Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
proof of $\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$ without solving $\int\frac{1}{x^2+1}dx$ So I know $\int\frac{1}{x^2+1}dx=\arctan(x)+c$
Now when i tried to integral $\frac{1}{x^2+1}$ in a different way i got this:
$$\frac{1}{x^2+1}=\frac{1}{(x-i)(x+i)}=\frac{A}{x-i}+\frac{B}{x+i}\\\implies1=Ax+Ai+Bx-Bi=x(A+B)+i(A-B)\\\implies
\begin{cases}
A+B=0
\\[2ex]
A-B=\frac{1}{i}=-i
\end{cases}\\\implies2A=-i\implies A=\frac{-i}{2}\implies B=\frac{i}{2}$$
So in the end i get:$$\int\frac{1}{x^2+1}dx=\int\frac{i}{2(x+i)}-\frac{i}{2(x-i)}dx=\frac{i}{2}\int\frac{1}{(x+i)}-\frac{1}{(x-i)}dx$$
After integrating this i get $\frac{i}{2}(\ln(x+i)-\ln(x-i)+C)$ after comparing this to $\arctan(x)$ at $x=0$ i find $C=-i\pi$
$\therefore~\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$
now my question is how can i prove that without using the integral
| By definition of $\ln$ for complex numbers:
$$\ln(x+i)=\ln|x+i|+i\operatorname{Arg}(x+i)$$
where $\operatorname{Arg}$ is the principal value of the $\arg$
It is clear that:
$$\operatorname{Arg}(x+i)=\begin{cases}\arctan\frac{1}{x}&\text{ if }x\ge0\\\pi+\arctan\frac{1}{x}&\text{ if }x<0\end{cases}$$
Analogously:
$$\ln(x-i)=\ln|x-i|+i\operatorname{Arg}(x-i)$$
In this case it is:
$$\operatorname{Arg}(x-i)=\begin{cases}-\arctan\frac{1}{x}&\text{ if }x\ge0\\-\pi-\arctan\frac{1}{x}&\text{ if }x<0\end{cases}$$
Moreover observe that:
$|x+i|=|x-i|$
Putting everything together you get:
$\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)=\begin{cases}\stackrel{x\ge0}=\frac{i}{2}(2i\arctan\frac{1}{x}-i\pi)&=-\arctan\frac{1}{x}+\frac{\pi}{2}=\\\stackrel{x<0}=\frac{i}{2}(2i\pi+2i\arctan\frac{1}{x}-i\pi)&=-\arctan\frac{1}{x}-\frac{\pi}{2}=\end{cases}=\arctan x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Show $x=0$ and $x=1$ are the only integer solutions I'm trying to show that the only solutions to $x^2-x+1=y^2$ are when $x=0$ and $x=1$. All I can think of is completing the square gives $x^2-x+1=(x-\frac{1}{2})^2+\frac{3}{4}$, which is clearly not a perfect square. Does this suffice? Should I use induction?
| Multiply by the equation by $4$ and complete the square for $x$. We have
\begin{eqnarray*}
(2x-1)^2+3=(2y)^2 \\
(2y+2x-1)(2y-2x+1)=3
\end{eqnarray*}
So Either
\begin{eqnarray*}
\cases{2y+2x-1=3 \\
2y-2x+1=1} \text{or}
\cases{2y+2x-1=-3 \\
2y-2x+1=-1}
\end{eqnarray*}
or
\begin{eqnarray*}
\cases{2y+2x-1=1 \\
2y-2x+1=3} \text{or} \cases{2y+2x-1=-1 \\
2y-2x+1=-3}
\end{eqnarray*}
so $(x,y)$ has solutions $\color{blue}{(1, \pm 1)}$ and $\color{green}{(0, \pm 1)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Finding real zeros of polynomial of third degree to solve inequality \begin{align}{}
& \dfrac{x^3+2 x^2}{2} < x+2 \\[18pt]
& x^3 + 2x^2 < 2x+4 \\
& x^3 + 2x^2 -2x - 4 < 0 \\
\end{align}
Let $p(x) = x^3 + 2x^2 -2x - 4$. Setting $p(x) = 0$
*
*Cauchy's bound tells me the real zeros should be in the range of $[-5, 5]$
*Rational zeros theorem gives me candidates to perform polynomial division with: $\{\pm1, \pm2, \pm4 \}$
*Descarte's rule of signs predicts 1 positive and either 2 or no negative real zeros.
Now, performing polynomial division of the form $p(x) \div (x - c)$, where c is each of the candidate rational zeros above, I can find $x = -2$ as a real zero, leaving polynomial $q(x) = x^2 - 2$, which is a quadratic, yielding $x = \pm \sqrt{2}$
Solving the inequality, by testing values on each interval, gives $(-\infty, -2) \cup (-\sqrt{2}, \sqrt{2})$ as the answer.
| Let $p(x) = x^3 + 2x^2 -2x - 4$. Setting $p(x) = 0$
*
*Cauchy's bound tells me the real zeros should be in the range of $[-5, 5]$
*Rational zeros theorem gives me candidates to perform polynomial division with: $\{\pm1, \pm2, \pm4 \}$
*Descarte's rule of signs predicts 1 positive and either 2 or no negative real zeros.
Now, performing polynomial division of the form $p(x) \div (x - c)$, where c is each of the candidate rational zeros above, I can find $x = -2$ as a real zero, leaving polynomial $q(x) = x^2 - 2$, which is a quadratic, yielding $x = \pm \sqrt{2}$
Solving the inequality, by testing values on each interval, gives $(-\infty, -2) \cup (-\sqrt{2}, \sqrt{2})$ as the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the following equation $ \frac{1}{2^x} + \frac{1}{4^x} + \frac{1}{6^x} + \frac{1}{8^x} = 1 $ . How can I solve the following equation?
$$
\dfrac{1}{2^x} +
\dfrac{1}{4^x} +
\dfrac{1}{6^x} +
\dfrac{1}{8^x} =
1
$$
| Write the equation as $2^{-x} + 4^{-x} + 6^{-x} + 8^{-x} = 1$. Define the function $g(x) = 2^{-x} + 4^{-x} + 6^{-x} + 8^{-x}$. Then, $g$ is a differentiable function, with $g'(x) = -2^{-x}\ln 2 - 4^{-x}\ln 4 - 6^{-x}\ln 6 - 8^{-x}\ln 8$, which is always negative. Hence, $g$ is a decreasing continuous function, so it's a one-one map. That is, there exists a single root of the above equation.
Furthermore, since $g(1) = \frac{25}{24} > 1$, and $g(2) = \frac{205}{576} < 1$, we have by the intermediate value theorem, that the root is between $1$ and $2$. Also, from the closeness of $\frac{25}{24}$ to $1$, we would be inclined to think that such a root is very close to $1$. Indeed, as far as Wolfram Alpha is concerned, the root is $1.03411$, and from our analysis, we confirm this as being the unique root.
I do not think there are purely algebraic ways of solving this problem (our method involved calculus).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2414987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that the expression $5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}$ is divisible by 19.
Prove that the expression
$$5^{2n+1} * 2^{n+2} + 3^{n+2} * 2^{2n+1}$$
is divisible by $19$.
I'll skip the basis step (as I have done last time) but I can conclude that it's only divisible by 19 for integers n ≥ 0 (or whole numbers).
II.
Assume that
$$5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}$$
is divisible by 19. Then,
$$5^{2k+3} * 2^{k+3} + 3^{k+3} * 2^{2k+3}$$
is divisible by 19.
Now this is where I get lost, I try to "dismember" the expression to get
$$5^{2k}* 5^3 * 2^k * 2^3 + 3^k * 3^3 * 2^{2k} * 2^3$$
I also try to get it similar to to the assumption to make use of the said assumption yielding
$$5^{2k}* 5 * 5^2 * 2^k * 2^2 * 2 + 3^k * 3^2 * 3 * 2^{2k} * 2 * 2^2$$
$$5^{2k+1} * 5^2 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$5^{2k+1} * 25 * 2^{k+2} * 2 + 3^{k+2} * 3 * 2^{2k+1} * 2^2$$
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
And this is where I get lost.. : (
Am I missing out something? Had I done it wrong? The number 19 is prime which makes it hard to handle for me. Thanks!
EDIT : After some pondering, I answered it this way :
$$50 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1}$$
I realized that 50 can be written as 38 + 12 (and 38 is a multiple of 19) Hence,
$$ 38 + 12 * 5^{2k+1} * 2^{k+2} + 12 * 3^{k+2} * 2^{2k+1} $$
Factoring out 12, I get :
$$ 38 + 12(5^{2k+1} * 2^{k+2} + 3^{k+2} * 2^{2k+1}) $$
38 is divisible by 19 and the long expression is divisible by 19 (per the assumption) and qed.
Is this correct ?
| First let's rewrite the expression so as to better highlight innate arithmetical structure.
$\ \ \quad\qquad\begin{align}
19\ \mid &\,\ 5^{2n+1} \cdot 2^{n+2} + 3^{n+2} \cdot 2^{2n+1}\\[.2em]
\iff\, \bmod 19\!:\, &\ \ \ \ \color{#0a0}{20}\cdot 50^n\, \equiv\, - \color{#0a0}{18}\cdot 12^n
\end{align}$
Now the induction is extremely simple if we view it arithmetically $\!\bmod 19,\,$ where it amounts to multiplying the base congruence $\,\color{#0a0}{20}\equiv \color{#0a0}{-18}\,$ by the $\color{#c00}n$'th power of $\,50\equiv 12.\,$ In induction format
$\ \ \quad\qquad\qquad\qquad\begin{align}
\color{#0a0}{20}\, \ &\equiv\, \color{#0a0}{-18}\qquad\qquad\ {\rm i.e.}\ \ \ P(0)\\[.3em]
\color{#0a0}{20}\cdot 50^{\large\color{#c00} n} &\equiv\, 12^{\large\color{#c00} n}(\color{#0a0}{-18})\qquad\! {\rm i.e.}\ \ \ P(\color{#c00}n)\\
\times\,\qquad 50\ \ &\equiv\, 12\\[.2em]
\hline
\!\!\Longrightarrow\ \ 20\cdot 50^{\large\color{#c00}{n+1}} &\equiv\, 12^{\large\color{#c00}{n+1}}(-18)\ \ \ {\rm i.e.}\ \ \ P(\color{#c00}{n\!+\!1})^{\phantom{|^|}}\!\!\!\!\!
\end{align}$
The final congruence is the product the two prior congruences using the Congruence Product Rule. In number theory we make such deductions by using Congruence Product and Power Rules. The power rule encapsulates such arithmetical inductions for convenient reuse. The inductive proofs in the other answers are in fact special cases of the proof of the Power Rule (e.g. see here where I highlight this in great detail).
We can simplify further using $\,\color{#0a0}{20\equiv 1,\ 18\equiv -1},\ 50\equiv 12\pmod{\!19}\,$ to obtain
$\qquad\qquad\qquad\qquad\qquad\begin{align}&\color{#0a0}{20}\cdot 50^n + \color{#0a0}{18}\cdot 12^n\\[.2em]
\equiv\ &\ \ 1\cdot 12^n\ \color{#0a0}{-\ 1}\cdot 12^n\end{align}$
which effectively completely encapsulates the induction in the Congruence Power Rule.
Remark $ $ Even if you don't know congruences you can still take advantage of these arithmetical simplifications by using a version of the Product Rule in divisibility form as below
$$\qquad\qquad\begin{align} {\rm mod}\,\ m\!:\, A\equiv a,\, B\equiv b&\ \ \,\Longrightarrow\,\ \ AB\equiv ab\qquad\bf\text{Congruence Product Rule}\\[3pt]
m\mid A-a,\ B-b&\,\Rightarrow\, m\mid AB-ab\qquad\bf\,\text{Divisibility$\ $ Product Rule}\\[4pt]
{\bf Proof}\quad (A-a)B+a(B&-b)\, = AB-ab\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
Evaluate $\displaystyle{\lim_{n\to \infty}}(1-\frac 12 +\frac 13 - \frac 14 + \cdots + \frac{1}{2n-1}-\frac{1}{2n}) $ I want to calculate this limit.
$$\lim_{n\to \infty}(1-\frac 12 +\frac 13 - \frac 14 + \cdots + \frac{1}{2n-1}-\frac{1}{2n}) $$
I tried to pair the terms of the sum in order to reduce each other but without any succes. By writing the sum:
$$\frac{1}{1*2} + \frac{1}{3*4}+\frac{1}{5*6}+\cdots+\frac{1}{2n(2n-1)}$$
I have not reached anything useful. Could you help me?
| Solution 1:
Prove by induction that
$$1-\frac 12 +\frac 13 - \frac 14 + \cdots + \frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+...+\frac{1}{2n}$$
Then use the fact that $\frac{1}{n+1}+...+\frac{1}{2n}=\frac{1}{n}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}$ is a Riemann sum.
Solution 2
$$\begin{align}
u_n &=1-\frac 12 +\frac 13 - \frac 14 + \cdots + \frac{1}{2n-1}-\frac{1}{2n}\\&=1+\frac 12 +\frac 13 + \frac 14 + \cdots + \frac{1}{2n-1}+\frac{1}{2n}-2 (\frac{1}{2}+...+\frac{1}{2n})\\&=\left(1+\frac 12 +\frac 13 + \frac 14 + \cdots + \frac{1}{2n-1}+\frac{1}{2n} -\ln (2n)\right)\\&\qquad-\left(\frac{1}{1}+\frac{1}{2}+\dots+\frac{1}{n}-\ln(n)\right)+\ln(2)
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Partial fraction decomposition of $\frac{1}{x^4-x^2}$ I have been having a debate over whether, when you factor the denominator into $x^2$, $x-1$, and $x+1$, you need a fraction that says $\frac{A}{x}$ and one that has $\frac{B}{x^2}$ or if you only need the fraction with $x^2$ as the denominator. When I worked it out, I was only able to get the correct answer when I didn't do the $\frac{A}{x}$ fraction, but maybe I made such a roundabout mistake that I happened to get the right thing the wrong way. When do you need two fractions and when don't you?
| Another way:
$$\frac {1}{x^4-x^2}=\frac {1}{x^2(x^2-1)}=\frac {x^2-(x^2-1)}{x^2(x^2-1)}=\frac 1{x^2-1}-\frac 1{x^2}.$$
And $$\frac 1{x^2-1}=\frac 12 \frac 2{(x-1)(x+1)}=\frac 12 \frac {(x+1)-(x-1)}{(x-1)(x+1)}=\frac 1{2(x-1)}-\frac 1{2(x+1)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Given a positive number $b$, under what conditions does there exist a rectangle with perimeter $2b$ and area $\frac{b}{2}$? Given a positive number $b$, under what conditions does there exist a rectangle with perimeter $2b$ and area $\frac{b}{2}$?
$$2l + 2w = 2b$$ $$l+w=b$$
and $$lw = \frac{b}{2}$$
I tried to use AGM inequality
$$lw = \frac{l+w}{2}$$
For that to be a AGM inequality the right side had to be squared and the equality will only hold if $l = w$. I don't know how to get the conditions from here.
| The answer is that there is always a solution whenever $$b\geq 2.$$ One example solution is given by
$$w \equiv \frac{b + \sqrt{b^2-2b}}{2}$$
$$\ell \equiv \frac{b - \sqrt{b^2-2b}}{2}$$
In this case, you can confirm that the sum is $2b$ and the product is $\frac{b}{2}$, exactly as required. The condition $b\geq 2$ ensures that the discriminant is positive so that the solutions are real.
1. You're asking for which values of $b$ there exists a rectangle with perimeter $2b$ and area $\frac{b}{2}$.
*Let $\ell$ and $w$ be the sides of such a rectangle. Then you're asking for which values of $b$ does there exist a solution $\langle \ell, w\rangle$ to the following system of equations:
$$\ell w =\frac{b}{2} \qquad \ell+w = b.$$
*We can eliminate one of the variables using one of the equations, say $\ell = b - w$. Then the question becomes for which values of $b$ does there exist a solution $\langle w\rangle$ to the equation
$$\begin{align*}(b-w)w &= \frac{b}{2}\\ bw - w^2 &= \frac{b}{2}\\w^2-bw+\frac{b}{2} &= 0\end{align*}$$
*This formula is quadratic in $w$, and hence it always has two solutions. In order to be a physical rectangle though, we must have that $w>0$ and also $\ell > 0$ (which requires that $b-w > 0$, i.e. $w<b$.)
Putting this all together, we ask for what values of $b$ does this quadratic equation have a solution $w$ where $0<w<b$.
*The solution in general is given by the quadratic formula:
$$w = \frac{b\pm\sqrt{b^2-2b}}{2}$$
Hence our question is: for which values of $b$ does at least one of the following conditions hold:
$$0 < \frac{b+\sqrt{b^2-2b}}{2} < b\qquad\text{(positive root)}$$
$$0 < \frac{b-\sqrt{b^2-2b}}{2} < b\qquad\text{(negative root)}$$
*With some arithmetic, we can show that each condition is actually equivalent to:
$$-b < \sqrt{b^2-2b} < b$$
*We are given that $b>0$, so assuming the quantity under the square root is also positive ($b^2 > 2b$, i.e. $b>2$), the condition $-b < \sqrt{b^2-2b}$ is always satisfied, and we only need to worry about whether
$$\begin{align*}\sqrt{b^2-2b} &< b\\
b^2 - 2b &< b^2 \\
-2b & < 0\\
b &> 0\\
\end{align*}$$
*We originally asked under which conditions a rectangle with the desired physical parameters existed. We converted this into a quadratic formula $w^2-bw + \frac{b}{2} = 0$ and sought solutions $w$ which were physically possible, meaning that $0<w<b$.
The first condition we imposed was that the discriminant $b^2 -2b$ would be positive so that the solutions $w$ wouldn't be imaginary. This is equivalent to requiring $b>2$.
It turns out that under that restriction, the quadratic formula always has a positive root less than $b$. (To see this, note that $b^2-2b$ is always less than $b^2$, so $\sqrt{b^2-2b}$ is always less than $b$, so $b-\sqrt{b^2-2b}$ is always positive.)
*Hence the answer is that there is always a solution whenever $$b\geq 2.$$ One example solution is given by
$$w \equiv \frac{b + \sqrt{b^2-2b}}{2}$$
$$\ell \equiv \frac{b - \sqrt{b^2-2b}}{2}$$
You know that each of these quantities is positive because $\sqrt{b^2-2b} < \sqrt{b^2} = b$, and of course their sum is $(2b)/2 = b$ and their product is
$$\frac{1}{4}[b^2 - (b^2-2b)] = \frac{b}{2}$$
exactly as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2420895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
Show that $\lfloor(1+\sqrt{3})^{2n-1}\rfloor$ is divisible by $2^n$
Show that $\lfloor(1+\sqrt{3})^{2n-1}\rfloor$ is divisible by $2^n$ where $n$ is a positive integer.
We have \begin{align*}\lfloor(1+\sqrt{3})^{2n-1}\rfloor &= (\sqrt{3}+1)^{2n-1}-(\sqrt{3}-1)^{2n-1}\\&=\dfrac{(4+2\sqrt{3})^n}{1+\sqrt{3}}+\dfrac{(4-2\sqrt{3})^n}{1-\sqrt{3}}\\&=2^n\left(\dfrac{(2+\sqrt{3})^n}{1+\sqrt{3}}+\dfrac{(2-\sqrt{3})^n}{1-\sqrt{3}}\right).\end{align*} Thus we must show that the second term is an integer. How can we continue?
| Hint: $1 \pm \sqrt{3}$ are the roots of $t^2-2t-2=0\,$, so $\;a_{n} = (1+\sqrt{3})^{n}+(1-\sqrt{3})^{n}$ is the solution to the recurrence $a_{n+2}=2a_{n+1}+2a_{n}\,$ with $a_0 = a_1 = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Limit of a sequence : $x_n = \frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \ldots + \frac{1}{n(n+1)(n+2)}$ Can someone help me with this problem?
Finding the limit $\lim_{n \to \infty}\ x_n$ where
$$x_n = \frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \ldots + \frac{1}{n(n+1)(n+2)},\quad n\in\mathbb{N}.$$
I don't have a clue how to do this.
| Hint. On may observe that
$$
\frac{1}{n(n+1)(n+2)}=\frac12\frac{2}{n(n+1)(n+2)}=\frac12\frac{(n+2)-n}{n(n+1)(n+2)}
$$ giving
$$
\frac{1}{n(n+1)(n+2)}=\frac{1}{2n(n+1)}-\frac{1}{2(n+1)(n+2)}
$$ the given sum is thus a telescoping one.
Hope you can finish it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$\sum_{n=1}^{2017}\left(\left((n+2)^4\bmod{(n+1)^4}\right)\bmod{4}\right)$
What's
$$\sum_{n=1}^{2017}\left(\left((n+2)^4\bmod{(n+1)^4}\right)\bmod{4}\right)$$
What have I tried?
$$(n+2)^4=n^4+8n^3+24n^2+32n+16$$
$$(n+1)^4=n^4+4n^3+6n^2+4n+1$$
Remainder:
$$4n^3+18n^2+28n+15$$
mod:
$$2n^2-1\pmod{4}$$
I can compute $\sum x^2$ but I don't know what to do with $$\sum_{n=1}^{2017}\left(2n^2-1\mod{4}\right)$$
| Hint: Compute the sum of even and odd $n$ separately. That is, $$\sum_{n=1}^{2b+1}{(2n^2-1\mod{4})} = \sum_{k=1}^{b}{(2(2k)^2-1\mod{4})} + \sum_{k=0}^{b}{(2(2k+1)^2-1\mod{4})}$$
Just a word of caution: If $a \mod{4} \in \{0,1,2,3\}$ then it is useful to use $2n^2+3$ instead of $2n^2-1$, especially when $n$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+...\to ?\;\;$ (Click here.) $$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+\dots\to~?$$
I tried like below
$$\frac{1}{1^2}+\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2+3^2}+\dots=\\\sum_{n=1}^{\infty}\frac{1}{1^2+2^2+3^2+\dots+n^2}=\\\sum_{n=1}^{\infty}\frac{1}{\frac{n(n+1)(2n+1)}{6}}=\\\sum_{n=1}^{\infty}\frac{6}{n(n+1)(2n+1)}=\\
$$
Then I can use the fraction ,but $$\frac{1}{n(n+1)(2n+1)}=\frac{1}{n}+\frac{1}{n+1}+\frac{-4}{2n+1}$$ This is ugly to turn into telescopic series .
Can you help me to find :series converge to ?
Thanks in advance .
| When I post it , I find an idea ...
$$\sum_{n=1}^{\infty}\frac{6}{n(n+1)(2n+1)}=\\
\sum_{n=1}^{\infty}\frac{6\times 2 \times 2}{(2n)(2n+2)(2n+1)}=\\
\sum_{n=1}^{\infty}\frac{24}{(2n)(2n+2)(2n+1)}=\\
\sum_{n=1}^{\infty}\frac{24}{1}(\frac{1}{(2n)(2n+1)(2n+2)})=\\
\sum_{n=1}^{\infty}\frac{24}{2}(\frac{1}{(2n)(2n+1)}-\frac{1}{(2n+1)(2n+2)})=\\
12\times \frac{1}{2\times 3}=2$$ but wolfram says $$\sum_{n=1}^{\infty}\frac{6}{n(n+1)(2n+1)}=6(3 - 4log(2))$$ Is there something I had missed ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
How many positive integral solutions for $\frac{xy}{x+y}= 2^4\cdot 3^5 \cdot 5^{4}$ Consider the equation:
$$\frac{xy}{x+y}= 2^4\cdot 3^5 \cdot 5^{4}$$
How many positive integral solutions are possible?
| We can rearrange to give $x = \dfrac{yT}{y-T}$, where $T:=2^4\cdot 3^5\cdot 5^4$
To have positive solutions then, we clearly need $y > T$ and by symmetry $x>T$ also.
