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To evaluate the given determinant Question: Evaluate the determinant
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|$
My answer:
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|= \left|
\begin{array}{cc} b^2c^2 & bc & c \\
c^2a^2 & ca & a \\
a^2b^2 & ab & b \\
\end{array}
\right| + \left|
\begin{array}{cc} b^2c^2 & bc & b \\
c^2a^2 & ca & c \\
a^2b^2 & ab & a \\
\end{array}
\right|= abc \left|
\begin{array}{cc} bc^2 & c & 1 \\
ca^2 & a & 1 \\
ab^2 & b & 1 \\
\end{array}
\right| +abc \left|
\begin{array}{cc} b^2c & b & 1 \\
c^2a & c & 1 \\
a^2b & a & 1 \\
\end{array}
\right|$
how do I proceed from here?
| $F=\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|$
$=\dfrac1{abc}\left|
\begin{array}{cc} ab^2c^2 & abc & a(b+c) \\
c^2a^2b & bca & b(c+a) \\
a^2b^2c & abc & c(a+b) \\
\end{array}
\right|$
$=\left|
\begin{array}{cc} ab^2c^2 &1& a(b+c) \\
c^2a^2b &1& b(c+a) \\
a^2b^2c &1& c(a+b) \\
\end{array}
\right|$
$R_3'=R_3-R_1,R_2'=R_2-R_1$
$F=\left|
\begin{array}{cc} ab^2c^2 &1& a(b+c) \\
abc^2(a-b) &0& -a(a-b) \\
-ab^2c(c-a) &0& b(c-a) \\
\end{array}
\right|$
$=(a-b)(c-a)\left|
\begin{array}{cc} ab^2c^2 &1& a(b+c) \\
abc^2 &0& -a \\
-ab^2c &0& b \\
\end{array}
\right|$
$=(a-b)(c-a)(-1)^{1+2}\cdot\left|
\begin{array}{cc} abc^2 & -a \\
-ab^2c & b \\
\end{array}
\right|$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Expanding an expression in a certain field If $\mathbb F_2$ is a field of characteristic $2$, then we have $x+x=y+y=z+z=0$ for all $x,y,z \in \mathbb F_2$. When I expand $(x+y)(y+z)(z+x)$, I get
\begin{align}
(x+y)(y+z)(z+x) &= xz^2+y^2z+yz^2+x^2y+x^2z+xy^2+2xyz \\
&= xz^2+y^2z+yz^2+x^2y+x^2z+xy^2+(x+x)yz \\
&= xz^2+y^2z+yz^2+x^2y+x^2z+xy^2.
\end{align}
I got $2xyz=0$ in $\mathbb F_2$, but how can I simplify my above expression even more?
| In $\mathbb{F}_2$, we also have $x^2=x$ and so on. So your expression will result in $0$. Another way to look at it is that at least two of $x,y$ or $z$ will take the same value, in which case at least one of $x+y, y+z, x+z$ will be $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An integral arising from Kepler's problem $\frac{1}{2\pi}\int_0^{2\pi}\ \frac{\sin^2\theta\ \text{d}\theta}{(1 + \epsilon\cos\theta)^2}$ I'm dealing with this integral in my spare time, since days and days, and it's really interesting. I'll provide to write what I tried until now, and I would really appreciate some help in understanding how to continue.
$$\Phi(\epsilon) = \frac{1}{2\pi}\int_0^{2\pi}\ \frac{\sin^2\theta\ \text{d}\theta}{(1 + \epsilon\cos\theta)^2}$$
Note: The integral arises in studying the Kepler's problem, which is why $\epsilon\in [0;\ 1]$.
I started with the passages to the complex plane:
$$\sin\theta = \frac{1}{2i}(z - z^{-1}) ~~~~~~~ \cos\theta = \frac{1}{2}(z + z^{-1}) ~~~~~~~ \text{d}\theta = \frac{\text{d}z}{iz}$$
thence
$$\begin{align*}
\Phi(\epsilon) & = \frac{1}{2\pi}\oint_{|z| = 1}\ \frac{-\frac{1}{4}(z - z^{-1})^2\ \text{d}z}{iz\left[1 + \epsilon\frac{(z + z^{-1}}{2}\right]^2}
\\\\
& = -\frac{1}{8\pi i}\oint_{|z| = 1}\ \frac{(z - z^{-1})^2\ \text{d}z}{z\left[\frac{2 + \epsilon(z + z^{-1})}{2}\right]^2}
\\\\
& = -\frac{1}{4\pi i}\oint_{|z| = 1}\ \frac{(z - z^{-1})^2\ \text{d}z}{z(2 + \epsilon(z + z^{-1})^2}
\\\\
& = -\frac{1}{4\pi i}\oint_{|z| = 1}\ \frac{\left(\frac{z^2 - 1}{z}\right)^2\ \text{d}z}{z\left(2 + \epsilon\frac{z^2+1}{z}\right)^2}
\\\\
& = -\frac{1}{4\pi i}\oint_{|z| = 1}\ \frac{(z^2-1)^2\ \text{d}z}{z(2z + \epsilon(z^2+1))^2}
\end{align*}$$
Poles et cetera
Poles are at $z_0 = 0$ and when $2z + \epsilon(z^2+1) = 0$, namely
$$z_1 = \frac{1+\sqrt{1 - \epsilon^2}}{\epsilon} ~~~~~~~~~~~ z_2 = \frac{1-\sqrt{1 - \epsilon^2}}{\epsilon}$$
Checking that $z_1\cdot z_2 = 1$ is ok, the question is now:
Question 1.: which one, aside $z_0$ lies into the unitary circle?
Question 2.: How to proceed?
| I am assuming $0 < \epsilon < 1$ in the following. The cases
$\epsilon = 0$ and $\epsilon = 1$ have to be handled separately,
see below.
There is a small error in your calculation, the root of
$2z + \epsilon(z^2+1) = 0$ are
$$
z_1 = \frac{-1-\sqrt{1 - \epsilon^2}}{\epsilon} \, , \quad z_2 = \frac{-1+\sqrt{1 - \epsilon^2}}{\epsilon}
$$
Question 1.: which one, aside $z_0$ lies into the unitary circle?
From
$$
1+\sqrt{1 - \epsilon^2} > 1 > \epsilon
$$
it follows that $z_1 < -1$ and consequently, $-1 < z_2 < 0$, i.e. $z_2$
is inside the unit disk and $z_1$ outside.
Question 2.: How to proceed?
You already have that
$$
\Phi(\epsilon) = -\frac{1}{4\pi i}\oint_{|z| = 1} \frac{(z^2-1)^2\ }{z(2z + \epsilon(z^2+1))^2} \, dz =
\frac{1}{2\pi i} \oint_{|z| = 1} f(z) \, dz
$$
with
$$
f(z) := -\frac{1}{2 \epsilon^2} \frac{(z^2-1)^2\ }{z(z-z_1)^2(z-z_2)^2}
$$
From the residue theorem it follows that
$$
\Phi(\epsilon) = \text{Res}(f, 0) + \text{Res}(f, z_2) \, .
$$
$f$ has a simple pole at $z = 0$, therefore
$$
\text{Res}(f, 0) = \lim_{z \to 0} z f(z) = -\frac{1}{2 \epsilon^2} \, .
$$
$f$ has a double pole at $z = z_2$.
One possible method to compute the residue is the
limit formula for higher order poles:
$$
\text{Res}(f, z_2) = \lim_{z \to z_2} \frac{d}{dz} \bigl((z-z_2)^2 f(z) \bigr) \, .
$$
Special cases: For $\epsilon = 0$ the integral becomes
$$
\Phi(0) = -\frac{1}{4\pi i}\oint_{|z| = 1} \frac{(z^2-1)^2}{2z^3} \, dz
= -\frac{1}{8\pi i}\oint_{|z| = 1} \bigl( z - \frac 2z + \frac{1}{z^3}\bigr) \, dz
$$
which can easily be computed using the residue theorem.
Alternatively,
$$
\Phi(0) = \frac{1}{2\pi}\int_0^{2\pi} \sin^2\theta \, d\theta
$$
can be computed directly.
For $\epsilon = 1$, the integral becomes
$$
\Phi(1) = -\frac{1}{4\pi i}\oint_{|z| = 1} \frac{(z-1)^2}{z(z+1)^2} \, dz
$$
which is infinite due to the singularity at $z=-1$. Alternatively,
$$
\Phi(1) = \frac{1}{2\pi}\int_0^{2\pi} \frac{\sin^2\theta}{(1 + \cos\theta)^2} \, d\theta
= \frac{1}{2\pi}\int_0^{2\pi} \frac{(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2})^2 }{(2 \cos^2 \frac{\theta}{2})^2} \, d\theta
= \frac{1}{2\pi}\int_0^{2\pi} \tan^2 \frac{\theta}{2}\, d\theta
$$
is infinite.
| {
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"url": "https://math.stackexchange.com/questions/1625048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculate this integral $\int \frac{x^2}{4x^4+25}dx$. I have to calculate this integral $\int \frac{x^2}{4x^4+25}dx$.
I dont have any idea about that.
I thought about parts integration.
Thanks.
| Hint:
$$\int \frac { x^{ 2 } }{ 4x^{ 4 }+25 } dx=\int { \frac { x^{ 2 } }{ 4x^{ 4 }-20{ x }^{ 2 }+25-20{ x }^{ 2 } } dx } =\int { \frac { { x }^{ 2 } }{ { \left( 2{ x }^{ 2 }-5 \right) }^{ 2 }-20{ x }^{ 2 } } dx } =\int { \frac { x^{ 2 }dx }{ \left( 2x^{ 2 }-2\sqrt { 5 } x-5 \right) \left( 2x^{ 2 }+2\sqrt { 5 } x-5 \right) } } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$? How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$ ?
$A = \left.\frac{s^2-s}{s^2+1} \right\vert_{s=-1} = \frac{1-(-1)}{1+1}=1$
| Notice, $$\frac{s-1}{s+1}\frac{s}{s^2+1}=\frac{A}{s+1}+\frac{Bs+C}{s^2+1}$$
$$s^2-s=(A+B)s^2+(B+C)s+(A+C)$$
comparing the corresponding coefficients on both the sides, one should get $$A+B=1\tag 1$$
$$B+C=-1\tag 2$$
$$A+C=0\tag 3$$
on solving (1), (2) & (3), one can easily get $\color{red}{A=1}, \color{blue}{B=0}, \color{red}{C=-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find the antiderivative of this function $\frac{1}{1+x^4}$? How to find the antiderivative of the function $f(x)=\frac{1}{1+x^4}$?
My math professor suggested to use the method of partial fractions, but it doesn't seem to work, because the denominator cannot be factored at all.
Attempting to integrate with other familiar methods (integration by parts, trigonometric substitution) didn't work out as well.
Does anyone have an idea what is the best approach to this problem?
| It's easy to see factorization in the first line.
\begin{align*}
\int \frac{1}{1+x^{4}}\mathrm{d}x &=\int \left ( \frac{\dfrac{\sqrt{2}}{4}x+\dfrac{1}{2}}{x^{2}+\sqrt{2}x+1}+\frac{-\dfrac{\sqrt{2}}{4}x+\dfrac{1}{2}}{x^{2}-\sqrt{2}x+1} \right )\mathrm{d}x\\
&=\int \left [ \frac{\dfrac{\sqrt{2}}{4}\left ( x+\dfrac{\sqrt{2}}{2} \right )+\dfrac{1}{4}}{\left ( x+\dfrac{\sqrt{2}}{2} \right )^{2}+\dfrac{1}{2}}+\frac{-\dfrac{\sqrt{2}}{2}\left ( x-\dfrac{\sqrt{2}}{2} \right )+\dfrac{1}{4}}{\left ( x-\dfrac{\sqrt{2}}{2} \right )^{2}+\dfrac{1}{2}} \right ]\mathrm{d}x \\
&=\frac{\sqrt{2}}{8}\ln\left | \frac{x^{2}+\sqrt{2}x+1}{x^{2}-\sqrt{2}x+1} \right |+\frac{\sqrt{2}}{4}\arctan\left ( \sqrt{2}x+1 \right )+\frac{\sqrt{2}}{4}\arctan\left ( \sqrt{2}x-1 \right )+C.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$ I saw some resolutions here like $\sqrt{x+\sqrt{x+\sqrt{x}}}- \sqrt{x}$, but I couldn't get the point to find
$\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.
I tried $\frac{1}{x}.(\sqrt{x+\sqrt{x+\sqrt{x}}})=\frac{\sqrt{x}}{x}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)=\frac{1}{\sqrt{x}}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)$ but now I got stuck. Could anyone help?
| For large $x$ you have $\sqrt x<x$ and so
$$
\sqrt{x+\sqrt{x+\sqrt{x}}}<\sqrt{x+\sqrt{2x}}<\sqrt{3x}.
$$
Since $\sqrt{x+\sqrt{x+\sqrt{x}}}/x<\sqrt{3x}/x\to0$ when $x\to\infty$, your limit is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$ If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$
Using the condition for common root, $$(3c-5b)(b-3a)=(c-5a)^2$$
$$3bc-9ac-5b^2+15ab=c^2+25a^2-10ac$$
$$25a^2+5b^2+c^2=15ab+3bc+ac$$
There is pattern in the equation. The coefficients of the middle terms and first terms of both sides are $(5\times 5,5)$ and $(5\times 3,3)$.
I tried to use Lagrange multipliers. But isnt there any simpler way to minimize $a+b+c$?
| The polynomial $x^2+3x+5$ is irreducible over $\mathbb{Q}$. Hence, if $x^2+3x+5$ and $ax^2+bx+c$, with $a,b,c\in\mathbb{Q}$, have a common root, they must be proportional. That is, $$ax^2+bx+c=a\left(x^2+3x+5\right)\,.$$
The problem would be more challenging if $x^2+3x+5$ is replaced by $x^2+3x+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve $z^4+2z^3+3z^2+2z+1 =0$ Solve $z^4+2z^3+3z^2+2z+1 =0$ with $z$: a complex variable.
Attempt at solving the problem:
We divide the polynom by $z^2$ and we get:
$z^2+2z+3+\dfrac{2}{z}+ \dfrac{1}{z^2}=0 $ $ $ We set $w=z+ \dfrac{1}{z}$
We now have $w^2+2w+5=0$
$\bigtriangleup = -16$
Let's find $\omega$ such that $\omega^2=-16$
We have $\omega=4i$
Therefore we have the 2 roots:
$w_ {1}=-1-2i$ and $ w_ {2}=-1+2i $
The issue is: I don't know how to find z
|
"I don't know how to find z"
Sure you do!
....
If you are correct in what you have done so far and you have
$z + \frac 1z = w$
And $w_1 = -1-2i$ and $w_2 = -1 + 2i$ then
you need to solve $z +\frac 1z = (-1-2i)$ or $z^2 +(1+2i)z + 1=0$
ANd $z + \frac 1z = (-1+2i)$ or $z^2 + (1-2i)z + 1 = 0$.
Both of which can be solved by quadratic equation to get the four solutions
$z = \frac {(-1\pm 2i) \pm\sqrt{(1\mp 2i)^2 -4}}{2}$
(And if you are worried about expressing $\sqrt{M}$ just convert to polar coordinates.)
That is.... if everything you've done is correct.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Matrix induction proof Given the following
$\lambda_{1}=\frac{1-\sqrt{5}}{2}$ and $\lambda_{2}=\frac{1+\sqrt{5}}{2}$
How do I prove this using induction:
$\begin{align*}
A^k=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
\lambda_2^{k-1}-\lambda_1^{k-1} & \lambda_2^{k}-\lambda_1^{k}\\
\lambda_2^{k}-\lambda_1^{k} & \lambda_2^{k+1}-\lambda_1^{k+1}
\end{array}\right),\,k>0
\end{align*}$
when $
A=\left(\begin{array}{cc}
0 & 1\\
1 & 1
\end{array}\right)\in\text{Mat}_{2}(\mathbb{R})
$.
I know how induction works and it holds for $k=1$. But I'm stuck at $k+1$.
My own work (Revision):
\begin{align*}
A^{k+1}=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
\lambda_2^{k-1}-\lambda_1^{k-1} & \lambda_2^{k}-\lambda_1^{k}\\
\lambda_2^{k}-\lambda_1^{k} & \lambda_2^{k+1}-\lambda_1^{k+1}
\end{array}\right)\cdot\left(\begin{array}{cc}
0 & 1\\
1 & 1
\end{array}\right)=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
\lambda_2^{k}-\lambda_1^{k} & \lambda_2^{k-1}-\lambda_1^{k-1}+\lambda_2^{k}-\lambda_1^{k}\\
\lambda_2^{k+1}-\lambda_1^{k+1} & \lambda_2^{k}-\lambda_1^{k}+\lambda_2^{k+1}-\lambda_1^{k+1}
\end{array}\right)
\end{align*}
So this is Fibonacci I guess? And therefore equivalent to:
\begin{align*}
\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
\lambda_2^{k}-\lambda_1^{k}& \lambda_2^{k+1}-\lambda_1^{k+1}\\
\lambda_2^{k+1}-\lambda_1^{k+1} & \lambda_2^{k+2}-\lambda_1^{k+2}
\end{array}\right)
\end{align*}
(Revision 2)
$A^k=P\Lambda^{k}P^{-1}$:
\begin{align*}
\left(\begin{array}{cc}
-\lambda_2& -\lambda_1\\
1 & 1
\end{array}\right)\cdot\left(\begin{array}{cc}
\lambda_1^k& 0\\
0 & \lambda_2^k
\end{array}\right)\cdot\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
-1& -\lambda_1\\
1 & \lambda_2
\end{array}\right)=\frac{1}{\sqrt{5}}\left(\begin{array}{cc}
\lambda_1^k\lambda_2-\lambda_1\lambda_2^k& \lambda_1^{k+1}\lambda_2-\lambda_1\lambda_2^{k+1}\\
\lambda_2^k-\lambda_1^k & \lambda_2^{k+1}-\lambda_1^{k+1}
\end{array}\right)
\end{align*}
Is this the right way?
| 1) A straightforward proof which is more natural than recursion, in my opinion (for a recursion proof see 2).)
Use diagonalization identity $A=P\Lambda P^{-1}$ from which $A^k=P\Lambda^kP^{-1} \ \ (1)$
where $\Lambda$ is the diagonal matrix diag$(\lambda_1,\lambda_2)$.
Here is an extension of my first explanation:
Indeed the columns of matrix $P$ are eigenvectors associated with $\lambda_1$ ans $\lambda_2$ in this order ; we can take
$P=\left(\begin{array}{cc}
-\lambda_2 & -\lambda_1\\
1 & 1
\end{array}\right)$ with
$P^{-1}=\dfrac{1}{\sqrt{5}}\left(\begin{array}{cc}
-1 &-\lambda_1 \\
1 &\lambda_2
\end{array}\right)$.
Plugging these expressions into formula (1) gives the answer.
2) @dk20, as you asked for a recursion proof, I add it to the previous text:
One wants to prove that, for any $k>1$, matrices
$A=\left(\begin{array}{cc}
0 & 1 \\
1 & 1
\end{array}\right)\left(\begin{array}{cc}
\lambda_2^{k-1}-\lambda_1^{k-1} & \lambda_2^{k}-\lambda_1^{k}\\
\lambda_2^{k}-\lambda_1^{k} & \lambda_2^{k+1}-\lambda_1^{k+1}
\end{array}\right)$
and $B=\left(\begin{array}{cc}
\lambda_2^{k}-\lambda_1^{k} & \lambda_2^{k+1}-\lambda_1^{k+1}\\
\lambda_2^{k+1}-\lambda_1^{k+1} & \lambda_2^{k+2}-\lambda_1^{k+2}
\end{array}\right)$
are identical.
It is clear that $A_{1j}=B_{1j}$ ($j=1,2$:coefficients of the first line).
Let us now prove that $A_{21}=B_{21}$ (bottom left coefficients), i.e.,
$\lambda_2^{k-1}-\lambda_1^{k-1} + \lambda_2^{k}-\lambda_1^{k}=\lambda_2^{k+1}-\lambda_1^{k+1}$
This equation is equivalent to the following one:
$\lambda_1^{k-1}(1+\lambda_1-\lambda_1^2)=\lambda_2^{k-1}(1+\lambda_2-\lambda_2^2)$
which is evidently fulfilled because it boils down to $0=0$ ; indeed, $\lambda_1$ and $\lambda_2$ are both roots of the quadratic equation $x^2-x-1=0$ ($\lambda_2$ is the "golden number").
The reason why $A_{22}=B_{22}$ is identical (change $k$ into $k+1$).
| {
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"timestamp": "2023-03-29T00:00:00",
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How to minimize $ab + bc + ca$ given $a^2 + b^2 + c^2 = 1$? The question is to prove that $ab + bc + ca$ lies in between $-1$ and $1$, given that $a^2 + b^2 + c^2 = 1$.
I could prove the maxima by the following approach. I changed the coordinates to spherical coordinates:
$a = \cos A \\
b = \sin A \cos B \\
c = \sin A \sin B$
Using that $\cos X + \sin X$ always lies between $- \sqrt 2$ and $\sqrt 2$ I proved that $a + b + c$ lies between $- \sqrt 3$ and $\sqrt 3$.
Using $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$ I could prove that $ab + bc + ca$ is maximum at $1$ but I can't prove the minima.
| From what you've done, $ab+bc+ca = \dfrac{(a+b+c)^2 - (a^2+b^2+c^2)}{2} \geq \dfrac{0 - 1}{2} = \dfrac{-1}{2}$, and this is the minimum value you sought. The minimum occurs when $a+b+c = 0, a^2+b^2+c^2 = 1$. To solve for $a,b,c$ you only need to find one solution of the system of $2$ equations above, then you are done. Since there are $3$ variables and only $2$ equations, you can take $c = \dfrac{1}{2}, \Rightarrow a, b$ are the solutions of the equation: $4x^2+2x-1 = 0$, thus $a = x = \dfrac{-1+\sqrt{5}}{4}, \Rightarrow b = -\dfrac{1}{2} - a = \dfrac{-1-\sqrt{5}}{4}$ by Viete's theorem on quadratic equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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The nature of roots of the quadratic equation $ax^2+(b-c)x-2b-c-a=0,$ If the expression $ax^2+2bx+c$, where $a$ is a non-zero real number, has the same sign as that of $a$ for every real value of $x$, then roots of the quadratic equation $ax^2+(b-c)x-2b-c-a=0$ are:
(A) real and equal
(B) real and unequal
(C) non-real having positive real part
(D) non-real having having negative real part
As the expression $ax^2+2bx+c$ has the same sign as that of $a$ for every real value of $x$, so if $a>0,$ then $4b^2-4ac<0$ and if $a<0$, then $4b^2-4ac>0$
To determine the nature of roots of the equation $ax^2+(b-c)x-2b-c-a=0, I found its discriminant
$\Delta =(b-c)^2+4a(2b+c+a)=b^2+c^2-2bc+8ab+4ac+4a^2$
Now I am not able to find the nature of roots of the equation.
| Since $ax^2+2bx+c$ has always the same sign as $a$ for any real $x$, it has no real roots, so $4b^2 - 4ac < 0$.
Now try writing
\begin{align}
(b-c)^2 + 4a(2b+c+a) &= (b-c)^2 + 4(2ab+ac+a^2) \\
&= (b-c)^2 + 4(2ab+b^2+a^2) + 4(ac - b^2) .
