Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculate the maximum possible area of the quadrilateral (functions and derivatives) The total area $A(x)$ in $m^2$ of a quadrilateral is given by:
$$A(x)= 2 \sqrt{ x^2 -16} +\frac 14\sqrt{68x^2-x^4-256},\;\; (4 < x < 8)$$
How do I calculate the maximum possible area of the quadrilateral.
As i have understood I need to find the derivative of A(x) then find the stationary points but I am struggling to first find the derivative and then find the stationary points.
Thanks in advance.
| It looks like your derivative you posted in the comment to the other hint is correct.
Hint: Find the common denominator and add the fractions. The stationary points will occur when the derivative is equal to zero, and the derivative will equal zero only when the numerator is equal to zero. What values of $x$ that are within your range make the numerator zero?
$$\dfrac{2x}{\sqrt{x^2-16}} + \dfrac{136x-4x^3}{8\sqrt{68x^2-x^4-256}} = \dfrac{2x}{\sqrt{x^2-16}} + \dfrac{34x-x^3}{2\sqrt{68x^2-x^4-256}}$$ $$=\dfrac{4x\sqrt{68x^2- x^4-256 }+(34x-x^3)\sqrt{x^2 - 16}}{2\sqrt{x^2 - 16}\sqrt{68x^2 -x^4 - 256}}$$
Clearly, the numerator equals zero when $x=0$. That's because $x = 0$ gives the minimum possible area! But you need the maximum area, and you need the solution such that $4\lt x \lt 8$. Put the numerator equal to zero, in a separate equation, and then solve for $x$.
$$4x\sqrt{68x^2 - x^4 - 256} + (34x - x^3)\sqrt{x^2 - 16} = 0$$ $$\iff 4\sqrt{68x^2 - x^4 - 256} + (34 - x^2)\sqrt{x^2 - 16} = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/682186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Rationale for a convention: Why use the semiperimeter in Heron's formula? Heron's formula says that the area of a triangle whose sides have lengths $a, b, c$ is $\sqrt{s(s-a)(s-b)(s-c)}$ where $s=(a+b+c)/2$ is the semiperimeter. It can also be stated by saying that the area is $\frac14\sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}$. Is there a substantial reason to prefer the first form, using the semiperimeter, over the second? It seems as if that's the only form I've seen in published sources.
(Either way, it's the simplest function of $a$, $b$, $c$ that is 2nd-degree homogeneous and is equal to $0$ whenever the three vertices are on a common line or at a common point.)
PS inspired by comments:
Let's compare lengths:
$$
\sqrt{s(s-a)(s-b)(s-c)}\text{ where }s=(a+b+c)/2\text{ is the semiperimeter}
$$
$$
\phantom{\frac{\vert}{}}\sqrt{s(s-a)(s-b)(s-c)}\text{ where $s$ is the semiperimeter}
$$
$$
\sqrt{s(s-a)(s-b)(s-c)}\text{ where }s=(a+b+c)/2
$$
$$
\frac 1 4 \sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}
$$
| The reason I think is simplicity:
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
Is much more simple than:
$$A=\frac{1}{4}\sqrt{(a+b+c)(b+c−a)(c+a−b)(a+b−c)}$$
I see in the comments that you say we have to add $\text{where } 2s=a+b+c$, in the standard form, and so the length of the two forms become the same. However, this is a useless argument. Continuing with the logic we must add, $\text{where a,b,c are lengths of sides of the triangle}.$ This, everyone will agree that is pointless, and will unnecessarily increase the length and apparent complexity of the formula.
The same goes with $s$ here, we do need to add $s$ is this and this. Anyone studying geometry will immediately recognize that $s$ stands for the semi-perimeter. For example, we would never recognize in a geometry question, $p$ as the perimeter, however when we see $s$, the first thought that comes is the semi-perimeter. Hence, I think, its a bit stubborn to add that part as you have done in the postscript, and then compare lengths.
And this said, the point is not on length but on how simple the formula is. The better form must be the one that allows more easily to do what the formula is about, that is calculating the area. Let us take an example with a $13, 14, 15$ triangle, and calculate using both forms as I would have done myself with a pen and paper:
$$
\begin{array}{c|lcr}
\text{Form 1} & \text{ Form 2} \\
\hline
{ s=\frac{13+14+15}{2}=21 \\
s - a=21 - 13=8 \\
s - b=21 - 14=7 \\
s - c=21 - 15=6 \\
\text{So, } A=\sqrt{21\cdot 7 \cdot 8 \cdot 6}=\sqrt{3 \cdot (7 \cdot 7) \cdot 2 \cdot 4 \cdot 2 \cdot 3} \\
=7\cdot 3\cdot 2\cdot 2=84 } & {a + b + c =13+14+15=42 \\
a + b - c =13+14-15=27-15=12 \\
a + c - b =13+15-14=28-14=14 \\
b + c - a =14+15-13=28-16=16 \\
\text{So, } A=\frac{1}{4}\sqrt{42\cdot 12 \cdot 14 \cdot 16}\\=\frac{1}{4}\sqrt{2 \cdot 21 \cdot 2 \cdot 6 \cdot 2 \cdot 7 \cdot 2 \cdot 8} \\=\sqrt{21 \cdot 6 \cdot 7 \cdot 8}=7\cdot 3\cdot 2\cdot 2=84}\\
\end{array}
$$
Even with small dimensions, we can see that the first form has an advantage. We can calculate $s$ easily and then $s-a$, $s-b$, .. as its just a subtraction of two terms. However, calculating $a + b - c$, $a + c - b$ .. are three terms calculation are very unrelated to each other. This becomes clearer when the sides are larger and variable. Thus, we can see that the standard form has advantages, its easier to use, write and state. There is one more important aspect.
There is an important area formula in terms of $s$, using the inradius, $A =rs$, where r is the inradius. If we draw the incircle, and the sides divided by the points of tangency can be conveniently expressed in terms of $s$. There are many examples where properties of triangles could be reprsented in terms of $s$, and without using $s$, it would be very untidy. As an example, the standard form allows a convenient expression of the inradius, whle the other form would give a very untidy expression:
$$r=\frac{A}{s}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$
Generalizations of the Herons formula, like the Brahmagupta's formula, or even the Bretschneider's formula, which gives us the area of a general convex quadrilateral:
$$A=\sqrt {(s-a)(s-b)(s-c)(s-d) - abcd \cdot \cos^2 \left(\frac{\alpha + \gamma}{2}\right)}$$
This is a very convenient and simple expression. However, if we replace $s$ by $\frac{a+b+c+d}{2}$, and rewrite the formula without introducing $s$:
$$A = \frac{1}{4}\sqrt{(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d) - 16abcd \cos^2 \left(\frac{\alpha + \gamma}{2}\right)}$$
This is an extremely unwieldy formula and unmanageable for general use. Thus, we can see that writing out the formula in full is a bad idea, which is apparent as we generalize the formula. On the other hand, the standard form, is easy to state and work with and more useful, and hence I think the preference over other forms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/685328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
An equation, where the solution does not exist, but on solving the equation we got a solution. why this is happening? The solution of the equation
$\sqrt{(x+1)} -\sqrt{(x-1)}= \sqrt{(4x-1)}$
is $\frac{5}{4}$,but when we put $x=\frac{5}{4}$ in the given equation, then it does not satisfy the equation.
Actually, if we take $f(x)=\sqrt{(x+1)} -\sqrt{(x-1)} -\sqrt{(4x-1)}$ then we can see that
$f(x)$ is defined when $x \geq 1$ and $f(1) \geq 0\mbox{ and }f'(x) \geq 0$ so, the function is monotone increasing and it will never appear zero.
so, my question is , In this type of equation where the solution actually does not exist, then why should we get this type of solution?
my solution procedure is,
$$
\begin{align}
\sqrt{(x+1)} -\sqrt{(x-1)}&= \sqrt{(4x-1)}\\
\implies 2x-2\sqrt{x^2-1}&=4x-1\\
\implies {-2}\sqrt{x^2-1}&= 2x-1\\
\implies 4(x^2-1)&=4x^2+1-4x\\ \implies x&=5/4
\end{align}$$
| Whenever we square, we immediately introduce extraneous root
Observe that
$\displaystyle\frac54$ is actually a root of $$\sqrt{x+1}=\sqrt{4x-1}-\sqrt{x-1}$$
Also, observe that $\displaystyle2x-1=-2\sqrt{x^2-1}\le0\implies 2x\le1\iff x\le\frac12$ for real $x$
But, $\displaystyle{\sqrt{x-1}}$ is not real unless $x\ge1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/685687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 0
} |
Let $p$ be prime and $\left(\frac{-3}p\right)=1$. Prove that $p$ is of the form $p=a^2+3b^2$
Let $p$ be prime and $\left(\frac{-3}p\right)=1$, where $\left(\frac{-3}p\right)$ is Legendre symbol. Prove that $p$ is of the form $p=a^2+3b^2$.
My progress:
$\left(\frac{-3}p\right)=1 \Rightarrow$ $\left(\frac{-3}p\right)=\left(\frac{-1}p\right)\left(\frac{3}p\right)=(-1)^{\frac{p-1}2}(-1)^{\left\lfloor\frac{p+1}6\right\rfloor}=1 \Rightarrow$ $\frac{p-1}2+\left\lfloor\frac{p+1}6\right\rfloor=2k$ I'm stuck here. This is probably not the way to prove that.
Also tried this way:
$\left(\frac{-3}p\right)=1$, thus $-3\equiv x^2\pmod{p} \Rightarrow$ $p|x^2+3 \Rightarrow$ $x^2+3=p\cdot k$
stuck here too.
Any help would be appreciated.
| Let $\omega\in\Bbb C$ satisfy $\omega^2+\omega+1$. Then $\Bbb Z[\omega]$ is a UFD, since $N(a+b\omega)=a^2-ab+b^2$ defines an Euclidean norm.
Since $\left(\frac{-3}{p}\right)=1$, there is an $s$ such that $p\mid s^2+3$. Working over $\Bbb Z[\omega]$ we have $$p\mid (s+1+2\omega)(s+1+2\omega^2)$$ If $p$ is prime we would have both $p\mid s+1$ and $p\mid 2$. This is impossible, so $p$ cannot be prime. Therefore we must have $p=\alpha\beta$, where $\alpha,\beta\in\Bbb Z[\omega]$. Taking norms, we get $$p^2=N(p)=N(\alpha)N(\beta)$$ Since neither $\alpha,\beta$ are units, we have $N(\alpha),N(\beta)\neq 1$. This forces $N(\alpha)=N(\beta)=p$. Let $\alpha=m+n\omega$, then $$p=N(\alpha)=m^2-mn+n^2$$ Without loss of generality we may assume that $n$ is even, as $p$ is an odd prime. We can then write:
$$4p=4m^2-4mn+n^2+3n^2\\4p=(2m-n)^2+3n^2\\p=\left(m-\frac{n}{2}\right)^2+3\left(\frac{n}{2}\right)^2$$ Choose $a=m-\frac{n}{2}$ and $b=\frac{n}{2}$ and we get $$p=a^2+3b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/685958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 2
} |
Approximating $\ln{\frac{x-y}{x+y}}$ For $x>>y$,
$$\ln{\frac{x-y}{x+y}} = \ln{\left[ x\left( 1-y/x \right) \right]} -\ln{\left[ x\left( 1+y/x \right) \right]} \approx -2\frac{y}{x}$$
However, the following does not work:
$$\ln{\frac{x-y}{x+y}} \approx \ln{\frac{x-y}{x}} = \ln{\left( 1-\frac{y}{x} \right)} \approx -\frac{y}{x}$$
Can anyone explain why the above approximation is not valid?
| The correct way to expand $\dfrac{x-y}{x+y}$ is using the binomial theorem.
$$ \frac{1}{x+y} = \frac1x\left(1+\frac{y}{x}\right)^{-1} = \frac1x\left(1-\frac{y}{x} + \ldots\right)$$
Hence
$$ \frac{x-y}{x+y} = \frac{x-y}{x}\left(1-\frac{y}{x} + \ldots\right) \approx 1 -2\frac{y}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/687037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Maclaurin series of $\frac{1}{1+\sin x}$ Find the terms through degree four of the Maclaurin series of $f(x)$.
$$f(x) = \frac{1}{1+\sin x}$$
My work:
The Maclaurin series for $\sin x$ up to degree $4$ is $x - \frac{x^3}{6} + \frac{x^5}{120}$
The Maclaurin series for $\frac{1}{1+x}$ up to degree $4$ is $1 - x + x^2 - x^3 + x^4$
I substituted $x - \frac{x^3}{6} + \frac{x^5}{120}$ for $x$ in $1 - x + x^2 - x^3 + x^4$
Did I do this right?
Plugging this into WolframAlpha, I get this: http://goo.gl/SKddyh
Which doesn't seem like the answer in the text: $1-x+x^2-\frac{5x^3}{6}+\frac{2x^4}{3}$
| You did it right. The answer on WolframAlpha is the same. Try typing "expand" in front on WA.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/692406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How can I prove the trigonometric Problem? How can I show the following trigonometric problem :
$$\frac{1}{3}\leq \frac{\sec^2\theta-\tan^2\theta}{\sec^2\theta+\tan^2
\theta}\leq 3$$
I have tried in the following way :
$$
\begin{align}
& \phantom{\Rightarrow}
\frac{\sec^2\theta-\tan^2\theta}{\sec^2\theta+\tan^2
\theta}-\frac{1}{3} \\[8pt]
& \Rightarrow \frac{1}{\sec^2\theta+\tan^2
\theta}-\frac{1}{3} \\[8pt]
& \Rightarrow \frac{\cos^2\theta}{1+\sin^2
\theta}-\frac{1}{3} \\[8pt]
& \Rightarrow \frac{1-\sin^2\theta}{1+\sin^2
\theta}-\frac{1}{3} \\[8pt]
& \Rightarrow \frac{2-4\sin^2\theta}{3(1+\sin^2
\theta)} \\[8pt]
& \Rightarrow \frac{2\cos 2\theta}{3(1+\sin^2
\theta)}
\end{align}
$$
How can I show $ \dfrac{2\cos 2\theta}{3(1+\sin^2
\theta)}\geq 0$ ?
| You have
$$
\frac 1 3 \le \frac{1-u}{1+u} \le 3
$$
where $u = \sin^2\theta$.
You cannot multiply through by $1+u$ because $1+u$ may be either positive or negative, depending on the value of $u$.
If $\dfrac{1-u}{1+u}\le 3$ then $\dfrac{1-u}{1+u}-3\le0$, so $\dfrac{1-u}{1+u}-\dfrac{3(1+u)}{1+u}\le 0$, and that becomes $\dfrac{-2-4u}{1+u} \le 0$. From that one gets $\dfrac{\frac 1 2 + u}{1+u}\ge0$. This changes signs at $-1/2$ and at $-1$, and we have
$$
\frac{\frac 1 2 + u}{1+u} \begin{cases} >0 & \text{if }u>-1/2, & \\ <0 & \text{if }-1/2>u>-1, \\ > 0 & \text{if } -1>u. \end{cases}
$$
Observe that you never have $-1>\sin^2\theta$, so you can ignore the third piece above. Then you need to ask: If $-1/2>\sin^2\theta>-1$, then what is $\sin\theta$ and then what is $\theta$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/694637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find this ODE solution $(y-x)\sqrt{x^2+1}\dfrac{dy}{dx}=(1+y^2)^{\frac{3}{2}}$ Find all solutions $y$ to the ODE
$$(y-x)\sqrt{x^2+1}\dfrac{dy}{dx}=(1+y^2)^{\frac{3}{2}}$$
My try:
$$(y-x)\dfrac{d(\sqrt{1+y^2})}{y}=\dfrac{(1+y^2)dx}{\sqrt{x^2+1}}$$
then
$$\dfrac{(y-x)\cdot d(\sqrt{1+y^2})}{y(1+y^2)}=\dfrac{d(\sqrt{1+x^2})}{x}$$
then I can't.Thank you for you help
| Another CAS gave this "nice" expression
$$x(y)=\frac{-2 c_1 y+2 c_1 \tan ^{-1}(y)-c_1^2-\tan ^{-1}(y)^2+2 y \tan ^{-1}(y)+1}{c_1^2
y-2 c_1 y \tan ^{-1}(y)-2 c_1-y+y \tan ^{-1}(y)^2+2 \tan ^{-1}(y)}$$ which simplifies if $y(0)=0$ but does not change to be more pleasant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving for $\sum_{n = 1}^{\infty} \frac{n^3}{8^n}$? I was trying to solve $ \displaystyle \sum_{n = 1}^{\infty} \frac{n^3}{8^n}$ and I found a way to solve it and I want if there are generalizations for, say, $\displaystyle \sum_{n=1}^{\infty} \frac{n^k}{a^n}$ in terms of $k$ and $a$. I would also like to know if there is a better way to solve it. Here's how I did it:
First I decomposed the series into the following sums:
$S_1 = \frac{1}{8} + \frac{1}{64} + \dots = \frac{\frac{1}{8}}{\frac{7}{8}}$
$S_2 = \frac{7}{64} + \frac{7}{512} + \dots = \frac{\frac{7}{64}}{\frac{7}{8}}$
$S_3 = \frac{19}{512} + \frac{19}{4096} + \dots = \frac{\frac{19}{512}}{\frac{7}{8}}$
And deduced that the sum can be written as $\frac{8}{7} \displaystyle \sum_{n = 1}^{\infty} \frac{3n^2 - 3n + 1}{8^n}$
$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{8^n}$ is easy to evaluate -- it's $\frac{1}{7} $by geometric series
$\displaystyle \sum_{n = 1}^{\infty} \frac{n}{8^n}$ can be evaluated in a whole host of ways to get an answer of $\frac{8}{49}$.
It remains to evaluate $\displaystyle \sum_{n = 1}^{\infty} \frac{n^2}{8^n}$, for which I took a similar approach as the cubics by decomposing it into many sums:
$T_1 = \frac{1}{8} + \frac{1}{64} + \dots = \frac{\frac{1}{8}}{\frac{7}{8}}$
$T_2 = \frac{3}{64} + \frac{3}{512} + \dots = \frac{\frac{3}{64}}{\frac{7}{8}}$
And so forth, coming to the conclusion that it is equal to $\frac{8}{7} \displaystyle \sum_{n = 1}^{\infty} \frac{2n-1}{8^n}$
Now, I used this information and the above values for $\displaystyle \sum_{n = 1}^{\infty} \frac{1}{8^n}$ and $\displaystyle \sum_{n = 1}^{\infty} \frac{n}{8^n}$ to get the sum as $\frac{776}{2401}$, which is confirmed by WA.
So, I would like to reiterate here: Is there a simpler way to compute this sum, and are there any known generalizations for this problem given an arbitrary $a$ in the denominator and arbitrary $k$ as the exponent in the numerator?
| There turns out to be a standard trick that applies here: let
$$ f(x) = \sum_{i=1}^{\infty} \frac{x^n}{8^n} $$
We can compute this sum because it is a geometric series. The neat idea, now, is that
$$ f'(x) = \sum_{i=1}^{\infty} \frac{n x^{n-1}}{8^n} $$
or alternatively,
$$ x f'(x) = \sum_{i=1}^{\infty} \frac{n x^n}{8^n} $$
Repeat a few times, then plug in $x=1$, and you get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/698598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Triangle Inequality question with fractions Given $a, b \in \mathbb{R}$, prove that
$$\frac{|a + b|}{1 + |a + b|} \le \frac{|a|}{1 + |a|} + \frac{|b|}{1+|b|}$$
When does equality hold?
The only useful thing I could get (using the triangle inequality) is:
$$\begin{align}1 + \left|\left(a + \frac{ab}{2}\right) + \left(b - \frac{ab}{2}\right)\right| &\le 1 + \left|a + \frac{ab}{2}\right| + \left|b - \frac{ab}{2}\right|\\
&\le 1 + |a| + |b| + |ab|\\
&=(1 + |a|)(1 + |b|)\end{align}$$
But it seems to be inapplicable! Any ideas?
| Because the function $f(x)=\frac{1}{1+\frac{1}{x}}=\frac{x}{1+x}$ is increasing, we have
$$\frac{|a+b|}{1+|a+b|} = \frac{1}{1+\frac{1}{|a+b|}} \le \frac{1}{1+\frac{1}{|a|+|b|}}=\frac{|a|}{1+|a|+|b|}+\frac{|b|}{1+|a|+|b|}\le\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$$
In the first inequality we have used the triangle inequality $|a+b|\le |a|+|b|$ and in the second inequality, the positivity of $|a|$ and $|b|$.
That is, equality holds if and only if $a=0$ or $b=0$ because otherwise the last inequality is strict.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/698869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why does the following nonlinear system have 21 solutions? I am curious as to why the following nonlinear system has 21 solutions (according to Wolfram Alpha).
$$y+xy^2-x^3+2xz^4=0 \\ -x-y^3-3x^2y+3yz^4=0 \\ -\frac{5}{2}y^2z^3-2x^2z^3-\frac{z^7}{2}=0$$
I see that there is one real and twenty complex solutions:
$(0,0,0)$
and
Is there any way to figure out the number of solutions by looking at the degrees of the equations? Like the highest degree of any term is $7$, and there are $3$ variables, so there are $7 \times 3 = 21$ solutions?
| When we do the algebra to exactly solve this system, it has $21$ distinct solutions $(x,y,z)$ (disregarding multiplicity). As previously commented, it helps to separate the cases $z=0$ and $z \neq 0$.
Case: $z = 0$
When $z = 0$, the third equation holds automatically. Substituting $z = 0$ in the first two equations gives a system of two cubic polynomial equations in two unknowns $x,y$:
$$ y + xy^2 - x^3 = 0 $$
$$ x + y^3 + 3x^2y = 0 $$
A classic approach to eliminating variables in polynomial systems computes resultants to find common roots of the equations. Let's motivate that computation in the simple case at hand.
Think of these (homogeneous) equations as respectively a quadratic and a cubic in $y$, with coefficients that happen to be polynomials in $x$, we can formulate the system in a matrix form as follows:
$$ \begin{pmatrix} 0 & x & 1& -x^3 \\ 1 & 0 & 3x^2 & x \end{pmatrix}
\begin{pmatrix} y^3 \\ y^2 \\ y \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
This, superficially at least, separates the dependence on $x$ from the dependence on $y$. If only the "coefficients" formed a square matrix, we could take its determinant and set that to zero as a necessary condition for a common root.
With just a bit of manipulation, this is what the Sylvester matrix does for us. Let's stagger the rows some and expand to a square $5\times 5$ matrix thusly:
$$ \begin{pmatrix} x & 1& -x^3 & 0 & 0 \\ 0 & x & 1& -x^3 & 0 \\ 0 & 0 & x & 1& -x^3
\\ 1 & 0 & 3x^2 & x & 0 \\ 0 & 1 & 0 & 3x^2 & x \end{pmatrix}
\begin{pmatrix} y^4 \\y^3 \\ y^2 \\ y \\ 1 \end{pmatrix}
= \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$
So a necessary condition for a root $(x,y)$ satisfying both equations to exist is for the determinant of this Sylvester matrix to be zero. That determinant is the resultant of the two polynomials with respect to variable $y$, and gives us this equation:
$$ x(16x^8 + 8x^4 + 1) = x(4x^4 + 1)^2 = 0 $$
So either $x= 0$ or $4x^4 + 1 = 0$:
$$ x = 0, 0.5 + 0.5i, 0.5 - 0.5i, -0.5 + 0.5i, -0.5 - 0.5i $$
Remarkably the corresponding $y$ values satisfy the same equation, so we can piece together the five pairs $(x,y)$ which satisfy the system with $z=0$, just as shown in the Question. Easily, if $x=0$, the system implies $y^3 = 0$ and $y=0$ (and conversely), thus accounting for the only real solution to the system. In the other cases $x$ and $y$ are then nonzero.
One final observation here provides that there are exactly four nonzero solution pairs $(x,y)$. The first cubic equation can now be rewritten:
$$ \frac{y}{x} = x^2 - y^2 $$
The absolute values $|x|=|y|$ are all equal (whatever roots are chosen), and it follows that $|x^2 - y^2| = 1$. Since $x^2,y^2 = \pm \frac{i}{2}$, we must have $x^2 = -y^2$. The rewritten cubic now tells us $y = ix$ iff $x^2 = \frac{i}{2}$ and $y = -ix$ iff $x^2 = -\frac{i}{2}$.
Case: $z \neq 0$
Dividing the third of the Question's original equations by $-z^3/2$ (since it is nonzero), we have:
$$ 5y^2 + 4x^2 + z^4 = 0 $$
Since the only real solution of this is the trivial one already noted, we will not find any further real solutions. However substituting $-z^4 = 5y^2 + 4x^2$ into the first two original equations gives once more a pair of cubic equations in two unknowns $x,y$:
$$ y - 9xy^2 - 9x^3 = 0 $$
$$ x + 16y^3 + 15x^2y = 0 $$
The resultants of these equations with respect to $x,y$ give us:
$$ 9y(144y^8 + 24y^4 + 1)= 9y(12y^4 + 1)^2 = 0 $$
$$ -x(729x^8 + 216x^4 + 16)= -x(27x^4 + 4)^2 = 0 $$
Discarding all but the nonzero roots, $x^4 = -\frac{4}{27}$ and $y^4 = -\frac{1}{12}$. Presumably if one checks, each choice of such root $x$ corresponds to one for $y$, so that we get four nonzero solution pairs $(x,y)$. However now $z^4 = -(5y^2 + 4x^2)$ will give us, for each nonzero solution pair $(x,y)$, four nonzero fourth roots for $z$. This is precisely what the Wolfram Alpha output cited in the Question says.