Define $v:=x-T, w:=y-T$, giving $v = \dfrac{T^2+Tw}{w} -T \implies vw=T^2$
Then the number of solutions of $vw=T^2 = 2^8\cdot 3^{10}\cdot 5^{8}$ is simply $9\times 11\times 9 = 891$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $n$ is a positive integer, then $3^{2^{n}}-1$ is divisible by $2^{n+2}$ Prove that if $n$ is a positive integer, then $3^{2^{n}}-1$ is divisible by $2^{n+2}$
I tried to prove this using induction so for the base case $n=1$ you get that $8 \mid 8$
For the inductive step assume $P(k): 2^{k+2}\mid 3^{2^{k}}-1$ is true for some $k \geq 1$ I want to show $P(k+1)$ is true then $2^{k+2+1}=( 3^{2^{k}}-1) * 2$ I want to somehow get to $3^{2^{k+1}}-1$ I just got stuck and I'm unsure if this is the right approach.
| For some integer $k$: $\ 3^{2^n}-1=2^{n+2}k$ and from this you want to derive, for some integer $m$: $\ 3^{2^{n+1}}-1=2^{n+3}m$.
$$3^{2^n}3^{2^n}-1=2^{n+3}m$$
$$(2^{n+2}k+1)(2^{n+2}k+1)-1=2^{n+3}m\Longrightarrow$$
$$2^{2n+4}k^2+2^{n+3}k+1-1=2^{n+3}m\Longrightarrow$$
$$2^{n+1}k^2+k=m$$
Because $k$ is an integer $m$ is an integer.
As a test, if $n=1$ then $k=\frac88=1$, so when $n=2$ then $m=2^2\cdot 1+1=5=\frac{80}{16}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding polynomial with root $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$, what is the degree of a root? a) In $\mathbb{R}$, $\sqrt{2}$ and $\sqrt{3}$ are algebric over $\mathbb{Q}$. Find the polynomial of degree $4$ over $\mathbb{Q}$ satisfiable by $\sqrt{2}+\sqrt{3}$
b) Wich is the degree of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$?
c) With is the degree of $\sqrt{2}\cdot \sqrt{3}$ over $\mathbb{Q}$?
I've found (a) here: Find the minimal polynomial of $\sqrt2 + \sqrt3 $ over $\mathbb Q$
For $b$, what is the degree of an element over $\mathbb{Q}$ and how to justify it?
| It is: $a=\sqrt{2}+\sqrt{2}$, then $a^2=(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}$. Hence $a^2-5=2\sqrt{6}$ and $(a^2-5)^2=24$, so
$a^4-10a^2+1=0$ and the polynomial is given by $X^4-10X^2+1\in\mathbb{Q}[X]$, which is irreducible.
Edit: So the degree of $\sqrt{2}+\sqrt{3}$ is $4$. It is the degree of the minimal polyomial.
c) $\sqrt{2}\cdot\sqrt{3}=\sqrt{6}$ and it is $X^2-6$ the minimal polynomial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2424632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$ Evaluate $$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx})$$
I tried the following:
$$\lim_{x\rightarrow \infty} (\sqrt{x^2 +ax} - \sqrt{x^2 +bx}) \cdot \frac{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
But ended up with
$$\lim_{x\rightarrow \infty} \frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}$$
Which I'm not sure what to do with.
| We may suppose $x>0$, so
\begin{align}\sqrt{x^2+ax}+\sqrt{x^2+bx}&=x\biggl(\sqrt{1+\frac ax}+\sqrt{1+\frac bx} \,\biggr)=x\biggl(1+\frac a{2x}+1+\frac b{2x} +o\Bigl(\frac1x\Bigr) \biggr)\\
&=2x+\frac{a+b}2+o(1),\end{align}
so that the denominator is equivalent at $\infty$ to $2x$, and
$$\sqrt{x^2 +ax} - \sqrt{x^2 +bx}=\frac{ax-bx}{\sqrt{x^2 +ax} + \sqrt{x^2 +bx}}\sim_\infty\frac{(a-b)x}{2x}=\frac{a-b}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Inequality : $\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)<\sum_{cyc}\frac{a^3+b^3}{c^2+ab}$
Let $\{a, b, c\} \subset \mathbb{R}^+$. Prove that
$$\displaystyle\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)<\displaystyle\sum_{cyc}\frac{a^3+b^3}{c^2+ab}$$
My work :
WLOG, let $\;c\geq b \geq a$, we have
$$\frac{b^3+c^3}{a^2+bc}\geq \frac{b^3+c^3}{b^2+bc}$$
Since $(b-c)(b-a) \leq 0$, we have
$$\frac{c^3+a^3}{b^2+ac}\geq \frac{c^3+a^3}{bc+ba}. $$
$\begin{eqnarray}
\sum_{cyc}\frac{a^3+b^3}{c^2+ab}&\geq &\frac{a^3+b^3}{c^2+ab} + \frac{b^3+c^3}{b^2+bc} + \frac{c^3+a^3}{bc+ba} \\
&\geq& \frac{a^3+b^3}{c^2+ab} + \frac{b^2-bc+c^2}{b} + \frac{c^2-ca+a^2}{c}\\
&\geq &\frac{a^3+b^3}{c^2+ab} + b-a+\frac{c^2}{b} + \frac{a^2}{c}=S
\end{eqnarray}$
Case 1 : $b-a \geq c-b$
we have $ \frac{b}{a} \geq \frac{c}{b}$, so $b^2\geq ac, \;2b \geq a+c$
so $$ S \geq \frac{a^3+b^3}{c^2+ab} +c-b+\frac{c^2}{b} + \frac{a^2}{c}$$
$$\geq \left(\frac{c^2}{b}+c\right)+\left(\frac{a^3+b^3}{c^2+ab}-b+ \frac{a^2}{c}\right)$$
Since $\frac{2}{9}\left(\frac{bc}{a}\right)=\frac{2}{9}\left(\frac{c^2a+b^2c-c^2a}{ab}\right)=\frac{2}{9}\left(\frac{c^2}{b}\right)+\frac{2}{9}\left(\frac{c(b^2-ca)}{ab}\right)\geq \frac{2}{9}\left(\frac{c^2}{b}\right)$
As $\frac{4}{9}a + \frac{4}{9}b + \frac{4}{9}c < \frac{8}{9}b + \frac{4}{9}\frac{c^2}{b}$, so $\frac{2}{9}\frac{ab}{c} < \frac{2}{9}\frac{abc}{c^2}< \frac{2}{9}c$
Similarly, $\frac{2}{9}\frac{ac}{b} < \frac{2}{9}c$
Thus, $\frac{4}{9}(a+b+c)+\frac{2}{9}\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right) < S $
Case 2 : $b-a \leq c-b$
we have $b^2\leq ac, \;2b \leq a+c$
so $ S \geq \left(\frac{c^2}{b}+c\right)+\left(\frac{a^3+b^3}{c^2+ab}-b+ \frac{a^2}{c}\right)\geq \left(\frac{2c^2}{a+c}+c\right)+\left(\frac{a^3+b^3}{c^2+ab}-b+ \frac{a^2}{c}\right)$
Since $b^2 \leq ac$, so $2ca \geq b(a+c)$
$\leftrightarrow c(2ca) \geq c(ab)+c(bc)$
$\leftrightarrow \frac{2c^2}{a+c} \geq \frac{bc}{a}$
Thus, $\frac{2c^2}{a+c}+c \geq \frac{bc}{a}+c > \frac{4}{9}(a+b+c) +\frac{2}{9}\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)$
By C-S,
$\frac{1}{2a+b}\leq \frac{2}{a}+ \frac{1}{b}$
Similarly, $\frac{1}{2b+c}\leq \frac{2}{b}+ \frac{1}{c}$, so
$\displaystyle\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)\leq \frac{4}{9}(a+b+c) +\frac{2}{9}\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)$
therefore,
$\displaystyle\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)<\displaystyle\sum_{cyc}\frac{a^3+b^3}{c^2+ab}$ $\blacksquare$
*
*Please help me check if there is any mistake in my work. Thank you.
*If you have different idea, please provide.
| By C-S
$$\sum_{cyc}ab\left(\frac{1}{2a+c}+\frac{1}{2b+c}\right)=\sum_{cyc}ab\left(\frac{1}{a+a+c}+\frac{1}{b+b+c}\right)\leq$$
$$\leq\sum_{cyc}\frac{ab}{9}\left(\frac{1^2}{a}+\frac{2^2}{a+c}+\frac{1^2}{b}+\frac{2^2}{b+c}\right)=$$
$$=\frac{2}{9}(a+b+c)+\frac{4}{9}\sum_{cyc}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=$$
$$=\frac{2}{9}(a+b+c)+\frac{4}{9}\sum_{cyc}\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)=\frac{2}{9}(a+b+c)+\frac{4}{9}(a+b+c)=\frac{2}{3}(a+b+c).$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{a^3+b^3}{c^2+ab}\geq\frac{2}{3}(a+b+c).$$
We'll prove a stronger inequality:
$$\sum_{cyc}\frac{a^3+b^3}{c^2+ab}\geq a+b+c.$$
Indeed, by C-S
$$\sum_{cyc}\frac{a^3+b^3}{c^2+ab}=\sum_{cyc}\frac{a^4}{c^2a+a^2b}+\sum_{cyc}\frac{b^4}{c^2b+b^2a}\geq$$
$$\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(c^2a+a^2b)}+\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(c^2b+b^2a)}=$$
$$=\frac{(a^2+b^2+c^2)^2}{2}\left(\frac{1}{a^2b+b^2c+c^2a}+\frac{1}{a^2c+b^2a+c^2b}\right)\geq$$
$$\geq\frac{(a^2+b^2+c^2)^2}{2}\cdot\frac{4}{\sum\limits_{cyc}(a^2b+a^2c)}=\frac{2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b+a^2c)}.$$
Thus, it remains to prove that
$$2(a^2+b^2+c^2)^2\geq(a+b+c)\sum_{cyc}(a^2b+a^2c)$$ or
$$\sum_{cyc}(2a^4-a^3b-a^3c+2a^2b^2-2a^2bc)\geq0,$$
which is true by Muirhead.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving equations which includes factorials I tried solve this question:
If $x$ things can be arranged in $m$ ways, $x-2$ things can be arranged in $n$ ways and $x-6$ things can be arranged in $p$ ways and $m = 30np$, then find $x$.
$m = 30np$
$\therefore x! = 30(x-2)!(x-6)!$
$\therefore x(x-1)(x-2)! = 30(x-2)!(x-6)!$
$\therefore x(x-1) = 30(x-6)!$
In the last equation, if you use try and error method, then you will find $6$ is answer for $x$...
But is there a way to find $x$ mathematically?
| $x(x-1) = 30(x-6)!$ Suppose $x > 6$
$\frac {x(x+1)}{x-6} = \frac {x^2 - 6x + 7x-42 + 42}{x-6} = x + 7 + \frac {42}{x-6} = 30(x-7)!$ Suppose $x > 7$.
$\frac {x(x+1)}{(x-6)(x-7)} = 1 + \frac {14}{x-7} + \frac{42}{(x-6)(x-7)} = 30(x-8)!$
But $1 + \frac {14}{x-7} + \frac{42}{(x-6)(x-7)}\le 1 + 14 + 21 = 36$ so $(x-8)! = 1$.
Or in other words $x \le 9$.
But $x(x+1) = 30(x-6)!$ so $5|x(x+1)$ so $5|x$ or $5|x-1$ so $x = 6$ or $x=5$. $x$ can't be $5$.
or
Basically $x(x-1) \approx x^2$ and $(x-6)! \approx (\frac x2)^{x-6}$ should indicate $x-6 \approx 2$. Knowing that $30|x(x-1)$ pinpoints it to $x = 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $2x^{2n-1}(1+x^{2n})^{\frac{1-n}{n}}$ I have to calculate the limit
$\lim_{n\to +\infty} 2x^{2n-1}(1+x^{2n})^{\frac{1-n}{n}}$
and to find out if the sequence of function converges uniformly.
I foud that the limit is $0$ if $|x|<1$ and $\pm\infty$ if $|x|>1$, but I'm not sure if I'm right.
Then I have to study if $f_n (x) = 2x^{2n-1}(1+x^{2n})^{\frac{1-n}{n}}$ converges uniformly. I tried with some estimations, but I can't find a $sup$ that converges if $n\to + \infty$
Thank you in advice
| Since the functions are odd, we can first assume $x \geq 0$.
$$\frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}} = \frac{2x^{2n-1}}{(1+x^{2n})^{\left(1-\frac{1}{n}\right)}} = \frac{2x^{2n}\frac{(1+x^{2n})^\frac{1}{n}}{x}}{1+x^{2n}} = \frac{2x^{2n}}{1+x^{2n}}\left(x^n + \frac{1}{x^n}\right)^\frac{1}{n}$$
If $x > 1$ then $\frac{1}{x^n} \le x^n$:
$$x = \sqrt[n]{x^n} \le \left(x^n + \frac{1}{x^n}\right)^\frac{1}{n} \leq (2x^n)^\frac{1}{n} = \sqrt[n]{2}\cdot x\xrightarrow{n\to\infty} x$$
$$\frac{2x^{2n}}{1+x^{2n}} = \frac{2}{\frac{1}{x^{2n}} + 1}\xrightarrow{n\to\infty} 2$$
So:
$$\lim_{n\to\infty} \frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}} = 2x$$
If $x = 1$:
$$\lim_{n\to\infty} \frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}} = 1$$
If $x < 1$, we have $\lim_{n\to\infty} 2x^{2n-1} = 0$ and $\lim_{n\to\infty} (1+x^{2n})^\frac{n-1}{n} = 1$. Thus:
$$\lim_{n\to\infty} \frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}} = 0$$
We can conclude:
$$\lim_{n\to\infty} \frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}} = \begin{cases}
0, & \text{if $x < 1$} \\
1, & \text{if $x = 1$} \\
2x, & \text{if $x > 1$} \\
\end{cases}$$
Expanding the domain to $\mathbb{R}$, we get:
$$\lim_{n\to\infty} \frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}} = \begin{cases}
2x, & \text{if $x < -1$} \\
-1, & \text{if $x = -1$} \\
0, & \text{if $|x| < 1$} \\
1, & \text{if $x = 1$} \\
2x, & \text{if $x > 1$} \\
\end{cases}$$
Functions $x \mapsto \frac{2x^{2n-1}}{(1+x^{2n})^\frac{n-1}{n}}$ are continuous for every $n \in \mathbb{N}$. Since the limit function is not continuous, the convergence cannot be uniform.
Edit:
Let's check if the convergence is uniform in $[0,a]$ for $a < 1$:
Let $f_n(x) = 2x^{2n-1}(1+x^{2n})^{\frac{1-n}{n}}$. The derivative is:
$$f_n'(x) = -4(n-1)x^{4n-2}(1+x^{2n})^{\left(-1+\frac{1-n}{n}\right)} + 2(2n-1)x^{2n-2}(1+x^{2n})^\frac{1-n}{n}$$
Rearranging $f'_n(x) = 0$ gives:
$$\frac{x^{2n}}{1+x^{2n}} = \frac{1}{2}\frac{2n-1}{n-1}\implies x^{2n}=1-2n < 0$$
Thus, $f_n$ has no stationary points. This implies that the supremum on $[0,a]$ is $|f(a)| = 2a^{2n-1}(1+a^{2n})^{\frac{1-n}{n}}$.
$$\|f_n - 0\|_{\infty,[0,a]} = \|f_n\|_{\infty,[0,a]} = 2a^{2n-1}(1+a^{2n})^{\frac{1-n}{n}} \xrightarrow{n\to\infty} 0$$
Thus, $f_n \xrightarrow{n\to\infty} 0$ uniformly on $[0,a]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I prove this inequality: $\frac{a^2}{3}+b^2+c^2>ab+bc+ca$? If $abc=1$ and $a^3>36$ prove that,
$\frac{a^2}{3}+b^2+c^2>ab+bc+ca$
I tried to use the general proof method.
$\frac{a^2}{3}+b^2+c^2-ab-bc-ca>0$
Symbolically notation:
$\frac{a^2}{3}+b^2+c^2-ab-bc-ca>0 \Rightarrow(x+y+z+...)^2>0$
But, after trying too much, I accepted the defeat.
| As an old (not too old) Olympiad participant, I love the questions of Proof. My reputation is very low right now.But, I think the proof is correct.
\begin{align}\frac{a^2}{3}+b^2+c^2-ab-bc-ca&=\frac{a^2}{4}+b^2+c^2-ab-bc-ca+\frac{a^2}{12}\\
&=\left({\frac a2-b-c}\right)^2+\frac{a^2}{12}-3bc\\
&=\left({\frac a2-b-c}\right)^2+\frac{a^2}{12}-\frac 3a\\
&=\left({\frac a2-b-c}\right)^2+\frac{a^3-36}{12a}\\
&>0
\end{align}
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Complete factorization of $64y^6 - 1$ If we use the difference of two cubes identity, then we have
$$64y^6 - 1 = (4y^2-1)(16y^4+4y^2+1) = (2y+1)(2y-1)(16y^4+4y^2+1)$$ which is the suggested solution of the textbook.
But if we use the difference of two squares identity, we then have
$$64y^6 - 1 = (8y^3+1)(8y^3-1) = (2y+1)(2y-1)(4y^2-2y+1)(4y^2+2y+1).$$
I then came up with the question: what is complete factorization? Of course we can do this trick:
$$16y^4+4y^2+1 = 16y^4+8y^2+1 - 4y^2 = (4y^2+1 - 2y)(4y^2+1+2y).$$
How do we know when to stop factorizing?
| The answer depends on the base ring/field you consider:
The complete factorisation over $\mathbf Z, \mathbf Q$ or $\mathbf R$ is indeed
$$(2x-1)(2x+1)(4x^2+2x+1)(4x^2-2x+1)$$
But on $\mathbf C$, you obtain a product of linear factors involving the sixth roots of unity:
\begin{align}
64x^6-1&=\prod_{k=1}^6\Bigl(2x-\mathrm e^{\tfrac{2ik\pi}{6}}\Bigr)\\
&=\underbrace{\Bigl(2x-\mathrm e^{\tfrac{i\pi}{3}}\Bigr)\Bigl(2x-\mathrm e^{\tfrac{5i\pi}{3}}\Bigr)}_{\textstyle4x^2+2x+1}\,\underbrace{\Bigl(2x-\mathrm e^{\tfrac{2i\pi}{3}}\Bigr)\Bigl(2x-\mathrm e^{\tfrac{4i\pi}{3}}\Bigr)}_{\textstyle4x^2-2x+1}\,\Bigl(\underbrace{2x-\mathrm e^{\tfrac{3i\pi}{3}}\rule[-1.5ex]{0pt}{1.5ex}}_{\textstyle2x+1}\Bigr)\,\Bigl(\underbrace{2x-\mathrm e^{\tfrac{6i\pi}{3}}\rule[-1.5ex]{0pt}{1.5ex}}_{\textstyle2x-1}\Bigr)
\end{align}\,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Show $r-s\sqrt{2}$ is a root if $r+s\sqrt{2}$ is
Given that $x = r+s\sqrt{2}$ is a solution to $x^2+ax+b=0$ for some $a,b,r,s\in \mathbb{Q}$ ($s\ne 0$), show that $x = r - s\sqrt{2}$ is also a solution.
If we let $f(x) = x^2+ax+b$, then:
$$f(r+s\sqrt{2})=r^2+(2s\sqrt{2}+a)r+(2s^2+as\sqrt{2}+b)=0 $$
We know the above must be true.
Then, substituting $x = r-s\sqrt{2}$ (or setting $s = -s$ in the above equation) gives:
$$f(r-s\sqrt{2})=r^2+(a-2s\sqrt{2})r+(2s^2-as\sqrt{2}+b) $$
$$ = f(r+s\sqrt{2})-2s\sqrt{2}(2r+a)$$
$$ =-2s\sqrt{2}(2r+a)$$
How can I show this is equal to $0$? I feel like I'm missing something very obvious, or perhaps have made a mistake in my substitutions somewhere.
Additionally, the result seems similar to the theorem in complex numbers (which I've been taught without proof) that $f(z) = 0\iff f(\bar{z}) = 0$. Here we're dealing with $\sqrt{2}\notin \mathbb{Q}$ rather than $i \notin \mathbb{R}$, so are there any simple proofs of the complex conjugate theorem that could be repurposed here?
| Putting $x = r+s\sqrt{n}$
where
$\sqrt{n}$ is irrational
(using $n$ for $2$
in the hope of generalizing)
into
$x^2+ax+b = 0$
gives
$\begin{array}\\
0
&=(r+s\sqrt{n})^2+a(r+s\sqrt{n})+b\\
&=r^2+2rs\sqrt{n}+s^2n+ar+as\sqrt{n}+b\\
&=r^2+s^2n+ar+b+s(2r+a)\sqrt{n}\\
\end{array}
$
Since
$\sqrt{n}$
is irrational,
we must have both
$r^2+s^2n+ar+b = 0$
and
$s(2r+a) = 0$.
Replicating the calculation above
but with $-s$ for $s$,
$(r-s\sqrt{n})^2+a(r-s\sqrt{n})+b
=r^2+s^2n+ar+b-s(2r+a)\sqrt{n}
=0
$
so
$r-s\sqrt{n}$
is also a root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2434958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $a,b,c$ are positive reals, then prove that $(a+b)(a+c)\ge 2 \sqrt {abc(a+b+c)}$
If $a,b,c$ are positive reals, then prove that $(a+b)(a+c)\ge 2 \sqrt {abc(a+b+c)}$
My tries:
By applying AM-GM inequality,
$$a+b\ge 2\sqrt
{ab}$$
and,
$$a+c\ge2\sqrt{ac}$$
and clearly LHS $=a^2+ac+ba+bc$, which is $\ge4a\sqrt{bc}$
Similarly, $(b+a)(b+c)\ge4b\sqrt{ac}$ and $(c+a)(c+b)\ge4c\sqrt{ab}$
What next?(I gave the Cauchy Schwarz tag because I do not know if this can be solved by Cauchy Schwarz inequality.)
| You have $$a(a+b+c) + bc \geq 2\sqrt{abc(a+b+c)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Asymptotic behaviour of $\frac{p_n-2}{p_n}$ I am interested in how fast this prime product approaches to 0:
$$
\prod_{n=2}^{N}\frac{p_n - 2}{p_n}
$$
Where $p_n$ is the nth prime number.
Numerical computation suggests that for $N > 1312$, $k = 0.5$
we have the following upper bound:
$$
\prod_{n=2}^{N}\frac{p_n - 2}{p_n} > \frac{1}{p_N^{k}}
$$
Can anyone prove this?
| We know that $n \ln n < p_n $ for all $n\geq 1$ and you are looking for lower bound so $\prod \limits_{n=2}^{N} (1-\frac{2}{p_n}) = \frac{1}{5} \prod \limits_{n=4}^{N} (1-\frac{2}{p_n}) \geq 0.2 \prod \limits_{n=4}^{N} (1-\frac{2}{n \ln n}) = 0.2 e^{\sum \limits_{n=4}^{N} \ln(1-\frac{2}{n \ln n})}$ and since $\ln(1-\frac{2}{t}) > -\frac{3}{t}$ for all $t>3.44$ and that $n \ln n > 3.44$ for all $n\geq 4$.
