Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to find the subgradient of $x \mapsto \max \left( (x+1)^2, (x-3)^2 \right) $ at $x=1$? I need help finding the subgradient of the following function at the point $x = 1$.
$$ \max \left( (x+1)^2, (x-3)^2 \right) $$
I think it's $$ [-4,4] $$ since that is the range between the left-hand and right-hand derivatives, but I can't prove it.
| We can observe that
$$f(x) = \max((x+1)^2, (x-3)^2) = \begin{cases} (x+1)^2 & x \geq 1 \\ (x-3)^2 & x < 1 \end{cases}$$
and this function passes through $(1,4)$. Consider the line
$$y = a(x-1) + 4$$
This line intersects the parabola $y = (x+1)^2$ at $(1,4)$ and $(a-3,(a-2)^2)$ and intersects the parabola $y = (x-3)^2$ at $(1,4)$ and $(a+5,(a+2)^2)$. As such, the line only intersects $f(x)$ twice if $a-3 > 1$ or $a+5 < 1$, or equivalently, if $a > 4$ or $a < -4$. The subgradients are the values of $a$ for which this line does not intersect the graph of $f$ twice, which would be $[-4,4]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4101455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Taylor Series $\sin(3x)$ centered at $a=\frac{\pi}{6}$ Using a table I got
\begin{array}{c|c|c}
n & f^{n}(x) & f^{n}\left(\frac{\pi}{6}\right) \\
\hline
0 & \sin(3x) & 1\\
1 & 3\cos(3x) & 0\\
2 & -9\sin(3x) & -9\\
3 & -27\cos(3x) & 0\\
4 & 81\sin(3x) & 81
\end{array}
From this table, I expanded $\sin(3x)$ centered at $a=\frac{\pi}{6}$, I get
\begin{align}
\require{cancel}
f(x) &= \frac{1}{0!}\left(x-\frac{\pi}{6}\right)^0+\frac{0}{1!}\left(x-\frac{\pi}{6}\right)^1-\frac{9}{2!}\left(x-\frac{\pi}{6}\right)^2+\frac{0}{3!}\left(x-\frac{\pi}{6}\right)^3+\frac{81}{4!}\left(x-\frac{\pi}{6}\right)^4+\cdots \\
&= \frac{1}{0!}\left(x-\frac{\pi}{6}\right)^0+\cancel{\frac{0}{1!}\left(x-\frac{\pi}{6}\right)^1}-\frac{9}{2!}\left(x-\frac{\pi}{6}\right)^2+\cancel{\frac{0}{3!}\left(x-\frac{\pi}{6}\right)^3}+\frac{81}{4!}\left(x-\frac{\pi}{6}\right)^4+\cdots \\
&= \frac{1}{0!}\left(x-\frac{\pi}{6}\right)^0-\frac{3^2}{2!}\left(x-\frac{\pi}{6}\right)^2+\frac{3^4}{4!}\left(x-\frac{\pi}{6}\right)^4-\frac{3^6}{6!}\left(x-\frac{\pi}{6}\right)^6
\end{align}
Is the following Taylor Series correct for the function $f(x)=\sin(3x)$ centered at $a=\frac{\pi}{6}$?
\begin{align}
f(x) &=\sum_{n=1}^{\infty}(-1)^n\frac{3^{2n-2}}{(2n-2)!}\left(x-\frac{\pi}{6}\right)^{2n-2}
\end{align}
| Yes. This seems to be the answer you'll get with/without the table. Just that without, you'd have to work out the values of different derivatives of f but you would ultimately arrive that result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proving if $a_{k}\ge a_{k-1}+1$ then $1+\frac{1}{a_{0}}(1+\frac{1}{a_{1}-a_{0}})...(1+\frac{1}{a_{n}-a_{0}})\le \prod_{k=0}^{n}(1+\frac{1}{a_{k}})$ I've worked on this problem with the sequence $a_{k}$ being the natural numbers, that is $a_{k}=a_{k-1}+1$ and $a_{0}=1$. Over the naturals, $\prod_{k=0}^{n}(1+\frac{1}{a_{k}})$ can be proven to be $n+1$.
I've been able to recognize the obvious pattern in $\prod_{k=2}^{n}(1+\frac{1}{k-1})$ to be equal to $n$, but I do not know how to prove this rigorously.
That said, over the naturals, the inequality reduces to $n+1 \le n+1$. That got me thinking if the sequence in question could be expanded with the condition $a_{k}\ge a_{k-1}+1$ and specifically if this would not lead to the situation corresponding with RHS being greater than LHS.
Thank you for your help.
|
Given reals $a_0 > 0$ and $ a_k \geq a_{k-1} + 1$, show that $$ 1 + \frac{1}{a_0}\prod_{i=1}^{n} ( 1 + \frac{1}{ a_i - a_0 }) \leq \prod_{i=0}^n ( 1 + \frac{1}{a_i}).$$
Claim 1: $ \prod_{i=0}^n (1 + \frac{1}{a_i}) \leq \frac{a_{n+1}}{a_0} $
Proof: This follows from the telescoping inequalities
$$ \prod_{i=0}^n (1 + \frac{1}{a_i}) = \prod_{i=0}^n \frac{a_i + 1}{a_i} \leq \prod_{i=0}^n \frac{ a_{i + 1} } { a_i} = \frac{a_{n+1}}{a_0}. $$
Equality holds iff $a_i = i+a_0$.
Claim 2: $ 1 + \frac{1}{a_0} \prod_{i=1}^{n} ( 1 + \frac{1}{ a_i - a_0 }) \leq \prod_{i=0}^n ( 1 + \frac{1}{a_i}).$
Proof by induction.
Base Case: $ 1 + \frac{1}{a_0} + \frac{1}{ a_0 (a_1 - a_0) }
\leq 1 + \frac{1}{a_0} + \frac{1}{a_1} + \frac{1}{ a_0 a_1 } \\
\Leftrightarrow a_1 \leq (a_0+1) (a_1 - a_0) \\
\Leftrightarrow a_0 ( a_1 - a_0 - 1) \geq 0 , $
which is true.
Induction step: Applying the induction hypothesis, we have
$$ 1 + \frac{1}{a_0} \prod_{i=1}^{k+1} ( 1 + \frac{1}{ a_i - a_0 }) \leq 1 + ( 1 + \frac{1}{a_{k+1} - a_0 } ) \times \left[ \prod_{i=0}^k (1 + \frac{1}{a_i} ) - 1 \right] \\
= ( 1 + \frac{1}{a_{k+1} - a_0 } ) \times \left[ \prod_{i=0}^k (1 + \frac{1}{a_i} ) \right] - \frac{1}{a_{k+1} - a_0} $$
We WTS that the RHS is $\leq ( 1 + \frac{1}{ a_{k+1} }) \prod_{i=0}^{k} ( 1 + \frac{1}{a_i}).$
Shifting terms around and factoring, this is equivalent to showing that
$$ \frac{a_0}{ (a_{k+1} - a_0 ) ( a_{k+1} ) } \times \left[ \prod_{i=0}^k ( 1 + \frac{1}{a_i})\right] \leq \frac{1}{ a_{k+1} - a_0 } .$$
This follows directly from the previous claim by replacing the term in the square brackets.
Again, equality holds iff $a_i = i+a_0$.
Note:
*
*The standard algebraic manipulations of the induction approach yielded the first claim directly. Assuming that this could be proved by induction in a direct manner, the first claim had to be true.
*The first claim could be strengthened to $ \prod_{i=0}^n (1 + \frac{1}{a_i}) \leq \frac{ a_{n} + 1 }{a_0}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4108966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Maximize $x + y$ with constraint $y \cdot x^2=15$. As the title says, I am asked to maximize the sum of two numbers $x + y$ when $x^2 \cdot y = 15$.
The thing is... I tried substitution but got nowhere, it seems $h(x) = x + \frac{15}{x^2}$ only has a minimum value, not a maximum.
I've been starting to think there is an error in the excercise. I tried Lagrange multiplicators next, having:
\begin{align}
f(x,y) &= x+y\\
g(x,y) &= x^2\cdot y - 15 = 0
\end{align}
Then:
\begin{align}
\nabla f = \lambda \nabla g
\end{align}
that gives:
\begin{align}
2xy &= 1\\
2x^2 &= 1\\
x^2\cdot y &= 15
\end{align}
then we find
\begin{align}
2xy &= 2x^2\\
x(2y - x) &= 0\\
x = 0 &\lor x = 2y
\end{align}
Now, from $x^2 \cdot y = 15$ we know $x,y>0$, then we only have one solution $x=2y$, applying that to $g(x)$ we get $x^2 \cdot \frac{x}{2} = 15 \implies x = \sqrt[3]{30}, y = \frac{\sqrt[3]{30}}{2}$.
Then the only point that I can use from Lagrange is $(\sqrt[3]{30}, \frac{\sqrt[3]{30}}{2})$. But what guarantee do I have that said point is the maximum of the sum of $x + y$?. Doesn't Lagrange only find extremes? How do I check this result?
As I said earlier, if I graph $h(x) = x + \frac{15}{x^2}$ in $R^2$ I find that $x=\sqrt[3]{30}$ is the value in the $x$ axis where the minimum value of $h(x)$ is, so this only feeds my suspicions about the problem having an error and the point needed being actually a minimum instead of a maximum.
Can anyone shed some light into this?
| There is definitely an error in this. You can see that from the logical framework, the product of two numbers being fixed means you can take one to 0 and the other to infinity on slider-scale. Then the sum goes to infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find angle between $y=\sin x$ and $y=\cos x$ at their intersection point. Find angle between $y=\sin x$ and $y=\cos x$ at their intersection point.
Intersection points are $\frac{\pi}{4}+\pi k$ and to find angle between them we need to compute derivatives at intersection points but then I can't combine them to get an answer which is $\arctan2\sqrt2$. Will be thankful for your help.
| If you think about these as parametrized curves $(t,\sin t)$ and $(t,\cos t)$, then the angle between them at a point $t_0$ is the angle between $(1,\cos t_0)$ and $(1,-\sin t_0)$. If $t_0 = \frac{\pi}{4} + \pi k$, then we get
$$\cos\theta = \frac{(1,\pm\frac{1}{\sqrt{2}})\cdot(1,\mp\frac{1}{\sqrt{2}})}{\frac{3}{2}} = \frac{1}{2}\cdot\frac{2}{3} = \frac{1}{3}.$$
This, of course, yields an angle of $\arccos\frac{1}{3}$, which is the same as $\arctan 2\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4116134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Finding $(a+b)^n + (a-b)^n$ when $b$ is a radical We aim to find the unit digit of the following expression: $(5+\sqrt{24})^{1234}+(5-\sqrt{24})^{1234}$.
I tried finding a pattern of summing two binomials by observing:
$(a+b)^2+(a-b)^2 = (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2) = 2a^2 + 2b^2$
$(a+b)^3+(a-b)^3 = (a^3 + 3a^2b + 3ab^2 + b^3) + (a^3 - 3a^2b + 3ab^2 - b^3) = 2a^3+6ab^2$
$(a+b)^4+(a-b)^4 = (a^4+4a^3b+6a^2b^2+4ab^3+b^4) + (a^4-4a^3b+6a^2b^2-4ab^3+b^4) = 2a^4+12a^2b^2+2b^4$
and so on. For $(5+\sqrt{24})^{n}+(5-\sqrt{24})^{n}$ where $n$ is even, I observed that every resulting term except for the final term $2b^n$ returns a unit digit of $0$, as every term consists $c\cdot 5^m$ where $c$ and $m$ are even.
Hence, the unit digit of $(5+\sqrt{24})^{1234}+(5-\sqrt{24})^{1234}$ equals to the unit digit of $\sqrt{24}^{1234}$, which is $24^{617}$. By the cyclicity of $4$, I concluded that the unit digit of $(5+\sqrt{24})^{1234}+(5-\sqrt{24})^{1234}$ is $4$.
Unfortunately, this was a part of a multiple choice problem, and $4$ was not an available option. What was my erroneous assumption, and what is a mathematical way to tackle this problem?
| Well that was careless of me! The unit digit would be the doubled value of $4$, as it is $2b^n$ that determines the unit digit, not $b^n$.
Hence, the unit digit of the expression would be $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4116460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximize $x^2y$ where $x^2+y^2=100$. (Looking for other approaches)
For positive numbers $x,y$ we have $x^2+y^2=100$. For which ratio of $x$ to $y$, the value of $x^2y$ will be maximum?
$1)\ 2\qquad\qquad2)\ \sqrt3\qquad\qquad3)\ \frac32\qquad\qquad4)\ \sqrt2$
It is a problem from a timed exam so I'm looking for alternative approaches to solve this problem quickly. Here is my approach:
We have $x^2=100-y^2$, so $x^2y=y(100-y^2)$. By differentiating with respect to $y$ and putting it equal to zero, we have:
$$(100-y^2)-2y^2=0\qquad\qquad y^2=\frac{100}3$$
Hence $x^2=\frac{200}3$ and $\frac{x^2}{y^2}=2$ therefore $\frac{x}{y}=\sqrt2$.
Is there another quick approach to solve it? (like AM-GM inequality or others)
| I think your solution is fine, the only change I would make it to point out that the constraint $x^2 + y^2 = 100$ can be changed to $x^2 + y^2 = z$ for any positive $z$ without changing the solution, since it is just ratios. But here is an alternative:
$$x^2 + y^2 = 1$$
$$r = x^2y$$
$$x = wy$$
Write r in terms of w:
$$r = (wy)^2y = w^2y^3$$
$$(wy)^2 + y^2 = 1 \implies y^2 = (1+w^2)^{-1}$$
Using the old "the critical points of $r$ are included in the critical points of $r^2$" trick:
$$r^2 = \dfrac{w^4}{(1 + w^2)^{3}} = \dfrac{1}{w^2 + 3 + 3w^{-2} + w^{-4}} > 0$$
$$d(r^2) = -r^{2}(2w - 6w^{-3} - 4w^{-5}) = 0$$
$$w^6 - 3w^{2} - 2 = 0$$
$$(w^2 - 2)(w^2+1)^2 = 0$$
$$w = \sqrt 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find the integral solution with minimum sum given the product is constant.
Let $x,y,z,n$ be positive integers such that $n$ has at least three prime divisors.
Find the minimum of $x+y+z$ such that $xyz=n$.
First, note that $x,y,z$ must be positive integers.
Consider $n=945$. Then its prime factorization is $3\cdot 3 \cdot 3 \cdot 5 \cdot 7$.
The following is my approach:
Choose two smallest integers in the prime factorization, which are $3,3$ and multiply them, we get $$9\cdot 3 \cdot 5 \cdot 7$$
Next choose two smallest integers among the numbers above, which are $3,5$ and multiply them, we get
$$9\cdot 15\cdot 7$$
So the minimum of $x+y+z$ should be $9+15+7=31$.
My question is: Is there a general solution for the minimum of $x+y+z$ in terms of $n$?
| You can solve the problem via integer linear programming as follows. Given prime factorization $n=\prod_p p^{n_p}$, there are $\prod_p (n_p+1)$ possible factors. For each factor $j=\prod_p p^{j_p}$, where $j_p \le n_p$, let nonnegative integer decision variable $x_j$ be the number of times the factor is used. The problem is to minimize $\sum_j j x_j$ subject to
\begin{align}
\sum_j x_j &= 3 \tag1 \\
\sum_j j_p x_j &= n_p &&\text{for all $p$} \tag2
\end{align}
Constraint $(1)$ selects three factors. Constraint $(2)$ uses each prime factor the correct number of times.
For example, $n=2160=2^4 3^3 5^1$ has $n_2=4$, $n_3=3$, and $n_5=1$, and there are $(4+1)(3+1)(1+1)=40$ factors. An optimal solution is $x_{12}=2$, $x_{15}=1$, with objective value $2 \cdot 12 + 1 \cdot 15 = 39$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4119377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $(x+1)(x+2)(x+3)(x+k) + 1$ is a perfect square then the value of $k$ is? If $(x+1)(x+2)(x+3)(x+k) + 1$ is a perfect square then the value of $k$ is ?
My approach:
The given expression can be written as $(y-1)(y)(y+1)(y+k-2) + 1$ where $y = x + 2$
This gives me $(y)(y^2 - 1)(y+k-2) = n^2 - 1$ something seems similar in LHS and RHS. But not sure how to proceed next? Should I put $(y)(y+k-2) = 1$ ? although it doesn't seem right for some reason also I dont think it leads anywhere.
Any hints/explanations are appreciated!
| So we must have for some quadratic $f$,
$$(x+1)(x+2)(x+3)(x+k) = (f+1)(f-1)$$
This implies the LHS can be split into two quadratics which differ only in the constant term, by $2$. Clearly $f$ is monic, and $f$'s linear term can only be among $(3x, 4x, 5x)$, as combining any two of the known three terms on LHS into a quadratic gives $(x+1)(x+2), (x+1)(x+3)$ or $(x+2)(x+3)$.
Corresponding to $(3x, 4x, 5x)$, the only possibilities for $k$ are $(0,2, 4)$, by requiring the remaining quadratic factor in LHS to also have same linear term. Now for each of these, the difference in constant term of the two quadratic factors is in order $(2, 1, 2)$, so the only solution is $k\in \{0, 4\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding values of $\theta$ for which $(\cos\theta,\sin\theta)$ lies inside the triangle formed by $x+y=2$, $x−y=1$, and $6x+2y−\sqrt{10}=0$
Find the range of values of $\theta$, such that $\theta\in[0,2\pi]$ for which $(\cos \theta,\sin \theta)$ lies inside the triangle formed by $x + y = 2, x − y = 1$ and $6x + 2y − \sqrt{10} =0$.
Not getting any hint of how can I get the range of those points is there any short method of doing so.
I have plotted the graph and then I found out that the line $6x + 2y − \sqrt{10} =0$ which forms one of the sides of the triangle has $O(0,0)$ and $P(\cos \theta, \sin\theta)$ on opposite sides so applied the formula of power of point.
So $L_3(O)\cdot L_3(P)<0$ and then solved to find out the range.
If there is any short method please tell me.
| Working from the diagram OP has posted, $D=(1,0)$ corresponds to $\theta=0$, so we only need to work out $E$. Now, $E$ is the intersection of $6x+2y=\sqrt{10}$ and $x^2+y^2=1$. In polar coordinates, these are $6r\cos\theta+2r\sin\theta=\sqrt{10}$ and $r=1$, so it comes down to solving $6\cos\theta+2\sin\theta=\sqrt{10}$. Noting that $6^2+2^2=40$, we divide both sides by $\sqrt{40}=2\sqrt{10}$ to get $${3\over\sqrt{10}}\cos\theta+{1\over\sqrt{10}}\sin\theta={1\over2}$$ Let $\eta=\arcsin{3\over\sqrt{10}}$ which implies $\cos\eta={1\over\sqrt{10}}$ and $\tan\eta=3$. Then we have $1/2=\sin\eta\cos\theta+\cos\eta\sin\theta=\sin(\theta+\eta)$. This is satisfied by $\theta+\eta=\pi/6$, but $\tan\eta=3>1$ implies $\eta>\pi/4$ and thus $\theta+\eta>\pi/4$, so we go to $\theta+\eta=5\pi/6$. Then, $$\theta={5\pi\over6}-\eta={5\pi\over6}-\arctan3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that the tangent plane to the surface $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1$ Show that the tangent plane to the surface $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1$$
At $(x_0,y_0,z_0)$ is given by $$\frac{x_{0}x}{a^{2}}-\frac{y_{0}y}{b^{2}}-\frac{z_{0}z}{c^{2}}=1$$
Then assume $a=b=c=1$ and show the tangent lines to the surface in the intersection with $x=x_0$ construct a cone. ($x_0 >1$).
Define $$F(x,y,z)=\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}-1$$
Then the tangent plane to the the surface at the given point is:
$$\frac{2x_{0}}{a^{2}}\left(x-x_{0}\right)-\frac{2y_{0}}{b^{2}}\left(y-y_{0}\right)-\frac{2z_{0}}{c^{2}}\left(z-z_{0}\right)=0$$
On the other hand the point is also on the surface so the equation is:
$$\frac{x_{0}x}{a^{2}}-\frac{y_{0}y}{b^{2}}-\frac{z_{0}z}{c^{2}}-\left(\frac{2x_{0}^{2}}{a^{2}}-\frac{2y_{0}^{2}}{b^{2}}-\frac{2z_{0}^{2}}{c^{2}}\right)=0=0$$
$$\frac{x_{0}x}{a^{2}}-\frac{y_{0}y}{b^{2}}-\frac{z_{0}z}{c^{2}}=1$$
Also if $a=b=c=1$ then the points on the intersection of the surface with $x=x_0$ satisfy the following relation:$$x_0^2-y^2-z^2=1$$
$$z^2+y^2=x_0^2-1\tag{1}$$
Now should I find the tangent lines to $(1)$? And show that they construct a cone?
| Here is what the second part of the question means though it is written a bit ambiguous.
The intersection of plane $x = x_0$ and surface $x^2 - y^2 - z^2 = 1$ (as $a = b = c = 1$) is a circle. Different values of $y, z$ lead to different level curves. For example, if $y = 0$, you get a hyperbola $x^2-z^2 = 1$. Similarly if $z = 0$, you get hyperbola $x^2-y^2 = 1$. You can think of it as rotating plane $y = 0$ about x-axis. The intersection curve of a generated plane this way, with the hyperboloid, is a hyperbola. For ease of working, we parametrize the surface as,
$r(x, \theta) = (x, \sqrt{x^2-1} \cos \theta, \sqrt{x^2-1} \sin \theta), 0 \leq \theta \leq 2\pi$
For any given $\theta = \alpha$, the level curve is then given by,
$r(x) = (x, \sqrt{x^2-1} \cos \alpha, \sqrt{x^2-1} \sin \alpha)$
$r'(x) = (1, \frac{x}{\sqrt{x^2-1}} \cos \alpha, \frac{x}{\sqrt{x^2-1}} \sin \alpha)$
Equation of tangent line to the curve at $x = x_0$ is $ \ r(x_0) + r'(x_0) t$. So,
$x = x_0 + t, y = \big(\sqrt{x_0^2-1} + \frac{x_0 t}{\sqrt{x_0^2-1}}\big) \cos \alpha, z = \big(\sqrt{x_0^2-1} + \frac{x_0 t}{\sqrt{x_0^2-1}}\big) \sin \alpha$
Substituting, $t = x - x_0$, please note $\sqrt{x_0^2-1} + \frac{x_0 t}{\sqrt{x_0^2-1}} = \frac{x_0}{\sqrt{x_0^2 - 1}} (x - \frac{1}{x_0})$
So $y^2 + z^2 = \frac{x_0^2}{x_0^2-1} (x - \frac{1}{x_0})^2$ and that is equation of a cone with its axis along x-axis and vertex at $(\frac{1}{x_0}, 0 , 0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving for the derivative of absolute value, gone wrong The absolute value $|x|$ can be represented as $\sqrt{x^{2}}$, as per this question.
Let $f(x) = |x| = \sqrt{x^{2}}$. Solving for $f'(x)$, let $u = x^{2}$. Then,
\begin{align*}\frac{df}{dx} &= \frac{df}{du}\cdot\frac{du}{dx} \\ &= \frac{d}{du}(\sqrt{u})\cdot\frac{d}{dx}(x^{2}) \\ &=\frac{1}{2\sqrt{u}}\cdot2x \\ &= \frac{x}{\sqrt{x^{2}}} \\ &= \frac{x}{|x|}\end{align*}
We can also see that $\sqrt{x^{2}} = \left(\sqrt{x}\right)^{2}$. Then, let $u = \sqrt{x}$. Solving for $f'(x)$,
\begin{align*}\frac{df}{dx} &= \frac{df}{du}\cdot\frac{du}{dx} \\ &= \frac{d}{du}(u^{2})\cdot\frac{d}{dx}(\sqrt{x}) \\ &= 2u\cdot \frac{1}{2\sqrt{x}} \\ &= \frac{\sqrt{x}}{\sqrt{x}} \\ &= 1\end{align*}
I think the problem here is by letting $u = \sqrt{x}$. What seems to be the problem?
| Yes. You are working over the reals, so how can $\sqrt{-1}$ be defined when you substitute $u=\sqrt x$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solution of the Diophantine equation
What are the possible triples (x,y,z) in positive integers such >that,
$$(x+y)^{2}+3x+y+1=z^{2}$$
I have used the inequality approach and many others but wasn't able to find an answer.
| Let $t:=x+y$ so that the Diophantine equation is equivalent to
$$t^2+t+2x+1=z^2.$$
In particular this shows that $z$ is odd, say $z=2s+1$. Multiplying by $4$ yields
$$(4s+2)^2=4z^2=4t^2+4t+8x+4=(2t+1)^2+8x+3,$$
and a bit of rearranging then shows that
$$(4s+2)^2-(2t+1)^2=8x+3.$$
Conversely, for any pair of integers $s$ and $t$ we have
$$(4s+2)^2-(2t+1)^2\equiv3\pmod{8},$$
which shows that the integral solutions are parametrized by
\begin{eqnarray*}
(x,y,z)
&=&\left(\frac{(4s+2)^2-(2t+1)^2-3}{8},t-\frac{(4s+2)^2-(2t+1)^2-3}{8},2s+1\right)\\
&=&\left(2s^2+2s-\frac{t^2+t}{2},-2s^2-2s+\frac{t^2-t}{2},2s+1\right).
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find out intersections of two absolute value functions? If we are asked to find out common points where two equations intersect each other, for example let us assume that two equations are $Y=12-x^2$ and $Y=4x$, we can solve it in the following way:
\begin{align*}4x&= 12-x^2\\
x^2+4x-12&=0\\
x^2+6x-2x-12&=0\\
x(x+6)-2(x+12)&=0\\
(x-2)(x+6)&=0
\end{align*}
So we have now two roots of $x$, $2$ and $-6$.
Plugging these values in one of the above equations we get two values for $y$, $8$ and $-24$.
So in $(2,8)$ and $(-6,-24)$ the above two lines intersect each other.
But what will we do if we are asked to find out the points where another two equations, $y=|x|$ and $y=|x^2-4|$ meet?
Is there any methodical approach which help find out the common points?
| In general, to find intersection between two curves $ y = f(x) $ and $ y = g(x) $,
you would look for solutions of $ f(x) = g(x) $ as you showed in your example.
To find the intersections between $ y = |f(x)| $ and $ y = |g(x)| $, you should thus solve $ |f(x)| = |g(x)| $, or equivalently, $ f^2(x) = g^2(x) $.
With $ f(x) = x $ and $ g(x) = x^2 - 4 $, look for solutions of
$ x^2 = (x^2 - 4)^2$, i.e. the roots of $ x^4 - 9x^2 + 16 $. You should find four of them:
\begin{align}
x_1 &= -\frac{1 + \sqrt{17}}{2} \approx -2.562\\
x_2 &= -\frac{-1 + \sqrt{17}}{2} \approx -1.562\\
x_3 &= \frac{-1 + \sqrt{17}}{2} \approx 1.562\\
x_4 &= \frac{1 + \sqrt{17}}{2} \approx 2.562
\end{align}
Then as you mentioned you can find the intersections by plugging the values into either $ |f(x)| $ or $ |g(x)| $ (it should yield the same results with both).
With your example, the curves therefore intersect at four points:
\begin{align}
P_1 &= (x_1, |f(x_1)|)\\
P_2 &= (x_2, |f(x_2)|)\\
P_3 &= (x_3, |f(x_3)|)\\
P_4 &= (x_4, |f(x_4)|)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4126996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$a_n=(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n}) $ converge Hi guys I have a problem that i need to prove that the sequence:
$$a_n=(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n}) $$ converges
I need to show its a monotone sequence with an upper and lower bound.
For lower bound, 0 seems to fit nicely.
I can also see that I always multiply by something bigger than one so its always increasing.
My only issue was finding an upper bound, with my calculator I figured that something around e is the upper bound but I can't prove it.
I attempted by induction:
$$ \left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{4}\right)\dots\left(1+\frac{1}{2^{n-1}}\right)\left(1+\frac{1}{2^{n}}\right)<4 $$
after multiplying both sides I got :
$$ \left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{4}\right)\dots\left(1+\frac{1}{2^{n-1}}\right)\left(1+\frac{1}{2^{n}}\right)\left(1+\frac{1}{2^{n+1}}\right)<4\left(1+\frac{1}{2^{n+1}}\right)<4\cdot2=8 $$
But it seems wrong and Im kind of lost.
Also tried using $$log_2a_n=log(3)-log(2)+log(5)-log(4)+log(9)-log(8)+\dots log(2^n+1)-log(2^n)$$
No matter how much I try, I just cant prove an upper bound, any suggestions?
| For the lower bound :
Using Bernoulli's inequality we have $x\geq 0$:
$$(1+x)^{\frac{1}{2}}\leq 1+\frac{1}{2}x$$
So :
$$(1+\frac{1}{2})^{\frac{1}{2}}(1+\frac{1}{4})^{\frac{1}{2}}\cdots(1+\frac{1}{2^n})^{\frac{1}{2}}< (1+\frac{1}{4})(1+\frac{1}{8})\cdots(1+\frac{1}{2^{n+1}})$$
Now we can conclude easily
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4129331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
In $\triangle{ABC}$, $\angle ABC=45^ \circ$. $X$ is a point on $BC$ such that $BX=\frac{1}{3}BC$ and $\angle AXC=60^ \circ$. Find $\angle ACB$. Problem
In $\triangle{ABC}$, $\angle ABC=45^ \circ$. $X$ is a point on $BC$ such that $BX=\frac{1}{3}BC$ and $\angle AXC=60^ \circ$. Find $\angle ACB$.
The problem looks easy. Though I couldn't solve it in an efficient way. Finally I solved it using trigonometry.
Trig solution
Let $BX = a$ units, then $BC = 3a$ and $XC = 3a-a= 2a$ units. $\angle AXC =60^ \circ$ and $\angle ABC= 45^ \circ$, then $\angle BAX= 60^ \circ -45^ \circ = 15^ \circ$.
Applying sine rule in $\triangle ABX$,
$$\frac{BX}{\sin \angle BAX} = \frac{AX}{\sin \angle ABC}$$
$$\implies \frac{a}{\sin 15°} = \frac{AX}{\sin 45°} \tag{1}$$
In $∆AXC$ , let $\angle ACB = \theta$, then $\angle XAC = (120 - \theta)$ and by sine rule,
$$\frac{XC}{\sin \angle XAC} = \frac{AX}{\sin \angle ACB}$$
$$\implies \frac{2a}{\sin (120 - \theta)} = \frac{AX}{\sin \theta} \tag{2}$$
Dividing $(1)$ by $(2)$,
$$\frac{\sin (120-\theta)}{2\sin 15^ \circ}= \frac{\sin \theta}{\sin 45°}$$
$$\implies 2\sin 15°\cdot\sin \theta = \sin 45°\cdot\sin (120-\theta)$$
$$\implies \frac{\sqrt{3}–1}{\sqrt 2}.\sin \theta = \frac{1}{\sqrt 2}.(\sin120°.\cos \theta - \cos 120°.\sin \theta).$$
$$\implies (\sqrt{3}–1).\sin \theta = \frac{\sqrt 3}{2}.\cos \theta + \frac{1}{2}.\sin \theta$$
$$\implies \tan \theta= 2+\sqrt 2$$
$$\implies \theta=75^ \circ$$
Thus, $\angle ACB = 75°$.
