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Basic question about modular arithmetic applied to the divisor sum function $\sigma(n)$ when $n=5p$ While studying the divisor sum function $\sigma(n)$ (as the sum of the divisors of a number) I observed that the following expression seems to be true always (1):
$\forall\ n=5p, p\in\Bbb P,\ p\gt 5,\ if\ d(5p)=4\ then\ \sigma(5p)=3,0\ (mod\ 9)$
Meaning that if the number $n=5k$ has only four divisors, so $5$ and other prime $p$ divide $n$, then the sum of the divisors is congruent to 3 or 0 modulo 9. I just observed that manually in the range $[1..400]$.
The divisors of a $5p$ number as the type above are $\{1,5,p,5p\}$
E.g.
$d(35) = \{1,5,7,35\}$
$d(55) = \{1,5,11,55\}$
And the sum of the divisors is $\sigma(5p)= 1+5+p+5p$, thus (2):
$(1+5+p+5p)\ mod\ 9 = 1 + 5 + ((p + 5p)\ mod\ 9) =$
$6 + ((p + 5p)\ mod\ 9) = 6 + (6p\ mod\ 9)$
... but from that point I am not sure how to conclude that for the case above (1) it will be $3$ or $0$.
I think that this must happen at (2) if (1) is true:
$6p\ mod\ 9 = 6 * (p\ mod\ 9) \equiv 3, 6\ (mod\ 9)$
and then is true only if:
$p\equiv 1,x?\ (mod\ 9)$
Any help or hint is very appreciated, or a counterexample if (1) is wrong and then the congruence is false, thank you!
| After reading the answers, I think I have found my own way to the solution. It is based on the property of the prime numbers, for $p\gt 3$ it belongs to Case 1: $1+6x$ or Case 2: $5+6x$. So replacing:
Case 1:
$1+5+p+5p = 1+5+(1+6x)+5(1+6x)=1+5+1+6x+5+30x = 12 + 36x = 12 (1 + 3x)$
And
$12(1+3x)\ mod\ 9 = 3 * (1 + (3\ or \ 6\ or\ 0\ (mod 9))$ so
$3(1+3)\ or\ 3(1+6)\ or\ 3(1+0)$ is respectively $12, 21, 3$ and $(mod\ 9)$ respectively $3,3,3$, so when $p=1+6x$ then $1+5+p+5p \equiv 3\ (mod\ 9)$
Case 2:
$1+5+p+5p = 1+5+(5+6x)+5(5+6x)=1+5+5+6x+25+30x = 36 + 36x = 36 (1 + x)$
And
$36(1+x)\ mod\ 9 = 0 * ((1+x)\ mod\ 9)$ so when $p=5+6x$ then $1+5+p+5p \equiv 0\ (mod\ 9)$
| {
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"url": "https://math.stackexchange.com/questions/1249399",
"timestamp": "2023-03-29T00:00:00",
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Finding $\int \frac{1+\sin x \cos x}{1-5\sin^2 x}dx$
Find $\int \frac{1+\sin x \cos x}{1-5\sin^2 x}dx$
I used a bit of trig identities to get: $\int \frac {2+\sin (2x)}{-4+\cos(2x)}dx$ and using the substitution: $t= \tan (2x)$ I got to a long partial fractions calculation which doesn't seem right.
Any hints on how to do it please?
| We have
$$\int\frac{1+\sin x\cos x}{1-5\sin^2x}\mathrm{d}x
=\int\frac{\mathrm{d}x}{1-5\sin^2x}+\int\frac{\sin x\cos x}{1-5\sin^2x}\mathrm{d}x$$
and
$$\int\frac{\sin x\cos x}{1-5\sin^2x}\mathrm{d}x
=-\frac{1}{10}\int\frac{\mathrm{d}\left(1-5\sin^2x\right)}{1-5\sin^2x}
=-\frac{1}{10}\ln\left(1-5\sin^2x\right).$$
Then,
$$\int\frac{\mathrm{d}x}{1-5\sin^2x}
=\int\frac{\mathrm{d}x}{\cos^2x-4\sin^2x}
=\int\frac{1}{1-4\tan^2x}\frac{\mathrm{d}x}{\cos^2x}
=\frac{1}{2}\int\frac{\mathrm{d}\left(2\tan x\right)}{1-\left(2\tan x\right)^2}
=\frac{1}{2}\tanh^{-1}\left(2\tan x\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the least degree Polynomial whose one of the roots is $ \cos(12^{\circ})$ Find the least degree Polynomial with Integer Coefficients whose one of the roots is $ \cos(12^{\circ})$
My Try: we know that $$\cos(5x)=\cos^5x-10\cos^3x\sin^2x+5\cos x\sin^4x$$ Putting $x=12^{\circ}$ and Converting $\sin$ to $\cos$ we have
$$\frac{1}{2}=x^5-10x^3(1-x^2)+5x(1-2x^2+x^4)$$ $\implies$
$$32x^5-40x^3+10x-1=0$$
Is this the Least degree? Please let me know.
| Without any quartic solving:
Set $y=2x$. The polynomial becomes
$$y^5-5y^3+5y-1=(y-1)(y^4+y^3-4y^2-4y+1)$$
As $2\cos\dfrac \pi{15}\neq 1$, it is a root of $\,f(y)=y^4+y^3-4y^2-4y+1$.
Let's show this polynomial is irreducible over $\mathbf Z$. Indeed, it has no integer root (the only possibilities are $1$ and $-1$, none of which is a root). If it were reducible, it would be as the product of two (irreducible) quadratic polynomials, hence it would also be reducible modulo $2$.
However, modulo $2$, we have:
$$f(y)=y^4+y^3+1.$$
It has no root in $\mathbf F_2$. Suppose we can write:
$$y^4+y^3+1=(y^2+ay+b)(y^2+a'y+b')$$
By identification, we obtain at once $bb'=1$, whence $b=b'=1$ and the system:
$$a+a'=1,\quad aa'=0,\quad a+a'=0,$$
which has no solution.
Thus the polynomial $y^4+y^3-4y^2-4y+1$ is irreducible, and the minimal polynomial of $\cos\dfrac \pi{15}$ is:
$$16x^4+8x^3-16x^2-8x+1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $f(x) = \sin^4 x+\cos^2 x\;\forall x\; \in \mathbb{R}\;,$ Then $\bf{Max.}$ and $\bf{Min.}$ value of $f(x)$
If $f(x) = \sin^4 x+\cos^2 x\;\forall x\; \in \mathbb{R}\;,$ Then $\bf{Max.}$ and $\bf{Min.}$ value of $f(x).$
My Solution:: Let $$\displaystyle y = \sin^4 x+\cos^2 x \leq \sin^2 x+\cos^2 x=1$$
And for Minimum, We take $$\displaystyle y = \sin^4 x+\cos^2 x=(1-\cos^2 x)^2+\cos^2 x$$
So $$\displaystyle y=\cos^4 x-\cos^2 x+1 = \left(\cos^2 x-\frac{1}{2}\right)^2+\frac{3}{4}\geq \frac{3}{4}$$
So We get $\displaystyle y=\sin^4 x+\cos^2 x\in \left[\frac{3}{4}\;,1\right]$
My question is How can we find Min. of $f(x)$ other then that method,
Something Like Using Inequality.,Thanks
| $$\cos^4x-\cos^2x=\cos^2x(\cos^2x-1)=-\sin^2x\cos^2x=-\dfrac{(2\sin x\cos x)^2}4=-\dfrac{\sin^22x}4$$
For real $x,0\le\sin^22x\le1\iff-1\ge-\sin^22x\ge0$
In case you don't know double angle formula, $(\sin x\pm\cos x)^2\ge0\iff-1\le2\sin x\cos x\le1$
| {
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"url": "https://math.stackexchange.com/questions/1252666",
"timestamp": "2023-03-29T00:00:00",
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Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$ Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$
I am stuck at understanding why the constraint $xy\gt 1$. Here is my work so far
let $\arctan x =a\implies x=\tan a$
let $\arctan y =b\implies y=\tan b$
therefore $\frac{x+y}{1-xy}=\frac{\tan a+\tan b}{1-\tan a\tan b}=\tan(a+b)$
$\implies a+b=\arctan \frac{\tan a+\tan b}{1-\tan a\tan b} $
$\implies \arctan x+\arctan y = \arctan\frac{x+y}{1-xy}$
I know $\pi$ is the period of $\tan x$ so $xy\gt 1$ constraint must have something to do with this. But I am not able to figure out how exactly these period and $xy\gt 1$ are related. Any help is appreciated. Thanks!
| I think this might be an idea for you.
Let $$f(x,y)= \arctan(x)+\arctan(y)-\arctan(\frac{x+y}{1-xy})$$ defined for $xy> 1$, then we have
$$ \frac{ \partial f}{\partial x }= \frac{1}{1+x^2} -\frac{(1-xy)+y(x+y)}{(1-xy)^2 +(x+y)^2 }= \frac{1}{1+x^2} -\frac{1+y^2}{(1+y^2)(1+x^2) }=0 $$
Similarly you can show that $ \frac{ \partial f}{\partial y}=0$. Hence our function is constant, i.e. $f(x,y)=C$. Now let $x=\tan(\frac{\pi}{3})= \sqrt{3}$ and $y=\tan(\frac{\pi}{4})=1$ then $xy \approx 1.732050807568877 > 1$. On the other hand $$f(x,y)=\frac{\pi}{3} +\frac{\pi}{4}- arctan \Big( \frac{ 1+\sqrt{3}}{1-\sqrt{3}} \Big) =\frac{\pi}{3} +\frac{\pi}{4} -(-\frac{5\pi}{12})= \pi $$ However $f(x,y)$ is constant, and so $f(x,y)= \pi$ for all $xy>1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solutions to $\frac{1}{a} + \frac{1}{b} = \frac{1}{100}$? I encountered this problem yesterday and successfully solved it. I'm interested in seeing other people's approach to solving this problem.
Problem: How many ordered pairs $(a, b)$ are solutions to $\frac{1}{a} + \frac{1}{b} = \frac{1}{100}$ where $a, b \in \mathbb{Z}$?
Edit: The solution is:
$49$
| This reminded me of problem 454 on project Euler which I wrote an algorithm for a year ago. (Here is the algorithm if anyone is interested)
So here is how I would solve this question.
Observe that
$$\frac{1}{2} = \frac{1}{3} + \frac{1}{6} \\ \frac{1}{3} = \frac{1}{4} + \frac{1}{12} \\ \frac{1}{4} = \frac{1}{5} + \frac{1}{20}$$
which can be generalized into
$$\frac{1}{a} = \frac{1}{a+1} + \frac{1}{a^2+a}$$
So immediately we get our first solution:
$$\frac{1}{100} = \frac{1}{101} + \frac{1}{10100}$$
Note the equation above can be further generalized into
$$\frac{1}{a} = \frac{1}{(a+n)} + \frac{n}{a^2+an}$$
in which case $n$ can be converted into 1 as long as $n$ is divisible by $a^2+an$. Thus to find all solutions to the equation is basically to find how many distinct $n$'s are there that are divisible by $a^2$ - how many factors do $a^2$ have.
Note: the question is asking for the number of ordered pairs, in which case the summation of 2 distinct fractions count as 2.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Prove there do not exist prime numbers $a$, $b$, and $c$ such that $a^3+b^3=c^3$ Prove there do not exist prime numbers $a$, $b$, and $c$ such that $a^3+b^3=c^3$.
From what I understand this proof requires a proof by contradiction or contrapositive...
| $a^3+b^3=(a+b)(a^2-ab+b^2)$ So we have
$(a+b)(a^2-ab+b^2)=c^3$
So $(a+b)=c$ and $(a^2-ab+b^2)=c^2$
Hence $(a+b)^2=a^2+2ab+b^2=a^2-ab+b^2$ which is a contradiction since it implies $3ab=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1261058",
"timestamp": "2023-03-29T00:00:00",
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Evaluating trigonometric limit: $\lim_{x \to 0} \frac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2}$
Evaluate $\lim_{x \to 0} \cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2} $
This is what I've tried yet:
$$\begin{align} & \cfrac{x(\tan 2x - 2\tan x)}{4\sin^4 x} \\
=&\cfrac{x\left\{\left(\frac{2\tan x}{1-\tan^2 x} \right) - 2\tan x\right\}}{4\sin^4 x}\\
=& \cfrac{2x\tan x \left(\frac{\tan^2 x}{1 - \tan^2 x}\right) }{4\sin^4 x} \\
=& \cfrac{x\tan^3 x}{2\sin^4 x (1-\tan^2 x)} \\
=& \cfrac{\tan^3 x}{2x^3\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)} \\
=& \cfrac{\frac{\tan^3 x}{x^3} }{2\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)}\end{align}$$
Taking limit of the above expression, we've :
$$\lim_{x\to 0} \cfrac{\frac{\tan^3 x}{x^3} }{2\left(\frac{\sin x}{x}\right)^4(1-\tan^2 x)} = \lim_{x\to 0} \cfrac{\cos^2x}{2\cos 2x} = \cfrac{1}{2} $$
Firstly, is my answer right or am I doing somewhere wrong?
Secondly, this seems a comparatively longer method than expected for objective type questions. I'm seeking for a shortcut method for such type of questions. Is there any method I should've preferred?
| Using power series, we only need two:
$$ \tan{x} = x + \frac{1}{3}x^3 + O(x^5), $$
and
$$ \cos{x} = 1- \frac{1}{2}x^2 + O(x^4). $$
Then
$$ x(\tan{2x}-2\tan{x}) = x\left( 2x+\frac{8}{3}x^3 - 2x - \frac{2}{3}x^3 + O(x^5) \right) = 2 x^4 + O(x^5), $$
and
$$ (1-\cos{2x})^2 = \frac{1}{4}(2x)^4 + O(x^6) = 4x^4 + O(x^6), $$
and then just divide to find the limit as $1/2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Write 100 as the sum of two positive integers
Write $100$ as the sum of two positive integers, one of them being a multiple of $7$, while the other is a multiple of $11$.
Since $100$ is not a big number, I followed the straightforward reasoning of writing all multiples up to $100$ of either $11$ or $7$, and then finding the complement that is also a multiple of the other. So then
$100 = 44 + 56 = 4 \times 11 + 8 \times 7$.
But is it the smart way of doing it? Is it the way I was supposed to solve it? I'm thinking here about a situation with a really large number that turns my plug-in method sort of unwise.
| $7x+11y=100$
$7x=100-11y$
$x=\frac{100-11y}7=14-2y+\frac{2+3y}7$
$a=\frac{2+3y}7$
$7a=2+3y$
$3y=-2+7a$
$y=\frac{-2+7a}3=-1+2a+\frac{1+a}3$
$b=\frac{1+a}3$
$3b=1+a$
$a=3b-1$
$y=\frac{-2+7(3b-1)}3=\frac{-9+21b}3=-3+7b$
$x=\frac{100-11(-3+7b)}7=\frac{133-77b}7=19-11b$
$\begin{matrix}
x\gt 0&\to&19-11b\gt 0&\to&11b\lt 19&\to&b\lt\frac{19}{11}&\to&b\le 1&\\
&&&&&&&&&b=1\\
y\gt 0&\to&-3+7b\gt 0&\to&7b\gt 3&\to&b\gt\frac37&\to&b\ge 1&\\
\end{matrix}$
$1$ is the only integral value of $b$ for which both $x$ and $y$ yield positive values, as required by the problem. So, $x=8$ and $y=4$.
| {
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"url": "https://math.stackexchange.com/questions/1265426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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find coefficient of $x^{50}$ Let $f(x)=\frac{1}{(1+x)(1+x^2)(1+x^4)}$then find the coefficient of term $x^{50}$ in $(f(x))^3$.I think that we can set $$(f(x))^3=\frac{a}{(1+x)^3}+\frac{b}{(1+x^2)^3}+\frac{c}{(1+x^4)^3}$$ and find a,b and c then use Taylor seri .
| We have $f(x)=\frac{1}{(1+x)(1+x^2)(1+x^4)}=\frac{1-x}{1-x^8}=(1-x)(1-x^8)^{-1}$ if $|x^8|<1$.
Now note that when $|x|<1$ then \begin{align*}
(1-x)^{-1}=&\sum\limits_{n\geq 0}x^n\\
-(1-x)^{-2}=&\sum\limits_{n\geq 1}nx^{n-1}\\
+2(1-x)^{-3}=&\sum\limits_{n\geq 2}n(n-1)x^{n-2}=\sum\limits_{n\geq 0}(n+2)(n+1)x^{n}\\
(1-x)^{-3}=&\sum\limits_{n\geq 0}\frac{(n+2)(n+1)}{2}x^{n}
\end{align*}
which means $f(x)^3=(1-x)^3(1-x^8)^{-3}=(1-3x+3x^2-x^3)\sum\limits_{n\geq 0}\frac{(n+2)(n+1)}{2}x^{8n}$
On the RHS, we have the terms of the forms $x^{8n}, x^{8n+1}, x^{8n+2}, x^{8n+3}$. Since none of $8n, 8n+1, 8n+3$ but $8n+2$ provide 50 (for $n=6$) so the required coefficient of $x^{50}$ is $3\frac{8.7}{2}=84$
| {
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"timestamp": "2023-03-29T00:00:00",
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compute improper integrals using integration by parts Compute
\begin{equation*}
\int_0^\infty \frac{\sin^4(x)}{x^2}~dx\text{ and }\int_0^\infty \frac{\sin (ax) \cos (bx)}{x}~dx.
\end{equation*}
For the first integral I tried letting $u = \sin ^4 x$ and $dv= \frac{1}{x^2}~dx$, which simplified to $\int_0 ^\infty \frac{4 \sin^3(x) \cos(x)}{x}~dx$ and applying integration by parts again with $u = \sin^3(x) \cos (x)$ and $dv = \frac{1}{x}$, I got
$$\left.\vphantom{\frac11}[\sin ^3 (x) \cos (x) \ln (x)] \right|_0^\infty - \int_0^\infty \ln (x) [3 \sin ^2 (2x) - \sin^4 (x)]~dx$$
The only problem is that it appears that the term $[\sin ^3 (x) \cos (x) \ln (x)] \big\vert_0^\infty$ evaluates to infinity and neither the $3\int_0^\infty \ln (x) \sin ^2 (2x)~dx$ term nor the $3\int_0^\infty \ln (x) \sin^4 (x)~dx$ converges.This looks like a dead end but the other choice of $u$ and $dv$ looks even worse, so I'm not sure how to proceed.
For the second integral I tried $u = \sin (ax) \cos (bx)$ and $dv = \frac{1}{x}$, which gives me $[\sin (ax) \cos (bx) \ln (x)] \big\vert_0^\infty + \int_0^\infty \frac{b-a}{2} \ln (x) \cos ((a-b)x)~dx - \int_0 ^ \infty \frac{b+a}{2} \ln (x) \cos ((a+b)x)~dx$, which also looks like a dead end because all three of the terms go to infinity.
Any help is appreciated.
| $$\begin{align}\int_0^{\infty} dx \frac{\sin^4{x}}{x^2} &= \int_0^{\infty} dx \frac{\sin^2{x}}{x^2} - \int_0^{\infty} dx \frac{\sin^2{x} \cos^2{x}}{x^2} \\ &= \int_0^{\infty} dx \frac{\sin^2{x}}{x^2} - \frac14 \int_0^{\infty} dx \frac{\sin^2{2 x}}{x^2} \\ &= \int_0^{\infty} dx \frac{\sin^2{x}}{x^2} - \int_0^{\infty} dx \frac{\sin^2{2 x}}{(2 x)^2}\\ &= \int_0^{\infty} dx \frac{\sin^2{x}}{x^2} - \frac12 \int_0^{\infty} du \frac{\sin^2{u}}{u^2}\\ &= \frac12 \int_0^{\infty} dx \frac{\sin^2{x}}{x^2}\end{align}$$
Now if only you could do that last integral...
| {
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For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$
$$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a).$$
Clearly, $$\left\{\det(A)\neq0\left|\begin{matrix}c\neq b\\a\neq c\\b\neq a\\a,b,c\neq 0\end{matrix}\right.\right\}\\$$
Is it sufficient to say that the matrix is invertible provided that all 4 constraints are met? Would Cramer's rule yield more explicit results for $a,b,c$ such that $\det(A)\neq0$?
| This is a transpose of the Vandermonde matrix: http://en.wikipedia.org/wiki/Vandermonde_matrix
According to the wikipedia page, the determinant of the matrix will be non-zero precisely when none of the elements $a, b$, or $c$ are the same.
| {
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Prove $(8k)^{8k}+(8k+1)^{8k+1}$ and $(8k+1)^{8k+1}+(8k+2)^{8k+2}$ are never perfect squares
Prove $$(8k)^{8k}+(8k+1)^{8k+1}\ \ \text{ and } \ \ \ (8k+1)^{8k+1}+(8k+2)^{8k+2}$$ are never perfect squares ($k\ge 1$).
mod $8$ gives $1$ for both, which is a quadratic residue, so doesn't solve it. Found in AoPS.
| We can say something a bit stronger. For $n$ a positive integer, $n^n+(n+1)^{n+1}$ cannot be a perfect square if $n$ is even or $n\equiv 1 \pmod{4}$.
For even $n$ assume $n=2x>0$ and
$$
(2x)^{2x}+(2x+1)^{2x+1}=u^2 \\
(2x+1)^{2x+1} = (u-(2x)^x)(u+(2x)^x)
$$
Let $A=u-(2x)^x,B=u+(2x)^x$ and $g=\gcd(A,B)$. Then
$$
g\mid B-A=2(2x)^x \\
g\mid (2x+1)^{2x+1} \\
\implies
g \mid \gcd\left(2(2x)^x,(2x+1)^{2x+1}\right)=1
$$
So $A,B$ have no common factor and $AB=(2x+1)^{2x+1}$ so we must have $A=a^{2x+1},B=b^{2x+1}$ for odd positive integers $a,b$ with $b\ge a+2$ and $ab=2x+1$. Then
$$
\begin{align}
2(2x)^x &= b^{2x+1}-a^{2x+1} \\
&= (b-a)\left(b^{2x}+b^{2x-1}a+\cdots+ba^{2x-1}+a^{2x}\right) \\
&> (b-a)(2x+1)(ab)^x && \text{by AM-GM} \\
&> 2(2x)^x
\end{align}
$$
a contradiction, so this is not possible with $n$ even.
For $n=1$ it is easily confirmed that $1^1+2^2=5$ is not a square.
For $1<n\equiv 1 \pmod 4$ assume $n=2x-1, x>2$ and
$$
(2x-1)^{2x-1}+(2x)^{2x}=u^2 \\
(2x-1)^{2x-1} = (u-(2x)^x)(u+(2x)^x)
$$
As before let $A=u-(2x)^x,B=u+(2x)^x$ and it follows that $\gcd(A,B)=1$, $A=a^{2x-1},B=b^{2x-1}$ for odd positive integers $a,b$ with $ab=2x-1=n$. Since $n\equiv 1\pmod{4}$, $a\equiv b \pmod{4}$ and so $b\ge a+4$.
Then
$$
\begin{align}
2(2x)^x &= b^{2x-1}-a^{2x-1} \\
&= (b-a)\left(b^{2x-2}+b^{2x-3}a+\cdots+ba^{2x-3}+a^{2x-2}\right) \\
&> (b-a)(2x-1)(ab)^{x-1} \\
&\ge 4(2x-1)^x \\
&=4(2x)^x\left(1-\frac{1}{2x}\right)^x \\
&> 2(2x)^x
\end{align}
$$
where the last step follows from $\left(1-\frac{1}{2t}\right)^t>\frac{1}{2}$ for all $t>1$ and the assumption $x>2$. Again we reach a contradiction, so it is not possible for numbers of this form to be perfect squares when $n\equiv 1 \pmod 4$.
| {
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"url": "https://math.stackexchange.com/questions/1275168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Number of ways to throw at most 14 with 4 dice - generating functions Determine the chance to throw at most 14 with 4 normal dice. I will set up the right generating function to determine the number of ways tot thow at most 14 with 4 normal dice and I need some help. I know that:
\begin{align}
x_1 + x_2 + x_3 + x_4 \leq 14 \qquad \text{with} \qquad 1 \leq x_i \leq 6.