\end{align}
| {
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Find the value of $x$ which is correct I have one exercise which is $$(x+2013)(x+2014)(x+2015)(x+2016)+1=0$$
I tag $A=x+2013$ or other for many ways but still can not find the first $x$ value. please help.
| If you let $y=x+2014$, then the equation becomes
$$(y-1)y(y+1)(y+2)+1=0 \Leftrightarrow (y^2+y-1)^2=0$$
So
$$y = \frac{-1\pm\sqrt{5}}{2} \Rightarrow x = \ldots$$
\begin{align}
(y-1)y(y+1)(y+2)+1 & = y^4+2 y^3-y^2-2 y+1 \\
{} & = \left( y^4+y^3-y^2 \right) + y^3-2y+1 \\
{} & = y^2 \left( y^2+y-1 \right) + \left(y^3+y^2-y\right) -y^2-y+1 \\
{} & = y^2 \left( y^2+y-1 \right) + y\left(y^2+y-1\right) -\left( y^2+y-1\right) \\
{} & = \left( y^2+y-1 \right)^2
\end{align}
\begin{align}
(y-1)y(y+1)(y+2)+1 & = y^4+2 y^3-y^2-2 y+1 \\
{} & = \left( y^4+2 y^3+y^2 \right) - 2y^2-2 y+1 \\
{} & = \left( y^2 + y \right)^2 - 2(y^2+y)+1 \\
{} & = \left( y^2+y-1 \right)^2
\end{align}
| {
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Image of a family of circles under $w = 1/z$ Given the family of circles $x^{2}+y^{2} = ax$, where $a \in \mathbb{R}$, I need to find the image under the transformation $w = 1/z$. I was given the hint to rewrite the equation first in terms of $z$, $\overline{z}$, and then plug in $z = 1/w$.
However, I am having difficulty doing this.
I completed the square in $x^{2}+y^{2}=ax$ to obtain $\left(x - \frac{a}{2} \right)^{2} + y^{2} = \left(\frac{a}{2} \right)^{2}$.
Then, given that $\displaystyle x = Re(z) = \frac{z+\overline{z}}{2}$ and $\displaystyle y = Im(z)= \frac{-(z-\overline{z})}{2i}$, I made those substitutaions and my equation became $\left( \frac{z+\overline{z}-a}{2}\right)^{2} - \left(\frac{\overline{z}-z}{2} \right)^{2} = \left(\frac{a}{2} \right)^{2}$.
Then, sbustituting in $z = \frac{1}{w}$, this became $\displaystyle \frac{\left(\frac{1}{w} + \frac{1}{\overline{w}} - a \right)^{2}}{4} - \frac{\left(\frac{1}{\overline{w}} - \frac{1}{w} \right)^{2}}{4} = \frac{a^{2}}{4}$.
Beyond this, my algebra gets very wonky.
Could someone please tell me what my final result should be? Knowing that would allow me to work backwards and then apply these methods to other problems (of which I have many to do!).
Thanks.
| Write $z=x+iy$, so $x^2+y^2=z\bar{z}$, and $x=\frac{z+\bar{z}}{2}$. Thus the circles can be described by
$$
z\bar{z}=a\frac{z+\bar{z}}{2}
$$
Upon doing $z=1/w$, you get
$$
\frac{1}{w\bar{w}}=\frac{a}{2}\frac{\bar{w}+w}{w\bar{w}}
$$
that becomes
$$
a\frac{\bar{w}+w}{2}=1
$$
Writing $w=X+iY$, you get
$$
aX=1
$$
More generally, if you have the circle $x^2+y^2+ax+by+c=0$, with $z=x+iy$ you get
$$
z\bar{z}+a\frac{z+\bar{z}}{2}+b\frac{z-\bar{z}}{2i}+c=0
$$
Upon changing $z=1/w$, the equation becomes
$$
\frac{1}{w\bar{w}}+a\frac{\bar{w}+w}{2w\bar{w}}
+b\frac{\bar{w}-w}{2iw\bar{w}}+c=0
$$
and removing $w\bar{w}$ from the denominator,
$$
1+a\frac{w+\bar{w}}{2}-b\frac{w-\bar{w}}{2i}+cw\bar{w}=0
$$
or, with $w=X+iY$,
$$
1+aX-bY+c(X^2+Y^2)=0
$$
So the circle becomes a straight line if $c=0$ (that is, it passes through the origin), otherwise it is transformed into a circle.
Similarly, the line $ax+by+c=0$ becomes
$$
a\frac{\bar{w}+w}{2w\bar{w}}+b\frac{\bar{w}-w}{2iw\bar{w}}+c=0
$$
and finally
$$
aX-bY+c(X^2+Y^2)=0
$$
so it becomes a circle if $c\ne0$, or a straight line through the origin otherwise.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the sequence of combinations contains an odd number of odd numbers
Let $n$ be an odd integer more than one. Prove that the sequence $$\binom{n}{1}, \binom{n}{2}, \ldots,\binom{n}{\frac{n-1}{2}}$$
contains an odd number of odd numbers.
I tried writing out the combination form as $$\frac{(2k+1)!}{(m!)((2k+1)-m)!}.$$ How do I use this to show that the sequence contains an odd number of odd numbers?
| Suppose $n=2k+1$
Note that $\binom{2k+1}{1}=\binom{2k+1}{2k}$, $\binom{2k+1}{2}=\binom{2k+1}{2k-1}$,...,$\binom{2k+1}{k}=\binom{2k+1}{k+1}$.
Thus
$$\binom{2k+1}{1}+\binom{2k+1}{2}+\cdots+\binom{2k+1}{k}+\binom{2k+1}{k+1}+\cdots+\binom{2k+1}{2k-1}+\binom{2k+1}{2k}=2^{2k+1}-2$$
From the above considerations,
$$2\binom{2k+1}{1}+2\binom{2k+1}{2}+\cdots+2\binom{2k+1}{k}=2^{2k+1}-2.$$ Therefore
$$\binom{2k+1}{1}+\binom{2k+1}{2}+\cdots+\binom{2k+1}{k}=2^{2k}-1 \tag{$*$}$$
If the number of odd terms in $\binom{2k+1}{1},\binom{2k+1}{2}, \ldots, \binom{2k+1}{k}$ is even, then the sum in $(*)$ is even. This contradiction shows that the number of odd terms is odd.
| {
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About Factorization I have some issues understanding factorization.
If I have the expression $x^{2}-x-7$ then (I was told like this) I can put this expression equal to zero and then find the solutions with the quadratic formula, so it gives me $x_{0,1}= 1 \pm 2\sqrt{2}$ then $$x^{2}-x-7 = (x-1-2\sqrt{2})(x-1+2\sqrt{2}).$$
That is correct I have checked it.
Now for the expression $3x^{2}-x-2$ if I do the same I have $x_{0} = 1$ and $x_1=\frac{-2}{3}$ so I would have $$3x^{2}-x-2 = (x-1)(x+\frac{2}{3})$$
but this is not correct since $(x-1)(x+\frac{2}{3}) = \frac{1}{3}(3x^{2}-x-2)$,
the correct factorization is $3x^{2}-x-2 = (3x+2)(x-1)$.
So I guess finding the roots of a quadratic expression is not sufficient for factorizing.
| $3(x - 1)(x + {2 \over 3}) = (x -1)(3x + 2) = (3x - 3)(x + {2 \over 3}) =...$ etc. are all valid factoring. The leading coefficient is just a constant.
And if $(x - 1)(x + {2 \over 3}) = 0$ then $3(x - 1)(x + {2 \over 3}) = 0 = (x - 1)(x + {2 \over 3}) $.
If you are concerned about going from roots to factoring think of it this way:
If the roots of $P(x) = ax^n + ...... + c$ are $r_1, .... ,r_n$ then the polynomial factors as $a(x - r_1)(x - r_2).....(x - r_3) = P(x)$ or $(x - r_1)(x - r_2).....(x - r_3) = P(x)/a$. When setting to 0 the $a$ doesn't matter as if $y = 0$ then $a*y = 0$ also no matter what the $a$ is.
Solving for 0, the $a$ gets lost. Going from the roots back, we only take $(x - r_i)$ so the resulting coefficient will always be $1$. The resulting polynomial will be $1/a$ of the original.
| {
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Prove that $\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$ Without using Mathematical Induction, prove that $$\frac{1}{n+1} + \frac{1}{n+3}+\cdots+\frac{1}{3n-1}>\frac{1}{2}$$
I am unable to solve this problem and don't know where to start. Please help me to solve this problem using the laws of inequality. It is a problem of Inequality.
Edit: $n$ is a positive integer such that $n>1$.
| $f(x) = 1/x$ is strictly convex, therefore
$$
\frac{1}{2n} < \frac 12 \left( \frac{1}{n+k} + \frac{1}{3n-k} \right)
$$
for $k = 1, ..., n-1$, or
$$
\frac{1}{n+k} + \frac{1}{3n-k} > \frac {1}{2n} + \frac {1}{2n}
$$
Combining terms pairwise from both ends of the sum shows that
$$
\frac{1}{n+1} + \frac{1}{n+3}+\dots+\frac{1}{3n-3} + \frac{1}{3n-1} >
\underbrace{\frac {1}{2n} + \frac {1}{2n} + \dots +\frac {1}{2n} + \frac {1}{2n}}_{n \text{ terms}}
= \frac 12.
$$
(If $n$ is odd then the middle term $ \frac {1}{2n}$ is not combined with another one.
But since $n> 1$ there is at least one "pair" to combine, which gives
the strict inequality.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{n \rightarrow \infty}\frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}$ Find:
$$\lim_{n \rightarrow \infty} \frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}$$
The sequence $\frac{1}{nx^2 \log{(1+ \frac{x}{n})}}=\frac{1}{x^3 \frac{\log{(1+ \frac{x}{n})}}{\frac{x}{n}}}$ converges pointwise to $\frac{1}{x^3}$. So if we could apply Lebesgue's Dominated Convergence Theorem, we have:
$\lim_{n \rightarrow \infty} \frac{1}{n} \int_{1}^{\infty} \frac{\mathrm dx}{x^2 \log{(1+ \frac{x}{n})}}=\lim_{n \rightarrow \infty} \int_{1}^{\infty} \frac{\mathrm dx}{x^3}=\frac{1}{2}$
I have a problem with finding a majorant. Could someone give me a hint?
| I think one could do this in a conceptually simpler way:
Since
$$
\frac{y}{1+y}<\log(1+y)<y
$$
your integrand is bounded as
$$
\frac{1}{x^3}<\frac{1}{nx^2\ln(1+x/n)}<\frac{1}{nx^2(x/n)/(1+x/n)}=\frac{1}{x^3}(1+x/n).
$$
By monotonicity, your integral satisfies
$$
\frac{1}{2}=\int_1^{+\infty}\frac{1}{x^3}\,\mathrm dx<\int_1^{+\infty}\frac{1}{nx^2\ln(1+x/n)}\,\mathrm dx<\int_1^{+\infty}\frac{1}{x^3}(1+x/n)\,\mathrm dx=\frac{1}{2}+\frac{1}{n}.
$$
Now squeeze.
| {
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Cauchy like inequality $(5\alpha x+\alpha y+\beta x + 3\beta y)^2 \leq (5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)$ Problem: Prove that for real $x, y, \alpha, \beta$,
$(5\alpha x+\alpha y+\beta x + 3\beta y)^2 \leq (5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)$.
I am looking for an elegant (non-bashy) solution.
It closely resembles Cauchy inequality but $\alpha y +\beta y$ part is creating a problem.
I also tried to define a suitable inner product but couldn't.
Since, the inequality is homogenous, (for non-zeros) it reduces to,
$(5mn+m+n+3)^2 \leq (5m^2+2m+3)(5n^2+2n+3)$, by putting $\alpha = \beta m$ and $x=n y$. But I couldn't take it furthur from here.
So, any hints, solutions (especially along lines of inner product) would be very welcome.
| By C-S $(5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)=$
$=\left(\left(\sqrt5\alpha+\frac{\beta}{\sqrt5}\right)^2+\frac{14\beta^2}{5}\right)\left(\left(\sqrt5x+\frac{y}{\sqrt5}\right)^2+\frac{14y^2}{5}\right)\geq$
$=\left(\left(\sqrt5\alpha+\frac{\beta}{\sqrt5}\right)\left(\sqrt5x+\frac{y}{\sqrt5}\right)+\frac{14\beta y}{5}\right)^2=(5\alpha x+\alpha y+\beta x + 3\beta y)^2$
| {
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Intergate $\int \frac{x}{(x^2-3x+17)^2}\ dx$
$$\int \frac{x}{(x^2-3x+17)^2}\ dx$$
My attempt:
$$\int \frac{x}{(x^2-3x+17)^2}\ dx=\int \frac{x}{\left((x-\frac{3}{2})^2+\frac{59}{4}\right)^2}\ dx$$
let $u=x-\frac{3}{2}$
$du=dx$
$$\int \frac{u+\frac{3}{2}}{\left((u)^2+\frac{59}{4}\right)^2}\ du$$
How can I continue from here?
| One has $$\int \frac{u+\frac{3}{2}}{u^2+\frac{59}{4}} du = \int \frac{u}{u^2+\frac{59}{4}}du + \int \frac{\frac{3}{2}}{u^2+\frac{59}{4}}du.$$ The first term can be computed be setting $v = u^2+\frac{59}{4}$ and the second thanks to the $\arctan$ function.
| {
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Probability: Finding the Number of Pears Given Two Scenarios
You have a bag containing 20 apples, 10 oranges, and an unknown number
of pears. If the probability that you select 2 apples and 2 oranges is
equal to the probability that you select 1 apple, 1 orange, and 2
pears, then what is the number of pears originally in the bag?
Using $n$ as the number of pears, I found the probability of selecting 2 apples and 2 oranges to be: $$\frac{\dbinom{20}{2}*\dbinom{10}{2}}{\dbinom{30+n}{4}}$$
seeing as the number of ways to choose two apples and 2 oranges (order shouldn't matter) would be given by $\dbinom{20}{2}*\dbinom{10}{2}$. I put this over the total number of possibilities, which was found by choosing $4$ fruits from a total of $30+n$ fruits.
As the problem stated, this equaled the probability of choosing 1 apple, 1 orange, and 2 pears, which would be:
$$\frac{\dbinom{20}{1}*\dbinom{10}{1}*\dbinom{n}{2}}{\dbinom{30+n}{4}}$$
And thus: $$\frac{\dbinom{20}{2}*\dbinom{10}{2}}{\dbinom{30+n}{4}}=\frac{\dbinom{20}{1}*\dbinom{10}{1}*\dbinom{n}{2}}{\dbinom{30+n}{4}}$$
$$\Rightarrow\dbinom{20}{2}*\dbinom{10}{2}=\dbinom{20}{1}*\dbinom{10}{1}*\dbinom{n}{2}$$
$$\Rightarrow\dbinom{n}{2}=42.75$$
which doesn't seem to be correct. Can anyone tell me where I went wrong? I believe it is most likely a conceptual mistake; is this not how you calculate the respective probabilities?
| It is good that you were alert. I believe everything you are doing is correct. If we continue, then
$$\binom{n}{2} = \frac{\binom{20}{2}\binom{10}{2}}{20(10)} = \frac{171}{4}.$$
I think it is ok to have a decimal number here.
This gives
\begin{align*}
\frac{n!}{2!(n-2)!} &= \frac{171}{4}\\
\implies \frac{n!}{(n-2)!} &= \frac{171}{2}\\
n(n-1) &= \frac{171}{2}\\
n^2-n+\frac{1}{4} &= \frac{171}{2}+\frac{1}{4}\\
\left(n-\frac{1}{2}\right)^2 &= \frac{343}{4}\\
\implies n &= \sqrt{\frac{343}{4}}+\frac{1}{2}\\
n&=9.76013
\end{align*}
which I feel is an appropriate approach. The numbers are usually chosen so that when you complete the square, you will end up with a whole number $n$, but I think there was an error.
| {
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What is the the integral of $\sqrt{x^a + b}$? How do you evaluate $\displaystyle\int\sqrt{x^a + b}\,\,\text{dx}$, where $a \neq 0$ and $a \neq 1$?
For example, how do you evaluate $\displaystyle\int\sqrt{x^2 + 1}\,\text{dx}$? If we let $u=x^2+1$, then $du=2x\,\text{dx}$. We cannot do this because there is no $2x$ in the original function. Of course you cannot let $u=\sqrt{x^2 + 1}$, because $du=\displaystyle\frac{x}{\sqrt{x^2+1}}\,\text{dx}$. There is no $\displaystyle\frac{x}{\sqrt{x^2+1}}$ in the original function. So how do we solve this?
How about $\displaystyle\int\sqrt{x^3 + 1}\,\text{dx}$?
| Kim Peek's "funny" hypergeometric solution is really the series solution near $x=0$. We have, for $|x^a/b| < 1$,
$$ \sqrt{x^a + b} = \sqrt{b} \sqrt{1 + x^a/b} =
\sqrt{b} \sum_{k=0}^\infty {1/2 \choose k} (x^a/b)^k$$
so integrating term-by-term
$$ \int \sqrt{x^a + b}\; dx = \sum_{k=0}^\infty {1/2 \choose k} \dfrac{x^{ak+1}}{(ak+1) \; b^{k-1/2}}$$
where $${1/2 \choose k} = \dfrac{\Gamma(3/2)}{\Gamma(k+1)\; \Gamma(3/2-k)}
= \dfrac{(2k)!}{(-4)^k (k!)^2 (1-2k)}$$
If you take the definition of the hypergeometric function as a power series, you get exactly this series.
Alternatively, for $|x^a/b| > 1$ you get a different series involving negative powers of $x$:
$$\sqrt{x^a + b} = x^{a/2} \sqrt{1 + b/x^a} = \sum_{k=0}^\infty {1/2 \choose k} b^k x^{(1/2 - k)a}$$
so that
$$ \int \sqrt{x^a + b}\; dx = \sum_{k=0}^\infty {1/2 \choose k} \dfrac{b^k x^{(1/2-k)a+1}}{(1/2-k)a + 1}$$
which also has a hypergeometric representation.
| {
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Distance between two circles on a cube I found this problem in a book on undergraduate maths in the Soviet Union (http://www.ftpi.umn.edu/shifman/ComradeEinstein.pdf):
A circle is inscribed in a face of a cube of side a. Another circle is circumscribed about a neighboring face of the cube. Find the least distance between points of the circles.
The solution to the problem is in the book (page 61), but I am wondering how to find the maximum distance between points of the circles and I cannot see how the method used there can be used to find this.
| I placed the inscribed circle on the top face (+y direction) and the circumscribed circle on the front face (+z direction). Their locus of points is
$$ \begin{align}
\vec{r}_1 & = \begin{bmatrix} r_1 \cos \theta_1 & \frac{a}{2} & r_1 \sin \theta_1 \end{bmatrix} \\
\vec{r}_2 & = \begin{bmatrix} r_2 \cos \theta_2 & r_2 \sin \theta_2 & \frac{a}{2} \end{bmatrix}
\end{align} $$
where $r_1 = \frac{a}{2}$, $r_2 = \frac{a}{\sqrt{2}}$ and $a$ is the side of the cube.
The distance (squared) is $$d^2 = \| \vec{r}_1 -\vec{r}_2 \|^2 $$ which is a function of $\theta_1$ and $\theta_2$.
It comes out as
$$ \frac{d^2}{a^2} = \frac{5}{4} - \frac{\sqrt{2} \cos\theta_1 \cos \theta_2 + \sin \theta_1 + \sqrt{2} \sin \theta_2}{2} $$
To minimize this you have to set $$\frac{\partial}{\partial \theta_1} \frac{d^2}{a^2} = 0$$ and at the same time $$\frac{\partial}{\partial \theta_2} \frac{d^2}{a^2} = 0$$
The system to solve is $$\begin{align} \frac{\sin \theta_1 \cos \theta_2}{\sqrt{2}} - \frac{\cos\theta_1}{2} & = 0 \\ \frac{\cos \theta_1 \sin \theta_2}{\sqrt{2}} - \frac{\cos\theta_2}{\sqrt{2}} & = 0 \end{align} $$
From the first equation $\cos \theta_2 = \frac{\cot \theta_1}{\sqrt{2}}$. When used in the second equation I get $\theta_1 = \arctan \sqrt{2}$.
The result I get was $$\frac{d^2}{a^2} = \frac{5}{4} - \frac{\sqrt{6}}{2} $$
| {
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"url": "https://math.stackexchange.com/questions/1650881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $a^5 + b^5 + c^5$
Suppose we have numbers $a,b,c$ which satisfy the equations
$$a+b+c=3,$$
$$a^2+b^2+c^2=5,$$
$$a^3+b^3+c^3=7.$$
How can I find $a^5 + b^5 + c^5$?
I assumed we are working in $\Bbb{C}[a,b,c]$. I found a reduced Gröbner basis $G$:
$$G = \langle a+b+c-3,b^2+bc+c^2-3b-3c+2,c^3-3c^2+2c+\frac{2}{3} \rangle$$
I solved the last equation for $c$ and got 3 complex values. When I plug into the 2nd equation $(b^2+bc+c^2-3b-3c+2)$ I get a lot of roots for $b$, and it is laborious to plug in all these values.
Is there a shortcut or trick to doing this? The hint in the book says to use remainders. I computed the remainder of $f = a^5 + b^5 + c^5$ reduced by $G$:
$$\overline{f}^G = \frac{29}{3}$$
How can this remainder be of use to me?
Thanks. (Note: I am using Macaualay2)
| Using just Macaulay2, you can do the following
Macaulay2, version 1.6.0.1
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases, PrimaryDecomposition,
ReesAlgebra, TangentCone
i1 : R=QQ[a,b,c]
o1 = R
o1 : PolynomialRing
i2 : i1=ideal(a+b+c-3,a^2+b^2+c^2-5,a^3+b^3+c^3-7)
2 2 2 3 3 3
o2 = ideal (a + b + c - 3, a + b + c - 5, a + b + c - 7)
o2 : Ideal of R
i3 : S=R/i1
o3 = S
o3 : QuotientRing
i4 : phi=map(S,R)
o4 = map(S,R,{- b - c + 3, b, c})
o4 : RingMap S <--- R
i6 : use R
o6 = R
o6 : PolynomialRing
i7 : phi(a^5+b^5+c^5)
29
o7 = --
3
o7 : S
(I deleted i5 and o5 as I made a typo in the input there)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
Obviously since this is a 5th degree polynomial, solving it is not going to be possible (or may be hard). However I think that factoring it to get $x^5 − x^3 + x − 2 = (x^2-x+1)(x^3+x^2-x-2)$ will help. We know both roots of the quadratic are complex, so we need only focus on the cubic $x^3+x^2-x-2$. How can we use this to show that the real root $a$ of it has $\lfloor a^6 \rfloor = 3$?
| We have $\sqrt[6]{3} \approx 1.2009$ and $\sqrt[6]{4} \approx 1.2599$. Let $f(x)=x^3+x^2-x-2$. Then $f(1.2) \approx -0.032$ and $f(1.25) \approx 0.2656$. So, $a$ must be between $1.2$ and $1.25$.
EDIT: As I said in the comments below, I see no way of showing that there is only one root using only precalculus. But, for completeness of my answer: If there were two roots, then the Mean Value Theorem would imply that the derivative is $0$ between the roots. But, $f'(x) = 3x^2+2x-1$. The roots of this are $x=-1$ and $x=\frac{2}{3}$. Using whichever method one likes, you can see that there is a local maximum at $x=-1$ and a local minimum at $x=\frac{2}{3}$. And, both $f(-1)$ and $f\left(\frac{2}{3}\right)$ are negative. So, there cannot be another root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How many integer-sided right triangles are there whose sides are combinations?
How many integer-sided right triangles exist whose sides are combinations of the form $\displaystyle \binom{x}{2},\displaystyle \binom{y}{2},\displaystyle \binom{z}{2}$?
Attempt:
This seems like a hard question, since I can't even think of one example to this. Mathematically we have,
$$\left(\dfrac{x(x-1)}{2} \right)^2+\left (\dfrac{y(y-1)}{2} \right)^2 = \left(\dfrac{z(z-1)}{2} \right)^2\tag1$$
where we have to find all positive integer solutions $(x,y,z)$.
