Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
A sum of series with the inverse squared central binomial coefficient A nice challenge by Cornel Valean:
Show that
$$2\sum _{n=1}^{\infty }\frac{2^{4 n}}{\displaystyle n^3 \binom{2 n}{n}^2}-\sum _{n=1}^{\infty }\frac{2^{4 n}}{\displaystyle n^4 \binom{2 n}{n}^2}+\sum _{n=1}^{\infty }\frac{2^{4 n} H_n^{(2)}}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}=\frac{\pi^3}{3}.$$
I have to say that I am not experienced in series involving squared central binomial coefficient, so I leave it for people who are experts in such series.
All approaches are appreciated. Thank you.
| An excellent answer was already given (the chosen one), but good to have more ways in place.
A solution by Cornel Ioan Valean
Instead of calculating all three series separately, we might try to calculate them all at once. So, we have that
$$2\sum _{n=1}^{\infty }\frac{2^{4 n}}{\displaystyle n^3 \binom{2 n}{n}^2}-\sum _{n=1}^{\infty }\frac{2^{4 n}}{\displaystyle n^4 \binom{2 n}{n}^2}+\sum _{n=1}^{\infty }\frac{2^{4 n} H_n^{(2)}}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}$$
$$=\sum _{n=1}^{\infty }\frac{2^{4n} (4n^2-1+n^2 H_n^{(2)})}{\displaystyle n^4 (2 n+1) \binom{2 n}{n}^2}=\sum _{n=1}^{\infty }\frac{2^{4n} (4-1/n^2+ H_n^{(2)})}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}$$
$$=\sum _{n=1}^{\infty }\frac{2^{4n}(4-1/n^2+ H_n^{(2)}\color{blue}{+(4 n^2-1) H_{n-1}^{(2)}}-\color{blue}{(4 n^2-1) H_{n-1}^{(2)}})}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}$$
$$=\sum _{n=1}^{\infty }\frac{2^{4n}(\color{red}{4n^2H_n^{(2)}}-\color{blue}{(4 n^2-1) H_{n-1}^{(2)}})}{\displaystyle n^2 (2 n+1) \binom{2 n}{n}^2}$$
$$=\sum _{n=1}^{\infty}\left(\frac{2^{4n+2}H_n^{(2)}}{\displaystyle (2n+1) \binom{2 n}{n}^2}-\frac{2^{4n}(2n-1)H_{n-1}^{(2)} }{\displaystyle n^2\binom{2 n}{n}^2}\right)$$
$$=\lim_{N\to\infty}\sum _{n=1}^{N}\left(\frac{2^{4n+3}H_n^{(2)}}{\displaystyle (n+1) \binom{2 n+2}{n+1}\binom{2 n}{n}}-\frac{2^{4n-1}H_{n-1}^{(2)} }{\displaystyle n\binom{2 n}{n}\binom{2 n-2}{n-1}}\right)$$
$$=\lim_{N\to\infty}\frac{2^{4N+3}H_N^{(2)}}{\displaystyle (N+1) \binom{2 N+2}{N+1}\binom{2 N}{N}}=\frac{\pi^3}{3},$$
where we used the asymptotic form of the central binomial coefficient, $\displaystyle \binom{2 N}{N}\sim \frac{4^N}{\sqrt{\pi N}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Prove that this number is divisible by 7 Without using induction, how can it be proved that 7 divides $3^{2n+1}+2^{n+2}$ for each $n\in\mathbb{N}$?
I tried to expand it using $\frac{x^{n+1}-1}{x-1}=1+x+..+x^n$ but I had no success.
It would be great if more than one proof is provided.
| \begin{eqnarray*}
\sum_{n=0}^{\infty} (3^{2n+1}+2^{n+2})x^n = \frac{3}{1-9x}+\frac{4}{1-2x} = \frac{ \color{red}{7} (1-6x)}{(1-9x)(1-2x)}.
\end{eqnarray*}
This function clearly has integer coefficients
\begin{eqnarray*}
\frac{ (1-6x)}{(1-9x)(1-2x)}=(1-6x) \left( 1 +9x+81x^2+ \cdots \right) \left( 1 +2x+4x^2+ \cdots \right).
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$(a+1)(b+1)(c+1)\leq4$ for triangle sides $a,b,c$ with $ab+bc+ac=1$
Given that $a,b,c$ are the lengths of the three sides of a triangle, and $ab+bc+ac=1$, the question is to prove $$(a+1)(b+1)(c+1)\leq4\,.$$
Any idea or hint would be appreciated.
This is Problem 6 of Round 1 of the BMO (British Mathematical Olympiad) 2010/2011, as can be seen here.
Remark. This question has been self-answered. Nevertheless, any new approach is always welcome!
| OK first let's expand the bracket
$(a+1)(b+1)(c+1)=abc+ab+ac+bc+a+b+c+1$.
Now we know that $ab+ac+bc=1$ so we actually need $abc+a+b+c+1 \leq 3$ or $abc+a+b+c \leq{2}$.
Since $a,b$ and $c$ form the sides of a triangle, we know that $a \leq b+c$ and $b \leq a+c$ and $c \leq a+b$.
I found it hard to progress from here and wondered if the result was actually true so did a thought experiment. Let us say $a,b$ and $c$ are all equal to $1/\sqrt{3}$. This would be an equilateral triangle and $ab+bc+ac=1/3+1/3+1/3=1$.
Then $(a+1)(b+1)(c+1)=abc+ab+ac+bc+a+b+c+1$=
$1/3 \sqrt{3}+1/3+1/3+1/3+1/\sqrt{3}+1/\sqrt{3}+1/\sqrt{3}+1=$
$1/3\sqrt{3}+1+\sqrt{3}+1$.
Which needs to be $\leq{4}$
Iff $1/3\sqrt{3} +\sqrt{3} \leq2$
iff $1/3+3 \leq 2\sqrt{3}$. Which is true.
Let's take another extreme case: $a$ and $b$ are just under $1$ and $c$ is close to $0$ then we can also have $ab+ac+bc=1$. Here $(a+1)(b+1)(c+1)$ will also be just under $4$ so I believe that the inequality is correct. I can show that we need $abc+a+b+c \leq{2}$ but don't know how to do that right now. I'll think about. But we haven't yet used the triangle inequalities so I suspect they are needed.
Not being able to finish it is killing me :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Check the convergence of the series $\displaystyle{\sum_{n=1}^{+\infty}\frac{\left (n!\right )^2}{\left (2n+1\right )!}4^n}$ I want to check if the following series converge or not.
*
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{\left (n!\right )^2}{\left (2n+1\right )!}4^n}$
I suppose we have to find here an upper bound and apply then the comparison test. But I don’t really have an idea which bound we could take. Could you give me a hint?
*
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2n}}$
We have a term that is a product of the form $\frac{2i-1}{2i}=1-\frac{1}{2i}$. To apply the comparison test we have to find an upper bound. Does it holds that $1-\frac{1}{2i}\leq \frac{1}{2}$ and so $$\prod_{i=1}^n\left (1-\frac{1}{2i}\right )\leq \prod_{i=1}^n \frac{1}{2}=\frac{1}{2^n}$$ Then taking the sum we get $$\sum_{n=1}^{+\infty}\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2n}\leq \sum_{n=1}^{+\infty} \frac{1}{2^n}=1$$ So from the comparison test the original sum must converge also.
Is everything correct?
*
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2\cdot 4\cdot 6\cdot \ldots \cdot 2n\cdot (2n+2)}}$
We have a term that is a product of the form $\frac{2i-1}{2i+2}$. Which upper bound could we use in this case?
| Some hints:
For first we can use Raabe's test
$$n\left(\frac{a_n}{a_{n+1}}-1 \right) = \frac{n}{2(n+1)}$$
For second
$$\frac{1}{2\sqrt{n}} \leqslant \frac{1}{2} \frac{3}{4} \cdots \frac{2n-1}{2n} \leqslant \frac{1}{\sqrt{2n}}\quad (1)$$
Proof:
For $n=1$ we have $\frac{1}{2} \leqslant \frac{1}{2} \leqslant \frac{1}{\sqrt{2}} $, so let's assume $n \geqslant 2$. We have
$$\frac{3}{4}>\frac{2}{3}, \frac{5}{6}>\frac{4}{5},\frac{7}{8}>\frac{6}{7}, \cdots, \frac{2n-1}{2n}>\frac{2n-2}{2n-1}$$
multiplication this inequalities gives
$$\frac{3}{4} \frac{5}{6} \cdots \frac{2n-1}{2n} > \frac{2}{3} \frac{4}{5} \cdots \frac{2n-2}{2n-1}$$
Now if we multiply left and right sides on left side, the we have
$$\left( \frac{3}{4} \frac{5}{6} \cdots \frac{2n-1}{2n} \right)^2 > \frac{1}{n} $$
Which is left side of (1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
The solution of the indefinite integral contains an error (I know correct answer), but I cannot find it There is an exercise on indefinite integral in some infinitesimal calculus book:
$$
\int \sqrt{x^{2} +1} \cdot dx
$$
The solution uses the first substitution x = sinh u and after some transformations the book gets the result:
$$
\frac{1}{2} \cdot \left( x\sqrt{x^{2} +1} +\ln\left( x+\sqrt{x^{2} +1}\right)\right) +C
$$
MAXIMA shows me a result:
$$
\[\frac{\operatorname{asinh}(x)}{2}+\frac{x\ \sqrt{{{x}^{2}}+1}}{2}\]
$$
It seems to be same, but the question is: the book says that there is an another solution via substitution with tan. I tried to solve it with the tan-substitution and got very different result. Please, show me where is my error:
$$
\int \sqrt{x^{2} +1} \cdot dx=\int \sqrt{\tan^{2} a +1} \cdot \frac{da}{\cos^{2} a} =\int \sqrt{\frac{1}{\cos^{2} a}} \cdot \frac{da}{\cos^{2} a} =\int \frac{da\cdot \cos a}{\cos^{4} a} =\int \frac{d(\sin a)}{\left(\cos^{2} a\right)^{2}} =
$$
$$
= \int \frac{d(\sin a)}{\left( 1\ -\ \sin^{2} a\right)^{2}} =\int \frac{dt}{\left( 1-t^{2}\right)^{2}} =\int \frac{e^{u} \cdot du}{\left( 1-e^{2\cdot u}\right)^{2}} =\frac{1}{2} \cdot \int \frac{2\cdot e^{u} \cdot du}{\left( 1-e^{2\cdot u}\right)^{2}} =\frac{1}{2} \cdot \int \frac{d\left( e^{2\cdot u}\right)}{\left( 1-e^{2\cdot u}\right)^{2}} =
$$
$$
= \frac{1}{2} \cdot \int \frac{dz}{( 1-z)^{2}} =\frac{1}{2} \cdot \int \frac{d( z-1)}{( z-1)^{2}} =\frac{1}{2} \cdot \int \frac{dv}{v^{2}} =-\frac{1}{2\cdot v} +C
$$
And then I'm trying to return back to the x through v -> z -> u -> t -> a -> x variables "back"-substitutions:
$$
= -\frac{1}{2\cdot ( z-1)} +C=-\frac{1}{2\cdot \left( e^{2\cdot u} -1\right)} +C=\frac{1}{2\cdot \left( 1-e^{2\cdot \ln t}\right)} +C=\frac{1}{2\cdot \left( 1-t^{2}\right)} +C=\frac{1}{2\cdot \left( 1-\sin^{2} a\right)} =
$$
$$
= \frac{1}{2\cdot \cos^{2} a} +C=\frac{1}{2\cdot \cos(\arctan x) \cdot \cos(\arctan x)} +C=\frac{1}{2\cdot \frac{1}{\sqrt{1+x^{2}}} \cdot \frac{1}{\sqrt{1+x^{2}}}} +C=\frac{1+x^{2}}{2} +C
$$
| You made a mistake going from $u$ to $z$: if $z = e^{2u}$, then $dz = 2e^{2u}\,du$, where you take it to be $2e^u\,du$. In fact, if you look at just what happens when you go from $t$ to $z$, you have replaced $t^2$ by $z$, but also replaced $dt$ by $dz$, where it should be $2\sqrt z\,dz$.
You could have stopped at $t$: once you have $\int \frac{dt}{(1-t^2)^2}$, you can take the partial fraction decomposition, and finish with no more substitutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3799770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $f(x)=\sin^{-1} (\frac{2x}{1+x^2})+\tan^{-1} (\frac{2x}{1-x^2})$, then find $f(-10)$ Let $x=\tan y$, then
$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$
Given answer is $0$
What’s wrong here?
| Let $\tan^{-1}\dfrac{2x}{1-x^2}=u\implies-\dfrac\pi2<u<\dfrac\pi2$
$\tan u=\dfrac{2x}{1-x^2}$
$\implies\sec u+\sqrt{1+\left(\dfrac{2x}{1-x^2}\right)^2}=\dfrac{1+x^2}{|1-x^2|}$
$\sin u=\dfrac{\tan u}{\sec u}=\text{sign of}(1-x^2)\cdot\dfrac{2x}{1+x^2}$
$\implies u=\sin^{-1}\left(\text{sign of}(1-x^2)\cdot\dfrac{2x}{1+x^2}\right)$
So if $1-x^2<0\iff x^2>1, u=\sin^{-1}\left(-\dfrac{2x}{1+x^2}\right)=-\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How to solve diffrential equations of Chemical Kinetics 3rd order reaction [Problem setting]
(ii) A chemical reaction is described by
$$
A+B+C \to^{k_1} D\\
D \to_{k_2} A+B+C
$$
If $x_1, x_2, x_3, x_4$ are the concentrations of A, B, C and D respectively, then write down the governing equations.
Hence deduce the equations for
$$
A+A+A \to^{k_1} A_3 \\
A_3 \to_{k_2} A + A + A \\
$$
Solve these equations and illustrate the solutions in a graph of the concentration of $A$ against $A_3$.
D.N.Burghers, M.S.Borrie, "Modelling with Differential Equations", 1990, p.134 Chapter6 Exercises 3.(ii)
[Solved halfway]
If $x$ is the concentration of $A$, $y$ of $A_3$,
\begin{eqnarray*}
-\frac{1}{3}\frac{dx}{dt}
&=& k_{1}x^3 - k_{2}y\\
-\frac{dy}{dt} &=& k_2y - k_1x^{3}
\end{eqnarray*}
where $k_1, k_2$ are the reaction rates.
rearrange,
\begin{eqnarray}
\left\{
\begin{array}{ll}
\frac{dx}{dt}
&=& 3k_{2}y - 3k_{1}x^3 & (1) \\
\frac{dy}{dt} &=& k_1x^{3} - k_2y & (2)
\end{array}
\right.
\end{eqnarray}
rearrange eq(1),
$$
\frac{dx}{dt} = 3k_2y - 3k_{1}x^3\\
y = \frac{1}{3k_2}(3k_{1}x^3 + \frac{dx}{dt})\\
\frac{dy}{dt} = \frac{1}{3k_2}(3k_{1}3x^2\frac{dx}{dt} + \frac{d^2x}{dt^2})
= \frac{3k_{1}x^2}{k_2}\frac{dx}{dt} + \frac{1}{3k_2}\frac{d^2x}{dt^2}
$$
substitute to (2) from it, derive differential equation.
\begin{eqnarray*}
\frac{dy}{dt} &=& k_1 x^3 - k_2y\\
\frac{3k_{1}x^2}{k_2}\frac{dx}{dt} + \frac{1}{3k_2}\frac{d^2x}{dt^2} &=& k_1 x^3 - k_2\frac{1}{3k_2}(3k_{1}x^3 + \frac{dx}{dt})\\
\frac{3k_{1}x^2}{k_2}\frac{dx}{dt} + \frac{1}{3k_2}\frac{d^2x}{dt^2} &=& k_1 x^3 - k_{1}x^3 -\frac{1}{3} \frac{dx}{dt}\\
9k_{1}x^2 \frac{dx}{dt} + \frac{d^2x}{dt^2} &=& -k_2 \frac{dx}{dt}\\
\frac{d^2x}{dt^2} + (9k_{1}x^2 + k_2)\frac{dx}{dt} &=& 0\\
\end{eqnarray*}
replace $p=dx/dt$,
$$
\frac{d^2 x}{dt^2}
= \frac{dp}{dt}
= \frac{dp}{dx}\frac{dx}{dt}
= \frac{dp}{dx} p
$$
substitute $p$,
$$
\frac{d^2x}{dt^2} + (9k_{1}x^2 + k_2)\frac{dx}{dt} = 0\\
\frac{dp}{dx} p + (9k_{1}x^2 + k_2)p = 0\\
$$
if $p \neq 0$,
$$
\frac{dp}{dx} p + (9k_{1}x^2 + k_2)p = 0\\
\frac{dp}{dx} + (9k_{1}x^2 + k_2) = 0\\
\frac{dp}{dx} = -(9k_{1}x^2 + k_2)\\
p = -3k_1x^3 -k_2x + c_0
$$
now have a first order differential equation and can separate the variables to give,
$$
\int \frac{dx}{3k_1x^3 + k_2x - c_0} = - \int dt\\
\frac{1}{3k_1} \int \frac{dx}{x^3 + \frac{k_2}{3k_1}x - \frac{c_0}{3k_1} } = -t+ c_1\\
$$
How can I solve the rest?
| From general principles of mass conservation, or the construction of the system by mass exchange terms that balance, you can easily see that
$$
3x+y=C=3x_0+y_0
$$
is a constant. Thus the system lives on the line $y=C-3x$. From the right side of the first equation we see that equilibrium points satisfy $k_1x^3-k_2y=0$. There is only one intersection point of the rising cubic $y=\frac{k_1}{k_2}x^3$ with the falling line $y=C-3x$, thus exactly one stationary point for the system.
Eliminating $y$ from the first ODE, one finds
$$
\dot x = -3(k_1x^3-k_2y)=-3(k_1x^3+3k_2x-k_2C)
$$
so that the $x$-derivative of the right side is negative for all $x>0$. This means that the stationary point is attracting or stable, all "physical" solutions converge to that point.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding remainder of $123^{456}$ divided by 88 using Chinese Remainder Theorem I tried using Chinese remainder theorem but I kept getting 19 instead of 9.
Here are my steps
$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equiv (-2)^9 \equiv -512 \equiv -6 \equiv 5 \pmod{11}\\
x_2 &= 123^{456} \equiv 123^0 \equiv 1 \pmod{8}\\
y_2 &= 1^{-1} \equiv 1 \pmod{8} \\
123^{456}
&\equiv \sum_{i=1}^2 x_i\times\frac{M}{m_i} \times y_i
\equiv 9\times\frac{88}{11}\times5 + 1\times\frac{88}{8} \times1 \equiv 371
\equiv 19 \pmod{88}
\end{split}
$$
| $y_1$ should've been the inverse of $8\pmod{11}$, not of $9\pmod{11}$, so $y_1=7$.
Similarly, $y_2$ should've been the inverse of $11\pmod 8$, not of $1\pmod 8$, so $y_2=3$.
Therefore, the result is: $9\times\frac{88}{11}\times \color{red}{7}+1\times\frac{88}{8}\times \color{red}{3}=537\equiv 9\pmod{88}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
} |
recurrence relation where $c_n = c_{n-1} + 2c_{n-2}$
A sequence $(c_n)$ is defined recursively as follows: $c_0 = 1, c_1 = 1, $ and $c_n = c_{n-1} + 2c_{n-2}$ for $n\geq 2$. We use $[x^n]g(x)$ to denote the coefficient of $x^n$ of the polynomial $g(x).$ Show that $c_{2n} = [x^{2n}] \dfrac{1}2\left(\dfrac{1}{1-x-2x^2}+\dfrac{1}{1+x-2x^2}\right)$ and that $\sum_{n\geq 0} c_{2n}x^n = \dfrac{1-2x}{1-5x+4x^2}.$ From this, one can deduce that $c_{2n} = 5c_{2n-2} - 4c_{2n-4}.$ Obtain a similar equation for $c_{2n}, c_{2n-4}$ and $c_{2n-8}.$
I know that $\dfrac{1}2\left(\dfrac{1}{1-x-2x^2}+\dfrac{1}{1+x-2x^2}\right) = \dfrac{1-2x^2}{1-5x^2+4x^4}$ and so if I show that $c_{2n} = [x^{2n}] \dfrac{1-2x^2}{1-5x^2+4x^4},$ I can replace $x^2$ with $x$ and get that $\sum_{n\geq 0} c_{2n}x^n = \dfrac{1-2x}{1-5x+4x^2}$. But I'm not sure how to show that $c_{2n} = [x^{2n}] \dfrac{1-2x^2}{1-5x^2+4x^4}.$ I don't think I'll need to compute the exact coefficient and it does not seem useful to manipulate the recurrence equation by substituting $n$ with $2n$. I tried showing that $(1-5x^2+4x^4) \sum_{n\geq 0} c_{2n} x^{2n} = 1-2x^2.$ Matching coefficients results in $c_0+(c_2-5c_0)x^2 + \sum_{n\geq 0} (c_{2n}-5c_{2n-2}-2c_{2n-4})x^{2n} = 1-2x^2,$ but it seems I'd need to prove something like $c_{2n}-5c_{2n-2}-2c_{2n-4}.$ I think figuring out how to come up with $\dfrac{1-2x}{1-5x+4x^2}$ should help me obtain a similar equation relating $c_{2n}, c_{2n-4}, c_{2n-8}$
| Using generating functions, I obtained the expression for $G(z) = \sum_n c_n z^n$:
$$
G(z) = \frac{c_0(1-z)+c_1z}{(1+z)(1-2z)}
$$
Now you need to expand the expression with partial fractions, you will obtain on RHS two expressions of the form
$$
G(z) = \lambda_1\sum_{z=0}^{\infty}(-1)^kz^k + \lambda_1\sum_{z=0}^{\infty}(-s_1)^kz^k
$$
By taking the coefficients for the term $z^n$ you will get your closed-form expression. Here you will need to find the constants $\lambda_1, \lambda_2$ using partial fractions, and $s_1$ by using Generalized binomial coefficient for $\frac{1}{1-2z}$. Can you handle from here?
EDIT: In the partial fraction step it's better to group $c_0 + (c_1-c_0)z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Expected number of days before all magical seeds become apple trees This is a question that I came across recently.
At the end of day $0$, $6$ magical seeds are planted. On each day following it, each seed has a chance to magically transform into an apple tree with a probability of $\frac{1}{2}$. The outcomes of seeds are independent of each another.
What is the expected number of days for all six seed to have become apple trees?
My solution:
$E(n)$ - number of expected days from the point that there is only n seed(s) left.
So, $E(1)$ - number of expected days for the last seed to grow.
$E(1) = 1 + \frac{1}{2} E(1) \,or \,E(1) = 2$. This we anyway know from a coin flip analogy.
$E(2) = 1 + \frac{2}{4} E(1) + \frac{1}{4} E(2) \,or\, E(2) = \frac{8}{3}$.
This comes from the fact that if at the end of a day, two seeds are left, I have 3 possible events - i) both seeds become trees the next day ($+1$ day). ii) one seed becomes tree and one seed is left (probability $\frac{2}{4}$). So we further add expected number of days for $E(1)$. iii) None of the seeds become a tree (probability $\frac{1}{4}$). So we add further expected number of days for $E(2)$.
Similarly, $E(3) = 1 + \frac{3}{8} E(1) + \frac{3}{8} E(2) + \frac{1}{8} E(3)$
$E(4) = 1 + \frac{4}{16} E(1) + \frac{6}{16} E(2) + \frac{4}{16} E(3) + \frac{1}{16} E(4)$
$E(4) = 1 + \frac{4}{16} E(1) + \frac{6}{16} E(2) + \frac{4}{16} E(3) + \frac{1}{16} E(4)$
$E(5) = 1 + \frac{5}{32} E(1) + \frac{10}{32} E(2) + \frac{10}{32} E(3) + \frac{5}{32} E(4) + \frac{1}{32} E(5)$
$E(6) = 1 + \frac{6}{64} E(1) + \frac{15}{64} E(2) + \frac{20}{64} E(3) + \frac{15}{64} E(4) + \frac{6}{64} E(5) + \frac{1}{64} E(6)$
This gives me an answer of $E(6) = \frac{55160}{13671}$. However the answer given is $(\log_2 6)$. I do not understand how the answer got into $\log$. When I calculate both, they are not same values.
Also, are there more generic and faster methods that I could use to get to the answer?
| Let $X_i$ be the r.v. corresponding to the number of the day where the $i$-th seed become an apple tree. Obviously, for all $i=1, ..., 6$ and for all $n \geq 1$,
$$\mathbb{P}(X_i = n) = \frac{1}{2^n}$$
What you are asked to compute is
$$\mathbb{E}(\max(X_i))$$
But for all integer $n \geq 1$,
$$\mathbb{P}(\max(X_i)=n) = \mathbb{P}(\max(X_i) \leq n) - \mathbb{P}(\max(X_i) \leq n-1) = \prod_{i=1}^6 \mathbb{P}(X_i \leq n) - \prod_{i=1}^6 \mathbb{P}(X_i \leq n-1)$$
$$=\left( 1 - \frac{1}{2^n} \right)^6-\left( 1 - \frac{1}{2^{n-1}} \right)^6$$
And $$\mathbb{E}(\max(X_i)) = \sum_{n=1}^{+\infty} n \left[\left( 1 - \frac{1}{2^n} \right)^6-\left( 1 - \frac{1}{2^{n-1}} \right)^6\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
For what values of $k$ is there a perfect $x^p$ in {$n, n+1, n+2, ... ,kn$}? I was thinking about this the other day, and I can't seem to find the answer.
It is a fairly simple proof by induction to show that for all $n ∈ ℕ$, there is a perfect square in ${n, n+1, n+2, ... , 2n}$.
I am trying to generalize this question. For what values of $k$ is there always perfect cube in ${{n, n+1, n+2,..., kn}}$?
More generally, for what values of $k$ is there always a perfect $x^p$ in ${n, n+1, n+2, ... , kn}$? I'm not sure how to approach a proof of this. Maybe it's way above me?
Thanks.
| For $n = 1$, since $1$ is a $p$'th power, any $k$ will do. For $n = 2$, the next largest $p$'th power if $2^p$, so
$$kn \ge 2^p \implies k \ge 2^{p-1} \tag{1}\label{eq1A}$$
I will show the minimum allowed value of $k = 2^{p-1}$ always works. For any $2 \le n \le 2^p$, this value of $k$ works. Also, if $n = m^p$ for any integer $m$, then this $k$ also works. Next, consider
$$m^p \lt n \lt (m+1)^p \tag{2}\label{eq2A}$$
for an integer $m \ge 2$.