\end{align}
\begin{align}
x_1 + x_2 + x_3 + x_4 + x' = 14 \qquad \text{with} \qquad 0 \leq x' \leq 10
\end{align}
So we have:
\begin{align}(x + x^2 + x^3 + x^4 + x^5 + x^6)^4 \cdot (0 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 + x^9 + x^{10}).
\end{align}
Does this make sense? I do not have to determine the coefficient of $x^{14}$ only the right generating function.
| The probability generating function for rolling a normal die is:
$G(t)=\frac{t}{6} + \frac{t^2}{6} + \frac{t^3}{6} + \frac{t^4}{6} + \frac{t^5}{6} + \frac{t^6}{6}$
To find the probability generating function for the sum of four independent dice then you have
$G_S(t)=\left(\frac{t}{6} + \frac{t^2}{6} + \frac{t^3}{6} + \frac{t^4}{6} + \frac{t^5}{6} + \frac{t^6}{6}\right)^4$
Where $S=X_1+X_2+X_3+X_4$
Since you want the probability of getting at most $14$ you want the sum of the coefficients of $t^4$ up to and including $t^{14}$ in the expansion. This is because the terms in the expansion represent mutually exclusive events.
I used Wolfram and got the probability to be $\frac{721}{1296}$
Hopefully I haven't made and error and have understood your question correctly.
| {
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"timestamp": "2023-03-29T00:00:00",
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Quadratic solutions puzzle The equation $x^2+ax+b=0$, where $a\neq b$, has solutions $x=a$ and $x=b$. How many such equations are there?
I'm getting $1$ equation as I can only find $a=b=0$ as an equation, which is not allowed.
$$x=\frac{±\sqrt{a^2-4 b}-a}2$$
$x=a$ or $b$ so these are the equations
$$a=\frac{\sqrt{a^2-4 b}-a}2$$
$$b=\frac{-\sqrt{a^2-4 b}-a}2$$
$$a=\frac{-\sqrt{a^2-4 b}-a}2$$
$$b=\frac{\sqrt{a^2-4 b}-a}2$$
The only solution for all of these is $a=b=0$, but is this right?
| You can do this by:
$$(x-a)(x-b)=x^2-(a+b)x+ab=x^2+ax+b$$
Equating constant terms gives $ab=b$ which means $a=1$ or $b=0$
Equating coefficients of $x$ gives $-a-b=a$ or equivalently $2a=-b$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding Number of Ordered Solutions to Equation $$ A \times B \times C \times D \times E \times F = 7 \times 10^7 $$
How can I find the number of ordered solutions for integers (I mean for integers $A,B,C,D,E,F$) so that they can satisfy the above equation?
| The factorization of $7 \times 10^7 $ is simply $2^7 \times 5^7 \times 7 $.
Suppose for now we restrict the search for positive integer solutions only.
For non-negative integers $x_i, y_i, z_i $, let $A = 2^{x_1} \cdot 5^{y_1} \cdot 7^{z_1}, B = 2^{x_2} \cdot 5^{y_2} \cdot 7^{z_2} , \ldots, F = 2^{x_6} \cdot 5^{y_6} \cdot 7^{z_6} $.
Then their product is $2^7 \times 5^7 \times 7^1 = 2^{\sum_{j=1}^6 x_j} \times 5^{\sum_{j=1}^6 x_j} \times 7^{\sum_{j=1}^6 x_j} $.
$ \Rightarrow \displaystyle \sum_{j=1}^6 x_j = 7, \sum_{j=1}^6 y_j = 7, \sum_{j=1}^6 z_j = 1 $.
By Stars and Bars, like TravisJ pointed out, we have $ {7+6 - 1 \choose 6-1} = {12 \choose 5} $ ordered solution for $<x_i>$ and $<y_i>$, and ${1+6 - 1 \choose 6-1} = 6 $ ordered solution for $<z_i>$.
Multiplying them together gives a total of $ {12 \choose 5} ^2 \times 6 = 3763584 $ ordered solution for positive integers $A$ to $F$.
If we drop the constraint for positive integers, then we could have:
Case 1: 6 Positive integers
Case 2: 4 positive integers, 2 negative integers
Case 3: 2 positive integers, 4 negative integers
Case 4: 6 Negative integers
Combining all these cases together yields a total of
$3763584 \times \left [ {6 \choose 0} + {6 \choose 2} + {6 \choose 4} + {6 \choose 6} \right ] = 120434688 $ ordered solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Conic reduction I'm trying to reduce this conic :
$x^2+y^2+2xy+x+y=0$
to a canonical form.
I started with finding the eigenvalues of the matrix associated to the quadratic form $x^2+y^2+2xy$
I found $z_1=2 , z_2=0$
and a basis for the diagonalized matrix $e_1= ({1\over \sqrt2},{1\over \sqrt2})$ $e_2=({1\over \sqrt2},{-1\over \sqrt2})$
The quadratic form can be rewritten in this base as : $2x^2$ but what should I do next? How can I rewrite the other terms? I know that the solution is a couple of parallel lines...
| Following the method blindly $x^2+y^2+2xy+x+y={\bf x}^T A{\bf x}+K{\bf x}=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0$,
we use $P=\begin{pmatrix}\frac1{\sqrt{2}}&\frac1{\sqrt{2}}\\\frac1{\sqrt{2}}&-\frac1{\sqrt{2}}\end{pmatrix}$.
Then ${\bf x}=P{\bf x}'$ gives ${\bf x}'^T P^TAP{\bf x}'+KP {\bf x}'=2x'^2+\sqrt{2}x'=\sqrt{2}x'(\sqrt{2}x'+1)=0$, two parallel lines, which we could also see directly since: $x^2+y^2+2xy+x+y=(x+y)^2+(x+y)=(x+y)(x+y+1)=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$a^{13} \equiv a \bmod N$ - proof of maximum $N$ From Fermat's Little Theorem, we know that $a^{13} \equiv a \bmod 13$. Of course $a^{13} \equiv a \bmod p$ is also true for prime $p$ whenever $\phi(p) \mid 12$ - for example, $a^{13} = a^7\cdot a^6 \equiv a\cdot a^6 = a^7 \equiv a \bmod 7$.
So far I have that the largest $N$ for which all $ a^{13} \equiv a \bmod N$, is $N = 2\cdot 3\cdot 5\cdot 7\cdot 13 = 2730$
Can someone either put together an elegant proof of this, or find and prove a different limit?
| Putting in $a=2$, we get that $N$ divides $2^{13} - 2 = 2 \cdot 3^2 \cdot 5 \cdot 7 \cdot 13$.
On the other hand, putting in $a=3$, we get that $N$ divides $3^{13} - 3 = 2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 13 \cdot 73$.
Hence $N$ must divide $2 \cdot 3 \cdot 5 \cdot 7 \cdot 13 = 2730$.
| {
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"timestamp": "2023-03-29T00:00:00",
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The box has minimum surface area Show that a rectangular prism (box) of given volume has minimum surface area if the box is a cube.
Could you give me some hints what we are supposed to do??
$$$$
EDIT:
Having found that for $z=\frac{V}{xy}$ the function $A_{\star}(x, y)=A(x, y, \frac{V}{xy})$ has its minimum at $(\sqrt[3]{V}, \sqrt[3]{V})$, how do we conclude that the box is a cube??
We have that $x=y$. Shouldn't we have $x=y=z$ to have a cube??
| Volume $V=xyz$ given. Area $A(x,y,z)=2(xy+yz+zx)$, to minimise, when $x,y,z>0$ and $xyz=V$.
Fact. If $a,b,c>0$, then $a+b+c\ge 3\sqrt[3]{abc}$, and equality holds if and only if $a=b=c$.
Proof. We set $X=\sqrt[3]{a}$, $Y=\sqrt[3]{b}$ and $Z=\sqrt[3]{c}$. Then the identity
$$
X^3+Y^3+Z^3-3XYZ=\frac{1}{2}(X+Y+Z)\big((X-Y)^2+(Y-Z)^2+(Z-X)^2\big),\tag{1}
$$
holds. This means that
$$
a+b+c-3\sqrt[3]{abc}=X^3+Y^3+Z^3-3XYZ\ge 0,
$$
as the right-hand-side of $(1)$ is non-negative, and the equality only if $X-Y=Y-Z=Z-X=0$ or $X=Y=Z=0$ or $a=b=c=0$. $\quad\Box$
Hence,
$$
\frac{A}{2}=xy+yz+zx\ge 3\sqrt[3]{xy\cdot yz\cdot zx}=3\sqrt[3]{x^2y^2z^2}=3V^{2/3}.
$$
and equality holds iff $xy=zx=yz$ or equivalently iff $x=y=z$.
Indeed, $A$ is minimised when $x=y=z$, and $A_{\mathrm{min}}=6V^{2/3}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Arc length of natural log function I am currently trying to find the arc length of $f(x)=ln(x)$, which involves the integral
$$\int \sqrt{1+\frac1{x^2}}dx$$
I managed to solve the integral correctly but I want to know if there is a simpler way as I ended up canceling two terms later on.
For reference this is what I did:
$$\int \sqrt{1+\frac1{x^2}}dx$$
$$\int \frac1x\sqrt{x^2+1}dx$$
$$u_1=\frac1x$$
$$du_1=-\frac{dx}{x^2}$$
$$v_1=\int \sqrt{x^2+1}dx$$
$$v_1=\frac12(sinh^{-1}(x)+x\sqrt{x^2+1})$$
$$\int \frac1x\sqrt{x^2+1}dx=\frac{sinh^{-1}(x)+x\sqrt{x^2+1}}{2x}+\frac12\int\frac{sinh^{-1}(x)}{x^2}dx+\int\frac1x\sqrt{x^2+1}dx$$
$$\frac12\int \frac1x\sqrt{x^2+1}dx=\frac{sinh^{-1}(x)+x\sqrt{x^2+1}}{2x}+\frac12\int\frac{sinh^{-1}(x)}{x^2}dx$$
$$\int \frac1x\sqrt{x^2+1}dx=\frac{sinh^{-1}(x)+x\sqrt{x^2+1}}x+\int\frac{sinh^{-1}(x)}{x^2}dx$$
$$u_2=sinh^{-1}(x)$$
$$du_2=\frac{dx}{\sqrt{x^2+1}}$$
$$v_2=\int\frac{dx}{x^2}$$
$$v_2=\frac1x$$
$$\int \frac1x\sqrt{x^2+1}dx=\frac{sinh^{-1}(x)}x+{\sqrt{x^2+1}}-\frac{sinh^{-1}(x)}x+\int\frac{dx}{x\sqrt{x^2+1}}$$
$$\int \frac1x\sqrt{x^2+1}dx={\sqrt{x^2+1}}+\int\frac{dx}{x\sqrt{x^2+1}}$$
$$t=\sqrt{1+x^2}$$
$$t^2-1=x^2$$
$$dt= \frac x{\sqrt{x^2+1}} dx$$
$$\int \frac1x\sqrt{x^2+1}dx={\sqrt{x^2+1}}+\int\frac x{x^2\sqrt{x^2+1}}dx$$
$$\int \frac1x\sqrt{x^2+1}dx={\sqrt{x^2+1}}+\int\frac{dt}{t^2-1}$$
$$\int \frac1x\sqrt{x^2+1}dx={\sqrt{x^2+1}}-tanh^{-1}(t)+C$$
$$\int \frac1x\sqrt{x^2+1}dx={\sqrt{x^2+1}}-tanh^{-1}(\sqrt{1+x^2})+C$$
Is there a simpler way to do this question instead of what I did?
| Standard ways to find this integral are to substitute $x=\tan\theta$ or $x=\sinh t$.
You can also let $\frac{1}{x}=\tan\theta$, so $x=\cot\theta$ and
$\displaystyle\int\sqrt{1+\frac{1}{x^2}}\;dx=\int\sec\theta\big(-\csc^{2}\theta\big)d\theta$. $\;\;$Then letting $u=\sec\theta, dv=-\csc^{2}\theta d\theta$ gives
$\displaystyle\sec\theta\cot\theta-\int\sec\theta d\theta=\csc\theta-\ln\lvert\sec\theta+\tan\theta\rvert+C=\sqrt{x^2+1}-\ln\bigg\vert\frac{\sqrt{x^2+1}+1}{x}\bigg\vert+C$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the 2x2 matrix given 2 equations. Find the $2 \times 2$ matrix ${A}$ such that ${A}^2 = {A}$ and
${A} \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$
I have tried to express A as a matrix with variables a,b,c, and d, but it gets too messy with the equations. Any help is appreciated.
| Notice that $A \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}$ and also $A \begin{pmatrix} 6 \\ 2 \end{pmatrix} = A^2 \begin{pmatrix} 7 \\ -1 \end{pmatrix} = A \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}$.
Thus $A \begin{pmatrix} 7 &6 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 6& 6 \\ 2& 2 \end{pmatrix}$.
And therefore $A = \begin{pmatrix} 6& 6 \\ 2& 2 \end{pmatrix}\begin{pmatrix} 7 &6 \\ -1 & 2 \end{pmatrix}^{-1} = \frac{1}{10}\begin{pmatrix} 9& 3 \\ 3& 1 \end{pmatrix}$.
| {
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Find the integer x: $x \equiv 8^{38} \pmod {210}$ Find the integer x: $x \equiv 8^{38} \pmod {210}$
I broke the top into prime mods:
$$x \equiv 8^{38} \pmod 3$$
$$x \equiv 8^{38} \pmod {70}$$
But $x \equiv 8^{38} \pmod {70}$ can be broken up more:
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod {10}$$
But $x \equiv 8^{38} \pmod {10}$ can be broken up more:
$$x \equiv 8^{38} \pmod 5$$
$$x \equiv 8^{38} \pmod 2$$
In the end,I am left with:
$$x \equiv 8^{38} \pmod 5$$
$$x \equiv 8^{38} \pmod 2$$
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod 3$$
Solving each using fermat's theorem:
*
*$x \equiv 8^{38}\equiv8^{4(9)}8^2\equiv64 \equiv 4 \pmod 5$
*$x \equiv 8^{38} \equiv 8^{1(38)}\equiv 1 \pmod 2$
*$x \equiv 8^{38} \equiv 8^{6(6)}8^2\equiv 64 \equiv 1 \pmod 7$
*$x \equiv 8^{38} \equiv 8^{2(19)}\equiv 1 \pmod 3$
So now, I have four congruences. How can i solve them?
| An alternate solution using the Chinese Remainder Theorem.
First off, $2, 3, 5$ and 7 are pairwise relatively prime thus we know the following system of congruences has a unique solution modulo $2\times3\times5\times7=210$.
\begin{cases} x \equiv 4 \pmod {5} \\ x \equiv 0 \pmod {2} \\ x \equiv 1 \pmod {7} \\ x \equiv 1 \pmod {3} \end{cases}
Start with the first congruence and turn it into an equation
$x\equiv 4 \pmod 5$
$\iff x = 5k + 4$ (by definition of congruence)
Plugging this in the second congruence:
$5k+4 \equiv 0\pmod {2}$
$\iff$
$5k \equiv 0 \pmod {2}$
To get rid of the 5 on the left hand side, multiply both sides of the congruence by the inverse of 5 mod 2 which is 1 by the extended euclidean algorithm. Thus:
$k \equiv 0 \pmod {2}$
$\iff$
$k = 2k'$
Which means $x = 5k + 4 = 5(2k') + 4 = 10k' + 4$
Now replacing in the third equation:
$10k' + 4 \equiv 1 \pmod {7}$
$\iff$
$10k' \equiv 4 \pmod {7}$
Multiply both sides by 5, the inverse of 10 mod 7.
$k' \equiv 6 \pmod {7}$
$\iff$
$k' = 7k'' + 6$
Which means $x = 10k' +4 = 10(7k'' + 6) + 4 = 70k'' + 64$
Finally replacing in the last equation:
$70k'' + 64 \equiv 1 \pmod {3}$
$\iff$
$70k' \equiv 0 \pmod {3}$
Multiply both sides by 1, the inverse of 70 mod 3.
$k' \equiv 0 \pmod {3}$
$\iff$
$k'' = 3k'''$
Thus
$x = 70k'' + 64 = 70(3k''') + 64 = 210k''' + 64$
$\iff$
$x \equiv 64 \pmod {210}$
| {
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"timestamp": "2023-03-29T00:00:00",
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For which values of $\alpha \in \mathbb{R}$, does the series $\sum_{n=1}^\infty n^\alpha(\sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1})$ converge? How do I study for which values of $\alpha \in \mathbb{R}$ the following series converges?
(I have some troubles because of the form [$\infty - \infty$] that arises when taking the limit.)
$$\sum_{n=1}^\infty n^\alpha(\sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1})$$
| First hint: $$\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1} = \sqrt{n+1}-\sqrt{n}+\sqrt{n-1}-\sqrt{n}$$
First term:
$$\sqrt{n+1}-\sqrt{n} = (\sqrt{n+1}-\sqrt{n})\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$$
Second term:
$$\sqrt{n-1}-\sqrt{n} = (\sqrt{n-1}-\sqrt{n})\frac{\sqrt{n-1}+\sqrt{n}}{\sqrt{n-1}+\sqrt{n}} = \frac{-1}{\sqrt{n-1}+\sqrt{n}}$$
Plug in the first and the second result into the original term:
$$\begin{align}
\frac{1}{\sqrt{n+1}+\sqrt{n}}-\frac{-1}{\sqrt{n-1}+\sqrt{n}}&=\frac{\sqrt{n-1}-\sqrt{n+1}}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n-1}+\sqrt{n})}\\
&=\frac{-2}{(\sqrt{n+1}+\sqrt{n})(\sqrt{n-1}+\sqrt{n})(\sqrt{n+1}+\sqrt{n-1})} \end{align}
$$
Thus $\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1} = O\left(\frac{1}{\sqrt{n}^3}\right)$. This should help you...
| {
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"timestamp": "2023-03-29T00:00:00",
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Dividing a Matrix into three parts The matrix $A$ is given by $$\left(\begin{array}{ccc}
1 & 2 & 1 \\
1 & 1 & 2 \\
2 & 3 & 1 \end{array} \right)$$
Given that $A^3$ can be expressed as $A^3$=$aA^2+bA+cI$, find the values of $a,b,c$. Furthermore, express the inverse $A^{-1}$ in the form $A^{-1}$=$dA^2+eA+fI$, where $d$,$e$, and $f$ are rational numbers
I was given this problem in a homework set and did not make much progress. I'm not sure as to how I should separate matrix $A$ into three components.
| Calculating $\det(A-\lambda I)=\det\begin{pmatrix}1-\lambda&2&1\\1&1-\lambda&2\\2&3&1-\lambda\end{pmatrix}$ gives
$\hspace{.3 in}(1-\lambda)^3+11-10(1-\lambda)=-(\lambda^3-3\lambda^2-7\lambda-2)$,
so by the Cayley-Hamilton theorem, $\;\;A^3-3A^2-7A-2I=0$
and therefore $\;\;A\big[\frac{1}{2}(A^2-3A-7I)\big]=I$.
Alternatively, using the idea suggested by hkmather802, we can find $A^2$ and $A^3$ to get the equation
$\begin{pmatrix}24&35&25\\25&36&29\\35&51&36\end{pmatrix}=a\begin{pmatrix}5&7&6\\6&9&5\\7&10&9\end{pmatrix}+b\begin{pmatrix}1&2&1\\1&1&2\\2&3&1\end{pmatrix}+c\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$.
Just using the first row, we get the linear system
$5a+b+c=24,\;\;7a+2b=35,\;\;6a+b=25$;
so solving the last two equations gives $a=3, b=7$; and substituting in the 1st equation gives $c=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Infinite solutions for $(\frac{n+1}{n})^a\cdot (\frac{m+1}{m})^b = 2$ Given $(\frac{n+1}{n})^a\cdot (\frac{m+1}{m})^b = 2$ where a, b, n, and m are all positive integers, are there infinitely many solutions $(a,b,n,m)$?
| Let's try this:
$(n+1)^{a}(m+1)^{b}=2\cdot n^{a}m^{b}$
Obviously either $(n+1)$ or $(m+1)$ must be even but not both. Take
$(n+1)=2^{j}\cdot r;(n+1)^{a}=2^{a\cdot j}r^{a}$
Now $n+1$ is relatively prime to $n$ so if $a\cdot j>1$ then $m=2^{k}s$
and we have
$a\cdot j-b\cdot k=1$
Thus $(a,b),(a,j),(j,k),(j,b)$ are all relatively prime.
When we ``reverse engineer'' this we find a problem. We are left
with:
$r^{a}\cdot(2^{k}s+1)^{b}=(2^{j}r-1)^{a}(s)^{b}$
Thus every factor of $r^{a}$ is a factor of $s^{b}$ and vise versa;
$r=s,a=b$ .
And the positive resolution of.
$a\cdot j-b\cdot k=1,a=b$ is
$a=b=1,j-k=1$
$\left(\frac{2^{j}}{2^{j}-1}\right)\left(\frac{2^{j-1}+1}{2^{j-1}}\right)=2$
$j=2$
Your solution
| {
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"url": "https://math.stackexchange.com/questions/1290495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Simplifying $\Big[\dfrac{5-\sqrt{a}}{5+\sqrt a}-\dfrac{\sqrt a+5}{\sqrt a-5}+2\Big]^{-2}$ Simplifying $$\Big[\dfrac{5-\sqrt{a}}{5+\sqrt a}-\dfrac{\sqrt a+5}{\sqrt a-5}+2\Big]^{-2}$$
When I try, the numerator cancels out to $0$, yet the answer sheet says $(25-a)^2/10000$. Where am I going wrong & how is getting $10 000$ there even possible?
| Let $s = \sqrt{a}$:
$$
\frac{5-s}{5+s}-\frac{s+5}{s-5}
= \frac{-(s-5)^2 - (s+5)^2}
{(5+s)(5-s)}
= \frac{(s-5)^2 + (s+5)^2}
{s^2-5^2}
= \frac{2s^2 + 2 \cdot 5^2}{s^2-5^2}
= \frac{2(a + 25)}{a-25}
$$
Can you finish the problem?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Math about Geometric series In a geometric series, the sum of $1^{st}$ term $+$ $2^{nd}$ term $+$ $3^{rd}$ term $= 38$,
the sum of $2^{nd}$ term $+ 4^{th}$ term $= 17 \frac{1}{3}$;
how to calculate the common ratio? ( it is for sure the answer is $r= \frac{2}{3}$ and the first term $= 18$)
| $$a(1+r+r^2) = 38\\ar+ar^3 = \frac{52}3 $$ eliminating $a$ leaves us with
$$\frac{1+r+r^2}{r+r^3} = \frac{57}{26}.$$ this can be rewritten as $$ 0=57r^3-26r^2+31r-26 = (3r-2)(19r^2+4r+13).$$ the quadratic equation $19r^2 + 4r+13=0 $ has no real roots, therefore $$r = \frac23, a = 18. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding the area under $\frac{3}{2x+1}$ and $3x-2$ Find the area between these curves.
$$y=\dfrac{3}{2x+1},\qquad y=3x-2;\qquad x=2\quad \text{et} \quad y=0 $$
Indeed,
I calculate the integral of the blue function between $1$ and $2$. Then, I will calculate the area of the triangle between the yellow line and the x-axis on the interval $\dfrac{2}{3}$ and $1$, which gives me:
$$\int _{\frac{3}{2}}^1\left(3x-2\right)dx+\int _1^2\frac{3}{2x+1}dx=\frac{3}{2}\ln \left(\frac{5}{3}\right)-\frac{7}{8}$$
Am I right in the way how to find calculate the area and if that so that $\frac{3}{2}\ln \left(\frac{5}{3}\right)-\frac{7}{8}$ is it correct?
| i think the first integral must be $$\int_{2/3}^{1}(3x-2)dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can I simplify $1\times 2 + 2 \times 3 + .. + (n-1) \times n$ progression? I have a progression that goes like this:
$$1\times 2 + 2 \times 3 + .. + (n-1) \times n$$
Is there a way I can simplify it?
| Yes.
\begin{align*}
1\cdot 2 + 2\cdot 3 + \dotsb + (n-1)\cdot n
&= 1\cdot(1+1) + 2\cdot(2+1) + \dotsb + (n-1)\cdot((n-1)+1) \\
&= 1^2 + 1 + 2^2 + 2 + \dotsb + (n-1)^2 + (n-1) \\
&= (1^2+2^2+\dotsb+(n-1)^2) + (1+2+\dotsb+(n-1))
\end{align*}
Both of those sums have well-known closed forms; see here, for example.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Distance between line and a point Consider the points (1,2,-1) and (2,0,3).
