{align*}
\right.$$
I got
$$\left \{
\begin{align*}
2x+3y &= 2n\pi \pm\frac{\pi}{3} \\
3x+2y &= 2m\pi \pm\frac{\pi}{6}
\end{align*}
\right.$$
The final answer in the book is
$x=\dfrac{1}{5}
\left[ (6m-4n)\pi \pm \dfrac{\pi}{2} \mp \dfrac{2\pi}{3} \right]$
Why is it $\mp \dfrac{2\pi}{3}$?
If for some value we had taken
$$2x+3y=2n\pi -\frac{\pi}{3}$$
and
$$3x+2y=2m\pi +\frac{\pi}{6}$$
The value of $x$ would be
$$\frac{1}{5}[(6m-4n)\pi+\frac{\pi}{2}+\frac{2\pi}{3}]$$
which is not included in the general solution.
| $1)$ You have:
$$2x+3y = 2n\pi \pm\frac{\pi}{3}$$
Multiply by $-2$ and get:
$$-4x-6y = -4n\pi \mp\frac{2\pi}{3} \quad (I)$$
$2)$ You have:
$$ 3x+2y= 2m\pi \pm\frac{\pi}{6}$$
Multiply by $3$ and get:
$$9x+6y= 6m\pi \pm \frac{\pi}{2}\quad (II)$$
$(I)+(II)$:
$$5x=(6m-4n)\pi \mp\frac{2\pi}{3}\pm \frac{\pi}{2}$$
PS: $\mp\frac{2\pi}{3}$ means that you got opposite signals when multiplied by a negative number, in that case it was $-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Computing $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$
What is $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$ ?
Here are a few remarks:
*
*Since $x\mapsto \frac{2^x}{x}$ is increasing when $x\geq 2$, one might be tempted to use the integral test. This fails: when doing so, one gets $a_n\leq \sum_{k=1}^n \frac{2^k}{k}\leq b_n$ where $a_n\sim \frac{2^n}{\ln (2)n}$ and $b_n\sim \frac{2^{n+1}}{\ln (2)n}$.
Unfortunately $b_n$ is too big and this estimate doesn't yield the limit.
*Here's my solution: since it's easy to sum $2^k$ and the difference $\frac{1}{k}-\frac{1}{k+1}$ is small, it's natural to try summation by parts: $$\begin{align} \sum_{k=1}^n \frac{2^k}{k}
&=\frac{S_n}{n+1}-1+\sum_{k=1}^n S_k \left(\frac{1}{k}-\frac{1}{k+1} \right)\quad \text{where} \; S_n=\sum_{k=0}^n 2^k\\
&= \frac{2^{n+1}}{n+1} + \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)} - \underbrace{1 - \sum_{k=1}^n\left(\frac{1}{k(k+1)}\right) - \frac{1}{n+1}}_{\text{bounded}}\\
\end{align}$$
Intuition suggests $\displaystyle \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}=o\left(\frac{2^n}n \right)$ but it's not immediate to prove. I had to resort to another summation by parts! Indeed
$$\begin{align}\small\sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}&= \small 2\left[ \frac{2^{n+1}}{n(n+1)} + 2\sum_{k=1}^n \left(\frac{2^{k+1}}{k(k+1)(k+2)}\right)-\frac 12 -2\sum_{k=1}^n \left(\frac{1}{k(k+1)(k+2)}\right) - \frac{1}{n(n+1)}\right]\\
&\small\leq \frac{2^{n+2}}{n(n+1)}+\frac{2^{n+2}}{n(n+1)(n+2)}\cdot n \\
&\small= o\left(\frac{2^n}n \right)
\end{align}$$
Hence $$\sum_{k=1}^n \frac{2^k}{k} = \frac{2^{n+1}}{n+1} + o\left(\frac{2^n}n \right)$$ and $$\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k} = 2$$
This solution is quite tedious and computational... That's why I'm looking for a shorter or smarter solution that avoids summation by parts (integration by parts is easy to perform on functions, it just gets quite heavy with series).
| Write the expression as
$$\frac{\sum_{k=1}^{n}2^k/k}{2^n/n}.$$
Note the denominator $\to \infty.$ That rings the Stolz-Cesaro bell, so consider
$$\frac{2^{n+1}/(n+1)}{2^{n+1}/(n+1) -2^n/n} = \frac{1}{1 -(n+1)/(2n)} \to \frac{1}{1/2} = 2.$$
By Stolz-Cesaro, the desired limit is $2.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove that $\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$ Prove that:
$$\tan 7°30' = \sqrt {6} - \sqrt {3} + \sqrt {2} - 2$$
My Work:
I guess that I have to use the formula :
$$\tan A = \frac {2 \tan(\frac {A}{2})}{1-\tan^2 (\frac {A}{2})}$$
But, I am not being able to use it. Please help me.
| Hint
$$\tan{(15^{\circ})}=2-\sqrt{3}$$
and let$x=\tan{7.5^{\circ}}$.we have
$$\dfrac{2x}{1-x^2}=2-\sqrt{3}\Longrightarrow (2-\sqrt{3})x^2+2x-(2-\sqrt{3})=0$$
so we have
$$x^2+2(2+\sqrt{3})x-1=0$$
then we have
$$(x+2+\sqrt{3})^2=1+(2+\sqrt{3})^2=8+4\sqrt{3}=2(\sqrt{3}+1)^2$$
then we have
$$x=\sqrt{2}(\sqrt{3}+1)-2-\sqrt{3}=\sqrt{6}+\sqrt{2}-2-\sqrt{3}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the following inequality $x^2+y^2+1>x\sqrt{y^2+1}+y\sqrt{x^2+1}$ Can anybody help me prove this inequality ?
| Hint
For $x \ge 0$ and $y \ge 0$ use
AM-GM
$$x^2+(y^2+1) \ge 2\sqrt{x^2(y^2+1)}=2x\sqrt{y^2+1}$$
$$(x^2+1)+y^2 \ge 2\sqrt{y^2(x^2+1)}=2y\sqrt{x^2+1}$$
Sum both and prove that the equality is not possible.
Also think about $x<0$ or $y<0$. That should be easy.
Can you finish?
| {
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Find all $n \in \mathbb{Z}$ such that $\sqrt{\frac{4n-2}{n+5}}$ is a rational number How does one approach this problem, where
all $\sqrt{\frac{4n-2}{n+5}}$ is a rational number when $n \in \mathbb{Z}$.
| Let $p,q$ be relatively prime positive integers such that $$\frac{4n-2}{n+5} = \frac{p^2}{q^2}$$
(note that $p\neq 0$ because $4n-2=0$ doesn't have integer solution.)
Then, we have $4n-2 = q^2k$, $n+5=p^2k$, $k\in\Bbb Z$. Now, $$4n-2 = 4(n+5) - 22\implies (4p^2-q^2)k = 22 \implies (2p-q)(2p+q)k = 22.$$
Notice that $2p-q\equiv2p+q\pmod 2$, and since $4\not\mid 22$, they must both be odd, implying that $$(2p-q)(2p+q)\mid 11$$
This leads to small number of cases:
*
*$(2p-q)(2p+q) = 11$
(We will come back to this case.)
*$(2p-q)(2p+q) = -11$
This would imply that $(2p-q)+(2p+q) = \pm 10$. Contradiction.
*$(2p-q)(2p+q) = 1$
This would imply that $2p-q=2p+q$, i.e. $q = 0$. Contradiction.
*$(2p-q)(2p+q) = -1$
This would imply that $(2p-q)+(2p+q) = 0$, i.e. $p = 0$. Contradiction.
Finally, we conclude that $(2p-q)(2p+q) = 11$, and since $2p-q<2p+q$, we have that $2p-q = 1$ and $2p-q=11$ (the other case when negative factors are chosen leads to negative $p$ and $q$). Solving the system we get $p=3$ and $q = 5$ (also, $k = 2$). This gives us
\begin{align}
4n-2 &= 50\\
n+5 &= 18
\end{align}
which implies that $n = 13$.
| {
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"url": "https://math.stackexchange.com/questions/2094027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Very interesting integral limit I found this interesting problem on AoPS forum but no one has posted an answer. I have no idea how to solve it.
$$
\int_0^\infty \sin(x^n)\,dx
$$
For all positive rationals $n>1$, $I_n$ denotes the integral as above.
If $P_n$ denotes the product
$$
P_n=\prod_{r=1}^{n-1}I_{\bigl(\!\frac{n}{r}\!\bigr)}\,,
$$
then evaluate the following limit $L$
$$
L=\lim_{n\to\infty}\bigl(\sqrt{n}\,P_n\bigr)^{\frac{1}{n}}
$$
| since
$$T_{\frac{r}{n}}=\int_{0}^{+\infty}\sin{(x^{\frac{r}{n}})}dx=\Gamma\left(\dfrac{r}{n}+1\right)\sin{\dfrac{\pi r}{2n}}=\dfrac{r}{n}\Gamma\left(\dfrac{r}{n}\right)\sin{\dfrac{\pi r}{2n}},r=1,2,,\cdots,n-1$$
proof see:3.1
Lemma $1$:
$$\prod_{i=1}^{n-1}\sin{\dfrac{i\pi}{2n}}=\dfrac{\sqrt{n}}{2^{n-1}}$$
Use this well known
$$z^{n-1}+z^{n-2}+\cdots+z+1=\prod_{k=1}^{n-1}(z-(z-z_{k})$$
let $z=1$ then we have
$$n=\prod_{i=1}^{n-1}\left|1-\cos{\dfrac{2k\pi}{n}}+i\sin{\dfrac{2k\pi}{n}}\right|$$
since
$$\left|1-\cos{\dfrac{2k\pi}{n}}+i\sin{\dfrac{2k\pi}{n}}\right|=2\sin{\dfrac{k\pi}{n}}$$
$$\prod_{i=1}^{n-1}\sin{\dfrac{i\pi}{n}}=\dfrac{n}{2^{n-1}}$$
and
$$\dfrac{2n}{2^{2n-1}}=\prod_{i=1}^{2n-1}\sin{\dfrac{i\pi}{2n}}=\left(\prod_{i=1}^{n-1}\sin{\dfrac{i\pi}{2n}}\right)^2$$
By done!
Lemma 2: $$f=\Gamma\left(\dfrac{1}{n}\right)\Gamma\left(\dfrac{2}{n}\right)\cdots\Gamma\left(\dfrac{n-1}{n}\right)=\dfrac{(2\pi)^{\frac{n-1}{2}}}{\sqrt{n}}$$
since
$$f^2=\left(\Gamma\left(\dfrac{1}{n}\right)\Gamma\left(\dfrac{n-1}{n}\right)\right)\cdot \left(\Gamma\left(\dfrac{2}{n}\right)\Gamma\left(\dfrac{n-2}{n}\right)\right)\cdots\left(\Gamma\left(\dfrac{n-1}{n}\right)\Gamma\left(\dfrac{1}{n}\right)\right)=\dfrac{\pi}{\sin{\frac{\pi}{n}}}\cdot\dfrac{2\pi}{\sin{\frac{2\pi}{n}}}\cdots\dfrac{\pi}{\sin{\dfrac{(n-1)\pi}{n}}}$$
so we have
$$f=\dfrac{(2\pi)^{\frac{n-1}{2}}}{\sqrt{n}}$$
so we have
$$P_{n}=\prod_{r=1}^{n-1}T_{\frac{r}{n}}=\prod_{r=1}^{n-1}\dfrac{r}{n}\Gamma\left(\dfrac{r}{n}\right)\sin{\dfrac{\pi r}{2n}}=\dfrac{(n-1)!}{n^{n-1}}\cdot\dfrac{\sqrt{n}}{2^{n-1}}\cdot \dfrac{(2\pi)^{\frac{n-1}{2}}}{\sqrt{n}}=\left(\dfrac{\sqrt{2\pi}}{2n}\right)^{n-1}(n-1)!$$
By stirling formula
we have
$$\lim_{n\to+\infty}(\sqrt{n}P_{n})^{\frac{1}{n}}=\dfrac{\sqrt{2\pi}}{2e}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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} |
Positive rational numbers with sum of their squares equal to $n$
Prove that for all positive integers $n$ there exist $n$ distinct, positive rational numbers with sum of their squares equal to $n$.
