Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
if $S=\sin x+2\sin (2x)+\cdots+n\sin nx$,$C=\cos x +2\cos (2x)+\cdots+n\cos (nx)$ if:
$S=\sin x+2\sin (2x)+\cdots+n\sin nx$,
$C=\cos x +2\cos (2x)+\cdots+n\cos (nx).$
prove that $4\sin^2 (x/2).S=(n+1)\sin (nx)-n\sin(nx+x)$
I can solve this easily using complex numbers(ie taking $C+iS$,which becomes an AGP) but was wondering if it can be solved using basic trig identities.
Source S.L.Loney plane trigonometry
Thank you!
| Easy to see that for $\sin\frac{x}{2}=0$ it's true.
But for $\sin\frac{x}{2}\neq0$ by the telescopic summation we obtain: $$S=-\left(\sum_{k=1}^n\cos{kx}\right)'=-\left(\frac{\sum\limits_{k=1}^n2\sin\frac{x}{2}\cos{kx}}{2\sin\frac{x}{2}}\right)'=$$
$$=-\left(\frac{\sum\limits_{k=1}^n\left(\sin\left(k+\frac{1}{2}\right)x-\sin\left(k-\frac{1}{2}\right)x\right)}{2\sin\frac{x}{2}}\right)'=-\left(\frac{\sin\left(n+\frac{1}{2}\right)x-\sin\frac{x}{2}}{2\sin\frac{x}{2}}\right)'=$$
$$=-\frac{\left(n+\frac{1}{2}\right)\cos\left(n+\frac{1}{2}\right)x\sin\frac{x}{2}-\frac{1}{2}\sin\left(n+\frac{1}{2}\right)x\cos\frac{x}{2}}{2\sin^2\frac{x}{2}}=$$
$$=\frac{\frac{1}{2}\left(\sin(n+1)x+\sin{nx}\right)-\left(n+\frac{1}{2}\right)(\sin(n+1)x-\sin{nx})}{4\sin^2\frac{x}{2}}=$$
$$=\frac{(n+1)\sin{nx}-n\sin(n+1)x}{4\sin^2\frac{x}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to think of factorising $x^7+x^2+1$ to $(x^2+x+1)(x(x-1)(x^3+1)+1)$ (Thales 2016) I was just doing the following question:
Find a prime number which divides the number $A=14^7+14^2+1$.
I solved it by finding the result which is $A=105413504+196+1=105413701$ and then trying out all prime numbers till I found that 211 divides it. However, obviously this is extremely tedious. I hence looked at the solution which says that $x^7+x^2+1=(x^2+x+1)(x(x-1)(x^3+1)+1)$ and from here by saying that $x=14$ we get the solution. However I can't seem to think of how to intuitively turn $x^7+x^2+1$ into $(x^2+x+1)(x(x-1)(x^3+1)+1)$. I realize that from the question it is obvious to go looking for factors of A and hence trying to factorize $14^7+14^2+1$, but I can't work out how to go about factorizing it, what are the steps which you need to take in order to factorize a given polynomial. Could you please explain to me how to go about factorizing such an expression and how to intuitively think of each step?
| You can "see" that $\omega = \exp \left( \frac{2i\pi}{3}\right)$ and $\omega^2= \exp \left( \frac{-2i\pi}{3}\right)$ are roots of $x^7+x^2+1$, therefore $x^7+x^2+1$ factorizes by $x^2+x+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
What is complex number z if $z^8+16z^4+256=0$? So far, I have set y to equal to $z^4$ and used the quadratic equation to solve $y = -8+8\sqrt{3}i$ or $-8-8\sqrt{3}i$. How do I determine the 8 different values of $z$?
| Note that $$z^8+16z^4+256$$
$$=z^8+32z^4+256-16z^4$$
$$=(z^4+16)^2-(4z^2)^2$$
$$=(z^4-4z^2+16)(z^4+4z^2+16)$$
$$=(z^4-4z^2+16)(z^2-2z+4)(z^2+2z+4)$$
Now considering $(z^4-4z^2+16)=0$, let $x=z^2$ then we have $$x^2-4x+16=0$$
$$x=\frac{4\pm\sqrt{-48}}{2}=\frac{4\pm 4i\sqrt{3}}{2}=2\pm2i\sqrt{3}$$
So you have $z=\pm\sqrt{2+2i\sqrt{3}}$ and $z=\pm\sqrt{2-2i\sqrt{3}}$ in this case. The other cases are quadratics.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Find the minimum of $P = (a - b)(b - c)(c - a)$
Given $a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = ab + bc + ca + 6$. Find the minimum of
$$P = (a - b)(b - c)(c - a)$$
My solution:
*
*We have:
$$a^2 + b^2 + c^2 = ab + bc + ca + 6$$
$$\implies 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca + 12$$
$$\implies (a - b)^2 + (b - c)^2 + (c - a)^2 = 12$$
*
*Using AM-GM Inequality, we have:
$$(a - b)^2 + (b - c)^2 + (c - a)^2 \geq 3 \sqrt[3]{((a - b)(b - c)(c - a))^2}$$
$$\implies 3 \sqrt[3]{P^2} \leq 12$$
$$\implies -8 \leq P \leq 8$$
*
*Therefore, $\min P = -8$
Is this solution correct? If not, then why?
| Minimize $P = (a-b)(b-c)(c-a)$
given $(a - b)^2 + (b - c)^2 + (c - a)^2 = 12$
WLOG, say $a \ge b \ge c, \, a - b = x, a - c = y$ where $x \ge 0, y \ge x$
which gives,
$P (x,y) = xy(x-y)$ ...(i)
$\, G(x,y) = x^2 + y^2 + (x-y)^2 - 12 = 0$ ...(ii)
Using Lagrange,
$Q(x,y) = P (x,y) + \lambda G(x,y) = xy(x-y) + \lambda (2x^2 + 2y^2 -2xy - 12)$
You get $\, 2xy-y^2 + \lambda(4x - 2y) = 0$ ..(iii)
$\, x^2-2xy + \lambda(4y - 2x) = 0$ ...(iv)
From (iii) and (iv),
You get $y = 2x, x = 2y$
Taking $y = 2x$ (as $y \ge x$) and substituting in (ii), you get $x = \sqrt 2$
From (i) that gives min $\, P = -4\sqrt2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Proving that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1\implies (a+1)(b+1)(c+1)\geq 64$ where $a,b,c>0$. I am trying to solve the following problem:
Let $a,b,c>0$ with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$. Prove that:
$$(a+1)(b+1)(c+1)\geq 64$$
So far, I have gotten that by AM-GM, $(a+1)\geq 2\sqrt{a}$, $(b+1)\geq 2\sqrt{b}$ and $(c+1)\geq 2\sqrt{c}$ so:
$$(a+1)(b+1)(c+1)\geq 8\sqrt{abc} \tag{1}$$
Then using AM-GM on $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, we get that:
\begin{align}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & \geq 3\times \frac{1}{\sqrt[3]{abc}} \\
1 & \geq 3\times \frac{1}{\sqrt[3]{abc}} \\
\frac{1}{3} & \geq \frac{1}{\sqrt[3]{abc}} \\
\frac{1}{27} & \geq \frac{1}{abc} \\
abc & > 27
\end{align}
Substituting this into $(1)$, we get:
$$(a+1)(b+1)(c+1)\geq 8\sqrt{27}$$
So I clearly went wrong somewhere, though I'm not sure why. It would be best if you could provide a solution using AM-GM or Muirhead's Inequality. The question seems so simple, where equality happens at $a=b=c=3$, though I can't prove this. Thank you in advance for your solutions and hints!
| Another way.
We can use a homogenization.
We need to prove $$\prod_{cyc}\left(a+\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)\geq\frac{64}{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^3}$$ or
$$\prod_{cyc}(a^2+ab+ac+bc)\geq64a^2b^2c^2$$ or
$$\prod_{cyc}((a+b)(a+c))\geq64a^2b^2c^2$$ or
$$\prod_{cyc}(a+b)\geq8abc$$ or
$$\prod_{cyc}(a+b)\geq8\prod_{cyc}\sqrt{ab}$$ or
$$\prod_{cyc}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)\geq8,$$ which is true by AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Bounds on $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}$ It's simple. What are the bounds on $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}$ as $a,b,c>0$. Thanks in advance!
Edit: I need bound that can actually be touched i.e. $\alpha$ and $\beta$ such that $\alpha\leq\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\leq\beta$
| There are no well defined bounds.
Let $S$ denote the given expression. Note first that
$$S > \frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=1.$$
On the other hand, with $a=1$, $c=\varepsilon$ and $b=\varepsilon^2$ we have
$$S=\frac{1}{1+\varepsilon^2} + \frac{2\varepsilon}{1+\varepsilon} \to 1$$
as $\varepsilon \downarrow 0$. Thus the largest lower bound for $S$ is $1$.
For an upper bound, assume wlog that $a \leq b,c$. Then
$$S \leq \frac{a}{a+b} + \frac{b}{b+a}+\frac{c}{c+a} < 2.$$
With $a=1$, $b=\varepsilon$ and $c=\varepsilon^2$ we have
$$S = \frac{2}{1+\varepsilon} + \frac{\varepsilon^2}{1+\varepsilon^2} \to 2$$
as $\varepsilon \downarrow 0$. Hence the smallest upper bound for $S$ is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $-1^2+3^2-5^2\mp ...+(2^n-1)^2=2^{2n-1}$ Playing with numbers, I construct following expression.
Can it be shown that
$$\sum_{i=1}^{2^{n-1}}(-1)^i(2i-1)^2=2^{2n-1}$$
attempt
We can construct following, using finite calculus as
$(-1)^2+3^2+7^2+...+(4k-5)^2 = \binom{k}1+8\binom{k}2+32\binom{k}3\quad\quad eq(1)$
$1^2+5^2+9^2+...+(4k-3)^2 = \binom{k}1+24\binom{k}2+32\binom{k}3\quad\quad eq(2)$
Let $4k-1=2^n-1$ so we can write above claim as
$eq(1)-eq(2)-1+(4k-1)^2=2^{2n-1}$
Which is equivalent to show
$(2^n-1)^2-16\binom{2^{n-2}}2-1= 2^{2n-1}$
Here I'm stuck. Thanks
| Using the Hockey-stick identity:
$$S(n)=\sum_{k=0}^{n-1} (2k+1)^2=\sum_{k=0}^{n-1} \left(8{k+1\choose 2}+1\right)=8{n+1\choose 3}+n=\frac{n(2n+1)(2n-1)}{3}$$
$$T(n)=\sum_{k=0}^{n-1} (4k+1)^2=\sum_{k=0}^{n-1} \left( 16k^2+8k+1\right)=\sum_{k=0}^{n-1} \left(32{k+1\choose2}-8{k+1\choose1}+9\right)\\=32{n+1\choose3}-8{n+1\choose2}+9n=\frac n3(16n^2-12n-1)$$
Let $N=2^{n-2}$, then
$$-1^2+3^2-5^2+\dots+(2^n-1)^2\\=-1^2+3^2-5^2+\dots+(4N-1)^2\\=S(2N)-2T(N)\\=8N^2\\=2^{2n-1}$$
To answer the comment:
$$1^2+3^2+5^2+\dots+(4N-1)^2=\sum_{k=0}^{2N-1} (2k+1)^2=S(2N)$$
$$1^2+5^2+\dots+(4N-3)^3=1^2+5^2+\dots+(4(N-1)+1)^3=\sum_{k=0}^{N-1}(4k+1)^2=T(N)$$
Now, the first row minus twice the second row yields:
$$-1^2+3^2-5^2+\dots+(4N-1)^2=S(2N)-2T(N)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3855368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Collatz Conjecture: If a non-trivial cycle exists, would the sum of powers of $2$ be less than $2n$? For the Collatz Conjecture, it seems to me that if $m$ is the sum of the powers of $2$ for a non-trivial $n$ cycle (where each $x_1, \dots, x_n $ is odd and $x_i > 1$), it follows that $m < 2n$
Is my reasoning correct?
Let:
*
*$\nu_2(x)$ be the 2-adic valuation of $x$
*$x_1, x_2, \dots, x_n$ be $n$ distinct integers such that:
*
*$x_{i+1} = \dfrac{3x_i + 1}{2^{\nu_2(3x_i+1)}}$
*$x_i > 1$
Observations:
*
*$\left(3 + \dfrac{1}{x_{i}}\right) = \left(\dfrac{x_{i+1}}{x_{i}}\right)2^{\nu_2(3x_{i} + 1)}$ since:
*
*$x_{i+1} = \dfrac{3x_{i}+1}{2^{\nu_2(3x_{i}+1)}}$
*$2^{\nu_2(3x_{i}+1)}x_{i+1} = 3x_{i} + 1$
*$\prod\limits_{i=1}^{n}\left(3 + \frac{1}{x_i}\right) = \left(\dfrac{x_{n+1}}{x_1}\right)\prod\limits_{i=1}^n2^{\nu_2(3x_i + 1)}$
This follows directly from the previous observation.
*
*$\left(3 + \dfrac{1}{x_{\text{max}}}\right)^{n} \le \left(\dfrac{x_{n+1}}{x_1}\right)\prod\limits_{i=1}^n2^{{\nu}_2(3x_i + 1)} \le \left(3 + \dfrac{1}{x_{\text{min}}}\right)^{n}$
This follows directly from the previous observation.
Claim:
If $n \ge 1$, $x_1, x_2, \dots, x_n$ forms a cycle, then $\sum\limits_{i=1}^n \nu_2(3x_i+1) < 2n$
Argument:
(1) Assume $x_1, x_2, \dots, x_n$ form a cycle such that $x_{i+n} = x_i$
(2) Let $m = \sum\limits_{i=1}^n \nu_2(3x_i+1)$ so that:
$$\left(3 + \dfrac{1}{x_{\text{max}}}\right)^{n} \le 2^{m} \le \left(3 + \dfrac{1}{x_{\text{min}}}\right)^{n}$$
(3) Clearly, $2^m > 3^n$ so that: $2^{\frac{m}{n}} > 3$
(4) It follows:
*
*$$3 + \frac{1}{x_{\text{max}}} \le 2^{\frac{m}{n}} \le 3 + \frac{1}{x_{\text{min}}}$$
*$$\frac{1}{x_{\text{max}}} \le 2^{\frac{m}{n}} - 3 \le \frac{1}{x_{\text{min}}}$$
*$$x_{\text{max}} \ge \frac{1}{2^{\frac{m}{n}} - 3} \ge x_{\text{min}}$$
(5) $2^{\frac{m}{n}} - 3 < 1$ since $x_{\text{min}} > 1$ and $2^{\frac{m}{n}} > 3$
(6) It follows:
*
*$$2^{\frac{m}{n}} < 2^2$$
*$$m < 2n$$
| [Correction: this argument contains a mistake]
Actually, a stronger result can be shown using the following argument:
Looking at both even and odd numbers in the cycle, it contains $n$ odd numbers and $m$ even numbers.
Since
$$ \frac{3^{m}}{2^{m+n}} < 1$$
it follows that
$$ 3^{m} < 2^{m+n} $$
$$ (\frac{3}{2})^m < 2^n $$
$$ m \cdot \ln (3/2) < n \cdot \ln 2 $$
$$ m < \frac{\ln 2}{\ln(3/2)}\cdot n < 1.71n $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3859842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
if $x_n=\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}$ prove
if $x_n=\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}$ prove that for $n\ge 2$ $$x_{n+1}-x_n<\frac{1}{n!}$$.
I think induction works best here. The case when $n=2$ is easy as $x_3-x_2=\sqrt{2+\sqrt[3]{3}}-\sqrt{2}=0.44<\frac{1}{2!}$.
However i am not able to proceed further ,the complex radicals are creating a lot of trouble
| Hint: use the fact that $$\sqrt{y}-\sqrt{x}=\frac{y-x}{\sqrt{y}+\sqrt{x}}$$
For example
$$x_{n+1}-x_n=\frac{\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n+1]{n+1}}}-\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}{\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n+1]{n+1}}}}+\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}}<\\
\frac{\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n+1]{n+1}}}-\sqrt[3]{3+\sqrt[4]{4+...\sqrt[n]{n}}}}{2}<...$$
Next $$\sqrt[3]{y}-\sqrt[3]{x}=\frac{y-x}{\sqrt[3]{y^2}+\sqrt[3]{yx}+\sqrt[3]{x}}$$
leading to
$$...<\frac{1}{2}\frac{\sqrt[4]{4+...\sqrt[n+1]{n+1}}-\sqrt[4]{4+...\sqrt[n]{n}}}{3}$$
And so on ...
General idea comes from the following identity $$y^n-x^n=(y-x)(y^{n-1}+y^{n-2}x+y^{n-3}x^2+...+yx^{n-2}+x^{n-1})$$
Replace $y\to \sqrt[n]{y}$, $x\to \sqrt[n]{x}$ and consider that both $y>1$ and $x>1$. As a result
$$\sqrt[n]{y}-\sqrt[n]{x}<\frac{y-x}{n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3868496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
System of equations from roots of polynomial I'm given the equation $3072x^4-2880x^3+840x^2-90x+3=0$ and told that its roots are $\alpha, \alpha r, \alpha r^2, \alpha r^3,$ for some $r\in \mathbb{R}$.
By considering the sum of the roots, the product, etc. I've found that \begin{gather}\alpha(1+r+r^2+r^3)=\frac{15}{16} \\ \alpha^2r(1+r+2r^2+r^3+r^4)=\frac{35}{128} \\ \alpha^3 r^3(1+r+r^2+r^3)=\frac{15}{512} \\ \alpha^4 r^6=\frac{1}{1024}\end{gather}
But this looks like a rather complex system and I can't see any obvious way solve this for $\alpha$ and $r$.
How can this system be solved?
EDIT
I can see that all of the denominators are powers of $2$, but I can't see how that will help me here.
| Answer :
$3072x^4-2880x^3+840x^2-90x+3=0$
Factoring by 3
$\Rightarrow $
$1024x^4-960x^3 +280x^2 - 30x+1=0 $
We can see $\frac{1}{2} $ is solution of the equation
$\Rightarrow $
$\frac{1024x^4-960x^3 +280x^2 - 30x+1}{x-\frac{1}{2}} $=0
$\Rightarrow $
$1024x^3 - 448 x^2 + 56 x - 2=0$
We can see $\frac{1}{4}$ is solution of the equation
$\Rightarrow $
$\frac{1024x^3 - 448 x^2 + 56 x - 2}{x-\frac{1}{4}} =0$
$\Rightarrow $
$1024x^2 - 192 x + 8=0$
$\triangle=192^2 - 32(1024)=4096$
$x_1=\frac{192-\sqrt{4096}}{2(1024)}$
And:
$x_2=\frac{192+\sqrt{4096}}{2 (1024) }$
$\Rightarrow $
$x_1=\frac{1}{16}$
$x_2=\frac{1}{8}$
No forget $\frac{1}{2} $also is a solution of the equation
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
For $a,b,c>0$ proving $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geqslant a + b + c + \frac{4(a - b)^2}{a + b + c}$ The problem with which I have a problem it's this:
For $a,b,c>0$ prove that
$$
\frac{a^2}{b} +
\frac{b^2}{c} +
\frac{c^2}{a} \geqslant
a + b + c +
\frac{4(a - b)^2}{a + b + c}
$$
Titu's Lemma and AM-GM work no good because this similar looking inequality is sharper.
After trying these, I decided to go for the following.
Here is my work:
Multiply $abc(a + b + c)$ to both the sides,
$$
(a + b + c)
(a^3c + ab^3 + bc^3) \geqslant abc(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) + 4abc (a - b) ^2
$$
After some work we are left to prove that:
$$
\sum_{cyc} {a^2b^3 + ab^4 - 2a^2 b^2 c} \geqslant abc(4 a^2 + 4 b^2 - 8ab)
$$
How to prove this or there is some better approach?
This inequality was the first in a list of many inequalities, but I don't think that means it is easy.
Thanks for help!
| The SOS proof. We have
$$\frac{a^2}{b} +
\frac{b^2}{c} +
\frac{c^2}{a} -
(a + b + c) -
\frac{4(a - b)^2}{a + b + c}$$
$$=\frac{a(b^2-2bc+ca)^2+b(c^2-ab)^2+c(a^2-2ab+bc)^2}{abc(a+b+c)} \geqslant 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3870314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding specific values of $n$ such that $n^2+3n+3$ is factorisable with certain constraints on the factors I received the following question during a Maths Olympiad:
Let $n$ be a positive integer. When is it possible to express $n^2+3n+3$ into the form $ab$ with $a$ and $b$ being positive integers, and such that the difference between $a$ and $b$ is smaller than $2\sqrt{n+1}$?
My work so far:
WLOG $a<b$. We can factor this quadratic into $(n+1)(n+2)+1$. This means both $n+1$ and $n+2$ cannot divide $n^2+3n+3$, and $a$ can at most be $n$. If $a=n$, then $b=\frac{n^2+3n+3}n=n+3+\frac3n>n+3$ and $b$ must be equal to or larger than $n+4$. The difference between $a$ and $b$ is at least $4$. We need $4<2\sqrt{n+1}\Rightarrow2<\sqrt{n+1}\Rightarrow4<n+1\Rightarrow n>3$.
Apart from this, $n^2+3n+3\equiv1,3\pmod 6$. It also seems that apart from $3$, the only primes that divide $n^2+3n+3$ are of the form $6k+1$. However this is conjecture and based on the few test cases I checked.
Quadratic residues might come in handy in this problem but I'm not sure.
How could I do this problem?
| As you suggested, WLOG, let $a \le b$, with an integer $c$ such that
$$b = a + c \tag{1}\label{eq1A}$$
The constraint then means
$$0 \le c \lt 2\sqrt{n + 1} \implies 0 \le c^2 \lt 4(n + 1) \tag{2}\label{eq2A}$$
We then get
$$\begin{equation}\begin{aligned}
a(a + c) & = n^2 + 3n + 3 \\
a^2 + ac & = n^2 + 3n + 3 \\
4a^2 + 4ac & = 4n^2 + 12n + 12 \\
4a^2 + 4ac + c^2 & = 4n^2 + 12n + 12 + c^2 \\
(2a + c)^2 & = 4n^2 + 12n + 12 + c^2
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Using \eqref{eq2A} gives
$$4n^2 + 12n + 12 \le (2a + c)^2 \lt 4n^2 + 12n + 12 + 4(n + 1) = 4n^2 + 16n + 16 \tag{4}\label{eq4A}$$
Note $(2n + 3)^2 = 4n^2 + 12n + 9$ is less than the lower boundary. Also, $(2n + 4)^2 = 4n^2 + 16n + 16$ is the exclusive upper boundary. This means there's no perfect square in that range, i.e., there's no integer $2a + c$ which satisfies \eqref{eq4A} and, thus, no integer $n$ matching the conditions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3872041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proof that $2(a^4+b^4+c^4)<(a^2+b^2+c^2)^2$ if and only if line segments of length $a$,$b$ and $c$ form a triangle Obviously I tried to somehow manipulate the triangle inequality
$a<b+c$
$a^2<(b+c)^2$
Similarly,
$b^2<(a+c)^2$ and $c^2<(a+b)^2$
After adding these together:
$a^2+b^2+c^2<(a+b)^2+(a+c)^2+(b+c)^2$
But it seems I even work in the wrong direction, cuz I have to find the lower bound on the sum of squares, not the one on the top.
So then I tried to rewrite the triangle inequality like that:
$c>a-b$
$b>a-c$
$a>b-c$
and work from here but I also didn't manage to get to the point, where I feel like I have this problem in my bag
| Hint. For $a,b,c\in\mathbb R^+$, we have $$a^4 + b^4 + c^4 - 2\cdot \left(a^2b^2 + b^2c^2 + a^2c^2\right)=(a+b+c)(a+b-c)(a+c-b)(a-b-c)$$ Therefore, your inequality is equivalent to $$(a+b+c)(a+b-c)(c+a-b)(a-b-c)<0\\\iff (a+b+c)(a+b-c)(c+a-b)(b+c-a)>0$$ Can you finish? Why is it trivial that the sides of a triangle fulfill this inequality? Why is there no other option, but $a,b,c$ to be sides of a triangle?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3872363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A hard integral $\int\frac{(x^4-1)\sqrt{x^4+1}}{x^8+1} \mathrm{d}x$ This integtare is very hard for me，I do not known how to deal with it.
$$\int\frac{(x^4-1)\sqrt{x^4+1}}{x^8+1} \mathrm{d}x$$
I use “mathematica” and obtain this
next the “maple” ：
the “maple” tell me it's a Elementary function
| Note
$$\frac2{x^8+1}=\frac1{x^4+1}\left( \frac1{x^4+\sqrt2x^2+1}+ \frac1{x^4-\sqrt2x^2+1}\right)
$$
Then
\begin{align}
&\int\frac{(x^4-1)\sqrt{x^4+1}}{x^8+1} \> dx \\
=& \frac12 \int\frac{\frac{x^4-1}{\sqrt{x^4+1}}dx}{x^4+\sqrt2x^2+1}
+\frac12 \int\frac{\frac{x^4-1}{\sqrt{x^4+1}}dx}{x^4-\sqrt2x^2+1}\\
=& \frac12\int\frac{d\sqrt{ x^2+\frac1{x^2}}}{x^2+\frac1{x^2}+\sqrt2}
+\frac12 \int\frac{d\sqrt{ x^2+\frac1{x^2}}}{x^2+\frac1{x^2}-\sqrt2}\\
=& \frac1{2\sqrt[4]2}\tan^{-1} \frac{\sqrt{ x^2+\frac1{x^2}} }{\sqrt[4]2 } - \frac1{2\sqrt[4]2}\coth^{-1} \frac{\sqrt{ x^2+\frac1{x^2}} }{\sqrt[4]2 }+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3873069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Using A Determinant to Solve a Linear System I recently asked how to determine the standard form equation for an ellipse,
$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where the ellipse has a given eccentricity
$e$ and passes through three points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$. I got a fantastic
answer here, but there's some magic in the answer I don't understand.
To recap the solution, I can start with the general ellipse (actually conic) equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, which has six unknowns ($A$, $B$, ..., $F$), and then take advantage of the constraints on the problem to set $B = 0$ and $A$ as a constant factor $(1-e)^2$ of $C$. This leave me with four unknowns ($A/C$, $D$, $E$, and $F$) which can be solved with three points using the system:
$$
\begin{bmatrix}
(1-e^2)x^2 + y^2 & x & y & 1\\
(1-e^2)x_1^2 + y_1^2 & x_1 & y_1 & 1\\
(1-e^2)x_2^2 + y_2^2 & x_2 & y_2 & 1\\
(1-e^2)x_3^2 + y_3^2 & x_3 & y_3 & 1
\end{bmatrix}
\begin{bmatrix}
A/C\\
D\\
E\\
F
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}
$$
The linked answer somehow converts this system to the determinant equation
$$
\begin{vmatrix}
(1-e^2)x^2+y^2 & x & y & 1 \\
(1-e^2)x_1^2+y_1^2 & x_1 & y_1 & 1 \\
(1-e^2)x_2^2+y_2^2 & x_2 & y_2 & 1 \\
(1-e^2)x_3^2+y_3^2 & x_3 & y_3 & 1
\end{vmatrix}=0
$$
which allows me to solve
$$
A/C = \begin{vmatrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{vmatrix}
\\D = -
\begin{vmatrix}
(1-e^2)x_1^2+y_1^2 & y_1 & 1 \\
(1-e^2)x_2^2+y_2^2 & y_2 & 1 \\
(1-e^2)x_3^2+y_3^2 & y_3 & 1
\end{vmatrix}
\\etc...
$$
I see the mechanics of computing $A/C$, $D$, ..., but I don't understand why I get to build the determinant formula from the original linear system. I thought I was maybe onto something with Cramer's rule, but that goes nowhere quickly. Can anyone explain or point me at an answer?
| At this stage:
$\begin{bmatrix}
(1-e^2)x^2 + y^2 & x & y & 1\\
(1-e^2)x_1^2 + y_1^2 & x_1 & y_1 & 1\\
(1-e^2)x_2^2 + y_2^2 & x_2 & y_2 & 1\\
(1-e^2)x_3^2 + y_3^2 & x_3 & y_3 & 1
\end{bmatrix}
\begin{bmatrix}
A/C\\
D\\
E\\
F
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}$
If this has a solution, either we have the trival solution: $A = D = E = F = 0$ or the matrix is "singular."
If the matrix is singular, then its determinant must be equal to 0. And $A/C, D, E, F$ are in the kernel of the matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Proving $abc-1+\sqrt\frac 2{3}\ (a-c)\ge 0$ The question is this:
If $a\ge b\ge c\ge 0$ and $a^2+b^2+c^2=3$, then prove that
$$abc-1+\sqrt\frac 2{3}\ (a-c)\ge 0$$
For my work on this inequality, I have proved already under constraints that it is true.
Proof for: $\sqrt{3}(bc - 1) + \sqrt{2}(1-c)\geqslant0.$
$$
\sqrt{3}abc +
\sqrt{2}a -
\sqrt{3} -
\sqrt{2}c
\geqslant 0
$$
$$
a\left(
\sqrt{3}bc +
\sqrt{2}
\right)
+ (-1)\left(
\sqrt{3} +
\sqrt{2}c
\right) \geqslant 0
$$
$$
(1 + 1)(a\left(
\sqrt{3}bc +
\sqrt{2}
\right)
+ (-1)\left(
\sqrt{3} +
\sqrt{2}c
\right)) \geqslant 0
$$
By Chebyshev,
$$
(a - 1)
(\sqrt{3}bc +
\sqrt{2} +
\sqrt{3} +
\sqrt{2}c
)\geqslant0
$$
$$
a \geqslant 1
$$
Chebyshev Inequality requires the sequences to be monotonous. As $a+1>0$, we need to have the other sequence also in the same order, hence the condition: $\sqrt{3}bc + \sqrt{2} \geqslant\sqrt{3} + \sqrt{2}c$. The sequences are $(a,-1)$ and $(\sqrt{3}bc + \sqrt{2} ,\sqrt{3} + \sqrt{2}c)$.