I find this hard to do. But here was my idea. Since we have $x^2(x-1)^2+y^2(y-1)^2 = z^2(z-1)^2$, we can try doing $x = y+1$. If we can prove there are infinitely many solutions to,
$$(y+1)^2y^2+y^2(y-1)^2 = z^2(z-1)^2\tag2$$
then we are done.
| Solving $(1)$ for $z$, we have,
$$z = \frac{1\pm\sqrt{1\pm4w}}{2}\tag3$$
where,
$$w^2 = (x^2-x)^2+(y^2-y)^2\tag4$$
It can be shown that $(4)$ has infinitely many integer solutions. (Update: Also proven by Sierpinski in 1961. See link given by MXYMXY, Pythagorean Triples and Triangular Numbers by Ballew and Weger, 1979.)
However, the problem is you still have to solve $(3)$. I found with a computer search that with $x<y<1000$, the only integers are $x,y,z = 133,\,144,\,165$, so,
$$\left(\dfrac{133(133-1)}{2} \right)^2+\left (\dfrac{144(144-1)}{2} \right)^2 = \left(\dfrac{165(165-1)}{2} \right)^2$$
P.S. If you're curious about rational solutions, then your $(1)$ and $(2)$ have infinitely many.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Prove that if $ 2^n $ divides $ 3^m-1 $ then $ 2^{n-2} $ divides $ m $ I got a difficult problem. It's kind of difficult to prove. Can you do it?
Let $ m,n\geq 3 $ be two positive integers. Prove that if $ 2^n $ divides $ 3^m -1$ then $ 2^{n-2} $ divides $ m $
Thanks :-)
| Because $n \geq 3$ we get $8 \mid 3^m-1$ and so $m$ must be even .
Let $m=2^l \cdot k$ with $k$ odd . Now use the difference of squares repeatedly to get :
$$3^m-1=(3^k-1)(3^k+1)(3^{2k}+1)\cdot \ldots \cdot (3^{2^{k-1} \cdot l}+1)$$
Each term of the form $3^s+1$ with $s$ even has the power of $2$ in their prime factorization exactly $1$ because: $$3^s+1 \equiv 1+1\equiv 2 \pmod{8}$$
Also $k$ is odd so :
$$3^k+1 \equiv 3+1 \equiv 4 \pmod{8}$$ has two factors of $2$ .
Finally the term $3^k-1 \equiv 3-1 \equiv 2 \pmod{8}$ has one factor of $2$ .
This means that $3^m-1$ has $1+2+l-1=l+2$ two's in his prime factorization .
But $2^n \mid 3^m-1$ so $n \leq l+2$ and then $l \geq n-2$ .
This means that $2^{n-2} \mid m$ as wanted .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Solve $3 = -x^2+4x$ by factoring I have $3 = -x^2 + 4x$ and I need to solve it by factoring. According to wolframalpha the solution is $x_1 = 1, x_2 = 3$.
\begin{align*}
3 & = -x^2 + 4x\\
x^2-4x+3 & = 0
\end{align*}
According to wolframalpha $(x-3) (x-1) = 0$ is the equation factored, which allows me to solve it, but how do I get to this step?
| \begin{align}x^2-4x+3&=x^2-3x-x+3\\
&=x(x-3)-(x-3)\\
&=(x-1)(x-3)
\end{align}
Note: You could also see that the sum of coefficients is zero, hence one root is $x=1$. Now divide the quadratic by $x-1$ to get the other factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate the limit $\lim_{x\to \infty}( \sqrt{4x^2+x}-2x)$
Evaluate :$$\lim_{x\to \infty} (\sqrt{4x^2+x}-2x)$$
$$\lim_{x\to \infty} (\sqrt{4x^2+x}-2x)=\lim_{x\to \infty} \left[(\sqrt{4x^2+x}-2x)\frac{\sqrt{4x^2+x}+2x}{\sqrt{4x^2+x}+2x}\right]=\lim_{x\to \infty}\frac{{4x^2+x}-4x^2}{\sqrt{4x^2+x}+2x}=\lim_{x\to \infty}\frac{x}{\sqrt{4x^2+x}+2x}$$
Using L'Hôpital $$\lim_{x\to \infty}\frac{1}{\frac{8x+1}{\sqrt{4x^2+x}}+2}$$
What should I do next?
| Hint :
$\displaystyle\lim_{x\to \infty}\frac{x}{\sqrt{4x^2+x}+2x}=\lim_{x\to \infty}\frac{1}{\sqrt{4+\frac{1}{x}}+2}$ , dividing numerator and denominator by $x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove $\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2$ if $a+b+c=0$ Found this lovely identity the other day, and thought it was fun enough to share as a problem:
If $a+b+c=0$ then show $$\frac{a^3+b^3+c^3}{3}\frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2.$$
There are, of course, brute force techniques for showing this, but I'm hoping for something elegant.
| Let $T_{m}$ be $a^m+b^m+c^m$.
Let $k=-ab-bc-ca$, and $l=abc$.
Note that this implies $a,b,c$ are solutions to $x^3=kx+l$.
Using Newton's Identity, note the fact that $T_{m+3}=kT_{m+1}+lT_{m}$(which can be proved by summing $x^3+kx+l$)
It is not to difficult to see that $T_{2}=2k$, $T_3=3l$, from $a+b+c=0$.
From here, note that $T_{4}=2k^2$ using the identity above.
In the same method, note that $T_{5}=5kl$.
From here, note $T_{7}=5k^2l+2k^2l=7k^2l$ from $T_{m+3}=kT_{m+1}+lT_{m}$ . Therefore, the equation simplifies to showing that $k^2l \times l=(kl)^2$, which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 7,
"answer_id": 4
} |
$\lim_{x\to 0} (2^{\tan x} - 2^{\sin x})/(x^2 \sin x)$ without l'Hopital's rule; how is my procedure wrong?
please explain why my procedure is wrong i am not able to find out??
I know the property limit of product is product of limits (provided limit exists and i think in this case limit exists for both the functions).
The actual answer for the given question is $\frac{1}{2}\log(2)$.
My course book has shown that don't use this step but has not given the reason.
AND Please TELL why i am WRONG
| One can rewrite:
$$\frac{2^{\tan x}-2^{\sin x}}{x^2\sin x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{\tan x - \sin x}{x^2 \sin x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{1-\cos x}{x^2 \cos x}=\frac{2^{\tan x}-2^{\sin x}}{\tan x - \sin x}\frac{\sin^2 x}{x^2}\frac{1}{(1+\cos x)\cos x}$$
$$=2^{\sin x}\cdot\frac{2^{\tan x-\sin x}-1}{\tan x - \sin x}\cdot \left(\frac{\sin x}{x}\right)^2\cdot\frac{1}{(1+\cos x)\cos x}\to1\cdot\ln 2\cdot 1^2\cdot \frac{1}{2\cdot1} = \frac{\ln 2}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
How to evaluate this limit? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to 0+}\left(\frac{\left[\ln\left(\frac{5+x^2}{5+4x}\right)\right]^6\ln\left(\frac{5+x^2}{1+4x}\right)}{\sqrt{5x^{10}+x^{11}}-\sqrt{5}x^5}\right)$$
| Let's try the elementary way. We have
\begin{align}
L &= \lim _{x \to 0^{+}}\left(\dfrac{\left[\log\left(\dfrac{5 + x^{2}}{5 + 4x}\right)\right]^{6} \log\left(\dfrac{5 + x^{2}}{1 + 4x}\right)}{\sqrt{5x^{10} + x^{11}} - \sqrt{5}x^5}\right)\notag\\
&= \lim _{x \to 0^{+}}\left(\dfrac{\left[\log\left(1 + \dfrac{x^{2} - 4x}{5 + 4x}\right)\right]^{6} \cdot\log 5}{\sqrt{5x^{10} + x^{11}} - \sqrt{5}x^5}\right)\notag\\
&= \log 5\lim _{x \to 0^{+}}\left(\dfrac{\left[\dfrac{\log\left(1 + \dfrac{x^{2} - 4x}{5 + 4x}\right)}{\dfrac{x^{2} - 4x}{5 + 4x}}\right]^{6}\left(\dfrac{x^{2} - 4x}{5 + 4x}\right)^{6}}{\sqrt{5x^{10} + x^{11}} - \sqrt{5}x^5}\right)\notag\\
&= \log 5\lim _{x \to 0^{+}}\left(\dfrac{1\cdot\left(\dfrac{x^{2} - 4x}{5 + 4x}\right)^{6}}{\sqrt{5x^{10} + x^{11}} - \sqrt{5}x^5}\right)\notag\\
&= \log 5\lim _{x \to 0^{+}}\left(\dfrac{x^{6}\left(\dfrac{x - 4}{5 + 4x}\right)^{6}}{x^{5}\{\sqrt{5 + x} - \sqrt{5}\}}\right)\notag\\
&= \log 5\lim _{x \to 0^{+}}\left(\dfrac{x - 4}{5 + 4x}\right)^{6}\cdot\frac{x}{\sqrt{5 + x} - \sqrt{5}}\notag\\
&= \left(\frac{4}{5}\right)^{6}\log 5\lim _{x \to 0^{+}}\frac{x(\sqrt{5 + x} + \sqrt{5})}{5 + x - 5}\notag\\
&= \left(\frac{4}{5}\right)^{6}2\sqrt{5}\log 5\notag\\
&= \frac{8192\sqrt{5}\log 5}{15625}\notag
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $I= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x$
$$I= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x$$
My Endeavour :
\begin{align}I&= \int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x\\ &= \int \frac{x}{\sqrt{1+ x^4}}\,\mathrm d x - \int \frac{1}{x\sqrt{1+ x^4}}\,\mathrm d x\end{align}
\begin{align}\textrm{Now,}\;\;\int \frac{x}{\sqrt{1+ x^4}}\,\mathrm d x &= \frac{1}{2}\int \frac{2x^3}{x^2\sqrt{1+ x^4}}\,\mathrm dx\\ \textrm{Taking}\,\,(1+ x^4)= z^2\,\,\textrm{and}\,\, 4x^3\,\mathrm dx= 2z\,\mathrm dz\,\, \textrm{we get} \\ &= \frac{1}{2}\int \frac{z\,\mathrm dz}{\sqrt{z^2-1}\, z}\\ &= \frac{1}{2}\int \frac{\mathrm dz}{\sqrt{z^2-1}}\\ &= \frac{1}{2}\ln|z+ \sqrt{z^2 -1}|\\ &= \frac{1}{2}\ln|\sqrt{1+x^4}+ x^2|\\ \textrm{Now, with the same substitution, we get in the second integral}\\ \int \frac{1}{x\sqrt{1+ x^4}}\,\mathrm d x &= \frac{1}{2}\int \frac{2x^3}{x^4\sqrt{1+ x^4}}\,\mathrm dx\\ &= \frac{1}{2}\int \frac{z\,\mathrm dz}{( z^2 -1)\;z} \\ &=\frac{1}{2}\int \frac{\mathrm dz}{ z^2 -1} \\ &=\frac{1}{2^2}\, \ln \left|\frac{ z+1}{z-1}\right| \\ &= \frac{1}{2^2}\, \ln \left|\frac{ \sqrt{1+ x^4}+1}{\sqrt{1+ x^4}-1}\right|\;.\end{align}
So, \begin{align}I&=\int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x \\ &=\frac{1}{2}\, \ln|\sqrt{1+x^4}+ x^2|- \frac{1}{2^2}\, \ln \left|\frac{ \sqrt{1+ x^4}+1}{\sqrt{1+ x^4}-1}\right| + \mathrm C\;.\end{align}
Book's solution:
\begin{align}I&=\int \frac{x^2 -1}{x\sqrt{1+ x^4}}\,\mathrm d x \\ &= \ln\left\{\frac{1+x^2 + \sqrt{1+x^4}}{x}\right\} + \mathrm C\;.\end{align}
And my hardwork's result is nowhere to the book's answer :(
Can anyone tell me where I made the blunder?
| From here and here we learn that (mistake 1)
\begin{align}
\int \frac{x}{\sqrt{1+ x^4}}dx&=\frac12\arcsin x^2\\
&=\frac12\ln(x^2+\sqrt{1+x^4})
\end{align}
Hence (for part 2 I basically take your solution multiplied by $-1$: mistake 2)
\begin{align}
I&=\frac12\ln x^2+\sqrt{1+x^4})-\frac14 \ln \frac{ \sqrt{1+ x^4}-1}{\sqrt{1+ x^4}+1}\\
&=\frac12\ln(x^2+\sqrt{1+x^4})-\frac14 \ln \frac{ \sqrt{1+ x^4}-1}{\sqrt{1+ x^4}+1}\frac{\sqrt{1+ x^4}+1}{\sqrt{1+ x^4}+1}\\
&=\frac12\ln(x^2+\sqrt{1+x^4})-\frac14 \ln \frac{ x^4}{(\sqrt{1+ x^4}+1)^2}\\
&=\frac12\ln(x^2+\sqrt{1+x^4})-\frac12 \ln \frac{ x^2}{\sqrt{1+ x^4}+1}\\
&=\frac12 \ln \frac{ (x^2+\sqrt{1+x^4})(\sqrt{1+ x^4}+1)}{x^2}\\
&=\frac12 \ln \frac{ 2(x^2+\sqrt{1+x^4})(\sqrt{1+ x^4}+1)}{2x^2}\\
&=\frac12 \ln \frac{ 2x^2\sqrt{1+ x^4}+2x^2+2(1+x^4)+2\sqrt{1+x^4}}{2x^2}\\
&=\frac12 \ln \frac{ (1+x^2+\sqrt{1+x^4})^2}{2x^2}\\
&=\ln \frac{ 1+x^2+\sqrt{1+x^4}}{x}-\frac12\ln 2\\
\end{align}
Note $\int \frac{dz}{z^2-1}=\frac12\ln\frac{z-1}{z+1}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
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Cauchy Residue Theorem Integral I have been given the integral $$\int_0^ {2\pi} \frac{sin^2\theta} {2 - cos\theta} d\theta $$ I have use the substitutions $z=e^{i\theta}$ |$d\theta = \frac{1}{iz}dz$ and a lot of algebra to transform the integral into this $$\frac{-i}{2} \oint \frac{1}{z^2}\frac{(z-1)^2}{z^2-4z+1}dz$$ In order to find the residues i further broke the integral into $$\frac{-i}{2} \oint \frac{1}{z^2}\frac{(z-1)^2}{(z+-r_1)(z-r_2)}dz$$ where $r_1 = 2+\sqrt{3}$ and $r_2 = 2-\sqrt{3}$ giving me three residues at $z=0|z=r_1|z=r_2 $
My question is where do I go from here? Thanks.
| There was an error in the original post. We have
$$\int_0^{2\pi}\frac{\sin^2(\theta)}{2-\cos(\theta)}d\theta=-\frac i2\oint_{|z|=1}\frac{(z^2-1)^2}{z^2(z^2-4z+1)}\,dz$$
There are two poles inside $|z|=1$. The first is a second order pole at $z=0$ and the second is a first order pole at $z=r_2$.
To find the reside of the first pole we use the general expression for the residue of a pole of order $n$
$$\text{Res}\{f(z), z= z_0\}=\frac{1}{(n-1)!}\lim_{z\to z_0}\left(\frac{d^{n-1}}{dz^{n-1}}\left((z-z_0)^nf(z)\right)\right)$$
Here, we have
$$\begin{align}
\text{Res}\left(\frac{-i(z^2-1)^2}{2z^2(z^2-4z+1)}, z= 0\right)&=\frac{1}{(2-1)!}\lim_{z\to 0}\left(\frac{d^{2-1}}{dz^{2-1}}\left((z-0)^2\frac{-i(z^2-1)^2}{2z^2(z^2-4z+1)}\right)\right)\\\\
\end{align}$$
To find the residue at $z=r_2$ we have simply
$$\text{Res}\left(\frac{-i(z^2-1)^2}{2z^2(z^2-4z+1)}, z= r_2\right)=\lim_{z\to r_2}\frac{-i(z^2-1)^2}{2z^2(z-r_1)}=-\frac{i}{2}\frac{(r_2^2-1)^2}{r_2^2(r_2-r_1)}$$
The rest is left as an exercise for the reader.
| {
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"timestamp": "2023-03-29T00:00:00",
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Make $2^8 + 2^{11} + 2^n$ a perfect square Can someone help me with this exercise? I tried to do it, but it was very hard to solve it.
Find the value of $n$ to make $2^8 + 2^{11} + 2^n$ a perfect square.
It is the same thing like $4=2^2$.
| Here is a very late answer since I just saw the problem:
By brute force, we may check that $n=12$ is the smallest possible integer such that $2^8+2^{11}+2^n$ is a perfect square. We also claim that this is the only integer.
To see why the above is true, let $2^8+2^{11}+2^n=2^8(1+2^3+2^k)=2^8(9+2^k), k \ge 4$. Now, we only need to find all integers $k \ge 4 $ such that $9+2^k$ is a perfect square. Let $m^2=9+2^k, m \ \in \mathbb{Z^+}$. Hence we have $2^k=(m-3)(m+3)$. Clearly, $m$ must be odd; otherwise, both $m-3$ and $m+3$ would be odd, contradiction. Hence, $2\mid m-3$ and $2 \mid m+3$. But we also claim it is impossible that both $m-3$ and $m+3$ divides $4$; otherwise, $4 \mid 2m \Rightarrow 2 \mid m$, which is a contradiction for $m$ odd. Since $m-3 < m+3$, we conclude that, for any integer $k \ge 4$, we must have $m-3=2$ and $m+3=2^{k-1}$. But the former already implies $m=5$, so we must have $k-1=3 \Rightarrow k=4$ as our only solution, which is precisely what we claimed earlier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to prove the following binomial identity How to prove that
$$\sum_{i=0}^n \binom{2i}{i} \left(\frac{1}{2}\right)^{2i} = (2n+1) \binom{2n}{n} \left(\frac{1}{2}\right)^{2n} $$
| Suppose we seek to verify that
$$\sum_{q=0}^n {2q\choose q} 4^{-q} =
(2n+1) {2n\choose n} 4^{-n}$$
using a method other than induction.
Introduce the Iverson bracket
$$[[0\le q\le n]]
= \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{w^q}{w^{n+1}} \frac{1}{1-w}
\; dw$$
This yields for the sum (we extend the sum to infinity because the
Iverson bracket controls the range)
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}} \frac{1}{1-w}
\sum_{q\ge 0} {2q\choose q} w^q 4^{-q}
\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}} \frac{1}{1-w}
\frac{1}{\sqrt{1-w}}
\; dw.$$
We have used the Newton binomial to obtain the square root.
Continuing we have
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{n+1}} \frac{1}{(1-w)^{3/2}}
\; dw.$$
Use the Newton binomial again to obtain
$${n+1/2\choose n}
= \frac{1}{n!} \prod_{q=0}^{n-1} (n+1/2-q)
= \frac{1}{2^n \times n!} \prod_{q=0}^{n-1} (2n+1-2q)
\\ = \frac{1}{2^n \times n!} \frac{(2n+1)!}{2^n n!}
= 4^{-n} \frac{(2n+1)!}{n!\times n!}
= 4^{-n} (2n+1) {2n\choose n}.$$
This is the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$
$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\rightarrow \infty$
So Using $\bf{A.M\geq G.M}\;,$ We get $$\frac{x+1+x+2+x+3+x+4+x+5}{5}\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$x+3\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\leq 3$$
and equality hold when $x+1=x+2=x+3=x+4=x+5\;,$ Where $x\rightarrow \infty$
So $$\lim_{x\rightarrow 0}\left[\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right]=3$$
Can we solve the above limit in that way, If not then how can we calculate it
and also plz explain me where i have done wrong in above method
Thanks
| $$\lim _{t\to 0}\left(\left[\left(\frac{1}{t}+1\right)\left(\frac{1}{t}+2\right)\left(\frac{1}{t}+3\right)\left(\frac{1}{t}+4\right)\left(\frac{1}{t}+5\right)\right]^{\frac{1}{5}}-\frac{1}{t}\right) = \lim _{t\to 0}\left(\frac{\sqrt[5]{1+15t+85t^2+225t^3+274t^4+120t^5}-1}{t}\right) $$
Now we use the Taylor's development at the first order
$$= \lim _{t\to 0}\left(\frac{1+3t-1+o(t)}{t}\right) = \color{red}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 2
} |
Find all possible values of $c^2$ in a system of equations. Numbers $x,y,z,c\in \Bbb R$ satisfy the following system of equations:
$$x(y+z)=20$$
$$y(z+x)=13$$
$$z(x+y)=c^2$$
Find all possible values of $c^2$.
To try to solve this, I expanded the equations:
$$xy+xz=20$$
$$yz+xy=13$$
$$xz+yz=c^2$$
Then I subtracted the first equation from the second one to get:
$$xz-yz=7$$
I added and subtracted this equation with the 3rd and got the following equations:
$$2yz=7-c^2$$
$$2xz=7+c^2$$
I then added these equations, factored out $2z$ and divided by $2$ to get:
$$z(x+y)=7$$
So, I found one possible value of $c^2 = 7$. How do I find the other values if they exist or how do I prove that there are no other values if they don't? Thanks!
| Let $c^2 = s$. Eliminating $x$ and $y$, you get an equation in $s$ and $z$:
$$ s^2 + 2 z^2 s - 66 z^2 - 49 $$
Thus
$$ z^2 = \dfrac{s^2 - 49}{66 - 2 s}$$
Since $z^2 \ge 0$, we need either $s \le -7$ or $7 \le s < 33$. This corresponds to $\sqrt{7} \le c < \sqrt{33}$. We then have
$$ \eqalign{y &= \dfrac{z (33 - c^2)}{c^2 + 7}\cr
x &= \dfrac{z(33 - c^2)}{c^2-7} \cr}$$
In particular, $c = \sqrt{7}$ is not possible. So the result is that
$\sqrt{7} < c < \sqrt{33}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Summing the terms of a series I have a really confusing question from an investigation. It states-
Find the value of:
$$\sqrt{1^3+2^3+3^3+\ldots+100^3}$$
How would I go about answering this??
| And if you don't know the formula
and don't need it exactly,
$\sum_{k=1}^{100} k^3
\approx \int_0^{100} x^3 dx
=\frac{100^4}{4}
$
so the result is
$\sqrt{\frac{100^4}{4}}
=\frac{100^2}{2}
=5000
$.
If you add in the
usual correction of
$\frac12 f(n)$,
the result is
$\sqrt{\frac{100^4}{4}+\frac12 100^3}
=\frac{100^2}{2}\sqrt{1+\frac{2}{100}}
\approx \frac{100^2}{2}(1+\frac{1}{100})
=5050
$.
Shazam!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof that $\sum_{i=0}^n 2^i = 2^{n+1} - 1$ $\sum_{i=0}^n 2^i = 2^{n+1} - 1$
I can't seem to find the proof of this. I think it has something to do with combinations and Pascal's triangle. Could someone show me the proof? Thanks
| Mathematical induction will also help you.
*
*(Base step) When $n=0$, $\sum_{i=0}^0 2^i = 2^0 = 1= 2^{0+1}-1$.
*(Induction step) Suppose that there exists $n$ such that $\sum_{i=0}^n 2^i = 2^{n+1}-1$. Then $\sum_{i=0}^{n+1}2^i=\sum_{i=0}^n 2^i + 2^{n+1}= (2^{n+1}-1)+2^{n+1}=2^{n+2}-1.$
Therefore given identity holds for all $n\in \mathbb{N}_0$.
