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How to evaluate $\iiint_{\mathbb{R}^3} \exp \left(-x^2 -y^2-z^2 \right) \mathrm{d}x \mathrm{d}y \mathrm{d}z$ How to evaluate
$$
\iiint_{\mathbb{R}^3} \exp \left( -x^2 -y^2 -z^2 \right) \mathrm{d}x \mathrm{d}y \mathrm{d}z,
$$
My Attempt:
If we use the transformation to spherical polar coordinates $R, \theta, \phi$, then we have
$$
\begin{align}
x &= R \sin \theta \cos \phi, \\
y &= R \sin \theta \sin \phi, \\
z &= R \cos \theta,
\end{align}
$$
where $0 \leq r < +\infty$, $0 \leq \theta \leq \pi$, and $0 \leq \phi < 2 \pi$.
And thus the Jacobian determinant
\begin{align}
&\frac{ \partial (x, y, z) }{ \partial (R, \theta, \phi ) } = \left\lvert \begin{matrix} \frac{\partial x}{\partial R } & \frac{\partial x }{ \partial \theta } & \frac{ \partial x }{ \partial \phi} \\ \frac{\partial y}{\partial R } & \frac{\partial y }{ \partial \theta } & \frac{ \partial y }{ \partial \phi} \\ \frac{\partial z }{\partial R } & \frac{ \partial z }{ \partial \theta } & \frac{ \partial z }{ \partial \phi} \end{matrix} \right\rvert \\
&= \left\lvert \begin{matrix} \sin \theta \cos \phi & R \cos \theta \cos \phi & - R \sin \theta \sin \phi \\ \sin \theta \sin \phi & R \cos \theta \sin \phi & R \sin \theta \cos \phi \\ \cos \theta & -R \sin \theta & 0 \end{matrix} \right\rvert \\
&= R^2 \sin \theta \left\lvert \begin{matrix} \sin \theta \cos \phi & \cos \theta \cos \phi & - \sin \phi \\ \sin \theta \sin \phi & \cos \theta \sin \phi & \cos \phi \\ \cos \theta & - \sin \theta & 0 \end{matrix} \right\rvert \\
&= R^2 \sin \theta \left( -\sin \phi \left\lvert \begin{matrix} \sin \theta \sin \phi & \cos \theta \sin \phi \\ \cos \theta & -\sin \theta \end{matrix} \right\rvert - \cos \phi \left\lvert \begin{matrix} \sin \theta \cos \phi & \cos \theta \cos \phi \\ \cos \theta & -\sin \theta \end{matrix} \right\rvert \right) \\
&= R^2 \sin \theta \left( -\sin^2 \phi \left\lvert \begin{matrix} \sin \theta & \cos \theta \\ \cos \theta & -\sin \theta \end{matrix} \right\rvert - \cos^2 \phi \left\lvert \begin{matrix} \sin \theta & \cos \theta \\ \cos \theta & -\sin \theta \end{matrix} \right\rvert \right) \\
&= R^2 \sin \theta \left( -\sin^2 \phi (-1) - \cos^2 \phi (-1) \right) \\
&= R^2 \sin \theta.
\end{align}
And, therefore we have
\begin{align}
\iiint_V &\exp \left( -x^2 -y^2 -z^2 \right) \, dx\, dy\, dz = \int_0^{2\pi} \int_0^\pi \int_0^\infty \exp\left( -R^2 \right)R^2 \sin \theta \, dR \, d\theta \, d\phi \\
&= 4 \pi \int_0^\infty R^2 \exp \left( - R^2 \right) \, dR \\
&= 4 \pi \left( - \frac{ R \exp \left( -R^2 \right) }{2 } \right)_{R=0}^{R=\infty} + 2 \pi \int_0^\infty \exp \left( -R^2 \right) \, dR \\
&= 2 \pi \int_0^\infty \exp \left( -R^2 \right) \, dR \\
&=
\end{align}
Is what I have done so far correct and clear enough? If so, then how to proceed from here? How to evaluate
$$
\int_0^\infty \exp \left( -R^2 \right) \, dR ?
$$
| The integral $\int_{0}^{\infty}e^{-R^2}dR$ is the Gaussian Integral.
It is very well known, the wikipedia link I have given above also contains a derivation. It equals-
$\int_{0}^{\infty}e^{-R^2}dR=\frac{\sqrt{\pi}}{2}$
So for your original integral,
$\begin{align}&\int\int\int_{\mathbb{R}^3}e^{-(x^2+y^2+z^2)}dxdydz\\&=2\pi\int_{0}^{\infty}e^{-R^2}dR\\&=2\pi\frac{\sqrt{\pi}}{2}\\&=\pi\sqrt{\pi}\\&=\pi^{\frac{3}{2}}\end{align}$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Can the substitution be different in this proof of a Fermat theorem? I was reading one of the Fermat theorem proof from the wikipedia
Specifically the theorem: " If $2^{k}+1$ is an odd prime, then $k$ is a power of 2."
In the proof it says:
Substituting $a=2^{r}$,$b=-1$,$b=-1$, $m = s$ and using that $s$ is
odd
I originally thought that we could have used $a=2^s$ and $m=r$ and it seemed to me it should work too, but I was wondering if I am misunderstanding something as that statement there using that s is odd I am not sure if it means something more and would create a problem with a different substitution
| If $k\in \Bbb Z^+$ and if $k$ is not a power of $2$ then $k\ge 3$ and $k$ has an odd divisor $p>1.$ So $k=k'p$ with $p$ odd and $p\ge 3$ and $k'\in \Bbb Z^+.$ Then, because $p$ is odd, we have $$2^k+1=2^{pk'}+1=(2^{k'})^p+1\equiv (-1)^p+1\equiv (-1)+1\equiv 0 \pmod {2^{k'}+1}$$ so $2^{k'}+1$ is a divisor of $2^k+1$ with $1<2^{k'}+1<2^{k'\cdot 3}+1\le 2^{k'p}+1=2^k+1$
so $2^k+1$ is not prime.
We can also see that $2^{k'}+1$ is a proper divisor of $2^k+1$ from the algebraic identity $x^p+1=(x+1)(x^{p-1}-x^{p-2}+-...+1)$ when $p$ is odd and $p\ge 3$. E.g
$x^3+1=(x+1)(x^2-x+1).$
$ x^5+1=(x+1)(x^4-x^3+x^2-x+1).$
$x^7+1=(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)$, et cetera.
For example with $x=2^{k'}$ and if $p=5.$ Then $x>1$ so $$x^4-x^3+x^2-x+1=(x^4-x^3)+(x^2-x)+1>1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4208507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}} + \frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$ Question: Simplify $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}} + \frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$
My Attempt:
On rationalizing both fractions separately and then adding (since denominator becomes $\sqrt{3}$) I got $$-2\sqrt{6}+2\sqrt{2+\sqrt{3}}+2\sqrt{2-\sqrt{3}}$$
However the given answer is $$\frac{\sqrt{6}}{3}$$
Even more confusing is that on inputting the problem into wolframalpha the solution is given as $$\sqrt{2}$$
I have broken my head over this for a couple of hours and I just can't find a solution. Hope someone can help.
| Hint
$$2-\sqrt3=\dfrac{(\sqrt3-1)^2}2$$
$$\implies\sqrt2-\sqrt{2-\sqrt3}=\dfrac{2-(\sqrt3-1)}{\sqrt2}$$
$$\implies\dfrac{2-\sqrt3}{\sqrt2-\sqrt{2-\sqrt3}}=\sqrt{\dfrac23}\cdot\dfrac{2-\sqrt3}{\sqrt3-1}$$
$$\dfrac{2+\sqrt3}{\sqrt2+\sqrt{2+\sqrt3}}=?$$
Now take the sum
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is my method correct to answering this question? Is there a quicker method to solve this question?
What is the smallest number that can be written as the sum of three, four and five consecutive numbers?
I encountered this question while doing my Math summer homework. I have tried to make progress on this question.
Sum of three consecutive numbers = $x + x+1 + x+2 = 3x+3$
Sum of four consecutive integers = $x + x+1 + x+2 + x+3 = 4x+6$
Sum of five consecutive integers = $x + x+1 + x+2 + x+3 + x+4 = 5x+10$
The number must be the lowest common multiple of $3x+3$ , $4x+6$ and $5x+10$, which is $60x + 30$.
Substituting $x = 0$ gives us the lowest positive, non-zero and whole number, which is $30$.
$$30 = 9 + 10 + 11\\
30 = 6 + 7 + 8 + 9\\
30 = 4 + 5 + 6 + 7 + 8$$
Is my answer correct? If not, where in my method have I produced an error? Is there an ever quicker method to solve this question?
| Your reasoning contains a subtle (essentially notational) error, which does not greatly affect the result you obtain. You say:
The number must be the lowest common multiple of $3x+3$, $4x+6$ and $5x+10$, which is $60x + 30$.
But the value of $x$ in each of $3x+3$, $4x+6$ and $5x+10$ is not the same; by your own results it is either $4$ or $6$ or $9$. So saying $60x+30$ and $x=0$ is not logically consistent: you did not mean to imply that the sum $3x+3=3$ was equal to $4x+6=6$ or $5x+10=10$.
The better reasoning goes as follows: One of the sums has $3$ as a factor, another has $2$ (but not $4$) as a factor, and the third has $5$ as a factor. Since they each equal the same sum, that sum must have $2,3,5$ as factors, and so must be an odd multiple of $30$. In your terms, $S=60k+30$ where $k$ is not identically the same as the $x$ in any of your sums.
$30$ itself is in fact a solution, and so are $90,150,\dots$, viz:
$$90=29+30+31=21+22+23+24=16+17+18+19+20 \\
150=49+50+51=36+37+38+39=28+29+30+31+32$$
The smallest such number is $30$, as you found.
As I get set to post this, I see that Erick Wong has made essentially the same objection.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Relationship between Chebyshev polynomials and square roots ($\sqrt{3}+\sqrt{2}=\frac{1}{\sqrt{T_1(5)-\sqrt{T_1(5)^2-1}}}$ etc.) (If my English is strange, I would appreciate it if you could correct it.)
There seems to be a property about the sum of square roots. (This is almost self-explanatory.)
let $ a,\ b,\ t \in \mathbb{N}^0,\ a\geq b$
$$ \sqrt{a}+\sqrt{b} =
\begin{cases}
A &(t\ mod\ 2 = 0)&\\
\frac{a-b}{A} &(t\ mod\ 2= 1)&
\end{cases}
$$
$$ A = \sqrt[2^t]{B+(-1)^{t}\sqrt{B^2-(a-b)^{2^t}}}$$
$$ B = \frac{(\sqrt{a}+\sqrt{b})^{2^t}+(\sqrt{a}-\sqrt{b})^{2^t}}{2}$$
Interestingly, $B$ seems to be equal to $T_{2^{t-1}}(a+b)$ when $a-b = 1$.
$T_n(x)$ is Chebyshev polynomials that is expressed by following equation.
$$T_n(x) = n\sum_{k=0}^{n}{(-2)^k\frac{(n+k-1)!}{(n-k)!(2k)!}(1-x)^k}$$
Originally, the range of $x$ is $-1$ to $1$, but this property can be seen when extended to natural numbers.
Please comment if you know anything.
| According to wiki, $T_n(x)$ can be expressed by the following equation too.
$$ T_n(x) = \frac{\left(x+\sqrt{x^2-1}\right)^n+\left(x-\sqrt{x^2-1}\right)^n}{2} $$
Hence,
$$ T_{2^{t-1}}(a+b) = \frac{(a+b+\sqrt{(a+b)^2-1})^{2^{t-1}}+(a+b-\sqrt{(a+b)^2-1})^{2^{t-1}}}{2} $$
On the other hands,
\begin{eqnarray*}
B &=& \frac{(\sqrt{a}+\sqrt{b})^{2^t}+(\sqrt{a}-\sqrt{b})^{2^t}}{2}\\
&=& \frac{(a+b+\sqrt{4ab})^{2^{t-1}}+(a+b-\sqrt{4ab})^{2^{t-1}}}{2}
\end{eqnarray*}
That is, if the square roots are equal, then $T_{2^{t-1}}(a+b)$ and $B$ are equal.
Therefore,
\begin{eqnarray*}
(a+b)^2-1 &=& 4ab\\
a^2+2ab+b^2-1 -4ab &=& 0\\
(a-b)^2 &=& 1\\
a-b &=&\pm1
\end{eqnarray*}
This indicates that $T_{2^{t-1}}(a+b)$ and $B$ are equal when $a-b =\pm1$.
For example, let $a=3,b=2$,
\begin{eqnarray*}
\sqrt{3}+\sqrt{2} &=& \frac{1}{\sqrt{T_1(5)-\sqrt{T_1(5)^2-1}}} = \frac{1}{\sqrt{5-\sqrt{24}}}\\
&=& \sqrt[4]{T_2(5)+\sqrt{T_2(5)^2-1}}= \sqrt[4]{49 +\sqrt{2400}}\\
&=& \frac{1}{\sqrt[8]{T_4(5)-\sqrt{T_4(5)^2-1}}} = \frac{1}{\sqrt[8]{4801-\sqrt{23049600}}}\\
&\vdots&
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Decomposing a large block $7\times 4 \times 5$ into small ones $1\times 1 \times 1,\ 2\times 2 \times 2,\ 3\times 3 \times 3$ Consider a block of size $7\times 4 \times 5$. We want to make this with cubes of size $1\times 1 \times 1,\ 2\times 2 \times 2,\ 3\times 3 \times 3$. If $x,\ y,\ z$ cubes are needed respectively, then what is the smallest number $x$ ?
Proof : Here $z$ must be in $\{0,1,2\}$. $(x,y,z) = (38,6,2),\ (33,10,1),\ (44,12,0)$ are eamples. Hence I think that $x=33$ is the smallest. How can we prove this ?
| As noted in the comments, there is a valid configuration with $(x,y,z)=(22,8,2)$.
Claim:$\;$Every valid configuration $(x,y,z)$ has $x\ge 22$.
Proof:
Suppose $(x,y,z)$ is a valid configuration.
Summing the volumes, we get $x+8y+27z=140$, hence $x\equiv 4-3z\;(\text{mod}\;8)$.
As you noted, we must have $z\in\{0,1,2\}$.
Place the large block so it has horizontal base $7{\times}4$ and height $5$, and regard it as comprised of $28$ vertical $1{\times}1{\times}5$ columns.
Then each of the $28$ vertical columns either contains a $1$-cube, or passes through the interior of exactly one $3$-cube.
It follows that $x\ge 28-9z$.
Consider $3$ cases . . .
Case $(1)$:$\;z=0$.
Then from $x\ge 28-9z$, we get $x\ge 28$.
Case $(2)$:$\;z=1$.
Then from $x\ge 28-9z$, we get $x\ge 19$, and from $x\equiv 4-3z\;(\text{mod}\;8)$,we get $x\equiv 1\;(\text{mod}\;8)$, hence $x\ge 25$.
Case $(3)$:$\;z=2$.
Now regard the large block as comprised of $35$ horizontal $1{\times}1{\times}4$ rows.
Since $z=2$, exactly $18$ of the $35$ horizontal rows pass through the interior of a $3$-cube.
Necessarily each of those $18$ rows also contains a $1$-cube, hence $x\ge 18$.
But then from $x\equiv 4-3z\;(\text{mod}\;8)$,we get $x\equiv 6\;(\text{mod}\;8)$, so $x\ge 22$.
Thus in all cases, we have $x\ge 22$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the measure of segment AB in triangle rectangle ABC For reference:
Given the triangle $ABC$, straight at $B$. The perpendicular bisector of $AC$ intersects at $P$
with the angle bisector of the outer angle $B$, then $AF \parallel BP$ ($F\in BC$) is drawn.
If $FC$ = $a$, calculate BP(x). (Answer: $\frac{a\sqrt2}{2})$
My progress:
Point P is on the circumcircle of ABC because the angle $\measuredangle ABP = 135^o$
$ Where~ AC = 2R\\
\triangle CBP \rightarrow
PC^2 = BP^2 + BC^2 - \sqrt2BCBP\\
but~PC = PA = R\sqrt{2} \text{(since P is in the bisector AC)}\\
2R^2 = BP^2 +BC^2 - \sqrt{2}BCBP\\
0= BP^2 - \sqrt{2}BCBP + BC^2 - 2R^2\\
0= (BP - \frac1{ \sqrt{2}}BC)^2 + \frac{BC^2}2 - 2R^2\\
0= (BP - \frac1{ \sqrt{2}}BC)^2 + \frac{BC^2-4R^2}2\\
0= (BP - \frac1{ \sqrt{2}}BC)^2 - \frac{AB^2}2\\
$
I can't find the relationship between BC, AB and a...
If anyone finds another way to solve by geometry I would be grateful
|
You have done most of the work by coming up with the nice construction. I will just mark point $Q$, draw chord $CQ$ and use it to find the answer.
As $BP$ and $AQ$ are parallel, $\angle AFB = \angle FBP = 45^0$
So $\angle CFQ = 45^0$ and as $\angle CQF = 90^0$, $CQ = \cfrac{a}{\sqrt2}$
But also note that, $\angle CAQ = \angle BAP = \angle A - 45^0$
So we must have, chord $BP$ = chord $CQ$.
$\therefore x = \cfrac{a}{\sqrt2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4217432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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For $1 \not= \alpha \in \mathbb{C}$ such that $\alpha^7 = 1$, evaluate $\alpha + \alpha^2 + \alpha^4.$ For $1 \not= \alpha \in \mathbb{C}$ such that $\alpha^7 = 1$, evaluate $\alpha + \alpha^2 + \alpha^4.$
My solution :
let $$p = \alpha + \alpha^2 + \alpha^4$$ and $$q = \alpha^3 + \alpha^5 + \alpha^6.$$
We know $$1 + \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 = 0,$$
$$p + q = \alpha + \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6$$
$$ = -1$$
and
$$pq = (\alpha + \alpha^2 + \alpha^4)(\alpha^3 + \alpha^5 + \alpha^6)$$
$$= \alpha^4 + \alpha^6 + \alpha^7 + \alpha^5 + \alpha^7 + \alpha^8 + \alpha^7 + \alpha^9 + \alpha^{10}$$
$$= 2.$$
Therefore, $p$ and $q$ are the two roots of the following equation :
$$x^2 + x + 2 = 0$$
and
$$p = \alpha + \alpha^2 + \alpha^4$$
$$= \frac{-1 ± \sqrt{7} i}{2}.$$
Would there be other ways of evaluating? I'm thinking of polar forms but not sure how to do this with it.
| Yet another way (albeit not too different from OP's), using that $\,|\alpha| = 1\,$, $\,\overline \alpha=\dfrac{1}{\alpha}\,$:
*
*$\require{cancel}\displaystyle \;p+\overline p = \alpha + \alpha^2+\alpha^4 + \overline \alpha + \overline \alpha^2 + \overline \alpha^4 =\alpha + \alpha^2+\alpha^4 + \frac{1}{\alpha}+\frac{1}{\alpha^2}+\frac{1}{\alpha^4} \\\displaystyle =\frac{\cancel{\alpha^5+\alpha^6+\alpha+\alpha^3+\alpha^2+1\color{red}{+\alpha^4}}-\color{red}{\alpha^4}}{\alpha^4}=\frac{-\alpha^4}{\alpha^4}=-1$
*$\displaystyle \;|p|^2=\cancel{|\alpha|^2}\,|1+\alpha+\alpha^3|^2=\left(1+\alpha+\alpha^3\right)\left(1+\overline\alpha+\overline\alpha^3\right)= \left(1+\alpha+\alpha^3\right)\left(1+\frac{1}{\alpha}+\frac{1}{\alpha^3}\right) \\\displaystyle = \frac{\left(1+\alpha+\alpha^3\right)\left(1+\alpha^2+\alpha^3\right)}{\alpha^3}=\frac{\cancel{1+\alpha^2+\alpha^3+\alpha}+\alpha^3+\cancel{\alpha^4}+\alpha^3+\cancel{\alpha^5+\alpha^6}}{\alpha^3}=2$
The two relations give the real part $\dfrac{-1}{2}$ and magnitude $\sqrt{2}$ of $\,p\,$, so $\,p = \dfrac{-1}{2} \pm i\,\sqrt{2 - \dfrac{1}{4}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "8",
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$\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}\geq 2(\sin\frac{A}{2}\sin\frac{B}{2}+\sin\frac{B}{2}\sin\frac{C}{2}+\sin\frac{A}{2}\sin\frac{C}{2})$ Let $A,B,C$ be the three angles of a triangle. Prove that:
$\sin\dfrac{A}{2}+\sin\dfrac{B}{2}+\sin\dfrac{C}{2}\geq 2(\sin\dfrac{A}{2}\sin\dfrac{B}{2}+\sin\dfrac{B}{2}\sin\dfrac{C}{2}+\sin\dfrac{A}{2}\sin\dfrac{C}{2})$
Here all I did:
$p=\dfrac{a+b+c}{2}$
$\sin \dfrac{A}{2} = \sqrt { \dfrac {(p-b)(p-c)}{bc}} $
$\Rightarrow \sin \dfrac{A}{2} \sin \dfrac{B}{2} = \dfrac{(p-c)}{c} \sqrt { \dfrac {(p-a)(p-b)}{ba}}$
$\Rightarrow 2 \sin \dfrac{A}{2} \sin \dfrac{B}{2} \le\dfrac{(p-c)}{c} \dfrac {2p-a-b}{\sqrt{ba}}$
$\Rightarrow 2 \sin \dfrac{A}{2} \sin \dfrac{B}{2} \le\dfrac{(p-c)}{c} \dfrac {c}{\sqrt{ba}}$
$\Rightarrow 2 \sin \dfrac{A}{2} \sin \dfrac{B}{2}\le \dfrac{(p-c)}{\sqrt{ba}}$
So we need to prove that : $\sum \dfrac{(p-c)}{\sqrt{ba}} \le \sum\sqrt { \dfrac {(p-b)(p-c)}{bc}} $. But I still have no idea (I'm not sure that's true either). I hope to get help from everyone. Thanks a lot
| Here's an alternative that take advantage of convexity of sine in $(0,\pi)$
$$
\begin{align}
\left(\sin{\frac{A}{2}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}}\right)\cdot 2\cdot\sin{\frac{A/2+B/2+C/2}{3}}&\geq\frac{2}{3}\left(\sin{\frac{A}{2}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}}\right)^{2}\\
&\geq 2\left(\sin{\frac{A}{2}\cdot\sin{\frac{B}{2}}+\sin{\frac{B}{2}}\cdot\sin{\frac{C}{2}}+\sin{\frac{C}{2}}}\cdot\sin{\frac{A}{2}}\right)
\end{align}
$$
Recall that $2\sin{\frac{A/2+B/2+C/2}{3}}=2\sin{\frac{\pi}{6}}=1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Write a linear system in a new basis thanks for spending some time helping :)
How exactly do you represent a linear system in a different basis?
Let's say the new basis B = {(1,1),(-1,1)}
How do I write the system
\begin{cases} x+y=7\\ 2x–4y=-16 \end{cases}
in the new basis?
| In terms of matrices your system is
$$\left[\begin{array}{cc}1&1\\2&-4\end{array}\right]
\left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}7\\-16\end{array}\right],$$
but if your basis change is
$$v_1=e_1+e_2,$$
$$v_2=-e_1+e_2,$$
then the old basis in terms of the new one is expressed as
$$e_1=\frac{1}{2}v_1-\frac{1}{2}v_2,$$
$$e_2=\frac{1}{2}v_1+\frac{1}{2}v_2.$$
The basis change associated matrix is
$B=\left[\begin{array}{cc}1&-1\\1&1\end{array}\right]$.
If the problem above can be written as
$Tu=w$
then by multiplying with $B^{-1}$ you will get the equivalent problem given by
$$(B^{-1}TB)B^{-1}u=B^{-1}w,$$
which is
$$\left[\begin{array}{cc}0&-3\\-2&-3\end{array}\right]
\left[\begin{array}{c}\bar x\\\bar y\end{array}\right]=
\left[\begin{array}{c}-\frac{9}{2}\\-\frac{23}{2}\end{array}\right],$$
where the new variables are
$$\bar x=\frac{1}{2}x+\frac{1}{2}y,$$
$$\bar y=-\frac{1}{2}x+\frac{1}{2}y.$$
So the system is
\begin{cases} 3\bar y=\frac{9}{2}\\ \\2\bar x+3\bar y=\frac{23}{2}\end{cases}
Addendum
The original problem's solution is
$u_0=\left[\begin{array}{c}2\\5\end{array}\right]$
and the transformed problem has the solution
$\left[\begin{array}{c}\frac{7}{2}\\\frac{3}{2}\end{array}\right]$, that perfectly coincides with $B^{-1}u_0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
In the figure, the polygons shown are regular. Find $"x"$. For reference:
my progress:
I marked the angles I could find but couldn't finish
$a_{i9} = \frac{180(n-2)}{n}=\frac{180(9-2}{9}=\angle140^\circ\\a_{i6}= \frac{180(6-2)}{6}=\angle120^\circ\\\angle ADB = \frac{360^\circ-2(140^\circ)}{2} = \angle 40^\circ\\\angle BJI = \frac{\angle 180^\circ-\angle 120^\circ}{2}=30^\circ\\a_{i5}= \frac{180(5-2)}{5}=\angle108^\circ\\\\
S_{ai5} = 180(n-2) = 180(5-2) =\angle 540^\circ \\
\angle HDE = \frac{540 - 3(140)}{2}=\angle 60^\circ$
| Lengthen $BA$ to the left and find the angles those it makes with $AM$ and $AL$. Then using them find $\angle MAL$ and $\angle LAN$. Then you can find $\angle ANL$ and eventually $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve in $\mathbf{N}$ the equation $9x^2+p=y^2$ In these days, I have been trying to solve this problem:
Let $p \in \mathbf{N}$ a positive large integer ($> 10^9$). Find all $x, y \in\mathbf{N}$ such that:
$$9x^2+p=y^2$$
The first approach that I have tried is the following. We know that:
$$(t+n)^2-t^2=t^2+2\cdot t \cdot n +n^2-t^2=2\cdot t \cdot n +n^2$$
Also, we can build every possible square $w_n$ greater than $p$ in this way:
$$w_n=\left\lfloor\sqrt{p}\right\rfloor^2+2\cdot \left\lfloor\sqrt{p}\right\rfloor \cdot n +n^2 = \left(n+\left\lfloor\sqrt{p}\right\rfloor^2\right)^2$$
For example, let $p=5$. Follows that: $w_1=\left(1+\left\lfloor\sqrt{5}\right\rfloor^2\right)^2=9$, $w_2=\left(2+\left\lfloor\sqrt{5}\right\rfloor^2\right)^2=16$ and so on.
Now, using this idea and applying to the first equation:
$$9x^2=\left(n+\left\lfloor\sqrt{p}\right\rfloor^2\right)^2-p$$
I am not allowed to apply Pell's equation because $9=3^2$ and calcultaing $\Delta$ in $x$ doesn't help anymore.
Another approach is based on Pell's equation. I thought to express $9x^2=8x^2+x^2$ and then:
$$8x^2+x^2+p=y^2\leftrightarrow 8x^2+p=y^2-x^2\leftrightarrow (y^2-x^2)-8x^2=p \leftrightarrow u^2-8x^2=p$$
But then, in order to generate all the solutions, I have to guess the first one (or one of them) that is pretty complicated for big $p$.
So, how can we do that? Are there any other solutions?
Thanks.
| First, reformulate the problem as the following:
$$
y^2 - 9x^2 = p \Rightarrow(y-3x)(y+3x)= p
$$
Now, for any given $p$, find its prime factors. Then, for any 2-partitions of them, solve a simple equation system.
To simplify some cases, suppose $p$ is factorized to $p_1 \times p_2$. Now, solve the following system:
$$
y-3x = p_1
$$
$$
y+3x = p_2
$$
So, $y = \frac{p_1 + p_2}{2}$, and $x = \frac{p_2 - p_1}{6}$. It gives us a heuristic to find potential answers more quickly. $p_2$ is in the form of $6k + p_1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Find the maximum of the value $F=x^3y+y^3z+z^3x$ let $x,y,z$ be real number.if $x+y+z=3$,show that
$$x^3y+y^3z+z^3x\le \dfrac{9(63+5\sqrt{105})}{32}$$
and the inequality $=$,then $x=?,y=?,z=?$
I can solve if add $x,y,z\ge 0$,also see: Calculate the maximum value of $x^3y + y^3z + z^3x$ where $x + y + z = 4$ and $x, y, z \ge 0$.
But for real $x,y,z$ I can't solve it
| It is not surprizing that Mathematica does not produce explicit expressions for $(x,y,z)$.
Using it with the restriction $x>y>z$, it gives that $y$ and $z$ are respectively the fourth and the first roots of the equation
$$56 t^6-336 t^5+378 t^4+819 t^3-2079 t^2+1512 t-351=0$$
Using @River Li's comment, $(x,y,z)$ are the solutions of the cubic
$$u^3-3u^2-\frac{3}{8} \left(3+\sqrt{105}\right)u+\frac{3}{112} \left(147+17 \sqrt{105}\right)=0$$ and, using the trigonometric method, they are
$$x=1+\sqrt{\frac{11+\sqrt{105}}{2} } \cos \left(\frac{1}{3} \cos
^{-1}\left(-\frac{91+9 \sqrt{105}}{7 \sqrt{2}
\left(11+\sqrt{105}\right)^{3/2}}\right)\right)=3.71382\cdots$$
$$y=1-\sqrt{\frac{11+\sqrt{105}}{2} } \sin \left(\frac{\pi
}{6}-\frac{1}{3} \cos ^{-1}\left(-\frac{91+9 \sqrt{105}}{7 \sqrt{2}
\left(11+\sqrt{105}\right)^{3/2}}\right)\right)=1.20642\cdots$$
$$z=1-\sqrt{\frac{11+\sqrt{105}}{2} } \sin \left(\frac{\pi
}{6}+\frac{1}{3} \cos ^{-1}\left(-\frac{91+9 \sqrt{105}}{7 \sqrt{2}
\left(11+\sqrt{105}\right)^{3/2}}\right)\right)=-1.92024\cdots$$
Trying to convert to radicals would be more than problematic.
Edit
After a series of mistakes, using @River Li's comments, the factorization of the sextic as the product of two cubic polynomials
$$(t^3 + b_2t^2 + b_1t + b_0)\,(t^3 + a_2t^2 + a_1t + a_0)$$gives
$$b_2=-3 \qquad \qquad b_1=-\frac{3}{8} \left(3-\sqrt{105}\right)\qquad \qquad b_0=\frac{3}{112} \left(147-17 \sqrt{105}\right)$$
$$a_2=-3 \qquad \qquad a_1=-\frac{3}{8} \left(3+\sqrt{105}\right)\qquad \qquad a_0=\frac{3}{112} \left(147+17 \sqrt{105}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$ How to find $\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$? Here is my attempt.
Put $f(x)=\frac{x}{\log x}(x^{\frac{1}{x}}-1)$ for $x>1$. Then
\begin{aligned}
\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1) &= \lim\limits_{x\rightarrow\infty} f(x) \\
&= \lim\limits_{x\rightarrow\infty}\frac{x^{\frac{1}{x}}-1}{\frac{\log x}{x}} \\
&= \lim\limits_{x\rightarrow\infty} \frac{x^{\frac{1}{x}-2}(1-\log x)}{\frac{x^2+\log x}{x^4}} \\
&= \lim x^{\frac{1}{x}}\lim x^2\lim\frac{1-\log x}{x^2+\log x} \\
&= 1 \cdot (+\infty)\cdot (0)\mbox{.}
\end{aligned}
This method does not work.
| Rewrite the limmand as
$$\frac{n}{\log n}\left(n^{\frac{1}{n}}-1\right) = \frac{n}{\log n}\left(e^{\frac{\log n}{n}}-1\right)$$
Then with $x = \frac{\log n}{n}$ we have
$$\lim_{x \to 0^+} \frac{e^x-1}{x} = 1 $$
the classic limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Differential equation $y' = (2y^2 + x)/(3y^2 + 5)$ (Apostol, section 8.28, ex. 30) Problem
This is from Apostol's Calculus book, section 8.28, exercise 30.
Let $y = f(x)$ be that solution of the differential equation
$$y' = \dfrac{2y^2 + x}{3y^2 + 5}$$
which satisfies the initial condition $f(0) = 0$.
(a) The differential equation shows that $f'(0) = 0$. Discuss whether $f$ has a relative maximum or minimum or neither at 0.
(b) Notice that $f'(x) \geq 0$ for each $x \geq 0$ and that $f'(x) \geq \frac{2}{3}$ for each $x \geq \frac{10}{3}$. Exhibit two positive numbers $a$ and $b$ such that $f(x) > ax - b$ for each $x \geq \frac{10}{3}$.
(c) Show that $x/y^2 \to 0$ as $x \to +\infty$.
(d) Show that $y/x$ tends to a finite limit as $x \to +\infty$ and determine this limit.
Solution attempt
Please, I would like to ask for verification of my attempt below.
(a) The second derivative is given by:
$$y'' = \dfrac{(3y^2 + 5)(4yy' + 1) - 6yy'(2y^2 + x)}{(3y^2 + 5)^2}$$
At zero, the value is:
$$f''(0) = \dfrac{5}{(5)^2} = \dfrac{1}{5}$$
Since this value is positive, we can conclude that the concavity of $f$ at 0 is up, so $f$ has a relative minimum at 0.
(b) Since $f'(x) \geq \frac{2}{3}$ for each $x \geq \frac{10}{3}$, in order to have the line $ax - b$ below $f(x)$, we can choose $a = \frac{2}{3}$. Also, we can choose $b$ such that $ax - b = 0$ at $x = \frac{10}{3}$ (since $f(x) \geq 0$ there). This gives:
$$\frac{2}{3} \cdot \frac{10}{3} - b = 0 \implies b = \frac{20}{9}$$
So, the numbers are $a = \frac{2}{3}$ and $b = \frac{20}{9}$.
(c) In item (b), we found that:
$$y > \frac{2}{3}x - \frac{20}{9}$$
for $x \geq \frac{10}{3}$. This implies:
$$\frac{3}{2} y + \frac{10}{3} > x$$
for $x \geq \frac{10}{3}$. Dividing both sides by $y^2$, this becomes:
$$\dfrac{3}{2y} + \dfrac{10}{3y^2} > \dfrac{x}{y^2}$$
Also, we know that $x/y^2 \geq 0$, so we have:
$$\dfrac{3}{2y} + \dfrac{10}{3y^2} > x/y^2 \geq 0$$
Since we know that $y > \frac{2}{3}x - \frac{20}{9}$, we have that $y \to +\infty$ as $x \to +\infty$, so the left-hand side of the left-hand inequality above tends to zero as $x \to \infty$:
$$\dfrac{3}{2y} + \dfrac{10}{3y^2} \to 0\text{ as }x \to \infty$$
Therefore, by the squeeze theorem, it follows that $x/y^2 \to 0$ as $x \to +\infty$.
