Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find all values of a for which the following system has a solution, no solution and infinitely many solution. Find all values of a for which the following system
$$\begin{cases}x+2y+z=a^2
\\x+y+3z=a
\\3x+4y+8z=8\end{cases}$$
has a solution, no solution and infinitely many solution.
I found the reduced row echelon form of this system which is:
$\begin{pmatrix}
1 & 0 & 0 & 4a^2+12a-40\\
0 & 1 & 0 & -a^2-5a+16\\
0 & 0 & 1 & -a^2-2a+8
\end{pmatrix}$
Does that mean this system has only one solution and there is no value for a which makes the system infinitely many solutions and no solution?
| Yes since the RREF is the following
$$\begin{pmatrix}
1 & 2 & 1 &a^2\\
1 & 1 & 3 & a\\
3 & 4 & 8 & 8
\end{pmatrix} \to \begin{pmatrix}
1 & 2 & 1 &a^2\\
0 & 1 & -2 & a^2-a\\
0 & 0 & 1 & 8-2a-a^2
\end{pmatrix}$$
by Rouché–Capelli theorem, we always have exacty one soution for the system.
| {
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"timestamp": "2023-03-29T00:00:00",
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Computation of 10th power of a matrix I have been given the following matrix $A$. I am required to find 10th power of $A$.
$$
A = \begin{pmatrix}
3 & 1 & 1 \\
2 & 4 & 2 \\
-1 & -1 & 1 \\
\end{pmatrix}
$$
I tried using Cayley Hamilton Theorem and found the characteristic polynomial as:
$t^3-8t^2+20t-16=0$.From here I find myself unable to compute the said power. Long division is daunting to me in this case. Please suggest how to go about it. The eigenvalues come out to be $2$( double) and $4$.
| The eigenvectors for this matrix come out to be
$$\left( \begin{array}{c} -1\\1\\0\end{array} \right),\left( \begin{array}{c} -1\\0\\1\end{array} \right), \left( \begin{array}{c} -1\\-2\\1\end{array} \right)$$
To diagonalize the matrix, if $A = SDS^{-1}$, then
$$S = \left( \begin{array}{c c c} -1&-1&-1\\1&0&-2\\0&1&1\end{array} \right), A = \left( \begin{array}{c} 2&0&0\\0&2&0\\0&0&4\end{array} \right)$$
Then $A^{10} = SD^{10}S^{-1}$, and finding 10th power of a diagonal matrix is trivial.
Can you take it from here?
| {
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Let f(x) = $x^2+ax+b,a,b \in R$. If $f(1)+f(2)+f(3)=0$, then the nature of the roots of the equation $f(x) =0$ is ..... Let $f(x) = x^2+ax+b$ for $a,b \in \mathbb{R}$. If $f(1)+f(2)+f(3)=0$, then the nature of the roots of the equation $f(x) =0$ is
(A) real
(B) imaginary
(C) real and distinct
(D) equal roots
My attempts:
\begin{align}
f(1) &= 1+a+b \\
f(2) &= 4+2a+b \\
f(3) &= 9+3a+b \\
f(1)+f(2)+f(3)
&= 1+a+b+4+2a+b+9+3a+b \\
0 &= 14+6a+3b
\end{align}
now how can we take it further about the nature of the roots , whether the roots of $f(x)=0$ is imaginary or real , please help, thanks...
| We have three real terms summing up to $0$.
They can't be all zero as a quadratic has at most two zeroes.
Hence at least one term is positive and at least one term is negative, hence the roots must be distinct real roots.
| {
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If $a+b+c=3$, then show $\sqrt{a^2 + ab +b^2}+ \sqrt{b^2 + bc +c^2}+\sqrt{c^2 + ac +a^2} \geq \sqrt{3}$ If $a+b+c=3$, and $a,b,c$ are positive real numbers, then show $\sqrt{a^2 + ab +b^2} + \sqrt{b^2 + bc +c^2} +\sqrt{c^2 + ac +a^2}\geq \sqrt{3}$
Normally when I do inequalities I try to first find where equality would be achieved, but in this case I have no idea where to start to find it.
From $a^2 - 2ab +b^2 \geq 0$ we can obviously get $a^2 -ab +b^2\geq 3ab$, and after substituting that, we can change the inequality to showing that $\sqrt{ab} + \sqrt{bc} +\sqrt{ac} \geq 1$, but clearly if a=b=c, then equality is not achieved. I wanted to try to use Cauchy-Schwarz or something of the sort, but those inequalities generally are used to find the upper bound of sums of roots.
| Use QM-AM to obtain \begin{align*}\sum_{cyc}\sqrt{a^2+ab+b^2}\geqslant \sum_{cyc}\frac{a+b+\sqrt{ab}}{\sqrt{3}}&=\frac{2(a+b+c)}{\sqrt{3}}+\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{\sqrt{3}}\\&=2\sqrt{3}+\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{\sqrt{3}}\\&>\sqrt{3}\end{align*}
| {
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Given the two functions $f(x)=x^2+3x+5$ and $g(x)=\sqrt{2-|x-4|}$ work out $g^{-1}$ (Sweden 1961) Given the two functions $f(x)=x^2+3x+5$ and $g(x)=\sqrt{2-|x-4|}$ work out $g^{-1}$ for some subdomain.
I have managed to prove that $g^{-1}$ exists in the following way:
$g(x)=\sqrt{6-x}$ with $x\in[4,6]$ we have that g is 1-1 in $[4,6]$ and hence $g^{-1}$ exists.
This is where I got stuck and couldn't continue. Could you please explain how to work out $g^{-1}$ and explain every step of your thought process as well as intuitively, how you though of each step?
| Why did you restrict to the domain $[4,6]$? It is not the complete domain of $g$.
the domain is all $x$ where $|x-4| \le 2$ so $0\le x-4 \le 2$ or $-2 \le x-4 \le 0$ so $4\le x \le 6$ or $2\le x \le 4$. So the domain is $[2,6]$
And we can easily see that $g$ is not one to one because for any $x \in (4,6]$ where $g(x) = \sqrt {2-|x-4|} =K$ we can have an $x' = 8-x\in [2,4)$ where $g(x') = \sqrt{2 -|(8-x)-4|}=\sqrt{2-|4-x|}=\sqrt{2-|x-4|}=K$. For example $g(5) =\sqrt{2-|5-4|} = \sqrt{2-|3-4|} = g(3)$ and $g(4.1)=\sqrt{2-|4.1-4|} = \sqrt{2-|4-3.9|}=g(3.9)$ etc.
So no inverse will exist.
To find the inverse (if it exists)
Let $y=g^{-1}(x)$ so $x = g(g^{-1}(x)) = g(y) = \sqrt{2-|y-4|}$ so we solve for $y$.
$x = \sqrt{2-|y-4|}$ so
$x^2 = 2-|y-4|$
If $y \ge 4$ then $x^2 = 2-(y-4)=6-y$ and $y= 6-x^2$.
If $y < 4$ then $x^2 = 2-(4-y)=y-2$ and $y=x^2+2$
But that is not a well defined function. If $6-x^2 > 4 > x^2 + 2$, that is to say if $x^2 < 2$ then we have no we to pick which value $y=g^{-1}(x)$ should be.
so the inverse does not exist and the function is not one to one.
| {
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Use proof by induction to show that for any positive integer $n\geq 2$ the following holds Use proof by induction to show that for any positive integer $n\geq 2$ the following holds:
$(1+\frac{1}{3})(1+\frac{1}{5})...(1+\frac{1}{2n-1})>\frac{\sqrt{2n+1}}{3}$
Proof: Base case: $n=2$. $LHS:\frac{4}{3}$; $RHS:\frac{\sqrt{5}}{3}$. Clearly, $\frac{4}{3}>\frac{\sqrt{5}}{3}$.
Next, we assume that for some positive integer $n\geq2$ the above statement holds.
Here is where I get consuded. Do I add the next term i.e $(1+\frac{1}{2n})$ to both sides? Or do I multiply the right hand side by that term to continue?
Any tips on how to tackle induction proofs like these? Thanks!
| Assuming that $(1+\frac{1}{3})(1+\frac{1}{5})...(1+\frac{1}{2n-1})>\frac{\sqrt{2n+1}}{3}$ consider $(1+\frac{1}{3})(1+\frac{1}{5})...(1+\frac{1}{2n-1})(1+\frac1 {2n+1})$. This quantity is greater than $\frac{\sqrt{2n+1}}{3} (1+\frac1 {2n+1})$. We want to show that $\frac{\sqrt{2n+1}}{3} (1+\frac1 {2n+1}) >\frac{\sqrt{2n+3}}{3}$. For this use the fact that $\sqrt{2n+1} (1+\frac1 {2n+1})=\sqrt {2n+1} +\frac 1 {\sqrt {2n+1}}$. Puting $x=2n+1$ you will now require the inequality $\sqrt x +\frac 1 {\sqrt x } >\sqrt {x+2}$ or $x+1 >\sqrt {x(x+2)}$. But $x(x+2)=x^{2}+2x < x^{2}+2x+1=(x+1)^{2}$ and you can finish the proof by taking square root on both sides.
| {
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Logaritmic equation with floor brackets $ \left \lfloor \frac{x-1}{2} -\left \lfloor \frac{x}{2} \right \rfloor \right \rfloor=\log_{10}x$ Solve equation
$ \left \lfloor \frac{x-1}{2} -\left \lfloor \frac{x}{2} \right \rfloor \right \rfloor= \log_{10}x$
I dont understand solution explained in the book.
$ \left \lfloor \frac{x-1}{2} -\left \lfloor \frac{x}{2} \right \rfloor \right \rfloor= \left \lfloor \frac{x}{2}- \left \lfloor \frac{x}{2} \right \rfloor -\frac{1}{2} \right \rfloor=\left \lfloor \left \{ \frac{x}{2} \right \}-\frac{1}{2} \right \rfloor$
I dont understand why is
$\left \lfloor \frac{x}{2}- \left \lfloor \frac{x}{2} \right \rfloor -\frac{1}{2} \right \rfloor=\left \lfloor \left \{ \frac{x}{2} \right \}-\frac{1}{2} \right \rfloor$
| $\{x\}$, the fractional part of x, is defined as $\{x\} = x - \lfloor x\rfloor$. (For example, $\{3.14\} = 0.14$, $\{-0.6\} = 0.4$, $\{1\} = 0$, etc. Thus, $\boxed{\frac{x}{2}-\left\lfloor\frac{x}{2}\right\rfloor = \bigg\{\frac{x}{2}\bigg\}.}$
| {
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Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
Let $x,y,z$ are the lengths of sides of a triangle such that $x+y+z=2$. Find the range of $xy+yz+xz-xyz$ .
What I Tried: I have tried by doing $(x+y+z)(x+y+z)=2*2 = 4.$
Also I got $2(xy+yz+xz)+(x^2+y^2+z^2)=4$.
But I think it is a wrong method to proceed...
| Let $F$ denote the function $xy + yz + xz - xyz$. Since $x + y + z = 2$, the area of the triangle by Heron's formula is
$$A = \sqrt{(1-x)(1-y)(1-z)} = \sqrt{F - 1}.$$
Hence the minimum value of $F$ is $1$, which is attained iff we allow triangles of zero area.
The maximum $F$ occurs for maximum $A$. If we're allowed to assume that the triangle with maximum area for a given perimeter is equilateral, the maximum $F$ occurs when $x = y= z= 2/3$, which gives $F = 28/27$ as others have found. If not, we can prove it as follows. Fix $x$ and write
$$F = yz(1 - x) + x(2 - x).$$
By the triangle inequality $x \le 1$, so $F$ is a maximum when $yz$ is a maximum. Since $y + z$ is fixed, the maximum $F$ for this $x$ occurs when $y = z$. Similarly for a fixed $y$ and a fixed $z$. Hence the maximum $F$ occurs for an equilateral triangle.
| {
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How to find $AL$ in a triangle with one side trisected? Here is the original diagram of the question:
Let $D$ and $E$ be the trisection points of $BC$. And let $K$ and $L$ be points on $AB$ and $AC$ such that $\angle BKE= \angle DLC = \alpha$. If $KB=16$, $AK=12$ and $LC=6$. Then find the length of $AL$.
The only thing I could think of that could use the fact that $\angle BKE$ and $\angle DLC$ are equal is the sine law. Also for convenience let $BD=DE=EC=y$.
$$\dfrac{2y}{\sin \alpha}=\dfrac{16}{\sin (\alpha + \angle B)}=\dfrac{KE}{\sin\angle B}$$
$$\dfrac{2y}{\sin \alpha}=\dfrac{6}{\sin{(\alpha + \angle C)}}=\dfrac{LD}{\sin \angle C}$$
Since I got $\sin \angle C$ and $\sin \angle B$ I thought of using sine rule again in the original triangle.
$$\dfrac{x+6}{\sin \angle B}=\dfrac{28}{\sin \angle C}$$
$$\begin{align*}
\dfrac{\sin (\alpha + \angle B)}{\sin ( \alpha + \angle C)} & =\dfrac{16}{6} \\
\implies \dfrac{\sin (\alpha + \angle B)+\sin ( \alpha + \angle C)}{\sin (\alpha + \angle B)-\sin ( \alpha + \angle C)} & = \dfrac{11}{5} \\
\implies 2 \cdot \tan \biggl(\alpha + \dfrac{B+C}{2} \biggr) \cot \biggl( \dfrac{B-C}{2} \biggr) & = \dfrac{11}{5}
\end{align*}$$
But I could not really get anywhere with these equations. Since the terms got very messy and I really could not simplify anything.
Would anyone please provide an elementary solution? Or perhaps give a hint on how to complete my approach?
| Extends $LD$ to meet $AB$ at $F$, extends $KE$ meets $AC$ at $G$.
Let $L'$ be on $FL$ such that $BL'\mathbin{\!/\mkern-5mu/\!} AL$. Similarly $CK'\mathbin{\!/\mkern-5mu/\!}AK$.
Then $BL'=\frac 12CL=3$, $CK'=\frac 12BK=8$.
Now since triangles $CK'G$ and $AKG$ are similar, we have $AG=AC\cdot \frac{AK}{AK-CK'}=3(x+6)$. Similarly $AF=28\frac x{x-3}$.
Now since triangles $AFL$ and $AGK$ are similar, we have $\frac {AF}{AL}=\frac {AG}{AK}$, then:
\begin{align}
\frac {28x}{x(x-3)}&=\frac {3(x+6)}{12}\\\\
4\cdot 28&=(x-3)(x+6)\\\\
112&=x^2+3x-18\\\\
0&=(x-10)(x+13)\\
\end{align}
Giving us $x=10$.
| {
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$\sqrt{1-x^2}$ is not differentiable at $x = 1$. Please give me an easier proof if exists.
Prove that $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
My proof is here:
Let $0 < h \leq 2$.
$\frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = \frac{\sqrt{2h-h^2}}{-h} = \frac{2h-h^2}{-h\sqrt{2h-h^2}} = \frac{h-2}{\sqrt{2h-h^2}}$.
So, $\lim_{h\to 0+} \frac{\sqrt{1-(1-h)^2}-\sqrt{1-1^2}}{-h} = -\infty$.
So, $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
I want an easier proof.
Is there any general theorem or general proposition to prove the above fact?
By the way, the following proposition is not true. $f(x) = x^{\frac{1}{3}}$ is not differentiable at $x = 0$ and $g(x) = x^3$ is differentiable at $x = 0$ but $f(g(x)) = x$ is differentiable at $x=0$.
If $f(x)$ is not differentiable at $x = a$ and $g(x)$ is differentiable at $x = b$ and $g(b) = a$, then $f(g(x))$ is not differentiable at $x = b$.
So, we cannot prove as follows:
$\sqrt{x}$ is not differentiable at $x = 0$. And $1-x^2$ is differentiable at $x = 1$. So, $\sqrt{1-x^2}$ is not differentiable at $x = 1$.
| $$\lim_{x\to1}\frac{\sqrt{1-x^2}-0}{x-1}=\lim_{x\to1}\sqrt{\frac{1+x}{1-x}}.$$
You can conclude.
| {
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Proof: For all real numbers $x$, If $x^2−5x+4≥0$, then either $x≤1$ or $x≥4$. Is this a valid technique? For all real numbers $x$, If $x^2−5x+4≥0$, then either $x≤1$ or $x≥4$. Is this a valid technique?
Instead of showing that either (x-4) and (x-1) are both positive or both negative, can I just do what I have written below instead?
$x^2−5x+4=(x-4)(x-1)≥0$
There are two possible situations: $x≥4$ or $x<4$.
if $x<4$ then $(x-4)<0$ and so we get $(x-1)≤ 0$ which implies that $x≤1$.
so either $x≤1$ or $x≥4$.
| Yes, it is valid. Well done!
Here I propose an alternative way to solve it:
\begin{align*}
x^{2} - 5x + 4 = \left(x^{2} - 5x + \frac{25}{4}\right) - \frac{25}{4} + 4 = \left(x - \frac{5}{2}\right)^{2} - \frac{9}{4} \geq 0
\end{align*}
which is equivalent to
\begin{align*}
\left|x - \frac{5}{2}\right|^{2} \geq \frac{9}{4} \Longleftrightarrow\left(x - \frac{5}{2} \geq \frac{3}{2}\right)\vee\left(x - \frac{5}{2} \leq -\frac{3}{2}\right)\Longleftrightarrow (x \geq 4)\vee(x\leq 1)
\end{align*}
Hopefully this helps.
| {
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How to solve $\cos x-\sin 3x=\cos 2x$? My attempt:
$$\begin{align} (\cos x- \cos 2x) - \sin 3x &= 0 \\
2\sin \frac{3x}{2}\sin \frac{x}{2} - 2 \sin \frac{3x}{2}\cos \frac{3x}{2}&=0\\
\sin\frac{3x} {2} \left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)&=0 \end{align}$$
Now either $\sin\frac{3x} {2} = 0$ or $\left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)=0$. Solving for the former,
$$\frac{3x}{2}=n\pi\rightarrow x = \frac{2n\pi}{3}$$
How can I solve $\left(\sin\frac{x} {2} - \cos\frac{3x} {2}\right)=0$? I know the elementary identities involving trigonometric ratios, but not complex numbers or calculus.
| This is an elementary equation $\cos a=\cos b$ under cover: just consider that $\sin a=\cos(\pi/2-a)$. Thus you get
$$
\cos\Bigr(\frac{\pi}{2}-\frac{x}{2}\Bigr)=\cos\frac{3x}{2}
$$
and you get the two families of solutions
$$
\frac{3x}{2}=\frac{\pi}{2}-\frac{x}{2}+2n\pi
\qquad\text{or}\qquad
\frac{3x}{2}=-\frac{\pi}{2}+\frac{x}{2}+2n\pi
$$
| {
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Prove $3\left(9-5\sqrt{3}\right) \sum \frac{1}{a} \geqslant \sum a^2+\frac32\cdot\frac{\left[(\sqrt3-2)(ab+bc+ca)+abc\right]^2}{abc}$ Let $a,\,b,\,c$ are positive real numbers satisfy $a+b+c=3.$ Prove that
$$3\left(9-5\sqrt{3}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant a^2+b^2+c^2 + \frac32 \cdot \frac{\left[(\sqrt3-2)(ab+bc+ca)+abc\right]^2}{abc}. \quad (1)$$
Note. From $(1)$ we get
$$3\left(9-5\sqrt{3}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant a^2+b^2+c^2.$$
It's was posted here.
My solution is write it as SOS
$$\sum \frac{\left[(9-4\sqrt3)c+ab\right](2c+\sqrt3-3)^2(a-b)^2}{24abc} \geqslant 0.$$
Any comments and solutions are welcome and appreciated
| $uvw$ kills this problem!
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that:
$$\frac{9(9-5\sqrt3)v^2}{w^3}\geq9u^2-6v^2+\frac{3(w^3-3(2-\sqrt3)v^2)^2}{2w^3}$$ or $f(w^3)\geq0,$ where
$$f(w^3)=6(9-5\sqrt3)u^4v^2-2(3u^2-2v^2)uw^3-3(w^3-3(2-\sqrt3)uv^2)^2.$$
But $$f''(w^3)=-6<0,$$ which says that $f$ is a concave function.
We know that the concave function gets a minimal value for an extreme value of $w^3$,
which by $uvw$ happens in the following cases.
*
*$w^3\rightarrow0^+$.
In this case our inequality is obviously true.
*Two variables are equal.
Let $b=a$ and $c=3-2a$, where $0<a<1.5.$
Thus, we obtain an inequality of one variable, which easy to check.
I got that it's enough to prove: $$(a-1)^2\left(a-\frac{3-\sqrt3}{2}\right)^2(6-2\sqrt3-a)\geq0,$$ which is obvious.
| {
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Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?
Let $n$ be an integer. If the tens digit of $n^2$ is 7, what is the units digit of $n^2$?
So $n^2 \equiv 7 \pmod{100}$? If this is the case then this can be written as $n^2 = 100k +7$, where $k \in \Bbb Z.$
Here one can see that no matter what the choice of $k$, the units digit will be $7$. Thus $n^2 \equiv 7 \pmod{10}.$ However this was wrong. The correct answer is $\textbf{6}.$
What am I doing wrong here? It seems that $n^2 \equiv 7 \pmod{100}$ doesn't hold. If the tens digit is $7$ should I have that $n^2 \equiv 7k \pmod{100}$, where $k$ represents the unit digit of $70$ and not a multiplication?
| The tens digit is $7$, not the units and you want to find the unit digit.
So if the unit digit is $x$ then the number ends with $7x$ and $\pmod{100}$ what you are trying to say is $n^2 \equiv 70 + x\pmod {100}$.
The way I would do this is let $n= 10k + a$ where $a,k$ are single digits. And the hundreds place don't affect the last two digits we might as well assume $n$ has only two digits.
$n^2 = 100k^2 + 20ak + a^2 = 100m + 70 + x$.
As $7$ is odd but $2ak$ is even so $a^2$ must be two digits and we carried an odd digit. if $a = 0,1,2,3,4,5,6,7,8,9$ then $a^2 = 0,1,4,9,14,25,36,49,64,81$.. So $a = 4$ or $6$.
So if $a = 4$ and $n=10k +4$ then $n^2 = 100k^2 + 80k+ 16$ and if $a=6$ and $n=10k+6$ then $n62 = 100k^2 + 120k + 36$. In either event $x = 6$.
We can have $8k +1\equiv 7$ and $k = 2,7$ or we can have $2k+3 \equiv 7$ and $k=2,7$. Notice $24,26,74,76$ when squared all end with $76$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3880913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Volume with spherical polar coordinates Determine the volume between the surface $z=\sqrt{4-x^2-y^2}$ and the area of the xy plane determined by $x^2+y^2\le 1,\ x+y>0,\ y\ge 0$.
I convert to spherical polar coordinates.
$$K=0\le r\le 1,\ 0\le \phi \le \frac{3\pi}{4},\ 0\le \theta \le 2\pi$$
$$\iiint_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi drd\phi d\theta$$
I can't figure out how to take $\int_{K} (\sqrt {4-r^2\sin^2\phi \cos^2\theta-r^2\sin^2\phi \sin^2\theta)}r^2\sin\phi dr$, which makes me think I made a mistake somewhere.
EDIT: Thanks for all the answers.
Now I understand how the limits of $\theta ,r,z$ works.
I don't fully understand where the function "disappear".
$\sqrt {4-x^2-y^2} =\sqrt {4-r^2}$
Why isn't it then:
$\int \int \int _{K} {\sqrt {4-r^2}rdzdrd\theta }$
| Area on XY plane is bound by $x^2 + y^2 \leq 1, y \geq 0, x + y \geq 0$
This is a sector of the circle $x^2 + y^2 \leq 1$ bound between positive $X$-axis and line $y = -x$ in the second quadrant. This comes from the fact that $y \geq 0$ so part of the circle in third and fourth quadrant of $XY$ plane is not included. $x + y \geq 0$ is true for quarter of the circle in the first quadrant as both $x$ and $y$ are positive. It is also true for part of the circle in the second quadrant above line $y = -x$ as $|y| \geq |x|$.
Now you are asked to find the volume between this area on XY plane and $z = \sqrt{4-x^2-y^2}$. So it is essentially a cylinder ($\frac{3}{8}$ cross section of a cylinder of radius $1$) cut out of the sphere of radius $2$ above $XY$ plane.
So here is how it will look in cylindrical coordinates -
$\displaystyle \int_{0}^{3\pi/4} \int_{0}^{1} \int_{0}^{\sqrt{4-r^2}} r \, dz \, dr \, d\theta$
| {
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How many 4 digit number can be formed by using 1,2,3,4,5,6,7 if at least one digit is repeated My approach for this was selecting 4 numbers from which one is being repeated
$$\boxed{A}\boxed{B}\boxed{C}\boxed{A} $$
$$\boxed{A}\boxed{B}\boxed{A}\boxed{A} $$
$$\boxed{A}\boxed{A}\boxed{A}\boxed{A} $$
and permitting them which gives 7$C_4$×4!/2! +7$C_4$×4!/3!+7$C_4$×4!/4! Since 2,3,4 no. Are being repeated numbers are being repeated
What they have done is
$7^4$-7$P_4$ subtracting cases where all 4 no. Are different from total cases. I agree but what is the flaw in my method
| The quick way
$$7^4 - \left[\binom{7}{4} \times 4!\right].$$
The hard way:
$$S_1 + S_2 + S_3 + S_4.$$
$S_1 = 7 =$ # ways all 4 numbers the same.
$S_2 = 7 \times 6 \times 4 =$ # ways 3 numbers the same.
$S_3 = \binom{7}{2}\times\binom{4}{2} =$ # ways of having two pairs of numbers.
$S_4 = 7 \times \binom{6}{2} \times 4 \times 3 =$ # of ways of having 2 of one number and two other numbers.
| {
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"url": "https://math.stackexchange.com/questions/3887103",
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prove $\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$
Prove if $a,b,c$ are positive $$\sum_\text{cyc}\frac{a+2}{b+2}\le \sum_\text{cyc}\frac{a}{b}$$
My proof:After rearranging we have to prove $$\sum_\text{cyc} \frac{b}{b^2+2b} \le \sum_\text{cyc} \frac{a}{b^2+2b}$$
As inequality is cyclic:
let $a\ge b\ge c$ then $$\frac{1}{a^2+2a}\le \frac{1}{b^2+2b}\le \frac{1}{c^2+2c}$$.The rest follows by rearrangement inequality.
The case $a\ge c\ge b$ is analogous.
Thus Proved!
Is it correct?...And any other alternative ways possible?
| I apologize for the first non-obvious proof and I give you a refinement
Hint :$a\geq b \geq c$
First prove that :
$$\frac{2(x-y)}{x+y+y^2}\leq \frac{x}{y}-\frac{x+2}{y+2}\quad (1)$$
For that multiply by $y(y+2)(x+y+y^2)$ , put in factor and it becomes :
$$2(x-y)^2\geq 0$$
Apply $(1)$ for $(a,b)$,$(b,c)$,$(c,a)$
Now we need to show :
$$\frac{2(a-b)}{a+b+b^2}+\frac{2(b-c)}{b+c+c^2}+\frac{2(c-a)}{a+c+a^2}\geq 0$$
Now introducing $f(c)$
$$\frac{2(b-c)}{b+c+c^2}+\frac{2(c-a)}{a+c+a^2}=f(c)$$
Using derivatives prove that $f(c)$ is decreasing when $c$ increases .
Now we put $b=c$ and the inequality becomes :
$$\frac{2(a-b)}{a+b+b^2}+\frac{2(b-a)}{a+b+a^2}\geq 0$$
Wich is obvious with the condition $a\geq b \geq c $
| {
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Irreducible factors of $x^q-1$ over $\mathbb{Z}_p$ The polynomial $x^3-1$ over $\mathbb{Z}_p$, where $p$ is a prime, factors as $(x-1)(x^2+x+1)$.