Edit: If you want to apply combinations and Pascal's triangle, observe
\begin{align}
2^0&=\binom{0}{0}\\
2^1&=\binom{1}{0}+\binom{1}{1}\\
2^2&=\binom{2}{0}+\binom{2}{1}+\binom{2}{2}\\
2^3&=\binom{3}{0}+\binom{3}{1}+\binom{3}{2}+\binom{3}{3}\\
\vdots&=\vdots\\
2^n&=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+\cdots+\binom{n}{n}
\end{align}
Hockey stick identity says that
$$
\sum_{i=r}^n \binom{i}{r}=\binom{n+1}{r+1}.
$$
and so
\begin{align}
\binom{0}{0}+\binom{1}{0}+\cdots+\binom{n}{0}&=\binom{n+1}{1}\\
\binom{1}{1}+\binom{2}{1}+\cdots+\binom{n}{1}&=\binom{n+1}{2}\\
\binom{2}{2}+\binom{3}{2}+\cdots+\binom{n}{2}&=\binom{n+1}{3}\\
\vdots&=\vdots\\
\binom{n}{n}&=\binom{n+1}{n+1}
\end{align}
Add all terms, then we get
\begin{align}
\sum_{i=0}^n 2^i &= \sum_{i=1}^{n+1} \binom{n+1}{i}\\
&=\sum_{i=0}^{n+1}\binom{n+1}{i}-1\\
&=2^{n+1}-1.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1677359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of sum of the series What would be the sum of following ?
$$\lim_{n\to\infty} \left[\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + \cdots + \frac{1}{(n+n)^{2}}\right]$$
I tried to turn it into integral :
$\displaystyle\int \frac{1}{(1+\frac{r}{n})^{2}}\frac{1}{n^{2}} $ but
I can't figure out how to deal with $\frac{1}{n^{2}}$
| $$\lim_{n\to\infty}0\le\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]\le\lim_{n\to\infty}\frac{n}{(n+1)^2}$$
$$\lim_{n\to\infty}0\le\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]\le\lim_{n\to\infty}\frac{\frac1n}{(1+\frac1n)^2}$$
$$0\le\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]\le0$$
$$\lim_{n\to\infty} [\frac{1}{(n+1)^{2}} + \frac{1}{(n+2)^{2}} + \frac{1}{(n+3)^{2}} + ... + \frac{1}{(n+n)^{2}}]=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Integrating $\int\frac{x^2-1}{(x^2+1)\sqrt{x^4+1}}\,dx$ I came across a question today...
Find $$\displaystyle\int\dfrac{x^2-1}{(x^2+1)\sqrt{x^4+1}}\,dx$$
How to do this? I tried to take $x^4+1=u^2$ but no result. Then I tried to take $x^2+1=\frac{1}{u}$, but even that didn't work. Then I manipulated it to $\int \dfrac{1}{\sqrt{1+x^4}}\,dx+\int\dfrac{2}{(x^2+1)\sqrt{1+x^4}}\,dx$, butI have no idea how to solve it.
Wolframalpha gives some imaginary result...but the answer is $\dfrac{1}{\sqrt2}\arccos\dfrac{x\sqrt2}{x^2+1}+C$
| Hint Divide the numerator and denominator by $x^2$, to get:
$$\int \frac{(1-\frac{1}{x^2})dx}{(x+\frac{1}{x})(\sqrt {(x+\frac{1}{x})^2-2} )}$$
Then put $x+\frac{1}{x}=t$
$$\int \frac{dt}{t(\sqrt{t^2-2})}$$
Which is easily taken care of by putting $t=\sqrt2 \sec\theta$,
$$\int \frac{(\sqrt2 \sec\theta\tan \theta)d\theta}{\sqrt2 \sec\theta \sqrt2 \tan\theta}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Square root of $\sqrt{1-4\sqrt{3}i}$ How can we find square root of the complex number
$$\sqrt{1-4\sqrt{3}i}?$$
Now here if I assume square root to be $a+ib$ i.e.
$a+ib=\sqrt{\sqrt{1-4\sqrt{3}i}}$, then after squaring both sides, how to compare real and imaginary part?
Edit: I observed
$\sqrt{1-4\sqrt{3}i}=\sqrt{4-3-4\sqrt{3}i}=\sqrt{2^2+3i^2-4\sqrt{3}i}=\sqrt{(2-\sqrt{3}i)^2}$ which made calculation easier.
| You need to first find the square root of $1 - 4 \sqrt{3}i$ using the same method: let $(c+di)^2 = 1 - 4 \sqrt{3} i$ and then compare real and imaginary parts to find $c$ and $d$ explicitly. This will give you two different answers. Then assume $(a+bi)^2 = c + di$, and compare real and imaginary parts to find $a$ and $b$ explicitly. This gives you two solutions for both solutions of $(c+di)^2 = 1 - 4 \sqrt{3} i$, so in total, you get four different answers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
How would one solve the following equation? This equation is giving me a hard time.
$$e^x(x^2+2x+1)=2$$
Can you show me how to solve this problem algebraically or exactly? I managed to solve it using my calculator with one of its graph functions. But I would like to know how one would solve this without using the calculator.
Highly appreciated,
Bowser.
| The answer given by Desmos for intersection of the two curves $y=e^x$ and $y=\frac {2}{(x+1)^2}$ is $\color{red}{x=0.249}$. Now we have
$$(x+1)^2=2e^{-1}\iff x^2+2x+1=2(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}-\frac{x^5}{60}+O(x^6))$$ hence $$1-4x-\frac{x^3}{3}+\frac{x^4}{12}-\frac{x^5}{60}+20\cdot O(x^6)=0$$
The first approximation $1-4x=0$ gives $\color{red}{x\approx 0.25}$
The second approximation $ 1-4x-\frac{x^3}{3}=0$ gives $\color{red}{x\approx 0.24872}$
And we can continue but we see that the first approach is already good enough.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Weird Inequality that seems to be true Is it true that:
$$\left (3x+\frac{4}{x+1}+\frac{16}{y^2+3}\right )\left (3y+\frac{4}{y+1}+\frac{16}{x^2+3}\right )\geq 81,\ \forall x,y\geq 0$$
I have proved that $3x+\frac{4}{x+1}+\frac{16}{x^2+3}= 9 +\frac{(x-1)^2 (3x^2+1)}{(x+1)(x^2+3)}, \ \forall x\geq 0$, but I did not succeed in proving the initial inequality.
Thanks, for the counterexample. This is for sure good:
$$\left (3x+\frac{4}{x+1}+\frac{8}{\sqrt{2(y^2+1)}}\right )\left (3y+\frac{4}{y+1}+\frac{8}{\sqrt{2(x^2+1)}}\right )\geq 81,\ \forall x,y\geq 0$$
| This inequality is false, e.g. $x=2.5$, $y=0.5$. You will obtain a value of $79.9901<81$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that the inequality $\sin^8(x) + \cos^8(x) \geq \frac{1}{8}$ is true for every real number. Prove that the inequality $\sin^8(x) + \cos^8(x) \geq \frac{1}{8}$ is true for every real number.
| $$\sin^8 x+\cos^8x \ge \frac 18;$$
$$ \left (\frac{1-\cos2x}{2} \right )^4+ \left (\frac{1+\cos2x}{2} \right )^4\ge \frac 18;$$
$$(1-\cos2x)^4+(1+\cos2x)^4 \ge2$$
$$1-4\cos2x+6\cos^22x-4\cos^32x+\cos^42x+$$
$$+1-4\cos2x+6\cos^22x-4\cos^32x+\cos^42x\ge 2$$
$$12\cos^22x+2\cos^42x \ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Establish the identity $\frac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta\$ Establish the identity:
$$\dfrac{\cot\theta + \sec\theta}{\cos\theta + \tan\theta} = \sec\theta \cot\theta$$
The first step I got was:
$$\sec\theta \cot\theta = \dfrac{\sec\theta \cot\theta\,\big(\cos\theta + \tan\theta\big)}{\cos\theta + \tan\theta}$$
Then it tells me to rewrite the factor $$\cos\theta + \tan\theta$$
in the numerator using reciprocal identities.
How would I do that?
Here is what the assignment looked like:
| Continuing from what you got:
$$\sec\theta \cot\theta = \dfrac{\sec\theta \cot\theta\,\big(\cos\theta + \tan\theta\big)}{\cos\theta + \tan\theta}$$
and since $\sec \theta \cos \theta = 1, \cot \theta \tan \theta = 1$, expand the brackets:
$$\sec\theta \cot\theta = \frac{\cot \theta + \sec \theta}{\cos \theta + \tan \theta}$$
There is no need to rewrite $\sec \theta \cot \theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Maximum minimum values in trigonometry
Find minimum value of $2\sin^2a+3\cos^2a$
Solving it we get $2+ \cos^2a$
Answer: $3$ (taking $\cos a$ as $-1$)
Why are we using the minimum cosine value as $-1$ instead of using the cosine as $0$?
This can make the minimum value as $2$.
| $$2 \sin^2a+3 \cos^2 a= 3\cos^2a+2-2\cos^2 a=\cos^2a+2$$
$$0\le \cos^2a \le 1 \Rightarrow 2\le \cos^2a+2 \le 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal
Question: Find a value of $n$ such that the coefficients of $x^7$ and $x^8$ are in the expansion of $\displaystyle \left(2+\frac{x}{3}\right)^{n}$ are equal.
My attempt:
$\displaystyle \binom{n}{7}=\binom{n}{8} $
$$ n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) \times 2^{n-7} \times (\frac{1}{3})^7= n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7) \times 2^{n-8} \times (\frac{1}{3})^8 $$
$$ \frac{6}{7!} = \frac{n-7}{40320} $$
$$ n-7 = 48 $$
$$ n=55 $$
| The coefficient of $x^7$ is
$$\binom{n}{7}\frac{2^{n-7}}{3^7}$$
And the coefficient of $x^8$ is
$$\binom{n}{8}\frac{2^{n-8}}{3^8}$$
Comparing them we get:
$$\binom{n}{8}=\binom{n}{7}\frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Integral $\int \sqrt{\frac{x}{2-x}}dx$ $$\int \sqrt{\frac{x}{2-x}}dx$$
can be written as:
$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx.$$
there is a formula that says that if we have the integral of the following type:
$$\int x^m(a+bx^n)^p dx,$$
then:
*
*If $p \in \mathbb{Z}$ we simply use binomial expansion, otherwise:
*If $\frac{m+1}{n} \in \mathbb{Z}$ we use substitution $(a+bx^n)^p=t^s$
where $s$ is denominator of $p$;
*Finally, if $\frac{m+1}{n}+p \in \mathbb{Z}$ then we use substitution
$(a+bx^{-n})^p=t^s$ where $s$ is denominator of $p$.
If we look at this example:
$$\int x^{\frac{1}{2}}(2-x)^{\frac{-1}{2}}dx,$$
we can see that $m=\frac{1}{2}$, $n=1$, and $p=\frac{-1}{2}$ which means that we have to use third substitution since $\frac{m+1}{n}+p = \frac{3}{2}-\frac{1}{2}=1$ but when I use that substitution I get even more complicated integral with square root. But, when I tried second substitution I have this:
$$2-x=t^2 \Rightarrow 2-t^2=x \Rightarrow dx=-2tdt,$$
so when I implement this substitution I have:
$$\int \sqrt{2-t^2}\frac{1}{t}(-2tdt)=-2\int \sqrt{2-t^2}dt.$$
This means that we should do substitution once more, this time:
$$t=\sqrt{2}\sin y \Rightarrow y=\arcsin\frac{t}{\sqrt{2}} \Rightarrow dt=\sqrt{2}\cos ydy.$$
So now we have:
\begin{align*}
-2\int \sqrt{2-2\sin^2y}\sqrt{2}\cos ydy={}&-4\int\cos^2ydy = -4\int \frac{1+\cos2y}{2}dy={} \\
{}={}& -2\int dy -2\int \cos2ydy = -2y -\sin2y.
\end{align*}
Now, we have to return to variable $x$:
\begin{align*}
-2\arcsin\frac{t}{2} -2\sin y\cos y ={}& -2\arcsin\frac{t}{2} -2\frac{t}{\sqrt{2}}\sqrt\frac{2-t^2}{2}={} \\
{}={}& -2\arcsin\frac{t}{2} -\sqrt{t^2(2-t^2)}.
\end{align*}
Now to $x$:
$$-2\arcsin\sqrt{\frac{2-x}{2}} - \sqrt{2x-x^2},$$
which would be just fine if I haven't checked the solution to this in workbook where the right answer is:
$$2\arcsin\sqrt\frac{x}{2} - \sqrt{2x-x^2},$$
and when I found the derivative of this, it turns out that the solution in workbook is correct, so I made a mistake and I don't know where, so I would appreciate some help, and I have a question, why the second substitution works better in this example despite the theorem i mentioned above which says that I should use third substitution for this example?
| Let me try do derive that antiderivative. You computed:
$$f(x)=\underbrace{-2\arcsin\sqrt{\frac{2-x}{2}}}_{f_1(x)}\underbrace{-\sqrt{2x-x^2}}_{f_2(x)}.$$
The easiest term is clearly $f_2$:
$$f_2'(x)=-\frac{1}{2\sqrt{2x-x^2}}\frac{d}{dx}(2x-x^2)=\frac{x-1}{\sqrt{2x-x^2}}.$$
Now the messier term. Recall that $\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}}$. So:
\begin{align*}
f_1'(x)={}&-2\frac{1}{\sqrt{1-\left(\sqrt{\frac{2-x}{2}}\right)^2}}\frac{d}{dx}\sqrt{\frac{2-x}{2}}=-2\frac{1}{\sqrt{1-\frac{2-x}{2}}}\cdot\frac{1}{\sqrt2}\frac{d}{dx}\sqrt{2-x}={} \\
{}={}&-2\sqrt{\frac2x}\cdot\frac{1}{\sqrt2}\cdot\frac{1}{2\sqrt{2-x}}\cdot(-1)=\frac{2}{\sqrt x}\frac{1}{2\sqrt{2-x}}=\frac{1}{\sqrt{2x-x^2}}.
\end{align*}
So:
$$f'(x)=f_1'(x)+f_2'(x)=\frac{x}{\sqrt{2x-x^2}}=\frac{x}{\sqrt x}\frac{1}{\sqrt{2-x}}=\frac{\sqrt x}{\sqrt{2-x}},$$
which is your integrand. So you were correct after all! Or at least got the correct result, but no matter how I try, I cannot find an error in your calculations.
As for the book's solution, take your $f$, and compose it with $g(x)=2-x$. You get the book's solution, right? Except for a sign. But then $g'(x)=-1$, so the book's solution is also correct: just a different change of variables, probably, though I cannot really guess which.
| {
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"answer_id": 1
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What's the formula for this series for $\pi$? These continued fractions for $\pi$ were given here,
$$\small
\pi = \cfrac{4} {1+\cfrac{1^2} {2+\cfrac{3^2} {2+\cfrac{5^2} {2+\ddots}}}}
= \sum_{n=0}^\infty \frac{4(-1)^n}{2n+1}
= \frac{4}{1} - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} + \cdots\tag1
$$
$$\small
\pi = 3 + \cfrac{1^2} {6+\cfrac{3^2} {6+\cfrac{5^2} {6+\ddots}}}
= 3 - \sum_{n=1}^\infty \frac{(-1)^n} {n (n+1) (2n+1)}
= 3 + \frac{1}{1\cdot 2\cdot 3} - \frac{1}{2\cdot 3\cdot 5} + \frac{1}{3\cdot 4\cdot 7} - \cdots\tag2
$$
$$\small
\pi = \cfrac{4} {1+\cfrac{1^2} {3+\cfrac{2^2} {5+\cfrac{3^2} {7+\ddots}}}}
= 4 - 1 + \frac{1}{6} - \frac{1}{34} + \frac {16}{3145} - \frac{4}{4551} + \frac{1}{6601} - \frac{1}{38341} + \cdots\tag3$$
Unfortunately, the third one didn't include a closed-form for the series. (I tried the OEIS using the denominators, but no hits.)
Q. What's the series formula for $(3)$?
| The third one should be obtained from $4.1.40$ in A&S p.68 using $z:=ix$ (from Euler I think not sure) :
$$-2\,i\,\log\frac{1+ix}{1-ix} = \cfrac{4x} {1+\cfrac{(1x)^2} {3+\cfrac{(2x)^2} {5+\cfrac{(3x)^2} {7+\ddots}}}}
$$
Except that the expansion of the function at $x=1$ is simply your expansion for $(1)$.
Some neat variants :
$$\varphi(x):=\int_0^{\infty}\frac{e^{-t}}{x+t}dt= \cfrac{1} {x+1-\cfrac{1^2} {x+3-\cfrac{2^2} {x+5-\cfrac{3^2} {x+7-\ddots}}}}$$
$$\text{the previous one was better for large $x$...}$$
$$\int_0^{\infty}e^{-t}\left(1+\frac tn\right)^n\,dt=1+ \cfrac{n} {1+\cfrac{1(n-1)} {3+\cfrac{2(n-2)} {5+\cfrac{3(n-3)} {7+\ddots}}}}$$
$$\sum_{k=0}^\infty\frac 2{(x+2k+1)^2}= \cfrac{1} {x+\cfrac{1^4} {3x+\cfrac{2^4} {5x+\cfrac{3^4} {7x+\ddots}}}}$$
$$\text{and thus $\dfrac{\zeta(2)}2$ for $x=1$ (Stieltjes)}$$
$$\text{The last one was obtained after division by $n$ at the limit $n=0$ :}$$
$$\begin{align}
\int_0^1\frac{t^{x-n}-t^{x+n}}{1-t^2}dx&=\sum_{k=0}^\infty\frac 1{x-n+2k+1}-\frac 1{x+n+2k+1}\\
&=\cfrac{n} {x+\cfrac{1^2(1^2-n^2)} {3x+\cfrac{2^2(2^2-n^2)} {5x+\cfrac{3^2(3^2-n^2)} {7x+\ddots}}}}\\
\end{align}$$
Your continued fraction appears too in a neat and recent book by Borwein, van der Poorten, Shallit, Zudilin "Neverending Fractions: An Introduction to Continued Fractions" at the end of the pages $167-169$ reproduced for convenience here (hoping there is no problem with that...) :
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Proof of an identity that relates hyperbolic trigonometric function to an expression with euclidean trigonometric functions.
Given a line $r$ and a (superior) semicircle perpendicular to $r$, and an arc $[AB]$ in the semicircle, I need to prove that
$$
\sinh(m(AB)) = \frac{\cos(\alpha)+\cos(\beta)}{\sin(\alpha)\sin(\beta)} \\
\cosh(m(AB)) = \frac{1 + \cos(\alpha)\cos(\beta)}{\sin(\alpha)\sin(\beta)}
$$
where $\alpha$ is $\angle A'OA$ and $\beta$ is $\angle B'OB$.
The argument $m(AB)$ of $sinh(m(AB))$ and $cosh(m(AB))$ above is the hyperbolic Cayley-Klein hyperbolic metric of a hyperbolic segment
$$
m(AB) = \ln \left| \frac{AA' \cdot BB'}{BA' \cdot AB'} \right|
$$
where $AB$ is the euclidean measure.
I tried writing the left side of both of them as the exponential definition. But it is so hard to manipulate the right side because always appear other segments like AO, BO and AB...
Thanks.
| Using the definition of the hyperbolic sine and hyperbolic cosine functions, we have
$$
\sinh m(AB)
= \frac{e^{m(AB)} - e^{-m(AB)}}{2}
= \frac{(AA'\cdot BB')^2 - (BA'\cdot AB')^2}{2(AA'\cdot BB')(BA'\cdot AB')} \\
\cosh m(AB)
= \frac{e^{m(AB)} + e^{-m(AB)}}{2}
= \frac{(AA'\cdot BB')^2 + (BA'\cdot AB')^2}{2(AA'\cdot BB')(BA'\cdot AB')} \\
$$
Using the law of cosines and letting $R$ be the radius of the circle, we have
$$
\begin{eqnarray}
AA'^2 &=& 2R^2(1 - \cos \alpha) \\
BB'^2 &=& 2R^2(1 - \cos \beta) \\
AB'^2 &=& 2R^2(1 + \cos \alpha) \\
BA'^2 &=& 2R^2(1 + \cos \beta) \\
\end{eqnarray}
$$
Combining these for convenience, we have
$$
\begin{eqnarray}
(AA' \cdot BB')^2 &=& 4R^4(1 - \cos \alpha - \cos \beta + \cos \alpha \cos \beta) \\
(BA' \cdot AB')^2 &=& 4R^4(1 + \cos \alpha + \cos \beta + \cos \alpha \cos \beta) \\
(AA' \cdot BB')^2 - (BA' \cdot AB')^2 &=& 8R^4(\cos \alpha + \cos \beta) \\
(AA' \cdot BB')^2 + (BA' \cdot AB')^2 &=& 8R^4(1 + \cos \alpha \cos \beta) \\
(AA' \cdot BB')(BA' \cdot AB') &=& 4R^4 \sin \alpha \sin \beta
\end{eqnarray}
$$
Plugging these back into the hyperbolic sine and hyperbolic cosine functions, we finally have
$$
\sinh m(AB) = \frac{8R^4(\cos \alpha + \cos \beta)}{2 \cdot 4R^4 \sin \alpha \sin \beta} = \frac{\cos \alpha + \cos \beta}{\sin \alpha \sin \beta} \\
\cosh m(AB) = \frac{8R^4(1 + \cos \alpha \cos \beta)}{2 \cdot 4R^4 \sin \alpha \sin \beta} = \frac{1 + \cos \alpha \cos \beta}{\sin \alpha \sin \beta} \\
$$
Which is what we wanted to prove.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does $a_{n} = \frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + ... + \frac{1}{\sqrt{n^2+2n-1}}$ converge? $a_{n} = \frac{1}{\sqrt{n^2+n}} + \frac{1}{\sqrt{n^2+n+1}} + ... + \frac{1}{\sqrt{n^2+2n-1}}$
and I need to check whether this sequence converges to a limit without finding the limit itself. I think about using the squeeze theorem that converges to something (I suspect '$1$').
But I wrote $a_{n+1}$ and $a_{n-1}$ and it doesn't get me anywhere...
| On the one hand,
$$a_n \ge \mbox{smallest summand} \times \mbox{number of summands}= \frac{1}{\sqrt{n^2+2n-1}}\times n .$$
To deal with the denominator, observe that
$$n^2+2n-1 \le n^2+2n+1=(n+1)^2.$$
On the other hand,
$$a_n \le \mbox{largest summand}\times \mbox{number of summands} = \frac{1}{\sqrt{n^2+n}}\times n.$$
To deal with the denominator, observe that
$$n^2+n \ge n^2$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1690092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove that: $[12\sqrt[n]{n!}]{\leq}7n+5$? How prove that: $[12\sqrt[n]{n!}]{\leq}7n+5$,$n\in N$ I know $\lim_{n\to \infty } (1+ \frac{7}{7n+5} )^{ n+1}=e$ and $\lim_{n\to \infty } \sqrt[n+1]{n+1} =1$.
| By AM-GM
$$\frac{1+2 + 3 + \cdots + n}{n} \ge \sqrt[n]{1 \times 2 \times 3 \times \cdots \times n}$$
$$\implies \frac{n+1}2 \ge \sqrt[n]{n!} \implies 6n+6 \ge 12\sqrt[n]{n!}$$
But $7n+5 \ge 6n+6$ for $n \ge 1$...
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Solving for $k$ when $\arg\left(\frac{z_1^kz_2}{2i}\right)=\pi$ Consider $$|z|=|z-3i|$$
We know that if $z=a+bi\Rightarrow b=\frac{3}{2}$
$z_1$ and $z_2$ will represent two possible values of $z$ such that $|z|=3$. We are given $\arg(z_1)=\frac{\pi}{6}$
The value of $k$ must be found assuming $\arg\left(\frac{z_1^kz_2}{2i}\right)=\pi$
My attempt:
We know $z_1=\frac{3\sqrt{3}}{2}+\frac{3}{2}i$ and $z_2=-\frac{3\sqrt{3}}{2}+\frac{3}{2}i$ by solving for $a$.
So let $z_3=\frac{z_1^kz_2}{2i}$ $$z_3=\frac{z_1^kz_2}{2i} = \frac{i\left(\frac{3\sqrt{3}}{2}+\frac{3}{2}i\right)^k\left(-\frac{3\sqrt{3}}{2}+\frac{3}{2}i\right)}{-2}$$
We know $$\arg(z_3)=\arctan\left(\frac{\operatorname{Im}(z_3)}{\operatorname{Re}(z_3)}\right)=\pi \Rightarrow \frac{\operatorname{Im}(z_3)}{Re(z_3)}=\tan(\pi)=0$$
This is the part where I get stuck; I assume that $\operatorname{Re}(z_3)\neq0$ and then make the equation $$\operatorname{Im}(z_3)=0$$
However, I am not sure on how to get the value of $k$ from this, or if I am in the right direction.