For $p = 1$, we have $k = 1$, with this working since each value is it's own first power. For $p = 2$, we have $k = 2$, which you've stated you can prove using induction. Thus, consider $p \ge 3$. Using \eqref{eq2A}, the Binomial theorem expansion with $x \gt 0$ giving $(1 + x)^p = 1 + px + \frac{p(p-1)}{2}x^2 + \ldots + x^p \gt 1 + px$, and $m \ge 2 \implies m + m \ge m + 2 \implies 2m - 1 \ge m + 1$, we then get
$$\begin{equation}\begin{aligned}
\left(2^{p-1}\right)n & \gt \left(2^{p-1}\right)m^p \\
& = \left(\frac{1}{2}\right)\left(2^{p}\right)m^p\left(\frac{(m+1)^p}{(m+1)^p}\right) \\
& = \left(\frac{1}{2}\right)\left(\frac{2m}{m+1}\right)^p(m+1)^p \\
& = \left(\frac{1}{2}\right)\left(1 + \frac{m - 1}{m+1}\right)^p(m+1)^p \\
& \gt \left(\frac{1}{2}\right)\left(1 + \frac{p(m - 1)}{m+1}\right)(m+1)^p \\
& \ge \left(\frac{1}{2}\right)\left(1 + \frac{3(m - 1)}{m+1}\right)(m+1)^p \\
& = \left(\frac{1}{2}\right)\left(\frac{4m - 2}{m+1}\right)(m+1)^p \\
& = \left(\frac{2m - 1}{m+1}\right)(m+1)^p \\
& \ge (m+1)^p
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Thus, $(m+1)^p \in [n,kn]$, confirming $k = 2^{p-1}$ (as well as any larger $k$) will always work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit of multivariable function $f(x,y) = {(x^2+y^2)}^{x^2y^2}$ $$f(x,y) = {(x^2+y^2)}^{x^2y^2}$$
I need to find the limit at (0,0) point
I applied the exponent rule and got $$e^{x^2y^2ln(x^2+y^2)}$$
and now with chain rule, I need to find the limit of $${x^2y^2ln(x^2+y^2)}$$
and how? :D
There isn't L'Hôpital's rule for multivariable function, right?
| We have that
$$ {(x^2+y^2)}^{x^2y^2}=e^{x^2y^2 \log(x^2+y^2)} \to 1$$
indeed
$$x^2y^2 \log(x^2+y^2)=(x^2+y^2) \log(x^2+y^2) \frac{x^2y^2}{x^2+y^2} \to 0\cdot 0=0 $$
since by $t=x^2+y^2 \to 0$ by standard limits
$$(x^2+y^2) \log(x^2+y^2) =t\log t \to 0$$
and since $x^2+y^2 \ge 2xy$
$$0\le\frac{x^2y^2}{x^2+y^2} \le \frac{x^2y^2}{2xy} =\frac12 xy \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Number of non negative integer solutions of $x+y+2z=20$
The number of non negative integer solutions of $x+y+2z=20$ is
Finding coefficient of $x^{20}$ in
$$\begin{align}
&\left(x^0+x^1+\dots+x^{20}\right)^2\left(x^0+x^1+\dots+x^{10}\right)\\
=&\left(\frac{1-x^{21}}{1-x}\right)^2\left(\frac{1-x^{11}}{1-x}\right)\\
=&\left(1-x^{21}\right)^2(1-x)^{-3}\left(1-x^{11}\right)
\end{align}$$
i.e.finding coefficient of $x^{20}$ in $$(1-x)^{-3}-x^{11}(1-x)^{-3}$$
Or, coefficient of $x^{20}$ in $(1-x)^{-3}-$ coefficient of $x^9$ in $(1-x)^{-3}=\binom{22}{20}-\binom{11}{9}=176$
The asnwer is given as $121$. What's my mistake?
EDIT (after seeing @lulu's comment):
Finding coefficient of $x^{20}$ in $$\begin{align}
&\left(x^0+x^1+...+x^{20}\right)^2\left(x^0+x^2+...+x^{20}\right)\\
=&\left(\frac{1-x^{21}}{1-x}\right)^2\left(\frac{1-x^{22}}{1-x^2}\right)\\
=&\left(1-x^{21})^2(1-x)^{-2}(1-x^{22})(1-x^2\right)^{-1}
\end{align}$$
i.e.finding coefficient of $x^{20}$ in $$(1-x)^{-2}(1-x^2)^{-1}$$
Not able to proceed next.
| Solving another way one has $0\le z \le10$ so we have in total the solutions of
$$x+y=20\space\space\text{ for } (x,y,0)\\x+y=18\space\space\text{ for } (x,y,1)\\x+y=16\space\space\text{ for } (x,y,2)\\x+y=14\space\space\text{ for } (x,y,3)\\x+y=12\space\space\text{ for } (x,y,4)\\x+y=10\space\space\text{ for } (x,y,5)\\x+y=8\space\space\text{ for } (x,y,6)\\x+y=6\space\space\text{ for } (x,y,7)\\x+y=4\space\space\text{ for } (x,y,8)\\x+y=2\space\space\text{ for } (x,y,9)\\x+y=0\space\space\text{ for } (x,y,10)$$
Thus we have
$$21+19+17+15+\cdots+5+3+1=121$$ solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Implicit Differentiation of $x+y= \arctan(y)$? I'm trying to verify that $x+y = \arctan(y)$ satisfies this differential equation: $1+(y^2)+(y^2)y'= 0$. To do so, I tried to differentiate $x+y = \arctan(y)$ to get $y'$, but only got so far:
$$1 = \left(\frac{1}{1+y^2}y' -1 \right) y'.$$
Now I'm not sure how I can isolate $y'$, or whether I am even taking the right first steps to solving the problem. Any hints/help about what direction I should take?
| You had differentiated correctly, but made an error when you re-arranged your equation:
$$ [ \ x \ + \ y \ ] \ ' \ \ = \ \ [ \ \arctan(y) \ ] \ ' \ \ \Rightarrow \ \ 1 \ + \ y' \ \ = \ \ \frac{1}{1 \ + \ y^2} · y' $$
$$ \Rightarrow \ \ 1 \ \ = \ \ \frac{1}{1 \ + \ y^2} · y' \ - \ y' \ \ \Rightarrow \ \ 1 \ \ = \ \ \left(\frac{1}{1+y^2} \ - \ 1 \right) · y' \ \ . $$
rather than $ \ \left(\frac{1}{1+y^2}y' -1 \right) y' \ \ . $ You will indeed verify the solution of the differential equation once you have the correct expression:
$$ 1 \ \ = \ \ \left(\frac{1}{1+y^2} \ - \ 1 \right) · y' \ \ \Rightarrow \ \ 1 \ - \ \left(\frac{1}{1+y^2} \ - \ 1 \right) · y' \ \ = \ \ 1 \ - \ \left(\frac{-y^2}{1+y^2} \right) · y' \ \ = \ \ 0 $$
[since $ \ 1 + y^2 \ \neq \ 0 \ \ , $ it is "safe" to multiply the equation through by it]
$$ \Rightarrow \ \ 1·(1+y^2) \ - \ (-y^2) · y' \ \ = \ \ 1 \ + \ y^2 \ + \ (y^2) · y'\ \ = \ \ 0 \ \ . $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving $(a+b+c) \Big(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Big) \leqslant 25$ For $a,b,c \in \Big[\dfrac{1}{3},3\Big].$ Prove$:$
$$(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) \leqslant 25.$$
Assume $a\equiv \text{mid}\{a,b,c\},$ we have$:$
$$25-(a+b+c) \Big(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\Big) =\dfrac{2}{bc} (10bc-b^2-c^2) +\dfrac{c+b}{abc} (a-b)(c-a)\geqslant 0.$$
I wish to find a proof with $a:\neq {\rm mid}\left \{ a, b, c \right \},$ or another proof$?$
Actually$,$ I also found a proof true for all $a,b,c \in \Big[\dfrac{1}{3},3\Big],$ but very ugly.
After clearing the denominators$,$ need to prove$:$
$$f:=22abc-a^2c-a^2b-b^2c-ab^2-bc^2-ac^2\geqslant 0$$
but we have$:$
$$f=\dfrac{1}{32} \left( 3-a \right) \left( 3-b \right) \Big( c-\dfrac{1}{3} \Big) +
\left( 3-a \right) \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) +\\+{
\frac {703}{32}}\, \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) \left(
c-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( 3-a \right) \left( 3-c
\right) \left( a-\dfrac{1}{3} \right) +\dfrac{1}{4} \left( 3-b \right) \left( 3-c
\right) \left( c-\dfrac{1}{3} \right) +\dfrac{5}{4} \left( 3-c \right) \left( c-\dfrac{1}{3}
\right) \left( a-\dfrac{1}{3} \right) +{\frac {49}{32}} \left( 3-c \right)
\left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) + \left( 3-b \right)
\left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) +\\+{\frac {21}{16}}\,
\left( 3-b \right) \left( a-\dfrac{1}{3} \right) \left( b-\dfrac{1}{3} \right) \\+\dfrac{5}{4}\,
\left( 3-a \right) \left( c-\dfrac{1}{3} \right) \left( a-\dfrac{1}{3} \right) +\dfrac{1}{32}
\, \left( 3-a \right) ^{2} \left( 3-c \right) +\dfrac{1}{4}\, \left( 3-b
\right) \left( b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{32} \left( 3-b \right) ^{2}
\left( a-\dfrac{1}{3} \right) +{\frac {9}{32}} \left( a-\dfrac{1}{3} \right) \left(
b-\dfrac{1}{3} \right) ^{2}+\dfrac{1}{4} \left( a-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) ^{
2}+\dfrac{1}{4} \left( b-\dfrac{1}{3} \right) \left( 3-b \right) ^{2}+{\frac {9}{32}}
\, \left( b-\dfrac{1}{3} \right) \left( c-\dfrac{1}{3} \right) ^{2}$$
So we are done.
If you want to check my decomposition$,$ please see the text here.
| Let $f(a,b,c)=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$. Note that $f$ is concave of each variable (if other variables are fixed). Hence, since concave on $I$ fucntion attains its maximum at endpoint of $I$ (here $I=[m,M]=\left[\frac{1}{3},3\right]$)
$$
\max_{(a,b,c)\in I^3} f=\max_{(a,b,c)\in\{m,M\}^3} f.
$$
Thus, we need only to compute these 8 values and choose the maximal one.
Details: consider any point $(a,b,c)$, fix $b$ and $c$ and consider $f$ as a function of $a$. We obtain
$$
f(a,b,c)\leq\max\{f(m,b,c),f(M,b,c)\},
$$
so we can assume that $a\in\{m,M\}$. Now repeat this argument for $b$ and $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Proving $6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$, where $x+y+z=0$ Question :
Let $x,y,z$ be reals satisfying $x+y+z=0$. Prove the following inequality:$$6(x^3+y^3+z^3)^2 \leq (x^2+y^2+z^2)^3$$
My Attempts :
It’s obvious that $x,y,z$ are either one negative and two positive numbers or one positive and two negative numbers.
In addition, putting $(x,y,z)$ and $(-x,-y,-z)$ into the inequality has the same outcome.
Therefore, without loss of generality, I suppose $x,y \geq 0$ and $z\leq0$.
$x+y=-z\\ \Longrightarrow (x+y)^2=z^2 \\ \Longrightarrow (x^2+y^2+z^2)^3=8(x^2+xy+y^2)^3$
is what I have got so far, and from here I can’t continue.
Am I on the right direction? Any suggestions or hints will be much appreciated.
| Scaling, one may assume that $z = -1$, so that $x + y = 1$. Then the desired inequality is
$$6(x^3 + y^3 - 1)^2 \leq (x^2 + y^2 + 1)^3$$
Since $x^3 + y^3 = (x + y)^3 - 3xy(x + y) = 1 - 3xy$ and $x^2 + y^2 + 1 = (x + y)^2 - 2xy + 1 = 2 - 2xy$, the desired inequality is
$$54(xy)^2 \leq (2 - 2xy)^3$$
Thus it makes sense to look at $xy = x(1 - x) = {1 \over 4} - (x+{1 \over 2})^2$, whose range is $(-\infty, {1 \over 4}]$. Letting $r = xy$, we need that for $r \leq {1 \over 4}$ we have
$$54r^2 \leq (2 - 2r)^3$$
However, the polynomial $54r^2 - (2 - 2r)^3$ can be directly factorized into $2(r + 2)^2(4r - 1)$, which is nonpositive in the domain $(-\infty, {1 \over 4}]$. Hence the inequality holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Solve differential equation: $y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$ I need to solve this differential equation: $y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$.
My attempt
I found that $y = x^2$ is a solution. Then I tried to put $y = x^2f(x)$, and solved this way:
$$2xf(x) + x^2f'(x) = xf^2(x) + 2xf(x) - x \implies x^2f'(x) = xf^2(x) - x \implies$$
$$xf'(x) = f^2(x) - 1 \implies \frac{df(x)}{f^2(x) - 1} = \frac{dx}{x} \implies$$
$$\frac{1}{2}\ln\left|\frac{f(x) - 1}{f(x) + 1}\right| = \ln|x| + C_* \implies \frac{f(x) - 1}{f(x) + 1} = Cx^2 \implies$$
$$f(x) = \frac{1 + Cx^2}{1 - Cx^2}$$
And we lost a solution $f(x) = -1$. So finally we have
$$y = x^2\frac{1 + Cx^2}{1 - Cx^2}, y = -x^2$$
Now I have 3 questions:
$\quad 1)$ Is my solution correct? I'm not sure that all solutions were found.
$\quad 2)$ When can we use particular solution to find all other solutions? I mean doing something like $y = g(x)h(x)$, where $g(x)$ is a particular solution.
$\quad 3)$ Is there an easier method to solve this equation?
| For question 3
$$y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$$
$$y'- 2\frac{y}{x} = \frac{y^2}{x^3} - x$$
$$\left ( \dfrac y {x^2} \right )'= \frac{y^2}{x^5}-\frac{1}{x} $$
It's separable.
$$\left ( \dfrac y {x^2} \right )'=\dfrac 1 x \left ( \frac{y^2}{x^4}-1 \right) $$
$$\dfrac {du}{u^2-1}=\dfrac {dx}{x}$$
Where $ u=\dfrac y {x^2}$
Integrate.
I got this
$$y(x)=x^2\dfrac {1+Kx^2}{1-Kx^2}$$
For $K=0$ you have the particular solution you find by inspection :$y=x^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3821061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determine all zeros of the polynomial $X^4 - 2X^3 - X^2 + 2X + 1 \in \mathbb C[X]$. This is Exercise 14 on page 110 of Analysis I by Amann and Escher.
The hint given is as follows: multiply the polynomial by $1/X^2$ and substitute $Y = X - 1/X$.
If I attempt this, I get the following:
\begin{align*}
X^4 - 2X^3 - X^2 + 2X + 1 &= 0\\
\Rightarrow X^2 - 2X - 1 + \frac{2}{X} + \frac{1}{X^2} &= 0.
\end{align*}
My problem is that I don't understand how to make the suggested substitution. I'm wondering if there is something obvious I'm missing.
I appreciate any help.
| $Y^2= X^2-2+\dfrac{1}{X^2}$. So $X^2-1+\dfrac{1}{X^2} = Y^2+1$. Then
$$X^2-2X-1+\dfrac{2}{X}+1/{X^2} = Y^2+1-\left(2X-\dfrac{2}{X}\right)=Y^2-2Y+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3822979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Computing $\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$ The problem is: $$\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$$
I factored the numerator to get: $\frac{(x-3)(x+5)}{|x+5|}$
How do i solve the rest?
| Another place where "factor out the big part" helps. Here, $|x|$ is growing without bound so $x^2$ is the big part in the numerator and $|x|$ is the big part in the denominator.
\begin{align*}
L &= \lim_{x \rightarrow -\infty} \frac{x^2 + 2x - 15}{|x+5|} \\
&= \lim_{x \rightarrow -\infty} \frac{x^2(1 + 2/x - 15/x^2)}{|x(1+5/x)|} \\
&= \lim_{x \rightarrow -\infty} \frac{x^2}{|x|} \frac{1 + 2/x - 15/x^2}{|1+5/x|}
\end{align*}
We should be able to see that the second fraction is going to $\frac{1+0-0}{|1+0|}$. (Making this easy to see is why you factor out the "big part".) The first fraction is always positive (positive over positive) once $x < 0$, so
$$ L = \lim_{x \rightarrow -\infty} (|x| \cdot 1) = \infty \text{.} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Reduced row echelon form of an augmented matrix is not unique I am given a sysetem of linear equations which after graphing, have no solution (the three lines intersect at different points). Now I am trying to prove this algebraically.
As an augmented matrix,
$$
\begin{bmatrix}
1 & -1 & 3 \\
1 & 1 & 1\\
2 & 3 & 6\\
\end{bmatrix}
$$
*
*$R_{1}-R_{2} \Rightarrow R_{2}$
$$
\begin{bmatrix}
1 & -1 & 3 \\
0 & -2 & 2\\
2 & 3 & 6\\
\end{bmatrix}
$$
continue from here in $(1)$ or $(2)$
$(1)$
*
*$R_{1}-\frac{1}{2}R_{3} \Rightarrow R_{3}$
$$
\begin{bmatrix}
1 & -1 & 3 \\
0 & -2 & 2\\
0 & -\frac{5}{2} & 0\\
\end{bmatrix}
$$
*
*$-\frac{1}{2}R_{2} \Rightarrow R_{2}$
$$
\begin{bmatrix}
1 & -1 & 3 \\
0 & 1 & -1\\
0 & -\frac{5}{2} & 0\\
\end{bmatrix}
$$
*
*$\frac{5}{2}R_{2} + R_{3} \Rightarrow R_{3}$
$$
\begin{bmatrix}
1 & -1 & 3 \\
0 & 1 & -1\\
0 & 0 & -\frac{5}{2}\\
\end{bmatrix}
$$
Then in RREF
$$
\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & -1\\
0 & 0 & -\frac{5}{2}\\
\end{bmatrix}
$$
$(2)$
*
*$-2R_{1}+R_{3} \Rightarrow R_{3}$
$$
\begin{bmatrix}
1 & -1 & 3 \\
0 & -2 & 2\\
0 & 5 & 0\\
\end{bmatrix}
$$
*
*$-\frac{1}{2}R_{2} \Rightarrow R_{2}$
$$
\begin{bmatrix}
1 & -1 & 3 \\
0 & 1 & -1\\
0 & 5 & 0\\
\end{bmatrix}
$$
*
*$-5R_{2} + R_{3} \Rightarrow R_{3}$
$$
\begin{bmatrix}
1 & -1 & 3 \\
0 & 1 & -1\\
0 & 0 & 5\\
\end{bmatrix}
$$
Then in RREF
$$
\begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & -1\\
0 & 0 & 5\\
\end{bmatrix}
$$
which is different from the RREF in $(1)$
Can someone explain why I end up with a different RREF? I thought all RREF are unique, but clearly not in this case. Of course as mentioned earlier, the system has no solutions and both augmented matrices show this but their RREF's are not unique still.
| Any RREF must have pivot 1 in each row. Your matrices do not satisfy this condition (look at row 3). So these are not RREFs. RREF of any matrix is unique. It is not a completely trivial statement, but it is a fact.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3827883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Difference equation: verify answer? I would like to solve the difference equation
$$ k_{n+1} - \frac{\omega_2+1}{\omega_1 +1 } k_n = \frac{\omega_1 -\omega_2}{2( \omega_1 +1)}$$
Where $\omega_1 , \omega_2 $ are fixed positive real numbers.
I obtained the solution
$$k_n = (k_0 - \tfrac12) \left( \frac{\omega_2 +1}{\omega_1+1} \right)^n + \frac12$$
In the following way:
If
\begin{equation}
k_{s+1}-ak_s = b
\end{equation}
Let
\begin{equation}
k_s = A a^s +B
\end{equation}
be an ansatz (sorry for changing from $n$ to $s$, they are the same)
Then,
\begin{align*}
k_{s+1}-ak_s &= b \\
A a^{s+1} +B - A a^{s+1} +aB &= b \\
B(1-a) &= b \\
B &= \frac{b}{1-a}
\end{align*}
So,
\begin{equation}
k_s = A a^s + \frac{b}{1-a}
\end{equation}
So, in this case,
\begin{align*}
a &=\frac{\omega_2+1}{\omega_1 +1 } \\
b &= \frac{\omega_2 -\omega_1}{2( \omega_1 +1)}
\end{align*}
Figuring out $B$,
\begin{align*}
1- a &=\frac{\omega_1 - \omega_2}{\omega_1+1} \\
b &= \frac{\omega_1 -\omega_2}{2( \omega_1 +1)}\\
\frac{1}{1- a }&=\frac{\omega_1+1} {\omega_1 - \omega_2}\\
\frac{b}{1- a }&=\frac{\omega_1+1} {\omega_1 - \omega_2} \cdot \frac{\omega_1 -\omega_2}{2( \omega_1 +1) }\\
&= \frac12
\end{align*}
I think that this is incorrect, but I can't see any mistake I made.
Thank you!
| Hoping that you do not mind, let me give a small trick for this kind of problems
You have
$$k_{s+1}-ak_s = b$$ and the problem would be easy if $b=0$.
So, let $k_s=x_s+C$, replace and expand
$$x_{s+1}+C -ax_s-a C = b$$ Select $C$ such that
$$C-aC=b \implies C=\frac b {1-a}\implies x_{s+1}-ax_s = 0\implies x_s=A a^{s-1}$$ Back to $k_s$
$$k_s=x_s+\frac b {1-a}=A a^{s-1}+\frac b {1-a}$$ and then your good result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3828614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can I solve $\lim_{(x,y)\to\ (0,0)} \frac{x^2y^2}{x^2+x^2y^2+y^2}$ by converting to polar coordinates? Is it correct to solve this problem like this?
$$\lim_{(x,y)\to\ (0,0)} \frac{x^2y^2}{x^2+x^2y^2+y^2} $$
$$\lim_{(x,y)\to\ (0,0)} \frac{1}{1+\frac{x^2+y^2}{x^2y^2}}$$
$$\frac{x^2+y^2}{x^2y^2}=\lim_{r\to\ 0} \frac{1}{r^2\cos^2\theta\sin^2\theta}=\infty\implies \lim_{(x,y)\to\ (0,0)} \frac{1}{1+\frac{x^2+y^2}{x^2y^2}}=0$$
| Directly by polar we obtain
$$\frac{x^2y^2}{x^2+x^2y^2+y^2}=\frac{r^2\cos^2\theta \sin^2\theta}{1+ r^2\cos^2\theta \sin^2\theta}\to \frac{0}{1+0}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Minimize $(x+y)(y+z)(z+x)$ given $xyz(x+y+z) = 1$ $x,y,z$ are positive reals and I am given $xyz(x+y+z) = 1$. Need to minimize $(x+y)(y+z)(z+x)$. Here is my approach.
Using AM-GM inequality
$$ (x+y) \geqslant 2 \sqrt{xy} $$
$$ (y+z) \geqslant 2 \sqrt{yz} $$
$$ (z+x) \geqslant 2 \sqrt{zx} $$
So, we have
$$ (x+y)(y+z)(z+x) \geqslant 8xyz $$
Also, I got
$$ \frac{x+y+z+(x+y+z)}{4} \geqslant \bigg[ xyz(x+y+z) \bigg] ^{1/4} $$
$$ \therefore x+y+z \geqslant 2 $$
But, I am stuck here. Any hints ?
| For $x=y=z=\frac{1}{\sqrt[4]3}$ we get a value $\frac{8}{\sqrt[4]{27}}.$
We'll prove that it's a minimal value.
Indeed, we need to prove that $$\prod_{cyc}(x+y)\geq\frac{8}{\sqrt[4]{27}}$$ or
$$27\prod_{cyc}(x+y)^4\geq4096x^3y^3z^3(x+y+z)^3.$$
Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that $$(9uv^2-w^3)^4\geq4096u^3w^9$$ or $f(w^3)\geq0,$ where
$$f(w^3)=(9uv^2-w^3)^4-4096u^3w^9.$$
But it's obvious that $f$ decreases, which says that it's enough to prove our inequality for a maximal value of $w^3$, which by $uvw$ happens for equality case of two variables.
Since the last inequality is symmetric and homogeneous, it's enough to assume $y=z=1$ and we need to prove that:
$$27(x+1)^8\geq256x^3(x+2)^3,$$ which is true by AM-GM:
$$27(x+1)^8=27(x^2+2x+1)^4=27\left(3\cdot\frac{x^2+2x}{3}+1\right)^4\geq$$
$$\geq27\left(4\sqrt[4]{\left(\frac{x^2+2x}{3}\right)^3\cdot1}\right)^4=256x^3(x+2)^3$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3829530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Real roots of $x^7+5x^5+x^3−3x^2+3x−7=0$? The number of real solutions of the equation,
$$x^7+5x^5+x^3−3x^2+3x−7=0$$
is
$$(A) 5 \quad (B) 7 \quad (C) 3 \quad (D) 1.$$
Using Descartes rule we may have maximum no. of positive real roots is $3$ and negative real root is $0.$ So there can be either $3$ real roots or $1$ real root but how to conclude what will be the no. of real roots exactly. Can you please help me?
| It is easy, by the rational root theorem, to find that $1$ is the only rational root. Then, dividing the polynomial by $x-1$ you get
$$
x^6+x^5+6 x^4+6 x^3+7 x^2+4 x+7
$$
that, by Decartes Rule, has zero positive roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the Eigenvectors $T\left(\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\right)=\left[\begin{array}{ll}d & b \\ c & a\end{array}\right]$ $V=$ $M_{2 \times 2}(\mathbb{R})$ y $T\left(\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\right)=\left[\begin{array}{ll}d & b \\ c & a\end{array}\right]$
I think the matrix associated to T is
$A= $$\left[\begin{array}{ll}0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\1 & 0 & 0 & 0 \end{array}\right]$
Then we can get the eigenvalues if we swap $R_4$ to $R_1$
$A`= $$\left[\begin{array}{ll}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \end{array}\right]$
Then the eigenvalue is 1 but from here Im stuck to find the eigenvectors or is something wrong with my process? thank you!
| The eigenvalues and eigenvectors of $T$ may be found directly from the given formula
$T \left ( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right ) = \begin{bmatrix} d & b \\ c & a \end{bmatrix}, \tag 1$
for we have
$T^2 \left ( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right ) = T \left ( \begin{bmatrix} d & b \\ c & a \end{bmatrix} \right ) = \begin{bmatrix} a & b \\ c & d\end{bmatrix}; \tag 2$
thus,
$T^2 = I, \tag 3$
or
$T^2 - I = 0; \tag 4$
if $\mu$ is an eigenvalue of $T$, that is, if
$TZ = \mu Z\tag 5$
for some
$0 \ne Z \in M_{2 \times 2}(\Bbb R), \tag 6$
then
$T^2Z = T(TZ) = T(\mu Z) = \mu TZ = \mu (\mu Z) = \mu^2Z, \tag 7$
whence
$(\mu^2 - 1)Z = \mu^2 Z - Z = T^2 Z - Z = (T^2 - I)Z = 0, \tag 8$
so in light of (6) we have
$\mu^2 - 1 = 0, \tag 9$
which implies
$\mu = \pm 1; \tag{10}$
now if
$\mu = 1, \tag{11}$
$\begin{bmatrix} d & b \\ c & a \end{bmatrix} = T \left (\begin{bmatrix} a & b \\ c & d \end{bmatrix} \right ) =
\mu \begin{bmatrix} a & b \\ c & d \end{bmatrix} =
\begin{bmatrix} a & b \\ c & d \end{bmatrix}, \tag{12}$
which forces
$a = d; \tag{13}$
an eigenmatrix for eigenvalue $1$ thus takes the form
$\begin{bmatrix} a & b \\ c & a \end{bmatrix}, \tag{14}$
where $a, b, c \in \Bbb R$ are arbitrary. It is now easy to see that the $1$-eigenspace of $T$ is of dimension $3$. On the other hand, when
$\mu = -1, \tag{15}$
$\begin{bmatrix} d & b \\ c & a \end{bmatrix} = T \left (\begin{bmatrix} a & b \\ c & d \end{bmatrix} \right ) =
\begin{bmatrix} -a & -b \\ -c & -d \end{bmatrix}, \tag{16}$
whence,
$a = - d \tag{16}$
$b = -b, \; c = -c ,\Longrightarrow b = c = 0; \tag{17}$
the eigenmatrix thus becomes
$\begin{bmatrix} a & 0 \\ 0 & -a \end{bmatrix}, \; a \in \Bbb R; \tag{18}$
it is clear that the $-1$ eigenspace of $T$ is of dimension $1$.
Since the sum of the dimensions of the $1$ and $-1$ eigenspaces is
$4 = \dim M_{2 \times 2}(\Bbb R), \tag{19}$
we conclude there are no more eigenvectors/eigenvalues to be had.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Iteration on sequences I am trying to find a general formula for this sequence, in order to find its limit (has to be this way):
$$X_{k+1} = \frac {X_k} 2 + \frac 1 {X_k}$$
I cannot get a grasp on it; after $2$ or $3$ iterations, the sequence gets huge and I don't see the pattern.