This solution is impossible without knowing the values of $\sin 15^ \circ$ and $\tan 75^ \circ$. And I find trigonometry boring. So, can this problem be solved in some other ways?
| Drop a perpendicular from point $C$ on $AX$ and let the feet of this perpendicular be $D$.
$DX=\frac {1}{2}XC=BX$ and thereafter $BD=DC$ and $AD=BD$ by simple angle chasing. Thus $D$ is the circumcentre of $\triangle ABC$ and $\angle ACB=90-15=75^{\circ}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4131869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $k$ is multiple of $3$, then $x_k$ is even.
Let $ \ (x_k) \in \mathbb{N}^{\mathbb{N}} \ $ be such that $ \ x_0=0 \ $, $ \ x_1=1 \ $ and $$ \ x_k = 3x^3_{k-1}+7x^5_{k-2} \ $$ $ \ \forall k \geq 2 \ $. Prove that $ \, x_k \, $ is even, for all $k$ multiple of $3$.
I tried to use induction with steps of size $3$:
$x_2 = 3 \ \Rightarrow \ x_3 = 3 \cdot 3^3 + 7 = 88 \ $ is even.
Let $k$ be a multiple of $3$ such that $ \, x_k \, $ is even. Then
$x_{k+3} = 3x^3_{k+2} + 7x^5_{k+1} = 3 (3x^3_{k+1}+7x^5_{k}) + 7(3x^3_{k}+7x^5_{k-1}) = \ ...$
I am trying to arrange this recurrence formula in a way to substitute $ \, x_k \, $ for $ \, 2m \, $ and obtaining $ \ x_{k+3} = 2 \cdot ( ...... ) \ $, but failing in the process.
| We could show, by modular congruence, that $x_{3m} \equiv 0 \pmod2$. We know that $7 \equiv 3 \equiv1 \pmod2$. We also know that $x^2 \equiv x \pmod2$ in the special case of mod 2 congruence: if $x$ is even, so will be $x^2$; if $x$ is odd, so will be $x^2$. Therefore, we can generalize to $x^n \equiv x \pmod2$. (*)
Finally, all of this translates to:
$$ x_k \equiv 3(x_{k-1})^3+7(x_{k-2})^5\equiv (x_{k-1})^3 + (x_{k-2})^5 \equiv x_{k-1}+x_{k-2} \pmod2 $$
Since $x_{k-1}=3(x_{k-2})^3+7(x_{k-3})^5$, we have that
$$x_{k-1} \equiv x_{k-2}+x_{k-3} \pmod2$$
Therefore,
$$x_k \equiv 2 x_{k-2}+x_{k-3} \equiv x_{k-3} \pmod2 \textrm{ (**)}$$
Using $k=3m$, we can rewrite this as
$$ x_{3m} \equiv x_{3(m-1)} \pmod2 $$
By induction, we see that $x_{3m} \equiv x_{0} \pmod2$ (***). Since $x_0 \equiv 0 \pmod2$, we have that $x_{3m}$ must be even.
Edit for further clarity:
(*):
Let $y$ be any natural number (including $0$). If $y$ is odd, then its remainder upon division by $2$ must be $1$ (by the definition of odd numbers: those which are not divisible by $2$). If $y$ is even, then its remainder upon division by $2$ must be $0$ (by the definition of even numbers: those which are divisible by $2$).
First, if $y$ is odd, we have the following congruence $\pmod2$:
$$y \equiv 1 \pmod2$$
Using the properties of modular congruence, the relation will still be valid upon multiplication of both sides by any integer. If we multiply both sides by $y$:
$$y^2 \equiv y \pmod2$$
Since $y$ is obviously still congruent to $1$, the above equiation can also be written as:
$$y^2 \equiv y \equiv 1 \pmod{2}$$
I can once again multiply all sides by $y$, obtaining:
$$y^3 \equiv y^2 \equiv y \equiv 1 \pmod{2}$$
By induction, then, we arrive at the conclusion that
$$y^n \equiv y \equiv 1 \pmod2$$
for any natural integer $n$.
Repeating the above but when $y$ is even, we would arrive at
$$y \equiv 0 \pmod2 \implies y^2 \equiv 0 \pmod2 \implies y^2 \equiv y \equiv 0 \pmod2$$
which would naturally lead us into
$$y^n \equiv y \equiv 0 \pmod2$$
So, (*) is valid for any natural $y$, as $y^n \equiv y \pmod2$ is valid when $y$ is even and when it's odd.
(**):
$2 \equiv 0 \pmod2$, which means $2x_{k-2} \equiv 0 \pmod2$.
(***):
Note that the congruence relation $x_{3m} \equiv x_{3(m-1)} \pmod2$ can be naturally extended to $x_{3m} \equiv x_{3(m-1)} \equiv x_{3(m-2)} \equiv \ldots \equiv x_{3} \equiv x_0 \pmod2$ by the use of induction. First, it has been already calculated that $x_3 \equiv x_0 \pmod2$ (given that $x_3=88$). Second, notice that $m$ is any given integer (such that $k-1=3m-1\geq2$, as the recurrence formula demands for both $x_k$ and $x_{k-1}$). Therefore, $m-1=m'$ and the modular congruence equiation could be easily re-obtained (as, again, the derivation of this congruence equation involved any given integer $m$ such that $k=3m\geq2$, not a specific $m$):
$$x_{3m'} \equiv x_{3(m'-1)} \pmod2$$
Substituting $m'=m-1$, we would have
$$x_{3(m-1)} \equiv x_{3(m-2)} \pmod2$$
Since $x_{3m'} \equiv x_{3(m'-1)} \pmod2$, the above translates to
$$x_{3m} \equiv x_{3(m-1)} \equiv x_{3(m-2)} \pmod2$$
I can repeat this induction up until $3m=3,m=1$, which means
$$x_{3m} \equiv \ldots \equiv x_3 \equiv x_0 \pmod2$$
Note: I can in fact impose that $x_{3m} \equiv x_0 \pmod2$ based on this induction alone. Remember, this recurrence relation requires that the term on the left (in $x_{3m} \equiv x_{3(m-1)} \pmod2$) respects $3m-1\geq2$, or $m\geq1$. It does not matter if the term on the right respects that condition or not (refer back to the derivation of the recurrence formula): what is truly important is that both $x_k$ and $x_{k-1}$ are greater than $2$. Therefore, $x_3 \equiv x_0 \pmod2$ is correct and does not depend on any previous calculation of the parity of $x_3$.
$$x_{3m} \equiv \ldots \equiv x_3 \equiv x_0 \pmod2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4133057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Identifying the equation of the surface containing all points that have a distance to the x-axis twice that to the yz-plane The distance $D_{1}$ of an arbitrary point P(x, y, z) to the x-axis can be obtained using the right angled triangle formed by the coordinates of the point:
$$y^2 + z^2 = D^2\\
D_{1} = \sqrt{y^2+z^2}$$
The distance $D_{2}$ from P to the yz-plane, $x = 0$, is found by:
$$D_{2} = \frac{|ax + by + cz + d|}{\sqrt{a^2 + b^2 + c^2}} = \frac{|1x + 0y + 0z + 0|}{\sqrt{1^2}} = |x|$$
Equating the two distances:
$$D_{1} = D_{2}\\
\sqrt{y^2+z^2} = 2|x|$$
The resulting equation is in the form of an elliptic cone, its axis being the x-axis:
$$y^2 + z^2 = 4x^2$$
Have I correctly identified the surface that contains all points twice the distance to the x-axis as to the yz-plane? What other methods could be used to find this surface?
| Let
$$y^2+z^2= r^2$$
$ r= x \cdot \tan \alpha = 2 x $ is a circular cone of semi-vertical angle $\alpha= \tan^{-1}2.$
The cone is swept out by rotation of $ y= 2x \text { or } z=2 x$ about the x-axis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4135528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find basis for space of integer columns such that $AX=B$ I have the following problem:
Determine all the integer solutions to the system of equations $AX=0$, when $A=\left[
\begin{matrix}
4 & 7 & 2 \\
2 & 4 & 6
\end{matrix} \right]$. Find a basis for the space of integer columns vectors $B$ such that $AX=B$.
I first diagonalize the matrix as follows and have to find $Q^{-1},P$ such that $A'=Q^{-1}AP$.
$$
\begin{gather*}
\left[
\begin{matrix}
4 & 7 & 2 \\
2 & 4 & 6
\end{matrix} \right]\xrightarrow{2r_2-r_1}\left[
\begin{matrix}
4 & 7 & 2 \\
0 & 1 & 10
\end{matrix} \right]\xrightarrow{r_1-7r_2}\left[
\begin{matrix}
4 & 0 & -68 \\
0 & 1 & 10
\end{matrix} \right]\xrightarrow{c_3+17c_1}\left[
\begin{matrix}
4 & 7 & 0 \\
0 & 1 & 10
\end{matrix} \right]\\
\xrightarrow{2r_2-r_1}\left[
\begin{matrix}
4 & 7 & 2 \\
0 & 1 & 10
\end{matrix} \right]
\xrightarrow{-c_3}\left[
\begin{matrix}
4 & 7 & 0 \\
0 & 1 & -10
\end{matrix} \right]\xrightarrow{c_3+10c_2}\left[
\begin{matrix}
4 & 0 & 0 \\
0 & 1 & 0
\end{matrix} \right]=A'
\end{gather*}
$$
Then we can form the matrix equivalent to applying the operations used to diagonalize the matrix.
$$
\begin{gather*}
Q^{-1}=\left[
\begin{matrix}
1 & -7 \\
0 & 1
\end{matrix} \right]\left[
\begin{matrix}
1 & 0 \\
-1 & 2
\end{matrix} \right]=\left[
\begin{matrix}
8 & -14 \\
-1 & 2
\end{matrix} \right] \\
P = \left[
\begin{matrix}
1 & 0 & 17 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{matrix} \right]
\left[
\begin{matrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{matrix} \right]
\left[
\begin{matrix}
1 & 0 & 0 \\
0 & 1 & 10 \\
0 & 0 & 1
\end{matrix} \right]=\left[
\begin{matrix}
1 & 0 & -17 \\
0 & 1 & 10 \\
0 & 0 & -1
\end{matrix} \right]
\end{gather*}
$$
The solution to $A'X'=0$ is given by $X'=[0,0,n]$ for $n\in\mathbb{Z}$. Then, $X=PX'=[-17n,10n,-n]^t$, which after I tested a few cases, it seems to be correct. So, it seems that so far so good.
But to find the basis I need to get $Q$, but when I try to invert $Q^{-1}$, I get that $Q=\left[
\begin{matrix}
1 & 7 \\
.5 & 4
\end{matrix} \right]\not\in \mathbb{Z}$, so that just means my calculation of $Q^{-1}$, and possibly $P$ was incorrect. So, how can one correctly calculate $Q^{-1}$ and $P$?
| You first need to swap $r_2$ with $r_1$ before you begin in order to use the algorithm listed in Artin (which I assume you are using as this is a question from the book). Look at step 1 of the proof listed. You must permute rows and columns to move a nonzero entry with smallest absolute value to the upper left corner. This could be done with column operations, but it's not as easy to compute the matrix multiplication after.
Therefore, the first elementary matrix you get, and last one you multiply to find $Q^{-1}$ is
$\pmatrix{0 & 1\\
1 & 0}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is the solution to the following correct? Graph the area that is bounded by $\sqrt{x} + \sqrt{y} = \sqrt{a}$
$y = +(\sqrt{a} - \sqrt{x})^2 $
First we find the zeros:
$$\sqrt{a} - \sqrt{x} = 0 \iff x = a$$
Also $a$ cannot be negative so we take integral bounds of lower bound of 0 and upper bound of a.
$$\frac{S}{2} = \int_{0}^{a} (\sqrt{a} - \sqrt{x})^2dx = \int_{0}^a (a + x - 2 \sqrt{ax}) = a\cdot a + a^2/2 -2\sqrt{a}\cdot a^{3/2}\frac{2}{3}) = a^2(1 + 1/2 - 2\cdot a\cdot\frac{2}{3}) = a^2(3/2 - a\cdot\frac{4}{3})
$$
Mistakes:
$$\frac{S}{2} \text{would have to be only } S$$
$$ \text{The second last line should be:}a\cdot a + a^2/2 -2\sqrt{a}\cdot a^{3/2}\frac{2}{3} = a^2(3/2-4/3)$$
| No, note that the graph of the function can only lie in the first quadrant. Hence what you marked $\frac S2$ would actually be just $S$.
Also note the calculation error in the second last line, $a^{\frac 12}.a^{\frac 32}=a^2$ and not $a^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4139020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the matrix of a linear endomorphism of $2\times 3$ matrices, given by $\xi \mapsto \psi \xi \varphi$ for $\psi$ and $\varphi$ given matrices I would like to solve a problem in which I'm asked to determine trace and determinant of a linear endomorphism $f:x \mapsto A_1xA_2$ where $$A_1 = \left(\begin{array}{cc}7 & 5\\ 0 & 3\end{array}\right)$$ and $$A_2 = \left(\begin{array}{ccc}1 & 0 & 0\\ -1 & 4 & 0\\2 & -3 & 2\end{array}\right)$$ are both real-valued matrices.
Edit
For those interested, I add my solution below, which I was able to find thanks to @Matthew Leingang's advice! Here it is:
I picked base $\{\left(\begin{array}{cc}1&0\\0&0\\0&0\end{array}\right),\left(\begin{array}{cc}0&1\\0&0\\0&0\end{array}\right),\left(\begin{array}{cc}0&0\\1&0\\0&0\end{array}\right),\left(\begin{array}{cc}0&0\\0&1\\0&0\end{array}\right),\left(\begin{array}{cc}0&0\\0&0\\1&0\end{array}\right),\left(\begin{array}{cc}0&0\\0&0\\0&1\end{array}\right),\} = \{x_1,x_2,x_3,x_4,x_5,x_6\}$
of the space of $3\times 2$- real valued matrices.
Then I computed $f(x_i) = A_1 x_i A_2$, $\forall i \in \{1, ...,6\}$ and reduced the obtained matrices in terms of linear combinations of $x_i$; explicitely I obtained:
*
*$L(x_1) = 7(x_1 - x_3 + 2x_5) + 5(x_2 - x_4 + 2x_6)$
*$L(x_2) = 3(x_2 - x_4 + 2x_6)$
*$L(x_3) = 7(4x_3 - 3x_5) + 5(4x_4 - 2x_6)$
*$L(x_4) = 3(4x_4 - 3x_6)$
*$L(x_5) = 2(7x_5 + 5x_6)$
*$L(x_6) = 6x_6$
Hence I conclude that $f$ can be represented with the matrix:
$$ [f] = \left(\begin{array}{cccccc}
7 & 0 & 0 & 0 & 0 & 0\\
5 & 3 & 0 & 0 & 0 & 0\\
- 7 & 0 & 28 & 0 & 0 & 0\\
- 5 & - 3 & 20 & 12 & 0 & 0\\
14 & 0 & - 21 & 0 & 14 & 0\\
10 & 6 & - 15 & - 9 & - 10 & 6\\
\end{array}\right)$$
Therefore $tr([f]) = 70$ and $\det([f]) = 592704$. Hope it helps! (and don't hesitate to tell me if you spot any mistake).
| First, the map $f$ is a map from $M_{2 \times 3}(\mathbb{R})$ to itself (and not $M_{3 \times 2}(\mathbb{R}))$. Instead of calculating some matrix representing $f$ directly, we can use the following trick:
Note that $A_1$ is upper triangular with distinct diagonal values, hence $A_1$ is diagonalizable with eigenvalues $7,3$. Choose (linearly independent) eigenvectors $v ,w$ such that $A_1 v = 7v, A_1 w = 3w$. Similarly, $A_2^T$ is upper triangular with distinct diagonal values, hence $A_2^T$ is also diagonalizable with eigenvalues $1,4,2$. Choose (linearly independent) eigenvectors $x,y,z$ such that $A_2^T x = x, A_2^T y = 4y, A_2^T z = 2z$.
Now, note that
$$ f \left( v \cdot x^T \right) = A_1 \cdot v \cdot x^T \cdot A_2 = \left( 7v \right) \cdot \left( A_2^T \cdot x \right)^T = \left( 7v \right) \cdot \left( x \right)^T = 7 v \cdot x^T, \\
f \left( v \cdot y^T \right) = \left( 7 v \right) \cdot \left( 4y \right)^T = 28 v \cdot y^T, \\
f \left( v \cdot z^T \right) = \left( 7v \right) \cdot \left( 2z \right)^T = 14 v \cdot z^T, \\
f \left( w \cdot x^T \right) = \left( 3w \right) \cdot \left( x \right)^T = 3 w \cdot x^T, \\
f \left( w \cdot y^T \right) = \left( 3w \right) \cdot \left( 4y \right)^T = 12 w \cdot y^T,\\
f \left( w \cdot z^T \right) = \left( 3w \right) \cdot \left( 2z \right)^T = 6 w\cdot z^T.$$
Hence, we found six eigenvectors of $f$ associated to the distinct eigenvalues $7,28,14,3,12,6$. This means that $f$ is diagonalizable and so
$$ \operatorname{trace}(f) = 7 + 28 + 14 + 3 + 12 + 6 = 70, \\
\det(f) = 7 \cdot 28 \cdot 14 \cdot 3 \cdot 12 \cdot 6 = 592704. $$
This argument obviously generalizes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4143145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Analysis of a seemingly-complicated algebraic identity This is a self-answered question. I post it here since it wasn't trivial for me.
Consider the equality
$$
2\frac{x^4-4 x^3+2 x^2 (y^2+2)-4 x y^2+y^2 (y^2+8)}{(x^2-2 x+y^2)^2}=2+\frac{4y^2}{(1-x)^2}, \tag{1}
$$
where $x,y$ are real numbers.
Mathematica claims that if this equality holds, then if $y \neq 0$, then
$$
x\in \{\pm\sqrt{2-y^2}, 2 \pm \sqrt{2-y^2}\}.
$$
Is there any chance we could verify this "by hand"?
(If $y=0$, then the equality holds for any $x \neq 2$.)
The motivation for considering this equality comes from this problem.
| Multiplying equation $(1)$ by $(x^2-2 x+y^2)^2$, we obtain
$$
2x^4-8 x^3+4x^2y^2+8x^2-8xy^2+2y^2(y^2+8)=
$$
$$2x^4+8x^2+2y^4-8x^3-8xy^2+4x^2y^2+\frac{4y^2(x^2-2 x+y^2)^2}{(1-x)^2},
$$
which after cancellations becomes
$$
16y^2=\frac{4y^2(x^2-2 x+y^2)^2}{(1-x)^2}.
$$
Thus, if $y \neq 0$, we can divide by $4y^2$, and get
$$
4=\big(\frac{x^2-2 x+y^2}{1-x}\big)^2,
$$
so
$$
x^2-2 x+y^2=\pm 2(1-x).
$$
If $x^2-2 x+y^2=2(1-x)$, then $x^2+y^2=2 \Rightarrow x=\pm \sqrt{2-y^2}$.
If $x^2-2 x+y^2=2(x-1)$, then solving the quadratic for $x$ one get $x=2 \pm \sqrt{2-y^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4145916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute $\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$ Compute the given line integral:
*
*$$\int_{\left(0,0,0\right)}^{\left(3,4,5\right)}\frac{xdx+ydy+zdz}{\sqrt{x^{2}+y^{2}+z^{2}}}$$
Let $P=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}},Q=\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}}, R=\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}$ then $$\color{blue}{\frac{\partial P}{\partial y}}=\frac{-xy}{(x^2+y^2+z^2)^{3/2}},\color{green}{\frac{\partial P}{\partial z}}=\frac{-xz}{(x^2+y^2+z^2)^{3/2}}$$$$\color{blue}{\frac{\partial Q}{\partial x}}=\frac{-xy}{(x^2+y^2+z^2)^{3/2}},\color{red}{\frac{\partial Q}{\partial z}}=\frac{-yz}{(x^2+y^2+z^2)^{3/2}}$$$$\color{green}{\frac{\partial R}{\partial x}}=\frac{-xz}{(x^2+y^2+z^2)^{3/2}},\color{red}{\frac{\partial R}{\partial y}}=\frac{-yz}{(x^2+y^2+z^2)^{3/2}}$$
So the vector field is conservative which means there is $f$ such that $\vec \nabla f=F$, I don't know how to continue.
| If you have checked the completeness then your expression is definitely of the form
$$\frac{\partial U}{\partial x}dx + \frac{\partial U}{\partial y}dy + \frac{\partial U}{\partial z}dz$$
for some (potential) function $U$. Hence
$$\begin{cases}
\dfrac{\partial U}{\partial x} = \dfrac{x}{\sqrt{x^2+y^2+z^2}} \\
\dfrac{\partial U}{\partial y} = \dfrac{y}{\sqrt{x^2+y^2+z^2}} \\
\dfrac{\partial U}{\partial z} = \dfrac{z}{\sqrt{x^2+y^2+z^2}}
\end{cases}$$
Integrating the first equality w.r.t $x$ we have
$$\int \dfrac{\partial U}{\partial x}dx = \int\dfrac{x}{\sqrt{x^2+y^2+z^2}}dx$$
substitute $t(x) = \sqrt{x^2+y^2+z^2}$ in the integral on the RHS. Then, $$dt = \dfrac{x}{\sqrt{x^2+y^2+z^2}}dx$$ and therefore
$$\begin{align}
U &= \int dt \\
&= t + C(y,z) \\
&= \sqrt{x^2+y^2+z^2} + C(y,z) \tag{1}\end{align}$$
Similarly,
$$U = \sqrt{x^2+y^2+z^2} + C(x,y) \tag{2}$$
$$U = \sqrt{x^2+y^2+z^2} + C(z,x) \tag{3}$$
Now, you show that $C(x,y) = \text{const}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4147367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Triple integral between the paraboloids $z = x^2 + y^2$ and $z = 4 - x^2 - 3 y^2$. I have to compute $\int_D f$, where $D$ is the region in ${\mathbb{R}}^3$ between the paraboloids $z = x^2 + y^2$ and $z = 4 - x^2 - 3 y^2$, and $f : D \to \mathbb{R}$ is given by $f(x , y , z) = x^2 + 2 y^2$.
My attempt: If I use cylindrical coordinates $(\rho , \theta , z) \in (0 , \infty) \times (0 , 2 \pi) \times \mathbb{R}$ given by
$$
\left\{
\begin{array}
xx = \rho \cos \theta \\
y = \rho \sin \theta \\
z = z
\end{array}
\right.
$$
The bounds of $z$ with the paraboloids are
$$
{\rho}^2 = x^2 + y^2 \leq z \leq 4 - x^2 - 3 y^2 \leq 4 - {\rho}^2 (1 + 2 {\sin}^2 \theta).
$$
(why do we have $x^2 + y^2 \leq z \leq 4 - x^2 - 3 y^2$ and not for example $x^2 + y^2 \geq z \geq 4 - x^2 - 3 y^2$ without drawing?). From this we deduce that
$$
x^2 + y^2 \leq 4 - x^2 - 3 y^2 \quad \Longrightarrow \quad x^2 + 2 y^2 \leq 2 \quad \Longrightarrow \quad \rho \leq \sqrt{\frac2{1 + 2 {\sin}^2 \theta}}.
$$
Finally the integral is
$$
\int_{\theta = 0}^{2 \pi} \left(\int_{\rho = 0}^{\sqrt{\frac2{1 + 2 {\sin}^2 \theta}}} \left(\int_{z = {\rho}^2}^{4 - {\rho}^2 (1 + 2 {\sin}^2 \theta)} {\rho}^3 (1 + 2 {\sin}^2 \theta) \, dz\right) \, d \theta\right) \, d \theta = \frac{32 \pi}{9 \sqrt{3}}.
$$
Does it look good?
| In your working, you have $\rho \leq \sqrt{\frac2{1 + 2 {\sin}^2 \theta}}$. That is not correct as you can see. It should be,
$\rho \leq \sqrt{\frac2{1 + {\sin}^2 \theta}}$. Also the same correction is needed in integrand, $x^2+2y^2 = \rho^2 (1+\sin^2\theta)$ and not $\rho^2 (1 + 2 \sin^2\theta)$.
Having said that, here is what I would suggest as a simpler working. You found the intersection of surfaces as $x^2+2y^2 = 2$. So use the following substitution,
$x = \sqrt2 \ r \cos\theta, y = r \sin\theta$ then the ellipse $x^2+2y^2 = 2$ transforms to a circle of radius $1$ centered at origin. Jacobian of transformation is $\sqrt2 \ r$.
So equations of paraboloids become,
$z = x^2 + y^2 = r^2 + r^2 \cos^2\theta; z = 4 - x^2 - 3y^2 = 4 - 2 r^2 - r^2 \sin^2\theta$
Integrand $x^2 + 2y^2 = 2 r^2$.
Now we evaluate the integral,
$\displaystyle \iiint_D (x^2+2y^2) \ dx \ dy \ dz = 2 \sqrt2 \int_0^{2\pi} \int_0^1 \int_{r^2 + r^2 \cos^2\theta}^{4 - 2 r^2 - r^2 \sin^2\theta} r^3 \ dz \ dr \ d\theta $
$ \displaystyle = \frac{4 \sqrt2 \ \pi}{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to deduce the number of elements in a finite group given the group generators? I am studying for my finals, and I came across a question I don't really understand.
We have a finite group $G$ with two generators: $a$ ($a^3=E$) and $b$ ($b^2=E$), where $E$ is the identity. We are also given the relation $(ab)^2=E$. Now the question asks to show that $G$ has $6$ elements that can be represented as
$$E, a, a^2, ab, a^2b, b$$
I am not sure how I can deduce the number of elements in my group from my generators?
Thanks!
| Here is a two-step approach. At first we derive some identities from the reduction rules which enable us to simplify representation of elements. Then we build a $6\times 6$ multiplication table from
\begin{align*}
A:=\{E,a,b,a^2,ab,a^2b\}\subset G\tag{1}
\end{align*}
and check if each row and each column contains all elements from $A$, from which we can conclude that
\begin{align*}\
\color{blue}{A=G}
\end{align*}
Reduction rules:
We know that $\{E,a,b,a^{-1},b^{-1}\}\subseteq G$, since $G$ is a group containing the neutral element $E$ and with each element $g\in G$ the inverse $g^{-1}\in G$, whereby we have to keep in mind, that not always two elements are necessarily pairwise distinct.
At least we take for granted according to the problem statement that $E,a$ and $b$ are pairwise different.
From $a^3=E$ we obtain by multiplication with $a^{-1}$ and similarly from $b^2=E$ we obtain by multiplication with $b^{-1}$
\begin{align*}
a^2&=a^2E=a^2\left(a a^{-1}\right)=a^3 a^{-1}=Ea^{-1}=a^{-1}\tag{2.1}\\
b&=bE=b\left(bb^{-1}\right)=b^2b^{-1}=Eb^{-1}=b^{-1}\tag{2.2}
\end{align*}
From $(ab)^2=E$ we obtain by multiplication with $a^{-1}$ from the left and by multiplication with $b^{-1}$ from the right
\begin{align*}
a^{-1}=a^{-1}E=a^{-1}(ab)^2=\left(a^{-1}a\right)bab=bab\tag{2.3}\\
b^{-1}=Eb^{-1}=(ab)^2b^{-1}=\left(aba\right)\left(bb^{-1}\right)=aba\tag{2.4}
\end{align*}
Combining (2.1) to (2.4) we obtain
\begin{align*}
a^2&=a^{-1}=bab\tag{3.1}\\
b&=b^{-1}=aba\tag{3.2}
\end{align*}
A Pause and a recheck:
Looking again at (1) we observe the set $A$ shows some asymmetries.
*
*We find $a^2$ in $A$ but not $b^2$. No problem. We know from (3.1) and (3.2) that $a^2=a^{-1}$ and we also know that $b^2=E\in A$ and $b=b^{-1}\in A$.
*We find $a^2b$ but not $ab^2$. No problem, since $ab^2=aE=a\in G$.
*We find $ab$, but not $ba$. This should be clarified. We obtain from (3.1) and (3.2)
\begin{align*}
ba&=baE=bab^2=(bab)b=a^2b\in G\\
\end{align*}
and we are now ready to build up the multiplication table.
Multiplication table:
We obtain
\begin{align*}
\begin{array}{c|cccccc}
\cdot &E&a&b&a^2&ab&a^2b\\
\hline
E&E&a&b&a^2&ab&a^2b\\
a&a&a^2&ab&E&a^2b&b\\
b&b&a^2b&E&ab&a^2&a\\
a^2&a^2&E&a^2b&a&b&ab\\
ab&ab&b&a&a^2b&E&a^2\\
a^2b&a^2b&ab&a^2&b&a&E
\end{array}
\end{align*}
Since each row and each column contains the elements from $A$ we conclude
\begin{align*}
\color{blue}{A=G}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4148591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
The equation $\tan x = \tan 2x \tan 4x \tan 8x$ In the question we have the equality
$$\tan 6^{\circ} \tan 42^{\circ} = \tan 12^{\circ} \tan 24^{\circ}$$
which is equivalent to
$$ \tan 6^{\circ} = \tan 12^{\circ} \tan 24^{\circ} \tan 48^{\circ}$$
This means that the equation
$$\tan x = \tan 2x \tan 4x \tan 8x$$
has the solution $x =6^{\circ} = \frac{\pi}{30}$. How to find all the solution of this equation?
| $\sin x=0$ is obviously a solution. We will assume for further solutions that $\sin x\neq 0.$
Now:
$$\begin{align}\sin 2x&=2\sin x\cos x\\\sin 4x&=4\sin x\cos x\cos2x\\
\sin 8x&=2\sin 4x\cos 4x\\&=8\sin x\cos x\cos 2x\cos 4x.\end{align}$$
So:
$$\begin{align}\frac{\sin x}{\cos x}&=\tan x\\&=\tan 2x\tan 4x\tan 8x\\&=\frac{64\sin^3 x\cos^3 x \cos 2x}{\cos 8x}
\end{align}$$
Cancelling $\sin x\neq 0$ and multiplying both sides by $\cos x\cos 8x$ you get:
$$\begin{align}\cos 8x &= 64\sin^2x\cos^4x\cos 2x\\
&=64(1-\cos^2 x)\cos^4x (2\cos^2x-1)
\end{align}$$
If $u=\cos x$ then:
$$\cos 8x = T_8(n)=128 u^8 - 256 u^6 + 160 u^4 - 32 u^2 + 1$$
where $T_n$ is the Chebyshev polynomial of the first kind, $T_n(\cos x)=\cos nx.$
Together, this means we have the equation:
$$q(u)=256 u^8 - 448 u^6 + 224 u^4 - 32 u^2 + 1=0$$
Wolfram Alpha helps me compute:
$$q(u)=\frac{T_{15}(u)T_1(u)}{T_3(u)T_5(u)}\tag 1$$
(I theorized this equality after playing with the numerical roots of $q.$)
Wolfram verified (1), but it is not really a satisfying answer for “why?”