I have tried another way but that was untrue. I have reached this far. The constraint $\sqrt{3}(bc - 1) + \sqrt{2}(1-c)\geqslant0$ isn't true always. Try $(a,b,c) = (\sqrt{3},0,0)$.
Thanks for extensions or other solutions too are welcome!
| Let $$f(a,b,c,\lambda)=abc-1+\sqrt{\frac{2}{3}}(a-c)+\lambda(a^2+b^2+c^2-3).$$
Thus, in the minimum point we need
$$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=\frac{\partial f}{\partial\lambda}=0,$$
which gives:
$$bc+\sqrt{\frac{2}{3}}+2\lambda a=ac+2\lambda b=ab-\sqrt{\frac{2}{3}}+2\lambda c=0.$$
Now, if $c=0$, so $$3=a^2+b^2\leq2a^2,$$ which gives $$a\geq\sqrt{\frac{3}{2}}$$ and $$abc-1+\sqrt{\frac{2}{3}}(a-c)=\sqrt{\frac{2}{3}}a-1\geq0.$$
Now, let $c>0$.
Thus, $$ \frac{bc+\sqrt{\frac{2}{3}}}{a}=\frac{ab-\sqrt{\frac{2}{3}}}{c}=\frac{ac}{b},$$ which gives
$$b^2c+b\sqrt{\frac{2}{3}}=a^2c$$ and $$b^2a-b\sqrt{\frac{2}{3}}=ac^2,$$ which after summing gives $$b^2=ac,$$ which with our condition gives $$a^2+ac+c^2=3$$ and we need to prove that
$$\sqrt{a^3c^3}+\sqrt{\frac{2}{3}}(a-c)\cdot\frac{a^2+ac+c^2}{3}\geq\sqrt{\left(\frac{a^2+ac+c^2}{3}\right)^3}.$$
Now, let $a=xc$ and $a^2+c^2=2uac.$
Thus, $x\geq1$ and $u\geq1$ and we need to prove that:
$$\sqrt{x^3}+\sqrt{\frac{2(x^2+1-2x)}{3}}\cdot\frac{x^2+x+1}{3}\geq\sqrt{\left(\frac{x^2+x+1}{3}\right)^3}$$ or
$$1+\sqrt{\frac{4(u-1)}{3}}\cdot\frac{2u+1}{3}\geq\sqrt{\left(\frac{2u+1}{3}\right)^3}$$ or
$$27+4(u-1)(2u+1)^2+12\sqrt{3(u-1)}(2u+1)\geq(2u+1)^3$$ or
$$(u-1)(4u^2-2u-11)+6\sqrt{3(u-1)}(2u+1)\geq0,$$ which is obvious for $4u^2-2u-11>0$ or $u>\frac{1+\sqrt{45}}{4}.$
Id est, it's enough to prove that $$6\sqrt{3}(2u+1)\geq\sqrt{u-1}(-4u^2+2u+11)$$ for $$1\leq u\leq \frac{1+\sqrt{45}}{4}.$$
Indeed, we need to prove that:
$$108(2u+1)^2\geq(u-1)(4u^2-2u-11)^2$$ or
$$229+355u+304u^2+68u^3+32u^4-16u^5\geq0$$ or
$$229+355u+304u^2+24u^3+24u^4+4u^3(11+2u-4u^2)\geq0$$ and we are done in this case.
Also, we need to check, what happens for $b=c$ and for $a=b$.
Two these cases lead to inequalities of one variable.
I hope there is a solution without LM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3874529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
When does this limit coincide with the derivative of the function? I am working with the following limit
$$\lim_{h\to 0}\frac{f\left(\frac{1}{c+1}+h\right)-f\left(\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1})}\right)}{h}$$
where $c$ is a constant with $0 \leq c \leq 1$. Note that $$\lim_{h \to 0}\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1})}=\frac{1}{c+1}$$
Under what conditions on the function f will this limit coincide with the derivative of $f$ at $\frac{1}{1+c}$ (assuming this exists):
$$f'\left(\frac{1}{1+c}\right)=\lim_{h \to 0}\frac{f\left(\frac{1}{c+1}+h\right)-f\left(\frac{1}{c+1}\right)}{h}$$
| By the MVT, there is $\xi$ between $\frac{1}{c+1}+h$ and $\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1})}$ such that
\begin{eqnarray}
&&f\left(\frac{1}{c+1}+h\right)-f\left(\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1})}\right)\\
&=&f'(\xi)\left[\left(\frac{1}{c+1}+h\right)-\left(\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1})}\right)\right].
\end{eqnarray}
Let
$$ g(h)=\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1)}}=\frac{1}{2+\sqrt{c\left(\frac{1}{\frac{1}{c+1}+h}-1\right)}(1-c^{-1)}} $$
and then $g(0)=\frac{1}{c+1}$ and
$$ g'(0)=-\frac{1}{\left[2+\sqrt{c\left(\frac{1}{\frac{1}{c+1}+h}-1\right)}(1-c^{-1)}\right]^2}\cdot\frac12 \frac{1-c^{-1}}{\sqrt{c\left(\frac{1}{\frac{1}{c+1}+h}-1\right)}}(-1)\frac{1}{(\frac{1}{c+1}+h)^2}\bigg|_{h=0}=\cdots$$
So
$$ g(h)=g(0)+g'(0)h+\cdots $$
and hence
$$\lim_{h->0}\frac{f\left(\frac{1}{c+1}+h\right)-f\left(\frac{1}{2+\sqrt{c\frac{1-\frac{1}{c+1}-h}{\frac{1}{c+1}+h}}(1-c^{-1})}\right)}{h}=-f'(\frac1{c+1})\dots.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
minimum value of $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$ if $x^2+y^2+z^2=1$ If $x^2+y^2+z^2=1$ what is the minimum value of $\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}$ for $x,y,z \gt 0$ ?
I would like to know if the minimum could be found using simpler ways.(like $AM \ge GM \ge HM$).
knowing $xy+yz+xz \geq \frac{-1}{2}$ for real $x,y,z$ might help.
| As mentioned in comments, the standard way to approach such constrained minimization problems is to use the method of Lagrange multipliers. There might be a simpler or more elegant solution, I don't know.
Let
$$s = \frac{xy}z + \frac{yz}x + \frac{xz}y$$
That is,
$$s = \frac{(xy)^2 + (yz)^2 + (xz)^2}{xyz}$$
Alternatively,
$$s = xyz\left(\frac1{x^2} + \frac1{y^2} + \frac1{z^2}\right)$$
Also note that
$$s^2=2 +
\left(\frac{xy}{z}\right)^2 + \left(\frac{yz}{x}\right)^2 + \left(\frac{xz}{y}\right)^2$$
(That form might lead to a simpler solution).
From symmetry considerations, it seems reasonable that $x=y=z=\sqrt3/3, s=\sqrt3$ is the solution. We can verify that it is a (local) stationary point of $s$ using first order differences.
The constraint equation is
$$x^2 + y^2 + z^2 = 1$$
so
$$2x\Delta x + 2y\Delta y + 2z\Delta z = 0$$
If we let $x=y=z=q=\sqrt3/3$, that reduces to
$$\Delta x + \Delta y + \Delta z = 0$$
or
$$\Delta z=-\Delta x - \Delta y$$
In what follows, we plug
$$\begin{align}\\
x=q+\Delta x\\
y=q+\Delta y\\
z=q+\Delta z\\
s=3q+\Delta s\\
\end{align}$$
into
$$s = \frac{(xy)^2 + (yz)^2 + (xz)^2}{xyz}$$
The aim is to show that $\Delta s=0$
First, note that
$$(q+\Delta x)^2=q^2+2q\Delta x$$
etc. And
$$(q^2+2q\Delta x)(q^2+2q\Delta y)\\
= q^4 + 2q^3(\Delta x + \Delta y)$$
So the numerator of $s$ is
$$\begin{align}\\
q^4 & + 2q^3(\Delta x + \Delta y)\\
+ q^4 & + 2q^3(\Delta y + \Delta z)\\
+ q^4 & + 2q^3(\Delta x + \Delta z)\\
= 3q^4 & + 2q^3(2\Delta x + 2\Delta y + 2\Delta z)\\
= 3q^4 &\\
\end{align}$$
And the denominator of $s$ is
$$\begin{align}\\
& (q + \Delta x)(q + \Delta y)(q + \Delta z)\\
= & (q^2 + q(\Delta x + \Delta y))(q + \Delta z)\\
= & q^3 + q^2(\Delta x + \Delta y + \Delta z)\\
= & q^3\\
\end{align}$$
Thus
$$3q + \Delta s = \frac{3q^4}{q^3} = 3q$$
Hence
$$\Delta s = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3880366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Interpolation polynomial of second degree - deriving coefficients formula
Prove that if $p(x)=\alpha x^2+ \beta x + \gamma$ is an interpolating polynomial of second degree of function $f$ (so $p(a)=f(a), p(b)=f(b), p(c)=f(c)$), passing through points $a<c<b$, then
$$
\alpha = \frac{(b-c)f(a)+(c-a)f(b)+(a-b)f(c)}{(b-c)(a-c)(a-b)} \\
\beta = - \left(\frac{(b^2-c^2)f(a)+(c^2-a^2)f(b)+(a^2-b^2)f(c)}{(b-c)(a-c)(a-b)}\right)
$$
I guess it should come from a set of equations:
$$
p(a) = f(a)=\alpha a^2 + \beta a + \gamma\\
p(b) = f(b)=\alpha b^2 + \beta b + \gamma\\
p(c) = f(c)=\alpha c^2 + \beta c + \gamma
$$
but I am completely clueless when solving it. I can't get neither $\alpha$ nor $\beta$ looking like in the thesis.
Could anyone give me a hand?
| Expand
$$\alpha=\frac{\begin{vmatrix}f(a)&a&1\\f(b)&b&1\\f(c)&c&1\\\end{vmatrix}}{\begin{vmatrix}a^2&a&1\\b^2&b&1\\c^2&c&1\\\end{vmatrix}}$$
$$\beta=\frac{\begin{vmatrix}a^2&f(a)&1\\b^2&f(b)&1\\c^2&f(c)&1\\\end{vmatrix}}{\begin{vmatrix}a^2&a&1\\b^2&b&1\\c^2&c&1\\\end{vmatrix}}$$
The denominators are Vandermonde.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3882427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Asymptotic behavior of tail series $A_n=\sum_{m\ge n+1} \frac{n!}{m!}$ invoked by $e$ Recalling a series, denote
$$
A_n=\sum_{m=n+1}^{\infty} {\frac{n!}{m!}}
$$
which is relative to the tail of $e$, and sometime emerges in some well-known limits like
$$
\lim_{n\to\infty} n\sin(2\pi e n!) = 2\pi
$$
as $n\to\infty$, the very first order of $A_n$ is trivial, since
$$
\frac1{n+1}<A_n<\frac1{n-1}
$$
so
$$
A_n \sim \frac1{n} + o(n^{-1})
$$
when it comes to the higher order, a possible method I used is using the Euler–Maclaurin formula with reciprocal gamma function, which may be an arguably inconvenient approach for this 'simple' series. Here, may I ask for other quick methods to its asymptotic form, some first items of which I found is (may not correct)
$$
A_n \sim \frac1{n} - \frac1{n^3} + \frac1{n^4} + o(n^{-4})
$$
where the squared item is happened to be missing.
| We can rewrite $A_n$ in multiple ways
$$
\eqalign{
& A_{\,n} = \sum\limits_{n + 1\, \le \,m} {{{n!} \over {m!}}} = \sum\limits_{0\, \le \,k}
{{{n!} \over {\left( {n + 1 + k} \right)!}}} = \cr
& = \sum\limits_{0\, \le \,k} {{1 \over {\left( {n + 1} \right)^{\,\overline {\,k + 1\,} } }}}
= \sum\limits_{0\, \le \,k} {n^{\,\underline {\, - \left( {k + 1} \right)\,} } } = \cr
& = {1 \over {n + 1}}\sum\limits_{0\, \le \,k} {{1 \over {\left( {n + 2} \right)^{\,\overline {\,k\,} } }}}
= {1 \over {n + 1}}{}_1F_{\,1} \left( {\left. {\matrix{ 1 \cr {n + 2} \cr } \;} \right|\;1} \right) = \cr
& = \Gamma \left( {n + 1} \right)\sum\limits_{0\, \le \,k} {{1 \over {\Gamma \left( {n + 1 + k + 1} \right)}}}
= \;e\;\gamma \left( {n + 1,1} \right) \cr}
$$
where:
*
*$n^{\,\underline {\,k\,} } ,\quad n^{\,\overline {\,k\,} } $ represent respectively the
Falling and Rising Factorial;
*${}_1F_{\,1}$ is the Confluent Hypergeometric Function;
*$\gamma(s,z)$ is the Lower Incomplete Gamma Function.
From the expression in the Rising Factorial, inverting $n$ into $1/z$ we get the asymptotics
$$
\begin{array}{l}
\frac{1}{{\left( {n + 1} \right)^{\,\overline {\,k + 1\,} } }}\quad \left| {\;z = } \right.\frac{1}{n}\quad = \frac{1}{{\left( {\frac{1}{z} + 1} \right)^{\,\overline {\,k + 1\,} } }} = \\
= \frac{1}{{\left( {\frac{1}{z} + 1} \right)\left( {\frac{1}{z} + 2} \right) \cdots \left( {\frac{1}{z} + k + 1} \right)}} = \\
= \frac{{z^{\,\left( {k + 1} \right)} }}{{\left( {z + 1} \right)\left( {2\,z + 1} \right) \cdots \left( {\left( {k + 1} \right)\,z + 1} \right)}}\quad \left| {\,\left| {\,z\,} \right| < \frac{1}{{k + 1}}} \right.\quad = \\
= z^{\,k + 1} \left( {\sum\limits_{0\, \le \,l_{\,1} \,} {\left( { - z} \right)^{\,\,l_{\,1} } } } \right)\left( {\sum\limits_{0\, \le \,\,\,l_{\,2} \,} {\left( { - 2\,z} \right)^{\,\,l_{\,2} } } } \right) \cdots \left( {\sum\limits_{0\, \le \,\,\,l_{\,k + 1} \,} {\left( { - \left( {k + 1} \right)\,z} \right)^{\,\,l_{\,k + 1} } } } \right) = \\
= z^{\,k + 1} \sum\limits_{0\, \le \,s\,} {\left( { - 1} \right)^{\,\,s} \left( {\sum\limits_{\scriptstyle \left\{ {\begin{array}{*{20}c}
{0\, \le \,l_{\,j} } \\
{l_{\,1} + l_{\,2} + \cdots l_{\,k + 1} = s} \\
\end{array}} \right. \atop
\scriptstyle \, } {\prod\limits_{1\, \le \,j\, \le \,k + 1} {j^{\,\,l_{\,j} } } } } \right)\;z^{\,\,s} } = \\
= z^{\,\,k + 1} \sum\limits_{0\, \le \,s\,} {\left( { - 1} \right)^{\,\,s} \left\{ \begin{array}{c}
s + \,k + 1 \\
\,k + 1 \\
\end{array} \right\}\;z^{\,\,s} } = \sum\limits_{0\, \le \,s\,} {\left( { - 1} \right)^{\,\,s - \,k - 1} \left\{ \begin{array}{c}
s \\
\,k + 1 \\
\end{array} \right\}\;z^{\,\,s} } \\
\end{array}
$$
or more simply
$$
\eqalign{
& {1 \over {\left( {n + 1} \right)^{\,\overline {\,k + 1\,} } }} = n^{\,\underline {\, - \left( {k + 1} \right)\,} } = \cr
& = \sum\limits_{0\, \le s\,} {\left( { - 1} \right)^{\,\,\, - \left( {k + 1} \right) - s}
\left[ \matrix{ - \left( {k + 1} \right) \cr s \cr} \right]\;n^{\,\,s} } = \cr
& = \sum\limits_{0\, \le s\,} {\left( { - 1} \right)^{\,\,\,k + 1 - s}
\left\{ \matrix{ - s \cr k + 1 \cr} \right\}\;n^{\,\,s} } = \cr
& = \sum\limits_{0\, \le s\,} {\left( { - 1} \right)^{\,\,\,k + 1 - s}
\left\{ \matrix{ s \cr k + 1 \cr} \right\}\;n^{\,\, - \,s} } \cr}
$$
Thus
$$
\eqalign{
& A_{\,n} = \sum\limits_{0\, \le s\,} {\left( {\sum\limits_{0\, \le \,k\,\left( { \le \,s - 1} \right)}
{\left( { - 1} \right)^{\,\,\,k + 1 - s} \left\{ \matrix{ s \cr k + 1 \cr} \right\}\;} } \right)n^{\,\, - \,s} } = \cr
& = \sum\limits_{0\, \le s\,} {{{c_{\,s} } \over {n^{\,s} }}} \cr
& c_{\,s} = 0,1,0, - 1, - 1,2, - 9,9,50, - 267,413,2180, \cdots \cr}
$$
But the Stirling Numbers are known to grow very rapidly.
Here in fact is a lin-log graph of the coefficients.
The expression in Incomplete Gamma instead gives
$$
\eqalign{
& A_{\,n} = \;e\;\gamma \left( {n + 1,1} \right) \sim \cr
& \sim {{\Gamma \left( {n + 1} \right)e^{\,n + 1} } \over {\sqrt {2\pi } \left( {n + 1} \right)^{\,n + 3/2} }}
\left( {1 + {{11} \over {12\left( {n + 1} \right)}} - {{23} \over {288\left( {n + 1} \right)^{\,2} }}
+ O\left( {{1 \over {n^{\,3} }}} \right)} \right) \cr}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3883120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proof on 3 equations crossing at one point $(a+b+c=0)$ WITHOUT determinant of the matrix So, as in title, I have to prove that if there are 3 linear equations:
*
*$ax + by + c = 0$
*$bx + cy + a = 0$
*$cx + ay + b = 0$
and those cross at one point, then $a + b + c = 0$. I know that's true because it is easy to do the proof using matrix determinants. However, in that case I was told that it is possible to do the proof using only linear algebra in matrix. I have no idea how to do it and I'm curious.
Thank you in advance for your effort.
edit:
I think I know the answer. You can correct me if I'm wrong. If we have a matrix of those equations it looks like this: r_1 = [a b c], r_2 =[b c a], r_3 = [c a b]. If we add r_1 and r_2 to r_3 we get a follownig equation: (a+b+c)x + (b+c+a)y + (c+a+b) = 0. The only way this equaton can be true is if a+b+c=0.
| Let $A$ be the matrix of the coefficients and $A|B$ the augmented matrix.
Overdeterminate system like this have solution only if $\text{rank}(A)=\text{rank}(A|B)$.
As $\text{rank}(A)=2$ must be $\text{rank}(A|B)=2$ which means that
$$\det(A|B)=\left|
\begin{array}{ccc}
a & b & -c \\
b & c & -a \\
c & a & -b \\
\end{array}
\right|=0$$
which is $$a^3-3 a b c+b^3+c^3=0\to (a+b+c) \left(a^2-a b-a c+b^2-b c+c^2\right)=0$$
satisfied when $a+b+c=0$.
$$.......................................$$
Alternatively
$$
\begin{cases}
ax + by + c = 0\\
bx + cy + a = 0\\
cx + ay + b = 0\\
\end{cases}
$$
Solve the first two equations
$$x=\frac{c^2-a b}{b^2-a c};\;y=\frac{a^2-b c}{b^2-a c}$$
plug in the third
$$\frac{a^3-3 a b c+b^3+c^3}{b^2-a c}=0\to a^3-3 a b c+b^3+c^3=0\\ (a+b+c) \left(a^2-a b-a c+b^2-b c+c^2\right)=0\to a+b+c=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I solve $(3x^2 \tan y- 2y^3/x^3 )dx + (4y^3 + x^3 \cos^2 y+ 3y^2/x^2 ) dy=0$ Somebody please help me to solve this Differential equation.
$\left(3x^2\tan (y)- \frac{2y^3}{x^3}\right)dx+ \left(4y^3 + \frac{x^3}{\cos^2y} + \frac{3y^2}{x^2}\right) dy=0 $
| $$\left(3x^2\tan (y)- \frac{2y^3}{x^3}\right)dx+ \left(4y^3 + \frac{x^3}{\cos^2y} + \frac{3y^2}{x^2}\right) dy=0$$
$$\left(\tan (y)dx^3+ y^3d\dfrac 1 {x^2}\right)+ \left(dy^4 +{x^3}d{\tan y} + \frac{dy^3}{x^2}\right) =0$$
$$d(\tan (y)x^3)+ d\dfrac {y^3} {x^2}+ dy^4 =0$$
Integrate.
$$x^3\tan (y)+ \dfrac {y^3} {x^2}+ y^4 =C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $C\frac{dC}{dr}\ + S\frac{dS}{dr}\ = (C^2 + S^2)\cos{\theta}$ Given $$C=1+r\cos{\theta}\ +\frac{r^2\cos{2\theta}}{2!}\ + \frac{r^3\cos{3\theta}}{3!}\ + \dotsb$$ and $$S = r\sin{\theta}\ + \frac{r^2\sin{2\theta}}{2!}\ + \frac{r^3\sin{3\theta}}{3!}\ + \dotsb$$
Show the following$$C\frac{dC}{dr}\ + S\frac{dS}{dr}\ = (C^2 + S^2)\cos{\theta}$$
I am currently solving the problems given in Differential Calculus for Beginners by Joseph Edwards. As a beginner I am completely clueless about the approach to the above question. I tried to find the answer keys to this book, but sadly non exist on the internet.
| $$
C\frac{dC}{dr}+S\frac{dS}{dr}=C\sum_{n=1}^\infty\frac{nr^{n-1}\cos n\theta}{n!}+S\sum_{n=1}^\infty\frac{nr^{n-1}\sin n\theta}{n!}\\
=C\sum_{n=1}^\infty\frac{ r^{n-1}\cos n\theta}{(n-1)!}+S\sum_{n=1}^\infty\frac{ r^{n-1}\sin n\theta}{(n-1)!}=C\sum_{n=0}^\infty\frac{ r^{n }\cos (n+1)\theta}{ n !}+S\sum_{n=0}^\infty\frac{ r^{n }\sin (n+1)\theta}{ n !}\\
=C\sum_{n=0}^\infty\frac{ r^ n }{ n !}(\cos n\theta\cos\theta-\sin n\theta\sin\theta )+S\sum_{n=0}^\infty\frac{ r^ n }{ n !}(\sin n\theta\cos\theta+\cos n\theta\sin\theta )\\
= C^2 \cos\theta-C S\sin\theta+SC\sin\theta+S^2 \cos\theta
=(C^2+S^2)\cos\theta
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
$(a^4+b^2) \leq (a^2+b^2)^2$ for all $a, b \in \mathbb{R}_+$? Let $a, b \in \mathbb{R}_+$.
Question. Is valid that
$$(a^4+b^2) \leq (a^2+b^2)^2?\tag{1}$$
I tried what follows:
$$(a^2+b^2)^2=a^4+2a^2b^2+b^4 \geq a^4+2a^2b^2.$$
For this, can I to conclude that $(1)$ holds?
| HINT
You can simply expand the RHS and subtract the LHS:
\begin{align*}
(a^{2} + b^{2})^{2} = a^{4} + 2a^{2}b^{2} + b^{4} \Rightarrow (a^{2} + b^{2})^{2} - a^{4} - b^{2} = 2a^{2}b^{2} + b^{4} - b^{2} = b^{2}(2a^{2} + b^{2} - 1)
\end{align*}
When the last expression is nonnegative?
Answer: outside the ellipse $2a^{2} + b^{2} = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$ Prove that $\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}}\gt1$, where a, b, c >0.
I tried,
$\sqrt{\frac{a}{a+b}}+\sqrt{\frac{b}{b+c}}+\sqrt{\frac{c}{c+a}} = \frac{a}{\sqrt{a}\sqrt{a+b}}+\frac{b}{\sqrt{b}\sqrt{b+c}}+\frac{c}{\sqrt{c}\sqrt{c+a}} \ge \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sqrt{a}\sqrt{a+b}+\sqrt{b}\sqrt{b+c}+\sqrt{c}\sqrt{c+a}}$.
Then $\sqrt{a}\sqrt{a+b} \le \sqrt{2}/2 (2a+(a+b))$.
This only gave me $\sqrt{2}/2 + \sqrt{2}(\frac{ab+bc+ca}{a+b+c})$.
| Another way.
$$\sum_{cyc}\sqrt{\frac{a}{a+b}}=\sum_{cyc}\frac{\sqrt{a}}{\sqrt{a+b}}>\sum_{cyc}\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}>\sum_{cyc}\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3887110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
To show that there are $3$ numbers between $n^2$ and $(n+1) ^2$ such that they satisfy a certain property. Here is the problem
Show that for any natural number $n$, one can find three distinct
natural numbers $a, b, c$ between $n^2$ and $(n + 1)^2$ such that $a^2 + b^2$ is divisible
by $c$.
It seems like here are several ways to express $a^2 + b^2$.
\begin{array}
aa^2 + b^2 &= (a - b) ^2 + 2ab \\
&= (a + b) ^2 - 2ab \\
&= \frac { (a + b) ^2 + (a - b)^2 }{2} \\
&= 2b^2 + (a + b)(a - b)
\end{array}
What the problem is literally saying: Find algebraic expressions $X,Y$ and $Z$ such that $n^2 \lt X,Y,Z \lt(n + 1)^2$ and $Y^2 + Z^2 = AX$, for some expression $A$. (Hence, $X|Y^2 + Z^2$)
In the original problem, $X$ is $c$, $Y$ and $Z$ are $b$ and $a$.
People with enough experience might know the answer at once, but is it possible to actually find the expressions from identities/relations? For example the relations that I have found above?
From my trying it is very difficult because one needs to search for expression that are restricted between $n^2$ and $(n + 1)^2$, and then the factoring. What I tried was just plugging random expressions to $Y$ and $Z$ and trying to find a suitable $X$ that suffices the factoring, but to no avail.
Please help, and thanks for that!
| There is in fact another way to do this:
Thm 1: Let $N$ be a sufficiently large integer. Then between $N$ and
$N +\lceil 2\sqrt{N} \rceil$ there are 3 integers $A$, $B$, $C$ such that
$C|(A^2+B^2)$.
To prove Thm 1, we make the following claim:
Claim 2: For each $M$ and $k$, the equation $(M-k)^2 \equiv_k (M+k)^2 \equiv_M k^2$.
We now finish the proof of Thm 1: So let $k_1,k_2$ be distinct positive integers both no larger than $\sqrt{N}$ such that $N+\sqrt{N} > k_1^2+k^2_2 > N$, picking
$k_1=\lfloor \sqrt{N} \rfloor$ and $k_2 = O(N^{1/4})$ should work.
Then let $C=k^2_1+k^2_2$, and then let $A=C+k_1$ and $B=C+k_2$. Then $A,B,C$ satisfy the conditions of Thm 1. $\surd$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3887734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Linear dependency in $\mathbb{R}^4$ Show that the three vectors are linearly dependent in $\mathbb{R}^4$.
$u=(0,3,1,-1)\ v=(6,0,5,1)\ w=(4,-7,1,3)$
They are linearly independent if $c_{1}u+c_{2}v+c_{3}w=0\Rightarrow c_{1}=c_{2}=c_{3}=0$
$$
c_{1}u+c_{2}v+c_{3}w\Rightarrow \begin{pmatrix}
0 & 6 &4 \\
3&0 &-7 \\
1 &5 &1 \\
-1 &1 &3
\end{pmatrix}
\Rightarrow
\begin{pmatrix}
1 &0 &-\frac{7}{3} \\
0 &1 &\frac{2}{3} \\
0 &0 &0 \\0
&0 &0
\end{pmatrix}
$$
*
*How does that show that the vectors are linearly dependent?
*Write each vector as a linear combination of the other two.
EDIT.
$$\begin{pmatrix}
0\\3
\\1
\\-1
\end{pmatrix}=a\begin{pmatrix}
6\\0
\\5
\\1
\end{pmatrix}\cdot b\begin{pmatrix}
4\\-7
\\1
\\3
\end{pmatrix}\Rightarrow a=\frac{2}{7},\ b=-\frac{3}{7}$$
Have i understood 2. correct?
| We have found, by row reduced echelon form (RREF), that the system $Ac=0$ has a not trivial solution (i.e. a solution with $c\neq 0$) that is precisely the definition for linear dependency for the columns of $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3889834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
One root common to $ax^2+2bx+c=0$ and $dx^2+2ex+f=0$
If three distinct numbers $a,b,c$ are in GP, and the equations $ax^2+2bx+c=0$ and $dx^2+2ex+f=0$ have a common root, then which of the following statements is correct? $1.$ $d,e,f$ are in GP. $2.$ $d,e,f$ are in AP. $3.$ $\frac da,\frac eb, \frac fc$ are in GP. $4.$ $\frac da,\frac eb, \frac fc$ are in AP.
My attempt:
Let $r$ be the common ratio in $a,b,c$. So, $b=ar, c=ar^2$.
So, the first equation becomes $ax^2+2arx+ar^2=0\implies x^2+2rx+r^2=0$.
Let $\alpha$ be the common root. So, $\alpha^2+2r\alpha+r^2=0$. Also, $d\alpha^2+2e\alpha+f=0\implies \alpha^2+2\frac ed\alpha+\frac fd=0$.
On comparing, I get $r=\frac ed, r^2=\frac fd\implies(\frac ed)^2=\frac fd\implies e^2=fd$.