Altogether the $z=0$ cases involved five solutions triples (only one real), and the $z \neq 0$ cases provide additionally $4\times 4 = 16$ solution triples, for a combined $21$ distinct solution points $(x,y,z)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/704281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
a quadratic equation for two unknown number find values of $p$ such that the equation $4x^2 + 3px - 2p = 0$ has? below are a few choices of the value p:
a) 2 real roots
b) 1 real roots
c) no roots or complex roots
so far i did for a) 2 real roots
$$(3p)^2 - 4(4)(-2p) > 0\\
9p^2 +32p > 0\\
9\left( p^2+\frac{32p}{9} \right) > 0\\
p^2+\frac{32p}{9} > 0\\
p^2 +\frac{32p}{9} + \left(\frac{32}{18}\right)^2 > 0\\
\left(p + \frac{32}{18} \right) ^2 > 0 \\
p + \frac{32}{18} > 0\\
p > -\frac{32}{18} $$
| The roots are given by the quadratic formula:
$$
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
= \frac{-3C \pm \sqrt{9C^2 + 32p}}{2a}
$$
If the thing under the square root is positive, you'll get two roots. Under what conditions will you get just one? Under what condition will you get none?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/706694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How does $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$? Good afternoon my wonderful friends!
Whenever I do this equation I set it up using the difference of two cubes, which is as follows:
$(a+b)^3 = (a+b)(a^2-ab+b^2) = a^3 + b^3$
Whenever I try to use this formula I always get:
$(x+h) x^2 - xh + h^2$
Simplify:
$x^3 - x^2h + h^2x + x^2h - xh^2 + h^3$
Simplify:
x^3 + h^3$
I don't understand were the three's come from in the final answer: $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$
| I don't know where you get "$a^2-ab+b^2$" from. Just use distributivity to carry out the multiplications by hand:
\begin{align}
(a+b)^2 &= (a+b)(a+b) = a(a+b)+b(a+b) \\&= a^2+ab+ba+b^2 = a^2+2ab+b^2
\end{align}
Thus,
\begin{align}
(a+b)^3 &= (a+b)(a+b)^2 = (a+b)(a^2+2ab+b^2) = a(a^2+2ab+b^2)+b(a^2+2ab+b^2) \\&= a^3+2a^2 b+ab^2 + ba^2 + 2ab^2+b^3 = a^3 + 3a^2b + 3ab^2 + b^3.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/708165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Closed Form for $\int_0^1 \frac{\log(x)}{\sqrt{1-x^2}\sqrt{x^2+2+2\sqrt{2}}}dx$ Is there a closed form for the following integral?
$$\int_0^1 \frac{\log(x)}{\sqrt{1-x^2}\sqrt{x^2+2+2\sqrt{2}}}dx$$
It is approximately equal to $-0.48878092308456029189008$.
Mathematica is unable to find a closed form for this integral.
| We let $$I(a,b)=\int^1_0\frac{x^{2a}}{\sqrt{1-x^2}\sqrt{x^2+b}}dx$$
so that OP's integral is equal to $\frac12\frac{d}{da}I(0,2+2\sqrt2)$.
According to Mathematica, we have $$I(a,b)=\frac{\sqrt\pi}{2\sqrt b}\frac{\Gamma(a+1/2)}{\Gamma(a+1)}{}_2F_1(\frac12,a+\frac12;a+1\mid-1/b)\\
=\frac{\sqrt\pi}{2\sqrt b}\frac{\Gamma(a+1/2)}{\Gamma(a+1)}(1+\frac1b)^{-1/2}{}_2F_1(\frac12,\frac12;a+1\mid\frac{1}{b+1})
=\frac{\sqrt\pi}{2\sqrt{b+1}}\frac{\Gamma(a+1/2)}{\Gamma(a+1)}{}_2F_1(\frac12,\frac12;a+1\mid\frac{1}{b+1}).$$
I'm currently looking for suitable transforms to relate $I(a,b)$ to $I(-a,b)$. It looks like DLMF15.8.4 will be a good candidate.
Edit: Using DLMF15.8.4, we have $$\sin(a\pi)\sqrt{b}I(a,b)=\cos(a\pi)I(-a,1/b)-b^aI(a,1/b).$$
Taking derivative with respect to $a$: $$\sin(a\pi)\sqrt{b}I_a(a,b)+\pi\cos(a\pi)\sqrt{b}I(a,b)=-\cos(a\pi)I_a(-a,1/b)-\pi\sin(a\pi)I(-a,1/b)-b^aI_a(a,1/b)+b^a\log bI(a,1/b).$$
Let $a=0$:
$$\pi\sqrt{b}I(0,b)=-I_a(0,1/b)-I_a(0,1/b)-\log bI(0,1/b).$$
Noting $$I(0,b)=\frac{K\left(\sqrt{\frac{1}{1+b}}\right)}{\sqrt{1+b}},$$ we have $$I_a(0,1/b)=-\frac12(\pi\sqrt{b}I(0,b)+\log bI(0,1/b))\\
=-\frac12\sqrt{\frac{b}{1+b}}\left(\pi K\left(\sqrt{\frac{1}{1+b}}\right)+\log b K\left(\sqrt{\frac{b}{1+b}}\right)\right).$$
Thus, we can get OP's integral by taking $b=1/(2+2\sqrt2)=\frac{\sqrt2-1}{2}$:
$$I=\frac12I_a(0,2+2\sqrt2)=-\frac{\sqrt2-1}{4}\left(\pi K\left(\sqrt{2\sqrt2-2}\right)+\log\left(\frac{\sqrt2-1}{2}\right) K\left(\sqrt2-1\right)\right)\\
=-\frac{\sqrt2-1}{4}\left(\pi K'\left(\sqrt2-1\right)-\log\left(2\sqrt2+2\right) K\left(\sqrt2-1\right)\right).$$
Now note that $\sqrt2-1=k_2$ is the second Elliptic Integral Singular Value. Using the value of $K(k_2)$ there, we conclude that
$$I=-\frac{\sqrt2-1}{4}\left(\pi K'\left(\sqrt2-1\right)-\log\left(2\sqrt2+2\right) K\left(\sqrt2-1\right)\right)\\
=-\frac{\sqrt2-1}{4}K(k_2)\left(\pi \frac{K'(k_2)}{K(k_2)}-\log\left(2\sqrt2+2\right)\right)\\
=-\frac{\sqrt2-1}{4}\frac{\Gamma(\tfrac18)\Gamma(\tfrac38)\sqrt{\sqrt2+1}}{2^{13/4}\sqrt{\pi}}\left(\sqrt{2}\pi-\log\left(2\sqrt2+2\right)\right)\\
=-\frac{\Gamma(\tfrac18)\Gamma(\tfrac38)\sqrt{\sqrt2-1}}{2^{21/4}\sqrt{\pi}}\left(\sqrt{2}\pi-\log\left(2\sqrt2+2\right)\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/709443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
Value of this definite integral $\int_{0}^{\infty} \frac{ \ln(x)}{x^2+2x+4} dx $ So I came across this question on brilliant.org and didn't know how to go about it:
$$\int_{0}^{\infty} \frac{ \ln(x)}{x^2+2x+4} dx $$
I tried to complete the squares in the denominator and then use a trigonometric substitution (with tan) but I didn't get anything beyond that.
How do I do it? Do we need to differentiate under the integral sign here?
| This CAN be done with real variables and no, you do not need to differentiate with respect to a parameter. Here's how:
$$I=\int_{0}^{\infty} \frac{\ln(x)}{x^2+2x+4} dx $$
$$I=\int_{0}^{\infty} \frac{\ln(x)}{{(x+1)}^2+3}dx $$
$$I=\frac{1}{3} \int_{0}^{\infty} \frac{\ln(x)}{ ({ \frac{x+1}{\sqrt{3}} })^2 + 1} dx $$
Now, set
$$\frac{x+1}{\sqrt{3}} = \tan(t)$$
Thus $t$ is between $\frac{\pi}{6}$ and $\frac{\pi}{2}$.
We now have
$$I=\frac{1}{\sqrt{3}} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \ln{ (\sqrt{3} \tan(t) - 1 )} dt $$
Now, use this property:
$$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$$
So, we have:
$$I= \frac{1}{\sqrt{3}} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \ln{ \left( \sqrt{3} \frac{ \sqrt{3} + \tan(t) }{\sqrt{3} \tan(t) - 1} - 1\right) }dt $$
On simplifying,
$$I=\frac{1}{\sqrt{3}} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \ln{\left( \frac{4}{\sqrt{3} \tan(t) - 1}\right)} dt $$
Thus,
$$I=\frac{1}{\sqrt{3}} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \ln{(4)} dt - I$$
$$I= \frac{\pi}{3 \sqrt{3} } \ln{2} $$
and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/709516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
A question on eigenvalues Let $A,B\in M_{2}(\mathbb{R})$ so that $A^2 = B^2 = I$. Which are eigenvalues of $AB$?
1) $1\pm \sqrt 3$
2)$3 \pm 2\sqrt2$
3)$\dfrac {1}{2},2$
4)$2 \pm 2\sqrt 3$
| It's easy to show that possibilities (1) and (4) are impossible. Note that the magnitude of the determinants for both $A$ and $B$ must be $1$, indeed
$$1 = \det(A^2) = \det(B^2) \implies |\det(A)| = |\det(B)| = 1$$
This necessarily implies that $|\det(AB)|=1$, i.e. the product of the eigenvalues must have magnitude $1$. This condition is satisfied by options (2) and (3), where the product of the eigenvalues is just $1$, but not by options (1) and (4).
Now let me show that both possibilities (2) and (3) are possible. For (3), let
$$A = \begin{pmatrix}\sqrt{2} & 1 \\ -1 & -\sqrt{2}\end{pmatrix},\ \ \ \ \ \text{and}\ \ \ \ \ \ B=\begin{pmatrix}\frac{1}{\sqrt{2}} & -1 \\ -\frac{1}{2} & -\frac{1}{\sqrt{2}}\end{pmatrix}$$
You can easily verify that $A$ and $B$ satisfy $A^2 = B^2 = I$. Their product is given by
$$AB = \begin{pmatrix}\frac{1}{2} & -\sqrt{2}-\frac{1}{\sqrt{2}} \\ 0 & 2\end{pmatrix}$$
which has eigenvalues $1/2$ and $2$, exactly as required. Let me now elaborate on how I found this example, and the same procedure will allow you to construct an example for case (2).
There is a rather well known parametrization for matrices satisfying $A^2 = I$ in the $2\times 2$ case. You can check that any $2\times 2$ matrix in the form of
$$A=\begin{pmatrix}x & y \\ \frac{1-x^2}{y} & -x\end{pmatrix}$$
for any $x,y\in\mathbb{R}$ will satisfy $A^2 = I$. This condition can be found in the wikipedia article for involutory matrices for example. To simplify the expression above, let's set $y=1$ in general. Then we try a product of the form
$$AB = \begin{pmatrix}a & 1 \\ 1-a^2 & -a\end{pmatrix}\begin{pmatrix}b & 1 \\ 1-b^2 & -b\end{pmatrix} = \begin{pmatrix}ab+1-b^2 & a-b\\ab^2 + (1-a^2)b - a & 1-a^2+ab\end{pmatrix}$$
where $a$ and $b$ are free to vary. The idea is to choose $a$ and $b$ so that we have our desired eigenvalues. However it is still rather difficult to find the eigenvalues we need. So let's try making the product matrix triangular so that we can read the eigenvalues right off the diagonal.
Making the product lower triangular requires $a=b$ which just gives the trivial solution $A=B$ and $AB=I$.
If we try making the product upper triangular, we find the condition $b = -\frac{1}{a}$ by solving the quadratic in $b$ found in the bottom left entry. The product then reduces to
$$AB = \begin{pmatrix}-\frac{1}{a^2} & a+\frac{1}{a} \\ 0 & -a^2\end{pmatrix}$$
This is almost what we need, except that the eigenvalues here are negative. To fix this, we simply take the negative of $B$ used in the expressions above, i.e.
$$B=-\begin{pmatrix}b & 1 \\ 1-b^2 & -b\end{pmatrix}$$
We've lucked out here since the solution now allows us to produce any pair of positive eigenvalues $\lambda_1$ and $\lambda_2$ such that $\lambda_1\lambda_2 = 1$.
Taking $a = \sqrt{2}$ gives a solution for case (3) which I've included above. Taking $a = \sqrt{3 + 2\sqrt{2}}$ will then give case (2) which I leave you to work out yourself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/713949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Derivatives of Logarithmic function Determine $f'(x)$ for $f(x) = ln(x + \sqrt{x^2 + 1})$
My handbook has the answer as $\displaystyle\frac{1}{\sqrt{x^2 + 1}}$ with no steps on how they got there.
I tried to get there, but somewhere I am getting things wrong. This is what I have
\begin{align}
f'(x) & = \displaystyle\frac{1}{x + \sqrt{x^2 + 1}}\cdot\tfrac{d}{dx}(x + \sqrt{x^2 + 1})\\
& = \displaystyle\frac{\tfrac{1}{x + \sqrt{x^2 + 1}} \cdot\tfrac{d}{dx}(x + \sqrt{x^2 + 1})}{x + \sqrt{x^2 + 1}}\\
& = \displaystyle\frac{\tfrac{1}{x + \sqrt{x^2 + 1}} \cdot (1 + \tfrac{d}{dx}(x^2 + 1)^{\tfrac{1}{2}})}{x + \sqrt{x^2 + 1}} \\
& = \displaystyle\frac{\tfrac{1}{x + \sqrt{x^2 + 1}} \cdot (1 + \tfrac{1}{2}(\tfrac{d}{dx}x^2 + 0)^{-\tfrac{1}{2}})}{x + \sqrt{x^2 + 1}} \\
& = \displaystyle\frac{\tfrac{1}{x + \sqrt{x^2 + 1}} \cdot (1 + \tfrac{1}{2}(2x)^{-\tfrac{1}{2}})}{x + \sqrt{x^2 + 1}} \\
& = \displaystyle\frac{\tfrac{1}{x + \sqrt{x^2 + 1}} \cdot (1 + \tfrac{1}{2\sqrt{2x}})}{x + \sqrt{x^2 + 1}} \\
& = \displaystyle\frac{\tfrac{1}{x + \sqrt{x^2 + 1}} + \tfrac{2\sqrt{2x}}{x + \sqrt{x^2 + 1}}}{x + \sqrt{x^2 + 1}} \\
\end{align}
Here I gave up, I have a mistake, but I can't figure out where
| Using the chain rule: set u = x + (x^2 + 1)^(1/2) ==> du/dx = 1 + x/(x^2 + 1)^(1/2) =
= (x + (x^2 + 1)^1/2)/(x^2 + 1)^1/2, and dy/du = 1/u = 1/(x + (x^2 + 1)^1/2). So:
dy/dx = dy/du*du/dx = 1/(x^2 + 1)^1/2.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/714599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Taylor Series of $ \frac{1}{1-x^2} $ about x=2 I am trying to form a taylor series of the following:
$ \frac{1}{1-x^2} $ about $x=2$
I tried factoring the equation such that it becomes the following:
$ \frac{1}{{(1+x)}{(1-x)}} $
I tried to substitute $ x = h + 2 $ into the equation and obtained the following after using partial fractions to decompose the result:
$ \frac{1}{2(h+3)} - \frac{1}{2(h+1)} $
I do not know how to proceed from here.
I know I can just compute all the derivatives of the expression and evaluate them. But this would be non-trivial. Could someone please advise me on how I could solve this question?
| Hint: Since $x = (x-2) + 2$, then $x^2 = [(x-2) + 2]^2 = (x-2)^2 + 4(x-2) + 4$ so
\begin{align*}
\frac{1}{1 - x^2} &= \frac{1}{1 - [(x-2)^2 + 4(x-2) + 4]} = \frac{1}{-(x-2)^2 -4(x-2)-3}\\
&= -\frac{1}{[(x-2) + 3][(x-2) + 1]} \, .
\end{align*}
Now use partial fractions to finish it off.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/714997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
I am not sure if the answer for the dividing rational expressions problem should be simplified $$ \frac{x}{x+2} \div \frac{1}{x^2 - 4} $$
-I am not sure if the answer for the problem (which is attached) would be $\frac{x^3-4x}{x+2}$ or if it could be simplified further.
| Remember that:
$$a \div \dfrac{1}{c}=ac$$
Therefore we can convert
$$\frac{x}{x+2} \div \dfrac{1}{x^2-4}$$
into:
$$\left(\frac{x}{x+2}\right)\left(x^2-4\right)$$
We can simplify this by factoring $x^2-4$ using the difference of squares formula, which is:
$$a^2-b^2=(a+b)(a-b)$$
So:
$$\left(\frac{x}{x+2}\right)\left(x^2-4\right)=\left(\frac{x}{x+2}\right)(x+2)(x-2)$$
Do you see that we can cancel $x+2$ out?
$$\require{cancel}{\left(\frac{x}{\cancel{x+2}}\right)\cancel{(x+2)}(x-2)}$$
Now we have:
$$x(x-2)$$
$$=x^2-2x$$
$$\displaystyle \boxed{\therefore \dfrac{x}{x+2} \div \dfrac{1}{x^2-4}=x^2-2x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/717238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the determinant without row expansion Show that the determinant of the matrix
\begin{bmatrix} 1& a& a^3\\
1& b& b^3\\
1& c& c^3\end{bmatrix}
is $(a-b)(b-c)(c-a)(a+b+c)$ without expanding.
I was able to get out $(a-b)(b-c)(c-a)$ but couldn't complete.
| Use row operations to simplify. In the process, the factorization drops right out.
\begin{align}
\det \begin{bmatrix}
1 & a & a^3 \\
1 & b & b^3 \\
1 & c & c^3
\end{bmatrix}
&= \det \begin{bmatrix}
1 & a & a^3 \\
0 & b-a & b^3-a^3 \\
0 & c-a & c^3-a^3
\end{bmatrix} \\
&= \det \begin{bmatrix}
1 & a & a^3 \\
0 & b-a & (b-a)(b^2+ab+b^2) \\
0 & c-a & (c-a)(c^2+ac+a^2)
\end{bmatrix} \\
&= (b-a)(c-a)\det \begin{bmatrix}
1 & a & a^3 \\
0 & 1 & b^2+ab+a^2 \\
0 & 1 & c^2+ac+a^2
\end{bmatrix} \\
&= (b-a)(c-a)\det \begin{bmatrix}
1 & a & a^3 \\
0 & 1 & b^2+ab+a^2 \\
0 & 0 & c^2-b^2+ac-ab
\end{bmatrix} \\
&= (b-a)(c-a)(c^2-b^2 + ac-ab)\det \begin{bmatrix}
1 & a & a^3 \\
0 & 1 & b^2+ab+a^2 \\
0 & 0 & 1
\end{bmatrix} \\
&= (b-a)(c-a)(c-b)(c+b+a) \\
&= (a-b)(b-c)(c-a)(a+b+c).
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/720589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do I solve this integral with hyperbolic functions? I was studying mechanics when I f ound a problem that lead to an integral that I can't solve. Basically the problem asked to find the period of oscillation function of the energy $E$ of a particle (mass $m$) in a field with a potential defined by:
$\displaystyle U(x) = -\frac{U_0}{\cosh^2(\alpha x)}$ with $-U_0 < E < 0$
From the differential equation $\displaystyle \frac{1}{2}m\dot{x}^2 -\frac{U_0}{\cosh^2(\alpha x)} = E$ you get the following integral for the period:
$\displaystyle T = 2\sqrt{2m} \int_0^{x_1} \frac{dx}{\sqrt{E + \frac{U_0}{\cosh^2(\alpha x)}}} $
Where $x_1$ is the positive solution of the equation $U(x) = E \implies \displaystyle x_1 = \frac{1}{\alpha} \cosh^{-1} \left( \sqrt{-\frac{U_0}{E}}\right)$.
The integral can then be written as:
$\displaystyle T = \frac{2\sqrt{2m}}{\alpha\sqrt{|E|}} \int_0^{\alpha x_1} \frac{dx}{\sqrt{1 + \frac{U_0}{E}\frac{1}{\cosh^2(x)}}}$
and that's where my solution stops... I can't find a good substitution to solve this. Thanks.
| When pulling $\sqrt{\lvert E\rvert}$ out of the denominator, you made a sign error. Since $E < 0$, you have
$$E + \frac{U_0}{\cosh^2 (\alpha x)} = (-E)\left(- 1 + \frac{-U_0/E}{\cosh^2 (\alpha x)}\right),$$
so the period would be
$$T = \frac{2\sqrt{2m}}{\alpha\sqrt{\lvert E\rvert}} \int_0^{\alpha x_1} \frac{dx}{\sqrt{\frac{-U_0/E}{\cosh^2 x} - 1}}.$$
Now let us write $c^2 = -U_0/E$ and $\gamma = \cosh^{-1} c = \alpha x_1$. Substituting $u = \cosh x$, we find
$$\begin{align}
\int_0^\gamma \frac{dx}{\sqrt{\left(\frac{c}{\cosh x}\right)^2-1}} &= \int_1^c \frac{1}{\sqrt{\left(\frac{c}{u}\right)^2-1}}\cdot \frac{du}{\sqrt{u^2-1}}\\
&= \int_1^c \frac{u\,du}{\sqrt{(c^2-u^2)(u^2-1)}}\\
&= \frac{1}{2}\int_1^{c^2} \frac{dv}{\sqrt{(c^2-v)(v-1)}}\tag{$v = u^2$}\\
&= \frac{1}{2} \int_1^{c^2} \frac{\sqrt{(c^2-v)(v-1)}}{c^2-1}\left(\frac{1}{v-1} + \frac{1}{c^2-v}\right)\,dv\\
&= \frac{1}{2} \int_1^{c^2} \sqrt{\frac{c^2-v}{v-1}} + \sqrt{\frac{v-1}{c^2-v}}\,dv.
\end{align}$$
Now we can substitute $t = \frac{c^2-v}{v-1}$ for the first summand, and $t = \frac{v-1}{c^2-v}$ for the second to obtain
$$\begin{align}
\int_0^\gamma \frac{dx}{\sqrt{\left(\frac{c}{\cosh x}\right)^2-1}} &= \int_0^\infty \frac{\sqrt{t}}{(t+1)^2}\,dt\tag{$\ast$}\\
&= B\left(\tfrac{1}{2},\tfrac{3}{2}\right)\\
&= \frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{3}{2}\right)}{\Gamma(2)}\\
&= \frac{\sqrt{\pi}\cdot \frac{1}{2}\sqrt{\pi}}{1}\\
&= \frac{\pi}{2},
\end{align}$$
and
$$T = \frac{\sqrt{2m}\cdot\pi}{\alpha\sqrt{\lvert E\rvert}}.$$
An alternative way to evaluate the integral $(\ast)$ is the residue theorem:
Choosing the branch of $\sqrt{z}$ with $\operatorname{Im}\sqrt{z} > 0$ on $\mathbb{C}\setminus [0,\infty)$ and a keyhole contour, the integrals over the circular arcs tend to $0$ in the limit, and $\lim\limits_{\varepsilon \downarrow 0} \sqrt{t+i\varepsilon} = \sqrt{t}$, $\lim\limits_{\varepsilon\downarrow 0} \sqrt{t-i\varepsilon} = -\sqrt{t}$ lead to
$$2\int_0^\infty \frac{\sqrt{t}}{(t+1)^2}\,dt = 2\pi i \operatorname{Res} \left(\frac{\sqrt{z}}{(z+1)^2}; -1\right) = 2\pi i \frac{1}{2\sqrt{-1}} = \pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/721129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $n^7+7$ can never be a perfect square. Prove that for a positive integer $n$, $n^7+7$ cannot be a perfect square.
I managed to show that $n \equiv 5 \pmod{8}$ or $n \equiv 9 \pmod{16}$. But nothing came from that so I presume another approach is needed???
Thanks for any help.
| Here is an alternative proof that is hopefully slightly simpler.
We will make use of a simple theorem whose proof is skipped here:
For any odd integer number $m$, we have $m^2\equiv 1(mod 4)$.
Assume on the contrary, $n^7+7$ is a perfect square, then there exists an integer $x$ such that $n^7+7=x^2$.
If $x$ is odd, $n$ would have to be even. And the left side would be $3 (mod 4)$ that contradicts the fact that the right side is $1 (mod 4)$ by the theorem above. So $x$ has to be even, and $n$ has to be odd. Further more, applying the above theorem on $n^6=(n^2)^3$, we know $n^7\equiv n (mod 4)$. Note the right hand size is $0 (mod 4)$ since $x$ is even, we must have $$n\equiv 1 (mod 4).$$
Let's denote $x=2b$ where $b$ is another positive integer. Now we get $n^7+7=4b^2$. Further utilizing the fact that $7=2^7-11^2$. We can rewrite the above equation as $n^7+2^7=4b^2+11^2$, or
$$(n+2)(n^6-2n^5+4n^4-8n^3+16n^2-32n+64)=4b^2+11^2.$$
Now, notice $n+2\equiv 3 (mod 4)$, it MUST have a prime factor $p$ with $p \equiv 3 (mod 4)$, and $p \mid (4b^2+11^2)$. We next consider two cases:
Case one: $11 \nmid b.$
Since $p \mid (4b^2+11^2)$, we know $p \neq 11$, $p \nmid b$, $4b^2 \equiv -11^2(mod p)$. Note $\frac{p-1}{2}$ is odd, ${\left(4b^2\right)}^{\frac{p-1}{2}} \equiv {\left(-11^2\right)}^{\frac{p-1}{2}}(mod p)$, ${(2b)}^{p-1}=-11^{p-1}(mod p)$. Applying Fermat's little theorem, we get $1 \equiv -1 (mod p)$, a contradiction.