So, $u=\cos x$ is a root of $q$ if and only if $\cos 15x=0$ but $\cos 5x\neq 0$ and $\cos 3x\neq 0.$
This amounts to $x=\frac{m\pi}{30}$ where $\gcd(m,30)=1.$
So, in $[-\pi,\pi],$ the solutions are $$\begin{align}x=0,&\pm\frac{\pi}{30},\pm\frac{29\pi}{30},\pm\frac{7\pi}{30},\pm\frac{23\pi}{30},\\
&\pm\frac{11\pi}{30},\pm\frac{19\pi}{30},\pm\frac{13\pi}{30},\pm\frac{17\pi}{30}
\end{align}$$
So, other than $0,$ these are the “primitive roots” of $\cos(15x)=0,$ or the $x$ such that $e^{ix}$ is a primitive $60$th root of unity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4149132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Is this Fourier Series correct I need to confirm the Leibniz series through using Fourier series. I have done what I believe is the correct Fourier series just want to know if I am on the right path? If it is correct I don't see any way to evaluate it that comes out as Leibniz.
Leibniz: \begin{equation}
\frac{\pi}{4}=\sum_{i=0}^\infty \frac{(-1)^i}{(2i+1)}
\end{equation}
Function:
$$f(x) = \begin{cases}
0, &\text{for}\ -\pi<x<0 \\
1, &\text{for}\ 0\leq x<\pi
\end{cases}
$$
Fourier:
\begin{equation}
b_n = \frac{1}{\pi}\int_{-\pi}^\pi f(x) \sin(xn) dx=\frac{1}{\pi}(\int_{0}^\pi 1 \sin(xn) dx+\int_{-\pi}^0 0 \sin(xn) dx)=\frac{1}{\pi}(1/n)(-1(-1)^n+1)+0)=\frac{1-(-1)^n}{\pi*n}
\end{equation}
\begin{equation}
a_n =\frac{1}{\pi}(\int_{0}^\pi 1 \cos(xn) dx +\int_{-\pi}^0 0 \cos(xn) dx) = \frac{1}{\pi}(0+0)=0
\end{equation}
\begin{equation}
a_0 = \frac{1}{\pi}(\int_{0}^\pi 1 dx +\int_{-\pi}^0 0 dx) = \frac{1}{\pi}(\pi-0+0)=1
\end{equation}
\begin{equation}
f(x) = \frac{a_0}{2}+\sum_{i=1}^\infty (a_n\cos(nx)+b_n\sin(nx)
=\frac{1}{2}+\sum_{i=1}^\infty (\frac{1-(-1)^n}{\pi*n}\sin(nx)
\end{equation}
| What you have obtained is that
$$
f(x) = \frac{1}{2} + \sum_{n=1}^{\infty}\frac{1 - (-1)^n}{\pi n} \sin(nx).
$$
Therefore, it follows that, for $x = \frac{\pi}{2}$,
$$
\begin{align}
f(\pi/2) = 1 &= \frac{1}{2} + \sum_{n=1}^{\infty}\frac{1 - (-1)^n}{\pi n} \sin(n\pi/2) \\
\text{using the note from below} \\
&= \frac{1}{2} + \sum_{k=1}^{\infty}\frac{1 - (-1)^{4k-3}}{\pi (4k-3)}\sin\left(\frac{\pi(4k-3)}{2}\right) + \sum_{k=1}^{\infty}\frac{1 - (-1)^{4k-2}}{\pi (4k-2)}\sin\left(\frac{\pi(4k-2)}{2}\right) + \sum_{k=1}^{\infty}\frac{1 - (-1)^{4k-1}}{\pi (4k-1)}\sin\left(\frac{\pi(4k-1)}{2}\right) + \sum_{k=1}^{\infty}\frac{1 - (-1)^{4k}}{\pi (4k)}\sin\left(\frac{\pi(4k)}{2}\right)\\
&= \frac{1}{2} + \sum_{k=1}^{\infty}\frac{1 - (-1)^{4k-3}}{\pi (4k-3)} - \sum_{k=1}^{\infty}\frac{1 - (-1)^{4k-1}}{\pi (4k-1)} \\
&= \frac{1}{2} + \frac{2}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{(4k-3)} - \sum_{k=1}^{\infty}\frac{1}{(4k-1)}\right) \\
&= \frac{1}{2} + \frac{2}{\pi}\left(\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2k-1}\right) \\
\text{shifting indices by 1, i.e. by letting $i = k-1$} \\
&= \frac{1}{2} + \frac{2}{\pi}\sum_{i=0}^{\infty}\frac{(-1)^{i}}{2i+1}.
\end{align}
$$
Equating the left and right hand sides then yields
$$
\frac{\pi}{4} = \sum_{i=0}^{\infty}\frac{(-1)^{i}}{2i+1},
$$
precisely as required.
Note:
In the above we used the fact that
$$
\sin\left(\frac{\pi n}{2}\right) =
\begin{cases}
1 & \text{if } n = 4k-3,\ k \in \mathbb{N} \\
0 & \text{if } n = 4k-2,\ k \in \mathbb{N} \\
-1 & \text{if } n = 4k-1,\ k \in \mathbb{N} \\
0 & \text{if } n = 4k,\ k \in \mathbb{N},
\end{cases}
$$
for example
$$
\sin\left(\frac{n\pi}{2}\right) = 1, 0, -1, 0, 1, 0, -1,...
$$
for
$$
n = 1,2,3,4,5,6,7...
$$
respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find surface area and centroid of a parabolic surface in $\mathbb{R}^{3}$ I have come to a problem in a multivariable calculus book that I am having problems with.
The problem can be stated :
Let $a \in \mathbb{R}$ with $a > 0$. Find the surface area of the part of a paraboloid given
by $z^{2} + x^{2} = 2ay$ that is cut out by a plane given by $y =a$. Then find the centroid
of the surface.
My attempt at computing the surface area is given below :
Let the surface be denoted $S$. We see :
\begin{equation}
y = a \text{ and } z^{2} + x^{2} = 2ay \Rightarrow z^{2} + x^{2} = 2a^{2}
\end{equation}
So intersection of $S$ with $y =a$ plane is a circle of radius $\sqrt{2}a$.
Let :
\begin{equation}
\require{cancel}
E = \left\{ (r,\theta) \in \mathbb{R}^{2} \; : \; r \in [0,\sqrt{2}a] \text{ and } \theta \in [-\pi,\pi] \right\}
\end{equation}
Now let's define a map from $E$ to $S$.
\begin{equation}
\Phi : \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}
\end{equation}
Let :
\begin{align}
x & = r\cos(\theta)\\
z & = r\sin(\theta)
\end{align}
So :
\begin{align}
z^{2} + x^{2} = 2ay & \Rightarrow y = \frac{z^{2} + x^{2}}{2a} \\
& \Rightarrow y = \frac{r^{2}\sin^{2}(\theta) + r^{2}\cos^{2}(\theta)}{2a}\\
& \Rightarrow y = \frac{r^{2}\cancel{(\sin^{2}(\theta) + \cos^{2}(\theta))}}{2a}\\
& \Rightarrow y = \frac{r^{2}}{2a}
\end{align}
So :
\begin{equation}
\Phi(r,\theta) = \left( r\cos(\theta) , \frac{r^{2}}{2a} , r \sin(\theta) \right)
\end{equation}
Where :
\begin{align}
x(r,\theta) & = r\cos(\theta)\\
y(r,\theta) & = \frac{r^{2}}{2a} \\
z(r,\theta) & = r\sin(\theta)
\end{align}
We see :
\begin{align}
x_{r} & = \cos(\theta)\\
y_{r} & = \frac{\cancel{2}r}{\cancel{2}a} = \frac{r}{a} \\
z_{r} & = \sin(\theta)
\end{align}
and :
\begin{align}
x_{\theta} & = -r \sin(\theta)\\
y_{\theta} & = 0\\
z_{\theta} & = r\cos(\theta)
\end{align}
Let :
\begin{align}
U & = x_{r}^{2} + y_{r}^{2} + z_{r}^{2} \\
& = \cos^{2}(\theta) + \frac{r^{2}}{a^{2}} + \sin^{2}(\theta)\\
& = \frac{r^{2}}{a^{2}} + \cancel{ \left( \sin^{2}(\theta) + \cos^{2}(\theta) \right)}\\
& = \frac{r^{2}}{a^{2}} + 1
\end{align}
and :
\begin{align}
V & = x_{\theta}^{2} + y_{\theta}^{2} + z_{\theta}^{2} \\
& = r^{2}\sin^{2}(\theta) + 0 + r^{2}\cos^{2}(\theta)\\
& = r^{2} \cancel{\left( \sin^{2}(\theta) + \cos^{2}(\theta) \right)}\\
& = r^{2}
\end{align}
and :
\begin{align}
W & = x_{r}x_{\theta} + y_{r}y_{\theta} + z_{r}z_{\theta}\\
& = \cos(\theta)(-r\sin(\theta)) + 0 + \sin(\theta)r\cos(\theta)\\
& = r\sin(\theta)\cos(\theta) - r\sin(\theta)\cos(\theta)\\
& = 0
\end{align}
so :
\begin{equation}
\sqrt{UV - W^{2}} = \sqrt{ r^{2}\left( \frac{r^{2}}{a^{2}} + 1 \right) } = r\sqrt{ \frac{r^{2}}{a^{2}} + 1 }
\end{equation}
We know :
\begin{align}
\text{Area}(S) & = \int\int_{E} \sqrt{UV - W^{2}} d(r,\theta)\\
& = \int\int_{E} r\left( \frac{r^{2}}{a^{2}} + 1 \right)^{\frac{1}{2}} d(r,\theta)\\
& = \int_{-\pi}^{\pi} \int_{0}^{\sqrt{2}a} r\left( \frac{r^{2}}{a^{2}} + 1 \right)^{\frac{1}{2}} dr d\theta
\end{align}
Let :
\begin{align}
u & = \frac{r^{2}}{a^{2}} + 1\\
\frac{du}{dr} & = \frac{2r}{a^{2}}
\end{align}
So :
\begin{align}
\text{Area}(S) & = \int_{-\pi}^{\pi} \int_{0}^{\sqrt{2}a} \left( \frac{a^{2}}{2} \right) \frac{du}{dr} u^{\frac{1}{2}} dr d\theta\\
& = \frac{a^{2}}{2} \int_{-\pi}^{\pi} \left( \int_{0}^{\sqrt{2}a} u^{\frac{1}{2}} \frac{du}{dr} dr \right) d\theta\\
& = \frac{a^{2}}{2} \int_{-\pi}^{\pi} \left( \int_{u(0)}^{u(\sqrt{2}a)} u^{\frac{1}{2}} du \right) d\theta\\
& = \frac{a^{2}}{2} \int_{-\pi}^{\pi} \left( \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \Bigr|_{u(0)}^{u(\sqrt{2}a)} \right) d\theta\\
& = \frac{\cancel{2}a^{2}}{\cancel{6}} \int_{-\pi}^{\pi} \left( u^{\frac{3}{2}} \Bigr|_{1}^{\frac{2\cancel{a^{2}}}{\cancel{a^{2}}} + 1} \right) d\theta\\
& = \frac{a^{2}}{3} \int_{-\pi}^{\pi} \left( u^{\frac{3}{2}} \Bigr|_{1}^{3} \right) d\theta\\
& = \frac{a^{2}}{3} \int_{-\pi}^{\pi} \left( 3^{\frac{3}{2}} - 1 \right) d\theta\\
& = \frac{a^{2}}{3} \int_{-\pi}^{\pi} (3 \sqrt{3} - 1 ) d\theta\\
& = \frac{a^{2}(3\sqrt{3} - 1)}{3} \int_{-\pi}^{\pi} d\theta\\
& = \frac{2\pi a^{2} (3\sqrt{3} - 1)}{3} \; \checkmark
\end{align}
Now my attempt at finding the centroid is given below :
We see that $S$ is symmetric w.r.t. the x and z coordinates. So :
\begin{align}
\overline{x} & = 0\\
\overline{z} & = 0
\end{align}
Let :
\begin{equation}
A = \text{Area}(S) = \frac{2\pi a^{2} (3 \sqrt{3} - 1) }{ 3 }
\end{equation}
We know :
\begin{equation}
\overline{y} = \frac{1}{A} \int\int_{E} y(r,\theta) \sqrt{UV - W^{2}} d(r,\theta)
\end{equation}
We know :
\begin{equation}
y(r,\theta) = \frac{r^{2}}{2a}
\end{equation}
and :
\begin{equation}
\sqrt{UV - W^{2}} = r\sqrt{ \frac{r^{2}}{a^{2}} + 1 }
\end{equation}
So :
\begin{align}
\overline{y} & = \frac{1}{A} \int\int_{E} \frac{r^{2}}{2a} r \sqrt{ \frac{r^{2}}{a^{2}} + 1 } d(r,\theta)\\
& = \frac{1}{2aA} \int\int_{E} r^{3} \left( \frac{r^{2}}{a^{2}} + 1 \right)^{\frac{1}{2}} d(r,\theta)
\end{align}
I'm not sure how to compute this integral. I suspect that maybe I did something wrong when I was finding the surface area or made a mistake in writing the expression for $\overline{y}$, but I don't see where the mistake is made. If what I have above is actually correct, then what is the best approach to evaluate the integral ?
| The surface area $A$ you found is correct.
The integral with respect to $\theta$ contributes a factor of $2\pi$. The remaining integral is simple. Start by integrating by parts:
$$\int x^3\sqrt{x^2+n^2}\,\mathrm dx = uv - \int v\,\mathrm du$$
Take $u=x^2$ and $\mathrm dv = x\left(x^2+n^2\right)^{\frac12}\,\mathrm dx$, so that $\mathrm du=2x\,\mathrm dx$ and $v=\frac13\left(x^2+n^2\right)^{\frac32}$.
$$\implies \int x^3\sqrt{x^2+n^2}\,\mathrm dx=\frac13x^2\left(x^2+n^2\right)^{\frac32}-\frac23\int x\left(x^2+n^2\right)^{\frac32}\,\mathrm dx$$
Then make a substitution, etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there a closed form to the sum $\sum_{k=1}^n \frac{k}{2k - 1}$ ? If there is, what would that be? I know that the sum of the first n natural numbers is $\frac{n(n + 1)}{2}$ and that the sum of the first n odd numbers is $n^2$, so I wondered what would happen if you had the ratio of the $n^{th}$ natural number to the $n^{th}$ odd number and took the summation of that.
FYI, I haven't hit college yet so my knowledge of sums is basic, I pretty much know nothing that's more advanced than the sums I've mentioned. I'm hoping for an explanation of how to find the closed form of $\sum_{k=1}^n \frac{k}{2k - 1}$ or a way to be certain that there is none.
| Starting with
$$\frac{k}{2k-1} = \frac{1}{2} \left( \frac{2k-1}{2k-1} + \frac{1}{2k-1} \right)$$
the sum is
$$S_n = \frac{n}{2} + \frac{1}{2} \sum_{k=1}^n \frac{1}{2k-1}$$
As noted in the comments, the harmonic numbers are defined to be
$$H_p= \sum_{k=1}^p \frac{1}{k}.$$
Here $p=2n-1$,
so
$$\begin{aligned}
H_{2n-1} &= 1 + \frac{1}{2}+ \frac{1}{3} + \cdots + \frac{1}{2n-1} \\&=\left(1+\frac{1}{3}+\cdots+\frac{1}{2n-1} \right)\,+\, \left( \frac{1}{2}+\frac{1}{4} + \cdots+\frac{1}{2n-2} \right)\\
&= \left( 1+\frac{1}{3}+\cdots+\frac{1}{2n-1}\right) \,+\,\frac{1}{2} H_{n-1}
\end{aligned} $$
and $$S_n = \frac{n}{2} + \frac{1}{2} \left(H_{2n-1}-\frac{1}{2}H_{n-1}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Help in solving the following Sturm-Liouville problem $-x^2y''-2xy'-\lambda y=0$ I have problems finding the eigenvalues and eigenfunctions of the equation:
$$-x^2y''-2xy'-\lambda y=0$$
The domain is $[1,\pi]$, with conditions $y(1)=y(\pi)=0$.
I have proved the values of $\lambda$ have to be nonnegative.
By making the subsitution $y=x^m$, I get to the condition:
$$m^2+m+\lambda=0$$
Which leads to the solutions:
$$y=ax^{-\frac{1+\sqrt{1-4\lambda}}{2}}+bx^{-\frac{1-\sqrt{1-4\lambda}}{2}}$$
I don't know if I'm going the right way about this or if I'm missing information about how to solve this type of problems.
| This is a regular Sturm-Liouville problem because it is non-singular on $[1,\pi]$. Start with a solution $y(1)=0$ and $y'(1)=1$ in order to normalize the problem at $x=1$. This gives
$$
y_{\lambda}(x) = \frac{1}{\sqrt{1-4\lambda}}\left(-x^{-\frac{1+\sqrt{1-4\lambda}}{2}}+x^{-\frac{1-\sqrt{1-4\lambda}}{2}}\right) \\
= \frac{1}{\sqrt{1-4\lambda}}\frac{1}{\sqrt{x}}\left(x^{\frac{\sqrt{1-4\lambda}}{2}}-x^{-\frac{\sqrt{1-4\lambda}}{2}}\right) \\
% = \frac{1}{\sqrt{1-4\lambda}}\frac{1}{\sqrt{x}}\left(e^{\ln(x)}\right)
% =\frac{1}{\sqrt{1-4\lambda}}\frac{1}{\sqrt{x}}x^{-\frac{-\sqrt{1-4\lambda}}{2}}\left(x^{\sqrt{1-4\lambda}}-1\right)
=\frac{1}{\sqrt{1-4\lambda}}\frac{1}{\sqrt{x}}2\sinh\left(\frac{\sqrt{1-4\lambda}}{2}\log(x)\right) \\
= -\frac{1}{\sqrt{1-4\lambda}}\frac{1}{\sqrt{x}}2\sin\left(i\frac{\sqrt{1-4\lambda}}{2}\log(x)\right)
$$
The eigenvalue equation requires us to find the zeros in $\lambda$ of an entire function:
$$
y_{\lambda}(\pi)=0.
$$
The eigenvalues satisfy
$$
i\frac{\sqrt{1-4\lambda}}{2}\log(\pi)=\pm n\pi \\
\sqrt{1-4\lambda}=\pm 2ni\pi/log(\pi) \\
1-4\lambda = -4n^2\pi^2/(\log(\pi))^2 \\
\lambda = \frac{1}{4}+n^2\pi^2/(\log(\pi))^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
if $u = \arccos \left(\frac{x+y}{\sqrt{x}+\sqrt{y}}\right)$ then $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y}$ Edit: this is a repost
if possible i need some sort of shorcut, 80 questions, 90 min to solve these type of questions i have only 1 min time solt for each question
Context: i've looked at PYQ's and this year mock tests and found out that this question was one of the most repeated ones and i'm having an exam next week and my teachers are not available
The answer i got
$x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{-x\left(\sqrt{x}+\sqrt{y}-(x+y)\frac{1}{2\sqrt{x}}\right)} {\sin^2(u)(\sqrt{x}+\sqrt{y})^2}- \frac{y\left(\sqrt{x}+\sqrt{y}-(x+y)\frac{1}{2\sqrt{y}}\right) } {\sin^2(u)\left(\sqrt{x}+\sqrt{y}\right)^2}$
The answer im getting is nowhere near the options that have been given
Options
1.$\frac{-1}{2}\sin(u)$
2. $\frac{-1}{2}\cot(u)$
3. $\frac{-1}{2}\tan(u)$
4. $\frac{-1}{2}\cos(u)$
| Write $\cos(u)=\frac{x+y}{\sqrt{x}+\sqrt{y}}$ and differentiate implicitly wrt $x$ and $y$. The RHS is symmetric in $x$ and $y$ so you only need to actually compute the derivative once. This gives
$$-\sin(u)\frac{\partial u}{\partial x}=\frac{1}{\sqrt{x}+\sqrt{y}}-\frac{1}{2\sqrt{x}}\frac{x+y}{(\sqrt{x}+\sqrt{y})^2},$$
$$-\sin(u)\frac{\partial u}{\partial y}=\frac{1}{\sqrt{x}+\sqrt{y}}-\frac{1}{2\sqrt{y}}\frac{x+y}{(\sqrt{x}+\sqrt{y})^2}.$$
Multiply the first equation by $x$ and the second by $y$ and add the equations to find
\begin{align}
-\sin(u)\left(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}\right)&=\frac{x+y}{\sqrt{x}+\sqrt{y}}-\frac{x+y}{\sqrt{x}+\sqrt{y}}\left(\frac{1}{2}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{1}{2}\frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}}\right)\\
&=\frac{1}{2}\cos(u)
\end{align}
where the last equality follows from the original equation. Divide by $-\sin(u)$ and you're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $x^2 - 10ax - 11b = 0$ has roots $c$ and $d$ and $x^2 - 10cx - 11d = 0$ has roots $a$ and $b$, then find $a + b + c + d$? If $x^2 - 10ax - 11b = 0$ have roots $c$ and $d$ and $x^2 - 10cx - 11d = 0$ have roots $a$ and $b$, then find $a + b + c + d$?
My attempt-
Using Vieta's formula, we get
$$ c+d=10a, a+b=10c $$
Because sum of roots $=\frac{-b}{a}$
Also using Vieta's formula we get
$$ cd=-11b, ab=-11d $$
Because product of roots $=\frac{c}{a}$
So now we need to solve these equations-
$$ c+d=10a, a+b=10c, cd=-11b, ab=-11d $$
So I solved it like this-
In the above line let each 4 equation be eq $1,2,3,4$.
Dividing eq $1$ by $2$ and cancelled $10$ in RHS because its both in numerator and denominator,
$$ \frac{c+d}{a+b}=\frac{a}{c} $$
Using componendo alone we get
$$ \frac{c+d+a+b}{a+b}=\frac{a+c}{c} $$
Putting all the values using eq $1$ and $2$,
$$ \frac{10a+10c}{ac}=\frac{a+c}{c} $$
Solving this we get $a=10$
Now doing mischiefs(some changes) with eq $1$,
$$d=10a-c =>d=100-c $$
Let the above equation be equation $5$.
Similarly doing some mischiefs with equation $2$ also
$$ b=10c-10 $$
Let above equation be equation $6$.
Now using equation $4$,
$$ ab=-11d $$
We know $a=10$ and using equation $5$ and $6$.
$$ 10(10c-10)=-11(100-c) $$
$$ 100c-100=-1100+11c $$
$$ 89c+1000=0 $$
$$ c=\frac{-1000}{89} $$
Now see adding equation $1$ and $2$
$$a+b+c+d=10a+10c$$
Thus we need to only find now $10(a+c)$.
$$ 10(a+c) $$
$$ 10(10+\frac{-1000}{89}) $$
But wait this is the wrong answer according to my book. What have I done wrong?
Here is the correct solution according to a website-
https://www.sarthaks.com/?qa=blob&qa_blobid=286792613098888701
Please help me with this question. Apologies for asking a homework question. Apologies for something missing in the question. Thanks for your time.
| In your approach, after doing componendo alone, you substitute $a+b = ac$, which should $10c$ and it shouldn't give you anything new. Instead, we can do as follows which is straightforward calculation.
$d = 10 a - c$ and $b = 10c - a$
$(ab)(cd) = (-11b)(-11d) = 121bd$
If $b=0$, then $d = 0 = a = c$ and same for $d = 0$, $a=0$ and $c=0$ cases
Otherwise, $ac = 121$
$ab = 10ac - a^2 = -11d = -110a + 11 c$
$a^2 -110a +11\frac{121}{a} - 1210 = 0$
$a^3 - 10*11a^2 - 10*11^2a + 11^3 = 0$
$(a^3+11^3) - 110a(a + 11) = 0$
$(a+11)(a^2 - 121a + 11^2) = 0$
$a$ has 3 solutions and find solution set for each case ($c = \dfrac{121}{a}$, $d = 10 a - c$ and $b = 10c - a$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4156661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Polynomial Graph Question Whose Solution Does Not Make Sense The question is:
How many roots of the equation $3x^4 +6x^3 + x^2 +6x +3$ are real?
The textbook's solution is:
Let $f(x) = 3x^4 +6x^3 + x^2 +6x +3$ .
Now $$f'(x) =12x^3 +18x^2 +2x +6\\
\implies f''(x) = 36x^2 +36x +2 ≠ 0 \; \forall x \in \mathbb{R} $$
So, graph of $f'(x)$ intersects $x$-axis only once.
Hence, $f(x)$ has only one turning point.
So $f(x)=0$ has maximum two real roots.
Now $f(0)=3>0$ and $f(-1)=-5<0$
So graph cuts $x$-axis between $-1$ and $1$.
Hence $f(x)=0$ has two real roots.
My Question:
How on earth does $f''(x)$ have no real solutions? Using the quadratic formula I get $2$ real negative roots. Am I missing something here? Also how does the fact that the graph cuts the $x$-axis between $1$ and $-1$ imply that there are $2$ real roots? I just started reading the portion in the textbook on Graphs Of Polynomial Functions and I am in $12$th grade so if you could, please explain in a little simple way thank you. Also sorry if the formatting is weird this is the first time I have asked a question here :) .
| Since you have sorted out the source of misunderstanding, here I offer an alternative solution, which quite surprisingly* allows you to find the exact roots of the polynomial while avoiding calculus. The method is (much) longer, but I find it interesting and I hope you will too.
(*I say surprisingly because mostly when solving quartic equations we don't have a hope of finding the exact roots without using the quartic formula.)
Alternative Solution
We begin with a key observation: when a polynomial's coefficients are mirror images of each other about the centre of the polynomial, then the roots of the polynomial come as pairs of reciprocals of each other. For example, the roots of the polynomial $2x^2+5x+2$ are $-2$ and $-0.5$ which are reciprocals of each other.
If you would like me to, I can prove this for you, but it may be a good exercise to try it for yourself.
Using this fact, we may now observe that your polynomial satisfies this property: the coefficients are unchanged if we 'reflect' them about the middle term.
Therefore, since the polynomial is a quartic it has $4$ roots and its roots must be of the form
$$-\alpha,-\frac{1}{\alpha},-\beta, -\frac{1}{\beta}.$$
Using the Factor theorem, we can now see that your polynomial can be factorised in the following way:
$$\begin{align}3x^4+6x^3+x^2+6x+3&\equiv 3(x+\alpha)\left(x+\frac{1}{\alpha}\right)(x+\beta)\left(x+\frac{1}{\beta}\right)\\
&\equiv 3\left(x^2+x\left(\alpha+\frac{1}{\alpha}\right)+1\right)\left(x^2+x\left(\beta+\frac{1}{\beta}\right)+1\right)\\
&\equiv 3x^4+3x^3\left(\alpha+\frac{1}{\alpha}+\beta+\frac{1}{\beta}\right)+3x^2\left(2+\left(\alpha+\frac{1}{\alpha}\right)\left(\beta+\frac{1}{\beta}\right)\right)\\
&\quad+3x\left(\alpha+\frac{1}{\alpha}+\beta+\frac{1}{\beta}\right)+3
\end{align}$$
Divide both sides by $3$ and for simplicity's sake let
$$a=\alpha+\frac{1}{\alpha},\quad b=\beta+\frac{1}{\beta}$$
and we find that
$$x^4+2x^3+\frac{1}{3}x^2+2x+1\equiv x^4+x^3(a+b)+x^2(2+ab)+x(a+b)+1.$$
What we really have here is a pair of simultaneous equations:
$$\begin{align}
a+b&=2\iff a=2-b\\
2+ab&=\frac{1}{3}
\end{align}$$
Replacing the $a$ in the $2$nd equation by the $a$ in the $1$st equation, we are left with a quadratic in $b$:
$$\begin{align}
2+b(2-b)=\frac{1}{3}&\implies b^2-2b-\frac{5}{3}=0\\
&\implies b=\frac{3\pm 2\sqrt6}{3}
\end{align}$$
The nice thing about this solution is that we then find that
$$a=\frac{3\mp 2\sqrt6}{3}.$$
Therefore, without loss of generality we can let
$$a=\frac{3+2\sqrt6}{3},\quad b=\frac{3- 2\sqrt6}{3}.$$
We now have $2$ quadratic equations that allow us to find the exact roots of the quartic polynomial!!
$$\begin{align}&\frac{3+2\sqrt6}{2}=\alpha+\frac{1}{\alpha}\implies \alpha^2-\alpha\left(\frac{3+ 2\sqrt6}{2}\right)+1=0\\
&\frac{3-2\sqrt6}{2}=\beta+\frac{1}{\beta}\implies \beta^2-\beta\left(\frac{3-2\sqrt6}{2}\right)+1=0
\end{align}$$
Now all you need to find the values of $-\alpha,-\frac{1}{\alpha},-\beta, -\frac{1}{\beta}$ is to solve these two quadratic equations, and indeed only one of them has real roots (which you can find exactly using the quadratic formula).
I hope that was helpful, or you at least found it interesting. Sorry for the length, I've just tried to make it is clear as possible. If you have any questions please don't hesitate to ask :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4168778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
My solution to $\int \frac{2x^2}{x^2+4}dx$ $$\int \frac{2x^2}{x^2+4}dx$$
Attempt. I have actually done this integral but I'm wondering if what I did was correct as I have a different answer from WolframAlpha. Here it is: $$\int \frac{2x^2}{x^2+4}dx$$
$$2\int \frac{x^2}{x^2+4}dx$$ $$2\int 1dx -2\int\frac{4}{x^2+4}dx$$ $$2x-2\int\frac{4}{x^2+4}dx$$ and here, while I now realise that ther right integral is just $\tan(\frac{x}{2})$, I continued. Let $x=2\tan(\theta)$: $$2x -8\int\frac{1}{4\tan^2(\theta)+4}d\theta$$ $$2x -2\int\frac{1}{\tan^2(\theta)+1}d\theta$$ $$2x -2\int\frac{1}{\sec^2(\theta)}d\theta$$ $$2x -2\int\cos^2(\theta)d\theta$$ $$2x -2\int\frac{\cos(2\theta)+1}{2}d\theta$$ $$2x-\int\cos(2\theta)d\theta-\theta$$ $$2x-\int\cos(2\theta)d\theta-\theta$$ $$2x-\frac12 \sin(2\theta)-\theta$$ $$2x-\frac12 \sin(2\arctan(\frac{x}{2}))-\arctan(\frac{x}{2})$$
Again, I don't really care if there was a shortcut way to this, I just want to know if it is correct or not.
| You are trying to integrate a rational function—a quotient of polynomials. If the degree of the numerator is greater than or equal to the degree of the denominator, then use polynomial long division to turn the improper fraction into a proper fraction. This will make life much easier. Here,
$$
\frac{2x^2}{x^2+4}=2-\frac{8}{4+x^2} \, .
$$
Now the problem has been reduced to evaluating
$$
\int\frac{dx}{4+x^2} \, .
$$
We can proceed in the following way:
$$
\int\frac{dx}{4+x^2} = \frac{1}{4}\int\frac{dx}{1+\left(\frac{x}{2}\right)^2} \, .
$$
By the chain rule,
$$
\frac{d}{dx}\left(\arctan\left(\frac{x}{2}\right)\right)=\frac{1}{1+\left(\frac{x}{2}\right)^2}\cdot\frac{1}{2} \tag{*}\label{*} \, ,
$$
and so
$$
\int\frac{dx}{4+x^2}=\frac{1}{4}\int\frac{dx}{1+\left(\frac{x}{2}\right)^2}=\frac{1}{2}\arctan\left(\frac{x}{2}\right)+C \, .
$$
The overall integral therefore equals
$$
\int 2-\frac{8}{4+x^2} \, dx = 2x -4\arctan\left(\frac{x}{2}\right) + C \, .
$$
The substitution you made is perfectly valid, but it is easier to directly apply the chain rule in reverse as I did in $\eqref{*}$. Substitutions should be reserved for more complicated integrals where it is not immediately clear how you can apply the chain rule in reverse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4170337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $(b + c) , (c + a) , (a + b)$ are in H.P. then find the relation between $\dfrac{a}{b+c} , \dfrac{b}{c+a} , \dfrac{c}{a+b}$ .