The polynomial $x^7-1$ over $\mathbb{Z}_{13}$ factors as, $(x-1)(x^2+3x+1)(x^2+5x+1)(x^2+6x+1)$.
All these $(x-1),(x^2+x+1),(x^2+3x+1),(x^2+5x+1),(x^2+6x+1)$ are irreducible factors.
Does a polynomial $x^q-1$ over $\mathbb{Z}_p$, where $p,q$ are primes ($q<p$), always factorize to $(x-1)$ times several quadratic factors of the form $(x^2+ax+1)$, where $a \in \mathbb{Z}_p$? (a quadratic factor with constant term "1"?)
Thanks a lot in advance.
| No. For example, $x^7 - 1 = (x-1)(x^6+x^5+ x^4+x^3+x^2+x+1)$ modulo $19$ as a product of irreducible factors. It is easy to see that $x-1$ is always a factor, but from that point on I think more specific conditions may be required on $p$ and $q$ ($q<p$ seems arbitrary in this respect) for this breakup to occur.
Note : Showing explicitly that the sixth degree polynomial is irreducible requires Galois theory.
We will use the demonstration in the comments for this purpose. We wish to show that $$
x^6+x^5+ x^4+x^3+x^2+x+1
$$
is irreducible modulo $19$. Note that a root of this polynomial can be found in a finite field extension of $\mathbb F_{19}$, which will be of the form $\mathbb F_{19^m}$ for some $m$.
However, any root of this polynomial will also satisfy $x^7 = 1$, therefore we are ask : for which $m$ is there a non-unit $x$ such that $x^7 = 1$ as elements of $\mathbb F_{19^m}$?
If there is such an element $x$, then $7$ must divide $19^m-1$. Conversely, because $7$ is prime, $7$ dividing $19^m-1$ implies the converse by Cauchy's theorem. Therefore, the question now is : which $m$ are such that $7$ divides $19^m-1$?
One can easily check that the set of all such $m$ is of the form $n\mathbb N$ for $n$ being the possible smallest value of $m$. Combining Euler's theorem with a few simple tests, one sees that $m$ can be any multiple of $6$.
So, any root of $x^6+x^5+\ldots+1 = 0$ must then lie in an extension of degree at least $6$. However, because the degree of this polynomial is also at most $6$, it follows that the root lies in an extension of degree $6$, and therefore $x^6+x^5+\ldots+1$ is its minimal polynomial, hence irreducible over $\mathbb F_{19}$.
| {
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Generalize the formula :$1=1, 3+5=8, 7+9+11=27, 13+15+17+19=64$ So the solution of this is $n^3$, as $1=1^3, 3+5=2^3, 7+9+11=3^3$
So I find that the $n+1 = m^2+3m+2 + (n_m)(m+1)$ where $n_m$ is the largest number in the previous equation and $m$ is the number of terms in last equation. How can I prove by induction that the solution is $n^3$, do I have to prove $(n+1)^3$ is equal to that or any other method or I prove that $(n+1)^3-n = 2m^2+2m+2+n_m$ (the differience between $(n+1)$ and $n$
| Let us find the general term of $1,3,7,13,\cdots$
As the next consecutive differences are $2,4,6,\cdots$
$$T_n=an(n-1)+b(n-1)(n-2)+cn(n-2)$$
$n=1\implies1=c(1-2),c =-1$
$n=2\implies3=2a$
$n=3,7=6a+2b+3c=3(3)+2b+3(-1),2b=1$
$$\implies T_n=\cdots=n^2-n+1$$
We need the sum of $n$ terms from $k=n^2-n+1$ to $(n+1)^2-(n+1)+1-1$ with the common difference $=2$
which is $$n(n^2-n+1)+\sum_{k=1}^n(2k)=n^3-n^2+n+\dfrac n2(2+(n-1)2)=?$$
| {
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Which positive integers $a$ and $b$ make $(ab)^2-4(a+b) $ a square of an integer? Which positive integers
$a$ and $b$ make
$(ab)^2-4(a+b)
$
a square of an integer?
I saw this in quora,
and found that
the only solutions with
$a \ge b > 0$
are
$(a, b, (ab)^2–4(a+b)) = (5, 1, 1)$
and $(3, 2, 16)$.
Another “solution” is
$a = b = 2, (ab)^2–4(a+b) = 0$.
My solution is
messy and computational,
and I wonder if
there is a more elegant solution.
Here is my solution.
Assume $a \ge b$
and write
$n^2 = (ab)^2–4(a+b)$ so $n < ab$.
Let $n = ab-k$
where $ab > k>0$ so
$(ab)^2–4(a+b) = (ab-k)^2 = (ab)^2–2kab+k^2$
or
$k^2–2kab+4(a+b) = 0$.
Then
$\begin{array}\\
k
&= \dfrac{2ab-\sqrt{4a^2b^2–16(a+b)}}{2}
\qquad \text{(use "-" since } k < ab)\\
&= ab-\sqrt{a^2b^2–4(a+b)}\\
&=(ab-\sqrt{a^2b^2–4(a+b)})\dfrac{ab+\sqrt{a^2b^2–4(a+b)}}{ab+\sqrt{a^2b^2–4(a+b)}}\\
&=\dfrac{4(a+b)}{ab+\sqrt{a^2b^2–4(a+b)}}\\
\end{array}
$
Therefore,
since $k \ge 1, 4(a+b) \ge ab$
so
$0 \ge ab-4(a+b) = ab-4(a+b)+16–16
=(a-4)(b-4)-16$
or $16 \ge (a-4)(b-4)$.
This gives a finite number of possible $a, b$,
all at least $4$.
Computation shows that
none of these are solutions.
To get the possible values of
$a$ and $n$ in terms of $b$
for any fixed $b$,
do this:
Since
$n^2
= a^2b^2-4(a+b)$,
$\begin{array}\\
b^2n^2
&= a^2b^4-4b^2a-4b^3\\
&= a^2b^4-4b^2a+4-4b^3–4\\
&=(b^2a-2)^2–4(b^3+1)\\
\end{array}
$
so
$4(b^3+1)
= (b^2a-2)^2-b^2n^2
= (b^2a-2-bn)(b^2a+bn)
$.
For each factorization
$r*s = 4(b^3+1)$, try
$r=b^2a-2-bn, s=b^2a-2+bn$.
This gives
$s-r=2bn$,
so if $2b$ divides $s-r$,
then
$n=\dfrac{s-r}{2b}$.
Adding $s$ and $r$,
$2b^2a-4=s+r$ so if
$2b^2$ divides $s+r+4$,
then $a = \dfrac{s+r+4}{2b^2}$.
This allows us to compute all solutions for any fixed value of b.
Running this for $1 \le b \le 16$
gives the solutions above.
For $a \ge b \ge 5$,
the restriction
$16 \ge= (a-4)(b-4)$
gives a finite set of possibilities which computation shows yields no additional solutions.
I sure would like to see
a more elegant solution.
Also,
this messy algebra
provides lots of opportunities
for errors.
| If $(ab)^2-4(a+b)$ is greater than $(ab-1)^2$ then it cannot be a square, since it is strictly between two consecutive squares. Hence
$$(ab)^2-4(a+b) \le (ab-1)^2=(ab)^2-2ab+1$$
$$2ab-4a-4b-1\le0$$
$$2(a-2)(b-2)=2ab-4a-4b+8\le 9$$
which, again, gives a finite set of possibilities to be checked.
WLOG suppose $a \ge b$. We need only consider the cases:
*
*$b=1,2$
*$b=3, a\le6$
*$b\ge 4, a < 3$ (this case contradicts $a\ge b$)
For $b=1$, $(ab)^2-4(a+b) = a^2-4a-4$. For $a\ge7$, $a^2-4a-4> a^2-6a+9=(a-3)^2$. But $a^2 -4a-4 < a^2-4a+4=(a+2)^2$. So we only need to check $1 \le a \le 6$.
For $b=2$, $(ab)^2-4(a+b) = 4a^2-4a-8$, which is a square only if $a^2-a-2$ is. For $a\ge4$, $a^2-a-2> a^2-2a+1 =(a-1)^2$. But $a^2-a-2<a^2$, so we only need to check $a=2,3$.
For $b=3$, $(ab)^2-4(a+b) = 9a^2-4a-12$, and we only need to check $3 \le a \le 6$.
| {
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A unique question in combinatorics and non-commutative variables. Let $x,y$ be two variables that satisfy:
$xy=yx+1$ (they are not commutative).
Find $(xy)^2 ,(xy)^3, (yx)^2, (yx)^3$ as a linear combination in terms of $y^jx^j$.
Then find a formula for $(xy)^n$ and $(yx)^n$.
After some calculations we get:
$(xy)^2=y^2x^2+3yx+1$.
$(xy)^3=y^3x^3+6y^2x^2+7yx+1$.
$(xy)^4=y^4x^4+10y^3x^3+25y^2x^2+15yx+1$.
$(yx)^2=y^2x^2+yx$.
$(yx)^3=y^3x^3+3y^2x^2+y$
Now, given $xy=yx+1$ we can use the binomial formula:
$(xy)^n=\sum_{k=0}^{n} {n \choose k} (yx)^k$.
Denote:
$(yx)^{n}=\sum a_{n,k} y^kx^k$.
Then,
$(yx+1)^n=(yx)(yx)^n=\sum {n\choose k} yxy^kx^k$.
After computing we get,
$(yx)^{n+1}= \sum a_{n,k} (ky^kx^k+y^{k+1}x^{k+1})$.
By comparing the coefficient of $[y^kx^k]$ in $(yx)^{n+1}$:
$a_{n,k}=ka_{n-1,k}+a_{n-1,k-1}$
| The following is valid for $n\geq 0$:
\begin{align*}
\color{blue}{(xy)^n=\sum_{k=0}^n{n+1\brace k+1}y^kx^k}\tag{1}
\end{align*}
where ${n\brace k}$ are the Stirling numbers of the second kind. We show (1) by deriving the recurrence relation
\begin{align*}
&{n+1\brace k}=k{n\brace k}+{n\brace k-1}\qquad\qquad n, k\geq 1\\
&{0\brace 0}=1,\qquad\quad {n\brace 0}={0\brace n}=0\qquad \ n\geq 1\tag{2}
\end{align*}
for the Stirling numbers of the second kind. We do this in two steps and start with the relation $xy=yx+1$. We get for $k\geq 1$:
\begin{align*}
\color{blue}{xy^k}&=(xy)y^{k-1}\\
&=(yx+1)y^{k-1}=yxy^{k-1}+y^{k-1}\\
&=y(yx+1)y^{k-2}=y^2xy^{k-2}+2y^{k-1}\\
&=y^3xy^{k-3}+3y^{k-1}\\
&=\cdots\\
&\,\,\color{blue}{=y^kx+ky^{k-1}}\tag{3}
\end{align*}
Now we consider
\begin{align*}
(xy)^n=\sum_{k=0}^na_{n,k}y^kx^k\qquad\qquad n\geq 0\tag{4}
\end{align*}
and derive a recurrence relation for $a_{n,k}, n,k\geq 0$ with the help of (3).
We obtain for $n\geq 0$:
\begin{align*}
\color{blue}{\sum_{k=0}^{n+1}a_{n+1,k}y^kx^k}&=(xy)^{n+1}=xy(xy)^n\\
&=xy\sum_{k=0}^na_{n,k}y^kx^k\tag{5}\\
&=\sum_{k=0}^na_{n,k}\left(xy^{k+1}\right)x^k\\
&=\sum_{k=0}^na_{n,k}\left((k+1)y^{k+1}x^k+y^{k+1}x^{k+1}\right)\tag{6}\\
&=\sum_{k=0}^na_{n,k}(k+1)y^kx^k+\sum_{k=1}^{n+1}a_{n,k-1}y^kx^k\tag{7}\\
&\,\,\color{blue}{=\sum_{k=0}^{n+1}\left((k+1)a_{n,k}+a_{n,k-1}\right)y^kx^y}\tag{8}
\end{align*}
Comment:
*
*In (5) we use the representation (4).
*In (6) we apply (3).
*In (7) we split the sum and shift the index of the right-hand sum to also have terms $y^kx^k$.
*In (8) we collect the sums by setting $a_{n,-1}=a_{n,n+1}=0, n\geq 0$.
We obtain from (8) the following recurrence relation for $a_{n,k}, n,k\geq 0$
\begin{align*}
a_{n+1,k}&=(k+1)a_{n,k}+a_{n,k-1}\\
\\
{n+2\brace k+1}&=(k+1){n+1\brace k+1}+{n+1\brace k}
\end{align*}
and since the boundary conditions in (2) can also be shown, the claim (1) follows.
Notes:
*
*We can easily derive from (1) for $n\geq 1$ the identity
\begin{align*}
\color{blue}{(yx)^n}&=y(xy)^{n-1}x\\
&=y\left(\sum_{k=0}^{n-1}{n\brace k+1}y^kx^k\right)x\\
&=\sum_{k=0}^{n-1}{n\brace k+1}y^{k+1}x^{k+1}\\
&\,\,\color{blue}{=\sum_{k=1}^n{n\brace k}y^kx^k}
\end{align*}
*A well-known instantiation of (1) is connected with the product rule of differentiation: $D(fg)=(Df)g+f(Dg)$. Considering the differential operator $(Df)(x)=\frac{d}{dx}f(x)$ and the multiplication operator $X$ with $(Xf)(x)=xf(x)$, we have
\begin{align*}
(xD)^n=\sum_{k=1}^n{n\brace k}x^kD^x
\end{align*}
See also chapter 4 in the referenced paper below.
*A nice paper providing many informations and insights around this theme is Combinatorial models of creation-annihilation by P. Blasiak and P. Flajolet.
| {
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If $2^{2k}-x^2\bigm|2^{2k}-1$ then $x=1$ This is the $y=2^k$ case of this question.
Suppose that $k\geq1$ and $0<x<2^k$ and $2^{2k}-x^2\bigm|2^{2k}-1$. Is it necessarily the case that $x=1$?
Equivalently: Suppose that there are two positive divisors of $2^{2k}-1$ which average to $2^k$. Is it necessarily the case that these two divisors are $2^k-1$ and $2^k+1$?
| I continue from Thomas Browning's (the author of the question) answer. We desire to show that
$$nx^2-4(n-1)y^2=1$$
has no solutions. Note that any solution must satisfy $\gcd(nx,y)=1$. We can rewrite the equation as
$$(nx)^2-4n(n-1)y^2=n,$$
so if
$$x^2-4n(n-1)y^2=n$$
has no solutions with $\gcd(x,y)=1$ then we're done. I'm going to prove that using the fact that
$$\frac xy\approx \sqrt{4n(n-1)}\approx 2n$$
and then squeezing the inequalities together and proving that they're too tight to hold. This corner of number theory is called Diophantine Approximation, and I happen to know about it. Start with
$$\sqrt{4n(n-1)}=[2(n-1);\overline{1,4(n-1)}]$$
This is easier to prove backwards. Let
$$t=2(n-1)+\frac 1{1+\frac 1{t+2(n-1)}}$$
and then it's easy to find that the positive solution is $t=\sqrt{4n(n-1)}$.
Also if
$$x^2-dy^2=n$$
then
$$\frac xy=\sqrt{d+\frac n{y^2}}=\sqrt{d}\sqrt{1+\frac n{dy^2}}$$
$$\frac xy-\sqrt{d}<\frac n{2\sqrt{d}y^2}$$
In our case $n>0$ and $d=4n(n-1)$ so
$$0<\frac xy-\sqrt{4n(n-1)}<\frac 1{4y^2\sqrt{1-1/n}}$$
Now from Hardy and Wright intro to number theory page 153:
Theorem 184. If
$$\left|\frac pq -x\right|<\frac 1{2q^2}$$
then $p/q$ is a convergent.
Note that when H&W say convergent they require it to be in lowest terms. Which is true of our previous expression, so $x/y$ is a convergent of $\sqrt{4n(n-1)}$. But the residues $x^2-dy^2$ left by a convergent $\frac xy$ to the continued fraction of $\sqrt d$ are periodic with the same period as the continued fraction itself. You can verify that when $d=4n(n-1)$ the residues are $1$ and $-4(n-1)$.
\begin{align*}
[2(n-1)]&=\frac{2(n-1)}1 &(2(n-1))^2-4n(n-1)1^2&=-4(n-1)\\
[2(n-1);1]&=\frac{2n-1}1 &(2n-1)^2-4n(n-1)1^2&=1\\
[2(n-1);1,4(n-1)]&=\frac{8n^2-10n+2}{4n-3} &(8n^2-10n+2)^2-4n(n-1)(4n-3)^2&=-4(n-1)\\
[2(n-1);1,4(n-1),1]&=\frac{8n^2-8n+1}{4n-2}&(8n^2-8n+1)^2-4n(n-1)(4n-2)^2&=1\\
[2(n-1);1,4(n-1),1,4(n-1)]&=\frac{32n^3-56n^2+26n-2}{16n^2-20n+5}&(\dots)^2-4n(n-1)(\dots)^2&=-4(n-1)
\end{align*}
So $n$ can never be a residue, therefore our equation has no solution.
| {
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"url": "https://math.stackexchange.com/questions/3903856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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$\int_{-1}^1{x^4}dx$ by substitution My student asked me: While solving $\int_{-1}^1{x^4}dx$, why can't we substitute as $t=x^2$
$$x^2 =t, 2xdx = dt, x=\pm \sqrt t, \int_{-1}^{1}(x^2 \cdot x \cdot x) dx = \int_1^1{\pm t
\sqrt{t}dt} =0 \ne \frac{1}{5}$$
I think the problem comes from $x = \pm \sqrt t$, but I can't properly explain why this happens. How can I prove that this substitution is wrong?
| Since for $r>0$, $\dfrac{\partial x^r}{\partial x}=r\times x^{r-1}$ and if $r-1+r=4$ then $r=\dfrac{5}{2}$.
Thus,
\begin{align}\int_{-1}^1 x^4\,dx&\overset{x\rightarrow x^4 \text{is even function}}=2\int_0^1 x^4 dx\\
&\overset{y=x^{\frac{5}{2}}}=2\times \frac{2}{5}\int_0^1 ydy\\
&=\frac{4}{5}\int_0^1 y dy\\
&=\frac{4}{5}\left[\frac{y^2}{2}\right]_0^1\\
&=\frac{4}{5}\times \frac{1}{2}\\
&=\boxed{\frac{2}{5}}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$ Solve fourth degree inequality $x^4 - (1 + (m + 1)^2)x^2 + (m + 1)^2 \geq 0$ where $m \in \mathbb{R}$.
I tried using substitution as follows:
$$ a = x^2 $$
$$ b = (m + 1)^2$$
Using the substitution:
$$ a^2 - (1 + b)a + b \geq 0$$
Substituting in the second degree equation:
$$ \frac{-(1 + b) + \sqrt{4 + (1 + b)^2}}{2} $$
Solving the binomial coefficients $(m + 1)^4$ and $(m + 1)^2$, I got:
$$ \frac{-(1 + (m + 1)^2) + \sqrt{m^4 + 4m^3 + 8m^2 + 8m + 8}}{2} $$
From here I don't know how to proceed...
| You made a mistake. From here
$$ a^2 - (1 + b)a + b \geq 0$$
you went here
$$ a=\frac{-(1 + b) + \sqrt{4 + (1 + b)^2}}{2} $$
While the correct solution is
$$a=\frac{-(1 + b) + \sqrt{ (1 + b)^2-4b}}{2} =\frac{-(1 + b) + \sqrt{ (1 - b)^2}}{2}$$
$$a_1=1;\;a_2=b$$
$$a\le 1\lor a\ge b$$
that is
$$x^2\le 1 \lor x^2\ge (m+1)^2$$
or
$$-1\le x \le 1 \lor x\ge| m+1|$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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About the divergence of an infinite series. Why is the series $\sum \frac{1}{n}$ not convergent, even though $$
\lim_{n \to \infty}\frac{1}{n}=0?$$
| $$
1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^n}=\\
=1+\frac{1}2+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\cdots+\frac{1}{8}\right)+\cdots+\left(\frac{1}{2^{n-1}+1}+\cdots+\frac{1}{2^n}\right)
$$
But
$$
\frac{1}{3}+\frac{1}{4}\ge \frac{2}{4}=\frac{1}{2}, \\
\frac{1}{5}+\cdots+\frac{1}{8} \ge \frac{4}{8}=\frac{1}{2},
$$
and in general
$$
\frac{1}{2^{n-1}+1}+\cdots+\frac{1}{2^n}\ge \frac{2^{n-1}}{2^n}=\frac{1}{2}
$$
and hence
$$
1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2^n}\ge 1+\frac{n-1}{2}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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System of equations involving 4 variables
If $$a + b = 6$$
$$ax + by = 10$$
$$ax^2 + by^2 = 24$$
$$ax^3 + by^3 = 62$$
then $$ax^4 + by^4 = ?$$
I got $$a(x-1) + b(y-1) = 4$$ $$ax(x-1) + by(y-1) = 14$$ $$ax^2(x-1) + by^2(y-1) = 38$$ by subtracting the given equations and also got $$a(x-1)^2 + b(y-1)^2 = 10$$ $$ax^2(x-1)^2 + by^2(y-1)^2 = 24$$ by further subtracting the three equations.
I don't think I'm going anywhere with this process and so I'm not sure how to approach this. Any help is appreciated.
| Define $A_i$ as follows
\begin{eqnarray*}
a+b=A_0 \\
ax+by=A_1 \\
ax^2+by^2=A_2 \\
ax^3+by^3=A_3 \\
ax^4+by^4=A_4 .
\end{eqnarray*}
Multiply the first & third equations & subtract the square of the second equation
\begin{eqnarray*}
A_0 A_2 -A_1^2 = ab(x-y)^2.
\end{eqnarray*}
Multiply the first & fourth equations & subtract the product of the second & third equations
\begin{eqnarray*}
A_0 A_3 -A_1 A_2 = ab(x+y)(x-y)^2 \\
(x+y) = \frac{A_0 A_3 -A_1 A_2}{A_0 A_2 -A_1^2}.
\end{eqnarray*}
Now multiply the first & fifth equations & subtract the square of the third equation
\begin{eqnarray*}
A_0 A_4 - A_2^2 = ab(x^2-y^2)^2 = ab(x+y)^2(x-y)^2 = \frac{(A_0 A_3 -A_1 A_2)^2}{A_0 A_2 -A_1^2}.
\end{eqnarray*}
Now plug in your values of $A_0,A_1,A_2,A_3$ and solve for $A_4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the inequality $\sqrt[3]{1+(3+x)\sqrt{x}+3x} - \sqrt[3]{1-(3+x)\sqrt{x}+3x} > x + a$ So am trying to solve this inequality, $\sqrt[3]{1+(3+x)\sqrt{x}+3x} - \sqrt[3]{1-(3+x)\sqrt{x}+3x} > x + a$, the problem is of course for values $0<a<1$, I tried working through it multiple times but the solutions i found, which they are $\sqrt{1-\sqrt{1-a}}<x<1 $ are the wrong ones when I try wolfram alpha, can anyone show me the right way to do it please?
| You obviously need $x > 0$ and further have
$$\sqrt[3]{(1+\sqrt x)^3} - \sqrt[3]{(1-\sqrt x)^3} > x+a$$
$\iff 1-a > (\sqrt x - 1)^2 \iff \sqrt{1-a} > |\sqrt x - 1|$
For $x \in (0, 1)$, we have this as
$ \sqrt{1-a} > 1-\sqrt x \implies x > (1-\sqrt{1-a})^2 $
For $x \geqslant 1$, we have this as
$ \sqrt{1-a} > \sqrt x - 1 \implies (1+\sqrt{1-a})^2 > x $
Thus the solution set is $x \in \left((1-\sqrt{1-a})^2, (1 + \sqrt{1-a})^2\right)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove a hard inverse trigonometry mathematical induction $\arctan\left(\frac{1}{2\left(1\right)^{2}}\right)+\arctan\left(\frac{1}{2\left(2\right)^{2}}\right)+\arctan\left(\frac{1}{2\left(3\right)^{2}}\right)+...+\arctan\left(\frac{1}{2\left(n\right)^{2}}\right)=\frac{\pi}{4}-\arctan\left(\frac{1}{2n+1}\right)$
I assume you need to do something like "take tan of both sides" but that doesn't really work for me given the summation on the LHS. Any hints?
| I'll assume you were able to prove the base case $n = 1$.
The inductive hypothesis is to assume $$\arctan \dfrac{1}{2\cdot1^2}+\cdots+\arctan \dfrac{1}{2\cdot n^2} = \dfrac{\pi}{4} - \arctan \dfrac{1}{2n+1}. \quad (1)$$
Then, you need to prove $$\arctan \dfrac{1}{2\cdot1^2}+\cdots+\arctan \dfrac{1}{2\cdot n^2} +\arctan \dfrac{1}{2\cdot (n+1)^2} = \dfrac{\pi}{4} - \arctan \dfrac{1}{2n+3}. \quad (2)$$
If you subtract $(1)$ from $(2)$, you get the equation $$\arctan \dfrac{1}{2\cdot (n+1)^2} = \arctan \dfrac{1}{2n+1} - \arctan \dfrac{1}{2n+3}. \quad (3)$$
So, after assuming the inductive hypothesis $(1)$, you need to prove equation $(3)$. Then you can add equation $(3)$ to equation $(1)$ to prove equation $(2)$, which completes the inductive step.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help solving the limit of the sequence: $\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$ Given the following sequence,
$$\left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) }$$
I am asked to determine to which $l \in \mathbb{R}$ the sequence converges.
This is what I have tried so far:
$$
\lim_{n} \left( \frac{2n^2 - 1}{2n^2 + 1} \right) ^ { \left( \frac{2n^3 - n}{n + 3} \right) } \implies
\left( \frac{2n^2 + 1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) } \implies
\left( \frac{2n^2}{2n^2 - 1} + \frac{1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) } \implies
\left( 1 + \frac{1}{2n^2 - 1} \right) ^ { - \left( \frac{2n^3 - n}{n + 3} \right) } \implies ? $$
My objective was to algebraically modify the sequence such that I could be able to reduce it to the form: $$\lim _{x\to +\infty }\left(1+{\frac {1}{x}}\right)^{x}=e$$
but seem to be stuck at the ? point.
Any suggestion? Thank you.
| Hint
$$=\left(\left(1-\dfrac2{2n^2+1}\right)^{-(2n^2+1)/2}\right)^{-\frac{2(2n^3-n)}{(2n^2+1)(n+3)}}$$
Alternatively
$$=\left(\dfrac{(1-1/2n^2)^{2n^2}}{(1+1/2n^2)^{2n^2}}\right)^{\dfrac{2n^3-n}{2n^2(n+3)}}$$
Now the numerator converges to $e^{-1}$
What about the denominator & the exponent?
| {
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"timestamp": "2023-03-29T00:00:00",
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Equation of Circle touching a straight line and passing through the centroid of a triangle and a particular point If the equation $3{x^2}y + 2x{y^2} - 18xy = 0$ represent the sides of a triangle whose centroid is G, then the equation of the circle which touches the line $x-y+1=0$ and passing through G and $(1,1)$ is | Start with the general form of a circle:
$$x^2+y^2 + ax+by+c = 0$$
It passes through $(1, 1)$ and $(2,3)$, so $a+b+c = -2$ and $2a+3b+c = -13$. Therefore $(2a + 3b + c) - 2(a+b+c) = b-c = -9 \Rightarrow b = c - 9$.