So we are left with $0.2 e^{-3 \sum \limits_{n=4}^{N} \frac{1}{n \ln n}}$
$\frac{1}{n \ln n}$ is monotonic deceasing function so we can bound the summation by integral, and we get that $0.2 e^{-3 \sum \limits_{n=4}^{N} \frac{1}{n \ln n}} \geq 0.2 e^{-3 \int \limits_{4}^{N} \frac{1}{n \ln n}dn}$
and $0.2 e^{-3 \int \limits_{4}^{N} \frac{1}{n \ln n}dn} = 0.2 e^{-3(\ln \ln N-\ln \ln 4)}$ since $\int \frac{1}{n \ln n} dn = \ln \ln n+C$
Now we are left with $0.2 e^{-3(\ln \ln N-\ln \ln 4)} = \frac{\ln^3(4)}{5 \ln^3(N)} \approx \frac{0.53284}{\ln ^3 N} > \frac{1}{p_N^{0.5}}$ for all $N>10^7$
Actually a very good approach using Dusart result about the sum $ \ln \ln x +B -\frac{1}{2 \ln^2 x}<\sum \limits_{p \leq x} \frac{1}{p} < \ln \ln x +B +\frac{1}{2 \ln^2 x}$ for $x>300$ where $B \approx 0.2614972128476427$ is Meissel–Mertens constant, one can bound your product by :
$\frac{0.814886}{\ln^2 p_N} <\prod \limits_{n=2}^{N} (1-\frac{2}{p_n}) < \frac{0.843639}{\ln^2 p_N}$
And the exact value is $c = \lim \limits_{N \to \infty} \ln^2 p_N \prod \limits_{n=2}^{N} (1-\frac{2}{p_n}) \approx 0.832417$
in this case $\prod \limits_{n=2}^{N} (1-\frac{2}{p_n}) > \frac{0.814886}{\ln^2 p_N} > \frac{1}{p_N^{0.5}}$ for all $N>1388$ , by computer checking for all $1388 \geq N \geq 1312$ one conclude that its true for all $N>1312$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2435929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to prove that this limit doesn't exits? The limit is $$ \lim_{(x,y)\to(0,0)} \frac{x\sin(y)-y\sin(x)}{x^2 + y^2}$$
My calculations: I substitute $y=mx$
\begin{align}\lim_{x\to 0} \frac{x\sin(mx)-mx\sin(x)}{x^2 + (mx)^2} &= \lim_{x\to 0} \frac{x(\sin(mx)-m\sin(x)}{x^2(1 + m^2)}\\ &= \lim_{x\to 0} \frac{1}{1+m^2}\bigg[\frac{\sin(mx)}{x}- \frac{m\sin(x)}{x}\bigg]\end{align}
Can I say that the limit $$ \lim_{x\to 0}\frac{\sin(mx)}{x}$$
doesn't exist because it depends on $m$, so the entire limit doesn't exist?
| I would just
throw in the
first terms of
the power series
and see what happens.
Since
$\sin(x)
= x-x^3/6+O(x^5)
$,
\begin{align}
\frac{x\sin(y)-y\sin(x)}{x^2 + y^2}
&=\frac{x(y-y^3/6+O(y^5))-y(x-x^3/6+O(x^5))}{x^2 + y^2}\\
&=\frac{xy-xy^3/6+O(xy^5)-yx+yx^3/6+O(x^5y))}{x^2 + y^2}\\
&=\frac{-(xy^3+yx^3)/6+O(xy^5)+O(x^5y))}{x^2 + y^2}\\
&=-\frac{xy(y^2+x^2)/6+O(xy^5)+O(x^5y))}{x^2 + y^2}\\
&=-\frac{xy}{6}+\frac{xy(O(y^4)+O(x^4))}{x^2 + y^2}
\end{align}
so
\begin{align}
\left|\frac{x\sin(y)-y\sin(x)}{x^2 + y^2}+\frac{xy}{6}\right|
&=\left|\frac{xy(O(y^4)+O(x^4))}{x^2 + y^2}\right|\\
&\le O(y^4)+O(x^4)\\
\end{align}
since
$0 \le (|x|-|y|)^2
=x^2+y^2-2|xy|$
so
$\frac{|xy|}{x^2+y^2}
\le \frac12$.
Therefore
the limit is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find $x-\frac{2x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}-\frac{2x^9}{9}+\cdots$ Find
$$S=x-\frac{2x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}-\frac{2x^9}{9}+\frac{x^{11}}{11}+\cdots=\sum_{n=0}^\infty\frac{x^{6n+1}}{6n+1}-\frac{2x^{6n+3}}{6n+3}+\frac{x^{6n+5}}{6n+5}$$
I could not find Regular pattern but I tried to differentiate $S$ as:
$$\frac{dS}{dx}=1-2x^2+x^4+x^6-2x^8+x^{10}+\cdots$$
any clue here?
| $$M=\frac x1+\frac{x^3}3+\frac{x^5}5+\frac{x^7}7+\frac{x^9}9+\frac{x^{11}}{11}+\cdots$$
is the Maclaurin series of $\tanh^{-1}x$ for $|x|<1$. The difference between this and $S$ is
$$M-S=\frac{3x^3}3+\frac{3x^9}9+\frac{3x^{15}}{15}+\dots$$
$$=\frac{(x^3)^1}1+\frac{(x^3)^3}3+\frac{(x^3)^5}5+\dots=\tanh^{-1}x^3$$
Thus
$$S=\tanh^{-1}x-\tanh^{-1}x^3,\ |x|<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2436975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that the perpendicular from the origin upon the straight line Prove that the perpendicular drawn from the origin upon the straight line joining the points $(c\cos \alpha, c\sin \alpha)$ and $(c\cos \beta, c\sin \beta)$ bisects the distance between them.
My Attempt:
Equation of the line joining the points $(c\cos \alpha, c\sin \alpha)$ and $(c\cos \beta, c\sin \beta)$ is:
$$y-c\sin \beta=\dfrac {c\sin \alpha - c\sin \beta}{c\cos \alpha- c\cos \beta} (x-c\cos \beta)$$
$$y-c\sin \beta =\dfrac {\sin \alpha - \sin \beta}{\cos \alpha - \cos \beta} (x-c\cos \beta)$$
$$x(\sin \alpha - \sin \beta)-y(\cos \alpha - \cos \beta)=c \sin \alpha. \cos \beta - c \cos \alpha. \sin \beta$$
$$x(\sin \alpha - \sin \beta) - y (\cos \alpha - \cos \beta)= c\sin (\alpha - \beta)$$
| Hint:
Using Prosthaphaeresis Formulas
The gradient of $$A(c\cos\alpha,c\sin\alpha);B(c\cos\beta,c\sin\beta)$$
is $$\dfrac{\sin\alpha-\sin\beta}{\cos\alpha-\cos\beta}=-\cot\dfrac{\alpha+\beta}2$$ assuming $\sin\dfrac{\alpha-\beta}2\ne0$ as for $\sin\dfrac{\alpha-\beta}2=0,\alpha\equiv\beta\pmod{2\pi}\implies A,B$ coincide.
The midpoint$(M)$ of $$(c\cos\alpha,c\sin\alpha);(c\cos\beta,c\sin\beta)$$
is $$\left(\dfrac{c(\cos\alpha+\cos\beta)}2,\dfrac{c(\sin\alpha+\sin\beta)}2\right)$$
So, the gradient of $O(0,0);M$ will be $$\dfrac{\dfrac{c(\sin\alpha+\sin\beta)}2-0}{\dfrac{c(\cos\alpha+\cos\beta)}2-0}=\tan\dfrac{\alpha+\beta}2$$
assuming $\cos\dfrac{\alpha-\beta}2\ne0$ as for $\cos\dfrac{\alpha-\beta}2=0,\alpha\equiv\pi+\beta\pmod{2\pi}\implies A,B$ becomes extremities of a diameter .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2437182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Prove this inequality $3(abc+4)\ge 5(ab+bc+ca)$
Let $\{a,b,c\}\subset\mathbb R$ such that $a+b+c=3$ and $abc\ge -4$. Prove that: $$3(abc+4)\ge 5(ab+bc+ca).$$
*) $ab+bc+ca<0$ This ineq is right
*) $ab+bc+ca\ge 0$ then in $ab,bc, ca$ at least a non-negative number exists assume is $ab$
$\Rightarrow \displaystyle f(ab)=(3c-5)ab+5c^2-15c+12$
+)$\displaystyle 3c-5 > 0\Rightarrow \displaystyle f \geq 5c^2-15c+12=5(c-\frac{3}{2})^2+\frac{3}{4} > 0$
+) $\displaystyle 3c-5 \leq 0$. And we have:
$\displaystyle \Rightarrow \frac{(3-c)^2}{4}+c(3-c) \geq ab+bc+ca \geq 0\displaystyle \Leftrightarrow -1 \leq c \leq \frac{5}{3}$
$\Rightarrow \displaystyle f \geq (3c-5)\frac{(3-c)^2}{4}+5c^2-15c+12 \geq 0$
$\displaystyle \Leftrightarrow (c-1)^2(c+1) \geq 0$
Help me to check up and post your solution. Thanks very much
| I think your solution is true and very nice.
My proof.
We can assume that $ab+ac+bc\geq0$, otherwise the inequality is obvious.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove a linear inequality of $w^3$,
which says it's enough to prove our inequality for an extreme value of $w^3$,
which happens in the following cases.
*
*$w^3=-4$.
In this case our inequality is obviously true;
*$b=a$ and $c=3-2a$.
Hence, $$a^2(3-2a)\geq-4$$ or
$$(a-2)(2a^2+a+2)\leq0,$$ which gives $a\leq2$ and we need to prove that
$$3a^2(3-2a)+12\geq5(a^2+2a(3-2a))$$ or
$$(a-1)^2(2-a)\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_0^{π/2} \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx$ Find the value of
$$\int_0^{\pi/2} \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx$$
I'm having a very tough time solving this question, can any body give me some hints?
| Note that after lettimg $x=\pi/2-t$, we obtain
$$I:=\int_{0}^{\frac{\pi}{2}} \frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx
=\int_{0}^{\frac{\pi}{2}} \frac{(\frac{\pi}{2}-t)\sin t\cos t}{\sin^4 t+\cos^4 t}\,dt=\int_{0}^{\frac{\pi}{2}} \frac{\frac{\pi}{2}\sin t\cos t}{\sin^4 t+\cos^4 t}\,dt-I$$
which implies that
\begin{align*}
I&=\frac{\pi}{4}\int_{0}^{\frac{\pi}{2}} \frac{\sin t\cos t}{\sin^4 t+\cos^4 t}\,dt\quad (s:=\sin(t))\\
&=\frac{\pi}{4}\int_{0}^{1}\frac{s}{s^4+(1-s^2)^2}\,ds
\\
&=\frac{\pi}{8}\int_{0}^{1}\frac{4s}{(2s^2-1)^2+1}\,ds
\\
&=\frac{\pi}{8}\left[\arctan(2s^2-1)\right]_0^1=\frac{\pi^2}{16}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2439244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Center of mass of a pyramid For an electromagnetic exercise, I need to find the center of mass of a pyramid. The pyramid is made of a square base of lenth a and four equilateral triangles with sides that mesure a long to. This is what I came up with for the moment but something seems to be wrong, when I try with some values, the results aren't right.
The summit of the pyramid is A, the four corners of the base are BCDE, the center of mass is o and the center of the base H. I used the fact that the center of mass is the center of the sphere going through all summits of the pyramid. p is the distance between H and O.
$$CH=\frac{a*\sqrt{2}}{2}, AC^2=AH^2+HC^2 \Rightarrow AH^2=a^2-\frac{a^2}{2}=\frac{a^2}{2}$$
O is the center of mass so OA=OC
$$OC^2=p^2*\frac{a.\sqrt{2}}{2}, OA=\frac{a^2}{2}-p$$
$$OC=OA \Rightarrow OC^2=OA^2$$
$$p^2*\frac{a.\sqrt{2}}{2}=\frac{a^4}{4}-a^2p+p^2$$
$$a^2p=\frac{a^4}{4}-\frac{a.\sqrt{2}}{2}$$
$$p=\frac{a^4-a*2\sqrt{2}}{4*a^2}$$
Is there a mistake because when I replace a with a value, the result doesn't seem right.
Thanks in advance.
Image of the pyramid with all the points
| First let's calculate the height $h$ of the pyramid. The height of the equilateral triangle of side $a$ is $\frac{a\sqrt{3}}{2}$ so by the Pythagorean theorem we get:
$$h = \sqrt{\left(\frac{a\sqrt{3}}{2}\right)^2 - \left(\frac{a}{2}\right)^2} = \frac{a}{\sqrt2}$$
Let's place the pyramid in the Cartesian coordinate system such that the center of the square base is at $(0,0,0)$ and the vertex is at $\left(0,0,\frac{a}{\sqrt2}\right)$. Since the pyramid is symmetrical around the $z$-axis, it is clear that the center of mass will be on the $z$-axis. We just have to calculate its $z$ coordinate. Let $\overline{z}$ be the coordinate, $M$ the mass of the pyramid and $\rho$ its density.
We have
$$\overline{z} = \frac{1}{M}\iiint_V z\cdot\rho(x,y,z)\,dV = \frac{\rho}{M}\iiint_V z\,dV = \frac1V \iiint_V z\,dv$$
since the density $\rho$ is constant.
To integrate over the volume of the pyramid, notice that for $z \in \left[0,\frac{a}{\sqrt2}\right]$, we have to integrate over the square $\left[-\left(\frac{a}2 - \frac{z}{\sqrt2}\right), \left(\frac{a}2 - \frac{z}{\sqrt2}\right)\right]^2$. You can see this from the similarity of the triangles $\Delta(0,0,0)\left(\frac{a}{2},0,0\right)\left(0,0,\frac{a}{\sqrt2}\right)$ and $\Delta(0,0,z)\left(x,0,z\right)\left(0,0,\frac{a}{\sqrt2}\right)$ and solving for $x$ from here.
\begin{align}
\iiint_V z\,dV &= \int_0^{\frac{a}{\sqrt2}} \int_{-\left(\frac{a}2 - \frac{z}{\sqrt2}\right)}^{\left(\frac{a}2 - \frac{z}{\sqrt2}\right)} \int_{-\left(\frac{a}2 - \frac{z}{\sqrt2}\right)}^{\left(\frac{a}2 - \frac{z}{\sqrt2}\right)} z \,dxdydz\\
&= \int_0^{\frac{a}{\sqrt2}} \int_{-\left(\frac{a}2 - \frac{z}{\sqrt2}\right)}^{\left(\frac{a}2 - \frac{z}{\sqrt2}\right)} z(a-z\sqrt2) \,dydz\\
&= \int_0^{\frac{a}{\sqrt2}} z(a-z\sqrt2)^2 \,dz\\
&= \frac{a^4}{24}
\end{align}
The volume is:
$$V = \frac{1}{3}Bh = \frac13 a^2 \cdot \frac{a}{\sqrt2} = \frac{a^3}{3\sqrt2}$$
Therefore,
$$\overline{z} = \frac1{V}\iiint_V z\,dV = \frac{a}{4\sqrt2}$$
The center of mass is at $\left(0,0,\frac{a}{4\sqrt2}\right) = \left(0,0,\frac{h}4\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots$ converges
Prove $$\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots$$ converges, where the sums are for all primes $p$.
I've found in this link that $$\frac{1}{2}\sum_p \frac{1}{p^2}+\frac{1}{3}\sum_p \frac{1}{p^3}+\cdots$$
converges to $K$, where $K<1$.
I know that $\sum_p \frac{1}{p^2}\le \sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6},$
because see Basel problem and specific values of Riemann zeta function.
Therefore also $$\sum_p \frac{1}{p^k}\le \sum_p \frac{1}{p^2}\le\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
for all $k\ge 3$, $k\in\mathbb Z.$
| Notice that
$$\sum_{p} \frac{1}{p^s}\lt\zeta(s)-1$$
so your sum is bounded from above by
$$
\begin{align}
\sum_{s=2}^\infty(\zeta(s)-1)
&=\sum_{s=2}^\infty \sum_{n=2}^\infty \frac{1}{n^s}\\
&=\sum_{n=2}^\infty \sum_{s=2}^\infty \frac{1}{n^s}\\
&=\sum_{n=2}^\infty\left(\frac{1}{1-\frac{1}{n}}-1-\frac{1}{n}\right)\\
&=\sum_{n=2}^\infty\left(\frac{1}{n-1}-\frac{1}{n}\right)\\[9pt]
&=1
\end{align}
$$
and so your sum is less than one, and must also trivially be positive, so it converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How to evaluate $\lim_{x\to0}\frac{\sin^2\left(\frac x2\right)-\frac{x^2}4}{e^{x^2}+e^{-x^2}-2}$? $$\begin{align*}
\lim_{x \to 0} \frac{\sin^2 \left(\frac{x}{2}\right) - \frac{x^2}{4}}{e^{x^{2}} + e^{-x^{2}} - 2} &\overset{L}{=} \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} + (-2x)e^{-x^{2}}} \\
&= \lim_{x \to 0} \frac{\sin \frac{x}{2} \cos \frac{x}{2} - \frac{1}{2}x}{2xe^{x^{2}} -2xe^{-x^{2}}} \\
&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{(2x)(2x)e^{x^{2}} - (2x)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{\frac{1}{2}\cos^2 \frac{x}{2} - \frac{1}{2}\sin^2 \frac{x}{2} - \frac{1}{2}}{4x^2 e^{x^{2}} + 4x^2 e^{-x^{2}}} \\
&\overset{L}{=} \lim_{x \to 0} \frac{\frac{1}{2}\left( -\sin \frac{x}{2} \cos \frac{x}{2} \right) - \frac{1}{2} \left( \sin \frac{x}{2} \cos \frac{x}{2} \right)}{(4x^2)(2x)e^{x^{2}} + (4x^2)(-2x)(e^{-x^{2}})} \\
&= \lim_{x \to 0} \frac{-\sin \frac{x}{2} \cos \frac{x}{2}}{8x^3e^{x^{2}} - 8x^3 e^{-x^{2}}} \\
\end{align*}$$
After evaluating the limit as $x \to 0$, I noticed that the problem comes up to be in an indeterminate form of $0/0$. I immediately utilized the L'Hospital Rule by differentiating both the numerator and denominator.
However, after using L'Hospital rule for 5-6 times, I noticed that the question will go through a loop of $0/0$ indeterminants.
In my second attempt,
I have tried multiplying $\exp(x^2)$ in both the numerator and denominator with hopes to balance out the $\exp(x^{-2})$. However, an indeterminant is $0/0$ still resulting.
Any help would be appreciated, thank you all!
| $$=\lim_{x\to0}e^{x^2}\cdot\dfrac{\left(\sin\dfrac x2\right)^2-\left(\dfrac x2\right)^2}{(e^{x^2}-1)^2}$$
$$=-\dfrac1{2^4}\lim_{x\to0}e^{x^2}\cdot\lim_{x\to0}\dfrac{\sin\dfrac x2+\dfrac x2}{\dfrac x2}\cdot\lim_{x\to0}\dfrac{\dfrac x2-\sin\dfrac x2}{\left(\dfrac x2\right)^3}\left(\dfrac1{\lim_{x\to0}\dfrac{e^{x^2}-1}{x^2}}\right)^2$$
$\lim_{x\to0}\dfrac{\sin\dfrac x2+\dfrac x2}{\dfrac x2}=\lim_{x\to0}\left(\dfrac{\sin\dfrac x2}{\dfrac x2}+1\right)=?+1$
$\lim_{x\to0}\dfrac{e^{x^2}-1}{x^2}=1$
For $I=\lim_{x\to0}\dfrac{\dfrac x2-\sin\dfrac x2}{\left(\dfrac x2\right)^3},$ set $x=2y$
and use Are all limits solvable without L'Hôpital Rule or Series Expansion, to find $6I=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2442967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Find the remainder when $f(x)$ divided by $(x^2 + x + 1)(x+1)$. When a polynomial $f(x)$ is divided by $x^2 + x + 1$ and $(x+1)^2$, the remainder are $x+5$ and $x-1$ respectively. Find the remainder when $f(x)$ is divided by $(x^2 + x + 1)(x+1)$.
First, I let the remainder be $Ax^2 + Bx +C$, then I try to find the values of $A$, $B$ and $C$. There is $3$ unknowns so we need three equations but I can only get two equations.
| $$p(x)\equiv (x-1)\pmod{(x+1)^2} $$
implies $p(x)\equiv (x-1) \equiv -2 \pmod{(x+1)}$. We know that $p(x)\equiv(x+5)\pmod{x^2+x+1}$, or
$$ p(x) = (x+5) + (x^2+x+1) r(x) $$
hence
$$ p(x) \equiv 4 + r(x)\pmod{(x+1)}$$
and $r(x)\equiv -6\pmod{(x+1)}$. It follows that
$$ p(x) = (x+5)+(x^2+x+1)(-6+(x+1)s(x)) $$
and
$$ p(x)\equiv \color{red}{(x+5)-6(x^2+x+1)} \pmod{(x+1)(x^2+x+1)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Is the inequality $\:\frac{r^2}s +\frac{s^2}r\geqslant 2\max\{r,s\}\:$ a valid one? I'm not sure if
$$\frac 12\left(\frac{r^2}s + \frac{s^2}r\right)\;\geqslant\;
\max\{r,s\}$$
holds true for all $\,r,s\in\mathbb R^{>0}$.
I think so $\,\ddot\smile\:$ but
Do you have a good proof for it?
$\ddot\frown\:$ The cyclic analogue in three variables
$$\max \big\{u,v,w\big\} \;\overset{\displaystyle ?}{\leqslant}\;
\frac 13\left(\frac{u^2}v + \frac{v^2}w + \frac{w^2}u\right)$$
certainly/already fails because evaluating the RHS at $\,(u,v,w)=(3,2,1)\,$ yields $\,2\tfrac{17}{18}$.
|
Your Claim is False See below: Indeed,
\begin{split}
&&\frac 12\left(\frac{r^2}s + \frac{s^2}r\right)\;\geqslant\;
\max\{r,s\}\\
&\Longleftrightarrow& \frac 12\left(\frac{r^2}s + \frac{s^2}{r}\right) \;\geqslant\; r \quad and \quad\frac 12\left(\frac{r^2}{s} + \frac{s^2}r\right)\;\geqslant\;s \\ &\Longleftrightarrow& \left(\frac{r^2}s + \frac{s^2}{r}\right) \;\geqslant\; 2r \quad and \quad\left(\frac{r^2}{s} + \frac{s^2}r\right)\;\geqslant\;2s \\& \Longleftrightarrow& \left(\frac{r}s + \frac{s^2}{r^2}\right) -2\;\geqslant\;
0 \quad and \quad\left(\frac{r^2}{s^2} + \frac{s}r\right)-2\;\geqslant\;
0\\& \Longleftrightarrow &f(t)\ge 0 \quad and \quad f(\frac{1}{t})\ge 0
\end{split}
Where $t=\frac{s}{r}$ and $f(t) = t^2 +\frac{1}{t} - 2$ , for $t>0$. Let study $f(t)$ for $t>0$.