If $(b + c) , (c + a) , (a + b)$ are in H.P. then $\dfrac{a}{b+c} , \dfrac{b}{c+a} , \dfrac{c}{a+b}$ are in $(i)$ A.P. $(ii)$ G.P. $(iii)$ H.P. $(iv)$ None of These.
What I Tried:- I have that $(b+c),(c+a),(a+b)$ are in H.P. .
$\implies \dfrac{1}{b+c}, \dfrac{1}{c+a}, \dfrac{1}{a+b}$ are in A.P. .
$\implies \dfrac{1}{c+a} - \dfrac{1}{b+c} = \dfrac{1}{a+b} - \dfrac{1}{c+a}.$
$\implies \dfrac{(b + c) - (c + a)}{(c + a)(b+c)} = \dfrac{(c+a) - (a+b)}{(a+b)(c+a)}.$
$\implies \dfrac{(b - a)}{(b + c)} = \dfrac{(c - b)}{(b + a)}.$
$\implies (b^2 - a^2) = (c^2 - b^2).$
$\implies 2b^2 = a^2 + c^2.$
So I was able to prove that $a^2 , b^2 , c^2$ are in A.P. , but I am not sure how to show the claim in the question.
Can anyone help me? Thank You.
| As , $a^2,b^2,c^2$ & $\frac{1}{(b+c)}, \frac{1}{(c+a)}, \frac{1}{(a+b)}$ are in $AP$ then , $$\frac{2}{(c+a)}=\frac{1}{(b+c)}+\frac{1}{(a+b)}$$ $$\frac{2b}{(c+a)}=\frac{b}{(b+c)}+\frac{b}{(a+b)}$$ $$\frac{2b}{(c+a)}=\frac{ba+\color{green}{b^2+b^2}+bc}{(a+b)(b+c)}$$
Since we have $a^2,b^2,c^2$ are in $AP$ , we have : $$2b^2=a^2+c^2$$
So , finally $$\frac{2b}{(c+a)}=\frac{\color{blue}{a^2+ab}+\color{red}{c^2+cb}}{(a+b)(b+c)}$$ $$\frac{2b}{(c+a)}=\frac{\color{blue}{a(a+b)}+\color{red}{c(b+c)}}{\color{blue}{(a+b)}\color{red}{(b+c)}}$$ $$\color{green}{\frac{2b}{(c+a)}=\frac{a}{(b+c)}+\frac{c}{(a+b)}}$$
We're DONE . Hence its in $AP$ .
Please consider UpVoting
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
A discussion about exponential diophantine equations and pythagorean triples
Do all pythagorean triples $(a,b,c)$ have the identity that the three
(exponential diophantine) equations
\begin{equation} a^x+b^y=c^z \end{equation} \begin{equation}
b^y+c^z=a^x \end{equation} \begin{equation} a^x+c^z=b^y \end{equation}
all have no more than one positive integer solution $(x,y,z)$?
This question is inspired by three exponential diophantine equations (1)$3^x+4^y=5^z$,(2) $4^y+5^z=3^x$, and (3) $3^x+5^z=4^y$.
Equations (1)(2)(3) all have no more than $1$ unique integer solution. Here, I'll give a proof for equation (3).
Find all positive integer solutions $(x,y,z)$ of the equation $3^x+5^z=4^y$.
$\because 3^x+5^z=4^y$, $\therefore (-1)^x+1^z\equiv 0\;\; (\operatorname{mod} 4)$. Also, since $x,y,z$ are all positive integers, so $x$ must be a multiple of $2$. Let $x=2a$. Then $3^{2a}+5^z=2^{2y}$, so we can get $$5^z=\left(2^y+3^a\right)\left(2^y-3^a\right).$$
Suppose $2^y+3^a=5^m, 2^y-3^a=5^n\;\;(m,n\in\mathbb{N}, m>n)$, subtracting the two equations, we get $2\cdot3^a=5^m-5^n$. If $n\geqslant1$, then $5\mid 5^m-5^n$, so $5\mid 2\cdot3^a$, which is a contradiction. Therefore, $n=0$.
We have $2^y=3^a+1$. For $a=1$, we get $y=2$. For $a\geqslant 2$, writing the equation in modulo $4$, we get $0\equiv (-1)^a+1$, so $a$ must be odd. Let $a=2k+1$, then $3^{2k+1}+1=2^a$. i.e. $3\cdot 9^k+1=2^y$. Writing the equation in modulo $8$, we get $3\cdot 1^k+1\equiv 0$. Which is another contradiction, so $(a,y)=(1,2)$, which leads us to $x=2a=2$.
Plugging in $x=2, y=2$ into the original equation (equation (3)), we get $5^z=7$. Therefore, $z$ is not an integer, so the exponential
diophantine equation $3^x+5^z=4^y$ doesn't have any solutions.
I'm curious that is the above statement (at the beginning of the question) true for all pythagorean triples. e.g. Does the exponential
diophantine equation $20^x+29^z=21^y$ have no more than one unique solution? Is there a general way to tackle with this problem?
Note:
*
*Again, I hope my question is not a duplication.
*$\mathbb{N}$ is the set of positive integers.
| It's an interesting question.
For the particular equation
$$
3^x+5^z=4^y
\qquad\qquad\;\;\;\,
$$
your attempt to prove that there are no positive integer solutions had flaws, as was noted in the comments.
However for that equation, there are, in fact, no positive integer solutions.
This can be shown as follows . . .
Let $\mathbb{N}$ denote the set of positive integers and let $R,S,T$ be defined as
\begin{align*}
R&=\{3^x\;\text{mod}\;20{\,\mid\,}x\in\mathbb{N}\}
\\[4pt]
S&=\{5^z\;\text{mod}\;20{\,\mid\,}z\in\mathbb{N}\}
\\[4pt]
T&=\{4^y\;\text{mod}\;20{\,\mid\,}y\in\mathbb{N}\}
\\[4pt]
\end{align*}
Then we get
\begin{align*}
R&=\{1,3,7,9\}
\qquad\qquad\;\,
\\[4pt]
S&=\{5\}
\\[4pt]
T&=\{4,16\}
\\[4pt]
\end{align*}
and it's then easily verified that there do not exist $r,s,t$ with $r\in R,\;s\in S,\;t\in T$ such that
$$
r+s\equiv t\;(\text{mod}\;20)
\qquad
$$
Thus the congruence
$$
3^x+5^z\equiv 4^y\;(\text{mod}\;20)
\;\;\,
$$
has no positive integer solutions, hence the equation
$$
3^x+5^z=4^y
\qquad\qquad\;\;\;
$$
has no positive integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Computing an infinite sum with Fourier series Let $F: \mathbb R \to \mathbb R$ which Fourier series is :
$$
F(x) = \frac{a_0}{2} + \sum^{\infty}_{k=1}a_n \cos\Bigl(\frac{n\pi x}{7}\Bigr),
$$
where $a_n = \frac{4}{7}\int^{7}_{0}e^{-4x}\cos\bigl(\frac{n\pi x}{7}\bigr)dx$
Use this Fourier series to compute the value of $\sum^{\infty}_{n=1}\frac{e^{-28}-(-1)^n}{784 + n^2 \pi^2}$
So far, I computed $a_n = 112\cdot \frac{(1-(-1)^n e^{-28})}{784 + n^2 \pi^2}$. I also tried to look at $F(14) = \frac{a_0}{2}+ \sum^{\infty}_{k=1}a_k$, which looks kind of similar to what I want, but haven't concluded anything so far.
| Computing $a_0$ and $a_n$ yields
$$F(x) = \frac{1 - e^{-28}}{14} + 112 \sum_{k=1}^\infty \frac{1-(-1)^n e^{-28}}{784 + n^2 \pi^2} \cos\left(\frac{n\pi x}7\right)$$
When $x=7$,
$$\begin{align*}
F(7) &= \frac{1 - e^{-28}}{14} + 112 \sum_{k=1}^\infty \frac{1-(-1)^n e^{-28}}{784 + n^2 \pi^2} (-1)^n \\[1ex]
&= \frac{1 - e^{-28}}{14} + 112 \sum_{k=1}^\infty \frac{(-1)^n- e^{-28}}{784 + n^2 \pi^2} \\[1ex]
\implies \sum_{k=1}^\infty \frac{e^{-28} - (-1)^n}{784 + n^2 \pi^2} &= \frac{1 - e^{-28}}{1568} - \frac{F(7)}{112}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all integer solutions to the equation $2^x + 3^y + 1 = 6^z$? Disclaimer: this is not homework.
I saw the following question randomly in an anime(!) and was curious.
For $z \leq 2$, I found the following solutions:
$x = 1, y = 1, z = 1$
$x = 2, y = 0, z = 1$
$x = 3, y = 3, z = 2$
$x = 5, y = 1, z = 2$
I entered the problem into Wolfram Alpha and it seems to confirm my suspicions that these are the only solutions.
Are there any solutions for $z > 2$? If not, how do we prove this?
| There are no solutions for $z \ge 3$. To prove this, we split the search into four cases.
Case I: $x \ge 3$. Then $2^x \equiv 0 \pmod{8}$ and $3^y \equiv 1 \ \text{or} \ 3 \pmod{8}$, and thus, $2^x+3^y+1 \equiv 2 \ \text{or} \ 4 \pmod{8}$. However, $6^z \equiv 0 \pmod{8}$. So there are no solutions in this case.
Case II: $x = 2$. The equation becomes $3^y+5 \equiv 6^z$. Since $z \ge 3$, we need $3^y = 6^z-5 \ge 6^3-5 = 211$, i.e., $y \ge 5$. But then $3^y+5 \equiv 2 \pmod{3}$ while $6^z \equiv 0 \pmod{3}$. So there are no solutions in this case.
Case III: $x = 1$. The equation becomes $3^y+3 \equiv 6^z$. Then, we have $3^y \equiv 1 \ \text{or} \ 3 \pmod{8}$, and so, $3^y+3 \equiv 4 \ \text{or} \ 6 \pmod{8}$. However, $6^z \equiv 0 \pmod{8}$. So there are no solutions in this case.
Case IV: $x = 0$. The equation becomes $3^y+2 \equiv 6^z$. Since $z \ge 3$, we need $3^y = 6^z-2 \ge 6^3-2 = 214$, i.e., $y \ge 5$. But then $3^y+2 \equiv 2 \pmod{3}$ while $6^z \equiv 0 \pmod{3}$. So there are no solutions in this case.
Therefore, there are no solutions for $z \ge 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4184007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating the integral $\int_{-\pi/4}^{\pi/4} 11 \sqrt{\sec^2\theta-1}\,d\theta$ Evaluating the integral $\int_{-\pi/4}^{\pi/4} 11 \sqrt{\sec^2\theta-1}\,d\theta$.
Recall that $\tan^2\theta+1=\sec^2\theta \iff \sec^2\theta-1=\tan^2\theta$. So,
\begin{equation*}
\begin{aligned}
\int_{-\pi/4}^{\pi/4} 11 \sqrt{\sec^2\theta-1}\,d\theta
& = 11 \int_{-\pi/4}^{\pi/4} \sqrt{\tan^2\theta}\,d\theta \\
& = 11 \int_{-\pi/4}^{\pi/4} \tan\theta\,d\theta \\
& = 11 \int_{-\pi/4}^{\pi/4} \dfrac{\sin\theta}{\cos\theta}\,d\theta \\
\end{aligned}
\end{equation*}
Let $u=\cos\theta$, $\,du=-\sin\theta\,du$.
Hence
\begin{equation*}
\begin{aligned}
& = -11 \int \dfrac{\,du}{u} \\
& = -11 \left. \ln|\cos \theta|\right]_{-\pi/4}^{\pi/4} \\
& = -11 (\ln|\cos (pi/4)| - \ln|\cos (-pi/4)|) \\
& = -11 (\ln|\dfrac{\sqrt{2}}{2}| - \ln|\dfrac{\sqrt{2}}{2}|) \\
\end{aligned}
\end{equation*}
which equals 0 but the textbook says the answer is $11\ln 2$. Where did I mess up?
| \begin{equation*}
\begin{aligned}
\int_{-\pi/4}^{\pi/4} 11 \sqrt{\sec^2\theta-1}\,d\theta
& = 11 \int_{-\pi/4}^{\pi/4} \sqrt{\tan^2\theta}\,d\theta \\
& = 11 \int_{-\pi/4}^{\pi/4} \lvert\tan\theta\rvert\,d\theta \\
& = 2\cdot11\int_{0}^{\pi/4} \tan\theta\,d\theta \\
& = 22[-\log(\cos \theta)]_{0}^{\pi/4}=22(-\log(\cos(\pi/4))+\log(\cos0)) \\
& = 22\frac{\log 2}{2}=11\log 2
\end{aligned}
\end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$ I am a student in Germany, and I prepare for Math Olympiad by solving math problems. I have been solving the following question, which took about 4 hours to solve.
Prove the following inequality without using calculator:
$$\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$$
Can you check my proof? It would be really grateful.
First, we can define function $f(x)$ as following:
$$f(x) = \sqrt[3]{x+1} - \sqrt[3]{x}\space(x > 0)$$
$$f(3) = \sqrt[3]{3+1} - \sqrt[3]{3} = \sqrt[3]{4} - \sqrt[3]{3}$$
$$f(2) = \sqrt[3]{2+1} - \sqrt[3]{2} = \sqrt[3]{3} - \sqrt[3]{2}$$
Then, we will differentiate $f(x)$ to check whether $f(x)$ is a decreasing function or not. $f '(x)$ must be a falling function if $f '(x)$ < 0.
$$f'(x) = \frac{1}{3\sqrt[3]{(x+1)^2}} - \frac{1}{3\sqrt[3]{x^2}}$$
Since the minuend is smaller than the subtrahend (minuend has a bigger denominator than the denominator of subtrahend), we can say $f '(x)$ is less than 0 which makes $f(x)$ a decreasing function. Falling function means that $f(a) > f(a+1)$. Substitute $a=2$ and we get:
$$f(2) > f(3)$$
$$\sqrt[3]{3} - \sqrt[3]{2} > \sqrt[3]{4} - \sqrt[3]{3}$$
$$\sqrt[3]{4} - \sqrt[3]{3} < \sqrt[3]{3} - \sqrt[3]{2}$$
Thank you for reading this text, but it would be more grateful if you check my solution, and comment my solution.
I wish you a beautiful day, and stay safe.
| I guess your question has been answered. Here is a different way to see it. Note that one has $$a^3-b^3=(a-b)(a^2+ab+b^2).$$ It follows that $$1=(\sqrt[3]{4}-\sqrt[3]{3})(\sqrt[3]{16}+\sqrt[3]{12}+\sqrt[3]{9})=(\sqrt[3]{3}-\sqrt[3]{2})(\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}).$$ It suffices now to show that $$\sqrt[3]{16}+\sqrt[3]{12}+\sqrt[3]{9}>\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}$$
$$\Leftrightarrow \sqrt[3]{16}+\sqrt[3]{12}>\sqrt[3]{6}+\sqrt[3]{4},$$ which is clear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
$8 \times 8$ checkerboard, how many distinct rectangles containing at least $4$ black squares?
An $8 \times 8$ checkerboard has alternating black and white squares. How many distinct rectangles, with sides on the grid lines of the checkerboard and containing at least $4$ black squares, can be drawn on the checkerboard?
I ended up getting$$3(2) + 5(2)(2) + (6(2)(2) - 2) + (5(2)(2) - 2) + (4(2)(2) - 2) + (3(2)(2) - 2) + (2(2)(2) - 2) + 1 = 97.$$ Is this correct?
EDIT: By distinct I mean not identical -- for example a $1 \times 4$ rectangle that begins with a black square would be considered the same as another $1 \times 4$ rectangle that begins with a black square, and those would be different from both (1) a $1 \times 4$ rectangle that begins with a white square and (2) a $4 \times 1$ rectangle.
| For rectangles, if there exists an $a\times b$ rectangle with $a\neq b$ such that it contains at least four black squares, then its transpose $b\times a$ will certainly be a distinct rectangle with at least four black squares that will fit on the chessboard. Similarly, any rectangle such that its transpose is a distinct rectangle will have the property that reversing the colors within the rectangle is possible by shifting it (for example, the only two $7\times 8$ rectangles on the board are both distinct, as one has a white square in the upper left corner while the other has a black square in the upper left corner). Thus for any $a\times b$ rectangle that has at least four white squares, reversing the colors (by shifting one row up or down as needed) will contain at least four black squares. This is true for any $a\times b$ rectangle that possesses at least four black squares except $1\times 7$ rectangles. A $1\times 7$ rectangle that possesses four black squares only contains three white squares.
Next, let's consider squares. The smallest $a\times a$ square that contains at least four black squares is the $3\times 3$ square. This contains either 5 black and 4 white or 4 black and 5 white. Its transpose will be nondistinct. But, shifting it one row or column will reverse the colors and be distinct. Thus, each such square will have a second "copy" except for the $8\times 8$ square because it cannot be shifted any rows or columns to reverse the colors.
This yields:
$1\times 7, 7\times 1$: 2 distinct, count each once
$1\times 8, 2\times 4^+, 3\times 4^+, 4\times 5^+, 5\times 6^+, 6\times 7^+, 7\times 8$: 21, count each four times.
$3\times 3, 4\times 4, 5\times 5, 6\times 6,7\times 7$: 5, count each twice.
$8\times 8$: Count once.
For this count, I am using $2\times 4^+$ to represent all distinct pairs where the second number is at least 4, thus $2\times 4, 2\times 5, 2\times 6, 2\times 7, 2\times 8$ are all represented by this shorthand.
This yields $2+21\cdot 4+5\cdot 2+1 = 97$, which is the answer you arrived at.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the minimal polynomial of $\alpha=\sqrt{3+2\sqrt{2}}$ over $\mathbb{Q}$ Question: Find the minimal polynomial of $\alpha=\sqrt{3+2\sqrt{2}}$ over $\mathbb{Q}$
Thoughts: the "standard" method of starting by squaring (twice) to get rid of the square roots, because I don't have a nice way of showing the resulting polynomial is irreducible. So..
Attempt: It would be great if I could get our $\alpha$ in the form $$(a+b)^2=a^2+2ab+b^2=\sqrt{3+2\sqrt{2}}.$$ So, $$3+2\sqrt{2}=(\sqrt{2})^2+2\sqrt{2}+1=(\sqrt{2}+1)^2.$$So,
$$\alpha=\sqrt{3+2\sqrt{2}}=\sqrt{2}+1.$$So,
$$(\alpha-1)^2=2\\
\alpha^2-2\alpha+1=2\\
\alpha^2-2\alpha-1=0.$$So let $f(x)=x^2-2x-1$. Since $f(x)$ is irreducible over $\mathbb{Q}$ by the Rational Roots Test (since it has degree $2$), $f(x)$ is monic, and $f(\alpha)=0$, we conclude that $f(x)$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$.
Does this look okay?
| OP's proof is good, and there is not much room left to simplify. The following are just alternatives.
*
*Let $\beta = \sqrt{3 - 2\sqrt{2}}$ and note that $\alpha\beta=1$ and $\alpha^2+\beta^2=6$. Given that $0 \lt \beta \lt \alpha$ it follows that $\alpha-\beta=\sqrt{(\alpha-\beta)^2}=\sqrt{\alpha^2+\beta^2-2\alpha\beta}=\sqrt{6-2}=2$.
Then $\alpha(-\beta)=-1$ and $\alpha+(-\beta)=2$, so $\alpha, -\beta$ are the roots of $x^2 - 2 x - 1\,$.
*The "standard" method can also work. Suppose "starting by squaring (twice) to get rid of the square roots", then $x^4 -6x^2+1=0=(x^2-1)^2-4x^2=(x^2-2x-1)(x^2+2x-1)$. Only the first quadratic factor has a root larger than $2$, which must be $\alpha$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find locus of mid-point of chord of circle I am working through a pure maths book as a hobby and am struggling with the very last part of the question below. I give the whole of the question just in case it proves relevant to the answer.
A circle passes through the points A, B and C which have coordinates $(0,3),(\sqrt3,0)$ and $(-\sqrt3,0$ respectively.
Find (i) the equation of the circle , (ii) the length of the minor arc BC, (iii) the equation of the circle on AB as diameter.
And then follows this part which I cannot do:
A line $y=mx-3$ of variable gradient m cuts the circle ABC in two points L and M. Find in cartesian form the equation of the locus of the mid-point of LM.
I have worked out the first parts to get:
(i) $x^2+y^2-2y-3=0$
(ii)$\frac{4\pi}{3}$
(iii)$x^2+y^2-3\sqrt x-3y=0; x^2+y^2+2y-3=0$
But I cannot get the very last part of (iii). I thought if I could find the coordinates of intersection I could then take the coordinates of the mid-point.
I have tried as follows:
Circle equation $x^2+y^2-2y-3=0 \rightarrow y=\frac{x^2+y^2-3}{2}$
At intersection $\frac{x^2+y^2-3}{2}=mx-3$
I then try to solve for this but in no way can I get the answer which the book says is $x^2+y^2+2y-3=0$
| We solve the system
$$\begin{cases}
x^2+y^2-2y-3=x^2+(y-1)^2-4=0\\
y=mx-3
\end{cases}$$
finding
$$x^2+(mx-3-1)^2-4=0$$
that is
$$(1+m^2)x^2-8mx+12=0\qquad(1).$$
Hence the locus of midpoints is given by
$$\frac{x_1+x_2}{2}=\frac{4m}{1+m^2},\quad\frac{y_1+y_2}{2}=\frac{4m^2}{1+m^2}-3=\frac{m^2-3}{1+m^2}.$$
The discriminant $\Delta$ of the quadratic equation (1) should be non-negative which yields the condition $|m|\geq \sqrt{3}$.
Note that for $m=0,\sqrt{3},-\sqrt{3}$, we get the points $(0,-3), (\pm\sqrt{3},0)$. Making a drawing, we guess that the locus is along a circle. Then we find the circle through the $3$ points (which are symmetric to $A,B,C$ with respect to the line $x=0$):
$$x^2+y^2+2y-3=0.$$
Finally, we verify that the whole locus of midpoints is along above circle:
$$x^2+y^2+2y-3=x^2+(y+1)^2-4=\left(\frac{4m}{1+m^2}\right)^2
+\left(\frac{2m^2-2}{1+m^2}\right)^2-4=0.$$
Notice that, because of the condition $|m|\geq \sqrt{3}$, the locus is not the whole circle, but just the $\color{blue}{\rm upper\ arc}$ with endpoints $(\pm\sqrt{3},0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Inequality proof $\prod_{cyc} (ab+1)$ Prove that if for positive reals $a,b,c$ with $a^2+b^2+c^2+ab+bc+ca \le 2$ and $a+b+c=1$ then
$$(ab+1)(bc+1)(ca+1)\ge ((1-a)(1-b)(1-c))^2.$$
I've tried expanding and i've noticed that $a+b+c=1$ and $a^2+b^2+c^2+ab+bc+ca\le 4$ imply $a^2+b^2+c^2 \le 3$ but i'm not sure how that helps...
| You don't need the additional constraint $a^2+b^2+c^2+ab+bc+ac\leq 2$.
Expanding on my comment, $a+b+c =1 \land a,b,c\geq 0 \implies 0\leq a,b,c \leq 1$. This also implies that
$$0 \leq 1-a, 1-b, 1-c \leq 1$$
Which further implies that
$$(1-a)(1-b)(1-c) \leq 1$$
Which again implies that
$$((1-a)(1-b)(1-c))^2 \leq 1 \quad (*)$$
However,
$$(1+ab)(1+bc)(1+ac) \geq 1\times 1 \times 1 =1 \quad (**)$$
Combining $*, **$, we get
$$(1+ab)(1+bc)(1+ac) \geq 1 \geq ((1-a)(1-b)(1-c))^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
if $S_n = (3 + \sqrt{5})^n + (3 - \sqrt {5})^n$ then show that $S_n$ is an integer by induction this is all that i have tried:
let $n=1$ so equation gives $(3 + \sqrt {5}) + (3 - \sqrt {5}) = 6$ which is an integer
so it is true for $n=1$
now let it be true for $k \ge n$ then we have:
$(3 + \sqrt {5})^k + (3 - \sqrt {5})^k$ is an integer
for k+1: $(3 + \sqrt {5})^k(3 + \sqrt {5}) + (3 - \sqrt {5})^k(3 - \sqrt {5})$
$= 3((3 + \sqrt {5})^k + (3 - \sqrt {5})^k) + \sqrt {5}((3 + \sqrt {5})^k - (3 - \sqrt {5})^k)$
$= (integer) + \sqrt {5}((3 + \sqrt {5})^k - (3 - \sqrt {5})^k)$
from here I started facing a problem.
I wasn't able to prove $\sqrt {5}((3 + \sqrt {5})^k - (3 - \sqrt {5})^k)$ an integer.
can someone please help me out?
(ONLY USING INDUCTION!)
| Use the fact that $$a^{n+1} + b^{n+1} = (a+b)(a^n + b^n) - ab(a^{n-1} + b^{n-1}),$$ so that if $a = 3 + \sqrt{5}$, $b = 3 - \sqrt{5}$, then $ab = 3^2 - 5 = 4$, hence $$S_{n+1} = 6S_n - 4S_{n-1}.$$ Consequently your induction base case should cover two consecutive values of $n$, and your induction step relies on the fact that the previous two values of $S$ are integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
What's more likely: $7$-digit number with no $1$'s or at least one $1$ among its digits?
A $7$-digit number is chosen at random. Which is more likely: the number has no $1$'s among its digits or the number has at least one $1$ among its digits?
Here's how I did it: The question is asking whether $8(9)^6$ (the number of those with no $1$'s among its digits) or $9(10)^6 - 8(9)^6$ (the number of those with at least one $1$ among its digits). Some tedious multiplying shows that $8(9)^6 = 4241528 < 4500000$, which demonstrates that $9(10^6) - 8(9)^6$ i.e. the number having at least one $1$'s among its digits is more likely.
However, I am wondering if there is a slicker way to get the answer without having to do any tedious multipication.
| Note that $8\cdot 9^6 < 9\cdot 10^6-8\cdot 9^6$ is equivalent to
$$2\cdot 8\cdot 9^6 < 9\cdot 10^6.$$
Getting rid of common factors on both sides by considering prime factorizations, we can reduce this as far as
$$3^{10} < 2^2 \cdot 5^6.$$
Now both sides are perfect squares, so by taking square roots this is equivalent to
$$3^5 < 2 \cdot 5^3.$$
This we might actually be able to calculate:
\begin{align}
3^5 &= 3\cdot 9^2 = 3\cdot 81 = 243, \\
2\cdot 5^3 &= 2\cdot 5\cdot 25 = 10\cdot 25 = 250.
\end{align}
This confirms the original inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How to find area $\triangle ABC$, where $AC=BD, CE=2, ED=1, AE=4$ and $\angle CAE=2 \angle DAB$ In a triangle $\triangle ABC$, $AC=BD, CE=2, ED=1, AE=4$ and $\angle CAE=2 \angle DAB$. Find area of $\triangle ABC$.
This is what it looks like, but how to justify the red ones or maybe there is another way to solve this problem?
| Take a Cartesian coordinate system where $\vec d=(0,0)$,
$\vec e=(1,0)$, $\vec c=(3,0)$ and $\vec b=(-b,0)$ with $b>0$,
$\vec a=(a_x,a_y)$. So $CE=2$ and $ED=1$ are satisfied.
$$ AC=BD \to
b^2=(3-a_x)^2+a_y^2
% \to
% b^2=9+a_x^2-6a_x+a_y^2
\to
b^2-9+6a_x=a_x^2+a_y^2
$$
$$ AE=4 \to
(a_x-1)^2+a_y^2=4^2.
\to
a_x^2-2a_x+1+a_y^2=16
\to
a_x^2+a_y^2=15+2a_x.
$$
$$
b^2=24-4a_x.
$$
$$
b=2\sqrt{6-a_x}.
$$
Vectors $\vec{AB}=(-b-a_x,-a_y)$,
$\vec{AD}=(-a_x,-a_y)$ and the dot product is
$$
\vec{AB}\cdot \vec{AD}=\cos\alpha |\vec{AB}||\vec{AD}|.
$$
$$
-a_x(-b-a_x)+a_y^2=\cos\alpha \sqrt{(-b-a_x)^2+a_y^2}\sqrt{a_x^2+a_y^2}.
$$
$$
+a_x b+15+2a_x=\cos\alpha \sqrt{b^2+2ba_x +15+2a_x}\sqrt{15+2a_x}.
$$
$$
2a_x \sqrt{6-a_x}+15+2a_x=\cos\alpha \sqrt{2ba_x +39-2a_x}\sqrt{15+2a_x}.
$$
Vector $\vec{AE}=(1-a_x,-a_y)$ and
$\vec{AC}=(3-a_x,-a_y)$, so the dot product is
$$
\vec{AE}\cdot \vec{AC}=\cos(2\alpha) |\vec{AE}||\vec{AC}|.
$$
$$
(1-a_x)(3-a_x)+a_y^2=\cos(2\alpha) \sqrt{(1-a_x)^2+a_y^2}\sqrt{(3-a_x)^2+a_y^2}
=\cos(2\alpha) 4 b
$$
$$
3-4a_x +15+2a_x
=\cos(2\alpha) 4 b
$$
$$
18-2a_x
=\cos(2\alpha) 4 b
$$
$$
9-a_x
=\cos(2\alpha) 2\cdot 2\sqrt{6-a_x}
=4\sqrt{6-a_x}(2\cos^2\alpha-1)
$$
Numerical solution of these 2 coupled equations for $\cos \alpha$ and $a_x$ converges
to $AC=BD=b=3$, $\alpha\approx 14.4775^\circ$, $a_x=3\frac34$, $a_y=3\sqrt{15}/4$. The area is half the product of baseline and altitude, $(b+3)a_y/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove that the limit of $f(x,y)=\dfrac{x^2 \sin(y) + y^2 \sin(x)}{x^2+y^2}$ is $0$ as $(x,y)$ approaches $(0,0)$? I need to show that $\displaystyle\lim_{(x,y)\rightarrow(0,0)}\dfrac{x^2 \sin(y) + y^2 \sin(x)}{x^2+y^2}$ exists. I know that it is equal to zero. Until now, all I know how to do is to prove using the $\epsilon-\delta$ definition. I tried to prove it by the following way:
Let $\delta>0$ be a real number such that $0<\sqrt{x^2+y^2}<\delta$. We have $0< x^2+y^2<\delta^2$.