What should I do in order to get the value of $k$?
| You already know that $\arg(z_1)=\frac{\pi}{6}$, and moreover $\arg(z_2)=\frac{5\pi}{6}$ and $\arg \left(\frac{1}{2i}\right)=\frac{-\pi}{2}$. Multiplying complex numbers results in adding their arguments (modulo $2\pi$) so you get the equation
$$\arg\left(\frac{z_1^kz_2}{2i}\right)=k\frac{\pi}{6}+\frac{5\pi}{6}-\frac{\pi}{2}=\frac{(k+2)\pi}{6}=\pi+2m\pi$$which gives you$$k=4+12m$$where $m\in \mathbb{Z}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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closed form for $I(n)=\int_0^1\left ( \frac{\pi}{4}-\arctan x \right )^n\frac{1+x}{1-x}\frac{dx}{1+x^2}$ $$I(n)=\int_0^1\left ( \frac{\pi}{4}-\arctan x \right )^n\frac{1+x}{1-x}\frac{dx}{1+x^2}$$
for $n=1$ I tried to use $\arctan x=u$
and by notice that $$\frac{1+\tan u}{1-\tan u}=\cot\left ( \frac{\pi}{4}-u \right )$$ then $\frac{\pi}{4}-u=y$ and got
$$I(1)=\int_0^{\pi/4}y\cot y dy$$
which equal to $$I(1)=\frac{1}{8}(4G +\pi \log 2)$$
Where $G$ is Catalan constant
so $$I(n)=\int_0^{\pi/4}x^n\cot x dx$$
but how to find the closed form for any n using real or complex analysis ?
and what about $I(2)$ it seems related to $\zeta(3)$ ?!!
| For $n=2$ we have, integrating by parts, $$I\left(2\right)=\int_{0}^{\pi/4}x^{2}\cot\left(x\right)dx=\frac{\pi^{2}}{16}\log\left(\frac{1}{\sqrt{2}}\right)-2\int_{0}^{\pi/4}x\log\left(\sin\left(x\right)\right)dx
$$ and now we can use the Fourier series of $\log\left(\sin\left(x\right)\right)$ $$\log\left(\sin\left(x\right)\right)=-\log\left(2\right)-\sum_{k\geq1}\frac{\cos\left(2kx\right)}{k},\,0<x<\pi$$ and so $$I\left(2\right)=\frac{\pi^{2}}{16}\log\left(\frac{1}{\sqrt{2}}\right)+\frac{\pi^{2}}{16}\log\left(2\right)+2\sum_{k\geq1}\frac{1}{k}\int_{0}^{\pi/4}x\cos\left(2kx\right)dx
$$ $$=\frac{\pi^{2}}{32}\log\left(2\right)+\pi\sum_{k\geq1}\frac{\sin\left(\frac{\pi k}{2}\right)}{4k^{2}}+\frac{1}{2}\sum_{k\geq1}\frac{\cos\left(\frac{\pi k}{2}\right)}{k^{3}}-\frac{1}{2}\zeta\left(3\right)
$$ and now since $$\cos\left(\frac{\pi k}{2}\right)=\begin{cases}
-1, & k\equiv2\,\mod\,4\\
1, & k\equiv0\,\mod\,4\\
0, & \textrm{otherwise}
\end{cases}
$$ we have $$\frac{1}{2}\sum_{k\geq1}\frac{\cos\left(\frac{\pi k}{2}\right)}{k^{3}}=\frac{1}{2}\sum_{k\geq1}\frac{\left(-1\right)^{k}}{8k^{3}}=-\frac{3}{64}\zeta\left(3\right)
$$ using the relation between the Dirichlet eta function and the Riemann zeta funcion. Similary, since $$\sin\left(\frac{\pi k}{2}\right)=\begin{cases}
-1, & k\equiv3\,\mod\,4\\
1, & k\equiv1\,\mod\,4\\
0, & \textrm{otherwise}
\end{cases}
$$ we have $$\sum_{k\geq1}\frac{\sin\left(\frac{\pi k}{2}\right)}{k^{2}}=\sum_{k\geq1}\frac{\left(-1\right)^{k-1}}{\left(2k-1\right)^{2}}=K
$$ where $K$ is the Catalan's constant. Finally we have $$I\left(2\right)=\frac{\pi^{2}}{32}\log\left(2\right)+\frac{\pi}{4}K-\frac{35}{64}\zeta\left(3\right).$$
Addendum: As tired notes, this method can be generalized for a general $n$. I will write only a sketch of the proof. Integrating by parts we have $$I\left(n\right)=\int_{0}^{\pi/4}x^{n}\cot\left(x\right)dx=\frac{\pi^{n}}{4^{n}}\log\left(\frac{1}{\sqrt{2}}\right)-n\int_{0}^{\pi/4}x^{n-1}\log\left(\sin\left(x\right)\right)dx
$$ and using the Fourier series we get $$I\left(n\right)=\frac{\pi^{n}}{4^{n}}\log\left(\frac{1}{\sqrt{2}}\right)+\frac{\pi^{n}}{4^{n}}\log\left(2\right)+n\sum_{k\geq1}\frac{1}{k}\int_{0}^{\pi/4}x^{n-1}\cos\left(2kx\right)dx$$ $$=\frac{\pi^{n}}{2^{2n+1}}\log\left(2\right)+\frac{n}{2^{n}}\sum_{k\geq1}\frac{1}{k^{n+1}}\int_{0}^{k\pi/2}y^{n-1}\cos\left(y\right)dy$$ and now the last integral can be calculated using an iterating integration by parts. We will get series very similar to the other case, which can be treated with the same techinques.
| {
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"url": "https://math.stackexchange.com/questions/1697651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Trigonometric inequality in sec(x) and csc(x) How can I prove the following inequality
\begin{equation*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right) \geq 3+%
\sqrt{2},~~~\forall x\in \left( 0,\frac{\pi }{2}\right) .
\end{equation*}%
I tried the following
\begin{eqnarray*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right) &\geq
&\left( 1+1\right) \left( 1+\frac{1}{\cos x}\right) \\
&=&2\left( 1+\frac{1}{\cos x}\right) \\
&\geq &2\left( 1+1\right) =4,
\end{eqnarray*}
but $4\leq 3+\sqrt{2}$.
| Expand the expression to get
$$\left(1+\frac{1}{\sin x}\right)\left(1+\frac{1}{\cos x}\right)=1+\frac{1}{\sin x}+\frac{1}{\cos x}+\frac{1}{\sin x\cos x}$$
Then using the identity $\sin x\cos x = \frac{1}{2}\sin 2x$, rewrite as
\begin{eqnarray*}
\left( 1+\frac{1}{\sin x}\right) \left( 1+\frac{1}{\cos x}\right) &=
& 1+\frac{1}{\sin x}+\frac{1}{\cos x}+\frac{2}{\sin 2x} \\
&\geq&1+1+1+2 \text{ for }x\in (0,\frac{\pi}{2}) \\
&= &5\\
&\geq & 3+\sqrt{2}\text{ since } 2\geq \sqrt{2}
\end{eqnarray*}
| {
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$a,b,c,d,e$ are positive real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the range of $e$.
$a,b,c,d,e$ are positive real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the range of $e$.
My book tells me to use tchebycheff's inequality
$$\left(\frac{a+b+c+d}{4}\right)^2\le \frac{a^2+b^2+c^2+d^2}{4}$$
But this not the Chebyshev's inequality given in wikipedia. Can someone state the actual name of the inequality so I can read more about it?
(I got $e\in\left[0,\frac{16}{5}\right]$ using the inequality)
| As @ChenJiang stated, its a case of cauchy's inequality
$$\left(\frac{a+b+c+d}{4}\right)^2\le \frac{a^2+b^2+c^2+d^2}{4}$$
$$(a+b+c+d)^2\le 4(a^2+b^2+c^2+d^2)$$
$$(8-e)^2\le 4(16-e^2)$$
$$5e^2-16e\le 0$$
$$e(5e-16)\le 0$$
$$\implies 0\le e\le \frac{16}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Use Cauchy Product to Find Series Representation of $[\ln(1 + x)]^2$ Problem:
Let $f(x) = [\ln(1 + x)]^2$. Use the series for the logarithm to compute that
\begin{align*}
f(x) = [\ln(1 + x)]^2 = \sum_{n = 2}^{\infty}(-1)^n\Bigg(\sum_{k = 1}^{n - 1} \frac{1}{(n - k)k}\Bigg) x^n.
\end{align*}
Use this to evaluate the 5th derivative of $f$ evaluated at $0$.
My attempt:
Using $\ln(1 + x) = \sum_{k = 1}^{\infty} (-1)^{k - 1} \frac{x^k}{k}$, the Cauchy product of $f(x) = (\ln(1 + x))(\ln(1 + x))$ gives us
\begin{align*}
f(x) &= \sum_{n = 1}^{\infty} \Bigg(\sum_{k = 1}^{n} \frac{(-1)^{n - k - 1}}{(n - k)} \Bigg(\frac{(-1)^{k - 1}}{k}\Bigg)\Bigg)x^n\\
&= \sum_{n = 1}^{\infty} \Bigg(\sum_{k = 1}^{n} \frac{(-1)^{n - 2}}{(n - k)k} \Bigg)x^n\\
&= \sum_{n = 1}^{\infty} \Bigg(\sum_{k = 1}^{n} \frac{(-1)^{n - 2}}{(n - k)k} \Bigg)x^n\\
&= \sum_{n = 1}^{\infty} (-1)^{n}\Bigg(\sum_{k = 1}^{n} \frac{1}{(n - k)k} \Bigg)x^n.\\
\end{align*}
This has the form of the solution but I'm having trouble shifting the indices. As for using it to find the 5th derivative, I just need to take the 5th derivative of the series and plug in $x = 0$? Thanks for your help!
| Note that for $-1<x< 1$ we have
$$\begin{align}
\log^2(1+x)&=\sum_{k=1}^\infty\sum_{m=1}^\infty\frac{(-1)^{k+m}x^{k+m}}{k\,m}\tag 1\\\\
&=\sum_{k=1}^\infty\sum_{n=k+1}^\infty\frac{(-1)^{n}x^{n}}{k\,(n-k)} \tag 2\\\\
&=\sum_{n=2}^\infty\sum_{k=1}^{n-1}\frac{(-1)^{n}x^{n}}{k\,(n-k)} \tag 3\\\\
\end{align}$$
as was to be shown!
NOTES:
In going from $(1)$ to $(2)$, we introduced a new index $n=k+m$. Then, with $m=n-k$, the lower limit for $n$ begins at $k+1$.
In going from $(2)$ to $(3)$, we interchanged the order of summation and note that $n\ge 2$ from $(2)$.
| {
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Linear system 2 unknowns There are $x$ white and $y$ black pearls and their ratio is $z$. If I add six black and six white pearles, the ratio doubles.
I did the following:
$ \frac{x+6}{y+6} = \frac{2x}{y}$ and then I get
$xy -6(2x-y)=0$
I can find solutions by guessing. Is there any other way?
ADDED:
Now I have to solve the problem for
$ \frac{c(x+y)+6}{x+y+12} = 2c $ and once again I am lost in factoring out variables.
| $\frac{x}{y} = z$ and $\frac{x+6}{y+6} = 2z$
$\implies x = zy$ and $\frac{zy+6}{y+6} = 2z$
$\implies 6 = z(y+12) \implies z = \frac{6}{y+12}$
$\implies x = zy = \frac{6y}{y+12}$
Since $x$ and $y$ are the number of pearls, they must be integers. That is
$(y+12) | (6y)$.
There are only 3 possible $y$ values by giving $x=2, x=3, x=4$. Other $x$ values gives non-integer or non-positive $y$'s.
Thus, all possible $(x,y)$ pairs are $(2,6), (3,12), (4,24)$
| {
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If $p>3$ and $p+2$ are twin primes then $6\mid p+1$ I have to prove that if $p$ and $p+2$ are twin primes, $p>3$, then $6\ |\ (p+1)$.
I figure that any prime number greater than 3 is odd, and therefore $p+1$ is definitely even, therefore $2\ |\ (p+1)$. And if I can somehow prove $3\ |\ (p+1)$, then I would be done. But I'm not sure how to do that. Help!
| The easy way.
Note that one of $p,p+1,p+2$ must be divisible by $3$, since they are three consecutive numbers, and since $p$ and $p+2$ are prime, that must be $p+1$. We can do the same to show that $p+1$ is divisible by $2$.
Looking modulo $6$.
We can look $\mod 6$. We see that
\begin{align}
6k+0\equiv 0\mod 6&\Rightarrow 6|6k+0\\
6k+1\equiv 1\mod 6&\Rightarrow \text{possibly prime}\\
6k+2\equiv 2\mod 6&\Rightarrow 2|6k+2\\
6k+3\equiv 3\mod 6&\Rightarrow 3|6k+3\\
6k+4\equiv 4\mod 6&\Rightarrow 2|6k+4\\
6k+5\equiv 5\mod 6&\Rightarrow \text{possibly prime}\\
\end{align}
So for a number to be prime it must be either of the form $6k+1$ or $6k-1$ (equivalent to $6k+5$ since $6k-1=6(k-1)+5$). So if you have two primes, $p$ and $p+2$, then $p=6k-1$ and $p+2=6k+1$ for some $k$; thus, $p+1=6k$ is a multiple of $6$.
| {
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Solve $ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $
Question: Solve $$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 0 \le x \le 360^{\circ} $$
My attempt:
$$ 6\sin^2(x)+\sin(x)\cos(x)-\cos^2(x)=5 $$
$$ 6(\frac{1}{2} - \frac{\cos(2x)}{2}) + \sin(x)\cos(x) -(\frac{1}{2} + \frac{\cos(2x)}{2}) = 5 $$
$$ 3 - 3\cos(2x)+ \sin(x)\cos(x) - \frac{1}{2} - \frac{\cos(2x)}{2} = 5$$
$$ \frac{7\cos(2x)}{2} - \sin(x)\cos(x) + \frac{5}{2} = 0 $$
$$ 7\cos(2x) - 2\sin(x)\cos(x) + 5 = 0 $$
$$ 7\cos(2x) - \sin(2x) + 5 = 0 $$
So at this point I am stuck what to do, I have attempted a Weierstrass sub of $\tan(\frac{x}{2}) = y$ and $\cos(x) = \frac{1-y^2}{1+y^2}$ and $\sin(x)=\frac{2y}{1+y^2} $ but I got a quartic and I was not able to solve it.
| HINT
Perhaps the other methods are easier but to continue where you left off,
Realize that $$\sin 2x=7\cos 2x+5$$
Use $$\sin^2 2x+\cos^2 2x=1$$
To make your last equation into a quadratic for $\cos 2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 2
} |
If a, b, c are three natural numbers with $\gcd(a,b,c) = 1$ such that $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$ then show that $a+b$ is a square.
If a, b, c are three natural numbers with $\gcd(a,b,c) = 1$ such that $$\frac{1}{a} + \frac{1}{b}= \frac{1}{c}$$ then show that $a+b$ is a perfect square.
This can be simplified to:
$$a+b = \frac{ab}{c}$$
Also, first few such examples of $(a,b,c)$ are $(12, 4, 3)$ and $(20, 5, 4)$. So, I have a feeling that $b$ and $c$ are consecutive.
I don't think I have made much progress.
Any help would be appreciated.
| To show this, we note that $c(a+b)=ab$. Now let $g$ be the gcd of $a$ and $b$, which need not necessarily be $1$. Denote $a=a'g$ and $b=b'g$ so that we get $c(a'+b') = a'b'g$.
Because $a' + b'$ is relatively prime to both $a'$ and $b'$, it follows that it divides $g$. But g also divides $c(a'+b')$. Further, note that $g$ is coprime to $c$, because it was the gcd of $a$ and $b$, so that gcd($g$,$c$)=gcd($a$,$b$,$c$)=1. It follows that $a'+b'=g$, and therefore that $a+b = (a'+b')g = g^2$.
For example, $\frac{1}{6} + \frac{1}{3} = \frac{1}{2}$ and $(3,6)=3$ and $3+6=9=3^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
If the tangent at the point $P$ of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the major axis and minor axis at $T$ and $t$ respectively If the tangent at the point $P$ of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the major axis and minor axis at $T$ and $t$ respectively and $CY$ is perpendicular on the tangent from the center,then prove that $Tt.PY=a^2-b^2$
Let the point $P$ be $(a\cos\theta,b\sin\theta)$,then the equation of the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1$.
It meets the major axis at $T(a\sec\theta,0)$ and the minor axis at $t(0,b\csc\theta)$
$Tt=\sqrt{a^2\sec^2\theta+b^2\csc^2\theta}$
Now I found $Y$,the foot of perpendicular from center $C$ of the ellipse to the tangent as $(\frac{\frac{\cos\theta}{a}}{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}},\frac{\frac{\sin\theta}{b}}{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}})$
$PY=\sqrt{(a\cos\theta-\frac{\frac{\cos\theta}{a}}{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}})^2+(b\sin\theta-\frac{\frac{\sin\theta}{b}}{\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2}})^2}$
I simplified this expression to get $PY=\frac{(a^2-b^2)\sin\theta\cos\theta}{(\frac{\cos^2\theta}{a^2}+\frac{\sin^2\theta}{b^2})a^2b^2}\sqrt{a^4\sin^2\theta+b^4\cos^2\theta}$ and $Tt=\sqrt{a^2\sec^2\theta+b^2\csc^2\theta}=\frac{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}}{\sin\theta \cos\theta}$
But $PY.Tt$ is not $a^2-b^2$I do not understand where have i gone wrong.Maybe i found $Y$ wrong,i used the formula for the foot of the perpendicular $(x',y')$ from $(x_1,y_1)$ on the line $Ax+By+C=0$ as $\frac{x'-x_1}{A}=\frac{y'-y_1}{B}=\frac{-(Ax_1+By_1+C)}{A^2+B^2}$
| Equation of $Tt$ is $bx\cos \theta + ay\sin \theta - ab = 0$. Hence $C{Y^2} = \frac{{ab}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$ and $C{P^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta $. Now,
$$P{Y^2} = C{P^2} - C{Y^2} = \frac{{ab}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }} - \left( {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } \right)$$
$$ = \frac{{{a^2}{b^2}\left( {{{\cos }^4}\theta + {{\sin }^4}\theta - 1} \right) + {{\sin }^2}\theta {{\cos }^2}\theta \left( {{a^4} + {b^4}} \right)}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$$
$$ = \frac{{{a^2}{b^2}\left( {{{\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right)}^2} - 2{{\sin }^2}\theta {{\cos }^2}\theta - 1} \right) + {{\sin }^2}\theta {{\cos }^2}\theta \left( {{a^4} + {b^4}} \right)}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$$
$$ = \frac{{{a^2}{b^2}\left( {1 - 2{{\sin }^2}\theta {{\cos }^2}\theta - 1} \right) + {{\sin }^2}\theta {{\cos }^2}\theta \left( {{a^4} + {b^4}} \right)}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$$
$$ = \frac{{{{\sin }^2}\theta {{\cos }^2}\theta {{\left( {{a^2} - {b^2}} \right)}^2}}}{{{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta }}$$
$$\Rightarrow PY = \frac{{\left( {{a^2} - {b^2}} \right)\sin \theta \cos \theta }}{{\sqrt {{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta } }}$$
Hence, $Tt \cdot PY=a^2-b^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1703222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can I calculate this limit $\lim_{(x,y)\to(0,0)} \frac{xy(1-cos(x^2+y^2))}{(x^2+y^2)^{\frac{5}{2}}}$? How can I calculate this limit $$\lim_{(x,y)\to(0,0)} \dfrac{xy(1-cos(x^2+y^2))}{(x^2+y^2)^{\frac{5}{2}}}$$ at the origin?
I tried to use the substitution $x ^2 + y^2=t$ but how can I evaluate the value of $xy$? I even tried to use polar coordinates but to avail.
| Outline: Note that $(x-y)^2\ge 0$, so $|xy|\le \frac{1}{2}(x^2+y^2)$. One can also get this from polar coordinates, for $|xy|=r^2|\cos\theta\sin\theta|=\frac{1}{2}r^2|\sin(2\theta)|\le \frac{r^2}{2}$.
Now you can comfortably let $t=x^2+y^2$. You will need to look at the behaviour of $1-\cos t$ near $0$. This can be taken care of mechanically by using L'Hospital's Rule. Or else one can use the beginning of the Maclaurin series for $\cos t$. Or else one can note that
$1-\cos t=\frac{1-\cos^2 t}{1+\cos t}=\frac{\sin^2 t}{1+\cos t}$.
The conclusion will be that the limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Expansion and factorisation I have a little problems with a few questions here and I need help.. Thanks ...
*
*Factorise completely
$$9x^4 - 4x^2 - 9x^2y^2 + 4y^2 $$
My workings ..
$$ (3x^2+2x)(3x^2-2x) - y^2 (9x^2-4) = (3x^2 + 2x)(3x^2 -2x) - y^2 (3x+2)(3x-2) $$
*Factorise $3x^2 + 11x - 20$ and , hence Factorise completely
$$11a - 11b - 20 + 3a^2 + 3b^2 - 6ab$$
My workings ...
$$ 11(a-b) - 20 + (3a-3b)(a-b)$$
*Evaluate the following by algebraic expansion of factorisation
(A) $78^2 + 312 + 4$
(B) $501^2 - 1002 + 1$
Thanks a lot !
| *
*$$9x^4 - 9x^2y^2 - 4x^2 + 4y^2$$
Group in paris such that; $9x^2(x^2-y^2) - 4(x^2 - y^2) = (x^2 - y^2)(9x^2 - 4)$
Then using the difference of 2 squares we get: $$(x-y)(x+y)(3x-2)(3x+2)$$
2.$3x^2 + 11x - 20$, factorise to $(x-5)(3x-4)$. As @mathlove pointed out $11(a-b) - 20 + 3(?)^2$, where $ ? = a-b$. Which is similar to the previous equation. So utilising it's result we get it factorised down to $(a-b-5)(3(a-b-4)$
*$78^2 + 312 + 4 = 78^2 + 316$, using the fact that $(a + b)^2 = a^2 + 2ab + b^2$, we get $(70 + 8)^2 + 316 = (4900 + 1120 + 64) + 316 = 6400$
*Apply the same reasoning as number 3
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
Question: Prove $\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$ $=$ $\frac{\tan(x) + \cot(x)}{\sec(x) + \csc(x)}$
My attempt:
$$\frac{\sec(x) - \csc(x)}{\tan(x) - \cot(x)}$$
$$ \frac{\frac {1}{\cos(x)} - \frac{1}{\sin(x)}}{\frac{\sin(x)}{\cos(x)} - \frac{\cos(x)}{\sin(x)}} $$
$$ \frac{\sin(x)-\cos(x)}{\sin^2(x)-\cos^2(x)}$$
$$ \frac{(\sin(x)-\cos(x))}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))} $$
$$ \frac{1}{\sin(x)+\cos(x)} $$
Now this is where I am stuck , I thought of multiplying the numerator and denominator by $$ \frac{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}}{\frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}} $$ but that did not work out well..
| multiply and divide $ \frac{1}{\sin{x}+\cos{x}}$ by $ \frac{1}{\sin{x}\cos{x}}$ then in the numerator substitute $1$ by $sin^2{x} + cos^2{x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Unknown Inequality $$ \left( \sqrt{3}\sqrt{y(x+y+z)}+\sqrt{xz}\right)\left( \sqrt{3}\sqrt{x(x+y+z)}+\sqrt{yz}\right)\left( \sqrt{3}\sqrt{z(x+y+z)}+\sqrt{xy}\right) \leq 8(y+x)(x+z)(y+z)$$
I can prove this inequality, but i need know if this inequaliy is known...
| We need to prove that
$$\prod\limits_{cyc}(a\sqrt{3(a^2+b^2+c^2)}+bc)\leq8\prod\limits_{cyc}(a^2+b^2)$$
which is true even for all reals $a$, $b$ and $c$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, and $abc=w^3$.