So, I am getting option $1$ as correct. But the answer is given as $4$. What's my mistake?
While this post indeed has a lot of good answers, my question was about my mistake in the method I followed. Scilife has answered that in the comments below.
| From the progression $ \ a \ , \ b = ar \ , \ c = ar^2 \ \ , $ we can write the first quadratic polynomial as
$$ ax^2 \ + \ 2bx \ + \ c \ \ = \ \ ax^2 \ + \ 2·ar·x \ + \ ar^2 \ \ , \ \ $$
for which we immediately see that it is a binomial-square with a "double zero" $ \ -r \ \ , $ as Scilife has observed. For the second polynomial, we label its other zero as $ \ s \ $ and may then express it as
$$ dx^2 \ + \ 2ex \ + \ f \ \ = \ \ d·(x + r)·(x - s) \ \ = \ \ dx^2 \ + \ d·(r - s)·x \ + \ (-d·r·s) \ \ . $$
Since the choice of $ \ s \ $ is unspecified, there is no general statement we can make about what sort of progression $ \ d \ , \ \frac12 d·(r - s) \ , \ -d·rs \ \ $ constitutes. To investigate the other choices for this question, we might directly construct the ratios
$$ \frac{e}{b} \ \ = \ \ \frac{\frac12·d \ · \ (r \ - \ s)}{a·r} \ \ = \ \ \left(\frac12 \ - \ \frac{s}{2r} \right)·\frac{d}{a} \ \ \ , \ \ \ \frac{f}{c} \ \ = \ \ \frac{-d· \ r ·s}{a·r^2} \ \ = \ \ - \frac{s}{r} ·\frac{d}{a} \ \ . $$
This form for the expressions then makes it evident that
$$ \frac{\frac{d}{a} \ + \ \frac{f}{c}}{2} \ = \ \frac12 · \left( \ \frac{d}{a} \ + \ \left[ \ - \frac{s}{r} ·\frac{d}{a} \ \right] \ \right) \ \ = \ \ \frac12 · \frac{d}{a} · \left( \ 1 \ - \ \frac{s}{r} \ \right) \ \ = \ \ \frac{e}{b} \ \ , $$
indicating that choice $ \ \mathbf{(4)} \ $ is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3892161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}=\sqrt{2 \pi }$ This question is the last part of a problem leading to proof of Stirling's approximation. I've already proved that $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}$ exists and that $\underset{n\to \infty }{\text{lim}}\frac{2^{4 n} (n!)^4}{((2 n)!)^2 \ (2 n+1)}=\frac{\pi }{2}$.
Hence, the question asks to assume $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}$ exists, and then asks to use $\underset{n\to \infty }{\text{lim}}\frac{2^{4 n} (n!)^4}{((2 n)!)^2 \ (2 n+1)}=\frac{\pi }{2}$ to show $\underset{n\to \infty }{\text{lim}}\frac{n!}{n^{n+\frac{1}{2}} \ e^{-n}}=\sqrt{2 \pi }$.
My attempt goes like this:
Since $\sqrt{x}$ is continuous, we can use $\sqrt{\underset{n\to \infty }{\text{lim}}f(n)}=\underset{n\to \
\infty }{\text{lim}}\sqrt{f(n)}$ to get $\underset{n\to \infty }{\text{lim}}\frac{2^{2 n} (n!)^2}{(2 n)! \
\sqrt{2 n+1}}=\sqrt{\frac{\pi }{2}}$.
Then we can eliminate $2^{2 n}n!$ to get $$\frac{2^{2 n} (n!)^2}{\sqrt{1+2 n} (2 n)!}=\frac{n!}{\sqrt{1+2 n} \left(n-\frac{1}{2}\right) \left(n-\frac{3}{2}\right) \cdots \
\frac{3}{2}\frac{1}{2}}$$
Then factor out $n^n$ and adjust $\sqrt{1+2n}$ to get $$\frac{2^{2 n} (n!)^2}{\sqrt{1+2 n} (2 n)!}=\frac{n!}{n^{n+\frac{1}{2}} \sqrt{2+\frac{1}{n}} \left(1-\frac{1}{2n}\right) \left(1-\frac{3}{2n}\right) \cdots \
\frac{3}{2n}\frac{1}{2n}}$$
The $\sqrt{2+\frac{1}{n}}$ factor will give a $\sqrt{2}$, so the remaining $\prod _k^n \left(1-\frac{2 k-1}{2 n}\right)$ must somehow relate to $\sqrt{2} e^{-n}$.
However, I'm not sure how to do this.
| I didn't use the given limit, but Stirling's approximation can solve this in one shot. You can rewrite the given limit as:
$$L = \lim_{n \to \infty}\frac{n!}{\left( \frac ne\right)^n \sqrt n}$$
Then, by stirling's approximation $n! = \sqrt{2\pi n}\ \left( \frac ne \right)^n + O(\frac{1}{n})$, the limit becomes
$$L = \frac{\sqrt{2\pi} \cdot n!}{n!} = \sqrt{2\pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3893379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find every point whose distance from each of the two coordinate axes equals its distance from the point $(4, 2)$ Here is my attempt:
Let $(x, y)$ be any such point.
Then the distance of $(x, y)$ from the $x$-axis is $\lvert y \rvert$, whereas the distance of that point from the $y$-axis is $\lvert x \rvert$. Thus we have the equalities
$$
\lvert x \rvert = \lvert y \rvert = \sqrt{ (x-4)^2+(y-2)^2}.
$$
From $\lvert x \rvert = \lvert y \rvert$, we obtain $y = \pm x$.
Thus we have the equations
$$
\sqrt{ (x-4)^2 + (\pm x -2)^2} = \lvert x \rvert,
$$
which implies
$$
(x-4)^2 + ( \pm x - 2)^2 = x^2,
$$
which simplifies to
$$
x^2 - 2(4 \pm 2)x + 20 = 0.
$$
Thus we have the following two quadratic equations
$$
x^2 - 12x + 20 = 0 \qquad \mbox{ and } \qquad x^2 -4x + 20 = 0,
$$
and the solutions of these quadratic equations are
$$
x = \frac{12 \pm 8 }{ 2 } \qquad \mbox{ and } \qquad x = \frac{ 4 \pm 8 \iota }{ 2 },
$$
that is,
$$
x = 10, 2 \qquad \mbox{ and } \qquad x = 2 \pm 4 \iota.
$$
We will of course only need the real values for our $x$.
Thus there are eight possible points satisfying the condition given in the problem, namely
$$
(10, 10), (10, -10), (-10, 10), (-10, -10), (2, 2), (2, -2), (-2, 2), (-2, -2).
$$
Is my solution correct in terms of the technique employed as well as the answers obtained? Or, are there any mistakes?
| Point lies on the bisector $(t,t)$
$$(t-4)^2+(t-2)^2=t^2$$
$$t^2-12 t+20=0\to t=2;\;t=10$$
Points are $(2,2);\;(10,10)$
The other possibility, $A$ on the other bisector $(t,-t)$, leads to the equation:
$$(-t-2)^2+(t-4)^2 =t^2$$
which has no real solutions.
$$$$
$$
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3896578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
qq, can I solve equations with complex numbers like that? So I have 2 examples:
$\frac{27|z|}{z}=\frac{1}{3}\bar z *z^3 \implies 27|z|=\frac{1}{3}\bar z *z^4 \implies 27|z|=\frac{1}{3} |z|^2*z^3 \implies \frac{27}{|z|}=\frac{1}{3} z^3 \implies \frac{27}{ \sqrt{a^2 + b^2}}=\frac{1}{3} (a+bi)^3 $
On the left hand side I see no imaginary part, therefore I know that: $ b = 0$ nad $ \frac{27}{ a}=\frac{1}{3} a^3 $
$8z(\bar z) = \bar z^5 \implies 8z|z|= \bar z ^5 \implies 8\frac{|z|^2}{\bar z}|z|= \bar z ^5 \implies 8|z|^3= \bar z ^6 \implies 8(\sqrt{a^2 + b^2})^3 = (a-bi)^6$
Same thing - so I know that: $ b = 0 $ , so: $ 8a^3 = a^6$
Is that ok?
| hint
Approach by polar expression with
$$z=re^{it}=r(\cos(t)+i\sin(t))$$
the equation becomes
$$27\frac{r}{re^{it}}=\frac 13.re^{-it}.r^3e^{3it}$$
or, after simplification
$$81=r^4e^{3it}=3^4e^{2ik\pi}$$
So, the solution of your equation is
$$z=3e^{i\frac{2k\pi}{3}}\; where\;\; k=0,1,2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to transform quadratic terms to something like $u^2+v^2$? I am told that by the substitution of $x = (a\cos \theta)u - (b\sin \theta) v$ and $y = (a\sin \theta)u + (b\cos \theta)v$, we can reduce $3x^2 + 2xy + 3y^2$ to $u^2 + v^2$.
I tried to plug in the substitutions, but I did not see how the quadratic term is reduce to $u^2 + v^2$. What I got was something complicated containing $a$, $b$, $\cos \theta$ and $\sin \theta$.
I am asked to reduce $3x^2 + 2xy + 3y^2 - x - 2y$ to $u^2 + v^2$ by similar substitutions. I do not know how to construct this substitution. Could you explain how do find this kind of substitutions in general?
| I think you are using a substitution more complicated than necessary. You also asked as to how such substitutions work in the general case, so I'll address that in this answer.
Consider the general bivariate quadratic polynomial,
$$f(x,y) = ax^2 + 2hxy + by^2 + 2gx + 2fy + c$$
There's a really neat way of writing this using matrices,
$$
f(x,y) =
\begin{pmatrix}x & y & 1\end{pmatrix}
\begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c\end{pmatrix}
\begin{pmatrix} x \\ y \\ 1\end{pmatrix} $$
Or like this,
$$
f(x,y) =
\begin{pmatrix}x & y\end{pmatrix}
\begin{pmatrix}a & h\\ h & b\end{pmatrix}
\begin{pmatrix}x \\ y\end{pmatrix} +
2\begin{pmatrix}g & f\end{pmatrix}
\begin{pmatrix}x \\ y\end{pmatrix}
+c
$$
which is even better as now the terms of degree $0, 1, 2$ are all separate. What we want to do given such a quadratic polynomial, is substitute the variables $(u, v)$ which will make the cross term $xy$ or $uv$ vanish. Note that we cannot always eliminate both the cross and linear terms.
Such a substitution is essentially a rotation of the axes of the plane from $x, y$ to $u, v$. So our problem reduces to finding the appropriate angle by which the axes should be rotated.
Consider the rotation matrix,
$$
R(\theta) =
\begin{pmatrix}
\cos \theta & \sin \theta\\
-\sin \theta & \cos \theta
\end{pmatrix}
$$
$R(\theta)$ is an orthonormal matrix with determinant 1, and hence, its inverse is simply its transpose. This can also be seen as the inverse operator must be a rotation by $-\theta$. So we get,
$$
R^{-1}(\theta) = R^{T}(\theta) = R(-\theta) =
\begin{pmatrix}
\cos \theta & -\sin \theta\\
\sin \theta & \cos \theta
\end{pmatrix}
$$
Now do the substitution
$$
\begin{pmatrix}u \\ v\end{pmatrix} =
R(\alpha)\begin{pmatrix}x \\ y\end{pmatrix}
$$
where $\alpha$ is the appropriate angle.
We get,
$$
f(u,v) =
\begin{pmatrix}u & v\end{pmatrix}
\underbrace{
R(\alpha)
\begin{pmatrix}a & h\\ h & b\end{pmatrix}
R^{-1}(\alpha)}_{M(\text{Let})}
\begin{pmatrix}u \\ v\end{pmatrix}
+
2\begin{pmatrix}g & f\end{pmatrix}
R^{-1}(\alpha)
\begin{pmatrix}u \\ v\end{pmatrix}
+c
$$
Doing the calculations, we find that,
$$
M =
\begin{pmatrix}
a\cos^2 \alpha + b\sin^2 \alpha + h\sin 2\alpha & h\cos 2\alpha - \dfrac{a-b}{2}\sin 2\alpha\\
h\cos 2\alpha - \dfrac{a-b}{2}\sin 2\alpha & a\sin^2 \alpha + b\cos^2 \alpha - h\sin 2\alpha
\end{pmatrix}
$$
where I've used the double angle identities $\cos 2t = \cos^2 t - \sin^2 t$ and $\sin 2t = 2\sin t \cos t$.
Therefore, to make the cross term vanish, the angle $\alpha$ must satisfy
$$h\cos 2\alpha - \dfrac{a-b}{2}\sin 2\alpha = 0$$
There are two cases:
*
*$\alpha = \dfrac{\pi}{4}$ when $a = b$
*$\tan 2\alpha = \dfrac{2h}{a-b}$
In your particular case, $a = b = 3$, so $\alpha = \dfrac{\pi}{4}$. The substitution is just $$(x, y) = \left(\dfrac{1}{\sqrt{2}}u - \dfrac{1}{\sqrt{2}}v, \dfrac{1}{\sqrt{2}}u + \dfrac{1}{\sqrt{2}}v \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How do I solve $\overline{(A \cap B) \cup (\overline{A} \cap C)} = (A \cap \overline{B}) \cup (\overline{A} \cap \overline{C})$? The question is to prove $\overline{(A \cap B) \cup (\overline{A} \cap C)} = (A \cap \overline{B}) \cup (\overline{A} \cap \overline{C})$ using laws of set theory.
I got stuck after the following steps:
$$
\overline{(A \cap B) \cup (\overline{A} \cap C)} = \overline{(A \cap B)} \cap \overline{(\overline{A} \cap C)}
\\ = (\overline{A} \cup \overline{B}) \cap (A \cup \overline{C})
\\ = ((\overline{A} \cup \overline{B}) \cap A) \cup ((\overline{A} \cup \overline{B}) \cap \overline{C})
\\ = (\overline{A} \cap A) \cup (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})
\\ = (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})
$$
Not sure what to do next. I somehow feel that the absorption law could be used here, but don't know how.
| Taking what you have done so far
$$\begin{align}
\overline{(A \cap B) \cup (\overline{A} \cap C)} & = \overline{(A \cap B)} \cap \overline{(\overline{A} \cap C)}\\
& = (\overline{A} \cup \overline{B}) \cap (A \cup \overline{C})\\
& = ((\overline{A} \cup \overline{B}) \cap A) \cup ((\overline{A} \cup \overline{B}) \cap \overline{C})\\
& = (\overline{A} \cap A) \cup (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})\\
& = (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C}) & \tag{1}
\end{align}$$
it looks fine.
Next you go as follows. We know from definition of union that
$$(\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \subseteq (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C})$$
Although
$$\begin{align}
x \in (\overline{B} \cap \overline {C}) & \implies x \in \overline{B} \wedge x \in \overline{C}\\
& \implies \big(x \in \overline{B} \wedge x \in \overline{C} \big) \wedge(\overline{A} \vee A)\\
& \implies \big(x \in \overline{B} \wedge x \in \overline{C} \wedge x \in \overline{A} \big) \vee \big( x \in \overline{B} \wedge x \in \overline{C} \wedge x \in A \big)\\
& \implies (x \in \overline{C} \wedge x \in \overline{A}) \vee (x \in \overline{B} \wedge x \in A)\\
& \implies (x \in \overline{C} \cap \overline{A}) \vee (x \in \overline{B}\cap A)\\
& \implies x \in (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C})
\end{align}$$
which means that $(\overline{B} \cap \overline{C}) \subseteq (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}).$ So we conclude that
$$\begin{align}
(\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C}) & \subseteq (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C})\\
& \subseteq (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C})
\end{align}$$
Continuing from $(1)$ we get the following final conclusion
$$\therefore (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \cup (\overline{B} \cap \overline{C}) = (\overline{B} \cap A) \cup (\overline{A} \cap \overline{C}) \quad \square$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral with square root + Trig $$ \int_0^{\pi} \sqrt{-3 \sin(x) - 3 \cos(x) + 5} ~~dx$$
I tried to use substitution:
$$ u = -3 \sin(x) -3 \cos(x) + 5 \\ du = -3 \cos(x) + 3 \sin(x) $$
But because $ ( \cos(x) )' = - \sin(x)$ it does not work that well (the signs are now flipped...)
I even tried to change the $\cos(x)$ into $\sin( x - \frac{\pi}{2} )$ but it didn't get me very far..
I also tried using this online tool: https://www.integral-calculator.com/
And it couldn't find the anti-derivative, but only gave the answer of:
$$ \approx 5.1363$$
I would appreciate your help, really any way possible to solve this ugly question would be appreciated! Thank you very much!
| Similar to @Andrew's comment make
$$\sin(x)+\cos(x)=\sqrt{2} \cos \left(x-\frac{\pi }{4}\right)=\sqrt{2} \left(1-2 \sin ^2\left(\frac{x }{2}-\frac{\pi}{8}\right)\right)$$ This makes
$$5-3\sin(x)-3\cos(x)=(5-3 \sqrt{2})+6 \sqrt{2} \sin ^2\left(\frac{x}{2}-\frac{\pi }{8}\right)$$
$$\sqrt{5-3\sin(x)-3\cos(x)}=a\,\sqrt{1+k\sin ^2\left(\frac{x}{2}-\frac{\pi }{8}\right)}$$ with $$a=\sqrt{5-3 \sqrt{2}}\qquad \text{and}\qquad k=\frac{6(6+5 \sqrt{2})}{7} $$
Now, let $\frac{x}{2}-\frac{\pi }{8}=t$ that is to say $x=2t+\frac{\pi }{4}$ to make the antiderivative
$$I=2a \int \sqrt{1+k \sin^2 (t)}\,dt =2a\,E(t|-k)$$ where appears the elliptic integral on fthe first kind.
Back to $x$, $a$ and $k$, using the bounds and simplifying, we the have
$$\color{red}{\int_0^{\pi} \sqrt{5-3 \sin(x) - 3 \cos(x) }\,dx =}$$
$$\color{red}{2 \sqrt{5-3 \sqrt{2}} \left(E\left(\frac{\pi }{8}|-\frac{6\left(6+5
\sqrt{2}\right)}{7} \right)+E\left(\frac{3 \pi }{8}|-\frac{6\left(6+5
\sqrt{2}\right)}{7} \right)\right)}$$ which is
$$5.1363328412422184497730638111478712316615873410060$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3899587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integration of hypergeometric function involving power and exponential function I would be grateful if someone can explain how to evaluate the following
$$
\int_{0}^{\infty} y^a exp(-y/b)\,_2F_1(1,1;2;-c y^2) dy
$$
where $a>0$, $b>0$, and $c>2$. I searched on special functions (using Gradestien book and others) but I could not find an answer.
| I do not know if a closed form exist.
Let $y=\frac{x}{\sqrt{c}}$ and $k=\frac{1}{b \sqrt{c}}$ to make the integral to be
$$ c^{-\frac{a+1}{2} }\int_{0}^{\infty}x^{a-2} e^{-k x} \log \left(x^2+1\right)\,dx$$ Let
$$J_a=\int_{0}^{\infty}x^{a-2} e^{-k x} \log \left(x^2+1\right)\,dx$$
For $J_2$ it is simple. Now appear the Meijer G function in all the cases I tried
$$J_1=\frac{G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
0,1 \\
0,0,0,\frac{1}{2}
\end{array}
\right)}{2 \sqrt{\pi }}\qquad \qquad J_3=\frac{2 G_{1,3}^{3,1}\left(\frac{k^2}{4}|
\begin{array}{c}
0 \\
0,0,\frac{3}{2}
\end{array}
\right)}{\sqrt{\pi } k^2}$$
$$J_4=\frac{4 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
0,1 \\
0,0,\frac{3}{2},2
\end{array}
\right)}{\sqrt{\pi } k^3}\qquad \qquad J_5=\frac{8 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
0,1 \\
0,0,2,\frac{5}{2}
\end{array}
\right)}{\sqrt{\pi } k^4}$$
$$J_6=\frac{16 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
0,1 \\
0,0,\frac{5}{2},3
\end{array}
\right)}{\sqrt{\pi } k^5}\qquad \qquad J_7=\frac{32 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
0,1 \\
0,0,3,\frac{7}{2}
\end{array}
\right)}{\sqrt{\pi } k^6}$$
$$J_8=\frac{64 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
0,1 \\
0,0,\frac{7}{2},4
\end{array}
\right)}{\sqrt{\pi } k^7}\qquad \qquad J_9=\frac{128 G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
0,1 \\
0,0,4,\frac{9}{2}
\end{array}
\right)}{\sqrt{\pi } k^8}$$ and, I hope, we could conjecture that for $a >3$
$$\color{blue}{J_a=\frac {2^{a-2}}{k^{a-1}\,\sqrt \pi}G_{2,4}^{4,1}\left(\frac{k^2}{4}|
\begin{array}{c}
0,1 \\
0,0,\frac{k-1}2,\frac{k}{2}
\end{array}
\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding a rational number in between two rationals Can anyone give me some hints or show me how to solve this problem?
Problem Find a rational number in between $\frac{9}{10}$ and $\frac{10}{11}$ which may be written in the form $\frac{m}{2^n}$ where $m$ is an integer and $n$ is a non negative integer.
I just do not know where to even start or what theorems I need to use to approach this. I would appreciate hints or guidance to solving these types of problems.
| Another possibility is to use the following well-known fact:
$$\frac{a}{c} < \frac{b}{d} \Longrightarrow \frac{a}{c} < \frac{a+b}{c+d} < \frac{b}{d}$$
A simple consequence of this is that
$$\frac{a}{c} < \frac{b}{d} \Longrightarrow \frac{a}{c} < \frac{ma+nb}{mc+nd} < \frac{b}{d} \quad (m,n > 0) $$
This means that for your question, we simply have to find multiples of $10$ and $11$ which add to a power of 2. The simplest one would be $1(10) + 2(11) = 32 = 2^5$, which gives
$$
\frac{9}{10} < \frac{1(9) + 2(10)}{1(10) + 2(11)} < \frac{10}{11} \\
\color{white}{\text{empty}}\\
\frac{9}{10} < \frac{29}{32} < \frac{10}{11}
$$
Trying for $64$ or $128$ just yields fractions equivalent to the above, but we can use $19(10) + 6(11) = 256 = 2^8$ to find another one:
$$
\frac{9}{10} < \frac{19(9) + 6(10)}{19(10) + 6(11)} < \frac{10}{11} \\
\color{white}{\text{empty}}\\
\frac{9}{10} < \frac{231}{256} < \frac{10}{11}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$ This is the $y=2^k$ case of this question.
Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?
Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?
| For $k\ge 1$ and $0<x<2^k$, suppose$$(2^{2k}-x^2) | (2^{2k}-1)$$for some $x>1$, and hence that $2^{2k}-x^2$ is composed only of the prime factors of $2^{2k}-1$ but lacks one or more of them (or perhaps contains all of the distinct factors but with one or more of them to a lesser power).
Thus let$$2^{2k}-1=(2^k-1)(2^k+1)=pqr\cdot stu$$and suppose, for some $x>1$, that$$2^{2k}-x^2=pq\cdot stu$$and hence divides $2^{2k}-1$.
Then since for $x\ge1$,$$2^{2k}-x^2=(2^k-x)(2^k+x)$$and hence the sum$$(2^k-x)+(2^k+x)=2^{k+1}$$then$$pqr+stu=pq+stu=2^{k+1}$$which is impossible.
Hence it is clear that removing one or more prime factors from either $2^k-1$ or $2^k+1$, while leaving the other addend intact, must make their sum less than $2^{k+1}$. Still less can their sum be $2^{k+1}$ if one or more prime factors are removed from both $2^k-1$ and $2^k+1$.
The remaining possibility is that removing one or more prime factors from both $2^k-1$ and $2^k+1$ and re-arranging the remaining prime factors, might yield a sum $=2^{k+1}$.
E.g. for $k=6$, $(2^k-1)(2^k+1)=63\cdot65=3^2\cdot7\cdot5\cdot13$, and$$3^2\cdot7+5\cdot13=2^{k+1}=2^7$$Removing one $3$-factor and re-arranging the four remaining distinct prime factors in the seven possible ways we get$$3\cdot7+5\cdot13=86$$$$3\cdot5+7\cdot13=106$$$$3\cdot13+5\cdot13=74$$$$3+7\cdot5\cdot13=458$$$$5+3\cdot7\cdot13=278$$$$7+3\cdot5\cdot13=202$$$$13+3\cdot5\cdot7=118$$Noteworthy here is that all sums are odd multiples of $2^1$.
For $k=10$, removing one of the $5$-factors from $2^{2k}-1=1023\cdot1025=3\cdot11\cdot31\cdot5^2\cdot41$, I find the fifteen possible sums of two addends containing the remaining five distinct primes are all odd multiples of $2^2$.
If it could be shown, then, that all such sums dividing $2^{2k}-1$ are odd multiples of some power of $2$, and hence not equal to $2^{k+1}$, it would follow that, for $k\ge1$ and $x>1$, $2^{2k}-x^2$ does not divide $2^{2k}-1$.
This of course is not a full answer, but it lays out a possible approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 4
} |
Infinite ways to represent $\pi$ as product of nested square roots of $2$ and $2^n$ and odd numbers Polygon inscribed in a circle leads to famous infinite product by Viete's formula.here
One way to represent that as infinite nested radical of 2 as follows
$$\pi = \lim_{n\to\infty}2^n \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}}$$ being total number of '2's inside the nested square roots is 'n'
After some manipulation in numbers some interesting findings I'm able to observe as follows
$$3\pi = \lim_{n\to\infty}2^n \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...\sqrt{2-\sqrt{2}}}}}}}$$
$$5\pi = \lim_{n\to\infty}2^n \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...\sqrt{2-\sqrt{2-\sqrt{2}}}}}}}}$$
In both these situations total number of '2's inside the nested square roots are 'n'.
As the n grows big each converge to respective multiples of $\pi$ more and more decimals
Python code to verify my findings
from decimal import*
getcontext().prec = 100
# nested sqrts of 2 & $\pi$, $3\pi$,$5\pi$
b = int(input("Enter number of iterations: "))
a = Decimal(2).sqrt()
for i in range(b):
a = Decimal(2 + a).sqrt()
a = Decimal(2 - a).sqrt()
Pi = 2 ** (b + 2) * a
print("\n\n Pi derived from Square as initial regular Polygon is:\n\n ", Pi)
a = Decimal(2).sqrt()
a = 2 - a
a = Decimal(a).sqrt()
b = int(input("Enter required number of cycles to calculate Pi: "))
for i in range(b):
a = 2 + a
a = Decimal(a).sqrt()
a = 2 - a
a = Decimal(a).sqrt()
Pi_3 = 2**(b+3)*a
C = Pi_3/3
print("\n\n",C)
a = Decimal(2).sqrt()
a = 2 - a
a = Decimal(a).sqrt()
a = 2 - a
a = Decimal(a).sqrt()
b = int(input("Enter required number of cycles to calculate Pi: "))
for i in range(b):
a = 2 + a
a = Decimal(a).sqrt()
#print("\n\n",a)
a = 2 - a
a = Decimal(a).sqrt()
Pi_5 = 2**(b+4)*a
C = Pi_5/5
print("\n\n",C)
# for practical purposes I have divided by 3 & 5 respectively (next to regular derivation by Viete's method) to get the value of $\pi$
Viete's formula is provable by geometry i.e. square becoming octagon then 16gon and so on ...
Is there any way to prove above 2 formulas by geometric shapes or some other methods? Are there similar infinite ways like those above?(with the help of nested square roots of 2 apart from equilateral triangle, regular Pentagon). Please comment. Thanks in advance
| Similarly to formulas, related to Viète's formula, in your formulas is considered a limit $\lim_{n\to\infty} 2^n\sqrt{2-a_n}$, where $a_n=\sqrt{2+a_{n-1}}$ with the initial conditions
$a_2=-\sqrt{2}$ for the first sequence and $a_3=-\sqrt{2-\sqrt{2}}$ for the second. Assuming $a_{n-1}=2\cos x_{n-1}$ for $x_{n-1}\in [0,\pi]$, we obtain $a_n=2\cos\tfrac {x_{n-1}}{2}$. Putting $x_n=\tfrac {x_{n-1}}2$ for all sufficiently big $n$, we obtain
$$\lim_{n\to\infty} 2^n\sqrt{2-a_n}=\lim_{n\to\infty} 2^n \sqrt{2-2\cos x_n}=\lim_{n\to\infty} 2^{n+1} \sin x_{n+1}.$$
For the first sequence we have $a_2=2\cos\tfrac{3\pi}4$, so $x_2=\tfrac{3\pi}4$ and
$$\lim_{n\to\infty} 2^n\sqrt{2-a_n}=\lim_{n\to\infty} 2^{n+1} \sin \left(\tfrac{3\pi}4\cdot \frac 1{2^{n-1}}\right)=3\pi.$$
For the second sequence we have $a_3=2\cos x_3=-\sqrt{2-\sqrt{2}}<0$. So $x_3\in \left(\tfrac {\pi}2, \pi\right) $ and
$4\cos^2 x_3=2-\sqrt{2}$. Thus $2\cos 2x_3=-\sqrt{2}$ and so $2x_3=\tfrac {5\pi}4$ and
$$\lim_{n\to\infty} 2^n\sqrt{2-a_n}=\lim_{n\to\infty} 2^{n+1} \sin \left(\tfrac{5\pi}8\cdot \frac 1{2^{n-2}}\right)=5\pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Hint for finding the sum: $\sum_{n=0}^{\infty}\frac{1}{4n^2-1} x^{2n+1}$? I know that $\sum_{n=0}^{\infty}\frac{1}{4n^2-1}$ is a telescoping sum, but $x^{2n+1}$ in the sum complicates it a bit.
| The sum is $-x+\dfrac {x^3}3+\dfrac{x^5}{15}+\dfrac{x^7}{35}+\dfrac{x^9}{63}+\cdots+\dfrac{x^{2n+1}}{4n^2-1}+\cdots$.