Now, since the line $x-y+1 =0 \Rightarrow y = x+1$ is tangent to the circle, substitute in $y = x+1$.
This equation gives us the possible values of $x$, so by setting $\Delta = 0$, we ensure there is only one (real) value of $x$ common to both the circle and line, which is what a tangent is:
$$x^2 + (x+1)^2 + ax + b(x+1) + c = 0$$
$$\Rightarrow 2x^2 + (a+b+2)x + (b+c+1) = 0$$
$$\Rightarrow B^2-4AC = 0 : (a+b+2)^2 - 8(b+c+1) = 0$$
$$\Rightarrow (-c)^2 - 8(b+c+1) = 0 \tag{$a+b+c = -2$}$$
$$\Rightarrow c^2 -8(c - 9 + c + 1) = 0 \tag{$b=c-9$}$$
which gives $(a,b,c) = (-9, 1, 8)$ as expected.
| {
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"url": "https://math.stackexchange.com/questions/3917805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving identity in infinite series I am trying to prove the equation
\begin{align*}
\sum_{n=0}^{\infty} a^n \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdots b_d^{c_d} = \prod_{i=1}^d \frac{1}{1 - a b_i},
\end{align*}
where we have that $n = c_1 + c_2 + ... c_d$. This is what I have tried so far:
\begin{align*}
\sum_{n=0}^{\infty} a^{c_1 + c_2 + \ldots + c_d} \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdots b_d^{c_d}
&= \sum_{n=0}^{\infty} a^{c_1} \cdot a^{c_2} \cdot \cdot \cdot a^{c_d} \sum_n b_1^{c_1} \cdot b_2^{c_2} \cdot \cdot \cdot b_d^{c_d} \\
&= \sum_{c_1}^{\infty} a^{c_1} \cdots \sum_{c_d}^{\infty} a^{c_d} \cdot \sum_{c_1}^{} b_1^{c_1} \cdots \sum_{c_d} b_d^{c_d} \\
&= \sum_{c_1}^{\infty} (a b_1)^{c_1} \cdots \sum_{c_d}^{\infty} (ab_d)^{c_d}.
\end{align*}
My doubts in here are whether it is even possible to split up the sum in this manner and then simply recombining it in the last line, since the product of two sums is not necessarily the sum of the product.
From here I would use
\begin{align*}
\sum_{k}^{\infty} x^k = \frac{1}{1-x},
\end{align*}
such that it would become
\begin{align}
\frac{1}{1-a b_1} \times \cdots \times\frac{1}{1 - a b_d} = \prod_{i = 1}^{d} \frac{1}{1-a b_i}.
\end{align}
I would like to ask where my errors are and if there is a better way to prove this identity. Thank you for your time.
| By induction on $d$.
$$\begin{aligned}
\frac{1}{1-ab_1}
\frac{1}{1-ab_2} &= \left(\sum_{n=0}^\infty (ab_1)^n\right)\left(\sum_{n=0}^\infty (ab_2)^n\right)\\
&=\sum_{n=0}^\infty \left(\sum_{l=0}^nb_1^lb_2^{n-l}\right) a^n\\
&=\sum_{n=0}^\infty \left(\sum_{c_1+c_2=n}b_1^{c_1} b_2^{c_2}\right) a^n
\end{aligned}$$
according to Cauchy product formula.
Now apply again Cauchy product formula to pass from $d$ to $d+1$:
$$\begin{aligned}
\prod_{i=1}^{d+1} \frac{1}{1 - a b_i} &=\left(\prod_{i=1}^{d} \frac{1}{1 - a b_i}\right)\frac{1}{1 - a b_{d+1}}\\
&= \left(\sum_{n=0}^\infty \left(\sum_{c_1+ \dots +c_d=n}b_1^{c_1} \dots b_d^{c_d}\right) a^n\right)\left(\sum_{n=0}^\infty (ab_{d+1})^n\right)\\
&= \sum_{n=0}^\infty \left(\sum_{l=0}^n \left(\sum_{c_1+ \dots +c_d=l}b_1^{c_1} \dots b_d^{c_d}\right)b_{d+1}^{n-l}\right)a^n\\
&=\sum_{n=0}^\infty \left(\sum_{c_1+\dots + c_{d+1}=n}b_1^{c_1} \dots b_{d+1}^{c_{d+1}}\right) a^n
\end{aligned}$$
| {
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Prove Inequalities $\frac{x(2+x)}{1+x} \leq f(x) \leq \frac{x(2-x)}{1-x}$ Using Calculus I am often asked to prove inequalities with natural logarithms. An example question I am stuck on is this:
Let $f(x) = \ln(\frac{1+x}{1-x})$ for $-1 < x < 1$. Prove that
$$\frac{x(2+x)}{1+x} \leq f(x) \leq \frac{x(2-x)}{1-x}$$ with equality iff $x=0$.
My general approach has been to try to find a function which upper and lower bounds the function $1/x$ such that the integral of the function will result in the upper and lower bound. I know that the definition of $\ln(x) = \int_{1}^{x} \frac{1}{t}dt$.
Example I want to prove (1)$f_1(x) < \ln(x) <f_2(x)$ . If I can find (2)$F_1(x) < 1/x < F_2(x)$, then by taking the integral I can get the original inequality. Problem is I find it difficult to come up with the (1) relationship. Espically when it gets complicated.
Are there better ways / tricks that can help me solve these proofs more easily? What would be your approach to this problem?
| Let
$$g(x)=\frac{x(2-x)}{1-x}- \ln\frac{1+x}{1-x}
$$
$$g’(x)= \frac{x(x^2-x+2)}{(1-x)^2(1+x)},\>\>\>
g’’(x) =\frac{2(3x^2+1)}{(1-x)^3(1+x)^2}
$$
Note $g(0)=g’(0)=0$ and $g’’(x)>0$, which means that $g(x)$ is convex with the minimum at $g(0)=0$. Thus, $g(x)\ge 0$, or
$$\ln\frac{1+x}{1-x} \le \frac{x(2-x)}{1-x}$$
Similarly, let
$$h(x)=\ln\frac{1+x}{1-x} - \frac{x(2+x)}{1+x}
$$
$$h’(x)= \frac{x(x^2+x+2)}{(1-x)(1+x)^2},\>\>\>h’’(x) =\frac{2(3x^2+1)}{(1-x)^2(1+x)^3}$$
Note $h(0)=h’(0)=0$ and $h’’(x)>0$, which means that $h(x)$ is convex with the minimum at $h(0)=0$. Thus, $h(x)\ge 0$, or
$$ \frac{x(2+x)}{1+x}\le \ln\frac{1+x}{1-x} $$
| {
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Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$ Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$
I couldn't do this question and hence I looked at the solution which goes as follows:
if $x=0$ then $x^2+2xy=0\ne5$ hence $x\ne0$
I state that $y=lx$. Hence the system becomes:
$x^2(1+2l)=5$, $x^2(l^2-3l)=-2$ which becomes $5l^2-11l+2=0$, hence $l=2$ or $l=\frac{1}{5}$.
Hence the possible solutions are:
$(1, 2), (-1,-2), (\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}), (-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}})$.
My question is why was I supposed to think of substituting $y=lx$? Why was this supposed to be intuitive and is there a more intuitive approach?
| Alternative approach.
Solve in real numbers the system of equations $x^2+2xy=5$, $y^2-3xy=-2$
$3x^2 + 6xy = 15~$ and $~2y^2 - 6xy = -4.$
Adding gives $3x^2 + 2y^2 = 11.$
Then, you have that $5 - x^2 = 2xy.$
Therefore, $25 - 10x^2 + x^4 = 4x^2y^2 = (2x^2)(11 - 3x^2) = 22x^2 - 6x^4.$
Therefore, $7x^4 - 32x^2 + 25 = 0.$
Applying the quadratic equation against $x^2$ gives that $x^2 \in \{1, \frac{25}{7}\}.$
Using that $2y^2 = 11 - 3x^2$ gives that
$y^2 \in \{4, \frac{1}{7}\}.$
Therefore, there are only 8 possible solutions to the equations:
$(x,y) \in \{(\pm 1, \pm 2), (\pm \frac{5}{\sqrt{7}}, \pm \frac{1}{\sqrt{7}})\}.$
Checking each of these 8 possibilities against the original equation gives final answer of:
$$(x,y) \in \left\{(1, 2), (-1, -2),
\left(\frac{5}{\sqrt{7}}, \frac{1}{\sqrt{7}}\right),
\left(-\frac{5}{\sqrt{7}}, -\frac{1}{\sqrt{7}}\right)\right\}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find centroid of shape given by a circle and a cardoid I'm stuck on finding the centroid of this shape (the shaded area):
With the sphere having a radius of $r=2$, and the cardioide being given by $r=1+\cos\phi$
I found the Area, which is
$A=\frac12(2\pi r^2-\frac32 \pi)=\frac{13}{4} \pi$
Now I don't know how to proceed further. Could anyone point me in the right direction?
Thank you :)
| There is a mistake in your area calculation -
$A = \frac12(\pi r^2-\frac32 \pi)=\frac{5}{4} \pi$ (you wrote $2\pi r^2 $ instead of $\pi r^2$ by mistake)
Now coming to finding centroid,
We know in polar coordinates, $x = r\cos \theta, y = r \sin \theta$ so the centroid ($\overline{x}, \overline {y}$) of the polar region is given by
$\displaystyle \overline{x} = \dfrac{2}{3} \frac {\int_{\alpha}^{\beta} r^3 \cos\theta \, \mathrm{d} \theta}{\int_{\alpha}^{\beta} r^2 \, \mathrm{d} \theta} \,$ (numerator is $r \cos \theta .(r^2 d\theta))$
Similarly, $\displaystyle \overline{y} = \dfrac{2}{3} \frac {\int_{\alpha}^{\beta} r^3 \sin\theta \, \mathrm{d} \theta}{\int_{\alpha}^{\beta} r^2 \, \mathrm{d} \theta}$
The denominator is nothing but $2A \,$ where $A$ is the area of the region which you have already found. Also consider the polar region as a collection of infinitely small triangles and that is how this formula.
In this case our shaded region is bound by $r = (1 + \cos \theta) \,$ below and $r = 2$ above. Please also note that both curves go between $0 \leq \theta \leq \pi$ as we are interested in shaded area above $X$-axis. So,
$\displaystyle \int_{\alpha}^{\beta} r^3 \cos\theta \, \mathrm{d} \theta = \displaystyle \int_{0}^{\pi} [2^3 - (1 + \cos \theta)^3] \cos\theta \, \mathrm{d} \theta = -\frac{15 \pi}{8}$
$\displaystyle \int_{\alpha}^{\beta} r^3 \sin\theta \, \mathrm{d} \theta = \displaystyle \int_{0}^{\pi} [2^3 - (1 + \cos \theta)^3] \sin\theta \, \mathrm{d} \theta = 12$
$\displaystyle \int_{\alpha}^{\beta} r^2 \, \mathrm{d} \theta \,$ is nothing but $2A$ and as you already found the area, our denominator is $\frac{5 \pi}{2}$.
So $\overline{x} = \frac{2}{3} \times \frac{-15 \pi/8}{5\pi/2} = -\frac{1}{2}$
$\overline{y} = \frac{2}{3} \times \frac{12}{5\pi/2} = \frac{16}{5\pi}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3922870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$? Consider the integral domain $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$?
I know the following elementary facts. We have
\begin{equation}
\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right] = \left\{ \frac{m + n \sqrt{5}}{2} : m, n \in \mathbb{Z} \text{ are both even or both odd} \right\}.
\end{equation}
For every $\frac{m + n \sqrt{5}}{2} \in \mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$, define its norm as usual:
\begin{equation}
N\left(\frac{m + n \sqrt{5}}{2}\right)=\frac{m^2-5n^2}{4}.
\end{equation}
Since $m, n$ are both even or both odd, it is easy to see that the norm is an integer. From this fact it is easily seen that $\frac{m + n \sqrt{5}}{2}$ is a unit of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ if and only if $m^2 - 5n^2=4$ or $m^2 - 5n^2=-4$. Now since $N(4+\sqrt{5})=11$ we easily get that $4+\sqrt{5}$ is an irreducible element of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. If $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ were a unique factorization domain, we could conclude that $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. But I do not know if $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ is a unique factorization domain. Does someone know if it is?
Thank you very much in advance for your attention.
| Yes, $\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$ is a UFD because it is norm-Euclidean.
| {
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Does anyone know of a much faster way to solve this polynomial question? A monic cubic $p(x) $ is divided by $(x^2 +x +1)$ and the remainder is $(2x+3)$. When $p(x)$ is divided by $x(x+3)$, the remainder is $5(x+1)$. Find $p(x)$
So the way I solved it was through using long division - as you can imagine very lengthy, and not very efficient in exam situations. I then equated the remainders with the given remainders. However, in an exam situation, would there be an even faster way? I am open to suggestions.
| This may not be a quicker way, but it is systematic and will work in all cases where the two moduli have no common factors. Hopefully, it also might make the use of the Extended Euclidean Algorithm with the Chinese Remainder Theorem more transparent.
We want
$$
\begin{align}
p(x)&\equiv2x+3&&\pmod{x^2+x+1}\tag{1a}\\
p(x)&\equiv5x+5&&\pmod{x^2+3x}\tag{1b}
\end{align}
$$
so we apply the Extended Euclidean Algorithm (as implemented in this answer and adapted to polynomials) to $x^2+3x$ and $x^2+x+1$:
$$
\begin{array}{c|c|c|c}
x^2+3x&1&0\\
x^2+x+1&0&1\\
2x-1&1&-1&1\\
\frac74&-\frac{2x+3}4&\frac{2x+7}4&\frac{2x+3}4
\end{array}\tag2
$$
Thus, we get
$$
(2x+7)\left(x^2+x+1\right)-(2x+3)\left(x^2+3x\right)=7\tag3
$$
So we have
$$
-\frac{(2x+3)\left(x^2+3x\right)}7\equiv\left\{
\begin{align}
&1&&\pmod{x^2+x+1}\\
&0&&\pmod{x^2+3x}
\end{align}
\right.\tag4
$$
and
$$
\frac{(2x+7)\left(x^2+x+1\right)}7\equiv\left\{
\begin{align}
&0&&\pmod{x^2+x+1}\\
&1&&\pmod{x^2+3x}\\
\end{align}
\right.\tag5
$$
Adding $2x+3$ times $(4)$ to $5x+5$ times $(5)$ yields one solution to $(1)$:
$$
\begin{align}
&\overbrace{(5x+5)\frac{(2x+7)\left(x^2+x+1\right)}7}^{\substack{0&\pmod{x^2+x+1}\\5x+5&\pmod{x^2+3x}}}\overbrace{-(2x+3)\frac{(2x+3)\left(x^2+3x\right)}7}^{\substack{2x+3&\pmod{x^2+x+1}\\0&\pmod{x^2+3x}}}\\
&=\frac{6x^4+31x^3+45x^2+53x+35}7\tag6
\end{align}
$$
The Chinese Remainder Theorem says that this is the solution mod $\left(x^2+x+1\right)\left(x^2+3x\right)$, so we can subtract $\frac67$ of the modulus to get
$$
\begin{align}
p(x)
&=\frac{6x^4+31x^3+45x^2+53x+35}7-\frac67\left(x^2+x+1\right)\left(x^2+3x\right)\\
&=\bbox[5px,border:2px solid #C0A000]{x^3+3x^2+5x+5}\tag7
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3927103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Number Theory : Solve the system of congruence $28x+17y\equiv 18 \pmod{41}$ and $31x+11y\equiv 35\pmod{41}$ Number Theory : Solve the system of congruence
(1) $28x+17y\equiv 18 \pmod{41}$
(2) $31x+11y\equiv 35\pmod{41}$
Attempt :
we know that equation (1) and (2) are in the same$\pmod{41}$. so we can use Modular arithmetic
lets multiply the first equation (1) by $31$ and the second equation(2) by $28$.
(1) $31\cdot(28x+17y)\equiv 31\cdot18 \pmod{41}$
(1) $868x+527y\equiv 558\pmod{41}$
(2) $28\cdot(31x+11y)\equiv 28\cdot35\pmod{41}$
(2) $868x+308y\equiv 980\pmod{41}$
So finally we can subtract equation (1) from (2) we get:
$219y\equiv -422\pmod{41}$
lets check the $\gcd(219,41)$ by Euclidian algorithm :
$219 = 41\cdot 5 + 14$
$41= 14\cdot 2 + 13$
$14= 13\cdot 1 + 1$
$\gcd(219,41)=1$
Hence, because the $\gcd$ is equal to $1$ we can find the Inverse and multiply the equation by the Inverse to find $y$.
$219y\equiv 1\pmod{41}$
$219a = 1+41k$ , the $41k$ must end with the digit of $8$ for $1+$digit $8$ will be $9$ so $k$ must be multiply of number with end digit of $8$.
I don't know how to continue from here .
| Let $p \in \Bbb N$ be a prime number.
Recall the following two facts from elementary number theory:
Let $p$ be a prime integer.
For all $u,v,w \in \Bbb Z$ satisfying $u\not\equiv 0 \pmod{p}$, $v\not\equiv 0 \pmod{p}$ and $w\not\equiv 0 \pmod{p}$,
$\tag 1 u \equiv v \pmod{p} \; \text{ iff } \; uw \equiv vw \pmod{p}$
If $a \in \Bbb N$ and $a\not\equiv 0 \pmod{p}$ and $a\not\equiv 1 \pmod{p}$ then there exist $q \in \Bbb Z$ and $r \in \Bbb N$ such that
$\tag 2 aq \equiv r \pmod{p} \; \text{ and } \; 1 \le r \lt p \; \text{ and } \; r \lt a$
The above theory is the counterpart to algorithms that compute the inverse
of non-zero elements in $(\mathbb{Z}/p\mathbb{Z})^\times$.
We are now ready to continue where the OP left off.
Since
$\quad 219y\equiv -422\pmod{41} \; \text{ iff } \; 14y\equiv 29\pmod{41}$
we are motivated to 'divide' (with a small negative residue) $41$ by $14$ and write ,
$\quad 41 = 14 \cdot 3 - 1$
and set $q = 3$.
Next,
$\quad 14y\equiv 29\pmod{41} \; \text{ iff } \; (3 \cdot 14) y\equiv 3 \cdot 29\pmod{41} \; \text{ iff } \; 1y\equiv 5 \pmod{41}$
and the OP's problem is solved in one step.
By being flexible with the sign of the residue, this is a one step solution. If you can only perform Euclidean division, this is what happens:
Presented with
$\quad 14y\equiv 29\pmod{41}$
we divide $41$ by $14$ and write ,
$\quad 41 = 14 \cdot 2 + 13$
and set $q = -2$.
Next,
$\quad 14y\equiv 29\pmod{41} \; \text{ iff } \; (-2 \cdot 14) y\equiv (-2 \cdot 29)\pmod{41} \; \text{ iff } \; 13 y\equiv 24 \pmod{41}$
and write
$\quad 41 = 13 \cdot 3 + 2$
and set $q = -3$
Next,
$\quad 13y\equiv 24\pmod{41} \; \text{ iff } \; (-3 \cdot 13) y\equiv (-3 \cdot 24)\pmod{41} \; \text{ iff } \; 2 y\equiv 10 \pmod{41}$
and write
$\quad 41 = 2 \cdot 20 + 1$
and set $q = -20$
Next,
$\quad 2 y\equiv 10 \pmod{41} \; \text{ iff } \; (-20 \cdot 2) y\equiv (-20 \cdot 10)\pmod{41} \; \text{ iff } \; 1y\equiv 5 \pmod{41}$
and the OP's problem is solved in $3$ steps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3935020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Justifying a solution for a system of equations in linear algebra I'm trying to understand how to solve systems of linear equations using matrices, however I'm faced with the following problem. Given the following system:
$$
\begin{cases}
x+2y+3z-3w=a \\
2x-5y-3x+12w=b \\
7x+y+8x+5w=c
\end{cases}
$$
What steps do I need to do to confirm and justify that it only has a single solution, which is:
$$
37a +13b = 9c
$$
Appreciate all hints! Thanks.
| You can get your answer using simple elementary row operations:
$$\left(
\begin{matrix}
1 & 2 & 3 & -3 & | & a\\
2 & -5 & -3 & 12 & | & b \\
7 & 1 & 8 & 5 & | & c \\
\end{matrix}
\right) >>{\text{$R2=R2-2R1, R3=R3-7R1$}}>>$$ $$\left(
\begin{matrix}
1 & 2 & 3 & -3 & | & a\\
0 & -9 & -9 & 18 & | & b-2a \\
0 & -13 & -13 & 26 & | & c-7a \\
\end{matrix}
\right)>>{\text{$R3=R3+(- \frac{13} {9} R2)$}}>>$$
$$\left(
\begin{matrix}
1 & 2 & 3 & -3 & | & a\\
0 & -9 & -9 & 18 & | & b-2a \\
0 & 0 & 0 & 0 & | & c-7a + [- \frac{13} {9}(b-2a)]\\
\end{matrix}
\right)$$
Now, notice that the last row of the matrix is a zero row, thus the system has a solution only if the right side of the equation is equal to zero. thus-
$$c-7a + [- \frac{13} {9}(b-2a)] =0$$
$$-\frac{37}{9}a-\frac{13}{9}b+c=0$$
$$-37a -13b +9c=0$$
$$37a+13b=9c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3935872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Lagrange's Multiplier for $f(x,y,z) = (xyz)^2$ Given a function:
$$f(x,y,z) = (xyz)^2$$
And the set: $$S = \{(x,y,z) : x^2 + y^2 + z^2 = a^2\} $$
Does $f$ have a maxima on $S$?
(Is this argument correct? : Yes, because f has a maxima and minima in S if S is compact and f is continuous. But here S is compact so we have a maxima)
Also, how do I tell if a function has maxima or minima on a set if the set is not compact?
I tried using the lagrange multiplier.
$f_x = 2xy^2z^2$, $f_y = 2yx^2z^2$, $f_z = 2zx^2y^2$, $f_\lambda = x^2 + y^2 + z^2 - a^2$
$$f_x + \lambda g_x = 0$$
$$f_y + \lambda g_y = 0$$
$$f_z + \lambda g_z = 0$$
$$f_\lambda = 0$$
However on solving this system of equations using substitution, I am unable to find a solution for (x,y,z). I did find one point though (0,0,0), for $a = 0$, Can I set $a$ to be $0$?
$$2xy^2z^2 = \lambda2x$$
$$2yx^2z^2 = \lambda2y$$
$$2zx^2y^2 = \lambda2z$$
So suppose $x = 0$ and $\lambda \not = 0$, we have $\lambda2y = 0$, so $y = 0$, similarly $z = 0$, $(0,0,0)$ satisfies $f_\lambda = 0$ for $a = 0$, so this is one point.
However, evaluating for other points, I am unable to simplify.
For $x \not = 0$, we have: $y^2z^2 = \lambda$
But how do I proceed from now?
Can I say by symmetry, if any one of $x$, $y$ or $z$ is $0$ then they all have to be $0$?
If so, then we have all non zero :
$$x^2z^2 = x^2y^2 = y^2z^2$$
which means $$ x^2 = y^2 = z^2$$
And then substituting this into constraint, we get 6 points of the form
$$( \pm\ a/\sqrt3, \pm\ a/\sqrt3, \pm\ a/\sqrt3)$$
Is this correct?
| Define : $\mathcal{L}(x,y,z,\lambda) = (xyz)^2 - \lambda\cdot(x^2 + y^2 + z^2 - a^2)$ for any $\lambda \in \mathbb{R}$ , and for any $x, y, z \in \mathbb{R}$ .
So, $~\mathcal{L}_x = 2xy^2z^2 - 2\lambda x~$ , $~\mathcal{L}_x = 2x^2yz^2 - 2\lambda y~$ , and $~\mathcal{L}_x = 2x^2y^2z - 2\lambda z~$ . Also, we have : $~\mathcal{L}_\lambda = a^2 - (x^2 + y^2 + z^2)~$ .
Equating these to $0$ , we get : $\lambda = x^2y^2 = y^2z^2 = z^2x^2~$ , and $~~a^2 = x^2 + y^2 + z^2$ .
Now, if any of $x, y, z$ is $0$ , then $f$ won't be maximized for sure, in fact it'll get minimized then. So , multiplying the first $3$ constraints by $z^2, x^2 , y^2$ respectively , we get $\lambda x^2 = \lambda y^2 = \lambda z^2$ , which implies that $f$ is maximized when $x^2 = y^2 = z^2 = k~$ (Say) .
Thus, by Lagrange Multiplier Theorem, using the 4th constraint, we get that $f$ is maximized when $a^2 = x^2 + y^2 + z^2 = 3k$ , i.e. $~x^2 = y^2 = z^2 = \frac{a^2}{3}$ .
Therefore, $f$ is maximized when $~(x,y,z) = \left(\pm\dfrac{| a |}{3} , \pm\dfrac{| a |}{3} , \pm\dfrac{| a |}{3} \right)$ , and the maximum value of will be $\dfrac{a^6}{27}$ .
In fact, you don't need a strong result like Lagrange Multiplier in order to solve this problem. Observe that by simple AM-GM Inequality, we have : $$(xyz)^2 = x^2\cdot y^2 \cdot z^2 \leqslant \left[\dfrac{x^2 + y^2 + z^2}{3}\right]^3 = \left(\dfrac{a^2}{3}\right)^3 = \dfrac{a^6}{27}$$ and the maximum is attained when $x^2 = y^2 = z^2 = \dfrac{a^2}{3}$ , which is certainly same as the result produced using Lagrange Multiplier. Hope it helps.
| {
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"url": "https://math.stackexchange.com/questions/3937945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Range of quadratic function using discriminant
Let $x^2-2xy-3y^2=4$. Then find the range of $2x^2-2xy+y^2$.
Let $2x^2-2xy+y^2=a$.
Then $ax^2-2axy-3ay^2=4a=8x^2-8xy+4y^2\implies (a-8)x^2-(2a-8)xy-(3a+4)y^2=0$.
We divide both side by $y^2$ and let $t=\frac{x}{y}$.
Then it implies $(a-8)t^2-(2a-8)t-(3a+4)=0$.
Since its discriminant is not negative, $\frac{\Delta}{4}\ge 0\implies a^2-7a-4\ge 0$. It gives us $a$ can have negative values like $-1$. But if clearly contracts $a-4=x^2+4y^2\ge 0\implies a\ge 4$. Where did I mistake?
| Others have pointed out where you went wrong. I just want to provide an alternative proof.
$$x^2-2xy-3y^2=4=(x+y)(x-3y)$$
Denote $z=x+y, w=x-3y$, then $zw=4, x=(3z+w)/4, y=(z-w)/4$.
We want to find the range of $$2x^2-2xy+y^2=\frac{1}{16} (5w^2+14wz+13z^2)=\frac 72+\frac{1}{16} (5w^2+13z^2) \tag 1$$ under the constraint $zw=4$.
Clearly $(1)$ can be as large as you want. And if you apply AM-GM you get the minimum
$$\frac 72+\frac{1}{16} (5w^2+13z^2) \ge \frac 72 + \frac{1}{16} 2 \sqrt{5} \sqrt{13}|wz|=\frac{7+\sqrt{65}}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3942812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Am I allowed to use this method? Question: solve
$\frac{d^2y}{dx^2}+\frac{dy}{dx}=2x+3$.
My attempt:
The general solution to the corresponding homogeneous DE is $y_g=c_1+c_2e^{-x}$.
Now to find the particular integral, rewrite the DE as $$(D^2+D)y=2x+3$$
We have $y_p=\frac{1}{D(D+1)} (2x+3)$.
