Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Is $f(x)= \frac{1}{x} - \frac{1}{e^x-1}$ monotonic? We have $\displaystyle f(x)= \frac{1}{x} - \frac{1}{e^x-1}$, additionally $f(0)=\frac{1}{2}$. Determine whether $f(x)$ is monotonic.
I tried to do this by checking if $f'(x)<0$, however it does not look very helpful. I don't know if it's tricky, or I am just blind to something obvious here.
Thanks for any hints in advance!
| Observe that
$$f(x) = \frac{1}{x} - \frac{1}{2} \coth \frac{x}{2} + \frac{1}{2}.$$
Hence
$$f'(x) = -\frac{1}{x^2} + \frac{1}{4} \mathrm{cosech}^2 \frac{x}{2}.$$
So it is sufficient to prove that $4\sinh^2 \frac{x}{2} \ge x^2$ for $x \ge 0$. This follows from $\sinh y \ge y$ for $y \ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Tangent line problems Problem 1
Find common tangent to the curve:
$y+x^2=-4$ and $x^2+y^2=4$.
My idea:
Let $t1... y=ax+b$ is a tangent line to the first curve.
Let $t2... y=cx+d$ is a tangent line to the second curve.
But we are seracing for common tangent so that mean that $a=c$ and $b=d$.
By using implicit differentiation we can find $a$ and $b$.
$$y+x^2=-4$$
$$y'+2x=0$$
$$y'=-2x$$
$$x^2+y^2=4$$
$$2x+2yy'=0$$
$$x+yy'=0$$
$$y'=\frac{-x}{y}$$
$$-2x=\frac{-x}{y} \to y=\frac{1}{2},$$$x$ is any real number
Problem 2
For which value of the coefficients $a$, $b$ and $c$ $\in$ R is the x-coordinate axis tangent to the curve?
$y=ax^2+bx+c$
My idea:
Curve and the tangent line must have just one common point. Our curve is a parabola and the number of dots on x axis is a number of solution of quadratic equation. So we need $a,b,c$ to be equal to $b^2-4ac=0$
Problem 3
Find a line that is tangent to the curve
$y = x^4 - 2x^3 - 3x^2 + 5x + 6$
in at least two points .
My idea:
Let that tangent be $t=ax+b$.
If we that line be a tangent line at two point to the curve equation $y(x)-t(x)=0$ must have at least two (not equal) solution.
$$x^4 - 2x^3 - 3x^2 + 5x + 6-(ax+b)=0$$
$$x^4 - 2x^3 - 3x^2 + 5x + 6=ax+b$$
$$a=5 , b=6$$
$$x^4 - 2x^3 - 3x^2=0$$ $$x_1=0, x_2=-1, x_3=3$$
Problem 4
I am trying to find the number of tangents to a curve that all pass through the origin. The curve's equation is $y=x^3+x^2−22x+20.$ I also need to find the equation of said tangents.
My work:
Let's use formula for tangent line:
$$y-y_0=y'(x_0)(x-x_0)$$
We know that $x=0$ and $y=0$.
Let's find $y'(x_0)$:
$y'(x_0)=3x_0^2+2x_0-22$
Plugin what we know:
$$-y_0=(3x_0^2+2x_0-22)*-x_0$$
We aslo know that $y_0=x_0^3+x_0^2-22x_0+20$
$$-x_0^3-x_0^2+22x_0-20=-3x_0^3-2x_0^2+22x_0$$
$$2x_0^3+x_0^2-20=0$$
Only real solution is 2. $x_0=2$ and $y_0=-12$.
Let tangent line be $y=kx+l$.
$k=y'(2)=-6$
$-12=-6*2+l\to l=0$
But we want l to be zero.
Our solution is $y=-6x$.
| Another solution for problem 3:
Let solve the more general problem:
Let $p(x)$ be a polynomial of degree $n\ge 4$. Find bitangents to the
curve $y=p(x)$.
Let $y-p(x_0)=p'(x_0)(x-x_0)$ be the equation of the tangent at the point $(x_0,p(x_0))$. It will be a bitangent if the abscissae equation for the intersection points of this line with the curve has another double root.
We'll write this equation using the Taylor's expansion of $p(x)$ at $x=x_0$:
$$p(x)=p(x_0)+p'(x_0)(x-x_0)+\frac{p''(x_0)}{2}(x-x_0)^2+\dots+\frac{p^{(n)}(x_0)}{n!}(x-x_0)^n.$$
The equation for the intersection points is thus
$$\frac{p''(x_0)}{2}(x-x_0)^2+\dots+\frac{p^{(n)}(x_0)}{n!}(x-x_0)^n$$
and the other intersection points abscissae are roots of
$$\frac{p''(x_0)}{2}+\frac{p'''(x_0)}{6}(x-x_0)+\dots+\frac{p^{(n)}(x_0)}{n!}(x-x_0)^{n-2}$$
So the abscissae of the other points of contact with the curve are the multiple roots of this equation which are different from $x_0$.
In the present case, as $n=4$, you get a quadratic equation, which must have a double root in $X=x-x_0$:
$$6x_0^2-6x_0-1+(4x_0-3)X+X^2=0,$$
whence the condition
$$(4x_0-3)^2-4(6x_0^2-6x_0-1)=-8x_0^2+13=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Derivation and tangent problem set Problem 1
On the curve $y=\frac{1}{1+x^2}$ find a point in which tangent line is parallel to the horizontal axis.
My idea:
Let's find $y'$.
$$y'=\frac{-2x}{(1+x^2)^2}$$
If we want a tangent line to be parallel to x-axis it must look like this $t...y=kx+l$ such that $k=0$.
Let's find for which $x$ we have that situation.
$$y'=\frac{-2x}{(1+x^2)^2}=0$$
$$x=0\to y=1$$
$$(x,y)=(0,1)$$
Problem 2
Determine the angle of the curve
$y^2 - 3x^2 + x + 1 = 0$ and
$xy^2 - 1 = 0$
intersect in the first quadrant
My work:
$y^2 - 3x^2 + x + 1 =xy^2 - 1$
One solution for $x>0$ and $y>0$ is $x=y=1$.
Let's now find derivative of easch function at that point.
$$2xyy'=-y^2$$
$$y'=\frac{-y^2}{2xy}=\frac{-1}{2}$$
$$2yy'=6x-1$$
$$y'=\frac{6x-1}{2y}=\frac{5}{2}$$
And the angle is $\alpha =arctg|\frac{k_1-k_2}{1+k_1*k_2}|$ where $k_1=\frac{-1}{2}$ and $k_2=\frac{5}{2}$
Problem 3
Find the angle of intersection of the tangents to the curve
$3x^2 - 6x + 2y^2 - 3 = 0$
from the point $T( 1,3 )$?
My work:
First (1,3) isn't on the curve.
We will need that tangent equation:
$$y-y_0=y'(x_0)(x-x_0)$$
From the task we know that $x=1$ and $y=3$.
Let's try to find $x_0$ and $y_0$.
Using implict differentation we will find $y'$.
$$6x-6+4yy'=0$$
$$3x-3+2yy'=0$$
$$2yy'=-3x+3$$
$$y'=\frac{-3x+3}{2y}$$
Plugin that in equation:
$$3-y_0=\frac{-3x_0+3}{2y_0}(1-x_0)$$
$$3-y_0=\frac{-3x_0+3}{2y_0}+\frac{-3x_0^2-3x_0}{2y_0}$$
$$3-y_0=\frac{-3x_0^2+3-6x_0}{2y_0}$$
$$6y_0-2y_0^2=-3x_0^2+3-6x_0$$
But I can't get something nice...Is my problem 1 and 2 ok and please help me with 3
| The solutions to problems 1) and 2) are correct.
For problem 3) note that you are searching the straight lines of equation $(y-3)=m(x-1)$ that are tangent to the curve $3x^2-6x+2y^2-3=0$. This means that you want the values of $m$ such that the system:
$$
\begin{cases}
(y-3)=m(x-1)\\
3x^2-6x+2y^2-3=0
\end{cases}
$$
has one double solution, and this is done if the discriminant of the second degree equation that we obtain substituting from the first to the second equation is null. This gives an equation ( of second degree) in $m$ and solving this equation you find the slopes of the tangent lines.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $N$ with $N^2=10^4M+N$ I'm a high school student in France and I participated in a math olympiad, and there was a question which I found impossible to solve. Maybe they are here some people who can help me:
We take a number $N$ which can be written by numbers $a,b,c$ and $d$ in
the same order.
so $N=a(10^3)+b(10^2)+c10+d$. And $M\in \mathbb{Z}$ with
$N^2=M(10^4)+N$
Find $M$.
| $$N^2-N=10^4 \cdot M$$
$$N(N-1)=2^4\cdot 5^4\cdot M$$
$N(N-1)=16\cdot 625\cdot M$
$\gcd(N,N-1)=1$
Then 1)$$N=2^4k; N-1=5^4l, k,l \in \mathbb N$$ or 2)$$N-1=2^4k; N=5^4l, k,l \in \mathbb N$$
1) $10^4\le N-1=5^4l \le 10^5 \Rightarrow 2\le l \le 15$ and $16|N$
2)$10^4\le N=5^4l \le 10^5 \Rightarrow 2\le l \le 15$ and $16|(N-1)$
1) $l=2, N-1=1250$, but $16 \not |N=1251$
$l=3, N-1=1875$, but $16 \not |N=1876$
and so on ...
$l=15, N-1=9375, N=9376=586 \cdot 16$
$$N^2=9376^2=87909376=10^4\cdot 8790+9376$$
$$M=8790$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1727583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that there are no positive integer solutions $(x, y)$ to the equation $x^2 - y^2 = 1$ Prove the following:
Theorem. There are no positive integer solutions $(x, y)$ to the equation
$x^2-y^2=1$.
| Here is another take.
$x^2-y^2=1$ implies $x^2=y^2+1$.
$x^2=y^2+1 > y^2$ implies $x>y$, that is, $x\ge y+1$.
But then $x^2 \ge (y+1)^2 = y^2+2y+1 \ge y^2+2+1 > y^2+1$, since $y \ge 1$.
Thus, $x^2 > y^2+1$, contradiction.
Another, more succinct way to express this is:
$$
y^2 < y^2+1 < (y+1)^2
$$
Thus, $y^2+1$ is strictly between two consecutive squares and so cannot be a square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $\sin y+\cos y=\frac{1}{2}$ then find $\frac{\sin^3y}{\cos^2y}+\frac{\cos^3y}{\sin^2y}$
If $$\sin y+\cos y=\frac{1}{2} \tag{1}$$ Then find $$x=\frac{\sin^3y}{\cos^2y}+\frac{\cos^3y}{\sin^2y} \tag{2}$$
Given that $$\sin y=\frac{1}{2}-\cos y$$
Squaring both sides we get
$$8\cos^2y-4\cos y-3=0$$
Hence
$$\cos y=\frac{1-\sqrt{7}}{4}$$ and so $$\sin y=\frac{1+\sqrt{7}}{4}$$
substituting the above values in Eq $(2)$ and using the Binomial Theorem we get
$$x=\frac{\sin^5y+\cos^5y}{\sin^2y \cos^2y}$$ so
$$x=\frac{\dfrac{\left(1-\sqrt{7}\right)^5+\left(1+\sqrt{7}\right)^5}{4^5}}{\dfrac{36}{256}}$$
Now by the Binomial theorem $$\left(1-\sqrt{7}\right)^5+\left(1+\sqrt{7}\right)^5=2\left(1+10 \times 7+5 \times 49\right)=632$$
So $$x=\frac{\dfrac{632}{4^5}}{\dfrac{36}{256}}=\frac{79}{18}$$
I feel this is a very lengthy approach; can I get a better approach?
|
Can i get any better approach
How about the following way?
Let $s=\sin y,c=\cos y$.
Squaring the both sides of $s+c=1/2$ gives
$$s^2+2sc+c^2=\frac 14\quad\Rightarrow\quad sc=-\frac 38.$$
Hence,
$$\begin{align}\frac{s^3}{c^2}+\frac{c^3}{s^2}&=\frac{s^5+c^5}{(sc)^2}\\\\&=\frac{(s^2+c^2)(s^3+c^3)-s^2c^2(s+c)}{(sc)^2}\\\\&=\frac{s^3+c^3-(sc)^2/2}{(sc)^2}\\\\&=\frac{(s+c)(s^2-sc+c^2)-(sc)^2/2}{(sc)^2}\\\\&=\frac{(1/2)(1-sc)-(sc)^2/2}{(sc)^2}\\\\&=\frac{79}{18}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Seeking combinatorial or group theoretic proof for permutation identity While working on another problem, I found the following combinatorial equality, but I got it analytically, and I'm curious to find a counting argument.
Fix $n$ a positive integer. For $n_1\leq n_2\leq \cdots \leq n_k$ with $\sum n_i=n$, let $s_{n_1,\dots,n_k}$ be the number of permutations in $\Sigma_n$ with sorted cycle lengths $n_1,n_2,\dots,n_k$.
Then show:
$$\sum_{n_1,\dots,n_k} 4^{k-1}s_{n_1,\dots,n_k}=n\cdot n!$$
where the sum is restricted to the case where all $n_i$ are odd.
I suppose we could just write $t_k$ as the number of permutations in $\Sigma_n$ composted of $k$ odd cycles, and write it as $\sum_{k} 4^{k-1}t_k = n\cdot n!$.
For example, for $n=5$, the possible permutations of signature $s_{5}=4!$, $s_{1,1,3}=2\binom{5}{2}=20$, $s_{1,1,1,1,1}=1$ so the sum is:
$$4^0\cdot24+4^2\cdot 20 + 4^4\cdot 1=600=5\cdot 5!$$
I have a proof of this, which is gross - it involves substituting the power series for $\theta=\arctan x$ into the power series for $\sin 4\theta$ and $\cos 4\theta$ (for odd and even $n$, respectively.) That seems unpleasant, so I am seeking a more direct combinatorial argument.
One thing I considered was the possibility of using the identity:
$$\sum_{n=0}^{m} n\cdot n! =(m+1)!-1$$
| Permutations with odd cycles only and cycle count marked have
the species
$$\mathfrak{P}(\mathcal{U}\mathfrak{C}_{=1}(\mathcal{Z})
+ \mathcal{U}\mathfrak{C}_{=3}(\mathcal{Z})
+ \mathcal{U}\mathfrak{C}_{=5}(\mathcal{Z})
+ \mathcal{U}\mathfrak{C}_{=7}(\mathcal{Z})
+ \cdots).$$
which yields the EGF
$$G(z,u) =
\exp\left(uz+
u\frac{z^3}{3}+
u\frac{z^5}{5}+
u\frac{z^7}{7}+\cdots\right)
\\ = \exp\left(u\sum_{k\ge 0}\frac{z^{2k+1}}{2k+1}\right)
\\ = \exp\left(u\log\frac{1}{1-z}
- u\sum_{k\ge 1} \frac{z^{2k}}{2k}\right)
\\ = \exp\left(u\log\frac{1}{1-z}
- u\frac{1}{2}\log\frac{1}{1-z^2}\right).$$
Now here we have $u=4$ so we continue with
$$\frac{1}{(1-z)^4}
\exp\left(2\log(1-z^2)\right)
= \frac{(1-z^2)^2}{(1-z)^4}
= \frac{(1+z)^2}{(1-z)^2}
\\ = (1+2z+z^2) \frac{1}{(1-z)^2}.$$
We get as our result the quantity
$$\frac{1}{4} n! [z^n] G(z, 4)
= \frac{1}{4} n!
\left({n+1\choose 1} + 2{n\choose 1}
+ {n-1\choose 1}\right)
\\ = \frac{1}{4} n!
(n+1+2n+n-1) = n! \times n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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Generating function for the partition function Could someone explain what is the reasoning behind the following equality? Or maybe direct me to a proof of the following equality?
$$\sum_{n=0}^{\infty}p(n)x^n = \prod_{k=1}^{\infty}(1-x^k)^{-1}$$ where $\sum_{n=0}^{\infty}p(n)x^n$ is the generating function for the partition function.
| For the basic generating functions in term of $(1 - x^k) ^ {-1}$, we can write them out as a infinite sum.
$(1 - x) ^ {-1}$ = $1 + x + x^2 + x^3 + ....$
$(1 - x^2) ^ {-1}$ = $1 + x^2 + x^4 + x^6....$
$(1 - x^3) ^ {-1}$ = $1 + x^3 + x^6 + x^9....$
and so on.
So $\prod(1 - x^k) ^ {-1}$ is just an infinite product of
($1 + x + x^2 + x^3 + ....$)( $1 + x^2 + x^4 + x^6....$)($1 + x^3 + x^6 + x^9....$)....
The for each power term $x^n$, there is a corresponding coefficient p(n). And we need to sum up all these power terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If the quadratic equation $x^2+(2-\tan\theta)x-(1+\tan\theta)=0$ has two integral roots, If the quadratic equation $x^2+(2-\tan\theta)x-(1+\tan\theta)=0$ has two integral roots,then find the sum of all possible values of $\theta$ in interval $(0,2\pi).$
The given quadratic equation is $x^2+(2-\tan\theta)x-(1+\tan\theta)=0$.
So $x=\frac{\tan\theta-2\pm\sqrt{\tan^2\theta+8}}{2}$
I am stuck here.I do not know what is the condition for two integer roots.
| Let $t=\tan \theta$. We need that $t^2+8$ be a perfect square, so $$t^2+8=s^2\qquad(*)$$
When $\theta\in (0,2\pi)$, the value $t$ can be all real but if $t$ and $s$ must be integers we have $s=t+h$ where $h$ is a positive integer. It follows $$2ht+h^2=8\qquad (**)$$ An obvious solution of $(*)$ is $t=1$ so for $(**)$ we have to solve $$2h+h^2=8$$ whose only solution is $h=2$ but for $t>1$ there are no solutions for $(**)$
So the only values to be considered are $\frac{\pi}{4}$ and $\frac{5\pi}{4}$ which correspond to $t=\tan \theta=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Fewest steps to show $\sin^2(s) = \frac{1}{2} - \frac{1}{2} \cos(2s)$? Is there a method for getting from this statement
$
\sin^2(s)
$
to this statement
$
\frac{1}{2}\ -\ \frac{1}{2}\cos(2s)
$
in fewer steps than the following?
$\begin{align}
\cos(2s) &= \cos^2(s) - \sin^2(s) \\
\frac{1}{2} \cos(2s) &= \frac{1}{2} \cos^2(s) - \frac{1}{2} \sin^2(s) \\
-\frac{1}{2} \cos(2s) &= -\frac{1}{2} \cos^2(s) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \frac{1}{2} - \frac{1}{2} \cos^2(s) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \frac{1}{2} ( 1 - \cos^2(s) ) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \frac{1}{2} ( \sin^2(s) ) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \sin^2(s) \\
\end{align}$
| You can certainly eliminate several of the steps. I might write it as
$$\cos 2s = \cos^2 s - \sin^2 s\tag{dbl angle formula}$$
$$\cos 2s = 1-2\sin^2 s\tag{$\cos^2$ in terms of $\sin^2$}$$
$$1-\cos 2s = 2 \sin^2 s\tag{subtr both sides from $1$}$$
$$\tfrac12 - \tfrac12\cos 2s = \sin^2 s\tag{halve both sides}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$ Prove that $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$
$$\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{\frac{2}{\cos\theta} +\frac{3\sin\theta}{\cos\theta}+5\sin\theta-7\cos\theta+5}{\frac{2\sin\theta}{\cos\theta} +\frac{3}{\cos\theta}+5\cos\theta+7\sin\theta+8}$$
$$=\frac{2+3\sin\theta+5\sin\theta\cos\theta-7\cos^2\theta+5\cos\theta}{2\sin\theta+3+5\cos^2\theta+7\sin\theta\cos\theta+8\cos\theta}$$
I am stuck here.I tried to factorize numerator and denominator but does not succeed.
| Let $\cos\theta=c,\sin\theta=s$. Then,
$$\begin{align}&s(2+3s+5sc-7c^2+5c)-(1-c)(2s+3+5c^2+7cs+8c)\\&=5c^3+5cs^2-5c+3c^2+3s^2-3\\&=5c(c^2+s^2-1)+3(c^2+s^2-1)\\&=0\end{align}$$
from which the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739433",
"timestamp": "2023-03-29T00:00:00",
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How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$? How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$ ?
I know the sum converges because it is a positive sum for every $n$ and it is smaller than $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ that converges and equals $1$. I need a direction...
| \begin{align*}
\frac{1}{n(n+3)} &= \frac{1}{3n}-\frac{1}{3(n+3)} \\
\sum_{n=1}^{\infty} \frac{1}{n(n+3)} &=
\sum_{n=1}^{\infty}
\left[ \frac{1}{3n}-\frac{1}{3(n+3)} \right] \\
&=\lim_{N\to \infty}
\left[
\sum_{n=1}^{N} \frac{1}{3n}-\sum_{n=1}^{N} \frac{1}{3(n+3)}
\right] \\
&=\sum_{n=1}^{3} \frac{1}{3n}-
\lim_{N\to \infty} \sum_{n=1}^{3} \frac{1}{3(N+n)} \\
&=\frac{11}{18}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine whether $x^2 - 14x + 30 \equiv 0$ mod 1615 is solvable. If so, find its solutions... Determine whether $x^2 - 14x + 30 \equiv 0\pmod{1615}$ is solvable. If so, find its solutions.
I assume the best way to solve this is via Chinese Remainder Theorem, but first i would have to break down the mod and if there is a solution then utilize C.R.T. No quite sure how to attack this with the right numbers, any hints/help are very appreciated.
| $$x^2-14x+30\equiv 0\pmod{5\cdot 17\cdot 19}$$
$$\iff (x-7)^2\equiv 19\pmod{5\cdot 17\cdot 19}$$
$$\iff \begin{cases}(x-7)^2\equiv 19\equiv 2^2\pmod{5}\\(x-7)^2\equiv 19\equiv 6^2\pmod{17}\\(x-7)^2\equiv 19\equiv 0^2\pmod{19}\end{cases}$$
$$\iff \begin{cases}x-7\equiv \pm 2\pmod{5}\\x-7\equiv \pm 6\pmod{17}\\x-7\equiv 0\pmod{19}\end{cases}$$
I used the fact that $5,17,19$ are prime and Euclid's Lemma: e.g., if $5\mid (x-7)^2-2^2=((x-7)+2)((x-7)-2)$, then $x-7\equiv \pm 2\pmod{5}$.