For $n = 1$ we can just take $1$. For $n = 2$ we can take $\left(\dfrac{1}{5}\right)^2+\left(\dfrac{7}{5}\right)^2 = 2$. How can we generalize this to any $n$?
| You can get $x^2+y^2=1$ infinitely many ways using Pythagorean triples, just make sure all the triples are distinct, then add the expressions. If $n$ is odd, use an initial $1$ not from a triple.
Edit 2: (previous idea didn't work) Suppose that
$$x=\frac{k^2-2k-1}{k^2+1},\ y=\frac{k^2+2k-1}{k^2+1}. \tag{1}$$
Then $x^2+y^2=2,$ and if $k$ is even both $x,y$ in (1) can be shown to be in lowest terms. This means we have infinitely many distinct solutions to $x^2+y^2=2,$ so that as @N.S suggests in his comment below we can get $n$ distinct rational squares with sum $n.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can the limit of this $F(x,y)$ is $0$ when it is taken along $y=x^2$ or $ x=y^2$?
Find $$\lim_{(x,y) \to (0,0)}\frac{3x^{2}y}{x^{2}+y^{2}}$$ if it
exists.
From my textbook, it said that the limits along the parabolas $$y=x^{2}\text{ and } x=y^{2}$$ also turn out to be 0.
I couldn't figure out why, as if we set $$x=y^{2}$$, for example
then $$\lim_{(x,y) \to (0,0)}\frac{3x^{4}}{x^{2}+(x^{2})^{2}}$$
$$\lim_{(x,y) \to (0,0)}\frac{3x^{4}}{x^{2}+x^{4}}$$
$$\lim_{(x,y) \to (0,0)}\frac{3}{0+1}$$
then the answer is supposed to be $3$? right?
| On $y=x^2$ we have
$$
\frac{3x^2y}{x^2+y^2}=\frac{3x^4}{x^2+x^4}=
\frac{3x^4}{x^2(1+x^2)}=\frac{3x^2}{1+x^2}
$$
which has limit $0$ when $x\to0$.
On $x=y^2$ we have
$$
\frac{3x^2y}{x^2+y^2}=\frac{3y^5}{y^4+y^2}=\frac{3y^5}{y^2(y^2+1)}
=\frac{3y^3}{y^2+1}
$$
which again has limit $0$ for $y\to0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to find the sum of this infinite series: $\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$ How do I find this particular sum?
$$\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$$
where $H_n = \sum_{k=1}^{n}\frac1k$.
This was given to me by a friend and I have absolutely no idea how to proceed as I have never done these questions before. If possible, please give a way out without using polylogarithmic functions or other non-elementary functions.
| recall: $\displaystyle H_a = \int_0^1 \frac{1-x^a}{1-x}\,\mathrm{d}x$, and integration by parts once we have
$$\int_0^1 x^{a-1} \ln (1-x)\,\mathrm{d}x = -\frac{H_a}{a}$$
Thus,
\begin{align*}&\sum\limits_{n=1}^{\infty} \frac{H_{n+a}}{n(n+a)} = -\sum\limits_{n=1}^{\infty} \frac{1}{n}\int_0^{1} x^{n+a-1} \ln (1-x)\,\mathrm{d}x= \int_0^{1} x^{a-1} \ln^{2}(1-x)\,\mathrm{d}x\\&= \lim\limits_{b \to 1}\frac{\partial^2}{\partial b^2}\int_0^{1} x^{a-1}(1-x)^{b-1}\,\mathrm{d}x= \lim\limits_{b \to 1} \frac{\partial^2}{\partial b^2} \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\\&= \lim\limits_{b \to 1}\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\left((\psi_0(b) - \psi_0(a+b))^2 + \psi_1(b) - \psi_1(a+b)\right)\\&= \frac{1}{a}\left((\gamma + \psi_0(a+1))^2 + \frac{\pi^2}{6} - \psi_1(a+1)\right)\end{align*}
Hence put $a=2$ we get
$$\sum\limits_{n=1}^{\infty} \frac{H_{n+2}}{n(n+2)} =\frac{1}{2}\left [ \left ( \gamma +\underset{\psi _{0}\left ( 3 \right )}{\underbrace{\frac{3}{2}-\gamma}} \right )^{2}+\frac{\pi ^{2}}{6}-\underset{\psi _{1}\left ( 3 \right )}{\underbrace{\left (\frac{\pi ^{2}}{6}-\frac{5}{4} \right )} }\right ]=\frac{7}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Integral related to Pythagoras theorem ${2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\mathrm dx=h^2$. Integral related to Pythagoras theorem
Triangle ABC is a right angle triangle, where Angle $ABC=90^o$.
$h$ is perpendicular to the hypotenuse AC and meet at angle ABC.
Where $a$ and $b$ are two small sides
How can I Show that h can be represented in term of this integral $(1)$
$${2\over \pi}\int_{0}^{\infty}{(ab)^3\over (a^2+b^2x^2)(b^2+a^2x^2)}\mathrm dx=h^2\tag1$$.
Any hints on this can be relate to Pythagoras theorem
Basic formulas : $AC^2=AB^2+BC^2$ and area, $A={bh\over 2}$
| Let
$$ I=\int^\infty_0 \frac{dx}{(a^2+b^2x^2)(b^2+a^2x^2)}.\tag{1}$$
Changing variable from $x\to\frac1{x}$ gives
$$ I=\int^\infty_0 \frac{x^2dx}{(a^2+b^2x^2)(b^2+a^2x^2)}.\tag{2}$$
So $a^2$(1)+$b^2$(2) gives
$$ a^2I+b^2I=\int^\infty_0 \frac{dx}{b^2+a^2x^2}=\frac{\pi}{2ab}$$
and hence
$$ I=\frac{\pi}{2ab(a^2+b^2)}.$$
So
$$ \frac{2}{\pi}\int^\infty_0 \frac{(ab)^3dx}{(a^2+b^2x^2)(b^2+a^2x^2)} = \frac{(ab)^3}{ab(a^2+b^2)}= \frac{a^2b^2}{a^2+b^2}=\frac{2}{\pi}(ab)^3I=\frac{a^2b^2}{a^2+b^2}=h.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of $\int\frac{dx}{x+ \sqrt{x^2-x+1}}$ Evaluate :
$$\frac{dx}{x+ \sqrt{x^2-x+1}}$$
After dividing and multiply with $x-\sqrt{x^2-x+1}$, I get
$x+\ln |x-1|-\int \frac{\sqrt{x^2-x+1}}{x-1}dx$.
Is $\int \frac{\sqrt{x^2-x+1}}{x-1}dx$ is integrable in terms of elementary functions?
| There is a substitution, called the Euler substitution, which solves your integral rather neatly. In your case we set $t = \sqrt{x^2 - x + 1} + x$; after subtracting $x$ from both sides and squaring, we get $x^2 -x + 1 = t^2 - 2xt + x^2$, which can be solved (after subtracting $x^2$ from both sides) to get $x = \frac{t^2 - 1}{2t - 1}$. In this case, $\frac{dt}{dx} = \frac{1}{\sqrt{x^2 - x + 1}}(2x-1) + 1$, and so
\begin{eqnarray} \int \frac{dx}{x + \sqrt{x^2 - x + 1}} = \int (\frac{1}{x + \sqrt{x^2 - x + 1}}\cdot\frac{1}{\frac{dt}{dx}})\frac{dt}{dx}dx \\= \int \frac{1}{t}\cdot\frac{1}{[\frac{1}{t - (t^2-1)/(2t-1)}\cdot(2(\frac{t^2-1}{2t-1}) - 1)] + 1} dt = \dots \\ \dots = \int\frac{t^2 - t + 1}{3t^3 - 3t^2} dt\end{eqnarray}
which is quite easy to solve directly by partial fractions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Partitions of $n$ with exactly 3 parts I'm learning about generating functions, and came across this question: find the generating function for the number of partitions of a number $n$, into exactly 3 parts. I just solved a problem where the condition was that each part is no greater than 3, but I'm kind of stuck on this one. Any ideas? Thanks!
| Hint: The following can be found somewhat more detailed in this answer.
A generating function for the number of partitions with exactly three parts is
\begin{align*}
&\frac{1}{(1-x)(1-x^2)(1-x^3)}-\frac{1}{(1-x)(1-x^2)}\\
&\qquad=\frac{1}{(1-x)(1-x^2)}\left(\frac{1}{1-x^3}-1\right)\\
&\qquad=\frac{x^3}{(1-x)(1-x^2)(1-x^3)}\\
&\qquad=x^3+x^4+2x^5+3x^6+4x^7+\color{blue}{5}x^8+7x^9\cdots
\end{align*}
Example: There are $\color{blue}{5}$ partitions of $8$ with three summands
\begin{align*}
8&=1+1+6\\
&=1+2+5\\
&=1+3+4\\
&=2+2+4\\
&=2+3+3
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Showing a vector belongs in the span Without using dimension, I have to show
From: question
$\overrightarrow{x} \in \mathbb{R}^3$ is also in span $\{v_1, v_2, n\}$ where $n = v_1 \times v_2$. Where $v_1, v_ 2\in \mathbb{R}^3$ as well.
We can let $x = \langle x_1, x_2, x_3 \rangle$, but that isn't very helpful.
I know the set is linearly independent, could that be used? Should I write their sum in the form of components?
| I'll use the notation from the previous answer, since I don't like subscripts. We are given two vectors $\mathbf{u}$ and $\mathbf{v}$, and we want to show that any given vector $\mathbf{x}$ is in the span of $\{\mathbf{u}, \mathbf{v}, \mathbf{n}\}$, where $\mathbf{n} = \mathbf{u} \times \mathbf{v}$.