(d) From item (c), we know that $x/y^2 \to 0$ as $x \to +\infty$. So:
$$\begin{aligned}
\lim_{x \to +\infty} y' &= \lim_{x \to +\infty} \dfrac{2y^2 + x}{3y^2 + 5} \\
&= \lim_{x \to +\infty} \dfrac{2 + x/y^2}{3 + 5/y^2} \\
&= \dfrac{2 + 0}{3 + 0} \\
&= \dfrac{2}{3}
\end{aligned}$$
So, as $x \to +\infty$, the function $y$ approaches a line of slope $\frac{2}{3}$. That is, $y$ approaches a line of the form $y=\frac{2}{3}x + C$, so $y/x \to \frac{2}{3}$ as $x \to +\infty$.
The argument in part (d) seems a bit informal; I'm not sure how to make it more rigorous.
| Based on Paul Sinclair's suggestion in the comments to the question, here is an attempt at a new argument for part (d).
We found that $\lim_{x \to +\infty} y' = 2/3$. This means that, given $\epsilon > 0$, there is a $N > 0$ such that, if $x > N$, then
$$\left|y' - \dfrac{2}{3}\right| < \epsilon$$
This implies
$$-\epsilon < y' - \dfrac{2}{3} < \epsilon \implies -\epsilon + \dfrac{2}{3} < y' < \epsilon + \dfrac{2}{3}$$
Integrating, we get:
$$\begin{aligned}
&\int_0^x \left(-\epsilon + \dfrac{2}{3}\right) dz < \int_0^x y' dz < \int_0^x \left(\epsilon + \dfrac{2}{3}\right) dz \\
\implies & \left(-\epsilon + \dfrac{2}{3}\right)x < y - y(0) < \left(\epsilon + \dfrac{2}{3}\right)x \\
\implies & \left(-\epsilon + \dfrac{2}{3}\right)x < y < \left(\epsilon + \dfrac{2}{3}\right)x
\end{aligned}$$
Dividing by $x$ (since $x \neq 0$), we get:
$$\begin{aligned}
&-\epsilon + \dfrac{2}{3} < \dfrac{y}{x} < \epsilon + \dfrac{2}{3} \\
\implies & -\epsilon < \dfrac{y}{x} - \dfrac{2}{3} < \epsilon \\
\implies & \left|\dfrac{y}{x} - \dfrac{2}{3}\right| < \epsilon
\end{aligned}$$
That is, we have shown that, given $\epsilon > 0$, there is a $N > 0$ such that, if $x > N$, then $|y/x - 2/3| < \epsilon$. That is,
$$\lim_{x \to +\infty} \dfrac{y}{x} = \dfrac{2}{3}$$
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proof of $\left(1+\frac1n\right)^n<3-\frac3n$ for $n\geq7$? We have the following inequality:
$$e_n:=\left(1+\frac1n\right)^n<3-\frac3n,$$
for $n\in\mathbb{N}$ and $n\geq7$.
My proof as follows: $e_1=2<2.75,e_2=2.25<2.75$, and for $\forall n\in\mathbb{N},n\geq2$,
\begin{align*}
\left(1+\frac{1}{n}\right)^n
&=2+\sum_{k=2}^{n}
\frac{1}{k!}\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)\\
&\leq2+\sum_{k=2}^{n}\frac{1}{k!}\\
&\leq2+\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{3^2}+\cdots+\frac{1}{3^{n-2}}\right)\\
&<2+\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{3^2}+\cdots+\frac{1}{3^{n-2}}+\cdots\right)\\
&=2.75.
\end{align*}
When $n\in\mathbb{N}$ and $n\geq12$,
$$\left(1+\frac{1}{n}\right)^n<2.75=3-\frac{3}{12}\leq3-\frac3n.$$
And for $n=7,8,9,10,11$, we can check the inequality directly (or use wolfram mathematica).
What I want to konw: Are there any elementary proofs of this inequality?
Any help and hints will welcome!
| This isn't exactly elementary but does the job. We wish to show that $f(x)<3$ for all real $x\ge7$ where $f(x)=(1+1/x)^x/(1-1/x)$. Its derivative is given by $$f'(x)=g(x)\left(x(x^2-1)\log\left(1+1/x\right)-x^2-1\right)$$ where $g(x)=(1+1/x)^x/((x-1)^2(x+1))$ is positive. Since $$x(x^2-1)\log(1+1/x)-x^2-1<(x^2-1)(x\log(1+1/x)-1)<0$$ for all $x>1$, we know that $f(x)$ is strictly decreasing for all $x>1$. The claim follows as $f(7)<3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
6-dimensional integral with strange result I am working through a physics paper where I encounter the following 6-dimensional integral:
$$ G(\mathbf{q}) = \frac{2}{(2\pi)^6 n^2} \int_{k \leq q_F} \int_{k' \leq q_F} \frac{\mathbf{q} \cdot (\mathbf{q} + \mathbf{k} -\mathbf{k}')}{|\mathbf{q} + \mathbf{k} -\mathbf{k}'|^2} \mathrm{d}^3 \mathbf{k} \mathrm{d}^3 \mathbf{k}'.$$
Here $\mathbf{q},\mathbf{k},\mathbf{k}'$ are vectors in $\mathbb{R}^3$ and $q,k,k'$ are their euclidean norms.
The authors present a solution to this intergal which is claimed to be exact, but sadly without any ansatz or idea to solve it. It reads:
\begin{align}
G(\mathbf{q}) = \frac{9}{32} \left( \frac{q}{q_F} \right)^2 \left\lbrace \frac{2}{105} \left[ 24 \left( \frac{q_F}{q} \right)^2 + 44 + \left( \frac{q}{q_F} \right)^2 \right] \\
- 2 \frac{q_F}{q} \left[ \frac{8}{35} \left( \frac{q_F}{q} \right)^2 - \frac{4}{15} + \frac{1}{6} \left( \frac{q}{q_F} \right)^2 \right] \ln \Bigg \vert \frac{q + 2q_F}{q - 2q_F} \Bigg \vert \\
+ \left( \frac{q}{q_F} \right)^2 \left[ \frac{1}{210} \left( \frac{q}{q_F} \right)^2 - \frac{2}{15} \right] \ln \Bigg \vert \frac{q^2 - 4q_F^2}{q^2} \Bigg \vert\right\rbrace
\end{align}
I find this result rather strange because of all the numerical fractions and I have never seen any result similar to this, which is why I don´t have any idea how to arrive at this result.
So far the only idea I had was to substitute with a relative coordinate $\mathbf{t} = \mathbf{k} -\mathbf{k}'$, but I have trouble finding out the new integral boundaries. Further I would take $\mathbf{q}$ parallel to the $z$-axis, such that I can use $\mathbf{q} \cdot \mathbf{k} = qk \cos\theta$ and use spherical coordinates. Also the definitions $n = \frac{q_F^3}{3\pi^2}$ and $\int_{k \leq q_F} \mathrm{d}^3 \mathbf{k} = \frac{4\pi}{3}q_F^3$ could be helpful. Maybe someone has done something similar before and knows how to approach an integral like this.
Any help or ideas would be greatly appreciated, thanks in advance!
| So for anyone still interested or coming across this later, here is the solution. A colleague and I finally managed to solve it. It is however quite long and tedious, so I will only present the main ideas and the end results.
We start with a transformation to $\mathbf{r} = \frac{\mathbf{k} + \mathbf{k}'}{2}$ and $\mathbf{t} = \mathbf{k} - \mathbf{k}'$ such that
\begin{align}
\mathbf{k} = \mathbf{r} + \frac{\mathbf{t}}{2} \text{ and } \mathbf{k}' = \mathbf{r}
- \frac{\mathbf{t}}{2} \,.
\end{align}
One can then verify that the absolute value of the determinant of the Jacobian is equal to $1$, so for the differentials we now have $\mathrm{d} \mathbf{k} \mathrm{d} \mathbf{k}' = \mathrm{d} \mathbf{r} \mathrm{d} \mathbf{t}$. From the integration bounds the conditions
$$|\mathbf{k}| = |\mathbf{r} + \frac{\mathbf{t}}{2}| \leq q_F \text{ and } |\mathbf{k}'| = |\mathbf{r} - \frac{\mathbf{t}}{2}| \leq q_F$$
follow. This can be visualized by two spheres with radius $q_F$ overlapping. $\mathbf{t}$ is then the vector connection their centers. $\mathbf{r}$ has its origin at the midpoint of the line connecting their centers and all possible $\mathbf{r} \pm \frac{\mathbf{t}}{2}$ are confined to the overlapping region. For a similar picture see Fetter, Walecka - Quantum theory of many-particle systems,2003 pg. 28. We then express the integration over $\mathrm{d} \mathbf{r}$ using step functions as
\begin{align}
\int \Theta(q_F - |\mathbf{r} + \tfrac{\mathbf{t}}{2}|)\Theta(q_F - |\mathbf{r} - \tfrac{\mathbf{t}}{2}|) \mathrm{d} \mathbf{r} = 2V \Theta(1-x)
\end{align}
where we have set $x = \frac{t}{2q_F}$ and $V$ is the volume of the spherical cap which makes up the overlapping region and has to be multiplied by $2$ because of symmetry. $V$ can be calculated using the formulas form https://en.wikipedia.org/wiki/Spherical_cap with $h = q_F - \frac{t}{2}$ and $a^2 = q_F^2 - \frac{t^2}{4}$, so the result of the integral in terms of $x$ is
\begin{align}
\frac{4 \pi}{3} q_F^3 \left(1 - \tfrac{3}{2} x + \tfrac{1}{2} x^3 \right) \Theta(1-x) \,.
\end{align}
The step function is there to account for values of $\frac{t}{2}>q_F$ in which case there would be no overlap, so everything would be equal to $0$. We can also plug in $n = \frac{q_F^3}{3\pi^2}$ to rewrite the prefactor, leaving us with the integral
\begin{align}
G(\mathbf{q}) = \frac{3}{8\pi q_F^3} \int \left( 1 - \tfrac{3}{2} x(t) + \tfrac{1}{2} x^3(t) \right) \Theta(1-x(t)) \frac{\mathbf{q} \cdot (\mathbf{q}+\mathbf{t})}{|\mathbf{q} + \mathbf{t}|^2} \mathrm{d} \mathbf{t} \,.
\end{align}
If we set $\mathbf{q}$ parallel to the $z$-axis we can make use of spherical coordinates and we arrive at
\begin{align}
G(\mathbf{q}) = \frac{3}{4q_F^3} \int_0^{2q_F} \left( 1 - \tfrac{3}{4q_F} t + \tfrac{1}{16q_F^3} t^3 \right) \int_0^\pi \frac{q^2 t^2 \sin\theta + q t^3 \cos\theta \sin\theta}{q^2 + t^2 + 2qt\cos\theta} d\theta dt \,.
\end{align}
Evaluating the angular part gives
\begin{align}
A(q,t) \equiv q t \ln \left| \frac{q+t}{q-t} \right| + \frac{t(q^2 + t^2)}{2q} \ln \left| \frac{q-t}{q+t} \right| + t^2 \,.
\end{align}
We now have to perform the $t$-integration over the product of $A(q,t)$ with the polynomial. As already said at the beginning this is a very long calculation, but it is elementary and does not require any special tricks, you just have to be really careful because there are going to be a lot of terms. I will present the solution of each part of the polynomial:
\begin{align}
\frac{3}{4q_F} \int_0^{2q_F} A(q,t) dt &= \left[ \frac{3}{4} \left( \frac{q}{q_F} \right) - \frac{3}{32} \left( \frac{q}{q_F} \right)^3 -\frac{3}{2} \left( \frac{q_F}{q} \right) \right] \ln \left| \frac{q + 2q_F}{q - 2q_F} \right| + \frac{3}{8} \left( \frac{q}{q_F} \right)^2 + \frac{3}{2} \\
- \frac{9}{16 q_F^4} \int_0^{2q_F} t A(q,t) dt &= - \frac{3}{80} \left( \frac{q}{q_F} \right)^4 \ln \left| \frac{q^2 - 4q_F^2}{q^2} \right| + \left[ - \frac{3}{4} \left( \frac{q}{q_F} \right) + \frac{9}{5} \left( \frac{q_F}{q} \right) \right] \ln \left| \frac{q + 2q_F}{q - 2q_F} \right| \\
& \quad - \frac{3}{20} \left( \frac{q}{q_F} \right)^2 - \frac{9}{5} \\
\frac{3}{64q_F^6} \int_0^{2q_F} t^3 A(q,t) dt &= \frac{3}{2240} \left( \frac{q}{q_F} \right)^6 \ln \left| \frac{q^2 - 4q_F^2}{q^2} \right| + \left[ \frac{3}{20} \left( \frac{q}{q_F} \right) - \frac{3}{7} \left( \frac{q_F}{q} \right) \right] \ln \left| \frac{q + 2q_F}{q - 2q_F} \right|\\
& \quad + \frac{3}{280} \left( \frac{q}{q_F} \right)^2 + \frac{3}{560} \left( \frac{q}{q_F} \right)^4 + \frac{3}{7} \,.
\end{align}
If we add up all terms and simplify we end up with
\begin{align}
G(q) &= \left[ \frac{3}{20} \left( \frac{q}{q_F} \right) - \frac{3}{32} \left( \frac{q}{q_F} \right)^3 -\frac{9}{70} \left( \frac{q_F}{q} \right) \right] \ln \left| \frac{q + 2q_F}{q - 2q_F} \right| \notag \\
& \quad + \left[ - \frac{3}{80} \left( \frac{q}{q_F} \right)^4 + \frac{3}{2240} \left( \frac{q}{q_F} \right)^6 \right] \ln \left| \frac{q^2 - 4q_F^2}{q^2} \right| \notag \\
& \quad + \frac{33}{140} \left( \frac{q}{q_F} \right)^2 + \frac{3}{560} \left( \frac{q}{q_F} \right)^4 + \frac{9}{70}
\end{align}
which after factorizing and grouping is exactly the desired equation. You can also check by expanding the expression of $G(\mathbf{q})$ given in the question.
| {
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"url": "https://math.stackexchange.com/questions/4226726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Compute $\frac 17$ in $\Bbb{Z}_3$
Compute $\frac 17$ in $\Bbb{Z}_3.$
We will have to solve $7x\equiv 1\pmod p,~~p=3.$
*
*We get $x\equiv 1\pmod 3.$
*Then $x\equiv 1+3a_1\pmod 9,$ so $7(1+3a_1)\equiv 1 \pmod 9$ basically lifting the exponent of $p=3,$ we get $1+3a_1\equiv 4\pmod 9\implies a_1\equiv 1\pmod 3.$
*So let $$x\equiv 1+3\cdot 1+3^2\cdot a_2 \pmod 2\implies 7(4+3^2\cdot a_2)\equiv 1\pmod {27}\implies 4+3^2\cdot a_2\equiv 4\pmod {27}\implies a_2\equiv 0 \pmod 3.$$
*So let $$ x\equiv 1+3\cdot 1+3^2\cdot 0+ 3^3\cdot a_3 \pmod {81}\implies 7(4+3^2\cdot 0+3^3\cdot a_3)\equiv 1\pmod {81}\implies 4+3^3\cdot a_3\equiv 58\pmod {81}\implies a_2\equiv 2 \pmod 3.$$
*So let $$ x\equiv 1+3\cdot 1+3^2\cdot 0+ 3^3\cdot 2+3^4\cdot a_4 \pmod {243}\implies 7(4+3^2\cdot 0+3^3\cdot 2+3^4\cdot a_4)\equiv 1\pmod {243}\implies 1+3+54+3^4\cdot a_4\equiv 139\pmod {243}\implies a_4\equiv 1 \pmod 3.$$
*So let $$ x\equiv 1+3\cdot 1+3^2\cdot 0+ 3^3\cdot 2+3^4\cdot 1+ 3^5\cdot a_5 \pmod {729}\implies 7(4+3^2\cdot 0+3^3\cdot 2+3^4\cdot 1+3^5\cdot a_5)\equiv 1\pmod {729}\implies 1+3+54+81\equiv 625\pmod {243}\implies a_5\equiv 2 \pmod 3.$$
I haven't worked out but I think $a_6$ is $0.$
So the sequence we are getting is $(a_0,a_1,a_2,a_3,a_4,\dots)=(1,1,0,2,1,2,\dots).$
But I am not sure if it's correct, since it's not being periodic. Any help?
| You are doing correct.
For a slightly more efficient approach, note that $7\mid 3^6-1$ by Fermat's little theorem. Then by noting that $3^6-1=7\cdot104$, we have
\begin{align*}
\frac{1}{7}
&= -\frac{104}{1 - 3^6} \\
&= -\sum_{n=0}^{\infty}104 \cdot 3^{6n} \\
&= 1 + \sum_{n=0}^{\infty} (3^{6} - 1)3^{6n} - \sum_{n=0}^{\infty}104 \cdot 3^{6n} \\
&= 1 + \sum_{n=0}^{\infty}(3^6-105)3^{6n}
\end{align*}
in $\mathbb{Z}_3$, the ring of $3$-adic integers. Then by using $3^6-105 = 212010_{(3)}$, it follows that
$$ \frac{1}{7} = \overline{212010}212011_{(3)} = \overline{021201}1_{(3)}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the value of the PM segment in the figure below? For reference:
Given a rectangle, ABCD (AB > BC) by B draws a perpendicular to AC which
intercepts "P" to CD, and an "M" the perpendicular to BD drawn by B.
$\space$ The prolongation of DM and BC intersect at Q.
Calculate PM if DQ =$ 17$ and BP = $9$. (answer: $4$)
My progresss:
$$\triangle BCP \sim \triangle QCD\\
\frac{BC}{CQ}=\frac{PC}{2PC}=\frac{PB}{DQ}\rightarrow\\
\frac{BC}{17}=\frac{1}{2}=\frac{9}{DQ} \therefore BC = \frac{17}{2}; DQ=18\\
\triangle DCQ: DC^2+CQ^2 = DQ^2 \rightarrow\\
DC^2+17^2 =18^2 \therefore DC =\sqrt35\\
\text{draw} ~MI \perp BQ \implies \triangle MBQ (\text{isosceles})\\
\triangle BMI \sim \triangle DCQ (AA):\\
BI = \frac{BQ}{2} = \frac{\frac{17}{2}+17}{2} =\frac{51}{4}\\\\
\frac{BI}{17}=\frac{BP+PM}{DQ}\rightarrow \frac{51}{68}=\frac{9+PM}{18} \therefore \boxed{ PM=4,5}$$
I didn't see where my mistake is..(book answer = 4)
|
Hints: If you show that In figure M is midpoint of QT and DM=PM, then we have:
In right angled triangle TBQ, BM=MQ so:
$17-PM=9+PM\Rightarrow PM=4$
| {
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A method to derive the quadratic formula I wonder if this is a valid method to derive the quadratic formula.
It is not an optimal method at all, but I am intrigued if it is a genuine way or has any gap.
The method starts from the equation whose coefficients are such that it is not necessary to use the quadratic formula to solve it,
$$(a^2 + b^2)x^2+ 2bcx+c^2 =0$$ becouse we can easily solve for x, having at the right side a quadratic form: $$a^2x^2 = -b^2x^2 - 2bcx - c^2 \implies a^2x^2 = -(bx + c)^2 $$
Obtaining the equation $(a^2 + b^2)x^2+ 2bcx+c^2 =0$ is trivial and the coefficient $a$ is squared so that $(a^2 + b^2)$ is only zero when $a$ or $b$ are zero. Of course $a,b,c \in \mathbb{R}$
First we find the roots of our special equation: $a^2x^2 = -(bx + c)^2 $
Square rooting both sides: $ax = \pm i(bx + c) $ where $ i=\sqrt{-1}$
Solving x we obtain: $ x = \dfrac{\pm ic}{a \mp ib}$
Now we are ready to derive the general quadratic formula. Note that the equation $(a^2 + b^2)x^2+ 2bcx+c^2 =0$ is just a quadratic equation $a'x^2 + b'x +c'$ so we can rearrange the coefficients:
$$a'=(a^2 + b^2)$$
$$b'=2bc$$
$$c'= c^2$$
Let's solve for $a$, $b$ and $c$
$$c'= c^2 \implies c = \pm \sqrt{c'} $$
$$b'=2bc \implies b= \frac{b'}{\pm 2\sqrt{c'}} $$
$$a'=(a^2 + b^2) \implies a = \pm \sqrt{a' - \frac{b'^2}{4c'} } $$
And now we substitute $a$, $b$ and $c$ in the above result $ x = \dfrac{\pm ic}{a \mp ib}$ and we obtain:
$$ x = \frac{\pm i \sqrt{c'}}{\pm \sqrt{a' - \frac{b'^2}{4c'} - \mp i \frac{b'}{2\sqrt{c'}} }}$$
Multiplying top and bottom by $\pm \sqrt{c'}$
$$ x = \frac{\pm i c'}{\pm \sqrt{c'} \sqrt{a' - \frac{b'^2}{4c'} - \mp i \frac{b'}{2} }}$$
Taking $4c'$ out of the square root and simplifying
$$ x = \frac{\pm 2 i c'}{\pm \sqrt{4a'c' - b'^2} \mp i b'}$$
Multiplying up and down by the conjugate of the denominator:
$$ x = \frac{(\pm 2 i c')(\pm \sqrt{4a'c' - b'^2} \mp i b')}{(\pm \sqrt{4a'c' - b'^2} \mp i b')(\pm \sqrt{4a'c' - b'^2} \pm i b')} \implies$$
$$ \implies x = \frac{(\pm 2 i c')(\pm \sqrt{4a'c' - b'^2} \mp i b')}{4a'c'}$$
Note that $ i \sqrt{4a'c' - b'^2} = \sqrt{b'^2 - 4a'c'}$ and $\pm i * \pm ib' = -b'$ So simplifying:
$$ x = \frac{-b'}{2a'} \pm \frac{\pm \sqrt{b'^2 - 4a'c'} }{2a'}$$
And, we've derived the formula.
| [Too long for a comment]
There is another way to derive the quadratic formula. But it is so complicated, it is almost barely usable.
Part 1
We begin with: $$x^2-px+q=0.$$ Let's assume this equation is of the form $A(x-\alpha)(x-\beta)$ for some constant $A$ and roots $\alpha$, $\beta$. So $$x^2-px+q=A(x-\alpha)(x-\beta)=Ax^2-A(\alpha+\beta)x+A\alpha\beta$$
Comparing coefficients, it follows that $A=1$, $\alpha+\beta=p$ and $\alpha\beta=q$. Now, with this information, we may analyse $\alpha^4+\beta^4$. Recall: $$(m+n)^4=\color{red}{m^4+n^4}+2mn(\color{green}{2m^2}+3mn\color{green}{+2n^2})$$
Let $m=\alpha$ and $n=\beta$. We wish to analyse the expression in red, but first we must evaluate the expression in green. Note: $$(\alpha+\beta)^2=\color{green}{\alpha^2+\beta^2}+2\alpha\beta$$ or, since $p=\alpha+\beta$ and $q=\alpha\beta$ we have: $$p^2=\alpha^2+\beta^2+2q$$ so $$\alpha^2+\beta^2=p^2-2q.$$ So now we have that, if: $(\alpha+\beta)^4=\color{red}{\alpha^4+\beta^4}+2\alpha\beta(2\alpha^2+3\alpha\beta+2\beta^2)$ then $$p^4=\alpha^4+\beta^4+2q(2p^2-q)$$ or $$\color{blue}{p^4}-4q\color{blue}{p^2}-\underbrace{(\alpha^4-2q^2+\beta^4)}_{\large (\alpha^2-\beta^2)^2}=0.$$
So now, we have solved a quadratic. Let's replace $\alpha$ with $u$ and $\beta$ with $v$. We have:
Let $x^2-Px-Q=0.$ If there exist $u$ and $v$ such that $$P=4uv\quad ,\quad Q=(u^2-v^2)^2$$ then $(u+v)^2$ is a root.
The quadratic equation can now be found by solving for $u$ and $v$, and then calculating $(u+v)^2$. The algebra becomes incredibly messy here.
Part 2
Expanding $Q$, we have: $$Q=u^4-2(\color{purple}{uv})^2+v^4=u^4-2\bigg(\color{purple}{\frac P4}\bigg)^2+v^4$$ $$\therefore\qquad u^4+v^4=\frac{P^2+8Q}8.$$ Now this is the equation of a super-ellipse, so we can solve for $u$ and $v$ by recalling the familiar trig identity $\cos^2\vartheta+\sin^2\vartheta=1$. Our equation is equivalent to $$\left(\cfrac{u^2}{\sqrt{\frac{P^2+8Q}8}}\right)^2+\left(\cfrac{v^2}{\sqrt{\frac{P^2+8Q}8}}\right)^2=1$$ or, given some simplification, $$u=\sqrt{\frac 12\cos\vartheta\sqrt{\frac{P^2}2+4Q}}\quad ,\quad v=\sqrt{\frac 12\sin\vartheta\sqrt{\frac{P^2}2+4Q}}$$ And now we ought to find $\vartheta$ such that $uv=\dfrac P4$. We have: $$uv=\frac 12\sqrt{\cos\vartheta\sin\vartheta\bigg(\frac{P^2}2+4Q\bigg)}=\frac P4$$ and, since $2\cos\vartheta\sin\vartheta=\sin(2\vartheta)$, then $$\frac P2=\sqrt{\frac 12\sin(2\vartheta)\bigg(\frac{P^2}2+4Q\bigg)}$$ or $$\sin(2\vartheta)=\frac{P^2}{P^2+8Q}=\lambda$$ or $$\vartheta=\frac 12\arcsin\lambda.$$ And now we use the identites, $$\begin{align}\cos\bigg(\frac 12\arcsin\lambda\bigg)&=\frac 12\Big(\sqrt{1+\lambda}+\sqrt{1-\lambda}\Big)\\ \sin\bigg(\frac 12 \arcsin\lambda\bigg)&=\frac 12\Big(\sqrt{1+\lambda}-\sqrt{1-\lambda}\Big)\end{align}$$ to finally derive (partially in terms of $\lambda$): $$u=\frac 12\sqrt{\frac P{\sqrt{2\lambda}}\Big(\sqrt{1+\lambda}+\sqrt{1-\lambda}\Big)}\quad ,\quad v=\frac 12\sqrt{\frac P{\sqrt{2\lambda}}\Big(\sqrt{1+\lambda}-\sqrt{1-\lambda}\Big)}.$$
Finally: $$(u+v)^2=\Bigg(\frac 12\sqrt{\frac P{\sqrt{2\lambda}}\Big(\sqrt{1+\lambda}+\sqrt{1-\lambda}\Big)}+\frac 12\sqrt{\frac P{\sqrt{2\lambda}}\Big(\sqrt{1+\lambda}-\sqrt{1-\lambda}\Big)}\Bigg)^2$$ $$=\frac P{4\sqrt{2\lambda}}\bigg(\sqrt{\sqrt{1+\lambda}+\sqrt{1-\lambda}}+\sqrt{\sqrt{1+\lambda}-\sqrt{1-\lambda}}\bigg)^2=\frac P{2\sqrt{2\lambda}}\Big(\sqrt{1+\lambda}+\sqrt{2\lambda}\Big)$$ $$=\frac P2\Bigg(1+\sqrt{\frac 12\bigg(1+\frac 1{\lambda}\bigg)}\Bigg)=\frac P2\bigg(1+\sqrt{1+\frac{4Q}{P^2}}\bigg)=\frac{P+\sqrt{P^2+4Q}}2$$
So now we can let $\alpha=\dfrac{P+\sqrt{P^2+4Q}}2$ since that is a root, and we know that $\alpha+\beta=P$.$$\therefore \beta=\frac{P-\sqrt{P^2+4Q}}2$$ and we finally have the form: $\dfrac{P\pm\sqrt{P^2+4Q}}2$. Now, substitute $P=-\dfrac BA$ and $Q=-\dfrac CA$.
| {
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"url": "https://math.stackexchange.com/questions/4231346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Transform ODE $(u-x)u_x + u + x = 0$ to polar coordinates According to Peter Olver in his book “Applications of Lie Groups to Differential Equations”, p. 104, Example 2.32, the ODE:
$$(u-x)u_x + u + x = 0 \tag{1}$$
is transformed in polar coordinates with $x = r\cos \theta$ and $u = r\sin\theta$ to:
$$\frac{dr}{d\theta} = r.$$
How do I transform $u_x=\partial_x(r\sin\theta)$ to the new coordinates? I tried the following, but the equation I got is very far from $\frac{dr}{d\theta} = r$. I rewrote $\partial_x$ in terms of $\partial_r$ and $\partial_\theta$ using the chain rule. Supposing $f = f(r,\theta)$ is a function of $r$ and $\theta$, we should have:
$$\begin{aligned} \partial_x f(r,\theta) &= (\partial_x r)\partial_r f + (\partial_x \theta)\partial_\theta f\\
&=\cos\theta\partial_r f + \left(\partial_x \arctan\frac ux\right) \partial_\theta f\\
&=\cos\theta\partial_r f + \left(\frac {1}{1+\theta^2}2\theta\frac{xu_x - u}{x^2}\right) \partial_\theta f.\end{aligned}$$
Thus:
$$\begin{aligned} u_x &=\cos\theta\partial_r (r\sin\theta) + \left(\frac {1}{1+\theta^2}2\theta\frac{xu_x - u}{x^2}\right) \partial_\theta (r\sin\theta)\\
&=\cos\theta\sin\theta + \left(\frac {2\theta}{1+\theta^2}\frac{xu_x - u}{(r\cos\theta)^2}\right) (-r\cos\theta + r_\theta \sin\theta)\\
&=\cos\theta\sin\theta - \left(\frac {2\theta}{1+\theta^2}\frac{r(u_x\cos\theta - \sin\theta)}{r\cos\theta}\right) + \left(\frac {2\theta}{1+\theta^2}\frac{\cos\theta u_x - \sin\theta}{r\cos^2\theta}\right) r_\theta \sin\theta.\end{aligned}$$
Rearranging:
$$\begin{aligned} u_x\left[1 + \frac{2\theta}{1+\theta^2}\left(1+\frac{r_\theta}{r}\tan\theta \right)\right] &= \cos\theta\sin\theta + \frac {2\theta}{1+\theta^2}\left(\tan\theta - \frac{r_\theta}{r}\tan^2\theta\right).\end{aligned}$$
So:
$$ u_x = \frac{\cos\theta\sin\theta + \frac {2\theta}{1+\theta^2}\left(\tan\theta - \frac{r_\theta}{r}\tan^2\theta\right)}{1 + \frac{2\theta}{1+\theta^2}\left(1+\frac{r_\theta}{r}\tan\theta \right)}$$
Then, rewriting (1) as $u_x = \frac{x+u}{x-u}$, I obtained:
$$\frac{\cos\theta\sin\theta + \frac {2\theta}{1+\theta^2}\left(\tan\theta - \frac{r_\theta}{r}\tan^2\theta\right)}{1 + \frac{2\theta}{1+\theta^2}\left(1+\frac{r_\theta}{r}\tan\theta \right)}=\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}.$$
| $$(u-x)u_x + u + x = 0 \tag{1}$$
$$-xu_x + u =- x -uu_x$$
$$\left(\dfrac u x \right)' =\dfrac 1 {x^2}(x+uu_x)$$
$$\left(\dfrac u x\right)' =\dfrac 1 {2x^2}(x^2+u^2)'$$
Multiply both sides by $ \dfrac {dx}{d\theta}$:
$$\dfrac {d}{d \theta}(\tan \theta )=\dfrac 1 {2r^2 \cos^2 \theta}\dfrac {dr^2}{d\theta }$$
$$\dfrac {dr}{d \theta}=r$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving for the value of $a,b$ in $f(x)=(ax+b)(x^5+1)-(5x+1)$ s.t. $(x^2+1)|f(x)$
$Q.$ If $f(x)=(ax+b)(x^5+1)-5x-1$ is divisible by $x^2+1$ . Then the value of $2a+3b$ $?$
MY APPROACH : We have , $(x^2+1)|f(x)$ then $(x-i)|f(x)$ and $(x+i)|f(x)$ .
So by Factor Theorem we have $f(-i)=0$ and $f(i)=0$ $$f(i)=-a+bi+(a-5)i+b-1=0$$ $$f(-i)=-a-bi-(a-5)i+b-1=0$$
By this I concluded that $a-b=-1$ . But this is not enough information to solve the problem .
| We have
$$\begin{align}&ax^6+bx^5+x(a-5)+(b-1)=P(x)(x^2+1)+mx+n \\\\
\implies &\begin{cases} mi+n=-a+bi+i(a-5)+(b-1)\\-mi+n=-a-bi-i(a-5)+(b-1)\end{cases}\\\\
\implies &\begin{cases} n=\frac{2(b-1)-2a}{2}=b-a-1\\m=\frac{2bi+2i(a-5)}{2i}=a+b-5\\
\end{cases}\\\\
\implies&\begin{cases}n=0,\thinspace b-a=1\\m=0,\thinspace a+b=5
\end{cases} \\\\
\implies &~~a=2, \thinspace b=3\\\\ \implies &~~2a+3b=13. \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $(f+g)(x)=4$ and $(f\circ g)(x)=7-4x$ What is the sum of the possible values for $g(2)$?
If $f$ and $g$ are polynomial functions so that $(f+g)(x)=4$ and
$(f\circ g)(x)=7-4x$, what is the sum of the possible values for
$g(2)$?
$1)2\qquad\qquad\qquad2)3\qquad\qquad\qquad3)-3\qquad\qquad\qquad4)-2$
To solve this problem First I realized that $f$ and $g$ should be linear polynomials in to have $(f\circ g)(x)=-4x+7$. Hence $f(x)=ax+b$ and $g(x)=cx+d$.
$$(f+g)(x)=(a+c)x+(b+d)=4$$
$$(f\circ g)(x)=(ac)x+(ad+b)=-4x+7$$
Hence, $$a+c=0$$ $$b+d=4$$ $$ac=-4$$ $$ad+b=7$$
Then got two cases:
First, $a=2\quad, c=-2,\quad d=3,\quad b=1\quad \Rightarrow g(x)=-2x+3\Rightarrow g(2)=-1$
Second, $a=-2\quad,c=2,\quad d=-1,\quad b=5\quad \Rightarrow g(x)=2x-1\Rightarrow g(2)=3$
So the first choice is correct.