Now use Chinese Remainder Theorem to find all the $4$ solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\gcd(80,8a^2+1)=1$ Show that $\gcd(80,8a^2+1)=1$
Let $\gcd(80,8a^2+1)=d$, then we have: $d|80a^2+10,80a^2\Rightarrow\ d|10$
So $d=1\ or\ 2\ or\ 5\ or\ 10$
Obviously $d$ can't be $2\ or\ 10$,but how can we show $d$ can't be 5??
|
$\gcd(80,8a^2+1)$
Factors of $80$ are $8$ and $10$.So,if $\gcd\neq0$ then $8a^2+1$ must have $8$(or $2$ if you consider prime factors) or $10$ as factor.But clearly $8a^2+1$ cannot have $8$(or $2$) or $10$ as factors.Hence they have no common factors.hence,$\gcd=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I show that $\frac {a^2}b + \frac {b^2}c + \frac {c^2}a \ge \frac {(a + b + c)(a^2 + b^2 + c^2)}{ab + bc + ca}?$ For positive real numbers $a, b, c$, show that
$$\frac {a^2}b + \frac {b^2}c + \frac {c^2}a \ge \frac {(a + b + c)(a^2 + b^2 + c^2)}{ab + bc + ca}.$$
I don't know how to solve this at all. Can you provide any hints?
| We have to prove that $\implies$
$\sum_{cyclic}\frac{a^2}{b}\geq{\frac{(a+b+c)(a^2+b^2+c^2)}{ab+bc+ca}}$
$\implies (ab+bc+ca)(\sum_{cyclic}\frac{a^2}{b})\geq(a+b+c)(a^2+b^2+c^2)$
Now expand both sides and many terms are cancelled out.Then we get the
inequality $\implies(\sum_{cyclic}\frac{ca^3}{b})\geq(ac^2+ba^2+cb^2)$
Now use the AM-GM inequality to
1.$(\frac{ca^3}{b}\;,\frac{ab^3}{c})$
2.$(\frac{ab^3}{c}\;,\frac{bc^3}{a})$
3.$(\frac{bc^3}{a}\;,\frac{ca^3}{b})$
Now the required inequality is proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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False Proof that $\sqrt{4}$ is Irrational Everyone with any basic knowledge of number theory knows the classic proof of the irrationality of $\sqrt{2}$. Curious about generalizations using elementary methods, I looked up the irrationality of $\sqrt{3}$, and found the following:
Say $ \sqrt{3} $ is rational. Then $\sqrt{3}$ can be represented as $\frac{a}{b}$, where a and b have no common factors.
So $3 = \frac{a^2}{b^2}$ and $3b^2 = a^2$. Now $a^2$ must be divisible by $3$, but then so must $a $ (fundamental theorem of arithmetic). So we have $3b^2 = (3k)^2$ and $3b^2 = 9k^2$ or even $b^2 = 3k^2 $ and now we have a contradiction.
Such a proof follows the same basic logic as the proof for $\sqrt{2}$, except for using the fundamental theorem of arithmetic to replace and generalize the trivial fact that $n$ is even if $n^2$ is even. However, when I apply this proof format to $\sqrt{4} $ (which is clearly an integer and thus rational) I get the following:
Say $ \sqrt{4} $ is rational. Then $\sqrt{4}$ can be represented as $\frac{a}{b}$, where a and b have no common factors.
So $4 = \frac{a^2}{b^2}$ and $4b^2 = a^2$. Now $a^2$ must be divisible by $4$, but then so must $a $ (fundamental theorem of arithmetic). So we have $4b^2 = (4k)^2$ and $4b^2 = 16k^2$ or even $b^2 = 4k^2 $, which implies that $b=4n$ by the fundamental theorem. Now we have a contradiction (since can note that both $a$ and $b$ are divisible by $4$ and we assumed they were coprime)
This proof is clearly false, yet I fail to see where it differs. Where does it do so?
| Just because $a^2$ is divisible by $4$, that doesn't mean $a$ is.
| {
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"url": "https://math.stackexchange.com/questions/1748624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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You cast a pair of dice. If you get a 6 and an 8 before 7 comes up twice, you win. What is the probability of winning? You cast a pair of dice. If you get a 6 and an 8 before 7 comes up twice, you win. What is the probability of winning?
What I tried was
\begin{align*}
P(X=6) & = \frac{5}{36}\\
P(X=8) & = \frac{5}{36}\\
P(X=7) & = \frac{6}{36} = \frac{1}{6}
\end{align*}
I try finding the probability of getting $6$ and $8$ and not $7$
$$P(X=6)+P(X=8)+P(X \neq 7) = \frac{5}{26} + \frac{5}{36} + \frac{5}{6} = 1.11$$
I stick since I know that the probability cannot be greater than one. The solution key says the answer is $0.5456$.
| Everything other than $6$, $7$, or $8$ is irrelevant. So we can imagine that we are repeating an experiment where the probability of $6$ is $5/16$, as is the probability of $8$, while $7$ has probability $6/16$.
Condition on the result of the first toss. Suppose it is a $6$. Then $6$'s become irrelevant, and the race is between $7$ and $8$. The first now has probability $\frac{6}{11}$ and the second has probability $\frac{5}{11}$. If we get an $8$, we win. If we get a $7$, then the probability we win is $\frac{5}{11}$. So the contribution to the probability is
$$\frac{5}{16}\left(\frac{5}{11}+\frac{6}{11}\frac{5}{11} \right).$$
There is the same contribution to the probability of winning from first toss is $8$.
Now we look at what happens if the first toss is $7$. If the next toss is $7$, we have lost. If it is $6$ or $8$, then our probability of winning is $\frac{5}{11}$. The contribution to the probability is
$$\frac{6}{16}\left(\frac{10}{16}\frac{5}{11}\right).$$
We conclude that the contribution from $6$ or $8$ first is $\frac{850}{(16)(121)}$, and the contribution from $7$ first is $\frac{300}{(256)(11)}$. Add. The official answer seems to be right.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate $\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}-xe^{\pi x}-1}{(\ln{(1+x)})^2}$? How to evaluate $\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}-xe^{\pi x}-1}{(\ln{(1+x)})^2}$?
So I think we expand to $x^2$ since the lowest term for $\ln(1+x)$ is $x$
Let $u=\arctan{(x)}$
$\lim\limits_{x\to 0}\frac{e^{\arctan{(x)}}-xe^{\pi x}-1}{(\ln{(1+x)})^2}=\lim\limits_{x\to 0}\frac{1+u+\frac{u^2}{2}+o(u^2)-(x+\pi x^2+o())-1}{x^2+o()}=\lim\limits_{x\to 0}\frac{1+x+o(x^3)+\frac{x^2+o()}{2}+o(u^2)-(x+\pi x^2+o())-1}{x^2+o()}=\frac12-\pi$
My answer is rather messy and likely incorrect. Could someone provide an easier way to solve such problem?
| Let's try to simplify the expression first. We have
\begin{align}
L &= \lim_{x \to 0}\frac{e^{\arctan x} - xe^{\pi x} - 1}{(\log(1 + x))^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{e^{\arctan x} - xe^{\pi x} - 1}{\left(\dfrac{\log(1 + x)}{x}\right)^{2}\cdot x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{e^{\arctan x} - \arctan x + \arctan x - xe^{\pi x} - 1}{1^{2}\cdot x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{e^{\arctan x} - \arctan x - 1}{x^{2}} + \lim_{x \to 0}\frac{\arctan x - xe^{\pi x}}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{e^{\arctan x} - \arctan x - 1}{(\arctan x)^{2}}\cdot\frac{(\arctan x)^{2}}{x^{2}} + \lim_{x \to 0}\frac{\arctan x - xe^{\pi x}}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{e^{\arctan x} - \arctan x - 1}{(\arctan x)^{2}} + \lim_{x \to 0}\frac{\arctan x - x + x - xe^{\pi x}}{x^{2}}\notag\\
&= \lim_{t \to 0}\dfrac{e^{t} - t - 1}{t^{2}} + \lim_{x \to 0}\frac{\arctan x - x}{x^{2}} + \lim_{x \to 0}\frac{1 - e^{\pi x}}{x}\text{ (putting }t = \arctan x)\notag\\
&= \lim_{t \to 0}\dfrac{e^{t} - 1}{2t} + \lim_{x \to 0}\frac{\arctan x - x}{x^{2}} - \lim_{x \to 0}\frac{e^{\pi x} - 1}{\pi x}\cdot \pi \text{ (using LHR for first limit)}\notag\\
&= \frac{1}{2} - \pi - \lim_{t \to 0}\frac{t - \tan t}{\tan^{2} t}\text{ (putting }t = \arctan x)\notag\\
&= \frac{1}{2} - \pi - \lim_{t \to 0}\frac{t - \tan t}{t^{2}}\cdot\frac{t^{2}}{\tan^{2}t}\notag\\
&= \frac{1}{2} - \pi - \lim_{t \to 0}\frac{t - \tan t}{t^{2}}\notag
\end{align}
The last limit for $(t - \tan t)/t^{2}$ as $t \to 0$ can be easily shown to be $0$ using the inequalities $$\sin t < t < \tan t$$ for $0 < t < \pi/2$. Hence the desired limit is $\dfrac{1}{2} - \pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Approximate roots of nonlinear equation (non-integer polynomial) In case of pulsating bubble arising from underwater explosion, bubble radius satisfies the following equation.
$x^3\dot{x}^{2} + x^3 + \frac{k}{x^{3(\gamma-1)}} = 1$
The minimum and maximum bubble radii can be obtained by substituting $\dot{x}=0$ in the above equation.
$x^3 + \frac{k}{x^{3(\gamma-1)}} = 1$
As per [1], for small $k$, the roots of the above equation are
$x_{0} \sim k^{\frac{1}{3(\gamma-1)}}\bigg[1+k^{\frac{1}{\gamma-1}}/3(\gamma-1)\bigg]$; minimum
$x_{m} \sim 1-\frac{k}{3}\bigg[1+\big(\gamma-\frac{2}{3}\big)k\bigg]$; maximum
I would like to know the derivation of these approximate roots.
Screenshot from the original source [1] has been included for more details.
[1] Hicks, A.N. 1972. The Theory of Explosion Induced Ship Whipping Motions. NCRE Report R579.
| Case $x$ maximum and $k$ small :
$x^3$ is large compared to $\frac{k}{x^{3(\gamma-1)}}$
$x^3=1-\frac{k}{x^{3(\gamma-1)}} \quad\to\quad x=\left(1-\frac{k}{x^{3(\gamma-1)}}\right)^{1/3} $
First approximate: $x\simeq 1 \quad\to\quad x\simeq \left(1-\frac{k}{1}\right)^{1/3} \simeq 1-\frac{k}{3}$
Second approximate: $x\simeq 1-\frac{k}{3} \quad\to\quad x\simeq\left(1-\frac{k}{(1-\frac{k}{3} )^{3(\gamma-1)}}\right)^{1/3}$
Expending to series of powers of $k$ leads to :
$$x\simeq 1-\frac{k}{3}-\frac{\gamma-\frac{2}{3}}{3}k^2$$
Case $x$ small :
$x^3$ is small compared to $\frac{k}{x^{3(\gamma-1)}}$
$\frac{k}{x^{3(\gamma-1)}}=1-x^3 \quad\to\quad x=\left(\frac{k}{1-x^3}\right)^{\frac{1}{3(\gamma-1)}}$
First approximate: $x\simeq 0 \quad\to\quad x\simeq\left(k\right)^{\frac{1}{3(\gamma-1)}}$
Second approximate: $x\simeq (k)^{\frac{1}{3(\gamma-1)}} \quad\to\quad x\simeq \left(\frac{k}{1-\left(k^{\frac{1}{3(\gamma-1)}} \right)^3}\right)^{\frac{1}{3(\gamma-1)}} \simeq (k)^{\frac{1}{3(\gamma-1)}} \left(1+(k)^{\frac{1}{(\gamma-1)}} \right)^{\frac{1}{3(\gamma-1)}} $
$$x\simeq (k)^{\frac{1}{3(\gamma-1)}} \left(1+ {\frac{1}{3(\gamma-1)}} (k)^{\frac{1}{(\gamma-1)}} \right) $$
| {
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"url": "https://math.stackexchange.com/questions/1750957",
"timestamp": "2023-03-29T00:00:00",
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Solving the Diophantine Equation $x^2 - y! = 2001$ and $x^2 - y! = 2016$ I had recently faced a problem:
Solve the Diophantine Equation $x^2 - y! = 2001$.
Solving it was quite easy. You show how $\forall y \ge 6$, $9|y!$ and since $3$ divides the RHS, it must divide the LHS and if $3|x^2 \implies 9|x^2$ and so the LHS is divisible by $9$ and the RHS is not. Contradiction. Hence, the only solution is $(45, 4)$.
That made me wonder, how we can solve the Diophantine Equation $x^2 - y! = 2016$. We cannot apply the same logic here. $2016$ is a multiple of $9$ and it is clear that $3|x$ and $9 \nmid x$. How do I proceed from here?
| Since $2016= 2^5 \cdot 3^2 \cdot 7$, we can use the same argument with $2$ or $7$ instead of $3$.
With $2$ we have:
If $y \ge 8$, then $2^5 \mid y!$ and so $2^5 \mid x^2$. This implies that $2^3 \mid x$ and, since $2^6 \mid y!$, we'd have $2^6 \mid 2016$, which is false.
Therefore, we only have to test $y \le 7$. The only solution is $x=84, y=7$.
With $7$ we have:
If $y \ge 14$, then $7^2 \mid y!$ and so $7^2 \mid x^2$. This implies that $7^2 \mid x$ and we'd have $7^2 \mid 2016$, which is false.
Therefore, we only have to test $y \le 13$. The only solution is $x=84, y=7$.
With $3$, the argument is similar to the one for $2$: If $y \ge 9$, then $3^4 \mid y!$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751081",
"timestamp": "2023-03-29T00:00:00",
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Let $\textbf{v} = (1, 1, 1, 1)$. Find a basis for... Let $\textbf{v}=(1,1,1,1)$. Find a basis for $\{\textbf{u}\in\Bbb{R}^4\ |\ \textbf{u}\cdot\textbf{v}=0\}$
How can I do this? In particular, I do not understand $\textbf{u}\cdot\textbf{v}=0.$
| Let
$$u=\begin{pmatrix}x\\y\\z\\w\end{pmatrix}$$
Then,
$$u\cdot v=0\Rightarrow x+y+z+w=0$$
$$w=-x-y-z$$
$$u=\begin{pmatrix}x\\y\\z\\-x-y-z\end{pmatrix}=x\begin{pmatrix}1\\0\\0\\-1\end{pmatrix}+y\begin{pmatrix}0\\1\\0\\-1\end{pmatrix}+z\begin{pmatrix}0\\0\\1\\-1\end{pmatrix}$$
The set
$$\left\{\begin{pmatrix}1\\0\\0\\-1\end{pmatrix},\begin{pmatrix}0\\1\\0\\-1\end{pmatrix},\begin{pmatrix}0\\0\\1\\-1\end{pmatrix}\right\}$$ span the space and are linearly independent. They form a basis.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to simplify $y = \sin(\frac{\arcsin(x)}{n}), n≥1$? $y = \sin(\frac{\arcsin(x)}{n}), n≥1$
I know that:
$\lim \limits_{x \to 0} \frac{x}{y} = n$
But I can't figure out what the curve of $x/y$ practically represents. Is there an obvious simple solution?
| Let $\theta = \arcsin x$ and we have to assume that $|x| \lt \dfrac{1}{\sqrt 2}$.
Then $\tan \theta = \dfrac{x}{\sqrt{1-x^2}}$ and $|\tan \theta| \lt 1$.
\begin{align}
\cos \dfrac{\theta}{n} + i \sin \dfrac{\theta}{n}
&= (\cos \theta + i \sin \theta)^{\frac 1n} \\
&= (\cos \theta)^{\frac 1n}(1 + i \tan \theta)^{\frac 1n} \\
&= x^{\frac 1n}\left(1 + i\dfrac{x}{\sqrt{1-x^2}}\right)^\frac 1n\\
&= x^{\frac 1n}\sum_{k=0}^\infty
\binom{1/n}{k} i^k\left(\dfrac{x}{\sqrt{1-x^2}}\right)^k \\
\cos \dfrac{\theta}{n}
&= x^{\frac 1n}\sum_{k=0}^\infty
(-1)^k\binom{1/n}{2k}\left(\dfrac{x^2}{1-x^2}\right)^k \\
\sin\dfrac{\theta}{n}
&= x^{\frac 1n} \dfrac{x}{\sqrt{1-x^2}} \sum_{k=0}^\infty
(-1)^k\binom{1/n}{2k+1}\left(\dfrac{x^2}{1-x^2}\right)^k \\
\end{align}
NOTES
We define $\binom zn$ where $z \in \mathbb R$ and $0 \le n \in \mathbb Z$ as follows
$(z)_n =
\begin{cases}
1 & \text{If $n = 0$.}\\
z(z-1)(z-2)\cdots(z-n+1) &\text{If $n \ge 1$.}
\end{cases}$
then $\binom zn = \dfrac{(z)_n}{n!}$
It can be shown that, if $|x| < 1$, then $(1 + x)^z = \sum_{k=0}^\infty \binom zk x^k$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove this inequality: $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$
Let $a$,$b$,$c$ be positive real numbers such that $abc=1$. Prove that
$$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$$
I tried various methods. But, couldn't solve it. It'd be great if anyone can help.
| You already have a proof using AM-GM and Muirhead. For a pure AM-GM proof, note that $\frac12 \left(\frac{a}b+\frac{b}a\right) \geqslant 1$ and so on, so half the LHS takes care of the $3$ on RHS.
For the rest, note you can cyclically sum the (weighted) AM-GMs:
$$\frac5{18} \left(\frac{a}b+\frac{a}c\right)+\frac2{18} \left(\frac{b}a+\frac{b}c\right)+\frac2{18} \left(\frac{c}a+\frac{c}b\right)\geqslant \frac{\sqrt[3]a}{\sqrt[6]{b c}}=\sqrt{a}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this inequality true for all k ? $\sum_{n=k}^{n=+\infty} \frac{1}{n^4} \leq (\sum_{n=k}^{n=+\infty} \frac{1}{n^2})^3$ Can it be generalized for other powers ? Wolfram seems to say it is true for k below 20000.
I stumbled upon it randomly when trying to approximate $\sum_{n=1}^{n=+\infty} \frac{1}{n^4}$.
My reasoning was :
$$\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2=\sum_{n=k}^{n=+\infty} \frac{1}{n^4} + (\text{double products}) \geq\sum_{n=k}^{n=+\infty} \frac{1}{n^4}$$
So
$$\sum_{n=1}^{n=+\infty} \frac{1}{n^4} \leq \sum_{n=1}^{n=k-1} \frac{1}{n^4}+\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2 \leq \left(\sum_{n=1}^{n=k-1} \frac{1}{n^4}\right)+\left(\frac{1}{k-\frac{1}{2}}\right)^2$$
where the last inequality comes from An inequality: $1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53$.
Then I noticed that, perhaps, I could raise the last term to the power of 3 instead of just 2, making the inequality stronger.
| Let's see what we can say about
comparing
$s_1(k)
=\sum_{n=k}^{\infty} \frac1{n^a}
$
with
$s_2(k)
=\left(\sum_{n=k}^{\infty} \frac1{n^b}\right)^c
$
for large enough $k$,
where
$a > 1$ and $b > 1$
so the sums converge.
Using the integral approximation,
$s_1(k)
=\sum_{n=k}^{\infty} \frac1{n^a}
\approx \int_k^{\infty} \frac{dx}{x^a}
=-\frac1{(a-1)x^{a-1}}\big|_k^{\infty}
=\frac1{(a-1)k^{a-1}}
$.
Therefore
$s_2(k)
=\left(\sum_{n=k}^{\infty} \frac1{n^b}\right)^c
\approx \left(\frac1{(b-1)k^{b-1}}\right)^c
=\frac1{(b-1)^c k^{c(b-1)}}
$,
so
$\dfrac{s_1(k)}{s_2(k)}
\approx \dfrac{\frac1{(a-1)k^{a-1}}}{\frac1{(b-1)^c k^{c(b-1)}}}
= \dfrac{{(b-1)^c k^{c(b-1)}}}{{(a-1)k^{a-1}}}
= \dfrac{{(b-1)^c }}{{(a-1)}}k^{c(b-1)-(a-1)}
$.
Therefore
if
$c(b-1) > a-1$,
then
$s_1(k)
> s_2(k)
$
for all large enough $k$;
if
$c(b-1) < a-1$,
then
$s_1(k)
< s_2(k)
$
for all large enough $k$;
If
$c(b-1) = a-1$,
then
$\dfrac{s_1(k)}{s_2(k)}
\approx \dfrac{{(b-1)^c }}{{(a-1)}}
$,
so the result
depends on this ratio.
For your case of
$a=4$ and $b=2$,
the key difference is
$c(b-1)-(a-1)
=c-3
$.
If
$c > 3$,
then
$s_1(k) > s_2(k)$
for large enough $k$;
if
$c < 3$,
then
$s_1(k) < s_2(k)$
for large enough $k$.
If $c=3$,
which is your case,
the ratio is
$\dfrac{(b-1)^c }{(a-1)}
=\dfrac{1}{3}
< 1
$,
so
$s_1(k)
\approx \frac13 s_2(k)
< s_2(k)
$
for large enough $k$,
which confirm's
Robert Israel's result
(good thing too,
because any result of mine
that differs from
a result of his
is probably wrong).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\int_0^\infty {dx\over (x^2+a^2)(x^2+b^2)}={\pi\over2ab(a+b)}$ Show that $$\int_0^\infty {dx\over (x^2+a^2)(x^2+b^2)}={\pi\over2ab(a+b)}$$ where $a,b>0$. I'm not sure how to simplify this. Any solutions or hints are greatly appreciated.
| You should take the partial fraction of $\dfrac{1}{(x^2+a^2)(x^2+b^2)}$.
This way, it is easier for you to integrate the expression and you should expect the $\arctan$ function in the numerator.
Recall:
$\dfrac{1}{(x^2+a^2)(x^2+b^2)}=\dfrac{Ax+B}{x^2+a^2}+\dfrac{Cx+D}{x^2+b^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve limit $\lim_{x\to 0} \frac{x - \sin (x)}{(x \sin (x))^{3/2}}$ Hi I must solve the next limit $\infty - $\infty$ usibg L'Hopital and Taylor series.
$\lim\limits_{x \to 0} \frac{x - \sin (x)}{(x \sin (x))^{3/2} }$
I tried to eliminate the root with multiplying by
$1 = \frac{(x \sin(x))^{\frac{1}{2}})}{(x \sin (x))^{\frac{1}{2}}} $
Byt whe I tried to apply L'Hopital the Sin function is very problematic
| You have with Taylor series around $0$ : $$\sin(x)=x-\frac{x^3}{6}+o(x^3)$$
$$\lim\limits_{x \to 0} \frac{x - \sin (x)}{(x \sin (x))^{3/2} }=\lim\limits_{x \to 0} \frac{x - x + \frac{x^3}{6}+o(x^3)}{(x (x+o(x))^{3/2} }=\lim\limits_{x \to 0} \frac{\frac{x^3}{6}+o(x^3)}{x^3 +o(x^3) }= \frac{1}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $6(\sin^{10}A+\cos^{10}A)-15(\sin^8A+\cos^8A)+10(\sin^6A+\cos^6A)-1=0$ Prove that
$$6(\sin^{10}A+\cos^{10}A)-15(\sin^8A+\cos^8A)+10(\sin^6A+\cos^6A)-1=0$$
Expression can be verified for different values of $A$ such as $\frac\pi4,\frac\pi2$ etc. But to prove it for general value of A?
| Set $x=\sin^2A$ and $y=\cos^2A$. Then $x+y=1$ and
\begin{align}
1&=(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\\
&=x^5+y^5+5xy(x^3+y^3)+10x^2y^2(x+y)\\
&=x^5+y^5+5xy(x+y)(x^2-xy+y^2)+10x^2y^2\\
&=x^5+y^5+5xy(x^2-xy+y^2)+10x^2y^2\\
&=x^5+y^5+5xy((x+y)^2-3xy)+10x^2y^2\\
&=x^5+y^5+5xy(1-3xy)+10x^2y^2\\
&=x^5+y^5-5x^2y^2+5xy
\end{align}
Similarly,
\begin{align}
1&=(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4\\
&=x^4+y^4+4xy(x^2+y^2)+6x^2y^2\\
&=x^4+y^4+4xy(1-2xy)+6x^2y^2\\
&=x^4+y^4-2x^2y^2+4xy
\end{align}
and
\begin{align}
1&=(x+y)^3=x^3+3x^2y+3xy^2+y^3\\
&=x^3+y^3+3xy(x+y)\\
&=x^3+y^3+3xy
\end{align}
Set $p=xy$; we have proved that
\begin{align}
x^5+y^5&=5p^2-5p+1\\
x^4+y^4&=2p^2-4p+1\\
x^3+y^3&=-3p+1
\end{align}
so finally your expression is
$$
6(5p^2-5p+1)-15(2p^2-4p+1)+10(-3p+1)-1=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $ab+ 2cb+\sqrt3 ac$? Three positive real numbers $a,b,c$ satisfy the equations $a^2+\sqrt3 ab+b^2=25$, $b^2+c^2=9$ and $a^2+ac+c^2=16$ .Then find the value of $ab+ 2cb+\sqrt3 ac$?