Then $y_p=\frac{1}{D+1}(x^2+3x)$.
Here's my question:
Am I allowed to use Taylor's expansion?
If this is allowed then I can find $y_p$ easily.
We would have $y_p=1-D+D^2-D^3+D^4-...(x^2+3x)$
and $y_p=x^2+3x-(2x+3)+2-0$
| $$(D^2+D)y=2x+3$$
It means:
$$y''+y'=2x+3$$
Not the equation you write in the first line.
$$\frac{d^2y}{dx^2}y+\frac{dy}{dx}y=2x+3$$
$$y''y+y'y=2x+3$$
And yes you have that:
$$\dfrac 1{1+D}=1-D+D^2-D^3.......$$
But you have:
$$y_p=\frac{1}{D(D+1)} (2x+3)$$
$$y_p=\left (\dfrac{1}{D}-\dfrac 1 {D+1} \right)(2x+3)$$
$$y_p=\dfrac{1}{D} (2x+3)-\dfrac 1 {D+1} (2x+3)$$
$$y_p=(x^2+3x)-\dfrac 1 {D+1} (2x+3)$$
Or as you did:
$$y_p=\dfrac 1{D+1}(x^2+3x)=x^2+3x -(2x+3)+2$$
$$\implies y_p=x^2+x-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3942993",
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"source": "stackexchange",
"question_score": "1",
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} |
$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$ I've tried to solve this limit:
$$\lim_{x \to 0} \frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$
Here,
$$\frac{\ln (1+x \arctan x)-e^{x^2}+1}{\sqrt{1+2x^4}-1}$$ $$\sim \frac{(x \arctan x-x^2)(\sqrt{1+2x^4}+1)}{2x^4}$$ $$\sim \frac{(x \arctan x-x^2)}{x^4}= \frac{ \arctan x-x}{x^3} \sim \frac{x-x}{x^3} \sim 0$$
But the final result should be $-\frac{4}{3}$.
Any help would be appreciated.
| I think the mistake is from the first step.Emm...in fact,$$\lim \frac{{\ln (1 + x\arctan x)}}{{\sqrt {1 + 2{x^4}} - 1}} = \frac{{\ln (1 + x\arctan x)(\sqrt {1 + 2{x^4}} + 1)}}{{2{x^4}}} = \infty , \\ \lim \frac{{{e^{{x^2}}} - 1}}{{\sqrt {1 + 2{x^4}} - 1}} = \frac{{({e^{{x^2}}} - 1)(\sqrt {1 + 2{x^4}} + 1)}}{{2{x^4}}} = \infty $$
However,when $${f_1} \sim \alpha ,{f_2} \sim \beta $$ but $$\frac{\alpha }{g} = \infty or\frac{\beta }{g} = \infty $$
you can't apply
$$\frac{{{f_1} - {f_2}}}{g} \sim \frac{{\alpha - \beta }}{g}$$
I don't know if I made it clear to you...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3944331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\lim_{x \to + \frac{1}{2}} [ \tan(\pi x^2)+(2x-1)4^x+6x-4 ] \cot(x-\frac{1}{2})$ I've tried to solve this limit:
$\lim_{x \to + \frac{1}{2}} \left [ \tan(\pi x^2)+(2x-1)4^x+6x-4 \right ] \cot(x-\frac{1}{2})$
the first parentesis should tend to $\frac{\sqrt{2}}{2}-1$ but I don't know how to procede.
The final result should be $2 \pi + 10$
| To make life easier, let $x=y+\frac 12$ to make the expression
$$\left(\left(4^{y+1}+6\right) y+\tan \left(\frac{1}{4} \pi (2
y+1)^2\right)-1\right) \cot (y)$$ and now, compose series one piece at the time
$$4^{y+1}=4 e^{y \log(4)}=4+4 y \log (4)+2 y^2 \log ^2(4)++O\left(y^3\right)$$
Now, after expansion of the argument, and using $\tan(a+b)=\cdots$
$$\tan \left(\frac{1}{4} \pi (2y+1)^2\right)=1+2 \pi y+2 \left(\pi +\pi ^2\right) y^2+O\left(y^3\right)$$
$$\left(4^{y+1}+6\right) y+\tan \left(\frac{1}{4} \pi (2
y+1)^2\right)-1=(10+2 \pi ) y+y^2 \left(2 \left(\pi +\pi ^2\right)+4 \log(4)\right)+O\left(y^3\right)$$
$$\cot(y)=\frac{1}{y}-\frac{y}{3}+O\left(y^3\right)$$ Make the product and the expression is
$$(10+2 \pi )+y \left(2 \pi\left(\pi +1\right)+8 \log (2)\right)+O\left(y^2\right)$$ which shows the limit and also how it is approached.
Edit for clarification about $\tan \left(\frac{1}{4} \pi (2y+1)^2\right)$
$$\frac{1}{4} \pi (2y+1)^2=\frac{\pi }{4}+\pi y(y+1)=\frac{\pi }{4}+a$$
$$\tan \left(\frac{\pi }{4}+a\right)=\frac{1+\tan(a)}{1-\tan(a)}=\frac {1+a+\frac{a^3}{3}+\frac{2 a^5}{15}+O\left(a^7\right) } {1-a-\frac{a^3}{3}-\frac{2 a^5}{15}+O\left(a^7\right) }$$
Long division
$$\tan \left(\frac{\pi }{4}+a\right)=1+2 a+2 a^2+\frac{8 a^3}{3}+\frac{10 a^4}{3}+\frac{64 a^5}{15}+\frac{244
a^6}{45}+O\left(a^7\right)$$ Now, replace $a=\pi y(y+1)$
$$\tan \left(\frac{1}{4} \pi (2y+1)^2\right)=1+2 \pi y+2 \left(\pi +\pi ^2\right) y^2+O\left(y^3\right)$$
| {
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Determinant and inverse matrix calculation of a special matrix Is there any smart way of calculating the determinant of this kind matrix?
\begin{pmatrix}
1 & 2 & 3 & \cdots & n \\
2 & 1 & 2 & \cdots & n-1 \\
3 & 2 & 1 & \cdots & n-2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n & n-1 & n-2 & \cdots & 1 \end{pmatrix}
I encountered this in a problem for the case $n=4$. I need to find the inverse matrix.
I doubt the idea of this problem is to calculate all the cofactors and then the inverse matrix in the usual way.
I see the pattern here and some recursive relations... but I am not sure if this helps for calculating the determinant of the existing matrix.
| Call your matrix $A$. Let $\{e_1,e_2,\ldots,e_{n-1}\}$ be the standard basis of $\mathbb R^{n-1}$. Let also $e=\sum_ie_i=(1,1,\ldots,1)^T$ and
$$
L=\pmatrix{1\\ -1&1\\ &\ddots&\ddots\\ &&-1&1}.
$$
Then $B:=LAL^T=\pmatrix{1&e^T\\ e&-2I_{n-1}}$. Using Schur complement, we obtain
$$
\det(A)=\det(B)=\det(-2I_{n-1})\left(1-e^T(-2I_{n-1})^{-1}e\right)=(-2)^{n-1}\frac{n+1}{2}.
$$
To find $A^{-1}$, partition $L$ as $\pmatrix{1&0\\ -e_1&L'}$. Then
\begin{aligned}
A^{-1}
&=L^TB^{-1}L\\
&=\pmatrix{1&-e_1^T\\ 0&L'^T}
\pmatrix{\frac{2}{n+1}&\frac{1}{n+1}e^T\\ \frac{1}{n+1}e&\frac{1}{2(n+1)}ee^T-\frac12I_{n-1}}
\pmatrix{1&0\\ -e_1&L'}\\
&=\pmatrix{\frac{1}{n+1}&\frac{1}{2(n+1)}e^T+\frac12e_1^T\\ \ast&\frac{1}{2(n+1)}e_ne^T-\frac12L'^T}
\pmatrix{1&0\\ -e_1&L'}\\
&=\pmatrix{\frac{-n}{2(n+1)}&\frac{1}{2(n+1)}e_n^T+\frac12e_1^T\\ \ast&\frac{1}{2(n+1)}e_ne_n^T-\frac12L'^TL'}.
\end{aligned}
By direct calculation, $L'^TL'$ is the symmetric tridiagonal matrix whose main diagonal is $(2,\ldots,2,1)$ and whose superdiagonal is a vector of $-1$s. Since $A^{-1}$ is symmetric, we finally get
$$
A^{-1}=\pmatrix{
\frac{-n}{2(n+1)}&\frac12&&&\frac{1}{2(n+1)}\\
\frac12&-1&\frac12\\
&\frac12&-1&\ddots\\
&&\ddots&\ddots&\frac12\\
\frac{1}{2(n+1)}&&&\frac12&\frac{-n}{2(n+1)}
}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3947667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How to evaluate $\int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}$ without using antiderivative? Someone gives a solution here:
\begin{align*} \int_0^{\frac{\pi}{2}} \frac{{\rm d}\alpha}{1+\cos\alpha\cos\beta}&=\int_0^{\frac{\pi}{2}} \sum_{n=0}^{\infty} (-\cos\alpha\cos\beta)^n{\rm d}\alpha\\ &=\sum_{n=0}^{\infty}(-\cos\beta)^n\int_0^{\frac{\pi}{2}} \cos^n\alpha{\rm d}\alpha\\ &=\sum_{n=0}^{\infty}(-\cos \beta) ^n\frac{\sqrt{\pi}\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)}{2\Gamma\left(\frac{n}{2}+1\right)}\\ &=\frac{\pi-2\arcsin\cos \beta}{2\sqrt{1-\cos^2\beta}}\\ &=\frac{\beta}{\sin \beta}. \end{align*}
Is it correct? How to obtain the fourth equlity?
| For any $\beta \textrm{ is not a multiple of }\pi$, we multiply both the numerator and denominator by $1-\cos \alpha \cos \beta.$
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{d \alpha}{1+\cos \alpha \cos \beta}
=& \int_0^{\frac{\pi}{2}} \frac{1-\cos \alpha \cos \beta}{1-\cos ^2 \alpha \cos ^2 \beta} d \alpha \\
=& \int_0^{\frac{\pi}{2}} \frac{\sec ^2 \alpha}{\sec ^2 \alpha-\cos ^2 \beta} d \alpha-\cos \beta \int_0^{\frac{\pi}{2}} \frac{\cos \alpha}{1-\cos ^2 \alpha \cos ^2 \beta}d\alpha \\
=& \int_0^{\frac{\pi}{2}} \frac{d(\tan \alpha)}{\tan ^2 \alpha+\sin ^2 \beta}-\int_0^{\frac{\pi}{2}} \frac{d\left(\sin \alpha \cos \beta\right)}{\sin ^2 \alpha \cos ^2 \beta+\sin ^2 \beta} \\
=& \frac{1}{\sin \beta}\left[\tan ^{-1}\left(\frac{\tan \alpha}{\sin \beta}\right)\right]_0^{\frac{\pi}{2}}-\frac{1}{\sin \beta}\left[\tan ^{-1}\left(\frac{\sin \alpha \cos \beta}{\sin \beta}\right)\right]_0^{\frac{\pi}{2}} \\
=&\frac{\pi}{2 \sin \beta}-\frac{1}{\sin \beta}\left(\frac{\pi}{2}-\beta\right) \\
=& \frac{\pi}{\sin \beta}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3948358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve integral $\int^1_0 \frac{1-x^2}{{(1+x^2)}\sqrt{1+x^4}}dx$ using subsituition $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$ Hi this question has been posed in my integration book where it has been asked to solve it using the given substitution or any other substitution, I've found identical question posted here {1} with different substitutions
The problem is when using the given substitution I end up with integral
$$\int^{\pi/4}_0 \frac{\sin{\theta}}{2x}d\theta $$
Solving for $x$ : $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$
According to WolframAlpha: $$x = \frac{|\csc{\theta}||1 \pm \sqrt{\cos{2\theta}}|}{\sqrt{2}} $$
I do know the sign for x is $+ve$ so I'll take $+ve$ solutions after opening the modulus, I do not know which sign to prefer in $ \pm \sqrt{\cos{2\theta}} $ term
Eitherway, I solved for both cases :
$$ \frac{1}{\sqrt{2}}\int^{\pi/4}_0\frac{\sin^2{\theta}}{1 + \sqrt{\cos{2\theta}}} d\theta -(I) $$
$$ \frac{1}{\sqrt{2}}\int^{\pi/4}_0\frac{\sin^2{\theta}}{1 - \sqrt{\cos{2\theta}}} d\theta -(II) $$
Which leads to :
$$ \frac{1}{\sqrt{2}}\int^{\pi/4}_0 1 \pm \sqrt{\cos{2\theta}} d\theta = \frac{\pi}{4\sqrt{2}}
\pm\frac{\Gamma(\frac{3}{4})^2}{2\sqrt{2}} $$
Here I cheated and used WolframAlpha as I was exhausted.
Both the cases lead to a similar answer with difference of a function unknown to me: $\pm\frac{\Gamma(\frac{3}{4})^2}{2\sqrt{2}}$
The correct answer is :
$$\frac{\pi}{4\sqrt{2}}$$
My main questions are:
*
*What am I doing wrong?
*How did author arrive at this ingenious substitution $\sqrt{1+x^4} = {(1+x^2)}\cos{\theta}$ , this doesn't appear to me trial n' error kind of substitution, I have spent hours doing algebraic transformation on this and no matter how you procced and plug the variable your integrand will only have one of these terms only $2x$, $1-x^2$, $1+x^2$ and $ \sqrt{1+x^4}$ and $2x$ being the simplest.
Related question: How do I integrate the following? $\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}$
| Divide top and bottom by $x^2$:
$$\int_0^1\frac{\frac{1}{x^2}-1}{\left(\frac{1}{x}+x\right)\sqrt{\frac{1}{x^2}+x^2}}dx$$
which suggests using the substitution $t = \frac{1}{x}+x$:
$$\int_2^\infty \frac{dt}{t\sqrt{t^2-2}} = \frac{1}{\sqrt{2}}\sec^{-1}\left(\frac{t}{\sqrt{2}}\right)\Biggr|_2^\infty = \frac{\pi}{4\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
A question related to $f(n)=n^4+n^2+1$ If $$f(n)=n^4+n^2+1$$ then we have to evaluate $$\frac{f\left(3\right)f\left(5\right)f\left(7\right)f\left(9\right)f\left(11\right)f\left(13\right)}{f\left(2\right)f\left(4\right)f\left(6\right)f\left(8\right)f\left(10\right)f\left(12\right)}$$ which, when run in Desmos, returns 61. But obviously we can't evaluate this with brute force. The farthest I have reached is that $$n^4+n^2+1=n^2(n+1)+\frac{n+1}{n+1}=(n^2+\frac{1}{n+1})(n+1)$$ Any hints?
| We note that $n^4+n^2+1 = (n^4+2n^2+1)-n^2 = (n^2+1)^2-n^2 = (n^2+n+1)(n^2-n+1)$
Writing the value of $(n^2+n+1) = g(n)$, the value of $g(n-1) = n^2-n+1$
So we get eg $f(n)=g(n)g(n-1)$
And $\frac{f(3)f(5)f(7)...f(13)}{f(2)f(4)f(6)...f(12)}$
is equal to $\frac{g(2)g(3)g(4)...g(12)g(13)}{g(1)g(2)g(3)g(4)...g(12)} = \frac{g(13)}{g(1)}$
This evaluates to $(13*14+1)/(1*2+1)$ or $\frac{183}{3} = 61$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3952369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Algebra problem with isomorphism Let $G=(-1,1)$ and the binary operation such that
$x*y=\frac{x\sqrt{1-y^2}+y\sqrt{1-x^2}}{\sqrt{1-(xy)^2+2xy\sqrt{(1-x^2)(1-y^2)}}}$.
Prove that (G,*) it is a isomorph group with the group (R,+).
I tried to find the bijective function $f:G\to R$ such that $f(x*y)=f(x)+f(y)$ but I could not.
| Let $x=\sin\alpha$, $y=\sin\beta$, so $\alpha,\beta\in(-\pi/2,\pi/2)$. (always do such substitution when there is a $\sqrt{1-x^2}$)
Then $x*y = \frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\sqrt{1 - \sin^2\alpha\sin^2\beta + 2\sin\alpha\sin\beta\cos\alpha\cos\beta}} = \frac{\sin(\alpha+\beta)}{\sqrt{\cos^2\alpha\cos^2\beta + \sin^2(\alpha+\beta)}} = \frac{t}{\sqrt{1+t^2}} = \tanh\mathrm{asinh}\,t = g(t) $ where $t = \frac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta} = \tan\alpha+\tan\beta = \frac{x}{\sqrt{1-x^2}} + \frac{y}{\sqrt{1-y^2}} = f(x)+f(y) $.
(I have replaced $1$ in the denominator by $(\sin^2\alpha+\cos^2\alpha)(\sin^2\beta+\cos^2\beta)$ because we need something symmetric with respect to $\alpha\leftrightarrow\beta$, and we need to get rid of that complicated term with the factor $2$; we did it by including it into $\sin^2(\alpha+\beta)$.)
So $x*y=g(f(x)+f(y))$, and since in fact $g(f(x))=x$, $f$ and $g$ indeed give the desired isomorphism.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3958305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Markov's Chain stationary matrix determination I have the following transition matrix $P$
\begin{equation}
P=
\begin{pmatrix}
1/2 & 1/3 & 1/6\\
1/4 & 3/4 & 0\\
1/5 & 2/5 & 2/5
\end{pmatrix}
\end{equation}
I have tried to calculate the stationary matrix as
\begin{equation}
P^n= C D^n C^{-1}
\end{equation}
With $D=C^{-1} P C$ the diagonal matrix. Calculating $C$ with the eigenvalues and eigenvectors
\begin{equation}
C=
\begin{pmatrix}
v_1 & w_1 & z_1\\
v_2 & w_2 & z_2\\
v_3 & w_3 & z_3
\end{pmatrix}
\end{equation}
where the eigenvectors are
\begin{align}
\vec{v} =\langle (v_1,v_2,v_3)\rangle &= \langle(1,1,1)\rangle \\
\vec{w} =\langle(w_1,w_2,w_3)\rangle &= \langle\left(\frac{2i\sqrt{39}+3}{6},\frac{-3i\sqrt{39}-7}{16},1\right)\rangle \\
\vec{v} =\langle(v_1,v_2,v_3)\rangle &= \langle\left(\frac{-2i\sqrt{39}+3}{6},\frac{3i\sqrt{39}-7}{16},1\right)\rangle \\
\end{align}
so the diagonal matrix is
\begin{equation}
D=
\begin{pmatrix}
1 & 0 & 0\\
0 & \frac{-i\sqrt{39}+39}{120} & 0\\
0 & 0 & \frac{i\sqrt{39}+39}{120}
\end{pmatrix}
\end{equation}
how can i calculate $D^{n}$? is the diagonalization correct?
I tried by other way using $BP=B$ where B is the final vector state, then
\begin{align}
[p_1 \quad p_2 \quad p_3] \begin{pmatrix}
1/2 & 1/3 & 1/6\\
1/4 & 3/4 & 0\\
1/5 & 2/5 & 2/5
\end{pmatrix} = [p_1 \quad p_2 \quad p_3]
\end{align}
using Gauss-Jordan elimination for solving this system it gives me infinite solutions.
I dont know if it is a problem with the transition matrix.
| Since $1$ is the dominant eigenvalue of the Markov Chain (the derivation of that fact can be found elsewhere, thus I leave that up to the reader to investigate if desired), all we need to solve is
$$ \begin{bmatrix} 1/2 & 1/4 & 1/5 \\1/3 & 3/4 & 2/5 \\ 1/6 & 0 & 2/5 \end{bmatrix} \begin{bmatrix} x \\y \\ z \end{bmatrix}= \begin{bmatrix} x \\y \\ z \end{bmatrix}$$
Solving gives infinite (why?) solutions as follows: $x=3.6z,y=6.4z$ with $z$ free. Using the constraint $x+y+z=1$ we find $z=18/55,y=32/55,z=1/11$ and thus we find for the Stationary matrix: $$ \begin{bmatrix} 18/55 & 18/55 & 15/55 \\32/55 & 32/55 & 32/55 \\ 1/11 & 1/11 & 1/11 \end{bmatrix}$$ The rudiments of the calculations is nothing but row operations (Gauss Elimination)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3959859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric problem (problem from a Swedish 12th grade ‘Student Exam’ from 1932) The following problem is taken from a Swedish 12th grade ‘Student Exam’ from 1932.
The sum of two angles are $135^\circ$ and the sum of their tangents are $5$. Calculate the angles.
Is there a shorter/simpler solution than the one presented below that I made some months ago? It seems rather ‘lengthy’.
Solution
Let the angles be $\alpha$ and $\beta$.
We have
$$
\left\{
\begin{aligned}
\alpha+\beta&=135°,\\
\tan(\alpha)+\tan(\beta)&=5.
\end{aligned}
\right.
$$
Since
$$
\tan(x)+\tan(y)
=\frac{\sin(x)}{\cos(x)}+\frac{\sin(y)}{\cos(y)}
=\frac{\sin(x)\cos(y)+\cos(x)\sin(y)}{\cos(x)\cos(y)}
=\frac{\sin(x+y)}{\cos(x)\cos(y)}
$$
we have
$$
5
=\tan(\alpha)+\tan(\beta)
=\frac{\sin(\alpha+\beta)}{\cos(\alpha)\cos(\beta)}
=\frac{\sin(135°)}{\cos(\alpha)\cos(\beta)}
=\frac{\frac{1}{\sqrt{2}}}{\cos(\alpha)\cos(\beta)}
$$
which gives
$$
\cos(\alpha)\cos(\beta)=\tfrac{1}{5\sqrt{2}}.
$$
Further
$$
-\tfrac{1}{\sqrt{2}}
=\cos(135°)
=\cos(\alpha+\beta)
=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)
=\tfrac{1}{5\sqrt{2}}-\sin(\alpha)\sin(\beta)
$$
which gives
$$
\sin(\alpha)\sin(\beta)
=\tfrac{1}{5\sqrt{2}}+\tfrac{1}{\sqrt{2}}
=\tfrac{6}{5\sqrt{2}}.
$$
Since $\alpha+\beta=135°$ we have
\begin{align*}
\sin(\alpha)\sin(\beta)
&
=\sin(\alpha)\sin(135°-\alpha)
\\&=\sin(\alpha)\bigl(\sin(135°)\cos(\alpha)-\cos(135°)\sin(\alpha)\bigr)
\\&=\sin(\alpha)\bigl(\tfrac{1}{\sqrt{2}}\cos(\alpha)+\tfrac{1}{\sqrt{2}}\sin(\alpha)\bigr)
\\&=\tfrac{1}{\sqrt{2}}\sin(\alpha)\bigl(\cos(\alpha)+\sin(\alpha)\bigr)
\\&=\tfrac{1}{\sqrt{2}}\bigl(\sin(\alpha)\cos(\alpha)+\sin^2(\alpha)\bigr)
\\&=\tfrac{1}{\sqrt{2}}\Bigl(\tfrac{1}{2}\sin(2\alpha)+\tfrac{1}{2}\bigl(1-\cos(2\alpha)\bigr)\Bigr)
\\&=\tfrac{1}{2\sqrt{2}}\bigl(\sin(2\alpha)+1-\cos(2\alpha)\bigr)
\end{align*}
why
\begin{gather*}
\tfrac{6}{5\sqrt{2}}=\tfrac{1}{2\sqrt{2}}\bigl(\sin(2\alpha)+1-\cos(2\alpha)\bigr)
\\\quad\Leftrightarrow\quad
\tfrac{12}{5}=\sin(2\alpha)+1-\cos(2\alpha)
\\\quad\Leftrightarrow\quad
\tfrac{7}{5}=\sin(2\alpha)-\cos(2\alpha)
\\\quad\Leftrightarrow\quad
\tfrac{7}{5}=\sqrt{2}\sin(2\alpha-\tfrac{\pi}{4})
\\\quad\Leftrightarrow\quad
\tfrac{7}{5\sqrt{2}}=\sin(2\alpha-\tfrac{\pi}{4})
\end{gather*}
which gives
\begin{gather*}
2\alpha-\tfrac{\pi}{4}=
\begin{cases}
\arcsin(\tfrac{7}{5\sqrt{2}})+n_12\pi\\
\pi-\arcsin(\tfrac{7}{5\sqrt{2}})+n_22\pi
\end{cases}
\\\quad\Leftrightarrow\quad
\alpha=
\begin{cases}
\tfrac{1}{2}\bigl(\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{4}+n_12\pi\bigr)\\
\tfrac{1}{2}\bigl(\pi-\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{4}+n_22\pi\bigr)
\end{cases}
\\\quad\Leftrightarrow\quad
\alpha=
\begin{cases}
\tfrac{1}{2}\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{\pi}{8}+n_1\pi\\
-\tfrac{1}{2}\arcsin(\tfrac{7}{5\sqrt{2}})+\tfrac{5\pi}{8}+n_2\pi
\end{cases}
\end{gather*}
where $\beta=135°-\alpha=\frac{3\pi}{4}-\alpha$ and $n_1,n_2\in\mathbb{Z}$, and vice versa since the problem is ‘symmetrical’.
The original exam
| Recall that $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$. Now, given $\alpha+\beta=135^{\circ}$ and $\tan(\alpha+\beta)=\tan 135^\circ=-1$, you get $\frac{5}{1-\tan\alpha\tan\beta}=-1$, i.e. $\tan\alpha\tan\beta=6$. Now, knowing the sum ($5$) and the product ($6$) of $\tan\alpha$ and $\tan\beta$, using Vieta formulas we conclude that $\tan\alpha$ and $\tan\beta$ are the solutions of the quadratic equation $x^2-5x+6=0$, i.e. $\tan\alpha$ and $\tan\beta$ are $2$ and $3$ in some order. This is now very easy to finish off: $\alpha=\arctan 2+n\cdot 180^{\circ}$ and $\beta=135^{\circ}-\alpha=\arctan 3-n\cdot 180^{\circ}$, or vice versa.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Is $\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$ a rational number? Is there a way to show that
$$\alpha=\sqrt[3]{7+5 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}$$
is a rational number?
I found $\alpha=3$ from doing simplifications. But, I would like to known a different approach.
| Anay's idea of cubing both sides looks promising, but there is a little subtlety in the method. Let's explore.
Start with
$\alpha=\sqrt[3]{7+5\sqrt2}+\sqrt[3]{20-14\sqrt2}$
Cube with the Binomial Theorem:
$\alpha^3=\color{blue}{7+5\sqrt2}+\color{brown}{3(\sqrt[3]{7+5\sqrt2})^2\sqrt[3]{20-14\sqrt2}+3\sqrt[3]{7+5\sqrt2}(\sqrt[3]{20-14\sqrt2})^2}+\color{blue}{20-14\sqrt2}$
Combine the terms that are the same color in the above expression:
$\alpha^3=\color{blue}{27-9\sqrt2}+\color{brown}{3\sqrt[3]{7+5\sqrt2}\sqrt[3]{20-14\sqrt2}(\sqrt[3]{7+5\sqrt2}+\sqrt[3]{20-14\sqrt2})}$
$\alpha^3=27-9\sqrt2+3\sqrt[3]{2\sqrt2}\alpha$
Render $\sqrt[3]{2\sqrt2}=\sqrt2$:
$\alpha^3-3\sqrt2\alpha-27+9\sqrt2=0$
Now suppose that $\alpha$ is rational in this final equation. Then the left side is rational only if $\sqrt2$ has a zero overall coefficient, meaning $3\alpha-9=0$ or $\alpha=3$ is the only possible rational root. So, we try this value and find that the equation holds. Thus $\alpha=3$ checks out.