First let's focus on the point $\mathbf{w} = p \mathbf{u} + q \mathbf{v}$ that is the foot of the perpendicular from $\mathbf{x}$ to the plane spanned by $\mathbf{u}$ and $\mathbf{v}$. The scalars $p$ and $q$ are unknown, as yet -- we have to find them. To get perpendicularity, we need $p$ and $q$ to satisfy the equations
$$(\mathbf{x} - \mathbf{w}) \cdot \mathbf{u} = (\mathbf{x} - p \mathbf{u} - q \mathbf{v}) \cdot \mathbf{u} = 0
$$
$$
(\mathbf{x} - \mathbf{w}) \cdot \mathbf{v} = (\mathbf{x} - p \mathbf{u} - q \mathbf{v}) \cdot \mathbf{v} = 0
$$
In other words, $p$ and $q$ that are solutions of the linear system
$$
p (\mathbf{u} \cdot \mathbf{u}) + q (\mathbf{u} \cdot \mathbf{v}) = \mathbf{x} \cdot \mathbf{u}
$$
$$
p (\mathbf{u} \cdot \mathbf{v}) + q (\mathbf{v} \cdot \mathbf{v}) = \mathbf{x} \cdot \mathbf{v}
$$
A solution exists because the determinant of this system is
$(\mathbf{u} \cdot \mathbf{v})(\mathbf{v} \cdot \mathbf{v}) - (\mathbf{u} \cdot \mathbf{v})^2$, which is positive unless $\mathbf{u}$ and $\mathbf{v}$ are linearly independent. Or, you can just see that the system has a solution by calculating it via Cramer's rule.
Since $\mathbf{x} - \mathbf{w}$ is orthogonal to both $\mathbf{u}$ and $\mathbf{v}$, it must be parallel to $\mathbf{n} = \mathbf{u} \times \mathbf{v}$. So, there is a scalar $k$ such that
$\mathbf{x} - \mathbf{w} = k \mathbf{n}$. But then
$$
\mathbf{x} = \mathbf{w} + k \mathbf{n} = p \mathbf{u} + q \mathbf{v} + k \mathbf{n}
$$
which means that $\mathbf{x}$ lies in the span of $\{\mathbf{u}, \mathbf{v}, \mathbf{n}\}$.
| {
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How do we evaluate $\int\frac{dx}{\sin^{4}{x}+1}$? How do we evaluate $$\displaystyle\int{\dfrac{dx}{\sin^{4}{x}+1}}?$$
don't know how to solve this, please help.
| By substituting $x=\arctan t$ the problem boils down to
$$ \int \frac{1+t^2}{(1+t^2)^2+t^4}\,dt = \int\left(\frac{1+i}{4}\cdot\frac{1}{t^2+\frac{1+i}{2}}+\frac{1-i}{4}\cdot\frac{1}{t^2+\frac{1-i}{2}}\right)\,dt $$
leading to:
$$ \frac{1}{2\sqrt{2}}\left(\sqrt{1-i}\,\arctan\frac{t}{\sqrt{\frac{1-i}{2}}}+\sqrt{1+i}\,\arctan\frac{t}{\sqrt{\frac{1+i}{2}}}\right) $$
that equals
$$ 2^{-7/4}\left[\sqrt{2+\sqrt{2}}\arctan\frac{t\sqrt{2+2\sqrt{2}}}{1-t^2\sqrt{2}}+\sqrt{2-\sqrt{2}}\,\text{arctanh}\frac{t\sqrt{2\sqrt{2}-2}}{1+t^2\sqrt{2}}\right]$$
by exploiting $\arctan(a)\pm\arctan(b)=\arctan\frac{a\pm b}{1\mp ab}$.
| {
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"url": "https://math.stackexchange.com/questions/2109901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding Limit with Taylor series Find the limit using Taylor-series , $f(x) =\frac{\ln(1+x^2)-x^2}{\sqrt{1+x^4}-1}$.
I calculated the limit of of the $\ln(1+x^2)$ which is equal to
${x^2} - \frac{2x^3}{3} + ...O(x^n)$
and $\ln(1+x^2) -x^2 $ = $\frac{-2x^3}{3} + ...$
but when i calculate $\sqrt{1+x^4}$ i get $0$ for first second and third derivative(when i plug in $0$) for the $4$rth derivative(after plugging in zero) i get $6$.
So Taylor series for $\sqrt{1+x^4}-1 = \frac{6x^4}{4!}+ ...$ and taking the next derivative gives again zero :( in simple i get this experession $\frac{-8x^3}{3x^4} + ..$ and $ \displaystyle{\lim_{x \to 0} f(x)}$. can someone tell me whether these steps are right or wrong?
| Hint
Use the generalized binomial theorem to show that $$\sqrt{1+x^4}=(1+x^4)^{\frac 12}=1+\frac{x^4}{2}-\frac{x^8}{8}+\cdots$$
Another way could be to consider Taylor series $$\sqrt{1+t}=1+\frac{t}{2}-\frac{t^2}{8}+O\left(t^3\right)$$ and replace $t$ by $x^4$.
| {
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Solve the equation $2\arcsin x=\arcsin(\frac{3}{4}x)$ $$2\arcsin x=\arcsin(\frac{3}{4}x)$$ so $x\in[-1,1]$
so we have:
$2\arcsin x=y\Rightarrow\sin\frac{y}{2}=x$ and $\arcsin x=y \Rightarrow \sin y=\frac{3}{4}x\Rightarrow\frac{4}{3}\sin y=x$ ,
$y\in[-\frac{\pi}{2},\frac{\pi}{2}]$
$$\sin\frac{y}{2}-\frac{4}{3}\sin y=0$$$$\sin\frac{y}{2}-\frac{4}{3}\cdot 2\sin\frac{y}{2}\cos\frac{y}{2}=0$$ $$\sin\frac{y}{2}\cdot (1-\frac{8}{3}\cos\frac{y}{2})=0$$
Is it done properly at this point?
1.$\sin \frac{y}{2}=0\Rightarrow x=0$
And what else?
| $\sin\frac{y}{2}\cdot (1-\frac{8}{3}\cos\frac{y}{2})=0$
$\sin\frac{y}{2} = 0$
So far so good
contining... suppose $\sin\frac{y}{2} \ne 0$
$(1-\frac{8}{3}\cos\frac{y}{2})=0\\
\frac{8}{3}\cos\frac{y}{2})=1\\
\cos\frac{y}{2}=\frac 38\\
\frac y2=\arccos\frac 38\\
x = \sin(\arccos\frac 38)\\
x = \sqrt {1 - \frac {9}{64}}\\
x = \frac {\sqrt{55}}{8}$
However, if this were the case.
since $\frac {\sqrt 2}{2}< \frac {\sqrt {55}}{8}< 1 \implies \frac {\pi}{4} < \arcsin \frac {\sqrt {55}}{8}< \frac {\pi}{2}\\2\arcsin \frac {\sqrt {55}}{8} > \frac {\pi}{2}$
while
$\arcsin \frac 34 \frac {\sqrt {55}}{8} < \frac {\pi}{2}$
$x = 0$ is the only solution.
One more consideration. Had we found a non-zero solution, then $-x$ would also be a solution, and the technique here managed to obliterate that possibility.
Cutting out all of the algebra:
$\forall x\ne 0, |2\arcsin x| > |\arcsin(\frac 34 x)|$
Which means that $2\arcsin x \ne \arcsin(\frac 34 x)$
unless $x = 0$
| {
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Question on proof of $1+2+\dots+n=\frac{n(n+1)}{2}$ by induction. I saw some video where it needs to prove $1+2+\dots+n=\frac{n(n+1)}{2}$ inductively. So it has to be true if $k=1$ and $k+1$ are true.
So, for $k=1$:
$$1=\frac{1(1+1)}{2}=\frac{1(2)}{2}=\frac{2}{2}=1$$
it is valid.
For $k+1$ here is the proof he does:
$$
\begin{align}
1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2} &(1)\\
\frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2} &(2)\\
&=\frac{k^2+2k+k+2}{2}&(3)\\
&=\frac{k^2+3k+2}{2}&(4)\\
&=\frac{(k+1)(k+2)}{2}&\text{factoring (4)}
\end{align}$$
Therefore this formula is valid for $k+1$.
But is this true? I think not. He is just undoing what he have just done. To prove it I think I need to do this:
$$
\begin{align}
1+2+\dots+k+k+1&=\frac{(k+1)(k+2)}{2}\\
\frac{k(k+1)}{2}+k+1&=\frac{(k+1)(k+2)}{2}\\
\frac{k(k+1)+2(k+1)}{2}&=\frac{(k+1)(k+2)}{2}\\
\frac{(k+1)(k+2)}{2}&=\frac{(k+1)(k+2)}{2}\\
(k+1)(k+2)&=(k+1)(k+2)\\
k+1&=k+1\\
k&=k\\
0&=0\text{ or }1=1
\end{align}$$
Therefore, $k+1$ is valid in this formula.
Is this right or am I making a mistake somewhere?
| Proving for sum =k+1
RhS= $\dfrac {(k)(k+1)}{2} + k + 1 $
Which gives
=$\dfrac {(k+1)(k+2)}{2}$
= $\dfrac {([k+1])([k+1] +1)}{2}$
Which means it hold for n and also n+1 !
| {
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Find all $n\in \mathbb{Z}^+ : 7|5^{6n}+5^n+2$ Question: We want to find all $n\in \mathbb{Z}^+$ such that $7|5^{6n}+5^n+2$.
Answer: We have that $7|5^{6n}+5^n+2 \iff 5^{6n}+5^n+2 \equiv 0 \bmod 7$.
But $\phi(7)=6\wedge\text{gcd}(5,7)=1$ and we can apply Fermat's theorem, so $5^6 \equiv1 \bmod 7\implies 5^{6n}\equiv1 \bmod 7 $. So, we have to find all positive integers $n$ such that
$$5^n+3\equiv0 \bmod 7 \iff 5^n \equiv 4 \bmod 7 \iff 2\cdot 5^n\equiv1 \bmod 7$$
I stuck in this point. Any help please?
Thank you in advance.
| From $5^n \equiv 4 (\bmod{7})$ write $(-2)^n \equiv 4 (\bmod{7})$ so that
$(-2)^{n-2} \equiv 1 (\bmod{7}).$ The order of $-2 (\bmod{7})$ is $6$, so $6\mid n-2$. Therefore $n =6k+2$ for any integer $k$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Equation involving fields Let $ K $ be a field and $f:K-\left \{ 0 \right \}\rightarrow K-\left \{ 0 \right \}$ a function with $f(f(x))=x^{-1},\forall x\in K-\left \{ 0 \right \}$ and $ f(1)\neq 1.$
Knowing that $ f^{2}(x)-f(x)+1=0 $ has an unique solution in $K-\left \{ 0 \right \} $, determine $ f(2)$.
I haven't found something useful yet.
| Suppose that $a$ is a root of the polynomial $x^2-x+1$.
Then $(x-a)$ divides the polynomial $x^2-x+1$.
This means that $(x-a)(x-a)=x^2-x+1$ (because the root is unique).
it follows that $x^2-2a+1=x^2-x+1$ and so $2a=1$ and $a^2=1$.
we conclude that $1=1^2=(2a)^2=4a^2=4$.
So the field must have characteristic $3$, and so $2=-1$.
Let $z=f(1)$. Notice $z^{-1}=f(f(z))=f(f(f(1))=f(1)=z$. It follows that $z^2=1$. The only two roots of the polynomial $x^2-1$ are $1$ and $-1$. We conclude $z=-1$.
So we have $f(1)=-1$, and since $1=f(f(1))=f(-1)$ we have $1=f(-1)=f(2)$.
| {
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Problem of Complex Numbers in Geometry using Roots of Unity Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \neq 1$. Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6$. Then $\alpha$ and $\beta$ are roots of the quadratic
$x^2 + px + q = 0$
for some integers $p$ and $q$. Find the ordered pair $(p,q)$.