Is there some way to find the desired value without actually finding values of $a,b,c$ or any other smart method to find $a,b,c$
| What a prowess, @Piquito ! Here is another way of treating this issue.
Its "pros": it is systematic. Its "cons": it necessitates a computer algebra system (I used Mathematica).
Let us give the following numbers to the 3 equations:
$$\begin{cases}a^2+\sqrt 3ab+b^2&=&25 \ \ &(1)\\b^2+c^2&=&9 \ \ \ \ \ &(2) \\a^2+ac+c^2&=&16 \ \ \ &(3)\end{cases}$$
Let us consider equations (1) and (3) as quadratics in $a$.
They must have a common root. This can be expressed by setting their resultant to $0$ (https://en.wikipedia.org/wiki/Resultant):
This gives:
$$81 - 66b^2 + b^4 + 41\sqrt{3}bc - \sqrt{3}b^3c - 7c^2 + 2b^2c^2 -
\sqrt{3}bc^3 + c^4=0 \ \ \ (4)$$
Now, we use constraint (2) meaning that point $(b/3,c/3)$ is on the unit circle, a constraint that we can translate into the following one (classical parameterization of the unit circle https://en.wikipedia.org/wiki/Tangent_half-angle_formula):
$$b = 3\dfrac{1 - t^2}{1 + t^2}, \ \ c = 3\dfrac{2 t}{1 + t^2} \ \ \ (5)$$
for a certain $t \in (-\infty,+\infty)$. Plugging (5) into (4) gives:
$$12 - 16\sqrt{3}t - 35t^2 + 16\sqrt{3}t^3 + 12t^4=0 \ \ \ (6)$$
This antipalindromic (https://en.wikipedia.org/wiki/Reciprocal_polynomial) 4th degree polynomial has four explicit (real) solutions
$$t=\dfrac{1}{12}(\pm9-4\sqrt{3}\pm\sqrt{273-72 \sqrt{3}}) \ \ \ (7)$$
(the two $\pm$ signs are independant: their four combinations are valid).
The objective of finding the value of
$$X = ab+ 2cb+\sqrt3 ac$$
is now within reach, because we are able to express it as an expression of $t$ alone, because:
*
*this is the case for $b$ and $c$ (formulas (5)).
*Concerning $a$, being the solution of quadratic equation (3), can be expressed as a function of $c$, itself function of $t$.
Then, it remains to check that, whatever the root chosen in (7), one gets the same result $X=24$. This is the case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Let $f(x)=x^5$. For $x_1>0$, let $p_1=(x_1,f(x_1))$.Draw a tangent at the point $p_1$ Let $f(x)=x^5$. For $x_1>0$, let $p_1=(x_1,f(x_1))$. Draw a tangent at the point $p_1$ and let it meet the graph again at point $p_2$. Then draw a tangent at $p_2$ and so on . Show that , the ratio $\frac{A(\bigtriangleup p_np_{n+1}p_{n+2})}{A( \bigtriangleup p_{n+1}p_{n+2}p_{n+3} )}$ is constant.
I don't think just taking the ratio of two triangles will help, something more need to be used.
| Expanding on my comment above, consider any point $P(a, a^5)$ other than the origin on the curve $f(x)$. The slope of the tangent at $P$ is $5a^4$. Hence the point of intersection is $Q(b, b^5)$, where $a \neq b$ and
$$5a^4 = \frac{b^5-a^5}{b-a} \implies b^3 + 2ab^2+3a^2b+4a^3 = 0$$
$$\implies r^3+2r^2+3r+4 = 0$$
where $r = b/a$. Clearly as a cubic this has at least one real root, and it must be negative. As the derivative $3r^2+4r+3$ has negative discriminant, it cannot have any more roots. It is also easy to check $r < -1$. Thus let $k = -r > 1$, and we have for any $P_n(x_n, x_n^5)$, the next point in sequence is $P_{n+1} (-kx_n, -k^5 x_n^5)$.
Now the ratio of areas you seek is
$$\frac{A(\bigtriangleup p_np_{n+1}p_{n+2})}{A( \bigtriangleup p_{n+1}p_{n+2}p_{n+3} )} = \frac{\begin{vmatrix} x_n & x_n^5 & 1 \\ x_{n+1} & x_{n+1}^5 & 1 \\
x_{n+2} & x_{n+2}^5 & 1 \end{vmatrix}
}{\begin{vmatrix} x_{n+1} & x_{n+1}^5 & 1 \\ x_{n+2} & x_{n+2}^5 & 1 \\
x_{n+3} & x_{n+3}^5 & 1 \end{vmatrix}
} = \frac{\begin{vmatrix}
x_n & x_n^5 & 1 \\
-kx_n & -k^5x_n^5 & 1 \\
k^2x_n & k^{10}x_n^5 & 1 \\
\end{vmatrix}
}{\begin{vmatrix}
-kx_n & -k^5x_n^5 & 1 \\
k^2x_n & k^{10}x_n^5 & 1 \\
-k^3x_n & -k^{15}x_n^5 & 1 \\
\end{vmatrix}
} = \frac1{k^6
}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What's the solution to this puzzle? I saw this on Instagram with no solution and was wondering what the answer is. I got $33$.
$$1+4=5$$
$$2+5=12$$
$$3+6=21$$
$$8+11=?$$
| 96
For each line add the two numbers and store it in D. Add D to the C from the previous step.
A+B=C D
1+4=5 5 (4+1 gives 5)
2+5=12 7 (5+2 is 7. Add 7 to 5 = 12 from previous step)
3+6=21 9 (6+3 is 9. Add 9 to 12 = 21 from previous step)
4+7=32 11 (4+7 is 11. Add 11 to 21 = 32 from previous step)
5+8=45 13
6+9=60 15
7+10=77 17
8+11=96 19
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Finding out a limit using Taylor series. So the limit is the following:
$$\lim_{x \to 0}{\frac{x^2-\frac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}}$$
Expansions for $\sin(x)$ and $\cos(x)$ are given:
$$\sin x = x-\frac{x^3}{3!} + \frac{x^5}{5!}-...+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!} + o(x^{2n})$$
$$\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-...+(-1)^n\frac{x^{2n}}{(2n)!}+o(x^{2n+1})$$
Here is what I tried:
$$\lim_{x->0}{\frac{x^2-\frac{x^6}{2}-x^2(1-\frac{x^4}{2}+\frac{x^8}{4!}+o(x^{2*5}))}{x^{10}+o(x^{10*2})}}=\lim_{x->0}{\frac{-\frac{x^{10}}{4!}-o(x^{12})}{x^{10}+o(x^{20})}}$$
This is where I am stuck. I figured that the problem occurs when expending $o()$. What am I missing here?
| The numerator is
$$x^2-\frac{x^6}2-x^2\Bigl(1-\frac{x^4}2+\frac{x^8}{24}+o(x^8)\Bigr)=-\frac{x^{10}}{24}+o(x^{10}),$$
hence
$$\frac{x^2-\cfrac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}\sim_0\frac{-\dfrac{x^{10}}{24}}{x^{10}}=-\frac1{24}.$$
*Editors Note: Sorry, can't leave comments yet. When you factor in the $-x^2$ you'll end with $-\frac{1}{24}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How to prove that $\sin x = x \, _{0}F_{1}(-;\frac{3}{2};-\frac{x^{2}}{4})$? How to prove that $\sin x = x \, _{0}F_{1}(-;\frac{3}{2};-\frac{x^{2}}{4})$?
where $_{0}F_{1}$ is the hypergeometric series?
| One may recall the definition of the generalized hypergeometric function
$$
{}_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;z): = \sum_{n=0}^\infty \frac{(a_1)_n\dots(a_p)_n}{(b_1)_n\dots(b_q)_n} \, \frac {z^n} {n!}
$$ which gives
$$
{}_0F_1\left(-;\frac{3}{2};-\frac{x^{2}}{4}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{\left(\frac32\right)_n} \, \frac {x^{2n}} {4^n \:n!}.
$$ Then one may check that
$$
\begin{align}
\frac1{\left(\frac32\right)_n\:4^n \:n!}&=\frac1{\left(\frac32\right)\left(\frac32+1\right)\left(\frac32+2\right)\cdots\left(\frac32+n-1\right)\:4^n \:n! }
\\\\&=\frac{2^n}{3\times 5\times 7 \cdots \times (2n+1)\:4^n \:n! }
\\\\&=\frac{2^n\times \color{blue}{2\times 4\times 6 \cdots \times (2n)} }{\color{blue}{2}\times3\times \color{blue}{4} \times5\times \color{blue}{6}\times7 \cdots \times \color{blue}{2n}\times (2n+1)\:4^n \:n! }
\\\\&=\frac{2^n\times 2^n\times n! }{(2n+1)!\:4^n \:n! }
\\\\&=\frac1{(2n+1)!}
\end{align}
$$ giving
$$
x\:{}_0F_1\left(-;\frac{3}{2};-\frac{x^{2}}{4}\right) = \color{blue}{\sum_{n=0}^\infty (-1)^n\frac {x^{2n+1}} {(2n+1)!}}=\color{red}{\sin x}
$$
as announced.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A Chinese Exam Question which is.....quite hard Let $f(x)=x^2-2x-3$, and $x_n$ be some sequence.
$x_1=2$, $x_n =$ the $x$ coordinate of the point of intersection of the $x$ axis and the line joining $P(4,5)$ and $Q_n(x_n, f(x_n))$.
Find an expression for $x_n$
Find $\lim \limits_{n\to\infty} Q_n(x_n,f(x_n))$.
| The slope of the line connecting points $(4,5)$ and $(x_n,y_n)$ is $m_n=\dfrac{5-y_n}{4-x_n}$
So the equation of the line is
$\begin{equation}
y-5=\dfrac{5-y_n}{4-x_n}(x-4)
\end{equation}$
$x_{n+1}$ will be the point on this line for which $y=0$, so (omitting the algebra)
$\begin{eqnarray}
x_{n+1}&=\dfrac{4x_n-4y_n}{5-yn}\\
&=\dfrac{4x_n^2-13x_n-12}{x_n^2-2x_n-8}
\end{eqnarray}$
In the limit we will have
$\begin{equation}
x=\dfrac{4x^2-13x-12}{x^2-2x-8}
\end{equation}$
or
$\begin{eqnarray}
x^3-6x^2+5x+12&=0\\
(x+1)(x-3)(x-4)&=0
\end{eqnarray}$
The sequence $\{x_n\}$ will never make it beyond 3 which is a zero of $f(x)$, so the limit of $Q_n$is $(3,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve the equation on the interval $\; 0 \le \theta \lt 2\pi $ Hey I have two questions for math exchange! Let me list them first and show you what I have tried. By the way one can not use a calculator on the test review!
Solve the equations on the interval $\; 0 \le \theta \lt 2\pi $:
*
*$2 \cos^2\theta - 3\cos \theta + 1 = 0 $
*$\cot \theta = 2 \cos \theta$
So for the first one I tried:
\begin{gather}
2 \cos^2\theta - 3\cos \theta + 1 = 0\\
(2\cos^2\theta - 1) \times (3\cos\theta - 1) = 0\\
\begin{aligned}
2 \cos^2 \theta -1 &= 0 &\text{ or }&& 3\cos\theta -1 =0\\
\cos^2\theta &= \dfrac{1}{2} &\text{ or }&& \theta = \cos^{-1}\left(\dfrac{1}{3}\right)
\end{aligned}
\end{gather}
After that I just got confused... I can't use a calculator so I do not know what to do next..
So The next problem was even more diffcult:
$$\cot \theta = 2 \cos \theta$$
And then immediately after that I got confused since there are two theta's, and I have absolutely no clue what to do next!
On the anwser key for test review, the correct anwsers are:
*
*$$2 \cos^2\theta - 3\cos \theta + 1 = 0$$
Answer:$\qquad\left\{ 0,\;\dfrac{\pi}{3} ,\; \dfrac{5\pi}{3} \right\}$
*$$\cot \theta = 2 \cos \theta$$
Answer : $\qquad\left\{\dfrac{\pi}{6} ,\; \dfrac{\pi}{2},\; \dfrac{5\pi}{6},\;\dfrac{3\pi}{2} \right\}$
Help would be greatly appreciated. Thankyou for reading!
| *
*$2\cos^2\theta-3\cos\theta+1=0$:
Use the change of variables $x=\cos\theta$, in order to write the equation as
$$
2x^2-3x+1=0
$$
Solving for $x$ yields
$$
x=1\quad or\quad x=\frac{1}{2}
$$
so in terms of $\theta$:
$$
\theta=0\quad or \quad \theta=\frac{\pi}{3}\quad or \quad \theta=-\frac{\pi}{3}
$$
*$\frac{\cos\theta}{\sin\theta}=2\cos\theta$:
Multiplying both terms by $\sin\theta$ yields
$$
\cos\theta=2\sin\theta\cos\theta\quad \Leftrightarrow\quad \sin(\frac{\pi}{2}-\theta)=\sin2\theta
$$
And solving for $\theta$ gives us
$$
\frac{\pi}{2}-\theta = 2\theta +2n\pi \quad or \quad \frac{\pi}{2}-\theta = \pi- 2\theta+2n\pi,
$$
that is:
$$
\theta\in \{\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}\}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Expressing the roots of a cubic as polynomials in one root All roots of $8x^3-6x+1$ are real. (*)
The discriminant of $8x^3-6x+1$ is $5184=72^2$ and so the splitting field of $8x^3-6x+1$ has degree $3$.
Therefore, all three roots can be expressed as polynomials in any one given root.
Indeed, if $a$ is a root, then the others are $2a^2-1$ and $-2a^2-a+1$.
This can be easily checked. But how can we find these expressions in the first place?
I've tried this: let $b,c$ be the other roots. Then from Vieta's formulas we get $$b+c=-a, \qquad bc=-\dfrac{1}{8a}$$
The corresponding quadratic has discriminant $3-3a^2$ but it is not at all clear that this is the square of a polynomial in $a$. (It is $(4 a^2+a-2
)^2=(b-c)^2$, as it should be.) I'm stuck here.
More generally, is there an algorithm that
given a cubic with integer coefficients and
having real splitting field of degree $3$,
expresses all three roots as polynomials in any one given root?
(*) From the triple-angle formula $\sin (3\theta) = - 4\sin^3\theta + 3\sin\theta$ when $\sin (3\theta) = 1/2$, these roots are $\sin(10^\circ)$, $\sin(50^\circ)$, $\sin(-70^\circ)$, but perhaps this is immaterial here.
| Based on the answers in this question, I'm able to finish the general case.
Let $x^3+px+q$ be a cubic with rational coefficients and having real splitting field of degree $3$. Let its roots be $a,b,c$. Then
$$b+c=-a, \qquad bc=-\dfrac{q}{a}$$
and $b,c$ are the roots of the quadratic
$$
h(x)=(x-b)(x-c)=x^2+ax-\dfrac{q}{a}
$$
The discriminant of $h$ is $b-c$. Now comes the nice idea from those answers:
$$
d=(a-b)(a-c)(b-c)=h(a)(b-c)
$$
where $d^2=-4p^3-27q^2$ is the discriminant of the original cubic. The hypothesis on the cubic implies that $d$ is rational. Therefore,
$$
b-c = \dfrac{d}{h(a)}
$$
To write this as a polynomial in $a$, we solve the linear system on $A,B,C$ implied by
$$
d=(Aa^2+Ba+C)h(a)=(Aa^2+Ba+C)(2a^2-\dfrac{q}{a})
$$
or
$$
ad=(Aa^2+Ba+C)(2a^3-q)=(Aa^2+Ba+C)(-2pa-3q)
$$
The solution is
$$
A=-\dfrac{6p}{d},
\quad
B=\dfrac{9q}{d},
\quad
C=-\dfrac{4p^2}{d}
$$
which gives
$$
b,c = -\dfrac{a}{2} \pm \dfrac{1}{2}(Aa^2+Ba+C)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Convert Integral Rectangular to Polar How can convert this problem
$$
\int_0^2 \int_x^\sqrt{8-x^2} \left(x^2+y^2\right)^{3/2} dydx
$$
I convert limits and funtion to polar cordinates as follows:
$$
\begin{split}
r^2 &= x^2+y^2\\
x = 2 &\to r \cos \theta = 2\\
x = 2 &\to r \cos \theta = 0\\
x = y &\to r \cos \theta = r \sin \theta \to 1 = \tan \theta\\
y = \sqrt{8 - x^2} &\to y^2 + x^2 = 8 \to r^2 = 8
\end{split}
$$
But i dont know how to rebuild de integral with your limits
| Do the usual and draw the region, and check that (fill in details)
$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\;\;,\;\;\frac\pi4\le t\le \frac\pi2\;,\;\;0\le r\le 2\sqrt2$$
so your integral becomes (don't forget the Jacobian!)
$$\int_0^{2\sqrt2}\int_{\pi/4}^{\pi/2}r^4\,dtdr=\frac\pi4\int_0^{2\sqrt2}r^4\,dr=\frac\pi{20}\cdot128\sqrt2=\frac{32\sqrt2\;\pi}5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1768584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding A 1-form on $R^2 - {\{(0,0)}\}$ I want to find a 1-form on $R^2 - {\{(0,0)}\}$ such that $w(Y) = 0$ and $w(X) = 1$. Here, $$X = -y\frac{\partial }{\partial x}
+ x\frac{\partial}{\partial y}\ \text{and}\ Y = x\frac{\partial }{\partial x}
+ y\frac{\partial}{\partial y}.$$
I let $w = adx+bdy$ be a 1-form and then, using the given conditions compute: $$(adx+bdy)(-y\frac{\partial }{\partial x}
+ x\frac{\partial}{\partial y}) = 1,\ \text{and similarly},\ (adx+bdy)(x\frac{\partial }{\partial x}
+ y\frac{\partial}{\partial y}) = 0.$$
I then obtain the following equations: $-ay+bx=1$, $ax+by=0$.
To solve for $a,b$, I then set it up as a matrix/vector system as
$$\underset{\begin{array}{c}\\ A \end{array}}%
{%
\begin{pmatrix}
x & y \\
-y & x%
\end{pmatrix}%
}%
\underset{\begin{array}{c}\\ x \end{array}}%
{%
\begin{pmatrix}
a \\
b%
\end{pmatrix}%
}=\underset{\begin{array}{c}\\ b \end{array}}%
{%
\begin{pmatrix}
0 \\
1%
\end{pmatrix}%
}$$
I then simply invert the matrix and obtain that $a$ = $\frac{-y}{x^2+y^2}$ and $b$ = $\frac{x}{x^2+y^2}$.
However, this seems to be incorrect - the solution in the back of the book gives $$a = \frac{x}{x^2+y^2}\ \text{and}\ b = \frac{y}{x^2+y^2}.$$ Can someone kindly let me know what I did wrong?
| Let $r = x^2 + y^2$. Note that if the values are $a = \frac{x}{r}$ and $b = \frac{y}{r}$, then we will have: $w = \frac{x}{r}$ $dx$ + $\frac{y}{r}$ $dy$. Then, the following follows: $ w(X) = \frac{1}{r} (-xy +xy)$ = $0$,$ w(Y) = \frac{1}{r} (x^2+y^2)$ = $\frac{1}{r} r = 1.$ This contradicts the conditions given ($w(X) =1$, $w(Y) =0$). Thus, the solution in the book can't be the 1-form on $R^2 -{\{(0,0)}\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if $n \in \mathbb{N}$ and $n \ge 2$, then $2^{n + 1} \le 3^n$. Prove that if $n \in \mathbb{N}$ and $n \ge 2$, then $2^{n + 1} \le 3^n$.
My method:
If $n = 2$, $2^{n + 1} \le 3^n$ then $2^3 \le 3^2$ is $8 \le 9$, which holds for $n = 2$.
$2^{k + 1} \le 3^k$ then $2^{k + 2} \le 3^{k + 1}$. Then
$2 \cdot 2 \cdot 2^k \lt 3 \cdot 3^k$
$4 \cdot 2^k \lt 3 \cdot 3^k$
Therefore, $4 \cdot 2^k \lt 3 \cdot 3^k$. Hence, $2^{n + 1} \le 3^n$ for $n \ge 2$.
Is there a problem with: $2^{k + 1} \le 3^k$ then $2^{k + 2} \le 3^{k + 1}$ or any other part of this proof?
| Formal proof by induction.
First, show that this is true for $n=2$:
$2^{2+1}<3^2$
Second, assume that this is true for $n$:
$2^{n+1}<3^n$
Third, prove that this is true for $n+1$:
$2^{n+2}=$
$2\cdot\color\red{2^{n+1}}<$
$2\cdot\color\red{3^n}<$
$3\cdot3^n=$
$3^{n+1}$
Please note that the assumption is used only in the part marked red.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Limit of $\sum_{r=1}^{n} \frac{r}{2^r}$ as $n\to \infty$
Consider the sequence $u_{n}=\sum\limits_{r=1}^{n} \frac{r}{2^r}$ with $n \ge 1$.
Then the limit of $u_{n}$ as $ n \rightarrow \infty$ is ?
I actually treated each term as an AGP and got $$u_{n}=2-\frac{1}{2^{r+1}}-\frac{n}{2^{r+1}}$$
But how to get the limit?
| This is a well-known result due to Nicole Oresme
\begin{align}
\sum_{n=1}^\infty \frac{n}{2^n}=2
\end{align}
Here is the proof
Consider the function
\begin{align*}
f(z)=\frac{1}{1-z}
\end{align*}
The power series expansion of $f(z)$ is
\begin{align}
f(z)=\sum_{n=1}^{\infty}z^n
\end{align}
Now take the derivative of $f$, which is
\begin{align*}
f'(z)=\frac{1}{(1-z)^2}=\sum_{n=1}^{\infty}nz^{n-1}
\end{align*}
Multiplying by $z$
\begin{align}
zf'(z)=\frac{z}{(1-z)^2}=\sum_{n=1}^{\infty}nz^{n}\tag{1}
\end{align}
Substituting $z=1/2$ in (1)
\begin{align}
\sum_{n=1}^\infty \frac{n}{2^n}=\frac{\frac{1}{2}}{\biggl(1-\frac{1}{2}\biggl)^2}=2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1773818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\displaystyle\lim _{x\to 0}\frac{\sqrt{1+x+x²}-1}{\sqrt{1+x}-\sqrt{1-x}}$ First I need to prove that this limit; $\displaystyle\lim _{x\to 0}\frac{\sqrt{1+x+x^2}-1}{\sqrt{1+x}-\sqrt{1-x}}$ converges, then I have to find its limit. Now I don't know how to prove that it converges (these epsilon proofs are still something that I'm trying to learn). And to actually find the limit; I tried to rewrite it by multiplying by $\displaystyle{\sqrt{1+x}+\sqrt{1-x}\over \sqrt{1+x}+\sqrt{1-x}}$ but I didn't get any further..