I wonder can we solve this problem with other approaches? ( Looing for approaches with less calculations and more elegance! )
| Using the condition that $f(x) + g(x) = 4$, $f(x) = ax + b, \ g(x) = -ax + (4-b)$, which results in:
$$f(g(x)) = f(-ax + 4 - b) = a(-ax+4-b) + b = 7 - 4x$$
hence $-a^2x = -4x \implies a = \pm 2$ and $a(4-b)+b = 7 \implies b = 1, 5$. Thus $g(x) = -2x+3$ or $2x-1$, so the sum of the possible values of $g(2)$ is $-1 + 3 = 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the limit of $\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$ How to find the limit of the following function?
$$\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$$
What I tried is as follows.
$$\tan^{-1} x = y \implies \tan y = x $$
$$\frac{x \tan^{-1} x}{1-\cos x}=\frac{\tan y \cdot y}{1-\cos(\tan y) }$$
But it didn't work.
Please Help me.
| Hint. From $\cos (A+B)=\cos A\cos B-\sin A\sin B$ with $B=A$ we have $\cos 2A=\cos^2A-\sin^2A=(1-\sin^2A)-\sin^2 A=1-2\sin^2 A.$ Hence $2\sin^2 A=1-\cos 2A.$ With $A=2x$ therefore $$2\sin^2(x/2)=1-\cos x.$$ So if $\cos x\ne 1$ then $$\frac {x\tan^{-1}x}{1-\cos x}=\frac {x\tan^{-1}x}{2\sin^2(x/2)}=$$ $$=\left(\frac {(x/2)}{\sin (x/2)}\right)^2\cdot 2\cdot \frac {\tan^{-1}x}{x}.$$ Remark: The "half-angle formulae" are usually presented as $$\frac {1-\cos x}{2}=\sin^2(x/2)$$ and $$\frac {1+\cos x}{2}=\cos^2(x/2).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Suppose $a_1=1$ and $a_n=\sqrt{n-a_{n-1}}$ $\forall n \in\mathbb{Z_{\ge 2}}$. What is the least integer $n$ for which $a_n>4$? ⌊$a_{100}$⌋? Suppose $a_1=1$ and $a_n=\sqrt{n-a_{n-1}}$ $\forall n \in\mathbb{Z_{\ge 2}}$. What is the least integer $n$ for which $a_n>4$? What is $\left \lfloor a_{100} \right \rfloor$, where $\left \lfloor u \right \rfloor$ is the greatest integer that is less than or equal to $u$?
As an example, the $n^\text{th}$ Fibonacci number can be found without finding the preceding $(n-1)$ numbers.
I am not sure if we can solve my problem efficiently, easily, and without a calculator.
Using a computer calculator, I found that
$\sqrt{19-\sqrt{18-\sqrt{17-\sqrt{\cdots-1}}}}<4$ and $\sqrt{20-\sqrt{19-\sqrt{18-\sqrt{17-\sqrt{\cdots-1}}}}}>4$
If we can not solve this easily without a calculator, then can we just get an upper limit and a lower limit "good guesses"?
Here, "good guesses" like when we try to find $\sqrt[3]{50}$, we know that $3^3=27$ and $4^3=64$, therefore, $\sqrt[3]{50}$ is a value between $3$ and $4$. So, to get a good approximation, we do not need to look at any value $\notin (3,4)$.
I am interested in finding an explicit method rather than guessing.
Your help would be appreciated. THANKS!
| Your second question
*
*It's easy to see that $\forall n \geqslant 2 \ , \ a_n \leqslant \sqrt{n}\ $ . So $a_{100} < \sqrt{100} = 10$ and $a_{99} < \sqrt{99} < 10 $.
*Then $\ \ a_{100} = \sqrt{100 -a_{99}} > \sqrt{100-10} > 9 \ $ .
*So $\ \fbox{$\ \left\lfloor a_{100} \right\rfloor = 9\ $}\ $ .
Your first question
*
*$\forall n \geqslant 2 \ , \ a_n<\sqrt{n}$
*$\forall n \geqslant 3 \ , \ a_n=\sqrt{n-a_{n-1}}>\sqrt{n-\sqrt{n-1}}$
*$\forall n \geqslant 4 \ , \ a_n=\sqrt{n-a_{n-1}}>\sqrt{n-\sqrt{n-1-\sqrt{n-2}}}$
*$\forall n \geqslant 5 \ , \ a_n=\sqrt{n-a_{n-1}}>\sqrt{n-\sqrt{n-1-\sqrt{n-2-\sqrt{n-3}}}}$
*We can show that $(a_n)_{n\geqslant 1}$ is increasing.
*$a_{19} < \sqrt{19-\sqrt{18-\sqrt{17}}} < 4$
*$a_{20} > \sqrt{20-\sqrt{19-\sqrt{18-\sqrt{17}}}} >4$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Writing $ \frac{1}{n}\sum_{i=0}^{n-1}{\frac{m}{m-i}} =\frac{m}{n}\left( H_n-H_{m-n} \right) $ Given the following series, where $m,n$ are integers such that $n<m$:
\begin{align}
\frac{1}{n}\sum_{i=0}^{n-1}{\frac{m}{m-i}}\\
=\frac{m}{n}\left( H_n-H_{m-n} \right)
\end{align}
We know that harmonic number $H_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + + \frac{1}{n}$. So,
\begin{align}
\frac{1}{n}\sum_{i=0}^{n-1}{\frac{m}{m-i}}\\
=\frac{m}{n}\sum_{i=0}^{n-1}{\frac{1}{m-i}}\\
=\frac{m}{n}(\frac{1}{m-0}+\frac{1}{m-1}+\frac{1}{m-2}+\cdots \frac{1}{(m-n)+1})\\
\end{align}
So,
\begin{align}
H_n=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots \frac{1}{n}\\
H_{m-n}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots \frac{1}{m-n}\\
\end{align}
Since $m>n$, how we can guarantee as well that $H_n - H_{m-n}$ is positive please? So, I see that difference here $H_n - H_{m-n} > 0$.
Problem 1: How we can proceed please to derive $\frac{m}{n}\left( H_n-H_{m-n} \right)$?
Edit: I made a mistake as it's $H_m - H_{m-n}$.
|
We obtain for positive integers $m>n$
\begin{align*}
\color{blue}{\frac{1}{n}\sum_{i=0}^{n-1}\frac{m}{m-i}}
&=\frac{m}{n}\sum_{i=0}^{n-1}\frac{1}{m-i}\\
&=\frac{m}{n}\sum_{i=0}^{n-1}\frac{1}{m-(n-1-i)}\tag{1}\\
&=\frac{m}{n}\sum_{i=0}^{n-1}\frac{1}{m-n+1+i}\\
&=\frac{m}{n}\sum_{i=m-n+1}^m\frac{1}{i}\tag{2}\\
&=\frac{m}{n}\left(\sum_{i=1}^m\frac{1}{i}-\sum_{i=1}^{m-n}\frac{1}{i}\right)\tag{3}\\
&\,\,\color{blue}{=\frac{m}{n}\left(H_m-H_{m-n}\right)}
\end{align*}
and the claim follows.
Comment:
*
*In (1) we change the order of summation: $i\to n-1-i$.
*In (2) we shift the index and start with $i=m-n+1$. To compensate this shift we substitute $i$ with $i-m+n-1$ within the scope of the sum.
*In (3) we write the sum as difference of Harmonic numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Improper integral $\int_0^{\infty} \frac{1}{(x+1)(\pi^2+\ln^2(x))}dx$ I am having trouble solving
$$\int_0^{\infty} \frac{1}{(x+1)(\pi^2+\ln^2(x))}dx$$
let $\ln(x)=u$ then $x du= dx $
$$\Rightarrow \int_0^{\infty} \frac{1}{(x+1)(\pi^2+\ln^2(x))}dx=\int_0^{\infty} \frac{x+1-1}{(x+1)(\pi^2+u^2)}du=\int_0^{\infty} \frac{1}{\pi^2+u^2}du-\int_0^{\infty} \frac{1}{(e^u+1)(\pi^2+u^2)}du$$
$$=\frac{1}{2}-\int_0^{\infty} \frac{1}{(e^u+1)(\pi^2+u^2)}du$$
How do you proceed from here?
I put it into an integral calculator and it stated "Antiderivative or integral could not be found."
How do you solve this ?
| Enforcing the substitution $x\mapsto e^{-x}$ reveals
$$\int_0^\infty \frac1{(x+1)(\pi^2+\log^2(x))}\,dx=\int_{-\infty}^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx$$
Then, note that
$$\begin{align}
\int_{-\infty}^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx&=\int_{-\infty}^0 \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx+ \int_0^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx\\\\
&=\int_0^\infty \frac{1}{(1+e^{x})(\pi^2+x^2)}\,dx+ \int_0^\infty \frac{1}{(1+e^{-x})(\pi^2+x^2)}\,dx\\\\
&=\int_0^\infty \frac{1}{\pi^2+x^2}\,dx\\\\
&=\frac12
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4242103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving a function is increasing in $n$. I am trying to prove that the function $\frac{n}{2n+1}$, defined for $n \in \mathbb{N}$, decreases in $\mathbb{N}$. I attempted it by induction, but I'm not convinced that I fully need induction. Why can I not prove that for an arbitrary $n$, $f(n) \leq f(n+1)$ and deduce that, because $n$ was arbitrary, this holds for all $n$? The only thing left out would be the base case, but. I'm not fully sure why I need it here.
Regardless, here is my attempt at the induction:
Let $f: \mathbb{N} \to \mathbb{R}$ be defined by $f(n) = \frac{n}{2n+1}$. We prove by induction on $n$ that $f$ is increasing in $n$. If $n = 1$, we notice that
\begin{align*}
f(1) = \frac{1}{3} \leq \frac{2}{5} = f(2).
\end{align*}
Suppose inductively that we have $f(n) \leq f(n+1)$ for some $n \geq 1$. So we have
$\frac{n}{2n+1} \leq \frac{n+1}{2n+3}$. First, we have
\begin{align*}
\frac{n+1}{2n+3} \leq \frac{n+3}{2n+3}.
\end{align*}
Furthermore, $2n + 5 \geq 2n + 3$, so $\frac{1}{2n + 5} \leq \frac{1}{2n+3}$, so $\frac{n+3}{2n + 3} \leq \frac{n+3}{2n+5}$. Therefore, it follows that
\begin{align*}
\frac{n+1}{2n+3} \leq \frac{n+3}{2n+3} \leq \frac{n+3}{2n + 5} = \frac{(n+2) + 1}{2(n+2) + 1},
\end{align*}
so $f(n+1) \leq f(n+2)$, which closes the induction
| $\dfrac{n}{2n+1} = \dfrac{1}{2} \left( \dfrac{2n}{2n+1} \right) = \dfrac{1}{2} \left( \dfrac{2n+1 - 1}{2n+1} \right) = \dfrac{1}{2} \left(1 - \dfrac{1}{2n+1} \right)$
Since $ \dfrac{1}{2n+1} $ decreases with increasing $n$, then $ \left( 1 - \dfrac{1}{2n+1} \right)$ is increasing with $n$, thus $\dfrac{n}{2n+1} $ increases with increasing $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
formular for sum of column in a Matrix I have a trivial problem. I need to create a vector, which has the sum of the columns in a AxB matrix. What would be the mathematical formula for that?
Thank you
| Hint: The following examples might be helpful.
\begin{align*}
\begin{pmatrix}a&b&c\\d&e&f\end{pmatrix}
\begin{pmatrix}1\\1\\1\end{pmatrix}
&=\begin{pmatrix}a+b+c\\d+e+f\end{pmatrix}\\
\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}a&b&c\\d&e&f\end{pmatrix}
&=\begin{pmatrix}a+d&b+e&c+f\end{pmatrix}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Problem in normalization inequalities Let $a,b,c>0$. Prove that: $$\dfrac{(a+b)^2(b+c)^2(c+a)^2}{abc} \ge \dfrac{64}{27}(a+b+c)^3$$
First solution:
$\bullet$ Since the inequality is homogeneous, we may normalize $a+b+c=3$, we need to prove:
$$(a+b)(b+c)(c+a)\ge8\sqrt{abc}$$
$\bullet$ Notice that, $a+b+c=3$, we may have $$(a+b)(b+c)(c+a)=(3-a)(3-b)(3-c)=3(ab+bc+ca)-abc \ge 3.\sqrt{3abc(a+b+c)}-abc=9\sqrt{abc}-abc$$
$\bullet$ So we need to prove $$9\sqrt{abc}-abc \ge 8\sqrt{abc}$$ or $$abc\le1$$
which is true by AM-GM: $$abc\le\dfrac{(a+b+c)^3}{27}=1$$
Second solution:
$\bullet$ Since the inequality is homogeneous, we may normalize $a+b+c=1$, we need to prove:
$$27(a+b)^2(b+c)^2(c+a)^2\ge64abc$$
$\bullet$ This solution is more convenient, by two famous inequalities $(a+b)(b+c)(c+a)\ge\dfrac{8}{9}(a+b+c)(ab+bc+ca)$ and $(ab+bc+ca)^2\ge3abc(a+b+c)$, the problem is solved
My question:
*
*We can normalize $a+b+c=k$ with any number $k$, but why the author find $k=3$ or $k=1$ and make the problem very simple.
*I have seen in many proofs for inequalities, the author normalize $abc=k;(a+b)(b+c)(c+a)=k;ab+bc+ca=k$,...Can you explain to me how to know which expression we need to normalize? Can you start from this post with a different normalization method, or can you give me some other examples?
*Is there a way to solve this problem without using the normalization method?
|
*Yes, of course.
For example, after full expanding we need to prove that:
$$\sum_{sym}(27a^4b^2-5a^4bc+27a^3b^3-30a^3b^2c-19a^2b^2c^2)\geq0,$$ which is true by Muirhead.
Also, we can prove that the inequality $$(a+b)^2(a+c)^2(b+c)^2\geq\frac{64}{27}abc(a+b+c)^3$$ is true for any reals $a$, $b$ and $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I find the Laurent series expansion How can I find the Laurent series expansion for
$$\frac{z^2-1}{z^2+1} $$ with $|z|>2$.
$$\frac{z^2-1}{z^2+1} =\frac{z^2+1-2}{z^2+1}=1-\frac2{z^2+1}$$
The fraction on the right-hand-side can be expressed as $$\frac1{z^2+1}=\frac i{2(z+i)}+\frac{-i}{2(z-i)}$$
also
$$\frac 1{z+i}=\frac1{z}\frac1{1-(-\frac i z)}=\sum_{k=-\infty}^{-1}(-1)^{-k+1}i^{-k+1}z^k $$
$$\frac1{z-1}=\frac 1 z \frac1{1-\frac i z }=\frac 1 z \sum_{k=0}^\infty (\frac i z)^k $$
And here I don't know how to combine these two series.
| Many exercises concerning the computation of Laurent expansions can simply be solved by making use of the geometric series
$$\sum_{k=0}^\infty x^k = \frac{1}{1-x}.$$
However, the important point is that the above identity is valid only for $x \in \mathbb{C}$ satisfying $|x| <1$. Since you are working under the condition that $|z|>2$, you thus cannot apply the identity directly, but it works if you somehow manage to replace $z$ by $1/z$.
We have
$$\frac{1}{z^2+1} = \frac{1}{z^2} \cdot \frac{1}{1+\frac{1}{z^2}} = z^{-2} \cdot \frac{1}{1 - (-\frac{1}{z^2})} = z^{-2} \cdot \sum_{k=0}^\infty (-z^{-2})^k = \sum_{k=0}^\infty (-1)^k z^{-2k-2}.$$
Hence
$$\frac{z^2-1}{z^2+1} = (z^2-1) \cdot \frac{1}{z^2+1} = z^2 \cdot \Big( \sum_{k=0}^\infty (-1)^k z^{-2k-2}\Big) - \Big( \sum_{k=0}^\infty (-1)^k z^{-2k-2}\Big)$$
$$= \Big( \sum_{k=0}^\infty (-1)^k z^{-2k}\Big) - \Big( \sum_{k=0}^\infty (-1)^k z^{-2k-2}\Big) = \Big( \sum_{k=-1}^\infty (-1)^{k+1} z^{-2k-2}\Big) - \Big( \sum_{k=0}^\infty (-1)^k z^{-2k-2}\Big)$$
$$= 1 + \sum_{k=0}^\infty ((-1)^{k+1} - (-1)^k) z^{-2k-2} = 1 - 2 \sum_{k=0}^\infty (-1)^k z^{-2k-2}.$$
The manipulations which I used (e.g., taking the difference of two (infinite) series) are valid because of the absolute convergence of the geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For any positive real numbers $a,b,c$, show that $\min\{(b-c)^2,(c-a)^2,(a-b)^2)\} \leq \frac{a^2+b^2+c^2}{5}$
For any positive real numbers $a,b,c$, show that $\min\{(b-c)^2,(c-a)^2,(a-b)^2)\} \leq \frac{a^2+b^2+c^2}{5}$
I managed to show this with a 3 instead of a 5 :
The inequality is symmetric in $a$, $b$, $c$ so we can assume $0<c\leq b \leq a$, so $M=\min\{(b-c)^2,(c-a)^2,(a-b)^2\} = \min\{(b-c)^2,(a-b)^2\}$ and $a-c=a-b+b-c \geq 2\min\{(a-b),(b-c)\}$ so $(a-c)^2\geq 4\min\{(a-b),(b-c)\}^2=4M$. Thus we have $6M \leq (b-c)^2+(c-a)^2+(a-b)^2=2(a^2+b^2+c^2-(ab+ac+bc)) \leq 2(a^2+b^2+c^2)$ so $M\leq \frac{a^2+b^2+c^2}{3}$.
| Hint: Show that for non-negative real $ a \leq b$,
$$ \min \{ a^2, (b-a)^2 \} \leq \frac{ a^2 + b^2 } { 5 },$$
by considering cases of $ b \geq 2a$ and $ a \leq b < 2a$.
Corollary: The original statement holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What is the measure of $\overset{\LARGE{\frown}}{AB}$ in the figure below? For reference:
In the figure calculate $\overset{\LARGE{\frown}}{AB}$ if $ \overset{\LARGE{\frown}}{BC} = 90^o$
My progress:
Relationships I found:
$FO$ is angle bissector
$\triangle OBC(isosceles):\measuredangle OCJ=\measuredangle OBJ=45^o\\
\measuredangle ADE = 90^o\\
\triangle DAJ(isosceles):\measuredangle DAJ = \measuredangle DJA\\
\measuredangle JAB = \measuredangle JCB = \frac{\measuredangle{JOB}}{2}$
| $Let BD = a ~and~ r = radius\\
AODJ~ is~ a ~square\\
BC = r\sqrt2\\
Power ~of~ Point~ Theorem :a(a+\sqrt2) = r^2 \rightarrow r^2-ar\sqrt{2}-a^2=0\\
\therefore r=\frac{a\sqrt{2}+-a\sqrt{6}}{2}\implies \frac{r}{a}=\frac{\sqrt{6}+\sqrt{2}}{2}\\
law ~of~ sines:\frac{r}{a}=\frac{sen(45+\frac{x}{2})}{sen(\frac{x}{2})}\\
\frac{\sqrt{6}+\sqrt{2}}{2}=\frac{\frac{\sqrt{2}}{2}(cos(\frac{x}{2})+sen(\frac{x}{2}))}{sen(\frac{x}{2})}\\
\sqrt{3}+1=\frac{(1+tg(\frac{x}{2}))}{tg(\frac{x}{2})}\\
k\sqrt{3}+k=1+k\implies k=\frac{\sqrt{3}}{3}\\
\therefore x = 60^o$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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What is the mistake in my reasoning in calculating limit where $x \to -\infty$ I had this question in an exam today:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x}
$$
My question is: I have tried to solve it and I also found two other methods to solve it, one in my textbook and one from somebody else. All three methods give different answers, and I can't see the issue with either of the methods which are not mine. I need help figuring out what the invalid operation(s) in each of the wrong methods are, and which 2 (or 3) methods are wrong.
I didn't solve it in the exam, but afterwards tried this:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \Bigg(\frac{\sqrt{x^2+4x}}{2x}\Bigg) - 1 \\
= \; \frac{1}{2}\lim \limits_{x \, \to \,-\infty} \; \Bigg(\frac{\sqrt{x^2+4x}}{x}\Bigg) - 1 \\
= \; \frac{1}{2}\lim \limits_{x \, \to \,-\infty} \; \Big(\sqrt{x^2+4x}\cdot \frac{1}{x}\Big) - 1 \\
= \; \frac{1}{2} \; \Big(\lim \limits_{x \, \to \,-\infty}\sqrt{x^2+4x} \, \cdot \, \lim \limits_{x \, \to \,-\infty}\frac{1}{x}\Big) - 1 \\
= \; \frac{1}{2} \; \Bigg(\sqrt{\lim \limits_{x \, \to \,-\infty}x^2+4x}\,\cdot \, \lim \limits_{x \, \to \,-\infty}\frac{1}{x}\Bigg) - 1 \\
= \; \frac{1}{2} \; \Bigg(\sqrt{\lim \limits_{x \, \to \,-\infty}x^2+4x}\,\cdot \, 0\Bigg) - 1 \\
= \; -1 \\
$$
I understand my mistake: I multiply an infinite amount with 0 and say it is equal to 0 in my last step.
However, our study guide has that exact example with the following solution:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2(1+\frac{4}{x})}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{|x|\sqrt{(1+\frac{4}{x})}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{-x\sqrt{(1+\frac{4}{x})}-2x}{2x} \; ; \; (x < 0) \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{-x(\sqrt{(1+\frac{4}{x})}+2)}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{-(\sqrt{(1+\frac{4}{x})}+2)}{2} \\
= \; \frac{-\Bigg(\sqrt{\lim \limits_{x \, \to \,-\infty} \; (1+\frac{4}{x})}+2\Bigg)}{2} \\
= \; \frac{-(\sqrt{1+0}+2)}{2} \\
= \; \frac{-(\sqrt{1}+2)}{2} \\
= \; \frac{-(1+2)}{2} \\
= \; -\frac{3}{2}
$$
To complicate matters even more: someone else solved the problem like this and got a third answer:
$$
\lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}-2x}{2x} \\
= \; \lim \limits_{x \, \to \,-\infty} \; \frac{\sqrt{x^2+4x}}{2x} - 1 \\
= \; \lim \limits_{x \, \to \,-\infty} \; \sqrt{\frac{x^2+4x}{4x^2}} - 1 \\
= \; \lim \limits_{x \, \to \,-\infty} \; \sqrt{\frac{1}{4} + \frac{1}{x}} - 1 \\
= \; \; \sqrt{\lim \limits_{x \, \to \,-\infty} \; \Big(\frac{1}{4} + \frac{1}{x}\Big)} - 1 \\
= \; \sqrt{\lim \limits_{x \, \to \,-\infty} \; \frac{1}{4} + \lim \limits_{x \, \to \,-\infty} \; \frac{1}{x}} - 1 \\
= \; \sqrt{\frac{1}{4} + 0} - 1 \\
= \; \sqrt{\frac{1}{4}} - 1 \\
= \; \frac{1}{2} - 1 \\
= \; -\frac{1}{2}
$$
| Exact explanations were made about the mistake you made in the comments. You missed that $x$ is negative.
You can observe this simple fact:
$$\begin{align}x=\begin{cases}-\sqrt{x^2},~\text{if}~ x<0\\\sqrt{x^2}, ~\text{if}~x>0\end{cases}\end{align}$$
This immediately reveals your mistake.
I would calculate this limit as follows:
$$\begin{align}\lim \limits_{x\to -\infty} \frac{\sqrt{x^2+4x}-2x}{2x}&=-\lim_{x \to\infty} \frac{\sqrt{x^2-4x}+2x}{2x}\\
&=-\lim_{x \to\infty} \left(\sqrt{\frac{x^2}{4x^2}-\frac{4x}{4x^2}}+1\right)\\
&=-\lim_{x \to\infty} \left(\sqrt{\frac 14-\frac{1}{x}}+1\right)\\
&=-\left(\frac 12+1\right)\\
&=-\frac 32.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding the area of a region in an isosceles triangle that contains squares of area $256$ and $49$ There was a quiz posted on F*cebook by someone. Here's the problem.
And here's my attempt:
First, as you can see there, I drew a line from one of the corner of the pink square to the one of the corner of the green square. The degree has been symbolized as alpha ($\alpha$).
$$\begin{align}
\alpha &= \tan^{-1}\left(\frac{16}{23}\right)\\
\end{align}$$
That means the degree between the line (almost diagonally to the green square) and the side line of the green square is $90-\alpha$. Let's say the side green square is $x$
We know $\cot(x)=\tan(90^{0} - x)$, so
$$\begin{align}
\tan(90^{0} - \alpha) = \cot(\alpha) &= \frac{16}{x}\\
x &= 16 \tan(\alpha)\\
&= 16 \tan\left(\tan^{-1}\left(\frac{16}{23}\right)\right)\\
&= \frac{16^2}{23} = \frac{256}{23}
\end{align}$$
Now, I assume (I can't tell reason, it's just my guesswork) that $x$ I've found earlier has the same length as the blue side one. I also assume those two blue are right triangles.
Finally, the last calculation is:
$$\begin{align}
& \left(\frac{1}{2}\cdot \frac{256}{23}\cdot 16\right) + \left(\frac{1}{2}\cdot 7 \left( \frac{256}{23} + 9\right)\right)\\
&= 159.5
\end{align}$$
The poster gave me cry emoji and didn't say anything, I conclude my answer is incorrect. Where's the mistake?
| Let $\alpha$ is the base angle. Then the area of one (bigger) blue triangle is $\frac{16^2}{2} \tan (2\pi-2\alpha)$, the area of the second blue triangle is $\frac{7^2}{2} \tan \alpha$. On the other hand, $\tan \alpha=\frac{16\tan (2\pi-2 \alpha)+9}{7}$ or (using double-angle formula and considering that $\tan (2\pi-2\alpha)=-\tan 2\alpha$) $$7\tan \alpha \cdot (1-\tan^2 \alpha)=-32 \tan \alpha-9 \tan^2 \alpha +9 \implies \tan \alpha =3, \tan 2\alpha=-\frac{3}{4}$$ Thus, the blue area is $8 \cdot 16 \cdot \frac{3}{4}+\frac{7^2 \cdot 3}{2}=169.5$
Your assumption is incorrect as the blue side is $12 \ne \frac{256}{23}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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finding the characteristic polynomial of $A$ given the characteristic polynomial of $A^2$ $A$ is a $3\times3$ matrix, the Jordan form of $A^2$ is $$
\begin{pmatrix}
25 & 1 & 0\\
0 & 25 & 0\\
0 & 0 & 1\\
\end{pmatrix}
$$
Find the characteristic polynomial, minimal polynomial and Jordan form of $A$ if $A$ has only negative eigenvalues.
I so far got to:
$P_{A^2}(X)=(x-25)^2(x-1)$, by cayley hamilton:
$P_{A^2}(A^2)=(A^2-25I)^2(A^2-I)=[(A-5I)(A+5I)]^2(A-1)(A+1)=0$
Therefore, $[(A-5I)(A+5I)]^2(A-1)(A+1)$ is divided by the minimal polynomial of $A$, and given all the eigenvalues of $A$ are negative meaning it is one of: $(x+1), (x+5), (x+5)(x+1), (x+5)^2(x+1), (x+5)^2$.
If $A$ is a diagonalizable matrix there is a matrix $M$ so $D=M^{-1}AM$ and $D$ is a diagonal matrix, however, by this equation we also get $D^2=M^{-1}A^2M$, of course $D^2$ is also diagonal meaning $A^2$ is diagonalizable in contradiction to the fact the Jordan form of $A^2$ is not diagonalizable.
Therefore, the minimal polynomial of $A$ is not a product of distinct linear factors, leaving us with $(x+1), (x+5), (x+5)^2(x+1), (x+5)^2$.
I can't figure out how to eliminate the other options, I know the first two options and the last one are incorrect because if the only eigenvalue of $A$ is $-1$ then the $25$ can't be an eigenvalue of $A^2$, I just don't know how to prove it (there is no proof in our book in relation to eigenvalues and power of matrices). Thanks.
| Let $\lambda_1, \lambda_2, \lambda_3$ the eigen values of $A$. $$P_A(X) = (x-\lambda_1)(x-\lambda_2)(x-\lambda_3)$$
Since the eigen values of $A$ are $25$, $25$, $1$. So you can assume that $\lambda_1^2=\lambda_2^2=25$ and $\lambda_3^2=1$. You can find the values of $\lambda_i$ since they are negative.
| {
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"url": "https://math.stackexchange.com/questions/4259064",
"timestamp": "2023-03-29T00:00:00",
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Expected value of the area of an annulus Consider the following problem:
Let $X,Y,Z,T$ be independent random variables with standard normal
distribution ($N(0,1)$). Consider the circles with center $(0,0)$ and radius
$(X,Y)$ and $(Z,T)$, respectively. Let $A$ be equal to the area of the
annulus defined by these circles (i.e. the region covered between the two concentric circles).
Calculate $\mathbb{E}(A)$ (the
expected value of the random variable $A$).
My attempt:
Since these random variables are independent, their joint probability density function can be written as:
$$ f(x,y,z,t) = \prod_{cycl} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} = \frac{1}{4\pi^2} e^{-\frac{1}{2}(x^2 + y^2 + z^2 +t^2)} $$
On the other hand, the area of the annulus can be calculated the following way:
$$ A= \pi |X^2+Y^2-Z^2-T^2| =: g(X,Y,Z,T) $$
So this means that
$$ \mathbb{E}(A) = \mathbb{E}(g(X,Y,Z,T)) = \iiiint\limits_{\mathbb{R}^4} g(\mathbf{x})f(\mathbf{x}) d\mathbf{x} = $$
$$ = \frac{1}{4\pi}\iiiint\limits_{\mathbb{R}^4} |x^2+y^2-z^2-t^2|e^{-\frac{1}{2}(x^2 + y^2 + z^2 +t^2)} \,dx\,dy\,dz\,dt = $$ $$= \frac{1}{2\pi} \iiiint\limits_{x^2+y^2>z^2+t^2} (x^2+y^2-z^2-t^2)e^{-\frac{1}{2}(x^2 + y^2 + z^2 +t^2)} \,dx\,dy\,dz\,dt =$$
$$ =\frac{1}{2\pi}\int\limits_{-\infty}^{\infty} e^{-\frac{x^2}{2}} \int\limits_{-\infty}^{\infty}e^{-\frac{y^2}{2}} \int\limits_{-\infty}^{\infty} e^{-\frac{z^2}{2}} \int\limits_{-\sqrt{x^2+y^2-z^2}}^{\sqrt{x^2+y^2-z^2}} (x^2+y^2-z^2-t^2)e^{-\frac{t^2}{2}} \,dt\,dz\,dy\,dx$$
The inner integral can be written as
$$\int\limits_{-\sqrt{x^2+y^2-z^2}}^{\sqrt{x^2+y^2-z^2}} (x^2+y^2-z^2-t^2)e^{-\frac{t^2}{2}} \,dt =$$
$$= (x^2+y^2-z^2)\int\limits_{-\sqrt{x^2+y^2-z^2}}^{\sqrt{x^2+y^2-z^2}} e^{-\frac{t^2}{2}} \,dt - \int\limits_{-\sqrt{x^2+y^2-z^2}}^{\sqrt{x^2+y^2-z^2}} t^2e^{-\frac{t^2}{2}} \,dt$$
which is fine, but I'm not able to evaluate it because of the limits...
Is my approach wrong? Thanks for any help in advance.
| The length of the radius is $U := \sqrt{X^2+Y^2}$ and $U^2=X^2+Y^2\sim\chi^2(2)$ therefore the random area is $A_{X,Y}=\pi U^2\sim \Gamma(1,2\pi)$. Same for $A_{Z,T}=\pi H^2$. The characteristic function of $\pi(U^2-H^2)$ is
$$E[e^{i\pi\theta(U^2-H^2)}]=E[e^{i\pi\theta U^2}]E[e^{-i\pi\theta H^2}]=(1-2i\pi\theta)^{-1}(1+2i\pi\theta)^{-1}=\frac{1}{1+4\pi^2\theta^2}$$
We manipulate this a bit:
$$\frac{1}{1+4\pi^2\theta^2}=\frac{1}{4\pi}\frac{\frac{1}{\pi}}{\frac{1}{4\pi^2}+\theta^2}$$
We get the antitransform (i.e. the density)
$$f_{\pi(U^2-H^2)}(y)=\frac{1}{4\pi}e^{-\frac{|y|}{2\pi}}$$
The expected area of the annulus is
$$E[|\pi(U^2-H^2)|]=\int_\mathbb{R}|y|f_{\pi(U^2-H^2)}(y)dy=2\int_{[0,\infty)}y\frac{1}{4\pi}e^{-\frac{|y|}{2\pi}}dy=2\pi$$
| {
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Solve $\lfloor x \rfloor ^2 + 9 \{x\}^2=6x-10, \forall x \in \mathbb{R}$ Solve $$\lfloor x \rfloor ^2 + 9 \{x\}^2=6x-10, \forall x \in \mathbb{R}$$
I tried to:
$$\lfloor x \rfloor ^2 + 9 (x-\lfloor x \rfloor)^2=6x-10$$
$$ 18 x \lfloor x \rfloor -10 \lfloor x \rfloor^2=(3x-1)^2 +9 $$
but I got stuck from here, or I started the exercise wrong? Thank you!
| Notice that
$$x\ge \lfloor x\rfloor>x-1$$
$$x^2\ge \lfloor x\rfloor ^2>x^2-2x+1$$
$$x^2+9>9\{x\}^2+\lfloor x\rfloor^2>x^2-2x+1$$
$$x^2+9>6x-10>x^2-2x+1$$
$$4+\sqrt5\approx 6.2>x>4-\sqrt5\approx 1.7$$
Hence $\lfloor x\rfloor\in\{2,3,4,5,6\}$. Now do casework based on each value of $\lfloor x\rfloor$.
| {
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Some intuition behind why we visit $\frac{2}{3}$ of Natural Numbers? Consider stating at $0$ on the number line of integers. Toss a fair coin and if heads advance $1$ space and if tails advance $2$ spaces.
*
*What number has the highest probability of being visited?
*What is the probability of visiting very large numbers roughly?
First part
The following recursive equation can be formed quite nicely conditioning on the first toss: $p_n = \frac{1}{2} p_{n-1} + \frac{1}{2}p_{n-2}$ (for $n \geq 3)$
In other words $p_n$ is the average of $p_{n-1}$ and $p_{n-2}$.
The average of $n$ values is less than or equal to the maximum of the values. Hence, for example, $10$ cannot be the most probably visited square because you visit either $8$ or $9$ with greater probability.
Likewise $9$ cannot be for the same logic continuing until we see $3$ cannot be. Hence the only candidates are $1$ or $2$.
The probability of visiting $1$ is $\frac{1}{2}$ and the probability of visiting $2$ is $\frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} = \frac{3}{4} $
Hence $2$ is the most probably visited square.