We have the possibility that even though $\alpha=3$ is a root of the cubic equation, it may not be the original sum of the cube roots with which we started. To check that, divide the cubic by $\alpha-3$ to get a quadratic equation whose remaining real roots are other possibilities for $\alpha$. This gives the factorization
$(\alpha-3)(\alpha^2+3\alpha+9-3\sqrt2)=0$
The discriminant of the quadratic factor is $-27+12\sqrt2=-\sqrt{729}+\sqrt{288}<0$. Thereby $\alpha=3$ is the only real root and must be the actual sum $\sqrt[3]{7+5\sqrt2}+\sqrt[3]{20-14\sqrt2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3962831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving a convoluted proof to an inequality: $x,y>0$ such that $x^2+y^2=1$. Prove that, $x^3+y^3 \geqslant \sqrt2 xy $ As the title described, I was trying to find an alternative proof to
If $x,y$ are positive numbers such that $x^2 + y^2=1$, prove that $x^3 + y^3 \geqslant \sqrt2 xy $.
Here's the proof that I've found (I'm sorry, I forgot where I got it):
Apply the Chebyshev's inequality on the tuplets $(x^2, y^2)$ and $\left( \frac1y, \frac1x\right)$, we have $$ \frac12 \left( \frac{x^2}y + \frac{y^2}x \right) \geqslant \frac{x^2 + y^2}2 \cdot \frac{1/y + 1/x}2 \quad \Rightarrow \quad \frac{x^2}y + \frac{y^2}x \geqslant \frac12 \left(\frac1x + \frac1y \right)$$ Apply AM-HM inequality on the tuplets $(x,y)$, we have $$ \frac{x+y}2 \geqslant \frac2{1/x + 1/y} \quad \Rightarrow \quad \frac1x + \frac1y \geqslant \frac4{x+y} $$ Apply Cauchy-Schwartz inequality on the tuplets $(x,y)$ and $(1,1)$, we have $$ x + y = x \cdot 1 + y\cdot 1 \leqslant \sqrt{x^2 + y^2} \sqrt2 = \sqrt2 $$ Combining these 3 inequalities above yield $$ \dfrac{x^2}y + \dfrac{y^2}x \geqslant \frac12 \cdot \frac4{x+y} \geqslant \frac12 \cdot \frac4{\sqrt2} = \sqrt2 $$ The result follows.
Now since I love to punish myself, I tried to find a harder proof as such:
We can let $(x,y) = (\cos\theta, \sin\theta) $, where $\theta \in (0, \tfrac\pi2) $. The inequality in question becomes $$ \begin{array} {l c l }
\cos^3 \theta + \sin^3 \theta &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(\cos^2 \theta - \sin \theta \cos \theta + \sin^2\theta) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
(\cos\theta + \sin\theta)(1 - \sin \theta \cos \theta ) &\geqslant &\sqrt2 \cos \theta \sin \theta \\
\cos\theta + \sin\theta &\geqslant & \cos \theta \sin \theta ( \sqrt2 + \cos \theta + \sin \theta) \\
\dfrac1{\sin\theta \cos\theta} &\geqslant & \dfrac{\sqrt2}{\cos \theta \sin \theta} + 1 \\
\end{array}$$ Apply Weierstrass substitution ($t = \tan\frac\theta2$, where $0<t<1$) yields $$ \dfrac{(1+t^2)^2}{2t(1-t^2)} \geqslant \dfrac{\sqrt2 (1+t^2)}{(1-t^2) +2t} + 1
$$ which simplifies to $$ - \dfrac{t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1}{ 2t(t-1)(t+1) (t^2 - 2t - 1)} \geqslant 0 $$ or $$ t^6 - 2\sqrt2 t^5 - 3t^4 -8t^3 + 3t^2 + 2\sqrt2 t - 1 \leqslant 0, \quad\quad\quad 0<t<1$$
Now how do I prove the sextic polynomial inequality above (which is true)?
| Here a relatively "natural" way to show the inequality. It avoids a time consuming (or rather time wasting) analysis of a quite specific 6th degree polynomial where the amount of learnings and insights to be expected doesn't seem to justify the effort.
To show is equivalently
$$\frac{x^2}y + \frac{y^2}x \geq \sqrt 2$$
\begin{eqnarray*}\frac{x^2}y + \frac{y^2}x
& \stackrel{\frac 1t \text{convex for }t>0}{\geq} & \frac 1{x^2y+y^2x}\\
& = & \frac 1{xy(x+y)}\\
& \stackrel{GM-AM: xy\leq \frac 12}{\geq} & \frac 2{x+y}\\
& \stackrel{C.S.:x+y\leq \sqrt 2}{\geq} & \frac 2{\sqrt 2} = \sqrt 2
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3965251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 10,
"answer_id": 0
} |
Find a basis of the generated subspace Consider the vectors
$$
v_1 =\begin{pmatrix}
1\\ 0\\ 0\\ 1
\end{pmatrix}, \quad v_2=\begin{pmatrix}
\sqrt{2}\\ 1\\ 2\\0
\end{pmatrix}, \quad v_3 =\begin{pmatrix}
\sqrt{2}+2\\ 1\\ 2\\ 2
\end{pmatrix}
$$
and let $S$ be the subspace generated by $v_1, v_2, v_3$. Find the dimension of $S$ and a basis for $S$.
${\bf My \ solution:}$ the dimension of $S$ is clearly $2$ since since $v_1, v_2, v_3$ are linearly dependent (or, better, the rank of the matrix whose columns are the three vectors is $2$). In order to find a basis I use the echelon form. It is
$$
\begin{pmatrix}
1 & \sqrt{2} &\sqrt{2}+2\\
0&1&1\\
0&2 &2\\
1&0&2
\end{pmatrix}\xrightarrow{}\begin{pmatrix}
1 & \sqrt{2} &\sqrt{2}+2\\
0&1&1\\
0&0&0\\
0&0&0
\end{pmatrix}
$$
which means that a basis for $S$ is given by the vectors $v_1$ and $v_2$.
Is my solution correct or am I missing something?
Thank you in advance!
${\bf EDIT:}$ My steps for row echelon form:
[
\begin{split}
&\begin{pmatrix}
1 & \sqrt{2} &\sqrt{2}+2\\
0&1&1\\
0&2 &2\\
1&0&2
\end{pmatrix}\xrightarrow{R_1-R_3}\begin{pmatrix}
1 & \sqrt{2} &\sqrt{2}+2\\
0&1&1\\
0&2&2\\
0&\sqrt{2} &\sqrt{2}
\end{pmatrix}\xrightarrow[R_3\leftrightarrow R_4]{R_3-2R_2}\\[5pt] &\begin{pmatrix}
1 & \sqrt{2} &\sqrt{2}+2\\
0&1&1\\
0&\sqrt{2}&\sqrt{2}\\
0&0&0
\end{pmatrix}\xrightarrow{R_3-\sqrt{2}R_2}\begin{pmatrix}
1 & \sqrt{2} &\sqrt{2}+2\\
0&1&1\\
0&0&0\\
0&0&0
\end{pmatrix}
\end{split}
]
| It's correct, however an overkill.
You observed correctly that $v_3=2v_1+v_2$, so they are linearly dependent, but clearly $v_1,v_2$ are not parallel, hence independent, so the dimension they span is indeed $2$.
But then any $2$ independent elements of the span form a basis, so just take $v_1$ and $v_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Expressing $ \frac{\cos x + \sin x}{\cos x - \sin x} $ as a single trigonometric function For a while now, I have been trying to get this expression into 1 term instead of the 2 term fraction.
$$
\frac{\cos x + \sin x}{\cos x - \sin x}
$$
The most I have gotten down to is get it to a single ratio, ie,
$$
\frac{1 + \tan x}{1 - \tan x}
$$
I have been looking for an identity or formula of some sort to factorize $ \cos x + \sin x $ and $ \cos x - \sin x $.
Any help will be appreciated.
Thanks
| $
\dfrac{\cos x + \sin x}{\cos x - \sin x} = \dfrac{1+ \tan x}{1 - \tan x}
$
Now, $\tan (x+y) = \dfrac{\tan x+ \tan y}{1 - \tan x \tan y}$
Therefore, $\tan (x+ \frac{\pi}{4}) = \dfrac{\tan x+ 1}{1 - \tan x}$
Therefore, $\dfrac{\cos x + \sin x}{\cos x - \sin x} = \tan(x+ \frac{\pi}{4})$
Is this what you meant by 1 term?.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3967860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
simplify fraction (stone drops off a cliff) How do I get from
$$ \frac {dt} {T} = \frac {dx} {gt} \sqrt { \frac {g} {2h} } $$
to
$$ \frac {dt} {T} = \frac {1} {2 \sqrt {hx} } dx $$
where $x(t) = \frac {1} {2} gt^2 $ and $ T = \sqrt { \frac {2h} {g}}$.
I'm currently struggling with Griffiths' Introduction to Quantum Mechanics. This is from a worked example on p.11-12 of the 2nd edition.
| We have
$$
x=\frac{1}{2}gt^2
\quad\Leftrightarrow\quad
gt^2=2x
$$
so
\begin{align*}
\frac{dt}{T}
&
=\frac{dx}{gt}\sqrt{\frac{g}{2h}}
=\sqrt{\frac{g}{g^2t^2\cdot2h}}\,dx
=\sqrt{\frac{1}{gt^2\cdot2h}}\,dx
\\&
=\sqrt{\frac{1}{2x\cdot2h}}\,dx
=\sqrt{\frac{1}{4hx}}\,dx
=\frac{1}{\sqrt{4hx}}\,dx
=\frac{1}{2\sqrt{hx}}\,dx.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A simple but nonsense infinite sums rule Assume $a>0$ and $0<r<1$ are given. The geometric series is known to be $\sum_0^{\infty} ar^n = \frac{a}{1-r}$. Now differentiate w.r.t $r$ to obtain $\sum_0^{\infty}anr^n = \frac{ar}{(1-r)^2}$. So far we're cool.
Now take the two series $$S_1 = \sum_n \frac{1}{2^n}$$ and $$S_2 = \sum_n \frac{n}{2^n}.$$ One would trivially expect $S_2>S_1$, because obviously $S_2$ has an additional $n$ multiplied by each term. To my surprise, this is not the case! Invoking the result of the previous paragraph, you get
$$S_1 = \frac{1}{1-1/2} = 2$$
and
$$ S_2 = \frac{1/2}{1/4} = 2$$
so that $S_1 = S_2$.
Does this make any sense to you? Because it sure makes no sense to me. Or am I losing it?
| Yes, you are right $S_1 = S_2 = 2$ and it might be somewhat astonishing the first time. But first let's have a look at your derivation which needs to be somewhat revised. We can take $a=1$ and consider
\begin{align*}
S_1(r)=\sum_{n=0}^\infty r^n=\frac{1}{1-r}\tag{1}
\end{align*}
Derivation with respect to $r$ gives
\begin{align*}
\frac{d}{dr}S_1(r)&=\frac{d}{dr}\frac {1}{1-r}=\frac{1}{(1-r)^2}\tag{2}
\end{align*}
Note we have numerator $1$ and not $r$. On the other hand we also obtain
\begin{align*}
\frac{d}{dr}S_1(r)&=\frac{d}{dr}\sum_{n=0}^\infty r^n=\sum_{n=0}^\infty nr^{n-1}=\sum_{n=1}^\infty nr^{n-1}\\
&=\sum_{n=0}^\infty (n+1)r^n\\
&=\sum_{n=0}^\infty nr^n+\sum_{n=0}^\infty r^n\\
&=\sum_{n=0}^\infty nr^n+\frac{1}{1-r}\\
\sum_{n=0}^\infty nr^n& = \frac{d}{dr}S_1(r) - \frac{1}{1-r}=\frac{1}{(1-r)^2}-\frac{1}{1-r}\tag{3}
\end{align*}
We obtain from (1) - (3) by substituting $r=\frac{1}{2}$:
\begin{align*}
\color{blue}{S_1\left(\frac{1}{2}\right)}&=\frac{1}{1-\frac{1}{2}}\,\color{blue}{=2}\\
\color{blue}{S_2}&=\sum_{n=0}^\infty \frac{n}{2^n}
=\frac{1}{\left(1-\frac{1}{2}\right)^2}-\frac{1}{1-\frac{1}{2}}=4-2\,\,\color{blue}{=2}
\\
\end{align*}
We observe the term with $n=0$ in $S_2$ is zero, but the numerator $n$ with index $n>0$ is a compensation for it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3969800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Problem proving $\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx$ I am trying to show that the value of the following integral is:
$$\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx=2^{n-1}\pi^{\frac{1}{2}}(2n+1)n!\delta_{mn}+2^n\pi^{\frac{1}{2}}(n+2)!\delta_{(n+2)m}+2^{n-2}\pi^{\frac{1}{2}}n!\delta_{(n-2)m},$$
where $H_n$ is the $n$-th Hermite polynomial.
My attempt is the following:
Suppose that the value of the integral is already known and that:
\begin{equation*}
\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx=\lambda
\end{equation*}
From the recurrence relation we have to:
\begin{equation*}
H_{n+1}(x)=2xH_{n}(x)-2nH_{n-1}(x)
\end{equation*}
Therefore, since it is valid for any $ n $ then it will also be valid for any $ m $, that is:
\begin{equation*}
H_{m+1}(x)=2xH_m(x)-2mH_{m-1}(x)
\end{equation*}
Solving for the terms $2xH_ {m} (x)$ and $2xH_ {n} (x)$ we have:
\begin{align}
2xH_ {n} (x) & = H_ {n + 1} (x) + 2nH_ {n-1} (x) \\
2xH_ {m} (x) & = H_ {m + 1} (x) + 2mH_ {m-1} (x)
\end{align}
And multiplying the \ textcolor {red} {\ textbf {equation 1}} by $ 2xH_m (x) $ and substituting the cleared value we have:
\begin{align*}
4x^2H_{n}(x)H_{m}(x)&=2xH_m(x)\left[H_{n+1}(x)+2nH_{n-1}(x)\right]\\
4x^2H_{n}(x)H_{m}(x)&=\left[H_{m+1}(x)+2mH_{m-1}(x)\right]\left[H_{n+1}(x)+2nH_{n-1}(x)\right]\\
4x^2H_{n}(x)H_{m}(x)&= H_{m+1}(x)H_{n+1}(x)+2nH_{m+1}(x)H_{n-1}(x)+2mH_{m-1}(x)H_{n+1}(x)+4mnH_{m-1}(x)H_{n-1}(x)\\
\end{align*}
And from what we have just found we can conclude that:
{\begin{align*}
4\lambda&=\int_{-\infty}^{\infty}e^{-x^2}\left[H_{m+1}(x)H_{n+1}(x)\right]dx+\int_{-\infty}^{\infty}e^{-x^2}2nH_{m+1}(x)H_{n-1}(x)dx+\int_{-\infty}^{\infty}2me^{-x^2}H_{m-1}(x)H_{n+1}(x)+\int_{-\infty}^{\infty}4mne^{-x^2}H_{m-1}(x)H_{n-1}(x)dx
\end{align*}
So going back to the integral that we already had:
\begin{align*}
\cfrac{1}{4}\times4\lambda&=\cfrac{1}{4}\times \sqrt{\pi}\left\{2^{n+1}(n+1)!\left[\delta_{(m+1)(n+1)}+2m\delta_{(m-1)(n+1)}\right]+2^{n-1}(n-1)!\left[\delta_{(m+1)(n-1)}2n+4mn\delta_{(m-1)(n-1)}\right]\right\}\\
\lambda&=\cfrac{\sqrt{\pi}}{4}\left\{2^{n+1}(n+1)!\left[\delta_{(m+1)(n+1)}+2m\delta_{(m-1)(n+1)}\right]+2^{n-1}(n-1)!\left[\delta_{(m+1)(n-1)}2n+4mn\delta_{(m-1)(n-1)}\right]\right\}\\
\end{align*}
And as it was supposed from the beginning:
\begin{equation*}
\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx=\lambda
\end{equation*}
Then finally it will come to:
$\int_{-\infty}^{\infty}x^2e^{-x^2}H_n(x)H_m(x)dx=\cfrac{\sqrt{\pi}}{4}\left\{2^{n+1}(n+1)!\left[\delta_{(m+1)(n+1)}+2m\delta_{(m-1)(n+1)}\right]+2^{n-1}(n-1)!\left[\delta_{(m+1)(n-1)}2n+4mn\delta_{(m-1)(n-1)}\right]\right\}$
But I can't find the way to give the answer $2^{n-1}\pi^{\frac{1}{2}}(2n+1)n!\delta_{mn}+2^n\pi^{\frac{1}{2}}(n+2)!\delta_{(n+2)m}+2^{n-2}\pi^{\frac{1}{2}}n!\delta_{(n-2)m}$
| Starting from here
$$
\cfrac{\sqrt{\pi}}{4}\left\{2^{n+1}(n+1)!\left[\delta_{(m+1)(n+1)}+2m\delta_{(m-1)(n+1)}\right]+2^{n-1}(n-1)!\left[\delta_{(m+1)(n-1)}2n+4mn\delta_{(m-1)(n-1)}\right]\right\}
$$
First, rewrite the $\delta$s so that they all have only $m$. We do this by noting that, for example, if $m-1 = n+1$, then $m = n+2$, so $\delta_{(m-1)(n+1)} = \delta_{m(n+2)}$. We can then also use $m\delta_{m(n+2)} = (n+2)\delta_{m(n+2)}$ to further simplify. This gives
$$
\cfrac{\sqrt{\pi}}{4}\left\{2^{n+1}(n+1)!\left[\delta_{mn}+2(n+2)\delta_{m(n+2)}\right]+2^{n-1}(n-1)!\left[\delta_{m(n-2)}2n+4n^2\delta_{mn}\right]\right\}
$$
Now expand the products:
$$
\sqrt{\pi}\left[2^{n-1}(n+1)!\delta_{mn}+2^{n}(n+2)(n+1)!\delta_{m(n+2)}+2^{n-2}n(n-1)!\delta_{m(n-2)}+2^{n-1}n^2(n-1)!\delta_{mn}\right]
$$
Finally, collect terms and simplify the factorials to get
$$
\sqrt{\pi}\left[2^{n-1}(2n+1)n!\delta_{mn}+2^{n}(n+2)!\delta_{m(n+2)}+2^{n-2}n!\delta_{m(n-2)}\right]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3972541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the integer solutions to $4^x - 9^y = 55$ I want to find the integer solutions of:
$$ 4^x - 9^y = 55$$
For now, I see that $x = 3, y = 1$ is an integer solution to the equation. How can I rigorously prove there are no other solutions for $x, y$ integers?
I tried to solve for $y$, but to no avail. WolframAlpha tells me that it is the only solution, but it doesn't provide an explanation.
| I like the explanations given in @MishaLavrov's answer, and I understand this only as a comment to his answer.
There, the second sentence introduces a bit obfuscation, I think: the bare idea of needing a crosscheck of $\{0,1,2,3,4,5\} \times
\{0,1,2,3\}$ insertions is somehow surprising (and misleading) after the introductory statement.
$$4^a - 9^b = 55 $$
This can be factored, and the factors of the lhs can be crosschecked with the two factorizings of the rhs, where the obvious sizes of the factors in the lhs are brought in agreement with the order of sizes of the factors of the rhs:
$$
(2^a+3^b)(2^a-3^b)= \begin{cases} (11) \cdot (5) & (a)\\ (55) \cdot (1) &(b) \end{cases} $$
Test $(a)$:
$$ (2^a+3^b)-(2^a-3^b)=2\cdot 3^b \overset?= 11-5 = 6 = 2\cdot 3^1 \implies b=1\\
(2^a+3^b)+(2^a-3^b)=2\cdot 2^a\overset?= 11+5 = 16 = 2 \cdot 2^3 \implies a=3$$
$\qquad \qquad$ both ways of summing the corresponding factors give valide solutions.
Test $(b)$:
$$ (2^a+3^b)-(2^a-3^b)=2\cdot 3^b\overset?= 55-1 = 54 = 2\cdot 3^3 \implies b=3\\
(2^a+3^b)+(2^a-3^b)=2\cdot 2^a\overset?= 55+1 = 56 = 2 \cdot 2^2 \cdot 7 \implies \text{error}
$$
$\qquad \qquad$ the second way of summing the corresponding factors does not agree with the problem.
Conclusion: We have only one solution: $4^3 - 9^1 = 55$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3973084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Some horrid integrals I am confused with $$(1) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} + 1}{x^3 + 1}}~~dx$$
Speaking as an example on (1).
The only thing I could do from here is only to do a u-sub:
$$ u = \ln(x) + 1 \\ x = e^{u-1} \\ du = \frac{1}{x} dx \\ dx = x (du) \Rightarrow dx = e^{u-1} (du) $$
And so this becomes:
$$ \int \sqrt{ \frac{u}{e^{3u-3} + 1} } ~~ du \\ $$
And from here it's basically a deadend.. I would appreciate if you could give some insights..
Interestingly, these similar integrals have interesting solutions:
$$(2) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x} +1}{x^3}}~~dx = \sqrt{2 e \pi}$$
$$(3) ~~~~ \int_{1/e}^{\infty} \sqrt{\frac{\ln{x}}{x^3 + 1}}~~dx = \text{a complex number ?!}$$
(2) - where and why does the pi comes into play? this looks like an interesting solution..
(3) - why is the solution a complex number if the area is right infront my eyes, and it is real just like every other areas? What is happening here?
Mystery solved about the complex number - the lower bound should be $1$ and not $1/e$ my bad! But this is not the main mystery ;)
Thank you :-)
| Concerning $(3)$ $$\int_{\frac 1e}^{\infty} \sqrt{\frac{\log{(x)}}{x^3 + 1}}\,dx $$ it is normal to get a complex result.
Around $x=\frac 1e$ we have
$$\sqrt{\frac{\log{(x)}}{x^3 + 1}}=i\, \sum_{n=0}^\infty a_n \left(x-\frac{1}{e}\right)^n$$ and the first coefficients are
$$a_0=\frac{ e^{3/2}}{\left(1+e^3\right)^{1/2}}\qquad a_1 =-\frac{e^{5/2} \left(4+e^3\right)}{2 \left(1+e^3\right)^{3/2}}\qquad a_2 =\frac{e^{7/2} \left(22-4 e^3+e^6\right)}{8 \left(1+e^3\right)^{5/2}}$$ So
$$\int_{\frac 1e}^{1} \sqrt{\frac{\log{(x)}}{x^3 + 1}}\,dx = i\, \sum_{n=0} ^\infty \frac {(e - 1)^{n + 1} } {(n+1) e^{n+1} }a_n$$ using the given terms, this leads to
$$\int_{\frac 1e}^{1} \sqrt{\frac{\log{x}}{x^3 + 1}}\,dx \sim 0.373129 \,i$$
In French, we have an expression which says "This, Sir, is the cause of your daughter's being dumb". In French, this simply means "This explains that !"
Edit
Consider
$$\sqrt{\frac{\log{(x)}}{x^3 + a}}=\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n} x^{-3 n-\frac{3}{2}} \sqrt{\log (x)} \,a^n$$ This gives
$$\int\sqrt{\frac{\log{(x)}}{x^3 + a}}\,dx=$$
$$\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n} \left(\frac{\sqrt{2 \pi } \text{erf}\left(\sqrt{3
n+\frac{1}{2}} \sqrt{\log (x)}\right)}{(6 n+1)^{3/2}}-\frac{2 x^{-3
n-\frac{1}{2}} \sqrt{\log (x)}}{6 n+1}\right)\,a^n$$
$$\int_1^\infty\sqrt{\frac{\log{(x)}}{x^3 + a}}\,dx=\sqrt{2\pi}\sum_{n=0}^\infty \frac{\binom{-\frac{1}{2}}{n}}{(6 n+1)^{3/2}}\,a^n$$
For $a=1$, computing the partial sums
$$S_p=\sqrt{2\pi}\sum_{n=0}^p\frac{\binom{-\frac{1}{2}}{n}}{(6 n+1)^{3/2}}$$
$$\left(
\begin{array}{cc}
p & S_p \\
10 & 2.45333 \\
20 & 2.45303 \\
30 & 2.45297 \\
40 & 2.45294 \\
50 & 2.45293 \\
60 & 2.45293 \\
70 & 2.45292
\end{array}
\right)$$
which is the result given by numerical integration (this number is not recognized by inverse symbolic calculators).
This sum is alternating and we have
$$\frac{S_{p+2}}{S_p}=1-\frac{4}{p}+\frac{51}{4 p^2}+O\left(\frac{1}{p^3}\right)$$ which explains the rather slow convergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3975665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 0
} |
Find all positive integer solutions for the following equation $5^a + 4^b = 3^c$:
Find all positive integer solutions for the following equation:
$$5^a + 4^b = 3^c$$
My first guess would be to study the equation in mod, but I tried modulo 3, 4, 5, and 9 and I can't find anything.
| Another route: Follow the reasoning in Aqua's answer up to $5^a=(3^d-2^b)(3^d+2^b)$.
Then ask if $3^d-2^b$ and $3^d+2^b$ share any factors. If $m\mid (3^d-2^b) \land m \mid (3^d+2^b)$, then $m \mid (3^d-2^b)+(3^d+2^b)=2\cdot3^d$, which means $m\in\{1,2,3^x\}$. But $2\not \mid (3^d-2^b)$ and $3\not \mid (3^d-2^b)$, so $m=1$. In other words, $\gcd((3^d-2^b),(3^d+2^b))=1$
But the product of the two terms is a power of $5$, so one term must be exactly $1$ and the other term is the power of $5$. Plainly the smaller term will be $1$, so we solve $(3^d-2^b)=1$, which has only two solutions: $b=d=1$, and by Mihailescu's theorem $d=2,b=3$
Substituting into the other term, we find $3^1+2^1=5^1$ and $3^2+2^3=17 \ne 5^d$. So there is only one solution to the original equation: $5^1+4^1=3^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3976046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Unable to reconstruct a new polynomial to find a given value Suppose $f(x)= x^3+2x^2+3x+3$ and has roots $a , b ,c$.
Then find the value of
$\left(\frac{a}{a+1}\right)^{3}+\left(\frac{b}{b+1}\right)^{3}+\left(\frac{c}{c+1}\right)^{3}$.
My Approach :
I constructed a new polynomial $g(x) = f\left(\frac{x^{\frac{1}{3}}}{1-x^{\frac{1}{3}}}\right)$ and then used the Vieta's formula for the sum of roots taken one at a time to solve the sum.
But then I realised that I won't be able to do so as $g(x)$ is not a polynomial anymore.
Can anyone help me please.
| We shall make use of the following well-known identity:
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz).$$
Now, using Vieta's Formula, $a+b+c=-2, ab+ac=bc=3$, and $abc=-3$.
Thus, by direct expansion, we have that
$(a+1)(b+1)(c+1)=-1.$
\begin{align}
\dfrac{a}{a+1}+\dfrac{b}{b+1} + \dfrac{c}{c+1} &= 3- \left(\dfrac{1}{a+1} +
\dfrac{1}{b+1} + \dfrac{1}{c+1} \right) \\
& = 3- \dfrac{(b+1)(c+1) + (a+1)(c+1) + (b+1)(a+1)}{-1} \\
& = 3 + bc+b+c+1 + ac+ a + c+1 + ab+ a +b+1 \\
& = 5.