Hence, it's obvious that the last inequality is equivalent yo $f(w^3)\leq0$, where $f$ is a convex function.
Hence, $f$ gets a maximal value for an extremal value of $w^3$.
We know that an equation $(x-a)(x-b)(x-c)=0$ or $x^3-3ux^2+3v^2x=w^3$ has three real roots $a$, $b$ and $c$.
Thus, a graph $y=w^3$ cross a graph $y=x^3-3ux^2+3v^2x$ in three points (maybe two of them coincide)
Let $u$ and $v^2$ be constants.
Hence, $w^3$ gets an extremal value for an equality case of two variables.
Id est, it remains to prove our inequality for $b=c=1$, which gives
$(x-1)^2(13x^2+34x+25)\geq0$, which is obvious. Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to determine if the following series converge or not? $\Sigma_{n=1}^{\infty} a_n $ where:
*
*$ a_n = \frac{1}{\ln(n)^{\ln(n)}}$
*$a_n = \frac{1}{n }-\ln\left( 1+\frac{1}{n}\right)$
in the first case, I really have no idea
in the second case, is it correct to say that for $ \frac{1}{n }-\ln\left( 1+\frac{1}{n}\right)$ is (by taylor expansion) $\frac{1}{2n^2}+O(\frac{1}{n^3})$ and therefore, by the limit comparison test converges?Is there any other way?
Thanks in advance
| For
2.
$\begin{array}\\
a_n
&= \frac{1}{n }-\ln\left( 1+\frac{1}{n}\right)\\
&= \frac{1}{n }-\int_1^{1+1/n} \frac{dx}{x}\\
&= \frac{1}{n }-\int_0^{1/n} \frac{dx}{1+x}\\
&= \int_0^{1/n} (1-\frac{1}{1+x})dx\\
&= \int_0^{1/n} (\frac{x}{1+x})dx\\
&< \int_0^{1/n} x\,dx\\
&= \frac{x^2}{2}|_0^{1/n}\\
&= \frac{1}{2n^2}\\
\end{array}
$
For a lower bound,
from the integral,
$a_n > 0$.
To be more precise,
$\begin{array}\\
a_n
&= \int_0^{1/n} (\frac{x}{1+x})dx\\
&\gt \frac1{1+1/n}\int_0^{1/n} x\,dx\\
&= \frac1{1+1/n}\frac{x^2}{2}|_0^{1/n}\\
&= \frac1{1+1/n}\frac{1}{2n^2}\\
&= \frac{1}{2n(n+1)}\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Determine whether $p_n$ is decreasing or increasing, if $p_{n+1} = \frac{p_n}{2} + \frac{1}{p_n}$ If $p_1 = 2$ and $p_{n+1} = \frac{p_n}{2}+ \frac{1}{p_n}$, determine $p_n$ is decreasing or increasing.
Here are the first few terms:
$$p_2 = \frac{3}{2}, p_3 = \frac{3}{4} + \frac{2}{3} = \frac{17}{12}, p_4 = \frac{17}{24} + \frac{12}{17} = \frac{577}{408}$$
The sequence seems decreasing to me so I tried to prove it by induction. Need to prove $p_k - p_{k+1} \gt 0$ for all n.
For n = 1, $p_1 - p_2 = 2- \frac{3}{2} = \frac{1}{2} \gt 0$ (true)
However when I tried to prove it for $k+1$, I ran into problems.
Assume $p_k - p_{k+1} \gt 0$ is true for n =k, then it must be also true for $n =k+1$.
$$p_{k+1} - p_{k+2} = \frac{p_k}{2} + \frac{1}{p_k} - \frac{p_{k+1}}{2} - \frac{1}{p_{k+1}} = \frac{p_k - p_{k+1}}{2} + (\frac{1}{p_k}-\frac{1}{p_{k+1}})$$
but $\frac{1}{p_k}-\frac{1}{p_{k+1}} \lt 0$
I don't know what to do from there.
| AM/GM. $\frac{x}{2}+\frac{1}{x}<x$ iff $x^2>2$. So it is enough to show that if $x^2>2$ then $(\frac{x}{2}+\frac{1}{x})^2>2$ or $\frac{x^2}{4}+\frac{1}{x^2}>1$. But by AM/GM $(\frac{x^2}{4}+\frac{1}{x^2})/2>\sqrt{\frac{1}{4}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
sum of the series $\frac{2^n+3^n}{6^n}$ from $n=1$ to $\infty$ Find the sum of the series $\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}=?$
My thoughts: find $\sum_{n=1}^{\infty} 2^n$, $\sum_{n=1}^{\infty} 3^n$ and $\sum_{n=1}^{\infty} 6^n$ (although I don't know how yet...)
Then, $\sum_{n=1}^{\infty} \frac{2^n+3^n}{6^n}= \frac{\sum_{n=1}^{\infty} 2^n}{\sum_{n=1}^{\infty} 3^n} + \frac{\sum_{n=1}^{\infty} 3^n}{\sum_{n=1}^{\infty} 6^n}$, am I on the right track? I really don't know how else to solve it.
Thanks.
| $$\sum_{i=1}^\infty \frac{2^n + 3^n}{6^n}=\sum_{i=1}^\infty (\frac{2}{6})^n + \sum_{i=1}^\infty (\frac{3}{6})^n=\frac{1}{2}+1=\frac{3}{2}$$
sums from forumla of geometric series $1/3^n$ and $1/2^n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability Problem - Finding a pdf Below is a problem I did. However, I did not come up with the answer in book.
I am thinking that I might have the wrong limits for the integral. I am hoping
somebody can point out what I did wrong.
Bob
Problem
Let $Y = \sin X$, where $X$ is uniformly distributed over $(0, 2 \pi)$. Find
the pdf of $Y$.
Answer:
\begin{eqnarray*}
P(Y<=y_0) &=& P( \sin x <= y_0 ) = P( x <= \arcsin( y_0) ) \\
P(Y<=y_0) &=& \int_0^{\arcsin(y_0)} \frac{1}{2 \pi} dx =
\frac{x}{2 \pi} \Big|_0^{\arcsin(y_0)} =
\frac{\arcsin(y_0)}{2 \pi} - \frac{0}{2 \pi} \\
P(Y<=y_0) &=& \frac{\arcsin(y_0)}{2 \pi} \\
F(Y) &=& \frac{\arcsin(Y)}{2 \pi} \\
f(y) &=& \frac{1}{2 \pi \sqrt{1 - y^2}} \\
\end{eqnarray*}
However, the book gets:
\begin{eqnarray*}
f(y) &=& \frac{1}{\pi \sqrt{1 - y^2}} \\
\end{eqnarray*}
I believe that book is right because
\begin{eqnarray*}
\int_{-1}^{1} \frac{1}{\pi \sqrt{1 - y^2} } dy &=&
\frac{\arcsin{y}}{\pi} \Big|_{-1}^{1} =
\frac{\frac{\pi}{2}}{\pi} - \frac{-\frac{\pi}{2}}{\pi} \\
\int_{-1}^{1} \frac{1}{\pi \sqrt{1 - y^2} } dy &=& 1 \\
\end{eqnarray*}
| your answer is correct indeed just a little mistake. in the first step:
$$P(Y \le y) = P( \sin x \le y ) \Rightarrow P(x_1 \le x \le x_2 )$$ as shown in figure below (sorry figure is for $y=\sin(x+\theta)$ but still is useful .just draw the figure for $\sin(x)$ in your imagination. also figure is from "Probability, Random Variables and Stochastic Processes" by Papoulis A. chapter 5 example 7. which is very similar to your problem).
since $\sin(x)$ intersects with each horizontal line two times. so we can now write the right answer as below:
$$P(Y \le y)= \frac{1}{2\pi} \int_0^{x_{1}}dx+\frac{1}{2\pi} \int_{x_{2}}^{2\pi}dx$$
and we have in mind that $x_2=\pi-\arcsin(y)=\pi-x_1$ so we get:
$$P(Y \le y)= \frac{1}{2\pi} \int_0^{x_{1}}dx+\frac{1}{2\pi} \int_{x_{2}}^{2\pi}dx = \frac{1}{2\pi}\times (x_{1}+2\pi-x_{2})=\frac{1}{2\pi}\times (x_{1}+\pi+x_{1}) \\ = \frac{1}{2\pi}\times (2x_{1}+\pi)=\frac{1}{2}+\frac{1}{\pi}\arcsin(y)$$
now by taking derivative with respect to $y$ we get:
$$f(y) = \frac{1}{\pi \sqrt{1 - y^2}}$$
EDIT
To answer the question in comments, when $y=a\sin(x+\theta)$ and $x$ is uniform on $(0,2\pi)$ and we want to find $F(y),f(y)$ i.e. CDF and PDF of $y$. We want to find $F(y)=P(Y\le y)$. To find where $Y\le y$ notice that according to the figure above, this happens in two intervals (when the sine figure is less than the $y$ line) which is $x\in (-\pi,x_0)$ and when $x \in (x_1,\pi)$. Therefore
$$P(Y \le y) = P((-\pi,x_0) \cup (x_1,\pi)) = P((-\pi,x_0)) + P((x_1,\pi))$$
The total length of this interval is $(x_0 - (-\pi)) + (\pi-x_1) = 2\pi + x_0 - x_1$ and $x_0 = \text{Arcsin}\left(\frac{y}{a}\right)-\theta , x_1 = \pi - \text{Arcsin}\left(\frac{y}{a}\right) - \theta $ which results from solving $y = a\sin(x+\theta)$. Substituting these in
$$P(Y \le y) = \frac{1}{2\pi}(2\pi + x_0 - x_1) = \frac{1}{2} + \frac{1}{\pi}\text{Arcsin}\left(\frac{y}{a}\right)$$.
Also please notice that since $x_1 = \pi - x_0 - 2\theta$ the length of the interval can also be written as $L = 2\pi + x_0 - x_1 = \pi + 2x_0 + 2\theta$ which results in the same solution.
| {
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"url": "https://math.stackexchange.com/questions/1712936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to find the stationary point of $f(x,y)=\sin x \sin y \sin (x+y)$ How to find the stationary point(s) of $f(x,y)=\sin x \sin y \sin (x+y)$
With $x,y\in(0,\pi)$
So far I have found $$\nabla f =(\color{red}{\sin x\cos (x+y)+\cos x\sin (x+y)\sin y},\color{blue}{\sin y\cos (x+y)+\cos y\sin (x+y)\sin x)}$$
So we need
$$(\sin x \cos (x+y)+\cos x \sin (x+y)) \sin y=0 \\ (\sin y \cos (x+y)+\cos y\sin (x+y))\sin x=0$$
Problem is I don't know how to solve this.
| $\left(\sin x \cos(x+y) + \cos x\sin(x+y)\right) \sin y = 0$ when $\sin x \cos(x+y) + \cos x\sin(x+y) = 0$ or when $\sin y = 0$. I trust that you know when $\sin y = 0$. What about that first one?
\begin{align}
\sin x \cos(x+y) + \cos x\sin(x+y) &= 0 \\
\sin x \cos(x+y) &= -\cos x\sin(x+y)\\
-\tan x &= \tan(x+y)\\
\tan(-x) &= \tan(x+y)
\end{align}
Because tangent is periodic with period $\pi$, then this equation is satisfied when $x+y = -x + n\pi$, where $n$ is any integer (provided there are no domain restriction violations). This is equivalent to $y = -2x + n\pi$.
The second equation you mentioned is nearly identical.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I find the vector $T\begin{pmatrix} 5 & 0 \\ -10 & -13 \end{pmatrix}$? I defined a function $T: M^R_{2x2} \rightarrow R_4[x]$ and I defined:
$T\begin{pmatrix} 2 & 3 \\ 1 & 0 \end{pmatrix} = x^2$
$T\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} = 3x - 4$
$T\begin{pmatrix} 0 & 2 \\ 4 & 5 \end{pmatrix} = 2x^2 - 7$
How can I find the following function if I know that all three vectors above are linearly independent:
$T\begin{pmatrix} 5 & 0 \\ -10 & -13 \end{pmatrix} = ?$
What is the method to handle such questions?
| Hint: You'll need to determine whether there exist $a,b,c$ for which
$$
\pmatrix{5&0\\-10&-13} =
a\pmatrix{2&3\\1&0} +
b\pmatrix{1&0\\0&2} +
c \pmatrix{0&2\\4&5}
$$
and, if such $a,b,c$ exist, find them.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\frac{y}{x}$ from $3x + 3y = yt = xt + 2.5x$ I need to find the ratio of
$$\frac{y}{x}$$
If given that
$$3x + 3y = yt = xt + 2.5x$$
So what I tried is:
$$t = \frac{3x + 3y}{y}$$
And then put it in the equation
$$\frac{x(3x + 3y)}{y} + 2.5x = \frac{(3x + 3y)}{y}y$$
$$\frac{x(3x + 3y)}{y} + 2.5x = 3x + 3y$$
$$\frac{3x^2}{y} + \frac{3yx}{y} + 2.5x = 3x + 3y$$
$$\frac{3x^2}{y} + 3x + 2.5x = 3x + 3y$$
$$\frac{3x^2}{y} + 2.5x = 3y$$
Here I got stuck. I didn't know how to find the ratio. Can someone help me?
| Assuming your calculations so far are correct (I didn't check), you are almost there. Divide both sides by $y$, you will get
$$\frac {3x^2} {y^2} + \frac {2.5x} y = 3,$$
a quadratic equation for $\frac xy$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the equation of the normal line with the minimum y-coordinate is $ y = \frac{-\sqrt{2}}{2}x + {1\over k}$
Question: The curve in the figure is the parabola $y=kx^2$ where $k>0$.
Several normal lines to this parabola are also shown. Consider the points in the first quadrant from which the normal lines are drawn. Notice that as the $x$ coordinate gets smaller, the $y$-coordinate of the intersection of the normal with the other arm of the parabola also decreases until it reaches a minimum, and then it increases. The normal line with the minimum $y$ coordinate is dotted.
$(a)$ Show that the equation of the normal to the parabola at a point $(x_0,y_0)$ is $y = {-x\over 2kx_0} + kx_0^2 + {1\over 2k}$
(b) Show that the equation of the normal line with the minimum y-coordinate is $ y = \frac{-\sqrt{2}}{2}x + {1\over k} $
What I have done:
$(a)$ Show that the equation of the normal to the parabola at a point $(x_0,y_0)$ is $y = {-x\over 2kx_0} + kx_0^2 + {1\over 2k}$
$$ f(x) = kx^2 $$
$$ f( x_{0}) = kx_{0}^2 $$
$$ f'(x) = 2kx $$
$$ f'(x_{0}) = 2kx_0 $$
$$ Normal = -1/m $$
$$ m= {-1\over 2kx_0} $$
$$ y-y_1 = m (x-x_1) $$
$$ y-kx_0^2 = {-1\over 2kx_0}(x-x_0) $$
$$ y = {-x\over 2kx_0} + kx_0^2 + {1\over 2k} $$
(b) Show that the equation of the normal line with the minimum y-coordinate is $ y = \frac{-\sqrt{2}}{2}x + {1\over k} $
$$ y = \frac{-\sqrt{2}}{2}x + {1\over k} $$
$$ {-1\over m} = \frac{-\sqrt{2}}{2} $$
$$ m = \sqrt{2} $$
$$ f'(x) = 2kx $$
$$ 2kx = \sqrt{2} $$
$$ x = {\sqrt{2}\over 2k} $$
$$ f({\sqrt{2}\over 2k}) = {1\over 2k}$$
$$ y-y_1 = m (x-x_1) $$
$$ y - {1\over 2k} = \frac{-\sqrt{2}}{2} (x-{\sqrt{2}\over 2k}) $$
$$ y = \frac{-\sqrt{2}}{2}x + {1\over k} $$
However I am stuck trying to prove that this is the minimum y coordinate.
I've attempted
$$ y = y$$
$$ kx^2 = {-x\over 2kx_0} + kx_0^2 + {1\over 2k}$$
$$ kx^2 + (\frac{1}{2kx_0})x -(kx_0^2 + \frac{1}{2k}) = 0$$
$$ x={-b\pm\sqrt{b^2-4ac} \over 2a}$$
$$ x={- (\frac{1}{2kx_0})\pm\sqrt{ (\frac{1}{4k^2x_0^2})-4(k)(-kx_0^2 - \frac{1}{2k})} \over 2k}$$
$$ x={- (\frac{1}{2kx_0})\pm\sqrt{ \frac{1}{4k^2x_0^2}-4(-k^2x_0^2 - \frac{k}{2k})} \over 2k}$$
$$ x={-\frac{1}{2kx_0}\pm\sqrt{ \frac{1}{4k^2x_0^2}+4k^2x_0^2 + 2} \over 2k}$$
$$ x={-\frac{1}{2kx_0}\pm\sqrt{ (2kx_0 + \frac{1}{2kx_0})^2} \over 2k}$$
$$ x={-\frac{1}{2kx_0}\pm { (2kx_0 + \frac{1}{2kx_0})} \over 2k}$$
$$ x = x_0 $$
$$ x = \frac{-1}{2k^2x_0} - x_0 $$
Now I am lost..
| You did almost all the work. From
$$ x={-\frac{1}{2kx_0}\pm { (2kx_0 + \frac{1}{2kx_0})} \over 2k}$$
the $x$ you are looking for is the one with the '$-$':
$$x={-\frac{1}{2kx_0}- { (2kx_0 + \frac{1}{2kx_0})} \over 2k}=-\frac{1}{2k^2x_0}-x_0$$
(the other one is simply $x=x_0$).
Now
$$y=k\left(-\frac{1}{2k^2x_0}-x_0\right)^2$$
so you just have to minimize
$$\frac{1}{2k^2x_0}+x_0$$
and it's easy to see that the minimum is at $x_0=\frac{1}{\sqrt2k}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the point on the cone closest to (1,4,0) Find the point on the cone $z^2=x^2+y^2$ nearest to the point $P(1,4,0)$.
This is a homework problem I've not made much headway on.
| This seems like an exercise in Lagrange Multipliers. You need to minimize the distance function $f(x,y,z) = (x-1)^2 + (y-4)^2 + z^2$ (which is the square distance from $(x,y,z)$ to $(1,4,0)$) subject to the constraint that $g(x,y,z) = x^2 + y^2 - z^2 = 0$. Any point which minimizes such $f$ subject to the constraint will satisfy $$\nabla f = \lambda \nabla g$$ for some $\lambda \in \mathbb R$. This gives us 4 equations: \begin{align*} 2(x-1) &= 2\lambda x \\ 2(y-4) &= 2\lambda y \\ 2z&= -2\lambda z \\ x^2 + y^2 - z^2 &= 0.\end{align*} The third equation gives $z = 0$ or $\lambda = -1$. If $z=0$, then the fourth equation gives $x=0, y= 0$. But then the first two equations couldn't be satisfied, so this is impossible. We conclude that $\lambda = -1$. In this case $x = 1/2$, $y=2$ so by the fourth equation $z = \pm\sqrt{17/4}$. By the symmetry in $z$, both sides of the $\pm$ work, so the points which minimize the distance are $$\left(\tfrac 1 2, 2, \sqrt{17/4}\right) \,\,\,\, \text{ and } \,\,\,\, \left(\tfrac 1 2, 2, -\sqrt{17/4}\right).$$ The distance is then $$d_{\text{min}} = \sqrt{\left(\tfrac 1 2 - 1 \right)^2 + (2-4)^2 + \left( \pm\sqrt{17/4}\right)^2}= \sqrt{\tfrac 1 4 + 4 + \tfrac{17}4} = \sqrt{\tfrac{34}{4}} = \tfrac{\sqrt{34}}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,$\log_8(kx^2+wx+f)=2.$ Let $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,$\log_8(kx^2+wx+f)=2.$If $k,w,f$ are relatively prime positive integers,then find the value of $k+w+f.$
The given equation is $\log_8(kx^2+wx+f)=2$ i.e. $kx^2+wx+f=64$
Since $x=\frac{1}{3}$ or $x=-15$ satisfies the equation,so
$\frac{k}{9}+\frac{w}{3}+f=64.............(1)$
$225k-15w+f=64............................(2)$
I need to find $k,w,f$ from these equations,but i need three equations to find three variables.I am stuck here.
| Eliminating $f$ gives $44k=3w$. Take $w=44,k=3$. Then we get $f=49$. Note that the general solution is $k=3h,w=44h,f=64-15h$, but the requirement that the numbers are relatively prime positive integers forces $h=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I factorize this numerator? $$\lim_{x\to -3} \frac 1{x+3} + \frac 4{x^2+2x-3}$$
I have the solution I just need to know how i turn that into:
$$\frac {(x-1)+4}{(x+3)(x-1)}$$
I know this might be really simple but I'm not sure how to factorise the numerator.
Thanks in advance!
| You may use common denominator technique:
$$
\frac 1{x+3} + \frac 4{x^2+2x-3} = \frac {1}{x+3} + \frac {4}{(x+3)(x-1)} = \frac {1 \times (x-1) + 4 \times 1}{(x+3)(x-1)} = \frac{x-1+4}{(x+3)(x-1)}
$$
Take a denominator which divisible to both denominators as the common denominator and divide that by each denominator separately, then multiply its quotient by its numerator.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac{1}{3}-\cos x$$
$$ 1 - \cos^2 x = \dfrac{1}{9} + \cos^2 x - \dfrac{2}{3}\cos x$$
$$ -2\cos^2 x + 2/3\cos x +\dfrac{8}{9}=0$$
$$ \boxed{\cos^2 x -\dfrac{1}{3}\cos x -\dfrac{4}{9} = 0}$$
Can I just substitute $\cos x$ by $z$ and solve as if it was a simple second degree equation and then obtain $x$ by taking the inverse cosine? I have tried to do this but I cannot get the right result. If I do this, I obtain the following results:
$$ z_1 = -0.520517 \longrightarrow x_1 = 121.4º\\
z_2= 0.8538509 \longrightarrow x_2 = 31.37º$$
I obtain $x$ from $z$ by taking the inverse cosine.
The correct result should be around 329º which corresponds to 4.165 rad. My question is if what I am doing is wrong because I have tried multiple times and I obtain the same result (or in the worst case, I have done the same mistake multiple times).
| An even simpler way is to substitute $x \rightarrow y - \frac{\pi}{4}$
You equation is now $$\sqrt{2} \sin(y) = \frac{1}{3}$$
The solutions are
$$\begin{aligned} x & = \arcsin \left( \frac{\sqrt{2}}{6} \right) - \frac{\pi}{4} + 2\pi n \\ x & = \arccos \left( \frac{\sqrt{2}}{6} \right) - \frac{\pi}{4} + 2 \pi n
\end{aligned}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding ALL solutions of the modular arithmetic equation $25x \equiv 10 \pmod{40}$ I am unsure how to solve the following problem. I was able to find similar questions, but had trouble understanding them since they did not show full solutions.