Case two: $11 \mid b.$
Let $b=11 \times a$, $a$ is an integer. We have
$$ (n+2)(n^6-2n^5+4n^4-8n^3+16n^2-32n+64)=(4a^2+ 1)\times 11^2. $$
We can manually verify that
$$ 11 \nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $$ for $n \equiv -5 (mod 11)$ to $n \equiv 5 (mod 11).$ Hence $11^2 \mid n+2.$ Note $11^2 \equiv 1(mod 4)$, we still have the claim that $\frac{n+2}{11^2}$ has a prime factor $p$ with $p \equiv 3(mod 4)$, and $p \mid 4a^2+1$.
$$ 4a^2 \equiv -1 (mod p)\\ {\left(4a^2\right)}^{\frac{p-1}{2}} \equiv -1 (mod p) \\ {\left(2a\right)}^{p-1} \equiv -1 (mode p) \\ 1 \equiv -1 (mod p),$$
a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/721834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Prove $2(x^4+y^4+z^4)+2xyz+7\ge 5(x^2+y^2+z^2)$ for $x, y, z \ge 0$
Let $x,y,z\ge 0$. Show that
$$2(x^4+y^4+z^4)+2xyz+7\ge 5(x^2+y^2+z^2).$$
my idea: let $$x+y+z=p,xy+yz+xz=q,xyz=r$$
since
$$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=p^2-2q$$
and
$$(xy+yz+xz)^2=x^2y^2+y^2z^2+x^2z^2+2xyz(x+y+z)$$
$$\Longrightarrow x^2y^2+y^2z^2+x^2z^2=q^2-2pr$$and
$$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+x^2z^2)=(p^2-2q)^2-2(q^2-2pr)$$
so
$$\Longleftrightarrow 2[(p^2-2q)^2-2(q^2-2pr)]+2r+7\ge 5(p^2-2q)$$
$$\Longleftrightarrow 2p^4-8p^2q+4q^2+8pr-5p^2+10q+2r+7\ge 0$$
then I can't
This link has a similar problem:
see this
Maybe this problem can use AM-GM inequality,But I can't.Thank you
| Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that $f(w^3)\geq0$, where
$$f(w^3)=2(81u^4-108u^2v^2+18v^4+12uw^3)+2w^3+7-5(9u^2-6v^2).$$
But, $f$ is a linear function, which says that it's enough to prove our inequality
for an extremal value of $w^3$.
We know that $x$, $y$ and $z$ are non-negative roots of the equation
$$(X-x)(X-y)(X-z)=0$$ or
$$X^3-3uX^2+3v^2X-w^3=0$$ or
$$w^3=X^3-3uX^2+3v^2X,$$
which says that a line $Y=w^3$ and the graph of $Y=X^3-3uX^2+3v^2X$
have three common points and $w^3$ gets an extremal value, when the line $Y=w^3$
is a tangent line to the graph $Y=X^3-3uX^2+3v^2X$,
which happens for equality case of two variables, or maybe $w^3=0$.
Id est, it's enough to check two cases.
*
*$w^3=0$.
Let $z=0$.
We need to prove that
$$2(x^4+y^4)+7\geq5(x^2+y^2)$$ or
$$2x^4-5x^2+3.5+2y^4-5y^2+3.5\geq0,$$
which is true because $5^2-4\cdot2\cdot3.5<0$;
*$y=x$.
We need to prove that
$$4x^4+2(z-5)x^2+2z^4-5z^2+7\geq0,$$
for which it's enough to prove that
$$(z-5)^2-4(2z^4-5z^2+7)\leq0$$ or
$$(z-1)^2(8z^2+16z+3)\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/722824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
} |
How to factor the polynomial $x^4-x^2-2x-1$? By taking advantage of a computer algebra system, I found out that
$$x^4 - x^2 - 2x - 1 = (x^2 + x + 1)(x^2 - x - 1)$$
However, I don't know a straightforward way to solve this by hand. This was in a high-schooler's homework set, so I am assuming it can be solved using elementary methods.
| Upon looking at this expression I immediately noticed that the $-x^2-2x-1$ could be simplified to $-(x+1)^2$. That is how you should be later on. For example, when you see $4x^2-4x+1$, you should right away notice that it equals $(2x-1)^2$
$$x^4-x^2-2x-1=x^4-(x+1)^2$$
$$=\left(x^2\right)^2-(x+1)^2$$
$$=(x^2+x+1)(x^2-(x+1)$$
$$=(x^2+x+1)(x^2-x-1)$$
$$\displaystyle \color{green}{\boxed{\therefore x^4-x^2-2x-1=(x^2+x+1)(x^2-x-1)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/723228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Two definite integrals These integrals are closely related since $\frac{\pi^2}{8}=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}$ and $G=\sum_{n\mathop=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^2}$. I'm not able to prove them though.
Show that
$$\int_0^1 \frac{1}{\sqrt{x^2-1}}\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)\,\mathrm{d}x =\frac{\pi^2}{8}$$
$$\int_1^{\infty} \frac{1}{x\sqrt{x^2-1}}\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)\,\mathrm{d}x =2G$$
where $G$ is Catalan's constant.
| Note that $\cosh^{-1}{x}=\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)$ and $\frac{d}{dx}\cosh^{-1}{x}=\frac{1}{\sqrt{x^2-1}}$. Substituting $u=\cosh^{-1}{x}$ into the first integral, we get:
$$\int_0^1 \frac{1}{\sqrt{x^2-1}}\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)\,\mathrm{d}x=\int_{\frac{i\pi}{2}}^{0}udu=\frac12u^2\big |_{\frac{i\pi}{2}}^0=-\frac12(\frac{i\pi}{2})^2=\frac{\pi^2}{8}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/724865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Showing a function $f$ has a zero I have been working on this problem and was wondering if anyone could give me some hints as to how to answer this. So far as $p$ is a polynomial it is continuous and $g$ is given as being continuous, and so by algebra of continuity $f$ is continuous, now does the solution include the use of the Intermediate Value Theorem?
Any help would be appreciated.
Let $f = p + g$ , where $p$ is an odd degree polynomial and $g:$ $\mathbb{R}$ $\rightarrow$ $\mathbb{R} $ is a bounded, continuous function. Show that f must have a zero.
| Suppose that $g$ is continuous and bounded by $M>0$ and $p(x)= x^n+\ldots+a_0$ where $n$ is odd. Then $f(x)=x^n+ a_{n-1}x^{n-1}\ldots+a_0+g(x)$ which may be written as $x^n\big(1+\ldots\frac{a_0}{x^n}+\frac{g(x)}{x^n}\big)$ when $x\not=0$, the basic idea is that when $|x|$ is big $x^n$ dominate and $f$ contain the sign of this.
More formally we need to estimate the quantity inside the parenthesis.
\begin{align}\bigg|\frac{a_{n-1}}{x}+\ldots+\frac{a_0}{x^n}+\frac{g(x)}{x^n}\bigg|\le\frac{|a_{n-1}|}{|x|}+\ldots+\frac{|a_0|}{|x^n|}+\frac{|g(x)|}{|x^n|}\\
\le\frac{|a_{n-1}|}{|x|}+\ldots+\frac{|a_0|}{|x^n|}+\frac{M}{|x^n|}\end{align}
Let choose $|x|\ge N=\max\{1,2^2|a_{n-1}|\ldots,2^{n+1}|a_0|,2^{n+2}M\}$
Thus
\begin{align}\frac{|a_{n-1}|}{|x|}+\ldots+\frac{|a_0|}{|x^n|}+\frac{M}{|x^n|}\le\frac{|a_{n-1}|}{|x|}+\ldots+\frac{|a_0|}{|x|}+\frac{M}{|x|}\\
\le\frac{1}{2^2}+\ldots+\frac{1}{2^{n+1}}+\frac{1}{2^{n+2}}\\
<\sum_{n=2}^\infty \frac{1}{2^n}=\frac{1}{2}\end{align}
Hence $\bigg|\frac{a_{n-1}}{x}+\ldots+\frac{a_0}{x^n}+\frac{g(x)}{x^n}\bigg|< \frac{1}{2}$ and so
$$\frac{1}{2}< 1+\frac{a_{n-1}}{x}+\ldots+\frac{a_0}{x^n}+\frac{g(x)}{x^n}$$
Then for $x\ge N$ we therefore have
$$0<\frac{x^n}{2}< x^n \bigg(1+\frac{a_{n-1}}{x}+\ldots+\frac{a_0}{x^n}+\frac{g(x)}{x^n}\bigg)=f(x)$$
and for $y\le -N$ we thus have
$$0>\frac{y^n}{2}>y^n \bigg(1+\frac{a_{n-1}}{y}+\ldots+\frac{a_0}{y^n}+\frac{g(y)}{y^n}\bigg)=f(y)$$
Since $f$ is continuous then is continuous on $[y,x]$ and since $f(y)<0<f(x)$. Hence by the IVT we have that there is a $c\in (y,x)$ such that $f(c)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/725079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Help with discrete math proof? I am having trouble proving the following:
If $x\in R$ and $x > 0$, then $x^4+1 \geq x^3+x$.
Work: I tried to rearrange the equation as $x^4-x^3-x+1 \geq 1$, but that does not really help. I also tried proof by cases where case 1 would be that x is irrational and case 2 would be that x is rational. However, that has not got me far either. I am not really sure how to approach this problem.
| Hint: try the rearrangement inequality.
details: if $x>1$:
$$
\frac 1x < 1\\
x<x^4\\
x + x^3 = 1\cdot x + \frac 1x \cdot x^4 \le \frac 1x \cdot x + 1\cdot x^4 = 1+x^4
$$
Otherwise:
$$
\frac 1x \ge 1\\
x\ge x^4\\
x + x^3 = 1\cdot x + \frac 1x \cdot x^4 \le \frac 1x \cdot x + 1\cdot x^4 = 1+x^4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/725447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove $\frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta} = \frac{2}{\sin\theta}$ How to prove:
$\frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta} = \frac{2}{\sin\theta}$
Please help.
| $$\sin^2\theta=1-\cos^2\theta=(1-\cos\theta)(1+\cos\theta)$$
$$\implies \frac{\sin\theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin\theta}$$
Now add the second term
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/726442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Finding $Ker(T_a)$ and $Im(T_a)$ there will be a matrix $A \in R^{2 x 2}$, $T_A:R^{2 x 2} \rightarrow R^{2 x 2}$
Define by $T_A(X)=$$
\begin{pmatrix}
1 & 2\\
0 & 3 \\
\end{pmatrix}
$$X-X$$
\begin{pmatrix}
1 & 2\\
0 & 3 \\
\end{pmatrix}
$$\forall X \in R^{2 x 2} $
Find $Ker(T_A)$ and it basis and $Im(T_A)$ and it basis.
As for $Ker(T_A)$ we need to find the matrix X so that $$
\begin{pmatrix}
1 & 2\\
0 & 3 \\
\end{pmatrix}X-X \begin{pmatrix}
1 & 2\\
0 & 3 \\
\end{pmatrix}=\begin{pmatrix}
0 & 0\\
0 & 0 \\
\end{pmatrix}
$$
So I got the following equations $$\begin{cases} a_1+2a_3-a_1=0\\
a_2+2a_4-2a_1-3a_2=0\\
3a_3-a_3=0\\
3a_4-2a_3+3a_4=0
\end{cases}
$$
There is the trival answer that is 0 for all a, but the answer is not \begin{pmatrix}
0 & 0\\
0 & 0 \\
\end{pmatrix}
| If we denote
$$X=\begin{pmatrix}
a & b\\
c & d \\
\end{pmatrix}$$
then
$$AX-XA=\begin{pmatrix}
2c & (2d-2b-2a)\\
2c & -2c \\
\end{pmatrix}=\begin{pmatrix}
0 & 0\\
0 & 0 \\
\end{pmatrix}\iff (c=0)\land (d=b+a)$$
hence
$$\operatorname{ker}(T_A)=\operatorname{span}\Bigg(I_2,\begin{pmatrix}
0 & 1\\
0 & 1 \\
\end{pmatrix}\Bigg)$$
and clearly we have
$$\operatorname{Im}(T_A)=\operatorname{span}\Bigg(\begin{pmatrix}
1 & 0\\
1 & -1 \\
\end{pmatrix},\begin{pmatrix}
0 & 1\\
0 & 0 \\
\end{pmatrix}\Bigg)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/726576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In Triangle, $\sin\frac A2\!+\!\sin\frac B2\!+\!\sin\frac C2\!-\!1\!=\!4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$ To prove
$$\sin\frac A 2+\sin\frac B 2+\sin\frac C 2-1=4\sin\frac{\pi -A}4\sin\frac{\pi -B}4\sin\frac{\pi-C}4$$
My approach :
$$
\begin{align}
\text{L.H.S.} & = \sin\frac{A}{2}+\sin \frac{B}{2}+\sin\frac{C}{2} -1 \\[8pt]
& = 2\sin\frac{A+B}{4}\cos\frac{A-B}{4}+\cos\left(\frac{\pi}{2}-\frac{C}{2}\right)-1 \\[8pt]
& = 2\sin\frac{\pi -C }{4}\cos\frac{A-B}{4} - 2\sin^2\left(\frac{\pi -C}{4} \right) \\[8pt]
& =2\sin\frac{\pi -C }{4}\left\{ \cos\frac{A-B}{4} - \sin\left(\frac{\pi -C}{4} \right)\right\}
\end{align}
$$
Unable to move further please help. thanks.
| Note that
$$\sin\frac{C}{2}=\cos\frac{A+B}{2}$$
and now use following formulas
$$\sin p+\sin q=2\sin\frac{p+q}2\cos\frac{p-q}2$$
$$\cos p-\cos q=-2\sin\frac{p+q}2\sin\frac{p-q}2$$
to obtain
$$\begin{align}\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}&=\sin\frac{A}{2}+\sin\frac{B}{2}+\cos\frac{A+B}{2}\\&=2\sin\frac{A+B}{4}\cos\frac{A-B}{4}+1-2\sin^2\frac{A+B}{4}\\
&=2\sin\frac{A+B}{4}\left(\cos\frac{A-B}{4}-\sin\frac{A+B}{4}\right)+1\\
&=2\sin\frac{A+B}{4}\left(\cos\frac{A-B}{4}-\cos\left(\frac{\pi}2-\frac{A+B}{4}\right)\right)+1\\
&=4\sin\frac{\pi-A}4\sin\frac{\pi-B}4\sin\frac{\pi-C}4+1\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/727852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
probability need help on correlation problem A deck of 52 cards is shuffled you are dealt 13 cards. Let $X$ and $Y$ denote, respectively, the number of aces and the number of spades in your hand. Show that $X$ and $Y$ are uncorrelated.
I try to write $X=X_1+X_2+\cdots+X_{13}$, such that $X_1\ldots X_{13}$ is the indicator of the $i^{th}$ card. But this doesn't work. please help.
| It will suffice to show that $E(XY)=E(X)E(Y)$. The expected number of Aces is $13\cdot \frac{4}{52}$, and the expected number of spades is $13\cdot\frac{13}{52}$. Thus $E(X)E(Y)=\frac{13}{4}$.
We now compute $E(XY)$. Note that $X$ and $Y$ are not independent.
For the calculation, we condition on $X$. The possibilities are $X=0,1,2,3,4$. For each of these values of $X$, we calculate $E(XY|X=x)$.
It is clear that $E(XY|X=0)=0$.
Next, we find $E(XY|X=1)$. Given that $X=1$, the probability that the lone Ace is the Ace of spades is $\frac{1}{4}$, and the probability it is another Ace is $\frac{3}{4}$.
If the lone Ace is the Ace of spades, we are choosing $12$ more cards from the $48$ non-Aces, and the expected number of spades is $1+12\cdot \frac{12}{48}$. If the Ace is a non-spade, the same reasoning shows that the expected number of spades is $12\cdot\frac{12}{48}$. Thus
$$E(XY|X=1)=\frac{1}{4}\cdot 4+\frac{3}{4}\cdot 3=\frac{13}{4}.$$
Next we deal with $X=2$. If there are $2$ Aces, the probability that one of them is the Ace of spades is $\frac{1}{2}$. The same reasoning as in the previous case shows that
$$E(XY|X=2)=2\cdot\frac{13}{4}=\frac{26}{4}.$$
Next comes $X=3$. If there are $3$ Aces, the probability they include the Ace of spades is $\frac{3}{4}$. If that is the case, we have $1$ spade for sure, and we are choosing $10$ cards from the $48$ non-spades. We obtain
$$E(XY|X=3)=3\cdot\left(\left(1+\frac{10}{4}\right)\frac{3}{4}+\frac{10}{4}\frac{1}{4} \right)=\frac{39}{4}.$$
Finally,
$$E(XY|X=4)=4\cdot \left(1+\frac{9}{4}\right)=\frac{52}{4}.$$
The probability of $k$ Aces is $\binom{4}{k}\binom{48}{13-k}/\binom{52}{13}$.
Finally, calculate. We get
$$E(XY)=\frac{13}{4}\cdot\frac{1}{\binom{52}{13}}\left(1\binom{4}{1}\binom{48}{12}+2\binom{4}{2}\binom{48}{11}+3\binom{4}{3}\binom{48}{10}+4\binom{48}{9} \right).$$
This simplifies to $\dfrac{13}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/729894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding max/min of multivariable function The following function
$f(x,y) = 3xy + \frac{6}{1 + x^2 + y^2 }$ within $\frac{1}{3} \leq x^2 +y^2 \leq 4$
I do partial differention
$\frac{\partial z}{\partial x} = 3y - \frac{12x}{1 + x^2 + y^2}$
$\frac{\partial z}{\partial y} = 3x - \frac{12y}{1 + x^2+ y^2 }$
I try to simplify and get an expression of just one variable setting
$\frac{\partial z}{\partial x} - \frac{\partial z}{\partial x} = 0$
getting that $x = y$
$3x - \frac{12x}{1 + x^2+x^2 }$
solving for $x = 0$, $x_1 = 0, x_2 = \frac{1}{\sqrt{2}}, x_3 = \frac{-1}{\sqrt{2}}$
which gives me
$f(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) = 3\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}} + \frac{6}{1 + \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 } = \frac{9}{2}$
second part of the problem is to try the outer and inner boundaries. I set for inner boundary:
$x = \frac{1}{\sqrt{3}}\cos{t}$
, $y = \frac{1}{\sqrt{3}}\sin{t}$
and outer
$x = 2\cos{t}$
, $y = 2sin{t}$
I then create a new function:
$h(t) = f(x, y) = f(2\cos{t}, 2\sin{t})$ differentiate and get $\sin{t} = \cos{t}$ for $ x = 0$
$f(2\frac{1}{\sqrt{2}},2\frac{1}{\sqrt{2}}) = 3\frac{2}{\sqrt{2}}\frac{2}{\sqrt{2}} + \frac{6}{1 + \left(\frac{2}{\sqrt{2}}\right)^2 + \left(\frac{2}{\sqrt{2}}\right)^2 } = \frac{36}{5}$
which actually yields the correct maximum.
for the inner boundary I get
$f(\frac{1}{\sqrt{3}}\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}}\frac{1}{\sqrt{2}}) = 5$
but neither 9/2 or 5 gets the correct minimum! I can't see what i'm doing wrong.
| This function is begging for the substitution $x=\sqrt r\cos\varphi$, $y=\sqrt r\sin\varphi$, because then you want to find the minimum or the maximum of
$$f(r,\varphi)=3r\cos\varphi\sin\varphi+\frac{6}{1+r}=\frac{3r}2\sin(2\varphi)+\frac6{1+r}$$
where $\dfrac13\leq r\leq4$. Note that
$$\frac6{1+r}-\frac{3r}2\le\frac6{1+r}+\frac{3r}2\sin(2\varphi)\le\frac6{1+r}+\frac{3r}2$$
with equalities if $\sin(2\varphi)=-1$ or $\sin(2\varphi)=1$.
You should be able to find the minimum of $\dfrac6{1+r}-\dfrac{3r}2$ and the maximum of $\dfrac6{1+r}+\dfrac{3r}2$ easily.
This can be done with almost no computing:
*
*$\dfrac6{1+r}-\dfrac{3r}2$ is obviously decreasing, so the minimum is for $r=4$.
*$\left(\dfrac6{1+r}+\dfrac{3r}2\right)''=\dfrac{12}{(1+r)^3}>0$, so the function is convex and you only have to check value for $r=\dfrac13$ or $r=4$. The maximum is for $r=4$, because even $$\frac6{1+0}+\frac{3\cdot0}2=6<\frac6{1+4}+6=\frac6{1+4}+\frac{3\cdot4}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/733127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Harder Trigonometry Identity ($\sec A+\csc A$) How do I prove:
$\sin A (1 + \tan A) + \cos A (1 + \cot A) = \sec A + \csc A$
I've tried expanding the brackets by multiplying sin A and cos A to the left hand side but to no avail. Where should I start from?
| $\tan A=\sin A/\cos A$, and $\cot A=\cos A/\sin A$, so that making this substitution, we have \begin{align}
\sin A\left(1+\frac{\sin A}{\cos A}\right)+\cos A\left(1+\frac{\cos A}{\sin A}\right) & =\sin A + \frac{\sin^2 A}{\cos A}+\cos A +\frac{\cos^2 A}{\sin A} \\
& =\sin A+\cos A+\frac{1-\cos^2 A}{\cos A}+\frac{1-\sin^2 A}{\sin A} \\
& = \sin A+\cos A+\sec A-\cos A+\csc A-\sin A \\
& = \sec A +\csc A
\end{align}
Where I also made use of $\cos^2 A+\sin^2 A=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/733708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Prove by induction $1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{2^n} \ge 1 + \frac {n}{2}$ Prove by induction $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots + \frac{1}{2^n} \ge 1 + \frac {n}{2}$
I can't explain in words how the left hand side of the equation is achieved soI shall provide an example.
When $n = 2$, $\frac{1}{2^2} = \frac{1}{4}$ so the left hand side will be $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$.
When $n = 3$, $\frac{1}{2^3} = \frac{1}{8}$, so the left hand side will be $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}$ and so forth.
I have gotten up to the point where it is calculating the algebra. However, the result ends up being $ 1 + \frac{[2^(k+1)*k] + 2}{2(2^{k+1}} \le 1+ \frac{k+1}{2}$ when the following should be $\ge$. What can I do to solve the question?
| Hint: Look at your example, appropriately grouped and then with terms made systematically smaller:
$$1+{1\over2}+\left({1\over3}+{1\over4}\right)+\left({1\over5}+{1\over6}+{1\over7}+{1\over8}\right)
\ge1+{1\over2}+\left({1\over4}+{1\over4}\right)+\left({1\over8}+{1\over8}+{1\over8}+{1\over8}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/735911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Demonstrate that the sequence is decreasing $$\left( 1-\frac{1}{n+1}\right)^n$$
How do I demonstrate that the sequence decreases? I tried using the newton binomial but I end up with a terribly complicated expression.
| This is a classic exercise.
Let
$$a_n= \left( 1-\frac{1}{n+1}\right)^n=\left( \frac{n}{n+1}\right)^n$$
$$\frac{a_n}{a_{n+1}} =\left( \frac{n}{n+1}\right)^n\left( \frac{n+2}{n+1}\right)^{n+1}=\left( \frac{n}{n+1}\right)^{n+1}\left( \frac{n+2}{n+1}\right)^{n+1}\frac{n+1}{n}$$
Now write
$$\left( \frac{n}{n+1}\right)^{n+1}\left( \frac{n+2}{n+1}\right)^{n+1}=\left( \frac{n^2+n}{(n+1)^2}\right)^{n+1}=\left( 1-\frac{1}{(n+1)^2}\right)^{n+1}$$
and use Bernoulli inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/736891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Help on this inequality If a,b,c are positive numbers, prove the inequality
$$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} ≥ \frac{3}{1+abc}
$$
| We need to prove that
$$\sum_{cyc}\frac{1+abc}{a(1+b)}\geq3$$ or
$$\sum_{cyc}\frac{1+abc+a+ab}{a(1+b)}\geq6$$ or
$$\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}\geq6$$ or
$$\sum_{cyc}\frac{1+a}{a(1+b)}+\sum_{cyc}\frac{b(1+c)}{1+b}\geq6,$$
which is AM-GM:
$$\sum_{cyc}\frac{1+a}{a(1+b)}+\sum_{cyc}\frac{b(1+c)}{1+b}\geq3\sqrt[3]{\prod_{cyc}\frac{1+a}{a(1+b)}}+3\sqrt[3]{\prod_{cyc}\frac{b(1+c)}{1+b}}=6.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/739945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate the integral Can some one provide me hint to evaluate the following integral.
$$
\int\csc^{2}\left(x\right)
\ln\left(\cos\left(x\right) - \sqrt{\vphantom{\largeA}\,\cos\left(2x\right)\,}\,\right)
\,{\rm d}x
$$.
| MAPLE returns
$$-\cot(x)\ln(\cos(x)-\sqrt{\cos(2x)})-\cot \left( x \right) +x-\sqrt { \left( -2\,
\left( \cos \left( x \right) \right) ^{2}+1 \right) \left( \left(
\cos \left( x \right) \right) ^{2}-1 \right) } \left( -1/2\,{\frac {-
2\, \left( \cos \left( x \right) \right) ^{3}+2\, \left( \cos \left(
x \right) \right) ^{2}+\cos \left( x \right) -1}{\sqrt { \left( -\cos
\left( x \right) -1 \right) \left( 2\, \left( \cos \left( x \right)
\right) ^{3}-2\, \left( \cos \left( x \right) \right) ^{2}-\cos
\left( x \right) +1 \right) }}}+1/2\,{\frac {-2\, \left( \cos \left(
x \right) \right) ^{3}-2\, \left( \cos \left( x \right) \right) ^{2}
+\cos \left( x \right) +1}{\sqrt { \left( 1-\cos \left( x \right)
\right) \left( 2\, \left( \cos \left( x \right) \right) ^{3}+2\,
\left( \cos \left( x \right) \right) ^{2}-\cos \left( x \right) -1
\right) }}} \right) \left( \sin \left( x \right) \right) ^{-1}{
\frac {1}{\sqrt {2\, \left( \cos \left( x \right) \right) ^{2}-1}}}$$
It seems you have a tricky integral on your hands. If I find a substitution that returns this I'll post
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/741826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Complex numbers system of equations problem with 5 variables Let $z_0$,$z_1$,$z_2$,$z_3$ and $z_4$ such that $z_i\in C$ that hold:
$$(1)|z_0|=|z_1|=|z_2|=|z_3|=|z_4|=1$$
$$(2)z_0+z_1+z_2+z_3+z_4=0$$
$$(3) z_0z_1+ z_1z_2+z_2z_3+z_3z_4+z_4z_0=0$$
Prove that the solutions $z_i$ of this equation lay on the corners of regular pentagon.