I am unable to find a good starting place for this problem. Should I start by working backwards using the roots?
| $$w^7-1=(w-1)(w^6+w^5+w^4+\dots+w+1) $$
So $$w^6+w^5+w^4+\dots+w+1 =0 \tag{1}$$
Now use Vieta's Formula.
First note that the sum of roots is $-p$, so we have that
$$-p=\alpha+\beta=w^6+w^5+w^4+w^3+w^2+w=-1$$
From $\text{(1)}$ Also,
$$q=(\omega + \omega^2 + \omega^4) \times (\omega^3 + \omega^5 + \omega^6)$$
Expanding,
$$q=w^4(w^{6}+w^{5}+w^{4}+w^{3}+w^{2}+w+1)+2w^7=2$$
So $p=1, q=2$.
| {
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Prove that $\sum\limits_{cyc} a^7 \geq \sum\limits_{cyc}a^4b^3$
Prove that $a^7+b^7+c^7\ge a^4b^3+b^4c^3+c^4a^3$
SOURCE : "A Brief Introduction to Olympiad Inequalities" by Evan Chen
It was one of the practice problems. Equality case is easy. I tried AM-GM and Muirhead, but could not seem to find a suitable proof for the inequality.
Any hint would be really helpful.
Thanks in advance !
^__^
| Using $\text{AM-GM}$ this can be solved. Note that by $\text{AM-GM}$ $$4a^7+3b^7=a^7+a^7+a^7+a^7+b^7+b^7+b^7 \ge 7a^4b^3$$$$4b^7+3c^7=b^7+b^7+b^7+b^7+c^7+c^7+c^7 \ge 7b^4c^3$$$$4c^7+3a^7=c^7+c^7+c^7+c^7+a^7+a^7+a^7 \ge 7c^4a^3$$
Adding the three and dividing by seven, we have $$a^7+b^7+c^7 \ge a^4b^3+b^4c^3+c^4a^3$$
| {
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How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas? How to find $\tan(-\frac{5\pi}{16})$ with half-angle formulas?
I tried the $\pm \sqrt{\frac{1-\cos{A}}{1+\cos{A}}}$ and $\frac{\sin{A}}{1+\cos{A}}$ but got stuck once there were square roots on top and bottom like $\frac{\sqrt{...}}{1-\sqrt{...}}.$
Using the cosine over cosine in square root I got up to
$$=-\sqrt{ \frac{ 1+\sqrt{\frac{1+\cos(5\pi/3)}{2}}}{1-\sqrt{\frac{1+\cos(5\pi/3)}{2}}} }$$
| The problem is straightforward, it's just a form which is more trouble to rationalize that it's worth. And there are several equivalent expressions depending upon which identities are used.
$$ \tan\left(-\frac{5\pi}{16}\right)=-\frac{\sin\left(\frac{5\pi}{8}\right)}{1+\cos\left(\frac{5\pi}{8}\right)}$$
and
$$ \sin\left(\frac{5\pi}{8}\right)=\sqrt{\frac{1-\cos(5\pi/4)}{2}}=\frac{1}{2}\sqrt{2+\sqrt{2}}$$
whereas
$$ \cos\left(\frac{5\pi}{8}\right)=-\sqrt{\frac{1+\cos(5\pi/4)}{2}}=-\frac{1}{2}\sqrt{2-\sqrt{2}}$$
Therefore
$$\tan\left(-\frac{5\pi}{16}\right)=\frac{\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}-2} $$
It takes many steps to rationalize this denominator so one may as well leave it in this form.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $ (100 a+10b+c)^2 =(a+b+c)^5.$ Find a three-digits number $\overline{abc}$ such that $\overline{abc}^2=(a+b+c)^5.$
It is easy to see that
$$
(a+b+c)^5 \leq 999^2 \implies a+b+c< \sqrt[5]{999^2}\leq 15
$$
and
$$
(100 a+10b+c)^2<15^5 \implies 100 a+10b+c>\sqrt{15^5} \leq 871.
$$
Also
$$
(100 a+10b+c)^2 =(a+b+c) \mod 2
$$
implies $a+b=0 \mod 2$.
Similarly
$$
(100 a+10b+c)^2 = c^2 = (a+b+c) \mod 5.
$$
But it not enough to find the solution $243$.
No more ideas.
| Let
$$x = \sqrt{a+b+c} = \sqrt[5]{\overline{abc}},\quad a\in \{1, 2\dots9\}, \quad b,c\in\{0,1\dots9\},$$
then
$$\begin{cases}
\sqrt{1} <= x < \sqrt{27}\\
\sqrt[5]{100} < x < \sqrt[5]{999}\\
x\in\mathbb N,
\end{cases}$$
$$\begin{cases}2.5 < x < 3.99\\
x\in\mathbb N,
\end{cases}$$
$$x=3,$$
$$\begin{cases}
\overline{abc} = x^5 = 243\\
a+b+c = x^2 = 9,
\end{cases}$$
$$\boxed{a=2,\quad b=4,\quad b=3}.$$
| {
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How to calculate a number of $x$ values such that $f(x)$ is a quadratic residue modulo $m$? Let $a, b$ be any fixed positive integers, $f(x) = (3x + 3a)^2 - b$
How to calculate a number of $x$ values such that $f(x)$ is a quadratic residue modulo $m$ without actually finding them?
Edit: I updated the question to make it more clear.
| Step 1: You can ignore the $-b$ since for each residue of $(3x+3a)^2-b$ there is a corresponding residue of $(3x+3a)^2$ whichi s just $b$ more than the former.
Step 2: Similarly you can ignore the $a$ since for each value of $(3x+3a)$ there is a value $(3y)$ with $y=x+a \pmod m$.
So we want to find the number of quadratic residues of the form $(3x)^2 \pmod m$.
If $m$ is a prime greater than $3$, then there are $\frac{m+1}{2}$ quadratic residues $r$ mod $m$, and for each one of those $q$, since $3$ has an inverse mod $m$,
$x=3^{-1}q\pmod m$ is a number such that $(3x)^2 = r.$ So if $m$ is an odd prime greater than $3$, then the answer is $\frac{m+2}{2}$.
If $m$ is the product of distinct primes all greater than $3$ then again we can find $x=3^{-1}\pmod m$ and the number of quadratic residues matches the number of possble values of $(3x)^2$; and that is the product of the number of quadratic residues for each factor. So for example, when $m = 5\cdot 11 \cdot 19$ there are
$3\cdot 6 \cdot 10$ possible values of $3x)^2\pmod m$.
If $m=3$ or $m=9$, then there of course is only one residue, namely $(3x)^2=0$.
Other cases can be solved as well, but counting the residues for an arbitrary prime power is a bit trickier.
| {
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Find all possible values of $x$ if $\ln(x) = \sin(x)$. My Mathematics Teacher gave me the following problem :
Find all the possible values of $x$ if $\ln(x) = \sin(x)$.
I tried graphing both $\ln(x)$ and $\sin(x)$. I found that they intersect at a single point $\approx (2.22, 0.8)$. However, I am unable to find a "mathematical" solution to this problem using the elementary properties of trigonometry and logarithms. I tried some Taylor expansions, but that did not help. Can anybody provide a hint as to what should be done ?
Also, what can be the possible values of $x$ is the problem is generalized as $\sin(x)=\log_b(x) ? $ One simple observation is that $0<x\leq b$ as maximum value of the sine function is 1. How to proceed ?
| As said in answers and comments, the solution requires numerical methods.
Since you graphed the functions, you noticed that they intersect close to $x=2$. Then, in order to approximate the solution, you could have used Taylor expansions (as you though about) but building them around $x=2$.
You have $$\log(x)=\log (2)+\frac{x-2}{2}-\frac{1}{8} (x-2)^2+\frac{1}{24}
(x-2)^3+O\left((x-2)^4\right)$$ $$\sin(x)=\sin (2)+(x-2) \cos (2)-\frac{1}{2} (x-2)^2 \sin (2)-\frac{1}{6} (x-2)^3 \cos
(2)+O\left((x-2)^4\right)$$
Limited to first order, we then have $$\log(x)-\sin(x)=(\log (2)-\sin (2))+(x-2) \left(\frac{1}{2}-\cos (2)\right)+O\left((x-2)^2\right)$$ leading to the solution $$x_{(1)}=\frac{2 (-1+\log (2)-\sin (2)+2 \cos (2))}{2 \cos (2)-1}\approx 2.23593$$ Limited to second order,we have $$\log(x)-\sin(x)=((\log (2)-\sin (2))+(x-2) \left(\frac{1}{2}-\cos (2)\right)+(x-2)^2
\left(\frac{\sin (2)}{2}-\frac{1}{8}\right)+O\left((x-2)^3\right)$$ and solving the quadratic would give $$x_{(2)}\approx 2.21872$$ while the "exact" solution given by Newton method would be $$x\approx 2.21911$$
Edit
May be, it would be nicer to build Taylor expansion around $x=\frac{2\pi}3$. This would give
$$\log(x)-\sin(x)=\left(\log \left(\frac{2 \pi }{3}\right)-\frac{\sqrt{3}}{2}\right)+\frac{(3+\pi )
}{2 \pi } \left(x-\frac{2 \pi }{3}\right)+\left(\frac{\sqrt{3}}{4}-\frac{9}{8 \pi
^2}\right) \left(x-\frac{2 \pi }{3}\right)^2+O\left(\left(x-\frac{2 \pi
}{3}\right)^3\right)$$ Limiting to first order, the solution would be
$$x_{(1)}=\frac{\pi \left(6+3 \sqrt{3}+2 \pi -6 \log \left(\frac{2 \pi }{3}\right)\right)}{3
(3+\pi )}\approx 2.22408$$ Using the second order, this would give $$x_{(2)}\approx 2.21901$$
| {
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How to find the minimal-polynomial of $\begin{pmatrix} 0&1&0&1\\1&0&1&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$
$E:=\begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix} A:= \begin{pmatrix} 0&1&0&1\\1&0&1&0\\0&0&0&0\\0&0&0&0\end{pmatrix} A^2:= \begin{pmatrix} 1&0&1&0\\0&1&0&1\\0&0&0&0\\0&0&0&0\end{pmatrix}$
My attempt:
The idea is to find a linear combination $lin(E+A+A^2+...+A^n) = A^{n+1}$
I tried it by taking the first column of each matrix and look for a solution:
(1) E|A
$x\begin{pmatrix} 1\\0\\0\\0\end{pmatrix} = \begin{pmatrix} 0\\1\\0\\0\end{pmatrix}$
This can obviously not work
(2) $lin(E+A) = A^2$
$x\begin{pmatrix} 1\\0\\0\\0\end{pmatrix} +y \begin{pmatrix} 0\\1\\0\\0\end{pmatrix} = \begin{pmatrix} 1\\0\\0\\0\end{pmatrix}$
Which works for $x=1, y=0.$
Now the minimalpolynomial should be $0\cdot E+ 1\cdot A + A^2 \rightarrow 0+x+x^2$, but it seems to be wrong in further calculations.
| The easiest way to find a minimal polynomial is to put the matrix in Jordan normal form. The minimal polynomial is $p(x) = \prod (x-\lambda)^{d_\lambda}$, where the product runs over the eigenvalues $\lambda$ and $d_\lambda$ is the size of the largest Jordan block for that eigenvalue (not the multiplicity of the eigenvalue).