Second part
Solving the recurrence relation formed in the first part (using the intials $p_1$ and $p_2$ ) yields: $p_n = \frac{2}{3} + \frac{1}{3} \cdot (-\frac{1}{2})^n$
This explains the periodicity noticed prior and gives us the long-run average probability of $\frac{2}{3}$
This number seems way to high for my intuition. I expected it to be $\frac{1}{2} $ can someone provide some intuition behind $\frac{2}{3}$?
| The intuition behind it is very simple: the average step length is $\frac32$. Therefore you would expect to traverse a stretch of length $n$ in about $\frac{n}{3/2}=\frac23 n$ steps. So about $\frac23$ of the integers in that stretch will be visited.
| {
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Finding inverse function of $y=\sqrt{x+\sqrt{2x-1}}$
If inverse of the function $y=\sqrt{x+\sqrt{2x-1}}$ be equal to
$y=ax^2+bx+c$, then what is the value of $a^2+b^2+c^2$?
$A) 3$
$B) 4$
$C) 5$
$D) 6$
I tried to isolate $x$ from the equation $y=\sqrt{x+\sqrt{2x-1}}$:
$$y^2=x+\sqrt{2x-1}$$
$$(y^2-x)^2=2x-1$$
$$y^4-2xy^2+x^2=2x-1$$
$$y^2(y^2-2x)=-x^2+2x-1$$
But I still have $-2x$ in the parenthesis of the LHS.
| Since we are told that the inverse function is quadratic, we know that the curve function $ \ y \ = \ \sqrt{x+\sqrt{2x-1}} \ \ $ is the "upper half" of a "horizontal" parabola (which is not difficult to verify). We can find one point on the curve by considering that its domain is $ \ x \ \ge \ \frac12 \ \ ; $ the coordinates of the vertex for the horizontal parabola are then $ \ x \ = \ \frac12 \ , \ y \ = \ \sqrt{\frac12 \ + \ 0} \ = \ \frac{\sqrt2}{2} \ \ . $ The vertex of the parabola corresponding to the inverse function is therefore $ \ x \ = \ \frac{\sqrt2}{2} \ , \ y \ = \ \frac12 \ \ , $ so we can write the "vertex form" of its curve equation as $ \ y \ = \ a·\left(x - \frac{\sqrt2}{2} \right)^2 + \frac12 \ \ . $
We can next choose any second point on the "horizontal" parabola that is convenient to calculate, say, $ \ x \ = \ 1 \ \Rightarrow \ y \ = \ \sqrt{1 + \sqrt{2-1}} \ = \ \sqrt2 \ \ , $ which tells us that $ \ (\sqrt2 \ , \ 1) \ $ is a point on the inverse-function parabola. Hence,
$$ 1 \ \ = \ \ a·\left(\sqrt2 - \frac{\sqrt2}{2} \right)^2 \ + \ \frac12 \ \ = \ \ a·\left( \frac{\sqrt2}{2} \right)^2 \ + \ \frac12 \ \ = \ \ \frac12·a \ + \ \frac12 \ \ \Rightarrow \ \ a \ = \ 1 \ \ . $$
The "standard form" of the inverse function is given by $ \ y \ = \ \left(x - \frac{\sqrt2}{2} \right)^2 + \frac12 \ = \ x^2 \ - \ \sqrt2·x \ + \ 1 \ \ , $ for which the correct choice is $ \ \mathbf{(B)} \ \ [ \ 1^2 + (-\sqrt2)^2 + 1^2 \ = \ 4 \ ] \ \ . $
| {
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"timestamp": "2023-03-29T00:00:00",
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The sum of $\frac{999}{1999} + \frac{999}{1999} \frac{998}{1998} + ...+ \frac{999}{1999} \frac{998}{1998}...\frac{1}{1001}$? Ive tried to represent $\frac{999}{1999}$ as $1 - \frac{1000}{1999}$ and so on, but it didnt lead me anywhere. I also tried to represent it as $\frac{999!}{1999!} \cdot \frac{1998!}{998!} + \frac{999!}{1999!} \cdot \frac{1997!}{997!} + \dots + \frac{999!}{1999!} \cdot \frac{1002!}{2!} + \frac{999!}{1999!} \cdot \frac{1001!}{1!} + \frac{999!}{1999!} \cdot \frac{1000!}{0!}$. Basically, it means that all that's left to do is calculate what $ \frac{1998!}{998!}+\frac{1997!}{997!} + \dots + \frac{1000!}{1!}$ is, but i am stuck and don't have any idea how to do that. Is there any other way to do it faster?
| It becomes easier if the terms are added from right to left.
The sum of the last two terms is
$$
\frac{999}{1999} \frac{998}{1998} \cdots \frac{3}{1003}\frac{2}{1002}\left(1 + \frac{1}{1001}\right) = \frac{999}{1999} \frac{998}{1998} \cdots \frac{3}{1003} \cdot \frac{2}{1001} \, .
$$
The sum of the last three terms is therefore
$$
\frac{999}{1999} \frac{998}{1998} \cdots \frac{4}{1004}\frac{3}{1003}\left( 1 + \frac{2}{1001}\right) =
\frac{999}{1999} \frac{998}{1998} \cdots \frac{4}{1004} \cdot \frac{3}{1001} \, .
$$
Continuing in this way one finally gets for the complete sum
$$
\frac{999}{1999} \left( 1 + \frac{998}{1001}\right) = \frac{999}{1001} \, .
$$
Another option is to write the sum as
$$
\sum_{k=1}^{999} \frac{999! (1999-k)!}{(999-k)! 1999!}
= \frac{1}{\binom{1999}{1000}} \sum_{k=1}^{999} \binom{1999-k}{1000} = \frac{\binom{1999}{1001}}{\binom{1999}{1000}}
= \frac{999}{1001} \, ,
$$
using the the hockey-stick identity.
| {
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Calculate the value of $ abc \cdot (a + b + c) $ Let $ a, b $ and $ c $ be numbers such that $$ \begin {cases} a ^ 2-ab = 1 \\ b ^ 2-bc = 1 \\ c ^ 2-ac = 1 \end {cases} $$Calculate the value of $ abc \cdot (a + b + c) $
Attempt: Can I solve the problem by symmetry?
| Another way.
We have $$a^2b-ab^2=b,$$ $$b^2c-c^2b=c$$ and $$c^2a-a^2c=a,$$ which gives $$\sum_{cyc}(a^2b-ab^2)=a+b+c$$ or $$(a-b)(a-c)(b-c)=a+b+c.$$
In another hand, $$\prod_{cyc}(a^2-ab)=1$$ or $$abc(a-b)(b-c)(c-a)=1,$$ which gives $$abc(a+b+c)=-1.$$
| {
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Show that $(I-\mathbf x\mathbf y^T)^{-1} = I-\frac{1}{\mathbf x^T\mathbf y- 1}\mathbf x\mathbf y^T$ Let $x,y\in\mathbb R^n$ and suppose that $x^Ty \neq 1$. Show that $(I-\mathbf x\mathbf y^T)^{-1} = I-\frac{1}{\mathbf x^T\mathbf y- 1}\mathbf x\mathbf y^T$. Note, I need to compute this directly not as some special case.
We need to show that two things:
*
*$(I-x y^T)(I-\frac{1}{x^T y- 1} x y^T) = I$, and
*$(I-\frac{1}{x^T y- 1} x y^T)(I-x y^T) = I$
By definition.
For (1) I have (EDIT 1)
$$\begin{equation}\begin{split}(I-x y^T)(I-\frac{1}{x^T y- 1} x y^T) &= I - \frac{1}{x^T y- 1} x y^T-x y^T + \frac{1}{x^T y- 1} x y^Tx y^T \\ &=I - \frac{1}{x^T y- 1} x y^T-x y^T\left(\frac{x^T y- 1}{x^T y- 1}\right) + \frac{1}{x^T y- 1} x y^Tx y^T \\ &= I - \frac{1}{x^T y- 1} \left[x y^T + (x^Ty-1)x y^T - x y^Txy^T\right] \\ &=I - \frac{1}{x^T y- 1} \left[x y^T + x^Tyx y^T-x y^T -x y^Txy^T\right] \\ &=I - \frac{1}{x^T y- 1}\left[x^Tyx y^T -x y^Txy^T\right] \\ &=I - \frac{1}{x^T y- 1}\left[x^Tyx y^T -x (y^Tx)y^T\right] \\ &=I - \frac{1}{x^T y- 1}\left[x^Tyx y^T -(y^Tx)x y^T\right]\\ &=I - \frac{1}{x^T y- 1}\left[x^Tyx y^T -x^Tyx y^T\right] \\ &= I.\end{split}\end{equation}$$
With the help in the comments! I think I got it.
And similar multiplication for (2)...
| For (1), we will use the fact that $\frac{x^Ty-1}{x^Ty-1} = 1$ and $x^Ty = y^Tx$. We also use the fact that $xy^Txy^T = (x(y^Tx)y^T) = (y^Tx)xy^T =y^Txxy^T =x^Tyxy^T $. Obser for (1) we have that
$$\begin{equation}\begin{split}\left(I-x y^T\right)\left(I-\frac{1}{x^T y- 1} x y^T\right) &= I - \frac{1}{x^T y- 1} x y^T-x y^T + \frac{1}{x^T y- 1} x y^Tx y^T \\ &=I - \frac{1}{x^T y- 1} x y^T-x y^T\left(\frac{x^T y- 1}{x^T y- 1}\right) + \frac{1}{x^T y- 1} x y^Tx y^T \\ &= I - \frac{1}{x^T y- 1} \left[x y^T + (x^Ty-1)x y^T - x y^Txy^T\right] \\ &=I - \frac{1}{x^T y- 1} \left[x y^T + x^Tyx y^T-x y^T -x^Tyxy^T\right] \\ &=I - \frac{1}{x^T y- 1}\left[x^Tyx y^T -x^Tyxy^T\right] \\ &=I.\end{split}\end{equation}$$
For (2) we have
$$\begin{equation}\begin{split}\left(I-\frac{1}{x^T y- 1} x y^T\right)\left(I-x y^T\right) &= I - xy^T - \frac{1}{x^T y- 1} x y^T+ \frac{1}{x^T y- 1} x y^Txy^T \\ &= I. \text{ by (1) calculation}\end{split}\end{equation}$$
Thus, $(I-x y^T)^{-1} = I-\frac{1}{x^T y- 1} x y^T$.
| {
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How to prove:$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sum_{cyc}{\sqrt{2(a^2+b^2)}}$ Problem: Let $a,b,c>0. $ Prove that: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sqrt{2(a^2+b^2)}+\sqrt{2(b^2+c^2)}+\sqrt{2(c^2+a^2)}$$
I have seen problem before, and I tried to prove: $$2(a+b+c)\ge\sqrt{2(a^2+b^2)}+\sqrt{2(b^2+c^2)}+\sqrt{2(c^2+a^2)}(!)$$
since by C-S inequality: $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c$. But (!) is not true.
Anyone can help me give a hint to solve this nice problem?. Thanks!
| Applying Cauchy-Schwarz inequality: $LHS = \sum_{\text{cyc}} \dfrac{a^2+b^2}{b}=\sum_{\text{cyc}}\dfrac{\left(\sqrt{a^2+b^2}\right)^2}{b}\ge\dfrac{\left(\sum_{\text{cyc}}\sqrt{a^2+b^2}\right)^2}{\sum_{\text{cyc}} a}\ge RHS\iff \sum_{\text{cyc}}\sqrt{a^2+b^2}\ge \sum_{\text{cyc}} \sqrt{2}a$, but this is true since $\sum_{\text{cyc}} \sqrt{a^2+b^2} \ge \sum_{\text{cyc}} \dfrac{a+b}{\sqrt{2}}= \sum_{\text{cyc}} \sqrt{2}a$ .
| {
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Prove $\frac{d}{\sqrt{a^2+b^2+c^2}}+\sqrt{3}\ge2\left(\sqrt{\frac{d}{a+b+c}}+\frac{\sqrt{ab+bc+ca}}{a+b+c}\right)$
Problem: Let $a,b,c,d>0.$ Prove that: $$\frac{d}{\sqrt{a^2+b^2+c^2}}+\sqrt{3}\ge2\left(\sqrt{\frac{d}{a+b+c}}+\frac{\sqrt{ab+bc+ca}}{a+b+c}\right)$$
It is my old teacher's problem.
My attempt: I guess a=b=c=d so using C-S inequality: $$R.H.S\le2\sqrt{\frac{d}{a+b+c}+\frac{ab+bc+ca}{(a+b+c)^2}}$$
But it seems weak to get the proof. I hope we can find a good solution for nice problem.
Thank you!
| We need to prove that:
$$\frac{d}{\sqrt{a^2+b^2+c^2}}+\sqrt3-\frac{2\sqrt{ab+ac+bc}}{a+b+c}\geq2\sqrt{\frac{d}{a+b+c}}$$ and since $$\sqrt3(a+b+c)\geq2\sqrt{ab+ac+bc},$$ by AM-GM we obtain:
$$\frac{d}{\sqrt{a^2+b^2+c^2}}+\sqrt3-\frac{2\sqrt{ab+ac+bc}}{a+b+c}\geq2\sqrt{\frac{d\left(\sqrt3-\frac{2\sqrt{ab+ac+bc}}{a+b+c}\right)}{\sqrt{a^2+b^2+c^2}}}.$$
Thus, it's enough to prove that:
$$\frac{\sqrt3-\frac{2\sqrt{ab+ac+bc}}{a+b+c}}{\sqrt{a^2+b^2+c^2}}\geq\frac{1}{a+b+c}.$$
Now, let $a^2+b^2+c^2=t(ab+ac+bc).$
Thus, we need to prove that:
$$\frac{\sqrt3-\frac{2}{\sqrt{t+2}}}{\sqrt{t}}\geq\frac{1}{\sqrt{t+2}}$$ or
$$\sqrt{3(t+2)}\geq2+\sqrt{t},$$ which is true by C-S:
$$\sqrt{3(t+2)}=\sqrt{(2+1)(2+t)}\geq2+\sqrt{t}.$$
| {
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$1-\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\frac{1}{11\cdot 3^5}+\frac{1}{13\cdot 3^6}--++\cdots.$ I want to evaluate the series
$$1-\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\frac{1}{11\cdot 3^5}+\frac{1}{13\cdot 3^6}--++\cdots.$$
I can rewrite this as
$$1+\sum_{n\geq 1} (-3)^{-3n}\left(\frac{3}{6n-1}+\frac{1}{6n+1}\right)$$
The answer should be $\frac{\ln7}{2}$. Although I can't see explicitly, since I get a log function, maybe it can be solved by differentiating some test function $f(x)$ and then substituting an appropriate number.
| For context and inspiration, let's remember that the linear factors in the power series for $\log(1+x)$ come from integrating a geometric series:
$$
\log(1+x) = \int_0^x \frac1{1+t}\,dt = \int_0^x \sum_{n=0}^\infty (-1)^n t^n \,dt = \sum_{n=0}^\infty (-1)^n \frac{t^{n+1}}{n+1}.
$$
So if we recognize our series as a special value of a power series, which turns out to be
$$
f(x) = 1-\sqrt3\biggl( \frac{x^5}5 + \frac{x^7}7 - \frac{x^{11}}{11} - \frac{x^{13}}{13} + \frac{x^{17}}{17} + \frac{x^{19}}{19} - \frac{x^{23}}{23} - \frac{x^{25}}{25} + \cdots \biggr)
$$
evaluated at $x=\frac1{\sqrt3}$, then we can work out what that power series is by writing it as an integral of a rational function:
\begin{align*}
f(x) &= 1 - \sqrt3 \int_0^x ( x^4 + x^6 - x^{10} - x^{12} + x^{16} + x^{18} - x^{22} - x^{24} + \cdots )\,dt \\
&= 1 - \sqrt3 \int_0^x (t^4+t^6)(1-t^6+t^{12}-t^{18}+\cdots)\,dt \\
&= 1 - \sqrt3 \int_0^x \frac{t^4+t^6}{1+t^6} \,dt = 1 - \sqrt3 \int_0^x \frac{t^4}{t^4-t^2+1} \,dt.
\end{align*}
Using partial fractions (well, telling my computer to use partial fractions), we obtain
\begin{align*}
f(x) &= 1 - \int_0^x \biggl( \sqrt3 + \frac{t-\sqrt3/2}{t^2-t\sqrt3+1} - \frac{t+\sqrt3/2}{t^2+t\sqrt3+1} \biggr) \,dt \\
&= 1 - x\sqrt3 - \frac{\log (x^2-x\sqrt3+1)}2 + \frac{\log (x^2+x\sqrt3+1)}2,
\end{align*}
and therefore the series we want is
\begin{align*}
f\biggl( \frac1{\sqrt3} \biggr) &= 1 - 1 - \frac{\log (1/3)}2 + \frac{\log (7/3)}2 = \frac{\log7}2
\end{align*}
as predicted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4279594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Cantor function integral. I'm trying to solve the following integral for a problem about fractals involving Cantor set:
$$\mathcal{I}=\int_{0}^{1}C\left(\sqrt{1-x^2}\right)dx$$
Where $C(x)$ denotes the Cantor ternary function.
I don't know a lot about measure theory and Lebesgue integrals, so I don't know what to do here. I tried the substitution $u=\sqrt{1-x^2}$, but it led me nowhere.
I've made a small program to numerically evaluate it, and it yielded $\mathcal{I}\approx0.7357895383$.
Is it possible to find a closed-form expression or a series expansion for this integral?
| Partial solution.
Put $t = \sqrt{1-x^2}$. Hence $x = \sqrt{1-t^2}$ and
$\mathcal{I} = \int_0^1 \frac{C(t) tdt}{\sqrt{1-t^2}}$.
Open Cantor set $A = \bigsqcup_{n \ge1, 1 \le k \le 2^{n-1}} (a_{nk}, b_{nk})$, where $(a_{11}, b_{11}) = (\frac13, \frac23)$, $(a_{21}, b_{21}) = (\frac19, \frac29)$, $(a_{22}, b_{22}) = (\frac79, \frac89)$, $(a_{31}, b_{31}) = (\frac1{27}, \frac2{27})$, $(a_{32}, b_{32}) = (\frac7{27}, \frac8{27})$, $(a_{33}, b_{33}) = (\frac{19}{27}, \frac{20}{27})$, $(a_{34}, b_{34}) = (\frac{25}{27}, \frac{26}{27})$ etc. Denote by $C_{ij}$ the value of $C$ on $(a_{nk}, b_{nk})$. It's easily seen that $C_{11} = \frac{1}2, C_{21} = \frac{1}4, C_{22} = \frac{3}4, C_{31} = \frac{1}8, C_{32} = \frac{3}8, C_{33} = \frac{5}8, C_{34} = \frac{7}8$ and
$C(t) = \frac{2k-1}{2^n}$ for $t \in (a_{nk}, b_{nk})$. Thus
$$\mathcal{I} = \int_A \frac{C(t)tdt}{\sqrt{1-t^2}} = \sum_{n \ge 1, 1 \le k \le 2^{n-1}} \int_{a_{nk}}^{b_{nk}} \frac{C(t)tdt}{\sqrt{1-t^2}} = \sum_{n \ge 1, 1 \le k \le 2^{n-1}} \int_{a_{nk}}^{b_{nk}} \frac{2k-1}{2^n}\frac{tdt}{\sqrt{1-t^2}} = $$
$$ = \sum_{n \ge 1, 1 \le k \le 2^{n-1}} \frac{2k-1}{2^n} \int_{a_{nk}}^{b_{nk}} \frac{tdt}{\sqrt{1-t^2}}$$
It's sufficient to paste formulas for $a_{nk}$ and $b_{nk}$ and find the sum.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving $\int_{0}^{2\pi}\frac{d\phi}{2+\sin(\phi)}$ with complex exponentials I’m working through the textbook A First Course in Complex Analysis and I’m trying to solve problem 4.30:
Compute the real integral
$$\int_{0}^{2\pi}\frac{d\varphi}{2+\sin(\varphi)}$$
by writing the sine function in terms of the complex exponential and making the substitution $z=e^{i\varphi}$ to turn the real integral into a complex integral.
I started by rewriting the sine function as $\frac{e^{i\varphi}-e^{-i\varphi}}{2i}$:
$$=\int_{0}^{2\pi} \frac{d\varphi}{2+\tfrac{e^{i\varphi}-e^{-i\varphi}}{2i}}=2i\int_{0}^{2\pi} \frac{d\varphi}{4i+e^{i\varphi}-e^{-i\varphi}}$$
Then I substituted $z=e^{i\varphi}$, as the problem asked, and got $dz=izd\varphi$, so $d\varphi=\tfrac{-idz}{z}$. I also defined my path as $\gamma:[0,2\pi]\rightarrow \mathbb{C}$ as $\gamma(\varphi)=e^{i\varphi}$
$$=2i\int_{\gamma} \frac{1}{4i+z-\tfrac{1}{z}} \frac{-idz}{z}=2\int_{\gamma} \frac{dz}{z^2+4iz-1}$$.
We can see that the inside of the integral is a rational function, so it should be holomorphic everywhere where it’s defined. The poles can then be calculated by the quadratic equation as $\pm \sqrt{3}i-2i$.
However, these poles clearly lie outside of the region enclosed by $\gamma$, the unit circle.
Since the inside of the integral is defined and holomorphic on all points on $\gamma$ and inside $\gamma$, and $\gamma$ is a closed curve, we have:
$$2\int_{\gamma} \frac{dz}{z^2+4iz-1}=2\cdot0=0$$
However, this is very obviously wrong, because the original integral is always positive and hence can’t be zero.
So my questions are:
*
*Where is my mistake?
*How would you solve this correctly?
| Here is another approach that relies on Calculus II.
We have
$$
I=\int_0^{2\pi}\frac{dx}{2+\sin x}=\int_{0}^{\pi}\frac{dx}{2+\sin x}+\int_{\pi}^{2\pi}\frac{dx}{2+\sin x}
$$
Setting $y=x-\pi/2, z=x-3\pi/2$ in the first and second integral, respectively, we get
\begin{eqnarray}
I&=&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dy}{2+\sin\left(y+\frac{\pi}{2}\right)}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dz}{2+\sin\left(z+\frac{3\pi}{2}\right)}\cr
&=&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dy}{2+\cos y}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dz}{2-\cos z}\cr
&=&\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dx}{2+\cos x}+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{dx}{2-\cos x}
\end{eqnarray}
Let $t=\tan(x/2)$. Since
$$
\cos x=\frac{1-t^2}{1+t^2}
$$
and
$$
dx=\frac{2dt}{1+t^2}
$$
we get
\begin{eqnarray}
I&=&\int_{-1}^{1}\frac{1}{2+\frac{1-t^2}{1+t^2}}\cdot\frac{2dt}{1+t^2}+
\int_{-1}^{1}\frac{1}{2-\frac{1-t^2}{1+t^2}}\cdot\frac{2dt}{1+t^2}\cr
&=&\int_{-1}^{1}\frac{2}{2(1+t^2)+1-t^2}dt+
\int_{-1}^{1}\frac{2}{2(1+t^2)-(1-t^2)}dt\cr
&=&4\int_{0}^{1}\frac{1}{3+t^2}dt+
4\int_{0}^{1}\frac{1}{1+3t^2}dt\cr
&=&4\int_{0}^{1}\frac{1}{3+t^2}dt+
\frac{4}{3}\int_{0}^{1}\frac{1}{\frac{1}{3}+t^2}dt\cr
&=&\frac{4}{\sqrt{3}}\left[\tan^{-1}\left(\frac{t}{\sqrt{3}}\right)\right]_0^1+\frac{4\sqrt{3}}{3}\left[\tan^{-1}(\sqrt{3}t)\right]_0^1\cr
&=&\frac{4}{\sqrt{3}}\left[\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)-\tan^{-1}(0)+\tan^{-1}(\sqrt{3})+\tan^{-1}(0)\right]\cr
&=&\frac{4}{\sqrt{3}}\left(\frac{\pi}{6}+\frac{\pi}{3}\right)\cr
&=&\frac{4}{\sqrt{3}}\cdot\frac{\pi}{2}\cr
&=&=\frac{2\pi}{\sqrt{3}}.
\end{eqnarray}
| {
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"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find $\angle CAD$ if $\triangle ABC$ is right angled at $B$, $\angle BAD = 30^\circ, \angle ADB = \angle ADC = 15^\circ$ Find angle $\theta$ in the below diagram.
This is a question that was brought to me by a high school student.
While I came up with a trigonometric solution and a synthetic solution, I am posting here to see more solutions that others come up with (esp. other synthetic solutions)
My immediate solution involved combination of a simple construction and trigonometry, basically knowing that $\tan 30^\circ = \dfrac{1}{\sqrt3}$ and $\tan 15^\circ = 2 - \sqrt3$.
We draw perp from $D$ to $AB$ extend and $BC$ extend. We also note that $\angle DBC = \angle DBE = 45^\circ$. If $AB = x, BE = y$, we find $x$ in terms of $y$. We next find $CF$ in terms of $y$ and subtracting from $y$ gives us $BC$ and we show $BC = x$.
Then in search of a synthetic solution, I drew a few more lines and as $FE$ is perpendicular bisector of $BD$, $BG = GD, BH = HD$.
So we see that $AB = BG$ and by A-A-S, $\triangle BHC \cong \triangle BHG$ which leads to $AB = BC$ and we have $\theta = 15^\circ$.
Look forward to more interesting solutions.
| Sine rule for $\Delta ABD$:
$$\frac{AB}{\sin 15^\circ}=\frac{BD}{\sin 30^\circ} \Rightarrow BD=\frac{AB}{2\sin 15^\circ}$$
Sine rule for $\Delta BCD$:
$$\frac{BC}{\sin 30^\circ}=\frac{BD}{\sin 105^\circ} \Rightarrow BC=\frac{BD}{2\sin 105^\circ}=\frac{BD}{2\cos 15^\circ}=\frac{AB}{4\sin 15^\circ\cos 15^\circ}=AB$$
Hence, the triangle $ABC$ is right angled and isosceles, implying $\theta=15^\circ$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4293587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
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Discrete calculus: is my proof about the difference of two consecutive powers correct? Theorem statement: $\displaystyle \Delta^dF_n(x) = \sum_{k=0}^{n-1} \binom{n-1}{k} \Delta^{d-1}F_{n-1-k}(x)(-1)^k$
Proof:
$$\begin{align} F_n(x) &= x^n =\Delta^0F(x) \\[2ex] \Delta^1F_n(x) &= x^n -(x-1)^n \\[2ex] &= x^n - \sum_{k=0}^n\binom nk x^{n-k}(-1)^k \\[2ex] &= \sum_{k=0}^{n-1}\binom{n-1}{k}x^{n-k}(-1)^{k} \\[2ex] &=x^{n-1} -(n-1)x^{n-2} + \ldots \pm (n-1)x^2 \mp 1 \\[2ex] &= \sum_{k=0}^{n-1}\binom{n-1}{k}\Delta^0F_{n-1-k}(x)(-1)^k\end{align}$$
$$\begin{align}\Delta^2F_n(x) &=\Delta^1F_n(x) -\Delta^1F_n(x-1) \\[2ex] &= \sum_{k=0}^{n-1}\binom{n-1}{k}x^{n-k}(-1)^k - \sum_{k=0}^{n-1}\binom{n-1}{k}(x-1)^{n-k}(-1)^k \\[2ex] & = x^{n-1}-(x-1)^{n-1}- (n-1)x^{n-2} - (n-1)(x-1)^{n-2} +\ldots \pm(n-1)x^2 -(n-1)(x-1)^2 \\[2ex] & = \sum_{k=0}^{n-1} \binom{n-1}{k}\Delta^1F_{n-1-k}(x)(-1)^k\end{align}$$
$$\begin{align} \Delta^3F_n(x) &= \Delta^2F_n(x) - \Delta^2F_n(x-1) \\[2ex] &= \sum_{k=0}^{n-1} \binom{n-1}{k}\Delta^1F_{n-1-k}(x)(-1)^k -\sum_{k=0}^{n-1} \binom{n-1}{k}\Delta^1F_{n-1-k}(x-1)(-1)^k \\[2ex] &= \sum_{k=0}^{n-1} \binom{n-1}{k} (\Delta^1F_{n-1-k}(x) - \Delta^1F_{n-1-k}(x-1))(-1)^k \\[2ex] &= \sum_{k=0}^{n-1} \binom{n-1}{k} \Delta^2F_{n-1-k}(x)(-1)^k \end{align}$$
$$\therefore \Delta^dF_n(x) = \sum_{k=0}^{n-1} \binom{n-1}{k} \Delta^{d-1}F_{n-1-k}(x)(-1)^k$$
Is this correct? If not, where did I go wrong?
If it is correct, what can I do with this? Where can I apply the derived identity?
I originally set out to prove that $\Delta^n F_n(x) = c \ \forall x$, but I'm not sure if I'm any closer to this now.
| There is a mistake early on:
\begin{align*}
x^n - (x - 1)^n &= x^n - \left(x^n - nx^{n-1} + \frac{n(n-1)}{2}x^{n-2} - \cdots \right) \\
&= nx^{n-1} - \frac{n(n-1)}{2}x^{n-2} + \frac{n(n-1)(n-2)}{6}x^{n-3} + \cdots \\
&= \sum_{i = 0}^{n-1} \binom{n}{i}(-1)^{n-i+1}x^i \\
\end{align*}
Whereas you said this is $x^{n-1} -(n-1)x^{n-2} + \ldots \pm (n-1)x^2 \mp 1 = (x - 1)^{n-1}$.
The easiest way to describe $\Delta^k$ of some polynomial is to write the polynomial in the basis $\{x^{\underline i} : i \ge 0\}$ where $x^{\underline i} = x(x-1)(x-2)\cdots(x-i+1)$. This is a nice basis for applying $\Delta$ since $\Delta(x^{\underline i}) = ix^{\underline{i - 1}}$.
The connection coefficients are the Stirling numbers of the second kind:
$$x^n = \sum_{k = 0}^n \begin{Bmatrix} n \\ k \end{Bmatrix} x^{\underline k}.$$
However, if all you care about is the identity $\Delta^n(x^n) = n!$ then it suffices that the coefficient of $x^{\underline n}$ is $1$ and the other coefficients are just some integers.
Now you can show (by induction, formally) that $\Delta^{k}(x^{\underline i}) = i(i - 1) \dots (i - k + 1)x^{\underline{i - k}} = i^{\underline k} x^{\underline{i - k}}$ and that this is $0$ if $k > i$.
It therefore follows that
$$\Delta^{n}(x^n) = \sum_{k = 0}^n \begin{Bmatrix} n \\ k \end{Bmatrix} \Delta^{n}(x^{\underline k}) = \begin{Bmatrix} n \\ n \end{Bmatrix} \Delta^{n}(x^{\underline n}) = n^{\underline n} x^{\underline 0} = n!.$$
Alternatively, you can prove this without the change of basis:
*
*$\Delta^0(x^0) = 0!, \Delta^1(x^1) = x - (x - 1) = 1!$
*Assume $\Delta^{i}(x^i) = i!$ for $i < n$
*Then $\Delta^{i + 1}(x^i) = 0$ for all $i < n$ and in particular, $\Delta^{n - 1}(x^i) = i$ except $i = n - 1$.
*Now apply $\Delta^{n-1}$ to the identity at the very top:
\begin{align*}
\Delta^{n-1}(\Delta^1(x^n)) &= \Delta^{n-1}\left(nx^{n-1} - \frac{n(n-1)}{2}x^{n-2} + \frac{n(n-1)(n-2)}{6}x^{n-3} + \cdots \right) \\
&= \Delta^{n-1}(nx^{n-1}) \\
&= n\Delta^{n-1}(x^{n-1}) \\
&= n(n-1)!
\end{align*}
by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4294063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Rationalizing $\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}$ in two ways gives different answers I have a doubt see when we rationalize denominator of expression $$\frac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}- \sqrt{1-\cos x}}$$
we get answer $$\frac{1+\sin x}{\cos x}$$ but when we rationalize numerator we get
$$\frac{\cos x}{1+\sin x}$$
How is this possible, because rationalizing means just multiplying by $1$?
| Putting $a$ for $\cos x$,
you have
(notice the MathJax)
$\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}
$
I now do the rationalizing
(using
$(u+v)(u-v)=u^2-v^2)$
and later substitute
$a = \cos(x)$
and use
$\sin^2(x)+\cos^2(x) = 1$.
I am doing this
in excruciating detail
so you can see
all the steps involved.
Once you understand these,
you should be able
to do this kind of thing
by yourself.
$\begin{array}\\
\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}
&=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}+\sqrt{1-a}}\\
&=\dfrac{1+a+2\sqrt{1+a}\sqrt{1-a}+1-a}{(1+a)-(1-a)}\\
&=\dfrac{2+2\sqrt{(1+a)(1-a)}}{2a}\\
&=\dfrac{1+\sqrt{1-a^2}}{a}\\
&=\dfrac{1+\sqrt{1-\cos^2(x)}}{\cos(x)}\\
&=\dfrac{1+\sin(x)}{\cos(x)}\\
\end{array}
$
I will be glad to answer
any questions you may have.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295216",
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"source": "stackexchange",
"question_score": "5",
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Find the limit given that $f(1)=1$, $f(x+y)=f(x)+f(y)+2xy$ and $f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4}$ Given the function $f: \mathbb{R}\to \mathbb{R}$ satisfies:
\begin{cases}
f(1)=1 \\
f(x+y)=f(x)+f(y)+2xy,\quad&\forall x,y \in \mathbb{R}\\
f\left(\frac{1}{x}\right)=\frac{f(x)}{x^4},&\forall x\neq0
\tag{1}\label{eqn1}
\end{cases}
Then find the limit of:
\begin{align}
L=\lim_{x \to 0} \frac{\frac{1}{e^{2f(x)}}-\sqrt[3]{1+f(x)}}{\ln\bigl(1+f(x)\bigr)}
\end{align}
My attempts
I can easily guess that $f(x)=x^2$, which totally satisfies \eqref{eqn1}. What I did was that I set $g(x)=f(x)-x^2$, then I got:
\begin{cases}
g(x)+g(y)=g(x+y)\\
g\left(\frac{1}{x}\right)x^4=g(x)
\tag{2}\label{eqn2}
\end{cases}
From \eqref{eqn1}, substitute $x=y=0$ we have $f(0)=0$
Now $g(1)=g(0)=0$, then I substitute $y=1-x$ in \eqref{eqn2} and got that
\begin{align}
g\left(\frac{1}{x}\right)x^4+g\left(\frac{1}{1-x}\right)(1-x)^4=0
\end{align}
Because the values of either $\frac1x$ or $\frac1{1-x}$ must lie between $0$ and $1$, I suspect that we can prove $g \equiv 0$.
Even if I have $f$ then, I'm still clueless about how to calculate the limit.
Any idea or solution or suggestion on the tools?