Any help is greatly appreciated!
| If $X_0=\sqrt 2$, then the sequence is constant, so we can assume $X_0\not = \sqrt 2$. We may also assume that $X_0>0$, as the negative case is the same.
Let $X_0=\sqrt 2\cdot\frac {r+1}{r-1} $, or $r=\frac{X_0+\sqrt2}{X_0-\sqrt2}$, then the closed form is $X_k=\sqrt 2\cdot\frac {r^{2^k}+1}{r^{2^k}-1}$. The way to get the closed form is as follows.
Dividing by $\sqrt 2$:
$$\frac {X_{k+1}}{\sqrt 2}=\frac 12(\frac {X_{k}}{\sqrt 2}+\frac {\sqrt 2}{X_{k}})$$
Substituting $\frac {X_{k}}{\sqrt 2}=S_k+1$:
$$S_{k+1}+1=\frac 12 \cdot(S_k+1+\frac 1{S_k+1})$$
$$S_{k+1}=\frac 12 \cdot(S_k-1+\frac 1{S_k+1})$$
$$S_{k+1}=\frac 12 \cdot\frac {S_k^2}{S_k+1}$$
Taking the inverse:
$$\frac 1{S_{k+1}}=2 \cdot\frac {S_{k}+1}{S_k^2}$$
$$\frac 1{S_{k+1}}= 2\cdot (\frac 1{S_{k}})^2+2\cdot \frac 1{S_{k}}$$
Completing the square:
$$\frac 1{S_{k+1}}+\frac 12=2\cdot(\frac 1{S_{k}}+\frac 12)^2$$
$$\frac 2{S_{k+1}}+1=(\frac 2{S_{k}}+1)^2$$
This means each term is the square of the previous term, which we can denote as $\frac 2{S_{k}}+1=r^{2^{k}}$. Substituting everything back into $X_k$ then we have the close form.
Proving that the closed form is actually correct:
\begin{align}
\frac {X_k}2+\frac 1{X_k}
&=\frac 1{\sqrt2} \cdot (\frac {r^{2^k}+1}{r^{2^k}-1}+\frac {r^{2^k}-1}{r^{2^k}+1})\\
&=\frac 1{\sqrt2} \cdot \frac {(r^{2^k}+1)^2+({r^{2^k}-1})^2}{(r^{2^k}-1)({r^{2^k}+1})} \\
&=\frac 1{\sqrt2}\cdot \frac {r^{2^{k+1}}+2r^{2^k}+1+r^{2^{k+1}}-2r^{2^k}+1}{r^{2^{k+1}}-1}\\
&=\sqrt 2\cdot \frac {r^{2^{k+1}}+1}{r^{2^{k+1}}-1}\\
&=X_{k+1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3838549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show $\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx=\frac{5\pi^3}{64}+\frac\pi{16}\ln^22$ Tried to evaluate the integral
$$I=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx$$
and managed to show that
\begin{align}
I &= \int_0^1 \frac{\ln^2x}{(x+1)^2+1} \, dx + \int_0^1 \frac{\ln^2x}{(x+1)^2+x^2} \, dx\\
&= \int_0^1 \frac{\ln^2x}{(x+1+i)(x+1-i)} \, dx
+ \int_0^1 \frac{\ln^2x}{(x+1+ix )(x+1-ix )} \, dx\\
&= -2\operatorname{Im}\operatorname{Li}_3\left(-\frac{1+i}2\right)
-2\operatorname{Im} \operatorname{Li}_3(-1-i)
\end{align}
which is equal to $ \frac{5\pi^3}{64}+\frac\pi{16}\ln^22$.
It is perhaps unnecessary, though, to resort to evaluation in complex space. I would like to work out an elementary derivation of this integral result.
| $$I(a)=\int_0^\infty\frac{x^{-a}}{x^2+2x+2}dx\overset{x\to 1/x}{=}\int_0^\infty\frac{x^a}{2x^2+2x+1}dx$$
$$=\Im\int_0^\infty\frac{(1+i)x^a}{1+(1+i)x}dx\overset{(1+i)x=u}{=}\Im \frac{1}{(1+i)^a}\int_0^\infty\frac{u^a}{1+u}du$$
$$=-\Im\frac{\pi\csc(a\pi)}{(1+i)^a}=2^{-a/2}\csc(a\pi)\sin(\frac{\pi}{4}a),$$
and your integral is $I''(0).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$
Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$
Hint: Start by dividing the whole equation by $a$
At first I have tried solving the equation without using the hint provided in my exercise and directly applying completing square, I get $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-c}{a}}$. So if I am to use the hint, I obtain the appropriate answer. But I wonder if I am asked the same question in my exam where the hint will not be provided then how am I supposed to answer.
I would like to know how one should approach this kind of question and how do I realise when to divide the whole equation with in this case $a$ or is there any other ways so that I can avoid dividing the whole equation by $a$.Thanks in advance for any help you are able to provide!
EDIT: Here's my steps. Please see where have I done wrong.
\begin{align}
ax^2+2bx+c&=0 \\
a\left[\left(x+\frac{b}{a}\right)^2-\frac{b^2}{a^2}\right] + c&=0 \\
a\left(x+\frac{b}{a}\right)^2-\frac{b^2}{a} + c&=0 \\
\left(x+\frac{b}{a}\right)^2&=\left(\frac{b^2}{a}-c\right)\left(\frac{1}{a}\right)\\
\left(x+\frac{b}{a}\right)^2&=\frac{a(b^2-c)}{a^2}\\
\left(x+\frac{b}{a}\right)^2&=\frac{b^2-c}{a} \\
x+\frac{b}{a}&=\pm\sqrt{\frac{b^2-c}{a}} \\
\implies x&=-\frac{b}{a}\pm\sqrt{\frac{b^2-c}{a}}\\
\end{align}
| HINT: Complete the square:
$$ax^2+2bx+c=0$$
$$a\left(x^2+\frac{2b}{a}x\right)+c=0$$
Try the rest from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$
$\cos^6x + \sin^6x = -3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = (-3/2)(2\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = (-3/2)(\sin^22x)$
How can I get it into $ 1 - (3/4)\sin^2(2x)$?
| It should be
$(\cos^2x+\sin^2x)^3=1$
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x\color{red}{=1}$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x=1$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x=1$
$\cos^6x + \sin^6x = 1-3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = 1-\dfrac3{\color{red}4}(4\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = 1-\dfrac3{\color{black}4}(2\cos x\sin x)^2$
$\cos^6x + \sin^6x = 1-\dfrac34(\sin^22x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 5
} |
Simplifying equation using trigonometric sum and differences identities Recently started learning trigonometric identities in school, having problems with this one. Tried solving using the sum and differences identities but keep having repeating everything. Thanks.
$\frac{-\cos(x)\sin(x)\pm \cos(y)\sin(y)}{\sin^2(y)-\cos^2(x)}=\tan(x\mp y)$
| We have
$$LHS = \frac{-\cos x \sin x + \cos y \sin y}{\sin^2y-\cos^2x} = \dfrac{1}{2}\dfrac{\sin(2y)-\sin(2x)}{\sin^2y-\cos^2y} = \dfrac{\sin(y-x)\cos(y+x)}{\sin^2y-\cos^2x} $$
$\sin(-z) = -\sin(z)$ So $\sin(y-x) = -\sin(x-y)$
$$\begin{align}& LHS = \dfrac{\sin(x-y)\cos(y+x)}{\color{blue}{\cos^2x-\sin^2y}} = \dfrac{\sin(x-y)\cos(x+y)}{\color{blue}{\cos(x+y)\cos(x-y)}}\\
& \ = \dfrac{\sin(x-y)}{\cos(x-y)} =\tan(x-y) =RHS\end{align}$$
Use $y = -y$ to get the other one.
For the blue highlighted part (a short proof)
$\cos(x+y)\cos(x-y) = \dfrac{1}{2}\left[\cos(2x)+\cos(2y)\right] = \dfrac{1}{2}\left[(2\cos^2x -1) + (1-2\sin^2y)\right]\\ = \cos^2x-\sin^2y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove or disprove the recursively defined sequence is convergent. The sequence $\{a_n\}$ is defined by $a_1=1, a_2=0$ and $a_{n+2}=a_{n+1}+\displaystyle\frac{a_n}{n^2}$ for $n\in \mathbb{N}$.
Since $\displaystyle\frac{1}{n^2}$ is summable, when $n$ is large, the sequence is something like $a_n=a_{n-1}+\displaystyle\sum_{i\leq n-2}\frac{a_i}{i^2}$, so I think the sequence should be convergent.
Then I want to use the Monotone convergent theorem, i.e. to show $\{a_n\}$ is monotonic and bounded.
For monotonic, it is easy to see that $\{a_n\}$ is increasing.
But for the upper bound, assuming $\{a_n\}$ converges and taking the limit $n\to \infty$ does not give any hints for me to find a suitable upper bound. I have also used computer programs to compute up to the 10000th term, but it seems that $\{a_n\}$ is still increasing, does not converges to a certain number.
So I wonder if it is convergent or not.
| Well this took longer than I thought. I feel like there must be an easier solution...
Claim 1: $a_n\le\sqrt n$ for all $n$. This holds for $n=1$ and $n=2$. Actually, we will want to assume $n\ge 3$ later, so we can also check $a_3=1\le\sqrt3$. Now if $a_n\le\sqrt n$ and $a_{n+1}\le\sqrt{n+1}$, then
$$a_{n+2}=a_{n+1}+\frac{a_n}{n^2}\le\sqrt{n+1}+\frac{\sqrt n}{n^2},$$
and it suffices to show $\sqrt{n+1}+\frac{\sqrt n}{n^2}\le \sqrt{n+2}$. Note
$$\sqrt{n+1}+\frac{\sqrt n}{n^2}\le\sqrt{n+1}+\frac{\sqrt{n+1}}{n^2}=\sqrt{n+1}\left(1+\frac{1}{n^2}\right)$$
and the inequality
$$\sqrt{n+1}\left(1+\frac{1}{n^2}\right)\le\sqrt{n+2}$$
is equivalent to
$$(n+1)\left(1+\frac{1}{n^2}\right)^2\le n+2.$$
With some elbow grease this is equivalent to
$$n^4\ge 2n^3+2n^2+n+1.$$
Now since $n\ge 3$,
$$n^4\ge 3n^3=2n^3+n^3\ge 2n^3+3n^2=2n^3+2n^2+n^2\ge 2n^3+2n^2+n+1.$$
This establishes Claim 1.
Claim 2: $a_n=\sum_{i=1}^{n-2}\frac{a_i}{i^2}$ for $n\ge 3$. This holds for $n=3$, and if $a_n=\sum_{i=1}^{n-2}\frac{a_i}{i^2}$, then
$$a_{n+1}=a_n+\frac{a_{n-1}}{(n-1)^2}=\frac{a_{n-1}}{(n-1)^2}+\sum_{i=1}^{n-2}\frac{a_i}{i^2}=\sum_{i=1}^{n-1}\frac{a_i}{i^2}.$$
Finishing up: we now have $a_n=\sum_{i=1}^{n-2}\frac{a_i}{i^2}\le\sum_{i=1}^{n-2}n^{-\frac32}.$ Pick your favorite way to show this is the partial sum of a convergent $p$-series, and we're done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3847186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Using Squeeze Theorem to compute $\lim_{(x,y)\to(0,0)} \frac{x²y}{x^2+xy+y^2} = 0$ Can you help me to show that $\lim_{(x,y)\to(0,0)} \frac{x²y}{x^2+xy+y^2} = 0$ with the squeeze theorem ?
I'm running out of ideas to bound $|\frac{x²y}{x^2+xy+y^2}|$. I was thinking to use $|xy|\leq |x^2+xy+y^2|$, which seems to be a right inequality but don't know to to show it.
| $2xy=(x+y)^2-x^2-y^2,$ so $$|f(x,y)|=\frac{x^2|y|}{|x^2+xy+y^2|}=\frac{2x^2|y|}{(x+y)^2 + x^2 + y^2} \le \frac{2x^2|y|}{x^2+y^2}= |x|\frac{2|xy|}{x^2+y^2} \le |x| \frac{x^2+y^2}{x^2+y^2}=|x|$$
where we also used $2|xy| \le x^2+y^2$ which follows from $(|x|-|y|)^2\ge 0$. Sending $(x,y)\to 0$ gives the result.
A now deleted comment informed me that there is an shorter proof:
$$ |f(x,y)|=\frac{x^2|y|}{|x^2+xy+y^2|}=\frac{2x^2|y|}{(x+y)^2 + x^2 + y^2} \le 2|y|.$$
This follows because $x^2 \le (x+y)^2 + x^2 + y^2$. Together with the above, we have the improved bound
$$ |f(x,y)| \le \min(|x|,2|y|).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3849458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx$. How can i evaluate
$$\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx$$
$$=\frac{\pi ^3}{6}-\frac{\pi ^3}{6\sqrt{2}}-4\pi +6\pi \ln \left(2\right)-\frac{\pi }{2\sqrt{2}}\ln ^2\left(2\sqrt{2}+3\right)-\sqrt{2}\pi \operatorname{Li}_2\left(2\sqrt{2}-3\right)$$
This is what I've done thus far.
\begin{align*}
&\int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx\\
&=\frac{\pi }{2}\int _0^{\pi }\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)\:dx\\[2mm]
&=\frac{\pi }{4}\int _0^{\pi }\sin \left(\frac{x}{2}\right)\operatorname{Li}_2\left(\cos \left(x\right)\right)\:dx+\frac{\pi }{4}\int _{\pi }^{2\pi }\sin \left(\frac{x}{2}\right)\operatorname{Li}_2\left(\cos \left(x\right)\right)\:dx\\[2mm]
&=\frac{\pi }{2}\int _0^{\pi }\sqrt{\frac{1+\cos \left(x\right)}{2}}\operatorname{Li}_2\left(-\cos \left(x\right)\right)\:dx=\pi \int _0^{\infty }\frac{\operatorname{Li}_2\left(\frac{t^2-1}{1+t^2}\right)}{\left(1+t^2\right)\sqrt{1+t^2}}\:dt\\[2mm]
&=\frac{\pi }{2}\int _1^{\infty }\frac{\operatorname{Li}_2\left(\frac{x-2}{x}\right)}{x\sqrt{x}\sqrt{x-1}}\:dx=\frac{\pi }{2}\int _0^1\frac{\operatorname{Li}_2\left(1-2x\right)}{\sqrt{1-x}}\:dx\\[2mm]
&=\pi \zeta \left(2\right)+\frac{\pi }{\sqrt{2}}\int _{-1}^1\frac{\sqrt{1+x}\ln \left(1-x\right)}{x}\:dx
\end{align*}
But I'm not sure how to proceed with either that polylogarithmic integral on the $5$th line nor the last one.
I'd appreciate any hints or ideas, thanks.
| $$\small \int _0^{\pi }x\sin \left(x\right)\operatorname{Li}_2\left(\cos \left(2x\right)\right)dx=\frac{\pi^3}{6}+6\pi \ln 2-4\pi +\frac{\pi}{\sqrt 2}\left(\operatorname{Li}_2(-(3+2\sqrt 2))-\operatorname{Li}_2(-(3-2\sqrt 2))\right)$$
$$\small\int_{-1}^1\frac{\sqrt{1+x}\ln(1-x)}{x}dx=6\sqrt 2 \ln 2 -4\sqrt 2+\operatorname{Li}_2(-(3+2\sqrt 2))-\operatorname{Li}_2(-(3-2\sqrt 2))$$
To show this result we can continue with the integral left at the end of the question body.
$$\int_{-1}^1\frac{\sqrt{1+x}\ln(1-x)}{x}dx=\int_{-1}^1 \frac{\ln(1-x)}{\sqrt{1+x}}dx+\int_{-1}^1 \frac{\ln(1-x)}{x\sqrt{1+x}}dx=\mathcal I+\mathcal J$$
$$\mathcal I=\int_{-1}^1 \frac{\ln(1-x)}{\sqrt{1+x}}dx \overset{1+x=t^2}=2\int_0^\sqrt 2\ln(2-x^2)dx=2\int_0^\sqrt 2 (x-\sqrt 2)'\ln(2-x^2)dx$$
$$\overset{IBP}=2\sqrt 2\ln 2 -4\int_0^\sqrt 2 \frac{x}{x+\sqrt 2}dx=6\sqrt 2 \ln 2 -4\sqrt 2$$
$$\mathcal J(t)=\int_{-1}^1 \frac{\ln(1-tx)}{x\sqrt{1+x}}dx\Rightarrow \mathcal J'(t)=-\int_{-1}^1 \frac{1}{(1-tx)\sqrt{1+x}}dx$$
$$\overset{1+x\to x^2}=-\frac{2}{t}\int_0^\sqrt 2\frac{1}{\frac{1+t}{t}-x^2}dx=-\frac{2\operatorname{arctanh}\left(\sqrt 2 \sqrt{\frac{t}{1+t}}\right)}{\sqrt t\sqrt{1+t}}$$
$$\mathcal J(0)=0\Rightarrow \mathcal J=-2\int_0^1 \frac{\operatorname{arctanh}\left(\sqrt 2 \sqrt{\frac{t}{1+t}}\right)}{\sqrt t\sqrt{1+t}}dt\overset{\frac{t}{1+t}=x^2}=-4\int_0^\frac{1}{\sqrt 2}\frac{\operatorname{arctanh}(\sqrt 2 x)}{1-x^2}dx$$
$$\overset{\sqrt 2 x\to x}=-4\sqrt 2 \int_0^1 \frac{\operatorname{arctanh} x}{2-x^2}dx\overset{x=\frac{1-t}{1+t}}=\int_0^1 \frac{\ln t}{3-2\sqrt 2+t}dt-\int_0^1\frac{\ln t}{3+2\sqrt 2 +t}dt$$
$$=\operatorname{Li}_2(-(3+2\sqrt 2))-\operatorname{Li}_2(-(3-2\sqrt 2))$$
Putting this togheter gives the announced result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3853327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this expression by the conjugate of the denominator.
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \right) = \lim _{x\to 2}\left(\dfrac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}\right) $$
But it still gives the indeterminate form $\frac{0}{0}$ .
But multiplying the expression by the conjugate of the demoninator and numerator we get
$$\lim _{x\to 2}\left(\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \cdot \dfrac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\right) $$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{1} \cdot \dfrac{1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{6-x-4}{3-x-1} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{2-x}{2-x} \cdot \dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}\right)$$
Now we can evaluate the limit:
$$\lim _{x\to 2}\left(\dfrac{\sqrt{3-2}+1}{\sqrt{6-2}+2}\right) = \dfrac{1}{2}$$
Taking this example, I would like to understand why rationalization was used. What did it change in the expression so the evaluation was possible? Especially, why multiplying by the numerator's and denominator's conjugate?
I am still new to limits and Calculus, so anything concerning concepts I'm missing is appreciated. I still couldn't understand how a limit supposedly tending to $\frac{0}{0}$ went to be $\frac{1}{2}$, I really want to understand it.
Thanks in advance for you answer.
| If you have an expression $$\frac{f(x)}{g(x)}$$where both $\lim_{x\to a}f(x)=0$ and $\lim{x\to a}g(x)=0$, you get a limit that looks like $\frac 00$. Now in this case, you write $$f(x)=(x-a)f_1(x)\\g(x)=(x-a)g_1(x)$$
Then
$$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim\frac{(x-a)f_1(x)}{(x-a)g_1(x)}=\lim_{x\to a}\frac{f_1(x)}{g_1(x)}$$
We can simplify the $x-a$ terms, because the expression is non zero, no matter how close we are, except at $x=a$. If your new limit has a solution, then you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 5
} |
Let $a_1,a_2,...,a_n>0$ be such that $\sum_{k=1}^n a_k=\frac{1}{2}n(n+1)$
Let $a_1,a_2,...,a_n>0$ be such that $\sum_{k=1}^n a_k=\frac{1}{2}n(n+1)$, then least value of
$$\sum_{k=1}^n\frac{(k^2-1)a_k+k^2+2k}{a_k^2+a_k+1}\text{ is,}$$
What I tried:
$$\sum_{k=1}^{n}k=\frac{1}{2}n(n+1)\implies k=a_k\tag{$\because\sum_{k=1}^n a_k=\frac{1}{2}n(n+1)$}$$
So replacing $a_k$ with $k$,
$$\begin{aligned}\require{cancel}
\sum_{k=1}^n\frac{(k^2-1)a_k+k^2+2k}{a_k^2+a_k+1}&=\sum_{k=1}^n\frac{(k^2-1)k+k^2+2k}{k^2+k+1}\\
&=\sum_{k=1}^{n}\frac{k^3-k+k^2+2k}{k^2+k+1}\\
&=\sum_{k=1}^{n}\frac{k^3+k+k^2}{k^2+k+1}\\
&=\sum_{k=1}^{n}\frac{k\color{red}{\bcancel{(k^2+k+1)}}}{\color{red}{\bcancel{(k^2+k+1)}}}\\
&=\sum_{k=1}^{n}k=\frac{1}{2}n(n+1)
\end{aligned}$$
Is this a correct method to solve this problem? because something feels off with this as the question asks for the least value but I get a direct result.
Please provide a correct method to solve if this is wrong
| Tips:
$$\frac{(k^2-1)a_k+k^2+2k}{a_k^2+a_k+1}=2k-a_k+\frac{(a_k+1)(a_k-k)^2}{a_k^2+a_k+1}\ge2k-a_k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Definite integral evaluation of $\frac{\sin^2 x}{2^x + 1}$
Compute
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{2^x + 1} \ dx.$$
So looking at the limits I first checked whether the function is even or odd by substituting $x = -x$ but this gave me:
$$\frac{\sin^2 x}{2^x + 1} \cdot 2^x$$
so the function is clearly neither odd or even and I don't know what to infer from this. I can't think of a good substitution either.
Any help would be appreciated!
| You did the right thing. Using your substitution, you have
$$
I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{2^x + 1} \ dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{2^x + 1} 2^x \ dx.
$$
Adding the two equivalent formulations you have
$$
2 I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{2^x + 1} (2^x +1) \ dx
= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\sin^2 x} \ dx = \int_0^{\frac{\pi}{2}} (1-\cos(2x))\ dx=\frac{\pi}{2}$$
which gives $I = \frac{\pi}{4}$ as the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Standard Form Ellipse Through Three Points and Parallel to X and Y Axes I want the general form $\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1$ for an ellipse with a specified eccentricity $e$ that passes through three (non-collinear) points $(x_1,y_1), (x_2, y_2), (x_3, y_3)$ and is parallel to the X and Y axes (i.e. major axis of ellipse parallel to the X axis and minor axis parallel to the Y axis).
I found this gem on Wikipedia:
$$
\frac{({\color{red}x} - x_1)({\color{red}x} - x_2) + {\color{blue}q}\;({\color{red}y} - y_1)({\color{red}y} - y_2)}
{({\color{red}y} - y_1)({\color{red}x} - x_2) - ({\color{red}y} - y_2)({\color{red}x} - x_1)} =
\frac{(x_3 - x_1)(x_3 - x_2) + {\color{blue}q}\;(y_3 - y_1)(y_3 - y_2)}
{(y_3 - y_1)(x_3 - x_2) - (y_3 - y_2)(x_3 - x_1)}\ .
$$
where ${\color{blue}q} = \frac{a^2}{b^2} = \frac{1}{1 - e^2}$, which I think is supposed to work, but a) converting this equation to standard form is a bear (and maybe isn't doable?), and b) seems to introduce $xy$ terms which leads me to believe the ellipse will be tilted with respect to the X and Y axes.
Is this the right equation to be working with? If so, is there a standard form of the equation? Is there a different/better way to accomplish the task?
P.S. Having the standard form is pretty important: I'm going to use this with a graphics app where knowing $x_0, y_0, a,$ and $b$ is required.
| Alternatively, the equation can be re-arranged in a compact form:
$$
\begin{vmatrix}
(1-e^2)x^2+y^2 & x & y & 1 \\
(1-e^2)x_1^2+y_1^2 & x_1 & y_1 & 1 \\
(1-e^2)x_2^2+y_2^2 & x_2 & y_2 & 1 \\
(1-e^2)x_3^2+y_3^2 & x_3 & y_3 & 1
\end{vmatrix}=0$$
where $e\ne 1$ and comparing with the general form
$$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$
Now,
\begin{align}
A &= (1-e^2) C \\ \\
B &= 0 \\ \\
C &=
\begin{vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1
\end{vmatrix} \\ \\
D &= -
\begin{vmatrix}
(1-e^2)x_1^2+y_1^2 & y_1 & 1 \\
(1-e^2)x_2^2+y_2^2 & y_2 & 1 \\
(1-e^2)x_3^2+y_3^2 & y_3 & 1
\end{vmatrix} \\ \\
E &=
\begin{vmatrix}
(1-e^2)x_1^2+y_1^2 & x_1 & 1 \\
(1-e^2)x_2^2+y_2^2 & x_2 & 1 \\
(1-e^2)x_3^2+y_3^2 & x_3 & 1
\end{vmatrix} \\ \\
F &= -
\begin{vmatrix}
(1-e^2)x_1^2+y_1^2 & x_1 & y_1 \\
(1-e^2)x_2^2+y_2^2 & x_2 & y_2 \\
(1-e^2)x_3^2+y_3^2 & x_3 & y_3
\end{vmatrix} \\
\end{align}
Re-arrange the equation as
$$A
\left( x+\frac{D}{2A} \right)^2+
C
\left( y+\frac{E}{2C} \right)^2=
\frac{D^2}{4A}+\frac{E^2}{4C}-F$$
implying the centre is
$$\left( -\frac{D}{2A}, -\frac{E}{2C} \right)$$
and the semi-axes
$$
(a,b)=
\left(
\sqrt{\frac{D^2}{4A^2}+\frac{E^2}{4AC}-\frac{F}{A}},
\sqrt{\frac{D^2}{4AC}+\frac{E^2}{4C^2}-\frac{F}{C}}
\right)$$
for $0 \le e<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3864927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$(\varepsilon, \delta)$ for continuity of a multivariable function
Define $f : \Bbb R^2 \to \Bbb R^2$, $$f(x,y) = \begin{cases}
0 & (x,y) = 0 \\
\frac{xy(x^2-y^2)}{x^2+y^2} & (x,y) \ne 0
\end{cases} $$ Determine if $f$ is continuous.
At $\Bbb R^2_{\ne 0}$ we have that $f$ is continuous. In order to see if it's continuous at the origin I was approaching it using epsilon-delta. We have that $|f(x,y) - f(0,0)| < \varepsilon$ whenever $\sqrt{(x-0)^2+(y-0)^2} = \sqrt{x^2+y^2} < \delta$.
Now $$|f(x,y)-f(0,0)| = \frac{|xy||x^2-y^2|}{|x^2+y^2|} \leqslant \frac{|xy||x^2+y^2|}{|x^2+y^2|} =|xy|.$$
However I'm losing $x^2+y^2$ here since that's what I would need for $\delta$. What am I missing in my approach here? Is there another way I could bound this?
| For all $(x,y)\in\mathbb{R}^2$ we have
$$
\left\lbrace
\begin{align*}
&2|xy|\le x^2+y^2\\[4pt]
&|x^2-y^2|\le x^2+y^2\\[4pt]
\end{align*}
\right.
$$
hence for all $(x,y)\in\mathbb{R}^2{\setminus}\{(0,0)\}$ we have
$$
2|f(x,y)|
=
\frac
{(2|xy|)(|x^2-y^2|)}
{x^2+y^2}
\le
\frac
{(x^2+y^2)(x^2+y^2)}
{x^2+y^2}
=
x^2+y^2
$$
which approaches $0$ as $(x,y)$ approaches $(0,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Invertibility of the $2\times 2$ symmetric matrix
Let $a, b, c \in \mathbb{R}$ \ $ \{0\} $ be nonzero, real constants. For which value of $a, b, c$ does the matrix $$A = \begin{pmatrix}a & b\\ b & c\end{pmatrix} $$
have two non-zero eigenvalues?