Its derivative is $-1+x^2+\dfrac{x^4}3+\dfrac{x^6}5+\dfrac{x^8}{7}+\cdots+\dfrac{x^{2n}}{2n-1}+\cdots.$
$=-1+x\left(x+\dfrac{x^3}3+\dfrac{x^5}5+\dfrac {x^7}7+\cdots+\dfrac{x^{2n-1}}{2n-1}+\cdots\right)$
$=-1+x\tanh^{-1}x$.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3921518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\int_0^\infty \frac{\cos(ax)}{(1+x^2)\sqrt{x}}dx$. $$
\mbox{How can I compute}\
\int_{0}^{\infty}\frac{\cos\left(ax\right)}{\left(1 + x^{2}\right)\,\sqrt{\,{x}\,}}\,\mathrm{d}x\ ?.
$$
I saw many questions about computations of that integral without the term $\,\sqrt{\,{x}\,}\,$ and that it can be solved by using various techniques ( contour integral, iterated integral, etc$\ldots$ ). But how can I compute my integral $?$.
| Let
$$
f(a) = \int_0^{+\infty} \frac{\cos(ax)}{(1+x^2)\sqrt{x}} \, \mathrm{d} x,
$$
First, we note that $f(a)=f(-a)$, hence $f$ is even and we will focus only on $a \ge 0$.
Second,
$$
\begin{split}
\left| f(a) \right| &\le \left|\int_0^{+\infty} \frac{\cos(ax)}{(1+x^2)\sqrt{x}} \, \mathrm{d} x \right|\\
&\le \int_0^{+\infty} \frac{\left|\cos(ax)\right|}{(1+x^2)\sqrt{x}} \, \mathrm{d} x \\
&\le \int_0^{+\infty} \frac{\mathrm{d} x}{(1+x^2)\sqrt{x}} = f(0)
\end{split}$$
With some math one finds that
$$
f(0) = \int_0^{+\infty} \frac{\mathrm{d} x}{(1+x^2)\sqrt{x}} =2\int_0^{+\infty} \frac{\mathrm dt}{1+t^4} = \frac{\pi}{\sqrt{2}} \approx 2.22
$$
(see here)
Now, if we differentiate twitce under integral symbol (we can do it because of Leibniz integral rule)
$$
\begin{split}
f''(a) &= -\int_0^{+\infty} \frac{\cos(ax) \cdot x^2}{(1+x^2)\sqrt{x}} \, \mathrm{d} x\\
&= -\int_0^{+\infty} \frac{\cos(ax) \left(x^2+1-1\right)}{(1+x^2)\sqrt{x}} \, \mathrm{d} x\\
&= - \left[\int_0^{+\infty} \frac{\cos(ax)}{\sqrt{x}} \, \mathrm{d} x -\int_0^{+\infty} \frac{\cos(ax) }{(1+x^2)\sqrt{x}} \, \mathrm{d} x \right] \\
&= f(a) - \int_0^{+\infty} \frac{\cos(ax)}{\sqrt{x}} \, \mathrm{d} x
\end{split}$$
Now,
$$
\begin{split}
\int_0^{+\infty} \frac{\cos(ax)}{\sqrt{x}} \, \mathrm{d} x &= \frac{1}{2}\int_\mathbb{R}\frac{\cos(ax)}{\sqrt{|x|}} \, \mathrm{d} x \\
&= \mathcal F \left(\frac{1}{\sqrt{|x|}}\right)(a) \\
&=\frac{1}{2} \sqrt{\frac{2\pi}{|a|}}\\
&= \sqrt{\frac{\pi}{2}}\sqrt{\frac{1}{|a|}}
\end{split}
$$
where we used $ \mathcal F \left(\frac{1}{\sqrt{|x|}}\right)(a) = \frac{1}{2} \sqrt{\frac{2\pi}{|a|}}$ (see Fourier Transform of $\frac{1}{\sqrt{|x|}}$)
So, the problem is now to solve
$$
\begin{cases}
f''(a) -f(a) = -\sqrt{\frac{\pi}{2}}\sqrt{\frac{1}{|a|}}\\
f(0) = \frac{\pi}{\sqrt 2}\\
\end{cases}
$$
From linear differential equations we know that a solution is
$$
f(a) = c_1 \mathrm{e}^a + c_2 \mathrm{e}^{-a} + f_\mathrm p(a)
$$
where $c_1$ and $c_2$ are real numbers and $f_\mathrm p$ is the complementary solution.
Wolfram Alpha helps us to know that
$$
\begin{split}
f(a) &= c_1 \mathrm{e}^a + c_2 \mathrm{e}^{-a} + \sqrt{\frac{\pi}{2}}\int_0^{a} \frac{\mathrm{e}^{-x+a}-\mathrm{e}^{x-a}}{2\sqrt{x}}\, \mathrm d x\\
&= c_1 \mathrm{e}^a + c_2 \mathrm{e}^{-a} + \sqrt{\frac{\pi}{2}}\int_0^{a} \frac{\sinh(a-x)}{\sqrt{x}}\, \mathrm d x
\end{split}
$$
Now, if we add the initial condition, we get
$$
c_1+c_2 = \frac{\pi}{\sqrt 2},
$$
hence, after renaming the constant $c_1 = C$, the solution is
$$
\begin{split}
f(a)&= C \mathrm{e}^a + \left(\frac{\pi}{\sqrt 2} - C\right) \mathrm{e}^{-a} + \sqrt{\frac{\pi}{2}}\int_0^{a} \frac{\sinh(a-x)}{\sqrt{x}}\, \mathrm d x\\
&= 2C \sinh(a) + \frac{\pi}{\sqrt 2} \mathrm{e}^{-a} + \sqrt{\frac{\pi}{2}}\int_0^{a} \frac{\sinh(a-x)}{\sqrt{x}}\, \mathrm d x
\end{split}
$$
If you find some other initial value (for example $f(1)$), then you can eliminate the constant $C$ too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Convergence of $\sum_{n=1}^{\infty} \left(\frac{1}{n} - \sin\left(\frac{1}{n} \right) \right)^{\alpha}$ I am trying to seek the real values of $\alpha$ for convergence of the following series :
$\sum_{n=1}^{\infty} \left(\frac{1}{n} - \sin\left(\frac{1}{n} \right) \right)^{\alpha}$
My trial:
Since $n>1$, $0<\frac{1}{n}<1$, thus we can expand sin function as
\begin{align}
\sin\left(\frac{1}{n} \right) = \frac{1}{n} - \frac{1}{3!} \frac{1}{n^3} + \cdots
\end{align}
Hence the series becomes
\begin{align}
\sum_{n=1}^{\infty} \left(\frac{1}{n} - \sin\left(\frac{1}{n} \right) \right)^{\alpha}
= \sum_{n=1}^{\infty} \left(\frac{1}{3!}\frac{1}{n^3} - \cdots\right)^{\alpha} \leq \frac{1}{3!}\sum_{n=1}^{\infty} \left(\frac{1}{n}\right)^{3\alpha}
\end{align}
Thus my guess is from p-test $3\alpha >1$, i.e., $\alpha>\frac{1}{3}$.
Is this approach admissible?
| Your intuition is correct for using the Taylor expansion of $\sin x$ around $x=0$. Note that for large enough $n$:
$$
{1\over n}-{1\over 6n^3}\le\sin {1\over n}\le {1\over n}-{1\over 5n^3}
$$
You have already done the lower bound. The upper bound is just on its way!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3923160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
inequality $\frac{2(x + y)^2}{2x^2 + y^2} \leq 3$ I'm looking for a "nice" way to show that the following inequality holds, i.e. without differentiating and determining the maximum:
$$
\frac{2(x + y)^2}{2x^2+ y^2} \leq 3 \quad \forall x,y \in \mathbb{R}
$$
It's rather easy to show
$$
\frac{4xy}{x^2 + y^2} \le 2
$$
and obviously
$$
\frac{2x^2 + 2y^2}{2x^2+y^2} \le 2
$$
but combining these two does not give me a sufficient low bound.
| Well $2x^2+y^2> 0$ and $2(x+y)^2\ge 0$ so
So $\frac{2(x + y)^2}{2x^2+ y^2} \leq 3\iff$
$2(x+y)^2 \leq3(2x^2 + y^2) \iff$
$2x^2 + 4xy +2y^2 \leq 6x^2 + 3y^2 \iff$
$0 \leq 4x^2 -4xy +y^2 \iff$
$0 \leq(2x -y)^2 \iff$ God's in his heaven and all's right in the world.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How to solve this quadratic congruence? $27w^2+20w+35 \equiv 0 \pmod{23}$ Given $27w^2+20w+35 \equiv 0 \pmod{23}$
How to solve this quadratic congruence?
All hints are welcome.
I completed the square ,so I obtained $27(x+\frac{10}{27})^2+\frac{845}{27} \equiv 0 \pmod{23} $
I need to get 2 solutions in the least residue system which are the correct answers: $w\equiv 6 \pmod{23}$ and $w \equiv 12 \pmod{23}$
| Since $27 \equiv 4$ we can write the equation as $4w^2 + 20w + 35 \equiv 0.$ Completing the square gives $(2w+5)^2 + 10 \equiv 0,$ i.e. $(2w+5)^2 \equiv -10.$ But $-10 \equiv -10+2\cdot 23=36=6^2,$ so $2w+5\equiv\pm 6,$ i.e. $2w=-5\pm 6.$
Case $+$: $2w=-5+6=1\equiv 1+23=24=2\cdot12$ so $w\equiv12.$
Case $-$: $2w=-5-6=-11\equiv -11+23=12=2\cdot6$ so $w\equiv6.$
Thus the solutions are $w=12$ and $w=6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3925246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove the determinant is $0$ This is Problem 16.17 from the book Exercises in Algebra by A. I. Kostrikin.
Prove that
$$
\left|\begin{array}{ccccc}
\dfrac{1}{2 !} & \dfrac{1}{3 !} & \dfrac{1}{4 !} & \cdots & \dfrac{1}{(2 k+2) !} \\
1 & \dfrac{1}{2 !} & \dfrac{1}{3 !} & \cdots & \dfrac{1}{(2 k+1) !} \\
0 & 1 & \dfrac{1}{2 !} & \cdots & \dfrac{1}{(2 k) !} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & \dfrac{1}{2 !}
\end{array}\right|=0, \quad k \in \mathbb{N}
$$
My Attempt: I tried to expand it by the first column, but it seemed to be more complicated when I did that. I also tried to add edges to the determinant(in the hope that it will be easier to calculate), but I still failed to work it out.
So, My Question is, how to calculate this determinant?
| We want to prove that the $(n-1)\times(n-1)$ matrix
$$
A_n=\pmatrix{
\dfrac{1}{2!} &\dfrac{1}{3!} & \cdots &\cdots & \cdots &\dfrac{1}{n!} \\
1 &\dfrac{1}{2!} &\dfrac{1}{3!} &\cdots &\cdots &\dfrac{1}{(n-1)!} \\
0 & 1 & \ddots & \ddots & \ddots & \vdots\\
\vdots & \ddots & \ddots & \ddots&\ddots & \vdots\\
\vdots & \ddots & \ddots &1& \dfrac{1}{2!} & \dfrac{1}{3!}\\
0 & \cdots & \cdots & 0 & 1 & \dfrac{1}{2!}
}
$$
is singular when $n\ge4$ is even. This can be proved by mathematical induction on $n$. The base case $n=4$ can be verified directly. In the inductive step, note that for any positive integer $m$,
$$
\sum_{k=0}^m\frac{(-1)^k}{k!}\frac{1}{(m-k)!}=\frac{1}{m!}\sum_{k=0}^m(-1)^k\binom{m}{k}=\frac{1}{m!}(1-1)^m=0.
$$
Move the first and the last summands out of the sum, we obtain
$$
\frac{1}{m!}+\sum_{k=1}^{m-1}\frac{(-1)^k}{k!}\frac{1}{(m-k)!}=\frac{(-1)^{m+1}}{m!}.
$$
Denote the $i$-th row of $A_n$ by $a_i$. The previous identity means that
\begin{aligned}
u&:=a_1+\sum_{k=1}^{n-1}\frac{(-1)^k}{k!}a_{k+1}\\
&=\left(-\frac{1}{2!},\,\frac{1}{3!},\,-\frac{1}{4!},\,\frac{1}{5!},\,\ldots,\,-\frac{1}{(n-2)!},\,\frac{1}{(n-1)!},\,\frac{1}{(n-1)!}-\frac{1}{n!}\right).
\end{aligned}
(Note that the last entry of $u$ is not $-\frac{1}{n!}$, because the last column of $A_n$ ends with $\frac{1}{2!}$, not $1$.) In other words, by some appropriate elementary row operations, we can modify the first row of $A$ to $u$. Therefore, by the multilinearity of the determinant function, $\det(A_n)$ remains unchanged if we replace the first row of $A_n$ by
$$
v:=\frac12(a_1+u)
=\left(0,\,\frac{1}{3!},\,0,\,\frac{1}{5!},\,\ldots,\,0,\,\frac{1}{(n-1)!},\,\frac{1}{2\times(n-1)!}\right).
$$
Now suppose $n\ge6$ is even. By Laplace expansion along $v$ and by induction assumption, we get
\begin{aligned}
\det(A_n)
&=-\frac{1}{3!}\det(A_{n-2})-\frac{1}{5!}\det(A_{n-4})-\cdots-\frac{1}{(n-1)!}\det(A_2)+\frac{1}{2\times(n-1)!}\\
&=-\frac{1}{(n-1)!}\det(A_2)+\frac{1}{2\times(n-1)!}\\
&=-\frac{1}{(n-1)!}\frac{1}{2!}+\frac{1}{2\times(n-1)!}\\
&=0.
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3933060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 0
} |
An Interesting Number Theory Problem: Solve $2520=(x+y+xy)^2+2xy+2y-3x, x, y \in \mathbb Z$ Solve:
$2520=(x+y+xy)^2+2xy+2y-3x$
Please help, I tried Solving SFFT and Quadratic, but couldn't end up with a result.
Edit: I solved this much.I factorized it
$((x+1)(y+1)-1)^2+2((x+1)(y+1)-1)-5x$
Then substituted $t=((x+1)(y+1)-1)$
so $t^2+2t-5x=2520$
and after this i am stuck
Edit 2: Solve for Integers
| $$2520=x^2+y^2+(xy)^2+2x^2y+2y^2x+2xy+2xy+2y-3x$$ so, if we group in terms of $y$ $$2520=y^2(x^2+2x+1)+y(2x^2+4x+2)+x^2-3x$$ so $$0=y^2\cdot(x+1)^2+2y(x+1)^2+x^2-3x-2520$$ so solving the quadratic equation in the usual way:
$$\Delta=4(x+1)^4-4(x+1)^2(x^2-3x)=4(x+1)^2(x^2+2x+1-x^2+3x+2520)$$ so we have found out that $$\Delta=(2x+2)^2\cdot(5x+2521)$$
but we know
$$y=\frac{-2(x+1)^2\pm\sqrt{\Delta}}{2(x+1)^2}=\pm\frac{\sqrt{\Delta}}{2(x+1)^2}-1=\pm\frac{2(x+1)\sqrt{5x+2521}}{2(x+1)^2}-1=\pm\frac{\sqrt{5x+2521}}{(x+1)}-1$$ so for $y$ to be an integer, $x+1$ must divide $\sqrt{5x+2521}$ and $5x+2521$ must be a perfect square. Let $5x+2521=k^2$ so $x=\frac{k^2-2521}{5}$ so $$\big(\frac{k^2-2521}{5}+1\big)^2\text{ divides } k$$ so $$(k^2-2516)^2\text{ divides } 25k$$ so $$(k^2-2516)^2\leq 25|k|$$ which gives us $k=\pm50,\pm51$. If $k=\pm 50$, then because $x=\frac{k^2-2521}{5}$, x wouldn't be an integer so contradiction. So $k=51$, so $x=\frac{51^2-2521}{5}=16$
Because $x=16$ wecan now see that because $y=\pm\frac{\sqrt{5x+2521}}{(x+1)}-1=\pm3-1$ so $y$ is $2$ or $-4$.
So the solutions are $(x,y)=(16,-4)$ and $(16,2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3934004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Two persons A and B toss a die. The person who first throws $5$ wins. What is the probability of A’s winning? It is to prove that if A begins, then
$$P(A_{wins})= \frac{6}{11}$$
Similarly, if B begins,
$$P(A_{wins})= \frac{5}{11}$$
Now, for the given problem, P= P(A wins)= P(A begins and wins or B begins and A wins)
But (A begins and wins) and (B begins and A wins) are mutually exclusive events.
So P= P(A begins and wins) + P(B begins and A wins)$= \frac{6}{11}+ \frac{5}{11}= 1.$
This means that A will surely win and B will surely lose, which is clearly incorrect. So what's the mistake in all of this?
| A begins and probability of A wins, it will be an infinity geometric series such that:
$$\frac{1}{6}+(\frac{5}{6})^2\frac{1}{6}+(\frac{5}{6})^4\frac{1}{6}+...=\frac{1}{6}(1+(\frac{5}{6})^2+(\frac{5}{6})^4+...)=\frac{1}{6}(\frac{1}{1-(\frac{5}{6})^2})=\frac{6}{11}$$
B begins and probability of A wins, it will be an infinity geometric series such that:
$$(\frac{5}{6})(\frac{1}{6})+(\frac{5}{6})^3\frac{1}{6}+(\frac{5}{6})^5\frac{1}{6}+...=\frac{5}{36}(1+(\frac{5}{6})^2+(\frac{5}{6})^4+...)=\frac{5}{36}(\frac{1}{1-(\frac{5}{6})^2})=\frac{5}{11}$$
$\frac{6}{11}+\frac{5}{11}=1$ does not mean A always wins.
We have an event and two possibilities
$$P(A_{\text{first and wins}})=1- P(B_{\text{second and loses}})=\frac{6}{11}$$
$$P(A_{\text{second and wins}})=1- P(B_{\text{first and loses}})=\frac{5}{11}$$
One of them always wins.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3937171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integration problem yields two solutions. But are they the same? I have an integration problem: $\int\frac{1}{x\sqrt {x^{12}-1}}dx$.
My attempt was to let $u=\sqrt {x^{12}-1}$. Solving it yields:
$$\frac{1}{6}\int\frac{1}{u^2+1}du$$
$$=\frac{1}{6}\tan^{-1}(u)+C$$
$$=\frac{1}{6}\tan^{-1}\sqrt {x^{12}-1}+C$$
However, my TA substituted $u=\frac{1}{x}$, and subsequently $v=u^6$. Solving it yields:
$$-\frac{1}{6}\int\frac{1}{\sqrt {1-v^2}}dv$$
$$=-\frac{1}{6}\sin^{-1}(v)+C$$
$$=-\frac{1}{6}\sin^{-1}\frac{1}{x^6}+C$$
I do not understand how the two answers are different but mathematically are correct. When I tried to draw a right-angled triangle using the first solution, I would have:
opposite = $\sqrt{x^{12}-1}$
adjacent = $1$
hypotenuse = $x^6$
So from the above, $\sin\theta=\frac{\sqrt{x^{12}-1}}{x^6}$ and $\theta=\arcsin\frac{\sqrt{x^{12}-1}}{x^6}$, which is not the same as $-\arcsin\frac{1}{x^6}$
May I know what is wrong with my deduction above?
Note: I also tried to differentiate both solutions. The first solution will yield back to the integrand. But the second solution does not. But strangely, if I use any arbitrary limits on the two solutions, both give me the same answer.
Thank you in advance.
| According to your consideration,
$\cos\theta = \frac{1}{x^6} \Rightarrow \theta = \cos^{-1}(\frac1{x^6})$
And $\cos^{-1}x = \frac\pi2-\sin^{-1}x$.
So, $\boxed{\theta = \frac\pi2 - \sin^{-1}\left(\frac{1}{x^6}\right)}$.
Now, $\frac{1}{6}\cdot\frac\pi2$ is just a constant and you have the required result as $\boxed{- \frac16\sin^{-1}\left(\frac{1}{x^6}\right) +k}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3937968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
express a matrix using Kronecker product How to express the following matrix as a Kronecker product of two $2\times 2$ matrices:
\begin{equation*}
A =
\begin{pmatrix}
0 & 0 & 0 & 1+j \\
0 & 0 & 1-j& 0 \\
0 & 1-j & 0& 0\\
1+j& 0& 0&0
\end{pmatrix}
\end{equation*}
| I am not sure this is what you were looking for...
$$
A =
\left(\begin{array}{c c}
0 & 1 \\
1 & 0
\end{array}\right)
\otimes
\left(\begin{array}{c c}
0 & 1 \\
1 & 0
\end{array}\right)
+
\left(\begin{array}{c c}
0 & 1 \\
-1 & 0
\end{array}\right)
\otimes
\left(\begin{array}{c c}
0 & j \\
-j & 0
\end{array}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3940070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find an orthogonal basis for the subspace described by $x − y − z = 0$. Find an orthogonal basis for the subspace described by $x − y − z = 0$.
First, find two linearly independent vectors in the span of the subspace above.
$v_1=\begin{pmatrix} 1\\ 1\\0 \end{pmatrix}$ $v_2=\begin{pmatrix} 1\\ 0\\1 \end{pmatrix}$
Then find the projection of $v_2$ on to $v_1$. this equals $\frac{v_2⋅v_1}{v_1⋅v_1}v_1=\begin{pmatrix} 1/2\\ 1/2\\0 \end{pmatrix}$
Now I need to find the orthogonal component. $\begin{pmatrix} 1\\ 0\\1 \end{pmatrix}-\begin{pmatrix} 1/2\\ 1/2\\0 \end{pmatrix}=\begin{pmatrix} 1/2\\ -1/2\\1 \end{pmatrix}$
Therefore, the orthogonal basis for the subspace is $\begin{pmatrix} 1\\ 1\\0 \end{pmatrix},\begin{pmatrix} 1/2\\ -1/2\\1 \end{pmatrix}$
Questions: Did I do this correctly? Do my steps make sense? Alternative solutions?
| Equation of the surface is $x-y-z=0$.
We have two independent variable here let us take them $x,y$ so we determine $z$ from this. By this you got $v_1=\begin{pmatrix} 1\\ 1\\0 \end{pmatrix} $, Now take assume another vector which is orthogonal to it, let it be $\begin{pmatrix} x\\ y\\z \end{pmatrix}$ so $x.1+y.1+z.0=0$ so $x+y = 0$,(the dot product must be zero) . Also it has to satisfy $x-y-z=0$. So we get $z=-2y, x=-y$ . Here only one independent variable so it is a straight line.
So we can take this vector as $v_2=\begin{pmatrix} 1\\ -1\\2 \end{pmatrix} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3941133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Using trigonometric substitution to integrate $\int\frac{x^3dx}{\sqrt{25-x^2}}$
I tried solving this calculus problem so many times but I never got an answer that at least looked similar to the choices in the image. Can someone help?
I don't understand where the Sine came from. This is the answer I get
$$\frac13(25-x^2)^{3/2} - 25\sqrt{25-x^2} + C $$
Edits are appreciated
| Integration by parts. Your result is equal to mine, BTW
$$\int \frac{x^3}{\sqrt{25-x^2}} \, dx=\int (-x^2)\,\frac{ -x}{\sqrt{25-x^2}}\,dx=$$
$$\int \frac{ -x}{\sqrt{25-x^2}}\,dx=\sqrt{25-x^2}+C$$
$$=-x^2\sqrt{25-x^2}-\int\left(-2x\sqrt{25-x^2}\right)\,dx=$$
$$=-x^2\sqrt{25-x^2}-\frac{2}{3} \left(25-x^2\right)^{3/2}+C=$$
$$=-x^2\sqrt{25-x^2}-\frac{2}{3}\sqrt{25-x^2}(25-x^2)+C=$$
$$=-\sqrt{25-x^2}\left(x^2+\frac{2}{3}(25-x^2)\right)+C=-\frac{1}{3} \sqrt{25-x^2} \left(x^2+50\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3941675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
A polynomial has the same remainder when divided by $x+k$ or $x-k$; what is $k$? Question
Given that $y = 3x^3 + 7x^2 - 48x + 49$ and that $y$ has the same remainder when it is divided by $x + k$ or $x - k$, find the possible values of $k$.
My attempt
Let $f(x) = 3x^3 + 7x^2 - 48x + 49$
$\text{Using Remainder Theorem,}$
\begin{align}
f(-k) &= f(k) \\
3(-k)^3 + 7(-k)^2 - 48(-k) + 49 &= 3(k)^3 + 7(k)^2 - 48(k) + 49 \\
-3k^3 - 7k^2 + 48k + 49 &= 3k^3 + 7k^2 - 48k + 49 \\
-3k^3 --3k^3 - 7k^2 - 7k^2 + 48k + 48k + 49 - 49 &= 0 \\
-6k^3 + 14k^2 + 96k &= 0 \\
\frac{-6k^3}{k} + \frac{14k^2}{k} + \frac{96k}{k} &= \frac{0}{k} \\
6k^2 + 14k - 96 &= 0
\end{align}
$\text{Comparing } 6k^2 + 14k - 96 = 0 \text{ with } ak^2 + bk + c = 0, a = 6, b = 14, c = -96$
\begin{align}
k = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a } &= \frac{ -(14) \pm \sqrt{(14)^2 - 4(6)(-96)} }{ 2(6) } \\
k &= 3 \text{ and } -5\frac{1}{3}
\end{align}
$\therefore k = 3, -5\frac{1}{3} \text{ or } 0 $
My answer is incorrect. The correct answer is $k = 0, 4 \text{ or } -4$
|
Your application of the Remainder Theorem does seem to be the approach for which the problem was designed. This is offered as a remark somewhat too long for a "comment", just to say a little about how your calculation can also be interpreted.
In writing the equation $ \ f(-k) \ = \ f(k) \ \ , $ we are in a sense asking for what values of $ \ x \ $ the function $ \ f(x) \ = \ 3x^3 + 7x^2 -48x + 49 \ \ $ [marked in blue in the graph above] "behaves" as an even-symmetry function. We can "resolve" our function into two "component" functions with even and odd symmetry, which is easily done for polynomial functions:
$$ f(x) \ \ = \ \ f_{even}(x) \ + \ f_{odd}(x) $$
$$ \rightarrow \ \ 3x^3 + 7x^2 -48x + 49 \ \ = \ \ (7x^2 + 49) \ + \ (3x^3 - 48x) \ \ ; $$
$ f_{even}(x) \ \ \text{and} \ \ f_{odd}(x) \ $ are marked in green and red, respectively. For the even function, by definition, $ \ f(-k) \ = \ f(k) \ \ $ is true for all real numbers $ \ k \ \ , $ whereas this is only true for an odd function at its $ \ x-$ intercepts, where $ \ f(-k) \ = \ -f(k) \ = \ 0 \ \ . $
So we seek those values of $ \ x = k \ $ at which $ \ f_{odd}(k) = 0 \ \ ; $ thus, we can simply write
$$ 3k^3 - 48k \ = \ 0 \ \ \Rightarrow \ \ 3k · (k^2 - 16) \ = \ 0 \ \ \Rightarrow \ \ k \ = \ 0 \ , \ -4 \ , \ 4 \ \ . $$
The odd-symmetry function, then, can be thought of as a "filter" that only "passes" the values of $ \ f(x) \ $ that possess even symmetry. We see in the graph above that these are the marked points at which $ \ f(x) = f_{even}(x) \ $ and $ \ f_{odd}(x) = 0 \ \ . $ The solution $ \ k = 0 \ $ is "trivial" in that $ \ f(-0) = f(0) \ $ must be true for a function that is continuous at $ \ x = 0 \ \ . $
[A separate conclusion that we can make is that, except at $ \ x = 0 \ \ , $ there is no value $ \ x = k \ $ for which our function can have $ \ f(-k) = -f(k) \ \ , $ since $ \ f_{even} \ $ is never equal to zero. More generally, for a polynomial function, there is no value of $ \ x = k \ $ other than $ \ x = 0 \ $ at which $ \ f(-k) = -f(k) \ $ if the constant term of the polynomial is positive.]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3943318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $a, b, c, d>0$ such that $a+b+c=1$, prove that $a^3+b^3+c^3+abcd\ge \min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$ If $a, b, c, d>0$, such that $a+b+c=1$, prove that: $$a^3+b^3+c^3+abcd\ge \min\left(\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right).$$
I tried solving it as follows:
$$a^3+b^3+c^3=3abc+1-3(ab+bc+ac).$$
From Schur we have that:
$$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c).$$
Hence $1+9abc\ge 4(ab+bc+ac)$.
This is as far as I got. I do not know how to introduce the $\min(\frac{1}{4}, \frac{1}{9}+\frac{d}{27})$ to my inequalities. Could you please explain to me how to solve it?
| This is an easy application of Lagrange as the first equality of $$3a^2+bcd=3b^2+acd=3c^2+abd=\lambda$$ gives $3(a+b)(a-b)=(a-b)cd$ so either $a=b$ or $3(a+b)=cd$. Thus there are two possibilities:
*
*$a=b=c$ so that $f(a,b,c)=1/9+d/27$;
*$a=b$ and $3(b+c)=ad$ so that $d=3/a-3$ since $c=1-2a$ giving $$f(a,b,c)=2a^3+(1-2a)^3+a^2(1-2a)\left(\frac3a-3\right)=3a^2-3a+1\ge\frac14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3945610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
The letter A appears an even number of times. How many different sequences could Dr. Lizardo have written down?
Dr. Lizardo writes down a sequence of $n$ letters, where each letter is A, B, or C, and the letter A appears an even number of times. How many different sequences could Dr. Lizardo have written down?