Any help is appreciated!
| Calculating the given limit is standard, when you prove that $ f ( x ) = x ^ 2 $ for all $ x \in \mathbb R $. Let's focus on the nontrivial part of the problem. Using $ g ( 1 ) = 0 $,
$$ g ( x + y ) = g ( x ) + g ( y ) \text , \tag 0 \label 0 $$
and
$$ g \left ( \frac 1 x \right ) = \frac { g ( x ) } { x ^ 4 } \text , \tag 1 \label 1 $$
you can see that for any $ x \in \mathbb R \setminus \{ - 1 , 0 \} $,
\begin{align*}
\frac { g \left ( x ^ 2 \right ) } { x ^ 4 ( x + 1 ) ^ 4 } + \frac { g ( x ) } { x ^ 4 ( x + 1 ) ^ 4 }
& = \frac { g \left ( x ^ 2 + x \right ) } { \left ( x ^ 2 + x \right ) ^ 4 } \\
& \stackrel { \eqref {1} } = g \left ( \frac 1 { x ^ 2 + x } \right ) \\
& \stackrel { \eqref {0} } = g \left ( \frac 1 x \right ) - g \left ( \frac 1 { x + 1 } \right ) \\
& \stackrel { \eqref {1} } = \frac { g ( x ) } { x ^ 4 } - \frac { g ( x + 1 ) } { ( x + 1 ) ^ 4 } \\
& \stackrel { \eqref {0} } = \frac { g ( x ) } { x ^ 4 } - \frac { g ( x ) + g ( 1 ) } { ( x + 1 ) ^ 4 } \\
& = \frac { \left ( 4 x ^ 3 + 6 x ^ 2 + 4 x + 1 \right ) g ( x ) } { x ^ 4 ( x + 1 ) ^ 4 } \text ,
\end{align*}
which implies
$$ g \left ( x ^ 2 \right ) = 2 x ^ 5 ( x + 1 ) ^ 4 \left ( x ^ 2 + 3 x + 1 \right ) g ( x ) \text . \tag 2 \label 2 $$
Using \eqref{0} and \eqref{2} you get
\begin{align*}
& & & 2 x ^ 5 ( x + 1 ) ^ 4 \left ( x ^ 2 + 3 x + 1 \right ) g ( x ) + 2 g ( x y ) + 2 y ^ 5 ( y + 1 ) ^ 4 \left ( y ^ 2 + 3 y + 1 \right ) g ( y ) \\
& & \stackrel { \eqref {2} } = & g \left ( x ^ 2 \right ) + 2 g ( x y ) + g \left ( y ^ 2 \right ) \\
& & \stackrel { \eqref {0} } = & g \left ( ( x + y ) ^ 2 \right ) \\
& & \stackrel { \eqref {2} } = & 2 ( x + y ) ^ 5 ( x + y + 1 ) ^ 4 \left ( ( x + y ) ^ 2 + 3 ( x + y ) + 1 \right ) g ( x + y ) \\
& & \stackrel { \eqref {0} } = & 2 ( x + y ) ^ 5 ( x + y + 1 ) ^ 4 \left ( ( x + y ) ^ 2 + 3 ( x + y ) + 1 \right ) \bigl ( g ( x ) + g ( y ) \bigr ) \text ,
\end{align*}
which for $ y = 1 $ shows that
$$ \Bigl ( ( x + 1 ) ^ 5 ( x + 2 ) ^ 4 \left ( x ^ 2 + 5 x + 5 \right ) - x ^ 5 ( x + 1 ) ^ 4 \left ( x ^ 2 + 3 x + 1 \right ) - 1 \Bigr ) g ( x ) = 0 \text . \tag 3 \label 3 $$
Therefore, $ g ( x ) = 0 $ for all $ x \in \mathbb R \setminus A $, where $ A $ is a finite set of real numbers consisting of $ - 1 $, $ 0 $, and finitely many roots of the polynomial appearing on the left-hand side of \eqref{3}. For any $ a \in A $, choosing a positive integer $ n $ large enough so that $ a + n > b $ for all $ b \in A $, you can use \eqref{0} to see that $ g ( a ) = 0 $, and hence $ g ( x ) = 0 $ for all $ x \in \mathbb R $, or equivalently $ f ( x ) = x ^ 2 $ for all $ x \in \mathbb R $.
| {
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"answer_count": 3,
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Permutations of the set $\{1,\ldots,n\}$ are there such that, no two of the numbers $1,2,3$ are consecutive
How many permutations of the set $\{1,\ldots,n\}$ are there such that no two of the numbers $1,2,3$ are consecutive?
My first thought was the inclusion-exclusion formula, like here, but it got messy, so I decided to work with partitions.
Let's consider $3$ places for the numbers $1,2,3$. Later, we can arrange them in $3!$ ways. Then, we're left with $n-3$ elements. The gaps between $1^{\mathrm{st}}$ and $2^{\mathrm{nd}}$ and $2^{\mathrm{nd}}$ and $3^{\mathrm{rd}}$ element should in no way be empty, so, we need partitions of the set $\{4,\ldots,n\}$ into $4,3$ and $2$ subsets. There are $S(n-3,4),S(n-3,3)$ and $S(n-3,2)$ of them respectively, where $S(n,k)$ is a Stirling number of the second kind. After the sizes of sets have been chosen, we can permute those remaining $n-3$ elements from the original set. If I'm not wrong, this should be equivalent to the problem of placing distinct objects (here numbers) into distinct boxes (here, gaps). Therefore, my answer is $\#=3!(n-3)!\left(S(n-3,4)+2S(n-3,3)+S(n-3,2)\right).$
Is this correct? Or did I over/undercount?
| I will solve the problem in two ways.
Method 1: We use the Inclusion-Exclusion Principle.
There are $n!$ permutations of the elements of the set $\{1, 2, 3, \ldots, n\}$. From these, we must exclude those permutations in which at least two of the numbers $1, 2, 3$ are consecutive.
Since there are only three numbers in the subset, there can be at most two pairs of consecutive numbers from the subset $\{1, 2, 3\}$ in the permutation, with two pairs occurring when all three of those numbers appear consecutively in the permutation.
A pair of consecutive numbers from the subset $\{1, 2, 3\}$: There are $\binom{3}{2}$ ways to select which two of those three numbers are consecutive. We now have $n - 1$ objects to permute, a block of two consecutive numbers and the other $n - 2$ numbers in the set $\{1, 2, 3, \ldots, n\}$. The $n - 1$ objects can be permuted in $(n - 1)!$ ways. The numbers in the block can be permuted in $2!$ ways. Hence, there are
$$\binom{3}{2}(n - 1)!2!$$
such permutations.
However, if we subtract that amount, we will have subtracted each permutation in which there are two pairs of consecutive numbers drawn from the subset $\{1, 2, 3\}$ twice, once when we designate the first two of the three consecutive numbers as the pair of consecutive numbers and once when we designate the last two of the three consecutive numbers as the pair of consecutive numbers. We only want to subtract such permutations once, so we must add them to the total.
Two pairs of consecutive numbers from the subset $\{1, 2, 3\}$: For this to occur, the three numbers in the subset must appear in consecutive positions. Consequently, we have $n - 2$ objects to permute, the block consisting of the numbers $1, 2, 3$ and the other $n - 3$ numbers in the set $\{1, 2, 3, \ldots, n\}$. The $n - 2$ objects can be permuted in $(n - 2)!$ ways. The numbers in the block can be permuted in $3!$ ways. Hence, there are
$$\binom{3}{3}(n - 2)!3!$$
such permutations.
By the Inclusion-Exclusion Principle, the number of admissible permutations is
$$n! - \binom{3}{2}(n - 1)!2! + \binom{3}{3}(n - 2)!3!$$
Method 2: We arrange the $n - 3$ numbers in the subset $\{4, 5, 6, \ldots, n\}$, then place the numbers $1, 2, 3$ in the spaces this creates.
There are $(n - 3)!$ ways to arrange the $n - 3$ numbers in the subset $\{4, 5, 6, \ldots, n\}$. This creates $n - 2$ spaces in which to place the numbers $1, 2, 3$, including $n - 4$ spaces between successive numbers in the permutation of the numbers in the subset $\{4, 5, 6, \ldots, n\}$ and two at the ends of the row. To ensure that no two of the three numbers $1, 2, 3$ are consecutive, we must choose three of these $n - 2$ spaces in which to place the numbers $1, 2, 3$, which can be done in $\binom{n - 2}{3}$ ways. The three numbers $1, 2, 3$ can be arranged in the selected spaces in $3!$ ways. Hence, the number of admissible permutations is
$$(n - 3)!\binom{n - 2}{3}3!$$
We show that the above methods yield equivalent results.
\begin{align*}
n! - \binom{3}{2}(n - 1)!2! + \binom{3}{3}(n - 2)!3! & = n! - 3 \cdot 2 \cdot (n - 1)! + 1 \cdot 6 \cdot (n - 2)!\\
& = n(n - 1)(n - 2)! - 6(n - 1)(n - 2)! + 6(n - 2)!\\
& = (n - 2)![n(n - 1) - 6(n - 1) + 6]\\
& = (n - 2)!(n^2 - n - 6n + 6 + 6)\\
& = (n - 2)!(n^2 - 7n + 12)\\
& = (n - 2)!(n - 3)(n - 4)
\end{align*}
\begin{align*}
(n - 3)!\binom{n - 2}{3}3! & = (n - 3)! \cdot \frac{(n - 2)!}{3!(n - 5)!} \cdot 3!\\
& = (n - 3)!(n - 2)(n - 3)(n - 4)\\
& = (n - 2)!(n - 3)(n - 4)
\end{align*}
so the results we obtained are equivalent.
Let's compare your result with the above result.
In order for no two of the numbers $1, 2, 3$ to be consecutive in a permutation of the set $\{1, 2, 3, \ldots, n\}$, $n$ must be at least $5$.
For $n = 5$ and $n = 6$, our answers agree.
However, if $n = 7$, my formula yields
$$(n - 2)!(n - 3)(n - 4) = 5! \cdot 4 \cdot 3 = 1440$$
Using the table of Stirling numbers of the second kind, we see that your result gives
\begin{align*}
3!(n - 3)![S(n - 3, 2) + S(n - 3, 3) + S(n - 3, 4)] & = 3!4![S(4, 2) + S(4, 3) + S(4, 4)]\\
& = 3!4!(7 + 6 + 1)\\
& = 2016
\end{align*}
so your method overcounts.
I suspect the error lies in permuting elements which are in different blocks. Once you separate the numbers $4, 5, 6, \ldots, n$, they can only be permuted within each block.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Determinant of matrix with $2$'s and the pattern $3\ 1\ 3$
Let for $n\ge 1, D_n$ be the determinant of the $n\times n$ matrix $A_n$ where the entries along the main diagonal (i.e. of the form $(i,i)$ for $1\leq i\leq n$) are all $3$, entries of the form $(i,i+1)$ are equal to $1$, entries of the form $(i, i+2)$ are equal to $3$, and all other entries are $2$. Find, with proof, a formula for the determinant of $D_n$.
For instance, the matrix $A_n$ for $n=4$ is shown below.
$$A_4 =\begin{pmatrix} 3 & 1 & 3 & 2 \\
2 & 3 & 1 & 3\\
2 & 2 & 3 & 1\\
2 & 2 & 2 & 3\end{pmatrix}.$$
I found the following formula after a bit of experimentation, but I wasn't able to prove it: $$D_n = 1 + 12\lfloor \frac{n}6\rfloor + \begin{cases}2,&\text{ if $n\equiv 1\bmod 6$}\\
6,&\text{ if $n\equiv 2\bmod 6$}\\
10, &\text{ if $n\equiv 3\bmod 6$}\\
12, &\text{ if $n\equiv 4 $ or $5\bmod 6$}\\
0,&\text{ otherwise}\end{cases}.$$
One can define $D_0 := 1$. I know $D_2 = 7$ and using row operations. So far, I've come up with the following sequence of operations, but I seem to be stuck after the last step. In the description below, $R_i$ represents the $i$th row and $C_i$ represents the ith column of $A_n$. Suppose $n > 2$.
Perform the operation $R_i \mapsto R_i - R_n$ for $2\leq i < n.$ Then the resulting matrix has all zeroes for entries of the form $(i, j)$ for $i < j$ that are not on the last row. The entries on the last row are $n-1$ $2$'s followed by a $3$. Also, entries along the main diagonal are all $1$'s and the only entry of the form $(i,i+1)$ where $2\leq i <n$ that does not equal $-1$ is $(n-1,n)$, and that entry equals $-2$. Entries of the form $(i,i+2)$ where $2\leq i < n-2$ are equal to one while entry $(n-2,n)$ equals $0$ if it exists.
Now perform the operations $C_i \mapsto C_i - C_1$ for $2\leq i < n$. This removes all the $2$'s on the last row except for the first, and the entries on the top row are $3, -2,0$, followed by $-1$'s and ending with $2$ (the sequence stops whenever $n$ is reached, so if $n=3$, the entries would be $3,-2,0$).
Finally, perform the operations $R_i \mapsto R_i + R_{i+1}$ for $2\leq i\le n-2$.
After performing these operations, for $n=5$, the following matrix is obtained:
$$\begin{pmatrix}
3& -2 & 0 & -1 & 2\\
0 & 1 & 0 & 0 & 2\\
0 & 0 & 1 & 0 & -2\\
0 & 0 & 0 & 1 & -2\\
2 & 0 & 0 & 0 & 3
\end{pmatrix}.$$
I can't seem to "zero out" the $2$ in the bottom left corner, despite trying various row and column operations. I think I should split the proof into cases based on the remainder when $n$ is divided by $6$, which is suggested by the formula at the beginning.
I tried considering other approaches such as finding eigenvalues and deriving a recurrence relation for $D_n$, but those approaches didn't seem to help much.
| Let $e$ be the vector of ones and $N$ be the full-sized nilpotent Jordan block. Since $A_n$ is a rank-one update of $B_n=I-N+N^2$, by Sylvester's determinant lemma we get
$$
\det(A_n)=\det(B_n+2ee^T)=\det(B_n)(1+2e^TB_n^{-1}e)=1+2e^TB_n^{-1}e.
$$
So, we need to find the sum of all entries in $B_n^{-1}$. Since $B_n$ is a polynomial in $N$, so must be its inverse. Let $B_n^{-1}=c_0I+c_1N+c_2N^2+\cdots+c_{n-1}N^{n-1}$. The equation $B_nB_n^{-1}=I$ gives the recurrence relation $c_0=c_1=1$ and $c_k=c_{k-1}-c_k$, whose solution is given by a repetition of the pattern $(1,1,0,-1,-1,0)$. So, when $n=6q+r$, we have
\begin{aligned}
B_n^{-1}
&=(I+N-N^3-N^4)+(N^6+N^7-N^9-N^{10})+\cdots\\
&\quad+(N^{6q-6}+N^{6q-5}-N^{6q-3}-N^{6q-2})\\
&\quad+
\begin{cases}
0&\text{if }r=0,\\
N^{6q}&\text{if }r=1,\\
N^{6q}+N^{6q+1}&\text{if }r=2,3,\\
N^{6q}+N^{6q+1}-N^{6q+3}&\text{if }r=4,\\
N^{6q}+N^{6q+1}-N^{6q+3}-N^{6q+4}&\text{if }r=5.\\
\end{cases}
\end{aligned}
Since the sum of all entries in $N^k+N^{k+1}-N^{k+2}-N^{k+3}$ is equal to $6$, we get
\begin{aligned}
e^TB_n^{-1}e
&=\begin{cases}
6q&\text{if }r=0,\\
6q+r&\text{if }r=1,\\
6q+r+(r-1)&\text{if }r=2,3,\\
6q+r+(r-1)-(r-3)&\text{if }r=4,\\
6q+r+(r-1)-(r-3)-(r-4)&\text{if }r=5\\
\end{cases}\\
&=\begin{cases}
n&\text{if }r=0,1,\\
n+1&\text{if }r=2,5,\\
n+2&\text{if }r=3,4.\\
\end{cases}
\end{aligned}
Thus
\begin{aligned}
\det(A_n)=1+2e^TB_n^{-1}e
&=\begin{cases}
2n+1&\text{if }r=0,1,\\
2n+3&\text{if }r=2,5,\\
2n+5&\text{if }r=3,4.\\
\end{cases}
\end{aligned}
PS. The form $B_n=I-N+N^2$ suggests that there might be a simpler solution that uses generating functions, but I am satisfied with my current answer here.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Why do I get incorrect solutions $ x = 0 $ and $ x = 2 $ for $ x + 1 + \sqrt{4x + 1} = 0 $? Here is my incorrect attempt at solving
$$ x + 1 + \sqrt{4x + 1} = 0. $$
Subtracting $ \sqrt{4x + 1} $ from both sides,
$$ x + 1 = -\sqrt{4x + 1} $$
Squaring both sides,
$$ x^2 + 2x + 1 = 4x + 1 $$
Subtracting $ 4x + 1 $ from both sides,
$$ x^2 - 2x = 0 $$
We have obtained
$$ x(x - 2) = 0 $$
It has two solutions: $ x = 0 $ and $ x = 2 $.
But if we substitute $ x = 0 $ in LHS of the original equation we get
$$ x + 1 + \sqrt{4x + 1} = 1 + \sqrt{1} = 2 $$
If we substitute $ x = 2 $ in LHS of the original equation we get
$$ x + 1 + \sqrt{4x + 1} = 3 + \sqrt{9} = 6 $$
Apparently I have made a mistake in some step that has led to this contradiction. Which step is incorrect in my solution above? It must be the squaring step that changes the minus sign to positive sign.
But squaring both sides is an often used step in many equations. What rules I need to keep in mind while solving such equations so that I do not get incorrect solution after squaring both sides?
| Be careful to square both side,
Make sure both positive...
.
Here my solution.,
Hope this could be referenced....
${x}+\mathrm{1}+\sqrt{\mathrm{4}{x}+\mathrm{1}}=\mathrm{0} \\ $
$\left(\sqrt{\mathrm{4}{x}+\mathrm{1}}\right)^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{4}{x}+\mathrm{1}}+\mathrm{3}=\mathrm{0} \\ $
$\left(\sqrt{\mathrm{4}{x}+\mathrm{1}}+\mathrm{3}\right)\left(\sqrt{\mathrm{4}{x}+\mathrm{1}}+\mathrm{1}\right)=\mathrm{0} \\ $
${No}\:{Solution} \\ $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Calculate the integral $\int_0^{\infty} x^{11} e^{-x^3}dx$ Calculate the integral $\int_0^{\infty} x^{11} e^{-x^3}dx$.
I wanted to approach this using the reduction formula. Here's what I have tried
$$\int x^2e^{-x^3}dx = -\frac{1}{3}e^{-x^3}$$
Then extending this to $n$
$$I_n = \int_0^{\infty}x^ne^{-x^3}dx $$
Using substitution I can get
$$u = x^{n-2}; du = (n-2)x^{n-3}; dv = x^2e^{-x^3}; v = -\frac{1}{3}e^{-x^3}$$
Plugging this in
$$I_n = \int_0^{\infty}x^ne^{-x^3}dx = \left[-\frac{x^{n-2}}{3}e^{-x^3} \right]_0^{\infty}+\frac{(n-3)}{3} \int_0^{\infty}x^{n-3}e^{-x^3}dx $$
$$\implies I_n = \int_0^{\infty}x^ne^{-x^3}dx =\frac{(n-3)}{3} \int_0^{\infty}x^{n-3}e^{-x^3}dx$$
$$=\frac{n-3}{3}I_{n-3}$$
Then starting from $n=11$ I get
$$\frac{8}{3} \cdot \frac{5}{3} \cdot\frac{2}{3} \int_0^{\infty} x^2e^{-x^3}dx$$
However this does not produce the answer which is $2$. How should I go about the reduction formula for this?
| Rather than trying to do a reduction formula on the original integrand, first perform the substitution $$y = x^3, \quad dy = 3x^2 \, dx.$$ Then for an integrand of the form $$f_n(x) = x^{3n+2} e^{-x^3}$$ we have
$$\int f_n(x) \, dx = \frac{1}{3}\int y^n e^{-y} \, dy.$$ This integral is now amenable to integration by parts with the usual choice $$u = y^n, \quad du = ny^{n-1} \, dy, \\ dv = e^{-y} \, dy, \quad v = -e^{-y}:$$
$$I_n(y) = \int y^n e^{-y} \, dy = -y^n e^{-y} + n \int y^{n-1} e^{-y} \, dy = -y^n e^{-y} + n I_{n-1}(y).$$
If we undo the initial substitution, this gives
$$\int x^{3n+2} e^{-x^3} \, dx = -\frac{1}{3}x^{3n} e^{-x^3} + \frac{n}{3} \int x^{3n-1} e^{-x^3} \, dx.$$
However, in your case, we simply have $n = 3$, and
$$\begin{align}
I_3(y) &= -y^3 e^{-y} + 3 I_2(y) \\
&= -y^3 e^{-y} - 3y^2 e^{-y} + 6 I_1(y) \\
&= -y^3 e^{-y} - 3y^2 e^{-y} - 6y e^{-y} + 6 I_0(y) \\
&= -e^{-y} (y^3 + 3y^2 + 6y + 6),
\end{align}$$
and in terms of $x$, this gives
$$\int f_3(x) \, dx = -\frac{1}{3}e^{-x^3}(x^9 + 3x^6 + 6x^3 + 6) + C.$$
So the definite integral is $$\int_{x=0}^\infty f_3(x) \, dx = 0 + \frac{6}{3} = 2.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
prove that $\int_0^\infty \frac{\sin^2 x-x\sin x}{x^3} \, dx= \frac{1}{2} - \ln 2$
Prove that
$$ \int_{0}^{\infty} \frac{\sin^2 x-x\sin x}{x^3} \, dx = \frac{1}{2} - \ln 2 .$$
Integration by parts gives
\begin{align*}
&\lim_{R\to \infty} \int_{0}^{R} \frac{\sin^2 x-x\sin x}{x^3} \, dx \\
&= \lim_{R\to \infty} \biggl( \int_{0}^{R} \frac{\sin^2x}{x^3} \, dx - \int_{0}^{R} \frac{\sin x}{x^2} \, dx \biggr)\\
&= \lim_{R\to\infty} \biggl( \frac{\sin^2 x}{-2x^2}\Biggr\rvert_{0}^{R} - \int_{0}^{R} \frac{\sin (2x)}{-2x^2} \, dx - \biggl(-\frac{\sin x}{x} \Biggr\rvert_{0}^{R} + \int_{0}^{R} \frac{\cos x}{x} \,dx \biggr) \biggr) \\
&= \lim_{R\to \infty} \biggl(\frac{1}{2} + \int_{0}^{2R} \frac{\sin u}{u^2/2} \, \Bigl(\frac{1}2 \, du\Bigr) - \biggl( 1 + \int_{0}^{R} \frac{\cos x}{x} \, dx \biggr)\biggr) \\
&\hspace{22em}\text{(using the substitution $u\mapsto 2x$)}\\
&= -\frac{1}{2} + \lim_{R\to \infty} \biggl(\int_{0}^{2R} \frac{\sin u}{u^2} \, du - \int_{0}^{R} \frac{\cos x}{x} \,dx \biggr)\\
&= \frac{1}{2} + \lim_{R\to\infty} \biggl(\int_{R}^{2R} \frac{\cos x}{x} \, dx \biggr)
\end{align*}
Thus it suffices to show that $\lim_{R\to\infty} \int_{R}^{2R} \frac{\cos x}{x} \, dx = \ln 2$. The Taylor series expansion of $\cos x$ is given by $\cos x = \sum_{i=0}^{\infty} \frac{(-1)^i x^{2i}}{(2i)!}$.
(If the step below (the one involving the interchanging of an infinite sum and integral) is valid, why exactly is it valid? For instance, does it use uniform convergence?)
The limit equals
$$
\lim_{R\to\infty} \int_{R}^{2R} \biggl( \frac{1}{x} + \sum_{i=1}^{\infty} \frac{(-1)^i x^{2i-1}}{(2i)!} \biggr) \, dx
= \ln 2 + \lim_{R\to\infty} \sum_{i=1}^{\infty} \biggl[\frac{(-1)^ix^{2i}}{(2i)(2i)!}\biggr]_{R}^{2R} . $$
But I don't know how to show $\lim_{R\to\infty} \lim_{R\to\infty} \sum_{i=1}^{\infty} \Bigl[\frac{(-1)^ix^{2i}}{(2i)(2i)!}\Bigr]_{R}^{2R} = -2 \ln 2$.
| $$I=\int \frac{\sin^2 (x)-x\sin (x)}{x^3}\, dx=\int \frac{\sin^2 (x)}{x^3}\, dx-\int \frac{\sin (x)}{x^2}\, dx$$
$$\int \frac{\sin^2 (x)}{x^3}\, dx=\frac 1 2\int \frac{1-\cos(2x)}{x^3}\,dx$$
$$\int \frac{1-\cos(2x)}{x^3}\,dx=-\frac 1{2x^2}-\int \frac{\cos(2x)}{x^3}\,dx$$
$$\int \frac{\cos(2x)}{x^3}\,dx=-\frac{\cos(2x)}{x^2}-\int\frac{\sin(2x)}{x^2}\,dx$$
$$\int\frac{\sin(2x)}{x^2}\,dx=-\frac{\sin(2x)}{x}+2\int \frac{\cos(2x)}x\,dx$$
$$\int \frac{\cos(2x)}x\,dx=\int \frac{\cos(y)}y\,dy=\text{Ci}(y)=\text{Ci}(2x)$$
Combining all the above
$$\color{red}{\int \frac{\sin^2 (x)}{x^3}\, dx=\text{Ci}(2 x)-\frac{\sin (x) (\sin (x)+2 x \cos (x))}{2 x^2}}$$
$$\int \frac{\sin (x)}{x^2}\, dx=-\frac{\sin (x)}{x}+\int \frac{\cos(x)}x\,dx$$
$$\color{red}{\int \frac{\sin (x)}{x^2}\, dx=-\frac{\sin (x)}{x}+\text{Ci}(x)}$$
So,
$$\color{blue}{I=-\text{Ci}(x)+\text{Ci}(2 x)-\frac{1}{4 x^2}+\frac{\cos (2 x)}{4 x^2}+\frac{\sin(x)}{x}-\frac{\sin (2 x)}{2 x}}$$ If $x\to \infty$ its value is $0$.
Now, for $x\to 0$, using expansion we have
$$\left(\log (2)-\frac{1}{2}\right)-\frac{x^2}{12}+\frac{13
x^4}{1440}+O\left(x^6\right)$$
$$J=\int_\epsilon^\infty \frac{\sin^2 (x)-x\sin (x)}{x^3}\, dx=\left(\frac{1}{2}-\log (2)\right)+\frac{\epsilon^2}{12}+O\left(\epsilon^4\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305593",
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"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
What's the measure of the $\angle BAC$ in the triangle below? For reference: In the right triangle $ABC$, right at $B$, the corner $AF$ is drawn such that $AB = FC$ and $\angle ACB = 2 \angle BAF$. Calculate $\angle BAC$.
My progress:
$\triangle ABF: cos(\frac{C}{2}) = \frac{x}{AF}\\
AF^2 = x^2+BF^2\\
\triangle AFC: Law ~of~ cosines:\\
AF^2 = x^2+AC^2-2.x.AC.cosC\\
\triangle ABC:\\
cos C = \frac{BC}{AC} =\frac{x+BC}{AC}\\
x^2+(x+BF)^2 = AC^2\\
Th.Stewart \triangle ABC:\\
AC^2.BF+x^3=AF^2BC+BC.x.BF$
...??
| Just a note: This is a method for reference, it approximates your angle answer:
We know that:
$\tan(\frac{c}{2}) = \frac{BF}{x}$
$\tan(90-c) = \frac{x+BF}{x} = 1 + \frac{BF}{x} = 1+\tan(\frac{c}{2})$
Using the relation between sine, cosine and tangent:
$\frac{\cos(c)}{\sin(c)}$ = $1+\frac{\sin(\frac{c}{2})}{\cos(\frac{c}{2})}$
Then using the half-angle formulae:
$\frac{\cos(c)}{\sin(c)}$ = $1\pm\frac{\sqrt{\frac{1-\cos(c)}{2}}}{\sqrt{\frac{1+\cos(c)}{2}}}$
$\frac{\cos(c)}{\sin(c)}$ = $1+\frac{\sqrt{\frac{1-\cos^2(c)}{4}}}{\frac{1+\cos(c)}{2}}$
Using Pythagorean-Trig identity:
$\frac{\cos(c)}{\sin(c)}$ = 1+$\frac{\sin(c)}{\cos(c)+1}$
$\cos^2(c)+\cos(c)$=$\sin^2(c) +\sin(c)\cos(c)+\sin(c)$
$\cos(2c)+\cos(c)-\sin(c)\cos(c)-\sin(c)=0$
$2\cos(2c)+2\cos(c)-\sin(2c)-2\sin(c)=0$
Because $c$ is real:
$(\cos(\frac{c}{2}))(-\sin(c)+2\cos(c)-1)=0$
$\sin(c)=2\cos(c)-1$
$\sqrt{1-\cos^2(c)} = 2\cos(c)-1$
$4\cos^2(c)-4\cos(c)+1=1-\cos^2(c)$
$(\cos(c))(5\cos(c)-4) = 0$
--> $\cos(c)=\frac{4}{5}$, given that $0<c<90$
$c \approx 37^o$
$180^o - 37^o - 90^o = 53^o$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
How to evaluate $I=\int_0^\pi\frac{\cos(\theta+\alpha)\sin(\theta)d\theta}{\sqrt{r^2+a^2-2ra\cos(\theta)}}$ Let's assume that $r>a>0$ and $0<\alpha<\pi$, to evaluate the following integral:
$$I=\int_0^\pi\frac{\cos(\theta+\alpha)\sin(\theta)d\theta}{\sqrt{r^2+a^2-2ra\cos(\theta)}}$$
I used a change of variable $u=\sqrt{r^2+a^2-2ra\cos(\theta)}$ to get:
$$I=\frac{1}{4r^2a^2}\int_{r-a}^{r+a}(\cos\alpha(r^2+a^2-u^2)-\sin\alpha \sqrt{(2ra)^2-(r^2+a^2-u^2)^2})du$$
Then I was stuck to integrate the complicated square root. Do you have an idea?
| For $\cos(\theta+\alpha) = \cos\theta\cos\alpha-\sin\theta\sin\alpha$
$$
I = \cos\alpha \int_{0}^{\pi} \frac{\cos\theta\sin\theta}{\sqrt{r^2+a^2-2ra\cos\theta}} \,\mathrm{d}\theta - \sin\alpha \int_{0}^{\pi} \frac{\sin^2\theta}{\sqrt{r^2+a^2-2ra\cos\theta}} \,\mathrm{d}\theta
$$
The first integral is trivial, here I focus on the second
$$
\begin{aligned}
\int_{0}^{\pi} \frac{\sin^2\theta}{\sqrt{r^2+a^2-2ra\cos\theta}} \,\mathrm{d}\theta
&= \int_{0}^{\pi} \frac{\sin^2\theta}{\sqrt{(r+a)^2-4ra\cos^2\frac{\theta}{2}}} \,\mathrm{d}\theta \quad \text{let }k^2=\frac{4ra}{(r+a)^2} \text{ and } \frac{\theta}{2}\mapsto\theta \\
&= \frac2{r+a} \int_{0}^{\pi/2} \frac{\sin^22\theta}{\sqrt{1-k^2\cos^2\theta}} \,\mathrm{d}\theta \quad \theta\mapsto\frac{\pi}{2}-\theta \\
&= \frac2{r+a} \int_{0}^{\pi/2} \frac{\sin^22\theta}{\sqrt{1-k^2\sin^2\theta}} \,\mathrm{d}\theta
\end{aligned}
$$
Integrating by parts gives
$$
\int_{0}^{\pi/2} \frac{\sin^22\theta}{\sqrt{1-k^2\sin^2\theta}} \,\mathrm{d}\theta = \frac4{k^2} \int_{0}^{\pi/2} \cos2\theta \sqrt{1-k^2\sin^2\theta} \,\mathrm{d}\theta
$$
where denote $I=\frac4{k^2}J$.
From another aspect, setting
$$
\sin^22\theta=A\cos2\theta(1-k^2\sin^2\theta) + B(1-k^2\sin^2\theta) + C
$$
or
$$
-4\sin^4\theta+4\sin^2\theta = 2k^2A\sin^4\theta - ((k^2+2)A+k^2B)\sin^2\theta + (A+B+C)
$$
comparing the coefficients, we solve
$$
A=-\frac2{k^2}, \quad B=-\frac{2(k^2-2)}{k^4}, \quad C=\frac{4(k^2-1)}{k^4}
$$
which suggests
$$
I=-\frac2{k^2}J + \frac{4(k^2-1)}{k^4}K(k^2) - \frac{2(k^2-2)}{k^4}E(k^2)
$$
Item of $J$ can be cancelled with $I=\frac4{k^2}J$, hence
$$
I = \int_{0}^{\pi/2} \frac{\sin^22\theta}{\sqrt{1-k^2\sin^2\theta}} \,\mathrm{d}\theta = \frac{8(k^2-1)}{3k^4}K(k^2) - \frac{4(k^2-2)}{3k^4}E(k^2)
$$
or plug-in $k^2$
$$
\int_{0}^{\pi} \frac{\sin^2\theta}{\sqrt{r^2+a^2-2ra\cos\theta}} \,\mathrm{d}\theta = -\frac{r+a}{3r^2a^2}\left((r-a)^2K\left(\frac{4ra}{(r+a)^2}\right) - (r^2+a^2)E\left(\frac{4ra}{(r+a)^2}\right)\right)
$$
the rest of calculation is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding $\int \frac{d x}{x+\sqrt{1-x^{2}}}$.
I have to calculate the following integral:
$$
\int \frac{d x}{x+\sqrt{1-x^{2}}}
$$
An attempt:$$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}} & \stackrel{x=\sin t}{=} \int \frac{\cos t}{\sin t+\cos t} d t \\
&=\int \frac{\cos t(\cos t-\sin t)}{\cos 2 t} d t
\end{aligned}
$$
I find the solution is
$$\frac{\ln{\left(x + \sqrt{1 - x^{2}} \right)}}{2} + \frac{\sin^{-1}{\left(x \right)}}{2}+C$$
How can I get this without trigonometric substitution?
| Let $y=\sqrt{1-x^2}$, then $ydy=-xdx$.