We have
$$f'(t) = 2t -\frac{1}{t^2} = 0\Longleftrightarrow t= \frac{1}{\sqrt[3]{2}}$$
where $$\lim_{t\to 0}f(t)=\lim_{t\to \infty}f(t)=\infty$$
Therefore $f(\frac{1}{\sqrt[3]{2}})$ is the only minimum of $f$ on $(0,\infty)$ that is,
$$\forall t>0,~~ f(t) \ge f(\frac{1}{\sqrt[3]{2}}) = 3*2^{-\frac{2}{3}} -2$$
But $$3*2^{-\frac{2}{3}} -2<0 $$
therefore there exists $0<a<1$ such that $f(a) = f(1) = 0$ and $f(t)< 0$ for $t\in(a,1)$ in particular
$$f(\frac{1}{\sqrt[3]{2}}) = 3*2^{-\frac{2}{3}} -2<0.$$
Moreover, we have, $$a=\frac{1+\sqrt{5}}{2}\qquad \text{that is,}\qquad f( \frac{-1+\sqrt{5}}{2}) = f(1)=0 $$
and $$f(t)< 0 \qquad \text{for}\qquad \frac{-1+\sqrt{5}}{2}<t<1.$$
Hence your claim is FALSE
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2446074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Summation of a series-combination of a GP and another series Find the sum of the series:
$$\frac{1}{2}\biggl(\frac{1}{2}\biggl)+\frac{2}{3}\biggl({\frac{1}{2}\biggl)}^2+\frac{3}{4}\biggl({\frac{1}{2}\biggl)}^3+\cdots+
n^{th}terms$$
I calculated the $n^{th}$ term to be $\frac{n}{n+1}\frac{1}{2^n}$
$$\sum_{k=1}^n \frac{k}{k+1}\frac{1}{2^k}=\sum_{k=1}^n \frac{1}{2^k}\biggl(1-\frac{1}{k+1}\biggl)=\sum_{k=1}^n \frac{1}{2^k}-\sum_{k=1}^n \frac{1}{2^k(k+1)} $$
The first summation is a GP. But then I could not solve the second summation as it was not telescoping. Please help.
| You have $\frac{1}{1-x} = \sum_{i=0}^\infty x^i$
So $-\ln(1-x) = \sum_{i=0}^\infty \frac{1}{i+1}x^{i+1}$
Then $$\frac{-\ln(1-x)}{x} = \sum_{i=0}^\infty \frac{1}{i+1}x^{i}$$
So, you have $$\frac{-\ln(\frac{1}{2})}{\frac{1}{2}} = 1 + \sum_{i=1}^\infty \frac{1}{i+1}\frac{1}{2^i}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proof verification: $\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$ Is either of the following methods correct?
Prove
$$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$
First Method
Preliminary Analysis:
We know that $a = 2$, $L= \frac{1}{2}$, and $f(x)= \frac{1}{x}$. By the precise definition of limit we have the following:
$$
0<|x-a|<\delta \implies |f(x)-L|<\varepsilon\\
0<|x-2|<\delta \implies \left|\frac{1}{x}-\frac{1}{2}\right|<\varepsilon \implies \left|\frac{2}{2x}-\frac{x}{2x}\right| <\varepsilon \implies \left|\frac{2-x}{2x}\right| <\varepsilon \\
\implies \left|\frac{-(-2+x)}{2x}\right| <\varepsilon \implies \left|\frac{x-2}{2x}\right| <\varepsilon \implies \frac{\left|x-2\right|}{\left|2x\right|} <\varepsilon \implies \left|x-2\right|<\varepsilon \left|2x\right|
$$
let $\delta = \varepsilon\left|2x\right|$ but we need to simplify it further because delta should be in terms of $\varepsilon$ only.
Assume $|x-a| < 1$
$$ |x-2| < 1 \implies -1 <x -2<1 \implies -1+2<x<1+2 \implies 1<x<3$$
Then we have to simplify $|2x|$ as well, which ends up being
$$ 2<2x<6 \implies -6<2x<6 \implies |2x| <6$$
Now consider the inequality we discovered, specifically, $\left|x-2\right|<\varepsilon \left|2x\right|$ this inequality is also valid for $\left|x-2\right|<\varepsilon\cdot6$.
Let $\delta = \min{\{1, \varepsilon\cdot6}\}$
Proof:
Given $\varepsilon > 0$ let $\delta = \min{\{1, \varepsilon\cdot6}\}$ if $ 0<|x-2|<\delta \implies |x-2|<1 \implies 1 <x < 3 \implies 2 < 2x < 6 \implies - 6 < 2x < 6 \\ \implies |2x| < 6$
We also have $|x - 2| < \varepsilon \cdot6$.
$$
\left|\frac{1}{x}-\frac{1}{2}\right|\implies \left|\frac{2}{2x}-\frac{x}{2x}\right| \implies \left|\frac{2-x}{2x}\right| \\
\implies \left|\frac{-(-2+x)}{2x}\right|\implies \left|\frac{x-2}{2x}\right| \implies \frac{\left|x-2\right|}{\left|2x\right|} < \frac{\varepsilon \cdot 6}{6} = \varepsilon
$$
By the precise definition of limit $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$
Second Method
Proof by Definition/Property:
By the direct substitution property if $f$ is a polynomial or a rational function and $a$ is in the domain of $f$, then
$$\lim_{x \to a} f(x) = f(a)$$
Then by the direct substituon property of limit: $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$
Are either of the above methods correct? (I am putting the question here as well in case someone misses it)
| proof-verification:
(1)Given $\varepsilon > 0$, let $\delta = \min{\{1, 6\varepsilon}\}$.
(2)if $ 0<|x-2|<\delta \implies |x-2|<1 \implies 1 <x < 3 \implies 2 < 2x < 6 \implies - 6 < 2x < 6 \\ \implies |2x| < 6$
(3)We also have $|x - 2| < \varepsilon \cdot6$.
$$
\left|\frac{1}{x}-\frac{1}{2}\right|\implies \left|\frac{2}{2x}-\frac{x}{2x}\right| \implies \left|\frac{2-x}{2x}\right| \\
\implies \left|\frac{-(-2+x)}{2x}\right|\implies \left|\frac{x-2}{2x}\right| \implies \frac{\left|x-2\right|}{\left|2x\right|} < \frac{\varepsilon \cdot 6}{6} = \varepsilon
$$
The writing of these line 2 and 3 are terrible. One should not write those confusing big "implication" arrows for doing simple algebra. Moreover, (3) is incorrect: $|x-2|<6\varepsilon$ and $|2x|<6$ do not imply that
$$
\frac{|x-2|}{|2x|}<\frac{6\varepsilon}{6}.
$$
By the precise definition of limit, $$\lim_{x\to2} \frac{1}{x} = \frac{1}{2}$$
You want to get the estimate like
$$
\frac{|x-2|}{|2x|}<\epsilon.
$$
The intuition is as follows. On the one hand, $|x-2|$ can be as small as possible if $x$ is close to $2$. One the other hand, when $x$ is close to $2$, the quantity $\frac{1}{|2x|}$ is bounded by some positive real number. Thus together, one can get $\frac{|x-2|}{|2x|}$ as small as one wants when $x$ is close to $2$.
Now we turn the intuition to a rigorous proof. Given $\epsilon>0$, let $\delta = \min\{1,\epsilon\}$. If $0<|x-2|<\delta$, then $1<x<3$, which implies that
$$
\frac{1}{|2x|}<1 \tag{1}
$$
On the other hand, by the definition of $\delta$, we also have
$$
|x-2|<\epsilon.\tag{2}
$$
Combining (1) and (2), we have the desired inequality
$$
\frac{|x-2|}{|2x|}<\epsilon.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2447907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to solve system of equation, $\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3$, $\sqrt{y-4}+\sqrt{z-4}=4\sqrt3$ and $ \sqrt{x-9}+\sqrt{z-9}=4\sqrt3$ . $$\begin{cases}\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3\\\sqrt{y-4}+\sqrt{z-4}=4\sqrt3\\\sqrt{x-9}+\sqrt{z-9}=4\sqrt3\end{cases}$$ I tried somthing,like go to the power of two , and change of variables... but it became more complicated . Is there an idea to solve this system of equation ?
Thanks in advance
| Hint:
Eliminate $y$ and $z$,
$$y=(4\sqrt3-\sqrt{x-1})^2+1,\\z=(4\sqrt3-\sqrt{x-9})^2+9$$
and
$$\sqrt{44+x-8\sqrt3\sqrt{x-1}}+\sqrt{44+x-8\sqrt3\sqrt{x-9}}=4\sqrt3.$$
By plotting, you can see that this equation has two real solutions. It is possible, by successive squarings and regroupings, to turn it to a polynomial. But this will be tedious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2451634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Positive integer solutions to $xy=4x+7y$ I think I have found all the positive integer solutions to $xy=4x+7y$. I did first was to make $y$ the subject of the expression, and then I
$$xy-7y=4x$$
$$y=\frac{4x}{x-7}$$
$$y=\frac{4}{x-7} * x$$
I started finding the divisors of $4$: $1,2,4$, and setting $x-7$ to be equal to these divisors. I then got the solution pairs: $(8, 32), (9, 18), (11, 11)$.
I also realised that if $x-7$ and $x$ were divisible by $7$, then $7$ could be 'cancelled out' from the expression. I substituted $x=14, 21, 35$, and got $3$ more solution pairs: $(14, 8)$, $(21, 6)$, $(35, 5)$. After this, I concluded that there might not be any more solutions, because $$\lim_{x\to \infty} (4*\frac{x}{x-7}) = 4*1=4,$$
and that means that when $x$ is another multiple of $7$, $y$ will never equal $4$.
However, how can I prove that there are no other solutions rigorously, without checking each number case by case? Is there a way to find the total number of integer solutions without knowing what they are?
| Let $\gcd(x,y) = k$ and let $x = ak; y = bk$ and therefore $\gcd(a,b) = 1$.
$abk^2 = 4ak + 7bk$
$abk = 4a + 7b$
So $a|7b$ so $a|7$ so $a = 7$ or $a=1$.
$b|4a$ so $b|4$ so $b=1,2$ or $4$.
$k = \frac 4b + \frac 7a$.
So there are $6$ solutions.
$a=1; b=1; k=11$ and $x=y = 11$. $121 = 44 + 77$
$a=1; b=2; k=9$ and $x=9;y=18$. $9*18 = 4*9 + 7*18$.
$a=1; b=4; k=8$ and $x=8;y=36$. $8*36 = 4*8 + 7*36$
$a=7; b=1; k=5$ and $x=35; y=5$. $5*35= 4*35+7*5$
$a=7; b=2; k=3$ and $x=21; y=6$. $6*21=4*21 + 7*6$
$a=7; b=4; k= 2$ and $x=14; y=8$. $8*14 = 4*24 + 7*8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Proving that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2y^3}{x^2+y^2}=0$ I need to show that $\lim\limits_{(x,y) \to (0,0)} \frac{x^2y^3}{x^2+y^2}=0$. My attempted method was showing that as the distance between $(x,y)$ and $(0,0)$ approaches $0$, so does the function in the limit. I tried to show $\lvert{\frac{x^2y^3}{x^2+y^2}}\rvert \leq \sqrt{x^2+y^2}$ for $(x,y) \neq 0$. I expanded the terms to get $\frac{x^2\lvert y^3\rvert}{x^2+y^2} \leq \frac{(x^2+y^2)\sqrt{x^2+y^2}}{x^2+y^2}$, or $(x^2+y^2)^\frac{3}{2}\geq x^2\lvert y^3\rvert$. I am unsure of how to prove this and whether this method is valid or would work in showing the existence of the above limit at all.
| A common approach, when you see squares as in the denominator, is conversion to polar coordinates: $x=r\cos t$, $y=r\sin t$, then
$$
\lim_{(x,y)\to(0,0)}\frac{x^2y^3}{x^2+y^2} = \lim_{r\to 0} \frac{(r\cos t)^2(r\sin t)^3}{(r\cos t)^2 + (r\sin t)^2}
= \lim_{r \to 0} \frac{r^5 \cos^2 t \sin^3 t}{r^2(1)}
= \cos^2 t\sin^3 t \lim_{r\to 0}r^3 = 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $(a^2 + 1)(b^2 + 1)(c^2 + 1) \ge (a + b)(b + c)(c + a)$ for $a, b, c \in \mathbb{R}$ How to prove that $(a^2 + 1)(b^2 + 1)(c^2 + 1) \ge (a + b)(b + c)(c + a)$ for $a, b, c \in \mathbb{R}$ ? I have tried AM-GM but with no effect.
| \begin{eqnarray*}
(abc-1)^2 + \frac{1}{2}\sum_{perms} a^2(b-1)^2 \geq 0.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Complete solution for a system of polynomial equations A research question led to the following system of polynomial equations.
\begin{align*}
& x^3 - x^2y + x^2z + x^2 - xy + xz + y^2 - yz - y = 0 \\
& x^3z - 3x^2y + 3xy^2 + xz^2 - y^3 + y^2 - 2yz = 0 \\
& x^4 - x^3y - 2x^3z + x^3 + x^2yz - x^2y + x^2z + xy^2 + xyz - xy - xz^2 - yz^2 = 0 \\
& x^4z + x^4 - x^3y + x^2yz - x^2y + 2x^2z^2 + x^2z - x^2 - xy^2z + xy^2 + xy - y^2z + z^3 - z = 0
\end{align*}
I used the poly_system function in SymPy (a Python library for symbolic mathematics) to solve this system, and it returned the following solutions:
$$[(-1, 0, 0), (0, 0, -1), (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 1, 0), (1, 2, 1), (2, 3, 0), (3, 6, 1)]$$
This was better than I had hoped, but I am concerned that the function may fail to find all solutions. How can I be certain that no other solutions exist?
| Using Maple, I took a "plex($x,y,z$)" Groebner basis of the ideal generated by the left sides of your equations. The result was
$$ \eqalign{&{z}^{5}-{z}^{3}
,\cr
&y{z}^{3}-y{z}^{2}
,\cr
&{y}^{2}{z}^{2}-{z}^{4}-{y}^{2}z-y{z}^{2}+yz+{z}^{2}
,\cr
&{y}^{3}z-7\,{z}^{4}-8\,{y}^{2}z+5\,y{z}^{2}+7\,yz+7\,{z}^{2}
,\cr
&{y}^{4}-18\,{z}^{4}-4\,{y}^{3}-23\,{y}^{2}z+27\,y{z}^{2}+3\,{y}^{2}+21
\,yz+18\,{z}^{2}
,\cr
&-3\,{z}^{4}+6\,x{z}^{2}+{y}^{3}-4\,{y}^{2}z+7\,y{z}^{2}-3\,{z}^{3}-4\,
{y}^{2}-yz+3\,{z}^{2}+3\,y+3\,z
,\cr
&6\,{z}^{4}+6\,xyz-{y}^{3}+{y}^{2}z-7\,y{z}^{2}+3\,{z}^{3}+4\,{y}^{2}-2
\,yz-6\,{z}^{2}-3\,y-3\,z
,\cr
&2\,x{y}^{2}-{y}^{3}+{y}^{2}z-{z}^{3}-2\,xy-2\,xz+y+z
,\cr
&3\,{z}^{4}+12\,{x}^{2}z-2\,{y}^{3}+5\,{y}^{2}z-11\,y{z}^{2}+6\,{z}^{3}
+8\,{y}^{2}-7\,yz-3\,{z}^{2}-6\,y-6\,z
,\cr
&6\,{x}^{2}y-{y}^{3}+{y}^{2}z+2\,y{z}^{2}-3\,{z}^{3}-6\,xy-6\,xz-2\,{y}
^{2}+4\,yz+3\,y+3\,z
,\cr
&-{z}^{4}+4\,{x}^{3}-{y}^{2}z+5\,y{z}^{2}-4\,{z}^{3}+4\,{x}^{2}-8\,xy+y
z+{z}^{2}+4\,z
\cr
}$$
From the first basis element, all solutions must have $z \in \{-1, 0, 1\}$.
If $z = -1$, the second basis element tells us $y=0$, and then the sixth gives us $x=0$.
If $z = 0$, the fifth basis element tells us $y \in \{0,1,3\}$. If $y=0$ we get $x=0$ or $x=-1$ from the 11th basis element. If $y=1$ we get $x=0$ or $x=1$ from the 10th basis element. If $y=3$ we get $x=2$ from the 8th basis element.
If $z=1$, the fourth basis element gives us $y = 0$, $2$ or $6$. If $y=0$, the 6th basis element gives us $x=0$. If $y=2$, the 6th basis element gives us $x=1$. If $y=6$, the 6th basis element gives us $x=3$.
Thus the listed solutions are the only ones.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Simplest way to solve radical equation $\sqrt{3x+1}-\sqrt{x+4}=1$ I have the following equation:
$$\sqrt{3x+1}-\sqrt{x+4}=1$$
I can get the answer $x=5$ through tedious and long algebraic manipulation with quite a few extraneous solutions. It's not elegant. Is there a simple, straightforward way to solve this equation?
| Condition: $x\geq -\frac{1}{3}$.
One has $\sqrt{3x+1} = \sqrt{x+4} + 1$, then $3x+1 = x+4 +1 + 2\sqrt{x+4}$.
So $x-2 = \sqrt{x+4}$. Note that $x \geq 0$.
Then $x^2-4x+4=x+4$, or $x^2-5x = 0$. So $x=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $5+4\cdot 5+4\cdot5^2+4\cdot5^3+4\cdot5^4+4\cdot5^5$
Evaluate
$$5+4\cdot 5+4\cdot5^2+4\cdot5^3+4\cdot5^4+4\cdot5^5.$$
The options are $5^6$, $5^7$, $5^8$, $5^9$, $5^{10}$.
I'm new to this site. I came across this question in an Olympiad foundation site. I have no idea how to solve it. Can I get the answer of this question. Thanks.
| $5 + 4.5 + 4.5^2+4.5^3+4.5^4+4.5^5 = 5+5(5-1)(1+5+5^2+5^3+5^4) = 5+5(5^5-1) = 5^6.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2453929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Probability that no $2 \times 2$ red square is formed Each square of a $3 × 3$ board is coloured either red or blue at random (each having probability $1/2$). Then find
the probability that there is no $2 × 2$ red square.
I was trying to calculate favorable cases using inclusion-exclusion principle.
Total ways=$2^9$
Now there can be $4-$ $2 \times 2$ square so I first subtracted $C(4,1) 2^5$ but once 2 of the 4 square colored then number of remaining square starts depending on position of two squares. How to tackle it? Is inclusion-exclusion correct way to proceed or is there a better method?
Given answer is $\frac{417}{512}$
| there is a $\frac {1}{16}$ chance a that any $2\times 2$ block is all red.
There are four $2\times 2$ blocks.
Now we must exclude the cases where we have double, triple or quadruple counted.
two $2\times 2$ blocks -- adjacent blocks $4\frac {1}{2^6}$ + opposite blocks $ 2\frac {1}{2^7}$
three $2\times 2$ blocks, $4\frac {1}{2^8}$
four $2\times 2$ blocks (all red) $\frac {1}{2^9}$
Chance that there is at least one $2\times 2$ block $= \frac 1{4} - \frac 1{16} - \frac 1{2^6} + \frac {1}{2^6} - \frac {1}{2^9} = \frac {95}{512}$
Chance that there no $2\times 2$ block $1 - \frac {95}{512} = \frac {417}{512}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2454763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I solve this integral using complex anlysis? $$\int_0^{\infty} \frac{\cos x }{(x^2+a^2)^2(x^2+b^2)}dx$$
I've tried changing $\cos x$ with $e^{iz}$ and $(x^2+a^2)^2(x^2+b^2)$ with $(z^2+a^2)^2(z^2+b^2)$ giving the integral $$\oint_\gamma \frac{e^{iz} }{(z^2+a^2)^2(z^2+b^2)}dz$$ over some contour and using partial fractions, but I did not get results.
I'm having problems finding the contour.
Some advices will be helpful.Thanks
| By parity of the integrand function (and its integrability)
$$ \int_{0}^{+\infty}\frac{\cos x}{(x^2+a^2)^2(x^2+b^2)}\,dx = \frac{1}{2}\,\text{Re}\int_{-\infty}^{+\infty}\frac{e^{ix}\,dx}{(x^2+a^2)^2 (x^2+b^2)} $$
equals, by the ML lemma and the residue Theorem,
$$ \frac{1}{2} \text{Re}\lim_{R\to +\infty}\oint\frac{e^{iz}\,dz}{(z^2+a^2)^2 (z^2+b^2)}= -\pi\text{ Im}\!\!\!\!\sum_{z_0\in\{ia,ib\}}\!\!\!\text{Res}\left(\frac{e^{iz}}{(z^2+a^2)^2 (z^2+b^2)},z=z_0 \right) $$
(assuming $a,b\in\mathbb{R}^+$ and $a\neq b$). A simple approach might be to recognize that by partial fraction decomposition the given integral only depends on the parametric integrals
$$ I_1(A) = \int_{0}^{+\infty}\frac{\cos x}{x^2+A}\,dx,\qquad I_2(A) = \int_{0}^{+\infty}\frac{\cos x}{(x^2+A)^2} $$
defined for $A\in\mathbb{R}^+$. We have $I_1(A)=\frac{\pi}{2\sqrt{A}}e^{-\sqrt{A}}$ and $I_2(A) = -\frac{d}{dA} I_1(A)$, from which it follows that $I_2(A)=\frac{\pi}{4A\sqrt{A}}(1+\sqrt{A})\,e^{-\sqrt{A}}$. Now it is enough to consider the cases $A=a^2$, $A=b^2$ and combine them to get the value of the original integral.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ rational? For this question,
Show the following irrational-looking expressions are actually rational numbers.
(a) $\sqrt{4+2\sqrt{3}}-\sqrt{3}$, and
(b) ...
I solved it as follows:
$$\begin{align} x &= \sqrt{4+2\sqrt{3}}-\sqrt{3},\\
x+\sqrt{3} &= \sqrt{4+2\sqrt{3}},\\
(x+\sqrt{3})^{2} &= (\sqrt{4+2\sqrt{3}})^{2},\\
\end{align}\\
x^{2}+2\sqrt{3}x-(1+2\sqrt{3}) = 0,\\
(x-1)(x+(1+2\sqrt{3}))=0.$$
My question is that, there are two numbers satisfying $x = \sqrt{4+2\sqrt{3}}-\sqrt{3}$, but one of them is irrational. Then, how can we say $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational as a whole?
| A classical trap: when you square the members of an equation, you introduce alien solutions.
Going from $$x+a=b$$ to $$x^2+2ax+a^2=b,$$
you introduce
$$x+a=-b$$
that has noting to do with the original problem.
In the given question, the irrational parts of $a$ and $b$ cancel each other in $b-a$, but not in $-b-a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Long Division of Proper Polynomials I took the derivative of function, yielding the result: $$\frac{-3x^4 - 4x^3 - 3x^2 + 18x + 6}{x^6 - 6x^3 + 9}$$ I know this to be correct and wanted to simplify the answer.
I've never attempted long division on an expression in which the highest order is in the denominator before but attempted:
$$
\require{enclose}
\begin{array}{r11}
-3x^{-2} - 4x^{-3} - 3x^{-4} ... \\[-3pt]
x^6 - 6x^3 + 9 \enclose{longdiv}{-3x^4 - 4x^3 - 3x^2 + 18x + 6}\kern-.2ex \\[-3pt]
\underline{-3x^4-18x+27x^{-2} } \\[-3pt]
-4x^3-3x^2+36x+6 \\[-3pt]
\underline{-4x^3-24-36x^{-3}} \\[-3pt]
-3x^2+36x-18-+27x^{-2}-36x^{-3} \\[-3pt]
...