I already found the fact that $ac \neq b^2$ but I am wondering if there is anything more.
| Note the characteristic polynomial of $A$ is
\begin{align}
P\left (\lambda\right ) &= \left |\lambda I - A \right |\\
&= \left (\lambda - a\right )\left (\lambda - c \right ) - b^2\\
&= \lambda^2 - \left (a + c\right )\lambda + \left (ac - b^2\right )
\end{align}
In order for $A$ to have two distinct non-zero eigenvalues, we must have $ac - b^2 \neq 0$ and $\Delta_\lambda \neq 0$. Hence, we have
\begin{align}
\left (a+c\right )^2 - 4\left (ac - b^2\right ) &\neq 0\\
\left (a - c\right )^2 + 4b^2 &\neq 0
\end{align}
Note that this means we cannot have $a = c$ and $b = 0$ simultaneously (i.e. $A$ cannot be a multiple of the identity).
Therefore, we require $b^2 \neq ac$, and $A \neq kI$ for some real $k$.
We can easily check to confirm that these conditions are sufficient.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3867407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Real part of function vanishes Consider the function
$$
f(a+bi) = e^{-bi} + e^{\left(\frac{a\sqrt3}{2} + \frac{b}{2}\right)i} + e^{\left(-\frac{a\sqrt3}{2} + \frac{b}{2}\right)i}.
$$
Using a Mathematica plot, I can see that the real part of $f$ vanishes on circles in the complex plane, see below. Is there a way to rigorously show this?
| They look like circles, but they are not. If you look very closely, they are a little bit flatened at their vertical extremes. I will present a far more convincing mathematical argument below.
Note that
$$
f(z) = e^{-Im(z)i} + e^{-Im(z\bar\omega)i} + e^{-Im(z\omega)i}.
$$
Then translate it to cartesian coordinates, obtaining
$$
f(a+bi) = e^{-bi} + e^{\left(\frac{a\sqrt3}{2} + \frac{b}{2}\right)i} + e^{\left(-\frac{a\sqrt3}{2} + \frac{b}{2}\right)i}.
$$
Then, the real part will be
\begin{align*}
Re(f(a+bi)) & = \cos(-b) + \cos\left(\frac{a\sqrt3}{2} + \frac{b}{2}\right) + \cos\left(-\frac{a\sqrt3}{2} + \frac{b}{2}\right) \\
& = \cos(b) + 2\cos\left(\frac{a\sqrt3}{2}\right)\cos\left(\frac{b}{2}\right).
\end{align*}
So we are looking for $(a,b)\in\mathbb R^2$ such that
$$
\cos(b) + 2\cos\left(\frac{a\sqrt3}{2}\right)\cos\left(\frac{b}{2}\right)=0.
$$
If we solve it for $a=0$, according to Wolfram alpha, we get
$$
b = 4\left(\pi n \pm \arctan\left(\sqrt{2\sqrt{3}-3}\right)\right)
$$
where the solutions closest to $0$ are
$$
b^{\pm} = \pm 4\arctan\left(\sqrt{2\sqrt{3}-3}\right).
$$
If we solve it for $b=0$, according to Wolfram alpha, we get
$$
a = \frac{4(3\pi n \pm \pi)}{3\sqrt{3}}
$$
where the solutions closest to $0$ are
$$
a^{\pm} = \pm \frac{4\pi}{3\sqrt{3}}.
$$
By the plot, it seems that the points $(a^{\pm},0)$, $(0,b^{\pm})$ are in a same circumpherence. Let us see that they are not.
Suppose the points $(a^{\pm},0)$ and $(0,b^{\pm})$ are in a circumpherence $C$ of equation $(x-x_0)^2 + (y-y_0)^2 = r^2$ (a priori, we are not even assuming that $C$ is centered at the origin).
Using that $(a^{\pm},0)$ are in $C$ and $a^-=-a^+\neq0$, we get
$$
\left\{\begin{array}{l} (a^+-x_0)^2 + y_0^2 = r^2 \\
(-a^+-x_0)^2 + y_0^2 = r^2\end{array}\right. \Rightarrow a^+x_0 = 0 \Rightarrow x_0=0,
$$
and substituting it, we get
$$
(a^+)^2 + y_0^2 = r^2.
$$
Analogously, we show that $y_0=0$ and
$$
(b^+)^2 = r^2.
$$
Putting all together, we conclude that $x_0=y_0=0$, so $C$ is centered at the origin, and its radius satisfies
$$
r= a^+ = b^+\ \ \Rightarrow\ \ \arctan\left(\sqrt{2\sqrt{3}-3}\right) = \frac{\pi}{3\sqrt{3}}.
$$
A simple computation (I used Wolfram alpha again) shows that the last equality is false.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3868839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{x\to0}\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$ I was trying to evaluate
$$\lim_{x\to0}\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$$
I have tried taking natural logarithm first:
$\lim_{x\to0}\frac{\ln(\sin x)-\ln x}{1-\cos x}=\lim_{x\to0}\frac{\frac{\cos x}{\sin x}-\frac{1}x}{\sin x}\quad\quad\text{(L'Hopital Rule)}\\
=\lim_{x\to0}\frac{\cos x}{x\sin x}-\frac{1}{x^2}\quad\quad(\lim_{x\to0}\frac{\sin x}x=1)\\
=\lim_{x\to0}\frac{\cos x-1}{x^2}\quad\quad\quad(\lim_{x\to0}\frac{\sin x}x=1)$
and after this I eventually have the limit equaling $-\frac{1}2$, which means that the original limit should be $\frac{1}{\sqrt{e}}$.
However, I graphed $f(x)=\left(\frac{\sin x}x\right)^\frac{1}{1-\cos x}$ on Desmos, and it turned out that the limit is approximately $0.7165313$, or $\frac{1}{\sqrt[3]{e}}$.
Therefore I think there's something wrong in my approach, but I couldn't find it. Any suggestions?
| $$\lim_{x\to0}\frac{\frac{\cos x}{\sin x}-\frac{1}x}{\sin x}=\lim_{x\to0}\frac{(x-\frac{x^3}{2!}+\frac{x^5}{4!}+\cdots)-(x-\frac{x^3}{3!}+\frac{x^5}{5!}+\cdots)}{x^3}=\frac{1}{6}-\frac{1}{2}=-\frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is the limiting sum of $\frac{1}{1\left(3\right)}+\frac{1}{3\left(5\right)}+\frac{1}{5\left(7\right)}+···+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$ I know $a$ (first term) is $\frac{1}{3}$, but I'm not sure what the common ratio would be from this. I assume it is the $\frac{1}{(2n-1)(2n+1)}$, but this is not a number. Any hints?
| $$S=\frac12\big(\frac{1}{2-1}-\frac{1}{2+1}+\frac{1}{4-1}-\frac{1}{4+1}+ \frac{1}{6-1}-\frac{1}{6+1}+\cdot\cdot\cdot+\frac{1}{2n-1}-\frac{1}{2n+1}\big) $$
Cancelling similar terms we finally get:
$$S= \frac12\big(1-\frac{1}{2n+1}\big)=\frac12\times\frac{2n}{2n+1}$$
$$\lim _{n→∞ } S=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluation of a telescoping sum I have come to a problem in a book on elementary mathematics that I don't understand the solution to. The problem has two parts :
a.) Factorize the expression $x^{4} + x^{2} + 1$
b.) Compute the value of the sum $\sum_{k=1}^{n} \frac{k}{k^{4} + k^{2} + 1}$ in terms
of $n \in \mathbb{N}$.
I was able to perform part (a.) to get :
\begin{equation}
x^{4} + x^{2} + 1 = (x^{2} + x + 1)(x^{2} - x + 1)
\end{equation}
I wasn't able to do part (b.), but in the answer key the first part of the solution is :
\begin{align}
\frac{k}{k^{4} + k^{2} + 1} & = \frac{k}{(k^{2} + k + 1)(k^{2} - k + 1)} \\
& = \frac{1}{2} \left( \frac{1}{k^{2} -k + 1} - \frac{1}{k^{2}+k+1} \right)
\end{align}
I can see that the first transformation above comes from the answer to part (a.). I do not understand how they did the second transformation. Could someone show me how to go from the second expression to the third above ?
| We can start from
$$ \frac{A}{k^{2} -k + 1} + \frac{B}{k^{2}+k+1}=\frac{A(k^{2}+k+1)+B(k^{2} -k + 1)}{(k^{2} -k + 1)(k^{2}+k+1)} = \frac{k}{(k^{2} + k + 1)(k^{2} - k + 1)} $$
from which we obtain $A=-B=\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3873819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to calculate the arc length of a difficult radical function I have been struggling with an arc length question, and I want to make sure I get this right. I have the function of:
\begin{align}
f(x) = \sqrt{7.2 (x-\frac {1}{7}}) - 2.023, [0.213, 0.127].
\end{align}
I have found the derivative of the function and set up my integral this way:
\begin{align}
I &= \int_{0.127}^{0.213} \sqrt{1 + \frac{12.96}{7.2x-\frac{7.2}{7}}}~dx
\end{align}
Letting A = 12.96 and simplifying:
\begin{align}
I &= \int_{0.127}^{0.213} \sqrt{\frac{7.2x-\frac{7.2}{7}+A}{7.2x-\frac{7.2}{7}}}~dx
\end{align}
$u=7.2x-\frac{7.2}{7}, du= 7dx, dx=\frac{du}{7}$:
\begin{align}
I &= \int_a^b \sqrt{\frac{{u}+A}{u}}~\frac{du}{7}
\end{align}
\begin{align}
I &= \frac{1}{7}\int_a^b \sqrt{\frac{{u}+A}{u}}~du
\end{align}
$u = C\tan^2v\\ du = 2C \tan v \sec^2 v ~ dv$
\begin{align}
I
&= \frac{1}{7}\int_{x=a}^{x=b} \sqrt{\frac{u + A}{u}}~du\\
&= \frac{1}{7}\int_{x=a}^{x=b} \sqrt{\frac{A(\tan^2 v + 1)}{A \tan^2 v}}~2A\tan v \sec^2 v ~ dv\\
&= \frac{2A}{7}\int_{x=a}^{x=b} \sqrt{\frac{\sec^2 v}{\tan^2 v}}\tan v \sec^2 v ~ dv; & \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \sqrt{\frac{\frac{1}{\cos^2 v}}{\frac{\sin^2 v}{\cos^2 v}}}\frac{\sin v}{\cos^3 v}~ dv; \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \sqrt{\frac{1}{\sin^2 v}}\frac{\sin v}{\cos^3 v}~ dv \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{\sin v}\frac{\sin v}{\cos^3 v}~ dv \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{\cos^3 v}~ dv \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{\cos v}{\cos^4 v}~ dv \\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{\cos v}{(1-\sin^2(v))^2} dv~
\end{align}
This is where I am stuck. Could I make a substitution such as:
$t = \sin v\\dt=\cos v\ dt\\\frac{dt}{cos\ v}=dv$
and then:
\begin{align}
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{\cos v}{(1-t^2)^2} \frac{dt}{cos\ v}~\\
&= \frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{(1-t^2)^2} dt~
\end{align}
which gives me an ordinary partial fractions integral.
Could I make this substitution or is it not possible because I would have 2 different variables in my integral, and if it's not possible, how else could I solve this integral?
| I actually found the answer to my question.
\begin{align}
& I=\frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{\cos^3 v}~ dv \\
& =\frac{2A}{7}\int_{x=a}^{x=b} sec^3v~ dv \\
& =\frac{A}{7}\int_{x=a}^{x=b} \sec v \tan v+ln \lvert\sec v + \tan v\rvert~ dv \\
\end{align}
The new bounds would be:
$arctan\frac{\sqrt{7.2*1.27-\frac{7.2}{7}}}{\sqrt{A}}$ and
$arctan\frac{\sqrt{7.2*0.213-\frac{7.2}{7}}}{\sqrt{A}}$
It was much simpler than I thought.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
cyclic rational inequalities $\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$ when $a+b+c=1$ I've been practicing for high school olympiads and I see a lot of problems set up like this:
let $a,b,c>0$ and $a+b+c=1$. Show that
$$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$$
Any problem that involves cyclic inequalities like these always stump me. I know I'm supposed to use Cauchy-Schwarz or AM-GM at some point, but I can never get to a place where this might be useful. My first instinct is to get common denominators and hope stuff simplifies, but I can never get farther than that. For example, in this problem I did the following:
$$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}$$
$$=\frac{(a^2+3)(b^2+3)+(b^2+3)(c^2+3)+(c^2+3)(a^2+3)}{(a^2+3)(b^2+3)(c^2+3)}$$
$$=\frac{a^2b^2+b^2c^2+c^2a^2+6(a^2+b^2+c^2)+27}{(a^2+3)(b^2+3)(c^2+3)}$$
but this is where I get stuck. I've tried using Cauchy-Schwarz on parts of this fraction to simplify it, but I can never get anything to work. How could you prove this inequality, and what are the important things to look out for in problems of this nature
| Tangent Line method helps.
Indeed, let $a=\frac{x}{3},$ $b=\frac{y}{3}$ and $c=\frac{z}{3}.$
Thus, $x+y+z=3$ and
$$\frac{27}{28}-\sum_{cyc}\frac{1}{a^2+3}=\sum_{cyc}\left(\frac{9}{28}-\frac{9}{x^2+27}\right)=\frac{9}{28}\sum_{cyc}\frac{x^2-1}{x^2+27}=$$
$$=\frac{9}{28}\sum_{cyc}\left(\frac{x^2-1}{x^2+27}-\frac{1}{14}(x-1)\right)=\frac{9}{392}\sum_{cyc}\frac{(x-1)^2(13-x)}{x^2+27}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3878235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Factoring $X^5 - X^4 - X^3 + X^2 + X$ into irreducible factors: Is my solution correct? I'm trying to reduce $f(x) = x^5 - x^4 - x^3 + x^2 + x$ in $F_3[X]$ in irreducible factors.
What I came up with was $(x^2 + x + 2)(x^2 - 2x + 2)(x)$
How I did it was I first found all the irreducible polynomials of degree $2$ in $F_3[X]$. Then I did long divisions on $f(x)$ with one of the polynomials I found (brute force), until found a factor. Which was $x^3 - 2x^2 + 2x$. Dividing it by $x$ gives $x^2 - 2x + 2$. And that's how I found the factors.
Is this a correct solution to the question? I'm especially having doubts about the last part of my solution where I just divide it by X. Hopefully this is okay to ask here.
| Your answer is right.
I like the following way: $$x^5-x^4-x^3+x^2+x=x(x^4-x^3-x^2+x+1)=$$
$$=x(x^4+2x^3-x^2-2x+1)=x(x^2+x-1)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3879184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Why is $\sec^2 (\tan^{-1} (\frac{x}{2})) = 1 + \tan^2 (\tan^{-1} (\frac{x}{2}))$ and not $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$?
Why is $\sec^2 (\tan^{-1} (\frac{x}{2})) = 1 + \tan^2 (\tan^{-1} (\frac{x}{2}))$ and not $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$?
If you replace $\sec^2$ with $\tan^2 + 1$, it should be $(1 + \tan^2 )(\tan^{-1} (\frac{x}{2}))$, right?
However, it seems that it is not multiplication here : $\sec^2 (\tan^{-1} (\frac{x}{2}))$, when the $1$ becomes independent of the $\tan^2$ somehow.
Could someone explain why?
| The identity
$$\sec^2\theta\equiv1+\tan^2\theta$$
holds for all values of $\theta$, not just some values- that's why it's an identity, as opposed to an equation. Replacing $\theta$ by $\tan^{-1}(\frac{x}{2})$ yields the correct identity:
$$\sec^2(\tan^{-1}(\frac{x}{2}))\equiv1+\tan^2(\tan^{-1}(\frac{x}{2}))\equiv1+\left(\tan(\tan^{-1}(\frac{x}{2}))\right)^2\equiv1+\frac{x^2}{4}$$
Edit
Thinking that $\sec^2\theta$ means $\sec^2\times\theta$ is meaningless- $\sec^2$ or indeed any other trigonometrical function has no meaning unless it has an argument, ie the value the that is an input into the trigonometical function (in your case $\tan^{-1}\frac{x}{2}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3884630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving a f is continuous at 1/3 Show that $f(x) = \frac{1}{5x}$ is continuous at $x = \frac{1}{3}$. I have to use an $\varepsilon - \delta$ proof? I am having trouble choosing my delta because I am confused as to how $|\frac{1}{x} - \frac13|$ can get to $|x - \frac{1}{3}|$?
| A function is continuous at a given point $x_0$ if for every $ϵ>0$ you can find some $δ>0$ such that $|x−x_0|<δ⟹|f(x)−f(x_0)|<ϵ.$
What is useful with these problems is to "solve them backwards". Start with
$$\left|\frac{1}{5x} - \frac{1}{15}\right| < \epsilon$$
What happens when $0<x<3$?
$$\frac{1}{5x} - \frac{1}{15} < \epsilon$$
$$\frac{1}{x} - \frac{1}{3} < 5 \epsilon$$
$$\frac{1}{x} < \frac{1}{3} + 5 \epsilon = \frac{1+15 \epsilon}{3}$$
$$x > \frac{3}{1+15\epsilon}$$
What happens when $x>3?$
$$-\frac{1}{5x} + \frac{1}{15} < \epsilon$$
$$-\frac{1}{x} + \frac{1}{3} < 5\epsilon$$
$$ \frac{1}{3} - 5\epsilon <\frac{1}{x}$$
Now we're gonna want to guarantee that the left side of this inequality is positive, so choose $\epsilon <\frac{1}{15}.$
$$ \frac{1-15\epsilon}{3} <\frac{1}{x}$$
$$ \frac{3}{1-15\epsilon} >x$$.
In conclusion, we see that when $\epsilon <\frac{1}{15}$ and $x>0,$ then $x$ lies in the range
$$\frac{3}{1+15\epsilon} < x < \frac{3}{1-15\epsilon}.$$
So
$$\frac{3}{1+15\epsilon} -3 < x -3 < \frac{3}{1-15\epsilon} -3.$$
Now think about how you can choose $\delta$ so that $|x-3| < \delta$ implies the above condition. (Hint: just make sure $\delta$ is the minimum of the two). Do you need to worry about $x<0$? (Hint: you could just choose $\delta$ to guarantee this doesn't happen).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3884941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Basis of Field F over $\mathbb{Q}$ I am self studying Field theory and got struck on this problem.
If $\;F=\mathbb{Q} \left(\sqrt{2}, \sqrt{3}\right),\;$ find
$\;\left[F:\mathbb{Q}\right]\;$ and a basis of $F$ over $\mathbb{Q}$.
I have proved that $\left[F:\mathbb{Q}\right] =4$, but there is a problem in basis elements.
My basis set is $\left\{a, b\sqrt{2}, c\sqrt{3}\right\}$ such that $a, b, c$ belongs to $\mathbb{Q}$. But what should be fourth element and why ?
I am not able to see .
Kindly help.
| A basis of $F$ over $\mathbb{Q}$ is $\;\left\{1,\sqrt{2},\sqrt{3},\sqrt{6}\right\},$ indeed for any element $\;x\in F\;$ there exist $\;r_1,\;r_2,\;r_3,\;r_4,\;r_5,\;r_6,\;r_7,\;r_8\in\mathbb{Q}\;$ such that
\begin{align}
x&=\dfrac{r_1+r_2\sqrt{2}+r_3\sqrt{3}+r_4\sqrt{6}}{r_5+r_6\sqrt{2}+r_7\sqrt{3}+r_8\sqrt{6}}\\
&=\dfrac{r_1+r_2\sqrt{2}+r_3\sqrt{3}+r_4\sqrt{6}}{r_5+r_6\sqrt{2}+r_7\sqrt{3}+r_8\sqrt{6}}\cdot\dfrac{r_5+r_6\sqrt{2}-r_7\sqrt{3}-r_8\sqrt{6}}{r_5+r_6\sqrt{2}-r_7\sqrt{3}-r_8\sqrt{6}}\\
&=\dfrac{\left(r_1+r_2\sqrt{2}+r_3\sqrt{3}+r_4\sqrt{6}\right)\left(r_5+r_6\sqrt{2}-r_7\sqrt{3}-r_8\sqrt{6}\right)}{\left(r_5+r_6\sqrt{2}\right)^2-\left(r_7\sqrt{3}+r_8\sqrt{6}\right)^2}\\
&=\dfrac{s_1+s_2\sqrt{2}+s_3\sqrt{3}+s_4\sqrt{6}}{s_5+s_6\sqrt{2}}\\
&=\dfrac{s_1+s_2\sqrt{2}+s_3\sqrt{3}+s_4\sqrt{6}}{s_5+s_6\sqrt{2}}\cdot\dfrac{s_5-s_6\sqrt{2}}{s_5-s_6\sqrt{2}}\\
&=\dfrac{\left(s_1+s_2\sqrt{2}+s_3\sqrt{3}+s_4\sqrt{6}\right)\left(s_5-s_6\sqrt{2}\right)}{\left(s_5+s_6\sqrt{2}\right)\left(s_5-s_6\sqrt{2}\right)}\\
&=q_11+q_2\sqrt{2}+q_3\sqrt{3}+q_4\sqrt{6}
\end{align}
where $\;q_1,\;q_2,\;q_3,\;q_4\in\mathbb{Q}\;$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3887179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the recurrence relation: $na_n = (n-4)a_{n-1} + 12n H_n$ I want to solve
$$ na_n = (n-4)a_{n-1} + 12n H_n,\quad n\geq 5,\quad a_0=a_1=a_2=a_3=a_4=0. $$
Does anyone have an idea, what could be substituted for $a_n$ to get an expression, which one could just sum up? We should use
$$ \sum_{k=0}^n \binom{k}{m} H_k = \binom{n+1}{m+1} H_{n+1} - \frac{1}{m+1} \binom{n+1}{m+1} $$
to simplify the result.
| Given by a CAS.
Isuppose that is missing the condition $a_3=0$.
Starting from $n=5$, the sequence
$$\left\{\frac{137}{25},\frac{1009}{150},\frac{17953}{2450},\frac{151717}{19600},\frac{
170875}{21168},\frac{1474379}{176400},\frac{3751927}{435600},\frac{20228477}{22869
00}\right\}$$ is not recognized by $OEIS$.
Being totally stuck, I used a CAS without conditions and got, after a lot of simplifications,
$$a_n=\frac{(-3 n^4+22 n^3-69 n^2+194 n+288+96 C) } {4 (n-3) (n-2) (n-1) n }+$$ $$\frac{12 (n-4) (n+1) \left(n^2-3 n+6\right) H_{n+1} } {4 (n-3) (n-2) (n-1) n }$$
For $n=0,1,2,3$, this leads to indeterminate forms. For $n=4$
$$a_4=C+\frac{25}{4}=0 \implies C=-\frac{25}{4}$$ leading to
$$a_n=(n-4)\frac{12 (n+1) \left(n^2-3 n+6\right) H_{n+1}-(n-3) \left(3 n^2-n+26\right) } {4 (n-3) (n-2) (n-1) n }$$ which is identical to what @Raffaele wrote in a comment.
Asymptotically,
$$a_n=3 \left(\gamma -\frac{1}{4}\right)+3 \log (n)+\frac{11}{2 n}-\frac{25}{4
n^2}+O\left(\frac{1}{n^3}\right)$$
I would really like to know how, starting from scratch, we could arrive to the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Show that three numbers form an arithmetic progression
The numbers $a,b$ and $c$ form an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.
We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{ } \forall \text{ }n\ge2$). I am stuck here and I would be very grateful if you could give me a hint.
| More hint:
\begin{align}&\quad(a^2+ab+b^2) + (b^2+bc + c^2) \\&= a^2+c^2 +2b^2 + b(a+c) \\&= a^2 + c^2 + 2b^2 + b(2b) \\&= a^2 + c^2 + 4b^2\\&=a^2+c^2+(a+c)^2\end{align}
In the worst case you can just substitute $b = \frac {a+c}2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3892856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Asymptotic estimate as $N \rightarrow \infty$ of $\sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\}$ Looking for the exact if possible, otherwise the asymptotic expansion and best estimate of the error terms as $N \rightarrow \infty$ of the two fractional sums $$\sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\}$$ My literature search has not found any examples similar to this. I have some material for the divisors terms such as $\left\lfloor{N/a}\right\rfloor^k$ and $\left\{{N/a}\right\}^k$. This is part of the calculation of the summations over the floor function of this argument.
From Benoit Cloitre. On the circle and divisor problem. November 2012 we have
$$\lim_{x \rightarrow \infty} \sum_{n = 1}^{x} \left\lfloor{\frac{x}{{n}^{2}}}\right\rfloor \sim
\zeta \left({2}\right) x
+ \zeta \left({\frac{1}{2}}\right) \sqrt{x}
+ O \left({{x}^{\theta}}\right)$$
Where $\theta = 1/4 + \epsilon$ is the estimated best error.
| Asymptotic is $(1 - \gamma) N$, where $\gamma$ is Euler–Mascheroni constant.
Proof
For any $x, y$:
$$
\begin{array}\\
\{x \pm y\} &= x \pm y - [x \pm y] \\
&= x \pm y - [[x] + \{x\} \pm [y] \pm \{y\}] \\
&= x - [x] \pm \{y\} - [\{x\} \pm \{y\}].
\end{array}
$$
Now set $x = \frac Nn, y = \frac N{n^2}$ to break the sum apart:
$$
\sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\}
= \underbrace{\sum\limits_{n = 1}^{N} \frac Nn}_{(1)}
- \underbrace{\sum\limits_{n = 1}^{N} \left[ \frac Nn \right]}_{(2)}
\pm \underbrace{\sum\limits_{n = 1}^{N} \left\{ \frac{N}{n^2} \right\} }_{(3)}
- \underbrace{\sum\limits_{n = 1}^{N} \left[ \left\{ \frac Nn \right\} \pm \left\{\frac{N}{n^2} \right\} \right]}_{(4)}.
$$
$(1)$ is Harmonic series, $(1) = N \ln N + \gamma N + \frac 12 + o(1)$.
$(2)$ is divisor summatory function, $(2) = N \ln N + N(2\gamma - 1) + O(\sqrt N)$.
$(3) = \underbrace{ \sum\limits_{n = 1}^{ \left[ \sqrt N \right]} \left\{ \frac{N}{n^2} \right\} }_{(3.1)}
+ \underbrace{ \sum\limits_{n = \left[ \sqrt N \right]+1}^{N} \left\{ \frac{N}{n^2} \right\} }_{(3.2)}.
$
$
(3.1) \leq \sum\limits_{n = 1}^{ \left[ \sqrt N \right]} 1 \leq \sqrt N.
$
$(3.2) = \sum\limits_{\left[ \sqrt N \right]+1}^{N} \frac{N}{n^2}
\leq N \cdot \sum\limits_{\left[ \sqrt N \right]+1}^{N} \frac{1}{n (n-1)}
= N \cdot \left( \sum\limits_{\left[ \sqrt N \right]+1}^{N} \frac{1}{n-1} - \frac{1}{n} \right)
= N \left( \frac{1}{\left[ \sqrt N \right]} - \frac{1}{N} \right)
\leq \frac {N}{\sqrt{N} + 1} - 1.
$
$(4) = O(\sqrt N)$. Proof is very technical and is written below.
Putting $(1)$, $(2)$, $(3)$, $(4)$ together, and leaving only leading asymptotic terms we have
$$
\sum\limits_{n = 1}^{N} \left\{{\frac{\left({n \pm 1}\right)}{{n}^{2}} N}\right\}
= (1 - \gamma) N + O(\sqrt N).
$$
Proving $(4) = O(\sqrt N)$
We want to show that $\sum\limits_{n = 1}^{N} \left[ \left\{ \frac Nn \right\} \pm \left\{\frac{N}{n^2} \right\} \right] = O(\sqrt N)$.
$$
\sum\limits_{1}^{N} [...] = \sum\limits_{1 \leq n \leq \frac{N}{\left[\sqrt N \right]} }[...] + \sum\limits_{ \frac{N}{\left[\sqrt N \right]} < n \leq N } [...], \\
$$
We split the sum in such a way, so that
*
*First part doesn't have too many summands.