The question:
Find ALL solutions (between $1$ & $40$) to the equation $25x \equiv 10 \pmod{40}$.
| Let's use the definition of congruence. $a \equiv b \pmod{n} \iff a = b + kn$ for some integer $k$. Hence, $25x \equiv 10 \pmod{40}$ means $$25x = 10 + 40k$$ for some integer $k$. Dividing each side of the equation $25x = 10 + 40k$ by $5$ yields $$5x = 2 + 8k$$
for some integer $k$. Thus,
$$5x \equiv 2 \pmod{8}$$
Since $\gcd(5, 8) = 1$, $5$ has a multiplicative inverse modulo $8$. To isolate $x$, we must multiply both sides of the congruence $5x \equiv 2 \pmod{8}$ by the multiplicative inverse of $5$ modulo $8$. To find the multiplicative inverse, we use the extended Eucldean algorithm.
\begin{align*}
8 & = 5 + 3\\
5 & = 3 + 2\\
3 & = 2 + 1\\
2 & = 2 \cdot 1
\end{align*}
Working backwards through this partial sequence of Fibonacci numbers to solve for $1$ as a linear combination of $5$ and $8$ yields
\begin{align*}
1 & = 3 - 2\\
& = 3 - (5 - 3)\\
& = 2 \cdot 3 - 5\\
& = 2(8 - 5) - 5\\
& = 2 \cdot 8 - 3 \cdot 5
\end{align*}
Therefore, $1 \equiv -3 \cdot 5 \pmod{8}$. Hence, $-3 \equiv 5^{-1} \pmod{8}$. Since $-3 \equiv 5 \pmod{8}$, we have $5 \equiv 5^{-1} \pmod{8}$. Thus, $5 \cdot 5x \equiv x \pmod{8}$. Hence,
\begin{align*}
5x & \equiv 2 \pmod{8}\\
5 \cdot 5x & \equiv 5 \cdot 2 \pmod{8}\\
x & \equiv 10 \pmod{8}\\
x & \equiv 2 \pmod{8}
\end{align*}
What remains is for you to find the solutions of the congruence $x \equiv 2 \pmod{8}$ such that $0 \leq x < 40$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does the sequence $\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}$ converge? I'm trying to determine if this sequence converges as part of answering whether it's monotonic:
$$
\left\{\frac{n!}{1\cdot 3\cdot 5\cdot ... \cdot (2n-1)}\right\}
$$
First, I tried expanding it a bit to see if I could remove common factors in the numerator and denominator:
$$
\left\{\frac{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot ...\cdot n}{1\cdot 3\cdot 5\cdot 7\cdot 9 \cdot ...\cdot (2n-1)}\right\}
$$
Second, I tried looking at elements of the sequence with common factors removed:
$$
1, \frac{2}{3}, \frac{2}{5}, \frac{2\cdot 4}{5\cdot 7}, \frac{2\cdot 4}{7\cdot 9}, ...
$$
Third, I tried looking at the elements again as fractions without simplifications:
$$
\frac{1}{1}, \frac{2}{3}, \frac{6}{15}, \frac{24}{105}, \frac{120}{945}, ...
$$
Last, I tried searching for similar questions on Stack Exchange and I found one for $\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ but I didn't understand how that might apply to my question. So, any hints would be much appreciated.
| The reciprocal of the term of interest is
$$\begin{align}
\frac{(2n-1)!!}{n!}&=\left(\frac{(2n-1)}{n}\right)\left(\frac{(2(n-1)-1)}{(n-1)}\right)\left(\frac{(2(n-2)-1)}{(n-2)}\right) \cdots \left(\frac{5}{3}\right)\left(\frac{3}{2}\right)\\\\
&=\left(2-\frac{1}{n}\right)\left(2-\frac{1}{n-1}\right)\left(2-\frac{1}{n-2}\right) \cdots \left(\frac{5}{3}\right)\left(\frac{3}{2}\right)\\\\
&\ge \left(\frac32\right)^{n-1}
\end{align}$$
Therefore, we see immediately that the limit of interest is $0$.
| {
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Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$ Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$
Answer:
This should be easy enough...
$f'(x) = 2x$
The tangent line in the point $(a, a^2)$ is $y - a^2 = 2 (x - a) \rightarrow y = 2x + a^2 - 2a$
The equation for the normal line is: $y - a^2 = -\frac{1}{2}(x - a) \rightarrow y = -\frac{1}{2}x + a^2 + \frac{1}{2}a$
Now to determine the point of intersection we just see when $f(x)$ and the normal line is equal, i.e.
$x^2 = -\frac{1}{2}x + a^2 + \frac{1}{2}a$
But this seems like a nonsense equation...
| The slope of the normal line is going to be: $-\frac{1}{2a}$.
set $g(x)=-\frac{1}{2a}x+(a^2+\frac{1}{2})$
You want to solve $g(x)=f(x)$.
$-\frac{1}{2a}x+(a^2+\frac{1}{2})=x^2$
$x^2+\frac{1}{2a}x+\frac{1}{16a^2}=(a^2+\frac{1}{2})+\frac{1}{16a^2}$
$x=\pm\sqrt{(a^2+\frac{1}{2})+\frac{1}{16a^2}}-\frac{1}{4a}$
Personally, I prefer vectors, but it amounts to the same thing:
$(a,a^2)+t[-2a,1]=(x,x^2)$
$a^2+t=(a(1-2t))^2$
And you can solve in this way to get a $t$, which I think is more intuitive, and less "symbol pushy"
| {
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"timestamp": "2023-03-29T00:00:00",
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What is $\arctan(x) + \arctan(y)$ I know $$g(x) = \arctan(x)+\arctan(y) = \arctan\left(\frac{x+y}{1-xy}\right)$$
which follows from the formula for $\tan(x+y)$. But my question is that my book defines it to be domain specific, by which I mean, has different definitions for different domains:
$$g(x) = \begin{cases}\arctan\left(\dfrac{x+y}{1-xy}\right), &xy < 1 \\[1.5ex]
\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x>0,\; y>0,\; xy>1 \\[1.5ex]
-\pi + \arctan\left(\dfrac{x+y}{1-xy}\right), &x<0,\; y<0,\; xy > 1\end{cases}$$
Furthermore, When I plot the function $2\arctan(x)$, it turns out that the book definition is correct. I don't understand how such peculier definition emerges. Thank you.
| Here is a straightforward (though long) derivation of the piece wise function description of $\arctan(x)+\arctan(y)$.
We will show that:
$\arctan(x)+\arctan(y)=\begin{cases}-\frac{\pi}{2} \quad &\text{if $xy=1$, $x\lt 0$, and $y \lt 0$} \\\frac{\pi}{2} \quad &\text{if $xy=1$, $x\gt 0$, and $y \gt 0$} \\ \arctan\left(\frac{x+y}{1-xy}\right) \quad &\text{if $xy \lt 1$} \\\arctan\left(\frac{x+y}{1-xy}\right)+\pi \quad &\text{if $xy \gt 1$ and $x \gt 0$, $y \gt 0$}\\ \arctan\left(\frac{x+y}{1-xy}\right)-\pi \quad &\text{if $xy \gt 1$ and $x \lt 0$, $y \lt 0$} \end{cases}$
working from these below statements:
*
*$\arctan$ is strictly increasing and $\arctan(0)=0$
*if $x \gt 0$ then $\arctan(x)+\arctan(\frac{1}{x})= \frac{\pi}{2}$ and if $x \lt 0$ then $\arctan(x)+\arctan(\frac{1}{x})=-\frac{\pi}{2}$
*$\text{dom}(\arctan)=\mathbb R$ and $\text{image}(\arctan)=(-\frac{\pi}{2},\frac{\pi}{2})$
Case 1: $xy=1$
Under these circumstances, you can derive that $x\gt 0$ and $y \gt 0 \implies \arctan(x)+\arctan(y)=\frac{\pi}{2}$. Alternatively, $x \lt 0$ and $y \lt 0 \implies \arctan(x)+\arctan(y) = -\frac{\pi}{2}$
Case 2: $xy \lt 1$
Under these circumstances, you can derive that $\arctan(x)+\arctan(y) \in (-\frac{\pi}{2},\frac{\pi}{2})$
Case 3: $xy \gt 1$
There are two subcases to consider. Either $x \gt 0$ and $y \gt 0$ OR $x \lt 0$ and $y \lt 0$.
If $x \gt 0$ and $y \gt 0$, the we argue as follows.
Because $x \gt 0$ and $xy \gt 1$, we necessarily have that $y \gt \frac{1}{x}$. Next, because $x \gt 0$, we must have that $\arctan(x)+\arctan(\frac{1}{x}) = \frac{\pi}{2}$. Given that $\arctan$ is a strictly increasing function, we then have that $\frac{\pi}{2}=\arctan(x)+\arctan(\frac{1}{x}) \lt \arctan(x)+\arctan(y)$. Finally, we know that for any $z \gt 0:\arctan(z) \in (0,\frac{\pi}{2})$.
With the above argument, we conclude that $\arctan(x)+\arctan(y) \in (\frac{\pi}{2}, \pi)$.
A similar argument would lead one to have that if $x \lt 0$ and $y \lt 0$, then $\arctan(x)+\arctan(y) \in (-\pi,-\frac{\pi}{2})$.
With these three cases established, we can now look at the identity:
$$\tan(\alpha+\beta)=\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$$
Letting $\alpha=\arctan(x)$ and $\beta=\arctan(y)$, we can rewrite the above identity as:
$$\tan(\arctan(x)+\arctan(y))=\frac{\tan(\arctan(x))+\tan(\arctan(y))}{1-\tan(\arctan(x))\tan(\arctan(y))}=\frac{x+y}{1-xy} \quad (\dagger_1)$$
From our previous work, we know that across all values of $x,y \in \mathbb R$, the sum $\arctan(x)+\arctan(y)$ can take on the values:
*
*$-\frac{\pi}{2}$ or $\frac{\pi}{2}$
*$(-\frac{\pi}{2},\frac{\pi}{2})$
*$(-\pi,-\frac{\pi}{2})$ or $(\frac{\pi}{2},\pi)$
Importantly, we know that $\tan$ is a periodic function (with period of $\pi$) defined on $\cdots (-\frac{3\pi}{2},-\frac{\pi}{2}), (-\frac{\pi}{2},\frac{\pi}{2}),(\frac{\pi}{2},\frac{3\pi}{2})\cdots \quad (\dagger_2)$
Further, by definition, $\arctan$ is the inverse function of $\tan$ when $\tan$ is restricted to $(-\frac{\pi}{2},\frac{\pi}{2})$.
With the above work in mind, how do we now precisely determine what $\arctan(x)+\arctan(y)$ equals?
Firstly, we know from Case 1 that if $x \lt 0, y\lt 0,$ and $xy=1$, then $\arctan(x)+\arctan(y)=-\frac{\pi}{2}$. Similarly, if $x \gt 0, y \gt 0,$ and $xy=1$, then $\arctan(x)+\arctan(y)=\frac{\pi}{2}$
Next, we can use the result from Case 2 and $(\dagger_1)$ to conclude that if $xy \lt 1$, then $\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right)$. This is because the argument of $\tan$, in the instance of Case 2, spans from $(-\pi/2,\pi/2)$, which implies that we are dealing with a $\tan$ function whose domain is restricted to $(-\pi/2,\pi/2)$. Therefore, when we take the $\arctan$ of $\tan(z)$, it simply computes to $z$.
So far, then, we have the piecewise function:
$\arctan(x)+\arctan(y)=\begin{cases}-\frac{\pi}{2} \quad &\text{if $xy=1$, $x\lt 0$, and $y \lt 0$} \\\frac{\pi}{2} \quad &\text{if $xy=1$, $x\gt 0$, and $y \gt 0$} \\ \arctan\left(\frac{x+y}{1-xy}\right) \quad &\text{if $xy \lt 1$}\end{cases}$
For the final part (i.e. consider when $xy \gt 1$), suppose $z=\arctan(x)+\arctan(y) \in (-\pi,-\frac{\pi}{2}) \cup (\frac{\pi}{2},\pi)$. Although it is tempting to take the $\arctan$ of both sides in $(\dagger_1)$, we must remember that $\arctan$ is only the inverse function of $\tan$ restricted to $(-\frac{\pi}{2},\frac{\pi}{2})$. It is therefore invalid to claim that $\arctan\circ \tan(z)=z$. However, what IS valid is to employ the fact that $\tan$ is periodic as identified in $(\dagger_2)$. Therefore, suppose $z \in (-\pi,-\frac{\pi}{2})$, which corresponds to $x \lt 0$ and $y \lt 0$. We know that $\tan(z)=\tan(z+\pi)$, where $z+\pi \in (0,\frac{\pi}{2})$. Returning to $(\dagger_1)$, let us then write:
$$\tan(z+\pi)=\tan(z)=\frac{x+y}{1-xy}$$
We can then apply $\arctan$ to $\tan(z+\pi)$ which yields:
$$z+\pi=\arctan\left(\frac{x+y}{1-xy}\right) \implies \arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right)-\pi$$
If $z \in (\frac{\pi}{2},\pi)$, which corresponds to $x \gt 0$ and $y \gt 0$ , a similar result applied to $z-\pi$ will yield:
$$z-\pi=\arctan\left(\frac{x+y}{1-xy}\right) \implies \arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right)+\pi$$
With this we can completely fill in our piecewise function as follows:
$\arctan(x)+\arctan(y)=\begin{cases}-\frac{\pi}{2} \quad &\text{if $xy=1$, $x\lt 0$, and $y \lt 0$} \\\frac{\pi}{2} \quad &\text{if $xy=1$, $x\gt 0$, and $y \gt 0$} \\ \arctan\left(\frac{x+y}{1-xy}\right) \quad &\text{if $xy \lt 1$} \\\arctan\left(\frac{x+y}{1-xy}\right)+\pi \quad &\text{if $xy \gt 1$ and $x \gt 0$, $y \gt 0$}\\ \arctan\left(\frac{x+y}{1-xy}\right)-\pi \quad &\text{if $xy \gt 1$ and $x \lt 0$, $y \lt 0$} \end{cases}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 3
} |
Estimate the bound of the sum of the roots of $1/x+\ln x=a$ where $a>1$ If $a>1$ then $\frac{1}{x}+\ln x=a$ has two distinct roots($x_1$ and $x_2$, Assume $x_1<x_2$). Show that $$x_1+x_2+1<3\exp(a-1)$$
First I tried to estimate the place of the roots separately. I have got that $x_1\leq \frac{1}{a}$ and $\exp(a-1)<x_2<\exp(a)$. Then I tried to think about $x_1 + x_2$ as a whole. Because $1/x_1+\ln x_1=a$ and $1/x_2+\ln x_2=a$. I tried to express $x_1+x_2$ as a function of $a$, But I failed. I have no idea to solve this problem. Please help me :)
| I can show the inequality when $a$ is close enough to $1$ (namely, $a\leq 1+ln(5/4)\approx 1.22$) or when $a$ is big enough (namely, $a \geq 1+\ln(5) \approx 2.6$). In the sequel $f(x)$ denotes $x+\ln(\frac{1}{x})$.
When $a$ is close to $1$. Let us put $w=\sqrt{e^{a-1}-1}$. The inequality then becomes $x_1+x_2\leq 2+3w^2$. Now I claim that when $w\in[0,\frac{1}{2}]$, one has
$$
\begin{array}{lcl}
x_1 & \leq & 1-\sqrt{2}w+\frac{4}{3}w^2 \\
x_2 & \leq & 1+\sqrt{2}w+\frac{5}{3}w^2 \\
\end{array}
$$
To check the first claim, let $g(w)=f(1-\sqrt{2}w+\frac{4}{3}w^2)-1-\ln(1+w^2)$. Then the derivative of $g$ is $g'(w)=\frac{w^3(\frac{4}{3}\sqrt{2}w-\frac{34}{9})}{(1+w^2)(1-\sqrt{2}w+\frac{4}{3}w^2)^2}<0$. So $g$ is decreasing on $(0,\frac{1}{2})$, whence $g(w)\leq g(0)=0$, proving the first claim.
To check the second claim, let $h(w)=f(1+\sqrt{2}w+\frac{5}{3}w^2)-1-\ln(1+w^2)$. Then the derivative of $h$ is $h'(w)=\frac{w^2(\frac{-5}{3}\sqrt{2}w^2-\frac{28}{9}w+sqrt(2)))}{(1+w^2)(1+\sqrt{2}w+\frac{5}{3}w^2)^2}$. So $h'$ has a unique zero in $(0,\frac{1}{2})$ ; since $h(0)=0$ and $h(\frac{1}{2})\approx 0.00091 >0$, we have $h(w)\geq 0$ for any $w\in(0,\frac{1}{2})$, proving the second claim.
When $a$ is big enough. Let us put $u=e^{a-1}$. The inequality then becomes $x_1+x_2+1\leq 3u$. Now I claim that when $u\geq 5$, one has
$$
\begin{array}{lcl}
x_1 & \leq & (3-e)u-1 \\
x_2 & \leq & eu \\
\end{array}
$$
To check the first claim, let $k(u)=f((3-e)u-1)-1-ln(u)$. Then the derivative of $k$ is $k'(w)=\frac{-1} {u((3-e)u-1)^2}<0$. So $k$ is decreasing on $(0,\infty)$, and since $k(5)\approx -1.05 <0$, this proves the first claim.
The second claim follows from the identity $f(eu)-1-ln(u)=\frac{1}{eu}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Taking inverse Fourier transform of $\frac{\sin^2(\pi s)}{(\pi s)^2}$ How do I show that
$$\int_{-\infty}^\infty \frac{\sin^2(\pi s)}{(\pi s)^2} e^{2\pi isx} \, ds = \begin{cases} 1+x & \text{if }-1 \le x \le 0 \\ 1-x & \text{if }0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases}$$
I know that $\sin^2(\pi s)=\frac{1-\cos(2\pi s)}{2}=\frac{1-(e^{2\pi i s}-e^{-2\pi i s}))/2}{2}$, so
$$\int_{-\infty}^\infty \frac{\sin^2(\pi s)}{(\pi s)^2} e^{2\pi isx} \, ds=2\int_0^\infty \frac{2e^{2\pi isx}-(e^{2\pi is(1+x)}+e^{2\pi i s(-1+x)})}{4\pi^2s^2} \, ds$$
I am also allowed to use the known identity
$$\int_{-\infty}^\infty \frac{1-\cos(a \pi x)}{(\pi x)^2} \, dx = |a|$$
for some real number $a$.
| Note that we can write
$$\begin{align}
\int_{-\infty}^\infty\frac{\sin^2(\pi s)}{(\pi s)^2}e^{i2\pi sx}\,ds&=\frac12\int_{-\infty}^\infty\frac{1-\cos(2\pi s)}{(\pi s)^2}e^{i2\pi sx}\,ds\\\\
&=\int_0^\infty \frac{1-\cos(2\pi s)}{(\pi s)^2}\,\cos(2\pi sx)\,ds\\\\
&=\int_0^\infty \frac{\cos(2\pi sx)-\frac12\left(\cos(2\pi s(x+1))+\cos(2\pi s(x-1))\right)}{(\pi s)^2}\,ds\\\\
&=\int_0^\infty \frac{\cos(2\pi sx)-1}{(\pi s)^2}\,ds\\\\
&+\frac12\int_0^\infty \frac{1-\cos(2\pi s(x+1))}{(\pi s)^2}\,ds\\\\
&+\frac12\int_0^\infty \frac{1-\cos(2\pi s(x-1))}{(\pi s)^2}\,ds\\\\
&=-|x|+\frac12|x+1|+\frac12|x-1|
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
What is $2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$ equal to? I came across this question while doing my homework:
$$\Large 2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty=?$$
$$\small\text{OR}$$
$$\large\prod\limits_{x=1}^{\infty} (2x)^{\frac{1}{4x}} = ?$$
My Attempt:
$\large 2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$
$\large \Rightarrow 2^{\frac{1}{4}}\cdot2^{\frac{2}{8}}\cdot2^{\frac{3}{16}}\cdot2^{\frac{4}{32}}\cdots\infty$
$\large \Rightarrow 2^{(\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \cdots \infty)}$
OR, $\large 2^{\space (\sum\limits_{x=1}^{\infty} \frac{x}{2^{x+1}})}$
$\cdots$ That's it ... I am stuck here ... It does not resemble any series I know...
How do you do it? Thanks!
| For every $x\in \mathbb R$ which $|x|\lt 1$, we have: $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$ From here: $$\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$$ Now, multiplying $x^2$ in both side we get that: $$\sum_{n=1}^\infty nx^{n+1}=\frac{x^2}{(1-x)^2}$$ And so: $$\sum_{n=1}^\infty n\left(\frac{1}{2}\right)^{n+1}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Limiting question:$\displaystyle \lim_{x\to\ 0} \frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x -\sin\ x}$ How do I find the value of $$\lim_{x\to\ 0} \frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x - \sin\ x}$$
in easy way.
| As Henry W; commented, Taylor series make thigs quite simple.
$$A=a \tan(x) \implies \log(A)=\tan(x)\log(a)=\Big(x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^6\right)\Big)\log(a)$$ $$A=e^{\log(a)}\implies A=1+x \log (a)+\frac{1}{2} x^2 \log ^2(a)+\frac{1}{6} x^3 \left(\log ^3(a)+2 \log
(a)\right)+O\left(x^4\right)$$ Soing the same with the second term of numerator $$B=a \sin(x) \implies B=1+x \log (a)+\frac{1}{2} x^2 \log ^2(a)+\frac{1}{6} x^3 \left(\log ^3(a)-\log
(a)\right)+O\left(x^4\right)$$ So the numerator is $$\frac{1}{2} x^3 \log (a)+O\left(x^4\right)$$ Using again the series for $\sin(x)$ and $\tan(x)$, the denominator is $$\frac{x^3}{2}+O\left(x^4\right)$$ which makes the ratio $$\frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x -\sin\ x}=\log (a)+O\left(x^1\right)$$ Using more terms would lead to $$\frac{a^{\tan\ x} - a^{\sin\ x}}{\tan\ x -\sin\ x}=\log (a)+x \log ^2(a)+O\left(x^2\right)$$ showing the limit and how it is approached.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve: $\int\frac{\sin 2x}{\sqrt{3-(\cos x)^4}}$ (and a question about $t=\tan \frac{x}{2}$) I tried substituting $$t=\tan \frac{x}{2}$$ but the nominator is $\sin {2x}$, so is there a way to get from $$\sin x=\frac{2x}{1+x^2}$$ to an expression with $\sin 2x$?
| $$\int\frac{\sin(2x)}{\sqrt{3-\cos^4(x)}}\space\text{d}x=$$
Use $\sin(2x)=2\sin(x)\cos(x)$:
$$2\int\frac{\sin(x)\cos(x)}{\sqrt{3-\cos^4(x)}}\space\text{d}x=$$
Substitute $u=\cos(x)$ and $\text{d}u=-\sin(x)\space\text{d}x$:
$$-2\int\frac{u}{\sqrt{3-u^4}}\space\text{d}u=$$
Substitute $s=u^2$ and $\text{d}s=2u\space\text{d}u$:
$$-\int\frac{1}{\sqrt{3-s^2}}\space\text{d}s=-\int\frac{1}{\sqrt{3}\sqrt{1-\frac{s^2}{3}}}\space\text{d}s=-\frac{1}{\sqrt{3}}\int\frac{1}{\sqrt{1-\frac{s^2}{3}}}\space\text{d}s=$$
Substitute $p=\frac{s}{\sqrt{3}}$ and $\text{d}p=\frac{1}{\sqrt{3}}\space\text{d}s$:
$$-\int\frac{1}{\sqrt{1-p^2}}\space\text{d}p=-\arcsin\left(p\right)+\text{C}=-\arcsin\left(\frac{s}{\sqrt{3}}\right)+\text{C}=$$
$$-\arcsin\left(\frac{u^2}{\sqrt{3}}\right)+\text{C}=-\arcsin\left(\frac{\cos^2(x)}{\sqrt{3}}\right)+\text{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
An inequality involving $\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$
$$\frac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}$$
Let $(x, y, z)$ be non-negative real numbers such that $x^2+y^2+z^2=2(xy+yz+zx)$.
Question: Find the maximum value of the expression above.
My attempt:
Since $(x,y,z)$ can be non-negative, we can take $x=0$, then equation becomes
$$y^2 + z^2=2xy$$
This implies that $(y-z)^2=0$.
So this implies that the required value is $$\frac{y^3 + z^3}{(y+z)(y^2 + z^2)}=\frac{1}{2}$$
But this wrong as the correct answer is $\frac{11}{18}$.
What is wrong with my method?
| Let $P$ be the expression we want to maximise.
Using the following notation: $S_1=x+y+z$, $S_2=xy+xz+yz$ and $S_3=xyz$.
From the hypothesis we get that, $S_1^2=4S2$.