\end{array}
$$
It seems that for every term I eliminate, I gain another of a lower order. Is there a problem with my method, do these two expressions not divide cleanly?
How can one tell whether the division of polynomials is possible?
| Note that the denominator is $(x^3-3)^2$, so you can try dividing by $x^3-3$. If you are looking for an exact factor there is a sneaky trick here - you can put $x^3=3$ to test divisibility, and this gives $-9x-12-3x^2+18x+6\neq 0$.
But you can still divide the numerator by $x^3-3$ to obtain $$-3x^4-4x^3-3x^2+18x+6=-(x^3-3)(3x+4)-3x^2+9x-6$$ which leaves your answer in a kind of partial fraction form as $$-\frac{3x+4}{x^3-3}-\frac{3x^2-9x+6}{(x^3-3)^2}=-\frac{3x+4}{x^3-3}-3\frac{(x-1)(x-2)}{(x^3-3)^2}$$
Note also that you can factorise $x^3-3$. If you take $t$ as the real cube root of $3$ then you have the real factorisation $(x-t)(x^2+tx+t^2)$ and if further $\omega^2+\omega+1=0$ so that $\omega$ is a complex cube root of unity then the factorisation can be completed as $(x-t)(x-\omega t)(x-\omega^2 t)$. You can use these to get a breakdown into more conventional partial fractions.
Whether any of this is useful to your purpose will depend on what you are trying to achieve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $\int_{0}^{1} \frac{x \log x \:dx}{1+x^2}$ Find the value of $$I=\int_{0}^{1} \frac{x \log x \:dx}{1+x^2}$$
My Try: I used Integration by parts
So
$$I=\frac{1}{2}\log x \times \log (1+x^2) \biggr\rvert_{0}^{1}-\int_{0}^{1}\frac{\log(1+x^2)}{2x}dx$$
So
$$I=\frac{-1}{2}\int_{0}^{1}\frac{\log(1+x^2)}{x}$$
Can we proceed from here?
| Since $\int_{0}^{1}x^n \log(x)\,dx = -\frac{1}{(n+1)^2}$ and $\frac{x}{1+x^2}=x-x^3+x^5-x^7+\ldots$ on $(0,1)$, by termwise integration we have
$$ \int_{0}^{1}\frac{x\log x}{1+x^2}\,dx = -\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{6^2}+\frac{1}{8^2}-\ldots = \color{blue}{-\frac{\pi^2}{48}}$$
as a consequene of the Basel problem. The exchange of $\sum$ and $\int$ is allowed by the dominated convergence Theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2465343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ Prove that $ \frac{x}{yz} + \frac{y}{xz} + \frac{z}{xy} \geq \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $ for $x,y,z \in \Bbb{R}$ and $xyz > 0 $.
I know that i can use the axioms of the real numbers, but i can't finde an usefull equivalent transformation
| Multiply through by $xyz$ and you get $$x^2 + y^2 + z^2 \geq xy + yz + zx.$$ Now note that $$(x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0$$ with equality only if $x = y = z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Solving systems of equation with unknown How do i solve this system?
$$
\left\{
\begin{array}{c}
ax+y+z=1 \\
x+ay+z=a \\
x+y+az=a^2
\end{array}
\right.
$$
Ive reduced to this form. How should i continue to get infinitely many solutions, no solution and unique?
$$ \left[
\begin{array}{ccc|c}
1&a&1&a\\
0&(a^2)-1&a-1&(a^2)-1\\
0&(a-1)(a-2)&0&a-1
\end{array}
\right] $$
| If $A$ is the matrix of the coefficients, then for Cramer's theorem the system has one unique solution if $\det A\ne 0$
$\det A=\left|
\begin{array}{lll}
a & 1 & 1 \\
1 & a & 1 \\
1 & 1 & a \\
\end{array}
\right|=a^3-3 a+2$
$\det A=0 \to (a-1)^2 (a+2)=0\to a_1=1;\;a_2=-2$
Then for $a\ne 1;\;a\ne 1$ the system has one and only one solution
$$x=-\frac{a+1}{a+2},\;y=\frac{1}{a+2},\;z=\frac{(a+1)^2}{a+2}$$
$$.$$
If $a=1$ the completed matrix $A|B$ of the system becomes
$
A|B=\left(
\begin{array}{lll|l}
1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 \\
1 & 1 & 1 & 1 \\
\end{array}
\right)
$
As $\text{rank } A= \text{rank } A|B=1$
the system has $\infty^{3-1}=\infty^2$ solutions given by the equation
$x+y+z=1$ whose solutions are $(t,u,1-t-u)$
they have two parameters which is linked to the $\infty^2$ solutions
$$
.
$$
If $a=-2$
$
A|B=\left(
\begin{array}{rrr|r}
-2 & 1 & 1 & 1 \\
1 & -2 & 1 & -2 \\
1 & 1 & -2 & 4 \\
\end{array}
\right)$
$\text{rank } A|B=3$ while $\text{rank }A=2$
they are different so the system is impossible, has no solutions.
Hope this helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2466418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $a+b+c=0$ prove that $ 3(a^2+b^2+c^2) \times (a^5+b^5+c^5) = 5(a^3+b^3+c^3) \times (a^4+b^4+c^4) $
If $a+b+c=0$, for $a,b,c \in\mathbb R$, prove
$$ 3(a^2+b^2+c^2) \times (a^5+b^5+c^5) = 5(a^3+b^3+c^3) \times (a^4+b^4+c^4) $$
I made this question as a more difficult (higher degree) version of this question. My idea was that algebraic brute force methods are easy to distinguish from more sophisticated ones if the degree of the terms in the question is higher.
The question was specifically made using the method from my answer to the linked question.
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, $u=0$ and
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)=-6v^2,$$
$$a^3+b^3+c^3=a^3+b^3+c^3-3abc+3abc=$$
$$=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc=3abc=3w^3,$$
$$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+a^2c^2+b^2c^2)=$$
$$=36v^4-2((ab+ac+bc)^2-2abc(a+b+c))=36v^4-18v^4=18v^4$$ and
$$a^5+b^5+c^5=(a^3+b^3+c^3)(a^2+b^2+c^2)-\sum_{cyc}(a^3b^2+a^3c^2)=$$
$$=(-6v^2)\cdot3w^3-(a+b+c)(a^2b^2+a^2c^2+b^2c^2)+abc(ab+ac+bc)=$$
$$=-18v^2w^3+3v^2w^3=-15v^2w^3.$$
Id est, $$3(a^2+b^2+c^2)(a^5+b^5+c^5)-5(a^3+b^3+c^3)(a^4+b^4+c^4)=$$
$$=3(-6v^2)(-15v^2w^3)-5\cdot3w^3\cdot18v^4=0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2469690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $-1$is a root for $ax^2+bx−3$, find $a^2+b^2$
Given: -1 is a root for $ax^2+bx-3$, with $a,b$ being positive primes, $x\in \Bbb R$.
Find: the numeric value for $a^2+b^2$.
Background: question asked in an entrance exam (Colégio Militar 2005).
My attempt: the other root is $3/a$ and by substitution we can easily find that $$a-b=3\ \ \text{or}\ \ a^2+b^2-2ab=9.$$
I got stuck at this point... how to get the value for $ab$? Hints please.
| Pure obfuscation.
Nothing about roots or algebra or the sum of $a^2 + b^2$ are relevant. That $-1$ is a root simply means $a(-1)^2 + b(-1)-3 = 0$ or in other words $a - b =3$ or $a = b+3$.
What does matter is that $a,b$ are positive primes and all primes except $2$ are odd. If they were both odd primes then $a -b$ would be an even number. So one of them is even so one of them is $2$. So the other is $2+3 = 5$.
So $a^2 + b^2 = 29$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2471382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving the cubic $x^3-x^2-2x+1 = 0$ Solving the cubic $x^3-x^2-2x+1 = 0$.
Using the Cubic Formula I get the following three solutions.
$x_1 = \frac{1}{3} - \frac{1}{3}\left(\frac{7}{2} + \frac{21}{2}i\sqrt{3} \right)^{1/3} - \frac{7}{3}\frac{1}{\left( \frac{7}{2} + \frac{21}{2}i\sqrt{3} \right)^{1/3}} \cong -1.2469796037174670610+2.10^{-20}i$
$x_2 = \frac{1}{3} + \frac{1}{6}\left( \frac{7}{2}+\frac{21}{2}i\sqrt{3} \right)^{1/3}(1+i\sqrt{3}) + \frac{7}{6}\frac{1-i\sqrt{3}}{\left( \frac{7}{2} + \frac{21}{2}i\sqrt{3} \right)^{1/3}} \cong .44504186791262880859 - 3.10^{-20}i$
$x_3 = \frac{1}{3} + \frac{1}{6}\left( \frac{7}{2}+\frac{21}{2}i\sqrt{3} \right)^{1/3}(1-i\sqrt{3}) + \frac{7}{6}\frac{1+i\sqrt{3}}{\left( \frac{7}{2} + \frac{21}{2}i\sqrt{3} \right)^{1/3}} \cong 1.8019377358048382525 + 3.10^{-20}i$
It is clear that the three solutions are all real solutions, but is there a way I can remove the complex components algebraically? My ultimate goal is to describe what the Galois group from this polynomial would like.
| Use $$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}.$$
Finally we obtain $$x_1=-2\cos\frac{2\pi}{7},$$
$$x_2=-2\cos\frac{4\pi}{7}$$ and
$$x_3=-2\cos\frac{6\pi}{7}.$$
Because
$$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=\frac{2\sin\frac{\pi}{7}\cos\frac{2\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{4\pi}{7}+2\sin\frac{\pi}{7}\cos\frac{6\pi}{7}}{2\sin\frac{\pi}{7}}=$$
$$=\frac{\sin\frac{3\pi}{7}-\sin\frac{\pi}{7}+\sin\frac{5\pi}{7}-\sin\frac{3\pi}{7}+\sin\frac{7\pi}{7}-\sin\frac{5\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac{1}{2},$$
which says $$x_1+x_2+x_3=1.$$
Now, $$x_1x_2+x_1x_3+x_2x_3=4\left(\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\cos\frac{6\pi}{7}+cos\frac{4\pi}{7}\cos\frac{6\pi}{7}\right)=$$
$$=2\left(\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}+cos\frac{4\pi}{7}+\cos\frac{2\pi}{7}\right)=-2$$ and
$$x_1x_2x_3=-8\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}\cos\frac{6\pi}{7}=-4\left(\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}\right)\cos\frac{6\pi}{7}=$$
$$=-2\left(1+\cos\frac{2\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{4\pi}{7}\right)=-2\left(1-\frac{1}{2}\right)=-1$$ and use the Viete's theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2471995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $AC$ WITHOUT using Cosine law...
In the following figure find side $AC$ WITHOUT using Cosine law:
There is a simple solution involving Cosine law,but is it possible to calculate $AC$ without using it??
| Let $DK$ be an altitude of $\Delta ABD$, $AC=x$ and $BD=y$.
Thus, $$DK=\frac{y\sqrt3}{2},$$
$$BK=\frac{y}{2}$$ and
$$BC=\sqrt{x^2-1}.$$
Thus, since $$\frac{AB}{BK}=\frac{AC}{CD}$$ and
$$\frac{BC}{DK}=\frac{AC}{AD},$$ we got the following system:
$$\frac{1}{\frac{y}{2}}=\frac{x}{1}$$ and
$$\frac{\sqrt{x^2-1}}{\frac{y\sqrt3}{2}}=\frac{x}{x+1}.$$
Thus, $$\frac{\sqrt{x^2-1}}{\frac{\sqrt3}{x}}=\frac{x}{x+1}$$ or
$$\sqrt{x^2-1}(x+1)=\sqrt3$$ or
$$x^4+2x^3-2x-4=0$$ or
$$(x+2)(x^3-2)=0$$ or
$$x=\sqrt[3]{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Cycles in $5×5$ Latin squares (solution found) One of my favorite puzzle formats is KenKen, in which you are to find an $n×n$ Latin square to be filled with whole numbers from 1 to $n$, given sums, differences, products or quotients of various groups of entries. The puzzles come in various sizes, including $5×5$.
When I solve a $5×5$ puzzle I always seem to find at the end that at least two rows are cyclic permutations of each other. Is this always true in $5×5$ Latin squares, or is there some weird coincidence going on?
Turns out I found the solution. Over 90% of 5×5 Latin squares actually have at least one pair of cyclically permuted rows. So, a 5×5 KenKen without such a cyclic permutations turns out to be relatively rare!
| A counter example where none of the rows are a cyclic permutation of another:
$$\begin{bmatrix}
1&2&3&4&5\\
2&3&5&1&4\\
3&5&4&2&1\\
4&1&2&5&3\\
5&4&1&3&2
\end{bmatrix}$$
But, it turns out that there are only $5$ reduced Latin squares out of $56$ total for $5×5$ squares. So, less than 10% of all $5×5$ solutions should be expected to lack such a cyclic row permutation. Hence such solutions are hard to find.
Here are the other four reduced squares with no cyclic row permutations:
$$\begin{bmatrix}
1&2&3&4&5\\
2&5&1&3&4\\
3&1&4&5&2\\
4&3&5&2&1\\
5&4&2&1&3
\end{bmatrix}$$
$$\begin{bmatrix}
1&2&3&4&5\\
2&5&4&1&3\\
3&4&2&5&1\\
4&1&5&3&2\\
5&3&1&2&4
\end{bmatrix}$$
$$\begin{bmatrix}
1&2&3&4&5\\
2&4&5&3&1\\
3&5&2&1&4\\
4&3&1&5&2\\
5&1&4&2&3
\end{bmatrix}$$
$$\begin{bmatrix}
1&2&3&4&5\\
2&4&1&5&3\\
3&1&5&2&4\\
4&5&2&3&1\\
5&3&4&1&2
\end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2474323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
(1) Sum of two factorials in two ways; (2) Value of $a^{2010}+a^{2010}+1$ given $a^4+a^3+a^2+a+1=0$. Question $1$:
Does there exist an integer $z$ that can be written in two different ways as $z=x!+y!$,where $x,y\in \mathbb N$ and $x\leq y$?
Answer: $0!=1!$ so $0!+2!=3=1!+2!$
Question $2$:
If $a^4+a^3+a^2+a+1=0$, find the value of $a^{2010}+a^{2010}+1$.
Answer: If $a^4+a^3+a^2+a+1=0$ then $$(a-1)(a^4+a^3+a^2+a+1)=0\implies a^5-1=0\implies a=1$$So the value of the required expression is $3$
I think that it is wrong as at $a=1$,$\frac{a^5-1}{a-1}$ is not defined.
Please give me some hints to solve thes problems?
| The first solution is wrong, because $3\neq z!$ for any $z$.
The second solution is wrong.
If $a=1$, then $a^4+a^3+a^2+a+1=1+1+1+1+1\neq 0$, so it is obvious you made a serious mistake in your second solution.
In particular, you multiplied the equation by $(a-1)$, which of course produced a solution of $a=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2474724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Prove that $2^n +1$ is divisible by $3$ for all positive integers $n$. I just want to know if I went on the right direction. With induction
Let $n=1$, then $2^1+1= 3$, which is divisible by $3$. Then show proof for
$n+1.$
$2^n+1=3k$
So we get $2^{n+1}+1, \rightarrow 2^n+2+1, \rightarrow 3k+3= 3(k+1)$. Thus $2^n+1$ is divisible by $3$.
Now if I wanted to show that $2^n+1$ is divisble by $3$, $\forall$ odd integers $n$.
Would it be with induction:
$n=1$, then $2+1=3$, and $3|3$.
Let $n=2k+1$, since n is odd, then we get $2^{2k+1}+1=3m$. Now we need to show for $k+1$.
We get: $2^{2k+2+1}+1=3m \rightarrow 2^{2k+1}*2^2+4-3 \rightarrow 4(2^{2k+1}+1)-3$
$\rightarrow 4(3m)-3 \rightarrow 3(4m-1)$, thus $2^n+1$ is divisible by $3$.
| $2^n+1$ is divisible by $3$ only when $n = 2k+1, (k∈\mathbb{Z}^*)$.
Use modulus:
For every odd number $n$, we have
$$2^n≡2 (\mod n)$$
and for every even number $n$, we have
$$2^n≡1 (\mod n)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
How to find $\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$? How to find $$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$$
My Try :
$$x^2+x-2=(x-1)(x+2)$$
$$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{(x-1)(x+2)}\cdot \frac{(\sqrt{x}+\sqrt{x+3})+3}{(\sqrt{x}+\sqrt{x+3})+3}=\\ =\lim_{x \to 1} \frac{2x+2\sqrt{x(x+3)}-6}{(x-1)(x+2)((\sqrt{x}+\sqrt{x+3})+3)}=\\\lim_{x \to 1} \frac{2(x-3+\sqrt{x(x+3)})}{(x-1)(x+2)((\sqrt{x}+\sqrt{x+3})+3)} $$
Now what ?
| Say $t=\sqrt{x}$, then
\begin{eqnarray}
% \nonumber to remove numbering (before each equation)
\lim_{t \to 1} \frac{t+\sqrt{t^2+3}-3}{t^4+t^2-2} &= &\lim_{t \to 1} \frac{t-1+\sqrt{t^2+3}-2}{(t^2+2)(t^2-1)} \\
&=& \lim_{t \to 1} \frac{1}{(t^2+2)(t+1)} + \frac{\sqrt{t^2+3}-2}{(t^2+2)(t^2-1)}\\
&=& {1\over 6}+\lim_{t \to 1} \frac{t^2-1}{(t^2+2)(t^2-1)(\sqrt{t^2+3}+2)}\\
&=& {1\over 6}+ {1\over 12}\\
&=& {1\over 4}
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2475585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that the limit as $(x,y,z)$ approaches $0$ of $(x^3+y^3+z^3)/(xyz)$ does not exist I am asked to show that
$$
\lim_{(x,y,z)\to(0,0,0)}\frac{x^3+y^3+z^3}{xyz}
$$
does not exist. I know that I need to use $x=at$, $y=bt$, and $z=bt$ but I don't understand what I should do after inserting them into the function.
| In
$\frac{x^3+y^3+z^3}{xyz}
$
let
$y = ax$
and
$z = bx$.
These values are on
a line through the origin.
It becomes
$\frac{x^3+(ax)^3+(bx)^3}{x(ax)(bx)}
=\frac{x^3+a^3x^3+b^3x^3}{abx^3}
=\frac{1+a^3+b^3}{ab}
$.
Therefore,
along this line,
the ratio depends
only on $a$ and $b$.
Therefore
the limit does not exist
because,
by choosing different
$a, b$ values,
the ration can take on
different values.
For example,
if $a=b=1$,
the ratio is $3$;
if $a=b=2$,
the ratio is $17/4$;
if $b=a$,
the ratio is $(1+2a^3)/a^2$;
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2479088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $x\in \Bbb R,$ solving $x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}$
Given: $x\in \mathbb R$, $x=\sqrt{1+\sqrt{1+\sqrt{1+x}}}$
Find: numeric value of $x$
Problem from a math contest. Sorry if it is a duplicate, but could not find anything similar using the search tool.
My attempt: I squared both sides, and developed the resulting expression, but I'm getting to nowhere.
Hints and/or answers please.
Is there any standard way to approach problems like these?
| We know that the Golden ratio $\phi = \frac{1+\sqrt5}{2}$ and $~\bar{\phi}= \frac{1 -\sqrt5}{2}$ satisfy the equation
$$x^2 = 1+x \Longleftrightarrow x= \sqrt {1 + x}~~x>0~ \implies x= f(x)$$ where, $f(x)= \sqrt {1 + x},~~x>0.$
This means that $ \phi $ is the only fix points of the function $f(x)= \sqrt {1 + x},~x>0$
But $$ \sqrt {1 + \sqrt {1 + \sqrt{1+x}}} = f\circ f\circ f(x)= f^3(x) .$$
Therefore your equation reduces to
$$x= f^3(x)$$
Which mean that $x$ is fix point of $f^3$.
On the other hand, $|f'(x)| =\frac{1}{2\sqrt{x+1}} \le \frac12$ then,
$$|(f^3(x))'| = 3|f'(x)\cdot f'(f^2)(x)| \le \frac 34<1$$
then $f^3$ is a contraction and hence, $f^3$ has a unique fix point satisfying $x=f^3(x)$ . Whereas,
$$ \color{red}{x= f^3(x) \implies f(x) = f^3(f(x)) \implies x= f(x) }$$
since we observse that $f(x)$ is also a point fix of $f^3$
by unicity? we get $x=f(x).$
But $\phi $ is the only positive number satisfying $x=f(x)$.
Hence, the only solution to $x=f^3(x)$ is $$x=\phi = \frac{1+\sqrt5}{2}~$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2480022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Find $\lim_{n \to \infty } \sqrt[3]{n^3+1} - \sqrt{n^2+1}$ Find
$$\lim_{n \to \infty} \sqrt[3]{n^3+1}-\sqrt{n^2+1}$$
I already tried to use the Sqeeze theorem on it, but I just was not able to find some reasonable upper series for it, only lower:
$$n\sqrt[3]{1+\frac{1}{n^3}}-n\sqrt{1+\frac{1}{n ^2}}$$
$$n\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$$
$$\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right) \leq n\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$$
Is there anyone who can give me a hint as to how to solve it?
| Find
$$\lim_{n \to \infty} \sqrt[3]{n^3+1}-\sqrt{n^2+1}$$
$$n\sqrt[3]{1+\frac{1}{n^3}}-n\sqrt{1+\frac{1}{n ^2}}$$
$$n\left(\sqrt[3]{1+\frac{1}{n^3}}-\sqrt{1+\frac{1}{n ^2}}\right)$$
$$n\left(\left({1+\frac{1}{n^3}}\right)^{1/3}-\left({1+\frac{1}{n^2}}\right)^{1/2}\right)$$
As $n \to \infty$,this can be approximated as
$$n\left(\left({1+\frac{1}{3n^3}+other terms}\right)-\left({1+\frac{1}{2n^2}}+other terms\right)\right)$$
$$n\left({\frac{1}{3n^3}-\frac{1}{2n^2}}+other terms\right)$$
$${\frac{1}{3n^2}-\frac{1}{2n}}+other terms$$
As $n \to \infty$, all terms become 0. So answer is 0.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2482681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
For which $n \in \mathbb{N}$ it is true that $x^2+x+1 | (x+1)^n-x^n-1$ I have to find such $n \in \mathbb{N}$ for which $x^2+x+1 | (x+1)^n-x^n-1$.