*In the second sum we have $n > \frac{N}{\left[\sqrt N \right]} \geq \sqrt N$, which means we can "drop" braces: $\left\{ \frac{N}{n^2} \right\} = \frac{N}{n^2}$.
*It will be convenient to work with the second sum later.
First sum is $O(\sqrt N)$ because the $[...]$ part equals either $-1$, $0$ or $1$:
$$
\left| \sum\limits_{1 \leq n \leq \frac{N}{\left[\sqrt N \right]}} [...] \right| \leq \sum\limits_{1 \leq n \leq \frac{N}{\left[\sqrt N \right]}} 1 = O(\sqrt{N}) .
$$
We'll split second sum even further, so that we can also "drop" braces for $\left\{ \frac Nn \right\}$:
$$
\sum\limits_{ \frac{N}{\left[\sqrt N \right]} < n \leq N} [...]
= \sum\limits_{k=1}^{\left[ \sqrt N \right] - 1} \sum\limits_{\frac{N}{k + 1} < n \leq \frac Nk} [...].
$$
Note, that $\frac{N}{k + 1} < n \leq \frac Nk \implies k \leq \frac Nn < k + 1 \implies \left\{ \frac Nn \right\} = \frac Nn - k$.
$$
[...]
= \left[ \left\{ \frac Nn \right\} \pm \left\{\frac{N}{n^2} \right\} \right]
= \left[ \frac Nn - k \pm \frac {N}{n^2} \right]
= \left[ N \frac{n \pm 1}{n^2} \right] - k.
$$
When "$\pm$" is "$+$", the $[...]$ is either $0$ or $1$. We want to find for how many $n$ it is $1$.
$$
\left[ N \frac{n + 1}{n^2} \right] - k = 1 \iff N \frac{n + 1}{n^2} \geq k + 1 \iff \frac{k+1}{N}n^2 - n - 1 \leq 0, \\
\text{where} \; n \in \left( \frac{N}{k+1}; \frac Nk \right].
$$
Solving quadratic inequality gives $n \in \left( \frac{N}{k+1}; \frac{N}{k+1} \frac{1 + \sqrt{1 + 4 \frac{k+1}{N}}}{2} \right] $. Length of this semi-interval is
$$
\frac{N}{k+1} \frac{1 + \sqrt{1 + 4 \frac{k+1}{N}}}{2} - \frac{N}{k+1}
= \frac{N}{k+1} \frac{-1 + \sqrt{1 + 4 \frac{k+1}{N}}}{2}
= \frac{N}{k+1} \frac{-1 + 1 + 4 \frac{k+1}{N}}{2 \left(1 + \sqrt{1 + 4 \frac{k+1}{N}} \right) }
= \frac{2}{1 + \sqrt{1 + 4 \frac{k+1}{N}}} < 1.
$$
This means that at most $1$ integer $n$ can be inside that semi-interval.
When "$\pm$" is "$-$", the logic is similar, in that case there can be at most $2$ integer $n$ for which $[...] \neq 0$.
Finally, for the second sum we have $$
\left| \sum\limits_{ \frac{N}{\left[\sqrt N \right]} < n \leq N} [...] \right|
\leq \sum\limits_{k=1}^{\left[ \sqrt N \right] - 1} 2
= O(\sqrt N).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
In $\mathbb{R}$ find the solutions of the equation $x^4-x^3-18x^2+3x+9=0$ I couldn't solve this question and hence looked at the solution which goes as follows:
$0$ is no a root of the equation. Hence:
$$x^2-x-18+\frac{3}{x}+\frac{9}{x^2}=0. \text{ So, }
x^2+\frac{9}{x^2}-(x-\frac{3}{x})-18=0$$
I state that $y=x-\frac{3}{x}$, so we have that $y^2=x^2+\frac{9}{x^2}-6$, in other words $x^2+\frac{9}{x^2}=y^2+6$. Hence the equation is written as $y^2+6-y-18=0$ so $y^2-y-12=0$, hence $y=4$ or $y=-3$ so $x=2\pm\sqrt{7}$, or $x=\frac{-3\pm\sqrt{21}}{2}$
Could you please explain to me why the solution author of the solution thought of originally dividing by the equation by x and after that substituting $x-\frac{3}{x}$ with $y$? Also if you can think of a more intuitive approach could you please show it?
| For the factorization, using the long, long way ,consider
$$(x^2+ax+b)(x^2+cx+d)-(x^4-x^3-18x^2+3x+9)=0$$ Expand and group terms to gat
$$(b d-9)+x (a d+b c-3)+x^2 (a c+b+d+18)+x^3 (a+c+1)=0$$ So, $c=-a-1$; replace
$$(b d-9)+x ((-a-1) b+a d-3)+x^2 ((-a-1) a+b+d+18))=0$$ So, $b=a^2+a-d-18$; replace
$$\left(d \left(a^2+a-d-18\right)-9\right)+x (-a (a (a+2)-2
d-17)+d+15)=0$$ So, $d=\frac{a^3+2 a^2-17 a-15}{2 a+1}$; replace
$$\frac {(a-3)(a+4)(a^4+2 a^3-23 a^2-24 a-3 )} {(1+2a)^2 }=0$$ The quartic does not show any real root : so $a=3\implies c=-4$ or $a=-4\implies c=3$. Choose the one you prefer and run back for $b$ and $d$.
Edit
I made a mistake when I wrote "the quartic does not show any real root". What I was supposed to write is that "the quartic does not show any rational root". This correction followed @Macavity's comment.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Prove the inequality $x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$. I want to prove $$x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 <0$$
for $x>0$ real number.
I tried supposing that $$x-4x \frac{x}{\sqrt{x}+x} +3x \left( \frac{x}{\sqrt{x}+x} \right)^2 -3 \left( \frac{x}{\sqrt{x}+x} \right)^2 \geq 0$$ holds to get a contradiction but I couldn't find the solution.
Can you help me proceed?
| Answer :
$x-4x\frac{x}{\sqrt{x} +x} +3x(\frac{x}{\sqrt{x} +x})^2 - 3(\frac{x}{\sqrt{x} +x})^2 =1-4\frac{x}{\sqrt{x} +x} +3(\frac{x}{\sqrt{x} +x})^2 - 3\frac{x}{(\sqrt{x} +x)^2 }= \frac{(\sqrt{x} +x)^2 - 4x(\sqrt{x} +x) +3x^2 - 3x}{(\sqrt{x} +x)^2} =\frac{x+2x\sqrt{x} +x^2 - 4x\sqrt{x} - 4x^2 +3x^2 - 3x}{(\sqrt{x} +x)^2} =\frac{-2x-2x\sqrt{x}}{(\sqrt{x} +x)^2}=\frac{-2\sqrt{x}}{\sqrt{x} +x} <0$
Finally :
$x-4x\frac{x}{\sqrt{x} +x} +3x(\frac{x}{\sqrt{x} +x})^2 - 3(\frac{x}{\sqrt{x} +x})^2 <0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find a side of smaller triangle
$\triangle ABC$ has side $\overline{AB}=160 cm$, $\overline{BC}=190 cm$, and $\overline{CA}=190 cm$.
Point $D$ is along side $\overline{AB}$ and $\overline{AD} = 100 cm$. Point $E$ is along side $\overline{CA}$.
Determine the length of $\overline{AE}$ if the area of $\triangle ADE$ is three-fifths the area of $\triangle ABC$.
Is my solution correct? Thanks in advance!
| As indicated in comment, you're assuming $\triangle AED$ is isosceles (i.e., $DE$ || $BC$) which need not be true.
If you know formula for area of a $\triangle$ is also given as
$$ \dfrac{1}{2}AB\cdot AC\sin A$$
then using this one can directly solve
$$ \dfrac{1}{2}AD\cdot AE\sin A = \dfrac{3}{5} \cdot\dfrac{1}{2}AB\cdot AC\sin A$$
$$ \Rightarrow AE = \dfrac{3}{5}\dfrac{AB\cdot AC}{AD}$$
without any use of trigonometry.
Otherwise,
$$ \dfrac{1}{2}AD\cdot h_2 = \dfrac{3}{5} \cdot\dfrac{1}{2}AB\cdot h_1$$
$$ \Rightarrow h_2 = \dfrac{24}{25}h_1$$
and since the two right triangles formed by $h_1$, $h_2$ and containing $\angle A$ are similar
$$ \dfrac{AE}{AC} = \dfrac{h_2}{h_1}$$
$$ \Rightarrow AE = \dfrac{24}{25}AC$$
Either should give $AE = 182.4 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given X+Y+Z = 0, find $\frac{(X^3+Y^3+Z^3)}{XYZ}$ The result can be found using the equation :
$(X^3+Y^3+Z^3) - 3XYZ = (X+Y+Z)(X^2 - XY +Y^2 - YZ +Z^2 - XZ)$
Since X+Y+Z = 0, the right side of the equation is equal to 0. Therefore $X^3+Y^3+Z^3 = 3XYZ$ and the answer to the problem is 3.
However what if we calculate $X^3+Y^3+Z^3$ as $(X+Y+Z)^3 - 3X^2Y - 3Y^2X - 3X^2Z - 3z^2X - 3Y^2Z - 3Z^2Y$. $(X+Y+Z)^3 = 0$, so $X^3+Y^3+Z^3$ can be replaced with $- 3X^2Y - 3Y^2X - 3X^2Z - 3Z^2X - 3Y^2Z - 3Z^2Y$.
$\frac{- 3X^2Y - 3Y^2X - 3X^2Z - 3Z^2X - 3Y^2Z - 3Z^2Y}{XYZ}$ = $\frac{-3X - 3Y}{Z} - \frac{-3X-3Z}{Y} - \frac{-3Y-3Z}{X}$
If X+Y+Z = 0, consequenty 3X+3Y+3Z = 0. Our expression can be rewritten as $\frac{-3Z}{Z} + \frac{-3Y}{Y} +\frac{-3X}{X}$, so the answer is -9.
Could you please tell me which way of solving this problem is right and why
| The first attempt is correct, assuming that $X, Y, Z \neq 0$. You can verify this with a simple example, say $X = Y = 1, Z = -2$. The mistake in the second attempt is due to the fact that you have missed a term in your expression of $X^3 + Y^3 + Z^3$. Specifically,
$$(X + Y + Z)^3 = X^3 + Y^3 + Z^3 + 3(X^2 Y + Y^2 X + X^2 Z + Z^2 X + Y^2 Z + Z^2 Y) + \color{blue}{6XYZ}$$
To avoid errors like the one you made, always check that the number of terms in your expansion is correct! We expect $3^3 = 27$ terms in the expansion of $(X + Y + Z)^3$, and your expansion only gave $21$. You can then plug this into your second method and you will see that you get the correct answer of $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Why does the series diverge? Why does the series $\sum_{n=1}^\infty(-1)^n n$ diverge?
Is it because the expanded series:
$ -1 + 2 -3 + 4 -5 + 6... $
is the difference between the infinite sum of the even and odd numbers which would be infinity?
| Method 1
Try grouping 2 terms at a time
$-1+2-3+4-5+...$
$=(-1+2)+(-3+4)+(-5+6)+...$
$=1+1+1+...$
$=∞$
Method 2
$-1=-1$
$-1+2=1$
$-1+2-3=-2$
$-1+2-3+4=2$
$-1+2-3+4-5=-3$
$-1+2-3+4-5+6=3$
As we can observe the sequence doesn't seem to(and will not) converge at a particular points on the number line
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3907996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integration of $\int \sqrt\frac{x-1}{x^5}\ dx$? How would you integrate $\int \sqrt\frac{x-1}{x^5}\ dx$?
I've tried to split it up to integrate by parts, with $u = \sqrt{x-1}$ and $v' = \frac{1}{x^5}$. Then I get $-\frac{2}{3}x^{-3/2}(x-1)^{1/2} + \frac{1}{3} \int x^{-3/2}(x-1)^{-1/2}$. It seems to me that I need to further pursue integration by parts on the second integral, but that would seem to throw me into an endless loop. How can I proceed?
I am given that the answer to this is $\frac{2}{3}(1-\frac{1}{x})^{3/2} + C$. What might be the fastest way to achieve this answer?
| Another approach:
Let $$I=\int \sqrt{\frac{x-1}{x^{5}}}dx$$so
\begin{eqnarray*}
I&=&\int \frac{\sqrt{x-1}}{x^{5/2}}dx\\
&=&2 \int \frac{\sqrt{u^{2}-1}}{u^4}du, \quad u=\sqrt{x}, du=\frac{1}{2\sqrt{x}}dx\\
&=&2\int\sin^{2}(s)\cos(s)ds, \quad u=\sec(s), du=\tan(s)\sec(s)ds\\
&=&2\int p^{2}dp, \quad p=\sin(s), dp=\cos(s)ds\\
&=&\frac{2p^{3}}{3}+C\\
&=&\frac{2}{3}\left(\frac{x-1}{x^{5}} \right)^{3/2}x^{6}+C
\end{eqnarray*}
Therefore, $$\boxed{\int \sqrt{\frac{x-1}{x^{5}}}dx=\frac{2}{3}\left(\frac{x-1}{x^{5}} \right)^{3/2}x^{6}+C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3908121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Let $a_{n+1}=\sqrt[3]{\frac{1+a_{n}}{2}}$ show that $0 \leq a_{n} \leq 1$ Let $0 \leq a_{1} \leq 1$ and $a_{n+1}=\sqrt[3]{\frac{1+a_{n}}{2}}$ for all $n \geq 1$.
Show that:
(a) $0 \leq a_{n} \leq 1$ for all n.
(b) $ a_{n+1} \geq a_{n}$ for all n.
I don't know if the following is correct.
My attempt.
By using induction:
(a) Show that $0 \leq a_{n} \leq 1$ for all n.
We have $0 \leq a_{1} \leq 1$ and we want to show that $0 \leq a_{2} \leq 1$ for all n.
$0 \leq a_{1} \leq 1 \Rightarrow \frac{1}{2} \leq \frac{a_{1}+1}{2} \leq 1 \Rightarrow \sqrt[3]{\frac{1}{2}} \leq \sqrt[3]{\frac{a_{1}+1}{2}} \leq \sqrt[3]1 \Rightarrow 0.8 \leq a_{2} \leq 1 \Rightarrow 0 \leq a_{2} \leq 1 $
Now we suppose that the relation is true for n.
We want to show that it is also true for n+1.
We have:
$0 \leq a_{n} \leq 1 \Rightarrow \frac{1}{2} \leq \frac{a_{n}+1}{2} \leq 1 \Rightarrow \sqrt[3]{\frac{1}{2}} \leq \sqrt[3]{\frac{a_{n}+1}{2}} \leq \sqrt[3]1 \Rightarrow 0.8 \leq a_{n+1} \leq 1 \Rightarrow 0 \leq a_{n+1} \leq 1 $
(b) Show that $ a_{n+1} \geq a_{n}$ for all n.
$ a_{n+1} \geq a_{n} \Rightarrow \sqrt[3]{\frac{1+a_{n}}{2}} \geq a_{n} \Rightarrow 1+a_{n} \geq 2a_{n}^3$
This is true because we know that $0 \leq a_{n} \leq 1$ and that $ a_{n} \geq a_{n}^3$.
| For (a) you don't need to prove the statement for $a_2$.
You can simplify the argument by proving first that $a_n\ge0$. This is true for $a_1$; suppose it is true for $a_n$; then $a_{n+1}=\sqrt[3]{(a_n+1)/2}\ge0$.
Now the other inequality, which you did well: if $a_n\le1$, then $(a_n+1)/2\le1$ and therefore $a_{n+1}=\sqrt[3]{(a_n+1)/2}\le1$.
For (b) the arrows are in the wrong direction! You need to prove that $a_{n+1}\ge a_n$ and you can't take this as an assumption. However
$$
a_{n+1}\ge a_n
\quad\text{if and only if}\quad
a_{n+1}^3\ge a_n^3
\quad\text{if and only if}\quad
\frac{a_{n}+1}{2}\ge a_n^3
$$
so we are reduced to proving that $2a_n^3\le a_n+1$.
Your argument is fine: $a_n\le1$ implies $a_n^3\le a_n$ and $a_n^3\le 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3910321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Showing $\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$ I found the following exercise reading my calculus notes:
If $x,y$ and $z$ are positive real numbers, show that $$\frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\geq 8$$
I've been trying to solve it for a while. However, I have no idea how to approach it. Any help is welcome.
| Note that
$(x-1)^2 \geq 0 \implies x^2+1 \geq 2x $
$(y-1)^2 \geq 0 \implies y^2+1 \geq 2y $
$(z-1)^2 \geq 0 \implies z^2+1 \geq 2z $
Multiplying these three inequalities you will get $(x^2+1)(y^2+1)(z^2+1)\geq 8xyz \implies \frac{(x^2+1)(y^2+1)(z^2+1)}{xyz}\ge8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3913377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Showing that the inflection points of $x\sin x$ lie on a certain curve
Show that the inflection points of $f(x)=x\sin x$ are on the curve $$y^2(x^2+4)=4x^2.$$
I checked the graph of each function, but it seems that $f$ has infinitely many inflection points. How should I proceed?
| An inflection point is where $f''$ changes sign,
so we need to find where $f''(x)=0$.
$f(x)=x\sin(x),
f'(x)=\sin(x)+x\cos(x),
f''(x)=\cos(x)+\cos(x)-x\sin(x)
=2\cos(x)-x\sin(x)
$.
So we want to show
that this satisfies
$y^2(x^2+4)=4x^2$.
If $f''(x)=0$ then
$2\cos(x)=x\sin(x)
$
or $x=2/\tan(x)$.
Since $f(x)=x\sin(x)$,
we want
$x^2\sin^2(x)(4/\tan^2(x)+4)=4x^2
$
or
$\sin^2(x)(1/\tan^2(x)+1)=1
$
or
$1=\sin^2(x)(\cos^2(x)/\sin^2(x)+1)
$
or
$1=\cos^2(x)+\sin^2(x)
$
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3913739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Derivation of bivariate Gaussian copula density The multivariate Gaussian copula density, derived here, is
$$c(u_1,\ldots,u_n;\Sigma)=|\Sigma|^{-\frac{1}{2}}\exp\!\left(-\frac{1}{2}x^{\top}(\Sigma^{-1}-I)x\right)$$
where $\Sigma$ is the covariance matrix, and $x=[\Phi^{-1}(u_1),\ldots,\Phi^{-1}(u_n)]^{\top}$.
The bivariate Gaussian copula density, based on the pair-wise correlation coefficient $\rho$, is
$$
c\left(u_{1}, u_{2} ; \rho\right)=\frac{1}{\sqrt{1-\rho^{2}}} \exp \left\{-\frac{\rho^{2}\left(x_{1}^{2}+x_{2}^{2}\right)-2 \rho x_{1} x_{2}}{2\left(1-\rho^{2}\right)}\right\}
$$
What is the derivation of the second formula from the first?
| Note that with standard normal marginals
$$\Sigma=\left[\begin{array}{cc} 1 & \rho \\ \rho & 1 \end{array}\right],\,\, |\Sigma| = 1 - \rho^2$$
and
$$\Sigma^{-1}= \frac{1}{1- \rho^2}\left[\begin{array}{cc} 1 & -\rho \\ -\rho & 1 \end{array}\right], \,\, \Sigma^{-1}-I= \frac{1}{1- \rho^2}\left[\begin{array}{cc} \rho^2 & -\rho \\ -\rho & \rho^2 \end{array}\right]$$
Hence,
$$- \frac{1}{2}\mathbf{x}^{\top}(\Sigma^{-1}-I)\mathbf{x} = \frac{-1}{2(1- \rho^2)}\left[\begin{array}{cc} x_1 & x_2 \end{array}\right]\left[\begin{array}{cc} \rho^2 & -\rho \\ -\rho & \rho^2 \end{array}\right]\left[\begin{array}{cc} x_1 \\ x_2 \end{array}\right] \\= \frac{-1}{2(1- \rho^2)}\left[\begin{array}{cc} x_1 & x_2 \end{array}\right]\left[\begin{array}{cc} \rho^2x_1 -\rho x_2 \\ -\rho x_1 + \rho^2 x_2 \end{array}\right] \\= -\frac{\rho^2 (x_1^2 +x_2^2)- 2\rho x_1 x_2 }{2(1-\rho^2)},$$
and, thus,
$$|\Sigma|^{-\frac{1}{2}}\exp\!\left(-\frac{1}{2}\mathbf{x}^{\top}(\Sigma^{-1}-I)\mathbf{x}\right) = \frac{1}{\sqrt{1- \rho^2}} \exp \left\{-\frac{\rho^2 (x_1^2 +x_2^2)- 2\rho x_1 x_2 }{2(1-\rho^2)} \right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to determine asymptotics for $\sum_{ab^2 < x} ab^2$ by summing separately over $a$ and $b$ I do not understand how to get asymptotics for the double sum
$$\sum_{ab^2 < x} ab^2$$
If I sum over $a$ first, I get
$$\sum_{b^2<x} b^2 \sum_{a < x/b^2} a = \frac12 \sum_{b^2<x} b^2 \frac{x^2}{b^4} \sim \frac{\zeta(2)}{2} x^2.$$
If I sum over $b$ first, I get
$$\sum_{a<x} a \sum_{b < \sqrt{x/a}} b^2 = \frac13 \sum_{a<x} a \left( \frac{x}{a} \right)^{3/2} \sim \frac{2}{3} x^2. $$
What is happening. Am I doing something forbidden in switching summations this way? I suppose it is a possible divergence problem, so I do I get precise asymptotics for this sum?
| TLDR: Your second method uses two asymptotics. The asymptotic approximation for the inner summation throws out "lower order terms" which end up mattering for outer summation.
For integers $N \ge 0$, we have $$\displaystyle\sum_{k = 1}^{N}k^2 = \dfrac{N(N+1)(2N+1)}{6}.$$ Combining this with the inequality $y-1 \le \lceil y \rceil-1 \le y$, we get $$\dfrac{(y-1)y(2y-1)}{6} \le \displaystyle\sum_{k = 1}^{\lceil y \rceil-1}k^2 \le \dfrac{y(y+1)(2y+1)}{6},$$ i.e.
$$\dfrac{1}{3}y^3-\dfrac{1}{2}y^2+\dfrac{1}{6}y \le \displaystyle\sum_{k < y}k^2 \le \dfrac{1}{3}y^3+\dfrac{1}{2}y^2+\dfrac{1}{6}y$$ for any real number $y \ge 1$. Applying this yields
$$\dfrac{x^{3/2}}{3a^{3/2}}-\dfrac{x}{2a}+\dfrac{x^{1/2}}{6a^{1/2}} \le \sum_{b < \sqrt{x/a}}b^2 \le \dfrac{x^{3/2}}{3a^{3/2}}+\dfrac{x}{2a}+\dfrac{x^{1/2}}{6a^{1/2}},$$ and thus, $$\sum_{a < x}\left[\dfrac{x^{3/2}}{3a^{1/2}}-\dfrac{x}{2}+\dfrac{x^{1/2}a^{1/2}}{6}\right] \le \sum_{a < x}a\sum_{b < \sqrt{x/a}}b^2 \le \sum_{a < x}\left[\dfrac{x^{3/2}}{3a^{1/2}}+\dfrac{x}{2}+\dfrac{x^{1/2}a^{1/2}}{6}\right].$$
However, note that $\displaystyle\sum_{a < x}\dfrac{x^{3/2}}{3a^{1/2}} \sim \dfrac{2}{3}x^2$, $\displaystyle\sum_{a < x}\dfrac{x}{2} \sim \dfrac{1}{2}x^2$, and $\displaystyle\sum_{a < x}\dfrac{x^{1/2}a^{1/2}}{6} \sim \dfrac{1}{9}x^2$. So the lower order terms that the approximation $\displaystyle\sum_{k < y}k^2 \sim \dfrac{1}{3}y^3$ drops end up contributing a significant amount to the double sum. Hence, all you know from the second method is that the double sum is asymptotically bounded between $\dfrac{5}{18}x^2$ and $\dfrac{23}{18}x^2$.
We can analyze the first method in the same manner. Using $$\dfrac{1}{2}y^2-\dfrac{1}{2}y \le \sum_{k < y}k \le \dfrac{1}{2}y^2+\dfrac{1}{2}y,$$ we get $$\dfrac{x^2}{2b^4}-\dfrac{x}{2b^2} \le \sum_{a < x/b^2}a \le \dfrac{x^2}{2b^4}+\dfrac{x}{2b^2},$$ and thus, $$\sum_{b < \sqrt{x}}\left[\dfrac{x^2}{2b^2}-\dfrac{x}{2}\right] \le \sum_{b < \sqrt{x}}b^2\sum_{a < x/b^2}a \le \sum_{b < \sqrt{x}}\left[\dfrac{x^2}{2b^2}+\dfrac{x}{2}\right].$$ As you already noted, $\displaystyle\sum_{b < \sqrt{x}}\dfrac{x^2}{2b^2} \sim \dfrac{\zeta(2)}{2}x^2$. The other term scales like $\displaystyle\sum_{b < \sqrt{x}}\dfrac{x}{2} \sim \dfrac{1}{2}x^{3/2} = o(x^2)$. Hence, the double sum does indeed scale like $\dfrac{\zeta(2)}{2}x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Limit of a product with growing number of factors I'm trying to solve the following limit:
$$L=\lim_{n\to\infty}\frac{(n^2-1)(n^2-2)\cdots(n^2-n)}{(n^2+1)(n^2+3)\cdots(n^2+2n-1)}=\lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1}$$
Since originally, the limit yields an $1^\infty$ indeterminate expression, my first idea was taking logarithms:
$$\log L = \lim_{n\to\infty}\log\left(\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1}\right)=\\
=\lim_{n\to\infty}\sum_{k=1}^{n}\log\left(\frac{n^2-k}{n^2+2k-1}\right)=\\=\lim_{n\to\infty}\sum_{k=1}^n\log\left(\frac{1-\frac{k}{n^2}}{1+\frac{2k-1}{n^2}}\right)=\\
=\lim_{n\to\infty}\sum_{k=1}^n\left(\log\left(1-\frac{k}{n^2}\right)-\log\left(1+\frac{2k-1}{n^2}\right)\right)=\\=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1-\frac{k}{n^2}\right)^{n^2}-\log\left(1+\frac{2k-1}{n^2}\right)^{n^2}\right)=\\
=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}-\log\left(1+\frac{1}{\frac{n^2}{2k-1}}\right)^{\frac{n^2}{2k-1}(2k-1)}\right)$$
Now if I could do the limit of the terms inside the sum I'd have:
$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}-\log\left(1+\frac{1}{\frac{n^2}{2k-1}}\right)^{\frac{n^2}{2k-1}(2k-1)}\right)=\\=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(\log e^{-k}-\log e^{2k-1}\right)=\\
=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(-k-2k+1\right)=\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\left(-3k+1\right)$$
and then the limit would be trivial. The question is: is that allowed? Why? If not, how could I proceed?
Thanks!
| Let's take the first part
$$\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)} = -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}}$$
as example. Note that $-\frac{n^2}{k}$ tends to $\infty$ for any $k=1,2,\cdots,n$, so we can use Taylor's expansion
$$\left(1+\frac{1}{x}\right)^x = e - \frac{e}{2x} + O\left(\frac{1}{x^2}\right) \quad (x \to \infty)$$
to obtain
$$\begin{aligned}
\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n\log\left(1+\frac{1}{-\frac{n^2}{k}}\right)^{-\frac{n^2}{k}(-k)}
&= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k\log\left[e+\frac{ek}{2n^2}+O\left(\frac{k^2}{n^4}\right)\right] \\
&= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n \left[k + k\log\left(1+\frac{k}{2n^2}+O\left(\frac{k^2}{n^4}\right)\right)\right] \\
&= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n \left[k + k\left(\frac{k}{2n^2}+O\left(\frac{k^2}{n^4}\right)\right)\right] \\
&= -\lim_{n\to\infty}\frac{1}{n^2}\sum_{k=1}^n k\left[1+O\left(\frac{1}{n}\right)\right],
\end{aligned}$$
where we have used $\frac{k}{n^2}=O\left(\frac{1}{n}\right)$. So the term $O(1/n)$ can be discarded.