What I have tried so far
The last letter is predefined in terms of "A or not" based on the first $n-1$ letters. If there are an odd number of As so far, the last one has to be A, otherwise, B or C. Therefore, there are $3^{n-1}$ ways to write the first $n-1$ letters. However, the last letter can either have 1 choice or 2 choices.
I am thinking maybe we need to do $S_n = 3^{n-1} + S_{n-1}$. However, I need a closed form, not a recurrance. How can I do this?
Thank you!
Note: $S_n = 3^{n-1} + S_{n-1}$ seems to be working for the first few values of $n$ (I tested through $n=4$), so finding a closed form of that recurrance would work as well.
| This is a bit more of a linear algebra proof which is probably not what you are seeking but I'd still like to share it anyways. Define two sequences $a_n$ being the number of length-$n$ sequences with even number of As and $b_n$ being the number of length-$n$ sequences with odd number of As. Clearly $a_0 = 1$ and $b_0 = 0$
Notice that the if you have an length-$(n-1)$ sequences with even number of As: there are two ways to add one character (either B or C) so that the new length-$n$ sequence has even As and there is one way to add one character (A) so that the new length-$n$ sequence has odd As.
Similarly if you have an length-$(n-1)$ sequences with odd number of As: there is one way to add one character (A) so that the new length-$n$ sequence has even As and there are two ways to add one character (B or C) so that the new length-$n$ sequence has odd As.
From here we can define a recurrence relation in terms of matrices
$$ \left( \begin{matrix} a_{n} \\ b_{n} \end{matrix} \right) =
\left( \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right)
\left( \begin{matrix} a_{n-1} \\ b_{n-1} \end{matrix} \right) =
\left( \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right)^n
\left( \begin{matrix} a_{0} \\ b_{0} \end{matrix} \right)$$
If we diagonalize this matrix we get
$$ \left( \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right) =
\left( \begin{matrix} -1 & 1 \\ 1 & 1 \end{matrix} \right)
\left( \begin{matrix} 1 & 0 \\ 0 & 3 \end{matrix} \right)
\frac{1}{2} \left( \begin{matrix} -1 & 1 \\ 1 & 1 \end{matrix} \right)$$
$$ \left( \begin{matrix} a_{n} \\ b_{n} \end{matrix} \right) =
\left( \begin{matrix} -1 & 1 \\ 1 & 1 \end{matrix} \right)
\left( \begin{matrix} 1 & 0 \\ 0 & 3^n \end{matrix} \right)
\frac{1}{2} \left( \begin{matrix} -1 & 1 \\ 1 & 1 \end{matrix} \right)
\left( \begin{matrix} 1 \\ 0 \end{matrix} \right) =
\frac{1}{2} \left( \begin{matrix} 3^n+1 \\ 3^n-1 \end{matrix} \right)$$
thus $a_n = \frac{3^n+1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3946649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Does this series that resembles the Maclaurin series of $\cosh x$ converge? Does this series converge?
$$\left(\frac{f_{i}^2}{2!} +\frac{f_{i}^4}{4!}\left(2\kappa c\right)+\frac{f_{i}^6}{6!}\left(2\kappa c\right)^2 + \frac{f_{i}^8}{8!}\left(2\kappa c\right)^3+ \cdots\right), \tag{1}$$
where $k, c \geq 0$. I am trying to establish a relation with the Maclaurin series of $\cosh x$:
$$ 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots=\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}. \tag{2}$$
| Multiply the series given by Eq. (1) by $\frac{2 \kappa c}{2\kappa c}$
$$\frac{1}{2\kappa c}\left(\frac{f_{i}^2}{2!}\left(2\kappa c \right) +\frac{f_{i}^4}{4!}\left(2\kappa c\right)2+\frac{f_{i}^6}{6!}\left(2\kappa c\right)^3 + \frac{f_{i}^8}{8!}\left(2\kappa c\right)^4+ \cdots\right), $$
Add a zero to the above expression
$$\frac{1}{2\kappa c}\left(1+\frac{f_{i}^2}{2!}\left(2\kappa c \right) +\frac{f_{i}^4}{4!}\left(2\kappa c\right)2+\frac{f_{i}^6}{6!}\left(2\kappa c\right)^3 + \frac{f_{i}^8}{8!}\left(2\kappa c\right)^4+ \cdots -1\right), $$
comparing with Eq. (2) of the question, we can establish the relation with $\cosh x$
$$\frac{1}{2\kappa c}\left(\sum_{n=0}^{\infty} \frac{\left[f_{i}\left(2\kappa c\right)^{\frac{1}{2}}\right]^{2n}}{2n!}- 1\right) =\frac{\cosh \left[f_{i}\left(2\kappa c\right)^{\frac{1}{2}} \right]-1}{2\kappa c}.$$
Which is the result given by @Sangchul Lee.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3950655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that : $\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx=\pi\sqrt{\phi^{-1}}$ I found it with help of Wolfram alpha .
$$\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx=\pi\sqrt{\phi^{-1}}$$
This integral admits an antiderivative (I can add it if so) so I can conclude that this result is true .I would like to see a proof with the use of complex integration or a 'real' proof . For that I think we can put $y=\frac{2}{x^2+1}$ and use integration by parts (perhaps?).For the proof with complex integration(if it exists I'm new for it ) can you detail the solution ?
Thanks!
| Here's a brute-force method:
$$
\begin{aligned}
\int{\arctan}\left( \frac{2}{x^2+1} \right) \,dx&=\int{\mathrm{arccot}}\left( \frac{1}{2}+\frac{x^2}{2} \right) \,dx\\
&=x\;\mathrm{arccot} \left( \frac{1}{2}+\frac{x^2}{2} \right) +\int{\frac{x^2}{\frac{5}{4}+\frac{x^2}{2}+\frac{x^4}{4}}}\,dx\\
&=x\;\mathrm{arccot} \left( \frac{1}{2}+\frac{x^2}{2} \right) -\frac{1}{2}\int{\frac{\sqrt{5}-x^2}{\frac{5}{4}+\frac{x^2}{2}+\frac{x^4}{4}}}\,dx+\frac{1}{2}\int{\frac{\sqrt{5}+x^2}{\frac{5}{4}+\frac{x^2}{2}+\frac{x^4}{4}}}\,dx\\
&=x\;\mathrm{arccot} \left( \frac{1}{2}+\frac{x^2}{2} \right) \\&+\frac{\displaystyle \int{\frac{\sqrt{2\left( -1+\sqrt{5} \right)}+2x}{-\sqrt{5}-\sqrt{2\left( -1+\sqrt{5} \right)}x-x^2}}\,dx}{\sqrt{2\left( -1+\sqrt{5} \right)}}+\frac{\displaystyle \int{\frac{\sqrt{2\left( -1+\sqrt{5} \right)}-2x}{-\sqrt{5}+\sqrt{2\left( -1+\sqrt{5} \right)}x-x^2}}\,dx}{\sqrt{2\left( -1+\sqrt{5} \right)}}\\&+\int{\frac{1}{\sqrt{5}-\sqrt{2\left( -1+\sqrt{5} \right)}x+x^2}}\,dx+\int{\frac{1}{\sqrt{5}+\sqrt{2\left( -1+\sqrt{5} \right)}x+x^2}}\,dx\\
&=x\;\mathrm{arccot} \left( \frac{1}{2}+\frac{x^2}{2} \right) \\&+\frac{\log \left( \sqrt{5}-\sqrt{2\left( -1+\sqrt{5} \right)}x+x^2 \right)}{\sqrt{2\left( -1+\sqrt{5} \right)}}-\frac{\log \left( \sqrt{5}+\sqrt{2\left( -1+\sqrt{5} \right)}x+x^2 \right)}{\sqrt{2\left( -1+\sqrt{5} \right)}}\\
&+\sqrt{\frac{2}{\sqrt{5}+1}} \arctan\left(\frac{2 x+\sqrt{2 \left(\sqrt{5}-1\right)}}{\sqrt{2 \left(\sqrt{5}+1\right)}}\right)\\&-\sqrt{\frac{2}{\sqrt{5}+1}} \arctan \left(\frac{\sqrt{2 \left(\sqrt{5}-1\right)}-2 x}{\sqrt{2 \left(\sqrt{5}+1\right)}}\right)
\\&\overset{\text{def}}{=} F(x)
\end{aligned}
$$
One can easily see that
$$
\underset{x\to \infty }{\text{lim}}x \text{ arccot}\left(\frac{x^2}{2}+\frac{1}{2}\right)=0
$$
and
$$
\begin{aligned}
&\lim_{x\to\infty} \frac{\log \left(x^2-\sqrt{2 \left(\sqrt{5}-1\right)} x+\sqrt{5}\right)}{\sqrt{2 \left(\sqrt{5}-1\right)}}-\frac{\log \left(x^2+\sqrt{2 \left(\sqrt{5}-1\right)} x+\sqrt{5}\right)}{\sqrt{2 \left(\sqrt{5}-1\right)}} \\&=\lim_{x\to\infty}\frac{\log \left(\frac{x^2-\sqrt{2 \left(\sqrt{5}-1\right)} x+\sqrt{5}}{x^2+\sqrt{2 \left(\sqrt{5}-1\right)} x+\sqrt{5}}\right)}{\sqrt{2 \left(\sqrt{5}-1\right)}}\\
&=0
\end{aligned}
$$
Also it is easy to see that
$$
\underset{x\to \infty }{\text{lim}}\arctan\left(\frac{\sqrt{2 \left(\sqrt{5}-1\right)}-2 x}{\sqrt{2 \left(\sqrt{5}+1\right)}}\right)=-\dfrac{\pi}{2}
$$
and
$$
\underset{x\to \infty }{\text{lim}}\arctan\left(\frac{2 x+\sqrt{2 \left(\sqrt{5}-1\right)}}{\sqrt{2 \left(\sqrt{5}+1\right)}}\right) = \dfrac{\pi}2
$$
Thus, combine the above results, from Newton-Leibniz formula, we get
$$
\int_{0}^{\infty}\arctan\left(\frac{2}{x^2+1}\right)dx= F(\infty) - F(0)=\sqrt{\frac{2}{\sqrt{5}+1}} \pi = \color{blue}{\pi\sqrt{\phi^{-1}}}
$$
As desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3951074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Non-numerical proof of an inequality
Let $ 0 < s < 1 $ be the unique solution to the equation $$ \frac{1}{2^s} + \frac{1}{6^s} + \frac{1}{12^s}=1, $$ and show that $$ \frac{1}{6^s}+\frac{1}{12^s} \geq \left(\frac{2}{7}\right)^s. $$
It is easy do this numerically; you can compute $ s \simeq 0.7584 $ with a calculator, and show that the desired inequality holds with this value plugged in.
I'm looking for a more general proof, maybe geometric, using some calculus, or using concavity of the function $ x \mapsto x^s $ for $ 0 < s < 1 $.
| Denote $a = \frac{1}{2^s}$ and $b = \frac{1}{3^s}$.
We have $a + a b + a^2 b = 1$ which results in $b = \frac{1 - a}{a + a^2}$.
Also, since $a^2 < b$, we have $\frac{1 - a}{a + a^2} > a^2$
or $\frac{(a^2 + 1)(1 - a - a^2)}{a + a^2} > 0$ which results in $1 - a - a^2 > 0$.
Since $x\mapsto x^s$ is concave on $x>0$, we have
$(\frac{6}{7})^s + (\frac{8}{7})^s \le 2(\frac{\frac{6}{7} + \frac{8}{7}}{2})^s = 2$
which results in $(\frac{2}{7})^s \le \frac{2}{4^s + 3^s}$.
It suffices to prove that $\frac{1}{6^s} + \frac{1}{12^s} > \frac{2}{4^s + 3^s}$ or
$ab + a^2 b > \frac{2}{\frac{1}{a^2} + \frac{1}{b}}$
or $\frac{(1-a)^2(a+1)(1 - a - a^2)}{a^4 + a^3 - a + 1} > 0$ which is true.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3951376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Antiderivative of $\sqrt{(x^2-a^2)(x^2-b^2)(x^2-c^2)}$ Is it possible to explicitly find the antiderivative of $\sqrt{(x^2-a^2)(x^2-b^2)(x^2-c^2)}$? It is rather straight-forward to find it when $a=0$, or $b=0$ or $c=0$. This is also easy in the case where $a=b$, or $b=c$, or $a=c$. But I have been struggling to do this in the general case. Any help is appreciated. Thanks!
| Hint:
$\sqrt{(x^2-a^2)(x^2-b^2)(x^2-c^2)}=\sqrt{(x+a)(x-a)(x+b)(x-b)(x+c)(x-c)}$
consult with Lauricella function
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3953366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find a cyclic sum Given that
$$ \frac{a}{bc-a^2}+\frac{b}{ca-b^2}+\frac{c}{ab-c^2}=0,\quad a,b,c\in \mathbb{R}, $$
find the value of
$$ \frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}. $$
I solved it via direct computation which is clearly not the best way to do. Is there any simple transformation I could perform or geometical meaning of this equation?
| From first equation (get rid of the denumerators) we get $$abc(a^2+b^2+c^2+ab+bc+ca)= a^2b^2(a+b)+b^2c^2(b+c)+c^2a^2(c+a)$$ which can be easly transforemd to $$abc(a^2+b^2+c^2+2ab+2bc+2ca)= (a^2b^2+b^2c^2+c^2a^2)(a+b+c)$$ so $$abc(a+b+c)^2= (a^2b^2+b^2c^2+c^2a^2)(a+b+c)$$
Now if $a+b+c\neq 0$ then we have $$\color{red}{abc(a+b+c)= a^2b^2+b^2c^2+c^2a^2}$$
But since \begin{align}a^2b^2 +b^2c^2&\geq 2b^2ac \\
b^2c^2 +c^2a^2&\geq 2c^2ab \\
c^2a^2 +a^2b^2&\geq 2a^2bc \\
\end{align}
we have always $$abc(a+b+c)\leq a^2b^2+b^2c^2+c^2a^2$$
with equality iff $a=b=c$. So, since we have equality sign in red equation, we deduce that $a=b=c$, but this can not hold since then fractions are not defined.
So $\boxed{a+b+c =0}$ and now I suppose it is not difficult to finish....
Let $E$ be the expression. Then $$E = b\Big(\frac{1}{(ca-b^2)^2}-\frac{1}{(bc-a^2)^2}\Big)+ c\Big(\frac{1}{(ab-c^2)^2} -\frac{1}{(bc-a^2)^2}\Big) $$ Value in each big bracket is $0$ so $E=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3955883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solutions of $2^{1978}(1+2^{4a-2030})=y(y+1)$
I am given that $a,y$ positive integers and
$$2^{1978}(1+2^{4a-2030})=y(y+1).$$
I want to maximize $a$.
I guessed that $a$ is maximized when $2^{1978}=y$ and hence $4a-2030=1978\implies a =1002$. However, there is the other case of when a factor of $1+2^{4a-2030}$ is "contained" in $2^{1978}$, that is: $(2^{1978}x, \frac{1+2^{4a-2030}}{x}) = (y,y+1) \text{ or }=(y+1,y)$. I have proven for the $(y,y+1)$ case that $2^{4a-30} \ge 2^{1978}x^2$ which imply that there may be infinite solutions and so there is no maximum.
| If we multiply given equation $$2^{1978}(1+2^{4a-2030})=y(y+1)$$
with $4$ and then add $1$ we get $$\boxed{2^{1980}+2^{4a-50}+1=(2y+1)^2}$$
Now suppose there exists $a>1002$, then $$1980<2a-24$$ so $$\color{red}{2^{1980}+2^{4a-50}+1} <(2^{2a-25}+1)^2$$
and since $$2^{4a-50}<\color{red}{2^{1980}+2^{4a-50}+1}$$
so $$2^{4a-50}<\color{red}{(2y+1)^2}<(2^{2a-25}+1)^2$$ which implies $$2^{2a-25}<2y+1<2^{2a-25}+1$$
and thus no solution. So $a\leq 1002$ and thus $a_{\max} = 1002$ (which clearly works).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3957030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Is there any analytical approximation to this integral? I have an integral:$$ I(x) = x\int_{x}^{\infty}K_{5/3}(\xi)\,\rm{d}\xi~,$$
where $K$ is modified Bessel function.
I need to evaluate this integral for tens of thousand different $x$ ultimately to use it for some numerical calculation. In the extreme limits of $x\gg 1$ or $x\sim 0$, the approximations are:
$$\lim_{x\rightarrow 0}I(x) = \frac{4\pi}{\sqrt{3}\Gamma(1/3)}\left(\frac{x}{2}\right)^{1/3}~,$$
and
$$\lim_{x\rightarrow \infty}I(x) = \sqrt{\frac{\pi x}{2}}e^{-x}~.$$
However, I would like to know if there are any approximate expressions $I_\text{approx}(x)$ valid for all $x$ such that $$\frac{|I_\text{approx}(x)-I(x)|}{I(x)}\lesssim 0.1~,$$
for all $x$.
| Using hypergeometric functions, the antiderivative exists. So,
$$I(x)=-\frac{\pi x}{\sqrt{3}}-\frac{2^{2/3} \sqrt{3} \pi x^{1/3} \,
_1F_2\left(-\frac{1}{3};-\frac{2}{3},\frac{2}{3};\frac{x^2}{4}\right)}{\Gamma
\left(-\frac{2}{3}\right)}-$$ $$\frac{81 x^{11/3} \Gamma \left(\frac{4}{3}\right) \,
_1F_2\left(\frac{4}{3};\frac{7}{3},\frac{8}{3};\frac{x^2}{4}\right)}{320\
2^{2/3}}$$
As series,
$$_1F_2\left(-\frac{1}{3};-\frac{2}{3},\frac{2}{3};u\right)=\sum_{n=0}^\infty \frac{\Gamma \left(-\frac{2}{3}\right)}{(1-3 n) n! \Gamma \left(n-\frac{2}{3}\right)}u^n$$
$$_1F_2\left(\frac{4}{3};\frac{7}{3},\frac{8}{3};u\right)=\sum_{n=0}^\infty \frac{4 \Gamma \left(\frac{8}{3}\right)}{(3 n+4) n! \Gamma \left(n+\frac{8}{3}\right)} u^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3957173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$ Solve for $x$ in
$$3x^3=({x^2+\sqrt{18}x+\sqrt{32}})(x^2-\sqrt{18}x-\sqrt{32})-4x^2$$
I'm really not good with 4th degree equations but since the 1st term in the RHS looked a simple (a+b)(a-b) application, I tried solving that but I'm really not able to reach to the final answer as it only gets more complicated...
can anyone please help me out?
The equation I got was:
$$x^4-3x^3-22x^2-48x-32=0$$
| Expand $$x^4-3 x^3-22 x^2-48 x-32=0$$
Try to factorize $$x^4-3 x^3-22 x^2-48 x-32=\left(a x+b+x^2\right) \left(c x+d+x^2\right)$$
RHS gives $$x^4+x^3 (a+c)+x^2 (a c+b+d)+x (a d+b c)+b d$$
thus we must have
$$
\begin{cases}
a+c=-3\\
ac+b+d=-22\\
ad+bc=-48\\
bd=-32\\
\end{cases}
$$
Final factorization should be
$$\left(x^2-6 x-8\right) \left(x^2+3 x+4\right)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3957453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$A^3$ is similar to $B^3$ but $A^2$ is not similar to $B^2$ Find an example of $A,B$ such that $A^3$ is similar to $B^3$, and $A^2$ is not similar to $B^2$, and $A$ is not similar to $B$.
I only have some clues that $p_{A^2}(x) \neq p_{B^2}(x)$ and $p_{A}(x) \neq p_{B}(x)$
I first wanted to start from $A^3 = S^{-1} B^3 S$
Then let $S$ and $B$ to be some arbitrary matrices, where $S$ is invertiable, so that I could obtain $A^3$ which is similar to $B^3$. Thus, $A = \sqrt[\leftroot{-3}\uproot{3}3]{A^3}$
However, I realize I don't know how to find square root or cube root of matrices.
| Let $\omega:=e^{\frac{2\pi i}{3}}$.
Then
$$
A=\begin{pmatrix}
\omega & 0\\
0 &\omega
\end{pmatrix}
\text{ and }
B=\begin{pmatrix}
\omega & 0\\
0 &1\\
\end{pmatrix}
$$
will do the trick.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $a^2 + b^2 +c^2 =4$ and $a^3 + b^3 + c^3 = 8$ then find $a^4+b^4+c^4$ This is a popular question, but I can find $a^4+b^4+c^4$ if I know $a+b+c, a^2+b^2+c^2, a^3 +b^3 +c^3$ or in special case $a+b+c=0$ we only need a more condition like $a+b+c$. I have tried in many times and can't find out the way to archive it.
| Well, it sounds logical that we need three equations, because three variables and stuff. This is wrong, though. Sometimes one equation can give us two variables, like $x^2+y^2=0\Rightarrow x=0,\;y=0$ (if we stay in $\mathbb R$, that is; but if we don't, then the original question can't be answered either). Same thing here.
Obviously†, $a^3+b^3+c^3\leqslant(a^2+b^2+c^2)^{3/2}$, and the equality is only possible when two of the variables are $0$. So $(0,0,2)$ and its permutations are indeed the only real solutions, from where you can easily find the answer.
So it goes.
†
$$(a^2+b^2+c^2)^3\geqslant \overbrace{a^6+b^6+c^6+\underbrace{a^4b^2+a^2b^4}_\text{apply AM-GM}+\underbrace{b^4c^2+b^2c^4}_\text{here too}+\underbrace{a^4c^2+a^2c^4}_\text{and here}}^\text{skipped quite a lot of terms, all of them quadratic}\geqslant \\\geqslant a^6+b^6+c^6+2a^3b^3+2b^3c^3+2a^3c^3=(a^3+b^3+c^3)^2,\text{ hence}\\
a^3+b^3+c^3\leqslant (a^2+b^2+c^2)^{3/2}=4^{3/2}=8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Minimum value of $\sin^{-1}x \cos^{-1}x$ Let us consider the function $~~f(x)= \sin^{-1}x \cos^{-1}x.$ I want to find the minimum value of $f$.
So, first I find the derivative $df$ of $f$ and put $df=0$ to get $\cos^{-1}x = \sin^{-1}x$ and from this critical point I got the maximum value of $f$ which is $\frac{\pi^2}{16}.$ But I want to find the minimum value of $f$.
Please help me to find this.
| As $-\dfrac\pi2\le\sin^{-1}x\le\dfrac\pi2$
and $0\le\cos^{-1}x\le\pi$
Clearly $x=-1$ yields the minimum value
Alternatively,
$$f(x)=\sin^{-1}x\cos^{-1}x=\left(\dfrac\pi2-\cos^{-1}x\right)\cos^{-1}x=\left(\dfrac\pi4\right)^2-\left(\cos^{-1}x-\dfrac\pi4\right)^2$$
Now $f(x)$ will be minimum if $\left(\cos^{-1}x-\dfrac\pi4\right)^2$ is maximum
$0\le\cos^{-1}x\le\pi\implies0-\dfrac\pi4\le\cos^{-1}x-\dfrac\pi4\le\pi-\dfrac\pi4$
$\implies\left(\cos^{-1}x-\dfrac\pi4\right)^2\le\left(\pi-\dfrac\pi4\right)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3960397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve the following differential equation: $(2x^3-x^2-y)dx-(2x^2y-x)dy=0$ I'd like your help with this differential equation:
$(2x^3-x^2-y)dx-(2x^2y-x)dy=0$
It seems not difficult, but I have no idea about it
Thanks a lot for help)
| Seek a total derivative $df=0$ by multiplying by an Ansatz $x^ay^b$ viz.$$f_x=2x^{a+3}y^b-x^{a+2}y^b-x^ay^{b+1},\,f_y=x^{a+1}y^b-2x^{a+2}y^{b+1}$$so$$0=\frac{f_{xy}-f_{yx}}{x^{a-1}y^{b-1}}=2bx^4-bx^3-(a+b+2)xy+2(a+2)x^2y^2.$$Equating coefficients gives $a=-2,\,b=0$, so$$df=(2x-1-y/x^2)dx+(1/x-2y)dy\implies f=x^2-x+y/x-y^2-C.$$The constraint therefore reduces to $x^2-x+y/x-y^2=C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3963893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to
If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.
Here's the proof that I've found (I'm sorry, I forgot where I got it):
Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.
Now since I love to punish myself, I tried to find a harder proof as such:
We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2) +2t} + 1
$$ which simplifies to $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$
Now how do I prove the sextic polynomial inequality above (which is true)?
| (I agree that most of the algebraic steps here are not obvious nor intuitive. It was a lot of wishful thinking that this method could work, in part because of the background of the problem.)
For the sextic inequality, from the graph, you know that $\sqrt{2} - 1 $ is a double root, so $ ( x - (\sqrt{2} - 1) )^2$ is a factor of the polynomial.
(Alternatively, we know from the work you did that that's the equality case.)
So, we have
$$\frac{ x^6 - 2\sqrt{2} x^5 - 3 x^4 - 8 x^3 + 3 x^2 + 2 \sqrt{2} x - 1} { ( x - (\sqrt{2} - 1) )^2} \\ = x^4 - 2x^3 - 2 ( 1 + \sqrt{2} ) x^2 - 2 ( 3 + 2 \sqrt{2} ) x - 2\sqrt{2} - 3.$$
We make the observation (by staring really hard) that the quartic factors as
$$ (x^2 - ( 2 + \sqrt{2} ) x - \sqrt{2} - 1 ) ( x^2 + \sqrt{2} x + \sqrt{2} + 1 ) $$
We can now easily show that on $ 0 < x < 1$, this value is negative. (First term is negative, second term is positive.)
Hence, the original expression is $ \leq 0$, with equality at $ \sqrt{2} - 1 $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 10,
"answer_id": 4
} |
Deciding whether given infinite series converges or diverges I am self-learning Real Analysis from Understanding Analysis by Stephen Abbott. I am having some trouble determining the convergence or divergence of the below infinite series. I have some initial thoughts which I have put down. Any inputs would be really helpful.
How do I build more skill at finding out if a given infinite series is convergent/divergent?
Exercise 2.7.2
(c) $1 - \frac{3}{4} + \frac{4}{6} - \frac{5}{8} + \frac{6}{10} - \frac{7}{12} + \ldots$
(d) $1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \frac{1}{9} + \ldots$
(e) $1 - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{4^2} + \frac{1}{5} - \frac{1}{6^2} + \frac{1}{7} - \frac{1}{8^2} + \ldots$
Proof.
(c) Let $(s_n)$ be the sequence of partial sums of the infinite series $\sum_{n=1}^{\infty}a_n$.
The general expression for the infinite series is,
\begin{align*}
1 + \sum_{n=1}^{\infty}(-1)^{n}\cdot\frac{n+2}{2n}
\end{align*}
I cannot apply the alternate series test, because
\begin{align*}
\lim_{n\to\infty}a_n = \lim_{n\to\infty}\frac{n+2}{2n} = \frac{1}{2} \ne 0
\end{align*}
I can't think of a known convergent or divergent series to compare with, in order to use the Alternate Series test.
(d) One observation is, this is not a rearrangement of the alternating harmonic series.
| (c) Your series is $\sum_{n=1}^\infty(-1)^{n+1}\frac{n+1}{2n}$ and, since you don't have $\lim_{n\to\infty}(-1)^{n+1}\frac{n+1}{2n}=0$, the series diverges.
(d) The sum of the first $3$ terms is greater than $1$. The sum of the first $6$ terms is greater than $1+\frac14$. The sum of the first $9$ terms is greater than $1+\frac14+\frac17$. So, your series diverges.
(e) The sum of the positive terms diverges, whereas the sum of the negative terms converges, and therefore this series diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3966123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$ This integral
$$\int_{0}^{1} \frac{3(x^3+x^2-x)+1}{3(x^2-x)+1}dx$$
is fabricated based on an interesting point which I may post later.
The question is: How would you do it?
| \begin{gather*}
I=\int ^{1}_{0}\left(\frac{3x^{3}}{3x( x-1) +1} +1\right) dx=\int ^{1}_{0}\frac{3x^{3} dx}{3x( x-1) +1} +1\\
Let\ I_{1} =\int ^{1}_{0}\frac{x^{3} dx}{3x( x-1) +1}\\
I_{1} =\int ^{1}_{0}\frac{( 1-x)^{3} dx}{3( 1-x)( 1-x-1) +1} =\int ^{1}_{0}\frac{( 1-x)^{3} dx}{3x( x-1) +1}\\
because\int ^{a}_{0} f( x) dx=\int ^{a}_{0} f( a-x) dx\\
Adding\ both,\\
2I_{1} =\int ^{1}_{0}\frac{x^{3} +( 1-x)^{3} dx}{3x( x-1) +1}\\
Apply\ a^{3} +b^{3} =( a+b)\left( a^{2} +b^{2} -ab\right)\\
2I_{1} =\int ^{1}_{0}\frac{3x( x-1) +1dx}{3x( x-1) +1} =1\\
I_{1} =\frac{1}{2}\\
I=3I_{1} +1=\frac{3}{2} +1=\frac{5}{2}\\
\\
\end{gather*}which I hope is the correct answer:)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3970899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Find variables from a multiplication of geometric sequences
Given $$f(x) = \sum_{i=0}^\infty a_ix^i$$
And $$f(x)(1+2x+2x^2+x^3) = \frac{1}{(1-x)^3}$$
Find the values of $a_0$, $a_1$, and $a_2$.
I try to expand them but it doesn't seem to lead me to any solution.