I have tried with insertion of complex numbers with property that $z_i=1\angle \phi_i$ with $\phi_i=i 360^°/5$ and $i\in\{0,1,2,3,4\}$
I am interested if I should use $z_i=1\angle ( \phi_i+\alpha)$ with $\alpha \in \{0,2\pi\}$
$$1\angle (\alpha)+ 1\angle ( \phi+\alpha)+ 1\angle ( 2\phi+\alpha)+ 1\angle ( 3\phi+\alpha)+ 1\angle ( 4\phi+\alpha)=0$$
$$ 1\angle (\phi+\alpha)+ 1\angle ( 3\phi+\alpha)+1\angle ( 5\phi+\alpha) +1\angle ( 7\phi+\alpha)+1\angle ( 4\phi+\alpha)=0$$
| Let us define $a,b,c,d$ as follows:
$$
a=\frac{z_0}{z_4},~b=\frac{z_1}{z_4},~c=\frac{z_2}{z_4},~d=\frac{z_3}{z_4}.
$$
The proposed equations are equivalent to the equations
$$
|a|=|b|=|c|=|d|=1\tag{1}
$$
$$
a+b+c+d=-1\tag{2}
$$
$$
a b+b c+ c d+ d+a=0\tag{3}
$$
The equations $(2)$ and $(3)$ can be written in the form
$$
\left\{\eqalign{\phantom{(1+b)}a + \phantom{(1+c)}d&\,=\,-c-b-1 \cr
(1+b)a + (1+c)d&\,=\,-bc }\right.\tag{4}
$$
Let us consider two cases:
*
*
$b=c$, in this case the substitution of the first equation in $(4)$ in the second
yields
$ (1+b)(1+2b)= b^2$ or equivalently $b^2+3b+1=0$, which is absurd since this equation has
only real roots of absolute value different from $1$ in contradiction with $(1)$.
*
$b\ne c$, here the system $(4)$ can be solved with respect to $a$ and $d$ and we get
$$
a=\frac{b+(c+1)^2}{b-c},\qquad d=\frac{(b+1)^2+c}{c-b}.\tag{5}
$$
Noting that $\overline{b}=1/b$ and $\overline{c}=1/c$, we conclude from $(5)$ that we have also
$$
\overline{a}=\frac{b (c+1)^2+c^2}{c (c-b)},\qquad \overline{d}=\frac{ b^2 + (1 + b)^2 c}{b (b - c)}.\tag{6}
$$
Now the equation $a\overline{a}=|a|^2=1$ becomes
$$
\left(\frac{b+(c+1)^2}{b-c}\right)\left(\frac{b (c+1)^2+c^2}{c (c-b)}\right)=1\,,
$$
which is equivalent to
$
(c^2+3c+1)(b^2+b(c^2+c+1)+c^2)=0
$,
but $c^2+3c+1\ne0$ since $|c|=1$, hence
$$b^2+b(c^2+c+1)+c^2=0\,.\tag{7}$$
In similar way, the equation $d\overline{d}=|d|^2=1$ becomes
$$
\left(\frac{(b+1)^2+c}{c-b}\right)\left(\frac{ b^2 + (1 + b)^2 c}{b (b - c)}\right)=1\,,
$$
which is equivalent to
$
(b^2+3b+1)(c^2+c(b^2+b+1)+b^2)=0
$.
So,
$$c^2+c(b^2+b+1)+b^2=0\,.\tag{8}$$
Subtracting $(8)$ and $(7)$ yields $(bc-1)(c-b)=0$, but we have seen that $c\ne b$, so
we must have $cb=1$ or equivalently
$$c =\frac{1}{b}=\overline{b}\,.\tag{9}$$
Replacing back in $(7)$ we get $b^2+c+1+b+c^2=0$ or equivalently
$$b^4+b^3+b^2+b+1=0\,.$$
Thus, $b$ is a fifth root of $1$ different from $1$, that is
$$b\in\{\omega,\omega^2,\omega^3,\omega^4\}\quad\hbox{ where
$\omega=\exp \left(\frac{2\pi i}{5}\right)\,$.}$$
It follows that $c=b^{-1}=b^4$ and $c^2=b^8=b^3$.
From $(5)$ we conclude that
$$\eqalign{
a&\,=\,\frac{b+b^3+2b^4+1}{b-b^4}\,=\,\frac{b^4\!-b^2}{b-b^4}\,=\,b^3\,,\cr
d&\,=\,b^2\,,}$$
where we used the identities $1+b+b^2+ b^3+b^4=0$ and $b^5=1$.
We get the solution $(z_0,z_1,z_2,z_3,z_4)$ with
$$
z_0=b^3z_4\,,~z_1=b z_4\,,~z_2=b^4 z_4\,,~z_3=b^2 z_4\,,
$$
which are the vertices of a regular pentagon in some order, (A regular pentagon, or a regular pentagonal star).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/741910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Solving the congruence $7x \equiv 41 \mod{13}$ I have to solve the following linear congruence:
$$7x \equiv 41 \mod{13}$$
The question where I got this from comes in two parts. The first is that it asks to find the set of the inverses of $7 \mod 13$, which turns out to be $[2]_{13}$ but, I found that solution by solving the linear diophantine equation $$7x - 13k - 41$$ which is just the original equation in disguise i.e. $7x \equiv 41 \mod{13}$.
Now, I'm terribly confused as to the solution of the congruence equation. Where am I going wrong and how should I proceed?
Thanks!
EDIT: From @Bill's hint
\begin{align}
7x &\equiv 41 \mod{13} \\
x &\equiv \frac{41}{7} \mod{13} \\
x &\equiv \frac{28}{7} \mod{13} \quad (?) \\
x &\equiv 4 \mod{13} \\
\end{align}
Then, this gives me the solution $[4]_{13}$.
| The OP states
... but, I found that solution by solving the linear diophantine equation
From this site I found
Here the equation is $7x\equiv 41 \pmod {13}$ or
$$\tag 1 7x\equiv 2 \pmod {13}$$
Here also, multiplication $7 \times x$ is repeated addition. So you can also get the inverse of ${[7]_{13}}^{-1}$ and/or the solution of $\text{(1)}$ using the same technique (might as well cycle through it all):
$\quad 7 \times 1 \equiv 0 + 7 \equiv 7 $
$\quad 7 \times 2 \equiv 7 + 7 \equiv 1 $
$\quad 7 \times 3 \equiv 1 + 7 \equiv 8 $
$\quad 7 \times 4 \equiv 8 + 7 \equiv 2 $
$\quad 7 \times 5 \equiv 2 + 7 \equiv 9 $
$\quad 7 \times 6 \equiv 9 + 7 \equiv 3 $
$\quad 7 \times 7 \equiv 3 + 7 \equiv 10$
$\quad 7 \times 8 \equiv 10 + 7 \equiv 4$
$\quad 7 \times 9 \equiv 4 + 7 \equiv 11$
$\quad 7 \times 10 \equiv 11 + 7 \equiv 5$
$\quad 7 \times 11 \equiv 5 + 7 \equiv 12$
$\quad 7 \times 12 \equiv 12 + 7 \equiv 6$
$\quad 7 \times 13 \equiv 6 + 7 \equiv 0 $
Looking over your work you can pick out both the inverse and the solution.
You can set up a spreadsheet to handle these problems and/or check your work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/744390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Several (advanced) L'Hospital problems Problems :
$$
\begin{align}
&\text{A}.\ \lim_{x\rightarrow1}(2-x)^{\tan\left(\frac{\pi x}{2}\right)}\\
&\text{B}.\ \lim_{x\rightarrow 0}\left(\cot x-\frac{1}{x}\right)\\
&\text{C}.\ \lim_{x\rightarrow0}\frac{\sin^{-1}x-x}{\tan^{-1}x-x}\\
&\text{D}.\ \lim_{x\rightarrow 1}(2-x)^{\tan\left(\frac{\pi x}{2}\right)}\\
&\text{E}.\ \lim_{x\rightarrow0}\left(a^x+b^x\right)^\frac{1}{x}
\end{align}
$$
Here are my solutions concerning A~E.
A. $$\ln y=\tan\left(\frac{\pi x}{2}\right)\ln(x-2)=\frac{\tan\left(\frac{\pi x}{2}\right)}{\frac{1}{\ln(2-x)}}$$
$$\lim_{x\rightarrow 1}\frac{\tan\left(\frac{\pi x}{2}\right)}{\frac{1}{\ln(2-x)}}=_H\lim_{x\rightarrow1}\frac{\frac{\pi}{2}\sec^2(\frac{\pi x}{2})}{x-2}=\infty$$so, $$A\rightarrow \infty$$
Is this conclusion right?
C.
$$=_H\lim_{x\rightarrow0}\frac{\frac{1}{\sqrt{1-x^2}}-1}{\frac{1}{1+x^2}-1}=_H\lim_{x\rightarrow0}\frac{\frac{-1}{2}\frac{1}{\sqrt{(1-x^2)^3}}(-2x)}{-\frac{2x}{(1+x^2)^2}}=\frac{1}{-2}=-\frac{1}{2}$$
| For example:
$$\lim_{x\to 0}\cot x-\frac1x=\lim_{x\to 0}\frac{x\cos x-\sin x}{x\sin x}\stackrel{\text{l'Hospital}}=\lim_{x\to 0}\frac{\cos x-x\sin x-\cos x}{\sin x+x\cos x}=$$
$$=-\lim_{x\to 0}\frac{\sin x}{\frac{\sin x}x+\cos x}=-\frac0{1+1}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/745427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to find maximum/minimum of $y=\frac{x(x^2-x+2)}{x^2-9}$? How can I find minimum and maximum of $y=\frac{x(x^2-x+2)}{x^2-9}$?
In other words, points, where $y'=0$.
My current steps are:
$$y=\frac{x(x^2-x+2)}{x^2-9}=x-1+\frac{11x-9}{x^2-9}$$
$$\frac{dy}{dx}=1+\frac{11(x^2-9)-2x(11x-9)}{(x^2-9)^2}$$
\begin{align}
y'=0 &\Rightarrow 1+\frac{11x^2-99-22x^2+18x}{(x^2-9)^2}=0 \\
&\Rightarrow\frac{11x^2-99-22x^2+18x+(x^2-9)^2}{(x^2-9)^2}=0 \\
&\Rightarrow 11x^2-99-22x^2+18x+x^4-18x^2+81=0 \\
&\Rightarrow x^4-29x^2+18x-18=0
\end{align}
Mathematica 8 gives me two complex and two real roots, but I have no idea how to solve 4th order polynomial equations, nor I am sure if I really have to do this, considering that nothing this ''complex'' should be done in second semester calculus homework.
| You can use partial fractions here $$\frac{11x-9}{x^2-9}=\frac A{x+3}+\frac B{x-3}=\frac 7{x+3}+\frac 4{x-3}$$ so that $$y=x-1+\frac 7{x+3}+\frac 4{x-3}$$ and $$y'=1-\frac 7{(x+3)^2}-\frac 4{(x-3)^2}$$
Then you should be able to sketch the functions to see what is happening, and to see that any real zeros of the derivative will be at local extrema.
Once your sketch shows you what is going on, you should be able to find suitable values of $x$ on either side of the zeros of $y'$ - the intermediate value theorem then tells you zeros exist (stay away from $x=\pm 3$), and you can use standard numerical methods to estimate the roots.
Splitting the function into simpler components greatly simplifies sketching, and sketching greatly helps if the situation looks a bit complicated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/745672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Compute integral $\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$ Compute the integral.
$$\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$$
The answer at the back of the book is
$$\frac{\pi}{4\sin(\frac{3\pi}{8})}$$
| Note
\begin{align}
\int_{-\infty}^\infty \frac{x^4}{1+x^8} dx
& =-\frac1{2i}\int_{-\infty}^\infty \left(\frac{1}{1+i x^4} - \frac{1}{1-i x^4} \right)dx \\
& = -\text{Im} \int_{-\infty}^\infty \frac{1}{1+i x^4}dx
\overset{t=i^{1/4}x} = -\text{Im} \left( e^{-i\frac\pi8}\int_{-\infty}^\infty \frac{1}{1+t^4}dt\right)\\
&= -\text{Im} \left( e^{-i\frac\pi8}\cdot \frac\pi{\sqrt2} \right)= \frac\pi{\sqrt2}\cdot\sin\frac\pi8= \frac\pi2\sqrt{1-\frac1{\sqrt2}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/747136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Solve $2\sqrt{(x-1)(x+2)}\ge|x+1|-2$ Can you please show me how can I solve this inequality. I would like to see how it can be done without the graph of the functions. Thank you!
$$2\sqrt{(x-1)(x+2)}\ge|x+1|-2$$
| If $|x+1|-2$ is negative or zero, the inequality is obvious, so assume $|x+1| >2$, so that $x>1$ or $x < (-3)$.
If $x>1$, then $|x+1|=x+1$ so that your inequality becomes $2\sqrt{(x-1)(x+2)} \geq x-1$, which is equivalent
to $4(x-1)(x+2) \geq (x-1)^2$, or $4(x+2) \geq x-1$, i.e. $3x+9 \geq 0$ which is clear.
If $x<(-3)$, then $|x+1|=-x-1$ so that your inequality becomes $2\sqrt{(x-1)(x+2)} \geq -x-3$, which is equivalent
to $4(x-1)(x+2) \geq (x+3)^2$, or $3x^2-2x-17 \geq 0$, which is true because
$3x^2-2x-17=16-(x+3)(11-3x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/748381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the real parameters $a$ and $b$ in limit Can you please help me solve this problem which requires to find the values of real parameters $a$ and $b$ so that the relation below is satisfied:
$$\lim_{x\to -\infty}(\sqrt {x^2+x+1}-ax+b)=0$$
Thank you very much!
| You have
$$
\lim_{x\to -\infty}\left(\sqrt {x^2+x+1}-ax+b\right)=0.
$$
Let $x=-y$, then you have
$$
\begin{align}
\lim_{-y\to -\infty}\left(\sqrt {(-y)^2+(-y)+1}-a(-y)+b\right)&=0\\
\lim_{y\to \infty}\left(\sqrt {y^2-y+1}+ay+b\right)&=0.\tag1
\end{align}
$$
Now, divide both sides of $(1)$ by $y$. It turns out to be
$$
\begin{align}
\lim_{y\to \infty}\left(\frac{\sqrt {y^2-y+1}+ay+b}{y}\right)&=0\\
\lim_{y\to \infty}\left(\sqrt {\frac{y^2-y+1}{y^2}}+a+\frac{b}{y}\right)&=0\\
\lim_{y\to \infty}\left(\sqrt {1-\frac{1}{y}+\frac{1}{y^2}}+a+\frac{b}{y}\right)&=0\\
1+a&=0\\
a&=\boxed{\Large\color{blue}{-1}}
\end{align}
$$
Hence, $(1)$ becomes
$$
\begin{align}
\lim_{y\to \infty}\left(\sqrt {y^2-y+1}-y+b\right)&=0\\
\lim_{y\to \infty}\left(\sqrt {y^2-y+1}-y\right)=-b\\
\lim_{y\to \infty}\left(\sqrt {y^2-y+1}-\sqrt{y^2}\right)=-b\\
\lim_{y\to \infty}\left(\left(\sqrt {y^2-y+1}-\sqrt{y^2}\right)\cdot\frac{\sqrt {y^2-y+1}+\sqrt{y^2}}{\sqrt {y^2-y+1}+\sqrt{y^2}}\right)=-b\\
\lim_{y\to \infty}\left(\frac{y^2-y+1-y^2}{\sqrt {y^2-y+1}+\sqrt{y^2}}\right)=-b\\
\lim_{y\to \infty}\left(\frac{-y+1}{\sqrt {y^2-y+1}+\sqrt{y^2}}\right)=-b.\tag2
\end{align}
$$
Now, divide the numerator and denominator part in $(2)$ by $y$, yield
$$
\begin{align}
\lim_{y\to \infty}\left(\frac{-1+\frac{1}{y}}{\sqrt {1-\frac{1}{y}+\frac{1}{y^2}}+\sqrt{1}}\right)&=-b\\
\frac{-1+0}{\sqrt {1-0+0}+\sqrt{1}}&=-b\\
-\frac{1}{2}&=-b\\
b&=\boxed{\Large\color{blue}{\frac{1}{2}}}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/749861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the limit of a function?? Determine the following limit:
$$\lim\limits_{x\to0}\frac{(1+x\cdot2^x)^{\frac{1}{x^2}}}{(1+x\cdot3^x)^{\frac{1}{x^2}}}$$
| $$\lim_{x\to0}\frac{(1+x\cdot2^x)^{\frac{1}{x^2}}}{(1+x\cdot3^x)^{\frac{1}{x^2}}}$$
$$=\lim_{x\to0}\left(\dfrac{1+x\cdot2^x}{1+x\cdot3^x}\right)^{\dfrac1{x^2}} =\lim_{x\to0}\left(1+\dfrac{x(2^x-3^x)}{1+x\cdot3^x}\right)^{\dfrac1{x^2}}$$
$$=\left(\lim_{x\to0}\left(1+\dfrac{x(2^x-3^x)}{1+x\cdot3^x}\right)^{\dfrac{1+x\cdot3^x}{x(2^x-3^x)}}\right)^{\lim_{x\to0}\dfrac{\dfrac{x(2^x-3^x)}{1+x\cdot3^x}}{x^2}}$$
Setting $\dfrac{1+x\cdot3^x}{x(2^x-3^x)}=n$ which $\to\infty$
the inner limit converges to $e$ as $\displaystyle \lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$
For $\displaystyle\lim_{x\to0}\dfrac{\dfrac{x(2^x-3^x)}{1+x\cdot3^x}}{x^2}$
$=\displaystyle\frac1{\lim_{x\to0}(1+x\cdot3^x)}\cdot\left(\lim_{x\to0}\frac{2^x-1}x-\lim_{x\to0}\frac{3^x-1}x\right)$
$\displaystyle=1\cdot(\ln2-\ln 3)=\ln\frac23$
Now we know, $\displaystyle e^{\ln a}=a$ for real $a>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/750346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why does $2^{n+1} + 2^{n+1} = 2^{n+2}$? Simple question, why does:
$2^{n+1} + 2^{n+1} = 2^{n+2}$ ?
Furthermore, why does this only work for powers of 2?
Thanks.
| $$2^{n+1} + 2^{n+1} = 2\cdot 2^{n+1} = 2^1 \cdot 2^{n+1} = 2^{n+1 + 1} = 2^{n+2}$$
We use the fact that $a^n\cdot a^m = a^{n+m}$.
Added: For larger bases, say we have an integer base $a$, then
$$\underbrace{a^{n+1}+a^{n+1} + \cdots + a^{n+1}}_{\large a \text{ terms } }= a\cdot a^{n+1} = a^{n+2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/750710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to closed the sum $\displaystyle \sum_{k=0}^n \dfrac{(-1)^k(2k+1)!!}{(n-k)!k!(k+1)!}$ How to closed the sum $\displaystyle S=\sum_{k=0}^n \dfrac{(-1)^k(2k+1)!!}{(n-k)!k!(k+1)!}$
I'm trying divide two cases $n$ odd and $n$ even. I predict that
$S=\begin{cases}\dfrac{1}{2^n\left[\left(\frac{n}{2}\right)!\right]^2}, & \quad \text{if $n$ even} \\ \\ \dfrac{-1}{2^n\left(\frac{n-1}{2}\right)!\left(\frac{n+1}{2}\right)!}, & \quad \text{if $n$ odd}\end{cases}$
or I can write
$\displaystyle S=\sum_{k=0}^n \dfrac{(-1)^k(2k+1)!!}{(n-k)!k!(k+1)!}= \dfrac{(-1)^n\left(\frac{2n-1-(-1)^n}{2}\right)!!}{n!\left(\frac{2n+1-(-1)^n}{2}\right)!!}$
Now I want you to help me prove it.
| Suppose we seek to evaluate
$$\sum_{k=0}^{n} \frac{(-1)^k}{2^k}
{2k+1\choose k} {n\choose n-k}$$
which is
$$\sum_{k=0}^{n} {n\choose k}
\frac{(-1)^k}{2^k} {2k+1\choose k}.$$
Introduce
$${2k+1\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2k+1}}{z^{k+1}} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1+z}{z}
\sum_{k=0}^n
{n\choose k}
\frac{(-1)^k}{2^k}
\frac{(1+z)^{2k}}{z^k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1+z}{z}
\left(1-\frac{(1+z)^2}{2z}\right)^n
\; dz
\\ = \frac{1}{2^n} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1+z}{z^{n+1}}
(-1-z^2)^n \; dz
\\ = \frac{(-1)^n}{2^n} \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1+z}{z^{n+1}}
(1+z^2)^n \; dz.$$
This is
$$\frac{(-1)^n}{2^n}
\left([z^n](1+z^2)^n + [z^{n-1}] (1+z^2)^n\right).$$
When $n$ is even the first term contributes and we obtain
$$\frac{(-1)^n}{2^n} {n\choose n/2}.$$
When $n$ is odd the second term contributes and we obtain
$$\frac{(-1)^n}{2^n} {n\choose (n-1)/2}.$$
Joining these two formulas we obtain
$$\frac{(-1)^n}{2^n} {n\choose \lfloor n/2\rfloor}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/750891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Volume of a parallelpiped from its sides and diagonals? If a parallelogram has sides of length $a$ and $b$, and diagonals of length $d$ and $e$, then we can find its area in the following way. By the polarization identity, we have $a b \cos\theta = \frac{1}{4}|d^2-e^2|$. Now the area is $a b \sin \theta$, which is equal to $\sqrt{a^2 b^2 - (a b \cos \theta)^2}$, or
$$\sqrt{a^2 b^2 - \frac{1}{16}(d^2-e^2)^2}.$$
This is not an exceptionally pretty formula, but maybe it's the best one can do.
How can such a formula be generalized to higher dimensions? In particular, what is the volume of a parallelpiped, given its edge and diagonal lengths?
| Let the parallelepiped with a vertex $O$ at the origin have vertices $A$, $B$, $C$. Write $D = B+C$; $E= C+A$; $F=A+B$; and $G=A+B+C$. Further, assign these names to the lengths of edges and diagonals:
$$a := |OA|\quad b := |OB| \quad c := |OC| \quad d := |AD| \quad e := |BE| \quad f := |CF| \quad g := |OG|$$
Note that a parallelepiped (and it properties) is completely determined by six quantities: the lengths of three edges meeting at a vertex, and the measures of the three angles between those edges. The total number of edge-lengths and diagonal-lengths is seven, so there's a dependency among these values, namely
$$4 \;(\; a^2 + b^2 + c^2\;) = d^2 + e^2 + f^2 + g^2$$
Although one diagonal (say, $g$) is unnecessary, it helps to simplify the volume formula a bit:
$$\begin{align}
32 V^2 &= 32 a^2 b^2 c^2 + d^2 e^2 f^2 + g^2 ( d^2 e^2 + e^2 f^2 + f^2 d^2 )\\
&- 2 g^2 ( a^2 d^2 + b^2 e^2 + c^2 f^2 ) - 2 ( a^2 e^2 f^2 + d^2 b^2 f^2 + d^2 e^2 c^2 )
\end{align}$$
In the case of a rectangular parallelepiped, for which $d^2=e^2=f^2=g^2=a^2+b^2+c^2$, the formula reduces to $V = a b c$, as expected.
I derived the formulas using coordinates
$$A = (a_x,0,0) \qquad B = (b_x, b_y, 0) \qquad C = (c_x, c_y, c_z)$$
Expressing the various distances in terms of these
$$a^2 = A\cdot A = a_x^2 \qquad b^2 = B\cdot B = b_x^2 + b_y^2 \qquad c^2 = C\cdot C = c_x^2 + c_y^2 + c_z^2$$
$$d^2 = \overrightarrow{AD}\cdot \overrightarrow{AD}=\dots \quad e^2 = \overrightarrow{BE}\cdot \overrightarrow{BE} =\dots \quad f^2 = \overrightarrow{CF}\cdot \overrightarrow{CF}=\dots$$
along with
$$V = a_x b_y c_z$$
I systematically eliminated $a_x$, $b_x$, $b_y$, $c_x$, $c_y$, $c_z$ from the system with the help of Mathematica's Resultant[] function. (Appropriate applications of the Law of Cosines should work just as well, but the Resultant process lets me crank through the equations without having to think too hard. :)
An $n$-dimensional parallelotope is determined by $\frac{1}{2}n(n+1)$ values. (These are the triangular numbers. You can see their relevance by noting the coordinatization done above considers vectors using $1$, $2$, $3$, ..., $n$ non-zero coordinates.) The figure has n characteristic edge-lengths and $2^{n-1}$ diagonals, so expressing volume in terms of these is certainly possible. With an increasing number of "extra" lengths, the resulting formula can take a variety of forms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/752839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Sum of Binomial Series of form $\binom{2000}{3k-1}$ Find the Value of $$ \binom{2000}{2}+\binom{2000}{5}+\binom{2000}{8}+\cdots+\binom{2000}{1997}+\binom{2000}{2000}$$
| You're off to a good start! In your post you defined quantities $X$ and $Y$ and found that $2X+Y=2^{2000}$. This post builds on that by giving you a second relation between $X$ and $Y$.