Note: An alternative solution is given here:
Lemma. Given $f(0) = 0$ and that finite $f'(0)$ exists, let
$$a_n = \sum_{k=1}^n f\left(\frac{k}{n^2}\right) = f\left(\frac{1}{n^2}\right) + f\left(\frac{2}{n^2}\right) + \cdots + f\left(\frac{n}{n^2}\right),$$
then
$$\lim_{n\to\infty} a_n = \frac{f'(0)}{2}.$$
Proof. First notice that
$$f'(0) = \lim_{x \to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x \to 0} \frac{f(x)}{x},$$
and using the definition of limit, $\forall \varepsilon > 0$, $\exists \delta > 0$, s.t. $\forall 0 < x < \delta$,
$$f'(0) - \varepsilon < \frac{f(x)}{x} < f'(0) + \varepsilon.$$
Particularly for $x > 0$, we have $(f'(0)-\varepsilon)x < f(x) < (f'(0)+\varepsilon)x$.
Now pick $N \in \mathbb{N}$, s.t. $N > \dfrac{1}{\delta}$. For $n > N$,
$$\frac{k}{n^2} \leq \frac{1}{n} < \frac{1}{N} < \delta, \quad k=1,2,\cdots,n,$$
and hence
$$(f'(0) - \varepsilon) \cdot \frac{k}{n^2} < f\left(\frac{k}{n^2}\right) < (f'(0) + \varepsilon) \cdot \frac{k}{n^2}, \quad k=1,2,\cdots,n.$$
Taking summation over $k$ gives
$$\frac{f'(0)-\varepsilon}{2} \cdot \frac{n+1}{n} < \sum_{k=1}^n f\left(\frac{k}{n^2}\right) < \frac{f'(0)+\varepsilon}{2} \cdot \frac{n+1}{n}.$$
Since $\dfrac{n+1}{n} \to 1$ as $n \to \infty$, then $\exists N_1 \in \mathbb{N}$, s.t. $\forall n > N_1$,
$$\frac{f'(0)-\varepsilon}{2} - \frac{\varepsilon}{2} < \sum_{k=1}^n f\left(\frac{k}{n^2}\right) < \frac{f'(0)+\varepsilon}{2} + \frac{\varepsilon}{2},$$
that is, $\dfrac{1}{2}f'(0)-\varepsilon < \sum\limits_{k=1}^n f\left(\frac{k}{n^2}\right) < \dfrac{1}{2}f'(0)+\varepsilon$. Therefore we have
$$\lim_{n\to\infty} a_n = \lim_{n\to\infty} \sum\limits_{k=1}^n f\left(\frac{k}{n^2}\right) = \frac{f'(0)}{2}.$$
Return to your problem, we have
$$I = \lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k-1} = \lim_{n\to\infty}\prod_{k=1}^n\frac{n^2-k}{n^2+2k} = \lim_{n\to\infty}\prod_{k=1}^n \frac{1-\frac{k}{n^2}}{1+2\frac{k}{n^2}} = \lim_{n\to\infty} e^{\sum_{k=1}^n f(k/n^2)},$$
where $f(x) = \ln\left(\frac{1-x}{1+2x}\right)$ and the second equality comes from
$$1 \leftarrow \left(1-\frac{1}{n^2}\right)^n \leq \prod_{k=1}^n \frac{n^2+2k-1}{n^2+2k} = \prod_{k=1}^n \left(1-\frac{1}{n^2+2k}\right) \leq 1.$$
Therefore $I = e^{\frac{f'(0)}{2}} = \boxed{e^{-\frac{3}{2}}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3919681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A question on floor function. Prove that for $n \in N$,
$[\sqrt{n} + \frac{1}{2} ] = [\sqrt{n-\frac{3}{4}} + \frac{1}{2}]$
where [.] is the greatest integer function.
My attempt: k < $\sqrt{n} + \frac{1}{2} $ < k+1 .
Obviously, $\sqrt{n-\frac{3}{4}} + \frac{1}{2}$ < $\sqrt{n} + \frac{1}{2} $,
so it would suffice to show that $\sqrt{n-\frac{3}{4}} + \frac{1}{2}$> k , but it turns out that the inequlity is very weak so i am not able to prove it.
Is there any way to do this using induction? Or am i trying the right thing?
Any help is appreciated , thanks!
| We have, where $k$ is the integer part of $\sqrt{n} + \frac{1}{2}$,
$$\begin{equation}\begin{aligned}
k & \lt \sqrt{n} + \frac{1}{2} \\
k - \frac{1}{2} & \lt \sqrt{n} \\
k^2 - k + \frac{1}{4} & \lt n
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Since $k$ and $n$ are positive integers, \eqref{eq1A} can actually be restated as being
$$\begin{equation}\begin{aligned}
k^2 - k + 1 & \le n \\
k^2 - k + \frac{1}{4} & \le n - \frac{3}{4} \\
\left(k - \frac{1}{2}\right)^2 & \le n - \frac{3}{4} \\
k - \frac{1}{2} & \le \sqrt{n - \frac{3}{4}} \\
k & \le \sqrt{n - \frac{3}{4}} + \frac{1}{2}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Since $\sqrt{n} + \frac{1}{2} \gt \sqrt{n - \frac{3}{4}} + \frac{1}{2}$, as you stated, this shows $k$ is also the integer part of $\sqrt{n - \frac{3}{4}} + \frac{1}{2}$, which means that
$$\left\lfloor \sqrt{n} + \frac{1}{2} \right\rfloor = \left\lfloor \sqrt{n - \frac{3}{4}} + \frac{1}{2} \right\rfloor \tag{3}\label{eq3A}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3921886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A question about floor functions and series The sequence $\{a_n\}^{∞}_{n=1} = \{2,3,5,6,7,8,10,...\}$
consists of all the positive integers that are not perfect squares.
Prove that $a_n= n+ [\sqrt{n} + \frac{1}{2}]$.
Well, I managed to prove that
$ m^2 < n+ [\sqrt{n} + \frac{1}{2}] < (m+1)^2 $,
where $[\sqrt{n} + \frac{1}{2}] = m$.
But is this enough to answer the question? Or do we also need to prove that $n+ [\sqrt{n} + \frac{1}{2}] $ can take all non-perfect square values. If yes, then how?
Any help is appreciated, thanks!
| The following is just a first step toward the answer.
From
$$
\eqalign{
& a_{\,n} = n + \left\lfloor {\sqrt n + {1 \over 2}} \right\rfloor \cr
& a_{\,n + 1} - a_{\,n}
= 1 + \left\lfloor {\sqrt {n + 1} + {1 \over 2}} \right\rfloor
- \left\lfloor {\sqrt n + {1 \over 2}} \right\rfloor \cr}
$$
since we have
$$
\eqalign{
& \left\lfloor {x - y} \right\rfloor = - \left\lceil {y - x} \right\rceil = \cr
& = \left\lfloor x \right\rfloor - \left\lfloor y \right\rfloor
+ \left\lfloor {\left\{ x \right\} - \left\{ y \right\}} \right\rfloor = \cr
& = \left\lfloor x \right\rfloor - \left\lfloor y \right\rfloor
- \left[ {\left\{ x \right\} < \left\{ y \right\}} \right] \cr}
$$
where $[P]$ denotes the Iverson bracket, then
$$
\eqalign{
& a_{\,n + 1} - a_{\,n} = 1 + \left\lfloor {\sqrt {n + 1} + {1 \over 2}} \right\rfloor
- \left\lfloor {\sqrt n + {1 \over 2}} \right\rfloor = \cr
& = 1 + \left\lfloor {\sqrt {n + 1} - \sqrt n } \right\rfloor
+ \left[ {\left\{ {\sqrt {n + 1} + {1 \over 2}} \right\}
< \left\{ {\sqrt n + {1 \over 2}} \right\}} \right] = \cr
& = 1 + \left[ {\left\{ {\sqrt {n + 1} + {1 \over 2}} \right\}
< \left\{ {\sqrt n + {1 \over 2}} \right\}} \right] \cr}
$$
Now the Iverson bracket will be one, and thus we will have a jump in $a_{\,n}$, when
$$
\left\{ {\sqrt {n + 1} + {1 \over 2}} \right\} < \left\{ {\sqrt n + {1 \over 2}} \right\}
$$
which for $1 \le n$ , which means $sqrt{n+1} -\sqrt{n} < 1/2$, can occur only when
$$
\sqrt n + {1 \over 2} < m < \sqrt {n + 1} + {1 \over 2}
$$
with $m$ an integer, that is
$$
\eqalign{
& \sqrt n < m - {1 \over 2} < \sqrt {n + 1} \quad \Rightarrow \cr
& \Rightarrow \quad n < m^{\,2} + {1 \over 4} - m < n + 1\quad \Rightarrow \cr
\quad \Rightarrow \cr
& \Rightarrow \quad n - {1 \over 4} < m\left( {m - 1} \right) < n + {3 \over 4}\quad \Rightarrow \cr
& \Rightarrow \quad n_ * = m\left( {m - 1} \right) \cr}
$$
Thus
$$
a_{\,n + 1} - a_{\,n} = 2\quad {\rm iff}\quad n = m\left( {m - 1} \right)
$$
It remains then to demonstrate that
$$
\forall q\;\exists m:\quad a_{\,n_{\, * } } + 1
= m\left( {m - 1} \right) + \left\lfloor {\sqrt {m\left( {m - 1} \right)}
+ {1 \over 2}} \right\rfloor + 1 = q^{\,2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the limit of $a_1 = 1 $ , $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$ $a_1 = 1 $ , $ a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n}$ , $n \in \mathbb{N}$
It is easy to prove that the limit exists: the boundary of that expression is $0$ and it is monotonically decreasing.
The problem is to actually find the limit (it is $0$) because if I take arithmetic of limits:
$ \lim_{x \to +\infty} a_{n+1} = \frac{\sqrt{1+ \lim_{n \to +\infty} a_n^2+1}}{\lim_{n \to +\infty} a_n}$
$g = \frac{\sqrt{1+g^2+1}}{g} \implies g^2 = {\sqrt{1+g^2}-1} \implies g^2 + 1 = {\sqrt{1+g^2}} \implies 0 = 0$
| You can prove it directly like this
$$a_{n+1} = \frac{\sqrt{1+a_n^2}-1}{a_n} < \frac{\sqrt{1+a_n^2+\frac{1}{4}a_n^4}-1}{a_n} = \frac{1+\frac{1}{2}a_n^2-1}{a_n} = \frac 12 a_n$$
Therefore $a_n \to 0$, $n\to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
An asymptotic about Integral of Legendre Polynomials I want to show asymptotics of the following integral involving Legendre Polynomial:
For $0<t<\theta<\frac\pi2$,
$$\Big|\int_0^\theta \frac{1}{\sqrt{\theta}+\sqrt{\theta-t}} \frac{1-P_n(\cos t)}{t}dt \Big| \leq C \theta^{-\frac12} \cdot \log (n\theta),$$ where $C$ is the constant, $P_n(x)$ is a Legendre Polynomial, $n$ is a positive integer.
I am trying to use Bernstein's inequality for trigonometric polynomials:
Since $P_n(1)=1$ and $|P_n(\cos t)|\leq 1$, $$|P_n(\cos t)-1|=|P_n(\cos t)-P_n(\cos 0)|\leq nt||P_n ||_{\infty}\leq nt.$$
Then I can only get the below estimate but I have no further idea $$\Big|\int_0^\theta \frac{1}{\sqrt{\theta}+\sqrt{\theta-t}} \frac{1-P_n(\cos t)}{t}dt \Big| \leq \int_0^\theta \frac{1}{\sqrt{\theta}+\sqrt{\theta-t}} \frac{|1-P_n(\cos t)|}{t}dt \leq \int_0^\theta \frac{1}{\sqrt{\theta}+\sqrt{\theta-t}} \frac{nt}{t}dt=O(n).$$
Any suggestions are welcome! Thank you for your help!
| First, suppose that $\frac{1}{n} <\theta <\frac{\pi}{2}$. We split the range of integration into two parts: $0<t<\frac{1}{n}$ and $\frac{1}{n}<t<\theta$, respectively. On the first interval, we use the known limiting relation between the Legendre polynomials and the Bessel function of the first kind $J_0$:
\begin{align*}
&\int_0^{1/n} {\frac{1}{{\sqrt \theta + \sqrt {\theta - t} }}\frac{{1 - P_n (\cos t)}}{t}dt} = \int_0^1 {\frac{1}{{\sqrt \theta + \sqrt {\theta - s/n} }}\frac{{1 - P_n \left( {\cos \left( {\frac{s}{n}} \right)} \right)}}{s}ds} \\ & \sim \int_0^1 {\frac{1}{{\sqrt \theta + \sqrt {\theta - s/n} }}\frac{{1 - J_0 (s)}}{s}ds} \le K\int_0^1 {\frac{1}{{\sqrt \theta + \sqrt {\theta - s/n} }} s ds} \le \frac{K}{2}\theta ^{ - 1/2}
\end{align*}
with a suitable positive constant $K$ and large $n$. On the remaining interval, we have
\begin{align*}
\left| {\int_{1/n}^\theta {\frac{1}{{\sqrt \theta + \sqrt {\theta - t} }}\frac{{1 - P_n (\cos t)}}{t}dt} } \right| & \le \theta ^{ - 1/2} \int_{1/n}^\theta {\left| {\frac{{1 - P_n (\cos t)}}{t}} \right|dt} \\ & \le 2\theta ^{ - 1/2} \int_{1/n}^\theta \frac{{dt}}{t} = 2\theta ^{ - 1/2} \log (n\theta ) .
\end{align*}
Thus, there is a constant $C >0$ such that
$$
\left| {\int_0^\theta {\frac{1}{{\sqrt \theta + \sqrt {\theta - t} }}\frac{{1 - P_n (\cos t)}}{t}dt} } \right| \le C \theta ^{ - 1/2} \max(1,\log (n\theta )).
$$
If $0 <\theta <\frac{1}{n}$, then
\begin{align*}
& \int_0^\theta {\frac{1}{{\sqrt \theta + \sqrt {\theta - t} }}} \frac{{1 - P_n (\cos t)}}{t}dt \le \theta ^{ - 1/2} \int_0^\theta {\frac{{1 - P_n (\cos t)}}{t}dt} \\ & \le \theta ^{ - 1/2} \int_0^{1/n} {\frac{{1 - P_n (\cos t)}}{t}dt} \sim \theta ^{ - 1/2} \int_0^1 {\frac{{1 - J_0 (s)}}{s}ds} \\ & \le K\theta ^{ - 1/2} \int_0^1 {sds} = \frac{K}{2}\theta ^{ - 1/2} .
\end{align*}
In summary, there is a constant $C >0$ such that
$$
\left| {\int_0^\theta {\frac{1}{{\sqrt \theta + \sqrt {\theta - t} }}\frac{{1 - P_n (\cos t)}}{t}dt} } \right| \le C \theta ^{ - 1/2} \max(1,\log (n\theta )).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show the inequality : $a^{(2(1-a))}+b^{(2(1-b))}+c^{(2(1-c))}+c\leq 1$ Claim :
Let $0.5\geq a \geq b \geq 0.25\geq c\geq 0$ such that $a+b+c=1$ then we have :
$$a^{(2(1-a))}+b^{(2(1-b))}+c^{(2(1-c))}+c\leq 1$$
To prove it I have tried Bernoulli's inequality .
For $0\leq x\leq 0.25$ we have :
$$x^{2(1-x)}\leq 2x^2$$
As in my previous posts we have the inequality $x\in[0,0.5]$ :
$$x^{2(1-x)}\leq 2^{2x+1}x^2(1-x)$$ applying this for each variables $a,b$ we want to show :
$$2^{2a+1}a^2(1-a)+2^{2b+1}b^2(1-b)+2c^2+c\leq 1$$
Now by Bernoulli's inequality we have:
$$2^{2x+1}\leq 2(1+2x)$$
Remains to show :
$$2(1+2a)a^2(1-a)+2(1+2b)b^2(1-b)+2c^2+c\leq 1\quad(1)$$
The function :
$$f(x)=2(1+2x)x^2(1-x)$$ is concave for $x\in [\frac{1}{8}+\frac{\sqrt{\frac{19}{3}}}{8},0.5]$
So we can use Jensen's inequality remains to show :
$$2\left(2(1+a+b)\left(\frac{a+b}{2}\right)^2\left(1-\left(\frac{a+b}{2}\right)\right)\right)+2c^2+c\leq 1$$
So it reduces to a one variable inequality and using derivatives it's not hard to show that :
$$g(c)=2f\left(\frac{1-c}{2}\right)+2c^2+c\leq 1$$
For $c\in[0,1-2\left(\frac{1}{8}+\frac{\sqrt{\frac{19}{3}}}{8}\right)]$
It shows the equality case $a=b=0.5$ and $c=0$ but inequality $(1)$ is false for the other equality case $a=0.5$ and $b=c=0.25$.
We have also the inequality for $x\in[0.25,0.5]$ (we can prove it using logarithm and then derivative)
$$x^{(2(1-x))}\leq x^22^{-5(x-0.25)(x-0.5)+1}$$
Using Bernoulli's inequality :
$$x^22^{-5(x-0.25)(x-0.5)+1}\leq 2(x^2+x^2(-5(x-0.25)(x-0.5)))$$
So Remains to show :
$$2(a^2+a^2(-5(a-0.25)(a-0.5)))+2(b^2+b^2(-5(b-0.25)(b-0.5)))+2c^2+c\leq 1\quad (2)$$
Question :
Have you a proof ? How to show $(2)$ ?
Thanks in advance !
| Some thoughts
We first give some auxiliary results (Facts 1 through 4). The proofs are not difficult and thus omitted.
Fact 1: If $\frac{1}{2} \ge x \ge \frac{1}{4}$, then
$x^{2(1-x)} \le \frac{528x^2+572x-93}{650}$.
Fact 2: If $0\le x \le \frac{1}{4}$, then $x^{2(1-x)} \le 3x^2 - 2x^3$.
(Hint: Use Bernoulli inequality.)
Fact 3: If $\frac{1}{2} \ge x \ge \frac{1}{4}$, then
$x^{2(1-x)} \le \frac{752x^2-24x-1}{320}$.
Fact 4: If $0\le x \le \frac{1}{4}$, then $x^{2(1-x)} \le 3x^2 - 4x^3$.
(Hint: Use Bernoulli inequality.)
Now, we split into two cases:
*
*$c \le \frac{1}{5}$:
By Facts 1-2, it suffices to prove that
$$\frac{528a^2+572a-93}{650} + \frac{528b^2+572b-93}{650} + 3c^2 - 2c^3 + c \le 1$$
or
$$650c^3-264a^2-264b^2-975c^2-286a-286b-325c+418 \ge 0.$$
It is verified by Mathematica.
*$\frac{1}{5} < c \le \frac{1}{4}$:
By Facts 1, 3, 4, it suffices to prove that
$$\frac{528a^2+572a-93}{650} + \frac{752b^2-24b-1}{320} + 3c^2 - 4c^3 + c \le 1$$
or
$$83200c^3-16896a^2-48880b^2-62400c^2-18304a+1560b-20800c+23841 \ge 0.$$
It is verified by Mathematica.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3927668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Determine the linear function $f: R \rightarrow R$ There's a linear function $f$ whose correspondence rule is $f(x)=|ax^2-3ax+a-2|+ax^2-ax+3$. State the values of the parameter $a$ that fully define the function $f$.
According to the problem condition the function $ f (x) $ is linear by
so the coefficient that accompanies $ x ^ 2 $ must be equal to 0, this is
meets when:
\begin{align*}
ax^2-3ax+a-2 & < 0\\
a(x^2-3x) &<2-a \\
\
a\underbrace{\left(x^2-3x+\frac{9}{4}\right)}_{
\left(x-\frac{3}{2}\right)^2≥ 0} &<2-a +\frac{9}{4}a\\
a \left(x-\frac{3}{2}\right)^2 &<2+\frac{5}{4}a\\
0 & <2+\frac{5}{4}a \\
-\frac{8}{5} & <a
\end{align*}
What else is missing to analyze?
| For any $a$, in order for the $x^2$ term to vanish, the inequality $ax^2-3ax+a-2\le0$ must hold for all values of $x$. Using your analysis, this is equivalent to
$$a\left(x-\frac32\right)^2\le2+\frac54a$$
If $a > 0$, there is some $x$ large/small enough such that LHS > RHS. Hence $a \le 0$.
Moreover, if $2 + \dfrac54 a < 0$, the above inequality fails for $x = \dfrac32$. Hence $a \ge -\dfrac 85$.
This gives the range $-\dfrac 8 5 \le a \le 0$; indeed in this range, LHS is nonpositive for all $x$ while RHS is nonnegative, so our initial inequality holds, and the function is linear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3933777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Efficient method to evaluate integrals with multiple roots in denominator $$ \int \frac{x^2-1}{x\sqrt{(x^2 + \beta x +1 )} \sqrt{(x^2 + \alpha x +1 )}} dx $$
I saw this question but I don't think the methods in answers are much applicable here.
My attempt:
$$ \int \frac{x^2-1}{x\sqrt{(x + \frac{\beta}{2})^2 + ( 1 - \frac{\beta^2}{4} )} \sqrt{(x + \frac{\alpha}{2})^2 + ( 1 - \frac{\alpha^2}{4} )}} dx= \frac{1}{\sqrt{ 1- \frac{\beta^2}{4} } \sqrt{1- \frac{\alpha^2}{4} }}\int \frac{x^2-1}{x\sqrt{\frac{(x + \frac{\beta}{2})^2}{ ( 1 - \frac{\beta^2}{4} )} +1} \sqrt{\frac{(x + \frac{\alpha}{2})^2}{ ( 1 - \frac{\alpha^2}{4} )} +1}} dx
$$
Not sure what's the best way to go now..
| Divide both numerator and denominator by $x^2$
$$\int \frac{1-\frac{1}{x^2}}{\sqrt{(x+\beta +\frac{1}{x})}\sqrt{(x+\alpha +\frac{1}{x})}}dx$$
which immediately suggests using the substitution $t=x+\frac{1}{x}$
$$\int \frac{dt}{\sqrt{\left(t+\frac{\alpha+\beta}{2}\right)^2-\left(\frac{\alpha-\beta}{2}\right)^2}} = \cosh^{-1}\left[\frac{t+\frac{\alpha+\beta}{2}}{\left|\frac{\alpha-\beta}{2}\right|}\right]$$
from a simple hyperbolic substitution. This makes the final answer
$$\cosh^{-1}\left[\frac{x+\frac{1}{x}+\frac{\alpha+\beta}{2}}{\left|\frac{\alpha-\beta}{2}\right|}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3934821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A question of Roots of unity
By considering the ninth roots of unity, show that: $\cos(\frac{2\pi}{9}) +
\cos(\frac{4\pi}{9}) + \cos(\frac{6\pi}{9}) + \cos(\frac{8\pi}{9}) = \frac{-1}{2}$.
I know how to find the roots of unity, but I am unsure as to how I can use them in finding the sum of these $4$ roots.
| Note that\begin{multline}\cos\left(\frac{2\pi}9\right)+\cos\left(\frac{4\pi}9\right)+\cos\left(\frac{6\pi}9\right)+\cos\left(\frac{8\pi}9\right)=\\=\frac12\left(e^{2\pi i/9}+e^{-2\pi i/9}+e^{4\pi i/9}+e^{-4\pi i/9}+e^{6\pi i/9}+e^{-6\pi i/9}+e^{8\pi i/9}+e^{-8\pi i/9}\right)\end{multline}But this is half the sum of all ninth roots of unity other than $1$. So, it's half the sum of the roots of$$x^8+x^7+x^6+x^5+x^4+x^3+x^2+x$$and that sum is $-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3937022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Is my method of solving equation correct? The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$
using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$
$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
$$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$
Got $\frac{32}{(a+b)^2}$ from the fact that $(a+b)^2(a+b)^3=32$ and $a+b=2$
$$a^5+b^5+5ab\biggl(\frac{32}{(2)^2}\biggr)+10a^2b^2(2)=32$$
$$a^5+b^5+40ab+20a^2b^2=32$$
From when I defined a and b earlier, I substitute and get
$$\left(\sqrt[5]{16+\sqrt{x}}\right)^5+\left(\sqrt[5]{16-\sqrt{x}}\right)^5+40\sqrt[5]{\left(16-\sqrt{x}\right)\left(16+\sqrt{x}\right)}+20\sqrt[5]{\left(16-\sqrt{x}\right)^2\left(16+\sqrt{x}\right)^2}=32$$
$$\require{cancel}\cancel{16}\cancel{+\sqrt{x}}+\cancel{16}\cancel{-\sqrt{x}}+40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=\cancel{32} 0$$
$$40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=0$$
$$20\biggl(2\sqrt[5]{256-x}+\sqrt[5]{\left(256-x\right)\left(256-x\right)\biggr)}=0$$
Then let $u=\sqrt[5]{256+{x}}$,
$$20(2u+u^2)=0$$
$$u(u+2)=0$$
$$u=0,-2$$
Substituting u to get x from $u=\sqrt[5]{256+{x}}$, I get $$x=\cancel{-288},256$$
However, since the original equation has a $\sqrt{x}$, which can't be negative, I eliminate $x=-288$, leaving just $$x=256$$ as my answer.
So, this is how I arrived on my answer. Did I perform any mathematical errors or any illegal mathematical maneuvers? Please let me know.
Thank you!
| When you went from :
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
to
$$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$
You incorrectly assumed that
$$a^3 + b^3 = (a+b)^3 = \biggl(\frac{32}{(a+b)^2}\biggr)$$
Therefore you cannot say that
$$a^3 + b^3 = \frac{32}{2^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3938150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
} |
Solving $x^d \equiv a \pmod{29}$ using primitive roots Can anyone please detail the general approach to questions of the form $x^d \equiv a \pmod{29}$? For example, Wolfram Alpha states that $x^5 \equiv 8 \pmod{29}$ has one solution, $x^4 \equiv 4 \pmod{29}$ has no solution, and $x^7 \equiv 12 \pmod{29}$ has many solutions, but I have no idea how I'd go about proving that no solutions exist, or that any solutions I find are comprehensive.
I know that $2$ is a primitive root $\bmod 29$, so $2^{28} \equiv 1 \pmod{29}$. For the first one I can manipulate to get $(2^{23})^5 \equiv 8 \pmod{29}$ to give $x \equiv 2^{23} \equiv 10 \pmod{29}$ as a solution, but this doesn't prove that no other solutions exist, does it? Also, I don't know how this approach would work with the third problem because $12$ cannot be written $2^n$.
Furthermore, I'm unsure of the general technique to show that the second is not soluble?
| Since $2$ is a primitive root modulo $29$, its multiplicative order is $28$, which means
$$2^y \equiv 2^z \pmod{29} \iff y \equiv z \pmod{28} \tag{1}\label{eq1A}$$
One way to solve your congruence equations is to first express the right hand sides as a congruent power of $2$. The first $2$ examples are quite easy. With the third one, i.e.,
$$x^7 \equiv 12 \pmod{29} \tag{2}\label{eq2A}$$
note $12 \equiv 3(4) \equiv (32)(4) \equiv 2^7 \pmod{29}$ (as J. W. Tanner's question comment states). Next, since $2$ is a primitive root and $x \not\equiv 0 \pmod{29}$, then there's an integer $1 \le a \le 28$ such that $x \equiv 2^a \pmod{29}$. Thus, \eqref{eq2A} becomes
$$\left(2^{a}\right)^7 \equiv 2^7 \pmod{29} \implies 2^{7a} \equiv 2^7 \pmod{29} \tag{3}\label{eq3A}$$
Using \eqref{eq1A}, this gives
$$7a \equiv 7 \pmod{28} \implies a \equiv 1 \pmod{4} \tag{4}\label{eq4A}$$
This gives multiple answers of $x \equiv 2^{4b + 1} \pmod{29}$ for $0 \le b \le 6$, i.e., $x \equiv 2, 3, 19, 14, 21, 17, 11 \pmod{29}$.