We know that $\left\vert x^2\sin(y)+y^2\sin(x)\right\vert\leq\left\vert x^2\sin(y)\right\vert+\left\vert y^2\sin(x)\right\vert=x^2\left\vert\sin(y)\right\vert+y^2\left\vert\sin(x)\right\vert\leq x^2+y^2$.
Thus
$$0<\left\vert x^2\sin(y)+y^2\sin(x)\right\vert\leq x^2+y^2<\delta^2$$
and dividing the inequality by $x^2+y^2>0$ we have
$$ 0<\dfrac{\left\vert x^2\sin(y)+y^2\sin(x)\right\vert}{x^2+y^2}<1<\dfrac{\delta^2}{x^2+y^2}. $$
I thought that this could be useful because $\dfrac{\left\vert x^2\sin(y)+y^2\sin(x)\right\vert}{x^2+y^2}=\left\vert\dfrac{x^2\sin(y)+y^2\sin(x)}{x^2+y^2}\right\vert$, but I don't know how to proceed.
| It is enough to show that the limit of $\frac{x^2 \sin y}{x^2+y^2}$ is $0$ ( the other half follows by symmetry). The modulus of this is at most $|\sin y|$, which is less than $|y|$, so we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4205040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Is it true that $\forall x>3, \exists yx$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 ?$
Is it true that $\forall x>3, \exists y<x$ and $\exists z>x$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 ?$
Here is my proof but I am stuck in the middle. And I am also wondering if I am on the right track?
Let $y = x - a$ for some real number $a$ that $a>0$, and $z = x + b$ for some real number $b$ that $b>0$.
Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 \Rightarrow \frac{1}{x}+\frac{1}{x-a}+\frac{1}{x+b}=1.$
After combining similar terms, I got:
$a = \frac{x(x^2+bx-3x-2b)}{x^2+(b-2)x-b}$, and $b = \frac{x(x^2-ax-3x+2a)}{-x^2+(a+2)x-a}$
For $a$ exist, $x^2+(b-2)x-b \neq 0$. And for $b$ exist, $-x^2+(a+2)x-a \neq 0$.
Let $f(x) = x^2+(b-2)x-b $ and $g(x) = -x^2+(a+2)x-a $.
By the formula of quadratic function: $x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$:
$f(x)\neq 0$ $\Rightarrow (b-2)^2+4b < 0$ ,and $g(x)\neq 0\Rightarrow(a+2)^2-4a<0$.
Since $(b-2)^2+4b \geq 0$,and $(a+2)^2-4a\geq0$ , no matter how we choose $a$ and $b$, $f(x)=0$ and $g(x)=0$ always have a solution.
Then I stuck at here, I understand that I need to prove $f(x)\neq0$ and $g(x)\neq0$ for all $x >3$ and $a,b$ have to be positive to make the statement true. Otherwise, it's false. Really hope to get some hints/help!! Thank you!
| It's easier if you substitute $a = 1/x, b=1/y, c=1/z$. I'll assume that we want $x,y,z > 0$. Then the conditions just say that $a,b,c$ are numbers between $0$ and $1$ so that $a + b + c = 1$ and $b$ is the biggest, $c$ is the smallest, and $a$ is in between. When is this possible?
You need $a < 1/2$ or else $b$ can't be bigger than $a$. But that's all you need: once you pick $a < 1/2$, just take $b$ to be $1/2$ and then take $c = 1 - a - b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$\epsilon$-$\delta$ proof of $\lim_{x \to 3} x^2 = 9$ I've been learning about $\epsilon$-$\delta$ proofs and attempted to come up with my own proof that
$$
\lim_{x \to 3} x^2 = 9
$$
exists (I did use some help from some textbooks). Is my proof valid and free of redundancies? My proof:
Scratch work:
Written formally, this is:
$$
\forall \epsilon > 0, \exists \delta > 0 \text{ s.t. } 0 < |x - 3| < \delta \implies |x^2 - 9| < \epsilon
$$
We start off by simplifying the conclusion of the implication:
$$
|x^2 - 9| = |x + 3|\cdot |x - 3|
$$
Now we have $|x + 3| \cdot |x - 3| < \epsilon$. Because $|x - 3| < \delta$, we have $|x - 3| \cdot |x + 3| < \delta |x + 3|$. In an attempt to find a suitable $\delta$, we let $\delta|x + 3| < \epsilon$. Solving for $\delta$, we get $\delta < \frac{\epsilon}{|x + 3|}$. There is a problem though: $\delta$ is defined in terms of $\epsilon$ and the randomly chose point $x$. The definition of $\delta$ can only depend on $\epsilon$. To get around this problem, we will have to estimate the size of $|x + 3|$.
We start of by assuming that $\delta < 1$, which implies that $|x - 3| < 1$. Solving for $x + 3$, we get $5 < x + 3 < 7$. We now know that $|x + 3| < 7$ when $\delta < 1$. Since $|x^2 - 9| < \delta|x + 3|$ and $|x + 3| < 7$ (if $\delta < 1$), we have $|x^2 - 9| < 7 \delta$. In an attempt to find a suitable $\delta$, we let $7 \delta < \epsilon$. Solving for $\delta$, we get $\delta < \frac{\epsilon}{7}$ when $\delta < 1$. So we have now deduced two restrictions: $\delta < 1$ and $\delta < \frac{\epsilon}{7}$. To satisfy both restrictions, we let $\delta < \min \left\{ 1, \frac{\epsilon}{7} \right\}$. When $\epsilon > 7$, then $\delta < 1$ and when $\epsilon < 7$, we have $\delta < \frac{\epsilon}{7}$. We can now write up the proof.
Proof:
For every $\epsilon > 0$ there exists a $0 < \delta < \min \left\{ 1, \frac{\epsilon}{7} \right\}$ such that $0 < |x - 3| < \delta \implies |x^2 - 9| < \epsilon$. There are two things that can happen: $\delta < 1$ and $\delta < \frac{\epsilon}{7}$.
Case 1 - $\delta < 1$:
The implication's hypothesis is $0 < |x - 3| < \delta$. Multiply both sides by $|x + 3|$ to get $0 < |x^2 - 9| < |x + 3| \delta$. Because $\delta < 1$, we know (from our scratchwork) that $|x + 3| < 7$ and $\delta$ is also less that $\frac{\epsilon}{7}$. This means that $0 < |x^2 - 9| < |x + 3| \delta < 7 \delta < [7 \frac{\epsilon}{7} = \epsilon]$.
Case 2 - $\delta < \frac{\epsilon}{7}$:
For $\delta$ to be less that $\frac{\epsilon}{7}$, we must have that $\frac{\epsilon}{7} < 1$ (as per the definition of $\min$). Because $\delta < \frac{\epsilon}{7}$ and $\frac{\epsilon}{7} < 1$ we have $\delta < 1$. Look to Case 1 for what follows after.
$\square{}$
| There is no need to consider several cases. Define $\delta=\min\left\{\frac\varepsilon7,1\right\}$. Then, if $|x-3|<\delta$, it follows from your computations that$$\left|x^2-9\right|=|x-3||x+3|<\frac\varepsilon7\times7=\varepsilon.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4207900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Prove $\int^{\infty}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ diverges First of all, I noticed that $\sin^2(\pi(x+\frac{1}{x}))$ "approaches" $\sin^2(\pi x)$ as $x \to \infty$. In other words, it seems that the area under one period of $\sin^2(\pi(x+\frac{1}{x}))$ should converge to the value of the area under one period of $\sin^2(\pi x)$ as $x \to \infty$.
The following is my best attempt at formalizing this idea:
Let $f(x) = \sin^2(\pi(x+\frac{1}{x}))$ and $\xi_n = \frac{n+\sqrt{n^2-4}}{2}$, where $n \geq 2$. Notice that $f(\xi_n)=0$.
Now, consider the limit of $\xi_{n+1} - \xi_{n}$ as $n \to \infty$:
$$\begin{align} \lim_{n\to\infty} \xi_{n+1}-\xi_n &= \lim_{n\to\infty}\frac{(n+1)+\sqrt{(n+1)^2-4}}{2}-\frac{n+\sqrt{n^2-4}}{2} \\ &= \frac{1}{2}\lim_{n\to\infty}1+\sqrt{(n+1)^2-4}-\sqrt{n^2-4} \\ &= \frac{1}{2}\lim_{n\to\infty}\frac{2+\frac{2}{n}+2\sqrt{1+\frac{2}{n}-\frac{3}{n^2}}}{\frac{1}{n}+\sqrt{1+\frac{2}{n}-\frac{3}{n^2}}+\sqrt{1-\frac{4}{n^2}}} \\ &= 1 \end{align}$$ What the result above tells us is that as x
increases, the period of $\sin^2(\pi(x+\frac{1}{x}))$ approaches the
period of $\sin^2(\pi x)$. Furthermore, $x+\frac{1}{x} \approx x$ as
$x\to \infty$, which implies that $\sin^2(\pi(x+\frac{1}{x}))$
"approaches" $\sin^2(\pi x)$ as $x \to \infty$. Using both of these
facts, we can claim that $\int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx$ approaches $\int^{1}_{0} \sin^2(\pi x) dx$ as $n \to \infty$.
We can also compute the value of $\int^{1}_{0} \sin^2(\pi x) dx$:
$$\begin{align} \int^{1}_{0} \sin^2(\pi x) dx &= \frac{1}{2}\int^{1}_{0}1-\cos(2\pi x)dx \\ &=\frac{1}{2}\{(1-\frac{1}{2\pi}\sin(2\pi))-(0-\frac{1}{2\pi}\sin(0))\} \\ &= \frac{1}{2}\end{align}$$ Thus, we can state that:
$$\lim_{n\to\infty} \int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx = \frac{1}{2}$$
Notice that we can write $\int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx$ as... $$\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx - \int^{\xi_{n}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$$ If we consider $\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ as a sequence, then $\lim_{n\to\infty} \int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx = \frac{1}{2}$
tells us that $\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ is
not a Cauchy sequence, and hence $\int^{\xi_{n+1}}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ is divergent as $n \to \infty$.
Thus, we can finally claim that $\int^{\infty}_{0} \sin^2(\pi(x+\frac{1}{x}))dx$ is divergent.
Now, the part of my argument which I find to be weak is:
What the result above tells us is that as $x$
increases, the period of $\sin^2(\pi(x+\frac{1}{x}))$ approaches the
period of $\sin^2(\pi x)$. Furthermore, $x+\frac{1}{x} \approx x$ as
$x\to \infty$, which implies that $\sin^2(\pi(x+\frac{1}{x}))$
"approaches" $\sin^2(\pi x)$ as $x \to \infty$. Using both of these
facts, we can claim that $\int^{\xi_{n+1}}_{\xi_n} \sin^2(\pi(x+\frac{1}{x}))dx$ approaches $\int^{1}_{0} \sin^2(\pi x) dx$ as $x \to \infty$.
How can I make this part more rigorous/stronger? I don't really know how to achieve any more rigor with my current level of mathematics.
| Alternative proof:
It suffices to prove that,
for any $M > 1$,
$\int_M^{M+4} \sin^2 (\pi(x + 1/x))\, \mathrm{d} x \ge \frac{1}{16}$.
Let $N \in (M + 1, M + 3)$ be a positive integer.
Let
$$x_1 = \frac{N}{2} + \frac{1}{8} + \frac{1}{8}\sqrt{16N^2 + 8N - 63}$$
and
$$x_2 = \frac{N}{2} + \frac{3}{8} +
\frac{1}{8} \sqrt{16N^2 + 24N -55}.$$
We have $x_2 - x_1 \ge \frac{1}{4}$,
and $M < x_1 < x_2 < M + 4$ (easy to prove), and
$$x_1 + \frac{1}{x_1} = N + \frac{1}{4},
\quad x_2 + \frac{1}{x_2} = N + \frac{3}{4}.$$
Since $x\mapsto x + 1/x$ is strictly increasing on $(1, \infty)$,
we have $N + 1/4 \le x + 1/x \le N + 3/4$ for all $x \in [x_1, x_2]$ and thus
$\sin^2 (\pi(x + 1/x)) \ge \frac{1}{4}$
for all $x \in [x_1, x_2]$.
Thus,
$\int_M^{M+4} \sin^2 (\pi(x + 1/x))\, \mathrm{d} x \ge \int_{x_1}^{x_2} \sin^2 (\pi(x + 1/x))\, \mathrm{d} x \ge \frac{1}{4}(x_2 - x_1) \ge \frac{1}{16}$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4211726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Necessary condition on $|x-1|$ to assure $|x^2 -1| < \frac{1}{2}$. Is my solution correct? How to get the tightest answer? Question: Determine a condition on $|x-1|$ that will assure that:-
$$|x^2 -1| < \frac{1}{2}$$
My solution:-
Let $f(x) = x^2$
$$\lim _{x \rightarrow 1} f(x) = \lim _{x \rightarrow 1} x^2 = 1$$
To find $\delta$ so that
$\: \: |f(x) - 1| < \epsilon = \frac{1}{2}\: \:$ whenever $\: \: 0 < |x-1| < \delta$
$$|f(x) - 1| < \epsilon = |x^2-1| = |x+1||x-1|$$
$$|f(x)-1| = |x+1||x-1|$$
$$|f(x)-1| < \delta \: |x+1| \tag {1}$$
$$$$
$$0 < |x-1| < \delta$$
$$|x-1| < \delta$$
$$-\delta < x-1 < \delta$$
$$-\delta +2 < x+1 < \delta + 2$$
$$-\delta-2 < x+1 < \delta + 2$$
$$|x+1| < \delta + 2 \tag{2}$$
Subsitutuing equation (2) in (1)
$$|f(x) -1| < \delta \: |x+1|$$
$$| f(x) - 1| < \delta \: (\delta + 2)$$
Solving $\delta \: (\delta + 2) = \frac{1}{2}$ for $\delta$ we get:
$$\delta = \frac{-2\pm \sqrt{6}}{2}$$
Since $\delta$ can only be positive, the necessary condition is
$$0 < |x-1| < \frac{-2+\sqrt{6}}{2} \approx 0.2247$$
| $$|x^2 - 1|<\frac{1}{2}$$
$$-\frac{1}{2} < x^2 - 1<\frac{1}{2}$$
$$1 -\frac{1}{2} < x^2 < 1+ \frac{1}{2}$$
$$\sqrt{ \frac{1}{2} }< x < \sqrt{\frac{3}{2}} \text{ or } -\sqrt{ \frac{3}{2} }< x <-\sqrt{\frac{1}{2}}
$$
$$\sqrt{ \frac{1}{2} }-1< x-1 < \sqrt{\frac{3}{2}}-1 \text{ or } -\sqrt{ \frac{3}{2} }-1< x-1 <-\sqrt{\frac{1}{2}}-1
$$
A necessary (but not sufficient) condition is $$|x-1| < \sqrt{\frac{3}{2}}+1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
For which ${n \in \mathbb{Z}^+_0}$ is the number ${2^n - 1}$ divisible by 7? This is how far I've gotten:
Let ${n=3k+r}$, where ${r=0,1}$ or $2$ (since ${0 \le r < 3}$).
Case ${r=0}$:
\begin{equation}
\begin{aligned}
2^{3k} &\equiv_7 1\\
\Rightarrow 2^{3k}-1 &\equiv_7 0\\
k &\in \mathbb{Z}^+_0\\
\end{aligned}
\end{equation}
For cases ${r=1}$ and ${r=2}$:
\begin{equation}
\begin{aligned}
2^{3k+r}-1 &\equiv_7 0\\
\Leftrightarrow 2^{3k} \cdot 2^r -1 &\equiv_7 0\\
\end{aligned}
\end{equation}
which gives
\begin{equation}
\begin{aligned}
2^0-1=0 &\equiv_7 0\\
2^1-1=1 &\equiv_7 1\\
2^2-1=3 &\equiv_7 3\\
\end{aligned}
\end{equation}
I know (by more or less messing around) that ${r=0,3}$ or $6$ for ${k \in \mathbb{Z}^+_0}$, but I'm getting nowhere in trying to prove it. And I'm not sure how $r$ actually can be 3 or 6 since $r$ should be smaller than 3. Right?
Any hints or guidings towards the right answer would be much appreciated. Thanks!
| Bear in mind if $a \equiv 1$ then $a\times k \equiv 1 \times k \equiv k$ and $a^m \equiv 1^m \equiv 1$.
So as you have figured we can writ any $n$ as $n = 3k + r$ where $r=0,1$ or $2$ and $k$ can be any natural number.
And we figure $2^3 \equiv 8 \equiv 1$ so........
$2^{3k+r}= 2^{3k}\times 2^r =(2^3)^k\times 2^r = 8^k \times 2^r \equiv 1^k\times 2^r \equiv 2^r \pmod 7$. Always.
And from that we can conclude that for all natural numbers we have one of the following $3$ cases:
*
*If $n \equiv 0 \pmod 3$ that is if there is a $k$ so $n = 3k + 0$ then
$2^{n} \equiv 2^0\equiv 1 \pmod 7$. And our case is always satisfied.
*If $n \equiv 1 \pmod 3$ that is if there is a $k$ so $n = 3k + 1$ then
$2^n \equiv 2^1 \equiv 2 \pmod 7$. And our case is never satisfied (but we we always have $n \equiv 1\pmod 7 \implies 2^n\equiv 2 \pmod 7$.)
*If $n \equiv 2 \pmod 3$ that is if there is a $k$ so $n = 3k + 2$ then
$2^n \equiv 2^2 \equiv 4\pmod 7$. And our case is never satisfied (be we always have $n\equiv 2\pmod 7\implies 2^n\equiv 4\pmod 7$. Always.)
So that's that.
if $n \equiv 0 \pmod 3$ then $2^n \equiv 1\pmod 7$. Always. And if $n \not \equiv 0 \pmod 3$ then $2^n \not \equiv 1 \pmod 7$. Never.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to write $f(z) = \frac{1}{\left(z-3\right)\left(z-4\right)\left(z-5\right)}$ as a Laurent Series I have the following function:
$f(z) = \frac{1}{\left(z-3\right)\left(z-4\right)\left(z-5\right)}$
Through partial fractions I got the values of A, B and C as below:
$=\frac{A}{(z-3)} + \frac{B}{(z-4)} + \frac{C}{(z-5)}$
$A = \frac{1}{2}$, $B=-1$, $C = \frac{1}{2}$
so altogether I got: $\frac{1}{2(z-3)} - \frac{1}{(z-4)} + \frac{1}{2(z-5)}$
Now, I need to find the Laurent series at $z=3$ in $ 1<|−3|<4$. I'm not too sure how to progress from here. My intuition is to find the power series at $\frac{1}{z+a}$ at $z=3$ but not too sure what to do after.
Any hints or anything would really help.
| That's a start! Now we need to find a series for each of the terms.
$\displaystyle \frac{1/2}{z-3}$ is the first series.
$\displaystyle \frac{-1}{z-4} = \frac{-1}{(z-3)-1} = -\frac{1}{z-3}\frac{1}{1-\frac{1}{z-3}}=-\frac{1}{z-3}\left(1+ \frac{1}{z-3}+ \frac{1}{(z-3)^2} + \frac{1}{(z-3)^3}+\cdots \right)$
Similarly for the third term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$x_1$ , $x_2$ are roots of $x=5-x^2$. Find the equation with roots $\frac1{(x_1+1)^3}$ and $\frac1{(x_2+1)^3}$.
Suppose $x_1$ , $x_2$ are the roots of the equation $x=5-x^2$.Then
$\dfrac1{(x_1+1)^3}$ and $\dfrac1{(x_2+1)^3}$ are roots of which
equation?
$1)125x^2+16x=1$
$2)125x^2=16x+1$
$3)125x^2=12x+1$
$4)125x^2+12x=1$
I solved this problem with the following approache,
I've denoted the roots of the original equation by $\alpha$ and $\beta$ rather than $x_1$, $x_2$ ,
$S=\alpha+\beta=-1$ and $P=\alpha\beta=-5$. We find $S'$ and $P'$ of the new equation,
$$P'=\dfrac1{[(\alpha+1)(\beta+1)]^3}=\dfrac1{(S+P+1)^3}=-\frac1{125}$$
$$S'=\dfrac{(\beta+1)^3+(\alpha+1)^3}{[(\alpha+1)(\beta+1)]^3}=-\frac1{125}\times\left((\alpha^3+\beta^3)+3(\alpha^2+\beta^2)+3(\alpha+\beta)+2\right)$$
$$=-\dfrac{(S^3-3PS)+3(S^2-2P)+3S+2}{125}=-\dfrac{-16+33-3+2}{125}=-\frac{16}{125}$$
Hence the new equation is $125x^2+16x-1=0$. And the answer is first choice.
This was a problem from a timed exam. So can you solve it with other approaches (preferably quicker one) ?
| Method-1
Given equation is $x^2+x-5=0$. It's root is $x_1$, hence $x_1^2+x_1-5=0$ $-(1)$
Also note that the equation we have to form has symmetric roots
Therefore think of an expression of type $p(x)=f(x)^2+f(x)-5$ such that $p(\frac{1}{(x_1+1)^3})=x_1^2+x_1+5=0$ which makes $\frac{1}{(x_1+1)^3}$ a root of $p(x)$
Such expression should have $f(\frac{1}{(x_1+1)^3})=x_1$
Clearly $\displaystyle f(x)=\frac{1}{(x)^{\frac{1}{3}}}-1$
Substituting $f(x)$ we get
$p(x)=5(x)^{\frac{2}{3}}+x^{\frac{1}{3}}-1$
And therefore our equation is
$5(x)^{\frac{2}{3}}+x^{\frac{1}{3}}=1$ whch is further simplified to $125x^2+16x=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Catalan constant's integral representation Catalan constant is known to have a rich source of integral identities, here is the formula I found:
$$ \int_0^\infty \frac{\sin^{-1}(\sin(x))}{x} \,dx \ =2G.$$
This can be proved by analyzing the function and using the identity:
$$ \frac{2G}{\pi}-\frac{1}{2}=\ln\left(\prod_{n=1}^\infty \frac{(4n-1)^{4n-1}}{(4n-3)^{2n-1}(4n+1)^{2n}}\right).$$
Is there any "neat" way to prove the formula directly without using the identity? It's really hard for me to prove the identity.
Update
I found my own way to prove this integral indentity:
Using Fourier series of triangle wave, with a little change in the coefficient, I got this equation:
$$\sin^{-1}(\sin(x))=\frac{4}{\pi}\sum_{n=0}^\infty (-1)^{n}\frac{\sin((2n+1)x)}{(2n+1)^{2}}$$
Divide both sides by $x$ and integrate:
$$\int_{0}^\infty \frac{\sin^{-1}(\sin(x))}{x}dx=\frac{4}{\pi}\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)^{2}}\int_{0}^\infty \frac{\sin((2n+1)x)}{x}dx$$
Using the fact that $\int_{0}^\infty \frac{\sin((2n+1)x)}{x}dx=\int_{0}^\infty \frac{\sin(x)}{x}dx=\frac{\pi}{2}$, we got:
$$\int_{0}^\infty \frac{\sin^{-1}(\sin(x))}{x}dx=2\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)^{2}}=2G$$
| Note that
\begin{align*}
\int_{0}^{\infty} \frac{\arcsin(\sin x)}{x} \, \mathrm{d}x
&= \frac{1}{2}\int_{-\infty}^{\infty} \frac{\arcsin(\sin x)}{x} \, \mathrm{d}x \\
&= \lim_{N\to\infty} \frac{1}{2}\int_{-N\pi}^{(N+1)\pi} \frac{\arcsin(\sin x)}{x} \, \mathrm{d}x \\
&= \lim_{N\to\infty} \frac{1}{2}\int_{0}^{\pi} \left( \sum_{n=-N}^{N} \frac{(-1)^n}{x + n \pi} \right) \arcsin(\sin x) \, \mathrm{d}x.
\end{align*}
Using the identity $ \lim_{N\to\infty} \sum_{n=-N}^{N} \frac{(-1)^n}{x + n \pi} = \frac{1}{\sin x}$, we get
\begin{align*}
\int_{0}^{\infty} \frac{\arcsin(\sin x)}{x} \, \mathrm{d}x
&= \frac{1}{2}\int_{0}^{\pi} \frac{\arcsin(\sin x)}{\sin x} \, \mathrm{d}x \\
&= \int_{0}^{\pi/2} \frac{x}{\sin x} \, \mathrm{d}x.
\end{align*}
Substituting $t=\tan(x/2)$, the last integral reduces to
\begin{align*}
\int_{0}^{\pi/2} \frac{x}{\sin x} \, \mathrm{d}x
&= \int_{0}^{\pi/2} \frac{x\sec^2(x/2)}{2\tan(x/2)} \, \mathrm{d}x \\
&= 2\int_{0}^{1} \frac{\arctan(t)}{t} \, \mathrm{d}t \\
&= 2\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_{0}^{1} t^{2n} \, \mathrm{d}t \\
&= 2\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} \\
&= 2G.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4219466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Using first principles find derivative of ln(sec(x)) The question is to use first principles only.
Thus I started with the same and got
$$
y = \ln(\sec(x))
$$
$$
\frac{dy}{dx} = \lim_{h\to 0} \frac{\ln(\sec(x+h)) - \ln(\sec(x))}{h}
$$
after this I do not understand how do I eliminate the $h$ in the denominator. I tried to implement $\ln(A) - \ln(B) = \ln\bigl(\frac{A}{B}\bigr)$ which ultimately led to
$$
\frac{dy}{dx} = \lim_{h\to 0} \frac{\ln\bigl(\frac{\sec(x+h)}{\sec(x)}\bigr)}{h}
$$
here I converted $\sec()$ to $\cos()$
$$
\frac{dy}{dx} = \lim_{h\to 0} \frac{\ln\bigl(\frac{\cos(x)}{\cos(x+h)}\bigr)}{h}
$$
Still I cannot proceed further.
| $$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(\sec(x+h)) -\ln(\sec(x))}{h}
$$
Using $\ln(A) - \ln(B) = \ln(\frac{A}{B})$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(\frac{\sec(x+h)}{\sec(x)})}{h}
$$
coverting $\sec(x)$ to $\cos(x)$ using $\cos(x) = \frac{1}{\sec(x)}$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(\frac{\cos(x)}{\cos(x+h)})}{h}
$$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x)}{\cos(x+h)}-1)}{h}
$$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x) - \cos(x+h)}{\cos(x+h)})}{h}
$$
multiplying and dividing by $\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x) - \cos(x+h)}{\cos(x+h)})}{h}\frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}
$$
repositioning
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x) - \cos(x+h)}{\cos(x+h)})}{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}\frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{h}
$$
seperating limit
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\ln(1+\frac{\cos(x) - \cos(x+h)}{\cos(x+h)})}{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}\lim_{h\to0} \frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{h}
$$
As $h$ approaches 0 so does $\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}$ as the numerator beging to approach $0$. ($\cos(x) - \cos(x)$)
let us assume $t = \frac{\cos(x)-\cos(x+h)}{\cos(x+h)}$ and hence t approaches $0$ when $h$ approaches $0$
hence equation turns out to be
$$
\frac{d}{dx}\ln\sec(x) = \lim_{t\to0} \frac{\ln(1+t)}{t} \lim_{h\to0} \frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{h}
$$
Using the standard limit $\lim_{x\to0} \frac{ln(x+1)}{x} = 1$
Therefore
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\frac{\cos(x)-\cos(x+h)}{\cos(x+h)}}{h}
$$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\cos(x)-\cos(x+h)}{h\cos(x+h)}
$$
Applying $\cos(A) - \cos(B) = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{-2\sin(\frac{2x+h}{2})\sin(\frac{-h}{2})}{h\cos(x+h)}
$$
Bringing the $-2$ down
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\sin(\frac{2x+h}{2})\sin(\frac{-h}{2})}{\frac{-h}{2}\cos(x+h)}
$$
rearranging
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\sin(\frac{2x+h}{2})}{\cos(x+h)} \lim_{h\to0} \frac{\sin(\frac{-h}{2})}{\frac{-h}{2}}
$$
Using standard limit $\lim_{x\to0} \frac{sin(x)}{x} = 1$
Therefore
$$
\frac{d}{dx}\ln\sec(x) = \lim_{h\to0} \frac{\sin(\frac{2x+h}{2})}{\cos(x+h)}
$$
Putting $h = 0$
$$
\frac{d}{dx}\ln\sec(x) = \frac{\sin(\frac{2x}{2})}{\cos(x)}
$$
$$
\frac{d}{dx}\ln\sec(x) = \frac{\sin(x)}{\cos(x)}
$$
$$
\frac{d}{dx}\ln\sec(x) = \tan(x)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4221017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Prove $\sum_{cyc}\frac{xy+1}{(x+y)^2}\geq 3$ when $x^2+y^2+z^2+(x+y+z)^2\leq 4$.
Let $x,y,z\in \Bbb{R}^+$ such that $x^2+y^2+z^2+(x+y+z)^2\leq 4$. Prove that $$\sum_{cyc}\frac{xy+1}{(x+y)^2}\geq 3.$$
As there are three fractions in the left side and a single term in the right side, I thought Cauchy-Schwarz might be of help. But from Cauchy-Schwarz, I got $$\sum_{cyc}\frac{xy+1}{(x+y)^2}\geq \frac{\left(\sum_{cyc}\sqrt{xy+1}\right)^2}{x^2+y^2+z^2+(x+y+z)^2}.$$
Then we have to prove that $$\left(\sum_{cyc}\sqrt{xy+1}\right)^2\geq12.$$
But this doesn't seem to work. So, how to solve the problem?
| We have
\begin{align*}
\sum_{\mathrm{cyc}} \frac{xy + 1}{(x + y)^2}
&\ge \sum_{\mathrm{cyc}} \frac{xy + \frac{x^2 + y^2 + z^2 + (x + y + z)^2}{4}}{(x + y)^2}\\
&= \sum_{\mathrm{cyc}} \frac{2xy + x^2 + y^2 + z^2 + xy + yz + zx}{2(x + y)^2}\\
&= \sum_{\mathrm{cyc}} \frac{(x + y)^2 + (y + z)(z + x)}{2(x + y)^2}\\
&= \frac{3}{2} + \sum_{\mathrm{cyc}} \frac{(y + z)(z + x)}{2(x + y)^2}\\
&\ge \frac32 + 3\sqrt[3]{\prod_{\mathrm{cyc}}\frac{(y + z)(z + x)}{2(x + y)^2}}\\
&= 3.