So the expression we want to maximise is:
$P=\dfrac{x^3+y^3+z^3}{(x+y+z)(x^2+y^2+z^2)}=\dfrac{S_1^3-3S_1S_2+3S3}{2S_1S_2}$
Then, simplify it using the hypothesis, in a way such that we only get in terms of $S_1$ and $S_3$ :
$P=\dfrac{1}{2}+\dfrac{6S_3}{S_1^3}$
Now, consider a polynomial with roots $x,y,z$. Ofcourse it is $F(x)=x^3-S_1x^2+S_2x-S_3$. Then, for the roots to be real, the discriminant of this polynomial must be $\geq 0$, so:
$S_1^2S_2^2-4S_2^3-4S_1^3S_3+18S_1S_2S_3-27S_3^2 \geq 0$
Again, using the hypothesis, we get:
$\dfrac{1}{2}S_1^3-27S_3 \geq 0$
Hence, $\dfrac{S_3}{S_1^3} \leq \dfrac{1}{54}$
Finally, $P \leq \dfrac{1}{2}+6\left(\dfrac{1}{54}\right)=\boxed{\dfrac{11}{18}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Given three coordinates (a,b,c), (d,e,f), and (l,m,n), what is the center of the circle in the 3D plane (h,k,i) that contains these three points. I have tried the following:
$$(a-h)^2+(b-k)^2+(c-i)^2=r^2$$
$$(d-h)^2+(e-k)^2+(f-i)^2=r^2$$
$$(l-h)^2+(m-k)^2+(n-i)^2=r^2$$
Subtracted equation 2 from 1, equation 3 from equation 2, and equation 3 from equation 1 to get:
$$2(a-d)h+2(b-e)k+2(c-f)i=a^2-d^2+b^2-e^2+c^2-f^2$$
$$2(d-l)h+2(e-m)k+2(f-n)i=d^2-l^2+e^2-m^2+f^2-n^2$$
$$2(a-l)h+2(b-m)k+2(c-n)i=a^2-l^2+b^2-m^2+c^2-n^2$$
Closer inspection revealed to me that the left side formed a 3 x 3 matrix and the left a 3 by 1 matrix.
Tried solving for $(h,k,i)$ by taking the inverse of the left side matrix and multiplying that by the right side 3 by 1 matrix, but it turns out the left side matrix's rows are dependent, so the determinant of the left side matrix is 0 (singular matrix); hence, no inverse.
Please guide me as to solve this problem.
Thank you in advance.
| For points $(a,b,c)$ and $(d,e,f)$, their perpendicular bisector can be found by:
$$\begin{align*}
(x-a)^2 + (y-b)^2+(z-c)^2 &= (x-d)^2 + (y-e)^2 + (z - f)^2\\
a^2-2ax+b^2-2by+c^2 -2cz &= d^2 - 2dx + e^2 - 2ey + f^2 - 2fz\\
2(a-d)x +2(b-e)y + 2(c-f)z &= a^2+b^2+c^2-d^2-e^2-f^2
\end{align*}$$
Do the same and find the perpendicular bisector of $(d,e,f)$ and $(l,m,n)$.
For three points $(a,b,c)$, $(d,e,f)$ and $(l,m,n)$, their common plane $px+qy+r = z$ can be found by:
$$\begin{pmatrix}
a&b&1\\d&e&1\\l&m&1
\end{pmatrix}\begin{pmatrix}
p\\q\\r
\end{pmatrix} = \begin{pmatrix}
c\\f\\n
\end{pmatrix}$$
This might have some problem if their common plane is perpendicular with the $xy$-plane, but you can always change to use $x$ or $y$ to be the subject of the equation and try to solve again. You are using a computer anyway.
Now you have three equations in the form of $tx + uy + vz = w$, and you should be able to solve for their intersection $(x,y,z) = (h,k,i)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solve $\cos 2x - 3\sin x - 1 = 0$ using addition formula
Solve $\cos 2x - 3\sin x - 1 = 0, \quad 0^{\circ} \le x \le 360^{\circ}$
\begin{align} \cos 2x - 3\sin x - 1 = 0
&\iff 1 - 2\sin^2 x - 3\sin x - 1 = 0 \\
&\iff- 2\sin^2 x - 3\sin x = 0 \\
&\iff2\sin^2 x + 3\sin x =0\\
&\iff\sin x(2\sin x + 3) =0 \\
&\iff\sin x = 0 \lor 2 \sin x + 3= 0
\end{align}
I could go on but the book gives the answer, $0^{\circ}, 180^{\circ}, 360^{\circ}$ and I am mystified as to where these answers have come from.
| $\sin x =0 \Leftrightarrow x =\pi n (0^{\circ}, 180^{\circ}, 360^{\circ})$
$2\sin x=-3 \Rightarrow \sin x = -\frac 32 - $impossible
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rolling a die until two rolls sum to seven Here's the question:
You have a standard six-sided die and you roll it repeatedly, writing
down the numbers that come up, and you win when two of your rolled
numbers add up to $7$. (You will almost surely win.) Necessarily, one of the
winning summands is the number rolled on the winning turn. A typical
game could go like this: $1, 1, 4, 5, 3$; you win on the 5th turn
because $3 + 4 = 7$. How many turns do you expect to play?
Here's what I've tried:
We seek $E(N)$ where $N$ is a random variable counting the number of turns it takes to win. Then $N \ge 2$, and $$E(N) = \sum_{n=2}^\infty n P(N=n) = \sum_{n=1}^\infty P(N > n).$$
I want to find either $P(N=n)$, the probability that I win on the $n$th turn, or $P(N > n)$, the probability that after $n$ turns I still haven't won. Note that $P(N = 1) = 0$. Let $X_k$ be the number rolled on the $k$th turn. Then $$P(N = 2) = P(X_1 + X_2 = 7) = \sum_{x=1}^6 P(X_1 = x)P(X_2 = 7-x) = 6\cdot \frac{1}{6}\cdot\frac{1}{6} = \frac{1}{6}.$$
So far so good. To compute $P(N > 3)$ I let $A_{i, j} = \{\omega \in \{1, \dotsc, 6\}^3 : w_i + w_j = 7\}$ and used the inclusion-exclusion principle and symmetry to find $$|A_{1,2} \cup A_{2,3} \cup A_{1,3}| = 3|A_{1,2}| - 3|A_{1,2}\cap A_{1,3}| = 90$$
so $P(N > 3) = \frac{126}{216} = \frac{7}{12}$. This is the probability that no two of three dice sum to seven. Similarly, I found $P(N > 4)$ to be $\frac{77}{216}$.
I don't see how to generalize the above. I also thought that $$P(N > n) = P(X_i + X_j \ne 7 \text{ for all }1 \le i\ne j \le n) = (1 - P(X_i + X_j = 7))^{\binom{n}{2}} = \left(\frac{5}{6}\right)^{n(n-1)/2}$$
but that's false because the events are not independent.
I also tried $$P(N = n) = P(X_n = 7 - X_k \text{ for some } 1 \le k < n \text{ and }N \ne n - 1)$$
where that last clause is shorthand for "and the previous rolls did not secure your victory".
This yields the recursion $p_n = (1-(5/6)^{n-1})(1-p_{n-1})$, $p_1 = 0$, which didn't agree with my previously computed probabilities. (Perhaps I made an error.)
| While not exactly stated, it seems that you win only if the sum of consecutive numbers sums to 7.
Obviously you can't win on the first roll. For every roll thereafter, you have a $1/6$ chance of winning on the next roll, (and a $5$ in $6$ chance of the game continuing).
Now, $P(N=2) = \frac{1}{6}$, $P(N=3) = \frac{5}{6}\times\frac{1}{6}$,
$P(N=4) = \left(\frac{5}{6}\right)^2\times\frac{1}{6}$,
so
$P(N=n) = \left(\frac{5}{6}\right)^{n-2}\times\frac{1}{6}$.
Thus $$E(N) = \sum P(N=n)n = 2\times\frac{1}{6} + 3\times\frac{5}{6}\times\frac{1}{6} + 4\times\left(\frac{5}{6}\right)^2\times\frac{1}{6} + \dotsb =
\frac{1}{6} \sum_{n=0}^{\infty}(n+2)\left(\frac{5}{6}\right)^n.$$
I don't know if you need to derive this, but,
$$\sum_{n=1}^{\infty} n x^{n-1} = \sum_{n=0}^{\infty}(n+1) x^n = \frac{1}{1-x}\sum_{n=0}^\infty x^n = \frac{1}{(1-x)^2}$$
so
$$\frac{1}{6}\sum_{n=0}^{\infty}(n+2)\left(\frac{5}{6}\right)^n = \frac{1}{6}\sum_{n=0}^\infty(n+1)\left(\frac{5}{6}\right)^n + \frac{1}{6}\sum_{n=0}^\infty \left(\frac{5}{6}\right)^n$$
which equals
$$\frac{1}{6}\times\frac{1}{\left(1-\frac{5}{6}\right)^2} + \frac{1}{6}\times\frac{1}{1-\frac{5}{6}} = 7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 2
} |
Roots of a Quartic (Vieta's Formulas)
Question: The quartic polynomial $x^4 −8x^3 + 19x^2 +kx+ 2$ has four distinct real roots denoted
$a, b, c,d$ in order from smallest to largest. If $a + d = b + c$ then
(a) Show that $a + d = b + c = 4$.
(b) Show that $abcd = 2$ and $ad + bc = 3$.
(c) Find $ad$ and $bc.$
(d) Find $a, b, c, d$ and $k$.
My attempt:
$$ x^4 −8x^3 + 19x^2 +kx+ 2 $$
With Vieta's formulas;
$$ a+b+c+d = 8 $$
$$ab+ac+ad+bc+bd+cd = 19$$
$$abc + abd + acd + bcd = -k$$
$$ abcd = 2 $$
(a) Show that $a + d = b + c = 4$
As $$ a+b+c+d = 8 $$
but $b+c = a+d $
$$ 2a+2d = 8 $$
$$ a+d = 4 $$
Hence $$ a+d = b+c = 4 $$
(b) Show that $abcd = 2$ and $ad + bc = 3$
As $$ abcd = 2 $$
and
$$ab+ac+ad+bc+bd+cd = 19$$
$$ ad+bc + a(b+c) + d(b+c) = 19 $$
$$ ad+bc + (a+d)(b+c) = 19 $$
$$ ad+bc + (4)(4) = 19 $$
$$ ad+bc + 16 = 19 $$
$$ ad +bc = 3 $$
(c) Find $ad$ and $bc.$
Given $ad +bc = 3$ and $ abcd = 2 $
Hence $$ bc = \frac {2}{ad}$$
$$ad +bc = 3$$
$$ ad + \frac {2}{ad} = 3 $$
Let $ad = z $
$$ z^2 - 3z + 2 = 0 $$
$$ ad = 2 $$
$$ bc = 1 $$
(d) Find $a, b, c, d$ and $k$.
Given
$$abc + abd + acd + bcd = -k$$
$$ ad(b+c)+ bc(a+d) = - k $$
But $b+c=a+d=4$
$$ 4ad+ 4bc = - k $$
$$ 4(ad+bc) = -k $$
$$ 4(3) = -k $$
$$ k = -12 $$
Now this is the part which I am stuck on.. How do I find $a,b,c,d$?
| Hint If you know that $ad = 2$ and $a + d = 4$, then $$(x - a)(x - d) = x^2 - (a + d) x + ad = x^2 - 4 x + 2 ,$$ so finding $a, d$ is just finding the roots of that quadratic. Of course, finding $b, c$ is analogous.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Express $1/(x-1)$ in the form $ax^2+bx+c$
Let $x$ be a root of $f=t^3-t^2+t+2 \in \mathbb{Q}[t]$ and $K=\mathbb{Q}(x)$. Express $\frac{1}{x-1}$ in the form $ax^2+bx+c$, where $a,b,c\in \mathbb{Q}$.
I have proved that $f$ is the minimal polynomial of $x$ over $\mathbb{Q}$ but I am stuck showing the above claim. I tried writing $\frac{1}{x-1}=ax^2+bx+c$ and solve for $a,b,c$ but it didn't seem to work. Any idea?
| Using the Extended Euclidean Algorithm as implemented in this answer, we get
$$
\begin{array}{r}
&&x^2&1&-(x+2)/3\\\hline
1&0&1&-1&(1-x)/3\\
0&1&-x^2&x^2+1&(x^3-x^2+x+2)/3\\
x^3-x^2+x+2&x-1&x+2&-3&0\\
\end{array}
$$
which means that
$$
\left(\vphantom{x^2}x-1\right)\left(x^2+1\right)+\left(x^3-x^2+x+2\right)\cdot\left(-1\vphantom{x^2}\right)=-3
$$
Therefore,
$$
-\frac{x^2+1}3\equiv\frac1{x-1}\pmod{x^3-x^2+x+2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
f(x) is a function such that $\lim_{x\to0} f(x)/x=1$ $f(x)$ is a function such that $$\lim_{x\to0} \frac{f(x)}{x}=1$$ if
$$\lim_{x \to 0} \frac{x(1+a\cos(x))-b\sin(x)}{f(x)^3}=1$$
Find $a$ and $b$
Can I assume $f(x)$ to be $\sin(x)$ since $\sin$ satisfies the given condition?
| Hint. You may use the standard Taylor series expansions, as $x \to 0$,
$$
\begin{align}
\cos x&=1-\frac{x^2}2+O(x^4)\\
\sin x&=x-\frac{x^3}6+O(x^4)
\end{align}
$$ giving
$$
\begin{align}
\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}&=\frac{(1+a-b) x+\frac16 (-3 a+b) x^3+O(x^5)}{(f(x))^3}
\\\\&=\frac{(1+a-b) x+\frac169 (-3 a+b) x^3+O(x^5)}{x^3(1+\epsilon(x))^3}
\end{align}
$$ where, as $x \to 0$, we have used $f(x)=x(1+\epsilon(x))$ with $\epsilon(x) \to 0$.
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\lim\limits_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x}$. Is Wolfram wrong or is it me? What am I doing wrong?
My attempt
$$\begin{align}
\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x} &= \lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + x} \cdot \frac{\sqrt{x^2 + 3x} + \sqrt{x^2 + x}}{\sqrt{x^2 + 3x} + \sqrt{x^2 + x}} =\\
&= \lim_{x \to -\infty} \frac{2x}{\sqrt{x^2 + 3x} + \sqrt{x^2 + x}} =\\
&= \lim_{x \to -\infty} \frac{\frac1x \cdot 2x}{\sqrt{\frac{x^2}{x^2} + \frac{3x}{x^2}} + \sqrt{\frac{x^2}{x^2} + \frac{x}{x^2}}} = \\
&= \lim_{x \to -\infty} \frac{2}{\sqrt{1 + \frac3x} + \sqrt{1 + \frac1x}} = 1
\end{align}$$
Wolfram result
| On your way to the last line, you're tacitly assuming that, for example, $\frac1x\sqrt{x^2+x} = \sqrt{\frac{x^2}{x^2}+\frac{x}{x^2}}$. But that is only true when $x$ is positive!
When $x$ is negative, $\frac1x\sqrt{\cdots\vphantom{x}}$ will be negative, and thus it can never be written as a sum of square roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Tridiagonal matrix inner product inequality I want to show that there is a $c>0$ such that
$$
\left<Lx,x\right>\ge c\|x\|^2,
$$
for alle $x\in \ell(\mathbb{Z})$ where
$$
L=
\begin{pmatrix}
\ddots & \ddots & & & \\
\ddots & 17 & -4 & 0 & \\
\ddots & -4 & 17 & -4 & \ddots \\
& 0 & -4 & 17 & \ddots \\
& & \ddots & \ddots &\ddots
\end{pmatrix},
$$
is a tridiagonal matrix and
$$
x=
\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix}.
$$
I know that the following holds
\begin{align}
\left<Lx,x\right>&=\left<\begin{pmatrix}
\ddots & \ddots & & & \\
\ddots & 17 & -4 & 0 & \\
\ddots & -4 & 17 & -4 & \ddots \\
& 0 & -4 & 17 & \ddots \\
& & \ddots & \ddots &\ddots
\end{pmatrix}\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix},\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix}\right>\\
&=\left< -4\begin{pmatrix}
\vdots \\
x_{2} \\
x_1 \\
x_{0} \\
\vdots
\end{pmatrix} +17\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix} -4\begin{pmatrix}
\vdots \\
x_{0} \\
x_{-1} \\
x_{-2} \\
\vdots
\end{pmatrix},\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix} \right>\\
&=-4\left<\begin{pmatrix}
\vdots \\
x_{2} \\
x_1 \\
x_{0} \\
\vdots
\end{pmatrix},\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix} \right> + 17\left<\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix},\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix} \right> -4\left<\begin{pmatrix}
\vdots \\
x_{0} \\
x_{-1} \\
x_{-2} \\
\vdots
\end{pmatrix},\begin{pmatrix}
\vdots \\
x_{1} \\
x_0 \\
x_{-1} \\
\vdots
\end{pmatrix} \right>.
\end{align}
Hence,
$$
\left<Lx,x\right>=-4k+17\|x\|^2,
$$
where
$$
k=\sum_{j\in \mathbb{Z}}{x_j(x_{j+1}+x_{j-1})}.
$$
Obviously $\|x\|^2\ge 0$, but how can I choose $c$ such that the inequality holds? Here I get stuck, any hints?
| You can also finish your proof by noting that $k \le 2 \, \|x\|^2$ (by applying Hölder's inequality). Hence,
$$\langle L \, x , x \rangle \ge -4 \, k + 17 \, \|x\|^2 \ge 9 \, \|x\|^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1758470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What's the value of $x$ in the following equation?
So this is how I approached this question, the above equations could be simplified to :
$$a = \frac{4(b+c)}{b+c+4}\tag{1!}$$
$$b = \frac{10(a+c)}{a+c+10}\tag{2}$$
$$c=\frac{56(a+b)}{a+b+56}\tag{3}$$
From above, we can deduce that $4 > a$ since $\frac{(b+c)}{b+c+4} < 1$ similarly $10 > b, 56 > c$ so $a + b + c < 70$
Let, $$(a + b + c)k = 70\tag4$$
Now let, $$\alpha(b+c) = b+c+4\tag{1'}$$
$$\beta(a+c) = a+c+10\tag{2'}$$
$$\gamma( a+b ) = a+b+56\tag{3'}$$
Now adding the above 3 equations we get :
$$2(a+b+c) + 70 = a(\gamma + \beta) + c(\beta + \alpha) + b(\alpha + \gamma) \rightarrow (2 + k)(a+b+c) = a(\gamma + \beta) + c(\beta + \alpha) + b(\alpha + \gamma)$$
Now from above we see that coefficient of $a,b,c$ must be equal on both sides so, $$(2 + k) = (\alpha + \beta) = (\beta + \gamma) = (\alpha + \gamma)$$
Which implies $\beta = \gamma = \alpha = 1+ \frac{k}{2} = \frac{2 + k}{2}$,
Now from $(1)$ and $(1')$ we get $a = \frac{4}{\alpha} = \frac{8}{2+k}$ similarly from $(2),(2')$ and $(3),(3')$ we find, $b = \frac{20}{2+k}, c = \frac{112}{2+k}$
Thus from above we get $a+b+c = \frac{140}{2+k}$ and from $(4)$ we get: $\frac{140}{2+k} = \frac{70}{k}$ from which we can derive $k = 2$
Thus we could derive $a = 2, b = 5, c = 28$ but, the problem now is $a, b, c$ values don't satisfy equation $(4)$ above for $k =2$
Well so, where do I err ? And did I take the right approach ? Do post the solution about how you solved for $x$.
| Your deductions are wrong and that is what is misleading you. Integers can be both positive and negative.
If you solve equations (1), (2) and (3) simultaneously you can find a, b and c.
I did this to find $$a=3$$
$$b=5$$
$$c=7$$
You can then plug this into the forth equation given in the problem to solve for x.
$$x = \frac{abc}{ a + b + c} $$
which solves to give $$x=\frac{105}{15}=7$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1759327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Picking two random points on a disk I try to solve the following:
Pick two arbitrary points $M$ and $N$ independently on a disk $\{(x,y)\in\mathbb{R}^2:x^2+y^2 \leq 1\}$ that is unformily inside. Let $P$ be the distance between those points $P=d(M,N)$. What is the probabilty of $P$ being smaller than the radius of the disk?
Picking a specific point $s$ as the first point $M$ on a disk has probabilty $P(M=s)=0$ since we are looking at the real numbers. My intention, that is probably wrong, is that chosing the second point $N$ such that the distance between them is bigger than the radius of the disk, is as likely as being smaller.
I have absolutely no idea how to solve this.
| Let $R$ be the distance between $O$, the origin, and $M$. The probability that $R$ is less than or equal to a value $r$ is
$$P(R\le r) = \begin{cases}
\frac{\pi r^2}{\pi\cdot 1^2} = r^2, & 0\le r\le 1\\
1, &r>1\\
0, &\text{otherwise}
\end{cases}$$
The probability density function of $R$ is
$$f_R(r) = \frac{d}{dr}P(R\le r) = \begin{cases}
2r, & 0< r< 1\\
0, &\text{otherwise}
\end{cases}$$
If $OM = d$, $0\le d\le 1$, then the area of the intersection of the unit circles centred at $O$ and $M$ respectively is
$$\begin{align*}
A(d) &= 4\left(\frac12\cos^{-1}\frac d2\right) - 2\left(\frac12d\sqrt{1^2-\frac {d^2}{2^2}}\right)\\
&= 2\cos^{-1}\frac d2 - \frac 12 d\sqrt{2^2-d^2}\\
\frac{A(d)}{\pi\cdot 1^2}&=\frac{2}{\pi}\cos^{-1}\frac d2 - \frac{1}{2\pi} d\sqrt{2^2-d^2}
\end{align*}$$
The last line is the probability that $N$ is within a unit distance of $M$, as a function of distance $d = OM$.
Combining with the probability density function $f_R$ above, the required probability is
$$p = \int_0^1\left(\frac{2}{\pi}\cos^{-1}\frac r2 - \frac{1}{2\pi} r\sqrt{2^2-r^2}\right)\cdot 2r\ dr \approx 0.58650$$
(WolframAlpha)
$$\begin{align*}
I_1 &= \int_0^1 r\cos^{-1}\frac r2\ dr\\
&= \int_{\pi/2}^{\pi/3} 2\theta\cos\theta\cdot(-2)\sin\theta \ d\theta && (r = 2\cos \theta)\\
&= -2\int_{\pi/2}^{\pi/3}\theta\sin 2\theta \ d\theta\\
&= \int_{\pi/2}^{\pi/3}\theta\ d\cos2\theta\\
&= \left[\theta\cos2\theta\right]_{\pi/2}^{\pi/3} - \int_{\pi/2}^{\pi/3}\cos2\theta\ d\theta\\
&= \left[\theta\cos2\theta - \frac 12 \sin2\theta\right]_{\pi/2}^{\pi/3}\\
&= -\frac{\pi}{6} - \frac{\sqrt3}{4} +\frac\pi2\\
&= \frac\pi 3 - \frac{\sqrt3}{4}\\
\frac4\pi I_1 &= \frac 43 - \frac{\sqrt3}{\pi}
\end{align*}$$
$$\begin{align*}
I_2 &= \int_0^1 r^2\sqrt{2^2-r^2}\ dr\\
&= -16\int_{\pi/2}^{\pi/3} \cos^2\theta\sin^2\theta\ d\theta&&(r=2\cos \theta)\\
&= -4\int_{\pi/2}^{\pi/3} \sin^2 2\theta\ d\theta\\
&= -4\int_{\pi/2}^{\pi/3} \frac{1-\cos4\theta}{2}\ d\theta\\
&= -2\left[\theta-\frac{1}{4}\sin4\theta\right]_{\pi/2}^{\pi/3}\\
&= -2\left(\frac\pi3+\frac{\sqrt3}8-\frac\pi2\right)\\
&= \frac\pi3-\frac{\sqrt3}{4}\\
\frac1\pi I_2 &= \frac13-\frac{\sqrt3}{4\pi}
\end{align*}$$
$$p = 1-\frac{3\sqrt3}{4\pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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