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Find the equation of a circle tangent to two lines Find the equation of the circle tangent to the $x$-axis at the origin and tangent to the line $4x-3y+24=0$.
My Approach:
Let the equation of the required circle be:
$$x^2+y^2+2gx+2fy+c=0$$
Let the equation to the tangent at origin $(0,0)$ to the above circle be $gx+fy+c=0$.
Then, what should I do? please help me to continue.
Thanks in Advance.
| As the circle touches the x-axis at (0,0) then the centre of the circle must be on the y-axis and the y coordinate of the centre must be equal to the radius. So the centre is $(0,r)$. So the equation is:
$$x^2+(y-r)^2=r^2$$
Then you want to find the point of intersection of the circle and the straight line. Lets rearrange the straight line:
$$x=\frac{3y-24}{4}$$
Sub in:
$$\left(\frac{3y-24}{4}\right)^2+(y-r)^2=r^2$$
Multiple by $4^2$ to get rid of the fractions:
$$(3y-24)^2+16(y-r)^2=16r^2$$
Expand:
$$9y^2-144y+576+16y^2-32yr+16r^2=16r^2$$
Collect and sort by powers of $y$:
$$25y^2-16(2r+9)y +576=0$$
Now we expect there to be only one point of intersection so this quadratic should have only one solution. This occurs when the discriminant is zero.
$$(16(2r+9))^2-4\cdot25\cdot576=0$$
Pull out the $16^2$ term and factorise $576$ in preparation to divide out common factor.
$$256(2r+9)^2-4\cdot25\cdot64\cdot9=0$$
Divide by common factor of $256$.
$$(2r+9)^2-25\cdot9=0$$
Expand and multiply:
$$4r^2+36r+81-225=0$$
Collect:
$$4r^2+36r-144=0$$
Divide by 4:
$$r^2+9r-36=0$$
Factorise:
$$(r-3)(r+12)=0$$
Solve:
So $r=3$ or $r=-12$
Note that $r=-12$ is a valid solution. It means that the circle is below the axis where earlier we assumed it was above the axis (when we set the centre to $(0,r)$).
So the two solutions are: $x^2+(y-3)^2=3^2$ and $x^2+(y+12)^2=12^2$
Or in expanded form: $x^2+y^2-6y=0$ and $x^2+y^2+24y=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Divisibility Proof $8\mid (x^2 - y^2)$ for $x$ and $y$ odd $x,y \in\Bbb Z$. Prove that if $x$ and $y$ are both odd, then $8\mid (x^2 - y^2)$.
My Proof Starts:
Assume $x$ and $y$ are both odd. So, $x = 2k + 1$ and $y = 2l +1$ for some integers $k$ and $l$. Thus,
\begin{align}
x^2 - y^2 &= (2k + 1)^2 - (2l + 1)^2 \\
&= 4k^2 + 4k + 1 - (4l^2 + 4l + 1) \\
&= 4k^2 + 4k - 4l^2 - 4l
\end{align}
My two concerns:
1) Is this correct so far?
2) How would I deal with the “$8\;\mid$” part?
| You're correct so far. What you need to finish is this
Hint: $4u^2+4u = 8v$
Solution:
$ 4u^2+4u=4(u+1)u=8\binom{u+1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Factorization of $x^5-1$ over $\mathbb F_{19}$ I've come across this question Factorization of $x^5-1$ over $F_{11}$ and $F_{19}$. The answer was good but I don't understand how to actually solve it.
It says that i can split $x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)$. If I expand this I get $x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd$. So to my understanding this gives:
$a+c=1 \mod 19$
$ac+d+b=1 \mod 19$
$ad+bc=1 \mod 19$
$bd=1 \mod 19$
What I dont understand is how to solve this system of equations. Any pointers? Thank you!
| Paying some time on this problem in order to give an elementary proof for the O.P. one has the cyclotomic $x^4+x^3+x^2+x+1$ is irreducible over $\Bbb F_{19}$.
We have
$$a+c=1\Rightarrow (a,c)\in S_1\subset \Bbb F_{19}\text { x }\Bbb F_{19}\\bd=1\Rightarrow (b,d)\in S_2\subset \Bbb F_{19}\text { x }\Bbb F_{19}$$
$$S_2=\{(1,1),(2,-9),(3,-6),(4,5),(5,4),(6,-3),(7,-8),(8,-7),(9,-2),(-9,2),(-8,7),(-7,8),(-6,3),(-5,-4),(-3,6),(-2,9),(-1,-1)\}$$
$$ad+bc=1\Rightarrow\begin{cases}a+b^2c=b\\ad^2+c=d\end{cases}\Rightarrow\begin{cases}c(b+1)=1\\a(d+1)=1\end{cases}\qquad (*)$$ Then we can evaluate directly all the cases $0\le a\le 18$ using the explicit set $S_2$ and $(*)$ the following way:
Let $E=ac+b+d$; we need to have $E=1$
►$a=0,1$ is trivially discarded.
►$(a,c)=(2,-1)\Rightarrow d+1=10,b+1=-1\Rightarrow E=-2+9-2=5\ne 1$
►$(a,c)=(3,-2)\Rightarrow d+1=13,b+1=9\Rightarrow E=-6+12+8=14\ne 1$
►$(a,c)=(4,-3)\Rightarrow d+1=5,b+1=6\Rightarrow E=-12+4+5\ne 1$
An so on till
►$(a,c)=(18,2)\Rightarrow d+1=-1,b+1=10\Rightarrow E=36+9-2\ne 1$
This shows that $x^4+x^3+x^2+x+1$ does not factorize on $\Bbb F_{19}[x]$
$$***$$
For $\Bbb F_{11}$ the same method applies and we find at once for $(a,c)=(2,-1)$
$$x^4+x^3+x^2+x+1=(x+2)(x-3)(x-4)(x+6)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve for power series $y'' - 9y = 0$ I really need help with this, the solution from this equation is $y(x) = c_1 e^{3x} + c_2 e^{-3x}$. But I can't get to it, I obtain the next:
$$y(x) = \sum_{n=0}^{\infty}a_nx^n$$
$$y''(x) = \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2}$$
Then the coefficients must be $a_{2m} = \frac{9^m a_0}{(2m)!}$ and $a_{2m+1}= \frac{9^ma_1}{(2m+1)!}$
Substituting in the first equation I have:
$$y(x) = a_0 \sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} + a_1\sum_{m=0}^{\infty} \frac{3^{2m}x^{2m+1}}{(2m+1)!} $$
Since $e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n!}$, I think its obvious that the first part of the last equation is $a_0e^{3x}$, but in the second part I dont really know how to get $a_1e^{-3}$. I am wrong?
| Note that $a_0=y(0)=c_1+c_2$ and $a_1=y'(0)=3c_1-3c_2$. Then
\begin{align}
y(x) &= a_0 \sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} + a_1\sum_{m=0}^{\infty} \frac{3^{2m}x^{2m+1}}{(2m+1)!}\\ \ \\
&= (c_1+c_2) \sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} + (3c_1-3c_2)\sum_{m=0}^{\infty} \frac{3^{2m}x^{2m+1}}{(2m+1)!}\\ \ \\
&= c_1\,\left(\sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} + 3\sum_{m=0}^{\infty} \frac{3^{2m}x^{2m+1}}{(2m+1)!}\right)+c_2\,\left(\sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} - 3\sum_{m=0}^{\infty} \frac{3^{2m}x^{2m+1}}{(2m+1)!}\right)\\ \ \\
&= c_1\,\left(\sum_{m=0}^{\infty} \frac{(3x)^{2m}}{(2m)!} + \sum_{m=0}^{\infty} \frac{3^{2m+1}x^{2m+1}}{(2m+1)!}\right)+c_2\,\left(\sum_{m=0}^{\infty} \frac{(-3x)^{2m}}{(2m)!} \sum_{m=0}^{\infty} \frac{(-3x)^{2m+1}}{(2m+1)!}\right)\\ \ \\
&=c_1\,e^{3c}+c_2\,e^{-3x}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find minimal value $ \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$ without derivatives. Find minimal value of $ \sqrt {{x}^{2}-5\,x+25}+\sqrt {{x}^{2}-12\,\sqrt {3}x+144}$ without using the derivatives and without the formula for the distance between two points.
By using the derivatives I have found that the minimal value is $13$ at $$ x=\frac{40}{23}(12-5\sqrt{3}).$$
| If we complete the squares we get $$\sqrt{\left(x-{5\over 2}\right)^2 +{75\over 4}} + \sqrt{ (x-6\sqrt{3})^2 +36}.$$ This is the distance from the point $P(x,0)$ to the point $Q(5/2, -5\sqrt{3}/2)$ plus the distance from $P(x,0)$ to the point $R(6\sqrt{3},6).$ Choosing $x$ amounts to sliding $P$ along the $x$-axis. The shortest sum of distances will be the straight line segment from $Q$ to $R$. This length is $$\sqrt{ \left( 6+{5\sqrt{3}\over 2}\right)^2 +(6\sqrt{3}-5/2)^2}=13.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $d$ must be a perfect square
Prove that if $a,b,c,d$ are integers such that $$(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^2 = d,$$ then $d$ is a perfect square.
In order for $(a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c)^2$ to be an integer, either $a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ has to be the square root of a positive integer or $a+2^{\frac{1}{3}}b+2^{\frac{2}{3}}c$ must be a perfect square. How do we deal with those conditions?
| We expand and get
$$
d=(a+2^{1/3}b+2^{2/3}c)^2\\
=(a^2+4bc)+(2c^2+2ab)2^{1/3}+(b^2+2ac)2^{2/3}
$$
If $d$ is to be an integer, then from the expression above we must have $2c^2+2ab=b^2+2ac=0$. Assuming $a\neq0$, this means that
$$
b=-\frac{c^2}{a}\\
c=-\frac{b^2}{2a}
$$
Inserting one of these into the other, we get
$$
b=-\frac{\frac{b^4}{4a^2}}{a}=-\frac{b^4}{4a^3}\\
b(b^3+4a^3)=0
$$
so since $a\neq0$, and $\sqrt[3]4\notin \Bbb Q$, we must have $b=0$, which gives $c=0$, and $d=a^2$, a perfect square.
On the other hand, if $a=0$, we have $c^2=b^2=0$, so $d=0$. Whether you count this as a perfect square is up to your definitions, but I would say that it is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A better way to evaluate a certain determinant
Question Statement:-
Evaluate the determinant:
$$\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix}$$
My Solution:-
$$
\begin{align}
\begin{vmatrix}
1^2 & 2^2 & 3^2 \\
2^2 & 3^2 & 4^2 \\
3^2 & 4^2 & 5^2 \\
\end{vmatrix} &=
(1^2\times2^2\times3^2)\begin{vmatrix}
1 & 1 & 1 \\
2^2 & \left(\dfrac{3}{2}\right)^2 & \left(\dfrac{4}{3}\right)^2 \\
3^2 & \left(\dfrac{4}{2}\right)^2 & \left(\dfrac{5}{3}\right)^2 \\
\end{vmatrix}&\left[\begin{array}{11}C_1\rightarrow\dfrac{C_1}{1} \\
C_2\rightarrow\dfrac{C_2}{2^2}\\
C_3\rightarrow\dfrac{C_3}{3^2}\end{array}\right]\\
&=
(1^2\times2^2\times3^2)\begin{vmatrix}
1 & 0 & 0 \\
2^2 & \left(\dfrac{3}{2}\right)^2-2^2 & \left(\dfrac{4}{3}\right)^2-2^2 \\
3^2 & 2^2-3^2 & \left(\dfrac{5}{3}\right)^2-3^2 \\
\end{vmatrix} &\left[\begin{array}{11}C_2\rightarrow C_2-C_1 \\
C_3\rightarrow C_3-C_1\end{array}\right]\\
&=
(1^2\times2^2\times3^2)
\begin{vmatrix}
1 & 0 & 0 \\
2^2 & 2^2-\left(\dfrac{3}{2}\right)^2 & 2^2-\left(\dfrac{4}{3}\right)^2 \\
3^2 & 3^2-2^2 & 3^2-\left(\dfrac{5}{3}\right)^2 \\
\end{vmatrix}\\
&=(1^2\times2^2\times3^2)
\begin{vmatrix}
1 & 0 & 0 \\
2^2 & \dfrac{7}{4} & \dfrac{20}{9} \\
3^2 & 5 & \dfrac{56}{9} \\
\end{vmatrix}\\
&=(1^2\times2^2)
\begin{vmatrix}
1 & 0 & 0 \\
2^2 & \dfrac{7}{4} & 20 \\
3^2 & 5 & 56 \\
\end{vmatrix}\\
&=4\times(-2)\\
&=-8
\end{align}
$$
As you can see, my solution is a not a very promising one. If I encounter such questions again, so would you please suggest a better method which doesn't include this ridiculous amount of calculations.
| The direct formula for $3$ by $3$ determinants isn't so bad
$$\begin{vmatrix}
a & b & c \\
d & e & f \\
g & h & i \\
\end{vmatrix}=aei+bfg+cdh-ceg-bdi-afh$$
so
$$\begin{vmatrix}
1 & 4 & 9 \\
4 & 9 & 16 \\
9 & 16 & 25 \\
\end{vmatrix}=225+576+576-729-400-256=-8.$$
Row operations and other similar tricks tend to speed things up only when the matrix is $4$ by $4$ or larger.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
} |
Compute $\int^{\pi}_0\frac{1}{(5-3\cos x)^3}dx$ Compute $$\int^{\pi}_0\frac{1}{(5-3\cos x)^3}dx$$
Tried use substitution
$$\mu=5-3\cos x$$
But made the problem even complecated...Any help?
Thank you~
| $$I=\int^{\pi}_0\frac{1}{(5-3\cos x)^3}dx$$
$\cos(-x)=\cos(x)$, hence we can write:
$$
I=\frac{1}{2}\int^{\pi}_{-\pi}\frac{1}{(5-3\cos x)^3}dx=\frac{1}{2}\int^{2\pi}_{0}\frac{1}{(5+3\cos x)^3}dx
$$
Last expression can be calculated through methods of complex analysis:
$$
I=\frac{1}{2}\int_{|z|=1}\frac{1}{\left(5+\frac{3}{2}(z+\frac{1}{z})\right)^3}\frac{dz}{iz}=-i\frac{1}{2}\int_{|z|=1}\frac{z^2}{\left(5z+\frac{3}{2}(z^2+1)\right)^3}dz\to\\
I=-\frac{4i}{27}\int_{|z|=1}\frac{z^2}{\left(z^2+\frac{10}{3}z+1\right)^3}dz
$$
Roots of the denominator: $z_1=-\frac{1}{3},z_2=-3$. Obviously, within unit circle only $z_1$, then:
$$
I=2\pi i*\left(-\frac{4i}{27}\right)Res_{z=z_1}\frac{z^2}{\left(z^2+\frac{10}{3}z+1\right)^3}
$$
Here we have pole of order 3, hence:
$$
Res_{z=z_1}\frac{z^2}{\left(z^2+\frac{10}{3}z+1\right)^3}\to\\\frac{1}{2!}\frac{d^2}{dz^2}\frac{z^2}{(z-z_2)^3}|_{z=z_1}=\frac{1}{2!}\left(\frac{2}{(z_1-z_2)^3}-\frac{12z_1}{(z_1-z_2)^4}+\frac{12z^2_1}{(z_1-z_2)^5}\right)=\frac{1593}{16384}
$$
Thus,
$$
I=2\pi i*\left(-\frac{4i}{27}\right)*\frac{1593}{16384}=\frac{59}{2048}\pi
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve the differential equation $\frac{dy}{dx} + 3yx = 0$ for the values $x = 0$ when $y = 1$ - Solution Review
Solve the differential equation $\frac{dy}{dx} + 3yx = 0$; $x = 0$ when $y = 1$.
I solved this DE using the integration factor method. However, online calculators are giving me a different answer, where they instead used the separation of variables method.
Please review my solution and indicate if/where my reasoning is false, why it is false, how to fix it, and what the correct reasoning should be. Thank you.
My solution is as follows.
$\dfrac{dy}{dx} + 3xy = 0$
$e^{ \int 3x } dx = e^{ \frac{3x^2}{2} }$
$ \dfrac{dy}{dx} e^{ \frac{3x^2}{2} } + 3yxe^{ \frac{3x^2}{2} } = 0$
$ \dfrac{d}{dx} \left( y e^{ \frac{3x^2}{2}} \right) = \dfrac{dy}{dx} e^{ \frac{3x^2}{2} } + 3xye^{ \frac{3x^2}{2}} $
$ \therefore \dfrac{d}{dx} \left( y e^{ \frac{3x^2}{2}} \right) = 0$
$ \displaystyle\int \dfrac{d}{dx} \left( y e^{ \frac{3x^2}{2}} \right) dx + C = 0$
$ \Rightarrow y e^{ \frac{3x^2}{2}} + C = 0$
We want to solve for y. We know that $x = 0$.
$ \therefore ye^0 + C = 0$
$ \Rightarrow y = -C$
| We separate the variables
$\frac{dy}{y}=-3xdx$
which we integrate as follows
$ln(\frac{y}{C})=-\frac{3}{2}x^2$
and
$y=Ce^{-\frac{3}{2}x^2}$
if $x=0$, $y=1$ so $C=1$
and finally
$$y=e^{-\frac{3}{2}x^3}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Simplifying fraction of infinite series There is $\frac{1+\left( jx\right) ^{-1/5}\sum\limits_{k=2}^{\infty }\frac{%
\left( jx\right) ^{k/5}}{k!}}{1+\left( jx\right) ^{1/5}%
+\sum\limits_{k=2}^{\infty }\frac{\left( jx\right) ^{k/5}}{k!}}$
(where $x\epsilon \mathbb{R}$, $x>10$ and $j=\sqrt{-1}$).
It is $\lim\limits_{k\rightarrow \infty }\frac{\frac{\left( jx\right) %
^{\left( k\right) /5}}{k!}}{\frac{\left( jx\right) ^{\left( k-1\right) /5}}{%
\left( k-1\right) !}}=\lim\limits_{k\rightarrow \infty }\frac{1}{k}=0$,
i.e. the sum converges.
Is it ok (why or why not?) to approximate (in first order)
enumerator and denominator and say
$\frac{1+\left( jx\right) ^{-1/5}\sum\limits_{k=2}^{\infty }\frac{\left( jx\right) ^{k/5}}{k!}}{%
1+\left( jx\right) ^{1/5}+\sum\limits_{k=2}^{\infty }\frac{\left( jx\right) %
^{k/5}}{k!}}\approx \frac{1}{1+\left( jx\right) ^{1/5}}$ ?
And how can I simplify
$\frac{1+\left( jx\right) ^{-1/5}\sum\limits_{k=2}^{\infty }\frac{%
\left( jx\right) ^{k/5}}{k!}}{1+\left( jx\right) ^{1/5}+%
\sum\limits_{k=2}^{\infty }\frac{\left( jx\right) ^{k/5}}{k!}}$ ?
| We can use the power series representation of the exponential function
\begin{align*}
\exp(x)=\sum_{k=0}^\infty \frac{x^k}{k!}\qquad\qquad\qquad x\in\mathbb{C}
\end{align*}
in order to simplify the expression. We get
\begin{align*}
\sum_{k=2}^\infty \frac{(jx)^{(k/5)}}{k!}&=\sum_{k=2}^\infty \frac{\left((jx)^\frac{1}{5}\right)^k}{k!}
=\exp\left((jx)^\frac{1}{5}\right)-1-(jx)^{\frac{1}{5}}\tag{1}
\end{align*}
We obtain with the help of (1):
\begin{align*}
\frac{1+(jx)^{-\frac{1}{5}}\sum_{k=2}^\infty \frac{(jx)^{(k/5)}}{k!}}
{1+(jx)^{\frac{1}{5}}+\sum_{k=2}^\infty \frac{(jx)^{(k/5)}}{k!}}
&=\frac{1+(jx)^{-\frac{1}{5}}\left(\exp\left((jx)^\frac{1}{5}\right)-1-(jx)^{\frac{1}{5}}\right)}{1+(jx)^{\frac{1}{5}}+\left(\exp\left((jx)^\frac{1}{5}\right)-1-(jx)^{\frac{1}{5}}\right)}\\
&=\frac{(jx)^{-\frac{1}{5}}\left(\exp\left((jx)^\frac{1}{5}\right)-1\right)}{\exp\left((jx)^\frac{1}{5}\right)}\\
&=(jx)^{-\frac{1}{5}}\left(1-\exp\left(-(jx)^{\frac{1}{5}}\right)\right)\tag{2}\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1972945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
To find the range of $\sqrt{x-1} + \sqrt{5-x}$ To find the range of $\sqrt{x-1} + \sqrt{5-x}$.
I do not know how to start?
Thanks
| Clearly $1\le x\le5$
As $\dfrac{1+5}2=3, 1-3\le x-3\le5-3$
WLOG $x-3=2\cos2y,$ where $0\le2y\le\pi$
$\sqrt{x-1}=2\cos y$ and $\sqrt{5-x}=2\sin y$
Now $\sin y+\cos y=\sqrt2\sin\left(y+\dfrac\pi4\right)$ and $0\le y\le\dfrac\pi2$
$\implies\dfrac1{\sqrt2}\le\sin\left(y+\dfrac\pi4\right)\le1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1975431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
cube roots of unity Let $\omega \ne 1$ be a complex cube root of unity.
If $$(4 + 5\omega + 6\omega ^2)^{n^2 + 2} + (6 + 5\omega ^2 + 4\omega)^{n^2 + 2} + (5 + 6\omega + 4\omega ^2)^{n^2 + 2} = 0$$ then $n$ can be...
| The LHS of the given expression can be written thus:
$$\left( 4 + 5\omega + 6\omega ^2 \right)^{n^2 + 2}+ \left( (4 + 5\omega + 6\omega ^2)w \right)^{n^2 + 2}+\left( (4 + 5\omega + 6\omega ^2)w^2 \right)^{n^2 + 2}$$
$$=\left( 4 + 5\omega + 6\omega ^2 \right)^{n^2 + 2}.F \ \ \text{with} \ \ F=1+w^{n^2 + 2}+(w^2)^{n^2 + 2}$$
The first factor is never zero, being a power of a non-zero element.
Thus, it suffices to examine the cases where $F=0$.
As $n^2+2\equiv \begin{cases}2 \ (mod. 3) & when \ n \equiv 0 \ (mod. 3)\\
0 \ (mod. 3)& when \ n \equiv1 \ (mod. 3)\\ 0 \ (mod. 3)& when \ n \equiv 2 \ (mod. 3)\end{cases}\ \ $, thus
*
*if $n \ \equiv 0 \ $ mod. 3: $F=1+w^{n^2 + 2}+(w^{n^2 + 2})^2=\dfrac{1-(w^{n^2 + 2})^3}{1-w^{n^2 + 2}}=\dfrac{1-(w^3)^{n^2 + 2}}{1-w^{n^2 + 2}}=0$
*otherwise $F=1+1+1=3\neq 0$ iff $n=1,2 \ mod. 3$.