In this case, your matrix is diagonalizable, so all the Jordan blocks are $1 \times 1$. It has 4 eigenvectors with eigenvectors 0 (multiplicity 2), 1, and -1. The minimal polynomial is $p(x) = x(x-1)(x+1)$.
| {
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Simple yet confusing probability problem. A bag contains x green candies and y red candies. A candy is selected at random from the bag and it's color is noted. It is then replaced into the bag with an additional ten candies of the same color. A second candy is then randomly chosen. Find the probability that the second candy is red.
I think that the answer is y/(x+y+10) but am unsure.
Also where would you recommend to get more practice on probality problems
| We have x green candies and y red.
Case 1-
Green candy is picked in first attempt.
$\frac x{x+y}$
Now green candies x + 10.
Red candy is picked in second attempt.
$\frac {y}{x+y+10}$
Probability = $\frac {x}{x+y} × \frac {y}{x+y+10} $
Case 2-
Red candy is picked in first attempt.
Probability = $\frac y{x+y}$
Now red candies is y + 10.
Red candy is picked in second attempt.
$\frac {y+10}{x+y+10}$
Probability = $\frac {y}{x+y} × \frac {y+10}{x+y+10} $
Total Probability = $\frac {x}{x+y} × \frac {y}{x+y+10} +\frac {y}{x+y} × \frac {y+10}{x+y+10} $
| {
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How to prove $\ln(1+x)\ge x/(1+x/2)$ when $x > 0$? I think the equation establishes only when $x = 0$, so the condition should be x>=0. The first derivative of $\ln(1+x)-x/(1+x/2)$ is $0$ at $x=0$. So the inequality looks good to establish. Is there a better way to prove it?
| To show
$\ln(1+x)
\ge x/(1+x/2)
$.
Since
$\ln(1+x)
=\int_0^x \dfrac{dt}{1+t}
$,
this is equivalent to
$\dfrac1{x}\int_0^x \dfrac{dt}{1+t}
\ge \dfrac1{1+x/2}
$.
Since
$\dfrac1{1+x/2}$
is the value of
$\dfrac1{1+t}
$
at the midpoint,
I'll combine the two halves.
$\begin{array}\\
\int_0^x \frac{dt}{1+t}
&=\int_0^{x/2} \frac{dt}{1+t}+\int_{x/2}^x \frac{dt}{1+t}\\
&=\int_0^{x/2} \frac{dt}{1+t}+\int_0^{x/2} \frac{dt}{1+x-t}\\
&=\int_0^{x/2} dt\left(\frac{1}{1+t}+\frac{1}{1+x-t}\right)\\
&=\int_0^{x/2} dt\left(\frac{1+x-t+1+t}{(1+t)(1+x-t)}\right)\\
&=\int_0^{x/2} dt\left(\frac{2+x}{1+t+x-t+t(x-t)}\right)\\
&=(2+x)\int_0^{x/2} dt\left(\frac{1}{1+x+t(x-t)}\right)\\
&\ge(2+x)\int_0^{x/2} dt\left(\frac{1}{1+x+x^2/4}\right)
\qquad\text{since }t(x-t) \le x^2/4\\
&=\dfrac{2+x}{(1+x/2)^2}\int_0^{x/2} dt\\
&=\dfrac{x}{2}\dfrac{4(2+x)}{(2+x)^2}\\
&=\dfrac{2x}{x+2}\\
&=\dfrac{x}{1+x/2}\\
\end{array}
$
so that
$\dfrac1{x}\int_0^x \frac{dt}{1+t}
\ge \dfrac{1}{1+x/2}
$.
Note that
the only inequality used
is that
$t(x-t)
\le x^2/4$
for $0 \le t \le x/2$.
If we use
$t(x-t) \ge 0$,
we can get an inequality
going the other way:
$\begin{array}\\
\int_0^x \frac{dt}{1+t}
&=(2+x)\int_0^{x/2} dt\left(\frac{1}{1+x+t(x-t)}\right)\\
&\le(2+x)\int_0^{x/2} dt\left(\frac{1}{1+x}\right)\\
&=\dfrac{2+x}{1+x}\int_0^{x/2} dt\\
&=\dfrac{x(2+x)}{2(1+x)}\\
\end{array}
$
so that
$\ln(1+x)
\le \dfrac{x(2+x)}{2(1+x)}
$.
The difference of the two bounds is
$\begin{array}\\
\dfrac{x(2+x)}{2(1+x)}-\dfrac{2x}{x+2}
&=\dfrac{x(2+x)(x+2)-2x(2(1+x))}{2(1+x)(x+2)}\\
&=\dfrac{x(x^2+4x+4-(4+4x))}{2(1+x)(x+2)}\\
&=\dfrac{x^3}{2(1+x)(x+2)}\\
&\le \dfrac{x^3}{4}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
How many numbers smaller than $10^6$ contain exactly three '$9$'s and have an odd sum of digits? i came up with an idea that i choose
Even Even Even 9 9 9 - and i sort it in ways such that
all even are the same - $\frac{6!}{3!*3!}$
two are the same - $\frac{6!}{3!*2!}$
all are different - $\frac{6!}{3!}$
other one is that
Odd Odd Even 9 9 9
odd are the same - $\frac {6!}{3!*2!}$
all three are differnt - $\frac{6!} {3!}$
and my result is a sum of all this options.
Am i right?
| With your idea:
*
*Three even digits and the 9:
*
*The same even digit every time (5 choices: $0, 2, 4, 6, 8$ and permutation): $5 \cdot \frac{6!}{3! \cdot 3!}$
*Two even digits are the same: $5 \cdot 4 \cdot \frac{6!}{3! \cdot 2!}$
*The three even digits are different: $5 \cdot 4 \cdot 3 \cdot \frac{6!}{3!}$
*Two odd digits, one even (4 choices for the odd digit: $1, 3, 5, 7$):
*
*The same odd digit : $\underbrace{5}_{even} \cdot \underbrace{4}_{odd} \cdot \frac{6!}{3! \cdot 2!}$
*Odd digits are different: $\underbrace{5}_{even} \cdot \underbrace{4 \cdot 3}_{odd} \cdot \frac{6!}{3!}$
The answer is the sum of all cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2129311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Finding the area between a line and a curve
The two equations are $x+1$ and $4x-x^2-1$.
The answer is $\frac{1}{6}$, but I've done it 4 different times and gotten -$\frac{15}{2}$ each time.
My working:
*
*$x+1$ = $4x-x^2-1$
*$x^2-3x+2 = 0$
*$(x-1)(x-2)$ means $x=1$ or $x=2$
*$\int_1^2$ $3x-x^2$
*$[\frac{3x^2}{2}-\frac{x^3}{3}]_1^2$
*$\frac{3(1)^2}{2}-\frac{(1)^3}{3}$ = $\frac{3}{2}-\frac{1}{3}$
*$\frac{3(2)^2}{2}-\frac{(2)^3}{3}$ = $\frac{12}{2}-\frac{8}{3}$
*($\frac{3}{2}-\frac{1}{3}$)-($\frac{12}{2}-\frac{8}{3}$) = $-\frac{9}{2}-\frac{9}{3}$
*-$\frac{15}{2}$
| Hint: You need to be doing top curve minus bottom curve i.e;
$$(4x-x^2-1) - (x+1) = 3x-x^2-2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Mathematical Induction for an alternating series It's been a while since I have done a problem like this. I have this problem
$$\sum_{i=0}^n (-1)^{i+1} i^2 = \frac {(-1)^{n+1}n(n+1)}{2}$$
So I have gotten this far:
Base Case:
$$n=1$$
$$(-1)^2+1^2 = \frac{ (-1)^2 1(2) }{2}$$
$$ 1 = 1$$
Assume:
$$\sum_{i=0}^n (-1)^{i+1} i^2 = \frac {(-1)^{n+1}n(n+1)}{2}$$
Prove:
$n = n+1$
$$\sum_{i=0}^n+1 (-1)^ {i+1} i^2 = \frac {(-1)^{n+2}(n+1)(n+2)}{2} $$
I think that I can just replace the i's with (n+1) too
Which would be this
$$\sum_{i=0}^{n+1} (-1)^ {n+2} (n+1)^2 = \frac {(-1)^{n+2}(n+1)(n+2)}{2} $$
I don't remember where to go from here.
| If $n$ is even,
$$\begin{align}
\sum_{i=0}^n (-1)^{i+1}i^2&=-0^2+1^2-2^2+3^2-4^2+\cdots+(n-1)^2-n^2\\
&=(1-2)(1+2)+(3-4)(3+4)\cdots+((n-1)-n)((n-1)+n)\\\\
&=-(1+2+3+4+\cdots+(n-1)+n)\\
&=-\frac{n(n+1)}2\end{align}$$
If $n$ is odd,
$$\begin{align}
\sum_{i=0}^n (-1)^{i+1}i^2&=-0^2+1^2-2^2+3^2-4^2-\cdots-(n-1)^2+n^2\\
&=(-0+1)(0+1)+(-2+3)(2+3)+\cdots+(-(n-1)+n)((n-1)+n)\\\\
&=1+2+3+4+\cdots+(n-1)+n\\
&=\frac{n(n+1)}2\end{align}$$
Hence
$$\sum_{i=0}^n (-1)^{i+1}i^2=(-1)^{n+1}\frac {n(n+1)}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How do they got the solution for this $2\times 2$ matrix?
For $y'=\frac{1}{x}\begin{pmatrix}0 &1 \\ 2 & -1\end{pmatrix}y$ the solutions are given by $y_1=\begin{pmatrix}x \\ x\end{pmatrix},y_2=\begin{pmatrix}x^{-2} \\ -2x^{-2}\end{pmatrix}$ with $x\in \Bbb R,y\in \Bbb R^2$
I would like to know how they got this solutions.
I tried the following:$$det\frac{1}{x}\begin{pmatrix}0-\lambda &1 \\ 2 & -(1+\lambda)\end{pmatrix}=\frac{1}{x}(\lambda^2+\lambda-2)$$
$\Rightarrow \lambda_{1,2}=-\frac{1}{2}+/-\frac{3}{2}\Rightarrow\lambda_1=-2 , \lambda_2=1$
Than $$ker\begin{pmatrix}-1&1 \\ 2 & -2\end{pmatrix}\Rightarrow v_1=\begin{pmatrix}1 \\ 1\end{pmatrix}\Rightarrow y_1=e^x\begin{pmatrix}1 \\ 1\end{pmatrix}$$
also $$ker\begin{pmatrix}2&1 \\ 2 & 1\end{pmatrix}\Rightarrow v_1=\begin{pmatrix}1 \\ -2\end{pmatrix}\Rightarrow y_2=e^{-2x}\begin{pmatrix}1 \\ -2\end{pmatrix}$$
I guess this way is wrong
Edit:
$y_1'=\frac{1}{x}y_2 $ and $y_2'=\frac{1}{x}(2y_1-y_2)$
| Let $x = e^u$. Then we have
$$ \def\d#1#2{\frac{d#1}{d#2}}
\d yu = \d yx \cdot \d xu = y' \cdot x $$
Hence, your equation reads
$$ \d yu = \begin{pmatrix} 0 & 1 \\ 2 & -1 \end{pmatrix}y $$
Now solve this linear system as usual and resubstitue $e^u$ by $x$, giving the stated solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is $x^2 \equiv -1 \pmod{365}$ solvable? I know that a similar question already exists, but I have a different question to ask.