Noticing that
$$d(x-y)= dx-\left(-\frac{x}{y} d x\right)=\frac{(x+y)dx}{y} \Rightarrow \frac{d(x-y)}{x+y}=\frac{d x}{\sqrt{1-x^2}}, $$
we split the integral into 2 parts as below.
$$
\begin{aligned}
I &=\int \frac{d x}{x+y} \\
&\left.=\frac{1}{2} \int \frac{(d x+d y)+(d x-d y)}{x+y}\right]\\&= \frac{1}{2}\left[\int \frac{d(x+y)}{x+y} + \int \frac{d(x-y)}{x+y}\right] \\ &=\frac{1}{2}\left[\ln |x+y|+\int \frac{d x}{\sqrt{1-x^2}} \right] \\ &=\frac{1}{2}\left[\ln \left| x+\sqrt{1-x^2}\right|+\sin ^{-1} x\right]+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Find the minimum of $\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}$ Let $0<x,y<\frac {\pi}{2}$ such that $\sin (x+y)=\frac 23$, then find the minimum of
$$\frac{\sin x}{\cos y}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}+\frac{\cos y}{\sin x}$$
A) $\frac 23$
B) $\frac 43$
C) $\frac 89$
D) $\frac {16}{9}$
E) $\frac{32}{27}$
My attempts:
I think that the all possible answers are wrong. Because, by Am-Gm inequality we have
$$\frac{\sin x}{\cos y}+\frac{\cos y}{\sin x}+\frac {\cos x}{\sin y}+\frac{\sin y}{\cos x}≥2+2=4.$$
But, Wolfram Alpha gives us a different result : The Global Minimum doesn't exist. However, the local minimum must be $6.$
But, still the problem is not solved. Because Wolfram's graph shows that the minimum can be less than $6$.
I also tried
Let $$\sin x=a,\cos y=b,\cos x=c,\sin y=d$$ with
$$ab+cd=\frac 23≥2\sqrt{abcd}\implies abcd≤\frac 19\\ a^2+c^2=b^2+d^2=1 $$
then I need
$$\min \left(\frac ab+\frac ba+\frac cd+\frac dc\right)$$
But, I can't do anything from here. Finally, I attach the graph drawn by WA.
| You are correct. Let $$f(x,y) = g + \frac{1}{g} + h + \frac{1}{h}$$ where $$g(x,y) = \frac{\sin x}{\cos y}, \quad h(x,y) = \frac{\cos x}{\sin y}.$$ Then since $0 \le x, y \le \pi/2$, we must have $0 \le \sin x, \cos x, \sin y, \cos y \ge 1$, hence $g, h \ge 0$. Then AM-GM proves $g + 1/g \ge 2$, as well as $h + 1/h \ge 2$, thus $f \ge 4$, and this is without the added constraint $\sin (x+y) = \frac{2}{3}$. None of the answer choices is greater than $4$, so all are incorrect.
But also be advised that Wolfram Alpha is also misleading, because we have established that it is impossible for $f$ to be negative. Thus there must be a lower bound.
The actual minimum is $6$. To see this, let $c_1 = \arcsin \frac{2}{3} \approx 0.729728$, and $c_2 = \pi - c_1 \approx 2.41186$. These comprise the two possible values for $x + y$ in the square $[0,\pi/2]^2$. Then the transformation $u = x+y$, $v = x-y$ gives
$$f = 8\frac{\sin u \cos^2 v}{\cos 2v - \cos 2u} = 4 \frac{\sin u \cos^2 v}{ \cos^2 v + \sin^2 u - 1}$$ Then since $\sin u = 2/3$,
$$f = \frac{24\cos^2 v}{-5 + 9 \cos^2 v} = \frac{8}{3} + \frac{\frac{40}{27}}{t - \frac{5}{9}}$$ where $t = \cos^2 v$. Consider the range of $t$. Since $x + y \in \{c_1, c_2\}$ and $0 \le x, y \le \pi/2$, it follows that $0 \le |x-y| \le c_1$, and in turn, $$\frac{5}{9} \le t \le 1.$$ As $f$ is obviously a decreasing function of $t$ on this interval, the minimum is attained when $t$ is as large as possible, namely $t = 1$, corresponding to $v = 0$, or $x = y \in \{c_1/2, c_2/2\}$, and $$f = \frac{8}{3} + \frac{40/27}{4/9} = 6.$$
Here is a plot of $f(x,y)$ and the two curves for which $\sin (x+y) = \frac{2}{3}$. As you can see, they do have a global minimum, and as $f$ is never negative, neither are the curves.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4320041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
What´s the value of the area of the triangle below? For reference: The sides of an acute-angled triangle measure
$3\sqrt2$, $\sqrt{26}$ and $\sqrt{20}$.
Calculate the area of the triangle (Answer:$9$)
My progress...
Is there any way other than Heron's formula since the accounts would be laborious or algebraic manipulation for the resolution?
$p=\frac{\sqrt{18}+\sqrt{20}+\sqrt{26}}{2}\implies S_{ABC} =\sqrt{p(p-\sqrt{20})(p-\sqrt{26})(p-\sqrt{18})}$
|
If $a = \sqrt{18}, b = \sqrt{20}, c = \sqrt{26}$, first note that $a^2 + b^2 \gt c^2$ where $c$ is the longest side. That means it is an acute-angled triangle and the orthocenter is inside the triangle. So if we drop a perp from vertex $A$, then using Pythagoras,
$AD^2 = 20 - x^2 = 26 - (\sqrt{18} - x)^2$
$ \implies x = \sqrt 2$ and we get $AD = \sqrt{18}$
So, $A = \frac 12 a \cdot AD = 9$
By the way, if you notice that $26 - 8 = 20 - 2$ and $\sqrt{8} + \sqrt{2} = a$, you can get to altitude $AD = \sqrt{18} ~$ a bit faster.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4322288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Global extreme points of function wrt. two variables Hi I am currently working on a problem in theoretical chemistry and am struggling a bit. While discussing Born-Oppenheimer energy functions, I was asked to find all the extreme points and saddle points of the follwing function:
I honestly didn't encounter such a problem before; I tried looking stuff up and found something here:
https://www.massmatics.de/merkzettel/#!217:Globales_Extremum_-_mehrdimensionale_Funktion
(watch out, it's in German)
Basically, what I should do is find the gradient of the function's first derivatives and then construct a linear system of equations. Which shouldn't really be possible as it is a function of grade 4.
Furthermore, I should find the Hesse matrix, but I don't really know if it is also applicable here.
Is there another trick I could use on this function? I find it most peculiar that the terms are arranged in the way they are, maybe there is another, more easy method here?
Thanks in advance!
| We will use these extrema in two varaibles notes as a guide.
We have
$$U(x, y) = (x\left(\frac{x^4}{4}-\frac{x^3}{3}-x^2+3\right) \left(\frac{y^4}{4}-\frac{y^3}{3}-y^2+4\right)$$
Finding critical points
$$f_x = \left(x^3-x^2-2 x\right) \left(\frac{y^4}{4}-\frac{y^3}{3}-y^2+4\right) = 0\\ f_y = \left(\frac{x^4}{4}-\frac{x^3}{3}-x^2+3\right) \left(y^3-y^2-2 y\right) = 0$$
From the first, we have $x = -1, 0, 2$ and from the second, we have $y = -1,0,2$ to give us a total of nine critical points
as
$$(x, y) = (-1,-1),(-1,0),(-1,2),(0,-1),(0,0),(0,-2),(2,-1),(2,0),(2,2)$$
The Hessian determinant, $|H(x, y)|$ is given by
$\left|\left(
\begin{array}{cc}
\left(3 x^2-2 x-2\right) \left(\dfrac{y^4}{4}-\dfrac{y^3}{3}-y^2+4\right) & \left(x^3-x^2-2 x\right) \left(y^3-y^2-2 y\right) \\
\left(x^3-x^2-2 x\right) \left(y^3-y^2-2 y\right) & \left(\dfrac{x^4}{4}-\dfrac{x^3}{3}-x^2+3\right) \left(3 y^2-2 y-2\right) \\
\end{array}
\right)\right|$
We also have
$$f_{xx} (x, y) = \left(3 x^2-2 x-2\right) \left(\frac{y^4}{4}-\frac{y^3}{3}-y^2+4\right)$$
For the nine critical points, we find
*
*Saddles: $(-1,0), (0, -1), (0, 2), (2, 0)$
*Local Min: $(-1, 1), (-1, 2), (2, -1)$
*Local Max: $(0, 0)$
*Global Min: $(2,2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality in cyclic order : $\sum\frac{8}{(a+b)^2+4abc}+a^2+b^2+c^2\ge\sum\frac{8}{a+3}$ Prove:$$\sum_{cyc} \frac{8}{(a+b)^2+4abc}+a^2+b^2+c^2\ge\sum_{cyc} \frac{8}{a+3}$$ where $a,~b,~c$ are positive real numbers.
My thought: I think Holder's inequality will be used. But can't understand how to start? Please give any hint.
| Hint: Using A.M-G.M inequality,
$$(a+b)^2\le2(a^2+b^2)~~\text{and}~~ 4abc\le2c(a^2+b^2) $$
$$\therefore\; (a+b)^2+4abc\le 2(a^2+b^2)(c+1) $$
It remains to prove,
$$\sum_{cyc}\left( \frac{4}{(a^2+b^2)(c+1)}+\frac{a^2+b^2}{2}\right)\ge \sum_{cyc} \frac{8}{a+3}$$
$$ \frac{4}{(a^2+b^2)(c+1)}+\frac{a^2+b^2}{2}\ge {\frac{2\sqrt{2}}{\sqrt{c+1}}}$$
$$\frac{2+(a+1)}{2}\ge \sqrt{2(a+1)} \implies \frac{4}{\sqrt{2(a+1)}}\ge\frac{8}{a+3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the generating function of $f(n) = \sum_{k = 0}^n \binom{n}{k} (-1)^{n-k}C_{k}$ I want to find the generating function of $f(n) = \sum_{k = 0}^n \binom{n}{k} (-1)^{n-k}C_{k}$, where $C_k$ is the $k$-th Catalan number. So, using the definition of an ordinary generating function:
$$F(x)=\sum_{n \ge 0} \biggl(\sum_{k = 0}^n \binom{n}{k} (-1)^{n-k}C_{k} \biggr)x^n$$
recalling that: $$C(x) =\sum_{n \ge 0}C_nx^n = \frac{1- \sqrt{1-4x}}{2x}$$
The first idea was to see $F(x)$ as a product of two formal series and I have already seen a proof in this regard where they use the theorem of residues, yet I am looking for something less refined. Any idea?
| \begin{align}
F(x) &= \sum_{n \ge 0} \left(\sum_{k = 0}^n \binom{n}{k} (-1)^{n-k}C_{k} \right)x^n \\
&= \sum_{k \ge 0} (-1)^k C_{k} \sum_{n \ge k} \binom{n}{k} (-x)^n \\
&= \sum_{k \ge 0} (-1)^k C_{k} \frac{(-x)^k}{(1+x)^{k+1}} \\
&= \frac{1}{1+x}\sum_{k \ge 0} C_{k} \left(\frac{x}{1+x}\right)^k \\
&= \frac{1}{1+x}\cdot \frac{1-\sqrt{1-4\cdot\frac{x}{1+x}}}{2\cdot\frac{x}{1+x}} \\
&= \frac{1-\sqrt{\frac{1-3x}{1+x}}}{2x}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4335670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate some integrals with hypergeometric function There are some interesting integral problems:
$$\int_{0}^{\infty} \cos\left (\frac{x}{2} \right ) \left ( \cosh(x)-\frac{4x^{3/2}{}_1F_2\left ( 1;\frac{5}{4},\frac{7}{4};\frac{x^2}{4} \right ) }{3\sqrt{\pi} } \right )\text{d}x
=\frac{4}{5}.$$
$$\int_{0}^{\infty} e^{-x} \left ( \cosh(x)-\frac{4x^{3/2}{}_1F_2\left ( 1;\frac{5}{4},\frac{7}{4};\frac{x^2}{4} \right ) }{3\sqrt{\pi} } \right )\text{d}x
=\frac{3}{4}.$$
Where ${}_1F_2$ is generalized hypergeometric function.
My question is that:
*
*How to prove these closed expressions?
*Are there any generalizations?
| I present an approach using a well-known property of the Laplace Transform, which is typically a good approach for such integrals.
For the first integral:
$$\int_{0}^{+\infty} f\left(x\right) g\left(x\right) \, dx = \int_{0}^{+\infty} \left(\mathcal{L} f\right)\left(y\right)\left(\mathcal{L}^{-1} g\right)\left( y\right) \, dy$$
Letting $f\left(x\right) = x \cos \left(\frac{x}{2}\right)\left(\cosh (x) -\frac{4x^{3/2} {}_{1} F_{2} \left(1;\frac{5}{4},\frac{7}{4};\frac{x^2}{4}\right)}{3\sqrt{\pi}}\right)$ and $g(x) = \frac{1}{x}$:
$$\left( \mathcal{L} f \right) (y) = \frac{2\sqrt{2}(9-20y(y-i))}{(2y-i)^{3/2}(5-4y(y-i))^2}-\frac{2\sqrt{2}(-9+20y(y+i))}{(2y+i)^{3/2}(5-4y(y+i))^2}+\frac{1}{(-2y+(2+i))^2}+\frac{1}{(2y-(2-i))^2}+\frac{1}{(2y+(2-i))^2}+\frac{1}{(2y+(2+i))^2}$$
(I am not able to write the derivation of the Laplace transform currently, but I will update my answer with the derivation soon)
$$\left( \mathcal{L}^{-1} g \right) \left(y \right) = 1$$
This gives then:
$$I = \int_{0}^{\infty} \frac{2\sqrt{2}(9-20y(y-i))}{(2y-i)^{3/2}(5-4y(y-i))^2}-\frac{2\sqrt{2}(-9+20y(y+i))}{(2y+i)^{3/2}(5-4y(y+i))^2}+\frac{1}{(-2y+(2+i))^2}+\frac{1}{(2y-(2-i))^2}+\frac{1}{(2y+(2-i))^2}+\frac{1}{(2y+(2+i))^2}\,dy$$
$$\implies I = \frac{1}{2}\left(\frac{1}{-2y-(2-i)}+\frac{1}{-2y+(2-i)}+\frac{1}{-2y+(2+i)}-\frac{1}{2y+(2+i)}+\frac{4}{\sqrt{y+\frac{i}{2}}(-5+4y(y+i))}+\frac{4}{(-5+4y(y-i))\sqrt{y-\frac{i}{2}}}\right) \Bigg]_{0}^{\infty} =\frac{4}{5}$$
As was to be shown. The same idea works for your second integral.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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A composite function problem The question is:
Suppose $$f(x) = x^2+1,$$ $$g(x) = 3-x.$$
Find the values for $x$ for such that $$(g\circ f)(x) = (f \circ g)(x).$$
I tried banging my head for one hour but my answer doesn't match the one given by the book which is $1/\sqrt{2}$ and $-1/\sqrt{2}$.
I think the answer given in the book is wrong because I even tried putting the given answer in $(g\circ f)(x)$ and $(f \circ g)(x)$ and the two don't match up.
My answer: $$ x =\frac{3\pm\sqrt{-7}}{2} $$
| Given that $f(x) = x^2 + 1$ and $g(x) = 3 -x$, the composite function should look as follows for the LHS of the equation, hence,
$$3-(x^2 + 1)$$
RHS,
$$(3-x)^2 + 1$$
Thus,
$$ 3-(x^2 + 1) = (3-x)^2 + 1$$.
Simplifying we get,
$$2x^2-6x+8 = 0\\ x^2 - 3x + 4 = 0.$$
In order to find the value's of $x$ satisfying the equation we can use the quadratic formula, and you should get $\frac{3 \pm i\sqrt{7}}{2}$ where $i = \sqrt{-1}$. Meaning you are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4350747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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limit, quotient of roots How can I find
$$
\lim_{n \to \infty} \frac{\sqrt[3]{n + 2} - \sqrt[3]{n + 1}}{\sqrt{n + 2} - \sqrt{n + 1}} \sqrt[6]{n - 3}?
$$
If I multiply by $\sqrt{n + 2} + \sqrt{n + 1}$ I could get no divisor, but I cannot get any result on this way. On the other hand I could also transform this expression:
$$
\frac{(n + 2)^{1 / 3} \left(1 - {\left(\frac{n + 1}{n + 2}\right)}^{1 / 3}\right)}{(n + 2)^{1 / 2} \left(1 - {\left(\frac{n + 1}{n + 2}\right)}^{1 / 2}\right)} \sqrt[6]{n - 3} = {\left(\frac{n - 3}{n + 2}\right)}^{1 / 6} \frac{1 - {\left(\frac{n + 1}{n + 2}\right)}^{1 / 3}}{1 - {\left(\frac{n + 1}{n + 2}\right)}^{1 / 2}}.
$$
However I also cannot see how continue to see that the limit is $\frac 2 3$. Any help would be appreciated.
| Your idea of multiplying and dividing by $\sqrt{n+2}+\sqrt{n+1}$ (or by the conjugate) is a good one, and we can do the same thing for the conjugate of the numerator. Since the denominator had square roots, its conjugate came from the difference of squares formula
$$a-b = (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})$$
Similarly, since the numerator has cube roots, its conjugate must come from the difference of cubes formula
$$a-b = (\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a}^2+\sqrt[3]{a}\sqrt[3]{b}+\sqrt[3]{b}^2)$$
Applying these factors the expression, we obtain
$$\lim_{n\to\infty} \frac{\sqrt[3]{n + 2} - \sqrt[3]{n + 1}}{\sqrt{n + 2} - \sqrt{n + 1}} \cdot \frac{\sqrt{n+2}+\sqrt{n+1}}{\sqrt{n+2}+\sqrt{n+1}} \cdot \frac{\sqrt[3]{n+2}^2+\sqrt[3]{n+2}\sqrt[3]{n+1}+\sqrt[3]{n+1}^2}{\sqrt[3]{n+2}^2+\sqrt[3]{n+2}\sqrt[3]{n+1}+\sqrt[3]{n+1}^2} \cdot\sqrt[6]{n - 3} $$
$$= \lim_{n\to\infty} \frac{\sqrt{n+2}+\sqrt{n+1}}{\sqrt[3]{n+2}^2+\sqrt[3]{n+2}\sqrt[3]{n+1}+\sqrt[3]{n+1}^2}\cdot\sqrt[6]{n - 3}$$
Then pull factors of $n$ from their respective roots
$$\lim_{n\to\infty} \frac{\sqrt{1+\frac{2}{n}}+\sqrt{1+\frac{1}{n}}}{\sqrt[3]{1+\frac{2}{n}}^2+\sqrt[3]{1+\frac{2}{n}}\sqrt[3]{1+\frac{1}{n}}+\sqrt[3]{1+\frac{1}{n}}^2}\cdot\sqrt[6]{1 - \frac{3}{n}} \cdot \frac{n^{\frac{1}{2}}n^{\frac{1}{6}}}{n^{\frac{2}{3}}} = \frac{1+1}{1+1+1} = \frac{2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4351218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Ways to find basis of a span of matrices
let $W= $$\left(\begin{matrix}
b & -a & 3a+b \\
c & 0 & a+2c \\
-3c & c & b \\
\end{matrix} \right)
$ $ \in M_3( \Bbb R)$ and $a,b,c \in \Bbb R$ I would like to know ways to find the basis of W using basic methods so I can have more options while practicing.
for this case I first found the span $W = a$ $\left(\begin{matrix}
0 & -1 & 3 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{matrix} \right)
$
$+b \left(\begin{matrix}
1 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right)
$
$+c \left(\begin{matrix}
0 & 0 & 0 \\
1 & 0 & 2 \\
-3 & 1 & 0 \\
\end{matrix} \right)
$
so from here the span is $W=$Sp$\{$$\left(\begin{matrix}
0 & -1 & 3 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{matrix} \right)
$ ,
$\left(\begin{matrix}
1 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right)
$ ,
$\left(\begin{matrix}
0 & 0 & 0 \\
1 & 0 & 2 \\
-3 & 1 & 0 \\
\end{matrix} \right)
$ $\}$
and a basis needs to span and be linear independent so we will check linear independence by using coordinates vector by using the standard basis for $ M_3( \Bbb R)$ ($dim M_3( \Bbb R)=9$)
then using matrix elementary row operations for the following matrix
$\left(\begin{matrix}
0 & -1 & 3 & 0 & 0 & 1 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 &1 \\
0 & 0 & 0 & 1 & 0 & 2 & -3 & 1 & 0 \\
\end{matrix} \right)
$ $R_1 \iff R_2$
$\left(\begin{matrix}
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 &1 \\
0 & -1 & 3 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 2 & -3 & 1 & 0 \\
\end{matrix} \right)
$
we have 3 opening elements and no zero rows so it is linearly independent and we found the span so it is a basis for W.
this is the way I know and I am just wondering if there are any other methods (that are suitable for linear algebra 1 course)
Thank you!
| From the matrices
$$\left(\begin{matrix}
0 & -1 & 3 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{matrix} \right)
,\left(\begin{matrix}
1 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right)
,
\left(\begin{matrix}
0 & 0 & 0 \\
1 & 0 & 2 \\
-3 & 1 & 0 \\
\end{matrix} \right)
$$
you ought to choose a linear combination
$$K\left(\begin{matrix}
0 & -1 & 3 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{matrix} \right)
+L\left(\begin{matrix}
1 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{matrix} \right)
+M\left(\begin{matrix}
0 & 0 & 0 \\
1 & 0 & 2 \\
-3 & 1 & 0 \\
\end{matrix} \right)
$$
and equate to the zero matrix
$\left(\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} \right)
$ to deduce that the coefficients all $K,L,M$ are zero, which will imply that they are linearly independent.
This leads you to use
$$\left(\begin{matrix}
L & -K & 3K+L \\
M & 0 & K+2M \\
-3M & M & L \\
\end{matrix} \right)=
\left(\begin{matrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{matrix} \right),
$$
with that goal in mind.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ball drawing problem with a condition (with replacing and without replacing) You have $9$ balls in a bag: $4$ blue, $2$ purple, $3$ green. If you draw $3$ balls and I tell you that you've drawn at least $1$ blue, what is the probability that you've drawn $3$ blue balls?
*
*solve for both, replacing and not replacing the balls.
My attempts so far: using Bayes' rule we get $P(3\ blue | al1\ blue) = \frac{P(al1\ blue | 3\ blue) \cdot P(3\ blue)}{P(al1\ blue)}$, here "$al1$" is "at least one."
Hence, without replacing I get: $$P(3\ blue | al1 \, blue) = \frac{1 \cdot \left(\frac{4}{9} \cdot \frac{3}{8} \cdot\frac{2}{7}\right)}{\left(1-all\ non\ blue\right)}=\frac{1 \cdot \left(\frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7}\right)}{\left(1-\frac{5}{9} \cdot \frac{4}{8}\cdot\frac{3}{7}\right)}=\frac{2}{37}.$$
With replacing, the formula is the same but numerators and denominators don't decrease and we get $\dfrac{16}{151}$.
This doesn't feel right, what are more intuitive ways to
think about this problem and hence solve it?
As this doesn't feel right, I have tried taking another approach which feels much better yet I am not 100% sure: there are $3$ possibilities: the first, the second and the third ball was blue. In case of without replacing summing them up will be:
$$\frac{3\cdot2}{8\cdot7} + \frac{4\cdot2}{8\cdot7} + \frac{4\cdot3}{9\cdot8} = \frac{105}{252},$$
and in case of replacements, again, the approach is the same but numbers dont decrease
| As an alternative approach without replacement,
*
*there in a sense ${9 \choose 3} = 84$ equally likely ways of choosing three balls from nine
*${4 \choose 3}=4$ are all blue
*${5 \choose 3}=10$ are none blue
*$84-10=74$ have at least one blue
so the conditional probability you want has probability $\frac{4}{74}=\frac{2}{37}$ as you found.
With replacement,
*
*there in a sense $9^3 = 729$ equally likely ways of choosing the three balls from nine
*$4^3=64$ are all blue
*$5^3=125$ are none blue
*$729-125=604$ have at least one blue
so the conditional probability you want has probability $\frac{64}{604}=\frac{16}{151}$ as you also found.
| {
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"source": "stackexchange",
"question_score": "1",
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How do you solve the equation $2^{|x+1|}-|2^x-1| = 2^x+1$? | ... | is absolute value
$2^{|x+1|}-|2^x-1| = 2^x+1$, move $|2^x-1|$ to RHS: $2^{|x+1|} = 2^x+1+|2^x-1|$
Log base 2 each side:
$|x+1| = log_2 (2^x+1+|2^x-1|)$
Here we quickly observe: $-1$ is not a solution but $0$ is a solution. If you try negative numbers $-2$ is also a solution.
How do I find the value of $x$ (not intuitively like I did previously) for which the equation is true?
Thanks in advance for any help.
edit: Solved, thanks to all who replied. Split $2^{|x+1|} = 2^x+1+|2^x-1|$ into three cases: $x >= 0$ and $x < -1$ and $-1 <= x < 0$. Solution: $[0, \infty]$ ∪ ${-2}$
| If $|x+1|\ge 0$ the equation becomes
$$
2\cdot 2^x-|2^x-1|=2^x+1
$$
that is equivalent to
$$
\begin{cases}
2^x-1\ge0\\
2\cdot 2^x-2^x+1=2^x+1
\end{cases}
\quad \lor\quad
\begin{cases}
2^x-1<0\\
2\cdot 2^x+2^x-1=2^x+1
\end{cases}
$$
can you solve these two systems? and solve also the other case: for $|x+1|< 0$ ?
| {
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} |
Calculate Integral of intersection between two surfaces
Calculate
$$\int_{\partial F} 2xy\, dx + x^2 \, dy + (1+x-z) \, dz$$
for the intersection of $z=x^2+y^2$ and $2x+2y+z=7$. Go clockwise with respect to the origin.
My attempt:
I first calculate the intersection, which is pretty easy
$$z=x^2+y^2 \wedge 2x+2y+z=7 \Rightarrow \, (x+1)^2+(y+1)^2=7$$
which is a circle.
Then I find the curl of the vector field
\begin{align*}
F = \begin{pmatrix} 2xy \\ x^2 \\ 1+x-z\end{pmatrix} \quad \quad \text{curl }F =\begin{pmatrix} 0 \\ -1 \\ 0\end{pmatrix}
\end{align*}
By Stokes theorem I have...
\begin{align*}
\int_{\partial F} F \, ds = \iint_F \text{curl } F \,dA
\end{align*}
I was told that I have to look at the circle $(x+1)^2+(y+1)^2=7$ on the plane $z=7$. We can easily find a normal vector on this plane, $n=(0,0,1)$. So we see that only the $z$-component of the curl/rotation is relevant.
\begin{align*}
\iint_{(x+1)^2+(y+1)^2\leq 7} 0 \, dxdy = \int^{\sqrt{7}}_{r=0} \int^{2\pi}_{\varphi=0} 0 \, d\varphi dr = 0
\end{align*}
which is very likely wrong. What is my mistake?
| You should consider the surface $S$ given by the intersection of the cylinder $(x+1)^2+(y+1)^2\leq 7+2=9$ with the plane $2x+2y+z=7$. Then $\mathbf{n}=-(2,2,1)/3$ (the ellipse $\partial S$ is clockwise oriented) and by Stokes' theorem:
$$
\begin{align*}\int_{\partial S} \mathbf{F} \, d\mathbf{s} &= \iint_S \text{curl }(\mathbf{F}) \,d\mathbf{S}\\
&=\iint_{(x+1)^2+(y+1)^2\leq 9} (0,-1,0)\cdot(-(2,2,1)) \, dxdy\\
&=\iint_{(x+1)^2+(y+1)^2\leq 9} 2 \, dxdy\\
&=2\, \text{Area}(\{(x+1)^2+(y+1)^2\leq 9\})=2\cdot 9\pi=18\pi.
\end{align*}$$
| {
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"source": "stackexchange",
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Derive an upper bound for $\log(x)$ A paper has given that for $x_0 > 0$,
$$
\log x \leq \frac{x}{x_0} + \log x_0 - 1.
$$
The way I tried to prove this is:
For $|t| < 1$,
\begin{align*}
\log(1 - t) = -\sum_{k=1}^n \frac{t^k}{k}
\end{align*}
We can replace $t = 1 - \frac{x}{x_0}$,
\begin{align*}
\log x - \log x_0 & = - [(1 - \frac{x}{x_0}) + \frac{1}{2}(1 - \frac{x}{x_0})^2 + ...]\\
& = -1 + \frac{x}{x_0} - [\frac{1}{2}(1 - \frac{x}{x_0})^2 + \frac{1}{3}(1 - \frac{x}{x_0})^3 + ...]\\
& = -1 + \frac{x}{x_0} - c
\end{align*}
How can we show that $c$ is positive and what's the range of $x$ (is it $0 < x < 2x_0$)?
| Hint: $\log x$ is concave, so it remains below its tangent at $x_0$.
| {
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Let $f(x) = \sin^{-1}(\frac{2x}{1+x^2})$ Show that $f(x) = 2\tan^{-1}(x)$
Let $$f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) ~~ -\infty<x<\infty.$$
Show that,
(a) $f(x) = 2\tan^{-1}(x)$ for $-1\leq x \leq 1$ and
(b) $f(x) = \pi-2\tan^{-1}(x)$ for $x \geq 1.$
Proof: I started off by equating $$\sin^{-1}\left(\frac{2x}{x^2+1}\right)=2\tan^{-1}(x)$$
(a)
We wish to show that these are equal for $-1\leq x \leq 1$.
For this domain $\displaystyle -1 \leq\frac{2x}{x^2+1} \leq 1 \implies -\frac{\pi}{2} \leq \sin^{-1}\left(\frac{2x}{x^2+1}\right) \leq \frac{\pi}{2}$
$$\frac{2x}{x^2+1} = \sin(2\tan^{-1}(x))$$
The task is now to show $\sin(2\tan^{-1}(x))=\frac{2x}{x^2+1}$
$$2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x))=\frac{2x}{x^2+1}$$
For $-1 \leq x \leq 1 \implies -\frac{\pi}{4}\leq\tan^{-1}(x)\leq\frac{\pi}{4} \implies -\frac{\sqrt{2}}{2}\leq \sin(\tan^{-1}(x)) \leq \frac{\sqrt{2}}{2}$
Also, $\frac{\sqrt{2}}{2}\leq\cos(\tan^{-1}(x)) \leq 1$
$$2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x)) \iff 2(\frac{x}{\sqrt{x^2+1}})(\frac{1}{\sqrt{x^2+1}})= \frac{2x}{x^2+1}$$
Which was to be shown.
(b)
$\displaystyle x \geq 1 \implies 0<\frac{2x}{1+x^2}\leq 1 \implies 0< \sin^{-1}\left(\frac{2x}{x^2+1}\right) \leq \frac{\pi}{2}$
Hence,
We must show that $$\sin^{-1}(\frac{2x}{x^2+1}) = \pi - 2\tan^{-1}(x) \iff \frac{2x}{x^2+1} = \sin(\pi - 2\tan^{-1}(x))$$ for $x\geq 1$
$$\sin(\pi - 2\tan^{-1}(x))=\sin(\pi)\cos(2\tan^{-1}(x)) - \cos(\pi)\sin(2\tan^{-1}(x))=\sin(2\tan^{-1}(x))$$
$$\sin(2\tan^{-1}(x)) = 2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x))=\frac{2x}{x^2+1}$$
Which was to be demonstrated.
Note: This problem didn't flow like I thought it would. I had imagined that during some of the intermediate steps, I would be presented with the option of choosing an $f(x)$ or trig function value that would only be true in one of the intervals. But no such situation presented itself. Did I do something wrong? Did I overlook something?
| As already commented by @DatBoi that the proof is incorrect, I would suggest you a different approach.
$ f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) ~~ -\infty<x<\infty.$
a) Substitute $x= \tan\theta$
$\implies f(x)= \sin^{-1}(\sin2\theta)$
$\implies f(x) =2\theta\quad -\pi/2\leq2\theta\leq\pi/2$
$\implies f(x)= 2\tan^{-1}x \quad -\pi/2\leq2\ (\tan^{-1}x)\leq\pi/2$
$\implies f(x) = 2\tan^{-1}x \quad -\pi/4\leq\ (\tan^{-1}x)\leq\pi/4$
$\implies f(x)= 2\tan^{-1}x \quad -1\leq x\leq1$
Now, I hope you can do the second part yourself.
| {
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"source": "stackexchange",
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Hard Integral without Partial Fraction Decomposition So I was trying to use partial fraction decomposition on this problem, and I realized that it didn't work, as it is already in partial fraction decomposition form.
$\int{\frac{3x+4}{(x^2+5)^2}dx}=$
Well, $\frac{3x+4}{(x^2+5)^2}=\frac{Ax+B}{x^2+5}+\frac{Cx+D}{(x^2+5)^2}$
$(x^2+5)^2[\frac{3x+4}{(x^2+5)^2}]=(x^2+5)^2[\frac{Ax+B}{x^2+5}+\frac{Cx+D}{(x^2+5)^2}]$
$3x+4=(Ax+B)(x^2+5)+(Cx+D)$
$3x+4=Ax^3+5Ax+Bx^2+5B+Cx+D$
$3x+4=(A)x^3+(B)x^2+(5A+C)x+(5B+D)$
So, $A=0$, $B=0$, $5A+C=3$, and $5B+D=4$
So, $A=0$, $B=0$, $C=3$, and $D=4$
$\int{\frac{3x+4}{(x^2+5)^2}dx}=\int{\frac{(0)x+(0)}{x^2+5}+\frac{(3)x+(4)}{(x^2+5)^2}}dx$
$\int{\frac{3x+4}{(x^2+5)^2}dx}=\int{\frac{3x+4}{(x^2+5)^2}}dx.$
As this clearly doesn't work, I am wondering what the integral would be. WolframAlpha tells me that the integral is as follows:
$\int{\frac{3x+4}{(x^2+5)^2}dx}=\frac{1}{50}(\frac{5(4x-15)}{x^2+5}+4\sqrt{5}\tan^{-1}(\frac{x}{\sqrt{5}}))+c$
I am not sure how to get here. Any help would be greatly appreciated.
| Note that
\begin{align}
\left( \frac1{x^2+5}\right)’&= -\frac{2x}{(x^2+5)^2},\>\>\>\>\>
\left( \frac x{x^2+5}\right)’= \frac{10}{(x^2+5)^2}
- \frac{1}{x^2+5}
\end{align}
Their linear combination below yields
$$
\left( \frac {4x-15}{x^2+5}\right)’= \frac{10(3x+4)}{(x^2+5)^2}
-\frac{4}{x^2+5}
$$
Integrate both sides to obtain
\begin{align}
\int{\frac{3x+4}{(x^2+5)^2}dx}&=\frac1{10}
\frac {4x-15}{x^2+5}+\frac25\int\frac1{x^2+5}dx\\
&= \frac1{10}
\frac {4x-15}{x^2+5}+\frac2{5\sqrt5}\tan^{-1}\frac x{\sqrt5}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Suppose X$\sim$ Cauchy(0,1). Then what will be the distribution of $\frac{1-X}{1+X}$? In order to find distribution of $\frac{1-X}{1+X}$ below approach I followed,
Let,
\begin{align}
Y = \frac{1-X}{1+X}
\end{align}
Then, cdf of Y is
\begin{align}
F_{Y}(y) = P(Y \leq y)
\end{align}
\begin{align}
= P\left(\frac{1-X}{1+X} \leq y\right)
\end{align}
\begin{align}
= 1 - P\left(X < \frac{1-y}{1+y}\right)
\end{align}
\begin{align}
= 1 - \int_{-\infty}^{\frac{1-y}{1+y}} f(x) \,dx
\end{align}
\begin{align}
= 1 - \int_{-\infty}^{\frac{1-y}{1+y}} \frac{1}{\pi}\cdot \frac{1}{1+x^2} \,dx
\end{align}
\begin{align}
= 1 - \frac{1}{\pi}\cdot \left[tan^{-1}x\right]_{-\infty}^{\frac{1-y}{1+y}}
\end{align}
\begin{align}
F_{Y}(y) = \frac{1}{\pi}\cdot \left[-\frac{\pi}{2}+tan^{-1}\left({\frac{1-y}{1+y}}\right)\right]
= \frac{1}{2} -\frac{1}{\pi}.tan^{-1}\left({\frac{1-y}{1+y}}\right)
\end{align}
and then
\begin{align}
f_{Y}(y) = \frac{d F_{Y}(y)}{dy} = \frac{1}{\pi}\cdot \frac{1}{1+y^2}
\end{align}
But I have a little confusion here how to find range of Y from X? And CDF of Y is doesn't looks like cdf of a cauchy distribution.
| This fact is useful in thinking about your problem: If $U,V\sim N(0,1)$ are independent standard normals, their ratio $X=U/V$ has the standard Cauchy distribution. Then $Y=(1-X)/(1+X) = (V-U)/(V+U) = S/T$ where $S=(V-U)/\sqrt 2$ and $T=(V+U)/\sqrt 2$. But $S$ and $T$ are also independent standard normals, so $Y$ has the same distribution as $X$.