Edit: I know that it converges to $1/2$, but that's what wolfram alpha says :|
| The standard way is to transform the limit into
$$
\lim _{x\to 0}
\frac{(\sqrt{1+x+x^2}-1)(\sqrt{1+x+x^2}+1)}
{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}
\cdot
\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x+x^2}+1}
$$
that becomes
$$
\lim_{x\to0}
\frac{x+x^2}{2x}
\cdot
\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x+x^2}+1}
=
\lim_{x\to0}
\frac{1+x}{2}
\cdot
\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x+x^2}+1}
$$
The alternative is to use Taylor expansions:
\begin{gather}
\sqrt{1+x+x^2}=1+\frac{1}{2}(x+x^2)+o(x)=1+\frac{1}{2}x+o(x)
\\
\sqrt{1+x}=1+\frac{1}{2}x+o(x)
\\
\sqrt{1-x}=1-\frac{1}{2}x+o(x)
\end{gather}
So the limit is
$$
\lim_{x\to0}\frac{1+\frac{1}{2}x+o(x)-1}{(1+\frac{1}{2}x+o(x))-(1-\frac{1}{2}x+o(x))}=\frac{\frac{1}{2}x+o(x)}{x+o(x)}=\frac{1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Splitting fractions with a linear denominator: $\frac{2x-1}{x+2}$ How can $$\frac{2x-1}{x+2}$$ be split to give $$A-\frac{B}{x+2}$$
where $A$ and $B$ are integers?
The solution is $$2-\frac{5}{x+2}.$$
| $$\frac{2x-1}{x+2}=\frac{2(x+2)-5}{x+2}=\frac{2(x+2)}{x+2}-\frac{5}{x+2}=2-\frac{5}{x+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Evaluating the integral $\int \frac{x^2+x}{(e^x+x+1)^2}dx$
Evaluate $$\int \frac{x^2+x}{(e^x+x+1)^2}dx$$
I tried converting in the form of Quotient rule(seeing the square in the denominator), neither am I able to make the denominators' derivative in the numerator. Some hints would be great. Thanks.
| Let $\displaystyle u=\frac{x^2+x}{xe^x}=(x+1)e^{-x}$ and $\displaystyle dv=\frac{xe^x}{(e^x+x+1)^2}dx,\;$ so $\displaystyle du=-xe^{-x}dx$ and $\displaystyle v=\frac{e^x}{e^x+x+1}$.
Then $\displaystyle\int\frac{x^2+x}{(e^x+x+1)^2}dx=(x+1)e^{-x}\cdot\frac{e^x}{e^x+x+1}-\int-\frac{x}{e^x+x+1}dx=\frac{x+1}{e^x+x+1}+\int\frac{x}{e^x+x+1}dx$
$\displaystyle\hspace{.2 in}=\frac{x+1}{e^x+x+1}+\int\frac{e^x+x+1}{e^x+x+1}dx-\int\frac{e^x+1}{e^x+x+1}dx=\frac{x+1}{e^x+x+1}+x-\ln\big|e^x+x+1\big|+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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A regular tetrahedron is centered at the origin with vertices $ (0,0,1$) and $(a,0,b)$. All of its vertices satisfy $x^2+y^2+z^2=1$
Find the remaining two vertices with respect to $a$ and $b$.
| To solve everything from scratch will result in a too long answer,
so we assume these important properties to be known:
Image 1: Tetraeder in cube construction. (Source)
*
*The tetraeder in the task has the internal arm length $1$, meaning
the distance from origin to top vertex. Then the edge length between
two vertices is $L=2\sqrt{2/3}$.
(This corresponds to a cube length $s=2/\sqrt{3}$ in the above image)
*The internal angle between two arms is $\theta\approx 109^0$ with
$\arccos \theta = -1/3$.
(This is only used to check the following results)
For the second vertex $Q=(a,0,b)$ we have the constraints that it lies on the unit sphere around the origin and on the sphere around $P=(0,0,1)$ of radius $L$.
\begin{align}
1 &= x^2 + y^2 + z^2 \\
L^2 &= x^2 + y^2 + (z-1)^2
\end{align}
The intersection is the circle at height
$$
L^2 = 2-2 z \iff z = 1 - L^2/2 = 1-4/3 = -1/3
$$
with the equation
$$
x^2+y^2=1-1/9=(2\sqrt{2}/3)^2 \quad (*)
$$
which has radius $r = 2 \sqrt{2}/3$.
(Large version)
Image 2: Unit sphere constraint (red), $L$ sphere constraint (green). plane constraints $z=-1/3$ and $y = 0$.
Cut with the plane $y=0$ this leaves two choices:
$$
Q=\left( \pm2\sqrt{2}/3,0,-1/3 \right)
$$
Note that we can pick only one of these solutions and not both as feasible vertices, as the angle between them is not $\theta$.
We continue with the positive choice $Q_1$.
The constraints for $P$ and $A$ again lead to two choices for the third vertex $R$:
(Large version left image) (Large version right image)
Images 3 and 4: Unit sphere (purple) and $L$ sphere constraints for $P$ (red) and $Q$ (green). Intersection circles (yellow).
The intersection of unit sphere and $L$ sphere of $Q$
$$
1 = x^2 + y^2 +z^2 \\
L^2 = (x-2\sqrt{2}/3)^2 + y^2 + (z+1/3)^2
$$
gives
$$
L^2 = 8/3 = 1 -(4\sqrt{2}/3)x + 8/9 + (2/3)z + 1/9 \iff \\
1 = - 2\sqrt{2}x + z
$$
as equation of the plane where the intersection circle lies in.
Intersecting this plane with circle $(*)$ we get the equations
$$
1 = -2\sqrt{2} x -1/3 \iff
x = -\sqrt{2}/3
$$
and
$$
8/9 = 2/9 + y^2 \iff y = \pm \sqrt{6}/3
$$
which gives two feasible vertices
$$
R = (-\sqrt{2}/3, \pm \sqrt{6}/3, -1/3)
$$
It turns out that these are the remaing two vertices we are looking for.
(Large version)
Image 5: The resulting tetraeder $(P,Q_1, R_1, R_2)$. Note the two measured tetraeder angles $\theta$. Note the alternative tetraeder vertex $Q_2$.
Summary:
We found two tetraeder that fit the task.
$$
\begin{array}{l}
P = \left( 0, 0, 1 \right) \\
Q_1 = \left( 2\sqrt{2}/3,0,-1/3 \right) \\
R_! = \left( -\sqrt{2}/3, \sqrt{6}/3, -1/3 \right) \\
R_2 = \left( -\sqrt{2}/3, -\sqrt{6}/3, -1/3 \right)
\end{array}
\quad\quad
\begin{array}{l}
P = \left( 0, 0, 1 \right) \\
Q_2 = \left( -2\sqrt{2}/3,0,-1/3 \right) \\
R_3 = \left( \sqrt{2}/3, \sqrt{6}/3, -1/3 \right) \\
R_4 = \left( \sqrt{2}/3, -\sqrt{6}/3, -1/3 \right)
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to calculate limit? I'm puzzled with this limit. The answer is -0.5, but how to get it? $\lim_\limits{x \to \infty}1-x+\sqrt{\frac{x^3}{x+3}}$
| I've tried to multiply by conjugate not only part with a variable, but all the expression, and this way I calculate it without replace and L'Hospital's Rule.
$\lim_\limits{x \to \infty} 1-x+\sqrt{\frac{x^3}{x+3}}=\lim_\limits{x \to \infty}\frac{\frac{x^3}{x+3}-(x-1)^2}{\sqrt{\frac{x^3}{x+3}}+x-1} = \lim_\limits{x \to \infty}\frac{-x^2+5x-3}{(x+3)\left(\frac{\sqrt{x^3}+\sqrt{x+3}(x-1)}{\sqrt{x+3}}\right)}=\lim_\limits{x \to \infty}\frac{-x^2+5x-3}{\sqrt{x^4+3x^3}+(x+3)(x+1)}$
Now we can see easily that factor before $x^2$ is $-1$ in numerator and $2$ in denominator, so the result is $-\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Karatsuba multiplication for $x = 123 , y = 100$ Given the numbers $x = 123$ and $y = 100$ how to apply the Karatsuba algorithm to multiply these numbers ?
The formula is
xy=10^n(ac)+10^n/2(ad+bc)+bd
As I understand $n = 3$ (number of digits) and I tried writing the numbers as
x = 10*12+3 , y = 10*10 +0 thus a = 12 , b = 3 , c = 10 , d = 0
or
x = 100*1+23 , y = 100*1 +0 thus a = 1 , b = 23 , c = 1 , d = 0
I looked at some explanations of the algorithm and tried it successfully for other numbers , but I don't know how to solve it in this particular case.
Is this example part of a more general case of the algorithm (like 3-digit numbers)? I found a question that may be related (Karatsuba multiplication with integers of size 3) and from the answer I gather that it's impossible , so is it that Karatsuba can't multiply $2$ numbers of $3$-digits or is there a way to do this ?
| $xy=123\cdot 100$
$(12\cdot 10+3)(10\cdot 10+0)$
$xy=10^n(ac)+10^n/2(ad+bc)+bd$, where $n=2
a=12, b=3, c=10, d=0$
subcategory 1:
$ac=12\cdot 10 = 120$
$bd=3\cdot 0 = 0$
$ad+bc= (12+3)(10+0)=150-120-0 = 30..$
$120\cdot 10^2 + 30\cdot 10^1 + 0$
subcategory 2:
$12\cdot 10$
$a=1,
b=2,
c=1,
d=0$
$ac=1$
$bd=0$
$ad+bc=(a+b)(b+d)-ac-bd$
$ad+bc= (1+2)(1+0)-1-0 = 2$
$ac\cdot 10^2 + (ad+bc)\cdot 10 + bd =120..$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Let $1 + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots=s$, show that then $\sum_1^\infty\frac{1}{n^3}=\frac{8}{7}s$ Let $1 + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{7^3} + \dots=s$, show that then $\sum_1^\infty\frac{1}{n^3}=\frac{8}{7}s$.
This is the last part of a problem that I am working on. So far, we have shown that the cosine series for $x^2$ is
$x^2=\frac{\pi^2}{3}+4\sum_1^\infty (-1)^n\frac{\cos(nx)}{n^2}$
The sine series for $x^2$ is
$x^2=2\pi\sum_1^\infty(-1)^{n+1}\frac{\sin nx}{n}-\frac{8}{\pi}\sum_1^\infty\frac{\sin ((2n-1)x)}{(2n-1)^3}$
The sine series for x is
$x=2(\sin x -\frac{\sin 2x}{2}+ \frac{\sin 3x}{3}-\dots)$
and using the above we got that
$\frac{\pi^3}{32}=1 - \frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\dots$.
Not sure if any of this matters, I just don't know how to use this information to solve the last part. I don't really know if this information is needed at all, but I assume it is.
| \begin{align*}
\sum_1^{\infty} \frac 1 {n^3} &= \sum_{\text{even}} \frac 1 {n^3} + s \\
&= \sum_1^{\infty} \frac{1}{(2k)^3} + s\\
&= \frac 1 8 \sum_1^{\infty} \frac 1 {k^3} + s
\end{align*}
Rearrange for the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove this inequality in complex domain (5) Let $z_{1},z_{2},z_{3}\in C$
show that
$$|z_{1}+z_{2}+z_{3}|^2+|(z_{1}-z_{2})(z_{1}-z_{3})|+|(z_{2}-z_{3})(z_{2}-z_{1})|
+|(z_{3}-z_{1})(z_{3}-z_{2})|\le 3(|z_{1}|^2+|z_{2}|^2+|z_{3}|^2)$$
Iif prove following inequality,Use Cauchy-Schwarz inequality we have
$$(|z_{1}+z_{2}+z_{3}|^2\le (|z_{1}|+|z_{2}|+|z_{3}|)^2\le 3(|z_{1}|^2+|z_{2}|^2+|z_{3}|^2)$$
| Let $O$, $A$, $B$ and $C$ be points in the plain and $M$ is a gravity canter of $\Delta ABC$.
Hence,
$3(OA^2+OB^2+OC^2)=9OM^2+AB^2+AC^2+BC^2\geq$
$\geq9OM^2+AB\cdot AC+AB\cdot BC+AC\cdot BC$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Weird differential equation - Jacobi? In class we had differential equations of the type
$$y'=\frac{\left(Ax+By \right)y+ \alpha x + \beta y}{\left(Ax+By \right) x+ax+by},$$
where $A,\alpha,a,B,\beta,b$ are constants. The names of the constants were chosen in a way, so that the 3 different constants for each variable look a bit an a and b.
For example $$y'=\frac{(x-y)y-x-y}{(x-y)x+x+y}$$
I did not fully understand the method of solution. In class the professor called them "Jacobi Differtial calculus" (translated). I did not find anything suitable on the web, except Wolfram Mathworld http://mathworld.wolfram.com/JacobiDifferentialEquation.html There is at least a differential equation named Jacobi. But this does not seem to be the same as in my question. Any help?
My question:
Q1 What are DE's like those called?
Q2 How to solve them?
| For $y'=\dfrac{(x-y)y-x-y}{(x-y)x+x+y}$ ,
Let $u=\dfrac{y}{x}$ ,
Then $y=xu$
$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=\dfrac{(x-xu)xu-x-xu}{(x-xu)x+x+xu}$
$x\dfrac{du}{dx}=\dfrac{(1-u)xu-1-u}{(1-u)x+1+u}-u$
$x\dfrac{du}{dx}=\dfrac{(x-1)u-xu^2-1}{(1-x)u+x+1}-u$
$x\dfrac{du}{dx}=\dfrac{xu^2-(x-1)u+1}{(x-1)u-x-1}-u$
$x\dfrac{du}{dx}=\dfrac{xu^2-(x-1)u+1-(x-1)u^2+(x+1)u}{(x-1)u-x-1}$
$x\dfrac{du}{dx}=\dfrac{u^2+2u+1}{(u-1)x-u-1}$
$\dfrac{dx}{du}=\dfrac{(u-1)x^2-(u+1)x}{(u+1)^2}$
$\dfrac{dx}{du}+\dfrac{x}{u+1}=\dfrac{(u-1)x^2}{(u+1)^2}$
Luckily this becomes a Bernoulli ODE.
Let $v=\dfrac{1}{x}$ ,
Then $x=\dfrac{1}{v}$
$\dfrac{dx}{du}=-\dfrac{1}{v^2}\dfrac{dv}{du}$
$\therefore-\dfrac{1}{v^2}\dfrac{dv}{du}+\dfrac{1}{(u+1)v}=\dfrac{(u-1)}{(u+1)^2v^2}$
$\dfrac{dv}{du}-\dfrac{v}{u+1}=-\dfrac{(u-1)}{(u+1)^2}$
I.F. $=e^{-\int\frac{du}{u+1}}=e^{-\ln(u+1)}=\dfrac{1}{u+1}$
$\therefore\dfrac{d}{du}\left(\dfrac{v}{u+1}\right)=-\dfrac{(u-1)}{(u+1)^3}$
$\dfrac{v}{u+1}=-\int\dfrac{(u-1)}{(u+1)^3}du$
$\dfrac{1}{(u+1)x}=\int\left(-\dfrac{1}{(u+1)^2}+\dfrac{2}{(u+1)^3}\right)du$
$\dfrac{1}{(u+1)x}=\dfrac{1}{u+1}-\dfrac{1}{(u+1)^2}+c$
$\dfrac{1}{x}=\dfrac{u}{u+1}+c(u+1)$
$\dfrac{1}{x}=\dfrac{\dfrac{y}{x}}{\dfrac{y}{x}+1}+c\left(\dfrac{y}{x}+1\right)$
$\dfrac{xy}{x+y}+c(x+y)=1$
$\dfrac{xy}{x+y}-1=C(x+y)$
Now for $y'=\dfrac{(Ax+By)y+\alpha x+\beta y}{(Ax+By)x+ax+by}$ ,
Let $u=\dfrac{y}{x}$ ,
Then $y=xu$
$\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$
$\therefore x\dfrac{du}{dx}+u=\dfrac{(Ax+Bxu)xu+\alpha x+\beta xu}{(Ax+Bxu)x+ax+bxu}$
$x\dfrac{du}{dx}=\dfrac{(A+Bu)xu+\alpha+\beta u}{(A+Bu)x+a+bu}-u$
$x\dfrac{du}{dx}=\dfrac{Bxu^2+(Ax+\beta)u+\alpha}{(Bx+b)u+Ax+a}-u$
$x\dfrac{du}{dx}=\dfrac{Bxu^2+(Ax+\beta)u+\alpha-(Bx+b)u^2-(Ax+a)u}{(Bx+b)u+Ax+a}$
$x\dfrac{du}{dx}=\dfrac{-bu^2+(\beta-a)u+\alpha}{(A+Bu)x+a+bu}$
$\dfrac{dx}{du}=-\dfrac{(A+Bu)x^2+(a+bu)x}{bu^2+(a-\beta)u-\alpha}$
Which also luckily that this becomes a Bernoulli ODE.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Evaluating $\int _0^{\frac{\pi }{2}}\:\frac{\sqrt[3]{\sin^2\left(x\right)}}{\sqrt[3]{\sin^2\left(x\right)}+\sqrt[3]{\cos^2\left(x\right)}}dx$ how i evaluate this integral with integration by substitution $x=\frac{\pi}{2}-t$
$$\int _0^{\frac{\pi }{2}}\:\frac{\sqrt[3]{\sin^2\left(x\right)}}{\sqrt[3]{\sin^2\left(x\right)}+\sqrt[3]{\cos^2\left(x\right)}}dx$$
| We have $$\begin{align} I=\int_{0}^{\pi/2}\frac{\sqrt[3]{\sin^{2}\left(x\right)}}{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}}dx=
& \int_{0}^{\pi/2}\frac{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}-\sqrt[3]{\cos^{2}\left(x\right)}}{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}}dx
\\ =
& \frac{\pi}{2}-\int_{0}^{\pi/2}\frac{\sqrt[3]{\cos^{2}\left(x\right)}}{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}}dx.
\end{align}$$ And now since holds $$\int_{a}^{b}f\left(x\right)dx=\int_{a}^{b}f\left(a+b-x\right)dx
$$ we have $$I=\frac{\pi}{2}-\int_{0}^{\pi/2}\frac{\sqrt[3]{\cos^{2}\left(\frac{\pi}{2}-x\right)}}{\sqrt[3]{\sin^{2}\left(\frac{\pi}{2}-x\right)}+\sqrt[3]{\cos^{2}\left(\frac{\pi}{2}-x\right)}}dx=\frac{\pi}{2}-\int_{0}^{\pi/2}\frac{\sqrt[3]{\sin^{2}\left(x\right)}}{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}}dx
$$ hence $$\int_{0}^{\pi/2}\frac{\sqrt[3]{\sin^{2}\left(x\right)}}{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}}dx=\frac{\pi}{4}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Odd binomial sum equality has only trivial solution? Suppose $$\sum_{k\ {\rm odd}}^n {n \choose k} 2^{(k-1)/2} = \sum_{k\ {\rm odd}}^m {m \choose k} 2^{(k-1)/2} 3^{(m-k)/2}.$$
Does $m=n=1$?
Clearly $m \leq n$, and for every $n$ there is at most one $m$.
| Here we find a closed expression for the series and show that a solution with $m$ even is not possible.
We use the convention $\binom{n}{k}=0$ if $0\leq n < k$ and start with the left-hand side. We obtain
\begin{align*}
\sum_{k\ {\rm odd}}^n {n \choose k} 2^{(k-1)/2}&= \sum_{k=0}^n \binom{n}{2k+1} 2^k\tag{1}\\
&=\frac{1}{\sqrt{2}}\sum_{k=0}^n\binom{n}{2k+1}\left(\sqrt{2}\right)^{2k+1}\tag{2}\\
\end{align*}
Comment:
*
*In (1) we replace $k$ with $2k+1$ in the summands without changing anything since we add only zeros.
*In (2) we do a rearrangement which is helpful for further steps.
We consider
\begin{align*}
f(x)=\sum_{k=0}^n\binom{n}{k}x^k=(1+x)^n
\end{align*}
and get the odd part of $f(x)$ via
\begin{align*}
\frac{1}{2}\left(f(x)-f(-x)\right)=\sum_{k=0}^n\binom{n}{2k+1}x^{2k+1}
\end{align*}
We obtain from (2)
\begin{align*}
\sum_{k\ {\rm odd}}^n {n \choose k} 2^{(k-1)/2}&=\frac{1}{2\sqrt{2}}\left(f(\sqrt{2})-f(-\sqrt{2})\right)\\
&=\frac{1}{2\sqrt{2}}\left(\left(1+\sqrt{2}\right)^n-\left(1-\sqrt{2}\right)^n\right)\tag{3}
\end{align*}
Similarly we can transform the right-hand side:
\begin{align*}
\sum_{k\ {\rm odd}}^m {m \choose k} 2^{(k-1)/2} 3^{(m-k)/2}&=3^{(m-1)/2}\sum_{k\ {\rm odd}}^m {m \choose k} \left(\frac{2}{3}\right)^{(k-1)/2}\\
&=3^{m/2}\frac{1}{2\sqrt{2}}\left(f\left(\sqrt{\frac{2}{3}}\right)
-f\left(-\sqrt{\frac{2}{3}}\right)\right)\\
&=3^{m/2}\frac{1}{2\sqrt{2}}\left(\left(1+\sqrt{\frac{2}{3}}\right)^m-\left(1-\sqrt{\frac{2}{3}}\right)^m\right)\\
&=\frac{1}{2\sqrt{2}}\left(\sqrt{3}+\sqrt{2}\right)^m-\left(\sqrt{3}-\sqrt{2}\right)^m\tag{4}
\end{align*}
We conclude from (3) and (4) OPs equation is equivalent with
\begin{align*}
\left(1+\sqrt{2}\right)^n-\left(1-\sqrt{2}\right)^n
=\left(\sqrt{3}+\sqrt{2}\right)^m-\left(\sqrt{3}-\sqrt{2}\right)^m\tag{5}
\end{align*}
In case $m=n=1$ we obtain
\begin{align*}
\left(1+\sqrt{2}\right)-\left(1-\sqrt{2}\right)&=2\sqrt{2}\\
\\
\left(\sqrt{3}+\sqrt{2}\right)-\left(\sqrt{3}-\sqrt{2}\right)&=2\sqrt{2}
\end{align*}
and equality holds.
The left-hand side of (5) has the representation
\begin{align*}
\left(1+\sqrt{2}\right)^n&-\left(1-\sqrt{2}\right)^n\\
&=\sum_{k=0}^n\binom{n}{k}\left(\sqrt{2}\right)^k-\sum_{k=0}^n\binom{n}{k}(-1)^k\left(\sqrt{2}\right)^k\tag{6}\\
&=A\sqrt{2}\qquad\qquad \text{ with }A\in\mathbb{N}
\end{align*}
Since odd $k$ only provide contributions to the expression in (6) we observe that the resulting value is an integer multiple of $\sqrt{2}$.
The right-hand side has the representation
\begin{align*}
\left(\sqrt{3}+\sqrt{2}\right)^n&-\left(\sqrt{3}-\sqrt{2}\right)^n\\
&=\sum_{k=0}^n\binom{n}{k}\left(\sqrt{2}\right)^k\left(\sqrt{3}\right)^{n-k}
-\sum_{k=0}^n\binom{n}{k}(-1)^k\left(\sqrt{2}\right)^k\left(\sqrt{3}\right)^{n-k}\tag{7}\\
&=
\begin{cases}
B\sqrt{2}\qquad\qquad &\text{ with }B\in\mathbb{N} \text { if } n \text { is odd}\\
C\sqrt{6}\qquad\qquad &\text{ with }C\in\mathbb{N} \text { if } n \text { is even}\\
\end{cases}
\end{align*}
Again odd $k$ only provide contributions to the expression in (7). If $n$ is odd, $n-k$ is even and the resulting value is an integer multiple of $\sqrt{2}$. Otherwise, if $n$ is even we get additionally a factor $\sqrt{3}$ resulting in an integer multiple of $\sqrt{6}$.