How can I isolate a0, a1, and a2 to find them?
| $f(x)= a_0+a_1x+a_2x^2+O(x^3)$
$f(x)(1+2x+2x^2+x^3)$
$=a_0+(2a_0+a_1)x+(2a_0+2a_1+a_2) x^2
+O(x^3)$
$$\textrm{Again } f(x)(1+2x+2x^2+x^3)=\frac{1}{(1-x)^3}=1+3x+6x^2+O(x^3)$$
Comparing the like powers of x we get:$$a_0=1,a_1=1,a_2=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}>\frac{2}{3}$ Prove that $\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}>\frac{2}{3}$ for any $n \in N$
I used AM $\geq$ HM and got $$\frac{1}{n}+\frac{1}{n+1}+...+\frac{1}{2n}\geq\frac{(n+1)^2}{n+n+1+...+2n}=\frac{(n+1)^2}{\frac{3}{2}n(n+1)}>\frac{2}{3}$$
My question is whether this solution is correct and also why $n+n+1+...+2n = \frac{3}{2}n(n+1)$, since I used wolfram to compute that and cannot see it myself.
Any tips on how to compute sums like this or other possible solutions for this problem would be appreciated as well.
| That is correct, and $$n+(n+1) +(n+2) +\dots +(n+n) \\ =(n+1)\times n +(1+2+3+\dots +n) \\ = n(n+1) + \frac{n(n+1)}{2} \\ =\frac 32 n(n+1) $$
Alternatively, you can use the formula for the sum of an arithmetic progression $$S=\frac {\text{number of terms}}{2} (\text{first term + last term}) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Bijective association between the linear space of symmetric tensors and linear space of homogeneous polynomials example edit:
Given a symmetric tensor G there is a bijective mapping between the $d$ indexes and a vector of $K$ numbers representing the time each variable $x_t$ appears in the associated monomial. For example, imagine $G_{112}$ being a 3 dimensional tensor with every index spanning from 1 to 3 (so $G_{333}$ is the last element)
The element is associated with $G_{112}x_1^2x_2^1x_3^0 = G_{112}x^{f([1,1,2])}$ where $f([1,1,2]) = [2,1,0]$.
More precisely
\begin{equation}
p(x) = \sum G_ix^{f(i)}
\end{equation}
So i want to "see" this relation using an example so i took two 2-dimensional symmetric tensors like \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
which is associated with the homogeneous polynomial
\begin{equation}
p(x_1,x_2) =x^2_1+x^2_2
\end{equation}
and
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
which is associated with the homogeneous polynomial
\begin{equation}
q(x_1,x_2) = 2x_1x_2
\end{equation}
Now if i do the product
\begin{equation}
p(x_1,x_2)q(x_1,x_2) = 2x_1^3x_2 + 2x_1x_2^3
\end{equation}
i should obtain a homogeneous polynomial which is associated to a symmetric tensor, and that symmetric tensor is
\begin{equation}
\begin{pmatrix}
0 & \frac{1}{2} \\
\frac{1}{2} & 0
\end{pmatrix}
\begin{pmatrix}
\frac{1}{2} & 0 \\
0 & \frac{1}{2}
\end{pmatrix}
\begin{pmatrix}
\frac{1}{2} & 0 \\
0 & \frac{1}{2}
\end{pmatrix}
\begin{pmatrix}
0 & \frac{1}{2} \\
\frac{1}{2} & 0
\end{pmatrix}
\end{equation}
But i can also calculate the tensor product of the two original tensors which is 4-dimensional and is
\begin{equation}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
\end{equation}
They should be the same... What am i doing wrong? Can you give an example of this link between tensors and polynomials?
| This is because the tensor product of symmetric tensors is not necessarily a symmetric tensor. In your example, the rank-$4$ tensor
$$
T = \begin{pmatrix}
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} &
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}\\
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix} &
\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}\\
\end{pmatrix}
$$
is not symmetric. For example, the entries $T_{1112} = 1$ and $T_{1211} = 0$ are not equal.
There's another indication that simply taking the tensor product doesn't work, i.e. tensor products of symmetric tensors are not commutative, but multiplying polynomials certainly is.
I'm not familiar with this mapping between homogeneous polynomials and symmetric tensors that you have described, but I think what you want is probably the symmetric product, which means that after taking the tensor product you need to average all entries with indices in the same orbit w.r.t. permutation. Note that the symmetric product is also commutative, which is a good sign. Again taking the $1112$ set of indices as an example, we have $T_{1112} = 1$, $T_{1121} = 1$, $T_{1211} = 0$, $T_{2111} = 0$, so the symmetrized version should have $T_{(1112)} = \frac{1}{2}$. You can check that this gives you the same matrix as the one obtained from the product of the polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3973097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\int_0^{\pi} \frac{x \sin x}{2 + \cos x}dx$ Problem : $$\int_0^{\pi} \frac{x \sin x}{2 + \cos x}dx$$
I tried $x=\pi - t$ but this just made integrand more messy
because its denominator isn't $2 + \cos x$.
I want to try this without $\tan(x/2)=t$ (like symmetric or etc.)
Thanks for help.
| The present answer is quite applicable and straightforward but I want to interpret my way of solving this kind of question.
Let us start with the substitution: $$ \bbox[yellow] { 2+\cos (x)=\frac{4-\cos^2 (x)}{2-\cos (x)} }$$ This is followed by: $$\int_{0}^{\pi} \frac{(2-\cos (x))\space x \sin (x)}{4-\cos^2 (x)} dx = \int_{0}^{\pi} \frac{2x \sin (x)}{4-\cos^2 (x)} dx-\int_{0}^{\pi} \frac{x \cos (x) \sin (x)}{4-\cos^2 (x)} dx$$ The evaluation of the first part of the integral comes from the lemma: $$\int_{0}^{\pi} x f(\sin (x)) dx=\frac{\pi}{2} \int_{0}^{\pi} f(\sin (x))dx$$ If we set $f(\sin (x))$ to be $\frac{2\sin(x)}{3+\sin^2(x)}$: $$\int_{0}^{\pi} x f(\sin(x))dx=\frac{\pi}{2} \int_{0}^{\pi} \frac{2 \sin(x)}{3+\sin^2(x)} dx=\frac{\pi}{2} \int_{0}^{\pi} \frac{2\sin(x)}{4-\cos^2 (x)} dx$$ We can integrate it using the method of substitution: $$\begin{align} y=\frac{\cos(x)}{2}\rightarrow -\frac{\pi}{2} \int_{\frac{1}{2}} ^{-\frac{1}{2}} \frac{dy}{1-{y^2}}
&=\Biggl[-\frac{\pi}{2}\tanh^{-1}y\Biggl]^{-\frac{1}{2}}_{\frac{1}{2}} \\
&=-\frac{\pi}{2} \tanh^{-1}\Biggl(-\frac{1}{2}\Biggl)+\frac{\pi}{2}\tanh^{-1}\Biggl(\frac{1}{2}\Biggl) \\
&=\pi \tanh^{-1} \Biggl(\frac{1}{2}\Biggl) \\
&= \frac{\pi}{2} \ln(3) \end{align}$$
If we apply integration by parts for the second integral: $$\begin{align} \int_{0}^{\pi}\frac{x \sin(x) \cos(x)}{4-\cos^2(x)}dx
&=\Biggl[x\frac{\ln(8-2\cos^2(x))}{2}\Biggl]_{0}^{\pi}-\frac{1}{2}\int_{0}^{\pi} \ln(8-2\cos^2(x))dx \\
&= \frac{\pi}{2} \ln(6)-\frac{1}{2} \int_{0}^{\pi}\ln(8-2\cos^2(x))dx \\
&= \frac{\pi}{2} \ln(6)-\frac{1}{2} \int_{0}^{\pi} \ln(2)+\ln(2-\cos(x))+\ln(2+\cos(x)) dx \end{align}$$ The final result is: $$\begin{align}
& \frac{\pi}{2} \ln(3)-\frac{\pi}{2}\ln(6)+\frac{\pi}{2} \ln(2)+\frac{1}{2}\int_{0}^{\pi} \ln(2-\cos(x))dx+\frac{1}{2} \int_{0}^{\pi} \ln(2+\cos(x))dx \\
&= \pi \ln\Biggl(\frac{1+\sqrt{3}}{2}\Biggl)\space+\space \pi \ln\Biggl(\frac{1+\sqrt{3}}{2}\Biggl) \\
&= 2\pi \ln\Biggl(\frac{1+\sqrt{3}}{2}\Biggl) \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3973742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving a Complex Diophantine Equation with One Known I am struggling to find a way to implement Euclid's Algorithm in order to solve this diophantine equation. The $N$ will be known and the set of solutions I wish to find will be a set of decreasing $X$ solutions that satisfy $$x^2 - 4 y^2 = n.$$
When simplifying I have concluded that the solution could be as follows...
$$x^2 - 4 y^2 = (x - 2y) (x + 2y)$$ when $n = 12$,
$$12 = (x - 2y) (x + 2y)$$
would it be possible to use Euclid's Algorithm to solve for the GCD in decreasing order?
The poposed solutions are a set of $n=12$, $x = 4$ and $y = 1$.
| Say $12=(x+2y)(x-2y)=ab$, where $x$ and $y$ are positive integers,
with $x+2y=a, $ and $x-2y=b$. Then $a>b$, and $2x=a+b$.
That means $a+b$ must be even, so $a$ and $b$ are both odd or both even.
But $a$ and $b$ cannot be both odd, since $ab=12$ is even.
Therefore, $a$ and $b$ are both multiples of $2$.
Since $12=2\times2\times3=ab$, it follows that $a=2\times3$ and $b=2$,
so $x=(a+b)/2=4$ and $y=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do we solve the equation $3\sec^{2}(x) - 4 = 0$? I am confused about the answer to this question:
$$3\sec^{2}(x) - 4 = 0$$
I got the answer
$$x = \frac{\pi}{6} + 2n\pi\quad\text{or}\quad x = \frac{5\pi}{6} + 2n\pi$$
However, the book's answer is $\pi/6 + n\pi, 5\pi/6 + n\pi$
Isn't the period of a secant function $2\pi$? If so, why is the answer saying $\pi$?
| The proposed equation is equivalent to
\begin{align*}
3\sec^{2}(x) - 4 = 0 & \Longleftrightarrow 4\cos^{2}(x) - 3 = 0\\\\
& \Longleftrightarrow \cos(x) = \pm\frac{\sqrt{3}}{2}\\\\
& \Longleftrightarrow \cos(x) = \pm\cos\left(\frac{\pi}{6}\right)
\end{align*}
Based on it, we can conclude that
\begin{align*}
\cos(x) = \cos\left(\frac{\pi}{6}\right) \Longleftrightarrow x = \pm\frac{\pi}{6} + 2m\pi
\end{align*}
as well as
\begin{align*}
\cos(x) = -\cos\left(\frac{\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right) \Longleftrightarrow x = \pm\frac{5\pi}{6} + 2n\pi
\end{align*}
Gathering all solutions, we obtain the following solution set:
\begin{align*}
S = \left\{x\in\mathbb{R} \mid \left(x = \frac{\pi}{6} + m\pi\right)\vee\left(x = \frac{5\pi}{6} + n\pi\right)\right\}
\end{align*}
In order to understand it properly, notice that
\begin{align*}
\begin{cases}
-\dfrac{5\pi}{6} = \dfrac{\pi}{6} - \pi\\\\
-\dfrac{\pi}{6} = \dfrac{5\pi}{6} - \pi
\end{cases}
\end{align*}
Based on it, we can express the solution set as proposed in the book.
EDIT
Considering the comment of @HansEngler, we can also solve the proposed equation as follows:
\begin{align*}
3\sec^{2}(x) - 4 = 0 & \Longleftrightarrow 4\cos^{2}(x) - 3 = 0\\\\
& \Longleftrightarrow 2\cos(2x) - 1 = 0\\\\\
& \Longleftrightarrow \cos(2x) = \cos\left(\frac{\pi}{3}\right)\\\\
& \Longleftrightarrow 2x = \pm\frac{\pi}{3} + 2k\pi\\\\
& \Longleftrightarrow x = \pm\frac{\pi}{6} + k\pi
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3980608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Is there a general formula for the eccentricity $e$ of ellipsoid? For an ellipse of eccentricity $e$ the formulas are:
${x^2 \over a^2} + {y^2 \over b^2} = 1 \\ e = \sqrt {1-\left({b \over a} \right)^2}$
what about the "3D case"?
| In the aspect of measure of the boundary
Perimeter of an ellipse
\begin{align}
1 &= \frac{x^2}{a^2}+\frac{y^2}{b^2} \tag{$a \ge b$} \\
P &= 4aE(e_{ab}) \\
e_{ab} &= \frac{\sqrt{a^2-b^2}}{a} \\
\end{align}
Surface area of an ellipsoid
\begin{align}
1 &= \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \tag{$a \ge b \ge c$} \\
S &= 2\pi
\left[
c^2+\frac{bc^2}{\sqrt{a^2-c^2}}F(\theta,k)+b\sqrt{a^2-c^2} E(\theta,k)
\right] \\
\theta &= \cos^{-1} \frac{c}{a} \\
k &= \frac{a}{b} \sqrt{\frac{b^2-c^2}{a^2-c^2}} \\
&= \frac{e_{bc}}{e_{ac}} \\
k' &= \sqrt{1-k^2} \\
&= \frac{c}{b} \sqrt{\frac{a^2-b^2}{a^2-c^2}} \\
&= \frac{e_{ab}}{e_{ac}}
\end{align}
*
*For prolate spheroid
$$b=c \implies (k,k')=(0,1)$$
*
*For oblate spheroid
$$a=b \implies (k,k')=(1,0)$$
*
*$k$ or $k'$ is not quite well-defined for the case of sphere.
In the aspect of confocal system
Confocal conics
\begin{align}
1 &= \frac{x^2}{a^2+s}+\frac{y^2}{b^2+s} \\
e^2 &= \frac{a^2-b^2}{a^2+s}
\end{align}
Confocal quadrics
\begin{align}
1 &= \frac{x^2}{a^2+s}+\frac{y^2}{b^2+s}+\frac{z^2}{c^2+s} \\
\kappa^2 &= \frac{a^2-c^2}{a^2+s}
\end{align}
*
*For $s\to \infty$, $\kappa \to 0$ which close to a sphere.
*For $s\to -c$, $\kappa \to 1$ which shrinks to focal ellipse namely
$$0=z=\frac{x^2}{a^2-c^2}+\frac{y^2}{b^2-c^2}-1$$
*
*This can not analogous to paraboloids.
See also another posts of mine here and here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3982732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx$
\begin{equation}
\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx
\end{equation}
This is a MCQ question and there are 5 options to choose which are "A.$2\sqrt{3}\pi+9\ln 3$, B.$2\sqrt{7}\pi+8\ln 3$ , C.$2\sqrt{3}\pi+8\ln 3$ , D.$2\sqrt{2}\pi+2\ln 3$, E.Other solution."
How do you solve this integral?
This appears in MCQ test, so I think there should be a trick to solve this without using too much force. The test have 30 questions and 2 hours, so each question should be finished under 4mn ( I doubted it though). Here is my solution that I spend around 3h to solve it. ( Sorry for not using latex and sorry for the inconveniences )Photo
I appreciate any solution tricks. Thanks!
| Tou have done a good work and obtained the correct result.
Trying on my side, using as you did $s=\sin(x)$, we ned with
$$\int_{0}^{\frac{\pi}{2}} \frac{2304\cos x}{(\cos 4x-8\cos 2x+15)^2} \,dx=36 \int_{0}^1 \frac{ds}{ \left(s^4+s^2+1\right)^2}$$
$$s^4+s^2+1=\left(s^2-s+1\right) \left(s^2+s+1\right)$$ Using partial fraction decomposition
$$\frac{1}{ \left(s^4+s^2+1\right)^2}=\frac{1-s}{2 \left(s^2-s+1\right)}+\frac{s+1}{2 \left(s^2+s+1\right)}-$$ $$\frac{s}{4
\left(s^2-s+1\right)^2}+\frac{s}{4 \left(s^2+s+1\right)^2}$$ The first and the second are not difficult
$$I_1=\int\frac{1-s}{ s^2-s+1}\,ds=-\frac 12\Bigg[\int \frac{2s-1}{ s^2-s+1}ds -\int\frac{ds}{ s^2-s+1}\Bigg]$$
$$I_2=\int\frac{s+1}{s^2+s+1}ds=\frac 12\Bigg[\int \frac{2s+1}{ s^2+s+1}ds -\int\frac{ds}{ s^2+s+1}\Bigg]$$
Now, for the third and fourth antiderivatives which look like
$$J=\int \frac s{(s^2+as+1)^2} ds=\int \frac s{(s-r_1)^2(s-r_2)^2} ds$$ partial fraction decomposition again
$$\frac s{(s-r_1)^2(s-r_2)^2}=\frac{r_1}{(r_2-r_1)^2 (s-r_1)^2}+\frac{r_1+r_2}{(r_2-r_1)^3 (s-r_1)}-$$ $$\frac{r_1+r_2}{(r_2-r_1)^3
(s-r_2)}+\frac{r_2}{(r_2-r_1)^2 (s-r_2)^2}$$
This is a pure nightmare !
At the end, after recombining evrything, the antiderivative is
$$\frac{6 s\left(1-s^2\right)}{s^4+s^2+1}+4 \sqrt{3} \tan ^{-1}\left(\frac{16 s \left(1-s^2\right)}{3 \sqrt{3}}\right)+9 \log \left(\frac{s^2+s+1}{s^2-s+1}\right)$$ and, for the definite integral, your good result
$$2 \sqrt{3} \pi +9 \log (3)$$
All this work took me more than one hour. Be sure that I have been looking for tricks but ... no one came to my mind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
how to solve this into partial fractions I'm having a bit of a hard time putting this into partial fractions:
$$\frac{10}{x^2+2x+1+\pi^2}.$$
I know that the roots of the denominator are $-1 \pm i\pi$, but I dont know how to proceed on puting this in partial fractions. Can anyone here help me, please?
| \begin{gather*}
I=\frac{10}{x^{2} +2x+1+\pi ^{2}} =\frac{10}{( x+1+i\pi )( x+1-i\pi )}\\
=\frac{10}{2i\pi }\frac{2i\pi }{( x+1+i\pi )( x+1-i\pi )}\\
=\frac{10}{2i\pi }\frac{( x+1+i\pi ) -( x+1-i\pi )}{( x+1+i\pi )( x+1-i\pi )}\\
=\frac{10}{2i\pi }\left(\frac{1}{( x+1-i\pi )} -\frac{1}{( x+1+i\pi )}\right)
\end{gather*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Proof Check: $4x^2+3x+17$ is continuous by using a $\epsilon$, $\delta$ Argument Can anyone double check my following epsilon delta proof.
I want to prove that the following function is continuous with an Epsilon Delta Argument.
$$ f: x \in \mathbb R \mapsto (4x^2+3x+17) \in \mathbb R $$
So I started with
$$\left\lvert f(x)-f(y) \right\rvert =$$
$$= \left\lvert (4x^2+3x+17) - (4y^2+3y+17) \right\rvert$$
$$= \left\lvert (4x^2+3x)-(4y^2+3y) \right\rvert$$
$$= 3\left\lvert (4x^2+x)-(4y^2+y) \right\rvert$$
$$= 3\left\lvert (\frac {4}{3}x^2+x)-(\frac {4}{3}4y^2+y) \right\rvert $$
$$= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert$$
$$= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert$$
(Triangel inequality)
$$\le 3 \left(\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 \right\rvert + \left\lvert x-y \right\rvert \right)$$
Note that $\delta \le 1$
$$\le 3 \left(\frac {4}{3}\left\lvert x^2 - y^2 \right\rvert + \delta \right)$$
$$= 3 \left(\frac {4}{3}\left\lvert (x - y) \right\rvert\left\lvert (x +y) \right\rvert + \delta \right)$$
$\delta \le 1$
$$\le 3 \left(\frac {4}{3}\left\lvert x + y \right\rvert \delta + \delta \right)$$
Adding zero in form of y - y
$$= 3 \left(\frac {4}{3}\left\lvert x + y - y + y \right\rvert \delta + \delta \right)$$
$$= 3 \left(\frac {4}{3}\left\lvert x - y + 2y \right\rvert \delta + \delta \right)$$
Triangel inequality
$$\le 3 \left(\frac {4}{3}(\left\lvert x - y \right\rvert +\left\lvert 2y \right\rvert) \delta + \delta \right)$$
Note that $\delta \le 1$
$$\le 3 \left(\frac {4}{3}(\delta +\left\lvert 2y \right\rvert) \delta + \delta \right)$$
$$= 4\delta^2+4\left\lvert 2y \right\rvert \delta + 3\delta $$
$$= \delta(4\delta+4\left\lvert 2y \right\rvert + 3) $$
Note that $\delta \le 1$
$$\le \delta(4 +4\left\lvert 2y \right\rvert + 3) $$
$$\le \delta(\left\lvert 8y \right\rvert + 7) = \epsilon$$
Therefore $$\delta = \frac {\epsilon}{(\left\lvert 8y \right\rvert + 7)} \;\; with \; \delta \le1$$
Thanks in advance for the help, I really appreciate it. :)
| Your solution looks good. The following steps could be simplified a bit to make it more readable.
\begin{align}
\left\lvert f(x)-f(y) \right\rvert
&=
3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert
\quad (\textrm{why bother factoring out $3$?})\\
& \le 3 \left(\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 \right\rvert + \left\lvert x-y \right\rvert \right)\\
& \le 3 \left(\frac {4}{3}\left\lvert x^2 - y^2 \right\rvert + \delta \right)
\quad (\color{red}{\textrm{for $|x-y|<\delta$} })\\
&= 3 \left(\frac {4}{3}\left\lvert (x - y) \right\rvert\left\lvert (x +y) \right\rvert + \delta \right)\\
&= 3 \left(\frac {4}{3}\left\lvert x - y + 2y \right\rvert \delta + \delta \right)\\
&\le 3 \left(\frac {4}{3}(\left\lvert x - y \right\rvert +\left\lvert 2y \right\rvert) \delta + \delta \right)
\quad (\textrm{triangle inequality})\\
&\le 3 \left(\frac {4}{3}(\delta +\left\lvert 2y \right\rvert) \delta + \delta \right)\\
&=4(\delta +\left\lvert 2y \right\rvert) \delta + 3\delta
\\
&\le
4(1+\left\lvert 2y \right\rvert) \delta + 3\delta
\quad (\delta\le 1)\\
&\le (7+|8y|)\delta
\end{align}
Set $$\delta =\min(\frac {\epsilon}{\left\lvert 8y \right\rvert + 7} ,1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Combinatorial/ Binomial Identity I am finding difficulty in showing the following identity:
$$\sum_{k=1}^n {n-1\choose k-1} {2n-1\choose k}^{-1} = \frac{2}{n+1}.$$
Mainly because of the inverse part.
| We have by binomial coefficient algebra that
$$\sum_{k=1}^n {n-1\choose k-1} {2n-1\choose k}^{-1}
= \sum_{k=1}^n \frac{(n-1)!/(n-k)!/(k-1)!}
{(2n-1)!/(2n-1-k)!/k!}
\\ = {2n-1\choose n}^{-1}
\sum_{k=1}^n k \frac{(2n-1-k)!}{n! (n-k)!}
\\ = \frac{1}{n} {2n-1\choose n}^{-1}
\sum_{k=1}^n k {2n-1-k\choose n-k}
\\ = \frac{1}{n} {2n-1\choose n}^{-1}
[z^n] (1+z)^{2n-1} \sum_{k=1}^n k \frac{z^k}{(1+z)^k}.$$
Now the coefficient extractor enforces the range and we get
for the extractor only
$$[z^n] (1+z)^{2n-1} \sum_{k\ge 1} k \frac{z^k}{(1+z)^k}
= [z^n] (1+z)^{2n-1}
\frac{z/(1+z)}{(1-z/(1+z))^2}
\\ = [z^n] (1+z)^{2n-1}
\frac{z(1+z)}{(1+z-z)^2}
= [z^{n-1}] (1+z)^{2n}
= {2n\choose n-1}.$$
Restoring the scalar in front now yields
$$\frac{1}{n} {2n-1\choose n}^{-1} {2n\choose n-1}
= \frac{1}{n} 2n \frac{1}{n+1} = \frac{2}{n+1}.$$
This is the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3990786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the reaction of the table
A uniform rod of mass $m$ and length $2a$ is held inclined to the vertical at an angle $\alpha$ with its lower end in contact with a smooth horizontal table and is released from rest. Prove that when the inclination of the rod to the vertical is $\theta$, the reaction of the table is
$$\frac{mg(4-6\cos{\alpha}\cos{\theta}+3\sin^2{\theta})}{(1+3\sin^2{\theta})^2} $$
I got reaction of the table as $\dfrac{mg(4-6\cos{\alpha}\cos{\theta}+3\cos^2{\theta})}{(1+3\sin^2{\theta})^2} $ which is not what we need to prove but I think my answer is correct, If $\alpha=0$ and $\theta=0$ then rod is vertical so reaction must be $mg$ which is satisfy my answer but $\dfrac{mg(4-6\cos{\alpha}\cos{\theta}+3\sin^2{\theta})}{(1+3\sin^2{\theta})^2} $ is not satisfied
Can anyone help me for this? Am I correct?
| I got the same result as you.
The motion of the rod can be understood as translation of the center od mass (midpoint of rod) and rotation around the center of mass. As there are no horizontal forces acting on the center of mass, the center of mass only moves vertically. If we denote its height by $y$, the conservation of energy gives
$$\frac12 m\dot{y} + \frac12 I\omega^2 + mgy = mgy(0).$$
where the moment of inertia is $I = \frac{ma^2}3$ and $\omega = \dot{\theta}$ is angular rotation around the center of mass. We can note that $y = a\cos\theta$ so $\dot{y} = -a\omega\sin\theta$. Plugging this in we get
$$\frac12ma^2\omega^2\sin^2\theta + \frac16ma^2\omega^2 + mga\cos\theta=mga\cos\alpha $$
which implies
$$\omega = \sqrt{\frac{6g(\cos\alpha - \cos\theta)}{a(1+3\sin^2\theta)}}.$$
Now, since the table is smooth, the reaction force $\vec{N}$ acts only upwards. Its projection orthogonally to the rod is $N\sin\theta$. The resulting torque $Na\sin\theta$ drives the rotation of the rod. The rotational equation of motion is
$$N\sin\theta = I\dot{\omega}.$$
After some calculation, we get
$$\dot\omega = \sqrt{\frac{6g}{a}} \frac1{2\sqrt{\frac{\cos\alpha - \cos\theta}{1+3\sin^2\theta}}}\frac{(4-6\cos\alpha\cos\theta +3\cos^2\theta)\sin\theta}{2(1+3\sin^2\theta)^2} \omega = \frac{3g(4-6\cos\alpha\cos\theta +3\cos^2\theta)\sin\theta}{2a(1+3\sin^2\theta)^2}$$
so
$$N = I\frac{\dot\omega}{\sin\theta} = \frac{ma^2}3\frac{3g(4-6\cos\alpha\cos\theta +3\cos^2\theta)}{2a(1+3\sin^2\theta)^2} = \frac{mg(4-6\cos\alpha\cos\theta +3\cos^2\theta)}{(1+3\sin^2\theta)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Unifying the connections between the trigonometric and hyperbolic functions There are many, many connections between the trigonometric and hyperbolic functions, some of which are listed here. It is probably too optimistic to expect that a single insight could explain all of these connections, but is there a holistic way of seeing the parallels between $\sin$ and $\sinh$, $\cos$ and $\cosh$? Can all of these seemingly disparate connections be shown to be essentially the same, or at least very similar?
Geometric connections
*
*Sine and cosine parameterise the unit circle $x^2+y^2=1$, just as hyperbolic sine and cosine parameterise the 'unit hyperbola' $x^2-y^2=1$. Both circles and hyperbolas are conic sections.
*The sector of the circle connecting the points $(0,0)$, $(1,0)$, and $(\cos t,\sin t)$ has an area of $t/2$. The region of the hyperbola connecting the points $(0,0)$, $(1,0)$, and $(\cosh t,\sinh t)$ has an area of $t/2$. This can even be used to define the hyperbolic functions geometrically, and many authors do the same with the trigonometric functions.
*Sine and hyperbolic sine are odd, whereas cosine and hyperbolic cosine are even. But sine and cosine are periodic functions, unlike the hyperbolic counterparts.
*The analogue of the identity $\cos^2x+\sin^2x \equiv 1$ is $\cosh^2x-\sinh^2x \equiv 1$. The compound angle formulae are almost identical to their hyperbolic counterparts, save for a pesky minus sign:
\begin{align}
\sin(x+y) &= \sin(x)\cos(y)+\cos(x)\sin(y) \\
\sinh(x+y) &= \sinh(x)\cosh(y) + \cosh(x)\sinh(y) \\[4pt]
\cos(x+y) &= \cos(x)\cos(y) \color{red}{-} \sin(x)\sin(y) \\
\cosh(x+y) &= \cosh(x)\cosh(y) \color{blue}{+} \sinh(x)\sinh(y) \, .
\end{align}
*In general, given a trigonometric function, it is possible to write down the corresponding hyperbolic identity using Osborn's rule: replace every occurrence of $\cos$ with $\cosh$; replace every occurrence of $\sin$ by $\sinh$; but negate the product of two $\sinh$ terms.
Analytic connections
*
*$\sin$ is the unique solution to the initial value problem
\begin{align}
f''(x) &= \color{red}{-}f(x) \\
f'(0) &= 1 \\
f(0) &= 0 \, ,
\end{align}
and the corresponding initial value problem for $\sinh$ is the same, except $f''(x) = \color{blue}{+}f(x)$.
*Likewise, the initial value problem for $\cos$ is
\begin{align}
f''(x) &= \color{red}{-}f(x) \\
f'(0) &= 0 \\
f(0) &= 1 \, ,
\end{align}
and again we see a mysterious sign change for $\cosh$: $f''(x) = \color{blue}{+}f(x)$.
*It follows that the higher-order derivatives of $\sin$ and $\sinh$ form periodic sequences.