We want to examine if $x^2 \equiv -1 \pmod{365}$ has a solution.
My thought is: $365=5\cdot 73$. So,The congruence $x^2 \equiv -1 \pmod{365}$ has solution, if and only if, the congrueces $x^2 \equiv -1 \pmod 5$ and $x^2 \equiv -1 \pmod{73}$ has solutions. So, if we use Legendre's Symbol we have
*
*$x^2 \equiv -1 \pmod 5$ has solution $\iff (-1/5)=1 $ (and with simple calculations, indeed)
*$x^2 \equiv -1 \pmod{71}$ has solution $\iff (-1/73)=1 $
Now, can we conclude that the congruence $x^2 \equiv -1 \pmod{365}$ has solution?
And more general: If we have the congruence $x^2 \equiv a \pmod n$ with $n=p_1^{n_1}\cdots p_k^{n_k},\ \gcd(a,n)=1$, which is equivalent with the system $x^2 \equiv a {\pmod p_1^{n_1}},\ldots,x^2 \equiv a \pmod{p_k^{n_k}}$, can we conclude that the first has solution if and only if each one of $x^2\equiv a\pmod{p_i^{n_i}},\ \forall i=1,\ldots,k$ has solution?
Thank you.
| Yes. Let a solution to $x^2+1 \equiv 0 \pmod {5}$ be $r_{1}$, and let a solution to $x^2+1 \equiv 0 \pmod {73}$ be $r_{2}$
Then note that by CRT, we have that there exists such $x$ that $$x \equiv r_{1} \pmod {5}$$ $$x \equiv r_{2} \pmod {73}$$
Exists. Then note that for such $x$, $$x^2+1 \equiv 0 \pmod {5}$$ $$x^2+1 \equiv 0 \pmod {73}$$
Which gives $$x^2+1 \equiv 0 \pmod {365}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find the limit of the expression
Find
$$ \lim_{x\to\infty}\frac{\sqrt{1-\cos^3(1/x)}\cdot(3^{1/x}-5^{-1/x})}
{\log_2( 1+x^{-2} + x^{-3})}$$
Firstly I replace $x$ with $1/t$. Then $t$ tends to $0$. And then I fix numerator with formulas for limits considering $t \to 0$.
But I don't know what to do with denominator. Any help?
| Since it is the denominator term that is the source of concern, note that after the substitution $x=1/t$, we have
$$\begin{align}
\log_2(1+x^{-2}+x^{-3})&=\log_2(1+t^2+t^3)\\\\
&=\left(\frac{\log_2(1+t^2+t^3)}{t^2+t^3}\right)\,(t^2+t^3)\\\\
&=\left(\frac{\log(1+t^2+t^3)}{\log(2)\,(t^2+t^3)}\right)\,(t^2+t^3)
\end{align}$$
For the numerator, we see that
$$\begin{align}
\sqrt{1-\cos^3(1/x)}&=\sqrt{1-\cos^3(t)}\\\\
&=\sqrt{1-\cos(t)}\sqrt{1+\cos(t)+\cos^2(t)}\\\\
&=\sqrt{2\sin^2(t/2)}\sqrt{1+\cos(t)+\cos^2(t)}\\\\
&=\sqrt{2}\sin(t/2)\sqrt{1+\cos(t)+\cos^2(t)}
\end{align}$$
Additionally, we have
$3^{1/x}-5^{-1/x}=3^t\left(1-e^{-\log(15)t}\right)$$
Now, we can write
$$\begin{align}
\frac{\sqrt{1-\cos^3(1/x)}\left(3^{1/x}-5^{-1/x}\right)}{\log_2(1+x^{-2}+x^{-3})}&=\frac{\sqrt{1-\cos^3(t)}\left(3^{t}-5^{-t}\right)}{\log_2(1+t^{2}+t^{3})}\\\\
&=\sqrt{2}\sqrt{1+\cos(t)+\cos^2(t)}\frac{\sin(t/2)\left(3^{t}-5^{-t}\right)}{\log_2(1+t^{2}+t^{3})}\\\\
&=\color{blue}{\sqrt{2}(3^t)\sqrt{1+\cos(t)+\cos^2(t)}}\frac{\color{red}{\sin(t/2)}\color{green}{\left(1-e^{-\log(15)t}\right)}}{\color{purple}{\log_2(1+t^{2}+t^{3})}}\\\\
&=\color{blue}{\left(\sqrt{2}(3^t)\sqrt{1+\cos(t)+\cos^2(t)}\right)}\\\\
&\times \frac{\color{red}{\sin(t/2)}}{\color{orange}{t/2}}\\\\
&\times \frac{\color{gold}{t^2+t^3}}{\color{purple}{\frac{\log(1+t^2+t^3)}{\log(2)}}}\\\\
&\times \frac{\color{green}{1-e^{-\log(15)t}}}{\log(15)t}\\\\
&\times \color{orange}{\frac{t}{2}}\times {\log(15)t}\times\frac{1}{\color{gold}{t^2+t^3}}
\end{align}$$
Can you finish now?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Prove that $\frac{1}{2}\cdot \frac{3}{4} \cdots \frac{2n-1}{2n} \leq \frac{1}{\sqrt{3n+1}}$ without induction
Prove that $\frac{1}{2}\cdot \frac{3}{4} \cdots \frac{2n-1}{2n} \leq \frac{1}{\sqrt{3n+1}}$
I know this can be easily proved by induction. But I am looking for another approach. How do I prove this without induction?
Here this question exists - How does one prove that $\frac{1}{2}\cdot\frac{3}{4}\cdots \frac{2n-1}{2n}\leq \frac{1}{\sqrt{3n+1}}?$. But the only one solution there uses induction. But I am looking for solution other than induction.
| If we consider
$$a_n = \frac{(2n-1)!!}{(2n)!!} = \frac{1}{4^n}\binom{2n}{n}=\prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)\tag{1}$$
we have:
$$ a_n^2 = \frac{1}{4}\prod_{k=2}^{n}\left(1-\frac{1}{k}+\frac{1}{4k^2}\right) = \frac{1}{4n}\prod_{k=2}^{n}\left(1+\frac{1}{4k(k-1)}\right)\tag{2} $$
hence:
$$ 4n a_n^2 \leq \exp\sum_{k=2}^{n}\frac{1}{4k(k-1)} \leq \exp\left(\frac{1}{4}\right) \tag{3}$$
and:
$$ a_n \leq \sqrt{\frac{1}{4e^{-1/4}n}} \tag{4}$$
is a stronger inequality, since $4e^{-1/4}\approx 3+\frac{1}{9}$. No induction, just squaring and creative telescoping.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2136612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to prove in this trig problem i have to prove this
$$\frac{\cos 3x}{\sin 2x \sin 4x}+\frac{\cos 5x}{\sin 4x \sin 6x}+\frac{\cos 7x}{\sin 6x \sin 8x}+\frac{\cos 9x}{\sin 8x \sin 10x} = \frac{1}{2}\csc x(\csc 2x - \csc 10x)$$
i tried taking lcm but does not leads to anything. i believe i have to write numerator as something in terms of denominator which i do not know how. Help.
Thanks
| The right side looks like a result of a telescopic sum, so we'll try to prove
$$ \frac{\cos(nx)}{\sin((n-1)x)\sin((n+1)x)}=\frac{1}{2}\csc(x)(\csc((n-1)x)-\csc((n+1)x)) $$
Why this? It may not be true, but if it is, we have
$$\begin{align*}
& \frac{\cos 3x}{\sin 2x\sin 4x} + \frac{\cos 5x}{\sin 4x\sin 6x} +\frac{\cos 7x}{\sin 6x\sin8x} +\frac{\cos 9x}{\sin 8x\sin10x} \\
=& \frac{1}{2}\csc x\left(\csc 2x - \csc 4x + \csc 4x - \csc 6x + \csc 6x - \csc 8x + \csc 8x - \csc 10x \right) \\
=& \frac{1}{2}\csc x (\csc 2x-\csc 10x) \\
\end{align*}$$
So, we'll start operating the right side. We have
$$\begin{align*}
& \frac{1}{2}\csc(x)(\csc((n-1)x)-\csc((n+1)x)) \\
=& \frac{1}{2\sin (x)}\left( \frac{1}{\sin((n-1)x)}-\frac{1}{\sin((n+1)x)} \right) \\
=& \frac{1}{2\sin(x)}\cdot \frac{\sin((n+1)x)-\sin((n-1)x)}{\sin((n-1)x)\sin((n+1)x)} \\
=& \frac{\sin((n+1)x)-\sin((n-1)x)}{2\sin(x)\sin((n-1)x)\sin((n+1)x)} \\
=& \frac{(\sin(nx)\cos(x)+\sin(x)\cos(nx))-(\sin(nx)\cos(x)-\sin(x)\cos(nx))}{2\sin(x)\sin((n-1)x)\sin((n+1)x)} \\
=& \frac{2\sin(x)\cos(nx)}{2\sin(x)\sin((n-1)x)\sin((n+1)x)} \\
=& \frac{\cos(nx)}{\sin((n-1)x)\sin((n+1)x)}
\end{align*}$$
And we're ready
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$3$ numbers $x,y,z$ pairwise relatively prime, sum of any two is multiple of third, find $abc$
Three distinct positive integers $x,y,z$ are pairwise relatively prime, and the sum of any two is a multiple of the third one. Find $xyz$.
I don't know how to start.
| Suppose WLOG $x < y < z$. Then $z \mid x + y < 2 z$, and there are no other multiples of $z$ between $z$ and $2 z$. It follows that $x + y = z$.
Now $x \mid y + z = x + 2 y$, so $x \mid 2 y$, and since $\gcd(x, y) = 1$, we have $x \mid 2$, and thus $x = 1$ or $x = 2$.
Similarly $y \mid x + z = 2 x + y$, so as above $y = 1$ or $y = 2$. Since $x < y$, we must have $x = 1$, $y = 2$, and thus $z = x + y = 3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality trouble: $(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3$ The following inequality is exercise 1.8 from this book.
For any real $a,b,c$, prove the following $$(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3.$$
I've managed to prove this via brut-force and Muirhead's inequality (Very unsatisfying). However I'm having difficulty understanding the solution in the back of the book. I'm also interested if anyone else knows of any nice proofs of this inequality?
In particular, the solution at the back says the following (word for word):
If we can show that $\frac{27}{64}(a+b)^2(b+c)^2(c+a)^2\geq (ab+bc+ca)^2$, then the conclusion follows. Denote $S_1=a+b+c$, $S_2=ab+bc+ca$ and $S_3=abc$. We need to show that $27(S_1S_2-S_3)^2\geq 64S_2^3.$
It then goes onto prove the last inequality via cases. However I'm struggling to see how it is sufficient to prove either of those two inequalities.
We can rewrite $a^2+ab+b^2 = (\frac{a+b}{2})^2+\frac{3}{4}(a^2+b^2)$ and so by AM-GM it is sufficient to prove $$\sqrt{\frac{27}{64}(a+b)^2(b+c)^2(c+a)^2}\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)} \geq (ab+bc+ca)^3 .$$ However $$\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)} \geq (ab+bc+ca)^2 $$ is clearly wrong and so this does not justify the first claim in the solution. This does give me an idea of where the $\frac{27}{64}$ term comes from however. Other than this I'm a bit stuck and any help would be appreciated.