Thus the result is that the LHS of your formula is 0 iff $n=3k$ for some integer $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove the result of this integration? How to prove that?
$$\int_0^1 \frac{1 - e^{-t} - e^{-1/t}}{t}\ \text{d}t = \gamma$$
where $\gamma = 0.5772156649015328606065\ldots$ is the Euler-Mascheroni constant.
Additional question: is there a way to evaluate it via Residues Theorem too?
| Preliminary results.
Let us begin with a definition of the Euler-Mascheroni Constant
\begin{equation}
\gamma = \lim_{n \to \infty} H_{n} - \mathrm{ln}(n)
\label{eq:1}
\tag{1}
\end{equation}
where $H_{n}$ are the harmonic numbers defined as
\begin{equation}
H_{n} = \displaystyle\sum_{k=1}^{n} \frac{1}{k}
\label{eq:2}
\tag{2}
\end{equation}
Let
\begin{equation}
\int\limits_{0}^{1} \frac{1-(1-x)^{n}}{x} \mathrm{d} x = H_{n}
\label{eq:3}
\tag{3}
\end{equation}
Proof:
\begin{align}
\tag{a}
\int\limits_{0}^{1} \frac{1-(1-x)^{n}}{x} \mathrm{d} x & = \int\limits_{0}^{1} \frac{1-y^{n}}{1-y} \mathrm{d} y \\
& = \int\limits_{0}^{1} \frac{1}{1-y} \mathrm{d} y \, - \int\limits_{0}^{1} \frac{y^{n}}{1-y} \mathrm{d} y \\
\tag{b}
& = \int\limits_{0}^{1} \displaystyle\sum_{k=0}^{\infty} y^{k} \mathrm{d} y \,\, - \int\limits_{0}^{1} y^{n} \displaystyle\sum_{k=0}^{\infty} y^{k} \mathrm{d} y \\
& = \displaystyle\sum_{k=0}^{\infty} \frac{y^{k+1}}{k+1} \Big|_{0}^{1} \,\, - \displaystyle\sum_{k=0}^{\infty} \frac{y^{k+n+1}}{k+n+1} \Big|_{0}^{1} \\
& = (1 + \frac{1}{2} + \frac{1}{3} + \dots) \, - (\frac{1}{n+1} + \frac{1}{n+2} +\frac{1}{n+3} + \dots) \\
& = 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} = H_{n}
\end{align}
Notes:
a. Let $y=1-x$
b. Expand $$\frac{1}{1-y} = \sum\limits_{k=0}^{\infty} y^{k}$$
Main result.
\begin{align}
\gamma &= \int\limits_{0}^{1} \frac{1-\mathrm{e}^{-x}-\mathrm{e}^{-1/x}}{x} \mathrm{d}x \\
&= \int\limits_{0}^{1} \frac{1-\mathrm{e}^{-x}}{x} \mathrm{d}x - \int\limits_{0}^{1} \frac{\mathrm{e}^{-1/x}}{x} \mathrm{d}x \\
&= \int\limits_{0}^{1} \frac{1-\mathrm{e}^{-x}}{x} \mathrm{d} x \,\, - \int\limits_{1}^{\infty} \frac{\mathrm{e}^{-x}}{x} \mathrm{d} x
\label{eq:4}
\tag{4}
\end{align}
We made the substitution $z=1/x$, then switched variables back to $x$.
Make the substitution $x = \frac{y}{n}$ in equation \eqref{eq:3} to obtain
\begin{align}
H_{n} & = \int\limits_{0}^{n} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \\
& = \int\limits_{0}^{1} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \,\, + \int\limits_{1}^{n} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \\
& = \int\limits_{0}^{1} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \,\, + \int\limits_{1}^{n} \frac{1}{y} \mathrm{d} y \,\, - \int\limits_{1}^{n} \frac{(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \\
& = \int\limits_{0}^{1} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d} y \,\, + \mathrm{ln}(n) \, - \int\limits_{1}^{n} \frac{(1-\frac{y}{n})^{n}}{y} \mathrm{d} y
\label{eq:5}
\tag{5}
\end{align}
Now we invoke the limit defnition of the exponential function
\begin{equation}
\mathrm{e}^{\pm x} = \lim_{n \to \infty} \left(1 \pm \frac{x}{n} \right)^{n}
\label{eq:6}
\tag{6}
\end{equation}
rearrange equation \eqref{eq:5} and take $\lim_{n \to \infty}$, we have
\begin{equation}
\lim_{n \to \infty} \left(H_{n} - \mathrm{ln}(n)\right)
= \lim_{n \to \infty} \left(\int\limits_{0}^{1} \frac{1-(1-\frac{y}{n})^{n}}{y} \mathrm{d}y
- \int\limits_{1}^{n} \frac{(1-\frac{y}{n})^{n}}{y} \mathrm{d}y\right)
\end{equation}
The left hand side equals $\gamma$ by equation \eqref{eq:1} as does the right hand side by equations
\eqref{eq:6}, \eqref{eq:5}, and \eqref{eq:4}.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
$\sqrt{3}$ represented as continued fraction Am I incorrect in the assumption that I deal with $\sqrt{3}$ in the same way I would approach $\sqrt{2}$, by adding and subtracting $1$ such that:
$\sqrt{3}=1+\sqrt{3}-1=1+\cfrac{2}{1+\sqrt{3}}$
The table representation is $[1; 1,2]$.
This is what I have so far:
$\sqrt{3}=1+\cfrac{2}{2+\cfrac{2}{1+\sqrt{3}}}$
Which seems to lead me to $[1;2,2,...]$ any help would be appreciated, thanks
| If you obtain a rational approximation of $\sqrt{3}$, for example $\frac{362}{209}$ you'll find its continued fraction is $$1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{3}}}}}}}$$ That 3 looks extraneous, so we might form the conjecture that $$\sqrt{3}=1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\ddots}}}}}}$$
Then we can prove that by calling the RHS x
$$x=1+\frac{1}{1+\frac{1}{1+\color{red}{1+\frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\ddots}}}}}}}=1+\frac{1}{1+\frac{1}{1+x}}$$
Then by simple algebra we can verify that $x=\sqrt{3}$.
On a small digression, these rational approximations might seem hard to obtain but I found that with the chakravala method in a few minutes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
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} |
Finding the roots of any cubic with trigonometric roots.
Question: How would you find the roots of a cubic polynomial whose roots are only expressive in trigonometric forms?
I'm really confused on how you would solve it. Some examples:$$x^3+x^2-2x-1=0\\x_1=2\cos\frac {2\pi}7,x_2=2\cos\frac {4\pi}7,x_3=2\cos\frac {8\pi}7\tag{1}$$$$x^3-x^2-9x+1=0\\x_1=4\cos\frac {2\pi}7,x_2=4\cos\frac {4\pi}7+1,x_3=4\cos\frac {8\pi}7+1\tag{2}$$$$x^3+x^2-6x-7=0\\x_1=2\left(\cos\frac {4\pi}{19}+\cos\frac {6\pi}{19}+\cos\frac {10\pi}{19}\right)\\x_2=2\left(\cos\frac {2\pi}{19}+\cos\frac {14\pi}{19}+\cos\frac {16\pi}{19}\right)\\x_3=2\left(\cos\frac {8\pi}{19}+\cos\frac {12\pi}{19}+\cos\frac {20\pi}{19}\right)\tag{3}$$
With $(3)$ being a very famous relation with this problem.
Now that I know it's possible, I'm wondering if there is a simple way to find the roots of any cubic with trigonometric roots. And is it possible to use the method to find the roots of cubics such as $x^3+x^2-10x-8=0$?
| The strategy to solve a cubic equation is first to change it to the form $x^3+px+q=0$, i.e. to eliminate the squared term.
Cardano's method works well if there is only $1$ real root ($4px^3+27q^2>0$). If there are $3$ real roots ($4px^3+27q^2<0$), you can set $x=A\cos\theta$ ($A>0$). The equation becomes
$$A^3\cos^3\theta+pA\cos\theta+q=0.$$
We choose $A$ so that $A^3\cos^3\theta+pA\cos\theta$ is proportional to the expansion of $\cos 3\theta$ in function of $\cos\theta$. Remember $\cos 3\theta=4\cos^3\theta-3\cos\theta$. So we must have
$$\frac{A^3}4=\frac{pA}{-3}\iff A^2=\frac{-p}3\iff A=\sqrt{\frac{-p}3}\qquad\text{(we chose $A>0$)}.$$
We obtain
$$-\frac p3\sqrt{-\frac p3}(4\cos^3\theta-3\cos\theta)=-q\iff\cos3\theta=-\frac q{-\dfrac p3\sqrt{-\dfrac p3}}$$
There remains to solve this standard trigonometric equation.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Why is $\lim_{n \rightarrow \infty} \bigg( \bigg| \frac{n^{n^2} (n+2)^{(n+1)^2}}{(n+1)^{2n^2+2n+1}} \bigg| \bigg)=e$? I know that the following is the correct limit, but I have difficulties in seeing just why this is.
$$\lim_{n\to\infty}\left| \frac{n^{n^2} (n+2)^{(n+1)^2}}{(n+1)^{2n^2+2n+1}}\right|=e$$
| Starting with $2 n^2 + 2n + 1 = n^2 + (n+1)^2$ then
\begin{align}
\frac{n^{n^2} (n+2)^{(n+1)^2}}{(n+1)^{2n^2+2n+1}} &= \left(\frac{n}{n+1}\right)^{n^{2}} \, \left( \frac{n+2}{n+1} \right)^{(n+1)^{2}} \\
&= \frac{ \left(1 + \frac{1}{n+1} \right)^{(n+1)^{2}} }{ \left( 1 + \frac{1}{n} \right)^{n^{2}}} \\
&= \frac{e^{(n+1)^{2} \, \ln(a_{n+1})} }{ e^{n^{2} \, \ln(a_{n})} } \hspace{5mm} a_{n} = 1 + \frac{1}{n} \\
&= e^{n^{2} \, \ln\left(1 + \frac{2}{n}\right) + n \, \ln\left( 1 + \frac{1}{n+1} \right) } \, e^{\ln\left( \left(1 + \frac{1}{n+1}\right)^{n+1} \right)}
\end{align}
Now
\begin{align}
\lim_{n \to \infty} \left[\frac{n^{n^2} (n+2)^{(n+1)^2}}{(n+1)^{2n^2+2n+1}} \right] &= e^{0 + 0} \, e^{\ln(e)} = e.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Performing LU factorization on Ax = b. Can someone show me step by step how this works? Consider matrix
$$
A= \begin{bmatrix}
0 & -2 & 1 \\
2 & 1 & -1 \\
-2 & -2 & 1 \\
\end{bmatrix}
$$
and resulting vector :
b = (3
0
1)
Perform LU factorization with row swapping, indicating at each step the Gauss transform or pivot matrix used. Then solve Ax = b.
LU factorization is something I just cannot understand or wrap my head around. I would greatly appreciate it if someone could work this out and show the steps.
Thanks guys.
| We are given:
$$A = \begin{bmatrix} 0 & -2 & 1 \\ 2 & 1 & -1 \\ -2 & -2 & 1 \end{bmatrix},b = \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix}$$
We are asked to solve $Ax = b$ using the LU factorization with pivoting. Our approach will be:
*
*$(1.)$ Compute $PA = LU$, if $PA = LU$, then $LU x = Pb$, so find $L, U, P$
*$(2.)$ Solve $Ly = Pb$ for $y$ using Forward Substitution
*$(3.)$ Solve $U x = y$ for $x$ using Backward Substitution
Step (1.)
We let $U = A$ and write $P$ and $L$ as the $3~x~3$ identity matrix, so:
$$U = \begin{bmatrix} 0 & -2 & 1 \\ 2 & 1 & -1 \\ -2 & -2 & 1 \end{bmatrix}, ~L = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, ~P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
The first step of the factorization process is to determine the entry of largest magnitude in column $1$. This is the entry $2$ in row $2$. We therefore swap rows $1$ and $2$ of the matrices $U$ and $P$ to obtain:
$$U = \begin{bmatrix} 2 & 1 & -1\\ 0 & -2 & 1 \\ -2 & -2 & 1 \end{bmatrix}, ~L = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, ~P = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
We then subtract suitable multiples of row $1$ of $U$ from rows $2$ through $3$, to create zeros in the first column of $U$ below the diagonal. The negative of the multiples are stored in the subdiagonal entries of the first column of the matrix $L$. These operations are ($R2$ already has a zero in the first position, so we leave it alone): $ 1 \times R1 + R3 \rightarrow R3$. This gives the matrices:
$$U = \begin{bmatrix} 2 & 1 & -1\\ 0 & -2 & 1 \\ 0 & -1 & 0 \end{bmatrix}, ~L = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}, ~P = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
We now repeat this for $R2$ and $R3$. We notice that we already have the largest magnitude in $R22$, so no row swap is needed. We now want to get zeros in the second position and see we can take $-\dfrac{1}{2} \times R2 + R3 \rightarrow R3$ and again update that entry in $L$ with the negative and we get:
$$U = \begin{bmatrix} 2 & 1 & -1\\ 0 & -2 & 1 \\ 0 & 0 & -\dfrac{1}{2} \end{bmatrix}, ~L = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & \dfrac{1}{2} & 1 \end{bmatrix}, ~P = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$
Since $U$ is upper triangular and $L$ is lower triangular, we now have $PA = LU$ (you can check this by multiplying those out).
Can you now do steps $(2.)$ and $(3.)$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving Logarithm Equation Solve for x:$$x^2(\log_{10} x)^5 = 100$$
Here's what I've tried:
$(\log_{10} x)^5 = A \\ \log_{ \log x} A= 5 = \frac{\log_{10} A}{\log_{10} \log x}$
Not sure how to continue
| $$x^2(\log_{10}x)^{5}=10^2$$
$$x^{\frac{2}{5}}\log_{10}x=10^{\frac{2}{5}}$$
$$\log_{10}x^{x^\frac{2}{5}}=\log_{10}10^{10^\frac{2}{5}}$$
$$x^{x^\frac{2}{5}}=10^{10^\frac{2}{5}}$$
so the $$x=10$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$
$P(x)$ is a polynomial of degree 5 such that $P(x)-1$ is divisible by $(x-1)^3$ and $P(x)$ is divisible by $x^3$. Find $P(x)$.
No idea where to start, would $P(x)$ be of the form $x^3(Ax^2+Bx+C)$?
| Less bashy than Parcly Taxel's answer:
Since $P$ is divisible by $x^3$ and $P-1$ is divisible by $(x-1)^3$, we know that $P'$ is divisible by $x^2(x-1)^2$. Since $P'$ is degree $4$, it must be a constant multiple of this. Say $P'(x)=Ax^2(x-1)^2=Ax^4-2Ax^3+Ax^2$.
Then $P(x)$ is an antiderivative of this, namely $\frac{A}{5}x^5-\frac{A}{2}x^4+\frac{A}{3}x^3+B$. At $0$ this is $0$ and at $1$ this is $1$, so we get $B=0$ and $\frac{A}{30}=1$. Thus $P(x)=\frac{30}{5}x^5-\frac{30}{2}x^4+\frac{30}{3}x^3=6x^5-15x^4+10x^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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Sequence of fractions that converges to $\sqrt{n}$ Begin with any two positive integers $a,b$. Let
$$s_1 = \frac{a}{b}$$
and recursively we define $s_{i+1}$ from $s_i$ as follows: If ${s_i}$ is the fraction $\frac{a'}{b'}$, then set
$$s_{i+1} = \frac{a'+2b'}{a'+b'}$$
(Notice that the actual value of the $s_{i+1}$ does not depend on the exact way we choose to represent $s_i$ in fractions)
How do we show that $s_i \to \sqrt{2}$ as $i\to \infty$?
In general, sending $\frac{a'}{b'}$ to $\frac{a'+nb'}{a'+b'}$ supposedly leads to $s_i \to \sqrt{n}$. How might we prove this?
| In the general case dividing by $b'$ in the nominator and the denominator we have that:
$$s_{i+1} = \frac{s_i + n}{s_i + 1}$$
Now consider two cases, namely $s_i \le \sqrt{n}$ and $s_i > \sqrt{n}$. Now let WLOG $s_i \le \sqrt{n}$. Then expressing $s_{i+2}$ by $s_i$ we have that:
$$s_{i+2} = \frac{s_{i+1} + n}{s_{i+1} + 1} = \frac{\frac{s_i + n}{s_i + 1} + n}{\frac{s_i + n}{s_i + 1} + 1} = \frac{(n+1)s_i + 2n}{2s_i + (n+1)}$$
From this we have that $s_{i+2} \ge s_i \iff (n+1)s_i + 2n \ge 2s_i^2 + (n+1)s_i \iff n \ge s_i^2$, which is true. Also we have that $s_{i+2} \le \sqrt{n} \iff ((n+1) - 2\sqrt{n})s_i \le \sqrt{n}(n+1) - 2n$, but this is true as: $((n+1) - 2\sqrt{n})s_i \le ((n+1) - 2\sqrt{n})\sqrt{n} = \sqrt{n}(n+1) - 2n$. Therefore if $i$ is odd/even we have that the sequence of odd/even indexes is convergent. Let WLOG $i$ be odd, then we can write $\lim_{n \to \infty} s_{2i} = L$. Taking the limit from the previous equation we have:
$$L = \frac{(n+1)L + 2n}{2L + (n+1)} \iff 2L^2 + (n+1)L = (n+1)L + 2n \iff L = \sqrt{n}$$
Similarly we can prove the case when $s_i > \sqrt{n}$ and also that the subsequence with indexes of other parity is convergent to $\sqrt{n}$.
Finaly as both $s_{2i}$ and $s_{2i+1}$ converge to $\sqrt{n}$ we can conclude that $s_i$ converges to $\sqrt{n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solve for $x$ where $0\leq x\leq 360$ Solve
$$4\sin x \cdot \sin 2x \cdot \sin 4x =\sin 3x$$
My Attempt :
Here,
$$4\sin x \cdot \sin 2x \cdot \sin 4x=\sin 3x$$
$$4\sin x \cdot (2\sin x \cdot \cos x ). (4 \sin x \cdot \cos x \cdot \cos 2x)=\sin3x$$
$$32\sin^3 x \cdot \cos^2 x \cdot \cos2x=\sin3x$$
How should I proceed further?
| Since
$$ \sin a \cdot \sin b = \frac{1}{2} \left[ \cos (a - b) - \cos (a+b) \right] $$
$$ \sin a \cdot \cos b = \frac{1}{2} \left[ \sin (a + b) + \sin (a - b) \right] $$
and
$$ \sin a - \sin b = 2 \sin\left[ \frac{1}{2} (a-b) \right] \cos \left[ \frac{1}{2} (a+b) \right] $$
we can use these identities to solve
$$ 4 \sin x \cdot \sin 2x \cdot \sin 4x = \sin 3 x $$
$$ \Rightarrow \quad 2 \sin x \cdot \left[ \cos (-2x) - \cos 6x \right] = \sin 3x $$
$$ \Rightarrow \quad 2 \sin x \cdot \cos 2x - 2 \sin x \cdot \cos 6x = \sin 3x $$
$$ \Rightarrow \quad \sin 3x + \sin (-x) - (\sin 7x + \sin(-5x)) = \sin 3x $$
$$ \Rightarrow \quad \sin 3x - \sin x - \sin 7x + \sin 5x = \sin 3x $$
$$ \Rightarrow \quad - \sin x - \sin 7x + \sin 5x = 0 $$
$$ \Rightarrow \quad \sin x = \sin 5x - \sin 7x $$
$$ \Rightarrow \quad \sin x = 2 \sin (-x) \cos 6x $$
$$ \Rightarrow \quad \sin x = - 2 \sin x \cos 6x $$
$$ \Rightarrow \quad \cos 6x = - \frac{1}{2} $$
$$ \Rightarrow \quad 6x = \frac{2}{3} (3 \pi n \pm \pi) $$
$$ \Rightarrow \quad x = \frac{1}{9} (3 \pi n \pm \pi) $$
or
$$\sin x = 0 \quad \Rightarrow \quad x = n \pi \mathrm{.}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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The idea behind the sum of powers of 2 I know that the sum of power of $2$ is $2^{n+1}-1$, and I know the mathematical induction proof. But does anyone know how $2^{n+1}-1$ comes up in the first place.
For example, sum of n numbers is $\frac{n(n+1)}{2}$. The idea is that we replicate the set and put it in a rectangle, hence we can do the trick. What is the logic behind the sum of power of $2$ formula?
| Here's a geometric intuition for why $2^0 + 2^1 + 2^2 + \dots + 2^n = 2^{n+1} - 1$:
Here's the idea. Notice that the boxes for $1 + 2 + 4$ are right next to a single box of size $8$. They perfectly fill that box once you add the gold square in, so $1 + (1 + 2 + 4) = 8$, meaning $1 + 2 + 4 = 8 - 1$. Similarly, the boxes for $1 + 2 + 4 + 8 + 16 + 32$ are right above a single box of size $64$, and the smaller boxes, plus the gold box, completely fill the box for $64$. That means that $1 + (1 + 2 + 4 + 8 + 16 + 32) = 64$, or that $1 + 2 + 4 + 8 + 16 + 32 = 64 - 1$.
More generally, this gives an intuition for why $1 + (2^0 + 2^1 + \dots + 2^n) = 2^{n+1}$: the boxes for the smaller powers of two, plus one gold box, perfectly fill the next box.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
The least positive real number $k$ for which $k\left(\sqrt{(x+y)(y+z)(z+x)}\right)\geq x\sqrt{y}+y\sqrt{z}+z\sqrt{x}$
The least positive real number $k$ for which $$k\left(\sqrt{(x+y)(y+z)(z+x)}\right)\geq x\sqrt{y}+y\sqrt{z}+z\sqrt{x}$$
Where $x,y,z>0$
$\bf{My\; Try::}$ Here $$k\geq \frac{x\sqrt{y}+y\sqrt{z}+z\sqrt{x}}{\sqrt{(x+y)(y+z)(z+x)}}$$
Using $\bf{A.M\geq G.M}$ Inequality
$$(x+y)(y+z)(z+x)\geq 8xyz$$
How can i solve it after that, Help required, Thanks
| Let $x=y=z$.
Hence, $k\geq\frac{3}{2\sqrt2}$.
We'll prove that $k=\frac{3}{2\sqrt2}$ is valid.
Indeed, we need to prove that
$9(x+y)(y+z)(z+x)\geq8(x\sqrt{y}+y\sqrt{z}+z\sqrt{x})^2$.
By C-S $(x+y+z)(xy+yz+zx)\geq(x\sqrt{y}+y\sqrt{z}+z\sqrt{x})^2$.
Thus, it remains to prove that
$$9(x+y)(y+z)(z+x)\geq8(x+y+z)(xy+yz+zx)$$ or
$$\sum_{cyc}z(x-y)^2\geq0$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that, for large N, the proportion of positive integers n≤N...