Equivalently, one can represent the iid variables $U,V\sim N(0,1)$ as $U=R\cos\Theta, V=R\sin\Theta$, where $R,\Theta$ are independent, with $R$ Rayleigh distributed and $\Theta$ uniform on $[0,2\pi)$, so $X=\tan\Theta$ and $Y=(1-X)/(1+X)=\tan(\pi/4-\Theta)$. Since $\Theta$ is uniform, so is $\pi/4-\Theta$, so $Y$ has the same distribution as $X$.
| {
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Maximum value of $\sqrt{1-\sqrt{a_1}} + \sqrt{1-\sqrt{a_2}} + \cdots + \sqrt{1-\sqrt{a_n}}$ for $a_i \in [0,1], a_1+a_2+ \cdots +a_n = 2$
Given $n \ge 3$ real numbers $a_1, a_2, \cdots, a_n \in [0,1]$ such that $a_1 + a_2+ \cdots +a_n =2$, find the maximum value of:
$$P =\sqrt{1-\sqrt{a_1}} + \sqrt{1-\sqrt{a_2}} + \cdots + \sqrt{1-\sqrt{a_n}}.$$
So far my only guess is that the equality occurs when there are $n-x$ numbers equal to $0$ and $x$ numbers equal to $\dfrac{2}{x}$, then we have:
$$P \le n-x + x \sqrt{1-\sqrt{\frac{2}{x}}} = n-x + \sqrt{x^2-x\sqrt{2x}}$$
The function above reaches maximum at $x=3$, so we have:
$$P \le n-3 + \sqrt{9 -3\sqrt{6}} $$
And I didn't know how to proceed any further.
| Hint : use the EV method with :
$$f(x)=\sqrt{1-\sqrt{1-\frac{x}{x+1}}}$$
We have on $(0,\infty)$:
$$f'''(x)>0$$
With the constraint (for $0<x_i\leq 1$):
$$\sum_{i=1}^{n}\frac{1-x_i}{x_i}=2\,,\sum_{i=1}^{n}\left(\frac{1-x_i}{x_i}\right)^2=\operatorname{constant}$$
Now use corollary 1.4 (see http://emis.maths.adelaide.edu.au/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf )
| {
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Find $a, b, c$ such that $a^3+b^3+c^3-3abc=2017$.
Find all natural numbers $a, b, c$ such that $a\leq b\leq c$ and $a^3+b^3+c^3-3abc=2017$.
My Attempt
$$a^3+b^3+c^3-3abc=2017$$
$$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=2017*1$$
Now, $a+b+c$ can't be equal to $1$ as $a, b, c$ are natural numbers.
So, $$a+b+c=2017$$ $$a^2+b^2+c^2-ab-bc-ca=1$$
How should I proceed after this?
|
It was an easy question, just one important result and a bit of number theory was needed
| {
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Equalities with 2 given substitions Let $\alpha$ and $\beta$ be real numbers such that $\alpha + \beta = 1$ and $-\alpha\beta = 1$ Which of the following equalities hold?
A. $\alpha^2 + \beta^2 = 2$
B. $\alpha^3 + \beta^3 = 3$
C. $\alpha^4 + \beta^4 = 6$
D. $\alpha^5 + \beta^5 = 12$
E. $\alpha^6 + \beta^6 = 18$
My attempts of solving this problem:
A.$$ \alpha^2 + \beta^2 = 2 $$
$$\alpha^2 + (-2\ \alpha \cdot \beta) + \beta^2 + 2\ \alpha\cdot\beta$$
$$=\left(\alpha + \beta \right)^2 + \left( -2\ \alpha\beta \right) $$
We already know that $\alpha + \beta = 1$ and $-\alpha\beta = 1$ which implies the following equation :
$1^2 + 2 \cdot 1= 3$
So we can conclude A isn't the answer
B.
$$\alpha^3 + \beta^3 = \left(\alpha +\beta \right)\left(\alpha^2 - \alpha\beta + \beta^2 \right)$$
$$1\cdot \left(3 + 1)\right) =4 $$
So B is neither the right answer.
For the following multiple choices I can't get to the final answer. Could someone help me with this?
Kind regards in advance.
| Choice E. Use $\alpha^6+\beta^6 = (\alpha^3+\beta^3)^2 -2(\alpha\beta)^3 = 4^2 - 2(-1) = 18$.
| {
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"question_score": "2",
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Can $\mathbb{R}^3$ with Hadamard product be represented as matrices? It is known that $\mathbb{R}^2$ with Hadamard product, represented as pairs or numbers $(a,b)$ with element-wise operations is isomorphic to real matrices of the form $\left(
\begin{array}{cc}
\frac{a+b}{2} & \frac{a-b}{2} \\
\frac{a-b}{2} & \frac{a+b}{2} \\
\end{array}
\right)$, and also to the split-complex numbers.
But I wonder, whether it is possible to represent $\mathbb{R}^3$ with Hadamard product and element-wise operations as real matrices in a similar way?
| The following works for complex matrices. Let $\omega = \exp(i 2\pi/3)$ be a primitive cube root of $1$. Define
$$
H(a,b,c) :=
\begin{bmatrix}
\frac{a+b+c}{3} & \frac{a+\omega b+\omega^2 c}{3} &
\frac{a+\omega^2 b+\omega c}{3}
\\
\frac{a+\omega^2 b+\omega c}{3} & \frac{a+b+c}{3} &
\frac{a+\omega b+\omega^2 c}{3}
\\
\frac{a+\omega b+\omega^2 c}{3} & \frac{a+\omega^2 b+\omega c}{3} & \frac{a+b+c}{3}
\end{bmatrix}
$$
Then
$$
H(a,b,c)+ H(d,e,f) = H(a+d,b+e,c+f)
\\
H(a,b,c) H(d,e,f) = H(ad,be,cf)
$$
Probably we can adapt this to get real $6 \times 6$ matrices, replacing each complex number by a real $2\times 2$ matrix in the usual way.
| {
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"question_score": "3",
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Under these conditions, determine the polynomial in two variables. Let $f(x,y)=(x^2+y^2+2)^{xy}$ defined on $\mathbb R^2$ and let $g(x,y)$ be a degree $2$ polynomial in two variables.
Suppose $\displaystyle\lim_{(x,y)\to (0,0)} \dfrac{g(x,y)-f(x,y)}{x^2+y^2}=0$. Then, determine $g(x,y)$.
Here is what I did so far.
Let $g(x,y)=a_0 x^2+a_1 y^2 +a_3 xy + a_4 x+ a_5 y+ a_6.$
From supposition, $\lim_{(x,y)\to (0,0)}g(x,y)-f(x,y)=0$ has to hold. And
since $\lim g(x,y)=a_6, \lim f(x,y)=1$, I get $a_6=1.$
If I use polar coordinates,
\begin{align}
&\dfrac{g(x,y)-f(x,y)}{x^2+y^2}\\
&=\dfrac
{r^2(a_0 \cos^2 \theta + a_1 \sin^2 \theta +a_3 \cos\theta \sin \theta)+ r(a_4 \cos \theta + a_5 \sin \theta)+1-(r^2+2)^{r^2 \cos \theta \sin \theta}}
{r^2}\\
&=a_0 \cos^2 \theta + a_1 \sin^2 \theta +a_3 \cos\theta \sin \theta+\dfrac{1}{r}(a_4 \cos \theta + a_5 \sin \theta)+ \dfrac{1-(r^2+2)^{r^2 \cos \theta \sin \theta}}{r^2}
\end{align}
When $r\to 0$ with $\theta=0,$ this has to converge to $0$ so
$a_0+\dfrac{a_4}{r}+\dfrac{1-1}{r^2}\to 0$ as $r\to 0$.
Thus $a_4=a_0=0.$
Similarly, when $r\to 0$ with $\theta=\frac{\pi}{2},$ this has to converge to $0$ so
$a_1+\dfrac{a_5}{r}+\dfrac{1-1}{r^2}\to 0$ as $r\to 0$.
Thus $a_5=a_1=0$.
(I'm not sure the argument above is correct.)
Thus, all that is left is to determine $a_3.$
Now, $g(x,y)=a_3 xy +1$. I'm stacked here.
How can I determine $a_3$ ?
| Recall that for $t\to 0$, $e^t=1+t+O(t^2)$ and $\ln(1+t)=t+o(t)$.
Therefore,for $(x,y)\to (0,0)$,
\begin{align}(x^2+y^2+2)^{xy}&=\exp(xy\ln(x^2+y^2+2))\\
&=1+xy\ln(x^2+y^2+2)+O(x^2y^2)\\
&=1+xy\left(\ln(2)+\ln\left(1+\frac{x^2+y^2}{2}\right)\right)+o(x^2+y^2)\\
&=1+xy\left(\ln(2)+\frac{x^2+y^2}{2}+o(x^2+y^2)\right)+o(x^2+y^2)\\
&=1+\ln(2)xy+o(x^2+y^2).\end{align}
Hence $g(x,y)=1+\ln(2)xy$. In particular $a_3=\ln(2)$.
| {
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} |
How do we find the exact value of $\int_{0}^{\infty} \frac{\ln ^{n}\left(1+x^{2}\right)}{1+x^{2}} d x$, where $n\in \mathbb N?$ Latest Edit
By @KStarGamer’s help, I can finally find a reduction formula for $I_n$ as below: $$\boxed{I_n= 2 \ln 2 I_{n-1}+ (n-1)!\sum_{k=0}^{n-2} \frac{2^{n-k}-2}{k!}\zeta(n-k) I_k}
$$
where $n\geq 2.$
In my post, I started to evaluate $$I_1=\int_{0}^{\infty} \frac{\ln \left(1+x^{2}\right)}{1+x^{2}} d x =\pi \ln 2, $$
then I challenge myself on $$I_2=\int_{0}^{\infty} \frac{\ln ^{2}\left(1+x^{2}\right)}{1+x^{2}}dx$$
Again, letting $x=\tan \theta$ as for $I_1$ yields$$I_2=\int_{0}^{\frac{\pi}{2}} \ln ^{2}\left(\sec ^{2} \theta\right) d \theta= 4 \int_{0}^{\frac{\pi}{2}} \ln ^{2}(\cos x) dx $$
It’s very hard to deal with $\ln^2$ and I was stuck. Suddenly a wonderful identity came to my mind. $$
2\left(a^{2}+b^{2}\right)=(a+b)^{2}+(a-b)^{2},
\\$$
by which
$\displaystyle 2\left[\ln ^{2}(\sin x)+\ln ^{2}(\cos x)\right]=[\ln (\sin x)+\ln (\cos x)]^{2}+[\ln (\sin x)-\ln (\cos x)]^{2} ,\tag*{}\\ $
we have
$\displaystyle 4 L=\underbrace{\int_{0}^{\frac{\pi}{2}} \ln ^{2}\left(\frac{\sin 2 x}{2}\right)}_{J} d x+\underbrace{\int_{0}^{\frac{\pi}{2}} \ln ^{2}(\tan x) d x}_{K} \tag*{}\\ $
For the first integral, using $ \int_{0}^{\frac{\pi}{2}} \ln (\sin x) d x=-\dfrac{\pi}{2} \ln 2 $ yields
$ \begin{aligned}J &=\int_{0}^{\frac{\pi}{2}}[\ln (\sin 2 x)-\ln 2]^{2} d x \\&=\int_{0}^{\frac{\pi}{2}} \ln ^{2}(\sin 2 x) d x-2 \ln 2 \int_{0}^{\frac{\pi}{2}} \ln (\sin 2 x) d x +\frac{\pi \ln ^{2} 2}{2} \\& \stackrel{x\mapsto 2x}{=} \frac{1}{2} \int_{0}^{\pi} \ln ^{2}(\sin x) d x-\ln 2 \int_{0}^{\pi} \ln (\sin x) d x+\frac{\pi \ln ^{2} 2}{2} \\& \stackrel{symmetry}{=} L-\ln 2(-\pi \ln 2)+\frac{\pi \ln ^{2} 2}{2} \\&=L+\frac{3 \pi \ln ^{2} 2}{2}\end{aligned}\tag*{} \\$
For the second integral, letting $ y=\tan x $ and using my post yields
$ \displaystyle K=\int_{0}^{\infty} \frac{\ln ^{2} y}{1+y^{2}} d y=\frac{\pi^{3}}{8}, \tag*{} \\$
then
$ \displaystyle 4L=L+\frac{3 \pi \ln ^{2} 2}{2}+\frac{\pi^{3}}{8} \Rightarrow L=\frac{\pi^{3}}{24}+\frac{\pi \ln ^{2} 2}{2}\tag*{} $
Hence
$ \displaystyle \boxed{I_2=4L= \frac{\pi^{3}}{6}+2 \pi \ln ^{2} 2} \tag*{} $
My Question: Can I go further with $I_n$, where $n\geq 3$?
| With differentiation of $$\frac2\pi\int_{0}^{\frac\pi2} (2\cos t)^adt = \frac{\Gamma(a+1)}{\Gamma^2\left(\frac a2+1\right)}$$
in $a$, it can be shown that $J_k=\frac2\pi \int_{0}^{\frac\pi2}\ln^k(2\cos t)dt$ are the power-series coefficients of the exponential function below
$$
\exp\bigg( \sum_{k=1}^\infty \frac{(-1)^k}k (1-2^{1-k})\zeta(k)x^k\bigg)= \sum_{k=0}^\infty \frac{J_k}{k!}x^k
$$
which results in
\begin{align}
J_0 &= 1,\>\>\>\>\>
J_1= 0, \>\>\>\>\>
J_2=\frac12 \zeta(2),\>\>\>\>\>
J_3= -\frac32 \zeta(3),\>\>\>\>\>
J_4= \frac{57}{8} \zeta(4) \\
J_5 &= -\frac{15}2\zeta(2)\zeta(3) -\frac{45}2 \zeta(5),\>\>\>\>\>
J_6= \frac{45}2 \zeta^2(3)+ \frac{12375}{64} \zeta(6),\>\>\>\>\>
J_7= \cdots
\end{align}
Now, note that
\begin{align}
I_n & = \int_{0}^{\infty} \frac{\ln ^{n}(1+x^{2})}{1+x^{2}} d x\\
&= \int_{0}^{\frac{\pi}{2}} \ln ^{n}\sec ^{2} t\> d t= 2^{n-1}\pi\sum_{k=0}^n \binom nk (-1)^{k} \ln^{n-k}(2)\>J_{k}
\end{align}
Thus
\begin{align}
I_1 &= \pi\ln2 \\
I_2 &
= \pi [\zeta(2) +2\ln^22]\\
I_3
&=\pi [6\zeta(3) +6\ln2\> \zeta(2) +4\ln^32] \\
I_4
&= \pi[57\zeta(4)+48\ln2\>\zeta(3)+24\ln^22\>\zeta(2)+8\ln^42]\\
I_5
&= \pi[120\zeta(2)\zeta(3)+360\zeta(5)+570\ln2\>\zeta(4)\\
&\hspace{2cm}+240\ln^22\>\zeta(3)+80\ln^32\>\zeta(2)+16\ln^52]\\
I_6&=\cdots
\end{align}
| {
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Integrate : $\int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$ Integrate :
$$I = \int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$$
My attempt : substitute $\sin t = x-1$, $u = \tan \frac{t}{2}$
$$I = \int \frac{1}{x^{2} \sqrt{1-(x-1)^{2}}} dx = \int \frac{1}{(\sin t + 1)^{2} \cos t} \cos t dt$$
$$= \int \frac{1}{(\sin t + 1)^{2}} dt = \int \frac{2(u^{2}+1)}{(u+1)^{4}} du$$
$$= 2 \int \left(\frac{1}{(u+1)^{2}} - \frac{2}{(u+1)^{3}} + \frac{2}{(u+1)^{4}} \right)$$
$$= - \frac{2}{u+1} + \frac{2}{(u+1)^{2}} - \frac{4}{3(u+1)^{3}} + const.$$
$$= \frac{-2(3u^{2}+3u+2)}{3(u+1)^{3}} + const.$$
where
$$A = \sqrt{2x-x^{2}},~~ u = \sqrt{\frac{1-A}{1+A}}$$
However, Wolfram gives the following answer :
$$ - \frac{\sqrt{2x-x^{2}}~(x+1)}{3x^{2}} + const. $$
which is way more simple than mine. Is there any other way to integrate this? The $u$ in my answer is very complicated, I just couldn't change my answer into Wolfy's.
| $$I = \int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$$
$$I = \int \frac{1}{x^{3} \sqrt{\frac{2}{x}-1}} dx$$
Substitute $\displaystyle u=\frac{1}{x}$ to get
$$I=-\int\frac{u}{\sqrt{2u-1}}du$$
$$-2I=\int\frac{2u-1+1}{\sqrt{2u-1}}du$$
Can you finish it now?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of consecutive numbers with a rounded multiplier I know full well that the sum of the integers from $1$ to $n$ is $\frac{n * \left (n+1\right )}{2}$ but I need a formula that works for the rounded multiples of $\frac{9}{7}$ the integers from $1$ to $n$.
e.g.
$n$
$\lfloor\frac{9}{7}n\rceil$
cumulative sum
1
1
1
2
3
4
3
4
8
4
5
13
5
6
19
6
8
27
any ideas? i don't even know what to search for so i don't know how to tag this question.
| You’re adding all the numbers up to $\left\lfloor\frac97n\right\rceil$ except those with remainders $2$ and $7$ modulo $9$. So you have $\frac{\left\lfloor\frac97n\right\rceil\left(\left\lfloor\frac97n\right\rceil+1\right)}2$, but you need to subtract the missing numbers. There are $\left\lfloor\frac{n+5}7\right\rfloor$ missing numbers with remainder $2$, and they add up to $2\left\lfloor\frac{n+5}7\right\rfloor+9\cdot\frac{\left\lfloor\frac{n+5}7\right\rfloor\left(\left\lfloor\frac{n+5}7\right\rfloor-1\right)}2=\frac{\left\lfloor\frac{n+5}7\right\rfloor\left(9\left\lfloor\frac{n+5}7\right\rfloor-5\right)}2$, and similarly there are $\left\lfloor\frac{n+1}7\right\rfloor$ missing numbers with remainder $7$, and they add up to $7\left\lfloor\frac{n+1}7\right\rfloor+9\cdot\frac{\left\lfloor\frac{n+1}7\right\rfloor\left(\left\lfloor\frac{n+1}7\right\rfloor-1\right)}2=\frac{\left\lfloor\frac{n+1}7\right\rfloor\left(9\left\lfloor\frac{n+1}7\right\rfloor+5\right)}2$, so the sum is
$$
\frac12\left(\left\lfloor\frac97n\right\rceil\left(\left\lfloor\frac97n\right\rceil+1\right)-\left\lfloor\frac{n+5}7\right\rfloor\left(9\left\lfloor\frac{n+5}7\right\rfloor-5\right)\right)-\left\lfloor\frac{n+1}7\right\rfloor\left(9\left\lfloor\frac{n+1}7\right\rfloor+5\right)\;.
$$
| {
"language": "en",
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"source": "stackexchange",
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Let $M=2\times 3^2\times 5^3\times 7^4$. Let $a_1,a_2,...,a_{120}$ be the positive divisors of $M$. Calculate $\prod_{i=1}^{120} a_i$ Let $M=2\times 3^2\times 5^3\times 7^4$. Let $a_1,a_2,...,a_{120}$ be the positive divisors of $M$. Calculate $$\prod_{i=1}^{120} a_i$$.
My Attempt:
I am aware how to find the sum of these $120$ divisors.$$\sum_{i=1}^{120}a_i=(2^0+2^1)(3^0+3^1+3^2)(5^0+5^1+5^2+5^3)(7^0+7^1+7^2+7^3+7^4)$$But I have never come across a problem asking for product of divisors before.
In the product of divisors $2^0$ and $2^1$ will occur $(1+2)(1+3)(1+4)=60$ times each.
Similarly, $3^0$,$3^1$ and $3^2$ will occur $(1+1)(1+3)(1+4)=40$ times each.
$5^0, 5^1, 5^2, 5^3$ will occur $(1+1)(1+2)(1+4)=30$ times each.
$7^0, 7^1, 7^2, 7^3, 7^4$ will occur $(1+1)(1+2)(1+3)=24$ times each.
So product of divisors equals $$(2^0)^{60}.(2^2)^{60}.(3^0)^{40}.(3^1)^{40}.(3^2)^{40}.(5^0)^{30}.(5^1)^{30}.(5^2)^{30}.(5^3)^{30}.(7^0)^{24}.(7^1)^{24}.(7^2)^{24}.(7^3)^{24}.(7^4)^{24}=2^{60}.3^{120}.5^{180}.7^{240}$$
Is this correct.
Can there be a shorter or more direct way to do it.
| Let's order the $a_{1}, \ldots, a_{120}$ so that $a_{i} \cdot a_{121-i} = M$ for all $1 \leq i \leq 120.$ We can do this because an integer $d$ is a positive divisor of $M$ if and only if $\frac{M}{d}$ is a positive integer divisor of $M$ (and since $M$ is not a square, $d$ and $\frac{M}{d}$ are never equal).
Now, note that
$$\left(\prod_{i=1}^{120}a_{i}\right)^{2} = \left(\prod_{i=1}^{120}a_{i}\right)\left(\prod_{i=1}^{120}a_{121-i}\right),$$ because the second product on the right side is just a permutation of the first, and multiplication is commutative.
Then, we can write
$$\left(\prod_{i=1}^{120}a_{i}\right)\left(\prod_{i=1}^{120}a_{121-i}\right) = \prod_{i=1}^{120} a_{i}a_{121-i},$$ once again using the commutativity of multiplication. However, since $a_{i}a_{121-i} = M$ for all $i$, it follows that
$$\prod_{i=1}^{120} a_{i}a_{121-i} = \prod_{i=1}^{120} M = M^{120}.$$
So, since $$\left(\prod_{i=1}^{120}a_{i}\right)^{2} = M^{120},$$ we see that
$$\prod_{i=1}^{120}a_{i} = M^{60}.$$
Finally, since $M = 2 \times 3^{2} \times 5^{3} \times 7^{4}$, we have
$$\prod_{i=1}^{120}a_{i} = 2^{60} \times 3^{120} \times 5^{180} \times 7^{240},$$
which agrees with what you got.
| {
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Maximum and Minimum value of an implicit function For the real value of $x$, $f\left( x \right)$ satisfies $f{\left( x \right)^3} - f{\left( x \right)^2} - {x^2}f\left( x \right) + {x^2} = 0$. When the maximum value of $f(x)$ is $1$ and the minimum value of $f(x)$ is $0$, what is the value of $f\left( { - \frac{4}{3}} \right) + f\left( 0 \right) + f\left( {\frac{1}{2}} \right) = \_\_\_\_\_$
My approach is as follow, as it is an implicit function we need to find the roots is $f(x)$.
We end up getting $(f(x)-1)(f(x)+x)(f(x)-x)=0$, so we end up getting three function viz.
$f(x)=1$; $f(x)=-x$ & $f(x)=x$ but how do we proceed further
| $y=f(x). y^3-y^2-x^2y+x^2=0$. Max $y$ is $1$. Min $y$ is $0$. Find $f(-4/3)+f(0)+f(1/2)$
$g(x,y)= (y-x)(y+x)(y-1)=0$.
The constraint requires only certain $y$ values for a given $x$.
$x=0\implies y\in\{0,1\}$
$x=1/2\implies y^3-y^2-y/4+1/4=0.\implies (2y-1)(2y+1)(y-1)=0\implies y\in\{1/2,-1/2,1\}$
$x=-4/3\implies y^3-y^2-16y/9+16/9=0\implies (9y^2-16)(y-1)=(3y-4)(3y+4)(y-1)=0$
$\implies y \in \{4/3, -4/3,1\}$
I'm getting only 18 possibilities, but those should be pared down. Not sure how to use the min and max values. $f(x)\equiv1$ doesn't allow a minimum value of $0$, so the $1$'s can be ignored.
Consistent definition of $f(x)$ requires consistently using $f(x)\equiv x$ or $f(x)\equiv -x$. So, combinations are reduced still further to $\{-5/6, 5/6\}$
| {
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Closed form of $\sum_{i=0}^{\infty} x^i (i!(i+1)!)^{-1/2}$ I am looking for a closed form of this series
$$
\sum\limits_{i=0}^{\infty} \frac{x^i}{\sqrt{i!(i+1)!}},\:x \geq 0.
$$
The main problem is that this series contains roots of factorials. But many special functions just contain factorials, not their roots. Is it possible to express the sum of this series in some well-known (or little known) functions?
| You will not find closed form, but using
$$n^{-s} = \frac{1}{\Gamma(s)} \int_0^\infty t^{s-1} e^{-nt} \, {\rm d}t$$
for $s=1/2$ (it should work for general $s$) and substituting $e^{-t} \rightarrow t$, you can bound your sum above $$f(x)=\sum_{n=0}^\infty \frac{x^n}{n!\sqrt{n+1}} = \frac{1}{\sqrt{\pi}}\int_0^1 \frac{e^{xt}}{\sqrt{-\log(t)}} \, {\rm d}t \stackrel{t=1-\tau}{=} \frac{e^x}{\sqrt{\pi}}\int_0^1 \frac{e^{-x\tau}}{\sqrt{-\log(1-\tau)}} \, {\rm d}\tau \\ \leq \frac{e^x}{\sqrt{\pi}}\int_0^1 \frac{e^{-x\tau}}{\sqrt{\tau}} \, {\rm d}\tau \stackrel{x\tau=u^2}{=} \frac{e^x}{\sqrt{x}} \frac{2}{\sqrt{\pi}} \int_0^{\sqrt{x}} e^{-u^2} \, {\rm d}u \leq \frac{e^x}{\sqrt{x}}\, ,$$
where $-\log(1-t) \geq t$ was used.
You can find better approximations, by expanding
$$\frac{1}{\sqrt{-\ln(1-t)}} = \frac{1}{\sqrt{t}} - \frac{\sqrt{t}}{4} - \frac{7t^{3/2}}{96} + O\left(t^{5/2}\right)$$
about $t=0$, and which leads to
$$f(x)= \frac{e^x}{\sqrt{x}} \left( 1 - \frac{1}{8x} - \frac{7}{128x^2} + O\left(1/x^3\right) \right) {\rm erf}\left( \sqrt{x} \right) + \frac{1}{\sqrt{\pi}} \left( \frac{31}{96x} + \frac{7}{64x^2} \right) \\
= \frac{e^x}{\sqrt{x}} \left( 1 - \frac{1}{8x} - \frac{7}{128x^2} + O\left(1/x^3\right) \right) \, .$$
Furthermore I played around via try and error an looked for smooth elementary functions that match $f(x)=1+\frac{x}{\sqrt{2}}+O(x^2)$ up to first order about $x=0$ and till second order asymptotically. My ansatz was
$$f(x)=e^x \sqrt{\frac{1+c/x+1/x^2}{x+a+b/x+1/x^2}}$$
and Taylor expanding about $x=0$ and $x=\infty$ gives the coefficients
$$a=\frac{4\sqrt{2}}{3}-\frac{7}{8} \\
b=\frac{\sqrt{2}}{3} + \frac{7}{8} \\
c=\frac{4\sqrt{2}}{3} - \frac{9}{8}$$
which approximates $f(x)$ reasonably well and bounds it from above. You can do much better by increasing the order inside the square-root from $1/x^2$ to $1/x^3$ and using the two new degrees of freedom to match up to third order about $x=0$. The expressions for the coefficients become more complicated though.
Here is a graph:
| {
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How do we parametrize the curve determined by the intersection of $x^{2} + y^{2} + z^{2} = 1$ and $x + y + z = 1$ and find it arc length? How do we parametrize the curve determined by the intersection of $x^{2} + y^{2} + z^{2} = 1$ and $x + y + z = 1$?
Should I make a new equation $x^{2} + y^{2} + z^{2} = x + y + z$?
Or do I need to use spherical coordinates to parametrize $x^{2} + y^{2} + z^{2} = 1$ ?
Also, I want to find the arc length of this parametric equation.
| Yet another answer.
The unit sphere in $\mathbb{R}^n$ can be described by $\|x\|^2 = 1$, and the equivalent plane by $e^T x = 1$ where $e=(1,...,1)$. Let $Q$ be an orthogonal rotation such that $Qe = \sqrt{n}e_1$, where $e_1 = (1,0,...,0)$.
Using the transformation $y=Qx$ we get the equivalent equations $\|y\|^2 = 1$ and $e_1^Ty = {1 \over \sqrt{n}}$. Equivalently
$y_2^2+\cdots+ y_n^2 = 1-{1 \over n}$, so we see that this is an $n-1$ dimensional $\sqrt{1-{1 \over n}}$ sphere centered at $({1 \over \sqrt{n}},0,...,0)$.
Mapping back we see that the intersection is parameterised by
${1 \over n} e + \sum_{k=2}^n y_kQ^T e_k$, where $y_2^2+\cdots+ y_n^2 = 1-{1 \over n}$.
For $n=3$ we can parameterise this as
${1 \over 3}(1,1,1) + {\sqrt{2 \over 3}}\cos \theta \ u +{\sqrt{2 \over 3}}\sin \theta \ v$ where ${1 \over \sqrt{3}} (1,1,1),u,v$ are orthonormal.
| {
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How to decompose $\frac{1}{(1 + x)(1 - x)^2}$ into partial fractions Good Day.
I was trying to decompose $$\frac{1}{(1 + x)(1 - x)^2}$$ into partial fractions.
$$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{B}{(1 - x)^2}$$
$$1 = A(1 - x)^ 2 + B(1 + x)$$
Substitute $x = 1$, $$B = \frac{1}{2}$$
Substitute $x = -1$, $$A = \frac{1}{4}$$
However, $$\frac{1}{4(1 + x)} + \frac{1}{2 (1 - x)^2}$$ doesn't seem to equal $$\frac{1}{(1 + x)(1 - x)^2}$$
How do we decompose this into partial fractions?
Thanks
| Using the hint given by @GEdgar, notice that
$$\frac{1}{(1+x)(1-x)^{2}}=\frac{A}{1-x}+\frac{B}{(1-x)^{2}} +\frac{C}{1+x}.$$
Solving, we get
$$1=A(x^{2}-1)+B(x+1)+C(x-1)^{2}$$
Equality coefficients, we have
$$\begin{cases}-A+B+C=1,\\ 0A+B-2C=0,\\A+0B+C=0\end{cases}$$
Solving the system equations, we get
$$\left(A,B,C\right)=\left(-\frac{1}{4},\frac{1}{2}, \frac{1}{4}\right).$$
Therefore,
$$\frac{1}{(1+x)(1-x)^{2}}=\frac{-1/4}{1-x}+\frac{1/2}{(1-x)^{2}} +\frac{1/4}{1+x}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How to show $\int_0^{\infty} \frac{\sin^3 x}{\cosh x\>+\>\cos x}\frac{dx}x=\frac{\pi}{8}$ I would like to evaluate the integral below$$\int_0^{\infty} \frac{\sin^3 x}{\cosh x+\cos x}\frac{dx}x $$
which I found to be $\frac \pi8$ numerically. I was able to evaluate a similarly looking, yet simpler, integral
$$\int_0^{\infty} \frac{\sin x}{\cosh x+\cos x}\frac{dx}x =\frac\pi4$$
by writing the integrand in the Frullani format and apply the theorem. However, it does not work in this case and I am not sure if they are related. Any solution is appreciated.
| Since
$$
\frac{\sin(x)}{\cosh(x)+\cos(x)}=2\sum_{n\ge 1}(-1)^{n-1}e^{-nx}\sin(nx
)$$
(shown here), we can combine the above with $\sin^2(x)\sin(nx) = \frac14 \left[2 \sin(n x) + \sin((2 - n) x) - \sin((2 + n) x)\right]$ and $\int_0^\infty e^{-ax}\frac{\sin{(bx)}}{x}\mathrm{d}x = \arctan\left(\frac{b}{a}\right)$ to get
\begin{align*}
I &= \frac{1}{2}\sum_{n\ge 1}(-1)^{n-1}\int_{0}^{\infty}\frac{e^{-nx}}{x}\left[2 \sin(n x) + \sin((2 - n) x) - \sin((2 + n) x)\right] \, \mathrm{d}x\\
& = \frac{1}{2}\sum_{n\ge 1}(-1)^{n-1} \left[\frac\pi2 + \arctan\left(\frac{2-n}{n} \right) - \arctan\left(\frac{2+n}{n} \right)\right]\\
& = \frac{1}{2}\sum_{n\ge 1}(-1)^{n-1} \left[\arctan(n+1) - \arctan(n-1)\right]\\
& = \frac{1}{2}\left[-\arctan(0) + \arctan(1) \right]\\
& = \frac{\pi}{8}
\end{align*}
as desired.