Let $\omega=e^{2\pi i/3}$. So $\omega^3=1$. Then by the binomial theorem, $$(1+\omega)^{2000}=\binom{2000}{0}+\binom{2000}{1}\omega+\binom{2000}{2}\omega^2+\ldots+\binom{2000}{2000}\omega^2$$
Now we compare the real parts of each side.
$$\Re((1+\omega)^{2000})=\Re(\binom{2000}{0}+\binom{2000}{1}\omega+\binom{2000}{2}\omega^2+\ldots+\binom{2000}{2000}\omega^2)$$
$$1+\omega=\frac{1+\sqrt3 i}{2}=e^{\pi i/3}$$ so $(1+\omega)^{2000}=\omega$, and we know that $\Re (\omega)=\Re (\omega^2)=\frac{-1}{2}$. Plugging in above, we see that
$$\frac{-1}{2}=\binom{2000}{0}-\frac{1}{2}\binom{2000}{1}-\frac{1}{2}\binom{2000}{2}+\binom{2000}{3}+\ldots-\frac{1}{2}\binom{2000}{2000}$$
This gives us a second equation, $\frac{-1}{2}=X-\frac{Y}{2}-\frac{X}{2}$.
Now you have 2 equations in 2 unknowns, and you just need to solve for $X$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/754825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Need help with basic factoring equation I'm just trying to brush up on my factoring of quadratic equations.
$$\frac1{x+3} + \frac1{x^2 + 5x +6}$$
$$\frac1{x+3} + \frac1{(x+2)(x+3)}$$
$$\frac{(x+2)(x+3) + (x+3)}{(x+2)(x+2)(x+3)}$$
Then I'm stuck. I think you can factor out the $(x+2)(x+3)$ from top and bottom?
| Between your second and third step it can be improved: since you have
$
\frac1{x+3} + \frac1{(x+2)(x+3)}
$
in the second step, you want to give all fractions a common denominator - the least common denominator. Since both terms have a $x+3$ in the denom. already, multiply the first term by $\frac{x+2}{x+2}$, and add the fractions to get
$$
\frac{x+2}{(x+2)(x+3)} + \frac 1{(x+2)(x+3)} = \frac{x+3}{(x+2)(x+3)} = \frac{1}{x+2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/755869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$ If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$
This can be easily done by calculas but is there any way to do do this by algebra
| If $a=b=\frac{1}{\sqrt2}$ then $a+b+\frac{1}{ab}=\sqrt2+2$.
We'll prove that it's a minimal value.
Indeed, let $a+b=2u$ and $ab=v^2$.
Hence, the condition gives $4u^2-2v^2=1$,
which says that $v^2=\frac{4u^2-1}{2}$ and $4u^2=1+2v^2\leq1+2u^2$, which gives $u\leq\frac{1}{\sqrt2}$
and we need to prove that
$$2u+\frac{2}{4u^2-1}\geq\sqrt2+2$$ or
$$(1-\sqrt2u)(2\sqrt2-1+4u+2(1-2u^2))\geq0$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/756702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
How to find answer to the sum of series $\sum_{n=1}^{\infty}\frac{n}{2^n} $ I have put his on wolfram and obtained answer as follows:
$\sum_{n=1}^{\infty}\frac{n}{2^n} = 2$
And the series is convergent too because $\lim_{n\to\infty} \frac {n}{2^n} = 0$
However I am wondering if there is a convenient way to solve this; I don't think you can represent it by a geometric progression either. So how do we have to do it on paper?
| In this answer we go for the 'one trick pony' approach - all we know is that for $k \ge 0$,
$\tag 1 \sum_{n=k}^{\infty}\frac{1}{2^n} = 2^{1-k}$
We decompose the summands of $\sum_{n=1}^{\infty}\frac{n}{2^n}$ in a natural/straightforward manner, arranging these numbers into a table:
$$\begin{pmatrix}
\frac{1}{2} & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \dots \\
0 & \frac{1}{4} & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \dots \\
0 & 0 & \frac{1}{8} & \frac{1}{16} & \frac{1}{32} & \dots \\
0 & 0 & 0 & \frac{1}{16} & \frac{1}{32} & \dots \\
0 & 0 & 0 & 0 & \frac{1}{32} & \dots \\
0 & 0 & 0 & 0 & 0 & \dots \\
. \\
. \\
. \\
\end{pmatrix}$$
Using $\text{(1)}$ we add up the entries in each row,
$$\begin{pmatrix}
1 \\
\frac{1}{2} \\
\frac{1}{4} \\
\frac{1}{8} \\
\frac{1}{16} \\
. \\
. \\
. \\
\end{pmatrix}$$
And now we add up the entries of our column vector, giving
$\quad \text{ANS: } 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Joint distribution probabilities I have a question that is similar to the following(made up here):
The construction of a tower of cards is done is two stages, procrastination and the actual building. The time in minutes needed to complete each stage are independent discrete random variables, X and Y, with probability functions;
$f_X(x) = \frac{7}{10}$ if $ x = 2, \frac{3}{10}$ if $x = 3$, and $0$ otherwise.
$f_Y(x) = \frac{2}{5}$ if $x = 3, \frac{2}{5}$ if $x = 4, \frac{1}{5}$ if $x = 5$ $0$ otherwise
What is the probability the task took more than six minutes to complete?
Now I haven't dealt with joint distribution problems before. But I can see 3 scenarios that yield more than 6 minutes of time elapsed. $f_X(2) $ then $f_Y(5)$ or $f_X(3)$ then $f_Y(4)$ or $f_X(3)$ then $f_Y(5)$
Can I simply then take $(\frac{7}{10}*\frac{1}{5} + \frac{3}{10}*\frac{2}{5} + \frac{3}{10}*\frac{1}{5})$? This seems right at $.32$. Furthermore if I do the other three scenarios I get a total probability of one, which increases my confidence with it once again. Any hints or confirmation? Thank you for your time
| This explicit computation may help:
\begin{align}
P[X+Y > 6]
&= P[X +Y > 6, X=2] + P[X +Y > 6, X=3] \\
&= P[Y = 5, X = 2] + P[ Y \geq 4, X=3 ] \\
&= P[Y = 5]P[ X = 2] + P[ Y \geq 4]P[ X=3 ] \\
&= \left(\frac{1}{5} \right)\left(\frac{7}{10} \right) +
\left(\frac{3}{5} \right)\left(\frac{3}{10} \right) \\
&= \frac{8}{25}.
\end{align}
So, you did in fact make the right computations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/757338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Express each function in the form $u(x,y) + iv (x,y)$ I was doing some homework with complex numbers and I'm stuck with these two, I hope that someone can solve these and clear it up for me. Thank you.
*
*ln(1+z)
*z/(3+z)
Samples,
| For the first one a function $w=re^{i\theta+2\pi n}$ satisfies $$\ln w=\ln re^{i\theta}=\ln r+i(\theta+2\pi n)$$Now $w=1+z=1+x+iy$. So $$r=\sqrt{(1+x)^2+y^2} \ \ \text{and} \ \ \theta=\arctan\{\frac{y}{1+x}\}$$
For the second one use $$\frac{1}{u+iv}=\frac{1}{u+iv}\cdot \frac{u-iv}{u-iv}=\frac{u-iv}{u^2+v^2}=\frac{u}{u^2+v^2}-i\frac{v}{u^2+v^2}$$ So,
$$\frac{z}{3+z}=\frac{x+iy}{3+x+iy}
\\ =\frac{x+iy}{3+x+iy} \cdot \frac{3+x-iy}{3+x-iy}
\\ =\frac{3x+x^2+y^2+i3y}{(3+x)^2+y^2}
\\ =\frac{3x+x^2+y^2}{(3+x)^2+y^2}+i \frac{3y}{(3+x)^2+y^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/758052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A function $f$ has the property that $f(x+y)=f(x)+f(y)+3xy$. If $f(1)=2$, what is $f(8)$? A function $f$ has the property that $f(x+y)=f(x)+f(y)+3xy$. If f(1)=2, what is f(8)?
I would like to try to tackle this problem but I need somewhere to start as I really have no idea at all on how to start.
At first I assumed that perhaps $f$ was a polynomial function, like a quadratic, but the $3xy$ term in the equation says otherwise.
Update -Thanks for everyone's help
This is what everyone has suggested. Instead of trying to find out the function, just use the f(1) value to find f(8):
$f(1+1)= f(2) = f(1) + f(1) +3(1)(1) = 7$
$f(4) = f(2) + f(2) + 3(2)(2) = 26$
$f(8) = f(4) + f(4) + 3(4)(4) = 100$
| Setting $y=x$, we get
$$f(2x) = 2f(x) + 3x^2$$
Hence, if $x=2^n$, we get
$$f(2^{n+1}) = 2f(2^n) + 3 \cdot 2^{2n} = 2(2f(2^{n-1}) + 3 \cdot 2^{2n-2}) + 3 \cdot 2^{2n} = 4 f(2^{n-1}) + 3(2^{2n-1} + 2^{2n})$$
In general, we have
$$f(2^{n+1}) = 2^{k+1} f(2^{n-k}) + 3 \left(2^{2n} + 2^{2n-1} + \cdots +2^{2n-k}\right) = 2^{k+1}f(2^{n-k}) + 3 \cdot 2^{2n-k} \left(2^{k+1}-1\right)$$
Setting $k=n$, we get
$$f(2^{n+1}) = 2^{n+1} f(1) + 3 \cdot 2^n \cdot \left(2^{n+1}-1\right)$$
Setting $n=2$, gives you the answer.
A more direct way is to recognize that
$$f(x+1) = f(x) + f(1) + 3x$$
Similarly, we get
\begin{align}
f(x) & = f(x-1) + f(1) + 3(x-1)\\
f(x-1) & = f(x-2) + f(1) + 3(x-2)\\
f(x-2) & = f(x-3) + f(1) + 3(x-3)\\
\vdots & = \vdots\\
f(2) & = f(1) + f(1) + 3 \cdot 1
\end{align}
Using telescopic summation we get
$$f(x+1) = f(1) + x f(1) + 3 \sum_{k=1}^x k$$
Hence,
$$f(x+1) = (x+1)f(1) + \dfrac{3x(x+1)}2$$
Since $f(1)$ is given as $2$, we get
$$f(x+1) = 2(x+1) + \dfrac{3x(x+1)}2 = \dfrac{(x+1)(3x+4)}2$$
Hence,
$$f(x) = \dfrac{x(3x+1)}2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/759837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$, then what values can $t$ take?
If $9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$, then what values can $t$ take?
This is what I have done:
Let $y = 3^x$
$$9^{x+1} + (t^2 - 4t - 2)3^x + 1 > 0$$ $$\implies9y^2 + (t^2 - 4t - 2)y + 1 > 0$$
For the LHS to be greater than zero, $b^2 - 4ac$ has to be $\lt 0$, since coefficient of $y^2$ is greater than $0$(which will give us an upward opening parabola).
$$(t^2 - 4t - 2)^2 - 36< 0$$
The answer that I get is different from the correct answer. The correct answer to this is $t \in \mathbb{R} - \{2\}$. What did I do wrong?
| Given $$9^{x+1}+(t^2-4t-2)3^x+1>0$$
We can write it as $$3^x\bigg[9\cdot 3^x+(t^2-4t-2)+\frac{1}{3^x}\bigg]>0$$
So $$\underbrace{3^x}_{>0} \bigg[\bigg(3\cdot 3^{\frac{x}{2}}-\frac{1}{3^{\frac{x}{2}}}\bigg)^2+(t-2)^2\bigg]>0$$
So for Above statement is True $\forall \ x \; \mathbb{R},$ If $t\in \mathbb{R}-\{2\}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/760560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How find the $AP+\frac{1}{2}BP$ minmum value An equilateral triangle $ABC$ such $$AB=BC=AC=2a>0$$
A circle $O$ is inscribed in triangle $ABC$,and the point $P$ on the circle $O$.
Find the minimum $$AP+\dfrac{1}{2}BP$$
My idea: let $$A(-a,0),B(a,0),O(0,\dfrac{\sqrt{3}}{3}a)$$
then the circle equation is $$
x^2+(y-\dfrac{\sqrt{3}a}{6})^2=\dfrac{1}{12}a^2$$
let $P(x,y)\;$ , then
$$|PA|+\dfrac{1}{2}|PB|=\sqrt{(x+a)^2+y^2}+\dfrac{1}{2}\sqrt{(x-a)^2+y^2}$$
where $$
x^2+(y-\dfrac{\sqrt{3}a}{6})^2=\dfrac{1}{12}a^2$$
then I can't.Thank you
| Some ideas:
Taking it from where you left it and correcting the coordinates of $\;O\;$ , we get the circle's equation is
$$x^2+\left(y-\frac a{2\sqrt3}\right)^2=\frac{a^2}{12}\;\;\;\;(I)$$
Observe that since the $\;x$-axis is tangent to the circle, the absolute value of the center's $\;y$-coordinate equals the circle's radius...
Now, with your notation
$$|PA|+\frac12|PB|=\sqrt{(x+a)^2+y^2}+\frac12\sqrt{(x-a)^2+y^2}\;\;\;\;(II)$$
Thus, we want to minimize (II) contiioned to (I) , and thus Lagrange's Multipliers may help here...or
$$(I)\implies x=\pm\sqrt{\frac{a^2}{12}-\left(y-\frac a{2\sqrt3}\right)^2}$$
and substitute in (II) to have a one-variable extrema problem (first derivative and stuff)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/762836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Showing a function has first and second partials everywhere and continuity of the partials Let $f \left( \begin{array}{ccc}
x \\
y \end{array} \right)= \begin{cases}
xy\dfrac{x^2-y^2}{x^2+y^2} & \mbox{if $(x,y) \neq (0,0)$}\\
0 & \mbox{if $(x,y) = (0,0)$} \end{cases}. $
I have this function and I want to show it's first and second partials exist everywhere, the first partials are continuous and the seconds are not. Computing the partials is one thing but what Do i do about continuity?
Here are the first partials:
away from $0$
$$D_1(f) = \large\frac{4x^2y^3 + x^4y-y^5}{(x^2+y^2)^2}$$
$$D_2(f) = \large\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}$$
| I'll handle $D_1(f)$.
Let $(x,y)\neq (0,0)$.
Note that
$$\begin{align}
|D_1(f)(x,y)|&=\left|\dfrac {y\left(x^4+4x^2y^2-y^4\right)}{\left(x^2+y^2\right)^2}\right|\\
&\leq \dfrac {y\left(x^4+4x^2y^2+y^4\right)}{\left(x^2+y^2\right)^2}\\
&=\dfrac{y\left(\left(x^2+y^2\right)^2+2x^2y^2\right)}{\left(x^2+y^2\right)^2}\\
&=y+\dfrac{2x^2y^3}{\left(x^2+y^2\right)^2}.\\
\end{align}$$
Thus. to prove that $\lim \limits_{(x,y)\to (0,0)}\left(D_1(f)(x,y)\right)=0$, it is enough to prove that $$\lim \limits_{(x,y)\to (0,0)}\left(y+\dfrac{2x^2y^3}{\left(x^2+y^2\right)^2}\right)=0.$$
In turn, to prove this, it enough to prove that $$\lim \limits_{(x,y)\to (0,0)}\left(y\right)=0 \land \lim \limits_{(x,y)\to (0,0)}\left(\dfrac{2x^2y^3}{\left(x^2+y^2\right)^2}\right)=0.$$
This is a consequence of
$$\dfrac{2x^2y^3}{\left(x^2+y^2\right)^2}=\dfrac{2x^2y^3}{x^4+2x^2y^2+y^4}\leq \dfrac{2x^2y^3}{2x^2y^2}=y\substack{(x,y)\to (0,0)\\\huge\longrightarrow} 0.$$
Therefore, if $D_1(f)(0,0)=0$, then $D_1(f)$ is continuous. If $D_1(f)(0,0)\neq 0$, it is discontinuous at the origin.
Since $D_1(f)(y,x)=-D_2(f)(x,y)$, the same conclusions follow immediately for $D_2(f)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/763519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Greatest prime factor of $4^{17}-2^{28}$ I have seen the solution to this problem.
What is the greatest prime factor of $ \ 4^{17} - 2^{28} \ $?
Answer: 7
$$ 4^{17}-2^{28} \ = \ 2^{34}-2^{28} \ = \ 2^{28} \ (2^6-1) \ = \ 2^{28} \ \cdot \ 63 \ = \ 2^{28}\ \cdot \ 3^2 \ \cdot \ 7 $$
I understand the problem up to this part: $2^{28} \ (2^6-1) \ $ .
Why did this occur? I can't explain it.
Thanks!
| You are asking how you went from $2^{34} - 2^{28} = 2^{28}(2^6-1)$? You are factoring out $2^{28}$ from both of the numbers. $2^{34} - 2^{28}=(2^{34} - 2^{28})*\frac{2^{28}}{2^{28}}=\frac{2^{34} - 2^{28}}{2^{28}}*2^{28}=(2^6-1)*2^{28}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Expression generating $\left( \frac{3}{10}, \, \frac{3}{10} + \frac{33}{100}, \, \frac{3}{10} + \frac{33}{100} + \frac{333}{1000}, \dots \right)$ I'm looking for a closed-form expression (in terms of $n$), that will give the sequence
$$
(s_n) = \left( \frac{3}{10}, \, \frac{3}{10} + \frac{33}{100}, \, \frac{3}{10} + \frac{33}{100} + \frac{333}{1000}, \dots \right).
$$
Can anyone think of one? I made a related post to this question several minutes ago but I realized I was interpreting the sequence wrong.
| Built on my answer to the OP's previous post of the similar question, then:
$s_n = \displaystyle \sum_{k=1}^n \left(\dfrac{1}{3} - \dfrac{1}{3\cdot 10^k}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove inequality $(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$ I am trying to prove the following inequality:
$$(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$$
for all $a>0, b>0$.
Does anyone know how to prove it?
Thanks a lot in advance!
| The equality holds for $a=b$, so we can just consider the case for $a>b$. In this case, to show this inequality, it is equivalent to show that
$$4\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\leq(\ln(a)-\ln(b)).$$
As $a>b>0$, we can suppose that $a=e^x$ and $b=e^y$ where $x>y,\ x,y\in\mathbf{R}$. So we have to show that
$$4\frac{e^\frac{x}{2}-e^\frac{y}{2}}{e^\frac{x}{2}+e^\frac{y}{2}}\leq(x-y),$$
or
$$4\frac{e^\frac{x-y}{2}-1}{e^\frac{x-y}{2}+1}\leq(x-y).$$
Now, denote $\frac{x-y}{4}=z$, the problem becomes to show that for $z>0$ the following inequality holds
$$\frac{e^{2z}-1}{e^{2z}+1}\leq{z}.$$
Obviously, this holds for $z\geq{1}$. So we only consider the case for $0<z<1$, and we have that
$$\frac{e^{2z}-1}{e^{2z}+1}=\frac{e^{z}-e^{-z}}{e^{z}+e^{-z}}=\frac{2(z+\frac{z^3}{3!}+\frac{z^5}{5!}+...)}{2(1+\frac{z^2}{2!}+\frac{z^4}{4!}+...)}\leq{z}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/765738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Find the axis of rotation of a rotation matrix by inspection (NOT by solving $Kv=v$) $$K=\
\begin{pmatrix}
0 & 0 & 1\\
-1 & 0 & 0\\
0 & -1 & 0
\end{pmatrix}$$
Find the axis of rotation for the rotation matrix $K$ by INSPECTION.
This is from my other thread click here to view it
Everything you see below is me finding the axis of rotation by solving $Kv=v$. Just to show you how much working it requires:
Noting that the axis of rotation consists of vectors that remain unmoved. That is a vector $v$ satisfying $Kv = v$. Or, $Kv - Iv=0$ where $I$ is the $3\times3$ identity matrix. For matrix $K$ after solving the homogeneous equations given by $(K-I)v=0$ and showing the working:
$(K-I)v=0$
So
$$K-I=\
\begin{pmatrix}
0 & 0 & 1\\
-1 & 0 & 0\\
0 & -1 & 0
\end{pmatrix}-\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}=\
\begin{pmatrix}
-1 & 0 & 1\\
-1 & -1 & 0\\
0 & -1 & -1
\end{pmatrix}$$
therefore
$$\begin{pmatrix}
-1 & 0 & 1\\
-1 & -1 & 0\\
0 & -1 & -1
\end{pmatrix}v=0$$
writing out the components for $v$ gives
$$\begin{pmatrix}
-1 & 0 & 1\\
-1 & -1 & 0\\
0 & -1 & -1
\end{pmatrix}\begin{pmatrix}
x \\
y \\
z
\end{pmatrix}=0$$
Multiplying out gives three equations
$-x+z=0$
$-x-y=0$
$-y-z=0$
Since
$$
v=\begin{bmatrix}x\\y\\z\end{bmatrix}
$$
Here's the solution parametrically in terms of $x$
\begin{align*}
z&= x\\
y&=-x\\
x&=x
\end{align*}
Hence the axis of rotation is given by the line
$$
\begin{bmatrix}
x\\-x\\x
\end{bmatrix}=x\begin{bmatrix}1\\-1\\1\end{bmatrix}\quad x\in\Bbb R
$$
That is, the axis of rotation is
$$
\operatorname{Span}\left\{\begin{bmatrix}1\\-1\\1\end{bmatrix}\right\}
$$
As you can see this was a lot of work so i would be so grateful if someone could please explain in simple english how to get the answer: $$
\operatorname{Span}\left\{\begin{bmatrix}1\\-1\\1\end{bmatrix}\right\}
$$
by using Inspection?
Many thanks to all that helped so far particularly Brian Fitzpatrick in the last thread
| Neglect the sign for the moment and think of
$$
K=\
\begin{pmatrix}
0 & 0 & 1\\
-1 & 0 & 0\\
0 & -1 & 0
\end{pmatrix}
$$
as a permutation matrix: $(1\to 3), (2\to 1), (3\to 2)$ or shorter $(312)$. So you permute your coordinate axes. This points to a rotation axis along one of the vectors $(\pm 1,\pm 1,\pm 1)^T$. Apply this set of vectors on $K$ to get
$$
\begin{pmatrix}
0 & 0 & 1\\
-1 & 0 & 0\\
0 & -1 & 0
\end{pmatrix}
\begin{pmatrix}
\color{red}{\pm 1} \\ \color{blue}{\pm 1} \\ \pm 1
\end{pmatrix}=
\begin{pmatrix}
\pm 1 \\ \color{red}{\mp 1} \\ \color{blue}{\mp 1}
\end{pmatrix}
$$
So you choose $\color{red}{+ 1},\color{blue}{- 1}$ and $+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/766565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Infinite series representation of elliptic integrals $F(k,\phi)$ and $E(k,\phi).$ Here is my working. For a natural number $n,$$$2^{2n-1}\left(-1\right)^n\sin^{2n}\phi=\tfrac{1}{2}\left(-1\right)^n\binom{2n}{n}+\sum\limits_{r=0}^{n-1}\binom{2n}{r}\left(-1\right)^r\cos 2\left(n-r\right)\phi.$$
According to the binomial theorem,
$$\frac{1}{\sqrt{1-x^2}}=1+\tfrac{1}{2}x^2+\tfrac{3}{8}x^4+\cdots+\tfrac{\left(2n-1\right)!!}{\left(2n\right)!!}x^{2n}+\cdots$$ and upon substituting $k\sin\phi$ for $x$ and integrating we get
$$\phi+\sum\limits_{n=1}^{\infty}\frac{\left(2n-1\right)!!k^{2n}}{\left(2n\right)!!2^{2n-1}}\left\lbrace\frac{\phi}{2}\binom{2n}{n}+\sum\limits_{r=0}^{n}\binom{2n}{r}\frac{\left(-1\right)^{n+r}\sin 2\left(n-r\right)\phi}{2\left(n-r\right)}\right\rbrace=F(k,\phi)?$$
Similarly, since $\sqrt{1-x^2}$ has a series expansion with opposite sign past the first term, and the general term has $\left(2n-3\right)!!$ in place of $\left(2n-1\right)!!,$
$$\phi-\sum\limits_{n=1}^{\infty}\frac{\left(2n-3\right)!!k^{2n}}{\left(2n\right)!!2^{2n-1}}\left\lbrace\frac{\phi}{2}\binom{2n}{n}+\sum\limits_{r=0}^{n}\binom{2n}{r}\frac{\left(-1\right)^{n+r}\sin 2\left(n-r\right)\phi}{2\left(n-r\right)}\right\rbrace=E(k,\phi)?$$
I would like to know if somebody could confirm or correct these formulae, since I only have some tabulated series in a book to go on and these are not listed as far as I can see. I believe the series for $F(k,\tfrac{\pi}{2})$ gives the following expression:
$$\int^{\tfrac{\pi}{2}}_0\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}=\frac{\pi}{2}\left\lbrace1+\left(\frac{1}{2}\right)^2k^2+\left(\frac{1\times 3}{2\times 4}\right)^2k^4+\left(\frac{1\times 3\times 5}{2\times 4\times 6}\right)^2k^6+\cdots\right\rbrace,$$ and the second gives
$$\int^{\tfrac{\pi}{2}}_0\sqrt{1-k^2\sin^2\phi}d\phi=\frac{\pi}{2}\left\lbrace 1-\left(\frac{1}{2}\right)^2k^2-\left(\frac{1\times 3}{2\times 4}\right)^2\frac{k^4}{3}-\left(\frac{1\times 3\times 5}{2\times 4\times 6}\right)^2\frac{k^6}{5}-\cdots\right\rbrace$$ which I think is correct, but I would like somebody to confirm the general formula, ie. the series representations for the elliptic integrals if possible. I cannot find any explicit representations of these integrals as series. I'm aware this is a pretty simple question but I just need a confirmation of either being right or wrong, no need for fancy answers even though they're welcome.
| Assuming that your last line is (you have a typo in the integrand), I just confirm that your development is perfectly correct $$F(k,\pi/2)=\int^{\pi/2}_0\frac{d\phi}{\sqrt{1-k^2\sin^2 \phi}}=\frac{K\left(\frac{k^2}{k^2-1}\right)}{\sqrt{1-k^2}}$$ for which the Taylor expansion built at $k=0$ effectively leads to $$F(k,\pi/2)=\frac{\pi}{2}\left\lbrace1+\left(\frac{1}{2}\right)^2k^2+\left(\frac{1\times 3}{2\times 4}\right)^2k^4+\left(\frac{1\times 3\times 5}{2\times 4\times 6}\right)^2k^6+\cdots\right\rbrace$$
Nice work and well done !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/769937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding matching roots If ${4 + \sqrt{2}}$ is one root of a quadratic equation given by ${x^2 - Px + Q =0}$ where P and Q are rational numbers then find the missing root. The answer is ${4 - \sqrt{2}}$.