$x^2+x+1$ has 2 complex roots: $x_1=-((-1)^{(1/3)})$, $x_2=(-1)^{(2/3)}$ so I tried to divide $(x+1)^n-x^n-1$ by $(x-x_1)$ and then $(x-x_2)$ but it was too difficult. Any other hints?
| As $(x-1)(x^2+x+1)=x^3-1,$ the roots of $x^2+x+1=0$ are complex cube roots$(w,w^2)$ of unity
So, $w^2+w+1=0$
Let $w=e^{2\pi i/3}\implies w^2=e^{-2\pi i/3},-w^2=e^{\pi i(1-2/3)}$
Now using About Euler's formula $e^{ix}=\cos x+i\sin x$,
$$f(w)=(w+1)^n-w^n-1=(-w^2)^n-w^n-1=(e^{\pi i/3})^n-(e^{2\pi i/3})^n-1$$
$$=\cos\dfrac{n\pi}3+i\sin\dfrac{n\pi}3-1-\cos\dfrac{2n\pi}3-i\sin\dfrac{2n\pi}3$$
We need $f(w)=0$
Equate the imaginary & the real parts -
$$\sin\dfrac{2n\pi}3=\sin\dfrac{n\pi}3$$
$\implies$ either $(i)\dfrac{2n\pi}3\equiv\dfrac{n\pi}3\pmod{2\pi}$
or $n\equiv0\pmod6$
But then $\cos\dfrac{n\pi}3-1-\cos\dfrac{2n\pi}3\ne0$
or $(ii)\dfrac{2n\pi}3\equiv\pi-\dfrac{n\pi}3\pmod{2\pi}$
$\implies n\equiv1\pmod2$
Again, $\cos\dfrac{n\pi}3-1-\cos\dfrac{2n\pi}3=\cos\dfrac{n\pi}3-2\cos^2\dfrac{n\pi}3=-\cos\dfrac{n\pi}3\left(2\cos\dfrac{n\pi}3-1\right)$
I hope yo can take it from here!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$F(2n-1) = F(n-1)^2 + F(n)^2$, where $F(i) $ is the $i$'th Fibonacci number, for all natural numbers greater than $1$ I'm proceeding by induction, but I have no idea how to do the inductive step. The Fibonacci sequence takes on the definition:
$F(0)=0$
$F(1) = 1$
$F(2) = 1$
So far, my proof is as follows:
Base Case:
$F(2) = 1 = F(1)^2 + F(0)^2$. This case holds
Inductive Step: We assume the claim holds for $n$ and show it holds for $n+1$...
$F(2n+1) = F(2n)+F(2n-1)=F(2n) + F(n-1)^2+F(n)^2$
I am stuck on how to reduce this expression to $F(n)^2 + F(n+1)^2$...
Edit: Thank you for the responses. I realize I made an egregious error in defining the fibonacci sequence. I have corrected it to the definition we use in my class.
Edit by DS : Now changed the indexing to the standard convention of the Fibonacci numbers.
| Yet another approach. By setting $M=\small\begin{pmatrix}1 & 1 \\ 1 & 0\end{pmatrix}$ we have
$$ \begin{pmatrix} F_{n+2} \\ F_{n+1}\end{pmatrix} = M \begin{pmatrix} F_{n+1} \\ F_{n}\end{pmatrix},\qquad M^n=\begin{pmatrix}F_{n+1} & F_n\\ F_{n} & F_{n-1}\end{pmatrix} $$
and $F_{2n+1}$ is the top-left element of $M^{2n}=(M^n)^2$, where
$$ \begin{pmatrix}F_{n+1} & F_n\\ F_{n} & F_{n-1}\end{pmatrix}^2 = \begin{pmatrix}\color{blue}{F_{n+1}^2+F_n^2} & F_n(F_{n-1}+F_{n+1})\\ F_n(F_{n-1}+F_{n+1}) & F_n^2+F_{n-1}^2\end{pmatrix} $$
leads to
$$ F_{2n+1}=F_{n+1}^2+F_n^2,\qquad \boxed{F_{2n\color{red}{-1}}=F_n^2+F_{n-1}^2.}$$
By Binet's explicit formula we have $F_{2n}=F_n L_n$, where Lucas numbers $L_n$ are defined through $L_0=2,L_1=1$ and $L_{n+2}=L_{n+1}+L_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2484884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Trigonometry Problems Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac{1}{2}$. The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.
My student left me with a list of questions just now (he also said that he wanted to be helped at school :P)
| \begin{align} \frac{\sin x}{\sin y} &= 3 \tag{1}\label{1} \\ \frac{\cos x}{\cos y} &= \frac{1}{2} \tag{2}\label{2} \end{align}
\begin{align} \frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} &=\frac{p}{q} \tag{3}\label{3} \end{align}
Find $p+q$.
\begin{align}
\frac{\sin 2x}{\sin 2y}
&=
\frac{2\sin x\cos x}{2\sin y\cos y}
=\frac32
\tag{4}\label{4}
\end{align}
\begin{align}
\frac{\sin x}{\sin y} &=
\frac{6\cos x}{\cos y} = 3
\tag{5}\label{5}
\end{align}
\begin{align}
\frac{\sin^2 x}{\sin^2 y} &=
\frac{36\cos^2 x}{\cos^2 y} = 9
\tag{6}\label{6}
\\
\frac{\sin^2 x+36\cos^2 x}{\sin^2 y+\cos^2 y}
&= 9
\tag{7}\label{7}
\\
35\cos^2 x
&= 8
\\
2\cos^2 x
&= \frac{16}{35}
\tag{8}\label{8}
\\
2\cos^2 x-1
&=
\frac{16}{35}-1
\tag{9}\label{9}
\\
\cos 2x&=-\frac{19}{35}
\tag{10}\label{10}
\end{align}
Similarly,
\begin{align}
\frac{\sin x}{6\sin y} &=
\frac{\cos x}{\cos y} = \frac12
\tag{11}\label{11}
\end{align}
\begin{align}
\frac{\sin^2 x}{36\sin^2 y} &=
\frac{\cos^2 x}{\cos^2 y} = \frac14
\tag{12}\label{12}
\\
\frac{\sin^2 x+\cos^2 x}{36\sin^2 y+\cos^2 y}
&= \frac14
\tag{13}\label{13}
\end{align}
\begin{align}
35\sin^2 y+1&=4
\tag{14}\label{14}
\\
\sin^2 y&=\frac{3}{35}
\tag{15}\label{15}
\\
1-2\sin^2y&=1-\frac{6}{35}
\tag{16}\label{16}
\\
\cos2y&=\frac{29}{35}
\tag{17}\label{17}
.
\end{align}
So,
\begin{align}
\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}
&=
\frac32-\frac{19}{29}=\frac{49}{58}=\frac{p}{q}.
\end{align}
$p+q=49+58=107$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2490794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\lim\limits_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}$
Calculate
$$\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}.$$
My Attempt :
$$=\lim_{x\to 3} \frac {\cos (\pi x)+1}{(x+5)^{\frac {1}{3}} -2}
=\lim_{x\to 3} \frac {1-(\pi x)/2! +1}{(5)^{1/3} + \dfrac {x}{3\cdot 5^{2/3}} -\dfrac {x^2}{45\cdot 5^{2/3}} -2}.$$
| $$\lim_{x\to 3} \dfrac {\cos (\pi x)+1}{(x+5)^{1/3} -2}=\frac{\pi^2}{2}\lim_{x\rightarrow3}\left(\frac{\sin^2\frac{\pi(x-3)}{2}}{\frac{\pi^2(x-3)^2}{4}}\cdot\frac{(x-3)^2}{\sqrt[3]{x+5}-2}\right)=$$
$$=\frac{\pi^2}{2}\lim_{x\rightarrow3}(x-3)\left(\sqrt[3]{(x+5)^2}+2\sqrt[3]{x+5}+4\right)=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Solving $a_{n+2}-6a_{n+1}+8a_n=27n^2+18$
Solve the recurrence relation
$$a_{n+2}-6a_{n+1}+8a_n=27n^2+18$$
I found the homogenous solutions which are $r=2,4$ meaning that
$$a_n=C_12^n+C_24^n$$. I'm not sure what to do after that.
| Write $a_{n+2} - 6a_{n+1} + 8a_n = (a_{n+2} - 4a_{n+1})- 2(a_{n+1} - 4a_n) = 27n^2+18$. This suggests that you put $b_n = a_n - 4a_{n-1}$, then you have: $b_{n+2}-2b_{n+1} = 27n^2 + 8$. So, you now have: $b_n = (b_n - 2b_{n-1})+ 2(b_{n-1}-2b_{n-2})+ 4(b_{n-2} - 2b_{n-3})+\cdots +2^{n-2}(b_2-2b_1) + 2^{n-1}b_1 = (27(n-2)^2+8)+2(27(n-3)^2+8)+4(27(n-4)^2+8)+\cdots+2^{n-2}((a_2-4a_1)-(a_1-4a_0))+ 2^{n-1}(a_1-4a_0)$. This sum is easy to evaluate. Repeat this trick again for the $a_n$ and you can solve it without using the characteristic equation. Its a bit long but is ...fun...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2491641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find an f(x) that fits specific critera (Polynomial Regression) I need a function (I don't care how messy it is) that when:
$f(1) = 1$ <--this has $1!$ different terms
$f(2) = a-b$ <--this has $2!$ different terms
$f(3) = a^2b-a^2c-b^2a+b^2c+c^2a-c^2b$ <--this has $3!$ different terms
$f(4) = a^3b^2c-a^3b^2d-a^3c^2b+a^3c^2d+a^3d^2b-a^3d^2c-b^3a^2c+b^3a^2d+b^3c^2a-b^3c^2d-b^3d^2a+b^3d^2c+c^3a^2b-c^3a^2d-c^3b^2a+c^3b^2d+c^3d^2a-c^3d^2b-d^3a^2b+d^3a^2c+d^3b^2a-d^3b^2c-d^3c^2a+d^3c^2b$ <--this has $4!$ different terms
$f(5) = a^4b^3c^2d-a^4b^3c^2e-a^4b^3d^2c+a^4b^3d^2e+a^4b^3e^2c-a^4b^3e^2d-a^4c^3b^2d+a^4c^3b^2e+a^4c^3d^2b-a^4c^3d^2e-a^4c^3e^2b+a^4c^3e^2d+a^4d^3b^2c-a^4d^3b^2e-a^4d^3c^2b+a^4d^3c^2e+a^4d^3e^2b-a^4d^3e^2c-a^4e^3b^2c+a^4e^3b^2d+a^4e^3c^2b-a^4e^3c^2d-a^4e^3d^2b+a^4e^3d^2c+b^4a^3c^2d-$ <--this would have $5!$ different terms (If I typed all 120 of them)
This function lists all of the unique combinations of the letters and the powers used. Hopefully you can see the pattern between the what is stated above, I just can't find a way to describe it.
This function would be a necessary part for working out a formula for 2-dimensional polynomial regression.
I do realize that there are multiple answers for this and any help for this would be much appreciated.
| Seems to be
$$
\eqalign{
& f\left( n \right) = \sum\limits_{\left\{ {j_1 , \cdots ,j_n } \right\} \in S\;even} {\left( {x_{\,j_{\,1} } - x_{\,j_{\,2} } } \right)x_{\,j_{\,3} } ^2 x_{\,j_{\,4} } ^3 \cdots x_{\,j_{\,n} } ^{n - 1} } = \cr
& = \sum\limits_{\left\{ {j_1 , \cdots ,j_n } \right\} \in S\;even} {\left( {x_{\,j_{\,1} } - x_{\,j_{\,2} } } \right)\prod\limits_{3\, \le \,k\, \le \,n} {x_{\,j_{\,k} } ^{k - 1} } } \quad |\;2 \le n \cr}
$$
where $S\, \text{even}$ denotes the set of the permutations of $\{1,2,\cdots,n\}$
which are even (even number of inversions).
That is, we denote $a,b,c, ...$ as $x_1,x_2,..$.
Then for example, in the case of $n=3$ we have $3!$ permutations, half of which are even and half odd. So the expression above becomes
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2495384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why does $3^{16} \times 7^{-6}$ become $\frac{3^{16}} {7^{6}}$? I was doing an exercise on exponents:
$$\begin{align}
\left(3^{-8} \times 7^3\right)^{-2} &= \left(3^{-8}\right)^{-2}\times \left(7^3\right)^{-2} \\
&= 3^{16} \times 7^{-6} \\
&= \frac{3^{16}} {7^{6}} \\
\end{align}$$
Why did $7^{-6}$ turn to $7^{6}$? More generally, why does a negative exponent turn positive when moved to the denominator? Would appreciate kindergarten language ;-D
| The question has thoroughly been answered but I do want to contribute with this table that I like.
Begin with this table:
\begin{array}{ccl}
7^5 & = & 7\cdot 7\cdot 7\cdot 7\cdot 7
\\
7^4 & = & 7\cdot 7\cdot 7\cdot 7
\\
7^3 & = & 7\cdot 7\cdot 7
\\
7^2 & = & 7\cdot 7
\\
7^1 & = & 7
\end{array}
Going up:
*
*The previous line is multiplied by $7$ on the right hand side.
*The exponent is increased by $1$ on the left hand side.
Going down:
*
*The previous line is divided by $7$ on the right hand side.
*The exponent is decreased by $1$ on the left hand side.
What would be a natural extension? To keep doing this pattern both up and down! And thus:
\begin{array}{ccl}
7^5 & = & 7\cdot 7\cdot 7\cdot 7\cdot 7 \phantom{\dfrac1{7^3}}
\\
7^4 & = & 7\cdot 7\cdot 7\cdot 7\phantom{\dfrac1{7^3}}
\\
7^3 & = & 7\cdot 7\cdot 7\phantom{\dfrac1{7^3}}
\\
7^2 & = & 7\cdot 7 \phantom{\dfrac1{7^3}}
\\
7^1 & = & 7\phantom{\dfrac1{7^3}}
\\
7^0 & = & \dfrac77 = 1\phantom{\dfrac1{7^3}}
\\
7^{-1} & = & \dfrac17 \phantom{\dfrac1{7^3}}
\\
7^{-2} & = & \dfrac{1/7}7 = \dfrac1{7^2}
\\
7^{-3} & = & \dfrac{1/7^2}7 = \dfrac1{7^3}
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2502263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 2
} |
Find sum of all positive real numbers $x$ such that $\sqrt[3]{2+x} + \sqrt[3]{2-x}$ becomes an integer.
Find sum of all positive real numbers $x$ such that $\sqrt[3]{2+x} + \sqrt[3]{2-x}$ becomes an integer.
I thought Euler's identity may help,so letting $a=\sqrt[3]{2+x}, b=\sqrt[3]{2-x},c=-y$ ($y \in \Bbb Z)$ we have : $a+b+c=0$ leading to :$(2+x)+(2-x)+(-y^3)=3\sqrt[3]{2+x} \sqrt[3]{2-x}(-y)$, but at this point I think sum of such $x's$ can't be found.
| $f(x)=\sqrt[3]{2+x} + \sqrt[3]{2-x}$
$$f'(x)=\frac{\sqrt[3]{2-x}^2-\sqrt[3]{x+2}^2}{3 \sqrt[3]{2-x}^2 \sqrt[3]{x+2}^2}$$
$f'(x)=0$ at $x=0$
as $f'(x)>0$ for $x<0$ and $f'(x)<0$ for $x>0$ then $x=0$ is a maximum
and $f(0)=2\sqrt[3]{2}$
$f(x)$ can be integer if $f(x)=2$ that is $x=\dfrac{10}{3 \sqrt{3}}$
or if $f(x)=1$ that is $x=\sqrt{5}$
So the sum of positive $x$ such that $f(x)$ is integer is
$\dfrac{10}{3 \sqrt{3}}+\sqrt{5}$
Hope this can be useful
Edit
Without derivatives
proof that
$$\sqrt[3]{2-x}+\sqrt[3]{x+2}\leq 2 \sqrt[3]{2},\;\forall x\in\mathbb{R}\quad(*)$$
$$\sqrt[3]{2-x}\leq 2 \sqrt[3]{2} - \sqrt[3]{x+2}$$
$$\left(\sqrt[3]{2-x}\right)^3\leq \left(2 \sqrt[3]{2} - \sqrt[3]{x+2}\right)^3$$
$$2-x\leq -x+14-12\sqrt[3]{4}\sqrt[3]{x+2}+6 \sqrt[3]{2} \sqrt[3]{(x+2)^2}$$
$$-12\leq -12\sqrt[3]{4}\sqrt[3]{x+2}+6 \sqrt[3]{2} \sqrt[3]{(x+2)^2}$$
$$\sqrt[3]{2} \sqrt[3]{(x+2)^2}-2\sqrt[3]{4}\sqrt[3]{x+2}+2\geq 0$$
Set $\sqrt[3]{x+2}=w$
$$\sqrt[3]{2} w^2-2\sqrt[3]{4}w+2\geq 0$$
which is verified for all $w\in\mathbb{R}$
and so is the $(*)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find a formula for $\sum\limits_{n=0}^N(-1)^n\frac{(2n+1)^3}{(2n+1)^4+4}$ I found this sum in the mathematial induction chapter of The art of Computer Programming and i have no idea how to solve it.
$\dfrac{1^3}{1^4+4}-\dfrac{3^3}{3^4+4} + ... +\dfrac{(-1)^n(2n+1)^3}{(2n+1)^4+4} $
I tried writing it as $\dfrac{1^3}{1*1^3+4}-\dfrac{3^3}{3*3^3+4} + ... +\dfrac{(-1)^n(2n+1)^3}{(2n+1)*(2n+1)^3+4} $
and then writing it as
$\dfrac{1}{1*1^3+4}-\dfrac{3+5}{3*3^3+4} +...+\dfrac{(-1)^n(((2n+1)^2-(2n+1)+1)+...+((2n+1)^2+(2n+1)-1))}{(2n+1)*(2n+1)^3+4} $
but did not know how to continue.
I also tried writing it as
$\dfrac{1}{1+\dfrac{4}{1^3}}-\dfrac{1}{3+\dfrac{4}{3^3}} + ... +\dfrac{(-1)^n}{2n+1+\dfrac{4}{(2n+1)^3}} $ but without succes.
|
Keyword: Concatenation.
First note that, for every $x$, $$\frac{4x^3}{x^4+4}=\frac{4x^3}{(x^2+2)^2-4x^2}=\frac{x^2}{(x-1)^2+1}-\frac{x^2}{(x+1)^2+1}$$ hence the partial sum $S_{2N+1}$ of $2N+1$ terms is such that $$4S_{2N+1}=\frac45+\sum_{n=1}^N\left(\frac{4(4n+1)^3}{(4n+1)^4+4}-\frac{4(4n-1)^3}{(4n-1)^4+4}\right)$$ that is, $$4S_{2N+1}=\frac45+U_N-V_N$$
where $$U_N=\sum_{n=1}^N\frac{(4n+1)^2}{(4n)^2+1}+\frac{(4n-1)^2}{(4n)^2+1}=\sum_{n=1}^N\frac{2((4n)^2+1)}{(4n)^2+1}=2N$$ and $$V_N=\sum_{n=1}^N\frac{(4n+1)^2}{(4n+2)^2+1}+\frac{(4n-1)^2}{(4n-2)^2+1}$$ Thus, $$V_N=\frac95-\frac{(4N+3)^2}{(4N+2)^2+1}+\sum_{n=1}^N\frac{(4n+1)^2}{(4n+2)^2+1}+\frac{(4n+3)^2}{(4n+2)^2+1}$$ that is, $$V_N=\frac95-\frac{(4N+3)^2}{(4N+2)^2+1}+\sum_{n=1}^N\frac{2((4n+2)^2+1)}{(4n+2)^2+1}=\frac95-\frac{(4N+3)^2}{(4N+2)^2+1}+2N$$ Coming back to $S_{2N+1}$, one gets $$4S_{2N+1}=\frac45-\frac95+\frac{(4N+3)^2}{(4N+2)^2+1}=\frac{8N+4}{(4N+2)^2+1}$$ hence, finally,
$$S_{2N+1}=\frac{2N+1}{4(2N+1)^2+1}$$
The even numbered sums $S_{2N}$ are solvable by a similar treatment.
Fun fact: $$\lim_{N\to\infty}S_N=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2503245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Derivatives in $\mathbb{R}^n$
Question: Let $~f : \mathbb{R}^n \setminus \{0\} \to \mathbb{R}^n$, $x \mapsto \frac{x}{x_1^2 + \dots + x_n^2}$.
Show that $~f$ is differentiable at every point $x \in \mathbb{R}^n \setminus \{0\}$ and compute the Jacobian. Then
provide evidence of whether it is possible to extend $f$ to a
continuous function on $\mathbb{R}^n$.
What I have so far is that we can let $\mathbf{f} = (f_1, \dots, f_n)$, where
$$f_1 = \frac{x_1}{x_1^2 + \dots + x_n^2},~~f_2 = \frac{x_2}{x_1^2 + \dots + x_n^2},\dots,~~f_n = \frac{x_n}{x_1^2 + \dots + x_n^2}.$$
Since these are rational functions the partials exist, and from $(x_1, \dots, x_n) \neq (0, \dots, 0)$ we have that $x_1^2 + \dots + x_n^2 \neq0$. Meaning, all functions $\mathbf{f}$ are also continuous (not positive this exactly holds?). Hence, we have partials of form
$$\begin{align*}
\frac{\partial f_1}{\partial x_1} &= \frac{(x_1^2 + \dots +x_n^2) - 2x_1(x_1)}{(x_1^2 + \dots + x_n^2)^2} = \frac{x_1^2 + \dots +x_n^2 - 2x_1^2}{(x_1^2 + \dots + x_n^2)^2} = \frac{x_2^2 + \dots +x_n^2 - x_1^2}{(x_1^2 + \dots + x_n^2)^2} \\
\frac{\partial f_1}{\partial x_2} &= \frac{-2x_2(x_1)}{(x_1^2 + \dots + x_n^2)^2} = \frac{-2x_1x_2}{(x_1^2 + \dots + x_n^2)^2} \\
& \vdots \\
\frac{\partial f_1}{\partial x_n} &= \frac{-2x_n(x_1)}{(x_1^2 + \dots + x_n^2)^2} = \frac{-2x_1x_n}{(x_1^2 + \dots + x_n^2)^2},
\end{align*}
$$
and so on for each $f_1, \dots, f_n$.
Then we can construct the Jacobian such that
$$D_{\mathbf{f}}(\mathbf{x}) =
\begin{pmatrix}
\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \dots & \frac{\partial f_1}{\partial x_n} \\
\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \dots & \frac{\partial f_2}{\partial x_n} \\
\vdots & \vdots & \ddots & \vdots \\
\frac{\partial f_n}{\partial x_1} & \frac{\partial f_n}{\partial x_2} & \dots &\frac{\partial f_n}{\partial x_n}
\end{pmatrix}.$$
Is this correct up until this point? How do I extend this $f$ to a continuous function on $\mathbb{R}^{n}$ (if possible). It seems we already have shown for $\mathbb{R}^n$?
| Continuous extension is not possible. For suppose $f(0)=\alpha$, then $f(x)\rightarrow\alpha$ as $x\rightarrow 0$. But $|f(x)|=\left|\dfrac{x}{|x|^{2}}\right|=\dfrac{1}{|x|}\rightarrow\infty$ as $x\rightarrow 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2505342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $a + b = 20$, then what is the maximum value of $ab^2$? Knowing that
$$a + b = 20$$
and $a,b \in \mathbb{Z}_{+}$
What is the maximum value that $ab^2$ can take?
If $b$ is even then so is $a$. Hence, let $b = 2k$ and $a = 20 - 2k$, where $k \in \{1, 2, 3, \cdots, 9\}$
$$ab^2 = (20 - 2k)(2k)^2 = 80k^2 - 8k^3$$
The value of $k$ which maximizes $ab^2$ is therefore $k = 7$ which gives $ab^2 = 1176$.
However this only shows that this is the maximum for even values of $a$ and $b$. It turns out that the maximum of $ab^2$ is when both are odd.
How do I solve this question since my approach is clearly not very elegant?
My solution:
$$a = 20 - b \Longrightarrow ab^2 = (20 - b)b^2$$
Differentiate to find the maximum value of $b$
$$40b - 3b^2 = b(40-3b) = 0$$
$$\Longrightarrow b = \left\lfloor \frac{40}{3} \right\rfloor$$ $$\Longrightarrow b = \left\lceil \frac{40}{3} \right\rceil$$
Therefore $a, b = 7, 13$ or $a, b = 6, 14$. Inspecting these $2$ cases yields the former.