\end{align}
$$\dfrac{a}{a+1} \cdot \dfrac{b}{b+1} \cdot \dfrac{c}{c+1} = \dfrac{abc}{-1}=3.$$
\begin{align}
\dfrac{a}{a+1}\cdot \dfrac{b}{b+1} + \dfrac{a}{a+1}\cdot \dfrac{c}{c+1} + \dfrac{b}{b+1}\cdot \dfrac{c}{c+1} & = \dfrac{3(c+1)}{c} + \dfrac{3(b+1)}{b} + \dfrac{3(a+1)}{a} \\
& = 3 \left(3+ \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{a} \right) \\
& = 3 \left(3 + \dfrac{ab+bc+ca}{abc}\right) \\
& = 6.
\end{align}
Let $A=\dfrac{a}{a+1}, B=\dfrac{b}{b+1} , C=\dfrac{c}{c+1}$. Then,
\begin{align}
A^3 + B^3 + C^3 & = (A+B+C)(A^2+B^2+C^2-AB-BC-CA)+3ABC \\
& = 5 \left((A+B+C)^2-2(AB+BC+CA)-AB-BC-CA \right) + 3ABC& \\
& = 5 (25 -3(6))+3(3) \\
& = \mathbf{44}.
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/3976572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
A Negative binomial problem $P(X \ge 5)$ equals?
Consider a sequence of independent Bernoulli trials with the
probability of success in each trial being $\dfrac{1}{3}$. Let $X$
denote the number of trials required to get the second success. Then
$P(X \ge 5)$ equals.
$A=\dfrac{3}{7}$
$B=\dfrac{16}{27}$
$C=\dfrac{2}{3}$
$D=\dfrac{9}{13}$
This is a problem of negative binomial $r=2$
$P(X=x)= {x+r-1 \choose r-1}p^rq^x ;\ \ x=0,1,2..$
$P(X\ge5)=1-P(X < 5) \implies 1-(P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4))$
This is very time consuming how do I save time on this problem.
How do I utilize CDF of negative binomial distribution in this problem?
${\displaystyle k\mapsto 1-I_{p}(k+1,\,r),}$ the regularized incomplete beta function
| Easy is $B:\frac{16}{27}$
$$P(X\geq 5)=1-P(X\leq 4)=1-\binom{2-1}{2-1}\Bigg(\frac{1}{3}\Bigg)^2\cdot\Bigg(\frac{2}{3}\Bigg)^0-\binom{3-1}{2-1}\Bigg(\frac{1}{3}\Bigg)^2\cdot\Bigg(\frac{2}{3}\Bigg)^1-\binom{4-1}{2-1}\Bigg(\frac{1}{3}\Bigg)^2\cdot\Bigg(\frac{2}{3}\Bigg)^2=$$
$$=1-\frac{3}{27}-\frac{4}{27}-\frac{4}{27}=\frac{16}{27}$$
Note that there are at least 2 parametrizations of NBinomial. I used the one counting the trials before s successes whose pmf is
$$P(X=n)=\binom{n-1}{s-1}\theta^s(1-\theta)^{n-s}$$
Where the support is $n=s,s+1,s+2,...$
| {
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"timestamp": "2023-03-29T00:00:00",
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How many tangent lines to the curve $y = \frac{x}{x+1}$ pass through the point $(1,2)$? The question:
How many tangent lines to the curve $y = \frac{x}{x+1}$ pass through the point $(1,2)$? At which points do these tangent lines touch the curve?
My attempt:
To find the slope of the tangent line to the curve, we need to get the first derivative of $y = \frac{x}{x+1}$, which is $y' = \frac{1}{(x+1)^2}$.
So, the line equation of the tangent line is $y = m(x-1) + 2$, where $m = \frac{1}{(x+1)^2}$ for the given $x$.
We also know that the point where the tangent line intersects the curve must be both on the tangent line and the curve, so using that,
$$\frac{x}{x+1} = \frac{x-1}{(x+1)^2} + 2$$
$$\frac{x-1}{(x+1)^2} + 2 - \frac{x}{x+1} = 0$$
$$\frac{x - 1 + 2x^2 + 4x + 2 - x^2}{(x+1)^2} = 0$$
$$\frac{x^2 + 5x + 1}{(x+1)^2} = 0$$
And I'm stuck. I'd appreciate if you can point out where I'm wrong and how to better approach this problem.
| For a fraction $\frac{a}{b}$ to be equal to 0, we must have $a=0$. In this case we have $$x^2+4x+1=0$$ and the result follows.
EDIT:
Note that $$\frac{x-1}{(x+1)^2}+2-\frac{x}{x+1}=\frac{x^2+4x+1}{(x+1)^2}$$
You might wanna check your steps.
| {
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"timestamp": "2023-03-29T00:00:00",
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No. of ways of selecting 3 squares when they do not lie in same row, column, or diagonal
Total number of ways of selecting 3 small squares on a normal chess board so that they don’t belong to the same row, column or diagonal line, is equal to:
No. of ways of selecting 3 squares when they do not lie in same row or same column $=(64×49×36)×\frac{1}{3!}=18816$
Total ways of selecting the squares when they lie in the same diagonal line $=2({8 \choose 3} + {7 \choose 3} +{6 \choose 3}+{5 \choose 3} + {4 \choose 3}+ {3 \choose 3})=392$
But this only counts the cases when all three are in the same diagonal. How to count the cases when two of the three lie in the same diagonal?
| I will count the number of ways of selecting three named positions $X,Y,Z$ such that no two are in the same row or column under various conditions. The counts differ slightly according to the parity of $n$ and so I will assume that $n$ is even.
$A$. No other conditions.
$B$. $X$ and $Y$ are on the same diagonal.
$C$. $X,Y$ and $Z$ are on the same diagonal.
$D$. $XY$ and $XZ$ are on perpendicular diagonals.
Inclusion/exclusion
$B-C-2D$ counts the number of ways that $X$ and $Y$ are on the same diagonal and that $Z$ is not on the same diagonal as either.
$C+3D$ counts the number of ways that each of $X,Y,Z$ is on a diagonal with at least one other.
Then the total number of possibilities with no two of $X,Y,Z$ on the same diagonal is $$A-3(B-C-2D)-(C+3D)=A-3B+2C+3D.$$ Each configuration occurs $3!$ times because of the naming of the positions and so the answer requested in the post is given by $$\frac{1}{6}(A-3B+2C+3D).$$
A
$$A=n^2(n-1)^2(n-2)^2.$$
For $n=8$ this is $112896$.
B
The diagonals with one orientation are of lengths $1,2,...,n-1,n,n-1, ...,2,1$. Therefore the number of ways of choosing two positions on one of these diagonals is$${2 \choose 2} + {3 \choose 2} +...+{n-1 \choose 2}+{n \choose 2} + {n-1 \choose 2}+... {3 \choose 2}+{2\choose 2}=\frac{1}{6}n(n-1)(2n-1).$$ This must be multiplied by the number of orientations, $2$, the number of ways of naming $X/Y$, also $2$, and the number of ways of choosing the position of $Z$, $(n-2)^2$.
$$B=\frac{2}{3}n(n-1)(n-2)^2(2n-1) .$$
For $n=8$ this is $20160$.
C
The number of ways of choosing three positions on one of the sets of parallel diagonals is$${3 \choose 3} +...+{n-1 \choose 3}+{n \choose 3} + {n-1 \choose 3}+... {3 \choose 3}=\frac{1}{12}n(n-1)^2(n-2).$$
This must be multiplied by the number of orientations, $2$, and the number of ways of naming $X,Y,Z$, $6$. $$C=n(n-1)^2(n-2).$$ For $n=8$ this is $2352$.
D
The condition for no two of $X,Y,Z$ to be on the same row or column is simply that the legs of the L-shape must be unequal. There are then $8$ different orientations of the L-shape and $2$ ways of choosing $Y$ and $Z$.
I have not found a neat formula for the number of L-shapes with a particular orientation but it should be a reasonable challenge to find one. The shapes are however easy to count and for $n=8$ the number is $100$ (split into $49$ for one colour of squares and $51$ for the other).
$$D=1600.$$
Grand total
$$\frac{1}{6}(112896-3\times 20160+2\times 2352+3\times 1600)=10320.$$
Addendum
The formula for $D$ appears to be $\frac{1}{6}n(n-2)(5n^2-16n+8)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Evaluate $\int\frac{x^4}{(x^2-1)^3}dx$ I tried:$$I=\int\frac{x^4}{(x^2-1)^3}dx=\int\frac{(x^2-1)(x^2+1)+1}{(x^2-1)^3}dx=\int\frac{x^2+1}{(x^2-1)^2}+\frac{1}{(x^2-1)^3}dx$$
For first fraction we can write it as $\frac{1}{x^2-1}+\frac{2}{(x^2-1)^2}$. therefor we have:
$$I=\frac12\ln\left|\frac{x-1}{x+1}\right|+\int\frac{2}{(x^2-1)^2}dx+\int\frac{1}{(x^2-1)^3}dx$$
I don't know how to evaluate remaining integrals.
| Integrate by parts to reduce the remaining integrals
$$\int \frac{dx}{(x^2-1)^3}=\int \frac1{4x^3}d\left( \frac{-x^4}{(x^2-1)^2}\right)= -\frac14 \frac{x}{(x^2-1)^2}-\frac34 \int \frac{dx}{(x^2-1)^2}
$$
$$\int \frac{dx}{(x^2-1)^2}=\int \frac1{2x}d\left( \frac{-x^2}{x^2-1}\right)= -\frac12 \frac{x}{x^2-1}-\frac12\int \frac{dx}{x^2-1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3981505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the period of this sinusoid? I'm stuck trying to find the period of this sinusoid and would really like some pointers to different ways to approach this problem.
$$x(t) = \cos(\frac{4\pi t}{5})\sin^2(\frac{8\pi t}{3})$$
I would like to turn the product into a sum and do something about that $2$ in the exponent of $\sin$
$$x(t) = \cos(\frac{4\pi t}{5})\sin^2(\frac{8\pi t}{3}) = \cos(\frac{4\pi t}{5})\left[ 1 - \cos(\frac{16 \pi t}{3}) \right]\frac{1}{2} $$
$$ = \frac{1}{2}\left[ \cos(\frac{4\pi t}{5}) + \cos(\frac{4\pi t}{5}) \cos(\frac{16 \pi t}{3})\right]$$
$$ = \frac{1}{2}\left[ \cos(\frac{4\pi t}{5}) + \frac{1}{2}\cos(\frac{4\pi t}{5} + \frac{16 \pi t}{3}) \cos(\frac{16 \pi t}{3} - \frac{4\pi t}{5})\right]$$
So the sum of cosines will have a period that is the LCM of of the periods of the individuals terms we added. The angular frequencies are $\frac{4\pi}{5}, \frac{92\pi}{15}, \frac{68\pi}{15}$.
Converting this to periods -> we get $\frac{5}{2}, \frac{15}{46}, \frac{15}{34}$, but the answer is $\frac{30}{4}$
| To get the LCM, we want the smallest $T$ which is an integer multiple of the three periods. That is, we want integers $n_1,n_2,n_3$ such that $$T=\frac{5}{2}n_1=\frac{15}{46}n_2=\frac{15}{34}n_3$$ with $T$ as small as possible.
To this end we first focus on the middle equality, which rearranges to $23n_1=3n_2$. Since $23$ and $3$ are prime, this requires that $23$ divides $n_2$ and $3$ divides $n_1$. Hence there must be an integer $k$ such that $n_1=3k$ and $n_2=23k$.
With this, the third quantity of the period equality becomes redundant and we have $$T=\frac{15}{2}k=\frac{15}{34}n_3.$$
From this we deduce the value of $n_3$ and therefore know $T$ in terms of $k$. From this we deduce the smallest such value and conclude. (I leave the reader to fill in the details.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Justifying $\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$ for large $x$
I need to find a way to justify the inequality $$\frac{5x^2 + 3x + 1}{x^3 - x^2 - 1} < \frac{10}{x}$$ that holds for large values of x.
As an instance, I will justify an example inequality just to show you the intended strategy. Consider the inequality
$$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5}{x}.$$ For example, $2x^2 - 7$ can be written as $x^2 + (x^2 - 7)$. Because $x^2 - 7$ is a positive value for all $x < -\sqrt{7}$, removing it from the denominator makes the absolute value of the fraction greater. Also note that when $x < -\frac{21}{10}$, the numerator $|5x + 21| < 5|x|$, and this happens for all $x < -\sqrt{7}$. So $$\left|\frac{5x + 21}{2x^2 - 7}\right| < \frac{5|x|}{x^2} = \frac{5}{x}$$ as long as $x < -\sqrt{7}$.
| Hint:
Set
$$
f(x)=\frac{5x^3 + 3x^2 + x}{x^3 - x^2 - 1}.
$$
Then you have
$$
\lim_{x\to+\infty}f(x)=5\tag{1}
$$
Now use the limit definition to show that $f(x)<10$ for sufficiently large $x$. So you get
$$
\frac{f(x)}{x}<\frac{10}{x}
$$
for sufficiently large $x$.
Notes: The limit (1) implies that
$$
|f(x)-5|<1
$$
for sufficiently large $x$. So you have
$$
4<f(x)<6<10.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximum number of non-congruent triangles with side lengths integers and less than $9$
Find the maximum number of non-congruent triangles whose side lengths are integers and less than $9$.
I tried using brute force to count all the triangles but there are so many triangles that can be formed and I counted about $60$ of them but after that I gave up. Is there any systematic approach to question?
| We do some systematic counting.
First the isosceles (including equilateral) ones. If the base is $1$, other two sides can be any from $1$ to $8$. If base is $2$, other two can be any from $2$ to $8$. If base is $3$ - $2$ to $8$. Base $4$ - $3$ to $8$. Base $5$ - $3$ to $8$. Base $6$ - $4$ to $8$. Base $7$ - $4$ to $8$. Base $8$ - $5$ to $8$. Total $$8+7+7+6+6+5+5+4=48$$
For scalene, we write the triplet of side lengths in decreasing order respecting triangle inequality.
*
*$[8,7,(6 \,\text{to}\, 2)]. [8,6,(5\,\text{to}\, 3)]. [8, 5, 4]. [8,4,✗].$
*$[7,6,(5 \,\text{to}\, 2)]. [7,5,(4 \,\text{to}\, 3)]. [7, 4, ✗].$
*$[6,5,(4 \,\text{to}\, 2)]. [6,4,3]. [6, 3, ✗].$
*$[5,4,(3 \,\text{to}\, 2)]. [5,3,✗].$
*$[4,3,2]. [4,2,✗].$
*$[3,2,✗].$
Sum : $5+3+1+4+2+3+1+2+1=22$
Grand total : $\boxed{48+22=70}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\sin (\cot ^{-1} (1+x)) = \cos (\tan^{-1} (x))$ Let $cot^{-1} (1+x)=a$
Then $x =1-\cot a$
So $$\sin a =\cos (\tan^{1}(1-\cot a))$$
$$a=\frac{\pi}{2}- \tan^{-1} (1-\cot a)$$
$$\cot a =\frac 12$$
So $x=\frac 12$, but given answer is $-\frac 12$. What am I doing wrong?
| $\sin (\cot ^{-1} (1+x)) = \cos (\tan^{-1} (x)).$
Let $\cot^{-1} (1+x)=A \implies \cot A =1+x.$ So $LHS=\sin A=
\frac{1}{\sqrt{2+2x+x^2}}$ Let $\tan B=x \implies RHS=\cos B =\frac{1}{\sqrt{1+x^2}}$. Finally $LHS=RHS \implies 2+2x+x^2=1+x^2 \implies x=-1/2.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $(G,*)$ be a group and $a,b \in G$ with $a*b = b^5*a^3$. Show that $|ba^{-1}| = |b^3a^3|$. Let $(G,*)$ be a group and $a,b \in G$ with $a*b = b^5*a^3$. Show that
$|ba^{-1}| = |b^3a^3|$.
attempt:
Write $a*b=ab$. Let $a,b \in G$.
Then, $ab=b^5a^3$. Note that
$ba^{-1} = a^{-1}b^5a^2$,
$ab^{-1} = b^2(b^3a^3)b^{-2}$, and
$b^3a^3 = b^{-2}(ab^{-1})b^2$.
Denote the identity element of $G$ by $e_G$.
Fact: for all $a \in G$,
$|a| = |a^{-1}|$.
Let $|ba^{-1}| = m$ and $|b^3a^3| = p$.
To show $m \mid p$:
\begin{align*}
(b^3a^3)^p &= e_G \\
(b^{-2}(ab^{-1})b^2)^p &= e_G \\
b^{-2}(ab^{-1})^pb^2 &= e_G \\
b^{-2}((ba^{-1})^{-1})^pb^2 &= e_G \\
((ba^{-1})^{-1})^p &= e_G
\end{align*}
By fact, $|(ba^{-1})^{-1}| = m$.
Hence, $m \mid p$ as desired.
To show $p \mid m$:
\begin{align*}
(ba^{-1})^m &= e_G \\
((ab^{-1})^{-1})^m &= e_G \\
((b^2(b^3a^3)b^{-2})^{-1})^m &= e_G \\
((b^2(b^3a^3)b^{-2})^m)^{-1} &= e_G
\end{align*}
Here I got a little bit confuse.
Does it true that in:
\begin{equation*}
((b^2(b^3a^3)b^{-2})^m)^{-1} = e_G
\end{equation*}
implies
$((b^2(b^3a^3)b^{-2})^m) = e_G$? (This is come from fact.) If yes, then
$|b^3a^3| = m$. Hence, $p \mid m$.
And thus, $m=p$.
| Claim 1: For any $x,y \in G$ we have $|xyx^{-1}|=|y|$
Proof: Exercise
Claim 2: For any $x \in G$ we have $|x|=|x^{-1}|$
Proof: Exercise
Now, from your last step of actual manipulation we have, $b^3a^3=b^{-2}(ab^{-1})b^2$.
So applying Claim 1 twice, we have, $|b^3a^3|=|b^{-2}(ab^{-1})b^2|=|b^{-1}(b^{-1}(ab^{-1})b)b|=|b^{-1}(ab^{-1})b|=|ab^{-1}|$
Now applying Claim 2, we get, $|b^3a^3|=|ab^{-1}|=|(ab^{-1})^{-1}|=ba^{-1}$
| {
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Primitive roots of a finite field I understand the definition of a primitive root of integers; however, I am quite confused trying to find the primitive roots of $\frac{\mathbb{Z}/3\mathbb{Z}[x]}{(x^3-x+\overline{1})}$. I know that $\frac{\mathbb{Z}/3\mathbb{Z}[x]}{(x^3-x+\overline{1})}$ has elements of the form $ax^2+bx+c$, with $a,b,c \in \mathbb{Z}/3\mathbb{Z}[x]$, so it has $27$ elements, and $\mathbb{F}_{27} \simeq \frac{\mathbb{Z}/3\mathbb{Z}[x]}{(x^3-x+\overline{1})}$.
Also, it is not clear to me if I must find an element of order 26 or an element of order $\varphi(27) = 18$, where $\varphi$ is the Euler Totient Function.
Could someone give me a tip?
| As Leoli1 commented, you have to find an element of order $26$,
since the multiplicative group of $\mathbb F_{27}$ has order $26$.
You may consider this brute force, but using $x^3\equiv x-1$ we have
$x^4\equiv x^2-x, $ $x^5\equiv -x^2+x-1$, $x^6\equiv x^2+x+1$, $x^7\equiv x^2+2x+2$, $x^8\equiv2x^2-1$,
$x^9\equiv x+1, x^{10}\equiv x^2+x$, $x^{11}\equiv x^2+x-1$, $x^{12}\equiv x^2-1$, and $x^{13}\equiv-1$.
It follows that $x$ is a primitive root.
(I could have taken a short cut: $x^{12}\equiv (x^4)^3\equiv (x^2-x)^3\equiv(x^2)^3-x^3= x^6-x^3$
$=x^3(x^3-1)\equiv (x-1)(x-2)\equiv (x-1)(x+1)=x^2-1$.)
| {
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Integral of $1/x^2$ without power rule I was wondering if it was possible to evaluate the following integral without using the power rule for negative exponents
\begin{equation*}
\int \frac{1}{x^2} \; dx
\end{equation*}
When using integration by parts, you end up with the same integral in the rhs so it seems out of luck
\begin{equation*}
\int \frac{1}{x^2} \; dx = \frac{\ln x}{x} + \int \frac{\ln x}{x^2} \; dx
\end{equation*}
This question is inspired by this blackpenredpen's video
Using integration by parts with $u = \frac{1}{x^2}$ and $v = x$ is NOT accepted. If you write
\begin{equation*}
\int \frac{1}{x^2} \; dx = \frac{1}{x} + 2 \int \frac{1}{x^2} \; dx
\end{equation*}
you are still implicitly using the power rule to compute the derivative of $\frac{1}{x^2}$ so it is not correct per the rules.
The same rules apply for $u$-substitution which implicitly use the power rule. See the following with $u = \frac{1}{x}$ such that
\begin{equation*}
\int \frac{1}{\left(\frac{1}{x}\right)^2} \left(\frac{1}{x}\right)' \; dx = \int \frac{1}{u^2} \; du
\end{equation*}
and here the power rule is also considered used to compute the derivative of $\frac{1}{x}$ although it can be subject to discussion (geometric proof, limit definition of derivative, etc...)
| $$ \frac{1}{x^2} = e^{\ln \left ( \frac{1}{x^2} \right ) } = e^{\ln \left ( x^{-2} \right ) } = e^{-2\ln (x ) } = (e^{-2})^{\ln(x)} $$
$$ \text{when} ~~a \in \mathbb{R^+}: ~~~\int a^{\ln(x)} dx = \int \frac1x x^{\ln(a) + 1} dx \\ u=\ln x \Rightarrow du = \frac1x dx \\ = \int e^{(\ln(a) + 1)u} \\v = ( \ln(a) + 1)u \Rightarrow du = \frac{1}{( \ln(a) + 1)} dv \\ = \frac{1}{ \ln(a) + 1} \int e^v dv = \frac{1}{\ln(a) + 1} e^v+C= \frac{1}{\ln(a) + 1}e^{(\ln(a) + 1)u} + C = \frac{e^{(\ln(a) + 1)u}}{\ln(a) + 1} + C = \frac{x^{\ln( a ) + 1}}{\ln(a) + 1} + C $$
Substituting $a = e^{-2}$ we get:
$$ \frac{x^{-2 + 1}}{-2 + 1} + C = - \frac{1}{x} + C$$
| {
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"answer_id": 7
} |
Find the derivative of $f(x)=x^{x^{\dots}{^{x}}}$.
Find the derivative of $f(x)$:
$$f(x)=x^{x^{\dots}{^{x}}}$$
Let $n$ be the number of overall $x's$ in $f(x)$. So for $n=1$, $f(x)=x$. I then tried to determine a pattern by solving for the derivative from $n=1$ to $n=5$. Here's what I got:
\begin{align}
n = 2 \Longrightarrow f(x) &= x^x \\
f'(x) &= \frac{d}{dx}\left(e^{x\ln \left(x\right)}\right) \\
&= e^{x\ln \left(x\right)}\frac{d}{dx}\left(x\ln \left(x\right)\right) \\
&= e^{x\ln \left(x\right)}\left(\ln \left(x\right)+1\right) \\
&= x^x\left(\ln \left(x\right)+1\right)
\end{align}
\begin{align}
n = 3 \Longrightarrow f(x) &= x^{x^{x}} \\
f'(x) &= \frac{d}{dx}\left(e^{x^x\ln \left(x\right)}\right) \\
&= e^{x^x\ln \left(x\right)}\frac{d}{dx}\left(x^x\ln \left(x\right)\right) \\
&= e^{x^x\ln \left(x\right)}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right) \\
&= x^{x^x}\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)
\end{align}
\begin{align}
n = 5 \Longrightarrow f(x) &= x^{x^{x^{x^{x}}}} \\
f'(x) &= ... \\
&= x^{x^{x^{x^x}}}\left(x^{x^{x^x}}\ln \left(x\right)\left(x^{x^x}\ln \left(x\right)\left(x^x\ln \left(x\right)\left(\ln \left(x\right)+1\right)+x^{x-1}\right)+x^{x^x-1}\right)+x^{x^{x^x}-1}\right)
\end{align}
However, I am not sure if I see a pattern here that can help solve the question.
| $$
f(x) = x^{f(x)}\\
\ln(f(x))=f(x) \ln(x)\\
\frac{1}{f(x)}f'(x) = \frac{f(x)}{x} + \ln(x)f'(x),\\
f'(x)=\left(\frac{1}{f(x)}-\ln(x)\right)^{-1}\frac{f(x)}{x},\\
f'(x)=\left(\frac{f(x)}{1-\ln(x)f(x)}\right)\frac{f(x)}{x}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3992667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Failing a basic integration exercise; where did I go wrong? (This is a basic calculus exercise gone wrong where I need some feedback to get forward.)
I've attempted to calculate an integral by first integrating it by parts and then by substituting. The result I got is not correct though. Can I get a hint about where started to make mistakes?
Here are my steps:
$$\int \sqrt{x}\cdot\sin\sqrt{x}\,dx$$
*
*Applying integration by parts formula
$$\int uv' \, dx = uv - \int u'v \, dx$$
where $u=\sqrt{x}$, $v'=\sin\sqrt{x}$, $u'=\frac{1}{2\sqrt{x}}$ and $v=-\cos\sqrt{x}$, resulting in:
$$\int \sqrt{x}\cdot\sin\sqrt{x}\,dx = \sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{x}} \cdot -\cos\sqrt{x} \, dx$$
*Next, I tried substituting $x$ with $g(t)=t^2$ in the remaining integral, i. e. replacing each $x$ in the integral with $t^2$ and the ending $dx$ with $g'(t)=2t\,dt$. I would later bring back the $x$ by substituting $t=\sqrt{x}$ after integration. Continuing from where we left by substituting:
$$\sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{x}} \cdot -\cos\sqrt{x} \, dx$$
$$\Longrightarrow \sqrt{x} \cdot -\cos\sqrt{x} - \int \frac{1}{2\sqrt{t^2}} \cdot -\cos\sqrt{t^2}\cdot2t \, dt$$
*Then I pulled out the constant multipliers from the integral:
$$\sqrt{x} \cdot -\cos\sqrt{x} - \frac{1}{2} \cdot -1 \cdot 2 \int \frac{1}{\sqrt{t^2}} \cdot \cos\sqrt{t^2}\cdot t \, dt$$
which turned out to eliminate each other (resulting in just $\cdot1$), so we end up with:
$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \int \frac{1}{\sqrt{t^2}} \cdot \cos\sqrt{t^2}\cdot t \, dt$$
*Reducing the integrand by reducing $\frac{1}{\sqrt{t^2}}\Rightarrow\frac{1}{t}$, which again is eliminated by multiplying with the integrand's $t$, and reducing $\cos\sqrt{t^2}\Rightarrow\cos t$. Therefore resulting in:
$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \int \cos t \, dt$$
where the integral can be solved as $\int \cos t \, dt = \sin t + C$. So now we are at:
$$\sqrt{x} \cdot -\cos\sqrt{x} \cdot \sin t + C$$
The correct answer however is
$$\int \sqrt{x} \sin\sqrt{x} \, dx = 4 \sqrt{x} \sin\sqrt{x} - 2 (x - 2) \cos\sqrt{x} + C$$
So something somewhere in my process went horribly wrong. What?
| After the substitution $u = \sqrt{x}$, the integral becomes $2\int u^2 \sin(u) \ du$ as Adam Rubinson has said. We can proceed using tabular integration:
$$\begin{array}{c|c} u^2 & \sin(u) \\ \hline 2u & -\cos(u) \\ \hline \ 2 & -\sin(u) \\ \hline \ 0 & \cos(u)\end{array}$$
where we differentiate on the left and integrate on the right, and multiply the terms diagonally (the signs alternate).