*If we solve the initial value problems shown above, we obtain the exponential forms of all $4$ functions:
\begin{align}
\sin x &= \frac{e^{\color{green}{i}x}-e^{-\color{green}{i}x}}{2\color{green}{i}} \quad{} \cos x = \frac{e^{\color{\green}{i}x}+ e^{-\color{green}{i}x}}{2} \\[3pt]
\sinh x &= \frac{e^{x}-e^{-x}}{2} \quad{} \cosh x = \frac{e^x + e^{-x}}{2} \, .
\end{align}
*All $4$ functions are analytic, and their Taylor series bear a striking resemblance to each other:
\begin{align}
\sin x &= x \color{red}{-} \frac{x^3}{3!} + \frac{x^5}{5!} \color{red}{-} \frac{x^7}{7!} + \ldots \\[4pt]
\sinh x &= x \color{blue}{+} \frac{x^3}{3!} + \frac{x^5}{5!} \color{blue}{+} \frac{x^7}{7!} + \ldots \\[4pt]
\cos x &= 1 \color{red}{-} \frac{x^2}{2!} + \frac{x^4}{4!} \color{red}{-} \frac{x^6}{6!} + \ldots \\[4pt]
\cosh x &= 1 \color{blue}{+} \frac{x^2}{2!} + \frac{x^4}{4!} \color{blue}{+} \frac{x^6}{6!} + \ldots
\end{align}
*And Euler's formula
$$
e^{ix} = \cos x + i \sin x
$$
is replaced by the underwhelming
$$
e^x = \cosh x + \sinh x \, .
$$
| The connection between the trigonometric and hyperbolic functions becomes more intimate when one introduces split-complex numbers: the numbers of the form $a+bj$, where $j^2=1$ and $j\ne \pm 1$.
Using Taylor series you can easily find that
$$
e^{jx} = \cosh x + j\sinh x , \quad
\sinh x = \frac{e^{jx}-e^{-jx}}{2j}, \quad \cosh x = \frac{e^{jx} + e^{-jx}}{2}.
$$
One can understand $\cos x$ and $\sin x$ as the real and imaginary parts of the complex exponential $e^{ix}$. Likewise, $\cosh x$ and $\sinh x$ are the real and 'imaginary' parts of the split-complex exponential $e^{jx}$.
The analogy can be summarized as follows:
$$
\begin{align}
& &&\textbf{Complex numbers} && \textbf{Split-complex numbers} \\
&\text{Imaginary unit:} && i^2 = -1, && j^2 = 1, \\
&\text{Numbers:} && z = a+bi, && z = a+bj, \\
&\text{Modulus:} && |z| = \sqrt{a^2+b^2}, && |z| = \sqrt{a^2-b^2} \; (\text{for } a^2 \ge b^2), \\
&\text{Euler's formula:} && e^{ix} = \cos x + i \sin x, && e^{jx} = \cosh x + j \sinh x, \\
&\text{Pythagorean theorem:} && |e^{ix}| = \cos^2 x + \sin^2 x =1, && |e^{jx}| = \cosh^2 x - \sinh^2 x = 1, \\
&\text{Geometric meaning:} && \text{Unit circle,} && \text{Unit hyperbola (right branch)}. \\
\end{align}
$$
The connection is in the correspondence:
$$
\begin{pmatrix} \text{complex} \; i \\ \sin, \; \cos \end{pmatrix} \leftrightarrow \begin{pmatrix} \text{split-complex} \; j \\ \sinh, \; \cosh \end{pmatrix}.
$$
One can use imagination to say that $\sin$, $\cos$ and complex numbers come from the elliptic world, while $\sinh$, $\cosh$ and split-complex numbers are their counterparts from the hyperbolic world.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3999593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 6,
"answer_id": 1
} |
Find the coefficient of $x^{16}$ in $(1 + x + x^2 + x^3 + x^4 + x^5)^4$ We are supposed to find the coefficients of it, I wanted to know if my approach is right here. The final answers seems a bit iffy.
$$(1+x+\dots+x^5)^4=\left(\frac{1-x^6}{1-x}\right)^4=(1-x^6)^4(1-x)^{-4}$$
Using the binomial theoerem, I got the following:
$$
(1-x^6)^4=\sum_{k\ge0}(-1)^k\binom{4}{k}x^{6k}
$$
and using the negative binomial theorem,
$$
(1-x)^{-4}=\sum_{k\ge0}(-1)^k\binom{-4}{k}x^k=\sum_{k\ge0}\binom{4+k-1}{k}x^k
$$
So the $x^{16}$ coefficient is
$$
\binom{4}{0}\binom{8+16-1}{16}-\binom{4}{1}\binom{8+10-1}{10}+\binom{4}{2}\binom{8+4-1}{4}\
$$
$$
= 1(245157)-4(19448)+6(330)\
$$
$$
= 245157-77792+1980\
= 169345
$$
Not too sure if I did this right. I think my math may be off somewhere but can't figure out where.
| This is of course :-) intended to find out the answer to:
You have a bag of $16$ (identical) frogs, and wish to distribute them to $4$ of your wizard friends. However each wizard only has enough pocket space to carry $5$ frogs. In how many ways can you distribute the frogs?
And the answer by inclusion-exclusion is
$\binom {16+4-1}{4-1} - \binom{4}{1}\binom {16-6+4-1}{4-1} + \binom{4}{2}\binom {16-12+4-1}{4-1} = \binom {19}{3} - 4\binom {13}{3} + 6\binom {7}{3} = 969 -1144+210 = 35$
Edit to add: As Doug M mentions in comments to the main question, this is also the number of ways to roll $20$ on four distinct dice, since choosing $4$ times from $\{0,1,2,3,4,5\}$ to make a total of $16$ can be done the same number of ways as choosing $4$ times from $\{1,2,3,4,5,6\}$ to make $20$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Find $\lim_{k\rightarrow \infty} A^k x$ Let $\displaystyle A=\left(\begin{matrix}.5&1\\0&.75\end{matrix}\right),x=\left(\begin{matrix}-2\\1\end{matrix}\right)$.
*
*Find $\displaystyle \lim_{k\rightarrow \infty} A^k x$
*Find $\displaystyle \lim_{k\rightarrow \infty} \frac 1 {\Vert A^k x\Vert}A^k x$
I found the eigenvalues and eigenvectors to be $\displaystyle \lambda_1=.5, v_1=\left(\begin{matrix}1\\0\end{matrix}\right),\lambda_2=.75,v_2=\left(\begin{matrix}4\\1\end{matrix}\right)$.
x can be expressed as $x=-6v_1+v_2$, then $\displaystyle Ax=-6(.5)\left(\begin{matrix}1\\0\end{matrix}\right)+.75\left(\begin{matrix}4\\1\end{matrix}\right)=\left(\begin{matrix}-6(.5)+.75(4)\\.75\end{matrix}\right)$
Then $\displaystyle \lim_{k\rightarrow \infty} A^k x=\lim_{k\rightarrow \infty} \left(\begin{matrix}-6(.5)^k+(.75)^k(4)\\(.75)^k\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$ ?
And what is $\displaystyle \lim_{k\rightarrow \infty} \frac 1 {\Vert A^k x\Vert}A^k x=\lim_{k\rightarrow \infty} \left(\begin{matrix}\frac{-6(.5)^k+(.75)^k(4)}{\sqrt{(-6(.5)^k+(.75)^k(4))^2+((.75)^k)^2}}\\\frac{(.75)^k}{\sqrt{(-6(.5)^k+(.75)^k(4))^2+((.75)^k)^2}}\end{matrix}\right)=\left(\begin{matrix}\infty\\\infty\end{matrix}\right)$?
| Since you already did the first part, I will rewrite the second part in terms of the eigenvector decomposition
$$\lim_{k\to\infty} \frac{\left(\frac{1}{2}\right)^k\cdot(-6v_1)+\left(\frac{3}{4}\right)^k\cdot(v_2)}{\left|\left(\frac{1}{2}\right)^k\cdot(-6v_1)+\left(\frac{3}{4}\right)^k\cdot(v_2)\right|} \equiv \lim_{k\to\infty}\frac{av+bw}{|av+bw|}$$
The two vectors are not orthogonal, and the magnitude of the sum of non orthogonal vectors is given by
$$|av+bw| = \sqrt{(av+bw)\cdot(av+bw)} = \sqrt{a^2|v|^2+2abv\cdot w + b^2|w|^2}$$
Plugging in the vectors we have, this simplifies to
$$\sqrt{a^2+8ab+17b^2} = |b|\sqrt{\left(\frac{a}{b}\right)^2+ 8\frac{a}{b}+17}$$
Now why did we pull out $b$ instead of $a$ when the same formula could have applied either way? It's because $\frac{a}{b}\to 0$, thus we can take the limit without a L'Hopital situatuon occurring. Meaning the limit simplifies to
$$\lim_{k\to\infty} \frac{\frac{a}{b}v + w}{\sqrt{\left(\frac{a}{b}\right)^2+ 8\frac{a}{b}+17}} \to \frac{w}{\sqrt{17}} = \frac{1}{\sqrt{17}}\begin{pmatrix} 4 \\ 1 \end{pmatrix}$$
because $b>0$. In general you can see that the second limit will approach the unit eigenvector with the largest (by absolute value) eigenvalue.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The segment joining the vertices of altitudes A triangle $ABC$ with sides $AB=15,BC=14,AC=13$ is given. $AA_1=12$ and $BB_1$ are heights. Find $A_1B_1$.
I was thinking about the Ptolemy's theorem: $$A_1B_1\cdot CH=B_1C\cdot HA_1+B_1H\cdot CA_1,$$ where $H$ is the intersection of the heights (the orthocenter). We can use it because $HA_1CB_1$ is cyclic: $$\measuredangle HB_1C+\measuredangle HA_1C=180^\circ.$$ I was able to find that $CA_1=5$. I am not sure this is the easiest solution, because we have to find $5$ more segments. Any thoughts on the problem will be appreciated. Thank you!
| The area of the triangle is, by Heron's formula, $$|\triangle ABC| = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(6)(7)(8)} = 84,$$ thus $$AA_1 = \frac{2(84)}{14} = 12, \quad BB_1 = \frac{2(84)}{13} = \frac{168}{13},$$ and by the Pythagorean theorem, $$CA_1 = 5, \quad CB_1 = \frac{70}{13}.$$ Since by the Law of Cosines we have $$\cos \angle C = \frac{13^2 + 14^2 - 15^2}{2(13)(14)} = \frac{5}{13},$$ we obtain
$$(A_1 B_1)^2 = (CA_1)^2 + (CB_1)^2 - 2(CA_1)(CB_1) \cos \angle C = \left(\frac{75}{13}\right)^2,$$ and the desired length is $75/13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4008643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Determinant Identity: Elegant Solution Please. Can anyone provide a proof of this identity:
$$\begin{vmatrix}
(b+c)^2 &a^2&a^2\\
b^2&(a+c)^2&b^2\\
c^2&c^2&(a+b)^2\\
\end{vmatrix}=2abc(a+b+c)^3 ?$$
Ideally I would like a series of row column operations resulting in the formula.
No brute force calculations. Note that letting $a=0$ we get two rows equal. So $abc$ is a factor. But I have unable to produce this by a simple series of operations, I mean to get one row or column all of whose entries are divisible by $a$. Getting divisibility by $(a+b+c)^2$ is also not hard just subtract one column from the other two. After that however I have been unable to make any further progress. Hopefully someone here is smarter than I.
| $$\begin{vmatrix}
(b+c)^2 &a^2&a^2\\
b^2&(a+c)^2&b^2\\
c^2&c^2&(a+b)^2\\
\end{vmatrix}= \begin{vmatrix}
(b+c)^2-a^2 &0&a^2\\
0&(a+c)^2-b^2&b^2\\
c^2-(a+b)^2&c^2 - (a+b)^2&(a+b)^2\\
\end{vmatrix}$$
$$\begin{vmatrix}
(b+c)^2-a^2 &0&a^2\\
0&(a+c)^2-b^2&b^2\\
c^2-(a+b^2)&c^2 - (a+b)^2&(a+b)^2\\
\end{vmatrix} =(a+b+c)^2\begin{vmatrix}
b+c-a &0&a^2\\
0&a+c-b&b^2\\
c-a-b& c-a-b&(a+b)^2\\
\end{vmatrix}$$
$$=(a+b+c)^2\begin{vmatrix}
b+c-a &0&a^2\\
0&a+c-b&b^2\\
-2b& -2a& 2ab\\
\end{vmatrix} (R_3 \to R_3 -(R_1+R_2))$$
$$=\frac{(a+b+c)^2}{ab}\begin{vmatrix}
a(b+c-a) &0&a^2\\
0&b(a+c-b)&b^2\\
-2ab& -2ab& 2ab\\
\end{vmatrix}$$
After this just do $R_{1} \to R_1 + R_3$ and $R_{2} \to R_2 + R_3$
$$=\frac{(a+b+c)^2}{ab}\begin{vmatrix}
ab+ac &a^2&a^2\\
b^2&ba+ac&b^2\\
0& 0& 2ab\\
\end{vmatrix} ={(a+b+c)^2}\begin{vmatrix}
b+c &a&a\\
b&a+c&b\\
0& 0& 2ab\\
\end{vmatrix}$$
$$={(a+b+c)^3}\begin{vmatrix}
1 &a&a\\
1&a+c&b\\
0& 0& 2ab\\
\end{vmatrix} = {(a+b+c)^3}\begin{vmatrix}
1 &a&a\\
0&c&b-a\\
0& 0& 2ab\\
\end{vmatrix}$$
This gives the required value of $2abc(a+b+c)^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4015504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Let ABCD be a tetrahedron of volume 1 and M,N,P,Q,R,S on AB,BC,CD,DA,AC,BD s.t. MP,NQ,RS are concurrent. Then the volume of MNRSPQ is less than 1/2. So far, I have denoted
$$\frac{MA}{MB}=a,\frac{BN}{NC}=b,\frac{CP}{PD}=c,\frac{DQ}{QA}=d,\frac{AR}{RC}=e,\frac{BS}{SD}=f$$
and then I have computed the volumes of the "small" tetrahedra
$$V_1=V_{AMRQ}=\frac{ae}{(a+1)(e+1)(d+1)};$$
$$V_2=V_{BMSN}=\frac{bf}{(b+1)(f+1)(a+1)};$$
$$V_3=V_{CRNP}=\frac{c}{(c+1)(e+1)(b+1)};$$
$$V_4=V_{DPQS}=\frac{d}{(d+1)(c+1)(f+1)}.$$
Also, by Menelaus theorem for tetrahedra, one can deduce that
$$abcd=1;~~~ af=ce; ~~~b=fde.$$
It remains to show that $$V_1+V_2+V_3+V_4\geq \frac{1}{2}.$$
The problem has a well known analogue in 2-D: for a triangle $ABC$ of area 1 and $AM, BN, CP$ concurrent cevians, the area of the triangle $MNP$ is less than 1/4.
Later edit/update:
I have managed to rewrite $V_k$ as follows:
$$V_1={\frac {a}{ \left( a+1 \right) \left( d+1 \right) \left( cd+1
\right) }};$$
$$V_2={\frac {b}{ \left( a+1 \right) \left( b+1 \right) \left( ad+1
\right) }};$$
$$V_3= {\frac {c}{ \left( c+1 \right) \left( b+1 \right) \left( ab+1
\right) }};$$
$$V_4= {\frac {d}{ \left( c+1 \right) \left( d+1 \right) \left( bc+1
\right) }},$$
Plus, we still have $abcd=1$.
| Let $X$ be the intersection of the three segments $MP$, $NQ$, $RS$.
Since $ABCD$ is non-degenerate, its volume is one, the points $A,B,C,D$ are not in a plane, so we can uniquely write
$$
X = aA+bB+cC+dD\ , \qquad a+b+c+d=1\ .
$$
Then grouping pairs of vertices, we obtain formulas for $M,N,P,Q,R,S$ in terms of these weights, corresponding to the groupings:
$$
\begin{aligned}
X
&=
(a+b)\cdot \underbrace{\frac {aA+bB}{a+b}}_{M} +
(c+d)\cdot \underbrace{\frac {cC+dD}{c+d}}_{P}
\\[2mm]
&=
(a+c)\cdot \underbrace{\frac {aA+cC}{a+c}}_{R} +
(b+d)\cdot \underbrace{\frac {bB+dD}{b+d}}_{S}
\\[2mm]
&=
(a+d)\cdot \underbrace{\frac {aA+dD}{a+d}}_{Q} +
(b+c)\cdot \underbrace{\frac {bB+cC}{b+c}}_{N} \ .
\end{aligned}
$$
(They really know in Romania vectors in 3D in the VIII.th class. The relations
the children could have been written involve the vectors $OX$, $OA$, $OB$, $OCV$, $OD$ instead of the above.)
So $M$ is situated on $AB$ such that $M=\frac 1{a+b}(aA+bB)$, which gives
$AM:AB=b:(a+b)$. We write similar relations for $AR:AC=c:(a+c)$ and $AQ:AD=d:(a+d)$. So the volume $[AMRQ]$, seen as a fraction of the volume $[ABCD]=1$
is
$$
[AMRQ]
=
\frac{AM}{AB}\cdot
\frac{AR}{AC}\cdot
\frac{AQ}{AD}\cdot
[ABCD]
=\frac{bcd}{(a+b)(a+c)(a+d)}\cdot 1\ .
$$
So the given geometric inequality is equivalent to the algebraic inequality
$$
\sum\frac{bcd}{(a+b)(a+c)(a+d)}
\ge
\frac 12\ .
$$
The sum has four terms obtained by acting with the powers of the cyclic permutation $(a,b,c,d)$ on the written term.
I could find quickly only the brute force proof. We multiply with the common denominator and have to show the domination:
$$
\underbrace{\sum a^3b^2c}_{4\cdot 3\cdot 2=24\text{ terms}} +
2\underbrace{\sum a^2b^2c^2}_{4\text{ terms}}
\ge
2\underbrace{\sum a^3bcd}_{4\text{ terms}} +
4\underbrace{\sum a^2b^2cd}_{6\text{ terms}}
\ .
$$
The domination is clear, the monominal type $(3,2,1,0)$
dominates $(3,1,1,1)$ and $(2,2,1,1)$. And $(2,2,2,0)$ dominates $(2,2,1,1)$.
So one can stop here.
But for the convenience of the reader, let us write this explicitly. First of all
$$
b^2c+c^2b+b^2d+d^2b + c^2d+d^2c\ge 6bcd\ .
$$
(Arithmetic-geometric mean.) We multiply with $a^3$ and sum. It remains (i.e. is sufficient) to show
$$
\frac 23
\underbrace{\sum a^3b^2c}_{4\cdot 3\cdot 2=24\text{ terms}} +
2\underbrace{\sum a^2b^2c^2}_{4\text{ terms}}
\ge
4\underbrace{\sum a^2b^2cd}_{6\text{ terms}}
\ .
$$
Now use $a^3b^2c+ab^2c^3\ge 2a^2b^2c^2$ to estimate downwards the LHS with a scalar multiple of $\sum a^2b^2c^2$, to have a simpler LHS, so
$$
\begin{aligned}
\frac 23
\underbrace{\sum a^3b^2c}_{4\cdot 3\cdot 2=24\text{ terms}} +
2\underbrace{\sum a^2b^2c^2}_{4\text{ terms}}
&\ge
\frac 23\cdot 6
\underbrace{\sum a^2b^2c^2}_{4\text{ terms}} +
2\underbrace{\sum a^2b^2c^2}_{4\text{ terms}}
=
6
\underbrace{\sum a^2b^2c^2}_{4\text{ terms}}
\\
&\ge
4\underbrace{\sum a^2b^2cd}_{6\text{ terms}}
\ .
\end{aligned}
$$
The last inequality follows from $a^2b^2c^2+a^2b^2d^2\ge 2a^2b^2cd$.
$\square$
Note: The problem was proposed by Flavian Georgescu, Bucharest, at that time college student (elev), so maximally 18yo, and it should be doable with the aid of "thin air" available in the 8.th class (14yo students) of the Romanian mathematical matter from the school book. I suppose Titu or a mean inequality was the solution of the author, if he did it as above, else i see only the possibility of a geometrical argument using $[XMRQ]$ and the similar tetraedra.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Where is the mistake in a point closest to cross section of two planes and a line problem The problem:
Find the point that lies on the line that is the intersection of two planes:
plane one: $$3x-2y+z = 2$$
plane two: $$x+y+2z = 4$$
The point also needs to be the same distance away from points $A(1,-1,-1),B(1,-1,3)$
My try:
First I found the direction of the line that is the intersection of two planes. Here I used cross product between $(3,-2,1)$ and $(1,1,2)$, which are the normals of the two planes. I got $s = (-1,-1,1)$ That is the direction of the intersection line.
What I did next was I solved the system of equations: $$3x-2y+z-2 = 0\\x+y+2z-4 = 0$$
I got: $z = 7y +2$ and $x = \frac{-5y}{3}$
Here I chose $y$ to be equal to $1$ $\implies y = 1 \implies x = \frac{-5}{3} \implies z = 9$
So the equation of the intersection line is: $(1,\frac{-5y}{3}, 9) + t(-1,-1,1) = (x,y,z)$
Then I wanted to find the plane where all the points that have the same distance from $A(1,-1,-1),B(1,-1,3)$ are located. it is trivial this plane's normal will be the direction vector $AB$, thus the normal is $n=(0,0,4)$
Then I wanted to position the plane($O$...origin) : $$OA + \frac{1}{2}AB = (1,-1,-1) +\frac{1}{2}(1,-1,3) = (\frac{3}{2},\frac{-3}{2},\frac{1}{2}) = \text{\{This is the point to position the plane\}}$$
So the equation of this plane is: $4z = 4*\frac{1}{2}=2$
I got $z_0$ form the last coordinate from the point.
Then I Solved for intersection between the first line(intersection between plane one and two) and this last plane. I got the wrong answer: $x = \frac{19}{2}, y = \frac{41}{6}, z = \frac{1}{2}$ I got this results by inserting parametrically defined $z$ into the plane equation. My answer is wrong.
The correct solution is: $T(1,1,1)$
Where did I go wrong?
| I would have done this a slightly different way. The two planes are
3x−2y+z=2 and x+y+2z=4. From the equation of the second plane, x= 4- y- 2z. Replace x in the first equation by that: 3(4- y- 2z)- 2y+ z= 12- 3y- 6z- 2y+ z= 12- 5y- 5z= 2 so 5y+ 5z= 10 and y+ z= 2 so z= 2- y and x= 4- y- 2(2- y)= 4- y- 4+ 2y= y.
Taking y= t, parametric equations for the line of intersection are x= t, y= t, z= 2- at. Any point on that line of intersection is of the form (x, y, z)= (t, t, 2- t).
The distance, squared, from A(1, -1, -1) to (t, t, 2- t) is (t- 1)^2+ (t+ 1)^2+ (3- t)^2.
The distance, squared, from B(1, -1, 3) to (t, t, (2- t)) is (t- 1)^2+ (t+ 1)^2+ (1+ t)^2.
For a point on the line equally distant from A an(td B, we must have (t- 1)^2+ (t+ 1)^2+ (3- t)^2= (t- 1)^2+ (t+ 1)^2+ (1+ t)^2.
Those first two squares on the left cancel the first two on the right leaving (3- t)^2= (1+ t)^2. Taking the square root of both sides 3- t= 1+ t so 2t= 2 and t= 1. Taking 3- t= -(1+ t)= -1- t does not give a solution.
So the point on the line equidistant from A and B is (t, t, 2- t)= (1, 1, 2- 1)= (1, 1, 1).
Check: Is (1, 1, 1) on the plane 3x- 2y+ z= 2? Yes,3- 2+ 1= 2.
Is (1, 1, 1) on the plane x+ 2y+ z= 4? Yes 1+ 2+ 1= 4.
The distance, squared, from (1, 1, 1) to A(1,-1,- 1) is (1- 1)^2+ (1-(-1))^2+ (1- (-1))^2= 0+ 4+ 4= 8.
The distance, squared, from (1, 1, 1) to B(1, -1, 3) is (1- 1)^2+ (1- (-1))^2+ (1- 3)^2= 0+ 4+ 4= 8.
Yes, (1, 1, 1) is equally distant from A and B.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Solve for natural numbers: $(2a+3)(3a+4) = 35^b$ Find $a$ and $b$ be natural numbers, such that $(2a + 3)(3a + 4) = 35^b$
One can see $a=1, b=1$ is a solution. The challenge is to prove there are no other solutions.
My intuition is that $2a + 3 = 5^b$, and $3a + 4 = 7^b$ - otherwise, $5$ would divide both terms, therefore their difference - so, $5 | a + 1, 7| a+ 1\implies 35 | a + 1,$ then LHS would not be divisible by $35$, contradiction. So, the first parenthesis is $5^b$ and second is $7^b$.
| hint
$$3(2a+3)-2(3a+4)=1$$
so, By Bezout's Theorem, $ 2a+3 $ and $ 3a+4 $ are relatively primes.
on the other hand
$$(2a+3)(3a+4)=5^b.7^b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} $ I have to find for which values of $a \in \Bbb N, a \ne 0$ the following limit exists and it is finite:
$$\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} $$
Applying L'Hôpital's rule:
$$\frac{1-\cos (ax)}{x^2}\sim \frac{\sin (ax) \cdot a}{2x}\sim \frac{a^2}{2}$$
Then $$ \frac{\cos (\pi \cdot \frac{1-\cos (ax)}{x^2})}{x^2} \sim \frac{\cos (\pi \cdot \frac{a^2}{2})}{x^2}.$$
$$\cos (\pi \cdot \frac{a^2}{2})=0 \implies a^2=1+2k, \quad k \in \Bbb N$$
and in this case the limit is $0$.
In the book the suggested solution is $a^2=1+2k$ for which the limit is $$ \frac{(-1)^k \cdot (2k+1)^2\cdot\pi}{24}$$
but I don't understand this solution.
Trying to solve the limit with $a^2=1+2k$:
$$ \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} \sim \frac{-\sin (\pi \cdot \frac{1-\cos ax}{x^2}) \cdot \pi \cdot \frac{\sin (ax) a x^2 - (1-\cos (ax)) 2x}{x^4}}{2x} $$$$ = \sin \left( \frac{\pi}{2}+k \pi\right) \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4}$$$$= (-1)^k \cdot \pi \cdot \frac{(1-\cos(ax))2- \sin(ax) ax}{2 \cdot x^4} $$
| This part of what you wrote is incorrect:
$$\frac{\cos (\pi \cdot \frac{1-\cos (ax)}{x^2})}{x^2} \sim \frac{\cos (\pi \cdot \frac{a^2}{2})}{x^2}$$
You're taking the limit of the top of the fraction, but not the bottom. Instead, you need to consider under what circumstances the limit of the entire fraction might exist. Of course, this can only happen if the top tends towards zero, i.e. if
$$\cos (\pi \cdot \frac{a^2}{2}) = 0$$
So the limit might exist if $a^2 = 1 + 2k$ for some $k \in \mathbb N$. Next, assuming $a$ is one of these values, compute the limit of the entire fraction using L'Hospital's rule:
$$
\lim_{x\to 0} \frac{\cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{x^2} = \lim_{x\to 0} \frac{\frac{d}{dx} \space \cos \left(\pi \cdot \frac{1-\cos (ax)}{x^2}\right)}{\frac{d}{dx} x^2} $$
$$ = \lim_{x\to 0} \frac{-\pi \left(\frac{a\sin(ax)x - 2(1 - \cos(ax))}{x^3}\right)\sin(\pi\cdot\frac{1 - \cos(ax)}{x^2})}{2x}
$$
$$
= \left( \lim_{x\to 0} \frac{-\pi\cdot(a \sin(ax) x - 2\cdot(1 - \cos(ax)))}{2x^4} \right) \left( \lim_{x\to 0} \sin\left(\pi \frac{1 - \cos(ax)}{x^2}\right) \right)
$$
You already know the second limit is $\sin\left(\pi\cdot\frac{a^2}{2}\right) = \pm 1$ (depending on the particular value of $a$). The first limit is straightforward. Just apply L'Hospital's Rule repeatedly until you have your answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve system of non-linear equations $\frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}= \frac{1}{y^2}+\frac{1}{(y-x)^2}-\frac{1}{(1-y)^2}=0$ Whilst solving a problem, the following system of non-linear equations arose:
$$0= \frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}$$
$$0= \frac{1}{y^2}+\frac{1}{(y-x)^2}-\frac{1}{(1-y)^2}$$
I have tried to solve it via brute force and by trying to add them both in a way that could give me the answer, but I still haven't got it.
I can't manage to find the intersection of the two parametric equations that these equations give. I know that I can find approximate solutions with the Wolfram Alpha tool, but I would like to know, if any, an analytical or strong numerical way to proceed.