Conclusion: OPs equation is not valid if $m$ is even.
The case $m$ odd needs further investigations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Showing that $\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}L_{2^n+1}}\right)=\frac{3}{\phi^2}$ Infinite product
$F_{n}:=[1,1,2,3,5,8,\cdots]$ and
$L_{n}:=[1,3,4,7,\cdots]$
for $n=1,2,3,\cdots$ respectively.
$\frac{1+\sqrt5}{2}=\phi$
Show that,
$$\prod_{n=1}^{\infty}\left(1+\frac{1}{F_{2^n+1}L_{2^n+1}}\right)=\frac{3}{\phi^2}$$
We took the idea from this site
Expand the product
(1)
$$\left(1+\frac{1}{F_{3}L_{3}}\right)\cdot\left(1+\frac{1}{F_{5}L_{5}}\right)\cdot\left(1+\frac{1}{F_{9}L_{9}}\right)\cdots=\frac{3}{\phi^2}$$
$F_{n}=\frac{\phi^n-(-\phi)^{-n}}{\sqrt5}$
$L_{n}=\phi^n+(-\phi)^{-n}$
$F_{n}L_{n}=\frac{\phi^{2n}-(-\phi)^{-2n}}{\sqrt5}=F_{2n}$
Rewrite (1)
$$\left(1+\frac{1}{F_{6}}\right)\cdot\left(1+\frac{1}{F_{10}}\right)\cdot\left(1+\frac{1}{F_{18}}\right)\cdots=\frac{3}{\phi^2}$$
Still doesn't help much to get from LHS to RHS.
We have $F_{2n}=F^2_{n+1}-F^2_{n-1}$
I have substituted in, but the formula seem too messy and more complicated.
Can anybody please give a hand?
| Use $$F_{4k+2}+1=F_{2k+2}L_{2k}\tag{1}$$ with $k=2^{n-1}$. Also note $$\prod_{k=1}^{n}L_{2^{k}}=\prod_{k=1}^{n}\phi^{2^k}+(-\phi)^{-2^{k}}=\frac{\phi^{2^{k+1}}-(-\phi)^{-2^{k+1}}}{\phi^2-(-\phi)^{-2}}=\frac{F_{2^{k+1}}}{F_{2}}\tag{2} $$ From $\text{(1)}$, note that $$\left(1+\frac{1}{F_{6}}\right)\cdot\left(1+\frac{1}{F_{10}}\right)\cdot\left(1+\frac{1}{F_{18}}\right)\cdots \left(1+\frac{1}{F_{2^{k+1}+2}}\right)=\frac{F_{4}L_{2}L_{4} \dots L_{2^k}}{F_{2^{k+1}+2}} $$ Using $\text{(2)}$, we have $$\frac{F_{4}L_{2}L_{4} \dots L_{2^k}}{F_{2^{k+1}+2}}=\frac{3F_{2^{k+1}}}{F_{2^{k+1}+2}}$$
But $$\lim_{k \to \infty}\frac{3F_{2^{k+1}}}{F_{2^{k+1}+2}}=\frac{3}{\phi^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Number theory contradiction I was solving this question
How many pairs of integers $(a,b)$ that satisfies the following conditions:
*
*$1\le a,b\le 42$
*$a^9\equiv b^7\pmod{43}$
and I ran into a contradiction:
We know by Fermat's Little Theorem that $a^{42} \equiv 1 \pmod{43}$ and $b^{42} \equiv 1 \pmod{43}$. Thus, we make the given condition in the problem equivalent to $a^{54} \equiv b^{42} \equiv 1 \pmod{43}$. Then, $a^{12} \equiv 1 \pmod{43}$.
I don't see the problem in my reasoning here. We have $a^{54} \equiv 1 \pmod{43}$ is equivalent to saying $a^9\equiv b^7\pmod{43}$ since $43$ is prime, but $6^{54} \equiv 1 \pmod{43}$ while $6^9 \equiv 1 \pmod{43}$ and $6^9 \equiv b^7 \pmod{43}$ isn't true for all $b$ as $a^{54} \equiv 1 \pmod{43}$ suggests. So how is it not equivalent?
| 3 is a cyclical generator for $Z_{43}$
Every number in between 1 and 42 can be expressed as $3^n$
$a= 3^n, b = 3^m\\
a^9 \equiv 3^{9n} \mod 43\\
b^7 \equiv 3^{7m} \mod 43$
the only numbers that overlap are $3^{21} \equiv 42$ and $3^{42} \equiv 1.$
$a^9\equiv b^7\equiv 1 \mod 43$
$a = 1, 4, 16,21, 41,35, 11,$ and $b = 1,6,36.$ or
$a^9\equiv b^7\equiv 42 \mod 43$
$a = 42, 39, 27,22, 2,8, 32,$ and $b = 42, 37,7.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding an antiderivative $\newcommand{\dx}{\,\mathrm dx}$
I need to find the following:
$$\int \sqrt{\frac{1-x}{1+x}} \cdot \frac 1x \dx$$
Firstly, it has to be that $x\in (-1, 0)\cup (0,1]$. From this, it is implied that $1+x>0$. Also, it holds that $1-x \ge 0$.
What I did is that I multiplied both numerator and denominator by $\sqrt{1+x}$. Thus, we have:
$$ \int \frac{\sqrt{1-x^2}}{1+x}\cdot \frac 1x\dx \begin{array}[t]{l}
\displaystyle\, \,\overset{x = \sin t}{=} \int \frac{\cos t}{1+\sin t}\cdot \frac{1}{\sin t}\cdot \cos t \, \mathrm dt\\[3ex]
\quad = \displaystyle \int \frac{\cos^2 t}{(1+\sin t)\sin t}\,\mathrm dt\\[3ex]
\quad = \displaystyle \int \frac{1-\sin^2 t}{(1+\sin t)\sin t}\, \mathrm dt\\[3ex]
\quad = \displaystyle \int \frac{(1-\sin t)(1+\sin t)}{(1+\sin t)\sin t}\, \mathrm dt = \displaystyle \int \frac{1}{\sin t} - 1\, \mathrm dt.
\end{array}$$
Thus, the final answer is:
$$\log\left| \tan \left(\frac{\arcsin x}{2}\right)\right| - \arcsin x + C,$$
since $\int \frac{1}{\sin t} \, \mathrm dt = \log\left| \tan \left(\frac {t}{2}\right) \right| + c.$
The answer given in the textbook is:
$$ \log\left|\frac{\sqrt{1-x}-\sqrt{1+x}}{\sqrt{1-x}+\sqrt{1+x}}\right| + 2\arctan \left(\sqrt{\frac{1-x}{1+x}}\right) +constant$$
I have derived that the arguments in the logarithms are the same, but it seems difficult to connect $\arcsin x $ with $2\arctan \left(\sqrt{\frac{1-x}{1+x}}\right).$
| Let $x=\cos2y\implies0\le2y\le\pi$
$\sqrt{\dfrac{1-x}{1+x}}=\tan y$
$\implies2\arctan\sqrt{\dfrac{1-x}{1+x}}=2y=\arccos x=\dfrac\pi2-\arcsin x$
But why don't we try with
$$\sqrt{\dfrac{1-x}{1+x}}=u\implies x=\dfrac{1-u^2}{1+u^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1797857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating the definite integral $\int_0^3 \sqrt{9- x^2} \, dx$ I have been having a problem with the following definite integral:
$$\int_0^3 \sqrt{9- x^2} \, dx
$$
I am only familiar with u-substitution and am positive that it can be done with only that. Any help would be appreciated.
| Here's a quick method: If $y = \sqrt{9-x^2}$ then $y^2 = 9-x^2$ so $x^2+y^2=9$. If you know that $x^2 + y^2 = 9$ is the equation of a circle of radius $3$ centered at $(0,0)$, that means $y^2=9-x^2$ is also an equation of that circle, as is $y = \pm\sqrt{9-x^2}$. So $y = \sqrt{9-x^2}$ (without $\text{“}\pm\text{''}$) is the top half of the circle. So the integral is just the area of one quarter of the circle. It you know the area is $\pi r^2$ and $r=3$, that tells you the answer.
(Of course, this won't work if the point is to prove that $\pi r^2$ is the area.)
Another method:
\begin{align}
x & = 3\sin\theta \\
dx & = 3\cos\theta\,d\theta \\
\sqrt{9-x^2} & = \sqrt{9 - 9\sin^2\theta} = \sqrt{9} \sqrt{1-\sin^2\theta} = 3\cos\theta \\[6pt]
& \text{As $x$ goes from $0$ to $3$ then $\sin\theta$ goes from $0$ to $1$,} \\
& \text{so $\theta$ goes from $0$ to $\pi/2$.}
\end{align}
Hence the integral becomes
\begin{align}
& 9\int_0^{\pi/2} \cos^2\theta\,d\theta = 9\int_{\pi/2}^0 \cos^2\left(\frac\pi2 - \zeta\right) \,(-d\zeta) & & (\text{The substitution is $\theta=\frac\pi2-\zeta$}) \\[10pt]
= {} & 9 \int_{\pi/2}^0 (\sin^2\zeta) \,(-d\zeta) & & \Big(\text{since $\cos\left(\frac\pi2 - \zeta\right)= \sin\zeta$}\Big) \\[10pt]
= {} & 9 \int_0^{\pi/2} (\sin^2\zeta)\,d\zeta \\[10pt]
= {} & 9 \int_0^{\pi/2} \sin^2\theta\,d\theta.
\end{align}
So we have $$ 9 \int_0^{\pi/2} \cos^2\theta\,d\theta = 9 \int_0^{\pi/2} \sin^2\theta\,d\theta. $$
But on the other hand
$$
9 \int_0^{\pi/2} \cos^2\theta\,d\theta + 9 \int_0^{\pi/2} \sin^2\theta\,d\theta = 9 \int_0^{\pi/2} 1\,d\theta \qquad\qquad(\text{since } \cos^2\theta+\sin^2\theta =1.)
$$
The latter integral comes to $9\pi/2$, so each of the two integrals separately must be half of that, or $9\pi/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Writing answer for $2\cos x+\sin2x=0$ $$2\cos x + \sin 2x = 0\implies
\cos x = 0 \; \&\; \sin x = -1$$
So, the solution my book provides is
$x = π/2 + 2nπ\; \&\; x = 3π/2 + 2nπ$
Why is $3π/2$ (for $\cos$) not included in the general solution set?
|
$$2\cos x+\sin2x=0\\2 \cos x+ 2 \sin x \cos x=0 \\ \to 2 \cos x (1+\sin x)=0\\ \to \\ sinx=-1 ,cosx=0$$
With respect to figure $$\begin{cases}cos=0 & x = \pm \frac{\pi}{2}+2k\pi\\sinx=-1 & x=-\frac{\pi}{2}+2k\pi\end{cases} $$ and $$x=\pm \frac{\pi}{2} +2k\pi$$ include $x=\frac{3\pi}{2} ,\frac{-3\pi}{2},\frac{\pi}{2},\frac{-\pi}{2},...$so we can rewrite it as $\frac{3\pi}{2}+2k\pi \\or\\ \frac{-\pi}{2}+2k\pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{x+y+z}{3}+\frac{3}{\frac1x+\frac1y+\frac1z}\geq5\sqrt[3]{\frac{xyz}{16}}$ Let $x,y,z>0$. Prove that
$$\frac{x+y+z}{3}+\frac{3}{\frac1x+\frac1y+\frac1z}\geq5\sqrt[3]{\frac{xyz}{16}}$$
On the left-hand side we have arithmetic and harmonic means, while on the right geometric mean. The AM-GM-HM inequality states that AM $\geq$ GM $\geq$ HM, but here HM is on the wrong side of the inequality.
| Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, we need to prove that $f(u)\geq0,$ where
$$f(u)=u+\frac{w^3}{v^2}-\frac{5w}{\sqrt[3]{16}}.$$
Id est, it's enough to prove our inequality for the minimal value of $u$, which happens for equality case of two variables.
Since our inequality is symmetric and homogeneous, we can assume $y=z=1$ and $z=16a^3$
and we need to prove that:
$$\frac{16a^3+2}{3}+\frac{48a^3}{32a^3+1}\geq5a$$ or
$$(4a-1)^2(32a^4+16a^3-24a^2+a+2)\geq0$$ or
$$(4a-1)^2(2(4a^2-1)^2+a(4a-1)^2)\geq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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If $\sin^{-1}x + \sin^{-1}y=\sin^{-1}(x \sqrt{1-y^{2}}+ y \sqrt{1-x^{2}})$, then what is the area represented by the locus of point $(x, y)$? If $\sin^{-1}x + \sin^{-1}y=\sin^{-1}(x \sqrt{1-y^{2}}+ y \sqrt{1-x^{2}})$, then what is the area represented by the locus of point $(x, y)$? I'm totally blank about this question so please explain clearly and state all the basic steps!
| $$\alpha=sin^{-1}x\to sin\alpha=x \,\,\,\,\,and\,\,cos\,\alpha=\sqrt{1-sin^2 \alpha}=\sqrt{1-x^2}$$
$$\beta=sin^{-1}y\to sin\beta=y \,\,\,\,\,and\,\,cos\,\beta=\sqrt{1-sin^2\beta}=\sqrt{1-y^2}$$
let $t=sin^{-1}x+sin^{-1}y=\alpha+\beta$ , then
$$sin(t)=sin(\alpha+\beta)=sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)$$
$$sin(t)=x\sqrt{1-y^2}+\sqrt{1-x^2}y$$
$$t=sin^{-1}(x\sqrt{1-y^2}+\sqrt{1-x^2}y)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum and minimum values of $\sin^2\theta+\sin^2\phi$ when $\theta+\phi=\alpha$ Find the maximum and minimum values of $\sin^2\theta+\sin^2\phi$ when $\theta+\phi=\alpha$(a constant).
$\theta+\phi=\alpha\implies\phi=\alpha-\theta$
$\sin^2\theta+\sin^2\phi=\sin^2\theta+\sin^2(\alpha-\theta)$
Let $f(\theta)=\sin^2\theta+\sin^2(\alpha-\theta)$
$f'(\theta)=2\sin\theta\cos\theta-2\sin(\alpha-\theta)\cos(\alpha-\theta)$
Putting $f'(\theta)=0$ gives $\sin2\theta=\sin2(\alpha-\theta)$
$2\theta=2\alpha-2\theta\implies \alpha=2\theta$
If $\alpha=2\theta$,then by $\theta+\phi=\alpha$ gives $\phi=\theta$
I am stuck here,the answer given is maximum $1+\cos\alpha$ and minimum $1-\cos\alpha$.But i have found only one critical value(when $\phi=\theta$) and that too i cannot decide whether it will give maximum or minimum value.
Please help.
| Write
\begin{eqnarray}
\sin^2 \phi + \sin^2 \theta &=& \sin^2 \phi + 1-\cos^2 \theta \\
&=& 1+(\sin \phi - \cos \theta) (\sin \phi + \cos \theta) \\
&=& 1+(\sin \phi - \sin ( { \pi \over 2} - \theta)) (\sin \phi + \sin ( { \pi \over 2} - \theta)) \\
&=& 1+4 \cos ({ \phi -\theta + { \pi \over 2}\over 2} ) \sin ({ \phi +\theta - { \pi \over 2}\over 2} ) \sin ({ \phi -\theta + { \pi \over 2}\over 2} ) \cos ({ \phi +\theta - { \pi \over 2}\over 2} ) \\
&=& 1+ \sin (\phi -\theta + { \pi \over 2}) \sin ( \phi +\theta - { \pi \over 2} ) \\
&=& 1 - \sin (\phi -\theta + { \pi \over 2}) \cos \alpha
\end{eqnarray}
We can choose $\phi -\theta$ to have arbitrary values while maintaining
the relationship $\phi+\theta = \alpha$, and so the above can be written as
$1-\sin t \cos \alpha$, which can be easily extremized,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Solve first order partial diferential equation Consider the Cauchy problem:
$$\left\{\begin{array}{lll}
x^2\partial_x u+y^2\partial_yu=u^2\\
u(x,2x)=1
\end{array}\right.$$
It is easy to show that the characteristic equations are given by:
$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{px^2+qy^2}=\frac{dp}{2pu-2xp}=\frac{dq}{2qu-2yq}=dt$$
By the Lagrange-Charpit's method, we need a first integral $\Psi$ different to $\Phi=x^2p+y^2q-u^2$.
How I can find $\Psi$? Maybe we should get directly the characteristic curves of the previous system?
Many thanks!
| Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x^2$ , letting $x(1)=-1$ , we have $-\dfrac{1}{x}=t$
$\dfrac{dy}{dt}=y^2$ , we have $-\dfrac{1}{y}=t+y_0=-\dfrac{1}{x}+y_0$
$\dfrac{du}{dt}=u^2$ , we have $\dfrac{1}{u}=-t+f(y_0)=\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)$ , i.e. $u(x,y)=\dfrac{1}{\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)}$
$u(x,2x)=1$ :
$\dfrac{1}{\dfrac{1}{x}+f\left(\dfrac{1}{x}-\dfrac{1}{2x}\right)}=1$
$\dfrac{1}{x}+f\left(\dfrac{1}{2x}\right)=1$
$f\left(\dfrac{1}{2x}\right)=1-\dfrac{1}{x}$
$f(x)=1-\dfrac{1}{\dfrac{1}{2x}}=1-2x$
$\therefore u(x,y)=\dfrac{1}{\dfrac{1}{x}+1-2\left(\dfrac{1}{x}-\dfrac{1}{y}\right)}=\dfrac{1}{1+\dfrac{2}{y}-\dfrac{1}{x}}=\dfrac{xy}{xy+2x-y}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof that $3^n | 2^{3^n} + 1$ Question:
Proof by induction that $3^n | 2^{3^n} + 1$.
Attempt: $$ 2^{3^{n+1}} + 1 = 2^{3^n} 2^3 + 1 = 2^{3^n} 2^3 + 1 + 2^3 - 2^3 =
2^3( 2^{3^n} + 1 ) + 1 -2^3$$
And the first is $3^n |$ but second I don't know how to proof that.
| Hint: Use the fact that $$(2^{3^n})^3 + 1 = \left(2^{3^n} +1\right) \left((2^{3^n})^2- 2^{3^n} + 1\right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Which is the explanation of the identity $\sum_{k=0}^n {n \choose k}k^2 = 2^{n-2}n(n+1)$? A friend of mine put me a problem some time ago, and after trying to do it, I finally surrendered. I looked for the answer online so maybe I could guess why is it like it is, but I just can't understand anything about it. Could someone explain, rather using logic, why is the identity as it is?
$$\sum_{k=0}^n {n \choose k}k^2 = 2^{n-2}n(n+1)$$
| By definition,
$$\binom nk=\frac{n!}{k!(n-k)!},$$ so that
$$\binom nkk=\frac{n!}{(k-1)!(n-k)!}==\frac{n(n-1)!}{(k-1)!((n-1)-(k-1))!}=\binom {n-1}{k-1}n.$$
This trick allows you to "absorb" the factor $k$ inside the binomial number.
The trick doesn't work with $k^2$ because the denominator doesn't have a second factor $k$, but it does work with $k^2-k$:
$$\binom nkk(k-1)=\frac{n!}{(k-2)!(n-k)!}==\frac{n(n-1)(n-2)!}{(k-2)!((n-2)-(k-2))!}=\binom {n-2}{k-2}n(n-1).$$
Adding these two results and summing over $k$,
$$S=2^{n-1}n+2^{n-2}n(n-1)=2^{n-2}n(n+1).$$
Check with $n=5$.
$$\begin{align}k&\to\color{blue}1\cdot0+\color{blue}5\cdot1+\color{blue}{10}\cdot2+\color{blue}{10}\cdot3+\color{blue}5\cdot4+\color{blue}1\cdot5=2^4\cdot 5=80\\
k^2-k&\to\color{blue}1\cdot0\cdot\bar1+\color{blue}5\cdot1\cdot0+\color{blue}{10}\cdot2\cdot1+\color{blue}{10}\cdot3\cdot2+\color{blue}5\cdot4\cdot3+\color{blue}1\cdot5\cdot4=2^3\cdot 5\cdot4=160\end{align}$$
and summing,
$$k^2\to\color{blue}1\cdot0^2+\color{blue}5\cdot1^2+\color{blue}{10}\cdot2^2+\color{blue}{10}\cdot3^2+\color{blue}5\cdot4^2+\color{blue}1\cdot5^2=2^3\cdot 5\cdot6=240.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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For the general ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ,show that the midpoints of the chords lie on a straight line.
Question: A collection of parallel chords connect pairs of points on an ellipse, as shown
For the general ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, show that the midpoints of the chords lie on a straight line.
What I have tried/attempted:
Say that a line in the form $y=mx+c$ intersects the ellipse
so
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
$$ \Leftrightarrow x^2b^2 + y^2a^2 = a^2b^2 $$
$$ \Leftrightarrow x^2b^2 + (mx+c)^2a^2 = a^2b^2 $$
$$\Leftrightarrow x^2b^2 + (m^2x^2+2mcx+c^2)a^2=a^2b^2 $$
$$ \Leftrightarrow x^2b^2 + a^2m^2x^2+2mca^2x+c^2a^2 - a^2b^2 = 0 $$
$$ \Leftrightarrow (b^2+a^2m^2)x^2 + (2mca^2)x + (c^2a^2-a^2b^2) = 0 $$
$$\Leftrightarrow (b^2+a^2m^2)x^2 + (2mca^2)x + a^2(c^2-b^2) = 0 $$
Now I am stuck should I be doing something with this equation I have formed? Or is there a more easy geometrical approach to this question I am missing.
| $$\Leftrightarrow (b^2+a^2m^2)x^2 + (2mca^2)x + a^2(c^2-b^2) = 0 $$
By the midpoint of a parabola, the $x$-value of the midpoint of the intersection of the line and the ellipse will be at
$$\frac{-2mca^2}{2(b^2+a^2m^2)} = \frac{-mca^2}{b^2+a^2m^2}$$
Note that the $y$ value of the midpoint is linearly dependent on the $x$ value of the midpoint $(y=mx+c)$. Therefore, this expression shows that for any fixed values of $a$, $b$, and $m$, the collection of midpoints for ranging $c$ values is linearly dependent on the variable $c$.
For a concrete example, take the ellipse $\frac{x^2}{4} +\frac{y^2}{9}=1$ intersected by lines with slope $m=3$. The $x$ coordinate for the midpoint of the chords will be
$$x_c=\frac{-mca^2}{b^2+a^2m^2}=\frac{-3\cdot c\cdot4}{9+4\cdot 9}=\frac{-4c}{15}$$
and we can find the corresponding $y$ coordinates of the midpoint by plugging it into our linear equation
$$y_c=mx+c=3\cdot \frac{-4c}{15}+c=\frac{1}{5}c$$
So we have the collection of midpoints $(\frac{-4c}{15},\frac{1}{5}c)$ which is obvious to show lies on a straight line for changing values of $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is the sum of the prime factors of $2^{16}-1$? I know $2^{10}=1024$ and $2^6=64$, but it seems they are not very helpful in solving this problem. There must be a trick to solve the problem in an easy way.