Next, with your second example,
$$x^5 \equiv 8 \equiv 2^3 \pmod{29} \tag{5}\label{eq5A}$$
as done before, let $x \equiv 2^a \pmod{29}$ to get
$$2^{5a} \equiv 2^3 \pmod{29} \implies 5a \equiv 3 \pmod{28} \implies a \equiv 23 \pmod{28} \tag{6}\label{eq6A}$$
Thus, $x \equiv 2^{23} \pmod{29}$ is the answer, as J. W. Tanner's question comment also indicates.
With your final example of
$$x^4 \equiv 4 \equiv 2^2 \pmod{29} \implies x^2 \equiv \pm 2 \tag{7}\label{eq7A}$$
note $2$ is not a quadratic root modulo $29$ (which requires, as shown in this table that $p \equiv 1, 7 \pmod{8}$ but $29 \equiv 5 \pmod{8}$), and also $-2$ is not a quadratic residue (since that requires $p \equiv 1, 3 \pmod{8}$). In addition, using the method I show above, gives
$$2^{4a} \equiv 2^2 \pmod{29} \implies 4a \equiv 2 \pmod{28} \implies 2a \equiv 1 \pmod{14} \tag{8}\label{eq8A}$$
However, it's not possible to have an even value be equivalent to an odd value with an even modulo (e.g., since it would require $14 \mid 2a - 1$), so there are no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3943839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$z$ is a complex number such that $z^7=1$, where $z\not =1$. Find the value of $z^{100}+z^{-100} + z^{300}+z^{-300} + z^{500}+z^{-500}$ Let $z=e^{i\frac{2\pi}{7}}$
Then the expression, after simplification turns to
$$2[\cos \frac{200\pi}{7} +\cos \frac{600 \pi}{7} +\cos \frac{1000\pi}{7}]$$
How do I solve from here?
| Since $z^7=1$ we have that it's equal to
$$z^2+z^5+z^6+z+z^3+z^4$$
$$=z^6+z^5+z^4+z^3+z^2+z$$ because $z^{7k+1}=z, z^{7k+2}=z^2,...$. This is equal to $-1$ since $z^7-1=(z-1)(z^6+..+1)=0$ and thus $z^6+..+1=0$ noting that $z -1\neq 0$.
Also we can solve using sum of geometric progression.
$\frac{a(r^n-1)}{r-1}=\frac{z(z^6-1)}{z-1}=\frac{z^7-z}{z-1}=\frac{1-z}{z-1}=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3945696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
For what a will this series converge? $\sum_{n=1}^{\infty}n^a\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$
I tried to use the D'Alembert ratio test and it didn't quite work out. The series inside is telescoping but I don't know if that information will be useful. Can someone help?
| We have
$$
\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} = \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n}\sqrt{n+1}} = \frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n})}{\sqrt{n}\sqrt{n+1}(\sqrt{n+1} + \sqrt{n})} \sim \frac{1}{2n^{3/2}}.
$$
Hence $a_n = n^{\alpha} \Bigl( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \Bigr) \sim \frac{1}{2n^{3/2-a}}$. The sum $\sum_n a_n$ is convergent iff $\sum_n \frac{1}{n^{3/2-a}}$ is convergent. Using integral test for convergence we get that $\sum_n \frac{1}{n^{3/2-a}}$ is convergent iff $3/2-a > 1$.
So, the sum $\sum_n a_n$ is convergent iff $a<\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$xy''-(2x^2+1)y'=x^5e^{x^2}$ $xy''-(2x^2+1)y'=x^5e^{x^2}$
Hi there,
needs help with the second order ODE.
I've tried (for the homogenous part) to substitute $\frac{dy}{dx} =u$ to get $ u'-(2x+\frac{1}{x})u=0$
solving for the first non trivial solution
$y_1=C_1 e^{x^2}$
however with $y_2$ and the particular solution i stuck.
asks for help.
| $$y'(x)=u(x)$$
$$x u'(x)-\left(2 x^2+1\right) u(x)=0$$
$$u(x)=c_1 e^{x^2} x$$
$$y(x)=\int c_1 e^{x^2} x\,dx=\frac{c_1 e^{x^2}}{2}+c_2$$
To solve the non homogeneous equation $$xy''-(2x^2+1)y'=x^5e^{x^2}$$
we guess the particular solution as $y_1(x)=(a x^4+b x^3+c x^2+d x+e)e^{x^2}$
$$y_1'(x)=2 e^{x^2} x \left(a x^4+b x^3+c x^2+d x+e\right)+e^{x^2} \left(4 a x^3+3 b x^2+2 c x+d\right)=\\=e^{x^2} \left(2 a x^5+4 a x^3+2 b x^4+3 b x^2+2 c x^3+2 c x+2 d x^2+d+2 e x\right)$$
$$y_1''(x)=2 e^{x^2} \left(2 a x^6+9 a x^4+6 a x^2+2 b x^5+7 b x^3+3 b x+2 c x^4+5 c x^2+c+2 d x^3+3 d x+2 e x^2+e\right)$$
So plugging this in the equation we get
$$8 a x^5+6 b x^4+ x^3 (8 a+4 c)+x^2 (3 b+2 d)-d\equiv x^5e^{x^2}$$
$a=\frac18;\;c=-\frac{1}{4}$ so the particular solution is
$$y_1=\frac{1}{8} e^{x^2} x^2 \left(x^2-2\right)$$
The general solution of the DE is
$$y=\frac{c_1 e^{x^2}}{2}+c_2+\frac{1}{8} e^{x^2} x^2 \left(x^2-2\right)$$
$$y=\frac{1}{8} e^{x^2} \left(x^4-2 x^2+4 c_1\right)+c_2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve $\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$ Solve the equation,
$$\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$$
We have $$y^3\frac{d^2y}{dx^2}+y^4=1$$
I tried using change of dependent variable
Let $z=y^3\frac{dy}{dx}$
Then we get
$$y^3\frac{d^2y}{dx^2}+3y^2\left(\frac{dy}{dx}\right)^2=\frac{dz}{dx}$$
But i could not get an equation completely involving $z,x$
| Another possibility
Switch variables to make
$$-\frac {x''}{[x']^3}+y=\frac 1 {y^3}$$ Reduction of order $p=x'$ gives a separable equation
$$-\frac {p'}{p^3}=\frac 1 {y^3}-y\implies p=\pm\frac{y}{\sqrt{c_1 y^2-y^4-1}}$$ Now
$$x+c_2=\pm \int \frac{y}{\sqrt{c_1 y^2-y^4-1}}\,dy=\pm \frac 12 \int \frac{dt}{\sqrt{c_1 t-t^2-1}}$$ and you will arrive to some arctangent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find curve at the intersection of two level surfaces. Let C be the curve at the intersection of two level surfaces M(x, y, z) = 5 and N(x,y,z) = 0, passing through a point P (1,1,1). Let $M(x, y, z) = 2x^2 + y^2 + 2z^2$ and $N(x, y, z) = xy-z$. Find curve C in the parametric form <x(t), y(t), z(t)>.
I am quite stuck on this one. I could let $z = xy$ and then plugging that into M I'd have $2x^2 + y^2 + 2x^2y^2 = 5$, and I have no idea what would be good ways to parametrize that.
Another way is to let $y=z/x$, in which case I end up with $2x^4 + (2z^2-5)x^2+z^2 = 0$, but again, I'm quite stuck. Any help would be really appreciated!
Or, if it is impossible to find the curve C, the question I'm working on actually asks for the tangent line to curve C at P if x'(0) = 3. If this makes things easier, how exactly is using this tangent more useful?
| In the equation $2x^2 + y^2 + 2 x^2 y^2 = 5$ put $x = r \cos t$, $y = \sqrt{2} \,r \sin t$, get $r = \frac{\sqrt{5} }{\sqrt{1 + \sqrt{1 + 5\sin^2 2 t} } }$, and the parametrization
$$\left( \frac{\sqrt{5} \cos t}{\sqrt{1 + \sqrt{1 + 5\sin^2 2 t} } } , \frac{\sqrt{10}\sin t}{\sqrt{1 + \sqrt{1 + 5\sin^2 2 t} } } \right)$$
Note that $t= \tan^{-1} (\frac{1}{\sqrt{2}})$ maps to $(1,1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solving $\cos(40+\theta) = 3\sin(50+\theta)$ Question
Solve the following equation for $0≤\theta≤180$
$$
\cos(40+\theta) = 3\sin(50+\theta)
$$
Hint: $\cos(40) = \sin(50)$
Solution
$$
\cos(40+\theta) = 3\sin(50+\theta)
$$
$$
\cos(40)\cos\theta -\sin(40)\sin\theta = 3(\sin(50)\cos\theta +\sin\theta \cos(50))
$$
$$
\sin(50)\cos\theta-\cos(50)\sin\theta = 3(\sin(50)\cos\theta+\sin\theta \cos(50))
$$
$$
2\sin(50)\cos\theta+4\sin\theta \cos(50) = 0
$$
What do I do next?
| Observe that $\cos(\theta)=0$ cannot give a solution. Rearranging from your last line as suggested by comments, we get $\tan(-50)=2\tan(\theta)$. Recall that $\tan(-150)=\dfrac1{\sqrt 3}$ and the trigonometric identity $\tan(3\phi)
=\dfrac{3\tan(\phi)-\tan^3(\phi)}{1-3\tan^2(\phi)}$. So $\dfrac{6\tan(\theta)-8\tan^3(\theta)}{1-12\tan^2(\theta)}=\dfrac1{\sqrt 3}\implies \dfrac1{\sqrt 3}-6\tan(\theta)-4\sqrt3\tan^2(\theta)+8\tan^3(\theta)=0$. We transform it into a monic depressed cubic in $\tan(\theta)$ as follows:
$\dfrac1{8\sqrt 3}-\dfrac34\tan(\theta)-\dfrac{\sqrt3}2\tan^2(\theta)+\tan^3(\theta)=0$. Let $x+\dfrac{\sqrt3}6=\tan(\theta),$ then the equation may be rewritten as $f(x)=x^3-x-\dfrac{\sqrt3}9=0$.
The discriminant of $f(x)$ is then $\delta^2=-4(-1)^3-27(-\dfrac{\sqrt3}9)^2=3$, and by Cardano formula, $\chi=\sqrt[3]{\dfrac{\sqrt3}{18}+\sqrt{\dfrac{-3}{108}}}+\sqrt[3]{\dfrac{\sqrt3}{18}-\sqrt{\dfrac{-3}{108}}}=\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}+\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}$ is a root of $f$.
Observe that $f(\dfrac{\sqrt3}3)\gt 0,f(1)\lt 0, f(2)\gt 0, $ so by intermediate value theorem, $f$ has at least two real roots. However since $f(x)\in \Bbb R[x]$, if $\alpha$ is a non real root of $f$, then so is its complex conjugate $\bar{\alpha}\ne\alpha$. Therefore $f$ has three real roots, and from this we see that $\chi=\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}+\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}\in \Bbb R$.
Write $p=\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}, q=\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}, $ then $\chi=p+q\implies \chi^3=(p+q)^3=p^3+q^3+3pq(p+q)$. Now, $p^3+q^3=\dfrac {\sqrt 3}9, pq=\sqrt[3]{(\dfrac{\sqrt 3}{18})^2+\dfrac 1{36}}$. Suppose that $p+q<0$, then $|p+q|^3\lt 3pq|p+q|\implies (p+q)^2=p^2+2pq+q^2\lt 3pq \implies p^2+q^2\lt pq$, but by A.M.-G.M.inequality we have $p^2+q^2\gt 2|pq|=2pq$, which gives $0\lt 2pq\lt p^2+q^2\lt pq$, a contradiction.
Hence $\chi=p+q\gt 0$ and $\tan(\theta)=\chi+\dfrac{\sqrt3}6=\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}+\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}+\dfrac{\sqrt3}6\gt 0$. Since $\arctan(\phi)\gt 0\iff \phi \gt 0, \theta=\arctan(\sqrt[3]{\dfrac{\sqrt3}{18}+\dfrac{\sqrt{-1}}6}+\sqrt[3]{\dfrac{\sqrt3}{18}-\dfrac{\sqrt{-1}}6}+\dfrac{\sqrt3}6) \gt 0$, and such $\theta$ is a desired solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How did this shorter method of finding $x^r$ in equation come?
Q)What is coefficient of $x^{30}$ in $$\left(\frac{1-x^{11}}{1-x}\right)^3\left(\frac{1-x^{21}}{1-x}\right)$$
I did this question by expanding $\frac{1}{1-x}$ and multiplying and got a large equation which in the end came to the correct solution of 1111.However, in the book the following steps are provided:
=Coefficient of $x^{30}$ in $(1-x^{11})^3(1-x^{21})(1-x)^{-4}$
=Coefficient of $x^{30}$ in $(1-3x^{11}+3x^{22})(1-x^{21})(1-x)^{-4}$
=Coefficient of $x^{30}$ in $(1-3x^{11}+3x^{22}-x^{21})(1-x)^{-4}$Here,(according to my understanding) the terms greater than $x^{30} $are excluded as only the terms lower than $x^{30}$ can give the desired result(Expansion of $\frac{1}{1-x}=1+x+x^2+x^3+....$). However, in the next step, it equates the above to:
$$\frac{33!}{30!3!} +(-3)\frac{22!}{19!3!}+(3)\frac{11!}{8!3!}+(-1)\frac{12!}{9!3!}$$which also provides the correct solution. How did this equation come?
| $(1-x)^{-n}=1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+...$
Basically, the coefficient of $x^r$ in $(1-x)^{-n} = \binom{n+r-1}{r}$ where n is a natural number
In your case $n=4$
What we want is:
(Coefficient of $x^{30}$ in $(1-x)^{-4}) -(3 \cdot $ Coefficient of $x^{19}$ in $(1-x)^{-4})+(3 \cdot$ Coefficient of $x^8$ in $(1-x)^{-4}) -($ Coefficient of $x^9$ in $(1-x)^{-4}$)
You can take it from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\frac1{1+a}+\frac1{1+b}+\frac1{1+c}\leq\frac14(a+b+c+3)$ if $abc=1$ I found the following exercise in a problem book (with no solutions):
Given $a,b,c>0$ such that $abc=1$ prove that
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\leq\frac{a+b+c+3}{4}$$
I tried AM-GM for the fraction on the LHS but got stuck from there.
| Let $f(x) = \frac{1}{1 + x} - \frac{x+1}{4} + \frac{1}{2}\ln x$.
We have $f'(x) = \frac{(1-x)(x^2+x+2)}{4x(1+x)^2}$.
Thus, $f'(x) > 0$ on $(0, 1)$, and $f'(x) < 0$ on $(1, \infty)$.
Thus, $f(x)$ is strictly increasing on $(0, 1)$, and strictly decreasing on $(1, \infty)$.
Also $f(1) = 0$. Thus, $f(x) \le 0$ on $(0, \infty)$.
Thus, $f(a) + f(b) + f(c) \le 0$ which results in $\frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c}
\le \frac{a+b+c+3}{4}$ (using $\ln (abc) = 0$).
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to find the least in $E^{\circ}=\frac{5S^g}{162}+\frac{C^\circ}{50}+\frac{2\pi^2}{360}\textrm{rad}$? The problem is as follows:
Let $S^{\circ}$, $C^{g}$ and $R\,\textrm{rad}$ the measures of a positive angle in sexagesimal, centesimal and radian degrees respectively such as:
$$E^{\circ}=\frac{5S^g}{162}+\frac{C^\circ}{50}+\frac{2\pi^2}{360R}\textrm{rad}$$
Using this information find the least $E$.
The alternatives in my book are as follows:
$\begin{array}{ll}
1.&\textrm{6}\\
2.&\textrm{8}\\
3.&\textrm{10}\\
4.&\textrm{12}\\
\end{array}$
How exactly should I find the least angle here?.
So far what I remember is that the proportions between each unit are the same in the sense of:
$\frac{S}{360}=\frac{C}{400}=\frac{R}{2\pi}$
Using this information you may find a relationship between the variables.
Plugin these in the above equation I'm getting:
However I find confusing why the degree symbols for $S$ and $C$ appears swapped then I figured that what it might be intended could be that the digits for the angle are represented by the letter but the degree is contrary to what its unit mentions.
For $\frac{5S^g}{162}$:
$\frac{5S^g}{162}\times\frac{360^\circ}{400^g}=\frac{1S}{36}^{\circ}$
For $\frac{C^\circ}{50}$:
Since;
$\frac{S}{360}=\frac{C}{400}$
$C=\frac{10}{9}S$
$\frac{C^\circ}{50}=\frac{10S^\circ}{9}\times \frac{1}{50}=\frac{1S^\circ}{45}$
For: $\frac{2\pi^2}{360R}\textrm{rad}$
Since this case $R$ matches with the degree symbol $\textrm{rad}$ then I would apply the equation:
$\frac{S^{\circ}}{180}=\frac{R\rad}{\pi}$
$R \textrm{rad}=\frac{S^\circ \pi}{180}$
$\frac{2\pi^2}{360R}\textrm{rad}=\frac{2\pi^2 \times 180}{360S^\circ \pi}=\frac{\pi}{S^\circ}$
Then:
$E^{\circ}=\frac{5S^g}{162}+\frac{C^\circ}{50}+\frac{2\pi^2}{360R}\textrm{rad}$
$E^{\circ}=\frac{1S}{36}^{\circ}+\frac{1S^\circ}{45}+\frac{\pi}{S^\circ}$
$E^{\circ}=\frac{S^2+20\pi}{20S}$
Then I assume that this latter expression must me minimized but would it be used calculus for this?
If so:
$E^{\circ}\,'=\frac{1}{20}-\frac{\pi}{S^2}$
Equating this to zero:
$E^{\circ}=2\sqrt{5\pi}$
But I don't know if this is what it was intended?.
Can someone help me here?. I'm stuck, why it doesn't matches any of the alternatives given or could it be that I made a mistake?. Please help.
| While it is true that $R$ rad $=\dfrac {S^\circ \pi}{180^\circ} $, it is not true that $\dfrac {2\pi^2}{360R}$rad $= \dfrac {2\pi^2 \times 180^\circ}{360S^\circ\pi}$; the latter is actually equal to $\dfrac {2\pi^2}{360R \text{ rad}}$. To resolve this, we consider the numerical value of $R$ and the conversion of rad to degrees (sexagesimal) seperately:
$$R = \frac {S \pi}{180}, \quad 1 \text{ rad }=\frac {180}{\pi}^\circ$$
Now we have:
$$\frac {2\pi^2}{360R} \text{ rad } = \frac {2\pi^2 \times 180}{360S\pi} \times \frac {180^\circ}{\pi}=\frac {180^\circ}{S}$$
$$E^\circ = \frac{5S^g}{162}+\frac{C^\circ}{50}+\frac{2\pi^2}{360R}\textrm{rad} = \frac {S^\circ}{36} + \frac {S^\circ}{45} + \frac {180^\circ }S = \left(\frac S{20}+\frac{180}S\right)^\circ$$
To minimize this we may use calculus, or better yet, the 2-term form of AM $\ge$ GM:
$$a+b\ge 2\sqrt{ab}$$
$$\frac S{20}+\frac{180}S\ge2\sqrt{\frac S{20}\times \frac {180}S} =6$$
Hence the minimum of $E^\circ$ is $6^\circ$ (which occurs at $S = 60$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3961092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A relationship between $a$ and $b$ related to algorithm Rock(1)-Scissor(0)-Paper(2)
This is related to algorithm Rock(1)-Scissor(0)-Paper(2)
I need to find a relationship between $a$ and $b$ so that
$$\begin{matrix}
a & b \\
1 & 0\\
0 & 2\\
2 & 1
\end{matrix}$$
is always true and
$$\begin{matrix}
a & b \\
1 & 2\\
0 & 1\\
2 & 0
\end{matrix}$$
is always false. I had read a book about this involving to binary and computer. I'm quite sure that we must do some analytics among
$$\begin{matrix}
01 & 00 \\
00 & 10
\end{matrix}$$
and
$$\begin{matrix}
00 & 10 \\
10 & 01
\end{matrix}$$
and
$$\begin{matrix}
10 & 01 \\
01 & 00
\end{matrix}$$
What should I do next ?? And I wanna know more about this knowledge, thanks a real lot !
Edit: another solution that is
$$a- b+ 2\equiv 0\mod 3$$
|
I want some new observations for this problem like system of modulo-equations
This answer shows three systems.
Let us define $f(a,b):=pa+qa^2+rb+sb^2$ where $p,q,r,s$ are integers. (This is because $x^3\equiv x\pmod 3$ always holds.)
We want $f(a,b)$ to satisfy
$$f(1,0)\equiv f(0,2)\equiv f(2,1)\pmod 3,$$
i.e.
$$p+q\equiv -r+s\equiv -p+q+r+s\pmod 3,$$
i.e.
$$r\equiv -p+q\pmod 3,\qquad s\equiv -q\pmod 3$$
So, $f(a,b)$ can be written as
$$f(a,b)\equiv pa+qa^2+(-p+q)b-qb^2\pmod 3$$
*
*For $(p,q)\equiv (0,1)\pmod 3$, the only solutions of $f(a,b)\equiv a^2+b-b^2\equiv 1\pmod 3$ are $(a,b)\equiv (1,0),(0,2),(2,1),(1,1),(2,0)\pmod 3$.
*For $(p,q)\equiv (1,1)\pmod 3$, the only solutions of $f(a,b)\equiv a+a^2-b^2\equiv 2\pmod 3$ are $(a,b)\equiv (1,0),(0,2),(2,1),(0,1),(2,2)\pmod 3$.
*For $(p,q)\equiv (1,2)\pmod 3$, the only solutions of $f(a,b)\equiv a-a^2+b+b^2\equiv 0\pmod 3$ are $(a,b)\equiv (1,0),(0,2),(2,1),(0,0),(1,2)\pmod 3$.
From these observations, we get the following three systems whose solutions are $(a,b)=(1,0),(0,2),(2,1)$.
(1)$$\begin{cases}a^2+b-b^2\equiv 1\pmod 3
\\\\a+a^2-b^2\equiv 2\pmod 3\end{cases}$$
(2)$$\begin{cases}a^2+b-b^2\equiv 1\pmod 3
\\\\a-a^2+b+b^2\equiv 0\pmod 3\end{cases}$$
(3)$$\begin{cases}a+a^2-b^2\equiv 2\pmod 3
\\\\a-a^2+b+b^2\equiv 0\pmod 3\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Remainder of Polynomial Division of $(x^2 + x +1)^n$ by $x^2 - x +1$ I am trying to solve the following problem:
Given $n \in \mathbb{N}$, find the remainder upon division of $(x^2 + x +1)^n$ by $x^2 - x +1$
the given hint to the problem is:
"Compute $(x^2 + x +1)^n$ by writing $x^2 + x +1 = (x^2 - x +1) + 2x$. Then, use the uniqueness part of the division algorithm."
If I take $a = x^2 - x +1$ I have
$$(x^2 + x +1)^n = (a + 2x)^n = a^n + \binom{n}{1}a^{n-1} 2x+ \binom{n}{2}a^{n-2} (2x)^2 + \dots + (2x)^n$$
but how do I proceed further?
| You're trying to divide by $a$ with remainder. You almost have it, since in your expression for $(a+2x)^n$ all terms but the final $(2x)^n$ are multiples of $a$ so can be dropped in getting the remainder. You now can focus just on $(2x)^n.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Find an expression for $n$ in terms of $x$, $y$, and $z$. At a certain rate of compound interest, $100$ will increase to $200$ in $x$ years, $200$ will increase to $300$ in $y$ years, and $300$ will increase to $1,500$ in $z$ years. If $600$ will increase to $1,000$ in $n$ year, find an expression for $n$ in terms of $x$, $y$, and $z$.
I know:
$600(1+i)^n = 1000$
I wrote:
$200 = 100(1+i)^x$
$300 = 100(1+i)^{x+y}$
$1500 = 100(1+i)^{x+y+z}$
Also, I know that :
$600=100(1+i)^{2x+y}$
Hence:
$1000=100((1+i)^{x+y+z}-(1+i)^{x+z}-(1+i)^x)$
$1000=100(1+i)^{2x+y}((1+i)^{-x+z}-(1+i)^{-x}-(1+i)^{-x-y})$
$1000=600((1+i)^{-x+z}-(1+i)^{-x}-(1+i)^{-x-y})$
$(1+i)^n=(1+i)^{-x+z}-(1+i)^{-x}-(1+i)^{-x-y}$
I need to get $n$ as a function of $x$,$y$ and $z$
I have a problem with my last line.
Does anyone knows a faster way to get the solution ?
| Let's make it clean and simple. Instead of $1+i$, I'll put $q$ as the rate. And I'll cancel out all the unnecessary terms. So we have
$$
2 = q^x \qquad \frac{3}{2} = q^y \qquad 5 = q^z
$$
Now we want to make the ratio $\frac{10}{6}$ with these. We can 'massage' the expression a bit:
$$
\frac{10}{6}
= \frac{2 \cdot 5}{ \left( \frac{3}{2} \right)\cdot 2 \cdot 2}
= \frac{5}{ \left( \frac{3}{2} \right)\cdot 2} = 5 \cdot \left( \frac{3}{2} \right)^{-1} \cdot (2)^{-1}
$$
Now you can just plug in the known values and you're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many positive integral values of $n$, less than $100$, are there such that $n^3+72$ is completely divisible by $n+7$? How many positive integral values of $n$, less than $100$, are there such that $n^3+72$ is completely divisible by $n+7$ ?