And I'm a bit confused on how to derive that answer. I remember for complex conjugate roots the answer would be the same digits with the signs flipped. Is it the same concept even though it doesn't have any imaginary numbers?
| $$x^2-Px+Q=0\\
\implies x=\dfrac{P\pm\sqrt{P^2-4Q}}{2}=\dfrac{P}{2}\pm\dfrac{\sqrt{P^2-4Q}}{2}$$
One solution, given $x=4+\sqrt{2}$ is $P=8,Q=14$. Then, you have the two solutions $4-\sqrt{2},4+\sqrt{2}$.
In general, however, if $\dfrac{P}{2}+\dfrac{\sqrt{P^2-4Q}}{2}=a+\sqrt{b}$, a solution of $x^2-Px+Q=0$, then $\dfrac{P}{2}-\dfrac{\sqrt{P^2-4Q}}{2}=a-\sqrt{b}$, another solution of $x^2-Px+Q=0$, since:
$$\left(x-\dfrac{P}{2}-\dfrac{\sqrt{P^2-4Q}}{2}\right)\left(x-\dfrac{P}{2}+\dfrac{\sqrt{P^2-4Q}}{2}\right)=\\
x^2-\dfrac{Px}{2}+\dfrac{x\sqrt{P^2-4Q}}{2}-\dfrac{Px}{2}+\dfrac{P^2}{4}-\dfrac{P\sqrt{P^2-4Q}}{4}-\dfrac{x\sqrt{P^2-4Q}}{2}+\dfrac{P\sqrt{P^2-4Q}}{4}-\dfrac{{P^2-4Q}}{4}=\\
x^2-Px+0+0+0+Q=
x^2-Px+Q=0$$
We use the fact that the rational numbers are closed under multiplication.
This is our original equation. Hence,
Theorem: If a solution of $x^2-Px+Q=0$ is $\dfrac{P}{2}+\dfrac{\sqrt{P^2-4Q}}{2}$, then the other solution is $\dfrac{P}{2}-\dfrac{\sqrt{P^2-4Q}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/770050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Diophantine equation. Three. Diophantine equation. $X^2+Y^2=qZ^3$
I wonder at what values of the coefficient $q$ equation has a solution.
And of course I wonder how she looks like a formula describing their solutions.
For the special case when $X^2+Y^2=Z^3$ You can get a basic formula.
Has the solutions:
$X=2k^6+8tk^5+2(7t^2+8qt-9q^2)k^4+16(t^3+2qt^2-tq^2-2q^3)k^3+$
$+2(7t^4+12qt^3+6q^2t^2-28tq^3-9q^4)k^2+8(t^5+2qt^4-2q^3t^2-5tq^4)k+$
$+2(q^6-4tq^5-5q^4t^2-5q^2t^4+4qt^5+t^6)$
.................................................................................................................................................
$Y=2k^6+4(3q+t)k^5+2(9q^2+16qt+t^2)k^4+32qt(2q+t)k^3+$
$+2(-9q^4+20tq^3+30q^2t^2+12qt^3-t^4)k^2+4(-3q^5-tq^4+10q^3t^2+6q^2t^3+5qt^4-t^5)k-$
$-2(q^6+4tq^5-5q^4t^2-5q^2t^4-4qt^5+t^6)$
.................................................................................................................................................
$Z=2k^4+4(q+t)k^3+4(q+t)^2k^2+4(q^3+tq^2+qt^2+t^3)k+2(q^2+t^2)^2$
$q,t,k$ - What are some integers any sign. After substituting the numbers and get a result it will be necessary to divide by the greatest common divisor. This is to obtain the primitive solutions.
| For the equation: $$X^2+qY^2=Z^3$$
You can write this simple solution:
$$X=(p^2+qs^2)((p^4-q^2s^4)t^3-3(p^2+qs^2)^2kt^2+3(p^4-q^2s^4)tk^2-(p^4-6qp^2s^2+q^2s^4)k^3)$$
$$Y=2ps(p^2+qs^2)((p^2+qs^2)t^3-3(p^2+qs^2)tk^2+2(p^2-qs^2)k^3)$$
$$Z=(p^2+qs^2)((p^2+qs^2)t^2-2(p^2-qs^2)tk+(p^2+qs^2)k^2)$$
$q - $The ratio is given for the problem.
$p,s,t,k - $ integers asked us.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/772409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Partial fraction decomposition of a complicated rational function Find the partial fraction decomposition of the rational function $\displaystyle \frac{2x^3+7x+5}{(x^2+x+2)(x^2+1)}$
I have tried dividing first but keep running into problem after problem, please help.
| Step 1: Find $a, b, c, d$ such that: $\dfrac{2x^3 + 7x + 5}{(x^2 + x + 2)(x^2 + 1)} = \dfrac{ax + b}{x^2 + x + 2} + \dfrac{cx + d}{x^2 + 1}$
Step 2: Rewrite:
$\dfrac{ax + b}{x^2 + x + 2} + \dfrac{cx+d}{x^2 + 1} = \dfrac{1}{2}\cdot \dfrac{a(2x+1)}{x^2 + x + 2} + \dfrac{1}{2}\cdot\dfrac{2b - a}{(x + \frac{1}{2})^2 + (\frac{\sqrt{7}}{2})^2} + \dfrac{c}{2}\cdot \dfrac{2x}{x^2 + 1} + d\cdot\dfrac{1}{x^2+1}$.
Step 3: Make the right substitution and complete answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/773479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving Identities Can you help me prove this identity:
$$\frac{\sin x}{2\csc x}\left(\tan^2 x + \sin^2 x + \frac{\sin^2x}{\tan^2x}\right)= \frac{1}{2} \tan^2x$$
I have tried working on the left side and changing everything to sine and cosine, but I can't seem to get it down to only $\frac{1}{2} \tan^2 x$.
| Our identity that we need to prove:
$$\frac{\sin x}{2\csc x}\left(\tan^2 x + \sin^2 x +\frac{\sin^2 x}{\tan^2 x}\right)=\frac 12 \tan^2 x$$
Remember that $\dfrac{\sin^2 x}{\tan^2 x}=\cos^2 x$. This follows from the equality $\tan^2 x=\dfrac{\sin^2 x}{\cos^2 x}$.
$$\frac{\sin x}{2\csc x}\left(\tan^2 x+\sin^2 x+\cos^2 x\right)=\frac 12\tan^2 x$$
Remember that $\sin^2 x+\cos^2 x=1$
$$\frac{\sin x}{2\csc x}\left(\tan^2 x+1\right)=\frac 12\tan^2 x$$
Remember that $\tan^2 x+1=\sec^2 x$
$$\frac{\sin x}{2\csc x}\cdot \sec^2 x=\frac 12\tan^2 x$$
Remember that $\csc x =\dfrac{1}{\sin x}$ and that $\sec^2 x =\dfrac{1}{\cos^2 x}$.
$$\frac{\sin x}{\left(2\cdot\dfrac{1}{\sin x}\right)}\cdot \frac 1{\cos^2 x}=\frac 12\tan^2 x$$
$$\frac 12\cdot \sin x \cdot\sin x\cdot \frac 1{\cos^2 x}=\frac 12\tan^2 x$$
$$\frac 12\cdot \frac{\sin^2 x}{\cos^2 x}=\frac 12\tan^2 x$$
Finally, remember that $\tan^2 x=\dfrac{\sin^2 x}{\cos^2 x}$.
$$\frac 12\tan^2 x=\frac 12\tan^2 x \, \, \, \, \, \checkmark$$
$$\color{green}{\therefore \frac{\sin x}{2\csc x}\left(\tan^2 x + \sin^2 x +\frac{\sin^2 x}{\tan^2 x}\right)=\frac 12 \tan^2 x}$$
Hope I helped
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/773620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Can $x^3+x^2+1=0$ be solved using high school methods? I encountered the following problem in a high-school math text, which I wasn't able to solve it: $x^3 + x^2 + 1 = 0$ Am I missing something here, or is indeed a more advanced method necessary to solve this particular cubic?
| Let $x=au+b$. Then $$x^3+x^2+1= a^3u^3+3a^2bu^2+3ab^2u+b^3+a^2u^2+2abu+b^2+1$$ $$=a^3u^3+(3a^2b+a^2)u^2+(3ab^2+2ab)u+(b^3+b^2+1)$$We want $3b+1=0\implies b=-1/3$ and then $$b^3+b^2+1=\frac{-1+3+27}{27}=\frac{29}{27}$$ and $$3ab^2+2ab=a(1/3-2/3)=-\frac{a}{3}$$ So clearly denominators we get $$(3au)^3-3(3au)+29=0$$ Now let $m+n=3au$ so that $$m^3+n^3+3mn(m+n)-3(m+n)+29=0$$ or $$m^3+n^3+3(m+n)(mn-1)+29=0$$ and require that $mn=1$ so that we get $m^3+n^3=-29$ and $m^3n^3=1^3=1$. We have two numbers whose sum and product are known, so they solve the quadratic $y^2+29y+1=0$ which has roots (with help from a calculator) $$\frac{-29\pm 3\sqrt{93}}{2}$$ $m$ and $n$ are the cube roots of this. We never needed to specify $a$ so take it as $1$. We finally get $$x=u+b={1\over 3}(-1+m+n)={1\over 3}\left(-1+\sqrt[3]{\frac{-29+ 3\sqrt{93}}{2}}+\sqrt[3]{\frac{-29- 3\sqrt{93}}{2}}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/774737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Question in fraction (not simple ) I have a question and its answer but I don't know how can i solve
$$\frac {37}{13} = 2+ \frac {1}{x+\frac{1}{5+\frac{1}{y}}} $$
the answer $ x =1, y=2$
Could any one explain how to solve this ?? please
| \begin{align*}
\frac {37}{13}
&= 2+ \frac {1}{x+\frac{1}{5+\frac{1}{y}}} \\
\frac {11}{13}
&= \frac {1}{x+\frac{1}{5+\frac{1}{y}}} \\
\frac {13}{11}
&= x+\frac{1}{5+\frac{1}{y}} \\
\frac {13 - 11x}{11}
&= \frac{1}{5+\frac{1}{y}} \\
\frac {11}{13 - 11x}
&= 5+\frac{1}{y} \\
\frac {-54 + 55x}{13 - 11x}
&= \frac{1}{y} \\
\frac {13 - 11x}{-54 + 55x}
&= y
\end{align*}
Assuming $x$ and $y$ must be integers, note that
$$
\gcd(13 - 11x, -54 + 55x) = \gcd(13-11x, 11) = \gcd(13, 11) = 1
$$
Hence
$$
-54 + 55x = \pm 1 \implies x = 1
$$
Then
$$
y = \frac{13 - 11}{-54 + 55} = 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/776154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Are these two summations equivalent? Is $\sum_{y=2}^{\infty} (\frac{1}{y})(1-p)^{y-1}$ equivalent to $\sum_{y=1}^{\infty} (\frac{1}{y})(1-p)^{y}$ ?
| We'll go ahead and evaluate both the sums to make things more interesting. Recall that for $|x| < 1$,
$$\frac{1}{1-x} = 1 + x + x^2 + x^3 + ...$$
Integrating both sides, we get
$$-\ln(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ...$$
Substitute $x = 1-p$ (I'm assuming $| 1- p| < 1$). Then,
$$-\ln p = (1 - p) + \frac{1}{2}(1-p)^2 + \frac{1}{3}(1-p)^3 + \frac{1}{4}(1-p)^4 + \dots$$
So it follows that:
$$B = \sum_{y = 1}^{\infty} \frac{1}{y}(1-p)^y = -\ln p$$
Moving on, we subtract $(1-p)$ from both sides of our earlier result. This gives us:
$$-\ln p - (1-p) = \frac{1}{2}(1-p)^2 + \frac{1}{3}(1-p)^3 + \frac{1}{4}(1-p)^4 +\frac{1}{5}(1-p)^5 + \dots$$
$$\frac{-\ln p - (1-p)}{1-p} = \frac{1}{2}(1-p) + \frac{1}{3}(1-p)^2 + \frac{1}{4}(1-p)^3 +\frac{1}{5}(1-p)^4 + \dots$$
So we have:
$$A = \sum_{y = 2}^{\infty} \frac{1}{y}(1-p)^{y-1} = \frac{-\ln p - (1-p)}{1-p}$$
The rest is trivial. If $A = B$, then
$$\frac{-\ln p - (1-p)}{1-p} = -\ln p$$
$$\frac{\ln p + (1-p)}{1-p} = \ln p$$
$$\ln p + (1-p) = (1 - p) \ln p$$
$$\ln p + 1 - p = \ln p - p\ln p$$
$$-p\ln p = 1 - p$$
whose only (real) solution to this is when $p = 1$ (why? I'll leave this to you).
Hence, the two are equal if and only if $p = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/776780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the full Taylor expansion of the following: I need to find the full Taylor expansion of $$f(x)=\frac{1+x}{1-2x-x^2}$$
Any help would be appreciated. I'd prefer hints/advice before a full answer is given. I have tried to do partial fractions\reductions. I separated the two in hopes of finding a known geometric sum but I could not.
Edit: I guess you could say that I did not have the.... insight to take the path with the partial decomposition mentioned. I have done some work (I had to go to the gym that is why it took a while)
$$\frac{1+x}{1-2x-x^2}=\frac{1}{2(\sqrt{2}-x-1)}-\frac{1}{2(\sqrt{2}+x+1)}$$ I am going to work with this to go further.
I got to this:
$$\frac{1}{2}\left(\sum_{n=0}^\infty\frac{x^n}{(\sqrt{2}-1)^{n+1}}+\sum_{n=0}^\infty\frac{x^n}{(-\sqrt{2}-1)^{n+1}}\right) $$ But I think this is wrong for some reason.
Edit: Figured it out.
$$\begin{align*}
\implies\frac{1+x}{1-2x-x^2}&=\frac{1}{2(\sqrt{2}-x-1)}-\frac{1}{2(\sqrt{2}+x+1)} \\[2mm]
&=\frac{1}{2}\left(\frac{1}{a-x}-\frac{1}{b+x}\right) \mbox{where $a=\sqrt{2}-1$ and $b=\sqrt{2}+1$}. \\[2mm]
&=\frac{1}{2}\left(\frac{1}{a} \frac{1}{1-\frac{x}{a}}-\frac{1}{b}
\frac{1}{1-\frac{x}{-b}}\right) \\[2mm]
&=\frac{1}{2}\left(\frac{1}{a}\sum_{n=0}^\infty \frac{1}{a^n}x^n-\frac{1}{b}\sum_{n=0}^\infty\frac{1}{(-b)^n}x^n\right) \\[2mm]
&=\frac{1}{2}\left(\frac{1}{\sqrt{2}-1}\sum_{n=0}^\infty \frac{1}{(\sqrt{2}-1)^n}x^n-\frac{1}{\sqrt{2}+1}\sum_{n=0}^\infty\frac{1}{(-\sqrt{2}-1)^n}x^n\right) \\[2mm]
&=\frac{1}{2}\left(\sum_{n=0}^\infty\frac{x^n}{(\sqrt{2}-1)^{n+1}}+\sum_{n=0}^\infty\frac{x^n}{(-\sqrt{2}-1)^{n+1}}\right) \\
&=1+3x+7x^2+17x^3+\ldots
\end{align*}$$
| Here is an alternate approach. Suppose that
$$
\frac{1+x}{1-2x-x^2}=\sum_{k=0}^\infty a_kx^k
$$
Multiply by $1-2x-x^2$ to get
$$
\begin{align}
1+x
&=\underbrace{a_0}_{1}+\underbrace{(a_1-2a_0)}_1x+\sum_{k=2}^\infty\underbrace{(a_k-2a_{k-1}-a_{k-2})}_{0}x^k
\end{align}
$$
Equating coefficients of $x^k$ gives
$$
a_0=1,a_1=3,\text{ and }a_k=2a_{k-1}+a_{k-2}\text{ for }k\ge2
$$
Therefore,
$$
\frac{1+x}{1-2x-x^2}=1+3x+7x^2+17x^3+41x^4+99x^5+\dots
$$
If we wish a formula for $a_n$ we can solve the recurrence
$$
a_k=2a_{k-1}+a_{k-2}
$$
using the roots of $x^2-2x-1=0$, which are $1\pm\sqrt2$, and $a_0=1,a_1=3$ to get
$$
a_k=\frac{(1+\sqrt2)^{k+1}+(1-\sqrt2)^{k+1}}{2}
$$
which is the integer closest to $\dfrac{(1+\sqrt2)^{k+1}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/777535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
proving $\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\leq 2(\frac{b}{a}+\frac{c}{b}+\frac{a}{c})$ For $a+b+c=1; a,b,c>0$, prove that
$$
\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}
\leq 2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right).
$$
| since $$\dfrac{1+a}{1-a}=\dfrac{-(1-a)+2}{1-a}=-1+\dfrac{2}{1-a}=-1+\dfrac{2}{b+c}=-1+\dfrac{2a+2b+2c}{b+c}=1+\dfrac{2a}{b+c}$$
or $$\dfrac{1+a}{1-a}=\dfrac{a+b+c+a}{b+c}=\dfrac{b+c+2a}{b+c}=1+\dfrac{2a}{b+c}$$
so
$$\Longleftrightarrow 2\sum_{cyc}\dfrac{b}{a}-\sum_{cyc}\dfrac{1+a}{1-a}=2\sum_{cyc}\dfrac{b}{a}-\sum_{cyc}\dfrac{2a}{b+c}-3=2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}-\dfrac{2a}{b+c}-\dfrac{2b}{a+c}+\dfrac{2c}{a+b}\right)-3$$
$$\Longleftrightarrow \sum_{cyc}\dfrac{ab}{c(a+b)}\ge\dfrac{3}{2}$$
By Cauchy-Schwarz inequality we have
$$\sum_{cyc}\dfrac{ab}{c(a+b)}\sum_{cyc}abc(a+b)\ge (\sum_{cyc}ab)^2$$
$$\Longleftrightarrow \dfrac{(ab+bc+ac)^2}{abc(a+b)+abc(b+c)+abc(a+c)}\ge\dfrac{3}{2}$$
$$\Longleftrightarrow (ab+bc+ac)^2\ge 3abc$$
Becuase $$(ab+bc+ac)^2\ge 3abc(a+b+c)=3abc$$
This AM-GM inequality.By done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/777956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How many ways can you build a football/soccer team? I'm reading a book on the history of soccer. There are several ways to position a team, for example, four defenders, four midfielders and two forwards is the common 4-4-2. How many ways can you build a team, given that there are 10 players that move because the keeper is always in the same position?
I made some schemes:
10
1-9
1-1-8
1-1-1-7
1-1-1-1-6
1-1-1-1-1-5
1-1-1-1-1-1-4
1-1 1-1-1-1-1-3
1-1-1-1-1-1-1-1-2
1-1-1-1-1-1-1-1-1-1
1-2-7
1-2-1-6
1-2-1-1-5
1-2-1-1-1-4
1-2-1-1-1-1-3
1-2-1-1-1-1-1-2
1-21-1-1-1-1-1-1
1-3-6
1-3-1-5
Then I thought of another scheme, one that would group all possible combinations of diagrams from the number of lines that are willing players on the court.
L1 10
L2 1-9,9-1,2-8,8-2,3-7,7-3,6-4,4-6,5-5
L3
1-1-8,1-8-1,8-1-1,1-2-7,1-7-2, 2-7-1,2-1-7,7-2-1, 7-1-2,1-3-6,1-6-3 .....
I can not find the formula to translate this, if it is correct.
In looking for an answer I found this site. Could you guide me to a book or article to solve my problem?
| For variety, here's another approach.
Your initial approach looks like you settled on the following way to enumerate all ways to make a list of positive integers that adds to 10:
*
*Write down all lists that start with 1 followed by a list that adds to 9
*Write down all lists that start with 2 followed by a list that adds to 8
*...
where you repeat the same approach to write down all lists that add to 9, and so forth.
If $f(n)$ is the number of lists that add to $n$, then your approach suggests
$$ f(n) = f(n-1) + f(n-2) + \ldots + f(1) + 1 $$
We can simplify this recurrence by comparing $f(k)$ with $f(k+1)$: they share most of the same terms, so if we subtract them, most of them cancel out:
$$ f(k+1) - f(k) = f(k) $$
and thus
$$ f(k+1) = 2 f(k) $$
This recurrence is very easy to solve: we're just doubling $f(k)$ every time we move to the next value of $k$. Because $f(1) = 1$, we thus conclude $f(n) = 2^{n-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/780341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
integration of $\int \frac{1}{x+ i\:y}\mathrm{d} x$ I can't seem to find where this result comes from
$$
\int \frac{1}{x+ i\:y}\mathrm{d} x = \frac{\ln(x^2
+y^2)}{2} - i \: \arctan \left( \frac{x}{y} \right)
$$
by my calculation the result should be
$$
\int \frac{1}{x+ i\:y}\mathrm{d} x = \frac{\ln(x^2
+y^2)}{2} + i \: \arctan \left( \frac{y}{x} \right)
$$
as
$$
\frac{\mathrm{d} }{\mathrm{d} x} \ln (x+a) = \frac{1}{x+a} \Rightarrow \int \frac{1}{x+ a}\mathrm{d} x = \ln (x+a)
\\
\ln (x+a) = \ln (|x+a| e^{i \arg{(x+a)}}) = \ln \left( \sqrt{x^2+y^2} \right) + i \underbrace{\arg{(x+a)}}_{\arctan \left( \frac{y}{x} \right)}
$$
| There are different ways to express the same function, and if you compute an unspecified primitive (antiderivative) of some function, the integration constants further enlarge the number of possibilities to express the same family of functions in different ways.
Here, the relations
$$\arctan \frac{1}{\phi} = \int_0^{\frac{1}{\phi}}\frac{dt}{1+t^2} = \int_{\infty}^\phi \frac{1}{1+u^{-2}}\biggl(-\frac{1}{u^2}\biggr)\,du = \int_\phi^\infty \frac{du}{1+u^2} = \frac{\pi}{2}-\arctan\phi$$
for $\phi > 0$ and $\arctan\frac{1}{\psi} = -\arctan\frac{1}{\lvert\psi\rvert} = -\bigl(\frac{\pi}{2}-\arctan \lvert\psi\rvert\bigr) =-\frac{\pi}{2} - \arctan\psi$ for $\psi < 0$ show that the two functions $\arctan\frac{y}{x}$ and $-\arctan\frac{x}{y}$ only differ by a constant on each connected component of their common domain, hence both results are correct(1).
(1) Logically, there is also the possibility that both are incorrect when the two differ only by a locally constant function, but in this case, the results are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/782414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Pie Integral $\int_0^1 \log\frac{(x+\sqrt{1-x^2})^2}{(x-\sqrt{1-x^2})^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.$ Hi I am trying to show this astonishing result$$
\int_0^1 \log\frac{\big(x+\sqrt{1-x^2}\big)^2}{\big(x-\sqrt{1-x^2}\big)^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.
$$
Note we can to use $\ln(a/b)=\ln a-\ln b$ but that didn't help me much . After this I obtained integrals of the form
$$
\int_0^1 \log \big[\big(x\pm\sqrt{1-x^2}\big)^2\big] \frac{x\, dx}{1-x^2}
$$
which I am not sure how to handle. Thanks.
| I hope it is not too late. Let
$$ I(a)=\int_0^1 \log\frac{\big(x+a\sqrt{1-x^2}\big)^2}{\big(x-\sqrt{1-x^2}\big)^2} \frac{x\, dx}{1-x^2}, -1\le a\le 1. $$
Clearly $I(-1)=0$ and
\begin{eqnarray}
I'(a)&=&2\int_0^1 \frac{x}{(x+a\sqrt{1-x^2})\sqrt{1-x^2}}dx\\
&=&2\int_0^{\frac{\pi}{2}}\frac{\sin t}{\sin t+a\cos t}dt.