So the maximum value $ab^2$ can take is $1183$.
| If $a$ and $b$ are required to be positive, but not integers, then the AM-GM says that
$$
\frac{a+\frac b2+\frac b2}3\geq\sqrt[3]{a\cdot\frac b2\cdot\frac b2}
$$
with equality iff $a=\frac b2=\frac b2$. For $a,b$ integers, you just have to try the integers closest to this equality (i.e. $a=6$ and $a=7$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Checking that $2+\sqrt3$ is a cube root of $26 + 15 \sqrt3$ I am trying to show that $$\sqrt[3]{26 + 15 \sqrt{3}} = 2 + \sqrt{3}$$
My idea is to find the cube roots of $z=26 + 15\sqrt{3}$ via De Moivre's formula.
So $r=\sqrt{26^2 + (15\sqrt{3})^2} = \sqrt{1351}$, and $\theta = \tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)$.
Thus, $$z^{1/3} = (1351)^{2/3} \left(\cos \left[ \frac{\tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)}{3}+ \frac{2k\pi}{3} \right] + i\sin\left[ \frac{\tan^{-1} \big(\frac{15\sqrt{3}}{26} \big)}{3} + \frac{2k\pi} {3}\right]\right)$$ for $k = 0,1,2.$
This does not seem like I am heading in the right direction. Any clues or hints would be greatly appreciated. Thank you!
| You can just cube it, like the comments have suggested$$(2+\sqrt3)^3=8+12\sqrt3+18+3\sqrt3=26+15\sqrt3$$
If you want to start off with the left-hand side and gradually work your way to the right-hand side, the easiest way is to assume$$\sqrt[3]{26+15\sqrt3}=a+b\sqrt3$$Cube both sides via the binomial theorem and compare the coefficients.$$\begin{align*} & a^3+9ab^2=26\\ & 3a^2b+3b^3=15\end{align*}$$Solving gives you $a,b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2506923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Equation of tangents from external point to a circle I have point $(p,q)$ and circle $ x^2 + y^2 + 2gx + 2fy + c = 0$.
I'm aware you could do substitute in $y = mx+c$ and solve a quadratic or you could use $ y-(-f) = m(x-(-g)) \sqrt{1+m^2}$ where m is the slope of the line and $(-g, - f)$ is the coordinates of the centre. But are there other ways? perhaps using calculus/vectors/complex numbers
maybe take point $(4,-5)$ and circle $ x^2 + y^2 -6x + -4y + 4 = 0$. as example
| $x^2 + y^2 -6x +4y + 4 = 0$
centre $C(3;\;-2)$ radius $r=3$
$P(4;\;-5)$
Write the equation of the generic line passing through $P$
$y+5=m(x-4)$
adjust it in normal form
$\mathscr{F}:mx -y -4m-5=0$
Calculate the distance $d(m)$ from $\mathscr{F}$ to the centre $C$ of the circle
$$d(m)=\frac{|3m+2-4m-5|}{\sqrt{m^2+1}}$$
The line $\mathscr{F}$ is tangent to the circle if the distance $d(m)$ is equal to the radius
$$d(m)=3\to \frac{|-3-m|}{\sqrt{m^2+1}}=3$$
square both sides
$$9+6m+m^2=9m^2+9\to m_1=0;\;m_2=\frac34$$
the two tangents are
$y=-5;\;y=\frac{3}{4}x-8$
Second method
Works with any conic
Find the equation of the polar line of $P$ wrt the circle
Substitute
$$x^2\to x_px;\;y^2\to y_py;\;x\to\frac{x+x_p}{2};\;y\to \frac{y+y_p}{2}$$
$$4x-5y-6\,\frac{x+4}{2}+4\,\frac{y-5}{2}+4=0$$
$p:x-3y=18$
The intersection points of this line with the circle are the tangent points of the tangents from $P$ to the circle, namely $K(3,-5);\;H(4.8,-4.4)$
Hope this helps
$$...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2507739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Maximum and minimum values of function
For any number $c$ we let $f_c(x)$ be the smaller of the two numbers
$(x-c)^2$ and $(x-c-2)^2$. Then we define $g(c)=\int_0^1f_c(x)dx$.
Find the maximum and minimum values of $g(c)$ if $-2\le c \le 2$.
So the maximum/minimum values are either $g(\pm2)$ (the endpoints) or $g(c)$ where $g'(c)=0$.
$$g(2)=\int_0^1f_2(x)dx=\int_0^1\min((x-2)^2,(x-4)^2)dx\\=\int_0^1(x-2)^2dx=\frac{(1-2)^3}{3}-\frac{(0-2)^3}{3}=\frac{7}{3}$$
The last step I was thinking was because on the interval $[0,1]$, $|x-2|$ is always smaller than $|x-4|$.
In the same way, $$g(-2)=\int_0^1f_{-2}(x)dx=\int_0^1\min((x+2)^2,x^2)dx\\=\int_0^1x^2dx=\frac{1}{3}$$
So $7$ thirds and $1$ third for the endpoints. When the derivative is zero, I am having trouble. I am struggling to understand how the derivative and antiderivative of $g$ and $f_c$ would operate. Thanks.
Problem Source: James Stewart, Calculus, Integrals Chapter
| It can be shown that $g$ has a symmetry in $c=-\frac 12$.
$\displaystyle g(c)=\int_0^1 \min\left((x-c)^2,(x-c-2)^2\right)\mathop{dx}\quad$ let substitute $u=1-x$
$\displaystyle\phantom{g(c)}=\int_1^0 \min\left((1-u-c)^2,(1-u-c-2)^2\right)(-\mathop{du})\\\displaystyle\phantom{g(c)}=\int_0^1 \min\left((u+c-1)^2,(u+c+1)^2\right)\mathop{du}=g(-c-1)$
So we need only to study $g$ on the interval $[-\frac 12,2]$.
$f_c(x)=\min((x-c)^2,(x-c)^2-4(x-c)+4)=(x-c)^2+4\min(0,1+c-x)$
Let's have $\begin{cases}\displaystyle g_0(c)=\int_0^1(x-c)^2\mathop{dx}=c^2-c+\frac 13\\\displaystyle g_1(c)=4\int_{c+1}^1 (1+c-x)\mathop{dx}=-2c^2\end{cases}$
For $c\in[0,2]\implies g(c)=g_0(c)$
with a minimum in $g(\frac 12)=\frac 1{12}$ and a maximum $g(2)=\frac 73$.
For $c\in[-\frac 12,0]\implies g(c)=g_0(c)+g_1(c)$
with a maximum $g(-\frac 12)=\frac 7{12}$ and a minimum $g(0)=\frac 13$.
So overall the extrema of $g$ for $c\in[-2,2]$ are
*
*maximum $g(2)=\frac 73$
*minimum $g(-\frac 32)=g(\frac 12)=\frac 1{12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2508766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Minimum Value of expression 2 What is the minimum value of $$(2x+3y)(8/x + 3/y)$$ when $x$ and $y$ are positive?
I have tried expanding the equation but I did not get anything
I got : $$25 + \frac{6(4y^2 +x^2)}{xy}$$ but I don't know how to answer after this.
I have also tried AM-GM but I couldn't manipulate the variables to cancel out and get the inequality.
This is from Sipnayan 2017
| You were on the right track:
$$\begin{align}(2x+3y)\left(\frac{8}{x} + \frac{3}{y}\right) &= 25+24 \frac{y}{x}+6\frac{x}{y}\\
&=25+12\left(\frac{2y}{x}+\frac{x}{2y}\right) \\
&\geq 25+12 \times 2 \\
&=49\end{align}$$
and notice that $x=2,y=1$ achieves this bound, so the minimum value is $49$.
The bound $\frac{2y}{x} + \frac{x}{2y} \geq 2$ can be proved from AM-GM, or from letting $t = \frac{2y}{x}$ and considering the function $f(t) = t + \frac{1}{t}$ and using calculus.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2509102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How to solve recurrence relations using generating functions that do not start at $a_0$? For example,
$$
a_{n+1} - 2a_n = 2, \\ \textrm{for } n \geq 1, a_1 = 1
$$
I tried using
$$
\sum_{n=1}^\infty a_{n+1}x^{n+1} - \sum_{n=1}^\infty 2a_n x^{n+1} = \sum_{n=1}^\infty 2x^{n+1}
$$
if we let $f(x) = \sum_{n=1}^\infty a_n x^n$
$$
f(x) - a_1 x - 2x f(x) = \frac{2x^2}{1-x}
$$
and I get
$$
f(x) = \frac{x^2 + x}{(1-x)(1-2x)} = \frac{A}{1-x} + \frac{B}{1-2x}
$$
but the partial fraction decomposition is not giving me a solution. Also, using method of undetermined coefficients I got the right answer which is
$$
a_n = 2^{n-1}3 - 2
$$
edit1: typos
| Should the fraction on the RHS be
$$ \frac{2x^2}{1-x}\text{?}$$
edit: You should end up with
$$f(x) = \frac{2x^2}{(1-2x)(1-x)} + \frac{x}{(1-2x)}$$
which you can simplify and do partial fractions with and get the desired answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2510261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the simplest way to compute :$\int^{\pi}_0\bigl(\frac{\sin(x)}{5-4\cos(x)}\bigr)^2dx=\frac{\pi}{24}$ I am trying to compute the following integral.
$$\int^{\pi}_0\biggl(\frac{\sin(x)}{5-4\cos(x)}\biggr)^2dx$$
My attempt was to expand the integrand and make use of the standard change of variables $t =\tan x/2$. But it turns out to be a lengthy and exhausting computations. I can put all details here since it is not pleasant.
at the end I got the answer:
$$\int^{\pi}_0\biggl(\frac{\sin(x)}{5-4\cos(x)}\biggr)^2dx =\frac{\pi}{24}$$
Maybe it is not correct so do not trust this result at 100%
I would like to know if there is an easiest way or trick that quickly leads to the answer?
| Tha tangent half-angle substitution is not so bad since using $t=\tan(\frac x2)$
$$I=\int\biggl(\frac{\sin(x)}{5-4\cos(x)}\biggr)^2dx=\int\frac{8 t^2}{\left(t^2+1\right) \left(9 t^2+1\right)^2}\,dt$$ Using partial fractions
$$\frac{8 t^2}{\left(t^2+1\right) \left(9 t^2+1\right)^2}=\frac{9}{8 \left(9 t^2+1\right)}-\frac{1}{\left(9 t^2+1\right)^2}-\frac{1}{8
\left(t^2+1\right)}$$ The second term needs to be integrated by parts. So, we have
$$\int\frac{dt}{\left(9 t^2+1\right)}=\frac{1}{3} \tan ^{-1}(3 t)$$
$$\int\frac{dt}{\left(9 t^2+1\right)^2}=\frac{t}{2 \left(9 t^2+1\right)}+\frac{1}{6} \tan ^{-1}(3 t)$$
$$\int\frac{dt}{
\left(t^2+1\right)}=\tan ^{-1}( t)$$ making
$$I=-\frac{t}{2 \left(9 t^2+1\right)}-\frac{1}{8} \tan ^{-1}(t)+\frac{5}{24} \tan ^{-1}(3 t)$$ Integrating between $0$ and $\infty$, the first term is $0$ and you are left with $$\frac 5 {24}\frac \pi 2-\frac 1 {8}\frac \pi 2=\frac \pi {24}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Partial fractions decomposition of ${\frac{2x}{(x+2)^2}}$ Express in partial fraction form: $\displaystyle{\frac{2x}{(x+2)^2}}$
I think is $\displaystyle{\frac{2x}{(x+2)^{2}} = \frac{A}{x+2}+\frac{B}{(x+2)^2}}$
However when identifying $A$ and $B$, I'm not sure how to calculate A.
E.g. $$2x = A\cdot (x+2) + B$$
Substitute $x=-2$
$2\cdot(-2)$ = $A\cdot (2-2) +B$
$-4 = B$
In other questions there is always another factor to multiply by at this stage.
| For instance, you can multiply both sides by $(x+2)^2$ in
$$\frac{2x}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}$$
then
$$(x+2)^2\frac{2x}{(x+2)^2}=(x+2)^2\frac{A}{x+2}+(x+2)^2\frac{B}{(x+2)^2}$$
now simplify
$$(x+2)^2\frac{2x}{(x+2)^2}=(x+2)\cdot A+(x+2)^2\frac{B}{(x+2)^2}$$
and you can take the limit as $x\to 2$ both sides to show that:
$$ B = \underset{x\to -2}{\lim}{\ (x+2)^{2}\frac{2x}{(x+2)^2}}=-4. $$
for $A$ take before the limit the derivative w.r.t $x$ in both sides to get
$$ A = \underset{x\to -2}{\lim}{\ \frac{d}{dx}{2x}=2. }$$
Recall that, if $\ \displaystyle{f(x)=\frac{p(x)}{(x-r)^{k}}}$ and $(x-r)$ doesn't divide $p(x)$ you have:
$$f(x)=\frac{p(x)}{(x-r)^{k}}=\sum_{s=1}^{k}\frac{A_{s}}{(x-r)^{s}}$$
if you multiply by $(x-r)^{k}$ in both sides you will have:
$$ (x-r)^{k}f(x)=p(x)=\sum_{s=1}^{k}A_{s}\cdot(x-r)^{k-s}$$
and from that you have:
$$ A_{k}=\underset{x\to r}{\lim}{\ (x-r)^{k}f(x)} $$
and the others
$$A_{i}=\frac{1}{(k-i)!}\underset{x\to r}{\lim}{\ \frac{d^{k-i}}{dx^{k-i}}\left((x-r)^{k}f(x)\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2513766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 4
} |
Recurrence relation $a_{n} = 2n(a_{n-1} + 3^n(n!)$, $ a_{0} = 1$ I am facing a problem with solving this recurrence relations.
$a_{n} = 2n(a_{n-1} + 3^n(n!))$, $ a_{0} = 1$
This is what I got so far.
$a_{n}^c = 2n(a_{n-1}^c = 2n(2(n-1)a_{n-2}^c = 2^2(n)(n-1)a_{n-2}^c = 2^2 n(n-1)(2(n-2)a_{n-3}) = 2^3 n(n-1)(n-2) = .... 2^n(n!)(a_{0}^c) = c_{1}2^n n!$
| Consider making use of the exponential generating function
$$
\sum_{n=0}^\infty a_n \frac{x^n}{n!}.
$$
Multiply your relation with $x^n/n!$ and sum over all $n \geq 1$:
$$
\sum_{n=1}^\infty a_n \frac{x^n}{n!} = \left(\sum_{n=1}^\infty 2na_{n-1}\frac{x^n}{n!} + \sum_{n=1}^\infty 2n3^nn! \frac{x^n}{n!}\right)\\
= 2x\sum_{n=1}^\infty a_{n-1} \frac{x^{n-1}}{(n-1)!} + 2\sum_{n=1}^\infty n(3x)^n \\
= 2x\sum_{n=1}^\infty a_n \frac{x^n}{n!} + 2x + \frac{6x}{(1-3x)^2}.
$$
Note that we made use of the initial condition $a_0 = 1$ in the above. Rearranging the terms we get that
$$
\sum_{n=1}^\infty a_n\frac{x^n}{n!} = \frac{2x}{1-2x} + \frac{6x}{(1-2x)(1-3x)^2}
$$
Using partial fractions we may write
$$
\frac{6x}{(1-2x)(1-3x)^2} = 6x\left(\frac{3}{(1-3x)^2} + \frac{4}{1-2x}- \frac{6}{1-3x}\right) \\
= 6\sum_{n=0}^\infty n 3^nx^n + 24x\sum_{n=0}^\infty2^nx^n - 36x\sum_{n=0}^\infty3^nx^n \\
= \sum_{n=1}^\infty 6(3^n(n-2)+2^{n+1})x^n.
$$
The other terms is
$$
\frac{2x}{1-2x} = 2x\sum_{n=0}^\infty 2^nx^n = \sum_{n=0}^\infty 2^{n+1}x^{n+1} = \sum_{n=1}^\infty 2^n x^n
$$
Collecting everything we finally get that
$$
a_n = n!\left(13\cdot2^n + 2\cdot3^{n+1}(n-2)\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2514112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show $ f(m) = \left( 1 - \frac{1}{m^2} \right)^{m(m-1)/2} $ is lower-bounded by $\frac{1}{2}$ for $m \ge 1$ It is also possible to assume $m$ is an integer if this is useful.
One way I can see to solve the problem is to show
(1) $f(m)$ is monotonically decreasing, which can be done either by showing $f'(m) \leq 0$ or showing $f(m+1) \leq f(m)$ for all $m \ge 1$ .
(2) The limit as $m \to \infty$ of $f(m)$ is $\frac{1}{\sqrt{e}} > \frac{1}{2}$.
Showing $(2)$ is simple using $\lim_{n\to \infty} \left(1 + \frac{1}{n} \right)^n = e$. Showing $(1)$ is quite annoying using the derivative, I've reduced showing $(1)$ using the other approach to showing $\left(1 - \frac{1}{(m+1)^2} \right)^{m+1} \leq \left(1 - \frac{1}{m^2} \right)^{m-1}$.
| $$
\begin{align}
\frac{\left(1-\frac1{m^2}\right)^{m-1}}{\left(1-\frac1{(m+1)^2}\right)^{m+1}}
&=\left(\frac{m^2}{m^2-1}\frac{(m+1)^2}{(m+1)^2-1}\right)\left(\frac{m^2-1}{m^2}\frac{(m+1)^2}{(m+1)^2-1}\right)^m\tag1\\
&=\left(\frac{m(m+1)}{(m+2)(m-1)}\right)\left(\frac{(m-1)(m+1)^3}{m^3(m+2)}\right)^m\tag2\\
&=\left(\frac{m(m+1)}{(m+2)(m-1)}\right)\left(1-\frac{2m+1}{m^3(m+2)}\right)^m\tag3\\
&\ge\left(\frac{m(m+1)}{(m+2)(m-1)}\right)\left(1-\frac{2m+1}{m^2(m+2)}\right)\tag4\\
&=\left(\frac{m(m+1)}{(m+2)(m-1)}\right)\left(\frac{(m-1)(m^2+3m+1)}{m^2(m+2)}\right)\tag5\\
&=\frac{(m+1)(m^2+3m+1)}{m(m+2)^2}\tag6\\
&=\frac{m^3+4m^2+4m+1}{m^3+4m^2+4m}\tag7\\[9pt]
&\gt1\tag8
\end{align}
$$
Explanation:
$(1)$: move a factor from the numerator and denominator to the front
$(2)$: factor
$(3)$: write second factor as $1-x$
$(4)$: Bernoulli's Inequality
$(5)$: factor
$(6)$: combine factors
$(7)$: multiply
$(8)$: numerator is $1$ greater than the denominator
Thus, the sequence is decreasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Area of a square inside a square created by connecting point-opposite midpoint
Square $ABCD$ has area $1cm^2$ and sides of $1cm$ each.
$H, F, E, G$ are the midpoints of sides $AD, DC, CB, BA$ respectively.
What will the area of the square formed in the middle be?
I know that this problem can be solved by trigonometry by using Area of triangle ($\frac{1}{2}ab\sin{c}$) but,
is there another method or visual proof?
| Here's a solution using analytic geometry which doesn't require any particular insight:
Set up a coordinate system such that $B = (0, 0)$ and $C = (1, 0)$. Then the equation of line $BH$ is $y - 2x = 0$, and the equation of line $CG$ is $2y + x = 1$. Therefore, their intersection is $(\frac{1}{5}, \frac{2}{5})$. Similarly, the equation of line $ED$ is $y - 2x = -1$, so the intersection of lines $ED$ and $CG$ is $(\frac{3}{5}, \frac{1}{5})$. This gives that the side of the inner square is the distance between these two points, which is
$$\sqrt{\left(\frac{1}{5} - \frac{3}{5}\right)^2 + \left(\frac{2}{5} - \frac{1}{5}\right)^2} = \frac{1}{\sqrt{5}}.$$
Therefore, the area of the square is the square of this length, or $\frac{1}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2518341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "61",
"answer_count": 15,
"answer_id": 2
} |
A small trouble understanding the AMGM inequality usage for some basic minimization problems? I'm reading Byrne's "A first course in optimization". After the presentation and proof of the AMGM inequality, the author gives two examples:
Here's the first:
I guess I understand what was done, it seems he writes in a convenient form with the AMGM inequality:
$$\left(\frac{3\cdot12}{x} \cdot\frac{3\cdot18}{y} \cdot 3\cdot xy\right)^{\frac{1}{3}}\leq \frac{1 }{3} \left( \frac{3\cdot12}{x} +\frac{3\cdot18}{y} + 3\cdot xy\right)$$
Which in turn, yields:
$$\left(3^3\cdot 12 \cdot 18\right)^{\frac{1}{3}}=\left(3^3 2^3 3^3\right)^{\frac{1}{3}}=18\leq \frac{12}{x} +\frac{18}{y} + xy$$
From here, I know that for all possible choices of $x,y$, then $18$ is the smaller value, but I'm not sure on how to find the actual $x,y$. Given that he mentions that the smaller value for $\frac{1}{3} f(x,y)$ is $6$, then I guess he did the following:
$$\frac{1}{3} f(x,y)+\frac{1}{3} f(x,y)+\frac{1}{3} f(x,y) \leq 6+6+6=18$$
$$\overbrace{\left(\frac{4}{x} +\frac{6}{y} + \frac{xy}{3}\right)}^{= \;6}+\overbrace{\left(\frac{4}{x} +\frac{6}{y} + \frac{xy}{3}\right)}^{=\;6}+\overbrace{\left(\frac{4}{x} +\frac{6}{y} + \frac{xy}{3}\right)}^{=\;6}\leq 6+6+6=18$$
$$\overbrace{\left(\frac{4}{x} +\frac{4}{x} + \frac{4}{x}\right)}^{= \;6}+\overbrace{\left(\frac{6}{y} +\frac{6}{y} + \frac{6}{y}\right)}^{= \;6}+\overbrace{\left(\frac{xy}{3} +\frac{xy}{3} + \frac{xy}{3}\right)}^{= \;6}=\frac{12}{x} +\frac{18}{y} + xy\leq 18$$
And it seems that solving each of them individually yields the desired result, I am confused at why reorganizing $3$ sums whose total is $6$ in this way (supposing it is what actually was done) yields exactly the the minimal solution.
My doubt in the following example may stem from the same doubt I had before:
Here, he talks about a "constant sum" which is perhaps something very similar to what was done before, but I don't understand what is this "constant sum" nor why it is needed, I see that the second centered equation is the same as the first but I don't understand the workings of this "constant sum". I have tried to rewrite the equation in several other ways (on paper) to check if I could uncover something but I couldn't. Could you clarify?
| For the first :
I'm not sure on how to find the actual $x,y$
"(The smallest value occurs when the three terms are equal. Therefore,) each is equal to $6$" means that $\frac{12}{x}=6,\frac{18}{y}=6$ and $xy=6$.
For the second :
I don't understand what is this "constant sum" nor why it is needed
"The terms $x,y$ and $72-3x-4y$ do not have a constant sum" means that $x+y+(72-3x-4y)=72-2x-3y$ is not a constant.
"but the terms $3x, 4y$ and $72-3x-4y$ do have a constant sum" means that $3x+4y+(72-3x-4y)=72$ is a constant.