(a) Find a vector equation of the line through these points in parametric form.
(b) Find the distance between this line and the point (1,0,1). (Hint: Use the parametric form of the equation and the dot product)
I have solved (a), Forming:
Vector equation: (1,2,-1)+t(1,-2,4)
x=1+t
y=2-2t
z=-1+4t
However, I'm a little stumped on how to solve (b).
| Using this formula and your computations from (a), we get the expression for distance $d$:
$$
d =
\frac{\left\| \
\
\left(\
\begin{bmatrix}
1 \\ 2 \\ -1
\end{bmatrix}
-
\begin{bmatrix}
1 \\ 0 \\ 1
\end{bmatrix}
\
\right)
\times
\,
\begin{bmatrix}
1 \\ -2 \\ 4
\end{bmatrix}
\ \
\right\|}
{
\begin{Vmatrix}
1 \\ -2 \\ 4
\end{Vmatrix}
}
=
\frac{\left\| \
\
\begin{bmatrix}
0 \\ 2 \\ -2
\end{bmatrix}
\times
\begin{bmatrix}
1 \\ -2 \\ 4
\end{bmatrix}
\ \
\right\|}
{
\sqrt{1 + 4 + 16}
}
=
$$
$$
=
\frac{1}{\sqrt{21}}
\begin{Vmatrix}
\vec{\mathbf{i}} & \vec{\mathbf{j}} & \vec{\mathbf{k}} \\
0 & 2 & -2 \\
1 & -2 & 4
\end{Vmatrix}
=
\frac{1}{\sqrt{21}}
\begin{Vmatrix}
2 \cdot 4 - (-2) \cdot (-2) \\
(-2)\cdot 1 - 0 \cdot 4 \\
0 \cdot (-2 ) - 1 \cdot 2
\end{Vmatrix}
=
\frac{1}{\sqrt{21}}
\begin{Vmatrix}
4 \\
-2 \\
-2
\end{Vmatrix}
=
$$
$$
=
\frac{\sqrt{16+4+4}}{\sqrt{21}} = \sqrt{\frac{24}{21}}= \sqrt{\frac{8}{7}} = \frac{2\sqrt{2}}{\sqrt{7}} \approx 1.069
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$3^x + 4^y = 5^z$ This is an advanced high-school problem.
Find all natural $x,y$, and $z$ such that
$3^x + 4^y = 5^z$.
The only obvious solution I can see is $x=y=z=2$. Are there any other solutions?
| First mod the whole thing by 3, then $1^y = 2^z$, so $z = 2z_2$ for some $z_2\in\mathbb{N}$. So $3^x = 5^{2z_2} - 2^{2y} = (5^{z_2} + 2^y)(5^{z_2} - 2^y)$.
If $5^{z_2} - 2^y\ne 1$, then by modding each factor by 3 we find that $z_2\ne y \mod 3$ and $z_2 = y \mod 3$, contradiction.
We can invoke Mihăilescu's theorem to find that $x = y = z = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find h in terms of r A sphere and a cylinder have equal volumes. The sphere has a radius 3r.
The cylinder has radius 2r and height h.
Find h in terms of r.
I'm only 15, someone walk me through this as simple as possible :)
| Volume of sphere $= \frac 43 \pi R^3 = \frac 43 \pi (3r)^3$
Volume of cylinder $= \pi R^2 h = \pi(2r)^2h$
These two volumes should be equal so equate them:
$$\frac 43 \pi (3r)^3 = \pi(2r)^2h$$
Now let's solve for $h$ in terms of $r$.
First notice that $(3r)^3 = 3^3r^3 = 27r^3$. Likewise $(2r)^2 = 2^2r^2 = 4r^2$. Therefore
$$\frac 43 \pi (3r)^3 = \pi(2r)^2h \\ \implies \frac 43 \pi \color{red}{(27r^3)} = \pi\color{red}{(4r^2)}h \\ \implies \frac 43 \color{red}{(27r^3)}\pi = \color{red}{(4r^2)}\pi h \\ \implies \color{blue}{\frac{4\cdot 27}{3}}r^3\pi = 4r^2\pi h$$
Now before we go on, notice that there is a $4r^2\pi$ on both sides. So let's divide both sides by $4r^2\pi$:
$$\require{cancel}\frac{4\cdot 27}{3}r^3\pi = 4r^2\pi h \\ \implies \frac {27}{3}\color{blue}{(4r^2\pi)}r = \color{blue}{(4r^2\pi)} h \\ \implies \frac{\frac {27}{3}\cancel{\color{blue}{(4r^2\pi)}}r}{\cancel{\color{blue}{(4r^2\pi)}}} = \frac{\cancel{\color{blue}{(4r^2\pi)}} h}{\cancel{\color{blue}{(4r^2\pi)}}} \\ \implies \frac {27}{3}r=h \\ \implies 9r=h$$
So your answer is $\require{enclose}\enclose{box} {h=9r}$. This is an expression for $h$ in terms of $r$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $ x=\frac{\sin^3 t}{\sqrt{\cos 2t}}$ and $y = \frac{\cos^3 t}{\sqrt{\cos 2t}}\;,$ Then $ \frac{dy}{dx}$ in terms of $t$ If $\displaystyle x=\frac{\sin^3 t}{\sqrt{\cos 2t}}$ and $\displaystyle y = \frac{\cos^3 t}{\sqrt{\cos 2t}}\;,$ Then $\displaystyle \frac{dy}{dx}$ in terms of $t$
$\bf{My\; Try::}$ Using The Formula $\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$
Now Given $$\displaystyle x= \frac{\sin^3 t}{\sqrt{\cos 2t}}.$$ So $$\displaystyle \frac{dx}{dt} = \frac{\sqrt{\cos 2t}\cdot 3\sin^2 t \cdot \cos t-\sin^3 t \cdot \frac{1}{2\sqrt{\cos 2t}}\cdot -\sin 2t \cdot 2t}{\cos 2t}$$
So we get $$\displaystyle \frac{dx}{dt} = \frac{\cos 2t \cdot 3\sin^2 t \cdot \cos t+\sin^3 t \cdot \sin 2t \cdot t}{\cos 2t \cdot \sqrt{\cos 2t}}$$
Similarly Given $$\displaystyle y=\frac{\cos^3 t}{\sqrt{\cos 2t}}.$$ So $$\displaystyle \frac{dy}{dt} = \frac{-\sqrt{\cos 2t}\cdot 3\cos^2 t \cdot \sin t-\cos^3 t\cdot \frac{1}{2\sqrt{\cos 2t}}\cdot -\sin 2t \cdot 2t}{\cos 2t}$$
So we get $$\displaystyle \frac{dy}{dt} = \frac{-\cos 2t \cdot 3\cos^2 t \cdot \sin t+\cos^3 t\cdot \sin 2t\cdot t}{\cos 2t \cdot \sqrt{\cos 2t}}$$
So $$\displaystyle \frac{dy}{dx} = \frac{-3\cos^2 t \cdot \sin t\cdot \cos 2t+\cos^3 t\cdot \sin 2t\cdot t}{\cos 2t \cdot 3\sin^2 t \cdot \cos t+\sin^3 t \cdot \sin 2t \cdot t}$$
But this method seems complicated.
Is there is any other method by which we can calculate $\displaystyle \frac{dy}{dx}$?
| You could simplify the intermediate calculations using reduction formulas of the powers of sine and cosine functions $$x=\frac{\sin ^3(t)}{\sqrt{\cos (2 t)}}$$ $$\frac{dx}{dt}=\frac{\sin (2 t) \sin ^3(t)}{\cos ^{\frac{3}{2}}(2 t)}+\frac{3 \sin ^2(t) \cos
(t)}{\sqrt{\cos (2 t)}}=\frac{\sin ^2(t) (2 \cos (t)+\cos (3 t))}{\cos ^{\frac{3}{2}}(2 t)}$$ $$y=\frac{\cos ^3(t)}{\sqrt{\cos (2 t)}}$$ $$\frac{dy}{dt}=\frac{\sin (2 t) \cos ^3(t)}{\cos ^{\frac{3}{2}}(2 t)}-\frac{3 \sin (t) \cos
^2(t)}{\sqrt{\cos (2 t)}}=\frac{\sin (t)-\sin (5 t)}{4 \cos ^{\frac{3}{2}}(2 t)}$$ $$\frac{dy}{dx}=-\cot (3 t)$$
This was just done using your steps (I suppose some mistakes since $t$ terms cannot occur during the derivations).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve Inequality for $ |x| $ Given $$\big|\frac{(x-2)}{(x+3)}\big| < 4,$$ solve for $x.$
\ My solution
$$|x - 2| < 4|x + 3|$$
Since,
$ |x - 2| \ge |x| - |2| $ and
$ |x + 3| \le |x| + |3| $ according to triangle inequality;
$|x| - |2| < 4|x| + 4|3| $
$-14 < 3|x|$
$|x| > \frac{-14}{3}$
Is this the final answer?
| from the graph $$f(x) = \frac{x-2}{x+3} = 1-\frac5{x+3}$$ you can see that it has vertical asymptote $x = 3$ and a horizontal one $y = 1.$ the function $f$ is decreasing on $\-infty, -3)$ and increasing on $-3, \infty).$
solving $f(x) = 4$ gives $x = -14/3$ and $f(x) = -4$ gives $x = -2.$
therefore $$ \left|\frac{x-2}{x+3} \right| < 4 \text{ for } -\infty < x < -14/3, -2 < x < \infty.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the point of intersection of two tangents to a circle
19. A circle $C$, of radius $r$, passes through the points $A (a, 0)$, $A_{1} (-a, 0)$ and $B (0, b)$, where $a$ and $b$ are positive and are not equal; a circle $C_{1}$, of radius $r_{1}$, passes through $A, B$ and $B_{1} (0, -b)$. Prove that the centre of $C$ is $\displaystyle\left(0, \frac{b^{2} - a^{2}}{2b}\right)$ and that $r_{1}/r = b/a$.
Find the point of intersection of the tangents to the circle $C$ at $A$ and $A_{1}$.
I have completed the two proofs, but have gone wrong with the final part of the question.
I have said:
The centre of $C$ is $\left(0, \dfrac{b^{2} - a^{2}}{2b}\right)$.
The gradient of the radius to $A$ is:
$$\left.\left(\dfrac{b^{2} - a^{2}}{2b} - 0\right)\right/(0 - a) = \dfrac{b^{2} - a^{2}}{-2ab} = \dfrac{a^{2} - b^{2}}{2ab}.$$
$\therefore$ The equation of the tangent to $C$ at $A$ is:
$$y = -\dfrac{2ab}{a^{2} - b^{2}}(x - a).$$
The gradient of the radius to $A_{1}$ is: $\dfrac{b^{2} - a^{2}}{2ab}$.
$\therefore$ The equation of the tangent to $C$ at $A_{1}$ is:
$$y = -\dfrac{2ab}{b^{2} - a^{2}}(x + a).$$
$\therefore$ At the point of intersection:
$y = -\dfrac{2ab}{a^{2} - b^{2}}(x - a) \qquad (1)$
$y = -\dfrac{2ab}{b^{2} - a^{2}}(x + a) \qquad (2)$
I chose to eliminate $y$:
$-\dfrac{2ab}{a^{2} - b^{2}}(x - a) = -\dfrac{2ab}{b^{2} - a^{2}}(x + a)$
$(x - a)(b^{2} - a^{2}) = (x + a)(a^{2} - b^{2})$
$xb^{2} - xa^{2} - ab^{2} + a^{3} = xa^{2} - xb^{2} + a^{3} - ab^{2}$
$2xb^{2} - 2xa^{2} = 0$
$x = 0$
Sub $x$ into (2)
$y = -\dfrac{2a^{2}b}{b^{2} - a^{2}}$
But, the book of knowledge saith:
$(0, 2a^{2}b/(b^{2} - a^{2})).$
Which presumably means I've made a mistake with a sign somewhere (but may indicate a more significant misunderstanding); but I'm afraid I simply can't spot what I've done wrong - any ideas?
|
Compare the above drawings and determine who's answer is more correct.
| {
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"url": "https://math.stackexchange.com/questions/1305401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
limits and infinity I'm having trouble wrapping my head around some of the 'rules' of limits. For example,
$$
\lim_{x\to \infty} \sqrt{x^2 -2} - \sqrt{x^2 + 1}
$$
becomes
$$
\sqrt{\lim_{x\to \infty} (x^2) -2} - \sqrt{\lim_{x\to \infty}(x^2) + 1}
$$
which, after graphing, seems to approach zero. My question is how do you know for sure the answer is zero without graphing? Thanks!
| Hint
$$\sqrt{x^2 -2} - \sqrt{x^2 + 1}=(\sqrt{x^2 -2} - \sqrt{x^2 + 1})\cdot \frac{\sqrt{x^2 -2} + \sqrt{x^2 + 1}}{\sqrt{x^2 -2} +\sqrt{x^2 + 1}}=\frac{-3}{\sqrt{x^2 -2} + \sqrt{x^2 + 1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Conditional entropy This is really annoying me: Two players A and B play a best-of-three squash match and the first player to win two games wins the match. A wins a game with probability $p$, B wins with probability $1-p$. Let $X$ be the number of games won by A, let $Y$ be the number of matches played. Show $H(Y|X=x)$ takes its maximum value when $p=\frac{1}{2}$ for all $x$.
I think that $$P(Y=2|X=0)=0$$ and $$P(Y=3|X=1)=1,$$ so $H(Y|X=0)=0$ and $H(Y|X=1)=0$. Then $$P(Y=2|X=2)=\frac{p^2}{p^2+p^2(1-p)}=\frac{1}{2-p}$$ and $$P(Y=3|X=2)=\frac{p^2(1-p)}{p^2+p^2(1-p)}=\frac{1-p}{2-p}$$and therefore $$H(Y|X=2)=\frac{log(2-p)}{2-p}-\frac{1-p}{2-p}log\left(\frac{1-p}{2-p}\right)$$$$=log(2-p) + \left(\frac{1}{2-p}-1\right)log(1-p).$$ Differentiating this gives $$\frac{log(1-p)}{(2-p)^2}$$ which is non-zero at $p=\frac{1}{2}$. Where have I gone wrong?
| $\mathsf P(Y=2\mid X=0)$ is the probability that two games in a best out of three match were played, given that player A won no games. It is in face certain that player B will have won two games in a row if Player A won none of the games in a best of three match. Let's examine the possible outcomes and their probability measures.
$$\begin{array}{ccc| c c | l}
~ & ~ & ~ & X & Y & P(X=x, Y=y)
\\ \hline A & A & ~ & 2 & 2 & p^2
\\ A & B & A & 2 & 3 & p^2(1-p)
\\ A & B & B & 1 & 3 & p(1-p)^2
\\ B & A & A & 2 & 3 & p^2(1-p)
\\ B & A & B & 1 & 3 & p(1-p)^2
\\ B & B & ~ & 0 & 2 & (1-p)^2
\end{array}$$
Thus we have:
$$\begin{align}\mathsf P(Y=2 \mid X=0) & = 1
\\ \mathsf P(Y=3 \mid X=1) & = 1
\\ \mathsf P(Y=2 \mid X=2) & = \frac{1}{3-2p} \tag{$\bigstar$}
\\ \mathsf P(Y=3 \mid X=2) & = \frac{2-2p}{3-2p}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Finding a basis for the nullspace $A$=\begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & 2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix}
I need to find the null space for this matrix. After performing row operations $R_3 + 3R_1$, then $R_{new row 3}$ + $R_2$, I got the reduced row echelon form. (the variable I chose were $x$, $y$, $z$, $w$). I identified the pivot variables as $x$ and $y$, after expressing both of these in terms of the free variables $z$ and $w$, I got $$\begin{bmatrix}
x\\
y\\
z\\
w\\
\end{bmatrix} = z \begin{bmatrix}
23/2\\
4\\
1\\
0\\
\end{bmatrix} + w\begin{bmatrix}
-11/2\\
-2\\
0\\
1\\
\end{bmatrix}$$, and then chose both the column vectors alongside $z$ and $w$ as the basis for the nullspace. However the answer given is $$\begin{bmatrix}
23\\
8\\
2\\
0\\
\end{bmatrix}\begin{bmatrix}
9\\
4\\
0\\
2\\
\end{bmatrix}$$. What am I doing wrong?
| If matrix
$$
\begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & 2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix}
$$
Row reduced form of matrix is
$$
\begin{bmatrix}1 & 0 & -\frac{23}{2} & 0\\ 0 & 1 & -4 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 &0 &0\end{bmatrix}
$$
So basis is
$$
\begin{bmatrix}23 \\ 8 \\ 2 \\ 0\end{bmatrix}
$$
If matrix is
$$
\begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & -2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix}
$$
Row reduced form of matrix is
$$
\begin{bmatrix}1 & 0 & -\frac{23}{2} & -\frac{9}{2}\\ 0 & 1 & -4 & -2\\ 0 & 0 & 0 & 0\\ 0 & 0 &0 &0\end{bmatrix}
$$
So basis is
$$
\begin{bmatrix}23 \\ 8 \\ 2 \\ 0\end{bmatrix}
\begin{bmatrix}9 \\ 4 \\ 0 \\ 2\end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\sin(2x)-\cos(2x)-\sin(x)+\cos(x)=0$ I would like to share a trigonometry question here. Wonder is there another way to solve it or not.
$\sin(2x)-\cos(2x)-\sin(x)+\cos(x)=0$
$2\cos(x)\sin(x)-(1-2\sin^2(x))-\sin(x)+\cos(x)=0$
$2\cos(x)\sin(x)-1+2\sin^2(x)-\sin(x)+\cos(x)=0$
$2\sin^2(x)+2\cos(x)\sin(x)-\sin(x)+\cos(x)-1=0$
$(2\sin(x)+1)(\sin(x)+\cos(x)-1)=0$
So, we have $$2\sin(x)+1=0 \space\text{and}\sin(x)+\cos(x)-1=0$$
$2\sin(x)+1=0$
$\sin(x)=-\frac{1}{2}$
$x=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n$
$\sin(x)+\cos(x)-1=0$ which answered in my yesterday post $\cos x+\sin x=1$
and get $x=2n\pi,2n\pi+\frac{\pi}{2}$
Combine all the solutions, $$x=2\pi n,x=\frac{7\pi}{6}+2\pi n,x=2\pi n+\frac{\pi}{2}, x=\frac{11\pi}{6}+2\pi n$$
| If $\cos A-\sin A=\cos B-\sin B$
As $\cos A-\sin A=\cdots=\sqrt2\cos\left(A+\dfrac\pi4\right),$
$\cos\left(A+\dfrac\pi4\right)=\cos\left(B+\dfrac\pi4\right)$
$\implies A+\dfrac\pi4=2m\pi\pm\left(B+\dfrac\pi4\right)$ where $m$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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When does this equation have a solutions in integers: $ z ^ x + \bar z ^ y = 1 $? Let $ z $ be a complex number such that $ z = \alpha + i \beta $, where $ \alpha$ and $\beta$ are integers. Let $ \bar z $ be the complex conjugate of $ z $, and let $ x $ and $ y $ be integers.
When does the following equation have a solutions in integers?
$$ z ^ x + \bar z ^ y = 1 $$
| First of all, if you accept that $0^0=1$, then there are some trivial solutions:
Let $z$ be $0$ and let one of $x$ and $y$ be $0$ and the other one be any
positive integer, and you will have $z^x+\bar z^y=0+1=1$. It's easy to see that
if one of $x$, $y$ and $z$ is equal to $0$, then the above cases are the only
possible solutions.
So, from now on, we'll assume that none of $x$, $y$ and $z$
is equal to $0$. Note that in this case, we'll have $\left|z\right|\geq1$,
because $\alpha$ and $\beta$ are integers and at least one of them is nonzero.
Next, we will show that if $z^x+\bar z^y=1$, then $\left|z\right|\leq2$.
*
*If $x>0$ and $y>0$ then $\left|z\right|^{-2x}\leq1$. Also by
$z^x+\bar z^y=1$ we get $\bar z^x+z^y=\bar 1=1$. By multiplying these equations
we have $\left|z\right|^{2x}+\left|z\right|^{2y}+z^x\bar z^y+\bar z^x z^y=1$.
Now suppose that $x\leq y$ (the case $y\leq x$ is similar). From the last
equation we have $\left|z\right|^{2x}+\left|z\right|^{2y}+\left|z\right|^{2x}
(\bar z^{y-x}+z^{y-x})=1$ that yields $1+\left|z\right|^{2y-2x}+
(\bar z^{y-x}+z^{y-x})=\left|z\right|^{-2x}$ which is equivalent to
$1+\left|z\right|^{2y-2x}+2\Re(z^{y-x})=\left|z\right|^{-2x}$. Now because
$\alpha$ and $\beta$ are integers, the left side of the last equation is an
integer. But the right side is no more than $1$ and so it's an integer if and
only if $\left|z\right|=1$. That gives $\left|z\right|\leq2$.
*If $x>0$ and $y<0$ then $\left|z\right|^x\geq\left|z\right|$ and
$\left|z\right|^y\leq1$. So, by triangle inequality, we get
$\left|z\right|\leq\left|z\right|^x\leq\left|z\right|^y+\left|z^x+
\bar z^y\right|\leq2$. The case $y>0$ and $x<0$ is similar.
*If $x<0$ and $y<0$ then $\left|z\right|^{-x}\geq\left|z\right|$ and
$\left|z\right|^{-y}\geq\left|z\right|$. Now, because $z^x+\bar z^y=1$, we have
$z^{-x}+\bar z^{-y}=z^{-x}\bar z^{-y}$ and by triangle inequality we get
$\left|z\right|^{-(x+y)}=\left|z^{-x}\bar z^{-y}\right|=
\left|z^{-x}+\bar z^{-y}\right|\leq\left|z\right|^{-x}+\left|z\right|^{-y}$
which gives $(\left|z\right|^{-x}-1)(\left|z\right|^{-y}-1)\leq1$. Therefore,
$(\left|z\right|-1)^2\leq1$ which yields $\left|z\right|\leq2$.
Next, we show that $\left|z\right|\neq1$. If $\left|z\right|=1$ then $z=\pm1$ or
$z=\pm i$. In the former case, $z^x+\bar z^y$ is an even integer and can't be
equal to $1$. In the latter case, if $x$ and $y$ have the same parity, then
$z^x+\bar z^y$ is an even multiple of $i$, and if they have different parities,
then $z^x+\bar z^y$ will have a nonzero imaginary part. So in either case, the
equation $z^x+\bar z^y=1$ won't hold. So the only possible values for $z$ are
$\pm2$, $\pm2i$, $\pm(1+i)$ and $\pm(1-i)$.
Now, we're ready for determining all the possible solutions.
*
*If $x>0$ and $y>0$ then $\left|z\right|$ must be equal to $1$ as we proved before. But we also proved this is impossible.
*If $x>0$ and $y<0$ then $\left|z\right|^y<1$ because $\left|z\right|^y
\neq1$. So in the inequality $\left|z\right|^y\leq2$ that was derived before,
the equality can't hold. Therefore $z$ is equal to $\pm(1+i)$ or $\pm(1-i)$ and
we have $\left|z\right|=\sqrt{2}$. Now, if $x\geq2$ then $2\leq\left|z\right|^x
\leq\left|z^x+\bar z^y\right|+\left|z\right|^y\leq1+\frac{1}{\sqrt{2}}$ which
leads to a contradiction. So the only possible value for $x$ is $1$. This means
$\Im(z^x)=\pm1$. On the other hand, $\left|\Im(\bar z^y)\right|\leq
\left|z\right|^y\leq\frac{1}{\sqrt{2}}$ so $\Im(z^x+\bar z^y)\neq0$ and
therefore there are no possible solutions in this case either. The case $x<0$
and $y>0$ is similar.