\end{align*}
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right)$
What is the limit of $\,\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right)$ ?
My thought was $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\left(\frac{1}{\sin^2 x} - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\left(\frac{1}{x^2} - \frac{1}{x^2} \right) = 0$. I used $\lim\limits_{x\to 0} \frac{\sin x}{x} = 1$.
But $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\frac{x^2-\sin^2x}{x^2\sin^2x}$ and if I apply L'Hospital's Rule four times at the second expression, I get $\lim\limits_{x\to 0}\,\left(\frac{8\cos 2x}{24\cos 2x - 32\sin 2x - 8x^2\cos 2x}\right) = \frac {1}{3}$.
What am I missing?
| As noticed in the comments and given answers, in general, we can’t take the limit only for a single part of the expression with some exceptions as discussed in detail here:
*
*Analyzing limits problem Calculus (tell me where I'm wrong).
Indeed in this case, proceeding by Taylor’s expansion we have
$$\sin^2 x=\left(x-\frac{x^3}6+o(x^3)\right)^2=x^2-\frac{x^4}{3}+o(x^4)$$
pointing out the others terms which play a role in the determining the value for the limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integers solutions to the equation $x^4 + 3y^4 = z^2$ I would like to find the integer solutions to the equation $x^4 + 3y^4 = z^2$.
I know this has infinitely solutions, because I noticed if $x=y$, we have $4x^4 = z^2 \Rightarrow 2x^2 = z$ (for positive integers). So for instance $(1,1,2), (2,2, 8), ...$ are solutions.
How to find all the integer solutions to this equation?
| The below equation has solution shown below:
$x^4 + 3y^4 = z^2$
$x=3m^2-n^2$
$y^2=4mn(3m^2+n^2)$
$z=(9m^4+18m^2n^2+n^4)$
RHS of $(y^2)$ is made a square at $(m.n)=(12,3)$ &
we get after removing common factors:
$(x,y,z)=(47,28,2593)$
Also for, $(m,n)=(1,1)$ we get:
$(x,y,z)=(1,2,7)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
The probability that two randomly selected subsets of the set $\{1 , 2, 3, 4 , 5\}$ have exactly two elements in their intersection , is The probability that two randomly selected subsets of the set $\{1 , 2, 3, 4 , 5\}$ have exactly two elements in their intersection is $\frac{135}{2^9}$.
In this problem Find the probability that two randomly selected subsets of $\{1,2,3,4,5\}$ have exactly 2 elements common in their intersection , it is said that the sample space has $2^{10}$ elements.
I can not understand how sample space contains $2^{5}\times 2^5$ elements. It should have $2^5 \choose 2$ elements. It has $\binom{2^5}{2} + 2^5$ elements if we allow two chosen subsets being identical.
| Method 1: Let's say the sets are $A$ and $B$. You have four choices for each of the five elements in the set $\{1, 2, 3, 4, 5\}$: place an element in both sets $A$ and $B$, place it only in set $A$, place it only in set $B$, or place it in neither subset. Hence, the sample space has $4^5$ elements.
Method 2: There are $2^5$ ways to select set $A$ since we can choose to include or not include each element of the set $\{1, 2, 3, 4, 5\}$ in set $A$. For each such choice, there are $2^5$ ways to select subset $B$. Since the choice of sets $A$ and $B$ is independent, there are $2^5 \cdot 2^5 = 2^{10}$ ways to choose the two subsets of the set $\{1, 2, 3, 4, 5\}$.
What is wrong with your approach?
While you did not explain your reasoning, it appears that the term
$$\binom{2^5}{2}$$
is supposed to count the number of ways of choosing two different subsets of the set $\{1, 2, 3, 4, 5\}$, while $2^5$ is supposed to count the number of ways of choosing two identical subsets.
Method 3: There are $2^5$ ways of choosing the subset $A$ and one way to select set $B$ so that it is the same as set $A$. Thus, the term $2^5$ makes sense.
For each of the $2^5$ ways we can select set $A$, there are $2^5 - 1$ ways to select set $B$ such that set $B$ is different from set $A$. Hence, there are $2^5(2^5 - 1)$ ways to pick two different subsets.
Thus, we again obtain
$$2^5 + 2^5(2^5 - 1) = 2^5 + 2^{10} - 2^5 = 2^{10}$$
elements in the sample space.
Note: The problem with using $\binom{2^5}{2}$ for the number of ways of selecting two different subsets is that there are two orders in which we could pick the same pair of distinct subsets. Notice that
$$2\binom{2^5}{2} + 2^5 = 2 \cdot \binom{32}{2} + 2^5 = 2 \cdot \frac{32 \cdot 31}{2} + 32 = 32 \cdot 31 + 32 = 992 + 32 = 1024 = 2^{10}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $x^3+y^2$ and $x^2+y^3$ are integers, show that $x$, $y$ are integers
Given rational numbers $x$, $y$ such that $x^3+y^2$ and $x^2+y^3$ are integers, prove that $x$, $y$ are integers.
For this problem I don't even know how to start tackling it. I tried various ways:
*
*Letting $x = \frac{a}{b}$, $y = \frac{c}{d}$, which only makes it more complicated
*Doing operations on them: $x^3+y^2+x^2+y^3=x^2(x+1)+y^2(y+1)$, which I have no idea how to continue;
$(x^3+y^2)(x^2+y^3)=x^5+y^5+x^3y^3+x^2y^2$, and it's also probably too complex to break down
$x^3+y^2-x^2-y^3=(x-y)(x^2+xy+y^2)+(y-x)(x+y)$ seems more plausible, but I also can't do anything with this.
I would really appreciate any way of tackling this problem, because I have spent a while thinking about this problem.
EDIT: I have found a solution to this problem (this was edited 2 days after this was posted, so there were also answers before this):
Suppose $x = \frac{a}{b}$, $y = \frac{c}{d}$ in which $a$, $b$, $c$, $d$
are integers and $\gcd(a,b) = \gcd(c,d) = 1$. Then we have
$x^2+y^3=\frac{a^2}{b^2}+\frac{c^3}{d^3}=\frac{a^2d^3+b^2c^3}{b^2d^3}$,
so:
$b^2|a^2d^3+b^2c^3$, which means $b^2|a^2d^3$, and $\gcd(a,b)=1$ so
$b^2|d^3$, and
$d^3|a^2d^3+b^2c^3$, which means $d^3|b^2$, so $b^2=d^3$.
Doing the same to $x^3+y^2$, we get $b^3=d^2$. From that we get
$b^5=d^5$, so $b=d$. Substituting to $b^2=d^3$, we get $b^2=b^3$,
hence $b=d=1$, which implies that $x$ and $y$ are integers.
| Let $x^3+y^2=a$ and $x^2+y^3=b$, where $a,b \in\mathbb Z$, then we have
$$(a-x^3)^3=(b-x^2)^2 \\ x^9-3ax^6+x^4+3a^2x^3-2bx^2-(a^3-b^2)=0$$
and
$$(a-y^2)^2=(b-y^3)^3\\
y^9-3by^6+y^4+3b^2y^3-2ay^2+(a^2-b^3)=0.$$
Then the Rational root theorem immediately tells us $x \mid a^3-b^2$ and $y \mid a^2-b^3$. This means $x,y\in\mathbb Z.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 3,
"answer_id": 1
} |
Two methods of finding out $\frac{d^2y}{dx^2}$ when $\frac{dy}{dt} = 4t-2$ and $\frac{dx}{dt} = 2t + 3$ Find $\frac{d^2y}{dx^2}$ given that $\frac{dy}{dt} = 4t-2$ and $\frac{dx}{dt} = 2t + 3$ at $t=2$
I tried two methods for this question, both of which give me different answers. Which one is wrong and why so?
Method $1$: $\frac{dy}{dt} = 4t-2$ hence $\frac{d^2y}{dt^2} = 4$, similarly $\frac{d^2x}{dt^2} = 2$ therefore, $\frac{d^2y}{dx^2} = 2$
Method $2$: $\frac{dy}{dx} = \frac{4t-2}{2t+3}$ therefore $\frac{d^2y}{dx^2} = \frac{(2t+3)4 - (4t-2)2}{(2t+3)^2}.\frac{dt}{dx} = \frac{16}{(2t+3)^2}.\frac{1}{2t+3} = \frac{16}{7^3}$
I can't understand why one method is wrong while the other is right, however they have very different answers. What am I doing wrong?
| The formula $\frac{d^2y}{dx^2} = \frac{d^2y/dt^2}{d^2 x/dt^2}$ is wrong. A correct formula for $\frac{d^2y}{dx^2}$ in terms of the derivatives of $y$ and $x$ with respect of $t$ is the following:
$$\frac{d^2y}{dx^2}=\frac{dt}{dx}\frac{d}{dt}\left(\frac{dy/dt}{dx/dt}\right)=
\frac{d^2y/dt^2\cdot dx/dt-dy/dt\cdot d^2x/dt^2}{(dx/dt)^3}.$$
Therefore
$$\frac{d^2y}{dx^2}=\frac{4(2t+3)-(4t-2)2}{(2t+3)^3}=\frac{16}{(2t+3)^3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove there is infinite number of 3 consecutive numbers which are sum of 2 squares.
Prove there is infinite number of 3 consecutive numbers which are sum of 2 squares of ($0\notin $) natural numbers. Example: $$72 = 6^2+6^2$$
$$73= 8^2+3^2$$
$$74 =7^2+5^2$$
I could only find 232,233 and 234 is next such triple. I was also thinking about to implement $$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc) ^2$$ but don't know how.
| If we have a solution where $ n-1 = a^2 + b^2, n = c^2 + d^2, n+1 = e^2 + f^2 $, with all variables non-zero, then
*
*$n$ has to be odd, because $\{n-1, n, n+1\} \equiv \{0, 1, 2\} \pmod{4}$.
*$n^2 -1 = (ae+bf)^2 + (af - be)^2$ -> If either of these are 0, then $n^2 = 1 + m^2$ which easily leads to a contradiction.
*$ n^2 = (c^2 - d^2)^2 + (2cd)^2$. -> Since $n$ is odd, so $ c \neq d$ and thus $c^2 - d^2 \neq 0$.
*$n^2 + 1 = n^2 + 1$.
Thus, we have our 3 consecutive integers
So, starting with your seed of $ n = 73$, an infinite family exists.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Determine the maximum of $f(x) = x + \sqrt{4-x^2}$ without calculus I was given this problem in a Calc BC course while we were still doing review, so using derivatives or any sort of calculus was generally forbidden. We were only doing review because a lot of people hadn't taken Calc AB due to scheduling issues so the teacher felt he should at least cover some of what would be covered in AB for a while.
Of course using calculus this problem is quite trivial. $f'(x) = x/\sqrt{4-x^2}$, set this equal to zero and the maximum occurs at ($\sqrt2$, $2\sqrt2$). But again, this method was not allowed.
I immediately recognized the $\sqrt{4-x^2}$ as a semicircle with maximum value at 2, so I knew that because x was being added, the value must be greater than 2 at the maximum. Also, the x value must be less than two.
I then attempted to rearrange the equation such that it could be written in polar coordinates but, letting $y=f(x)$ caused $x^2$ to cancel out and I was left with $y^2 = 2x\sqrt{4-x^2}$ which only seems to complicate the equation further.
I've been stuck at this point for a while as I haven't thought too much about the problem, but my teacher dismissed the problem as he didn't realize he had assigned it. Regardless, this was in the pre-calculus portion of our textbook, so I assume that there has to be a way to solve it (our textbook only gives solutions for odd problems).
| $\text {Method}-\text{1a}$
$$\begin{align}&u=x+\sqrt {4-x^2},\thinspace u-x≥0\\ \implies &(u-x)^2=4-x^2\\ \implies &u^2-2ux+2x^2-4=0\\ \implies &2x^2-2ux+(u^2-4)=0\\ \implies &\Delta=u^2-2(u^2-4)≥0\\
\implies &u^2≤8\\
\implies &|u|≤2\sqrt 2\\ \implies &\max \left\{u\right\}=2\sqrt 2. \end{align}$$
$\text {Method}-\text{1b}$
\begin{align}&u=x+\sqrt {4-x^2},\thinspace u-x≥0\\ \implies &(u-x)^2=4-x^2\\ \implies &u^2-2ux+2x^2-4=0\\
\implies &2x^2-2ux+(u^2-4)=0\\
\implies &2\left(x-\frac u2\right)^2+\frac{u^2-8}{2}=0\\
\implies &u^2-8≤0\\
\implies &\max \left\{u\right\}=2\sqrt 2. \end{align}
$\text {Method}-\text{2a}$
$$\begin{align}&u^2=4+2x\sqrt {4-x^2}\\
\implies &x\sqrt{4-x^2}=a, \thinspace a=\frac{u^2-4}{2}\\
\implies &\frac{a^2}{x^2}+x^2=4, \thinspace x≠0 \\
\implies &\left(\frac ax-x\right)^2+2a=4 \\
\implies &\left(\frac ax-x\right)^2=4-2a≥0\\
\implies &a=\frac {u^2-4}{2}≤2\\ \implies &u^2≤8\\ \implies &\max \left\{u\right\}=2\sqrt 2.
\end{align}$$
$\text {Method}-\text{2b}$
$$\begin{align}&u^2=4+2x\sqrt {4-x^2},\thinspace x≠0\\
\implies &x\sqrt{4-x^2}=a, \thinspace a=\frac{u^2-4}{2}\\
\implies &\frac{a^2}{x^2}+x^2=4≥2\sqrt{\frac{a^2}{x^2}\times x^2}=2a \\
\implies &a=\frac {u^2-4}{2}≤2\\
\implies &\max \left\{u\right\}=2\sqrt 2.
\end{align}$$
$\text {Method}-3$
$$\begin{align}&\frac {u^2-4}{2}=\sqrt{x^2(4-x^2)},\thinspace x≥0\\
\implies &\frac {u^2-4}{2}=\sqrt{4x^2-x^4}\\
\implies &\sqrt{4-(x^2-2)^2}≤2\\
\implies &\frac {u^2-4}{2}≤2\\
\implies &\max \left\{u\right\}=2\sqrt 2.
\end{align}$$
$\text {Method}-4$
$$\begin{align}&u^2-4=2\sqrt{x^2(4-x^2)}, \thinspace x≥0\\
\implies &x^2+(4-x^2)≥2\sqrt{x^2(4-x^2)}\\
\implies &u^2-4≤4\\
\implies &\max \left\{u\right\}=2\sqrt 2.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 9,
"answer_id": 5
} |
how to prove $\sum_{cyc}(1-a_{1}+a_{1}a_{2})^2\ge\frac{n}{2}$ let $a_{i}\in [0,1]$,prove or disprove
$$f_{n}=(1-a_{1}+a_{1}a_{2})^2+(1-a_{2}+a_{2}a_{3})^2+\cdots+(1-a_{n}+a_{n}a_{1})^2\ge\dfrac{n}{2}$$
I can only prove $n=3$.
$$f_{3}=\sum_{cyc}[1-a_{1}(1-a_{2})]^2=3-2\sum_{cyc}a_{1}(1-a_{2})+\sum_{cyc}a^2_{1}(1-a_{2})^2$$
let $$t=a_{1}(1-a_{2})+a_{2}(1-a_{3})+a_{3}(1-a_{1})=1-(1-a_{1})(1-a_{2})(1-a_{3})-a_{1}a_{2}a_{3}\le 1$$
and
$$\sum_{cyc}a^2_{1}(1-a_{2})^2=(\sum_{cyc}a_{1}(1-a_{2}))^2-2\sum_{cyc}a_{1}a_{2}(1-a_{1})(1-a_{3})\ge t^2-\dfrac{1}{4}\sum_{cyc}a_{2}(1-a_{3})=t^2-\dfrac{1}{4}t$$
so we have
$$f_{3}\ge 3-2t+t^2-\dfrac{1}{4}t\ge\dfrac{3}{2}$$
But for $n$ I can't,maybe use induction to prove it?
| A proof for $n=4$.
Let $a_1=\frac{1}{a+1},$ $a_2=\frac{1}{b+1},$ $a_3=\frac{1}{c+1}$ and $a_4=\frac{1}{d+1},$ where $a\geq0$, $b\geq0,$ $c\geq0$ and $d\geq0.$
Thus, $$\prod_{cyc}(a+1)^2\left(\sum_{cyc}(1-a_1+a_1a_2)^2-2\right)=$$
$$=\prod_{cyc}(a+1)^2\left(\sum_{cyc}\left(\frac{a}{a+1}+\frac{1}{(a+1)(b+1)}\right)^2-2\right)=2+2a^2b^2c^2d^2+$$
$$+\sum_{cyc}(2a^2b^2c^2d+a^2b^2c^2+2a^2b^2cd+2a^2b^2c+a^2b^2+2a^2bc-2a^2cd+2a^2b+a^2+2ab+2a)\geq$$
$$\geq\sum_{cyc}(a^2b^2c^2-2a^2cd+a^2)=\sum_{cyc}a^2(cd-1)^2\geq0.$$
A proof for $n=5$ we can get by the same way, but it's more complicated.
I found a nice proof for $n=4$.
By using the previous substitutions and by AM-GM we obtain:
$$\sum_{cyc}(1-a_1+a_1a_2)^2=\sum_{cyc}\left(\frac{ab+a+1}{(a+1)(b+1)}\right)^2\geq$$
$$\geq2\left(\frac{(ab+a+1)(cd+c+1)}{(a+1)(b+1)(c+1)(d+1)}+\frac{(bc+b+1)(da+d+1)}{(b+1)(c+1)(d+1)(a+1)}\right)$$ and it's enough to prove that:
$$(ab+a+1)(cd+c+1)+(bc+b+1)(da+d+1)\geq(a+1)(b+1)(c+1)(d+1),$$ which is $$abcd+1\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
If $a^{2}+b^{2} \leq 4$, prove that $a+b \leq 4$ \begin{equation}
\text { If } a^{2}+b^{2} \leq 4 \text { prove that } a+b \leq 4 \text { }
\end{equation} What I have tried:
\begin{equation}
a^{2} \leq a^{2}+b^{2} \leq 4 \text { and } b^{2} \leq a^{2}+b^{2} \leq 4
\end{equation}
\begin{equation}
|a| \leq 2 \text { and }|b| \leq 2
\end{equation}
\begin{equation}\text { So } a+b
\text { can get the maximum value, then } a \geq 0 \text { and } b \geq 0 \text { }
\end{equation}\begin{equation}
\text { } 0 \leq a \leq 2 \text { and } 0 \leq b \leq 2 \text {}
\end{equation}
\begin{equation}
(a-b)^{2}=a^{2}+b^{2}-2 a b \geq 0
\end{equation}\begin{equation}
a^{2}+b^{2} \geq 2 a b \text { So at this step I am not certain what to do next. }
\end{equation}
| Applying the RMS-AM inequality :
$$ ^2 \ + \ ^2 \ ≤ \ 4 \ \ \Rightarrow \ \ \frac{+}{2} \ ≤ \ \sqrt{\frac{^2 + ^2}{2}} \ ≤ \ \sqrt2 \ \ \Rightarrow \ \ + \ ≤ \ 2 · \sqrt2 \ < \ 4 \ \ . $$
Another argument: the circle of radius $ \ 2 \ $ centered on the origin and its interior, as described by $ \ x^2 + y^2 \ \le \ 4 \ \ , $ lies entirely "below" the line $ \ x + y = \ 4 \ \ . $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Are there positive integers $x$, $y$ and prime numbers $p$ so that $\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{p}$ I have a solution for this, but I'm not really sure about that:
We have: $\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{p}$, multiplying both sides by $\sqrt[3]{x^2}-\sqrt[3]{xy}+\sqrt[3]{y^2}$ we get $x+y=\sqrt[3]{p}(\sqrt[3]{x^2}-\sqrt[3]{xy}+\sqrt[3]{y^2})$ so $\sqrt[3]{px^2}-\sqrt[3]{pxy}+\sqrt[3]{py^2}$ is an integer.
Suppose $\sqrt[3]{px^2}$, $\sqrt[3]{pxy}$, and $\sqrt[3]{py^2}$ are all integers, we can easily see that $p|x$ and $p|y$.
Let $x^2=p^2a^3$, $y^2=p^2b^3$ where $a$ and $b$ are positive integers. Then we can see that $a$ and $b$ also have to be perfect squares since $x=p\sqrt{a^3}$ and $y=p\sqrt{b^3}$ are integers. Since $\gcd(2,3)=1$, we can let $x^2=p^2a'^6$, $y^2=p^2b'^6$, or $x=pa'^3$, $y=pb'^3$ where $a'$ and $b'$ are integers. Subbing that to the original equation we get $\sqrt[3]{pa'^3}+\sqrt[3]{pb'^3}=\sqrt[3]{p}$, so $\sqrt[3]{p}(a'+b')=\sqrt[3]{p}$. But $a'+b'>1$, so $\sqrt[3]{p}(a'+b')>\sqrt[3]{p}$, hence a contradiction.
I can see I made a lot of assumptions here, like "Let $x^2=p^2a^3$, $y^2=p^2b^3$", or assuming all $\sqrt[3]{px^2}$, $\sqrt[3]{pxy}$, and $\sqrt[3]{py^2}$ are integers. Are there any problems with my work or is it good to go? And moreover, do you have a better solution than this? I appreciate your time and effort for this and thanks a lot in advance.
| Recall the algebraic identity:
$$a^3+b^3+c^3 - 3abc = \frac12(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$$
Whenever $a+b+c = 0$, we have $a^3+b^3+c^3 = 3abc$.
Substitute $a,b,c$ by $\sqrt[3]{p}, -\sqrt[3]{x}, -\sqrt[3]{y}$ and let $K = p - x - y$, we find
$$K = p - x - y = 3\sqrt[3]{pxy} > 0$$
Since $x,y > 0$, $K$ is a positive integer smaller than $p$.
Taking cube on the equality on the left and using the condition $p$ is a prime, we have
$$K^3 = 27pxy \implies p | K^3 \implies p | K$$
This contradicts with above fact that $K$ is a positive integer smaller than $p$.
As a result, equation $\sqrt[3]{x} + \sqrt[3]{y} = \sqrt[3]{p}$ has not positive integer solutions for any prime $p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Evaluate : $S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7}+\frac{1}{9\cdot10\cdot11}+\cdots$ Evaluate:$$S=\frac{1}{1\cdot2\cdot3}+\frac{1}{5\cdot6\cdot7} + \frac{1}{9\cdot10\cdot11}+\cdots$$to infinite terms
My Attempt:
The given series$$S=\sum_{i=0}^\infty \frac{1}{(4i+1)(4i+2)(4i+3)} =\sum_{i=0}^\infty \left(\frac{1}{2(4i+1)}-\frac{1}{4i+2}+\frac{1}{2(4i+3)}\right)=\frac{1}{2}\sum_{i=0}^\infty \int_0^1 \left(x^{4i}-2x^{4i+1}+x^{4i+2}\right) \, dx$$
So,$$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4} + \frac{x^2}{1-x^4}\right)dx=\frac{1}{2} \int_0^1 \left(\frac{1+x^2}{1-x^4}-\frac{2x}{1-x^4}\right)\,dx = \frac{1}{2} \int_0^1 \left(\frac{1}{1-x^2}-\frac{2x}{1-x^4}\right)\,dx$$
$$=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\int_{0}^{1}\frac{2x}{1-x^4}dx=\frac{1}{2}\int_{0}^1\frac{1}{1-x^2}dx-\frac{1}{2}\int_{0}^{1}\frac{1}{1-y^2}dy=0(y=x^2)$$
which is obviously absurd since all terms of $S$ are positive.
But if I do like this then I am able to get the answer, $$S=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1-x^4}-\frac{2x}{1-x^4}+\frac{x^2}{1-x^4}\right)dx=\frac{1}{2}\int_{0}^{1}\frac{(1-x)^2}{1-x^4}dx=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{1+x}-\frac{x}{1+x^2}\right)dx=\frac{\ln2}{4}$$
What is wrong with the previous approach
| Thinking about generalized harmonic numbers
$$\frac{1}{(4 i+1) (4 i+2) (4 i+3)}=\frac{1}{2 (4 i+1)}+\frac{1}{2 (4 i+3)}-\frac{1}{(4i+2)}$$
$$S_n=\sum_{i=0}^n\frac{1}{(4 i+1) (4 i+2) (4 i+3)}$$
$$\sum_{i=0}^n \frac{1}{ (4 i+1)}=\frac{1}{4} \left(\psi \left(n+\frac{5}{4}\right)-\psi \left(\frac{1}{4}\right)\right)$$
$$\sum_{i=0}^n \frac{1}{ (4 i+3)}=\frac{1}{4} \left(\psi \left(n+\frac{7}{4}\right)-\psi
\left(\frac{3}{4}\right)\right)$$
$$\sum_{i=0}^n \frac{1}{ (4 i+2)}=\frac{1}{4} \left(\psi \left(n+\frac{3}{2}\right)-\psi \left(\frac{1}{2}\right)\right)$$
$$S_n=\frac{1}{4} \left(H_{2 n+\frac{3}{2}}-H_{n+\frac{1}{2}}\right)$$ Using the asymptotics, then
$$S_n=\frac{\log (2)}{4}-\frac{1}{128 n^2}+O\left(\frac{1}{n^3}\right)$$
Thinking about the gaussian hypergeometric function
$$\sum_{i=0}^\infty\frac{x^{2i}}{(4 i+1) (4 i+2) (4 i+3)}=\frac{3 x \, _2F_1\left(\frac{1}{4},1;\frac{5}{4};x^2\right)+x \,
_2F_1\left(\frac{3}{4},1;\frac{7}{4};x^2\right)-3 \tanh ^{-1}(x)}{6 x}$$ and, if $x \to 1$, the limit.
Just, for the fun, notice that for $x=0$, the result is already $\frac 16=0.1667$ not "so far" from $\frac{\log(2)}4=0.1733$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
} |
Prove that: $a^{12}+b^{12}+c^{12} \geq \frac{2049}{8}\cdot a^4b^4c^4 $ Let $a,b,c$ be real and $a+b+c=0.$
Prove that:$$a^{12}+b^{12}+c^{12} \geq \frac{2049}{8}\cdot a^4b^4c^4$$
I know we can make the following:
Let $bc\geq0$ thus we need to prove
$$b^{12}+c^{12}+(b+c)^{12}\geq \frac{2049}{8} b^4c^4(b+c)^4$$or
$$\frac{(2b+c)^2(b+2c)^2(b-c)^2(4b^6+12b^5c+99b^4c^2+178b^3c^3+99b^2c^4+12bc^5+4c^6)}{8}\geq 0\text{,}$$which is true.
But I'm trying to find another solution. Can be?
| Let $ab\leq0$.
Thus, by AM-GM we obtain:
$$a^{12}+b^{12}+c^{12}-\frac{2049}{8}a^4b^4c^4=$$
$$=3a^4b^4c^4+2(ab+ca+bc)^6-24a^2b^2c^2(ab+ac+bc)^3-\frac{2049}{8}a^4b^4c^4=$$
$$=\frac{1}{8}(16(ab+ac+bc)^6-192(ab+ac+bc)^3a^2b^2c^2-2025a^4b^4c^4)=$$
$$=\frac{1}{8}(4(ab+ac+bc)^3-75a^2b^2c^2)(4(ab+ac+bc)^3+27a^2b^2c^2)=$$
$$=\frac{1}{8}(4(a^2+ab+b^2)^3+75a^2b^2c^2)(4(a^2+ab+b^2)^3-27a^2b^2c^2)=$$
$$=\frac{1}{8}(4(a^2+ab+b^2)^3+75a^2b^2c^2)\left(4\left((a+b)^2+2\left(-\frac{ab}{2}\right)\right)^3-27a^2b^2c^2\right)\geq$$
$$\geq\frac{1}{8}(4(a^2+ab+b^2)^3+75a^2b^2c^2)\left(4\left(3\sqrt[3]{(a+b)^2\left(-\frac{ab}{2}\right)^2}\right)^3-27a^2b^2c^2\right)=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Area enclosed between $2$ curves
Question: The area enclosed by the curves $$ y^2 = x \ \text{and} \ y^2 = \ 3x - 1, \ \ \ \ \ \text{where} \ \ 0\leq \ x\leq \frac{1}{2} \ $$
My work:
To begin with I re-arranged the curves in terms of $y$ to the power $1$
$$y_{1} = \sqrt{x}$$
and
$$y_{2} = \sqrt{3x-1}$$
and then proceeded to take the integral for both over the interval $0\leq \ x\leq \frac{1}{2}$
$$f(x_{1}) = \frac{1}{3\sqrt{2}}$$
and
$$f(x_{2}) = \frac{5^{\frac{3}{2}}}{9\sqrt{2}} - \frac{2}{9}$$
And therefore the area between the curves is $\displaystyle\int f(x_{1}) - f(x_{2})$ right?
Is this the correct method? Because the answer that was provided to me was $\frac{2\sqrt{2}}{9}$.
| I would integrate with respect to y.
$x = y^2\\
x = \frac {y^2}{3} + \frac 13$
The curves intersect at $(\frac 12, \frac {\sqrt {2}}{2})$ and $(\frac 12, -\frac{\sqrt {2}}{2})$
$\int_{-\frac {\sqrt {2}}{2}}^{\frac {\sqrt 2}{2}} \frac {y^2}{3} + \frac 13 - y^2\ dy$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4238371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sum of series with changing ratio I have a series of the type
$$\frac{1}{2} + \frac{1}{2} \cdot \frac{3}{4} + \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} + \dots $$
The series has a changing ratio $= \frac{2k-1}{2k}$.
The ratio in itself is increasing but always less than $1$.
Q. Is the sum of the series finite? What is the sum of the series?
I have programmed it in Matlab, but the result has not converged even after $10^{13}$ terms.
The difference keeps on decreasing, but the rate of decrease in difference is also decreasing.
Any reference to literature is also appreciated.
Thank you for your time !
| By Stirling's approximation, $n!\approx\sqrt{2\pi n}\left(\frac ne\right)^n$. On the other hand,\begin{align}\frac12\times\frac34\times\cdots\times\frac{2n-1}{2n}&=\frac{(2n-1)!!}{(2n)!!}\\&=\frac{(2n-1)!!\times2\times4\times\cdots\times(2n)}{2\times4\times\cdots\times(2n)\times2\times4\times\cdots\times(2n)}\\&=\frac{(2n)!}{2^{2n}(n!)^2}.\end{align}Putting all this together, you get that\begin{align}\frac12\times\frac34\times\cdots\times\frac{2n-1}{2n}&\approx\frac{\sqrt{4\pi n}\left(\frac{2n}e\right)^{2n}}{2^{2n}2\pi n\left(\frac ne\right)^{2n}}\\&=\frac1{\sqrt{\pi n}}.\end{align}Since the series $\displaystyle\sum_{n=1}^\infty\frac1{\sqrt{\pi n}}$ diverges, then so does your series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4240109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Number of matrices $A$ in $M_2(\mathbf{Z}/3\mathbf{Z})$ satisfying $A^{50}=I.$
Find the number of $2$x$2$ matrices $A$ with entries in $\mathbf{Z}/3\mathbf{Z}$ satisfying $A^{50}=I$.