\end{eqnarray}
Define
$$ A=\int_0^{\frac{\pi}{2}}\frac{\sin t}{\sin t+a\cos t}dt, B=\int_0^{\frac{\pi}{2}}\frac{\cos t}{\sin t+a\cos t}dt $$
and then
$$ A+aB=\frac{\pi}{2}, B-aA=-\log |a|.$$
Thus $A=\frac{\frac{\pi}{2}+a\log |a|}{1+a^2}$. However $\frac{a\log |a|}{1+a^2}$ is an odd function of $a$ and hence
$$ I(1)=2\int_{-1}^1\frac{\frac{\pi}{2}+a\log |a|}{1+a^2}da=2\int_{-1}^1\frac{\frac{\pi}{2}}{1+a^2}da=\frac{\pi^2}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/782964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
} |
Why do the even Bernoulli numbers grow so fast? Question is in the title.
We have:
$$B_{2n} \sim (-1)^{n-1} 4 \sqrt {\pi n} \left( \frac n {\pi e} \right)^{2n}$$
| Recall the generating function
$$f(z) = \sum_{q\ge 0} B_q \frac{z^q}{q!} = \frac{z}{e^z-1}.$$
We can treat it as if it were rational as the poles are countable with
no accumulation point, so we may use the techniques of extracting coefficients from rational generating functions, with the asymptotics being given by inverse powers of the dominant poles.
The pole at $z=0$ is cancelled by the factor $z$ in the numerator. The
nearest two poles are at $\pm 2\pi i,$ both at the same distance from zero ($2\pi$).
Computing the residues we obtain
$$\mathrm{Res}(f(z); z=\pm 2\pi i) = \pm 2\pi i$$
and from this we get the singular decomposition
$$\frac{2\pi i}{z-2\pi i} - \frac{2\pi i}{z+2\pi i}.$$
Turn this into geometric series to get
$$\frac{1}{z/(2\pi i)-1} - \frac{1}{z/(2\pi i)+1}
= - \frac{1}{1-z/(2\pi i)} - \frac{1}{1+z/(2\pi i)}
\\= -\sum_{q\ge 0} \frac{z^q}{(2\pi i)^q}
-\sum_{q\ge 0} (-1)^q \frac{z^q}{(2\pi i)^q}
= -2 \sum_{q\ge 0} \frac{z^{2q}}{(2\pi i)^{2q}}
\\ = -2 \sum_{q\ge 0} (-1)^q \frac{z^{2q}}{(2\pi)^{2q}}
= 2 \sum_{q\ge 0} (-1)^{q+1} \frac{z^{2q}}{(2\pi)^{2q}}.$$
Extracting coeffcients and since $B_{2n} = (2n)! [z^{2n}] f(z)$ we find that
$$B_{2n} \sim (2n)! \times 2 \times
\frac{(-1)^{n+1}}{(2\pi)^{2n}}.$$
Using Stirling's formula this becomes
$$\sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n}
\times 2 \times
\frac{(-1)^{n+1}}{(2\pi)^{2n}}.$$
This is
$$4 \times (-1)^{n+1} \times \sqrt{\pi n}
\left(\frac{n}{\pi e}\right)^{2n}.$$
Addendum. To justify rigorously the above asymptotics we need to
verify that there isn't an additional entire component not accounted
for in the complete singular decomposition.
To do this introduce the sum
$$g(w) =
\sum_{q\ge 1}\left(\frac{1}{w/q-1}-\frac{1}{w/q+1}\right)
= 2 \sum_{q\ge 1} \frac{1}{w^2/q^2-1}.$$
This can be evaluated using a standard technique from complex
variables and setting
$$G(z) = 1/(w^2/z^2-1)\times\pi\cot(\pi z)$$
we have
$$g(w) =
-
\left(
\mathrm{Res}(G(z); z=0)
+\mathrm{Res}(G(z); z=w)
+\mathrm{Res}(G(z); z=-w)
\right).$$
Computing the residues we get
$$g(w) = \pi w \cot(\pi w).$$
Now we claim that
$$f(z) = -\frac{1}{2}z + \sum_{q\ge 1}
\left(\frac{2\pi i q}{z-2\pi i q}-\frac{2\pi i q}{z+2\pi i q}\right).$$
The sum is $g(z/(2\pi i))$ so we get
$$-\frac{1}{2}z + \frac{z}{2i} \cot\left(\frac{z}{2i}\right)
= -\frac{1}{2}z + \frac{z}{2i} \times i \times
\frac{e^{z/2}+e^{-z/2}}{e^{z/2}-e^{-z/2}}
\\= \frac{1}{2} z
\left(-1 + \frac{e^z+1}{e^z-1}\right)
= \frac{1}{2}z \frac{2}{e^z-1} = \frac{z}{e^z-1}.$$
This shows rigorously that the entire part is $$-\frac{1}{2}z$$ and
hence makes no contribution to $B_{2n}$ where $n\ge 1$ and the above
singular decomposition is justified.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof via induction $1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$ (b) Prove that for every integer $n \ge 1$, $$1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$$
This is the second part of a two part question. Part (a) was the following:
Write the sum: $1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2)$ using summation notation.
It was simple enough : $\sum k(k+2)$.
For this question, the base case $(n=1)$ holds, as $1\cdot(1+2) = 3 = (1\cdot2\cdot9)/6$.
Induction step: Assume the above holds for all $n = k$, prove that it holds for all $n = k+1$
I'm a bit lost from here, help?
| Assume result is true for $n=k$ i.e.
$$S_k=\frac {k(k+1)(2k+7)}6$$
Then, adding the next term gives:
$$\begin{align}
S_{k+1}&=S_k+(k+1)(k+2)\\
&=\frac {k(k+1)(2k+7)}6+(k+1)(k+3)\\
&=\frac{k+1}6 \left( k(2k+7)+6(k+3)\right)\\
&=\frac{k+1}6\left( 2k^2+13k+18\right)\\
&=\frac{(k+1)(k+2)(2k+9)}6\\
&=\frac{(\overline{k+1})(\overline{k+1}+1)(2\overline{k+1}+7)}6\\
\end{align}$$
i.e. result is also true for $n=k+1$.
As shown in the question, the result is true for $n=1$.
Hence, by induction, the result is true for all positive integer $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/785355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Interior point of $\Delta\,ABC$ if $P(\lambda,2)$ is an interior point of $\Delta\,ABC$ formed by the lines
$$x+y=4$$ $$3x-7y=8$$ $$4x-y=31$$ Find $\lambda$
My Idea: The vertices of $\Delta ABC$ are
$A(\frac{18}{5},\frac{2}{5})$ $\:$ $B(7,-3)$ and $C(\frac{209}{25},\frac{61}{25}) $
So using areas of $\Delta ABC$, $\Delta APC$,$\Delta ABP$,$\Delta PBC$ we have
$$\begin{vmatrix}
&\frac{18}{5} & \frac{2}{5} & 1\\
& 7 & {-3}& 1\\
&\frac{209}{25} &\frac{61}{25} &1
\end{vmatrix}=\begin{vmatrix}
&\frac{18}{5} & \frac{2}{5} & 1\\
& \lambda & {2}& 1\\
&\frac{209}{25} &\frac{61}{25} &1
\end{vmatrix}+\begin{vmatrix}
&\frac{18}{5} & \frac{2}{5} & 1\\
& 7 & {-3}& 1\\
&\lambda &2 &1
\end{vmatrix}+\begin{vmatrix}
&\lambda & 2 & 1\\
& 7 & {-3}& 1\\
&\frac{209}{25} &\frac{61}{25} &1
\end{vmatrix}$$ from which we get $\lambda$
Is there any smarter way to do this..
| Another way is rough plot all these three lines with $y=2$ With seeing the points of intersection and a rough idea of their positions you will be able to directly tell the range of $\lambda$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/785895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate Gauss-like Integral Evaluate Integral
$$\int_0^\infty e^{-ay^{2}-\frac{b}{y^2}}dy $$
Where a and b are real and positive.
This integral is eerily similar to the Gaussian integral
$$\int_0^\infty e^{-\alpha x^2}dx = \frac{1}{2} \sqrt{\frac{\pi}{\alpha}}$$
This is an integral I have come across as a step in a problem doing some homework for Advanced Statistics... Not sure where to begin.
| This answer is taken from my answer here.
$$
\begin{align}
\int_{v=0}^\infty \exp\left(-av^2-\frac{b}{v^2}\right)\,dv&=\int_{v=0}^\infty \exp\left(-a\left(v^2+\frac{b}{av^2}\right)\right)\,dv\\
&=\int_{v=0}^\infty \exp\left(-a\left(v^2-2\sqrt{\frac{b}{a}}+\frac{b}{av^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dv\\
&=\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dv\\
&=\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\
\end{align}
$$
The trick to solve the last integral is by setting
$$
I=\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv.
$$
Let $t=-\frac{1}{v}\sqrt{\frac{b}{a}}\;\rightarrow\;v=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dv=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$, then
$$
I_t=\sqrt{\frac{b}{a}}\int_{t=0}^\infty \frac{\exp\left(-a\left(-\frac{1}{t}\sqrt{\frac{b}{a}}+t\right)^2\right)}{t^2}\,dt.
$$
Let $t=v\;\rightarrow\;dt=dv$, then
$$
I_t=\int_{t=0}^\infty \exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.
$$
Adding the two $I_t$s yields
$$
2I=I_t+I_t=\int_{t=0}^\infty\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)\exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.
$$
Let $s=t-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;ds=\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)dt$ and for $0<t<\infty$ is corresponding to $-\infty<s<\infty$, then
$$
I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}}.
$$
Thus
$$
\begin{align}
\int_{v=0}^\infty \exp\left(-av^2-\frac{b}{v^2}\right)\,dv&=\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\
&=\frac12\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Summation of Fibonacci numbers $F_n$ with $n$ odd vs. even Compare the summation below:
$$\begin{align}
\smash[b]{\sum_{i=1}^n F_{2i-1}}&=F_1+F_3+F_5+\cdots+F_{2n-1}\\
&=1+2+5+\cdots+F_{2n-1}\\
&=F_{2n}\\
\end{align}
$$
with this one:
$$\begin{align}
\smash[b]{\sum_{i=1}^n F_{2i}}&=F_2+F_4+F_6+\cdots+F_{2n}\\
&=1+3+8+\cdots+F_{2n}\\
&=F_{2n+1}-1\\
\end{align}$$
When I first discovered these patterns I was amazed. Naively I had thought that an every-other-number sum of Fibonacci numbers would be the same pattern whether the parity of their indices was odd or even, but I was wrong! Why is the above true, where the summation of odd-indexed Fibonacci numbers is another Fibonacci number, but the even-indexed sum is a Fibonacci number minus 1?
| A clean way to see this is by using generating functions. Define $F(z) = \sum_{n \ge 0} F_n z^n$, take the recurrence:
$$
F_{n + 2} = F_{n + 1} + F_n \qquad F_0 = 0, F_1 = 1
$$
Multiply by $z^n$, sum over all valid values for $n$, i.e., $n \ge 0$, and recognize the resulting sums:
$$
\frac{F(z) - F_0 - F_1 z}{z^2}
= \frac{F(z) - F_0}{z} + F(z)
$$
Solving:
$$
F(z) = \frac{z}{1 - z - z^2}
$$
We also have, if $A(z) = \sum_{n \ge 0} a_n z^n$ then:
\begin{align}
\sum_{n \ge 0} a_{2 n} z^{2 n}
&= \frac{A(z) + A(-z)}{2} \\
\sum_{n \ge 0} a_{2 n + 1} z^{2 n + 1}
&= \frac{A(z) - A(-z)}{2} \\
\sum_{n \ge 0} \left( \sum_{0 \le k \le n} a_k \right) z^n
&= \frac{A(z)}{1 - z}
\end{align}
So, for even/odd Fibonacci numbers:
\begin{align}
F_e(z)
&= \sum_{n \ge 0} F_{2 n} z^n \\
&= \frac{F(z^{1/2}) + F(- z^{1/2})}{2} \\
&= \frac{z}{1 - 3 z + z^2} \\
F_o(z)
&= \sum_{n \ge 0} F_{2 n + 1} z^n \\
&= \frac{F(z^{1/2}) - F(- z^{1/2})}{2 z^{1/2}} \\
&= \frac{1 - z}{1 - 3 z + z^2} \\
\end{align}
\begin{align}
\sum_{n \ge 0} \left( \sum_{0 \le k \le n} F_{2 n} \right) z^n
&= \frac{F(z^{1/2}) + F(- z^{1/2})}{2 (1 - z)} \\
&= \frac{z}{(1 - z) (1 - 3z + z^2)} \\
&= \frac{1 - z}{1 - 3 z + z^2} - \frac{1}{1 - z}
\end{align}
The first term is the generating function of the odd Fibonacci numbers, the second one is the generating function of the sequence of ones. Comparing coefficients:
$$
\sum_{0 \le k \le n} F_{2 n} = F_{2 n + 1} - 1
$$
Similarly, as $F_0 = 0$:
\begin{align}
\sum_{n \ge 0} \left( \sum_{0 \le k \le n} F_{2 n + 1} \right) z^n
&= \frac{F(z^{1/2}) - F(- z^{1/2})}{2 z^{1/2} (1 - z)} \\
&= \frac{1}{1 - 3z + z^2} \\
&= \frac{F_e(z) - F_0}{z}
\end{align}
The last expression corresponds to the even Fibonacci numbers shifted by one:
$$
\sum_{0 \le k \le n} F_{2 n + 1} = F_{2 n + 2}
$$
Note that we didn't need any premonition on what the sums would turn out to be.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Is this proof by induction correct? Prove by induction that for all $n\in\mathbb N$, $3\mid n^3+3n^2+2n$.
$$P(1) = (1)^3+3(1)^2+2(1) = 6$$
Which is clearly divisble by $3$. Therefore, $P(1)$ is true.
Assume $P(1),\ldots,P(n)$ and show $P(n+1)$ is true.
$$(n+1)^3+ 3(n+1)^2+ 2(n+1) = n^3+6n^2+11n+6$$
This can be rewritten: $(n^3+3n^2+2n)+3(n^2+3n+2)$.
First term is divisible by $3$ because $P(n)$ is true.
The second term is divisible by $3$ because it is a multiple of $3$.
Thus, their sum is divisible by $3$.
Is there anything I am missing?
| Your proof looks solid, but I might suggest an easier proof:
For all $n \in \mathbb{N}$, we have $n \equiv 0, 1, 2 \pmod{3}$. If $n \equiv 0 \pmod{3}$, we are done.
If $n \equiv 1 \pmod{3}$, then we have $n^3 + 3n^2 + 2n \equiv (1+2) \equiv 0 \pmod{3}$.
If $n \equiv 2 \pmod{3}$, then we have $n^3 + 3n^2 + 2n \equiv (2 + 1) \equiv 0 \pmod{3}$.
We conclude that $3|(n^3 + 3n^2 + 2n)$ for all natural numbers $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why are these two given expressions equivalent? $6\cos^2(\frac {x}{2})-7\cos(\frac {x}{2})+2=0$ is equivalent to the expression $(3\cos(\frac {x}{2})-2)(2\cos(\frac {x}{2})-1) = 0$
Could someone explain the steps involved in evaluating the first expression to equal the second expression?
I noticed the expression in this question here, and I was curious how this could be achieved.
| The answer comes from factoring.
When you foil the two parentheticals from the second expression, their products can be seen to equal the first.
We treat $\cos(\frac {x}{2})$ much like we would the variable $x$ within any given expression.
If given the expression $2x^2-3x+1 = 0$, we are able to factor this to show that $(-2x+1)(-x+1) = 0$.
Given this particular expression: $6\cos^2(\frac {x}{2})-7\cos(\frac {x}{2})+2=0$, we are able factor this to show that $(3\cos(\frac {x}{2})-2)(2cos(\frac {x}{2})-1) = 0$.
When foiled, it is shown that:
First: $3\cos(\frac {x}{2}) \cdot (2\cos(\frac {x}{2}) = 6\cos^2(\frac {x}{2})$
Second: The product of $3\cos(\frac {x}{2}) \cdot -1 = -3\cos(\frac {x}{2})$ and the product of $2cos(\frac {x}{2}) \cdot -2 = -4cos(\frac {x}{2})$, and their sum equals $-7\cos(\frac {x}{2})$
Third: $-2 \cdot -1 = 2$
Final, we add the first, second, and third steps together, and their sums are show to be equivalent to the initial expression: $6\cos^2(\frac {x}{2})$ [first] $-7\cos(\frac {x}{2})$[second]$+2$[third]$=0$.
And the result is [Final].
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/790358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
calculating the sum of a series I was wondering if there is a way to use complex-analysis to solve the sum of the following series? (just like you can use it to solve integrals of some kinds (using fourier's transform for integrals).
$$\sum^{\infty}_{n=0}\frac{1}{n^2+25}$$
and
$$\sum^{\infty}_{n=0}\frac{(-1)^n}{n^2+25}$$
And if not, how to solve them otherwise?
thanks in advance!
| We can evaluate the two sums using following expansions of trigonometry functions
which can be proved using Mittag Leffler's theorem:
$$
\frac{\cos x}{\sin x} = \lim_{N\to\infty}\sum_{n=-N}^N \frac{1}{x+n\pi}
\quad\text{ and }\quad
\frac{1}{\sin x} = \lim_{N\to\infty}\sum_{n=-N}^N \frac{(-1)^n}{x+n\pi}
$$
Let's look at the expansion at the left first. Substitute $x$ by $\pi y i$, move the term for $n = 0$ and group the terms for $+n$ and $-n$ togather, we get
$$\frac{\cos(\pi y i)}{\sin(\pi y i)} - \frac{1}{\pi y i} = \frac{1}{\pi}\sum_{n=1}^\infty \left(\frac{1}{y i + n} + \frac{1}{yi - n}\right)
= \frac{2 y}{\pi i}\sum_{n=1}^\infty\frac{1}{y^2+n^2}$$
This leads to
$$\sum_{n=0}^\infty \frac{1}{y^2+n^2} = \frac{1}{y^2} + \frac{\pi}{2y}\left( \frac{\cosh(\pi y)}{\sinh(\pi y)} - \frac{1}{\pi y}\right)
= \frac{1}{2y^2} + \frac{\pi}{2y}\frac{\cosh(\pi y)}{\sinh(\pi y)}
$$
The evaluation of the second sum is pretty similar, you get something like
$$
\sum_{n=0}^\infty \frac{(-1)^n}{y^2+n^2} = \frac{1}{2y^2} + \frac{\pi}{2y} \frac{1}{\sinh(\pi y)}
$$
As a result, one get
$$
\sum_{n=0}^\infty \frac{1}{25+n^2} = \frac{1}{50} + \frac{\pi}{10}\frac{\cosh(5\pi)}{\sinh(5\pi)}
\quad\text{ and }\quad
\sum_{n=0}^\infty \frac{(-1)^n}{25+n^2} = \frac{1}{50} + \frac{\pi}{10}\frac{1}{\sinh(5\pi)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/792817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a$ Hi I am trying to solve this integral $$
I:=\int_0^1 \log\left(\frac{1+ax}{1-ax}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}=\pi\arcsin\left(a\right),\qquad
\left\vert a\right\vert \leq 1.
$$
It gives beautiful result for $a = 1$
$$
\int_0^1 \log\left(\frac{1+ x}{1-x}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}
=\frac{\pi^2}{2}.
$$
I tried to write
$$
I=\int_0^1 \frac{\log(1+ax)}{x\sqrt{1-x^2}}dx-\int_0^1 \frac{\log(1-ax)}{x\sqrt{1-x^2}}dx
$$
If we work with one of these integrals we can write
$$
\sum_{n=1}^\infty \frac{(-1)^{n+1} a^n}{n}\int_0^1 \frac{x^{n-1}}{\sqrt{1-x^2}}dx-\sum_{n=1}^\infty \frac{a^n}{n}\int_0^1 \frac{x^{n-1}}{\sqrt{1-x^2}}dx,
$$
simplifying this I get an infinite sum of Gamma functions. which i'm not sure how to relate to the $\arcsin$ Thanks.
| It is not necessary to use complex analysis or to use power series to compute. From @Random Variable, we have
$$ I'(a) =\int_{-1}^1 \frac{dx}{(1+ax)\sqrt{1-x^2}}=\int_{-1}^1 \frac{1}{1+ax}d\arcsin x. $$
Let $x=\sin t$. Then
$$ I'(a) =\int_{-\pi/2}^{\pi/2} \frac{1}{1+a\sin t}dt. $$
Let $u=\tan\frac{t}{2}$. Then
\begin{eqnarray}
I'(a)&=&2\int_{-1}^{1} \frac{1}{u^2+1+2au}du=2\int_{-1}^{1} \frac{1}{(u+a)^2+1-a^2}du\\
&=&\frac{2}{\sqrt{1-a^2}}(\arctan\frac{1+a}{\sqrt{1-a^2}}-\arctan\frac{-1+a}{\sqrt{1-a^2}})\\
&=&\frac{2}{\sqrt{1-a^2}}(\arctan\frac{1+a}{\sqrt{1-a^2}}+\arctan\frac{1-a}{\sqrt{1-a^2}})\\
&=&\frac{2}{\sqrt{1-a^2}}\frac{\pi}{2}\\
&=&\frac{\pi}{\sqrt{1-a^2}}.
\end{eqnarray}
But $I(0)=0$ and so $ I(a)=\pi\arcsin a. $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/795493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 0
} |
How should I solve integrals of this type? The general form of the integral I want to solve is:
$$ \int e^{bx}\sin(ax) dx$$
Euler's formula has a nice connection, but the i makes it too complicated.
Doing it by parts doesn't seem to get me anywhere.
Do you have any tips for how to begin solving this?
| By using the exponential version of sine:
\begin{align}
\int e^{bx} \ \sin(ax) \ dx &= \frac{1}{2i} \int \left( e^{(b+ai)x} - e^{(b-ai)x}
\right) \ dx \\
&= \frac{1}{2i} \left[ \frac{e^{(b+ai)x}}{b+ai} - \frac{e^{(b-ai)x}}{b-ai} \right] \\
&= \frac{1}{2i (b+ai)(b-ai)} \left[ (b-ai) e^{(b+ai)x} - (b+ai)e^{(b-ai)x}
\right] \\
&= \frac{1}{2i(b^{2}+a^{2})} \left[ b \ e^{bx} \left(e^{ai x} - e^{-ai x} \right) - (ai) e^{bx} \left( e^{ai x} + e^{-ai x} \right) \right] \\
&= \frac{e^{bx}}{b^{2} + a^{2}} \left[ b \sin(ax) - a \cos(ax) \right]
\end{align}
By using integration by parts:
\begin{align}
\int e^{bx} \ \sin(ax) \ dx &= \frac{1}{b} e^{bx} \sin(ax) - \frac{a}{b} \int
e^{bx} \ \cos(ax) \ dx \\
&= \frac{1}{b} e^{bx} \sin(ax) - \frac{a}{b^{2}} e^{bx} \cos(ax) - \frac{a^{2}}{b^{2}} \int e^{bx} \ \sin(ax) \ dx
\end{align}
which is
\begin{align}
\left( 1 + \frac{a^{2}}{b^{2}} \right) \int e^{bx} \ \sin(ax) \ dx = \frac{1}{b} e^{bx} \sin(ax) - \frac{a}{b^{2}} e^{bx} \cos(ax)
\end{align}
or
\begin{align}
\int e^{bx} \ \sin(ax) \ dx = \frac{e^{bx}}{b^{2} + a^{2}} \left[ b \sin(ax) - a \cos(ax) \right]
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/796730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
How does WolframAlpha simplify sine and cosine? When I feed WolframAlpha an expression like $\sin({\pi\frac{2}{3}})$, it correctly prints that this is equal to $\frac{\sqrt3}{2}$, instead of the decimal expansion $0.866025403\ldots$.
Perhaps it has a lookup table for common fractions of $\pi$. Or is it likely to be more sophisticated? Does it solve the convergence of the series expansion for $\sin(x)$?
| It must be that $\sin\left(\dfrac{p}{q}\pi\right)$ is algebraic. To see why check out this question.
I am almost certain that W|A doesn't use power series unless the value is very small simply because it would be too slow to calculate the value of arbitrary trig functions using power series. It is more likely that there is a certain class of rational numbers such as $\frac{1}{3}$ where the forms of $\sin(\frac{\pi}{3})$ and $\cos(\frac{\pi}{3})$ are known and then formulae such as the double angle formula gives results for other rational numbers such as $\sin({\frac{2\pi}{3}})$. The result is then simplified and sent to the user.
This is only a conjecture as I do not have access to any Mathematica source code.