Edit: I realised straight after posting that $a^2+ab+b^2 \neq (\frac{a+b}{2})^2+\frac{3}{4}(a^2+b^2)$. So in fact my work is nonsense and can be ignored.
| We use the so called uvw method for proving inequality involving symmetric polynomials of $3$ variables.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Hence, our inequality it's $kw^6+A(u,v^2)w^3+B(u,v^2)\geq0$,
where $A$ and $B$ polynomials such that $A(0,0)=B(0,0)=0$.
Let $\theta^3=1$, where $\theta\neq1$ and $(a,b,c)=(1,\theta,\theta^2)$.
Thus, $a^2+ab+b^2=a^2+ac+c^2=b^2+bc+c^2=u^2=v^2=0$, but $w^3\neq0$,
which says that $k=0$.
Thus, our inequality is a linear inequality of $w^3$,
which says that it's enough to prove our inequality for an extremal value of $w^3$,
which happens for equality case of two variables.
Since our inequality is homogeneous and even degree, we can assume $b=c=1$,
which gives $$3(a^2+a+1)^2\geq(2a+1)^3$$ or
$$(a-1)^2(3a^2+4a+2)\geq0,$$
which is obvious.
Done!
| {
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Prove that this is an abelian group
$G$ is a group. $f:\:G\to G,\:a\mapsto a^3$ is a group monomorphism. Prove that $G$ is abelian.
My idea is that it's sufficient to prove that $(aba^{-1}b^{-1})^3=e$. So
$$
(aba^{-1}b^{-1})^3=a^3b^3a^{-3}b^{-3}
$$
and I am stuck. Or
$$
(aba^{-1}b^{-1})^3=a^3(ba^{-1}b^{-1})^3=a^3ba^{-3}b^{-1}
$$
but this kind of effort does not seem to be of any help.
| Here's my version (revised) . . .
Since the map $\:x\mapsto x^3$ is a monomorphism,
\begin{align*}
&\hspace{1pt}(ab)^3 = a^3b^3,\text{ for all }a,b \in G.\tag{1}\\[6pt]
&\hspace{1pt}\text{If }a,b \in G,\text{ and }a^3 = b^3,\text{ then }a = b.\tag{2}\\
\end{align*}
Let $a,b \in G.$
\begin{align*}
\text{Then}&&a^3b^3a^{-3}&=(aba^{-1})^3&&[\text{by} (1)]\\
&&&= (aba^{-1})(aba^{-1})(aba^{-1})&&\\
&&&= ab^3a^{-1}&&\\[8pt]
\text{But then}&&a^3b^3a^{-3} &= ab^3a^{-1}&&\\
\implies&& a^2b^3&=b^3a^2&&\\[8pt]
\text{So we have}&&a^2b^3&=b^3a^2,\,\text{ for all }a,b \in G.&&\tag{3}\\
&&&\text{(thus, squares commute}&&\\
&&&\text{ with cubes)}&&\\[8pt]
\text{Then}&&(a^2b)^3 &= a^6b^3&&[\text{by} (1)]\\
&&&= b^3a^6&&[\text{by} (3)]\\
&&&= (ba^2)^3&&[\text{by} (1)]\\[8pt]
\text{But then}&&(a^2b)^3 &= (ba^2)^3&&\\
\implies&& a^2b&=ba^2&&[\text{by} (2)]\\[8pt]
\text{So we have}&&a^2b&=ba^2,\,\text{ for all }a,b \in G.&&\tag{4}\\
&&&\text{(thus, squares commute}&&\\
&&&\text{ with everything)}&&\\[8pt]
\text{Then}&&(ab)^6 &=((ab)^2)^3&&\\
&&&= ((ab)(ab))^3&&\\
&&&= (ab)^3(ab)^3&&[\text{by} (1)]\\
&&&= a^3b^3a^3b^3&&[\text{by} (1)]\\
&&&=a(a^2)b^3a^3(b^2)b&&\\
&&&=ab^5a^5b&&[\text{by} (4)]\\[8pt]
\text{But then}&&(ab)^6 &= ab^5a^5b&&\\
\implies&& a(ba)^5b &=ab^5a^5b&&\\
\implies&& (ba)^5 &=b^5a^5&&\\
\implies&& (ba)^3(ba)^2 &=b^5a^5&&\\
\implies&& b^3a^3(ba)^2 &=b^5a^5&&[\text{by} (1)]\\
\implies&& a^3(ba)^2 &=b^2a^5&&\\
\implies&& a^3(ba)^2 &=a^5b^2&&[\text{by} (4)]\\
\implies&& (ba)^2 &=a^2b^2&&\\
\implies&& (ba)^2 &=b^2a^2&&[\text{by} (4)]\\
\implies&& baba &=b^2a^2&&\\
\implies&& ab &=ba&&\\[8pt]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Proving a summation result using strong induction I was recently taught strong induction in class and overall I'm still learning the topic. I am currently stuck on a problem and I don't even know where to begin with it. A starting place or hint would be great.
| Fix $k$ and apply induction on $n$. So, let us assume that $$\sum_{j=1}^\color{blue}nj(j+1)(j+2)\dots(j+k-1)=\frac{\color{blue}n(\color{blue}n+1)(\color{blue}n+2)\dots(\color{blue}n+k)}{k+1}.$$ Then
$$\begin{align}
&\sum_{j=1}^{\color{red}{n+1}}j(j+1)(j+2)\dots(j+k-1)\\
&=[(n+1)(n+2)(n+3)\dots(n+1+k-1)]+\sum_{j=1}^nj(j+1)(j+2)\dots(j+k-1)\\
&=[(n+1)(n+2)(n+3)\dots(n+k)]+\sum_{j=1}^nj(j+1)(j+2)\dots(j+k-1)\\
&=\overbrace{[(n+1)(n+2)(n+3)\dots(n+k)]}+\frac{n\overbrace{(n+1)(n+2)(n+3)\dots(n+k)}}{k+1}\\
&=(n+1)(n+2)(n+3)\dots(n+k)\cdot \left[1+\frac{n}{k+1}\right]\\
&=(n+1)(n+2)(n+3)\dots(n+k)\cdot \left[\frac{k+1+n}{k+1}\right]\\
&=(n+1)(n+2)(n+3)\dots(n+k)\cdot \left[\frac{n+k+1}{k+1}\right]\\
&=\frac{(n+1)(n+2)(n+3)\dots(n+k)(n+k+1)}{k+1}\\
&=\frac{\color{red}{(n+1)}\cdot[\color{red}{(n+1)}+1)]\cdot[\color{red}{(n+1)}+2]\dots[\color{red}{(n+1)}+k]}{k+1}
\end{align}$$
Hope this help.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute the following without the calculator
$$4\left(5+3\sqrt2\over 2\right)^4-16\left(5+3\sqrt2\over 2\right)^3-17\left(5+3\sqrt2\over 2\right)^2+27\left(5+3\sqrt2\over 2\right)-3$$
Please solve the following equation without using calculator.
Substituting $\left(5+3\sqrt2\over 2\right)$ to x must be the first step, but then I don't know how to factor it.
| Let $\alpha=\displaystyle\frac{5+\sqrt{3}}2, \beta=\displaystyle\frac{5-\sqrt{3}}2$.
$$\alpha+\beta=5; \alpha\beta=\frac 74$$
$\alpha,\beta$ are roots of the quadratic
$$x^2-(\alpha+\beta)+\alpha\beta=0\\
x^2-5x+\frac 74=0\\
4x^2-20x+7=0$$
i.e.
$$\overbrace{4\alpha^2-20\alpha+7}^{f(\alpha)}=0$$
Now consider the following:
$$\begin{array}
&&&&&\\
\alpha^2\cdot f(\alpha):&4\alpha^4&-2 0\alpha^3 &+7\alpha^2 & & &=0\\
\alpha\cdot f(\alpha): & &\;\;\;4\alpha^3&-20\alpha^2 &+7\alpha & &=0\\
- f(\alpha): & & &-4\alpha^2 &+20\alpha&-7&=0\\
+4: & & & & &+4&=4\\
\hline
\text{Adding}: &4\alpha^4&-16\alpha^3 &-17\alpha^2 &+27\alpha &-3&=\color{red}4\\
\hline
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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How to use powers on matrices In the questions compute $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6$ and $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{99}$, how would you solve these?
| $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}= 2 \begin{pmatrix} \cos \frac {\pi}{6} & -\sin\frac {\pi}{6} \\ \sin\frac {\pi}{6} & \cos \frac {\pi}{6} \end{pmatrix}$
$\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6= 2^6 \begin{pmatrix} \cos \frac {\pi}{6} & -\sin\frac {\pi}{6} \\ \sin\frac {\pi}{6} & \cos \frac {\pi}{6} \end{pmatrix}^6$
It is worth the exercise to see what happens when you multiply matrices that can be put into this form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2148061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$ Given that $a, b$ and $c$ are the sides of a triangle.
How to prove that $a^2 + b^2 + c^2 \lt 2(ab + bc + ca)$?
Maybe any hint? Am I going to wrong direction?
$$2(ab + bc + ca)-a^2 + b^2 + c^2>0$$
$$2ab + 2bc + 2ca-a^2 + b^2 + c^2>0$$
$$2b(a+c) + 2ca-a^2 + b^2 + c^2>0$$
...?
| Note that three sides $a,b,c$ in a triangle satisfy
$$ a<b+c, b<a+c, c<a+b. $$
So one has
$$ a^2<a(b+c), b^2<b(a+c), c^2<c(a+b). $$
Adding these three inequalities gives
$$ a^2+b^2+c^2<2ab+2bc+2ca. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Evaluating $\prod_{r=1}^{\infty} \frac{\sin \frac {a}{2^r}}{\tan^2 \frac {a}{2^r} \tan \frac {a}{2^{r-1}}+\tan \frac {a}{2^{r}}}$
The question is to evaluate $$\prod_{r=1}^{\infty} \frac{\sin \frac {a}{2^r}}{\tan^2 \frac {a}{2^r} \tan \frac {a}{2^{r-1}}+\tan \frac {a}{2^{r}}}$$
I could rewrite the denominator as $$\tan\frac {a}{2^r}\left(\tan \frac {a}{2^r}\tan \frac{a}{2^{r-1}}+1\right)$$which is same as $$\tan \frac{a}{2^r}-\tan\frac {a}{2^{r-1}}$$And so the product transforms to $$\prod_{r=1}^{\infty} \frac{\sin \frac {a}{2^r}}{\tan \frac{a}{2^r}-\tan\frac {a}{2^{r-1}}}$$ I have no idea on how to proceed after this. Any help is appreciated. Thanks.
| Lets start with:
$\tan \frac {a}{2^{r-1}} = \frac {2\tan \frac {a}{2^{r}}}{1-\tan^2 \frac {a}{2^{r}}}$
and that gets everything in terms of $\frac {a}{2^{r}}$
$u = \frac {a}{2^{r}}$
$\frac {\sin u}{\tan^2 u\frac {2 \tan u}{1-\tan^2 u}+\tan u}\\
\frac {\sin u (1-\tan^2 u)}{\tan u(1 +\tan^2 u)}\\
\frac {\sin u (1-\tan^2 u)}{\tan u(\sec^2 u)}\\
\cos^2 \cot u\sin u(1-\tan^2 u)\\
\cos u (\cos^2u-\sin^2 u)\\
\cos u (\cos 2u)$
$\prod_\limits{r=1}^{\infty} \cos \frac {a}{2^{r}} \cos \frac {a}{2^{r-1}} = \cos a\prod_\limits{r=1}^{\infty} \cos^2 \frac {a}{2^{r}}\\
\frac {\sin^2 a}{a^2} \cos {a}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can I find all the matrices that commute with this matrix? I would like to find all the matrices that commute with the following matrix
$$A = \begin{pmatrix}2&0&0\\ \:0&2&0\\ \:0&0&3\end{pmatrix}$$
I set $AX = XA$, but still can't find the solutions from the equations.
| >>> from sympy import *
>>> A = diag(2,2,3)
>>> X = MatrixSymbol('X',3,3)
>>> Matrix(A*X - X*A)
Matrix([
[ 0, 0, -X[0, 2]],
[ 0, 0, -X[1, 2]],
[X[2, 0], X[2, 1], 0]])
If $\rm A X = X A$, then $x_{13} = x_{23} = x_{31} = x_{32} = 0$. The other five entries are unconstrained.