Show that, for large $N$, the proportion of positive integers $n ≤ N$
(a) not divisible by $2^7$is $1 − \frac{1}{2^7}$
(b) not divisible by any of $2^7, 3^4, 5^3, 7^2, 11^2, 13^2, 17^2, 19^2, 23^2$ is $(1-\frac{1}{3^4})(1-\frac{1}{2^7})\cdots(1-\frac{1}{23^2}) ≈ .931.$
For part a, the number of naturals $i\leq n$ that are divisible by $p$ (call this count $c$) satisfies $\frac{n}{p}-1\lt c\lt \frac{n}{p}+1$, so the density $d = \lim_{n\rightarrow\infty}\frac{c}{n}$ satisfies $\frac{1}{p}-\frac{1}{n}\lt d\lt \frac{1}{p}+\frac{1}{n}$ for all $n$; therefore we must have $d=\frac{1}{p}$
How do I apply this to part (b)?
| It is pure probability at that point. not divisible by an of a,b,c,d, or e is the same as (not divisible by a and not divisible by b and ...) which are independent for large N so you have P(not divisible by a)*P(not divisible by b)P(not divisble by c)...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Coefficient of a term using binomial theorem I was just wondering how would I find the coefficient of any term let's say $x^3$ in the expansion of $(x^2+2x+2)^{10}$ using binomial expansion or any other technique. Please let me know if this can be found directly using a shortcut if any.
| We can also apply the binomial theorem twice in order to determine the coefficient of $x^3$ in $(x^2+2x+10)^{10}$.
It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. This way we can write e.g.
\begin{align*}
[x^k](1+x)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
[x^3](x^2+2x+2)^{10}&=[x^3]\sum_{k=0}^{10}\binom{10}{k}x^{2k}(2x+2)^{10-k}\tag{1}\\
&=\sum_{k=0}^{1}\binom{10}{k}[x^{3-2k}]\sum_{j=0}^{10-k}\binom{10-k}{j}(2x)^j2^{10-k-j}\tag{2}\\
&=\sum_{k=0}^1\binom{10}{k}\binom{10-k}{3-2k}2^{10-k}\tag{3}\\
&\binom{10}{0}\binom{10}{3}2^{10}+\binom{10}{1}\binom{9}{1}2^9\tag{4}\\
&=1\cdot120\cdot1024+10\cdot9\cdot 512\\
&=168960
\end{align*}
Comment:
*
*In (1) we apply the binomial formula
*In (2) we observe that only $k=0$ and $k=1$ may contribute via $x^{2k}$ something for the coefficient of $x^3$. So, we change the upper limit of the sum to $k=1$. We apply the coefficient of operator rule
\begin{align*}
[x^p]x^qA(x)=[x^{p-q}]A(x)
\end{align*}
and we also expand the inner binomial according to the binomial formula.
*In (3) we select the coefficient of $x^{3-2k}$ by selecting the summand with $j=3-2k$.
*In (4) we expand the sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that $5\cdot10^n+10^{n-1}+3$ is divisible by 9 Prove by induction the following:
$5*10^n+10^{n-1}+3$ is divisible by 9
Base case: $n=1$ $5*10+10^{1-1}+3=5*10+10^0+3=50+1+3=54$
$9|54=6$
Inductive Hypothesis: If $k$ is a natural number such that $9|5*10^k+10^{k-1} +3$
Inductive step:
Show that $S_k$ is true $\Rightarrow$ $S_{k+1}$ is true
$S_{k+1}$: $9|5*10^{k+1}+10^{k} +3$
$9|10(5*10^{k+1}+10^{k} +3)$
$9|5*10^{k+2}+10^{k+1} +10*3$
$9|5*(10^{k+1}*10^1)+(10^{k}*10^1) +(9+1)*3$
$9|5*(10^{k+1}*(9+1))+(10^{k}*10*(9+1)) +(9+1)*3$
$9|5(9*10^{k+1}+10^{k+1})+9*10^k+10^k+((9*3)+(1*3))$
This is were I am stuck for the last day try to figure out what move next would speed up the inductive proof as I have a feeling it can be finished up. Anyone help me see what I am unable to find.
| Note that if for some $n \geq 1$ the proposition is true then $5\cdot 10^{n} + 10^{n-1} + 3 = 9k$ for some $k$; hence
$$
5\cdot 10^{n+1} + 10^{n} + 3 = 90k - 27.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Inequality based on AM/GM Inequality Find the greatest value of $x^3y^4$
If $2x+3y=7 $ and $x≥0, y≥0$.
(Probably based on weighted arithmetic and geometric mean)
| $y$ is a function of $x$. That is, $y=(7-2x)/3.$ And $\frac {dy}{dx}=y'=-2/3.$
Let $f(x)=x^3y^4.$ Then $f'(x)=3x^2 y^4+4x^3y^3y'=3x^2y^4-(8/3)x^3y^3.$
For $xy\ne 0$ we have $f(x)>0$ and $f'(x)=f(x)(\frac {3}{x}-\frac {8}{3y}).$
So for $xy\ne 0$ we have $f'(x)>0\iff \frac {3}{x}-\frac {8}{3y}>0\iff 9y>8x\iff21-6x>8x \iff 3/2>x.$ Similarly for $xy\ne 0$ we have $f'(x)<0\iff 3/2<x.$
And we have $xy=0\implies f'(x)=0.$
So $f(x)$ is increasing for $x<3/2$ and is decreasing for $3/2<x<7/2.$ ($7/2$ is the upper limit of the domain of $f,$ as $y=(7-2x)/3\geq 0.$) Therefore, since $f(x)$ is continuous, $\max f=f(3/2).$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
} |
binomial congruence $\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}\equiv 0~or (-2)\pmod p$
Let $p\ge 5$ be a prime number. Show that $$\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}\equiv 0 \text{ or } -2\pmod p .$$
Examples:
If $p=5$, then
$$f=\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}=\binom{2}{1}+\binom{4}{2}=8\equiv -2\pmod 5 .$$
If $p=7$, then
$$f=\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}=\binom{2}{1}+\binom{4}{2}+\binom{6}{3}=28\equiv 0\pmod 7 .$$
If $p=11$, then
$$f=\sum_{i=1}^{\frac{p-1}{2}}\binom{2i}{i}=\binom{2}{1}+\binom{4}{2}+\binom{6}{3}+\binom{8}{4}+\binom{10}{5}=350\equiv -2 \pmod{11} .$$
| My solution:\begin{align*}&\sum_{i=0}^{\frac{p-1}{2}}\binom{2i}{i}=\sum_{i=0}^{\frac{p-1}{2}}\binom{2i}{i}\left(\dfrac{1}{2}+\frac{1}{2}\right)^{i}
=\sum_{i=0}^{\frac{p-1}{2}}\binom{2i}{i}\sum_{j=0}^{i}\binom{i}{j}\dfrac{1}{2^i}\\
&=\sum_{i=0}^{\frac{p-1}{2}}\sum_{j=0}^{i}\dfrac{(2i)!}{2^i\cdot i!\cdot j!(i-j)!}\\
&=\sum_{i=0}^{\frac{p-1}{2}}\sum_{j=0}^{i}\dfrac{(2i-1)!!}{j!(i-j)!}\\
&\equiv\sum_{i=0}^{\frac{p-1}{2}}\sum_{j=0}^{i}\dfrac{(-1)^i(p-1)(p-3)(p-5)\cdots(p-2i+1)}{j!(i-j)!}\\
&=\sum_{i=0}^{\frac{p-1}{2}}\sum_{j=0}^{i}\dfrac{(-1)^i2^i\left(\frac{p-1}{2}\right)!}{\left(\frac{p-1-2i}{2}\right)!j!(i-j)!}\\
&=\sum_{i=0}^{\frac{p-1}{2}}\sum_{j=0}^{i}(-2)^i\binom{\frac{p-1}{2}}{i}\binom{i}{j}\\
&=\sum_{i=0}^{\frac{p-1}{2}}(-4)^i\binom{\frac{p-1}{2}}{i}\\
&=(1-4)^{\frac{p-1}{2}}\\
&=(-3)^{\frac{p-1}{2}}\\
&\equiv \pm 1\pmod p
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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} |
$\lim _{x \to a} \frac{x^n-a^n}{x-a}$ when $n$ is irrational. Question on the theorem :$$\lim _{x \to a} \frac{x^n-a^n}{x-a}=na^{n-1}$$
Is it true when $n$ is irrational ? is there a proof ?
Ex:
$$\lim _{x \to a} \frac{x^{\sqrt{2}}-a^{\sqrt{2}}}{x-a}=\sqrt{2}a^{\sqrt{2}-1}$$
| Just another way to do it.
Define $$x=a(1+y)\implies A=\frac{x^n-a^n}{x-a}=\frac{a^n(1+y)^n-a^n}{ay}=a^{n-1}\frac{(1+y)^n-1}{y}$$ Apply the generalized binomial theorem or Taylor series around $y=0$ $$(1+y)^n=1+n y+\frac{n(n-1) }{2} y^2+O\left(y^3\right)$$ which makes $$\frac{(1+y)^n-1}{y}=n+\frac{n(n-1)}{2} y+O\left(y^2\right)$$ $$A=a^{n-1}\left( n+\frac{n(n-1)}{2} y+O\left(y^2\right)\right)$$ and $y\to 0$; then the limit $n a^{n-1}$.
If you want to go further, replace $y=\frac{x-a}a$ to get $$A=n a^{n-1}+\frac{n(n-1)}{2}a^{n-2}(x-a)+O\left((x-a)^2\right)$$ which shows the limit and how it is approached.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding range of this expression If a line makes angles $\alpha,\beta,\gamma$ with positive axes,then the range of $\sin\alpha\sin\beta + \sin\beta\sin\gamma +\sin\gamma\sin\alpha$ is? I am a noob in finding range , so please help me from beginning.
| Take a unit vector $\mathbf u$ parallel to the line: its components will be $(cos\alpha , cos\beta , cos\gamma)$,
with $0 \leqslant \alpha , \; \beta ,\; \gamma \leqslant \pi$, and $cos^2\alpha + cos^2\beta + cos^2\gamma=1$.
We shall note that the $3$ angles are not independent (there are $2$ degrees of freedom on the unit sphere),
and also that we cannot take two of them as independent variables, since their ranges are interrelated.
For instance there cannot be two angles null.
So the best practicable way is to resort to spherical coordinates, e,g.
$$
\begin{gathered}
- \pi < \phi = \text{longitude} \leqslant \pi \hfill \\
- \pi /2 \leqslant \theta = \text{latitude} \leqslant \pi /2 \hfill \\
\end{gathered}
$$
with which we will have
$$
\left\{ \begin{gathered}
\cos \alpha = \cos \theta \cos \phi \hfill \\
\cos \beta = \cos \theta \sin \phi \hfill \\
\cos \gamma = \sin \theta \hfill \\
\end{gathered} \right.\quad \left\{ \begin{gathered}
\sin \alpha = \sqrt {1 - \cos ^{\,2} \theta \cos ^{\,2} \phi } \hfill \\
\sin \beta = \sqrt {1 - \cos ^{\,2} \theta \sin ^{\,2} \phi } \hfill \\
\sin \gamma = \sqrt {1 - \sin ^{\,2} \theta } \hfill \\
\end{gathered} \right.
$$
Note that
$$
\begin{gathered}
0 \leqslant \sin \alpha = \cos \left( {\pi /2 - \alpha } \right)\quad \left| {\;0 \leqslant \alpha \leqslant \pi } \right. \hfill \\
\cos ^{\,2} \alpha + \cos ^{\,2} \beta + \cos ^{\,2} \gamma = 1\quad \Rightarrow \quad \sin ^{\,2} \alpha + \sin ^{\,2} \beta + \sin ^{\,2} \gamma = 2 \hfill \\
\end{gathered}
$$
so the sines are non-negative, and we can take the plus sign for the square root.
Therefore we can work to find the range of the function
$$
\begin{gathered}
f(\theta ,\phi ) = \sin \alpha \sin \beta + \;\sin \beta \sin \gamma + \;\sin \gamma \sin \alpha = \hfill \\
= \sqrt {\left( {1 - \cos ^{\,2} \theta \cos ^{\,2} \phi } \right)\left( {1 - \cos ^{\,2} \theta \sin ^{\,2} \phi } \right)} + \hfill \\
+ \sqrt {\left( {1 - \cos ^{\,2} \theta \sin ^{\,2} \phi } \right)\left( {1 - \sin ^{\,2} \theta } \right)} + \hfill \\
+ \sqrt {\left( {1 - \cos ^{\,2} \theta \cos ^{\,2} \phi } \right)\left( {1 - \sin ^{\,2} \theta } \right)} \hfill \\
\end{gathered}
$$
or, better, just work with
$$
g(\theta ,\phi ) = \sin \alpha + \sin \beta + \;\sin \gamma
$$
since
$$
g(\theta ,\phi )^2 = 2 + 2f(\theta ,\phi )
$$
and find that
$$
1 \leqslant f(\theta ,\phi ) \leqslant 2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The number of solutions of $x^2+2016y^2=2017^n$
The number of solutions of $x^2+2016y^2=2017^n$ is $k$. Write $k$ with $n$.
For $n = 1$, the only solution is $(1,1)$. For $n = 2$, it gets more complicated. Taking the equation modulo $2016$ we find that $x^2 \equiv 1 \pmod{2016}$. How do we continue?
Solving for $y$ we get $$y = \pm \sqrt{\dfrac{2017^n-x^2}{2016}}.$$
| If $x^2+2016y^2$ is divisible by $2017$, then taking $\mod 2017$ shows that $x\equiv\pm y\mod2017$. Then
$\left(\dfrac{x\pm2016y}{2017}\right)^2+2016\left(\dfrac{y\mp x}{2017}\right)^2=\dfrac{x^2+2016y^2}{2017}$,
shown just by expanding out. We can choose the sign of each of those so that the numbers are actual integers, not just rationals. Now in the case of $x^2+2016y^2=2017^n$, we can just use this to whittle our way down to $1$.
Now there's another way of looking at this construction. What I did was I took $x+y\sqrt{-2016}$ and noted that multiplying it by its conjugate gives you $x^2+2016y^2$. And we also determined that one of $\frac{x+y\sqrt{-2016}}{1+\sqrt{-2016}}$ and $\frac{x+y\sqrt{-2016}}{1-\sqrt{-2016}}$ can be written in the form $c+d\sqrt{-2016}$ where $c$ and $d$ are integers. And we also determined that continuing to divide what we get by the appropriate choice between $1\pm\sqrt{-2016}$, we eventually get down to either $1$ or $-1$.
So $x+y\sqrt{-2016}=\pm(1+\sqrt{-2016})^u(1-\sqrt{-2016})^v$ with $u+v=n$. This gives all such $x$ and $y$, and each $x$ and $y$ are determined by exactly one sign and pair $(u,v)$. This last statement is not difficult to prove; to do so, write out two different representations of a number in that form, divide out by the common factors and sign, and obtain $(1+\sqrt{-2016})^r=\pm(1-\sqrt{-2016})^r$ for some $r$. Then multiply both sides by $(1+\sqrt{-2016})^r$; on one side you get $(1+\sqrt{-2016})^{2r}$ which is not divisible by $2017$, but on the other side you get $\pm2017^r$, which is. So any representation must be unique.
So we can get any pair $x$ and $y$ that work from the equation
$x+y\sqrt{-2016}=\pm(1+\sqrt{-2016})^u(1-\sqrt{-2016})^v$
but the question (I assume) was asking about nonnegative (or positive) $x$ and $y$. Swapping $u$ and $v$ changes the sign of $y$, and swapping the sign in front changes both $x$ and $y$. And if $n$ is even then $u=v$ gives $x=2017^{\frac{n}{2}}$ and $y=0$. Otherwise, we can take $u>v$ and the choice of sign to be positive, and then just take the absolute value of the coefficients. So that should give us $\frac{n+1}{2}$ if $n$ is odd, and $\frac{n+2}{2}$ if you allow that exceptional $b=0$ case, or $\frac{n}{2}$ if not.
That is, there are $\left\lfloor\frac{n+2}{2}\right\rfloor$ pairs $(x, y)$ if you allow $y=0$, or $\left\lfloor\frac{n+1}{2}\right\rfloor$ if you don't.
| {
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"url": "https://math.stackexchange.com/questions/2010085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $\int\limits_{0}^{2\pi}\frac{a\cos x -1}{(a^2+1-2a\cos x)^{3/2}}dx.$ $$\int\limits_{0}^{2\pi}\frac{a \cos x -1}{(a^2+1-2a \cos x)^{3/2}}dx
= 2\int\limits_{0}^{\pi}\frac{a \cos x -1}{(a^2+1-2a\cos x)^{3/2}}dx.$$
If a<1, this integral doesn't converge. How to evaluate it for any other a? I think it can be expressed as elliptic integral $I(a^2)$ or calculated using series, but stuck using both ways.
| Note that $$I\left(a\right)=2\int_{0}^{\pi}\frac{a\cos\left(x\right)-1}{\left(a^{2}+1-2a\cos\left(x\right)\right)^{3/2}}dx
$$ $$=\frac{d}{da}\left(-2a\int_{0}^{\pi}\frac{1}{\sqrt{a^{2}+1-2a\cos\left(x\right)}}dx\right)
$$ and, since $a>1$, $$\int_{0}^{\pi}\frac{1}{\sqrt{\left(a+1\right)^{2}-2a-2a\cos\left(x\right)}}dx=\int_{0}^{\pi}\frac{1}{\sqrt{\left(a+1\right)^{2}-2a\left(1+\cos\left(x\right)\right)}}dx
$$ $$\stackrel{x=2u}{=}\int_{0}^{\pi/2}\frac{1}{\sqrt{\left(a+1\right)^{2}-2a\left(1+\cos\left(2u\right)\right)}}du=\frac{2}{a+1}\int_{0}^{\pi/2}\frac{1}{\sqrt{1-\frac{4a}{\left(a+1\right)^{2}}\cos^{2}\left(u\right)}}du$$ $$\stackrel{u\rightarrow\pi/2-u}{=}\frac{2}{a+1}\int_{0}^{\pi/2}\frac{1}{\sqrt{1-\frac{4a}{\left(a+1\right)^{2}}\sin^{2}\left(u\right)}}du
=\frac{2}{a+1}K\left(\frac{4a}{\left(a+1\right)^{2}}\right)$$ where $K(z)$ is the complete elliptic integral of the first kind. Hence we have $$I\left(a\right)=\frac{d}{da}\left(\frac{-4a}{a+1}K\left(\frac{4a}{\left(a+1\right)^{2}}\right)\right)=\color{red}{\frac{2\left((a+1)E\left(\frac{4a}{\left(a+1\right)^{2}}\right)-\left(a-1\right)K\left(\frac{4a}{\left(a+1\right)^{2}}\right)\right)}{a^{2}-1}}$$ for $a>1,$ where $E(z)$ is complete elliptic integral of the second kind.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2011319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluation of $\int x^{26}(x-1)^{17}(5x-3)dx$
Evaluation of $$\int x^{26}(x-1)^{17}(5x-3) \, dx$$
I did not understand what substution i have used so that it can simplify,
I have seems it is a derivative of some function.
Help me, Thanks
| Sorry for the informal writing am rushing out.
Hope it helps!
$$
(x - 1)^{17} = \sum \left(
\begin{array}{c}
17 \\
r
\end{array}
\right) x^{17-r}(-1)^r
\\
x^{26}(x - 1)^{17} = \sum \left(
\begin{array}{c}
17 \\
r
\end{array}
\right) x^{43-r}(-1)^r
\\
x^{26}(x - 1)^{17}(5x−3) = 5\sum \left(
\begin{array}{c}
17 \\
r
\end{array}
\right) x^{45-r}(-1)^r\space- 3\sum\left(
\begin{array}{c}
17 \\
r
\end{array}
\right) x^{44-r}(-1)^r
\\
=5x^{45} - \left(\sum\left[
5\left( \begin{array}{c}
17 \\
r+1
\end{array}\right) - 3\left( \begin{array}{c}
17 \\
r
\end{array}\right)
\right]x^{44-r}(-1)^r\right)
+3x^{27}
\\
\int x^{26}(x - 1)^{17}(5x−3) = 5\int x^{45} - \left(\sum\left[
5\left( \begin{array}{c}
17 \\
r+1
\end{array}\right) - 3\left( \begin{array}{c}
17 \\
r
\end{array}\right)
\right]\int x^{44-r}(-1)^r\right)
\\+3\int x^{27} + c
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Multivariable limit - perhaps a trickier problem I am stuck on. I am trying to solve the following limit:
$\lim_{(x,y) \to (0,0)} \frac{x^4y^4}{(x^2 + y^4)^3}$
(This is a more challenging problem from Folland Calculus, it seems).
I am pretty sure this limit does not exist (however, this is just a guess, and I am not 100% sure.)
I am trying to approach via the paths y = mx and x = my but... they don't seem to work. I have tried simpler cases such as x = 0 and y = 0.
Any help appreciate.
| Use polar coordinates :
$$\frac{x^4y^4}{(x^2+y^4)^3} = \frac{r^8\cos^4\theta\sin^4\theta}{r^6(\cos^2\theta+r^2\sin^4\theta)^3} = r^2\frac{\cos^4\theta\sin^4\theta}{(\cos^2\theta+r^2\sin^4\theta)^3}$$
Now the problem remaining is : is the fraction bounded or not ? You have, for $\theta\ne\frac\pi2\mod\pi$ :
$$\frac{\cos^4\theta\sin^4\theta}{(\cos^2\theta+r^2\sin^4\theta)^3} = \frac{\cos^4\theta\sin^4\theta}{\cos^6\theta\left(1+r^2\frac{\sin^4\theta}{\cos^2\theta}\right)^4}$$
So if you go to $0$ following the curve of equation $r=\cos\theta$, for example (this is a circle tangent to the $y$ axis at the origin), you have :
$$\frac{x^4y^4}{(x^2+y^4)^3} = \cos^2\theta\frac{\cos^4\theta\sin^4\theta}{\cos^6\theta(1+\sin^4\theta)}\xrightarrow[\theta\to\frac\pi2]{}\frac12\ne0$$
So your function is not continuous at the origin, as you suspected.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to prove that $f(x)$ is equal to $ \frac{1}{2}(\exp(x) -\exp(-x)) $? $$f(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$$
Since we know that $ \frac{1}{2}(\exp(x) - \exp(-x)) $ is same as
$$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{x^{n}}{n!} - \sum_{n=0}^{\infty}\frac{(-x)^{n}}{n!}\right)$$
we can place our $f(x)$ in this form. But first we can do more simplifying, we can write that form as $\displaystyle \frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n}-(-x)^n}{n!}$. In my book I found next steps, we are placing our $f(x)$ in this last form and we get
$$\frac{1}{2}\left(\sum_{n=0}^{\infty}\frac{x^{2n+1}-(-x)^{2n+1}}{(2n+1)!} + \frac{x^{2n}-(-x)^{2n}}{(2n)!}\right)$$
The part which I don't understand is, from where do we get that $\displaystyle\frac{x^{2n}-(-x)^{2n}}{(2n)!}$ ?
| They split the sum in an even and an odd part. In general
$$
\sum_{n = 0}^{+\infty} a_n = \sum_{n=0,\;n\; {\rm even}}^{+\infty}a_n + \sum_{n=0,\;n\; {\rm odd}}^{+\infty}a_n = \sum_{n=0}^{+\infty}a_{2n} + \sum_{n=0}^{+\infty}a_{2n+1}
$$
This is particularly useful here because $1-(-1)^{2n} = 1-1 = 0$ and $1 -(-1)^{2n+1} = 1 + 1 = 2$.