*If $x<0$ and $y<0$ and $\left|z\right|=2$ then
$\left|z\right|^x\leq\frac{1}{2}$ and $\left|z\right|^y\leq\frac{1}{2}$ and
equality holds iff $x=-1$ and $y=-1$ respectively. By triangle inequality we get
$\left|z^x+\bar z^y\right|\leq\left|z\right|^x+\left|z\right|^y\leq1$ so the
equalities must hold. By simple calculation, we see that only the case $z=2$
gives a solution.
If $x<0$ and $y<0$ and $\left|z\right|=\sqrt{2}$ then
$\left|z\right|^x\leq\frac{1}{\sqrt{2}}$ and
$\left|z\right|^y\leq\frac{1}{\sqrt{2}}$. If $x$ is equal to $-1$ then $y$ must
be equal to $-1$ too. Because $\left|\Re(z^x)\right|$ would be equal to
$\frac{1}{2}$ but if $y=-2$ then $\Re(\bar z^y)=0$ and if $y<-2$ then
$\left|\Re(\bar z^y)\right|\leq\left|z\right|^y=\frac{1}{2\sqrt{2}}$ and in
either case $z^x+\bar z^y\neq1$ (Similarly if $y=-1$ then $x=-1$). Because
$\Re(z^{-2})=\Re(\bar z^{-2})=0$ the case $x=y=-2$ is impossible. If $x\leq-2$
and $y\leq-3$ then because $\left|z^x\right|+\left|\bar z^y\right|\leq
\frac{1}{2}+\frac{1}{2\sqrt{2}}<1$, we won't have a solution and the case
$x\leq-3$ and $y\leq-2$ is similar. So in the only possible solutions, we have
$x=y=-1$. By simple calculation, we see that only $z=1\pm i$ gives a solution.
Hence, the only nontrivial solutions are $(x,y,z)=(-1,-1,2)$ and
$(x,y,z)=(-1,-1,1\pm i)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1308206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim\limits_{n\to\infty }\frac{1+\frac12+\frac13+\cdots+\frac1n}{1+\frac13+\frac15+\cdots+\frac1{2n+1}}$ I need to compute: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}$.
My Attempt: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}=\lim_{n\rightarrow \infty }\frac{2s}{s}=2$.
Is that ok?
Thanks.
| I don't know if there's still any interest in this question from more than a year ago, but here's an elementary solution:
The numerator in the problem is $H_n=1+\frac12+\frac13+\dots+\frac{1}{n},$ the $n^{\text{th}}$ harmonic number.
The denominator in the problem is
\begin{align}
1+\frac13+\frac15+\dots+\frac1{2n+1}&=H_{2n+1}-\big(\frac12+\frac14+\frac16+\dots+\frac{1}{2n}\big)
\\&=H_{2n+1}-\frac12 \big(1+\frac12+\frac13+\dots+\frac{1}{n}\big)
\\&=H_{2n+1}-\frac12 H_n.
\end{align}
Next note that
\begin{align}
H_{2n+1}-H_n&=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\dots+\frac{1}{2n+1}
\\&\le\frac{1}{n+1}+\frac{1}{n+1}+\frac{1}{n+1}+\dots+\frac{1}{n+1} \scriptsize{\quad(n+1\text{ terms})}
\\&=1.
\end{align}
Of course, $H_{2n+1}-H_n$ is always positive, being a sum of one or more positive fractions, so we have
$$0\lt H_{2n+1}-H_n \le 1$$
for all non-negative integers $n.$
Since $H_n\to\infty$ as $n\to\infty,$ it follows that
$$
\lim_{n\to\infty}\frac{H_{2n+1}-H_n}{H_n}=0.
$$
We can now solve the original problem:
\begin{align}
\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}&=\lim_{n\to\infty}\frac{H_n}{H_{2n+1}-\frac12 H_n}
\\&=\lim_{n\to\infty}\frac{H_n}{\frac12 H_n+(H_{2n+1}-H_n)}
\\&=\lim_{n\to\infty}\frac{1}{\frac12+\frac{H_{2n+1}-H_n}{H_n}}
\\&=\frac{1}{\frac12+\lim_{n\to\infty}\frac{H_{2n+1}-H_n}{H_n}}
\\&=\frac{1}{\frac12+0}
\\&=2.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
} |
Partial Fraction Using the Long Method I've been trying to express $$\frac{x^2-13}{x^3-7x+6}$$ as a partial fraction, and, so far I have arrived at $$x^2-13=A(x-2)(x+3)+B(x-1)(x+3)+C(x-1)(x-2)$$.
From this onwards, I substituted different values of x to remove unknowns and find the values of the unknowns individually. Yet, in my textbook,
$$1= B+C \tag 1$$ (for the coefficients of the squared x term)
$$0=A+2B-3C \tag 2$$ (for the coefficients of the non-squared terms)
$$-13=-6A-3B+2C\tag 3$$ (for the constants)
What I have trouble with is understanding how 1) is derived, for, from my perspective, it should be $$1=A+B+C$$. Also, I was wondering if these three equations can be solved simultaneously by using Gaussian elimination, or simply, simultaneously solve two of the equations and then solve the remaining equation form he remaining unknowns.
| As you have noticed, since $x^3-7x+6= (x-1)(x^2+x-6)=(x-1)(x-2)(x+3) $ we can write $$\frac{x^2-13}{x^3-7x+6}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x+3}$$ to get $$x^2-13=A(x-2)(x+3)+B(x-1)(x+3)+C(x-1)(x-2).$$ Then, to solve this equation for $A, B, C$, it is easier to plug in $x=1,2,-3$, i.e., the roots of denominator, to obtain:
$$x=1 \Rightarrow -12=-4A$$
$$x=2 \Rightarrow -9=5B$$
$$x=-3 \Rightarrow -4=20C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
| Define the $2$-adic valuation $\nu_2(r)$ of a nonzero rational number $r = \frac{p}{q}$ to be the number of times $2$ divides $p$ minus the number of times $2$ divides $q$. The $2$-adic valuation of the square of a rational number is even. But the $2$-adic valuation of $2$ is odd. Hence $2$ is not the square of a rational number.
This argument generalizes with no difficulty to the following: if $n$ is a positive integer, then $\sqrt[k]{n}$ is rational iff the $p$-adic valuation $\nu_p(n)$ of $n$ is always divisible by $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 16
} |
Equality with dilogarithms During some calculations with definite integrals I happened to get the equality
\begin{eqnarray}
2\, \textrm{Li}_2(-\frac{1}{2}) - 2 \, \textrm{Li}_2(\frac{1}{4})+ 2\, \textrm{Li}_2(\frac{2}{3})=
3 \log^2 2 - \log^2 3
\end{eqnarray}
Does this follow from some well known equalities for dilogarithms?
| I'm going to use the following 3 identities along with the known value $\text{Li}_{2} \left(\frac{1}{2} \right) = \frac{\pi^{2}}{12} - \frac{1}{2} \log^{2}(2)$:
$$\text{Li}_{2}(1-z) = - \text{Li}_{2} \left(1- \frac{1}{z} \right) - \frac{1}{2} \log^{2} (z) , \quad z \notin (-\infty,0] \tag{1}$$
$$\text{Li}_{2}(z) = - \text{Li}_{2} \left(\frac{1}{z} \right) - \frac{\pi^{2}}{6} - \frac{1}{2} \log^{2}(-z) \tag{2}, \quad z \notin [0,1)$$
$$ \text{Li}_{2}(z) = -\text{Li}_{2}(-z) + \frac{1}{2} \text{Li}_{2}(z^{2}) \tag{3}$$
Using $(1)$, which I think is sometimes referred to as Landen's identity, we have
$$\text{Li}_{2} \left(\frac{2}{3} \right) = \text{Li}_{2} \left(1- \frac{1}{3} \right) = - \text{Li}_{2} (-2) - \frac{1}{2} \log^{2} (3).$$
Then using $(2)$ (i.e., the inversion formula),
$$ \text{Li}_{2} \left(\frac{2}{3} \right) = \text{Li}_{2} \left(- \frac{1}{2} \right) + \frac{\pi^{2}}{6} + \frac{1}{2} \log^{2}(2) - \frac{1}{2} \log^{2}(3).$$
And using $(3)$ we have
$$ \text{Li}_{2} \left(- \frac{1}{2} \right) = -\text{Li}_{2} \left(\frac{1}{2} \right) + \frac{1}{2} \text{Li}_{2} \left(\frac{1}{4} \right) = -\frac{\pi^{2}}{12} + \frac{1}{2} \log^{2}(2) + \frac{1}{2} \text{Li}_{2} \left(\frac{1}{4} \right).$$
Therefore,
$$ \begin{align} &2 \text{Li}_{2} \left(- \frac{1}{2} \right) -2 \text{Li}_{2} \left(\frac{1}{4} \right) + 2 \text{Li}_{2} \left(\frac{2}{3} \right) \\ &= 4 \text{Li}_{2} \left(-\frac{1}{2} \right) - 2 \text{Li}_{2} \left(\frac{1}{4} \right)+ \frac{\pi^{2}}{3} + \log^{2}(2) - \log^{2}(3) \\ &= -\frac{\pi^{2}}{3} + 2 \log^{2}(2) + 2 \text{Li}_{2} \left(\frac{1}{4} \right) - 2 \text{Li}_{2} \left(\frac{1}{4} \right) + \frac{\pi^{2}}{3} + \log^{2}(2) - \log^{2}(3) \\ &= 3 \log^{2}(2) - \log^{2}(3). \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Integral of Trigonometric Identities $$\int(\sin(x))^3(\cos(2x))^2dx$$
I can write $$\sin^3(x)=\sin(x)(1-\cos^2(x)=\sin(x)-\sin(x)\cos^2(x)$$
for $$\cos^2(2x)=(1-\sin^2(x))^2=1-4\sin^2(x)+4\sin^4(x)$$
after simplifying the Trig identities i get:
$$\int(sin^3(x)-4sin^5(x)+4sin^7(x))dx$$
so i need to know how to go further :)
| take your integrand $$\sin^3 x-4\sin^5 x+4\sin^7 x $$ and factor the $\sin x$ so you get $$ \sin x\left(\sin^2 x-4\sin^4 x+4\sin^6 x\right) = \sin x\left( (1-\cos^2 x)-4(1-\cos^2 x)^2+4(1-\cos^2 x)^3 \right)$$
now make the substitution $$u = \cos x , \quad du = -\sin x \,dx $$ so you have $$\int\left( \sin^3 x-4\sin^5 x+4\sin^7 x \right)\, dx = -\int\left(1-u^2-4(1-u^2)^2+4(1-u^2)^3\right)\, du$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find $(f^{-1})'(a) = f(x) = 2x^3 + 3x^2+7x+4, a=4 $ Find $(f^{-1})'(a) = f(x) = 2x^3 + 3x^2+7x+4, a=4 $
How do I go about solving a problem like this? What are the steps?
| first you need to find $x$ such that $f(x) = 4.$ you know there is exactly one such $x$ because $f$ has an inverse. by trial and error, you find that $x = 0.$ so $f$ sends $0$ to $4.$ the slope of $f$ at $x = 0$ is $f'(0) = 7.$ therefore the slope of $f^{-1}$ at the point $y = 4$ is $\left(f^{-1}\right)'(4) = \frac17,$ the reciprocal of the slope of $f$ at the point $x = 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
with inequality $\frac{y}{xy+2y+1}+\frac{z}{yz+2z+1}+\frac{x}{zx+2x+1}\le\frac{3}{4}$ Let $x,y,z\ge 0$, show that
$$\dfrac{y}{xy+2y+1}+\dfrac{z}{yz+2z+1}+\dfrac{x}{zx+2x+1}\le\dfrac{3}{4}$$
I had solve
$$\sum_{cyc}\dfrac{y}{xy+y+1}\le 1$$
becasuse After some simple computations, it is equivalent to
$$(1-xyz)^2\ge 0$$
| Because by C-S $$\sum_{cyc}\frac{y}{xy+2y+1}\leq\frac{1}{4}\sum_{cyc}y\left(\frac{1^2}{xy+y}+\frac{1^2}{y+1}\right)=\frac{1}{4}\sum_{cyc}\left(\frac{1}{x+1}+\frac{y}{y+1}\right)=\frac{3}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Convolution of binomial coefficients As part of a (SE) problem I've been working on, I came up with this expression:
$$
\sum_{i=0}^M\binom{M-1+i}{i}\binom{M+i}{i}
$$
I'd like to get a closed form for this, but after a considerable amount of time searching my references and online sources (not to mention the time I've spent bashing this into other equally opaque equivalences), I've come up empty. Does anyone have a clue? I'll be happy to link to the original if asked, but the expression more or less tells the story.
| Remark. The Iverson bracket that was used in the first answer is
not quite appropriate as the infinite series it was substituted into
does not converge in a neighborhood of zero.
The appropriate form of the Iverson bracket here is
$$[[0\le q\le M]] =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{z^{q}}{z^{M+1}}\frac{1}{1-z} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{M+1}}\frac{1}{1-z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{M}}{w^{M+1}}
\sum_{q\ge 0} {q+M-1\choose M-1} (1+w)^q z^q
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{M+1}}\frac{1}{1-z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{M}}{w^{M+1}}
\frac{1}{(1-z(1+w))^M}
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{M+1}}\frac{1}{1-z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{M}}{w^{M+1}}
\frac{1}{(1-z-zw)^M}
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{M+1}}\frac{1}{(1-z)^{M+1}}
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{M}}{w^{M+1}}
\frac{1}{(1-zw/(1-z))^M}
\; dw\; dz.$$
Extracting the residue at $w=0$ we obtain
$$\sum_{q=0}^M {M\choose M-q} {M-1+q\choose q} \frac{z^q}{(1-z)^q}.$$
Substituting this into the integral in $z$ yields
$$\sum_{q=0}^M {M\choose M-q} {M-1+q\choose q}
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{M-q+1}}\frac{1}{(1-z)^{M+q+1}} \; dz$$
which is
$$\sum_{q=0}^M {M\choose M-q} {M-1+q\choose q}
{M+q+M-q\choose M+q}
\\= \sum_{q=0}^M {M\choose q} {M-1+q\choose M-1}
{2M\choose M+q}.$$
This is the same as what we obtained in the first version and we may
continue as before.
Addendum.
We can also evaluate this using the negative of the residue at
$w=(1-z)/z$ of the inner integral. This requires the derivative
(use Leibniz' rule)
$$\frac{1}{(M-1)!}\left(\frac{(1+w)^M}{w^{M+1}}\right)^{(M-1)}
\\ = \frac{1}{(M-1)!}
\sum_{q=0}^{M-1}
{M-1\choose q} \frac{M! \times (1+w)^{M-q}}{(M-q)!}
(-1)^{M-1-q}
\frac{(M+M-1-q)!}{M! \times w^{M+1+M-1-q}}
\\= (-1)^{M-1} \sum_{q=0}^{M-1} {M\choose q} (1+w)^{M-q}
(-1)^q {2M-1-q\choose M} \frac{1}{w^{2M-q}}.$$
Evaluation yields
$$(-1)^{M-1} \sum_{q=0}^{M-1} {M\choose q} \frac{1}{z^{M-q}}
(-1)^q {2M-1-q\choose M} \frac{z^{2M-q}}{(1-z)^{2M-q}}.$$
Re-write the double integral as follows:
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2M+1}}\frac{1}{1-z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{M}}{w^{M+1}}
\frac{1}{((1-z)/z-w)^M}
\; dw\; dz
\\ = \frac{(-1)^M}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{2M+1}}\frac{1}{1-z}
\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{M}}{w^{M+1}}
\frac{1}{(w-(1-z)/z)^M}
\; dw\; dz.$$
Substitute the evaluated derivative into this integral to get
$$- \sum_{q=0}^{M-1} {M\choose q}
(-1)^q {2M-1-q\choose M}
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{M+1}}\frac{1}{(1-z)^{2M+1-q}} \; dz
\\ = - \sum_{q=0}^{M-1} {M\choose q}
(-1)^q {2M-1-q\choose M}
{2M-q+M\choose M}
\\ = - \sum_{q=0}^{M-1} {M\choose q}
(-1)^q {2M-1-q\choose M}
{3M-q\choose M}.$$
We still need the value of the residue at infinity of the integral in
$w$ using the following formula:
$$\mathrm{Res}_{z=\infty} h(z)
= \mathrm{Res}_{z=0}
\left[-\frac{1}{z^2} h\left(\frac{1}{z}\right)\right]$$
which in the present case yields
$$- \mathrm{Res}_{w=0}
\frac{1}{w^2} \frac{(1+1/w)^M}{1/w^{M+1}}
\frac{1}{(1-z/(1-z)/w)^M}
\\ = - \mathrm{Res}_{w=0}
\frac{1}{w} (1+w)^M
\frac{w^M}{(w-z/(1-z))^M}
\\ = - \mathrm{Res}_{w=0}
(1+w)^M
\frac{w^{M-1}}{(w-z/(1-z))^M}.$$
This is zero by inspection and since the residues sum to zero we may
conclude that the sum has the alternate representation
$$\sum_{q=0}^{M} {M\choose q}
(-1)^q {2M-1-q\choose M}
{3M-q\choose M}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Integrate $\int_{0}^{\pi/2}x\sqrt{\tan{x}}\log{(\sin{x})}\,\mathrm dx$ When trying to solve this problem: How to Integrate $ \int^{\pi/2}_{0} x \ln(\cos x) \sqrt{\tan x}\,dx$
I found his sister integral has an interesting closed form provided my calculation is correct. I use an ugly series to find it. Can you use other methods to evaluate the integral? Such as Gamma function or residue method?
$$\int_{0}^{\pi/2}x\sqrt{\tan{x}}\log{\sin{x}}\,\mathrm dx=-\frac{\pi\sqrt{2}}{48}\big(\pi^2+12\pi \log{2}+24\log^2{2}\big) $$
| Substitute $t = \sqrt{\tan x}$
\begin{align}
I=&\int_{0}^{\pi/2}x\sqrt{\tan{x}}\ln({\sin x})\ dx\\
= &\int_0^\infty \frac{t^2 \ln\frac{t^4}{1+t^4}\ \tan^{-1}t^2}{1+t^4}dt\\
=&\int_0^\infty \int_0^1 \frac{2y \ t^4 \ln\frac{t^4} {1+t^4}}{(1+t^4)(1+y^4 t^4)}dy \ dt\\
=&\int_0^1 \frac{2y}{1-y^4} \int_0^\infty\bigg(\frac{\ln\frac{t^4} {1+t^4}}{1+y^2t^4}- \frac{\ln\frac{t^4} {1+t^4}}{1+t^4} \bigg)dt \ dy
\end{align}
Utilize
$$J(a)= \int_0^\infty \frac{a \ln\frac{t^4} {1+t^4}}{1+a^4 t^4}dt
= -\frac\pi{\sqrt2}\left[\tan^{-1}a+\ln(1+a)+\frac12\ln(1+a^2) \right]
$$
to obtain
\begin{align}
I= &\int_0^1 \frac{2 [J(y)-yJ(1)] }{1-y^4}dy \\
=&\ \sqrt2\pi \int_0^1 \frac{\frac\pi4y-\tan^{-1}y}{1-y^4}
+\frac{y\ln2 -\ln(1+y)}{1-y^4}
+\frac{y\ln2 -\ln(1+y^2)}{2(1-y^4)}\ dy\\
=&\ \sqrt2\pi \bigg[ \left(G-\frac{\pi^2}{16}-\frac\pi4\ln2\right)+\left( \frac{\pi^2}{12} -\frac\pi4\ln2 -\ln^22\right)\\
&\>\>\>\>\>\>\>\>\>\>\>\>\>
+\frac12\left( 2G+ \frac{\pi^2}{8} -\pi\ln2-\ln^22\right)\bigg]\\
=& \ \frac{\pi}{\sqrt2}\left( G+ \frac{\pi^2}{24}-\frac\pi2\ln2
-\frac34\ln^22\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
trigonometric inequality bound? Suppose that one wants to determine an upper bound of the trigonometric expression
$$
a\sin(x)+b\cos(x),
$$
where $a,b\in\mathbb{R}$. My instinct is to proceed as follows:
$$
a\sin(x)+b\cos(x)\leq |a|+|b|,
$$
which is correct. If one were to drop the modulus signs, would that make the bound incorrect?
| We can write $a\sin x+b\cos x=C \sin(x+y)$ where $C=\sqrt{a^2+b^2}$ and $y=\arctan (b/a)$ with the arctangent function accounting for the individual signs of $a$ and $b$ (see Note).
Thus, the maximum is $C=\sqrt{a^2+b^2}$.
NOTE:
We can see that this is correct using only the addition angle formula for the sine function along with the identity $\sin^2 x+\cos^2 x=1$. Write
$$C\sin(x+y)=(C\cos y)\sin x+(C\sin y)\cos x \implies a=C\cos y \,\,\text{and,}\,\, b=C\sin y$$
Solving for $C$ we have $C^2=a^2+b^2\implies C=\sqrt{a^2+b^2}$.
Solving for $-\pi\le y\le \pi$ we have $\tan y=b/a \implies y=\arctan (b/a)$ for $a>0$, $y=\arctan(b/a)+\pi$ for $a<0$ and $b>0$, and $y=\arctan(b/a)-\pi$ for $a<0$ and $b<0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int_{\pi/6}^{\pi/2} (\frac{1}{2} \tan\frac{x}{2}+\frac{1}{4} \tan\frac{x}{4}+ \cdots+\frac{1}{2^n} \tan\frac{x}{2^n}+\cdots) dx$ $$f(x)=\frac{1}{2} \tan\frac{x}{2}+\frac{1}{4} \tan\frac{x}{4}+....+\frac{1}{2^n} \tan\frac{x}{2^n}+...$$
Check the function $f(x)$ is continuous on $[\frac{\pi}{6},\frac{\pi}{2}]$ and evaluate $\displaystyle\int_{\pi/6}^{\pi/2} f(x)$.
| I didn't manage to finish this problem, and I would post it as a comment, but it's too long and I figured that it might help someone to help the OP find an answer, so I have posted my solution thus far.
So we begin by noting that $\tan(x)$ is continuous in the region $[-\frac{\pi}{2},\frac{\pi}{2}]$. We further note that:
$$I_{n}=\frac{1}{2^{n}}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\tan\left(\frac{x}{2^{n}}\right)\:\mathrm{d}x$$
Can be simplified by using the substitution $u_{n} = \frac{x}{2^{n}}, \implies \mathrm{d}u_{n} = \frac{\mathrm{d}x}{2^{n}}$, we thus can write:
$$I_{n} = \int_{\frac{\pi}{3 \cdot 2^{n+1}}}^{\frac{\pi}{2^{n+1}}}\tan\left(u\right)\:\mathrm{d}u$$
And we have that $\forall n > 1 : [\frac{\pi}{3 \cdot 2^{n+1}},\frac{\pi}{2^{n+1}}]\subset[-\frac{\pi}{2},\frac{\pi}{2}]$, thus the function is a sum of continuous functions over the domain of integration and is thus continuous itself.
We note that:
$$I_{n} = \left[-\ln\left(\cos(u)\right)\right]_{\frac{\pi}{3 \cdot 2^{n+1}}}^{\frac{\pi}{2^{n+1}}} = \ln\left(\cos\left(\frac{\pi}{3 \cdot 2^{n+1}}\right)\right)-\ln\left(\cos\left(\frac{\pi}{2^{n+1}}\right)\right)$$
Using log rules:
$$I_{n} = \ln\left(\frac{\cos\left(\pi/(3\cdot 2^{n+1})\right)}{\cos(\pi/2^{n+1})}\right) = \ln\left(\frac{1}{2\cos\left(\frac{\pi}{3 \cdot 2^{n}}\right)-1}\right) = -\ln\left(2\cos\left(\frac{\pi}{3\cdot 2^{n}}\right)-1\right)$$
It is useful to check that our integral doesn't blow up at large $n$:
$$\lim_{n\to\infty}I_{n} = 0$$
We note that:
$$\int_{-\frac{\pi}{6}}^{\frac{\pi}{3}}f(x)\:\mathrm{d}x = \sum_{n=1}^{N}I_{n} = -\ln\left(\prod_{n=1}^{N}\left(2\cos\left(\frac{\pi}{3\cdot 2^{n}}\right)-1\right)\right)$$
Now, I wish to make use of the fact that:
$$\prod_{n=1}^{N}\cos\left(\theta_{n}\right) = \frac{1}{2^{N}}\sum_{e \in S}\cos\left(e_{1}\theta_{1} + \cdots + e_{N}\theta_{N}\right)$$
And:
$$\sum_{n=1}^{\infty}\frac{\pi}{3 \cdot 2^{n}} = \frac{\pi}{3}$$
It is interesting to note that the dominating term in the product would be $\cos(\pi/3) = \frac{1}{2}$ and that $-\ln\left(2^{-1}\right) = \ln(2)$, which is the answer obtained by following math110's method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing that $(a^2-a+1)(b^2-b+1)(c^2-c+1) \leq 7$ How can I show that $(a^2-a+1)(b^2-b+1)(c^2-c+1) \leq 7$ given that $a+b+c = 3?$
Attempt: Setting $x=2a-1, y=2b-1, z=2c-1,$ we obtain that
$[(2a-1)^2+3][(2b-1)^2+3][(2c-1)^2+3] \leq 448,$
$s=x+y+z = 3, q=xy+yz+zx,$ and $p=xyz.$
So $(x^2+y^2+z^2) = 9-2q,$ and $x^2y^2+y^2z^2+z^2x^2 = q^2-6p.$
It suffices to prove that $9(9-2q)+3(q^2-6p)+p^2+27 \leq 448,$ or equivalently, that $p^2-18p+3q^2-18q-340 \leq 0.$
Further note that $p \leq 1$ (by AM-GM).
How can I conclude?
| The LHS is convex in each of the variables $a,b,c$ and $a, b,c \in [0,3]$ with $a+b+c=3$. So the LHS gets maximised when $a,b,c \in \{0,3\}$.
It isn't hard to verify that the maximum is when any two variables are zero and the other $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of a function with exponentiation In my assignment I have to calculate the following limit:
$$\lim_{x \to 0} \left(1+\frac {1-\cos x} {x} \right)^\frac{1}{x}$$
According to wolfram alpha, the result is $\sqrt{e} $.
However in my calculations I got a different one. Can you please let me know where did I get it wrong?
Here's my solution:
First we represent the argument differently:
$$\lim_{x \to 0} \left(1+\frac {1-\cos x} {x} \right)^\frac{1}{x}= \lim_{x \to 0} \left(\left(1+\frac {1-\cos x} {x} \right)^x\right)^ {\frac{1}{x}\cdot {x}} $$
Now we use the heine theorem and write the following:
$$\lim_{ n \to \infty} \left(\left(1+\frac {1-\cos n} {n} \right)^n\right)^ {\frac{1}{n}\cdot {n}} = \lim_{ n \to \infty} \left(\left(1+\frac {1}{n}-\frac{\cos n}{n} \right)^n\right)^ {\frac{1}{n}\cdot {n}}, $$ where $\frac {1}{n}-\frac{\cos n}{n} \to 0$. Therefore $\lim_{ n \to \infty}\left(1+\frac {1}{n}-\frac{\cos n}{n} \right)^n\to e. $
Since $\frac{1} {n}\cdot n \to 1$, the whole argument approaches
$$e^1=e$$
Where am I wrong here?
P. S. another approach to solving this is welcome.
Thanks,
Alan
| The best approach is to take logs. If $L$ is the desired limit then we have
\begin{align}
\log L &= \log\left(\lim_{x \to 0}\left(1 + \frac{1 - \cos x}{x}\right)^{1/x}\right)\notag\\
&= \lim_{x \to 0}\log\left(1 + \frac{1 - \cos x}{x}\right)^{1/x}\text{ (by continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{1}{x}\log\left(1 + \frac{1 - \cos x}{x}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x}\log\left(1 + \frac{1 - \cos^{2} x}{x(1 + \cos x)}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x}\log\left(1 + \frac{\sin^{2}x}{x^{2}}\cdot\frac{x}{(1 + \cos x)}\right)\notag\\
&= \lim_{x \to 0}\frac{1}{x}\cdot\dfrac{\log\left(1 + \dfrac{\sin^{2}x}{x^{2}}\cdot\dfrac{x}{(1 + \cos x)}\right)}{\dfrac{\sin^{2}x}{x^{2}}\cdot\dfrac{x}{(1 + \cos x)}}\cdot\dfrac{\sin^{2}x}{x^{2}}\cdot\dfrac{x}{(1 + \cos x)}\notag\\
&= \lim_{x \to 0}\dfrac{\log\left(1 + \dfrac{\sin^{2}x}{x^{2}}\cdot\dfrac{x}{(1 + \cos x)}\right)}{\dfrac{\sin^{2}x}{x^{2}}\cdot\dfrac{x}{(1 + \cos x)}}\cdot1\cdot\dfrac{1}{(1 + \cos x)}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\dfrac{\log\left(1 + \dfrac{\sin^{2}x}{x^{2}}\cdot\dfrac{x}{(1 + \cos x)}\right)}{\dfrac{\sin^{2}x}{x^{2}}\cdot\dfrac{x}{(1 + \cos x)}}\notag\\
&= \frac{1}{2}\lim_{t \to 0}\frac{\log(1 + t)}{t}\text{ (putting }t = \dfrac{\sin^{2}x}{x^{2}}\cdot\dfrac{x}{(1 + \cos x)})\notag\\
&= \frac{1}{2}\notag
\end{align}
It is now clear that $L = e^{1/2} = \sqrt{e}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Summation of series in powers of x with certain combinations as coefficients How can I find the sum: $$\sum_{k=0}^{n} \binom{n-k}{k}x^{k}$$
Edit: The answer to this question is: $$\frac{{(1+\sqrt{1+4x})}^{n+1}-{(1-\sqrt{1+4x})}^{n+1}}{2^{n+1}\sqrt{1+4x}}$$ I don't know how to arrive at this answer.
| Introduce the generating function
$$f(z) = \sum_{n\ge 0} z^n \sum_{k=0}^n {n-k\choose k} x^k.$$
This becomes
$$f(z) = \sum_{k\ge 0} x^k
\sum_{n\ge k} z^n {n-k\choose k}
\\ = \sum_{k\ge 0} x^k
\sum_{n\ge 0} z^{n+k} {n\choose k}
= \sum_{k\ge 0} x^k
\sum_{n\ge k} z^{n+k} {n\choose k}
\\ = \sum_{k\ge 0} x^k
\sum_{n\ge 0} z^{n+2k} {n+k\choose k}
= \sum_{k\ge 0} x^k z^{2k}
\sum_{n\ge 0} z^{n} {n+k\choose k}
\\ = \sum_{k\ge 0} x^k z^{2k}
\frac{1}{(1-z)^{k+1}}
= \frac{1}{1-z} \frac{1}{1-xz^2/(1-z)}
\\ = \frac{1}{1-z-xz^2}
= -\frac{1/x}{z^2+z/x-1/x}.$$
Solving $z^2+z/x-1/x$ we obtain $$\rho_{1,2}
= \frac{-1 \pm\sqrt{1+4x}}{2x}$$
and we have that
$$f(z) = -\frac{1}{x}\frac{1}{(z-\rho_1)(z-\rho_2)}
\\ = -\frac{1}{x}
\frac{1}{\rho_1-\rho_2}
\left(\frac{1}{z-\rho_1}-\frac{1}{z-\rho_2}\right)
\\ = -\frac{1}{x}
\frac{1}{\rho_1-\rho_2}
\left(\frac{1}{\rho_1}\frac{1}{z/\rho_1-1}
-\frac{1}{\rho_2}\frac{1}{z/\rho_2-1}\right).$$
Extracting coefficients from this we obtain
$$-\frac{1}{x}
\frac{x}{\sqrt{1+4x}}
\left(-\frac{1}{\rho_1^{n+1}}
+\frac{1}{\rho_2^{n+1}}\right).$$
Now since $\rho_1\rho_2 = -1/x$ this becomes
$$-\frac{1}{\sqrt{1+4x}}
\left(-(-1)^{n+1} x^{n+1} \rho_2^{n+1}
+(-1)^{n+1} x^{n+1} \rho_1^{n+1}\right).$$
Observe that
$$(-1)^{n+1} x^{n+1} \rho_{1,2}^{n+1}
= \frac{(1\mp\sqrt{1+4x})^{n+1}}{2^{n+1}}$$
so this finally becomes
$$-\frac{1}{\sqrt{1+4x}}
\frac{(1-\sqrt{1+4x})^{n+1}-(1+\sqrt{1+4x})^{n+1}}
{2^{n+1}}
\\ = \frac{1}{\sqrt{1+4x}}
\frac{(1+\sqrt{1+4x})^{n+1}-(1-\sqrt{1+4x})^{n+1}}
{2^{n+1}}.$$
Alternate derivation of the generating function.
Introduce
$${n-k\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n-k}}{z^{k+1}} dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z}
\sum_{k=0}^n \frac{x^k}{(1+z)^k z^k} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{(1+z)^{n}}{z}
\frac{x^{n+1}/(1+z)^{n+1}/z^{n+1}-1}
{x/(1+z)/z-1} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{n+1}
\frac{x^{n+1}/(1+z)^{n+1}/z^{n+1}-1}
{x-z-z^2} \; dz.$$
We omit the part that does not contribute to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} (1+z)^{n+1}
\frac{x^{n+1}/(1+z)^{n+1}/z^{n+1}}
{x-z-z^2} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{x^{n+1}}{x-z-z^2} \; dz.$$
This is
$$[z^n] \frac{x^{n+1}}{x-z-z^2}
= x [z^n] x^{n} \frac{1}{x-z-z^2}
\\ = x [z^n] \frac{1}{x-zx-z^2x^2}
= [z^n] \frac{1}{1-z-z^2x}.$$
We have the same generating function as above, QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Using the triangle inequality to bound $\frac{x^3 + 3x + 1}{10-x^3}$ for $|x+1|<2$ How do I use the triangle inequality to bound the function $$f(x) = \frac{x^3 + 3x + 1}{10 - x^3}$$
on the interval $|x+1|<2$? I understand how the triangle inequality works, but using fractions with triangle inequality is confusing me.
| Hint.
Use $\vert \vert x \vert - \vert y \vert \vert \le \vert x-y \vert$ with $y=-1$. You'll get $\vert x \vert < 3$ with your hypothesis. Hence you can bound
$$\vert x^3 + 3x + 1 \vert \le \vert x \vert^3 + 3\vert x \vert + 1 < 37$$
Also for $x \le 0$ you have $-x^3 \ge 0$. Therefore for $x \le 0$: $0 \le \frac{1}{10 - x^3} \le \frac{1}{10}$. And for $x \ge 0$ as $|x+1|<2$ you have $0 \le x <1$ and also $0 \le x^3 <1$ hence $10-x^3 \ge 9$.
Finally you get:
$$\left\vert {(x^3 + 3x + 1)\over(10 - x^3)}\right\vert < \frac{37}{9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral ${\large\int}_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}$ How to prove the following conjectured identity?
$$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{\sqrt[4]6}{3\sqrt\pi}\Gamma^2\big(\tfrac14\big)\tag1$$
It holds numerically with precision of at least $1000$ decimal digits.
Are there any other integers under the radical except $7$ and $1$ that result in a nice closed form?
| This is a partial answer to the second question. Mathematica could evaluate
$$\int_0^\infty\frac{dx}{\sqrt[4]{a+\cosh x}},$$
in term of the following Appell function:
$$
F_1\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{5}{4},\sqrt{a^2-1}-a,\frac{1}{\sqrt{a^2-1}-a}\right).
$$
For $a=0$ and $a=1$ there is a closed-form of this Appell function, so we get
$$\int_0^\infty\frac{dx}{\sqrt[4]{\cosh x}} = \frac{4\sqrt{\pi}\,\Gamma\left(\frac{9}{8}\right)}{\Gamma\left(\frac{5}{8}\right)}$$
and
$$\int_0^\infty\frac{dx}{\sqrt[4]{1+\cosh x}} = \frac{\Gamma^2\left(\frac{1}{4}\right)}{2^{3/4}\sqrt{\pi}}.$$
Numerically I've got your conjectured form for $a=7$ too:
$$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{\sqrt[4]6}{3\sqrt\pi}\Gamma^2\left(\tfrac14\right),$$
or in term of elliptic $K$ function:
$$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{4\sqrt[4]2\sqrt[4]3}{3}K\left(\tfrac{\sqrt{2}}{2}\right).$$
A related, somehow generalized question is here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 4,
"answer_id": 0
} |
High School Trigonometry ( Law of cosine and sine) I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine:
Let $\alpha$, $\beta$ and $\gamma$ be the angles of arbitrary triangle with sides a, b and c respectively. Then $${b - 2a\cos\gamma \over a\sin\gamma} + {c-2b\cos\alpha \over b\sin\alpha} + {a - 2c\cos\beta \over c\sin\beta}$$ is equal to (answer is zero but I need steps).
| Step 1:
Write cosines relations for a triangle:
$$a^2=b^2+c^2-2bc \cos(A)$$
$$b^2=a^2+c^2-2ac \cos(B)$$
$$c^2=a^2+b^2-2ab \cos(C)$$
Step 2:
Write sinus Area relation for the triangle:
$$Area(ABC)=\frac{ab \sin(C)}{2}=\frac{bc \sin(A)}{2}=\frac{ac \sin(B)}{2}$$
Step 3:
Write cosines relations as you have in your question:
$$a^2-b^2=c(c-2b \cos(A))$$
$$c-2b \cos(A)=\frac{a^2-b^2}{c}$$
$$b^2-c^2=a(a-2c \cos(B))$$
$$a-2c \cos(B)=\frac{b^2-c^2}{a}$$
$$c^2-a^2=b(b-2a \cos(C))$$
$$b-2a \cos(C)=\frac{c^2-a^2}{b}$$
Step 4:
Combine with sinus Area relation
$$c-2b \cos(A)=\frac{a^2-b^2}{c}$$
$$\frac{c-2b \cos(A)}{b \sin(A) }=\frac{a^2-b^2}{c b \sin(A) }$$
$$a-2c \cos(B)=\frac{b^2-c^2}{a}$$
$$\frac{a-2c \cos(B)}{c \sin(B) }=\frac{b^2-c^2}{ac \sin(B)}$$
$$b-2a \cos(C)=\frac{c^2-a^2}{b}$$
$$\frac{b-2a \cos(C)}{a \sin(C) }=\frac{c^2-a^2}{ab \sin(C)}$$
Step 5:
Combine with sinus Area relation
$$\frac{c-2b \cos(A)}{b \sin(A) }=\frac{a^2-b^2}{2.Area(ABC) }$$
$$\frac{a-2c \cos(B)}{c \sin(B) }=\frac{b^2-c^2}{2.Area(ABC)}$$
$$\frac{b-2a \cos(C)}{a \sin(C) }=\frac{c^2-a^2}{2.Area(ABC)}$$
Step 6:
Add them
$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=\frac{a^2-b^2}{2.Area(ABC) }+\frac{b^2-c^2}{2.Area(ABC)}+\frac{c^2-a^2}{2.Area(ABC)}$$
$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=\frac{a^2-b^2+b^2-c^2+c^2-a^2}{2.Area(ABC) }$$
$$\frac{c-2b \cos(A)}{b \sin(A) }+\frac{a-2c \cos(B)}{c \sin(B) }+\frac{b-2a \cos(C)}{a \sin(C) }=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 7
} |
If $f(x)=8x^3+3x$ , $x\in\mathbb{R}$, how do I find $\lim_{x \to \infty}\frac {f^{-1}(8x)-f^{-1}(x)}{x^{1/3}}$? Let $f: \mathbb R \to \mathbb R$ be defined as $f(x)=8x^3+3x$. Then $f$ is continuous , strictly increasing, and
$\lim _{x\to \infty}f(x)=\infty , \lim_{x \to -\infty}f(x)=-\infty$ , so $f$ is injective and surjective . How do I compute
$$\lim_{x \to \infty}\dfrac {f^{-1}(8x)-f^{-1}(x)}{x^{1/3}}?$$
| I got something different than the one in the prevous answer:
$f^{-1}(8x)=z$ and $f^{-1}(x)=y$ so
$8x=8z^{3}+3z$ and $x=8y^{3}+3y$
Therefore,
$8z^{3}\left ( 1+\frac{3}{z^{2}} \right )=8x\Rightarrow z\left ( 1+\frac{3}{z^{2}} \right )^{1/3}=x^{1/3}$ and similarly
$2y\left ( 1+\frac{3}{8y^{2}} \right )^{1/3}=x^{1/3}$.
Now, substituting, then factoring and cancelling $x^{1/3}$, we get
$\dfrac{z-y}{x^{1/3}}=\dfrac{\dfrac{x^{1/3}}{\left ( 1+\dfrac{3}{z^{2}} \right )^{1/3}}-\left (\dfrac{ x^{1/3}}{2\left ( 1+\dfrac{3}{8y^{2}} \right )^{1/3}}\right )}{x^{1/3}}=\\\dfrac{1}{\left ( 1+\dfrac{3}{z^{2}} \right )^{1/3}}-\left (\dfrac{1}{2\left ( 1+\dfrac{3}{8y^{2}} \right )^{1/3}}\right )$.
As $x\rightarrow \infty $, $z\rightarrow \infty$ so upon taking limits, we obtain $1-\frac{1}{2}=\frac{1}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.” This site
states:
Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$
“The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.”
In other words, according to Example $1$: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}.$$
Should:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}$$
not be:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^3 +(n + 1)^3}{2^3}$$
as everything in the left-hand side is cubed?
| Argument:
Every $k^3$ is the sum of $k$ consecutive odd numbers.
E.g.: $8 = 3+5$, $\quad 27 = 7+9+11$, $\quad 64 = 13+15+17+19$.
The sum of $n$ consecutive $k^3$ numbers, starting from $k = 1$, is the sum of
$n(n+1)/2$ consecutive odd numbers.
But the sum of a number of consecutive odd numbers is the square of
that number.
Therefore, the sum of $n$ consecutive $k^3$ starting with $1$ is $[n(n+1)/2]^2$.
Elaboration:
It is well known that $k^2$ is the sum of $k$ consecutive odd numbers starting from $1$. But $k^3$, $k^4$, and $k^p$ generally, are also the sum of $k$ consecutive odd numbers. This is because $k$ numbers whose arithmetic mean is $k$ sum to $k^2$; $k$ numbers whose mean is $k^2$ sum to $k^3$, et cetera. What is unique about the sum of cubes is that it is a sum of consecutive odd numbers with no gaps and no repetitions. E.g., the sum of the first five cubes is:
$$\underbrace{1^3}_{1} +\underbrace{2^3}_{3+5} +\underbrace{3^3}_{7+9+11} +\underbrace{4^3}_{13+15+17+19} +\underbrace{5^3}_{21+23+25+27+29}.$$
Therefore the sum is a square. But what square? In summing the first five cubes we have summed the first $1+2+3+4+5 = 15$ odd numbers, giving us $15^2$. And $15$ is the fifth triangle number. Hence generally, the sum of the first $n$ cubes is the square of the $n$th triangle number, or $[n(n+1)/2]^2$, i.e. the square of the sum of the first $n$ integers.
With sums of squares, there is repetition because all the sums of odd numbers start from $1$:
$$1^2 +2^2 +3^2 +4^2 = 1 +(1+3) +(1+3+5) +(1+3+5+7).$$
With fourth and higher powers there is no repetition, but there are gaps between the sets of odd numbers being summed:
$$1^4 +2^4 +3^4 +4^4 = 1 +(7+9) +(25+27+29) +(61+63+65+67).$$
Only sums of cubes are sums of consecutive odd numbers starting from $1$,
and therefore have this special connection with squares.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
How to compute the derivative of $\sqrt{x}^{\sqrt{x}}$? I know have the final answer and know I need to use the natural log but I'm confused about why that is.
Could someone walk through it step by step?
| Let $y= \sqrt{x}^{\sqrt x}$. Then $\ln(y) = \sqrt{x} \ln(\sqrt x) = \frac{1}{2} \sqrt x \ln(x)$. So,
$$
\frac{d}{dx} \ln(y) = \frac{d}{dx} \left(\frac{1}{2}\sqrt x\ln(x)\right)
$$
$$
\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left(\frac{\sqrt x}{x} + \frac{1}{2\sqrt x}\ln(x) \right)
$$
So,
$$
\frac{dy}{dx} = y \cdot \frac{1}{2} \left(\frac{\sqrt x}{x} + \frac{1}{2\sqrt x}\ln(x) \right) = \frac{1}{2}\sqrt{x}^{\sqrt x}\left(\frac{\sqrt x}{x} + \frac{1}{2\sqrt x}\ln(x) \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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List the elements of the field $K = \mathbb{Z}_2[x]/f(x)$ where $f(x)=x^5+x^4+1$ and is irreducible Since $\dim_{\mathbb{Z}_2} K = \deg f(x)=5$, $K$ has $2^5=32$ elements. So constructing the field $K$, I get:
\begin{array}{|c|c|c|}
\hline \text{polynomial} & \text{power of $x$} & \text{logarithm} \\\hline
0 & 0 & -\infty \\\hline
1 & 1 & 0 \\\hline
x & x & 1 \\\hline
x^2 & x^2 & 2 \\\hline
x^3 & x^3 & 3 \\\hline
x^4 & x^4 & 4 \\\hline
x^4+1 & x^5 & 5 \\\hline
x^4+x+1 & x^6 & 6 \\\hline
x^4+x^2+x+1 & x^7 & 7 \\\hline
x^4+x^3+x^2+x+1 & x^8 & 8 \\\hline
x^3+x^2+x+1 & x^9 & 9 \\\hline
x^4+x^3+x^2+x & x^{10} & 10 \\\hline
x^3+x^2+1 & x^{11} & 11 \\\hline
x^4+x^3+x & x^{12} & 12 \\\hline
x^2+1 & x^{13} & 13 \\\hline
x^3+x & x^{14} & 14 \\\hline
x^4+x^2 & x^{15} & 15 \\\hline
x^4+x^3+1 & x^{16} & 16 \\\hline
x+1 & x^{17} & 17 \\\hline
x^2+x & x^{18} & 18 \\\hline
x^3+x^2 & x^{19} & 19 \\\hline
x^4+x^3 & x^{20} & 20 \\\hline
\end{array}
Have I constructed the field correctly? I'm sure the claim that $K$ has $32$ elements is true, yet when I actually construct the field, I only get $22$ elements as $x^{21} = x^0 = 1$.
| The reason you have a problem is that $x^5+x^4+1$ is not irreducible in $\mathbb{Z}_2[x]$.
Thus, if $R=\mathbb{Z}_2[x]/(x^5+x^4+1)$, then $R^\times$ is not (as you would have expected if the polynomial were irreducible) a cyclic group of order $31$, so the fact that $x^{21}\equiv 1\bmod (x^5+x^4+1)$ is not surprising at all.
In fact, $x^5+x^4+1=(x^2+x+1)(x^3+x+1)$ in $\mathbb{Z}_2[x]$, so that
$$\mathbb{Z}_2[x]/(x^5+x^4+1)\;\cong\;\mathbb{Z}_2[x]/(x^2+x+1)\oplus \mathbb{Z}_2[x]/(x^3+x+1)\;\cong\;\mathbb{F}_4\oplus\mathbb{F}_8$$
so that for unit groups,
$$\mathbb{Z}_2[x]/(x^5+x^4+1)^\times\;\cong\;\mathbb{F}_4^\times\oplus\mathbb{F}_8^\times\;\cong\;\mathbb{Z}/3\mathbb{Z}\oplus\mathbb{Z}/7\mathbb{Z}\;\cong\;\mathbb{Z}/21\mathbb{Z}$$
Again, the ring $R=\mathbb{Z}_2[x]/(x^5+x^4+1)$, although having $32$ elements, is not a field, and therefore the unit group $R^\times$ is smaller than $R\setminus\{0\}$.
Even though you (completely coincidentally) chose a polynomial $x^5+x^4+1$ such that $x\in R^\times$, and such that $R^\times$ was cyclic, and such that $x$ was in fact a generator for $R^\times$, it could not help the fact that since $R^\times$ has $21$ elements and $R\setminus\{0\}$ has $31$ elements, you cannot get all the elements of $R\setminus\{0\}$ by successive powers of $x$.
Now, as to how to list elements.
In general, if $f\in \mathbb{Z}_p[x]$ is any polynomial of degree $n$, whether it is irreducible or not, then the $p^n$ elements of $\mathbb{Z}_p[x]/(f)$ are easily produced as the cosets
$$a_0+a_1x+\cdots+a_{n-1}x^{n-1}+(f),\qquad a_0,a_1,\ldots,a_{n-1}\in\mathbb{Z}_p$$
Of course, $f$ is irreducible $\iff$ $\mathbb{Z}_p[x]/(f)$ is a field, but the elements of $\mathbb{Z}_p[x]/(f)$ are always simple to write down. It's the choice of $f$ that determines the additive and multiplicative structure.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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problem in solving this problem from olympiad(use of invariant) Start with the set $\{3, 4, 12\}$. In each step you may choose two of the numbers $a$, $b$
and replace them by $0.6a − 0.8b$ and $0.8a + 0.6b$. Can you reach $\{4, 6, 12\}$
in finitely many steps:
Invariant here is that $a^2+b^2$ remains constant. Till here I am good
This part of the solution i didn't understand.
Since $a^2 +b^2 +c^2= 3^2 +4^2 +12^2 =13^2$ ,
the point $(a, b, c)$ lies on the sphere around $O$ with radius $13$. Because $4^2 +6^2 +12^2=
14^2$ , the goal lies on the sphere around $O$ with radius $14$. The goal cannot be reached.
please explain
Will it make any difference if z-points are different
| Well, $c$ does not change, so since $a^2+b^2$ does not change and $c$ does not change either we conclude $a^2+b^2+c^2$ does not change. In $(3,4,12)$ We have $a^2+b^2+c^2=169=13^2$.
On the other hand in $(4,6,12)$ we have $a^2+b^2+c^2=196=14^2$. Since $196$ is different from $169$ we cannot reach the desired target.
You can understand the proof without talking about spheres, it is redundant, they where just using the fact that the points $(x,y,z)$ on the sphere are those in which $\sqrt{x^2+y^2+z^2}=13$
| {
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"url": "https://math.stackexchange.com/questions/1330240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to evaluate $\int { \frac { dx }{ { 2x }^{ 4 }+{ 2x }^{ 2 }+1 } }$? How to integrate something like
$\int { \frac { dx }{ { 2x }^{ 4 }+{ 2x }^{ 2 }+1 } }$
I know the method of partial fractions.
But what if the denominator is not factorizeable like the one in the question?
| First pull out any constant coefficient of $x^4$ to get a monic polynomial.
In $$ x^4 + B x^2 + C $$ with $$ B^2 - 4 C < 0, $$ so that $C > 0,$ we get
$$ x^4 + B x^2 + C = (x^2 + \lambda x + \sqrt C) (x^2 - \lambda x + \sqrt C), $$
where
$$ \lambda = \sqrt {2 \sqrt C - B} $$
is also real.
The original problem was edited to $2 x^4 + 2 x^2 + 1 = 2 \left(x^4 + x^2 + \frac{1}{2} \right);$
$$ x^4 + x^2 + \frac{1}{2} = \left(x^2 + \sqrt{\sqrt 2 - 1} \; x + \frac{1}{\sqrt 2} \right) \left(x^2 - \sqrt{\sqrt 2 - 1} \; x + \frac{1}{\sqrt 2} \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the area remaining after flipping a rectangle inside a rectangle Let $r$ be the inside rectangle of base $b$ and height $h$.
Let $R$ be the outside rectangle of base $B$ and height $H$
The dimensions of $r$ and $R$ are related in the following way:
I want to find the area left when you flip $r$ against the walls of the outside triangle until its back in it's original position, here is a diagram of the area left after a few flips (orange region):
The pink rectangle indicates where the rectangle was before we started flipping it.
| The untouched area for $b=h$ (ie, when $B H$ is a square), is $$\text{ area}=b^2 \left(-2 \sqrt{3}-\sqrt{7}+\frac{2 \pi }{3}+7-4 \csc ^{-1}\left(\frac{4}{\sqrt{7}}\right)\right):$$
Here, $B:H=1:1.$ It is likely this is either optimal, or close to optimal.
The area $=0$ for $h\geq\left(\sqrt{7}-2\right) b$ - or, of course, for $b\geq\left(\sqrt{7}-2\right) h:$
Here, $B:H=1:\left(14+3 \sqrt{7}\right)/19$
Some Mathematica code to play with (example implementation: areaBH[2,2] and manipulateBH[2,2]).
Update
The total untouched area can therefore be calculated with:
Area $=$
If $\left(\sqrt{7}-2\right) b<h< \left(\sqrt{3}-1\right),$
\begin{align}{r}
\frac{1}{2} \left(h \left(-\sqrt{4 b^2+3 h^2}-2 \sqrt{b (3 b-4 h)}+4 h\right)+4 \left(b^2+h^2\right) \times\\
\left(\tan ^{-1}\left(\frac{\sqrt{b (3 b-4 h)}}{b+2 h}\right)-\tan ^{-1}\left(\frac{h}{\sqrt{4 b^2+3 h^2}}\right)\right)+2 b h-b \sqrt{b (3 b-4 h)}\right)\\
\end{align}
If $\left(\sqrt{3}-1\right) b<h<b/(\sqrt{3}-1),$
\begin{align}
&\frac{1}{2} \left(-b \left(\sqrt{3 b^2+4 h^2}-6 h\right)-h \left(\sqrt{4 b^2+3 h^2}+2 \left(\sqrt{3}-2\right) h\right)+4 \left(b^2+h^2\right) \times\\ \left(-\tan ^{-1}\left(\frac{h}{\sqrt{4 b^2+3 h^2}}\right)+\cot ^{-1}\left(\frac{b}{\sqrt{3 b^2+4 h^2}}\right)-\tan ^{-1}\left(\frac{b+\sqrt{3} h}{\sqrt{3} b-h}\right)+\tan ^{-1}\left(\frac{\sqrt{3} h-b}{\sqrt{3} b+h}\right)\right)-2 \left(\sqrt{3}-2\right) b^2\right)\\
\end{align}
If $b/(\sqrt{3}-1)<h<b/(\sqrt{7}-2),$
\begin{align}
b \left(-\frac{1}{2} \sqrt{3 b^2+4 h^2}-\sqrt{h (3 h-4 b)}+h\right)+2 \left(b^2+h^2\right) \left(\cot ^{-1}\left(\frac{b}{\sqrt{3 b^2+4 h^2}}\right)-\tan ^{-1}\left(\frac{2 b+h}{\sqrt{h (3 h-4 b)}}\right)\right)+2 b^2-\frac{1}{2} h \sqrt{h (3 h-4 b)}\\
\end{align}
otherwise, $0.$
Anyhow, the area that the OP asked for can be found with the functions areaBH[b,h] (which is slower) or fBH[b,h] (the formula given above) for any $b$ and $h.$
eg fBH[N[100],80] gives $470.616$ almost immediately, whereas fBH[100,80]//FullSimplify gives $-200 \left(82 \sqrt{3}+8 \sqrt{37}+5 \sqrt{139}+\frac{164 \pi }{3}-284-164 \cot ^{-1}\left(\frac{9}{\sqrt{5843-80 \sqrt{5143}}}\right)\right),$ which takes a little longer to calculate.
Note
If $\dfrac{\text{area}}{BH}$ is plotted instead of just $\text{area},$ the maximum value returns to $h=b$ as expected:
With[{b = 1}, Plot[fBH[b, h]/((2 h + b) (2 b + h)), {h, b (Sqrt@7 - 2),
b/(Sqrt@7 - 2)}, Filling -> Axis, Axes -> False, Frame -> True,
GridLines -> {{b, b (Sqrt@7 - 2), b/(Sqrt@7 - 2)}, {0, fBH[b, b]/((2 b + b)
(2 b + b))}}]]
so $b=h$ is indeed optimal, when scaled properly against $BH.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$ How can we find the value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$?
Note that $\sin^{-1}$ is the inverse sine function.
I'm asking for the solution $x$ for this equation.
Please work out the solution.
| My answer is similar to @juantheron.
First we know that the equation is valid for $0\le\theta\le\dfrac{\pi}2,$ for some $\theta$ in a right triangle. Now taking the cosine of both sides we have:
$$x=\cos(\sin^{-1}(x)+\sin^{-1}(1-x).$$
Using the angle sum formula for cosine and the Pythagorean Theorem, we get
\begin{align}
x&=\cos(\sin^{-1}(x))\cos(\sin^{-1}(1-x))-\sin(\sin^{-1}(x))\sin(\sin^{-1}(1-x))\\
&=\sqrt{1-x^2}\sqrt{1-(1-x)^2}-x(1-x)\\
&=\sqrt{(1-x^2)(2x-x^2)}-x(1-x)\\
2x-x^2&=\sqrt{(1-x^2)(2x-x^2)}\\
\end{align}
Now, noting that $x=0$ is a solution, the remaining solutions come from
\begin{align}
\sqrt{2x-x^2}&=\sqrt{1-x^2}\\
2x&=1\\
x&=\dfrac12.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/1333609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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solve system equation: $ a \cdot b = 3 \cdot a-b+1, b \cdot c = 3 \cdot b - c + 1, c \cdot a = 3 \cdot c - a + 1$ I want to solve this system of equations but i'm stuck.
Here is it:
$$ a \cdot b = 3 \cdot a - b + 1 $$
$$ b \cdot c = 3 \cdot b - c + 1 $$
$$ c \cdot a = 3 \cdot c - a + 1 $$
| Hint: If $a\neq-1\wedge b\neq-1\wedge c\neq -1$
$$\begin{cases}b=\frac{3a+1}{1+a}\\c=\frac{3b+1}{1+b}\\a=\frac{3c+1}{1+c}\end{cases}$$
What happens if you substitute (1) in (2) and, after that, (2) in (3)?
| {
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Fourier sine transform of $\frac{1}{2}+\frac{1-x^2}{4x}\ln\vert\frac{1+x}{1-x}\vert$ Show that
$$
\int_0^{\infty} kF(k)\sin(ka)\,dk = \frac{\pi}{2}aG(a)
$$
where
$$
F(x) = \frac{1}{2}+\frac{1-x^2}{4x}\ln\vert\frac{1+x}{1-x}\vert
$$
and
$$
G(x) = \frac{\sin x-x\cos x}{x^4}
$$
EDIT: The source can be found here. One should notice that the function $F(x)$ is not continuous at $x=1$.
EDIT2: The integral below may be of some help.
$$
\ln\vert\frac{a+x}{a-x}\vert = 2\int_0^{\infty}\frac{\sin at\sin xt}{t}\,dt
$$
EDIT3: Here is a mathematica code of my question:
F[x_] := (1/2 + (1 - x^2)/(4 x)*Log[Abs[(1 + x)/(1 - x)]])xSin[a*x];
Integrate[F[x], {x, 0, Infinity}, Assumptions -> a > 0]
The result is:
([Pi] (-a Cos[a] + Sin[a]))/(2 a^3)
EDIT4: By virtue of Bessel function $J_{\frac{1}{2}}(z)=\sqrt{\frac{2}{\pi z}}\sin(z)$, the integral identity turns to be
$$
\int_0^{\infty} k^{\frac{3}{2}}F(k)J_{\frac{1}{2}}(ka)\,dk = \sqrt{\frac{\pi}{2}a}G(a)
$$
In this occasion, some refs are useful:
(1) Lin Q.G.: Infinite integrals involving Bessel functions by an improved approach of contour integration and the residue theorem.
(2) Lucas S.K.: Evaluating infinite integrals involving Bessel functions of arbitrary order.
| The integral can be written as
\begin{align}
I&=\frac{1}{2}\int_0^{\infty} k\sin(ka)\left[1+\frac{1-k^2}{2k}\ln\left|\frac{1+k}{1-k}\right|\right]\,dk\\
&=\frac{1}{2}\int_0^{\infty} \left( 1-k^2 \right)\sin(ka)\left[\frac k{1-k^2}+\frac{1}{2}\ln\left|\frac{1+k}{1-k}\right|\right]\,dk\\
&=\frac{1}{2}\left( 1+\frac{d^2}{da^2} \right)\int_0^{\infty} \sin(ka)\left[\frac k{1-k^2}+\frac{1}{2}\ln\left|\frac{1+k}{1-k}\right|\right]\,dk
\end{align}
Now, recognizing that
\begin{equation}
\frac k{1-k^2}+\frac{1}{2}\ln\left|\frac{1+k}{1-k}\right|=\frac{1}{2}\frac{d}{dk}\left( k\ln\left|\frac{1+k}{1-k}\right| -2\right)
\end{equation}
The function $g(k)= k\ln\left|\frac{1+k}{1-k}\right| -2$ is is such that $g'(k)$ is integrable, $g(k)\sim 2k^{-2}/3$ for $k\to\infty$ and $g(k)\sim -2$ for $k\to 0$.
It comes
\begin{align}
I&=\frac{1}{4}\left( 1+\frac{d^2}{da^2} \right)\int_0^{\infty} \sin(ka)\frac{d}{dk}\left( k\ln\left|\frac{1+k}{1-k}\right| -2\right)\,dk\\
&=-\frac{1}{4}\left( 1+\frac{d^2}{da^2} \right)a\int_0^{\infty} \cos(ka)\left[k\ln\left|\frac{1+k}{1-k}\right| -2\right]\,dk\\
&=-\frac{1}{4}\left( 1+\frac{d^2}{da^2} \right)a\frac{d}{da}\int_0^{\infty} \sin(ka)\left[\ln\left|\frac{1+k}{1-k}\right|-\frac{2}{k}\right] \,dk
\end{align}
We know (GR 17.33.35) that
\begin{equation}
\ln\left|\frac{1+k}{1-k}\right|=2\int_0^\infty\sin t\sin kt \frac{dt}{t}
\end{equation}
and
\begin{equation}
\int_0^{\infty} \frac{2\sin(ka)}{k}dk=\pi
\end{equation}
thus
\begin{equation}
I=-\frac{1}{2}\left( 1+\frac{d^2}{da^2} \right)a\frac{d}{da}\left[\int_0^{\infty} \sin(ka)\,dk\int_0^\infty\sin t\sin kt \frac{dt}{t}-\pi\right]
\end{equation}
The contribution of the constant term in the bracket vanishes as it does not depend on $a$. The double integral is the sine transform of the sine transform of $\sin t/t$. Finally,
\begin{align}
I&=-\frac{\pi}{4}\left( 1+\frac{d^2}{da^2} \right)a\frac{d}{da}\left( \frac{\sin a}{a} \right)\\
&=\frac{\pi}{2}\frac{\sin a -a\cos a}{a^3}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Finding an isomorphism between polyomial quotient rings Let $F_1 = \mathbb{Z}_5[x]/(x^2+x+1)$ and $F_2 = \mathbb{Z}_5[x]/(x^2+3)$. Note neither $x^2+x+1$ nor $x^2+3$ has a root in $\mathbb{Z}_5$, so that each of the above are fields of order 25, and hence they are isomorphic from elementary vector space theory, however I'm tasked with constructing an explicit isomorphism between the two fields, so I applied the technique I've learned here and elsewhere as follows.
First, let $\phi: F_1 \rightarrow F_2$ by $x \mapsto ax+b$ (as the base field is fixed it suffices to find the image of $x$). Then we should have that $\phi(x^2+x+1) = x^2 + 3$, and the left hand side reduces to $\phi(x)^2 + \phi(x) + 1$ as $\phi$ is a homomorphism. Simplifying, we have $(ax+b)^2 + (ax+b) + 1 = a^2x^2 + 2abx + b^2 + ax+b+1 = a^2x^2 + (2ab+a)x + (b^2+b+1) = x^2+3$.
However, this appears to be a problem. Notably, since $a^2 = 1$ mod 5, we have $a = 1,4$. Picking the former, we have $2b+1 = 0$ and thus $b = 2$, a problem as $2^2+2+1 = 7 \neq 1$ mod 5. Trying $a=4$, we have $3b+4 = 0$ so $b = 2$, and again the same problem.
Are there flaws in my technique or arithmetic that I don't see? If not, what other options do I have to attempt to construct the isomorphism?
| HINT:
You want $(\mathbb{Z}/5)[x]/(x^2+x+1) \to (\mathbb{Z}/5)[x]/(x^2+3)$, $x \mapsto a x + b$, so $x^2+x+1 \mapsto (ax+b)^2+ (ax + b) +1$, and you want the latter a multiple of $x^2+3$. Divide $(ax+b)^2+ (ax + b) +1$ by $x^2+3$ and you get the remainder
$$a\,(2 b+1)\, x + (b^2 + b+1 - 3 a^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the last two digits of $33^{100}$
Find the last two digits of $33^{100}$
By Euler's theorem, since $\gcd(33, 100)=1$, then $33^{\phi(100)}\equiv 1 \pmod{100}$. But $\phi(100)=\phi(5^2\times2^2)=40.$
So $33^{40}\equiv 1 \pmod{100}$
Then how to proceed?
With the suggestion of @Lucian:
$33^2\equiv-11 \pmod{100}$ then $33^{100}\equiv(-11)^{50}\pmod{100}\equiv (10+1)^{50}\pmod{100}$
By using the binomial expansion, we have:
$33^{100}\equiv (10^{50}+50\cdot 10^{49}+ \cdots + 50\cdot 10+1)\pmod{100}$
$\implies 33^{100}\equiv (50\cdot 10+1)\pmod{100}\equiv 01 \pmod{100}$
| You can use fast exponentiation: modulo $100$
$$33^2\equiv -11,\quad 33^4\equiv 21,\quad 33^8\equiv 441\equiv 41,\quad 33^{16}\equiv1681\equiv -19$$
whence $\,33^{20}\equiv -19\cdot 21 =-(20-1)(20+1)\equiv 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Find the least positive angle satisfying the trigonometric equation $\sin^3 x+\sin^3 2x+\sin^3 3x=(\sin x+\sin 2x+\sin 3x)^3$. I did solve the question, but my method is highly tedious. I combined the sin and then opened the cubic.... Is there some trick? Something I am missing? Thanks.
| For $\sin ax+\sin bx=0$
Method $\#1:$
$\sin ax+\sin bx=0\iff\sin ax=-\sin bx=\sin(-bx)$ as $\sin(-A)=-\sin A$
$\implies ax=n\pi+(-1)^n(-bx)$ where $n$ is any integer
If $n$ is even $=2m$(say), $ax+2m\pi-bx\iff x=\dfrac{2m\pi}{a+b}$
Similarly for odd $n=2m+1$(say)
Method $\#2:$
Using Prosthaphaeresis Formula, $$\sin ax+\sin bx=2\sin\dfrac{(a+b)x}2\cos\dfrac{(a-b)x}2$$
If $\sin\dfrac{(a+b)x}2=0, \dfrac{(a+b)x}2=p\pi$
and if $\cos\dfrac{(a-b)x}2=0,\dfrac{(a-b)x}2=\dfrac{(2q+1)\pi}2$ where $p,q$ are integers
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $x=(9+4\sqrt{5})^{48}=[x]+f$ . Find $x(1-f)$.
If $x=(9+4\sqrt{5})^{48}=[x]+f$, where $[x]$ is defined as integral part of $x$
and $f$ is a fraction, then $x(1-f)$ equals .
$\color{green}{a.)\ 1} \\
b.)\ \text{less than}\ 1 \\
c.)\ \text{more than}\ 1 \\
d.)\ \text{between}\ 1 \text{and }\ 2 \\
e.)\ \text{none of these}\ \\ $
This question looks scary from the get go.
I tried to go with pattern
$(9+4\sqrt{5})^{1}\approx 17.94=17+0.94 \implies 17(1-0.94)=1.02\\
(9+4\sqrt{5})^{2}\approx 321.99=321+0.99 \implies 321(1-0.99)=3.21\\ $
i don't know if i interpreted the question correctly .
I look for a short and simple way.
I have studied maths upto $12$th grade. Thanks.
| Consider
$$P=(9+4\sqrt 5)^{48}+(9-4\sqrt 5)^{48}.$$
Note that $P$ is an integer.
Now we have $0\lt 9-4\sqrt 5\lt 1$. Hence we have
$$0\lt (9-4\sqrt 5)^{48}\lt 1.$$
Hence, we have
$$x=(9+4\sqrt 5)^{48}=P-1+1-(9-4\sqrt 5)^{48}.$$
This implies that $f=1-(9-4\sqrt 5)^{48}$.
Thus, we have
$$x(1-f)=(9+4\sqrt 5)^{48}(9-4\sqrt 5)^{48}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
$ \lim_{n\to+\infty} \frac{1\times 3\times \ldots \times (2n+1)}{2\times 4\times \ldots\times 2n}\times\frac{1}{\sqrt{n}}$
Knowing that :
$$I_n=\int_0^{\frac{\pi}{2}}\cos^n(t) \, dt$$
$$I_{2n}=\frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}\times\dfrac{\pi}{2}\quad \forall n\geq 1$$
$$I_{n}\sim \sqrt{\dfrac{\pi}{2n}}$$
Calculate:
$$
\lim_{n\to+\infty} \frac{1\times 3\times \ldots \times (2n+1)}{2\times 4\times \ldots\times 2n}\times\dfrac{1}{\sqrt{n}}$$
Indeed,
$$I_{2n}=\frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}\times\dfrac{\pi}{2}\quad \forall n\geq 1 \\
\frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}=\dfrac{2}{\pi}\times I_{2n}$$
then
$$
\begin{align*}
\frac{1\times 3\times \cdots \times (2n+1)}{2\times 4\times \cdots\times 2n}\times\dfrac{1}{\sqrt{n}}&=\dfrac{2}{\pi}\times I_{2n}\times (2n+1)\times\dfrac{1}{\sqrt{n}}\\
&=\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{\sqrt{n}} \times \sqrt{ \dfrac{2n\times I^{2}_{2n}}{2n} } \\
&=\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ 2n\times I^{2}_{2n} } \\
\end{align*}
$$
or $$(2n)I^{2}_{2n}\sim \dfrac{\pi}{2}$$
then
\begin{align*}
\frac{1\times 3\times \cdots \times (2n+1)}{2\times 4\times \cdots\times 2n}\times\dfrac{1}{\sqrt{n}}&=\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ 2n\times I^{2}_{2n} } \\
&\sim\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ \dfrac{\pi}{2} } \\
\end{align*}
$$
I'm stuck here
i think that i can go ahead
\begin{align*}
\frac{1\times 3\times \cdots \times (2n+1)}{2\times 4\times \cdots\times 2n}\times\dfrac{1}{\sqrt{n}}&\sim \dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ \dfrac{\pi}{2} } \\
&\sim \dfrac{2}{\pi}\times\dfrac{2n}{ \sqrt{2}\times n} \times \sqrt{ \dfrac{\pi}{2} } \\
&\sim \frac{ 2\sqrt{\pi} }{\pi}=\dfrac{2}{\sqrt{\pi}}
\end{align*}
am i right ? if that so is there any other way
| We are given
$$I_{2n}=\frac{\pi}{2}\frac{(2n-1)!!}{(2n)!!} \tag 1$$
and
$$I_n\sim \sqrt{\frac{\pi}{2n}} \tag 2$$
From $(2)$ is trivial to see that
$$I_{2n}\sim \frac12 \sqrt{\frac{\pi}{n}} \tag 3$$
Then, using $(1)$ and $(3)$, we find that
$$\begin{align}
\frac{(2n+1)!!}{\sqrt{n}(2n)!!}&=\frac{(2n+1)(2n-1)!!}{\sqrt{n}(2n)!!}\\\\
&=\frac{2n+1}{\sqrt{n}}\left(\frac{(2n-1)!!}{(2n)!!}\right)\\\\
&=\frac{2n+1}{\sqrt{n}}\left(\frac{2}{\pi}I_{2n}\right)\\\\
&\sim \frac{2n+1}{\sqrt{n}}\left(\frac{2}{\pi}\frac12\sqrt{\frac{\pi}{n}}\right)\\\\
&=\frac{2n+1}{n\sqrt{\pi}}\\\\
&\to \frac{2}{\sqrt{\pi}}\,\,\text{as}\,\,n\to \infty
\end{align}$$
Thus, we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\frac{(2n+1)!!}{\sqrt{n}(2n)!!}=\frac{2}{\sqrt{\pi}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Differentiate the Function $f(x)= \sqrt{x} \ln x$ Differentiate the Function $f(x)= \sqrt{x} \ln x$
| $f(x) = \sqrt{x} \ln x$. Using the product rule we have that $$f'(x) = \ln x \cdot \frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x} + \sqrt{x}\cdot\frac{\mathrm{d}}{\mathrm{d}x} \ln x$$
Hence $$f'(x) = \frac{\ln x}{2\sqrt{x}} + \frac{\sqrt{x}}{x}.$$
Further simplification results in $$f'(x) = \frac{\ln x}{2\sqrt{x}} + \frac{2}{2\sqrt{x}} = \frac{\ln x + 2}{2\sqrt{x}}$$
Doing it your way: $$\ln f(x) = \ln\left(\sqrt{x} \ln x\right) = \frac{1}{2}\ln x + \ln (\ln x).$$
Then differentiating both sides with respect to $x$ yields $$\frac{f'(x)}{f(x)} = \frac{1}{2x} + \frac{1}{x\ln x}.$$
Multiplying both sides by $f(x)$ gives
$$f'(x) = \sqrt{x} \ln x \left(\frac{1}{2x} + \frac{1}{x \ln x}\right) = \frac{\ln x + 2}{2\sqrt{x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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A proof by induction and trigonometry Do you know how to prove that $\displaystyle\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$ using induction?
I have tried with $n = 1$ which gives $\cos \frac{x}{2} = \frac{\sin(nx)}{(2\sin1/2x)}$
I am not sure on how to expand with the trigonometric formulas.
With $n= p+1$ I get LHS: $\cos(2(n+1)-1)$ which I summaries to $\cos(2n+1)$ which should be $\cos 2n \cos 1-\sin 2n \sin1$ plus the RHS $\frac{\sin(nx)}{(2\sin 1/2x)}$
RHS $p+ 1 = \frac{\sin(n+1x)}{(2\sin 1/2x)}$
Any ideas on how to proceed would be very helpful.
| I see this is quite an old question but here is a solution that assumes only basic high school compound angle formulas:
If we assume that $\displaystyle\sum_{k=1}^n \cos\left(\frac{(2k-1)x}{2}\right)=\frac{\sin(nx)}{2\sin\left(\frac x 2\right)}$ then it follows that:
$$\begin{align}
\sum_{k=1}^n \cos\left(\frac{(2k-1)x}{2}\right)+\cos\left(\frac{(2n+1}{2}\right)&=\frac{\sin(nx)}{2\sin\left(\frac x 2\right)}+\cos\left(\frac{(2n+1)x}{2}\right)\\
\sum_{k=1}^{n+1} \cos\left(\frac{(2k-1)x}{2}\right)&=\frac{\sin nx + 2\sin\left(\frac x 2\right)\cos\left(nx+\frac x 2\right)}{2\sin\left(\frac x 2\right)}\\
&=\frac{\sin nx + 2\sin\left(\frac x 2\right)\left[\cos nx\cos\left(\frac x 2\right)-\sin nx\sin\left(\frac x 2\right)\right]}{2\sin\left(\frac x 2\right)}\\
&=\frac{\sin nx \left[1-2\sin^2\left(\frac x 2\right)\right]+2\sin\left(\frac x 2\right)\cos\left(\frac x 2\right)\cos nx}{2\sin\left(\frac x 2\right)}\\
&=\frac{\sin nx\cos x+\sin x\cos nx}{2\sin\left(\frac x 2\right)}\\
&=\frac{\sin(nx+x)}{2\sin\left(\frac x 2\right)}=\frac{\sin\left((n+1)x\right)}{2\sin\left(\frac x 2\right)}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve $(z^1+z^2+z^3+z^4)^3$ using Pascals Triangle? In an exercise it seems I must use Pascal's triangle to solve this $(z^1+z^2+z^3+z^4)^3$. The result would be $z^3 + 3z^4 + 6z^5 + 10z^ 6 + 12z^ 7 + 12z^ 8 + 10z^ 9 + 6z^ {10} + 3z^ {11} + z^{12}$. But how do I use the triangle to get to that result? Personally I can only solve things like $(x+y)^2$ and $(x+y)^3$.
Thanks for any tips that may be given.
| $$(z+z^2+z^3+z^4)^3 = z^3\cdot\left(\frac{1-z^4}{1-z}\right)^3=z^3(1-3z^4+3z^8-z^{12})\sum_{n\geq 0}\binom{n+2}{2}z^n$$
hence:
$$\begin{eqnarray*}(z+z^2+z^3+z^4)^3&=&(z^3-3z^7+3z^{11}-z^{15})\sum_{n\geq 0}\binom{n+2}{2}z^n\\&=&\sum_{n\geq 3}\binom{n-1}{2}z^n-3\sum_{n\geq 7}\binom{n-5}{2}+3\sum_{n\geq 11}\binom{n-9}{2}z^n-\sum_{n\geq 15}\binom{n-13}{2}z^n\\&=&\sum_{n= 3}^{12}\binom{n-1}{2}z^n-3\sum_{n=7}^{12}\binom{n-5}{2}z^n+3\sum_{n=11}^{12}\binom{n-9}{2}z^n\\&=&\frac{1}{2}\left(\sum_{n= 3}^{6}(n-1)(n-2)z^n-2\sum_{n=7}^{10}(n-4)(n-11)z^n+\sum_{n=11}^{12}(n-13)(n-14)z^n\right)\end{eqnarray*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Solve $n(n+1)(n+2)=6m^3$ in positive integers
How to find all positive integers $m,n$ such that $n(n+1)(n+2)=6m^3$ ?
I can see that $m=n=1$ is a solution, but is it the only solution ?
| The paper Rational approximation to algebraic numbers of small height: the Diophantine equation $\left|ax^n-by^n\right|=1$ by Michael A. Bennett (J. reine angew. Math. 535 (2001),1–49) shows that the equation in the title has at most one solution in positive integers $x,y$ for given positive integers $a,b,n$ with $n\ge3$.
It follows that $x=y=1$ is the only solution both of $\left|2x^3-y^3\right|=1$ and of $\left|2x^3-3y^3\right|=1$.
From now on, $n$ is the $n$ in the question, not in the paper.
$n$, $n+1$ and $n+2$ have no factors other than $2$ in common. It follows that except for factors of $2$ and one factor of $3$ to cancel the factor of $3$ in the factor $6$, $n$, $n+1$ and $n+2$ must all be perfect cubes. This allows us to exclude non-trivial solutions by case analysis:
If $n+1$ is even, it does not contain the single uncubed factor of $3$, since otherwise $n$ and $n+2$ would both be perfect cubes. It has a single uncubed factor of $2$ to cancel the factor of $2$ in the factor $6$, since $n$ and $n+2$ are odd. Thus $n+1=2x^3$, and one of $n$ and $n+2$ must be a perfect cube $y^3$, and with $\left|2x^3-y^3\right|=1$ it follows that $x=y=1$.
If $n$ and $n+2$ are even, one of them, say, $n$, is divisible by $2$ but not by $4$. It cannot also contain the single uncubed factor of $3$, since otherwise the other two would have to be perfect cubes. Thus $n=2x^3$, and $n+1$ is either $y^3$ or $3y^3$. Again it follows that $x=y=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Quadratic polynomials describe the diagonal lines in the Ulam-Spiral I'm trying to understand why is it possible to describe every diagonal line in the Ulam-Spiral with an quadratic polynomial $$2n\cdot(2n+b)+a = 4n^2 + 2nb +a$$ for $a, b \in \mathbb{N}$ and $n \in 0,1,\ldots$.
It seems to be true but why?
Wikipedia says: "The pattern also seems to appear even if the number at the center is not 1 (and can, in fact, be much larger than 1). This implies [WHY?] that there are many integer constants b and c such that the function: $4n^2+bn+c$ as $n$ counts up $\{1, 2, 3, ...\}$, a number of primes that is large by comparison with the proportion of primes among numbers of similar magnitude."
I can't find a source with a detailed explanation.
I found these equations:
So here is the solution:
\begin{align*}
y_t - y_{t+1} - (y_{t+1} - y_{t+2}) &= 8\\
y_t - 2y_{t+1} + y_{t+2} &= 8\\
y_{t+2} - 2y_{t+1} + y_t &= 8
\end{align*}
1) We solve $y_{t+2} - 2y_{t+1} + y_t = 0.$
Let $y_t = A\beta^t$ \begin{align*}
A\beta^{t+2} - 2A\beta^{t+1} + A\beta^t &= 0\\
A\beta^{t}\cdot (\beta^2 - 2\beta + 1) &= 0
\end{align*}
$\beta^2 - 2\beta + 1 = 0$ has two identical solutions $\beta_{1,2} = 1$.
So with $A_1$ and $A_2t$ we get $$y_t = A_1 + A_2t.$$
2) $1 + a_1 + a_2 = 0$ and $a_1 = -2$ so let $y_t = ct^2$
\begin{align*}
c\cdot(t+2)^2 - 2c\cdot(t+1)^2 + ct^2 &= 8\\
c\cdot\big(t^2+4t+4 - 2\cdot(t^2+2t+1) + t^2\big) &= 8\\
c\cdot(t^2+4t+4 - 2t^2-4t-2 + t^2) &= 8\\
2c &= 8\\
c &= 4
\end{align*}
So $y_t = 4t^2$
3) The complete solution is $$y_t = 4t^2 + A_2t + A_1.$$
The "exclusion lines" seem to be interesting too:
$$4n^2+n$$ $$4n^2+3n$$ $$4n^2+3n-1$$ $$4n^2-n$$ seem not to have any primes at all.
Useful website I found a bit late: http://ulamspiral.com
| Note that the first "ring" of numbers has just 1 number in it, the next ring has $9-1=8$ numbers, then $25-9=16$, $49-25=24$, $81-49=32$, and so on. These numbers (aside from the first) are increasing by 8. When you start somewhere and go out along a diagonal, with each step you increase by 8 more than you did with the previous step. That is, the second difference of the sequence as you go out along a diagonal is a constant 8; $a_{n+2}-2a_{n+1}+a_n=8$. And the general solution of that difference equation ("recurrence relation") is $a_n=4n^2+bn+c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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Find the measurement of line BD So I was trying to find the measurement of $BD$ I drew green lines to make myself some angles, the measurement $3$ is from the point A to C, If only I can line $AE$ or $CE$ then I will just use the cosine law to get $BE$ and $DE$, however I can't figure it out.
| Hint: Let's call $H_B$ the foot of the line from $B$ to $AC$ and $H_D$ the foot of the line from $D$ to $AC$
Then you have $AH_B=AB \cos 30$, $BH_B=AB \sin 30$
and $CH_D=CD \cos 60$, $DH_D=CD \sin 60$
Thales gives us: $\dfrac{EB}{ED}=\dfrac{EH_B}{EH_D}=\dfrac{BH_B}{DH_D}$
But $H_BH_D=AC-AH_B-CH_D=H_BE+EH_D$
You can play with everything from now on...
EDIT: completing the work...
$H_BH_D=AC-AH_B-CH_D=3-AB \cos 30 -CD \cos 60=3-\dfrac{3\sqrt 3}{4}-2*\dfrac 12=\dfrac{8-3\sqrt 3}{4}$
$H_BE=\dfrac{H_BH_D}{1+\dfrac{DH_D}{BH_B}}=\dfrac{\dfrac{8-3\sqrt 3}{4}}{1+\dfrac{CD\sin 60}{AB \sin 30}}=\dfrac{\dfrac{8-3\sqrt 3}{4}}{1+\dfrac{4\sqrt 3}{3}}=\dfrac{24-9\sqrt 3}{12+16\sqrt 3}=\dfrac 34 \times \dfrac{8-3\sqrt 3}{3+4\sqrt 3}$
And $EH_D=\dfrac{4\sqrt 3}{3}EH_B=\sqrt 3 \times \dfrac{8-3\sqrt 3}{3+4\sqrt 3} $
$EB^2=BH_B^2+EH_B^2=\dfrac{9}{16}\times (1+(\dfrac{8-3\sqrt 3}{3+4\sqrt 3})^2)$
$ED^2=3\times (1+(\dfrac{8-3\sqrt 3}{3+4\sqrt 3})^2)$
$BD=\dfrac 14 \times (3+4\sqrt 3)\times \sqrt{1+(\dfrac{8-3\sqrt 3}{3+4\sqrt 3})^2}=\dfrac 12 \times \sqrt{37-6\sqrt 3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Product of cosines: $ \prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right) $
Evaluate
$$ \prod_{r=1}^{7} \cos \left({\dfrac{r\pi}{15}}\right) $$
I tried trigonometric identities of product of cosines, i.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)] $$
but I couldn't find the product.
Any help will be appreciated.
Thanks.
| I like the answer by @Steven Gregory because of the way the dominos fall and it seems the only one presented that a precalculus student could hope to find. Using the reflection and multiplication formulas for the gamma function a fairly compact proof is possible.
$$\begin{align}\prod_{r=1}^7\cos\frac{r\pi}{15} & = \prod_{r=1}^7\sin\left(\frac{\pi}2-\frac{r\pi}{15}\right) = \prod_{r=0}^6\sin\left(\frac{\pi}2-\frac{7\pi}{15}+\frac{r\pi}{15}\right) \\
& = \prod_{r=0}^6\sin\left(\frac{\pi}{30}+\frac{r\pi}{15}\right) = \prod_{r=0}^6\frac{\pi}{\Gamma\left(\frac1{30}+\frac{r}{15}\right)\Gamma\left(\frac{29}{30}-\frac{r}{15}\right)} & \tag{1} \\
& = \frac{\pi^7}{\prod_{r=0}^6\Gamma\left(\frac1{30}+\frac{r}{15}\right)\prod_{s=8}^{14}\Gamma\left(\frac1{30}+\frac{s}{15}\right)} = \frac{\pi^7\,\Gamma\left(\frac1{30}+\frac7{15}\right)}{\prod_{r=0}^{14}\Gamma\left(\frac1{30}+\frac{r}{15}\right)} & \tag{2} \\
& = \frac{\pi^7\,\Gamma\left(\frac12\right)}{(2\pi)^{(15-1)/2}15^{\frac12-15\left(\frac1{30}\right)}\Gamma\left(15\left(\frac1{30}\right)\right)} & \tag{3} \\
& = \frac1{2^7}
\end{align}$$
$(1)$ Using the reflection formula for the gamma function: $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$.
$(2)$ Reordering one product and multiplying and dividing by $\Gamma\left(\frac1{30}+\frac7{15}\right)$.
$(3)$ Using the multiplication formula for the gamma function: $$\prod_{k=0}^{n-1}\Gamma\left(x+\frac{k}n\right)=(2\pi)^{(n-1)/2}n^{\frac12-nx}\Gamma(nx)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 5,
"answer_id": 4
} |
Determine whether $\sum \frac{2^n + n^2 3^n}{6^n}$ converges For the series $$\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n},$$ I was thinking of using the root test? so then I would get $(2+n^2/n+3)/6$ but how do I find the limit of this?
| $$\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n}=\sum_{n=1}^{\infty}\left(\frac{1}{3^n}+\frac{n^2}{2^n}\right),$$
where
$$\sum_{n=1}^{\infty}\frac{1}{3^n}=\frac{1}{1-\frac{1}{3}}-1=\frac{1}{2}$$
and, by ratio test,
$$\frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}}=\frac{1}{2}\left(1+\frac{1}{n}\right)^2 \to \frac{1}{2}<1 \;\; \text{ as } \;\; n \to \infty,$$
meaning that $\sum_{n=1}^{\infty}\frac{n^2}{2^n}$ is absolutely convergent. Therefore, the whole series is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
How do you take the integral of $\int \frac{\mathrm{d}y}{y(3-y)}$? How do you evaluate the following? $$\int \frac{\mathrm{d}y}{y(3-y)}$$
I have looked at the solution and I don't understand how they are taking the integral of this? They go from:
$$\int \frac{\mathrm{d}y}{y(3-y)}$$
to
$$\int \frac13 \left(\frac1y + \frac{1}{3-y}\right)\, \mathrm{d}y$$
and how they get to this second step confuses me. Where did they get this extra $\frac13$ from?
| Partial fraction decomposition gives:
\begin{align*}
\dfrac{1}{y (3-y)} & = \dfrac{A}{y} + \dfrac{B}{3-y}\\
1&=A(3-y) + By
\end{align*}
Looking at the $y$ coefficients and the constant coefficients, we obtain the following two equations:
$$0 = A - B $$
$$1 = 3A$$
Which solve to give $A = B = \dfrac{1}{3}$.
Hence
$$\dfrac{1}{y (3-y)} = \dfrac{1}{3y} + \dfrac{1}{3(3-y)}$$
or
$$\dfrac{1}{y (3-y)} = \dfrac{1}{3} \left(\dfrac{1}{y} + \dfrac{1}{3-y}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to rewrite $\pi - \arccos(x)$ as $2\arctan(y)$? I get the following results after solving the equation $\sqrt[4]{1 - \frac{4}{3}\cos(2x) - \sin^4(x)} = -\,\cos(x)$, :
$$
x_{1} = \pi - \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z}\\
x_{2} = \pi + \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z}
$$
Wolfram Alpha, here the link, instead, gives the following results:
$
x_{1} = 2\pi n - 2\arctan(\sqrt{5 + 2\sqrt{6}}) : n \in \mathbb{Z}\\
x_{2} = 2\pi n + 2\arctan(\sqrt{5 + 2\sqrt{6}}) : n \in \mathbb{Z}
$
Now, supposing that my solutions are correct, this means that there must be a relation between:
$\pi - \arccos(\frac{\sqrt{6}}{3})$ and $- 2\arctan(\sqrt{5 + 2\sqrt{6}})$
or between:
$\pi + \arccos(\frac{\sqrt{6}}{3})$ and $+ 2\arctan(\sqrt{5 + 2\sqrt{6}})$
or viceversa. But, given the solutions I have found, how can I prove that they are effectively the same as the solutions Wolfram found? Mathematically?
P.S.: I have found out, by looking at the graphs of $y_{1} = \pi - \arccos(\frac{\sqrt{6}}{3})$ and $y_{2} = 2\arctan(\sqrt{5 + 2\sqrt{6}})$ e.g., that:
$\pi - \arccos(\frac{\sqrt{6}}{3}) = 2\arctan(\sqrt{5 + 2\sqrt{6}})$
Then, of course:
$\pi - \arccos(\frac{\sqrt{6}}{3}) + 2\pi = 2\arctan(\sqrt{5 + 2\sqrt{6}})+ 2\pi \\
\pi - \arccos(\frac{\sqrt{6}}{3}) + 4\pi = 2\arctan(\sqrt{5 + 2\sqrt{6}})+ 4\pi \\
... every\,\,360°n, n \in \mathbb{Z}$
And that:
$\pi + \arccos(\frac{\sqrt{6}}{3}) = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ 2\pi\\
\pi + \arccos(\frac{\sqrt{6}}{3}) + 2\pi = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ 4\pi\\
\pi + \arccos(\frac{\sqrt{6}}{3}) + 4\pi = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ 6\pi\\
...
\pi + \arccos(\frac{\sqrt{6}}{3}) + (n - 2)\pi = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ n\pi\\
$
So we can say that $\pi + \arccos(\frac{\sqrt{6}}{3}) + (n - 2)\pi$ differs from $-2\arctan(\sqrt{5 + 2\sqrt{6}}) + n\pi$ by just one lap ($2\pi$), otherwise they can be safely considered the same (for all integers).
So how to rewrite $\arccos$ in terms of $\arctan$?
Thanks for the attention!
| You can see in the figure the angle $\alpha$ corresponding to $\arccos(\frac{\sqrt{6}}{3})$ and the tangent of the same $\alpha$ from which you can easily deduce your own conclusion.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Is this formula true for $n\geq 1$:$4^n+2 \equiv 0 \mod 6 $? Is this formula true for $n\geq 1$:$$4^n+2 \equiv 0 \mod 6 $$.
Note :I have tried for some values of $n\geq 1$ i think it's true such that
:I used the sum digits of this number:$N=114$,$$1+1+4\equiv 0 \mod 6,1²+1²+4²\equiv 0 \mod 6,1^3+1^3+4^3\equiv 0 \mod 6,\cdots $$ ?
Thank you for any help
| For a less number theoretic approach, let $S_n$ be the statement that $4^n + 2$ is divisible by 6. Clearly $S_1$ is true since $4^1 + 2 = 6$. Now assuming that $4^k + 2$ is divisible by 6,
\begin{align*}
4^{k+1} + 2 &= 4\cdot4^n + 2 \\
&= 4\cdot4^n + 4\cdot2 - 6 \\
&= 4(4^n + 2) - 6,
\end{align*}
we immediately have that $4^{k+1} + 2$ is divisible by 6, so the induction is complete.
The other answers are far more elegant, however, and you should lean towards using modular arithmetic to solve these problems.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$
Let $0\le a\le b\le c,abc=1$, then show that
$$a+b^2+c^3\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}$$
Things I have tried so far:
$$\dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}=\dfrac{b^2c^3+ac^3+ab^2}{bc^2}$$
Since $abc=1$, it suffices to prove that
$$c+b^3c^2+bc^5\ge b^2c^3+ac^3+ab^2$$
then the problem is solved. I stuck in here.
| Write
$$a = \frac{p^3}{p q r},\, b= \frac{q^3}{p q r}, \,
c= \frac{r^3}{p q r}
$$
with $0 < p\le q \le r$ and substitute to get the equivalent inequality:
$$
p q^8 r^4 + p^5 q^3 r^5 + q r^{12}-p^6 q^7 - p q^5 r^7 - p^5 r^8 \ge 0$$
Now write $p=u,\, q=u+v, \,r=u+v+w$, with $u>0$, $\ v,w\ge 0$ and substitute to get an expression in $u$,$v$,$w$ with all coefficients positive, hence $\ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Possible values of z? Let $f:[-2,2]\to \mathbb {R} $ where $$f(x)=x^3+2(\sin x)^5+3(\tan x)^7+\left\lfloor\frac{x^2+1}{z}\right\rfloor $$ is an odd function then what are possible values of $z$?
$\lfloor\cdot\rfloor $ is the floor function.
| We know that the functions $x^n$ for $n$ odd, $\sin x$ and $\tan x$ are all odd functions. Furthermore, if $h$ and $g$ are odd, then $h(g(-x)) = h(-g(-x))=-h(g(-x))$, so the composition of two odd functions is again odd.
And we can therefore conclude that $x^3+ 2 (\sin x)^5 + 3 ( \tan x)^7$ is an odd function. So for $f$ to be odd, we want $y(x)=[\frac{x^2+1}{z}]$ to be odd.
Since $x^2+1$ is non negative, we have for positive $z$ that $[\frac{x^2+1}{z}] \ge 0$ for all $x \in [-2,2]$. And for negative $z$ we have $y(x) \le 0$. So for $y$ to be odd, we must find $z$ such that $[\frac{x^2+1}{z}] =0$.
We have $x^2+1 \in [1,5]$, so we want $z$ such that $[1/z,5/z] \subset [0, 1)$. So if $z >5$ we have $[1/z,5/z] \subset [0,1)$.
Thus for $z >5 $ we have $f$ is odd.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $a^2+b^2$ is a multiple of three, then a and b are multiples of three I have attempted to prove the above. I am uncertain about the correctness of my proof:
Both numbers have to be multiples of three, i.e. $3a+3b=3n$, $\ 3(a+b)=3n$
It is not possible to arrive at an integer that is a multiple of three without adding two integers that are multiples of three.
Assumption: Suppose that $b$ is not a multiple of three, then it can be expressed as $(3v \pm 1)$, therefore we have:
\begin{align*} a^2+(3v \pm 1)^2=3n\\ a^2= 3n-9v^2 \mp 6v -1\\ a=\sqrt{3(n-3v^2\mp2v-\frac{1}{3})}
\end{align*}
which is not a multiple of three? (or is it).