MY WORK :-
Well we know that $a+b$ divides $a^3+b^3$ thus $n+7$ divides $n^3+7^3$ thus it divides $n^3+343$. Now if $n+7$ divides $n^3+72$ then it divides their difference. Thus $n+7$ divides $343-72=271$. Now the factor of $271$ are $1$ and $271$ only. So if $n\lt100$ then there are no values possible; hence answer should be $0$. However in answer key it's said $1$. I want to know why it's $1$ ??
| Do the polynomial division with remainder:
$${n^3+72\over n+7}=n^2-7n+49-{271\over n+7}\ .$$
Now $n^2-7n+49$ is an integer for all $n$. Therefore it remains to check, for which $n\in[100]$ the number $271$ is divisible by $n+7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3966917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? How do we show $1-\cos x\ge\frac{x^2}3$ for $|x|\le1$? My first idea was to write $$1-\cos x=\frac12\left|e^{{\rm i}x}-1\right|^2\tag1,$$ which is true for all $x\in\mathbb R$, but I don't have a suitable lower bound for the right-hand side at hand.
| The positive root of the equation $1-\cos x = \frac{x^2}{3}$ is approximately $2.16$
so the inequality is valid on a larger interval, say $[-2, 2]$. To show this consider
$$\frac{1-\cos x}{\frac{x^2}{3}} = \frac{2 \sin^2 \frac{x}{2}}{\frac{x^2}{3}}= \frac{3}{2} \left(\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^2$$
Now, the function $t\mapsto \sin t$ is concave on $[0, \pi]$, so $\frac{\sin t}{t}$ is decreasing on $[0, \pi]$. Therefore, on the interval $[-2, 2]$ we have $\frac{\sin \frac{x}{2}}{\frac{x}{2}} \ge \sin 1$ and so
$$\frac{1-\cos x}{x^2/3} \ge \frac{3}{2} \sin^2 1= 1.06.. > 1$$
On the interval $[-1,1]$ the inequality is
$$1- \cos x \ge 2 \sin^2\frac{1}{2}\cdot x^2$$
where $2 \sin^2 \frac{1}{2} = 0.45969\ldots$
and this is the best estimate on $[-1,1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
Disagreement over the arc length of an ellipse I'm reading Elliptic Functions and Elliptic Integrals by Prasolov and Solovyev. On page $53$, it reads:
The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ can be given parametrically by the formulas $x=a\cos \varphi$, $y=b\sin\varphi$. The differential $dl$ of the length of an arc on the ellipse is equal to $\sqrt{dx^2+dy^2}=d\varphi\sqrt{a^2\cos ^2\varphi +b^2\sin ^2\varphi}$. If $a=1$ and $b=\sqrt{1-k^2}$, then $dl=d\varphi \sqrt{1-k^2\sin ^2\varphi}$. In this case the length of the arc on the ellipse between the end point of the small half axis, $B$, and the point $M=(\cos\varphi ,b\sin\varphi )$ is equal to $E (\varphi )=\int_0^{\varphi}\sqrt{1-k^2\sin ^2\psi}\, d\psi .$
I think there is an error. My approach is the following:
The upper half of the ellipse (there are the points $B$ and $M$) can be represented by $y=b\sqrt{1-x^2}$. The arc length, measured from the point $B=(0,b)$ to an arbitrary point in the first quadrant $M$ in terms of the horizontal component of $M$ is
$$s=\int_0^x \sqrt{\frac{1-k^2t^2}{1-t^2}}\, dt$$
where $b=\sqrt{1-k^2}$. The substitution $u=\arcsin t$ gives
$$s=\int_0^{\arcsin x}\sqrt{1-k^2\sin ^2 u}\, du.$$
The ellipse is parametrized by $x=\cos\varphi$, $y=b\sin\varphi$, therefore if $M=(\cos\varphi ,b\sin\varphi )$, then
$$\begin{align}s&=\int_0^{\arcsin \cos\varphi}\sqrt{1-k^2\sin ^2 u}\, du\\&=\int_0^{\frac{\pi}{2}-\varphi}\sqrt{1-k^2 \sin ^2 u}\, du.\end{align}$$
So the required arc length should be $E\left(\frac{\pi}{2}-\varphi\right)$, not $E(\varphi )$. A considerable amount of the following theorems in the book is "proved" assuming $E (\varphi )$ for the arc length, which seems a bit worrying. Maybe I'm missing something.
| If $x=a\cos \varphi$ and $y=b\sin\varphi$ then
$$
(dx)^2+(dy)^2 = a^2\sin^2\varphi\, d\varphi +b^2\cos^2\varphi\, d\varphi,
$$
contrary to the book's claim that it would be
$a^2\cos^2\varphi\, d\varphi +b^2\sin^2\varphi\, d\varphi.$
Yet $x=a\cos \varphi$ and $y=b\sin\varphi$ also implies that
$(x,y) = (a,0)$ when $\varphi = 0$ and $(x,y) = (0,b)$ when $\varphi = \frac\pi2,$ which implies that to integrate the curve from $(0,b)$ to an arbitrary point on the ellipse you should choose $\frac\pi2$ and not $0$ as the fixed end of your integral.
So altogether the book is not making sense.
But if we make just one change -- instead of $x=a\cos\varphi$ and $y=b\sin\varphi$,
let $x=a\sin\varphi$ and $y=b\cos\varphi$ --
then $(x,y) = (0,b)$ when $\varphi=0,$ it therefore makes sense to use $0$ as the fixed end of the integral when integrating the curve length from $(0,b)$ to an arbitrary point on the ellipse, and
$$
(dx)^2+(dy)^2 = a^2\cos^2\varphi\, d\varphi +b^2\sin^2\varphi\, d\varphi
$$
as claimed in the book.
So I will guess that the original intention was to set
$x=a\sin\varphi$ and $y=b\cos\varphi,$ but sometime between the original conception
of the integration and the time when the book was typeset, someone mistakenly wrote
the more usual formulas $x=a\cos\varphi$ and $y=b\sin\varphi$ instead of the particular formulas that are correct for this particular problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integral of functions of several variables The function is defined from $[-1,1]\times[-1,1]$ to $\Bbb R$, given by $f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}$ when $(x,y)≠0$ and $f(0,0)=0$.
I could find that the function is not continuous at origin, so not differentiable and also partial derivatives does not exist at origin.
But how do we evaluate the integral of the function over the given domain. Being function of several variable how do we deal with the point of discontinuity? Can we convert it to polar coordinates and apply residue theorem? Do we have any other methods?
Any help would be appreciated. Thanks in advance.
| This is a (somewhat?) standard example of how rearranging an iterated integral makes it converge conditionally to either of two different numbers.
\begin{align}
& \int_0^1 \left( \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dy \right) \, dx = +\frac \pi 4. \\[10pt]
& \int_0^1 \left( \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dx \right) \, dy = -\frac \pi 4. \\[10pt]
& \iint\limits_{0\,<\,y\,<\,x\,<\,1} \frac{x^2-y^2}{(x^2+y^2)^2} \, d(x,y) = +\infty. \\[10pt]
& \iint\limits_{0\,<\,x\,<\,y\,<\,1} \frac{x^2-y^2}{(x^2+y^2)^2} \, d(x,y) = -\infty. \\[10pt]
& \iint\limits_{\varepsilon\,<\,x\,<\,1 \\ \varepsilon\,<\,y\,<\,1} \frac{x^2-y^2}{(x^2+y^2)^2} \, d(x,y) = 0.
\end{align}
Only when the integrals of the positive and negative parts are both infinite can the values of the two iterated integrals differ.
Let $y = x\tan \theta$, so that $dy = x\sec^2\theta\,d\theta$ and
$x^2 + y^2= x^2\sec^2\theta$, and as $y$ goes from $0$ to $1$
then $\theta$ goes from $0$ to $\arctan(1/x)$. Then
\begin{align*}
\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \, dy
= {} & \int_0^{\arctan(1/x)}
\frac{x^2 - x^2 \tan^2\theta}{(x^2 + x^2\tan^2\theta)^2} \big( x\sec^2\theta\,d\theta\big) \\[10pt]
= {} & \frac 1 x \int_0^{\arctan(1/x)} \frac{1-\tan^2 \theta}{\sec^2 \theta} \, d\theta \\[10pt]
= {} & \frac 1 x \int_0^{\arctan(1/x)} (\cos^2\theta-\sin^2\theta) \, d\theta \\[10pt]
= {} & \frac 1 x \int_0^{\arctan(1/x)} \cos(2\theta) \, d\theta \\[10pt]
= {} & \frac 1 {2x} \sin\left(2\arctan \frac 1 x\right) \\[10pt]
= {} & \frac 1 x \sin\left(\arctan \frac 1 x \right) \cos\left( \arctan \frac 1 x \right) \\[10pt]
= {} & \frac 1 x \cdot \frac 1 {\sqrt{1+x^2}} \cdot \frac x {\sqrt{1+x^2}} = \frac 1 {1+x^2}. \\[10pt]
\text{And then}
& \int_0^1 \frac{dx}{1+x^2} = \frac \pi 4.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Convergence of $\sum^\infty_{n=0} \frac{\cos(n + \frac{1}{n^2})}{n \cdot \ln(n^2 + 1)}$ Is my idea correct? Convergence:
$$\sum^\infty_{n=0}\frac{\cos \left(n + \frac{1}{n^2} \right)}{n \cdot \ln \left( n^2 + 1 \right)}$$
Edit (I tried to follow the idea written by @Daniel Fischer):
$$\sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2})}{n\cdot ln(n^2 + 1)} = \sum_{n=0}^{\infty} \frac{cos(n)}{n \cdot ln(n^2 + 1)} + \sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2}) - cos(n)}{n \cdot ln(n^2 + 1)}$$
$\sum_{n=0}^{\infty} \frac{cos(n)}{n \cdot ln(n^2 + 1)}$ is convergent because of Dirichlet test and How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?
$\sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2}) - cos(n)}{n \cdot ln(n^2 + 1)} = \sum_{n=0}^{\infty} \frac{-2 \left(sin \left( \frac{2n+\frac{1}{n^2}}{2} \right) \cdot sin \left( \frac{\frac{1}{n^2}}{2} \right) \right)}{n \cdot ln \left(n^2 + 1 \right)} = \sum_{n=0}^{\infty} \frac{-2 \left(sin \left( n+\frac{1}{2n^2} \right) \cdot sin \left( \frac{1}{2n^2} \right) \right)}{n \cdot ln \left(n^2 + 1 \right)}$
Then:
$$\sum_{n=0}^{\infty} \left| \frac{-2 \left(sin \left( n+\frac{1}{2n^2} \right) \cdot sin \left( \frac{1}{2n^2} \right) \right)}{n \cdot ln \left(n^2 + 1 \right)} \right| \leq \sum_{n=0}^{\infty} \left| 2 \cdot \frac{ \frac{1}{2n^2} }{n \cdot ln \left(n^2 + 1 \right)} \right| \leq \sum_{n=0}^{\infty} \left| \frac{ 1 }{n^3 \cdot ln \left(n^2 + 1 \right)} \right| $$
For $n \geq 3$:
$$\sum_{n=0}^{\infty} \left| \frac{ 1 }{n^3 \cdot ln \left(n^2 + 1 \right)} \right| \leq \sum_{n=0}^{\infty} \left| \frac{1}{n^2} \right|$$
Therefore we know that $\sum_{n=0}^{\infty} \frac{cos(n + \frac{1}{n^2}) - cos(n)}{n \cdot ln(n^2 + 1)}$ it is also convergent. Is that correct?
| Continuing from the remark made in the comments,
$$\cos\biggl(n + \frac{1}{n^2}\biggr) = \cos n + \biggl(\cos \biggl(n + \frac{1}{n^2}\biggr) - \cos n\biggr).$$
The second term on the right decays like $1/n^2$, since cosine is Lipschitz continuous (prove this using the cosine sum identity). The partial sums of the first term on the right are uniformly bounded; to see this, it is convenient to remember cosine is the real part of the complex exponential,
$$ \cos n = \operatorname{Re} e^{in}. $$
Thus the partial sum of cosine is just the real part of a nice geometric series,
$$ \sum_{n = 0}^m \cos n = \operatorname{Re} \sum_{n = 0}^m e^{in} = \operatorname{Re} \frac{1 - e^{i(m + 1)}}{1 - e^i}. $$
The right hand side has uniformly bounded modulus in $m$, so we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $a_n=100a_{n−1}+134$ , find least value of $n$ for which $a_n$ is divisible by $99$
Let $a_{1}=24$ and form the sequence $a_{n}, n \geq 2$ by $a_{n}=100 a_{n-1}+134 .$ The first few terms are
$$
24,2534,253534,25353534, \ldots
$$
What is the least value of $\mathrm{n}$ for which $\mathrm{a}_{\mathrm{n}}$ is divisible by $99 ?$
So, In this post,the present answer there get to the result that $a_n \equiv a_1 + (n - 1)35 \equiv 35n - 11 \equiv 0 \pmod{99} \tag{1}\label{eq1A}$
But after that what i did is to multiply both sides by $9$ in the congruence
$35n - 11 \equiv 0 \pmod{99}$
$315n \equiv 0 \pmod{99}$
$18n \equiv 0 \pmod{99}$
dividing by $18$ ,we get
$n \equiv 0 \pmod{11}$ but I did not get the other part of the congruence which was arrived in that answer by this process ?
thankyou
| You need to solve $35n \equiv 11 \pmod{99}$. This can be done via a variation of the (Extended) Euclidean Algorithm.
$$
\begin{array}{ll}
99n \equiv 0 \pmod{99} & (1)\\
35n \equiv 11 \pmod{99} & (2)\\
-6n \equiv -33 \pmod{99} & (3) = (1)-3\times(2)\\
-n \equiv -187 \equiv -88 \pmod{99} & (4) = (2) + 6\times(3)
\end{array}
$$
Therefore $n\equiv 88 \pmod{99}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Prove that if $(a + b + c)(ab + ac + bc) = abc$ then sum of some two numbers equals $0$ Prove that if $(a + b + c)(ab + ac + bc) = abc$ then sum of some two numbers equals $0$.
Without loss of generality let's suppose that $a=0$ then $(b + c)bc = 0 \Rightarrow b+c = 0$ or
$bc = 0 \Rightarrow b=0 \lor c = 0$ and $a+b=0 \lor a+c=0$ respectively. Q.E.D.
Now let's suppose $abc \not = 0$ and $a+b+c \not = 0$ then from initial equation we can get the following one that must hold $\dfrac{1}{a+b+c} = \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}$
As $a,b,c$ are some arbitrary numbers (but not non-negative integers for example) I've got completely stuck here
| Hint:
$$(a + b + c)(ab + ac + bc) - abc=(a + b)(b + c)(c + a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3983909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $x^3 + \frac{1}{x^3}$ Question
$x^2 + \frac{1}{x^2}=34$ and $x$ is a natural number. Find the value of $x^3 + \frac{1}{x^3}$ and choose the correct answer from the following options:
*
*198
*216
*200
*186
What I have did yet
I tried to find the value of $x + \frac{1}{x}$. Here are my steps to do so:
$$x^2 + \frac{1}{x^2} = 34$$
$$\text{Since}, (x+\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}$$
$$\Rightarrow (x+\frac{1}{x})^2-2=34$$
$$\Rightarrow (x+\frac{1}{x})^2=34+2 = 36$$
$$\Rightarrow x+\frac{1}{x}=\sqrt{36}=6$$
I have calculated the value of $x+\frac{1}{x}$ is $6$. I do not know what to do next. Any help will be appreciated. Thank you in advanced!
| You wrote
$$(x+\frac{1}{x}) = x^2 + 2 + \frac{1}{x^2}.$$
But it should read
$$(x+\frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}.$$
Then we obtain $x+\frac{1}{x}=6.$
From $x^2 + \frac{1}{x^2}=34$ we obtain
$(1) \quad x^3+\frac{1}{x}=34 x$
and
$(2) \quad x+\frac{1}{x^3}=\frac{34}{x}.$
If we add (1) and (2) we get
$$x^3+\frac{1}{x^3}=33(x+\frac{1}{x}).$$
Ans so
$$x^3+\frac{1}{x^3}=6 \cdot 33=198.$$
Remark : if $x^2 + \frac{1}{x^2}=34$, then $x$ can not be a natural number !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solving a linear system of equations use row operations Can somebody help me find the simplest way to use elementary operations of multiplying one row by another and scalar multiplication of a row to find the solution to the system of equations?
$$2x_1+4x_2-x_3=7$$ $$x_1+x_2-x_3=0$$ $$3x_1-2x_2+3x_3=8$$
Every time I solve this it feels like it could have been done simpler... Thanks!
My attempt:
I first exchanged row 1 and row 2 to get:
$$x_1+x_2-x_3=0$$ $$2x_1+4x_2-x_3=7$$ $$3x_1-2x_2+3x_3=8$$
Then I did $-2R_1+R_2 \rightarrow R_2$ and $-3R_1+R_3 \rightarrow R_3$ to get:
$$x_1+x_2-x_3=0$$ $$0x_1+2x_2+x_3=7$$ $$0x_1-5x_2+6x_3=8$$
But now I can't use row 2 to eliminate the $x_2$ in row 3 without introducing fractions. I realize i could use back substitution, but I want to use row operations. Was there a different way I could have carried out the row operations to make this nicer? Thank you!
| We are given
$$\begin{pmatrix}
2 & 4 & -1\\
1 & 1 & -1\\
3 & -2 & 3
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=
\begin{pmatrix}
7\\
0\\
8
\end{pmatrix}$$
Take $R_1 - 2R_2 \to R_2$ and $-3R_1+2R_3 \to R_3$ to form
$$\begin{pmatrix}
2 & 4 & -1\\
0 & 2 & 1\\
0 & -16 & 9
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=
\begin{pmatrix}
7\\
7\\
-5
\end{pmatrix}$$
Then let $8R_2 + R_3 \to R_3$ to obtain
$$\begin{pmatrix}
2 & 4 & -1\\
0 & 2 & 1\\
0 & 0 & 17
\end{pmatrix}
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}=
\begin{pmatrix}
7\\
7\\
51
\end{pmatrix}$$
We then find that $x_1=1,x_2=2,x_3=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3986688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is $A$ a perfect square? Consider
$$A=a^2+2ab^2+b^4-4bc-4b^3,$$
where $a,b,c\in\mathbb{Z}$ and $b\neq0$ such that $b|a$ and $b|c$, so $b|A$.
Now I want to know that Is $A$ a perfect square?
| No, $A$ is not a perfect square for such $a,b,c$ values. As a counterexample, take $a=b=c=1$, for which $A=-4$.
EDIT: The answer below corresponds to the original question "Can $A$ be a perfect square?".
Yes, it can be 0. Let $a=bn$ and $c=bm$. Then:
$$\begin{align}
A&=a^2+2ab^2+b^4-4bc-4b^3\\
&=b^2n^2+2b^3n+b^4-4b^2m-4b^3\\
&=b^2 \big((n+b)^2-4(m+b)\big)
\end{align}$$
So any values that make $(n+b)^2-4(m+b)$ a square are valid ones. So, for instance, any $n=-b$ and $m=-b$ are enough, but you can find many others.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$ A=\begin{pmatrix} 0 & 2 & 0\\ 0 & 0 & 3 \\ 0 & 0 & 0 \end{pmatrix}$, calculate $e^A$ I have encounter a question in my book , it was For $ A=\begin{pmatrix}
0 & 2 & 0\\
0 & 0 & 3 \\
0 & 0 & 0
\end{pmatrix}$, calculate $e^A$
My solution way : I tried to find its eigenvalues , so i found that the only eigenvalue is $0$ and the eigenspace is $(1,0,0)$ .Hence , it cannot be diagonalized.
Then , i tried to use taylor exponential and it gives me $ A=\begin{pmatrix}
1 & 2 & 0\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .However the answer is $ A=\begin{pmatrix}
1 & 2 & 3\\
0 & 1 & 3 \\
0 & 0 & 1
\end{pmatrix}$ .
What am i missing ,can you help me?
| Note that $$A^2=\begin{pmatrix}
0 & 0 & 6\\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix},$$ which is where the $3$ in the upper-right corner comes from.
Since $A^3$ is just the $3\times 3$ matrix of zeroes, $e^A=I+A+\frac12 A^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Uniform convergence of a functional series $\sum\limits _{n=1}^{\infty}\frac{x\sin(xn)}{n+x}$ Here's the question: Check series for uniform convergence: $$\sum\limits _{n=1}^{\infty}\frac{x\sin(nx)}{n+x}$$
a) $ x\in E_{1} = (0;\pi) ;$
b) $ x\in E_{2} = (\pi;+\infty) .$
For a) I used the fact that $ \left| \sum\limits_{n=1}^{N}x\sin(nx) \right| \le \left| x\sum\limits_{n=1}^{N}\sin(nx) \right| \le \left| \displaystyle{\frac{x}{\sin(\frac{x}{2})}} \right| \le \pi$ , that means we have bounded partial sums. Sequence $ \{\displaystyle{\frac{1}{n+x}}\} $ is monotone and uniformly goes to 0 for every value of $ x\in E_{1} $ as n goes to $\infty$. Thus, series is umiform convergent on $ E_{1} = (0;\pi) $ . Question is, can I perform the same thing setting partial sums as $ \left| \sum\limits_{n=1}^{N}\sin(nx) \right| $ and sequence as $ \{\displaystyle{\frac{x}{n+x}}\} $ which tells us that it converges to $1$ and finally saying that it converges not uniformly?
| For part (b), consider a sequence $x_n = \frac{\pi}{4n} + 2n\pi \in E_2$. For $n < k \leqslant 2n$ we have
$$\frac{1}{\sqrt{2}} = \sin \left(\frac{\pi}{4}+ 2n^2 \pi \right) = \sin nx_n < \sin k x_n < \sin 2nx_n = \sin \left(\frac{\pi}{2}+ 2n^2 \pi \right) = 1$$
Thus,
$$ \left|\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} \right| =\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} > n \cdot \frac{x_n}{2n + x_n}\cdot \frac{1}{\sqrt{2}} = \frac{2n^2\pi + \frac{\pi}{4}}{\sqrt{2}(2n + 2n\pi + \frac{\pi}{4n})} $$
Since, the RHS tends to $+\infty$ as $n \to \infty$, the series cannot be uniformly convergent on $E_2$, by violation of the uniform Cauchy criterion.
Edit (9/16/2022):
The argument above is flawed because with $x_n = \frac{\pi}{4n} + 2n\pi$, we should have $\frac{\pi}{4} +2n^2\pi < kx_n \leqslant \frac{\pi}{2} + 2n^2\pi + 2n^2\pi$ and we don't have $\sin kx_n > 0$ for all $n< k\leqslant 2n$.
It should be amended as follows.
Consider a sequence $x_n = \frac{\pi}{4n} + 2\pi \in E_2$. For $n < k \leqslant 2n$ we have $\frac{\pi}{4} < \frac{\pi k}{4n}\leqslant \frac{\pi}{2}$ and
$$\frac{1}{\sqrt{2}} = \sin \left(\frac{\pi}{4}\right) < \sin \frac{\pi k}{4n} =\sin \left(\frac{\pi k}{4n}+ 2k\pi\right) = \sin kx_n < \sin \left(\frac{\pi}{2} \right) = 1$$
Thus,
$$ \left|\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} \right| =\sum_{k = n+1}^{2n}\frac{x_n \sin kx_n}{k+x_n} > n \cdot \frac{x_n}{2n + x_n}\cdot \frac{1}{\sqrt{2}} = \frac{2n\pi + \frac{\pi}{4}}{\sqrt{2}(2n + 2\pi + \frac{\pi}{4n})} $$
Since the RHS tends to $\frac{\pi}{\sqrt{2}}\neq 0$ as $n \to \infty$, the series cannot be uniformly convergent ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Closed form for $\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$ How to get a closed form for the following sum:
$$\sum_{k=0}^{n}\left(-1\right)^{k}F_{k}^{2}$$
Where $F_k$ is the $k$th Fibonacci number.
I tried to get a recurrence relation,but I failed and the closed form is given by $$\frac{\left(-1\right)^{n}F_{2n+1}-\left(2n+1\right)}{5}$$
| First let's take a look at the power series $ \sum\limits_{n\geq 0}{F_{n}^{2}x^{n}} $, it is easy to find what its radius of convergence would be using the ratio test, but for now let's denote it $ R<1 $, and let's define a function $ f:x\mapsto\sum\limits_{n=0}^{+\infty}{F_{n}^{2}x^{n}} $.
We have for all $ x\in\left(0,R\right) $ : \begin{aligned} \frac{1}{1-x}\sum_{n=0}^{+\infty}{F_{n}^{2}x^{n}}=\sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{F_{k}^{2}}\right)x^{n}}&=\sum_{n=0}^{+\infty}{F_{n}F_{n+1}x^{n}}\\&=\sum_{n=0}^{+\infty}{\left(F_{n+1}^{2}-F_{n}^{2}-\left(-1\right)^{n}\right)x^{n}}\\ \frac{f\left(x\right)}{1-x}&=\frac{f\left(x\right)}{x}-f\left(x\right)-\frac{1}{1+x}\\ \iff f\left(x\right)&=\frac{x\left(1-x\right)}{\left(1+x\right)\left(x^{2}-3x+1\right)}\end{aligned}
Now : \begin{aligned} \sum_{n=0}^{+\infty}{\left(\sum_{k=0}^{n}{\left(-1\right)^{n-k}F_{k}^{2}}\right)x^{n}}&=\left(\sum_{n=0}^{+\infty}{\left(-1\right)^{n}x^{n}}\right)\left(\sum_{n=0}^{+\infty}{F_{n}^{2}x^{n}}\right)\\ &=\frac{x\left(1-x\right)}{\left(1+x\right)^{2}\left(x^{2}-3x+1\right)}\\ &=\frac{1-x}{5\left(x^{2}-3x+1\right)}+\frac{1}{5\left(x+1\right)}-\frac{2}{5\left(x+1\right)^{2}} \\ &=\frac{1}{5}\sum_{n=1}^{+\infty}{F_{n}^{2}x^{n-1}}+\frac{1}{5}\sum_{n=0}^{+\infty}{\left(-1\right)^{n}x^{n}}-\frac{2}{5}\sum_{n=1}^{+\infty}{n\left(-1\right)^{n}x^{n-1}}\\ &=\frac{1}{5}\sum_{n=0}^{+\infty}{\left(F_{n+1}^{2}+\left(-1\right)^{n}\left(2n+3\right)\right)x^{n}}\end{aligned}
Thus, for all $ n\in\mathbb{N} $ : $$ \sum_{k=0}^{n}{\left(-1\right)^{k}F_{k}^{2}}=\frac{\left(-1\right)^{n}F_{n+1}^{2}+2n+3}{5} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to get the value of $2\left |\csc x \right | \sin x + 3\left | \cos y\right|\sec y$ given two constrains? The problem is as follows:
Given:
$\sqrt{\cos x}\cdot \sin y > 0$
and,
$\tan x\cdot \sqrt{\cot y} < 0$
Find:
$B=2\left |\csc x \right | \sin x + 3\left | \cos y\right|\sec y$
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\textrm{-1}\\
2.&\textrm{5}\\
3.&\textrm{1}\\
4.&\textrm{-5}\\
\end{array}$
I'm confused on exactly how to use the given clues to solve the problem?
To me the source of confusion is how to use the absolute value in the question?
From the first given expression I'm getting this, assuming squaring both sides of the equation will not modify its order:
$\left(\sqrt{\cos x}\cdot \sin y\right)^2 > 0^2$
$\cos x \cdot \sin^2 y > 0$
$\left(\tan x\cdot \sqrt{\cot y}\right )^2 < 0^2$
$\tan^2 x\cdot \cot y < 0^2$
But the thing is this where I'm stuck, where to go from here?. There isn't known a relationship between those angles. If so then I believe trigonometric identities could be used to simplify the expression. Therefore, should these expressions be divided or what?.}
Can someone help me here on what should be done? and more importantly why?. It would help me the most is an answer which would explain how does absolute value is used here?.
| $a<0$ doesn't necessarily mean $a^2<0$.
For example, $-2<0$ but $(-2)^2 = 4 >0$.
Now,
$\begin{align}B &= 2|\csc x|\sin x + 3 |\cos y|\sec y = 2 \frac{\sin x}{|\sin x|}+3\frac{|\cos y|}{\cos y}\\
\\& = 2\text{ sgn}(\sin x) + 3\text{ sgn}(\cos y)\end{align}$
Then, $\sqrt{\cos x} \sin y > 0 \Rightarrow \sin y > 0$ and $\cos x > 0$ as $\cos$ is under the square root.
Similarly, $\cot y>0$ and $\tan x<0$.
So we have,
$\sin y >0, \cot y = \frac{\cos y}{\sin y} > 0 \Rightarrow \boxed{\cos y >0}$
$\cos x>0, \tan x = \frac{\sin x}{\cos x}<0 \Rightarrow \boxed{\sin x<0}$
Thus,
$B = 2(-1) + 3(1) = 1$
Edit (based on comment)
As $\sin x<0 \Rightarrow |\sin x| = -\sin x$ and $\cos y >0 \Rightarrow |\cos y| = \cos y$.
So,
$\begin{align}B = 2 \frac{\sin x}{|\sin x|}+3\frac{|\cos y|}{\cos y} = 2 \frac{\sin x}{-\sin x}+3\frac{\cos y}{\cos y} = -2 + 3 = 1\end{align}$
Clarification for abs value:
Consider some numerical values. Let $x=1$, so $|x|=1=x$.
Again let $x=−1$. Now what is it's absolute value? It's again 1, right? $1=−(−1)$ or $|x|=−x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3994917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits