Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Show every prime $p\equiv 1,3$ (mod 8) can be written as $p=x^2+2y^2$
Let $K=\mathbb{Q}(\sqrt{-2})$. Show that $\mathcal{O}(K)$ is a principal ideal domain. Deduce that every prime $p\equiv 1, 3$ (mod 8) can be written as $p = x^2 + 2y^2$ with $x, y \in \mathbb{Z}$.
As $−2$ is squarefree $6\equiv 1$ (mod 4) we have $\mathcal{O}(K) = \mathbb{Z}[
\sqrt{2}$]. The
discriminant is $\Delta$ = −8. The degree $n = 2$. The signature is $(0, 2)$. Thus the Minkowski bound is
$$ B_k = \frac{2!}{2^2}=\frac{4}{\pi}\times \sqrt{8} = \frac{4\sqrt{2}}{\pi}<2$$
Hence $Cl(K)$ is generated by the empty set of ideal classes and so $Cl(K) = \{1\}$. So this means $\mathcal{O}(K)$ is a principal ideal domain I believe...
Ok, now if we let $p \equiv 1$ or $3$ (mod 8). By quadratic reciprocity, $−2 ≡ \alpha^2$
(mod p) for
some integer $\alpha$. Thus
$$X^2 + 2 ≡ (X + \alpha)(X − \alpha) \quad\text{(mod}~~ p).$$
Ok, now I am slightly stuck, can we apply some theorem here? Not sure if what above is correct to get to the desired result
| An elementary proof starting from scratch. If $p\equiv 1\pmod{8}$, in $\mathbb{F}_p^*$ there is an element with order $8$, which we may call $\alpha$. Since $\alpha^4+1=0$ we have $(\alpha+\alpha^{-1})^2=0 $ and $-2$ is a quadratic residue $\pmod{p}$. If $p\equiv 3\pmod{8}$ we may consider the degree of the splitting field of $\Phi_8(x)=x^4+1$ over $\mathbb{F}_p$, which is given by the least $k$ such that $8\mid(p^k-1)$, i.e. by $2$. So we have $\alpha\in\mathbb{F}_{p^2}$ with order $8$, and by Frobenius automorphism one of the quadratic factors of $x^4+1$ is given by
$$(x-\alpha)(x-\alpha^3)=x^2-(\alpha+\alpha^3)-1$$
and the other quadratic factor is given by
$$(x-\alpha^5)(x-\alpha^7) = x^2-(\alpha^5+\alpha^7)-1.$$
On the other hand $(\alpha+\alpha^3)^2 = \alpha^2+\alpha^6-2 = -2$, so $-2$ is a quadratic residue in this case, too.
Let $u\in\{1,\ldots,\frac{p-1}{2}\}$ be such that $u^2+2\equiv 0\pmod{p}$. We have $u^2+2 = kp $ for some $k\in\{1,\ldots,\lfloor\frac{p}{4}\rfloor\}$. Let $v$ be the minimum between $u\pmod{k}$ and $(k-u)\pmod{k}$. We have $v^2+2=kq$ for some $q<k$ and by the Lagrange-Brahmagupta-Fibonacci identity
$$ (a^2+2b^2)(c^2+2d^2)= (ac+2bd)^2 + 2(bc-ad)^2 $$
we have
$$ (u^2+2)(v^2+2) = (uv+2)^2+2(v-u)^2 =k^2 pq.$$
On the other hand both $uv+2$ and $v-u$ are multiples of $k$, so by starting with a representation of a multiple of $p$ as $A^2+2B^2$ we got a representation as $A^2+2B^2$ of a smaller multiple
$$ \left(\frac{uv+2}{k}\right)^2+\left(\frac{v-u}{k}\right)^2 = qp $$
and the trick can be iterated again, ultimately leading to a representation of $p$ as $A^2+2B^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find center of rational circle When a circle is defined by three points on its perimeter, it's possible to find its center $(h,k)$ using:
$$
h=\frac{\left| \begin{matrix}x_1^2+y_1^2 & y_1 & 1 \\x_2^2+y_2^2 & y_2 & 1 \\ x_3^2+y_3^2 & y_3 & 1\end{matrix} \right|}{2*\left| \begin{matrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right|},
k=\frac{\left| \begin{matrix}x_1 & x_1^2+y_1^2 & 1 \\x_2 & x_2^2+y_2^2 & 1 \\ x_3 & x_3^2+y_3^2 & 1\end{matrix} \right|}{2*\left| \begin{matrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right|}
$$
(Taken from here)
But this assumes that the points are defined using real numbers.
How should this formula be adapted when the points are defined using rationals with a common denominator? (i.e.: a point is defined using $(a/w, b/w)$).
I noticed that the denominator is the same as when calculating the orientation of three points. In that situation the 1's are replaced by the $w$'s. Is this also true for finding the center? Since the numerator contains squares, what are the implications then for the last column?
Update:
Based on @saulspatz answer I think the following is correct:
My actual situation is as follows (only for $h$):
$$
h=\frac{\left| \begin{matrix}\frac{x_1}{w_1}^2+\frac{y_1}{w_1}^2 & \frac{y_1}{w_1} & 1 \\\frac{x_2}{w_2}^2+\frac{y_2}{w_2}^2 & \frac{y_2}{w_2} & 1 \\ \frac{x_3}{w_3}^2+\frac{y_3}{w_3}^2 & \frac{y_3}{w_3} & 1\end{matrix} \right|}{2*\left| \begin{matrix}\frac{x_1}{w_1} & \frac{y_1}{w_1} & 1 \\ \frac{x_2}{w_2} & \frac{y_2}{w_2} & 1 \\ \frac{x_3}{w_3} & \frac{y_3}{w_3} & 1 \end{matrix} \right|}
$$
When I now multiply each row by its corresponding $w$ I have to correct the determinant by dividing by the same factor:
$$
h=\frac{\frac{1}{w_1^2w_2^2w_3^2}\left| \begin{matrix}x_1^2+y_1^2 & y_1w_1 & w_1^2 \\x_2^2+y_2^2 & y_2w_2 & w_2^2 \\ x_3^2+y_3^2 & y_3w_3 & w_3^2\end{matrix} \right|}{\frac{2}{w_1w_2w_3}\left| \begin{matrix}x_1 & y_1 & w_1 \\ x_2 & y_2 & w_2 \\ x_3 & y_3 & w_3 \end{matrix} \right|}
$$
I now multiply by $w_1^2w_2^2w_3^2$, which leaves:
$$
h=\frac{\left| \begin{matrix}x_1^2+y_1^2 & y_1w_1 & w_1^2 \\x_2^2+y_2^2 & y_2w_2 & w_2^2 \\ x_3^2+y_3^2 & y_3w_3 & w_3^2\end{matrix} \right|}{2w_1w_2w_3\left| \begin{matrix}x_1 & y_1 & w_1 \\ x_2 & y_2 & w_2 \\ x_3 & y_3 & w_3 \end{matrix} \right|}
$$
| If you multiply a column of a determinant by $c$ the determinant is also multiplied by $c$. So if we say that the points are $(\frac{x_k}{w}, \frac{y_k}{w})$ the values would become $$
h=\frac{\frac{1}{w^3} \left|\begin{matrix}x_1^2+y_1^2 & y_1 & 1 \\x_2^2+y_2^2 & y_2 & 1 \\ x_3^2+y_3^2 & y_3 & 1\end{matrix} \right|}{\frac{2}{w^2}\left| \begin{matrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right|},
k=\frac{\frac{1}{w^3}\left| \begin{matrix}x_1 & x_1^2+y_1^2 & 1 \\x_2 & x_2^2+y_2^2 & 1 \\ x_3 & x_3^2+y_3^2 & 1\end{matrix} \right|}{\frac{2}{w^2}\left| \begin{matrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right|}$$
Now after multiplying numerator and denominator by $w^3$, you could bring the the $w$ inside the determinant by multiplying the last column by $w$. That would make the last column $w$, but only in the denominators.
| {
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Partial fraction problem in $\int \frac{dx}{x^2(x^2+1)}$ I'm stuck with this integral:
$$\int \frac{dx}{x^2(x^2+1)}$$
I know that it can be solved in a simple way adding $x^2-x^2$ in the numerator, but I have troubles with partial fraction decomposition:
$$\frac{1}{x^2(x^2+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+1}$$
Then we get
$$1=Ax^4+Ax^2+Bx^3+Bx+Cx^4+Dx^3$$
But then I Find that
$$A+C=0$$
$$B+D=0$$
$$A=0$$
$$B=0$$
So everything is zero and nothing is equal to $1$. I know that this is inconsistent and there are for sure some mistakes, but I can't find out where they are. Thanks for your help.
| How did you get powers of $x^4$ in your expansion for the numerator?
$$\begin{align}\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+1}&=\frac{Ax(x^2+1)+B(x^2+1)+(Cx+D)x^2}{x^2(x^2+1)}\\
&=\frac{(A+C)x^3+(B+D)x^2+Ax+B}{x^2(x^2+1)}
\end{align}$$
Yielding equations:
$$\begin{align}B&=1\\A&=0\\B+D&=0\\A+C&=0\end{align}$$
| {
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If a,b, and c is real, and following the equation, solve for
If $a,b,c$ are real and aren't 0, that complete the following equations
$a^2 + a = b^2$
$b^2 + b = c^2$
$c^2 + c = a^2$
Solve for $(2a+c)(2b+a)(2c+b)$,
What I've done:
*
*Adding all of the equation of it and substact with $(a^2 + b^2 + c^2)$ of both side of the equation
$a+b+c=0$
*Substituting $a = -(b+c); b=-(c+a); c=-(a+b)$ to equation $(1),(2),(3)$ respectively:
$c^2 -b -c + 2bc= 0$
$a^2 - c -a + 2ac = 0$
$b^2 -b -a +2ab = 0$
*Expanding the equation into $4a^2c+2a^2b+9acb+4ab^2+2ac^2+4c^2b+2cb^2$
then re-arranged by it's constant
$4(a^2c+ab^2+c^2b)+2(a^2b+ac^2+cb^2)+9abc$
focused on
$2(a^2b+ac^2+cb^2)$
which can be written as
$2((a)ab+(c)ac+(b)cb)$
and subsituting $a, b, c$ that inside parentheses
$2(-(b+c)(ab)-(a+b)ac-(a+b)cb)
-2(ab^2 + a^2c + cb^2 +3abc)$
removing $3abc$, and $Cb^2$ from the parentheses
$-2(ab^2 + a^2c) - 6abc -cb^2$
so I get
$(4-2)(a^2c+ab^2+cb^2)+9abc-6abc$
$2(a^2c+ab^2+cb^2)+3abc$
But that's seems I can go to, I didn't find anything that could be useful anymore
| Adding all equations we get
$$a+b+c=0$$ and now we can eliminate the other variables, we obtain: $$3a^4-3a^2+a=0$$ and so on.
With $$c=-a-b$$ we get
$$b^2+b=(a+b)^2$$ and
$$(a+b)^2-(a+b)=a^2$$
From the equation $$b^2+b=a^2+b^2+2ab$$ we get
$$a^2+2ab-b=0$$ and $$b=\frac{a^2}{1-2a}$$ Can you proceed?
And we get $a$ from
$$a^2+a=\left(\frac{a^2}{1-2a}\right)^2$$
| {
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Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$
Prove that $9^n+4^{n+1}$ is a multiple of $5$ for all $n \in \mathbb{N}$
Proof. So i'm going to prove this by induction. The first case when $n=1$ is trivial since:
$$9+16=25,$$
implying $ 5 \mid 25$.
Now we need to show is divisible when $n=k+1$. We will use this later on, but $$9^k+4^{k+1}=5k.$$
The
n,
$$\begin{align}9^{k+1}+4^{k+2}
&= 9 \cdot 9^k+4 \cdot 4^{k+1} \\
&= 9 \cdot 9^k+(9-5) \cdot 4^{k+1} \\
&=9 \cdot 9^k+9 \cdot 4^{k+1} -5 \cdot 4^{k+1} \\
&= 9(9^k + 4^{k+1})-5 \cdot 4^{k+1} \\
&=\underbrace{9(5k)-5 \cdot 4^{k+1}}_{\text{This is where the $5k$ comes in}} \\
&= 5(9k-4^{k+1}),\end{align}$$
thus, the original expression a multiple of $5$.
Is my induction correct?
Edit: I see several answers that took a different approach, all is welcome it really helps me see it in a different way. Thank You!
| By Induction:
$n=0$: $9^0+4^1=5$ , divisible by $5$.
Hypothesis:
Assume : $9^n+4^{ n+1}$ is divisible by $5$.
Step:
$9^{n+1}+4^{n+2} =9 \cdot 9^{n} +4 \cdot 4^{n+1}=$
$(5+4)9^n +4 \cdot 4^{n+1} =$
$ 5 \cdot 9^n +4(9^n+4^{n+1})=$
$ 5 \cdot 9^n +4\cdot 5c= 5(9^n+4c)$,
where $c \in \mathbb{Z^+}$, since $(9^n+4^{n+1})$
is by hypothesis divisible by $5$,
| {
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Find all integer $t$ such that $\frac{t^2+9}{t+9}$ is an integer Find all integer $$ t \geq 0 $$ such that $$ \frac{t^2+9}{t+9} =k $$ k is integer.
How to solve this?
I've tried to solve this as an equation with parameter
$$
t^2-kt-9k+9=0 \\
D= k^2-36k+36\\
t_1=\frac{k-\sqrt{k^2-36k+36}}{2} \\
t_2=\frac{k+\sqrt{k^2-36k+36}}{2}
$$
so we need $$\sqrt{k^2-36k+36}$$ to be an int and also $$k-\sqrt{k^2-36k+36} \\
k+\sqrt{k^2-36k+36}$$ must be even... So I'm kinda stuck
| Your discriminant is wrong. You should get:
$$D=k^2-4(9-9k)=k^2\color{red}+36k\color{red}-36$$
The easier approach is to note that $\frac{t^2-81}{t+9}=t-9$ is an integer, so $$\frac{t^2+9}{t+9}=\frac{t^2-81}{t+9}+\frac{90}{t+9}$$ is an integer if and only if $\frac{90}{t+9}$ is an integer.
If you must use the discriminant method...
You only really need the discriminant to be a square, since if the polynomial has rational roots, then the roots are integers, since the polynomial has integer coefficients and lead coefficient $1.$
Completing the square, you need $D=(k+18)^2-(18^2+36)=(k+18)^2-360=n^2$ for some $n$, or $(k+18-n)(k+18+n)=360$.
Any factorization $90=AB$ gives us $k+18-n=2A, k+18+n=2B$, or $k=A+B-18$ and $n=B-A$ and $t=\frac{-k\pm n}{2}$ and this gives $t=A-9$ or $t=B-9.$
(Why did we use factorizations of $90$ rather than $360?$)
More geneally, if $f$ is a polynomial with integer coefficients, and $a$ is an integer such that $f(a)\neq 0$ then the set of $t$ such that:
$$\frac{f(t)}{t-a}$$
can be determined as follows. Let $u=t-a$ then:
$$\frac{f(t)}{t-a}=\frac{f(u+a)}{u}.$$ And $u$ divides $f(u+a)$ if and only if $u$ divides $f(a).$ So $t+a\mid f(t)$ if any only if $t+a\mid f(a).$
So write all the integer factors of $f(a)$ and then add $a$ to each of them.
In your example, $f(t)=t^2+9, a=-9,$ and $f(a)=90.$
| {
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If $a_n = a_{n-1}a_{n-3} + a_{n-2}^2$, then is it always divisible by $a_{n-4}$? The Somos Sequences are neat because they are integral yet contain a division operation.
If we consider the numerator $a_n = a_{n-1}a_{n-3} + a_{n-2}^2$ of the Somos-4 sequence, $a_1 = a_2 = a_3 = 1$, then is it always divisible by $a_{n-4}$?
Note that this is not obvious, since integrality in the orignal Somos-4 sequence implies that $a_{n-4}$ divides the numerator, $a_{n-4}$ and the numerator each built with respect to that sequence's previous elements and not with respect to the simpler sequence above.
| No. Working out the first few terms of the recurrence $a_n = a_{n-1} a_{n-3} + a_{n-2}^2$, we get
$$
\begin{array}{c|c}
n & a_n \\ \hline
1 & 1\\
2 & 1\\
3 & 1\\
4 & 2\\
5 & 3\\
6 & 7\\
7 & 23\\
8 & 118\\
9 & 1355\\
\end{array}
$$
from which we see $a_{5} = 3$ and $a_9 = 1355$, but $3 \nmid 1355$.
In fact for $n \leq 50$, $a_{n-4} \mid a_n$ only for $n=5,6,7,8$.
| {
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Evaluate $\lim_{x\to 0} \frac {1-(\cos 2x)^3(\cos 5x)^5(\cos 7x)^7(\sec 4x)^9(\sec 6x) ^{11}}{x^2}$
Evaluate $$\lim_{x\to 0} \frac {1-(\cos 2x)^3(\cos 5x)^5(\cos 7x)^7(\sec 4x)^9(\sec 6x) ^{11}}{x^2}$$
Now I can apply L'Hospital rule twice but believe it or not it would seriously be a very tedious task to do. I also tried writing $\cos 2x=\frac {e^{i2x}+e^{-i2x}}{2}$ and so on but couldn't continue due to large powers. Moreover I don't see any standard limits popping out. The only sequence I could notice was in the powers of the trigonometric functions which follow the series $3,5,7,9,11$
Any hints would be appreciated
| This is what one calls a crafted problem especially designed to intimidate students. One of the easiest approaches is to use the Taylor series expansion (presented in another answer).
However the problem can be solved in step by step manner using the standard limits $$\lim_{x\to 0}\frac{1-\cos x} {x^2}=\frac{1}{2},\lim_{x\to a} \frac{x^n-a^n} {x-a} =na^{n-1}\tag{1}$$ Dividing the first limit by $\cos x$ and noting that $\cos x\to 1$ as $x\to 0$ we can see that $$\lim_{x\to 0}\frac{1-\sec x}{x^2}=-\frac{1}{2}\tag{2}$$ Next we apply the usual algebraic technique of splitting to express the numerator $$1-abc\dots=1-a+a-abc\dots=1-a+a(1-bc\dots)$$ and note that each of $a, b, c\dots$ tends to $1$ as $x\to 0$. Thus the desired limit $L$ is equal to $$\lim_{x\to 0}\frac{1-\cos^32x}{x^2}+\lim_{x\to 0}\cos^32x\cdot\frac{1-\cos^55x\dots} {x^2}$$ The first limit in above expression is evaluated by writing it as $$\lim_{x\to 0}\frac{1-\cos^32x}{1-\cos 2x}\cdot\frac{1-\cos 2x}{(2x)^2}\cdot 4=3\cdot\frac{1}{2}\cdot 4=6$$ using standard limits $(1)$.
Thus $$L=6+\lim_{x\to 0}\frac{1-\cos^55x}{x^2}+\lim_{x\to 0}\cos^55x\cdot\frac{1-\cos^77x\dots}{x^2}$$ The first limit above evaluates to $5(1/2)25=125/2$ and thus $$L=\frac{137}{2}+\lim_{x\to 0}\frac{1-\cos^77x}{x^2}+\lim_{x\to 0}\cos^77x\cdot\frac{1-\sec^94x\dots}{x^2}$$ so that $$L=\frac{137}{2}+\frac{343}{2}+\lim_{x\to 0}\frac{1-\sec^94x}{x^2}+\lim_{x\to 0}\sec^94x\cdot\frac{1-\sec^{11}6x}{x^2}$$ or $$L=240-9\cdot\frac{1}{2}\cdot 16-11\cdot\frac{1}{2}\cdot 36=-30$$ using limits $(1),(2)$.
| {
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Find the highest power of prime which divides a factorial. Find the highest power of $10$ that divides $50!$.
I know by De-Polignac's formula ,the highest power of $5$ that divides $50!$ is $12$ & the highest power of $2$ that divides $50!$ is $47$.
I am able to find this because $5$ & $2$ are primes but here $10$ is not prime but $10=5 \times 2$ so what can we say about the power of 10
My observation say that it may be $47-12$
Please correct me
| As you found:
The highest power of $2$ that divides $50!$ is $\left\lfloor \frac{50}{2} \right \rfloor + \left\lfloor \frac{50}{2^2} \right \rfloor + \left\lfloor \frac{50}{2^3} \right \rfloor + \cdots$ which evaluates as $ 25+12+6+3+1$ $= 47$, each term being half-rounded-down of the previous one.
The highest power of $5$ that divides $50!$ is $\left\lfloor \frac{50}{5} \right \rfloor + \left\lfloor \frac{50}{5^2} \right \rfloor = 10+2= 12$, similarly dividing through by $5$ repeatedly.
So $50! = k\cdot 2^{47}\cdot 5^{12}$, where $k$ is not divisible by either $2$ or $5$. Clearly we can also write this as $50! = k\cdot 2^{35}\cdot 10^{12}$ - and again $k$ is coprime to both $2$ and $10$.
So the result is that $50!$ is divisible by $10^{12}$ but not by $10^{13}$, since here the powers of $5$ determine how many powers of $10$ can be extracted from $50!$.
| {
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Showing that $\frac{1}{2\pi}\int^{2\pi}_{0}(2\cos\theta)^{2n}d\theta=\frac{(2n)!}{n!n!}$ As the title suggests, I would like to show that: $$\frac{1}{2\pi}\int^{2\pi}_{0}(2\cos\theta)^{2n}d\theta=\frac{(2n)!}{n!n!}$$ for every positive integer $n$.
How can I go about doing that? I understand the that there are already solutions for the problem: $\frac{1}{2\pi}\int^{2\pi}_{0}(\cos\theta)^{2n}d\theta= {2n \choose n} \frac{\pi}{2^{2n-1}}$. Is there any modifications I can do to obtain the desired solution?
| Integrating by parts
\begin{align*}
\int_0^{2\pi}\cos^{2n}\theta d\theta&=\Big[\cos^{2n-1}\theta\sin\theta\Big]_0^{2\pi}-\int_0^{2\pi}\sin\theta(2n-1)\cos^{2n-2}\theta(-\sin\theta) d\theta\\
&=(2n-1)\int_0^{2\pi}(\cos^{2n-2}\theta-\cos^{2n}\theta)d\theta\\
&=(2n-1)\int_0^{2\pi}(\cos^{2n-2}\theta-\cos^{2n}\theta)d\theta\\
\end{align*}
So, we have
\begin{align*}
\int_0^{2\pi}\cos^{2n}\theta d\theta&=\frac{2n-1}{2n}\int_0^{2\pi}\cos^{2(n-1)}\theta d\theta\\
&=\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\int_0^{2\pi}\cos^{2(n-2)}\theta d\theta\\
&=\dots\\
&=\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-5}{2n-4}\cdots\frac{1}{2}\cdot\int_0^{2\pi}d\theta\\
&=\left(\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdot\frac{2n-5}{2n-4}\cdots\frac{1}{2}\right)\left(\frac{2n}{2n}\cdot\frac{2n-2}{2n-2}\cdot\frac{2n-4}{2n-4}\cdots\frac{2}{2}\right)\left(2\pi\right)\\
&=\;\frac{2\pi(2n)!}{2^{2n}(n!)^2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2742664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Integral $\int \frac {dx}{\sin^2 x + \tan^2x}$
I am trying to find the following integral:
$$\int \frac {dx}{\sin^2 x + \tan^2x}$$
I have tried the common thing to do when encountering rational functions that contains rational functions and converting everything in terms of $\tan \frac{x}{2}$ then substituting it.
$$\tan\frac{x}{2}=t\Rightarrow \sin x=\frac{2t}{1+t^2},\ \cos x=\frac{1-t^2}{1+t^2},\ dx=\frac{2}{1+t^2}dt$$
$$\Rightarrow \int \frac{dx}{\sin^2 x + \frac{\sin^2 x}{\cos^2 x}}=\int \frac{\frac{2}{1+t^2}}{\left(\frac{2t}{1+t^2}\right)^2+\left(\frac{2t}{1-t^2}\right)^2}dt$$
However I am stuck. Is there perhaps an easier way to approach it?
| $$\dfrac1{\sin^2x+\tan^2x}=\dfrac{\cos^2x}{\sin^2x(1+\cos^2x)}$$
Method$\#1:$
$$=\dfrac{\cos^2x+1-1}{\sin^2x(1+\cos^2x)}=\csc^2x-\dfrac1{(1-\cos^2x)(1+\cos^2x)}$$
Now $$\dfrac2{(1-\cos^2x)(1+\cos^2x)}=\dfrac1{1-\cos^2x}+\dfrac1{1+\cos^2x}$$
Finally $$\dfrac1{1+\cos^2x}=\dfrac{\sec^2x}{2+\tan^2x}$$
Method$\#2:$
$$\dfrac{\cos^2x}{\sin^2x(1+\cos^2x)}=\dfrac c{(1-c)(1+c)}=\dfrac A{1-c}+\dfrac B{1+c}$$ using Partial Fraction Decomposition where $c=\cos^2x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2745598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral values of $k$ for which polynomial has integral roots. How integral values of $k$ exists such that all roots of the polynomial:
$$f(x)=x^3-(k-3)x^2-11x+(4k-8)$$ are also integers.
Could someone please provide me some direction to proceed in this. First I thought first root would be obtained by hit and trial but it is not the case here. I also tried rewriting $x^3-(k-3)x^2-11x+(4k-8)=0$ as $x^3+3x^2-11x-8=k(x^2-4)$ and solve the question graphically but didn't succeed in that. Please give some direction
| Hint:
Since $$k ={x^3+3x^2-11x-8\over x^2-4}\implies x^2-4 \mid x^3+3x^2-11x-8$$
Since also $$x^2-4\mid (x+3)(x^2-4) = x^3+3x^2 -4x-12$$
we get
$$ x^2-4\mid ( x^3+3x^2-11x-8)-( x^3+3x^2-4x-12)= -7x+4$$
thus $$x-2\mid -7x+4\implies x-2 \mid -7x+4+7(x-2) = -10$$
Finally we have $x-2\in \{-10,-5,-2,-1,1,2,5,10\}$ so $x\in \{-8,-3,0,1,3,4,7,12\}$
Now, of course not all these are good...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Taylor series $\sqrt{1-2x+x^2+o(x^3)}$ I need to solve this asymptotic expansion for $x\to0$
$$\sqrt{1-2x+x^2+o(x^3)}$$
This is the expansion to use:
$$\sqrt{1+t} = 1+ \frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}+o(x^3)$$
let:
$$t = -2x+x^2+o(x^3)$$
My question is, should I stop to the first order or keep going?
How can I decide when to stop?
I have tried first and second order and I get:
$$\sqrt{1+t}= 1+ \frac{-2x+x^2+o(x^3)}{2}+o(-2x+x^2+o(x^3) = 1-x+\frac{1}{2}x^3+o(x^3)$$
The second grade:
$$\sqrt{1+t}= 1+ \frac{-2x+x^2+o(x^3)}{2} - \frac{\big(-2x+x^2+o(x^3)\big)^2}{8} +o\big(-2x+x^2+o(x^3)\big)^2 = 1-x+4x^2+\frac{1}{2}x^3-4x^3-\frac{1}{8}x^4+x^4+o(x^4)$$
| You are being told to work in the ring $\Bbb Q[[x]]/(x^3)$, and to find there a square root of $1-2x+x^2$. I think that the answer is obvious.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Precise definition of Multivariable limit I have encountered this limit: $\lim_{(x,y) to (0,0)} \frac{x^2+3y-7}{2+x^3-5y^2}$. We can substitute and find the limit to be $\frac{-7}{2}$since it is not a difficult limit, however I am trying to prove it rigorously and am getting stuck. I’m starting out with for all $\epsilon > 0$ there exists $\delta> 0$ such that $| \frac{x^2+3y-7}{2+x^3-5y^2}+\frac{7}{2}|<\epsilon $ whenever $\delta>\sqrt{x^2+y^2}$. I expanded the first absolute value and got $\frac{1}{2}|\frac{7x^3+2x^2-35y^2+6y}{x^3-5y^2+2}|$, but I cannot get past here to find any inequalities. I think it should be easy to do, since it’s a really easy limit there’s nothing undefined, but I’m stuck. Thank you.
| Note that
$$
\delta > \sqrt{x^2+y^2} \implies \delta > \lvert x\rvert,\, \delta>\lvert y\rvert
$$
Then $\delta > \sqrt{x^2+y^2}$ implies that
$$
\lvert 7x^3+2x^2-35y^2+6y \rvert < 7\delta^3 + 2\delta^2 +35\delta^2+6\delta = 7\delta^3 + 37\delta^2+6\delta
$$
Now, suppose without loss of generality that $\delta < 1/5$ (I picked this small number pretty arbitrarily, any small enough number works here) then
$$
\lvert x^3-5y^2+2 \rvert \geq 2 - \delta^3 - 5\delta^2 \geq 2-\frac{1}{5^3}-5\frac{1}{5^2} \geq 1
$$
Combining the inequalities we have (and noting that $\delta < 1/5$) we see that
$$
\frac{1}{2}\left\lvert \frac{7x^3+2x^2-35y^2+6y}{x^3-5y^2+2} \right\rvert < \frac{7\delta^3 + 37\delta^2+6\delta}{2} \leq \frac{7\delta + 37\delta+6\delta}{2} = 25\delta
$$
So, given $\epsilon >0$ picking $\delta = \min \{\epsilon /50, 1/5\}$ will give you the inequality
$$
\frac{1}{2}\left\lvert \frac{7x^3+2x^2-35y^2+6y}{x^3-5y^2+2} \right\rvert<\epsilon
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Invariant Factors of a complete tripartite graph Let $K_{4,4,12}$ be a complete tripartite graph on 20 vertices. Considering its adjacency matrix with a convenient labeling,its nonzero eigenvalues are -4,-8,12. Now the invariant factors are 4,4,24. How did they find the invariant factors in detail, please?
| The version of the paper I found says that the invariant factors are $4,4,24$; we get $1,9,36$ for the second example $K_{2,9,9}$.
In any case, here is how you get there. These are not the invariant factors of the adjacency matrix (there's a result about those elsewhere in the paper) but of the quotient matrix, which in the case of $K_{4,4,12}$ is
$$\begin{bmatrix} 0 & 4 & 12 \\ 4 & 0 & 12 \\ 4 & 4 & 0\end{bmatrix}.$$
(In general, it has zeroes along the diagonal and otherwise the $(i,j)$ entry is the size of the $j^{\text{th}}$ part of the complete multipartite graph.)
To find the invariant factors, we reduce the matrix to a diagonal matrix in which each diagonal entry divides the next, by doing integer-invertible row and column operations, like so:
\begin{align}
\begin{bmatrix} 0 & 4 & 12 \\ 4 & 0 & 12 \\ 4 & 4 & 0\end{bmatrix} & \leadsto \begin{bmatrix} 4 & 0 & 12 \\ 0 & 4 & 12 \\ 4 & 4 & 0\end{bmatrix} & \text{(swap a pivot into $(1,1)$)} \\
& \leadsto \begin{bmatrix} 4 & 0 & 12 \\ 0 & 4 & 12 \\ 0 & 4 & -12\end{bmatrix} & \text{(subtract row 1 from row 3)} \\
& \leadsto \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 12 \\ 0 & 4 & -12\end{bmatrix} & \text{(subtract $3\times$ column 1 from column 3)} \\
& \leadsto \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 12 \\ 0 & 0 & -24\end{bmatrix} & \text{(subtract row 2 from row 3)} \\
& \leadsto \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & -24\end{bmatrix} & \text{(subtract $3\times$ column 2 from column 3)} \\
& \leadsto \begin{bmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 24\end{bmatrix} & \text{(multiply the third row by $-1$)} \\\end{align}
In general, see the algorithm on Wikipedia.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limits of a product $N$ is a positive integer.
I want to calculate this limit, but i couldn't get anywhere when i tried.
$$P[n]=\left(1+\frac {1}{n^2}\right)\left(1+\frac {2}{n^2}\right)\cdots\left(1+\frac {n-1}{n^2}\right)$$ as $n\to \infty$
I tried to apply $\ln()$ at both sides to transform into a sum etc..
tried to use functions to limit the superior and inferior intervals of the function like this:
$$\exp\left(\left(n-1\right)\ln\left(1+\frac1{n^2}\right)\right)<P[n]<\exp\left((n-1)\ln\left(1+\frac{n-1}{n^2}\right)\right)$$
The only thing I get is that:
$1<\lim(P[n])<e$
Can anyone help me to solve this?
| $$\left(1+\frac {j}{n^2}\right)\left(1+\frac {n-j}{n^2}\right)=1+\frac{1}{n}+\frac{j(n-j)}{n^4}$$
By AM-GM we have
$$\sqrt{j(n-j)}\leq \frac{n}{2} $$
and hence
$$1+\frac{1}{n} \leq \left(1+\frac {j}{n^2}\right)\left(1+\frac {n-j}{n^2}\right) \leq 1+\frac{1}{n}+\frac{1}{4n^2}$$
Therefore
$$\left(1+\frac{1}{n}\right)^{\frac{n-1}{2}} \leq P[n] \leq \left( 1+\frac{1}{n}+\frac{1}{4n^2} \right)^\frac{n}{2}$$
Now
$$\lim_n \left(1+\frac{1}{n}\right)^{\frac{n-1}{2}}=\left( \left(1+\frac{1}{n}\right)^{n} \right)^{\frac{n-1}{2n}}=\sqrt{e}$$
And
$$\lim_n \left( 1+\frac{1}{n}+\frac{1}{4n^2} \right)^\frac{n}{2} =\lim_n \left(\left( 1+\frac{4n+1}{4n^2} \right)^\frac{4n^2}{4n+1}\right)^\frac{4n+1}{8n} =\sqrt{e}$$
Therefore
$$\lim_n P[n]=\sqrt{e}$$
| {
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"url": "https://math.stackexchange.com/questions/2753914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proving Hermite's Identity in a different approach To prove
$$S=\left [x \right]+\left [x+\frac{1}{n} \right]+\left [x +\frac{2}{n}\right]+\cdots+\left [x +\frac{n-1}{n}\right]=\left [nx \right]$$
using and starting with $$x-1 \lt \left [x \right]\le x \tag{1}$$ we have
$$x+\frac{1}{n}-1 \lt \left [x +\frac{1}{n}\right] \le x+\frac{1}{n} \tag{2}$$
$$x+\frac{2}{n}-1 \lt \left [x +\frac{2}{n}\right] \le x+\frac{2}{n} \tag{3}$$
$$x+\frac{3}{n}-1 \lt \left [x +\frac{3}{n}\right] \le x+\frac{3}{n} \tag{4}$$
and so on till
$$x+\frac{n-1}{n}-1 \lt \left [x +\frac{n-1}{n}\right] \le x+\frac{n-1}{n} \tag{n}$$
Adding all we get
$$nx+\frac{n(n-3)}{2} \lt S \le nx+\frac{n(n-1)}{2}$$
Now how can we prove that between $nx+\frac{n(n-3)}{2}$ and $nx+\frac{n(n-1)}{2}$ there is only on integer which is $\left [nx \right]$?
| Here is a proof I came up with
about a year ago:
For all $x$ which are real numbers, prove that $\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$
The basic idea is that
you have to see
how far $x$ is from
$\lfloor x \rfloor$
and,
if $d = x-\lfloor x \rfloor$,
what $\lfloor nd \rfloor$ is.
This allows you do get the sum.
| {
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"url": "https://math.stackexchange.com/questions/2754876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Without using the Rule of Sarrus, prove that: Without using the Rule of Sarrus, prove that:
$$\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|=3abc-a^3-b^3-c^3$$
My Approach:
$$LHS=
\left|
\begin{matrix}
(b+c)&(a-b)&a \\
(c+a)&(b-c)&b \\
(a+b)&(c-a)&c \\
\end{matrix}\right|$$
$$C_1\to C_1+C_2$$
$$=
\left|
\begin{matrix}
(c+a)&(a-b)&a \\
(a+b)&(b-c)&b \\
(b+c)&(c-a)&c \\
\end{matrix}\right|$$
$$C_1\to C_1-C_3$$
$$=
\left|
\begin{matrix}
c&(a-b)&a \\
a&(b-c)&b \\
b&(c-a)&c \\
\end{matrix}\right|$$
How do I complete the rest?
| \begin{align*}
\left|
\begin{matrix}
c&(a-b)&a \\
a&(b-c)&b \\
b&(c-a)&c \\
\end{matrix}\right|&= c\left|\begin{matrix}
(b-c)&b \\
(c-a)&c \\
\end{matrix}\right|-a\left|\begin{matrix}(a-b)&a\\(c-a)&c \\\end{matrix}\right|+b\left|\begin{matrix}(a-b)&a\\(b-c)&b \\\end{matrix}\right|\\
&=c[bc-c^2-bc+ab]-a[ac-bc-ac+a^2]+b[ab-b^2-ab+ac]\\
&=-c^3+abc+abc-a^3-b^3+abc\\
&=3abc-a^3-b^3-c^3.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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British Maths Olympiad (BMO) 2002 Round 1 Question 3 Proof without Cauchy-Schwarz? The question states:
Let $x,y,z$ be positive real numbers such that $x^2 + y^2 + z^2 = 1$
Prove that
$x^2yz + xy^2z + xyz^2 ≤ 1/3$
I have a proof of this relying on the fact that:
$x^2/3 +y^2/3 + z^2/3 \geq (x+y+z)^3/9 $ (A corollary of C-S I believe)
Is there an elementary proof without this fact (or C-S in general)?
| Here is my approach.
Using AM-GM,
$\begin{align} x^2+y^2+z^2 &\ge 3(xyz)^{2/3} \\ \implies 1 &\ge 3(xyz)^{2/3} \\ \implies (xyz)^{2/3} &\le \dfrac{1}{3} \\ \implies xyz \le \dfrac{1}{3\sqrt{3}} \tag 1 \end{align}$
Again, using AM-GM,
$\begin{align} \dfrac{x^3+y^3+z^3}{3} &\ge 3xyz \\ \implies x^3+y^3+z^3-3xyz &\ge 0 \\ \implies (x+y+z)(x^2+y^2+z^2 -xy-yz-zx) &\ge 0\end{align}\tag*{}$
Since $x,y,z \in \mathbb R^+ $, $x+y+z \ge 0 $.
$\begin{align}\therefore \, x^2+y^2+z^2 -xy-yz-zx &\ge 0 \\ \implies -x^2-y^2-z^2 +xy+yz+zx &\le 0 \\ \implies -2x^2-2y^2-2z^2 +2xy+2yz+2zx &\le 0 \\ \implies x^2+y^2+z^2 +2xy+2yz+2zx &\le 3(x^2+y^2+z^2) \\ \implies (x+y+z)^2 &\le 3 \\ \implies x+y+z &\le \sqrt{3} \tag 2 \end{align}$
Multiplying $(1)$ and $(2)$, we can get the desired result.
| {
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"source": "stackexchange",
"question_score": "11",
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Evaluating $\int \sqrt{\frac{\cos x - \cos^3 x}{1-\cos^3 x}}dx$
Evaluate $\int \sqrt{\frac{\cos x - \cos^3 x}{1-\cos^3 x}}dx$
My attempt:
$I=\int \sqrt{\frac{\cos x - \cos^3 x}{1-\cos^3 x}}dx=\int \sqrt{\frac{\cos x(1-\cos^2 x)}{1-\cos^3 x}}dx=\int \sqrt{\frac{\cos x\sin^2 x}{1-\cos^3 x}}dx=\int \sin x\sqrt{\frac{\cos x}{1-\cos^3 x}}dx=\int\sqrt{\frac{\cos x}{1-\cos^3 x}}(\sin x)dx=-\int \sqrt{\frac{\cos x}{1-\cos^3 x}}(-\sin x)dx$
Let $z=\cos x$
$\therefore dz=(-\sin x)dx$
$\therefore I=-\int\sqrt{\frac{z}{1-z^3}}dz$
I cannot understand how to proceed further. Please help.
| Hint: Substitute $u=z^{3/2}$. You should then be able to recognise the integral you get.
How could you have come up with this yourself? Well note that you have $\sqrt z dz$ on top (which is the same up to a constant as $d(z^{3/2})$), and you have $1-z^3$ on the denominator, whereas you're probably more used to seeing things of the form $1-z^2$ on the denominator. So the substitution $z^{3/2}$ has the effect of transforming the denominator into something you recognise, while also getting rid of the squareroot on the numerator.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Calculate residue in every pole of $\frac{z^2}{(z-2)^2(\cos{z}-1)^3}$ The problem is as simple as the title suggests. Of course, the formula for the residue of an order $n$ pole involving $n-1$ derivatives could be applied, but the computation will be extremely long, as there are poles of order $6$ in the function:
$\frac{z^2}{(z-2)^2(\cos{z}-1)^3}$
So I was trying to follow the way of finding out the Laurent series, but I am not sure how to proceed (I will need to compute the inverse of the series of $(\cos{z}-1)^3$ and I am not sure at all that this is an easy task.
Are there other approaches that I should explore?
Thanks a lot!!
| Let
$$f(z) = \frac {z^2} {(z - 2)^2} =
1 + \frac 4 {z - 2} + \frac 4 {(z - 2)^2}, \\
g(z) = \frac 1 {(\cos z - 1)^3}.$$
To find the expansion of $g$, take the product of the series of $1/g$ and of $g$ and equate the coefficients:
$$(a z^6 + b z^8 + c z^{10} + O(z^{12}))
(A z^{-6} + B z^{-4} + C z^{-2} + O(1)) = \\
a A + (a B + b A) z^2 + (a C + b B + c A) z^4 + O(z^6) = \\
1 + O(z^6) \Rightarrow \\
A = \frac 1 a, B = - \frac b {a^2}, C = \frac {b^2 - a c} {a^3}.$$
The series coefficients of $g$ at other poles $z_k = 2 \pi k$ are the same:
$$g(z) = -\frac 8 {(z - z_k)^6} -\frac 2 {(z - z_k)^4} -
\frac 4 {15 (z - z_k)^2} + O(1), \\
\operatorname{Res}_{z = z_k} f(z) g(z) =
-8[(z - z_k)^5]f(z) - 2[(z - z_k)^3]f(z) - \frac 4 {15} [(z - z_k)^1]f(z) = \\
\frac {8 (\pi k)^5 - 32 (\pi k)^4 + 63 (\pi k)^3 - 47 (\pi k)^2 + 8 \pi k + 45}
{30 (\pi k - 1)^7}.$$
The remaining residue is
$$\operatorname{Res}_{z = 2} f(z) g(z) = \frac {(3 \cot 1 - 1) \csc^6 1} 2.$$
The residue at $z = \infty$ is not defined, because $z = \infty$ is not an isolated singularity.
The sum of all residues is zero. This can be seen by considering the sequence of integrals of $f(z) g(z)$ over the circles with radii $(2 k + 1) \pi$. The sequence tends to zero, therefore the partial sums of the residues also tend to zero.
| {
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"question_score": "2",
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Find a matrix $N$ s.t. $N^2=H$ Given the below matrix, find a matrix $N$ s.t. $N^2=H$
$$H=\begin{bmatrix}2&i&2&1+i\\-i&2&-2i&1-i\\2&2i&5&2+2i\\1-i&1+i&2-2i&3\end{bmatrix}$$
I solved for the characteristic polynomial, which is $(x-1)^3(x-9)$
Any help will be appreciated.
| Your matrix is diagonalizable. More precisely, if$$M=\begin{pmatrix}1+i & -1-i & -2 & -i \\ 1-i & 0 & 0 & 1 \\ 2+2 i & 0 & 1 & 0 \\ 2 & 1 & 0 & 0\end{pmatrix},$$then$$M^{-1}.H.M=\begin{pmatrix}9&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}.$$So, an answer to your question is$$M.\begin{pmatrix}3&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}.M^{-1}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Express the given polar equation in simplest rectangular form: $r = 4 + 3cosθ$ Express the given polar equation in simplest rectangular form: $r = 4 + 3cosθ$
My attempt
Multiplying r on both sides, we get
$r^2=4r+3 rcos\theta$
since, $x=rcos\theta$ and $y=rsin\theta$
and $r^2=x^2+y^2$
we get $x^2+y^2=4r+3x$
can anyone please explain after this step..
| $$r^2 - 4 r = 3 x$$
$$x^2 +y^2 - 4\sqrt{x^2 +y^2} = 3 x$$
which represents a cardoid.. it is better to leave it that way.
if you try to get rid of radical sign you end up with another equation representing another symmetrical cardoid, not really what you wanted to represent at the outset:
$$x^2 +y^2 + 4\sqrt{x^2 +y^2} = 3 x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Without using a calculator, is $\sqrt[8]{8!}$ or $\sqrt[9]{9!}$ greater? Which is greater between $$\sqrt[8]{8!}$$ and $$\sqrt[9]{9!}$$?
I want to know if my proof is correct...
\begin{align}
\sqrt[8]{8!} &< \sqrt[9]{9!} \\
(8!)^{(1/8)} &< (9!)^{(1/9)} \\
(8!)^{(1/8)} - (9!)^{(1/9)} &< 0 \\
(8!)^{(9/72)} - (9!)^{8/72} &< 0 \\
(9!)^{8/72} \left(\left( \frac{8!}{9!} \right)^{(1/72)} - 1\right) &< 0 \\
\left(\frac{8!}{9!}\right)^{(1/72)} - 1 &< 0 \\
\left(\frac{8!}{9!}\right)^{(1/72)} &< 1 \\
\left(\left(\frac{8!}{9!}\right)^{(1/72)}\right)^{72} &< 1^{72} \\
\frac{8!}{9!} < 1 \\
\frac{1}{9} < 1 \\
\end{align}
if it is not correct how it would be?
| I would compare it by using logarithms:
$$\ln a = \frac{1}{8}(\ln1 + ...+\ln8)$$
$$\ln b = \frac{1}{9}(\ln1 + ...+\ln8+\ln9)=\frac{8}{9} \ln a + \frac{1}{9}\ln9$$
$$\ln b-\ln a=\frac{1}{9}(\ln9-\ln a)=\frac{1}{9} \ln\frac{9}{a}$$
$a$ is obviously less than 8 so:
$$\frac{9}{a} \gt 1$$
$$\ln\frac{9}{a}\gt0$$
$$\ln b - \ln a \gt 0$$
$$b \gt a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2762306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 10,
"answer_id": 1
} |
What is $\sqrt[4]{-1}$ So, we all know that $\sqrt{-1}=i$, that $\sqrt{-2}=\sqrt2i$, and so on. And that, for example, $\sqrt[3]{-27}=-3$. But I was wondering; what would $\sqrt[4]{-1}$ be? Since all $n^{th}$ roots of $-1$ that are odd result in $-1$ (Because $-1^{1,3,5,7,9,\dots}=-1$), I wondered what variation of $i$ would even roots of $-1$ give? (Like $\sqrt{-1},\sqrt[4]{-1},\sqrt[6]{-1},\dots$)
| We will use Euler's Identity, $$e^{i\pi} + 1 = 0,\tag1$$ or in particular, $$\begin{align} e^{i\theta} &= \cos\theta + i\sin\theta \\ &= \text{cis}\,\theta.\end{align}\tag2$$
From $(1)$, we get that $e^{i\pi} = -1$, thus $$\begin{align} \left(e^{(i\pi)/2}\right)^{1/2} &= e^{(i\pi)/4} \\ &= \sqrt [4] {-1}.\end{align}$$ Now, using $(2)$, by substituing $\theta = \dfrac{\pi}{4}$, we obtain the following result: $$\begin{align} \sqrt [4] {-1} &= \cos \left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right) \\ \\ &= \frac{1}{\sqrt{2}} + i\left(\frac{1}{\sqrt{2}}\right) \\ \\ &= \boxed{ \ \frac{1}{\sqrt{2}}\left(1+i\right). \ }\end{align}$$ See what pattern you can find now with the sequence, $\sqrt{-1}, \sqrt [4] {-1}, \sqrt [6] {-1},\ldots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How can one prove that $\sqrt{ 2} \cdot \sqrt{ 3} = \sqrt{ 6}$? I have already proved that $\sqrt{2}\cdot \sqrt{2}=2$ so I hope I can now use $\sqrt{3}\cdot \sqrt{3}=3$ and the same for 6.
The exercise comes from Stillwell: Mathematics and its History. The other exercises have not been complicated so probably the solution to proving $\sqrt{2}\cdot \sqrt{2}=2$ I borrowed from Martin K was way over the top. But it cannot be adapted, I think, to this one where the roots are not of the same number.
Here is my attempt:
$r^2=2, s^2=3$.
$r\cdot r=2$ implies $r=\sqrt{2}$ and $s\cdot s=3$
implies $s=\sqrt{3}$
I also know that $\sqrt{6}\cdot \sqrt{6}=6$.
$r^2\cdot s^2=6$ so $(rs)^2=6$
This implies that $(\sqrt{2}\sqrt{3})^2=\sqrt{6}\cdot \sqrt{6}$
Which I am not sure about, but hope, implies that $\sqrt{2}\sqrt{3}=\sqrt{6}$
| Let $\sqrt{2}\sqrt{3} = x$.
Substitute $\sqrt{2} = \dfrac{2}{\sqrt{2}}$ and $\sqrt{3} = \dfrac{3}{\sqrt{3}}$. Then we should have: $$\left(\frac{2}{\sqrt{2}}\right)\left(\frac{3}{\sqrt{3}}\right) = \sqrt{6}.\tag1$$ Note that the $LHS$ (Left Hand Side of the equation) is equal to $\dfrac{6}{\sqrt{2}\sqrt{3}}$. It follows, then, that: $$\begin{align} \frac{6}{\sqrt{2}\sqrt{3}} &=\sqrt{6} \\ &\Downarrow \\ \frac{6}{x} &= \sqrt{6}.\end{align}$$ For you to have made your substitutions in Eq. $(1)$, it must follow that $x = \sqrt{6}$.
$$\therefore \sqrt{2}\sqrt{3} = \sqrt{6}.\tag*{$\bigcirc$}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant? Why is the difference between these two functions a constant?
$$f(x)=\frac{2x^2-x}{x^2-x+1}$$
$$g(x)=\frac{x-2}{x^2-x+1}$$
Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant.
Of course I can calculate it is true: the difference is $2$, but my intuition is still completely off here. So, who can provide some intuitive explanation of what is going on here? Perhaps using a graph of some kind that shows what's special in this particular case?
Thanks!
BACKGROUND: The background of this question is that I tried to find this integral:
$$\int\frac{x dx}{(x^2-x+1)^2}$$
As a solution I found:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{2x^2-x}{3\left(x^2-x+1\right)}+C$$
Whereas my calculusbook gave as the solution:
$$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{x-2}{3\left(x^2-x+1\right)}+C$$
I thought I made a mistake but as it turned out, their difference was constant, so both are valid solutions.
| I'm not sure anyone is speaking to your observation that the two numerators have different degrees.
Let's flip this around the other way:
\begin{align*}
\frac{x-2}{x^2-x+1} + 2 &= \frac{x-2}{x^2-x+1} + 2\frac{x^2-x+1}{x^2-x+1} \\
&= \frac{2x^2 -x}{x^2-x+1} \text{.}
\end{align*}
That is, we started with a thing having a linear numerator and added a constant to it. But when we brought the constant to have a common denominator, it picked up a degree two factor. Then the addition was forced to produce a degree two sum.
To sum up, in the context of rational functions, when you add constants, you are adding polynomials having the degree of the denominator to the polynomials in the numerators. So constants effectively have "degree two in the numerator" in your example.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2764985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "54",
"answer_count": 16,
"answer_id": 7
} |
How to get the expected value of these numbers?
So for the E(x) of (a) I got 1 $\cdot$ ($\frac{1}{5} $) + 3 $\cdot$ ($\frac{2}{5} $)+ 9 $\cdot$ ($\frac{2}{5} $) = 9.8
Why is the answer showing E(x)= 5?
Also To compute the Variance, according to the formula, E($X^2$) - E$(X)^2$ shouldn't the answer be $\frac{181}{5} $ - $\frac{625}{5}$ ?
Many thanks in advance !
|
So for the E(x) of (a) I got $1 \cdot \tfrac 15 + 3 \cdot \tfrac25+ 9 \cdot\tfrac25 = 9.8$
Why is the answer showing E(x)= 5?
Because $\tfrac {1+ 6+18}5 =\tfrac {25}5$ not $\tfrac {49}5$ (...how did you even?)
Also To compute the Variance, according to the formula, E($X^2$) - E$(X)^2$ shouldn't the answer be $\frac{181}{5} $ - $\frac{625}{5}$ ?
$$\dfrac{1+2\cdot 3^2+2\cdot 9^2}{5}-\left(5\right)^2=\dfrac{181}5-\dfrac {125}5=\dfrac{56}5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2765795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Conceptual Idea of Functional Reduction I have a doubt regarding functional reduction(In mathematics, reduction refers to the rewriting of an expression into a simpler form).
I have a function, let me call it as $f(x)$ and I have rewritten it to a simpler form let me call it as $g(x)$. Can we say both $f(x)$ and $g(x)$ are equivalent? From a rough over view, we may argue that both render same result for the values of $x$.
Assume I have a function $f(x) = \frac{x^2-9}{x-3}$ . Clearly it is not defined at $x=3$. But if I rewrite it as $g(x) = x+3$ you can say that it is defined on $x=3$. Since $g(x)$ and $f(x)$ both don't agree on common domain and range can we say both are equivalent? Is it conceptually fair to argue such point? Then what is the point of reduction if we gain some domain which was not the original property of the function before reduction.?
| Equal functions agree on domain and values.
When one proves identities, one only requires the (restricted) functions to agree on their mutual domain, the intersection of their domains. If we do not make this relaxation, many identities become false. For instance, "$1 = \frac{\sin z}{\sin z}$" is undefined on the right for $z$ an integer multiple of $\pi$. If we use ideas from calculus, we can "fix this" -- the right-hand side has finite limits which equal the left-hand side at the points where the right-hand side is undefined.
You are right to keep the change of domain in mind. Although the above is what we do for identities, it is not what we do when solving equations. Consider $\sin x \cos x = \sin x$. If we "divide both sides by $\sin x$", we do exactly the sort of manipulation we were describing above. However, the integer multiples of $\pi$ that we remove from the domains of both sides actually contain solutions, which our reduced equation has "forgotten". So there are two ways to go.
First, we can choose to not reduce our domain by neither multiplying nor dividing both sides by something that can ever be zero...
\begin{align}
\sin x \cos x &= \sin x \\
\sin x \cos x - \sin x &= 0 \\
\sin x (\cos x - 1) &= 0 \\
\sin x = 0 \text{ or }& \cos x = 1 \\
x = \sin^{-1}(0) + 2 \pi k \text{ or }& x = \pi - \sin^{-1}(0) + 2 \pi k \\
\text{ or }& x = \pm \cos^{-1}(1) + 2 \pi k, \text{ for any integer $k$} \\
x = 2 \pi k \text{ or }& x = \pi + 2 \pi k \\
\text{ or }& x = \pm \pi/2 + 2 \pi k, \text{ for any integer $k$}
\end{align}
Second, we can allow ourselves to alter the domain, but must then check for solutions in the set we removed. To reduce writing, we reuse the particular solution sets found above.
\begin{align}
\sin x \cos x &= \sin x \\
\cos x &= 1 &&\text{ -- Check } \sin x = 0 \text{ in } \sin x \cos x = \sin x \text{.} \\
x &= \pm \pi/2 + 2 \pi k, \text{ for any integer $k$}
\end{align}
and we check
\begin{align}
\sin x &= 0 \\
x &= 2 \pi k \text{ or } x = \pi + 2 \pi k, \text{ for any integer $k$}.
\end{align}\begin{align}
x &= 2 \pi k: & \sin (2 \pi k) \cos (2 \pi k) &\overset{?}{=} \sin (2 \pi k) \\
&& 0 \cdot 1 &\overset{\checkmark}{=} 0 \text{ and }\\
x &= \pi + 2 \pi k: & \sin (\pi + 2 \pi k) \cos (\pi + 2 \pi k) &\overset{?}{=} \sin (\pi + 2 \pi k) \\
&& 0 \cdot -1 &\overset{\checkmark}{=} 0 \text{.}
\end{align}
So the solution set is $x = \pm \pi/2 + 2 \pi k$, $x = 2 \pi k$, or $x = \pi + 2 \pi k$, for any integer $k$. Notice that this second method has significantly more cognitive load for both the writer and the reader -- goals are stored for later use. I usually recommend the first method because of this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the value of $\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $?
What is the value of
$\sin 1 ^\circ \sin3^\circ\sin5^\circ \sin 7^\circ \sin 9^\circ \cdots \sin 179^\circ $ ?
The question is indeed intriguing. We could start by condensing it using $\sin \theta = \sin (180-\theta)$, This reduces the problem as the products till $89^\circ$. But that doesn't help proceed.
Thanks in advanced.
|
The product of the chords in this figure is
$(2\cos \frac {2\pi}{5})(2\cos \frac {\pi}{5})(2\cos 0)(2 \cos -\frac {\pi}{5})(2\cos -\frac {2\pi}{5}) = 2^5\prod_\limits {n=-2}^2 \cos \frac {n\pi}{5}$
If we map this figure to the complex plane the product of those lengths = $|(1+e^{\frac {\pi i}{5}})(1+e^{\frac {3\pi i}{5}})(1+e^{\frac {5\pi i}{5}})(1+e^{\frac {7\pi i}{5}})(1+e^{\frac {7\pi i}{5}})|$
Note: $(z+e^{\frac {\pi i}{5}})(z+e^{\frac {3\pi i}{5}})(z+e^{\frac {5\pi i}{5}})(z+e^{\frac {7\pi i}{5}})(z+e^{\frac {7\pi i}{5}}) = z^5 + 1$
Evaluated at $z= 1$
$2^5\prod_\limits {n=-2}^2 \cos \frac {n\pi}{5} = 2\\
\prod_\limits {n=-2}^2 \cos \frac {n\pi}{5} = 2^{-4}$
And this generalizes:
$\prod_\limits {n=1}^k \cos \frac {(2n-1)\pi}{2k} = 2^{-(k-1)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2766630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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computing probabilities of an negative binomial random variable Would appreciate if you could check if I answered the questions correctly:
$X$ is a Negative Binomial random variable with the parameters $\frac{1}{2}$ and $r =1,2,3,\ldots$.
$Y$ is a random variable that is defined as: $Y =
\begin{cases}
X, & x \le r+1 \\
X-1, & x \ge r+2
\end{cases}$
(1) Find $\Pr\{ Y = r+1 \}$
The result I got: $r\,(\frac{1}{2})^{r+2}$. Since we need, according to the definition of Negative Binomial, to calculate the probability of trials to make $r$ success, then we can calculate it by multiplying the parameters ($1/2$ and $r$) times $r+1+1$ to cover it all.
(2) Find $\mathbb{E}[Y]$
The result I obtained is $2r-1-\frac{r+2}{2^r}$. I calculated it using $r/p$ while trying to cover all of $Y$.
Hope I did this one correctly. Would appreciate your comments or corrections.
| Your description in (1) and the fact that you quoted $\mathbb{E}[X] = r/p$ for (2), I take it that you adopt the total number of trials for the definition of $X$ the Negative Binomial.
\begin{align*}
\Pr\bigl\{ Y = r+1 \bigr\} &= \overbrace{\Pr\bigl\{X = r+1 \bigr\} }^{Y=X} + \overbrace{ \Pr\bigl\{ X = r+2\bigr\} }^{Y=X-1} \qquad\text{, by definition of $Y$} \\
&= {r\choose r-1} p^rq + {r+1\choose r-1} p^r q^2 \Bigg|_{p=q=1/2}\\
&= r \cdot \frac1{ 2^{r+1}} + \frac{(r+1)r}{2!} \cdot \frac1{ 2^{r+2}} \\
&= \frac{ 4r }{ 2^{r+3}} + \frac{ r^2 + r }{ 2^{r+3}}\\
&= \frac{ r(r+5) }{ 2^{r+3}}
\end{align*}
Similarly, the expectation splits into two parts.
\begin{align*}
\mathbb{E}[ Y ] &= \Bigl\{ \text{contribution from $X \leq r+1$} \Bigr\} + \Bigl\{ \text{from $X \geq r+2$} \Bigr\} \\
&= \sum_{k=r}^{r+1} \Bigl[ \color{magenta}k \cdot \Pr\bigl\{ X = k \bigr\} \Bigr] + \Biggl\{ \mathbb{E}[ X-1 ] - \sum_{k=r}^{r+1} \Bigg[ (\color{magenta}k-1) \cdot \Pr\bigl\{ X = k \bigr\} \Bigg] \Biggr\} \\
\end{align*}
After canceling the $\color{magenta}{\text{magenta }k}\,$, we have
\begin{align*}
\mathbb{E}[ Y ] &= \sum_{k=r}^{r+1} \Bigl[ 1 \cdot \Pr\bigl\{ X = k \bigr\} \Bigr] + \mathbb{E}[ X-1 ]\\
&= \frac1{ 2^r} + \frac{ r }{ 2^{r+1}} + \frac{r}p\Bigg|_{p = 1/2} - 1 \\
&= 2r-1 + \frac{ r+2 }{ 2^{r+1}}
\end{align*}
Your own answer is just off by a minus sign.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve the integration How to find $$ \frac{\int_0^{\pi}x^3\log(\sin x)\,dx}{\int_0^{\pi} x^2 \log(\sqrt{2} \sin x)\,dx} $$
I couldn’t resolve it by using integration by parts.
| $\newcommand{\intd}{\,\mathrm{d}}$
Alternatively, and quite similarly I might add, this can be done using the substitution $u=a+b-x$ where $a$ and $b$ are the bounds of the integration. It will be important to note that $\sin(\pi-x)=\sin x$, looking at the numerator we have $I_1=\int_0^{\pi}x^3\log(\sin x)\,\mathrm{d}x$, making the substitution $u=\pi-x$ and switching the variable back to $x$ we get $$I_1=\int_0^\pi(\pi-x)^3\log(\sin x)\,\mathrm{d}x=\int_0^\pi(\pi^3-3\pi^2x+3\pi x^2-x^3)\log(\sin x)\,\mathrm{d}x$$ Adding $I_1+I_1$ we get $$2I_1=\int_0^\pi(\pi^3-3\pi^2x+3\pi x^2)\log(\sin x)\,\mathrm{d}x$$ We now have 3 resulting integrals which will be computed with the same method. Consider $I_2=\int_0^\pi x\log(\sin x)\intd x$, making the same substitution we have $I_2=\int_0^\pi(\pi-x)\log(\sin x)\intd x$ which implies $2I_2=\int_0^\pi \pi\log(\sin x)\intd x$ so $I_2=\frac\pi2\int_0^\pi\log(\sin x)\intd x$. Rewriting our expression for $I_1$ we obtain
\begin{align}
\frac{\pi^3}{2}\int_0^\pi\log(\sin x)\intd{x}-\frac{3\pi^2}{2}\int_0^\pi x\log(\sin x)\intd{x}+\frac{3\pi}{2}\int_0^\pi x^2\log(\sin x)\intd x
\end{align}
Let $I_3=\int_0^\pi\log(\sin x)\intd x$, this implies $I_2=\frac\pi2I_3$ so \begin{align}
I_1&=\frac{\pi^3}{2}I_3-\frac{3\pi^2}{2}\cdot\frac\pi2I_3+\frac{3\pi}{2}\int_0^\pi x^2\log(\sin x)\intd x\\
&=-\frac{\pi^3}{4}I_3+\frac{3\pi}{2}\int_0^\pi x^2\log(\sin x)\intd{x}
\end{align}
By a similar fashion in the denominator we have
\begin{align}
\int_0^\pi x^2\log(\sqrt2\sin x)\intd x&=\int_0^\pi \frac{x^2}{2}\log 2+x^2\log(\sin x)\intd{x}\\
&=\frac{\pi^3}{6}\log2+\int_0^\pi x^2\log(\sin x)\intd{x}
\end{align}
All that is left to compute is $I_3$ which will be done by use of the same substitution. We have that since $I_3=\int_0^\pi\log(\sin(\pi-x))\intd{x}$ that $$I_3=2\int_0^{\frac\pi2}\log(\sin x)\intd{x}=2\int_0^{\frac\pi2}\log(\cos x)\intd{x}$$
From this it follows that
\begin{align}
I_3&=\int_0^\pi\log(\sin x)\intd{x}=2\int_0^{\frac\pi2}\log(\sin x)\intd{x}\\
&=\int_0^{\frac\pi2}\log(\sin x)\intd{x}+\int_0^{\frac\pi2}\log(\cos x)\intd{x}\\
&=\int_0^{\frac\pi2}\log(\sin x\cos x)\intd{x}=\int_0^{\frac\pi2}\log\left(\frac{\sin 2x}{2}\right)\intd{x}\\
&=\int_0^{\frac\pi2}\log(\sin 2x)\intd{x}-\frac\pi2\log2\\
&\overbrace{=}^{u=2x}\frac12\int_0^\pi\ln(\sin u)\intd u-\frac\pi2\log2\\
&=\frac12I_3-\frac\pi2\log2
\end{align}
So $I_3=-\pi\log2$. Now we have condensed the expression down to
\begin{align*}\frac{\int_0^\pi x^3\log(\sin x)\intd{x}}{\int_0^\pi x^2\log(\sqrt2\sin x)\intd{x}}&=\frac{-\frac{\pi^3}{4}\cdot-\pi\log2+\frac{3\pi}{2}\int_0^\pi x^2\log(\sin x)\intd{x}}{\frac{\pi^3}{6}\log2+\int_0^{\pi}x^2\log(\sin x)\intd{x}}\\
&=\frac{\frac{\pi^4}{4}\log2+\frac{3\pi}{2}\int_0^\pi x^2\log(\sin x)\intd{x}}{\frac{\pi^3}{6}\log2+\int_0^\pi x^2\log(\sin x)\intd{x}}
\end{align*}
By inspection we see that the numerator is $\frac{3\pi}{2}$ times the denominator so the final answer is $\boxed{\frac{3\pi}{2}}$. Interestingly enough, although my approach did not involve Fourier series nor $\zeta(3)$ the final steps to the answer are remarkably similar. It is also important to note that since $x^2$ is even we could not make the same $\pi-x$ substitution to solve for that integral.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integration of $\frac{1}{1+x^2+x^4+\cdots +x^{2m}}$ Please calculate $$A_m=\int _{-\infty}^{\infty}\frac{dx}{1+x^2+x^4+\cdots +x^{2m}}$$
My idea is $$A_m=\int \frac{1-x^2}{1-x^{2m+2}}\: dx$$ so I tried to use residue theorem to upper half circle with radius $R$.
But, I cannot calculate it.
| Let $\gamma$ be the upper half circle $C$ with radius $R$ centered at $0$ and the segment from $A=(-R,0)$ to $B=(R,0)$. Clearly, inside $\gamma$, the function $f(z)=\frac{z^2-1}{z^{2m+2}-1}$ has $m$ poles $z=\omega_k\equiv=e^{\frac{\pi i k}{m+1}}=\omega_1^k$, $k=1,2,\cdots,m$ and on $AB$, $f(x)$ has two removable singular points $-1$ and $1$. For large $R$, using $\omega_k^{2m+2}=1$ and
$$ \frac{1+e^{ix}}{1-e^{ix}}=i\cot(\frac x2),$$
one has
\begin{eqnarray}
\int_\gamma f(z)\: dz&=&2\pi i\sum_{k=1}^m\text{Res}(f(z),z=\omega_k)\\
&=&2\pi i\sum_{k=1}^m\frac{\omega_k^2-1}{(2m+2)\omega_k^{2m+1}}\\
&=&\frac{\pi i}{m+1}\sum_{k=1}^m\bigg[\omega_k^{-2m+1}-\omega_k^{-2m-1}\bigg]\\
&=&\frac{\pi i}{m+1}\sum_{k=1}^m\bigg[\omega_k^{3}-\omega_k\bigg]\\
&=&\frac{\pi i}{m+1}\sum_{k=1}^m\bigg[\omega_1^{3k}-\omega_1^k\bigg]\\
&=&\frac{\pi i}{m+1}\bigg[\omega_1^3\frac{1-\omega_1^{3m}}{1-\omega_1^3}-\omega_1\frac{1-\omega_1^{m}}{1-\omega_1}\bigg]\\
&=&\frac{\pi i}{m+1}\bigg[\frac{1+\omega_1^{3}}{1-\omega_1^3}-\frac{1+\omega_1}{1-\omega_1}\bigg]\\
&=&-\frac{\pi i}{m+1}\bigg[\cot(\frac{3\pi}{2(m+1)})-\cot(\frac{\pi}{2(m+1)})\bigg].
\end{eqnarray}
On $C$,
$$\bigg|\int_C \frac{z^2-1}{z^{2m+2}-1}\: dz\bigg|\le\int_C \bigg|\frac{z^2-1}{z^{2m+2}-1}\bigg|\: |dz|\le \int_C\frac{|z|^2+1}{|z|^{2m+2}-1}\: |dz|=\frac{R^2+1}{R^{2m+2}-1}2\pi R\to 0$$
as $R\to\infty$. On $AB$
$$ \int_{-R}^Rf(z)dz=\int_{-R}^Rf(x)dx\to\int_{-\infty}^\infty f(\infty)dx $$
as $R\to\infty$.
Thus
$$\int _{-\infty}^{\infty}f(x)dx=-\frac{\pi}{m+1}\bigg[\cot(\frac{3\pi}{2(m+1)})-\cot(\frac{\pi}{2(m+1)})\bigg].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2770635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $a\in\mathbb{R}, a\neq 0$ such that the parabolas of the functions $f$ and $g$ are tangent.
Find $a\in\mathbb{R}, a\neq 0$ such that the parabolas of the functions $f$ and $g$ are tangent.
$f_a(x)=ax^2-(a+2)x-1$.
$g_a(x)=x^2-x-a$.
At first what I tried is find their derivative and just make them equal, but it's nowhere near the good answer...
What I tried:
since they must be tangent I thought their derivative should be equal so we have:
$f_a'(x)=2ax-(a+2)$ and $g_a'(x)=2x-1$ by making them equalling we get $a = 1$ and $a = -1$ at the same time... which is completely wrong (that's why I didn't added this).
Following B.Metha's advice we have:
$f_a(x) - g_a(x) = (a-1)x^2-(a+1)x+a-1.$
| $f_a(x)
=ax^2-(a+2)x-1
$,
$g_a(x)=x^2-x-a
$.
Their tangents are
$f_a'(x)
=2ax-(a+2)
$,
$g_a'(x)=2x-1
$.
The tangents have
the same slope when
$2ax-(a+2)
=2x-1
$
or
$2x(a-1) = a+1
$
or
$x = \frac{a+1}{2(a-1)}$
(if $a \ne 1$).
If $a=1$ then
$f_a=g_a$
so the curves are identical.
They intersect when
$ax^2-(a+2)x-1
=x^2-x-a
$
or
$(a-1)x^2-(a+1)x+a-1
=0
$.
To have the same slope there,
we must have
$\begin{array}\
0
&=(a-1)x^2-(a+1)x+a-1\\
&=(a-1)(\frac{a+1}{2(a-1)})^2-(a+1)(\frac{a+1}{2(a-1)})+a-1\\
&=\frac{(a+1)^2}{4(a-1)}-\frac{(a+1)^2}{2(a-1)}+a-1\\
&=-\frac{(a+1)^2}{4(a-1)}+a-1\\
&=\frac{4(a-1)^2-(a+1)^2}{4(a-1)}\\
&=\frac{3a^2-10a+3}{4(a-1)}\\
\end{array}
$
so
$3a^2-10a+3
=0$
or
$a
=\dfrac{10\pm\sqrt{100-36}}{6}
=\dfrac{10\pm\sqrt{64}}{6}
=\dfrac{10\pm 8}{6}
=\dfrac{18, 2}{6}
=3, \frac13
$.
For
$a=3$,
$x = \frac{4}{2(2)}
=1
$
and for
$a=\frac13$,
$x
=\frac{\frac43}{2(-\frac23)}
=-1
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Property of Bernstein polynomial $\sum_{k=0}^n k^2 B_{n,k} (x) $ $= n(n-1)x^2+nx$ Actually, I have to show:
$$ \sum_{k=0}^n B_{n,k} (x) =1 $$
$$ \sum_{k=0}^n \frac{k}{n} B_{n,k} (x) =x $$
$$ \sum_{k=0}^n \frac{k^2}{n^2} B_{n,k} (x) =x^2+ \frac{1}{n}x(1-x) $$
So the first two claims were easy, but I failed to show the last one.
Remark: $B_{n,k}$ = $\frac{n!}{k!(n-k)!} x^k(1-x)^{n-k}$.
$ \sum_{k=0}^n \frac{k^2}{n^2} B_{n,k} (x) $ = $\frac{1}{n^2} \sum_{k=0}^n k^2 B_{n,k} (x) $ . Is there a trick for $\sum_{k=0}^n k^2 B_{n,k} (x) $ ? Found in another question that $\sum_{k=0}^nk^2B_{n,k}(x)=n(n-1)x^2+nx $, but why?
| Evaluating the identity in the title we seek
$$\sum_{k=0}^n k^2 {n\choose k} x^k (1-x)^{n-k}.$$
This is a polyonomial of degree $n$ in $x$ from which we may
extract coefficients with $0\le q\le n$:
$$[x^q] \sum_{k=0}^n k^2 {n\choose k} x^k (1-x)^{n-k}
= \sum_{k=0}^q k^2 {n\choose k} [x^{q-k}] (1-x)^{n-k}
\\ = \sum_{k=0}^q k^2 {n\choose k} (-1)^{q-k} {n-k\choose q-k}
\\ = \sum_{k=2}^q k(k-1) {n\choose k} (-1)^{q-k} {n-k\choose q-k}
+ \sum_{k=1}^q k {n\choose k} (-1)^{q-k} {n-k\choose q-k}
\\ = n(n-1) \sum_{k=2}^q {n-2\choose k-2} (-1)^{q-k} {n-k\choose q-k}
+ n \sum_{k=1}^q {n-1\choose k-1} (-1)^{q-k} {n-k\choose q-k}.$$
We get for the first piece
$$n(n-1) \sum_{k=2}^q {n-2\choose k-2} (-1)^{q-k}
[z^{q-k}] (1+z)^{n-k}
\\ = n(n-1) [z^q] \sum_{k=2}^q {n-2\choose k-2} (-1)^{q-k}
z^k (1+z)^{n-k}.$$
Now here we may extend $k$ beyond $q$ as there is no contribution to
the coefficient extractor:
$$n(n-1) [z^q] \sum_{k=2}^n {n-2\choose k-2} (-1)^{q-k}
z^k (1+z)^{n-k}
\\ = n(n-1) [z^q] (-1)^q z^2 (1+z)^{n-2}
\sum_{k=2}^n {n-2\choose k-2} (-1)^{k-2}
z^{k-2} (1+z)^{-(k-2)}
\\ = n(n-1) [z^q] (-1)^q z^2 (1+z)^{n-2}
\left(1-\frac{z}{1+z}\right)^{n-2}
= n(n-1) [z^q] (-1)^q z^2.$$
This is $$n(n-1) \times [[q=2]].$$
We get for the second piece
$$n [z^q] \sum_{k=1}^n {n-1\choose k-1} (-1)^{q-k}
z^k (1+z)^{n-k}
\\ = - n [z^q] (-1)^q z (1+z)^{n-1}
\sum_{k=1}^n {n-1\choose k-1} (-1)^{k-1}
z^{k-1} (1+z)^{-(k-1)}
\\ = - n [z^q] (-1)^q z (1+z)^{n-1}
\left(1-\frac{z}{1+z}\right)^{n-1}
= - n [z^q] (-1)^q z.$$
This is $$n \times [[q=1]].$$
Collecting the two contributions we get
$$n(n-1)x^2 + nx$$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2772803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Show that the equation $x^4 + 3 y^4 = 131$ has no integer solutions. I have seen the result demonstrated by considering the equation$\pmod{5}$, but was wondering if this proof, working in $\mathbb{Z}_3$, is also valid, since it seems to require less working:
$$\forall x \in \mathbb{Z}_3, x = \{0,1,2\} .$$
So, for any $x$, $x^4 = \{0,1\}$. Also, in $\mathbb{Z}_3$ we note that $3=0$, $3y^4=0$, for any $y$.
Since $131=2$, the equation $x^4 + 3 y^4 = 131$ has no solutions in $\mathbb{Z}_3$.
By definition, for any integers $x,y$
$$x^4 + 3 y^4 \equiv 131 \pmod{3} \iff x^4 + 3 y^4 -131=3k$$
for some integer $k$. Letting $k=0$, the result follows. $\square$
| I imagine if you've actually seen this done more than once via $\mod 5$ it is not because $\mod 5$ is easy, but because it is interesting.
By Fermat's little Theorem $a^4 \equiv 0, 1\mod 5$ and determined by whether $5$ divides $a$ or not. So $x^4 + 3y^4\equiv 0, 1, 3,4\mod 5$ and which value will uniquely determine which of $x,y$ are divisible by $5$ and which are not. So $x^4 + 3y^4 \equiv 1 \mod 5\iff 5\not \mid x$ and $5\mid y$ (and as $3*5^4 > 131$ $y = 0$ and $x^4 = 131$).
That we can do that is certainly interesting and informative. Far more so then the final result we are trying to prove (As $3*5^4 > 131$ $y = 0$ and $x^4 = 131$ so there is no solution).
But doing it $\mod 3$ is certainly easier and more intuitive. But there is nothing particularly interesting or informative about $x^4 +3y^4 \equiv x^4 \equiv 0, 1\not \equiv 131 \mod 3$... at least nothing that can't be demonstrated by other problems.
....
More interesting would be to find all integer solutions to $x^4 + 3y^4 = 1956$.
$\mod 2$ we get $x$ are both even or odd but as $16\not \mid 1956$ they are both odd. Not very useful.
$\mod 3$ we get $x \equiv 0 \mod 3$. Somewhat useful, maybe we can try $\mod 9$ and get that $3y^4 \equiv 3 \mod 9$ so $3\not \mid y$. but that's not that useful.
So far we have $x,y$ are both odd. $3|x$ and $3\not \mid y$. That's... not much.
But $\mod 5$ we get $x^4 + 3y^4 \equiv 1 \mod 5$ so $x^4 \equiv 1\mod 5$ and $y^4 \equiv 0 \mod 5$. So $5\not \mid x$ and $5\mid y$. And as $3*5^4 = 1875 < 1956 < 3*10^4 = 30000$ we have $x^4 = 1956;y=0$ of $x^4 = 1956-1875=81; |y| = 5$. $x^4 = 1956$ is not possible. $x^4 = 81$ means $|x| =3$.
so $(\pm 3, \pm 5)$ are the only solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2773302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $(y_n)$ has a finite limit when $n \rightarrow \infty$. Let a real sequence $(a_n)$ defined by:
$x_1$ = 1; $x_n = \displaystyle\frac{2n}{(n-1)^2} \displaystyle\sum_{i=1}^{n-1} x_i$, $n=2,3,4, \dots$
For each integers $n$ let
$y_n = x_{n+1} - x_n$
Prove that $(y_n)$ has a finite limit when $n \rightarrow \infty$.
My attempt: We have
$x_{n+1} = \displaystyle\frac{2(n+1)}{n^2} \displaystyle\sum_{i=1}^{n} x_i = \displaystyle\frac{2(n+1)}{n^2}\left(\frac{2n}{(n-1)^2}+1\right) \left(\displaystyle\frac{2(n-1)}{(n-2)^2}+1\right) \dots 5 x_1$.
Hence, when $n \rightarrow \infty$, $(x_n)$ has a finte limit $\Rightarrow$ $(y_n)$ has a finite limit.
Is that right? Thank all!
| Not really an answer, just a clarification: $(x_n)_n$ indeed does not converge.
We have:
\begin{align}
x_{n+1} &= \frac{2(n+1)}{n^2}\left(\frac{2n}{(n-1)^2} + 1\right)\left(\frac{2(n-1)}{(n-2)^2} + 1\right)\cdots\left(\frac{2\cdot 2}{1^2} + 1\right)\\
&= \frac{2(n+1)}{n^2}\cdot\frac{2n + (n-1)^2}{(n-1)^2}\frac{2(n-1) + (n-2)^2}{(n-2)^2}\cdots\frac{2\cdot 2 + 1^2}{1^2}\\
&= \frac{2(n+1)}{n^2}\cdot\frac{n^2+1}{(n-1)^2}\frac{(n-1)^2 + 1}{(n-2)^2}\cdots\frac{2^2+1}{1}\\
&= \frac{2(n+1)}{n^2}\cdot\prod_{k=2}^n \frac{k^2+1}{(k-1)^2}\\
&\ge \frac{2(n+1)}{n^2}\cdot\prod_{k=2}^n \frac{k^2}{(k-1)^2}\\
&\ge \frac{2(n+1)}{n^2}\cdot\left(\frac{n!}{(n-1)!}\right)^2\\
&= \frac{2(n+1)}{n^2} \cdot n^2\\
&= 2(n+1)
\end{align}
so $(x_n)_n$ is unbounded.
However,
\begin{align}
y_n &= x_{n+1}-x_n\\
&= \frac{2(n+1)}{n^2}\cdot\prod_{k=2}^n \frac{k^2+1}{(k-1)^2} - \frac{2n}{(n-1)^2}\cdot\prod_{k=2}^{n-1} \frac{k^2+1}{(k-1)^2}\\
&= \left(\frac{2(n+1)}{n^2}\frac{n^2+1}{(n-1)^2} - \frac{2n}{(n-1)^2}\right)\cdot\prod_{k=2}^{n-1} \frac{k^2+1}{(k-1)^2}\\
&= \frac2{(n-1)^2}\left(1 + \frac1n + \frac1{n^2}\right)\cdot\prod_{k=2}^{n-1} \frac{k^2+1}{(k-1)^2}\\
\end{align}
which seems to converge to around $3.67608$, but it is difficult to bound the above product. Namely
$$\prod_{k=2}^{n-1} \frac{k^2+1}{(k-1)^2} \le \prod_{k=2}^{n-1} \frac{k^2+k}{(k-1)^2} = \prod_{k=2}^{n-1} \frac{k(k+1)}{(k-1)^2} = \frac12 \frac{(n-1)!n!}{(n-2)!} = \frac12 (n-1)^2n$$
not being good enough.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Writing $\frac{1}{x+5}$ as $\frac{A}{x}+\frac{B}{5}$ Is this possible? (in this case specifically, but also in general for any fraction with addition/subtraction in the denominator).
I thought this would be easy, but upon trying it and thinking some more
*
*theres 2 unknowns and one equation, so we probably can't identify a unique $A$ and $B$?
For example, I feel like the simplest manipulation is to multiple both sides by $5$ and move the $A$ term over, which gives
$$
B=\frac{5}{x+5} - \frac{5A}{x} =_{A=1} \frac{-25}{x(x+5)}
$$
which is not the same as if we assumed $A=2$
| Note that
$\frac{A}{x}+\frac{B}{5}=\frac{5A+Bx}{5x}=\frac{1}{x+5}\implies 5Ax+25A+Bx^2+5Bx=5x$$
then for $B=0$
$$5Ax+25A=5x \implies A=\frac{x}{x+5}$$
and for $A=0$
$$Bx^2+5Bx=5x\implies B=\frac{5}{x+5}$$
which are trivial solutions and then it is not possible obtain the equality for $A,B$ constants.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2780045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculate the determinant of this $5 \times 5$ matrix
Calculate the determinant of the matrix $$A=\begin{pmatrix} \sin\alpha
& \cos\alpha & a\sin\alpha & b\cos\alpha & ab \\
-\cos\alpha & \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2 \\ 0
& 0 & 1 & a^2
& b^2 \\ 0 & 0 & 0 & a
& b \\ 0 & 0 & 0 & -b
& a \end{pmatrix} \text{ with } (\alpha,a,b \in \mathbb{R})$$
I have trouble solving the determinant.. But what is immediately visible are those zeroes in the matrix, just one more zero is needed such that this matrix is a triangular matrix (the element $a_{54}$ must be zero for this but it is $-b$ instead). If it was zero we could just multiply the diagonal and the product would be our determinant.
I have tried various ways to form this matrix such that $a_{54}$ is zero but the way I formed harmed the matrix and I got a wrong determinant as solution : /
As example, I have multiplied row $4$ with $b$, multiply row $5$ with $a$ and then add row $4$ to row $5$. Because I multiplied row $5$ with $a$, I need to divide the determinant by $a$ at the end.
So then I have the matrix
$$\begin{pmatrix}
\sin\alpha & \cos\alpha & a\sin\alpha & b\cos\alpha & ab\\
-\cos\alpha & \sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2\\
0 & 0 & 1 & a^2 & b^2\\
0 & 0 & 0 & a & b\\
0 & 0 & 0 & 0 & a^2+b^2
\end{pmatrix}$$
$$\text{Thus }\det = \frac{\sin\alpha \cdot \sin\alpha \cdot 1 \cdot a \cdot (a^2+b^2)}{a}=\sin^2\alpha \cdot (a^2+b^2)$$
But this is wrong and I don't see how to get the correct determinant...?
| If you calculate along the first column,
\begin{align}
|A|&= \sin\alpha \begin{vmatrix}
\sin\alpha & -a^2\sin\alpha & b^2\cos\alpha & a^2b^2 \\
0 & 1 & a^2 &
b^2 \\ 0 & 0 & a &
b \\ 0 & 0 & -b &
a \end{vmatrix}
+\cos\alpha\begin{vmatrix}
\cos\alpha & a\sin\alpha & b\cos\alpha & ab \\
0 & 1 & a^2
& b^2 \\ 0 & 0 & a
& b \\ 0 & 0 & -b
& a \end{vmatrix} \\ \ \\
&=(\sin^2\alpha+\cos^2\alpha)\,\begin{vmatrix} 1 & a^2
& b^2 \\ 0 & a
& b \\ 0 & -b
& a \end{vmatrix} \\ \ \\
&=\begin{vmatrix} 1 & a^2
& b^2 \\ 0 & a
& b \\ 0 & -b
& a \end{vmatrix} \\ \ \\
&=a^2+b^2.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Prove the equality: $\det\left[\begin{smallmatrix} -2a &a+b &a+c \\ b+a&-2b &b+c \\ c+a&c+b &-2c \end{smallmatrix}\right] = 4(a+b)(b+c)(c+a)$
i have to prove:
$$\begin{vmatrix}
-2a &a+b &a+c \\
b+a&-2b &b+c \\
c+a&c+b &-2c
\end{vmatrix} = 4(a+b)(b+c)(c+a)$$
I have tried many calculations between the rows and columns of the determinant to get to the answer i want to solve the exercise however none of them gave me something right. Can someone help?
| First of all, note that the determinant is going to be a polynomial in a, b, and c of degree 3. Secondly, note that the polynomial is cyclic.
Substitute $a = -b$, to get,
$$\begin{vmatrix}
-2a &0 &c+a \\
0&2a &c-a \\
c+a&c-a &-2c
\end{vmatrix} = -2a(-4ac - (c-a)^2 + (c+a)^2) = 0$$
This shows that $a+b$ is factor of the polynomial.
Since $a+b$ is a factor and the polynomial is cyclic, $b+c, c+a$ are also factors. And since the degree of the polynomial is 3, there is at most only a constant factor left. Let the constant factor be $k$.
Thereby,
$$\begin{vmatrix}
-2a & a+b &c+a \\
a+b& -2b &b+c \\
c+a& b+c &-2c
\end{vmatrix} = k(a+b)(b+c)(c+a)$$
Put $a = 0, b = 1, c = 1$ to get,
$$\begin{vmatrix}
0 & 1 & 1 \\
1& -2 & 2 \\
1& 2 &-2
\end{vmatrix} = k(1)(2)(1) \Rightarrow k = 4$$
Thus,
$$\begin{vmatrix}
-2a & a+b &c+a \\
a+b& -2b &b+c \\
c+a& b+c &-2c
\end{vmatrix} = 4(a+b)(b+c)(c+a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2782450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inverse element in field extension $\mathbb{Q}(\sqrt{2},\sqrt{3})$ Determine $(2\sqrt{2}-3\sqrt{3}+5)^{-1}$ in $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$, which is of degree $4$ over $\mathbb{Q}$ and for $a= \sqrt{2} + \sqrt{3}$ follows $a^4-10a^2+1=0$.
I did calculate inverse elements in field extensions of degree $2$ over $\mathbb{Q}$, but am somehow stuck with this problem, any help is very much appreciated!
| You can carry out the "rationalise the denominator" in two steps.
First multiply $\frac{1}{5+2\sqrt{2}-3\sqrt{3}}$ by $\frac{5+2\sqrt{2}+3\sqrt{3}}{5+2\sqrt{2}+3\sqrt{3}}$ and get $\frac{5+2\sqrt{2}+3\sqrt{3}}{6+20\sqrt{2}}$.
Now multiply by by $\frac{6-20\sqrt{2}}{6-20\sqrt{2}}$ and get
$\frac{(5+2\sqrt{2}+3\sqrt{3})(6-20\sqrt{2})}{-764}$.
Multiply out the numerator and you're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2784929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving a fourth degree equation Well, I want to find the roots of the equation:
$$x^4+6x^3-9x^2-162x-243=0\tag1$$
And Wolfram Alpha tells me that I can factor this and get:
$$(x^2-3x-9)(x^2+9x+27)=0\tag2$$
Then it will be easy to solve... But my question is, how can I go from $(1)$ to $(2)$?
| To solve $x^4+6x^3−9x^2−162^x−243=0$,
first try and factor, if possible: $(x^2−3x−9)(x^2+9x+27)=0$
So since this factors into to quadratics, we can use quadratic formula and some other simplification methods such as completing the square.
The roots of $x^2-3x-9$ are the same as the roots of $x^2-x-1$, multiplied by three since the absolute value of coefficients strictly increases by a power of $3$. This all amounts to $(3/2)(1±√5)$ being the roots of $x^2-3x-9$, and as for $x^2+9x+27$, one notices the increasing powers of $9$, and therefore a root of $x^2+9x+27$ is equivalent to a root of $x^2+x+1$, multiplied by nine. This amounts to $(9/2)(-1±i√3)$ being the roots of $x^2+9x+27$.
The complete list of solutions to
$x^4+6x^3−9x^2−162^x−243=0$
is
$(9/2)(-1+i√3)$
$(9/2)(-1-i√3)$
$(3/2)(1+√5)$
$(3/2)(1-√5)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2785079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
The sum of angles in a triangle is greater than 180? So basically the question goes:
Let $ABC$ be a triangle and that $\cos A=-\frac 3 5$ and $\cos B=\frac 5 {13}$, find $\cos C$.
What I did was I get the value of $A$ and $B$ by calculating the arccosine of $-3/5$ and $5/13$ and the sum of both of these angles was already greater than $180$, how can I calculate $\cos C$?
| I suspect you're meant to calculate $\cos C$ by using the cosine of sums rule, not by computing inverse cosines.
With that in mind, and on the off chance that the minus sign on $-3/5$ is an error, and the cosine of angle $A$ is really meant to be $3/5$, I'll point out
$$
\arccos 3/5 \approx 53.13 \\
\arccos 5/13 \approx 67.38
$$
which leaves $C \approx 180-53.13-67.38 = 59.49$. And indeed, we see that (assuming the minus sign is spurious)
\begin{align}
\cos(A+B) & = \cos A \cos B - \sin A \sin B \\
& = \frac{3}{5} \times \frac{5}{13} - \frac{4}{5} \times \frac{12}{13} \\
& = \frac{15}{65} - \frac{48}{65} = -\frac{33}{65}
\end{align}
That would mean that $\cos C = 33/65$ (because $A+B+C = 180$), and sure enough,
$$
\arccos 33/65 \approx 59.49
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2785189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Show $\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$ is equivalent to $1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$ for $\lvert x\rvert < 1$ I have been asked to show that $$\frac{1}{1-\frac{x}{3-x}+\frac{x}{4-x}}$$ is equivalent to writing $$1+\frac{1}{2}\left(\frac{1}{2-x}-\frac{3}{6-x}\right)$$
From here I just tried to work out the bottom of the first fraction which I found to be $\frac{(3-x)(4-x)-x}{(3-x)(4-x)}$ now taking the reciprocal gives me $\frac{(3-x)(4-x)}{(3-x)(4-x)-x}$.
I did try factorising the bottom to get to $\frac{(3-x)(4-x)}{(x-6)(x-2)}$ and then using partial fractions gets me to $\frac{-15}{2(x-6)}+\frac{1}{2(x-2)}$ which is definitely not where I want to be.
I feel like I need to work from $\frac{(3-x)(4-x)}{(3-x)(4-x)-x}$ and somehow pull out a $1$ here but not entirely sure how.
Would really appreciate if anyone could help me.
Thank you.
| We have $$\frac 1 {1 - \frac {x}{3-x} + \frac x{4-x}}=\frac {x^2-7x+12}{x^2-8x+12}=1+\frac {x} {x^2-8x+12}=1+\frac 1 2 \left(\frac 1 {2-x} - \frac 3 {6-x}\right)$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\tan\left(\sum_{r=1}^{\infty} \arctan\left(\frac{4}{4r^2 +3}\right)\right)$
$$\tan\left(\sum_{r=1}^{\infty} \arctan\left(\dfrac{4}{4r^2
+3}\right)\right)= ? $$
I wrote it in the form:
$$\tan\left(\sum_{r=1}^{\infty} \arctan\left(\dfrac{\dfrac43}{\dfrac{4r^2}{3}
+1}\right)\right)$$
and tried to use:
$$\arctan x- \arctan y = \arctan\left(\dfrac{x-y}{1+xy}\right)$$ but that trick doesn't help here. How to go about solving this problem then?
| You are on the right track. Just note that
$$\frac{4}{4r^2+3} =\frac{1}{r^2+3/4}=\frac{1}{1+r^2-(1/2)^2}=\frac{(r+1/2)-(r-1/2)}{1+(r+1/2)(r-1/2)}$$
and therefore
$$\arctan\left(\frac{4}{4r^2+3}\right)=\arctan\left(r+\frac{1}{2}\right)- \arctan\left(r-\frac{1}{2}\right)\\=\arctan\left((r+1)-\frac{1}{2}\right)- \arctan\left(r-\frac{1}{2}\right).$$
Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
A polynomial inequality in three variables I have a polynomial $P(A,B,C)$ where $A,B,C \in \mathbb{R}$ and $A>0,B>0$.
$$P(A,B,C)=C^2+A^2-2CA+4AB-2B+C-A$$
When will
*
*$P(A,B,C)>0$
*$P(A,B,C)=0$
*$P(A,B,C)<0$
I have tried factorizing in various ways but none of them is very helpful. Any ideas here?
My attempt:
$P(A,B,C)=(C-A)^2+(C-A)+2B(2A-1)=(C-A)(C-A+1)+2B(2A-1)$
which leads to nowhere
| Let $x=C-A$, then $P(A,B,C)=x^2+x+2B(2A-1) \equiv f(x)$.
$$\Delta=1^2-4(2B)(2A-1)=1-8B(2A-1)$$
*
*$\Delta < 0 \implies f(x)>0$,
$$1<8B(2A-1) \implies 2A-1> \frac{1}{8B} \implies A> \frac{8B+1}{16B}$$
*
*$\Delta \ge 0 \land f(x) < 0$,
$$\frac{-1-\sqrt{1-8B(2A-1)}}{2} < C-A < \frac{-1+\sqrt{1-8B(2A-1)}}{2}$$ where $0<A \le \dfrac{8B+1}{16B}$.
*$\Delta \ge 0 \land f(x) = 0$,
$$C-A=\frac{-1 \pm \sqrt{1-8B(2A-1)}}{2}$$ where $0<A \le \dfrac{8B+1}{16B}$.
*$\Delta \ge 0 \land f(x) > 0$,
$$\left| C-A+\frac{1}{2} \right|> \frac{\sqrt{1-8B(2A-1)}}{2}$$ where $0<A \le \dfrac{8B+1}{16B}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to show that this integral is correct? How can one show that
$$\int_0^{\pi/2}\cos\left(\frac{x}{2}\right)\ln\left[\frac{1}{\alpha} \tan(x) \tan\left(\frac{x}{2}\right)\right] \sqrt{\sin(x) \tan \left(\frac{x}{2}\right)} \, \mathrm dx=-\frac{\ln(\alpha)}{\sqrt{2}}$$
assume $\alpha\ge1$.
I can't see how to simplify $\tan(x)\tan(x/2)$ and $\sin(x)\tan(x/2)$.
| Use properties of trig functions to get
\begin{equation*}
\sin(x)\tan\left(\frac{x}{2}\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\tan\left(\frac{x}{2}\right)=2\sin^2\left(\frac{x}{2}\right).
\end{equation*}
Now we can simplify the integral to
\begin{equation*}
\int_0^{\pi/2} \frac{1}{\sqrt{2}}\sin(x)\left(\ln \tan(x)\tan\left(\frac{x}{2}\right)-\ln\alpha\right) dx.
\end{equation*}
It is easy to verify that
\begin{equation*}
\int_0^{\pi/2} \frac{1}{\sqrt{2}}\sin(x)(-\ln\alpha)dx=-\frac{\ln \alpha}{\sqrt{2}}.
\end{equation*}
which is your answer. Now we only need to show
\begin{equation}\label{x}
\int_0^{\pi/2} \sin(x) \ln \tan(x)\tan\left(\frac{x}{2}\right) dx=0.
\end{equation}
Let $t=\cos(x)$, we can draw a right triangle with sides $1,t,\sqrt{1-t^2}$ and draw a bisector of the angle $x$. Then
\begin{equation*}
\tan(x)=\frac{\sqrt{1-t^2}}{t},\quad\tan\left(\frac{x}{2}\right)=\frac{\sqrt{1-t^2}}{1+t}
\end{equation*}
and the integral becomes
\begin{equation*}
\int_0^1 \ln \left(\frac{1-t}{t}\right)dt=\int_0^1 \ln(1-t)dt-\int_0^1 \ln tdt=0
\end{equation*}
by substitution $u=1-t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
Integrate $\int \frac{dx}{\sqrt{(x-a)(b-x)}}$
Integrate $\int \frac{dx}{\sqrt{(x-a)(b-x)}}$ where $b>a$
My Attempt
$$
\begin{align}
&\int \frac{1}{\sqrt{(x-a)(b-x)}}dx=\int\frac{dx}{\sqrt{-x^2+(a+b)x-ab}}\\
&=\int\frac{dx}{\sqrt{-(x^2-2.\frac{a+b}{2}x+ab+\frac{(a+b)^2}{4}-\frac{(a+b)^2}{4})}}\\
&=\int\frac{dx}{\sqrt{\frac{a^2+b^2+2ab}{4}-ab-\big[x-\frac{a+b}{2}\big]^2}}=\int\frac{dx}{\sqrt{\frac{(a-b)^2}{4}-\big(x-\frac{a+b}{2}\big)^2}}\\
&=\sin^{-1}\frac{x-\frac{a+b}{2}}{\frac{b-a}{2}}+C\color{red}{=\sin^{-1}\frac{2x-(a+b)}{b-a}+C}
\end{align}
$$
My reference shows the solution $I=2\sin^{-1}\sqrt{\frac{x-a}{b-a}}$, yet why am I getting a different solution or both the same ?
Thanx@lab bhattacharjee for the hint.
$$
\begin{align}
\sin^{-1}\frac{2x-(a+b)}{b-a}+C&=\frac{\pi}{2}-\cos^{-1}\frac{2x-(a+b)}{b-a}+C\\&=\frac{\pi}{2}-\pi+\cos^{-1}\frac{(a+b)-2x}{b-a}+C_1\\
&=\cos^{-1}\frac{(a+b)-2x}{b-a}+C_2
\end{align}
$$
Let,
$$
y=\cos^{-1}\frac{(a+b)-2x}{b-a}\implies\cos y=\frac{(a+b)-2x}{b-a}\\
2\sin^2\frac{y}{2}=1-\cos y=1-\frac{(a+b)-2x}{b-a}=\frac{b-a-a-b+2x}{b-a}=\frac{2x-2a}{b-a}\\
\sin^2\frac{y}{2}=\frac{x-a}{b-a}\implies \sin\frac{y}{2}=\sqrt{\frac{x-a}{b-a}}\\
\frac{y}{2}=\sin^{-1}\sqrt{\frac{x-a}{b-a}}\implies\boxed{y=2\sin^{-1}\sqrt{\frac{x-a}{b-a}}}
$$
| Hint:
From your answer if $\sin^{-1}\sqrt{\dfrac{x-a}{b-a}}=y$
$\sin y=\sqrt{\dfrac{x-a}{b-a}}$
$\implies\cos2y=1-2\sin^2y=\dfrac{a+b-2x}{b-a}$
$2y=\cos^{-1}\dfrac{a+b-2x}{b-a}=\dfrac\pi2-\sin^{-1}\dfrac{a+b-2x}{b-a}=\dfrac\pi2+\sin^{-1}\dfrac{2x-a-b}{b-a}$
So, you have reached to the right answer.
See also: Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
Alternatively,
As $(x-a)(b-x)=-\{x^2-(a+b)x+ab\}=\dfrac{(a-b)^2}4-\left(x-\dfrac{a+b}2\right)^2$
choose $x-\dfrac{a+b}2=\dfrac{(a-b)}2\cos2t$
$$2x=a+b+(a-b)(\cos^2t-\sin^2t)=2(a\cos^2t+b\sin^2t)$$
| {
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"answer_count": 2,
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Local truncation error of Crank-Nicolson for PDE $u_t+au_x = 0$
Exercise 4:
The Crank-Nicolson scheme for $u_t + a u_x = 0$ is given by
$$ \frac{U_{j,n+1}-U_{j,n}}{\Delta t} + \frac{a}{2}\frac{D_xU_{j,n}}{2\Delta x} + \frac{a}{2}\frac{D_xU_{j,n+1}}{2\Delta x} = 0 .$$
Show that the LTE is given by
$$ \mathcal{L}_\Delta u = au_{xxx} \left(\frac{1}{6} + \frac{p^2}{12}\right) {\Delta x}^2 + O({\Delta x}^3,{\Delta t}^3) , $$
where $p = a{\Delta t}/{\Delta x}$. Find the amplification factor and find the conditions for stability.
I am just trying to work out the LTE of the Crank-Nicolson scheme, however i do not get the same answers the book. Here is my working if anyone could have a look and tell me what i am doing wrong, thank you.
The scheme $ u_j^{n+1} = u_j^n -\frac{1}{4}(u_{j+1}^n -u_{j-1}^{n}+ u_{j+1}^{n+1}-u_{j-1}^{n+1})$ rewrites as
$$
\frac{(u(x,t+\Delta t) - u(x.t)}{\Delta t} + \frac{a}{4 \Delta x} [ u(x+\Delta x,t) - u(x-\Delta x ,t) + u(x+ \Delta x, t+ \Delta t)- u ( x-\Delta x , t+\Delta t)
$$
Expanding using Taylor series, I get
$$
u_t + \frac{1}{2} u_{tt} \Delta t + \frac{1}{6}u_{ttt} \Delta t^{2} + O(\Delta t^{3})
$$
and similarly for the $x$ and $t$. I get
$$
\frac{a}{4 \Delta x} [ u(x,t) + u_x \Delta + \frac{1}{2} u_{xx} \Delta x^2 + \frac{1}{6} u_{xxx} \Delta x^{3} + O(\Delta x ^{4})- (u(x,t)-u_{x} \Delta x......)]
$$
after some simplification i get the even terms remaining..
$$
\frac{a}{2}u_{x} + \frac{au_{xxx}\Delta x^{2}}{12} + O(\Delta x^{3})
$$
similarly for the last expansion. And $t$ and $x$ together, i get
$$
\frac{a}{2}u_{x} + \frac{a}{2} u_{xt} \Delta t + o(\Delta t^{2}) + \frac{a}{12}u_{xxx} \Delta x^{2}
$$
Putting all these terms back into the equation and using
$u_{t} - = -au_{x}$,
$u_{tt} = -au_{xt}$, I am left with $\frac{1}{6} u_{ttt} \Delta t^{2} + \frac{1}{6} a u_{xxx} \Delta x^{2} + O( \Delta t^{3} \Delta x^{3})$.
However the answer in the book is $ au_{xxx}(\frac{1}{6} + \frac{p^2}{12} )\Delta x^{2}$.
| Let us write the Crank-Nicolson method for the linear advection equation $u_t + au_x = 0$ by averaging forward and backward Euler time-integration and by using centered spatial differences:
$$
\frac{u_{j}^{n+1} - u_j^n}{\Delta t} = -\frac{a}{2}\left(\frac{u_{j+1}^n - u_{j-1}^n}{2\Delta x} + \frac{u_{j+1}^{n+1} - u_{j-1}^{n+1}}{2\Delta x}\right) ,
\qquad
u_j^n \simeq u(j\Delta x, n\Delta t) .
\tag{1}
$$
Thus, the scheme reads
$
u_{j}^{n+1} = u_j^n - \frac{1}{4} p\big(u_{j+1}^n - u_{j-1}^n + u_{j+1}^{n+1} - u_{j-1}^{n+1}\big) = 0
$,
where $p = {a \Delta t}/{\Delta x}$ denotes the Courant number. To perform the local truncation error analysis, we need to expand the smooth function $u$ as a Taylor series in two variables in the vicinity of $(x, t)$. Doing so at the third order, we have
\begin{aligned}
v &= u + h u_x + k u_t
+ \tfrac{1}{2} \left( h^2 u_{xx} + 2 hk u_{xt} + k^2 u_{tt}\right) \\
&+ \tfrac{1}{6} \left( h^3 u_{xxx} + 3 h^2 k u_{xxt} + 3 h k^2 u_{xtt} + k^3 u_{ttt}\right)
+ O(h^4 + k^4)
\end{aligned}
with the notation $v(x,t) = u(x+h, t+k)$ for the steps $h \in \lbrace{0, \pm\Delta x}\rbrace$ and $k \in \lbrace{0,\Delta t}\rbrace$. Assuming that $u$ satisfies the PDE $u_t = -au_x$, we replace time-derivatives by spatial derivatives as follows
$$
v = u + (h-ak)u_x
+ \tfrac{1}{2} ( h-ak)^2 u_{xx}
+ \tfrac{1}{6} ( h-ak)^3 u_{xxx}
+ O(h^4 + k^4) .
$$
For a fixed value of the Courant number, we obtain the following expressions to be substituted in the numerical scheme $(1)$:
\begin{aligned}
v_{j}^{n+1} &= u - p\Delta x\,u_x + \tfrac{1}{2}(p\Delta x)^2 u_{xx} - \tfrac{1}{6} (p\Delta x)^3 u_{xxx} + O(\Delta x^4) \\
v_j^n &= u \\
v_{j\pm 1}^{n} &= u \pm \Delta x\,u_x + \tfrac{1}{2} \Delta x^2 u_{xx} \pm \tfrac{1}{6} \Delta x^3 u_{xxx} + O(\Delta x^4) \\
v_{j\pm 1}^{n+1} &= u + (\pm 1 -p)\Delta x\, u_x
+ \tfrac{1}{2} ( \pm 1-p)^2\Delta x^2 u_{xx}
+ \tfrac{1}{6} ( \pm 1 -p)^3\Delta x^3 u_{xxx} + O(\Delta x^4)
\end{aligned}
The divided finite-difference operators become
\begin{aligned}
\frac{v_{j}^{n+1} - v_j^n}{\Delta t} &= {-a} u_x + \tfrac{1}{2}pau_{xx} \Delta x - \tfrac{1}{6} p^2 au_{xxx} \Delta x^2 + O(\Delta x^3) \, , \\
\frac{v_{j+1}^{n} - v_{j-1}^n}{2\Delta x} &= u_x + \tfrac{1}{6} u_{xxx} \Delta x^2 + O(\Delta x^3) \, , \\
\frac{v_{j+1}^{n+1} - v_{j-1}^{n+1}}{2\Delta x} &= u_x - pu_{xx}\Delta x + \tfrac{1}{6} (1 + {3}p^2) u_{xxx} \Delta x^2 + O(\Delta x^3)
\end{aligned}
leading to the local truncation error
$a u_{xxx} \big({\tfrac{1}{6}} + \tfrac{1}{12} p^2\big) \Delta x^2 + O(\Delta x^3)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Necessary condition for a special polynomial of degree $5$ to have a special quadratic factor I am trying to solve the problem no. 41(i) on page $33$, Higher Algebra by Barnard and Child.
The problem statement is:
If $ax^5+bx^2+c$ has a factor of the form $x^2+px+1$, prove that
$$(a^2-c^2)(a^2-c^2+bc)=a^2b^2.$$
By synthetic division, I have found the remainder of $ax^5+bx^2+c$ upon division by $x^2+px+1$ is
$$(ap^4-3ap^2-bp+a) x +(ap^3-2ap-b+c).$$
As $x^2+px+1$ is a factor of $ax^5+bx^2+c$, the memainder must be indentically zero. That is:\begin{align*}
ap^4-3ap^2-bp+a&=0 \tag{1}\\
ap^3-2ap-b+c&=0 \tag{2}
\end{align*}
Then subtracting $(1)$ from $(2)\times p$ gives
$$ap^2+cp-a=0.$$
I do not know how to eliminate $p$ from the equations. I do not even know if this approach will solve it. And I have not been able yet to think of other approach.
If it is easy, then please give me some hint. Otherwise a solution, please!
| Denote $f_1(t) = at^4 - 3at^2 - bt + a$, $f_2(t) = at^3 - 2at - (b - c)$, then $f_1(p) = f_2(p) = 0$. Therefore, defining\begin{align*}
f_3(t) &= t f_2(t) - f_1(t) = at^2 + ct - a,\\
f_4(t) &= a(tf_3(t) - f_2(t)) - c f_3(t) = (a^2 - c^2)t + ab,\\
f_5(t) &= (a^2 - c^2)^2 f_3(t) - a (f_4(t))^2 + 2a^2b f_4(t) - c(a^2 - c^2) f_4(t)\\
&= a(a^2 b^2 - (a^2 - c^2)(a^2 - c^2 + bc)),
\end{align*}
then$$
f_1(p) = f_2(p) = f_3(p) = f_4(p) = f_5(p) = 0 \Longrightarrow a(a^2 b^2 - (a^2 - c^2)(a^2 - c^2 + bc)) = 0,
$$
and $a ≠ 0$ implies$$
(a^2 - c^2)(a^2 - c^2 + bc) = a^2 b^2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2797638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving the system $5(\sin x + \sin y) = 1$ and $5(\sin 2x + \sin 2y) = 1$ To find the general solution $(x,y)$ satisfying the system of equations
\begin{align} 5(\sin x + \sin y) &= 1 \\ 5(\sin 2x + \sin 2y) &= 1
\end{align}
I applied $\sin C + \sin D$ rule and then divided these two equations, then I am stuck at
$$\cos\frac{x-y}{2} = 2\cos \frac{x+y}{2}\cos(x-y).$$
I do not know what to do further.
| First,$$
\begin{cases}
\sin x + \sin y = \dfrac{1}{5}\\
\sin 2x + \sin 2y = \dfrac{1}{5}
\end{cases} \Longrightarrow \begin{cases}
\sin \dfrac{x + y}{2} \cos \dfrac{x - y}{2} = \dfrac{1}{10} & (1)\\
\sin(x + y) \cos(x - y) = \dfrac{1}{10} & (2)
\end{cases}.
$$
Denote $u = \cos \dfrac{x - y}{2}$, $v = \sin \dfrac{x + y}{2}$, then (1) and (2) imply$$
\begin{cases}
uv = \dfrac{1}{10}\\
(2u^2 - 1) v \sqrt{1 - v^2} = \pm \dfrac{1}{20}
\end{cases} \Longrightarrow \begin{cases}
uv = \dfrac{1}{10} & (3)\\
(2u^2 - 1)^2 v^2 (1 - v^2) = \dfrac{1}{400} & (4)
\end{cases}.
$$
Since $v = \dfrac{1}{10u}$ by (3), eliminating $v$ from (4) yields$$
(2u^2 - 1)^2\left( 4u^2 - \frac{1}{25} \right) = u^4\\
\Longrightarrow 16 (u^2)^3 - \left( \frac{4}{25} + 17 \right) (u^2)^2 + \left( \frac{4}{25} + 4 \right) u^2 - \frac{1}{25} = 0.
$$
Note that $-π < y \leqslant x \leqslant π$ implies $l$ Now, there are six solutions to the last equation (see WA), i.e. $u_1, \cdots, u_6$. For each $u_k$, there is $v_k = \dfrac{1}{10u_k}$. Note that $\cos \dfrac{x + y}{2} = \pm \sqrt{1 - v^2}$. If $(2u_k^2 - 1) v_k > 0$, then$$
\frac{1}{20} = (2u_k^2 - 1) v_k \cos \frac{x + y}{2} \Longrightarrow \cos \frac{x + y}{2} = \sqrt{\smash[b]{1 - v_k^2}}.
$$
Otheriwise $(2u_k^2 - 1) v_k < 0$, then$$
\frac{1}{20} = (2u_k^2 - 1) v_k \cos \frac{x + y}{2} \Longrightarrow \cos \frac{x + y}{2} = -\sqrt{\smash[b]{1 - v_k^2}}.
$$
Thus for each $k$,$$
\begin{cases}
\cos \dfrac{x - y}{2} = u_k,\ \sin \dfrac{x - y}{2} = \pm \sqrt{\smash[b]{1 - u_k^2}}\\
\sin \dfrac{x + y}{2} = v_k,\ \cos \dfrac{x + y}{2} = ε_k \sqrt{\smash[b]{1 - v_k^2}}
\end{cases},
$$
where $ε_k \in \{\pm 1\}$ is determined as above. Since$$
\begin{cases}
\sin x = \sin \dfrac{x + y}{2} \cos \dfrac{x - y}{2} + \cos \dfrac{x + y}{2} \sin \dfrac{x - y}{2}\\
\cos x = \cos \dfrac{x + y}{2} \cos \dfrac{x - y}{2} - \sin \dfrac{x + y}{2} \sin \dfrac{x - y}{2}\\
\sin y = \sin \dfrac{x + y}{2} \cos \dfrac{x - y}{2} - \cos \dfrac{x + y}{2} \sin \dfrac{x + y}{2}\\
\cos y = \cos \dfrac{x + y}{2} \cos \dfrac{x - y}{2} + \sin \dfrac{x + y}{2} \sin \dfrac{x - y}{2}
\end{cases},
$$
there are twelve possible solutions $(\sin x, \cos x, \sin y, \cos y)$. After computing them numerically, it turns out that there are six distinct solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
trigonometric limit with integral: $\lim_{\alpha\to 0}\int^{\alpha}_{0}\frac{dx}{\sqrt{\cos x -\cos \alpha}}$ I've got following limit to calulate:
$$\lim_{\alpha\to 0}\int^{\alpha}_{0}\frac{dx}{\sqrt{\cos x -\cos \alpha}}$$
Is there any way to bound it by some "more convenient" limit?
| Assume $a>0$ and rewrite
$$
\cos x-\cos a=2\sin\Big(\frac{a+x}{2}\Big)\sin\Big(\frac{a-x}{2}\Big).
$$
Estimate: $t-\frac{t^3}{6}\le\sin t\le t$, $t\ge 0$ gives for $0\le x\le a$
$$
\underbrace{\frac{a^2-x^2}{2}\Big(1-\frac{(a+x)^2}{24}\Big)\Big(1-\frac{(a-x)^2}{24}\Big)}_{\ge\ \frac{a^2-x^2}{2}\big(1-\frac{a^2}{6}\big)\big(1-\frac{a^2}{24}\big)}\le 2\sin\Big(\frac{a+x}{2}\Big)\sin\Big(\frac{a-x}{2}\Big)\le\frac{a^2-x^2}{2},
$$
hence
$$
\frac{\sqrt{2}}{\sqrt{a^2-x^2}}\le\frac{1}{\sqrt{\cos x-\cos a}}\le\frac{\sqrt{2}}{\sqrt{a^2-x^2}}\frac{1}{\sqrt{1-\frac{a^2}{6}}}\frac{1}{\sqrt{1-\frac{a^2}{24}}}.
$$
Integrate:
$$
\frac{\pi}{\sqrt{2}}\le\int_0^a\frac{1}{\sqrt{\cos x-\cos a}}\,dx\le\frac{\pi}{\sqrt{2}}\frac{1}{\sqrt{1-\frac{a^2}{6}}}\frac{1}{\sqrt{1-\frac{a^2}{24}}}\to\frac{\pi}{\sqrt{2}},\quad a\to 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove the theorem of Nicomachus by induction Prove the theorem of Nicomachus(AD.100) by induction:
$$
1^3 = 1,\ 2^3 = 3+ 5,\ 3^3 = 7 + 9 + 11,\ 4^3 = 13 + 15 + 17 + 19,\ ...
$$
My approach: from looking at the above pattern you can tell there is something of the following sort:
$$ 2^{n-1} + q = n ^3,$$
where $q$ is odd s.t. $q = 2k + 1.$
| If induction is not mandatory,
let the $r$th term$(T_r)$ of $1,3,7,13$ be $$a(r-1)(r-2)+b(r-1)+c$$
$r=1\implies 1=c$
$r=2\implies3=b+c\iff b=2$
$r=3\implies7=2a+2b+c\iff a=1$
$$T_r=(r-1)(r-2)+2(r-1)+1=r^2-r+1$$
Sum of $n$ term $$\sum_{m=\frac{n^2-n+1}2-1}^{\frac{n^2+n-1}2-1}(2m+1)=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
How to factor $16x^2-y^2+8y-16$ so I get $(4x+y-4)(4x-y+4)$? I tried completing the square, making it equal to zero, and trying to think of it as a circle, but I still can't prove that these two are equivalent. I tried to factor this in Wolfram Alpha, and it spits out that answer. But I don't know how else to get to it.
| I f you want to be systematic, add in a variable $z$ to make this homogeneous,
$$ 16 x^2 - y^2 - 16 z^2 + 8 y z $$
with result (below)
$$ 16 x^2 - y^2 - 16 z^2 + 8 y z = 16 x^2 - (y-4z)^2 = (4x+y-4z)(4x-y+4z) $$
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left(
\begin{array}{rrr}
16 & 0 & 0 \\
0 & - 1 & 4 \\
0 & 4 & - 16 \\
\end{array}
\right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$
$$ P_{j-1} E_j = P_j $$
$$ E_j^{-1} Q_{j-1} = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = \left(
\begin{array}{rrr}
16 & 0 & 0 \\
0 & - 1 & 4 \\
0 & 4 & - 16 \\
\end{array}
\right)
$$
==============================================
$$ E_{1} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 4 \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$ P_{1} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 4 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; Q_{1} = \left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & - 4 \\
0 & 0 & 1 \\
\end{array}
\right)
, \; \; \; D_{1} = \left(
\begin{array}{rrr}
16 & 0 & 0 \\
0 & - 1 & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
$$
==============================================
$$ P^T H P = D $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 4 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
16 & 0 & 0 \\
0 & - 1 & 4 \\
0 & 4 & - 16 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 4 \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
16 & 0 & 0 \\
0 & - 1 & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
$$
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & - 4 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
16 & 0 & 0 \\
0 & - 1 & 0 \\
0 & 0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 1 & - 4 \\
0 & 0 & 1 \\
\end{array}
\right)
= \left(
\begin{array}{rrr}
16 & 0 & 0 \\
0 & - 1 & 4 \\
0 & 4 & - 16 \\
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $x^5+x^{-5}$ given the value of $x^2+x^{-2}$.
Find $x^5+\dfrac1{x^5}$ in its simplest form given that $x^2+\dfrac1{x^2}=a$ for $a,x>0$.
Attempt:
We write $$x^2+\frac1{x^2}=a\implies x^4-ax^2+1=0\implies x^5=ax^3-x$$ and $$\frac1{x^2}=a-x^2\implies \frac1{x^4}=a^2-2ax^2+x^4\implies \frac1{x^5}=\frac{a^2}x-2ax+x^3$$ so $$\begin{align}x^5+\frac1{x^5}&=(1+a)x^3-(1+2a)x+a^2\cdot\frac1x\\&=(1+a)x^3-(1+2a)x+a^2(ax-x^3)\\&=(1+a-a^2)x^3-(1+2a-a^3)x\\\implies x^5+\frac1{x^5}&=(a^2-a-1)x(a+1-x^2)\end{align}$$ But is this in its simplest form?
| Observe that $$(X+1/X)^2=a+2$$. This gives you $X+1/X$.
Now note that $$(X+1/X)^3=X^3+1/X^3+3(X+1/X)$$ and obtain $X^3+1/X^3$. Finally $$(X^2+1/X^2)(X^3+1/X^3)=X^5+1/X^5+(X+1/X)$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 4
} |
Solutions of $x^4+2tx^2+1=0$
I am looking for all the solutions of $x^4+2tx^2+1=0$
Is there any "quick" way to figure out all the complex-solutions $\pi_i(t)$ where $i=1,...4$ ?
I got $\pi_1(t)=\sqrt{-t+\sqrt{t^2-1}}\sqrt{t-\sqrt{t^2-1}}\sqrt{-t+\sqrt{1-t^2}}$ but apparently I am too "stupid" to verify my calculation. I always end up making mistakes (hopefully) in my calculation.
Would appreciate any help
| $$x^4+2tx^2+1=0$$
Substitute $y=x^2$
$$y^2+2ty+1=0$$
Now use the formula $$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
So, here $a=1,b=2t,c=1$
$$y=\frac{-t\pm \sqrt{4t^2-4(1)(1)}}{2(1)}$$
$$y=\frac{-t\pm \sqrt{4t^2-4}}{2}$$
$$y=-t\pm \sqrt{t^2-1}$$
Substitute back $y=x^2$
$$x^2=-t\pm \sqrt{t^2-1}$$
$$x=\pm\sqrt{-t\pm \sqrt{t^2-1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
2013 AIME I #5: Funky Real Root The real root of the equation $8x^3-3x^2-3x-1=0$ can be written in the form $\frac{\sqrt[3]{a}+\sqrt[3]{b}+1}{c}$, where $a$, $b$, and $c$ are positive integers. Find $a+b+c$.
I noticed that there was a $3x^2-3x-1$ present and I decided to add $(x+1)^3$ to get rid of this. When I did this, I ended up with $x=1/(9^{1/3}-1)$. I am not sure how to rationalize this crazy mess.
| What I will do for most cubic equations:
Step 1:
If the equation $ax^3+bx^2+cx+d=0$ has $b=0$ then skip this step. Otherwise, let $y=x+\dfrac{b}{3a}$.
In this case:
$$8x^3-3x^2-3x-1=0$$
Let $y=x+\dfrac{b}{3a}=x-\dfrac{1}{8}\Rightarrow x=y+\dfrac{1}{8}$.
\begin{equation}\begin{aligned}
8x^3-3x^2-3x-1=0 &\Leftrightarrow 8\left(y+\dfrac{1}{8}\right)^3-3\left(y+\dfrac{1}{8}\right)^2-3\left(y+\dfrac{1}{8}\right)-1=0 \\
&\Leftrightarrow 8\left(y^3+\dfrac{3}{8}y^2+\dfrac{3}{64}y+\dfrac{1}{512}\right) -3\left(y^2+\dfrac{1}{4}y+\dfrac{1}{64}\right)-3y-\dfrac{3}{8}-1=0\\
&\Leftrightarrow 8y^3-\dfrac{27}{8}y-\dfrac{45}{32}=0\\
\end{aligned}\end{equation}
The last equation now has $b=0$.
Step 2:
After having the last equation in the form $ax^3+cx+d=0$, write it again as $x^3+\dfrac{c}{a}x+\dfrac{d}{a}=0$ (if $a\ne 1$) and find real numbers $z,t$ that satisfy ${\begin{cases}z^3+t^3=\dfrac{d}{a}\\zt=\dfrac{-c}{3a}\end{cases}}$
This is why:
$8y^3-\dfrac{27}{8}y-\dfrac{45}{32}=0$
$\Leftrightarrow y^3-\dfrac{27}{64}y-\dfrac{45}{256}=0$
If you can find $z,t\in\mathbb{R}$ so that ${\begin{cases}z^3+t^3=\dfrac{-45}{256}\\zt=\dfrac{27}{192}\end{cases}}\Leftrightarrow {\begin{cases}z^3+t^3=\dfrac{-45}{256}\\z^3t^3=\dfrac{729}{262144}\end{cases}}\Leftrightarrow {\begin{cases}z=\sqrt[3]{\dfrac{-81}{512}}\\t=\sqrt[3]{\dfrac{-9}{512}}\end{cases}}\Leftrightarrow {\begin{cases}z=\dfrac{\sqrt[3]{-81}}{8}\\t=\dfrac{\sqrt[3]{-9}}{8}\end{cases}}$ then
$y^3-3yzt+z^3+t^3=0$
$\Leftrightarrow (y+z+t)(y^2+z^2+t^2-yz-zt-ty)=0$
$\Leftrightarrow (y+z+t)2(y^2+z^2+t^2-yz-zt-ty)=0$
$\Leftrightarrow (y+z+t)((y-z)^2+(z-t)^2+(t-y)^2)=0$
$\Leftrightarrow y=-z-t$ because you will find that $z\ne t$.
$\Rightarrow x=\dfrac{1}{8}-z-t=\dfrac{1}{8}-\dfrac{\sqrt[3]{-81}}{8}-\dfrac{\sqrt[3]{-9}}{8}=\dfrac{-\sqrt[3]{-81}-\sqrt[3]{-9}+1}{8}=\dfrac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
[Calc II]: Shell Method - Dealing with offsets So I'm told to use the shell method for a triangle bounded by $(1,1)$, $(1,2)$ and $(2,1)$. The function values, I've found, are then $y = 3 - x$, $y = 1$ and $x = 1$. I'm rotating it around the $y$-axis. I'm having trouble finding my values for $h(x)$ and $ρ(x)$ though. The two offsets are really screwing with me. Any help understanding how to deal with offsets using the shell method?
I know that the $a$ and $b$ for the definite integral should be $1$ and $2$, respectively. I've tried setting $h(x) = 3 - x$ and $ρ(x) = x$, but it does not give me the correct answer of $4\pi/5$.
| Since we are rotating around the $y$-axis, the method of shells says that the volume is
$$ V = \int_a^b A(x)\, dx $$
where $A(x)$ is the cross-sectional area given by
$$ A(x) = 2\pi\left(\textrm{radius}\right)\left(\textrm{height}\right). $$
Since we are rotating around $x=0$ and the triangle is on the first quadrant, the radius is $\rho(x) = x$. Now, vertically (in the $y$-direction) the triangle is bounded by $y_\textrm{lower}=1$ and $y_\textrm{upper}=3-x$. Consequently, the height is
$$ h(x) = y_\textrm{upper} - y_\textrm{lower} = 3-x-1 = 2-x. $$
Since the integration variable is $x$, the limits $a, b$ is determine by looking at the $x$-direction; it should be clear that $a = 1$ and $b = 2$. Hence,
\begin{align*}
V & = \int_1^2 2\pi x(2-x)\, dx \\
& = 2\pi\int_1^2 2x - x^2\, dx \\
& = 2\pi\left(x^2 - \frac{x^3}{3}\right)\bigg|_1^2 \\
& = 2\pi\left[\left(2^2 - \frac{2^3}{3}\right) - \left(1^2 - \frac{1^3}{3}\right)\right] \\
& = 2\pi\left[\left(4 - \frac{8}{3}\right) - \left(1 - \frac{1}{3}\right)\right] \\
& = 2\pi\left[\frac{4}{3} - \frac{2}{3}\right] \\
& = \frac{4\pi}{3}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial fraction of $\frac{s}{(s^2+2s+2)(s^2-2s+2)}$ I am trying to find the partial fraction of:
$$\frac{s}{(s^2+2s+2)(s^2-2s+2)}$$
I started off with:
$$\frac{A(s^2-2s+2)}{s^2+2s+2} + \frac{B(s^2+2s+2)}{s^2-2s+2}$$
After that I get the following equations:
$A+B = 0$; $-2A+2B =1$
Giving: $A=-B$, and secondly $4B=1$, Hence:
$A=-\frac{1}{4}$ and $B=\frac{1}{4}$
But is this correct?
| Since the factors in the denominator can't be factored over in the real numbers further, the best you can do is$$\frac {s}{(s^2+2s+2)(s^2-2s+2)}=\frac {As+B}{s^2+2s+2}+\frac {Cs+D}{s^2-2s+2}$$
Multiply both sides by the denominator to clear the fractions to get the equation$$s=(As+B)(s^2-2s+2)+(Cs+D)(s^2+2s+2)$$
Now let replace any occurrence of $s^2$ with $2s-2$, simplify, and repeat the process with replacing $s^2$ with $-2s-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can these definite integrals be expressed in closed form? In my research, I recently came across the integral
$$
I(a,b) = \int_{\theta_1}^{\theta_2} \frac{\sin^2 \theta}{(a \sin \theta - b \cos \theta)^2}\,\mathrm{d}\theta,
$$
where the range of integration $[\theta_1, \theta_2]$ is defined slightly differently in the two cases I'm interested in:
*
*Case I: $[\theta_1, \theta_2] = \{ \theta \in [0, \pi]: a \sin \theta - b \cos \theta > 1 \}$
*Case II: $[\theta_1, \theta_2] = \{ \theta \in [0, \pi]: a \sin \theta - b \cos \theta > b \}$
The constants $a$ and $b$ are positive, but other than that they are arbitrary.
In the case $b = 1$, these two cases are the same, the lower bound of integration is $\theta_1 = 2 \tan^{-1} (1/a)$, the upper bound is $\theta_2 = \pi$, and the integral evaluates to
$$
I(a,1) = \frac{a + (a^2 - 1) \tan^{-1} a}{(a^2 + 1)^2}
$$
(thanks to Mathematica.)
However, I haven't been able to find nearly so nice an expression for the integrand in the cases where $b \neq 1$. The bounds of integration can be found by solving a quadratic equation for $\cos \theta$, but the expressions aren't "nice". Mathematica is little help, since it has trouble finding an antiderivative that works in both Quadrant I and Quadrant II. (It also throws imaginary numbers into the results; they cancel out when you simplify the resulting expressions, but are rather unsightly all the same.)
If all else fails, I can numerically integrate this; but I'm wondering if there are any slick mathematical substitutions I can use to make my life easier.
| $a\sin\theta + b\cos\theta = \sqrt {a^2+b^2}\cos (\theta -\arctan \frac {a}{b})$
$\frac {1}{a^2+b^2} \int \sin^2\theta\sec^2(\theta - \arctan \frac {a}{b}) \ d\theta\\
\phi = \theta - \arctan \frac {a}{b}\\
\frac {1}{a^2+b^2} \int \sin^2(\phi+\arctan \frac {a}{b})\sec^2\phi \ d\theta\\
\frac {1}{a^2+b^2} \int (\sin\phi\cos\arctan \frac {a}{b} + \cos\phi\sin\arctan \frac {a}{b})^2\sec^2 \phi d\phi\\
\sin\arctan \frac {a}{b} = \frac {a}{\sqrt {a^2+b^2}},\cos\arctan \frac {a}{b} = \frac {b}{\sqrt {a^2+b^2}}\\
\frac {1}{a^2+b^2} \int \frac {b^2}{a^2+b^2}\sin^2\phi + \frac {2ab}{a^2+b^2}\sin\phi\cos\phi + \frac {a^2}{a^2+b^2}\cos^2\phi)\sec^2 \phi d\phi\\
\frac {1}{a^2+b^2} \int \frac {b^2}{a^2+b^2}\tan^2\phi + \frac {2ab}{a^2+b^2}\tan\phi + \frac {a^2}{a^2+b^2}d\phi\\
\frac {1}{a^2+b^2} \int \frac {b^2}{a^2+b^2}\sec^2\phi + \frac {2ab}{a^2+b^2}\tan\phi + \frac {a^2-b^2}{a^2+b^2}d\phi\\
\frac {1}{(a^2+b^2)^2} (b^2\tan\phi + 2ab \ln |\sec\phi| + (a^2-b^2)\phi)\\
\frac {1}{(a^2+b^2)^2} (b^2\tan(\theta_2 - \arctan \frac ab) - b^2\tan(\theta_1 - \arctan \frac ab) + 2ab \ln |\sec(\theta_2 - \arctan \frac ab)| - 2ab \ln |\sec(\theta_1 - \arctan \frac ab)| + (a^2-b^2)(\theta_2-\theta_1))$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding $\lim\limits_{n→∞}\left(\frac{\sqrt{n^2+n}-1}n\right)^{2\sqrt{n^2+n}-1}$ I have the limit
$$\lim_{n\to\infty}\left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}.$$
I simplified the limit by using $(f(x)-1)\cdot g(x)$, where $f(x)= \sqrt{n^2+n}-1$ and $g(x) = 2\sqrt{n^2+n}- 1$ since it is in the form of $1^\infty$. I ended up getting $$\frac{(3+2n)(n-(\sqrt{n^2-n})+1}{n},$$ which tends to $\infty$.
Please help how to proceed further or a different way to solve it.
| First of all, note that the $-1$ in the exponent is irrelevant, i.e.,
$$\lim_{n\to\infty}\left(\sqrt{n^2+n}-1\over n \right)^{2\sqrt{n^2+n}-1}
=\lim_{n\to\infty}\left(\left(\sqrt{n^2+n}-1\over n \right)^{2\sqrt{n^2+n}}\over\sqrt{1+{1\over n}}-{1\over n}\right)
=\lim_{n\to\infty}\left(\sqrt{n^2+n}-1\over n \right)^{2\sqrt{n^2+n}}$$
Now, abbreviating $\sqrt{n^2+n}$ to $s$ (for "$s$quare root") when convenient, and noting that $s\to\infty$ as $n\to\infty$, we have
$$\begin{align}
\left(\sqrt{n^2+n}-1\over n \right)^{2\sqrt{n^2+n}}
&=\left(\sqrt{n^2+n}-1\over\sqrt{n^2+n} \right)^{2\sqrt{n^2+n}}\left(\sqrt{n^2+n}\over n \right)^{2\sqrt{n^2+n}}\\
&=\left(1-{1\over s} \right)^{2s}\left(n^2+n\over n^2 \right)^{\sqrt{n^2+n}}\\
&=\left(\left(1-{1\over s} \right)^s\right)^2\left(\left(1+{1\over n} \right)^n\right)^{\sqrt{1+1/n}}\\
&\to (e^{-1})^2\cdot (e^1)^1=e^{-1}
\end{align}$$
(Note, in the limiting step, we use both the product theorem for limits and the theorem that $f(n)^{g(n)}\to L^\ell$ if $f(n)\to L$ and $g(n)\to\ell$ and $L^\ell$ is not of the indeterminate form $0^0$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2811250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Prove identity: $\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $ $$\sin \alpha= \frac{2\tan \frac{\alpha}{2}}{1+\tan^2 \frac{\alpha}{2}} $$
I am having a problem proving this identity. I write tan like $\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}$ and the squared one in the same way.
I eventually get $$\frac {2\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2}} $$
And I am stuck...
| In your last formula
*
*Your nominator is $\sin x$ due to $\sin 2x = 2\sin x \cos x$, written for $\frac{x}{2}$.
*Your denominator is $1$, due to $\sin^2 x + \cos^2 x = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2812197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of $10$ digit numbers such that every digit that appears appears exactly twice I need to find the number of $10$-digit numbers in which every digit that appears appears exactly twice.
I have attempted to solve this but am not sure if I am not overcounting something.
The only problem is that a $0$ cannot stand on the leftmost position.
The first digit, say $i$, can be chosen in $9$ ways. Then, there are $9$ remaining places for the second $i$.
Now, we have $9$ possible remaining digits and $8$ remaining places. We need to choose $4$ digits, we can do so in $\binom{9}{4}$ ways.
Now, the remaining places must be divided into two-element subsets, which can be done in ${8\brace 2}$ ways (Stirling numbers of the second kind).
Finally, we only need to distribute these $4$ numbers between these subsets, in $4!$ ways.
And so the final answer is:
$$9 \cdot 9 \cdot \binom{9}{4} \ {8 \brace 2} \cdot 4!$$
Is it the correct answer?
| Since the leading digit cannot be $0$, there are $9$ ways to pick the leading digit. There are $9$ ways to pick the other location for the leading digit. There are $\binom{9}{4}$ ways to select which four of the other nine digits appear in the number. There are $\binom{8}{2}$ ways to select the locations of the smallest of those numbers, $\binom{6}{2}$ ways to select the locations of the next smallest of those numbers, $\binom{4}{2}$ ways to select the location of the third smallest of those numbers, and $\binom{2}{2}$ ways to select the location of the largest of those numbers. Hence, the number of $10$-digit positive integers in which every digit that appears appears exactly twice is
$$\binom{9}{1}\binom{9}{1}\binom{9}{4}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
As Joriki pointed out in the comments, ${8 \brace 2}$ is the number of ways of placing $8$ objects in two non-empty subsets, not the number of ways of placing eight objects in four subsets of size $2$, which is
$$\frac{1}{4!}\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
If you replace your factor of ${8 \brace 2}$ by the above factor, your answer would agree with mine.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Characteristic Polynom of a Matrix for the complete Graph, I need to derive a formula for the characteristic polynom of a complete graph. Here are some example of the adjacency Matrices:
$$
K_2=
\begin{pmatrix}{}
0 & 1 \\
1 & 0 \\
\end{pmatrix}
$$
$$
\chi(K_2)=(-1 + x) (1 + x)
$$
$$
K_3=
\begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0 \\
\end{pmatrix}
$$
$$
\chi(K_3)=-(-2 + x) (1 + x)^2
$$
$$
K_4=\left(
\begin{array}{cccc}
0 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0 \\
\end{array}
\right)$$
$$
\chi(K_4)=(-3 + x) (1 + x)^3
$$
$$
K_5=
\left(
\begin{array}{ccccc}
0 & 1 & 1 & 1 & 1 \\
1 & 0 & 1 & 1 & 1 \\
1 & 1 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 & 1 \\
1 & 1 & 1 & 1 & 0 \\
\end{array}
\right)$$
$$
\chi(K_5)=-(-4 + x) (1 + x)^4
$$
I see the pattern:
$$
\chi(K_n)=(-1)^{n}(-(n-1) + x) (1 + x)^{(n-1)}
$$
Although I have no idea how to proove it. Any ideas?
Application:
From the closed formula one can derive the number of closed walks in the graph
| If you add the identity then you get a rank-one matrix whose characteristic polynomial is $$\chi(K_n + I) = (-1)^n x^{n-1} (x - n).$$ The factor $(x-n)$ coming from the lone eigenvalue $n$ with eigenvector $(1,1,...1).$ When passing from $K_n+I$ to $K_n$ you substitute $x \mapsto x+1$ and get the formula you want.
| {
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"answer_id": 0
} |
Use sum and product of roots to prove type of roots. Suppose $α$ and $β$ are the roots of $x^2-px+q$. Then
$(a)$ Find the quadratic equation whose roots are $α^2+β^{-2}$ and $β^2+α^{-2}$.
$(b)$ Prove that if $(p,q) \in \mathbb{R}$ then this equation will have equal roots if $p = 0$ and $p^2=4q$.
I have tried to solve part (b) several times but I seem to be heading nowhere…there is a lot of algebra involved and I seem to get lost in the forest every time I attempt it.
Note that: "this equation"in (b) refers to the quadratic equation derived in (a)
| For part a we have that
$(a)$ Find the quadratic equation whose roots are $α^2+β^{-2}$ and $β^2+α^{-2}$
*
*$p'=α^2+β^{-2}+β^2+α^{-2}=α^2+β^2+\frac{α^2+β^2}{α^2\beta^2}=p^2-2q+\frac{p^2-2q}{q^2}$
*$q'=(α^2+β^{-2})(β^2+α^{-2})=α^2β^2+2+\frac{1}{α^2\beta^2}=q^2+2+\frac1{q^2}$
then
$$x^2-(p^2-2q+\frac{p^2-2q}{q})x+q^2+2+\frac1{q^2}=0$$
and for part (b) we need $p'^2=4q'$ that is
$$p'^2=\left(p^2-2q+\frac{p^2-2q}{q^2}\right)^2=(p^2-2q)^2\left(1+\frac1{q^2}\right)^2=(p^2-2q)^2\left(1+\frac2{q^2}+\frac1{q^4}\right)$$
$$4q'=4\left(q^2+2+\frac1{q^2}\right)=4q^2\left(1+\frac2{q^2}+\frac1{q^4}\right)$$
which implies
$$(p^2-2q)^2=4q^2\iff(p^2-2q-2q)(p^2-2q+2q)=0 \iff p^2=4q \,\lor\, p=0$$
| {
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"answer_id": 0
} |
How to compute $\sum_n(2n - \sqrt{n^2+1}-\sqrt{n^2-1})$? How to compute $\sum_n (2n - \sqrt{n^2+1}-\sqrt{n^2-1})$?
I tried two ways:
1. \begin{align*}
(2n - \sqrt{n^2+1}-\sqrt{n^2-1})
&= n - \sqrt{n^2+1} + n -\sqrt{n^2-1} \\
&= \frac{1}{n+\sqrt{n^2-1}}-\frac{1}{n-\sqrt{n^2+1}},
\end{align*}
but I don't know how to do later.
2. \begin{align*}
(2n - \sqrt{n^2+1}-\sqrt{n^2-1})
&= 2n - \frac{(\sqrt{n^2+1} + \sqrt{n^2-1})}{1} \\
&= 2n - \frac{2}{\sqrt{n^2+1} - \sqrt{n^2-1}},
\end{align*}
but I don't know how to do later too.
| Let us assume that you need to compute the infinite summation with
$$a_n=2n - \sqrt{n^2+1}-\sqrt{n^2-1}$$ For large values of $n$, rewrite $$a_n=n\left(2- \sqrt{1+\frac 1{n^2}}- \sqrt{1-\frac 1{n^2}}\right)$$ and use the binomial expansion or Taylor series to get
$$a_n=\frac{1}{4 n^3}+\frac{5}{64 n^7}+O\left(\frac{1}{n^{11}}\right)$$ So,
$$\sum_{n=1}^\infty a_n\approx\sum_{n=1}^p a_n+\frac 14 \sum_{n=p+1}^\infty \frac{1}{ n^3}=\sum_{n=1}^p a_n-\frac{1}{8}\psi ^{(2)}(p+1)\approx \sum_{n=1}^p a_n +\frac 1 {8 p^2}$$
If we use $p=10$, the first summation is $\approx 0.635843$, the correction term is $0.00125$ making a total of $0.637093$ while the infinite summation looks to be $\approx 0.636974$. For sure, if we increase $p$, we shall get closer and closer.
Now, the question is : what is this number ?
| {
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Trigonometric version of second difference I am reading an article
Hall P, Turlach BA. On the estimation of a convex set with corners. IEEE Transactions on Pattern Analysis and Machine Intelligence. 1999 Mar;21(3):225-34
and stuck at an identity used there (p. 228). It can be written as follows:
$$
p(\theta_i)\sin(\theta_j - \theta_k) + p(\theta_j)\sin(\theta_k - \theta_i) + p(\theta_k)\sin(\theta_i - \theta_j) = -\frac{1}{2}(\theta_i - \theta_j)(\theta_j - \theta_k)(\theta_k - \theta_i)\{p(\theta_j) + p''(\theta_j)\} + O\{(\theta_k - \theta_i)^4\},
$$
where $-\pi < \theta_i < \theta_j < \theta_k < \pi$ and $p:(-\pi, \pi) \to \mathbb{R}$ is a function with three bounded derivatives on $[\theta_i, \theta_k]$.
Authors call this idenitity as trigonometric version of second difference. In the paper $p(\theta)$ is a support function of the convex figure on the plane, but the idenitity does not require this property.
I have tried to prove this idenitity, but failed. Here are my steps:
*
*We can write the identity in the equivalent form, by dividing both parts by $(\theta_i - \theta_j)(\theta_j - \theta_k)(\theta_k - \theta_i)$ and multiplying by $-1$:
$$
\frac{1}{2}\{p(\theta_j) + p''(\theta_j)\} = \frac{p(\theta_i)\sin(\theta_k - \theta_j) - p(\theta_j)\sin(\theta_k - \theta_i) + p(\theta_k)\sin(\theta_j - \theta_i)}{(\theta_j - \theta_i)(\theta_k - \theta_j)(\theta_k - \theta_i)} + O(\theta_k - \theta_i),
$$
*Substitute sines with first two parts of Taylor series as $\sin x = x - \dfrac{x^3}{6} + O(x^5)$:
$$
\frac{1}{2}\{p(\theta_j) + p''(\theta_j)\} = \frac{p(\theta_i)(\theta_k - \theta_j) - p(\theta_j)(\theta_k - \theta_i) + p(\theta_k)(\theta_j - \theta_i)}{(\theta_j - \theta_i)(\theta_k - \theta_j)(\theta_k - \theta_i)} - \frac{1}{6} \frac{p(\theta_i)(\theta_k - \theta_j)^3 - p(\theta_j)(\theta_k - \theta_i)^3 + p(\theta_k)(\theta_j - \theta_i)^3}{(\theta_j - \theta_i)(\theta_k - \theta_j)(\theta_k - \theta_i)} + O(\theta_k - \theta_i),
$$
*We can write a second divided difference in classic form without sines:
$$
\frac{1}{2}p''(\zeta) = \frac{p(\theta_i)(\theta_k - \theta_j) - p(\theta_j)(\theta_k - \theta_i) + p(\theta_k)(\theta_j - \theta_i)}{(\theta_j - \theta_i)(\theta_k - \theta_j)(\theta_k - \theta_i)}
$$
for some $\zeta \in [\theta_i, \theta_k]$ (see e.g. Bakhvalov N.S. Numerical methods: analysis, algebra, ordinary differential equations. MIR Publishers, 1977. P. 50).
Hence,
$$
\frac{1}{2}p''(\theta_j) = \frac{p(\theta_i)(\theta_k - \theta_j) - p(\theta_j)(\theta_k - \theta_i) + p(\theta_k)(\theta_j - \theta_i)}{(\theta_j - \theta_i)(\theta_k - \theta_j)(\theta_k - \theta_i)} + O(\theta_k - \theta_i),
$$
because $p''(\theta_j) = p''(\zeta) + O(\theta_k - \theta_i)$.
*After the subtraction of the second identity from the first one, we obtain:
$$
\frac{p(\theta_j)}{2} = - \frac{1}{6} \frac{p(\theta_i)(\theta_k - \theta_j)^3 - p(\theta_j)(\theta_k - \theta_i)^3 + p(\theta_k)(\theta_j - \theta_i)^3}{(\theta_j - \theta_i)(\theta_k - \theta_j)(\theta_k - \theta_i)} + O(\theta_k - \theta_i)
$$
The last identity does not seem to be true for any arbitrary function $p(\theta)$. If we try to take $\theta_i = \phi, \theta_j = 2 \phi, \theta_k = 3 \phi$, then
$$
\frac{p(\theta_j)}{2} = - \frac{1}{6} \frac{p(\theta_i)\phi^3 - 8 p(\theta_j)\phi^3 + p(\theta_k)\phi^3}{2 \phi^3} + O(\phi),
$$
or, equivalently,
$$
p(\theta_j) = \frac{p(\theta_i) + p(\theta_k)}{2} + O(\phi),
$$
and this is not true in the general case.
I need to prove the initial identity. Could someone find a mistake in my calculations?
| $\def\peq{\mathrel{\phantom{=}}{}}$Since $p'''$ is bounded, then\begin{align*}
p(θ_i) &= p(θ_j) + (θ_i - θ_j) p'(θ_j) + \frac{1}{2} (θ_i - θ_j)^2 p''(θ_j) + O((θ_i - θ_j)^3), \\
p(θ_k) &= p(θ_j) + (θ_k - θ_j) p'(θ_j) + \frac{1}{2} (θ_k - θ_j)^2 p''(θ_j) + O((θ_k - θ_j)^3).
\end{align*}
Also,\begin{align*}
\sin(θ_i - θ_j) &= (θ_i - θ_j) - \frac{1}{6} (θ_i - θ_j)^3 + O((θ_i - θ_j)^4), \\
\sin(θ_j - θ_k) &= (θ_j - θ_k) - \frac{1}{6} (θ_j - θ_k)^3 + O((θ_j - θ_k)^4), \\
\sin(θ_k - θ_i) &= (θ_k - θ_i) - \frac{1}{6} (θ_k - θ_i)^3 + O((θ_k - θ_i)^4).
\end{align*}
Note that $|θ_i - θ_j| \leqslant |θ_k - θ_i|$, $|θ_j - θ_k| \leqslant |θ_k - θ_i|$. After combining terms with respect to $p(θ_j)$, $p'(θ_j)$, $p''(θ_j)$,\begin{align*}
&\peq p(θ_i) \sin(θ_j - θ_k) + p(θ_j) \sin(θ_k - θ_i) + p(θ_k) \sin(θ_i - θ_j) \\
&= A_0 p(θ_j) + A_1 p'(θ_j) + A_2 p''(θ_j) + O((θ_k - θ_i)^4),
\end{align*}
where\begin{align*}
A_0 &= (θ_j - θ_k) + (θ_k - θ_i) + (θ_i - θ_j) \\
&\peq - \frac{1}{6} ((θ_j - θ_k)^3 + (θ_k - θ_i)^3 + (θ_i - θ_j)^3) \\
&= -\frac{1}{6} ((θ_j^3 - 3θ_j^2 θ_k + 3θ_j θ_k^2 - θ_k^3) + (θ_k^3 - 3θ_k^2 θ_i + 3θ_k θ_i^2 - θ_i^3) \\
&\peq + (θ_i^3 - 3θ_i^2 θ_j + 3θ_i θ_j^2 - θ_j^3)) \\
&= -\frac{1}{2} (-θ_i^2 θ_j + θ_i θ_j^2 - θ_j^2 θ_k + θ_j θ_k^2 - θ_k^2 θ_i + θ_k θ_i^2) \\
&= -\frac{1}{2} (θ_i - θ_j)(θ_j - θ_k)(θ_k - θ_i),
\end{align*}$$
A_1 = (θ_i - θ_j)(θ_j - θ_k) + (θ_k - θ_j)(θ_i - θ_j) = 0,
$$\begin{align*}
A_2 &= \frac{1}{2} (θ_i - θ_j)^2 (θ_j - θ_k) + \frac{1}{2} (θ_k - θ_j)^2 (θ_i - θ_j) \\
&= -\frac{1}{2} (θ_i - θ_j)(θ_j - θ_k)(θ_k - θ_i).
\end{align*}
Therefore,\begin{align*}
&\peq p(θ_i) \sin(θ_j - θ_k) + p(θ_j) \sin(θ_k - θ_i) + p(θ_k) \sin(θ_i - θ_j) \\
&= -\frac{1}{2} (θ_i - θ_j)(θ_j - θ_k)(θ_k - θ_i) (p(θ_j) + p''(θ_j)) + O((θ_k - θ_i)^4).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Division of $f(x):= x^6+x^4+x^3-x^2-1$ by $g(x):=x^4+x^3-1$ seems not so correct Let's say $f(x):= x^6+x^4+x^3-x^2-1$ and $g(x):=x^4+x^3-1$. Given that degree $(f(x))$ $>$ degree $(g(x))$, we can divide $f$ by $g$. The issue here is, continuous division of polynomial (starting from the higher powers) we have learnt in junior classes, at the end of the day, we get the remainder as $(-x^3-x+1)$ (quotient being $x^2-x+2$) but remainders can't be negative right? So, if we make the quotient $x^2-x+1$, I think the remainder obtained will be more suitable.
| The division is correct indeed we have that
$$(x^4+x^3-1)(x^2-x+2)=x^6-x^5+2x^4+x^5-x^4+2x^3-x^2+x-2=$$
$$=x^6+x^4+2x^3-x^2+x-2=(x^6+x^4+x^3-x^2-1)+(x^3+x-1)$$
and therefore
$$x^6+x^4+x^3-x^2-1=(x^4+x^3-1)(x^2-x+2)+(-x^3-x+1)$$
| {
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"url": "https://math.stackexchange.com/questions/2822281",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Short notation for a column vector where each element is the mean of each row of a given matrix I have a matrix $A$ defined as:
$$
A =
\begin{bmatrix}
a_{1,1} & a_{1,2} & \ldots & a_{1,m} \\
a_{2,1} & a_{2,2} & \ldots & a_{1,m} \\
\vdots & \vdots & \ddots & a_{1,m} \\
a_{n,1} & a_{n,2} & \ldots & a_{n,m}
\end{bmatrix}
$$
I have also a column vector $b$ defined as:
$$
b =
\begin{bmatrix}
b_{1} \\
b_{2} \\
\vdots \\
b_{n}
\end{bmatrix}
= \frac{1}{m}
\begin{bmatrix}
a_{1,1} + a_{1,2} + \ldots + a_{1,m} \\
a_{2,1} + a_{2,2} + \ldots + a_{1,m} \\
\vdots \\
a_{n,1} + a_{n,2} + \ldots + a_{n,m}
\end{bmatrix}
= \frac{1}{m}
\begin{bmatrix}
\sum_{i=1}^m a_{1,i} \\
\sum_{i=1}^m a_{2,i} \\
\vdots \\
\sum_{i=1}^m a_{n,i}
\end{bmatrix}
$$
My question is: there is a way to define $b$ without using the elements of $A$ (e.g. $a_{1,1}, a_{1,2}, \ldots, a_{n,m}$) but only performing some operations on directly on A?
| I think
$$b = \dfrac{1}{m}\begin{bmatrix}
a_{1,1} & a_{1,2} & \ldots & a_{1,m} \\
a_{2,1} & a_{2,2} & \ldots & a_{1,m} \\
\vdots & \vdots & \ddots & a_{1,m} \\
a_{n,1} & a_{n,2} & \ldots & a_{n,m}
\end{bmatrix} \underbrace{\begin{bmatrix} 1\\1\\ \vdots \\ 1\\1\end{bmatrix}}_{m\times1}=\dfrac{1}{m}A\,\vec{1}$$
is the most compact way.
| {
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How to compute integral on calculator? So i have this integral tending from 2(on top) to 1(on bottom ) of
$$\int_1^2x\ln(x^2 +3)x\,dx$$
If I turn it into $x^2\ln(x^2 +3)\,dx$, with integration by parts I get
$$\frac{16}{3}\,\ln(2) + \frac{49}{9}.$$
I was wondering, how is it possible that my TI nspire CAS does NOT recognize my integral? If I compute it, both with and without intervals, it gives me the same exact integral or a really weird equation with $\tan$ to the power of $-1.$
So I was wondering if any of you knew what could possibly be the problem. I tried typing it as the original integral, which is $x\ln(x^2 +3)x\,dx$, and also like this: $x^2\ln(x^2 +3)\,dx.$ Didn't get the answer above.
Any help would be really useful. :)
I am new here so mathjax is kind of a challenge to me... any help also on formatting the integral properly would be nice. :)
| By parts
$$\begin{align}\int x^2\log(x^2+3)\,dx&=\frac{x^3}3\log(x^2+3)-\int\frac{2x^4}{3(x^2+3)}dx\\
&=\frac{x^3}3\log(x^2+3)-\frac23\int\left(x^2-3+\frac9{x^2+3}\right)dx\\
&=\frac{x^3}3\log(x^2+3)-\frac{2x^3}{9}+2x-6\int\frac1{(\sqrt3t)^2+3}d(\sqrt3t)\\
&=\frac{x^3}3\log(x^2+3)-\frac{2x^3}{9}+2x-\frac63\sqrt3\arctan t\\
&=\frac{x^3}3\log(x^2+3)-\frac{2x^3}{9}+2x-2\sqrt3\arctan\frac x{\sqrt3}
.\end{align}$$
| {
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"source": "stackexchange",
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A geometric inequality for a triangle ABC I have to prove that:
$ \frac {a^2}{w_a^2} + \frac {b^2}{ w_b^2} +\frac {c^2}{ w_c^2} \ge 4,$
for the sides $a,b,c$ of the triangle $ABC,$ $ w_a, w_b, w_c $ the angle bisectors and $s$ its semiperemeter. We have for the angle bisectors of the triangle $ABC$:
$ w_a = \frac {2}{b+c} \sqrt{b c s (s-a)},$ $ w_b = \frac {2}{c+a} \sqrt{c a s (s-b)},$ $ w_c = \frac {2}{a+b} \sqrt{a b s (s-c)}.$
Then, we have:
$ \frac {a^2}{w_a^2} + \frac {b^2}{ w_b^2} +\frac {c^2}{ w_c^2} = \frac {1}{4 s} [ \frac { a^2 (b+c)^2}{b c (s-a)} + \frac { b^2 (a+c)^2}{a c (s-b)} + \frac { c^2 (a+b)^2}{a b (s-c)}].$
So, we have to prove
$ \frac{ a^2 (b+c)^2}{b c (s-a)} + \frac { b^2 (a+c)^2}{a c (s-b)} + \frac { c^2 (a+b)^2}{a b (s-c)} \ge 16 s $
Then I stuck, thank you.
| Let $m_a$, $m_b$ and $m_c$ be medians of the triangle to sides $a$, $b$ and $c$ respectively.
Thus, by C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{a^2}{w_a^2}\geq\sum_{cyc}\frac{a^2}{m_a^2}=4\sum_{cyc}\frac{a^2}{2b^2+2c^2-a^2}=4\sum_{cyc}\frac{a^4}{2a^2b^2+2a^2c^2-a^4}\geq$$
$$\geq\frac{4(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2+2a^2c^2-a^4)}\geq\frac{4(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2+a^4+c^4-a^4)}=\frac{4(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2+a^4)}=4.$$
| {
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Find the acute angle between the surfaces $xy^2z = 3x + z^2 $ and $3x^2-y^2+2z=1$ at the point $(1,-2,1)$ Find the acute angle between the surfaces $xy^2z = 3x + z^2 $ and $3x^2-y^2+2z=1$ at the point $(1,-2,1)$
Angle between curves at a point is given by the angle between their tangent planes at the point.
$$f(x,y,z):= 3x +z^2 -xy^2z$$
$\nabla f(1,-2,1) = \langle 3-y^2z,-2xyz,2z-xy^2\rangle_{(1,-2,1)} = \langle -1,4,-2 \rangle$
Equation of tangent plane to $xy^2z = 3x + z^2 $ will be
$-x + 4y -2z + d=0$ Putting $(1,-2,1)$ and solving for $d$ we have
$$x-4y+2z=11 \; \; \; (1)$$
Also, a nice way to write equation of tangent plane to curve $ax^2 + by^2 + cz^2 + 2ux + 2vy + 2wz + d =0$ at $P(x_0,y_0,z_0)$ would be:
$$ax\cdot x_0 + by\cdot y_0 + cz\cdot z_0 + u(x+x_0) + v(y+y_0) + w(z+z_0) +d=0$$
Hence tangent plane to $3x^2-y^2+2z=1$ will be $3x(1) -y(-2)+(z+1)=1$
$$ \Rightarrow 3x +2y+z=0 \; \; \; (2)$$
One of the reason for posting this is, how'd you follow this method to write equation of tangent plane to the first curve $xy^2z = 3x + z^2 $? Basically, is it possible to extend this method to equations where degree is greater than $2$ and contains terms such as $xyz$?
Now, angle between tangent planes is angle between their normals,
Direction ratios of normal to $(1)$ and $(2)$ respectively are
$a=\langle 1,-4,2\rangle$ and $b=\langle 3,2,1\rangle $
$\Rightarrow \theta= \arccos(\frac{a\cdot b}{|a||b|}) = \arccos(\frac{-3}{7\sqrt{6}})$
Problem here is, how'd I know if this is acute or not, i.e, if this is the answer that I was looking for ?
| Did you go ahead and take arccos$\left(\frac{-3}{7\sqrt{6}}\right)$? I get about 100$^o$. So your answer is no. The acute angle is 180 - 100 or about 80$^o$.
| {
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Show that the substitution $x=X-1$ and $y=Y+3$ turns $\frac{dy}{dx}=...$ into a homogenous equation. $$\frac{dy}{dx}=\frac{4x-y+7}{2x+y-1}$$
After substitution of the given variable I get $$\frac{dY}{dX} = \frac{4-\frac{Y}{X}}{2+\frac{Y}{X}}$$ which seems to give a homogenous equation. (There is no answer given for this part of the question)
But then the next question says find the particular solution with $x=0$ and $y=3$ giving your answer in the form $f(x,y)=c$. Then after this point I get an answer that is completely different to the answer given in the book.
Let $Y=vX$. Then $\frac{dY}{dX} = X\frac{dv}{dX}+v$
Replace, then $$X\frac{dv}{dX}+v=\frac{4-v}{2+v}$$
Making the Integrals
$$\int \frac{1}{\frac{4-v}{2+v}-v}dv=\int\frac{1}{X}dX$$
$$\frac{3\ln(1-v)}{5}-\frac{2\ln(4+v)}{5}-\ln(X)=c$$
Replacing everything back to $y$ and $x$.
$$\frac{3\ln(1-\frac{y-3}{x+1})}{5}-\frac{2\ln(4+\frac{y-3}{x+1})}{5}-\ln(x+1)=c$$
According to this $c=-\frac{2\ln(4)}{5}$
Even before the replacement stages the answer is already completely different to that of the answer given.
What did I do wrong?
Just for reference, answer given is $(x-y+4)^3(4x+y+1)^2 = 16$.
| Note that LHS integral should give
$$ \int \frac{2+v}{(v+4)(1-v)}dv = -\frac35 \ln(1-v) - \frac25 \ln(4+v) $$
After fixing the sign error, you can transform your answer to match the one given.
$$ 3\ln\left(1-\frac{y-3}{x+1}\right) + 2\ln\left(4+\frac{y-3}{x+1}\right) + 5\ln(x+1) = -5c $$
$$ 3\ln \left(\frac{x-y+4}{x+1} \right) + 2\ln\left(\frac{4x+y+1}{x+1}\right) + 5\ln(x+1) = -5c $$
Using the $\log$ properties, this simplifies to
$$ 3\ln (x-y+4) + 2\ln(4x+y+1) = -5c $$
Taking the exponential gives
$$ (x-y+4)^3(4x+y+1)^2 = e^{-5c} $$
and the given initial point gives $e^{-5c}=16$
| {
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How to get Jordan Form above the diagonal and not below it? Let $ M=
\begin{bmatrix}
0 & 0 & 0 \\
1 & 1 &1 \\
-1 & -1 & -1\\
\end{bmatrix}$ $\in M_{3}(\mathbb R^{3})$
I found that $\chi_{M}=X^{3}$ and $\mu_{M}=X^{2}$, since $\dim_{\mathbb R}\operatorname{Im}(M-0\cdot\operatorname{Id})=1$
So $$\ker(M-0\cdot\operatorname{Id})= \ker\begin{bmatrix}
0 & 0 & 0 \\
1 & 1 &1 \\
-1 & -1 & -1\\
\end{bmatrix}=\left\{\begin{bmatrix}
-1\\
1 \\
0 \\
\end{bmatrix}, \begin{bmatrix}
-1\\
0 \\
1 \\
\end{bmatrix} \right\}\quad(*)$$
Now define $(M-0\cdot\operatorname{Id})=N$.
It follows:
$0 \subset \ker N \subset \mathbb R^{3}$. We can find a basis of the complement to $(*)$ in $\mathbb R^{3}$, say $\begin{bmatrix}
1\\
0 \\
0 \\
\end{bmatrix}$. Now,
$N\begin{bmatrix}
1\\
0 \\
0 \\
\end{bmatrix}=\begin{bmatrix}
0\\
1 \\
-1 \\
\end{bmatrix} \in \ker N$. Now we find an element in $\ker N$ such that $\{\begin{bmatrix}
0\\
1 \\
-1 \\
\end{bmatrix}, v_{1}\}$ is a basis of $\ker N$. Let's choose $v_{1}=\begin{bmatrix}
-2\\
1 \\
1 \\
\end{bmatrix}$. So our Jordan basis is $$\left\{\begin{bmatrix}
1\\
0 \\
0 \\
\end{bmatrix},
\begin{bmatrix}
0\\
1 \\
-1 \\
\end{bmatrix},
\begin{bmatrix}
-2\\
1 \\
1 \\
\end{bmatrix}\right\}$$
Problem: For $S=\begin{bmatrix}
1 & 0 & -2\\
0 & 1 & 1\\
0 & -1 & 1\\
\end{bmatrix}$
and $J=S^{-1}MS$
, I only get a Jordan Matrix $J$ with ones below the diagonal, I am aware that this is also the Jordan Form, nonetheless in our lectures we state that $J$ has ones above the diagonal. For peace of mind, I would like to know I am not getting ones above the diagonal but rather below the diagonal?
| You must construct the Jordan basis more carefully.
You already determined that the Jordan form consists of one $2 \times 2$ block and one $1 \times 1$ block, i.e. $\dim \ker M = 2$ and $\dim \ker M^2 = 3$.
Write a table like this $$
\begin{array}{ccc}
\ker M^2 \,\dot-\,\ker M^1 & f_2\\
\ker M^1 \,\dot-\,\ker M^0 & f_1 & f_3\\
\end{array}
$$
Pick a vector $f_2 \in \ker M^2 \setminus \ker M^1$, such as $f_2 = \pmatrix{1 \\ 0 \\ 0}$ and define $f_1 = Mf_2 = \pmatrix{0 \\ 1 \\ -1}$.
Pick a vector $f_3 \in \ker M$ linearly independent with $f_1$, such as $f_3 = \pmatrix{-1 \\ 1 \\ 0}$.
Now $$\{f_1, f_2, f_3\} = \left\{\pmatrix{0 \\ 1 \\ -1}, \pmatrix{1 \\ 0 \\ 0}, \pmatrix{-1 \\ 1 \\ 0}\right\}$$ is a Jordan basis for $M$, with the Jordan form above the diagonal. Indeed: $$Mf_1 = 0$$ $$Mf_2 = f_1$$ $$Mf_3 = 0$$
Put those vectors in a matrix $S$ and we have $$J = S^{-1}MS = \pmatrix{0 & 1 & -1 \\ 1 & 0 & 1 \\ -1 & 0 & 0}^{-1}\pmatrix{0 & 0 & 0 \\ 1 & 1 & 1 \\ -1 & -1 & -1}\pmatrix{0 & 1 & -1 \\ 1 & 0 & 1 \\ -1 & 0 & 0} = \pmatrix{0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0}$$
| {
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Expected number of correct guess in a game Question is as follow.
There is a bag. In the bag, there are $a$ red cubes and $b$ blue cubes. Assume that she knows exactly how many cubes for each of the colors before the draw. Mary is going to draw all the cubes one by one out of the bag randomly. For every turn, she will make a guess on the color before drawing the cube. Find the expected number of correct guess in the game.
Trial:
I can solve for a simpler case, that is when she doesn't know the number of cubes of each color. But if she does, then I can figure out her strategy, that is she is going to guess the color which is greater in number in the bag. Also, I tried the find the probability of getting one correct only but cant gets a success because it seems to depend on the previous results.
| We find the following recurrence:
$$E(a,b) = \begin{cases}
\frac{bE(a,b-1) + a(E(a-1,b)+1)}{a+b}&\text{if } a\ge b,\\
E(b,a)&\text{if }b>a,\\
0&\text{if }a=b=0.
\end{cases}$$
These cases are pretty obvious: the first says "we gain one point if we guess right", the second says "we can pick the cube that is more common and it doesn't matter which that is", and the third says "you can't score any points on an empty bag"
From here we can begin to find easy ones: It's clear that $E(a,0) = a$, for instance. But it gets harder from there - each entry depends on the ones before it in ugly ways. I present here the values for $a,b \le 5$.
$$\begin{array}{c|cccccc}
\ &0&1&2&3&4&5\\
\hline
0 & 0 & 1 & 2&3&4&5\\
1&1&1\frac{1}{2}&2\frac{1}{3}&3\frac{1}{4}&4\frac{1}{5}&5\frac{1}{6}\\
2&2&2\frac{1}{3}&2\frac{5}{6}&3\frac{3}{5}&4\frac{7}{15}&5\frac{8}{21}\\
3&3&3\frac{1}{4}&3\frac{3}{5}&4\frac{1}{10}&4\frac{29}{35}&5\frac{37}{56}\\
4&4&4\frac{1}{5}&4\frac{7}{15}&4\frac{29}{35}&5\frac{23}{70}&6\frac{2}{63}\\
5&5&5\frac{1}{6}&5\frac{8}{21}&5\frac{37}{56}&6\frac{2}{63}&6\frac{67}{126}
\end{array}$$
From this a few other closed forms for special cases are apparent:
*
*for $(1,n)$, it's $n+\frac{1}{n+1}$
*for $(2,n)$ with $n>0$, it's $n+\frac{2(n+3)}{(n+1)(n+2)}$
There are sure to be others but I don't know them.
| {
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Problem related to quadratic equation If $x$ is a real number then what is the greatest value of
$$ f(x)=2(a-x)\left(x+\sqrt{x^2+b^2}\right)\;\;? $$
I cannot figure out how to convert this into a quadratic equation to find out the greatest value of $f(x).$
From what I've read, the answer is given by $a^2+b^2$.
| let $\sqrt{x^2+b^2}+x=y,$ then $\displaystyle \sqrt{x^2+b^2}-x = \frac{b^2}{y}$ and $\displaystyle 2x=y-\frac{b^2}{y}$
$\displaystyle f(x) = (2a-2x)(x+\sqrt{b^2+y^2}) = (2a-y-\frac{b^2}{y})\cdot y = (2ay-y^2+b^2) $
$f(y)= (a^2+b^2)-(y-b)^2\leq (a^2+b^2)$ equality hold when $y=b$
| {
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Proof that squares are divisible by 3 when their sum is In this proof, they write $3|a^2+b^2 \implies 3|a$, $3|b$. I tried using the same proof used to prove $3|a^2 \implies 3|a$, where $3$ being prime and writing $a^2 = a\cdot a$ suggests that $a$ is divisible by $3$. I'm not sure how to prove the $3|a^2+b^2$ case, though.
E9. There is no quadruple of positive integers $(x, y, z, u)$ satisfying $$x^2 + y^2 = 3(z^2 + u^2).$$
Solution. Suppose there is such a quadruple. We choose the solution with the smallest $x^2 + y^2$. Let $(a, b, c, d)$ be the chosen solution. Then $$a^2 + b^2 = 3(c^2 + d^2) \implies 3|a^2 + b^2 \implies 3|a, 3|b \implies a = 3a_1, b = 3b_1,\\a^2 + b^2 = 9(a^2_1 + b^2_1) = 3(c^2 + d^2) \implies c^2 + d^2 = 3(a^2_1 + b^2_1).$$
We have found a new solution $(c, d, a_1, b_1)$ with $c^2 + d^2 \lt a^2 + b^2$. Contradiction.
We have used the fact that $3|a^2 + b^2 \implies 3|a, 3|b$. Show this yourself. We will return to similar examples when treating infinite descent.
| In $\mathbb{Z}/3$, $a^2=1$, or $a^2=0$, $b^2=1$ or $0$ implies that $a^2+b^2=0$ if and only if $a^2=b^2=0$.
| {
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Grid walk with restrictions on some points
Given a $10\times10$ grid (each axis enumeration is $0,1,2,..,9$) a grid walk is started at point $(0,0)$ and finished at point $(9,9)$. Each move can be either to the right or up. How many paths are possible if points $(3,3), (6,3), (6,6)$ must not be crossed?
I know that the total number of possible paths is ${9+9\choose 9} = 48620$.
Inclusions/exclusion principle can be used. $(6,3)$ can be visited in:
$$
{6+3\choose 3}{9+9-(6+3)\choose 9-3}=7056
$$
Similar calculations can be made for the other two points. Where I'm having trouble is calculating crossing of $2$ and $3$ points simultaneously. How do I do that?
| If a path passes through $(3,3)$ and then $(6,3)$, it must travel only horizontally from $(3,3)$ to $(6,3)$. This gives us
$$\binom{3+3}{3}\cdot 1\cdot \binom{9+9-6-3}{9-3}$$
total paths, where the first binomial coefficient is the number of walks from $(0,0)$ to $(3,3)$, the $1$ is the number of walks from $(3,3)$ to $(6,3)$, and the second binomial coefficient is the number of walks from $(6,3)$ to $(9,9)$.
If a path passes through $(3,3)$ and then $(6,6)$, we will have
$$\binom{3+3}{3}\cdot \binom{6+6-3-3}{6-3} \cdot \binom{9+9-6-6}{9-6}$$
Where the first coefficient counts paths from $(0,0)$ to $(3,3)$, the second counts paths from $(3,3)$ to $(6,6)$, and the third counts paths from $(6,6)$ to $(9,9)$.
Using these two examples, you can calculate the number of paths passing through $(6,3)$ and $(6,6)$.
Now we must count the number of paths going through all three. First, the walk must pass through $(3,3)$, which can be done in this many ways:
$$\binom{3+3}{3}$$
Then it must pass through $(6,3)$, which can only be done $1$ way (by going horizontally). Then it must go through $(6,6)$, which can be done only $1$ way (by going vertically). Then, from $(6,6)$ to $(9,9)$, we have
$$\binom{9+9-6-6}{9-6}$$
Which gives us a total of
$$\binom{3+3}{3}\cdot 1\cdot 1\cdot \binom{9+9-6-6}{9-6}$$
Now, after calculating these binomial coefficients, you can do your inclusion-exclusion!
| {
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Proving $\pi = 48\tan^{-1}\frac{1}{18} + 32 \tan^{-1}\frac{1}{57} - 20\tan^{-1}\frac{1}{239}$ The below equation represents $\pi$ to some decimals using tangent inverse. I need to prove that the left hand side of the equation equals the right hand side.
$$
\pi = 48\tan^{-1}\frac{1}{18} + 32 \tan^{-1}\frac{1}{57} - 20\tan^{-1}\frac{1}{239}
$$
| To prove given formula, we'll use identity (see here)
$$
\arctan\frac{a_1}{b_1} + \arctan\frac{a_2}{b_2} = \arctan\frac{a_1 b_2+a_2b_1}{b_1b_2-a_1a_2}.
\tag{1}
$$
Denote $$\arctan\frac{1}{b} = f(b).\tag{2}$$
Then given formula will have form
$$
f(1) = 12 f(18) + 8 f(57) - 5 f(239).\tag{3}
$$
The proof will be based on $2$-term-RHS identities which can be easily checked via $(1)$.
a): since $f(239)=4f(5)-f(1)$ (see Machin's Formula), after substituting $f(239)$ to $(3)$ we get equivalent formula to prove:
$$
f(1) = 5f(5)-3f(18)-2f(57).\tag{4}
$$
b): since $f(18)=f(5)-f(7)$, we get new identity to prove:
$$
f(1)=2f(5)+3f(7)-2f(57).\tag{5}
$$
c): since $f(57)=f(7)-f(8)$, we came to new identity to prove:
$$
f(1) = 2f(5)+f(7)+2f(8).\tag{6}
$$
d):
since $f(8)=f(3)-f(5)$, we came to $2$-term-RHS formula:
$$
f(1)=2f(3)+f(7),\tag{7}
$$
e):
and since $f(7)=f(2)-f(3)$, we came to another well-known Machin-like (Euler's) formula:
$$
f(1)=f(2)+f(3).\tag{8}
$$
So, given formula can be obtained from $(8)$: each step replaces argument to more appropriate two arguments (and replace $5$ at the final step). The sketch is:
$(2),3 \longrightarrow \color{red}{3},\color{red}{7}$
$(3),7 \longrightarrow \color{red}{5},7,\color{red}{8}$
$5,7,(8) \longrightarrow 5,\color{red}{7},\color{red}{57}$
$5,(7),57 \longrightarrow \color{red}{5},\color{red}{18},57$
$(5),18,57 \longrightarrow \color{red}{1},18,57,\color{red}{239}$.
| {
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Use the Gauss elimination method in order to find $u$ in $\mathbb{R}^2$ such that $T(u)=\left(\begin{matrix} 3 & 1& 4 \end{matrix}\right)^T$ Consider the linear map $T:\mathbb{R}^2->\mathbb{R}^3$ with
$$T\left(\begin{matrix} x \\ y \end{matrix}\right)=\left(\begin{matrix} x-2y \\ 2x+y \\ -5x+8y \end{matrix}\right)$$ Use the Gauss elimination method in order to find $u$ in $\mathbb{R}^2$ such that $T(u)=\left(\begin{matrix} 3 \\ 1 \\ 4 \end{matrix}\right)$
My try:
$$\left(\begin{matrix} 1 & -2 & 3 \\ 2 & 1 & 1 \\ -5 & 8 & 4 \end{matrix}\right)_{R_2->R_2-2R_1}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 5 & -5 \\ -5 & 8 & 4 \end{matrix}\right)_{R_3->R_3+5R_1}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 5 & -5 \\ 0 & -2 & 19 \end{matrix}\right)_{R_2->\frac{R_2}{5}}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & -2 & 19 \end{matrix}\right)_{R_3->R_3+2R_2}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 17 \end{matrix}\right)_{R_3->\frac{R_3}{17}}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}\right)_{R_2->R_2+R_3}$$
$$=\left(\begin{matrix} 1 & -2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right)_{R_1->R_1-3R_3}$$
$$=\left(\begin{matrix} 1 & -2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right)_{R_1->R_1+2R_2}$$
$$=\left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right)$$
Then how to find $u$?
| You just what to eliminate in the first two columns.
You are fine through here.
$\begin{array}{cc|c} 1 & -2 & 3 \\ 0 & 1 & -1 \\ 0 & -2 & 19 \end{array}$
But then then you hit the problem. The vector you have been asked to find is not in the column space of the matrix. Your problem has no solution.
| {
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How to express $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ into $\frac{1}{2}\sqrt{5+3\sqrt{3}+2\sqrt{12+7\sqrt{3}}}$ I am just wondering how to express $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ into $\frac{1}{2}\sqrt{5+3\sqrt{3}+2\sqrt{12+7\sqrt{3}}}$ (This one is taken from Wolfram BTW)
Clearly, $\frac{\frac{1}{2}}{\sqrt[4]{\frac{3}4}-\sqrt[4]{\frac{1}{4}}}$ can be rewritten as $\frac{1}{2(\sqrt[4]{\frac{3}{4}}-\sqrt[4]{\frac{1}{4}})}$, but what should I do next?
I am a little inexperienced in manipulating surds, so some helps are very much appreciated
Thank you!
| We need to prove that
$$(\sqrt[4]3-1)^2(5+3\sqrt3+2\sqrt[4]3(2+\sqrt3))=2$$ or
$$2(5+3\sqrt3+2\sqrt[4]3(2+\sqrt3))=(1+\sqrt[4]3+\sqrt3+\sqrt[4]{27})^2,$$
which is obvious after full expanding.
| {
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Express the following complex numbers in the form of $a+bi$ I'm struggling with the following exercises:
I tried to use the reasoning as follows:
$$(a+bi)^n=(re^{\theta i})^n=r^ne^{\theta in}=r^n(\cos(\theta n)+i\sin(\theta n))$$
So for the first one I did:
$$2^{1/6}(\cos(-\frac{\pi}{3} \cdot \frac{1}{6})+i\sin(-\frac{\pi}{3} \cdot \frac{1}{6}))$$
But it gives me a decimal solution, that's why I'm not sure about the solution.
And for the second one, I was attempting to do the same but when I was calculating $r$ of $(1+\sqrt{-3}i)^{50}$, that is, the modulus of the complex number:
$$r=\sqrt{1^2+(\sqrt{-3})^2}=\sqrt{1-3}$$
which doesn't exist.
Any idea?
Thank you.
| We can calculate the norm of $z$ in this way:
$|z|^6=|z^6|=|1-\sqrt{3}i|=2$ so you have that
$|z|= 2^\frac{1}{6}$
Now you can define $y:= \frac{z}{2^\frac{1}{6}}$ and then
*
*$|y|=1$
*$y^6=\frac{1}{2}-\frac{\sqrt{3}}{2}i$
You can write $y$ in trigonometric form:
$y=cos(\theta)+i sin(\theta)$
and so
$\frac{1}{2}-\frac{\sqrt{3}}{2}i =y^6=cos(6\theta)+i sin(6\theta)$
Then you must impose the conditions:
$cos(6\theta)=\frac{1}{2}$
$sin(6\theta)=-\frac{\sqrt{3}}{2}$
and the solution is
$\theta=\frac{-\pi}{36}+\frac{k\pi}{3}$
for $k\in\{-1,-2,0,1,2,3\}$
for example for $k=0$ you have that
$y=cos(\frac{\pi}{36})-i sin(\frac{\pi}{36})$
and so
$z=2^{\frac{1}{6}}(cos(\frac{\pi}{36})-i sin(\frac{\pi}{36}))$
For the second
$(1+\sqrt{-3}i)^{50}=(1-\sqrt{3})^{50}=c$
that is a real number
$(1+i)^{100}=2^{50}(cos(\frac{100\pi}{4})+ i sin(\frac{100\pi}{4}))=$
$
2^{50}(-1+ i 0)=-2^{50}$
So the number is real and is
$-\frac{1}{c}2^{50}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Solutions of $x^2 -xy + y^2 \equiv 0 \pmod{43}$
$x^2 -xy + y^2 \equiv 0 \pmod{43}$
We could transform this into
$\frac{x^3 + y^3}{x+y} \equiv 0 \pmod{43}$
and thus
$x^3 + y^3 \equiv 0 \pmod{43}$
From what I could manage so far, I know that if $x = 3k$ , $y= 3k+1$, as $43 = 3k+1$.
Also there are no solutions for $43$ itself, $84$ and $127$.
How should I proceed?
| We have $x \equiv -y$ iff $x \equiv 0$ iff $y \equiv 0$.
Otherwise, as you have argued, $x^2 - xy + y^2 \equiv 0$ iff $x^3 + y^3 \equiv 0$ and this happens iff $(xy^{-1})^3 \equiv -1$.
So the solutions are $x=yu$, where $u$ is an element of order $6$ mod $43$, that is, $u=7$ or $u=37$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find solution to this ODE with powerseries I am working on solving this ODE
$$ (1+x^2)y'' + xy'-y =0 $$
with $1. y(0)=0,y'(0)=1 \\
2. y(0)=1, y'(0)=1$
with $$ y= \sum_{n=0}^{ \infty} a_n x^n , y'= \sum_{n=1}^{ \infty}na_n x^{n-1}, y''=\sum_{n=2}^{ \infty} n(n-1)a_n x^{n-2}$$
plugged into the equation follows:
$(1+x^2) \sum_{n=2}^{ \infty} n(n-1)a_n x^{n-2}+x \sum_{n=1}^{ \infty}na_n x^{n-1}-\sum_{n=0}^{ \infty} a_n x^n =0 $
$ \leftrightarrow \sum_{n=2}^{ \infty} n(n-1)a_n x^{n}+\sum_{n=2}^{ \infty} n(n-1)a_n x^{n-2}+ \sum_{n=1}^{ \infty} na_n x^{n}-\sum_{n=0}^{ \infty} a_n x^n =0$
$ \leftrightarrow 2a_2-a_0 + \sum_{n=1}^{ \infty} (n(n-1)a_n + (n+2)(n+1)a_{n+2}+ na_n - a_n) x^n = 0 $
you get that
$ a_{n+2}= \frac{a_n(1-n^2)}{(n+2)(n+1)} $
let now be $ a_0=1 $, $a_1=0 $ so that $a_2= \frac12, $( $2a_2 - a_0 = 0 \leftrightarrow a_2= \frac12 )$
for uneven $n$ such as $a_3, a_5,..$ it equals zero.
but for even $n$ i can not find any regularity:
$ n=2 , a_4 = \frac{a_2 (-3)}{12}= -\frac{1}{8}$
$ n=4 , a_6= \frac{a_4 (-15)}{30}= \frac{1}{16} $
$ n=6 , a_8= \frac{a_6 (-35)}{56}= - \frac{5}{128} $
i am totally stuck and do not know how to proceed :(
I appreciate any help of you guys !
| For (1), you should get $a_n = 0$ for all $n \ge 2$, i.e. $y(x) = x$.
For (2), the solution turns out to be $y(x) = x + \sqrt{1+x^2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$xy\frac{dy}{dx} = \sqrt{ x^2- y^2 - x^2y^2 -1}$ Solve the differential equation -
$$xy\frac{dy}{dx} = \sqrt{ x^2- y^2 - x^2y^2 -1}$$
I tried various methods such as putting $y=vx$, squarring both sides and then manipulating the expression but couldnt solve it. Any hint will be helpful. Thanks
| $$xy\frac{dy}{dx} = \sqrt{ x^2- y^2 - x^2y^2 -1}$$
Let $X=x^2$ and $Y=y^2$
$$X\frac{dY}{dX} = \sqrt{ X- Y - XY -1}$$
Let $u(X)=\sqrt{ X- Y - XY -1}$
$$Y=\frac{X-1-u^2}{X+1}$$
$$X\;Y' =X\left(\frac{1}{X+1}-\frac{2u\,u'}{X+1}-\frac{X}{(X+1)^2}+\frac{1}{(X+1)^2} +\frac{u^2}{(X+1)^2}\right)=u$$
$$u\,u'+\frac{X+1}{2X}u-\frac{u^2}{2(X+1)}=\frac{1}{X+1}$$
Let $v(X)=\frac{1}{u(X)}$
$$v'=-\frac{1}{X+1}v^3+\frac{X+1}{2X}v^2-\frac{1}{2(X+1)}v$$
This is an Abel's ODE of the first kind.
Many of them are not analytically solvable in terms of standard elementary and/or special functions. This is probably the case. For deeper investigation, see : https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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integral $\int_{0}^{1}\left( \left\lfloor{\frac{2}{x}} \right\rfloor-2 \left\lfloor{\frac{1}{x}} \right\rfloor \right)dx$ Evaluate $$\int_{0}^{1}\left( \left\lfloor{\frac{2}{x}} \right\rfloor-2 \left\lfloor{\frac{1}{x}} \right\rfloor \right)dx$$
My Attempt
$$I_{1}=\int_{0}^{1}\left\lfloor{\frac{2}{x}} \right\rfloor dx$$
Put $x=2t$
$$I_{1}=\int_{0}^{\frac{1}{2}}2\left\lfloor{\frac{1}{t}} \right\rfloor dt=\int_{0}^{\frac{1}{2}}2\left\lfloor{\frac{1}{x}}\right\rfloor dx$$
let $$I_{2}=\int_{0}^{1} \left\lfloor{\frac{1}{x}} \right\rfloor dx=\int_{0}^{\frac{1}{2}}\left\lfloor{\frac{1}{x}}\right\rfloor dx+\int_{\frac{1}{2}}^{1}\left\lfloor{\frac{1}{x}}\right\rfloor dx$$
Given integral
$$
\begin{align}
\int_{0}^{1}\left(\left\lfloor{\frac{2}{x}}\right\rfloor-2\left\lfloor{\frac{1}{x}}\right\rfloor \right)dx&=I_{1}-2I_{2}\\
&=2\int_{0}^{\frac{1}{2}}\left\lfloor{\frac{1}{x}}\right\rfloor dx-2\int_{0}^{\frac{1}{2}}\left\lfloor{\frac{1}{x}}\right\rfloor dx-2\int_{\frac{1}{2}}^{1}\left\lfloor{\frac{1}{x}}\right\rfloor dx\\
&=-2\int_{\frac{1}{2}}^{1}\left\lfloor{\frac{1}{x}}\right\rfloor dx\\
&=(-2)(1)=-2.
\end{align}
$$
But answer given is $\ln(\frac{4}{e})$
What mistake am I making?
| $$\begin{align*}\int_0^1 \left( \left\lfloor \dfrac{2}{x} \right\rfloor - 2\left\lfloor \dfrac{1}{x} \right\rfloor \right)dx & = \sum_{n=1}^\infty \left[\int_{\tfrac{1}{n+1}}^{\tfrac{2}{2n+1}}\left( \left\lfloor \dfrac{2}{x} \right\rfloor - 2\left\lfloor \dfrac{1}{x} \right\rfloor \right)dx + \int_{\tfrac{2}{2n+1}}^{\tfrac{1}{n}}\left( \left\lfloor \dfrac{2}{x} \right\rfloor - 2\left\lfloor \dfrac{1}{x} \right\rfloor \right)dx \right] \\ & = \sum_{n=1}^\infty \left[ \left( \dfrac{2}{2n+1}-\dfrac{1}{n+1} \right)(2n+1-2n) + \left(\dfrac{1}{n}-\dfrac{2}{2n+1} \right)(2n-2n) \right] \\ & = \sum_{n=1}^\infty \left(\dfrac{2}{2n+1}- \dfrac{1}{n+1} \right) = \ln\left( \dfrac{4}{e} \right)\end{align*}$$
The mistake you made was assuming the integral could be evaluated directly over intervals of length 1/2 (because neither $I_1$ nor $I_2$ is finite). Instead, you could use an infinite number of intervals (each time that $\left\lfloor \dfrac{2}{x} \right\rfloor > 2\left\lfloor \dfrac{1}{x} \right\rfloor$, which occurs when $\dfrac{1}{n+1} < x < \dfrac{2}{2n+1}$ for any positive integer $n$ because on that interval, $2n+1=\left\lfloor \dfrac{2}{x} \right\rfloor < 2n+2$ and $2n=2\left\lfloor \dfrac{1}{x}\right\rfloor < 2n+2$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find all $z$ for which $z^2+2z+2$ is real positive. Find all $z$ for which $z^2+2z+2$ is real positive.
My Attempt :
Let $z=x+iy, (x,y\in R)$:
$$z^2+2z+2= (x^2-y^2+2ixy) +2(x+iy)+2$$
$$=(x^2-y^2+2x+2)+i(2xy+2y)$$
| You want the real part to be positive and the imaginary part to be $0$
$$ (x^2-y^2+2x+2)+i(2xy+2y)$$
$$2xy+2y=0 \implies 2y(x+1)=0$$
Thus $y=0$ or $x=-1$
For $y=0$ we get $x^2+2x+2=(x+1)^2+1 >0$
For $ x=-1 $ we need $1-y^2>0$ which implies $|y|<1$
Thus the solution is either $ y=0$ or ($|y|<1$ and $x=-1$ )
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 5
} |
If $\sin x +\sin 2x + \sin 3x = \sin y\:$ and $\:\cos x + \cos 2x + \cos 3x =\cos y$, If $\sin x +\sin 2x + \sin 3x = \sin y\:$ and $\:\cos x + \cos 2x + \cos 3x =\cos y$, then $x$ is equal to
(a) $y$
(b) $y/2$
(c) $2y$
(d) $y/6$
I expanded the first equation to reach $2\sin x(2+\cos x-2\sin x)= \sin y$, but I doubt it leads me anywhere. A little hint would be appreciated. Thanks!
| If $z = e^{ix} = \cos x + i \sin x$
Then we have $z + z^2 + z^3 = e^{iy} = w$ (say)
Divide by $z^2$ to see that
$$z + \frac{1}{z} + 1 = \frac{w}{z^2}$$
The left side is real and thus
$$w = az^2$$
Since $|w| = |z| = 1$ we must have that $|a| = 1$
Thus $$w = \pm z^2$$
This gives rise to two equations:
$$z + z^3 = 0$$
and
$$z + 2z^2 + z^3 = 0$$
I will leave the rest to you.
And as others said, be careful that $e^{ix}$ is periodic and for the question to make sense you might need to put bounds on $x,y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
$\lim_{n \rightarrow \infty} ( 1 - \frac{2}{2.3}) ( 1 - \frac{2}{3.4}).......(1-\frac{2}{(n+1).(n+2)})$ Evaluate
$\lim_{n \rightarrow \infty} \left( 1 - \frac{2}{2\cdot3}\right) \left( 1 - \frac{2}{3\cdot4}\right)\ldots\left( 1 - \frac{2}{(n+1)(n+2)}\right)$
My attempts : i know that $1 - \frac {2}{k(k+1)} = \frac{(k+2)(k-1)}{k(k+1)}$
After that I'm not able to proceed further.
Any hints/solution will be apprecaited.
Thanks you and thanks in advance.
| Hint:
\begin{eqnarray*}
\frac{ 1\times 4}{ 3\times 2}\times \frac{ 2\times 5}{ 4\times 3}\times \frac{ 3 \times 6}{ 5 \times 4}\times \cdots
\end{eqnarray*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Are $1+p^3+p^6$ and $1+p^4+p^8$ coprime? What are the primes $p$ for which $1+p^3+p^6$ and $1+p^4+p^8$ are coprime?
I know it is true for $p=2$ and $p=3$ and not true for any $p \equiv 1 \mod 6$. I conjecture that it true for all primes $p \equiv 5 \mod 6$.
Any counterexample $> 10^8$.
This is relevant to OEIS sequence A046685.
| Let $p$ be an integer.
Suppose $\gcd(1+p^3+p^6,1+p^4+p^8) = u > 1$.
\begin{align*}
\text{Then}\;\;&1+p^3+p^6\equiv 0\;(\text{mod}\;u)\\[4pt]
\implies\;&(p^3-1)(p^6+p^3+1)\equiv 0\;(\text{mod}\;u)\\[4pt]
\implies\;&p^9-1\equiv 0\;(\text{mod}\;u)\\[4pt]
\implies\;&p^9\equiv 1\;(\text{mod}\;u)\\[10pt]
\text{Similarly}\;\;&1+p^4+p^8\equiv 0\;(\text{mod}\;u)\\[4pt]
\implies\;&(p^4-1)(p^8+p^4+1)\equiv 0\;(\text{mod}\;u)\\[4pt]
\implies\;&p^{12}-1\equiv 0\;(\text{mod}\;u)\\[4pt]
\implies\;&p^{12}\equiv 1\;(\text{mod}\;u)\\[10pt]
\text{Then}\;\;&
\begin{cases}
p^{12}\equiv 1\;(\text{mod}\;u)\\[4pt]
p^9\equiv 1\;(\text{mod}\;u)\\
\end{cases}\\[4pt]
\implies\;&p^3\equiv 1\;(\text{mod}\;u)\\[4pt]
\implies\;&p^6\equiv 1\;(\text{mod}\;u)\\[4pt]
\implies\;&1+p^3+p^6\equiv 3\;(\text{mod}\;u)\\[4pt]
\implies\;&0\equiv 3\;(\text{mod}\;u)\\[4pt]
\implies\;&u=3\\[4pt]
\implies\;&p^3\equiv 1\;(\text{mod}\;3)\\[4pt]
\implies\;&p\equiv 1\;(\text{mod}\;3)\\[4pt]
\end{align*}
It follows that $1+p^3+p^6$ and $1+p^4+p^8$ are relatively prime unless $p\equiv 1\;(\text{mod}\;3)$, in which case, their $\gcd$ is $3$.
For the case where $p$ is prime, $p\equiv 1\;(\text{mod}\;3)$ is equivalent to $p\equiv 1\;(\text{mod}\;6)$, hence $1+p^3+p^6$ and $1+p^4+p^8$ are relatively prime unless $p\equiv 1\;(\text{mod}\;6)$, in which case, their $\gcd$ is $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2846047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 1
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How do you evaluate limit of $\frac{\sqrt{a+x} - \sqrt a}{x\sqrt a(\sqrt{a+x})}$ when $x$ tends to $0$? I tried rationalization method where we multiply both the numerator and denominator with appropriate opposite factor of numerator. But I could only get $\frac{1}{\sqrt {a^2+ax}(2 \sqrt {a + \sqrt x)}}$. But the final solution in textbook says it should be $\frac{1}{2a^{3/2}}$. Please help.
| Easier solution: Write $y=\sqrt{x+a}$ and $b=\sqrt{a}$, then
\begin{align}\lim_{x \to 0}\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a}\sqrt{a+x}} &= \lim_{y \to b}\frac{y-b}{by(y^2-b^2)}\\&=\lim_{y \to b}\frac{1}{by(y+b)}\\
&= \frac{1}{2b^3}=\frac{1}{2a\sqrt{a}}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum of all numbers greater than $10000$ formed by using all digits $0,1,2,4,5$ Find the sum of all numbers greater than $10,000$ formed by using all digits $0,1,2,4,5$ and no digit being repeated in any number.
I could find that the number of such numbers are $96$ but I could not find their sum.
| Let's keep placeholders for where we will keep the digits :
[] [] [] [] []
Now, in the first box i.e: ten thousands place, we can put any number in the following set $ \{1,2,4,5 \}$ and after that have 4! ways of filling in the remaining numbers. So, there are 4! numbers with 1 as first number, 4! numbers with 2 as first number.. till 5. So, the sum of digits in ten thousand place is given as:
$$ 4!(1+2+4+5) \cdot 10^4$$
Similarly, suppose we fix some digit in the thousand place, then we have $ 3 \cdot 3!$ way of arranging the rest of numbers (remember we can't keep zero in first place). Hence the sum of no.s in thousand place is given as:
$$ 3 \cdot 3! ( 1 + 2 + 4 +5) \cdot 10^3$$
And, similar results for hundreds , tens and ones place. Now, to get the whole number, we just need to add up the parts of it's expanded form. Hence,
$$ \text{number} = 4!(1+2+4+5) \cdot 10^4 + 3 \cdot 3! (1+2+4+5) \cdot 10^3 + 3 \cdot 3! (1+2+4+5) \cdot 10^2 + 3 \cdot 3! (1+2+4+5) \cdot 10^1 + 3 \cdot 3! (1+2+4+5) \cdot 10^0 = 3119976$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Range of $a$ in Trigonometric equation
If the inequality $\sin^2 x+a\cos x+a^2>1+\cos x$ hold for any $x\in \mathbb{R}.$ Then range of $a$ is
Try: $1-\cos^2 x+a\cos x+a^2-1-\cos x>0$
$$\cos^2 x+(1-a)\cos x-a^2<0$$
$$4\cos^2 x+4(1-a)\cos x-4a^2<0$$
$$\bigg(2\cos x+(1-a)\bigg)^2-4a^2-(1-a)^2<0$$
Could some help me how to solve further, Thanks
| Given task allows suitable algebraic approach, because it is equivalent to the system of
\begin{cases}
\cos x =t\\
t\in[-1,1]\\
t^2+(1-a)t-a^2<0,\tag1
\end{cases}
wherein the parabola $P(t,a) = t^2+(1-a)t-a^2$ has the positive discriminant
$$D=(1-a)^2+4a^2.\tag2$$
The roots of $P(t,a)$ are
$$t_1 = \dfrac{a-1 - \sqrt D}2,\quad t_2 = \dfrac{a-1 + \sqrt D}2\tag3$$
wherein
$$P(t,a) = (t-t_1)(t-t_2)\tag4$$
is negative iff
$$t\in(t_1,t_2).\tag5$$
Condition $(5)$ is satisfied for all $t\in[-1,1]$ iff
$\mathbf{[-1,1] \subset (t_1,t_2)},$
or $t_1<-1$ and $t_2>1.$
This leads to the system in the form of
\begin{cases}
a-1-\sqrt{(a-1)^2+4a^2} < -2\\[4pt]
a-1 +\sqrt{(a-1)^2+4a^2} > 2,\tag6
\end{cases}
$$\sqrt{(a-1)^2+4a^2} > \max(a+1,\,3-a).\tag7$$
Since
$$\max(a+1,3-a) =
\begin{cases}
3-a,\quad\text{if}\quad a\in(-\infty,1)\\[4pt]
a+1,\quad\text{if}\quad a\in[1,\infty)\tag8
\end{cases}$$
is positive, then the inequality $(7)$ can be presented in the form of
$$\genfrac{[}{.}{0}{0}
{a\in(-\infty,1),\quad \sqrt{5a^2-2a+1} > 3-a}
{a\in[1,+\infty),\quad \sqrt{5a^2-2a+1} > a+1.}\tag9$$
Then
$$\genfrac{[}{.}{0}{0}
{a\in(-\infty,1),\quad 5a^2-2a+1 > a^2-6a+9}
{a\in[1,+\infty),\quad 5a^2-2a+1 > a^2+2a+1,}$$
$$\genfrac{[}{.}{0}{0}
{a\in(-\infty,1),\quad a^2+a-2 > 0}
{a\in[1,+\infty),\qquad\,\, a^2-a > 0,}$$
$$\genfrac{[}{.}{0}{0}
{a\in(-\infty,1),\quad (a+2)(a-1) > 0}
{a\in[1,+\infty),\qquad\,\, a(a-1) > 0\quad,}\tag{10}$$
$$\color{green}{\mathbf{a\in (-\infty,-2) \cup (1,+\infty).}}\tag{11}$$
Solution of $(7)$ can be illustrated, using Wolfram Alpha.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the probability that exactly one box remains empty Five identical(indistinguishable) balls are to be randomly distributed into $4$ distinct boxes. What is the probability that exactly one box remains empty?
we can choose the empty box in $4$ ways.
We should fit the $5$ balls now into $3$ distinct boxes in $^7C_5$ ways.
The correct answer must be: $\frac37$ (mcq)
Can you tell me how should I proceed then?
| I couldn't see how the previous answer came up with a solution so I did it differently.
$0$ boxes empty has $1,1,1,2$ box contents times $\frac{5!}{2!}$ ball arrangements $= \frac{4!}{3!}\cdot \frac{5!}{2!} = 240$
$1$ box empty has $1,1,3$ and $1,2,2$ box contents and $\frac{5!}{3!}$ and $\frac{5!}{2!\cdot 2!}$ ball arrangements $= (\frac{3!}{2!}\cdot \frac{5!}{3!}+\frac{3!}{2!}\cdot \frac{5!}{2!\cdot 2!})\cdot ^4C_1 = 600$
$2$ boxes empty has $1,4$ and $2,3$ box contents and $\frac{5!}{4!}$ and $\frac{5!}{3!\cdot 2!}$ ball arrangements $=(2!\cdot \frac{5!}{4!}+2!\cdot \frac{5!}{3!\cdot 2!})\cdot ^4C_2 = 180$
$3$ boxes empty has all $5$ in any one of $4$ boxes $= 4$
$P(1 \text{empty}) = \frac{600}{240+600+180+4} = \frac{600}{1024} = .58594$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $f(n)=\sum_{k=1}^n \frac n {n^2+k^2}$ is strictly increasing I want to show that the function $f:\mathbb N \to \mathbb Q$ defined by$$f(n)=\sum_{k=1}^n \frac n {n^2+k^2}$$
is strictly increasing. Could I have a hint on how to go about this? I have tried playing around with $f(n+1)-f(n)$ but it's all so messy with the different denominators that I'm not sure what to do.
| $$
\begin{align}
&\sum_{k=1}^{n+1}\frac{n+1}{k^2+(n+1)^2}-\sum_{k=1}^n\frac{n}{k^2+n^2}\tag1\\
&=\frac1{2n+2}+\sum_{k=1}^n\left(\frac{n+1}{k^2+(n+1)^2}-\frac{n}{k^2+n^2}\right)\tag2\\
&=\frac1{2n+2}-\sum_{k=1}^n\frac{n(n+1)-k^2}{\left(k^2+(n+1)^2\right)\left(k^2+n^2\right)}\tag3\\
&=\frac1{2n+2}-\frac1{n(n+1)}\sum_{k=1}^n\frac{1-\frac{k^2}{n(n+1)}}{\left(1+\frac{k^2}{(n+1)^2}\right)\left(1+\frac{k^2}{n^2}\right)}\tag4\\
&\gt\frac1{2n+2}-\frac1{n(n+1)}\sum_{k=1}^n\frac{1-\frac{k(k-1)}{n^2}}{\left(1+\frac{k^2}{n^2}\right)\left(1+\frac{(k-1)^2}{n^2}\right)}\tag5\\
&=\frac1{2n+2}-\frac1{n+1}\sum_{k=1}^n\left[\frac{\frac{k}{n}}{1+\frac{k^2}{n^2}}-\frac{\frac{k-1}{n}}{1+\frac{(k-1)^2}{n^2}}\right]\tag6\\[9pt]
&=0\tag7
\end{align}
$$
Explanation:
$(2)$: separate the $k=n+1$ term from the left sum and combine the rest
$(3)$: algebra
$(4)$: algebra
$(5)$: $\frac{k}{n+1}\gt\frac{k-1}{n}$ and $\frac{1-xy}{\left(1+x^2\right)\left(1+y^2\right)}$ is strictly decreasing in $x$ when $x,y\in[0,1]$
$(6)$: algebra
$(7)$: telescoping series sums to $\frac12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Spherical coordinates in surface integrals I am stuck on the following problem
Evaluate :
$$I=\iint x^2 y^2 z dS $$
where S is the positive side of lower half of the sphere $x^2 + y^2 + z^2 = a^2$
I tried using spherical coordinates and their jacobians but cannot seem to find the answer which is $$I= \frac {2\pi a^7} {105}$$
| Substitute $z=-\sqrt{a^2-x^2-y^2}$ and $dS=\sqrt{1+z_x^2+z_y^2}dA$. which turn $$I=-a\iint_{R} x^2y^2dA.$$ Use polar co-ordinate.$x=r\cos\theta, y=r\sin \theta$. $dxdy=rdrd\theta$. $r$ varies from $0$ to $a$ and $\theta$ varies from $0$ to $2\pi$. Which turn $I=-a\int_{0}^{2\pi}\sin^2\theta \cos^2\theta d\theta \int_{0}^a r^5 dr=\frac{-a^7}{6}\int_{0}^{2\pi}\sin^2\theta \cos^2\theta d\theta =\frac{-a^7}{24}\int_{0}^{2\pi}\sin^2 2\theta d\theta=\frac{-a^7}{48}\int_{0}^{2\pi}({1-\cos4\theta}) d\theta=\frac{-a^7\pi}{24}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2853473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int\frac{(1+x) \lfloor(1-x+x^2)(1+x+x^2)+x^2\rfloor}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$ If $$F(x) = \int\frac{(1+x) \lfloor(1-x+x^2)(1+x+x^2)+x^2\rfloor}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$$then find the value of $\lfloor F(99)- F(3)\rfloor$
My attempt : I could only reduce the portion of the numerator within $\lfloor\rfloor$ the floor function.
$$F(x) = \int\frac{(1+x)\lfloor1+2x^2 + x^4\rfloor}{1+2x+3x^2+4x^3+3x^4+2x^5+x^6}dx$$
I could not understand how to reduce further. Any help will be gratefully acknowledged.
| After some factorisation, what we're left to figure out is:
$$ I= \left \lfloor \int_3^{99} \frac{\lfloor (1+x^2)^2 \rfloor}{(1+x)(1+x^2)^2} dx \right \rfloor $$
But first, for $g(x) = (1+x^2)^2$, let's study $ \lfloor g(x) \rfloor$ and
$ \frac{\lfloor g(x) \rfloor}{g(x)}$.
Notice that for some positive integer $n$, $ \frac{\lfloor g(x) \rfloor}{g(x)}$ has a jump discontinuity at every $ x=\sqrt{\sqrt{n+1} -1}$ and that $ g(x)-1 \leq \lfloor g(x) \rfloor \leq g(x) $.
Consequently, for the interval $ \sqrt{\sqrt{n} -1} \leq x < \sqrt{\sqrt{n+1} -1}$, we have
$$ \frac{n}{n+1} < \frac{\lfloor (1+x^2)^2 \rfloor}{(1+x^2)^2} \leq 1 $$
So, since $ \frac{n}{n+1}$ increases as $n$ increases, $ \frac{99}{100} \leq \frac{\lfloor (1+x^2)^2 \rfloor}{(1+x^2)^2} \leq 1$ for all $ x \geq 3$.
Therefore,
$$ \int_3^{99} \frac{99}{100(1+x)} dx < I < \int_3^{99} \frac{1}{1+x} dx $$
$$ \frac{99}{100} \ln(25) < I < \ln(25) $$
$$ \implies I \approx 3.2 $$
$$ \implies \lfloor I \rfloor = 3 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $2+2\cos(2\pi\theta) \leq 4\exp(-2\|\theta \|^2)$. Let $\theta \in \mathbb{R}$, and let $\|\theta\|$ denote the distance to the integer nearest to $\theta$. I want to prove $$2+2\cos(2\pi\theta) \leq 4\exp(-2\|\theta \|^2).$$ From the double angle formula we arrive at the equivalent $$|\cos(\pi \theta) | \leq e^{-\|\theta\|^2},$$ and since both sides are 1-periodic, we need only check that $$0 \leq e^{-\theta^2} - \cos(\pi \theta) \qquad \text{for $\theta \in \left[0,\frac{1}{2}\right]$},$$ and $$0 \leq e^{-(1-\theta)^2} + \cos(\pi \theta) \qquad \text{for $\theta \in \left[\frac{1}{2},1\right]$}.$$ It seems like these aren't as easy as "take the derivative a few times." Any suggestions?
| In fact, because both sides of $|\cos\pi\theta|\le e^{-\Vert\theta\Vert^2}$ are symmetric around $\frac12$ ($f(\theta)=f(1-\theta)$), we only need to check $0\le e^{-\theta^2}-\cos\pi\theta$ for $\theta\in[0,1/2]$. This can be rewritten as
$$e^{-\theta^2}\ge\cos\pi\theta\tag1$$
The idea of the proof of $(1)$ is to find two polynomials $p(x),q(x)$ with $\cos\pi x\le q(x)\le p(x)\le e^{-x^2}$ with $x\in[0,1/2]$ and then prove $q(x)\le p(x)$ over the same range of $x$. These polynomials are truncations of the Maclaurin series of the two sides of $(1)$, to $x^2$ for the LHS and to $x^4$ for the RHS, so we have
$$e^{-x^2}\ge 1-\frac{x^2}2\ge1-\frac{(\pi x)^2}2+\frac{(\pi x)^4}{24}\ge\cos\pi x$$
$$-\frac{x^2}2\ge-\frac{(\pi x)^2}2+\frac{(\pi x)^4}{24}$$
$$-\frac12\ge-\frac{\pi^2}2+\frac{\pi^4x^2}{24}$$
$$\frac{\pi^2-1}2\cdot\frac{24}{\pi^4}\ge x^2$$
$$x\le\sqrt{\frac{\pi^2-1}2\cdot\frac{24}{\pi^4}}=1.045\dots$$
Since this contains $[0,1/2]$, $q(x)\le p(x)$ over the range in question and $(1)$ is proved.
| {
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"url": "https://math.stackexchange.com/questions/2856213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determine all possible integers $x$ and $y$ s.t. $3x + 7y \equiv 14 \pmod{28}$ and $x + 3y \equiv 8 \pmod{28}$. Determine all possible ints $x$ and $y$ that satisfy $3x + 7y \equiv 14 \pmod {28}$ and $x + 3y \equiv 8 \pmod {28}$. The answer should be in terms of $x\equiv r$ and $y \equiv s$ where $r$ and $s$ are remainders.
Here is what I have tried. Am I headed in the right direction?
$8\cdot (3x + 7y) \equiv 14 \pmod {28}$ and $14 \cdot (x + 3y) \equiv 8 \pmod {28}$
$24x + 56y \equiv 112 \pmod{ 28}$ and $14x + 42y \equiv 112 \pmod {28}$
$24x + 56y \equiv 14x + 42y \pmod {28}$
This is where I start to get unsure,
$24x + 28y + 28y \equiv 14x + 14y + 28y \pmod {28}$
$24x \equiv 14x + 14y \pmod {28}$
$48x \equiv 28x + 28y \pmod {28}$
So then...?
$20x \equiv 1 \pmod {28}$?
Stuck here. Is this even the right idea?
| We have
\begin{eqnarray*}
3x+7y \equiv 14 \pmod{28} \\
x+3y \equiv 8 \pmod{28}.
\end{eqnarray*}
Multiply the second equation by $3$ and then subtract the first
\begin{eqnarray*}
3x+9y \equiv 24 \pmod{28} \\
2y \equiv 10 \pmod{28}.
\end{eqnarray*}
So
\begin{eqnarray*}
y &\equiv 5 \pmod{28} \\
x & \equiv 21 \pmod{28}
\end{eqnarray*}
or\begin{eqnarray*}
y &\equiv 19 \pmod{28} \\
x & \equiv 7 \pmod{28}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2857002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Roots of $f$ and $f'$ for $1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$ I have this question from an admission exams. Given that $$f(x)=1+\sum_{k=0}^{100}\frac{(-1)^{k+1}}{(k+1)!}x(x-1)(x-2)\cdots (x-k)$$ find $S(f(x))-S(f'(x))$ where $S$ denotes the sum of the real roots for $f(x)$ respectively $f'(x)$
Here is how I tried, since for $x=1,2,\ldots,100, 101$ the sum part vanishes little by little and
$$f(1)=1-1=0$$
$$f(2)=1-2+\frac{1}{2!}2(2-1)=0$$
$$f(3)=1-3+\frac{1}{2!}3(3-1)-\frac{1}{3!}3(3-2)(3-1)=\binom{3}{0}-\binom{3}{1}+\binom{3}{2}-\binom{3}{3}$$ and we can see a pattern in the above equation so that we can rewrite $f(3)$ as $(1-1)^3=0\,$ and indeed that $f(101)=(1-1)^{101}=0$
Since the polynomial is also of order $101$, the roots are $x=1,2, \ldots, 101$ giving:$$S(f(x))=\sum_{j=1}^{101}j =5151$$ Is this correct? And how can $S(f'(x))$ be evaluated ? This doesnt seem so obvious.. Also now we can rewrite $$f(x)=-\frac{1}{101!}(x-1)(x-2)\cdots(x-101)$$
| We have
$$f(x) = a_{101}x^{101} + a_{100}x^{100} + \cdots$$
and we know that $a_{101} = -\frac1{101!}$. Vieta's formulas give
$$5151 = \text{ sum of roots of } f = -\frac{a_{100}}{a_{101}}$$
so $a_{100} = \frac{5151}{101!}$.
The derivative is
$$f'(x) = 101a_{101}x^{100} + 100a_{100}x^{99} + \cdots$$
so Vieta's formulas give
$$\text{ sum of roots of } f' = -\frac{100a_{100}}{101a_{101}} = -\frac{100\cdot \frac{5151}{101!}}{101\cdot \frac{-1}{101!}} = \frac{515100}{101}= 5100$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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