Here is an example. Suppose you know that $$\sin\left(\dfrac{\pi}{2}\right) = 1 \quad\text{and}\quad \cos\left(\dfrac{\pi}{2}\right) = 0.$$
We know that $$\sin^2(\theta) = \dfrac{1-\cos(2\theta)}{2}.$$
It must be that:
$$\sin^2\left(\dfrac{\pi}{4}\right) = \dfrac{1}{2} \quad\text{and}\quad \cos^2\left(\dfrac{\pi}{4}\right) = 1-\dfrac{1}{2} = \dfrac{1}{2}$$
$$\sin\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2} \quad\text{and}\quad \cos\left(\dfrac{\pi}{4}\right) = \dfrac{\sqrt{2}}{2}.$$
Continuing...
$$\sin\left(\dfrac{\pi}{8}\right) = \dfrac{\sqrt{{2-\sqrt{2}}}}{2} \quad\text{and}\quad \cos\left(\dfrac{\pi}{8}\right) = \dfrac{\sqrt{{2+\sqrt{2}}}}{2}$$
$$\sin\left(\dfrac{\pi}{16}\right) = \dfrac{\sqrt{2-\sqrt{{2+\sqrt{2}}}}}{2} \quad\text{and}\quad \cos\left(\dfrac{\pi}{16}\right) = \dfrac{\sqrt{2+\sqrt{{2+\sqrt{2}}}}}{2} $$
Which agrees with W|A.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/797844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int_0^{\pi/4}\log \tan x \frac{\cos 2x}{1+\alpha^2\sin^2 2x}dx=-\frac{\pi}{4\alpha}\text{arcsinh}\alpha$ Hi I am trying to prove this
$$
I:=\int_0^{\pi/4}\log\left(\tan\left(x\right)\right)\,
\frac{\cos\left(2x\right)}{1+\alpha^{2}\sin^{2}\left(2x\right)}\,{\rm d}x
=-\,\frac{\pi}{4\alpha}\,\text{arcsinh}\left(\alpha\right),\qquad \alpha^2<1.
$$
What an amazing result this is! I tried to write
$$
I=\int_0^{\pi/4} \log \sin x\frac{\cos 2x}{1+\alpha^2\sin^2 2x}-\int_0^{\pi/4}\log \cos x \frac{\cos 2x}{1+\alpha^2\sin^2 2x}dx
$$
and played around enough here to realize it probably isn't the best idea.
Now back to the original integral I, we can possibly change variables $y=\tan x$ and
re-writing the original integral to obtain
$$
\int_0^{\pi/4}\log \tan x \frac{\cos 2x}{1+{\alpha^2}\big(1-\cos^2 (2x)\big)}dx=\int_0^1 \log y \frac{1-y^2}{1+y^2}\frac{1}{1+{\alpha^2}\big(1-(\frac{1-y^2}{1+y^2})^2\big)}\frac{dy}{1+y^2}.
$$
Simplifying this we have
$$
I=\int_0^1\log y \frac{1-y^2}{1+y^2}\frac{(1+y^2)^2}{(1+y^2)^2+4\alpha^2y^2}\frac{dy}{1+y^2}=\int_0^1\log y \frac{1-y^2}{(1+y^2)^2+4\alpha^2y^2}dy
$$
Another change of variables $y=e^{-t}$ and we have
$$
I=-\int_0^\infty \frac{t(1-e^{-2t})}{(1+e^{-2t})^2+4\alpha^2 e^{-2t}} e^{-t}dt
$$
but this is where I am stuck...How can we calculate I? Thanks.
| I use on Ron's results from integrating by parts:
$$\alpha I(\alpha) = - \int_0^{\pi/4} dx \frac{\arctan{(\alpha \sin{2 x})}}{\sin{2 x}},$$
As an alternative way to complete the problem, use the method differentiating under the integral sign:
$$\frac{d}{d\alpha}(\alpha I(\alpha)) = -\int_{0}^{\pi/4}\frac{dx}{\alpha^2\sin^2{2x}+1}=-\frac{\pi}{4\sqrt{1+\alpha^2}}\\
\implies \alpha I(\alpha) = -\frac{\pi}{4}\int_{0}^{\alpha}\frac{d\tilde\alpha}{\sqrt{1+\tilde\alpha^2}} = -\frac{\pi}{4}\sinh^{-1}{\alpha}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/798227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\lim\limits_{n \rightarrow \infty} \left(1-\frac{1}{n^2}\right)^n= 1.$ Prove that $\lim\limits_{n \rightarrow \infty} \left(1-\frac{1}{n^2}\right)^n= 1.$
I need to show that there exists $N \in \mathbb{N}: \forall n \geq N : \left|(1-\frac{1}{n^2})^n-1\right| \lt \epsilon$ $,\:\:\forall \epsilon \gt 0.$
Since $(1-\frac{1}{n^2})^n \leq 1$ $\forall n \in \mathbb{N}_+ \Rightarrow \left|(1-\frac{1}{n^2})^n-1\right| = 1 - (1-\frac{1}{n^2})^n$
Therefore I need to solve $1 - \left(1-\frac{1}{n^2}\right)^n < \epsilon$ for $n$.
Unfortunately I cannot solve this explicitly for n. The best I can do is:
$$\log(1-\epsilon) \lt n \log\left(1-\frac{1}{n^2}\right)$$
Is there a way to solve this for $n$ explicitly? How can I prove this if this is if it is not solvable for $n$?
| Consider the expansion of $(1-\frac{1}{n^2})^n.$
$$\begin{align}\left(1-\frac{1}{n^2}\right)^n &= 1 - n\dfrac{1}{n^2}+\dfrac{n(n-1)}{2}\dfrac{1}{n^4} - \cdots \\ &= 1-\dfrac{1}{n}+\dfrac{n-1}{2n^3}+\cdots\\&\geq1-\dfrac{1}{n}\end{align}$$
We can construct the inequality $$1-\dfrac{1}{n} \leq \left(1-\frac{1}{n^2}\right)^n \leq 1$$
Subtract $1$ and then change the order of inequality
$$-\dfrac{1}{n} \leq \left(1-\frac{1}{n^2}\right)^n -1 \leq 0$$
$$0 \leq 1- \left(1-\frac{1}{n^2}\right)^n \leq \dfrac{1}{n}$$
After this you can just choose $N = \left\lceil\dfrac{1}{\epsilon}\right\rceil.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/800219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
The $n$'th derivative of $x^x$ I want to know the $n$'th derivative of $f(x)=x^x$.
Then, I'll calculate $f(0)$ with Taylor expansion of $f(x)$ on $a=1$.
Here is my answer, but it is unfinished.
The derivative of $f(x)=x^x$
$$\begin{align}
f'(x)&=x^x(\log x+1)\\
f''(x)&=x^x(\log x+1)^2+x^{x-1}\\
f'''(x)&=x^x(\log x+1)^3+3x^{x-1}(\log x+1)-x^{x-2}\\[5pt]
f(x)^{(4)}&=x^x(\log x+1)^4+4x^{x-1}(\log x+1)^2-4x^{x-2}(\log x+1)+3x^{x-2}+2x^{x-3}\\
f(x)^{(5)}&=x^x(\log x+1)^5+10x^{x-1}(\log x+1)^3-10x^{x-2}(\log x+1)^2+15x^{x-2}(\log x+1)\\&\quad+10x^{x-3}(\log x+1)-10x^{x-3}-6x^{x-4}\\
f(x)^{(6)}&=x^x(\log x+1)^6+15x^x(\log x+1)^4-20x^{x-2}(\log x+1)^3+45x^{x-2}(\log x+1)^2\\&\quad+30x^{x-3}(\log x+1)^2-50x^{x-3}(\log x+1)+15x^{x-3}-46x^{x-4}(\log x+1)\\&\quad+40x^{x-4}+24x^{x-5}
\end{align}$$
Taylor expansion of $f(x)=x^x$ in $a=1$
$$\begin{align}
f(x)&=\sum_{i=0}^{n-1}\frac{f^{(i)}(1)}{i!}\\[5pt]
&\qquad=\frac1{0!}+\frac1{1!}(x-1)+\frac2{2!}(x-1)^2+\frac3{3!}(x-1)^3+\frac8{4!}(x-1)^4+\frac{12}{5!}(x-1)^5\\&\qquad+\frac{54}{6!}(x-1)^6+\cdots
\end{align}$$
| For $n\in\{0,1,2,\dotsc\}$, we have
\begin{equation}\label{power-exp-deriv-eq}
(x^x)^{(n)}=n!x^{x-n}\sum_{k=0}^{n} x^{k} \sum_{j=0}^{k}\Biggl[\sum_{q=0}^{n-k} \frac{s(q+j,j)}{(q+j)!} \binom{j}{n-k-q}\Biggr]\frac{(\ln x)^{k-j}}{(k-j)!},
\end{equation}
where $s(n,k)$ for $n\ge k\ge0$ denotes the Stirling numbers of the first kind. Consequently, we have
\begin{equation}\label{power-exp-taylor-ser}
x^x=\sum_{n=0}^\infty\Biggl[\sum_{k=0}^{n} \sum_{q=k}^{n} \frac{s(q,k)}{q!} \binom{k}{n-q}\Biggr](x-1)^n, \quad |x-1|<1.
\end{equation}
For more information and their proofs, please see the site https://mathoverflow.net/a/439188 and the paper
*
*Jian Cao, Feng Qi, and Wei-Shih Du, Closed-form formulas for the $n$th derivative of the power-exponential function $x^x$, Symmetry 15 (2023), no. 2, Article 323, 13 pages; available online at https://doi.org/10.3390/sym15020323.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/802256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
How prove that there exists $m$ such that $|f(m)|\le \dfrac{\sqrt{b^2-4c}}{2}$ Let the function $$f(x)=x^2+bx+c,\qquad b^2-4c>0$$ Assume that $x_{1},x_{2}$ are the roots of $f(x)$ and $|x_{1}-x_{2}|\ge 1$.
Show that: There exists an integer $m$, such that
$$|f(m)|\le \dfrac{\sqrt{b^2-4c}}{2}$$
My idea: Since $$x_{1}+x_{2}=-b,x_{1}x_{2}=c$$
and $$f(x)=(x-x_{1})(x-x_{2})$$
so
$$|x_{1}-x_{2}|=\sqrt{(x_{1}+x_{2})^2-4x_{1}x_{2}}=\sqrt{b^2-4c}$$
so
$$|f(m)|=|(m-x_{1})(m-x_{2})|\le \dfrac{(x_{1}-x_{2})^2}{4}$$
Then I can't.Thank you very much.
| We consider the following equivalent problem.
For $ d \geq 1 $, define $ g(x) = x^2 + dx$. We want to show that there exists real numbers $ j \leq 0 \leq k$ with $k-j \geq 1$ such that for all $ j \leq x \leq k$, we have
$$ - \frac{d}{2} \leq g(x) \leq \frac{d}{2} . $$
Proof of Equivalent: To see why this is equivalent, is because we can go from $g(x)$ to $f(x)$ by a horizontal translation, and the translated range of $[j,k]$ guarantees us that we have an integer value.
Proof of new problem: With this simplified version, we can easily solve the inequality.
For $ 1 \leq d \leq 2$, the solution set is
$$ \frac{1}{2} ( -d - \sqrt{d(d+2)}) \leq x \leq \frac{1}{2}( \sqrt{d(d+2)}-d).$$
Verify that $k-j = \sqrt{d(d+2)} \geq \sqrt{3} \geq 1 $.
For $ 2 < d$, the solution set (which contains 0) is
$$\frac{1}{2} ( \sqrt{ d(d-2)} -d) \leq x \leq \frac{1}{2} ( \sqrt{ d(d+2) } -d ) .$$
Verify that $ k-j = \frac{1}{2} ( \sqrt{ d(d+2) } - \sqrt{d(d-2)}) = \frac{2d}{ \sqrt{d^2+2d} + \sqrt{d^2+2d}} > \frac{2d}{2d} = \geq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the general term of the recursive relation $a_{n+1}=\frac{1}{n+1}\sum\limits_{k=0}^n a_k 2^{2n-2k+1}$ I need to find the general term of the recursive relation $a_{n+1}=\frac{1}{n+1}\sum\limits_{k=0}^n a_k 2^{2n-2k+1}$
I know it's the central binomial sequence but I can't find a way to show it.
$a_0=1$
| We are given the recurrence relation
$$a_{n+1}=\frac{1}{n+1}\sum_{k=0}^{n}a_{k}2^{2n-2k+1}.$$
Then,
$$a_{n}=\frac{1}{n}\sum_{k=0}^{n-1}a_{k}2^{2n-2k-1}.$$
Multiplying the relation for $a_n$ on both sides by $\frac{2^2n}{n+1}$,
$$\frac{2^2n}{n+1}a_{n}=\frac{1}{n+1}\sum_{k=0}^{n-1}a_{k}2^{2n-2k+1}.$$
Subtracting this from the original recurrence relation, we get
$$a_{n+1}-\frac{2^2n}{n+1}a_{n} = \frac{1}{n+1}a_{n}2^{2n-2n+1} = \frac{2\,a_{n}}{n+1}\\
\implies a_{n+1} = \frac{2(2n-1)}{n+1}a_{n}.$$
Now we have a simple linear recurrence relation which may be solved by more familiar methods.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/804079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why this representation of circle is valid? A line passing through two distinct points $P_1(x_1,y_1),P_2(x_2,y_2)$ can be expressed by
$$\det\left| \begin{array}{ccc}
x-x_1&y-y_1 \\
x_2-x_1&y_2-y_1 \\ \end{array} \right|=0$$
Since line is set $A=\{a P_1+bP_2|a+b=1\}$ and it corresponds to determinant properties.
A circle passing three(non-colinear) points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ can be expressed by $$\det\left| \begin{array}{ccc}
(x-x_1)^2+(y-y_1)^2 & (x-x_1) & (y-y_1) \\
(x_2-x_1)^2+(y_2-y_1)^2 & (x_2-x_1) & (y_2-y_1) \\
(x_3-x_1)^2+(y_3-y_1)^2 & (x_3-x_1) & (y_3-y_1) \end{array} \right|=0$$
or $$\det\left| \begin{array}{ccc}
x^2+y^2 & x & y&1 \\
x_1^2+y_1^2 & x_1 & y_1&1 \\
x_2^2+y_2^2 & x_2 & y_2&1 \\
x_3^2+y_3^2 & x_3 & y_3&1 \\ \end{array} \right|=0$$
Please explain simple and neat as possible(like in determinant sense).
| Consier the set of equations $$Ax^2+Ay^2+Bx+Cy+D=0 \tag{1} $$
$$Ax_1^2+Ay_1^2+Bx_1+Cy_1+D=0 \tag{2} $$
$$Ax_2^2+Ay_2^2+Bx_2+Cy_2+D=0 \tag{3} $$
$$Ax_3^2+Ay_3^2+Bx_3+Cy_3+D=0 \tag{4} $$
this is a $4\times4$ linear system and it must satisfy,
$$\left( \begin{array}{ccc}
x^2+y^2 & x & y&1 \\
x_1^2+y_1^2 & x_1 & y_1&1 \\
x_2^2+y_2^2 & x_2 & y_2&1 \\
x_3^2+y_3^2 & x_3 & y_3&1 \\ \end{array} \right) \left( \begin{array}{ccc}
A\\
B \\
C \\
D \\ \end{array} \right)=0 \tag{5}$$
taking determinant you get your required condition. Also you can subtract the set of equations $1\to 4$ to reduce it into $3\times 3$ system and get the other result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/808506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Improper Integral $\int_{1}^{\infty} \sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) dx$ I need to calculate this improper integral.
$$\int_{1}^{\infty} \sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) dx$$
How do I prove that $$ \sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) $$ has an asymptotic equivalence with: $$ \frac{1}{\sqrt{x}} $$ for $x\rightarrow \infty$
And by the p-test that it diverges?
| Yes, as $x\rightarrow \infty\;$ you have the asmptotic relations
$$\sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right)
\sim \sin \left( \sin\left( \frac {1}{\sqrt{x}} \right) \right)
\sim \sin\left( \frac {1}{\sqrt{x}} \right)
\sim \frac {1}{\sqrt{x}}
$$
because for small $z\rightarrow 0\;$ you have $\sin(z) \sim z;\;$and therefore the integral
$$\int_{1}^{\infty} \sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) dx$$ diverges. You can get some better estimate with a CAS e.g.
$$\sin \left( \sin\left( \frac {1}{\sqrt{x}+1} \right) \right) =
\sqrt\frac{1}{x}-\frac{1}{x}+\frac{2}{3}\left(\frac{1}{x}\right)^{3/2}+O\left(\frac{1}{x^2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Laurent series for $\frac1{z^2+1}$ I have this problem:
Find the Laurent series around $z=0$, for $\dfrac{10}{(z+2)(z^2+1)}$ in the region $1<|z|<2$.
I did partial fractions and found this: $\dfrac{2}{z+2}-\dfrac{2z-4}{z^2+1}$,
then I have to know what's the Laurent series for $\dfrac{1}{z^2+1}$ to solve the problem.
Do you know that series? Thank you.
| Expanding geometrically,
$$\frac{2}{z+2} = \frac{1}{ 1 + \frac z 2} = \sum_{k=0}^\infty (-1)^k \left( \frac z 2\right)^k $$
Similarly,
$$\frac{2z-4}{z^2 + 1} = \frac{2z-4}{z^2}\cdot \frac 1 {1 + \frac{1}{z^2}} = \frac{2z-4}{z^2} \sum_{k=0}^{\infty}(-1)^k \left( \frac 1 {z^2}\right)^k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Let $a$ and $b$ be relatively prime integers. Show that $\gcd(a+b,a^2-3ab+b^2)= 1$ or $5$ Let $a$ and b be relatively prime integers. Show that $\gcd(a+b,a^2-3ab+b^2)=1$ or $5$
Proof: $s\mid (a+b)$ and $s\mid (a^2-3ab+b^2)$ implies $s\mid(a+b)^2=a^2+b^2+2ab$ and $s\mid (a^2-3ab+b^2)$
implies $s\mid((a+b)^2-(a^2-3ab+b^2))=5ab$.
From a previous question I posted I figure that the I am on the wrong track. Can someone help me with the next step?
| Note that $$a^2-3ab+b^2=(a+b)^2-5ab$$ So if $p$ is a prime, and such that $p|(a+b)$ and $p|(a+b)^2-5ab$ then $p|5ab$. Now $p\not | ab$ since if $p|a$, say then from $p|a+b$ follows $p|b$ contradiction the fact the $a$ and $b$ are relatively prime therefore $p|5$. The same argument shows that $gcd =5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Maximize the given variable under the following constraint Find the greatest integer $a$ such that $a^5 + 5^5$ is divisible by $a - 5$
| Make a polynomial (euclidean) division of $a^5+5^5$ by $a-5$ : since
$$
a^5+5^5=(a^5-5^5)+2(5^5)=(a-5)(a^4 + 5a^3 + 25a^2 + 125a + 625)+2(5^5)
$$
If $a$ is a solution, then $a-5$ must divide $2(5^5)$. It is then easy to see
that the largest solution is $a=2(5^5)+5=6255$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/811308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Diophantine equation with cubes. Interested in the solution in general Diophantine equations of the form:
$X^3+Y^3+Z^3=3XYZ+q$
$q$ - what some integer.
Solutions similar equations can be written.
Since this equation is easy, as it is quite symmetrical.
$X^3+Y^3+Z^3-3XYZ=R^3$
Such a solution can write.
$X=sp(p+s)$
$Y=s(2p^2+s^2)$
$Z=p(p^2+2s^2)$
$R=p^3+s^3$
And solutions can be written:
$X=s(9p^2+9ps+10s^2)$
$Y=s(6p^2+12ps+7s^2)$
$Z=3p^3+3p^2s+15ps^2+7s^3$
$R=3(p+2s)(p^2-ps+s^2)$
Whether there are any thoughts how to solve this equation?
At first I thought to use for solving Pell's equation, but I think that you can do without.
| You can use the following identity:
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$$
Fixed integer q has a finite number of factorial expansions. So we get a finite number of systems of two equations:
$$x+y+z = a$$
$$x^2+y^2+z^2-xy-xz-yz = b$$
for all integers $a$ and $b$ such that $ab=q$. Excluding the variable z we obtain the quadratic equation in x and y.
CALCULATIONS:
Let $u = x + y$ and $v=x y$. Then we have $z = a - u$ and
$$x^2+y^2+z^2-xy-xz-yz = u^2 - 2v + (a-u)^2 - v - (a-u)u =$$
$$u^2 - 3v + a^2 - 2au + u^2 - au +u^2 = 3u^2 - 3au - 3v + a^2$$
Thus, we have a equation in integers u and v:
$$ 3(u^2 - au) - 3v = b - a^2$$
It`s something like Pell's equation as you said. But it is more simple and we can express v from u:
$$ 3v = 3(u^2 - au) - (b - a^2)$$
Next step. Let suppose that (x,y) is an integer solution of original equation then because $u = x + y$ and $v=x y$ there is integer m such that $u^2-4v=m^2$ (this is a discriminant of a square equation $(\lambda - x)(\lambda - y) = \lambda^2-u\lambda+v=0$).
So $4v = u^2 - m^2$ and we have the following transformations of the equation:
$$ 3v = 3(u^2 - au) - (b - a^2)$$
$$12(u^2 - au)-3*4v = 4(b-a^2)$$
$$12(u^2 - au)-3(u^2-m^2)=4(b-a^2)$$
$$9u^2 - 12au + 3m^2=4(b-a^2)$$
$$9u^2 - 12au + 4a^2 + 3m^2=4b$$
$$(3u-2a)^2 + 3m^2=4b$$
Thus, we have that the number of solution for each fixed a and b is finite! So for each parameter q we can find all solutions (x,y,z) of the original equation $x^3+y^3+z^3 = 3xyz + q$ by the following formulas:
$$x=\frac{u-m}{2}$$
$$y=\frac{u+m}{2}$$
$$z=a-u$$
where u and m form a solution of the equation:
$$(3u-2a)^2 + 3m^2=4b$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/811456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}} dx$? How to calculate?
$$\int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}\, \mathrm dx$$
I try to let $x=\cos^2 t$, then
$$\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}=\tan\frac t2,\; dx=-2\sin t\cos t\,\mathrm dt $$
so
$$\int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}} \mathrm dx=-2 \int\tan\frac t2\sin t\cos t\,\mathrm dt$$
Thanks a lot!
| Alternate method:
$$ I = \int \sqrt{\frac{1- \sqrt{x}}{1 + \sqrt{x}}}\ dx$$
Putting $x = \cos^{2}{2t}$ so that $dx = -4\cos(2t) \sin(2t)\ dt$ and $t = \frac12 \cos^{-1}({\sqrt x})$
$$\begin{align}\implies I & = \int \frac{\sin(t)}{\cos(t)}\cdot -4\cos(2t) \sin(2t)\ dt\\ & = -4\int \frac{\sin t}{\cos t}[2\cos^2t - 1]\cdot 2 \sin t \cos t \ dt\\& = -8 \int\sin^2(t)[2\cos^2t - 1]\ dt\\& = -8 \int \frac{\sin^22t}{2} - \sin^2t \ dt\\& = -8 \int \frac{1 - \cos4t}{4} - \frac{1- \cos2t}{2} \ dt\\& = 2 \int \cos4t - 1\ dt + 4\int 1- \cos2t \ dt\\& = \frac{\sin 4t}{2} - 2t + C_1 + 4t - 2\sin 2t+ C_2\\& = \sin(2t) \cos(2t) + 2t - 2\sin 2t+ C\\& = \sqrt{x - x^2} + \cos^{-1}\sqrt{x}- 2\sqrt{1-x} + C\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/813079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
How to find $\lim_{n\rightarrow 0} \cos(\frac{\pi}{n} \sin n \cos n)$ and $\lim_{n\rightarrow 0} \frac{n}{\sin(2n) - \cos(\frac{n}{2}) +1}$ How can I determine these limits:
$$ a) \lim_{n\rightarrow 0} \cos(\frac{\pi}{n} \sin n \cos n)$$
$$ b) \lim_{n\rightarrow 0} \frac{n}{\sin(2n) - \cos(\frac{n}{2}) +1}$$
Note I cannot use l'Hospital. But I know that $\lim_{n\rightarrow 0} \frac{\sin n}{n} = 1$ and $\lim_{n\rightarrow 0} \frac{\cos n -1}{n} = 0$
I already found the limit of $\lim_{n\rightarrow 0} \frac{\sin(3n) \sin(2n)}{n^2}$:
$$\lim_{n\rightarrow 0} \frac{\sin(3n) \sin(2n)}{n^2}
= \lim_{n\rightarrow 0} \frac{\sin(3n)}{n} \lim_{n\rightarrow 0} \frac{ \sin(2n)}{n} =
3 \lim_{n\rightarrow 0} \frac{\sin(3n)}{3n} 2\lim_{n\rightarrow 0} \frac{ \sin(2n)}{2n} = 3 \cdot 2 = 6$$
Can I use a similar approach here?
| \begin{align}
(a)\quad\lim_{n\rightarrow 0} \cos\left(\frac{\pi\sin n \cos n}{n}\right)&=\cos\left(\lim_{n\rightarrow 0}\frac{\pi\sin n \cos n}{n}\right)\\
&=\cos\left(\lim_{n\rightarrow 0}\frac{\pi\sin n \cos n}{n}\right)\\
&=\cos\left(\pi\lim_{n\rightarrow 0}\cos n\frac{\sin n}{n}\right)\\
&=\cos(\pi)\\
&=-1.
\end{align}
Note that, it is valid to move the limit since cosine function is continuous.
\begin{align}
(b)\quad\lim_{n\rightarrow 0} \dfrac{\dfrac nn}{\dfrac{2\sin2n}{2n} -\left(\dfrac{ \cos\frac{n}{2}-1}{2\cdot\frac{n}{2}}\right)}=\frac{1}{2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/813769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How does $\frac{t^2}{t+1}$ equal $t-1+\frac{1}{t+1}$? I do the long division:
1: t+1 goes into $t^2$ t times
2: Subtract $t^2$ + 1 from $t^2$ and get -1
3: Answer: t - $\frac{1}{t+1}$
Am I missing something here?
| Polynomial division will also give you the answer quickly:
$$\begin{array}{cccccccc}
& & & t & & -1 & & \\
(t+1)& | & & t^2 & & +0t & &+0 \\
& & -( & t^2 & & +t &) & \\
& & & & & -t & & +0 \\
& & & &-(&-t & -1 & )\\
& & & & & & +1\\
\end{array}$$
$$\textrm{Result : } \boxed{t- 1 + \dfrac{1}{t+ 1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/814410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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.