Note: Out of all the solutions ($8$ it seems, $4$ real and $4$ complex), I only need the one that takes real values for $x$ and $y$, with $0<x<y<1$.
| Take the cue of $0<x<y<1$ and observe that $x+y=1$ satisfies both equations
$$\frac{1}{x^2}-\frac{1}{(y-x)^2}-\frac{1}{(1-x)^2}=
\frac{1}{y^2}+\frac{1}{(y-x)^2}-\frac{1}{(1-y)^2}=0$$
which leads to $\frac{1}{x^2}-\frac{1}{(1-2x)^2}-\frac{1}{(1-x)^2}=0$, or
$$x^4+6x^3-11x^2+6x -1=0$$
Factorize
$$[x^2+3(1+\sqrt2)x-(1+\sqrt2)][x^2+3(1-\sqrt2)x-(1-\sqrt2)]=0$$
to obtain the solution in the domain $(0,1)$
$$x= \frac12\left( \sqrt{31+22\sqrt2}-3\sqrt2-3\right)
$$
$$ y= \frac12\left(- \sqrt{31+22\sqrt2}+3\sqrt2+5\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Exponent equation with common power Solve for $x$ in
$$\log_{2}(2^{x-1}+3^{x+1}) = 2x-\log_{2}(3^x)$$
I simplified by doing:
$$\log_{2}(3^x \cdot 2^{x-1} + 3^{2x+1}) = 2x$$
$$\frac{6^x}{2} + 3^{2x+1} = 2^{2x}$$
$$6^x + 2 \cdot 3^{2x+1} = 2^{2x+1}$$
Where do I go from here? I tried moving more numbers around, but I haven't been able to get any closer to solving for x. Any help is appreciated.
| Put $3^x=a$ and $2^x=b$. Then
$$6a^2+ab-2b^2=0$$
$$\Rightarrow 6a^2+4ab-3ab-2b^2=0$$
$$\Rightarrow (3a+2b)(2a-b)=0$$
Now $3a+2b$ is always positive. So $2a-b=0$.
Therefore, $$2\cdot 3^x = 2^x$$
$$\Rightarrow x=\log_{2/3} 2=\frac{1}{\log_2 (2/3)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4023597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Algebraic inequality $\sum \frac{x^3}{(x+y)(x+z)(x+t)}\geq \frac{1}{2}$ The inequality is
$$\frac{x^3}{(x+y)(x+z)(x+t)}+\frac{y^3}{(y+x)(y+z)(y+t)}+\frac{z^3}{(z+x)(z+y)(z+t)}+\frac{t^3}{(t+x)(t+y)(t+z)}\geq \frac{1}{2},$$
for $x,y,z,t>0$.
It originates from a 3-D geometry problem involving volumes of tetrahedra etc. Actually, it is equivalent with that problem (see Let ABCD be a tetrahedron of volume 1 and M,N,P,Q,R,S on AB,BC,CD,DA,AC,BD s.t. MP,NQ,RS are concurrent. Then the volume of MNRSPQ is less than 1/2.).
The three variables simpler case
$$\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+x)(y+z)}+\frac{z^2}{(z+x)(z+y)}\geq \frac{3}{4}$$
can be proved using the Cauchy-Schwarz inequality in "Engel's form".
I have tried variants of Holder type inequalities, until now unsuccessfully.
| This is a solution using Carlson inequality: if $x_{ij}(1\leq i\leq m,1\leq j\leq n)$ are non-negative reals, then
$$
\prod_{s=1}^m\sum_{t=1}^nx_{st}\geq\left(\sum_{t=1}^n\left(\prod_{s=1}^mx_{st}\right)^{\frac{1}{m}}\right)^{m}
$$
Proof. According to the Carlson inequality,
$$
\begin{align}
&\sum\frac{x^3}{(x+y)(x+z)(x+t)}\cdot\sum(x+y)\cdot\sum(x+z)(x+t)\\
\geq&\left(\sum x\right)^3\\
\because&\sum(x+y)=2(x+y+z+t),\sum(x+z)(x+t)=(x+y+z+t)^2\\
\therefore&\sum\frac{x^3}{(x+y)(x+z)(x+t)}\cdot2(x+y+z+t)\cdot(x+y+z+t)^2\geq(x+y+z+t)^3
\end{align}
$$
One can now easily observe that the original inequality is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4026381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Find all $x,y,z \in \mathbb R$ Such that $\frac{1}{x}+y+z = \frac{1}{y}+x+z = \frac{1}{z}+x+y=3 $ Find $x,y,z \in \mathbb R$ Such that :
$$\frac{1}{x}+y+z = \frac{1}{y}+x+z = \frac{1}{z}+x+y =3$$
My Attempt:
I’ve turned this into a system of equations :
$$\cases{\frac{1}{x}+y+z =3 \\ \frac{1}{y}+x+z =3 \\ \frac{1}{z}+x+y=3 } \iff \cases{1+xy+zx=3x \\1+xy+zy=3y \\1+xz+yz=3z}$$
Multiplying some equations by $-1$ and adding them together we get:
$$z(y-x)=3(y-x)\iff z=3$$
Notice that $y\ne x$.
You can play a little bit with $x,y$ ’s values, you will end up with :
$$(x,y,z) \in \{(3,\frac{-1}{3},3), (-3, \frac{1}{3},3)\}$$
Edit: user pointed out in comments that the first triple of my solution doesn’t work in all cases, but i don’t know why.
And the values of $x,y,z$ Can swap places because of the symmetry in the equations.
My question is what would happen if $x=y$.
And there is one more thing to notice is that one of the obvious solutions is $$x=y=z=1$$
Thank you.
| If $x = y$,
$$\dfrac{1}{z} + x + y = 2x + 1/z = 3 \implies z = \dfrac{1}{3-2x}$$
Hence $x + 1/y + z = x + 1/x + 1/(3-2x) = 3$. Multiply this term by $x(3-2x)$ for $x \notin \{0, 3/2\}$ and after rearranging you will get the cubic equation
$$2x^3 - 9x^2 + 10x - 3 =(x-1)(2x^2 - 7x + 3) = 0$$
$\implies x = 1$, $x = 3$ or $ x = 0.5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4027723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving $z^4 + 4i\bar{z} = 0$ in detail I asked this question here Solving $z^4 + 4i\bar{z} = 0$
Though I accepted the answer, later I found out that I and WA don't agree on the result.
What I did:
Find the absolute value of $z$ by solving $|z^4| = |-4i\bar{z}|$
Getting $|z|^4 = 4|z|$ and hence $|z| = 0$ or $|z| = 4^{\frac{1}{3}}$
Leaving the trivial solution $z=0$ aside, we have $|z| = 4^{\frac{1}{3}}$
Now back to the original equation, multiplying it by $z$ we get:
$z^5 = -4i|z|^2$ and hence $z^5 = -4i4^{\frac{2}{3}}$ hence $z^5 = 4^{\frac{5}{3}}\operatorname{cis}(-\frac{\pi}{2})$
Getting the roots $z_k = 4^{\frac{1}{3}}\operatorname{cis}(\frac{-\frac{\pi}{2} + 2\pi k}{5})$ where $k=0,1,2,3,4$
So we get the multiplication of the complex roots to be:
$z_0\cdot z_1\cdot z_2\cdot z_3\cdot z_4 = 4^{\frac{5}{3}}\operatorname{cis}(\frac{-\pi}{10} + \frac{3\pi}{10} + \frac{7\pi}{10} + \frac{11\pi}{10} + \frac{15\pi}{10}) = 4^{\frac{5}{3}}\operatorname{cis}(\frac{35\pi}{10}) = -4^{\frac{5}{3}}i=-2^{\frac{10}{3}}i$
But when calculating with WA I get $-2^{\frac{5}{2}}i$.
What is the correct result?
| Wolframalpha returns the product of the roots of $\displaystyle\frac{z^4 + 4 i z^*}z = 0$ as
$$-10.0794 i,$$ which indeed actually equals the $$-2^{\frac{10}{3}}i$$ that you obtained.
(First dividing the equation by $z$ was to discard the trivial root $z=0,$ as desired.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $\prod_{k=0}^n |x-k| \le (n-1)!/2$ for $1 \le x \le n-1$.
Let $n \ge 3$, $x \in \Bbb R$ such that $1 \le x \le n-1$. Show that $\prod_{k=0}^n |x-k| \le (n-1)!/2$.
For $n=3$, $1 \le x \le 2$ we want to show $|x(x-1)(x-2)(x-3)|=x(x-1)(2-x)(3-x) \le 1$.
We see that some large bounds are easy to guess:
$$
x(x-1)(2-x)(3-x) \le 2 \times 1 \times 1 \times 2=4
$$
but not precise enough. Another try:
$$
x(x-1)(2-x)(3-x) \le x(3-x) \le 9/4
$$
the last inequality is from the study of $f(x)=x(3-x)$.
Same problem.
For $n=3$ and $1 \le x \le 2$ let $0 \le y=x-1 \le 1$.
We get using AM-GM
\begin{align}
|x(x-1)(x-2)(x-3)| &= |(y+1)(y)(y-1)(y-2)|\\
&=y(1-y^2)(2-y)\\
&\le (\frac{3-y^2}{3})^3\\
&\le 1
\end{align}
For the general case I tried to use the same approach:
let $1 \le x \le n-1$ ie. $0 \le y = (x-1)/(n-2) \le 1$. We have
\begin{align}
\prod_{k=0}^n |x-k|
&=
\prod_{k=0}^n |(n-1)y +1-k|\\
&=
\prod_{k=0}^l |(n-1)y +1-k| \prod_{k=l+1}^n |(n-1)y +1-k|\\
&=
\prod_{k=0}^l ((n-1)y +1-k) \prod_{k=l+1}^n (k-1-(n-1)y)\\
\end{align}
where $l = \lfloor (n-1)y+1\rfloor$. Applying AM-GM doesn't really help here. Suppose $y<1$ ie. $l<n$, we see that
$$
\prod_{k=0}^l ((n-1)y +1-k) \le \prod_{k=0}^l (n-k)=\frac{n!}{(n-l-1)!}
$$
and
$$
\prod_{k=l+1}^n (k-1-(n-1)y) \le \prod_{k=l}^{n-1} (k) = \frac{(n-1)!}{(l-1)!}
$$
So
$$
\prod_{k=0}^n |x-k|\le
\frac{(n-1)!}{(l-1)!}\frac{n!}{(n-l-1)!}
$$
The bound is too big...
I also tried this approach: let $f(x)= \prod_{k=0}^n (x-k)$, $f'(x)=f(x)(\sum_{k=0}^n \frac{1}{x-k})$.
I'm trying to find informations about $x$ such that $f'(x)=0$. Let $x \in \Bbb R - [|0,n|]$. So $\sum_{k=0}^n \frac{1}{x-k}=0$. But I'm stuck here.
My questions are:
*
*Does someone have a hint or a proof?
*Can someone give me some advice on how to handle such problems during an exam?
Thanks :)
| I would like to solve a more general problem than this problem and that problem.
First, we find all local maxima of the function $f(x) =\prod_{k=0}^n |x-k|$ in the interval $0 \le x \le n$.
For each interval $[p,p+1)$ with $p\in \{0,...,n-1\}$ we prove that there exists 1 local maximum.
We guess the value of $v_0,v_1,...,v_n$ such that there exists $x_p \in [p,p+1)$ satisfying two following conditions
$$v_0x = v_1(x-1) =...=v_n(x-k) \tag{1}$$
$$v_0+v_1+...+v_n=0\tag{2}$$
Solve these 2 conditions, we can have a solution (attention, this is not a unique solution, but we need only 1) as follows
$$v_k=\frac{1}{x_p-k}$$
with $x_p \in [p,p+1)$ satisfying this equation
$$\sum_{0 \le k \le n}\frac{1}{x_p-k} = 0 \tag{3}$$
Note: For information, it's easy to prove that the equation (3) has exactly one root in each interval $[p,p+1)$ where $p\in \{0,...,n-1\}$.
Now, we apply the AM-GM inequality for $v_k(x-k)$ with $0 \le k \le n$ (we notice that for $k >p$, $v_k$ is negative, then the terms $v_k(x-k)$ are all positive).
\begin{align}
f(x) &=\prod_{0 \le k \le n}|x-k| \\
&= \frac{1}{\prod_{0 \le k \le n}|v_k|} \prod_{0 \le k \le n}v_k(x-k) \\
&\le \frac{1}{\prod_{0 \le k \le n}|v_k|} \left(\frac{\sum_{0 \le k \le n}v_k(x-k)}{n+1} \right)^{n+1} = \frac{1}{\prod_{0 \le k \le n}|v_k|k} \left(\frac{-\sum_{0 \le k \le n}kv_k}{n+1} \right)^{n+1}
\end{align}
We note that,;from (1) and (3), we have
\begin{align}
-\sum_{0 \le k \le n}kv_k &= -\sum_{0 \le k \le n}\frac{k}{x_p-k} \\
&= n+1 - x_p\sum_{0 \le k \le n}\frac{1}{x_p-k} \\
&= n+1
\end{align}
Hence,
$$f(x) \le \frac{1}{\prod_{0 \le k \le n}|v_k|} \left(\frac{-\sum_{0 \le k \le n}kv_k}{n+1} \right)^{n+1} = \prod_{0 \le k \le n}|x_p-k|\left(\frac{n+1}{n+1} \right)^{n+1} = \prod_{0 \le k \le n}|x_p-k|$$
The local maximum in $x \in [p,p+1)$ occurs when $x = x_p$ where $x_p \in [p,p+1)$ is the solution of (3).
Second, we prove that $f(x_0) \ge f(x_1) \ge ...\ge f(x_{[\frac{n}{2}]})$ and $f(x)$ is a symmetric function at $x = \frac{n}{2}$ (in other words, $f(x) = f(n-x)$).
It's easy to prove the second statement about the symmetry.
For the first statement, for $x \in [p,p+1)$ with $p\in \{0,...,[\frac{n}{2}]-1\}$, we have
$$ f(x+1) = \prod_{0 \le k \le n}|(x+1)-k| = \frac{x+1}{|x-n|}\prod_{0 \le k \le n}|x-k| = \frac{x+1}{|x-n|} f(x)$$
As $x \in [p,p+1)$ with $p\in [0,...,([\frac{n}{2}]-1)]$, we have then $x< \frac{n+1}{2}$, so $\frac{x+1}{|x-n|} <1$ or $f(x+1) < f(x)$. We deduce that
$$\max_{x \in [p+1,p+2)}f(x) = \max_{x \in [p,p+1)}f(x+1) < \max_{x \in [p,p+1)}f(x)$$
or for all $p\in \{0,...,[\frac{n}{2}]-1\}$ $$f(x_{p})>f(x_{p+1}) \tag{4}$$
Finally, we solve your problem, which corresponds to the case: find the maximum of $f(x)$ in the interval $ 1 \le x \le n-1$ (another particular case is where $0 \le x \le n$ was solved here ).
From (4), we have the function $f(x)$ attains its maximum value $f(x_1)$ in $x \in [1,n-1]$ when $x=x_1$ where $x_1 \in (1,2)$ is the root of the equation (3).
$$f(x) \le f(x_1) = \prod_{0 \le k \le n}|x_1-k| = x_1(x_1-1)(2-x_1) \prod_{3 \le k \le n}(k-x_1) \le \left(\frac{x_1 + (x_1-1) + 2(2-x_1)}{3} \right)^3 \frac{1}{2}\prod_{3 \le k \le n}(k-1) = \frac{(n-1)!}{2}$$
Because $x_1$ doesn't satisfy the condtion of AM-GM inequality ($x_1 = x_1-1 = 2(2-x_1)$), we conclude that $$f(x) < \frac{(n-1)!}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4031976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Let $a,b,c$ be positive integers such that $a^3+b^3=2^c.$ Show that $a=b$.
Let $a,b,c$ be positive integers such that $$a^3+b^3=2^c.$$ Show that $a=b$.
I have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=2^c =2^x\cdot2^y$$ now it can only be that $a$ and $b$ are both odd or even since they sum to an even number. Thus if $a$ and $b$ are both odd I have that $a^2$ is odd, $ab$ is odd and $b^2$ is odd. This would imply that $a^2-ab+b^2 = 2^y =1$ which in turn implies that $a^3+b^3 = a+b \implies a=b.$ The problem I have is that if I would have considered that both $a$ and $b$ are even I would have gotten that $a=2t, b=2k$ from where $$8t^3+8k^3=2^c \implies t^3+k^3=2^{c-3}$$ but I couldn't deduce anything from here why cannot $a$ and $b$ be even?
| HINT:
*
*For the case where $a,b$ both even, factor out the largest power of $2$ from both of them. You will either reduce to exactly one of $a'$, $b'$ odd [which you already noted is impossible] or both $a,b$ odd. To elaborate a bit more, let $j$ be the largest integer s.t. $2^j$ divides both $a$ and $b$. Then write $2^ja'=a$ and $2^jb'=b$, with at least one of $a',b'$ odd.
Then $$(a')^3+(b')^3 = 2^{3j}2^{c-3j},$$ so as at least one of $a',b'$ is odd, then by your reasoning above, both $a'$ and $b'$ must be odd.
*For the case where $a,b$ both odd: You noted that this implies the equation
$$(a+b)(a^2-ab+b^2)= 2^c.$$ So then both $(a^2-ab+b^2)$ and $(a+b)$ must be powers of $2$. However, $a^2-ab+b^2$ is odd [why is this] so this gives $a^2-ab+b^2=1$, because $1$ is the only odd power of $2$. So this gives $(a+b)=(a^3+b^3)$ [why is this] which
is only possible [assuming $a,b$ positive integers] for $a,b = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Euler's Beta function for positive integers derivation I am aware that a proof for the Euler's Beta function for positive integers has already been asked, but I have tried to derive it myself without using the gamma function, and my result doesn't match. Any ideas?
My derivation (I used $n=sm+d$ and $t=\left(1-x\right)^m$):
\begin{align*}
B\left(m,n\right)&=\int_{0}^{1}{x^{m-1}\left(1-x\right)^{n-1}dx} \\
&=\frac{1}{m}\int_{0}^{1}{\left(1-x\right)^{n-1}dx^m} \\
&=\frac{1}{m}\int_{1}^{0}{\left(1-x\right)^{n-1}d\left(1-x\right)^m} \\
&=-\frac{1}{m}\int_{0}^{1}{\left(1-x\right)^{sm+d-1}d\left(1-x\right)^m} \\
&=-\frac{1}{m}\int_{1}^{0}{t^{\frac{sm+d-1}{m}}dt} \\
&=\frac{1}{m}\int_{0}^{1}{t^{\frac{sm+d-1}{m}}dt} \\
&=\frac{1}{m}\left(\frac{m}{sm+m+d}1^{\frac{sm+m+d-1}{m}}-\frac{m}{sm+m+d}0^{\frac{sm+m+d-1}{m}}\right) \\
&=\frac{1}{sm+m+d-1} \\
&=\frac{1}{m+n-1} \\
\end{align*}
Correct result:
$$B\left(m,n\right)=\frac{\left(m−1\right)!\cdot\left(n−1\right)!}{\left(m+n−1\right)!}$$
| Ok, I've found a solution:
let $a_i=\int_{0}^{1}{\left(m+i-1\right)x^{m+i-2}\left(1-x\right)^{n-i}dx}$
let $u_i=\left(1-x\right)^{n-i},du_i=\left(i-n\right)\left(1-x\right)^{n-i-1}dx$
let $v_i=x^{m+i-1},dv_i=\left(m+i-1\right)x^{m+i-2}dx$
\begin{align*}
a_i&=\int_{0}^{1}{\left(m+i-1\right)x^{m+i-2}\left(1-x\right)^{n-i}dx}\\
&=\int_{0}^{1}{u_idv_i}\\
&=\left[u_iv_i\right]_0^1+\int_{0}^{1}{v_idu_i}\\
&=\left[\left(1-v_i^{\frac{1}{m+i-1}}\right)^{n-i}v_i\right]_0^1-\int_{0}^{1}{x^{m+i-1}\left(i-n\right)\left(1-x\right)^{n-i-1}dx}\\
&=0-\left(i-n\right)\int_{0}^{1}{x^{m+i-1}\left(1-x\right)^{n-i-1}dx}\\
&=\frac{n-i}{m+i}\int_{0}^{1}{\left(m+i\right)x^{m+i-1}\left(1-x\right)^{n-i-1}dx}\\
&=\frac{n-i}{m+i}a_{i+1}\\
&=a_{n-1}\prod_{j=1}^{n-2}{\frac{n-j}{m+j}}\\
&=\frac{m!\left(n-1\right)!}{\left(m+n-2\right)!}a_{n-1}
\end{align*}
\begin{align*}
B\left(m,n\right)&=\int_{0}^{1}{x^{m-1}\left(1-x\right)^{n-1}dx}\\
&=\frac{1}{m}\int_{0}^{1}{mx^{m-1}\left(1-x\right)^{n-1}dx}\\
&=\frac{1}{m}a_1\\
&=\frac{1}{m}\frac{m!\left(n-1\right)!}{\left(m+n-2\right)!}a_{n-1}\\
&=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\int_{0}^{1}{\left(m+n-2\right)x^{m+n-3}\left(1-x\right)dx}\\
&=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\left[\int_{0}^{1}{\left(m+n-2\right)x^{m+n-3}dx}-\int_{0}^{1}{\left(m+n-2\right)x^{m+n-2}dx}\right]\\
&=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\left(\left[x^{m+n-2}\right]_0^1-\left[\frac{m+n-2}{m+n-1}x^{m+n+1}\right]_0^1\right)\\
&=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\left(1-\frac{m+n-2}{m+n-1}\right)\\
&=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}-\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-2\right)!}\frac{m+n-2}{m+n-1}\\
&=\frac{\left(m-1\right)!\left(n-1\right)!\left(m+n-1\right)}{\left(m+n-2\right)!}-\frac{\left(m-1\right)!\left(n-1\right)!\left(m+n-2\right)}{\left(m+n-1\right)!}\\
&=\frac{\left(m-1\right)!\left(n-1\right)!\left(m+n-1-m-n+2\right)}{\left(m+n-2\right)!}\\
&=\frac{\left(m-1\right)!\left(n-1\right)!}{\left(m+n-1\right)!}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4033811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve this limit $\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)$ \begin{align}
\lim\limits_{x\to 1}\left(\dfrac{2017}{1-x^{2017}} -\dfrac{2018}{1-x^{2018}}\right)
&= \lim\limits_{x\to 1}\dfrac{2017(x^{2017} + \dots + 1) - 2018(x^{2016} + \dots + 1)}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}\\\\
&= \lim\limits_{x\to 1}\dfrac{2017x^{2017} - x^{2016} - \dots - 1}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}
\end{align}
Did I do the right way with those step above?
I think next step is to separate:
$2017x^{2017} = \underbrace{x^{2017} + \dots + x^{2017}}_{\text{2017 addends}}$
Then combine with the rest and factorize $(1-x)$,
but it has a little confused after factorizing.
Please help me!!?
| You're on the right track. You can continue by dividing out the problematic factor $1-x$. What does long division of the numerator by $1-x$ give?
Hint:
If $P(x)=(x-1)Q(x)$ then $P'(x)=Q(x)+(x-1)Q'(x)$, so $Q(1)=P'(1)$.
Details:
Denote the polynomial in the numerator of the fraction by $P(x)$, so that$$P(x)=2017x^{2017}-x^{2016}-\ldots-x-1.$$By polynomial longdivision there exist polynomials $Q(x)$ and $R(x)$ with integer coefficients such that $P(x)=(x-1)Q(x)+R(x)$, where $\deg R(x)<\deg(x-1)$. That is to say $R(x)$ is constant, so $R(x)=R$ for some integer $R$. Plugging in $x=1$ shows that $$P(1)=(1-1)Q(1)+R(1)=R(1)=R,$$and it is not hard to see that $P(1)=0$. This means $P(x)=(x-1)Q(x)$ and so$$\lim\limits_{x\to 1}\dfrac{P(x)}{(1-x)(x^{2016} + \dots +1)(x^{2017} + \dots +1)}=\lim\limits_{x\to 1}\dfrac{Q(x)}{(x^{2016} + \dots +1)(x^{2017} + \dots +1)}.$$Here the hint comes in handy; we have\begin{eqnarray*}Q(1)&=&P'(1)=2017^2-2016-\ldots-2-1\\&=&2017^2-\frac12\cdot2017\cdot2016=2017\cdot1009.\end{eqnarray*}It follows that$$\lim\limits_{x\to 1}\dfrac{-Q(x)}{(x^{2016} + \dots +1)(x^{2017} + \dots +1)}=\frac{-2017\cdot1009}{2017\cdot2018}=-\frac12.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
If positive reals $a$ and $b$ satisfy $a\sqrt{a}+b\sqrt{b}=183, a\sqrt{b}+b\sqrt{a}=182$, find $\frac{9}{5}(a+b)$. Question from Math Olympiad:
Suppose $a, b$ are positive real numbers such that $a\sqrt{a} + b\sqrt {b} = 183$ and $a\sqrt{b} + b\sqrt {a} = 182$. Find $\frac{9}{5}(a+b)$.
My approach:
$a\sqrt{a} + b\sqrt {b} = 183$
$a\sqrt{b} + b\sqrt {a} = 182$
Therefore $(\sqrt{a} + \sqrt{b})(a+b) = 365$
I only see two integral possibilities:
$1 \times 365 $ and $5 \times 73 $
And on putting either of them, I am getting $657$ and $131.4$, whereas the answer is an integral $73$.
Please help me with the general method to go about the same.
This is the picture from the source.
| Hint : If you make the difference of the two equations, you get
$$(\sqrt{a}-\sqrt{b})(a-b)=0$$
Probably you can conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4036786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Rotation composition in $R^3$ and linear mappings I'm trying to understand why the result of a composition of linear transformations (two rotations) it is different from mapping the base vectors from $R^3$ to $R^3$ twice.
I'll explain.
I have a mapping: ϕ that maps vectors in $R^3$ to $R^3$ as a linear combinations of the canonical base vectors (e1,e2,e3).
The transformation matrix associated with ϕ is the following:
\begin{bmatrix}\cos(a)&0&\sin(a)\\0&1&0\\-\sin(a)&0&\cos(a)\end{bmatrix}
That is the classic counterclockwise rotation with respect to the e2 axis.
Now I want to map the canonical base vectors (e1,e2,e3) using ϕ.
$ϕ(\mathbf{e1}) = \cos(a) \bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) - \sin(a) \bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} \cos(a) \\ 0 \\ -\sin(a) \end{smallmatrix} \bigr)$
$ϕ(\mathbf{e2}) = 0 \bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr) + 1 \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr)$
$ϕ(\mathbf{e3}) = \sin(a) \bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + \cos(a) \bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr)$
After that, I want to perform a counterclockwise rotation around the rotated base vector e3 (e.g. ϕ(e3)).
To do this, i'm using this transformation matrix:
\begin{bmatrix}\cos(b)&-\sin(b)&0\\\sin(b)&\cos(b)&0\\0&0&1\end{bmatrix}
And I map the "old" rotated basis onto the new basis using Ψ:
$Ψ(ϕ(\mathbf{e1})) = \cos(b) \bigl( \begin{smallmatrix} \cos(a) \\ 0 \\ -\sin(a) \end{smallmatrix} \bigr) + \sin(b) \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} \cos(a)\cos(b) \\ \sin(b) \\ -\sin(a)\cos(b) \end{smallmatrix} \bigr)$
$Ψ(ϕ(\mathbf{e2})) = -\sin(b) \bigl( \begin{smallmatrix} \cos(a) \\ 0 \\ -\sin(a) \end{smallmatrix} \bigr) + \cos(b) \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} -\cos(a)\sin(b) \\ \cos(b) \\ \sin(a)\sin(b) \end{smallmatrix} \bigr)$
$Ψ(ϕ(\mathbf{e3})) = 0 \bigl( \begin{smallmatrix} \cos(a) \\ 0 \\ -\sin(a) \end{smallmatrix} \bigr) + 0 \bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr) + 1 \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} \sin(a) \\ 0 \\ \cos(a) \end{smallmatrix} \bigr)$
Assuming that what is written is true, now I have tried to do the verification, checking if the composition of the two rotations actually gives me the same result.
Therefore:
\begin{equation}
\begin{bmatrix}\cos(b)&-\sin(b)&0\\\sin(b)&\cos(b)&0\\0&0&1\end{bmatrix}*\begin{bmatrix}\cos(a)&0&\sin(a)\\0&1&0\\-\sin(a)&0&\cos(a)\end{bmatrix}=\begin{bmatrix}\cos(a)\cos(b)&-\sin(b)&\sin(a)\cos(b)\\\sin(b)\cos(a)&\cos(b)&\sin(a)\sin(b)\\-\sin(a)&0&\cos(a)\end{bmatrix}
\end{equation}
If now I tried to make a linear mapping transforming the canonical basic vectors with this composition of rotations, I would find a different result from the one found before.
Therefore (calling τ this linear mapping):
τ(e1) = \cos(a)\cos(b) $\bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr)$ + \sin(b)\cos(a) $\bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr)$ - \sin(a) $\bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} \cos(a)\cos(b) \\ \sin(b)\cos(a) \\ -\sin(a) \end{smallmatrix} \bigr)$
τ(e2) = -\sin(b) $\bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr)$ + \cos(b) $\bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr)$ + 0 $\bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} -\sin(b) \\ \cos(b) \\ 0 \end{smallmatrix} \bigr)$
τ(e3) = \sin(a)\cos(b) $\bigl( \begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix} \bigr)$ + \sin(a)\sin(b) $\bigl( \begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix} \bigr)$ + \cos(a) $\bigl( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \bigr)$ = $\bigl( \begin{smallmatrix} \sin(a)\cos(b) \\ \sin(a)\sin(b) \\ \cos(a) \end{smallmatrix} \bigr)$
Clearly:
Ψ(ϕ(e1)) ≠ τ(e1), Ψ(ϕ(e2)) ≠ τ(e2), Ψ(ϕ(e3)) ≠ τ(e3)
When instead I should have expected the same result...
I noticed that by reversing the order of the matrix product, I get the same result instead... I can't understand why.
| Note that$$\phi(e_1)=(\cos(a),0,-\sin(a))=\cos(a)e_1-\sin(a)e_3.$$Therefore\begin{align}\psi\bigl(\phi(e_1)\bigr)&=\psi\bigl(\cos(a)e_1-\sin(a)e_3\bigr)\\&=\cos(a)\psi(e_1)-\sin(a)\psi(e_3).\end{align}There is no reason for you to assume that this is equal to$$\cos(b)\phi(e_1)+\sin(b)\phi(e_2)+0\times\phi(e_3),$$which is what you did.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4037942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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.