Propositions as far as I know:
*
*The number of $2$x$2$ matrices with entries in $\mathbf{Z}/3\mathbf{Z}$ is $81$.
*$A^{50}=I$ implies that the eigenvalues of $A$ are roots of $\lambda^{50}=1.$
*In $\mathbf{Z}/3\mathbf{Z}$, if $A$ is diagonalizable and $\det(A)\neq 0$, then $A^{50}=I.$
*$A^{50}=I$ implies $\det(A)\neq 0$
Using prop. $3$ I get the following $14$ matrices satisfying $A^{50}=I$ (at least $14$ matrices, maybe
more).
But I'm not sure whether the condition in prop. 3 is necessary? Note that $\textrm{char} (\mathbf{Z}/3\mathbf{Z})\neq 0$.
| Consider: A matrix $A=\begin{pmatrix}a & b \\ c & d \end{pmatrix}$ is not a unit exactly if $ad - bc=0$. How many possibilities does this have?
*
*$a=0,b=0$ (9 possibilies)
*$d=0,c=0$ and $a\neq 0$ or $b\neq 0$ (8 possibilities)
*$a=0,c=0$ and $b\neq0, d\neq 0$ (4 possibilities)
*$d=0,b=0$ and $a\neq0,c\neq 0$ (4 possibilities)
*$ad=1,bc=1$ (4 possibilities)
*$ad=2,bc=2$ (4 possibilities)
for a total of $9+8+4+4+4+4=33$ possibilities.
Thus we have $3^4 - 33 = 48$ matrices that are units. The multiplicative order of such a matrix thus has to divide $48$, so it divides $50$ if and only if it divides the $\gcd(48,50)=2$.
Thus we only need to consider matrices with multiplicative order 1 or 2.
Order 1 is trivially only possible for $E$. So we only need to solve:
$$ A^2 = E$$
(this also includes $A=E$).
This is equivalent to
$$ a^2+bc = 1 = cb+d^2 $$
$$ ab+bd = 0 = ca+dc $$
Thus if $a=0$ then $bc=1$ (thus $b=c=\pm1$), $d=0$. So we get
$\begin{pmatrix} 0 & \pm 1 \\ \pm 1& 0 \end{pmatrix}$ ($\pm$ linked, 2 possibilites)
If $a=\pm1$ then $a^2=1$, so $bc=0$, thus $d=\pm1$. Then either $b=0$ or $c=0$. Assume $b=0$ ($c=0$ analogue). So $0=ca+dc$ if either $c=0$ or $a+d=0$. Thus either $c=0$ or $a=-d$. This gives us
*
*$\begin{pmatrix} \pm1 & 0 \\ 0& \pm1 \end{pmatrix}$ ($\pm$ not linked, 4 possibilites)
*$\begin{pmatrix} \pm 1 & 0 \\ c & \mp 1 \end{pmatrix}$ ($c=1,2$, $\pm$ linked, 4 possibilities)
*$\begin{pmatrix} \pm 1 & b \\ 0& \mp 1 \end{pmatrix}$ (similar 4 possibilities)
So we have a total of $2+4+4+4=14$ such matrices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A formula for $1^m+2^m+3^m+\ldots+n^m$ using binomial coefficients It is known that
$$
\sum_{k=1}^{n}k^1=\binom{n+1}{2}
$$
and
$$
\sum_{k=1}^{n}k^2=\binom{n+1}{2}+2\binom{n+1}{3}
$$
Is there a formula for
$$
\sum_{k=1}^{n}k^m,
$$
where $m$ is a positive integers, using binomial coefficients of the form $\binom{n+1}{k}$ ?
| Note
$$
x^m = \sum_{k=0}^m S_{m,k}\binom{x}{k}
\tag1$$
where
$$
S_{m,k} = S_2(m,k) k!
$$
and $S_2(m,k)$ is the Stirling number of the second kind.
For $m \ge 1$ we have $S_2(m,0) = 0$, so
$$
x^m = \sum_{k=1}^m S_{m,k}\binom{x}{k}
\tag2$$
For example
$$
x^5 = \binom{x}{1} +30\binom{x}{2} +150\binom{x}{3} +240\binom{x}{4} +120\binom{x}{5}
$$
From $(2)$ we get
$$
\sum_{x=1}^{n-1} x^m = \sum_{x=1}^n \sum_{k=1}^m S_{m,k}\binom{x}{k}
= \sum_{k=1}^m S_{m,k} \sum_{x=1}^n \binom{x}{k}
\tag3$$
Now
$$
\sum_{x=1}^n \binom{x}{k} = \binom{n+1}{k+1}\qquad\text{if } k \ge 1
$$
so from $(3)$ we have our answer: if $m \ge 1$ then
$$
\sum_{x=1}^{n} x^m = \sum_{k=1}^m S_{m,k} \binom{n+1}{k+1}
\tag4$$
Recall:
$S_{m,k} = S_2(m,k) k!$
where $S_2(m,k)$ is the Stirling number of the second kind.
For example,
$$
\sum_{x=1}^n x^5 = \binom{n+1}{2} +30\binom{n+1}{3} +150\binom{n+1}{4} +240\binom{n+1}{5} +120\binom{n+1}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using logarithmic differentiation or otherwise, differentiate $y = (x −1) (x − 2) (x − 3$)? And show that $y' = 3x^2-12x+11$ So far this is my method:
\begin{align*}
y = (x-1)(x-2)(x-3) & \Longleftrightarrow \ln(y) = \ln(x-1) + \ln(x-2) + \ln(x-3)\\\\
& \Longleftrightarrow \frac{y'}{y} = \frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3}\\\\
& \Longleftrightarrow
y' = y\left(\frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3}\right)
\end{align*}
| You are almost done.
Recall $y = (x-1)(x-2)(x-3)$.
So $y{'} = (x-1)(x-2)(x-3)(\frac{1}{x-1}+\frac{1}{x-2}+\frac{1}{x-3}) = (x-2)(x-3)+(x-1)(x-3)+(x-1)(x-2) = 3x^2-12x+11.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find incremental n-determinant with all ones on main diagonal and (almost) similar rows? So, the question is in the title.
$$\begin{vmatrix}
1 & 2 & 3 & 4 & \cdots & n - 1 & n \\
1 & 1 & 3 & 4 & \cdots & n - 1 & n \\
1 & 2 & 1 & 4 & \cdots & n - 1 & n \\
1 & 2 & 3 & 1 & \cdots & n - 1 & n \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 2 & 3 & 4 & \cdots & 1 & n \\
1 & 2 & 3 & 4 & \cdots & n - 1 & 1 \\
\end{vmatrix}$$
This determinant equals
$$(-1)^{n}\cdot(n - 1)!$$
But how can you find it?
| As mentioned by nejimban in the comments, we can make the matrix upper triangular and use the fact that the determinant is invariant under row operations. However, it should be $(-1)^{n-1}(n-1)!$ instead of $(-1)^n(n-1)!$
$$\begin{vmatrix}
1 & 2 & 3 & 4 & \cdots & n - 1 & n \\
1 & 1 & 3 & 4 & \cdots & n - 1 & n \\
1 & 2 & 1 & 4 & \cdots & n - 1 & n \\
1 & 2 & 3 & 1 & \cdots & n - 1 & n \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 2 & 3 & 4 & \cdots & 1 & n \\
1 & 2 & 3 & 4 & \cdots & n - 1 & 1 \\
\end{vmatrix}\!\begin{aligned}
\ \overset{r_2-r_1}{\longrightarrow} \
\end{aligned}\begin{vmatrix}
1 & 2 & 3 & 4 & \cdots & n - 1 & n \\
0 & -1 & 0 & 0 & \cdots & 0 & 0 \\
1 & 2 & 1 & 4 & \cdots & n - 1 & n \\
1 & 2 & 3 & 1 & \cdots & n - 1 & n \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 2 & 3 & 4 & \cdots & 1 & n \\
1 & 2 & 3 & 4 & \cdots & n - 1 & 1 \\
\end{vmatrix}\!\begin{aligned}
\ \overset{r_3-r_1}{\longrightarrow} \
\end{aligned}$$
$$\begin{vmatrix}
1 & 2 & 3 & 4 & \cdots & n - 1 & n \\
0 & -1 & 0 & 0 & \cdots & 0 & 0 \\
0 & 0 & -2 & 0 & \cdots & 0 & 0 \\
1 & 2 & 3 & 1 & \cdots & n - 1 & n \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 2 & 3 & 4 & \cdots & 1 & n \\
1 & 2 & 3 & 4 & \cdots & n - 1 & 1 \\
\end{vmatrix}\!\begin{aligned}
\ \overset{}{\longrightarrow} \
\end{aligned}\begin{vmatrix}
1 & 2 & 3 & 4 & \cdots & n - 1 & n \\
0 & -1 & 0 & 0 & \cdots & 0 & 0 \\
0 & 0 & -2 & 0 & \cdots & 0 & 0 \\
0 & 0 & 0 & -3 & \cdots & 0 & 0 \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
0 & 0 & 0 & 0 & \cdots & 2-n & 0 \\
0 & 0 & 0 & 0 & \cdots & 0 & 1-n \\
\end{vmatrix}\!\begin{aligned}
\ \
\end{aligned}$$
and since the determinant of a triangular matrix is the product of its diagonal elements we have
$$\begin{vmatrix}
1 & 2 & 3 & 4 & \cdots & n - 1 & n \\
1 & 1 & 3 & 4 & \cdots & n - 1 & n \\
1 & 2 & 1 & 4 & \cdots & n - 1 & n \\
1 & 2 & 3 & 1 & \cdots & n - 1 & n \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
1 & 2 & 3 & 4 & \cdots & 1 & n \\
1 & 2 & 3 & 4 & \cdots & n - 1 & 1 \\
\end{vmatrix}$$
$$=1\cdot(-1)\cdot(-2)\cdot(-3)\cdot\cdot\cdot(2-n)(1-n)=(-1)^{n\color{red}{-1}}\cdot(n-1)!$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If Alice gives Bob $m$ candies, then he'll have $n$ times her candies; if Bob gives Alice $n$ candies, then she'll have $m$ times his candies. I came up with a seemingly innocent problem of recreational mathematics by myself. It goes likes this.
Alice and Bob have some different amount of candies ($>1$ each). If Alice gives Bob $m$ candies, then Bob has $n$ times the candies remaining to Alice, and if Bob gives Alice $n$ candies, then Alice has $m$ times the candies remaining to Bob.
How many candies have Alice and Bob?
I expected infinite solutions, but running a simple Python code (up to 200 iterations) gave me only 4 answers:
$$\begin{array}{cccc}
\text{A} & \text{B} & m & n \\
\hline
7 & 5 & 3 & 2 \\
14 & 4 & 8 & 2 \\
11 & 5 & 7 &3 \\
11 & 7 & 8 & 5
\end{array}$$
Apparently the highest amount of candies that Alice/Bob can have is 14.
Does this limit actually exist? If so, how can I prove that?
| If we solve the system of equations $$\begin{cases}b+m = n(a-m) \\ a+n = m(b-n)\end{cases}$$ for $a,b$ in terms of $m,n$ we get $a = m+1 + \frac{m^2+m+n+1}{mn-1}$ and $b = n+1 + \frac{n^2+m+n+1}{mn-1}$. By symmetry, we may assume $m \le n$.
We want $\frac{m^2+m+n+1}{mn-1}$ to be an integer. This is easiest to achieve when $\frac{m^2+m+n+1}{mn-1} = 1$. In this case, the equation $m^2+m+n+1 = mn-1$ gives us $n = m + 2 + \frac{4}{m-1}$, which only works out when $m-1$ is $1$, $2$, or $4$. These correspond to $(m,n) = (2,8)$, $(3,7)$, and $(5,8)$.
It remains to check the cases where $\frac{m^2+m+n+1}{mn-1} \ge 2$. Since $m^2 \le mn$, this requires $m+n+1\ge mn-2$, or $(m-1)(n-1) \le 4$. This is true when $m=1$, or for the pairs $(m,n) = (2,2)$, $(2,3)$, $(2,4)$, $(2,5)$, and $(3,3)$. Moreover, when $m=1$, we have $a = 3 + \frac{4}{n-1}$, limiting us to $n=2$, $3$, and $5$.
We can check these finitely many cases and get only the following possibilities for $(a,b)$: $(4, 14)$, $(5, 7)$, $(5, 11)$, $(6, 6)$, and $(7, 11)$ (together with the symmetric cases where $a>b$). Some of these work with multiple values of $m$ and $n$: for example, $a=b=6$ works either with $m=n=2$ or with $m=n=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Find $\int\frac{x+1}{x^2+x+1}dx$ Find
$\int\frac{x+1}{x^2+x+1}dx$
$\int \frac{x+1dx}{x^2+x+1}=\int \frac{x+1}{(x+\frac{1}{2})^2+\frac{3}{4}}dx$
From here I don't know what to do.Write $(x+1)$ = $t$?
This does not work.Use partial integration?I don't think it will work here.
And I should complete square then find.
| Write $x+\frac{1}{2} = t.$ Then,
$$\frac{x+1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}} = \frac{x+\frac12 + \frac12}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}} = \frac{t}{t^2+\frac{3}{4}} + \frac{\frac12}{t^2+\frac{3}{4}}.$$
Can you integrate these two terms now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
The Residual of $\frac{1}{(z-1)(z+2)z}$ I am calculating the residual of $f(z)=\frac{1}{(z-1)(z+2)z}$ on $z=1$
Since it is a first order pole, It is easy to know that $Res(z=1)=\lim_{z\to1}(z-1)f(z)=\frac{1}{3}$
But when I calculate the Laurent Series:
$f(z)=\frac{1}{(z-1)(z+2)z}=\frac{1}{z-1}(\frac{1}{z}-\frac{1}{z+2})\frac{1}{2}$
and $\frac{1}{z}=\frac{1}{(z-1)+1}=\frac{1}{z-1}\frac{1}{1+\frac{1}{z-1}}=\frac{1}{z-1}\{1-\frac{1}{z-1}+\frac{1}{(z-1)^2}+\dots\}$
$\frac{1}{z+2}=\frac{1}{(z-1)+3}=\frac{1}{3}\frac{1}{1+\frac{z-1}{3}}=\frac{1}{3}\{1-\frac{z-1}{3}+\frac{(z-1)^2}{3^2}-\dots\}$
Therefore:$f(z)=\frac{1}{2}\{\frac{1}{(z-1)^2}-\frac{1}{(z-1)^3}+\dots\}-\frac{1}{6}\{\frac{1}{z-1}-\frac{1}{3}+\frac{z-1}{3^2}-\dots\}$
Then the coefficient of $\frac{1}{z-1}$ is $-\frac{1}{6}$. I remember that the residual of $z=z_0$ is actually the the coefficient $a_{-1}$ in the Laurent Series $f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$. What mistakes did I make?
| The series you wrote for $\frac 1 z$ is valid only when $|\frac 1 {z-1}| <1$ or $|z-1| >1$. The correct expansion of $\frac 1z$ is $1-(z-1)+(z-1)^{2}+\cdots$ valid for $|z-1|<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Summation notation $ 1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \ldots + \frac{1}{1+2+3 + \ldots +n} + \ldots $ How to write the sum
$$
1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \ldots + \frac{1}{1+2+3 + \ldots +n} + \ldots
$$
in summation (∑) notation.There are 2 series here, one the entire 1 + (1/1+2)... series and the other one in denominator of each term. This summation is a sub-part of a Induction problem.
| $$\sum_{i=1}^n \frac{1}{\sum_{j=1}^i j}$$
You can calculate $a_i=\sum_{j=1}^i j$ first, then represent $\frac{1}{a_i}=b_{i}-b_{i-1}$, then $\sum_{i=1}^n (b_i-b_{i-1})=b_n-b_0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $n=2^{10}\times 3^5$. Find number of divisors of $n^2$ which are less than $n$ but do not divide $n$
If $N=2^{10}\times 3^5$. Find number of divisors of $n^2$ which are less than $n$ but do not divide $n$
My solution:
$$n^2=2^{20}\times 3^{10}$$
Factors of $n^2=\left(20+1\right)(10+1)=21\times 11=231$
Factors which are greater than or equal to $n$ can be found as we have choices for power of 2 $\in [10,11,\dots 20]$ and that of 3 $\in[5,6,\dots10]$
We get the factors of $n^2$ which are greater than or equal to $n$ as $11\times 6=66$
Factors of $n$ which are less than $n=(10+1)(5+1)-1=65$ ($-1$ as $n$ has been counted in the above condition)
$\therefore$ factors which satisfy the condition are $231-66-65=100$
But the answer is $50$. What is the correct way to solve it and what is the error in my solution?
| Hints:
*
*How many factors of $n$ are there?
*How many factors of $n^2$ are there? OP stated 231.
*How many factors of $n^2$ are there, which are $\leq n$?
$\frac{231+1}{2} = 116$. (But, why?)
*Show that the factors of $n^2$ that are $\leq n$, but not factors of $n$, is the difference between 2 of the above values.
$116 -66 = 50.$ (But, why?)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integral of weird function I am trying to solve the following integral
\begin{equation}
{\cal I}=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}\frac{2\tau}{1+\omega^{2}\tau^{2}}\frac{\sin^{2}\left[\left(\omega+\Omega\right)t/2\right]}{\left[\left(\omega+\Omega\right)/2\right]^{2}}
\end{equation}
for finite $\Omega$. I could only solve it for $\Omega=0$, yielding
\begin{equation}
{\cal I}=2 \left[\tau^{2}\left(e^{-t/\tau}-1\right)+\tau \thinspace t\right]
\end{equation}
I also tried to solve using the fact below
\begin{equation}
{\cal I}=\int_{0}^{t}\int_{0}^{t}dt_{1}dt_{2} e^{-|t_1-t_2|/\tau}e^{i\Omega\left(t_{1}-t_{2}\right)}=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}\frac{2\tau}{1+\omega^{2}\tau^{2}}\frac{\sin^{2}\left[\left(\omega+\Omega\right)t/2\right]}{\left[\left(\omega+\Omega\right)/2\right]^{2}}
\end{equation}
although this did not help much. Any thoughts?
| After some minor simplification, we have for the antiderivative
$$I=\frac{4\tau}\pi\int \frac{\sin ^2\left(\frac{1}{2} t (\omega +\Omega )\right)}{\left(\tau ^2 \omega
^2+1\right) (\omega +\Omega )^2}\,d\omega$$ Using $x=\frac{t (\omega +\Omega )}2$ gives
$$\int \frac{\sin ^2\left(\frac{1}{2} t (\omega +\Omega )\right)}{\left(\tau ^2 \omega
^2+1\right) (\omega +\Omega )^2}\,d\omega=\frac {t^3}{8\tau^2}\int \frac{\sin^2(x) }{x^2(x^2+ax+b) }\,dx$$ with $a=-t \Omega$ and $b=\frac{t^2 \left(\tau ^2 \Omega ^2+1\right)}{4 \tau ^2}$.
Let $r$ and $s$ be the complex roots of the quadratic and use partial fraction decomposition
$$\frac 1{x^2(x-r)(x-s)}=\frac{r+s}{r^2 s^2 x}+\frac{1}{r^2 (r-s) (x-r)}-\frac{1}{s^2 (r-s)
(x-s)}+\frac{1}{r s x^2}$$ which make that we face two kinds of integratels
$$J=\int \frac{\sin^2(x) }{x^2} \,dx=\text{Si}(2 x)+\frac{\cos (2 x)-1}{2 x}$$
$$K_c=\int \frac{\sin^2(x) }{x-c} \,dx=-\frac{1}{2} (\cos (2 c) \text{Ci}(2 x-2 c)+\sin (2 c) \text{Si}(2 c-2 x)-\log
(x-c))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Limit of a function with 2 variables of the indeterminate form 0/0 For my vector calculus course, I need to solve this limit: $$\lim_{(x,y)\rightarrow (0,0)} \frac{1-\cos(x^2y)}{x^6+y^4}$$
By replacing (x,y) by (0,0), I noticed that the limit was of the form $\frac{0}{0}$ and decided to Taylor expand the numerator which gave me this inequality: $$\left | 1-\cos(x^2y) \right |\leq\frac{x^4y^2}{2}$$
Then I rewrote my limit using the inequality I just found: $$\left | \frac{1-\cos(x^2y)}{x^6+y^4} \right |\leq \frac{x^4y^2}{2(x^6+y^4)}\leq \frac{x^4y^2}{x^6+y^4}$$
Seeing that I was stuck, I decided to rewrite my expression in polar coordinates: $$\left | \frac{1-\cos(x^2y)}{x^6+y^4} \right |\leq\frac{(r\cos(\theta))^4(r\sin(\theta))^2}{(r\cos(\theta))^6+(r\sin(\theta))^4}$$
By making some simplifications, I get: $$\frac{(r\cos(\theta))^4(r\sin(\theta))^2}{(r\cos(\theta))^6+(r\sin(\theta))^4}=\frac{r^2\cos^4(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}$$
And when I take the limit of my new expression when r tends to 0, I get: $$\lim_{r\to 0} \frac{r^2\cos^4(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)} = \frac{0}{\sin^4(\theta)}=0$$ unless $\theta=k\pi$ where $k \epsilon \mathbb{Z}$
First, I wanted to know if my proof was correct and if it is not, where was my mistake. Second, I wanted to know if there was a proof without the condition that $\theta=k\pi$.
Thank you in advance for your answers,
IsaacM
| One possible variant to do is, that knowing $1-\cos \alpha=2\sin^2 \frac{\alpha}{2}$ we have
$$\frac{1-\cos(x^2y)}{x^6+y^4} = \frac{2\sin^2 \frac{x^2y}{2}}{x^6+y^4}$$
So we can consider
$$\frac{x^4y^2}{x^6+y^4} \leqslant \frac{x^4y^2}{2|x|^3y^2}=\frac{1}{2}x\cdot\text{sign}(x)\to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4259404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving $3z^2 + (2+3i)z + (5i-5) = 0$ In a Precalculus book, Complex numbers chapter, I am given the following exercise:
Solve the equation $3z^2 + (2+3i)z + (5i-5) = 0$, giving your answers in the form $z = a + bi$
From the way that the exercise is shown, I thought about using the quadratic formula to solve it. Starting with the discriminant, I found
$$\Delta = (2+3i)^2 - 4 \cdot 3 \cdot (5i-5) = 55-48i$$
If I could find the root of that complex number, I could finish this exercise. But how could I do that? (Wolfram gives be $8-3i$ as an answer for that root)
(by the way, this exercise comes before the root finding method for complex numbers)
| Use de Moivre's formula: $\Delta = 55 - 48i$ is the complex number with magnitude $\sqrt{55^2 + (-48)^2} = 73$ and whose argument $\theta$ satisfies $\cos \theta = \frac{55}{73}, \sin\theta = -\frac{48}{73}$.
Therefore, $\sqrt{\Delta}$ is the complex number with magnitude $\sqrt{73}$, and whose argument is either $\frac{1}{2} \theta$ or $\frac{1}{2} \theta + \pi$. Now, since $\theta$ is in the fourth quadrant, if we take the choice of $\theta$ where $-\frac{\pi}{2} < \theta < 0$, then $\frac{1}{2} \theta$ is also in the fourth quadrant. Now,
$$\cos\left(\frac{1}{2} \theta\right) = \sqrt{\frac{1 + \cos{\theta}}{2}} = \sqrt{\frac{64}{73}} = \frac{8}{\sqrt{73}}$$
and
$$\sin\left(\frac{1}{2} \theta\right) = -\sqrt{\frac{1 - \cos{\theta}}{2}} = -\sqrt{\frac{9}{73}} = -\frac{3}{\sqrt{73}}.$$
Therefore, one of the square roots is $\sqrt{73} \cdot ( \cos(\theta/2) + i \sin(\theta/2) ) = 8 - 3i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove $ \sum_{n=1}^{2q-1}\frac{n}{q}\sin\left(\frac{\pi np}{q}\right)=-\cot\left(\frac{\pi p}{q}\right)-\csc\left(\frac{\pi p}{q}\right) $ For $p$ and $q$ are positive integers, $p < q$.
How to prove this identity?
$$
\sum_{n=1}^{2q-1}\frac{n}{q}\sin\left(\frac{n\pi p}{q}\right)=-\cot\left(\frac{\pi p}{q}\right)-\csc\left(\frac{\pi p}{q}\right)=-\cot\left(\frac{\pi p}{2q}\right)
$$
This identity is comprised by two parts.
$$
\sum_{n=1}^{q-1}\frac{2n}{q}\sin\left(\frac{2n\pi p}{q}\right)=-\cot\left(\frac{\pi p}{q}\right)\\
\sum_{n=1}^{q}\frac{(2n-1)}{q}\sin\left(\frac{(2n-1)\pi p}{q}\right)=-\csc\left(\frac{\pi p}{q}\right)
$$
I found these identities during the calculations when proving the other identity in my previous question by comparison. I am curious how to prove the identities in general ways.
I also found some other identities all by comparison like:
For $q$ is odd
$$
\sum_{n=1}^{(q-1)/2}\frac{2n}{q}\sin\left ( \frac{2n\pi p}{q} \right )=\frac{(-1)^{p-1}}{2}\csc\left( \frac{\pi p}{q} \right ) \\
\sum_{n=1}^{(q-1)/2}\frac{(2n-1)}{q}\sin\left ( \frac{(2n-1)\pi p}{q} \right )=\frac{(-1)^{p-1}}{2}\cot\left( \frac{\pi p}{q} \right ) \\
\sum_{n=1}^{(q-1)/2}2\sin\left ( \frac{2n\pi p}{q} \right )=\cot\left( \frac{\pi p}{q} \right )+(-1)^{p-1}\csc\left( \frac{\pi p}{q} \right ) \\
\sum_{n=1}^{(q-1)/2}2\sin\left ( \frac{(2n-1)\pi p}{q} \right )=\csc\left( \frac{\pi p}{q} \right )+(-1)^{p-1}\cot\left( \frac{\pi p}{q} \right )
$$
For $q$ is even
$$
\sum_{n=1}^{q/2}\frac{2n}{q}\sin\left ( \frac{2n\pi p}{q} \right )=\frac{(-1)^{p-1}}{2}\cot\left( \frac{\pi p}{q} \right ) \\
\sum_{n=1}^{q/2}\frac{(2n-1)}{q}\sin\left ( \frac{(2n-1)\pi p}{q} \right )=\frac{(-1)^{p-1}}{2}\csc\left( \frac{\pi p}{q} \right ) \\
\sum_{n=1}^{q/2}2\sin\left ( \frac{2n\pi p}{q} \right )=\cot\left( \frac{\pi p}{q} \right )+(-1)^{p-1}\cot\left( \frac{\pi p}{q} \right ) \\
\sum_{n=1}^{q/2}2\sin\left ( \frac{(2n-1)\pi p}{q} \right )=\csc\left( \frac{\pi p}{q} \right )+(-1)^{p-1}\csc\left( \frac{\pi p}{q} \right )
$$
How to prove them in general ways?
| Thanks to Mr. metamorphy's answer. I rearrange the proof to make it suitable for my question.
Let $\omega =e^{2\pi pi/q}$
\begin{align}
(\omega-1)\sum_{n=1}^{q}n\omega^n&=(\omega-1)\left(\sum_{n=1}^{q}\omega^n+\sum_{n=2}^{q}\omega^n+\sum_{n=3}^{q}\omega^n+\cdots \sum_{n=q}^{q}\omega^n \right) \\
&=\omega(\omega^q-1)+\omega^2(\omega^{q-1}-1)+\omega^3(\omega^{q-2}-1)-1)+\cdots+\omega^q(\omega-1) \\
&=q\omega^{q+1}-\sum_{n=1}^{q}\omega^n=q\omega,\quad\text{since }\omega^q=1\text{ and }\sum_{n=1}^{q}\omega^n=0 \\
\sum_{n=1}^{q}\frac{n}{q}\omega^n&=\frac{\omega}{\omega-1},\quad\sum_{n=1}^{q}\frac{n}{q}\omega^{-n}=\frac{\omega^{-1}}{\omega^{-1}-1} \\
\sum_{n=1}^{q}\frac{2n}{q}\sin\left(\frac{2n\pi p}{q}\right)&=\sum_{n=1}^{q}\frac{2n}{q}\left(\frac{\omega^n-\omega^{-n}}{2i}\right)=\frac{1}{i}\left(\frac{\omega}{\omega-1}-\frac{\omega^{-1}}{\omega^{-1}-1}\right) \\
&=-\frac{1}{i}\left(\frac{1+\omega}{1-\omega}\right)=-\cot\left(\frac{\pi p}{q}\right) \\
\sum_{n=1}^{q}\frac{2n}{q}\cos\left(\frac{2n\pi p}{q}\right)&=\sum_{n=1}^{q}\frac{2n}{q}\left(\frac{\omega^n+\omega^{-n}}{2}\right)=\frac{\omega}{\omega-1}+\frac{\omega^{-1}}{\omega^{-1}-1} \\
&=\frac{\omega-1}{\omega-1}=1
\end{align}
Replace $q$ with $2q$,
$$
\sum_{n=1}^{2q}\frac{n}{q}\sin\left(\frac{n\pi p}{q}\right)=-\cot\left(\frac{\pi p}{2q}\right)=-\cot\left(\frac{\pi p}{q}\right)-\csc\left(\frac{\pi p}{q}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Approxiamating Arc Length with Chords and Taylor Series ($S=\frac{1}{3}(8c-C)(1+\epsilon)$) The problem I am working on comes directly from Chapter 3 of Advanced Calculus by Edwin Bidwell Wilson. A chapter that deals with Taylor series.
The problem in full is :
*If S is a circular arc and C its chord and v is the chord of half the arc, prove $$S= \frac{1}{3}(8v-C)(1+\epsilon)$$ where $$\epsilon \approx S^4/7680R^4$$
Here is a picture that helps visualize how I see the problem.