So your expression becomes
\begin{eqnarray}
\frac{1}{2}\left(\sum_{n=0}^{+\infty}\frac{x^n}{n!} + \sum_{n=0}^{+\infty}\frac{(-1)^nx^n}{n!}\right) &=& \frac{1}{2}\sum_{n=0}^{+\infty}\frac{(1 -(-1)^n)x^n}{n!} \\
&=& \frac{1}{2}\sum_{n=0}^{+\infty}\frac{(1 -(-1)^{2n})x^{2n}}{(2n)!} + \frac{1}{2}\sum_{n=0}^{+\infty}\frac{(1 -(-1)^{2n+1})x^{2n+1}}{(2n+1)!} \\
&=& 0 + \frac{2}{2}\sum_{n=0}^{+\infty}\frac{x^{2n+1}}{(2n+1)!}\\
&=& \sum_{n=0}^{+\infty}\frac{x^{2n+1}}{(2n+1)!}
\end{eqnarray}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find minimal polynomial for finite field How can we find minimal polynomials for $\alpha $ in $GF(2^{n})$.
What is the general approach to find minimal polynomials. I know about minimal polynomials they are monic etc. In particular i want to know about primitive polynomials of $GF(32)$.
| There are $\frac{\phi(2^n-1)}{n}$ primitive polynomials of degree $n$ over $GF(2)$. For $n=5$ we have $30/5=6$ primitive polynomials, namely
$$
x^5+x^2+1,\; x^5+x^3+1,\; x^5+x^3+x^2+x+1,\;x^5+x^4+x^2+x+1,\;
$$
$$
x^5+x^4+x^3+x+1,\;x^5+x^4+x^3+x^2+1.\;
$$
Indeed, the product of all of them, together with the factor $(x+1)$ gives $x^{31}-1$. Using a factorisation algorithm, e.g., the Berlekamp algorithm we obtain the factorization of $x^{31}-1$ over $GF(2)$.
Remark: All irreducible polynomials in $GF(2)[x]$ of degree $2, 3, 5$ are
primitive.
References: See here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Derivative with quotient rule Trying to get $\dfrac{d}{dx}\left[\frac{20e^x}{(e^x+4)^2}\right]$.
After quotient rule: $$f'(x)=20\dfrac{e^x(e^{2x}+8e^x+16)-e^x(2e^x+8e^x)}{(e^x+4)^4}\\\\\\= 20\dfrac{e^{3x}+8e^{2x}+16e^x-2e^{2x}-8e^x}{(e^x+4)^4}\\\\=20\dfrac{e^{3x}+6e^{2x}+8e^x}{(e^x+4)^4}\\\\=\dfrac{20e^x(e^x+4)(e^x+2)}{(e^x+4)^4}\\\\=\dfrac{20e^x(e^x+2)}{(e^x+4)^3}$$
But Wolframalpha says it should be: $-\dfrac{20e^x(e^x-4)}{(e^x+4)^3}$
Where is my mistake?
| Using:
$$
\frac{d}{dx}(e^x+4)^2=2e^x(e^x+4)
$$
we have
$$
f'(x)=\frac{20e^x(e^x+4)^2-20\cdot 2e^x(e^x+4)}{(e^x+4)^4}=\frac{20 e^x(e^x+4)(e^x+4-2e^x)}{(e^x+4)^4}=
$$
$$
=\frac{20e^x(-e^x+4)}{(e^x+4)^3}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $\frac12 3^{2-p}(x+y+z)^{p-1}\le\frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}$
Showing $\frac12 3^{2-p}(x+y+z)^{p-1}\le\frac{x^p}{y+z}+\frac{y^p}{x+z}+\frac{z^p}{y+x}$ with $p>1$, and $x,y,z$ positive
By Jensen I got taking $p_1=p_2=p_3=\frac13$, (say $x=x_1,y=x_2,z=x_3$)
$\left(\sum p_kx_k\right)^p\le\sum p_kx^p$ which means;
$3^{1-p}(x+y+z)^p\le x^p+y^p+z^p$
and if I assume wlog $x\ge y\ge z$ then $\frac{1}{y+z}\ge\frac{1}{x+z}\ge\frac{1}{x+y}$
so these $2$ sequences are similarly ordered, and by rearrangement inequality the rest follows ?
I should use power means, but any other solution is also appreciated.
| Indeed, Jensen works.
Let $x+y+z=3$.
Hence, we need to prove that $\sum\limits_{cyc}f(x)\geq0$, where $f(x)=\frac{x^p}{3-x}-\frac{1}{2}$.
$f''(x)=\frac{x^{p-2}\left((p-2)(p-1)x^2-6p(p-2)x+9p(p-1)\right)}{(3-x)^3}$.
If $p=2$ so $f''(x)>0$.
If $p>2$ so since $f''(0)>0$ and $\lim\limits_{x\rightarrow3^-}f''(x)>0$ and $\frac{3p}{p-1}>3$, we see that $f''(x)>0$.
If $1<p<2$ so since $f''(0)>0$ and $\lim\limits_{x\rightarrow3^-}f''(x)>0$, we see that $f''(x)>0$.
Thus, your inequality follows from Jensen.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is the sequence $a_{n}=\prod\limits_{i=1}^{n}\left(1+\frac{i}{n^2}\right)$ decreasing?
let $$a_{n}=\left(1+\dfrac{1}{n^2}\right)\left(1+\dfrac{2}{n^2}\right)\cdots \left(1+\dfrac{n}{n^2}\right)$$
since $$a_{1}=2,a_{2}=\dfrac{15}{8},a_{3}=\dfrac{1320}{729}\cdots $$
I found
$$a_{1}>a_{2}>a_{3}>\cdots$$
I conjecture
$\{a_{n}\}$ be monotone decreasing?
remark: It is well known this following limts
$$\lim_{n\to+\infty}a_{n}=e^{\lim_{n\to\infty}\sum_{i=1}^{n}\ln{\left(1+\frac{i}{n}\right)}}=e^{1/2}$$But the result seems to speculation regarding the inequality of useless,so how to prove inequality
$$a_{n}>a_{n+1},\forall n\in N^{+}$$
| Taking logarithms is often helpful when dealing with products, so let's try that here too. Grouping terms in the difference of the logarithms, we obtain
$$\log a_n - \log a_{n+1}
= \sum_{k = 1}^n \Biggl(\log\biggl(1 + \frac{k}{n^2}\biggr) - \log \biggl(1 + \frac{k}{(n+1)^2}\biggr)\Biggr) - \log \biggl(1 + \frac{1}{n+1}\biggr).$$
Now we want to bound the differences $\log \bigl(1 + \frac{k}{n^2}\bigr) - \log \bigl(1 + \frac{k}{(n+1)^2}\bigr)$ below, and $\log \bigl(1 + \frac{1}{n+1}\bigr)$ above in such a way that we obtain the desired inequality $\log a_n - \log a_{n+1} > 0$. We note that for $0 \leqslant x < y$ we have
$$\log (1 + y) - \log (1 + x) > \frac{y-x}{1+y}$$
by the mean value theorem. Using this for $x_k = \frac{k}{(n+1)^2},\, y_k = \frac{k}{n^2}$, we get
\begin{align}
\sum_{k = 1}^n \bigl(\log (1 + y_k) - \log (1 + x_k)\bigr)
&> \sum_{k = 1}^n \frac{y_k - x_k}{1+y_k} \\
&= \biggl(\frac{1}{n^2} - \frac{1}{(n+1)^2}\biggr) \sum_{k = 1}^n \frac{k}{1+ y_k} \\
&= \frac{2n+1}{(n+1)^2}\sum_{k = 1}^n \frac{y_k}{1+y_k} \\
&> \frac{2n+1}{(n+1)^2}\sum_{k = 1}^n \bigl(y_k - y_k^2\bigr) \\
&= \frac{2n+1}{(n+1)^2}\biggl(\frac{n+1}{2n} - \frac{(n+1)(2n+1)}{6n^3}\biggr) \\
&= \frac{2n+1}{(n+1)^2}\biggl(\frac{1}{2} + \frac{1}{6n} - \frac{1}{2n^2} - \frac{1}{6n^3}\biggr) \\
&= \frac{1}{(n+1)^2}\biggl(n + \frac{1}{2} + \frac{1}{3} -\frac{5}{6n} - \frac{1}{3n^2} - \frac{1}{6n^3}\biggr) \\
&= \frac{1}{(n+1)^2}\biggl((n+1) - \frac{1}{2} + \frac{1}{3} - \frac{5}{6n} - \frac{1}{3n^2} - \frac{1}{6n^3}\biggr),
\end{align}
and an easy verification shows that the final expression is greater than
$$\frac{1}{n+1} - \frac{1}{2(n+1)^2} + \frac{1}{3(n+1)^3} > \log \biggl(1 + \frac{1}{n+1}\biggr)$$
for $n \geqslant 5$.
The remaining cases are verified by hand:
$$a_1 = 2 > a_2 = \frac{15}{8} > a_3 = \frac{440}{243} > a_4 = \frac{14535}{8192} > a_5 = \frac{3420144}{1953125}.$$
So indeed the sequence $(a_n)$ is strictly decreasing.
| {
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"url": "https://math.stackexchange.com/questions/2022413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How can we show $\cos^6x+\sin^6x=1-3\sin^2x \cos^2x$? How can we simplify $\cos^6x+\sin^6x$ to $1−3\sin^2x\cos^2x$?
One reasonable approach seems to be using $\left(\cos^2x+\sin^2x\right)^3=1$, since it contains the terms $\cos^6x$ and $\sin^6x$. Another possibility would be replacing all occurrences of $\sin^2x$ by $1-\cos^2x$ on both sides and then comparing the results.
Are there other solutions, simpler approaches?
I found some other questions about the same expression, but they simplify this to another form: Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here?, Alternative proof of $\cos^6{\theta}+\sin^6{\theta}=\frac{1}{8}(5+3\cos{4\theta})$ and Simplify $2(\sin^6x + \cos^6x) - 3(\sin^4 x + \cos^4 x) + 1$. Perhaps also Prove that $\sin^{6}{\frac{\theta}{2}}+\cos^{6}{\frac{\theta}{2}}=\frac{1}{4}(1+3\cos^2\theta)$ can be considered similar. The expression $\cos^6x+\sin^6x$ also appears in this integral Find $ \int \frac {\tan 2x} {\sqrt {\cos^6x +\sin^6x}} dx $ but again it is transformed to a different from then required here.
Note: The main reason for posting this is that this question was deleted, but I still think that the answers there might be useful for people learning trigonometry. Hopefully this new question would not be closed for lack of context. And the answers from the deleted question could be moved here.
| If $a+b+c=0\;,$ Then $a^3+b^3+c^3=3abc$
Now write Identity as $$\sin^2 x+\cos^2 x+(-1) = 0\;,\text{ Then }\sin^6 x+\cos^6 +(-1)^3 = -3\sin^2 x\cos^2 x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2023931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Conjecture for the value of $\int_0^1 \frac{1}{1+x^{p}}dx$ While browsing the post Is there any integral for the golden ratio $\phi$?, I came across this nice answer,
$$ \int_0^\infty \frac{1}{1+x^{10}}dx=\frac{\pi\,\phi}5$$
it seems the general form is just
$$p \int_0^\infty \frac{1}{1+x^{p}}dx=\color{blue}{\frac{\pi}{\sin\big(\tfrac{\pi}{p}\big)}}$$
I wondered about
$$\int_0^\color{red}1 \frac{1}{1+x^p}dx=\,?$$
Mathematica could find messy closed-forms for $p=5,7$. After some laborious simplification,
$$5\int_0^1 \frac{1}{1+x^5}dx=\frac{\pi\sqrt{\phi}}{5^{1/4}}+\ln2+\sqrt{5}\ln\phi$$
Question 1: In general, is it true that for any $p$ ,
$$2p\,\int_0^1 \frac{1}{1+x^p}dx=\color{blue}{\frac{\pi}{\sin\big(\tfrac{\pi}{p}\big)}}+2\ln2-\psi\big(\tfrac{1}{p}\big)+\psi\big(\tfrac{p-1}{2p}\big)+\psi\big(\tfrac{p+1}{2p}\big)-\psi\big(\tfrac{p-1}{p}\big)$$
where $\psi(z)$ is the digamma function?
Note: The four digammas, implemented in Mathematica as PolyGamma[z], can be expressed as a sum of cosines x logarithms for odd $p=2m+1$. Let $k=\frac{2n-1}{p}\pi$, then,
$$-\psi\big(\tfrac{1}{p}\big)+\psi\big(\tfrac{p-1}{2p}\big)+\psi\big(\tfrac{p+1}{2p}\big)-\psi\big(\tfrac{p-1}{p}\big)=-4\sum_{n=1}^m \cos (k)\ln\big(\sin\tfrac{k}{2}\big)$$
Question 2: For even $p$, can we can also avoid the digamma by using cosines and logarithms?
| The following close-form
holds for any integer $p\ge 2$
\begin{align}
\int_0^1 \frac1{1+x^p}dx
= \frac2p \sum_{k=1}^{[\frac p2]} ( \theta_k \sin2\theta_k + \cos2\theta_k \ln \cos\theta_k)
\end{align}
where $\theta_k= \frac{p-2k+1}{2p}\pi $. In particular
\begin{align}
& \int_0^1 \frac1{1+x^5} dx= \frac{\pi\sqrt{\phi}}{5^{5/4}}+\frac15\ln2+\frac1{\sqrt5}\ln\phi\\
& \int_0^1 \frac1{1+x^6} dx= \frac\pi6 +\frac1{2\sqrt3}\ln(2+\sqrt3) \\
& \int_0^1 \frac1{1+x^7} dx
=\frac\pi{14}\csc \frac{\pi}7-\frac27\left(\frac{\ln\cos\frac{3\pi}7}{\sec\frac{\pi}7} +\frac{\ln\cos\frac{2\pi}7}{\sec\frac{3\pi}7} -\frac{\ln\cos\frac{\pi}7}{\sec\frac{2\pi}7} \right)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule) How to solve these limits using a formula for logarithm limit(without applying L'Hopitale rule)
$$ \lim_{x \to 0} \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3} $$
$$ \lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]}$$
I suppose in the second I may not take into account arctan and sin as sinx approximately equals x
| The second you've got down pat.
$$\lim_{x \to 0} \frac{\arctan 2x}{\sin[2 \pi(x+10)]} = \lim_{x \to 0} \frac{\arctan 2x}{x}\frac{x}{\sin(2\pi x)} = \frac{1}{\pi}$$
The first is a little trickier. Use the binomial theorem to get...
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan(x)} - \sqrt{1 + \sin(x)}}{x^3} = \lim_{x \to 0} \frac{\tan x - \sin x}{2x^3} = \lim_{x \to 0} \frac{\sin x }{x} \cdot \frac{1 - \cos x}{x^2} \cdot \frac{1}{2\cos x} = \frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find intersection of 2 linear equations I have two tables with data, both describing linear equations
Table 1:
x = 60, y = 0
x = 61, y = 1
x = 62, y = 2
x = 63, y = 3
x = 64, y = 4
x = 65, y = 5
x = 66, y = 6
x = 67, y = 7
x = 68, y = 8
x = 69, y = 9
x = 70, y = 10
Table 2:
x = 64, y = 0
x = 65, y = 2
x = 66, y = 4
x = 67, y = 6
x = 68, y = 8
x = 69, y = 10
Part 1:
In $y=mx+b$ for each table, I want to know which is the $m$, the $x$, and the $b$
Part 2:
I want to find the point where these two lines intersect. The intersection is clear to see by just looking at the data, but I would like the answer to be solved for.
Thanks!
|
Part 1:
For Table 1, we use the fact that the slope is the change in $y$ divided by the change in $x$. Or,$$m=\frac {y_2-y_1}{x_2-x_1}\tag1$$
Randomly choosing two points, say $(70,10)$ and $(69,9)$ and plugging them in, we get the slope as$$m=\frac {10-9}{70-69}=1$$
Thus, we now have $y=x+b$ for $b$ is the $y$-intercept. To find $b$, we plug in any point. Doing so, we get $b=-60$. Therefore, the values in Table $1$ can be represented by $\boxed{y=x-60}$.
For Table 2, we do the same thing that we did in table $1$ to get $\boxed{y=2x-128}$. (See if you can work through it!)
Part 2:
To find the intersection point, you need to find a point $(x,y)$ that satisfies both equations. Therefore, we have the system of equation$$\begin{align*} & y=x-60\\ & y=2x-128\end{align*}\tag2$$
Substituting $y$ with $x-60$, we get the solutions as\begin{align*} & x-60=2x-128\\ & \implies 68=x\end{align*}
Plugging $x=68$ into the first equation of $(2)$ gets us $y$ as $8$. Thus, their intersection point is $\boxed{(68,8)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Algebraic proof that if $a>0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $
Prove that if $a>0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $.
Using the fact that $a \gt 0$, multiply by $a$ on both sides and get everything to one side we have; $a^{11}-a^{10}-a+1 \geq 0$. By factoring $(a^{10}-1)(a-1) \geq 0 $.
I am not sure how to proceed any further.
| We need to proof $1+a^9\le\frac{1}{a}+a^{10}$. You already proof that is sufficient and necessary that $(a-1)(a^{10}-1)\ge 0$.
This last inequality is true (when $a>0$) because:
$(a-1)(a^{10}-1)=(a-1)^2(a^9+a^8+a^7+a^6+a^5+a^4+a^3+a^2+a+1)$
And is trivial that $a^9+a^8+a^7+a^6+a^5+a^4+a^3+a^2+a+1> 0$, because $a>0$, and always $(a-1)^2\ge 0$, multiplying this two inequalities we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $\int_{-1}^{+1} (1-x^2)^n dx = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$ I'am trying to prove that
$$\int_{-1}^{+1} (1-x^2)^n\:\mathrm{d}x = \frac{2^{2n+1}(n!)^2}{(2n+1)!}$$
Here what I have done so far. We know that:
\begin{equation}
\sin^{2}x + \cos^{2}x = 1
\end{equation}
Let $x = \sin\alpha$ so $\mathrm{d}x = \cos\alpha\:\mathrm{d}\alpha$
\begin{equation}
x^{2} = \sin^{2}\alpha
\end{equation}
\begin{equation}
\cos^{2}x = 1 -\sin^{2}x
\end{equation}
We have $x = 1 \Rightarrow \alpha = \frac{\pi}{2}$ and $x = -1 \Rightarrow \alpha = - \frac{\pi}{2}$, so we replace the original integral by:
$$\int_{-1}^{1} (x-1^2)^n\:\mathrm{d}x = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n+1}\alpha\:\mathrm{d}\alpha = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^{2n}\alpha) \cos\alpha\:\mathrm{d}\alpha$$
Now integrating by parts,
$$u = \cos^{2n}\alpha \quad \Rightarrow\:\mathrm{d}v = \cos\alpha\:\mathrm{d}\alpha \\
\mathrm{d}u = -2n\cos^{2n-1}\alpha\sin\alpha\:\mathrm{d}\alpha \quad \Rightarrow v = \sin\alpha \\
\int udv = uv - \int vdu$$
Then,
\begin{align}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos^{2n}\alpha) \cos\alpha\:\mathrm{d}\alpha &= \underbrace{\cos^{2n}\alpha \sin\alpha \Big|_{-\frac{\pi}{2}}^{\frac{\pi}{2}}}_{\text{= 0}} + 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n-1}\alpha \sin^{2} \alpha\:\mathrm{d}\alpha \\[10pt]
&= 2n \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n-1}\alpha (1-\cos^2\alpha)\:\mathrm{d}\alpha \\
&= 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n-1}\alpha\:\mathrm{d}\alpha - 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n+1}\alpha\:\mathrm{d}\alpha
\end{align}
Now we see that
$$(2n+1)\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2n+1}\alpha = 2n\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} cos^{2n-1}\alpha\:\mathrm{d}\alpha$$
I am stuck here, it's almost the same integral as the original one. Any suggestions or corrections will be very welcome. Thanks.
| If I try to prove it for $n=1$ I get:
$$\int_{-1}^1(1-x^2)dx=(x-x^3/3)|_{-1}^{1}=2-2/3=4/3\ne\frac{3!1!^2}{2^3}=\frac{3}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2033851",
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Proof: $\sum^n_{i = 0} {n\choose i} F_{i+m}$ is Fibonacci number I'm trying to solve the following problem:
Show
$$\sum^n_{i = 0} {n\choose i} F_{i+m}$$
is Fibonacci number.
I know many properties of binomial symbol and Fibonacci numbers but I have no idea how to start proving given formula.
| The task is to express
$$S_{n,m} = \sum_{q=0}^n {n\choose q} F_{q+m}$$
as a Fibonacci number. Using the generating function of the Fibonacci
numbers we find that
$$F_{q+m} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{q+m+1}} \frac{z}{1-z-z^2} \; dz.$$
We thus obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m+1}}
\frac{z}{1-z-z^2}
\sum_{q=0}^n {n\choose q} \frac{1}{z^q}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{m}}
\frac{1}{1-z-z^2}
\left(1+\frac{1}{z}\right)^n
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+m}}
\frac{1}{1-z-z^2} (1+z)^n
\; dz.$$
Using the golden ratio $$\varphi = \frac{1+\sqrt{5}}{2}$$ this becomes
$$-\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+m}}
\frac{1}{(z-1/\varphi)(z+\varphi)} (1+z)^n
\; dz.$$
Residues sum to zero so from the two simple poles we get
$$S_{n,m} - \frac{1}{1/\varphi+\varphi}
\varphi^{n+m} \left(1+\frac{1}{\varphi}\right)^n
- \frac{1}{-\varphi-1/\varphi}
\frac{1}{(-\varphi)^{n+m}} \left(1-\varphi\right)^n = 0.$$
Hence
$$S_{n,m} = \frac{1}{\varphi + 1/\varphi}
\left( \varphi^m (1+\varphi)^n
- \left(-\frac{1}{\varphi}\right)^m
\left(-\frac{1}{\varphi}+1\right)^n \right).$$
By definition we have $1+\varphi = \varphi^2$ and
$1+(-1/\varphi) = (-1/\varphi)^2$ so this becomes
$$S_{n,m} = \frac{1}{\varphi + 1/\varphi}
\left( \varphi^{m+2n}
- \left(-\frac{1}{\varphi}\right)^{m+2n}\right).$$
This is Binet's formula evaluated at $m+2n$ and we get the result
$$\bbox[5px,border:2px solid #00A000]{ F_{m+2n}.}$$
Remark. We have used the fact that
$$\mathrm{Res}_{z=\infty} \frac{1}{z^{n+m}}
\frac{1}{1-z-z^2} (1+z)^n
\\ = -\mathrm{Res}_{z=0} \frac{1}{z^2} z^{n+m}
\frac{1}{1-1/z-1/z^2} \left(1+\frac{1}{z}\right)^n
\\ = -\mathrm{Res}_{z=0} z^{m}
\frac{1}{z^2-z-1} (1+z)^n = 0.$$
| {
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Prove $4a^2+b^2+1\ge2ab+2a+b$ Prove $4a^2+b^2+1\ge2ab+2a+b$
$4a^2+b^2+1-2ab-2a-b\ge0$
$(2)^2(a)^2+(b)^2+1-2ab-2a-b\ge0$
Any help from here? I am not seeing how this can be factored
| As a function of $a$, the minimum of the quadratic $4a^2+b^2+1-(2ab+2a+b)$ is $\frac34 (b - 1)^2 \ge 0$.