What is the sum of the prime factors of $2^{16}-1$?
| Using $a^2-1=(a-1)(a+1)$ we have:
$$2^{16}-1=(2^8-1)(2^8+1)=(2^4-1)(2^4+1)(2^8+1)=15\cdot 17\cdot 257=3\cdot 5\cdot 17\cdot 257$$
So, the answer is $3+5+17+257=282$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Subtracting $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$ $\frac{(x+3)}{(x^2-1)} - \frac{(x-2)}{(x^2+2x+1)}$
To solve the problem I first dissembled the equation on the denominator
$ \frac{(x+3)}{(x-1)*(x+1)} - \frac{(x-2)}{(x+1)^2}$
I multiplied the denominator together and to do this, I think I have to multiply the top part as well right? This is where i get confused, I forgot how to do this problem as its been a long time. How would i go on to solve this?
| $$ \frac{x+3}{(x-1)(x+1)} - \frac{x-2}{(x+1)^2}=\frac{(x+3)\color{red}{(x+1)}-(x-2)\color{red}{(x-1)}}{(x-1)(x+1)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Recurrence Relation with two parameters and Summation This is a recurrence relation with two parameters which came up in a problem I was trying to solve.
Given
$$\begin{align}&A_n=pB_{n-1};\qquad &&B_n=q(A_{n-1}+B_{n-1})\\
&A_4=p; \qquad &&B_4=q; \\
&p+q=1;&&(0< p,q< 1)\end{align}$$
evaluate
$$\sum_{n=4}^\infty n(A_n+B_n)$$
From the above it can be seen that
$$A_n+B_n=q(A_{n-2}+B_{n-2})+pqB_{n-2}$$
and also that
$$B_n-B_{n-1}=qA_{n-1}-A_n$$
but it is not clear how these can help with the required summation. Logically one should first attempt to express $A_n, B_n$ explicitly in terns of $n$.
Altenratively the original recurrence equation can be represented as
$$\left(\begin{matrix}A_{n}\\B_n\end{matrix}\right)
=\left(\begin{matrix}0&p\\1-p&1-p\end{matrix}\right)
\left(\begin{matrix}A_{n-1}\\B_{n-1}\end{matrix}\right)$$
Suggestions on how to proceed would be appreciated.
| From your own calculations, notice that you may write
$$\begin{pmatrix} A_n \\ B_n \end{pmatrix} = \begin{pmatrix} 0 & p \\ q & q \end{pmatrix} \begin{pmatrix} A_{n-1} \\ B_{n-1} \end{pmatrix} = \begin{pmatrix} 0 & p \\ q & q \end{pmatrix}^2 \begin{pmatrix} A_{n-2} \\ B_{n-2} \end{pmatrix} = \dots = \begin{pmatrix} 0 & p \\ q & q \end{pmatrix}^{n-4} \begin{pmatrix} A_4 \\ B_4 \end{pmatrix} \ \forall n \ge 4 .$$
Let $M = \begin{pmatrix} 0 & p \\ q & q \end{pmatrix}$.
Notice that $A_n + B_n$ can be identified with the $1 \times 1$ matrix
$$\begin{pmatrix} A_n + B_n \end{pmatrix} = \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} A_n \\ B_n \end{pmatrix} = \begin{pmatrix} 1 & 1 \end{pmatrix} M^{n-4} \begin{pmatrix} A_4 \\ B_4 \end{pmatrix} ,$$
therefore, if you place your sum inside a $1 \times 1$ matrix, it becomes
$$(A_4 + B_4) + \sum \limits _{n \ge 5} n \begin{pmatrix} 1 & 1 \end{pmatrix} M^{n-4} \begin{pmatrix} A_4 \\ B_4 \end{pmatrix} = (1) + \begin{pmatrix} 1 & 1 \end{pmatrix} \sum \limits _{n \ge 5} n M^{n-4} \begin{pmatrix} A_4 \\ B_4 \end{pmatrix} = \\
(1) + \begin{pmatrix} 1 & 1 \end{pmatrix} \sum \limits _{k \ge 1} (k+4) M^k \begin{pmatrix} A_4 \\ B_4 \end{pmatrix} = \\
(1) + \begin{pmatrix} 1 & 1 \end{pmatrix} \sum \limits _{k \ge 1} k M^k \begin{pmatrix} A_4 \\ B_4 \end{pmatrix} + 4 \begin{pmatrix} 1 & 1 \end{pmatrix} \sum \limits _{k \ge 1} M^k \begin{pmatrix} A_4 \\ B_4 \end{pmatrix} .$$
The usual formulae for numbers are valid for matrices too, i.e.
$$\sum \limits _{k \ge 1} M^k = M (1-M)^{-1}, \qquad \sum \limits _{k \ge 1} k M^k = M (1-M)^{-2}$$
therefore your sum (as a matrix) becomes
$$(1) + \begin{pmatrix} 1 & 1 \end{pmatrix} M (1 - M)^{-1} \begin{pmatrix} A_4 \\ B_4 \end{pmatrix} + 4 \begin{pmatrix} 1 & 1 \end{pmatrix} M (1 - M)^{-2} \begin{pmatrix} A_4 \\ B_4 \end{pmatrix} .$$
It is easy to compute $(1-M)^{-1} = \frac 1 {p^2} \begin{pmatrix} p & p \\ q & 1 \end{pmatrix}$, so $(1-M)^{-2} = \frac 1 {p^4} \begin{pmatrix} p & p^2 + p \\ pq + q & pq + 1 \end{pmatrix}$, whence everything follows (honestly, I don not feel like performing matrix multiplications, it seems to me that you are more than able to finish this).
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$ If $$x=(a+\sqrt {a^2+b^3})^\frac {1}{3} + (a-\sqrt {a^2+b^3})^\frac {1}{3}$$ then prove that $x^3+3bx=2a$.
By observing the given question, I thought about cubing on both sides. But it becomes quiet vague and complex.
Can anyone help me with a simpler proof?
| Hint: $$(x+y)^3=x^3+y^3+3x^2y+3xy^2=x^3+y^3+3xy(x+y)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\int_{1/10}^{1/2}\left(\frac{\sin{x}-\sin{3x}+\sin{5x}}{\cos{x}+\cos{3x}+\cos{5x}}\right)^2dx$ How can I evaluate the following integral?
$$\int_{1/10}^{1/2}\left(\frac{\sin{x}-\sin{3x}+\sin{5x}}{\cos{x}+\cos{3x}+\cos{5x}}\right)^2dx$$
I notice that the numerator and the denominator are very similar. So my direction is to evaluate this integral by substitution. However, I cannot find a suitable substitution so that the integral can be nice.
| Using product-to-sum formula, we have
$$
\begin{aligned}
2 \cos x(\sin x-\sin 3 x+\sin 5 x) = & \sin 2 x-(\sin 4 x+\sin 2 x)+(\sin 6 x+\sin 4 x) \\
= & \sin 6 x \cdots (1)
\end{aligned}
$$
and
$$
\begin{aligned}
2 \sin x(\cos x+\cos 3 x+\cos 5 x) = & \sin 2 x+(\sin 4 x-\sin 2 x)+(\sin 6 x-\sin 4 x) \\
= & \sin 6 x \cdots (2)
\end{aligned}
$$
$(1)\div(2)$ yields
$$
\frac{\sin x-\sin 3 x+\sin 5 x}{\cos x+\cos 3 x+\cos 5 x}=\tan x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find an expression for $\frac{dy}{dx}$ in terms of $x$ and $y$ and verify that $P$ is a stationary point. A curve is defined by the equation $$2y+e^{2x}y^2=x^2+\frac{2}{e}$$
Find an expression for $\frac{dy}{dx}$ in terms of $x$ and $y$
\begin{align}
2y+e^{2x}y^2 & = x^2+\frac{2}{e} \\
2\frac{dy}{dx}+2e^{2x}y^2+2e^{2x}y\frac{dy}{dx} & = 2x \\
\frac{dy}{dx}+e^{2x}y^2+e^{2x}y\frac{dy}{dx} & = x \\
\frac{dy}{dx}\left(1+e^{2x}y\right) & = x-e^{2x}y^2 \\
\frac{dy}{dx} & = \frac{x-e^{2x}y^2}{1+e^{2x}y} \\
\end{align}
Verify that $P$ $(1, \frac{1}{e})$ is a stationary point on the curve.
Stationary point when $\frac{dy}{dx}=0$
$$\frac{x-e^{2x}y^2}{1+e^{2x}y}$$
\begin{align}
& = \frac{(1)-e^{2(1)}(\frac{1}{e})^2}{1+e^{2(1)}(\frac{1}{e})} \\
& = \frac{1-e^{2}e^{-2}}{1+e^{2}e^{-1}} \\
& = \frac{1}{1+e} \\
\end{align}
$$\frac{1}{1+e}\neq0$$
Thanks
| You have a mistake in the final part because:
$$
e^2e^{-2}=1
$$
so
$$
1-e^2e^{-2}=0
$$
and
$$
\frac{1-e^{2}e^{-2}}{1+e^{2}e^{-1}}=\frac{0}{1+e}=0
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of k when $f(x) = x^3 -4x^2 + 6x + k$ If $ \ \ a ,b \ \ $and $\ \ c \ \ $ are roots of $ \ \ f(x) = x^3 -4x^2 + 6x + k \ \ $
and $ \frac{1}{a^2 + b^2} + \frac{1}{b^2 + c^2} + \frac{1}{a^2 + c^2} = 1 \ \ \ \ $then find the value of $ \ k $
I have to tell you honestly that I cannot solve this question or find the way to get value of $ \ k$
I started with $(a + b +c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ac) \ \ $ $\Rightarrow a^2 + b^2 + c^2 = 4$
then $a^3 +b^3 + c^3 = (a+b+c)(a^2 + b^2 +c^2) - (ab+bc+ac)(a+b+c) + 3abc$
$ \ \ \ \ \ \ \ \ a^3 +b^3 + c^3 = -8 + 3k$
apply to $(a+b+c)^3 = a^3 + b^3 + c^3 + 3(a+b)(b+c)(a+c)$
$ \ \ \ \ \ \ \ \ \ \ \ \ \ 72 -3k = 3(a+b)(b+c)(a+c)$
finally , I don't know how relation between $ \frac{1}{a^2 + b^2} + \frac{1}{b^2 + c^2} + \frac{1}{a^2 + c^2} = 1 \ \ $ and $(a+b)(b+c)(a+c)$
Please tell me the way to solve this question correctly.
thank you in advance.
| HINTS to find $k^2$: Let $x=a^2+b^2$, $y=b^2+c^2$, $z=c^2+a^2$. Then $xy+yz+zx=(a^2+b^2+c^2)^2$ and you also know $xyz$ and $x+y+z$. Thus you know everything about $x,y,z$. Now $8a^2b^2c^2=(x+y-z)(x-y+z)(-x+y+z)$ and the R.H.S. is a symmetric polynomial in $x,y,z$, so you can find $a^2b^2c^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Proof of Leibniz $\pi$ formula I found the following proof online for Leibniz's formula for $\pi$:
$$\frac{1}{1-y}=1+y+y^2+y^3+\ldots$$
Substitute $y=-x^2$:
$$\frac{1}{1+x^2}=1-x^2+x^4-x^6+\ldots$$
Integrate both sides:
$$\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots$$
Now plug in $1$ for $x$:
$$\frac{\pi}{4}=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\ldots$$
The thing I am confused about is that the Taylor expansion for $\frac{1}{1-y}$ only works for $-1<y<1$. Why is this still a legitimate proof? At the end, we compute the integral of both sides to $y=-1$.
| Yes, you need an extra step. You need to know this: If
$$
\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots
$$
holds for $-1 < x < 1$ and the series on the right side converges at $x=1$, then the equation also holds at $x=1$.
Arthur beat me to doing it.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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Calculate equation of lines tangent to $\frac{x^2}{x-1}$ but also going through $(2,0)$ As the question states, I want to calculate the equations of two lines tangent to $\frac{x^2}{x-1}$, but also going through point $(2,0)$. Here's what I did:
Suppose there is some point at which the line touches the curve, $a$, which will obviously obey the equation $\left(a,\frac{a^2}{a-1}\right)$. The derivative is calculated as
\begin{align*}
f(x) = \frac{x^2}{x-1} = x^2(x-1)^{-1} \Rightarrow f'(x) = 2x(x-1)^{-1} -x^2(x-1)^{-2} = \frac{x^2-2}{(x-1)^2} \Rightarrow f'(a) = \frac{a^2-2}{(a-1)^2}.
\end{align*}
Filling this into the equation we get
\begin{align*}
y - \left(\frac{a^2}{a-1}\right) = \frac{a^2-2}{(a-1)^2}\left(x-a\right).
\end{align*}
Obviously this equation will contain the point $(2,0)$, so we fill it in and obtain the answer:
\begin{align*}
\frac{a^2}{a-1} = \frac{a^2-2}{(a-1)^2}\left(a-2\right) \Rightarrow a^2 +2a-4=0 \Rightarrow (a+1)^2=5 \Rightarrow a = -1\pm\sqrt 5.
\end{align*}
The equations are getting so complicated I'm pretty sure I'm doing it wrong, but I'm not quite sure what I'm doing wrong so I'm wondering if StackExchange could offer some input on my method?
| There's a simpler way to solve: take a line through $(2,0)$ with variable slope $t$:
$$y=t(x-2),$$
and write the equation for the intersection points with the curvbe has a double root:
$$t(x-2)=\frac{x^2}{x-1}\iff x^2=t(x-1)(x-2)\iff (1-t)x^2+3tx-2t=0.$$
There's a double root if and only if
$$\Delta=9t^2+8t(1-t)=t(t+8)=0,\enspace \text{i.e.}\quad t=0,\,-8.$$
Furthermore, the double root is equal to $\;x=\dfrac{3t}{2(t-1)}=0,\,\dfrac 43.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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evaluate if integral converge: $ \int_2^\infty \frac{2}{x(x+1)(x-1)} dx $ Evaluate if the following integral converges:
$$
\int_2^\infty \frac{2}{x(x+1)(x-1)} dx
$$
Here I go:
$$ 2\int_2^\infty \frac{1}{x(x+1)(x-1)} dx $$
Partial fractions of $\dfrac{1}{x(x+1)(x-1)}$ :
$$\frac{1}{x(x+1)(x-1)} =\frac{A}{x}+\frac{B}{(x+1)} + \frac{C}{(x-1)} = \frac{-1}{x}+\frac{1/2}{(x+1)} + \frac{1/2}{(x-1)} $$
Back to the integral:
$$2(-\int_2^\infty\frac{1}{x}+\frac{1}{2}\int_2^\infty\frac{1}{(x+1)} + \frac{1}{2}\int_2^\infty\frac{1}{(x-1)})dx$$
$$\lim_{a\to2..{b\to\infty}}2[(-\ln(x)+\frac{\ln(x+1)}{2} + \frac{\ln(x-1)}{2}]_a^b$$
$$2[(-\ln(b)+\frac{\ln(b+1)}{2} + \frac{\ln(b-1)}{2}]-2[(-\ln(2)+\frac{\ln(3)}{2} + \frac{\ln(1)}{2}] $$
$$(-\infty+\infty+\infty) + (\ln(3/4)) $$
So, the integral does not converge? I am unsure about the last part, if anyone can confirm me the answer or explain why it's wrong... thank you guys!
| The integral function is non-negative over the integration range and behaves like $\frac{1}{x^3}$ for large values of $x$, hence it is integrable for sure. Partial fraction decomposition gives:
$$\frac{2}{(x-1)x(x+1)} = \frac{1}{x(x-1)}-\frac{1}{x(x+1)}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{2}{x}+\frac{1}{x+1}\right)\tag{1}$$
hence:
$$ \int_{2}^{M}\frac{2\,dx}{(x-1)x(x+1)}=\log\left(\frac{4}{3}\right)+\log\left(1-\frac{1}{M^2}\right)\tag{2} $$
and by letting $M\to +\infty$:
$$ \int_{2}^{+\infty}\frac{2\,dx}{(x-1)x(x+1)}=\color{red}{\log\left(\frac{4}{3}\right)}.\tag{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Convert $16x^2+56x-80=0$ to the form of $(\text{something})^2=D$ Sorry about this very basic question, I want to convert the
equation $16x^2+56x-80=0$ to the form $(\text{something})^2=D$, I know that the answer is $(4x+7)^2-129$, but how can I convert this without usnig calculator
| You are "completing the square". The firs thing I would do is factor "16" out of the first two terms: $16(x^2+ \frac{7}{2}x)- 80= 0$. The reason I did that I that I know that a "perfect square" is of the form $(x+ a)^2= x^2+ 2ax+ a^2$. Compare that to $x^2+ \frac{7}{2}x$. The first two terms will be the same if $2a= \frac{7}{2}$ or $a= \frac{7}{4}$. In that case, $a^2= \left(\frac{7}{4}\right)^2= \frac{49}{16}$. We can get a "perfect square" by adding $\frac{49}{16}$. Of course, in order not to change the value, we have to subtract that also:
$16(x^2+ \frac{7}{2}x+ \frac{49}{16}- \frac{49}{16})- 80= 16(x^2+ \frac{7}{2}x+ \frac{49}{16})- 49- 80= 16(x- \frac{7}{4})^2- 129$.
Putting the "16" backinside the square will give $(4(x- \frac{7}{4}))^2= (4x- 7)^2$ so we have $(4x- 7)^2- 129= 0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help in proving an inequality Show that $$a^4 + b^4\ge\frac{1}{8}$$ if $a+b=1.$
| $\text {let: } k = \frac {(b-a)}{2}\\
\text {then: } b = \frac {(b+a)}{2} + \frac {(b-a)}{2} = \frac 12 + k\\
a = \frac {(b+a)}{2} - \frac {(b-a)}{2} = \frac 12 - k$
$a^4 + b^4\\
(\frac 12 - k)^4 + (\frac 12 + k)^2\\
[(\frac 12)^4 - 4(\frac 12)^3k + 6(\frac 12)^2k^2 -4(\frac 12)k^3 + k^4] + [(\frac 12)^4 + 4(\frac 12)^3k + 6(\frac 12)^2k^2 +4(\frac 12)k^3 + k^4]\\
(\frac 12)^4 + 6(\frac 12)^2k^2 + k^4\\
\frac 18 + 6(\frac 12)^2k^2 + k^4$
$k^2>0, k^4 >0$ for all k.
$\frac 18 + 6(\frac 12)^2k^2 + k^4 > \frac 18$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: $|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$
$$|a\sin x+b \cos x|\leq \sqrt{a^2+b^2}$$
I have tried: $$|a\sin x+b \cos x|\leq |a+b|\leq \sqrt{a^2+b^2}$$
enough to prove: $$|a+b|\leq \sqrt{a^2+b^2}$$
But I can find how to continue from here
| Take two vectors: $v_1 = (a,b)$ and $v_2 = (\sin x, \cos x)$. Their scalar product is $(v_1, v_2) = |v_1||v_2|cos(\phi) = a \sin x + b \cos x$ where $\phi$ is the angle between $v_1$ and $v_2$, but $|v_1| = \sqrt{a^2 + b^2}, |v_2| = 1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A certain unique rotation matrix One can find that the matrix
$A=\begin{bmatrix}
-\dfrac{1}{3} & \dfrac{2}{3} & \dfrac{2}{3} \\
\dfrac{2}{3} & -\dfrac{1}{3} & \dfrac{2}{3} \\
\dfrac{2}{3} & \dfrac{2}{3} & -\dfrac{1}{3} \\
\end{bmatrix} $
is at the same time $3D$ rotation matrix and for it the sum of entries in every column (row) is constant (here $-\dfrac{1}{3}+ \dfrac{2}{3} + \dfrac{2}{3} = 1)$.
The same is true if we change the order of columns in it.
For example:
$A_1=\begin{bmatrix}
\dfrac{2}{3} & \dfrac{2}{3} & -\dfrac{1}{3} \\
-\dfrac{1}{3} & \dfrac{2}{3} & \dfrac{2}{3} \\
\dfrac{2}{3} & -\dfrac{1}{3} & \dfrac{2}{3} \\
\end{bmatrix} $ $A_2=\begin{bmatrix}
\dfrac{2}{3} & -\dfrac{1}{3} & \dfrac{2}{3} \\
\dfrac{2}{3} & \dfrac{2}{3} & -\dfrac{1}{3} \\
-\dfrac{1}{3} & \dfrac{2}{3} & \dfrac{2}{3} \\
\end{bmatrix} $
Questions:
Is it any systematic way to find other non-trivial (without $0$ and $1$) rotation matrices with this property?
Especially it is interesting whether the above rotation matrices are the only ones with rational entries ?- maybe someone knows other rotation matrices exist where the sum of entries is constant..
and...
Can it be proved in some way that if the sum of entries in columns for a rotation matrix is constant then it should be equal to the length of column vectors?
| The three row vectors of the matrix define a triorthonormal frame, so that the tips of the vectors belong to a unit sphere. The sum of entries being a constant corresponds to a plane in the first octant, $x+y+z=c$, which intersects the sphere along a (small) circle.
You can take any three points on that circle that form an equilateral triangle and they fulfill the conditions on the rows (in addition to orthogonality). The transform is a rotation around the first octant bissector.
As the transform is orthogonal, the transpose of the matrix corresponds to the inverse rotation, also a rotation around the bissector, so that the condition on the rows also holds.
By the Rodrigues formula,
$$M=\cos(\theta)\left(\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}\right)+\frac{1-\cos(\theta)}3\left(\begin{matrix}1&1&1\\1&1&1\\1&1&1\end{matrix}\right)+\frac{\sin(\theta)}{\sqrt3}\left(\begin{matrix}\ \ 0&\ \ 1&-1\\-1&\ \ 0&\ \ 1\\\ \ 1&-1&\ \ 0\end{matrix}\right).$$
There is a simple infinity of solutions. Rational ones occur for $\sin(\theta)=0$ or $\sin(\theta)=\pm\frac{\sqrt3}2$, but this brings nothing new. Other ones are found when $\cos(\theta)$ and $\dfrac{\sin(\theta)}{\sqrt3}$
are simultaneously rational ($p^2+3q^2=\cos^2(\theta)+\sin^2(\theta)=1$).
| {
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Evaluate an increasing sum of binomial coefficients: $\sum_{k=1}^nk\binom{m+k}{m+1}$ I've been working on a problem and got to a point where I need the closed form of
$$\sum_{k=1}^nk\binom{m+k}{m+1}.$$
I wasn't making any headway so I figured I would see what Wolfram Alpha could do. It gave me this:
$$\sum_{k=1}^nk\binom{m+k}{m+1} = \frac{n((m+2)n+1)}{(m+2) (m+3)}\binom{m+n+1}{ m+1}.$$
That's quite the nasty formula. Can anyone provide some insight or justification for that answer?
| You can prove this by induction.
Here is the induction step:
$$
\begin{align*}
\sum_{k=1}^{n+1} k \binom{m+k}{m+1} &= \frac{n((m+2)n+1)}{(m+2)(m+3)}\binom{m+n+1}{m+1} + (n+1)\binom{m+n+1}{m+1} \\
&=\frac{(m+n+2)(m(n+1)+2n+3)}{(m+2)(m+3)} \binom{m+n+1}{m+1} \\
&=\frac{(n+1)(m(n+1)+2n+3)}{(m+2)(m+3)} \cdot \frac{(m+n+2)!}{(m+1)!(n+1)!} \\
&= \frac{(n+1)((m+2)(n+1)+1)}{(m+2)(m+3)}\cdot \binom{m+n+2}{m+1}.