It is needed to have
$$\small\frac{1}{12}(3x)(4y)(72-3x-4y)\le \frac{1}{12}\times\left(\frac{3x+4y+(72-3x-4y)}{3}\right)^3=\frac{1}{12}\times 24^3\ \ (\text{$=$ a constant})$$
The equality occurs when $3x=24,4y=24$ and $72-3x-4y=24$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve a quartic equation using trigonometric power identitiy? According to this post, we can solve a cubic equation
$$t^3+pt+q=0$$
by the trigonometric identity
$$4\cos^3\theta-3\cos\theta-\cos3\theta=0$$
So I've tried to solve the quartic equation
$$t^4+pt^2+qt+r=0$$
using the identity
$$8\cos^4\theta-4\cos2\theta-\cos4\theta-3=0$$
$$8\cos^4\theta-4(2\cos^2\theta-1)-\cos3\theta\cos\theta+\sin3\theta\sin\theta-3=0$$
$$8\cos^4\theta-8\cos^2\theta-\cos3\theta\cos\theta+1+\sin3\theta\sin\theta=0$$
If we let $t:=A\cos\theta$, then we have
$$A^4\cos^4\theta+A^2p\cos^2\theta+Aq\cos\theta+r=0$$
or
$$8\cos^4\theta+\frac{8p}{A^2}\cos^2\theta+\frac{8q}{A^3}\cos\theta+\frac{8r}{A^4}=0$$
now $\frac{8p}{A^2}=-8$ implies that $A=\sqrt{-p}$ and we need to find $\theta$ in the following system
$$\begin{cases}\cos3\theta=-\frac{8q}{A^3}\\1+\sin3\theta\sin\theta=\frac{8r}{A^4}\end{cases}$$
What would you think on solving this system? Is it possible to solve this system? I couldn't find anything on the web about the solving a quartic equation by trigonometric power identity, so I don't know if this way works ...
| The second way to get solutions in trigonometric form (without using trigonometric power indentity). Represent root's of quadratic equation $$ax^2+bx+c=0$$ in trigonometric form
$$x=cos\left(\frac12 arccos\left(\frac {b^2-4ac-2a^2}{2a^2}\right)+\frac {\pi n} 2 \right)-\frac {b}{2a}$$ where $n=0,1$. Then rewrite $$t^4+pt^2+qt+r=0$$ in form of $$\left(t^2+\alpha t+\beta \right)\left(t^2-\alpha t+\frac {r}{\beta} \right)$$ Now, we can find $\alpha$, $\beta$, solving system of equations $$\begin{cases} \alpha^2(\alpha^2+a)^2-4c\alpha-b^2=0\\ \beta^2+\frac {b}{\alpha} \beta-c=0 \end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $ \prod_{n=3}^\infty ( 1 - \binom{n}{2}^{-1} )$
$$ \prod_{n=3}^\infty \left ( 1 - \frac{1}{\binom{n}{2}} \right)$$
Expanding it out doesn't give any good patterns, I'm totally stuck.
The answer is 1/3 [according to WolframAlpha:
https://www.wolframalpha.com/input/?i=product+from+3+to+infinity+(1+-+1%2F((n)+comb+2)+)
| Note that we have
$$\begin{align}
\prod_{n=3}^N \left(1-\frac{1}{\binom{n}{2}}\right)&=\prod_{n=3}^N \left(1-\frac{2}{n(n-1)}\right)\\\\
&=\prod_{n=3}^N\frac{n^2-n-2}{n(n-1)}\\\\
&=\prod_{n=3}^N \frac{n+1}{n}\frac{n-2}{n-1}\\\\
&=\prod_{n=3}^N \frac{n+1}{n}\prod_{n=3}^N \frac{n-2}{n-1}\\\\
&=\frac{N+1}{3}\frac{1}{N-1}
\end{align}$$
Now let $N\to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2523437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Exponents Simplification I need assistance with simplifying this rational expression.
The expression is $$\frac{1}{\sqrt{29 + 12\sqrt{5}}}.$$
The answer, for reference, is $$\frac{2\sqrt{5} - 3}{11}.$$
Thank you!
| Hint:
$$ 29 + 12\sqrt{5} = 3^2 + 2^2\cdot 5 + 2\cdot 3\cdot 2\sqrt{5} = (3 + 2\sqrt{5})^2 $$
Thus
$$ \sqrt{29+12\sqrt{5}} = 3 + 2\sqrt{5} $$
If you can't easily see this, here's an algebraic approach: Suppose $\exists a,b \in \mathbb{Z}$ such that
$$ \sqrt{29+12\sqrt{5}} = a + b\sqrt{5} $$
Squaring both sides gives
$$ 29 + 12\sqrt{5} = a^2 + 5b^2 + 2ab\sqrt{5} $$
Since $a$ and $b$ are integers, it must be true that
$$ a^2 + 5b^2 = 29, \ 2ab = 12 $$
Solving this gives $(a,b) = (3,2), \ (-3,-2)$. We pick the solution pair that satisfies $a + b\sqrt{5} > 0$, which is $(3,2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2525637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\int_{-\infty}^{\infty} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$ is convergent.
Show that $$\int_{-\infty}^{\infty} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent.
$$\int_{0}^{\infty} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent by comparison test with $\int_{0}^{\infty} \dfrac{1}{x^4} dx$.
How do i show that $$\int_{-\infty}^{0} \dfrac{x^2}{(x^2+1)^2(x^2+2x+2)} \ dx$$ is convergent?
| Another way:
$$
0\le\frac{x^2}{(x^2+1)^2(\underbrace{(x+1)^2+1)}_{\ge 1}}\le\frac{x^2}{(x^2+1)^2}\le\frac{x^2+1}{(x^2+1)^2}=\frac{1}{x^2+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
To evaluate $\lim_{x \to 0^-}({\frac{\tan x}{x}})^\frac{1}{x^3}$ Evaluate $$\lim_{x \to 0^-}({\frac{\tan x}{x}})^\frac{1}{x^3}$$
I tried taking log on both sides and then using L'Hospital rule but its giving complex results.Are there any simpler methods to approach this?
| Equation $(4)$ in this answer says that
$$
\lim_{x\to0}\frac{\tan(x)-x}{x^3}=\frac13
$$
Therefore,
$$
\begin{align}
\left(\frac{\tan(x)}x\right)^{1/x^3}
&=\left(1+\frac{x^2}3+o\!\left(x^2\right)\right)^{1/x^3}\\
&=\left(\left(1+\frac{x^2}3+o\!\left(x^2\right)\right)^{1/x^2}\right)^{1/x}
\end{align}
$$
as $x\to0$, we can make $\left(1+\frac{x^2}3+o\!\left(x^2\right)\right)^{1/x^2}$ as close to $e^{1/3}\gt1$ as we wish. Thus,
$$
\begin{align}
\lim_{x\to0^-}\left(\left(1+\frac{x^2}3+o\!\left(x^2\right)\right)^{1/x^2}\right)^{1/x}
&=\lim_{x\to0^-}\left(e^{1/3}\right)^{1/x}\\
&=0
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Value of $e$ given that $\frac{a+b+c+d+e}{5} = 2 = \sqrt{\frac{a^2 + b^2 + c^2 + d^2 + e^2}{5}}$ I have five real numbers $a,b,c,d,e$ and their arithmetic mean is $2$. I also know that the arithmetic mean of $a^2, b^2,c^2,d^2$, and $e^2$ is $4$. Is there a way by which I can prove that the range of $e$ (or any ONE of the numbers) is $[0,16/5]$. I ran across this problem in a book and am stuck on it. Any help would be appreciated.
| \begin{align*}
&(a - 2)^2 + (b-2)^2 + (c-2)^2 + (d-2)^2 + (e - 2)^2\\[4pt]
&= (a^2 + b^2 + c^2 + d^2 +e^2) - 4(a + b + c + d +e) + 20\\[4pt]
&= 20 - 4(10) + 20\\[4pt]
&= 0\\[10pt]
&\;\text{hence}\\[10pt]
&\;a = b = c = d = e = 2\\[4pt]
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2529262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Compute $\sum\limits_{n=0}^\infty \frac{1}{a+(2n+1)^2},$, $a>0$ using residue theorem I need to compute the series $$\sum_{n=0}^\infty \frac{1}{a+(2n+1)^2},$$ where $a>0$ is constant. I know that this can be done using the residue theorem, but I don't really understand how this method works? Which function $f$ should I apply the residue theorem on?
| Replacing $a$ by $a^2$, From here Where Residue Theorem is used, We have, $$\begin{align}\frac{\pi}{a} \coth(\pi a) =\sum_{-\infty}^{\infty} \frac{1}{n^{2}+a^{2}}
&= \sum_{-\infty}^\infty \frac{1}{a^2+(2n)^2} +\sum_{-\infty}^\infty \frac{1}{a^2+(2n+1)^2} \\
&= \color{blue}{\frac{1}{4}}\sum_{-\infty}^\infty \frac{1}{\color{blue}{\left(\frac{a}{2}\right)^2}+n^2} +\color{blue}{2}\sum_{n=0}^\infty \frac{1}{a^2+(2n+1)^2} \\
&= \color{blue}{\frac{\pi}{2a}\coth(\frac{\pi a}{2})} +\color{blue}{2}\sum_{n=0}^\infty \frac{1}{a^2+(2n+1)^2}\end{align}$$
It is worth noticing that,$$\sum_{-\infty}^\infty \frac{1}{a^2+(2n+1)^2}=\sum_{n=0}^\infty \frac{1}{a^2+(2n+1)^2}+\sum_{n=1}^\infty \frac{1}{a^2+(2n-1)^2} \\=2\sum_{n=0}^\infty \frac{1}{a^2+(2n+1)^2}$$
We deduce from the previous step that:
$$\color{red}{ \sum_{n=0}^\infty \frac{1}{a^{\color{blue}2}+(2n+1)^2} =\frac{\pi}{4a}\left( 2\coth(\pi a)- \coth(\frac{\pi a}{2})\right)}$$
or
$$\color{blue}{ \sum_{n=0}^\infty \frac{1}{a +(2n+1)^2} =\frac{\pi}{4\sqrt{a}}\left( 2\coth(\pi \sqrt{a})- \coth(\frac{\pi \sqrt{a}}{2})\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2530744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_1^\infty \frac{1}{x(x^2+1)}\ dx$ $$I=\int_1^\infty \frac{1}{x(x^2+1)}\ dx$$
I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.
$$\frac{1}{x(x^2+1)} = \frac{1}{2} \left(\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i}\right)$$
So
$$I = \frac{1}{2}\int_1^\infty\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i} \ dx \\ = \frac{1}{2}\left[\log |x| - \log(|x+i|) - \log(|x-i|)\right]_1^\infty
$$
which evaluates to be $\infty$.
| Your partial fraction decomposition is incorrect. If you still want to use complex numbers, it's
$$ \frac{1}{x(1+x^2)} = \frac{1}{x} - \frac{1}{2(x+i)} - \frac{1}{2(x-i)} $$
Then
$$ \int_1^{\infty} f(x) \ dx = \left.\left(\ln |x| - \frac{1}{2}\ln |x+i| -
\frac{1}{2} \ln |x-i|\right)\right|_1^{\infty} = \left. \ln \left|\frac{x}{\sqrt{x^2+1}} \right|\right|_1^{\infty} = \frac{1}{2}\ln 2 $$
Note that
$$ \lim_{x\to \infty} \frac{x}{\sqrt{x^2+1}} = \lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x^2}}} = 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
$a,b,c$ such that $ax^2+bx+c \equiv 0$ (mod $3$) holds for all $x$? Today I wrote a computer program that finds whole positive $a,b,c$ such that $ax^2+bx+c \equiv 0$ (mod $n$) holds for all $x$. For $n=2$ I get several results, such as $x^2+x$, $3x^2+x$ and more. However, I was not able to find $a,b,c$ for the case $n=3$.
(Why) Are there no $a,b,c$ for $n=3$?
Remark: Of course $\text{gcd}(a,b,c) = 1$ and $a,b,c \in \mathbb{Z}$ and $a \not = 0$ or $b \not = 0$ or $c \not = 0$.
| What conditions are needed in order to conclude that the familiar quadratic formula holds, saying that if $ax^2+bx+c=0$ and $a\ne0,$ then $x= \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\text{ ?}$
One needs $2a$ to have a multiplicative inverse, regardless of which non-zero element $a$ is. Thus if $n=6,$ for example, and $a=3,$ you get $2a=0$ and that has no multiplicative inverse; thus dividing by $2a$ makes no sense.
But when $n=3$ and $a\ne 0$, then either $a=1$ and $2a=2,$ and $2$ is its own multiplicative inverse, or $a=2$ and $2a=1,$ and $1$ is its own multiplicative inverse. Thus the quadratic formula holds, so a quadratic equation cannot have more than two roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2535268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solution of differential equation in Trigonometric functions of $x$ and $y$ If $$\frac{dy}{dx} +\frac{\cos x(3\cos y-7\sin x-3)}{\sin y(3\sin x-7\cos y+7)}=0$$
Attempt: $(3\sin x\sin y-7\cos y \sin y+7\sin y)dy+(3\cos x\cos y-7\sin x\cos -3\cos x)dx =0$
could some help me how to solve it , thanks
| Hint: $$\dfrac{\sin y}{\cos x}\dfrac{dy}{dx}=-\dfrac{d(\cos y)}{d(\sin x)}=-\dfrac{3\cos y-7\sin x-3}{3\sin x-7\cos y+7}$$ Now use a suitable variable transformation to get a familiar equation as in example 4.14.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2535531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the integral $\int_{-1}^1\exp\left(ax+b\sqrt{1-x^2}\right)\,dx$ I wonder if there is a closed form expression for the following integral with respect to $x$:
\begin{equation}
\int_{-1}^1\exp\left(ax+b\sqrt{1-x^2}\right)\,dx
\end{equation}
where $a,b$ are some constant. Thanks so much! Any help would be greatly appreciated!
| $$\int_{-1}^1 \exp \left(a x+b \sqrt{1-x^2}\right) \, dx=\int_{-1}^1 \exp (a x) \exp \left(b \sqrt{1-x^2}\right) \, dx=\int_{-1}^1
\exp (a x) \sum _{n=0}^{\infty } \frac{\left(b \sqrt{1-x^2}\right)^n}{n!} \, dx=\sum _{n=0}^{\infty } \int_{-1}^1 \frac{\exp (a
x) \left(b \sqrt{1-x^2}\right)^n}{n!} \, dx=\sum _{n=0}^{\infty } \frac{2^{\frac{1}{2}-\frac{n}{2}} a^{-\frac{1}{2}-\frac{n}{2}}
b^n \pi I_{\frac{1+n}{2}}(a)}{\Gamma \left(\frac{1}{2}+\frac{n}{2}\right)}=\sqrt{\frac{2}{a}} \pi \sum _{n=0}^{\infty }
\frac{(2 a)^{-\frac{n}{2}} b^n I_{\frac{1+n}{2}}(a)}{\Gamma \left(\frac{1}{2}+\frac{n}{2}\right)}$$
For integral to solve I'm use a CAS.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2535772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
AM>HM Problem $\frac{1}{n+1}+...+\frac{1}{3n+1}>1$ I am having difficulty solving one of the problems from "Problems in Mathematical Analysis I" - W. J. Kaczor;M. T. Nowak .
It's a problem 1.2.5 b), and it goes like this:
1.2.5. For $n \in \mathbb{N}$, verify the following claims:
$$\tag{b} \qquad \dfrac{1}{n + 1} + \dfrac{1}{n + 2} + \dfrac{1}{n + 3} + \ldots + \dfrac{1}{3n + 1} \, > \, 1$$
In solutions it says: "Use the arithmetic-harmonic mean inequality!"
I tried to apply it on whole inequality but got:
\begin{align}
& \dfrac{\frac{1}{n+1}+\ldots+\frac{1}{3n+1}}{2n}\, >\, \dfrac{2n}{n+1+\ldots+3n+1} \\
\implies & \frac{1}{n+1}+\ldots+\frac{1}{3n+1}>\frac{8n^2}{2n(n+1+3n+1)} \\
\implies & \frac{1}{n+1}+\ldots+\frac{1}{3n+1}>\frac{2n}{2n+1}
\end{align}
| $$\frac { \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 3n+1 } }{ 2n+1 } >\frac { 2n+1 }{ n+1+n+2+...+3n+1 } \\ \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 3n+1 } >\frac { { \left( 2n+1 \right) }^{ 2 } }{ \frac { 3n+1+n+1 }{ 2 } \cdot \left( 2n+1 \right) } =\frac { 4n+2 }{ 4n+2 } =1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2538082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Show that $\tan^{-1} (\tfrac{1}{8} ) \geq \frac{1}{\sqrt{65}} $ Here I'd like to check that:
$$ \frac{1}{8} \geq \tan^{-1} (\tfrac{1}{8} ) \geq \frac{1}{\sqrt{65}} $$
The numbers do check out if one uses a calculator. They were using the Taylor series:
*
*$\frac{1}{\sqrt{65}} \;\;\;\;\;= 0.12403 $
*$\tan^{-1} \frac{1}{8} = 0.12435 $
*$\frac{1}{8} = 0.125$
The Taylor series for tangent could lead to an approximation. If $0 < x < \frac{\pi}{4}$ I believe we have:
$$ x - \frac{x^3}{3} < \tan^{-1} x < x $$
I do not recommend either of these, as a first recourse, but it came up in discussion last time.
These numbers might be somewhat mysterious (and they are). I got these using complex numbers. I said that:
\begin{eqnarray*}
5 &=& 1^2 + 2^2 \\
13 &=& 2^2 + 3^2
\end{eqnarray*}
And there are angles associated to these sum of squares $\theta_5 = \tan^{-1} \frac{1}{2}$ and $\theta_{13} = \tan^{-1} \frac{2}{3} $
then, using the difference of arctangents formula:
$$ \tan^{-1} \frac{2}{3} - \tan^{-1} \frac{1}{2} = \tan^{-1} \left[
\frac{ \frac{2}{3} - \frac{1}{2}}{1 + \frac{2}{3} \times \frac{1}{2}} \right] = \tan^{-1} \tfrac{1}{8}$$
and this leads to the left side inequalty above.
| Suppose $\theta = \tan^{-1}\frac18$. Then $\theta = \sin^{-1}\frac{1}{\sqrt{65}}$. Thus $\sin \theta = \frac{1}{\sqrt{65}}$. But $\theta > \sin\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simplify $\sqrt{1-\sin^4x}-\sqrt{1-\cos^4x}$
Simplify
$$\sqrt{1-\sin^4x}-\sqrt{1-\cos^4x}$$
My Try
$$A:=\sqrt{1-\sin^4x}-\sqrt{1-\cos^4x}\\A^2=1-\sin^4x-1+\cos^4x-2\sqrt{(1-\sin^4x)(1-\cos^4x)}\\A^2=\cos^4x-\sin^4x-2\sqrt{(\sin x\cos x)^2(1+\sin^2x)(1+\cos^2x)}$$
Now what ?
| \begin{align}
\sqrt{1-\sin^{4} x} - \sqrt{1-\cos^{4} x} & = \sqrt{(1-\sin^{2} x)(1+\sin^{2} x)} - \sqrt{(1-\cos^{2} x)(1+\cos^{2} x)} \\
& = \sqrt{(cos^{2} x)(1+\sin^{2} x)} - \sqrt{(sin^{2} x)(1+\cos^{2} x)} \\
& = \lvert \cos x\rvert\sqrt{1+\sin^{2} x} - \lvert \sin x\rvert\sqrt{1+\cos^{2} x} \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2541734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the real number $x$ such that $\det A=0$ $$A= \begin{bmatrix} 1 & x & x^2 & x^3\\ x & x^2 & x^3 & 1 \\ x^2 & x^3 & 1 & x \\ x^3 & 1 & x&x^2 \end{bmatrix}$$
I don't know how to approach this. I'd like to figure this out step-by-step so any suggestions would be greatly appreciated. Thanks in advance!
| By performing column operations we can evaluate the determinant to be $(x^4-1)^3$. So, the roots are $1$, $-1$, $i$ and $-i$.
Try $C_2 \to C_2 - xC_1$ and $C_4 \to C_4 - xC_3$. This will easily simplify the determinant into:
$\det A= \left| \begin{matrix} 1 & 0 & x^2 & 0\\ x & 0 & x^3 & 1 - x^4\\ x^2 & 0 & 1 & 0 \\ x^3 & 1-x^4 & x& 0\end{matrix}\right|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Bounds for change of variable U and V are independent and uniformly distributed on (0,1). Find the joint density of $X=\frac{\sqrt{U}}{\sqrt{U}+\sqrt{V}}$ and $Y={\sqrt{U}+\sqrt{V}}$.
I've been able to find without a hitch that $f_{XY}(x, y)=4(1-x)xy^3$ for some values of X and Y. My textbook confirmed that this is correct. My problem comes from evaluation the values of X and Y over which the joint pdf takes a positive value. I reasoned that, since $0<u<1$ and $0<v<1$, that $0<x<1$ and $0<y<2$. Integrating the pdf over these values of x and y, however, does not yield one. The textbooks says that the bounds are $0<x<1$ and $0<y<\min\left(\frac{1}{x}, \frac{1}{1-x}\right)$, which indeeds integrate to 1. How were these bounds for y obtained? I have a feeling that the answer might be connected to the fact that $X=\frac{\sqrt{U}}{Y}$, but I am unable to explain why.
| $X=\frac{\sqrt{U}}{\sqrt{U}+\sqrt{V}}$
$X = \frac{\sqrt{U}}{Y}$
$ XY = \sqrt{U}$
$ \sqrt{U}$ implies $0\le xy \le 1$ implies $0\le y \le \frac{1}{x}\tag 1$
and
$X=\frac{\sqrt{U}}{\sqrt{U}+\sqrt{V}} = \frac{\sqrt{U}+\sqrt{V}-\sqrt{V}}{\sqrt{U}+\sqrt{V}} $
$X = 1-\frac{\sqrt{V}}{\sqrt{U}+\sqrt{V}} = 1-\frac{\sqrt{V}}{Y}$
$(1-X)Y = \sqrt{V}$
$\sqrt{V}$ implies $0\le (1-x)y \le 1$ implies $0\le y \le \frac{1}{1-x}\tag 2$
Combining (1) and (2)
We get
$0\le y \le min(\frac{1}{x},\frac{1}{1-x})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Sum of series of fractions: $\frac{4}{1!}+ \frac{8}{2!}+ \frac{14}{3!} + \frac{22}{4!}+\cdots$
Find the sum of the series:
$$\frac{4}{1!}+ \frac{8}{2!}+ \frac{14}{3!} + \frac{22}{4!}+\cdots$$
I am not able to understand how to proceed. The numerator terms have nothing in common which might result in an AP or GP.
Please guide me with the approach.
| $a_n = \frac{n^2 + n + 2}{n!}\\
\sum a_n = \sum\frac{n^2 + n + 2}{n!}\\
\sum_\limits{n=1}^\infty\frac{n^2 - n}{n!} + \sum_\limits{n=1}^\infty\frac{2n}{n!} + \sum_\limits{n=1}^\infty \frac {2}{n!}\\$
$\sum_\limits{n=1}^\infty\frac{n^2 - n}{n!} = 0 + \sum_\limits{n=2}^\infty\frac{1}{(n-2)!} = \sum_\limits{n=0}^\infty\frac{1}{n!} = e\\
\sum_\limits{n=1}^\infty\frac{2n}{n!} = 2\sum_\limits{n=1}^\infty\frac{1}{(n-1)!} = 2e\\
\sum_\limits{n=1}^\infty \frac {2}{n!} = 2e- 2$
$5e-2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2544075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.