As mentioned before, $(a+b) \ne 3m \ $ if either $a$ or $b$ is not a multiple of 3, in which case assumption that $b$ is not a multiple of 3 is false. And hence it is a multiple of three, so is $a$.
| If $a\ne 0\ (\ mod\ 3)$, then $a^2\equiv 1\ (\ mod\ 3) $
If $a\equiv 0\ (\ mod\ 3)$, then $a^2\equiv 0\ (\ mod\ 3)$
If $b\ne 0\ (\ mod\ 3)$, then $b^2\equiv 1\ (\ mod\ 3) $
If $b\equiv 0\ (\ mod\ 3)$, then $b^2\equiv 0\ (\ mod\ 3)$
So, the only way to get $a^2+b^2\equiv 0\ (\ mod\ 3)$ is $a\equiv b\equiv 0\ (\ mod\ 3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$A$ and $B$ similar if $A^2=B^2=0$ and dimension of range $A$ and $B$ are equal Suppose $A$ and $B$ are linear transformations on finite dimensional vector space $V$,s.t. $A,B\neq 0$ and $A^2=B^2=0$. Suppose the dimension of range $A$ and $B$ are equal, can $A$ and $B$ be similar?
| that is surely possible. Take for example $A= \begin{pmatrix}0 & 0 \\ 1& 0 \end{pmatrix}$ and $ B=\begin{pmatrix}0 & 1 \\ 0& 0 \end{pmatrix}$ and then it holds that
$$
A^2=B^2=0
$$
and
$$
A=SBS^{-1} \text{ with } S=\begin{pmatrix}0 & 1 \\ 1& 0 \end{pmatrix}
$$
so $A$ and $B$ are indeed similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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The maximum and minimum values of the expression Here is the question:find the difference between maximum and minimum values of $u^2$ where $$u=\sqrt{a^2\cos^2x+b^2\sin^2x} + \sqrt{a^2\sin^2x+b^2\cos^2x}$$
My try:I have just normally squared the expression and got
$u^2=a^2\cos^2x+b^2\sin^2x + a^2\sin^2x+b^2\cos^2x +2\sqrt{a^2\cos^2x+b^2\sin^2x} \sqrt{a^2\sin^2x+b^2\cos^2x}$
$u^2=a^2+b^2 +2\sqrt{a^2\cos^2x+b^2\sin^2x} .\sqrt{a^2\sin^2x+b^2\cos^2x}$
I am not getting how to solve the irrational part,so how should we do it.Is there some general way to solve such questions?
| Assume WLOG $a > b > 0$, $A = \sqrt{a^2\cos^2x+b^2\sin^2x}, B = \sqrt{a^2\sin^2x+b^2\cos^2x}\Rightarrow A^2+B^2 = a^2+b^2\Rightarrow u^2 = (1\cdot A+1\cdot B)^2\leq (1^2+1^2)(A^2+B^2)=2(a^2+b^2)\Rightarrow u^2_{max} = 2(a^2+b^2)$. To find $u^2_{min}$, you need to find the min of $(a^2\cos^2x+b^2\sin^2x)(a^2\sin^2x+b^2\cos^2x)=f(\cos^2 x)=(a^2-(a^2-b^2)t)(b^2+(a^2-b^2)t), t = \cos^2x, 0 \leq t \leq 1=f(p) = (a^2-p)(b^2+p), p = (a^2-b^2)t, 0 \leq p \leq a^2-b^2\to f(p) = a^2b^2 + (a^2-b^2)p - p^2\Rightarrow f'(p) = a^2-b^2 - 2p=0 \iff p = \dfrac{a^2-b^2}{2}\Rightarrow f\left(\dfrac{a^2-b^2}{2}\right)=\dfrac{(a^2+b^2)^2}{4}$. At end points $p = 0, a^2-b^2, f(0) = a^2b^2, f(a^2-b^2) = a^2b^2$. Thus $u^2_{min} = a^2+b^2 + 2\sqrt{a^2b^2}=(a+b)^2$, since $a^2b^2 \leq \dfrac{(a^2+b^2)^2}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How can I prove the following equality I have the following equality :
$$I_1=-\frac{ab}{2\pi}\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=\frac{a}{a+b}-\frac{1}{2}$$
$$I_2=-\frac{ab}{2\pi}\int_0^\pi \frac{\sin(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=0$$
where $0 <b \leq a$.
I used the residues but I could not prove this equality
| One trick to facilitate analysis is to write
$$\sin^2x=\frac{1-\cos 2x}{2}$$
and
$$\cos^2x=\frac{1+\cos 2x}{2}$$
Thus,
$$a^2\sin^2(t)+b^2\cos^2(t)=\frac{b^2+a^2}{2}+\frac{b^2-a^2}{2}\cos 2t$$
For the integral of interest, we can write
$$\begin{align}
I_1&=-\frac{ab}{2\pi}\int_0^{\pi} \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt\\\\
&=-\frac{ab}{4\pi}\int_0^{2\pi}\frac{\cos t}{\frac{b^2+a^2}{2}+\frac{b^2-a^2}{2}\cos t}dt\\\\
&=-\frac{ab}{2\pi(b^2-a^2)}\int_0^{2\pi}\frac{\cos t}{\frac{b^2+a^2}{b^2-a^2}+\cos t}dt\\\\
&=-\frac{ab}{(b^2-a^2)}+\frac{ab(b^2+a^2)}{2\pi(b^2-a^2)^2}\int_0^{2\pi}\frac{1}{\frac{b^2+a^2}{b^2-a^2}+\cos t}dt
\end{align}$$
Can you finish now?
SPOLIER ALERT
SCROLL OVER SHADED AREA TO SEE ANSWER
Starting with the last term $I_1=-\frac{ab}{(b^2-a^2)}+\frac{ab(b^2+a^2)}{2\pi(b^2-a^2)^2}\int_0^{2\pi}\frac{1}{\frac{b^2+a^2}{b^2-a^2}+\cos t}dt$, we move to the complex plane by letting $z=e^{it}$. Then, $I_1$ becomes $$\begin{align}I_1&=-\frac{ab}{(b^2-a^2)}+\frac{ab(b^2+a^2)}{2\pi(b^2-a^2)^2}\oint_C \frac{-2i}{z^2+2\frac{b^2+a^2}{b^2-a^2}\,z+1}dz\\\\&=-\frac{ab}{(b^2-a^2)}+\frac{ab(b^2+a^2)}{2\pi(b^2-a^2)^2}\left(2\pi i \frac{-2i}{-2\frac{2ab}{a^2-b^2}}\right)\\\\&=-\frac{ab}{(b^2-a^2)}-\frac{a^2+b^2}{2(a^2-b^2)}\\\\&=\frac12 -\frac{a}{a+b}\end{align}$$which differs from the posted answer by a factor of $-1$. For $a=2$ and $b=1$, Wolfram Alpha confirms the value of the integral as $-1/6$ agreeing with the answer herein, whereas the posted answer is $+1/6$.
For $I_2$, enforce the substitution $t= \pi -x$. Then, we see
$$\begin{align}
I_2&=\int_0^{\pi} \frac{\sin(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt\\\\
&=\int_{\pi}^{0} \frac{\sin 2(\pi-x)}{a^2\sin^2(\pi-x)+b^2\cos^2(\pi-x)}(-1)dx\\\\
&=\int_{0}^{\pi} \frac{-\sin (2x)}{a^2\sin^2(x)+b^2\cos^2(x)}dx\\\\
&=-I_2
\end{align}$$
Thus, we have $I_2=-I_2$ which, of course, implies $I_2=0$ as expected!
| {
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"timestamp": "2023-03-29T00:00:00",
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How to integrate $5^{5^{5^x}}5^{5^x}5^x$? How to integrate $5^{5^{5^x}}5^{5^x}5^x$ ?
I guess it's simple but can't figure it out !
| Let $u=5^{5^x}$, then $x = \frac{\frac{\ln(u)}{\ln(5)}}{\ln{5}}$ and $du = 5^{x+5^x}\ln^2(5)\hspace{.1cm} dx \implies dx = \frac{5^{-x-5^x}}{\ln^2(5)} \hspace{.1cm} du$. Thus we transform
$ \begin{align}
\int 5^{5^{5^x}} 5^{5^x} 5^x \hspace{.1cm} dx &= \int 5^uu5^x \frac{5^{-x-5^x}}{\ln^2(5)} \hspace{.1cm} du \\
&= \int \frac{u5^{u-5\frac{\frac{\ln(u)}{\ln(5)}}{\ln{5}}}}{\ln^2(5)} \hspace{.1cm} du \\
&= \frac{1}{\ln^2(5)} \int u5^{u-5\frac{\frac{\ln(u)}{\ln(5)}}{\ln{5}}} \hspace{.1cm} du \\
&= \frac{1}{\ln^2(5)} \int 5^u \hspace{.1cm} du \\
&= \frac{1}{\ln^2(5)} \frac{5^u}{\ln(5)} \\
&= \frac{1}{\ln^2(5)} \frac{5^{5^{5^x}}}{\ln(5)} \\
&= \frac{5^{5^{5^x}}}{\ln^3(5)}
\end{align}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Value of an expression with cube root radical What is the value of the following expression?
$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}$$
| Start by noticing that $(2 + \sqrt{5})^{3} = 38 + 17 \sqrt{5}$ and $(2 - \sqrt{5})^{3} = 38 - 17 \sqrt{5}$. Now,
\begin{align}
\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38} &= \sqrt[3]{(2 + \sqrt{5})^{3}} - \sqrt[3]{(\sqrt{5} - 2)^{3}} \\
&= (2 + \sqrt{5}) - (-2 + \sqrt{5}) \\
&= 4.
\end{align}
What seems to have happened is that most answers presented here follow the pattern:
\begin{align}
\sqrt[3]{\ 17\sqrt{5}+38} + \sqrt[3]{17\sqrt{5}-38} &= \sqrt[3]{(2 + \sqrt{5})^{3}} + \sqrt[3]{(\sqrt{5} - 2)^{3}} \\
&= (2 + \sqrt{5}) + (-2 + \sqrt{5}) \\
&= 2 \sqrt{5}.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Nature of the roots of quadratic equation Here is the problem that I need to prove:
If $x$ is real and $\displaystyle{\ p = \frac{3(x^2+1)}{(2x-1)}}$, prove that $\ p^2-3(p+3) \geq 0$
Here is what I did:
\begin{align*}
p(2x-1)=3(x^2+1) \\
3x^2 - 2px + (p+3)=0 \\
b^2 - 4ac = 4(p^2-3(p+3))
\end{align*}
By inspection I can see that $p^2 > -3(p+3)$ for almost all values of $p \ $, therefore
$p^2-3(p+3) > 0 $. However, the question asks to show that $\ p^2-3(p+3) \geq 0$
If I make $p^2 = 3(p+3)$ I can find roots and so $\ p^2-3(p+3) = 0$, when $\displaystyle{p = \frac{3 \pm \sqrt{45}}{2}}$. Therefore $\ p^2-3(p+3) \geq 0$
Having done this, how can I mathematically show that $p^2$ is never $<$ than $3(p+3)$? Because I am not satisfied with just saying that by inspection $p^2$ is greater than $3(p+3)$.
Thank you
| I think you are overthinking this.
To show that $x$ satisfies the quadratic equation $3x^2−2px+(p+3)=0$, you have two cases :
(1) $b^2−4ac > 0$ :
Therefore $4(p^2−3(p+3)) > 0$ i.e. $p^2−3(p+3) > 0$
(2) $b^2−4ac = 0$ :
Then $4(p^2−3(p+3)) = 0$ i.e. $p^2−3(p+3) = 0$
The third case $b^2−4ac < 0$ here is irrevelant since $x$ is supposed to be a solution.
No need to compute anything here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\sum\limits_{i=1}^n \frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$, for all $x_i>0.$ Can you prove the following new inequality? I found it experimentally.
Prove that, for all $x_1,x_2,\ldots,x_n>0$, it holds that
$$\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod\limits _{j=1}^nx_j}} \ge
1\,.$$
| The inequality $\displaystyle\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$ is trivial given the claim below. Of course, the equality occurs if and only if $x_1=x_2=\ldots=x_n$.
Claim: For
every $i=1,2,\ldots,n$, we have $\displaystyle\frac{x_i}{\sqrt[n]{x_i^n+\left(n^n-1\right)\,\prod\limits_{j=1}^n\,x_j}}
\geq
\frac{x_i^{1-\frac{1}{n^n}}}{\sum\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}}$. The equality holds if and only if
$x_1=x_2=\ldots=x_n$.
Proof: The required inequality is equivalent to $$\left(\sum_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^n-x_i^{n\left(1-\frac{1}{n^n}\right)}\geq \left(n^n-1\right)\,x_i^{-\frac{1}{n^{n-1}}}\,\prod_{j=1}^n\,x_j\,.$$
Note that the expansion of $\left(\sum\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^n$ consists of $n^n$ terms of the form $x_{j_1}^{1-\frac{1}{n^n}}x_{j_2}^{1-\frac{1}{n^n}}\cdots x_{j_n}^{1-\frac{1}{n^n}}$, where $j_1,j_2,\ldots,j_n\in\{1,2,\ldots,n\}$. The product of these terms is equal to $$\left(\prod\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^{n^n}\,.$$ If we take the term $x_i^{n\left(1-\frac{1}{n^n}\right)}$ out of the product, we get the product of $n^n-1$ terms from the expansion of $\left(\sum\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^n-x_i^{n\left(1-\frac{1}{n^n}\right)}$, which is then equal to
$$\frac{\left(\prod\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^{n^n}}{x_i^{n\left(1-\frac{1}{n^n}\right)}}=x_i^{-\frac{1}{n^{n-1}}\left(n^n-1\right)}\,\prod\limits_{j=1}^n\,x_j^{n^n-1}\,.$$
By the AM-GM Inequality,
$$\frac{\left(\sum\limits_{j=1}^n\,x_j^{1-\frac{1}{n^n}}\right)^n-x_i^{n\left(1-\frac{1}{n^n}\right)}}{n^n-1}\geq \left(x_i^{-\frac{1}{n^{n-1}}\left(n^n-1\right)}\,\prod\limits_{j=1}^n\,x_j^{n^n-1}\right)^{\frac{1}{n^n-1}}=x_i^{-\frac{1}{n^{n-1}}}\,\prod\limits_{j=1}^n\,x_j\,,$$
which is what we want. Hence, the claim is true. The equality case happens, due to the AM-GM Inequality, if and only if $x_1=x_2=\ldots=x_n$.
How did I get the exponent $1-\frac{1}{n^n}$?
I assumed it was $k$ at first, and the desired inequality was equivalent to
$$\left(\sum\limits_{j=1}^n\,x_j^{k}\right)^n-x_i^{nk}\geq \left(n^n-1\right)\,x_i^{n(k-1)}\,\prod\limits_{j=1}^n\,x_j\,.$$
Then, the last inequality read
$$\frac{\left(\sum\limits_{j=1}^n\,x_j^{k}\right)^n-x_i^{nk}}{n^n-1}\geq \left(x_i^{-nk}\,\prod\limits_{j=1}^n\,x_j^{n^nk}\right)^{\frac{1}{n^n-1}}\,,$$
the right-hand side of which I wanted to equal $x_i^{n(k-1)}\,\prod\limits_{j=1}^n\,x_j$. Therefore, $\frac{n^nk}{n^n-1}=1$ and $n(k-1)=-\frac{nk}{n^n-1}$, both of which gave me $k=1-\frac{1}{n^n}$.
| {
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"url": "https://math.stackexchange.com/questions/1370084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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Finding $\frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ where $a, b, c$ are the roots of a cubic equation, without solving the cubic equation itself Suppose that we have a equation of third degree as follows:
$$
x^3-3x+1=0
$$
Let $a, b, c$ be the roots of the above equation, such that $a < b < c$ holds. How can we find the answer of the following expression, without solving the original equation?
$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a}
$$
| As I've suggested in the comment yesterday,
let $x=2m\cos y\implies(2m\cos y)^3-(2m\cos y)+1=0\ \ \ \ (1)$
As $\cos3y=4\cos^3y-3\cos y,$
$2m^3(\cos3y+3\cos y)-(2m\cos y)+1=0$
$\iff2m^3\cos3y+2m\cos y(m^2-1)+1=0\ \ \ \ (2)$
WLOG choose $m^2-1=0\iff m=\pm1$
Let $m=1$
$(1)$ reduces to $8\cos^3y-6\cos y+1=0 \ \ \ \ (3)$
and $(2)\implies\cos3y=-\dfrac12\implies3y=360^\circ n\pm120^\circ$ where $n$ is any integer
$\implies y=120^\circ n+40^\circ$ where $n\equiv-1,0,1\pmod3$
So, the roots of $(3)$ are
$\cos(-80^\circ)=\cos80^\circ$ $\cos40^\circ,\cos160^\circ=\cos(180^\circ-20^\circ)=-\cos20^\circ<0$
Clearly, $\cos40^\circ>\cos80^\circ>0>-\cos20^\circ$
$\implies c=2\cos40^\circ, b=2\cos80^\circ, a=2\cos160^\circ$
$\implies\dfrac ab=\dfrac{2\cos160^\circ}{2\cos80^\circ}=\dfrac{2\cos^280^\circ-1}{\cos80^\circ}=2\cos80^\circ-\dfrac1{\cos80^\circ}$
$\implies\sum_{\text{cyc}}\dfrac ab=2\sum_{\text{cyc}}\cos80^\circ-\sum_{\text{cyc}}\dfrac1{\cos80^\circ}$
Using Vieta's formula on $(3),\sum_{\text{cyc}}\cos80^\circ=0,$
$\cos40^\circ\cos80^\circ+\cos40^\circ\cos160^\circ+\cos80^\circ\cos160^\circ=\dfrac{-6}8$
and $\cos40^\circ\cos80^\circ\cos160^\circ=-\dfrac18$
and $\sum_{\text{cyc}}\dfrac1{\cos80^\circ}=\dfrac{\cos40^\circ\cos80^\circ+\cos40^\circ\cos160^\circ+\cos80^\circ\cos160^\circ}{\cos40^\circ\cos80^\circ\cos160^\circ}=\cdots=6$
Won't you try with $m=-1?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Is it possible to evaluate this binomial sum? Would it be possible to evaluate this sum?
$$\sum_{k=0}^{N/2}k\binom{N+1}{k},$$
where $N$ is even? I know that the sum
$$\sum_{k=0}^{N+1}k\binom{N+1}{k}=2^N(N+1)$$ (by http://mathworld.wolfram.com/BinomialSums.html, equation $(21)$), but I can't figure out a way to evaluate this sum from just $k=0$ to $N/2$.
| Since for $k$ such that $1 \leq k \leq N + 1$, we have
$$k\binom{N + 1}{k} = k \frac{(N + 1)!}{k! (N + 1 - k)!} = \frac{(N + 1)!}{(k - 1)!(N + 1 - k)!} = (N + 1)\binom{N}{k - 1}.$$
If $N$ is even, $\binom{N}{0} = \binom{N}{N}, \binom{N}{1} = \binom{N}{N - 1}, \binom{N}{2} = \binom{N}{N - 2}, \ldots, \binom{N}{N/2 - 1} = \binom{N}{N/2 + 1}$ implies that
\begin{align*}
& \sum_{k = 1}^{N/2}\binom{N}{k - 1} = \binom{N}{0} + \binom{N}{1} + \cdots + \binom{N}{N/2 - 1} \\
= &\frac{\binom{N}{0} + \cdots + \binom{N}{N} - \binom{N}{N/2} }{2} \\
= & \frac{1}{2}\sum_{k = 0}^N \binom{N}{k} - \frac{1}{2}\binom{N}{N/2} \\
= & 2^{N - 1} - \frac{1}{2}\binom{N}{N/2}.
\end{align*}
Hence
$$\sum_{k = 0}^{N/2} k\binom{N + 1}{k} = (N + 1)2^{N - 1} - \frac{1}{2}(N + 1)\binom{N}{N/2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given $\tan A + \tan B = 3x$ and $\tan A \tan B = 2x^2$, find $\tan A - \tan B$ Given $$\tan A + \tan B = 3x$$ and $$\tan A \tan B = 2x^{2},$$ how can one find $\tan A - \tan B$? I have tried substitution, but failed to find the answer.
Edit: Can this problem be solved using the formulas for sums and differences of tangents?
| We know that $$(a-b)^2=a^2+b^2-2ab=(a+b)^2-4ab$$ $$\implies a-b=\pm \sqrt{(a+b)^2-4ab}$$ Now, we have
$$\tan A-\tan B=\sqrt{(\tan A+\tan B)^2-4\tan A\tan B}$$ $$=\sqrt{(3x)^2-4(2x^2)}$$ $$=\sqrt{x^2}$$$$=\pm x=|x|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
Evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
How to evalute this equation without using calculator?
| Add the first two
$$\frac 1{1+\sqrt3+\sqrt2} + \frac 1{1+\sqrt3-\sqrt2}=2\frac{1+\sqrt3}{(1+\sqrt3)^2-2}=2\frac{1+\sqrt3}{2+2\sqrt3}=1.$$
Similarly, add the last two
$$2\frac{1-\sqrt3}{2-2\sqrt3}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
remainder of $a^2+3a+4$ divided by 7
If the remainder of $a$ is divided by $7$ is $6$, find the remainder when $a^2+3a+4$ is divided by 7
(A)$2$ (B)$3$ (C)$4$ (D)$5$ (E)$6$
if $a = 6$, then $6^2 + 3(6) + 4 = 58$, and $a^2+3a+4 \equiv 2 \pmod 7$
if $a = 13$, then $13^2 + 3(13) + 4 = 212$, and $a^2+3a+4 \equiv 2 \pmod 7$
thus, we can say that any number, $a$ that divided by 7 has remainder of 6, the remainder of $a^2 + 3a + 4$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
| $a^2 + 3a + 4 \equiv a^2 - 4a + 4 \equiv (a-2)^2 \pmod 7$
If $a\equiv b \pmod 7$, then $a^2 + 3a + 4 \equiv (b-2)^2 \pmod 7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Modular Quadratic Equation I'm trying to solve that equation:
$x^2-3x-5\equiv0\pmod{343}$
I've completed the square as follows:
$x^2-3x-5 \equiv x^2+340x-5\equiv(x+170)^2-170^2-5\pmod{343}\\
(x+170)^2 \equiv 93\pmod{343}\\
y^2 \equiv 93 \pmod{343}$
But I have no idea how to move on. How can I use the fact that $343=7^3$?
| HINT:
Start with $x^2-3x-5\equiv0\pmod7\iff0\equiv x^2-3x-5+7=(x-1)(x-2)$
If $x-1\equiv0\pmod7, x=7a+1$ where $a$ is any integer
Now $x^2-3x-5=(7a+1)^2-3(7a+1)-5=49a^2-7a-7$
which $\equiv0\pmod{7^2}\iff7|(a+1)\implies a\equiv-1\pmod7, a=7b-1$ where $b$ any integer
$\implies x=7a+1=7(7b-1)+1=49b-6$
Now $x^2-3x-5=(49b-6)^2-3(49b-6)-5\equiv49(-15b+1)\pmod{7^3}$
which will be $\equiv0\iff-15b+1\equiv0\pmod7\iff b\equiv1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Application of Jensen's inequality to $x^x+y^y+z^z$ Claim: If $x, y, z >0$ and $x+y+z = 3\pi, $ then $x^x + y^y + z^z > 81.$
My attempt: Let $f(w) = w^w$, so $f$ is convex on $(0, \infty).$ By Jensen's inequality, $f(x\frac{x}{3\pi}+ y\frac{y}{3\pi} + z\frac{z}{3\pi}) \leq \frac{x}{3\pi}f(x) + \frac{y}{3\pi}f(y) + \frac{z}{3\pi}f(z) < x^x + y^y + z^z. $
Could anyone advise me how to continue from here?
| You rather just want to use that $$\frac{1}{3}f(x)+\frac{1}{3}f(y)+\frac{1}{3}f(z) \geq f\left(\frac{1}{3}x+\frac{1}{3}y+\frac{1}{3}z\right)=f(\pi)=\pi^\pi > 3^3 =27$$
Multiply by 3 to get the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
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