\end{align*}
$$
| {
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Sum of $1^2+3^2+\cdots+(2n+1)^2$ Have trouble with proof. I've been working through a question on Courant's What is Mathematics?
This is the question: Prove $1^2+3^2+\cdots+(2n+1)^2=\frac{n(n+1)(2n+1)(2n+3)}{3}$. I called this $S_{(2n+1)^2}$.
What I've been given as a hint is that I can use the sum $1+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ (I called this $S_{n^2}$) in some way.
So here's my line of thought:
What do I need to do to $S_{n^2}=1^2+2^2+3^2+\cdots+n^2$ to get $S_{(2n+1)^2}=1^2+3^2+\cdots+(2n+1)^2$?
When I compared them, I thought that $S_{n^2}-(2^2+4^2+\cdots+(2n)^2) = S_{(2n+1)^2}$.
I defined $2^2+4^2+\cdots+(2n)^2$ as $S_{(2n)^2}$.
After a bit of head-scratching I managed to figure out $S_{(2n)^2}=2^2(0)+2^2(1)^2+2^2(2)^2+\cdots+2^2(n)^2=4(1^2+2^2+3^2+\cdots+n^2)=4S_{n^2} =\frac{2}{3}n(n+1)(2n+1)$
Now is where my confusion starts. So from my logic above I went forward with $S_{n^2}-S_{(2n)^2}=S_{(2n+1)^2}$ which gave me $-\frac{1}{2}(n(n+1)(2n+1))$ which isn't $S_{(2n+1)^2}$.
I don't know where I really went wrong. One of my main suspicions is that $S_{n^2}-S_{(2n)^2}=S_{n^2}-4S_{n^2}=-3S_{n^2}\neq S_{(2n+1)^2}$. But I don't see why.
| Using your hint, here is one way to go:
$$1^2 + 3^2 +\cdots +(2n+1)^2 = (1^2 + 2^2 + 3^2 + 4^2+\cdots +(2n+1)^2) - (2^2 + 4^2 + 6^2+\cdots+ (2n)^2)\\
(1^2 + 2^2 + 3^2 + 4^2+\cdots+ (2n+1)^2) - 4(1^2 + 2^2 + 3^2+\cdots+ n^2)$$
$$\frac{(2n+1)(2n+2)(4n+3)}{6} - \frac{4(n)(n+1)(2n+1)}{6}$$
Alternatively, you could use a proof by induction.
Base case:
$$n = 1\\
1^2 + 3^2 = \frac {1\cdot2\cdot3\cdot5}{3} = 10$$
Suppose:
$$1^2+3^2+\cdots+(2n+1)^2=\frac{n(n+1)(2n+1)(2n+3)}{3}$$
We will show that:
$$1^2+3^2+\cdots+(2n+1)^2 + (2n+3)^2=\frac{(n+1)(n+2)(2n+3)(2n+5)}{3}$$
By our hypothesis:
$$\frac{n(n+1)(2n+1)(2n+3)}{3} + (2n+3)^2=\frac{(n+1)(n+2)(2n+3)(2n+5)}{3}$$
And simplify.
| {
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"timestamp": "2023-03-29T00:00:00",
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Q27 from AMC 2012(Senior): Five consecutive integers that sum to a perfect square, and the three middle terms sum to a perfect cube.
Five consecutive integers $p,q,r,s,t$,each less than $10000$, produce a sum which is a perfect square, while the sum of $q,r,s$ is a perfect cube.What is the value of $ \sqrt{p+q+r+s+t}$ ?
What I have tried so far:
Let $p=r-2$
$p+q+r+s+t =5r $
$5r=x^2 $
$q+r+s =3r =y^3 $
$x^2 -y^3 =2r $
So,the only perfect squares which are divisible by $5$ are the multiples of $5$: $25,100,225,400...$
I also observed that $100-25=75,225-100=125$,where $125-75=50$.Trying that for $225-100=125,400-225=175$ where $175-125=50$
Then,for the perfect cube which are divisible by 3 and must be less than the perfect squares.
$30^3,60^3,90^3,120^3....$
And here is were I got stuck at...
Is the concept I'm using correct?
| Since $\frac{3}{5}x^2=y^3$ is a cube, the number of factors of 5 in $x^2$ must be even and one more than a multiple of 3. So choices are $4$ and $10$. Also the number of factors of 3 in $x^2$ must be even and one less than a multiple of $3$, so choices are $2$ and $8$. Now, $\frac{x^2}{5} \le 10000$, so $x$ must be a multiple of $3\cdot 5^2 = 75$. And in fact $x=75$ (so $r = \frac{75^2}{5}=1125$) works.
| {
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Does this wrong cancellation of $B$ work for $\overline{AB}/ \overline{BC}=A/C$? My teacher says that wrong cancellation of $B$ for the fraction$$\frac{\overline{AB}}{\overline{BC}}=\frac{A}{C}$$ will work for some numbers. I see some trivial cases when $A=B=C$, but are there more of them?
$A$, $B$ and $C$ are some digits.
| Interesting question! I've solved this before! Let $\overline{AB}=10a+b$, $\overline{BC}=10b+c$, we get$$\frac{10a+b}{10b+c}=\frac{a}{c}$$ and $$b(10a-c)=9ac$$Note that $a$, $b$ and $c$ must be nonzero digits. $c=a$ is our trivial case, which leads to $a=b=c$. For $a \neq c$ let $d=c-a$. Its clear that $d$ cannot take $\pm9$. Equation becomes $$b(9a-d)=9a(a+d) \\ 9a^2+9a(d-b)+bd=0$$It's immediate that $9|bd$ and since $d\neq \pm9$ only possible values for $b$ are $3$, $6$ and $9$.
For $b=3$ $$c=\frac{10a}{3a+1}$$and we must have $3a+1|10(3a+1)-3(10a)=10$, which gives $a=3$ only and this is one of trivial solutions.
For $b=6$ $$c=\frac{20a}{3a+2}$$and we must have $3a+2|20(3a+2)-3(20a)=40$, which gives $a=1, 2, 6$ and leads to $(a, b, c)=(1, 6, 4), (2, 6, 5)$ and trivial solution $(6, 6, 6)$.
For $b=9$ $$c=\frac{10a}{a+1}$$and we must have $a+1|10(a+1)-10a=10$, which gives $a=1, 4, 9$ and leads to $(a, b, c)=(1, 9, 5), (4, 9, 8)$ and trivial solution $(9, 9, 9)$.
We are done.
| {
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Fake proof that $\frac{e^x-1}{e^x+1}=e^x$, via integrating $\operatorname{sech} x$ in two ways We start with the integral:
$$\int \text{sech}(x)dx$$
Method 1
\begin{align}
\int \text{sech}(x)dx & = \int\frac{2}{e^x+e^{-x}}dx \\
&= \int\frac{2e^x}{e^{2x}+1}dx
\end{align}
Using the substitution $u=e^x$,
\begin{align}
\int \text{sech}(x)dx & = \int\frac{2}{u^2+1}du \\
&= 2\text{ arctan}(u)+c \\
&= 2\text{ arctan}(e^x)+c
\end{align}
Method 2
\begin{align}
\int \text{sech}(x)dx & = \int\frac{\text{cosh}(x)}{\text{cosh}^2(x)}dx \\
& = \int\frac{\text{cosh}(x)}{\text{sinh}^2(x)+1}dx
\end{align}
Using the substitution $u=\text{sinh}(x)$,
\begin{align}
\int \text{sech}(x)dx & = \int\frac{1}{u^2+1}du \\
&= \text{arctan}(\text{sinh}(x))+c \\
&= 2\text{ arctan}(\text{tanh}(\frac{x}{2}))+c \\
&= 2\text{ arctan}(\frac{e^x-1}{e^x+1})+c
\end{align}
Thus, we obtain:
$$\frac{e^x-1}{e^x+1}=e^x$$
However, I see no reason why they are equal. Did I do something wrong in the calculation?
| Because $2\arctan\left(\frac{e^x-1}{e^x+1}\right)=2\left(\arctan(e^x)-\frac\pi4\right)=2\arctan(e^x)+C'$.
The results differ by a constant.
| {
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Choosing a value so a line and circle intersect a one, two, and no points Let l be a line and C be a circle.
$y=x+d$, where $d$ is to be determined.
$C=x^2+y^2=4$
*
*Pick a value for $d$ so that l and C intersect at one point.
*Pick a value for $d$ so that l and C intersect at two distinct points.
*Pick a value for $d$ so that l and C do not intersect.
Am I able to use this: How do I calculate the intersection(s) of a straight line and a circle??
Graphically, I could answer B and C. I'm hoping there's an algebraic way to solve all three parts.
| For intersection points both equations
$$
y = x + d \\
x^2 + y^2 = 2^2
$$
must hold. Inserting the first into the second we get
$$
y = x + d
$$
and
$$
2^2 = x^2 + (x+ d)^2 = 2 x^2 + 2xd + d^2 \iff \\
2 = x^2 + x d + d^2/2 = (x + d/2)^2 - d^2/4 + d^2/2 \iff \\
2 - d^2/4 = (x + d/2)^2 \iff \\
x = \frac{-d \pm \sqrt{8 - d^2}}{2}
$$
Which gives the solutions
$$
P =
\left(
\frac{-d \pm \sqrt{8 - d^2}}{2},
\frac{-d \pm \sqrt{8 - d^2}}{2} + d
\right)
$$
The number of real solutions is determined by the argument of the square root
$$
\Delta(d) = 8 - d^2
$$
If $\Delta(d) = 0$ or $d = \pm\sqrt{8} = \pm 2\sqrt{2}$ there is one solution.
If $\Delta(d) > 0$ or $\lvert d \rvert < 2\sqrt{2}$ there are two solutions.
If $\Delta(d) < 0$ or $\lvert d \rvert > 2\sqrt{2}$ there is no (real) solution.
You can fiddle with it here.
| {
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Prove that $\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}>10^{-19.76}$
Prove that $\frac{1}{4\cdot 1976^3}-\frac{1}{16\cdot 1976^7}>10^{-19.76}$ without using a calculator.
I rearraged to get $4 \cdot 1976^4-1 > 10^{-19.76} \cdot 16 \cdot 1976^7$ and so we have to prove that $4 \cdot 1976^4-1-10^{-19.76} \cdot 16 \cdot 1976^7>0$. How should I do that?
| Seems really simple - am I missing something?
$a = 4\times 1976^3 < 4\times 2000^3 = 32\times 10^9$
$\implies 1/a > \frac{1}{32}10^{-9} > 3\times 10^{-11}$
$b= 16\times 1976^7 > 10^{22}$ and $1/b <10^{-22}$
$\implies \frac{1}{a} - \frac{1}{b} > 3 \times 10^{-11} - 10^{-22} >10^{-19.76}$ (howsoever you interpret that exponent)
| {
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Number of positive unequal integer solutions of $x+y+z+w=20$ What is the number of positive different integer solutions of $x+y+z+w=20$, where $x,y,z,w$ are all different and positive?
It would be nice if coding is not used. I am given the answer $552$.
| Just for comparison with Brian Scott's solution, here is the longhand count solution.
As is observed in several places above, it is sufficient to find the number of solutions with $x<y<z<w$ and multiply by 24, because each such solution can be permuted to give 24 solutions to the original question.
Next note that $1+2+3+4=10$ is the smallest total for 4 unequal positive integers, so we are not going to get many solutions. Now $4+5+6+7>20$, so $x=1,2$ or 3. Since $3+4+5+6=18$ the only possible solutions for $x=3$ are $(3,4,6,7),(3,4,5,8)$.
Similarly for $x=2$ we must have $y=3,4$ or 5. $y=5$ gives only $(2,5,6,7)$. $y=4$ gives only $(2,4,6,8),(2,4,5,9)$. $y=3$ gives only $(2,3,4,11),(2,3,5,10),(2,3,6,9),(2,3,7,8)$.
Finally in a similar way $x=1$ implies $y=2,3,4,5$ and gives solutions $(1,2,3,4),(1,2,4,13),(1,2,5,12),(1,2,6,11),(1,2,7,10),(1,2,8,9)$ and $(1,3,4,12),(1,3,5,11),(1,3,6,10),(1,3,7,9)$ and $(1,4,5,10),(1,4,6,9),(1,4,7,8)$, and $(1,5,6,8)$.
Total 23, hence 552 to question.
--------- shorter version ------
I have spelt this out in full detail, but note that since only the count is required setting out all those cases with $z+w=k$ is not really needed. It is enough to say something like:
$x=3$ implies $y=4$ and 2 solutions
$x=2$ implies: $y=3$, 4 solutions; or $y=4$, 2 solutions; or $y=5$, 1 solution
$x=1$ implies: $y=2$, 6 solutions; or $y=3$, 4 solutions; or $y=4$, 3 solutions; or $y=5$, 1 solution.
Total 23 solutions. Times 4! gives 552 to the original question.
----------- comment ----------
Teachers often give simple questions for you to practice methods like generating functions. But simple questions can often be answered faster by more elementary methods. Often fastest is best!
| {
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Evaluate $\lim_{n\to\infty}\left(\frac{3}{2}\cdot\frac{5}{3}\cdot\dots\frac{2n+1}{n+1}\right)^\frac{1}{n}$ What's the value of $$\lim_{n\to\infty}\left(\frac{3}{2}\cdot\frac{5}{3}\cdot\dots\frac{2n+1}{n+1}\right)^\frac{1}{n}?$$
I've tried the AM-GM inequality with no luck.
Also tried to right the inequality as:
$$\left(\left(2-\frac{1}{2}\right)\left(2-\frac{1}{3}\right)\dots\left(2-\frac{1}{n+1}\right)\right)^\frac{1}{n}.$$
Would be happy to hear any hints/proofs.
| Clearly the expression is $a_{n}^{1/n}$ where $a_{n} = \dfrac{(2n + 1)!}{n!(n + 1)!2^{n}}$ and hence $$\frac{a_{n + 1}}{a_{n}} = \frac{(2n + 3)!}{2^{n + 1}(n + 1)!(n + 2)!}\cdot\frac{2^{n}n!(n + 1)!}{(2n + 1)!} = \frac{(2n + 3)}{(n + 2)} \to 2$$ so desired limit is also $2$. We have used the following standard result:
If $\{a_{n}\}$ is a sequence of positive terms such that $a_{n + 1}/a_{n} \to L$ then $a_{n}^{1/n} \to L$.
One can also use Stirling's approximation for $n!$ but proof of Stirling's approximation is not easy.
| {
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eliminate the parameters Given:
$x = \frac{1}{2} \cos(\theta)$ and $y = 2\sin(\theta)$
Part a)
solving the first one for theta:
1) multiply both sides by $2$:
$$2x = \cos(\theta)$$
2) divide both sides by $\cos (\arccos)$:
$$\cos^{-1}(2x) = \theta$$
Part b)
Now plugging this into the second equation for theta:
1) plug in theta:
$$y = 2\sin(\cos^{-1}(2x))$$
2) Simplify:
$$y = 2 \sqrt{1-x^2}$$
$$\frac{1}{2} y=\sqrt{1-x^2}$$
3) Squaring both sides gives:
$$\frac{1}{4} y^2 = 1-x^2$$
$$\frac{1}{4} y^2 +x^2 = 1$$
I have laid it out in parts and subsections of each part please refer to the appropriate part and subsection where a mistake was made. One area where I notice there may be a mistake is in part b section 2 where I substitute $2\sin(\cos^{-1}(2x))$ with the square root. $\sqrt{1-x^{2}}$ is for $\sin(\cos^{-1}(x))$ I am not sure how the $2$ in $2\sin$ and the $2$ in $\cos^{-1}(2x)$ affect the substitution.
| This is an idea:
$$\begin{cases}x=\frac12\cos\theta\\{}\\y=2\sin\theta\end{cases}\implies \cos^2\theta+\sin^2\theta=4x^2+\left(\frac y2\right)^2=1$$
and you have an ellipse
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed form of the series.
Write the following series in closed form
$$\frac{1}{a-1}+\frac{2}{a-2}+\frac{2}{a-3}+\frac{1}{a-4}+\frac{1}{a-5}+\frac{2}{a-6}+\frac{2}{a-7}+\frac{1}{a-8}+\frac{1}{a-9}+\frac{2}{a-10}+\frac{2}{a-11}+\cdots+\frac{1}{a-4p+3}+\frac{2}{a-4p+2}+\frac{2}{a-4p+1}+\frac{1}{a-4p},$$
where $a$ is any positive number.
I tried to split this series into 4 sub series as follows
$$\frac{1}{a-1}+\frac{1}{a-5}+\frac{1}{a-9}+\cdots+\frac{1}{a-4p+3}+$$
$$+\frac{2}{a-2}+\frac{2}{a-6}\frac{2}{a-10}+\cdots+\frac{2}{a-4p+2}+$$
$$+\frac{2}{a-3}+\frac{2}{a-7}+\frac{2}{a-11}+\cdots+\frac{2}{a-4p+1}+$$
$$+\frac{1}{a-4}+\frac{1}{a-8}+\frac{1}{a-12}+\cdots+\frac{1}{a-4p}$$
but still unable to obtain the result.
| You have splitted the series in $4$ subseries. We see that for each sub series $(a-j)(a-i)=4,\forall j>i$. So we get the following series for every sub series:
$$(1):\sum_{i=0}^{p-1} \frac{1}{a-(4i+1)}\\(2):\sum_{i=0}^{p-1}\frac{2}{a-(4i+2)}\\(3):\sum_{i=0}^{p-1}\frac{2}{a-(4i+3)}\\(4):\sum_{i=0}^{p}\frac{1}{a-4i}.$$ Together in one expression we get:$$\frac{1}{a-4p}+\sum_{i=0}^{p-1} \left(\frac{1}{a-(4i+1)}+\frac{2}{a-(4i+3)}+\frac{2}{a-(4i+2)}+\frac{1}{a-4i}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $2^{x^2 + x} + \log_2x=2^{x+1}$
Solve the equation $2^{x^2 + x} + \log_2x=2^{x+1}$ where $x$ is real.
I tried to use derivatives, without success.
It's obvious that $x=1$ is solution.
Also, if $x \gt 1$ then $2^{x^2 + x} \gt 2^{x+1}$ and $\log_2x \gt 0$ therefore the equation has no solutions $x \gt 1$.
| It's clear that $x=1$ is a solution. Let's look at two cases: $x>1$ and $x<1$.
If $x>1$, then $\log_2x>0$ and $x^2+x>x+1$, which implies that $2^{x^2+x}>2^{x+1}$. It's impossible.
If $0<x<1$, then $\log_2x<0$ and $x^2+x<x+1$, which implies that $2^{x^2+x}>2^{x+1}$. It's impossible.
In summary, we only have one solution $x=1$.
| {
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"source": "stackexchange",
"question_score": "1",
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Find $\lambda$ and $\theta$ such that it validates this matricial equation Find $\lambda$ and $\theta$ such that it validates the matricial equation
$$ \left( \begin{array}{cc}
1 & 2 \\
2 & 3
\end{array} \right)
%
\left( \begin{array}{cc}
\cos \theta \\
\sin \theta
\end{array} \right) = \lambda\left( \begin{array}{cc}
\cos \theta \\
\sin \theta
\end{array} \right)
$$
What I have tried:
$$\left\{ \begin{array}{ll}
\cos \theta + 2\sin \theta &= \lambda \cos \theta\\
2\cos \theta + 3\sin \theta &= \lambda \sin \theta.\end{array} \right. $$
$$\left\{ \begin{array}{ll}
\ -2\cos \theta -4 \sin \theta &= -2\lambda \cos \theta\\
2\cos \theta + 3\sin \theta &= \lambda \sin \theta.\end{array} \right. $$
$$-\sin \theta = -2\lambda \cos \theta + \lambda \sin \theta$$
$$ (\lambda + 1)\sin \theta = 2 \lambda \cos \theta$$
I can´t find $\lambda$ and $\theta$
| For given $\theta$ this is an eigenvalue problem.
The characteristic equation of your matrix is $p(\lambda) = (1-\lambda)(3 - \lambda) -4 = \lambda^2 - 4\lambda -1$
And so the eigenvalues are
$ \lambda = \dfrac{4 \pm \sqrt{16+4}}{2} = 2 \pm \sqrt{5}$
It can be shown (though I used computer algebra) that the eigenvectors are
$$v_1 = \begin{bmatrix} 1\\1 \end{bmatrix} v_2 = \begin{bmatrix} \dfrac{2}{1+ \sqrt{5}} \\ \dfrac{2}{1-\sqrt{5}}\end{bmatrix} $$
The polar coordinate representation of a vector allows you to find the angle. Since there are two vectors, there should be two $\theta$, and so
$$\theta_i = \tan ^{-1}(\dfrac{y_i}{x_i})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Express $1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}$ in a simplifed form I need to express $$1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}$$ in a simplified form.
So I used the identity $$(1+x)^n=1 + \binom{n}{1}x + \binom{n}{2}x^2 + \dotsb + \binom{n}{n}x^n$$
Now on integrating both sides and putting $x=1$.
I am getting $$\frac{2^{n+1}}{n+1}$$ is equal to the given expression.But the answer in my book is $$\frac{2^{n+1}-1}{n+1}.$$
Where does that -1 term in the numerator come from?
| Note that $$\binom{n+1}{k+1}=\frac{n+1}{k+1}\binom{n}{k}$$
Therefore,
\begin{align}
\sum_{k=0}^{n}\frac{1}{k+1}{n\choose k}&=\frac{1}{n+1}\sum_{k=0}^{n}\frac{n+1}{k+1}{n\choose k}
\\&=\frac{1}{n+1}\sum_{k=0}^{n}{{n+1}\choose {k+1}}
\\&=\frac{1}{n+1}\left [\sum_{k=0}^{n+1}{{n+1}\choose k}-1\right ]
\\&=\frac{1}{n+1}\left (2^{n+1}-1\right )
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Prove $\int_{\frac{\pi}{20}}^{\frac{3\pi}{20}} \ln \tan x\,\,dx= - \frac{2G}{5}$ Context: This question
asks to calculate a definite integral which turns out to be equal to $\displaystyle 4 \, \text{Ti}_2\left( \tan \frac{3\pi}{20} \right) -
4 \, \text{Ti}_2\left( \tan \frac{\pi}{20} \right),$ where $\text{Ti}_2(x) = \operatorname{Im}\text{Li}_2( i\, x)$ is the
Inverse Tangent Integral function.
The source for this integral is this question on brilliant.org.
In a comment, the OP
claims that the closed form can be further simplified to $-\dfrac\pi5 \ln\left( 124 - 55\sqrt5 + 2\sqrt{7625 - 3410\sqrt5} \right) + \dfrac85 G$.
How can we prove that?
I have thought about using the formula $$\text{Ti}_2(\tan x) = x \ln \tan x+ \sum_{n=0}^{\infty} \frac{\sin(2x(2n+1))}{(2n+1)^2}. \tag{1}$$
but that only mildly simplifies the problem.