Hence $2\int u^2 \sin(u) \ du = 2 \left(-u^2 \cos(u) - -2u \sin(u) + 2\cos(u) \right) + C$, which is:
$$2 \left( -x \cos( \sqrt x )+2\sqrt x \sin(\sqrt x) + 2\cos(\sqrt x)\right)+C = 4 \sqrt{x} \sin(\sqrt x) - 2(x-2)\cos (\sqrt x) +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3994724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
What are the symmetric elements of $SU(3)$? Is there any simple way to know what are the matrices $M\in\operatorname{SU}(3)$ s.t $M=M^t$? For exemple, for $\operatorname{SU}(2)$ it is easy to verify that the symmetric elements are either diagonal or
$$\left(\begin{array}{cc}
0 & i \\
i & 0
\end{array}\right).$$
| I assume you are asking about the triplet representation, judging from your doublet paradigm for SU(2), which should not be "either-or", but rather
$$
i~ \begin{pmatrix}
\sin\phi & \cos\phi \\
\cos\phi & - \sin\phi
\end{pmatrix} .
$$
I gather you are looking for "seat-of-the-pants" easy, so your first look should be at the elements generated by Gell-Mann's symmetric generators, {$\lambda_3, \lambda_8, \lambda_1, \lambda_4, \lambda_6$}, given the generic formula for the triplet rep group element,
$$
\begin{align}
\exp(i\theta H) ={}
&\left[-\frac{1}{3} I\sin\left(\varphi + \frac{2\pi}{3}\right) \sin\left(\varphi - \frac{2\pi}{3}\right) - \frac{1}{2\sqrt{3}}~H\sin(\varphi) - \frac{1}{4}~H^2\right]
\frac{\exp\left(\frac{2}{\sqrt{3}}~i\theta\sin(\varphi)\right)}
{\cos\left(\varphi + \frac{2\pi}{3}\right) \cos\left(\varphi - \frac{2\pi}{3}\right)} \\[6pt]
& {} + \left[-\frac{1}{3}~I\sin(\varphi) \sin\left(\varphi - \frac{2\pi}{3}\right) - \frac{1}{2\sqrt{3}}~H\sin\left(\varphi + \frac{2\pi}{3}\right) - \frac{1}{4}~H^{2}\right]
\frac{\exp\left(\frac{2}{\sqrt{3}}~i\theta \sin\left(\varphi + \frac{2\pi}{3}\right)\right)}
{\cos(\varphi) \cos\left(\varphi - \frac{2\pi}{3}\right)} \\[6pt]
& {} + \left[-\frac{1}{3}~I\sin(\varphi) \sin\left(\varphi + \frac{2\pi}{3}\right) - \frac{1}{2\sqrt{3}}~H \sin\left(\varphi - \frac{2\pi}{3}\right) - \frac{1}{4}~H^2\right]
\frac{\exp\left(\frac{2}{\sqrt{3}}~i\theta \sin\left(\varphi - \frac{2\pi}{3}\right)\right)}
{\cos(\varphi)\cos\left(\varphi + \frac{2\pi}{3}\right)}
\end{align}
$$
generated by a traceless 3×3 Hermitian matrix H, normalized as $\operatorname{tr}( H^2) =2$, where
$$\varphi \equiv \frac{1}{3}\left[\arccos\left(\frac{3\sqrt{3}}{2}\det H\right) - \frac{\pi}{2}\right].$$
You notice that the linear combinations from this 5-generator set, squared, preserve the symmetry under transposition, so this subspace generates elements in your set.
Exploring "coincidences" where special coefficients annihilate terms involving antisymmetric generators might be a formidable problem, however.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) - \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $
Show $ ( \sin(\theta) - \cos(2\theta))^2 = 1+ \sin(\theta) - \sin(3 \theta) - \frac{1}{2}\cos(2 \theta) + \frac{1}{2}\cos(4 \theta) $
If I would start from the right expression I would be able to reduce it. But I'm not used to do the reverse.
I can see that Wolframalpha expands the left expression in many ways, one of which is the one I want, but are there any trick to do this expansion by hand?
This question is linked to this one.
| We'll use the following identities:
$\cos^2(\theta) = \dfrac{1+\cos(2\theta)}{2}$
$\sin^2{\theta} = \dfrac{1-\cos(2\theta)}{2}$
$\sin \theta = \dfrac{e^{i\theta} - e^{-i\theta}}{2i}$
$\cos \theta = \dfrac{e^{i\theta} + e^{-i\theta}}{2}$
$$
(\sin(\theta) - \cos(2\theta))^2 = \sin^2(\theta) + \cos^2(2\theta) - 2 \sin(\theta) \cdot \cos(2\theta)
$$
$$
\sin(\theta) \cdot \cos(2\theta) = \dfrac{e^{i\theta} - e^{-i\theta}}{2i} \cdot \dfrac{e^{2i\theta} + e^{-2i\theta}}{2} \\
= \dfrac{e^{3i\theta} + e^{-i\theta} - e^{i\theta} - e^{-3i\theta}}{4i} \\
= \dfrac{e^{3i\theta} - e^{-3i\theta}}{4i} - \dfrac{e^{i\theta} - e^{-i\theta}}{4i} \\
= \frac1{2} \sin(3\theta) - \frac1{2} \sin(\theta)
$$
As such,
$$
(\sin(\theta) - \cos(2\theta))^2 = \dfrac{1-\cos(2\theta)}{2} + \dfrac{1+\cos(4\theta)}{2} + \frac1{2} \sin(3\theta) - \frac1{2} \sin(\theta)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Finding the limit of $a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2})$ for arbitrary $a_{0}$ and $a_{1}$. Here is the problem statement:
Let $a$, $b$ $\in$ $\Bbb{R}$. A sequence $(a_{n})_{n \in \Bbb{N}}$ is defined recursively by
$$a_{0}:=a, \qquad a_{1}:=b, \qquad a_{n} = \frac{1}{2}(a_{n-1}+a_{n-2}) \quad \text{for} \quad n \geq 2.$$
Prove that $\lim_{n\to\infty}a_{n}$ exists and compute its value.
Now I have proved that the limit exists. My idea was to use the Monotone Convergence Theorem as follows:
Proof: Without loss of generality, assume $a<b$. By inspection of the first two terms of $a_{n}$, we see that
$$a_{3} = \frac{1}{2}(a_{2}+a_{1})=\frac{1}{2}\left[\frac{1}{2}(a_{1}+a_{0}) + a_{1}\right] = \frac{1}{2}(a_{1}+a_{0})+\frac{1}{4}(a_{1}-a_{0})=a_{2}+\frac{1}{4}(a_{1}-a_{0})>a_{2} \quad (\because a_{1}>a_{0}).$$
Therefore we need to show that the sequence is increasing, so our inductive hypothesis is $a_{n}<a_{n+1}$.
$$a_{n}<a_{n+1}\implies a_{n+1}=\frac{1}{2}(a_{n}+a_{n-1})<\frac{1}{2}(a_{n+1}+a_{n})=a_{n+2}.$$
Thus $a_{n}$ is increasing for $n\geq 2$. Now we need to show that it is bounded. By inspection, $a_{3}=\frac{1}{2}(a_{2}+a_{1})<\frac{1}{2}(a_{1}+a_{1}) = a_{1}$, and $a_{4}<a_{1}$ and so on. So again assume by induction $a_{n}<a_{1}$ and we have
$$a_{n+1}=\frac{1}{2}(a_{n}+a_{n-1})<\frac{1}{2}(a_{1}+a_{1}) = a_{1}.$$
Thus $a_{n}$ is bounded above. Hence by the Monotone Convergence Theorem, sequence is convergent and $\lim a_{n}$ exists.$\quad\square$
However, I'm stuck in finding the limit. Any hints would be appreciated.
| $$
\begin{bmatrix}
a_{n}\\
a_{n-1}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
a_{n-1}\\
a_{n-2}
\end{bmatrix}
$$
$$
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & \frac{1}{2}
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
1 & -1
\end{bmatrix}
$$
$$
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix}^{n}
=
\begin{bmatrix}
1 & 1\\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & \frac{1}{2}^{n}
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
1 & -1
\end{bmatrix}
$$
$$
\begin{bmatrix}
a_{n}\\
a_{n-1}
\end{bmatrix}
=
\begin{bmatrix}
\frac{1}{2} & \frac{1}{2}\\
0 & 1
\end{bmatrix} ^ {n}
\begin{bmatrix}
a_{1}\\
a_{0}
\end{bmatrix}
=
\begin{bmatrix}
1 & 1\\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
0 & \frac{1}{2}^{n}
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
1 & -1
\end{bmatrix}
\begin{bmatrix}
a_{1}\\
a_{0}
\end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Deriving what $\cos(A+2B)$ is We know that $\cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$ , but I don't understand with the $2B$. Would it just become $\cos(A+2B) = \cos(A)\cos(2B) - \sin(A)\sin(2B)$?
Thank you for any help you can give me.
| Further expansion is possible with repeated use of the angle addition identities.
$$\begin{align}
\cos (a + 2b) &= \cos a \cos 2b - \sin a \sin 2b \\
&= \cos a (\cos b \cos b - \sin b \sin b) - \sin a (\sin b \cos b + \cos b \sin b) \\
&= \cos a (\cos^2 b - \sin^2 b) - \sin a (2 \sin b \cos b) \\
&= \cos a \cos^2 b - \cos a \sin^2 b - 2 \sin a \sin b \cos b.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3997084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
I have to show that the sum of this double series is $\frac{1}{2}$ i have to solve this double series. i tried it, but i am not sure, that it is enough.
$$\sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \left(\left(\frac{1}{k+1} \cdot \left(\frac{k}{k+1}\right)^{i}\right) - \left(\frac{1}{k+2} \cdot \left(\frac{k+1}{k+2}\right)^{i}\right)\right)$$
$$= \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \frac{(1)^{i}}{k+1} - \frac{(1)^{i}}{k+2} $$
$$= \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{k+1} - \frac{1}{k+2} $$
this is a telescoping series. Because of that, the Series is convergence to $\lim_{n\to\infty} \frac{1}{2} - \frac{1}{n+2} = \frac{1}{2}$
Is this solution right and enough?
| Here is my attempt.
Since all terms are positive, we are allowed to swap the two summations.
$$\sum_{i=1}^\infty \sum_{k=1}^{\infty} \left( \frac{1}{k+1} (\frac{k}{k+1})^i - \frac{1}{k+2} (\frac{k+1}{k+2})^i \right) = $$
$$= \lim_{N \to \infty }\sum_{k=1}^N \sum_{i=1}^{\infty} \left( \frac{1}{k+1} (\frac{k}{k+1})^i - \frac{1}{k+2} (\frac{k+1}{k+2})^i \right) = $$
$$= \lim_{N \to \infty } \left( \sum_{k=1}^N \frac{1}{k+1} \sum_{i=1}^{\infty} (\frac{k}{k+1})^i - \sum_{k=1}^N \frac{1}{k+2} \sum_{i=1}^{\infty} (\frac{k+1}{k+2})^i \right) = $$
$$= \lim_{N \to \infty } \left( \sum_{k=1}^N \frac{1}{k+1} \sum_{i=1}^{\infty} (\frac{k}{k+1})^i - \sum_{k=1}^N \frac{1}{k+2} \sum_{i=1}^{\infty} (\frac{k+1}{k+2})^i \right) = $$
$$= \lim_{N \to \infty } \left( \sum_{k=1}^N \frac{k}{k+1} - \sum_{k=1}^N \frac{k+1}{k+2} \right) = $$
$$= \lim_{N \to \infty } \left( \frac{1}{2} - \frac{N+1}{N+2} \right) = - \frac{1}{2}$$
I don't know where I did a minus sign mistake. :/
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4000278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Analyse convergence of: $\displaystyle \sum^{\infty}_{n=3} \left (\sqrt[3]{n^3 + 2} - \sqrt{n^2 + 7}\right)$
Analyse convergence of:
$$\displaystyle \sum^{\infty}_{n=3} \left(\sqrt[3]{n^3 + 2} - \sqrt{n^2 + 7}\right )$$
I know that all elements are negative, so I need to take $-1$ out and I will have only positive elements and then I can use comparison criterion. However, I don't know how to do the first step. I need to make it rational.
| If you are able to prove that both series$$\sum_{n=3}^\infty\sqrt[3]{n^3+2}-n\quad\text{and}\quad\sum_{n=3}^\infty\sqrt{n^2+7}-n\tag1$$converge, or that one of them converges and the other one diverges, then you're done.
Now, note that\begin{align}\sqrt[3]{n^3+2}-n&=\sqrt[3]{n^3+2}-\sqrt[3]{n^3}\\&=\frac2{\sqrt[3]{n^3+2}^2+\sqrt[3]{n^3+2}\sqrt[3]{n^3}+\sqrt[3]{n^3}^2}\\&=\frac2{\sqrt[3]{n^3+2}^2+\sqrt[3]{n^3+2}\,n+n^2}\end{align}and therefore$$\lim_{n\to\infty}\frac{\sqrt[3]{n^3+2}-n}{\frac1{n^2}}=\frac23$$and therefore the first series from $(1)$ converges.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
A weird contour integral calculation I have function $\int_{-1}^{1}\frac{\sqrt{1-x^2}}{1+x^2}dx$.
Here to use residue thm, I rewrite the integral as $\int_{-1}^{1}\frac{\sqrt{1-z^2}}{1+z^2}dz$ with the poles $z=i$ and $-i$. However there is a given condition $-1<x<1$, so it means $z^2<1$. This is the problem because when the poles are at $|z|=1$. It's outside the boundary. So I have no idea about how I can calculate this integral
Here's my Residues
$\operatorname{Res}_{z=i}[f(z)]=\lim_{z\to i}\frac{\sqrt{1-z^2}}{i(1-zi)}dz=\frac{\sqrt{2}}{2i}$
$\operatorname{Res}_{z=-i}[f(z)]=\lim_{z\to -i}\frac{\sqrt{1-z^2}}{i(1+zi)}dz=\frac{\sqrt{2}}{2i}$
But I guess these are wrong
Edit: since I need a closed contour i replaced $x=cos\theta$ and $dx=-sin\theta d\theta$
And my integral is now $-\frac{1}{2}\oint \frac{\sqrt{1-cos^2\theta}}{cos^20+cos^2\theta}(-sin\theta) d\theta$=
$-\frac{1}{2}\oint \frac{-sin^2\theta}{cos^20+cos^2\theta}d\theta$
| Time for the dog-bone contour (dumbbell contour), which works perfectly.
Indeed, consider the contour $\mathcal{C}$ given as follows:
$\hspace{12em}$
With the principal branch cut, as the radius/band-width of $\mathcal{C}$ goes to zero,
\begin{align*}
\oint_{\mathcal{C}} \frac{i\sqrt{z-1}\sqrt{z+1}}{i(z^2+1)} \, dz
&\longrightarrow
\int_{-1-0^+i}^{1-0^+i} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^2+1} \, dz
- \int_{-1+0^+i}^{1+0^+i} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^2+1} \, dz \\\\
&\quad = 2\int_{-1}^{1} \frac{\sqrt{1-x^2}}{x^2+1} \, dx.
\end{align*}
On the other hand, Residue Theorem tells that
\begin{align*}
\oint_{\mathcal{C}} \frac{i\sqrt{z-1}\sqrt{z+1}}{z^4+1} \, dz
&= - 2\pi i \sum_{z_k \ : \ z_k^2 + 1 = 0} \underset{z=z_k}{\mathrm{Res}} \, \frac{i\sqrt{z-1}\sqrt{z+1}}{z^2+1} \\\\
&= 2\pi (\sqrt{2}-1)
\end{align*}
(In order to use Residue Theorem, consider a large circle, apply Residue Theorem to the region enclosed by this circle and $\mathcal{C}$, and then let the radius of the circle go to $\infty$.) Therefore
$$ \int_{-1}^{1} \frac{\sqrt{1-x^2}}{x^2+1} \, dx = \pi (\sqrt{2}-1) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4003763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determine the partial derivative of $\frac{\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}}{\cos ^2\left(x^2y^2\right)}$.
Determine the partial derivative of $f(x)$:
$$f(x,y)=\frac{\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}}{\cos ^2\left(x^2y^2\right)}$$
Here's what I have so far:
\begin{align}
\frac{\partial \:}{\partial \:x}\left(\frac{\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}}{\cos ^2\left(x^2y^2\right)}\right) &= \frac{\frac{\partial \:}{\partial \:x}\left(\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}\right)\cos ^2\left(x^2y^2\right)-\frac{\partial \:}{\partial \:x}\left(\cos ^2\left(x^2y^2\right)\right)\left(\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}\right)}{\left(\cos ^2\left(x^2y^2\right)\right)^2} \\
&= \frac{\left(\frac{2\tan \left(\frac{1}{x^2+1}\right)}{x}-\frac{2x\ln \left(y^2x^2\right)\sec ^2\left(\frac{1}{x^2+1}\right)}{\left(x^2+1\right)^2}+\frac{2x}{3\left(x^2+y^2\right)^{\frac{2}{3}}}\right)\cos ^2\left(x^2y^2\right)-\left(-2y^2x\sin \left(2y^2x^2\right)\right)\left(\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}\right)}{\left(\cos ^2\left(x^2y^2\right)\right)^2} \\
&= \frac{\cos ^2\left(x^2y^2\right)\left(\frac{2\tan \left(\frac{1}{x^2+1}\right)}{x}-\frac{2x\ln \left(x^2y^2\right)\sec ^2\left(\frac{1}{x^2+1}\right)}{\left(x^2+1\right)^2}+\frac{2x}{3\left(x^2+y^2\right)^{\frac{2}{3}}}\right)+2xy^2\sin \left(2x^2y^2\right)\left(\ln \left(x^2y^2\right)\tan \left(\frac{1}{x^2+1}\right)+\sqrt[3]{x^2+y^2}\right)}{\cos ^4\left(x^2y^2\right)}
\end{align}
However, I am not sure if this answer is correct since I get a different answer on emathhelp website.
| The only issue is that $ \frac{\partial}{\partial x} \cos^2(x^2 y^2)$ should be equal to
$$- 4 x y^2 \cos(x^2 y^2) \sin(x^2 y^2) $$
and with that correction it is good.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4005546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Any positive integer greater than $11$ is a nonnegative linear combination of $5$ and $4$. My solution Let $n\in\mathbb{Z}^{+}$, then there exists $k\in\mathbb{Z}_0^+$, such that $n=5k + i, i\in\{0,1,2,3,4\}$. Now analyzing by cases we have:
*
*If $i=0$, then
\begin{align*}
n = 5k \Rightarrow n = 5k + 4(0).
\end{align*}
*If $ i = 1 $, then
\begin{align*}
n & = 5k + 1 \\
& = 5k-5(3) +5(3) +1 \\
& = 5(k-3) + 15 + 1 \\
& = 5(k-3) +16 \Rightarrow n = 5(k-3) +4(4).
\end{align*}
*If $ i = 2 $, then
\begin{align*}
n & = 5k + 2 \\
& = 5k-5(2) +5(2) +2 \\
& = 5(k-2) + 10 + 2 \\
& = 5(k-2) +12 \Rightarrow n = 5(k-2) +4(3).
\end{align*}
*If $i=3$, then
\begin{align*}
n & = 5k + 3 \\
& = 5k-5 + 5 + 3 \\
& = 5(k-1) +8 \Rightarrow n = 5(k-1) +4(2).
\end{align*}
*If $i=4$, then
\begin{align*}
n = 5k + 4 \Rightarrow n = 5k + 4(1).
\end{align*}
Thus, every positive number can be expressed as a linear combination of $5$ and $4$. Now using that $n>11$, so we have:
\begin{align*}
n &> 11 \\
5k + i &> 5(2) +1 \\
5k-5(2) &> 1-i \\
5 (k-2) &> 1-i \\
k-2 &> \frac{1-i}{5} \\
k &> 2+\frac{1-i}{5}.
\end{align*}
So by increasing over the values that $ i $ takes, we have:
\begin{align*}
k &> 2+ \frac{1-i}{5} \geq 2+ \frac{1-0}{5}\\
k &> 2 + 0.2 = 2.2
\end{align*}
But $k\in\mathbb{Z}_0^+ \Rightarrow k \geq 3 \Rightarrow n \geq 15 $. Thus we have that every positive integer greater than or equal to $15$ is a non-negative linear combination of $5$ and $4$.
Finally, let's look at the cases that are still unverified, which are $12$, $13$ and $14$.
\begin{align*}
12 &= 5(0) +4(3) \\
13 &= 5(1) +4(2) \\
14 &= 5(2) +4(1).
\end{align*}
Therefore, any positive integer greater than $11$ is a nonnegative linear combination of $5$ and $4$.
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.
| We proceed by induction on $n.$ Observe that $12 = 3 \cdot 4 + 0 \cdot 5$ is a non-negative linear combination of $4$ and $5.$ We will assume inductively that any integer $n \geq 12$ can be written as a non-negative linear combination $n = 4x + 5y$ of $4$ and $5.$ Given that $x \geq 1,$ we have that $$4(x - 1) + 5(y + 1) = 4x - 4 + 5y + 5 = 4x + 5y + 1 = n + 1$$ is a non-negative linear combination of $4$ and $5,$ as desired. On the other hand, if we have that $x = 0,$ then $n = 5y.$ By hypothesis that $n \geq 12,$ we must have that $n \geq 15$ (because $n$ is a multiple of $5$) so that $y \geq 3.$ Consequently, we find that $$4 \cdot 4 + 5(y - 3) = 16 + 5y - 15 = 5y + 1 = n + 1$$ is a non-negative linear combination of $4$ and $5.$ QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4006307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
General Inequality Problem. The question goes like this :
Let $a,b,c,d$ be positive reals and given that $a+b+c+d=1$.
Prove that: $$6(a^3+b^3+c^3+d^3) \geqslant a^2+b^2+c^2+d^2 + \frac{1}{8}$$
My approach goes like this:
I wrote $a^3+b^3$ as $(a+b)(a^2+b^2-ab)$
Similarly $c^3+d^3$ as $(c+d)(c^2+d^2-cd)$
Although I am not sure if its the correct method, I tried reducing the powers.
Then, I got:
$6(a+b)(a^2+b^2-ab) -a^2-b^2 \geqslant c^2+d^2 - 6(c+d)(c^2+d^2-cd) +\frac{1}{8} $
I tried substituting $c,d$ in terms of $a,b$ but the calculation is lengthier than I expected. Please check if my reasoning is correct, and help me solve this in a shorter method. Thanks!
| Use Tangent Line Method
$$6x^3-x^2 \geqslant \frac{5x-1}{8}, \quad \forall x > 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4007820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
compute $\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}$
The question is:
$$
\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}
\quad \gamma:r(t)=(t,2t^2-2), \quad -1\leq t\leq 1
$$
I have tried to solve it like this:
since $Q_x=P_y$ it's potential vector field but the singularity is at origin, first i thought that i could evaluate it from $(-1,0)$ to $(1,0)$ along $x$ axis but the singularity is the problem, how should i proceed in this case?
Any suggestion would be great Thanks
| $$
\int_\gamma\frac{-y^2dx+2xy\ dy}{x^2+y^4}=\int_\gamma\frac{-y^2}{x^2+y^4}\,dx+\int_\gamma\frac{2xy}{x^2+y^4}\,dy$$
$$\gamma(t)=(t,2 t^2-2);\;t\in[-1,1];\;\gamma'(t)=(1, 4 t)$$
Substitute
$$\int_{-1}^1 -\frac{\left(2 t^2-2\right)^2}{\left(2 t^2-2\right)^4+t^2}\cdot 1\,dt+\int_{-1}^1 \frac{2 t \left(2 t^2-2\right)}{\left(2 t^2-2\right)^4+t^2}\cdot(4t)\,dt=$$
$$=\int_{-1}^1 \frac{4 \left(3 t^4-2 t^2-1\right)}{16 t^8-64 t^6+96 t^4-63 t^2+16}\,dt$$
Integrating is quite hard
denominator $16 t^8-64 t^6+96 t^4-63 t^2+16=(16 t^8-64 t^6+96 t^4-64 t^2+16)+t^2=16 \left(t^2-1\right)^4+t^2$
numerator can be factored $4 \left(3 t^4-2 t^2-1\right)=4\left(t^2-1\right) \left(3 t^2+1\right)$
and integral can be written as
$$\int_{-1}^1\frac{4\left(t^2-1\right) \left(3 t^2+1\right)}{16 \left(t^2-1\right)^4+t^2}\,dt$$
Now divide numerator and denominator by $16 \left(t^2-1\right)^4$
$$\int_{-1}^1\frac{\frac{4\left(t^2-1\right) \left(3 t^2+1\right)}{16 \left(t^2-1\right)^4}}{1+\frac{t^2}{16 \left(t^2-1\right)^4}}\,dt=\int_{-1}^1\frac{\frac{ 3 t^2+1}{4 \left(t^2-1\right)^3}}{1+\frac{t^2}{16 \left(t^2-1\right)^4}}\,dt$$
Now substitute $\frac{t}{4\left(t^2-1\right)^2}=u$
extremes become $t=-1\to u=-\infty;\;t=1\to u=+\infty$
while differentiating we get $\frac{3 t^2+1}{4 \left(t^2-1\right)^3}\,dt=-du$
Therefore the integral becomes
$$-\int_{-\infty}^{\infty} \frac{du}{1+u^2}=-\left[\arctan u\right]_{-\infty}^{\infty}=-\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4011290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Inequality Induction Proof, how should I proceed? I have been trying to prove inequalities using induction to no avail.
For example,
Prove the following:
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{n}{2^{n}}<2$
Base Case:
$\frac{1}{2^{1}} = 2$
$\frac{1}{2}<2\;$ Which is true.
We assume, for n=k, that
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{k}{2^{k}}<2\;\;\;$ is true.
As induction step with n=k+1, we need to prove that
$\frac{1}{2^{1}}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+...+\frac{k+1}{2^{k+1}}<2\;\;\;$
How should I proceed from here?
| I finally managed to prove it with induction, which was the goal.
I followed the suggestion to prove that $\;\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+...+\frac{k}{2^{k}} = 2-\frac{k+2}{2^{k}}\;$
for k+1 we have that
$\;\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+...+\frac{k}{2^{k}}+\frac{k+1}{2^{k+1}} = 2-\frac{k+3}{2^{k+1}}\;$
It follows from the inductive hypothesis that
$2-\frac{k+2}{2^{k}}+\frac{k+1}{2^{k+1}} = 2-\frac{k+3}{2^{k+1}}\;$
$2-\frac{2k+4}{2^{k+1}}+\frac{k+1}{2^{k+1}} = 2-\frac{k+3}{2^{k+1}}\;$
$2+\frac{-3-k}{2^{k+1}} = 2-\frac{k+3}{2^{k+1}}\;$
$2-\frac{k+3}{2^{k+1}} = 2-\frac{k+3}{2^{k+1}}\;$
Which verifies that $\;\frac{1}{2^{1}}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+...+\frac{k}{2^{k}}$ is 2 minus something else, which in turn means that it is less than 2 thus proving the original statement.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4013103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
find the maximum value of $m$ where $3^m$ be a factor of $A_{1395}$.
Consider the number $A_{1395}=\underbrace{333\cdots3}_{1395}$. what is
the maximum value of $m$ where $3^m$ be a factor of $A_{1395}$.