We can also vectorize $\rm A X = X A$, which yields the following homogeneous linear system
$$\left( (\mathrm I_3 \otimes \mathrm A) - (\mathrm A \otimes \mathrm I_3) \right) \mbox{vec} (\mathrm X) = 0_9$$
or,
$$\begin{bmatrix}0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{bmatrix} \mbox{vec} (\mathrm X) = \begin{bmatrix} 0\\0\\0\\0\\0\\0\\0\\0\\0\end{bmatrix}$$
Again, we conclude that $x_{13} = x_{23} = x_{31} = x_{32} = 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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How to find all solutions for $x^3=6x+6$ Could anyone help me to find how to find all solution for $x^3=6x+6$?
| Use Cardano's method:
Set $y=u+v$. As you have two unknowns instead of one, you can add a constraint on $u,v$, in order to simplify the equation. This equation becomes
$$(u+v)^3-6(u+v)-6=u^3+v^3+(u+v)(3uv-6)-6=0.$$
Adding the condition $3uv=6$, i.e. $uv=2$, you obtain the equation $u^3+v^3-6=0$, whence the system
$$\begin{cases}u^3+v^3=6\\u^3v^3=8\end{cases}\iff u^3, v^3\;\text{ are the roots of the quadratic equation}\quad t^2-6t+8=0.$$
Now $t^2-6t+8=(t-3)^2-9+8$, so $u^3, v^3=3\pm1=2,4$ and finally
$$u,v=\sqrt[3]{2},\; \sqrt[3]{4},\enspace\text{whence}\quad y=\sqrt[3]{2}+\sqrt[3]{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$ if :
$$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$$
then $a+b=?$
Without the use of the L'Hôspital's Rule
My Try :
$$\lim_{ x \to 0 }\left( \frac{ax+bx^3+\sin 3x}{x^3} \right)=0$$
$$\lim_{ x \to 0 }\left( \frac{x(a+bx^2)+\sin 3x}{x^3} \right)=0$$
$$\lim_{ x \to 0 }x(a+bx^2)+\sin 3x=0 $$
now ?
| We need to use the limit $$\lim_{x\to 0}\frac{x-\sin x} {x^{3}}=\frac{1}{6}\tag{1}$$ which can be easily established either via L'Hospital's Rule or Taylor's theorem. Now using $(1)$ we have $$\frac{\sin 3x - 3x}{x^{3}}\to - \frac{9}{2}\tag{2}$$ and hence $$\frac{\sin 3x}{x^{3}}+\frac{a}{x^{2}}+b\to 0\tag{3}$$ is equivalent to (via subtracting equation $(2)$ from equation $(3)$) $$\frac{3+a}{x^{2}}+b\to \frac{9}{2}\tag{4}$$ or $$\frac{3+a+bx^{2}}{x^{2}}\to \frac{9}{2}\tag{5}$$ It is now clear that $3+a+bx^{2}\to 0$ so that $a=-3$. And then $b=9/2$ follows from the penultimate equation. Note that everything apart from limit $(1)$ is based on algebra of limits.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2160584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Proof that expression is positive if conditions is met Is there a way to show that this expression is always positive as long as $b>0$ and $r>x$ ? Assume $r>0$ and $x>0$
\begin{equation}
b-1+\sqrt{(1+b)^2-\frac{4rb}{x}}
\end{equation}
It's simple with $b>1$ but I can't figure it out if it is true for lower values.
| This is false. Consider $b = \frac{1}{10}, x = 1, r = \frac32$. Clearly, we have $b,x,r>0$ and $r>x$.
$$\sqrt{(1+b)^2-\frac{4rb}{x}} = \sqrt{\left(1+\frac{1}{10}\right)^2-\frac{4(3/2)(1/10)}{1}} = \sqrt\frac{121-60}{100} = \frac{\sqrt{61}}{10}$$
$$b-1+\sqrt{(1+b)^2-\frac{4rb}{x}} = -\frac9{10} + \frac{\sqrt{61}}{10} = \frac{\sqrt{61}-\sqrt{81}}{10} < 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the sequence of Fibonacci quotients is Cauchy Consider the sequence defined by $a_1 = 1$ and $a_n =\frac{1}{1 + a_{n−1}}$
for all $n ≥ 2$. In general, this sequence can also be described as the sequence
of quotients of succesive terms in the Fibonacci sequence.
Prove that $(a_n)_{n\in \Bbb N}$ is a Cauchy sequence.
| Given the Fibonacci sequence where $a_{n+1} = a_n + a_{n-1}$, we have with $b_n = a_n/a_{n-1},$
$$\begin{align} |b_{n+1} - b_{n}| &= \left|\frac{a_{n+1}}{a_n}- \frac{a_{n}}{a_{n-1}} \right| \\ &= \left|\frac{a_{n+1}a_{n-1} - a_n^2}{a_n a_{n-1}} \right| \\ &= \left|\frac{a_na_{n-1} + a_{n-1.}^2 - a_n a_{n-1} - a_n a_{n-2}}{a_{n-1}^2 + a_{n-1}a_{n-2}} \right|. \end{align}$$
Note that the sequence is increasing and $a_{n-1}^2 + a_{n-1}a_{n-2} > 2 a_{n-1}a_{n-2}.$
Hence,
$$\begin{align} |b_{n+1} - b_{n}| &< \left|\frac{a_{n-1}^2 - a_{n}a_{n-2}}{2 a_{n-1}a_{n-2}}\right| \\ &= \frac{1}{2}\left|\frac{a_n}{a_{n-1}} - \frac{a_{n-1}}{a_{n-2}}\right| \end{align}.$$
By induction we can show that
$$|b_{n+1} - b_{n}| < \left(\frac{1}{2} \right)^{n-2}\left|\frac{a_2}{a_1} - \frac{a_1}{a_0} \right| = \left(\frac{1}{2} \right)^{n-2},$$
and
$$|b_{n+k} - b_{n}| < \sum_{j=0}^{k-1} (1/2)^{n-2 +j} = (1/2)^{n-2} \frac{1 - (1/2)^k}{1 - 1/2} < (1/2)^{n-3}.$$
Since the RHS converges to $0$ as $n \to \infty$ the sequence is Cauchy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rule for multiplying rational numbers (follow-up) I was thinking about the rule where
$\space\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$ and why it worked, and I found this question: Rule for multiplying rational numbers.
The best answer was:
Did:
One knows that $a/b$ may be defined as the number $x$ such that
$bx=a$, and that $c/d$ may be defined as the number $y$ such that
$dy=c$.
If one wants the multiplication on these objects to be associative and
commutative as it is on the integers, one should ask that
$ac=(bx)(dy)=(bd)(xy)$ hence that the object $xy$ fits the definition
of $(ac)/(bd)$.
And I was thinking, if in the end we get $\frac{(xy)(bd)}{bd}=\frac{ac}{bd}$, how do we know how to simplify the two $(bd)$s here? Don't we need the property itself $\space\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$ to know how to do that?
$$\frac{(xy)(bd)}{bd}=\frac{(xy)\cdot (bd)}{1\cdot (bd)}=\frac{xy}{1}\cdot\frac{bd}{bd}=\frac{xy}{1}\cdot 1=\frac{xy}{1}=xy$$
So do we just assume the property is true inductively (not mathematically inductively), or am I missing something?
PS: As a beginner, I may say some ridiculous things, excuse me for that.
| A completely formal proof depends on the particular axioms and definitions you work with, but as far as the intuition of it goes, consider the following points.
*
*The fraction $\frac{1}{b}$ denotes the multiplicative inverse of $b \ne 0\,$, which is to say that $x=\frac{1}{b}$ is the unique solution to $b \cdot x = 1\,$. The fraction $\frac{a}{b}$ can be thought of as shorthand for $a \cdot \frac{1}{b}\,$.
*Multiplying together $b \cdot \frac{1}{b}=1$ and $d \cdot \frac{1}{d}=1$ gives $b \cdot \frac{1}{b} \cdot d \cdot \frac{1}{d}=1\,$. Since multiplication is commutative, this can be rewritten as $(b \cdot d) \cdot (\frac{1}{b} \cdot \frac{1}{d})=1\,$. But the latter means that $\frac{1}{b} \cdot \frac{1}{d}$ is the multiplicative inverse of $b\cdot d\,$, which proves that $\frac{1}{b} \cdot \frac{1}{d} = \frac{1}{b \,\cdot\, d}\,$.
Therefore $\;\frac{a}{b}\cdot \frac{c}{d} \stackrel{\;(1)\;}{=} a \cdot \frac{1}{b} \cdot c \cdot \frac{1}{d} = (a \cdot c) \cdot (\frac{1}{b} \cdot \frac{1}{d}) \stackrel{\;(2)\;}{=} (a \cdot c) \cdot \frac{1}{b \,\cdot\, d} \stackrel{\;(1)\;}{=} \frac{a \,\cdot c}{b \,\cdot\, d}\,$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that the sum of pythagorean triples is always even Problem: Given $a^2 + b^2 = c^2$ show $a + b + c$ is always even
My Attempt, Case by case analysis:
Case 1: a is odd, b is odd. From the first equation,
$odd^2 + odd^2 = c^2$
$odd + odd = c^2 \implies c^2 = even$
Squaring a number does not change its congruence mod 2.
Therefore c is even
$ a + b + c = odd + odd + even = even$
Case 2: a is even, b is even. Similar to above
$even^2 + even^2 = c^2 \implies c$ is even
$a + b + c = even + even + even = even$
Case 3:
One of a and b is odd, the other is even
Without loss of generality, we label a as odd, and b as even
$odd^2 + even^2 = c^2 \implies odd + even = c^2 = odd$
Therefore c is odd
$a + b + c = odd + even + odd = even$
We have exhausted every possible case, and each shows $a + b + c$ is even. QED
Follow Up:
Is there a proof that doesn't rely on case by case analysis?
Can the above be written in a simpler way?
| $c^2 = a^2 + b^2 = (a+b)^2 - 2ab$.
$2ab = (a+b)^2 - c^2$
$2ab = (a+b+c)(a+b-c)$
Let $n = a+b+c$, and the above becomes:
$2ab = n(n-2c)$
So the right-hand side must be even, but since $n-2c$ is odd when $n$ is odd, $n$ must be even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.