Equivalent formulations include:
$$\, \text{Ti}_2\left( \tan \frac{3\pi}{20} \right) -
\, \text{Ti}_2\left( \tan \frac{\pi}{20} \right) \stackrel?= \frac{ \pi}{20} \ln \frac{ \tan^3( 3\pi/20)}{\tan ( \pi/20)} + \frac{2 G}{5} \tag{2}$$
$$ \sum_{n=0}^{\infty} \frac{\sin \left(\frac{3\pi}{10}(2n+1) \right)- \sin \left(\frac{\pi}{10}(2n+1)\right)}{(2n+1)^2} \stackrel?=\
\frac{2G}{5} \tag{3}$$
$$\int_{\pi/20}^{3\pi/20} \ln \tan x\,\,dx \stackrel?= - \frac{2G}{5} \tag{4}$$
A related similar question is this one.
| We can use the same technique of the linked answer for proving the claim. We first prove that $$2\int_{0}^{\pi/20}\log\left(\tan\left(5x\right)\right)dx=\int_{\pi/20}^{3\pi/20}\log\left(\tan\left(x\right)\right)dx.\tag{1}$$Let $$I=\int_{0}^{\pi/20}\log\left(\tan\left(5x\right)\right)dx$$ using the identity
$$\tan\left(\left(2n+1\right)x\right)=\tan(x)\prod_{k=1}^{n}\tan\left(\frac{k\pi}{2n+1}+x\right)\tan\left(\frac{k\pi}{2n+1}-x\right)$$
we have $$\begin{align}I=
& \int_{0}^{\pi/20}\log\left(\tan\left(x\right)\right)dx+\int_{0}^{\pi/20}\log\left(\tan\left(\frac{\pi}{5}-x\right)\right)dx
\\ +
& \int_{0}^{\pi/20}\log\left(\tan\left(\frac{\pi}{5}+x\right)\right)dx+\int_{0}^{\pi/20}\log\left(\tan\left(\frac{2\pi}{5}-x\right)\right)dx
\\ +
& \int_{0}^{\pi/20}\log\left(\tan\left(\frac{2\pi}{5}+x\right)\right)dx
\\ =
& \int_{0}^{\pi/20}\log\left(\tan\left(x\right)\right)dx+\int_{3\pi/20}^{\pi/5}\log\left(\tan\left(x\right)\right)dx
\\ +
&\int_{\pi/5}^{\pi/4}\log\left(\tan\left(x\right)\right)dx+\int_{7\pi/20}^{2\pi/5}\log\left(\tan\left(x\right)\right)dx
\\ +
&\int_{2\pi/5}^{9\pi/20}\log\left(\tan\left(x\right)\right)dx.
\end{align} $$ So we have $$I=\int_{0}^{\pi/20}\log\left(\tan\left(x\right)\right)dx+\int_{3\pi/20}^{\pi/4}\log\left(\tan\left(x\right)\right)dx+\int_{7\pi/20}^{9\pi/20}\log\left(\tan\left(x\right)\right)dx\tag{2}
$$ and in the last integral of $(2)$ if we put $x\rightarrow\frac{\pi}{2}-x
$ and recalling the identity $\tan\left(\frac{\pi}{2}-x\right)=\frac{1}{\tan\left(x\right)}
$, we get $$\begin{align}I=
& \int_{0}^{\pi/20}\log\left(\tan\left(x\right)\right)dx+\int_{3\pi/20}^{\pi/4}\log\left(\tan\left(x\right)\right)dx-\int_{\pi/20}^{3\pi/20}\log\left(\tan\left(x\right)\right)dx
\\ =
& \int_{0}^{\pi/4}\log\left(\tan\left(x\right)\right)-2\int_{\pi/20}^{3\pi/20}\log\left(\tan\left(x\right)\right)dx
\\ =
& 5\int_{0}^{\pi/20}\log\left(\tan\left(5x\right)\right)dx-2\int_{\pi/20}^{3\pi/20}\log\left(\tan\left(x\right)\right)dx
\end{align}$$ so finally we have $(1)$. Hence $$\begin{align}
\int_{\pi/20}^{3\pi/20}\log\left(\tan\left(x\right)\right)dx= & 2\int_{0}^{\pi/20}\log\left(\tan\left(5x\right)\right)dx \\
= & \frac{2}{5}\int_{0}^{\pi/4}\log\left(\tan\left(x\right)\right)dx \\ = & \color{red}{-\frac{2}{5}G}
\end{align}$$ as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to use the generalized binomial theorem to produce the power series of $(1-x)^{1/2}$ I am trying to see how to get from $\sqrt{1-x}$ to the power series $\displaystyle\sum_{m=0}^\infty\frac{-1}{2m-1}\,{2m \choose m}\,\frac{x^m}{4^m}$, ideally using the generalized binomial theorem.
I see that the Taylor expansion evaluated at $0$ is,
$f(x)=(1-x)^{1/2}$; $f(0)=1$; and the first coefficient of the power series is $1$.
$f^1(x)= - \frac{1}{2}(1-x)^{-1/2}$; $f^1(0)=-1/2$; and the second coefficient, $-\frac{x}{2}$.
$f^2(x) = -\frac{1}{4}(1-x)^{-3/2}$; $f^2(0)=-1/4$; and the third coefficient, $-\frac{x^2}{8}$.
$f^3(x)=-\frac{3}{8}(1-x)^{-5/2}$; $f^3(0)=-3/8$; and the fourth coefficient, $-\frac{3\cdot x^3}{8\cdot3!}=-\frac{x^3}{16}$.
It is also likely that ${2m\choose m}=\frac{2m!}{m!m!}$ may be part of the derivation, but I don't see a straightforward link.
| Let $f(x)=(1-x)^{1/2}$. Then, differentiating we have
$$\begin{align}
f^{(1)}(x)&=(-1)\left(\frac12\right) (1-x)^{-1/2}\\\\
f^{(2)}(x)&=(-1)^2\left(\frac12\right) \left(-\frac12\right) (1-x)^{-3/2} \\\\
f^{(3)}(x)&=(-1)^3\left(\frac12\right) \left(-\frac12\right) \left(-\frac32\right) (1-x)^{-5/2}\\\\
f^{(4)}(x)&=(-1)^4\left(\frac12\right) \left(-\frac12\right) \left(-\frac32\right)\left(-\frac52\right) (1-x)^{-7/2}\\\\
\vdots\\\\
f^{(n)}(x)&=-\frac{(2n-3)!!}{2^n}(1-x)^{-(2n-1)/2} \\\\
\end{align}$$
Therefore, $f(x)$ has the series expansion
$$f(x)=1-\sum_{n=1}^\infty \frac{(2n-3)!!}{2^n\,n!}\,x^n \tag 1$$
We can express the double factorial term in terms of single factorials by writing
$$\begin{align}
(2n-3)!!&=(2n-3)(2n-5)(2n-7)\cdots(5)(3)(1)\\\\
&=\frac{(2n-3)!}{(2n-4)(2n-6)(2n-8)\cdots (6)(4)(2)}\\\\
&=\frac{(2n-3)!}{2^{n-2}(n-2)!}\\\\
&=\frac{(2n)!}{(2n-1)2^n\,n!}\\\\
&=\frac{n!}{(2n-1)2^n}\binom{2n}{n} \tag 2
\end{align}$$
Substituting $(2)$ into $(1)$ yields
$$\begin{align}
f(x)&=1-\sum_{n=1}^\infty \frac{1}{4^n(2n-1)}\binom{2n}{n}\,x^n\\\\
&=-\sum_{n=1}^\infty \frac{1}{4^n(2n-1)}\binom{2n}{n}\,x^n
\end{align}$$
as was to be shown!
| {
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"url": "https://math.stackexchange.com/questions/1849779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can this upper bound for $\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$ be improved?
I would like to find the smallest possible upper bound for the following sum of prime radicals (OEIS A062048):
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P$
This is my attempt. It is too big, I am sure there is a better way of calculating this, but my knowledge is not very good (the questions are at the end of the explanation):
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor = \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_n} \rfloor$
by Bertrand's postulate $\lfloor \sqrt{p_{i+1}} \rfloor \lt \lfloor \sqrt{2 \cdot p_i} \rfloor$ so it is possible to replace in a first step $\lfloor \sqrt{3} \rfloor$ by $\lfloor \sqrt{2 \cdot 2} \rfloor$:
$\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_n} \rfloor \lt \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{2 \cdot 2} \rfloor + ... + \lfloor \sqrt{p_n} \rfloor$
then recursively doing the same with the next prime $\lfloor \sqrt{5} \rfloor \lt \lfloor \sqrt{2\cdot 3} \rfloor \lt \lfloor \sqrt{2\cdot (2 \cdot 2)} \rfloor$, etc. the result is:
$\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_i} \rfloor... \lt \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{2^2} \rfloor + \lfloor \sqrt{2^3} \rfloor + ... + \lfloor \sqrt{2^n} \rfloor =$
$=\sum_{i=1}^{n \ even}2^{\frac{n}{2}}+\sqrt{2}\cdot \sum_{i=1}^{n \ odd}2^{\frac{n-1}{2}}=$
$\sqrt{2} \lt 2$:
$\sum_{i=1}^{n \ even}2^{\frac{n}{2}}+\sqrt{2}\cdot \sum_{i=1}^{n \ odd}2^{\frac{n-1}{2}} \lt 2 \cdot \sum_{i=1}^n 2^{\frac{n}{2}} $
Finally, $\sum_{i=1}^n 2^i = (2^{i+1}-1)-2^0 = 2^{i+1}-2$, thus:
$2 \cdot \sum_{i=1}^n 2^{\frac{n}{2}} \lt 2 \cdot (2^{(\frac{n}{2}+1)}-2)=2^{(\frac{n}{2}+2)}-4$
So finally the expression would be:
$\sum_{i=1}^n \lfloor \sqrt{p_i} \rfloor, p_i \in \Bbb P \lt 2^{(\frac{n}{2}+2)}-4$
I would like to ask the following questions:
*
*Are the calculations correct or there are gaps / errors?
*Is there a better approach? Thank you!
UPDATE 2016/7/17: Thank you very much for the answers, for the sake of completeness I have sketched the solutions everybody kindly provided, compared with $\sum_{i=1}^{n}i=\frac{n \cdot (n+1)}{2}$. The reason is that this sequence is expected to grow slower than the sequence of the sum of the first $n$ elements, so it is a good expression to compare with. The only solution that was too big was mine, kindly improved by mathlove, that is why I wanted to ask for better values really under the value of the sum of the first $n$ natural numbers.
Blue: $\sum_{i=1}^{n}i=\frac{n \cdot (n+1)}{2}$
Red (mathlove) = $\sum_{i=1}^{n}\sqrt{p_i}\lt \sum_{i=1}^{n}2^{\frac i2}=\frac{2^{\frac{n+1}{2}}-\sqrt 2}{\sqrt 2-1}$
Yellow (Marco Cantarini) = $n\sqrt{n\log\left(n\right)+n\log\left(\log\left(n\right)\right)}-\frac{2\left(n-1\right)\sqrt{n-1}}{3\sqrt{\log\left(2\right)}}+1$
Green (LeGrandDODOM) (leading term only): $\displaystyle\frac 23 n^{\frac 32}\sqrt{\ln n}$
|
$\lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + ... + \lfloor \sqrt{p_i} \rfloor... \lt \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{2^2} \rfloor + \lfloor \sqrt{2^3} \rfloor + ... + \lfloor \sqrt{2^n} \rfloor =$
$=\sum_{i=1}^{n \ even}2^{\frac{n}{2}}+\sqrt{2}\cdot \sum_{i=1}^{n \ odd}2^{\frac{n-1}{2}}=$
Then, you used $\sqrt{2} \lt 2$, but I think you don't need to do so because
$$\sqrt 2\times 2^{\frac{n-1}{2}}=2^{\frac{n}{2}}$$
so the sum can be written as
$$\sum_{i=1}^{n}2^{\frac i2}\tag1$$
(By the way, it seems that you have
$$\sum_{i=1}^n 2^{\frac{i}{2}} =2^{(\frac{n}{2}+1)}-2$$
but this is incorrect. See when $n=1$.)
Finally, from $(1)$,
$$\sum_{i=1}^{n}\sqrt{p_i}\lt \sum_{i=1}^{n}2^{\frac i2}=\frac{2^{\frac{n+1}{2}}-\sqrt 2}{\sqrt 2-1}$$
| {
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"url": "https://math.stackexchange.com/questions/1850686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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$'a'$ for which roots of one equation lie between roots of other equation. If the range of values of $'a'$ for which the roots of the equation $x^2-2x-a^2+1=0$ lie between the roots of the equation $x^2-2(a+1)x+a(a-1)=0$ is $(p,q)$, find the value of $(q+\frac{1}{p^2})$
Could someone give me slight hint as I can't understand how to initiate.
| $x^2-2x-a^2+1=0$. Quadratic equation yields $x = \frac{2\pm \sqrt{4 - 4(a^+ 1)}}{2}$
$x^2-2(a+1)x+a(a-1)=0$. Q.E. yields $x = \frac{2(a+1) \pm \sqrt{(2(a+1))^2 - 4a(a-1)}}{2}$
We are told $\frac{2(a+1) - \sqrt{(2(a+1))^2 - 4a(a-1)}}{2}< \frac{2- \sqrt{4 - 4(a^+ 1)}}{2}$ and $\frac{2+ \sqrt{4 - 4(a^+ 1)}}{2} <\frac{2(a+1) + \sqrt{(2(a+1))^2 - 4a(a-1)}}{2}$
We work that out to figure that $p \le a \le q$. Then we evaluate $p + 1/q^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate the limit of $\frac{\sin \lfloor x+1\rfloor }{ \lfloor x+1\rfloor }$ at $x=-1$ $$f(x):=\begin{cases}\frac{ \sin \lfloor x+1\rfloor }{ \lfloor x+1\rfloor } & \lfloor x+1\rfloor \ne0 \\
0 & \lfloor x+1\rfloor=0 \ \end{cases}$$
Then at $x=-1$ find the limit
My work
$$\lfloor x+1\rfloor=0$$
$$0\le\ x+1\lt1$$
$$f(x):=\begin{cases}\frac{ \sin \lfloor x+1\rfloor }{ \lfloor x+1\rfloor } & \lfloor x+1\rfloor \ne0 \\
0 & -1\le x \lt 0 \ \end{cases}$$
$$LHL=\lim_{x\to -1^-}\frac{\sin(-1)}{-1}$$
$$RHL=\lim_{x\to -1^+}\frac{\sin(0)}{0}$$
$$LHL=\sin1$$
$RHL= \text{Not defined}$
But the answer say$$ RHL=0$$
Please tell me why I am wrong.
| Read carefully the definition: the function is defined at the right of $-1$.
Make your life simpler and change $x+1$ into $x$, so you need the limits at $0$ of
$$
g(x)=\begin{cases}
\dfrac{\sin\lfloor x\rfloor}{\lfloor x\rfloor} & \text{if $\lfloor x\rfloor\ne0$}
\\[6px]
0 & \text{if $\lfloor x\rfloor=0$}
\end{cases}
$$
For $0<x<1$, we have $\lfloor x\rfloor=0$, so $g(x)=0$ and therefore
$$
\lim_{x\to0^+}g(x)=0
$$
For $-1<x<0$, we have $\lfloor x\rfloor=-1$ and so $g(x)=\frac{\sin(-1)}{-1}=\sin 1$; therefore
$$
\lim_{x\to0^-}g(x)=\sin1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$
Prove that: $x^2+y^2+z^2=2xyz$ has no answer over $\Bbb{N}$
$$LHS=(x+y+z)^2-2(xy+yz+xz)=2xyz \implies (x+y+z)^2=2(xy+yz+xz)+2xyz$$
now what??
| We use a classical infinite descent argument.
Note that the right-hand side is even, so the left-hand side must be. It follows that two of $x$, $y$, $z$ are odd and the third even, or all three are even.
But two odd and one even is impossible, for then the right-hand side is divisible by $4$ and the left-hand side is not.
Thus $x=2x_1$, $y=2y_1$, $z=2z_1$ for some integers $x_1$, $y_1$, $z_1$.
Substituting we get $x_1^2+y_1^2+z_1^2=4x_1y_1z_1$.
Repeat the argument. We find that $x_1=2x_2$, and so on, with $x_2^2+y_2^2+z_2^2=8x_2y_2z_2$.
Continue. We conclude that $x$, $y$, $z$ are each divisible by arbitrarily high powers of $2$, so are all $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $f(x)$ satisfying the functional equation $x^2{f(x)} +f(1- x) =2x -x^4$ A function $f(x)$ satisfies the functional equation $x^2{f(x)} +f(1- x) =2x -x^4$ for all real $x$. Then $f(x)$ is given by.
My work
$$x^2{f(x)} +f(1- x) =2x -x^4$$
Replacing $x$ $by$ $1- x$
$$(1-x)^2{f(1 - x)} +f(x) =2x -x^4 + 4x^3 -6x^2 + 1$$
Subtracting them
$${f(x)}(1-x^2) +{f(1- x)}((1-x)^2 - 1) = 4x^3 -6x^2 + 1$$
What should I do next ?
| It is easier to solve a much more general problem:
Consider the functional equation
a(x) f(x) + b(x) f (1-x) = g(x).
Call a^* (x) = a(1-x) etc. and the equation becomes the two equations.
a f + b f^* = g
a^* f^* + b^* f = g^*. which is a 2-by-2 linear system:
Solving for f by inverting a 2-by-2 matrix gives
f(x) = (1/ (a a^* - b b^)) ( a^ g - b^* g^*).
This works for all functions a,b,g as long as a a^* - b b^* is not zero.
The given problem has a(x) = x^2, b(x) = 1, and g(x) = 2x-x^4.
Plugging these in gives
f(x) = (1/ (x^2(1-x)^2 -1)) ( (1-x)^2 g(x) - g(1-x) )
= (1/ (x^2(1-x)^2 -1)) ( (1-x)^2 (2x-x^4) - 2(1-x) + (1-x)^4)
which simplifies to 1-x^2.
| {
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"url": "https://math.stackexchange.com/questions/1854000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $10^{340} < \frac{5^{496}}{1985}$
Prove that $10^{340} < \dfrac{5^{496}}{1985}$.
I said since $2^{13} < 10^{4}$, we see that $5 = \dfrac{10}{2} > 10^{\frac{9}{13}}$ and so $10^{340} < \dfrac{10^{343.38}}{1985} <\dfrac{5^{496}}{1985}$. But then I have to estimate $\log_{10}{2} < 0.31$, which is pretty computational. Is there an easier way?
| $$2^{340}\cdot 5^{340} \lt \dfrac{5^{496}}{1985}$$
is equivalent to
$$2^{340}\lt \frac{5^{496-340}}{5\times 397}=\frac{5^{155}}{397}$$
which is equivalent to
$$397\cdot 2^{340}\lt 5^{155}$$
Since $397\lt 1024=2^{10}$, it is sufficient to prove that
$$2^{350}\lt 5^{155}$$
which is equivalent to
$$2^{70}\lt 5\cdot 5^{30}$$
which is equivalent to
$$5\gt \left(\frac{2^7}{5^3}\right)^{10}=(1.024)^{10}$$
Now, $(1.024)^{10}\lt (1.1)^{10}=(1.21)^5\lt 1.3\times (1.3)^4\lt 1.3\times (1.7)^2=3.757\lt 5$.
| {
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"url": "https://math.stackexchange.com/questions/1855146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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which $X$ can satisfy $XSX^2=D$? In the question, $X$, $S$, and $D$ are symmetric positive definite (SPD) matrices.
Just for an example, $X=S^{-\frac{1}{2}}(S^{\frac{1}{2}}DS^{\frac{1}{2}})^{\frac{1}{2}}S^{-\frac{1}{2}}$ is the solution to $XSX=D$.
$X^2=XX$, $X^{\frac{1}{2}}X^{\frac{1}{2}}=X$, and $X^{-\frac{1}{2}}X^{-\frac{1}{2}}=X^{-1}$. $X^{\frac{1}{2}}$ and $X^{-\frac{1}{2}}$ are also symmetric positive definite (SPD) matrices.
| If $X$ is a solution, $SX=X^{-1}DX^{-1}$ must be positive definite. Therefore $X$ must commute with $S$. But then $XSX^2=D$ gives $X^3=S^{-1}D$. So, a solution exists if and only if $S$ commutes with $D$ and $X=S^{-1/3}D^{1/3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I prove that $\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$ How do I prove that
$$\sqrt{20+\sqrt{20+\sqrt{20}}}-\sqrt{20-\sqrt{20-\sqrt{20}}} \approx 1$$
without using the calculator?
| In general, it holds that $$\sqrt{n(n-1)+\sqrt{n(n-1)+\sqrt{n(n-1)}}}=n-\frac{1}{8n^2}+O\left(\frac1{n^3}\right)$$ and that $$\sqrt{n(n-1)-\sqrt{n(n-1)-\sqrt{n(n-1)}}}=(n-1)+\frac{1}{8n^2}+O\left(\frac1{n^3}\right)\,$$ for all $n\geq 1$. Hence, their difference is $$1-\frac1{4n^2}+O\left(\frac1{n^3}\right)\,.$$ In particular, for $n=5$, the difference should be about $1-\dfrac1{100}=0.99$. This is quite close to the actual value of $0.9872649...$.
From $\sqrt{1+x}=1+\frac{1}{2}x+O\left(x^2\right)$, we have $\sqrt{1-\frac1n}=1-\frac1{2n}+O\left(\frac{1}{n^2}\right)$. This means $$\sqrt{n(n-1)}=n\,\sqrt{1-\frac{1}{n}}=n\,\Biggl(1-\frac1{2n}+O\left(\frac{1}{n^2}\right)\Biggr)=n-\frac{1}{2}+O\left(\frac{1}{n}\right)\,,$$ which can also be written as $$\sqrt{n(n-1)}=(n-1)+\frac{1}{2}+O\left(\frac{1}{n}\right)\,.$$ Ergo, $$\begin{align}\sqrt{n(n-1)+\sqrt{n(n-1)}}&=\sqrt{n^2-\frac{1}{2}+O\left(\frac{1}{n}\right)}&=&n\,\sqrt{1-\frac{1}{2n^2}+O\left(\frac{1}{n^3}\right)}\\&=n\,\Biggl(1-\frac{1}{4n^2}+O\left(\frac{1}{n^3}\right)\Biggr)&=&n-\frac{1}{4n}+O\left(\frac{1}{n^2}\right)\,.\end{align}$$ Similarly, $$\sqrt{n(n-1)-\sqrt{n(n-1)}}=(n-1)-\frac{1}{4n}+O\left(\frac{1}{n^2}\right)\,.$$ Hence, $$\begin{align}\sqrt{n(n-1)+\sqrt{n(n-1)+\sqrt{n(n-1)}}}&=\sqrt{n^2-\frac{1}{4n}+O\left(\frac{1}{n^2}\right)}\\&=n\,\sqrt{1-\frac{1}{4n^3}+O\left(\frac{1}{n^4}\right)}\\&=n\,\Biggl(1-\frac{1}{8n^3}+O\left(\frac{1}{n^4}\right)\Biggr)\\&=n-\frac{1}{8n^3}+O\left(\frac{1}{n^3}\right)\,.\end{align}$$ Likewise, $$\sqrt{n(n-1)-\sqrt{n(n-1)-\sqrt{n(n-1)}}}=(n-1)+\frac{1}{8n^2}+O\left(\frac{1}{n^3}\right)\,.$$ In fact, we can prove by induction on $k$ that $$f^+_k\big(n(n-1)\big)=n-\frac{1}{2^kn^{k-1}}+O\left(\frac{1}{n^k}\right)$$ and $$f^-_k\big(n(n-1)\big)=(n-1)-\frac{(-1)^k}{2^kn^{k-1}}+O\left(\frac{1}{n^k}\right)\,,$$ where $f^+_k(x):=\sqrt{x+f^+_{k-1}(x)}$ with $f^+_0(x):=0$ and $f^-_k(x):=\sqrt{x-f^-_{k-1}(x)}$ with $f^-_0(x):=0$ for all $x\geq 0$ and for each $k=1,2,3,\ldots$.
| {
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"url": "https://math.stackexchange.com/questions/1859741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 5
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