More precisely,
$$4a^2+b^2+1-(2ab+2a+b)=\left(2a-\frac{b+1}{2}\right)^2+\frac34 (b - 1)^2 \ge 0$$
| {
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Binomial theorem coefficient question Use the binomial theorem to find the coefficient for $x^3$ on both sides of the expansion of:
$(1+x)^3$$(1+x)^3$ $=$ $(1+x)^6$
i. Hence show $(_3C_0)^2+(_3C_1)^2+(_3C_2)^2+(_3C_3)^2$ $=$ $_6C_3$
ii. Use the same argument with $(1+x)^n(1+x)^n = (1+x)^{2n}$ to prove
$\sum_{k=0}^n(_nC_k)^2=_{2n}C_n$
Thank you.
| It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series. We also use the notation $\binom{n}{k}$ instead of $_nC_k$.
We obtain
\begin{align*}
[x^3](1+x)^3(1+x)^3&=[x^3]\left(\sum_{j=0}^3\binom{3}{j}x^j\right)\left(\sum_{k=0}^3\binom{3}{k}x^k\right)\tag{1}\\
&=\sum_{j=0}^3\binom{3}{j}[x^{3-j}]\sum_{k=0}^3\binom{3}{k}x^k\tag{2}\\
&=\sum_{j=0}^3\binom{3}{j}\binom{3}{3-j}\tag{3}\\
&=\sum_{j=0}^3\binom{3}{j}^2\tag{4}\\
\end{align*}
on the other hand we obtain
\begin{align*}
[x^3](1+x)^3(1+x)^3&=[x^3](1+x)^6\\
&=[x^3]\sum_{j=0}^6\binom{6}{j}x^j\tag{5}\\
&=\binom{6}{3}
\end{align*}
Comment:
*
*In (1) we apply the binomial theorem twice.
*In (2) we use the linearity of the coefficient of operator and apply the rule
\begin{align*}
[x^p]x^qA(x)=[x^{p-q}]A(x)
\end{align*}
to the summands of the left series.
*In (3) we select the coefficient of $x^{3-j}$ from the right series.
*In (4) we use the symmetry $\binom{n}{k}=\binom{n}{n-k}$.
*In (5) we again apply the binomial theorem and select the coefficient of $x^3$.
We conclude
\begin{align*}
\sum_{j=0}^3\binom{3}{j}^2=\binom{6}{3}\\
\end{align*}
The calculation to show
\begin{align*}
\sum_{j=0}^n\binom{n}{j}^2=\binom{2n}{n}\\
\end{align*}
can be done analogously.
| {
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Convergence of $x_{n+2} = \sqrt{x_{n+1}} + \sqrt{x_{n}}$
Let $x_0,x_1> 0$. Prove that the sequence defined by $x_{n+2} = \sqrt{x_{n+1}} + \sqrt{x_{n}}$ converges.
Here's my solution: it's easy to prove by induction that $$\forall n, 0<\min(4,u_0,u_1)\leq u_n\leq \max(4,u_0,u_1)$$
The sequence is therefore bounded with $0<\liminf u_n \leq \limsup u_n<\infty$.
By considering a subsequence that converges to $\limsup u_n$, you get two limit points $l_1$ and $l_2$ such that $\limsup u_n = \sqrt{l_1}+\sqrt{l_2}$. Since $\limsup u_n$ is the greatest limit point, this yields $\limsup u_n\leq 2 \sqrt{\limsup u_n}$ which in turn implies $\limsup u_n \leq 4$.
A similar reasoning proves $\liminf u_n \geq 4$, hence $$\liminf u_n = \limsup u_n = 4$$ and the sequence converges to $4$.
Do you know any less advanced proof that an undergraduate might think of ? Preferably something that relies on monotony or an auxillary sequence.
| If there is a limit, then $L=2\sqrt{L}$ so that $L=4$. Now
$$
x_{n+2}-4=\sqrt{x_{n+1}}-2+\sqrt{x_{n}}-2
=\frac{x_{n+1}-4}{\sqrt{x_{n+1}}+2} + \frac{x_{n}-4}{\sqrt{x_{n}}+2}
$$
If one can additional prove $x_k\ge 1$ for all $k$ beforehand, then
$$
|x_{n+2}-4|\le\frac13(|x_{n+1}-4|+|x_{n}-4|)
$$
To get a recursive bound, one can try to bound a combined expression of the type
\begin{align}
|x_{n+2}-4|+a|x_{n+1}-4|
&\le \frac{3a+1}3|x_{n+1}-4|+\frac13|x_{n}-4|
\\
&\le \frac{b}{3}\left(\frac{(3a+1)}b·|x_{n+1}-4|+\frac1{ab}a|x_{n}-4|\right)
\end{align}
To finish this bound one needs $a$ and $b$ to satisfy $3a+1\le b$ and $1\le ab$ which is possible for $\frac73\le b<3$ and for $b=\frac73$ requires $a\le\frac49$, $a\ge \frac37$. Using the larger bound results in
\begin{align}
|x_{n+2}-4|+\frac49|x_{n+1}-4|\le\frac79\left(|x_{n+1}-4|+\frac49·|x_{n}-4|\right)
\end{align}
giving linear convergence at the geometric rate $\dfrac79$. I.e.,
$$
|x_{n+1}-4|\le |x_{n+1}-4|+\frac49·|x_{n}-4|\le \left(\frac79\right)^n\left(|x_1-4|+\frac49·|x_0-4|\right)
$$
| {
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"question_score": "4",
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Question about Euler proof of divergence of the sum of the reciprocals of the primes. Consider the Euler proof :
$$\ln\sum_{n = 1}^{\infty}\frac{1}{n} = -\ln\sum_{p}\frac{1}{1-p^{-1}} = \sum_{p}\left(\frac{1}{p} + \frac{1}{2p^2} + \frac{1}{3p^3} + \dots\right) = A + \frac{1}{2}B + \frac{1}{3}C + \dots = A + K,$$ where $K < 1$.
My question is : why $K < 1$. Is it true because of $\frac{1}{2}B < \frac{1}{2}$, then $\frac{1}{3}C < \frac{1}{4}$ etc. So we could estimate our residue using sum of power of two?
| Yes, you are right. For all $d\geq 2$
$$\frac{1}{d}\sum_p\frac{1}{p^d}<\frac{1}{2^{d-1}}.$$
Infact, for $d\geq 2$,
$$\sum_p\frac{1}{p^d}\leq \frac{1}{2^d}+\sum_{n\geq 3}\frac{1}{n^d}< \frac{1}{2^d}+\int_{x\geq 2}\frac{dx}{x^d}= \frac{1}{2^d}+\frac{2}{(d-1)2^d}\leq \frac{3}{2^{d}}$$
which implies that
$$\frac{1}{d}\sum_p\frac{1}{p^d}\leq\frac{3}{d2^d}<\frac{1}{2^{d-1}}.$$
| {
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Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $ Find $\sin \theta$ if $\tan \theta +\sec \theta =1.5 $
$$\tan \theta +\sec \theta =1.5 $$
$$2\tan \theta +2\sec \theta =3 $$
$$2\sec \theta =3-2\tan \theta$$
$$4\sec^2 \theta =(3-2\tan \theta)^2$$
$$4+4\tan^2 \theta =9-12\tan \theta+4\tan^2 \theta$$
So I get $$\tan \theta = \frac{5}{12}$$
Thus $$\sin\theta=\frac{5}{13}$$
But If I do like this , $$\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta} =\frac{3}{2}$$
$$ 2\sin\theta +2=3\cos \theta$$
$$ (2\sin\theta +2)^2=9\cos^2 \theta$$
$$ 4\sin^2\theta+8\sin\theta +4=9-9\sin^2\theta$$
$$13\sin^2\theta+8\sin\theta-5=0$$
Therefore I get two answers $$\sin\theta=\frac{5}{13} , \sin\theta =-1$$
What is the reason behind this ? Why am I getting two answers in one method and one in another ?
| You have
$$
\sin\theta+1=\frac{3}{2}\cos\theta
$$
and so
$$
3\cos\theta=2(1+\sin\theta)
$$
Note that $\cos\theta\ge0$, so $-\pi/2\le\theta\le\pi/2$. We can square to get
$$
9-9\sin^2\theta=4+8\sin\theta+4\sin^2\theta
$$
so
$$
13\sin^2\theta+8\sin\theta-5=0
$$
and therefore
$$
\sin\theta=\frac{5}{13}
\qquad\text{or}\qquad
\sin\theta=-1
$$
However, $\sin\theta=-1$ implies $\theta=-\pi/2$, where $\tan\theta$ is undefined.
The only solution in $(-\pi,\pi]$, since $\cos\theta\ge0$, is
$$
\theta=\arcsin\frac{5}{13}
$$
The other one, that is, $\pi-\arcsin\frac{5}{13}$ must be discarded because it corresponds to a negative cosine.
Add integral multiples of $2\pi$, if you're interested in all solutions.
With the first method you don't get the spurious solution $\sin\theta=-1$ because the tangent is not defined there to begin with.
On the other hand, the equation $\sin\theta=5/13$ has two solutions in $(-\pi,\pi]$, one of them introduced by squaring and you should check both, unless you discard it a priori like I did.
| {
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"source": "stackexchange",
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} |
Prove the following trignometric inequality for all real $x$ Prove the following trignometric inequality for all $x \in \Bbb R$
$$x^2 \sin(x) + x \cos(x) + x^2 + {\frac 12} >0$$
take $x$ in the form of radians.
This particular question is the seventh question of the 1995 Indian RMO.
| There may be more efficient ways but this is what I have:
If it is positive (i.e. nonnegative) then certainly when we multiply it by a positive function it will stay positive... to this end, multiply it by $\sin^2(x)\cos^2(x)$ this gives
$$x^2\sin^3\cos^2(x)+x\sin^2(x)\cos^3(x)+x^2\sin^2(x)\cos^2(x)+\frac{1}{2}\sin^2(x)\cos^2(x)$$
Now this is positive provided that $x^2\cos^2(x)(\sin^2(x)-\sin^3(x))$ is positive and similarly if $\frac{1}{2}\cos^2(x)\sin^2(x) - x\sin^2(x)\cos^3(x)$ is positive. I'll show that the first term is positive. This first term is positive only if $g(x)=\sin^2(x)-\sin^3(x)$ is positive.
Upon taking the derivative we get that this function has roots precisely when $x=n\pi$ for $n\in \mathbb{Z}$. If we compute the second derivative we get
$$g''(x) = 3\sin^3(x)-2\sin^2(x)+2\cos^2(x)-6\sin(x)\cos^2(x)$$
and at $x=n\pi$ we have that $g''(x)=2>0$, therefore the function is always positive and likewise we can check the other term $\frac{1}{2}\cos^2(x)\sin^2(x) - x\sin^2(x)\cos^3(x)$ is positive.
| {
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Probability that 2 numbers will occur exactly 4 times. A die is rolled 10 times, what is the probability that 2 numbers will occur exactly 4 times? So I know that the probability of rolling any one number exactly 4 times would be
$$\frac{10!}{4!6!}\cdot \left(\frac{1}{6}\right)^4 \cdot\left(\frac{5}{6}\right)^6,$$
so would the probability that two numbers occur exactly four times be?
$$\frac{10!}{8!2!}\cdot\left(\frac{1}{6}\right)^4\cdot\left(\frac{1}{6}\right)^4\cdot\left(\frac{4}{6}\right)^2$$
| No, you have ignored two effects. The first is that you seem to be trying to calculate the probability that two specific numbers occur four times each, and the problem asks for any two numbers. The second is that once one number is given to have occured exactly four times, the odds of the other number also doing so are different that the original odds.
The easy way to do this is to ask how many ways to arrange to get two four-timers are there, and divide by $6^{10}$. So:
*
*There are $\binom{6}{2}$ ways to choose the two numbers.
*There are $\binom{10}{4}$ ways to choose which of the dice landed on the specific first number.
*There are then $\binom{6}{4}$ ways to choose which of the dice landed on the specific second number.
*There are then $4^2$ ways to choose the rolls of the remaining two dice.
The net probability is
$$
\frac{1}{6^{10}} \cdot \frac{6!}{2!4!}\cdot\frac{10!}{6!4!}\cdot \frac{6!}{4!2!}\cdot 4^2
$$
| {
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Solving basic system of equations $x, y$ are positive reals such that $y\sqrt{x^2 - y^2} = 48$ and $x + y + \sqrt{x^2 - y^2} = 24$. How to find $x$ and $y$? Squaring equations leads to complicated equations with polynomials of high degrees. Is there any way to omit it and solve it smarter?
| Call $x^2-y^2=z^2 \quad (1)$ then:
$yz=48 \quad(2)$ and $x+y+z=24 \quad(3)$
From $(1)$ we have:
$$x^2=y^2+z^2=(y+z)^2-2yz$$
and using $(2)$ and $(3)$ we get:
$$x^2=(24-x)^2-2.48 \Rightarrow x=10$$
and then $y=8$ or $y=6$
| {
"language": "en",
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How to express $ lcm \big( 1, 2, \dots, n \big)$ in terms of factorials $1!, 2!, 3!, \dots, n!$? I observed that if you arrange the factorials up to 18 in a certain way it is possible to get the LCM of all numbers of to 18. With some effort I could show:
\begin{eqnarray*} \frac{18! \times 3! }{9! \times 6! }
&=& \frac{10 \times 11 \times 12 \times 13 \times 14 \times 15 \times 16 \times 17 \times 18}{6 \times 5 \times 4} \\ \\
&=& \frac{2 \times 5 \times 11 \times 2^2 \times 3 \times 13 \times 2 \times 7 \times 3 \times 5 \times 2^4 \times 17 \times 2 \times 3^2}{3 \times 2 \times 5 \times 2^2} \\ \\
&=&
2^6 \times 3^3 \times 5 \times 7 \times 11 \times 13 \times 17
\\ \\
&=& lcm \big( 1,2,3,4,5, \dots, 17 , 18\big)
\end{eqnarray*}
How general is this? I thought maybe I could use the Mobius function. I may have left out some values
$$ \prod n!^{\; \mu(n/d)} = lcm \big( 1, 2, \dots, n \big)$$
where $\mu(n)$ is the Möbius function.
To check my work:
$$ lcm \big( 1, 2, \dots, 18 \big) = 2^4 \times 3^2 \times 5 \times 7 \times 11 \times 13 \times 17 $$
so my conjecture does not look right as stated. By the way
$$ 18! = 2^{16}\times 3^{8} \times 3^5 \times 5^3 \times 7 \times 9 \times 11 \times 13 \times 17$$
what about
$$ lcm\big(1,2,\dots,18\big)\times
lcm\big(1,2,\dots,9\big)\times lcm\big(1,2,\dots,6\big)\times
lcm\big(1,2,3,4\big)\times lcm\big(1,2,3\big)^2 \times lcm\big(1,2\big)^3 $$
in fact
\begin{eqnarray*} lcm \big( 1, 2, \dots, 9 \big) &=& 2^3 \times 3^2 \times 5 \times 7 \\
lcm \big( 1, 2, \dots, 6 \big) &=& 2^2 \times 3 \times 5 \\
lcm \big( 1, 2, 3,4 \big) &=& 2^2 \times 3 \\
lcm \big( 1, 2, 3 \big) &=& 2 \times 3 \\
lcm \big( 1, 2 \big) &=& 2 \\
\end{eqnarray*}
so I can conjecture the formla in one direction, interesting in its own right:
$$ n! = \prod_{k \leq n} lcm \big( 1,2,\dots, \frac{n}{k}\big)$$
| $lcm \big( 1,2,3, \dots, n\big) = \prod_{p \leq n} p^i$ where p is prime and $i$ such that $p^i \le n$ and $p^{i+1}>n$
Using Arthurs comment:
$lcm \big( 1,2,3, \dots, n\big) = \prod_{p \leq n}\frac{(p^i)!}{(p^i-1)!}$
For example
$lcm \big( 1,2,3, \dots, 18\big) = \frac{17! \times 13! \times 11! \times 7! \times 5! \times 9! \times 16!}{ 16! \times 12! \times 10! \times 6! \times 4! \times 8! \times 15! }$
So yes, this is always possible, but I don't know whether there is some more... meaningful way of finding such arepresentation.
Also your second statement: $n! = \prod_{k \leq n} lcm \big( 1,2,\dots, \frac{n}{k}\big)$ is disproven by $n=18$, as your general formula does not include the square/to the power of 3 you use for the two smallest $lcm$s (which seems kinda arbitrary) and I don't know why you'd include $lcm \big( 1, 2, 3,4 \big)$ since $4 \nmid 18$
| {
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How does one get that $1^3+2^3+3^3+4^3+\cdots=\frac{1}{120}$? While watching interesting mathematics videos, I found one of the papers of Srinivasa Ramanujan to G.H.Hardy in which he had written $1^3+2^3+3^3+4^3+\cdots=\frac{1}{120}$.
The problem is that every term on the left is more than $\frac{1}{120}$ yet the sum is $\frac{1}{120}$. How is that ???
I know that there are many and much more interesting things presented by Ramanujan (like $1-1+1-1+1...=\frac{1}{2}$ and $1+2+3+4.....=\frac{-1}{12}$) but for now I am interested in the summation in the title.
Any idea/hint is heartily welcome. Thanks.
Here is the video I'm talking about.
| I think I have rediscovered it (after watching the video you linked and read wiki biography of Ramanujan).
Start with $$
\frac{1}{x+1} =1 -x +x^2 -x^3 +-\cdots, \quad |x| <1.
$$ and differentiate to get $$
-\frac{1}{(x+1)^2} =-1 +2x -3x^2 +4x^3 -+\cdots, \quad |x| <1. \\
\frac{2}{(x+1)^3} =2 \cdot 1 -3 \cdot 2x +4 \cdot 3 x^2 -+\cdots, \quad |x| <1. \\
-\frac{6}{(x+1)^4} =-3 \cdot 2 \cdot 1 +4 \cdot 3 \cdot 2x -5 \cdot 4 \cdot 3 x^2 +-\cdots, \quad |x| <1.
$$
Take a magic mushroom and, ignoring $|x| <1$, let us take $x=1$ in each one. $$
\frac{1}{2} =1 -1 +1 -1 +- \cdots \\
-\frac{1}{4} =-1 +2 -3 +4 -+ \cdots \\
\frac{1}{4} =2 \cdot 1 -3 \cdot 2 +4 \cdot 3 -+ \cdots \\
-\frac{3}{8} =-3 \cdot 2 \cdot 1 +4 \cdot 3 \cdot 2 -5 \cdot 4 \cdot 3 +- \cdots
$$ Or more formally, $$
\sum_{m=1}^\infty (-1)^{m+1} m =\frac{1}{4}. \\
\sum_{m=1}^\infty (-1)^{m+1} m (m+1) =\frac{1}{4}. \\
\sum_{m=1}^\infty (-1)^{m+1} m (m+1) (m+2) =\frac{3}{8}.
$$
But notice $$ \begin{align}
\sum_{m=1}^\infty (-1)^{m+1} m^2
=&\sum_{m=1}^\infty (-1)^{m+1} m (m+1) -\sum_{m=1}^\infty (-1)^{m+1} m \\
=&\frac{1}{4} -\frac{1}{4} =0.
\end{align}
$$ and $$ \begin{align}
\sum_{m=1}^\infty (-1)^{m+1} m^3
=&\sum_{m=1}^\infty (-1)^{m+1} m (m+1) (m+2) -3\sum_{m=1}^\infty (-1)^{m+1} m^2 -2\sum_{m=1}^\infty (-1)^{m+1} m \\
=&\frac{3}{8} -3 \cdot 0 -2 \cdot \frac{1}{4}
=-\frac{1}{8} \quad \quad \ldots \spadesuit
\end{align}
$$
On the other hand, $$
\zeta(-3) :=1^3 +2^3 +3^3 +\cdots \\
2^4 \zeta(-3) =2 \cdot 2^3 +2 \cdot 4^3 +2 \cdot 6^3 +\cdots \\
$$ Subtract them, aligning the 2nd, 4th, 6th term like Ramanjunan did in his notebooks (shown in the video). $$
-15 \zeta(-3) =1^3 -2^3 +3^3 -+\cdots \quad \quad \ldots \heartsuit
$$ $\heartsuit$ and $\spadesuit$ together give us: (Hold your breath.) $$
\sum_{m=1}^\infty m^3 =\frac{1}{120}.
$$
Recently I also found a proof of Riemann conjecture, but the answer box is too narrow for me to type all that down.
P.s. seriously, I think Ramanujan's effort is sort of finding an interpretation of divergent series so that they have a real value, while their manipulation to be still consistent to our usual notion of arithmetics: arranging, addition, expanding, etc.?
Maybe this can be compared to the attempt to define quaternion as an extension of complex numbers, while inevitably discarding commutative law?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\frac{1}{b-c},\frac{1}{c-a},\frac{1}{a-b}$ are in arithmetic progression under given condition. Suppose $(b-c)^2,(c-a)^2,(a-b)^2$ are in arithmetic progression.
Then show that $\frac{1}{b-c},\frac{1}{c-a},\frac{1}{a-b}$ are also in an arithmetic progression.