$a) 2\quad\quad b)3\quad\quad c)4\quad\quad d)5\quad\quad e)6$
Here is my attempt :
We have $A_{1395}=3\times \underbrace{111\dots1}_{1395}$ sum of the digits of $\underbrace{111\cdots1}_{1395}$ is $1395$ which is a multiply of $9$. therefor this number is dividable to $9$. so we conclude $A_{1395}$ has the factor $3^3$ . to see whether it has more factor of $3$ or not I divided it to $27$. after doing some steps of long division I got:
$$\underbrace{11\cdots1}_{1395}÷27=4115226\cdots$$
Unfortunately I couldn't see any pattern as the result. so I don't know how to check this number has more factor of $3$ or not.
| Write using the binomial theorem,
\begin{align}
999\cdots 9 &= 10^{1395} -1 \\
&= (1+9)^{1395} -1 \\
&= \left(\array{1395\\1}\right)\cdot 9 + \left(\array{1395\\2}\right)\cdot9^2 + \cdots 9^{1395}
\end{align}
But $1395 = 3^2 \cdot 5 \cdot 31$. So, the first term is divisible by $3^4$ and each term thereafter is divisible by $3^5$ meaning the sum is divisible $3^4$ and not by $3^5$. Accordingly, $333\cdots 3$ is divisible by $3^3$ but not $3^4$. The answer is therefore $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Factoring $a^4+b^4+(a-b)^4$ I'm trying to factor $$a^4+b^4+(a-b)^4$$ so the result would be $2(a^2-ab+b^2)^2$ but I can't get that.
I rewrite it as:
$$a^4+b^4+(a-b)^4=(a^2+b^2)^2-2a^2b^2+(a-b)^4=(a^2-\sqrt2 ab+b^2)(a^2+\sqrt2 ab+b^2)+(a-b)^4$$
But I can't use difference of squares anymore because $(a-b)^4$ is not negative.
| Hint: $(a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4018565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
The center of the circumcircle lies on a side of a triangle Consider a triangle $ABC$. Let the angle bisector of angle $A$ be $AP,P\in BC$. $BP=16,CP=20$ and the center of the circumcircle of $\triangle ABP$ lies on the segment $AC$. Find $AB$.
$$AB=\dfrac{144\sqrt5}{5}$$
By Triangle-Angle-Bisector Theorem $$\dfrac{BP}{PC}=\dfrac{AB}{AC}=\dfrac{16}{20}=\dfrac{4}{5}\\ \Rightarrow AB=4x, AC=5x.$$ The cosine rule on $ABC$ gives $$BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cdot\cos\alpha \\ \iff 1296=41x^2-40x^2\cos\alpha,$$ where $\measuredangle A=\alpha.$ Is any of this helpful for the solution? Any help would be appreciated. Thank you in advance!
| Alternative solution using neither trigonometry nor power of a point:
You already figured out that $OP \parallel AB$, so draw two perpendicular lines $OQ$ and $BR$.
Let the radii $OA = OB = OP = 10x$.
Since $\triangle ABC \sim \triangle OPC$, we have $AB = \frac{20+16}{20} \cdot OP = 18x$, so $AQ = BQ = 9x$.
Therefore, $OQ = \sqrt{OA^2 - AQ^2} = \sqrt{19} x$, so $BR = \sqrt{19} x$.
$RP = OP - OR = 10x - 9x = x$.
$BP = \sqrt{BR^2 + RP^2} = 2 \sqrt{5} x$.
Given that $BP = 16$, we conclude that $x = \frac{16}{2\sqrt{5}} = \frac85 \sqrt{5}$.
Finally, $AB = 18 \cdot \frac85 \sqrt{5} = \frac{144}{5} \sqrt{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4020885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Divergence of $\lim_{N\to +\infty}[ \int_{1}^{N+1} f(x) dx -\sum_{n=1}^{N}f(n)] $ Divergence of $\lim_{N\to +\infty}[ \int_{1}^{N+1} f(x) dx -\sum_{n=1}^{N}f(n)] $ where $f(x)= sin(log_e x)(\frac{1}{x^{a}}-\frac{1}{x^{1-a}})$,$ 0<a<1/2$
My try -
$\lim_{N\to +\infty}[ \int_{1}^{N+1} sin(log_e x)(\frac{1}{x^{a}}-\frac{1}{x^{1-a}})dx - \sum_{n=1}^{N} sin ( log _en) (\frac{1}{n^{a}}-\frac{1}{n^{1-a}})] $ where $0<a<1/2$
Define S=$\lim_{N\to +\infty}[ \int_{1}^{N+1} sin(log_e x)(\frac{1}{x^{a}}-\frac{1}{x^{1-a}})dx - \sum_{n=1}^{N} sin ( log _en) (\frac{1}{n^{a}}-\frac{1}{n^{1-a}})] $
If $a=1/2$ then S is convergent to 0.
If $a<1/2$ then, $(\frac{1}{x^{a}}-\frac{1}{x^{1-a}})>0$ $ \forall x\geq 1$
and $(\frac{1}{n^{a}}-\frac{1}{n^{1-a}})>0$ $ \forall n\geq 1$.
Please prove that S is divergent when $0<a<1/2$
| *
*Integral $I(a)=\int_{1}^{N+1} \sin(\ln x)(\frac{1}{x^{a}}-\frac{1}{x^{1-a}})dx$ can be evaluated - just make a change $x=e^t$:
$I(a)=\int_{0}^{\ln(N+1)} \sin(t)(e^{-ta}-e^{(a-1)t})e^tdt=\int_{0}^{\ln(N+1)} \sin(t)(e^{t(1-a)}-e^{at})dt$ and $\sin(t)=\frac{e^{it}-e^{-it}}{2i}= \Im({e}^{it})$
Integral with exponents can be easily taken.
*Consider the integral as a sum: $\sum_{k=1}^{N}\int_{\ln(k)}^{\ln(k+1)} \sin(t)(e^{t(1-a)}-e^{at})dt$
So, you get a series with N term as $a_N=\left(\int_{\ln(N)}^{\ln(N+1)} \sin(t)(e^{t(1-a)}-e^{at})dt-\sin(\ln{N})(\frac{1}{N^{a}}-\frac{1}{N^{1-a}})\right)$ and you can investigate its behaviour at $N\to\infty$.
Let's consider $a_N$:
After evaluating the integral we get:
$a_N=\frac{1-a}{1+(1-a)^2}[(N+1)^{1-a}\sin(\ln(N+1))-N^{1-a}\sin(\ln(N))]-\frac{1}{1+(1-a)^2}[(N+1)^{1-a}\cos(\ln(N+1))-N^{1-a}\cos(\ln(N))]-\frac{a}{1+a^2}[(N+1)^{a}\sin(\ln(N+1))-N^{a}\sin(\ln(N))]+\frac{1}{1+a^2}[(N+1)^{a}\cos(\ln(N+1))-N^{a}\cos(\ln(N))]-[\sin(\ln{N})\frac{1}{N^{a}}]+[\sin(\ln{N})\frac{1}{N^{1-a}}]$
Now, let's consider, for example
$\frac{a}{1+a^2}[(N+1)^{a}\sin(\ln(N+1))-N^{a}\sin(\ln(N))]=\frac{a}{1+a^2}[N^a(1+\frac{1}{N})^{a}\sin(\ln{N}+\ln(1+\frac{1}{N}))-N^{a}\sin(\ln(N))]$
Expanding the sine of the sum, as well as using Taylor series expansion for large N we get
$\frac{aN^a}{1+a^2}[(1+\frac{a}{N}+O(\frac{1}{N^2}))\left(\sin(\ln{N})\cos((\frac{1}{N})+O(\frac{1}{N^2}))+\cos(\ln{N})\sin((\frac{1}{N})+O(\frac{1}{N^2}))\right)-\sin(\ln(N))]=\frac{aN^a}{1+a^2}[(1+\frac{a}{N}+O(\frac{1}{N^2}))\left(\sin(\ln{N})\cos(\frac{1}{N})+O(\frac{1}{N^2})+\cos(\ln{N})\sin(\frac{1}{N})+O(\frac{1}{N^2})\right)-\sin(\ln(N))]=\frac{a}{1+a^2}N^a\left(\frac{a\sin(\ln{N})}{N}+\frac{\cos(\ln{N})}{N}+O(\frac{1}{N^2})\right)$
Finally, evaluating all terms we get
$a_N=N^{-a}\left(\frac{(1-a)^2\sin(\ln{N})}{1+(1-a)^2}+\frac{(1-a)\cos(\ln{N})}{1+(1-a)^2}-\frac{(1-a)\cos(\ln{N})}{1+(1-a)^2}+\frac{\sin(\ln{N})}{1+(1-a)^2}+O(\frac{1}{N^2})\right)+N^{a-1}\left(-\frac{a^2\sin(\ln{N})}{1+a^2}-\frac{a\cos(\ln{N})}{1+a^2}+\frac{a\cos(\ln{N})}{1+a^2}-\frac{\sin(\ln{N})}{1+a^2}+O(\frac{1}{N^2})\right)-$$-\sin(\ln{N})\frac{1}{N^{a}}+\sin(\ln{N})\frac{1}{N^{1-a}}$
After grouping all terms we see that they cancel each other, so we get
$a_N=N^{a}O(\frac{1}{N^2})+N^{1-a}O(\frac{1}{N^2})=O(\frac{1}{N^{2-a}})+O(\frac{1}{N^{1+a}})$
But these series converges absolutely $\Rightarrow$ $a_N$ converges.
Please check my calculations.
| {
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"url": "https://math.stackexchange.com/questions/4021701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show $\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx= \int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$ I accidentally found out that the two integrals below
$$I_1=\int_0^\infty \frac{\tan^{-1}x^2}{1+x^2} dx,\>\>\>\>\>\>\>I_2=\int_0^\infty \frac{\tan^{-1}x^{1/2} }{1+x^2}dx$$
are equal in value. In fact, they can be evaluated explicitly. For example, the first one can be carried out via double integration, as sketched below.
\begin{align}
I_1&=\int_0^\infty \left(\int_0^1 \frac{x^2}{1+y^2x^4}dy\right)\frac{1}{1+x^2}dx\\
&= \frac\pi2\int_0^1 \left( \sqrt{\frac y2}+ \frac1{\sqrt{2y} }-1\right)\frac{1}{1+y^2}dy=\frac{\pi^2}8
\end{align}
Similarly, the second one yields $I_2=\frac{\pi^2}8$ as well.
The evaluations are a bit involved, though, and it seems an overreach to prove their equality this way, if only the following needs to be shown
$$\int_0^\infty \frac{\tan^{-1}x^2-\tan^{-1}x^{1/2} }{1+x^2}dx=0$$
The question, then, is whether there is a shortcut to show that the above integral vanishes.
| By the known formula for the difference of two inverse tangents, your last integral is
$$
\int_0^{ + \infty } {\frac{{\tan ^{ - 1} \left( {\frac{{x^2 - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}dx} = \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{x^2 - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}dx} + \int_1^{ + \infty } {\frac{{\tan ^{ - 1} \left( {\frac{{x^2 - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}dx} .
$$
Taking $y=1/x$ in the last integral gives
\begin{align*}
& \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{x^2 - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}dx} + \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{y^{1/2} - y^2 }}{{1 + y^{5/2} }}} \right)}}{{1 + y^2 }}dy} \\ & = \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{x^2 - x^{1/2} }}{{1 + x^{5/2} }}} \right)}}{{1 + x^2 }}dx} - \int_0^1 {\frac{{\tan ^{ - 1} \left( {\frac{{y^2 - y^{1/2} }}{{1 + y^{5/2} }}} \right)}}{{1 + y^2 }}dy} = 0.
\end{align*}
| {
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Impossible integral?
$$\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2$$
I saw this problem in a calculus exam. $f$ is assumed to be continuous. Using the Fundamental Theorem of Calculus, I calculated $f(t)= 3t^2+8t+8 $. But when I integrate $f$ it gives me $\int_{0}^{x^2-1} f(t) dt= x^6+x^4+3x^2-5$:
$$\frac {d}{dx}\int_{0}^{x^2-1} f(t) dt= \frac {d}{dx}[x^6+x^4+3x^2]$$
$$f(x^2-1)2x = 6x^5+4x^3+6x$$
$$f(x^2-1) = 3x^4+2x^2+3$$
$$f(x^2-1) = 3(x^2-1)^2+8(x^2-1)+8$$
$$f(t) = 3t^2+8t+8$$
$$\int_{0}^{x^2-1} 3t^2+8t+8 dt= x^6+x^4+3x^2-5$$
This means that there is no continuous function that satisfies the equation of the problem?
| You are correct that this does mean there are no continuous functions satisfying the equation. (There's a quicker way to see this: set $x=1$, then $LHS=0$, but $RHS=5$.)
| {
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Show that $2^n<2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$ Is there a way to show that (with integer $n>2$) $$2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$$
I tried to figure a way with derivative or by looking how both side are growing, but I have some trouble with the ceiling function.
| With $n+k= {\lceil n \log_23\rceil}<n \log_23+1$ we can use the Rhin bound (page 160: $|µ_1\log 2+µ_2\log 3|\geq H^{-13.3}$ with $H=max(|µ_1|,|µ_2|)$)
$$|(n+k) \log2 - n \log3|>\frac 1{(n+k)^{13.3}}>\frac1{(n \log_23+1)^{13.3}}>\frac 1{n^a}$$
If we choose $a=15$ the above is true for $n>41$
Now (growth of $x^a$ vs $a^x$), $$\frac 1{n^a}>\frac{1}{(\frac{3}{2})^n}$$
is true for $n>\frac{-aW_{-1}(\frac{-\log\frac{3}{2}}{a})}{log(\frac{3}{2})}$ or $n>196$ with chosen $a=15$ ($W$ is the productlog), and we also have (using $x>log(1+x)$)
$$\frac{1}{(\frac{3}{2})^n}>\log(1+\frac{1}{(\frac{3}{2})^n})=\log(1+\frac{2^n}{3^n})$$
So for $n>196$ we have $$|(n+k)\log2 - n \log3|=\log(\frac{2^{n+k}}{3^n})>\log(1+\frac{2^n}{3^n})$$
or
$$\frac{2^{n+k}}{3^n}>1+\frac{2^n}{3^n}$$
$$2^{\lceil n \log_23\rceil}-3^n>2^n$$
Similarly we have
$$|n \log3-(n+k-1) \log2|>\frac 1{(n+k-1)^{13.3}}>\frac1{(n \log_23)^{13.3}}>\frac 1{n^a}$$
which also holds for $a=15$, and additionally we have:
$$\frac{1}{(\frac{3}{2})^n}=\frac{1}{2^{n\log_2\frac{3}{2}}}>\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}}>\log(1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}})$$
So for $n>196$ we have
$$|n \log3-(n+k-1) \log2|=\log(\frac{2\cdot 3^n}{2^{n+k}})>\log(1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}})$$
or
$$\frac{2\cdot 3^n}{2^{n+k}}>1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}}$$
$$2\cdot 3^n>2^{n+k}+2^n$$
$$2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$$
This leads to ($n>196$):
$$\begin{array}{|c|}\hline 2^n<2^{\lceil n \log_23\rceil}-3^n<3^n-2^n\\\hline\end{array}$$
and with manual checking, the left inequality holds except for $n$ in $\{1,3,5\}$ and the right inequality holds except for $n$ in $\{1,2\}$
| {
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Trigonometric system with different coefficients I have a trigonometric system as below and I want to solve it for $x$.
\begin{gather}
A\cos(x)+B\cos(2x)+C\sin(2x) = D \\
-C\cos(2x)-A\sin(x)+B\sin(2x) = F
\end{gather}
Can anyone suggest a solution?
Regards.
| Note $(1)$ the first equation and $(2)$ the second. Multiplying $(1)$ by $C$ and $(2)$ by $B$ the sum of two resulting equations give $$A\cos(\theta-x)+\cos(2x)=\frac{BD-CF}{B^2+C^2}=M$$ Similarly you get $$A\sin(\theta-x)+\sin(2x)=\frac{CD-BF}{B^2+C^2}=N$$ where $\sin(\theta)=\dfrac CB$. Now taking the square and adding you get
$$A^2(1)=(M-\cos(2x))^2+(N-\sin(2x))^2$$ so you have
$$M\cos(2x)+N\sin(2x)=\frac{M^2+N^2+1-A^2}{2}$$ which is a standard equation you can solve. In fact you do have
$$\sin(\phi+x)=\frac{M^2+N^2+1-A^2}{2(M^2+N^2)}$$ where $\sin(\phi)=\dfrac{M}{M^2+N^2}$.
| {
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Finding minimum of value $x^2+y^2+z^2+t^2$
Suppose $x,y,z,t$ are real numbers which holds for these equations:
numbers:
${ \begin{cases}{x^2+6y+2t=2} \\ {y^2+2t+y=-14} \\
{z^2+4x+2y=-27} \\ {t^2+2z+y=5}\end{cases} }$
What is the minimum
value of $x^2+y^2+z^2+t^2$ ?
$$1)12\quad\quad\quad\quad\quad\quad2)18\quad\quad\quad\quad\quad\quad3)34\quad\quad\quad\quad\quad\quad4)38\quad\quad\quad\quad\quad\quad5)\text{none}$$
$$$$
I added all the equations together:
$$x^2+y^2+z^2+t^2+4x+10y+4t+2z=-34$$
$$(x^2+4x+4)+(y^2+10y+25)+(z^2+2z+1)+(t^2+4t+4)=0$$
$$(x+2)^2+(y+5)^2+(z+1)^2+(t+2)^2=0$$
Since each squares are non negative, each of them should be zero, therefore $x=-2,\quad y=-5,\quad z=-1,\quad t=-2.$
But when I put these numbers in the original equations neither of them holds for the equations. so can we conclude there is not exist such real numbers $x,y,z,t$ that holds for the equations and therefore the answer is $5)\text{none}$. am I right?
| I have both checked your solution, and confirmed it via WolframAlpha. No real solutions exist for this system of equations, and hence, the answer should be none.
Here is the link to the WolframAlpha computation.
| {
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Solving multivariate quadratic equations over the integers I am looking for a method (if it exists) to solve over the integers the following sum of squares equation:
$$ x_1^2 + x_2^2+x_3^2 + \cdots + x_n^2 = m,$$ with $m \in \mathbb{N}.$
Someone has any idea about books, articles dealing with this kind of problem?
Thanks in advance!
| If $m\in\mathbb{N}$, then $x_n\in\mathbb{N}$ meaning that any $x$-values will work. If $m$ is a perfect square, the are also infinite solutions because the Pythagorean theorem works for any number of dimensions.
For every Pythagorean triple, there is a Pythagorean quadruple where $C$ of the triple is replaced by $(A,B)$ of some other triple, and the pattern continues indefinitely. To explain, let's use a backwards example. In the triple
$(5,12,13),$ we can replace $5$, with $(3,4)$ because $3^2+4^2=5^2 \implies 3^2+4^2+12^2=13^2.$
To find triples in the forward direction, we need only consider that there is a side A for ever odd number greater than $1$. If we define Euclid's formula as
$ \quad A=m^2-k^2\quad B=2mk \quad C=m^2+k^2\quad$ we can solve the $C$-function for $k$ and find that any $m$-value that yields an integer $k$ provides the $(m,k)$ needed for a Pythagorean triple.
Here is how an $A=5$ triple is found to replace the $C$-value in $(3,4,5)$
\begin{equation}
A=m^2-k^2\implies k=\sqrt{m^2-A}\\
\text{for}\qquad \lfloor\sqrt{A+1}\rfloor \le m \le \frac{A+1}{2}
\end{equation}
The lower limit ensures $k\in\mathbb{N}$ and the upper limit ensures $m> k$.
$$A=5\implies \lfloor\sqrt{5+1}\rfloor=2\le m \le \frac{5+1}{2} =3\\
\land
\quad m\in\{3\}\implies k \in\{2\} $$
$$F(3,2)=(5,12,13)$$
A similar process will find triples for $A=13$ and so on indefinitely to form n-tuples such as where
$$3^2+4^2+12^2+84^2+132^2+12324^2+6900^2+1428^2=14197^2\\
\text{ or }\quad
15^2+8^2+144^2+404^2+93744^2=93745^2$$
| {
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Find the coefficient of $x^{100}$ in $\frac{x^4}{\prod_{i=1}^{4}(1-x^{i})}$ Find the coefficient of $x^{100}$ in $$\frac{x^4}{\prod_{i=1}^{4}(1-x^{i})}$$
This problem came from finding total number of 4-partition of an integer 100.
One uses mathematica and it maybe easy, but I want to find more Mathematical proof.
f[m_, n_] :=
Coefficient[Series[x^m/Product[(1 - x^i), {i, 1, m}], {x, 0, n}], x,
n];
f[4, 100]
$7153$ is obtained. More general:
FullSimplify[
SeriesCoefficient[x^4/Product[(1 - x^i), {i, 1, 4}], {x, 0, n}],
Assumptions -> Element[n, PositiveIntegers]]
$$
\frac{1}{288} \left(-32 U_n\left(-\frac{1}{2}\right)+(n+1) \left(2 n
(n+2)+9 (-1)^n-13\right)+36 \cos \left(\frac{\pi n}{2}\right)\right)$$
| Partial fractions:
$$\begin{split}
\frac{288x^4}{(1-x)(1-x^2)(1-x^3)(1-x^4)}
&=\frac{288x^4}{(1-x)^4(1+x)^2(1+x+x^2)(1+x^2)}\\
&=\frac{12}{(1-x)^4} - \frac{12}{(1-x)^3} + \frac{13}{(1-x)^2} \\
&\quad + \frac{9}{(1+x)^2} - \frac{32}{1+x+x^2} + \frac{36}{1+x^2} \\
&=\frac{12}{(1-x)^4} - \frac{12}{(1-x)^3} + \frac{13}{(1-x)^2} \\
&\quad + \frac{9}{(1+x)^2} - \frac{32(1-x)}{1-x^3} + \frac{36(1-x^2)}{1-x^4}
\end{split}$$
Taking coefficients:
$$\begin{split}
[x^n]\frac{288x^4}{(1-x)(1-x^2)(1-x^3)(1-x^4)} &=
12\binom{n+3}{4}-12\binom{n+2}{3}+13\binom{n+1}{2} \\
&\quad + 9(-1)^n\binom{n+1}{2} -32([3\mid n]-[3\mid(n-1)]) + 36([4\mid n]-[4\mid(n-2)])
\end{split}$$
| {
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Find the whole number solutions for W+X+Y+Z = 15 where W, X, Y, Z ≤ 6 I worked out a solution but don't know if its the right one. Is this the right way to approach the problem? Any help would be appreciated.
First, the number of non-negative integer solutions for $W+X+Y+Z = 15$ can be calculated using stars and bars:
$$\binom{n+k-1}{n} = \binom{15+4-1}{15}$$
Now, when value of W exceeds 6; i.e., for value of 7 violets the rule of the upper limit for W.
That means the number of violations W can have is among 15-7 = 8 identical items in 4 distinct bins.
Equation for violations becomes W+X+Y+Z = 15-7 = 8
As a result, total number of violations for W = C (n+k-1, k-1) = C (8+4-1, 4-1) = C (11, 3)
⸫ Total number of violations for all 4 bins when 1 bin cross upper limit = C (11, 3) × 4
Since, number of violations 8 is >n/2=15/2=7.5; we need to add in the subtracted repetitions.
When 2 bins cross above upper limit; W+X+Y+Z = 15-7-7 = 1
⸫ Total number of repetitions for all 4 bins when 2 bin cross upper limit = C (1+4-1, 4-1) × 4 = C (4, 3) × 4
When 3 bins cross above upper limit; W+X+Y+Z = 15-7-7-7 = -6 i.e., not possible.
Therefore, the number of integer solutions for $W+X+Y+Z = 15$ where $W, X, Y, Z \leq 6$ is $$\binom{18}{3} – \binom{11}{3} \times 4 + \binom43\times 4$$
| There is a mistake in your last term. The answer should be $180$ and not $172$. Also you can simplify the working.
There are multiple ways to tackle the problem -
*
*Solve as is using P.I.E what you did
*Simplify using change of variable and solve
*Solve using generating function
a) Using first method (how you did; I have used boxes and balls analogy in my working),
the answer should be $\displaystyle {18 \choose 3} - {4 \choose 1} {11 \choose 3} + {4 \choose 2} {4 \choose 1} = 180$
The second term is where we choose one of the boxes to have $7$ balls and then distribute rest $8$ balls in any of the boxes.
The third term is where we choose two of the boxes to have $7$ balls each and then put one remaining ball in any of the boxes. This term is wrong in your working.
b) Using second method,
We substitute $a = 6 - w, b = 6 - x, c = 6- y, d = 6 - z$.
So we have, $a + b + c + d = 9 \ $ where $0 \leq a, b, c, d \leq 6$.
Only one of them can have more than $6$ balls. So the answer is
${12 \choose 3} - {4 \choose 1} {5 \choose 3} = 180$
c) Using generating function
Find coefficient of $x^{15}$ in $(1+x+x^2+x^3+x^4+x^5+x^6)^4$ which is indeed $180$. I will leave it to work through it if this is a method which is of interest to you.
| {
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Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 roots
Prove that equation $x^5-5x^3+4x-1=0$ has exactly 5 root.
By using intermediate value theorem, I can show that $x^5-5x^3+4x-1=0$ has at least one root in each following intervals: $(-2,-1.5),\ (-1.5,-1),\ (-1,0.5),\ (0.5,1),\ (1,3)$. So , it has exactly 5 roots.
But I wonder that there is some logical ways we can find intervals that contain roots without guessing?
| Given $f(x)=x^5-5x^3+4x-1$, differentiate first:
$$ f'(x) = 5x^4 - 15x^2 +4 . $$
Setting the derivative equal to $0$, we have
$$ 0 = 5x^4 - 15x^2 +4 . $$
This is quadratic in $x^2$, so we can use the quadratic formula:
$$ x^2 = \frac{15 \pm \sqrt{145}}{10} . $$
Since $\sqrt{145}<15$, $x^2>0$, so our solutions are
$$ x = \pm \sqrt{\frac{15 \pm \sqrt{145}}{10}} . $$
So, we have four distinct local extrema for $f$. If there are five distinct roots, each must be on one of the following intervals:
$$ \left(-\infty, -\sqrt{\frac{15+\sqrt{145}}{10}}\right) , $$
$$ \left(-\sqrt{\frac{15+\sqrt{145}}{10}}, -\sqrt{\frac{15-\sqrt{145}}{10}}\right) , $$
$$ \left(-\sqrt{\frac{15-\sqrt{145}}{10}}, \sqrt{\frac{15-\sqrt{145}}{10}}\right) , $$
$$ \left(\sqrt{\frac{15-\sqrt{145}}{10}}, \sqrt{\frac{15+\sqrt{145}}{10}}\right) , $$
$$ \left(\sqrt{\frac{15+\sqrt{145}}{10}}, \infty\right) . $$
| {
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Inequality with quadratic and maximum Let $x,y$ be real numbers in $[0,1]$ such that $$x \ge y \geq \max\left\{x^2, \frac{2x}{3}, \frac{1+x}{4}, \frac{1}{3}\right\}.$$ Is it true that $x^2+2y \ge 2x$?
In order to prove this, the term $x^2$ is currently on the "wrong" side, between the condition and the needed inequality. Using a linear combination of the conditions $y\geq \dots$ is not sufficient.
| nope not true...
let $x = 0.6, y=0.4, \max\left\{(0.6)^2, \frac{2*0.6}{3}, \frac{1+0.6}{4}, \frac{1}{3}\right\}=\max\left\{0.36, 0.4, 0.4, 0.33\right\} =0.4\\$
$\therefore x \geq y \geq \max\left\{x^2, \frac{2x}{3}, \frac{1+x}{4}, \frac{1}{3}\right\}$
holds when x = 0.6 and y = 0.4. Now check if $x^2+2y \geq 2x \implies 1.16 \geq 1.2$ which is false! Hence it is not true that $x^2+2y \geq 2x$
| {
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.