Clearly this problem begs the question of where do I start. I can tell you I started by working with the fact that
$$\theta = 2\arcsin(C/2R)$$
thus
$$S = R\theta = 2R\arcsin(C/2R)$$
using the taylor series of $\arcsin$ gives us
$$S = 2R \sum_{n=0}^{\infty} \frac{(C/2R)^{2n+1}}{2n+1}{2n \choose n}\frac{1}{4^n}$$
You could also derive two pieces of information using the equations presented:
$$R = \frac{4h^2+c^2}{8h}$$
and
$$ h = \sqrt{v^2 - C^2/4}$$
to get
$$R = \frac{v^2}{2\sqrt{v^2-C^2/4}}$$
which gets us
$$S = C \sum_{n=0}^{\infty} \frac{(\frac{C^2}{v^2}(v^2-C^2/4))^{n}}{2n+1}{2n \choose n}\frac{1}{4^n}$$
might be stuck .
| $v=2R\sin\frac{\theta}{4}=2R\left(\frac{\theta}{4}-\frac{\theta^3}{384}+\frac{\theta^5}{122880}+o(\theta^5)\right)$
$C=2R\sin\frac{\theta}{2}=2R\left(\frac{\theta}{2}-\frac{\theta^3}{48}+\frac{\theta^5}{3840}+o(\theta^5)\right)$
$8v-C=2R\left(\frac{3\theta}{2}-\frac{\theta^5}{5120}+o(\theta^5)\right)$
$\frac{1}{1+\epsilon}=\frac{8v-C}{3S}=\frac{2R\left(\frac{3\theta}{2}-\frac{\theta^5}{5120}+o(\theta^5)\right)}{3R\theta}=1-\frac{\theta^4}{7680}+o(\theta^4)$
$1+\epsilon=\frac{1}{1-\frac{\theta^4}{7680}+o(\theta^4)}=1+\frac{\theta^4}{7680}+o(\theta^4)$
$\epsilon=\frac{\theta^4}{7680}+o(\theta^4)\approx \frac{\theta^4}{7680}=\frac{S^4}{7680R^4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4264043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can trigonometric substitution be used to solve this integral? The integral in question is $$\int \frac{x+1}{9x^2+6x+5}dx.$$ I first completed the square in the denominator giving $(3x+1)^2+4$ and proceeded to perform a $u$-substitution with $u = 3x+1$, $du=3~dx$, and $x=(u-1)/3$. After simplifying, I was left with $$\frac{1}{9}\int\frac{u+2}{u^2+4}du.$$ It is at this point I used trigonometric substitution with $u = 2\tan\theta$ and $du = 2\sec^2\theta~d\theta$ (I'm aware the integral can be written as $\frac 1 9\int\frac{u}{u^2+4}du+\frac 1 9\int\frac{2}{u^2+4}du$ and solved this way). After performing the trig substitution, I was left with $$\frac 1 9\int(\tan\theta + 1)~d\theta = \frac 1 9 \ln|\sec\theta|+\frac 1 9 \theta + C.$$ Rewriting everything in terms of $x$ gave me $$\frac 1 9\ln\left(\frac{(3x+1)^2+4}{2}\right)+\frac{1}{9}\tan^{-1}\left(\frac 1 2(3x+1)\right)+ C,$$ which is incorrect. The correct answer is $$\frac{1}{18}\ln\left(9x^2+6x+5\right)+\frac{1}{9}\tan^{-1}\left(\frac 1 2(3x+1)\right)+ C.$$ What went wrong with my trig substitution?
| $$\begin{align*}
\int \frac{x+1}{9x^2+6x+5}dx
&=\int\frac{x+1}{(3x+1)^2+2^2}dx\\ (u = 3x+1)\\
& = \frac{1}{9}\int\frac{u+2}{u^2+2^2}du\\ (u = 2\tan\theta)\\
&=\frac{1}{9}\int\frac{(2\tan\theta+2)(2\sec^2\theta)}{2^2(1+\tan^2\theta)}d\theta\\
&=\frac{1}{9}\int(\tan\theta+1)d\theta\\
& = \frac{1}{9}\ln{|\sec\theta|} + \frac{1}{9}\theta + C_0
\end{align*}$$
Now, $|x| = \sqrt{x^2}$
$\implies \ln|\sec\theta| = \ln{(\sqrt{\sec^2{\theta}})} = \frac{1}{2}\ln{\sec^2{\theta}} = \frac{1}{2}\ln{(1+\tan^2{\theta})}$
Which for the above:
$$\begin{align*}
\frac{1}{9}\ln{|\sec\theta|} + \frac{1}{9}\theta + C_0
& = \frac{1}{9}\times\frac{1}{2}\ln{(1+\tan^2{\theta})}+ \frac{1}{9}\theta + C_0\\
& = \frac{1}{9}\times\frac{1}{2}\ln\left(1+\left(\frac{3x+1}{2}\right)^2\right) + \tan^{-1}\left(\frac{3x+1}{2}\right)+C_0\\
& = \frac 1{18}\ln(9x^2 + 6x +5) + \tan^{-1}\left(\frac{3x+1}{2}\right)+(C_0 - \frac 1{18}\ln(4))\\
& = \frac 1{18}\ln(9x^2 + 6x +5) + \tan^{-1}\left(\frac{3x+1}{2}\right)+C
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4269166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to calculate sum, if the recurrence is given? Let $a_n=4a_{n-1}-a_{n-2}$, for $n\geq 3$. Prove that $\sum\limits_{n=1}^\infty \cot^{-1}{a_n^2}=\frac{\pi}{12}$, if $a_1=2$, $a_2=8$.
This recurrent formula is easy to solve. Characteristic equation $x^2-4x+1=0$ has two roots, $2\pm \sqrt{3}$. Therefore general solution is $a_n=a(2+\sqrt{3})^n+b(2-\sqrt{3})^n$, where a and b can be obtained from $a_1=2$ and $a_2=8$. But, I have no idea what to do next. How to calculate sum $\sum\limits_{n=1}^\infty \cot^{-1}{a_n^2}$?
Any help is welcome. Thanks in advance.
| Lemma: $\frac{a_{n+1}^2+a_n^2}{1+a_na_{n+1}}=4$
solving the recurrence,
$a_n=\frac{1}{\sqrt{3}}(a^n-1/a^n)$ where $a=2+\sqrt{3}$
$a_{n+1}^2+a_n^2=\frac{1+a^2}{a}(a^{2n+1}+1/a^{2n+1})-4 =4(a^{2n+1}+1/a^{2n+1}-1)$
$1+a_n a_{n+1}=3+(a^{2n+1}+1/a^{2n+1})-(a+1/a)=a^{2n+1}+1/a^{2n+1}-1$
using the lemma,
$\frac{a_{n+1}^2+a_n^2}{1+a_na_{n+1}}=4=\frac{a_{n+2}+a_n}{a_{n+1}}$
$a_{n+2}+a_n=\frac{a_{n+1}^3+a_n^2 a_{n+1}}{1+a_na_{n+1}}$
$a_{n+2}=\frac{a_{n+1}^3-a_n}{1+a_na_{n+1}}$
then continue with Martin R's soln.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4270022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find another sum of squares for $3^{12}-6^6+2^{12}$ I have a question about factorization of number $3^{12}-6^6+2^{12}$. By completing the square one can show that$$3^{12}-6^6+2^{12} = (3^6-2^6)^2+6^6 = 665^2+216^2$$ If we can find another representation of this number as sum of squares then may try Euler's factorization method. But how to find another sum of squares in the simplest way? Maybe other methods work?
| We have
\begin{eqnarray*}
665^2+216^2 = 488881 = 37 \times 13213 = (1^2+6^2)(a^2+b^2)=(a+6b)^2+(6a-b)^2.
\end{eqnarray*}
Now solve $a+6b=665,6a-b=216$ gives $a=53,b=102$ now swap $a$ and $b$ to get another representation as the sum of two squares.
$$ 420^2+559^2 =488881=665^2+216^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4270611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the partial sum $\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$ Evaluate the partial sum $$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)}$$
What I have tried:
Calculate the partial fractions which are (for sake of brevity) :
$$\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{n(n+2)}$$
So we get:
$$\sum_{n=1}^k \frac{1}{n(n+1)(n+2)} = \sum_{n=1}^k \left(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{n(n+2)}\right)$$
Then calculating a few numbers for $n$ we get:
$$\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{6} \right) + \left(\frac{1}{4} - \frac{1}{3} + \frac{1}{8} \right) + \left(\frac{1}{6} - \frac{1}{4} + \frac{1}{10}\right) . . . \left(\frac{1}{2n} - \frac{1}{n+1} + \frac{1}{n+2}\right)$$
The first two fractions cancel out in the first bracket and we're left with $\frac{1}{6}$, as for the second bracket the first fraction is cancelled out by the second fraction in the third bracket.
I have noticed that the first fractrion so $\frac{1}{2n}$ cancel out by every even term in the denominator for $-\frac{1}{n+1}$ so the equation becomes:
$$\left(-\frac{1}{2n+1}+\frac{1}{n+2}\right) = \left(\frac{n-1}{(2n+1)(n+2)} \right)$$
Have I approached this correctly? I would greatly appreciate some assistance on any improvements!
| This is a two step partial fraction decomposition
$$S_k=\sum_{n=1}^k \frac{1}{(n)(n+1)(n+2)}=\frac{1}{2}\sum_{n=1}^k \frac{1}{(n)(n+1)}-\frac{1}{(n+1)(n+2)}\tag{1}$$
$$S_k=\frac{1}{2}\sum_{n=1}^k \left(\frac{1}{(n)}-\frac{1}{(n+1)}\right)-\left(\frac{1}{(n+1)}-\frac{1}{(n+2)}\right)=\frac{1}{2}\sum_{n=1}^k \left(\frac{1}{(n)}-\frac{2}{(n+1)}+\frac{1}{(n+2)}\right)\tag{2}$$
Then you can write (2) in terms of the Harmonic Function $H_k=\sum_{n=1}^k \frac{1}{n}$
$$S_k=\frac{1}{2}\left( H_k-2\left( H_k+\frac{1}{k+1}-1\right)+\left( H_k+\frac{1}{k+1}+\frac{1}{k+2}-\frac{3}{2}\right) \right)\tag{3}$$
Which can immediately be simplified further as all the $H_k$ terms cancel out.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4272062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Show that the following inequality holds $\sqrt[3]{2}+\sqrt2>2$ Show that the following inequality holds $$\sqrt[3]{2}+\sqrt2>2$$
I have just studied cubic roots and we know that if $a<b$ then $\sqrt[3]{a}<\sqrt[3]{b}$ $(a,b\in\mathbb{R}).$ Here though we also have the square root $\sqrt[2]2$. How are we supposed to approach the problem? Thank you. The approximate values are $$\sqrt[3]{2}\approx1.26;\sqrt2\approx1.41$$
So using @rtybase suggestion we have to show $\sqrt[3]2+\sqrt2>2$. We know that $\sqrt{2}>1$. Can we subtract the second inequality from the first? Is subtracting inequalities even a thing? Then we will get $\sqrt[3]2>1=\sqrt[3]{1}$ which is true.
| We have:
$$\begin{array}{l}
\sqrt[3]{2} > 2 - \sqrt{2} \iff 2 > (2 - \sqrt{2})^3\iff \\
\iff 2 > 8 - 3\cdot 2^2\cdot \sqrt{2}+3\cdot 2\cdot (\sqrt{2})^2- 2\sqrt{2}=20-14\sqrt{2}\iff \\ \iff14\sqrt{2} > 18\iff 7\sqrt{2} > 9\iff 98 > 81 \end{array}
$$,
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Convergence of $a_{n+1}:=\sqrt{a_n+k}$ $a_0=\sqrt k$
$a_{n+1}:=\sqrt{a_n+k}$
I am trying to show the convergence by monotone convergence theorem and I can show its monotonicity by observing the square of each term but I am stuck at proving its boundedness. I feel it should be bounded as it increases slower and slower.
| What is $a_1$?
Assume $x_0$ is positive root of $x^2-x-k=0$ (Assume $k\gt 0$)
If $a_1 \gt x_0$, we could prove $x_0\lt a_{n+1}\lt a_n$
since $a_2 = \sqrt{a_1+k}$, so $a_1^2-a_2^2=a_1^2-a_1-k\gt 0$, so $a_2\lt g_1$
and $(a_1+\sqrt{a_1+k})(a_2^2-a_2-k)=(a_1+\sqrt{a_1+k})(a_1-\sqrt{a_1+k})=a_1^2-a_1-k\gt 0$, so $a_2^2-a_2-k\gt 0$, so $a_2\gt x_0$.
So our assumption holds for n=2
Similarly, assuming $x_0\lt a_{m+1} \lt a_m$
We have $a_{m+1}^2-a_{m+2}^2=a_{m+1}^2-a_{m+1}-k \ gt 0$, so $a_{m+2} \lt a_{m+1}$
and $(a_{m+1}+\sqrt{a_{m+1}+k})(a_{m+2}^2-a_{m+2}-k)=a_{m+1}^2-a_{m+1}-k \ gt 0$,
so $a_{m+2}^2-a_{m+2}-k \ gt 0$, so $a_{m+2} \gt x_0$
So $x_0\lt a_{n+1}\lt a_n$ is true for any n.
Similarly for $a_1 \lt x_0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4280708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximizing $a^2+b^2+c^2+d^2$ with given constraints The following problem is from a local contest which ended today:
Let $a,b,c,d$ be positive real numbers such that $$(a+b)(c+d)=143\\ (a+c)(b+d)=150\\ (a+d)(b+c)=169$$ Find the maximum value of $a^2+b^2+c^2+d^2$.
Here are my workings:
We have $$\tag{1}ac+bc+bd+ad=143$$ $$\tag{2}{ab+bc+cd+ad=150}$$ $$\tag{3}{ab+bd+ac+cd=169}$$
Summing $(1)$, $(2)$ and $(3)$, we have $$2(ab+bc+cd+ac+ad+bd)=462$$
We have $$\begin{align} a^2+b^2+c^2+d^2 &=(a+b+c+d)^2-2(ab+bc+cd+ac+ad+bd)\\ &= (a+b+c+d)^2-462\end{align}$$
So we have to maximize $a+b+c+d$.
I can't proceed from here. Minimizing the expression seems easy by Cauchy-Schwarz or AM-GM but I don't know how to maximize this.
| As @CalvinLin pointed out,
under the conditions, $a^2 + b^2 + c^2 + d^2$ is not bounded from above.
Explanation:
Let $t\in (0, 1)$ and
$$a = \frac{t}{2} + \frac{Q}{4t},
\quad b = -\frac{t}{2} + \frac{Q}{4t},
\quad c = \frac{143t}{Q} + \frac{19}{2t},
\quad d = \frac{143t}{Q} - \frac{19}{2t}$$
where $Q = \sqrt{ {t}^{4}+638\,{t}^{2}+361} - \sqrt{{t}^{4}+66\,{t}^{2}+361}$.
It is not difficult to verify
that
$a, b, c, d > 0$, and
$(a + b)(c + d) = 143$, and $(a + c)(b + d) = 150$,
and $(a + d)(b + c) = 169$.
However, $a^2 + b^2 + c^2 + d^2 = t^2 + 176 + \frac{361}{t^2} \to \infty$ as $t \to 0^{+}$.
Remarks: Explain how to obtain the above $a, b, c, d$.
Consider the system of equations
\begin{align*}
(a + b)(c + d) = 143, \tag{1}\\
(a + c)(b + d) = 150, \tag{2}\\
(a + d)(b + c) = 169. \tag{3}
\end{align*}
(1) gives
$d = \frac{143}{a + b} - c$.
[(2) - (3)] gives
$(b-a)c + ad - db + 19 = 0$
which results in $c = \frac{81a - 62b}{a^2 - b^2}$.
We see that if $a > b$ and $a-b \to 0^{+}$, then $c \to \infty$. So, we let $a - b = t$.
Then we have
$$4\,{t}^{2}{b}^{4}+8\,{t}^{3}{b}^{3}+ \left( 5\,{t}^{4}-352\,{t}^{2}-
361 \right) {b}^{2}+ \left( {t}^{5}-352\,{t}^{3}-361\,t \right) b-88\,
{t}^{4}+5022\,{t}^{2}
= 0.$$
We can solve $b$ in closed form. Indeed, dividing both sides by $4t^2$,
we have
$${b}^{4}+2\,t{b}^{3}+ \left( -88+ \frac{5}{4}t^2 -{\frac {361}{4t^2}} \right) {b}^{2}+ \left( -{\frac {361}{4t}} + \frac14{t}^{3}-88
\,t \right) b+{\frac {2511}{2}}-22\,{t}^{2} = 0.
$$
We eliminate the term $b^3$ by letting $b = z - t/2$:
$${z}^{4}+ \left( -{\frac {361}{4t^2}}-88-\frac14{t}^{2} \right) {z
}^{2}+{\frac {20449}{16}} = 0.$$
Luckily, it can be solved in closed form.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4282379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Without superior math, can we evaluate this limit? We all knew, with $$\lim_{x\to 0}\frac{\sin x - x}{x(1-\cos x)}$$ we can use L'Hôpital's rule or Taylor series to eliminate undefined form. But without all tools, by only using high school knowledge, how can we evaluate this limit? It seems difficult to transform numerator, any idea?
Thank you!
| I want to thank everyone for your help, after some hard work, i found the answer, i post it here for all of you (in case you all need reference later):
$$\lim_{x\to 0}\frac{\sin x - x}{x(1-\cos x)}=\lim_{x\to 0}\frac{\sin x - x}{2x\sin^2{\frac{x}{2}}}=\lim_{x\to 0}\frac{2(\sin x - x)}{x^3}\frac{\frac{x^2}{4}}{\sin^2{\frac{x}{2}}}=\lim_{x\to 0}\frac{2(\sin x - x)}{x^3}$$because $$\lim_{x\to 0}\frac{\frac{x^2}{4}}{\sin^2{\frac{x}{2}}}=1$$ when x approach to 0. Now, let $$L=\lim_{x\to 0}\frac{2(\sin x - x)}{x^3}(*)$$ use this indentity $$\sin x = 3\sin \frac{x}{3} - 4\sin^3\frac{x}{3}$$ $$=>L=\lim_{x\to 0}\frac{2(3\sin \frac{x}{3} - 4\sin^3\frac{x}{3} - x)}{x^3}=\lim_{x\to 0}\left(\frac{6}{27}\left(\frac{\sin\frac{x}{3}-\frac{x}{3}}{(\frac{x}{3})^3}\right)-\frac{8}{27}\right)$$ Now we can replace $$\lim_{x\to 0}\frac{\sin\frac{x}{3}-\frac{x}{3}}{(\frac{x}{3})^3}=\frac{L}{2}$$by comparing to (*), finally we have $$L=\frac{6}{27}\frac{L}{2}-\frac{8}{27}$$ solve this easy equation we conclude $$\lim_{x\to 0}\frac{\sin x - x}{x(1-\cos x)}=L=\frac{-1}{3}$$
And we have solved this limit without using L'Hopital's rule or Taylor series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4283735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Request for proof: Regularity of pi's continued fractions
Notice
In this post, a continued fraction is represented as follows
$$ a + \cfrac{1^2}{b+\cfrac{3^2}{b+\cfrac{5^2}{\ddots}}} = a +K^\infty_{k=1}\frac{(2k-1)^2}{b} $$
When I was checking the continued fractions of pi in wolframalpha, I noticed the following regularity.
\begin{eqnarray*}
\frac{4}{\pi} &=& 1 + K^\infty_{k=1}\frac{(2k-1)^2}{2} \\
\pi &=& 3 + K^\infty_{k=1}\frac{(2k-1)^2}{6} \\
\frac{16}{\pi} &=& 5 + K^\infty_{k=1}\frac{(2k-1)^2}{10}
\end{eqnarray*}
So I was wondering what the value of $ 7 + K^\infty_{k=1}\frac{(2k-1)^2}{14} $ would be. Some numerical investigation yielded the following speculative results.
The statement from here on out is unproven.
\begin{eqnarray*}
\frac{9\pi}{4} = 7 + K^\infty_{k=1}\frac{(2k-1)^2}{14} \\
\frac{256}{9\pi} = 9 + K^\infty_{k=1}\frac{(2k-1)^2}{18} \\
\frac{225\pi}{64} = 11 + K^\infty_{k=1}\frac{(2k-1)^2}{22} \\
\frac{1024}{25\pi} = 13 + K^\infty_{k=1}\frac{(2k-1)^2}{26} \\
\frac{1225\pi}{256} = 15 + K^\infty_{k=1}\frac{(2k-1)^2}{30} \\
\end{eqnarray*}
The numbers appearing on LHS are all square numbers, indicating that there is some regularity.
If you know of any proof of this or related information, please let me know.
Thank you.
| All these continued fractions (and many others..) can be derived from a formula known to Wallis and Euler. I do not know how (and if..) they proved it. It was also proven by Stieltjes and Ramanujan.
These Continued fractions can be evaluated using Euler's Differental Method. In this way one can prove the Wallis-Euler formula's which state that:
$4k+1+\cfrac{1^2}{2(4k+1)+\cfrac{3^2}{2(4k+1)+\cfrac{5^2}{2(4k+1)+...}}}=\cfrac{2k+1}{W(k)}\cfrac{4}{\pi}$
and
$4k+3+\cfrac{1^2}{2(4k+3)+\cfrac{3^2}{2(4k+3)+\cfrac{5^2}{2(4k+3)+...}}}=(2k+1)\cdot{W(k)}\cdot{\pi}$
with the Wallis product $W(k)=\cfrac{1\cdot3}{2\cdot2}\cdot\cfrac{3\cdot5}{4\cdot4}\cdot\cfrac{5\cdot7}{4\cdot4}....\cfrac{(2k-1)\cdot)2k+1)}{2k\cdot2k}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4285749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can we show $\frac{1-x^n}{1-c^n} + \left(1-\frac{1-x}{1-c}\right)^n \leq 1 $ for all $n \in \mathbb{N}$, $0 \leq c \leq x \leq 1$, $c \neq 1$?
How can we show $$\frac{1-x^n}{1-c^n} + \left(1-\frac{1-x}{1-c}\right)^n \leq 1 $$ for all $n \in \mathbb{N}$, $0 \leq c \leq x
\leq 1, c \neq 1$?
The context is this probability problem, but of course this problem might be of independent interest to inequality enthusiasts.
In my attempt, I have graphed the inequality on Desmos. We might note that the derivative changes sign in the range $c < x < 1$ so it may be unlikely differentiation would be of help.
| If $x=c$ or $n=1$ then inequality holds. So consider $x > c, n > 1$:
$$\frac{1-x^n}{1-c^n}+\left(\frac{x-c}{1-c}\right)^n\leq 1\Leftrightarrow
1-x^n+(1-c^n)\left(\frac{x-c}{1-c}\right)^n\leq 1-c^n\Leftrightarrow$$
$$(1-c^n)\left(\frac{x-c}{1-c}\right)^n\leq x^n-c^n\Leftrightarrow
\frac{1-c^n}{(1-c)^n} \leq \frac{x^n-c^n}{(x-c)^n}$$
Consider $f(x)=\frac{x^n-c^n}{(x-c)^n}$, then $$f'(x)=\frac{n x^{n-1}}{(x-c)^n}-\frac{n(x^n-c^n)}{(x-c)^{n+1}}=\frac{nc(c^{n-1}-x^{n-1})}{(x-c)^{n+1}}<0$$
$$f'(x)<0, x\leq 1 \Rightarrow f(1)\leq f(x)\Rightarrow \frac{1-c^n}{(1-c)^n} \leq \frac{x^n-c^n}{(x-c)^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4287733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Eliminate ${\theta, \varphi}$ from $\tan\theta+\tan\varphi=a$, $\sec\theta+\sec\varphi=b$, $\csc\theta+\csc\varphi=c$ Here is the problem:
Eliminate $\theta, \varphi$ from the equations
$$\tan\theta+\tan\varphi=a$$
$$\sec\theta+\sec\varphi=b$$
$$\csc\theta+\csc\varphi=c$$
What I have so far.
$$\tan \theta=a-\tan\varphi$$
$$\sec \theta=b-\sec\varphi$$
$$\csc \theta=c-\csc\varphi$$
And using
$$\tan \theta \csc\theta =\sec\theta$$ gives us
$$c\tan\varphi+a\csc\varphi-2\sec \varphi=ac-b$$
And also
$$\tan^2\theta+1=\sec^2\theta$$
gives
$$2a\tan\varphi-2b\sec\varphi=a^2-b^2$$
Thus we have two linear relations. But I cannot think of any more properties to use. Things like $\sin^2+\cos^2=1$ give a big mess. So what are the end steps ?
Update: The method used in this question: How to eliminate $\theta$ & $\phi$ from above equations
will work and solves the problem.
Following that method gives me
$$[(b^2-a^2)^2+4a^2]a^2c=4a^2[c(a^2+b^2)-2ab]$$
What a fantastic relation !
| In the meantime I have found an answer. By request I post it here.
Thanks to @AakashM for pointing out some mistakes in the solution, now corrected.
Write the equations as
$$\sin\theta\cos\varphi+\sin\varphi\cos\theta=
a\cos\theta\cos\varphi$$
$$\cos\theta+\cos\varphi=b\cos\theta\cos\varphi$$
$$\sin\theta+\sin\varphi=c\sin\theta\sin\varphi$$
Now define
$$\alpha=\frac{\theta+\varphi}{2}, \beta=\frac{\theta-\varphi}{2}$$
and so
$$4\sin\alpha\cos\alpha=a(\cos2\beta+\cos2\alpha)$$
$$4\cos\alpha\cos\beta=b(\cos2\beta+\cos2\alpha)$$
$$4\sin\alpha\cos\beta=c(\cos2\beta-\cos2\alpha)$$
Dividing the first and second,
$$\frac{\sin\alpha}{\cos\beta}=\frac{a}{b}$$
so
$$b\sin\alpha=a\cos\beta$$
Further subtracting the first and second,
$$4\cos\alpha(\sin\alpha-\cos\beta)=(a-b)
(\cos2\beta+\cos2\alpha)$$
$$2\cos\alpha(\sin\alpha-\cos\beta)=(a-b)
(\cos^2\beta-\sin^2\alpha)$$
$$2\cos\alpha(\sin\alpha-\cos\beta)=(a-b)
(\cos\beta-\sin\alpha)(\cos\beta+\sin\alpha)$$
So cancelling $\sin\alpha-\cos\beta$
we get
$$2\cos\alpha=(b-a)
(\cos\beta+\sin\alpha)$$
Now multiply by $a$ and convert the $\cos\beta$ to $\sin\alpha$ using the previous formula.
$$2a\cos\alpha=(b-a)
(a\cos\beta+a\sin\alpha)$$
$$2a\cos\alpha=(b-a)
(b\sin\alpha+a\sin\alpha)$$
$$2a\cos\alpha=(b^2-a^2)\sin\alpha$$
Finally we take the last equation in the form
$$2\sin\alpha\cos\beta=c(\cos^2\beta+\sin^2\alpha-1)$$
Now multiply by $a^2$ and convert the $\cos\beta$ as before,
$$2ab\sin^2\alpha=c(b^2\sin^2\alpha+a^2\sin^2\alpha-a^2)$$
$$2ab\sin^2\alpha=c((a^2+b^2)\sin^2\alpha-a^2)$$
So $$[c(a^2+b^2)-2ab]\sin^2\alpha=a^2c$$
Now to finish we use
$$(2a\cos\alpha)^2+(2a\sin\alpha)^2=4a^2$$
and substituting a previous equation,
$$((b^2-a^2)\sin\alpha)^2+(2a\sin\alpha)^2=4a^2$$
So
$$[(b^2-a^2)^2+4a^2]\sin^2\alpha=4a^2$$
and multiplying by $[c(a^2+b^2)-2ab]$,
$$[(b^2-a^2)^2+4a^2]a^2c=4a^2[c(a^2+b^2)-2ab]$$
our final relation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4288989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
If $3 \mid a^2+b^2$, then show that $9 \mid ab$.
If $3 \mid a^2+b^2$, then show that $9 \mid ab$.
If $3$ divides $a^2+b^2$ I have that $a^2+b^2 \equiv 0 \pmod{3}$. Looking at the squares it seems that starting from $9$ for every third I’ll get a remainder $0$. Also squares mod $3$ are only $0$ or $1$ so I must have $a^2 \equiv 0 \pmod{3}$ and $b^2 \equiv 0 \pmod{3} $, but still how does this imply the result we’re after?
| $a^2\equiv0\pmod3$ means $3\mid a^2$, which implies $3\mid a$ (by Euclid's lemma)
;
similarly $3\mid b$, so $9\mid ab$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4293873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
how can I solve the following equation (without complex numbers)? $$
\sqrt{45-x^2} = 3- x^2
$$
$$
\sqrt{13-x^2} = 7 - x^2
$$
I have tried the Quadratic formula. I always came up with the solution of $\{-3, 3\}$ by the first one and $\{-3, 3\}$ by the second one which is both wrong. Does this happen because I have to pay attention to a rule when substituting in a root, which I don't know, or how should I arrive at the solution that the first has none according to the graph and the second has the solution set according to the graph $\{-2,2\}$? I search a lot in the internet but there was no solution.
| Consider the first equation:
A general strategy would be to square both sides:
$$\sqrt{45-x^2} = 3- x^2 \implies 45-x^2 = (3-x^2)^2 = 9+x^4-6x^2 \iff x^4-5x^2-36=0$$
Using $a=x^2$, the equation becomes $a^2-5a-36=0$ with $a\ge 0$. The Quadratic Formula indeed say that the solutions are $a=9$ and $a=-4$. Since $a\ge 0$, only $a=9$ is solution, so $x=\pm 3$.
But! The first step of computation ($\sqrt{45-x^2}=3-x^2 \implies 45-x^2=(3-x^2)^2$) is an implication, not an equivalence, so it is important to check that $3$ and $-3$ are indeed solution. As you check, they are not, so the equation has no real solution.
What are $3$ and $-3$ doing here? The equation $45-x^2=(3-x^2)^2$ is actually equivalent to $3-x^2=\pm\sqrt{45-x^2}$ (assuming $45-x^2\ge 0$). $4$ and $-3$ are solutions of the "twin" equation $3-x^2=-\sqrt{45-x^2}$.
Other method to see this equation has no solution: $3-x^2=\sqrt{45-x^2}$ implies that $3-x^2\ge 0$ so $x^2\le 3$, hence $3-x^2\le 3$. This in turn implies that $45-x^2\ge 42$ so $\sqrt{45-x^2}\ge\sqrt{42}>3\ge 3-x^2$ so the equation has no real solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that: $\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{21}{2}$ Given 3 positive real numbers $x, y, z$ satisfies $xy+yz+xz=1$. Prove that: $$\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{21}{2}$$
(Only use AM-GM, Cauchy-Schwarz inequalities)
My progress:
Till now, I have not made much of a progress besides finding out that $x^2+y^2+z^2\geq1$ and $xyz\leq\frac{\sqrt{3}}{9}$ and turn the LHS into $$\frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(x+y)(y+z)}}+\frac{z}{\sqrt{(x+z)(y+z)}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$$
| Hint :
If we have :
$$f\left(x\right)=\frac{x}{\sqrt{x^{2}+1}}+\frac{1}{x^{2}}$$
And define :
$$h\left(x,y\right)=f\left(\frac{1-xy}{x+y}\right)-\frac{21}{2}xy\left(1+C\frac{\frac{2}{21}-\frac{2}{63xy}}{x^{2}+y^{2}+\left(\frac{1-xy}{x+y}\right)^{2}}\right)$$
Then with the constraint $0<xy\le1$ and $6<C<7$ a judicious constant try to show:
$$h\left(x,y\right)\ge0$$
Now use it three times .
For $0<x\leq 1$ and a end see Wolfram Alpha transformation
I recognize it's very ugly but all the good way are complete see other answer .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.