Please help.
| Since it is give that $(b-c)^2,(c-a)^2,(a-b)^2$ are in AP. So, you get that :
$$2(c-a)^2=(b-c)^2+(a-b)^2$$
$$\implies2c^2+2a^2-4ac=2b^2+c^2+a^2-2bc-2ab+2ac$$
$$\implies c^2+a^2-2ac=2b^2-2bc-2ab+2ac $$
$$\implies (c-a)^2=2(b^2-bc-ab+ac)$$
$$\implies (c-a)^2=-2(ab+bc-b^2-ac)$$
$$\implies (a-c)(c-a)=2(ab+bc-b^2-ac)$$
$$\implies (a-c)(c-a)=2(a-b)(b-c)$$
$$\implies \frac{(a-c)}{(a-b)(b-c)}=\frac{2}{(c-a)}$$
$$\implies \frac{(a-b)+(b-c)}{(a-b)(b-c)}=\frac{2}{(c-a)}$$
$$\implies \frac{1}{(b-c)}+\frac{1}{(a-b)}=\frac{2}{(c-a)}$$
That is what we needed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2058741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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if I so decide the limit? $\lim_{n\to +\infty}{\sqrt n * (\sqrt{n+1} - \sqrt{n-1})}$ $$ \lim_{n\to +\infty}{\sqrt n * (\sqrt{n+1} - \sqrt{n-1})} = \lim_{n\to +\infty}{\frac{\sqrt n * (\sqrt{n+1} - \sqrt{n-1})*(\sqrt{n+1} + \sqrt{n-1})}{(\sqrt{n+1} + \sqrt{n-1})}} = \lim_{n\to +\infty}{\frac{\sqrt n * (n + 1 - n + 1)}{(\sqrt{n+1} + \sqrt{n-1})}} = \lim_{n\to +\infty}{\frac{2 \sqrt n}{(\sqrt{n+1} + \sqrt{n-1})} } = \lim_{n\to +\infty}{\frac{2*\frac{\sqrt n}{\sqrt n}} {\frac{\sqrt{n+1}}{\sqrt n}+\frac{\sqrt{n-1}}{\sqrt n}}} = \lim_{n\to +\infty}{\frac{2}{\sqrt{1+\frac{1}{n}}+\sqrt{1 - \frac{1}{n}}}} = \frac{2}{\sqrt{1+0}+\sqrt{1-0}} = 1 $$
Can i do that? Divide the numerator and denominator by $$\sqrt n$$
| All your moves are fine. To gain more intuition for what is happening you might enjoy noting that to compute
$\lim_{n \to \infty} \left(\sqrt{ n^2 + n} - \sqrt{n^2 - n}\right),\;$ we see
$$\sqrt{n^2 + n} = \sqrt{\left(n+\frac{1}{2}\right)^2 - \frac{1}{4}}$$
$$\sqrt{n^2 + n} = \sqrt{\left(n-\frac{1}{2}\right)^2 - \frac{1}{4}}$$
The first term grows like $\left(n+ \frac{1}{2}\right)$ and the second like $\left(n-\frac{1}{2}\right)$ as $n \to \infty$, so their difference approaches $1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you find the real solutions to these simultaneous equations? I am looking for all real $(a,b,c)$ that satisfy the following
\begin{equation}
\left\{
\begin{array}{l}2a + a^2b = b\\
2b + b^2c = c\\
2c + c^2a = a\\
\end{array}
\right.
\end{equation}
I know that $a=b=c = 0$ is the only real solution to the problem I know of but I don't know how to prove it.
I was also given the hint, substitute $a = \tan(x)$.
| Rearrange
\begin{equation}
\left\{
\begin{array}{cccc}
\tan y &=& b &=& \dfrac{2a}{1-a^2} \\
\tan z &=& c &=& \dfrac{2b}{1-b^2} \\
\tan x &=& a &=& \dfrac{2c}{1-c^2} \\
\end{array}
\right.
\end{equation}
\begin{equation}
\left\{
\begin{array}{cccc}
\tan y &=& \tan 2x \\
\tan z &=& \tan 2y \\
\tan x &=& \tan 2z \\
\end{array}
\right.
\end{equation}
\begin{equation}
\left\{
\begin{array}{ccc}
y &=& p\pi+2x \\
z &=& q\pi+2y \\
x &=& r\pi+2z \\
\end{array}
\right.
\end{equation}
$$
\begin{pmatrix}
-2 & 1 & 0 \\
0 & -2 & 1 \\
1 & 0 & -2
\end{pmatrix}
\begin{pmatrix} x \\ y \\ z \end{pmatrix}=
\begin{pmatrix} p\pi \\ q\pi \\ r\pi \end{pmatrix}
$$
$$\begin{pmatrix} x \\ y \\ z \end{pmatrix}=
\begin{pmatrix}
-\frac{4}{7} & -\frac{2}{7} & -\frac{1}{7} \\
-\frac{1}{7} & -\frac{4}{7} & -\frac{2}{7} \\
-\frac{2}{7} & -\frac{1}{7} & -\frac{4}{7}
\end{pmatrix}
\begin{pmatrix} p\pi \\ q\pi \\ r\pi \end{pmatrix}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Can an indefinite integral evaluate to zero? Problem : Evaluate $\displaystyle\int \frac{\sin x+\cos x}{\cos^2 x+\sin^4 x} \, dx $
I evaluated in the following way, and somehow got zero :
$$I=\int \frac{\sin x+\cos x}{\cos^2 x+\sin^2 x(1-\cos^2 x)} \, dx$$
$$I=\int \frac{\sin x}{\cos^2 x+\sin^2 x(1-\cos^2 x)}dx +\int \frac{\cos x}{\cos^2 x+\sin^2 x(1-\cos^2 x)} \,dx $$
$$I=\int \frac{\sin x}{1-\sin^2 x\cos^2 x} \, dx +\int \frac{\cos x}{1-\sin^2 x\cos^2 x} \, dx$$
$$I=I_1+I_2$$
Substituting $\cos x=t$ in $I_1$ and $\sin x=u$ in $I_2$
This gives :
$$I_1=-\int \frac{1}{1-(1-t^2)t^2} \, dt $$
and
$$I_2=\int \frac{1}{1-u^2(1-u^2)} \, du $$
Since $I_1=-I_2$, $I=0$
My textbook gives me the answer :
$$I=\frac{1}{2\sqrt{3}}\log\left(\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right) + \tan^{-1}(\sin x-\cos x) + C$$
which seems to have involved the substitution $\sin x-\cos x=t$. I tried to simplify the denominator to make it a function of $(\sin x-\cos x)$ but I couldn't.
Could you please explain why my method did not work and how do I proceed to obtain the answer given ?
| Let $$I =\int\frac{\sin x+\cos x}{\cos^2 x+\sin^4 x}dx$$
Now we can write $\cos^2 x+\sin^4 x = \sin^4 x-\sin^2 x+1 = \cos^4 x-\cos^2 x+1$
So $$I = \int\frac{\sin x}{\cos^4 x-\cos^2 x+1}dx+\int\frac{\cos x}{\sin^4 x-\sin^2 x+1}dx$$
Now put $\cos x=t\;,$ Then $\sin xdx = -dt$ and put $\sin x=u\;,$ Then $\cos dx = du$
So $$I = -\underbrace{\int\frac{1}{t^4-t^2+1}dt}_{J}+\underbrace{\int\frac{1}{u^4-u^2+1}du}_{K}$$
Now Let $$J = \int\frac{1}{t^4-t^2+1}dt = \frac{1}{2}\int\frac{(t^2+1)-(t^2-1)}{t^4-t^2+1}dt$$
So $$J =\frac{1}{2}\int\frac{t^2+1}{t^4-t^2+1}dt+\frac{1}{2}\int\frac{t^2-1}{t^4-t^2+1}dt$$
So $$J = \frac{1}{2}\int\frac{1+\frac{1}{t^2}}{\left(t-\frac{1}{t}\right)^2+1}+\frac{1}{2}\int\frac{1-\frac{1}{t^2}}{\left(t+\frac{1}{t}\right)^2+3}dt$$
So we get $$J = \tan^{-1}\left(\frac{t^2-1}{t}\right)+\frac{1}{2\sqrt{3}}\ln \left|\frac{t^2-\sqrt{3}t+1}{t^2+\sqrt{3}t+1}\right|$$
Same calculation for $J$
So we get $$I = -\tan^{-1}\left(\frac{\cos^2 x-1}{\cos x}\right)-\frac{1}{2\sqrt{3}}\ln \left|\frac{\cos^2 x-\sqrt{3}\cos x+1}{\cos^2 x+\sqrt{3}\cos x+1}\right|+\tan^{-1}\left(\frac{\sin^2 x-1}{\sin x}\right)+\frac{1}{2\sqrt{3}}\ln \left|\frac{\sin^2 x-\sqrt{3}\sin x+1}{\sin^2 x+\sqrt{3}\sin x+1}\right|+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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$x^3 = x$ for all $x \in R$, where $R$ is a ring. Prove that $6x = 0$ for all $x \in R$. I am very confused about this problem. Because, clearly 6x = 0 only has one solution, 0. But I can make
$x^3 = x$
$x^3 -x=0$ (Adding additive inverse of x to both sides)
$x(x^2-1)=0$ (Distributive property of the ring)
$x(x+1)(x-1) = 0$ (Distributive property of the ring again)
which gives $x=0,1,-1$ as solutions.
Not also I don't really know how to prove this problem, it confuses me how this is possible? Is it because the ring could not be an integral domain?
| $(x+1)^3=x^3+3x^2+3x+1=x+1=x+3x^2+3x+1$implies that $3x^2+3x=0$. $3(x^2+x)=0$.
$3((x+1)^2+x+1)=3(x^2+2x+1+x+1)=3(x^2+x)+6x+6=0$.
$(1+1)^3=1+1$, $8=2$ implies $6=0$, so $6x+6=6x=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is there a property similar to $ x^2 = (x-1)(x+1)+1$ for $ x^3$? I am looking for a way to decompose $x^3$ in a similar way.
| $$\small \begin{vmatrix} x+n & 1 & 0 & 0 & \ddots & 0 & 0 & 0 & 0\\ -n & x+n-2 & 2 & 0 & \ddots & 0 & 0 & 0 & 0\\ 0 & -n+1 & x+n-4 & 3 & \ddots & 0 & 0 & 0 & 0\\ 0 & 0 & -n+2 & x+n-6 & \ddots & 0 & 0 & 0 & 0\\ \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots & \ddots\\ 0 & 0 & 0 & 0 & \ddots & x-n+6 & n-2 & 0 & 0\\ 0 & 0 & 0 & 0 & \ddots & -3 & x-n+4 & n-1 & 0\\ 0 & 0 & 0 & 0 & \ddots & 0 & -2 & x-n+2 & n\\ 0 & 0 & 0 & 0 & \ddots & 0 & 0 & -1 & x-n \end{vmatrix} = x^{n+1}$$
It goes without saying that $n$ is obviously an integer, since the determinant is always polynomial in $x$. By setting $n=1$, your decomposition can be recovered.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the coefficient of $x^{10}$ in $(1+x^2-x^3)^8$ The coefficient of $x^{10}$ in $(1+x^2-x^3)^8$. I tried to factor it into two binomials but it became way to long to solve by hand.
| The pertinent part of the polynomial expansion can be done by hand without too much fuss:
$$\begin{align}
(1+x^2-x^3)^8&=(1+x^2)^8-8x^3(1+x^2)^7+{8\choose2}x^6(1+x^2)^6-{8\choose3}x^9(1+x^2)^5+\cdots\\
&=(1+x^2)^5((1+x^2)^3-8x^3(1+x^2)^2+28x^6(1+x^2)-56x^9)+\cdots\\
&=(1+x^2)^5((1+3x^2+3x^4+x^6)-8x^3(1+2x^2+x^4)+28x^6(1+x^2)-56x^9)+\cdots\\
&=(1+5x^2+10x^4+10x^6+5x^8+x^{10})(1+3x^2-8x^3+3x^4-16x^5+29x^6-8x^7+28x^8-56x^9)+\cdots\\
&=\cdots+(5\cdot28+10\cdot29+10\cdot3+5\cdot3+1\cdot1)x^{10}+\cdots\\
&=\cdots+(140+290+30+15+1)x^{10}+\cdots\\
&=\cdots+476x^{10}+\cdots
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Use the sum identity and double identity for sine to find $\sin 3x$. Q. Use the sum identity and double identity for sine to find $\sin 3x$.
$$
\begin{align}
\sin 3x &= \sin (2x + x)\\
&=\sin 2x \cos x + \cos 2x \sin x \\
&= (2\sin x \cos x) \cos x + (1 - 2\sin^2 x) \sin x\\
&=2\sin x \cos^2 x + \sin x - 2\sin^3 x \\
&=2\sin x (1 - \sin^2 x) + \sin x - 2\sin^3 x\\
&= " 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x \\
&=3\sin x - 4\sin^3 x"
\end{align}
$$
The part of the problem I'm having trouble with is in quotations.
My question:
is how does $\sin x - 2\sin^3 x = 4\sin^3 x$?
I see it as this $\sin x - 2\sin^3 x = 2\sin^{3-1} x = 4\sin x$.
| With a hopefully obvious notation,
$$s_3=s_2c+c_2s=2sc^2+(1-2s^2)s=2s(1-s^2)+(1-2s^2)s=3s-4s^3.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Simplify $\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$ Simplify::
$$\frac {2^{n(1-n)}\cdot 2^{n-1}\cdot 4^n}{2\cdot 2^n\cdot 2^{(n-1)}}$$
My Attempt:
\begin{align}
&\frac {2^{n-n^2}\cdot 2^{n-1}\cdot 2^{2n}}{2\cdot 2^n\cdot 2^{n-1}}\\
&=\frac {2^{n-n^2+n-1+2n}}{2^{1+n+n-1}} \\
&=\frac {2^{4n-n^2-1}}{2^{2n}}
\end{align}
I could not move on. Please help me to continue.
| Since that is a quotient you can rewrite it as $$2^{(4n-n^2-1)-2n}= 2^{-(n-1)^2}$$
Notice indeed that $$(a-b)^2=a^2-2ab+b^2$$ so that you have $$(n-1)^2 = n^2-2n+1$$ so clearly also $$-(n-1)^2 = -n^2+2n-1$$ and by rewriting the middle term you get $$-(n-1)^2 = -n^2+4n-2n-1$$ because clearly $4n-2n =2n$ and as you can probably see this is the same as $$-(n-1)^2 = (-n^2+4n-1)-2n$$ by commutativity of addition
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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find range of $a^2+b^2$ without trigonometric substution given $7a^2-9ab+7b^2=9$ and $a,b$ are real no. then find range of $a^2+b^2$ without trigonometric substution
from $7a^2-9ab+7b^2=9$
$\displaystyle ab = \frac{7(a^2+b^2)-9}{9}$
put into inequality $\displaystyle a^2+b^2 \geq 2ab$
$\displaystyle a^2+b^2 \geq \frac{14(a^2+b^2)-18}{9}$
$\displaystyle a^2+b^2 \leq \frac{18}{5}$
i wan,t be able to find minimum,could some help me with this
| If $a = 0$ or $b = 0$, then $a^2+b^2 = \dfrac{9}{7}$, and this is one of the many values in the range of $a^2+b^2$. So assume $ab \neq 0$, then put $x = \dfrac{a}{b} > 0$ ( we can assume $a, b > 0$ ), and rewrite the equation by dividing both sides by $ab$ to have: $\dfrac{9}{ab} = 7\left(x+\dfrac{1}{x}\right)-9 = 7u-9, u = x+\dfrac{1}{x}\implies a^2+b^2 = ab\left(\dfrac{a}{b}+\dfrac{b}{a}\right) = ab\left(x+\dfrac{1}{x}\right) = abu = \dfrac{9u}{7u-9}$. Thus consider $f(u) = \dfrac{9u}{7u-9}, u \ge 2\implies f'(u) = \dfrac{9(7u-9)-7(9u)}{(7u-9)^2}= \dfrac{-81}{(7u-9)^2} < 0\implies f(u) \le f(2)= \dfrac{18}{5}$. Thus $(a^2+b^2)_{\text{max}} = \dfrac{18}{5} $. Note that the OP has made an edit to his answer, and I thought his first answer to the minimum value was incorrect, but due to trusting his own answer, I didn't validate it but it was confirmed by the 3rd answer to this question that it was correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Confusion in fraction notation $$a_n = n\dfrac{n^2 + 5}{4}$$
In the above fraction series, for $n=3$ I think the answer should be $26/4$, while the answer in the answer book is $21/2$ (or $42/4$). I think the difference stems from how we treat the first $n$. In my understanding, the first number is a complete part and should be added to fraction, while the book treats it as part of fraction itself, thus multiplying it with $n^2+5$.
So, I just want to understand which convention is correct.
This is from problem 6 in exercise 9.1 on page 180 of the book Sequences and Series.
Here is the answer sheet from the book (answer 6, 3rd element):
*
*$3,8,15,24,35$
*$\dfrac{1}{2},\dfrac{2}{3},\dfrac{3}{4},\dfrac{4}{5},\dfrac{5}{6}$
*$2, 4, 8, 16 \text{ and } 32$
*$-\dfrac{1}{6},\dfrac{1}{6},\dfrac{1}{2},\dfrac{5}{6},\dfrac{7}{6}$
*$25,-125,625,-3125,15625$
*$\dfrac{3}{2},\dfrac{9}{2},\dfrac{21}{2},21,\dfrac{75}{2}$
*$65, 93$
*$\dfrac{49}{128}$
*$729$
*$\dfrac{360}{23}$
*$3, 11, 35, 107, 323$; $3+11+35+107+323+...$
*$-1,\dfrac{-1}{2},\dfrac{-1}{6},\dfrac{-1}{24},\dfrac{-1}{120}$; $-1+(\dfrac{-1}{2})+(\dfrac{-1}{6})+(\dfrac{-1}{24})+(d\frac{-1}{120})+...$
*$2, 2, 1, 0, -1$; $2+2+1+0+(-1)+...$
*$1,2,\dfrac{3}{5},\dfrac{8}{5}$
| Here is my attempt at a helpful rule:
The expression $x\frac{y}{z}$ always means $x\times\frac{y}{z}$
except when $x,y,$ and $z$ are all integers written in decimal notation; then it means $x+\frac{y}{z}$.
So $n\frac{n^2+5}{4}$ means $n\times\frac{n^2+5}{4}$, but $3\frac14$ means $3+\frac14$.
| {
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"url": "https://math.stackexchange.com/questions/2074265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 7,
"answer_id": 4
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Find the value of $\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx $ Problem :
Find the value of $\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx$.
My approach :
\begin{align}
&\int^1_{-1} x \ln(1^x +2^x +3^x +6^x)\,dx \\
=& \ln(1^x +2^x +3^x +6^x) \frac{x^2}{2} - \int^1_{-1} \frac{1}{1^x+2^x+3^x+6^x}(2^x\log2+3^x \log3+6^x \log 6 )\frac{x^2}{2} \,dx
\end{align}
[By using by parts]
Is it the correct method of solving this, please suggest, will be of great help, thanks.
| $$1+2^x+3^x+6^x=(1+2^x)(1+3^x)$$
Now $\ln(AB)=\ln A+\ln B$
Apply $\displaystyle\int_p^qf(x)\ dx=\int_p^qf(p+q-x)\ dx$
in $I=\displaystyle\int_{-1}^1x\ln(1+a^x)\ dx=\int_{-1}^1(-x)\ln(1+a^{-x})\ dx=\int_{-1}^1(-x)\{\ln(1+a^x)-x\ln a\}dx$
$\implies\displaystyle I=-I+\ln a\int_{-1}^1x^2\ dx$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum\limits_{cyc}\frac{a}{a+b}\geq1+\frac{3\sqrt[3]{a^2b^2c^2}}{2(ab+ac+bc)}$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\geq1+\frac{3\sqrt[3]{a^2b^2c^2}}{2(ab+ac+bc)}$$
I tried C-S, uvw, BW and more, but without success.
| @Michael Rosenberger Thanks, here is my $0.02:
Write the whole inequality in terms of $ab, ac, bc$:
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}=\frac{ac}{ac+bc}+\frac{ab}{ab+ac}+\frac{bc}{bc+ab}$$
Then use the following inequality proved here - as a Lemma inside Andreas's answer:
$$ \frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+a} \geq 1+\frac{3\sqrt[3]{a \, b \ c}}{2(a+b+c)}$$
for $a\leftarrow ac$, $b\leftarrow ab$, $c\leftarrow bc$
That's all.
| {
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"url": "https://math.stackexchange.com/questions/2075297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Find $\lim\limits_{x\to\pi/4} \frac{1-\tan(x)^2}{\sqrt{2}*\cos(x)-1}$ without using L'Hôpital's rule. Find $$\lim_{x\to\pi/4} \frac{1-\tan(x)^2}{\sqrt{2}\times \cos(x)-1}$$ without using L'Hôpital's rule.
I can solve it using L'Hôpital's rule, but is it possible to solve it without using L'Hôpital's rule?
| You can try the following:
\begin{align}
\frac{1-\tan^2({x})}{\sqrt{2} \cos{x}-1}\frac{\sqrt{2} \cos{x}+1}{ \sqrt{2} \cos{x}+1 }&=\frac{1-\frac{\sin^2{x}}{\cos^2{x}}}{2 \cos^2{x}-1} (\sqrt{2} \cos{x}+1)\\ &=\frac{\cos^2{x}-\sin^2{x}}{\cos^2{x} (2\cos^2{x}-\sin^2{x}-\cos^2{x}) }(\sqrt{2} \cos{x}+1)\\
&=\frac{\sqrt{2} \cos{x}+1}{\cos^2x}.