Note that to re-write the arctangent expressions from equations $2\to 3$ we combine $\arctan(\alpha) - \arctan(\beta) = \arctan\left(\frac{\alpha - \beta}{1 + \alpha\beta} \right)$ and $\arctan(x) + \arctan\left( \frac1x\right)= \frac\pi2,\ x>0$. Thus
\begin{align*}
\frac\pi2 + \arctan\left(\frac{2-n}{n} \right) - \arctan\left(\frac{2+n}{n} \right) &=\frac\pi2 + \arctan\left(\frac{-n^2}{2}\right) \\
&\overset{\color{blue}{n^2/2 >0}}{=} \arctan\left(\frac{2}{n^2}\right) \\
&=\arctan(n+1) - \arctan(n-1)
\end{align*}
also exploiting the fact that $\arctan(x)$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4401894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the error in my workings for the proof of $\left(\frac{x}{x+2y} + \frac{y}{y+2z} + \frac{z}{z+2x}\right)\ge 1$ Here are my workings...
By Cauchy-Schwarz,
$$\left(\frac{x}{x+2y} + \frac{y}{y+2z} + \frac{z}{z+2x}\right)\left( \frac{x+2y}{x} + \frac{y+2z}{y} + \frac{z+2x}{z}\right) ≥ (1+1+1)^2$$
$$\left(\frac{x}{x+2y} + \frac{y}{y+2z} + \frac{z}{z+2x}\right)(3+2(y/x + z/y + x/z))≥ 9 \tag{1}$$
However, by AM-GM,
$y/x + z/y + x/z ≥ 3,$ meaning that there is no maximum value for $y/x + z/y + x/z.$
Hence, $3+2(y/x + z/y + x/z)$ has no maximum value and we can't find the minimum value of $\left(\frac{x}{x+2y} + \frac{y}{y+2z} + \frac{z}{z+2x}\right)$ using equation $(1)$?
Please help me spot the error and enlighten me on what's the correct way to solve it (Using AM-GM or Cauchy-Schwarz inequality)
| Observe,$$\sum_{\text{cyclic}}\frac{x}{x+2y}=\sum_{\text{cyclic}}\frac{x^2}{x^2+2xy}$$
Using Titu's Lemma,
$$\sum_{\text{cyclic}}\frac{x^2}{x^2+2xy}\ge\frac{(x+y+z)^2}{x^2+y^2+z^2+2(xy+yz+zx)}=1$$
$$\therefore\;\; \frac{x}{x+2y}+\frac{y}{y+2z}+\frac{z}{z+2x}\ge 1 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4405673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the power of the matrix. Let $A = \left( {\begin{array}{*{20}{c}}
0&1&1\\
1&0&1\\
1&1&0
\end{array}} \right)$.
I want to find $A^k,$ where $k \in N$. So far I calculated $A^2, A^3, A^4,...$ but I can not see the general formula for $A^k$. Here are $A^2, A^3, A^4, A^5$.
Not sure if this leads to anything but I found the general formula for $B^k$, where $B = \left( {\begin{array}{*{20}{c}}
1&1&1\\
1&1&1\\
1&1&1
\end{array}} \right)$.
${B^k} = \left( {\begin{array}{*{20}{c}}
{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\
{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\
{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}
\end{array}} \right)$
Thanks in advance.
| Note that you get $$A^2=A+2I$$
Therefore $A^n$ is a linear combination of $A$ and $I\ $, i.e. $$ A^n=a_nA+i_nI$$
You get $A^{n+1}=a_nA^2+i_nA=a_n(A+2I)+i_nA=(a_n+i_n)A+2a_nI$
The system to solve is $\begin{cases}a_{n+1}=a_n+i_n\\i_{n+1}=2a_n\end{cases}\ $ and reporting for $a_{n+2}$ gives $\begin{cases}a_{n+2}=a_{n+1}+2a_n\\ a_0=0\\ a_1=1\end{cases}$
This solves to $a_n=\alpha\,2^n+\beta\,(-1)^n$ and initial conditions give $\ \alpha=-\beta=\frac 13$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Finding the all possible values of $x$ such that $\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$ Find possible value of $x$ such that
$$\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$$
Progress: what I did was to consider a case when $x^2 -1 < 1$ $(xy < 1)$ and $3x>-1$ $(xy > -1)$ and then apply $\tan^{-1}(x) \pm tan^{-1}(y)$ identity and got the range.
But is it correct to say we got all the possible value of $x$ or we need to consider all possibilities of $x^2 - 1 > 1$ , $3x < -1$ etc?
As what my book did was just did the first case and then left it without telling what about the solutions from other cases of $tan^{-1}(x) \pm \tan^{-1}(y)$.
| I advise you to apply the identity
\begin{align*}
\tan(a + b + c) = \frac{\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c)}{1 - \tan(a)\tan(b) - \tan(a)\tan(c) - \tan(b)\tan(c)}
\end{align*}
so that you arrive at the following equation:
\begin{align*}
\frac{3x - x(x^{2} - 1)}{1 - (x^{2} - 1) - 2x^{2}} = 3 & \Longleftrightarrow \frac{x^{3} - 4x}{3x^{2} - 2} = 3\\\\
& \Longleftrightarrow x^{3} - 9x^{2} - 4x + 6 = 0\\\\
& \Longleftrightarrow (x^{3} + x^{2}) - 10(x^{2} + x) + 6(x + 1) = 0\\\\
& \Longleftrightarrow x^{2}(x^{2} + 1) - 10x(x + 1) + 6(x + 1) = 0\\\\
& \Longleftrightarrow (x^{2} - 10x + 6)(x + 1) = 0
\end{align*}
Hence the solution set is a subset of $S = \{5\pm\sqrt{19},-1\}$.
EDIT
It remains to determine if there are extraneous roots.
In order to do so, you can check out WolframAlpha (for example).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the maximum value of:$3\sqrt[3]{\dfrac{c^2-3a^2}{6}}-2\sqrt{\dfrac{a^2+b^2+c^2-ab-bc-ca}{3}}$ Let $a,b,c \in R$, find the maximum value of:$$3\sqrt[3]{\dfrac{c^2-3a^2}{6}}-2\sqrt{\dfrac{a^2+b^2+c^2-ab-bc-ca}{3}}$$
I have the solution:
$$5a^2+2b^2+c^2-2(ab+bc+ca)=5(a-\frac{b+c}{5})^2+\frac{1}{5}(3b-2c)^2\geqslant 0$$
which is equivalent to
$$ a^2+b^2+c^2-ab-bc-ca\geqslant \frac{c^2-3a^2}{2}$$
or
$$ 2\sqrt{\frac{a^2+b^2+c^2-ab-bc-ca}{3}}\geqslant 2\sqrt{\frac{c^2-3a^2}{6}}$$
Now, let $x=\sqrt[6]{\dfrac{c^2-3a^2}{6}}$ and the problem is find the maximum value of $P=3t^2-2t^3$, which is easy.
My question:
*
*Is there any other solution to this problem?
*I try to test it on Wolfram Alpha, the result is that the maximum value does not exist. Why ? Is Wolfram Alpha's algorithm wrong or does it use a special calculation method?
| Alternate approach:
$a^2+b^2+c^2-ab-bc-ca$ takes its max value when $b =\frac{a+c}2$. So $a^2+b^2+c^2-ab-bc-ca=\frac 34(a-c)^2$, so we aim to maximize
$$3\sqrt[3]{\dfrac{c^2-3a^2}{6}}- |c-a|$$
Let $c-a=t$, so $c^2-3a^2=(a+t)^2-3a^2=-2a^2+2at+t^2$. This takes maximum if $a=t/2$, and $c^2-3a^2=3/2 t^2$. Therefore, we aim to maximize
$$3\sqrt[3]{\dfrac{t^2}{4}}-|t|$$
And this takes the maximum when $t=2$, and the maximum value is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4410244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the value of an expression using the roots of a given polynomial
Let $a,$ $b,$ $c,$ $d,$ and $e$ be the distinct roots of the equation $x^5 + 7x^4 - 2 = 0.$ Find
\begin{align*} &\frac{a^4}{(a - b)(a - c)(a - d)(a - e)} + \frac{b^4}{(b - a)(b - c)(b - d)(b - e)} \\ &\quad + \frac{c^4}{(c - a)(c - b)(c - d)(c - e)} + \frac{d^4}{(d - a)(d - b)(d - c)(d - e)} \\ &\quad + \frac{e^4}{(e - a)(e - b)(e - c)(e - d)}. \end{align*}
I started by using Vieta's formulas to obtain the following system of equations:
\begin{cases}
&\sum_{\text{cyc}} a = -7, \\
&\sum_{\text{cyc}} ab = 0, \\
&\sum_{\text{cyc}} abc = 0, \\
&\sum_{\text{cyc}} abcd = 0, \\
&abcde = 2.
\end{cases}
Now let $f_a(x) = (x - b)(x - c)(x - d)(x - e)$, then $f_a(x) = x^4 - x^3\sum_{\text{cyc}} b + x^2 \sum_{\text{cyc}} bc -x \sum_{\text{cyc}}bcd + bcde$. By using the above system of equations, we can eventually simplify it down and see that $f_a(a) = a^4 + (7a^3 + a^4)4$. Now there are a ton of options for what we can do, but now the sum we desire to find is given by $$\sum_{\text{cyc}} \frac{a^4}{a^4 + (7a^3 + a^4)4}.$$
Using that $a^5 + 7a^4 - 2 = 0$, we can simplify further to find that $f_a(a)a = 10 - \frac{14}{a+7}$ and $\frac{2a}{a+7} = a^5$ so $$\sum_{\text{cyc}} \frac{a^4}{a^4 + (7a^3 + a^4)4} \cong \sum_{\text{cyc}} \frac{a}{5a+28}.$$
Now this should be simpler, but after thinking about it I don't really see a good way to proceed. What can I do? Did I drive into the weeds, or am I still on the misty road and I just don't see the end?
| Hint: Turns out Vieta isn't needed. Irrespective of which polynomial they are roots of, as long as $a,b,c,d,e$ are distinct,
$$\sum_{cyc} \frac{a^4}{(a-b)(a-c)(a-d)(a-e)}=1$$
Will expand on the hint in case you can't complete the proof...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{2^{ab}-1}{2^a-1} $ is composite number let $b>a>1$ be postive integers,show that
$$\dfrac{2^{ab}-1}{2^a-1} $$ is composite number
I try use $$(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)$$
so $x-1|x^n-1$.then let $x=2^a,n=b$ we have
$$\dfrac{2^{ab}-1}{2^a-1}=2^{(b-1)a}+2^{(b-2)a}+\cdots+2^a+1$$
but How to prove $2^{(b-1)a}+2^{(b-2)a}+\cdots+2^a+1$ is composite ？
| You also need $a >1$.
There is another less direct way to do this. Let us write $A=2^a-1$, $B=2^b-1$, and $C=2^{ab}-1$. Then both $A$ and $B$ divide $C$ so write $C=BD$ for some integer $D$. Also, the inequality $a <b$ gives the inequality $A<B$. Now, $a \ge 2$ also gives
$$B^2 < 2^{2b} -1$$ $$\le 2^{ab}-1 = C,$$ so in particular $B^2< C$, so $D$ must satisfy $D >B$ and [as $A < B$] in particular $A < D$.
Now suppose $p=\frac{C}{A}=\frac{BD}{A}$ for some prime $p$. We show next that this is impossible. First note that for $\frac{C}{A}=\frac{BD}{A}=p$ with $p$ prime, to be true, the prime $p$ must divide either $B$ or $D$.
Case 1: If $p$ divides $B$ then $B \ge p$ and as $D \ge B > A$, it follows then that $\frac{D}{A}>1$, and thus:
$$\frac{C}{A} = \frac{BD}{A} \ge \left(p × \frac{D}{A}\right) > p,$$ and in particular $\frac{C}{A} > p$, so we arrive at a contradiction.
Case 2: If $p$ divides $D$ then $D \ge p$, and as $B > A$, it follows then that $\frac{B}{A} > 1$, and thus:
$$\frac{C}{A} = \frac{BD}{A} \ge p ×\frac{B}{A} > p,$$ and in particular $\frac{C}{A} > p$, so we arrive here at a contradiction as well.
Note that this breaks down if $a=b$ [Case 2 in particular], or if $a=1$ [in which case $A=1$ and $D=1$]. Indeed, try $a=3$, $b=3$, and then try $a=1$, $b=5$. So you do indeed need $a<b$ and $a >1$.
You could do this another way. Use $C=BD$, with $A<\min\{B,D\}$, as established above. Then that $\frac{C}{A}=\frac{BD}{A}$ is an integer means $A$ can be factored $A=A_1A_2$ with $A_1|B$ and $A_2|D$ [one of $A_1,A_2$ may be $1$ and the other $A$]. Then $B$ can be written $B=A_1B'$, and furthermore, since $B>A \ge A_1$, it follows that $B' > 1$. Likewise, $D$ can be written $D=A_2D'$, and since $D>A \ge A_2$, it follows that $D'>1$.
Then:
$$\frac{C}{A}=\frac{BD}{A}$$ $$= \frac{(A_1B')(A_2D')}{A_1A_2} = B'D'.$$
As $B'$ and $D'$ are both integers greater than $1$ gives $B'D'$ composite, and thus, $\frac{C}{A} =$ $\frac{BD}{A}= $ $B'D'$ composite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\int_{0}^{2\pi}f(x)\cos(kx)dx \geq 0$ for every $k \geq 1$ given that $f$ is convex. Given $f: [0, 2\pi] \to \mathbb{R}$ convex function, prove that for every $k\geq1$
\begin{align}
\int_{0}^{2\pi}f(x) \cos (kx)dx \geq 0
\end{align}
I am completely stumped. What I have tried to do is return the query for $k=1$, and for that value of $k$ try to write the integral from $0$ to $2\pi$ as a sum of four integrals from $0$ to $\dfrac{\pi}{2}$ and use the the theorem for first derivative monotony. No luck so far.
Any help would be much appreciated.
Edit 1: I saw the link here about a similarly asked topic. However, this process gets the general case as I perceive it and I am really supposed to use the method described above. I will try this and come back with a definitive answer.
Edit 2: I cannot use the $f''(x) \geq 0$ argument due to the simple fact that I have not been formally taught it as part of the class.
Edit 3: Final proof, with thanks to the contributors below.
Let's start by setting $A = \frac{\pi}{2}$ and $B = \frac{3\pi}{2}$. It follows from basic trigonometry that:
\begin{align}
&\cos x \geq 0, \ x \in [A,B] \ \text{and}\\
&\cos x \leq 0, \ x \in [0, A] \cap [B, 2\pi].
\end{align}
And we also set $L$ to be the line segment such that $L(A) = f(A), \ L(B) = f(B)$. We will prove a basic property of said line in regards to the convex function $f$.
*
*I can take for granted that (we proved this in class)
\begin{align}
L(x) = \dfrac{x-A}{B-A}f(B) + \dfrac{B-x}{B-A}f(A)
\end{align}
so for $x \in [A,B]$ there exists $\lambda \in [0,1]$ such that: $x = \lambda A + (1-\lambda) B$. Taking the aforementioned expression and replacing it on $L(x)$ we get (I omit trivial algebra)
\begin{align}
L(x) = (1-\lambda) f(B) + \lambda f(A).
\end{align}
Since $f$ is convex, we can write
\begin{align}
&f(\lambda A + (1-\lambda) B) \leq \lambda f(A) + (1-\lambda) f(B)\\
\implies &f(x) \leq L(x), \ \forall \ x \in [A,B] \ \text{and} \ \lambda \in [0,1].
\end{align} $\blacksquare$
*We assume that there exists $x \in [0,A]: \ f(x) < L(x)$. Then there exists $A \in [x, B] \ \text{and} \ \lambda \in [0,1]: \ A = \lambda x + (1-\lambda) B$. Then
\begin{align}
&L(A) = \dfrac{A-x}{B-x}f(B) + \dfrac{B-A}{B-x}f(x), \ A \in [x,B]\\
\implies &L(A) = (1-\lambda)L(B)+\lambda L(x)\\
\implies &L(A) = (1-\lambda)f(B) + \lambda L(x).
\end{align}
Then assuming that $L(x) > f(x)$ we get
\begin{align}
L(A) > (1-\lambda) f(B) + \lambda f(x) \geq f(A), \ \text{assuming convexity}.
\end{align}
Because $L(A) = f(A)$ the above inequality becomes $L(A) > f(A)$ which is a contradiction.
$\blacksquare$
In the same spirit, for $x \in [B, 2\pi]$ doing the exact same replacements and applying the exact same principles we also get
\begin{align}
&f(\lambda B + (1-\lambda) 2 \pi) \leq \lambda f(B) + (1-\lambda) f(2\pi)\\
\implies &f(x) \leq L(x).
\end{align}
$\blacksquare$
We then set $g(x) = f(x) - L(x)$ and from (1) and (2) above it is safe to assume that $\cos x$ and $g(x)$ will have the same sign in the whole domain, that is:
\begin{align}
&g(x) \geq 0, \ \text{where} \ \cos x \geq 0\\
&g(x) \leq 0, \ \text{where} \ \cos x \geq 0.
\end{align}
*We have
\begin{align}
\int_0^{2\pi} g(x) \cos x dx = \int_0^{\frac{\pi}{2}} g(x) \cos x dx + \int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} g(x) \cos x dx + \int_{\frac{3\pi}{2}}^{2\pi}g(x) \cos x dx \geq 0,
\end{align}
because
\begin{align}
&\text{in} \left[0, \frac{\pi}{2} \right], \ \cos x \geq 0 \implies g(x) \geq 0\\
&\text{in} \left[\frac{\pi}{2}, \frac{3 \pi}{2} \right], \ \cos x \leq 0 \implies g(x) \leq 0\\
&\text{in} \left[\frac{3 \pi}{2}, 2 \pi \right], \ \cos x \geq 0 \implies g(x) \geq 0.\\
\end{align}
*We have that
\begin{align}
&\int_0^{2 \pi}\cos x dx = 0 \ \text{trivial}\\
&\int_0^{2 \pi}x \cos x dx = \int_0^{2 \pi}x (\sin x)' dx = \left[ x \sin x \right]_0^{2\pi} - \int_0^{2 \pi} \sin x dx = 0.
\end{align}
It then follows that
\begin{align}
\int_0^{2 \pi} f(x) \cos x dx = \int_0^{2 \pi} g(x) \cos x dx \geq 0
\end{align}
which was previously proven. For the case $k=1$, the proof is over.
For $k>1$, we have:
\begin{align}
\int_0^{2\pi}f(x) \cos (kx) dx = \sum_{i=0}^{k-1} \int_{\frac{2\pi i}{k}}^{\frac{2\pi (i+1)}{k}} f(x) \cos (kx) dx.
\end{align}
We perform the change of variable
\begin{align}
x = \dfrac{y+2\pi i}{k}\\
\implies \begin{cases}dx = \dfrac{1}{k}dy\\ x = \dfrac{2\pi i}{k} \to y=0\\ x = \dfrac{2\pi (i+1)}{k} \to y = 2\pi \end{cases}.
\end{align}
So the above sum becomes
\begin{align}
\sum_{i=0}^{k-1} \dfrac{1}{k} \int_{0}^{2\pi}f \left(\dfrac{y+2 \pi i}{k} \right) \cos (y+2\pi i)dy.
\end{align}
Because $f$ is convex in $[0, 2\pi]$ there exists $\lambda \in [0,1]$ such that:
\begin{align}
\theta f\left(\frac{y_1 + 2\pi i}{k}\right)
+ (1 - \theta)f\left(\frac{y_2 + 2\pi i}{k}\right)
&\ge f\left(\theta\frac{y_1 + 2\pi i}{k}
+ (1 - \theta) \frac{y_2 + 2\pi i}{k}\right)\\
&= f\left(\frac{\theta y_1 + (1 - \theta)y_2 + 2\pi i}{k}\right).
\end{align}
Having performed the change of variables:
\begin{align}
x_1 = \dfrac{y_1 + 2\pi i}{k} \ \text{and} \ x_2 = \dfrac{y_2 + 2\pi i}{k}.
\end{align}
So $f \left( \dfrac{y + 2\pi i}{k}\right)$ convex on $[0, 2\pi]$. Using the result from $k=1$ we have
\begin{align}
\int_0^{2\pi} f \left( \dfrac{y + 2\pi i}{k} \right)\cos y dy \geq 0.
\end{align}
$\blacksquare$
| As demonstrated in the other answers, it suffices to prove the statement for $k=1$, that is
$$
I = \int_{0}^{2\pi}f(x) \cos (x) \, dx \ge 0 \, .
$$
We can split the integral into four parts over the intervals $[0, \pi/2]$, $\pi/2, \pi]$, $[\pi , 3\pi/2]$ and $[3\pi/2, 2 \pi|$ and substitute $x = \pi - y$, $x=\pi + y$, $x =2\pi -y$ in the second, third, and fourth integral, respectively. This gives
$$
I = \int_0^{\pi/2} \bigl( f(x) - f(\pi -x) - f(\pi + x) + f(2\pi - x)\bigr) \cos(x) \, dx \, .
$$
So order to prove $I \ge 0$ it suffices to show that
$$ \tag{$*$}
f(\pi -x) + f(\pi + x) \le f(x) + f(2\pi - x)
$$
for $0 \le x \le \pi/2$.
And that follows directly from the convexity condition, applied to $x < \pi - x< 2 \pi-x$:
$$
f(\pi - x) \le \frac{\pi}{2 \pi - 2x}f(x) + \frac{\pi - 2x}{2 \pi - 2x}f(2\pi - x) \, ,
$$
and to $x < \pi + x < 2 \pi-x$:
$$
f(\pi + x) \le \frac{\pi-2x}{2 \pi - 2x}f(x) + \frac{\pi}{2 \pi - 2x}f(2\pi - x) \, .
$$
Adding these two inequalities gives exactly $(*)$, and that finishes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Probability of spelling IDAHO correctly I'm working with a problem stated as follows:
A sign reads IDAHO. Two letters are removed and put back at random, each
equally likely to be put upside down as in the correct orientation. What is the
probability that the sign still reads IDAHO?
So, we form our events as:
*
*$A = \{$IDAHO is spelt correctly$\}$
*$B = \{$2 "symmetric" characters are chosen$\}$
*$C = \{$1 "symmetric" and 1 "unsymmetric" character is chosen$\}$
*$D = \{$2 "unsymmetric" characters are chosen$\}$
Where symmetric means that no matter of orientation, the character looks just like it used to before.
Notice, $B,C,D$ form a partition of our sample space. Hence, the the total law of probability gives us:
$$P(A) = P(A|B)P(B) + P(A|C)P(C) + P(A|D)P(D) $$
We trivially know that $P(A|B) = 1/2$, since in choosing two symmetric letters.
Furthermore, $P(B) = \frac{\begin{pmatrix}
3 \\2
\end{pmatrix}}{\begin{pmatrix}
5 \\2
\end{pmatrix}} = 3/10$
Then, we have $P(A|C) = 2/4$, since in choosing 1 symmetric letter and 1 unsymmetric, we have to get the correct position on the unsymmetric letter, whereas the symmetric letter can be switched in orientation without changing the word. Meaning we have 2 out of a total 4 possible scenarios. 4 comes from the fact that we can choose to place the symmetric letter in 2 positions, and orient it in 2 ways, however, these are equivalent. Then, the latter unsymmetric letter has already been given a position, but can be oriented in 2 distinct ways.
Furthermore, $P(C) = \frac{\begin{pmatrix}
3 \\2
\end{pmatrix}\begin{pmatrix}
2 \\1
\end{pmatrix}}{\begin{pmatrix}
5 \\2
\end{pmatrix}} = 3/5$
Lastly, $P(A|D) = 1/6$, since we only have one scenario in which it can become the same word, and the total scenarios are given as we choose a position for the first letter (2), then we choose orientation (2), the latter letter has already been given a position and orientation can be choosen (2).
Also $P(D) = \frac{\begin{pmatrix}
2 \\
2\end{pmatrix}}{\begin{pmatrix}
5 \\2
\end{pmatrix}} = 1/10$
Finally, we have $P(A) = 1/2 \cdot 3/10 + 1/2 \cdot 3/5 + 1/6 \cdot 1/10 = 7/15$, which according to the answer sheet is wrong ($5/16$).
Does anyone of you guys see the mistake. I've tried to find it and gone through my argumentation but can't find the mistake.
Thanks.
| This answer identifies the errors in the question’s reasoning, mainly by restating the advice in Aaron Montgomery’s answer and Acccumulation’s comment, then uses all of Acccumulation’s advice to recalculate $P(A)$.
$P(A|B)$ is indeed $\frac{1}{2}$, but let us write it as $\frac{4}{8}$ instead, for consistency with the other probabilities and to follow Acccumulation’s advice.
For $P(A|C)$ and $P(A|D)$, the numbers of favourable outcomes are indeed $2$ and $1$ respectively, but the total numbers of outcomes are incorrect. By your own reasoning, they are both $2 \cdot 2 \cdot 2 = 8$.
One small error was not mentioned by the other users: the expression for $P(C)$ should say $\begin{pmatrix} 3 \\ 1 \end{pmatrix}$ instead of $\begin{pmatrix} 3 \\ 2 \end{pmatrix}$; it so happens that these have the same value.
\begin{align}
P(A) &= P(A|B)P(B) + P(A|C)P(C) + P(A|D)P(D) \\
&= \frac{4}{8} \cdot \frac{\begin{pmatrix} 3 \\ 2 \end{pmatrix}}{\begin{pmatrix} 5 \\ 2 \end{pmatrix}} + \frac{2}{8} \cdot \frac{\begin{pmatrix} 3 \\ 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix}}{\begin{pmatrix} 5 \\ 2 \end{pmatrix}} + \frac{1}{8} \cdot \frac{\begin{pmatrix} 2 \\ 2 \end{pmatrix}}{\begin{pmatrix} 5 \\ 2 \end{pmatrix}} \\
&= \frac{4 \begin{pmatrix} 3 \\ 2 \end{pmatrix} + 2 \begin{pmatrix} 3 \\ 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \end{pmatrix} + \begin{pmatrix} 2 \\ 2 \end{pmatrix}}{8 \begin{pmatrix} 5 \\ 2 \end{pmatrix}} \\
&= \frac{4 \cdot 3 + 2 \cdot 6 + 1}{8 \begin{pmatrix} 5 \\ 2 \end{pmatrix}} \\
&= \frac{12 + 12 + 1}{8 \begin{pmatrix} 5 \\ 2 \end{pmatrix}} \\
&= \frac{25}{80} \\
&= \frac{5}{16}. \\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Compute $\sum_{k=0}^{n}{\frac{\binom{n}{k}}{k+1}}$ Compute the sum:
$$\sum_{k=0}^{n}{\frac{\binom{n}{k}}{k+1}}$$
I tried to solve it:
$$\sum_{k=0}^{n}{\frac{(k+1)(k+2)\binom{n+2}{k+2}}{(n+1)(n+2)(k+2)}}=\sum_{k=0}^{n}{\frac{(k+1)\binom{n+2}{k+2}}{(n+1)(n+2)}}=\frac{1}{(n+1)(n+2)}\sum_{k=0}^{n}(k+1)\binom{n+2}{k+2}$$
But from this point I'm stuck.
Thank you!
| You should start by showing that:
$$ \frac{n\choose k}{k+1} = \frac{1}{n+1} {n+1\choose k+1} $$
Then notice you're summing over $k$, so you're free to take this $\frac{1}{n+1}$ out of the sum as a constant factor:
$$ \sum_{k=0}^n \frac{n\choose k}{k+1} = \frac{1}{n+1}\sum_{k=0}^n {n+1\choose k+1} $$
Finally,
$$ \sum_{k=0}^n {n+1\choose k+1} = \left[\sum_{k=0}^{n+1} {n+1\choose k+1}\right] - {n+1\choose n+1}$$
and you should be able to evaluate these two terms, by standard results about binomial coefficients.
1. The first term is the sum over row $(n+1)$ of Pascal's triangle, which gives $2^{n+1}$.
2. The second term is just $1$.
So the final answer is: $$ \sum_{k=0}^n \frac{n\choose k}{k+1} = \frac{2^{n+1} - 1}{n+1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4417043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Cuboid with inner and outer Cubes A cuboid with integer side lengths consists of cubes with side length 1. Such a cube is called an inner cube if none of its sides are visible from the outside, otherwise, i.e. if at least one side is visible from the outside, the cube is called an outer cube. The $3 \times 4 \times 7$ cuboid shown consists of 10 inner and 74 outer cubes.
For which a is there an $a \times b \times c$ cuboid with $a \le b$ and $a \le c$, with the same number of inner and outer cubes?
I have:
Number inner cubes: $(a-2)(b-2)(c-2)$
Number of outer cubes: $abc - (a-2)(b-2)(c-2)$
$(a-2)(b-2)(c-2) = abc - (a-2)(b-2)(c-2)$
$a=4+(8c+8b-16)/(bc-4c-4b+8)$
How can I determinate $a, b$ and $c$???
| We will show that the set of all solutions are finite (solutions with length greater than $2$), and we give some bounds to find all of the solutions.
Without lose of generality we can assume $2 < a \leq b \leq c$ (Note that if $a=1, 2$ then every cube is a visible cube). We will show that there are only finitely many natural numbers that are satisfying $2(a-2)(b-2)(c-2)=abc$, with $2 < a \leq b \leq c$.
Lemma (1): If $10 \leq d$, then $d < \sqrt[3]2(d-2)$.
(Hint: Prove that the statement holds for $d=10$, and then, by induction, prove that if the statement holds for $d=r$, then it holds for $d=r+1$.)
Lemma (2): Let $2 < a \leq b \leq c$ be integers that are satisfying $2(a-2)(b-2)(c-2)=abc$, then $a\leq 9$.
Proof: Suppose on contrary that $10 \leq a \leq b \leq c$, then Lemma (1) implies that
$$a.b.c < \sqrt[3]2(a-2).\sqrt[3]2(b-2).\sqrt[3]2(c-2)=2(a-2)(b-2)(c-2)=abc,$$
which is an obvious contradiction.
So, we'e proved that $a\in \{3, 4, \cdots, 9 \}$.
In the following we will consider two separated cases. The argument for the First case is due to the answer by "Jean Marie":
First case ($a=3 , 4$): In this case we have $2(a-2)\leq a$, and also clearly we have $b-2 < b$ and $c-2 < c$. Since all of them are positive, we can conclude that
$$2(a-2)(b-2)(c-2) < (a)(b)(c),$$
So we can conclude that there are no solution for $a=3, 4$.
Second case ($a=5, 6, 7, 8, 9$): In this case we will give some bounds for $b$, to show that there are only finitly many solutions.
Remark: Let's (ignore $t=3, 4$ and) fix some $t\in \{5, 6, \cdots, 9 \}$, and suppose that $a=t$, and let $\lambda_t=\frac{2(t-2)}{t}$ (You can easily verify that $1 < \lambda_t$.) Now consider these simple manipulation (we just set $a=t$!):
$$2(a-2)(b-2)(c-2)=abc \Leftrightarrow 2(t-2)(b-2)(c-2)=tbc \Leftrightarrow \lambda_t(b-2)(c-2)=bc,$$
so we are going to deal with equation $\lambda_t(b-2)(c-2)=bc$. [end of Remark]
Lemma (3): If $\frac{4\lambda_t}{\lambda_t -1} \leq d$, then $d < \sqrt[2]\lambda_t(d-2)$.
Proof: First verify that $\frac{4\lambda_t}{\lambda_t -1} \leq d$
implies $4 <\frac{4\lambda_t}{\lambda_t -1} \leq d$, so both of $d$ and $d-2$ are positive numbers (to obtain this we used $1 < \lambda_t$). Now consider the following:
$$\frac{4\lambda_t}{\lambda_t -1} \leq d \Rightarrow
4\lambda_t d \leq (\lambda_t -1) d^2 \Rightarrow
4\lambda_t d < (\lambda_t -1) d^2 +4\lambda_t \Rightarrow
$$
$$
\Rightarrow
d^2 < \lambda_t d^2 - 4\lambda_t d +4\lambda_t \Rightarrow
d^2 < \lambda_t (d-2)^2 \Rightarrow d < \sqrt[2]\lambda_t(d-2),
$$
and we are done.
Lemma (4): Let $4 < t=a \leq b \leq c$ be integers that are satisfying $2(a-2)(b-2)(c-2)=abc$, then $t \leq b< \frac{4\lambda_t}{\lambda_t -1}$.
Proof: To prove it, first we will consider the above remark, so we can consider the equation $\lambda_t(b-2)(c-2)=bc$, and we will switch into this equation; i.e we are dealing with solutions of this last equation satisfying $4 < t \leq b \leq c$, and we will prove that $ b < \frac{4\lambda_t}{\lambda_t -1}$.
Suppose on contrary that $\frac{4\lambda_t}{\lambda_t -1} \leq b \leq c$, then Lemma (3) implies that
$$b.c < \sqrt[2]\lambda_t(b-2) . \sqrt[2]\lambda_t(c-2)=\lambda_t (b-2)(c-2)=bc,$$
which is an obvious contradiction.
So, we'e proved that $ b < \frac{4\lambda_t}{\lambda_t -1}$, and we are done.
So we are dealing with finite sets of paired integers $(a, b)$ (with $2<a$), and if we suppose that $a$ and $b$ are given constants, then the equation turns into a linear equation containing just one variable $c$. Also, the reader can check this method works for higher dimentional cubes.
Also, note that the bounds in my solution are not sharp, but they are so small, and perhaps one can find all solution just by hand-based-calculation in less than 1 hour.
Algorithm:
(I): Fix some $t \in \{ 5, 6, \cdots, 9\}$, and set $a=t$.
(II): Now fix some $t \leq s < 4 + \frac{4t}{t-4}=8+\frac{16}{t-4}$, and set $b=s$.
(III): Now consider the equation $2(a-2)(b-2)(c-2)=abc$, and replace $a=t$ and $b=s$, and solve for the variable $c$, which would be equal $c=\frac{4(t-2)(s-2)}{2(t-2)(s-2)-ts}=2+\frac{2ts}{2(t-2)(s-2)-ts}$. If $c=2+\frac{2ts}{2(t-2)(s-2)-ts}$ is a natural number, then $(a, b, c)$ is a solution, otherwise it is not a solution.
Remark: Considering the bounds given in parts (I) and (II) in the algorithm, we have just to consider the following pairs:
*
*$t=5$: In this case $5 \leq s < 24$, and there are at most $19$ potential answers with $(a, b)=(t, s)$.
*$t=6$: In this case $6 \leq s < 16$, and there are at most $11$ potential answers with $(a, b)=(t, s)$.
*$t=7$: In this case $7 \leq s < 13+\frac{1}{3}$, and there are at most $7$ potential answers with $(a, b)=(t, s)$.
*$t=8$: In this case $8 \leq s < 12$, and there are at most $4$ potential answers with $(a, b)=(t, s)$.
*$t=9$: In this case $9 \leq s < 11+\frac{1}{5}$, and there are at most $3$ potential answers with $(a, b)=(t, s)$ (There are not any pairs in this case).
I think all the answers are listed in the answer by "Michael Hoppe" (I didn't verify all of the possibilites and also his answer).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Fractions in Questions and Answers
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