\end{align}
Now take the limit as usual.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2075487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
prove that: $\sqrt{2}=e^{1-{2K\over \pi}}\prod\limits_{n=1}^{\infty}\left({4n-1\over 4n+1}\right)^{4n}e^2$ show that
$$\sqrt{2}=e^{1-{2K\over \pi}}\prod_{n=1}^{\infty}\left({4n-1\over 4n+1}\right)^{4n}e^2$$
where K is the catalan's constant; $K=0.9156 ...$
My try:
take the ln
$${1\over2}\ln{2}=\left(1-{2K\over \pi}\right)\sum_{n=1}^{\infty}\ln{\left({4n-1\over 4n+1}\right)^{4n}}e^2$$
$${1\over2}\ln{2}=\sum_{n=1}^{\infty}\ln{\left({4n-1\over 4n+1}\right)^{4n}}e^2-
{2K\over \pi}\sum_{n=1}^{\infty}\ln{\left({4n-1\over 4n+1}\right)^{4n}}e^2$$
we know that
$${1\over 2}\ln{2}=\sum_{n=1}^{\infty}\ln{\left(4n-1\over 4n+1\right)}+\sum_{n=1}^{\infty}\ln{\left(4n+1\over 4n+2\right)}$$
sub: then we got
$$\sum_{n=1}^{\infty}\ln{\left(4n+1\over 4n+2\right)}=\sum_{n=1}^{\infty}\ln{\left({4n-1\over 4n+1}\right)^{4n-1}}e^2-
{2K\over \pi}\sum_{n=1}^{\infty}\ln{\left({4n-1\over 4n+1}\right)^{4n}}e^2$$
Anyway I am stuck, any help please. I tried and looked everywhere on wolfram can't find any similiar infinite product to simplify this further.
| Consider
\begin{align}
\sum_{n=1}^{\infty}4n \log\left({4n-1\over 4n+1}\right)+2 &=-
\sum_{n=1}^{\infty}4n \sum_{k=0}^\infty \frac{-2}{(2k+1)(4n)^{2k+1}}+2\\
&=-\sum_{k=1}^\infty \frac{2}{(2k+1)4^{2k}}\sum_{n=1}^{\infty}\ \frac{1}{n^{2k}}\\
&=-\sum_{k=1}^\infty \frac{2\zeta(2k)}{(2k+1)4^{2k}}\\
&= \frac{2K}{\pi}-1+\frac{\log(2)}{2}
\end{align}
Hence finally we have
$$\prod_{n=1}^{\infty}\left({4n-1\over 4n+1}\right)^{4n} e^{2} = \sqrt{2} \mathrm{exp} \left\{ \frac{2K}{\pi}-1 \right\}$$
ADDENDUM
We prove the last series using the generating function
$$\pi\;x\;\cot(\pi\;x)-1=-2\sum_{k=1}^\infty \zeta(2k)\;x^{2k}$$
By integration
$$4\pi\int^{1/4}_0\;x\;\cot(\pi\;x)\,dx -1=-2\sum_{k=1}^\infty \zeta(2k)\frac{x^{2k+1}}{(2k+1) 4^{2k}}$$
Note that
\begin{align}
\int^z_0 x\pi \cot(\pi x) \, dx
&=z\log(\sin\pi z)-\int^z_0 \log(\sin\pi x) dx\\
&=z\log(2\sin\pi z)-\int^z_0 \log(2\sin\pi x) dx\\
&=z\log(2\sin\pi z)-\frac{1}{2\pi }\int^{2\pi z}_0 \log\left(2\sin\frac{x}{2}\right) dx\\
&=z\log(2\sin\pi z)+\frac{\mathrm{cl}_2(2\pi z)}{2\pi}\\
\end{align}
Hence
$$ 4 \int^{1/4}_0 x\pi \cot(\pi x) \, dx -1= 4 \left(\frac{\log(2\sin\pi /4)}{4}+\frac{\mathrm{cl}_2(\pi/2)}{2\pi}\right)-1 = \frac{\log 2}{2}+\frac{K}{2\pi}-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Does $\sum \frac{a_n}{1+n a_n}$ converge if $a_n = \frac{1}{\sqrt{n}}$ if $n$ is a perfect square and $a_n = \frac{1}{n^2}$ otherwise? Given the sequence $\left\{ a_n \right\}$, where
$$a_n = \begin{cases} \frac{1}{\sqrt{n}} \ \mbox{ if } n \mbox{ is a perfect square} \\ \frac{1}{n^2} \ \mbox{ otherwise}, \end{cases}$$
does the series $$\sum \frac{a_n}{1 + n a_n}$$ converge or diverge?
My effort:
If $n = m^2$, then we have
$$\frac{a_n}{1+na_n} = \frac{ \frac{1}{m} }{ 1 + m^2 \frac{1}{m} } = \frac{1}{m(m+1)} = \frac{1}{\sqrt{n} (1 + \sqrt{n})}, $$
and otherwise
$$ \frac{a_n}{1+na_n} = \frac{ \frac{1}{n^2}}{1+ n \frac{1}{n^2}} = \frac{1}{n^2 + n}. $$
Thus we have
$$ \frac{a_n}{1+na_n} = \begin{cases} \frac{1}{\sqrt{n} (1 + \sqrt{n})} \ \mbox{ if } n \mbox{ is a perfect square} \\ \frac{1}{n( 1 + n)} \ \mbox{ otherwise}. \end{cases} $$
What next? How to proceed from here?
An afterthought:
If $n$ is a perfect square, then we note that
$$\frac{a_n}{1+na_n} = \frac{1}{\sqrt{n}(1+ \sqrt{n})} \geq \frac{1}{2n},$$
| Hint.
$$\sum_{n=1}^\infty\frac{a_n}{1+na_n}=\sum_{k=1}^\infty\frac{a_{k^2}}{1+k^2 a_{k^2}}+\sum_{n\text{ is not a square}} \frac{a_n}{1+na_n}$$
From what you've done we know
$$\sum_{k=1}^\infty\frac{a_{k^2}}{1+k^2 a_{k^2}}=\sum_{k=1}^\infty\frac 1{k(k+1)}<\infty$$
and
$$\sum_{n\text{ is not a square}} \frac{a_n}{1+na_n}=\sum_{n\text{ is not a square}}\frac 1{n(n+1)}<\sum_{n=1}^\infty\frac 1{n(n+1)}<\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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On the convergence of the sequence $\,a_n=\displaystyle\sum_{k=1}^{n-1}\left(\frac{n}{k(n-k)}\right)^{2}$ Let the sequence
$$
a_n=\sum_{k=1}^{n-1}\left(\frac{n}{k(n-k)}\right)^{2}, \quad n\ge 2.
$$
Show that:
A. $\,a_n\le 4$.
B. $\,\{a_n\}_{n\in\mathbb N}$ is decreasing, and hence converging.
C. Find $\lim_{n\to\infty}a_n$.
Clearly, B implies A.
The first thing that comes to one's mind is to make $\,\{a_n\}_{n\in\mathbb N}$ look like an approximation of a definite integral. However, we do not have the right power of $n$.
Numerical experiments suggest that $\lim_{n\to\infty}=3.2898\ldots$
| $$\begin{align}
a_n= & \sum_{k=1}^{n-1}\left[\frac{n}{k(n-k)}\right]^{2} \\
= & \sum_{k=1}^{n-1}\left[\frac{(n-k)+k}{k(n-k)}\right]^{2} \\
= & \sum_{k=1}^{n-1}\left[\frac{1}{k}+\frac{1}{n-k}\right]^{2} \\
= & \sum_{k=1}^{n-1}\left[\frac{1}{k^{2}}+\frac{1}{(n-k)^{2}}+\frac{2}{k(n-k)}\right] \\
= & \sum_{k=1}^{n-1}\frac{1}{k^{2}}+ \sum_{k=1}^{n-1}\frac{1}{(n-k)^{2}}+2 \sum_{k=1}^{n-1}\frac{1}{k(n-k)} \\
= & \sum_{k=1}^{n-1}\frac{1}{k^{2}}+ \sum_{k=1}^{n-1}\frac{1}{(n-k)^{2}}+\frac{2}{n} \sum_{k=1}^{n-1}\left[\frac{1}{k}+\frac{1}{n-k}\right] \\
= & \sum_{k=1}^{n-1}\frac{1}{k^{2}}+ \sum_{k=1}^{n-1}\frac{1}{(n-k)^{2}}+\frac{2}{n} \sum_{k=1}^{n-1}\frac{1}{k}+\frac{2}{n} \sum_{k=1}^{n-1}\frac{1}{n-k} \\
\end{align}$$
Now carefully note that the terms in the sum $\sum_\limits{k=1}^{n-1}\frac{1}{k^{2}}$ are same as those in $\sum_\limits{k=1}^{n-1}\frac{1}{(n-k)^{2}}$ only in the reverse order. Same also holds for $\sum_\limits{k=1}^{n-1}\frac{1}{k}$ and $\sum_\limits{k=1}^{n-1}\frac{1}{n-k}$ .
So we can write that
$$\begin{align}
a_n= & \sum_{k=1}^{n-1}\frac{2}{k^{2}}+\frac{2}{n} \sum_{k=1}^{n-1}\frac{2}{k} \\
= & 2 \sum_{k=1}^{n-1}\left[\frac{1}{k^{2}}+\frac{2}{nk}\right] \\
\end{align}$$
And as $n\to\infty$ , $\sum_{k=1}^{n-1}\frac{1}{k^{2}} \to \frac{\pi^2}{6}$ and $\sum_{k=1}^{n-1}\frac{2}{nk} \to 0$.
So the limit of $a_n$ is $\frac{\pi^2}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Limit question related to integration Find the limit,
$$L=\lim_{n\to \infty}\int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx$$
My try:
$$ \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}xdx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}(1-x)dx< \int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx< \int_{0}^{\frac{1}{2}}2^{\frac{1}{n}}(1-x)dx+ \int_{\frac{1}{2}}^{1}2^{\frac{1}{n}}xdx$$
Now taking the limit I get that,
$$\frac{1}{4}<L<\frac{3}{4}$$
But, how can I get the exact answer!!
This is Problem 11941 from the American Mathematical Monthly.
| Hint/Intuition:
For $x,y\ge0, \lim_{n\to\infty} \left(x^n+y^n\right)^{1/n}=\max\{x,y\}$.
Alternative Approach:
$$I=\int_{0}^{1}(x^n+(1-x)^n)^{\frac{1}{n}}dx=2\int_{0}^{1/2}(x^n+(1-x)^n)^{\frac{1}{n}}dx$$
$$=2\int_0^{1/2}(1-x)\left(1+\left(\frac{x}{1-x}\right)^n\right)^{1/n}$$
Let $u=\frac{x}{1-x}\implies dx=\frac{du}{(1+u)^2}$
$$I=2\int_0^1\frac{1}{1+u}(1+u^n)^{1/n}\frac{du}{(1+u)^2}$$
Note that for $y>0, 1<(1+y)^{1/n}<1+\frac{y}{n}$, so:
$$2\int_0^1\frac{du}{(1+u)^3}\le I \le 2\int_0^1\left(1+\frac{u^n}{n}\right)\frac{du}{(1+u)^3}$$
Now:
$$0\le I - 2\int_0^1\frac{du}{(1+u)^3}\le \frac{2}{n}\int_0^1\frac{u^n\,du}{(1+u)^3}\le \frac{2}{n}\int_0^1\frac{du}{(1+u)^3}=\frac{3}{4n}$$
So, the integral converges to $2\int_0^1\frac{du}{(1+u)^3}=\frac{3}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2079437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can you prove that $\frac{a+b}{ab+1}$ is real if $|a|=1$, $|b|=1$, and $ab\ne-1$? Can you prove that $\dfrac{a+b}{ab+1}$ is real if $|a|=1$, $|b|=1$, and that $ab$ isn't equal to $-1$?
| $$\dfrac{a+b}{ab+1}= \dfrac{(a+b)(1+\overline{a}\overline{b})}{(ab+1)(1+\overline{a}\overline{b})}=\dfrac{(a+b)(1+\overline{a}\overline{b})}{|ab+1|^2}=\dfrac{a+b+|a|^2\overline{b}+\overline{a}|b|^2}{|ab+1|^2}=\dfrac{a+b+\overline{b}+\overline{a}}{|ab+1|^2}=\dfrac{2Re(a+b)}{|ab+1|^2} \in \Bbb R$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2081984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find two variables $a,b \in {\bf Z}$ that the matrix $A$ is orthogonal I have to find two variables $a,b \in {\bf Z}$ that the given $n \times n$ matrix A becomes orthogonal.
\begin{equation*}
A =
\begin{pmatrix}
1&2 \\
a&b
\end{pmatrix}
\end{equation*}
I know that a $n \times n$ matrix is called orthogonal if $A^TA$ $=$ $id$ which means:
\begin{equation*}
\begin{pmatrix}
1&a \\
2&b
\end{pmatrix}
\cdot
\begin{pmatrix}
1&2 \\
a&b
\end{pmatrix}
=
\begin{pmatrix}
1+a^2&2+ab \\
2+ab&4+b^2
\end{pmatrix}
\stackrel{?}{=}
\begin{pmatrix}
1&0 \\
0& 1
\end{pmatrix}
\end{equation*}
| The system of equations has no solution at all, since $1+a^2=1$ implies $a=0$ and hence $2+ab=2\neq 0$. hence $A^TA\neq I_2$ for all $a,b$. This is, given the computation in the question, much more immediate than to use that the norm of a column vector must be $1$ for an orthogonal matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
| If $x = a+b$, then $x^3 = (3ab)x + (a^3+b^3)$.
(Conversely, if $x^3 - px - q = 0 $ then we may find $a,b$, such that $3ab = p, a^3+b^3=q$.
using Cardano's method for solving the cubic equation.)
Letting $w = e^{\frac {2i \pi}{3}} $ denote a cube root of unity, notice that the numbers $a+b, wa + w^2b$, and $w^2a + wb$ all satisfy the same cubic equation, $x^3 = 3abx + (a^3+b^3)$, and in general the solutions to the equation
$x^3 = 3w^iabx + (a^3+b^3) $ are given by $x = w^ja + w^{i-j}b$, for $j=0,1,2$.
There are thus $9$ possible values of $(2+\sqrt{5})^{1/3} + (2-\sqrt{5})^{1/3}$, depending on which cube roots you take, and they each satisfy one of $3$ cubic equations. But only one of these values, namely $1$, is real, let alone rational. Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 3
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Which integration formula for $\frac{1}{a^2-x^2}$ is correct? In my book integration formula for $\frac{1}{x^2-a^2}$ is given as $\frac{1}{2a}\ln(\frac{x-a}{x+a})$.
From the above formula we can write the formula for integration of $\frac{1}{a^2-x^2}$ as $-\frac{1}{2a}\ln(\frac{x-a}{x+a}) = \frac{1}{2a}\ln(\frac{x+a}{x-a}) $ [as we know $\ln(\frac{1}{x})=-\ln(x)$]
But, in my book the integration of $\frac{1}{a^2-x^2}$ is written as $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $.
We know that the integration in the second case cannot be $\frac{1}{2a}\ln(\frac{x+a}{x-a}) $ and $\frac{1}{2a}\ln(\frac{x+a}{{a-x}}) $ at the same time. Which formula is the correct one and why?
| Let $f(x)=\frac{1}{x^2-a^2}$.
$f$ is continuous at $(-\infty,-|a|),(-|a|,|a|)$ and $(|a|,+\infty)$.
in each interval
$$2af(x)=\frac{1}{x-a}-\frac{1}{x+a}$$
and $$\int f(x)dx=\frac{1}{2a}\ln(\frac{|x-a|}{|x+a|}).$$
the final expression depends on which interval $J$ we want the antiderivative.
for example, if $a>0$ and $J=(-a,a)$, we have
$$\int f(x)dx=\frac{1}{2a}\ln( \frac{a-x}{x+a} )+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Induction with floor, ceiling $n\le2^k\implies a_n\le3\cdot k2^k+4\cdot2^k-1$ for $a_n=a_{\lfloor\frac{n}2\rfloor}+a_{\lceil\frac{n}2\rceil}+3n+1$ Via induction I need to prove an expression is true. the expression is:
$n \leq 2^k \longrightarrow a_n \leq 3 \cdot k2^k + 4 \cdot
2^k-1$
for all $n,k \in \mathbb{Z^+}$
$a_n$ is a recursive function where
$a_1 = 3$
$a_n = a_{\lfloor \frac{n}{2} \rfloor} + a_{\lceil \frac{n}{2} \rceil} + 3n+1$
I am stuck at the point where I need to prove that $P(n+1)$ is true
$a_{\lfloor \frac{n+1}{2} \rfloor} + a_{\lceil \frac{n+1}{2} \rceil} + 3(n+1) + 1 \leq 3 \cdot k 2^k + 4 \cdot 2^k -1 $
It is the fact that I don't know how to rewrite the floor and ceiling functions to something else.
Can someone give me some ideas how to proceed with this?
| Hint: Your induction should be for $k$, not for $n$
$P(k): n \leq 2^k \Longrightarrow a_n \leq 3 \cdot k2^k + 4 \cdot
2^k-1$
For $k=1$ it is trivial
Suppose $P(k)$ is true and prove that $P(k+1)$ is true
$P(k+1): n \leq 2^{k+1} \Longrightarrow a_n \leq 3 \cdot (k+1)2^{k+1} + 4 \cdot
2^{k+1}-1$
But if $n\le 2^{k+1}$ then both $\lfloor n/2\rfloor\le 2^k$ and $\lceil n/2\rceil\le 2^k$. Next, write the recurrence and apply the induction hypothesis.
$a_n = a_{\lfloor \frac{n}{2} \rfloor} + a_{\lceil \frac{n}{2} \rceil} + 3n+1 \le 2\cdot(3 \cdot k2^k + 4 \cdot
2^k-1)+3n+1\le 3 \cdot (k+1)2^{k+1} + 4 \cdot
2^{k+1}-1$
After simplifications, the last inequality is equivalent with $n\le 2^{k+1}$, which is true by assumption.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Absolute value rational inequality I just stumbled upon this particular question and cannot answer
$$\left\lvert \frac{2x+1}{x-2} \right\rvert<1$$
I know that there are some rules in order to answer this Problem
$\frac{2x+1}{x-2} $ if only the $\frac{2x+1}{x-2}≥ 0 $
and
$\frac{(-)2x+1}{x-2}$ if only the $\frac{2x+1}{x-2}< 0 $
but I just can't find the answer because everything what i've learnt and saw on YouTube mixed up, and I can't tell which one is correct.
Can anybody give me an Explanation?
*Ps, I'm sorry, I'm not really familiar with MathJax
btw, I have seen another Problem on this site which is also Absolute value rational inequality, but I don't really understand.
| $|\frac {2x+1}{x-2}|<1$
To get rid of the absolute value:
$-1< \frac {2x+1}{x-2}<1$
When you mulitiply through by $x-2$ it is going to flip the direction of the inequalities if $x-2 < 0$
Suppose (x-2) > 0
$2-x < 2x+1 <x-2$ and $(x-2) > 0$
$2-x < 2x+1$ and $2x+1 <x-2$ and $(x-2) > 0$
$\frac 13 < x$ and $x <-3$ and $x > 2$
not compatible!
Suppose $(x-2) < 0$
$2-x > 2x+1 >x-2$ and $(x-2) < 0$
$2-x > 2x+1$ and $2x+1 > x-2$ and $(x-2) < 0$
$\frac 13 > x$ and $x > -3$ and $(x-2) < 0$
$x\in(-3 , \frac 13)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine whether the following sequence is increasing or decreasing $\frac{n^2+2n+1}{3n^2+n}$ Determine whether the following sequence is increasing or decreasing:
$$\frac{n^2+2n+1}{3n^2+n}$$
I'm not sure whether my solution is correct:
$$\frac{n^2+2n+1}{3n^2+n}=\frac{n(n+2)+1}{n(3n+1)}=\frac{n+2}{3n+1}+\frac{1}{n(3n+1)}.$$
Let's prove $\frac{n+2}{3n+1}$ is a decreasing sequence.
$$a_n>a_{n+1} \Leftrightarrow \frac{n+2}{3n+1}>\frac{n+3}{3n+4}\Leftrightarrow(n+2)(3n+4)>(n+3)(3n+1)\Leftrightarrow3n^2+10n+8>3n^2+10n+3\Leftrightarrow 8>3$$
So $\frac{n+2}{3n+1}$ is a decreasing sequence and we know that $\frac{1}{n(3n+1)}$ is also decreasing so our given sequence is a decreasing sequence as a sum of $2$ decreasing sequences.
| HINT Find the difference (an+1 - an) and study the sign of this difference. If it is positive, the sequence is increasing, otherwise it is decreasing.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Proving inequality for $n\in \mathbb{N}$ How can we prove the following: For all $x>0 $ and $n\in \mathbb{N}$,
$$\frac{x^n}{1+x+x^2+...+x^{2n}}\leq\frac{1}{2n+1}.$$
I was wondering if someone can help me. Thanks.
| As suggested by @fonfonx in the comments, you can write
\begin{align}
\frac{x^n}{1+x+x^2+...+x^{2n}} &= \frac{1}{\frac{1}{x^n}+\frac{1}{x^{n-1}}+\ldots+\frac{1}{x}+1+x+\ldots+x^{n}}\\
&= \frac{1}{1+(\frac{1}{x}+x)+(\frac{1}{x^2}+x^2)+\ldots+(\frac{1}{x^n}+x^n)}\\
&\leq\frac{1}{1+2+2+\ldots+2}\\
&=\frac{1}{1+2n}\\
\end{align}
where we have used the fact that $\frac{1}{x^i}+x^i \geq 2$ for $x>0$, which follows immediately from $(x^i-1)^2 \geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ac}{1+a+c}\geq \frac{3}{2}$ for $a$, $b$, $c$ positive and $1+a+b+c=2abc$
Given $1+a+b+c = 2abc$ and positivity of real numbers $a,b,c$, we are asked to prove that $$\frac{ab}{1+a+b}+\frac{bc}{1+b+c}+\frac{ac}{1+a+c}\geq \frac{3}{2}$$
If $d=a+b+c$ I got as far as to simplify the inequality into $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d-a}+\dfrac{1}{d-b}+\dfrac{1}{d-c}\geq 3$$
From $a=\frac{1+b+c}{2bc-1}$
I also can prove that $$ab+ac+bc\geq \frac{3}{2}$$
But cannot manage to get to the desired result.
| We need to prove next equivalent inequality $$\sum\limits_{cyc}\left(\frac{ab}{1+a+b}+1\right)\geq\frac{9}{2}$$ or
$$\sum\limits_{cyc}\frac{(a+1)(b+1)(c+1)}{(1+a+b)(1+c)}\geq\frac{9}{2}$$
Then by Cauchy-Schwarz inequality $$\sum\limits_{cyc}\frac{(a+1)(b+1)(c+1)}{(1+a+b)(1+c)}\geq\frac{9\prod\limits_{cyc}(a+1)}{\sum\limits_{cyc}(1+a+b)(c+1)}=$$
$$=\frac{9\prod\limits_{cyc}(a+1)}{\sum\limits_{cyc}(a+2ab+1+2a)}=\frac{9}{2}$$
Now, we have done!
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
How many rational solutions does $(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2$ have?
How many rational solutions does $(x^2+4x+8)^2+3x(x^2+4x+8)+2x^2$ have?
I don't know how to start...
| We can still use the classic "complete square" trick to do this without having to use the "double" variables as shown by the first answer. To this end, we have: $\left((x^2+4x+8) + \dfrac{3x}{2}\right)^2 - \dfrac{9x^2}{4}+2x^2=0\implies \left((x^2+4x+8)+\dfrac{3x}{2}\right)^2 = \left(\dfrac{x}{2}\right)^2$. Now using the familiar formula $a^2 = b^2 \implies a = \pm b$ to settle the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Sum of binomial coefficients involving $n,p,q,r$
Sum of binomial product $\displaystyle \sum^{n}_{r=0}\binom{p}{r}\binom{q}{r}\binom{n+r}{p+q}$
Simplifying $\displaystyle \frac{p!}{r!\cdot (p-r)!} \cdot \frac{q!}{r!\cdot (q-r)!}\cdot \frac{(n+r)!}{(p+q)! \cdot (n+r-p-q)!}$.
Could some help me with this, thanks
| The proposed identity can be derived (putting $r=s$) from this more general one
$$
\sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( \matrix{
m - \left( {r - s} \right) \cr
k \cr} \right)\left( \matrix{
n + \left( {r - s} \right) \cr
n - k \cr} \right)\left( \matrix{
r + k \cr
m + n \cr} \right)} = \left( \matrix{
r \cr
m \cr} \right)\left( \matrix{
s \cr
n \cr} \right)\quad \quad \left| \matrix{
\,{\rm 0} \le {\rm integer \, }m,n \hfill \cr
\;{\rm real}\;r,s \hfill \cr} \right.
$$
reported in "Concrete Mathematics" - pag. 171, and which is demonstrated through the following passages:
$$
\begin{gathered}
\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
k \\
\end{gathered} \right)\left( \begin{gathered}
n + \left( {r - s} \right) \\
n - k \\
\end{gathered} \right)\left( \begin{gathered}
r + k \\
m + n \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\begin{subarray}{l}
\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right) \\
\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right)
\end{subarray}} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
k \\
\end{gathered} \right)\left( \begin{gathered}
n + \left( {r - s} \right) \\
n - k \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n - j \\
\end{gathered} \right)\left( \begin{gathered}
k \\
j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\begin{subarray}{l}
\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right) \\
\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right)
\end{subarray}} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
k \\
\end{gathered} \right)\left( \begin{gathered}
k \\
j \\
\end{gathered} \right)\left( \begin{gathered}
n + \left( {r - s} \right) \\
n - k \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\begin{subarray}{l}
\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right) \\
\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right)
\end{subarray}} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
j \\
\end{gathered} \right)\left( \begin{gathered}
m - \left( {r - s} \right) - j \\
k - j \\
\end{gathered} \right)\left( \begin{gathered}
n + \left( {r - s} \right) \\
n - k \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
j \\
\end{gathered} \right)\left( \begin{gathered}
m + n - j \\
n - j \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
j \\
\end{gathered} \right)\left( \begin{gathered}
m + n - j \\
m \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m + n - j \\
\end{gathered} \right)} = \hfill \\
= \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,m + n} \right)} {\left( \begin{gathered}
m - \left( {r - s} \right) \\
j \\
\end{gathered} \right)\left( \begin{gathered}
r \\
m \\
\end{gathered} \right)\left( \begin{gathered}
r - m \\
n - j \\
\end{gathered} \right)} = \hfill \\
= \left( \begin{gathered}
r \\
m \\
\end{gathered} \right)\left( \begin{gathered}
s \\
n \\
\end{gathered} \right) \hfill \\
\end{gathered}
$$
consisting in:
*
*Vandermonde de-convolution
*shift of the 4th binomial
*"trinomial revision"
*sum on $k$ by Vandermonde convolution
*symmetry
*"trinomial revision"
*sum on $j$ by Vandermonde convolution
| {
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"url": "https://math.stackexchange.com/questions/2088564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
$a\sqrt{b} + b\sqrt{c} + c\sqrt{d} + d\sqrt{a} \le 4$ I've got stuck at this problem which I found some days ago in a book about inequalities.
If $a, b, c, d ∈ [0, +\infty)$ and $a+b+c+d=4$, then prove
$$a\sqrt{b} + b\sqrt{c} + c\sqrt{d} + d\sqrt{a} \le 4$$
I thought about Cauchy-Buniakowsky-Schwartz inequality but it didn't worked:
$$(a^2 + b^2 + c^2 + d^2)(b+c+d+a) \ge (a\sqrt{b} + b\sqrt{c} + c\sqrt{d} + d\sqrt{a})^2$$
I'm looking for some hints, not the entire solution.
Thanks!
| Because by AM-GM $\sum\limits_{cyc}a\sqrt{b}\leq\sum\limits_{cyc}\frac{a+ab}{2}=2+\frac{(a+c)(b+d)}{2}\leq4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$?
$\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$.
I think this is a simple exercise, but I get this:
$(n+1)^2+1=n^2+2n+2$.
$n^2+2n+2 = (n^2+1)+(2n+1)$
then $\gcd(n^2+1, (n+1)^2+1)=\gcd(n^2+1, 2n+1)$
and $\displaystyle n^2+1 = \frac{n(2n+1)}{2}+\left(-\frac{n}{2}+1\right)$
then $\gcd(n^2+1, 2n+1)=\gcd(2n+1, \frac{n}{2}-1)$.
But gcd is on integer numbers and $\dfrac{n}{2}-1$ is not always an integer number, so, I need some help?
| Let $d=\gcd(n^2+1, (n+1)^2+1)$
So $d$ divides their difference
$d|2n+1$ so by multiplication with $n$ one gets $d|2n^2+n$
But $d|2n^2+2$ so $d|n-2$ so $d|2n-4$ so $d|5$
By considering all the residues modulo $5$ of $n^2+1$ and $(n+1)^2+1$ we can see that $\gcd$ is $5$ if $n\equiv2$ mod $5$, and $1$ otherwise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Principal ideal by considering a product How can I show by considering the product of $\zeta_7^3 + \zeta_7^2 +1$ with $\zeta_7^3 + \zeta_7 +1$ that $(2, \zeta_7^3 + \zeta_7^2 +1 )$ is a principal ideal of $O_{Q(\zeta_7)}$?
| Let $\zeta=\zeta_7$. Multiply the numbers together. Doing so gives you the following:
\begin{align*}
(\zeta^3+\zeta^2+1)(\zeta^3+\zeta+1)&=\zeta^6+\zeta^5+\zeta^3+\zeta^4+\zeta^3+\zeta+\zeta^3+\zeta^2+1\\&=(\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1)+2\zeta^3\\
&=2\zeta^3
\end{align*}
$\zeta$ is a unit in this ring, so $2\in (\zeta^3+\zeta^2+1)$. In other words, $(2,\zeta^3+\zeta^2+1)=(\zeta^3+\zeta^2+1)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
then maximum and minimum value of $x+y$ if $x,y\in R$ and $x^3+y^3=2\;,$ then maximum and minimum value of $x+y$
using $\displaystyle \frac{x^3+y^3}{2}\geq \left(\frac{x+y}{2}\right)^3$
So $(x+y)^3\leq 2^3$ so $x+y\leq 2$
could some help me to find minimum value, thanks
| Note that
Since
$$
3x^2+3y^2y'=0
$$
we have
$$
y'=-\frac{x^2}{y^2}
$$
Therefore,
$$
\begin{align}
0
&=(x+y)'\\[6pt]
&=1+y'\\
&=1-\frac{x^2}{y^2}\\
\end{align}
$$
At $x=y=1$, we get a maximum of $2$.
$x=-y$ doesn't happen, but $\frac xy\to-1$ as $x\to\pm\infty$. Since $xy\le\frac{x^2+y^2}2$, we have $x^2-xy+y^2\ge\frac{x^2+y^2}2$. Therefore,
$$
\begin{align}
x+y
&=\frac{x^3+y^3}{x^2-xy+y^2}\\
&\le\frac4{x^2+y^2}\\[6pt]
&\to0
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2094891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Number of positive integer solutions of $\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}$ for distinct primes $p$ and $q$
Let $p$ and $q$ be distinct primes. Then find the number of positive integer solutions of the equation $$\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}$$
We get $pq=\frac{xy}{x+y}$
Now $x+y$ must divide $xy$ as L.H.S. is a positive integer with two prime factors but how do we make sure the same on R.H.S. ?
Given options are $3$ or $4$ or $8$ or $9$.
| Since $x$ and $y$ are positive, we have $\frac1x<\frac1{pq}$ and $\frac1y<\frac1{pq}$, which implies $x>pq$ and $y>pq$. This suggests the substitution $x:=pq+a$ and $y:=pq+b$ with positive integers $a$ and $b$.
Under this substitution, the given equation can be rewritten into the equivalent equation $ab=p^2q^2$. As $p^2q^2$ has nine positive divisors ($1$, $p$, $p^2$, $q$, $pq$, $p^2q$, $q^2$, $pq^2$, and $p^2q^2$), there are exactly nine positive integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2095007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Solve the equation ${\sqrt{4x^2 + 5x+1}} - 2{\sqrt {x^2-x+1}} = 9x-3$ I tried factoring the expression inside the square root, but that does not seem to help. Squaring the equation makes it even more terrible.
Can anyone provide a hint about what should be done?
| Rewrite your equation so that there is only one square root on each side. We get
$$\sqrt{4x^2+5x+1}=2\sqrt{x^2-x+1}+9x-3.$$
Squaring both sides we get
$$
\begin{align}
4x^2+5x+1&=4(x^2-x+1)+4(9x-3)\sqrt{x^2-x+1}+(81x^2-54x+9)\\
&=85x^2-58x+13+4(9x-3)\sqrt{x^2-x+1}\end{align}.$$
We get
$$81x^2-63x+12=-4(9x-3)\sqrt{x^2-x+1}.$$
Dividing by $3$, we get
$$27x^2-21x+4=-4(3x-1)\sqrt{x^2-x+1}.$$
Apply factoring at the LHS, we get
$$(9x-4)(3x-1)=-(3x-1)\cdot 4\sqrt{x^2-x+1}$$
Hence,
$$(9x-4)(3x-1)+(3x-1)\cdot 4\sqrt{x^2-x+1}=0.$$
Thus,
$$(3x-1)\cdot\Big[9x-4)+ 4\sqrt{x^2-x+1}\Big]=0.$$
Either
$x=\frac{1}{3}$ or
$$9x-4=-4\sqrt{x^2-x+1}.$$
Squaring again to both sides, we get
$$81x^2-72x+16=16(x^2-x+1).$$
We get
$$65x^2-56x=0.$$
Either $x=0$ or $x=\frac{56}{65}.$ Only $x=\frac{1}{3}$ satisfies the original equation.
| {
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"answer_count": 4,
"answer_id": 1
} |
Inequality related to sum of reciprocals: $\sum_{k=1}^{n} {\frac {1} {k^2}} > \frac {3n}{2n+1}$?
For every integer $n>1$, prove that :
$\sum_{k=1}^{n} {\frac {1} {k^2}} > \frac {3n}{2n+1}$
I don't seem to find any clue on how to relate the left side of the inequality to the right side.
I tried a little bit of AM-GM on the set {$\frac {1}{1^2}, \frac {1}{2^2},..,\frac {1}{n^2}$} :
$\sum_{k=1}^{n} {\frac {1} {k^2}} \geq n(\frac {1} {{(n!)}^2})^{1/n}$
Is that something helpful?
How to proceed ?
| for $n=2$ we have we have $$1+\frac{1}{4}>\frac{6}{5}$$ this is true since we have $$25>24$$
now we assume that is true:
$$\sum_{k=1}^n\frac{1}{k^2}>\frac{3n}{2n+1}$$
and we have to prove that
$$\sum_{k=1}^{n+1}\frac{1}{k^2}>\frac{3(n+1)}{2(n+1)+1}$$
we show that
$$\sum_{k=1}^{n+1}\frac{1}{k^2}>\frac{3n}{2n+1}+\frac{1}{(n+1)^2}>\frac{3(n+1)}{2(n+1)+1}$$
this is true since $$\frac{3n}{2n+1}+\frac{1}{(n+1)^2}-\frac{3(n+1)}{2(n+1)+1}={\frac {n \left( 2+n \right) }{ \left( 2\,n+1 \right) \left( n+1
\right) ^{2} \left( 2\,n+3 \right) }}
>0$$ since $n>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\int_{-1}^{1} (1-x^2)^n dx = \frac{(n!)^2 2^{2n + 1}}{(2n+1)!}$ Prove that :
$$\int_{-1}^{1} (1-x^2)^n dx = \frac{(n!)^2 2^{2n + 1}}{(2n+1)!}$$
| The trick is induction together with integration by parts. Denote our integral by $I_n$. Then
$$
I_{n+1}=\int_{-1}^1(1-x^2)^n(1-x^2)\text{d}x=I_n-\int_{-1}^1 x^2(1-x^2)^n\text{d}x.
$$
Integrating by parts by $u=x$ and $v'=x(1-x^2)^n$ we arrive at the recurrence $$I_{n+1}=\frac{2n+2}{2n+3}I_n.$$
Therefore $$I_{n+1}=\frac{2(n+1)\cdot 2(n+1)(n!)^2\cdot 2^{2n+1}}{2(2n+3)(n+1)(2n+1)!}=\frac{\bigl((n+1)!\bigr)^2\cdot 2^{2n+3}}{(2n+3)!},$$ which is the induction step. I leave to you the verification of the formula for $n=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to express $\sqrt{2-\sqrt{2}}$ in terms of the basis for $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ I've already shown that $\sqrt{2-\sqrt{2}}\in\mathbb{Q}(\sqrt{2+\sqrt{2}})$ but I want to write $\sqrt{2-\sqrt{2}}=a+b\alpha+c\alpha^2+d\alpha^3$ where $\alpha=\sqrt{2+\sqrt{2}}$. By squaring $\sqrt{2-\sqrt{2}}$ I get a system of four equations with four variables, but everything I try (Sage, for instance) takes too long to solve it. Is there a simpler way to write it in terms of the basis?
| There's a trigonometry trick for this particular case.
Let $\alpha=\sqrt{2+\sqrt{2}}$; then
$$
\alpha=2\sqrt{\frac{1+\sqrt{2}/2}{2}}=2\cos\frac{\pi}{8}
$$
whereas
$$
\beta=\sqrt{2-\sqrt{2}}=2\sqrt{\frac{1-\sqrt{2}/2}{2}}=2\sin\frac{\pi}{8}=
2\cos\left(\frac{\pi}{2}-\frac{\pi}{8}\right)=
2\cos\frac{3\pi}{8}
$$
Now
$$
\cos3t=4\cos^3t-3\cos t
$$
so
$$
\beta=2\left(4\frac{\alpha^3}{8}-3\frac{\alpha}{2}\right)=
\alpha^3-3\alpha
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Computing $ \lim \limits_{x \rightarrow \infty} \left(1+\frac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$ For $a \geq 0 $, the following limit
$$ L = \lim_{x \rightarrow \infty} \left(1+\dfrac{1}{x} \right)^{\sqrt{ax^2 +bx+c}}$$
can be computed by applying L'Hopital rule as follows
$$L = \exp \left(\lim_{x \rightarrow \infty} \frac{\log \left(1+\dfrac{1}{x} \right)}{\frac{1}{\sqrt{ax^2 +bx+c}}}\right) = \exp \left(\lim_{x \rightarrow \infty} \dfrac{\frac{-1}{x(x+1)}}{\frac{-(2ax+b)}{2(ax^2 +bx+c)^{3/2}}}\right) =$$
$$ = \exp \left(\lim_{x \rightarrow \infty} \dfrac{2(ax^2 +bx+c)^{3/2}}{x(x+1)(2ax+b)}\right) = \exp \left(\lim_{x \rightarrow \infty} \dfrac{2\left(a+ \frac{b}{x}+\frac{c}{x^2}\right)^{3/2}}{1\cdot\left(1+\frac{1}{x}\right)\left(2a+\frac{b}{x}\right)}\right) = \exp(\sqrt{a}).$$
I am wondering if exists another method to compute this limit.
Thanks for any hint!
| I unusually use this general method:
Let $\displaystyle \lim_{x \to x_o}u(x)=1$ and $\displaystyle \lim_{x\to x_o}V(x)=\infty$, then
$\displaystyle \lim_{x \to x_o}u^V=\lim_{x \to x_o}\Big[ \Big(1+(u-1) \Big)^{\frac{1}{u-1}}\space \Big]^{(u-1)V}=\Big[ \lim_{x\to x_o}\Big( 1+(u-1)\Big)^{\frac{1}{u-1} \space} \Big]^{\displaystyle\lim_{x\to x_o}(u-1)V}=e^{\displaystyle\lim_{x\to x_o}(u-1)V}$
Now let's apply what we got above to our problem. Here $ \{ u(x) = 1+\frac{1}{x}\} \to 1 $ and $\{V(x)=\sqrt{ax^2+bx+c}\} \to \infty$ as $x\to \infty$.
$\displaystyle \lim_{x\to \infty}\left(1+\frac{1}{x} \right)^{\sqrt{ax^2+bx+c}}=[1^\infty]=\exp \left( \lim_{x\to \infty}\left( 1+\frac{1}{x}-1\right)\sqrt{ax^2+bx+c}\right)=\exp \left( \lim_{x\to \infty}\frac{1}{x}\sqrt{ax^2+bx+c}\right)=\exp \left(\lim_{x\to\infty}\sqrt{a+\frac{b}{x}+\frac{c}{x^2}}\right)=\exp \sqrt a=e^{\sqrt a}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Calculate volume between two geometric figures I have a figure C that is defined as the intersection between the sphere $x^2+y^2+z^2 \le 1 $ and the cyllinder $x^2+y^2 \le \frac{1}{4}$.
How should i calculate the volume of this figure?
| The given cylinder and the given sphere share the sections given by $z=\pm\frac{\sqrt{3}}{2}$. If $|z|\leq\frac{\sqrt{3}}{2}$, the area of the section is $\frac{\pi}{4}$. If $\frac{\sqrt{3}}{2}\leq|z|\leq 1$, the area of the section is $\pi(1-z^2)$.
Integrating $1$ on sections, we get that the volume is given by:
$$ \frac{\pi}{4}\sqrt{3}+2\pi\int_{\sqrt{3}/2}^{1}(1-z^2)\,dz =\color{red}{\frac{\pi}{6}(8-3\sqrt{3})}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find the limit $\lim _{ x \to 0} \frac{\sqrt[3]{1+6x+3x^2+3x^3+3x^4}-\sqrt[4]{1+8x+4x^2+4x^3-2x^4}}{6x^2}$ I have been trying to find the limit,
$$\lim _{ x \to 0} \frac{\sqrt[3]{1+6x+3x^2+3x^3+3x^4}-\sqrt[4]{1+8x+4x^2+4x^3-2x^4}}{6x^2}$$
and sort of succeeded. But my $0$ answer doesn't converge with what Wolfram says which is $1/3$. Therefore, I would really appreciate it, if you could give me the right answer and drop a hint on the solution.
| Without Taylor nor the generalized Binomial theorem:
Use the identity
$$a^{12}-b^{12}=\\
(a-b)(a^{11}+a^{10}b^{1}+a^{9}b^{2}+a^{8}b^{3}+a^{7}b^{4}+a^{6}b^{5}+a^{5}b^{6}+a^{4}b^{7}+a^{3}b^{8}+a^{2}b^{9}+ab^{10}+b^{11}).$$
By multiplying/dividing by the conjugate "dodecanomial", the numerator becomes the polynomial
$$(1+6x+3x^2+3x^3+3x^4)^4-(1+8x+4x^2+4x^3-2x^4)^3=\\
(1+24x+228x^2+\cdots81x^{16})-(1+24x+204x^2+\cdots-8x^{12}),$$
while the denominator is a sum of $12$ rational expressions that tend to $1$. So after simplification, the limit is
$$\frac{228-204}{6\cdot12}=\frac13.$$
Alternatively, you can pull a factor $(1+2x)$ from both radicals and get
$$\frac{\sqrt[3]{1+\frac{-9x^2-5x^3+3x^4}{(1+2x)^3}}-\sqrt[4]{1+\frac{-20x^2-28x^3-18x^4}{(1+2x)^4}}}{6x^2}.$$
It is not hard to show that the Taylor development will yield
$$\frac{(1-\frac93x^2+\cdots)-(1-\frac{20}4x^2+\cdots)}{6x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Finding the 5th order taylor polynomial of a function without brute force I am currently stuck on this 5th order taylor polynomial question for almost an hour now, it's quite difficult if you brute force it but there was a hint given today by my tutor that there is an easier way to solve for the 5th taylor polynomial of this function below if you do some substitutions. Should I be substituting $x^2$ for $x$ and then find the 5th taylor polynomial and then substituting $x^2$ back in at the end or am I completely off with this suggestion?:$$f(x) = \frac{\sin(x^2)}{x^2}$$
| Building on the hint $u=x^2$.
$\sin(u)=u-\frac{u^3}{3!}+\frac{u^5}{5!}$
$\frac{\sin(u)}{u}=1-\frac{u^2}{3!}+\frac{u^4}{5!}$
Now back substituting in $x^2$
$\frac{\sin(x^2)}{x^2}=1-\frac{x^4}{3!}+\frac{x^8}{5!}+...$
Edit
Using $x=x^2$ as the substitution as op had.
$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...$
$\frac{\sin(x)}{x}=1-\frac{x^2}{3!}+\frac{x^4}{5!}+...$
Now back substituting $x^2$
$\frac{\sin(x^2)}{x^2}=1-\frac{(x^2)^2}{3!}+\frac{(x^2)^4}{5!}...$
It doesn't matter what you pick as your substitution. It may get confusing if you use the substitution $x=x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$a+b+c+ab+bc+ca+abc=1000$ find the minimum value of $a+b+c$ Given $a+b+c+ab+bc+ca+abc=1000$.
Find the minimum value of $a+b+c$.
Now we are considering $a$, $b$, $c$ to be integers and here in lies the pertinent problem as I could not get an answer in integers but in fractions.
| I am assuming here $a=b=c$ since $1000$ is the cube of $10$.
$$(a+1)(b+1)(c+1) = 1000$$
now, $(a+1)^3 = 1000$;
$$a+1 = 10\iff
a=9$$
so, $a+b+c = 3*9 = 27$.
Just that we can get the integer value but I cannot say $27$ is the minimum value for $a+b+c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Binomial theorem expansion. Why do the powers flip in certain circumstances? Okay given an expression like $(x+y)^5$
we have terms
$x^ky^{n-k}$
So we have ${5 \choose 0} x^0y^5 + {5 \choose 1} x^1y^4+ {5 \choose 2} x^2y^3+ {5 \choose 3} x^3y^2+ {5 \choose 4} x^4y^1+ {5 \choose 5} x^5$
However when we have $(3x^2+y)^5$
we supposedly have $x^{n-k}y^k$ and we get:
${5 \choose 0} (3x^2)^5y^0+ {5 \choose 1} (3x^2)^4y^1+{5 \choose 2} (3x^2)^3y^2+{5 \choose 3} (3x^2)^2y^3 + {5 \choose 4} (3x^2)y^4+ {5 \choose 5} (3x^2)^0y^5$
which ends up as:
$243x^{10}y^0+ 625x^8y+27x^6y^2+9x^4y^3+3x^2y^4+y^5$
somehow my coefficients are off for the second term : it's apparently $5$ times $3^4$. How in the heck does that arise do we always multiply coefficients by the power n, in this case 5 and secondly my biggest question is why on earth is n-k on x instead of y?
What would happen in a situation where we have $(4x^3+5y^2)^3$ which term would get n-k and would I have to multiply some coefficients by 3?
What if we had $(x^2+7y^3)^2$? would n-k go on the $7y^3$?
| Expansion of the expression $\left(x+y\right)^n$ gives :
$$\left(x+y\right)^n=\sum_{k=0}^n{n\choose k}x^ky^{n-k}$$
but you can also write it as :
$$\left(x+y\right)^n=\sum_{k=0}^n{n\choose k}x^{n-k}y^k$$
simply because addition and multiplication are commutative.
So you can put the exponent $n-k$ to whatever term you wish and put the exponent $k$ to the other one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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